Calculus/Print version

Welcome to the Wikibook of

Calculus

beginning with
Prerequisites and the Disciplines
of Differentiation and Integration,
and continuing with
Further Studies.

Mathematics is a universal study. It can be studied for its own sake as pure mathematics, or it can be applied to the natural world and to humanity as in the study of statistics. In the natural world, the nautilus shell presents one of the finest natural examples of a logarithmic spiral.

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Printable Version

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Contents

Precalculus

Contents

• Functions
• Exercises

Limits

Contents

• Introduction to Limits
• Finite Limits
• Infinite Limits
• Continuity
• Formal Definition of the Limit
• Exercises

Differentiation

Basics of Differentiation

• Differentiation - An Introduction
• More differentiation rules - More rules for differentiation
• Higher Order derivatives - An introduction to second power derivatives
• Implicit Differentiation
• Exercises

Applications of Derivatives

• Newton's Method
• Related Rates
• Kinematics
• Optimization
• Exercises

Important Theorems

• Extreme Value Theorem
• Rolle's Theorem
• Mean Value Theorem

Integration

Basics of Integration

• Integration

Integration techniques

• Integration techniques
• Further integration techniques
• Improper integrals
• Exercises

Applications of Integration

• Area
• Volume
• Volume of solids of revolution
• Arc length
• Surface area
• Work
• Centre of mass
• Pressure and force
• Probability

Parametric Equations

Contents

• Introduction to Parametric Equations
• Differentiation and Parametric Equations
• Integration and Parametric Equations

Polar Equations

Contents

• Introduction to Polar Equations
• Differentiation and Polar Equations
• Integration and Polar Equations

Sequences and Series

Basics

• Sequences
• Series

Calculus and Series

• Taylor series
• Power series
• Exercises

Multivariable & Differential Calculus

Contents

• Vectors
• Multivariable Calculus
• Ordinary Differential Equations
• Partial Differential Equations

Extensions

Further Analysis

• Systems of Ordinary Differential Equations

Formal Theory of Calculus

• Formal limits
• Real numbers
• Complex numbers
• Theory of Sequences

References

Contents

• Table of Trigonometry
• Summation notation
• Tables of Derivatives
• Tables of Integrals

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Introduction

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Introduction

Calculus is a tool used almost everywhere in the modern world to describe change and motion. Its use is widespread in science, engineering, medicine, business, industry, and many other fields. Just as algebra introduces students to new ways of thinking about arithmetic problems (by way of variables, equations, functions and graphs), calculus introduces new ways of thinking about algebra problems (considering, for example, how the height of a point moving along a graph changes as its horizontal position changes).

For a different perspective, have a look at the Wikipedia entry for Calculus.

What is calculus?

Calculus is the branch of mathematics dealing with instantaneous rates of change of continuously varying quantities. For example, consider a moving car. It is possible to create a function describing the displacement of the car (where it is located in relation to a reference point) at any point in time as well as a function describing the velocity (speed and direction of movement) of the car at any point in time. If the car were traveling at a constant velocity, then algebra would be sufficient to determine the position of the car at any time; if the velocity is unknown but still constant, the position of the car could be used (along with the time) to find the velocity.

However, the velocity of a car cannot jump from zero to 35 miles per hour at the beginning of a trip, stay constant throughout, and then jump back to zero at the end. As the accelerator is pressed down, the velocity rises gradually, and usually not at a constant rate (i.e., the driver may push on the gas pedal harder at the beginning, in order to speed up). Describing such motion and finding velocities and distances at particular times cannot be done using methods taught in pre-calculus, but it is not only possible but straightforward with calculus.

Calculus has two basic applications: differential calculus and integral calculus. The simplest introduction to differential calculus involves an explicit series of numbers. Given the series (42, 43, 3, 18, 34), the differential of this series would be (1, -40, 15, 16). The new series is derived from the difference of successive numbers which gives rise to its name "differential". Rarely, if ever, are differentials used on an explicit series of numbers as done here. Instead, they are derived from a series of numbers defined by a continuous function which are described later.

Integral calculus, like differential calculus, can also be introduced via series of numbers. Notice that in the previous example, the original series can almost be derived solely from its differential. Instead of taking the difference, however, integration involves taking the sum. Given the first number of the original series, 42 in this case, the rest of the original series can be derived by adding each successive number in its differential (42, 42+1, 43+(-40), 3+15, 18+16). Note that knowledge of the first number in the original series is crucial in deriving the integral. As with differentials, integration is performed on continuous functions rather than explicit series of numbers, but the concept is still the same. Integral calculus allows us to calculate the area under a curve of almost any shape; in the car example, this enables you to find the displacement of the car based on the velocity curve. This is because the area under the curve is the total distance moved, as we will soon see.

Why learn calculus?

Calculus is essential for many areas of science and engineering. Both make heavy use of mathematical functions to describe and predict physical phenomena that are subject to continual change, and this requires the use of calculus. Take our car example: if you want to design cars, you need to know how to calculate forces, velocities, accelerations, and positions. All require calculus. Calculus is also necessary to study the motion of gases and particles, the interaction of forces, and the transfer of energy. It is also useful in business whenever rates are involved. For example, equations involving interest or supply and demand curves are grounded in the language of calculus.

Calculus also provided important tools in understanding functions and has led to the development of new areas of mathematics including real and complex analysis, topology, and non-euclidean geometry.

What is involved in learning calculus?

Learning calculus, like much of mathematics, involves two parts:

• Understanding the concepts: You must be able to explain what it means when you take a derivative rather than merely apply the formulas for finding a derivative. Otherwise, you will have no idea whether or not your solution is correct. Drawing diagrams, for example, can help clarify abstract concepts.

• Symbolic manipulation: Like other branches of mathematics, calculus is written in symbols that represent concepts. You will learn what these symbols mean and how to use them. A good working knowledge of trigonometry and algebra is a must, especially in integral calculus. Sometimes you will need to manipulate expressions into a usable form before it is possible to perform operations in calculus.

What you should know before using this text

There are some basic skills that you need before you can use this text. Continuing with our example of a moving car:

• You will need to describe the motion of the car in symbols. This involves understanding functions.

• You need to manipulate these functions. This involves algebra.

• You need to translate symbols into graphs and vice verse. This involves understanding the graphing of functions.

• It also helps (although it isn't necessarily essential) if you understand the functions used in trigonometry since these functions appear frequently in science.

Scope

The first three chapters of this textbook cover the topics taught in a typical high school or first year college course. The first chapter, Precalculus, reviews those aspects of functions most essential to the mastery of Calculus, introduces the concept of the limit process. It also discusses some applications of limits and proposes using limits to examine slope and area of functions. The next two chapters, differentiation and integration, apply limits to calculate derivatives and integrals. The Fundamental Theorem of Calculus is used, as are the essential formulae for computation of derivatives and integrals without resorting to the limit process. The second and third chapters include articles that apply the concepts previously learned to calculating volumes, and so on as well as other important formulae.

The remainder of the central Calculus chapters cover topics taught in higher level Calculus topics: multivariable calculus, vectors, and series (Taylor, convergence, divergence).

Finally, the other chapters cover the same material, using formal notation. They introduce the material at a much faster pace, and cover many more theorems than the other two sections. They assume knowledge of some set theory and set notation.

Functions

Previous: Precalculus Current: Calculus/Print version Next: Limits Back to Calculus

Functions

Classical understanding of functions

To provide the classical understanding of functions, a function can be thought of as a machine. Machines take in raw materials, change them in a predictable way, and give out a finished product. The kinds of functions we consider here, for the most part, take in a real number, change it in a formulaic way, and give out another real number. A function is usually named f or g or something similar except when it is a special, named function (such as the pi function, the Riemann Zeta function, or the random variable function). A function is always defined as "of a variable" which tells the reader what to replace in the function.

For instance, $f(x) = 3x+2 \$ tells the reader:

• The function f is a function of x.
• To evaluate the function at a certain number, replace the x with the desired number.
• Replacing x with that number in the right side of the function will produce the function's output for that certain input.
• In English, the definition of $f \$ is interpreted, "Given a number, $f$ will return two more than the triple of that number."

Thus, if we want to know the value (or output) of the function at 3:

$f(x) = 3x+2 \$

$f(3) = 3(3)+2 \$ We evaluate the function at x = 3.

$f(3) = 9+2 = 11 \$ The value of $f \$ at 3 is 11.

Note that $f(3) \$ means the value of the dependent variable when $x \$ takes on the value of 3. So we see that the number 11 is the output of the function when we give the number 3 as the input. We refer to the input as the argument of the function (or the independent variable), and to the output as the value of the function at the given argument (or the dependent variable). A good way to think of it is the dependent variable $f(x) \$ 'depends' on the value of the independent variable $x \$. This is read as "the value of f of three is eleven", or simply "f of three equals eleven".

Notation

Functions are used so much that there is a special notation for them. The notation is somewhat ambiguous, so familiarity with it is important in order to understand the intention of an equation or formula.

Though there are no strict rules for naming a function, it is standard practice to use the letters $f$, $g$, and $h$ to denote functions, and the variable $x$ to denote an independent variable. $y$ is used for both dependent and independent variables.

When discussing or working with a function $f$, it's important to know not only the function, but also its independent variable $x$. Thus, when referring to a function $f$, you usually do not write $f$, but instead $f(x)$. The function is now referred to as "$f$ of $x$". The dependent variable is adjacent to the independent variable (in parenthesis). This is useful for indicating the value of the function at a particular value of the independent variable. For instance, if

$f(x)=7x+1\,$,

and if we want to use the value of $f$ for $x$ equal to $2$, then we would substitute 2 for $x$ on both sides of the definition above and write

$f(2)=7(2)+1=14+1=15\,$

This notation is more informative than leaving off the independent variable and writing simply '$f$', but can be ambiguous since the parentheses can be interpreted as multiplication. Consistency in notation greatly improves the readability of mathematical text.

Modern understanding of functions

The formal definition of a function states that a function is actually a rule that associates elements of one set called the domain of the function, with the elements of another set called the range of the function. For each value we select from the domain of the function, there exists exactly one corresponding element in the range of the function. The definition of the function tells us which element in the range corresponds to the element we picked from the domain. Classically, the element picked from the domain is pictured as something that is fed into the function and the corresponding element in the range is pictured as the output. Since we "pick" the element in the domain for whose corresponding element in the range we want to find, we have control over what element we pick and hence this element is also known as the "independent variable". The element mapped in the range is beyond our control and is "mapped to" by the function. This element is hence also known as the "dependent variable", for it depends on which independent variable we pick. Since the elementary idea of functions is better understood from the classical viewpoint, we shall use it hereafter. However, it is still important to remember the correct definition of functions at all times.

To make it simple, for the function $f(x)$, all of the $x$ values constitute the domain, and all of the values $f(x)$ ($y$ on the x-y plane) constitute the range.

Remarks

The following arise as a direct consequence of the definition of functions:

1. By definition, functions can take as many "inputs" at a time as desired but return only one "output", corresponding to that set of input. While one set of inputs cannot correspond to more than one output, the same output may correspond to more than one set of inputs. This is interpreted graphically as the vertical line test: a line drawn parallel to the axis of the dependent variable (normally vertical) through the graph of a function will intersect that function only once. Equivalently, this has an algebraic (or formula-based) interpretation. We can always say if $a = b$, then $f(a) = f(b)$, but if we only know that $f(a) = f(b)$ then we can't be sure that $a= b$.
2. Each function has a set of values, the function's domain, which it can accept as input. Perhaps this set is all positive real numbers; perhaps it is the set {pork, mutton, beef}. This set must be implicitly/explicitly defined in the definition of the function. You cannot feed the function an element that isn't in the domain, as the function is not defined for that input element.
3. Each function has a set of values, the function's range, which it can output. This may also be the set of real numbers, in which case the function is termed as a "real-valued" function. It may be the set of positive integers or even the set {0,1}. This set, too, must be implicitly/explicitly defined in the definition of the function.

This is an example of an expression which fails the vertical line test.

The vertical line test

The vertical line test is a systematic test to find out if an expression can serve as a function. Simply graph the expression and draw a vertical line along each point in the domain of the relation. If any vertical line ever touches the relation for more than one value, then the expression is not a function; if the line always touches only one value, then the expression is a function.

(There are a lot of useful curves, like circles, that aren't functions (see picture). Some people call these graphs with multiple intercepts, like our circle, "multi-valued functions"; they would refer to our "functions" as "single-valued functions".)

Important functions

$f(x)=a_n x^n + a_{n-1}x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0$ --Polynomial function

$f(x)=x\,$ --Identity function Takes an input and gives it as output.

$g(x)=c\,$ --Constant function Takes an input, ignores it, and always gives the constant c as an output. A polynomial of the zeroth degree

$g(x)=0\,$ --Zero function A special case of the above. A polynomial of the undefined degree.

$f(x)=mx+b\,$ --Linear function (function of a line)--polynomial of the first degree. Takes an input, multiplies by m and adds b

$\operatorname{sgn}(x) = \left\{ \begin{matrix} -1 & : & x < 0 \\ 0 & : & x = 0 \\ 1 & : & x > 0. \end{matrix} \right.$ -- The signum function Determines the sign of the argument x.

$f(x)=ax^2+bx+c \,$ -- The quadratic function. A particular case of the more general polynomial function.

Example functions

Some more simple examples of functions have been listed below.

$h(x)=\left\{\begin{matrix}1,&\mbox{if }x>0\\-1,&\mbox{if }x<0\end{matrix}\right.$ Gives 1 if input is positive, -1 if input is negative. Note that the function only accepts negative and positive numbers, not $0$. Mathematics describes this condition by saying $0$ is not in the domain of the function.

$g(y)=y^2\,$ Takes an input and squares it. $g(z)=z^2\,$ Exactly the same function, rewritten with a different independent variable. This is perfectly legal and sometimes done to prevent confusion (e.g. when there are already too many uses of $x$ or $y$ in the same paragraph.)

$f(y)=\left\{\begin{matrix}\int_{-y}^{y}e^{x^2}\,dx,&\mbox{if }y>0\\0,&\mbox{if }y\le0\end{matrix}\right.$ Takes an input and uses it as boundary values for an integration.

It is possible to replace the independent variable with any mathematical expression, not just a numeral. For instance, if the independent variable is itself a function of another variable, then it could be replaced with the value of that function. This is called composition, and is discussed next.

Manipulating functions

Functions can be manipulated in the same manner as any other variable; they can be added, multiplied, raised to powers, etc. For instance, let

$f(x)=3x+2\,$ and

$g(x)=x^2\,$.

Then

$\begin{align} f+g &= (f+g)(x)\\ &= f(x)+g(x)\\ &= (3x+2)+(x^2)\\ &= x^2+3x+2\, \end{align}$,

$\begin{align} f-g &= (f-g)(x)\\ &= f(x)-g(x)\\ &= (3x+2)-(x^2)\\ &= -x^2+3x+2\, \end{align}$,

$\begin{align} f\times g &= (f\times g)(x)\\ &= f(x)\times g(x)\\ &= (3x+2)\times(x^2)\\ &= 3x^3+2x^2\, \end{align}$,

$\begin{align} \frac{f}{g} &= \left(\frac{f}{g}\right)(x)\\ &= \frac{f(x)}{g(x)}\\ &= \frac{3x+2}{x^2}\\ &= \frac{3}{x}+\frac{2}{x^2} \end{align}$.

Composition of functions

However, there is one particular way to combine functions which cannot be done with other variables. The value of a function $f$ depends upon the value of another variable $x$; however, that variable could be equal to another function $g$, so its value depends on the value of a third variable. If this is the case, then the first variable is a function $h$ of the third variable; this function ($h$) is called the composition of the other two functions ($f$ and $g$). Composition is denoted by

$f\circ g=(f\circ g)(x)=f(g(x))$.

This can be read as either "f composed with g" or "f of g of x."

For instance, let

$f(x)=3x+2\,$ and

$g(x)=x^2\,$.

Then

$\begin{align} h(x) &= f(g(x))\\ &= f(x^2)\\ &= 3(x^2)+2\\ &= 3x^2+2\, \end{align}$.

Here, $h$ is the composition of $f$ and $g$. Note that composition is not commutative:

$f(g(x))=3x^2+2\,$, and

$\begin{align} g(f(x)) &= g(3x + 2)\\ &= (3x + 2)^2\\ &= 9x^2+12x+4\, \end{align}$

so $f(g(x))\ne g(f(x))\,$.

Composition of functions is very common, mainly because functions themselves are common. For instance: addition, multiplication, etc., can be represented as functions of more than one independent variable:

$\operatorname{plus}(x,y)=x+y$,

$\operatorname{times}(x,y)=x\times y$, etc.

Thus, the expression $2 \times 3 + 4$ is a composition of functions:

$2\times 3+4$	= $\operatorname{times}(2,3)+4$
	= $\operatorname{plus}( \operatorname{times}(2, 3), 4)$.

Since the function times equals $6$ if $x=2$ and $y=3$, then

$\operatorname{plus}(\operatorname{times}(2,3),4)= \operatorname{plus}(6,4)$.

Since the function plus equals $10$ if $x=6$ and $y=4$, then

$(2\times 3)+4=\operatorname{plus}(\operatorname{times}(2,3),4)=\operatorname{plus}(6,4)=10$.

Transformations

Transformations are a type of function manipulation that are very common. They consist of multiplying, dividing, adding or subtracting constants to either the argument or the output. Multiplying by a constant is known as Dilation and adding a constant is called Translation. Here are a few examples:

$f(2\times x) \,$ Dilation

$f(x+2)\,$ Translation

$2\times f(x) \,$ Dilation

$2+f(x)\,$ Translation

Examples of horizontal and vertical translations

Examples of horizontal and vertical dilations

Translations and Dilations can be either horizontal or vertical. Examples of both vertical and horizontal translations can be seen at right. The red graphs represent functions in their 'original' state, the solid blue graphs have been translated (shifted) horizontally, and the dashed graphs have been translated vertically. Dilations are demonstrated in a similar fashion. The function

$f(2\times x) \,$

has had its input doubled. One way to think about this is that now any change in the input will be doubled. If I add one to x, I add two to the input of the function, so it will now change twice as quickly. However, this is a horizontal dilation by $\frac{1}{2}$ because the distance to the y-axis has been halved. A vertical dilation, such as

$2\times f(x) \,$

is slightly more straightforward. In this case, you double the output of the function. The output represents the distance from the x-axis, so in effect, you have made the graph of the function 'taller'. Here are a few basic examples where a is any postive constant:

Original Graph	$f(x)\,$	Reflection through origin	$-f(-x)\,$
Horizontal Translation by a units right	$f(x-a)\,$	Horizontal Translation by a units left	$f(x+a)\,$
Horizontal Dilation by a factor of a	$f(x\times \frac{1}{a}) \,$	Vertical Dilation by a factor of a	$a\times f(x) \,$
Vertical Translation by a units down	$f(x)-a\,$	Vertical Translation by a units up	$f(x)+a\,$
Reflection over x-axis	$-f(x)\,$	Reflection over y-axis	$f(-x)\,$

Domain and Range

Notation

The notation used to denote intervals is very simple, but sometimes ambiguous because of the similarity to ordered pair notation:

Meaning	Interval Notation	Set Notation
All values greater than or equal to $a$ and less than or equal to $b$	$\left[a,b\right]$	$\left\{x:a\le x\le b\right\}$
All values greater than $a$ and less than $b$	$\left(a,b\right)$	$\left\{x:a < x < b\right\}$
All values greater than or equal to $a$ and less than $b$	$\left[a,b\right)$	$\left\{x:a\le x < b\right\}$
All values greater than $a$ and less than or equal to $b$	$\left(a,b\right]$	$\left\{x:a < x\le b\right\}$
All values greater than or equal $a$.	$\left[a,\infty\right)$	$\left\{x:x\ge a\right\}$
All values greater than $a$.	$\left(a,\infty\right)$	$\left\{x:x > a\right\}$
All values less than or equal to $a$.	$\left(-\infty,a\right]$	$\left\{x:x\le a\right\}$
All values less than $a$.	$\left(-\infty,a\right)$	$\left\{x:x < a\right\}$
All values.	$\left(-\infty,\infty\right)$	$\left\{x: x\in\mathbb{R}\right\}$

Note that $\infty$ must always be unbounded (that is, have no inclusive bracket, but instead have an exclusive parentheses). This is because $\infty$ is not a number, and therefore cannot be in our set. $\infty$ is really just a symbol that makes things easier to write, like the intervals above.
Note: ( is also denoted by ], and ) by [, i.e., (a,b) is the same as ]a,b[, and [a,b) is [a,b[. This is source of funny misunderstandings.

Domain

The domain of the function is the interval from -1 to 1

The domain of a function is the set of all points over which it is defined. More simply, it represents the set of x-values which the function can accept as input. For instance, if

$f(x)=\sqrt{1-x^2}$

then $f(x)$ is only defined for values of $x$ between $-1$ and $1$, because the square root function is not defined (in real numbers) for negative values. Thus, the domain, in interval notation, is $\left[-1,1\right]$. In other words,

$f(x) \mbox{is defined for } x\in [-1,1], \operatorname{ or } \{x:-1\le x\le 1\}$.

The range of the function is the interval from 0 to 1

Range

The range of a function is the set of all values which it attains (i.e. the y-values). For instance, if:

$f(x)=\sqrt{1-x^2}$,

then $f(x)$ can only equal values in the interval from $0$ to $1$. Thus, the range of $f$ is $\left[0,1\right]$.

One-to-one Functions

A function $f(x)$ is one-to-one (or less commonly injective) if, for every value of $f$, there is only one value of $x$ that corresponds to that value of $f$. For instance, the function $f(x)=\sqrt{1-x^2}$ is not one-to-one, because both $x=1$ and $x=-1$ result in $f(x)=0$. However, the function $f(x)=x+2$ is one-to-one because for every possible value of $f(x)$, there is exactly one corresponding value of $x$. Any function that looks like $f(x)=x^3+ax$, where $a\in \left[0,\infty\right)$, is one-to-one. Note that if you have a one-to-one function and translates/dilates it, it remains one-to-one (Of course you can't multiply $x$ or $f$ by a zero factor).

If you know what the graph of a function looks like, it is easy to determine whether or not the function is one-to-one. If every horizontal line intersects the graph in at most one point, then the function is one-to-one. This is known as the Horizontal Line Test.

Inverse functions

Function $f(x)$ has an inverse function if and only if $f(x)$ is one-to-one. For $f(x)$ and $g(x)$ such that $g(x)$ is the inverse function of $f(x)$:

$g(f(x)) = f(g(x)) = x\$.

For example, the inverse of $f(x)=x+2$ is $g(x)=x-2$. The function $f(x)=\sqrt{1-x^2}$ has no inverse.

Notation

The inverse function of $f$ is denoted as $f^{-1}(x)$. The inverse of a function $f^{-1}(x)$ is defined as the function that follows this rule

$f(f^{-1}(x))=f^{-1}(f(x)) = x$:

To determine $f^{-1}(x)$ when given a function f, substitute $f^{-1}(x)$ for $x$ and substitute $x$ for $f(x)$. Then solve for $f^{-1}(x)$, provided that it is also a function.

Example: Given $f(x) = 2x - 7$, find $f^{-1}(x)$.

Substitute $f^{-1}(x)$ for $x$ and substitute $x$ for $f(x)$., then solve for $f(x)$:

$f(x) = 2x - 7\,$

$x = 2[f^{-1}(x)] - 7\,$

$x + 7 = 2[f^{-1}(x)]\,$

$\frac{x + 7}{2} = f^{-1}(x)\,$

To check your work, confirm that $f(f^{-1}(x)) = x$:

$f(f^{-1}(x)) =$

$f \left (\frac{x + 7} {2}\right ) = {}$

$2\left (\frac{x + 7}{2}\right ) - 7 = (x+7) - 7 = x$

Graphing Functions

Graph of y=2x

It is sometimes difficult to understand the behavior of a function given only its definition; a visual representation or graph can be very helpful. A graph is a set of points in the Cartesian plane, where each point ($x$,$y$) indicates that $f(x)=y$. In other words, a graph uses the position of a point in one direction (the vertical-axis or y-axis) to indicate the value of $f$ for a position of the point in the other direction (the horizontal-axis or x-axis).

Functions may be graphed by finding the value of $f$ for various $x$ and plotting the points ($x$, $f(x)$) in a Cartesian plane. Since functions that you will deal with are generally continuous (see below), the parts of the function between the points can be approximated by drawing a line or curve between the points. Extending the function beyond the set of points is also possible, but becomes increasingly inaccurate.

Plotting points like this is laborious. Fortunately, many functions' graphs fall into general patterns. For a simple case, consider functions of the form

$f(x)={a}x \,$

The graph of $f$ is a single line, passing through ($0$,$0$) and $(1,a)$. Thus, after plotting the two points, a straightedge may be used to draw the graph as far as is needed.

Continuity

Most functions that you will deal with are not a random scattering of points. Rather, the graphs have one or more curves, which do not have sudden gaps in them, and may be drawn without lifting your pencil. Later, the principle of continuity will be defined formally using the concept of limits.

Algebraic manipulation

Purpose of review

This section is intended to review algebraic manipulation.

Rules of arithmetic and algebra

The following rules are always true (see: field theory).

• Addition
  • Commutative Law: $a+b=b+a \,$.
  • Associative Law: $(a+b)+c=a+(b+c)\,$.
  • Additive Identity: $a+0=a\,$.
  • Additive Inverse: $a+(-a)=0\,$.
• Subtraction
  • Definition: $a-b = a+(-b)\,$.
• Multiplication
  • Commutative law: $a\times b=b\times a\,$.
  • Associative law: $(a\times b)\times c=a\times (b\times c)\,$.
  • Multiplicative Identity: $a\times 1=a\,$.
  • Multiplicative Inverse: $a\times \frac{1}{a}=1,\qquad a \neq 0\,$
  • Distributive law: $a\times (b+c)=a\times b+a\times c\,$.
• Division
  • Definition: $\frac{a}{b}=a\times \frac{1}{b},\qquad b \neq 0\,$.

The above laws are true for all $a$, $b$, and $c$, whether $a$, $b$, and $c$ are numbers, variables, functions, or other expressions. For instance, although

$\frac{(x+2)(x+3)}{x+3}$	= $(((x+2)\times (x+3))\times (\frac{1}{x+3}))$
	= $((x+2)\times ((x+3)\times (\frac{1}{x+3})))$
	= $((x+2)\times (1)),\qquad x \neq -3 \,$
	= $x+2, \qquad x \neq -3\,$

is much longer than simply cancelling $x+3$ out in both the numerator and denominator, it is important to understand the longer method. Occasionally people do the following, for instance, which is incorrect:

$\frac{2\times (x + 2)}{2}=\frac{2}{2}\times \frac{x+2}{2}=\frac{x+2}{2}$.

The correct way is

$\frac{2\times (x + 2)}{2}=\frac{2}{2}\times \frac{x+2}{1}=1 \times \frac{x+2}{1}=x+2$,

where the number $2$ cancels out in both the numerator and the denominator.

Limits

Previous: Calculus Current: Calculus/Print version Next: Finite Limits Back to Calculus

Intuitive Look

A limit looks at what happens to a function when the input approaches a certain value. The general notation for a limit is as follows:

$\quad\lim_{x\to a} f(x) = L$

This is read as "The limit of f(x) as x approaches a is L". How we can determine whether a limit exists for f(x) at a and what the limit is is a technical point that we'll take up later. For now, we'll look at it from an intuitive standpoint.

Let's say that the function that we're interested in is $f(x)=x^2$, and that we're interested in its limit as x approaches 2. We can write the limit that we're interested in using the above notation as follows:

$\quad\lim_{x\to 2} x^2$

One way to try to evaluate what this limit is would be to choose values near 2, compute f(x) for each, and see what happens as they get closer to 2. This is implemented as follows:

$x$	1.7	1.8	1.9	1.95	1.99	1.999
$f(x)=x^2$	2.89	3.24	3.61	3.8025	3.9601	3.996001

Here we chose numbers smaller than 2, and approached 2 from below. We can also choose numbers larger than 2, and approach 2 from above:

$x$	2.3	2.2	2.1	2.05	2.01	2.001
$f(x)=x^2$	5.29	4.84	4.41	4.2025	4.0401	4.004001

We can see from the tables that as x grows closer and closer to 2, f(x) seems to get closer and closer to four, regardless of whether we approach 2 with x from above or from below. For this reason, we feel reasonably confident that the limit of $x^2$ as x approaches 2 is 4, or written in limit notation,

$\quad\lim_{x\to 2} x^2=4$

Now let's look at another example. Suppose we're interested in the function $f(x)=\frac{1}{x-2}$, and its behavior as x approaches 2. Here's the limit in limit notation:

$\quad\lim_{x\to 2} \frac{1}{x-2}$

Just as before, we can compute function values as x approaches 2 from below and from above. Here's a table, approaching from below:

$x$	1.7	1.8	1.9	1.95	1.99	1.999
$f(x)=\frac{1}{x-2}$	-3.333	-5	-10	-20	-100	-1000

And here from above:

$x$	2.3	2.2	2.1	2.05	2.01	2.001
$f(x)=\frac{1}{x-2}$	3.333	5	10	20	100	1000

In this case, the function doesn't seem to be approaching any value as x approaches 2. In this case we would say that the limit doesn't exist.

Both of these examples may seem trivial, but consider the following function:

$f(x) =\left\{\begin{matrix} x^2 & \mbox{if } x\neq 2 \\ \mbox{undefined} & \mbox{if } x=2\end{matrix}\right.$ OR $\begin{matrix} f(x) &=& \frac{x^2(x-2)}{x-2} \quad \end{matrix}$

Now, we see that these functions are completely identical; not just "almost the same," but actually, in terms of the definition of a function, completely the same. However, we are more comfortable with the second one algebraically, because it is easier to work with.

In algebra, we would simply say that we can cancel the term $(x-2)$, and then we have the function $f(x)=x^2$. This, however, would be a bit dishonest; the function that we have now is not really the same as the one we started with, because it is defined at x=2, and our original function was, specifically, not defined at x=2. In algebra we were willing to ignore this difficulty because we had no better way of dealing with this type of function. Now, however, in calculus, we can introduce a better, more correct way of looking at this type of function. What we want is to be able to say that, even though at $x=2$ the function doesn't exist, it works almost as though it does, and it's 4. It may not get there, but it gets really, really close. The only question that we have is: what do we mean by close?

Informal definition of a limit

As the precise definition of a limit is a bit technical, it is easier to start with an informal definition, and we'll explain the formal definition later.

We suppose that a function f is defined for x near c (but we do not require that it be defined when x=c).

Definition: (Informal definition of a limit) We call L the limit of f(x) as x approaches c , i.e. f(x) becomes close to L when x is close (but not equal) to c. When this holds we write $\lim_{x \to c} f(x) = L$ or $f(x) \to L \quad \mbox{as} \quad x \to c.$

Notice that the definition of a limit is not concerned with the value of f(x) when x=c (which may exist or may not). All we care about is the values of f(x) when x is close to c, on either the left or the right (i.e. less or greater).

Limit rules

Now that we have defined, informally, what a limit is, we will list some rules that are useful for manipulating a limit. These will all be proven, or left as exercises, once we formally define the fundamental concept of the limit of a function.

Suppose that $\lim_{x\to c} f(x) =L$ and $\lim_{x\to c} g(x) =M$ and that k is constant. Then

$\begin{align} \lim_{x\to c} k f(x) &=& k \cdot \lim_{x\to c} f(x) &=& k L \\ \lim_{x\to c} [f(x) + g(x)] &=& \lim_{x\to c} f(x) + \lim_{x\to c} g(x) &=& L + M \\ \lim_{x\to c} [f(x) - g(x)] &=& \lim_{x\to c} f(x) - \lim_{x\to c} g(x) &=& L - M \\ \lim_{x\to c} [f(x) \cdot g(x)] &=& \lim_{x\to c} f(x) \cdot \lim_{x\to c} g(x) &=& L\cdot M \\ \lim_{x\to c} {f(x) \over g(x)} &=& {\lim_{x\to c} f(x) \over \lim_{x\to c} g(x)} &=& {L \over M} & \hbox{as long as } M\neq 0 \end{align}$

Notice that in the last rule we need to require that M is not equal to zero (otherwise we would be dividing by zero which is an undefined operation).

These rules are known as identities; they are scalar multiplication, addition, subtraction, multiplication, and division of limits. (A scalar is a constant, and we say that when you multiply a function by a constant, you are performing scalar multiplication.)

Using these rules we can deduce some others. First, the constant rule states that if f(x) = b is constant for all x then the limit as x approaches c must be equal to b. In other words

$\lim_{x\to c} b = b$.

Second, the identity rule states that if f(x) = x then the limit of f as x approaches c is equal to c. That is,

$\lim_{x\to c} x = c$.

Third, using the rule for products many times we get that

$\lim_{x\to c} f(x)^n = \left(\lim_{x\to c} f(x) \right)^n$ for a positive integer n.

This is called the power rule.

Examples

Example 1 Find the limit $\lim_{x\to 2} {4x^3}.$

We need to simplify the problem, since we have no rules about this expression by itself. We know from the identity rule above that $\lim_{x\to 2} {x} = 2$. By the power rule, $\lim_{x\to 2} {x^3} = 2^3 = 8$. Lastly, by the scalar multiplication rule, we get $\lim_{x\to 2} {4x^3} = 4 \cdot 8=32$

Example 2

Find the limit $\lim_{x\to 2} 4x^3 + 5x +7.$

To do this informally, we split up the expression, once again, into its components. As above,$\lim_{x\to 2} 4x^3=32$.

Also $\lim_{x\to 2} 5x = 5\cdot\lim_{x\to 2} x = 5\cdot2=10$

and $\lim_{x\to 2} 7 =7$. Adding these together gives

$\lim_{x\to 2} 4x^3 + 5x +7 = 32 + 10 +7 =49.$

Example 3

Find the limit: $\lim_{x\to 2}$ $(4x^3 + 5x +7) \over (x-4)(x+10).$

From the previous example the limit of the numerator is $\lim_{x\to 2} 4x^3 + 5x +7 =49.$ The limit of the denominator is

$\lim_{x\to 2} (x-4)(x+10) = \lim_{x\to 2} (x-4) \cdot \lim_{x\to 2} (x+10) = (2-4)\cdot(2+10)=-24.$

As the limit of the denominator is not equal to zero we can divide which gives

$\lim_{x\to 2}$ $(4x^3 + 5x +7) \over (x-4)(x+10)$ $= - {49 \over 24}$.

Example 4

Find the limit: $\lim_{x\to 4}$ $(x^4 - 16x + 7) \over (4x-5).$

We apply the same process here as we did in the previous set of examples;

$\lim_{x\to 4}$ $(x^4 - 16x + 7) \over (4x-5)$ $=$ $\lim_{x\to 4} (x^4 - 16x + 7) \over \lim_{x\to 4} (4x-5)$ $=$ $( \lim_{x\to 4} (x^4) - \lim_{x\to 4} (16x) + \lim_{x\to 4} (7) ) \over \lim_{x\to 4} (4x) - \lim_{x\to 4} (5)$

We can evaluate each of these; $\lim_{x\to 4} (x^4) = 256.$ $\lim_{x\to 4} (16x) = 64.$ $\lim_{x\to 4} (7) = 7.$ $\lim_{x\to 4} (4x) = 16.$ $\lim_{x\to 4} (5) = 5.$ $= {199 \over 11}$

Example 5

Evaluate the limit $\lim_{x\to 0}$ $1-\cos(x) \over x$.

To evaluate this seemingly complex limit, be aware of your sine and cosine identities. We will also have to use two new facts. First, if f(x) is a trigonometric function (that is, one of sine, cosine, tangent, cotangent, secant or cosecant) and is defined at a, then $\lim_{x\to a} f(x) = f(a)$. Second, $\lim_{x\to 0}$ $\sin(x) \over (x)$ $= 1$.

To evaluate the limit, recognize that $1 - \cos(x)$ can be multiplied by $1+cos(x)$ to obtain $(1-cos^2(x))$ which, by our trig identities, is $sin^2(x)$. So, multiply the top and bottom by $1+cos(x)$ (This is allowed because it is identical to multiplying by one). This is a standard trick for evaluating limits of fractions; multiply the numerator and the denominator by a carefully chosen expression which will make the expression simplify somehow. In this case, we should end up with:

$\lim_{x\to 0}$ $1-\cos(x) \over x$ $=$ $1-\cos(x) \over x$ $\cdot$ $1 \over 1$ $=$ $1-\cos(x) \over x$ $\cdot$ $1 + \cos(x) \over 1+ \cos(x)$ $=$ $[(1 - \cos(x)) \cdot 1] + [(1 - \cos(x)) \cdot \cos(x)] \over x \cdot (1+ \cos(x))$ $=$ $1 - \cos(x) + \cos(x) - \cos^2(x) \over x \cdot (1+ \cos(x))$ $=$ $1 - \cos^2(x) \over x \cdot (1+ \cos(x))$ $=$ $\sin^2(x) \over x \cdot (1+ \cos(x))$ $=$ $\sin(x) \cdot \sin(x) \over x \cdot (1+ \cos(x))$ $=$ $\sin(x) \over x$ $\cdot$ $\sin(x) \over (1+ \cos(x))$

Your next step shall be to break this up into $\lim_{x\to 0}$$\sin(x) \over (x)$$\cdot$$\lim_{x\to 0}$$\sin(x) \over (1+\cos(x))$ by the limit rule of multiplication. By the fact mentioned above, $\lim_{x\to 0}$$\sin(x) \over (x)$$= 1$.

Next, $\lim_{x\to 0}$$\sin(x) \over (1+\cos(x))$ $=$ $\lim_{x\to 0}\sin x \over \lim_{x\to 0} (1+\cos x)$ $=$ $0 \over 1 + \cos 0$ $= 0$.

Thus, by multiplying these two results, we obtain 0.

We will now present an amazingly useful result, even though we cannot prove this yet. We can find the limit of any polynomial or rational function, as above, as long as that rational function is defined at c (so we are not dividing by zero). More precisely, c must be in the domain of the function.

Limits of Polynomials and Rational functions If f is a polynomial or rational function that is defined at c then $\lim_{x \rightarrow c} f(x) = f(c)$

We already learned this for trigonometric functions, so we see that it is easy to find limits of polynomial, rational or trigonometric functions wherever they are defined. In fact, this is true even for combinations of these functions; thus, for example, $\lim_{x\to 1} (\sin (x^2) + 4\cos^3(3x-1)) = \sin 1^2 + 4\cos^3 (3(1)-1)$.

Finding limits

Now, we will discuss how, in practice, to find limits. First, if the function can be built out of rational, trigonometric, logarithmic and exponential functions, then if a number c is in the domain of the function, then the limit is simply the value of the function at c.

If c is not in the domain of the function, then in many cases (as with rational functions) the domain of the function includes all the points near c, but not c. An example would be if we wanted to find $\lim_{x\to 0} \frac{x}{x}$, where the domain includes all numbers besides 0. In that case, we want to find a similar function, except with the hole filled in. The limit of this function at c will be the same, as can be seen from the definition of a limit. The function is the same as the previous except at a point c. The limit definition depends on f(x) only at the points where x is close to c but not equal to it. When $x \ne c$, this condition doesn't hold, and so the limit at c does not depend on the value of the function at c. Therefore, the limit of the new function is the same as of the previous function. And since the domain of our new function includes c, we can now (assuming it's still built out of rational, trigonometric, logarithmic and exponential functions) just evaluate the function at c as before.

In our example, this is easy; canceling the xs gives 1, which equals x/x at all points except 0. Thus, we have $\lim_{x\to 0}\frac{x}{x} = \lim_{x\to 0} 1 = 1$. In general, when computing limits of rational functions, it's a good idea to look for common factors in the numerator and denominator.

Lastly, note that the limit might not exist at all. There are a number of ways in which this can occur:

"Gap": There is a gap (more than a point wide) in the function where the function is not defined. As an example, in

$f(x) = \sqrt{x^2 - 16}$

f(x) does not have any limit when -4 ≤ x ≤ 4. There is no way to "approach" the middle of the graph. Note also that the function also has no limit at the endpoints of the two curves generated (at x = -4 and x = 4). For the limit to exist, the point must be approachable from both the left and the right. Note also that there is no limit at a totally isolated point on the graph.

"Jump": If the graph suddenly jumps to a different level, there is no limit. This is illustrated in the floor function (in which the output value is the greatest integer not greater than the input value).

Asymptote: In

$f(x) = {1 \over x^2}$

the graph gets arbitrarily high as it approaches 0, so there is no finite limit. In this case we say the limit is infinite.

Infinite oscillation: These next two can be tricky to visualize. In this one, we mean that a graph continually rises above and below a horizontal line. In fact, it does this infinitely often as you approach a certain x-value. This often means that there is no limit, as the graph never approaches a particular value. However, if the height (and depth) of each oscillation diminishes as the graph approaches the x-value, so that the oscillations get arbitrarily smaller, then there might actually be a limit.

The use of oscillation naturally calls to mind the trigonometric functions. An example of a trigonometric function that does not have a limit as x approaches 0 is

$f(x) = \sin {1 \over x}.$

As x gets closer to 0 the function keeps oscillating between -1 and 1. It is also true that sin(1/x) oscillates an infinite number of times on the interval between 0 and any positive value of x. The sine function, sin(x), is equal to zero whenever x=kπ, where k is a positive integer. Between each value of k, sin(x) oscillates between 0 and -1 or 0 and 1. Hence, sin(1/x)=0 for every x=1/(kπ). In between consecutive pairs of these values, 1/(kπ} and 1/[(k+1)π], sin(1/x) oscillates from 0, to -1, to 1 and back to 0 and so on. We may also observe that there are an infinite number of such pairs, and they are all between 0 and 1/π. There are a finite number of such pairs between any positive value of x and 1/π which implies that there must be infinitely many between x and 0. From our reasoning we may conclude that as x approaches 0 from the right, the function sin(1/x) does not approach any value. We say that the limit as x approaches 0 from the right does not exist. (In contrast, in the case of a "jump", the limits from each side did exist. Since they weren't equal, though, the regular limit didn't exist.)

Incomplete graph: Let us consider two examples.

First, let f be the constant function f(q)=2 where we specify that q is only allowed to be a rational number. Let $q_0$ be some rational number. Then $\lim_{q\to q_0} f(q) = 2$, since, for q close to $q_0$, f(q) is close to (in fact equals) 2.

Now let g be the similar-looking function defined on the entire real line, but we change the value of the function based on whether x is rational or not.

$g(x)=\left\{\begin{matrix} 2, & \mbox{if }x \mbox{ is rational} \\ 0, & \mbox{if }x \mbox{ is irrational} \end{matrix}\right.$

Now g has a limit nowhere! For let x be a real number; we show that g can't have a limit at x. No matter how close we get to x, there will be rational numbers (where g will be 2) and irrational numbers (where g will be 0). Thus g has no limit at any real number!

Using limit notation to describe asymptotes

Now consider the function

$g(x) = \frac {1}{x^2}.$

What is the limit as x approaches zero? The value of g(0) does not exist; the value of g(0) is not defined.

$\qquad g(0) = \frac {1}{0^2}$

Notice that we can make the function g(x) as large as we like, by choosing a small x, as long as x ≠ 0. For example, to make g(x) equal to one trillion, we choose x to be 10^-6. In this case, we say we can make g(x) arbitrarily large (as large as we like) by taking x to be sufficiently close to zero, but not equal to zero. And we express it algebraically as follows

$\lim_{x\to 0} g(x) = \lim_{x\to 0} \frac {1}{x^2} = \infty$

Note that the limit does not exist at x = 0.

Example: Finding Vertical and Horizontal Asymptotes

Use limits to find all vertical and horizontal asymptotes of the function $f(x) = \frac{1}{x+2} \$

The graph of $f(x) = \frac{1}{x+2} \$ has a vertical asymptote if f(x) goes to infinity. Therefore, we can determine any vertical asymptotes by setting the denominator equal to 0.

$x + 2 = 0 \$

$x + 2 - 2 = 0 - 2 \$

$x = -2 \$

Notice that $f(-2) = \frac{1}{0}$. Therefore, f(x) has a vertical asymptote at x = -2.

The graph of $f(x) = \frac{1}{x+2} \$ has a horizontal asymptote if f(x) converges on a value at infinity. We can find any horizontal asymptotes by finding $\lim_{x \to \infin} f(x) \ .$

$\lim_{x \to \infin} \frac{1}{x+2} = \frac{1}{\infin + 2} = \frac{1}{\infin} = 0$

Therefore f(x) has a horizontal asymptote at y = 0.

Application to calculus

To see the power of the limit, let's consider a moving car. Suppose we have a car whose position is linear with respect to time (that is, a graph plotting the position with respect to time will show a straight line). We want to find the velocity. This is easy to do from algebra, we just take the slope, and that's our velocity.

But unfortunately (or perhaps fortunately if you are a calculus teacher), things in the real world don't always travel in nice straight lines. Cars speed up, slow down, and generally behave in ways that make it difficult to calculate their velocities.

Now what we really want to do is to find the velocity at a given moment (the instantaneous velocity). The trouble is that in order to find the velocity we need two points, while at any given time, we only have one point. We can, of course, always find the average speed of the car, given two points in time, but we want to find the speed of the car at one precise moment.

Here is where the basic trick of differential calculus comes in. We take the average speed at two moments in time, and then make those two moments in time closer and closer together. We then see what the limit of the slope is as these two moments in time are closer and closer, and as those two moments get closer and closer, the slope comes out to be closer and closer to the slope at a single instant.

External links

• Online interactive exercises on limits
• Tutorials for the calculus phobe

Continuity

Defining Continuity

We are now ready for a formal definition of continuity, which was introduced in our review of functions. The definition is simple:

Definition of Continuity at a point If $f(x)$ is defined on an open interval containing c. Then $f(x)$ is said to be continuous at c if and only if $\lim_{x \rightarrow c} f(x) = f(c)$

Note that for f to be continuous at c, the definition requires three conditions:

1. that f is defined at c, so f(c) exists,
2. the limit as x approaches c exists, and
3. the limit and f(c) are equal.

If any of these do not hold then f is not continuous at c.

Notice how this relates to the intuitive idea of continuity. To be continuous, we want the graph of the function to approach the actual value of the function. This can be done by graphing the function itself. If the function is uniformly "smooth" (i.e. no "gaps", breaks, or sharp turns/corners) within an interval, then it is also continuous.

Definition of Continuity of a Function A function is said to be "continuous" if it is continuous at every point c in its domain.

Discontinuities

A discontinuity is a point where a function is not continuous.

Removable discontinuities

For example, the function $f(x) = \frac {x^2-9} {x-3}$ is considered to have a removable discontinuity at x = 3. It is discontinuous at that point because the fraction then becomes $\frac{0}{0}$ which is undefined. Therefore the function fails the very first definition of continuity.

Why do we call it removable though? Because, if we modify the function slightly, we can eliminate the discontinuity and make the function continuous.

To make the function f(x) continuous, we have to simplify f(x) so that $f(x) = \frac {x^2-9} {x-3}$ $=$ $\frac {(x+3)(x-3)} {(x-3)}$ $=$ $\frac {x+3} {1} \cdot \frac {x-3} {x-3}$ $=$ $\frac {x+3} {1} \cdot 1$ $= x + 3$.

As long as x ≠ 3, we can simplify the function f(x) to get a new function g(x) where

$g(x) = \left\{\begin{matrix} x + 3, & \mbox{if }x \ne 3 \\ 6, & \mbox{if }x = 3 \end{matrix}\right.$

Note that the function g(x) is not the same as the original function f(x), because g(x) has the extra point (3, 6). g(x) is now defined for x=3, the only thing keeping f(x) from being continuous. In fact, this kind of simplification is always possible with a removable discontinuity in a rational function. When the denominator is not 0, we can divide through to get a function which is the same. When it is 0, this new function will be identical to the old except for new points where previously we had division by 0.

Jump Discontinuities

Unfortunately, there are those discontinuities which cannot be removed from a function. Consider this k(x):

$k(x) = \left\{\begin{matrix} 1, & \mbox{if }x = 0 \\ -1, & \mbox{if }x \ne 0 \end{matrix}\right.$

k(x) is certainly defined for all points c, and the limit as x approaches any c even exists, the problem then lies in the fact that the limit and the function do not always equal each-other. Specifically

$\lim_{x\rightarrow 0}k(x)=-1$ But k(x) actually equals 1 at x=0, so...

$\lim_{x\rightarrow 0}k(x)\ne k(0)$ And the function is discontinuous.

Sadly there is no way to remove these discontinuities without fatally altering the function. There are no algebraic transformations to be applied and the function simply has to be accepted as discontinuous, ruling out many operations that will be introduced throughout calculus. These discontinuities are called nonremovable or essential discontinuities.

One-Sided Continuity

Just as a function can have a one-sided limit, a function can be continuous from a particular side.

Intermediate value theorem (IVT)

The intermediate value theorem is a very important theorem in calculus and analysis. It says:

If a function is continuous on a closed interval [a,b], then for every value k between f(a) and f(b) there is a value c on [a,b] such that f(c)=k.

Application: bisection method

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A few steps of the bisection method applied over the starting range [a₁;b₁]. The bigger red dot is the root of the function.

The bisection method is the simplest and most reliable algorithm to find roots to an equation.

Suppose we want to solve the equation f(x) = 0. Given two points a and b such that f(a) and f(b) have opposite signs, we know by the intermediate value theorem that f must have at least one root in the interval [a, b] as long as f is continuous on this interval. The bisection method divides the interval in two by computing c = (a+b) / 2. There are now two possibilities: either f(a) and f(c) have opposite signs, or f(c) and f(b) have opposite signs. The bisection algorithm is then applied recursively to the sub-interval where the sign change occurs. In this way we home in to a small sub-interval containing the root. The mid point of that small sub-interval is usually taken as the root.

Differentiation

What is differentiation?

Differentiation is a method that allows us to find a function that relates the rate of change of one variable with respect to another variable.

Informally, we may suppose that we're tracking the position of a car on a two-lane road with no passing lanes. Assuming the car never pulls off the road, we can abstractly study the car's position by assigning it a variable, x. Since the car's position changes as the time changes, we say that x is dependent on time, or x = x(t). But this is not enough to study how the car's position changes as the time changes; so far, we only have a way to tell where the car is at a specific time. Differentiation allows us to study $dx / dt$, which is the mathematical expression for how the car's position changes with respect to time.

Formally, we are able to find the slope at any point of a non-linear function (compare with determining the slope of a linear function: quite easy). Suppose we have determined the position of a particle at any time t modeled by the function:

$x = 3t^2 \,$

Differentiating this would give us the amount of change in distance with respect to time (a rate, or speed).

The Definition of Slope

On a Line

Three lines with different gradients

The slope of a line, also called the gradient of the line, is a measure of its inclination. A line that is horizontal has slope 0, a line from the bottom left to the top right has a positive slope, a line from the top left to the bottom right has a negative slope.

Gradient can be defined in two (equivalent) ways. The first way is to express it as how much the the line climbs for a given "step" horizontally. We denote a step in a quantity using a delta (Δ) symbol. Thus, a step in x is written as Δx. We can therefore write this definition of gradient as:

$\mbox{Gradient}=\frac{\Delta y}{\Delta x}$

Alternatively, we can define gradient as the "tangent function" of the line:

$\mbox{Gradient}=\tan\left( \alpha \right),$

where $\alpha$ is the angle between the line to the horizontal (measured clockwise). Those who know how the tangent function is generated (opposite side over adjacent side) will be able to spot the equivalence here.

If we have two points on a line, $P \left(x_1,y_1 \right)$ and $Q \left( x_2,y_2 \right)$, the step in x from P to Q is given by:

$\Delta x = x_2 - x_1\,$

Likewise, the step in y from P to Q is given by:

$\Delta y = y_2 - y_1\,$

This leads to the very important result below.

The definition of slope, m, between two points $(x_1, y_1)$ and $(x_2, y_2)$ on a line is $m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1}$.

On a Function

Most functions we are interested in are not straight lines (although they can be). We cannot define a gradient of a curved function in the same way as we can for a line. In order for us to understand how to find the gradient of a function at a point, we will first have to cover the idea of tangency. A tangent is a line which just touches a curve at a point, such that the angle between them at that point is zero. Consider the following four curves and lines:

(i)	(ii)
(iii)	(iv)

1. The line L crosses, but is not tangent to C at P.
2. The line L crosses, and is tangent to C at P.
3. The line L crosses C at more than one point, but is tangent to C at P.
4. There are many lines that cross C at P, but none are tangent. In fact, this curve has an undefined tangent at 'P.

A secant is a line drawn though two points on curve. We can construct a definition of the tangent as the limit of a secant of the curve drawn as the separation between the points tends to zero. Consider the diagram below.

As the distance h tends to zero, the secant line becomes the tangent at the point x₀. The two points we draw our line through are:

$P \left( x_0, f\left( x_0 \right) \right)$

and

$Q \left( x_0+h, f\left( x_0+h \right) \right)$

As a secant line is simply a line and we know two points on it, we can find its slope, m, from before:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Substituting in the points on the line,

$m = \frac{f\left( x_0+h \right) - f\left( x_0 \right)}{\left(x_0 + h \right) - x_0}.$

This simplifies to

$m = \frac{f\left( x_0+h \right) - f\left( x_0 \right)}{h}.$

This expression is called the difference quotient. Note that h can be postive or negative - it is perfectly valid to take a secant with a secondary point to the left.

Now, to find the slope of the tangent, m₀ we let h be zero. We cannot simply set it to zero as this would imply division of zero by zero which would yield an undefined result. Instead we must find the limit of the above expression as h tends to zero:

$m_0 = \lim_{h \to 0}\left[\frac{f\left( x_0+h \right) - f\left( x_0 \right)}{h}\right]$

Exercises

1. Find the slope of the tangent to the curve y=x² at x=1.

The Slope at a Point

Consider the formula for average velocity in the x-direction, $\frac{\Delta x}{\Delta t}$. This formula can be used to find approximate results for speed, but is rarely exact. To correct this we look at the change in position as the change in time approaches 0. Mathematically this is written as: $\lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$, where d denotes change, x denotes distance, and t denotes time. Compare the operator d with Δ. The delta should be familiar to you from studying slope. They both indicate a difference between two numbers, however d denotes an infinitesimal difference.

(Note that the letter s is often used to denote distance, which would yield $\frac{ds}{dt}$. The letter d is often avoided in denoting distance due to the ambiguity in the expression $\frac{dd}{dt}$.)

If a function f(x) is plotted on an (x,y) Cartesian coordinate system, differentiation will yield a function which describes the rate of change of y with respect to x.

The rate of change at a specific point is called the instantaneous rate of change. The line with a slope equal to the instantaneous rate of change and that only touches the graph at that specific point is known as a tangent.

Historically, the primary motivation for the study of differentiation was the tangent line problem: for a given curve, find the slope of the straight line that is tangent to the curve at a given point. The word tangent comes from the Latin word tangens, which means touching. Thus, to solve the tangent line problem, we need to find the slope of a line that is "touching" a given curve at a given point. But what exactly do we mean by "touching"?

The solution is obvious in some cases: for example, a line $y = m x + c$ is its own tangent; the slope at any point is $m$. For the parabola $y = x^2$, the slope at the point $(0,0)$ is $0$ (the tangent line is flat). In fact, at any vertex of any smooth function the slope is zero, because the slope of the tangent lines on either side of the point are opposite signs.

But how can you find the slope of, say, $y = \sin x + x^2$ at $x = 1.5$?

The easiest way to find slopes for any function is by differentiation. Differentiation results in another function whose value for any value $x$ is the slope of the original function at $x$. This function is known as the derivative of the original function, and is denoted by either a prime sign, as in $f'(x)\$ (read "f prime of x"), the quotient notation, $\frac{df}{dx}$ or $\frac{d}{dx}\left[ f\right]$, which is more useful in some cases, or the differential operator notation, $D_x[f(x)]$, which is generally just written as $D f(x)$.

Most of the time the brackets are not needed, but are useful for clarity if we are dealing with something like $D (fg)$ for a product.

Example: If$\ f(x) = 3x + 5$, $\ f'(x) = 3$, no matter what $x$.

Example: If $f(x) = \left|x\right|$ (the absolute value function) then $f'(x) = \left\{ \begin{matrix} -1, & x < 0 \\ \operatorname{undefined}, & x = 0 \\ 1, & x > 0 \end{matrix} \right. .$

Here, $f(x)$ is not smooth (though it is continuous) at $x=0$ and so the limits $\lim_{x \to 0^{+}} f'(x)$ and $\lim_{x \to 0^{-}} f'(x)$ (the limits as 0 is approached from the right and left respectively) are not equal. f'(0) is said to be undefined, and so f'(x) has a discontinuity at 0. This sort of point of non differentiability is called a cusp. Functions may also not be differentiable because they go to infinity at a point, or oscillate infinitely frequently.

The Definition of the Derivative

The derivative is the formula $m_{tan}$ (slope of tangent line) on a curve at a specific point.

An example

Draw a curve defined as $y=3x^2$ and select any point on it. We select the point at which x=4; what is the slope at this point? We can do it "the hard (and imprecise) way", without using differentiation, as follows, using a calculator and using small differences below and above the given point:

When $x=3.999$, $y=47.976003$.

When $x=4.001$, $y=48.024003$.

Then the difference between the two values of x is $\Delta x=0.002$.

Then the difference between the two values of y is $\Delta y=0.048$.

Thus, the slope $= \frac{\Delta y}{\Delta x} = 24$ at the point of the graph at which $x=4$.

Now using differentiation rules (see below) to solve this problem again, when $y=3x^2$ then the slope at any point of that curve is found by evaluating $y'=6x$.

Our x is 4, so that $y' = \frac{\Delta y}{\Delta x} = 6 * 4 = 24$ again. No need for a calculator!

$f'(x) = \lim_{\Delta x \to 0}\left[\frac{f(x+\Delta x)-f(x)}{\Delta x}\right]$

This is the definition of the derivative. If the limit exists we say that f is differentiable at x and its derivative at x is $f'(x)$. A visual explanation of this formula is that the slope of the tangent line is the limit of the slope of a secant line when the difference of the points ($\Delta x$) tends to zero.

Example

Let us try this for a simple function:

$f(x)\,$	= $\frac{x}{2}$
$f'(x)\,$	= $\lim_{\Delta x \to 0}\left(\frac{\frac{x+\Delta x}{2} - \frac{x}{2}}{\Delta x}\right)$
	= $\lim_{\Delta x \to 0}\left(\frac{1}{2}\right)$
	= $\frac{1}{2}$

This is consistent with the definition of the derivative as the slope of a function.

Sometimes, the slope of a function varies with $x$. This is demonstrated by the function $f(x)=x^2$,

$f(x)\,$	= $x^2\,$
$f'(x)\,$	= $\lim_{\Delta x \to 0}\left[\frac{(x+\Delta x)^2-x^2}{\Delta x}\right]$
	= $\lim_{\Delta x \to 0}\left(\frac{x^2+2x\Delta x+\Delta x^2-x^2}{\Delta x}\right)$
	= $\lim_{\Delta x \to 0}\left(\frac{2x\Delta x+\Delta x^2}{\Delta x}\right)$
	= $\lim_{\Delta x \to 0}\left(2x+\Delta x\right)$
	= $2x$

Understanding the Derivative Notation

The derivative notation is special and unique in mathematics. The most common use of derivatives you'll run into when first starting out with differentiating is the Leibniz notation, expressed as $\frac{dy}{dx}$. You may think of this as "rate of change in y with respect to x". You may also think of it as "infinitesimal value of y divided by infinitesimal value of x". Either way is a good way of thinking. Often, in an equation, you will see just $\frac{d}{dx}$, which literally means "derivative with respect to x". You may safely assume that it is the equivalent of $\frac{dy}{dx}$ for now.

As you advance through your studies, you will see that $dy$ and $dx$ can act as separate entities that can be multiplied and divided (to a certain degree). Eventually you will see derivatives such as $\frac {dx} {dy}$, which sometimes will be written $\frac{d}{dy}$. Or, you may see a derivative in polar coordinates marked as $\frac{d\theta}{dr}$.

All of the following are equivalent for expressing the derivative of $y = x^{2}$

• $\frac{dy}{dx} = 2x$
• $\frac{d}{dx} x^{2} = 2x$
• $dy = 2x dx \$
• $f '(x) = 2x \$
• $D(f(x)) = 2x \$

Exercises

Using the definition of the derivative find the derivative of the function $f(x)=2x+3$

Using the definition of the derivative find the derivative of the function $f(x)=x^3$. Now try $f(x)=x^4$. Can you see a pattern? In the next section we will find the derivative of $f(x)=x^n$ for all n.

The text states that the derivative of $\left|x\right|$ is not defined at $x = 0$. Use the definition of the derivative to show this.

Graph the derivative to $y=4x^2$ on a piece of graph paper without solving for $dy/dx$. Then, solve for $dy/dx$ and graph that; compare the two graphs.

Use the definition of the derivative to show that the derivative of $\sin x$ is $\cos x$. Hint: Use a suitable sum to product formulae and the fact $\lim_{t \rightarrow 0}\frac{\sin t}{t}=1$

Differentiation rules

The process of differentiation is tedious for large functions. Therefore, rules for differentiating general functions have been developed, and can be proved with a little effort. Once sufficient rules have been proved, it will be possible to differentiate a wide variety of functions. Some of the simplest rules involve the derivative of linear functions.

Derivative of a Constant Function

For any fixed real number $c$,

$\frac{d}{dx}\left[c\right]=0$

Intuition

The function f(x) = c is a horizontal line, which has a constant slope of zero. Therefore, it should be expected that the derivative of this function is zero, regardless of the value of x.

Proof

From the definition of a derivative:

$\lim_{\Delta x \to 0}\left[\frac{f(x+\Delta x)-f(x)}{\Delta x}\right]$

Let $f(x) = c$. Then $f(x+\Delta x) = c$ because there are no x's in the function with which to plug in $x + \Delta x$. Therefore:

$\frac{d}{dx}\left[c\right] = \lim_{\Delta x \to 0} \left[\frac{(c)-c}{\Delta x}\right] = 0$

Example

$\frac{d}{dx}\left[3\right]=0$

Example

$\frac{d}{dx}\left[z\right]=0$

Derivative of a Linear Function

For any fixed real numbers $m$ and $c$,

$\frac{d}{dx}\left[mx+c\right]=m$

The special case $\frac{dx}{dx} = 1$ shows the advantage of the $\frac{d}{dx}$ notation -- rules are intuitive by basic algebra, though this does not constitute a proof, and can lead to misconceptions to what exactly $dx$ and $dy$ actually are.

Constant multiple and addition rules

Since we already know the rules for some very basic functions, we would like to be able to take the derivative of more complex functions and break them up into simpler functions. Two tools that let us do this are the constant multiple rules and the addition rule.

The Constant Rule

For any fixed real number $c$,

$\frac{d}{dx}\left[cf(x)\right] = c \frac{d}{dx}\left[f(x)\right]$

The reason, of course, is that one can factor $c$ out of the numerator, and then of the entire limit, in the definition.

Example

We already know that

$\frac{d}{dx}\left[x^2\right]=2x$.

Suppose we want to find the derivative of $3x^2$

$\frac{d}{dx}\left[3x^2\right]$	= $3\frac{d}{dx}\left[x^2\right]$
	= $3\times2x\,$
	= $6x\,$

Another simple rule for breaking up functions is the addition rule.

The Addition and Subtraction Rules

$\frac{d}{dx}\left[f(x)\pm g(x)\right]= \frac{d}{dx}\left[f(x)\right]\pm\frac{d}{dx}\left[g(x)\right]$

Proof

From the definition:

$\lim_{\Delta x \to 0}\left[\frac{[f(x+\Delta x) \pm g(x + \Delta x) - [f(x) \pm g(x)]}{\Delta x}\right]$

$= \lim_{\Delta x \to 0} \left[\frac{[f(x+\Delta x) - f(x)] \pm [g(x + \Delta x) - g(x)]}{\Delta x}\right]$

$= \lim_{\Delta x \to 0} \left[\frac{[f(x+\Delta x) - f(x)]}{\Delta x}\right] \pm \lim_{\Delta x \to 0} \left[\frac{[g(x+\Delta x) - g(x)]}{\Delta x}\right]$

By definition then, this last term is $\frac{d}{dx} \left[f(x)\right] + \frac{d}{dx}\left[g(x)\right]$

Example

What is:

$\frac{d}{dx}\left[3x^2+5x\right]$	= $\frac{d}{dx}\left[3x^2+5x\right]$
	= $\frac{d}{dx}\left[3x^2\right]+\frac{d}{dx}\left[5x\right]$
	= $6x+\frac{d}{dx}\left[5x\right]$
	= $6x+5\,$

The fact that both of these rules work is extremely significant mathematically because it means that differentiation is linear. You can take an equation, break it up into terms, figure out the derivative individually and build the answer back up, and nothing odd will happen.

We now need only one more piece of information before we can take the derivatives of any polynomial.

The Power Rule

$\frac{d}{dx}\left[x^n\right]=nx^{n-1}, x\ne0$

For example, in the case of $x^2$ the derivative is $2x^1=2x$ as was established earlier. This rule is actually in effect in linear equations too, since $x^{n-1}=x^0$ when n=1, and of course, any real number or variable to the zero power is one.

This rule also applies to fractional and negative powers. Therefore

$\frac{d}{dx}\left[\sqrt x \right]$	= $\frac{d}{dx}\left[ x^{1/2}\right]$
	= $\frac 1 2 x^{-1/2}$
	= $\frac 1 {2\sqrt x}$

Since polynomials are sums of monomials, using this rule and the addition rule lets you differentiate any polynomial. A relatively simple proof for this can be derived from the binomial expansion theorem.

Derivatives of polynomials

With these rules in hand, you can now find the derivative of any polynomial you come across. Rather than write the general formula, let's go step by step through the process.

$\frac{d}{dx}\left[6x^5+3x^2+3x+1\right]$

The first thing we can do is to use the addition rule to split the equation up into terms:

$\frac{d}{dx}\left[6x^5\right]+\frac{d}{dx}\left[3x^2\right]+\frac{d}{dx}\left[3x\right]+\frac{d}{dx}\left[1\right].$

We can immediately use the linear and constant rules to get rid of some terms:

$\frac{d}{dx}\left[6x^5\right]+\frac{d}{dx}\left[3x^2\right]+3+0.$

Now you may use the constant muliplier rule to move the constants outside the derivatives:

$6\frac{d}{dx}\left[x^5\right]+3\frac{d}{dx}\left[x^2\right]+3.$

Then use the power rule to work with the individual monomials:

$6\left(5x^4\right)+3\left(2x\right)+3.$

And then do some algebra to get the final answer:

$30x^4+6x+3.\,$

These are not the only differentiation rules. There are other, more advanced, differentiation rules, which will be described in a later chapter.

Exercises

• Find the derivatives of the following equations:

$f(x) = 42$

$f(x) = 6x + 10$

$f(x) = 2x^2 + 12x + 3$

• Use the definition of a derivative to prove the derivative of a constant function, of a linear function, and the constant rule and addition or subtraction rules.

Extrema and Points of Inflexion

The four types of extrema.

Maxima and minima are points where a function reaches a highest or lowest value, respectively. There are two kinds of extrema (a word meaing maximum or minimum): global and local, sometimes referred to as "absolute" and "relative", respectively. A global maximum is a point that takes the largest value on the entire range of the function, while a global minimum is the point that takes the smallest value on the range of the function. On the other hand, local extrema are the largest or smallest values of the function in the immediate vicinity.

All extrema look like the crest of a hill or the bottom of a bowl on a graph of the function. A global extremum is always a local extremum too, because it is the largest or smallest value on the entire range of the function, and therefore also its vicinity. It is also possible to have a function with no extrema, global or local: y=x is a simple example.

At any extremum, the slope of the graph is necessarily zero, as the graph must stop rising or falling at an extremum, and begin to fall or rise. Becuase of this, extrema are also commonly called stationary points or turning points. Therefore, the first derivative of a function is equal to zero at extrema. If the graph has one or more of these stationary points, these may be found by setting the first derivative equal to zero and finding the roots of the resulting equation.

The function f(x)=x³, which contains a point of inflexion at the point (0,0).

However, a slope of zero does not guarantee a maximum or minimum: there is a third class of stationary point called a point of inflexion. Consider the function

$f \left(x \right) = x^3$.

The derivative is

$f^\prime \left(x \right) = 3 x^2$

The slope at x=0 is 0. We have a slope of zero, but while this makes it a stationary point, this doesn't mean that it is a maxima or minima. Looking at the graph of the function you will see that x=0 is neither, it's just a spot at which the function flattens out. True extrema require the a sign change in the first derivative. This makes sense - you have to rise (positive slope) to and fall (negative slope) from a maximum. In between rising and falling, on a smooth curve, there will be a point of zero slope - the maximum. A minimum would exhibit similar properties, just in reverse.

Good (B and C, green) and bad (D and E, blue) points to check in order to classify the extremum (A, black). The bad points lead to an incorrect classification of A as a minimum.

This leads to a simple method to classify a stationary point - plug x values slightly left and right into the derivative of the function. If the results have opposite signs then it is a true maximum/minimum. You can also use these slopes to figure out if it is a maximum or a minimum: the left side slope will be positive for a maximum and negative for a minimum. However, you must exercise caution with this method, as, if you pick a point too far from the extremum, you could take it on the far side of another extremum and incorrectly classify the point.

The Extremum Test

A more rigourous method to classify a stationary point is called the extremum test. As we mentioned before, the sign of the first derivative must change for a stationary point to be a true extremum. Now, the second derivative of the function tells us the rate of change of the first derivative. It therefore follows that if the second derivative is positive at the stationary point, then the gradient is increasing. The fact that it is a stationary point in the first place means that this can only be a minimum. Conversely, if the second derivative is negative at that point, then it is a maximum.

Now, if the second derivative is zero, we have a problem. It could be a point of inflexion, or it could still be an extremum. Examples of each of these cases are below - all have a second derivative equal to zero at the stationary point in question:

• $y = x^{3}$ has a point of inflexion at $x=0$
• $y = x^{4}$ has a minimum at $x=0$
• $y =-x^{4}$ has a maximum at $x=0$

However, this is not an insoluble problem. What we must do is continue to differentiate until we get, at the (n+1)th derivative, a non-zero result at the stationary point:

$f^{\prime} \left(x \right)=0, \,f^{\prime \prime} \left(x \right)=0,\, \ldots ,f^{\left(n\right)} \left(x \right)=0,\,f^{\left(n+1\right)} \left(x \right)\ne 0$

If n is odd, then the stationary point is a true extremum. If the (n+1)th derivative is positive, it is a minimum; if the (n+1)th derivative is negative, it is a maximum. If n is even, then the stationary point is a point of inflexion.

As an example, let us consider the function

$f \left( x \right) = -x^4$

We now differentiate untill we get a non-zero result at the stationary point at x=0 (assume we have already found this point as usual):

$f^\prime \left( x \right) = -4x^3$

$f^{\prime \prime} \left( x \right) = -12x^2$

$f^{\prime \prime \prime} \left( x \right) = -24x$

$f^{\left(4\right)} \left( x \right) = -24$

Therefore, (n+1) is 4, so n is 3. This is odd, and the fourth derivative is negative, so we have a maximum. Note that none of the methods give can tell you if this is a global extremum or just a local one. To do this, you would have to set the function equal to the height of the extremum and look for other roots.

See the "Sales Example" below for an application of these principles.

More Differentiation Rules

Chain Rule

We know how to differentiate regular polynomial functions. For example:

$\frac{d}{dx}(3x^3 - 6x^2 + x) = 9x^2 - 12x + 1$

However, we've not yet explored the derivative of an unexpanded expression. If we are given the function $f(x)=(x^2+5)^2$, we currently have no choice but to expand it and find the derivative of each part and then add the parts back together:

$f(x)=(x^2+5)^2 \, \!$

$f(x)=x^4 + 10x^2 + 25 \, \!$

$f '(x) = 4x^3 + 20x \, \!$

However, there is a useful rule known as the chain rule. The function above ($f(x)=(x^2+5)^2$) can be consolidated into two nested parts $f(x)=u^2$, where $u =m(x)= (x^2+5)$. Therefore:

if

$g(u) = u^2 \, \!$ and

$u = m(x)=x^2+5 \, \!$

Then:

$f(x) = g(m(x)) \, \!$

Then

$f '(x) = g '(m (x))m'(x) \, \!$

The chain rule states that if we have a function of the form $y(u(x))$ (i.e. $y$ can be written as a function of $u$ and $u$ can be written as a function of $x$) then:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Chain Rule If a function F(x) is composed to two differentiable functions g(x) and m(x), so that F(X)=g(m(x)), then F(x) is differentiable and, $F '(x)= g'(m(x))m'(x) \, \!$

We can now investigate the original function:

$\frac{dy}{du} = 2u$

$\frac{du}{dx} = 2x$

Therefore

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 2u\cdot 2x = 2(x^2+5)(2x) = 4x^3 + 20x$

This can be performed for more complicated equations. If we consider:

$\frac{d}{dx} \sqrt{1 + x^2}$

and let y=√u and u=1+x², so that dy/du=1/2√u and du/dx=2x, then, by applying the chain rule, we find that

$\frac d {dx} \sqrt{1 + x^2} = \frac {1} {2 \sqrt {1 + x^2}}\cdot 2x = \frac {x} \sqrt {1 + x^2}$

Product and Quotient Rules

When we wish to differentiate a more complicated expression such as:

$h(x) = (x^2+5)^5 \cdot (x^3 + 2)^3$

our only way (up to this point) to differentiate the expression is to expand it and get a polynomial, and then differentiate that polynomial. This method becomes very complicated and is particularly error prone when doing calculations by hand. It is advantageous to find the derivative of h(x) using just the functions f(x) = (x²+5)⁵ and g(x) = (x³ + 2)³ and their derivatives.

Derivatives of products (Product rule) $\frac{d}{dx}\left[ f(x) \cdot g(x) \right] = f'(x) \cdot g(x)+f(x) \cdot g'(x)\,\!$

What this rule basically means is that if one has a function that is the product of two functions, then all one has to do is differentiate the first function, multiply it by the other un- differentiated function, add that to the first function undifferentiated multiplied by the differentiated second function. For Example: If one were to take the function
$H(x) = (x+2)^3 \cdot 2(x+1)^2$,

its derivative would not be
$3(x+2)^2 \cdot 4(x+1)$.

Instead it would be
$3(x+2)^2 \cdot 2(x+1)^2 + (x+2)^3 \cdot 4(x+1)$

Another way of approaching this is if one were to have a function that was a product of the two functions A and B
$h(x) = A \cdot B$ Its derivative would be
$h'(x) = A' \cdot B + A \cdot B'$

Proving this rule is relatively straightforward, first let us state the equation for the derivative:

$\frac{d}{dx} \left[ f(x) \cdot g(x) \right] = \lim_{h \to 0} \frac{ f(x+h)\cdot g(x+h) - f(x) \cdot g(x)}{h}$

We will then apply one of the oldest tricks in the book--adding a term that cancels itself out to the middle:

$\frac{d}{dx} \left[ f(x) \cdot g(x) \right] = \lim_{h \to 0} \frac{ f(x+h)\cdot g(x+h) \mathbf{- f(x) \cdot g(x+h) + f(x) \cdot g(x+h)} - f(x) \cdot g(x)}{h}$

Notice that those terms sum to zero, and so all we have done is add 0 to the equation.

Now we can split the equation up into forms that we already know how to solve:

$\frac{d}{dx} \left[ f(x) \cdot g(x) \right] = \lim_{h \to 0} \left[ \frac{ f(x+h)\cdot g(x+h) - f(x) \cdot g(x+h) }{h} + \frac{f(x) \cdot g(x+h) - f(x) \cdot g(x)}{h} \right]$

Looking at this, we see that we can separate the common terms out of the numerators to get:

$\frac{d}{dx} \left[ f(x) \cdot g(x) \right] = \lim_{h \to 0} \left[ g(x+h) \frac{ f(x+h) - f(x) }{h} + f(x) \frac{g(x+h) - g(x)}{h} \right]$

Which, when we take the limit, becomes:

$\frac{d}{dx} \left[ f(x) \cdot g(x) \right] = f(x) \cdot g'(x) + g(x) \cdot f'(x)$, or the mnemonic "one D-two plus two D-one"

This can be extended to 3 functions:

$\frac{d}{dx}[fgh] = f(x) g(x) h'(x) + f(x) g'(x) h(x) + f'(x) g(x) h(x) \,$

For any number of functions, the derivative of their product is the sum, for each function, of its derivative times each other function.

For quotients, where one function is divided by another function, the equation is more complicated but it is simply a special case of the product rule.

$\frac{f(x)}{g(x)} = f(x) \cdot g(x)^{-1}$

Then we can just use the product rule and the chain rule:

$\frac{d}{dx} \frac{f(x)}{g(x)} = f'(x) \cdot g(x)^{-1} - f(x) \cdot g'(x) \cdot g(x)^{-2}$

We can then multiply through by 1, or more precisely: g(x)² / g(x)², which cancels out into 1, to get:

$\frac{d}{dx} \frac{f(x)}{g(x)} = \frac{f'(x) \cdot g(x)}{g(x)^2} - \frac{ f(x) \cdot g'(x) }{g(x)^{2}}$

This leads us to the so-called "quotient rule":

Derivatives of quotients (Quotient Rule) $\frac{d}{dx} \left[{f(x)\over g(x)}\right] = \frac{f'(x) \cdot g(x) - f(x) \cdot g'(x)}{g(x)^2}\,\!$

Which some people remember with the mnemonic "low D-high minus high D-low over the square of what's below."

Remember: the derivative of a product/quotient is not the product/quotient of the derivatives. (That is, differentiation does not distribute over multiplication or division.) However one can distribute before taking the derivative. That is $\frac{d}{dx}\left((a+b)\times(c+d)\right) = \frac{d}{dx}\left(ac+ad+bc+bd\right)$

Implicit Differentiation

Generally, one will encounter functions expressed in explicit form, that is, y = f(x) form. You might encounter a function that contains a mixture of different variables. Many times it is inconvenient or even impossible to solve for y. A good example is the function $y^2 + 2yx + 3 = 5x \,$. This function is too cumbersome to isolate y alone. Therefore, one can utilize implicit differentiation to find the derivative. To do so, consider y to be a nested function that is defined implicitly by x. You need to employ the chain rule whenever you take the derivative of a variable with respect to a different variable i.e. d/dx (the derivative with respect to x) of x is 1 d/dx (the derivative with respect to x) of y is dy/dx

Remember

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Therefore

$\frac{d}{dx} (y^3) = \frac{d}{dy} (y^3) \frac{dy}{dx}=3y^2 \cdot \frac{dy}{dx} \$

Examples

$xy \,=1$
can be solved as:
$y=\frac{1}{x}$
then differentiated:
$\frac{dy}{dx}=-\frac{1}{x^2}$
However, it can also be differentiated like this: $\frac{d}{dx}[xy]=\frac{d}{dx}[1]$
$x\frac{dy}{dx}+y=0$ (use the product rule)
$\frac{dy}{dx}=-\frac{y}{x}$ (solve for dy/dx)
Note that, if we substitute $y=\frac{1}{x}$ into $\frac{dy}{dx}=-\frac{y}{x}$, we end up with $\frac{dy}{dx}=-\frac{1}{x^2}$ again.

• Find the derivative of $y^2 + x^2$ $= 25$ with respect to x.

You are seeking $dy \over dx$.

Take the derivative of each side of the equation with respect to x.

$\frac{d (y^2 + x^2)}{dx} = \frac{d(25)}{dx}$

$2y \cdot \frac{dy}{dx} + 2x = 0$

$2y \cdot \frac{dy}{dx} = -2x$

$\frac{dy}{dx} = -\frac{x}{y}$

Exponential, logarithmic, and trigonometric functions

Exponential

To determine the derivative of an exponent requires use of the symmetric difference equation for determining the derivative:

$\frac{d}{dx} f(x) = \lim_{h \to 0} \frac{f(x+h) - f(x - h)}{2h}$

First we will solve this for the specific case of an exponent with a base of e and then extend it to the general case with a base of a where a is a positive real number.

First we set up our problem using f(x) = e^x:

$\frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x+h} - e^{x-h}}{2h}$

Then we apply some basic algebra with powers (specifically that a^{b + c} = a^b a^c):

$\frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x} e^{h} - e^{x} e^{-h} }{2h}$

Treating e^x as a constant with respect to what we are taking the limit of, we can use the limit rules to move it to the outside, leaving us with:

$\frac{d}{dx} e^x = e^x \cdot \lim_{h \to 0} \frac{e^{h} - e^{-h} }{2h}$

A careful examination of the limit reveals a hyperbolic sine:

$\frac{d}{dx} e^x = e^x \cdot \lim_{h \to 0} \frac{\sinh(h)}{h}$

The limit of sin(h)/h as h approaches 0 is equal to 1, leaving us with:

Derivative of the exponential function $\frac{d}{dx}e^x = e^x\,\!$

in which $f'(x) = f(x)$.

Now that we have derived a specific case, let us extend things to the general case. Assuming that a is a positive real constant, we wish to calculate:

$\frac{d}{dx}a^x$

One of the oldest tricks in mathematics is to break a problem down into a form that we already know we can handle. Since we have already determined the derivative of e^x, we will attempt to rewrite a^x in that form.

Using that e^ln(c) = c and that ln(a^b) = b · ln(a), we find that:

$a^x = e^{x \cdot \ln(a)}$

Thus, we simply apply the chain rule:

$\frac{d}{dx}e^{x \cdot \ln(a)} = \left[ \frac{d}{dx} x\cdot \ln(a) \right] e^{x \cdot \ln(a)}$

In which we can solve for the derivative and substitute back with e^{x · ln(a)} = a^x to get:

Derivative of the exponential function $\frac{d}{dx}a^x = \ln\left(a\right)a^x\,\!$

Logarithms

Closely related to the exponentiation is the logarithm. Just as with exponents, we will derive the equation for a specific case first (the natural log, where the base is e), and then work to generalize it for any logarithm.

First let us create a variable y such that:

$y = \ln\left(x\right)$

It should be noted that what we want to find is the derivative of y or $\frac{dy}{dx}$.

Next we will put both sides to the power of e in an attempt to remove the logarithm from the right hand side:

$e^y = x$

Now, applying the chain rule and the property of exponents we derived earlier, we take the derivative of both sides:

$\frac{dy}{dx} \cdot e^y = 1$

This leaves us with the derivative:

$\frac{dy}{dx} = \frac{1}{e^y}$

Substituting back our original equation of x = e^y, we find that:

Derivative of the Natural Logarithm $\frac{d}{dx}\ln\left(x\right) = \frac{1}{x}\,\!$

If we wanted, we could go through that same process again for a generalized base, but it is easier just to use properties of logs and realize that:

$\log_b(x) = \frac{\ln(x)}{\ln(b)}$

Since 1 / ln(b) is a constant, we can just take it outside of the derivative:

$\frac{d}{dx}\log_b(x) = \frac{1}{\ln(b)} \cdot \frac{d}{dx} \ln(x)$

Which leaves us with the generalized form of:

Derivative of the Logarithm $\frac{d}{dx}\log_b\left(x\right) = \frac{1}{x\ln\left(b\right)}\,\!$

Trigonometric Functions

Sine, Cosine, Tangent, Cosecant, Secant, Cotangent. These are functions that crop up continuously in mathematics and engineering and have a lot of practical applications. They also appear in more advanced mathematics, particularly when dealing with things such as line integrals with complex numbers and alternate representations of space like spherical and cylindrical coordinate systems.

We use the definition of the derivative, i.e.,

$f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$,

to work these first two out.

Let us find the derivative of sin x, using the above definition.

$f(x) = \sin{x} \, \!$ $f'(x) = \lim_{h \to 0}{\sin(x+h)-\sin{x} \over h}$ Definition of derivative $= \lim_{h \to 0}{\cos(x)\sin(h)+\cos(h)\sin(x) - sin(x) \over h}$ trigonometric identity $= \lim_{h \to 0}{\cos(x)\sin(h)+(\cos(h) - 1)\sin(x) \over h}$ factoring $= \lim_{h \to 0}{\cos(x)\sin(h) \over h}$ $+\lim_{h \to 0}{(\cos(h) - 1)\sin(x) \over h}$ separation of terms $= \cos{x} \, \! \times 1 + \sin{x} \, \! \times 0$ application of limit $= \cos{x} \, \!$ solution

Now for the case of cos x

$f(x) = \cos{x} \, \!$ $f'(x) = \lim_{h \to 0}{\cos(x+h)-\cos{x} \over h}$ Definition of derivative $= \lim_{h \to 0}{\cos(x)\cos(h)-\sin(h)\sin(x) - cos(x) \over h}$ trigonometric identity $= \lim_{h \to 0}{\cos(x)(\cos(h) - 1)-\sin(x)\sin(h) \over h}$ factoring $=\lim_{h \to 0}{\cos(x)(\cos(h) - 1) \over h}$ $- \lim_{h \to 0}{\sin(x)\sin(h) \over h}$ separation of terms $= \cos{x} \, \! \times 0 - \sin{x} \, \! \times 1$ application of limit $= -\sin{x} \, \!$ solution

Therefore we have established

Derivative of Sine and Cosine $\frac{d}{dx} \sin(x) = \cos(x)\,\!$ $\frac{d}{dx} \cos(x) = -\sin(x)\,\!$

To find the derivative of the tangent, we just remember that:

$\tan(x) = \frac{\sin(x)}{\cos(x)}$

which is a quotient. Applying the quotient rule, we get:

$\frac{d}{dx} \tan(x) = \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}$

Then, remembering that $\cos^2(x) + \sin^2(x) = 1$, we simplify:

$\frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)}$	$=\frac{1}{\cos^2(x)}$
	$=\sec^2(x)\,$

Derivative of the Tangent $\frac{d}{dx} \tan(x) = \sec^2(x)\,\!$

For secants, we just need to apply the chain rule to the derivations we have already determined.

$\sec(x) = \frac{1}{\cos(x)}$

So for the secant, we state the equation as:

$\sec(x) = \frac{1}{u}$

$u(x) = \cos(x)$

Take the derivative of both equations, we find:

$\frac{d}{dx} \sec(x) = \frac{-1}{u^2} \cdot \frac{du}{dx}$

$\frac{du}{dx} = -\sin(x)$

Leaving us with:

$\frac{d}{dx} \sec(x) = \frac{\sin(x)}{\cos^2(x)}$

Simplifying, we get:

Derivative of the Secant $\frac{d}{dx} \sec(x) = \sec(x) \tan(x)\,\!$

Using the same procedure on cosecants:

$\csc(x) = \frac{1}{\sin(x)}$

We get:

Derivative of the Cosecant $\frac{d}{dx} \csc(x) = -\csc(x) \cot(x)\,\!$

Using the same procedure for the cotangent that we used for the tangent, we get:

Derivative of the Cotangent $\frac{d}{dx} \cot(x) = -\csc^2(x) \,\!$

Inverse Trigonometric Functions

Arcsine, arccosine, arctangent. These are the functions that allow you to determine the angle given the sine, cosine, or tangent of that angle.

First, let us start with the arcsine such that:

$y=\arcsin(x)$

To find dy/dx we first need to break this down into a form we can work with:

$x = \sin(y)$

Then we can take the derivative of that:

$1 = \cos(y) \cdot \frac{dy}{dx}$

...and solve for dy / dx:

$\frac{dy}{dx} = \frac{1}{\cos(y)}$

At this point we need to go back to the unit triangle. Since y is the angle and the opposite side is sin(y) (which is equal to x), the adjacent side is cos(y) (which is equal to the square root of 1 minus x², based on the pythagorean theorem), and the hypotenuse is 1. Since we have determined the value of cos(y) based on the unit triangle, we can substitute it back in to the above equation and get:

Derivative of the Arcsine $\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1-x^2}}\,\!$

We can use an identical procedure for the arccosine and arctangent:

Derivative of the Arccosine $\frac{d}{dx} \arccos(x) = \frac{-1}{\sqrt{1-x^2}}\,\!$

Derivative of the Arctangent $\frac{d}{dx} \arctan(x) = \frac{1}{1+x^2}\,\!$

Exercises

Given the above rules, practice differentiation on the following.

1. $\frac{d}{dx}[(x^3 + 5)^{10}] \, \!$
2. $\frac{d}{dx}[x^3 + 3x] \, \!$
3. $\frac{d}{dx}[(x+4)\cdot (x+2)\cdot (x-3)]$
4. $\frac{d}{dx}[\frac{x+1}{3x^2}]$
5. $\frac{d}{dx}[3\cdot x^3]$
6. $\frac{d}{dx}[sin x \cdot x^4]$
7. $\frac{d}{dx}[2^x]$
8. $\frac{d}{dx}[e^{x^2}]$
9. $\frac{d}{dx}[ e^{2^x} ]$
10. $\frac{d}{dx}[ x^x ]$

Answers

Click arrow to show answers

1. $30 x^{2} \cdot ( x^3 + 5 )^{9}$
2. $3\cdot x^2+3$
3. $(x - 3)\cdot (x + 2) + (x + 4)\cdot (x + 2) + (x - 3)\cdot (x + 4)$
4. $-\frac{x+2}{3x^3}$
5. $9\cdot x^2$
6. $cos x \cdot x^4 + sin x \cdot 4x^3$
7. $\ln(2) \cdot 2^x$
8. $2x \cdot e^{x^2}$
9. $\ln(2) \cdot 2^x \cdot e^{2^x}$
10. $[\ln(x) + 1] \cdot x^x$

External links

• Online interactive exercises on derivatives
• Visual Calculus - Interactive Tutorial on Derivatives, Differentiation, and Integration

High Order Derivatives

Calculus/Double Derivatives

Newton's Method

Newton's Method

In Calculus, Newton's Method (also called the Newton-Raphson method) is a recursive algorithm for approximating the root of a differentiable function. We know simple formulas for finding the roots of linear and quadratic equations, and there are also more complicated formulae for cubic and quartic equations. At one time it was hoped that there would be formulas found for equations of quintic and higher-degree, though it was later shown by Neils Henrik Abel that no such equations exist. The Newton-Raphson method is a method for approximating the roots of polynomial equations of any order. In fact the method works for any equation, polynomial or not, as long the function is differentiable in a desired interval.

Newton's Method Let $f(x)$ be a differentiable function. Select a point $x_1$ based on a first approximation to the root, arbitrarily close to the function's root. To approximate the root you then recursively calculate using: $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$ As you recursively calculate, the $x_{n}$'s become increasingly better approximations of the functions root. For n number of approximations, $x_n = x_0 - \sum_{i=0}^n \frac{f(x_i)}{f'(x_i)}$

Examples

Find the root of the function $f(x) = x^2 \$.

$x_1 = f(2) = 4 \$

$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 2$

$x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 1$

$x_4 = x_3 - \frac{f(x_3)}{f'(x_3)} = \frac{1}{2}$

$x_5 = x_4 - \frac{f(x_4)}{f'(x_4)} = \frac{1}{4}$

$x_6 = x_5 - \frac{f(x_5)}{f'(x_5)} = \frac{1}{8}$

$x_7 = x_6 - \frac{f(x_6)}{f'(x_6)} = \frac{1}{16}$

$x_8 = x_7 - \frac{f(x_7)}{f'(x_7)} = \frac{1}{32}$

As you can see $x_n$ is gradually approaching zero (which we know is the root of f(x)). One can approach the function's root with arbitrary accuracy.

Answer: $f(x) = x^2 \$ has a root at $x = 0 \$.

Notes

This method fails when $f'(x) = 0$. In that case, one should choose a new starting place. Occasionally it may happen that $f(x) = 0$ and $f'(x) = 0$ have a common root. To detect whether this is true, we should first find the solutions of $f'(x) = 0$, and then check the value of $f(x)$ at these places.

Newton's method also may not converge for every function, take as an example:

$f(x) = \begin{cases} \sqrt{x-r}, & \mbox{for} x \ge r \\ -\sqrt{r-x}, & \mbox{for} x \le r \end{cases}$

For this function choosing any $x_1 = r - h$ then $x_2 = r + h$ would cause successive approximations to alternate back and forth, so no amount of iteration would get us any closer to the root than our first guess.

See Also

• Wikipedia:Newton-raphson method
• Wikipedia:Abel-Ruffini theorem

Related Rates

Introduction

One useful application of derivatives is as an aid in the calculation of related rates. What is a related rate? In each case in the following article the related rate we are calculating is a derivative with respect to some value. We compute this derivative from a rate at which some other known quantity is changing. Given the rate at which something is changing, we are asked to find the rate at which a value related to the rate we are given is changing.

Process for solving related rates problems  :

• Write out any relevant formulas and information.
• Take the derivative of the primary equation.
• Solve for the desired variable.
• Plug-in known information and simplify.

Related Rates

As stated in the introduction, when doing related rates, you generate a function which compares the rate of change of one value with respect to change in time. For example, velocity is the rate of change of distance over time. Likewise, acceleration is the rate of change of velocity over time. Therefore, for the variables for distance, velocity, and acceleration, respectively x, v, and a, and time, t:

$v = \frac{dx}{dt}$

$a = \frac{dv}{dt}$

Using derivatives, you can find the functions for velocity and acceleration from the distance function. This is the basic idea behind related rates: the rate of change of a function is the derivative of that function with respect to time.

Common Applications

Filling Tank

This is the easiest variant of the most common textbook related rates problem: the filling water tank.

• The tank is a cube, with volume 1000L.
• You have to fill the tank in ten minutes or you die.
• You want to escape with your life and as much money as possible, so you want to find the smallest pump that can finish the task.

We need a pump that will fill the tank 1000L in ten minutes. So, for pump rate p, volume of water pumped v, and minutes t:

$p = \frac{dv}{dt}$

Newton's Second Law

Using calculus, Sir Issac Newton was able to develop his Second Law of Motion. Let's play around with them. First, we know that velocity is the derivative of the location function, and acceleration is the derivative of velocity. Newton derived the following equations for mass m, velocity v, force F and momentum p:

$p = mv$

therefore, if Force is a change in momentum,

$F = \frac{dp}{dt}$

Examples

Related rates can get complicated very easily.

Example 1:

A cone with a circular base is being filled with water. You need to find a formula which will find the rate with which water is pumped.

• Write out any relevant formulas or pieces of information.

$V = \frac{1}{3} \pi r^2 h$

• Take the derivative of the equation above with respect to time. Remember to use the Chain Rule and the Product Rule.

$V = \frac{1}{3} \pi r^2h$

$\frac{dV}{dt} = \frac{\pi}{3} \left( r^2 \cdot \frac{dh}{dt} + 2rh \cdot \frac{dr}{dt} \right)$

Answer: $\frac{dV}{dt} = \frac{\pi}{3} \left( r^2 \cdot \frac{dh}{dt} + 2rh \cdot \frac{dr}{dt} \right)$

Example 2:

A spherical hot air balloon is being filled with air. The volume is changing at a rate of 2 cubic feet per minute. 
How is the radius changing with respect to time when the radius is equal to 2 feet?

• Write out any relevant formulas and pieces of information.

$V_{sphere} = \frac{4}{3} \pi r^3$

$\frac{dV}{dt} = 2$

$r = 2 \$

• Take the derivative of both sides of the volume equation with respect to time.

$V = \frac{4}{3} \pi r^3$

$\frac{dV}{dt}$	=	$\frac{4}{3} \cdot 3 \cdot \pi r^2 \cdot \frac{dr}{dt}$
	=	$4 \pi r^2 \cdot \frac{dr}{dt}$

• Solve for $\frac{dr}{dt}.$

$\frac{dr}{dt} = \frac{1}{4 \pi r^2} \cdot \frac{dV}{dt}$

• Plug-in known information.

$\frac{dr}{dt} = \frac{1}{16 \pi} \cdot 2$

Answer: $\frac{dr}{dt} = \frac{1}{8 \pi}$ ft/min.

Example 3:

An airplane is attempting to drop a box onto a house. The house is 300 feet away in horizontal distance and 
400 feet in vertical distance. The rate of change of the horizontal distance with respect to time is the same 
as the rate of change of the vertical distance with respect to time. How is the distance between the box and 
the house changing with respect to time at the moment? The rate of change in the horizontal direction with 
respect to time is -50 feet per second. 

Note: Because the vertical distance is downward in nature, the rate of change of y is negative. Similarly, the horizontal distance is decreasing, therefore it is negative (it is getting closer and closer).

The easiest way to describe the horizontal and vertical relationships of the plane's motion is the Pythagorean Theorem.

• Write out any relevant formulas and pieces of information.

$x^2 + y^2 = s^2 \$ (where s is the distance between the plane and the house)

$x = 300 \$

$y = 400 \$

$s = \sqrt{x^2 + y^2} = \sqrt{300^2 + 400^2} = 500 \$

$\frac{dx}{dt} = \frac{dy}{dt} = -50$

• Take the derivative of both sides of the distance formula with respect to time.

$x^2 + y^2 = s^2 \$

$2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 2s \cdot \frac{ds}{dt}$

• Solve for $\frac{ds}{dt}$.

$\frac{ds}{dt} = \frac{1}{2s} \left( 2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} \right)$

• Plug-in known information

$\frac{ds}{dt}$	=	$\frac{1}{2(500)} \left[ 2(300) \cdot (-50) + 2(400) \cdot (-50) \right]$
	=	$\frac{1}{1000} (-70000)$
	=	$-70 \$ ft/s

Answer: $\frac{ds}{dt} = -70$ ft/sec.

Example 4:

Sand falls onto a cone shaped pile at a rate of 10 cubic feet per minute.  The radius of the pile's base
is always 1/2 of its altitude.  When the pile is 5 ft deep, how fast is the altitude of the pile increasing?

• Write down any relevant formulas and information.

$V = \frac{1}{3} \pi r^2 h$

$\frac{dV}{dt} = 10$

$r = \frac{1}{2} h \$

$h = 5 \$

Substitute $r = \frac{1}{2} h$ into the volume equation.

$V \$	=	$\frac{1}{3} \pi r^2 h$
	=	$\frac{1}{3} \pi h \cdot \left( \frac{h^2}{4} \right )$
	=	$\frac{1}{12} \pi h^3$

• Take the derivative of the volume equation with respect to time.

$V = \frac{1}{12} \pi h^3$

$\frac{dV}{dt} = \frac{1}{4} \pi h^2 \cdot \frac{dh}{dt}$

• Solve for $\frac{dh}{dt}$.

$\frac{dh}{dt} = \frac{4}{\pi h^2} \cdot \frac{dV}{dt}$

• Plug-in known information and simplify.

$\frac{dh}{dt}$	=	$\frac{4}{\pi (5)^2} \cdot 10$
	=	$\frac{8}{5 \pi}$ ft/min

Answer: $\frac{dh}{dt} = \frac{8}{5 \pi}$ ft/min.

Example 5:

A 10 ft long ladder is leaning against a vertical wall.  The foot of the ladder is being pulled away from
the wall at a constant rate of 2 ft/sec.  When the ladder is exactly 8 ft from the wall, how fast is 
the top of the ladder sliding down the wall?

• Write out any relevant formulas and information.

Use the Pythagorean Theorem to describe the motion of the ladder.

$x^2 + y^2 = l^2 \$ (where l is the length of the ladder)

$l = 10 \$

$\frac{dx}{dt} = 2$

$x = 8 \$

$y = \sqrt{l^2 - x^2} = \sqrt{100-64} = \sqrt{36} = 6 \$

• Take the derivative of the equation with respect to time.

$2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0$ ($l^2 \$ so $\frac{dl}{dt} = 0$.)

• Solve for $\frac{dy}{dt}$.

$2x \cdot \frac{dx}{dt} + 2y \cdot \frac{dy}{dt} = 0$

$2y \cdot \frac{dy}{dt} = -2x \cdot \frac{dx}{dt}$

$\frac{dy}{dt} = - \frac{x}{y} \cdot \frac{dx}{dt}$

• Plug-in known information and simplify.

$\frac{dy}{dt}$	=	$\left( - \frac{8}{6} \right) (-2)$
	=	$\frac{8}{3}$ ft/sec

Answer: $\frac{dy}{dt} = \frac{8}{3}$ ft/sec.

Exercises

Problem Set

Here's a few problems for you to try:

1. A spherical balloon is inflated at a rate of 100 $ft^3$/min. Assuming the rate of inflation remains constant, how fast is the radius of the balloon increasing at the instant the radius is 4 ft?
2. Water is pumped from a cone shaped reservoir (the vertex is pointed down) 10 ft in diameter and 10 ft deep at a constant rate of 3 $ft^3$/min. How fast is the water level falling when the depth of the water is 6 ft?
3. A boat is pulled into a dock via a rope with one end attached to the bow of a boat and the other end held by a man standing 6 ft above the bow of the boat. If the man pulls the rope at a constant rate of 2 ft/sec, how fast is the boat moving toward the dock when 10 ft of rope is out?

Solution Set

1. $\frac{25}{16\pi} \frac{ft}{min}$
2. $\frac{1}{3\pi} \frac{ft}{min}$
3. $\frac{5}{2}\frac{ft}{sec}$

Kinematics

Introduction

Kinematics or the study of motion is a very relevant topic in calculus.

This section uses the following conventions:

• $x(t) \$ represents the position equation
• $v(t) \$ represents the velocity equation
• $a(t) \$ represents the acceleration equation

Differentiation

Average Velocity and Acceleration

Average velocity and acceleration problems use the algebraic definitions of velocity and acceleration.

• $v_{avg} = \frac{\Delta x}{\Delta t}$
• $a_{avg} = \frac{\Delta v}{\Delta t}$

Examples

Example 1:

A particle's position is defined by the equation $x(t) = x^3 - 2x^2 + x \$. Find the
average velocity over the interval [2,7].

• Find the average velocity over the interval [2,7]:

$v_{avg} \$	=	$\frac{x(7) - x(2)}{7-2}$
	=	$\frac{252 - 2}{5}$
	=	$50 \$

Answer: $v_{avg} = 50 \$.

Instantaneous Velocity and Acceleration

Instantaneous velocity and acceleration problems use the derivative definitions of velocity and acceleration.

• $v(t) = \frac{dx}{dt}$
• $a(t) = \frac{dv}{dt}$

Examples

Example 2:

A particle moves along a path with a position that can be determined by the function $x(t) = 4t^3 + e^t \$. 
Determine the acceleration when $t = 3$.

• Find $v(t) = \frac{ds}{dt}.$

$\frac{ds}{dt} = 12t^2 + e^t$

• Find $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}.$

$\frac{d^2s}{dt^2} = 24t + e^t$

• Find $a(3) = \frac{d^2s}{dt^2} |_{t=3}$

$\frac{d^2s}{dt^2} |_{t=3}$	=	$24(3) + e^3 \$
	=	$72 + e^3 \$
	=	$92.08553692... \$

Answer: $a(3) = 92.08553692... \$

Integration

• $x_2 - x_1 = \int_{t_1}^{t_2} v(t) dt$
• $v_2 - v_1 = \int_{t_1}^{t_2} a(t) dt$

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Optimization

Introduction

Optimization is the use of Calculus in the real world. Let us assume we are a pizza parlor and wish to maximize profit. Perhaps we have a flat piece of cardboard and we need to make a box with the greatest volume. How does one go about this process?

Obviously, this requires the use of maximums and minimums. We know that we find maximums and minimums via derivatives. Therefore, one can conclude that Calculus will be a useful tool for maximizing or minimizing (also known as "Optimizing") a situation.

Examples

Example 1:

A box manufacturer desires to create a box with a surface area of 100 inches squared. 
What is the maximum size volume that can be formed by bending this material into a box? 
The box is to be closed.  The box is to have a square base, square top, and rectangular sides.

• Write out known formulas and information

$A_{base} = x^2 \$

$A_{side} = x \cdot h \$

$A_{total} = 2x^2 + 4x \cdot h = 100$

$V = l \cdot w \cdot h = x^2 \cdot h$

• Eliminate the variable h in the volume equation

$2x^2 + 4xh = 100 \$

$x^2 + 2xh = 50 \$

$2xh = 50 - x^2 \$

$h = \frac{50 - x^2}{2x}$

$V \$	=	$(x^2) \left( \frac{50 - x^2}{2x} \right)$
	=	$\frac{1}{2} (50x - x^3)$

• Find the derivative of the volume equation in order to maximize the volume

$\frac{dV}{dx} = \frac{1}{2} (50-3x^2)$

• Set $\frac{dV}{dx} = 0$ and solve for $x \$

$\frac{1}{2} (50-3x^2) = 0$

$50-3x^2 = 0 \$

$3x^2 = 50 \$

$x = \pm \frac{\sqrt{50}}{\sqrt{3}}$

• Plug-in the x value into the volume equation and simplify

$V \$	=	$\frac{1}{2} \left[ 50 \cdot \sqrt{\frac{50}{3}} - \left( \sqrt{\frac{50}{3}} \right) ^3 \right]$
	=	$68.04138174.. \$

Answer: $V_{max} = 68.04138174.. \$

Sales Example

A small retailer can sell n units of a product for a revenue of r(n)=8.1n and at a cost of c(n)=n³-7n²+18n, with all amounts in thousands. How many units does it sell to maximize its profit?

The retailer's profit is defined by the equation p(n)=r(n) - c(n), which is the revenue generated less the cost. The question asks for the maximum amount of profit which is the maximum of the above equation. As previously discussed, the maxima and minima of a graph are found when the slope of said graph is equal to zero. To find the slope one finds the derivative of p(n). By using the subtraction rule p'(n)=r'(n) - c'(n):

$p(n)\,$	= $r(n) - c(n)\,$
$p'(n)\,$	= $\frac{d}{dn}\left[8.1n\right]-\frac{d}{dn}\left[n^3-7n^2+18n\right]\,$
	= $-3n^2+14n-9.9\,$

Therefore, when $-3n^2+14n-9.9\,=0$ the profit will be maximized or minimized. Use the quadratic formula to find the roots, giving {3.798,0.869}. To find which of these is the maximum and minimum the function can be tested:

p(0.869) = - 3.97321, p(3.798) = 8.58802

Because we only consider the functions for all n >= 0 (i.e. you can't have n=-5 units), the only points that can be minima or maxima are those two listed above. To show that 3.798 is in fact a maximum (and that the function doesn't remain constant past this point) check if the sign of p'(n) changes at this point. It does, and for n greater than 3.798 P'(n) the value will remain decreasing. Finally, this shows that for this retailer selling 3,798 units would return a profit of $8,588.02.

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Integration

Definition of the Integral

Figure 1

Figure 2

The rough idea of defining the area under the graph of f is to approximate this area with a finite number of rectangles. Since we can easily work out the area of the rectangles we get an estimate of the area under the graph. If we use a larger number of rectangles we expect a better approximation, and the limit as we approach an infinite number of rectangles will give the exact area.

Suppose first that f is positive and a<b. We pick an integer n and divide the interval [a,b] into n subintervals of equal width (see Figure 2). As the interval [a,b] has width b-a each subinterval has width $\Delta x = \frac{b-a}{n}.$ We denote the endpoints of the subintervals by $x_0,x_1,\ldots,x_n$ so

$x_i = a + i \Delta x \mbox{ for } i=0,1,\ldots, n.\,$

Figure 3

Now for each $i=1,\ldots,n$ pick a sample point $x_i^*$ in the interval $[x_{i-1},x_{i}]\!$ and consider the rectangle of height $f(x_i^*)$ and width $\Delta x$ (see Figure 3). The area of this rectangle is $f(x_i^*)\Delta x$. By adding up the area of all the rectangles for $i=1,\ldots,n$ we get that the area S is approximated by

$A_n= f(x_1^*) \Delta x + f(x_2^*) \Delta x + \cdots + f(x_n^*) \Delta x.$

A more convenient way to write this is with the summation notation as

$A_n = \sum_{i=1}^{n} f(x_i^*)\Delta x.$

For each number n we get a different approximation. As n gets larger the width of the rectangles gets smaller which yields a better approximation (see Figures 4 and 5). In the limit as $A_n$ as n tends to infinity we get the area of S.

Definition of the Definite Integral Suppose f is a continuous function on [a,b] and $\Delta x=\frac{b-a}{n}$. Then the definite integral of f between a and b is $\int_{a}^{b} f(x)\ dx = \lim_{n \to \infty} A_n= \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x.$
where $x_i^*$ are any sample points in the interval $[x_{i-1},x_{i}]$.

It is a fact that if f is continuous on [a,b] then this limit always exists and does not depend on the choice of the points $x_i^*\in[x_{i-1},x_{i}]$. For instance they may be evenly spaced, or distributed ambiguously throughout the interval. The proof of this is technical and is beyond the scope of this section.

Notation When considering the expression $\int_{a}^{b} f(x)\ dx$ the function f is called the integrand and the interval [a,b] is the interval of integration. Also a is called the lower limit and b the upper limit of integration.

One important feature of this definition is that we also allow functions which take negative values. If f(x)<0 for all x then $f(x_i^*)<0$ so $f(x_i^*)\Delta x<0$. So the definite integral of f will be strictly negative. More generally if f takes on both positive an negative values then $\int_a^b f(x)dx$ will be the area under the positive part of the graph of f minus the area under the graph of the negative part of the graph (see Figure 6). For this reason we say that $\int_a^b f(x) dx$ is the signed area under the graph.

Figure 6

Figure

The area spanned by $f(x)$

A geometrical proof that anti-derivative gives the area

Suppose we have an function F(x) which returns the area between x and some unknown point u. (Actually, u is the first number before x which satisfies F(u) = 0, but our solution is independent from u, so we won't bother ourselves with it.) We don't even know if something like F exists or not, but we're going to investigate what clue do we have if it does exists.

We can use F to calculate the area between a and b, for instance, which is obviously F(b)-F(a); F is something general. Now, consider a rather peculiar situation, the area bounded at $x$ and $x+\Delta x$, in the limit of $\Delta x \to 0$. Of course it can be calculated by using F, but we're looking for another solution this time. As the right border approaches the left one, the shape seems to be an infinitesimal rectangle, with the height of f(x) and width of $\Delta x$. So, the area reads:

$\mbox{Infinitesimal area} = \lim_{\Delta x \to 0} f(x) \Delta x$

Of course, we could use F to calculate this area as well:

$\mbox{Infinitesimal area} = \lim_{\Delta x \to 0} F(x + \Delta x) - F(x)$

By combining these eqautions, we have

$\lim_{\Delta x \to 0} F(x + \Delta x) - F(x) = \lim_{\Delta x \to 0} f(x) \Delta x$

If we divide both sides by $\Delta x$, we get

$\lim_{\Delta x \to 0} \frac{F(x + \Delta x) - F(x)}{\Delta x} = f(x)$

which is an interesting result, because the left-hand side is the derivative of F with respect to x. This remarkable result doesn't tell us what F itself is, however it tells us what the derivative of F is, and it is f.

Independence of Variable

It is important to notice that the variable x did not play an important role in the definition of the integral. In fact we can replace it with any other letter, so the following are all equal: $\int_a^b f(x) dx = \int_a^b f(t) dt=\int_a^b f(u) du = \int_a^b f(w) dw.$ Each of these is the signed area under the graph of f between a and b.

Left and Right Handed Riemann Sums

We could have decided to choose all our sample points $x_i^*$ to be on the right hand side of the interval $[x_{i-1},x_{i}]$ (see Figure 7). Then $x_i^*=x_{i}$ for all i and the approximation that we called $A_n$ for the area becomes

$A_n = \sum_{i=1}^{n} f(x_{i})\Delta x.$

This is called the right-handed Riemann sum, and the integral is the limit

$\int_{a}^{b} f(x)\ dx = \lim_{n \to \infty} A_n= \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i) \Delta x.$

Alternatively we could have taken each sample point on the left hand side of the interval. In this case $x_i^*=x_{i-1}$ (see Figure 8) and the approximation becomes

$A_n = \sum_{i=1}^{n} f(x_{i-1})\Delta x.$

Then the integral of f is

$\int_{a}^{b} f(x)\ dx = \lim_{n \to \infty} A_n= \lim_{n \to \infty} \sum_{i=1}^{n} f(x_{i-1}) \Delta x.$

The key point is that, as long as f is continuous, these two definitions give the same answer for the integral.

Figure 7

Figure 8

Example 1

In this example we will calculate the area under the curve given by the graph of $f(x) = x$ for x between 0 and 1. First we fix an integer n and divide the interval $[0,1]$ into n subintervals of equal width. So each subinterval has width

$\Delta x = \frac{1}{n}.$

To calculate the integral we will use the right-handed Riemann Sum. (We could have used the left-handed sum instead, and this would give the same answer in the end). For the right-handed sum the sample points are

$x_i^* = 0 + i\Delta x = \frac{i}{n} \quad i=1,\ldots,n$

Notice that $f(x_i^*) = x_i^* = \frac{i}{n}$. Putting this into the formula for the approximation,

$A_n = \sum_{i=1}^n f(x_{i}^*) \Delta x = \sum_{i=1}^n f(i/n)\Delta x = \sum_{i=1}^n \frac{i}{n} \cdot \frac{1}{n} = \frac{1}{n^2} \sum_{i=1}^n i.$

Now we use the formula

$\sum_{i=1}^n i = \frac{n(n+1)}{2}$

to get

$A_n = \frac{1}{n^2} \frac{n(n+1)}{2} = \frac{n(n+1)}{2n^2}.$

To calculate the integral of f between 0 and 1 we take the limit as n tends to infinity,

$\int_0^1 f(x) dx = \lim_{n\to \infty} \frac{n(n+1)}{2n^2} = \frac{1}{2}.$

Example 2

Next we show how to find the integral of the function $f(x) =x^2$ between x=a and x=b. This time the interval [a,b] has width b-a so

$\Delta x = \frac{b-a}{n}.$

Once again we will use the right-handed Riemann Sum. So the sample points we choose are

$x_i^* = a + i\Delta x = a + \frac{i(b-a)}{n}.$

Thus

$A_n\,$	$= \sum_{i=1}^n f(x_{i}^*) \Delta x$
	$=\sum_{i=1}^n f\left(a+\frac{(b-a)i}{n}\right)\Delta x$
	$=\frac{b-a}{n} \sum_{i=1}^n \left(a+\frac{(b-a)i}{n}\right)^2$
	$=\frac{b-a}{n} \sum_{i=1}^n \left( a^2 + \frac{2a(b-a)i}{n} + \frac{(b-a)^2i^2}{n^2} \right)$

We have to calculate each piece on the right hand side of this equation. For the first two,

$\sum_{i=1}^n a^2 = a^2 \sum_{i=1}^n 1 = na^2$

$\sum_{i=1}^n \frac{2a(b-a)i}{n} = \frac{2a(b-a)}{n} \sum_{i=1}^n i = \frac{2a(b-a)}{n}\cdot \frac{n(n+1)}{2}.$

For the third sum we have to use a formula

$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$

to get

$\sum_{i=1}^n \frac{(b-a)^2i^2}{n^2} = \frac{(b-a)^2}{n^2}\frac{n(n+1)(2n+1)}{6}.$

Putting this together

$A_n = \frac{b-a}{n} \left(na^2 + \frac{2a(b-a)}{n}\cdot \frac{n(n+1)}{2} + \frac{(b-a)^2}{n^2}\frac{n(n+1)(2n+1)}{6}\right).$

Taking the limit as n tend to infinity gives

$\int_a^b x^2 dx$	$= (b-a)\left(a^2 + a(b-a) + \frac{1}{3}(b-a)^2\right)$
	$=(b-a)\left( a^2 + ab - a^2 + \frac{1}{3}(b^2 - 2ab + a^2)\right)$
	$=\frac{1}{3}(b-a)(b^2+ab+a^2)$
	$=\frac{1}{3}(b^3-a^3).$

Basic Properties of the Integral

The Constant Rule

From the definition of the integral we can deduce some basic properties. We suppose that f and g are continuous on [a,b].

Integrating Constants
If c is constant then $\int_a^b c dx = c(b-a).$

When $c>0$ and $a<b$ this integral is the area of a rectangle of height c and width b-a which equals c(b-a).

Example

$\int_1^3 9 dx = 9(3-1)=9*2 = 18.$

$\int_{-2}^6 11 dx = 11(6-(-2))=11*8 = 88.$

$\int_{2}^{17} 0 dx = 0*(17-2) =0.$

Constant Rule
$\int_a^b c f(x) dx = c \int_a^b f(x) dx.$

When f is positive, the height of the function cf at a point x is c times the height of the function f. So the area under cf between a and b is c times the area under f. We can also give a proof using the definition of the integral, using the constant rule for limits,

$\int_a^b cf(x) dx = \lim_{n\to \infty} \sum_{i=1}^n cf(x_i^*)=c\lim_{n\to \infty} \sum_{i=1}^n f(x_i^*)=c\int_a^b f(x)dx.$

Example We saw in the previous section that

$\int_0^1 x dx = \frac{1}{2}$.

Using the constant rule we can use this to calculate that

$\int_0^1 3x dx = 3\int_0^1 x dx = 3.\frac{1}{2} = \frac{3}{2},$

$\int_0^1 -7x dx = -7\int_0^1 x dx = (-7).\frac{1}{2} = -\frac{7}{2}.$

Example We saw in the previous section that

$\int_a^b x^2 dx = \frac{1}{3}(b^3-a^3)$. We can use this and the constant rule to calculate that

$\int_1^3 2x^2 dx = 2\int_1^3 x^2 dx = 2.\frac{1}{3}.(3^3-1^3) = \frac{2}{3}(27-1) = \frac{52}{3}.$

The addition and subtraction rule

Addition and Subtraction Rules of Integration
$\int_a^b (f(x) + g(x)) dx = \int_a^b f(x) dx + \int_a^b g(x) dx.$
$\int_a^b (f(x) - g(x)) dx = \int_a^b f(x) dx - \int_a^b g(x) dx.$

As with the constant rule, the addition rule follows from the addition rule for limits:

$\int_a^b (f(x)+g(x)) dx$	=	$\lim_{n\to \infty} \sum_{i=1}^n f(x_i^*) + g(x_i^*)$
	=	$\lim_{n\to \infty} \sum_{i=1}^n f(x_i^*)+\lim_{n\to \infty} \sum_{i=1}^n g(x_i^*)$
	=	$\int_a^b f(x)dx+\int_a^b g(x)dx.$

The subtraction rule can be proved in a similar way.

Example From above $\int_1^3 9 dx = 18$ and $\int_1^3 2x^2 dx = \frac{52}{3}$ so

$\int_1^3 (2x^2 + 9)dx = \int_1^3 2x^2 dx + \int_1^3 9 dx = \frac{52}{3} + 18 = \frac{106}{3},$

$\int_1^3 (2x^2 - 9)dx = \int_1^3 2x^2 dx - \int_1^3 9 dx = \frac{52}{3} - 18 = -\frac{2}{3}.$

Example

$\int_0^2 4x^2 + 14 dx = 4\int_0^2 x^2 dx + \int_0^2 14 dx = 4 \cdot \frac{1}{3}(2^3-0^3) + 2 \cdot 14 = \frac{32}{3} + 28 = \frac{116}{3}.$

The Comparison Rule

Comparison Rule

• Suppose $f(x)\ge 0$ for all x in [a,b]. Then

$\int_a^b f(x) dx \ge 0.$

• Suppose $f(x)\ge g(x)$ for all x in [a,b]. Then

$\int_a^b f(x) dx \ge \int_a^b g(x) dx.$

• Suppose $M\ge f(x)\ge m$ for all x in [a,b]. Then

$M(b-a)\ge \int_a^b f(x) dx \ge m(b-a).$

If $f(x)\ge 0$ then each of the rectangles in the Riemann sum to calculate the integral of f will be above the y axis, so the area will be non-negative. If $f(x)\ge g(x)$ then $f(x)-g(x)\ge 0$ and by linearity of the integral we get the second property. Finally if $M\ge f(x)\ge m$ then the area under the graph of f will be greater than the area of rectangle with height m and less than the area of the rectangle with height M (see Figure 9). So

$M(b-a)=\int_a^b M \ge \int_a^b f(x) dx \ge \int_a^b m = m(b-a).$

Figure 9

Linearity with respect to endpoints

Additivity with respect to endpoints Suppose $a<c<b$. Then

$\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx.$

Again suppose that f is positive. Then this property should be interpreted as saying that the area under the graph of f between a and b is the area between a and c plus the area between c and b (see Figure 8)

Figure 8

Extension of Additivity with respect to limits of integration
When $a=b$ we have that $\Delta x=\frac{b-a}{n}=0$ so

$\int_a^a f(x) dx = 0.$

Also in defining the integral we assumed that a<b. But the definition makes sense even when b<a in which case $\Delta x = \frac{1}{n}(b-a)$ so has changed sign. This gives

$\int_b^a f(x) dx = -\int_a^b f(x) dx.$

With these definitions,

$\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx$

whatever the order of a,b,c.

Fundamental Theorem of Calculus

Statement of the Fundamental Theorem

Suppose that f is continuous on [a,b]. We can define a function F by

$F(x)= \int_{a}^{x} f(t)\ dt \mbox{ for } x \mbox{ in } [a,b].$

Fundamental Theorem of Calculus Part I Suppose f is continuous on [a,b] and F is defined by

$F(x)= \int_{a}^{x} f(t)\ dt.$

Then F is differentiable on (a,b) and for all $x\in(a,b)$,

$F'(x) = f(x).\,$o

Now recall that F is said to be an antiderivative of f if $F'(x) = f(x)\,.$.

Fundamental Theorem of Calculus Part II Suppose that f is continuous on [a,b] and that F is any antiderivative of f. Then

$\ \int_{a}^{b} f(x)\ dx=F(b)-F(a).$

Figure 1

Note: a minority of mathematicians refer to part one as two and part two as one. All mathematicians refer to what is stated here as part 2 as The Fundamental Theorem of Calculus.

Proofs

Proof of Fundamental Theorem of Calculus Part I

Suppose x is in (a,b). Pick $\Delta x$ so that $x+\Delta x$ is also in (a, b). Then

$F(x) = \int_{a}^{x} f(t) dt$

and

$F(x+\Delta x) = \int_{a}^{x+\Delta x} f(t) dt$.

Subtracting the two equations gives

$F(x + \Delta x) - F(x) = \int_{a}^{x + \Delta x} f(t) dt - \int_{a}^{x} f(t) dt.$

Now

$\int_{a}^{x + \Delta x} f(t) dt = \int_{a}^{x} f(t) dt + \int_{x}^{x + \Delta x} f(t) dt$

so rearranging this we have

$F(x + \Delta x) - F(x) = \int_{x}^{x + \Delta x} f(t) dt.$

According to the mean value theorem for integration, there exists a c in [x, x + Δx] such that

$\int_{x}^{x + \Delta x}f(t) dt = f(c) \Delta x$.

Notice that c depends on $\Delta x$. Anyway what we have shown is that

$F(x + \Delta x) - F(x) = f(c) \Delta x \,$,

and dividing both sides by Δx gives

$\frac{F(x + \Delta x) - F(x)}{\Delta x} = f(c)$.

Take the limit as $\Delta x\to 0$ we get the definition of the derivative of F at x so we have

$F'(x) = \lim_{\Delta x \to 0} f(c).$.

To find the other limit, we will use the squeeze theorem. The number c is in the interval [x, x + Δx], so x≤ c ≤ x + Δx. Also, $\lim_{\Delta x \to 0} x = x$ and $\lim_{\Delta x \to 0} x + \Delta x = x$. Therefore, according to the squeeze theorem,

$\lim_{\Delta x \to 0} c = x$.

As f is continuous we have

$F'(x) = \lim_{\Delta x\to 0} f(c)= f\left(\lim_{\Delta x \to 0} c\right) = f(x)$

which completes the proof.

Proof of Fundamental Theorem of Calculus Part II

Define $P(x) = \int_a^x f(t) dt.$ Then by the Fundamental Theorem of Calculus part I we know that P is differentiable on (a,b) and for all $x\in (a,b)$

$P'(x) = f(x).\,$

So P is an antiderivative of f. Now we were assuming that F was also an antiderivative so for all $x\in (a,b)$,

$P'(x) = F'(x) = f(x).\,$

A consequence of the Mean Value Theorem is that this implies there is a constant C such that for all $x\in (a,b)$,

$P(x) = F(x) + C.\,$,

and as P and F are continuous we see this holds when x=a and when x=b as well. Since we know that P(a)=0 we can put x=a into the equation to get 0=F(a) +C so C=-F(a). And putting x=b gives

$\int_a^b f(t) dx = P(b) = F(b) + C = F(b) - F(a).$

Integration of Polynomials

Using the power rule for differentiation we can find a formula for the integal of a power using the Fundamental Theorem of Calculus. Let $f(x) =x^n$. We want to find an antiderivative for f. Since the differentation rule for powers lowers the power by 1 we have that

$\frac{d}{dx} x^{n+1} = (n+1)x^n.$

As long as $n+1\neq 0$ we can divide by n+1 to get

$\frac{d}{dx} \left(\frac{x^{n+1}}{n+1}\right)= x^n = f(x).$

So the function $F(x) = \frac{x^{n+1}}{n+1}$ is an antiderivative of f. If a,b>0 then F is continuous on [a,b] and we can apply the Fundamental Theorem of Calculus we can calculate the integral of f to get the following rule.

Power Rule of Integration I
$\int_a^b x^n dx = \bigg[{x^{n+1} \over {n+1}}\bigg]_a^b = \frac{b^{n+1}-a^{n+1}}{n+1}.$ as long as $n\neq -1$ and $a,b>0.$

Notice that we allow all values of n, even negative or fractional. If n>0 then this works even if a or b are negative.

Power Rule of Integration II
$\int_a^b x^n dx = \bigg[{x^{n+1} \over {n+1}}\bigg]_a^b = \frac{b^{n+1}-a^{n+1}}{n+1}.$ as long as $n>0.$

Example

To find $\int_1^2 x^3 dx$ we raise the power by 1 and have to divide by 4. So $\int_1^2 x^3 dx = \bigg[\frac{x^4}{4}\bigg]_1^2 = \frac{2^4}{4} - \frac{1^4}{4} = \frac{15}{4}.$

Example

The power rule also works for negative powers. For instance $\int_1^3 \frac{1}{x^3} dx = \int_1^3 x^{-3} dx = \bigg[\frac{x^{-2}}{-2}\bigg]_1^3 = \frac{1}{-2}\left(3^{-2}-1^{-2}\right) = -\frac{1}{2}\left(\frac{1}{3^2} - 1\right) =-\frac{1}{2}\left(\frac{1}{9}-1\right) = \frac{1}{2} \cdot \frac{8}{9} = \frac{4}{9}.$

Example

We can also use the power rule for fractional powers. For instance $\int_0^5 \sqrt x dx = \int_0^5 x^{\frac{1}{2}} dx = \bigg[\frac{x^{\frac{3}{2}}}{\frac{3}{2}} dx\bigg]_0^5 = \frac{2}{3}\left(5^{\frac{3}{2}} - 0^{\frac{3}{2}}\right)=\frac{2}{3}\left(5^{\frac{3}{2}}\right)$

Example

Using the linearity rule we can now integrate any polynomial. For example

$\int_0^3 (3x^2 + 4x +2) dx = \bigg[x^3 + 2x^2 + 2x\bigg]_0^3 = 3^3 + 2 \cdot 3^2 + 2 \cdot 3-0 = 27+18+6=51.$

Indefinite Integrals

The Fundamental Theorem of Calculus tells us that if f is continuous then the function $F(x) = \int_a^x f(t) dt$ is an antiderivative of f (i.e. $F'(x) = f(x)$. However it is not the only antiderivative. We can add any constant to F without changing the derivative.

We write $F(x)=\int f(x) dx+C$ if the derivative of F is $F'(x) = f(x)\,.$

Example

Since the derivative of $x^4$ is $4x^3$ the general antiderivative of $4x^3$ is $x^4$ plus a constant. Thus

$\int 4x^3 dx = x^4 + C.$

Example: Finding antiderivatives

Let us return to the previous example, that of $6x^2$. How would we go about finding the integral of this function? Recall the rule from differentiation that

$\frac{d}{dx} x^n = n x^{n-1}\$

In our circumstance, we have:

$\frac{d}{dx} x^3 = 3 x^2\$

This is a start! We now know that the function we seek will have a power of 3 in it. How would we get the constant of 6? Well,

$2 \frac{d}{dx} x^3= 2 \times 3 x^2 =6x^2.\$

Thus, we say that $2x^3$ is an antiderivative of $6x^2$.

Basic Properties of Indefinite Integrals

Constant Rule for indefinite integrals
If c is constant then $\int c f(x) dx = c \int f(x) dx.$

Sum/Difference Rule for indefinite integrals

$\int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx.$

$\int f(x) - g(x) dx = \int f(x) dx - \int g(x) dx.$

Indefinite integrals of Polynomials

Since

$\frac{d}{dx} \frac{1}{n+1}x^{n+1} = x^{n}\$

we have the following rule for indefinite integrals.

Power rule for indefinite integrals If $n\neq -1 ,$ then

$\int x^n {dx}= \frac{1}{n+1} x^{n+1} +C.$

Integral of the Inverse function

Since

$\frac{d}{dx} \ln x = \frac{1}{x}$

We know that

$\int \frac{dx}{x} = \ln \left| x \right| +C$

Note that the polynomial integration rule does not apply when the exponent is -1. This technique of integration must be used instead. Since the argument of the natural logarithm function must be positive (on the real line), the absolute value signs are added around its argument to ensure that the argument is positive.

Integral of Sine and Cosine

In this section we will concern ourselves with determining the integrals of the sin and cosine function.

Recall that

$\frac{d}{dx}\sin{x}= \cos{x} \,$

$\frac{d}{dx}\cos{x}= -\sin{x}. \,$

So sin x is an antiderivative of cos x and -cos x is an antiderivative of sin x. Hence we get the following rules for integrating sin x and cos x.

$\int \cos{x}\ dx= \sin{x} +C$

$\int \sin{x}\ dx= -\cos{x} + C$

We will find how to integrate more complicated trigonometric functions in the chapter on Further integration techniques.

Integral of the Exponential function

Since

$\frac{d}{dx} e^x= e^x$

we see that $e^x$ is its own antiderivative. Perhaps a more useful definition of this rule can be given as:

$\frac{d}{dx} e^{f(x)} = f'(x) \cdot e^{f(x)}$

hence:

$\frac {d}{dx} e^x = (1)e^x$

Where the exponent (x) is differentiated to give a value of 1

Simplified: $\frac {d}{dx} e^x = (1)e^x$

Becomes: $\frac {d}{dx} e^x = e^x$

So the integral of an exponential function can be found thusly:

$\int e^x dx = e^x + C$

Integration Rules

$\int n dx = nx + C$

$\int x^n dx = \frac {x^{n+1}}{n+1} + C, n \ne -1$

$\int e^x dx = e^x + C$

$\int a^x dx = \frac {a^x}{\ln \left( a \right)} + C$

$\int \frac {1}{x} dx = \ln \left| x \right| + C, x > 0$

$\int \sin {x} dx = -\cos {x} + C$

$\int \cos {x} dx = \sin {x} + C$

$\int \tan {x} dx = -\ln \left| \cos{x} \right| + C$

$\int \csc {x} dx = -\ln \left| \csc{x} + \cot {x} \right| + C$

$\int \sec {x} dx = \ln \left| \sec{x} + \tan {x} \right| + C$

$\int \cot {x} dx = \ln \left| \sin{x} \right| + C$

$\int \sec^2 {x} dx = \tan {x} + C$

$\int \csc^2 {x} dx = -\cot {x} + C$

$\int \sec{x} \tan {x} dx = \sec {x} + C$

$\int \csc{x} \cot {x} dx = -\csc {x} + C$

$\int \frac {1}{\sqrt {a^2 - b^2 x^2}} dx = \frac{1}{b} \arcsin {\frac {b x}{a}} + C$

$\int \frac {1}{a^2 + b^2 x^2} dx = \frac {1}{a b} \arctan {\frac {b x}{a}} + C$

$\int \frac {1}{x \sqrt {x^2 - a^2}} dx = \frac {1}{a} \arcsec {\frac {\left| x \right|}{a}} + C$

The Substitution Rule

Suppose that we want to find $\int (\cos(x^2)\cdot x) dx.$

The Fundamental theorem of calculus tells us that we want to find an antiderivative of the function:

$f(x)= \cos (x^2)\cdot x$

Since $\sin(x)$ differentiates to $\cos(x)$ as a first guess we might try the function $\sin (x^2)$. But by the Chain Rule

$\frac{d}{dx} \sin (x^2) = \cos(x^2) \cdot \frac{d}{dx} x^2 = \cos(x^2) \cdot 2x = 2x \cos(x^2)$

which is almost what we wanted apart from the fact that there is an extra factor of 2 in front. But this is easily dealt with because we can divide by any constant so

$\frac{d}{dx}\frac{\sin (x^2)}{2} = \frac{1}{2} \cdot \frac{d}{dx} \sin (x^2) = \frac{1}{2} \cdot 2 \cos(x^2)x=x \cos(x^2)=f(x).$

So using the Fundamental Theorem of Calculus, $\int x \cos(x^2) dx = \frac{\sin(x^2)}{2} + C.$

In fact this technique will work for more general integrands. Suppose u is a differentiable function. Then to evaluate $\int u'(x) cos(u(x)) dx$ we just have to notice that by the Chain Rule

$\frac{d}{dx} \sin(u(x)) = \cos(u(x)) \frac{du}{dx} = u'(x) \cos(u(x)).$

As long at $u'$ is continuous the Fundamental Theorem applies and tells us that

$\int \cos(u(x)) u'(x) dx = \sin(u(x)) + C.$

Now the right hand side of this equation is just the intgral of $\cos(u)$ but with respect to u. If we write u instead of u(x) this becomes $\int \cos(u(x)) u'(x) dx = \sin(u) du + C = \int \cos(u) du.$

So for instance if $u(x) = x^3$ we have worked out that

$\int (\cos(x^3)\cdot 3x^2)dx = \sin(x^3) + C.$

Now there was nothing special about using the cosine function in the discussion above, and it could be replaced by any other function. Doing this gives us the substitution rule for indefinite integrals.

Substitution rule for indefinite integrals Assume u is differentiable with continuous derivative and that f is continuous on the range of u. Then $\int f(u(x)) \frac{du}{dx} dx = \int f(u) du.$

Notice that it looks like you can cancel in the expression $\frac{du}{dx} dx$ to leave just a $du$. This does not really make any sense as $\frac{du}{dx}$ is not a fraction, but is a good way to remember the substitution rule.

There is a similar rule for definite integrals, but we have to change the endpoints.

Substitution rule for definite integrals Assume u is differentiable with continuous derivative and that f is continuous on the range of u. Suppose $c=u(a)$ and $d=u(b)$. Then $\int_a^b f(u(x)) \frac{du}{dx} dx = \int_c^d f(u) du.$

Examples

Consider the integral

$\int_{0}^2 x \cos(x^2+1) \,dx$

By using the substitution u = x² + 1, we obtain du = 2x dx and

$\int_{0}^2 x \cos(x^2+1) \,dx$	$= \frac{1}{2} \int_{0}^2 \cos(x^2+1) 2x \,dx$
	$= \frac{1}{2} \int_{1}^{5}\cos(u)\,du$
	$= \frac{1}{2}(\sin(5)-\sin(1)).$

Note how the lower limit x = 0 was transformed into u = 0² + 1 = 1 and the upper limit x = 2 into u = 2² + 1 = 5.

Proof of the substitution rule

We will now prove the substitution rule for definite integrals. Let F be an anti derivative of f so

$F'(x) = f(x)$. By the Fundamental Theorem of Calculus

$\int_c^d f(u) du = F(d) - F(c).$

Next we define a function $G$ by the rule

$G(x) = F(u(x))\,.$

Then by the Chain rule G is differentiable with derivative

$G'(x) = F'(u(x)) u'(x)= f(u(x)) u'(x)\,.$

Integrating both sides with respect to x and using the Fundamental Theorem of Calculus we get

$\int_a^b f(u(x)) u'(x)dx =\int_a^b G'(x) dx = G(b) - G(a).$

But by the definition of F this equals

$G(b) - G(a) = F(u(b)) - F(u(a)) = F(d) - F(c) = \int_c^d f(u) du.$

Hence

$\int_a^b f(u(x)) u'(x)dx = \int_c^d f(u) du.$

which is the substitution rule for definite integrals.

Integration of even and odd functions

Recall that a function f is called odd if it satisfies $f(-x) = -f(x)$ and is called even if $f(-x) = f(x).$

Suppose f is a continuous odd function then for any a,

$\int_{-a}^a f(x) dx =0.$

If f is a continuous even function then for any a,

$\int_{-a}^a f(x) dx = 2 \int_0^a f(x)dx.$

Caution: For improper integrals (e.g. if a is infinity, or if the function approaches infinity at 0 or a, etc.), the first equation above is only true if $\int_0^a f(x) dx$ exists. Otherwise the integral is undefined, and only the Cauchy principal value is 0.

Suppose f is an odd function and consider first just the integral from -a to 0. We make the substitution u=-x so du=-dx. Notice that if x=-a then u=a and if x=0 then u=0. Hence $\int_{-a}^0 f(x) dx = - \int_a^0 f(-u) du= \int_0^a f(-u) du.$ Now as f is odd, $f(-u) = -f(u)$ so the integral becomes $\int_{-a}^0 f(x) dx = - \int_0^a f(u) du.$ Now we can replace the dummy variable u with any other variable. So we can replace it with the letter x to give $\int_{-a}^0 f(x) dx = - \int_0^a f(u) du = - \int_0^a f(x) dx.$

Now we split the integral into two pieces

$\int_{-a}^a f(x) dx = \int_{-a}^0 f(x) dx+\int_{0}^a f(x) dx = -\int_0^a f(x) dx + \int_0^a f(x) dx =0.$

The proof of the formula for even functions is similar, and is left as an exercise.

Integration by Parts

Integration by parts for indefinite integrals Suppose f and g are differentiable and their derivatives are continuous. Then

$\int f(x)g'(x) dx = f(x)g(x) - \int f'(x) g(x) dx.$

If we write u=f(x) and v=g(x) then using the Leibnitz notation du=f'(x) dx and dv=g'(x) dx and the integration by parts rule becomes

$\int u dv = uv - \int v du.$

For definite integrals the rule is essentially the same, as long as we keep the endpoints.

Integration by parts for definite integrals Suppose f and g are differentiable and their derivatives are continuous. Then

$\int_a^b f(x)g'(x) dx = \bigg[f(x)g(x)\bigg]_a^b - \int_a^b f'(x) g(x) dx$

$=f(b)g(b)-f(a)g(a) - \int_a^b f'(x) g(x) dx$.

This can also be expressed in Leibniz notation.

$\int_a^b udv = \bigg[uv\bigg]_a^b - \int_a^b v du.$

Example Find

$\int x\cos (x) \,dx$

Here we let:

$u = x$, so that $du = dx$,

$dv = \cos(x)dx$ , so that $v = \sin(x)$.

Then:

$\int x\cos (x) \,dx$	$= \int u \,dv$
	$= uv - \int v \,du$

$\int x\cos (x) \,dx = x\sin (x) - \int \sin (x) \,dx$

$\int x\cos (x) \,dx = x\sin (x) + \cos (x) + C$

where C is an arbitrary constant of integration.

Example

$\int x^2 e^x\,dx$

In this example we will have to use integration by parts twice.

Here we let

$u = x^2$, so that $du= 2x dx$,

$dv= e^xdx$, so that $v =e^x$.

Then:

$\int x^2e^x \,dx$	$= \int u \,dv$
	$= uv - \int v \,du$

$\int x^2e^x \,dx = x^2e^x - \int 2xe^x\,dx= x^2 e^x - 2\int xe^x\,dx.$

Now to calculate the last integral we use integration by parts again. Let

$u = ''x$, so that $du= dx$,

$dv= e^xdx$, so that $v =e^x$

and integrating by parts gives

$\int xe^x \,dx = xe^x - \int e^x\,dx= x e^x - e^x.$

So in the end

$\int x^2e^x \,dx = x^2 e^x - 2(x e^x - e^x)=x^2 e^x - 2x e^x + 2e^x=e^x(x^2-2x+2).$

Example Find

$\int \ln (x) \,dx.$

The trick here is to write this integral as

$\int \ln (x) \cdot 1 \,dx.$

Now let

$u = ln(x)$ so $du = 1/x dx$,

$v = x$ so $dv = 1dx$.

Then using integration by parts,

$\int \ln (x) \,dx$	$= x \ln (x) - \int \frac{x}{x} \,dx$
	$= x \ln (x) - \int 1 \,dx$

$\int \ln (x) \,dx = x \ln (x) - {x} + {C}$

$\int \ln (x) \,dx = x ( \ln (x) - 1 ) + C$

where, again, C is an arbitrary constant.

Example Find $\int \arctan(x) dx.$

Again the trick here is to write the integrand as $\arctan(x) = \arctan(x) \cdot 1$. Then let

u = arctan(x); du = 1/(1+x²) dx

v = x; dv = 1·dx

so using integration by parts,

$\int \arctan (x) \,dx$	$= x \arctan (x) - \int \frac{x}{1 + x^2} \,dx$
	$= x \arctan (x) - {1 \over 2} \ln \left( 1 + x^2 \right) + C.$

Example Find :$\int e^{x} \cos (x) \,dx$ This example uses integration by parts twice. First let,

u = e^x; thus du = e^xdx

dv = cos(x)dx; thus v = sin(x)

so

$\int e^{x} \cos (x) \,dx = e^{x} \sin (x) - \int e^{x} \sin (x) \,dx$

Now, to evaluate the remaining integral, we use integration by parts again, with

u = e^x; du = e^xdx

v = -cos(x); dv = sin(x)dx

Then

$\int e^{x} \sin (x) \,dx$	$= -e^{x} \cos (x) - \int -e^{x} \cos (x) \,dx$
	$= -e^{x} \cos (x) + \int e^{x} \cos (x) \,dx$

Putting these together, we get

$\int e^{x} \cos (x) \,dx = e^{x} \sin (x) + e^x \cos (x) - \int e^{x} \cos (x) \,dx$

Notice that the same integral shows up on both sides of this equation. So we can simply add the integral to both sides to get:

$2\int e^{x} \cos (x) \,dx = e^{x} ( \sin (x) + \cos (x) )$

$\int e^{x} \cos (x) \,dx = \frac{e^{x} ( \sin (x) + \cos (x) )}{2}$

Further Integration Techniques

We have already considered two major techniques we can use to solve integration problems, that of the method of substitution, and integration by parts.

Here we make use of these techniques to help us integrate complicated integrals.

Partial fractions

First, an example. Suppose we want to find $\int {3x+ 1 \over x^2+x} dx$. One way to do this is to simplify the integrand by finding constants $A$ and $B$ so that

${3x+ 1 \over x^2+x}={3x+ 1 \over x(x+1)}= {A \over x}+ {B \over x+1}.$

This can be done by cross multiplying the fraction which gives ${3x+1\over x(x+1)} = {{A(x+1) + Bx} \over {x(x+1)}}.$ As both sides have the same denominator we must have $3x+1 = A(x+1)+Bx$. This is an equation for $x$ so must hold whatever value $x$ is. If we put in $x=0$ we get $1 = A$ and putting $x=-1$ gives $-2=-B$ so $B=2$. So we see that

$\frac{3x+ 1}{x^2+x} = \frac{1}{x} + \frac{2}{x+1} \$

Returning to the original integral

$\int \frac{3x+1}{x^2+x} dx$	=	$\int \frac{dx}{x} + \int \frac{2}{x+1} dx$
	=	$\ln \left| x \right| + 2 \ln \left| x+1 \right| + C$

Rewriting the integrand as a sum of simpler fractions has allowed us to reduce the initial integral to a sum of simpler integrals. In fact this method works to integrate any rational function.

Method of Partial Fractions:

• Step 1 Use long division to ensure that the degree of $P(x)$ less than the degree of $Q(x)$.
• Step 2 Factor Q(x) as far as possible.
• Step 3 Write down the correct form for the partial fraction decomposition (see below) and cross multiply to find the constants.

To factor Q(x) we have to write it as a product of linear factors (of the form $ax+b$) and irreducible quadratic factors (of the form $ax^2+bx+c$ with $b^2-4ac<0$).

Some of the factors could be repeated. For instance if $Q(x) = x^3-6x^2+9x$ we factor $Q(x)$ as

$Q(x) = x(x^2-6x+9) = x(x-3)(x-3)=x(x-3)^2.$

It is important that in each quadratic factor we have $b^2-4ac<0$, otherwise it is possible to factor that quadratic piece further. For example if $Q(x) = x^3-3x^2 - 2x$ then we can write

$Q(x) = x(x^2-3x+2) = x(x-1)(x+2)$

We will now show how to write $P(x)/Q(x)$ as a sum of terms of the form

${A \over (ax+b)^k}$ and ${Ax+B \over (ax^2+bx+c)^k}.$

Exactly how to do this depends on the factorization of $Q(x)$ and we now give four cases that can occur.

Case (a) Q(x) is a product of linear factors with no repeats.

This means that $Q(x) = (a_1x+b_1)(a_2x+b_2)...(a_nx+b_n)$ where no factor is repeated and no factor is a multiple of another.

For each linear term we write down something of the form ${A \over (ax+b)}$, so in total we write

${P(x) \over Q(x)} = {A_1 \over (a_1x+b_1)} + {A_2 \over (a_2x+b_2)} + \cdots + {A_n \over (a_nx+b_n)}$

Example 1

Find $\int {1+x^2 \over (x+3)(x+5)(x+7)}dx$

Here we have $P(x)=1+x^2, Q(x)=(x+3)(x+5)(x+7)$ and Q(x) is a product of linear factors. So we write

$\frac{1+x^2}{(x+3)(x+5)(x+7)}=\frac{A}{x+3}+\frac{B}{x+5}+\frac{C}{x+7}$

Multiply both sides by the denominator

$1+x^2=A(x+5)(x+7)+B(x+3)(x+7)+C(x+3)(x+5)$

Substitute in three values of x to get three equations for the unknown constants,

$\begin{matrix} x=-3 & 1+3^2=2\cdot 4 A \\ x=-5 & 1+5^2=-2\cdot 2 B \\ x=-7 & 1+7^2=(-4)\cdot (-2) C \end{matrix}$

so $A=5/4, B=-13/2, C=25/4$, and

$\frac{1+x^2}{(x+3)(x+5)(x+7)}=\frac{5}{4x+12} -\frac{13}{2x+10} +\frac{25}{4x+28}$

We can now integrate the left hand side.

$\int \frac{1+x^2 \, dx}{(x+3)(x+5)(x+7)}= \frac{5}{4} \ln|x+3| - \frac{13}{2}\ln|x+5|+ \frac{25}{4}\ln|x+7|+ C$

Case (b) Q(x) is a product of linear factors some of which are repeated.

If $(ax+b)$ appears in the factorisation of $Q(x)$ k-times. Then instead of writing the piece ${A \over (ax+b)}$ we use the more complicated expression

${A_1\over ax+b} + {A_2\over (ax+b)^2} + {A_3\over (ax+b)^3} + \cdots + {A_k\over (ax+b)^k}$

Example 2

Find $\int {1 \over (x+1)(x+2)^2}dx$

Here P(x)=1" and "Q(x)=(x+1)(x+2)² We write

$\frac{1}{(x+1)(x+2)^2}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}$

Multiply both sides by the denominator $1= A(x+2)^2+B(x+1)(x+2)+C(x+1)$

Substitute in three values of x to get 3 equations for the unknown constants,

$\begin{matrix} x=0 & 1= 2^2A +2B+C \\ x=-1 & 1=A \\ x=-2 & 1= -C \end{matrix}$

so A=1, B=-1, C=-1, and

$\frac{1}{(x+1)(x+2)^2}=\frac{1}{x+1}-\frac{1}{x+2}-\frac{1}{(x+2)^2}$

We can now integrate the left hand side. $\int \frac{1}{(x+1)(x+2)^2} dx= \ln \frac{x+1}{x+2} + \frac{1}{x+2} +C$

Case (c) Q(x) contains some quadratic pieces which are not repeated.

If $(ax^2+bx+c)$ appears we use ${Ax+B \over (ax^2+bx+c)}.$

Case (d) Q(x) contains some repeated quadratic factors.

If $(ax^2+bx+c)$ appears k-times then use

${A_1x+B_1 \over (ax^2+bx+c)} + {A_2x+B_2 \over (ax^2+bx+c)^2} + {A_3x+B_3 \over (ax^2+bx+c)^3} + \cdots + {A_kx+B_k \over (ax^2+bx+c)^k}$

Trigonometric Substitution

If the integrand contains a single factor of one of the forms $\sqrt{a^2-x^2} \mbox{ or } \sqrt{a^2+x^2} \mbox{ or } \sqrt{x^2-a^2}$ we can try a trigonometric substitution.

• If the integrand contains $\sqrt{a^2-x^2}$ let $x = a \sin \theta$ and use the identity $1-\sin^2 \theta = \cos^2 \theta.$
• If the integrand contains $\sqrt{a^2+x^2}$ let $x = a \tan \theta$ and use the identity $1+\tan^2 \theta = \sec^2 \theta.$
• If the integrand contains $\sqrt{x^2-a^2}$ let $x = a \sec \theta$ and use the identity $\sec^2 \theta-1 = \tan^2 \theta.$

Sine substitution

This substitution is easily derived from a triangle, using the Pythagorean Theorem.

If the integrand contains a piece of the form $\sqrt{a^2-x^2}$ we use the substitution

$x=a\sin \theta \quad dx=a \cos \theta d\theta$

This will transform the integrand to a trigonometic function. If the new integrand can't be integrated on sight then the tan-half-angle substitution described below will generally transform it into a more tractable algebraic integrand.

Eg, if the integrand is √(1-x²),

$\begin{matrix} \int_0^1 \sqrt{1-x^2} dx & = & \int_0^{\pi/2} \sqrt{1-\sin^2 \theta} \cos \theta \, d\theta \\ & = & \int_0^{\pi/2} \cos^2 \theta \, d\theta \\ & = & \frac{1}{2} \int_0^{\pi/2} 1+ \cos 2\theta \, d\theta \\ & = & \frac{\pi}{4} \end{matrix}$

If the integrand is √(1+x)/√(1-x), we can rewrite it as

$\sqrt{\frac{1+x}{1-x}} = \sqrt{\frac{1+x}{1+x}\frac{1+x}{1-x}} =\frac{1+x}{\sqrt{1-x^2}}$

Then we can make the substitution

$\begin{matrix} \int_0^a \frac{1+x}{\sqrt{1-x^2}} dx & = & \int_0^\alpha \frac{1+\sin \theta}{\cos \theta} \cos \theta \, d\theta & 0 <a < 1 \\ & = & \int_0^\alpha 1+ \sin \theta \, d\theta & \alpha = \sin^{-1} a \\ & = & \alpha + \left[ - \cos \theta \right]_0^\alpha & \\ & = & \alpha + 1 - \cos \alpha & \\ & = & 1+ \sin^{-1} a - \sqrt{1-a^2} & \\ \end{matrix}$

Tangent substitution

File:Trig Sub Triangle 2.png

When the integrand contains a piece of the form $\sqrt{a^2+x^2}$ we use the substitution

$x = a \tan \theta \quad \sqrt{x^2+a^2} = a \sec \theta \quad dx = a \sec^2 \theta d\theta$

E.g, if the integrand is (x²+a²)^-3/2 then on making this substitution we find

$\begin{matrix} \int_0^z \left( x^2+a^2 \right)^{-\frac{3}{2}}dx & = & a^{-2} \int_0^\alpha \cos \theta \, d\theta & z>0 \\ & = & a^{-2} \left[ \sin \theta \right]_0^\alpha & \alpha = \tan^{-1} (z/a) \\ & = & a^{-2} \sin \alpha & \\ & = & a^{-2} \frac{z/a}{\sqrt{1+z^2/a^2}} & = \frac{1}{a^2} \frac{z}{\sqrt{a^2+z^2}} \\ \end{matrix}$

If the integral is

$I= \int_0^z \sqrt{x^2+a^2} \quad z>0$

then on making this substitution we find

$\begin{matrix} I & = & a^2 \int_0^\alpha \sec^3 \theta \, d\theta & & & \alpha = \tan^{-1} (z/a) \\ & = & a^2 \int_0^\alpha \sec \theta \, d\tan \theta & & & \\ & = & a^2 [ \sec \theta \tan \theta ]_0^\alpha & - & a^2 \int_0^\alpha \sec \theta \tan^2 \theta \, d\theta & \\ & = & a^2 \sec \alpha \tan \alpha & - & a^2 \int_0^\alpha \sec^3 \theta \, d\theta & + a^2 \int_0^\alpha \sec \theta \, d\theta \\ & = & a^2 \sec \alpha \tan \alpha & - & I & + a^2 \int_0^\alpha \sec \theta \, d\theta \\ \end{matrix}$

After integrating by parts, and using trigonometric identities, we've ended up with an expression involving the original integral. In cases like this we must now rearrange the equation so that the original integral is on one side only

$\begin{matrix} I & = & \frac{1}{2}a^2 \sec \alpha \tan \alpha & + & \frac{1}{2}a^2 \int_0^\alpha \sec \theta \, d\theta \\ & = & \frac{1}{2}a^2 \sec \alpha \tan \alpha & + & \frac{1}{2}a^2 \left[ \ln \left( \sec \theta + \tan \theta \right) \right]_0^\alpha \\ & = & \frac{1}{2}a^2 \sec \alpha \tan \alpha & + & \frac{1}{2}a^2 \ln \left( \sec \alpha + \tan \alpha \right) \\ & = & \frac{1}{2}a^2 \left( \sqrt{1+\frac{z^2}{a^2}} \right) \frac{z}{a} & + & \frac{1}{2}a^2 \ln \left( \sqrt{1+\frac{z^2}{a^2}}+\frac{z}{a} \right) \\ & = & \frac{1}{2}z\sqrt{z^2+a^2} & + & \frac{1}{2}a^2 \ln \left(\frac{z}{a} + \sqrt{1+\frac{z^2}{a^2}} \right) \\ \end{matrix}$

As we would expect from the integrand, this is approximately z²/2 for large z.

Secant substitution

This substitution is easily derived from a triangle, using the Pythagorean Theorem.

If the integrand contains a factor of the form $\sqrt{x^2-a^2}$ we use the substitution

$x = a \sec \theta \quad dx = a \sec \theta \tan \theta d\theta \quad \sqrt{x^2-a^2} = \tan \theta.$

Example 1

Find $\int_1^z \frac{\sqrt{x^2-1}}{x}dx.$

$\begin{matrix} \int_1^z \frac{\sqrt{x^2-1}}{x}dx & = & \int_1^\alpha \frac{\tan \theta }{\sec \theta }\sec \theta \tan \theta \,d\theta & z>1 \\ & = & \int_0^\alpha \tan^2 \theta \, d\theta & \alpha = \sec^{-1} z \\ & = & \left[ \tan \theta -\theta \right]_0^\alpha & \tan \alpha = \sqrt{\sec^2 \alpha -1} \\ & =& \tan \alpha -\alpha & \tan \alpha = \sqrt{z^2-1} \\ & =& \sqrt{z^2-1} - \sec^{-1} z & \\ \end{matrix}$

Example 2

Find $\int_1^z \frac{\sqrt{x^2-1}}{x^2} dx.$

$\begin{matrix} \int_1^z \frac{\sqrt{x^2-1}}{x^2}dx & = & \int_1^\alpha \frac{\tan \theta}{\sec^2 \theta}\sec \theta \tan \theta \, d\theta & z>1 \\ & = & \int_0^\alpha \frac{\sin^2 \theta}{\cos \theta} d\theta & \alpha = \sec^{-1} z \\ \end{matrix}$

We can now integrate by parts

$\begin{matrix} \int_1^z \frac{\sqrt{x^2-1}}{x^2}dx & = & -\left[ \tan \theta \cos \theta \right]_0^\alpha + \int_0^\alpha \sec \theta \, d\theta \\ & = & -\sin \alpha +\left[ \ln (\sec \theta + \tan \theta ) \right]_0^\alpha \\ & = & \ln (\sec \alpha + \tan \alpha ) - \sin \alpha \\ & = & \ln (z+ \sqrt{z^2-1} ) - \frac{\sqrt{z^2-1}}{z}\\ \end{matrix}$

Trigonometric integrals

Powers of Sine and Cosine

We will give a general method to solve generally integrands of the form cos^m (x)sinⁿ(x). First let us work through an example.

$\int (\cos^3{x})(\sin^2{x})\,dx$

Notice that the integrand contains an odd power of cos. So rewrite it as

$\int (\cos^2{x})(\sin^2{x}) \cos{x}\,dx$

We can solve this by making the substitution u = sin(x) so du = cos(x) dx. Then we can write the whole integrand in terms of u by using the identity

cos(x)² = 1 - sin²(x)=1-u².

So

$\begin{matrix} \int (\cos^3{x})(\sin^2{x})\,dx &=&\int (\cos^2{x})(\sin^2{x}) \cos{x}\,dx\\ &=&\int (1-u^2)u^2\,du\\ &=&\int u^2\,du - \int u^4\,du\\ &=&{1\over 3} u^3+{1\over 5}u^5 + C\\ &=&{1\over 3} \sin^3{x}- {1\over 5} \sin^5{x} + C \end{matrix}.$

This method works whenever there is an odd power of sine or cosine.

To evaluate $\int (\cos^m{x})(\sin^n{x})\,dx$ when either m or n is odd.

• If m is odd substitute u=sin x and use the identity cos²x = 1 - sin²x=1-u².
• If n is odd substitute u=cos x and use the identity sin²x = 1 - cos²x=1-u².

Example

Find $\int_0^{\pi/2} \cos^{40}(x)\sin^3(x) dx$.

As there is an odd power of sin we let $u=\cos x$ so du = - sin(x)dx. Notice that when x=0 we have u=cos(0)=1 and when $x=\pi/2$ we have $u=\cos(\pi/2) = 0$.

$\begin{matrix} \int_0^{\pi/2} \cos^{40}(x)\sin^3(x) dx &=& \int_0^{\pi/2} \cos^{40}(x)\sin^2(x) \sin(x) dx \\ &=& -\int_{1}^{0} u^{40} (1-u^2) du \\ &=&\int_{0}^{1} u^{40} (1-u^2) du\\ &=& \int_{0}^{1} u^{40} - u^{42} du \\ &=& [\frac{1}{41}u^{41} - \frac{1}{43}u^{43}]_0^1 \\ &=& \frac{1}{41}-\frac{1}{43}. \end{matrix}$

When both m and n are even things get a little more complicated.

To evaluate $\int (\cos^m{x})(\sin^n{x})\,dx$ when both m and n are even.
Use the identities sin²x = 1/2 (1- cos 2x) and cos²x = 1/2 (1+ cos 2x).

Example

Find $\int \sin^2 x\cos^4 x\,dx.$

As sin²x = 1/2 (1- cos 2x) and cos²x = 1/2 (1+ cos 2x) we have

$\int \sin^2 x\cos^4 x\,dx = \int \left( {1 \over 2}(1 - \cos 2x) \right) \left( {1 \over 2}(1 + \cos 2x) \right)^2 \,dx,$

and expanding, the integrand becomes

$\frac{1}{8} \left(\int 1 - \cos^2 2x + \cos 2x- \cos^3 2x \,dx\right).$

Using the multiple angle identities

$\begin{matrix} I & = & \frac{1}{8} \left( \int 1 \, dx - \int \cos^2 2x \, dx + \int \cos 2x \,dx -\int \cos^3 2x \,dx \right) \\ & = & \frac{1}{8} \left( x - \frac{1}{2} \int (1 + \cos{4x})\,dx + \frac{1}{2}\sin{2x} -\int \cos^2 2x \cos 2x \,dx\right) \\ & = & \frac{1}{16} \left( x + \sin{2x} + \int \cos{4x} \,dx -2 \int(1-\sin^2 2x)\cos 2x \,dx\right) \\ \end{matrix}$

then we obtain on evaluating

$I=\frac{x}{16}-\frac{\sin 4x}{64} + \frac{\sin^3 2x }{48}+C$

Powers of Tan and Secant

To evaluate $\int (\tan^m{x})(\sec^n{x})\,dx$.

1. If n is even and $n\ge 2$ then substitute u=tan x and use the identity sec²x = 1 + tan²x.
2. If n and m are both odd then substitute u=sec x and use the identity tan²x = sec²x-1.
3. If n is odd and m is even then use the identity tan²x = sec²x-1 and apply a reduction formula to integrate $\sec^j x dx\,.$

Example 1

Find $\int \sec^2 x dx$.

There is an even power of $\sec x$. Substituting $u=\tan x$ gives $du = \sec^2 x dx$ so

$\int \sec^2 x dx = \int du = u+C = \tan x + C.$

Example 2

Find $\int \tan x dx$.

Let $u=\cos x$ so $du=-\sin x dx$. Then

$\begin{matrix} \int \tan x dx &=& \int \frac{\sin x}{\cos x} dx \\ &=& \int \frac{-1}{u} du \\ &=& -\ln |u| + C \\ &=& -\ln |\cos x | + C\\ &=& \ln |\sec x| +C. \end{matrix}$

Example 3

Find $\int \sec x dx$.

The trick to do this is to multiply and divide by the same thing like this:

$\begin{matrix} \int \sec x dx &=& \int \sec x \frac{\sec x + \tan x}{\sec x + \tan x} dx \\ &=& \int \frac{\sec^2 x + \sec x \tan x}{\sec x+ \tan x} \end{matrix}.$

Making the substitution $u= \sec x + \tan x$ so $du = \sec x\tan x + \sec^2 x dx,$

$\begin{matrix} \int \sec x dx &=& \int \frac{1}{u} du\\ &=& \ln |u| + C \\ &=& \ln |\sec x + \tan x| + C \end{matrix}.$

More trigonometric combinations

For the integrals $\int \sin{nx}\cos{mx}\,dx$ or $\int \sin{nx}\sin{mx}\,dx$ or $\int \cos{nx}\cos{mx}\,dx$ use the identities

• $\sin{a}\cos{b} = {1\over 2}(\sin{(a+b)}+\sin{(a-b)}) \,$
• $\sin{a}\sin{b} = {1\over 2}(\cos{(a-b)}-\cos{(a+b)}) \,$
• $\cos{a}\cos{b} = {1\over 2}(\cos{(a-b)}+\cos{(a+b)}) \,$

Example 1

Find $\int \sin{3x}\cos{5x}\,dx.$

We can use the fact that sin a cos b=(1/2)(sin(a+b)+sin(a-b)), so

$\sin{3x}\cos{5x}=(\sin{8x}+\sin{(-2x)})/2 \,$

Now use the oddness property of sin(x) to simplify

$\sin{3x}\cos{5x}=(\sin{8x}-\sin{2x})/2 \,$

And now we can integrate

$\begin{matrix} \int \sin{3x}\cos{5x}\,dx & = & \frac{1}{2} \int \sin{8x}-\sin{2x}dx \\ & = & \frac{1}{2}(-\frac{1}{8}\cos{8x}+\frac{1}{2}\cos{2x}) +C \\ \end{matrix}$

Example 2

Find:$\int \sin x \sin 2x \,dx$.

Using the identities

$\sin x \sin 2x= \frac{1}{2} \left( \cos (-x)-\cos (3x) \right) = \frac{1}{2} (\cos x -\cos 3x).$

Then

$\begin{matrix} \int \sin{x}\sin{2x}\,dx & = & \frac{1}{2} \int (\cos{x}-\cos{3x})\,dx \\ & = & \frac{1}{2}(\sin{x}-\frac{1}{3}\sin{3x}) + C \end{matrix}$

Reduction formula

A reduction formula is one that enables us to solve an integral problem by reducing it to a problem of solving an easier integral problem, and then reducing that to the problem of solving an easier problem, and so on.

For example, if we let

$I_n = \int x^n e^x\,dx$

Integration by parts allows us to simplify this to

$I_n = x^ne^x - n\int x^{n-1}e^x\,dx=$

$I_n = x^ne^x - nI_{n-1} \,\!$

which is our desired reduction formula. Note that we stop at

$I_0 = e^x \,\!$.

Similarly, if we let

$I_n = \int_0^\alpha \sec^n \theta \, d\theta$

then integration by parts lets us simplify this to

$I_n = \sec^{n-2}\alpha \tan \alpha - (n-2)\int_0^\alpha \sec^{n-2} \theta \tan^2 \theta \, d\theta$

Using the trigonometric identity, tan²=sec²-1, we can now write

$\begin{matrix} I_n & = & \sec^{n-2}\alpha \tan \alpha & + (n-2) \left( \int_0^\alpha \sec^{n-2} \theta \, d\theta - \int_0^\alpha \sec^n \theta \, d\theta \right) \\ & = & \sec^{n-2}\alpha \tan \alpha & + (n-2) \left( I_{n-2} - I_n \right) \\ \end{matrix}$

Rearranging, we get

$I_n=\frac{1}{n-1}\sec^{n-2}\alpha \tan \alpha + \frac{n-2}{n-1} I_{n-2}$

Note that we stop at n=1 or 2 if n is odd or even respectively.

As in these two examples, integrating by parts when the integrand contains a power often results in a reduction formula.

Rational functions in Sine and Cosine

The "tan half angle" substitution

Another useful change of variables is

$t=\tan (x/2) \,$

With this transformation, using the double-angle trig identities,

$\sin x =\frac{2t}{1+t^2} \quad \cos x = \frac{1-t^2}{1+t^2} \quad \tan x = \frac{2t}{1-t^2} \quad dx=\frac{2dt}{1+t^2}$

This transforms a trigonometric integral into a algebraic integral, which may be easier to integrate.

For example, if the integrand is 1/(1 + sin x ) then

$\begin{matrix} \int_0^{\pi/2} \frac{dx}{1 + \sin x} & = & \int_0^1 \frac{2dt}{(1+t)^2} \\ & = & \left[ -\frac{2}{1+t} \right]_0^1 \\ & = & 1 \\ \end{matrix}$

This method can be used to further simplify trigonometric integrals produced by the changes of variables described earlier.

For example, if we are considering the integral

$I = \int_{-1}^1 \frac{\sqrt{1-x^2}}{1+x^2} dx$

we can first use the substition x= sin θ, which gives

$I = \int_{-\pi/2}^{\pi/2} \frac{ \cos^2 \theta}{1+ \sin^2 \theta} d\theta$

then use the tan-half-angle substition to obtain

$I = \int_{-1}^1 \frac{(1-t^2)^2}{1+6t^2+t^4}\frac{2dt}{1+t^2}$

In effect, we've removed the square root from the original integrand. We could do this with a single change of variables, but doing it in two steps gives us the opportunity of doing the trigonometric integral another way.

Having done this, we can split the new integrand into partial fractions, and integrate.

$\begin{matrix} I & = & \int_{-1}^1 \frac{2-\sqrt{2}}{t^2+3-\sqrt{8}}dt + \int_{-1}^1 \frac{2+\sqrt{2}}{t^2+3+\sqrt{8}}dt - \int_{-1}^1 \frac{2}{1+t^2}dt \\ & = &\frac{4-\sqrt{8}}{\sqrt{3-\sqrt{8}}} \tan^{-1} (\sqrt{3+\sqrt{8}}) + \frac{4+\sqrt{8}}{\sqrt{3+\sqrt{8}}} \tan^{-1}(\sqrt{3-\sqrt{8}}) - \pi \\ \end{matrix}$

This result can be further simplified by use of the identities

$3 \pm \sqrt{8}= \left( \sqrt{2} \pm 1 \right)^2 \quad \tan \left( \sqrt{2} \pm 1 \right) = \left( \frac{1}{4} \pm \frac{1}{8} \right)\pi$

ultimately leading to

$I=(\sqrt{2}-1)\pi \,$

In principle, this approach will work with any integrand which is the square root of a quadratic multiplied by the ratio of two polynomials. However, it should not be applied automatically.

E.g, in this last example, once we deduced

$I = \int_{-\pi/2}^{\pi/2} \frac{ \cos^2 \theta}{1+ \sin^2 \theta} d\theta$

we could have used the double angle formulae, since this contains only even powers of cos and sin. Doing that gives

$I = \int_{-\pi/2}^{\pi/2} \frac{1+\cos 2\theta}{3 - \cos 2\theta} d\theta = \frac{1}{2} \int_{-\pi}^\pi \frac{1+\cos \phi}{3 - \cos \phi} d\phi$

Using tan-half-angle on this new, simpler, integrand gives

$\begin{matrix} I & = & \int_{-\infty}^\infty \frac{1}{1+2t^2} \frac{dt}{1+t^2} \\ &= & \int_{-\infty}^\infty \frac{2dt}{1+2t^2} - \int_{-\infty}^\infty \frac{dt}{1+t^2} \end{matrix}$

This can be integrated on sight to give

$I = \frac{4}{\sqrt{2}}\frac{\pi}{2}-2\frac{\pi}{2} = (\sqrt{2}-1) \pi$

This is the same result as before, but obtained with less algebra, which shows why it is best to look for the most straightforward methods at every stage.

A more direct way of evaluating the integral I is to substitute t = tan θ right from the start, which will directly bring us to the line

$I = \int_{-\infty}^\infty \frac{1}{1+2t^2} \frac{dt}{1+t^2}$

above. More generally, the substitution t = tan x gives us

$\sin x =\frac{t}{\sqrt{1+t^2}} \quad \cos x = \frac{1}{\sqrt{1+t^2}} \quad dx=\frac{dt}{1+t^2}$

so this substitution is the preferable one to use if the integrand is such that all the square roots would disappear after substitution, as is the case in the above integral.

Alternate Method

In general, to evaluate integrals of the form

$\int \frac{A + B \cos x + C \sin x}{a + b \cos x + c \sin x} \, dx$,

it is extremely tedious to use the aforementioned "tan half angle" substitution directly, as one easily ends up with a rational function with a 4th degree denominator. Instead, we may first write the numerator as

$A + B \cos x + C \sin x \equiv p (a + b \cos x + c \sin x) + q \frac{d}{dx} (a + b \cos x + c \sin x) + r$.

Then the integral can be written as

$\int \left(p+\frac{q \frac{d}{dx} (a + b \cos x + c \sin x)}{a + b \cos x + c \sin x} + \frac{r}{a + b \cos x + c \sin x} \right) \, dx$

which can be evaluated much more easily.

Example

Evaluate $\int \frac{\cos x + 2}{\cos x + \sin x} \, dx$.

Let

$\cos x + 2 \equiv p (\cos x + \sin x) + q \frac{d}{dx} (\cos x + \sin x) + r$.

Then

$\cos x + 2 \equiv p (\cos x + \sin x) + q (-\sin x + \cos x) + r$

$\cos x + 2 \equiv (p+q) \cos x + (p-q) \sin x + r$.

Comparing coefficients of cos x, sin x and the constants on both sides, we obtain

$\left\{\begin{matrix} p+q & = & 1 \\ p-q & = & 0 \\ r & = & 2 \end{matrix} \right.$

yielding p = q = 1/2, r = 2. Substituting back into the integrand,

$\int \frac{\cos x + 2}{\cos x + \sin x} \, dx = \int \frac{1}{2} \, dx + \frac{1}{2} \int \frac{d(\cos x + \sin x)}{\cos x + \sin x} + \int \frac{2}{\cos x + \sin x} \, dx$.

The last integral can now be evaluated using the "tan half angle" substitution described above, and we obtain

$\int \frac{2}{\cos x + \sin x} \, dx = \sqrt{2} \ln \left| \frac{\tan \frac{x}{2} - 1 + \sqrt{2}}{\tan \frac{x}{2} - 1 - \sqrt{2}} \right| + C$.

The original integral is thus

$\int \frac{\cos x + 2}{\cos x + \sin x} \, dx = \frac{x}{2} + \frac{1}{2} \ln |\cos x + \sin x| + \sqrt{2} \ln \left| \frac{\tan \frac{x}{2} - 1 + \sqrt{2}}{\tan \frac{x}{2} - 1 - \sqrt{2}} \right| + C$.

Irrational functions

Integration of irrational functions is more difficult than rational functions, and many cannot be done. However, there are some particular types that can be reduced to rational forms by suitable substitutions.

Type 1

Integrand contains $\sqrt[n]{\frac{ax+b}{cx+d}}$

Use the substitution $u=\sqrt[n]{\frac{ax+b}{cx+d}}$.

Example

Find $\int \frac{1}{x} \sqrt{\frac{1-x}{x}} \, dx$.

Type 2

Integral is of the form $\int \frac{Px+Q}{\sqrt{ax^2+bx+c}} \, dx$

Write $Px+Q$ as $Px+Q = p \frac{d}{dx} (ax^2+bx+c) + q$.

Example

Find $\int \frac{4x-1}{\sqrt{5-4x-x^2}} \, dx$.

Type 3

Integrand contains $\sqrt{a^2-x^2}$, $\sqrt{a^2+x^2}$ or $\sqrt{x^2-a^2}$

This was discussed in "trigonometric substitutions above". Here is a summary:

1. For $\sqrt{a^2-x^2}$, use $x = a \sin \theta$.
2. For $\sqrt{a^2+x^2}$, use $x = a \tan \theta$.
3. For $\sqrt{x^2-a^2}$, use $x = a \sec \theta$.

Type 4

Integral is of the form $\int \frac{1}{(px+q) \sqrt{ax^2+bx+c}} \, dx$

Use the substitution $u=\frac{1}{px+q}$.

Example

Find $\int \frac{1}{(1+x)\sqrt{3+6x+x^2}} \, dx$.

Type 5

Other rational expressions with the irrational function $\sqrt{ax^2+bx+c}$

1. If $a>0$, we can use $u=\sqrt{ax^2+bx+c} \pm \sqrt{a}x$.
2. If $c>0$, we can use $u=\frac{\sqrt{ax^2+bx+c} \pm \sqrt{c}}{x}$.
3. If $ax^2+bx+c$ can be factorised as $a(x-\alpha)(x-\beta)$, we can use $u=\sqrt{\frac{a(x-\alpha)}{x-\beta}}$.
4. If $a<0$ and $ax^2+bx+c$ can be factorised as $-a(\alpha-x)(x-\beta)$, we can use $x=\alpha\cos^2\theta+\beta\sin^2\theta$,$/theta+\beta.$

Improper Integrals

In a definite integral ($\int_a^b f(x)\,dx$) the function has defined intervals and the function itself is continuous. In this section, we deal with integrals of functions where the interval is infinite (type I) or the function has infinite discontinuity in the intervals [a,b] (type II).

Type I: Infinite Integrals
An integral with infinite region includes ±$\infty$ included in the interval such as $\int_{-\infty}^\infty\ f(x)\,dx$. We cannot simply find the antiderivative and plug in $\infty$. We can however rewrite the integral using a limit. Let $\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{b\rightarrow \infty} \int_1^b \frac{1}{x^2}\,dx$
Now this represents a definite integral so we can find the antiderivative and see if the integral converges. $\lim_{b\rightarrow \infty} \int_1^b \frac{1}{x^2}\,dx = \lim_{b\rightarrow \infty} \bigg[-\frac{1}{x}\bigg]_1^b = \lim_{b\rightarrow \infty} -\frac{1}{b} + 1 = 1$
We can now define the the type 1 integral:
(a) If there is some value $b$ where $b \ge a$ and $\int_a^b f(x)\,dx$ exists, then
$\int_a^\infty f(x)\,dx = \lim_{b\rightarrow \infty} \int_a^b f(x)\,dx$

(b) If there is some value $a$ where $a \le b$ and $\int_a^b f(x)\,dx$ exists, then
$\int_{-\infty}^b f(x)\,dx = \lim_{a\rightarrow {-\infty}} \int_a^b f(x)\,dx$

(c) We can also define $\int_{-\infty}^\infty f(x)\,dx$ as

$\int_{-\infty}^\infty f(x)\,dx = \int_{-\infty}^a f(x)\,dx + \int_a^\infty f(x)\,dx$ assuming that both integrals converge.

**note that if the limits fail to exist, we say that the integral diverges and if the improper integrals yield a finite solution, the integral converges.

Lets look at an example: Evaluate the integral if it converges. $\int_{-\infty}^\infty xe^{x^{-2}}\,dx$
$\int_{-\infty}^\infty xe^{x^{-2}}\,dx = \lim_{a\rightarrow{-\infty}} \int_a^0 xe^{x^{-2}}\,dx + \lim_{b\rightarrow\infty} \int_0^b xe^{x^{-2}}\,dx$ Use the chain rule to find the antiderivative with $u=-x^2, dx=-\frac{du}{2x}$

$\lim_{a\rightarrow\infty} \int_a^0 xe^{x^{-2}}\,dx = \frac{-e^{(0)^{-2}}}{2} - \lim_{a\rightarrow{-\infty}} \frac{-e^{(a)^{-2}}}{2} = (-\frac{1}{2}) - (0)$

$\lim_{b\rightarrow\infty} \int_0^b xe^{x^{-2}}\,dx = \lim_{b\rightarrow\infty} \frac{-e^{(b)^{-2}}}{2} - \frac{-e^{(0)^{-2}}}{2} = (0) + \frac{1}{2}$

$\frac{1}{2} - \frac{1}{2} = 0$ This shows that the integral converges to 0.

Type II: Infinite Discontinuity
Integrating a function that contains a vertical asymptote.

Area

Introduction

Finding the area between two curves, usually given by two explicit functions, is often useful in calculus.

In general the rule for finding the area between two curves is

$A = A_{top} - A_{bottom} \$ or

If f(x) is the upper function and g(x) is the lower function

$A = \int_a^b[f(x)-g(x)]\, dx$

This is true whether the functions are in the first quadrant or not.

Area between two curves

Suppose we are given two functions y₁=f(x) and y₂=g(x) and we want to find the area between them on the interval [a,b]. Also assume that f(x)≥ g(x) for all x on the interval [a,b]. Begin by partitioning the interval [a,b] into n equal subintervals each having a length of Δx=(b-a)/n. Next choose any point in each subinterval, x_i*. Now we can 'create' rectangles on each interval. At the point x_i*, the height of each rectangle is f(x_i*)-g(x_i*) and the width is Δx. Thus the area of each rectangle is [f(x_i*)-g(x_i*)]Δx. An approximation of the area, A, between the two curves is

$A := \sum_{i=1}^{n} [f(x_{i}^{*})-g(x_{i}^{*})]\Delta x$.

Now we take the limit as n approaches infinity and get

$A = \lim_{n \to \infty} \sum_{i=1}^{n} [f(x_{i}^{*})-g(x_{i}^{*})]\Delta x$

which gives the exact area. Recalling the definition of the definite integral we notice that

$A = \int_a^b[f(x)-g(x)]\,dx$.

This formula of finding the area between two curves is sometimes known as applying integration with respect to the x-axis since the rectangles used to approximate the area have their bases lying parallel to the x-axis. It will be most useful when the two functions are of the form y₁=f(x) and y₂=g(x). Sometimes however, one may find it simpler to integrate with respect to the y-axis. This occurs when integrating with respect to the x-axis would result in more than one integral to be evaluated. These functions take the form x₁=f(y) and x₂=g(y) on the interval [c,d]. Note that [c,d] are values of y. The derivation of this case is completely identical. Similar to before, we will assume that f(y)≥ g(y) for all y on [c,d]. Now, as before we can divide the interval into n subintervals and create rectangles to approximate the area between f(y) and g(y). It may be useful to picture each rectangle having their 'width', Δy, parallel to the y-axis and 'height', f(y_i*)-g(y_i*) at the point y_i*, parallel to the x-axis. Following from the work above we may reason that an approximation of the area, A, between the two curves is

$A := \sum_{i=1}^{n} [f(y_{i}^{*})-g(y_{i}^{*})]\Delta y$.

As before, we take the limit as n approaches infinity to arrive at

$A = \lim_{n \to \infty} \sum_{i=1}^{n} [f(y_{i}^{*})-g(y_{i}^{*})]\Delta y$,

which is nothing more than a definite integral, so

$A = \int_c^d[f(y)-g(y)]\,dy$.

Regardless of the form of the functions, we basically use the same formula.

Volume

Volumes

In this section we will learn how to find the volume of a shape. The procedure is very similar to calculating the Area. The basic procedure is:

• Partition the shape in
• Calculate basal area of every partition
• Multiply by height of the partition
• Sum up all the volumes

So, given a function $f(x)$ that gives us the basal area at a given height x, we can write it up as follows:

$\sum_{i=1}^{n} f(x_{i})\Delta x$ Now limit it to infinity:

$\lim_{n \to \infty} \sum_{i=1}^{n} f(x_{i})\Delta x$ This is a Riemann's Sum, so we can rewrite it as:

$\int_a^b f(x)\,dx$

Examples

Calculate the volume of a square pyramid of base b and height h.

The basal shape is a square, and depends on the height x at which it is taken. For simplicity, we will consider an inverted pyramid, so that $f(x)=\left(\frac{b}{h}x\right)^2$. We integrate in the proper range (0 to h):

$\int_0^h \left(\frac{b}{h}x\right)^2\,dx = \left(\frac{b}{h}\right)^2\int_0^h x^2\,dx = \left(\frac{b}{h}\right)^2 \frac{h^3}{3} = \frac{b^2\cdot h}{3}$

Volume of solids of revolution

Revolution solids

A solid is said to be of revolution when it is formed by rotating a given curve against an axis. For example, rotating a circle positioned at (0,0) against the y-axis would create a revolution solid, namely, a sphere.

Calculating the volume

Calculating the volume of this kind of solid is very similar to calculating any volume: we calculate the basal area, and then we integrate through the height of the volume.

Say we want to calculate the volume of the shape formed rotating over the x-axis the area contained between the curves $f(x)$ and $g(x)$ in the range [a,b]. First calculate the basal area:

$|\pi f(x)^2 - \pi g(x)^2|$

And then integrate in the range [a,b]:

$\int_a^b |\pi f(x)^2 - \pi g(x)^2|\,dx = \pi \int_a^b |f(x)^2-g(x)^2|\,dx$

Alternatively, if we want to rotate in the y-axis instead, $f$ and $g$ must be invertible in the range [a,b], and, following the same logic as before:

$\pi \int_a^b |{f^{-1}(x)}^2-{g^{-1}(x)}^2|\,dx$

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Arc length

Suppose that we are given a function f and we want to calculate the length of the curve drawn out by the graph of f. If the graph were a straight line this would be easy -- the formula for the length of the line is given by pythagoras' theorem. And if the graph were a polygon we can calculate the length by adding up the length of each piece.

The problem is that most graphs are not polygons. Nevertheless we can estimate the length of the curve by approximating it with straight lines. Suppose the curve C is given by the formula y=f(x) for $a\le x\le b$. We divide the interval [a,b] into n subintervals with equal width $\Delta x$ and endpoints $x_0,x_1,\ldots,x_n$. Now let $y_i=f(x_i)$ so $P_i=(x_i,y_i)$ is the point on the curve above $x_i$. The length of the straight line between $P_i$ and $P_{i+1}$ is

$|P_iP_{i+1}| = \sqrt{ (y_{i+1} - y_{i})^2 + (x_{i+1}-{x_i})^2}.$

So an estimate of the length of the curve C is the sum

$\sum_{i=0}^{n-1} |P_iP_{i+1}|$

As we divide the interval [a,b] into more pieces this gives a better estimate for the length of C. In fact we make that a definition.

Definition (Length of a Curve)

The length of the curve y=f(x) for $a\le x\le b$ is defined to be $L = \lim_{n\to \infty} \sum_{i=0}^{n-1} |P_{i+1}P_i|.$

The Arclength Formula

Suppose that $f'$ is continuous on [a,b]. Then the length of the curve given by $y=f(x)$ between a and b is given by

$L = \int_a^b \sqrt{ 1+ (f'(x))^2 } dx$

And in Leibniz notation

$L = \int_a^b \sqrt{ 1+ \left(\frac{dy}{dx}\right)^2 } dx$

Proof: Consider $y_{i+1} - y_i= f(x_{i+1}) -f(x_i).$ By the Mean Value Theorem there is a point $z_i$ in $(x_{i+1},x_i)$ such that

$y_{i+1} - y_i= f(x_{i+1}) -f(x_i)= f'(z_i) (x_{i+1}-x_i)\,$.

So

$|P_iP_{i+1}|\,$	=	$\sqrt{ (y_{i+1} - y_{i})^2 + (x_{i+1}-x_i)^2}$
	=	$\sqrt{ (f'(z_i))^2 (x_{i+1}-x_i)^2 + (x_{i+1}-x_i)^2 }$
	=	$\sqrt{ (1+(f'(z_i))^2) (x_{i+1}-x_i)^2 }$
	=	$\sqrt{ (1+(f'(z_i))^2)} \Delta x.$

Putting this into the definiton of the length of C gives

$L=\lim_{n\to \infty} \sum_{i=0}^{n-1} \sqrt{ (1+(f'(z_i))^2)} \Delta x.$

Now this is the definition of the integral of the function $g(x) = \sqrt{1+(f'(x))^2}$ between a and b (notice that g is continuous because we are assuming that $f'$ is continuous). Hence

$L=\int_a^b \sqrt{1+(f'(x))^2} dx$

as claimed.

Arclength of a parametric curve

For a parametric curve, that is, a curve defined by $x=f(t)$ and $y=g(t)$, the formula is slightly different:

$L = \int_a^b \sqrt{ (f'(t))^2 + (g'(t))^2 }\,dt$

Proof: The proof is analogous to the previous one: Consider $y_{i+1} - y_i= g(t_{i+1}) -g(t_i)$ and $x_{i+1} - x_i= f(t_{i+1}) -f(t_i)$. By the Mean Value Theorem there are points $c_i$ and $d_i$ in $(t_{i+1},t_i)$ such that

$y_{i+1} - y_i= g(t_{i+1}) -g(t_i)= g'(c_i) (t_{i+1}-t_i)\,$ and

$x_{i+1} - x_i= f(t_{i+1}) -f(t_i)= f'(d_i) (t_{i+1}-t_i)\,$.

So

$|P_iP_{i+1}|\,$	=	$\sqrt{ (y_{i+1} - y_{i})^2 + (x_{i+1}-x_i)^2}$
	=	$\sqrt{ (g'(c_i))^2 (t_{i+1}-t_i)^2 + (f'(d_i))^2 (t_{i+1}-t_i)^2 }$
	=	$\sqrt{ (f'(d_i))^2)+(g'(c_i))^2) (t_{i+1}-t_i)^2 }$
	=	$\sqrt{ (f'(d_i))^2 + (g'(c_i))^2} \Delta t.$

Putting this into the definiton of the length of the curve gives

$L=\lim_{n\to \infty} \sum_{i=0}^{n-1} \sqrt{ (f'(d_i))^2 + (g'(c_i))^2 } \Delta t.$

This is equivalent to:

$L=\int_a^b \sqrt{ (f'(t))^2 + (g'(t))^2}\,dt$

Surface area

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Suppose we are given a function f and we want to calculate the surface area of the function f rotated around a given line. The calculation of surface area of revolution is related to the arc length calculation.

If the function f is a straight line, other methods such as surface area formulas for cylinders and conical frustra can be used. However, if f is not linear, an integration technique must be used.

Recall the formula for the lateral surface area of a conical frustrum:

$A = 2 \pi r l \,$

where r is the average radius and l is the slant height of the frustrum.

For y=f(x) and $a\le x\le b$, we divide [a,b] into subintervals with equal width Δx and endpoints $x_0,x_1,\ldots,x_n$. We map each point $y_i= f(x_i) \,$ to a conical frustrum of width Δx and lateral surface area $A_i \,$.

We can estimate the surface area of revolution with the sum

$A = \sum_{i=0}^{n} A_i$

As we divide [a,b] into smaller and smaller pieces, the estimate gives a better value for the surface area.

Definition (Surface of Revolution)

The surface area of revolution of the curve y=f(x) about a line for $a\le x\le b$ is defined to be

$A = \lim_{n\to \infty} \sum_{i=0}^{n} A_i$

The Surface Area Formula

Suppose f is a continuous function on the interval [a,b] and r(x) represents the distance from f(x) to the axis of rotation. Then the lateral surface area of revolution about a line is given by

$A = 2 \pi \int_a^b r(x) \sqrt{ 1+ (f'(x))^2 } dx$

And in Leibniz notation

$A = 2 \pi \int_a^b r(x) \sqrt{ 1+ \left(\frac{dy}{dx}\right)^2 } dx$

Proof:

$A \$	=	$\lim_{n \to \infty} \sum_{i=1}^n A_i$
	=	$\lim_{n \to \infty} \sum_{i=1}^n 2 \pi r_i l_i$
	=	$2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i l_i$

As $n \rightarrow \infty$ and $\Delta x \rightarrow 0$, we know two things:

1. the average radius of each conical frustrum $r_i$ approaches a single value

2. the slant height of each conical frustrum $l_i$ equals an infitesmal segment of arc length

From the arc length formula discussed in the previous section, we know that

$l_i = \sqrt{ 1+ (f'(x_i))^2 }$

Therefore

$A \$	=	$2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i l_i$
	=	$2 \pi \lim_{n \to \infty} \sum_{i=1}^n r_i \sqrt{ 1+ (f'(x_i))^2} \Delta x$

Because of the definition of an integral $\int_a^b f(x) dx = \lim_{n \to \infty} \sum_{i=1}^n f(c_i) \Delta x_i$, we can simplify the sigma operation to an integral.

$A = 2 \pi \int_a^b r(x) \sqrt{ 1+ (f'(x))^2 } dx$

Or if f is in terms of y on the interval [c,d]

$A = 2 \pi \int_c^d r(y) \sqrt{ 1+ (f'(y))^2} dy$

Work

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$W = \int F\,dr=\int ma\,dr=\int m \frac{dv}{dt}dr=m\int \frac{dr}{dt}dv=m\int v dv=\frac{1} {2} mv^2=\Delta E_k$

Centre of mass

Calculus/Centre of mass

Pressure and force

Calculus/Pressure and force

Probability

Calculus/Probability

The logarithm defined as an integral

Calculus/The logarithm defined as an integral

Calculus/Integration/Applications/Contents

Series

Introduction

An arithmetic series is the sum of a sequence of terms with a common difference. A geometric series is the sum of terms with a common ratio. For example, an interesting series which appears in many practical problems in science, engineering, and mathematics is the geometric series $r + r^2 + r^3 + r^4 + ...$ where the $...$ indicates that the series continues indefinitely. A common way to study a particular series (following Cauchy) is to define a sequence consisting of the sum of the first $n$ terms. For example, to study the geometric series we can consider the sequence which adds together the first n terms:

$S_n(r) = \sum_{i=1}^{n} r^i.$

Generally by studying the sequence of partial sums we can understand the behavior of the entire infinite series.

Two of the most important questions about a series are

• Does it converge?
• If so, what does it converge to?

For example, it is fairly easy to see that for $r > 1$, the geometric series $S_n(r)$ will not converge to a finite number (i.e., it will diverge to infinity). To see this, note that each time we increase the number of terms in the series, $S_n(r)$ increases by $r^{n+1}$, since $r^{n+1} > 1$ for all $r > 1$ (as we defined), $S_n(r)$ must increase by a number greater than one every term. When increasing the sum by more than one for every term, it will diverge.

Perhaps a more surprising and interesting fact is that for $|r| < 1$, $S_n(r)$ will converge to a finite value. Specifically, it is possible to show that

$\lim_{n \to \infty} S_n(r) = \frac{r}{1-r}.$

Indeed, consider the quantity

$(1-r)S_n(r) = (1-r)\sum_{i=1}^{n} r^n = \sum_{i=1}^{n} r^n - \sum_{i=2}^{n+1} r^n = r - r^{n+1}$

Since $r^{n+1}\to 0$ as $n\to \infty$ for $|r|<1$, this shows that $(1-r)S_n(r)\to r$ as $n\to \infty$. The quantity $1-r$ is non-zero and doesn't depend on $n$ so we can divide by it and arrive at the formula we want.

We'd like to be able to draw similar conclusions about any series.

Unfortunately, there is no simple way to sum a series. The most we will be able to do in most cases is determine if it converges. The geometric and the telescoping series are the only types of series in which we can easily find the sum of.

Convergence

It is obvious that for a series to converge, the a_n must tend to zero (because sum of any inifinite terms is infinity, except when the sequence approaches 0), but even if the limit of the sequence is 0, is not sufficient to say it converges.

Consider the harmonic series, the sum of 1/n, and group terms

$\begin{matrix} \sum_1^{2^m} \frac{1}{n} &= 1+ \frac{1}{2}+ & \frac{1}{3}+\frac{1}{4} +\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+ & \ldots & + \sum_{1+2^{n-1}}^{2^n} \frac{1}{p} \\ & > \frac{3}{2}+ & \frac{1}{4}2+ & \frac{1}{8}4+ & \ldots & + \frac{1}{2^n}2^{n-1} \\ & = \frac{3}{2}+ & \frac{1}{2} + & \frac{1}{2} + & \ldots & + \frac{1}{2} \quad (m \mbox{ terms}) \end{matrix}$

As m tends to infinity, so does this final sum, hence the series diverges.

We can also deduce something about how quickly it diverges. Using the same grouping of terms, we can get an upper limit on the sum of the first so many terms, the partial sums.

$1+\frac{m}{2} \le \sum_1^{2^m} \frac{1}{n} \le 1+m$

or

$1+\frac{\ln_2 m}{m}\le \sum_1^m \frac{1}{n} \le 1+ \ln_2 m$

and the partial sums increase like log m, very slowly.

Notice that to discover this, we compared the terms of the harmonic series with a series we knew diverged.

This is a convergence test (also known as the direct comparison test) we can apply to any pair of series.

• If b_n converges and |a_n|≤|b_n| then a_n converges.
• If b_n diverges and |a_n|≥|b_n| then a_n diverges.

There are many such tests, the most important of which we'll describe in this chapter.

Absolute convergence

Theorem: If the series of absolute values, $\sum_{n=1}^\infty \left| a_n \right|$, converges, then so does the series $\sum_{n=1}^\infty a_n$

We say such a series converges absolutely.

Proof:

Let $\epsilon>0$

According to the Cauchy criterion for series convergence, exists $N$ so that for all $N<m,n$:

$\sum_{k=n}^{m}|a_{k}|<\epsilon$

We know that:

$|\sum_{k=n}^{m}a_{k}|\leq\sum_{k=n}^{m}|a_{k}|$

And then we get:

$|\sum_{k=n}^{m}a_{k}|\leq\sum_{k=n}^{m}|a_{k}|<\epsilon$

Now we get:

$|\sum_{k=n}^{m}a_{k}|<\epsilon$

Which is exactly the Cauchy criterion for series convergence.

$Q.E.D$

The converse does not hold. The series 1-1/2+1/3-1/4 ... converges, even though the series of its absolute values diverges.

A series like this that converges, but not absolutely, is said to converge conditionally.

If a series converges absolutely, we can add terms in any order we like. The limit will still be the same.

If a series converges conditionally, rearranging the terms changes the limit. In fact, we can make the series converge to any limit we like by choosing a suitable rearrangement.

E.g, in the series 1-1/2+1/3-1/4 ..., we can add only positive terms until the partial sum exceeds 100, subtract 1/2, add only positive terms until the partial sum exceeds 100, subtract 1/4, and so on, getting a sequence with the same terms that converges to 100.

This makes absolutely convergent series easier to work with. Thus, all but one of convergence tests in this chapter will be for series all of whose terms are postive, which must be absolutely convergent or divergent series. Other series will be studied by considering the corresponding series of absolute values.

Ratio test

For a series with terms a_n, all positive, if

$\lim_{n \to \infty } \frac{a_{n+1}}{a_n} = r$

then

• the series converges if r<1
• the series diverges if r>1
• the series could do either if r=1, the test is not conclusive in this case.

E.g, suppose

$a_n=\frac{n!n!}{(2n)!}$

then

$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{(2n+1)(2n+2)}=\frac{n+1}{4n+2} \to \frac{1}{4}$

so this series converges.

Integral test

If f(x) is a monotonically decreasing, always positive function, then the series

$\sum_{n=1}^\infty f(n)$

converges if and only if the integral

$\int_1^\infty f(x)dx$

converges.

E.g, consider f(x)=1/x^p, for a fixed p.

• If p=1 this is the harmonic series, which diverges.
• If p<1 each term is larger than the harmonic series, so it diverges.
• If p>1 then

$\begin{matrix}\int_1^\infty x^{-p}dx & = & \lim_{s \to \infty}\int_1^s x^{-p}dx & \\ & = & \lim_{s \to \infty } \left. \frac{-1}{(p-1)x^{p-1}} \right|^s_1 & \\ & = & \lim_{s \to \infty } \left( \frac{1}{p-1}-\frac{1}{(p-1)s^{p-1}} \right) & =\frac{1}{p-1} \end{matrix}$

The integral converges, for p>1, so the series converges.

We can prove this test works by writing the integral as

$\int_1^\infty f(x)dx=\sum_{n=1}^\infty \int_n^{n+1} f(x)dx$

and comparing each of the integrals with rectangles, giving the inequalities

$f(n) \ge \int_n^{n+1} f(x)dx \ge f(n+1)$

Applying these to the sum then shows convergence.

Limit Comparison

• If b_n converges, and the limit

$\lim_{n \to \infty} \frac{a_n}{b_n}$

exists and is not zero, then a_n converges

• If c_n diverges, and

$\lim \frac{|a_n|}{c_n} > 0$

then a_n diverges

Example:

$a_n = n^{-\frac{n+1}{n}}$

For large n, the terms of this series are similar to, but smaller than, those of the harmonic series. We compare the limits.

$\lim \frac{|a_n|}{c_n} = \lim \frac{n}{n^{\frac{n+1}{n}}} = \lim \frac {1}{n^{\frac {1}{n}}}=1>0$

so this series diverges.

Alternating series

If the signs of the a_n alternate,

$a_n=(-1)^n |a_n| \,$

then we call this an alternating series. The series sum converges provided that

$\lim_{n \to \infty}a_n=0$ and $\ |a_{n+1}| < |a_n|$.

The error in a partial sum of an alternating series is smaller than the first omitted term.

$\left| \sum_{n=1}^\infty a_n - \sum_{n=1}^m a_n \right| < |a_{m+1}|$

Geometric series

The geometric series can take either of the following forms

$\sum_{n=0}^\infty ar^n$ or $\sum_{n=1}^\infty ar^{n-1}$

As you have seen at the start, the sum of the geometric series is

$\ S_n = \frac{a}{1-r}\quad\mbox{ for } \left| r \right| < 1$.

Telescoping series

$\sum_{n=0}^\infty (b_n - b_{n+1})$

Expanding (or "telescoping") this type of series is informative. If we expand this series, we get:

$\sum_{n=0}^k (b_n - b_{n+1})= (b_0 - b_1) + (b_1 - b_2) + ... + (b_{k-1} - b_k)$

Additive cancellation leaves:

$\sum_{n=0}^k (b_n - b_{n+1})= b_0 - b_k$

Thus,

$\sum_{n=0}^\infty (b_n - b_{n+1}) = \lim_{k \to \infty} \sum_{n=0}^k (b_n - b_{n+1}) = \lim_{k \to \infty} (b_0 - b_k) = b_0 - \lim_{k \to \infty} b_k$

and all that remains is to evaluate the limit.

There are other tests that can be used, but these tests are sufficient for all commonly encountered series.

Theory

Taylor Series

sin(x) and Taylor approximations, polynomials of degree 1, 3, 5, 7, 9, 11 and 13.

In mathematics, the Taylor series of an infinitely often differentiable real (or complex) function f defined on an open interval (a-r, a+r) is the power series

$\sum_{n=0}^{\infin} \frac{f^{(n)}(a)}{n!} (x-a)^{n}$

Here, n! is the factorial of n and f ⁽ⁿ⁾(a) denotes the nth derivative of f at the point a. If this series converges for every x in the interval (a-r, a+r) and the sum is equal to f(x), then the function f(x) is called analytic. To check whether the series converges towards f(x), one normally uses estimates for the remainder term of Taylor's theorem. A function is analytic if and only if a power series converges to the function; the coefficients in that power series are then necessarily the ones given in the above Taylor series formula.

If a = 0, the series is also called a Maclaurin series.

The importance of such a power series representation is threefold. First, differentiation and integration of power series can be performed term by term and is hence particularly easy. Second, an analytic function can be uniquely extended to a holomorphic function defined on an open disk in the complex plane, which makes the whole machinery of complex analysis available. Third, the (truncated) series can be used to approximate values of the function near the point of expansion.

The function e^-1/x² is not analytic: the Taylor series is 0, although the function is not.

Note that there are examples of infinitely often differentiable functions f(x) whose Taylor series converge, but are not equal to f(x). For instance, for the function defined piecewise by saying that f(x) = exp(−1/x²) if x ≠ 0 and f(0) = 0, all the derivatives are zero at x = 0, so the Taylor series of f(x) is zero, and its radius of convergence is infinite, even though the function most definitely is not zero. This particular pathology does not afflict complex-valued functions of a complex variable. Notice that exp(−1/z²) does not approach 0 as z approaches 0 along the imaginary axis.

Some functions cannot be written as Taylor series because they have a singularity; in these cases, one can often still achieve a series expansion if one allows also negative powers of the variable x; see Laurent series. For example, f(x) = exp(−1/x²) can be written as a Laurent series.

The Parker-Sockacki theorem is a recent advance in finding Taylor series which are solutions to differential equations. This theorem is an expansion on the Picard iteration.

List of Taylor series

Several important Taylor series expansions follow. All these expansions are also valid for complex arguments x.

Exponential function and natural logarithm:

$e^{x} = \sum^{\infin}_{n=0} \frac{x^n}{n!}\quad\mbox{ for all } x$

$\ln(1+x) = \sum^{\infin}_{n=1} \frac{(-1)^{n+1}}n x^n\quad\mbox{ for } \left| x \right| < 1$

Geometric series:

$\frac{1}{1-x} = \sum^{\infin}_{n=0} x^n\quad\mbox{ for } \left| x \right| < 1$

Binomial series:

$(1+x)^\alpha = \sum^{\infin}_{n=0} C(\alpha,n) x^n\quad\mbox{ for all } \left| x \right| < 1\quad\mbox{ and all complex } \alpha$

Trigonometric functions:

$\sin x = \sum^{\infin}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}\quad\mbox{ for all } x$

$\cos x = \sum^{\infin}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n}\quad\mbox{ for all } x$

$\tan x = \sum^{\infin}_{n=1} \frac{B_{2n} (-4)^n (1-4^n)}{(2n)!} x^{2n-1}\quad\mbox{ for } \left| x \right| < \frac{\pi}{2}$

$\sec x = \sum^{\infin}_{n=0} \frac{(-1)^n E_{2n}}{(2n)!} x^{2n}\quad\mbox{ for } \left| x \right| < \frac{\pi}{2}$

$\arcsin x = \sum^{\infin}_{n=0} \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1}\quad\mbox{ for } \left| x \right| < 1$

$\arctan x = \sum^{\infin}_{n=0} \frac{(-1)^n}{2n+1} x^{2n+1}\quad\mbox{ for } \left| x \right| < 1$

Hyperbolic functions:

$\sinh x = \sum^{\infin}_{n=0} \frac{1}{(2n+1)!} x^{2n+1}\quad\mbox{ for all } x$

$\cosh x = \sum^{\infin}_{n=0} \frac{1}{(2n)!} x^{2n}\quad\mbox{ for all } x$

$\tanh x = \sum^{\infin}_{n=1} \frac{B_{2n} 4^n (4^n-1)}{(2n)!} x^{2n-1}\quad\mbox{ for } \left| x \right| < \frac{\pi}{2}$

$\sinh^{-1} x = \sum^{\infin}_{n=0} \frac{(-1)^n (2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1}\quad\mbox{ for } \left| x \right| < 1$

$\tanh^{-1} x = \sum^{\infin}_{n=0} \frac{1}{2n+1} x^{2n+1}\quad\mbox{ for } \left| x \right| < 1$

Lambert's W function:

$W_0(x) = \sum^{\infin}_{n=1} \frac{(-n)^{n-1}}{n!} x^n\quad\mbox{ for } \left| x \right| < \frac{1}{e}$

The numbers B_k appearing in the expansions of tan(x) and tanh(x) are the Bernoulli numbers. The C(α,n) in the binomial expansion are the binomial coefficients. The E_k in the expansion of sec(x) are Euler numbers.

Multiple dimensions

The Taylor series may be generalised to functions of more than one variable with

$\sum_{n_1=0}^{\infin} \cdots \sum_{n_d=0}^{\infin} \frac{\partial^{n_1}}{\partial x^{n_1}} \cdots \frac{\partial^{n_d}}{\partial x^{n_d}} \frac{f(a_1,\cdots,a_d)}{n_1!\cdots n_d!} (x_1-a_1)^{n_1}\cdots (x_d-a_d)^{n_d}$

History

The Taylor series is named for mathematician Brook Taylor, who first published the power series formula in 1715.

Constructing a Taylor Series

Several methods exist for the calculation of Taylor series of a large number of functions. One can attempt to use the Taylor series as-is and generalize the form of the coefficients, or one can use manipulations such as substitution, multiplication or division, addition or subtraction of standard Taylor series (such as those above) to construct the Taylor series of a function, by virtue of Taylor series being power series. In some cases, one can also derive the Taylor series by repeatedly applying integration by parts. The use of computer algebra systems to calculate Taylor series is common, since it eliminates tedious substitution and manipulation.

Example 1

Consider the function

$f(x)=\ln{(1+\cos{x})} \,,$

for which we want a Taylor series at 0.

We have for the natural logarithm

$\ln(1+x) = \sum^{\infin}_{n=1} \frac{(-1)^{n+1}}{n} x^n = x - {x^2\over 2}+{x^3 \over 3} - {x^4 \over 4} + \cdots \quad\mbox{ for } \left| x \right| < 1$

and for the cosine function

$\cos x = \sum^{\infin}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n} = 1 -{x^2\over 2!}+{x^4\over 4!}- \cdots \quad\mbox{ for all } x\in\mathbb{C}.$

We can simply substitute the second series into the first. Doing so gives

$\left(1 -{x^2\over 2!}+{x^4\over 4!}-\cdots\right)-{1\over 2}\left(1 -{x^2\over 2!}+{x^4\over 4!}-\cdots\right)^2 +{1\over 3}\left(1 -{x^2\over 2!}+{x^4\over 4!}-\cdots\right)^3-\cdots$

Expanding by using multinomial coefficients gives the required Taylor series. Note that cosine and therefore $f$ are even functions, meaning that $f(x)=f(-x)$, hence the coefficients of the odd powers $x$, $x^3$, $x^5$, $x^7$ and so on have to be zero and don't need to be calculated. The first few terms of the series are

$\ln(1+\cos x)=\ln 2-{x^2\over 4}-{x^4\over 96}-{x^6\over 1440} -{17x^8\over 322560}-{31x^{10}\over 7257600}-\cdots$

The general coefficient can be represented using Faà di Bruno's formula. However, this representation does not seem to be particularly illuminating and is therefore omitted here.

Example 2

Suppose we want the Taylor series at 0 of the function

$g(x)=\frac{e^x}{\cos x}\,.$

We have for the exponential function

$e^x = \sum^\infty_{n=0} {x^n\over n!} =1 + x + {x^2 \over 2!} + {x^3 \over 3!} + {x^4 \over 4!} +\cdots$

and, as in the first example,

$\cos x = 1 - {x^2 \over 2!} + {x^4 \over 4!} - \cdots$

Assume the power series is

${e^x \over \cos x} = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots$

Then multiplication with the denominator and substitution of the series of the cosine yields

$\begin{align} e^x &= (c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots)\cos x\\ &=\left(c_0 + c_1 x + c_2 x^2 + c_3 x^3 + c_4x^4 + \cdots\right)\left(1 - {x^2 \over 2!} + {x^4 \over 4!} - \cdots\right)\\ &=c_0 - {c_0 \over 2}x^2 + {c_0 \over 4!}x^4 + c_1x - {c_1 \over 2}x^3 + {c_1 \over 4!}x^5 + c_2x^2 - {c_2 \over 2}x^4 + {c_2 \over 4!}x^6 + c_3x^3 - {c_3 \over 2}x^5 + {c_3 \over 4!}x^7 +\cdots \end{align}$

Collecting the terms up to fourth order yields

$=c_0 + c_1x + \left(c_2 - {c_0 \over 2}\right)x^2 + \left(c_3 - {c_1 \over 2}\right)x^3+\left(c_4+{c_0 \over 4!}-{c_2\over 2}\right)x^4 + \cdots$

Comparing coefficients with the above series of the exponential function yields the desired Taylor series

$\frac{e^x}{\cos x}=1 + x + x^2 + {2x^3 \over 3} + {x^4 \over 2} + \cdots$

Power Series

The study of power series is aimed at investigating series which can approximate some function over a certain interval.

Motivations

Elementary calculus (differentiation) is used to obtain information on a line which touches a curve at one point (i.e. a tangent). This is done by calculating the gradient, or slope of the curve, at a single point. However, this does not provide us with reliable information on the curve's actual value at given points in a wider interval. This is where the concept of power series becomes useful.

An example

Consider the curve of y = cos(x), about the point x = 0. A naïve approxomation would be the line y = 1. However, for a more accurate approximation, observe that cos(x) looks like an inverted parabola around x = 0 - therefore, we might think about which parabola could approximate the shape of cos(x) near this point. This curve might well come to mind:

$y = {1 - x^2 \over 2}$

In fact, this is the best estimate for cos(x) which uses polynomials of degree 2 (i.e. a highest term of x²) - but how do we know this is true? This is the study of power series: finding optimal approximations to functions using polynomials.

Definition

A power series is a series of the form

a₀x⁰ + a₁x¹ + ... + a_nxⁿ

or, equivalently,

$\sum_{j=0}^n a_jx^j$

Radius of convergence

When using a power series as an alternative method of calculating a function's value, the equation

$f(x)=\sum_{j=0}^n a_jx^j$

can only be used to study f(x) where the power series converges - this may happen for a finite range, or for all real numbers.

The size of the interval (around its center) in which the power series converges to the function is known as the radius of convergence.

An example

$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$ (a geometric series)

this converges when | x | < 1, the range -1 < x < +1, so the radius of convergence - centered at 0 - is 1. It should also be observed that at the extremities of the radius, that is where x = 1 and x = -1, the power series does not converge.

Another example

$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$

Using the ratio test, this series converges when the ratio of successive terms is less than one:

$\lim_{n \to \infty} \left|\frac{x^\left(n+1\right)}{\left(n+1\right)!}\frac{n!}{x^n}\right|<1$

$\lim_{n \to \infty} \left|\frac{x^n x^1}{n! \left(n+1\right)}\frac{n!}{x^n}\right|<1$

or $\lim_{n \to \infty}\left|\frac{x}{n+1}\right|<1$

which is always true - therefore, this power series has an infinite radius of convergence. In effect, this means that the power series can always be used as a valid alternative to the original function, e^x.

Abstraction

If we use the ratio test on an arbitary power series, we find it converges when

$\lim \frac{|a_{n+1}x|}{|a_n|} <1$

and diverges when

$\lim \frac{|a_{n+1}x|}{|a_n|} >1$

The radius of convergence is therefore

$r=\lim \frac{|a_n|}{|a_{n+1}|}$

If this limit diverges to infinity, the series has an infinite radius of convergence.

Differentiation and Integration

Within its radius of convergence, a power series can be differentiated and integrated term by term.

$\frac{d}{dx} \sum_{j=0}^\infty a_jx^j = \sum_{j=0}^\infty (j+1)a_{j+1}x^j$

$\int \sum_{j=0}^\infty a_jz^j dz = \sum_{j=1}^\infty \frac{a_{j-1}}{j}x^j$

Both the differential and the integral have the same radius of convergence as the original series.

This allows us to sum exactly suitable power series. For example,

$\frac{1}{1+x}=1-x+x^2-x^3+ \ldots$

This is a geometric series, which converges for | x | < 1. Integrating both sides, we get

$\ln (1+x) = x-\frac{x^2}{2}+\frac{x^3}{3} \ldots$

which will also converge for | x | < 1. When x = -1 this is the harmonic series, which diverges'; when x = 1 this is an alternating series with diminishing terms, which converges to ln 2 - this is testing the extremities.

It also lets us write power series for integrals we cannot do exactly:

$e^{-x^2}=\sum (-1)^n \frac{x^{2n}}{n!}$

The left hand side can not be integrated exactly, but the right hand side can be.

$\int_0^z e^{-x^2}dx=\sum \frac{(-1)^n z^{2n+1}}{(2n+1)n!}$

This gives us a power series for the sum, which has an infinite radius of convergence, letting us approximate the integral as closely as we like.

Further reading

• "Decoding the Rosetta Stone" article by Jack W. Crenshaw 2005-10-12

Multivariable & Differential Calculus

See also Algebra/Vectors

---

Two-Dimensional Vectors

Introduction

In most mathematics courses up until this point, we deal with scalars. These are quantities which only need one number to express. For instance, the amount of gasoline used to drive to the grocery store is a scalar quantity because it only needs one number: 2 gallons.

In this unit, we deal with vectors. A vector is a directed line segment -- that is, a line segment that points one direction or the other. As such, it has an initial point and a terminal point. The vector starts at the initial point and ends at the terminal point, and the vector points towards the terminal point. A vector is drawn as a line segment with an arrow at the terminal point:

A single vector with coordinate axes.

The same vector can be placed anywhere on the coordinate plane and still be the same vector -- the only two bits of information a vector represents are the magnitude and the direction. The magnitude is simply the length of the vector, and the direction is the angle at which it points. Since neither of these specify a starting or ending location, the same vector can be placed anywhere. To illustrate, all of the line segments below can be defined as the vector with magnitude $4\sqrt{2}$ and angle 45 degrees:

Multiple locations for the same vector.

It is customary, however, to place the vector with the initial point at the origin as indicated by the blue vector. This is called the standard position.

Component Form

In standard practice, we don't express vectors by listing the length and the direction. We instead use component form, which lists the height (rise) and width (run) of the vectors. It is written as follows:

$\begin{pmatrix} \mathrm{run} \\ \mathrm{rise}\end{pmatrix}$

Vector with rise and run measurements.

From the diagram we can now see the benefits of the standard position: the two numbers for the terminal point's coordinates are the same numbers for the vector's rise and run. Note that we named this vector u. Just as you can assign numbers to variables in algebra (usually x, y, and z), you can assign vectors to variables in calculus. The letters u, v, and w are usually used, and either boldface or an arrow over the letter is used to identify it as a vector.

When expressing a vector in component form, it is no longer obvious what the magnitude and direction are. Therefore, we have to perform some calculations to find the magnitude and direction.

Magnitude

$|\mathbf{u}| = \sqrt{u_x^2+u_y^2}$

where $u_x$ is the width, or run, of the vector; $u_y$ is the height, or rise, of the vector. You should recognize this formula as pythagorean's theorem. It is -- the magnitude is the distance between the initial point and the terminal point.

The magnitude of a vector can also be called the norm.

Direction

$\tan \theta = \frac{u_y}{u_x}$ Vector triangle with ux, uy, and theta labeled.

where $\theta$ is the direction of the vector. This formula is simply the tangent formula for right triangles.

Vector Operations

For these definitions, assume:

$\mathbf{u} = \begin{pmatrix} u_x \\ u_y \end{pmatrix}$ $\mathbf{v} = \begin{pmatrix} v_x \\ v_y \end{pmatrix}$

Vector Addition

Graphically, adding two vectors together places one vector at the end of the other. This is called tip-to-tail addition: The resultant vector, or solution, is the vector drawn from the initial point of the first vector to the terminal point of the second vector when they are drawn tip-to-tail:

Tip-to-tail addition

Fig: $\begin{pmatrix} 4 \\ 6 \end{pmatrix} + \begin{pmatrix} 1 \\ -3 \end{pmatrix} = \begin{pmatrix} 5 \\ 3 \end{pmatrix}$

Numerically:

$\mathbf{u} + \mathbf{v} = \begin{pmatrix} u_x + v_x \\ u_y + v_y \end{pmatrix}$

Scalar Multiplication

Graphically, multiplying a vector by a scalar changes only the magnitude of the vector by that same scalar. That is, multiplying a vector by 2 will "stretch" the vector to twice its original magnitude, keeping the direction the same.

Scalar multiplication Fig: $2 \begin{pmatrix} 3 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ 6 \end{pmatrix}$

Numerically, you calculate the resultant vector with this formula:

$c \mathbf{u} = \begin{pmatrix} c u_x \\ c u_y \end{pmatrix}$, where c is a constant scalar.

As previously stated, the magnitude is changed by the same constant:

$|c \mathbf{u}| = c |\mathbf{u}|$

Since multiplying a vector by a constant results in a vector in the same direction, we can reason that two vectors are parallel if one is a constant multiple of the other -- that is, that $\mathbf{u} || \mathbf{v}$ if $\mathbf{u} = c \mathbf{v}$ for some constant c.

We can also divide by a non-zero scalar by instead multiplying by the reciprocal, as with dividing regular numbers:

$\frac{\mathbf{u}}{c} = \frac{1}{c} \mathbf{u}, {c}\ne 0$

Dot Product and Perpendicular Vectors

The dot product, sometimes confusingly called the scalar product, of two vectors is given by:

$\mathbf{u} \cdot \mathbf{v} = u_x v_x + u_y v_y$

or alternatively:

$\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta$

where $\theta$ is the angle difference between the two vectors. If this angle is 90 degrees or ${\pi \over 2}$ (if the two vectors are orthogonal to each other), that is the vectors are perpendicular, then the dot product is 0.

The dot product of a vector with itself is given by:

$\mathbf{u} \cdot \mathbf{u} = |\mathbf{u}||\mathbf{u}| \cos 0 = |\mathbf{u}|^2$

Unit Vectors

A unit vector is a vector with a magnitude of 1. The unit vector of u is a vector in the same direction as $\mathbf{u}$, but with a magnitude of 1:

Unit vector.

The process of finding the unit vector of u is called normalization. As mentioned in scalar multiplication, multiplying a vector by constant C will result in the magnitude being multiplied by C. We know how to calculate the magnitude of $\mathbf{u}$. We know that dividing a vector by a constant will divide the magnitude by that constant. Therefore, if that constant is the magnitude, dividing the vector by the magnitude will result in a unit vector in the same direction as $\mathbf{u}$:

$\mathbf{w}=\frac{\mathbf{u}}{|\mathbf{u}|}$, where $\mathbf{w}$ is the unit vector of $\mathbf{u}$

Standard Unit Vectors

There are two special unit vectors. i points one unit directly right in the x direction, and j points one unit directly up in the y direction:

$\mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$

$\mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$

Using the scalar multiplication and vector addition rules, we can then express vectors in a different way:

$\begin{pmatrix} x \\ y \end{pmatrix} = x \mathbf{i} + y \mathbf{j}$

If we work that equation out, it makes sense. Multiplying x by i will result in the vector $\begin{pmatrix} x \\ 0 \end{pmatrix}$. Multiplying y by j will result in the vector $\begin{pmatrix} 0 \\ y \end{pmatrix}$. Adding these two together will give us our original vector, $\begin{pmatrix} x \\ y \end{pmatrix}$. Expressing vectors using i and j is called standard form.

Projections

Sometimes it is necessary to find the part of a vector parallel to another which is known as the projection of a vector $\mathbf{u}$ on $\mathbf{v}$ (notationally displayed as $\mathbf{pr_v}\mathbf{u}$).

To determine the projection we first note that $\mathbf{pr_v}\mathbf{u}$ is proportional to $\mathbf{v}$. In other words:

$\mathbf{pr_v}\mathbf{u} = \mathbf{c}\mathbf{v}$

$|\mathbf{pr_v}\mathbf{u}| = |\mathbf{c}\mathbf{v}|$

$|\mathbf{pr_v}\mathbf{u}| = \mathbf{c}|\mathbf{v}|$

Also, with the angle θ, and because of the definition of the cosine:

$|\mathbf{pr_v}\mathbf{u}| = \cos{\theta}|\mathbf{u}|$

By the dot product:

$\cos{\theta}= (\mathbf{u} \cdot \mathbf{v})/(|\mathbf{u}||\mathbf{v}|)$

By substituting this for $\mathbf{cos}$θ and then the new expression for $\mathbf{pr_v}\mathbf{u}$ into the equation relating it for c, we find that:

$\mathbf{c} = (\mathbf{u} \cdot \mathbf{v})/(|\mathbf{v}|^2)$

Which, finally, substituting this c in the oringinal vector-form equation gives us the widely used definition for $\mathbf{pr_v}\mathbf{u}$:

$\mathbf{pr_v}\mathbf{u} = (\mathbf{u} \cdot \mathbf{v})/(|\mathbf{v}|^2) \mathbf{v}$

Polar coordinates

Polar coordinates are an alternative two-dimensional coordinate system, which is often useful when rotations are important. Instead of specifying the position along the x and y axes, we specify the distance from the origin, r, and the direction, an angle θ.

Looking at this diagram, we can see that the values of x and y are related to those of r and θ by the equations

$\begin{matrix} x=r \cos \theta & r = \sqrt{x^2+y^2} \\ y=r \sin \theta & \tan \theta = \frac{y}{x} \end{matrix}$

Because tan^-1 is multivalued, care must be taken to select the right value.

Just as for Cartesian coordinates the unit vectors that point in the x and y directions are special, so in polar coordinates the unit vectors that point in the r and θ directions are also special.

We will call these vectors $\hat{\mathbf{r}}$ and $\hat{\boldsymbol{\theta}}$, pronounced r-hat and theta-hat. Putting a circumflex over a vector this way is often used to mean the unit vector in that direction.

Again, on looking at the diagram we see,
$\begin{matrix} \mathbf{i}= \hat{\mathbf{r}} \cos \theta -\hat{\boldsymbol{\theta}} \sin \theta & \hat{\mathbf{r}} = \frac{x}{r}\mathbf{i}+ \frac{y}{r}\mathbf{j} \\ \mathbf{j}= \hat{\mathbf{r}} \sin \theta +\hat{\boldsymbol{\theta}} \cos \theta & \hat{\boldsymbol{\theta}} = -\frac{y}{r}\mathbf{i} + \frac{x}{r}\mathbf{j} \end{matrix}$

Three-Dimensional Coordinates and Vectors

Basic definition

Two-dimensional Cartesian coordinates as we've discussed so far can be easily extended to three-dimensions by adding one more value: 'z'. If the standard (x,y) coordinate axes are drawn on a sheet of paper, the 'z' axis would extend upwards off of the paper.

Similar to the two coordinate axes in two-dimensional coordinates, there are three coordinate planes in space. These are the xy-plane, the yz-plane, and the xz-plane. Each plane is the "sheet of paper" that contains both axes the name mentions. For instance, the yz-plane contains both the y and z axes and is perpendicular to the x axis.

Therefore, vectors can be extended to three dimensions by simply adding the 'z' value.

$\mathbf{u} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$

To facilitate standard form notation, we add another standard unit vector:

$\mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$

Again, both forms (component and standard) are equivalent.

$\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = 1 \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}$

Magnitude: Magnitude in three dimensions is the same as in two dimensions, with the addition of a 'z' term in the radicand.

$| \mathbf{u} | = \sqrt{u_x^2 + u_y^2 + u_z^2}$

Three dimensions

The polar coordinate system is extended into three dimensions with two different coordinate systems, the cylindrical and spherical coordinate systems, both of which include two-dimensional or planar polar coordinates as a subset. In essence, the cylindrical coordinate system extends polar coordinates by adding an additional distance coordinate, while the spherical system instead adds an additional angular coordinate.

Cylindrical coordinates

a point plotted with cylindrical coordinates

The cylindrical coordinate system is a coordinate system that essentially extends the two-dimensional polar coordinate system by adding a third coordinate measuring the height of a point above the plane, similar to the way in which the Cartesian coordinate system is extended into three dimensions. The third coordinate is usually denoted h, making the three cylindrical coordinates (r, θ, h).

The three cylindrical coordinates can be converted to Cartesian coordinates by

$\begin{align} x &= r \, \cos\theta \\ y &= r \, \sin\theta \\ z &= h. \end{align}$

Spherical coordinates

A point plotted using spherical coordinates

Polar coordinates can also be extended into three dimensions using the coordinates (ρ, φ, θ), where ρ is the distance from the origin, φ is the angle from the z-axis (called the colatitude or zenith and measured from 0 to 180°) and θ is the angle from the x-axis (as in the polar coordinates). This coordinate system, called the spherical coordinate system, is similar to the latitude and longitude system used for Earth, with the origin in the centre of Earth, the latitude δ being the complement of φ, determined by δ = 90° − φ, and the longitude l being measured by l = θ − 180°.

The three spherical coordinates are converted to Cartesian coordinates by

$\begin{align} x &= \rho \, \sin\phi \, \cos\theta \\ y &= \rho \, \sin\phi \, \sin\theta \\ z &= \rho \, \cos\phi. \end{align}$

$r = \sqrt{x^2 + y^2 + z^2},$

$\theta\ = \arctan{\frac{y}{x}},$

$\phi\ = \arccos{\frac{z}{r}},$

Cross Product

The cross product of two vectors is a determinant:

$\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix}$

and is also a pseudovector.

The cross product of two vectors is orthogonal to both vectors. The magnitude of the cross product is the product of the magnitude of the vectors and the sin of the angle between them.

$|\mathbf{u} \times \mathbf{v}|=|\mathbf{u}||\mathbf{v}| \sin \theta$

This magnitude is the area of the parallelogram defined by the two vectors.

The cross product is linear and anticommutative. For any numbers a and b,

$\mathbf{u} \times \left( a\mathbf{v} +b\mathbf{w} \right) = a \mathbf{u} \times \mathbf{v} +b\mathbf{u} \times \mathbf{w} \quad \mathbf{u} \times \mathbf{v} = - \mathbf{v} \times \mathbf{u}$

If both vectors point in the same direction, their cross product is zero.

Triple Products

If we have three vectors we can combine them in two ways, a triple scalar product,

$\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$

and a triple vector product

$\mathbf{u} \times (\mathbf{v} \times \mathbf{w})$

The triple scalar product is a determinant

$\mathbf{u} \cdot(\mathbf{v} \times \mathbf{w}) = \begin{vmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{vmatrix}$

If the three vectors are listed clockwise, looking from the origin, the sign of this product is positive. If they are listed anticlockwise the sign is negative.

The order of the cross and dot products doesn't matter.

$\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w}$

Either way, the absolute value of this product is the volume of the parallelepiped defined by the three vectors, u, v, and w

The triple vector product can be simplified

$\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) =(\mathbf{u} \cdot \mathbf{w}) \mathbf{v} -(\mathbf{u} \cdot \mathbf{v}) \mathbf{w}$

This form is easier to do calculations with.

The triple vector product is not associative.

$\mathbf{u} \times (\mathbf{v} \times \mathbf{w}) \ne (\mathbf{u} \times \mathbf{v}) \times \mathbf{w}$

There are special cases where the two sides are equal, but in general the brackets matter. They must not be omitted.

Three-Dimensional Lines and Planes

We will use r to denote the position of a point.

The multiples of a vector, a all lie on a line through the origin. Adding a constant vector b will shift the line, but leave it straight, so the equation of a line is,

$\mathbf{r}=\mathbf{a}s+\mathbf{b}$

This is a parametric equation. The position is specified in terms of the parameter s.

Any linear combination of two vectors, a and b lies on a single plane through the origin, provided the two vectors are not colinear. We can shift this plane by a constant vector again and write

$\mathbf{r}=\mathbf{a}s+\mathbf{b}t+\mathbf{c}$

If we choose a and b to be orthonormal vectors in the plane (i.e unit vectors at right angles) then s and t are Cartesian coordinates for points in the plane.

These parametric equations can be extended to higher dimensions.

Instead of giving parametric equations for the line and plane, we could use constraints. E.g, for any point in the xy plane z=0

For a plane through the origin, the single vector normal to the plane, n, is at right angle with every vector in the plane, by definition, so

$\mathbf{r} \cdot \mathbf{n}=0$

is a plane through the origin, normal to n.

For planes not through the origin we get

$(\mathbf{r}- \mathbf{a}) \cdot \mathbf{n}=0 \quad \mathbf{r} \cdot \mathbf{n}=a$

A line lies on the intersection of two planes, so it must obey the constraint for both planes, i.e

$\mathbf{r} \cdot \mathbf{n}=a \quad \mathbf{r} \cdot \mathbf{m}=b$

These constraint equations con also be extended to higher dimensions.

Vector-Valued Functions

Vector-Valued Functions are functions that instead of giving a resultant scalar value, give a resultant vector value. These aid in the create of direction and vector fields, and are therefore used in physics to aid with visualizations of electric, magnetic, and many other fields. They are of the following form:

$\mathbf{F(t)} = \begin{pmatrix} \mathbf{a_1(t)} \\ \mathbf{a_2(t)} \\ \mathbf{a_3(t)} \\ \mathbf{.} \\ \mathbf{.} \\ \mathbf{a_n(t)} \end{pmatrix}$

Introduction

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Limits, Derivatives, and Integrals

Put simply, the limit of a vector-valued function is the limit of its parts.

Proof:

Suppose $\lim_{t \rightarrow c} \mathbf{F}(t) = \mathbf{L} = \begin{pmatrix} \mathbf{a_1} \\ \mathbf{a_2} \\ \mathbf{a_3} \\ \mathbf{.} \\ \mathbf{.} \\ \mathbf{a_n} \end{pmatrix}$

Therefore for any $\epsilon > 0$ there is a $\phi > 0$ such that

$0 < |t-c| < \phi \implies |\mathbf{F}(t) - \mathbf{L}| < \epsilon$

But by the triangle inequality $|a_1| \le |\mathbf{F}| \le |a_1| + |a_2| + |a_3| + ... + |a_n|$ $|a_1(t) - a_1| \le |\mathbf{F}(t) - \mathbf{L}|$

So

$0 < |t-c| < \phi \implies |a_1(t) - a_1| < \epsilon$

Therefore $\lim_{t \rightarrow c} a_1(t) = a_1$ A similar argument can be used through parts a_n(t)

Now let $\lim_{t \rightarrow c} \mathbf{F}(t) = \mathbf{L} = \begin{pmatrix} \mathbf{a_1} \\ \mathbf{a_2} \\ \mathbf{a_3} \\ \mathbf{.} \\ \mathbf{.} \\ \mathbf{a_n} \end{pmatrix}$ again, and that for any ε>0 there is a corresponding φ>0 such 0<|t-c|<φ implies

$|a_n(t) - a_n| < {\epsilon \over n}$

Then

$0 < |t-c| < \phi \implies |\mathbf{F}(t) - \mathbf{L}| \le {\epsilon_1 \over n} + ... + {\epsilon_n \over n} = \epsilon$

therefore!:

$\lim_{t \rightarrow c} \mathbf{F}(t) = \mathbf{L} = \begin{pmatrix} \mathbf{a_1} \\ \mathbf{a_2} \\ \mathbf{a_3} \\ \mathbf{.} \\ \mathbf{.} \\ \mathbf{a_n} \end{pmatrix} = \begin{pmatrix} \lim_{t \rightarrow c} \mathbf{a_1(t)} \\ \lim_{t \rightarrow c} \mathbf{a_2(t)} \\ \lim_{t \rightarrow c} \mathbf{a_3(t)} \\ \mathbf{.} \\ \mathbf{.} \\ \lim_{t \rightarrow c} \mathbf{a_n(t)} \end{pmatrix}$

From this we can then create an accurate definition of a derivative of a vector-valued function:

$\mathbf{F}'(t)=\lim_{h \rightarrow 0} {\mathbf{F}(t+h) - \mathbf{F}(t) \over h} = \begin{pmatrix} \mathbf{a_1(t)} \\ \mathbf{a_2(t)} \\ \mathbf{a_3(t)} \\ \mathbf{.} \\ \mathbf{.} \\ \mathbf{a_n(t)} \end{pmatrix}$

$= \lim_{h \rightarrow 0} {\begin{pmatrix} \mathbf{a_1(t + h)} \\ \mathbf{a_2(t + h)} \\ \mathbf{a_3(t + h)} \\ \mathbf{.} \\ \mathbf{.} \\ \mathbf{a_n(t + h)} \end{pmatrix} - \begin{pmatrix} \mathbf{a_1(t)} \\ \mathbf{a_2(t)} \\ \mathbf{a_3(t)} \\ \mathbf{.} \\ \mathbf{.} \\ \mathbf{a_n(t)} \end{pmatrix} \over h}$

$= \begin{pmatrix} \lim_{h \rightarrow 0} {a_1(t+h) - a_1(t) \over h} \\ \lim_{h \rightarrow 0} {a_2(t+h) - a_2(t) \over h} \\ \lim_{h \rightarrow 0} {a_3(t+h) - a_3(t) \over h} \\ \mathbf{.} \\ \mathbf{.} \\ \lim_{h \rightarrow 0} {a_n(t+h) - a_n(t) \over h} \end{pmatrix}$

The final step was accomplished by taking what we just did with limits.

By the Fundamental Theorem of Calculus integrals can be applied to the vector's components.

In other words: the limit of a vector function is the limit of its parts, the derivative of a vector function is the derivative of its parts, and the integration of a vector function is the integration of it parts.

Velocity, Acceleration, Curvature, and a brief mention of the Binormal

Assume we have a vector-valued function which starts at the origin and as its independent variables changes the points that the vectors point at trace a path.

We will call this vector $\mathbf{r}(t)$, which is commonly known as the position vector.

If $\mathbf{r}$ then represents a position and t represents time, then in model with Physics we know the following:

$\mathbf{r}(t+h) - \mathbf{r}(t)$ is displacement. $\mathbf{r}'(t) = \mathbf{v}(t)$ where $\mathbf{v}(t)$ is the velocity vector. $|\mathbf{v}(t)|$ is the speed. $\mathbf{r}''(t) = \mathbf{v}'(t) = \mathbf{a}(t)$ where $\mathbf{a}(t)$ is the acceleration vector.

The only other vector that comes in use at times is known as the curvature vector.

The vector $\mathbf{T}(t)$ used to find it is known as the unit tangent vector, which is defined as $\frac{\mathbf{v}(t)}{|\mathbf{v}(t)|}$ or shorthand $\mathbf{\hat v}$.

The vector normal $\mathbf{N}$ to this then is $\frac{\mathbf{T}'(t)}{|\mathbf{v}(t)|}$.

We can verify this by taking the dot product

$\mathbf{T} \cdot \mathbf{N} = 0$

Also note that $|\mathbf{v}(t)| = {ds \over dt}$

and

$\mathbf{T}(t) = \frac{v}{|v|} = \frac{\frac{dr}{dt}}{\frac{ds}{dt}} = {dr \over ds}$

and

$\mathbf{N} = \frac{\mathbf{T}'(t)}{|\mathbf{v}(t)|} = \frac{\frac{dT}{dt}}{\frac{ds}{dt}} = {dT \over ds}$

Then we can actually verify:

${d \over ds} ( \mathbf{T} \cdot \mathbf{T} )= {d \over ds} (1)$

${dT \over ds} \cdot \mathbf{T} + \mathbf{T} \cdot {dT \over ds} = 0$

$2 * \mathbf{T} \cdot {dT \over ds} = 0$

$\mathbf{T} \cdot {dT \over ds} = 0$

$\mathbf{T} \cdot \mathbf{N} = 0$

Therefore $\mathbf{N}$ is perpendicular to $\mathbf{T}$

What this gives rise to is the Unit Normal Vector $\frac{\frac{dT}{ds}}{|\frac{dT}{ds}|}$ of which the top-most vector is the Normal vector, but the bottom half $(|\frac{dT}{ds}|)^{-1}$ is known as the curvature. Since the Normal vector points toward the inside of a curve, the sharper a turn, the Normal vector has a large magnitude, therefore the curvature has a small value, and is used as an index in civil engineering to reflect the sharpness of a curve (clover-leaf highways, for instance).

The only other thing not mentioned is the Binormal that occurs in 3-d curves $\mathbf{T} \times \mathbf{N} = \mathbf{B}$, which is useful in creating planes parallel to the curve. < Calculus/Outline

In your previous study of calculus, we have looked at functions and their behaviour. Most of these functions we have examined have been all in the form

f(x) : R → R,

and only occasional examination of functions of two variables. However, the study of functions of several variables is quite rich in itself, and has applications in several fields.

We write functions of vectors - many variables - as follows:

f : R^m → Rⁿ

and f(x) for the function that maps a vector in R^m to a vector in Rⁿ.

Before we can do calculus in Rⁿ, we must familiarise ourselves with the structure of Rⁿ. We need to know which properties of R can be extended to Rⁿ

Topology in Rⁿ

We are already familar with the nature of the regular real number line, which is the set R, and the two-dimensional plane, R². This examination of topology in Rⁿ attempts to look at a generalization of the nature of n-dimensional spaces - R, or R²³, or Rⁿ.

Lengths and distances

If we have a vector in R², we can calculate its length using the Pythagorean theorem. For instance, the length of the vector (2, 3) is

$\sqrt{3^2+2^2}=\sqrt{13}$

We can generalize this to Rⁿ. We define a vector's length, written |x|, as the square root of the sum of the squares of each of its components. That is, if we have a vector x=(x₁,...,x_n),

$|\mathbf{x}| = \sqrt{x_1^2+x_2^2+\cdots+x_n^2}$

Now that we have established some concept of length, we can establish the distance between two vectors. We define this distance to be the length of the two vectors' difference. We write this distance d(x, y), and it is

$d(\mathbf{x}, \mathbf{y}) = |\mathbf{x} - \mathbf{y}| = \sqrt{\sum{(x_i-y_i)^2}}$

This distance function is sometimes referred to as a metric. Other metrics arise in different circumstances. The metric we have just defined is known as the Euclidean metric.

Open and closed balls

In R, we have the concept of an interval, in that we choose a certain number of other points about some central point. For example, the interval [-1, 1] is centered about the point 0, and includes points to the left and right of zero.

In R² and up, the idea is a little more difficult to carry on. For R², we need to consider points to the left, right, above, and below a certain point. This may be fine, but for R³ we need to include points in more directions.

We generalize the idea of the interval by considering all the points that are a given, fixed distance from a certain point - now we know how to calculate distances in Rⁿ, we can make our generalization as follows, by introducing the concept of an open ball and a closed ball respectively, which are analogous to the open and closed interval respectively.

an open ball

$B(\mathbf{a}, r)$

is a set in the form { x ∈ Rⁿ|d(x, a) < r}

an closed ball

$\overline{B}(\mathbf{a}, r)$

is a set in the form { x ∈ Rⁿ|d(x, a) ≤ r}

In R, we have seen that the open ball is simply an open interval centered about the point x=a. In R² this is a circle with no boundary, and in R³ it is a sphere with no outer surface. (What would the closed ball be?)

Boundary points

If we have some area, say a field, then the common sense notion of the boundary is the points 'next to' both the inside and outside of the field. For a set, S, we can define this rigorously by saying the boundary of the set contains all those points such that we can find points both inside and outside the set. We call the set of such points ∂S

Typically, when it exists the dimension of ∂S is one lower than the dimension of S. e.g the boundary of a volume is a surface and the boundary of a surface is a curve.

This isn't always true; but it is true of all the sets we will be using.

A set S is bounded if there is some positive number such that we can encompass this set by a closed ball about 0. --> if every point in it is within a finite distance of the origin, i.e there exists some r>0 such that x is in S implies |x|<r.

Curves and parametrizations

If we have a function f : R → Rⁿ, we say that f's image (the set {f(t) | t ∈ R} - or some subset of R) is a curve in Rⁿ and f is its parametrization.

Parametrizations are not necessarily unique - for example, f(t) = (cos t, sin t) such that t ∈ [0, 2π) is one parametrization of the unit circle, and g(t) = (cos at, sin at) such that t ∈ [0, 2π/a) is a whole family of parametrizations of that circle.

Collision and intersection points

Say we have two different curves. It may be important to consider

• when the two curves cross each other - where they intersect
• when the two curves hit each other at the same time - where they collide.

Intersection points

Firstly, we have two parametrizations f(t) and g(t), and we want to find out when they intersect, this means that we want to know when the function values of each parametrization are the same. This means that we need to solve

f(t) = g(s)

because we're seeking the function values independent of the times they intersect.

For example, if we have f(t) = (t, 3t) and g(t) = (t, t²), and we want to find intersection points:

f(t) = g(s)

(t, 3t) = (s, s²)

t = s and 3t = s²

with solutions (t, s) = (0, 0) and (3, 3)

So, the two curves intersect at the points (0, 0) and (3, 3).

Collision points

However, if we want to know when the points "collide", with f(t) and g(t), we need to know when both the function values and the times are the same, so we need to solve instead

f(t) = g(t)

For example, using the same functions as before, f(t) = (t, 3t) and g(t) = (t, t²), and we want to find collision points:

f(t) = g(t)

(t, 3t) = (t, t²)

t = t and 3t = t²

which gives solutions t = 0, 3 So the collision points are (0, 0) and (3, 9).

We may want to do this to actually model physical problems, such as in ballistics.

Continuity and differentiability

If we have a parametrization f : R → Rⁿ, which is built up out of component functions in the form f(t) = (f₁(t),...,f_n(t)), f is continuous if and only if each component function is also.

In this case the derivative of f(t) is

a_i = (f₁′(t),...,f_n′(t)). This is actually a specific consequence of a more general fact we will see later.

Tangent vectors

Recall in single-variable calculus that on a curve, at a certain point, we can draw a line that is tangent to that curve at exactly at that point. This line is called a tangent. In the several variable case, we can do something similar.

We can expect the tangent vector to depend on f′(t) and we know that a line is its own tangent, so looking at a parametrised line will show us precisely how to define the tangent vector for a curve.

An arbitary line is f(t)=at+b, with  :f_i(t)=a_it+b_i, so

f_i′(t)=a_i and

f′(t)=a, which is the direction of the line, its tangent vector.

Similarly, for any curve, the tangent vector is f′(t).

Angle between curves

We can then formulate the concept of the angle between two curves by considering the angle between the two tangent vectors. If two curves, parametrized by f₁ and f₂ intersect at some point, which means that

f₁(s)=f₂(t)=c,

the angle between these two curves at c is the angle between the tangent vectors f₁′(s) and f₂′(t) is given by

$\arccos{\mathbf{f}_1^\prime(s)\cdot\mathbf{f}_2^\prime(t) \over |\mathbf{f}_1^\prime(s)||\mathbf{f}_2^\prime(t)|}$

Tangent lines

With the concept of the tangent vector as being analogous to being the gradient of the line in the one variable case, we can form the idea of the tangent line. Recall that we need a point on the line and its direction.

If we want to form the tangent line to a point on the curve, say p, we have the direction of the line f′(p), so we can form the tangent line

x(t)=p+t f′(p)

Different parametrizations

One such parametrization of a curve is not necessarily unique. Curves can have several different parametrizations. For example, we already saw that the unit circle can be parametrized by g(t) = (cos at, sin at) such that t ∈ [0, 2π/a).

Generally, if f is one parametrization of a curve, and g is another, with

f(t₀) = g(s₀)

there is a function u(t) such that u(t₀)=s₀, and g'(u(t)) = f(t) near t₀.

This means, in a sense, the function u(t) "speeds up" the curve, but keeps the curve's shape.

Surfaces

A surface in space can be described by the image of a function f : R² → Rⁿ. f is said to be the parametrization of that surface.

For example, consider the function

f(α, β) = α(2,1,3)+β(-1,2,0)

This describes an infinite plane in R³. If we restrict α and β to some domain, we get a parallelogram-shaped surface in R³.

Surfaces can also be described explicitly, as the graph of a function z = f(x, y) which has a standard parametrization as f(x,y)=(x, y, f(x,y)), or implictly, in the form f(x, y, z)=c.

Level sets

The concept of the level set (or contour) is an important one. If you have a function f(x, y, z), a level set in R³ is a set of the form {(x,y,z)|f(x,y,z)=c}. Each of these level sets is a surface.

Level sets can be similarly defined in any Rⁿ

Level sets in two dimensions may be familiar from maps, or weather charts. Each line represents a level set. For example, on a map, each contour represents all the points where the height is the same. On a weather chart, the contours represent all the points where the air pressure is the same.

Limits and continuity

Before we can look at derivatives of multivariate functions, we need to look at how limits work with functions of several variables first, just like in the single variable case.

If we have a function f : R^m → Rⁿ, we say that f(x) approaches b (in Rⁿ) as x approaches a (in R^m) if, for all positive ε, there is a corresponding positive number δ, |f(x)-b| < ε whenever |x-a| < δ, with x ≠ a.

This means that by making the difference between x and a smaller, we can make the difference between f(x) and b as small as we want.

If the above is true, we say

• f(x) has limit b at a
• $\lim_{\mathbf{x}\rightarrow\mathbf{a}} \mathbf{f}(\mathbf{x}) = \mathbf{b}$
• f(x) approaches b as x approaches a
• f(x) → b as x → a

These four statements are all equivalent.

Rules

Since this is an almost identical formulation of limits in the single variable case, many of the limit rules in the one variable case are the same as in the multivariate case.

For f and g, mapping R^m to Rⁿ, and h(x) a scalar function mapping R^m to R, with

• f(x) → b as x → a
• g(x) → c as x → a
• h(x) → H as x → a

then:

• $\lim_{\mathbf{x}\rightarrow\mathbf{a}} (\mathbf{f} + \mathbf{g}) = \mathbf{b} + \mathbf{c}$
• $\lim_{\mathbf{x}\rightarrow\mathbf{a}} (h\mathbf{f}) = H\mathbf{b}$

and consequently

• $\lim_{\mathbf{x}\rightarrow\mathbf{a}} (\mathbf{f}\cdot\mathbf{g}) = \mathbf{b}\cdot\mathbf{c}$
• $\lim_{\mathbf{x}\rightarrow\mathbf{a}} (\mathbf{f}\times\mathbf{g}) = \mathbf{b}\times\mathbf{c}$

when H≠0

• $\lim_{\mathbf{x}\rightarrow\mathbf{a}} ({\mathbf{f}\over h}) = {\mathbf{b}\over H}$

Continuity

Again, we can use a similar definition to the one variable case to formulate a definition of continuity for multiple variables.

If f : R^m → Rⁿ, f is continuous at a point a in R^m if f(a) is defined and

$\lim_{\mathbf{x}\rightarrow\mathbf{a}} \mathbf{f}(\mathbf{x}) = \mathbf{f}(\mathbf{a})$

Just as for functions of one dimension, if f, g are both continuous at p, f+g, λf (for a scalar λ), f·g, and f×g are continuous also. If φ : R^m → R is continus at p, φf, f/φ are too if φ is never zero.

From these facts we also have that if A is some matrix which is n×m in size, with x in R^m, a function f(x)=A x is continuous in that the function can be expanded in the form x₁a₁+...+x_ma_m, which can be easily verified from the points above.

If f : R^m → Rⁿ which is in the form f(x) = (f₁(x),...,f_n(x) is continuous if and only if each of its component functions are a polynomial or rational function, whenever they are defined.

Finally, if f is continuous at p, g is continuous at f(p), g(f(x)) is continuous at p.

Special note about limits

It is important to note that we can approach a point in more than one direction, and thus, the direction that we approach that point counts in our evaluation of the limit. It may be the case that a limit may exist moving in one direction, but not in another.

Differentiable functions

We will start from the one-variable definition of the derivative at a point p, namely

$\lim_{x\rightarrow p} {f(x)-f(p) \over x-p} = f'(p)$

We can't divide by vectors, so this defintion can't be immediately extended to the multiple variable case. However, we can divide by the absolute value of a vector, so lets rewrite this definition in terms of absolute values

$\lim_{x\rightarrow p} \frac{\left|f(x)-f(p)-f'(p)(x-p)\right|}{\left| x-p\right|} = 0$

after pulling f'(p) inside and putting it over a common denominator.

So, how can we use this for the several-variable case?

If we switch all the variables over to vectors and replace the constant,(which performs a linear map in one dimension) with a matrix (which is also a linear map), we have

$\lim_{\mathbf{x}\rightarrow\mathbf{p}} {|\mathbf{f}(\mathbf{x})-\mathbf{f}(\mathbf{p})-A(\mathbf{x}-\mathbf{p})| \over |\mathbf{x}-\mathbf{p}|} = 0$

If this limit exists for some f : R^m → Rⁿ, and there is a matrix A which is m×n, we refer to this matrix as being the derivative and we write it as D_p f.

A point on terminology - in referring to the action of taking the derivative, we write D_p f, but in referring to this matrix itself, it is known as the Jacobian matrix and is also written J_p f. More on the Jacobian later.

Properties

There are a number of important properties of this formulation of the derivative.

Affine approximations

If f is differentiable at p for x close to p, |f(x)-(f(p)+A(x-p))| is small compared to |x-p|, which means that f(x) is approximately equal to f(p)+A(x-p).

We call an expression of the form g(x)+c affine, when g(x) is linear and c is a constant. f(p)+A(x-p) is an affine approximation to f(x).

Jacobian matrix and partial derivatives

The Jacobian matrix of a function is in the form

$\left(J_\mathbf{p} \mathbf{f}\right)_{ij} = \left.{\partial f_i \over \partial x_j}\right|_\mathbf{p}$

for a f : R^m → Rⁿ, J_p f' is a m×n matrix.

The consequence of this is that if f is differentiable at p, all the partial derivatives of f exist at p.

However, it is possible that all the partial derivatives of a function exist at some point yet that function is not differentiable there.

Continuity and differentiability

Furthermore, if all the partial derivatives exist, and are continuous in some neighbourhood of a point p, then f is differentiable at p. This has the consequence that for a function f which has its component functions built from continuous functions (such as rational functions, differentiable functions or otherwise), f is differentiable everywhere f is defined.

We use the terminology continuously differentiable for a function differentiable at p which has all its partial derivatives existing and are continuous in some neighbourhood at p.

Rules of taking Jacobians

If f : R^m → Rⁿ, and h(x) : R^m → R are differentiable at 'p':

• $J_\mathbf{p} (\mathbf{f}+\mathbf{g}) = J_\mathbf{p} \mathbf{f} + J_\mathbf{p} \mathbf{g}$
• $J_\mathbf{p} (h\mathbf{f}) = hJ_\mathbf{p} \mathbf{f} + \mathbf{f}(\mathbf{p}) J_\mathbf{p} h$
• $J_\mathbf{p} (\mathbf{f}\cdot \mathbf{g}) = \mathbf{g}^T J_\mathbf{p} \mathbf{f} + \mathbf{f}^T J_\mathbf{p}\mathbf{g}$

Important: make sure the order is right - matrix multiplication is not commutative!

Chain rule

The chain rule for functions of several variables is as follows. For f : R^m → Rⁿ and g : Rⁿ → R^p, and g o f differentiable at p, then the Jacobian is given by

$\left( J_{\mathbf{f}(\mathbf{p})} \mathbf{g}\right) \left( J_\mathbf{p} \mathbf{f}\right)$

Again, we have matrix multiplication, so one must preserve this exact order. Compositions in one order may be defined, but not necessarily in the other way.

Alternate notations

For simplicity, we will often use various standard abbreviations, so we can write most of the formulae on one line. This can make it easier to see the important details.

We can abbreviate partial differentials with a subscript, e.g,

$\partial_x h(x,y)= \frac{\partial h}{\partial x} \quad \partial_x \partial_y h= \partial_y \partial_x h$

When we are using a subscript this way we will generally use the Heaviside D rather than ∂,

$D_x h(x,y)= \frac{\partial h}{\partial x} \quad D_x D_y h= D_y D_x h$

Mostly, to make the formulae even more compact, we will put the subscript on the function itself.

$D_x h= h_x \quad h_{xy}=h_{yx}$

If we are using subscripts to label the axes, x₁, x₂ …, then, rather than having two layers of subscripts, we will use the number as the subscript.

$h_1 = D_1 h = \partial_1 h = \partial_{x_1}h = \frac{\partial h}{\partial x_1}$

We can also use subscripts for the components of a vector function, u=(u_x, u_y, u_y) or u=(u₁,u₂…u_n)

If we are using subscripts for both the components of a vector and for partial derivatives we will separate them with a comma.

$u_{x,y}=\frac{\partial u_x}{\partial y}$

The most widely used notation is h_x. Both h₁ and ∂₁h are also quite widely used whenever the axes are numbered. The notation ∂_xh is used least frequently.

We will use whichever notation best suits the equation we are working with.

Directional derivatives

Normally, a partial derivative of a function with respect to one of its variables, say, x_j, takes the derivative of that "slice" of that function parallel to the x_j'th axis.

More precisely, we can think of cutting a function f(x₁,...,x_n) in space along the x_j'th axis, with keeping everything but the x_j variable constant.

From the definition, we have the partial derivative at a point p of the function along this slice as

${\partial \mathbf{f} \over \partial x_j} = \lim_{t\rightarrow 0} {\mathbf{f}(\mathbf{p}+t\mathbf{e}_j) - \mathbf{f}(\mathbf{p}) \over t}$

provided this limit exists.

Instead of the basis vector, which corresponds to taking the derivative along that axis, we can pick a vector in any direction (which we usually take as being a unit vector), and we take the directional derivative of a function as

${\partial \mathbf{f} \over \partial \mathbf{d}} = \lim_{t\rightarrow 0} {\mathbf{f}(\mathbf{p}+t\mathbf{d}) - \mathbf{f}(\mathbf{p}) \over t}$

where d is the direction vector.

If we want to calculate directional derivatives, calculating them from the limit definition is rather painful, but, we have the following: if f : Rⁿ → R is differentiable at a point p, |p|=1,

${\partial \mathbf{f} \over \partial \mathbf{d}} = D_\mathbf{p} \mathbf{f}(\mathbf{d})$

There is a closely related formulation which we'll look at in the next section.

Gradient vectors

The partial derivatives of a scalar tell us how much it changes if we move along one of the axes. What if we move in a different direction?

We'll call the scalar f, and consider what happens if we move an infintesimal direction dr=(dx,dy,dz), using the chain rule.

$\mathbf{df}=dx\frac{\partial f}{\partial x} + dy\frac{\partial f}{\partial y}+dz\frac{\partial f}{\partial z}$

This is the dot product of dr with a vector whose components are the partial derivatives of f, called the gradient of f

$\operatorname{grad} \mathbf{f} = \nabla \mathbf{f} = \left(\frac{\partial \mathbf{f}(\mathbf{p})}{\partial x_1},\cdots, \frac{\partial \mathbf{f}(\mathbf{p})}{\partial x_n}\right)$

We can form directional derivatives at a point p, in the direction d then by taking the dot product of the gradient with d

${\partial \mathbf{f}(\mathbf{p}) \over \partial \mathbf{d}} =\mathbf{d} \cdot \nabla \mathbf{f}(\mathbf{p})$.

Notice that grad f looks like a vector multiplied by a scalar. This particular combination of partial derivatives is commonplace, so we abbreviate it to

$\nabla = \left( \frac{\partial }{\partial x}, \frac{\partial }{\partial y}, \frac{\partial }{\partial z}\right)$

We can write the action of taking the gradient vector by writing this as an operator. Recall that in the one-variable case we can write d/dx for the action of taking the derivative with respect to x. This case is similar, but ∇ acts like a vector.

We can also write the action of taking the gradient vector as:

$\nabla = \left( \frac{\partial }{\partial x_1}, \frac{\partial }{\partial x_2}, \cdots \frac{\partial }{\partial x_n}\right)$

Properties of the gradient vector

Geometry

• Grad f(p) is a vector pointing in the direction of steepest slope of f. |grad f(p)| is the rate of change of that slope at that point.

For example, if we consider h(x, y)=x²+y². The level sets of h are concentric circles, centred on the origin, and

$\nabla h = (h_x,h_y) = 2(x,y)= 2 \mathbf{r}$

grad h points directly away from the origin, at right angles to the contours.

• Along a level set, (∇f)(p) is perpendicular to the level set {x|f(x)=f(p) at x=p}.

If dr points along the contours of f, where the function is constant, then df will be zero. Since df is a dot product, that means that the two vectors, df and grad f, must be at right angles, i.e the gradient is at right angles to the contours.

Algebraic properties

Like d/dx, ∇ is linear. For any pair of constants, a and b, and any pair of scalar functions, f and g

$\frac{d}{dx} (af+bg)= a\frac{d}{dx}f + b\frac{d}{dx}g \quad \nabla (af+bg) = a \nabla f + b \nabla g$

Since it's a vector, we can try taking its dot and cross product with other vectors, and with itself.

Divergence

If the vector function u maps Rⁿ to itself, then we can take the dot product of u and ∇. This dot product is called the divergence.

$\operatorname{div}\, \mathbf{u} = \nabla \cdot \mathbf{u} = \frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \cdots \frac{\partial u_n}{\partial x_n}$

If we look at a vector function like v=(1+x²,xy) we can see that to the left of the origin all the v vectors are converging towards the origin, but on the right they are diverging away from it.

Div u tells us how much u is converging or diverging. It is positive when the vector is diverging from some point, and negative when the vector is converging on that point.

Example:

For v=(1+x², xy), div v=3x, which is positive to the right of the origin, where v is diverging, and negative to the left of the origin, where v is converging.

Like grad, div is linear.

$\nabla \cdot \left( a\mathbf{u}+b\mathbf{v} \right) = a\nabla \cdot \mathbf{u} + b\nabla \cdot \mathbf{v}$

Later in this chapter we will see how the divergence of a vector function can be integrated to tell us more about the behaviour of that function.

To find the divergence we took the dot product of ∇ and a vector with ∇ on the left. If we reverse the order we get

$\mathbf{u} \cdot \nabla = u_x D_x + u_y D_y + u_z D_z$

To see what this means consider i·∇ This is D_x, the partial differential in the i direction. Similarly, u·∇ is the the partial differential in the u direction, multiplied by |u|

Curl

If u is a three-dimensional vector function on R³ then we can take its cross product with ∇. This cross product is called the curl.

$\operatorname{curl}\,\mathbf{u} = \nabla \times \mathbf{u} =\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ D_x & D_y & D_z \\ u_x & u_y & u_z \end{vmatrix}$

Curl u tells us if the vector u is rotating round a point. The direction of curl u is the axis of rotation.

We can treat vectors in two dimensions as a special case of three dimensions, with u_z=0 and D_zu=0. We can then extend the definition of curl u to two-dimensional vectors

$\operatorname{curl}\,\mathbf{u} = D_y u_x-D_x u_y$

This two dimensional curl is a scalar. In four, or more, dimensions there is no vector equivalent to the curl.

Example:
Consider u=(-y, x). These vectors are tangent to circles centred on the origin, so appear to be rotating around it anticlockwise.

$\operatorname{curl}\,\mathbf{u} = D_y(-y)-D_x x= -2$

Example
Consider u=(-y, x-z, y), which is similar to the previous example.

$\operatorname{curl}\,\mathbf{u} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ D_x & D_y & D_z \\ -y & x-z & y \end{vmatrix} =2\mathbf{i} + 2\mathbf{k}$

This u is rotating round the axis i+k

Later in this chapter we will see how the curl of a vector function can be integrated to tell us more about the behaviour of that function.

Product and chain rules

Just as with ordinary differentiation, there are product rules for grad, div and curl.

• If g is a scalar and v is a vector, then

the divergence of gv is

$\nabla \cdot (g\mathbf{v})=g \nabla \cdot \mathbf{v} + (\mathbf{v} \cdot \nabla) g$

the curl of gv is

$\nabla \times (g\mathbf{v}) = g(\nabla \times \mathbf{v})+ (\nabla g) \times \mathbf{v}$

• If u and v are both vectors then

the gradient of their dot product is

$\nabla ( \mathbf{u} \cdot \mathbf{v} ) = \mathbf{u} \times (\nabla \times \mathbf{v} ) + \mathbf{v} \times (\nabla \times \mathbf{u} ) + (\mathbf{u} \cdot \nabla) \mathbf{v} + (\mathbf{v} \cdot \nabla) \mathbf{u}$

the divergence of their cross product is

$\nabla \cdot ( \mathbf{u} \times \mathbf{v} ) = \mathbf{v}\cdot ( \nabla \times \mathbf{u} ) - \mathbf{u}\cdot ( \nabla \times \mathbf{v} )$

the curl of their cross product is

$\nabla \times ( \mathbf{u} \times \mathbf{v} ) = (\mathbf{v} \cdot \nabla ) \mathbf{u} - (\mathbf{u} \cdot \nabla) \mathbf{v} + \mathbf{u}(\nabla \cdot \mathbf{v}) - \mathbf{v}(\nabla \cdot \mathbf{u})$

We can also write chain rules. In the general case, when both functions are vectors and the composition is defined, we can use the Jacobian defined earlier.

$\left. \nabla \mathbf{u}(\mathbf{v}) \right|_{\mathbf{r}}= \mathbf{J}_{\mathbf{v}} \left. \nabla \mathbf{v} \right|_{\mathbf{r}}$

where J_u is the Jacobian of u at the point v.

Normally J is a matrix but if either the range or the domain of u is R¹ then it becomes a vector. In these special cases we can compactly write the chain rule using only vector notation.

• If g is a scalar function of a vector and h is a scalar function of g then

$\nabla h(g) = \frac{dh}{dg} \nabla g$

• If g is a scalar function of a vector then

$\nabla = (\nabla g) \frac{d}{dg}$

This substitution can be made in any of the equations containing ∇

Second order differentials

We can also consider dot and cross products of ∇ with itself, whenever they can be defined. Once we know how to simplify products of two ∇'s we'll know out to simplify products with three or more.

The divergence of the gradient of a scalar f is

$\nabla^2 f(x_1,x_2,\ldots x_n) = \frac{\partial^2 f}{\partial x_1^2}+ \frac{\partial^2 f}{\partial x_2^2}+\ldots+\frac{\partial^2 f}{\partial x_n^2}$

This combination of derivatives is the Laplacian of f. It is commmonplace in physics and multidimensional calculus because of its simplicity and symmetry.

We can also take the Laplacian of a vector,

$\nabla^2 \mathbf{u}(x_1,x_2,\ldots x_n) = \frac{\partial^2 \mathbf{u}}{\partial x_1^2}+ \frac{\partial^2 \mathbf{u}}{\partial x_2^2}+ \ldots + \frac{\partial^2 \mathbf{u}}{\partial x_n^2}$

The Laplacian of a vector is not the same as the divergence of its gradient

$\nabla (\nabla \cdot \mathbf{u})- \nabla^2 \mathbf{u}= \nabla \times (\nabla \times \mathbf{u})$

Both the curl of the gradient and the divergence of the curl are always zero.

$\nabla \times \nabla f=0 \quad \nabla \cdot (\nabla \times \mathbf{u})=0$

This pair of rules will prove useful.

Integration

We have already considered differentiation of functions of more than one variable, which leads us to consider how we can meaningfully look at integration.

In the single variable case, we interpret the definite integral of a function to mean the area under the function. There is a similar interpretation in the multiple variable case: for example, if we have a paraboloid in R³, we may want to look at the integral of that paraboloid over some region of the xy plane, which will be the volume under that curve and inside that region.

Riemann sums

When looking at these forms of integrals, we look at the Riemann sum. Recall in the one-variable case we divide the interval we are integrating over into rectangles and summing the areas of these rectangles as their widths get smaller and smaller. For the multiple-variable case, we need to do something similar, but the problem arises how to split up R², or R³, for instance.

To do this, we extend the concept of the interval, and consider what we call a n-interval. An n-interval is a set of points in some rectangular region with sides of some fixed width in each dimension, that is, a set in the form {x∈Rⁿ|a_i ≤ x_i ≤ b_i with i = 0,...,n}, and its area/size/volume (which we simply call its measure to avoid confusion) is the product of the lengths of all its sides.

So, an n-interval in R² could be some rectangular partition of the plane, such as {(x,y) | x ∈ [0,1] and y ∈ [0, 2]|}. Its measure is 2.

If we are to consider the Riemann sum now in terms of sub-n-intervals of a region Ω, it is

$\sum_{i; S_i \subset \Omega} f(x^*_i)m(S_i)$

where m(S_i) is the measure of the division of Ω into k sub-n-intervals S_i, and x^*_i is a point in S_i. The index is important - we only perform the sum where S_i falls completely within Ω - any S_i that is not completely contained in Ω we ignore.

As we take the limit as k goes to infinity, that is, we divide up Ω into finer and finer sub-n-intervals, and this sum is the same no matter how we divide up Ω, we get the integral of f over Ω which we write

$\int_\Omega f$

For two dimensions, we may write

$\int\int_\Omega f$

and likewise for n dimensions.

Iterated integrals

Thankfully, we need not always work with Riemann sums every time we want to calculate an integral in more than one variable. There are some results that make life a bit easier for us.

For R², if we have some region bounded between two functions of the other variable (so two functions in the form f(x) = y, or f(y) = x), between a constant boundary (so, between x = a and x =b or y = a and y = b), we have

$\int_a^b\,\int_{f(x)}^{g(x)} h(x,y)\,dy dx$

An important theorem (called Fubini's theorem) assures us that this integral is the same as

$\int\int_\Omega f$.

Order of integration

In some cases the first integral of the entire iterated integral is difficult or impossible to solve, therefore, it can be to our advantage to change the order of integration.

$\int_a^b\,\int_{f(x)}^{g(x)} h(x,y)\,dx dy$

$\int_c^d\,\int_{e(y)}^{f(y)} h(x,y)\,dy dx$

As of the writing of this, there is no set method to change an order of integration from dxdy to dydx or some other variable. Although, it is possible to change the order of integration in an x and y simple integration by simply switching the limits of integration around also, in non-simple x and y integrations the best method as of yet is to recreate the limits of the integration from the graph of the limits of integration.

In higher order integration that can't be graphed, the process can be very tedious. For example, dxdydz can be written into dzdydx, but first dxdydz must be switched to dydxdz and then to dydzdx and then to dzdydx (but since 3-dimensional cases can be graphed, doing this would be seemingly idiotic).

Parametric integrals

If we have a vector function, u, of a scalar parameter, s, we can integrate with respect to s simply by integrating each component of u seperately.

$\mathbf{v}(s)=\int \mathbf{u}(s)\,ds \Rightarrow v_i(s)=\int u_i(s)\,ds$

Similarly, if u is given a function of vector of parameters, s, lying in Rⁿ, integration with respect to the parameters reduces to a multiple integral of each component.

Line integrals

In one dimension, saying we are integrating from a to b uniquely specifies the integral.

In higher dimensions, saying we are integrating from a to b is not sufficient. In general, we must also specify the path taken between a and b.

We can then write the integrand as a function of the arclength along the curve, and integrate by components.

E.g, given a scalar function h(r) we write

$\int_C h(\mathbf{r}) \, d\mathbf{r} = \int_C h(\mathbf{r}) \frac{d\mathbf{r}}{ds}\,ds = \int_C h(\mathbf{r}(s)) \mathbf{t}(s) \, ds$

where C is the curve being integrated along, and t is the unit vector tangent to the curve.

There are some particularly natural ways to integrate a vector function, u, along a curve,

$\int_C \mathbf{u}\, ds \quad \int_C \mathbf{u} \cdot d\mathbf{r} \quad \int_C \mathbf{u} \times d\mathbf{r} \quad \int_C \mathbf{u} \cdot \,\mathbf{n} ds$

where the third possibility only applies in 3 dimensions.

Again, these integrals can all be written as integrals with respect to the arclength, s.

$\int_C \mathbf{u} \cdot d\mathbf{r} = \int_C \mathbf{u} \cdot \mathbf{t} \, ds \quad \int_C \mathbf{u} \times d\mathbf{r} = \int_C \mathbf{u} \times \mathbf{t}\, ds \quad$

If the curve is planar and u a vector lieing in the same plane, the second integral can be usefully rewritten. Say,

$\mathbf{u}=u_t \mathbf{t}+u_n \mathbf{n} + u_b \mathbf{b}$

where t, n, and b are the tangent, normal, and binormal vectors uniquely defined by the curve.

Then

$\mathbf{u} \times \mathbf{t}= -\mathbf{b} u_n + \mathbf{n} u_b$

For the 2-d curves specified b is the constant unit vector normal to their plane, and u_b is always zero.

Therefore, for such curves,

$\int_C \mathbf{u} \times d\mathbf{r} = \int_C \mathbf{u} \cdot \,\mathbf{n} ds$

Green's Theorem

Let C be a piecewise smooth, simple closed curve that bounds a region S on the Cartesian plane. If two function M(x,y) and N(x,y) are continuous and their partial derivatives are continuous, then

$\int \int_S (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) dA = \oint_C M dx + N dy = \oint_C \mathbf{F} \cdot d\mathbf{r}$

In order for Green's theorem to work there must be no singularities in the vector field within the boundaries of the curve.

Green's theorem works by summing the circulation in each infinitesimal segment of area enclosed within the curve.

Inverting differentials

We can use line integrals to calculate functions with specified divergence, gradient, or curl.

• If grad V = u

$V(\mathbf{p})=\int_{\mathbf{p}_0}^{\mathbf{p}} \mathbf{u} \cdot d\mathbf{r} + h(\mathbf{p})$

where h is any function of zero gradient and curl u must be zero.

• If div u = V

$\mathbf{u}(\mathbf{p}) =\int_{\mathbf{p}_0}^{\mathbf{p}} V \, d\mathbf{r} + \mathbf{w}(\mathbf{p})$

where w is any function of zero divergence.

• If curl u = v

$\mathbf{u}(\mathbf{p}) = \frac{1}{2}\int_{\mathbf{p}_0}^{\mathbf{p}} \mathbf{v} \times d\mathbf{r} + \mathbf{w}(\mathbf{p})$

where w is any function of zero curl.

For example, if V=r² then

$\operatorname{grad} V = 2(x, y, z) = 2 \mathbf{r}$

and

$\begin{matrix} \int_{\mathbf{0}}^{\mathbf{r}} 2\mathbf{u} \cdot d\mathbf{u} & = & \int_{\mathbf{0}}^{\mathbf{r}} 2 \left( udu + vdv+ wdw \right) \\ & = & \left[ u^2 \right]_{\mathbf{0}}^{\mathbf{r}} + \left[ v^2 \right]_{\mathbf{0}}^{\mathbf{r}} + \left[ w^2 \right]_{\mathbf{0}}^{\mathbf{r}} \\ & = & x^2 + y^2 + z^2 = r^2\\ \end{matrix}$

so this line integral of the gradient gives the original function.

Similarly, if v=k then

$\mathbf{u}(\mathbf{p}) = \int_{\mathbf{p}_0}^{\mathbf{p}} \mathbf{k} \times d\mathbf{r}$

Consider any curve from 0 to p=(x, y', z), given by r=r(s) with r(0)=0 and r(S)=p for some S, and do the above integral along that curve.

$\begin{matrix} \mathbf{u}(\mathbf{p}) & = & \int_0^S \mathbf{k} \times \frac{d\mathbf{r}}{ds} \, ds \\ & = & \int_0^S \left( \frac{dr_x}{ds} \mathbf{j}-\frac{dr_y}{ds} \mathbf{i} \right) \, ds \\ & = & \mathbf{j} \int_0^S \frac{dr_x}{ds} \, ds - \mathbf{i} \int_0^S \frac{dr_y}{ds} \, ds \\ & = & \mathbf{j} [r_x(s)]^S_0 - \mathbf{i} [r_y(s)]^S_0 \\ & = & p_x\mathbf{j} - p_y \mathbf{i} = x\mathbf{j} - y \mathbf{i} \end{matrix}$

and curl u is

$\frac{1}{2}\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ D_x & D_y & D_z \\ -y & x & 0 \end{vmatrix} = \mathbf{k} = \mathbf{v}$

as expected.

We will soon see that these three integrals do not depend on the path, apart from a constant.

Surface and Volume Integrals

Just as with curves, it is possible to parameterise surfaces then integrate over those parameters without regard to geometry of the surface.

That is, to integrate a scalar function V over a surface A parameterised by r and s we calculate

$\int_A V(x,y,z) \, dS = \int \int_A V(r,s) \det J \, dr ds$

where J is the Jacobian of the tranformation to the parameters.

To integrate a vector this way, we integrate each component seperately.

However, in three dimensions, every surface has an associated normal vector n, which can be used in integration. We write dS=ndS.

For a scalar function, V, and a vector function, v, this gives us the integrals

$\int_A V \, \mathbf{dS} \quad \int_A \mathbf{v} \cdot \mathbf{dS} \quad \int_A \mathbf{v} \times \mathbf{dS}$

These integrals can be reduced to parametric integrals but, written this way, it is clear that they reflect more of the geometry of the surface.

When working in three dimensions, dV is a scalar, so there is only one option for integrals over volumes.

Gauss's divergence theorem

We know that, in one dimension,

$\int_a^b Df dx = f|_a^b$

Integration is the inverse of differentiation, so integrating the differential of a function returns the original function.

This can be extended to two or more dimensions in a natural way, drawing on the analogies between single variable and multivariable calculus.

The analog of D is ∇, so we should consider cases where the integrand is a divergence.

Instead of integrating over a one-dimensional interval, we need to integrate over a n-dimensional volume.

In one dimension, the integral depends on the values at the edges of the interval, so we expect the result to be connected with values on the boundary.

This suggests a theorem of the form,

$\int_V \nabla \cdot \mathbf{u} \, dV = \int_{\partial V} \mathbf{n} \cdot \mathbf{u} dS$

This is indeed true, for vector fields in any number of dimensions.

This is called Gauss's theorem.

There are two other, closely related, theorems for grad and curl:

• $\int_V \nabla u \, dV = \int_{\partial V} u \mathbf{n} dS$,

• $\int_V \nabla \times \mathbf{u} \, dV = \int_{\partial V} \mathbf{n} \times \mathbf{u} dS$,

with the last theorem only being valid where curl is defined.

Stokes' curl theorem

These theorems also hold in two dimensions, where they relate surface and line integrals. Gauss's divergence theorem becomes

$\int_S \nabla \cdot \mathbf{u} \, dS = \oint_{\partial S} \mathbf{n} \cdot \mathbf{u} ds$

where s is arclength along the boundary curve and the vector n is the unit normal to the curve that lies in the surface S, i.e in the tangent plane of the surface at its boundary, which is not necessarily the same as the unit normal associated with the boundary curve itself.

Similarly, we get

$\int_S \nabla \times \mathbf{u} \, dS = \int_C \mathbf{n} \times \mathbf{u} ds \quad (1)$,

where C is the boundary of S

In this case the integral does not depend on the surface S.

To see this, suppose we have different surfaces, S₁ and S₂, spanning the same curve C, then by switching the direction of the normal on one of the surfaces we can write

$\int_{S_1+S_2} \nabla \times \mathbf{u} \, dS = \int_S \nabla \times \mathbf{u} \, dS - \int_S \nabla \times \mathbf{u} \, dS \quad (2)$

The left hand side is an integral over a closed surface bounding some volume V so we can use Gauss's divergence theorem.

$\int_{S_1+S_2} \nabla \times \mathbf{u} \, dS = \int_V \nabla \cdot \nabla \times \mathbf{u} \, dV$

but we know this integrand is always zero so the right hand side of (2) must always be zero, i.e the integral is independent of the surface.

This means we can choose the surface so that the normal to the curve lying in the surface is the same as the curves intrinsic normal.

Then, if u itself lies in the surface, we can write

$\mathbf{u}=\left( \mathbf{u} \cdot \mathbf{n} \right) \mathbf{n} + \left( \mathbf{u} \cdot \mathbf{t} \right) \mathbf{t}$

just as we did for line integrals in the plane earlier, and substitute this into (1) to get

$\int_S \nabla \times \mathbf{u} \, dS = \int_C \mathbf{u} \cdot d\mathbf{r}$

This is Stokes' curl theorem < Calculus/Outline

Ordinary differential equations involve equations containing:

• variables
• functions
• their derivatives

and their solutions.

In studying integration, you already have considered solutions to very simple differential equations. For example, when you look to solving

$\int f(x) \,dx=g(x)$

for g(x), you are really solving the differential equation

$g'(x) = f(x) \,$

Notations and terminology

The notations we use for solving differential equations will be crucial in the ease of solubility for these equations.

This document will be using three notations primarily:

• f' to denote the derivative of f
• D f to denote the derivative of f
• ${df \over dx}$ to denote the derivative of f (for separable equations).

Terminology

Consider the differential equation

$3 f^{\prime \prime}(x)+5xf(x)=11$

Since the equation's highest derivative is 2, we say that the differential equation is of order 2.

Some simple differential equations

A key idea in solving differential equations will be that of Integration

Let us consider the second order differential equation (remember that a function acts on a value.)

$f''(x)=2 \,$

How would we go about solving this?. It tells us that on differentiating twice, we obtain the constant 2 so, if we integrate twice, we should obtain our result.

Integrating once first of all:

$\int f''(x) \,dx = \int 2 \,dx$

$f'(x)=2x+C_1 \,$

We have transformed the apparently difficult second order differential equation into a rather simpler one, viz.

$f'(x)=2x+C_1 \,$

This equation tells us that if we differentiate a function once, we get $2x+C_1$. If we integrate once more, we should find the solution.

$\int f'(x) \,dx = \int 2x+C_1 \,dx$

$f(x)=x^2+C_1x+C_2 \,$

This is the solution to the differential equation. We will get $f''=2 \,$ for all values of $C_1$ and $C_2$.

The values $C_1$ and $C_2$ are related to quanities known as initial conditions.

Why are initial conditions useful? ODEs (ordinary differential equations) are useful in modelling physical conditions. We may wish to model a certain physical system which is initially at rest (so one initial condition may be zero), or wound up to some point (so an initial condition may be nonzero and be say 5 for instance) and we may wish to see how the system reacts under such an initial condition.

When we solve a system with given initial conditions, we substitute them after our process of integration.

Example

When we solved $f''=2 \,$ say we had the initial conditions $f'(0)=3 \,$ and $f(0)=2 \,$. (Note, initial conditions need not occur at f(0)).

After we integrate we make subsitutions:

$f'(0)=20(0)+C_1 \,$

$3=C_1 \,$

$\int f'(x) \,dx = \int 2x+3 \,dx$

$f(x)=x^2+3x+C_2 \,$

$f(0)=0^2+3(0)+C_2 \,$

$2=C_2 \,$

$f(x)=x^2+3x+2 \,$

Without initial conditions, the answer we obtain is known as the general solution or the solution to the family of equations. With them, our solution is known as a specific solution.

Basic first order DEs

In this section we will consider four main types of differential equations:

• separable
• homogeneous
• linear
• exact

There are many other forms of differential equation, however, and these will be dealt with in the next section

Separable equations

A separable equation is in the form (using dy/dx notation which will serve us greatly here)

${dy \over dx} = f(x)/g(y)$

Previously we have only dealt with simple differential equations with g(y)=1. How do we solve such a separable equation as above?

We group x and dx terms together, and y and dy terms together as well.

$g(y)\ dy = f(x)\ dx$

Integrating both sides with respect to y on the left hand side and x on the right hand side:

$\int g(y)\,dy=\int f(x)\,dx+C$

we will obtain the solution.

Worked example

Here is a worked example illustrating the process.

We are asked to solve

${dy \over dx} = 3x^2y$

Separating

${dy \over y} = (3x^2)\,dx$

Integrating

$\int {dy \over y} = \int 3x^2\,dx$

$\ln{y}=x^3+C \,\!$

$y=e^{x^3+C}$

Letting $k = e^C$ where k is a constant we obtain

$y=ke^{x^3}$

which is the general solution.

Verification

This step does not need be part of your working, but if you have time, you can verify your answer by differentiation. We obtained

$y=ke^{x^3}$

as the solution to

${dy \over dx} = 3x^2y$

Differentiating the solution,

${dy \over dx} = 3kx^2e^{x^3}$

Since $y=ke^{x^3}$, we can write

${dy \over dx} = 3x^2y$

We see that we obtain our original differential equation, so we can confirm our working as being correct.

Homogeneous equations

A homogeneous equation is in the form

${dy \over dx} = f(y/x)$

This looks difficult as it stands, however we can utilize the substitution

$v = {y \over x}$

so that we are now dealing with F(v) rather than F(y/x).

Now we can express y in terms of v, as y=xv and use the product rule.

The equation above then becomes, using the product rule

${dy \over dx} = v+x{dv \over dx}$

Then

$v+x{dv \over dx} = f(v)$

$x{dv \over dx} = f(v)-v$

${dv \over dx} = {f(v)-v \over x}$

which is a separable equation and can be solved as above.

However let's look at a worked equation to see how homogeneous equations are solved.

Worked example

We have the equation

${dy \over dx} = {y^2 + x^2 \over yx}$

This does not appear to be immediately separable, but let us expand to get

${dy \over dx} = {y^2 \over yx} + {x^2 \over yx}$

${dy \over dx} = {x \over y} + {y \over x}$

Substituting y=xv which is the same as substituting v=y/x:

${dy \over dx} = 1/v + v$

Now

$v+x{dv \over dx} = 1/v + v$

Cancelling v from both sides

$x{dv \over dx} = 1/v$

Separating

$v\, dv = dx/x$

Integrating both sides

${1 \over 2}v^2+C= \ln(x) \,$

${1 \over 2}\left({y \over x}\right)^2= \ln(x)-C$

$y^2 = 2x^2 \ln(x) - 2Cx^2 \,$

$y = x\sqrt{2 \ln(x) - 2C}$

which is our desired solution.

Linear equations

A linear first order differential equation is a differential equation in the form

$a(x){dy \over dx} + b(x)y=c(x)$

Multiplying or dividing this equation by any non-zero function of x makes no difference to its solutions so we could always divide by a(x) to make the coefficient of the differential 1, but writing the equation in this more general form may offer insights.

At first glance, it is not possible to integrate the left hand side, but there is one special case. If b happens to be the differential of a then we can write

$a(x){dy \over dx} + b(x)y = a(x){dy \over dx} + y{da \over dx} = {d \over dx}a(x)y$

and integration is now straightforward.

Since we can freely multiply by any function, lets see if we can use this freedom to write the left hand side in this special form.

We multiply the entire equation by an arbitary, I(x), getting

$aI{dy \over dx} + bIy=cI$

then impose the condition

$\frac{d}{dx}aI = bI$

If this is satisified the new left hand side will have the special form. Note that multiplying I by any constant will leave this condition still satisfied.

Rearranging this condition gives

$\frac{1}{I}\frac{dI}{dx} = \frac{b-\frac{da}{dx}}{a}$

We can integrate this to get

$\ln I(x) = \int \frac{b(z)}{a(z)}dz - \ln a(x) + c \quad I(x)=\frac{k}{a(x)}e^{\int \frac{b(z)}{a(z)}dz}$

We can set the constant k to be 1, since this makes no difference.

Next we use I on the original differential equation, getting

$e^{\int \frac{b(z)}{a(z)}dz}{dy \over dx} + e^{\int \frac{b(z)}{a(z)}dz} \frac{b(x)}{a(x)}y =e^{\int \frac{b(z)}{a(z)}dz}\frac{c(x)}{a(x)}$

Because we've chosen I to put the left hand side in the special form we can rewrite this as

${d \over dx}(ye^{\int \frac{b(z)}{a(z)}dz}) = e^{\int \frac{b(z)}{a(z)}dz}\frac{c(x)}{a(x)}$

Integrating both sides and dividing by I we obtain the final result

$y = e^{-\int \frac{b(z)}{a(z)}dz} \left(\int e^{\int \frac{b(z)}{a(z)}dz}\frac{c(x)}{a(x)}dx + C\right)$

We call I an integrating factor. Similar techniques can be used on some other calclulus problems.

Example

Consider

$\frac{dy}{dx} + y \tan x = 1 \quad y(0)=0$

First we calculate the integrating factor.

$I=e^{\int \tan z dz} = e^ {\ln \sec x} = \sec x$

Multiplying the equation by this gives

$\sec x \frac{dy}{dx} + y \sec x \tan x = \sec x$

or

$\frac{d}{dx} y\sec x = \sec x$

We can now integrate

$y = \cos x \int_0^x \sec z \, dz = \cos x \ln (\sec x + \tan x)$

Exact equations

An exact equation is in the form

f(x, y) dx + g(x, y) dy = 0

and, has the property that

D_x f = D_y g

(If the differential equation does not have this property then we can't proceed any further).

As a result of this, if we have an exact equation then there exists a function h(x, y) such that

D_y h = f and D_x h = g

So then the solutions are in the form

h(x, y) = c

by using the fact of the total differential. We can find then h(x, y) by integration

Basic second and higher order ODE's

The generic solution of a n^th order ODE will contain n constants of integration. To calculate them we need n more equations. Most often, we have either

boundary conditions, the values of y and its derivatives take for two different values of x

or

initial conditions, the values of y and its first n-1 derivatives take for one particular value of x.

Reducible ODE's

1. If the independent variable, x, does not occur in the differential equation then its order can be lowered by one. This will reduce a second order ODE to first order.

Consider the equation:

$F\left(y,\frac{dy}{dx},\frac{d^2y}{dx^2}\right)=0$

Define

$u=\frac{dy}{dx}$

Then

$\frac{d^2y}{dx^2}=\frac{du}{dx}=\frac{du}{dy}\cdot\frac{dy}{dx}=\frac{du}{dy}\cdot u$

Substitute these two expression into the equation and we get

$F\left(y,u,\frac{du}{dy}\cdot u\right)$=0

which is a first order ODE

Example

Solve

$1+2y^2\operatorname{D}^2y=0$

if at x=0,  y=Dy=1

First, we make the substitution, getting

$1+2y^2 u \frac{du}{dy}=0$

This is a first order ODE. By rearranging terms we can separate the variables

$udu=-\frac{dy}{2y^2}$

Integrating this gives

$u^2/2=c+1/2y$

We know the values of y and u when x=0 so we can find c

$c=u^2/2-1/2y=1^2/2-1/(2\cdot 1)=1/2-1/2=0$

Next, we reverse the substitution

$\frac{dy}{dx}^2=u^2=\frac{1}{y}$

and take the square root

$\frac{dy}{dx}=\pm \frac{1}{\sqrt{y}}$

To find out which sign of the square root to keep, we use the initial condition, Dy=1 at x=0, again, and rule out the negative square root. We now have another separable first order ODE,

$\frac{dy}{dx}=\frac{1}{\sqrt{y}}$

Its solution is

$\frac{2}{3}y^\frac{3}{2}= x+d$

Since y=1 when x=0, d=2/3, and

$y=\left(1 + \frac{3x}{2} \right)^\frac{2}{3}$

2. If the dependent variable, y, does not occur in the differential equation then it may also be reduced to a first order equation.

Consider the equation:

$F\left(x,\frac{dy}{dx},\frac{d^2y}{dx^2}\right)=0$

Define

$u=\frac{dy}{dx}$

Then

$\frac{d^2y}{dx^2}=\frac{du}{dx}$

Substitute these two expressions into the first equation and we get

$F\left(x,u,\frac{du}{dx}\right)$=0

which is a first order ODE

Linear ODEs

An ODE of the form

$\frac{d^ny}{dx^n}+a_1(x)\frac{d^{n-1}y}{dx^{n-1}}+ ... +a_n y=F(x)$

is called linear. Such equations are much simpler to solve than typical non-linear ODEs. Though only a few special cases can be solved exactly in terms of elementary functions, there is much that can be said about the solution of a generic linear ODE. A full account would be beyond the scope of this book

If F(x)=0 for all x the ODE is called homogeneous

Two useful properties of generic linear equations are

1. Any linear combination of solutions of an homogenous linear equation is also a solution.
2. If we have a solution of a nonhomogenous linear equation and we add any solution of the corresponding homogenous linear equation we get another solution of the nonhomogenous linear equation

Variation of constants

Suppose we have a linear ODE,

$\frac{d^ny}{dx^n}+a_1(x)\frac{d^{n-1}y}{dx^{n-1}}+ ... +a_n y=0$

and we know one solution, y=w(x)

The other solutions can always be written as y=wz. This substitution in the ODE will give us terms involving every differential of z upto the n^th, no higher, so we'll end up with an n^th order linear ODE for z.

We know that z is constant is one solution, so the ODE for z must not contain a z term, which means it will effectively be an n-1^th order linear ODE. We will have reduced the order by one.

Lets see how this works in practice.

Example

Consider

$\frac{d^2y}{dx^2}+\frac{2}{x}\frac{dy}{dx}-\frac{6}{x^2}y=0$

One solution of this is y=x², so substitute y=zx² into this equation.

$\left( x^2\frac{d^2z}{dx^2}+2x\frac{dz}{dx}+2z\right) +\frac{2}{x} \left( x^2\frac{dz}{dx}+2xz \right) -\frac{6}{x^2}x^2 z=0$

Rearrange and simplify.

$x^2 D^2 z + 6xD z=0$

This is first order for Dz. We can solve it to get

$z=A x^{-5} \quad y=A x^{-3}$

Since the equation is linear we can add this to any multiple of the other solution to get the general solution,

$y=A x^{-3} + B x^2$

Linear homogenous ODE's with constant coefficents

Suppose we have a ODE

$(D^n+a_1 D^{n-1}+ ... + a_{n-1}D+a_0)y=0$

we can take an inspired guess at a solution (motivate this)

$y=e^{px}$

For this function Dⁿy=pⁿy so the ODE becomes

$(p^n+a_1 p^{n-1}+ ... + a_{n-1}p+a_0)y=0$

y=0 is a trivial solution of the ODE so we can discard it. We are then left with the equation

$p^n+a_1 p^{n-1}+ ... + a_{n-1}p+a_0)=0$

This is called the characteristic equation of the ODE.

It can have up to n roots, p₁, p₂ … p_n, each root giving us a different solution of the ODE.

Because the ODE is linear, we can add all those solution together in any linear combination to get a general solution

$y=A_1 e^{p_1 x} +A_2 e^{p_2 x} + ... + A_n e^{p_n x}$

To see how this works in practice we will look at the second order case. Solving equations like this of higher order uses the exact same principles; only the algebra is more complex.

Second order

If the ODE is second order,

$D^2 y + bDy+cy=0$

then the characteristic equation is a quadratic,

$p^2+bp+c=0$

with roots

$p_{\pm}=\frac{-b \pm \sqrt{b^2-4c}}{2}$

What these roots are like depends on the sign of b²-4c, so we have three cases to consider.

1/ b² > 4c

In this case we have two different real roots, so we can write down the solution straight away.

$y=A_{+}e^{p_{+}}+A_{-}e^{p_{-}}$

2/ b² < 4c

In this case, both roots are imaginary. We could just put them directly in the formula, but if we are interested in real solutions it is more useful to write them another way.

Defining k²=4c-b², then the solution is

$y=A_{+}e^{ikx-\frac{bx}{2}}+A_{-}e^{-ikx-\frac{bx}{2}}$

For this to be real, the A's must be complex conjugates

$A_{\pm}=A e^{\pm ia}$

Make this substitution and we can write,

$y=A e^{-bx/2}\cos (kx+a)$

If b is positive, this is a damped oscillation.

3/ b² = 4c

In this case the characteristic equation only gives us one root, p=-b/2. We must use another method to find the other solution.

We'll use the method of variation of constants. The ODE we need to solve is,

$D^2 y -2pDy+p^2y=0$

rewriting b and c in terms of the root. From the characteristic equation we know one solution is $y=e^{px}$ so we make the substitution $y=ze^{px}$, giving

$(e^{px}D^2z+2pe^{px}Dz+p^2e^{px}z)-2p(e^{px}Dz+pe^{px}z)+p^2e^{px}z=0$

This simplifies to D²z=0, which is easily solved. We get

$z=Ax+B \quad y=(Ax+B)e^{px}$

so the second solution is the first multiplied by x.

Higher order linear constant coefficent ODE's behave similarly: an exponential for every real root of the characteristic and a exponent multiplied by a trig factor for every complex conjugate pair, both being multiplied by a polynomial if the root is repeated.

E.g, if the characteristic equation factors to

$(p-1)^4(p-3)(p^2+1)^2=0$

the general solution of the ODE will be

$y=(A+Bx+Cx^2+Dx^3)e^x + Ee^{3x}+ F \cos (x+a) +Gx \cos(x+b)$

The most difficult part is finding the roots of the characteristic equation.

Linear nonhomogenous ODEs with constant coefficents

First, let's consider the ODE

$Dy-y=x$

a nonhomogenous first order ODE which we know how to solve.

Using the integrating factor e^-x we find

$y=c e^{-x} +1 -x$

This is the sum of a solution of the coresponding homogenous equation, and a polynomial.

Nonhomogeneous ODE's of higher order behave similarly.

If we have a single solution, y_p of the nonhomogeneous ODE, called a particular solution,

$(D^n+a_1 D^{n-1} + \cdots + a_n)y=F(x)$

then the general solution is y=y_p+y_h, where y_h is the general solution of the homogeneous ODE.

Find y_p for an arbitary F(x) requires methods beyond the scope of this chapter, but there are some special cases where finding y_p is straightforward.

Remember that in the first order problem y_p for a polynomial F(x) was itself a polynomial of the same order. We can extend this to higher orders.

Example:

$D^2y+y=x^3-x+1$

Consider a particular solution

$y_p=b_0+b_1 x+b_2 x^2 + x^3$

Substitute for y and collect coefficients

$x^3 + b_2 x^2 +(6+b_1)x +(2b_2+b_0)=x^3-x+1$

So b₂=0, b₁=-7, b₀=1, and the general solution is

$y=a \sin x + b \cos x + 1 -7x + x^3$

This works because all the derivatives of a polynomial are themselves polynomials.

Two other special cases are

$F(x)=P_n e^{kx} \quad y_p(x)=Q_n e^{kx}$

$F(x)=A_n \sin kx +B_n \cos kx \quad y_p(x)=P_n \sin kx +Q_n \cos kx$

where P_n,Q_n,A_n, and B_n are all polynomials of degree n.

Making these substitutions will give a set of simultaneous linear equations for the coefficients of the polynomials.

Non-Linear ODEs

If the ODE is not linear, first check if it is reducible. If it is neither linear nor reducible there is no generic method of solution. You may, with sufficient ingenuity and algebraic skill, be able to transform it into a linear ODE.

If that is not possible, solving the ODE is beyond the scope of this book By Full Prof. Meysam Soltani

First order

Any partial differential equation of the form

$h_1 \frac {\partial u}{\partial x_1} + h_2 \frac {\partial u}{\partial x_2} \cdots + h_n \frac {\partial u}{\partial x_n} = b$

where h₁, h₂ … h_n, and b are all functions of both u and Rⁿ can be reduced to a set of ordinary differential equations.

To see how to do this, we will first consider some simpler problems.

Special cases

We will start with the simple PDE

$u_z(x,y,z)=u(x,y,z) \quad (1)$

Because u is only differentiated with respect to z, for any fixed x and y we can treat this like the ODE, du/dz=u. The solution of that ODE is ce^z, where c is the value of u when z=0, for the fixed x and y

Therefore, the solution of the PDE is

$u(x, y, z)=u(x,y,0) e^z$

Instead of just having a constant of integration, we have an arbitary function. This will be true for any PDE.

Notice the shape of the solution, an arbitary function of points in the xy, plane, which is normal to the 'z' axis, and the solution of an ODE in the 'z' direction.

Now consider the slightly more complex PDE

$a_x u_x +a_y u_y + a_z u_z=h(u) \quad (2)$

where h can be any function, and each a is a real constant.

We recognise the left hand side as being a·∇, so this equation says that the differential of u in the a direction is h(u). Comparing this with the first equation suggests that the solution can be written as an arbitary function on the plane normal to a combined with the solution of an ODE.

Remembering from Calculus/Vectors that any vector r can be split up into components parallel and perpendicular to a,

$\mathbf{r}= \mathbf{r}_\perp + \mathbf{r}_\| = \left( \mathbf{r}-\frac{(\mathbf{r} \cdot \mathbf{a})\mathbf{a}}{|\mathbf{a}|^2} \right) + \frac{(\mathbf{r} \cdot \mathbf{a})\mathbf{a}}{|\mathbf{a}|^2}$

we will use this to split the components of r in a way suggested by the analogy with (1).

Let's write

$\mathbf{r} = (x,y,z) = \mathbf{r}_\perp + s\mathbf{a} \quad s = \frac{\mathbf{r} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}}$

and substitute this into (2), using the chain rule. Because we are only differentiating in the a direction, adding any function of the perpendicular vector to s will make no difference.

First we calculate grad s, for use in the chain rule,

$\nabla s = \frac{\mathbf{a}}{a^2}$

On making the substitution into (2), we get,

$h(u)= \mathbf{a} \cdot \nabla s \frac{d}{ds} u(s)= \frac{\mathbf{a} \cdot \mathbf{a}}{\mathbf{a} \cdot \mathbf{a}} \frac{d}{ds} u(s)= \frac{du}{ds}$

which is an ordinary differential equation with the solution

$s=c(\mathbf{r}_\perp) + \int^u \frac{dt}{h(t)}$

The constant c can depend on the perpendicular components, but not upon the parallel coordinate. Replacing s with a monotonic scalar function of s multiplies the ODE by a function of s, which doesn't affect the solution.

Example:

$u(x,t)_x=u(x,t)_t$

For this equation, a is (1, -1), s=x-t, and the perpendicular vector is (x+t)(1, 1). The reduced ODE is du/ds=0 so the solution is

u=f(x+t)

To find f we need initial conditions on u. Are there any constraints on what initial conditions are suitable?

Consider, if we are given

• u(x,0), this is exactly f(x),
• u(3t,t), this is f(4t) and f(t) follows immediately
• u(t³+2t,t), this is f(t³+3t) and f(t) follows, on solving the cubic.
• u(-t,t), then this is f(0), so if the given function isn't constant we have a inconsistancy, and if it is the solution isn't specified off the initial line.

Similarly, if we are given u on any curve which the lines x+t=c intersect only once, and to which they are not tangent, we can deduce f.

For any first order PDE with constant coefficients, the same will be true. We will have a set of lines, parallel to r=at, along which the solution is gained by integrating an ODE with initial conditions specified on some surface to which the lines aren't tangent.

If we look at how this works, we'll see we haven't actually used the constancy of a, so let's drop that assumption and look for a similar solution.

The important point was that the solution was of the form u=f(x(s),y(s)), where (x(s),y(s)) is the curve we integrated along -- a straight line in the previous case. We can add constant functions of integration to s without changing this form.

Consider a PDE,

$a(x,y)u_x+b(x,y)u_y=c(x,y,u)$

For the suggested solution, u=f(x(s),y(s)), the chain rule gives

$\frac{du}{ds} = \frac{dx}{ds}u_x+ \frac{dy}{ds}u_y$

Comparing coefficients then gives

$\frac{dx}{ds}=a(x,y) \quad \frac{dy}{ds}=b(x,y) \quad \frac{du}{ds}=c(x,y,u)$

so we've reduced our original PDE to a set of simultaneous ODE's. This procedure can be reversed.

The curves (x(s),y(s)) are called characteristics of the equation.

Example: Solve $yu_x=xu_y$ given u=f(x) for x≥0 The ODE's are

$\frac{dx}{ds}=y \quad \frac{dy}{ds}=-x \quad \frac{du}{ds}=0$

subject to the initial conditions at s=0,

$x(0)=r \quad y(0)=0 \quad u(0)=f(r) \quad r \ge 0$

This ODE is easily solved, giving

$x(s)=r \cos s \quad y(s)=r \sin s \quad u(s)=f(r)$

so the characteristics are concentric circles round the origin, and in polar coordinates u(r,θ)=f(r)

Considering the logic of this method, we see that the independance of a and b from u has not been used either, so that assumption too can be dropped, giving the general method for equations of this quasilinear form.

Quasilinear

Summarising the conclusions of the last section, to solve a PDE

$a_1(u,\mathbf{x}) \frac {\partial u}{\partial x_1} + a_2(u,\mathbf{x}) \frac {\partial u}{\partial x_2} \cdots + a_n(u,\mathbf{x}) \frac {\partial u}{\partial x_n} = b(u,\mathbf{x})$

subject to the initial condition that on the surface, (x₁(r₁,…,r_n-1, …x_n(r₁,…,r_n-1), u=f(r₁,…,r_n-1) --this being an arbitary paremetrisation of the initial surface--

• we transform the equation to the equivalant set of ODEs,

$\frac{dx_1}{ds}=a_1 \ \ldots \ \frac{dx_n}{ds}=a_n \quad \frac{du}{ds}=b$

subject to the initial conditions

$x_i(0)=f(r_1, \ldots ,r_{n-1}) \quad u=f(r_1,r_2, \ldots r_{n-1})$

• Solve the ODE's, giving x_i as a function of s and the r_i.
• Invert this to get s and the r_i as functions of the x_i.
• Substitute these inverse functions into the expression for u as a function of s and the r_i obtained in the second step.

Both the second and third steps may be troublesome.

The set of ODEs is generally non-linear and without analytical solution. It may even be easier to work with the PDE than with the ODEs.

In the third step, the r_i together with s form a coordinate system adapted for the PDE. We can only make the inversion at all if the Jacobian of the transformation to Cartesian coordinates is not zero,

$\begin{vmatrix} \frac{\partial x_1}{\partial r_1}&\cdots &\frac{\partial x_1}{\partial r_{n-1}} & a_1 \\ \vdots & \ddots & & \vdots \\ \frac{\partial x_n}{\partial r_1}&\cdots &\frac{\partial x_n}{\partial r_{n-1}} & a_n \end{vmatrix} \ne 0$

This is equivalent to saying that the vector (a₁, &hellip:, a_n) is never in the tangent plane to a surface of constant s.

If this condition is not false when s=0 it may become so as the equations are integrated. We will soon consider ways of dealing with the problems this can cause.

Even when it is technically possible to invert the algebraic equations it is obviously inconvenient to do so.

Example

To see how this works in practice, we will
a/ consider the PDE,

$uu_x+u_y+u_t=0$

with generic initial condition,

$u=f(x,y) \mbox{ on } t=0$

Naming variables for future convenience, the corresponding ODE's are

$\frac{dx}{d\tau}=u \quad \frac{dy}{d\tau}=1 \quad \frac{dz}{d\tau}=1 \quad \frac{du}{d\tau}=0 \quad$

subject to the initial conditions at τ=0

$x=r \quad y=s \quad t=0 \quad u=f(r,s)$

These ODE's are easily solved to give

$x=r+f(r,s)\tau \quad y=s+\tau \quad t=\tau \quad u=f(r,s)$

These are the parametric equations of a set of straight lines, the characteristics.

The determinant of the Jacobian of this coordinate transformation is

$\begin{vmatrix} 1+\tau\frac{\partial f}{\partial r} & \tau\frac{\partial f}{\partial s} & f \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{vmatrix}=1+\tau\frac{\partial f}{\partial r}$

This determinant is 1 when t=0, but if f_r is anywhere negative this determinant will eventually be zero, and this solution fails.

In this case, the failure is because the surface $sf_r=-1$ is an envelope of the characteristics.

For arbitary f we can invert the transformation and obtain an implicit expression for u

$u=f(x-tu,y-x)$

If f is given this can be solved for u.

1/ $f(x,y)=ax$, The implicit solution is

$u=a(x-tu) \Rightarrow u=\frac{ax}{1+at}$

This is a line in the u-x plane, rotating clockwise as t increases. If a is negative, this line eventually become vertical. If a is positive, this line tends towards u=0, and the solution is valid for all t.

2/ f(x,y)=x², The implicit solution is

$u=(x-tu)^2 \Rightarrow u=\frac{1+2tx-\sqrt{1+4tx}}{2t^2}$

This solution clearly fails when $1+4tx<0$, which is just when $sf_r=-1$ . For any t>0 this happens somewhere. As t increases this point of failure moves toward the origin.

Notice that the point where u=0 stays fixed. This is true for any solution of this equation, whatever f is.

We will see later that we can find a solution after this time, if we consider discontinuous solutions. We can think of this as a shockwave.

3/ $f(x,y)=\sin (xy)$
The implicit solution is

$u(x,y,t)=\sin \left( (x-tu)(y-x) \right)$

and we can not solve this explitly for u. The best we can manage is a numerical solution of this equation.

b/We can also consider the closely related PDE

$uu_x+u_y+u_t=y$

The corresponding ODE's are

$\frac{dx}{d\tau}=u \quad \frac{dy}{d\tau}=1 \quad \frac{dz}{d\tau}=1 \quad \frac{du}{d\tau}=y \quad$

subject to the initial conditions at τ=0

$x=r \quad y=s \quad t=0 \quad u=f(r,s)$

These ODE's are easily solved to give

$x=r+\tau f+\frac{1}{2}s\tau^2+\frac{1}{6}\tau^3 \quad y=s+\tau \quad t=\tau \quad u=f+s\tau+\frac{1}{2}\tau^2$

Writing f in terms of u, s, and τ, then substituting into the equation for x gives an implicit solution

$u(x,y,t)=f(x-ut+\frac{1}{2}yt^2-\frac{1}{6}t^3,y-t)+yt-\frac{1}{2}t^2$

It is possible to solve this for u in some special cases, but in general we can only solve this equation numerically. However, we can learn much about the global properties of the solution from further analysis

Characteristic initial value problems

What if initial conditions are given on a characteristic, on an envelope of characteristics, on a surface with characteristic tangents at isolated points?

Discontinuous solutions

So far, we've only considered smooth solutions of the PDE, but this is too restrictive. We may encounter initial conditions which aren't smooth, e.g.

$u_t = c u_x \quad u(x,0)= \left\{ \begin{matrix} 1, & x \ge 0 \\ 0, & x < 0 \end{matrix} \right.$

If we were to simply use the general solution of this equation for smooth initial conditions,

$u(x,t) = u(x + ct, 0) \,$

we would get

$u(x,t) = \left\{ \begin{matrix} 1, & x+ct \ge 0 \\ 0, & x+ct < 0 \end{matrix} \right.$

which appears to be a solution to the original equation. However, since the partial differentials are undefined on the characteristic x+ct=0, so it becomes unclear what it means to say that the equation is true at that point.

We need to investigate further, starting by considering the possible types of discontinuities.

If we look at the derivations above, we see we've never use any second or higher order derivatives so it doesn't matter if they aren't continuous, the results above will still apply.

The next simplest case is when the function is continuous, but the first derivative is not, e.g |x|. We'll initially restrict ourselves to the two-dimensional case, u(x, t) for the generic equation.

$a(x,t)u_x + b(x,t)u_t = c(u,x,t) \quad (1)$

Typically, the discontinuity is not confined to a single point, but is shared by all points on some curve, (x₀(s), t₀(s) )

Then we have

$\begin{matrix} x>x_0 & \lim_{x \rarr x_0} = u_+ \\ x<x_0 & \lim_{x \rarr x_0} = u_- \end{matrix}$

We can then compare u and its derivatives on both sides of this curve.

It will prove useful to name the jumps across the discontinuity. We say

$[u]=u_+-u_- \quad [u_x]=u_{x+}-u_{x-} \quad [u_t]=u_{t+}-u_{t-}$

Now, since the equation (1) is true on both sides of the discontinuity, we can see that both u₊ and u_-, being the limits of solutions, must themselves satisfy the equation. That is,

$\begin{matrix} a(x,t)u_{+x} + b(x,t)u_{+t} & = & c(u_+,x,t) \\ a(x,t)u_{-x} + b(x,t)u_{-t} & = & c(u_-,x,t) \end{matrix} \mbox{ where } \begin{matrix}x=x_0(s) \\ t=t_0(s)\end{matrix}$

Subtracting then gives us an equation for the jumps in the differentials

$a(x,t)[u_x] + b(x,t)[u_t] = 0 \,$

We are considering the case where u itself is continuous so we know that [u]=0. Differentiating this with respect to s will give us a second equation in the differential jumps.

$\frac{dx_0}{ds}[u_x] + \frac{dx_0}{dt}[u_t] =0$

The last two equations can only be both true if one is a multiple of the other, but multiplying s by a constant also multiplies the second equation by that same constant while leaving the curve of discontinuity unchanged, hence we can without loss of generality define s to be such that

$\frac{dx_0}{ds}=a \quad \frac{dt_0}{ds}=b$

But these are the equations for a characteristic, i.e discontinuities propagate along characteristics. We could use this property as an alternative definition of characteristics.

We can deal similarly with discontinuous functions by first writing the equation in conservation form, so called because conservation laws can always be written this way.

$(au)_x+(bu)_t=a_x u + b_t u +c \quad (1)$

Notice that the left hand side can be regarded as the divergence of (au, bu). Writing the equation this way allows us to use the theorems of vector calculus.

Consider a narrow strip with sides parallel to the discontinuity and width h

We can integrate both sides of (1) over R, giving

$\int_R (au)_x+(bu)_t \, dx dt = \int_R (a_x + b_t) u +c \, dx dt$

Next we use Green's theorem to convert the left hand side into a line integral.

$\oint_{\partial R} au dt - bu dx = \int_R (a_x + b_t) u +c \, dx dt$

Now we let the width of the strip fall to zero. The right hand side also tends to zero but the left hand side reduces to the difference between two integrals along the part of the boundary of R parallel to the curve.

$\int au_+dt-bu_+dx - \int au_-dt-bu_-dx = 0$

The integrals along the opposite sides of R have different signs because they are in opposite directions.

For the last equation to always be true, the integrand must always be zero, i.e

$\left( a \frac{dt_0}{ds} - b \frac{dx_0}{ds} \right) [u] = 0$

Since, by assumption [u] isn't zero, the other factor must be, which immediately implies the curve of discontinuity is a characteristic.

Once again, discontinuities propagate along characteristics.

Above, we only considered functions of two variables, but it is straightforward to extend this to functions of n variables.

The initial condition is given on an n-1 dimensional surface, which evolves along the characteristics. Typical discontinuities in the initial condition will lie on a n-2 dimensional surface embedded within the initial surface. This surface of discontinuity will propagate along the characteristics that pass through the initial discontinuity.

The jumps themselves obey ordinary differential equations, much as u itself does on a characteristic. In the two dimensional case, for u continuous but not smooth, a little algebra shows that

$\frac{d[u_x]}{ds}=[u_x]\left( \frac{\partial c}{\partial u} + a\frac{b_x}{b}-a_x \right)$

while u obeys the same equation as before,

$\frac{du}{ds}=c$

We can integrate these equations to see how the discontinuity evolves as we move along the characteristic.

We may find that, for some future s, [u_{x] passes through zero. At such points, the discontinuity has vanished, and we can treat the function as smooth at that characteristic from then on.}

_{Conversely, we can expect that smooth functions may, under the righr circumstances, become discontinuous.}

_{To see how all this works in practice we'll consider the solutions of the equation}

_{$u_t+uu_x=0 \quad u(x,0)=f(x)$}

_{for three different initial conditions.}

_{The general solution, using the techniques outlined earlier, is}

_{$u=f(x-tu) \,\!$}

_{u is constant on the characteristics, which are straight lines with slope dependent on u.}

_{First consider f such that}

_{$f(x)= \left\{ \begin{matrix} 1 & x>a \\ \frac{x}{a} & a \ge x > 0 \\ 0 & x \le 0 \end{matrix} \right. \quad a>0$}

_{While u is continuous its derivative is discontinuous at x=0, where u=0, and at x=a, where u=1. The characteristics through these points divide the solution into three regions.}

_{All the characteristics to the right of the characteristic through x=a, t=0 intersect the x-axis to the right of x=1, where u=1 so u is 1 on all those characteristics, i.e whenever x-t>a.}

_{Similarly the characteristic through the origin is the line x=0, to the left of which u remains zero.}

_{We could find the value of u at a point in between those two characteristics either by finding which intermediate characteristic it lies on and tracing it back to the initial line, or via the general solution.}

_{Either way, we get}

_{$f(x)= \left\{ \begin{matrix} 1 & x-t>a \\ \frac{x}{a+t} & a+t \ge x > 0 \\ 0 & x \le 0 \end{matrix} \right.$}

_{At larger t the solution u is more spread out than at t=0 but still the same shape.}

_{We can also consider what happens when a tends to 0, so that u itself is discontinuous at x=0.}

_{If we write the PDE in conservation form then use Green's theorem, as we did above for the linear case, we get}

_{$[u]\frac{dx_0}{ds}=\frac{1}{2}[u^2]\frac{dt_0}{ds}$}

_{[u²] is the difference of two squares, so if we take s=t we get}

_{$\frac{dx_0}{dt}=\frac{1}{2}(u_-+u_+)$}

_{In this case the discontinuity behaves as if the value of u on it were the average of the limiting values on either side.}

_{However, there is a caveat.}

_{Since the limiting value to the left is u- the discontinuity must lie on that characteristic, and similarly for u+; i.e the jump discontinuity must be on an intersection of characteristics, at a point where u would otherwise be multivalued.}

For this PDE the characteristic can only intersect on the discontinuity if

$u_- > u_+ \,$

If this is not true the discontinuity can not propogate. Something else must happen.

The limit a=0 is an example of a jump discontinuity for which this condition is false, so we can see what happens in such cases by studying it.

Taking the limit of the solution derived above gives

$f(x)= \left\{ \begin{matrix} 1 & x>t \\ \frac{x}{t} & t \ge x > 0 \\ 0 & x \le 0 \end{matrix} \right.$

If we had taken the limit of any other sequence of initials conditions tending to the same limit we would have obtained a trivially equivalent result.

Looking at the characteristics of this solution, we see that at the jump discontinuity characteristics on which u takes every value betweeen 0 and 1 all intersect.

At later times, there are two slope discontinuities, at x=0 and x=t, but no jump discontinuity.

This behaviour is typical in such cases. The jump discontinuity becomes a pair of slope discontinuities between which the solution takes all appropriate values.

Now, lets consider the same equation with the initial condition

$f(x)= \left\{ \begin{matrix} 1 & x \le 0 \\ 1-\frac{x}{a} & a \ge x > 0 \\ 0 & x > a \end{matrix} \right. \quad a>0$

This has slope discontinuities at x=0 and x=a, dividing the solution into three regions.

The boundaries between these regions are given by the characteristics through these initial points, namely the two lines

$x=t \quad x=a$

These characteristics intersect at t=a, so the nature of the solution must change then.

In between these two discontinuities, the characteristic through x=b at t=0 is clearly

$x=\left( 1- \frac{b}{a} \right) t + b \quad 0 \le b \le a$

All these characteristics intersect at the same point, (x,t)=(a,a).

We can use these characteristics, or the general solution, to write u for t<a

$u(x,t)= \left\{ \begin{matrix} 1 & x \le t \\ \frac{a-x}{a-t} & a \ge x > t \\ 0 & x > a \end{matrix} \right. \quad a >t \ge 0$

As t tends to a, this becomes a step function. Since u is greater to the left than the right of the discontinuity, it meeets the condition for propogation deduced above, so for t>a u is a step function moving at the average speed of the two sides.

$u(x,t)= \left\{ \begin{matrix} 1 & x \le \frac{a+t}{2} \\ 0 & x > \frac{a+t}{2} \end{matrix} \right. \quad t \ge a \ge 0$

This is the reverse of what we saw for the initial condition previously considered, two slope discontinuities merging into a step discontinuity rather than vice versa. Which actually happens depends entirely on the initial conditions. Indeed, examples could be given for which both processes happen.

In the two examples above, we started with a discontinuity and investigated how it evolved. It is also possible for solutions which are initially smooth to become discontinuous.

For example, we saw earlier for this particular PDE that the solution with the initial condition u=x² breaks down when 2xt+1=0. At these points the solution becomes discontinuous.

Typically, discontinuities in the solution of any partial differential equation, not merely ones of first order, arise when solutions break down in this way and progogate similarly, merging and splitting in the same fashion.

Fully non-linear PDEs

It is possible to extend the approach of the previous sections to reduce any equation of the form

$F(x_1,x_2, \ldots ,x_n,u,u_{x_1},u_{x_2}, \ldots , u_{x_n})=0$

to a set of ODE's, for any function, F.

We will not prove this here, but the corresponding ODE's are

$\frac{dx_i}{d\tau}=\frac{\partial F}{\partial u_i} \quad \frac{du_i}{d\tau} = -\left( \frac{\partial F}{\partial x_i} + u_i \frac{\partial F}{\partial u} \right) \quad \frac{du}{d\tau}=\sum_{i=1}^n u_i \frac{\partial F}{\partial u_i}$

If u is given on a surface parameterised by r₁…r_n then we have, as before, n initial conditions on the n, x_i

$\tau=0 \quad x_i=f_i(r_1,r_2, \ldots ,r_{n-1})$

given by the parameterisation and one initial condition on u itself,

$\tau=0 \quad u=f(r_1,r_2, \ldots ,r_{n-1})$

but, because we have an extra n ODEs for the u_i's, we need an extra n initial conditions.

These are, n-1 consistency conditions,

$\tau=0 \quad \frac{\partial f}{\partial r_i}= \sum_{j=1}^{n-1} u_i \frac{\partial f_i}{\partial r_j}$

which state that the u_i's are the partial derivatives of u on the initial surface, and one initial condition

$\tau=0 \quad F(x_1,x_2, \ldots ,x_n,u,u_1,u_2, \ldots , u_n)=0$

stating that the PDE itself holds on the initial surface.

These n initial conditions for the u_i will be a set of algebraic equations, which may have multiple solutions. Each solution will give a different solution of the PDE.

Example

Consider

$u_t=u_x^2+u_y^2, \quad u(x,y,0)=x^2+y^2$

The initial conditions at τ=0 are

$\begin{matrix} x=r & y=s & t=0 & u=r^2+s^2 \\ u_x=2r & u_y=2s & u_t=4(r^2+s^2) & \end{matrix}$

and the ODE's are

$\begin{matrix} \frac{dx}{d\tau}=-2u_x & \frac{dy}{d\tau}=-2u_y & \frac{dt}{d\tau}=1 & \frac{du}{d\tau}=u_t-2(u_x^2+u_y^2)\\ \frac{du_x}{d\tau}=0 & \frac{du_y}{d\tau}=0 & \frac{du_t}{d\tau}=0 & \end{matrix}$

Note that the partial derivatives are constant on the characteristics. This always happen when the PDE contains only partial derivatives, simplifying the procedure.

These equations are readily solved to give

$x=r(1-4\tau ) \quad y=s(1-4\tau ) \quad t=\tau \quad u=(r^2+s^2)(1-4\tau )$

On eliminating the parameters we get the solution,

$u=\frac{x^2+y^2}{1-4t}$

which can easily be checked. abc

Second order

Suppose we are given a second order linear PDE to solve

$a(x,y)u_{xx}+b(x,y)u_{xy}+c(x,y)u_{yy} = d(x,y)u_x+e(x,y)u_y +p(x,y)u +q(x,y) \quad (1)$

The natural approach, after our experience with ordinary differential equations and with simple algebraic equations, is attempt a factorisation. Let's see how for this takes us.

We would expect factoring the left hand of (1) to give us an equivalent equation of the form

$a(x,y)(D_x+\alpha_+(x,y)D_y)(D_x+\alpha_-(x,y)D_y)u \,$

and we can immediately divide through by a. This suggests that those particular combinations of first order derivatives will play a special role.

Now, when studying first order PDE's we saw that such combinations were equivalent to the derivatives along characteristic curves. Effectively, we changed to a coordinate system defined by the characteristic curve and the initial curve.

Here, we have two combinations of first order derivatives each of which may define a different characteristic curve. If so, the two sets of characteristics will define a natural coordinate system for the problem, much as in the first order case.

In the new coordinates we will have

$D_x+\alpha_+(x,y)D_y=D_r \quad D_x+\alpha_-(x,y)D_y =D_s$

with each of the factors having become a differentiation along its respective characteristic curve, and the left hand side will become simply u_rs giving us an equation of the form

$u_{rs}=A(r,s)u_r+B(r,s)u_s+C(r,s)u+D(r,s) \,$

If A, B, and C all happen to be zero, the solution is obvious. If not, we can hope that the simpler form of the left hand side will enable us to make progress.

However, before we can do all this, we must see if (1) can actually be factorised.

Multiplying out the factors gives

$u_{xx}+\frac{b(x,y)}{a(x,y)}u_{xy}+\frac{c(x,y)}{a(x,y)}u_{yy}= u_{xx}+(\alpha_++\alpha_-)u_{xy}+\alpha_+\alpha_-u_{yy}$

On comparing coefficients, and solving for the α's we see that they are the roots of

$a(x,y)\alpha^2+b(x,y)\alpha+c(x,y)=0 \,$

Since we are discussing real functions, we are only interested in real roots, so the existence of the desired factorisation will depend on the discriminant of this quadratic equation.

• $\mbox{If } b(x,y)^2 > 4 a(x,y)c(x,y)$

then we have two factors, and can follow the procedure outlined above. Equations like this are called hyperbolic

• $\mbox{If } b(x,y)^2 = 4 a(x,y)c(x,y)$

then we have only factor, giving us a single characteristic curve. It will be natural to use distance along these curves as one coordinate, but the second must be determined by other considerations.

The same line of argument as before shows that use the characteristic curve this way gives a second order term of the form u_rr, where we've only taken the second derivative with respect to one of the two coordinates. Equations like this are called parabolic

• $\mbox{If } b(x,y)^2 < 4 a(x,y)c(x,y)$

then we have no real factors. In this case the best we can do is reduce the second order terms to the simplest possible form satisfying this inequality, i.e u_rr+u_ss

It can be shown that this reduction is always possible. Equations like this are called elliptic

It can be shown that, just as for first order PDEs, discontinuities propagate along characteristics. Since elliptic equations have no real characteristics, this implies that any discontinuities they may have will be restricted to isolated points; i.e, that the solution is almost everywhere smooth.

This is not true for hyperbolic equations. Their behaviour is largely controlled by the shape of their characteristic curves.

These differences mean different methods are required to study the three types of second equation. Fortunately, changing variables as indicated by the factorisation above lets us reduce any second order PDE to one in which the coefficients of the second order terms are constant, which means it is sufficient to consider only three standard equations.

$u_{xx}+u_{yy}=0 \quad u_{xx}-u_{yy}=0 \quad u_{xx}-u_y=0$

We could also consider the cases where the right hand side of these equations is a given function, or proportional to u or to one of its first order derivatives, but all the essential properties of hyperbolic, parabolic, and elliptic equations are demonstrated by these three standard forms.

While we've only demonstrated the reduction in two dimensions, a similar reduction applies in higher dimensions, leading to a similar classification. We get, as the reduced form of the second order terms,

$a_1\frac{\partial^2 u}{\partial x_1^2}+ a_2\frac{\partial^2 u}{\partial x_2^2} + \cdots + a_n\frac{\partial^2 u}{\partial x_n^2}$

where each of the a_is is equal to either 0, +1, or -1.

If all the a_is have the same sign the equation is elliptic

If any of the a_is are zero the equation is parabolic

If exactly one of the a_is has the opposite sign to the rest the equation is hyperbolic

In 2 or 3 dimensions these are the only possibilities, but in 4 or more dimensions there is a fourth possibility: at least two of the a_is are positive, and at least two of the a_is are negative.

Such equations are called ultrahyperbolic. They are less commonly encountered than the other three types, so will not be studied here.

When the coefficients are not constant, an equation can be hyperbolic in some regions of the xy plane, and elliptic in others. If so, different methods must be used for the solutions in the two regions.

Elliptic

Standard form, Laplace's equation:

$\nabla^2 h=0$

Quote equation in spherical and cylindrical coordinates, and give full solution for cartesian and cylindrical coordinates. Note averaging property Comment on physical significance, rotation invariance of laplacian.

Hyperbolic

Standard form, wave equation:

$\nabla^2 h=c^2 h_{tt}$

Solution, any sum of functions of the form

$h=f(\mathbf{k}\cdot \mathbf{x}-\omega t) \quad \omega=|\mathbf{k}|c$

These are waves. Compare with solution from separating variables. Domain of dependance, etc.

Parabolic

The canonical parabolic equation is the diffusion equation:

$\nabla^2 h=h_t$

Here, we will consider some simple solutions of the one-dimensional case.

The properties of this equation are in many respects intermediate between those of hyperbolic and elliptic equation.

As with hyperbolic equations but not elliptic, the solution is well behaved if the value is given on the initial surface t=0.

However, the characteristic surfaces of this equation are the surfaces of constant t, thus there is no way for discontinuities to propagate to positive t.

Therefore, as with elliptic equations but not hyberbolic, the solutions are typically smooth, even when the initial conditions aren't.

Furthermore, at a local maximum of h, its Laplacian is negative, so h is decreasing with t, while at local minima, where the Laplacian will be positive, h will increase with t. Thus, initial variations in h will be smoothed out as t increases.

In one dimension, we can learn more by integrating both sides,

$\begin{matrix} \int_{-a}^b h_t dt & = & \int_{-a}^b h_{xx} dx \\ \frac{d}{dt} \int_{-a}^b h \,dt & = & \left[ h_x \right]_{-a}^b \end{matrix}$

Provided that h_x tends to zero for large x, we can take the limit as a and b tend to infinity, deducing

$\frac{d}{dt} \int_{-\infty}^\infty h \,dt$

so the integral of h over all space is constant.

This means this PDE can be thought of as describing some conserved quantity, initially concentrated but spreading out, or diffusing, over time.

This last result can be extended to two or more dimensions, using the theorems of vector calculus.

We can also differentiate any solution with respect to any coordinate to obtain another solution. E.g if h is a solution then

$\nabla^2 h_x = \partial_x \nabla^2 h = \partial_x \partial_t h =\partial_t h_x$

so h_x also satisfies the diffusion equation.

Similarity solution

Looking at this equation, we might notice that if we make the change of variables

$r = \alpha x \quad \tau =\alpha^2 t$

then the equation retains the same form. This suggests that the combination of variables x²/t, which is unaffected by this variable change, may be significant.

We therefore assume this equation to have a solution of the special form

$h(x,t)=f(\eta) \mbox{ where } \eta = \frac{x}{t^{1 \over 2}}$

then

$h_x = \eta_x f_{\eta} = t^{-{1\over 2}}f_{\eta} \quad h_t = \eta_t f_{\eta} = -\frac{\eta}{2t}f_{\eta}$

and substituting into the diffusion equation eventually gives

$f_{\eta \eta} + \frac{\eta}{2} f_{\eta}=0$

which is an ordinary differential equation.

Integrating once gives

$f_{\eta} = A e^{-\frac{\eta^2}{4}}$

Reverting to h, we find

$\begin{matrix} h_x & = & \frac{A}{\sqrt{t}} e^{-\frac{\eta^2}{4}} \\ h & = & \frac{A}{\sqrt{t}} \int_{-\infty }^x e^{-s^2/4t} ds +B \\ & = & A \int_{-\infty }^{x/2\sqrt{t}} e^{-z^2} dz +B \end{matrix}$

This last integral can not be written in terms of elementary functions, but its values are well known.

In particular the limiting values of h at infinity are

$h(-\infty ,t)=B \quad h(\infty ,t)= B + A \sqrt{\pi } ,$

taking the limit as t tends to zero gives

$h= \left\{ \begin{matrix} B & x<0 \\ B + A \sqrt{\pi } & x>0 \end{matrix} \right.$

We see that the initial discontinuity is immediately smoothed out. The solution at later times retains the same shape, but is more stretched out.

The derivative of this solution with respect to x

$h_x = \frac{A}{\sqrt{t}} e^{-x^2/4t}$

is itself a solution, with h spreading out from its initial peak, and plays a significant role in the further analysis of this equation.

The same similiarity method can also be applied to some non-linear equations.

Separating variables

We can also obtain some solutions of this equation by separating variables.

$h(x,t)=X(x)T(t) \Rightarrow X^{\prime \prime}T=X\dot{T}$

giving us the two ordinary differential equations

$\frac{d^2X}{dx^2}+k^2 X=0 \quad \frac{dT}{dt} = -k T$

and solutions of the general form

$h(x,t)= A e^{-kt} \sin (kx+\alpha) \,$

Systems of Ordinary Differential Equations

We have already examined cases where we have a single differential equation and found several methods to aid us in finding solutions to these equations. But what happens if we have two or more differential equations that depend on each other? For example, consider the case where

$D_t x(t) = 3y(t)^2 + x(t)t$

and

$D_t y(t) = x(t) + y(t)$

Such a set of differential equations is said to be coupled. Systems of ordinary differential equations such as these are what we will look into in this section.

First order systems

A general system of differential equations can be written in the form

$D_t \mathbf{x} = \mathbf{F}(\mathbf{x}, t)$

Instead of writing the set of equations in a vector, we can write out each equation explicitly, in the form:

$D_t x_1 = F_1(x_1, \ldots, x_n, t)$

$\vdots\!\;$

$D_t x_i = F_i(x_1, \ldots, x_n, t)$

If we have the system at the very beginning, we can write it as:

$D_t \mathbf{x} = \mathbf{G}(\mathbf{x}, t)$

where

$\mathbf{x} = (x(t), y(t)) = (x,y)$

and

$\mathbf{G}(\mathbf{x}, t) = (3y^2+xt,x+y)$

or write each equation out as shown above.

Why are these forms important? Often, this arises as a single, higher order differential equation that is changed into a simpler form in a system. For example, with the same example,

$D_t x(t) = 3y(t)^2 + x(t)t$

$D_t y(t) = x(t) + y(t)$

we can write this as a higher order differential equation by simple substitution.

$D_t y(t) - y(t) = x(t)$

then

$D_t x(t) = 3y(t)^2 + (D_t y(t) - y(t))t$

$D_t x(t) = 3y(t)^2 + t D_t y(t) - t y(t)$

Notice now that the vector form of the system is dependent on t since

$\mathbf{G}(\mathbf{x}, t) = (3y^2+xt,x+y)$

the first component is dependent on t. However, if instead we had

$\mathbf{H}(\mathbf{x}) = (3y^2+x,x+y)$

notice the vector field is no longer dependent on t. We call such systems autonomous. They appear in the form

$D_t \mathbf{x} = \mathbf{H}(\mathbf{x})$

We can convert between an autonomous system and a non-autonomous one by simply making a substitution that involves t, such as y=(x, t), to get a system:

$D_t \mathbf{y} = (\mathbf{F}(\mathbf{y}), 1) = (\mathbf{F}(\mathbf{x}, t), 1)$

In vector form, we may be able to separate F in a linear fashion to get something that looks like:

$\mathbf{F}(\mathbf{x}, t) = A(t)\mathbf{x} + \mathbf{b}(t)$

where A(t) is a matrix and b is a vector. The matrix could contain functions or constants, clearly, depending on whether the matrix depends on t or not.

Formal limits

Formal definition of limit (optional)

In preliminary calculus, the definition of a limit is probably the most difficult concept to grasp. If nothing else, it took some of the most brilliant mathematicians 150 years to arrive at it.

The intuitive definition of a limit is adequate in most cases, as the limit of a function is the function of the limit. But what is our meaning of "close"? How close is close? We consider the following limit of a function:

$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac {\sin(x)} {x}.$

You will have noted that f(0) = 0/0, and this is undefined. But the limit does exist and it equals 1. But how do we convert that intuition into mathematical sense?

We call L the limit of f(x) as x approaches c if for every positive number ε, there exists a number δ such that

$\left| f(x) - L \right| < \epsilon$

whenever

$0 < \left| x - c \right| < \delta.$

Understanding the parallels between the two definitions above is important: instead of saying ' f(x) approximately equals L ', the formal definition says that 'the difference between f(x) and L is less than any number epsilon'.

$\lim_{x \to c} f(x) = L$ if and only if $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L$

$\lim_{x \to c^-} f(x)$ means as x approaches c from the left, and $\lim_{x \to c^+} f(x)$ as x approaches c from the right.

See how to make formal proofs.

Notice that we have not yet defined ' x approaches c ' yet; we will discuss this later.

Limits are often written

$\lim_{x \rightarrow c} f(x)$

which may be read "the limit of f(x) as x approaches c".

Examples

1) What is the limit of f(x) = x + 7 as x approaches 4?

There are two steps to answering such a question; first we must determine the answer -- this is where intuition and guessing is useful, as well as the informal definition of a limit. Then, we must prove that the answer is right. For this problem, the answer happens to be 11. Now, we must prove it using the definition of a limit:

Informal: 11 is the limit because when x is roughly equal to 4, f(x) = x + 7 approximately equals 4 + 7, which equals 11.

Formal: We need to prove that no matter what value of ε is given to us, we can find a value of δ such that

$\left| f(x) - 11 \right| < \epsilon$

whenever

$\left| x - 4 \right| < \delta.$

For this particular problem, letting δ equal ε works (see choosing delta for help in determining the value of delta to use). Now, we have to prove

$\left| f(x) - 11 \right| < \epsilon$

given that

$\left| x - 4 \right| < \delta = \epsilon.$

Since |x - 4| < ε, we know
|f(x) - 11|  = |x + 7 - 11| = |x - 4| < ε, which is what we wished to prove.

2) What is the limit of f(x) = x² as x approaches 4?

Formal: Again, we pull two things out of thin air; the limit is 16 (use the informal definition to find the limit of f(x)), and δ equals √(ε+16) - 4. Note that δ is always positive for positive ε. Now, we have to prove
$\left| x^2 - 16 \right| < \epsilon$
given that
$\left| x - 4 \right| < \delta = \sqrt{\epsilon + 16} - 4$.

We know that |x + 4| = |(x - 4) + 8| ≤ |x - 4| + 8  < δ + 8 (because of the triangle inequality), thus

$\begin{matrix} \left| x^2 - 16 \right| & = & \left| x - 4 \right| \cdot \left| x + 4 \right| \\ \\ \ & < & (\delta) \cdot (\delta + 8) \\ \\ \ & = & (\sqrt{16 + \epsilon} - 4) \cdot (\sqrt{16 + \epsilon} + 4) \\ \\ \ & = & (\sqrt{16 + \epsilon})^2 - 4^2 \\ \\ \ & = & \epsilon \end{matrix}.$

3) Show that the limit of sin(1/x) as x approaches 0 does not exist.

Suppose the limit exists and is l. We will proceed by contradiction. Assume that $l\neq 1$, the case for $l=1$ is similar. Choose $\epsilon = l - 1$, then for every $\delta > 0$, there exists a large enough n such that $0 < x_0 = \frac{1}{\pi /2 + 2\pi n} < \delta$, but $|sin(1/x_0) - l|=|1-l|=\epsilon$ a contradiction.

The function sin(1/x) is known as the topologist's curve.

4) What is the limit of $x sin(1/x)$ as x approaches 0?

It is 0. For every $\epsilon > 0$, choose $\delta = \epsilon$ so that for all x, if $0 < |x| < \delta$, then $|x sin x - 0| <= | x | < \epsilon$ as required.

5) Prove that the limit of 1/x as x approaches 0 does not exist?

6) Prove that the limit of f(x)={x, when x is rational; 0, when x is irrational} as x approaches c for ALL (c not equal to zero, c is real number) does not exist?

Real numbers

< Calculus/Outline

Fields

You are probably already familiar with many different kinds of numbers from your past experience. Some of the commonly used types of numbers are

• Natural numbers, usually denoted with an N, are the numbers 0,1,2,3,...
• Integers, usually denoted with a Z, are the positive and negative natural numbers: ...-3,-2,-1,0,1,2,3...
• Rational numbers, denoted with a Q, are fractions of integers (excluding division by zero): -1/3, 5/1, 0, 2/7. etc.
• Real numbers, denoted with a R, are constructed and discussed below.

Note that these different classes of numbers have different properties. In the integers for example, any number always has an additive inverse: for any integer x, there is another integer t such that x+t=0. This should not be terribly surprising: from basic aritematic we know that t=-x. Try to prove to yourself that the natural numbers do not have an additive inverse for all numbers.

In mathematics, it is useful to note the important properties that each of these collections has. The rational numbers, which will be of primary concern in constructing the real numbers, have the properties

There exists a number 0 such that for any other number a, 0+a=a+0=a

For any two numbers a and b, a+b is another number

For any three numbers a,b, and c, a+(b+c)=(a+b)+c

For any number a there is another number -a such that a+(-a)=0

For any two numbers a and b, a+b=b+a

For any two numbers a and b,a*b is another number

There is a number 1 such that for any number a, a*1=1*a=a

For any two numbers a and b, a*b=b*a

For any three numbers a,b and c, a(bc)=(ab)c

For any three numbers a,b and c, a(b+c)=ab+ac

For every number a there is another number a^-1 such that aa^-1=1

When presented as a list as above, these probably seem quite intimidating. However on careful inspection, these properties, you should find, are nothing more than basic facts from artimetic, carefully constructed. Any collection of numbers (and operations + and * on those numbers) which satisfies the above properties is called a field. The properties above are usually called field axiom. As an exercise, determine if the integers form a field, and if not, which field axiom(s) they violate.

Even though the list of field axioms is quite extensive, it does not explore the properties of the rational numbers fully. Rational numbers also have an ordering.' A total ordering must satisfy several properties: for any numbers a, b, and c

if a ≤ b and b ≤ a then a = b (antisymmetry)

if a ≤ b and b ≤ c then a ≤ c (transitivity)

a ≤ b or b ≤ a (totality)

To familiarize yourself with these properties, try to show that (a) natural numbers, integers and rational numbers are all totally ordered and more generally (b) convince yourself that any collection of rational numbers are totally ordered (note that the inegers and natural are both collections specific rational numbers).

Finally, it is useful to note one characterization of the rational numbers: every rational number has a decimal expansion which is either repeating or terminating. The proof of this fact is omitted, however it follows from the definition of each rational number being a fraction, and thus by performing long division, the remainder at any stage can only take on positive integer values smaller than the denominator, of which there are finitely many.

Constructing the Real Numbers

There are two additional tools which are needed for the construction of the real numbers: the upper bound and the least upper bound. Definition A collection of numbers E is bounded above if there exists a number m such that for all x in E x≤m. Any number m which satisfies this condition is called an upper bound of the set E.

Definition If a collection of numbers E is bounded above with m as an upper bound of E, and all other upper bounds of E are bigger than m, we call m the least upper bound or supremum of E, denoted by sup E.

Note that many collections of rational numbers do not have a least upper bound which is also rational, although some do. Suppose the the numbers 5 and 10/3 are, together, taken to be E. The number 10/3 is not only an upper bound of E, it is a least upper bound. In general, there are many upper bounds (12, for instance, is an upper bound of the collection above), but it can have at most one upper bound.

Consider the collection of numbers $\{3, 3.1, 3.14, 3.141, 3.1415, \dots\}$: You may recognize these decimals as the first few digits of pi. Note however, that since each decimal terminates, each number in this collection is a rational number. This collection has many upper bounds. The number 4, for instance, is an upper bound. There is no least upper bound, at least not in the rational numbers. Try to convince yourself of this fact by attempting to construct such a least upper bound: (a) why does pi not work as a least upper bound (hint: pi does not having a repeating or terminating decimal expansion), (b) what happens if the proposed supremum is equal to a pi up to some decimal place, and zeros after (c) if the proposed supremum is bigger than pi, can you find a smaller upper bound which will work?

In fact, there are many such collections of rational numbers which do not have a least upper bound which is rational. We define, however the real numbers to any number which is a least upper bound of a collection of rationals.

Properties of Real Numbers

The reals are well ordered.

For all reals; a, b, c

Either b>a, b=a', or b<a.

If a<b and b<c then a<c

Also

b>a implies b+c>a+c

b>a and c>0 implies bc>ac

b>a implies -a>-b

Upper bound axiom

Every non-empty set of real numbers which is bounded above has a supremum.

The upper bound axiom is necessary for calculus. It is not true for rational numbers.

We can also define lower bounds in the same way.

Definition A set E is bounded below if there exists a real M such that for all x∈E x≥M Any M which satisfies this condition is called an lower bound of the set E

Definition If a set, E, is bounded below, M is an lower bound of E, and all other lower bounds of E are less than M, we call M the greatest lower bound or inifimum of E, denoted by inf E

The supremum and infimum of finite sets are the same as their maximum and minimum.

Theorem

Every non-empty set of real numbers which is bounded below has an infimum.

Proof:

Let E be a non-empty set of of real numbers, bounded below

Let L be the set of all lower bounds of E

L is not empty, by definition of bounded below

Every element of E is an upper bound to the set L, by definition

Therefore, L is a non empty set which is bounded above

L has a supremum, by the upper bound axiom

1/ Every lower bound of E is ≤sup L, by definition of supremum

Suppose there were an e∈E such that e<sup L

Every element of L is ≤e, by definition

Therefore e is an upper bound of L and e<sup L

This contradicts the definition of supremum, so there can be no such e.

If e∈E then e≥sup L, proved by contradiction

2/ Therefore, sup L is a lower bound of E

inf E exists, and is equal to sup L, on comparing definition of infinum to lines 1 & 2

Bounds and inequalities, theorems: $A \subseteq B \Rightarrow \sup A \le \sup B$ $A \subseteq B \Rightarrow \inf A \ge \inf B$ $\sup A \cup B = \max (\sup A, \sup B)$ $\inf A \cup B = \min (\inf A, \inf B)$

Theorem: (The triangle inequality)

$\forall a,b,c \in \R \quad |a-b| \le |a-c|+|c-b|$

Proof by considering cases

If a≤b≤c then |a-c|+|c-b| = (c-a)+(c-b) = 2(c-b)+(b-a)>b-a = |b-a|

Exercise: Prove the other five cases.

This theorem is a special case of the geometric theorem

The sum of two sides of a triangle is greater than or equal to the third side

and useful whenever we need to manipulate inequalities and absolute values

Theory of Sequences

A sequence is an ordered list of objects (or events). Like a set, it contains members (also called elements or terms), and the number of terms (possibly infinite) is called the length of the sequence. Unlike a set, order matters, and the exact same elements can appear multiple times at different positions in the sequence.

For example, (C, R, Y) is a sequence of letters that differs from (Y, C, R), as the ordering matters. Sequences can be finite, as in this example, or infinite, such as the sequence of all even positive integers (2, 4, 6,...).

An infinite sequence of real numbers (in blue). This sequence is neither increasing, nor decreasing, nor convergent. It is however bounded.

Examples and notation

There are various and quite different notions of sequences in mathematics, some of which (e.g., exact sequence) are not covered by the notations introduced below.

A sequence may be denoted (a₁, a₂, ...). For shortness, the notation (a_n) is also used.

A more formal definition of a finite sequence with terms in a set S is a function from {1, 2, ..., n} to S for some n ≥ 0. An infinite sequence in S is a function from {1, 2, ...} (the set of natural numbers without 0) to S.

Sequences may also start from 0, so the first term in the sequence is then a₀.

A finite sequence is also called an n-tuple. Finite sequences include the empty sequence ( ) that has no elements.

A function from all integers into a set is sometimes called a bi-infinite sequence, since it may be thought of as a sequence indexed by negative integers grafted onto a sequence indexed by positive integers.

Types and properties of sequences

A subsequence of a given sequence is a sequence formed from the given sequence by deleting some of the elements without disturbing the relative positions of the remaining elements.

If the terms of the sequence are a subset of an ordered set, then a monotonically increasing sequence is one for which each term is greater than or equal to the term before it; if each term is strictly greater than the one preceding it, the sequence is called strictly monotonically increasing. A monotonically decreasing sequence is defined similarly. Any sequence fulfilling the monotonicity property is called monotonic or monotone. This is a special case of the more general notion of monotonic function.

The terms non-decreasing and non-increasing are used in order to avoid any possible confusion with strictly increasing and strictly decreasing, respectively. If the terms of a sequence are integers, then the sequence is an integer sequence. If the terms of a sequence are polynomials, then the sequence is a polynomial sequence.

If S is endowed with a topology, then it becomes possible to consider convergence of an infinite sequence in S. Such considerations involve the concept of the limit of a sequence.

Sequences in analysis

In analysis, when talking about sequences, one will generally consider sequences of the form

$(x_1, x_2, x_3, ...)\,$ or $(x_0, x_1, x_2, ...)\,$

which is to say, infinite sequences of elements indexed by natural numbers. (It may be convenient to have the sequence start with an index different from 1 or 0. For example, the sequence defined by x_n = 1/log(n) would be defined only for n ≥ 2. When talking about such infinite sequences, it is usually sufficient (and does not change much for most considerations) to assume that the members of the sequence are defined at least for all indices large enough, that is, greater than some given N.)

The most elementary type of sequences are numerical ones, that is, sequences of real or complex numbers.

Summation notation=

Summation Notation

Summation notation allows an expression that contains a sum to be expressed in a simple, compact manner. The greek letter sigma, Σ, is used to denote the sum of a set of numbers.

Example

$\sum_{i=2}^7 i^2 = 2^2 + 3^2 + 4^2 + 5^2 + 6^2+7^2$

Let f be a function and N,M are integers with $N<M$. Then

$\sum_{i=N}^M f(i) = f(N) + f(N+1) + f(N+2) + \cdots + f(M).$

We say N is the lower limit and M is the upper limit of the sum.

Notice that we can replace the letter i with any other variable. For this reason i is referred to as a dummy variable. So

$\sum_{i=1}^4 i = \sum_{j=1}^4 j = \sum_{\alpha=1}^4 \alpha = 1 + 2 + 3 + 4$

Conventionally we use the letters i, j, k, m for dummy variables.

Examples

$\sum_{i=1}^5 i = 1 + 2 + 3 + 4 + 5$

Here, the dummy variable is i, the lower limit of summation is 1, and the upper limit is 5.

Example

Sometimes, you will see summation signs with no dummy variable specified, e.g

$\sum_1^4 i^3 =100$

In such cases the correct dummy variable should be clear from the context.

You may also see cases where the limits are unspecified. Here too, they must be deduced from the context.

Common Summations

$\sum_{i=1}^n c = c + c + ... + c = nc , c\in\mathbb{R}$

$\sum_{i=1}^n i = 1 + 2 + 3 + ... + n = {n(n+1)\over 2}$

$\sum_{i=1}^n i^2 = 1^2 + 2^2 + 3^2 + ... + n^2 = {n(n+1)(2n+1)\over 6}$

$\sum_{i=1}^n i^3 = 1^3 + 2^3 + 3^3 + ... + n^3 = {n^2(n+1)^2 \over 4}$

Tables of Integrals

Rules

• $\int cf(x)\,dx = c\int f(x)\,dx$
• $\int f(x)+g(x)\,dx = \int f(x)\,dx+ \int g(x)\,dx$
• $\int f(x)-g(x)\,dx = \int f(x)\,dx- \int g(x)\,dx$
• $\int u\,dv\ = uv - \int v\,du$

Powers

• $\int dx = x+C$
• $\int a\,dx = ax+C$
• $\int x^n\,dx = \frac{1}{n+1}x^{n+1}+C\qquad\mbox{ if }n\ne-1$
• $\int x^{-n}\,dx = \frac{1}{-n+1}x^{-n+1}+C\qquad\mbox{ if }n\ne1$
• $\int {1\over x}\,dx = \ln|x|+C$
• $\int \frac{1}{ax+b}\,dx = {1 \over a}\ln|ax+b|+C\qquad\mbox{ if }a\ne 0$

Trigonometric Functions

Basic Trigonometric Functions

• $\int \sin{x}\,dx = -\cos{x} + C$
• $\int \cos{x}\,dx = \sin{x} + C$
• $\int \tan{x}\,dx = \ln \left |{\sec{x}} \right | + C$
• $\int \sin^2{x}\,dx = \frac{1}{2} x - \frac{1}{4} \sin{2x} + C$
• $\int \cos^2{x}\,dx = \frac{1}{2} x + \frac{1}{4} \sin{2x} + C$
• $\int \tan^2{x}\,dx = \tan(x) - x + C$

Reciprocal Trigonometric Functions

• $\int \sec{x}\,dx = \ln \left |{\sec{x}}+\tan x \right | + C = \ln \left | \tan{\left( \frac1 2 x +\frac1 4 \pi \right) }\right |+C$
• $\int \csc{x}\,dx = - \ln \left |{\csc x + \cot x} \right | + C=\ln \left | \tan \left(\frac1 2 x \right) \right |+C$
• $\int \cot{x}\,dx = \ln \left |{\sin{x}} \right | + C$

• $\int \sec^2 kx\,dx = \frac1 k \tan{kx} + C$
• $\int \csc^2 kx\,dx = -\frac1 k \cot kx + C$
• $\int \cot^2 kx\,dx = -x-\frac1 k \cot kx + C$

• $\int \sec{x}\tan{x}\,dx = \sec x + C$
• $\int \csc^n kx \cot kx\,dx = -\frac{1}{ka} \csc^n ax + C$
• $\int \sec x \csc x\,dx =\ln \left | \tan x \right | + C$

Inverse Trigonometric Functions

• $\int {1\over \sqrt{1-x^2}}\,dx = \mbox{arcsin}(x) + C$
• $\int {1\over \sqrt{a^2-x^2}}\,dx = {1\over a}\mbox{arcsin}(x/a) + C \qquad\mbox{ if }a\ne 0$
• $\int {1\over 1+x^2}\,dx = \mbox{arctan}(x) + C$
• $\int {1\over a^2+x^2}\,dx = {1\over a}\mbox{arctan}(x/a) + C \qquad\mbox{ if }a\ne 0$

Exponential and Logarithmic Functions

• $\int e^x \,dx = e^x + C$
• $\int e^{ax} \,dx = {1\over a}e^{ax} + C \qquad\mbox{ if }a\neq 0$
• $\int a^x \,dx = {1\over \ln a}a^x + C \qquad\mbox{ if }a>0, a\neq 1$
• $\int \ln x \,dx = x\ln x-x + C$

Inverse Trigonometric Functions

• $\int \mbox{arcsin}(x) \,dx = x\,\mbox{arcsin}(x) + \sqrt{1-x^2} + C$
• $\int \mbox{arccos}(x) \,dx = x\,\mbox{arccos}(x) - \sqrt{1-x^2} + C$
• $\int \mbox{arctan}(x) \,dx = x\,\mbox{arctan}(x) - {1\over 2}\ln(1+x^2) + C$

Tables of Derivatives

General Rules

$\frac{d}{dx}(f + g)= \frac{df}{dx} + \frac{dg}{dx}$

$\frac{d}{dx}(cf)= c\frac{df}{dx}$

$\frac{d}{dx}(fg)= f\frac{dg}{dx} + g\frac{df}{dx}$

$\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{g\frac{df}{dx} - f\frac{dg}{dx}}{g^2}$

Powers and Polynomials

• $\frac{d}{dx} (c) = 0$
• $\frac{d}{dx}x=1$
• $\frac{d}{dx}x^n=nx^{n-1}$
• $\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt x}$
• $\frac{d}{dx}\frac{1}{x}=-\frac{1}{x^2}$
• $\frac{d}{dx}(c_n x^n + c_{n-1} x^{n-1} + c_{n-2}x^{n-2} + \cdots +c_2x^2 + c_1 x + c_0) = n c_n x^{n-1} + (n-1) c_{n-1} x^{n-2} + (n-2) c_{n-2}x^{n-3} + \cdots + 2c_2x+ c_1$

Trigonometric Functions

$\frac{d}{dx} \sin (x)= \cos (x)$

$\frac{d}{dx} \cos (x)= -\sin (x)$

$\frac{d}{dx} \tan (x)= \sec^2 (x)$

$\frac{d}{dx} \cot (x)= -\csc^2 (x)$

$\frac{d}{dx} \sec (x)= \sec (x) \tan (x)$

$\frac{d}{dx} \csc (x) = -\csc (x) \cot (x)$

Exponential and Logarithmic Functions

• $\frac{d}{dx} e^x =e^x$
• $\frac{d}{dx} a^x =a^x \ln (a)\qquad\mbox{if }a>0$
• $\frac{d}{dx} \ln (x)= \frac{1}{x}$
• $\frac{d}{dx} \log_a (x)= \frac{1}{x\ln (a)}\qquad\mbox{if }a>0, a\neq 1$
• $(f^g)' = \left(e^{g\ln f}\right)' = f^g\left(f'{g \over f} + g'\ln f\right),\qquad f > 0$
• $(c^f)' = \left(e^{f\ln c}\right)' = f' c^f \ln c$

Inverse Trigonometric Functions

• $\frac{d}{dx} \mbox{arcsin x} = \frac{1}{\sqrt{1-x^2}}$
• $\frac{d}{dx} \mbox{arccos x} = -\frac{1}{\sqrt{1-x^2}}$
• $\frac{d}{dx} \mbox{arctan x} = \frac{1}{1+x^2}$
• ${d \over dx} \arcsec x = { 1 \over |x|\sqrt{x^2 - 1}}$
• ${d \over dx} \arccot x = {-1 \over 1 + x^2}$
• ${d \over dx} \arccsc x = {-1 \over |x|\sqrt{x^2 - 1}}$

Hyperbolic and Inverse Hyperbolic Functions

${d \over dx} \sinh x = \cosh x$

${d \over dx} \cosh x = \sinh x$

${d \over dx} \tanh x = \mbox{sech}^2\,x$

${d \over dx} \,\mbox{sech}\,x = -\tanh x\,\mbox{sech}\,x$

${d \over dx} \,\mbox{coth}\,x = -\,\mbox{csch}^2\,x$

${d \over dx} \,\mbox{csch}\,x = -\,\mbox{coth}\,x\,\mbox{csch}\,x$

${d \over dx} \sinh^{-1} x = { 1 \over \sqrt{x^2 + 1}}$

${d \over dx} \cosh^{-1} x = {-1 \over \sqrt{x^2 - 1}}$

${d \over dx} \tanh^{-1} x = { 1 \over 1 - x^2}$

${d \over dx} \mbox{sech}^{-1}\,x = { 1 \over x\sqrt{1 - x^2}}$

${d \over dx} \mbox{coth}^{-1}\,x = {-1 \over 1 - x^2}$

${d \over dx} \mbox{csch}^{-1}\,x = {-1 \over |x|\sqrt{1 + x^2}}$

Table of Trigonometry

Definitions

• $\tan(x)= \frac{\sin x}{\cos x}$
• $\sec(x)= \frac{1}{\cos x}$
• $\cot(x)= \frac{\cos x}{\sin x}= \frac{1}{\tan x}$
• $\csc(x)= \frac{1}{\sin x}$

Pythagorean Identities

• $\sin^2 x + \cos^2 x =1 \$
• $1+\tan^2(x)= \sec^2 x \$
• $1+\cot^2(x)= \csc^2 x \$

Double Angle Identities

• $\sin(2 x)= 2\sin x \cos x \$
• $\cos(2 x)= \cos^2 x - \sin^2 x \$
• $\tan(2x) = \frac{2 \tan (x)} {1 - \tan^2(x)}$
• $\cos^2(x) = {1 + \cos(2x) \over 2}$
• $\sin^2(x) = {1 - \cos(2x) \over 2}$

Angle Sum Identities

$\sin \left(x+y\right)=\sin x \cos y + \cos x \sin y$

$\sin \left(x-y\right)=\sin x \cos y - \cos x \sin y$

$\cos \left(x+y\right)=\cos x \cos y - \sin x \sin y$

$\cos \left(x-y\right)=\cos x \cos y + \sin x \sin y$

$\sin x+\sin y=2\sin \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)$

$\sin x-\sin y=2\cos \left( \frac{x+y}{2} \right) \sin \left( \frac{x-y}{2} \right)$

$\cos x+\cos y=2\cos \left( \frac{x+y}{2} \right) \cos \left( \frac{x-y}{2} \right)$

$\cos x-\cos y=-2\sin \left( \frac{x+y}{2} \right)\sin \left( \frac{x-y}{2} \right)$

$\tan x+\tan y=\frac{\sin \left( x+y\right) }{\cos x\cos y}$

$\tan x-\tan y=\frac{\sin \left( x-y\right) }{\cos x\cos y}$

$\cot x+\cot y=\frac{\sin \left( x+y\right) }{\sin x\sin y}$

$\cot x-\cot y=\frac{-\sin \left( x-y\right) }{\sin x\sin y}$

Product-to-sum identities

$\cos\left (x\right ) \cos\left (y\right ) = {\cos\left (x + y\right ) + \cos\left (x - y\right ) \over 2} \;$

$\sin\left (x\right ) \sin\left (y\right ) = {\cos\left (x - y\right ) - \cos\left (x + y\right ) \over 2} \;$

$\sin\left (x\right ) \cos\left (y\right ) = {\sin\left (x + y\right ) + \sin\left (x - y\right ) \over 2} \;$

$\cos\left (x\right ) \sin\left (y\right ) = {\sin\left (x + y\right ) - \sin\left (x - y\right ) \over 2} \;$

Acknowledgements and Further Reading

Acknowledgements

Pieces of this book have been taken from wikipedia.

Contributors

In alphabetical order (by surname or display name):

• Aaron Paul (AKA Grimm)
• "Professor M." (no user page available)

Chaotic llama

• User:Cronholm144
• User:Fephisto
• User:Juliusross
• User:Stranger104
• User:Whiteknight

Further Reading

• Elements of the Differential and Integral Calculus by William Granville, a wikitext
• Elementary Calculus: An Approach Using Infinitesimals a free text
• First-Year Calculus Notes a GPL'd text
• Difference Equations to Differential Equations: An Introduction to Calculus another GPL'd text
• Calculus without Limits (Lecture Notes for Applied Calculus) by Karl Heinz Dovermann

License

Version 1.2, November 2002

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The "Title Page" means, for a printed book, the title page itself, plus such following pages as are needed to hold, legibly, the material this License requires to appear in the title page. For works in formats which do not have any title page as such, "Title Page" means the text near the most prominent appearance of the work's title, preceding the beginning of the body of the text.

A section "Entitled XYZ" means a named subunit of the Document whose title either is precisely XYZ or contains XYZ in parentheses following text that translates XYZ in another language. (Here XYZ stands for a specific section name mentioned below, such as "Acknowledgements", "Dedications", "Endorsements", or "History".) To "Preserve the Title" of such a section when you modify the Document means that it remains a section "Entitled XYZ" according to this definition.

The Document may include Warranty Disclaimers next to the notice which states that this License applies to the Document. These Warranty Disclaimers are considered to be included by reference in this License, but only as regards disclaiming warranties: any other implication that these Warranty Disclaimers may have is void and has no effect on the meaning of this License.

2. VERBATIM COPYING

You may copy and distribute the Document in any medium, either commercially or noncommercially, provided that this License, the copyright notices, and the license notice saying this License applies to the Document are reproduced in all copies, and that you add no other conditions whatsoever to those of this License. You may not use technical measures to obstruct or control the reading or further copying of the copies you make or distribute. However, you may accept compensation in exchange for copies. If you distribute a large enough number of copies you must also follow the conditions in section 3.

You may also lend copies, under the same conditions stated above, and you may publicly display copies.

3. COPYING IN QUANTITY

If you publish printed copies (or copies in media that commonly have printed covers) of the Document, numbering more than 100, and the Document's license notice requires Cover Texts, you must enclose the copies in covers that carry, clearly and legibly, all these Cover Texts: Front-Cover Texts on the front cover, and Back-Cover Texts on the back cover. Both covers must also clearly and legibly identify you as the publisher of these copies. The front cover must present the full title with all words of the title equally prominent and visible. You may add other material on the covers in addition. Copying with changes limited to the covers, as long as they preserve the title of the Document and satisfy these conditions, can be treated as verbatim copying in other respects.

If the required texts for either cover are too voluminous to fit legibly, you should put the first ones listed (as many as fit reasonably) on the actual cover, and continue the rest onto adjacent pages.

If you publish or distribute Opaque copies of the Document numbering more than 100, you must either include a machine-readable Transparent copy along with each Opaque copy, or state in or with each Opaque copy a computer-network location from which the general network-using public has access to download using public-standard network protocols a complete Transparent copy of the Document, free of added material. If you use the latter option, you must take reasonably prudent steps, when you begin distribution of Opaque copies in quantity, to ensure that this Transparent copy will remain thus accessible at the stated location until at least one year after the last time you distribute an Opaque copy (directly or through your agents or retailers) of that edition to the public.

It is requested, but not required, that you contact the authors of the Document well before redistributing any large number of copies, to give them a chance to provide you with an updated version of the Document.

4. MODIFICATIONS

You may copy and distribute a Modified Version of the Document under the conditions of sections 2 and 3 above, provided that you release the Modified Version under precisely this License, with the Modified Version filling the role of the Document, thus licensing distribution and modification of the Modified Version to whoever possesses a copy of it. In addition, you must do these things in the Modified Version:

A. Use in the Title Page (and on the covers, if any) a title distinct from that of the Document, and from those of previous versions (which should, if there were any, be listed in the History section of the Document). You may use the same title as a previous version if the original publisher of that version gives permission.

B. List on the Title Page, as authors, one or more persons or entities responsible for authorship of the modifications in the Modified Version, together with at least five of the principal authors of the Document (all of its principal authors, if it has fewer than five), unless they release you from this requirement.

C. State on the Title page the name of the publisher of the Modified Version, as the publisher.

D. Preserve all the copyright notices of the Document.

E. Add an appropriate copyright notice for your modifications adjacent to the other copyright notices.

F. Include, immediately after the copyright notices, a license notice giving the public permission to use the Modified Version under the terms of this License, in the form shown in the Addendum below.

G. Preserve in that license notice the full lists of Invariant Sections and required Cover Texts given in the Document's license notice.

H. Include an unaltered copy of this License.

I. Preserve the section Entitled "History", Preserve its Title, and add to it an item stating at least the title, year, new authors, and publisher of the Modified Version as given on the Title Page. If there is no section Entitled "History" in the Document, create one stating the title, year, authors, and publisher of the Document as given on its Title Page, then add an item describing the Modified Version as stated in the previous sentence.

J. Preserve the network location, if any, given in the Document for public access to a Transparent copy of the Document, and likewise the network locations given in the Document for previous versions it was based on. These may be placed in the "History" section. You may omit a network location for a work that was published at least four years before the Document itself, or if the original publisher of the version it refers to gives permission.

K. For any section Entitled "Acknowledgements" or "Dedications", Preserve the Title of the section, and preserve in the section all the substance and tone of each of the contributor acknowledgements and/or dedications given therein.

L. Preserve all the Invariant Sections of the Document, unaltered in their text and in their titles. Section numbers or the equivalent are not considered part of the section titles.

M. Delete any section Entitled "Endorsements". Such a section may not be included in the Modified Version.

N. Do not retitle any existing section to be Entitled "Endorsements" or to conflict in title with any Invariant Section.

O. Preserve any Warranty Disclaimers.

If the Modified Version includes new front-matter sections or appendices that qualify as Secondary Sections and contain no material copied from the Document, you may at your option designate some or all of these sections as invariant. To do this, add their titles to the list of Invariant Sections in the Modified Version's license notice. These titles must be distinct from any other section titles.

You may add a section Entitled "Endorsements", provided it contains nothing but endorsements of your Modified Version by various parties--for example, statements of peer review or that the text has been approved by an organization as the authoritative definition of a standard.

You may add a passage of up to five words as a Front-Cover Text, and a passage of up to 25 words as a Back-Cover Text, to the end of the list of Cover Texts in the Modified Version. Only one passage of Front-Cover Text and one of Back-Cover Text may be added by (or through arrangements made by) any one entity. If the Document already includes a cover text for the same cover, previously added by you or by arrangement made by the same entity you are acting on behalf of, you may not add another; but you may replace the old one, on explicit permission from the previous publisher that added the old one.

The author(s) and publisher(s) of the Document do not by this License give permission to use their names for publicity for or to assert or imply endorsement of any Modified Version.

5. COMBINING DOCUMENTS

You may combine the Document with other documents released under this License, under the terms defined in section 4 above for modified versions, provided that you include in the combination all of the Invariant Sections of all of the original documents, unmodified, and list them all as Invariant Sections of your combined work in its license notice, and that you preserve all their Warranty Disclaimers.

The combined work need only contain one copy of this License, and multiple identical Invariant Sections may be replaced with a single copy. If there are multiple Invariant Sections with the same name but different contents, make the title of each such section unique by adding at the end of it, in parentheses, the name of the original author or publisher of that section if known, or else a unique number. Make the same adjustment to the section titles in the list of Invariant Sections in the license notice of the combined work.

In the combination, you must combine any sections Entitled "History" in the various original documents, forming one section Entitled "History"; likewise combine any sections Entitled "Acknowledgements", and any sections Entitled "Dedications". You must delete all sections Entitled "Endorsements."

6. COLLECTIONS OF DOCUMENTS

You may make a collection consisting of the Document and other documents released under this License, and replace the individual copies of this License in the various documents with a single copy that is included in the collection, provided that you follow the rules of this License for verbatim copying of each of the documents in all other respects.

You may extract a single document from such a collection, and distribute it individually under this License, provided you insert a copy of this License into the extracted document, and follow this License in all other respects regarding verbatim copying of that document.

7. AGGREGATION WITH INDEPENDENT WORKS

A compilation of the Document or its derivatives with other separate and independent documents or works, in or on a volume of a storage or distribution medium, is called an "aggregate" if the copyright resulting from the compilation is not used to limit the legal rights of the compilation's users beyond what the individual works permit. When the Document is included in an aggregate, this License does not apply to the other works in the aggregate which are not themselves derivative works of the Document.

If the Cover Text requirement of section 3 is applicable to these copies of the Document, then if the Document is less than one half of the entire aggregate, the Document's Cover Texts may be placed on covers that bracket the Document within the aggregate, or the electronic equivalent of covers if the Document is in electronic form. Otherwise they must appear on printed covers that bracket the whole aggregate.

8. TRANSLATION

Translation is considered a kind of modification, so you may distribute translations of the Document under the terms of section 4. Replacing Invariant Sections with translations requires special permission from their copyright holders, but you may include translations of some or all Invariant Sections in addition to the original versions of these Invariant Sections. You may include a translation of this License, and all the license notices in the Document, and any Warranty Disclaimers, provided that you also include the original English version of this License and the original versions of those notices and disclaimers. In case of a disagreement between the translation and the original version of this License or a notice or disclaimer, the original version will prevail.

If a section in the Document is Entitled "Acknowledgements", "Dedications", or "History", the requirement (section 4) to Preserve its Title (section 1) will typically require changing the actual title.

9. TERMINATION

You may not copy, modify, sublicense, or distribute the Document except as expressly provided for under this License. Any other attempt to copy, modify, sublicense or distribute the Document is void, and will automatically terminate your rights under this License. However, parties who have received copies, or rights, from you under this License will not have their licenses terminated so long as such parties remain in full compliance.

10. FUTURE REVISIONS OF THIS LICENSE

The Free Software Foundation may publish new, revised versions of the GNU Free Documentation License from time to time. Such new versions will be similar in spirit to the present version, but may differ in detail to address new problems or concerns. See http://www.gnu.org/copyleft/.

Each version of the License is given a distinguishing version number. If the Document specifies that a particular numbered version of this License "or any later version" applies to it, you have the option of following the terms and conditions either of that specified version or of any later version that has been published (not as a draft) by the Free Software Foundation. If the Document does not specify a version number of this License, you may choose any version ever published (not as a draft) by the Free Software Foundation.