Calculus. Contents. Paul Sutcliffe. Office: CM212a.

Transcription

1 Calculus Paul Sutcliffe Office: CM212a. Books One and several variables calculus, Salas, Hille & Etgen. Calculus, Spivak. Mathematical methods in the physical sciences, Boas. Contents 1 Functions Functions, domain and range The graph of a function Graphs and transformations Even and odd functions Piecewise functions Operations with functions Inverse functions Elementary functions imits and continuity Definitions Results about limits and continuity Classification of discontinuities imits as x The intermediate value theorem

2 3 Differentiation Derivative as a limit Rules of differentiation Derivatives of higher order The chain rule Hôpital s rule Implicit differentiation Boundedness and monotonicity Critical points Min-Max problems Rolle s theorem The mean value theorem The inverse function rule Partial derivatives Integration Indefinite integrals Definite integrals The fundamental theorem of calculus The logarithm and exponential function revisited Standard integrals Integration by parts Integration by substitution Integration of rational functions using partial fractions Integration using a recurrence relation Definite integrals using even and odd functions Differential equations First order separable ODE s First order homogeneous ODE s First order linear ODE s First order exact ODE s Bernoulli equations Second order linear constant coefficient ODE s The homogeneous case

3 5.6.2 The inhomogeneous case Initial and boundary value problems Systems of first order linear ODE s Taylor series Taylor s theorem and Taylor series agrange form for the remainder Calculating limits using Taylor series Fourier series Calculating the Fourier coefficients Parseval s theorem Half range Fourier series Complex Fourier series Functions of several variables Taylor s theorem for functions of two variables Chain rule Implicit differentiation Double integrals Rectangular regions Beyond rectangular regions Polar coordinates Change of variables and the Jacobian The Gaussian integral

4 1 Functions Notation: Symbol s.t. iff Denotes for every there exists such that if and only if Open interval, (a, b) = {x : a < x < b}. Closed interval, [a, b] = {x : a x b}. Half-open interval, (a, b] = {x : a < x b}, etc. Semi-infinite interval, (a, ) = {x : x > a}, etc. The set of real numbers, (, ), is denoted by R. For two sets A and B then A B is the union, A B is the intersection and A\B is the set of all elements in A that are not in B. 1.1 Functions, domain and range Defn: A function f is a correspondence between two sets D (called the domain) and C (called the codomain), that assigns to each element of D one and only one element of C. The notation to indicate the domain and codomain is f : D C. For x D we write f(x) to denote the assigned element in C, and call this the value of f at x, or the image of x under f, where x is called the argument of the function. Extending the above notation we write f : D C : x f(x) ie. f is a function from D to C that associates x in D to f(x) in C. The set of all images is called the range of f and our notation for the domain and range is Dom f and Ran f = {f(x) : x Dom f}. 4

5 Figure 1: Illustration of the domain, codomain and range of a function. In (almost all of) this course we shall only deal with real-valued functions of a real variable, meaning that our functions assign real numbers to real numbers ie. both the domain and codomain are subsets of R. There are various ways of representing a function, but perhaps the most familiar is through an explicit expression. Eg. f : R R given by f(x) = x 2, x R. In this case Dom f = R and Ran f = [0, ). We say that f maps the real line onto [0, ). Eg. f(x) = 2x + 4, x [0, 6]. Here Dom f = [0, 6] and Ran f = [2, 4]. If the domain of a function f is not explicitly given, then it is taken to be the maximal set of real numbers x for which f(x) is a real number. Eg. f(x) = 2x + 4. Here Dom f = [ 2, ) and Ran f = [0, ). Eg. f(x) = 1/(1 x). Here Dom f = R\{1} = (, 1) (1, ) and Ran f = R\{0}. In addition to explicit expressions, a function f may be represented by other means, for example, as a set of ordered pairs (x, y), where x Dom f and y Ran f. 5

6 Note: y(x) is another common notation for a function. The element x in the domain is called the independent variable and the element y in the range is called the dependent variable. This notation is often used if the function is defined by an equation in two variables, including differential equations (see later). The function f(x) = x 2 is defined by the equation y = x 2, as we have simply written y = f(x). However, not all equations will define a function. Eg. The equation y 2 = x does NOT define a function y(x), because for x < 0 there are no real solutions for y, and for x > 0 there are two solutions for y, whereas a function must assign only one element of the range for each element of the domain. 1.2 The graph of a function Defn: The graph of a function f is the set of all points (x, y) in the xy-plane with x Dom f and y = f(x), ie. graph f = {(x, y) : x Dom f and y = f(x)}. The graph of a function over the interval [a, b] is the portion of the graph where the argument is restricted to this interval. Note: If you are asked to graph a function, but no interval is given, then try to choose an appropriate interval that includes all the interesting behaviour eg. turning points. To indicate on a graph exactly which points are included we use a closed circle to denote an included point, so if it is at the end of a curve segment this denotes that this end is a closed interval. We use an open circle to denote an excluded point associated with an open interval. 6

7 Figure 2: The graph of f(x) = cos x over [ 2π, 2π]. The graph of a function is a curve in the plane, but not every curve is the graph of a function. The following simple test determines whether a curve is the graph of a function. The vertical line test. If any vertical line intersects the curve more than once then the curve is not the graph of a function, otherwise it is. The proof is obvious, given the defining property of a function that only one element in the range is associated with an element in the domain. Figure 3: The vertical line test applied to a cubic curve and a circle. In the above figure the first curve is the graph of a function; in fact the cubic function f(x) = x 3 x 2 1. The second curve is not the graph of a function as the vertical line shown intersects the curve twice. In fact the curve is given by the equation y 2 = 4 (x 1) 2 ie. the circle of radius 2 with centre (x, y) = (1, 0). Note that even when a curve is not the graph of 7

8 a function, some sections of the curve may be graphs of functions. For the above circle example the section given by y 0 is the graph of the function f(x) = 4 (x 1) 2 with Dom f = [ 1, 3]. 8

9 1.3 Graphs and transformations The graph of a function f : R R can be shifted, reflected and rescaled by applying some simple transformations to the function and its argument, as follows: translation: f(x) + a shifts the graph of f(x) up by a units. f(x + b) shifts the graph of f(x) to the left by b units. reflection: f(x) reflects the graph of f(x) about the x-axis. f( x) reflects the graph of f(x) about the y-axis. f(x) reflects the portions of the graph of f(x) that are below the x-axis about the x-axis. scaling: λf(x) is a vertical stretch of the graph of f(x) if λ > 1 and a vertical contraction if 0 < λ < 1. f(µx) is a horizontal contraction of the graph of f(x) if µ > 1 and a horizontal stretch if 0 < µ < 1. Figure 4: Graph of f(x) = x 2 2x + 2, together with f(x) and f( x). Combining these transformations in turn is a way to create graphs from the graph of a simpler starting function. As an example, the figure shows how 9

10 Figure 5: Graph of f(x) = sin x, together with 2f(x) and f(2x). to obtain the graph of the function f(x) = (x 1) 2 1 over [ 1, 3] by starting from the function g(x) = x 2. Figure 6: Graphs of x 2, (x 1) 2, (x 1) 2 1, (x 1) 2 1, 2 (x 1) 2 1, 2+2 (x 1) 2 1. Here are some tips for graphing the reciprocal 1/f(x) from the graph of f(x). If f(x) is increasing (decreasing) then 1/f(x) is decreasing (increasing). The graphs of f(x) and 1/f(x) intersect iff f(x) = ±1. Figure 7: Graph of f(x) = 1 + x 2, together with 1/f(x). 10

11 1.4 Even and odd functions Defn. A function f is even if f(x) = f( x) ± x Dom f. Defn. A function f is odd if f(x) = f( x) ± x Dom f. The graph of an even function is symmetric under a reflection about the y-axis. The graph of an odd function is symmetric under a rotation by 180 about the origin (equivalent to a reflection in the y-axis followed by a reflection in the x-axis). Figure 8: Graphs of the even function x 2 and the odd function x 3. All functions f : R R can be written as the sum of an even and an odd function f(x) = f even (x) + f odd (x), where f even (x) = 1 2 f(x) f( x) and f odd(x) = 1 2 f(x) 1 2 f( x). This decomposition is often useful, as is the ability to spot an even or odd function as it can simplify some calculations. Eg. f(x) = (1 + x) sin x with f even (x) = x sin x and f odd (x) = sin x. 11

12 1.5 Piecewise functions Some functions are defined piecewise ie. different expressions are given for different intervals in the domain. Eg. The absolute value (or modulus) function { x if x 0 f(x) = x = x if x < 0. Figure 9: The graph of x over [ 1, 1]. Eg. f(x) = { x if 0 x < 1 x 1 if 1 x 2. In this example Dom f = [0, 2] but the function has a discontinuity at x = 1 (more about this later). Figure 10: The graph of a piecewise function. A step function is a piecewise function which is constant on each piece. An example is the Heaviside step function H(x) defined by { 0 if x < 0 H(x) = 1 if x > 0. Note that with this definition Dom H = R\{0}. It is sometimes convenient to extend the domain to R by defining the value of H(0) (some obvious candidates are 0, 1 2, 1). Figure 11: The Heaviside step function. 12

13 1.6 Operations with functions Given two functions f and g we can define the following: the sum is (f + g)(x) = f(x) + g(x), with domain Dom f Dom g. the difference is (f g)(x) = f(x) g(x), with domain Dom f Dom g. the product is (fg)(x) = f(x)g(x), with domain Dom f Dom g. the ratio is (f/g)(x) = f(x)/g(x), with domain (Dom f Dom g)\{x : g(x) = 0}. the composition is (f g)(x) = f(g(x)), with domain {x Dom g : g(x) Dom f}. Note that f g and g f are usually different functions. Eg. f(x) = sin x, g(x) = x 2 then (f g)(x) = sin(x 2 ) but (g f)(x) = sin 2 x. 1.7 Inverse functions Defn. A function f : D C is surjective (or onto) if Ran f = C, ie. if y C x D s.t. f(x) = y. Eg. f : R R given by f(x) = 2x + 1 is surjective. Eg. f : R R given by f(x) = x 2 is not surjective, since any negative number in the codomain is not the image of an element in the domain. Defn. A function f : D C is injective (or one-to-one) if x 1, x 2 D with x 1 x 2 then f(x 1 ) f(x 2 ). Eg. f(x) = 2x + 1 is injective since f(x 1 ) = f(x 2 ) implies x 1 = x 2. 13

14 Eg. f(x) = x 2 is not injective since f(x) = f( x) and x x if x 0. The following simple test can be applied to see if a function is injective from its graph. The horizontal line test. If no horizontal line intersects the graph of f more than once then f is injective, otherwise it is not. Eg. The function f(x) = x 3 3x is surjective as Ran f = R but it is not injective as the horizontal line y = 1 intersects the graph of f at 3 points (see Figure 12). Figure 12: Graph of f(x) = x 3 3x and the horizontal line y = 1. Defn. A function f : D C is bijective if it is both surjective and injective. Eg. From above we have seen that the function f(x) = 2x + 1 is bijective. Theorem of inverse functions. A bijective function f admits a unique inverse, denoted f 1, such that (f 1 f)(x) = x = (f f 1 )(x). It is clear from the definition that Dom f 1 = Ran f and Ran f 1 = Dom f. As the inverse undoes the effect of the function an equivalent definition is f(x) = y iff f 1 (y) = x. 14

15 Eg. We have seen that f(x) = 2x + 1 is bijective, so lets find its inverse. To simplify notation first write y = f 1 (x). Using the property x = f(f 1 (x)), we have x = f(y) = 2y + 1 and hence y = 1 2 (x 1) = f 1 (x). Note that given an injective function we can make a bijective function by taking the codomain equal to the range, and then an inverse exists. Eg. We have seen that f(x) = x 2 is not an injective function if Dom f = R, but it is injective if we take Dom f = [0, ). With this choice we can take the codomain equal to Ran f = [0, ) and f is now a bijective function. The inverse is f 1 (x) = x with Dom f 1 = Ran f = [0, ). Warning: Dont confuse the notation for the inverse with the same notation for the reciprocal. 1.8 Elementary functions Polynomials. inear functions are polynomials of degree one and take the general form f(x) = ax + b. Here Dom f = Ran f = R. The graph is a straight line with slope a and y-intercept (0, b), which is the point on the graph that intersects the y-axis. Quadratic functions are polynomials of degree two and take the general form f(x) = ax 2 + bx + c. Here Dom f = R, but the range is restricted. The graph is an upward parabola if a > 0 and a downward parabola if a < 0. By completing the square we can write f(x) = a(x + b 2a )2 + c b2 4a and see 2 that the vertex of the parabola is at ( b 2a, c b2 4a ). Hence, if a > 0 then the 2 range is given by Ran f = [c b2 4a, ) and if a < 0 then Ran f = (, c b2 2 4a ]. 2 Eg. f(x) = 2x 2 + 8x + 5 = 2(x 2) so Ran f = (, 13]. 15

16 Cubic functions are polynomials of degree three and take the general form f(x) = ax 3 + bx 2 + cx + d, with Dom f = Ran f = R. The graph of a typical example is shown in Figure 12 for the cubic function f(x) = x 3 3x. A cubic has at most two turning points (where the graph changes direction) and in the example these are at x = ±1. From the above examples it should be clear that for a polynomial function of arbitrary degree n, the domain is always R, and the range is also R if n is odd, but is restricted if n is even. Square root functions. These take the form f(x) = g(x) for some function g(x). By definition the square root is a non-negative number, so this can yield restrictions on the range. There can also be restrictions on the domain, since the square root is only defined if the argument g(x) is non-negative. Eg. f(x) = x has Dom f = Ran f = (, 0]. Rational functions. A rational function is a ratio of two polynomials f(x) = p(x)/q(x). The domain is given by Dom f = R\{x : q(x) = 0}. Eg. f(x) = 2x/(x 2 1) has Dom f = R\{±1}. The simplified form of a rational function is obtained by removing any common factors in the numerator and denominator. Eg. f(x) = 2x 4 x 2 4 = 2(x 2) (x + 2)(x 2) = 2 x + 2 if x 2. Note that we had to exclude the point x = 2 in the last equality to avoid a division by zero. Hence in this example the original rational function and its simplified form 2/(x + 2) are not quite the same function as they have different domains, being R\{±2} and R\{ 2} respectively. 16

17 If, in simplified form, q(a) = 0 then x = a is a vertical asymptote of the graph of f, which is a vertical line given by the equation x = a. Figure 13: Graph of f(x) = 2x/(x 2 1) and the vertical asymptotes x = ±1. There can also be horizontal asymptotes though a correct treatment requires the concept of a limit, which we have not yet studied. Roughly, a horizontal asymptote involves considering the behaviour of the function when x is large. For rational functions a horizontal asymptote exists only if the degree of the numerator is not larger than the degree of the denominator. In this case we can write f(x) = p 0 + p 1 x p n x n q 0 + q 1 x q n x n and the horizontal asymptote is the line y = p n /q n. In the example of Figure 13 we have n = 2 and p 2 = 0, q 2 = 1, so the horizontal asymptote is the line y = 0, which the graph approaches when x is large. A rational function is called a proper rational function if the degree of the numerator is less than the degree of the denominator. A rational function can always be written as the sum of a polynomial and a proper rational function by performing long division. 17

18 Figure 14: Graph of f(x) = 3x 2 /(2x 2 50), showing the vertical asymptotes x = ±5 and the horizontal asymptote y = 3/2. Eg. f(x) = (1 + x 6 )/(x 3 x 2 ) is not a proper rational function. x 3 + x 2 + x + 1 x 3 x 2 x x 6 x 5 x x 5 x 4 x x 4 x 3 x x 3 x 2 x This yields the required decomposition 1 + x 6 x 3 x = 2 x3 + x 2 + x x2 + 1 x 3 x 2. 18

19 Exponential and logarithm functions. An exponential function is any function of the form f(x) = b x where b (called the base) is a positive real number not equal to 1. For all values of the base Dom f = R and Ran f = (0, ) and the graph is continuous and passes through the point (0, 1). However, there are two types of behaviour: If b > 1 the graph is increasing and has a horizontal asymptote y = 0 along the negative x-axis. If 0 < b < 1 the graph is decreasing and has a horizontal asymptote y = 0 along the positive x-axis. Figure 15: Graph of f(x) = 2 x and g(x) = ( 1 2 )x = 1 2 x = 1/f(x). Two important properties of exponential functions are b x b y = b x+y and (b x ) y = b xy. There is a special value of the base (called the natural base) b = e = , where e is called Euler s number and is transcendental (like π) ie. it is not the root of a polynomial equation with rational coefficients. One way to calculate e is via the sum e = ! + 1 2! + 1 3! + 1 4! +... The function f(x) = e x is usually referred to as the exponential function. 19

20 An exponential function f(x) = b x is injective and so admits an inverse. To determine this write y = f 1 (x) so that f(y) = x ie. b y = x. This is the defining relation of a logarithm function g(x) = log b x = f 1 (x) which is simply the name given to the inverse of an exponential function f(x) = b x. Explicitly, y = log b x iff b y = x. The domain of a logarithm function is the range of an exponential function and hence is (0, ). The range of a logarithm function is the domain of an exponential function and is thus R. Notation: I shall use the notation log x as the shorthand for the logarithm log e x, which is known as the natural logarithm. Warning: This notation is not universal as some people use ln x to denote the natural logarithm and instead use log x to denote log 10 x. A logarithm grows very slowly for x large and positive (more slowly than a linear function) and has a vertical asymptote at x = 0 (see Figure 16). Figure 16: The natural logarithm log x and the vertical asymptote at x = 0. Some important properties of logarithms are: log b 1 = 0 from b 0 = 1. log b b x = x from the definition as the inverse of an exponential. log b (αβ) = log b α + log b β from b α b β = b α+β. 20

21 log b (α/β) = log b α log b β from b α /b β = b α β. log b (x r ) = r log b x from (a x ) r = a xr. Trigonometric functions. Defn. A function f(x) is periodic if p > 0 s.t. f(x + p) = f(x) x. The smallest such value of p is called the period of f. The simplest periodic functions are the trigonometric functions sin x and cos x with period 2π. In both cases the domain is R and the range is [ 1, 1]. The function tan x = sin x/ cos x has period π and the domain is R\{x = ( n)π for integer n} with associated vertical asymptotes x = ( n)π. The range is R. Figure 17: The trigonometric functions sin x, cos x and tan x. It is expected that you know the values of the trigonometric functions at the particular values of the argument shown in the table, plus those related by the symmetries apparent in the plots shown in Figure 17. radians 0 π/6 π/4 π/3 π/2 degrees sin cos It is also expected that you know the important trigonometric identities: 21

22 sin 2 x + cos 2 x = 1. addition formulae sin(x ± y) = sin x cos y ± cos x sin y cos(x ± y) = cos x cos y sin x sin y double angle formulae sin 2x = 2 sin x cos x cos 2x = 2 cos 2 x 1 = 1 2 sin 2 x half-angle formulae sin 2 x = cos 2x cos 2 x = cos 2x The reciprocals of the trigonometric functions have their own notation: csc x = 1/ sin x, sec x = 1/ cos x, cot x = 1/ tan x = cos x/ sin x. The above formulae lead to related formulae for these functions eg. sec 2 x = 1 + tan 2 x. The trigonometric functions are not injective over their full domains, but are injective over suitably restricted domains (given in the table), so that an inverse exists. f(x) Dom f Ran f f 1 (x) Dom f 1 Ran f 1 sin x [ π/2, π/2] [ 1, 1] sin 1 x or arcsin x [ 1, 1] [ π/2, π/2] cos x [0, π] [ 1, 1] cos 1 x or arccos x [ 1, 1] [0, π] tan x ( π/2, π/2) R tan 1 x or arctan x R ( π/2, π/2) Figure 18: Graphs of the functions sin 1 x, cos 1 x and tan 1 x. 22

23 Hyperbolic functions. Hyperbolic functions, with domain R, are defined in terms of the exponential function via sinh x = 1 2 (ex e x ), cosh x = 1 2 (ex + e x ), tanh x = sinh x/ cosh x. Figure 19: Graphs of the functions sinh x, cosh x and tanh x. From the properties of the exponential function it is easy to show (exercise) that some basic properties include sinh 0 = 0 and cosh 0 = 1. If x > 0 then 0 < sinh x < cosh x. sinh x is an odd function and cosh x is an even function. Addition formulae sinh(x ± y) = sinh x cosh y ± cosh x sinh y cosh(x ± y) = cosh x cosh y ± sinh x sinh y. Double argument formulae sinh(2x) = 2 sinh x cosh x cosh(2x) = sinh 2 x + cosh 2 x. 23

24 cosh 2 x sinh 2 x = 1. The reciprocals of hyperbolic functions have their own notation: cosech x = 1/sinh x, sech x = 1/cosh x, coth x = 1/tanh x. The above formulae lead to related formulae for these functions eg. sech 2 x = 1 tanh 2 x. 24

25 2 imits and continuity 2.1 Definitions Rough idea #1: A function f(x) has a limit at x = a if f(x) is close to whenever x is close to a. Rough idea #2: ets think about the function f(x) near x = a. If you approximate f(x) by a constant then you will make an error given by f(x). Suppose ε > 0 is the εrror at which you become unhappy with your approximation, that is, you are happy as long as f(x) < ε. The important issue for a limit is whether I can keep you happy simply by making sure that x stays within a δistance δ > 0 of a, that is by restricting to 0 < x a < δ. Note that we dont even care about what happens exactly at x = a. If I can always find a δistance δ that keeps you happy, no matter how small you choose your εrror ε, then we say that f(x) has a limit as x tends to a. Figure 20: The idea of a limit. 25

26 Defn: f(x) has a limit as x tends to a if We then write ε > 0 δ > 0 s.t f(x) < ε when 0 < x a < δ. lim f(x) = or equivalently f(x) as x a. x a If there is no such then we say that no limit exists. The limit does not require that f(a) is equal to or even that f(a) exists. This is because of the inequality 0 < x a in the definition. There is a lot more of this ε and δ stuff in Analysis I, where proofs concerning limits are considered. In this course we shall be less formal and deal only with methods to calculate limits or see that no limit exists. Rough idea #3: If f(a) exists then it would seem that this is a good candidate for the limit. This naive view turns out to be correct only if the graph of f(x) is a single unbroken curve with no holes or jumps, at least in a small region around x = a. Defn. A function f(x) is continuous at the point x = a if the following three properties all hold: f(a) exists. lim x a f(x) exists. lim x a f(x) is equal to f(a). Defn. A function f(x) is continuous on a subset S of its domain if it is continuous at every point in S. Defn. A function f(x) is continuous if it is continuous at every point in its domain. Eg. f(x) = x 2 is continuous and lim x 2 x 2 = f(2) = 4. 26

27 Eg. The following function has a discontinuity at x = 0 x 2 if 1 x < 0 f(x) = 1 if x = 0 x 2 if 0 < x 1 As mentioned earlier, to indicate on a graph exactly which points are included we use a closed circle to denote an included point, so if it is at the end of a curve segment this denotes that this end is a closed interval. We use an open circle to denote an excluded point associated with an open interval. This notation is demonstrated in Figure 21 where the graph of the above function is presented. Figure 21: A graph illustrating the use of open and closed circles. The limit exists at x = 0. This function is not continuous at x = 0 but the limit exists at this point and lim x 0 f(x) = 0 1 = f(0). Eg. The following function also has a discontinuity at x = 0 (see Figure 23 for its graph) { 1 if x 0 f(x) = x 2 if x > 0 In this case no limit exists at x = 0. This is clear because if we consider x < 0 then to get arbitrarily close to the function for x arbitrarily close to 0 would require the limit to be 1. However, if we consider x > 0 then to get arbitrarily close to the function for x arbitrarily close to 0 would require the limit to be 0. These two requirements are incompatible and so no limit exists at x = 0. 27

28 Figure 22: A graph illustrating the use of open and closed circles. No limit exists at x = 0. Eg. For the function f(x) = sin(1/x) no limit exists at x = 0 because as x approaches zero this function oscillates between 1 and -1 over smaller and smaller intervals. In particular, in any given interval 0 < x < δ it is always possible to find values x 1 and x 2 s.t. f(x 1 ) = 1 and f(x 2 ) = 1, thus f(x) cannot remain close to any given constant. Figure 23: A graph of the function sin(1/x). 2.2 Results about limits and continuity There are some simple facts about limits that follow from the definition: The limit is unique. If f(x) = g(x) (except possibly at x = a) on some open interval containing a then lim x a f(x) = lim x a g(x). If f(x) K on either an interval (a, b) or an interval (c, a) 28

29 and if lim x a f(x) = then K. (A similar result holds by replacing all with ). If lim x a f(x) = and lim x a g(x) = M then (i) lim x a (f(x) + g(x)) = + M (ii) lim x a (f(x)g(x)) = M (iii) if M 0 then lim x a (f(x)/g(x)) = /M. Note: In Durham the results (i),(ii),(iii) are sometimes known as the Calculus Of imits Theorem (COT). However, this seems to be a rather grand name for these results and you will not find this name in any books. Pinching Theorem. If g(x) f(x) h(x) for all x a in some open interval containing a and lim x a g(x) = lim x a h(x) = then lim x a f(x) =. There are also some simple facts about continuous functions that follow from the definition and the above results: If f(x) and g(x) are continuous functions then so are f(x)+g(x), f(x)g(x) and f(x)/g(x). All polynomial, rational, trigonometric and hyperbolic functions are continuous. If f(x) is continuous then so is f(x). Eg. All the following functions are continuous: 2x 3 + x + 7, 3x/(x 1), (1 + x 2 )/ sin x, tan x. Note: Although tan x is continuous, it is not continuous on the interval [0, π], because this interval includes the point x = π/2 which is not in the domain of tan x. Eg. lim x π/2 x 2 sin x = (π/2) 2 sin(π/2) = π 2 /4. Eg. lim x 0 1 x does not exist. The value of f(x) can be made greater than any 29

30 finite constant by taking x sufficiently close to zero, so no limit exists. Eg. Calculate lim x 3 2x 2 18 x 3. 2x 2 18 x 3 = 2(x + 3)(x 3) x 3 = 2(x + 3) if x 3. The value of the function at x = 3 is irrelevant in defining the limit as x 3 so 2x 2 18 lim x 3 x 3 = lim x 3 2(x + 3) = 12. Eg. Calculate lim x 5 x 25 x 25. x 5 x 25 = ( x 5)( x + 5) (x 25)( x + 5) = x 25 (x 25)( x + 5) = 1 if x 25. x + 5 x 5 lim x 25 x 25 = lim 1 = 1 x 25 x Eg. Calculate lim x 0 x 2 sin( 1 x ). As 1 sin( 1 x ) 1 then x2 x 2 sin( 1 x ) x2. Also lim x 0 x 2 = 0 = lim x 0 x 2. Hence by the pinching theorem lim x 0 x 2 sin( 1 x ) = 0. Figure 24: Graphs of f(x) = x 2 sin(1/x) and the bounding functions g(x) = x 2 and h(x) = x 2. Two important trigonometric limits that you may assume to be true and use are: sin x 1 cos x lim = 1 and lim = 0. x 0 x x 0 x 30

31 The graphs in Figure 25 provide some evidence to support these results. They can be proved by applying the pinching theorem but it requires a bit of work. For example, the first result requires the fact that sin x < x < tan x for 0 < x < π/2. Figure 25: Graphs of sin x and 1 cos x. As x = 0 is not in the domain of either function there x x should really be open circles at x = 0 on both graphs. Eg. Calculate lim x 0 sin(2x) x. sin(2x) lim x 0 x = lim x 0 2 sin(2x) 2x = lim u 0 2 sin(u) u = 2 1 = 2. The above used the change of variable u = 2x and then the first of the two important trigonometric limits given above. Eg. Calculate lim x 0 tan x x. tan x lim x 0 x sin x = lim x 0 x cos x = Eg. Calculate lim x 0 1 cos x x 2. For π < x < π 1 cos x x 2 = ( lim x 0 (1 cos x)(1 + cos x) x 2 (1 + cos x) )( sin x lim x x 0 ) 1 = 1 1 = 1. cos x ( ) 2 sin x 1 = x 1 + cos x Hence by using the first important trigonometric limit we have that {( ) 2 } 1 cos x sin x 1 lim = lim = (1) 2 1 x 0 x 2 x 0 x 1 + cos x =

32 2.3 Classification of discontinuities If a function f(x) is not continuous at a point x = a then we say it has a discontinuity at x = a. There are different types of discontinuity and these are best classified by considering how the function behaves on each side of the point x = a. This motivates the definition of the following one-sided limits. Defn: f(x) has a right-sided limit + = lim x a + f(x) as x tends to a from above if ε > 0 δ > 0 s.t f(x) + < ε when 0 < x a < δ. Note that the difference between this definition and the definition of the limit is the removal of the modulus sign in x a. This means that we only need to worry about points to the right of a when requiring the function to be close to +. Similarly, we have the definition Defn: f(x) has a left-sided limit = lim x a f(x) as x tends to a from below if ε > 0 δ > 0 s.t f(x) < ε when 0 < a x < δ. Here we only need to worry about points to the left of a when requiring the function to be close to. It should be obvious that = lim x a f(x) exists iff + and both exist and are equal. In which case = + =. There are 3 types of discontinuity: (i) Removable discontinuity. In this case exists but f(a). The discontinuity can be removed to make the continuous function { f(x) if x a g(x) = if x = a. 32

33 Eg. The following function (Figure 26i) has a removable discontinuity at x = 0, { x 2 if x 0 f(x) = 1 if x = 0. Removing this discontinuity yields the continuous function g(x) = x 2. Figure 26: 3 types of discontinuities at x = 0 : (i) removable; (ii) jump; (iii) infinite. (ii) Jump discontinuity. In this case both + and exist but +. Eg. The following function (Figure 26ii) has a jump discontinuity at x = 0, { 1 if x 0 f(x) = x 2 if x > 0. In this example + = 0 1 =. Eg. The signum function 1 if x < 0 sgn(x) = 0 if x = 0 1 if x > 0 has a jump discontinuity at x = 0. In this example + = 1 1 =. 33

34 (iii) Infinite discontinuity. In this case at least one of + or does not exist. Eg. The function f(x) = 1/x (Figure 26iii) has an infinite discontinuity at x = 0. In this example neither + nor exist. 2.4 imits as x. So far we have only been concerned with the limit of a function f(x) as x approaches a finite point a. However, it is also possible to define a limit as x. Rough idea #4: A function f(x) has a limit as x if f(x) can be kept arbitrarily close to by making x sufficiently large. Defn: f(x) has a limit as x tends to if We then write ε > 0 S > 0 s.t f(x) < ε when x > S. lim f(x) = or equivalently f(x) as x. x If there is no such then we say that no limit exists. The corresponding definition associated with lim x f(x) = should be obvious and is given by Defn: f(x) has a limit as x tends to if ε > 0 S < 0 s.t f(x) < ε when x < S. 1 Eg. lim x x = 0. This should be clear because 1 x can be made as close to zero as required by making x sufficiently large. An easy way to calculate 34

35 limits as x is to first make the substitution u = 1/x and then use the fact that if lim x f(x) exists then lim x f(x) = lim u 0 + f(1/u). This has transformed the limit into one that we are already familiar with. Eg. lim x 1 x = lim u 0 + u = 0. Eg. Calculate lim x x cos(1/x)+2 x. By using the substitution u = 1/x we have that x cos(1/x) + 2 lim x x = lim u u cos u u = lim (cos u + 2u) = cos(0) + 0 = 1. u 0 + Our earlier discussion of a horizontal asymptote can now be made more precise, in that the graph of a function f(x) will have a horizontal asymptote y = if lim x f(x) =. Eg. 2x + 3 lim x x + 5 = lim x x x = = 2. 1 Here we have used the result that lim x x = 0, together with the earlier facts about limits. We see that calculating this limit shows that the graph of this rational function has the horizontal asymptote y = 2, reproducing our earlier observations about the horizontal asymptotes of rational functions. 2.5 The intermediate value theorem The intermediate value theorem states that if f(x) is continuous on [a, b] and u is any number between f(a) and f(b) (ie. either f(a) < u < f(b) or f(b) < u < f(a)) then (at least one) c (a, b) s.t. f(c) = u. Eg. f(x) = sin x is continuous on [0, π/2] and f(0) = 0 < 1 2 < 1 = f(π/2) so by the intermediate value theorem there is (at least) one point x in (0, π/2) s.t. sin x = 1 2. It is important that f(x) is continuous throughout the interval, otherwise the theorem does not apply. 35

36 Figure 27: An illustration of the intermediate value theorem. Eg. f(x) = sgn(x) 1+x has f( 1) = < 1 5 < 1 2 = f(1) but there is no x in ( 1, 1) s.t. f(x) = 1 5. The intermediate value theorem does not apply because f(x) is not continuous at x = 0. There is a jump discontinuity at x = 0 with lim x 0 + f(x) = 1 1 = lim x 0 f(x). An application of the intermediate value theorem is locating the zeros of a function. If the function f(x) is continuous on [a, b] and we know that either f(a) < 0 < f(b) or f(b) < 0 < f(a) then by the intermediate value theorem the equation f(x) = 0 has at least one root between a and b. Eg. The function f(x) = x 2 2 is continuous on [1, 2] with f(1) = 1 < 0 and f(2) = 2 > 0, so there is at least one root of the equation x 2 2 = 0 in (1, 2) (of course the root is x = 2 = ). A repeated iterated application of this approach gives the bisection method, which can be used to locate the roots of a wide variety of equations to any desired accuracy. 36

37 3 Differentiation 3.1 Derivative as a limit Geometrically the derivative f (a) of a function f(x) at x = a is equal to the slope of the tangent to the graph of f(x) at x = a. Figure 28: The graph of a function f(x), the tangent at x = a and the secant through the points on the graph at x = a and x = a + h. A secant is a line that intersects a curve at two points (the part of the secant that is between the two intersection points is called a chord). Consider a secant that intersects the graph of f(x) at the two (x, y) points (a, f(a)) and (a+h, f(a+h)). The slope of this secant is given by the difference quotient (f(a + h) f(a))/h. As the two points approach each other ie. as h decreases, the secant approaches the tangent at x = a. The tangent is obtained in the limit as h 0 and the derivative at a is the slope of the secant in this limit ie. f (a) = lim h 0 f(a + h) f(a) h providing this limit exists. If this limit exists then we say that f(x) is differentiable at x = a. Eg. Use the limit definition of the derivative to calculate f (a) for f(x) = x 2. f (a) = lim h 0 f(a + h) f(a) h = lim h 0 (a + h) 2 a 2 h = lim h 0 2ah + h 2 h = lim h 0 (2a+h) = 2a. 37

38 This is, of course, the expected result from the knowledge that f (x) = 2x. Eg. Use the limit definition of the derivative to calculate f (π) for f(x) = sin x. f (π) = lim h 0 f(π + h) f(π) h = lim h 0 sin(π + h) sin π h = lim h 0 sin π cos h + cos π sin h h sin h = lim = 1 h 0 h where the final equality uses the earlier important trigonometric limit result. This is, of course, the expected answer from the knowledge that f (x) = cos x giving f (π) = cos π = 1. If f (a) exists for all a in Dom f we say that f(x) is differentiable and then f (a) defines a function f (x) called the derivative. Eg. Use the limit definition of the derivative to calculate the derivative of the function f(x) = x cos x. ( = lim h 0 f (x) = lim h 0 f(x + h) f(x) h x cos x cos h 1 h = lim h 0 (x + h) cos(x + h) x cos x h (x + h)(cos x cos h sin x sin h) x cos x = lim h 0 h x sin x sin h ) +cos x cos h sin x sin h = x sin x+cos x h where we have used both of the earlier important trigonometric limit results. The fact that the derivative is equal to the slope of the tangent of the graph allows us to determine a Cartesian equation for the tangent by calculating the derivative. Explicitly, the tangent line at the point (a, f(a)) is given by y = f(a) + f (a)(x a). Eg. For the function f(x) = 4x 3 x, find a Cartesian equation for the tangent to the graph at the point (1, 3). f (x) = 12x 2 1 so f (1) = 11. The tangent line is given by 38

39 y = f(1) + f (1)(x 1) = (x 1) = 11x 8 A necessary condition for a function to be differentiable at a point is that it is continuous at that point, but this is not a sufficient condition. Geometrically, there are two ways that a function can fail to be differentiable at a point where it is continuous: (i). The tangent line is vertical at that point. Eg. The function f(x) = x 1/3 is continuous in R but it is not differentiable at x = 0. The difference quotient at x = 0 is f(0 + h) f(0) h = h1/3 h = 1 h 2/3 and no limit exists as h 0 because the above grows without bound. (ii). There is no tangent line at that point. Eg. The function f(x) = x is continuous in R but it is not differentiable at x = 0. The difference quotient at x = 0 is { 1 if h < 0 f(0 + h) f(0) h = h h = 1 if h > 0 The left and right limits therefore both exist lim f(0+h) f(0) h 0 + h = 1 and lim f(0+h) f(0) f(0+h) f(0) h 0 h = 1 but these are not equal, so lim h 0 h does not exist. 3.2 Rules of differentiation The following rules of differentiation are easy to prove by considering the appropriate difference quotients. If f(x) and g(x) are differentiable at x then so are the following: 39

40 sum rule: f(x) + g(x) with derivative f (x) + g (x). product rule: f(x)g(x) with derivative f (x)g(x) + f(x)g (x). and providing g(x) 0 then so are the following: reciprocal rule: 1 g(x) with derivative g (x) (g(x)) 2. quotient rule: f(x) g(x) with derivative f (x)g(x) f(x)g (x) (g(x)) 2. Eg. Differentiate u(x) = (3x 2 1)/(x 4 + 3x + 1) u (x) = (3x2 1) (x 4 + 3x + 1) (3x 2 1)(x 4 + 3x + 1) (x 4 + 3x + 1) 2 = 6x(x4 + 3x + 1) (3x 2 1)(4x 3 + 3) = 6x5 + 18x 2 + 6x (12x 5 + 9x 2 4x 3 3) (x 4 + 3x + 1) 2 (x 4 + 3x + 1) Derivatives of higher order = 6x5 + 4x 3 + 9x 2 + 6x + 3 (x 4 + 3x + 1) 2. If f(x) is differentiable then its derivative f (x) may also be differentiable, in which case we denote its derivative by f (x) ie. the second derivative of f(x). Similarly the third derivative is f (x) and in general the n th derivative is f (n) (x) so that f (3) (x) = f. This notation is due to agrange ( ), but other notations are also common. f (x) = df dx, f (x) = d2 f dx 2, f (n) (x) = dn f dx n is due to eibniz ( ). f (x) = Df(x), f (x) = D 2 f(x), f (n) (x) = D n f(x) is due to Euler ( ). 40

41 Finally, if the independent variable represents time, say f(t), then f (t) = f and f (t) = f is due to Newton ( ). As we have seen, the derivative df dx is the slope of the tangent to the graph of f(x). This means that the derivative df dx measures the rate of change of the dependent variable f with respect to changes in the independent variable x. In the case that the independent variable is time t, the derivative measures the rate of change with time, so if the dependent variable, say X(t) represents the position of an object confined to a line then Ẋ is the velocity of the object, with Ẋ the speed, and Ẍ is the acceleration of the object. The product rule for differentiation is just the first case of the more general eibniz rule: If f(x) and g(x) are both differentiable n times then so is the product f(x)g(x) with D n (fg) = n k=0 ( ) n (D k f)(d n k g) k where ( ) n k = n! k!(n k)! is the binomial coefficient familiar from Pascal s triangle Figure 29: The binomial coefficients arranged into Pascal s triangle. Eg. Calculate D 3 (x 2 sin x) using D 3 (fg) = f(d 3 g)+3(df)(d 2 g)+3(d 2 f)(dg)+ (D 3 f)g. f = x 2, Df = 2x, D 2 f = 2, D 3 f = 0. g = sin x, Dg = cos x, D 2 g = sin x, D 3 g = cos x. 41

42 D 3 (x 2 sin x) = x 2 cos x (3)2x sin x + (3)2 cos x + 0 sin x = (6 x 2 ) cos x 6x sin x. 3.4 The chain rule To handle the differentiation of composite functions (f g)(x) = f(g(x)) we turn to the following theorem: The chain rule theorem states that if g(x) is differentiable at x and f(x) is differentiable at g(x) then the composition (f g)(x) is differentiable at x with (f g) (x) = f (g(x))g (x). Recall that prime denotes differentiation with respect to the argument so in eibniz notation the above formula may be written more crudely as d df f(g(x)) = dx dg where we need to be aware that on the right hand side the argument of the first factor is g(x) and the argument of the second factor is x. This form is useful as a mnemonic (think of cancelling the dg factor) but it is no more than this. Eg. Calculate d (( x + 1 ) 3 ). dx x In terms of the above notation g(x) = x + 1 x and f(x) = x 3 giving g (x) = 1 1 x and f (x) = 3x 4 thus 2 (( d x + 1 ) 3 ) ( = f (g(x))g (x) = 3 x + 1 ) 4 (1 1x ). dx x x 2 dg dx Eg. d dt cos(t2 ) = 2t sin(t 2 ). 42

43 3.5 Hôpital s rule Recall that the calculus of limits theorem states that if lim x a f(x) = and f(x) lim x a g(x) = M 0 then lim x a g(x) = M. Clearly this result is not applicable to the situation where = M = 0, which is called an indeterminant form. We have already seen how to deal with situations like this directly, but an alternative (and often easier) approach is sometimes available by making use of the following. Hôpital s rule If lim x a f(x) = lim x a g(x) = 0 and lim x a f (x) g (x) f(x) lim x a g(x) = lim f (x) x a g (x). exists then f Note: The condition that lim (x) x a g (x) exists contains the hidden conditions that f(x) and g(x) are differentiable on I = (a h, a) (a, a + h) for some h > 0, with g (x) 0 x I. Proof: We shall only consider the proof for the slightly easier situation in which f(x) and g(x) are both differentiable at x = a and g (a) 0. In this case f(x) lim x a g(x) = lim f(x) f(a) x a g(x) g(a) = lim x a f(x) f(a) x a g(x) g(a) x a = lim x a f(x) f(a) x a g(x) g(a) lim x a x a = f (a) g (a) = lim f (x) x a g (x). We can apply Hôpital s rule to calculate the two important trigonometric limits from earlier. Eg. Calculate lim x 0 sin x x. f(x) = sin x and g(x) = x are both differentiable. Also lim x 0 sin x = sin 0 = 0 and lim x 0 x = 0. Furthermore, f (x) = cos x and g (x) = 1 0. Thus Hôpital s rule applies to give sin x lim x 0 x = lim cos x x 0 1 = cos 0 = 1. 43

44 Eg. Calculate lim x 0 1 cos x x. f(x) = 1 cos x and g(x) = x are both differentiable. Also lim x 0 (1 cos x) = 1 cos 0 = 0 and lim x 0 x = 0. Furthermore, f (x) = sin x and g (x) = 1 0. Thus by Hôpital s rule 1 cos x lim x 0 x = lim x 0 sin x 1 = sin 0 = 0. If applying Hôpital s rule yields another indeterminant form then Hôpital s rule can be reapplied to this form and so on. Eg. Calculate lim x 0 2 sin x sin(2x) x sin x. 2 sin x sin(2x) lim x 0 x sin x = lim x 0 2 cos x 2 cos(2x) 1 cos x = lim x 0 2 cos x + 8 cos(2x) cos x 3.6 Implicit differentiation = lim x 0 2 sin x + 4 sin(2x) sin x = = 6. So far we have been differentiating functions defined explicitly in terms of an independent variable. Even if an explicit expression is not available then we may still be able to obtain an expression for the derivative if we know that a differentiable function satisfies a particular equation. This is the process of implicit differentiation, which involves differentiating an equation and using the chain rule. Eg. Consider a function y(x) that satisfies the equation x 2 + y 2 = 1. Differentiating this equation with respect to x gives 2x + 2y dy dx = 0 and hence dy dx = x/y. In particular, the function y = 1 x 2 satisfies the given equation and the above result yields dy dx = x/ 1 x 2, which we could have obtained directly by differentiating the explicit expression. 44

45 Eg. Assume that x(t) is a differentiable function that satisfies the equation cos(t x) = (2t + 1) 3 x. Express dx dt in terms of x and t. Differentiating both sides of the equation with respect to t gives sin(t x)(1 dx dt ) = 3(2t + 1)2 2x + (2t + 1) 3 dx dt and hence dx dt = 6x(2t + 1)2 + sin(t x) sin(t x) (2t + 1) Boundedness and monotonicity The following definitions apply to a function f(x) defined in some interval I. Defn: If a constant k 1 s.t. f(x) k 1 x in I we say that f(x) is bounded above in I and we call k 1 an upper bound of f(x) in I. Furthermore, if a point x 1 in I s.t. f(x 1 ) = k 1 we say that the upper bound k 1 is attained and we call k 1 the global maximum value of f(x) in I. Similarly, we have the following. Defn: If a constant k 2 s.t. f(x) k 2 x in I we say that f(x) is bounded below in I and we call k 2 a lower bound of f(x) in I. Furthermore, if a point x 2 in I s.t. f(x 2 ) = k 2 we say that the lower bound k 2 is attained and we call k 2 the global minimum value of f(x) in I. Defn: f(x) is bounded in I if it is both bounded above and bounded below in I ie. if a constant k s.t. f(x) k x in I. If no interval I is specified then it is taken to be the domain of the function. Eg. cos x is bounded (in R) because cos x 1 x R. Both these upper and lower bounds are attained as cos 0 = 1 and cos π = 1. The global maximum value in R is 1 and the global minimum value in R is 1. Eg. Consider f(x) = sgn(x) 1+x 2 for 1 x 1. Then f(x) is bounded in [ 1, 1] because f(x) 1 for all x in [ 1, 1] but neither of the bounds is attained, so it has no global maximum value and no global minimum value in [ 1, 1]. 45

46 Eg. On [0, π/2) the function tan x is bounded below but not bounded above. A condition that guarantees the existence of both a global maximum and minimum value (called extreme values) is provided by the following: The extreme value theorem states that if f is a continuous function on a closed interval [a, b] then it is bounded on that interval and has upper and lower bounds that are attained ie. points x 1 and x 2 in [a, b] s.t. f(x 2 ) f(x) f(x 1 ) x [a, b]. Defn: f(x) is monotonic increasing in [a, b] if f(x 1 ) f(x 2 ) for all pairs x 1, x 2 with a x 1 < x 2 b. Defn: f(x) is strictly monotonic increasing in [a, b] if f(x 1 ) < f(x 2 ) for all pairs x 1, x 2 with a x 1 < x 2 b. Obvious definitions apply with increasing replaced by decreasing. Eg. sgn(x) is monotonic increasing in [ 1, 1] but not strictly. Eg. x 2 is strictly monotonic increasing in [0, b] for any b > Critical points Defn: We say that f(x) has a local maximum at x = a if f(a) f(x) x (a h, a + h). h > 0 s.t. Remark: If f(x) has a local maximum at x = a and is differentiable at this point then f (a) = 0. Proof: As f(x) is differentiable at a then f(x) f(a) lim x a x a = lim x a + f(x) f(a) x a = f (a). 46

47 f(x) has a local maximum at a implies f(x) f(a) x a 0 for x (a h, a) hence f f(x) f(a) (a) = lim 0. x a x a Similarly, f(x) has a local maximum at a also implies f(x) f(a) x a 0 for x (a, a + h) hence f f(x) f(a) (a) = lim 0. x a + x a Putting these two together gives f (a) 0 f (a) and thus f (a) = 0. Defn: We say that f(x) has a local minimum at x = a if f(x) f(a) x (a h, a + h). h > 0 s.t. Remark: If f(x) has a local minimum at x = a and is differentiable at this point then f (a) = 0. Defn: f(x) has a stationary point at x = a if it is differentiable at x = a with f (a) = 0. Defn: An interior point x = a of the domain of f(x) is called a critical point if either f (a) = 0 or f (a) does not exist. Every local maximum and local minimum of a differentiable function is a stationary point but there may be other stationary points too. Eg. f(x) = x 3 has f (0) = 0 so x = 0 is a stationary point but it is neither a local maximum nor a local minimum because f(x) > f(0) for x (0, h) but f(x) < f(0) for x ( h, 0). Defn: If f(x) is twice differentiable in an open interval around x = a with f (a) = 0 and if f (x) changes sign at x = a then we say that x = a is a 47

48 point of inflection. Figure 30: The graph of f(x) = x 3 indicating the point of inflection at x = 0. Eg. f(x) = x 3 has f (x) = 6x so f (0) = 0. As f (x) < 0 for x < 0 and f (x) > 0 for x > 0 then f (x) changes sign at x = 0 so it is a point of inflection. The first derivative test Suppose f(x) is continuous at a critical point x = a. (i). If h > 0 s.t. f (x) < 0 x (a h, a) and f (x) > 0 x (a, a + h) then x = a is a local minimum. (ii). If h > 0 s.t. f (x) > 0 x (a h, a) and f (x) < 0 x (a, a + h) then x = a is a local maximum. (iii). If h > 0 s.t. f (x) has a constant sign x a in (a h, a + h) then x = a is a not a local extreme value (minimum/maximum). Eg. f(x) = x is continuous but f (x) 0 so there are no stationary points. The derivative does not exist at x = 0 so this is a critical point. f (x) < 0 for x ( 1, 0) and f (x) > 0 for x (0, 1) so by the first derivative test x = 0 is a local minimum. In fact f(0) = 0 is a global minimum as x 0. Eg. f(x) = x 4 2x 3 has f (x) = 4x 3 6x 2 = 2x 2 (2x 3). The only critical points are x = 0, 3/2. Now f (x) < 0 for x < 3/2 and f (x) > 0 for x > 3/2. 48

49 Thus f (x) has constant sign on both (sufficiently close) sides of x = 0 so this is not a local extreme value. However, f (x) changes sign from negative to positive as x passes through x = 3/2 so this is a local minimum. The second derivative test Suppose f(x) is twice differentiable at x = a with f (a) = 0. (i). If f (a) > 0 then x = a is a local minimum. (ii). If f (a) < 0 then x = a is a local maximum. Note that the theorem doesn t say anything about the case f (a) = 0. Eg. f(x) = 2x 3 9x x has f (x) = 6x 2 18x + 12 = 6(x 2 3x + 2) = 6(x 1)(x 2) giving stationary points at x = 1 and x = 2. Furthermore, f (x) = 6(2x 3) giving f (1) = 6 < 0 and f (2) = 6 > 0 so by the second derivative test x = 1 is a local maximum and x = 2 is a local minimum. Note that the first derivative test is more general than the second derivative test as it does not require the function to be differentiable at the critical point. Determining all critical points and asymptotes is the key to sketching the graph of a function. Eg. Sketch the graph of the function f(x) = 4x 5 4x 5 lim x ± x 2 1 = lim x ± 1 x x 2 1. ( ) ( ) 4 5/x = 0 = x 2 2 so y = 0 is a horizontal asymptote along both the positive and negative x-axis. There are vertical asymptotes at x = ±1. f (x) = 4(x2 1) 2x(4x 5) (x 2 1) 2 = 4x2 + 10x 4 (x 2 1) 2 = (4x 2)(x 2) (x 2 1) 2 so there are stationary points at x = 1 2, 2 with f(1 2 ) = 4 and f(2) = 1. 1 x f

50 As x passes through 1 2 then f (x) changes sign from negative to positive so this is a local minimum. As x passes through 2 then f (x) changes sign from positive to negative so this is a local maximum. Figure 31: Graph of f(x) = 4x 5 x 2 1. If a function is defined on an interval then extreme values can occur at the endpoints of the interval. Here an endpoint x = c is a point at which the function is defined but where the function is undefined either to the left or right of this point. For example, if the domain of a function is [a, b] then a and b are both endpoints. Defn: If c is an endpoint of f(x) then f has an endpoint maximum at x = c if f(x) f(c) for x sufficiently close to c. The obvious similar definition applies for an endpoint minimum. If a function is differentiable sufficiently close to an endpoint then examining the sign of the derivative can determine if there is an endpoint maximum or minimum. If f(x) is continuous on an interval [a, b] then the global extreme values in this interval are attained at either critical points or endpoints, so we can determine them by examining all the possibilities. 50

51 Eg. Find the global extreme values of f(x) = 1 + 4x x4 for x [ 1, 3]. Figure 32: Graph of f(x) = 1 + 4x x4 f (x) = 8x 2x 3 = 2x(4 x 2 ) which is zero at x = 0, 2; recall we only consider x [ 1, 3]. Critical point: f(0) = 1 and f(2) = 9 End points: f( 1) = 9 2 and f(3) = 7 2. The smallest of these four values is 7 2, which is the global minimum, whereas the largest is 9 which is the global maximum. Eg. Find the global extreme values of x 2 2 x + 2 for x [ 1 2, 2]. First of all we can write { x 2 + 2x + 2 if 1 2 f(x) = x < 0 x 2 2x + 2 if 0 x 2 f (x) = 2x + 2 for x [ 1 2, 0) so no critical points here. f (x) = 2x 2 for x (0, 2] so a critical point at x = 1 with f(1) = 1. lim x 0 + f (x) = 2 2 = lim x 0 f (x) hence f (x) does not exist at x = 0, making this a critical point with f(0) = 2. Endpoints: f( 1 2 ) = 5 4 and f(2) = 2. Thus the global minimum is 1 and the global maximum is 2. 51

52 Figure 33: Graph of a piecewise function with two quadratic pieces. 3.9 Min-Max problems Determining extreme values (minima and maxima) is the subject of optimization, which has many practical applications. We can use the ideas we have studied so far if we can express the quantity to be optimized as a function of one variable. Eg. Determine the ratio of the side lengths of a rectangle of fixed perimeter, so that its area is as large as possible. et the side lengths be x and y. The fixed perimeter is P = 2x + 2y and the area is A = xy. By substituting y = 1 2 P x into A we get A(x) = 1 2 P x x2 as a function of one variable. The physically allowed values are 0 x 1 2 P and A(x) vanishes at both end points. A (x) = 1 2 P 2x = 0 if x = 1 4P. As A (x) = 2 < 0 this is a local maximum with A( 1 4 P ) = 1 16 P 2 and is also the global maximum since there are no other critical points. With x = 1 4 P then y = 1 2 P x = 1 4P so the ratio of side lengths is 1 ie. the optimal rectangle is a square. Eg. What length of fencing is required to create two adjacent rectangular fields of the same width and total area 15, 000 m 2? Use length units of m and area units of m 2. et x be the common width and y the total height. The amount of fencing is F = 3x + 2y, from the 52

53 perimeter plus the extra dividing fence. The total area is = xy. Thus F (x) = 3x /x is the function to be minimized for x > 0. F (x) = /x 2 = 0 if x = 100. Then F (100) = = 600. Also F (x) = 60000/x 3 > 0 and the stationary point is a global minimum. The required length of fencing is 600m Rolle s theorem Rolle s theorem states that if f is differentiable on the open interval (a, b) and continuous on the closed interval [a, b], with f(a) = f(b), then there is at least one c (a, b) for which f (c) = 0. Proof: By the extreme value theorem x 1 and x 2 in [a, b] s.t. f(x 1 ) f(x) f(x 2 ) x [a, b]. If x 1 (a, b) then x 1 is a local minimum and f (x 1 ) = 0 so we are done. If x 2 (a, b) then x 2 is a local maximum and f (x 2 ) = 0 so we are done. The only case that is left is if both x 1 and x 2 are endpoints, a, b. But since f(a) = f(b) then in this case f(x 1 ) = f(x 2 ) = f(a) so the above bound becomes f(a) f(x) f(a) x [a, b]. Thus f(x) = f(a) is constant in the interval and f (x) = 0 x [a, b]. Figure 34: Illustration of Rolle s theorem. An obvious corollary of Rolle s theorem is that if f(x) is differentiable at every point of an open interval I then each pair of zeros of the function in I is separated by at least one zero of f (x). This is simply the special case 53

54 of Rolle s theorem with a and b taken to be the the zeros of f(x) so that f(a) = f(b) = 0. An application of the above result is to provide a limit on the number of distinct real zeros of a function. Eg. Show that f(x) = 1 7 x7 1 5 x6 + x 3 3x 2 x has no more than 3 distinct real roots. Figure 35: Graph of f(x) = 1 7 x7 1 5 x6 + x 3 3x 2 x showing 3 zeros. Suppose the required result is false and there are (at least) 4 distinct real roots of f(x). Then by Rolle s theorem there are (at least) 3 distinct real roots of f (x). Then by applying Rolle s theorem to f (x) there are (at least) 2 distinct real roots of f (x). However, f (x) = 6x 5 6x 4 + 6x 6 = 6(x 1)(x 4 + 1) has only one real root (at x = 1). This contradiction proves the required result The mean value theorem The mean value theorem states that if f is differentiable on the open interval (a, b) and continuous on the closed interval [a, b], then there is at least one c (a, b) for which f (c) = f(b) f(a). b a 54

55 Figure 36: Illustration of the mean value theorem. Geometrically this theorem states that there is at least one point in the interval at which the tangent to the curve is parallel to the line joining the two points (b, f(b)) and (a, f(a)) at the ends of the interval. The geometry makes the result clear but here is the proof. Proof: Define the function g(x) = (b a)(f(x) f(a)) (x a)(f(b) f(a)). Then g(a) = g(b) = 0 and as g(x) satisfies the requirements of Rolle s theorem then c (a, b) for which g (c) = 0. As g (x) = (b a)f (x) (f(b) f(a)) then setting x = c yields the required result f (c) = f(b) f(a) b a. Note that Rolle s theorem is just a special case of the mean value theorem with f(a) = f(b). The mean value theorem can be used to prove some obvious results relating monotonicity to the sign of the derivative. Here is an example: Suppose f(x) is continuous in [a, b] and differentiable in (a, b) with f (x) 0 throughout this interval. Then f(x) is monotonic increasing in (a, b). Proof: If a x 1 < x 2 b then by the mean value theorem c (a, b) s.t. f (c) = f(x 2) f(x 1 ) x 2 x 1. However, since f (x) 0 in (a, b) then f (c) 0 which imples that f(x 2 ) f(x 1 ) ie. f is monotonic increasing. 55

56 Similar obvious results and proofs follow for the cases of monotonic decreasing and strictly monotonic increasing/decreasing The inverse function rule The derivative of the inverse of a function can be obtained from the following: The inverse function rule states that if f(x) is continuous in [a, b] and differentiable in (a, b) with f (x) > 0 throughout this interval then its inverse function g(y) (recall g(f(x)) = x) is differentiable for all f(a) < y < f(b) with g (y) = 1/f (g(y)). Note: A similar theorem exists for the case f (x) < 0. Eg. f(x) = sin x has f (x) = cos x > 0 for x ( π/2, π/2). Its inverse function g(y) = sin 1 y is therefore differentiable in ( 1, 1) with g (y) = 1 f (g(y)) = 1 cos g(y) = 1 1 sin 2 g(y) but y = f(g(y)) = sin(g(y)) hence g (y) = 1. Thus we have shown that 1 y 2 d dx sin 1 x = 1 1 x 2. Eg. f(x) = tan x has f (x) = sec 2 x > 0 for x ( π/2, π/2). With this domain then Ran f = R hence the domain of the inverse function g(y) = tan 1 (y) is R. By the inverse function rule g (y) = 1 f (g(y)) = 1 sec 2 (g(y)) = tan 2 (g(y)) = y 2 where we have used y = f(g(y)) = tan(g(y)). Thus we have shown that d dx tan 1 x = x 2. 56

57 3.13 Partial derivatives So far we have only considered functions of a single variable eg. f(x). ater we shall study functions of two variables eg. f(x, y). For this situation we shall need the concept of a partial derivative. The partial derivative of f with respect to x is written as f x and is obtained by differentiating f with respect to x, keeping y fixed ie. y may be treated as a constant in performing the partial differentiation with respect to x. Eg. f(x, y) = x 2 y 3 sin x cos y + y has f x = 2xy3 cos x cos y. Similarly, the partial derivative of f with respect to y is written as f y and is obtained by differentiating f with respect to y, keeping x fixed. Eg. f(x, y) = x 2 y 3 sin x cos y + y has f y = 3x2 y 2 + sin x sin y + 1. Partial derivatives are defined in terms of the following limits (which we assume exist) f (x, y) = lim x h 0 f(x + h, y) f(x, y), h f (x, y) = lim y h 0 f(x, y + h) f(x, y). h An alternative notation for partial derivatives is to write f x for f x etc. By applying partial differentiation to these partial derivatives we may obtain the second order partial derivatives 2 f x = ( ) f 2 f, 2 x x y = ( ) f 2 f, 2 y y y x = ( ) f 2 f, y x x y = ( ) f. x y For the earlier example f(x, y) = x 2 y 3 sin x cos y + y we obtain 2 f x 2 = 2y3 + sin x cos y, 2 f y x = 6xy2 + cos x sin y, 2 f y 2 = 6x2 y + sin x cos y, 2 f x y = 6xy2 + cos x sin y. 57

58 Note the equality, in this example, of the two mixed partial derivatives 2 f x y = 2 f y x. It can be shown that this result is true in general, if f and all first and second order partial derivatives are continuous. An alternative notation for partial derivatives is etc 2 f x 2 = f xx 58

59 4 Integration 4.1 Indefinite integrals Defn: A function F (x) is called an indefinite integral or antiderivative of a function f(x) in the interval (a, b) if F (x) is differentiable with F (x) = f(x) throughout (a, b). We then write F (x) = f(x) dx. Eg. cos x dx = sin x in R. Eg. 1 x 2 dx = 1 x Eg. sgnx dx = x in R\{0}. in R\{0}. Note1: If F (x) is an indefinite integral of f(x) in (a, b) then so is F (x) + c for any constant c. In applications of integration it is important to include this arbitrary constant. We therefore write eg. cos x dx = sin x + c. Note2: If F 1 (x) and F 2 (x) are both indefinite integrals of f(x) in (a, b) then F 1 (x) F 2 (x) = c for some constant c. Defn: We say that f(x) is integrable in (a, b) if it has an indefinite integral F (x) in (a, b) that is continuous in [a, b]. Eg. cos x is integrable in any finite interval (a, b) because sin x is continuous in R. 1 Eg. x is not integrable in (0, 1) because its indefinite integral 1 2 x continuous at x = 0. is not Eg. sgn x is integrable in (0, 1) because its indefinite integral x is continuous in [0, 1]. 59

60 4.2 Definite integrals Defn: A subdivision S of [a, b] is a partition into a finite number of subintervals [a, x 1 ], [x 1, x 2 ],..., [x n 1, b] where a = x 0 < x 1 < x 2 <... < x n = b. The norm S of the subdivision is the maximum of the subinterval lengths a x 1, x 1 x 2,.., x n 1 b. (Thus a small value of S means that the interval [a, b] has been chopped up into small pieces.) The numbers z 1, z 2,..., z n form a set of sample points from S if z j [x j 1, x j ] for j = 1,.., n. Defn: Suppose that f(x) is a function defined for x [a, b]. The Riemann sum is n R = (x j x j 1 )f(z j ). j=1 The Riemann sum is equal to the sum of the (signed) areas of rectangles Figure 37: An illustration of a Riemann sum. of height f(z j ) and width x j x j 1. Here signed means that the areas of rectangles below the x-axis are counted negatively. If f(x) is continuous in [a, b] and S is small then we expect R to be a good approximation to the (signed) area under the graph of f(x) above the interval [a, b]. This turns out to be correct and the error in the approximation can be reduced to zero by taking the limit in which S tends to zero. This leads to the definition of the definite integral as b a f(x) dx = lim S 0 R and it can be shown that this limit exists if f(x) is continuous in [a, b]. 60

61 By definition we set a b f(x) dx = b a f(x) dx and therefore a a f(x) dx = 0. Figure 38: The definite integral is the signed area under the curve: b a f(x) dx = A 1 A 2 +A 3. There are a number of fairly obvious properties of the definite integral that are not too hard to prove. In the following let f(x) and g(x) both be integrable in (a, b). (i) inearity If λ, µ are any constants then λf(x) + µg(x) is integrable in (a, b) with b ( ) b b λf(x) + µg(x) dx = λ f(x) dx + µ g(x) dx. a (ii) If c [a, b] then b a f(x) dx = c a f(x) dx + b c f(x) dx. (iii) If f(x) g(x) x (a, b) then b a f(x) dx b a g(x) dx. (iv) If m f(x) M x [a, b] then m(b a) b a a f(x) dx M(b a). 4.3 The fundamental theorem of calculus So far we dont have any connection between indefinite and definite integrals. This is provided by the following. a 61

62 The fundamental theorem of calculus If f(x) is continuous on [a, b] then the function F (x) = x a f(t) dt defined for x [a, b] is continuous on [a, b] and differentiable on (a, b) and is an indefinite integral of f(x) on (a, b) ie. F (x) = d dx x a f(t) dt = f(x) throughout (a, b). Furthermore if F (x) is any indefinite integral of f(x) on [a, b] then b a f(t) dt = F (b) F (a) = [ F (x)] b a. We shall sketch the important points of the proof. For a x < x + h < b we have that F (x + h) F (x) = x+h a f(t) dt x a f(t) dt = x+h x f(t) dt where we have used property (ii). et m(h) and M(h) denote the minimum and maximum values of f(x) on the interval [x, x + h]. Then by property (iv) we have that m(h)h x+h x f(t) dt M(h)h. Thus F (x + h) F (x) m(h) M(h). h Since f(x) is continuous on [x, x+h] we have that lim h 0 + m(h) = lim h 0 + M(h) = f(x) and so by the pinching theorem F (x + h) F (x) lim h 0 + h = f(x). 62

63 A similar argument applies to the limit from below and together they give F (x) = lim h 0 F (x + h) F (x) h = f(x) which proves that F (x) is an indefinite integral of f(x). The proof of the first part of the theorem is completed by showing that F (x) is continuous from the right at x = a and from the left at x = b. Both these follow by a simple consideration of the limit of the relevant quotient. Finally, to prove the last part of the theorem the key observation is that any indefinite integral F (x) is related to F (x) by the addition of a constant. The fundamental theorem of calculus provides a simple rule for differentiating a definite integral with respect to its limits. Eg. d dx x sin 2 t dt = sin 2 x. We can combine this result with the chain rule if the limit is a more complicated expression d x 2 ( ) 1 Eg. dx 1 + e dt = 1 d t 1 + e x2 dx x2 = 2x. 1 + e x The logarithm and exponential function revisited We have already seen one definition of the logarithm function (as the inverse of the exponential function) but here is an alternative equivalent definition. The function 1 x is continuous for x (0, ) so we can define log t = t 1 1 x dx for t (0, ). By the fundamental theorem of calculus, this is differentiable for t (0, ) with d dt log t = 1 t. Thus log t is strictly monotonic increasing in (0, ) and it follows that log t is positive when t > 1, zero when t = 1, and negative when 0 < t < 1. 63

64 Figure 39: The logarithm function in terms of areas under the curve y = 1 x. In Figure 39 we illustrate the logarithm function in terms of areas under the curve y = 1 x. For t 1 > 1 then log t 1 is the area A 1 and for 0 < t 2 < 1 then log t 2 is minus the area A 2. As we have seen, the indefinite integral 1 x dx = log x is only valid for x > 0 because this is the domain of the function log x. However, we can extend this relation for the indefinite integral to all x 0 as 1 x dx = log x. It is easy to prove this result by considering the derivative of log( x) for x < 0. As we have already seen, the inverse of the function f(x) = log x is the exponential function g(y) = e y = exp y. We can therefore use the inverse function rule to calculate the derivative of the exponential function d dy ey = g 1 (y) = f (g(y)) = 1 1 = g(y) = e y. g(y) Thus we have shown that d dx ex = e x x R. If n is a positive integer then x n has the obvious definition of x multiplied by itself n times. Also, x n = 1/x n. We define x 1 n as the inverse function of y n in the interval (0, ), which has x 1 n > 0 and (x n) 1 n = x x (0, ). 64

65 If q is an integer then we define x q n = (x n) 1 q for x > 0. From this definition we have that x q 1 n = exp(log(x n ) q q ) = exp(q log(x 1 n )) = exp( n n log(x q 1 n )) = exp( n log(x 1 n ) n ) = exp( q log x). n Thus for any rational number r = q/n we have that for x > 0 x r = exp(r log x). If a is irrational (eg. a = 2) then we take this formula as a definition of x a, x a = exp(a log x). There are some important results concerning limits as x for the logarithm and exponential functions. To derive these results we begin with the following, emma 1: x 0, e x 1 + x Proof: Consider f(x) = e x (1 + x) then f(0) = 0 and f (x) = e x 1 0. Hence f(x) is monotonic increasing in [0, ) so f(x) 0 for all x 0. emma 2: x 0, and for any positive integer n, e x n j=0 xj /j!. Note that the case n = 1 corresponds to emma 1, and the proof for general n is similar to the proof of emma 1. Result 1 For any constant a > 0 log x lim = 0. x x a Roughly speaking this says that log x grows more slowly than any power of x. Proof: Put x = e y then log x lim x x a y = lim y e. ay 65

66 For y > 0 then by emma 2 with n = 2 As lim y 2 a 2 y 0 y e ay y 1 + ay a2 y y a2 y = 2 2 a 2 y = 0 then by the pinching theorem lim y y log x = 0 = lim eay x x. a Result 2 For any constant a > 0 x a lim x e = 0. x Roughly speaking this says that any power of x grows more slowly than e x. Proof is left as an exercise. Hint: similar to the proof of Result 1, using emma 2 and the pinching theorem. Result 3 For any constant a ( lim 1 + a x = e x x) a. Proof (for the case a > 0): Recall the definition of the derivative of a function f(x) at x = b. f f(b + h) f(b) (b) = lim. h 0 h Apply this to f(x) = log x so that f (x) = 1 x and take b = 1. log(1 + h) log(1) log(1 + h) 1 = lim = lim. h 0 h h 0 h With the change of variable h = a x this becomes log(1 + a x 1 = lim ) x a x ie. Since e x is continuous at a we have that e a = lim exp(x log(1 + a )) = lim x x x 66 a = lim x (x log(1 + a x )). ( 1 + a x) x.

67 4.5 Standard integrals A closed form expression is a function obtained from elementary functions by a finite sequence of elementary operations. Eg. e x2 + sin(1 + x 2 ) x 4 + e cos x is a closed form expression. A basic problem in integration is given a closed form expression f(x) can we find a closed form expression for its indefinite integral F (x) = f(x) dx? In general this is not possible, for example, e x2 dx does not have a closed form expression. The game is to try and reduce f(x) dx to some standard integrals by using a number of rules of integration eg. integration by parts, substitution,... Some standard integrals that you are expected to know include (we have seen some of these already but it is an exercise to prove them all by differentiating the right hand side in each case) 67

68 x a dx = xa+1 a c, for constant a 1 1 dx = log x + c x e x dx = e x + c sin x dx = cos x + c cos x dx = sin x + c sec 2 x dx = tan x + c cosec 2 x dx = cotx + c x dx = 2 tan 1 x + c 1 dx = 1 x 2 sin 1 x + c 1 dx = 1 + x 2 sinh 1 x + c 4.6 Integration by parts f(x)g (x) dx = f(x)g(x) f (x)g(x) dx. Eg. Eg. x 2 e 3x dx = 1 3 x2 e 3x 2 3 xe3x dx = 1 3 x2 e 3x xe3x + 9 e3x dx = 1 3 x2 e 3x 2 9 xe3x e3x + c. x cos x dx = x sin x sin x dx = x sin x+cos x +c. 68

69 Eg. Eg. Hence 1 log x dx = (log x)1 dx = (log x)x x dx = x log x x+c = x(log x 1)+c. x e x cos x dx = e x sin x e x sin x dx = e x sin x+e x cos x e x cos x dx. e x cos x dx = 1 2 ex (sin x+cos x) +c. 4.7 Integration by substitution If F (x) = f(x) dx then F (u(x)) = f(u(x)) u (x) dx. In more compact notation, by making the substitution from x to u(x) we have f(x) dx = f(u) du, where du = u (x) dx. Eg. Calculate x 2+3x dx. 4 3 Make the substitution u = 2 x2 then du = 6x dx and ( ) ( x x dx = u du = = tan 1 ( 2 x2 ) + c. ) u 2 du = tan 1 u +c Eg. Calculate tan x dx. Put u = cos x then du = sin x dx and sin x tan x dx = cos x dx = 1 du = log u + c = log cos x + c. u Eg. Calculate π 2 0 cos3 x dx. Put u = sin x then du = cos x dx. When x = 0 then u = 0 and when x = π/2 then u = 1. Therefore π 2 0 cos 3 x dx = π 2 0 cos x(1 sin 2 x) dx = (1 u 2 ) du = [ u 1 3 u3 ] 1 0 = = 2 3.

70 Eg. Calculate 1 (x 2 +4) dx. 2 Put x = 2 tan u then dx = 2 sec 2 u du and 1 (x 2 + 4) dx = 2 sec 2 u du (tan 2 2 (4 tan 2 u + 4) = u + 1) du 2 8(tan 2 u + 1) = du 2 8(tan 2 u + 1) 1 1 = 8 cos2 u du = 16 (1+cos 2u) du = 1 16 (u+1 2 sin 2u) = 1 (u+sin u cos u) 16 = 1 ( u + tan u ) = 1 ( u + tan u ) 16 sec 2 u tan 2 = 1 ( tan 1 x2 x ) u x2 4 1 Thus (x 2 + 4) dx = 1 ( tan 1 x x ) + c. 4 + x 2 For a general function f(x) the formula f (x) dx = log f(x) + c f(x) is a simple application of integration by substitution. Simply put u = f(x) so that du = f (x) dx. Then f (x) 1 f(x) dx = du = log u + c = log f(x) + c. u Eg. x x 4 + 4x + 7 dx = 1 4 log x4 + 4x c. 4.8 Integration of rational functions using partial fractions Given a rational function, first of all decompose it into a polynomial and a proper rational function. As the integration of a polynomial is trivial we can now restrict attention to the integration of proper rational functions. Every proper rational function can be written as a sum of partial fractions which take the form A Bx + C and (x α) k (x 2 + βx + γ) l 70

71 where the quadratic that appears in the above has no real roots. The number and type of partial fractions that appear depends on the factors in the denominator of the rational function. Each factor in the denominator of the form (x α) k fraction of the form generates a partial A 1 (x α) + A 2 (x α) A k (x α) k. Each factor in the denominator of the form (x 2 + βx + γ) l generates a partial fraction of the form B 1 x + C 1 (x 2 + βx + γ) + B 2x + C 2 (x 2 + βx + γ) B lx + C l (x 2 + βx + γ) l. The coefficients A k, B l, C l can be found directly by comparison with the rational function either by the substitution of specific values of x or by comparing coefficients of x. The integration is then completed by integrating the partial fraction. Eg. Calculate 2x x 2 x 2 dx. 2x x 2 x 2 = 2x (x 2)(x + 1) = a x 2 + b x + 1. To determine a and b multiply the above by (x 2)(x + 1) to give 2x = a(x + 1) + b(x 2). Putting x = 2 yields 4 = 3a and putting x = 1 yields 2 = 3b hence 4 2 2x x 2 x 2 = 3 x x + 1. Integrating this expression gives the required integral 4 2 2x x 2 x 2 dx = 3 x x + 1 = 4 3 log x log x c. 3 71

72 Eg. Calculate 2x 2 +3 x 3 2x 2 +x dx. 2x x 3 2x 2 + x = 2x2 + 3 x(x 1) 2 = a x + b x 1 + c (x 1) 2 As 2x = a(x 1) 2 + bx(x 1) + cx then putting x = 0 gives a = 3, while x = 1 gives 5 = c. Comparing coefficients of x 2 gives 2 = a + b thus b = 1. 2x x 3 2x 2 + x dx = x 1 x dx = 3 log x log x 1 (x 1) 2 x 1 +C. Eg. Calculate 3x 4 +x 3 +20x 2 +3x+31 (x+1)(x 2 +4) 2 dx. 3x 4 + x x 2 + 3x + 31 (x + 1)(x 2 + 4) 2 = a x bx + c x dx + f (x 2 + 4) 2 It is an exercise to show that a = 2, b = 1, c = 0, d = 0, f = 1. 3x 4 + x x 2 + 3x dx = (x + 1)(x 2 + 4) 2 x x x (x 2 + 4) dx 2 = 2 log 1 + x log(x2 + 4) 1 8 ( x ) x Integration using a recurrence relation 1 16 tan 1 x 2 + C. Eg. Calculate 1 0 x3 e x dx. Define I n = 1 0 xn e x dx for all integer n 0. 1 [ ] 1 1 I n+1 = x n+1 e x dx = x n+1 e x (n + 1)x n e x dx = e (n + 1)I n. 0 0 The recurrence relation I n+1 = e (n + 1)I n can be used to calculate I n for any positive integer n from the starting value I 0 = 1 0 ex dx = [e x ] 1 0 = e 1. I 1 = e I 0 = e (e 1) = 1, I 2 = e 2I 1 = e 2(1) = e 2, I 3 = e 3I 2 = e 3(e 2) = 6 2e is the required integral. 0 72

73 Eg. Calculate tan 4 x dx. Define F n (x) = tan n x dx for all integer n 0. F n+2 (x) + F n (x) = tan n x (1 + tan 2 x) dx = tan n x sec 2 x dx Put u = tan x then du = sec 2 x dx F n+2 (x) + F n (x) = u n du = un+1 n + 1 = tann+1 x n + 1. The recurrence relation F n+2 (x) = 1 n+1 tann+1 (x) F n (x) can be used to calculate the required integral for n even from the starting value F 0 (x) = 1 dx = x and for n odd from the starting value F1 (x) = log cos x. In particular, F 2 (x) = tan x x and F 4 (x) = 1 3 tan3 x tan x + x thus tan 4 x dx = 1 3 tan3 x tan x + x + c Definite integrals using even and odd functions The definite integral of an odd function over a symmetric interval is zero. If f odd (x) is an integrable odd function on [ a, a] then Proof: a a a f odd (x) dx = 0 = f odd ( x) dx + a a 0 a f odd (x) dx = 0. 0 a f odd (x) dx = f odd (u) du + a 0 a 0 f odd (x) dx (f odd ( x) + f odd (x)) dx = 0. A similarly argument shows that if f even (x) is an integrable even function on [ a, a] then a a f even (x) dx = 2 f even (x) dx. a 0 73

74 Recall that any function f(x) can be decomposed as a sum of even and odd functions f(x) = f even (x)+f odd (x). This can be a useful technique to calculate definite integrals over symmetric intervals as a a f(x) dx = a a f even (x) + f odd (x) dx = 2 a 0 f even (x) dx. Eg. Calculate 1 1 e x x 4 cosh x dx. For f(x) = ex x 4 cosh x we have f even(x) = 1 2 f(x) f( x) = [ ] e x x 4 cosh x dx = 2 0 x 4 = 2 5 x5 0 = 2 5. x 4 (e x +e x ) cosh x = x 4. Thus Eg. Calculate π 4 π 4 (1+2x) 3 cos x 1+12x 2 dx. For f(x) = (1+2x)3 cos x 1+12x we have 2 f even (x) = 1 2 f(x) f( x) = (cos x){(1+2x)3 +(1 2x) 3 } Thus π 4 π 4 2(1+12x 2 ) = (cos x)(2+24x2 ) (1 + 2x) 3 π [ ] π cos x 4 4 dx = 2 cos x dx = 2 sin x x (1+12x 2 ) = cos x. = = 2. 74

75 5 Differential equations A differential equation is an equation that relates an unknown function (the dependent variable) to one or more of its derivatives. The order of a differential equation is the order of the highest derivative that appears in the equation. A function that satisfies the differential equation is called a solution and the process of finding a solution is called solving the differential equation. If the dependent variable depends on only a single unknown variable (the independent variable) then the differential equation is called an ordinary differential equation (ODE). We shall deal only with ODE s. Differential equations for functions with several independent variables and involving partial derivatives are called partial differential equations (PDE s). These are generally much harder to solve and wont be discussed in this course. We shall usually discuss ODE s using y to denote the dependent variable and x for the independent variable, so solving the ODE involves finding y(x). The generic first order ODE may be written in the form dy dx = f(x, y) ( ) Generically this will have a one-parameter family of solutions ie. the general solution will contain an arbitrary constant. Assigning a particular value to this constant gives a particular solution. A particular solution will be singled out by the assignment of an initial value ie. requiring y(x 0 ) = y 0, for given x 0 and y 0. In this case the ODE is called an initial value problem (IVP). To solve an IVP, first solve the ODE to find the general solution and then determine the value of the constant so that the IVP is also satisfied. Depending upon the form of the function f(x, y) there are various methods that can be used to solve certain kinds of ODE s. We shall discuss these in turn. 75

76 5.1 First order separable ODE s The ODE ( ) is separable if f(x, y) = X(x)Y (y). It can be solved by direct integration as 1 Y (y) dy = X(x) dx. Eg. Solve the IVP y = xe x e y, dy dx = xey x with y(0) = 0. e y dy = xe x dx, e y = xe x + e x dx = xe x e x +c y = log(e x (1 + x) c). y(0) = 0 = log(1 c), so c = 0. y = log(e x (1 + x)) = x log(1 + x). Eg. Solve y dy = 2 y = 2x(1 + y2 ) (1 + x 2 ) 2. 2x (1 + x 2 ) dx, 2 tan 1 y = x2 c, y = tan ( ) x 2+c. Eg. Solve the IVP 1 + y dy = y y = x2 y y, y(3) = y (x 2 1) dx, y + log y = x3 3 x + c. Put x = 3 and y = 1 1 = 6 + c, c = 5, hence y + log y = x3 3 x 5. In this case the final solution can only be written in implicit form ie. it cannot be written in an explicit form y =... where the right hand side is a function of x only. 76

77 5.2 First order homogeneous ODE s The ODE ( ) is homogeneous if f(tx, ty) = f(x, y) t R. In this case the substitution y = xv produces a separable ODE for v(x). Eg. Solve f(tx, ty) = t2 y 2 t 2 x 2 txty y = y2 x 2. xy = y2 x 2 xy = f(x, y) hence homogeneous. Put y = xv then y = v + xv and the ODE becomes v+xv = x2 v 2 x 2 = v 1 x 2 v v, v = 1 xv, v dv = 1 x dx, v 2 = log x +log c, 2 v = ± 2 log c x, y = xv = ±x 2 log c x 5.3 First order linear ODE s The ODE ( ) is linear if f(x, y) = p(x)y + q(x). In this case the ODE can be solved in terms of the integrating factor The solution is given by I(x) = e p(x) dx. y = 1 I(x) I(x)q(x) dx. Proof: We need to show that y satisfies the ODE y + py = q. The first step is to observe that I = e p dx p = Ip. With y = 1 I Iq dx we have y = I I Iq dx I Iq = I I y + q = Ip I y + q = py + q. Eg. Solve the IVP y 2 x y = 3x3, y( 1) = 2. 77

78 In the above notation p = 2 x and q = 3x3. ( ) ( I = exp p dx = exp 2 ) ( x dx = exp ) 2 log x = 1 x 2. y = 1 I Iq dx = x 2 1 x 23x3 dx = x 2 ( ) 3 3x dx = x 2 2 x2 + c = 3 2 x4 + cx 2. Using y( 1) = 2 we have 2 = c hence c = 1 2 giving y = 3 2 x x First order exact ODE s An alternative form in which to write a first order ODE is M(x, y) dx + N(x, y) dy = 0. ( ) Rearranging gives dy dx = M(x, y) N(x, y) so this agrees with the form ( ) with f(x, y) = M(x,y) N(x,y). Note that a given f(x, y) does not correspond to a unique choice of M(x, y) and N(x, y) as only their ratio determines the ODE, so there is a freedom to multiply both M and N by the same arbitrary function. For any function g(x, y) the total differential dg is defined to be dg = g g dx + x y dy. You may think of dg as the small change in the function g that arises in moving from the point (x, y) to (x + dx, y + dy) where dx and dy are both small. In particular, if dg is identically zero then g is a constant. The ODE ( ) is called exact if there exists a function g(x, y) s.t. the left hand side of ( ) is equal to the total derivative dg ie. M = g x and N = g y. 78

79 In this case the ODE says that dg = 0 hence g = constant and this yields the solution of the ODE. The equality of the mixed partial derivatives 2 g y x = 2 g x y requires that an exact equation satisfies M y = N x. This results in the following test for exactness The ODE M(x, y) dx + N(x, y) dy = 0 is exact iff M y = N x. Eg. Show that (3e 3x y + e x ) dx + (e 3x + 1) dy = 0 is an exact equation and hence solve it. In this example M = 3e 3x y + e x and N = e 3x + 1. M y = 3e3x = N x hence this equation is exact. From M = g x we have g x = 3e3x y + e x giving g = e 3x y + e x + φ(y) where the usual constant of integration is replaced by an arbitrary function of integration φ(y) because of the partial differentiation. To determine this function we use g y = N, g y = e3x + 1 = e 3x + φ, so φ = 1, giving φ = y. Finally we have that g = e 3x y + e x + y = constant = c. In this case we can write the solution in explicit form y = (c e x )/(e 3x + 1). As an exercise check that this does indeed solve the starting ODE y = (3e 3x y + e x )/(e 3x + 1). Note that this example is also a linear ODE so could also have been solved using an integrating factor. Consider an ODE M dx + N dy = 0 that is not exact ie. M y N x. By multiplying the ODE by a function I(x, y) we can get an equivalent representation of the ODE as m dx + n dy = 0 where m = MI and n = NI. If this equation is exact ie. m y = n x then I is called an integrating factor for the original ODE. Eg. Show that x is an integrating factor for (3xy y 2 ) dx+x(x y) dy = 0. 79

80 M = 3xy y 2 so M y = 3x 2y, but N = x2 xy so N x = 2x y. Therefore M y N x and the ODE is not exact. However, with the integrating factor I = x we get m = MI = 3x 2 y y 2 x and n = NI = x 3 x 2 y so now m y = 3x2 2yx = n x which is exact. We can now solve this exact equation using the above method. g x = m = 3x2 y y 2 x giving g = x 3 y 1 2 y2 x 2 + φ(y). g y = n = x3 x 2 y = x 3 x 2 y + φ so φ = 0 and we can take φ = 0. This gives the final solution g = constant = c ie. x 3 y 1 2 y2 x 2 = c. 5.5 Bernoulli equations A Bernoulli equation is a nonlinear ODE of the form y + p(x)y = q(x)y n where n 0, 1, otherwise the equation is simply linear. These ODE s can be solved by first performing the substitution v = y 1 n, which converts the equation to a linear ODE for v(x). Eg. Solve y 2y x = x2 y 2. This is a Bernoulli equation with n = 2 hence put v = 1 y, which gives v = y y. Dividing the ODE by y 2 yields y 2 y 2 2 xy = x2 which in terms of v is v 2v x = x2 ie the linear ( equation ) v + 2 x v = x2. We solve this with an integrating factor I = exp 2 x dx = exp(2 log x) = x 2 as v = 1 x x 2 x 2 dx = 1 2 x ( x5 + c 5 ). So finally y = 1 v = 5x2 x 5 + c. 5.6 Second order linear constant coefficient ODE s The general form of a second order linear constant coefficient ODE is α 2 y + α 1 y + α 0 y = φ(x) ( ) 80

81 where α 2 0, α 1, α 0 are constants (the constant coefficients) and φ(x) is an arbitrary function of x. The ODE is still second order and linear if these constants are replaced by functions of x, but then the ODE is not so easy to solve The homogeneous case We first restrict to the case φ(x) = 0 ie. α 2 y + α 1 y + α 0 y = 0 ( ) in which case the second order ODE is called homogeneous. The first point to note is that if y 1 and y 2 are any two solutions of the homogeneous ODE ( ) then so is any arbitrary linear combination y = Ay 1 + By 2, where A and B are constants. Proof: y = Ay 1 + By 2 so y = Ay 1 + By 2 and y = Ay 1 + By 2. Hence α 2 y + α 1 y + α 0 y = α 2 (Ay 1 + By 2) + α 1 (Ay 1 + By 2) + α 0 (Ay 1 + By 2 ) = = A(α 2 y 1 + α 1 y 1 + α 0 y 1 ) + B(α 2 y 2 + α 1 y 2 + α 0 y 2 ) = = 0. Note that the linearity of the ODE is crucial for this result. The general solution of the ODE ( ) contains two arbitrary constants (because the ODE is second order). If y 1 and y 2 are any two independent particular solutions then the general solution is given by y = Ay 1 +By 2, with A and B the two arbitrary constants. The task is therefore to find two independent particular solutions. To find a particular solution, look for one in the form y = e λx. Putting this into ( ) yields the characteristic (or auxiliary) equation α 2 λ 2 + α 1 λ + α 0 = 0. (Char) There are 3 cases to consider, depending upon the type of roots of (Char). 81

82 (i). Distinct real roots. If (Char) has real roots λ 1 λ 2 then y 1 = e λ 1x and y 2 = e λ 2x are the two required particular solutions and the general solution is y = Ae λ 1x + Be λ 2x. Eg. Solve y y 2y = 0. Then (Char) is λ 2 λ 2 = 0 = (λ 2)(λ + 1) with roots λ = 2, 1. The general solution is y = Ae 2x + Be x. (ii). Repeated real root. (Char) has a repeated real root if α1 2 = 4α 0 α 2 ie. when α 0 = α2 1 4α 2. (Char) then reduces to ( α 2 λ 2 +α 1 λ+ α2 1 = 0 = α 2 λ+ α ) 2 1 with λ 1 = α 1 the real double root. 4α 2 2α 2 2α 2 Thus (Char) has produced only one solution y 1 = e λ 1x. However, as we now show, y = xe λ 1x is also a solution in this case. The original ODE ( ) now takes the form α 2 y + α 1 y + α2 1 4α 2 y = 0 = α 2 (y 2λ 1 y + λ 2 1y). With y = xe λ 1x we have y = e λ 1x +λ 1 xe λ 1x and y = 2λ 1 e λ 1x +λ 2 1xe λ 1x. Hence y 2λ 1 y + λ 2 1y = 2λ 1 e λ 1x + λ 2 1xe λ 1x 2λ 1 (e λ 1x + λ 1 xe λ 1x ) + λ 2 1xe λ 1x = 0. y 1 = e λ 1x and y 2 = xe λ 1x are the two required particular solutions and the general solution is y = Ae λ 1x + Bxe λ 1x. Eg. Solve 2y 12y +18y = 0. Then (Char) is 2λ 2 12λ+18 = 0 = 2(λ 3) 2 with double root λ = 3 The general solution is y = Ae 3x + Bxe 3x. 82

83 (iii). Complex roots. If the roots of (Char) are complex then they come in a complex conjugate pair ie. λ 1 = α+iβ and λ 2 = α iβ. We therefore have the two solutions y 1 = e λ 1x = e αx e iβx and y 2 = e λ 2x = e αx e iβx, but these are not the solutions we are looking for, since they are complex and we want real solutions. However, we have seen that we can take any linear combination of these solutions (even with complex coefficients) and it will still be a solution. In particular the combinations Y 1 = 1 2 (y 1 + y 2 ) = 1 2 eαx (e iβx + e iβx ) = e αx cos(βx) Y 2 = 1 2i (y 1 y 2 ) = 1 2i eαx (e iβx e iβx ) = e αx sin(βx) are both real and provide the two particular solutions we want. The general solution is an arbitrary real linear combination of Y 1 and Y 2 ie. y = Ae αx cos(βx) + Be αx sin(βx) = e αx (A cos(βx) + B sin(βx)) with A and B real. Recall that α and β are both real and are the real and imaginary parts of the complex root of (Char). Eg. Solve y 6y + 10y = 0. Then (Char) is λ 2 6λ + 10 = 0 with roots λ = 6± = 6±2i 2 = 3±i = α±iβ. Hence α = 3 and β = 1 and the general solution is y = e 3x (A cos x + B sin x) The inhomogeneous case We now return to the general inhomogeneous ODE α 2 y + α 1 y + α 0 y = φ(x) ( ) where φ(x) is a function that is not identically zero. The general solution to ( ) is obtained as a sum of two parts y = y CF + y P I 83

84 where y CF is called the complementary function and is the general solution of the homogeneous version of the ODE ie. ( ) which is obtained by removing the φ(x) term. This part of the solution contains the two arbitrary constants and we have already seen how to find this. y P I is called the particular integral and is any particular solution of ( ). We haven t yet seen any methods to find this. However, before we look at a method, let us first show that the combination y = y CF + y P I is indeed a solution of ( ). We have that y = y CF + y P I and y = y CF + y P I so put these into the left hand side of ( ) to get α 2 y + α 1 y + α 0 y = α 2 (y CF + y P I) + α 1 (y CF + y P I) + α 0 (y CF + y P I ) = α 2 y CF + α 1 y CF + α 0 y }{{ CF } = 0 by ( ) + α 2 y P I + α 1 y P I + α 0 y }{{ P } I = φ(x). = φ(x) by ( ) To find a particular integral we shall use the method of undetermined coefficients, which can be applied if φ(x) takes certain forms such as polynomials, exponentials or trigonometric functions, including sums and products of these. The idea is to try an appropriate form of the solution that contains some unknown constant coefficients which are then determined by explicit calculation and the requirement that the try yields a solution. The table below lists the kind of terms that can be dealt with if they appear in φ(x) (they can come with any constant coefficients in front) together with the form to try for y P I, containing undetermined constant coefficients a i. Term in φ(x) e γx x n cos(γx) sin(γx) Form to try for y P I a 1 e γx a 0 + a 1 x a n x n a 1 cos(γx) + a 2 sin(γx) a 1 cos(γx) + a 2 sin(γx) The motivation for the above is that the form of the try when differentiated zero, once or twice, should have a similar functional form to φ(x), as there is then a chance that the left hand side of ( ) could equal the right hand 84

85 side. If φ(x) contains sums/products of the kind of terms above then try sums/products of the listed forms for the try. Special case rule: If the suggested form of the try for y P I is contained in y CF then we know that this form wont work as the left hand side of ( ) will then be identically zero and so cant equal φ(x). In this case first multiply the suggested form of the try by x to get the correct try. In some special cases this rule may need to be applied twice as x times the suggested try may also be contained in y CF. Eg. Solve y y 2y = 7 2x 2. From earlier we already know that y CF = Ae 2x + Be x. To find y P I we try the form y = a 0 + a 1 x + a 2 x 2 to simplify the notation for the calculation we have dropped the P I subscript here. Now y = a 1 + 2a 2 x and y = 2a 2, so putting these into the ODE gives 2a 2 (a 1 + 2a 2 x) 2(a 0 + a 1 x + a 2 x 2 ) = 7 2x 2 = 2a 2 a 1 2a 0 + x( 2a 2 2a 1 ) 2a 2 x 2. Comparing coefficients of x 2, x 1, x 0 gives 2a 2 = 2, (a 2 = 1), 2a 2 2a 1 = 0, (a 1 = 1), 2a 2 a 1 2a 0 = 7, (a 0 = 2). We have found y P I = 2 x + x 2 and the general solution is y = y CF + y P I ie. y = Ae 2x + Be x 2 x + x 2. Eg. Solve y + 4y + 4y = 5e 3x. First find y CF by solving y + 4y + 4y = 0. (Char) is λ 2 + 4λ + 4 = 0 = (λ + 2) 2 with λ = 2 repeated, hence y CF = e 2x (A + Bx). For y P I try y = a 1 e 3x so y = 3a 1 e 3x and y = 9a 1 e 3x. Put these into the ODE 9a 1 e 3x + 12a 1 e 3x + 4a 1 e 3x = 5e 3x = 25a 1 e 3x so a 1 = 1 5 giving y P I = 1 5 e3x. The general solution is y = y CF + y P I ie y = e 2x (A + Bx) e3x. Eg. Solve y 2y + 2y = 10 cos(2x). First find y CF by solving y 2y + 2y = 0. (Char) is λ 2 2λ + 2 = 0 with λ = 1 ± i. Hence y CF = e x (A cos x + B sin x). For y P I try y = a 1 cos(2x) + a 2 sin(2x) so y = 2a 1 sin(2x) + 2a 2 cos(2x) and 85

86 y = 4a 1 cos(2x) 4a 2 sin(2x). Put these into the ODE y 2y + 2y = 4a 1 cos(2x) 4a 2 sin(2x) 2( 2a 1 sin(2x) + 2a 2 cos(2x)) + 2(a 1 cos(2x) + a 2 sin(2x)) = 10 cos(2x) = ( 2a 1 4a 2 ) cos(2x) + ( 2a 2 + 4a 1 ) sin(2x). Comparing coefficients of sin(2x) and cos(2x) gives 2a 2 + 4a 1 = 0, (a 2 = 2a 1 ), 2a 1 4a 2 = 10, 10a 1 = 10, (a 1 = 1, a 2 = 2). Thus y P I = cos(2x) 2 sin(2x). The general solution is y = e x (A cos x + B sin x) cos(2x) 2 sin(2x). Eg. Solve y y 2y = 6e x. From earlier we already know that y CF = Ae 2x + Be x. To find y P I we first think that we should try the form y = a 1 e x but note that this is contained in y CF so instead we try y = a 1 xe x. Then y = a 1 e x a 1 xe x and y = 2a 1 e x + a 1 xe x. Put into ODE 2a 1 e x +a 1 xe x (a 1 e x a 1 xe x ) 2a 1 xe x = 6e x = 3a 1 e x so a 1 = 2 and y P I = 2xe x. The general solution is y = Ae 2x + Be x 2xe x. Eg. Solve y y 2y = e x (8 sin(3x) 14 cos(3x)). From earlier we already know that y CF = Ae 2x + Be x. For y P I try y = e x (a 1 cos(3x) + a 2 sin(3x)). It is an exercise to show that a 1 = 1, a 2 = 1. The general solution is then y = Ae 2x + Be x + e x (cos(3x) sin(3x)). Eg. Solve y y 2y = 5 + 9x 2x cos x + 2 sin x. From earlier we already know that y CF = Ae 2x + Be x. For y P I try y = a 0 + a 1 x + a 2 x 2 + a 3 x 3 + a 4 cos x + a 5 sin x. It is an exercise to show that a 0 = 1, a 1 = 0, a 2 = 3 2, a 3 = 1, a 4 = 1, a 5 = 1. The general solution is then y = Ae 2x + Be x x2 + x 3 cos x sin x Initial and boundary value problems The values of the two arbitrary constants in the general solution of a second order ODE are fixed by specifying two extra requirements on the solution 86

87 and/or its derivative. An initial value problem (IVP) is when we require y(x 0 ) = y 0 and y (x 0 ) = δ for given constants x 0, y 0, δ. In this case the two constraints are given at the same value of the independent variable. A boundary value problem (BVP) is when we require y(x 0 ) = y 0 and y(x 1 ) = y 1 for given constants x 0 x 1, y 0, y 1. In this case the two constraints are given at two different values of the independent variable. The way to solve an IVP or BVP is to first find the general solution of the ODE and then determine the values of the two constants in this solution so that the extra conditions are satisfied. Eg. Solve the IVP y y 2y = 7 2x 2, y(0) = 5, y (0) = 1. From earlier we already know that the general solution of the ODE is y(x) = Ae 2x + Be x 2 x + x 2. Therefore y(0) = A + B 2 = 5, so A + B = 7. Now y (x) = 2Ae 2x Be x 1 + 2x, giving y (0) = 2A B 1 = 1, so 2A B = 2. The solution of these two equations for A and B is A = 3, B = 4. Therefore the solution of the IVP is y = 3e 2x + 4e x 2 x + x 2. Eg. Solve the BVP 4y + y = 0, y(0) = 1, y(π) = 2. (Char) is 4λ = 0 with roots λ = ± i 2. The general solution is y = A cos( x 2 ) + B sin(x 2 ). y(0) = A = 1 and y(π) = B = 2. Hence the solution of the BVP is y = cos( x 2 ) + 2 sin(x 2 ). 87

88 5.7 Systems of first order linear ODE s A system of n coupled first order linear ODE s for n dependent variables can be written as a single n th order linear ODE for a single dependent variable by eliminating the other dependent variables. An associated IVP is obtained if all the values of the dependent variables are specified at a single value of the independent variable. Eg. Solve the pair of first order linear ODE s y = y + z, z = 8y + 5z for y(x), z(x) by first finding a second order ODE for y(x). From the first equation z = y + y hence z = y + y. Substituting these expressions into the second equation gives y + y = 8y + 5(y + y) ie. the second order ODE y 4y + 3y = 0. We can now solve this ODE using our earlier methods. (Char) is λ 2 4λ + 3 = 0 = (λ 3)(λ 1) with roots λ = 1, 3. y = Ae x + Be 3x and then z = y + y = 2Ae x + 4Be 3x. Hence the general solution is y = Ae x + Be 3x, z = 2Ae x + 4Be 3x. We can extend the problem to an IVP by requiring that y(0) = 1, z(0) = 2. From the general solution y(0) = A + B = 1 and z(0) = 2A + 4B = 2 with solution A = 3, B = 2. Hence the solution of the IVP is y = 3e x 2e 3x, z = 6e x 8e 3x. 88

89 6 Taylor series 6.1 Taylor s theorem and Taylor series Taylor s theorem states that if f(x) has n + 1 continuous derivatives in an open interval I that contains the point x = a, then x I ( n f (k) ) (a) f(x) = (x a) k + R n (x) k! where R n (x) = 1 n! x a k=0 Proof: Fix x I then x a f (t) dt = f(x) f(a), so (x t) n f (n+1) (t) dt is called the remainder. f(x) = f(a) + x a f (t) dt. If we think of the integrand in the above as f (t) 1 then we can perform an integration by parts where we choose the constant of integration to be x so that 1 dt = (t x). This gives [ ] x x f(x) = f(a) + f (t)(t x) f (t)(t x) dt = f(a) + f (a)(x a) + a a x a f (t)(x t) dt. Note that this proves the theorem for n = 1. To prove the theorem for n = 2 we perform another integration by parts on this equation, x f(x) = f(a) + f (a)(x a) + f (t)(x t) dt a ] x = f(a) + f (a)(x a) [f (x t)2 (t) + 2 a = f(a) + f (a)(x a) + f (a) 2 (x a)2 + x a x f (x t)2 (t) a 2 f (x t)2 (t). 2 dt 89

90 This proves the theorem for n = 2. Continuing in this way the theorem follows after performing an integration by parts a total of n times. More formally, this can be written in the form of a proof by induction. The combination P n (x) = f(x) R n (x) is a polynomial in x of degree n called the n th order Taylor polynomial of f(x) about x = a P n (x) = f(a) + f (a)(x a) + f (a) 2! (x a) f (n) (a) (x a) n. n! If the term about x = a is not included when referring to the n th order Taylor polynomial of f(x), then by default this is taken to mean the choice a = 0, as this is the most common case. The Taylor polynomial P n (x) is an approximation to the function f(x). Generically, it is a good approximation if x is close to a and the approximation improves with increasing order n. The remainder provides an exact expression for the error in the approximation. Eg. Calculate the n th order Taylor polynomial of e x. f(x) = e x, f (x) = e x,..., f (k) (x) = e x so f (k) (0) = 1 giving P n (x) = 1 + x + x2 2! xn n! = n Figure 40 shows the graph of e x and its Taylor polynomials of order 0 to 3. k=0 x k k!. Figure 40: The function e x and its Taylor polynomials of order 0 to 3. 90

91 If f(x) is infinitely differentiable on an open interval I that contains the point x = a and in addition for each x I we have that lim n R n (x) = 0, then we say that f(x) can be expanded as a Taylor series about x = a and we write f (k) (a) f(x) = (x a) k. k! k=0 Infinite series and discussions/tests concerning their convergence are covered in the Analysis I course. Note that the n th order Taylor polynomial is obtained by taking just the first (n + 1) terms of the Taylor series (where we count terms even if they happen to be zero). Eg. From our earlier calculation we have that the Taylor series of e x is e x = k=0 x k k! = 1 + x + x2 2! + x3 3! + where denotes an infinite number of terms with increasing powers of x. Note that if we set x = 1 in the above Taylor series then we recover the formula presented earlier for Euler s number e = k=0 1 k! = ! + 1 2! + 1 3! + Eg. Calculate the Taylor series of sin x. f(x) = sin x, f(0) = 0, f (x) = cos x, f (0) = 1, f (x) = sin x, f (0) = 0, f (x) = cos x, f (0) = 1, f (4) (x) = sin x so now the pattern repeats and we see that for all non-negative integers k we have that f 2k (0) = 0 and f 2k+1 (0) = ( 1) k. 91

92 The Taylor series of sin x is therefore given by ( 1) k sin x = (2k + 1)! x2k+1 = x x3 3! + x5 5! x7 7! + k=0 Observe that only odd powers of x appear in the Taylor series of sin x as this is an odd function. A similar calculation (exercise) yields the Taylor series of cos x ( 1) k cos x = (2k)! x2k = 1 x2 2! + x4 4! x6 6! + k=0 in which only even powers of x appear, as cos x is an even function. To calculate the Taylor series of sinh x observe that for all non-negative integer k we have that f(x) = sinh x satisfies { f (k) sinh x if k is even (x) = cosh x if k is odd thus f (2k) (0) = 0 and f (2k+1) (0) = 1 giving 1 sinh x = (2k + 1)! x2k+1 = x + x3 3! + x5 5! + x7 7! + k=0 A similar calculation (exercise) yields the Taylor series of cosh x 1 cosh x = (2k)! x2k = 1 + x2 2! + x4 4! + x6 6! + k=0 These two results can also be obtained directly from the Taylor series of e x by noting that cosh x and sinh x are the even and odd parts of e x. As log x is not defined at x = 0 it does not make sense to consider the Taylor series of log x about x = 0. Instead we consider the Taylor series of log x about x = 1. 92

93 f(x) = log x, f(1) = 0, f (x) = 1 x, f (1) = 1, f (x) = 1 x, f (1) = 1, f (x) = 2 2 x, f (1) = 2, 3 f (4) (x) = 3 2 x, f (4) (1) = 3 2 and in general for k a positive integer 4 f (k) (x) = ( 1) k 1 (k 1)!, f (k) (1) = ( 1) k 1 (k 1)! x k Thus the Taylor series of log x about x = 1 is ( 1) k 1 (k 1)! log x = (x 1) k ( 1) k 1 = (x 1) k k! k k=1 This result is more often expressed in terms of the variable X = x 1, when it becomes log(1 + X) = ( 1) k 1 k=1 k X k. If we now rename the variable X as x we get ( 1) k 1 log(1 + x) = x k = x x2 k 2 + x3 3 x4 4 + k=1 This result is not valid at x = 1, and this is to be expected because the logarithm is not defined when its argument is zero. A careful examination of the requirement lim n R n (x) = 0 reveals that x must satisfy 1 < x 1 for the above Taylor series to be valid (ie. for the series to converge). k=1 6.2 agrange form for the remainder There is a more convenient expression for the remainder term in Taylor s theorem. The agrange form for the remainder is R n (x) = f (n+1) (c) (n + 1)! (x a)n+1, for some c (a, x). To prove this expression for the remainder we will first need to prove the following lemma: emma et h(t) be differentiable n + 1 times on [a, x] with h (k) (a) = 0 for 0 k n and h(x) = 0. Then c (a, x) s.t. h (n+1) (c) = 0. 93

94 Proof: h(a) = 0 = h(x) so by Rolle s theorem c 1 (a, x) s.t. h (c 1 ) = 0. h (a) = 0 = h (c 1 ) so by Rolle s theorem c 2 (a, c 1 ) s.t. h (c 1 ) = 0. Repeating this argument a total of n times we arrive at h (n) (a) = 0 = h (n) (c n ) so by Rolle s theorem c n+1 (a, c n ) s.t. h (n+1) (c n+1 ) = 0. This proves the lemma with c = c n+1 (a, x). Proof of the agrange form of the remainder: Consider the function [ ] f(x) Pn (x) h(t) = f(t) P n (t) (t a) n+1. (x a) n+1 By construction h(x) = 0. Also dk dt k (t a) n+1 is zero when evaluated at t = a for 0 k n. Furthermore, by definition of the Taylor polynomial P n (t) we have that f (k) (a) = P (k) n for 0 k n. Hence h (n) (a) = 0 for 0 k n. h(t) therefore satisfies the conditions of the lemma and we have that c (a, x) s.t. h (n+1) (c) = 0. As P n (t) is a polynomial of degree n then P n (n+1) (t) = 0. Also, dn+1 dt (t a) n+1 = (n + 1)! hence n+1 [ f(x) 0 = h (n+1) (c) = f (n+1) Pn (x) (c) (n + 1)! (x a) n+1 Rearranging this expression gives the required result f(x) = P n (x) + f (n+1) (c) (n + 1)! (x a)n+1. ]. (a) One use of the agrange form of the remainder is to provide an upper bound on the error of a Taylor polynomial approximation to a function. Suppose that f (n+1) (t) M, t in the closed interval between a and x. Then R n (x) M x a n+1 (n+1)! provides a bound on the error. 94

95 Eg. Show that the error in approximating e x by its 6 th order Taylor polynomial is always less than throughout the interval [0, 1]. In this example f(x) = e x and n = 6 so we first require an upper bound on f (7) (t) = e t for t [0, 1]. As e t is monotonic increasing and positive then e t e 1 = e < 3. Thus, in the above notation, we may take M = 3. As a = 0 we now have that R 6 (x) < 3 x 7 7! 3 7! for x [0, 1]. Evaluating 3 7! = < and the required result has been shown. 6.3 Calculating limits using Taylor series Defn: et n be a positive integer. We say that f(x) = o(x n ) (as x 0) if lim x 0 f(x) x n = 0. In particular, if α is any non-zero constant then αx m = o(x n ) iff m > n. Egs. 3x 4 = o(x 3 ), 3x 4 = o(x 2 ), 3x 8 x 6 = o(x 5 ), o(x 8 ) + o(x 5 ) = o(x 5 ). If we know that a Taylor series is valid in a suitable open interval, then it may be useful in calculating certain limits. It can be shown that the Taylor series we have already seen for sin x and cos x are valid for all x R. Although we have not proved this result you may assume that it is true and use it in calculating limits. As examples, we shall now calculate the two important trigonometric limits that we saw earlier. Eg. Use Taylor series to calculate lim x 0 sin x x. From the Taylor series of sin x we have that sin x = x x3 3! + o(x 4 ). Hence sin x lim x 0 x = lim x x3 3! + o(x 4 ) x 0 x Eg. Use Taylor series to calculate lim x 0 1 cos x x. = lim x 0 (1 x2 3! + o(x3 )) = lim x 0 (1 + o(x)) = 1. 95

96 From the Taylor series of cos x we have that cos x = 1 x2 2 + o(x3 ). Hence 1 cos x lim x 0 x x2 2 = lim + o(x3 ) x 0 x = lim x 0 x o(x3 ) x = lim x 0 ( x 2 +o(x2 )) = 0. 96

97 7 Fourier series In our study of Taylor series and Taylor polynomials we have seen how to write and approximate functions in terms of polynomials. If the function we are interested in is periodic, then it is more appropriate to use trigonometric functions rather than polynomials. This is the topic of Fourier series, which we shall describe in this chapter. Fourier series are important in a wide range of areas and applications. In particular, they play a vital role in the solution of certain partial differential equations. Examples of this application will be studied in the Dynamics course. Consider a function f(x) of period 2 which is given on the interval (, ). The functions cos nπx nπx and sin, with n any positive integer, are also periodic with period 2. It therefore seems reasonable to try and write f(x) in terms of these trigonometric functions. Note that a constant is trivially a periodic function (for any period) so we can also include a constant term. We therefore aim to write f(x) = a ( a n cos nπx + b n sin nπx ) (F S) n=1 where a 0, a n, b n are constants labelled by the positive integer n, and are called the Fourier coefficients of f(x). (It is just a convenient notation to call the constant term a 0 /2.). If these Fourier coefficients are such that this series converges then it is called the Fourier series of f(x). Eg. The function f(x) = 2 cos 2 πx πx + 3 sin has period 2. Its Fourier series contains only a finite number of terms as 2 cos 2 πx πx 2πx πx + 3 sin = 1 + cos + 3 sin. Therefore a 0 = 2, a 2 = 1, b 1 = 3, and all the other Fourier coefficients are zero. In general an infinite number of Fourier coefficients may be non-zero and we need a method to determine these from the given function f(x). 97

98 7.1 Calculating the Fourier coefficients In order to derive formulae for the Fourier coefficients we first need to prove some identities for integrals of trigonometric functions. In the following let m and n be positive integers. (i) (ii) (iii) (iv) (v) 1 1 cos nπx sin nπx sin mπx cos mπx sin mπx dx = 0. dx = 0. nπx cos dx = 0. nπx cos dx = nπx sin dx = { 0 if m n 1 if m = n. { 0 if m n 1 if m = n. The expression that occurs on the right hand side of the last two formulae arises so often that it is useful to introduce a shorthand notation for this. The Kronecker delta δ mn is defined to be { 0 if m n δ mn = 1 if m = n. So, for example, we may write 1 It is trivial to prove (i) : nπx cos sin mπx dx = [ nπ nπx sin dx = δ mn. ] nπx sin = 2 nπ sin(nπ) = 0. (ii) and (iii) are true by inspection, as they are both integrals of odd functions over a symmetric interval. We can prove (iv) and (v) by using cos(x ± y) = cos x cos y sin x sin y. First assume m n then 1 cos mπx nπx cos dx = 1 2 ( (m n)πx cos + cos 98 ) (m + n)πx dx

99 Now if m = n 1 = 1 [ 1 (m n)πx sin + 1 (m + n)πx sin 2π m n m + n cos mπx ] = 0. nπx cos dx = 1 ( 1+cos 2nπx ) dx = 1 [ x+ ] 2nπx sin = nπ This proves (iv) and a similar calculation (exercise) proves (v). We say that the set of functions {1, cos nπx nπx, sin } form an orthogonal set on [, ] because if φ, ψ are any two different functions from this set then φψ dx = 0. 1 We now use the identities (i),.., (v) to derive expressions for the Fourier coefficients in (F S). First of all, by integrating (F S) we have 1 f(x) dx = 1 { a0 2 + n=1 ( a n cos nπx + b n sin nπx using (i) and (ii). Thus we have derived an expression for a 0 a 0 = 1 f(x) dx. et m be a positive integer and multiply (F S) by cos mπx 1 cos mπx f(x) dx = 1 { a0 2 + ( a n cos nπx +b n sin nπx n=1 = a n δ mn = a m, n=1 Thus we have derived an expression for a n )} dx = a 0 and integrate to give )} cos mπx dx where we have used the identities (i), (iii), (iv). a n = 1 cos nπx f(x) dx. 99

100 Note that if we extend this relation to n = 0 then it reproduces the correct expression given above for a 0. Similarly, multiply (F S) by sin mπx 1 = sin mπx f(x) dx = 1 b n δ mn = b m, n=1 and integrate to give { a0 2 + n=1 Thus we have derived an expression for b n ( a n cos nπx +b n sin nπx )} sin mπx dx where we have used the identities (ii), (iii), (v). b n = 1 sin nπx f(x) dx. Eg. Calculate the Fourier series of the function f(x) = x for 1 < x < 1. It is to be understood here that f(x) is a periodic function with period 2. Its graph is shown in Figure 41. Figure 41: The graph of f(x) = x for 1 < x < 1, with period 2. We apply the above formulae with = 1. a 0 = 1 1 x dx = 2 [ ] x dx = x 2 = 1. 0 For n > 0 a n = x cos(nπx) dx = 2 x cos(nπx) dx 0 100

101 {[ ] 1 x 1 = 2 nπ sin(nπx) 0 0 = 2 n 2 π Furthermore, for n > 0 } 1 sin(nπx) dx = 2 [ ] 1 cos(nπx) nπ n 2 π 2 0 2(cos(nπ) 1) = 2 n 2 π 2(( 1)n 1). b n = 1 1 x sin(nπx) dx = 0 because this is the integral of an odd function over a symmetric interval. This demonstrates a general point that if f(x) is an even function on the interval (, ) then all b n = 0 and the Fourier series contains only cosine terms (plus a constant term). This is called a cosine series. Putting all this together we have the Fourier (cosine) series f(x) = n=1 2 n 2 π 2(( 1)n 1) cos(nπx) = π 2 ( cos(πx) cos(3πx) cos(5πx) + ) Note that in this example a 2n = 0 and a 2n 1 = 4 π 2 (2n 1) 2, so this Fourier (cosine) series could also be written as f(x) = π 2 n=1 cos((2n 1)πx) (2n 1) 2. To see how the Fourier series approaches the function f(x) define the partial sum S m (x) = a m ( a n cos nπx + b n sin nπx ). n=1 In this example, S 1 (x) = π cos(πx), S 2 3 (x) = π (cos(πx) cos(3πx)), S 5 (x) = π (cos(πx) cos(3πx) cos(5πx)), etc. 101

102 In Figure 42 we plot the graph of f(x) together with the partial sums S 1 (x), S 5 (x), S 11 (x). This helps to demonstrate that, in this case, lim m S m (x) = f(x). Figure 42: The graph of f(x) = x for 1 < x < 1, together with the partial sums S 1 (x), S 5 (x), S 11 (x). Eg. Calculate the Fourier series of the function f(x) = x for π < x < π. It is to be understood here that f(x) is a periodic function with period 2π. Its graph is shown in Figure 43. Note that the function is not continuous at the points x = (2p + 1)π, p Z, where there are jump discontinuities. Figure 43: The graph of f(x) = x for π < x < π, with period 2π. We apply the above formulae with = π. For n 0 a n = 1 π π π x cos(nx) dx = 0 because this is the integral of an odd function over a symmetric interval. 102

103 This demonstrates a general point that if f(x) is an odd function on the interval (, ) then all a n = 0 and the Fourier series contains only sine functions. This is called a sine series. For n > 0 b n = 1 π x sin(nx) dx = 1 {[ x ] π π } π π π n cos(nx) 1 + cos(nx) dx π π n = 1 { 2π [ ] π } 1 π n cos(nπ)+ n sin(nx) = 2 2 π n cos(nπ) = 2 n ( 1)n = 2 n ( 1)n+1. Putting all this together we have the Fourier (sine) series ( 2 f(x) = n ( 1)n+1 sin(nx) = 2 sin x 12 ) sin(2x)+13 sin(3x) 14 sin(4x)+ n=1 In Figure 44 we plot the graph of f(x) = x for π < x < π, together with Figure 44: The graph of f(x) = x for π < x < π, together with the graphs of the partial sums S 1 (x), S 6 (x), S 50 (x). the graphs of the partial sums S 1 (x), S 6 (x), S 50 (x). These graphs demonstrate that as more terms of the Fourier series are included it becomes an increasingly accurate approximation to f(x) inside the interval x ( π, π). However, notice what happens at the points x = ±π, where f(x) is not continuous. At these points the Fourier series converges to 0, which is the midpoint of the jump. Note the oscillations around the point of discontinuity, where the Fourier series under/overshoots. This is called the Gibbs phenomenon and the amount of under/overshoot tends to a constant (in fact about 9%) rather than dying away as more terms are included, 103

104 but the under/overshoot region becomes more localized. In the limit of an infinite number of terms the undershoot and overshoot occur at exactly the same point and cancel each other out. In general, we would like to know what happens to the partial sum S m (x) = a m ( a n cos nπx + b n sin nπx ) n=1 for all values of x in the limit as m, and how this is related to f(x). The following theorem provides an answer. Dirichlet s theorem et f(x) be a periodic function, with period 2, such that on the interval (, ) it has a finite number of extreme values, a finite number of jump discontinuities and f(x) is integrable on (, ). Then its Fourier series converges for all values of x. Furthermore, it converges to f(x) at all points where f(x) is continuous and if x = a is a jump discontinuity then it converges to 1 2 lim x a f(x) lim x a + f(x). 7.2 Parseval s theorem We shall now derive a relation between the average of the square of a function and its Fourier coefficients. Parseval s theorem If f(x) is a function of period 2 with Fourier coefficients a n, b n then 1 2 (f(x)) 2 dx = 1 4 a (a 2 n + b 2 2 n). Proof: 1 (f(x)) 2 dx = 1 {[ a0 2 + ( a n cos nπx +b n sin nπx n=1 104 n=1 )][ a0 2 + m=1 ( a m cos mπx +b m sin mπx )]} dx

105 = a n=1 m=1 (a n a m δ mn + b n b m δ mn ) = a (a 2 n + b 2 n). n=1 Parseval s theorem is useful in several contexts. One application is in finding the sum of certain infinite series. Eg. Calculate n=1 1 n by applying Parseval s theorem to the Fourier series 2 of f(x) = x for π < x < π. From earlier we calculated that the Fourier coefficients are given by a n = 0 and b n = 2 n ( 1)n+1. Hence by Parseval s theorem 1 2π π π x 2 dx = 1 2π [ x 3 3 ] π π = π2 3 = 1 2 n=1 4 n 2. Thus 1 n = π Certain infinite sums can also be obtained by evaluating a Fourier series at a particular value of x. Eg. Calculate 1 n=1 (2n 1) by evaluating the Fourier series of f(x) = x for 2 1 < x < 1, at the value x = 0. From earlier we calculated the Fourier (cosine) series to be f(x) = π 2 n=1 cos((2n 1)πx) (2n 1) 2. As f(x) = x satisfies the conditions of Dirichlet s theorem and is continuous for all x then, by Dirichlet s theorem, evaluating the Fourier series at x = 0 gives f(0) = 0. Hence 0 = π 2 n=1 n=1 1 (2n 1) 2. Thus 1 (2n 1) = π n=1 105

106 7.3 Half range Fourier series A half range Fourier series is a Fourier series defined on an interval (0, ) rather than the interval (, ). Given a function f(x), defined on the interval (0, ), we obtain its half range sine series by calculating the Fourier sine series of its odd extension { f(x) if 0 < x < f o (x) = f( x) if < x < 0 The Fourier coefficients of f o (x) are a n = 0 and b n = 1 f o (x) sin nπx dx = 2 0 f(x) sin nπx This gives the half range sine series for f(x) on (0, ) as f(x) = n=1 b n sin nπx. Given a function f(x), defined on the interval (0, ), we obtain its half range cosine series by calculating the Fourier cosine series of its even extension { f(x) if 0 < x < f e (x) = f( x) if < x < 0 The Fourier coefficients of f e (x) are b n = 0 and a n = 1 f e (x) cos nπx dx = 2 0 f(x) cos nπx This gives the half range cosine series for f(x) on (0, ) as f(x) = a n=1 a n cos nπx. For a given (physical) problem on an interval (0, ) it is usually the boundary conditions that determine whether an odd extension and a half sine Fourier series or an even extension and a half cosine Fourier series is more appropriate. dx. dx. 106

107 7.4 Complex Fourier series Fourier series take a much simpler form if written in terms of complex variables. Starting with the Fourier series f(x) = a ( a n cos nπx + b n sin nπx ) we use the relations cos nπx to rewrite this as f(x) = a n=1 where we have defined for n > 0 c n = 1 2 (a n ib n ) = 1 2 and for n < 0 c n = 1 2 (a n+ib n ) = 1 2 n=1 = 1 inπx 2 (e ( (a n ib n )e inπx +e inπx ) and sin nπx ) + (an + ib n )e inπx = c 0 = a 0 2 = 1 f(x) dx 2 f(x)(cos nπx f(x)(cos nπx = i inπx 2 (e e inπx ) n= c n e inπx nπx i sin ) dx = 1 f(x)e inπx 2 nπx +i sin ) dx = 1 2 f(x)e inπx. Note that all 3 cases (n = 0, n > 0, n < 0) can be written as a single compact formula c n = 1 2 f(x)e inπx, with the Fourier series f(x) = n= c n e inπx 107

108 8 Functions of several variables In this chapter we shall consider functions of two variables, say f(x, y). The extension to functions of more than two variables should be obvious. We have already seen partial derivatives, defined in terms of the limits f (x, y) = lim x h 0 f(x + h, y) f(x, y), h f (x, y) = lim y h 0 We have also introduced the total differential df as df = f f dx + x y dy. The gradient of f(x, y) is defined to be the vector ( f f = x, f ), y f(x, y + h) f(x, y). h where the symbol is called nabla. f points in the direction of the greatest rate of increase of the function f(x, y). A point (x, y) at which f vanishes, that is f = (0, 0), is called a stationary point. Eg. f = 1 2 (1 + x2 + y 2 ). f = (x, y) points along the radial half-line from the origin through the point (x, y). It is easy to see that moving out along a radial line is the direction of the greatest rate of increase of f since f = 1 2 (1 + r2 ), in terms of polar coordinates x = r cos θ, y = r sin θ. The only stationary point is (x, y) = (0, 0), and in fact f(x, y) attains its global minimum at this point since f(0, 0) = 1 2 and clearly f(x, y) 1 2. Below we shall see how to extend some of the concepts we have seen for functions of one variable to the two variable case. 8.1 Taylor s theorem for functions of two variables Suppose f(x, y) and all its partial derivatives up to and including order n + 1 are continuous throughout an open rectangular region centred at the point 108

109 (a, b). Then for all (x, y) in this region [ f(x, y) = f(a, b) + (x a) f ] (a, b) + (y b) f (a, b) x y + 1 [ (x a) 2 2 f 2 x (a, b) + 2(x a)(y b) 2 f n! n k=0 ( n k x y (a, b) + (y f b)2 2 y ) (x a) k (y b) n k n f x k y n k (a, b) + R n(x, y) where R n (x, y) is the agrange form of the remainder given by R n (x, y) = 1 (n + 1)! n+1 ( ) n + 1 k k=0 (x a) k (y b) n+1 k n+1 f (α, β) x k yn+1 k with (α, β) a point on the line segment joining (a, b) to (x, y). If lim n R n (x, y) = 0 for all (x, y) in this region then the n limit of the above expression is the Taylor series of f(x, y) about (a, b) in this region. Note that the n th order Taylor series is a polynomial in powers of (x a) and (y b) up to order n. Eg. Taylor expand f(x, y) = e x log(1 + y) about the origin (0, 0) up to and including terms of quadratic order and give an expression for the remainder term. f x = ex log(1 + y) = 0 at (x, y) = (0, 0). f = 1 at (x, y) = (0, 0). y = ex 1+y 2 f x = e x log(1 + y) = 0 at (x, y) = (0, 0). 2 2 f y = ex 2 (1+y) = 1 at (x, y) = (0, 0). 2 2 f x y = ex 1+y 3 f x 3 = 1 at (x, y) = (0, 0). = e x log(1 + y), 3 f y 3 = 2ex (1+y) 3, f(x, y) = y (2xy y2 ) + R 2 (x, y) with 3 f x 2 y = ex 1+y, 3 f x y 2 = ex (1+y) 2 ] (a, b)

110 ( R 2 (x, y) = 1 6 x 3 e α log(1 + β) + 3x 2 y eα eα 1+β 3xy2 (1+β) + y 3 2 2eα (1+β) ). 3 Here (α, β) is a point on the line segment joining (0, 0) to (x, y). 8.2 Chain rule Suppose f(x, y) but in addition x and y are both functions of the independent variable t. By substituting the t dependence of x(t) and y(t) directly into f(x(t), y(t)) we obtain a function f(t), with a derivative df dt. We may ask how this derivative is related to the partial derivatives f f x and x. The answer is obtained by taking the expression for the total derivative df = f f x dx + y dy and dividing by dt to obtain the chain rule df dt = f x dx dt + f y Eg. f(x, y) = x 2 y and x = cos t, y = sin t. Obtain df dt as a function of t by using the chain rule and check the result by expressing f in terms of t and differentiating directly. By the chain rule df dt = f x dx dt + f y dy dt. dy dt = 2xydx dt +x2dy dt = 2xy sin t+x2 cos t = 2 cos t sin 2 t+cos 3 t. By direct substitution f = cos 2 t sin t and df dt = 2 cos t sin2 t + cos 3 t. Suppose f(x, y) but this time x and y are both functions of two variables u and v ie. x(u, v) and y(u, v). Then by substituting these expressions into f(x(u, v), y(u, v)) we obtain f as a function of u and v and we may ask how the partial derivatives f f f u and v are related to the partial derivatives x and. The answer is again provided by the chain rule, which in this case reads f y f u = f x x u + f y f, and y u v = f x x v + f y y v. 110

111 Eg. f(x, y) = sin(x 2 y 2 ) and x = u + v, y = u v. Obtain f f u and v in terms of u and v by using the chain rule and check the result by expressing f directly in terms of u, v and then calculating the partial differentials. By the chain rule f u = f x x u + f y y u = 2x cos(x2 y 2 )(1) 2y cos(x 2 y 2 )(1) = 2(u + v) cos(4uv) 2(u v) cos(4uv) = 4v cos(4uv). f v = f x x v + f y y v = 2x cos(x2 y 2 )(1) 2y cos(x 2 y 2 )( 1) = 2(u + v) cos(4uv) + 2(u v) cos(4uv) = 4u cos(4uv). By direct substitution f = sin(4uv) thus f u = 4v cos(4uv) and f v = 4u cos(4uv). In the above examples it turned out to be easier to calculate the required derivatives by direct substitution. However, there are situations in which this is not always possible and then the chain rule comes into its own. 8.3 Implicit differentiation Suppose f(x, y) = x y and x, y are functions of t given implicitly by x 2 +y 2 = t 2 and x sin t = ye y. In this case it is not possible to explicitly solve for x(t) and y(t) so the chain rule needs to be used to obtain an expression for df dt. df dt = f x dx dt + f y dy dt = dx dt dy dt. To obtain these required terms we differentiate the given relations with respect to t to get 2x dx dt + 2ydy dt = 2t, and dx dt sin t + x cos t = dy dt ey (1 + y). These two equations can now be solved (exercise) for dx dt the above chain rule formula to get an expression for df dt and put into in terms of x, y, t. and dy dt 111

112 Given an implicit expression for a function, the value of its partial derivatives can be calculated using the chain rule. Eg. Suppose the function z(x, y) satisfies the equation 4x 2 y + z 5 x 3yz 2xy = 0. Calculate the value of z x at (x, y, z) = (1, 1, 1). First of all, it is easy to check that (x, y, z) = (1, 1, 1) indeed satisfies the given equation. Now differentiate the equation with respect to x keeping y fixed to get 8xy + z 5 + 5xz 4 z z 3y 2y = 0. x x Setting x = y = z = 1 this becomes z x = 0 thus at (1, 1, 1) we find that z x = 7/ Double integrals Rectangular regions Recall that for a function f(x) the definite integral b a f(x) dx is the signed area under the curve y = f(x) between x = a and x = b. This interpretation of the definite integral can be generalized for a function of two variables to a volume under a surface. Given a function f(x, y) and a region D in the (x, y) plane, the double integral f(x, y) dxdy D is the signed volume of the region between the surface z = f(x, y) and the region D in the z = 0 plane (see Figure 45). In a similar manner to our earlier construction of the definite integral, in terms of the limit of a Riemann sum of areas of rectangles, we can define the double integral as the limit of a Riemann sum of volumes of cuboids. For simplicity, consider the case of a rectangular region D = [a 0, a 1 ] [b 0, b 1 ]. We construct a subdivision S of D by defining the points of a two-dimensional lattice as a 0 = x 0 < x 1 <... < x n = a 1 and b 0 = y 0 < y 1 <... < y m = b 1. The edge lengths of the lattice are equal to dx i = x i x i 1 for i = 1,..., n and dy j = y j y j 1 for j = 1,..., m. The norm of the subdivision S is given by the maximum of all the dx i and dy j. 112

113 Figure 45: The double integral is the signed volume of the region between the surface z = f(x, y) and the region D in the z = 0 plane. It is obtained as the limit of a Riemann sum of cuboid volumes. We introduce sample points (p i, q j ) [x i 1, x i ] [y j 1, y j ] inside each rectangle of the lattice. The area of the rectangle is dx i dy j and we associate to this rectangle a cuboid of height f(p i, q j ). The volume of all the cuboids is the Riemann sum n m R = f(p i, q j )dx i dy j i=1 j=1 and the double integral is the limit f(x, y) dxdy = lim R. S 0 D In the special case that f(x, y) is a constant, say f(x, y) = c, then f(x, y) dxdy = c area(d). D For a rectangular region D = [a 0, a 1 ] [b 0, b 1 ] a practical way to compute the double integral is to use the following iterated integral result (valid providing f is continuous on D) a1 ( b1 ) b1 ( a1 ) f(x, y) dxdy = f(x, y) dy dx = f(x, y) dx dy. a 0 b 0 b 0 a 0 D 113