Fourier Series Roughy speaking, a Fourier series expansion for a function is a representation of the function as sum of s and coes. Expresg a musica tone as a sum of a fundamenta tone and various harmonics is such a representation. So is a spectra decomposition of ight waves. The main Fourier series expansions that we use in this course are stated in the next section. We sha never prove that these expansions are correct. But in the section Vaidity of Fourier series we give an eementary partia argument that, hopefuy, wi convince you that the expansions are indeed correct. In the section Usefuness of Fourier Series we introduce one of the many ways that Fourier series are used in appications. The Main Fourier Series Expansions. We sha shorty state three Fourier series expansions. They are appicabe to functions that are piecewise continuous with piecewise continuous first derivative. In appications, most functions satisfy these reguarity requirements. We start with the definition of piecewise continuous. A function f(x is said to be piecewise continuous if it is continuous except for isoated jump discontinuities. In the exampe beow, f(x is continuous except for jump f(x f(+ [f(++f( ] f( f( 3 discontinuities at x and x.5. If a function f(x has a jump discontinuity at x, then the vaue of f(x as it enters and eaves the jump are f(x x x im f(x and f(x + im x<x x x x f(x x>x respectivey. If f were continuous at x, we woud have f(x f(x + f(x. At c Joe Fedman.. A rights reserved.
a jump, however, there is no a priori reation between f(x and f(x ±. In the exampe above, f( is we beow both f( and f(+. On the other hand, it is fairy common for the vaue of f at the jump x to be precisey at the midpoint of the jump. That is f(x [f(x ++f(x ]. In the exampe, this is the case at x.5. Theorem (Fourier Series Let f(x be piecewise continuous with piecewise continuous first derivative. a Let f( f(. Then b k ( { f(x if f is continuous at x f(x++f(x otherwise for a x if and ony if b k f(x( dx. b Let f ( f (. Then a + for a x if and ony if a k c Let f be periodic of period. Then a + [ ak cos ( for a x if and ony if a k a k cos ( { f(x if f is continuous at x f(x++f(x otherwise f(xcos( dx. +bk ( { ] f(x if f is continuous at x f(x++f(x otherwise f(xcos ( dx and bk f(x ( dx Remark. One consequence of the above theorem is that if you are tod that, for exampe, β k ( for a x then, by part a with f(x, it is necessary that β k f(x ( dx for a k,, This is used repeatedy in ug Fourier series to sove differentia equations. c Joe Fedman.. A rights reserved.
Exampe. Consider the function f(x { if x if < x f(x Appy part b of the Fourier series Theorem with this f and. Then x a k f(xcos ( dx cos ( { if k dx if k kπ ( { if k if k kπ ( kπ if k if k > and k is even if k is odd and by part b of the Theorem ( (k / kπ + ( p p (p+π cos( (p+πx { if x if x if < x We have summed over a odd k by setting k p+ and summing over p,,,. There is a Java demo that shows the graphs of + p ( p (p+π cos( (p+πx for many vaues of. x Usefuness of Fourier Series. In this course, we use Fourier series as a too for soving partia differentia equations. This is just one of its many appications. To give you a preiminary taste of this appication, we now briefy consider one typica partia differentia equation probem. It arises in studying a vibrating string. Suppose that a vibrating string has its ends tied to nais at x y and x, y. Denote by y(x,t the ampitude of the string at position x and time t. If c Joe Fedman.. A rights reserved. 3
there are no externa forces, no damping and the ampitude is sma, y t (x,t c y x (x,t for a x, t ( y(,t for a t ( y(,t for a t (3 y(x, f(x for a x (4 y t (x, g(x for a x (5 A detaied derivation of this system of equations is provided in another set of notes. The first equation is the ewton s aw of motion appropriate for the current situation; the next two equations impose the requirements that the ends of the string are tied to nais; the fina two equations specify the initia position and and speed of the string. We are assuming that the initia position and speed f(x and g(x are given functions. The unknown in the probem is the ampitude y(x,t. For each fixed t, y(x,t is a function of the one variabe x. This function vanishes at x and x and thus, by part a of the Fourier series theorem has an expansion y(x,t b k (t ( The soution y(x,t is competey determined by the, as yet unknown, coefficients b k (t. Furthermore these coefficients can be found by substituting y(x,t b k(t ( into the above five requirements on y(x,t. First the PDE (: y t (x,t c y x (x,t k(t ( + b k π c [ ] b k(t+ k π c b k (t ( b k (t ( This says that, for each fixed t, the function, viewed as a function of x, has Fourier series expansion [ ] b k (t+ k π c b k (t (. Our Fourier series theorem then forces b k(t+ k π c b k (t for a k,t ( Because si for a integers k, conditions ( and (3 are satisfied by y(x,t c Joe Fedman.. A rights reserved. 4
b k(t ( regardess of what bk (t are. Finay substituting into (4 and (5 gives y(,t y t (,t By uniqueness of Fourier coefficients, once again, b k ( b k ( b k ( ( f(x b k (( g(x f(x ( dx (4 g(x ( dx (5 For each fixed k, equations (, (4 and (5 constitute one second order constant coefficient ordinary differentia equation and two initia conditions for the unknown function b k (t. You aready know how to sove constant coefficient ordinary differentia equations. Denoting α k the soution of (, (4 and (5 is and the soution of (, (, (3, (4, (5 is y(x,t f(x ( dx βk b k (t α k cos ( kπct + β k kπc ( kπct [ α k cos ( kπct + β k kπc ( kπct g(x ( dx ] ( The interpretation of this formua is provided in another set of notes. Vaidity of Fourier Series. It is beyond the scope of this course to give a fu justification for the Fourier series theorem. However, with just a itte high schoo trigonometry, we can justify a formua that is usefu in its own right and that aso provides arbitrariy good approximations to the Fourier series expansions. For concreteness, we sha just consider Fourier series that is part a of the theorem. The other parts are simiar. Imagine some appication in which we have c Joe Fedman.. A rights reserved. 5
to measure some function f(x obeying f( f(. For exampe, f(x might be the ampitude at some fixed time of a vibrating string. Because we can ony make finitey many measurements, we cannot determine f(x for a vaues of x. Suppose that we measure f(x for x, (,, where is some fixed integer. Define F ( n f ( n Because we do not know f(x for a x we cannot compute b k f(x( dx. But we can get a Riemann sum approximation to it ug ony x s for which f(x is known. A we need to do is divide the domain of integration up into intervas each of ength /. On the interva n x (n+ we approximate f(x ( ( by f n ( kπ n. f(x ( This gives B ( k n f( ( n kπ n n ( F ( n x ( Define More generay, et F, F,,, F be + numbers obeying F F. B k m F n ( mkπ We sha show that the Fourier series representation for the F n s F n B k ( ( is reay true. To do so we just evauate the right hand side by substituting in the definition c Joe Fedman.. A rights reserved. 6
of the B k s. B k ( m m m F m { F m ( mkπ F m ( mkπ ( mkπ ( ( The vaidity of ( is then an immediate consequence of part b of ( } Lemma. a { if cosθ cos(kθ ( cosθ+ θθ cosθ if cosθ b Let n and m be integers with n,m. Then ( mkπ { ( if n m if n m Proof: a If cosθ then θ is an integer mutipe of π so that cos(kθ for a integers k and cos(kθ, as caimed. If cosθ, we must show cos(kθ ( cosθ+ θθ cos θ or equivaenty (just moving the ( cosθ to the eft hand side cos(kθ+ cos(θ + or equivaenty (just cross mutipying by ( cos θ ( cosθ+ θθ cos θ ( cosθcos(kθ+( cosθcos(θ θθ c Joe Fedman.. A rights reserved. 7
The eft hand side is ( cosθ+ cosθ + cosθ + ( cosθcos(kθ+( cosθcos(θ cos(kθ cos(kθ cosθcos(kθ+cos(θ cosθcos(θ cos ( (k +θ cos ( (k θ +cos(θ cos( ( θ cos( ( +θ by the trig identity cosθcosφ cos(θ +φ+ cos(θ φ. Writing out the three sums on three separate rows ( cosθ+ cosθ ( cosθcos(kθ+( cosθcos(θ +cosθ +cos(θ+ +cos(( θ+cos(( θ cos(θ cos(( θ cos(( θ cos(θ cosθ cos(θ cos(( θ cos( ( θ cos( ( +θ cos(( θ+cos(θ cos((+θ (θ(θ which is exacty the desired identity. b Ug θφ cos(θ φ cos(θ+φ ( mkπ ( cos ( (n mkπ cos ( (n+mkπ We wish to appy the identity of part a twice, the first time with θ (n mπ time with θ (n+mπ. Observe that with n,m we have and the second π < ( π (n mπ ( π < π cos ( (n mπ if and ony if n m < π (n+mπ ( π < π cos ( (n mπ c Joe Fedman.. A rights reserved. 8
and ( (n mπ cos ( (n mπ ( (n+mπ cos ( (n+mπ ce ( m ( m. Appying part a twice Subtracting gives which is the desired resut. ( n m cos ( { (n mkπ if n m ( n m if n m cos ( (n+mkπ ( n m cos ( (n mkπ cos ( { (n+mkπ if n m if n m We remark that trig identities are often ceaner ooking and easier to derive ug exponentias of compex numbers. This is indeed the case for Fourier series formuae, which are just big trig identities. See the notes on compex numbers and exponentias. c Joe Fedman.. A rights reserved. 9