f(x + T ) = f(x), for all x. The period of the function f(t) is the interval between two successive repetitions.

Transcription

1 Fourier Series. Itroductio Whe the Frech mathematicia Joseph Fourier (768-83) was tryig to study the flow of heat i a metal plate, he had the idea of expressig the heat source as a ifiite series of sie ad cosie fuctios. Although the origial motivatio was to solve the heat equatio, it later became obvious that the same techiques could be applied to a wide array of mathematical ad physical problems. I this course, we will lear how to fid Fourier series to represet periodic fuctios as a ifiite series of sie ad cosie terms. A fuctio f(x) is said to be periodic with period T, if f(x + T ) = f(x), for all x. The period of the fuctio f(t) is the iterval betwee two successive repetitios.. Defiitio of a Fourier Series Let f be a bouded fuctio defied o the [, π] with at most a fiite umber of maxima ad miima ad at most a fiite umber of discotiuities i the iterval. The the Fourier series of f is the series f(x) = a + a cos x + a cos x + a 3 cos 3x b si x + b si x + b 3 si 3x +... where the coefficiets a ad b are give by the formulae a = π a = π b = π f(x)dx f(x)cos(x)dx f(x)si(x)dx

2 .3 Why the coefficiets are foud as they are We wat to derive formulas for a, a ad b. To do this we take advatage of some properies of siusoidal sigals. We use the followig orthogoality coditios: Orthogoality coditios (i) The average value of cos(x) ad si(x) over a period is zero. cos(x)dx = si(x)dx = (ii) The average value of si(mx)cos(x) over a period is zero. si(mx)cos(x)dx = (iii) The average value of si(mx)si(x) over a period, { π if m = si(mx)si(x)dx = otherwise (iv) The average value of cos(mx)cos(x) over a period, π if m = = cos(mx)cos(x)dx = π if m = if m Remark. The followig trigoometric idetities are useful to prove the above orthogoality coditios: cos A cos B = [cos(a B) + cos(a + B)] si A si B = [cos(a B) cos(a + B)] si A cos B = [si(a B) + si(a + B)]

3 Fidig a We assume that we ca represet the fuctio by f(x) = a + a cos x + a cos x + a 3 cos 3x b si x + b si x + b 3 si 3x +... () Multiply () by cos(x), ad itegrate from to π ad assume it is permissible to itegrate the series term by term. f(x) cos(x)dx = a cos(x) + a cos x cos(x)dx + a cos x cos(x)dx b si x cos(x)dx + b si x cos(x)dx +... = a π, because of the above orthogoality coditios a = π f(x) cos(x)dx Similarly if we multiply () by si(x), ad itegrate from to π, we ca fid the formula for the coefficiet b. To fid a, simply itegrate () from to π.

4 .4 Fourier Series Defiitio. Let f(x) be a π-periodic fuctio which is itegrable o [, π]. Set a = π a = π b = π the the trigoometric series a + f(x)dx f(x)cos(x)dx f(x)si(x)dx (a cos(x) + b si(x)) = is called the Fourier series associated to the fuctio f(x). Remark. Notice that we are ot sayig f(x) is equal to its Fourier Series. Later we will discuss coditios uder which that is actually true. Example. Fid the Fourier coefficiets ad the Fourier series of the squarewave fuctio f defied by f(x) = { if x < if x < π ad f(x + π) = f(x) Solutio: So f is periodic with period π ad its graph is:

5 Usig the formulas for the Fourier coefficiets we have a = π = π ( = π (π) = a = π f(x)dx dx + π f(x) cos(x)dx dx) = π ( dx + cos(x)dx) π = π ( + [ si(x) ] π ) = (si π si ) π = b = π = π ( f(x) si(x)dx dx + π ] π ) = π ( [ cos(x) si(x)dx) = (cos π cos ) π = π (( ) ), sice cos π = ( ). {, if is eve = π, if is odd

6 The Fourier Series of f is therefore f(x) = a + a cos x + a cos x + a 3 cos 3x b si x + b si x + b 3 si 3x +... = si x + si x + si 3x + si 4x + si 5x +... π 3π 5π = + π si x + si 3x + si 5x π 5π Remark. I the above example we have foud the Fourier Series of the square-wave fuctio, but we do t kow yet whether this fuctio is equal to its Fourier series. If we plot + π si x + π si x + si 3x 3π + π si x + si 3x + si 5x+ 3π 5π + π si x + si 3x + si 5x + si 7x+ 3π 5π 7π + π si x + si 3x + si 5x + si 7x + si 9x + si x + si 3x, 3π 5π 7π 9π π 3π we see that as we take more terms, we get a better approximatio to the square-wave fuctio. The followig theorem tells us that for almost all poits (except at the discotiuities), the Fourier series equals the fuctio. Theorem (Fourier Covergece Theorem) If f is a periodic fuctio with period π ad f ad f are piecewise cotiuous o [, π], the the Fourier series is coverget. The sum of the Fourier series is equal to f(x) at all umbers x where f is cotiuous. At the umbers x where f is discotiuous, the sum of the Fourier series is the average value. i.e. [f(x+ ) + f(x )].

7 Remark. If we apply this result to the example above, the Fourier Series is equal to the fuctio at all poits except,, π. At the discotiuity, observe that f( + ) = ad f( ) = The average value is /. Therefore the series equals to / at the discotiuities.

8 .5 Odd ad eve fuctios Defiitio. A fuctio is said to be eve if A fuctio is said to be odd if f( x) = f(x) for all real umbers x. f( x) = f(x) for all real umbers x. Example. cos x, x, x are examples of eve fuctios. si x, x, x 3 are examples of odd fuctios. The product of two eve fuctios is eve, the product of two odd fuctios is also eve. The product of a eve ad odd fuctio is odd. Remark. If f is a odd fuctio the while if f is a eve fuctio the (i) If f(x) is odd, the f(x)dx =, f(x)dx = f(x)dx. f(x)cos(x) is odd hece a = ad f(x)si(x) is eve hece b = π f(x)si(x)dx π (ii) If f(x) is eve, the f(x)cos(x) is eve hece a = π f(x)si(x) is eve hece b = f(x)cos(x)dx ad

9 Example.. Let f be a periodic fuctio of period π such that f(x) = x for π x < π Fid the Fourier series associated to f. Solutio: So f is periodic with period π ad its graph is: We first check if f is eve or odd. Therefore, f( x) = x = f(x), a = b = π so f(x) is odd. f(x)si(x)dx Usig the formulas for the Fourier coefficiets we have b = π = π ([ x x si(x)dx cos x ] π π ( = π ( x [ cos π] + [si ] π ) = cos π { / if is eve = / if is odd The Fourier Series of f is therefore f(x) = b si x + b si x + b 3 si 3x +... cos x )dx) = si x si x + 3 si 3x 4 si 4x + si 5x

10 . Let f be a periodic fuctio of period π such that f(x) = π x for π x < π Solutio: So f is periodic with period π ad its graph is: We first check if f is eve or odd. f( x) = π ( x) = π x = f(x), so f(x) is eve. Sice f is eve, b = a = π f(x) cos(x)dx

11 Usig the formulas for the Fourier coefficiets we have a = π f(x) cos(x)dx = (π x ) cos(x)dx π = π ([ (π x si x] π π ) si x x dx) = π ([(π π si π ) (π ) si ] + = x si(x)dx π = π ([ cos x] π π x ( = x ([ cos π] + [si ] π π ) = 4 cos π { 4/ if is eve = 4/ if is odd It remais to calculate a. a = π cos x )dx) (π x )dx = π [π x x3 3 ]π = 4π 3 The Fourier Series of f is therefore f(x) = a + a cos x + a cos x + a 3 cos 3x +... b si x + b si x + b 3 si 3x +... x si x dx) = π 3 + 4(cos x 4 cos x + 9 cos 3x 6 cos 4x + cos 5x +...) 5