Section 1.6. Fourier Series Let f (x) be a function de…ned on the interval [¡L; L] : We call the series 1 a0 X ³ n¼ x n¼ x ´ + an cos + bn sin ; 2 L L n=1

where the coe¢cients an and bn are given by, for n = 0; 1; 2; :::; Z 1 L n¼ x an = f (x) cos dx L ¡L L Z 1 L n¼x bn = f (x) sin dx; L ¡L L

the Fourier series expansion of f (x) on the interval [¡L; L] ; and denote by 1 a0 X ³ n¼x n¼x ´ f (x) » + an cos + bn sin : (Fourier Series) 2 L L n=1 Example 6.1. Find the Fourier series for ½ 0; if ¡ ¼ · x · 0 f (x) = x; if 0 < x · ¼

on [¡¼ ; ¼] :

Solution: L = ¼: We computer the coe¢cients an and bn as follows. For n > 0; Z Z 1 L n¼x 1 ¼ an = f (x) cos dx = f (x) cos nxdx L ¡L L ¼ ¡¼ Z 1 ¼ = x cos nxdx ¼ 0 µ ¶ Z 1 ¼ sin nx = xd ¼ 0 n · ¸¼ Z ¼ 1 x sin nx 1 = ¡ sin nxdx ¼ n ¼n 0 0 1 1 = ¡ 2 (¡ cos nx)¼0 ¼n 1 1 = (cos n¼ ¡ 1) ¼ n2 (¡1)n ¡ 1 = : ¼n2 1

1 a0 = ¼

Z

¼

1 f (x) dx = ¼ ¡¼

Z

0

¼

xdx =

µ

11 2 x ¼2

¶¼

=

0

¼ 2

Z 1 ¼ bn = f (x) sin nxdx ¼ ¡¼ Z 1 ¼ = x sin nxdx ¼ 0 · ¸ 1 ¡x cos nx sin nx ¼ = + ¼ n n2 0 1 = [¡x cos nx]¼0 ¼n 1 (¡1)n (¡1)n+1 =¡ [¼ cos n¼] = ¡ = : ¼n n n Thus, the Fourier series is 1 a0 X ³ n¼x n¼x ´ f (x) » + an cos + bn sin 2 L L n=1 " # 1 ¼ X (¡1)n ¡ 1 n¼ x (¡1)n+1 n¼x = + cos + sin 4 n=1 ¼ n2 L n L

Example 6.2. Find the Fourier series for f (x) = jxj on [¡¼ ; ¼] : Solution: We shall computer an and bn : Z Z Z 1 ¼ 1 ¼ 1 0 a0 = jxj dx = xdx + (¡x) dx ¼ ¡¼ ¼ 0 ¼ ¡¼ ¼ ¼ = + =¼ 2 2

2

Z 1 ¼ n¼x an = jxj cos dx ¼ ¡¼ ¼ Z Z 1 ¼ 1 0 = x cos nxdx + (¡x) cos nxdx ¼ 0 ¼ ¡¼ Z ¼ Z 0 1 1 = xd (sin nx) ¡ xd (sin nx) n¼ 0 n¼ ¡¼ · ¸ Z ¼ Z 0 1 1 1 1 ¼ 0 = (x sin nx)0 ¡ sin nxdx ¡ (x sin nx)¡¼ ¡ sin nxdx n¼ n¼ 0 n¼ n¼ ¡¼ 1 1 1 1 =¡ (¡ cos nx)¼0 + (¡ cos nx)0¡¼ n¼ n n¼ n 1 1 = 2 (cos n¼ ¡ 1) ¡ 2 (1 ¡ cos n¼) : n¼ n¼ Thus an =

2 2 (cos n¼ ¡ 1) = 2 ((¡1)n ¡ 1) : 2 n¼ n¼

Z 1 ¼ n¼x bn = jxj sin dx ¼ ¡¼ ¼ Z Z 1 ¼ 1 0 = x sin nxdx + (¡x) sin nxdx: ¼ 0 ¼ ¡¼ The second integral in the above expression for bn is Z 1 0 (¡x) sin nxdx ¼ ¡¼ Z 0 y=¡x 1 = (y) sin n (¡y) d (¡y) ¼ ¼ Z 1 0 = y sin nydy ¼ ¼ Z 1 ¼ =¡ y sin nydy: ¼ 0 So bn = 0;

3

and consequently, a0 X ³ n¼x n¼ x ´ jxj » + an cos + bn sin 2 L L n=1 1

¼ X 2 = + ((¡1)n ¡ 1) cos nx: 2 2 n=1 n ¼ 1

Note that in the second example, the Fourier series expansion contains only term cos (n¼x=L) : We call it the Fourier Cosine series. This is generally true that when f (x) is even; i.e., f (¡x) = f (x) ; we have Z 1 L n¼ x bn = f (x) sin dx L ¡L L Z Z 1 L n¼ x 1 0 n¼ x = f (x) sin dx + f (x) sin dx L 0 L L ¡L L Z Z 1 L n¼ x 1 0 n¼ (¡y) = f (x) sin dx + f (¡y) sin d (¡y) L 0 L L L L Z Z 1 L n¼ x 1 0 n¼ y = f (x) sin dx + f (y) sin dy = 0; L 0 L L L L where in the third line above, we change variable x = ¡y: Thus, for even function f (x), the Fourier series is always a Cosine series. Similarly, if f (x) is an odd function, i.e., f (¡x) = ¡f (x) ; then for n > 0; Z 1 L n¼x an = f (x) cos dx L ¡L L Z Z 1 L n¼ x 1 0 n¼ x = f (x) cos dx + f (x) cos dx L 0 L L ¡L L Z Z 1 L n¼ x 1 0 n¼ (¡y) = f (x) cos dx + f (¡y) cos d (¡y) L 0 L L L L Z Z 1 L n¼ x 1 0 n¼ y = f (x) cos dx + f (y) cos dy = 0: L 0 L L L L In this case, the Fourier series becomes f (x) »

a0 X n¼ x + bn sin 2 L n=1 1

4

that contains only sin (n¼ x=L) : We call it the Fourier Sine series. Some other terminologies. ² f (x) is called piecewise smooth if f (x) has continuous derivative everywhere except for …nite many points. For instance, f (x) = jxj is di¤erentiable everywhere except for x = 0: ² f (x) is called periodic with the period 2L if f (x + 2L) = f (x) for any x: For instance, f (x) = sin x is periodic with period 2¼ : For any function f (x) de…ned only on [¡L; L] ; one can always extend it to all real numbers as a periodic function, as shown in the …gure below.

y

-L

L

x

This is equivalent to say that f (x) is periodic with the period 2L: When f (L) = f (¡L) ; the extended function is continuous at x = kL for k = §1; §2; ::: Otherwise, the extended function is discontinuous at x = kL: ² f (x ¡ 0) is the limit from the left. f (x + 0) is the limit from the right. Theorem. Let f (x) be a piecewise smooth function on [¡L; L] : Then its Fourier series 1 a0 X ³ n¼x n¼x ´ f (x) » + an cos + bn sin ; on [¡L; L] : 2 L L n=1

converges to f (x) at any point where f (x) is continuous, and to the average[f (x ¡ 0) + f (x + 0)] =2 at all any x where f (x) is discontinuous. In particular, if f (x) is smooth 5

everywhere on [¡L; L] and f (L) = f (¡L) : Then a0 X ³ n¼ x n¼ x ´ f (x) = + an cos + bn sin ; on [¡L; L] : 2 L L n=1 1

To see convergence of a Fourier series, we notice that, using trigonometric identities, k¼x k¼x ak cos + bn sin L µ ZL L ¶ µ Z L ¶ 1 k¼» k¼ x 1 k¼ » k¼ x = f (») cos d» cos + f (») sin d» sin L ¡L L L L ¡L L L µ ¶ Z L 1 k¼» k¼x k¼ » k¼x = f (») cos cos + sin sin d» L ¡L L L L L Z 1 L k¼ (»¡ x) = f (») cos d»: L ¡L L So the partial sum sequence ¶ n µ a0 X k¼x k¼x Sn = + ak cos + bn sin 2 L L k=1 Z L Z n X 1 1 L k¼ (» ¡ x) = f (») d» + f (») cos d» 2L ¡L L ¡L L k=1 " # Z n 1 L 1 X k¼ (» ¡ x) = f (») + cos d» L ¡L 2 k=1 L ·µ ¶ ¸ 1 Z sin m + (» ¡ x) 1 L 2 = f (») d» L ¡L 2 sin ((» ¡ x) =2) converges to f (x) at any continuous point of f (x) : Example 6.3. From Example 6.1 above, we know the Fourier series for ½ 0; if ¡ ¼ · x · 0 f (x) = on [¡¼ ; ¼ ] x; if 0 < x · ¼

6

is 1 a0 X ³ n¼ x n¼ x ´ f (x) » + an cos + bn sin 2 L L n=1 " # 1 ¼ X (¡1)n ¡ 1 n¼ x (¡1)n+1 n¼x = + cos + sin : 4 n=1 ¼ n2 L n L

The function has discontinuity point x = 0; and its periodic extension has discontinuity x = ¼ and x = ¡¼ : So the Fourier series is converges everywhere except for x = ¡¼; 0; ¼ : Thus " # 1 n n+1 X ¼ (¡1) ¡ 1 n¼ x (¡1) n¼x f (x) = + cos + sin on (¡¼; 0) and on (0; ¼) : 2 4 n=1 ¼n L n L From Example 6.2, the Fourier series for f (x) = jxj on [¡¼; ¼ ] is ¼ X 2 jxj » + ((¡1)n ¡ 1) cos nx: 2 2 n=1 n ¼ 1

According to Theorem, the Fourier series converges to jxj everywhere. Thus ¼ X 2 jxj = + ((¡1)n ¡ 1) cos nx on [¡¼ ; ¼] : 2 n=1 n2 ¼ 1

Any function on [0; L] can be extended to [¡L; L] either as an even function ½ f (x) on [0; L] F (x) = : f (¡x) on [¡L; 0] y

-L

L

7

x

or as an odd function: G (x) =

½

f (x) ¡f (¡x)

on [0; L] : on [¡L; 0]

y

-L

L

x

Note that for even function extension, F (¡L) = f (¡ (¡L)) = f (L) = F (L) So this extended function is periodic on [¡L; L] : For the odd function extension, there is a possible discontinuous point x = 0: We know that the Fourier series for F (x) and G (x) have the form, respectively, a0 X ³ n¼ x n¼x ´ F (x) » + an cos + bn sin 2 L L n=1 1

a0 X n¼x = + an cos ; on [¡L; L] ; 2 L n=1 1

a0 X ³ n¼x n¼x ´ G (x) » + an cos + bn sin 2 L L n=1 1

1 X n¼x = bn sin ; on [¡L; L] : L n=1

8

This leads to Theorem. Let f (x) be a piecewise smooth function on [0; L] : Then f (x) has Sine series expansion a0 X n¼ x + bn sin on (0; L) 2 L n=1 Z L 2 n¼x bn = f (x) sin dx; L 0 L

f (x) »

1

and Cosine series expansion 1 X n¼ x an cos on [0; L] L n=0 Z 2 L n¼ x an = f (x) cos dx: L 0 L

f (x) »

Both Since and Cosine series converge to f (x) at all points where f (x) is continuous, and to the average[f (x ¡ 0) + f (x + 0)] =2 at all points x where f (x) is discontinuous. Some interesting properties and advanced topics on the theory of the Fourier series. n n¼x o1 ² cos is a sequence of mutually orthogonal functions under L n=0 the inner product (dot product) ³ n¼x ´ ³ m¼ x ´ cos ¢ cos L L Z L n¼x m¼x def = cos cos dx L L ¡L ½ 0 if n 6= m = : L if n = m ² Riemann Lemma: for any smooth function f (x) Z L n¼ x lim f (x) cos dx = 0 n!1 ¡L L n n¼x o1 ² The same properties hold for sin L n=0 9

² There are many other sequence of orthogonal functions having these properties. For instance, Legendre polynomial on [0; 1] Ln (t) =

n X k=0

(n + k)! k 2 k (t ¡ 1) : (n ¡ k)! (k!) 2

Another expression is called Rodrigues formula Ln (t) = The expansion

¢n 1 dn ¡ 2 t ¡ 1 : 2n n! dtn

f (x) »

n X

cn Ln (t)

k=0

is called the Fourier series with respect to Legendre polynomials. ² More general, we called a sequence of functions f©n (x)g1 n=0 are mutually orthogonal with respect to a weight ½(x) if Z L ©n (x) ©m (x) ½(x) dx = 0 if n 6= m: ¡L Z L [©n (x)]2 ½(x) dx = Án > 0 ¡L

is

For any function f (x), its Fourier series expansion with respect to f©n (x)g1 n=0 f (x) »

1 X

cn ©n (x) :

k=0

The Fourier coe¢cient cn can be calculated as follows. Multiplying the above expansion by ©k (x) ½(x) for any …xed k : f (x) ©k (x) ½(x) »

1 X

cn ©n (x) ©k (x) ½(x)

k=0

and then integrate the both sides Z L Z 1 X f (x) ©k (x) ½(x) dx » cn ¡L

k=0

10

L

¡L

©n (x) ©k (x) ½(x) dx:

When n 6= k; the weighted orthogonality implies Z L ©n (x) ©k (x) ½(x) dx = 0: ¡L

So Z

L

¡L

f (x) ©k (x) ½(x) dx »

1 X

Z

cn

k=0

= ck

Z

L

L

©n (x) ©k (x) ½(x) dx ¡L

©k (x) ©k (x) ½(x) dx ¡L

= ck Ák or 1 ck = Ák

Z

L

f (x) ©k (x) ½(x) dx:

¡L

We call f©n (x)g1 n=0 is complete if #2 Z L" n X lim f (x) ¡ cn ©n (x) ½(x) dx = 0; n!1

¡L

k=0

for any function f (x) : In this case, we say the Fourier series Square-converges to f (x) ; and write Z L 1 X 1 f (x) = cn ©n (x) ; cn = f (x) ©n (x) ½(x) dx: Án ¡L k=0 Homework: 1. Find the Fourier series for f (x) =

½

x 2x

if 0 · x · 1 if ¡ 1 · x < 0

and determine at what point the Fourier series does not converges to f (x) ; if any. 2. Find the Fourier Sine series for x on [0; 1] 3. Find the Fourier CoSine series for x on [0; 1] :

11