Worksheet 27: Fourier Series

Worksheet 27: Fourier series
Full Fourier series: if f is a function on the interval [, ], then the corresponding series is f ( x) a0 + an cos(nx) + bn sin(nx); 2 n=1

an =
1 1 bn =
f (x) cos(nx) dx,

f (x) sin(nx) dx.
Fourier cosine and sine series: if f is a function on the interval [0, ], then the corresponding cosine series is a0 f (x) + an cos(nx); 2 n=1 an = 2

f (x) cos(nx) dx,

0
and the corresponding sine series is
f (x) bn = 2
n=1
bn sin(nx); f (x) sin(nx).

0
Convergence theorem for full Fourier series: if f is a piecewise dierentiable function on [, ], then its Fourier series converges at every point. The sum of the series is computed as follows: 1
1. Forget about what the function f looks like outside of the interval [, ]. After all, the formulas for the coecients only feature the values of f on that interval. be the 2. Continue f periodically from [, ] to the whole real line; let f resulting function. (x), if f is 3. The sum of the Fourier series at the point x is equal to f (x+) + f (x))/2. continuous at x; otherwise, it is equal to (f Convergence for Fourier cosine series: forget about what f looks like outside of [0, ], then extend f as an even function to [, ], then use the above algorithm. Same approach works for sine series, except that you extend f to [, ] as an odd function. 1.* This problem shows an alternative way of proving that the functions sin(kx), k Z, k > 0, form an orthogonal set in C [0, ]. (a) Assume that u and v are eigenfunctions of the following problem: u (x) + u(x) = 0, 0 < x < ; u(0) = u( ) = 0; v (x) + v (x) = 0, 0 < x < ; v (0) = v ( ) = 0, where and are two real numbers. Integrate by parts twice and use the boundary conditions to show that

u (x)v (x) dx =
0 0
u(x)v (x) dx.
Use the dierential equations satised by u and v to show that
( )
0
u(x)v (x) dx = 0.
(b) Take u(x) = sin(kx), v (x) = sin(lx), for k, l positive integers and k = l. Verify that these functions satisfy the conditions of part (a) for certain and , and conclude that
sin(kx) sin(lx) dx = 0.
0
Solution: (a) We have

u (x)v (x) dx = u (x)v (x)| x=0

0 0
u (x)v (x) dx; u (x)v (x) dx.

0
u(x)v (x) dx = u(x)v (x)| x=0

0
Since u(0) = u( ) = v (0) = v ( ), the boundary terms vanish and we get

u (x)v (x) dx =
0 0
u(x)v (x) dx.
Next, u (x) = u(x) and v (x) = v (x); substituting this into the equation above, we get

u(x)v (x) dx =
0
u(x)v (x) dx.
(b) The functions u(x) and v (x) satisfy the equations of (a) for = k 2 and = l2 ; therefore,
(k 2 l2 )
0
u(x)v (x) dx = 0.
Since k 2 = l2 , the functions u and v are orthogonal. 2. Find the Fourier sine series for the function f (x) = x( x), 0 < x < . Solution: Integrate by parts: 2 2 = k bk =
x( x) sin(kx) dx
0
x( x) d(cos(kx))
0
2 2 = x( x) cos(kx)|x=0 + ( 2x) cos(kx) dx k k 0 2 2x d(sin(kx)) = k 2 0 2 4 = ( 2 x ) sin( kx ) | + sin(kx) dx x=0 k 2 k 2 0 4 4 = 3 cos(kx)| [1 (1)k ]. x=0 = k k 3
Therefore, bk = 0 for k even and bk = 8/(k 3 ) for k odd; the Fourier series is 8 f (x)
j =1
sin((2j 1)x) . (2j 1)3
3. Using the previous problem, nd the formal solution to the problem 2u u = , 0 < x < , t > 0; t x2 u(0, t) = u(, t) = 0, t > 0; u(x, 0) = x( x), 0 < x < . Answer:
u(x, t) =
j =1
8 2 e(2j 1) t sin((2j 1)x). 3 (2j 1)
4. Find the Fourier cosine series for the function f (x) = x, 0 < x < . Solution: We calculate a0 = ak = 2 2 = k
0
2
0
x dx = ,
0
( x) cos(kx) dx x d(sin(kx))
0
2 = k
sin(kx) dx =
2 [1 (1)k ]. 2 k
The corresponding Fourier series is f ( x) 4 + 2
j =1
cos((2j 1)x) . (2j 1)2
5. Find the full (sine and cosine) Fourier series for the function f (x) = Solution: We calculate a0 = 1 1 bk = ak =
0, < x < 0; 1, 0 < x < .
dx = 1,
0
cos(kx) dx = 0, k > 0,
0
sin(kx) dx =
0
1 (1)k . k
Therefore, the Fourier series is 1 f ( x) + 2
j =1
2 sin((2j 1)x). 2j 1
6. Assume that f (x) is an odd function on the interval [, ]. Explain why the full Fourier series of f consists only of sines (in other words, why the coecients next to the cosines are all zero). Solution: We have ak = 1
f (x) cos(kx) dx = 0,
since cos(kx) is even, f (x) is odd, the product f (x) cos(kx) is odd and thus its integral over [, ] is zero. 7. Using the Fourier series convergence theorem, nd the functions to which the series in problems 2, 4, and 5 converge. Sketch their graphs. Answers: For problem 2, the Fourier series converges to the 2 -periodic extension of the function g (x) = x( x), 0 x ; x( x), x 0.
For problem 4, the Fourier series converges to the 2 -periodic extension of the function |x| from the segment [, ]. 5
For problem 5, the Fourier series of the function 0, h(x) = 1, 1/2,
converges to the 2 -periodic extension < x < 0; 0 < x < ; x {, 0, }.
8. Assume that f is a function on the interval [0, ] whose graph is symmetric with respect to the line x = /2; in other words, f ( x) = f (x). If f (x)
k=1
bk sin(kx)
is the Fourier sine series of f , prove that bk = 0 for even k . (Hint: write out the formula for bk and make the change of variables y = x.) Solution: Making the change of variables y = x, we get bk = = = 2
f (x) sin(kx) dx
0
f ( y ) sin(k ( y )) dy
0
f (y )(1)k+1 sin(ky ) dy = (1)k+1 bk .

0
Therefore, for k even, bk = bk and thus bk = 0.

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Worksheet 27: Fourier series

f (x) cos(nx) dx,

f (x) sin(nx) dx.

f (x) cos(nx) dx,

and the corresponding sine series is

bn sin(nx); f (x) sin(nx).

u(x)v (x) dx.

Use the dierential equations satised by u and v to show that

Solution: (a) We have

u (x)v (x) dx = u (x)v (x)| x=0

u (x)v (x) dx; u (x)v (x) dx.

u(x)v (x) dx = u(x)v (x)| x=0

Since u(0) = u( ) = v (0) = v ( ), the boundary terms vanish and we get

u(x)v (x) dx.

u(x)v (x) dx.

sin((2j 1)x) . (2j 1)3

8 2 e(2j 1) t sin((2j 1)x). 3 (2j 1)

The corresponding Fourier series is f ( x) 4 + 2

cos((2j 1)x) . (2j 1)2

0, < x < 0; 1, 0 < x < .

Therefore, the Fourier series is 1 f ( x) + 2

For problem 5, the Fourier series of the function 0, h(x) = 1, 1/2,

converges to the 2 -periodic extension < x < 0; 0 < x < ; x {, 0, }.

f (y )(1)k+1 sin(ky ) dy = (1)k+1 bk .

Therefore, for k even, bk = bk and thus bk = 0.

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