Proving the Fourier Coefficients of f' Using Periodic Functions | Homework Help

[Kuma]

Homework Statement

Hi there.

The question is:

if f(x) is a periodic function with period 2pi and the derivative is f'(x) continuous on [-pi, pi]

show that the Fourier coefficients of f' are kbk and -kak if the Fourier coefficients of f are ak and bk respectively.

Homework Equations

The Attempt at a Solution

This is my attempt. Doesn't quite get there:

so ak' = 1/2pi int from -pi to pi of f'(x) cos kx dx

use parts:
f'(x) = du
f(x) = u
cos kx = v
-1/k sin kx = dv

i get

1/2pi[f(x) cos kx/k + 1/k int from -pi to pi f(x) sin kx]

of course the integral in the parts equation is just bk, and the f(x) cos kx/k evaluated from pi to -pi just becomes 2f(pi)cos kpi.

So then I'm left with.

1/pi[f(pi)cos k pi + (1/k)bk]

I guess I can write it as

[(-1)^k f(pi) + (1/k) bk]/pi

What did i do wrong??

 

[Dick]

Homework Statement

Hi there.

The question is:

if f(x) is a periodic function with period 2pi and the derivative is f'(x) continuous on [-pi, pi]

show that the Fourier coefficients of f' are kbk and -kak if the Fourier coefficients of f are ak and bk respectively.

Homework Equations

The Attempt at a Solution

This is my attempt. Doesn't quite get there:

so ak' = 1/2pi int from -pi to pi of f'(x) cos kx dx

use parts:
f'(x) = du
f(x) = u
cos kx = v
-1/k sin kx = dv

i get

1/2pi[f(x) cos kx/k + 1/k int from -pi to pi f(x) sin kx]

of course the integral in the parts equation is just bk, and the f(x) cos kx/k evaluated from pi to -pi just becomes 2f(pi)cos kpi.

So then I'm left with.

1/pi[f(pi)cos k pi + (1/k)bk]

I guess I can write it as

[(-1)^k f(pi) + (1/k) bk]/pi

What did i do wrong??

E.g. f(x) evaluated from -pi to pi is f(pi)-f(-pi). That's not 2f(pi).

 

[Kuma]

E.g. f(x) evaluated from -pi to pi is f(pi)-f(-pi). That's not 2f(pi).

isn't it f(pi) cos kpi - f(-pi) cos -kpi

I wasn't sure how to evaluate that, with k odd i think -2f(pi) and even its 2f(pi) isn't it?

 

[Dick]

isn't it f(pi) cos kpi - f(-pi) cos -kpi

I wasn't sure how to evaluate that, with k odd i think -2f(pi) and even its 2f(pi) isn't it?

No. f is periodic. So f(-pi)=f(pi) and cos(kpi)=cos(-kpi), right?

 

[Kuma]

Oh right. So then the first term is 0?
But then I'm still left with (1/k pi)(bk).

 

[Dick]

Oh right. So then the first term is 0?
But then I'm still left with (1/k pi)(bk).

In your integration by parts you've got v=cos(kx). Then dv=(-k*sin(kx))dx. Not 1/k. And the pi should have been included into definition of bk.