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.
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
f'(x) = du
f(x) = u
cos kx = v
-1/k sin kx = dv
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??