I'm kinda stuck on this problem (attached).. I know it works (from a previous class), but I forgot how to actually prove it.
Thank you!
If you have:
$\displaystyle S_n^{f}=a_0+\sum_{n=1}^{n} a_n \cos(nx) + \sum_{n=1}^{n} b_n \sin(nx)$
You know the first derivative of:
$\displaystyle a_n \cos(nx) = - a_n \cdot n \cdot \sin(nx)$
and so the second derivative will be
$\displaystyle -a_n \cdot n^2 \cdot \cos(nx)$
...
So the kth derivative will be:
$\displaystyle \pm a_n \cdot n^k \cdot \mbox{trig}(nx)$
We don't know if we get a sinus or a cosinus, that's the reason we wright trig(nx).
Do this also for the other sum.
Does this help you? ...