# Derivative of Fourier series

1. Oct 1, 2009

### caesius

We were given in a previous question,

$$s_{N}(x) = \frac{4}{\pi}\sum_{n=0}^{N-1}\frac{sin(2n+1)x}{2n+1}$$

1. The problem statement, all variables and given/known data

Show that
$$s'_{N}(x) = \frac{2sin(2Nx)}{\pi sinx}, x \neq l\pi$$

and

$$s'_{N}(x) = \frac{4N}{\pi}(-1)^l, x = l\pi$$

where l is any integer.

3. The attempt at a solution

Utterly stumped on this one, I'm aware it's not *NORMAL* differentiation, how exactly do you go about differentiating a series? We've never been taught that and I'm an attentive math student.

So I can't even start (and this is the last question), this is frustrating me...

Cheers

Last edited: Oct 1, 2009
2. Oct 1, 2009

### dalle

don't panic. all you have is a finite sum.
the formula $$(f+g)' = f' + g'$$ does work for finite sums!
the only problem left is finding the right trigonometric identity to use!

3. Oct 1, 2009

### caesius

But the function is w.r.t x, so my first thought was just to drop the x off. But that's wrong...

I don't get it, what's f and what's g?

Thanks for somewhere to start though

4. Oct 2, 2009

### dalle

dear ceasius, you are given a sequence of functions
$$s_1 = \frac{4}{\pi} \sin x , s_2 =\frac{4}{\pi} (\sin x + \frac{\sin (3 x}{3})), s_3 = \frac{4}{\pi} ( \sin x + \frac{\sin (3x)}{3} + \frac{\sin(5x)}{5}),..$$
your task is to prove that
$$\frac{d s_n(x)}{dx} =\frac{sin(2nx)}{\pi \sin x} ,x \ne \pi l$$
for every n. the way to do this is by using induction on n.
the case n=1 is simple
$$\frac{4}{\pi} \frac{d \sin x}{dx}=\frac{4}{\pi} \cos x$$
there is a trigonomtric formula that says
$$\sin(a) \cos(b) = \frac{1}{2}(\sin(a-b) + \sin (a+b))$$
apply this formula on $$\sin x \cos x$$, check for which x $$\sin x = 0$$ and you will get the right result. It's a good practice to try n=2
$$\frac{d s_2(x)}{dx}=\frac{4}{\pi}\frac{d(\sin(x)+\frac{\sin(3x)}{3})}{dx}=\frac{4}{\pi}(\frac{d\(sin(x)}{dx}+\frac{d\sin(3x)}{3 dx})=\frac{2 \sin(2x)}{\pi \sin x}+ \frac{4}{\pi} \cos(3x)$$
now apply the trigonometric formula on $$\sin x \cos(3x)$$ and you will get the right result.
I hope it's clear what's left to do

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