Proving the Derivative of a Fourier Series Using Induction

[caesius]

We were given in a previous question,

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

Homework Statement

Show that
<br /> s&#039;_{N}(x) = \frac{2sin(2Nx)}{\pi sinx}, x \neq l\pi<br />

and

<br /> s&#039;_{N}(x) = \frac{4N}{\pi}(-1)^l, x = l\pi<br />

where l is any integer.

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

 

[dalle]

don't panic. all you have is a finite sum.
the formula (f+g)&#039; = f&#039; + g&#039; does work for finite sums!
the only problem left is finding the right trigonometric identity to use!

 

[caesius]

don't panic. all you have is a finite sum.
the formula (f+g)&#039; = f&#039; + g&#039; does work for finite sums!
the only problem left is finding the right trigonometric identity to use!

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

 

[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