How can I derive the identity for this special Fourier series?

[poconnel]

can anyone give me a hint on deriving this identity:
sum(((p^n))/n)*sin(n*Q)= atan(2*p*sin(q)/(1-p^2)

n = 1 to infinity
p and q are polar coordinates

 

[poconnel]

$$\sum$$p$$^{n}$$sin(n*$$\o$$)=$$\\\\atan(\frac{2p*sin(\o)}{(1-p^{2})})$$

 

[Nchatooa]

Having difficulty with this Fourier series function. I know that it is an odd function...can anyone help me

 

[poconnel]

see attached

 

[LCKurtz]

Having difficulty with this Fourier series function. I know that it is an odd function...can anyone help me

It is poor practice to hijack a thread about one question by posing another question. You should start a new thread with your question.

 

[LCKurtz]

see attached

The function is periodic with period $2\pi$. Why are you integrating over a range of $3\pi$?

 

[poconnel]

The function is periodic with period $2\pi$. Why are you integrating over a range of $3\pi$?

I believe even though sin and cos are periodic over 2 pi a Fourier series is a representation over the given period and so the integrals must be taken over the period given. I forgot to multiply the integrals by $$\frac{1}{3\pi}$$

 

[LCKurtz]

No. You are given a function on $(-\pi,\pi)$. That function is extended periodically. The fact that you were asked to draw its graph over an interval of length $3\pi$ doesn't magically make the function periodic of that period.

The problem asks you to determine whether the given function is even, odd, or neither. If it is one of the first two, that suggests using a half-range expansion.

You are, as the saying goes, barking up the wrong tree.

 

[poconnel]

maybe this will work

 

[poconnel]

any help on my problem?

 

[LCKurtz]

What do you need help on? That last bmp looks like you have calculated the Fourier series correctly. Do you see what the answer to the original question is?

 

[poconnel]

$$\sum$$p$$^{n}$$sin(n*$$\o$$)=$$\\\\atan(\frac{2p*sin(\o)}{(1-p^{2})})$$

I was looking or hints on the proof of this identity

 

[poconnel]

$$\sum$$$$\frac{\rho}{n}$$$$\ast$$$$\rho$$$$^{n}$$sin$$\sigma$$=$$\frac{1}{2}$$$$\ast$$tan$$^{-1}$$$$\left($$$$\frac{2*\rho*sin(\sigma)}{1-\rho^{2}}$$$$\right)$$$$\right)$$

this is a Fourier series listed in most references but I can't derive it. Any help?

 

[poconnel]

can anyone give me a hint on deriving this identity:
sum(((p^n))/n)*sin(n*Q)= atan(2*p*sin(q)/(1-p^2)

n = 1 to infinity
p and q are polar coordinates

the actual identity is:

 

[Mute]

the actual identity is:

I would try the following: Write the left hand side as

$$\mbox{Im}\left(\sum_{n=1}^\infty \frac{z^{2n-1}}{2n-1}\right),$$

where "Im" means take the imaginary part of the result, and I've written $z = p \exp[i \phi]$.

Take a derivative of this with respect to z, evaluate the series, then integrate the result with respect to z. Impose the initial condition z = 0 the sum = 0 to fix the arbitrary constant, then write z in terms of p and phi again and finally take the imaginary part of the result.