Recent content by poconnel

the actual identity is:

\sum\frac{\rho}{n}\ast\rho^{n}sin\sigma=\frac{1}{2}\asttan^{-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?

I was looking or hints on the proof of this identity

any help on my problem?

maybe this will work

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}

see attached

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

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