let-f-x-log-cht-developp-f-at-fourier-serie-

Question Number 144699 by mathmax by abdo last updated on 28/Jun/21

let f (x) = \log (cht)

developp f at fourier serie

Answered by Olaf_Thorendsen last updated on 28/Jun/21

f is not periodic!

Answered by mathmax by abdo last updated on 28/Jun/21

sorry developp f at integr serie

Answered by mathmax by abdo last updated on 29/Jun/21

f (x) = \log (cht) \Rightarrow f^{^{'}} (x) = \frac{sht}{cht} = \frac{e^{t} - e^{- t}}{e^{t} + e^{t}}

=_{e^{t} = z} \frac{z - z^{- 1}}{z + z^{- 1}} = \frac{z^{2} - 1}{z^{2} + 1} = (z^{2} - 1) \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{2 n}

= \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{2 n + 2} - \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{2 n}

= \sum_{n = 0}^{\infty} {(- 1)}^{n} e^{(2 n + 2) t} - \sum_{n = 0}^{\infty} {(- 1)}^{n} e^{2 nt}

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \sum_{p = 0}^{\infty} \frac{{(2 n + 2) t}^{p}}{p!} - \sum_{n = 0}^{\infty} {(- 1)}^{n} \sum_{p = 0}^{\infty} \frac{{(2 nt)}^{p}}{p!}

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \sum_{p = 0}^{\infty} \frac{{(2 n + 2)}^{p}}{p!} t^{p} - \sum_{n = 0}^{\infty} {(- 1)}^{n} \sum_{p = 0}^{\infty} \frac{{(2 n)}^{p}}{p!} t^{p}