let f(x)=log(cht) developp f at fourier serie


Question and Answers Forum

All Questions Topic List
Relation and Functions Questions
Previous in All Question Next in All Question
Previous in Relation and Functions Next in Relation and Functions
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
	$f(x)=log(cht)⇒f^′ (x)=((sht)/(cht))=((e^t −e^(−t) )/(e^t +e^t )) =_(e^t =z) ((z−z^(−1) )/(z+z^(−1) ))=((z^2 −1)/(z^2 +1))=(z^2 −1)Σ_(n=0) ^∞ (−1)^n z^(2n) =Σ_(n=0) ^∞ (−1)^n z^(2n+2) −Σ_(n=0) ^∞ (−1)^n z^(2n) =Σ_(n=0) ^∞ (−1)^n e^((2n+2)t) −Σ_(n=0) ^∞ (−1)^n e^(2nt) =Σ_(n=0) ^∞ (−1)^n Σ_(p=0) ^∞ (({(2n+2)t}^p )/(p!))−Σ_(n=0) ^∞ (−1)^n Σ_(p=0) ^∞ (((2nt)^p )/(p!)) =Σ_(n=0) ^∞ (−1)^n Σ_(p=0) ^∞ (((2n+2)^p )/(p!))t^p −Σ_(n=0) ^∞ (−1)^n Σ_(p=0) ^∞ (((2n)^p )/(p!))t^p$
	$f (x) = \log (cht) \Rightarrow f^{^{'}} (x) = \frac{sht}{cht} = \frac{e^{t} - e^{- t}}{e^{t} + e^{t}}$ $=_{e^{t} = z} \frac{z - z^{- 1}}{z + z^{- 1}} = \frac{z^{2} - 1}{z^{2} + 1} = (z^{2} - 1) \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{2 n}$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{2 n + 2} - \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{2 n}$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} e^{(2 n + 2) t} - \sum_{n = 0}^{\infty} {(- 1)}^{n} e^{2 nt}$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \sum_{p = 0}^{\infty} \frac{{(2 n + 2) t}^{p}}{p!} - \sum_{n = 0}^{\infty} {(- 1)}^{n} \sum_{p = 0}^{\infty} \frac{{(2 nt)}^{p}}{p!}$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \sum_{p = 0}^{\infty} \frac{{(2 n + 2)}^{p}}{p!} t^{p} - \sum_{n = 0}^{\infty} {(- 1)}^{n} \sum_{p = 0}^{\infty} \frac{{(2 n)}^{p}}{p!} t^{p}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum