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

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{log}\left(\mathrm{cht}\right) \\ $$$$\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$

Answered by Olaf_Thorendsen last updated on 28/Jun/21

f is not periodic !

$${f}\:\mathrm{is}\:\mathrm{not}\:\mathrm{periodic}\:! \\ $$

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

sorry developp f at integr serie

$$\mathrm{sorry}\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie} \\ $$

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

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{log}\left(\mathrm{cht}\right)\Rightarrow\mathrm{f}^{'} \left(\mathrm{x}\right)=\frac{\mathrm{sht}}{\mathrm{cht}}=\frac{\mathrm{e}^{\mathrm{t}} −\mathrm{e}^{−\mathrm{t}} }{\mathrm{e}^{\mathrm{t}} \:+\mathrm{e}^{\mathrm{t}} } \\ $$$$=_{\mathrm{e}^{\mathrm{t}} \:=\mathrm{z}} \:\:\frac{\mathrm{z}−\mathrm{z}^{−\mathrm{1}} }{\mathrm{z}+\mathrm{z}^{−\mathrm{1}} }=\frac{\mathrm{z}^{\mathrm{2}} −\mathrm{1}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}=\left(\mathrm{z}^{\mathrm{2}} −\mathrm{1}\right)\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{z}^{\mathrm{2n}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{z}^{\mathrm{2n}+\mathrm{2}} −\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{z}^{\mathrm{2n}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{\left(\mathrm{2n}+\mathrm{2}\right)\mathrm{t}} \:−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{\mathrm{2nt}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \sum_{\mathrm{p}=\mathrm{0}} ^{\infty} \:\frac{\left\{\left(\mathrm{2n}+\mathrm{2}\right)\mathrm{t}\right\}^{\mathrm{p}} }{\mathrm{p}!}−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\sum_{\mathrm{p}=\mathrm{0}} ^{\infty} \:\frac{\left(\mathrm{2nt}\right)^{\mathrm{p}} }{\mathrm{p}!} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\sum_{\mathrm{p}=\mathrm{0}} ^{\infty} \:\frac{\left(\mathrm{2n}+\mathrm{2}\right)^{\mathrm{p}} }{\mathrm{p}!}\mathrm{t}^{\mathrm{p}} \:−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\sum_{\mathrm{p}=\mathrm{0}} ^{\infty} \:\frac{\left(\mathrm{2n}\right)^{\mathrm{p}} }{\mathrm{p}!}\mathrm{t}^{\mathrm{p}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com