f-z-j-z-z-2-j-2-z-0-lim-z-f-z-

Question Number 213555 by issac last updated on 08/Nov/24

f(z)=Σ_(j=−∞) ^∞ (z/(z^2 +j^2 )) , z∈(0,∞) lim_(z→∞) f(z)=??

$${f}\left({z}\right)=\underset{{j}=−\infty} {\overset{\infty} {\sum}}\:\frac{{z}}{{z}^{\mathrm{2}} +{j}^{\mathrm{2}} }\:,\:{z}\in\left(\mathrm{0},\infty\right) \\ $$$$\underset{{z}\rightarrow\infty} {\mathrm{lim}}\:{f}\left({z}\right)=?? \\ $$

Answered by lepuissantcedricjunior last updated on 09/Nov/24

f(z)=Σ_(j=−∞) ^(+∞) (z/(z^2 +j^2 ))=2Σ_(j=0) ^(+∞) (z/(z^2 +j^2 )) lim_(z→+∞) f(z)=lim_(n→+∞) (lim_(n→+∞) (2/z)Σ_(j=0) ^n (1/(1+((j/z))^2 ))) =lim_(z→+∞) (2∫_0 ^1 (dx/(1+x^2 ))) =lim_(z→+∞) ((𝛑/2)) =>lim_(z→+∞) f(z)=(𝛑/2)

$$\boldsymbol{{f}}\left(\boldsymbol{{z}}\right)=\underset{\boldsymbol{{j}}=−\infty} {\overset{+\infty} {\sum}}\frac{\boldsymbol{{z}}}{\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{j}}^{\mathrm{2}} }=\mathrm{2}\underset{\boldsymbol{{j}}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\boldsymbol{{z}}}{\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{j}}^{\mathrm{2}} } \\ $$$$\underset{\boldsymbol{{z}}\rightarrow+\infty} {\mathrm{lim}}\boldsymbol{{f}}\left(\boldsymbol{{z}}\right)=\underset{\boldsymbol{{n}}\rightarrow+\infty} {\mathrm{lim}}\left(\underset{\boldsymbol{{n}}\rightarrow+\infty} {\mathrm{lim}}\frac{\mathrm{2}}{\boldsymbol{{z}}}\underset{\boldsymbol{{j}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\boldsymbol{{j}}}{\boldsymbol{{z}}}\right)^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{\boldsymbol{{z}}\rightarrow+\infty} {\mathrm{lim}}\left(\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\boldsymbol{{dx}}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\underset{\boldsymbol{{z}}\rightarrow+\infty} {\mathrm{lim}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}\right) \\ $$$$=>\underset{\boldsymbol{{z}}\rightarrow+\infty} {\mathrm{lim}}\boldsymbol{{f}}\left(\boldsymbol{{z}}\right)=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$

Commented by mr W last updated on 09/Nov/24

Σ_(j=−∞) ^(+∞) (z/(z^2 +j^2 )) ≠ 2Σ_(j=0) ^(+∞) (z/(z^2 +j^2 ))

$$\underset{\boldsymbol{{j}}=−\infty} {\overset{+\infty} {\sum}}\frac{\boldsymbol{{z}}}{\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{j}}^{\mathrm{2}} }\:\neq\:\mathrm{2}\underset{\boldsymbol{{j}}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\boldsymbol{{z}}}{\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{j}}^{\mathrm{2}} } \\ $$

Answered by Berbere last updated on 10/Nov/24

f(z)=(1/z)+Σ_(j≥1) ((2z)/(z^2 +n^2 )) cotan(πz)=(1/z)+2Σ(z/(z^2 −n^2 ))⇒coan(it)=(1/(it))−2Σ((it)/(t^2 +n^2 )) ⇒icotan(it)=(1/t)+2Σ(t/(t^2 +n^2 )) =cotanh(z)=f(z)⇒lim_(z→∞) f(z)=1

$${f}\left({z}\right)=\frac{\mathrm{1}}{{z}}+\underset{{j}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{2}{z}}{{z}^{\mathrm{2}} +{n}^{\mathrm{2}} } \\ $$$${cotan}\left(\pi{z}\right)=\frac{\mathrm{1}}{{z}}+\mathrm{2}\Sigma\frac{{z}}{{z}^{\mathrm{2}} −{n}^{\mathrm{2}} }\Rightarrow{coan}\left({it}\right)=\frac{\mathrm{1}}{{it}}−\mathrm{2}\Sigma\frac{{it}}{{t}^{\mathrm{2}} +{n}^{\mathrm{2}} } \\ $$$$\Rightarrow{icotan}\left({it}\right)=\frac{\mathrm{1}}{{t}}+\mathrm{2}\Sigma\frac{{t}}{{t}^{\mathrm{2}} +{n}^{\mathrm{2}} } \\ $$$$={cotanh}\left({z}\right)={f}\left({z}\right)\Rightarrow\underset{{z}\rightarrow\infty} {\mathrm{lim}}{f}\left({z}\right)=\mathrm{1} \\ $$

Commented by issac last updated on 10/Nov/24

$$\left.{thx}!\:\::\right) \\ $$

Commented by York12 last updated on 11/Nov/24

What′s your instgram or facebook account , please sir

$$ \\ $$$$\mathrm{What}'\mathrm{s}\:\mathrm{your}\:\mathrm{instgram}\:\mathrm{or}\:\mathrm{facebook}\:\mathrm{account} \\ $$$$, \\ $$$$\mathrm{please}\:\mathrm{sir} \\ $$

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