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Question Number 140996 by mnjuly1970 last updated on 14/May/21
$Evaluation of :: Ω :=Σ_(n=0) ^∞ (((−1)^n )/(1+n^2 )) solution:: Ω:=1+Σ_(n=1) ^∞ (((−1)^n )/(n^2 −i^2 )) =(1/(2i)){Σ_(n=1) ^∞ (((−1)^n )/(n−i))−(((−1)^n )/(n+i))} :=1+(1/(2i)) (Φ−Ψ) where Φ:=Σ_(n=1) ^∞ (((−1)^n )/(n−i)) and Ψ :=Σ_(n=1) ^∞ (((−1)^n )/(n+i)) Φ:=Σ_(n=1) ^∞ (1/(2n−k)) −Σ_(n=1) ^∞ (1/(2n−1−i)) :=(1/2){Σ_(n=1) ^∞ (1/(n−(i/2)))−Σ_(n=1) ^∞ (1/(n−((1+i)/2)))} :=(1/2){ψ(1−((1+i)/2))−ψ(1−(i/2))} :=(1/2)(ψ(((1−i)/2))−ψ(1−(i/2))).... Ψ:=Σ_(n=1) ^∞ (((−1)^n )/(n+i)) =Σ_(n=1) ^∞ (1/(2n+i))−Σ_(n=1) ^∞ (1/(2n−1+i)) :=(1/2){Σ_(n=1) ^∞ (1/(n+(i/2)))−Σ_(n=1) ^∞ (1/(n+((i−1)/2)))} :=(1/2)(ψ(((1+i)/2))−ψ(1+(i/2))) .... Φ−Ψ:=(1/2){ψ(((1−i)/2))−ψ(((1+i)/2))} +(1/2){ψ(1+(i/2))−ψ(1−(i/2))} :=(1/2)(−πcotπ(((1−i)/2)))+(1/2)((2/i)−πcot(π(i/2))) :=−(π/2)tan(((πi)/2))−(π/2)cot(((πi)/2))−i :=−i−(π/(sin(πi)))=−i−((2iπ)/(e^(−π) −e^π )) :=−i+πicsch(π) .... Ω :=1+(1/(2i))(−i+πicsch(π))=(1/2)+(π/2) csch(π) ... Ω:=(1/2)+(π/2) csch(π)....✓✓✓$
$E v a l u a t i o n o f :: Ω := \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{1 + n^{2}}$ $s o l u t i o n ::$ $Ω := 1 + \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n^{2} - i^{2}} = \frac{1}{2 i} {\underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n - i} - \frac{{(- 1)}^{n}}{n + i}}$ $:= 1 + \frac{1}{2 i} (Φ - Ψ) w h e r e Φ := \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n - i}$ $a n d Ψ := \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + i}$ $Φ := \underset{n = 1}{\sum^{\infty}} \frac{1}{2 n - k} - \underset{n = 1}{\sum^{\infty}} \frac{1}{2 n - 1 - i}$ $:= \frac{1}{2} {\underset{n = 1}{\sum^{\infty}} \frac{1}{n - \frac{i}{2}} - \underset{n = 1}{\sum^{\infty}} \frac{1}{n - \frac{1 + i}{2}}}$ $:= \frac{1}{2} {ψ (1 - \frac{1 + i}{2}) - ψ (1 - \frac{i}{2})}$ $:= \frac{1}{2} (ψ (\frac{1 - i}{2}) - ψ (1 - \frac{i}{2})) . . . .$ $Ψ := \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n}}{n + i} = \underset{n = 1}{\sum^{\infty}} \frac{1}{2 n + i} - \underset{n = 1}{\sum^{\infty}} \frac{1}{2 n - 1 + i}$ $:= \frac{1}{2} {\underset{n = 1}{\sum^{\infty}} \frac{1}{n + \frac{i}{2}} - \underset{n = 1}{\sum^{\infty}} \frac{1}{n + \frac{i - 1}{2}}}$ $:= \frac{1}{2} (ψ (\frac{1 + i}{2}) - ψ (1 + \frac{i}{2})) . . . .$ $Φ - Ψ := \frac{1}{2} {ψ (\frac{1 - i}{2}) - ψ (\frac{1 + i}{2})}$ $+ \frac{1}{2} {ψ (1 + \frac{i}{2}) - ψ (1 - \frac{i}{2})}$ $:= \frac{1}{2} (- π c o t π (\frac{1 - i}{2})) + \frac{1}{2} (\frac{2}{i} - π c o t (π \frac{i}{2}))$ $:= - \frac{π}{2} t a n (\frac{π i}{2}) - \frac{π}{2} c o t (\frac{π i}{2}) - i$ $:= - i - \frac{π}{s i n (π i)} = - i - \frac{2 i π}{e^{- π} - e^{π}}$ $:= - i + π i c s c h (π) . . . .$ $Ω := 1 + \frac{1}{2 i} (- i + π i c s c h (π)) = \frac{1}{2} + \frac{π}{2} c s c h (π) . . .$ $Ω := \frac{1}{2} + \frac{π}{2} c s c h (π) . . . . ✓ ✓ ✓$
	Answered by qaz last updated on 14/May/21

	$\underset{n = 0}{\sum^{\infty}} \frac{{(- x)}^{n}}{1 + n^{2}} = \frac{1}{1 + {(x D)}^{2}} (\frac{1}{1 + x}) = y$ $\Rightarrow y + (x^{2} D^{2} + x D) y = \frac{1}{1 + x}$ $\Rightarrow y^{″} + \frac{1}{x} y^{'} + \frac{1}{x^{2}} y = \frac{1}{x^{2} (1 + x)}$ $s o l v e t h i s D E . . . .$
	Commented by mnjuly1970 last updated on 14/May/21

	$t h a n k y o u m r q a z . . .$
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