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Question Number 126440 by mnjuly1970 last updated on 20/Dec/20
... advanced calculus... prove that ::: Π_(n=2) ^∞ (1+(1/n^4 ) )=((cosh((√2) π)−cos((√2) π))/(4π^2 ))
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{advanced}\:\:{calculus}… \\ $$$$\:\:\:\:{prove}\:\:{that}\:::: \\ $$$$\:\:\:\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:\right)=\frac{{cosh}\left(\sqrt{\mathrm{2}}\:\pi\right)−{cos}\left(\sqrt{\mathrm{2}}\:\pi\right)}{\mathrm{4}\pi^{\mathrm{2}} } \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 20/Dec/20
Π_(n=2) ^∞ (1+(i/n^2 ))Π_(n=2) ^∞ (1−(i/n^2 )) =(1/((1−i^(3/2) )(1−(√i))))Π_(n=1) ^∞ (1−((((√i^3 ))^2 )/n^2 ))Π_(n=1) ^∞ (1−(((√i)/n))^2 ) =(1/(−(√i)(i+1))).((sin(i^(3/2) π))/(πi^(3/2) )).((sin((√i)π))/(π(√i))) =(1/(4π^2 ))(cosh((√2)π)−cos((√2)π)) (After simplifying)
$$\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{{i}}{{n}^{\mathrm{2}} }\right)\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{{i}}{{n}^{\mathrm{2}} }\right) \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{1}−{i}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)\left(\mathrm{1}−\sqrt{{i}}\right)}\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\frac{\left(\sqrt{{i}^{\mathrm{3}} }\right)^{\mathrm{2}} }{{n}^{\mathrm{2}} }\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}−\left(\frac{\sqrt{{i}}}{{n}}\right)^{\mathrm{2}} \right) \\ $$$$=\frac{\mathrm{1}}{−\sqrt{{i}}\left({i}+\mathrm{1}\right)}.\frac{{sin}\left({i}^{\frac{\mathrm{3}}{\mathrm{2}}} \pi\right)}{\pi{i}^{\frac{\mathrm{3}}{\mathrm{2}}} }.\frac{{sin}\left(\sqrt{{i}}\pi\right)}{\pi\sqrt{{i}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi^{\mathrm{2}} }\left({cosh}\left(\sqrt{\mathrm{2}}\pi\right)−{cos}\left(\sqrt{\mathrm{2}}\pi\right)\right)\:\:\:\:\left({After}\:{simplifying}\right) \\ $$
Commented by mnjuly1970 last updated on 20/Dec/20
thank you..
$${thank}\:{you}.. \\ $$
Answered by mathmax by abdo last updated on 21/Dec/20
((sinz)/z)=Π_(n=1) ^∞ (1−(z^2 /(n^2 π^2 ))) so Π_(n=2) ^∞ (1+(1/n^4 ))=(1/2)Π_(n=1) ^∞ (1+(1/n^4 ))=(1/2)Π_(n=1) ^∞ (1+(i/n^2 ))(1−(i/n^2 )) (z/π)=t ⇒((sin(πt))/(πt)) =Π_(n=1) ^∞ (1−(t^2 /n^2 )) ⇒Π_(n=1) ^∞ (1−(i/n^2 )) =Π_(n=1) ^∞ (1−((((√i))^2 )/n^2 ))=((sin(π(√i)))/(π(√i))) also Π_(n=1) ^∞ (1+(i/n^2 ))=Π_(n=1) ^∞ (1−((((√(−i)))^2 )/n^2 )) =((sin(π(√(−i))))/(π(√(−i)))) ⇒ Π_(n=2) ^∞ (1+(1/n^4 ))=(1/2)((sin(π(√i))sin(π(√(−i))))/(π^2 (√(−i^2 )))) =((sin(π(√i))sin(π(√(−i))))/(2π^2 )) sin(π(√i)) =((e^(i(π(√i))) −e^(−i(π(√i))) )/(2i)) =(1/(2i)){ e^(i(π((1/( (√2)))+(i/( (√2))))) −e^(−iπ((1/( (√2)))+(i/( (√2))))) } =(1/(2i))( e^(−(π/( (√2)))) (cos((π/( (√2))))+isin((π/( (√2))))−e^(π/( (√2))) (cos((π/( (√2))))−isin((π/( (√2))))) =(1/(2i))(cos((π/( (√2))))(e^(−(π/( (√2)))) −e^(π/( (√2))) ) +isin((π/( (√2))))(e^(−(π/( (√2)))) +e^(π/( (√2))) )) =−cos((π/( (√2))))sh((π/( (√2)))) +sin((π/( (√2))))ch((π/( (√2)))) sin(π(√(−i))) =((e^(i(π(√(−i)))) −e^(−iπ(√(−i))) )/(2i)) =((e^(iπ((1/( (√2)))−(i/( (√2))))) −e^(−iπ((1/( (√2)))−(i/( (√2))))) )/(2i)) =((e^(π/( (√2))) (cos((π/( (√2))))+isin((π/( (√2)))))−e^(−(π/( (√2)))) (cos((π/( (√2))))−isin((π/( (√2))))))/(2i)) =(1/(2i)) cos((π/( (√2))))(e^(π/( (√2))) −e^(−(π/( (√2)))) )+sin((π/( (√2))))((e^(π/( (√2))) +e^(−(π/( (√2)))) )/2) =cos((π/( (√2))))sh((π/( (√2))))+sin((π/( (√2))))ch((π/( (√2)))) rest to collect the calculus...
$$\frac{\mathrm{sinz}}{\mathrm{z}}=\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}−\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right)\:\mathrm{so} \\ $$$$\prod_{\mathrm{n}=\mathrm{2}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{4}} }\right)=\frac{\mathrm{1}}{\mathrm{2}}\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{4}} }\right)=\frac{\mathrm{1}}{\mathrm{2}}\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{i}}{\mathrm{n}^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{\mathrm{i}}{\mathrm{n}^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{z}}{\pi}=\mathrm{t}\:\Rightarrow\frac{\mathrm{sin}\left(\pi\mathrm{t}\right)}{\pi\mathrm{t}}\:=\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}−\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }\right)\:\Rightarrow\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}−\frac{\mathrm{i}}{\mathrm{n}^{\mathrm{2}} }\right) \\ $$$$=\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(\mathrm{1}−\frac{\left(\sqrt{\mathrm{i}}\right)^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }\right)=\frac{\mathrm{sin}\left(\pi\sqrt{\mathrm{i}}\right)}{\pi\sqrt{\mathrm{i}}}\:\mathrm{also} \\ $$$$\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{i}}{\mathrm{n}^{\mathrm{2}} }\right)=\prod_{\mathrm{n}=\mathrm{1}} ^{\infty} \left(\mathrm{1}−\frac{\left(\sqrt{−\mathrm{i}}\right)^{\mathrm{2}} }{\mathrm{n}^{\mathrm{2}} }\right)\:=\frac{\mathrm{sin}\left(\pi\sqrt{−\mathrm{i}}\right)}{\pi\sqrt{−\mathrm{i}}}\:\Rightarrow \\ $$$$\prod_{\mathrm{n}=\mathrm{2}} ^{\infty} \left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{4}} }\right)=\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{sin}\left(\pi\sqrt{\mathrm{i}}\right)\mathrm{sin}\left(\pi\sqrt{−\mathrm{i}}\right)}{\pi^{\mathrm{2}} \sqrt{−\mathrm{i}^{\mathrm{2}} }}\:=\frac{\mathrm{sin}\left(\pi\sqrt{\mathrm{i}}\right)\mathrm{sin}\left(\pi\sqrt{−\mathrm{i}}\right)}{\mathrm{2}\pi^{\mathrm{2}} } \\ $$$$\mathrm{sin}\left(\pi\sqrt{\mathrm{i}}\right)\:=\frac{\mathrm{e}^{\mathrm{i}\left(\pi\sqrt{\mathrm{i}}\right)} −\mathrm{e}^{−\mathrm{i}\left(\pi\sqrt{\mathrm{i}}\right)} }{\mathrm{2i}}\:=\frac{\mathrm{1}}{\mathrm{2i}}\left\{\:\mathrm{e}^{\mathrm{i}\left(\pi\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)\right.} −\mathrm{e}^{−\mathrm{i}\pi\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2i}}\left(\:\mathrm{e}^{−\frac{\pi}{\:\sqrt{\mathrm{2}}}} \:\:\left(\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{isin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)−\mathrm{e}^{\frac{\pi}{\:\sqrt{\mathrm{2}}}} \:\:\left(\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)−\mathrm{isin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\right)\right.\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2i}}\left(\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\left(\mathrm{e}^{−\frac{\pi}{\:\sqrt{\mathrm{2}}}} \:\:−\mathrm{e}^{\frac{\pi}{\:\sqrt{\mathrm{2}}}} \right)\:+\mathrm{isin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\left(\mathrm{e}^{−\frac{\pi}{\:\sqrt{\mathrm{2}}}} \:\:+\mathrm{e}^{\frac{\pi}{\:\sqrt{\mathrm{2}}}} \right)\right) \\ $$$$=−\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{sh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\:+\mathrm{sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{ch}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\mathrm{sin}\left(\pi\sqrt{−\mathrm{i}}\right)\:=\frac{\mathrm{e}^{\mathrm{i}\left(\pi\sqrt{\left.−\mathrm{i}\right)}\right.} −\mathrm{e}^{−\mathrm{i}\pi\sqrt{−\mathrm{i}}} }{\mathrm{2i}}\:=\frac{\mathrm{e}^{\mathrm{i}\pi\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)} −\mathrm{e}^{−\mathrm{i}\pi\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)} }{\mathrm{2i}} \\ $$$$=\frac{\mathrm{e}^{\frac{\pi}{\:\sqrt{\mathrm{2}}}} \:\left(\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{isin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\right)−\mathrm{e}^{−\frac{\pi}{\:\sqrt{\mathrm{2}}}} \left(\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)−\mathrm{isin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\right)}{\mathrm{2i}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2i}}\:\:\:\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\left(\mathrm{e}^{\frac{\pi}{\:\sqrt{\mathrm{2}}}} \:\:−\mathrm{e}^{−\frac{\pi}{\:\sqrt{\mathrm{2}}}} \right)+\mathrm{sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\frac{\mathrm{e}^{\frac{\pi}{\:\sqrt{\mathrm{2}}}} +\mathrm{e}^{−\frac{\pi}{\:\sqrt{\mathrm{2}}}} }{\mathrm{2}} \\ $$$$=\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{sh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{ch}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\mathrm{rest}\:\mathrm{to}\:\mathrm{collect}\:\mathrm{the}\:\mathrm{calculus}… \\ $$

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