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Question Number 126440 by mnjuly1970 last updated on 20/Dec/20

$$...advancedcalculus...$$$$provethat:::$$$$\underset{n=2}{\prod ^{\mathrm{\infty }}}(1+\frac{1}{n^4})=\frac{cosh\left(\sqrt{2}\pi \right)-cos\left(\sqrt{2}\pi \right)}{4{\pi }^2}$$$$$$

Answered by Dwaipayan Shikari last updated on 20/Dec/20

$$\underset{n=2}{\prod ^{\mathrm{\infty }}}(1+\frac{i}{n^2})\underset{n=2}{\prod ^{\mathrm{\infty }}}(1-\frac{i}{n^2})$$$$=\frac{1}{(1-i^{\frac{3}{2}})(1-\sqrt{i})}\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-\frac{{\left(\sqrt{i^3}\right)}^2}{n^2})\underset{n=1}{\prod ^{\mathrm{\infty }}}(1-{\left(\frac{\sqrt{i}}{n}\right)}^2)$$$$=\frac{1}{-\sqrt{i}(i+1)}.\frac{sin\left(i^{\frac{3}{2}}\pi \right)}{\pi i^{\frac{3}{2}}}.\frac{sin\left(\sqrt{i}\pi \right)}{\pi \sqrt{i}}$$$$=\frac{1}{4{\pi }^2}(cosh\left(\sqrt{2}\pi \right)-cos\left(\sqrt{2}\pi \right))\left(Aftersimplifying\right)$$

Commented by mnjuly1970 last updated on 20/Dec/20

$$thankyou..$$

Answered by mathmax by abdo last updated on 21/Dec/20

$$\frac{\mathrm{sinz}}{\mathrm{z}}=\prod _{\mathrm{n}=1}^{\mathrm{\infty }}(1-\frac{{\mathrm{z}}^2}{{\mathrm{n}}^2{\pi }^2})\mathrm{so}$$$$\prod _{\mathrm{n}=2}^{\mathrm{\infty }}(1+\frac{1}{{\mathrm{n}}^4})=\frac{1}{2}\prod _{\mathrm{n}=1}^{\mathrm{\infty }}(1+\frac{1}{{\mathrm{n}}^4})=\frac{1}{2}\prod _{\mathrm{n}=1}^{\mathrm{\infty }}(1+\frac{\mathrm{i}}{{\mathrm{n}}^2})(1-\frac{\mathrm{i}}{{\mathrm{n}}^2})$$$$\frac{\mathrm{z}}{\pi }=\mathrm{t}\Rightarrow \frac{\sin \left(\pi \mathrm{t}\right)}{\pi \mathrm{t}}=\prod _{\mathrm{n}=1}^{\mathrm{\infty }}(1-\frac{{\mathrm{t}}^2}{{\mathrm{n}}^2})\Rightarrow \prod _{\mathrm{n}=1}^{\mathrm{\infty }}(1-\frac{\mathrm{i}}{{\mathrm{n}}^2})$$$$=\prod _{\mathrm{n}=1}^{\mathrm{\infty }}(1-\frac{{\left(\sqrt{\mathrm{i}}\right)}^2}{{\mathrm{n}}^2})=\frac{\sin \left(\pi \sqrt{\mathrm{i}}\right)}{\pi \sqrt{\mathrm{i}}}\mathrm{also}$$$$\prod _{\mathrm{n}=1}^{\mathrm{\infty }}(1+\frac{\mathrm{i}}{{\mathrm{n}}^2})=\prod _{\mathrm{n}=1}^{\mathrm{\infty }}(1-\frac{{\left(\sqrt{-\mathrm{i}}\right)}^2}{{\mathrm{n}}^2})=\frac{\sin \left(\pi \sqrt{-\mathrm{i}}\right)}{\pi \sqrt{-\mathrm{i}}}\Rightarrow$$$$\prod _{\mathrm{n}=2}^{\mathrm{\infty }}(1+\frac{1}{{\mathrm{n}}^4})=\frac{1}{2}\frac{\sin \left(\pi \sqrt{\mathrm{i}}\right)\sin \left(\pi \sqrt{-\mathrm{i}}\right)}{{\pi }^2\sqrt{-{\mathrm{i}}^2}}=\frac{\sin \left(\pi \sqrt{\mathrm{i}}\right)\sin \left(\pi \sqrt{-\mathrm{i}}\right)}{2{\pi }^2}$$$$\sin \left(\pi \sqrt{\mathrm{i}}\right)=\frac{{\mathrm{e}}^{\mathrm{i}\left(\pi \sqrt{\mathrm{i}}\right)}-{\mathrm{e}}^{-\mathrm{i}\left(\pi \sqrt{\mathrm{i}}\right)}}{2\mathrm{i}}=\frac{1}{2\mathrm{i}}\{{\mathrm{e}}^{\mathrm{i}(\pi (\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}})}-{\mathrm{e}}^{-\mathrm{i}\pi (\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}})}\}$$$$=\frac{1}{2\mathrm{i}}({\mathrm{e}}^{-\frac{\pi }{\sqrt{2}}}(\cos \left(\frac{\pi }{\sqrt{2}}\right)+\mathrm{isin}\left(\frac{\pi }{\sqrt{2}}\right)-{\mathrm{e}}^{\frac{\pi }{\sqrt{2}}}(\cos \left(\frac{\pi }{\sqrt{2}}\right)-\mathrm{isin}\left(\frac{\pi }{\sqrt{2}}\right))$$$$=\frac{1}{2\mathrm{i}}(\cos \left(\frac{\pi }{\sqrt{2}}\right)({\mathrm{e}}^{-\frac{\pi }{\sqrt{2}}}-{\mathrm{e}}^{\frac{\pi }{\sqrt{2}}})+\mathrm{isin}\left(\frac{\pi }{\sqrt{2}}\right)({\mathrm{e}}^{-\frac{\pi }{\sqrt{2}}}+{\mathrm{e}}^{\frac{\pi }{\sqrt{2}}}))$$$$=-\cos \left(\frac{\pi }{\sqrt{2}}\right)\mathrm{sh}\left(\frac{\pi }{\sqrt{2}}\right)+\sin \left(\frac{\pi }{\sqrt{2}}\right)\mathrm{ch}\left(\frac{\pi }{\sqrt{2}}\right)$$$$\sin \left(\pi \sqrt{-\mathrm{i}}\right)=\frac{{\mathrm{e}}^{\mathrm{i}(\pi \sqrt{-\mathrm{i})}}-{\mathrm{e}}^{-\mathrm{i}\pi \sqrt{-\mathrm{i}}}}{2\mathrm{i}}=\frac{{\mathrm{e}}^{\mathrm{i}\pi (\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}})}-{\mathrm{e}}^{-\mathrm{i}\pi (\frac{1}{\sqrt{2}}-\frac{\mathrm{i}}{\sqrt{2}})}}{2\mathrm{i}}$$$$=\frac{{\mathrm{e}}^{\frac{\pi }{\sqrt{2}}}(\cos \left(\frac{\pi }{\sqrt{2}}\right)+\mathrm{isin}\left(\frac{\pi }{\sqrt{2}}\right))-{\mathrm{e}}^{-\frac{\pi }{\sqrt{2}}}(\cos \left(\frac{\pi }{\sqrt{2}}\right)-\mathrm{isin}\left(\frac{\pi }{\sqrt{2}}\right))}{2\mathrm{i}}$$$$=\frac{1}{2\mathrm{i}}\cos \left(\frac{\pi }{\sqrt{2}}\right)({\mathrm{e}}^{\frac{\pi }{\sqrt{2}}}-{\mathrm{e}}^{-\frac{\pi }{\sqrt{2}}})+\sin \left(\frac{\pi }{\sqrt{2}}\right)\frac{{\mathrm{e}}^{\frac{\pi }{\sqrt{2}}}+{\mathrm{e}}^{-\frac{\pi }{\sqrt{2}}}}{2}$$$$=\cos \left(\frac{\pi }{\sqrt{2}}\right)\mathrm{sh}\left(\frac{\pi }{\sqrt{2}}\right)+\sin \left(\frac{\pi }{\sqrt{2}}\right)\mathrm{ch}\left(\frac{\pi }{\sqrt{2}}\right)$$$$\mathrm{rest}\mathrm{to}\mathrm{collect}\mathrm{the}\mathrm{calculus}...$$

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