nice-calculus-prove-that-n-1-1-1-n-4-cosh-2pi-cos-2pi-4pi-2-

Question Number 134442 by mnjuly1970 last updated on 03/Mar/21

n i c e \dots . . c a l c u l u s \dots

p r o v e t h a t ::

ϕ = \underset{n = 1}{\prod^{\infty}} (1 + \frac{1}{n^{4}}) = \frac{c o s h (2 π) - c o s (2 π)}{4 π^{2}}

\dots \dots \dots \dots \dots \dots \dots . .

Answered by Dwaipayan Shikari last updated on 03/Mar/21

\underset{n = 1}{\prod^{\infty}} (1 + \frac{1}{n^{4}}) = \underset{n = 1}{\prod^{\infty}} (1 + \frac{i}{n^{2}}) (1 - \frac{i}{n^{2}}) = \frac{s i n h (π \sqrt{i}) s i n (π \sqrt{i})}{π^{2} i}

= \frac{s i n h (\frac{π}{\sqrt{2}} (1 + i)) s i n (\frac{π}{\sqrt{2}} (1 + i))}{π^{2} i} = - \frac{e^{\frac{π}{\sqrt{2}} + \frac{π}{\sqrt{2}} i} - e^{- \frac{π}{\sqrt{2}} - \frac{π}{\sqrt{2}} i}}{4 π^{2}} . (e^{\frac{π}{\sqrt{2}} i - \frac{π}{\sqrt{2}}} - e^{- \frac{π}{\sqrt{2}} i + \frac{π}{\sqrt{2}}})

= - \frac{e^{\sqrt{2} π i} - e^{\sqrt{2} π} - e^{- \sqrt{2} π} + e^{- \sqrt{2} π i}}{4 π^{2}} = \frac{e^{\sqrt{2} π} + e^{\sqrt{2} π} - e^{\sqrt{2} π i} - e^{- \sqrt{2} π i}}{4 π^{2}}

= \frac{c o s h (\sqrt{2} π) - c o s (\sqrt{2} π)}{2 π^{2}}

Commented by mnjuly1970 last updated on 03/Mar/21

v e r y n i c e \dots t h a n k s a l o t s i r . .