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

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

\dots a d v a n c e d c a l c u l u s \dots

p r o v e t h a t :::

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

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

\underset{n = 2}{\prod^{\infty}} (1 + \frac{i}{n^{2}}) \underset{n = 2}{\prod^{\infty}} (1 - \frac{i}{n^{2}})

= \frac{1}{(1 - i^{\frac{3}{2}}) (1 - \sqrt{i})} \underset{n = 1}{\prod^{\infty}} (1 - \frac{{(\sqrt{i^{3}})}^{2}}{n^{2}}) \underset{n = 1}{\prod^{\infty}} (1 - {(\frac{\sqrt{i}}{n})}^{2})

= \frac{1}{- \sqrt{i} (i + 1)} . \frac{s i n (i^{\frac{3}{2}} π)}{π i^{\frac{3}{2}}} . \frac{s i n (\sqrt{i} π)}{π \sqrt{i}}

= \frac{1}{4 π^{2}} (c o s h (\sqrt{2} π) - c o s (\sqrt{2} π)) (A f t e r s i m p l i f y i n g)

Commented by mnjuly1970 last updated on 20/Dec/20

t h a n k y o u . .

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

\frac{sinz}{z} = \prod_{n = 1}^{\infty} (1 - \frac{z^{2}}{n^{2} π^{2}}) so

\prod_{n = 2}^{\infty} (1 + \frac{1}{n^{4}}) = \frac{1}{2} \prod_{n = 1}^{\infty} (1 + \frac{1}{n^{4}}) = \frac{1}{2} \prod_{n = 1}^{\infty} (1 + \frac{i}{n^{2}}) (1 - \frac{i}{n^{2}})

\frac{z}{π} = t \Rightarrow \frac{\sin (π t)}{π t} = \prod_{n = 1}^{\infty} (1 - \frac{t^{2}}{n^{2}}) \Rightarrow \prod_{n = 1}^{\infty} (1 - \frac{i}{n^{2}})

= \prod_{n = 1}^{\infty} (1 - \frac{{(\sqrt{i})}^{2}}{n^{2}}) = \frac{\sin (π \sqrt{i})}{π \sqrt{i}} also

\prod_{n = 1}^{\infty} (1 + \frac{i}{n^{2}}) = \prod_{n = 1}^{\infty} (1 - \frac{{(\sqrt{- i})}^{2}}{n^{2}}) = \frac{\sin (π \sqrt{- i})}{π \sqrt{- i}} \Rightarrow

\prod_{n = 2}^{\infty} (1 + \frac{1}{n^{4}}) = \frac{1}{2} \frac{\sin (π \sqrt{i}) \sin (π \sqrt{- i})}{π^{2} \sqrt{- i^{2}}} = \frac{\sin (π \sqrt{i}) \sin (π \sqrt{- i})}{2 π^{2}}

\sin (π \sqrt{i}) = \frac{e^{i (π \sqrt{i})} - e^{- i (π \sqrt{i})}}{2 i} = \frac{1}{2 i} {e^{i (π (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})} - e^{- i π (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})}}

= \frac{1}{2 i} (e^{- \frac{π}{\sqrt{2}}} (\cos (\frac{π}{\sqrt{2}}) + isin (\frac{π}{\sqrt{2}}) - e^{\frac{π}{\sqrt{2}}} (\cos (\frac{π}{\sqrt{2}}) - isin (\frac{π}{\sqrt{2}}))

= \frac{1}{2 i} (\cos (\frac{π}{\sqrt{2}}) (e^{- \frac{π}{\sqrt{2}}} - e^{\frac{π}{\sqrt{2}}}) + isin (\frac{π}{\sqrt{2}}) (e^{- \frac{π}{\sqrt{2}}} + e^{\frac{π}{\sqrt{2}}}))

= - \cos (\frac{π}{\sqrt{2}}) sh (\frac{π}{\sqrt{2}}) + \sin (\frac{π}{\sqrt{2}}) ch (\frac{π}{\sqrt{2}})

\sin (π \sqrt{- i}) = \frac{e^{i (π \sqrt{- i)}} - e^{- i π \sqrt{- i}}}{2 i} = \frac{e^{i π (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}})} - e^{- i π (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}})}}{2 i}

= \frac{e^{\frac{π}{\sqrt{2}}} (\cos (\frac{π}{\sqrt{2}}) + isin (\frac{π}{\sqrt{2}})) - e^{- \frac{π}{\sqrt{2}}} (\cos (\frac{π}{\sqrt{2}}) - isin (\frac{π}{\sqrt{2}}))}{2 i}

= \frac{1}{2 i} \cos (\frac{π}{\sqrt{2}}) (e^{\frac{π}{\sqrt{2}}} - e^{- \frac{π}{\sqrt{2}}}) + \sin (\frac{π}{\sqrt{2}}) \frac{e^{\frac{π}{\sqrt{2}}} + e^{- \frac{π}{\sqrt{2}}}}{2}

= \cos (\frac{π}{\sqrt{2}}) sh (\frac{π}{\sqrt{2}}) + \sin (\frac{π}{\sqrt{2}}) ch (\frac{π}{\sqrt{2}})

rest to collect the calculus \dots