Advanced-calculus-prove-that-P-n-1-1-1-n-4-sin-pi-sinh-pi-pi-2-

Question Number 134435 by mnjuly1970 last updated on 03/Mar/21

\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 ::

P = \underset{n = 1}{\prod^{\infty}} (1 - \frac{1}{n^{4}}) = \frac{s i n (π) . s i n h (π)}{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^{2}}) \underset{n = 1}{\prod^{\infty}} (1 - \frac{1}{n^{2}}) = \frac{s i n π}{π} . \frac{s i n h (π)}{π} = \frac{s i n (π) s i n h (π)}{π^{2}} = 0

Commented by mnjuly1970 last updated on 03/Mar/21

t a y e b a l l a h s i r p a y a n \dots

Answered by mnjuly1970 last updated on 03/Mar/21

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

x := 1

∴ \frac{s i n (π)}{π} = \underset{n = 1}{\prod^{\infty}} (1 - \frac{1}{n^{2}}) \dots . . (1)

x : = i

\frac{s i n (π i)}{π i} \overset{(i^{2} = - 1)}{=} \prod_{n = 1} (1 + \frac{1}{n^{2}}) \dots . (2)

s i n (π i) = \frac{e^{- π} - e^{π}}{2 i} (*) \dots \dots . . (2^{*})

(2) a n d (*) :\Rightarrow \frac{s i n h (π)}{π} = \underset{n = 1}{\prod^{\infty}} (1 + \frac{1}{n^{2}}) \dots . 2^{*}

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