Advanced-Calculus-prove-that-determinant-i-n-0-1-x-2-2n-1-2-cos-pix-2-ii-n-0-1-x-2-2n-1-2-cosh

Question Number 135610 by mnjuly1970 last updated on 14/Mar/21

\dots . A d v a n c e d \dots \dots C a l c u l u s \dots .

p r o v e t h a t :

\begin{array}{cc} i :: \underset{n = 0}{\prod^{\infty}} (1 - \frac{x^{2}}{{(2 n + 1)}^{2}}) = c o s (\frac{π x}{2}) ✓ \\ i i :: \underset{n = 0}{\prod^{\infty}} (1 + \frac{x^{2}}{{(2 n + 1)}^{2}}) = c o s h (\frac{π x}{2}) ✓ ✓ \end{array}

\dots \dots \dots \dots .

Answered by Dwaipayan Shikari last updated on 14/Mar/21

c o s x = C (\frac{π}{2} - x) (\frac{π}{2} + x) (\frac{3 π}{2} - x) (\frac{3 π}{2} + x) \dots

c o s (x) h a s z e r o s a t = {\frac{π}{2}, - \frac{π}{2}, \frac{3 π}{2}, - \frac{3 π}{2} \dots .}

x = 0 \Rightarrow 1 = C \frac{π}{2} . \frac{π}{2} . \frac{3 π}{2} . \frac{3 π}{2} \dots \Rightarrow C = \frac{2}{π} . \frac{2}{π} . \frac{2}{3 π} . \frac{2}{3 π} \dots

c o s x = (1 - \frac{2 x}{π}) (1 + \frac{2 x}{π}) (1 - \frac{2 x}{3 π}) (1 + \frac{2 x}{3 π}) \dots .

c o s x = \underset{n = 0}{\prod^{\infty}} (1 - \frac{2 x^{2}}{π^{2} {(2 n + 1)}^{2}})

c o s (\frac{π}{2} x) = \underset{n = 0}{\prod^{\infty}} (1 - \frac{x^{2}}{{(2 n + 1)}^{2}})

c o s h (\frac{π}{2} x) = \underset{n = 0}{\prod^{\infty}} (1 + \frac{x^{2}}{{(2 n + 1)}^{2}}) x \to x i