prove-that-z-C-we-have-sinz-z-n-1-1-z-2-n-2-pi-2-

Question Number 67524 by mathmax by abdo last updated on 28/Aug/19

p r o v e t h a t \forall z \in C w e h a v e

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

Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

T h e c o m p l e m e n t f o r m u l a s g i v e u s \frac{π}{s i n (π z)} = Γ (z) Γ (1 - z)

N o w s i n (π z) = \frac{π}{Γ (z) Γ (1 - z)} = \frac{π}{- z Γ (z) Γ (- z)} c a u s e Γ (x + 1) = x Γ (x)

k n o w i n g t h a t \frac{1}{Γ (z)} = {z e}^{γ z} \underset{n = 1}{\prod^{\infty}} (1 + \frac{z}{n}) e^{- \frac{z}{n}}

\frac{1}{Γ (z) Γ (- z)} = {z e}^{γ z} \underset{n = 1}{\prod^{\infty}} (1 + \frac{z}{n}) e^{- \frac{z}{n}} . {(- z)}^{} e^{γ (- z)} \underset{n = 1}{\prod^{\infty}} (1 + \frac{- z}{n}) e^{- \frac{- z}{n}}

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

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

f i n a l l y w i t h w = π z

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