Prove-this-equation-2-x-1-x-1-2-1-2-2x-1-2-x-1-x-1-

Question Number 73974 by Raxreedoroid last updated on 17/Nov/19

Prove this equation

\frac{2^{x - 1} (x - \frac{1}{2})!}{\frac{1}{2}!} = \frac{(2 x - 1)!}{2^{x - 1} (x - 1)!}

Answered by mind is power last updated on 17/Nov/19

Γ (2 x) = (2 x - 1)!

β (x, x) = \frac{Γ (x) . Γ (x)}{Γ (2 x)}

β (x, x) = 2 \int_{0}^{\frac{π}{2}} \cos^{2 x - 1} (t) . \sin^{2 x - 1} (t) dt

= 2 \int_{0}^{\frac{π}{2}} . {(\sin (t) \cos (t))}^{2 x - 1} . dt

= 2 \int_{0}^{\frac{π}{2}} {(\frac{\sin (2 t)}{2})}^{2 x - 1} dt

= \frac{1}{2^{2 x - 2}} . \int_{0}^{\frac{π}{2}} . \sin^{2 x - 1} (2 t) dt

u = 2 t \Rightarrow β (x, x) = \frac{1}{2^{2 x - 2}} . \int_{0}^{π} \sin^{2 x - 1} (u) du = \frac{1}{2^{2 x - 2}} \int_{0}^{\frac{π}{2}} \sin^{2 x - 1} (u) du + \frac{1}{2^{2 x - 2}} \int_{\frac{π}{2}}^{π} \sin^{2 x - 1} (u) du

\int_{0}^{\frac{π}{2}} \sin^{2 x - 1} (u) du = \frac{1}{2} β (\frac{1}{2}, x)

\int_{\frac{π}{2}}^{π} \sin^{2 x - 1} (u) du = \int_{0}^{\frac{π}{2}} \sin^{2 x - 1} (u + \frac{π}{2}) d (u) = \frac{1}{2} β (x, \frac{1}{2}) = \frac{1}{2} β (\frac{1}{2}, x)

\Rightarrow \int_{0}^{π} \sin^{2 x - 1} (t) dt = β (x, \frac{1}{2})

\Rightarrow β (x, x) = \frac{1}{2^{2 x - 2}} . β (x, \frac{1}{2})

\Leftrightarrow \frac{Γ (x) . Γ (x)}{Γ (2 x)} = \frac{1}{2^{2 x - 2}} . \frac{Γ (x) . Γ (\frac{1}{2})}{Γ (x + \frac{1}{2})} \Leftrightarrow \frac{Γ (2 x)}{Γ (x) 2^{x - 1}} = \frac{Γ (x + \frac{1}{2})}{Γ (\frac{1}{2})} . 2^{x - 1}

Γ (x) = (x - 1)!, Γ (x + \frac{1}{2}) = (x - \frac{1}{2})!, Γ (2 x) = (2 x - 1)!

\Leftrightarrow \frac{2^{x - 1} . (x - \frac{1}{2})!}{(\frac{1}{2})!} = \frac{(2 x - 1)!}{(x - 1)! . 2^{x - 1}}

Commented by Raxreedoroid last updated on 17/Nov/19

My proof is the following

\frac{(2 n - 1)!}{2^{n - 1} (n - 1)!} = 1, 3, 15, 105, 945 \dots

\underset{k = 1}{\prod^{n - 1}} 2 k + 1 = 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \dots (2 n + 1)

let u = 2 k + 1

n = 1, \underset{k = 1}{\prod^{0}} u = 1

n = 2, \underset{k = 1}{\prod^{1}} u = 1 \cdot 3 = 3

n = 3, \underset{k = 1}{\prod^{2}} u = 1 \cdot 3 \cdot 5 = 15

n = 4, \underset{k = 1}{\prod^{3}} u = 1 \cdot 3 \cdot 5 \cdot 7 = 105

n = 5, \underset{k = 1}{\prod^{4}} u = 945

{It}^{'} s the same

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

\underset{k = 1}{\prod^{n - 1}} 2 = 2^{n - 1}

\underset{k = 1}{\prod^{n - 1}} x + \frac{1}{2} = \underset{k = 1 + \frac{1}{2}}{\prod^{n - 1 + \frac{1}{2}}} x = \frac{(n - \frac{1}{2})!}{(1 \frac{1}{2} - 1)!} = \frac{(n - \frac{1}{2})!}{(\frac{1}{2})!}

\underset{k = 1}{\prod^{n - 1}} 2 x + 1 = \frac{2^{x - 1} (x - \frac{1}{2})!}{(\frac{1}{2})!}

Commented by mind is power last updated on 17/Nov/19

yeah nice this for x \in N

Commented by Raxreedoroid last updated on 17/Nov/19

I am surprised becuase I proved this

in a simple way even though I {can}^{'} t understand

most of this

Commented by mind is power last updated on 17/Nov/19

if x \in N or x \in C ?