91Ó°ÊÓ

d

Here, we need to prove that and $\mathrm{Γ}\left(\frac{1}{2}-n\right)\mathrm{Γ}\left(\frac{1}{2}+n\right)=(-1{)}^n\mathrm{π}$and$(\mathrm{z}!)(-\mathrm{z})!=\frac{\mathrm{π}\mathrm{z}}{\sin \left(\mathrm{π}\mathrm{z}\right)}$.

Use the formula to prove the two questions.

$\mathrm{Γ}\left(p\right)\mathrm{Γ}(1-p)=\frac{\mathrm{π}}{\sin \left(\mathrm{π}p\right)}$

Use$p=\frac{1}{2}-n$

$\mathrm{Γ}\left(p\right)\mathrm{Γ}(1-p)=\frac{\mathrm{π}}{\sin \left(\frac{\mathrm{π}}{2}-\mathrm{π}n\right)}$

Use the sine formula to expand the denominator.

$\begin{array}{c}\sin (AÂ\pm B)=\sin A\cos BÂ\pm \cos A\sin B\\ \sin \left(\frac{\mathrm{Ï€}}{2}-n\mathrm{Ï€}\right)=\sin \frac{\mathrm{Ï€}}{2}\cos n\mathrm{Ï€}-\cos \frac{\mathrm{Ï€}}{2}\sin n\mathrm{Ï€}\end{array}$

Apply this identity to simplify further.

$\mathrm{Γ}\left(p\right)\mathrm{Γ}(1-p)=\frac{\mathrm{π}}{\sin \frac{\mathrm{π}}{2}\cos \left(n\mathrm{π}\right)-\cos \frac{\mathrm{π}}{2}\sin \left(n\mathrm{π}\right)}$

From trigonometric tables, use the following formulae.

$\begin{array}{c}\sin \frac{\mathrm{Ï€}}{2}=1\\ \cos \frac{\mathrm{Ï€}}{2}=0\\ \cos \left(n\mathrm{Ï€}\right)=(-1{)}^n\end{array}$

The equation can be simplified further.

$\mathrm{Γ}\left(p\right)\mathrm{Γ}(1-p)=(-1{)}^n\mathrm{π}$

The following relations between the Gamma function and factorials are known.

$\begin{array}{c}\mathrm{Γ}(p+1)=p!\\ \mathrm{Γ}(p+1)=p\mathrm{Γ}\left(p\right)\end{array}$

Use these formulae to simplify the left hand side of the equation in the question.

$\begin{array}{c}(z!)(-z)!=\mathrm{Γ}(z+1)\mathrm{Γ}(-z+1)\\ (z!)(-z)!=z\mathrm{Γ}\left(z\right)\mathrm{Γ}(1-z)\end{array}$

Simplify using established identities.

$\mathrm{Γ}\left(p\right)\mathrm{Γ}(1-p)=\frac{\mathrm{π}}{\sin \left(\mathrm{π}p\right)}$

Replace $p=z$.

$\begin{array}{l}\mathrm{Γ}\left(z\right)\mathrm{Γ}(1-z)=\frac{\mathrm{π}}{\sin \left(\mathrm{π}z\right)}\\ (z!)(-z)!=\frac{\mathrm{π}z}{\sin \left(\mathrm{π}z\right)}\end{array}$

Therefore, the equation is proved.

$(\mathrm{z}!)(-\mathrm{z})!=\frac{\mathrm{Ï€}\mathrm{z}}{\sin \left(\mathrm{Ï€}\mathrm{z}\right)}$