Question

Use equations (3.4) and (11.5) to show that $\mathit{\Gamma }\mathbf{\left(}\mathit{p}\mathbf{\right)}\mathbf{\square }{\mathit{p}}^{\mathbf{p}}{\mathit{e}}^{\mathbf{-}\mathbf{p}}\sqrt{\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{p}}}\left(1+\frac{1}{12p}+...\right)$.

Short answer

The equation has been proved.

Step-by-step solution

Given Information

The statement is$\Gamma \left(p\right)\square p^{\mathrm{p}}{e}^{-\mathrm{p}}\sqrt{\frac{2\mathrm{\pi }}{\mathrm{p}}}(1+\frac{1}{12\mathrm{p}}+...)$.

Definition ofthe sterling’s formula.

Sterling’s formula is used to simplify formulas involving factorial.

$\mathit{n}\mathbf{!}\mathbf{\square }{\mathit{n}}^{\mathbf{n}}{\mathit{e}}^{\mathbf{-}\mathbf{n}}\sqrt{\mathbf{2}\mathbf{\pi n}}$.

Prove the statement.

The statement is $\Gamma \left(p\right)\square p^{\mathrm{p}}{e}^{-\mathrm{p}}\sqrt{\frac{2\mathrm{\pi }}{\mathrm{p}}}(1+\frac{1}{12\mathrm{p}}+...)$.

The sterling’s formula is role="math" localid="1664882157502" $\Gamma n\square n^{\mathrm{n}}{e}^{-\mathrm{n}}\sqrt{2\mathrm{\pi n}}$.

The equation (11.5) is expressed as,

$\Gamma \left(n+1\right)\square n^{\mathrm{n}}{e}^{-\mathrm{n}}\sqrt{2\mathrm{\pi n}}\left(1+\frac{1}{12n}+\frac{1}{288n^2}+...\right)$

The equation (3.4) is expressed as,

$\Gamma \left(n+1\right)=n\Gamma \left(n\right)$

The value of $\Gamma \left(n\right)$becomes as follows.

$$\begin{gathered} \Gamma \left(n\right)=\frac{1}{n}\Gamma \left(n+1\right) \\ \Gamma \left(n\right)=n^{\mathrm{n}}{e}^{-\mathrm{n}}\sqrt{2\mathrm{\pi n}}\left(1+\frac{1}{12n}+\frac{1}{288n^2}+...\right) \end{gathered}$$

Hence, the statement has been proven.