Question

The following expression occurs in statistical mechanics:

$\mathit{P}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\left(np+u\right)\mathbf{!}\left(nq-u\right)\mathbf{!}}{\mathit{p}}^{\mathbf{n}\mathbf{p}\mathbf{+}\mathbf{u}}{\mathit{q}}^{\mathbf{n}\mathbf{q}\mathbf{-}\mathbf{u}}\mathbf{.}$

Use Stirling’s formula to show that

$\frac{\mathbf{1}}{\mathbf{P}}\mathbf{\square }{\mathit{x}}^{\mathbf{n}\mathbf{p}\mathbf{x}}{\mathit{y}}^{\mathbf{n}\mathbf{q}\mathbf{y}}\sqrt{\mathbf{2}\mathbf{\pi npqxy}}\mathbf{}\mathbf{,}\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{}\mathit{x}\mathbf{=}\mathbf{1}\mathbf{+}\frac{\mathbf{u}}{\mathbf{n}\mathbf{p}}\mathbf{,}\mathit{y}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathbf{u}}{\mathbf{n}\mathbf{q}}\mathbf{,}\mathit{p}\mathbf{+}\mathit{q}\mathbf{=}\mathbf{1}\mathbf{.}$

Hint: Show that${\mathbf{\left(}\mathit{n}\mathit{p}\mathbf{\right)}}^{\mathbf{n}\mathbf{p}\mathbf{+}\mathbf{u}}{\mathbf{\left(}\mathit{n}\mathit{q}\mathbf{\right)}}^{\mathbf{n}\mathbf{q}\mathbf{-}\mathbf{u}}\mathbf{=}{\mathit{n}}^n{\mathit{p}}^{\mathbf{n}\mathbf{p}\mathbf{+}\mathbf{u}}{\mathit{q}}^{\mathbf{n}\mathbf{q}\mathbf{-}\mathbf{u}}$.

Short answer

The statement has been proved.

Step-by-step solution

Given Information

The expression is $P=\frac{n!}{\left(np+u\right)!\left(nq-u\right)!}{p}^{np+u}{q}^{nq-u}$.

Definition of the sterling’s formula.

Sterling’s formula is used to simplify formulas involving factorial. $\mathit{n}\mathbf{!}\mathbf{~}{\mathit{n}}^{\mathbf{n}}{\mathit{e}}^{\mathbf{-}\mathbf{n}}\sqrt{\mathbf{2}\mathbf{\pi n}}$.

Prove the statement.

The expression is $P=\frac{n!}{\left(np+u\right)!\left(nq-u\right)!}{p}^{np+u}{q}^{nq-u}$.

The sterling’s formula is $n!~n^{\mathrm{n}}{e}^{-\mathrm{n}}\sqrt{2\mathrm{\pi n}}$.

Use Sterling’s formula in the expression.

The expression becomes as follows.

$$\begin{gathered} P=\frac{n!}{\left(np+u\right)!\left(nq-u\right)!}{p}^{np+u}{q}^{nq-u} \\ P=\frac{n^n{e}^{-n}\sqrt{2\mathrm{\pi n}}{p}^{np+u}{q}^{nq-u}}{{\left(np+u\right)}^{np+u}{e}^{-\left(np+u\right)}\sqrt{2\mathrm{\pi }\left(\mathrm{np}+\mathrm{u}\right)}{\left(\mathrm{nq}-\mathrm{u}\right)}^{\left(\mathrm{nq}-\mathrm{u}\right)}{\mathrm{e}}^{-\left(\mathrm{nq}-\mathrm{u}\right)}\sqrt{2\mathrm{\pi }\left(\mathrm{np}-\mathrm{u}\right)}}{p}^{np+u}{q}^{nq-u} \end{gathered}$$

The value (np)^np+u(nq)^nq-u becomes as follows.

$$\begin{gathered} {\left(np\right)}^{np+u}{\left(nq\right)}^{nq-u}=n^{np+u}{p}^{np+u}{n}^{nq-u}{q}^{nq-u} \\ {\left(np\right)}^{np+u}{\left(nq\right)}^{nq-u}=n^{n\left(p+q\right)}{p}^{np+u}{q}^{nq-u} \\ {\left(np\right)}^{np+u}{\left(nq\right)}^{nq-u}=n^n{p}^{np+u}{q}^{nq-u} \end{gathered}$$

Substitute the above value in the equation mentioned below.

$$\begin{gathered} P=\frac{n^n{e}^{-n}\sqrt{2\mathrm{\pi n}}{p}^{np+u}{q}^{nq-u}}{{\left(np\right)}^{np+u}{x}^{npx}{e}^{-\left(np+u\right)}\sqrt{2\mathrm{\pi npx}}{\left(nq\right)}^{\left(nq-u\right)}{y}^{nqy}{e}^{-\left(nq-u\right)}\sqrt{2\mathrm{\pi }\left(\mathrm{nqy}\right)}} \\ P=\frac{e^{-n}\sqrt{2\mathrm{\pi n}}}{x^{npx}{y}^{nqy}2\mathrm{\pi n}\sqrt{\mathrm{pxqy}}{\mathrm{e}}^{-\mathrm{n}}} \\ P=\frac{1}{x^{npx}{y}^{nqy}\sqrt{2\mathrm{\pi npxqy}}} \end{gathered}$$

Hence, the statement has been proven.