Question

Show that for integral n, m,

$\mathbf{1}\mathbf{/}\mathit{B}\mathbf{(}\mathit{n}\mathbf{,}\mathit{m}\mathbf{)}\mathbf{=}\mathit{m}\left(\begin{array}{l}n+m-1\\ n-1\end{array}\right)\mathbf{=}\mathit{n}\left(\begin{array}{l}n+m-1\\ m-1\end{array}\right)$

Hint: See Chapter 1, Section 13C, Problem 13.3.

Short answer

It has been proved that $1/B\left(n,m\right)=m\left(\begin{array}{l}n+m-1\\ n-1\end{array}\right)=n\left(\begin{array}{l}n+m-1\\ m-1\end{array}\right)$.

Step-by-step solution

Given information

Here, we need to prove that $1/B\left(n,m\right)=m\left(\begin{array}{l}n+m-1\\ n-1\end{array}\right)=n\left(\begin{array}{l}n+m-1\\ m-1\end{array}\right)$.

Definition of a Beta function

The beta function is defined as $\mathit{B}\mathbf{(}\mathit{p}\mathbf{,}\mathit{q}\mathbf{)}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{1}}{\mathbf{\int }}}{\mathit{x}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{)}}^{\mathbf{q}\mathbf{-}\mathbf{1}}\mathit{d}\mathit{x}$$B(p,q)=\underset{0}{\overset{1}{\int }}{x}^{p-1}{(1-x)}^{q-1}dx$.

Begin evaluating from the first fraction in the given expression

Write the first fraction in the given equation.

$m\left(\begin{array}{l}n+m-1\\ n-1\end{array}\right)$

Write the relation between Gamma functions and the Combination expression

Write the relation.

$\left(\begin{array}{l}p\\ n\end{array}\right)=\frac{\Gamma (p+1)}{n!\Gamma (p-n+1)}$

Combine the expressions and continue evaluation

Continue evaluating the expressions.

$m.\frac{\Gamma (n+m-1+1)}{(n-1)\Gamma (n+m-1-n+1+1)}=\frac{\Gamma (n+m)}{(n-1)\Gamma (m+1)}$

Use the identities of evaluating Gamma functions and continue calculations.

$\begin{array}{c}\Gamma (p+1)=p\Gamma \left(p\right)\\ m·\frac{\Gamma (n+m)}{(n-1)!m\Gamma \left(m\right)}=\frac{\Gamma (n+m)}{(m-1)!\Gamma \left(m\right)}\end{array}$

Again, use identities of Gamma functions to further simplify.

$\begin{array}{c}\Gamma \left(m\right)=(m-1)!\\ \frac{\Gamma (n+m)}{(m-1)!\Gamma \left(m\right)}=\frac{\Gamma (n+m)}{(n-1)!(m-1)!}\end{array}$

This gives the answer.

Repeat the same process for the second fraction in the expression

Start with the second fraction in the expression.

$n\left(\begin{array}{l}n+m-1\\ m-1\end{array}\right)$

Continue evaluating in the same way as before.

$n\left(\begin{array}{l}n+m-1\\ m-1\end{array}\right)=\frac{\Gamma (n+m)}{(n-1)!(m-1)!}$

Since both sides are equal, the given statement is proved.