Binomial theorem for $(1-x^2)^{n-\frac12}$

Solution 1:

According to the binomial series expansion we obtain \begin{align*} \color{blue}{\left(1-x^2\right)^{n-\frac{1}{2}}=\sum_{k=0}^{\infty} \binom{n-\frac{1}{2}}{k}(-1)^kx^{2k}\qquad\qquad |x|<1} \end{align*} If we prefer integral upper indices of the binomial coefficients we can also write \begin{align*} \color{blue}{\left(1-x^2\right)^{n-\frac{1}{2}}=\sum_{k=0}^\infty\binom{2n}{2k}\binom{2k}{k}\binom{n}{k}^{-1}\left(-\frac{1}{4}\right)^kx^{2k}}\tag{1} \end{align*}

We obtain (1) by writing \begin{align*} \color{blue}{\binom{n-\frac{1}{2}}{k}}&=\frac{1}{k!}\left(n-\frac{1}{2}\right)\left(n-\frac{3}{2}\right)\cdots\left(n-k+\frac{1}{2}\right)\tag{2.1}\\ &=\frac{1}{k!2^k}(2n-1)(2n-3)\cdots(2n-2k+1)\\ &=\frac{1}{k!2^k}\,\frac{(2n-1)!!}{(2n-2k-1)!!}\tag{2.2}\\ &=\frac{1}{k!2^k}\,\frac{(2n)!}{(2n)!!}\,\frac{(2n-2k)!!}{(2n-2k)!}\tag{2.3}\\ &=\frac{1}{k!2^k}\,\frac{(2n)!}{2^nn!}\,\frac{2^{n-k}(n-k)!}{(2n-2k)!}\tag{2.4}\\ &=\frac{1}{4^k}\,\frac{(2n)!}{n!}\,\frac{(n-k)!}{k!(2n-2k)!}\\ &=\frac{1}{4^k}\binom{2n}{2k}\frac{(2k)!(n-k)!}{n!k!}\\ &\,\,\color{blue}{=\frac{1}{4^k}\binom{2n}{2k}\binom{2k}{k}\binom{n}{k}^{-1}} \end{align*} and the claim (1) follows.

Comment:

• In (2.1) we use the definition $\binom{\alpha}{k}=\frac{1}{k!}\alpha(\alpha-1)\cdots(\alpha-k+1)$ with $\alpha\in\mathbb{C}$ and $k\in\mathbb{N}_{0}$.

• In (2.2) we use double factorials $(2n-1)!!=(2n-1)(2n-3)\cdots3\cdot 1$.

• In (2.3) we apply $(2n)!=(2n)!!(2n-1)!!$.

• In (2.4) we apply $(2n)!!=2^nn!$.