$n$-th derivative of $x^\alpha$ where $\alpha = m + 1/2$

Question

It is well-known that, for any real $\alpha$ and nonnegative integer $n$ $$ \frac{d^n x^\alpha}{dx^n} = \alpha(\alpha-1)\cdots(\alpha - n + 1) x^{\alpha - n} $$ I just found out that the coefficient is known as a falling factorial $$ \alpha(\alpha-1)\cdots(\alpha - n + 1) =: (\alpha)_n $$ where $(\alpha)_n$ is Pochhammer's symbol.

It seems that in the case $\alpha = 1/2$ and more generally $\alpha = m + 1/2$ where $m$ is a nonnegative integer, it is possible to express $(\alpha)_n$ with powers of 2 and standard (integer) factorials. For example: $$ (1/2)_4 = \frac12\bigg(-\frac12\bigg)\bigg(-\frac32\bigg)\bigg(-\frac52\bigg) =- \frac{1\times3\times5}{2^4} = -\frac{5!}{2^4\times(2\times4)} =- \frac{5!}{2^4\times2^2\times 2!} $$

Would anyone know a good reference to such formulas? I am guessing the cases $n>m$ and $n<m$ must be distinguished.

Thanks! p.

Answer 1

Your convention is a little different from what I am accustomed to; I usually denote Pochhammer symbols (rising factorials) by $(a)_n$, while falling factorials are $a^{(n)}$. That is,

$$\begin{align*} (a)_n&=\prod_{j=0}^{n-1}(a+j)\\ a^{(n)}&=\prod_{j=0}^{n-1}(a-j)=(-1)^n(-a)_n\end{align*}$$

The general relation you want is

$$\left(a+\frac12\right)_n=\frac{(2a)_{2n}}{4^n (a)_n}$$

and from similar considerations, one can derive

$$\left(a+\frac12\right)^{(n)}=\frac{(2a+1)^{(2n+1)}}{(2a-2n+1) 4^n a^{(n)}}$$

(which can be proven with e.g. the usual duplication formula for the gamma function).

As a reminder, $(1)_n=n^{(n)}=n!$.

Answer 2

Notice that:

\begin{align}\left(m+\frac12\right)_n&=\prod_{k=0}^{n-1}\left(m+\frac12-k\right)\\&=\prod_{k=0}^{n-1}\frac{\color{#BB4466}{2m+1-2k}}2\frac{\color{#BB4466}{2m-2k}}{2\color{#66AA99}{(m-k)}}\\&=\frac1{4^n}\left(\prod_{k=0}^{n-1}\frac1{\color{#66AA99}{m-k}}\right)\left(\prod_{k=-1}^{2n-2}\color{#BB4466}{(2m-k)}\right)\\&=\frac1{4^n}\frac{(m-n)!}{m!}\frac{(2m+1)!}{(2m-2n-3)!}\end{align}

provided that $m\ge n$ to avoid $0/0$. If $m<n$, we then re-interpret the factorials using

$$\frac{(m-n)!}{(2m-2n-3)!}=\frac1{(2m-n-3)_{(m-3)}}$$

where $(a)_0=1$ and $(a)_{n+1}=(a)_n(a-n)$ recursively defines the falling factorial to negative values if $0\le m\le3$.