# Binomial Coefficients with fractions: $\binom {m-\frac 12}m=\frac 1{2^{2m}}\binom {2m}m$.

From my earlier question here and the interesting solutions posted, we find interesting equivalents converting binomial coefficients with fractions to those without, e.g. $$\binom {m-\frac 12}m=\frac 1{2^{2m}}\binom {2m}m$$ and $$\binom {n+\frac 12}n=\frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$

Are there any "rules of thumb" for quickly converting a binomial coefficient with fractions into a binomial coefficient without fractions, adjusted with a coefficient as necessary?

Further edit:

The purpose of this question is not to derive the above (that has already been done elsewhere) but to ask if there is a handy rule of thumb for converting one form to another (with basis provided, of course).

Another example might be $$\binom {m-\frac 34}m$$ Perhaps one could consider a fractional binomial coefficient of the form $$\binom {m-\frac pq}m$$ and see if that can be converted into a binomial coefficient of integer parameters.

• the first fraction is equivalent to $$\frac{\Gamma \left(m+\frac{1}{2}\right)}{\sqrt{\pi } \Gamma (m+1)}$$ – Dr. Sonnhard Graubner Oct 29 '16 at 17:55
• The $\Gamma$-$Duplication\ Formula$. – Felix Marin Oct 30 '16 at 4:11
• Thanks for your comments, both. Looking at the question again, it appears that the first equation shown can be taken as the "rule of thumb". – hypergeometric Oct 30 '16 at 4:19

$$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ \begin{align} {m - 1/2 \choose m} & = {\pars{m - 1/2}! \over m!\pars{-1/2}!}= {\Gamma\pars{m + 1/2} \over m!\,\Gamma\pars{1/2}} \\[5mm] & = {1 \over m!\,\root{\pi}}\,\ \overbrace{{\root{2\pi}2^{1/2 - 2m}\,\Gamma\pars{2m} \over \Gamma\pars{m}}} ^{\ds{\color{#f00}{\large\S}}\,,\ \Gamma\pars{m + 1/2}}\,,\qquad\qquad\qquad \pars{~\Gamma\pars{1 \over 2} = \root{\pi}~} \\[5mm] & = {1 \over 2^{2m - 1}}\,{\pars{2m - 1}! \over m!\pars{m - 1}!} = {1 \over 2^{2m - 1}}\,{\pars{2m}!/\pars{2m} \over m!\pars{m!/m}} \\[5mm] & = {1 \over 2^{2m}}\,{\pars{2m}! \over m!\, m!} = \color{#f00}{{1 \over 2^{2m}}{2m \choose m}} \end{align} $$\ds{\color{#f00}{\large\S}:\ \Gamma\!-\!Duplication\ Formula}$$. See $$\ds{\mathbf{6.1.18}}$$ in Abramowitz & Stegun Table.

Note that there are several useful ways to express $$\ds{2m \choose m}$$:

$${2m \choose m} = 2^{2m}{m - 1/2 \choose m} = 2^{2m}\bracks{{-1/2 \choose m}\pars{-1}^{m}} = {-1/2 \choose m}\pars{-4}^{m} = {-1/2 \choose -1/2 - m}\pars{-4}^{m}$$

The other one is quite similar to this one.

• Thanks for your solution and the reference. Very useful. (+1) Also, very nice $\color{red}{\large\S}$! – hypergeometric Oct 30 '16 at 4:38
• @hypergeometric Thanks. You're welcome. – Felix Marin Oct 30 '16 at 4:40

In equation $(6)$ of this answer, the Euler-Maclaurin Sum Formula was used to derive the asymptotic formula $$\binom{n+\alpha}{n}=\frac{n^\alpha}{\Gamma(1+\alpha)}\left(1+\frac{\alpha+\alpha^2}{2n}-\frac{2\alpha+3\alpha^2-2\alpha^3-3\alpha^4}{24n^2}+O\!\left(\frac1{n^3}\right)\right)\tag{1}$$ by applying the idenity $$\binom{n+\alpha}{n}=\prod_{k=1}^n\left(1+\frac\alpha{k}\right)\tag{2}$$ Proven in equation $(11)$ of this answer, Gauss' Multiplication Formula says $$\prod_{k=0}^{n-1}\Gamma\!\left(x+\frac kn\right) =\sqrt{n2^{n-1}\pi^{n-1}}\frac{\Gamma(nx)}{n^{nx}}\tag{3}$$ So that \begin{align} \prod_{k=0}^1\binom{m+\frac k2}{m} &=\prod_{k=0}^1\frac{\Gamma\!\left(m+1+\frac k2\right)}{m!\,\Gamma\left(1+\frac k2\right)}\\ &=\frac{\Gamma(2m+2)}{m!^2\,2^{2m}\,\Gamma(2)}\\ &=\frac1{2^{2m}}\binom{2m+1}{1}\frac{(2m)!}{(m!)^2}\\ &=\frac{2m+1}{4^m}\binom{2m}{m}\tag{4} \end{align} and \begin{align} \prod_{k=0}^3\binom{m+\frac k4}{m} &=\prod_{k=0}^3\frac{\Gamma\!\left(m+1+\frac k4\right)}{m!\,\Gamma\left(1+\frac k4\right)}\\ &=\frac{\Gamma(4m+4)}{m!^4\,4^{4m}\,\Gamma(4)}\\ &=\frac1{4^{4m}}\binom{4m+3}{3}\frac{(4m)!}{(m!)^4}\\ &=\frac1{256^m}\binom{4m+3}{3}\binom{4m}{m}\binom{3m}{m}\binom{2m}{m}\tag{5} \end{align} so that $$\binom{m+\frac14}{m}\binom{m+\frac34}{m}=\frac{(4m+1)(4m+3)}{3\cdot64^m}\binom{4m}{m}\binom{3m}{m}\tag{6}$$ So we can compute the product $\binom{m+\frac14}{m}\binom{m+\frac34}{m}$ in terms of integer binomimals, but I don't know of a way to compute each in such terms.

• Very interesting solution. Thanks! (+1) – hypergeometric Oct 31 '16 at 2:45