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.

  • $\begingroup$ the first fraction is equivalent to $$\frac{\Gamma \left(m+\frac{1}{2}\right)}{\sqrt{\pi } \Gamma (m+1)}$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 29 '16 at 17:55
  • $\begingroup$ The $\Gamma$-$Duplication\ Formula$. $\endgroup$ – Felix Marin Oct 30 '16 at 4:11
  • $\begingroup$ 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". $\endgroup$ – 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.

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.