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I am working on finding that $\Gamma(x+\frac{1}{2})$ can be approximated with $\sqrt{x}\Gamma(x)$. I have tried many approximations, e.g., Stirling, Taylor series, etc, but I cannot find an appropriate solution. Can anyone give me any guidelines? Thanks!

Update: I forgot to mention that they almost match in Matlab simulations. e.g., plot([1:100],sqrt([1:100]),'r');hold on;plot([1:100],gamma([1:100]+1/2)./gamma([1:100]),'o')

Jack D'Aurizio
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pvm
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    Can you indicate the regime where the approximation is expected to be good. Is for small $x$, large $x$, around any particular point? What is the error term? – Sasha Oct 26 '16 at 15:24
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    Note that $\sqrt{x}\Gamma(x)$ is the geometric mean of $\Gamma(x)$ and $\Gamma(x+1)$. This combined with monotonicity is a reason why this approximation should at least make some sense even if it is not that good. What are you trying to get at? Are you trying to get an approximation to the error? – Ian Oct 26 '16 at 15:27
  • Perhaps it is of interest to you that according to Wikipedia, $\lim_{n \to \infty} \frac{\Gamma(n+\alpha)}{\Gamma(n)n^\alpha} = 1$ for all $\alpha \in \mathbb{R}$. Unfortunately it does not provide a source. – sTertooy Oct 26 '16 at 15:27
  • @Sasha 'x' is a natural number – pvm Oct 26 '16 at 15:31
  • @Ian I have the expression $\frac{\Gamma(x+1/2)}{\Gamma(x)}$ in my mathematical model and it would be very convenient to provide the $\sqrt{x}$ approximation that will make the model far more tractable – pvm Oct 26 '16 at 15:33
  • But in what regime? There are Stirling-type approximations available in the $x \to \infty$ regime. In the small $x$ regime it's less straightforward what to do. – Ian Oct 26 '16 at 15:42
  • @Ian The values that $x$ takes as a natural number should not exceed 100. So, it should be in the small $x$ regime. – pvm Oct 26 '16 at 15:54
  • Once you're larger than about 10 the approximation is already reasonable. Lower than that there is more discrepancy. For example, Gamma(3/2) is about 0.89 while Gamma(1) is 1, so the approximation is off by more than 10%. You can get a feel for why this would happen by thinking about a way to realize this approximation geometrically: it comes about by linearly interpolating the logarithm of $\Gamma(x)$ and then exponentiating. So if $\Gamma(x)$ looks basically like an exponential function on $[x,x+1]$ then this approximation is good, otherwise it's bad. – Ian Oct 26 '16 at 15:55
  • @Ian Thank you for the insights! If I find an appropriate solution I will let you know – pvm Oct 26 '16 at 16:34

2 Answers2

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A particular case of Gautschi's inequality gives $$\forall x>0,\qquad \sqrt{x}<\frac{\Gamma(x+1)}{\Gamma\left(x+\frac{1}{2}\right)}<\sqrt{x+1} \tag{1}$$ hence $$\frac{x}{\sqrt{x+1}}<\frac{\Gamma\left(x+\frac{1}{2}\right)}{\Gamma(x)}<\sqrt{x}. \tag{2}$$

Gautschi's inequality is a quite straightforward consequence of the log-convexity of the $\Gamma$ function.

A more accurate approximation comes from Stirling's inequality $$ \log\Gamma(x) = \left(x-\frac{1}{2}\right)\log x-x+\frac{1}{2}\log(2\pi)+\frac{O(1)}{x}\tag{3} $$ leading to: $$ \log\frac{\Gamma\left(x+\frac{1}{2}\right)}{\Gamma\left(x\right)} -\left(\frac{1}{2}\log x-\frac{1}{8x}\right) = \frac{O(1)}{x^3}\tag{4}$$

Essentially the same accuracy can also be recovered from the fact that

If $\{a_n\}_{n\geq 1}$ is defined through $$ a_n=\binom{2n}{n}\frac{\sqrt{\pi\left(n+\frac{1}{4}\right)}}{4^n} $$ $\{a_n\}_{n\geq 1}$ is an increasing sequence converging towards $1$.

This fact can be proved through Weierstrass products and creative telescoping, too.
The Legendre duplication formula then gives $$ \frac{\Gamma\left(x+\frac{1}{2}\right)}{\Gamma(x)}\sim \frac{x}{\sqrt{x+\frac{1}{4}}}. \tag{5}$$

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Jack D'Aurizio
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Stirling Asymptotic Expansion '$+$' Duplication Formula:

\begin{align} \ln\pars{\Gamma\pars{z + {1 \over 2}}} & \sim z\ln\pars{z} - z + {1 \over 2}\,\ln\pars{2\pi} - {1 \over 24z} + {7 \over 2880z^{3}} - {31 \over 40320z^{5}} + \mrm{O}\pars{1 \over z^{7}} \\[5mm] \ln\pars{\Gamma\pars{z}} & \sim \pars{z - \color{#f00}{1 \over 2}}\ln z - z + {1 \over 2}\ln\pars{2\pi} + \frac{1}{12z} - \frac{1}{360z^{3}} + {1 \over 1260z^{5}} + \mrm{O}\pars{1 \over z^{7}} \end{align}

$$ \ln\pars{\Gamma\pars{z + 1/2} \over \root{z}\Gamma\pars{z}} \sim -\,{1 \over 8z} + {1 \over 192z^{3}} + \mrm{O}\pars{1 \over z^{5}}\quad \mbox{as}\ \verts{z} \to \infty $$
Felix Marin
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