I am having trouble obtaining a lower bound for the following formula: $$ \ln\frac{\Gamma\left(\frac{x}{3}\right)}{\Gamma\left(\frac{x}{4}+1\right)\Gamma\left(\frac{x}{12}+1\right)}. $$ I tried using the well-known Stirling's approximation for the Gamma function, namely: $$ \Gamma(x)\approx\sqrt{\frac{2\pi}{x}}\left(\frac{x}{e}\right)^x $$ but it yields some very "ugly" and in fact unusable results. My question is: is there a better and simpler way to get this lower bound?
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$\begingroup$ How about recalling that $\log\Gamma$ is a convex function? $\endgroup$– Jack D'AurizioAug 18, 2015 at 22:12
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$\begingroup$ Sorry, but I have no idea what a convex function is. $\endgroup$– grzegorzmarciakowskiAug 18, 2015 at 22:18
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$\begingroup$ Have a look at en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem and en.wikipedia.org/wiki/Convex_function. $\endgroup$– Jack D'AurizioAug 18, 2015 at 22:34
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$\begingroup$ Do you have any restrictions on $x$? Is it small? Is it large? $\endgroup$– robjohn ♦Feb 18, 2020 at 15:26
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2 Answers
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By exploting the convexity of the $\log\Gamma$ function, it is not difficult to show that: $$ \log\left(\frac{\Gamma\left(\frac{x}{3}\right)}{\Gamma\left(\frac{x}{4}+1\right)\Gamma\left(\frac{x}{12}+1\right)}\right)\leq \left(\frac{\log 4}{3}-\frac{\log 3}{4}\right)x \tag{1}$$ holds for every $x$ big enough ($x\geq 3$ is fine). The RHS of $(1)$ is just the first term of the asymptotic expansion of the LHS at $x=+\infty$. Stirling's approximation/inequality provides the tighter inequality:
$$ \log\left(\frac{\Gamma\left(\frac{x}{3}\right)}{\Gamma\left(\frac{x}{4}+1\right)\Gamma\left(\frac{x}{12}+1\right)}\right)\leq \left(\frac{\log 4}{3}-\frac{\log 3}{4}\right)x+\log\left(6\sqrt{\frac{2}{\pi}}\right)-\frac{3}{2}\log x$$
that holds for every $x>0$. By computing an extra term, we have that the lower bound:
$$ LHS\color{red}{\geq} \left(\frac{\log 4}{3}-\frac{\log 3}{4}\right)x+\log\left(6\sqrt{\frac{2}{\pi}}\right)-\frac{3}{2}\log x\color{red}{-\frac{13}{12x}}\tag{2}$$
holds for every $x>0$.
answered Aug 18, 2015 at 23:17
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$\begingroup$ I may have misunderstood the question, but it seems this answer provides an upper bound, while the asker was looking for a lower bound? $\endgroup$– njuffaAug 19, 2015 at 6:16
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$\begingroup$ When I evaluate (1) numerically it seems to hold for $x \ge 3$, but the tighter bound in the next equation fails for $x = 3, 10, 20$. I double checked my code, but the high bound comes out as less than the function, e.g.
func(10) = -1.1800738578558259e-1, buthigh(10) = -1.6546909416432840e+0. I assumed $log(x)$ is the natural logarithm of $x$. $\endgroup$– njuffaAug 19, 2015 at 21:16 -
1$\begingroup$ Sorry for the false alarm. It works fine numerically, as-is. I misread "six times the square root of two over $\pi$" as the "sixth root of two over $\pi$". $\endgroup$– njuffaAug 19, 2015 at 21:25
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Let $u=12x$ then
\begin{align} \frac{\Gamma\left(\frac{x}{3}\right)}{\Gamma\left(\frac{x}{4}+1\right)\Gamma\left(\frac{x}{12}+1\right)}&=\frac{\Gamma\left(4u\right)}{3u^2\Gamma\left(3u\right)\Gamma\left(u\right)}\\ &=\frac{2^{8 u-\frac{3}{2}} 3^{-3 u-\frac{1}{2}}}{\sqrt{\pi } u^2}\times\frac{\Gamma \left(u+\frac{1}{4}\right) \Gamma \left(u+\frac{1}{2}\right) \Gamma \left(u+\frac{3}{4}\right)}{\Gamma (u) \Gamma \left(u+\frac{1}{3}\right) \Gamma \left(u+\frac{2}{3}\right)}\\ &=q(u)h(u) \end{align} i.e. $f(x)=q(x/12)h(x/12)$. You could easily check that $h(u)\sim \sqrt{u}$ hence for large enough $x$ \begin{align}\log f(x) &>\log q(x/12)\\ &=(\frac23x-\frac32)\log 2-(\frac14x+\frac12)\log3-2\log x+2\log12-\frac12\log\pi \end{align}
