17
$\begingroup$

The $\Gamma (x)$ function has just one minimum for $x>0$ . This result uses some properties of the gamma function:

  • $\Gamma ^{\prime \prime }(x)>0$ and $\Gamma (x)>0$ for all $x>0$
  • $\Gamma (1)=\Gamma (2)=1$.

Observing the following graph (created in SWP) of $y=\Gamma (x)$ this minimum is near $x=3/2$, but likely $\min \Gamma (x)\neq \Gamma \left( 3/2\right) =\dfrac{1}{2}\Gamma \left( 1/2\right) =\dfrac{1}{2}\sqrt{\pi }$.

alt text

I think that it is not possible to find analytically the exact value of $x_{\min }$, even by converting to an adequate problem in the interval $]0,1]$ and using the functional equation $\Gamma (x+1)=x\Gamma (x)$ and the reflection formula

$\Gamma (p)\Gamma (p-1)=\dfrac{\pi }{\sin px}\qquad $( $0\lt p\lt 1$)

Question:

a) Which is the best way to find $\min_{[1,2]}\Gamma (x)$ and does $x_{\min }$ lay in $[1,3/2]$ or in $[3/2,2]$?

b) Is there some useful series expansion of $\Gamma (x)$?

c) Which numeric method do you suggest?


Edit: Due to the shape of $\Gamma (x)$ I thought on the one-dimensional Davies-Swann-Campey method of direct search for unconstrained optimization, which approximates a function near a minimum by successive approximating quadratic polynomials.

$\endgroup$
  • 2
    $\begingroup$ A general rule of thumb in numerical computing: it's easier to compute (simple) roots of functions to the full precision of your environment than to compute extrema. $\endgroup$ – J. M. is a poor mathematician Aug 24 '10 at 22:18
  • $\begingroup$ wolframalpha.com/input/?i=min+Gamma%28x%29+from+0+to+3 $\endgroup$ – Memming May 9 '12 at 15:33
  • 1
    $\begingroup$ computation results in 1935 nature.com/nature/journal/v135/n3422/abs/135917b0.html $\endgroup$ – Memming May 9 '12 at 15:46
  • $\begingroup$ @Memming Thanks for the link. $\endgroup$ – Américo Tavares May 9 '12 at 15:48
  • $\begingroup$ So as it's not simple to compute the value analytically, it's worth stating that $\left(\frac{33}{20}\cdot\frac{\pi^4}{110}\ ,\ \frac{\pi^4}{110}\right)$ is an approximation to $(x,y)_{\text{min}}$ with an error of $\mathcal{O}(10^{-4})$ in the first and even less in the second variable. $\endgroup$ – Nikolaj-K Mar 1 '13 at 11:04
8
$\begingroup$

There indeed is no closed-form for the gamma function's minimum; what you can do instead, however, is to find the positive root of the digamma function (the logarithmic derivative of the gamma function), which should be available in your computing environment.

$\endgroup$
  • 1
    $\begingroup$ For instance, the computation in Mathematica goes something like x /. FindRoot[PolyGamma[x], {x, 1}, WorkingPrecision -> 20] which yields the result 1.4616321449683623413 . $\endgroup$ – J. M. is a poor mathematician Aug 24 '10 at 22:11
  • $\begingroup$ Thanks, you have answered to a), b) and confirmed that it is not possible to find analytically the gamma function's minimum! Since I have no access to Mathematica I will try to use PARI. $\endgroup$ – Américo Tavares Aug 24 '10 at 22:31
  • 1
    $\begingroup$ Américo: As for question b.), there are series expansions (see DLMF for instance), but none of them seem to be practical so I'd steer away from them. I believe PARI/GP has a digamma/polygamma function somewhere, just check the docs. $\endgroup$ – J. M. is a poor mathematician Aug 24 '10 at 22:42
  • $\begingroup$ @Mangaldan: I will check. Also I intend to study further the subject of series expansions for the present situation. $\endgroup$ – Américo Tavares Aug 24 '10 at 23:03
  • $\begingroup$ I'm actually trying to discourage you from looking at the series expansions (well okay, let's make an exception for Stirling); but if you think you can glean new insights, dig in: dlmf.nist.gov/5.7 $\endgroup$ – J. M. is a poor mathematician Aug 24 '10 at 23:17
4
$\begingroup$

According to MathWorld the minimum of the Gamma function for positive $x$ is 1.46163...; in particular I guess this is enough to deduce that it is smaller than $3/2$. You can follow the links along to find some references where this is proved.

$\endgroup$
  • $\begingroup$ Yes, it is enough. Thanks! In this case I have not checked before in MathWorld, although usualy I search many things there. $\endgroup$ – Américo Tavares Aug 24 '10 at 22:35

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.