Minimum of the Gamma Function $Gamma (x)$ for $x>0$. How to find $x_min$?What non-integer number has the smallest factorial?At which value (over $mathbbR^+$) is the gamma function strictly increasing?Explicit series for the minimum point of the Gamma function?Irrationality of $min$ and $arg,min$ of $Gamma|_[1, 2]$Relevance of the Gamma Function's local minimum at $xapprox 1.4616$How do you prove Gautschi's inequality for the gamma function?Limit for gamma functionCalculating the divisor, known to be small, of two Stirling approximations of the logarithmic Gamma function without overflowsBounding the Gamma FunctionMinimize the coefficient of asymptotic normalityDerivative of the Gamma functionOther forms for the derivative of the Gamma functionGeneralized gradient descent with constraintsNumerical methods for calculating derivative of gamma functionUsing the Weierstrass formula for the Gamma Function to find an expresision for $f(z)$

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Minimum of the Gamma Function $Gamma (x)$ for $x>0$. How to find $x_min$?


What non-integer number has the smallest factorial?At which value (over $mathbbR^+$) is the gamma function strictly increasing?Explicit series for the minimum point of the Gamma function?Irrationality of $min$ and $arg,min$ of $Gamma|_[1, 2]$Relevance of the Gamma Function's local minimum at $xapprox 1.4616$How do you prove Gautschi's inequality for the gamma function?Limit for gamma functionCalculating the divisor, known to be small, of two Stirling approximations of the logarithmic Gamma function without overflowsBounding the Gamma FunctionMinimize the coefficient of asymptotic normalityDerivative of the Gamma functionOther forms for the derivative of the Gamma functionGeneralized gradient descent with constraintsNumerical methods for calculating derivative of gamma functionUsing the Weierstrass formula for the Gamma Function to find an expresision for $f(z)$













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( 3/2right) =dfrac12Gamma left( 1/2right) =dfrac12sqrtpi $.



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 $Gamma (x+1)=xGamma (x)$ and the reflection formula



$Gamma (p)Gamma (p-1)=dfracpi sin pxqquad $( $0lt plt 1$)



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.










share|cite|improve this question











$endgroup$







  • 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 not a 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 $left(frac3320cdotfracpi^4110 , fracpi^4110right)$ is an approximation to $(x,y)_textmin$ with an error of $mathcalO(10^-4)$ in the first and even less in the second variable.
    $endgroup$
    – Nikolaj-K
    Mar 1 '13 at 11:04
















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( 3/2right) =dfrac12Gamma left( 1/2right) =dfrac12sqrtpi $.



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 $Gamma (x+1)=xGamma (x)$ and the reflection formula



$Gamma (p)Gamma (p-1)=dfracpi sin pxqquad $( $0lt plt 1$)



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.










share|cite|improve this question











$endgroup$







  • 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 not a 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 $left(frac3320cdotfracpi^4110 , fracpi^4110right)$ is an approximation to $(x,y)_textmin$ with an error of $mathcalO(10^-4)$ in the first and even less in the second variable.
    $endgroup$
    – Nikolaj-K
    Mar 1 '13 at 11:04














17












17








17


6



$begingroup$


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( 3/2right) =dfrac12Gamma left( 1/2right) =dfrac12sqrtpi $.



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 $Gamma (x+1)=xGamma (x)$ and the reflection formula



$Gamma (p)Gamma (p-1)=dfracpi sin pxqquad $( $0lt plt 1$)



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.










share|cite|improve this question











$endgroup$




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( 3/2right) =dfrac12Gamma left( 1/2right) =dfrac12sqrtpi $.



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 $Gamma (x+1)=xGamma (x)$ and the reflection formula



$Gamma (p)Gamma (p-1)=dfracpi sin pxqquad $( $0lt plt 1$)



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.







calculus approximation gamma-function numerical-methods special-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 29 '10 at 10:20







Américo Tavares

















asked Aug 24 '10 at 21:59









Américo TavaresAmérico Tavares

32.5k1181206




32.5k1181206







  • 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 not a 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 $left(frac3320cdotfracpi^4110 , fracpi^4110right)$ is an approximation to $(x,y)_textmin$ with an error of $mathcalO(10^-4)$ in the first and even less in the second variable.
    $endgroup$
    – Nikolaj-K
    Mar 1 '13 at 11:04













  • 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 not a 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 $left(frac3320cdotfracpi^4110 , fracpi^4110right)$ is an approximation to $(x,y)_textmin$ with an error of $mathcalO(10^-4)$ in the first and even less in the second variable.
    $endgroup$
    – Nikolaj-K
    Mar 1 '13 at 11:04








2




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 not a mathematician
Aug 24 '10 at 22:18




$begingroup$
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 not a 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




$begingroup$
wolframalpha.com/input/?i=min+Gamma%28x%29+from+0+to+3
$endgroup$
– Memming
May 9 '12 at 15:33




1




1




$begingroup$
computation results in 1935 nature.com/nature/journal/v135/n3422/abs/135917b0.html
$endgroup$
– Memming
May 9 '12 at 15:46




$begingroup$
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




$begingroup$
@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 $left(frac3320cdotfracpi^4110 , fracpi^4110right)$ is an approximation to $(x,y)_textmin$ with an error of $mathcalO(10^-4)$ in the first and even less in the second variable.
$endgroup$
– Nikolaj-K
Mar 1 '13 at 11:04





$begingroup$
So as it's not simple to compute the value analytically, it's worth stating that $left(frac3320cdotfracpi^4110 , fracpi^4110right)$ is an approximation to $(x,y)_textmin$ with an error of $mathcalO(10^-4)$ in the first and even less in the second variable.
$endgroup$
– Nikolaj-K
Mar 1 '13 at 11:04











2 Answers
2






active

oldest

votes


















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.






share|cite|improve this answer









$endgroup$








  • 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 not a 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 not a 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 not a 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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









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.






share|cite|improve this answer









$endgroup$








  • 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 not a 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 not a 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 not a mathematician
    Aug 24 '10 at 23:17















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.






share|cite|improve this answer









$endgroup$








  • 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 not a 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 not a 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 not a mathematician
    Aug 24 '10 at 23:17













8












8








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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 24 '10 at 22:09









J. M. is not a mathematicianJ. M. is not a mathematician

61.4k5152290




61.4k5152290







  • 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 not a 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 not a 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 not a mathematician
    Aug 24 '10 at 23:17












  • 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 not a 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 not a 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 not a mathematician
    Aug 24 '10 at 23:17







1




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 not a mathematician
Aug 24 '10 at 22:11




$begingroup$
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 not a 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




$begingroup$
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




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 not a mathematician
Aug 24 '10 at 22:42




$begingroup$
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 not a 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





$begingroup$
@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 not a mathematician
Aug 24 '10 at 23:17




$begingroup$
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 not a 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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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
















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.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    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














4












4








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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 24 '10 at 22:08









damianodamiano

1,7161010




1,7161010











  • $begingroup$
    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

















  • $begingroup$
    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
















$begingroup$
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





$begingroup$
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


















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