Minimum of the Gamma Function $Gamma (x)$ for $x>0$. How to find $x_{min}$?What non-integer number has the...
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Minimum of the Gamma Function $Gamma (x)$ for $x>0$. How to find $x_{min}$?
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$begingroup$
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 $min Gamma (x)neq Gamma left( 3/2right) =dfrac{1}{2}Gamma left( 1/2right) =dfrac{1}{2}sqrt{pi }$.
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)=dfrac{pi }{sin px}qquad $( $0lt plt 1$)
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
asked Aug 24 '10 at 21:59
Américo TavaresAmérico Tavares
32.5k1181206
$endgroup$
add a comment |
$begingroup$
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 $min Gamma (x)neq Gamma left( 3/2right) =dfrac{1}{2}Gamma left( 1/2right) =dfrac{1}{2}sqrt{pi }$.
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)=dfrac{pi }{sin px}qquad $( $0lt plt 1$)
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
asked Aug 24 '10 at 21:59
Américo TavaresAmérico Tavares
32.5k1181206
$endgroup$
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 $left(frac{33}{20}cdotfrac{pi^4}{110} , frac{pi^4}{110}right)$ 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
add a comment |
$begingroup$
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 $min Gamma (x)neq Gamma left( 3/2right) =dfrac{1}{2}Gamma left( 1/2right) =dfrac{1}{2}sqrt{pi }$.
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)=dfrac{pi }{sin px}qquad $( $0lt plt 1$)
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
asked Aug 24 '10 at 21:59
Américo TavaresAmérico Tavares
32.5k1181206
$endgroup$
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 $min Gamma (x)neq Gamma left( 3/2right) =dfrac{1}{2}Gamma left( 1/2right) =dfrac{1}{2}sqrt{pi }$.
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)=dfrac{pi }{sin px}qquad $( $0lt plt 1$)
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
calculus approximation gamma-function numerical-methods special-functions
asked Aug 24 '10 at 21:59
Américo TavaresAmérico Tavares
32.5k1181206
asked Aug 24 '10 at 21:59
Américo TavaresAmérico Tavares
32.5k1181206
edited Aug 29 '10 at 10:20
Américo Tavares
asked Aug 24 '10 at 21:59
Américo TavaresAmérico Tavares
32.5k1181206
asked Aug 24 '10 at 21:59
Américo TavaresAmérico Tavares
32.5k1181206
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 $left(frac{33}{20}cdotfrac{pi^4}{110} , frac{pi^4}{110}right)$ 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
add a comment |
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 $left(frac{33}{20}cdotfrac{pi^4}{110} , frac{pi^4}{110}right)$ 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
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 $left(frac{33}{20}cdotfrac{pi^4}{110} , frac{pi^4}{110}right)$ 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
$begingroup$
So as it's not simple to compute the value analytically, it's worth stating that $left(frac{33}{20}cdotfrac{pi^4}{110} , frac{pi^4}{110}right)$ 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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Aug 24 '10 at 22:09
J. M. is not a mathematicianJ. M. is not a mathematician
61.5k5152290
$endgroup$
1
$begingroup$
For instance, the computation in Mathematica goes something likex /. 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
|
show 3 more comments
$begingroup$
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.
answered Aug 24 '10 at 22:08
damianodamiano
1,7161010
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Aug 24 '10 at 22:09
J. M. is not a mathematicianJ. M. is not a mathematician
61.5k5152290
$endgroup$
1
$begingroup$
For instance, the computation in Mathematica goes something likex /. 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
|
show 3 more comments
$begingroup$
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.
answered Aug 24 '10 at 22:09
J. M. is not a mathematicianJ. M. is not a mathematician
61.5k5152290
$endgroup$
1
$begingroup$
For instance, the computation in Mathematica goes something likex /. 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
|
show 3 more comments
$begingroup$
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.
answered Aug 24 '10 at 22:09
J. M. is not a mathematicianJ. M. is not a mathematician
61.5k5152290
$endgroup$
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.
answered Aug 24 '10 at 22:09
J. M. is not a mathematicianJ. M. is not a mathematician
61.5k5152290
answered Aug 24 '10 at 22:09
J. M. is not a mathematicianJ. M. is not a mathematician
61.5k5152290
answered Aug 24 '10 at 22:09
J. M. is not a mathematicianJ. M. is not a mathematician
61.5k5152290
answered Aug 24 '10 at 22:09
J. M. is not a mathematicianJ. M. is not a mathematician
61.5k5152290
61.5k5152290
1
$begingroup$
For instance, the computation in Mathematica goes something likex /. 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
|
show 3 more comments
1
$begingroup$
For instance, the computation in Mathematica goes something likex /. 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
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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.
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– Américo Tavares
Aug 24 '10 at 22:31
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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.
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– Américo Tavares
Aug 24 '10 at 22:31
1
1
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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.
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– J. M. is not a mathematician
Aug 24 '10 at 22:42
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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.
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– J. M. is not a mathematician
Aug 24 '10 at 22:42
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@Mangaldan: I will check. Also I intend to study further the subject of series expansions for the present situation.
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– Américo Tavares
Aug 24 '10 at 23:03
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@Mangaldan: I will check. Also I intend to study further the subject of series expansions for the present situation.
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– Américo Tavares
Aug 24 '10 at 23:03
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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
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– J. M. is not a mathematician
Aug 24 '10 at 23:17
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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
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– J. M. is not a mathematician
Aug 24 '10 at 23:17
|
show 3 more comments
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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.
answered Aug 24 '10 at 22:08
damianodamiano
1,7161010
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Yes, it is enough. Thanks! In this case I have not checked before in MathWorld, although usualy I search many things there.
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– Américo Tavares
Aug 24 '10 at 22:35
add a comment |
$begingroup$
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.
answered Aug 24 '10 at 22:08
damianodamiano
1,7161010
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Aug 24 '10 at 22:08
damianodamiano
1,7161010
$endgroup$
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.
answered Aug 24 '10 at 22:08
damianodamiano
1,7161010
answered Aug 24 '10 at 22:08
damianodamiano
1,7161010
answered Aug 24 '10 at 22:08
damianodamiano
1,7161010
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
add a comment |
$begingroup$
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
add a comment |
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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.
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– J. M. is not a mathematician
Aug 24 '10 at 22:18
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wolframalpha.com/input/?i=min+Gamma%28x%29+from+0+to+3
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– Memming
May 9 '12 at 15:33
1
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computation results in 1935 nature.com/nature/journal/v135/n3422/abs/135917b0.html
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– Memming
May 9 '12 at 15:46
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@Memming Thanks for the link.
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– Américo Tavares
May 9 '12 at 15:48
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So as it's not simple to compute the value analytically, it's worth stating that $left(frac{33}{20}cdotfrac{pi^4}{110} , frac{pi^4}{110}right)$ 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.
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– Nikolaj-K
Mar 1 '13 at 11:04