Asymptotic of gamma function

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I came across a quetion:
Let $h$ go to zero. What is the asymptotic of $Gamma(x+o_p(h))$ where $xin(0,2)$? The difficulty is the limitation of x goes to zero.



Can I obtain $$Gamma(x+o_p(h))simGamma(x)$$
Any comments are welcomed.




fractions gamma-function gamma-distribution




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  • 3




    What is $o_p$? $
    – Szeto
    Jul 21 at 13:37














up vote
0
down vote

favorite












I came across a quetion:
Let $h$ go to zero. What is the asymptotic of $Gamma(x+o_p(h))$ where $xin(0,2)$? The difficulty is the limitation of x goes to zero.



Can I obtain $$Gamma(x+o_p(h))simGamma(x)$$
Any comments are welcomed.




fractions gamma-function gamma-distribution




share|cite|improve this question

















  • 3




    What is $o_p$? $
    – Szeto
    Jul 21 at 13:37












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I came across a quetion:
Let $h$ go to zero. What is the asymptotic of $Gamma(x+o_p(h))$ where $xin(0,2)$? The difficulty is the limitation of x goes to zero.



Can I obtain $$Gamma(x+o_p(h))simGamma(x)$$
Any comments are welcomed.




fractions gamma-function gamma-distribution




share|cite|improve this question













I came across a quetion:
Let $h$ go to zero. What is the asymptotic of $Gamma(x+o_p(h))$ where $xin(0,2)$? The difficulty is the limitation of x goes to zero.



Can I obtain $$Gamma(x+o_p(h))simGamma(x)$$
Any comments are welcomed.





fractions gamma-function gamma-distribution





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 21 at 12:51









Bernard

110k635103




110k635103









asked Jul 21 at 11:35









steven

296




296







  • 3




    What is $o_p$? $
    – Szeto
    Jul 21 at 13:37












  • 3




    What is $o_p$? $
    – Szeto
    Jul 21 at 13:37







3




3




What is $o_p$? $
– Szeto
Jul 21 at 13:37




What is $o_p$? $
– Szeto
Jul 21 at 13:37










1 Answer
1






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The series expansion of the function $Gamma(x)$ is :
$$Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$$
This formula is restricted at $x>0$ because $Gamma(xto 0)toinfty$.



The coefficients of the above Taylor series are computed as usual in terms of successive derivatives of $Gamma(x)$, which involves the polygamma functions $psi^(n)(x)$. The usual digamma function is $psi(x)=psi^(0)(x)$.



For $x=0$ and $epsilonto 0$ we have to look for asymptotic series :
$$Gamma(epsilon)sim frac1epsilon-gamma+frac112(6gamma^2+pi^2)epsilon+O(epsilon^3)$$
$gamma$ is the Euler-Mascheroni constant.
This is derived from the asymptotic series $(35)$ in http://mathworld.wolfram.com/GammaFunction.html






share|cite|improve this answer























  • Can you tell me the reference for the following result: $Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$, what $psi^(0)$ and $psi^(1)$?
    – steven
    Jul 21 at 16:40











  • functions.wolfram.com/GammaBetaErf/Gamma/06/01/04/01 functions.wolfram.com/GammaBetaErf/PolyGamma2
    – JJacquelin
    Jul 21 at 16:54










  • One can find the successive derivatives of $Gamma(x)$ from the definition of the polygamma functions $psi^(n)(x)=fracd^n+1dx^n+1ln(Gamma(x))$. mathworld.wolfram.com/PolygammaFunction.html . This allows to obtain the above Taylors series after some calculus.
    – JJacquelin
    Jul 21 at 17:10











  • I got it. Thanks for your kindly help.
    – steven
    Jul 22 at 1:22










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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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oldest

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up vote
1
down vote













The series expansion of the function $Gamma(x)$ is :
$$Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$$
This formula is restricted at $x>0$ because $Gamma(xto 0)toinfty$.



The coefficients of the above Taylor series are computed as usual in terms of successive derivatives of $Gamma(x)$, which involves the polygamma functions $psi^(n)(x)$. The usual digamma function is $psi(x)=psi^(0)(x)$.



For $x=0$ and $epsilonto 0$ we have to look for asymptotic series :
$$Gamma(epsilon)sim frac1epsilon-gamma+frac112(6gamma^2+pi^2)epsilon+O(epsilon^3)$$
$gamma$ is the Euler-Mascheroni constant.
This is derived from the asymptotic series $(35)$ in http://mathworld.wolfram.com/GammaFunction.html






share|cite|improve this answer























  • Can you tell me the reference for the following result: $Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$, what $psi^(0)$ and $psi^(1)$?
    – steven
    Jul 21 at 16:40











  • functions.wolfram.com/GammaBetaErf/Gamma/06/01/04/01 functions.wolfram.com/GammaBetaErf/PolyGamma2
    – JJacquelin
    Jul 21 at 16:54










  • One can find the successive derivatives of $Gamma(x)$ from the definition of the polygamma functions $psi^(n)(x)=fracd^n+1dx^n+1ln(Gamma(x))$. mathworld.wolfram.com/PolygammaFunction.html . This allows to obtain the above Taylors series after some calculus.
    – JJacquelin
    Jul 21 at 17:10











  • I got it. Thanks for your kindly help.
    – steven
    Jul 22 at 1:22














up vote
1
down vote













The series expansion of the function $Gamma(x)$ is :
$$Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$$
This formula is restricted at $x>0$ because $Gamma(xto 0)toinfty$.



The coefficients of the above Taylor series are computed as usual in terms of successive derivatives of $Gamma(x)$, which involves the polygamma functions $psi^(n)(x)$. The usual digamma function is $psi(x)=psi^(0)(x)$.



For $x=0$ and $epsilonto 0$ we have to look for asymptotic series :
$$Gamma(epsilon)sim frac1epsilon-gamma+frac112(6gamma^2+pi^2)epsilon+O(epsilon^3)$$
$gamma$ is the Euler-Mascheroni constant.
This is derived from the asymptotic series $(35)$ in http://mathworld.wolfram.com/GammaFunction.html






share|cite|improve this answer























  • Can you tell me the reference for the following result: $Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$, what $psi^(0)$ and $psi^(1)$?
    – steven
    Jul 21 at 16:40











  • functions.wolfram.com/GammaBetaErf/Gamma/06/01/04/01 functions.wolfram.com/GammaBetaErf/PolyGamma2
    – JJacquelin
    Jul 21 at 16:54










  • One can find the successive derivatives of $Gamma(x)$ from the definition of the polygamma functions $psi^(n)(x)=fracd^n+1dx^n+1ln(Gamma(x))$. mathworld.wolfram.com/PolygammaFunction.html . This allows to obtain the above Taylors series after some calculus.
    – JJacquelin
    Jul 21 at 17:10











  • I got it. Thanks for your kindly help.
    – steven
    Jul 22 at 1:22












up vote
1
down vote










up vote
1
down vote









The series expansion of the function $Gamma(x)$ is :
$$Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$$
This formula is restricted at $x>0$ because $Gamma(xto 0)toinfty$.



The coefficients of the above Taylor series are computed as usual in terms of successive derivatives of $Gamma(x)$, which involves the polygamma functions $psi^(n)(x)$. The usual digamma function is $psi(x)=psi^(0)(x)$.



For $x=0$ and $epsilonto 0$ we have to look for asymptotic series :
$$Gamma(epsilon)sim frac1epsilon-gamma+frac112(6gamma^2+pi^2)epsilon+O(epsilon^3)$$
$gamma$ is the Euler-Mascheroni constant.
This is derived from the asymptotic series $(35)$ in http://mathworld.wolfram.com/GammaFunction.html






share|cite|improve this answer















The series expansion of the function $Gamma(x)$ is :
$$Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$$
This formula is restricted at $x>0$ because $Gamma(xto 0)toinfty$.



The coefficients of the above Taylor series are computed as usual in terms of successive derivatives of $Gamma(x)$, which involves the polygamma functions $psi^(n)(x)$. The usual digamma function is $psi(x)=psi^(0)(x)$.



For $x=0$ and $epsilonto 0$ we have to look for asymptotic series :
$$Gamma(epsilon)sim frac1epsilon-gamma+frac112(6gamma^2+pi^2)epsilon+O(epsilon^3)$$
$gamma$ is the Euler-Mascheroni constant.
This is derived from the asymptotic series $(35)$ in http://mathworld.wolfram.com/GammaFunction.html







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 21 at 15:52


























answered Jul 21 at 15:45









JJacquelin

40k21649




40k21649











  • Can you tell me the reference for the following result: $Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$, what $psi^(0)$ and $psi^(1)$?
    – steven
    Jul 21 at 16:40











  • functions.wolfram.com/GammaBetaErf/Gamma/06/01/04/01 functions.wolfram.com/GammaBetaErf/PolyGamma2
    – JJacquelin
    Jul 21 at 16:54










  • One can find the successive derivatives of $Gamma(x)$ from the definition of the polygamma functions $psi^(n)(x)=fracd^n+1dx^n+1ln(Gamma(x))$. mathworld.wolfram.com/PolygammaFunction.html . This allows to obtain the above Taylors series after some calculus.
    – JJacquelin
    Jul 21 at 17:10











  • I got it. Thanks for your kindly help.
    – steven
    Jul 22 at 1:22
















  • Can you tell me the reference for the following result: $Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$, what $psi^(0)$ and $psi^(1)$?
    – steven
    Jul 21 at 16:40











  • functions.wolfram.com/GammaBetaErf/Gamma/06/01/04/01 functions.wolfram.com/GammaBetaErf/PolyGamma2
    – JJacquelin
    Jul 21 at 16:54










  • One can find the successive derivatives of $Gamma(x)$ from the definition of the polygamma functions $psi^(n)(x)=fracd^n+1dx^n+1ln(Gamma(x))$. mathworld.wolfram.com/PolygammaFunction.html . This allows to obtain the above Taylors series after some calculus.
    – JJacquelin
    Jul 21 at 17:10











  • I got it. Thanks for your kindly help.
    – steven
    Jul 22 at 1:22















Can you tell me the reference for the following result: $Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$, what $psi^(0)$ and $psi^(1)$?
– steven
Jul 21 at 16:40





Can you tell me the reference for the following result: $Gamma(x+epsilon)=Gamma(x)+Gamma(x)left(psi^(0)(x)right):epsilon+Gamma(x)left(left(psi^(0)(x)right)^2+psi^(1)(x)right) :fracepsilon^22+O(epsilon^3)$, what $psi^(0)$ and $psi^(1)$?
– steven
Jul 21 at 16:40













functions.wolfram.com/GammaBetaErf/Gamma/06/01/04/01 functions.wolfram.com/GammaBetaErf/PolyGamma2
– JJacquelin
Jul 21 at 16:54




functions.wolfram.com/GammaBetaErf/Gamma/06/01/04/01 functions.wolfram.com/GammaBetaErf/PolyGamma2
– JJacquelin
Jul 21 at 16:54












One can find the successive derivatives of $Gamma(x)$ from the definition of the polygamma functions $psi^(n)(x)=fracd^n+1dx^n+1ln(Gamma(x))$. mathworld.wolfram.com/PolygammaFunction.html . This allows to obtain the above Taylors series after some calculus.
– JJacquelin
Jul 21 at 17:10





One can find the successive derivatives of $Gamma(x)$ from the definition of the polygamma functions $psi^(n)(x)=fracd^n+1dx^n+1ln(Gamma(x))$. mathworld.wolfram.com/PolygammaFunction.html . This allows to obtain the above Taylors series after some calculus.
– JJacquelin
Jul 21 at 17:10













I got it. Thanks for your kindly help.
– steven
Jul 22 at 1:22




I got it. Thanks for your kindly help.
– steven
Jul 22 at 1:22












 

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