Application of Legendre's duplication formula The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Stirling-type formula for the logarithmic derivative of the Gamma functionVariations on the Stirling's formula for $Gamma(z)$Stirling's Approximation and Ramanujan's Proof of Bertrand's PostulateThe Duplication Formula for the Gamma Function by logarithmic derivatives.Asymptotic behaviour of generalized binomial coefficient $fracan(an-1)…(an-n+1)n!$Gamma duplication formula via HadamardDuplication formula for beta functionLegendre Gamma duplication formulaIntegrating $psi'$ to get $Gamma$On Ramanujan's proof of Bertrand's postulate (using stirling formula)
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Application of Legendre's duplication formula
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Stirling-type formula for the logarithmic derivative of the Gamma functionVariations on the Stirling's formula for $Gamma(z)$Stirling's Approximation and Ramanujan's Proof of Bertrand's PostulateThe Duplication Formula for the Gamma Function by logarithmic derivatives.Asymptotic behaviour of generalized binomial coefficient $fracan(an-1)…(an-n+1)n!$Gamma duplication formula via HadamardDuplication formula for beta functionLegendre Gamma duplication formulaIntegrating $psi'$ to get $Gamma$On Ramanujan's proof of Bertrand's postulate (using stirling formula)
$begingroup$
I am reading the book "Special functions: an introduction to classical functions of mathematical physics" by Nico M. Temme and I'm having trouble understanding how to find a constant (in page 62).
I understand why
$$
logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C,
$$
where $g$ is bounded for $tgt0$ and $C$ is a constant of integration, which has to be determined. I know that Legendre's duplication formula can be written in the form
$$
log frac2^2zGamma(z+1)Gamma(z+frac12)Gamma(2z+1) = frac12log pi.
$$
Now, the book asserts the following: Letting $ztoinfty$ and using Legendre's duplication formula we obtain $C=frac12log 2pi.$ But I don't understand how to use the duplication formula, when I let $ztoinfty$ the integral vanishes (since $g$ is bounded). Thus
$$
C=lim_ztoinfty logGamma(z+1)-(z+frac12)log z + z.
$$
Now, using Stirling's approximation, the above limit equals $frac12log 2pi$, but I don't want to use Stirling's approximation, I want to know how to use the duplication formula.
analysis gamma-function digamma-function
asked Mar 24 at 16:41
SebitasSebitas
707
$endgroup$
add a comment |
$begingroup$
I am reading the book "Special functions: an introduction to classical functions of mathematical physics" by Nico M. Temme and I'm having trouble understanding how to find a constant (in page 62).
I understand why
$$
logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C,
$$
where $g$ is bounded for $tgt0$ and $C$ is a constant of integration, which has to be determined. I know that Legendre's duplication formula can be written in the form
$$
log frac2^2zGamma(z+1)Gamma(z+frac12)Gamma(2z+1) = frac12log pi.
$$
Now, the book asserts the following: Letting $ztoinfty$ and using Legendre's duplication formula we obtain $C=frac12log 2pi.$ But I don't understand how to use the duplication formula, when I let $ztoinfty$ the integral vanishes (since $g$ is bounded). Thus
$$
C=lim_ztoinfty logGamma(z+1)-(z+frac12)log z + z.
$$
Now, using Stirling's approximation, the above limit equals $frac12log 2pi$, but I don't want to use Stirling's approximation, I want to know how to use the duplication formula.
analysis gamma-function digamma-function
asked Mar 24 at 16:41
SebitasSebitas
707
$endgroup$
$begingroup$
It has nothing obvious that $logGamma(z+1)=(z+frac12)log z - z + int_-infty^inftyg(t)e^-ztdt + C$ even less that $g(t) = 0$ for $t< 0$. And it depends on the tools you want to use : complex analysis, Fourier analysis, some theorems like duplication, reflection, Stirling
$endgroup$
– reuns
Mar 24 at 19:10
$begingroup$
@reuns I only want to use the duplication formula
$endgroup$
– Sebitas
Mar 24 at 19:14
$begingroup$
Once you know that $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ then the duplication formula gives $C = lim_z to inftylog 2^2z + logGamma(z+1)+log Gamma(z+frac12)-logGamma(2z+1)$. But again $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ has nothing obvious and you should probably look again at the steps they used to find it.
$endgroup$
– reuns
Mar 24 at 19:35
$begingroup$
I know it is not obvious, but after some trouble I was able to prove that $logGamma(z+1)=(z+frac12)log z-z + int_0^infty g(t)e^-ztdt + C$ but from this I really don't see how the duplication formula gives $C=lim_ztoinfty log 2^2z + log Gamma(z+1) + log Gamma(z+frac12)-logGamma(2z+1).$ Could you please elaborate that line?
$endgroup$
– Sebitas
Mar 24 at 20:24
add a comment |
$begingroup$
I am reading the book "Special functions: an introduction to classical functions of mathematical physics" by Nico M. Temme and I'm having trouble understanding how to find a constant (in page 62).
I understand why
$$
logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C,
$$
where $g$ is bounded for $tgt0$ and $C$ is a constant of integration, which has to be determined. I know that Legendre's duplication formula can be written in the form
$$
log frac2^2zGamma(z+1)Gamma(z+frac12)Gamma(2z+1) = frac12log pi.
$$
Now, the book asserts the following: Letting $ztoinfty$ and using Legendre's duplication formula we obtain $C=frac12log 2pi.$ But I don't understand how to use the duplication formula, when I let $ztoinfty$ the integral vanishes (since $g$ is bounded). Thus
$$
C=lim_ztoinfty logGamma(z+1)-(z+frac12)log z + z.
$$
Now, using Stirling's approximation, the above limit equals $frac12log 2pi$, but I don't want to use Stirling's approximation, I want to know how to use the duplication formula.
analysis gamma-function digamma-function
asked Mar 24 at 16:41
SebitasSebitas
707
$endgroup$
I am reading the book "Special functions: an introduction to classical functions of mathematical physics" by Nico M. Temme and I'm having trouble understanding how to find a constant (in page 62).
I understand why
$$
logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C,
$$
where $g$ is bounded for $tgt0$ and $C$ is a constant of integration, which has to be determined. I know that Legendre's duplication formula can be written in the form
$$
log frac2^2zGamma(z+1)Gamma(z+frac12)Gamma(2z+1) = frac12log pi.
$$
Now, the book asserts the following: Letting $ztoinfty$ and using Legendre's duplication formula we obtain $C=frac12log 2pi.$ But I don't understand how to use the duplication formula, when I let $ztoinfty$ the integral vanishes (since $g$ is bounded). Thus
$$
C=lim_ztoinfty logGamma(z+1)-(z+frac12)log z + z.
$$
Now, using Stirling's approximation, the above limit equals $frac12log 2pi$, but I don't want to use Stirling's approximation, I want to know how to use the duplication formula.
analysis gamma-function digamma-function
analysis gamma-function digamma-function
asked Mar 24 at 16:41
SebitasSebitas
707
asked Mar 24 at 16:41
SebitasSebitas
707
edited Mar 24 at 17:53
Sebitas
asked Mar 24 at 16:41
SebitasSebitas
707
asked Mar 24 at 16:41
SebitasSebitas
707
asked Mar 24 at 16:41
SebitasSebitas
707
707
$begingroup$
It has nothing obvious that $logGamma(z+1)=(z+frac12)log z - z + int_-infty^inftyg(t)e^-ztdt + C$ even less that $g(t) = 0$ for $t< 0$. And it depends on the tools you want to use : complex analysis, Fourier analysis, some theorems like duplication, reflection, Stirling
$endgroup$
– reuns
Mar 24 at 19:10
$begingroup$
@reuns I only want to use the duplication formula
$endgroup$
– Sebitas
Mar 24 at 19:14
$begingroup$
Once you know that $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ then the duplication formula gives $C = lim_z to inftylog 2^2z + logGamma(z+1)+log Gamma(z+frac12)-logGamma(2z+1)$. But again $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ has nothing obvious and you should probably look again at the steps they used to find it.
$endgroup$
– reuns
Mar 24 at 19:35
$begingroup$
I know it is not obvious, but after some trouble I was able to prove that $logGamma(z+1)=(z+frac12)log z-z + int_0^infty g(t)e^-ztdt + C$ but from this I really don't see how the duplication formula gives $C=lim_ztoinfty log 2^2z + log Gamma(z+1) + log Gamma(z+frac12)-logGamma(2z+1).$ Could you please elaborate that line?
$endgroup$
– Sebitas
Mar 24 at 20:24
add a comment |
$begingroup$
It has nothing obvious that $logGamma(z+1)=(z+frac12)log z - z + int_-infty^inftyg(t)e^-ztdt + C$ even less that $g(t) = 0$ for $t< 0$. And it depends on the tools you want to use : complex analysis, Fourier analysis, some theorems like duplication, reflection, Stirling
$endgroup$
– reuns
Mar 24 at 19:10
$begingroup$
@reuns I only want to use the duplication formula
$endgroup$
– Sebitas
Mar 24 at 19:14
$begingroup$
Once you know that $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ then the duplication formula gives $C = lim_z to inftylog 2^2z + logGamma(z+1)+log Gamma(z+frac12)-logGamma(2z+1)$. But again $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ has nothing obvious and you should probably look again at the steps they used to find it.
$endgroup$
– reuns
Mar 24 at 19:35
$begingroup$
I know it is not obvious, but after some trouble I was able to prove that $logGamma(z+1)=(z+frac12)log z-z + int_0^infty g(t)e^-ztdt + C$ but from this I really don't see how the duplication formula gives $C=lim_ztoinfty log 2^2z + log Gamma(z+1) + log Gamma(z+frac12)-logGamma(2z+1).$ Could you please elaborate that line?
$endgroup$
– Sebitas
Mar 24 at 20:24
$begingroup$
It has nothing obvious that $logGamma(z+1)=(z+frac12)log z - z + int_-infty^inftyg(t)e^-ztdt + C$ even less that $g(t) = 0$ for $t< 0$. And it depends on the tools you want to use : complex analysis, Fourier analysis, some theorems like duplication, reflection, Stirling
$endgroup$
– reuns
Mar 24 at 19:10
$begingroup$
It has nothing obvious that $logGamma(z+1)=(z+frac12)log z - z + int_-infty^inftyg(t)e^-ztdt + C$ even less that $g(t) = 0$ for $t< 0$. And it depends on the tools you want to use : complex analysis, Fourier analysis, some theorems like duplication, reflection, Stirling
$endgroup$
– reuns
Mar 24 at 19:10
$begingroup$
@reuns I only want to use the duplication formula
$endgroup$
– Sebitas
Mar 24 at 19:14
$begingroup$
@reuns I only want to use the duplication formula
$endgroup$
– Sebitas
Mar 24 at 19:14
$begingroup$
Once you know that $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ then the duplication formula gives $C = lim_z to inftylog 2^2z + logGamma(z+1)+log Gamma(z+frac12)-logGamma(2z+1)$. But again $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ has nothing obvious and you should probably look again at the steps they used to find it.
$endgroup$
– reuns
Mar 24 at 19:35
$begingroup$
Once you know that $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ then the duplication formula gives $C = lim_z to inftylog 2^2z + logGamma(z+1)+log Gamma(z+frac12)-logGamma(2z+1)$. But again $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ has nothing obvious and you should probably look again at the steps they used to find it.
$endgroup$
– reuns
Mar 24 at 19:35
$begingroup$
I know it is not obvious, but after some trouble I was able to prove that $logGamma(z+1)=(z+frac12)log z-z + int_0^infty g(t)e^-ztdt + C$ but from this I really don't see how the duplication formula gives $C=lim_ztoinfty log 2^2z + log Gamma(z+1) + log Gamma(z+frac12)-logGamma(2z+1).$ Could you please elaborate that line?
$endgroup$
– Sebitas
Mar 24 at 20:24
$begingroup$
I know it is not obvious, but after some trouble I was able to prove that $logGamma(z+1)=(z+frac12)log z-z + int_0^infty g(t)e^-ztdt + C$ but from this I really don't see how the duplication formula gives $C=lim_ztoinfty log 2^2z + log Gamma(z+1) + log Gamma(z+frac12)-logGamma(2z+1).$ Could you please elaborate that line?
$endgroup$
– Sebitas
Mar 24 at 20:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, define the two functions
$$ f(z) := (z + frac12)log(z)-z,quad
F(z) := f(z) + f(z - frac12) - f(2z). $$
Second, find that the power series expansion at infinity of
$ F(z) $ is
$$ F(z) = -2log(2) z - frac12 log(2) - frac18 z^-1 + O(z^-2). $$
The rest is a simple exercise.
answered Mar 24 at 19:58
SomosSomos
14.9k11337
$endgroup$
$begingroup$
Finally I understand why you defined such functions, thank you very much!
$endgroup$
– Sebitas
Mar 24 at 21:21
add a comment |
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1 Answer
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1 Answer
1
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$begingroup$
First, define the two functions
$$ f(z) := (z + frac12)log(z)-z,quad
F(z) := f(z) + f(z - frac12) - f(2z). $$
Second, find that the power series expansion at infinity of
$ F(z) $ is
$$ F(z) = -2log(2) z - frac12 log(2) - frac18 z^-1 + O(z^-2). $$
The rest is a simple exercise.
answered Mar 24 at 19:58
SomosSomos
14.9k11337
$endgroup$
$begingroup$
Finally I understand why you defined such functions, thank you very much!
$endgroup$
– Sebitas
Mar 24 at 21:21
add a comment |
$begingroup$
First, define the two functions
$$ f(z) := (z + frac12)log(z)-z,quad
F(z) := f(z) + f(z - frac12) - f(2z). $$
Second, find that the power series expansion at infinity of
$ F(z) $ is
$$ F(z) = -2log(2) z - frac12 log(2) - frac18 z^-1 + O(z^-2). $$
The rest is a simple exercise.
answered Mar 24 at 19:58
SomosSomos
14.9k11337
$endgroup$
$begingroup$
Finally I understand why you defined such functions, thank you very much!
$endgroup$
– Sebitas
Mar 24 at 21:21
add a comment |
$begingroup$
First, define the two functions
$$ f(z) := (z + frac12)log(z)-z,quad
F(z) := f(z) + f(z - frac12) - f(2z). $$
Second, find that the power series expansion at infinity of
$ F(z) $ is
$$ F(z) = -2log(2) z - frac12 log(2) - frac18 z^-1 + O(z^-2). $$
The rest is a simple exercise.
answered Mar 24 at 19:58
SomosSomos
14.9k11337
$endgroup$
First, define the two functions
$$ f(z) := (z + frac12)log(z)-z,quad
F(z) := f(z) + f(z - frac12) - f(2z). $$
Second, find that the power series expansion at infinity of
$ F(z) $ is
$$ F(z) = -2log(2) z - frac12 log(2) - frac18 z^-1 + O(z^-2). $$
The rest is a simple exercise.
answered Mar 24 at 19:58
SomosSomos
14.9k11337
answered Mar 24 at 19:58
SomosSomos
14.9k11337
answered Mar 24 at 19:58
SomosSomos
14.9k11337
answered Mar 24 at 19:58
SomosSomos
14.9k11337
14.9k11337
$begingroup$
Finally I understand why you defined such functions, thank you very much!
$endgroup$
– Sebitas
Mar 24 at 21:21
add a comment |
$begingroup$
Finally I understand why you defined such functions, thank you very much!
$endgroup$
– Sebitas
Mar 24 at 21:21
$begingroup$
Finally I understand why you defined such functions, thank you very much!
$endgroup$
– Sebitas
Mar 24 at 21:21
$begingroup$
Finally I understand why you defined such functions, thank you very much!
$endgroup$
– Sebitas
Mar 24 at 21:21
add a comment |
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$begingroup$
It has nothing obvious that $logGamma(z+1)=(z+frac12)log z - z + int_-infty^inftyg(t)e^-ztdt + C$ even less that $g(t) = 0$ for $t< 0$. And it depends on the tools you want to use : complex analysis, Fourier analysis, some theorems like duplication, reflection, Stirling
$endgroup$
– reuns
Mar 24 at 19:10
$begingroup$
@reuns I only want to use the duplication formula
$endgroup$
– Sebitas
Mar 24 at 19:14
$begingroup$
Once you know that $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ then the duplication formula gives $C = lim_z to inftylog 2^2z + logGamma(z+1)+log Gamma(z+frac12)-logGamma(2z+1)$. But again $logGamma(z+1)=(z+frac12)log z - z + int_0^inftyg(t)e^-ztdt + C$ has nothing obvious and you should probably look again at the steps they used to find it.
$endgroup$
– reuns
Mar 24 at 19:35
$begingroup$
I know it is not obvious, but after some trouble I was able to prove that $logGamma(z+1)=(z+frac12)log z-z + int_0^infty g(t)e^-ztdt + C$ but from this I really don't see how the duplication formula gives $C=lim_ztoinfty log 2^2z + log Gamma(z+1) + log Gamma(z+frac12)-logGamma(2z+1).$ Could you please elaborate that line?
$endgroup$
– Sebitas
Mar 24 at 20:24