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Looking for an explanation or insight on a form of the inverse of a restricted gamma function
$begingroup$
I posted this yesterday asking about how to find an inverse of a restricted gamma function. To put it concisely, I was looking at the gamma function in the positive reals after restricting it to be greater than its minimum to make it 1:1 where an inverse does exist (there's a graph in the last post so you can see what I mean better).
After doing more research into finding an inverse, I found this post/answer where the high precision answer is pretty much exactly what I want. The problem is that I don't understand it, how it's derived, where it came from, or anything. Does anybody have any insight on the derivation of this or where it came from or anything? Any help will be appreciated.
Thank you for your time.
analysis inverse gamma-function inverse-function
asked Feb 2 at 4:59
DoleseDolese
62
$endgroup$
add a comment |
$begingroup$
I posted this yesterday asking about how to find an inverse of a restricted gamma function. To put it concisely, I was looking at the gamma function in the positive reals after restricting it to be greater than its minimum to make it 1:1 where an inverse does exist (there's a graph in the last post so you can see what I mean better).
After doing more research into finding an inverse, I found this post/answer where the high precision answer is pretty much exactly what I want. The problem is that I don't understand it, how it's derived, where it came from, or anything. Does anybody have any insight on the derivation of this or where it came from or anything? Any help will be appreciated.
Thank you for your time.
analysis inverse gamma-function inverse-function
asked Feb 2 at 4:59
DoleseDolese
62
$endgroup$
add a comment |
$begingroup$
I posted this yesterday asking about how to find an inverse of a restricted gamma function. To put it concisely, I was looking at the gamma function in the positive reals after restricting it to be greater than its minimum to make it 1:1 where an inverse does exist (there's a graph in the last post so you can see what I mean better).
After doing more research into finding an inverse, I found this post/answer where the high precision answer is pretty much exactly what I want. The problem is that I don't understand it, how it's derived, where it came from, or anything. Does anybody have any insight on the derivation of this or where it came from or anything? Any help will be appreciated.
Thank you for your time.
analysis inverse gamma-function inverse-function
asked Feb 2 at 4:59
DoleseDolese
62
$endgroup$
I posted this yesterday asking about how to find an inverse of a restricted gamma function. To put it concisely, I was looking at the gamma function in the positive reals after restricting it to be greater than its minimum to make it 1:1 where an inverse does exist (there's a graph in the last post so you can see what I mean better).
After doing more research into finding an inverse, I found this post/answer where the high precision answer is pretty much exactly what I want. The problem is that I don't understand it, how it's derived, where it came from, or anything. Does anybody have any insight on the derivation of this or where it came from or anything? Any help will be appreciated.
Thank you for your time.
analysis inverse gamma-function inverse-function
analysis inverse gamma-function inverse-function
asked Feb 2 at 4:59
DoleseDolese
62
asked Feb 2 at 4:59
DoleseDolese
62
asked Feb 2 at 4:59
DoleseDolese
62
asked Feb 2 at 4:59
DoleseDolese
62
asked Feb 2 at 4:59
DoleseDolese
62
62
add a comment |
add a comment |
1 Answer
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$begingroup$
There is a magificent approximation. If you look at this question of mine which was about the solution of $$n!=a^n, 10^k$$
@robjohn proposed (with very detailed explanations)
$$
bbox[5px,border:2px solid #C0A000]{nsim eaexpleft(operatorname{W}left(frac k{ea}log(10)-frac1{2ea}log(2pi a)right)right)-frac12}tag{4}
$$
where $W(.)$ is Lambert function.
So, for your case, make $a=1$ and $k=frac{log (n!)}{log (10)}$ to get
$$n sim e expleft(Wleft(frac{2 log (n!)-log (2 pi )}{2 e} right)right)-frac 12$$
Just for the fun of it, try with $n=123$ the result would be $122.9999299$ !
If we know that the result should be an integer, then a good inverse of $Gamma(n+1)$ is given by
$$nsim eexpleft(operatorname{W}left(frac1{e}logleft(frac{Gamma(n)}{sqrt{2pi}}right)right)right)+frac12$$
answered Feb 2 at 7:11
Claude LeiboviciClaude Leibovici
125k1158135
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add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
There is a magificent approximation. If you look at this question of mine which was about the solution of $$n!=a^n, 10^k$$
@robjohn proposed (with very detailed explanations)
$$
bbox[5px,border:2px solid #C0A000]{nsim eaexpleft(operatorname{W}left(frac k{ea}log(10)-frac1{2ea}log(2pi a)right)right)-frac12}tag{4}
$$
where $W(.)$ is Lambert function.
So, for your case, make $a=1$ and $k=frac{log (n!)}{log (10)}$ to get
$$n sim e expleft(Wleft(frac{2 log (n!)-log (2 pi )}{2 e} right)right)-frac 12$$
Just for the fun of it, try with $n=123$ the result would be $122.9999299$ !
If we know that the result should be an integer, then a good inverse of $Gamma(n+1)$ is given by
$$nsim eexpleft(operatorname{W}left(frac1{e}logleft(frac{Gamma(n)}{sqrt{2pi}}right)right)right)+frac12$$
answered Feb 2 at 7:11
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
There is a magificent approximation. If you look at this question of mine which was about the solution of $$n!=a^n, 10^k$$
@robjohn proposed (with very detailed explanations)
$$
bbox[5px,border:2px solid #C0A000]{nsim eaexpleft(operatorname{W}left(frac k{ea}log(10)-frac1{2ea}log(2pi a)right)right)-frac12}tag{4}
$$
where $W(.)$ is Lambert function.
So, for your case, make $a=1$ and $k=frac{log (n!)}{log (10)}$ to get
$$n sim e expleft(Wleft(frac{2 log (n!)-log (2 pi )}{2 e} right)right)-frac 12$$
Just for the fun of it, try with $n=123$ the result would be $122.9999299$ !
If we know that the result should be an integer, then a good inverse of $Gamma(n+1)$ is given by
$$nsim eexpleft(operatorname{W}left(frac1{e}logleft(frac{Gamma(n)}{sqrt{2pi}}right)right)right)+frac12$$
answered Feb 2 at 7:11
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
There is a magificent approximation. If you look at this question of mine which was about the solution of $$n!=a^n, 10^k$$
@robjohn proposed (with very detailed explanations)
$$
bbox[5px,border:2px solid #C0A000]{nsim eaexpleft(operatorname{W}left(frac k{ea}log(10)-frac1{2ea}log(2pi a)right)right)-frac12}tag{4}
$$
where $W(.)$ is Lambert function.
So, for your case, make $a=1$ and $k=frac{log (n!)}{log (10)}$ to get
$$n sim e expleft(Wleft(frac{2 log (n!)-log (2 pi )}{2 e} right)right)-frac 12$$
Just for the fun of it, try with $n=123$ the result would be $122.9999299$ !
If we know that the result should be an integer, then a good inverse of $Gamma(n+1)$ is given by
$$nsim eexpleft(operatorname{W}left(frac1{e}logleft(frac{Gamma(n)}{sqrt{2pi}}right)right)right)+frac12$$
answered Feb 2 at 7:11
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
There is a magificent approximation. If you look at this question of mine which was about the solution of $$n!=a^n, 10^k$$
@robjohn proposed (with very detailed explanations)
$$
bbox[5px,border:2px solid #C0A000]{nsim eaexpleft(operatorname{W}left(frac k{ea}log(10)-frac1{2ea}log(2pi a)right)right)-frac12}tag{4}
$$
where $W(.)$ is Lambert function.
So, for your case, make $a=1$ and $k=frac{log (n!)}{log (10)}$ to get
$$n sim e expleft(Wleft(frac{2 log (n!)-log (2 pi )}{2 e} right)right)-frac 12$$
Just for the fun of it, try with $n=123$ the result would be $122.9999299$ !
If we know that the result should be an integer, then a good inverse of $Gamma(n+1)$ is given by
$$nsim eexpleft(operatorname{W}left(frac1{e}logleft(frac{Gamma(n)}{sqrt{2pi}}right)right)right)+frac12$$
answered Feb 2 at 7:11
Claude LeiboviciClaude Leibovici
125k1158135
answered Feb 2 at 7:11
Claude LeiboviciClaude Leibovici
125k1158135
answered Feb 2 at 7:11
Claude LeiboviciClaude Leibovici
125k1158135
answered Feb 2 at 7:11
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
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