Mixing
Finding an Inverse of Restricted Gamma Function
$begingroup$
I don't know/haven't used LaTeX yet but I'll do my best to keep it simple,
I'm working on my undergrad senior project and I'm trying to find an inverse function for f(x)=(x-1)! just in the positive reals. I was inspired to ask this question when in one of my probability classes my professor talked about how something like π! existed. Now, obviously, this isn't a 1:1 function so an inverse doesn't exist, but I first restricted the function just to x > 0 and then restricted it further after finding the minimum which is x=1.461632... or the positive root of the digamma function. You can see what I mean on this graph (the green one is what I'm trying to find the inverse of). After restricting the domain to x>1.461632... , the function is 1:1 and an inverse does exist.
This is where I'm stuck at.
I guess what I'm asking is that is there a way to find this inverse? I know that, for example, f^-1(120)=5 and f^-1(3(√π )/4)=2.5 but what of something like f^-1(25) or f^-1(e)? I've seen things like Stirling's Approximation and finding an inverse based off of that but I wanted to see if anybody else has any ideas of what I can do next.
Thank you for your time and let me know if you have any questions about my post.
factorial gamma-function inverse-function digamma-function
asked Jan 31 at 21:09
DoleseDolese
62
$endgroup$
add a comment |
$begingroup$
I don't know/haven't used LaTeX yet but I'll do my best to keep it simple,
I'm working on my undergrad senior project and I'm trying to find an inverse function for f(x)=(x-1)! just in the positive reals. I was inspired to ask this question when in one of my probability classes my professor talked about how something like π! existed. Now, obviously, this isn't a 1:1 function so an inverse doesn't exist, but I first restricted the function just to x > 0 and then restricted it further after finding the minimum which is x=1.461632... or the positive root of the digamma function. You can see what I mean on this graph (the green one is what I'm trying to find the inverse of). After restricting the domain to x>1.461632... , the function is 1:1 and an inverse does exist.
This is where I'm stuck at.
I guess what I'm asking is that is there a way to find this inverse? I know that, for example, f^-1(120)=5 and f^-1(3(√π )/4)=2.5 but what of something like f^-1(25) or f^-1(e)? I've seen things like Stirling's Approximation and finding an inverse based off of that but I wanted to see if anybody else has any ideas of what I can do next.
Thank you for your time and let me know if you have any questions about my post.
factorial gamma-function inverse-function digamma-function
asked Jan 31 at 21:09
DoleseDolese
62
$endgroup$
add a comment |
$begingroup$
I don't know/haven't used LaTeX yet but I'll do my best to keep it simple,
I'm working on my undergrad senior project and I'm trying to find an inverse function for f(x)=(x-1)! just in the positive reals. I was inspired to ask this question when in one of my probability classes my professor talked about how something like π! existed. Now, obviously, this isn't a 1:1 function so an inverse doesn't exist, but I first restricted the function just to x > 0 and then restricted it further after finding the minimum which is x=1.461632... or the positive root of the digamma function. You can see what I mean on this graph (the green one is what I'm trying to find the inverse of). After restricting the domain to x>1.461632... , the function is 1:1 and an inverse does exist.
This is where I'm stuck at.
I guess what I'm asking is that is there a way to find this inverse? I know that, for example, f^-1(120)=5 and f^-1(3(√π )/4)=2.5 but what of something like f^-1(25) or f^-1(e)? I've seen things like Stirling's Approximation and finding an inverse based off of that but I wanted to see if anybody else has any ideas of what I can do next.
Thank you for your time and let me know if you have any questions about my post.
factorial gamma-function inverse-function digamma-function
asked Jan 31 at 21:09
DoleseDolese
62
$endgroup$
I don't know/haven't used LaTeX yet but I'll do my best to keep it simple,
I'm working on my undergrad senior project and I'm trying to find an inverse function for f(x)=(x-1)! just in the positive reals. I was inspired to ask this question when in one of my probability classes my professor talked about how something like π! existed. Now, obviously, this isn't a 1:1 function so an inverse doesn't exist, but I first restricted the function just to x > 0 and then restricted it further after finding the minimum which is x=1.461632... or the positive root of the digamma function. You can see what I mean on this graph (the green one is what I'm trying to find the inverse of). After restricting the domain to x>1.461632... , the function is 1:1 and an inverse does exist.
This is where I'm stuck at.
I guess what I'm asking is that is there a way to find this inverse? I know that, for example, f^-1(120)=5 and f^-1(3(√π )/4)=2.5 but what of something like f^-1(25) or f^-1(e)? I've seen things like Stirling's Approximation and finding an inverse based off of that but I wanted to see if anybody else has any ideas of what I can do next.
Thank you for your time and let me know if you have any questions about my post.
factorial gamma-function inverse-function digamma-function
factorial gamma-function inverse-function digamma-function
asked Jan 31 at 21:09
DoleseDolese
62
asked Jan 31 at 21:09
DoleseDolese
62
asked Jan 31 at 21:09
DoleseDolese
62
asked Jan 31 at 21:09
DoleseDolese
62
asked Jan 31 at 21:09
DoleseDolese
62
62
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There is the algorithm by David W. Cantrell as described here. The idea is that we can use
$$g(x) = sqrt{2pi}{left(frac{x-1/2}{e}right)^{x-1/2}} - c$$
as an approximation of $Gamma(x)$. Here $c = sqrt{2pi}/e - Gamma(k)$ where $k$ is the positive zero of the derivative of $Gamma$. We can invert this function in terms of the Lambert $W$-function. First we set
$$L(x) = lnleft(frac{x+c}{sqrt{2pi}}right)$$
and then we get
$$AIG(y) := g^{-1}(y) = frac{L(x)}{Wleft(frac{L(x)}{e}right)} + 1/2.$$
As the arguments increase the error gets very small, consider following evaluations:
$$begin{array}{r|ll}
N & AIG((N-1)!) & text{error} \ hline
2 & hphantom{1}2.02 & hphantom{-}0.01 \
5 & hphantom{1}4.995 & -0.001 \
10 & hphantom{1}9.998 & -0.0002 \
20 & 19.9993 & -0.00004 \
end{array}$$
answered Jan 31 at 21:49
flawrflawr
11.7k32546
$endgroup$
add a comment |
$begingroup$
Bear in mind in reading this that I am an engineer, and this is just a non-rigorous sketch of what you could try.
Assuming you are allowed to use the Gamma Function integral defined for the positive reals only, then you can use this function to calculate the "Inverse Gamma Function" for the positive reals up to any desired cut off as long as you are willing utilize and prove the identity:
$$sin(pi z)=frac{pi z}{Gamma(1-z)Gamma(1+z)}tag{1}$$
where $Gamma(z)=int_0^{infty } e^{-t} t^{z-1} , dt$ for $z>0$
Rearranging (1) we have
$$frac{z}{Gamma(1+z)}=frac{ Gamma(1-z) sin(pi z)}{pi}=left(frac{1}{Gamma(z)}right)$$
You will immediately say that this only works for $0<z<1$ so it is not much use for your purpose, but we can write
$$frac{zprod _{k=1}^{n-1} (z+k)}{Gamma(n+z)}=frac{ Gamma(n-z) sin(pi (z+n-1))}{pi prod _{k=1}^{n-1} (z-k)} =left(frac{1}{Gamma(z)}right)$$
where n is a positve integer $ge 1$. Therefore you can now calculate your inverse Gamma Function with $z$ in the range $0<z<n$, without worrying about how to calculate negative factorials!
Edit 1:
Lets push this a bit further and define a Pochammer Function approximation to the Gamma Function $Gamma_P$
$$Gamma_P(z)=prod _{k=1}^{lfloor z-1rfloor } ((-lfloor z-1rfloor +z-1)+k)$$
where $lfloor zrfloor$ is the floor function on $z$ and
$$Gamma(z)=Gamma ((z-lfloor zrfloor )+1) Gamma_P(z)$$
which gives
$$frac{Gamma ((z-lfloor zrfloor )+1) Gamma_P(n-z) sin (pi (n+z-1))}{pi prod _{k=1}^{n-1} (z-k)}approxleft(frac{1}{Gamma(z)}right)tag{2}$$
with the true Gamma Function $Gamma ((z-lfloor zrfloor )+1)$ now just used in the range $1le ((z-lfloor zrfloor )+1) le 2$
(Note: Equation 2 is approximate, and $n$ can be set to $1$, simplifying the formula - not sure about the origin of the error here need to review this when I have time)
I've plotted $Gamma ((z-lfloor zrfloor )+1)$ alongside a very rough half sine wave approximation (in orange)
$$Gamma ((z-lfloor zrfloor )+1)approx1-left(1-frac{sqrt{pi }}{2}right) sin (pi (z-lfloor zrfloor ))$$
Finally a comparison between the approximate inverse Gamma Function using the half sine wave approximation and the real one
answered Feb 1 at 0:35
James ArathoonJames Arathoon
1,588423
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
oldest
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oldest
votes
$begingroup$
There is the algorithm by David W. Cantrell as described here. The idea is that we can use
$$g(x) = sqrt{2pi}{left(frac{x-1/2}{e}right)^{x-1/2}} - c$$
as an approximation of $Gamma(x)$. Here $c = sqrt{2pi}/e - Gamma(k)$ where $k$ is the positive zero of the derivative of $Gamma$. We can invert this function in terms of the Lambert $W$-function. First we set
$$L(x) = lnleft(frac{x+c}{sqrt{2pi}}right)$$
and then we get
$$AIG(y) := g^{-1}(y) = frac{L(x)}{Wleft(frac{L(x)}{e}right)} + 1/2.$$
As the arguments increase the error gets very small, consider following evaluations:
$$begin{array}{r|ll}
N & AIG((N-1)!) & text{error} \ hline
2 & hphantom{1}2.02 & hphantom{-}0.01 \
5 & hphantom{1}4.995 & -0.001 \
10 & hphantom{1}9.998 & -0.0002 \
20 & 19.9993 & -0.00004 \
end{array}$$
answered Jan 31 at 21:49
flawrflawr
11.7k32546
$endgroup$
add a comment |
$begingroup$
There is the algorithm by David W. Cantrell as described here. The idea is that we can use
$$g(x) = sqrt{2pi}{left(frac{x-1/2}{e}right)^{x-1/2}} - c$$
as an approximation of $Gamma(x)$. Here $c = sqrt{2pi}/e - Gamma(k)$ where $k$ is the positive zero of the derivative of $Gamma$. We can invert this function in terms of the Lambert $W$-function. First we set
$$L(x) = lnleft(frac{x+c}{sqrt{2pi}}right)$$
and then we get
$$AIG(y) := g^{-1}(y) = frac{L(x)}{Wleft(frac{L(x)}{e}right)} + 1/2.$$
As the arguments increase the error gets very small, consider following evaluations:
$$begin{array}{r|ll}
N & AIG((N-1)!) & text{error} \ hline
2 & hphantom{1}2.02 & hphantom{-}0.01 \
5 & hphantom{1}4.995 & -0.001 \
10 & hphantom{1}9.998 & -0.0002 \
20 & 19.9993 & -0.00004 \
end{array}$$
answered Jan 31 at 21:49
flawrflawr
11.7k32546
$endgroup$
add a comment |
$begingroup$
There is the algorithm by David W. Cantrell as described here. The idea is that we can use
$$g(x) = sqrt{2pi}{left(frac{x-1/2}{e}right)^{x-1/2}} - c$$
as an approximation of $Gamma(x)$. Here $c = sqrt{2pi}/e - Gamma(k)$ where $k$ is the positive zero of the derivative of $Gamma$. We can invert this function in terms of the Lambert $W$-function. First we set
$$L(x) = lnleft(frac{x+c}{sqrt{2pi}}right)$$
and then we get
$$AIG(y) := g^{-1}(y) = frac{L(x)}{Wleft(frac{L(x)}{e}right)} + 1/2.$$
As the arguments increase the error gets very small, consider following evaluations:
$$begin{array}{r|ll}
N & AIG((N-1)!) & text{error} \ hline
2 & hphantom{1}2.02 & hphantom{-}0.01 \
5 & hphantom{1}4.995 & -0.001 \
10 & hphantom{1}9.998 & -0.0002 \
20 & 19.9993 & -0.00004 \
end{array}$$
answered Jan 31 at 21:49
flawrflawr
11.7k32546
$endgroup$
There is the algorithm by David W. Cantrell as described here. The idea is that we can use
$$g(x) = sqrt{2pi}{left(frac{x-1/2}{e}right)^{x-1/2}} - c$$
as an approximation of $Gamma(x)$. Here $c = sqrt{2pi}/e - Gamma(k)$ where $k$ is the positive zero of the derivative of $Gamma$. We can invert this function in terms of the Lambert $W$-function. First we set
$$L(x) = lnleft(frac{x+c}{sqrt{2pi}}right)$$
and then we get
$$AIG(y) := g^{-1}(y) = frac{L(x)}{Wleft(frac{L(x)}{e}right)} + 1/2.$$
As the arguments increase the error gets very small, consider following evaluations:
$$begin{array}{r|ll}
N & AIG((N-1)!) & text{error} \ hline
2 & hphantom{1}2.02 & hphantom{-}0.01 \
5 & hphantom{1}4.995 & -0.001 \
10 & hphantom{1}9.998 & -0.0002 \
20 & 19.9993 & -0.00004 \
end{array}$$
answered Jan 31 at 21:49
flawrflawr
11.7k32546
answered Jan 31 at 21:49
flawrflawr
11.7k32546
answered Jan 31 at 21:49
flawrflawr
11.7k32546
answered Jan 31 at 21:49
flawrflawr
11.7k32546
11.7k32546
add a comment |
add a comment |
$begingroup$
Bear in mind in reading this that I am an engineer, and this is just a non-rigorous sketch of what you could try.
Assuming you are allowed to use the Gamma Function integral defined for the positive reals only, then you can use this function to calculate the "Inverse Gamma Function" for the positive reals up to any desired cut off as long as you are willing utilize and prove the identity:
$$sin(pi z)=frac{pi z}{Gamma(1-z)Gamma(1+z)}tag{1}$$
where $Gamma(z)=int_0^{infty } e^{-t} t^{z-1} , dt$ for $z>0$
Rearranging (1) we have
$$frac{z}{Gamma(1+z)}=frac{ Gamma(1-z) sin(pi z)}{pi}=left(frac{1}{Gamma(z)}right)$$
You will immediately say that this only works for $0<z<1$ so it is not much use for your purpose, but we can write
$$frac{zprod _{k=1}^{n-1} (z+k)}{Gamma(n+z)}=frac{ Gamma(n-z) sin(pi (z+n-1))}{pi prod _{k=1}^{n-1} (z-k)} =left(frac{1}{Gamma(z)}right)$$
where n is a positve integer $ge 1$. Therefore you can now calculate your inverse Gamma Function with $z$ in the range $0<z<n$, without worrying about how to calculate negative factorials!
Edit 1:
Lets push this a bit further and define a Pochammer Function approximation to the Gamma Function $Gamma_P$
$$Gamma_P(z)=prod _{k=1}^{lfloor z-1rfloor } ((-lfloor z-1rfloor +z-1)+k)$$
where $lfloor zrfloor$ is the floor function on $z$ and
$$Gamma(z)=Gamma ((z-lfloor zrfloor )+1) Gamma_P(z)$$
which gives
$$frac{Gamma ((z-lfloor zrfloor )+1) Gamma_P(n-z) sin (pi (n+z-1))}{pi prod _{k=1}^{n-1} (z-k)}approxleft(frac{1}{Gamma(z)}right)tag{2}$$
with the true Gamma Function $Gamma ((z-lfloor zrfloor )+1)$ now just used in the range $1le ((z-lfloor zrfloor )+1) le 2$
(Note: Equation 2 is approximate, and $n$ can be set to $1$, simplifying the formula - not sure about the origin of the error here need to review this when I have time)
I've plotted $Gamma ((z-lfloor zrfloor )+1)$ alongside a very rough half sine wave approximation (in orange)
$$Gamma ((z-lfloor zrfloor )+1)approx1-left(1-frac{sqrt{pi }}{2}right) sin (pi (z-lfloor zrfloor ))$$
Finally a comparison between the approximate inverse Gamma Function using the half sine wave approximation and the real one
answered Feb 1 at 0:35
James ArathoonJames Arathoon
1,588423
$endgroup$
add a comment |
$begingroup$
Bear in mind in reading this that I am an engineer, and this is just a non-rigorous sketch of what you could try.
Assuming you are allowed to use the Gamma Function integral defined for the positive reals only, then you can use this function to calculate the "Inverse Gamma Function" for the positive reals up to any desired cut off as long as you are willing utilize and prove the identity:
$$sin(pi z)=frac{pi z}{Gamma(1-z)Gamma(1+z)}tag{1}$$
where $Gamma(z)=int_0^{infty } e^{-t} t^{z-1} , dt$ for $z>0$
Rearranging (1) we have
$$frac{z}{Gamma(1+z)}=frac{ Gamma(1-z) sin(pi z)}{pi}=left(frac{1}{Gamma(z)}right)$$
You will immediately say that this only works for $0<z<1$ so it is not much use for your purpose, but we can write
$$frac{zprod _{k=1}^{n-1} (z+k)}{Gamma(n+z)}=frac{ Gamma(n-z) sin(pi (z+n-1))}{pi prod _{k=1}^{n-1} (z-k)} =left(frac{1}{Gamma(z)}right)$$
where n is a positve integer $ge 1$. Therefore you can now calculate your inverse Gamma Function with $z$ in the range $0<z<n$, without worrying about how to calculate negative factorials!
Edit 1:
Lets push this a bit further and define a Pochammer Function approximation to the Gamma Function $Gamma_P$
$$Gamma_P(z)=prod _{k=1}^{lfloor z-1rfloor } ((-lfloor z-1rfloor +z-1)+k)$$
where $lfloor zrfloor$ is the floor function on $z$ and
$$Gamma(z)=Gamma ((z-lfloor zrfloor )+1) Gamma_P(z)$$
which gives
$$frac{Gamma ((z-lfloor zrfloor )+1) Gamma_P(n-z) sin (pi (n+z-1))}{pi prod _{k=1}^{n-1} (z-k)}approxleft(frac{1}{Gamma(z)}right)tag{2}$$
with the true Gamma Function $Gamma ((z-lfloor zrfloor )+1)$ now just used in the range $1le ((z-lfloor zrfloor )+1) le 2$
(Note: Equation 2 is approximate, and $n$ can be set to $1$, simplifying the formula - not sure about the origin of the error here need to review this when I have time)
I've plotted $Gamma ((z-lfloor zrfloor )+1)$ alongside a very rough half sine wave approximation (in orange)
$$Gamma ((z-lfloor zrfloor )+1)approx1-left(1-frac{sqrt{pi }}{2}right) sin (pi (z-lfloor zrfloor ))$$
Finally a comparison between the approximate inverse Gamma Function using the half sine wave approximation and the real one
answered Feb 1 at 0:35
James ArathoonJames Arathoon
1,588423
$endgroup$
add a comment |
$begingroup$
Bear in mind in reading this that I am an engineer, and this is just a non-rigorous sketch of what you could try.
Assuming you are allowed to use the Gamma Function integral defined for the positive reals only, then you can use this function to calculate the "Inverse Gamma Function" for the positive reals up to any desired cut off as long as you are willing utilize and prove the identity:
$$sin(pi z)=frac{pi z}{Gamma(1-z)Gamma(1+z)}tag{1}$$
where $Gamma(z)=int_0^{infty } e^{-t} t^{z-1} , dt$ for $z>0$
Rearranging (1) we have
$$frac{z}{Gamma(1+z)}=frac{ Gamma(1-z) sin(pi z)}{pi}=left(frac{1}{Gamma(z)}right)$$
You will immediately say that this only works for $0<z<1$ so it is not much use for your purpose, but we can write
$$frac{zprod _{k=1}^{n-1} (z+k)}{Gamma(n+z)}=frac{ Gamma(n-z) sin(pi (z+n-1))}{pi prod _{k=1}^{n-1} (z-k)} =left(frac{1}{Gamma(z)}right)$$
where n is a positve integer $ge 1$. Therefore you can now calculate your inverse Gamma Function with $z$ in the range $0<z<n$, without worrying about how to calculate negative factorials!
Edit 1:
Lets push this a bit further and define a Pochammer Function approximation to the Gamma Function $Gamma_P$
$$Gamma_P(z)=prod _{k=1}^{lfloor z-1rfloor } ((-lfloor z-1rfloor +z-1)+k)$$
where $lfloor zrfloor$ is the floor function on $z$ and
$$Gamma(z)=Gamma ((z-lfloor zrfloor )+1) Gamma_P(z)$$
which gives
$$frac{Gamma ((z-lfloor zrfloor )+1) Gamma_P(n-z) sin (pi (n+z-1))}{pi prod _{k=1}^{n-1} (z-k)}approxleft(frac{1}{Gamma(z)}right)tag{2}$$
with the true Gamma Function $Gamma ((z-lfloor zrfloor )+1)$ now just used in the range $1le ((z-lfloor zrfloor )+1) le 2$
(Note: Equation 2 is approximate, and $n$ can be set to $1$, simplifying the formula - not sure about the origin of the error here need to review this when I have time)
I've plotted $Gamma ((z-lfloor zrfloor )+1)$ alongside a very rough half sine wave approximation (in orange)
$$Gamma ((z-lfloor zrfloor )+1)approx1-left(1-frac{sqrt{pi }}{2}right) sin (pi (z-lfloor zrfloor ))$$
Finally a comparison between the approximate inverse Gamma Function using the half sine wave approximation and the real one
answered Feb 1 at 0:35
James ArathoonJames Arathoon
1,588423
$endgroup$
Bear in mind in reading this that I am an engineer, and this is just a non-rigorous sketch of what you could try.
Assuming you are allowed to use the Gamma Function integral defined for the positive reals only, then you can use this function to calculate the "Inverse Gamma Function" for the positive reals up to any desired cut off as long as you are willing utilize and prove the identity:
$$sin(pi z)=frac{pi z}{Gamma(1-z)Gamma(1+z)}tag{1}$$
where $Gamma(z)=int_0^{infty } e^{-t} t^{z-1} , dt$ for $z>0$
Rearranging (1) we have
$$frac{z}{Gamma(1+z)}=frac{ Gamma(1-z) sin(pi z)}{pi}=left(frac{1}{Gamma(z)}right)$$
You will immediately say that this only works for $0<z<1$ so it is not much use for your purpose, but we can write
$$frac{zprod _{k=1}^{n-1} (z+k)}{Gamma(n+z)}=frac{ Gamma(n-z) sin(pi (z+n-1))}{pi prod _{k=1}^{n-1} (z-k)} =left(frac{1}{Gamma(z)}right)$$
where n is a positve integer $ge 1$. Therefore you can now calculate your inverse Gamma Function with $z$ in the range $0<z<n$, without worrying about how to calculate negative factorials!
Edit 1:
Lets push this a bit further and define a Pochammer Function approximation to the Gamma Function $Gamma_P$
$$Gamma_P(z)=prod _{k=1}^{lfloor z-1rfloor } ((-lfloor z-1rfloor +z-1)+k)$$
where $lfloor zrfloor$ is the floor function on $z$ and
$$Gamma(z)=Gamma ((z-lfloor zrfloor )+1) Gamma_P(z)$$
which gives
$$frac{Gamma ((z-lfloor zrfloor )+1) Gamma_P(n-z) sin (pi (n+z-1))}{pi prod _{k=1}^{n-1} (z-k)}approxleft(frac{1}{Gamma(z)}right)tag{2}$$
with the true Gamma Function $Gamma ((z-lfloor zrfloor )+1)$ now just used in the range $1le ((z-lfloor zrfloor )+1) le 2$
(Note: Equation 2 is approximate, and $n$ can be set to $1$, simplifying the formula - not sure about the origin of the error here need to review this when I have time)
I've plotted $Gamma ((z-lfloor zrfloor )+1)$ alongside a very rough half sine wave approximation (in orange)
$$Gamma ((z-lfloor zrfloor )+1)approx1-left(1-frac{sqrt{pi }}{2}right) sin (pi (z-lfloor zrfloor ))$$
Finally a comparison between the approximate inverse Gamma Function using the half sine wave approximation and the real one
answered Feb 1 at 0:35
James ArathoonJames Arathoon
1,588423
edited Feb 2 at 11:14
answered Feb 1 at 0:35
James ArathoonJames Arathoon
1,588423
answered Feb 1 at 0:35
James ArathoonJames Arathoon
1,588423
answered Feb 1 at 0:35
James ArathoonJames Arathoon
1,588423
1,588423
add a comment |
add a comment |
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