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| Axel Vogt... |
 Posted: Thu Nov 05, 2009 3:28 pm |
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At the MMA one finds the following identity for 1 < z < infinity
hypergeom([a, b],[c],z) =
exp(2*I*(c-a-b)*Pi)*(1/(1-z))^a*
hypergeom([a, c-b],[c],z/(z-1)) -
2*I*exp(I*(c-a-b)*Pi)*Pi*GAMMA(c)/GAMMA(a+b-c+1)/GAMMA(c-a)/GAMMA(c-b)*
hypergeom([a, b],[a+b-c+1],1-z)
which I was not aware of (usually the branch cut is excluded and I
am not used to rules how MMA treats branch cuts [Maple does counter
clockwise around branch point, which here is: from below towards
the real axis]).
Does anybody know where that comes from (or better: how to proof it)? |
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| Posted: Fri Nov 06, 2009 2:36 pm |
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Guest
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Axel Vogt schrieb:
[quote]
At the MMA one finds the following identity for 1 < z < infinity
hypergeom([a, b],[c],z) =
exp(2*I*(c-a-b)*Pi)*(1/(1-z))^a*
hypergeom([a, c-b],[c],z/(z-1)) -
2*I*exp(I*(c-a-b)*Pi)*Pi*GAMMA(c)/GAMMA(a+b-c+1)/GAMMA(c-a)/GAMMA(c-b)*
hypergeom([a, b],[a+b-c+1],1-z)
which I was not aware of (usually the branch cut is excluded and I
am not used to rules how MMA treats branch cuts [Maple does counter
clockwise around branch point, which here is: from below towards
the real axis]).
Does anybody know where that comes from (or better: how to proof it)?
[/quote]
Plotting for some specific parameters a,b,c suggests the formula to be
correct. It looks fairly useless, however, since the branch cut appears
twice. Another way of writing this is:
exp(I*(a+b-c)*Pi)*hypergeom([a, b],[c],z) -
exp(I*(c-a-b)*Pi)*(1/(1-z))^a*
hypergeom([a, c-b],[c],z/(z-1)) =
-2*I*Pi*GAMMA(c)/GAMMA(a+b-c+1)/GAMMA(c-a)/GAMMA(c-b)*
hypergeom([a, b],[a+b-c+1],1-z)
Substituting 1-z for z, where z<0:
exp(I*(a+b-c)*Pi)*hypergeom([a, b],[c],1-z) -
exp(I*(c-b)*Pi)*(-z)^(-a)*
hypergeom([a, c-b],[c],1-1/z) =
-2*I*Pi*GAMMA(c)/GAMMA(a+b-c+1)/GAMMA(c-a)/GAMMA(c-b)*
hypergeom([a, b],[a+b-c+1],z)
Further substituting 1/z for z, and c-b for b:
exp(I*(a-b)*Pi)*hypergeom([a, c-b],[c],1-1/z) -
exp(I*b*Pi)*(-z)^a*
hypergeom([a, b],[c],1-z) =
-2*I*Pi*GAMMA(c)/GAMMA(a-b+1)/GAMMA(c-a)/GAMMA(b)*
hypergeom([a, c-b],[a-b+1],1/z)
Linear combination gives:
(exp(I*(a-c)*Pi)-exp(I*(c-a)*Pi))*
hypergeom([a, b],[c],1-z) =
-2*I*Pi*GAMMA(c)/GAMMA(a+b-c+1)/GAMMA(c-a)/GAMMA(c-b)*
exp(-I*b*Pi)*hypergeom([a, b],[a+b-c+1],z)
-2*I*Pi*GAMMA(c)/GAMMA(a-b+1)/GAMMA(c-a)/GAMMA(b)*
exp(I*(c-a-b)*Pi)*(-z)^(-a)*hypergeom([a, c-b],[a-b+1],1/z)
as well as the same with c-b substituted for b, and 1/z for z. Somewhat
simplified:
hypergeom([a, b],[c],1-z) =
GAMMA(c)*GAMMA(a-c+1)/GAMMA(a+b-c+1)/GAMMA(c-b)*
exp(-I*b*Pi)*hypergeom([a, b],[a+b-c+1],z) +
GAMMA(c)*GAMMA(a-c+1)/GAMMA(a-b+1)/GAMMA(b)*
exp(I*(c-b)*Pi)*z^(-a)*hypergeom([a, c-b],[a-b+1],1/z)
This is more useful than the original formula as the branch cut appears
only once. Now you only have to find a proof for this one!
Martin. |
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| Axel Vogt... |
| Posted: Fri Nov 06, 2009 4:44 pm |
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Guest
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clicliclic at (no spam) freenet.de wrote:
[quote]Axel Vogt schrieb:
At the MMA one finds the following identity for 1 < z < infinity
hypergeom([a, b],[c],z) =
exp(2*I*(c-a-b)*Pi)*(1/(1-z))^a*
hypergeom([a, c-b],[c],z/(z-1)) -
2*I*exp(I*(c-a-b)*Pi)*Pi*GAMMA(c)/GAMMA(a+b-c+1)/GAMMA(c-a)/GAMMA(c-b)*
hypergeom([a, b],[a+b-c+1],1-z)
which I was not aware of (usually the branch cut is excluded and I
am not used to rules how MMA treats branch cuts [Maple does counter
clockwise around branch point, which here is: from below towards
the real axis]).
Does anybody know where that comes from (or better: how to proof it)?
Plotting for some specific parameters a,b,c suggests the formula to be
correct. It looks fairly useless, however, since the branch cut appears
twice. Another way of writing this is:
exp(I*(a+b-c)*Pi)*hypergeom([a, b],[c],z) -
exp(I*(c-a-b)*Pi)*(1/(1-z))^a*
hypergeom([a, c-b],[c],z/(z-1)) =
-2*I*Pi*GAMMA(c)/GAMMA(a+b-c+1)/GAMMA(c-a)/GAMMA(c-b)*
hypergeom([a, b],[a+b-c+1],1-z)
Substituting 1-z for z, where z<0:
exp(I*(a+b-c)*Pi)*hypergeom([a, b],[c],1-z) -
exp(I*(c-b)*Pi)*(-z)^(-a)*
hypergeom([a, c-b],[c],1-1/z) =
-2*I*Pi*GAMMA(c)/GAMMA(a+b-c+1)/GAMMA(c-a)/GAMMA(c-b)*
hypergeom([a, b],[a+b-c+1],z)
Further substituting 1/z for z, and c-b for b:
exp(I*(a-b)*Pi)*hypergeom([a, c-b],[c],1-1/z) -
exp(I*b*Pi)*(-z)^a*
hypergeom([a, b],[c],1-z) =
-2*I*Pi*GAMMA(c)/GAMMA(a-b+1)/GAMMA(c-a)/GAMMA(b)*
hypergeom([a, c-b],[a-b+1],1/z)
Linear combination gives:
(exp(I*(a-c)*Pi)-exp(I*(c-a)*Pi))*
hypergeom([a, b],[c],1-z) =
-2*I*Pi*GAMMA(c)/GAMMA(a+b-c+1)/GAMMA(c-a)/GAMMA(c-b)*
exp(-I*b*Pi)*hypergeom([a, b],[a+b-c+1],z)
-2*I*Pi*GAMMA(c)/GAMMA(a-b+1)/GAMMA(c-a)/GAMMA(b)*
exp(I*(c-a-b)*Pi)*(-z)^(-a)*hypergeom([a, c-b],[a-b+1],1/z)
as well as the same with c-b substituted for b, and 1/z for z. Somewhat
simplified:
hypergeom([a, b],[c],1-z) =
GAMMA(c)*GAMMA(a-c+1)/GAMMA(a+b-c+1)/GAMMA(c-b)*
exp(-I*b*Pi)*hypergeom([a, b],[a+b-c+1],z) +
GAMMA(c)*GAMMA(a-c+1)/GAMMA(a-b+1)/GAMMA(b)*
exp(I*(c-b)*Pi)*z^(-a)*hypergeom([a, c-b],[a-b+1],1/z)
This is more useful than the original formula as the branch cut appears
only once. Now you only have to find a proof for this one!
Martin.
[/quote]
Thank you, I lost you a bit through the substitutions ...
After substituting back does it result in
hypergeom([a, b],[c],x) =
GAMMA(c)*GAMMA(a-c+1)/GAMMA(a+b-c+1)/GAMMA(c-b)*
exp(-I*b*Pi)*hypergeom([a, b],[a+b-c+1],1-x)+
GAMMA(c)*GAMMA(a-c+1)/GAMMA(a-b+1)/GAMMA(b)*
exp(I*(c-b)*Pi)*(1-x)^(-a)*hypergeom([a, c-b],[a-b+1],1/(1-x))
with 1 < x < infinity ? Then it would be much better.
At least the right hand side satisfies the differential equation. |
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| ... |
| Posted: Fri Nov 06, 2009 5:26 pm |
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Axel Vogt schrieb:
[quote]
Thank you, I lost you a bit through the substitutions ...
After substituting back does it result in
hypergeom([a, b],[c],x) =
GAMMA(c)*GAMMA(a-c+1)/GAMMA(a+b-c+1)/GAMMA(c-b)*
exp(-I*b*Pi)*hypergeom([a, b],[a+b-c+1],1-x)+
GAMMA(c)*GAMMA(a-c+1)/GAMMA(a-b+1)/GAMMA(b)*
exp(I*(c-b)*Pi)*(1-x)^(-a)*hypergeom([a, c-b],[a-b+1],1/(1-x))
with 1 < x < infinity ? Then it would be much better.
At least the right hand side satisfies the differential equation.
[/quote]
Yes, my transformed z was restricted to -infinity < z < 0, so your x =
1-z will (once more) be restricted to 1 < x < infinity. My suggestion
would be to transform the 2F1(1/z) in the transformed formula to two
2F1(z)'s and see how it looks then (this is easy since z<0 is not on the
branch cut). To combine z (or 1-z) with 1/z in one formula usually means
trouble with the branch cuts.
Martin. |
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