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| Gerry... |
 Posted: Wed Oct 21, 2009 8:42 am |
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Guest
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Hi all,
does there exist a solution for the gamma ratio:
gamma(1/2-s)/gamma(1/2+s)
without using the zeta function!:-)
Now let s= a + b*Pi*I
For a >0 and integer and b = 0 i found that :
gamma(1/2-s)/gamma(1/2+s) = 1/4*(-4)^s*4^s*gamma(s)^2/gamma(2*s)^2
I was hoping that it would work also for b<>0 but...
it doesn't
Only the folowing gives a fixed error for any c>0
g1(s) = gamma(1/2-s)/gamma(1/2+s)
g2(s) = 1/4*(-4)^s*4^s*gamma(s)^2/gamma(2*s)^2
gx(s,j) = log( g2(s)/g1(s)/j-1/(2*j) )/ j /log(Pi);
err(j)= gx(c-(j+1)*Pi*I,j+1) - gx(c-1*Pi*I,j);
any comments are welcome
regards
Gerry |
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| Posted: Wed Oct 21, 2009 10:31 pm |
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Guest
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Gerry schrieb:
[quote]
does there exist a solution for the gamma ratio:
gamma(1/2-s)/gamma(1/2+s)
without using the zeta function!:-)
Now let s= a + b*Pi*I
For a >0 and integer and b = 0 i found that :
gamma(1/2-s)/gamma(1/2+s) = 1/4*(-4)^s*4^s*gamma(s)^2/gamma(2*s)^2
I was hoping that it would work also for b<>0 but...
it doesn't
Only the folowing gives a fixed error for any c>0
g1(s) = gamma(1/2-s)/gamma(1/2+s)
g2(s) = 1/4*(-4)^s*4^s*gamma(s)^2/gamma(2*s)^2
gx(s,j) = log( g2(s)/g1(s)/j-1/(2*j) )/ j /log(Pi);
err(j)= gx(c-(j+1)*Pi*I,j+1) - gx(c-1*Pi*I,j);
any comments are welcome
[/quote]
Why do you call the reformulation of your Gamma ratio a solution? It
still involves two Gamma functions and is more complicated than your
original expression! For integer s=n, your transformation is correct,
although the following amounts to the same, while being simpler:
Gamma(1/2-n)/Gamma(1/2+n) = (-4)^n/((2n-1)!!)^2
The double factorial p!! here denotes 2*4*6* ... *p if p is even, and
1*3*5* ... *p if p is odd (the latter case applies).
In general, the following equivalences can be used to transform Gamma
functions:
Gamma(x+1) = x*Gamma(x),
Gamma(x)*Gamma(1-x) = pi/sin(pi*x),
Gamma(1/2+x)*Gamma(1/2-x) = pi/cos(pi*x),
Gamma(2*x) = 2^(2*x-1)*Gamma(x)*Gamma(x+1/2)/sqrt(pi).
That's about all there is if you restrict yourself to at most three
Gamma functions per expression. The four expressions hold for any
complex argument; the second and third of them being equivalent.
When plotting your err(c,j) := gx(c-(j+1)*pi*#i, j+1) - gx(c-1*pi*#i, j)
versus c for arbitrary constant j (where c and j are real) I get the
impression that re(err(c,j)) is constant while im(err(c,j)) is a linear
sawtooth function with a downhill slope that crosses zero at all integer
c. Perhaps you want to know why this is so? Hmmm ...
Martin. |
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| Gerry... |
| Posted: Wed Oct 21, 2009 11:01 pm |
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On Oct 22, 6:31 am, cliclic... at (no spam) freenet.de wrote:
[quote]Gerry schrieb:
does there exist a solution for the gamma ratio:
gamma(1/2-s)/gamma(1/2+s)
without using the zeta function!:-)
Now let s= a + b*Pi*I
For a >0 and integer and b = 0 i found that :
gamma(1/2-s)/gamma(1/2+s) = 1/4*(-4)^s*4^s*gamma(s)^2/gamma(2*s)^2
I was hoping that it would work also for b<>0 but...
it doesn't
Only the folowing gives a fixed error for any c>0
g1(s) = gamma(1/2-s)/gamma(1/2+s)
g2(s) = 1/4*(-4)^s*4^s*gamma(s)^2/gamma(2*s)^2
gx(s,j) = log( g2(s)/g1(s)/j-1/(2*j) )/ j /log(Pi);
err(j)= gx(c-(j+1)*Pi*I,j+1) - gx(c-1*Pi*I,j);
any comments are welcome
Why do you call the reformulation of your Gamma ratio a solution? It
still involves two Gamma functions and is more complicated than your
original expression! For integer s=n, your transformation is correct,
although the following amounts to the same, while being simpler:
Gamma(1/2-n)/Gamma(1/2+n) = (-4)^n/((2n-1)!!)^2
The double factorial p!! here denotes 2*4*6* ... *p if p is even, and
1*3*5* ... *p if p is odd (the latter case applies).
In general, the following equivalences can be used to transform Gamma
functions:
Gamma(x+1) = x*Gamma(x),
Gamma(x)*Gamma(1-x) = pi/sin(pi*x),
Gamma(1/2+x)*Gamma(1/2-x) = pi/cos(pi*x),
Gamma(2*x) = 2^(2*x-1)*Gamma(x)*Gamma(x+1/2)/sqrt(pi).
That's about all there is if you restrict yourself to at most three
Gamma functions per expression. The four expressions hold for any
complex argument; the second and third of them being equivalent.
When plotting your err(c,j) := gx(c-(j+1)*pi*#i, j+1) - gx(c-1*pi*#i, j)
versus c for arbitrary constant j (where c and j are real) I get the
impression that re(err(c,j)) is constant while im(err(c,j)) is a linear
sawtooth function with a downhill slope that crosses zero at all integer
c. Perhaps you want to know why this is so? Hmmm ...
Martin.- Hide quoted text -
- Show quoted text -
[/quote]
All clear i found it :-)
For complex s we have the transformation:
gamma(1/2-s)/gamma(1/2+s)
=(-1)^(-s)*(-4)^s*4^s*gamma(s)*gamma(-2*s)/(gamma(-s)*gamma(2*s))
Can this be simplified ?
Thanks for the information Martin.
Gerry |
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| Gerry... |
| Posted: Wed Oct 21, 2009 11:31 pm |
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Guest
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On Oct 22, 11:01 am, Gerry <gerry... at (no spam) gmail.com> wrote:
[quote]On Oct 22, 6:31 am, cliclic... at (no spam) freenet.de wrote:
Gerry schrieb:
does there exist a solution for the gamma ratio:
gamma(1/2-s)/gamma(1/2+s)
without using the zeta function!:-)
Now let s= a + b*Pi*I
For a >0 and integer and b = 0 i found that :
gamma(1/2-s)/gamma(1/2+s) = 1/4*(-4)^s*4^s*gamma(s)^2/gamma(2*s)^2
I was hoping that it would work also for b<>0 but...
it doesn't
Only the folowing gives a fixed error for any c>0
g1(s) = gamma(1/2-s)/gamma(1/2+s)
g2(s) = 1/4*(-4)^s*4^s*gamma(s)^2/gamma(2*s)^2
gx(s,j) = log( g2(s)/g1(s)/j-1/(2*j) )/ j /log(Pi);
err(j)= gx(c-(j+1)*Pi*I,j+1) - gx(c-1*Pi*I,j);
any comments are welcome
Why do you call the reformulation of your Gamma ratio a solution? It
still involves two Gamma functions and is more complicated than your
original expression! For integer s=n, your transformation is correct,
although the following amounts to the same, while being simpler:
Gamma(1/2-n)/Gamma(1/2+n) = (-4)^n/((2n-1)!!)^2
The double factorial p!! here denotes 2*4*6* ... *p if p is even, and
1*3*5* ... *p if p is odd (the latter case applies).
In general, the following equivalences can be used to transform Gamma
functions:
Gamma(x+1) = x*Gamma(x),
Gamma(x)*Gamma(1-x) = pi/sin(pi*x),
Gamma(1/2+x)*Gamma(1/2-x) = pi/cos(pi*x),
Gamma(2*x) = 2^(2*x-1)*Gamma(x)*Gamma(x+1/2)/sqrt(pi).
That's about all there is if you restrict yourself to at most three
Gamma functions per expression. The four expressions hold for any
complex argument; the second and third of them being equivalent.
When plotting your err(c,j) := gx(c-(j+1)*pi*#i, j+1) - gx(c-1*pi*#i, j)
versus c for arbitrary constant j (where c and j are real) I get the
impression that re(err(c,j)) is constant while im(err(c,j)) is a linear
sawtooth function with a downhill slope that crosses zero at all integer
c. Perhaps you want to know why this is so? Hmmm ...
Martin.- Hide quoted text -
- Show quoted text -
All clear i found it :-)
For complex s we have the transformation:
gamma(1/2-s)/gamma(1/2+s)
=(-1)^(-s)*(-4)^s*4^s*gamma(s)*gamma(-2*s)/(gamma(-s)*gamma(2*s))
Can this be simplified ?
Thanks for the information Martin.
Gerry- Hide quoted text -
- Show quoted text -
[/quote]
Or better :
gamma(1/2-s)/gamma(1/2+s)
= 2^(4*s)*gamma(s)*gamma(-2*s)/(gamma(-s)*gamma(2*s))
Gerry |
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