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Guest
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 Posted: Thu Jun 09, 2005 3:06 pm Post subject:
hypergeometric GF
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What is the generating function satisfying
a(1)=1/(2*m-1);
a(n+1) = -(2*m-2+n)*n/(2*(2*m-1+2*n)*(1+n)) * a(n), n>=1
m,n positive integers? |
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G. A. Edgar
science forum Guru
Joined: 29 Apr 2005
Posts: 470
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| Posted: Thu Jun 09, 2005 5:20 pm Post subject:
Re: hypergeometric GF
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In article <1118336815.019647.41220@g49g2000cwa.googlegroups.com>,
<carlos@colorado.edu> wrote:
| Quote: | What is the generating function satisfying
a(1)=1/(2*m-1);
a(n+1) = -(2*m-2+n)*n/(2*(2*m-1+2*n)*(1+n)) * a(n), n>=1
m,n positive integers?
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Maple says,
| Quote: | E1 := a(n+1) = -(2*m-2+n)*n/(2*(2*m-1+2*n)*(1+n)) * a(n);
(2 m - 2 + n) n a(n) |
E1 := a(n + 1) = - -------------------------
2 (2 m - 1 + 2 n) (n + 1)
| Quote: | rsolve({E1,a(1)=1/(2*m+1)},a(n));
(-n - m) (1/2) n |
8 4 Pi (2 m - 1) GAMMA(2 m - 2 + n) (-1)
- ------------------------------------------------------
/ 1\
GAMMA(m) GAMMA|n + m - -| n (2 m + 1)
\ 2/
| Quote: | simplify(%);
(n + 1) (-n - m) (1/2) |
8 (-1) 4 Pi (2 m - 1) GAMMA(2 m - 2 + n)
------------------------------------------------------------
/ 1\
GAMMA(m) GAMMA|n + m - -| n (2 m + 1)
\ 2/
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/ |
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Daniel Lichtblau
science forum beginner
Joined: 12 May 2005
Posts: 48
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| Posted: Thu Jun 09, 2005 6:05 pm Post subject:
Re: hypergeometric GF
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carlos@colorado.edu wrote:
| Quote: | What is the generating function satisfying
a(1)=1/(2*m-1);
a(n+1) = -(2*m-2+n)*n/(2*(2*m-1+2*n)*(1+n)) * a(n), n>=1
m,n positive integers?
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In[65]:= InputForm[a[n] /. First[RSolve[{a[1]==1/(2*m-1),
a[n+1] == -(2*m-2+n)*n/(2*(2*m-1+2*n)*(1+n)) * a[n]}, a[n], n]]]
Out[65]//InputForm=
-(((-1)^n*4^(1 - n)*Gamma[1/2 + m]*Gamma[-2 + 2*m + n])/
((-1 + 2*m)*n*Gamma[-1 + 2*m]*Gamma[-1/2 + m + n]))
Daniel Lichtblau
Wolfram Research |
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Guest
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| Posted: Fri Jun 10, 2005 5:21 am Post subject:
Re: hypergeometric GF
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That is just a formula for the coefficients but not the generating
function. In Mathematica I tried
cn=-(2*m-2+n)*n/(2*(2*m-1+2*n)*(1+n));
{{g}}=GeneratingFunction[{a[n+1]==cn*a[n],a[1]==1/(2*m-1)},a[n],n,z];
but the answer is wrong. For small numeric instances of m, however,
the answers are easily found:
m=1 g=4*ArcSinh[Sqrt[z]/2]^2
m=2 g=(-240*(Sqrt[z*(4 + z)] - (4 + z)*ArcSinh[Sqrt[z]/2]) -
Sqrt[z^5*(4 + z)]*HypergeometricPFQ[{2, 2}, {7/2},
-z/4])/(60*Sqrt[z*(4 + z)])
What I would like to get is the general expression for arbitrary
integer m>0.
Thanks. |
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Daniel Lichtblau
science forum beginner
Joined: 12 May 2005
Posts: 48
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| Posted: Fri Jun 10, 2005 5:45 am Post subject:
Re: hypergeometric GF
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car...@colorado.edu wrote:
| Quote: | That is just a formula for the coefficients but not the generating
function. In Mathematica I tried
cn=-(2*m-2+n)*n/(2*(2*m-1+2*n)*(1+n));
{{g}}=GeneratingFunction[{a[n+1]==cn*a[n],a[1]==1/(2*m-1)},a[n],n,z];
but the answer is wrong. For small numeric instances of m, however,
the answers are easily found:
m=1 g=4*ArcSinh[Sqrt[z]/2]^2
m=2 g=(-240*(Sqrt[z*(4 + z)] - (4 + z)*ArcSinh[Sqrt[z]/2]) -
Sqrt[z^5*(4 + z)]*HypergeometricPFQ[{2, 2}, {7/2},
-z/4])/(60*Sqrt[z*(4 + z)])
What I would like to get is the general expression for arbitrary
integer m>0.
Thanks.
|
It's a beginning from which the rest can be attained. I'll work with
the form I showed. I assume the one given by G. Edgar is equivalent.
We'll use it to define a function of m and n.
InputForm[c[m_,n_] = a[n] /. First[RSolve[{a[1]==1/(2*m-1),
a[n+1] == -(2*m-2+n)*n/(2*(2*m-1+2*n)*(1+n))*a[n]}, a[n], n]]]
Now we'll use a pedestrian recursive definition of the same function
(on nonnegative integers).
b[m_,1] = 1/(2*m-1);
b[m_,n_] := b[m,n] =
-(2*m-2+(n-1))*(n-1)/(2*(2*m-1+2*(n-1))*n)*b[m,n-1]
A few numeric examples should demonstrate that these are in agreement.
In[11]:= b[5,7] - c[5,7]
Out[11]= 0
In[12]:= b[3,21] - c[3,21]
Out[12]= 0
In[16]:= InputForm[val = b[1/3,21] - c[1/3,21]]
Out[16]//InputForm=
5/4496856363183247 +
(Gamma[5/6]*Gamma[59/3])/(7696581394432*Gamma[-1/3]*
Gamma[125/6])
In[17]:= FunctionExpand[val]
Out[17]= 0
So I gather that what you wanted was not the closed form of a[m,n], but
rather a sum of said form over powers of x^n.
In[21]:= InputForm[Sum[c[m,n]*x^n, {n,1,Infinity}]]
Out[21]//InputForm=
(x*HypergeometricPFQ[{1, 1, -1 + 2*m}, {2, 1/2 + m}, -x/4])/(-1 + 2*m)
Daniel Lichtblau
Wolfram Research |
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Paul Abbott
science forum addict
Joined: 19 May 2005
Posts: 99
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| Posted: Fri Jun 10, 2005 5:50 am Post subject:
Re: hypergeometric GF
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In article <1118388086.398435.13560@g43g2000cwa.googlegroups.com>,
carlos@colorado.edu wrote:
| Quote: | That is just a formula for the coefficients but not the generating
function. In Mathematica I tried
cn=-(2*m-2+n)*n/(2*(2*m-1+2*n)*(1+n));
{{g}}=GeneratingFunction[{a[n+1]==cn*a[n],a[1]==1/(2*m-1)},a[n],n,z];
but the answer is wrong. For small numeric instances of m, however,
the answers are easily found:
m=1 g=4*ArcSinh[Sqrt[z]/2]^2
m=2 g=(-240*(Sqrt[z*(4 + z)] - (4 + z)*ArcSinh[Sqrt[z]/2]) -
Sqrt[z^5*(4 + z)]*HypergeometricPFQ[{2, 2}, {7/2},
-z/4])/(60*Sqrt[z*(4 + z)])
What I would like to get is the general expression for arbitrary
integer m>0.
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RSolve[{a[1] == 1/(2 m - 1),
a[n] == -(n-1) (2 m + n - 3) a[n-1]/(2 (2 (n-1)+2 m-1) n)}, a, n]
a[n_] = a[n] /. First[%]
gf[m_][z_] = Sum[a[n] z^n, {n, 1, Infinity}]
The answer is
z HypergeometricPFQ[{1, 1, 2m - 1}, {2, m + 1/2}, -z/4]/(2 m - 1)
and this yields you give above.
Cheers,
Paul
--
Paul Abbott Phone: +61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
The University of Western Australia (CRICOS Provider No 00126G)
AUSTRALIA http://physics.uwa.edu.au/~paul
http://InternationalMathematicaSymposium.org/IMS2005/ |
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Guest
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| Posted: Fri Jun 10, 2005 2:09 pm Post subject:
Re: hypergeometric GF
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That is perfect! I should have thought of using Sum rather
than GeneratingFunction. Thanks. |
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