An identity for generalized hypergeometric function
2
$begingroup$
I think the following identity is true,
$$
frac{4 (4 s+9)}{3 Gamma left(s+frac{5}{2}right) Gamma left(s+frac{7}{2}right)}-frac{16 (s+2)}{3 Gamma (s+3)^2}=frac{, _3F_2left(2,s+frac{5}{2},s+frac{7}{2};s+4,s+5;1right)}{Gamma (s+4) Gamma (s+5)}
$$
But I haven't found anything useful in the literature.
This identity is inspired by this question.
hypergeometric-function
asked Jan 15 at 0:16
ablmfablmf
2,58352452
$endgroup$
add a comment |
2
$begingroup$
I think the following identity is true,
$$
frac{4 (4 s+9)}{3 Gamma left(s+frac{5}{2}right) Gamma left(s+frac{7}{2}right)}-frac{16 (s+2)}{3 Gamma (s+3)^2}=frac{, _3F_2left(2,s+frac{5}{2},s+frac{7}{2};s+4,s+5;1right)}{Gamma (s+4) Gamma (s+5)}
$$
But I haven't found anything useful in the literature.
This identity is inspired by this question.
hypergeometric-function
asked Jan 15 at 0:16
ablmfablmf
2,58352452
$endgroup$
1
$begingroup$
Since $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=sum_{ngeq0}frac{(a_1)_n(a_2)_n(a_3)_n}{(b_1)_n(b_2)_n n!}$$ And since $(x)_n=frac{Gamma(x+n)}{Gamma(x)}$ you can factor: $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=frac{Gamma(b_1)Gamma(b_2)}{Gamma(a_1)Gamma(a_2)Gamma(a_3)}sum_{ngeq0}frac{Gamma(a_1+n)Gamma(a_2+n)Gamma(a_3+n)}{Gamma(b_1+n)Gamma(b_2+n) n!}$$ which will help you simplify things
$endgroup$
– clathratus
Jan 15 at 1:08
$begingroup$
This is only valid if s+5/2 is a nonpositive integer, right?
$endgroup$
– ablmf
Jan 15 at 14:43
$begingroup$
I think it's valid for $$text{Re},s>-frac52$$
$endgroup$
– clathratus
Jan 15 at 15:25
$begingroup$
Okay. For anyone who's interested in a proof, see Doron Zeilberger's book A=B, example 7.3.3.
$endgroup$
– ablmf
Jan 18 at 17:19
add a comment |
2
2
2
1
$begingroup$
I think the following identity is true,
$$
frac{4 (4 s+9)}{3 Gamma left(s+frac{5}{2}right) Gamma left(s+frac{7}{2}right)}-frac{16 (s+2)}{3 Gamma (s+3)^2}=frac{, _3F_2left(2,s+frac{5}{2},s+frac{7}{2};s+4,s+5;1right)}{Gamma (s+4) Gamma (s+5)}
$$
But I haven't found anything useful in the literature.
This identity is inspired by this question.
hypergeometric-function
asked Jan 15 at 0:16
ablmfablmf
2,58352452
$endgroup$
I think the following identity is true,
$$
frac{4 (4 s+9)}{3 Gamma left(s+frac{5}{2}right) Gamma left(s+frac{7}{2}right)}-frac{16 (s+2)}{3 Gamma (s+3)^2}=frac{, _3F_2left(2,s+frac{5}{2},s+frac{7}{2};s+4,s+5;1right)}{Gamma (s+4) Gamma (s+5)}
$$
But I haven't found anything useful in the literature.
This identity is inspired by this question.
hypergeometric-function
hypergeometric-function
asked Jan 15 at 0:16
ablmfablmf
2,58352452
asked Jan 15 at 0:16
ablmfablmf
2,58352452
asked Jan 15 at 0:16
ablmfablmf
2,58352452
asked Jan 15 at 0:16
ablmfablmf
2,58352452
asked Jan 15 at 0:16
ablmfablmf
2,58352452
2,58352452
1
$begingroup$
Since $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=sum_{ngeq0}frac{(a_1)_n(a_2)_n(a_3)_n}{(b_1)_n(b_2)_n n!}$$ And since $(x)_n=frac{Gamma(x+n)}{Gamma(x)}$ you can factor: $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=frac{Gamma(b_1)Gamma(b_2)}{Gamma(a_1)Gamma(a_2)Gamma(a_3)}sum_{ngeq0}frac{Gamma(a_1+n)Gamma(a_2+n)Gamma(a_3+n)}{Gamma(b_1+n)Gamma(b_2+n) n!}$$ which will help you simplify things
$endgroup$
– clathratus
Jan 15 at 1:08
$begingroup$
This is only valid if s+5/2 is a nonpositive integer, right?
$endgroup$
– ablmf
Jan 15 at 14:43
$begingroup$
I think it's valid for $$text{Re},s>-frac52$$
$endgroup$
– clathratus
Jan 15 at 15:25
$begingroup$
Okay. For anyone who's interested in a proof, see Doron Zeilberger's book A=B, example 7.3.3.
$endgroup$
– ablmf
Jan 18 at 17:19
add a comment |
1
$begingroup$
Since $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=sum_{ngeq0}frac{(a_1)_n(a_2)_n(a_3)_n}{(b_1)_n(b_2)_n n!}$$ And since $(x)_n=frac{Gamma(x+n)}{Gamma(x)}$ you can factor: $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=frac{Gamma(b_1)Gamma(b_2)}{Gamma(a_1)Gamma(a_2)Gamma(a_3)}sum_{ngeq0}frac{Gamma(a_1+n)Gamma(a_2+n)Gamma(a_3+n)}{Gamma(b_1+n)Gamma(b_2+n) n!}$$ which will help you simplify things
$endgroup$
– clathratus
Jan 15 at 1:08
$begingroup$
This is only valid if s+5/2 is a nonpositive integer, right?
$endgroup$
– ablmf
Jan 15 at 14:43
$begingroup$
I think it's valid for $$text{Re},s>-frac52$$
$endgroup$
– clathratus
Jan 15 at 15:25
$begingroup$
Okay. For anyone who's interested in a proof, see Doron Zeilberger's book A=B, example 7.3.3.
$endgroup$
– ablmf
Jan 18 at 17:19
1
1
$begingroup$
Since $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=sum_{ngeq0}frac{(a_1)_n(a_2)_n(a_3)_n}{(b_1)_n(b_2)_n n!}$$ And since $(x)_n=frac{Gamma(x+n)}{Gamma(x)}$ you can factor: $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=frac{Gamma(b_1)Gamma(b_2)}{Gamma(a_1)Gamma(a_2)Gamma(a_3)}sum_{ngeq0}frac{Gamma(a_1+n)Gamma(a_2+n)Gamma(a_3+n)}{Gamma(b_1+n)Gamma(b_2+n) n!}$$ which will help you simplify things
$endgroup$
– clathratus
Jan 15 at 1:08
$begingroup$
Since $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=sum_{ngeq0}frac{(a_1)_n(a_2)_n(a_3)_n}{(b_1)_n(b_2)_n n!}$$ And since $(x)_n=frac{Gamma(x+n)}{Gamma(x)}$ you can factor: $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=frac{Gamma(b_1)Gamma(b_2)}{Gamma(a_1)Gamma(a_2)Gamma(a_3)}sum_{ngeq0}frac{Gamma(a_1+n)Gamma(a_2+n)Gamma(a_3+n)}{Gamma(b_1+n)Gamma(b_2+n) n!}$$ which will help you simplify things
$endgroup$
– clathratus
Jan 15 at 1:08
$begingroup$
This is only valid if s+5/2 is a nonpositive integer, right?
$endgroup$
– ablmf
Jan 15 at 14:43
$begingroup$
This is only valid if s+5/2 is a nonpositive integer, right?
$endgroup$
– ablmf
Jan 15 at 14:43
$begingroup$
I think it's valid for $$text{Re},s>-frac52$$
$endgroup$
– clathratus
Jan 15 at 15:25
$begingroup$
I think it's valid for $$text{Re},s>-frac52$$
$endgroup$
– clathratus
Jan 15 at 15:25
$begingroup$
Okay. For anyone who's interested in a proof, see Doron Zeilberger's book A=B, example 7.3.3.
$endgroup$
– ablmf
Jan 18 at 17:19
$begingroup$
Okay. For anyone who's interested in a proof, see Doron Zeilberger's book A=B, example 7.3.3.
$endgroup$
– ablmf
Jan 18 at 17:19
add a comment |
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1
$begingroup$
Since $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=sum_{ngeq0}frac{(a_1)_n(a_2)_n(a_3)_n}{(b_1)_n(b_2)_n n!}$$ And since $(x)_n=frac{Gamma(x+n)}{Gamma(x)}$ you can factor: $$,_3F_2(a_1,a_2,a_3;b_1,b_2;1)=frac{Gamma(b_1)Gamma(b_2)}{Gamma(a_1)Gamma(a_2)Gamma(a_3)}sum_{ngeq0}frac{Gamma(a_1+n)Gamma(a_2+n)Gamma(a_3+n)}{Gamma(b_1+n)Gamma(b_2+n) n!}$$ which will help you simplify things
$endgroup$
– clathratus
Jan 15 at 1:08
$begingroup$
This is only valid if s+5/2 is a nonpositive integer, right?
$endgroup$
– ablmf
Jan 15 at 14:43
$begingroup$
I think it's valid for $$text{Re},s>-frac52$$
$endgroup$
– clathratus
Jan 15 at 15:25
$begingroup$
Okay. For anyone who's interested in a proof, see Doron Zeilberger's book A=B, example 7.3.3.
$endgroup$
– ablmf
Jan 18 at 17:19