Relationship between two hypergeometric functions
$begingroup$
Let $F(a, b, c, x)$ be the hypergeometric function. Suppose I can express $F(a, b, c, x)$ in finite terms, say in terms of $Gamma$ functions, for various $x$. Is there a way I can then deduce the value of $F(-a, b, c, x)$? For example, it is known that
$$Fbigg(frac{1}{2}, frac{1}{2}, 1, frac{1}{2}bigg) = frac{Gamma(1/4)^2}{2pi^{3/2}}.$$
Is there a way to use this result to deduce the value of $F(-1/2, 1/2, 1, 1/2)$? Likewise, it is known that
$$Fleft(frac{1}{12}, frac{5}{12}, 1, Big(frac{4}{85}Big)^3right) = frac{Gamma(1/7) Gamma(2/7) Gamma(4/7)}{8pi^2} frac{255^{1/4}}{7^{1/4}}.$$
Is there a way to use this result to deduce the value of $F(-1/12, 5/12, 1, 64/614125)$?
sequences-and-series power-series modular-forms hypergeometric-function elliptic-functions
edited Jan 23 at 10:48
Tito Piezas III
27.6k367176
asked Jul 13 '17 at 20:35
glebovgglebovg
7,15822147
$endgroup$
add a comment |
$begingroup$
Let $F(a, b, c, x)$ be the hypergeometric function. Suppose I can express $F(a, b, c, x)$ in finite terms, say in terms of $Gamma$ functions, for various $x$. Is there a way I can then deduce the value of $F(-a, b, c, x)$? For example, it is known that
$$Fbigg(frac{1}{2}, frac{1}{2}, 1, frac{1}{2}bigg) = frac{Gamma(1/4)^2}{2pi^{3/2}}.$$
Is there a way to use this result to deduce the value of $F(-1/2, 1/2, 1, 1/2)$? Likewise, it is known that
$$Fleft(frac{1}{12}, frac{5}{12}, 1, Big(frac{4}{85}Big)^3right) = frac{Gamma(1/7) Gamma(2/7) Gamma(4/7)}{8pi^2} frac{255^{1/4}}{7^{1/4}}.$$
Is there a way to use this result to deduce the value of $F(-1/12, 5/12, 1, 64/614125)$?
sequences-and-series power-series modular-forms hypergeometric-function elliptic-functions
edited Jan 23 at 10:48
Tito Piezas III
27.6k367176
asked Jul 13 '17 at 20:35
glebovgglebovg
7,15822147
$endgroup$
1
$begingroup$
My guess would be no, since $$(a)_{n}=(a)(a+1)cdots(a+n-1)$$ and $$(-a)_{n}=(-a)(-a+1)cdots(-a+n-1)=(-1)^{n}(a)(a-1)cdots(a-n+1)=(-1)^{n}a^{(n)}$$ are very different beasts. Here, $(a)_{n}$ is the rising factorial (a.k.a. Pochhammer symbol) and $a^{(n)}$ is the falling factorial.
$endgroup$
– parsiad
Jul 13 '17 at 21:23
$begingroup$
I guess no also. You asked about $F(-1/2, 1/2, 1, 1/2)$ which Mathematica evaluates to $sqrt{2}mathrm {EllipticE}[-1]/pi$ and I don't that can be expressed with $Gamma$ functions.
$endgroup$
– Somos
Jul 14 '17 at 1:10
$begingroup$
In your example, those $F$ values are $frac{2}{pi}K$ and $frac{2}{pi}E$ for $k^2=k'^2=frac{1}{2}$. Therefore $K=K'$, $E=E'$. Now use Legendre's $KE'+K'E-KK'=frac{pi}{2}$. Here that works, but that's a very special case of course.
$endgroup$
– ccorn
Jul 15 '17 at 13:02
2
$begingroup$
This is a particular case of your previous question. These hypergeometric series are essentially complete elliptic integrals and my answer (math.stackexchange.com/a/2354635/72031) shows that both $K, E$ can be expressed in terms of Gamma function and $pi$ for any singular modulus $k$. The hypergeometric function at the end of the answer is related to Ramanujan's theory of theta functions to alternative bases (Berndt calls this theories with signature $r$, classical case being $r=2$, and your hypergeometric geometric function belongs to $r=6$). The answer is yes for this case also.
$endgroup$
– Paramanand Singh
Jul 17 '17 at 7:49
$begingroup$
A minor correction in my previous comment : in the theory of signature $6$ the function ${}_{2}F_{1}(1/6,5/6,1,x)$ plays the role of complete elliptic integral $K$ and the role of $E$ is played by ${}_{2}F_{1}(-5/6,5/6,1,x)$ and it is this function which can be evaluated in terms of Gamma and $pi$ for singular moduli $k$ (or $x$). I doubt there is a connection between this function and $ F(-1/12,5/12,1,x)$ of your question.
$endgroup$
– Paramanand Singh
Jul 17 '17 at 17:27
add a comment |
$begingroup$
Let $F(a, b, c, x)$ be the hypergeometric function. Suppose I can express $F(a, b, c, x)$ in finite terms, say in terms of $Gamma$ functions, for various $x$. Is there a way I can then deduce the value of $F(-a, b, c, x)$? For example, it is known that
$$Fbigg(frac{1}{2}, frac{1}{2}, 1, frac{1}{2}bigg) = frac{Gamma(1/4)^2}{2pi^{3/2}}.$$
Is there a way to use this result to deduce the value of $F(-1/2, 1/2, 1, 1/2)$? Likewise, it is known that
$$Fleft(frac{1}{12}, frac{5}{12}, 1, Big(frac{4}{85}Big)^3right) = frac{Gamma(1/7) Gamma(2/7) Gamma(4/7)}{8pi^2} frac{255^{1/4}}{7^{1/4}}.$$
Is there a way to use this result to deduce the value of $F(-1/12, 5/12, 1, 64/614125)$?
sequences-and-series power-series modular-forms hypergeometric-function elliptic-functions
edited Jan 23 at 10:48
Tito Piezas III
27.6k367176
asked Jul 13 '17 at 20:35
glebovgglebovg
7,15822147
$endgroup$
Let $F(a, b, c, x)$ be the hypergeometric function. Suppose I can express $F(a, b, c, x)$ in finite terms, say in terms of $Gamma$ functions, for various $x$. Is there a way I can then deduce the value of $F(-a, b, c, x)$? For example, it is known that
$$Fbigg(frac{1}{2}, frac{1}{2}, 1, frac{1}{2}bigg) = frac{Gamma(1/4)^2}{2pi^{3/2}}.$$
Is there a way to use this result to deduce the value of $F(-1/2, 1/2, 1, 1/2)$? Likewise, it is known that
$$Fleft(frac{1}{12}, frac{5}{12}, 1, Big(frac{4}{85}Big)^3right) = frac{Gamma(1/7) Gamma(2/7) Gamma(4/7)}{8pi^2} frac{255^{1/4}}{7^{1/4}}.$$
Is there a way to use this result to deduce the value of $F(-1/12, 5/12, 1, 64/614125)$?
sequences-and-series power-series modular-forms hypergeometric-function elliptic-functions
sequences-and-series power-series modular-forms hypergeometric-function elliptic-functions
edited Jan 23 at 10:48
Tito Piezas III
27.6k367176
asked Jul 13 '17 at 20:35
glebovgglebovg
7,15822147
edited Jan 23 at 10:48
Tito Piezas III
27.6k367176
asked Jul 13 '17 at 20:35
glebovgglebovg
7,15822147
edited Jan 23 at 10:48
Tito Piezas III
27.6k367176
edited Jan 23 at 10:48
Tito Piezas III
27.6k367176
edited Jan 23 at 10:48
Tito Piezas III
27.6k367176
27.6k367176
asked Jul 13 '17 at 20:35
glebovgglebovg
7,15822147
asked Jul 13 '17 at 20:35
glebovgglebovg
7,15822147
asked Jul 13 '17 at 20:35
glebovgglebovg
7,15822147
7,15822147
1
$begingroup$
My guess would be no, since $$(a)_{n}=(a)(a+1)cdots(a+n-1)$$ and $$(-a)_{n}=(-a)(-a+1)cdots(-a+n-1)=(-1)^{n}(a)(a-1)cdots(a-n+1)=(-1)^{n}a^{(n)}$$ are very different beasts. Here, $(a)_{n}$ is the rising factorial (a.k.a. Pochhammer symbol) and $a^{(n)}$ is the falling factorial.
$endgroup$
– parsiad
Jul 13 '17 at 21:23
$begingroup$
I guess no also. You asked about $F(-1/2, 1/2, 1, 1/2)$ which Mathematica evaluates to $sqrt{2}mathrm {EllipticE}[-1]/pi$ and I don't that can be expressed with $Gamma$ functions.
$endgroup$
– Somos
Jul 14 '17 at 1:10
$begingroup$
In your example, those $F$ values are $frac{2}{pi}K$ and $frac{2}{pi}E$ for $k^2=k'^2=frac{1}{2}$. Therefore $K=K'$, $E=E'$. Now use Legendre's $KE'+K'E-KK'=frac{pi}{2}$. Here that works, but that's a very special case of course.
$endgroup$
– ccorn
Jul 15 '17 at 13:02
2
$begingroup$
This is a particular case of your previous question. These hypergeometric series are essentially complete elliptic integrals and my answer (math.stackexchange.com/a/2354635/72031) shows that both $K, E$ can be expressed in terms of Gamma function and $pi$ for any singular modulus $k$. The hypergeometric function at the end of the answer is related to Ramanujan's theory of theta functions to alternative bases (Berndt calls this theories with signature $r$, classical case being $r=2$, and your hypergeometric geometric function belongs to $r=6$). The answer is yes for this case also.
$endgroup$
– Paramanand Singh
Jul 17 '17 at 7:49
$begingroup$
A minor correction in my previous comment : in the theory of signature $6$ the function ${}_{2}F_{1}(1/6,5/6,1,x)$ plays the role of complete elliptic integral $K$ and the role of $E$ is played by ${}_{2}F_{1}(-5/6,5/6,1,x)$ and it is this function which can be evaluated in terms of Gamma and $pi$ for singular moduli $k$ (or $x$). I doubt there is a connection between this function and $ F(-1/12,5/12,1,x)$ of your question.
$endgroup$
– Paramanand Singh
Jul 17 '17 at 17:27
add a comment |
1
$begingroup$
My guess would be no, since $$(a)_{n}=(a)(a+1)cdots(a+n-1)$$ and $$(-a)_{n}=(-a)(-a+1)cdots(-a+n-1)=(-1)^{n}(a)(a-1)cdots(a-n+1)=(-1)^{n}a^{(n)}$$ are very different beasts. Here, $(a)_{n}$ is the rising factorial (a.k.a. Pochhammer symbol) and $a^{(n)}$ is the falling factorial.
$endgroup$
– parsiad
Jul 13 '17 at 21:23
$begingroup$
I guess no also. You asked about $F(-1/2, 1/2, 1, 1/2)$ which Mathematica evaluates to $sqrt{2}mathrm {EllipticE}[-1]/pi$ and I don't that can be expressed with $Gamma$ functions.
$endgroup$
– Somos
Jul 14 '17 at 1:10
$begingroup$
In your example, those $F$ values are $frac{2}{pi}K$ and $frac{2}{pi}E$ for $k^2=k'^2=frac{1}{2}$. Therefore $K=K'$, $E=E'$. Now use Legendre's $KE'+K'E-KK'=frac{pi}{2}$. Here that works, but that's a very special case of course.
$endgroup$
– ccorn
Jul 15 '17 at 13:02
2
$begingroup$
This is a particular case of your previous question. These hypergeometric series are essentially complete elliptic integrals and my answer (math.stackexchange.com/a/2354635/72031) shows that both $K, E$ can be expressed in terms of Gamma function and $pi$ for any singular modulus $k$. The hypergeometric function at the end of the answer is related to Ramanujan's theory of theta functions to alternative bases (Berndt calls this theories with signature $r$, classical case being $r=2$, and your hypergeometric geometric function belongs to $r=6$). The answer is yes for this case also.
$endgroup$
– Paramanand Singh
Jul 17 '17 at 7:49
$begingroup$
A minor correction in my previous comment : in the theory of signature $6$ the function ${}_{2}F_{1}(1/6,5/6,1,x)$ plays the role of complete elliptic integral $K$ and the role of $E$ is played by ${}_{2}F_{1}(-5/6,5/6,1,x)$ and it is this function which can be evaluated in terms of Gamma and $pi$ for singular moduli $k$ (or $x$). I doubt there is a connection between this function and $ F(-1/12,5/12,1,x)$ of your question.
$endgroup$
– Paramanand Singh
Jul 17 '17 at 17:27
1
1
$begingroup$
My guess would be no, since $$(a)_{n}=(a)(a+1)cdots(a+n-1)$$ and $$(-a)_{n}=(-a)(-a+1)cdots(-a+n-1)=(-1)^{n}(a)(a-1)cdots(a-n+1)=(-1)^{n}a^{(n)}$$ are very different beasts. Here, $(a)_{n}$ is the rising factorial (a.k.a. Pochhammer symbol) and $a^{(n)}$ is the falling factorial.
$endgroup$
– parsiad
Jul 13 '17 at 21:23
$begingroup$
My guess would be no, since $$(a)_{n}=(a)(a+1)cdots(a+n-1)$$ and $$(-a)_{n}=(-a)(-a+1)cdots(-a+n-1)=(-1)^{n}(a)(a-1)cdots(a-n+1)=(-1)^{n}a^{(n)}$$ are very different beasts. Here, $(a)_{n}$ is the rising factorial (a.k.a. Pochhammer symbol) and $a^{(n)}$ is the falling factorial.
$endgroup$
– parsiad
Jul 13 '17 at 21:23
$begingroup$
I guess no also. You asked about $F(-1/2, 1/2, 1, 1/2)$ which Mathematica evaluates to $sqrt{2}mathrm {EllipticE}[-1]/pi$ and I don't that can be expressed with $Gamma$ functions.
$endgroup$
– Somos
Jul 14 '17 at 1:10
$begingroup$
I guess no also. You asked about $F(-1/2, 1/2, 1, 1/2)$ which Mathematica evaluates to $sqrt{2}mathrm {EllipticE}[-1]/pi$ and I don't that can be expressed with $Gamma$ functions.
$endgroup$
– Somos
Jul 14 '17 at 1:10
$begingroup$
In your example, those $F$ values are $frac{2}{pi}K$ and $frac{2}{pi}E$ for $k^2=k'^2=frac{1}{2}$. Therefore $K=K'$, $E=E'$. Now use Legendre's $KE'+K'E-KK'=frac{pi}{2}$. Here that works, but that's a very special case of course.
$endgroup$
– ccorn
Jul 15 '17 at 13:02
$begingroup$
In your example, those $F$ values are $frac{2}{pi}K$ and $frac{2}{pi}E$ for $k^2=k'^2=frac{1}{2}$. Therefore $K=K'$, $E=E'$. Now use Legendre's $KE'+K'E-KK'=frac{pi}{2}$. Here that works, but that's a very special case of course.
$endgroup$
– ccorn
Jul 15 '17 at 13:02
2
2
$begingroup$
This is a particular case of your previous question. These hypergeometric series are essentially complete elliptic integrals and my answer (math.stackexchange.com/a/2354635/72031) shows that both $K, E$ can be expressed in terms of Gamma function and $pi$ for any singular modulus $k$. The hypergeometric function at the end of the answer is related to Ramanujan's theory of theta functions to alternative bases (Berndt calls this theories with signature $r$, classical case being $r=2$, and your hypergeometric geometric function belongs to $r=6$). The answer is yes for this case also.
$endgroup$
– Paramanand Singh
Jul 17 '17 at 7:49
$begingroup$
This is a particular case of your previous question. These hypergeometric series are essentially complete elliptic integrals and my answer (math.stackexchange.com/a/2354635/72031) shows that both $K, E$ can be expressed in terms of Gamma function and $pi$ for any singular modulus $k$. The hypergeometric function at the end of the answer is related to Ramanujan's theory of theta functions to alternative bases (Berndt calls this theories with signature $r$, classical case being $r=2$, and your hypergeometric geometric function belongs to $r=6$). The answer is yes for this case also.
$endgroup$
– Paramanand Singh
Jul 17 '17 at 7:49
$begingroup$
A minor correction in my previous comment : in the theory of signature $6$ the function ${}_{2}F_{1}(1/6,5/6,1,x)$ plays the role of complete elliptic integral $K$ and the role of $E$ is played by ${}_{2}F_{1}(-5/6,5/6,1,x)$ and it is this function which can be evaluated in terms of Gamma and $pi$ for singular moduli $k$ (or $x$). I doubt there is a connection between this function and $ F(-1/12,5/12,1,x)$ of your question.
$endgroup$
– Paramanand Singh
Jul 17 '17 at 17:27
$begingroup$
A minor correction in my previous comment : in the theory of signature $6$ the function ${}_{2}F_{1}(1/6,5/6,1,x)$ plays the role of complete elliptic integral $K$ and the role of $E$ is played by ${}_{2}F_{1}(-5/6,5/6,1,x)$ and it is this function which can be evaluated in terms of Gamma and $pi$ for singular moduli $k$ (or $x$). I doubt there is a connection between this function and $ F(-1/12,5/12,1,x)$ of your question.
$endgroup$
– Paramanand Singh
Jul 17 '17 at 17:27
add a comment |
1 Answer
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$begingroup$
In general, when you are given solutions for special values of $z$ in $text{F}(a,b,c,z)$ they are unique, unless otherwise specified. I have found three solutions in terms of gamma functions for $z=1/2$ here; none of them would allow you to find the specific answer that you seek.
As for your second case, I don't imagine it would be any different. However, I cannot say for certain as I am unable find anything like that. I did, however, ascertain that it is correct insofar as I was able to verify it numerically and from the gamma solution.
answered Jul 14 '17 at 21:04
Cye WaldmanCye Waldman
4,2112523
$endgroup$
add a comment |
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$begingroup$
In general, when you are given solutions for special values of $z$ in $text{F}(a,b,c,z)$ they are unique, unless otherwise specified. I have found three solutions in terms of gamma functions for $z=1/2$ here; none of them would allow you to find the specific answer that you seek.
As for your second case, I don't imagine it would be any different. However, I cannot say for certain as I am unable find anything like that. I did, however, ascertain that it is correct insofar as I was able to verify it numerically and from the gamma solution.
answered Jul 14 '17 at 21:04
Cye WaldmanCye Waldman
4,2112523
$endgroup$
add a comment |
$begingroup$
In general, when you are given solutions for special values of $z$ in $text{F}(a,b,c,z)$ they are unique, unless otherwise specified. I have found three solutions in terms of gamma functions for $z=1/2$ here; none of them would allow you to find the specific answer that you seek.
As for your second case, I don't imagine it would be any different. However, I cannot say for certain as I am unable find anything like that. I did, however, ascertain that it is correct insofar as I was able to verify it numerically and from the gamma solution.
answered Jul 14 '17 at 21:04
Cye WaldmanCye Waldman
4,2112523
$endgroup$
add a comment |
$begingroup$
In general, when you are given solutions for special values of $z$ in $text{F}(a,b,c,z)$ they are unique, unless otherwise specified. I have found three solutions in terms of gamma functions for $z=1/2$ here; none of them would allow you to find the specific answer that you seek.
As for your second case, I don't imagine it would be any different. However, I cannot say for certain as I am unable find anything like that. I did, however, ascertain that it is correct insofar as I was able to verify it numerically and from the gamma solution.
answered Jul 14 '17 at 21:04
Cye WaldmanCye Waldman
4,2112523
$endgroup$
In general, when you are given solutions for special values of $z$ in $text{F}(a,b,c,z)$ they are unique, unless otherwise specified. I have found three solutions in terms of gamma functions for $z=1/2$ here; none of them would allow you to find the specific answer that you seek.
As for your second case, I don't imagine it would be any different. However, I cannot say for certain as I am unable find anything like that. I did, however, ascertain that it is correct insofar as I was able to verify it numerically and from the gamma solution.
answered Jul 14 '17 at 21:04
Cye WaldmanCye Waldman
4,2112523
answered Jul 14 '17 at 21:04
Cye WaldmanCye Waldman
4,2112523
answered Jul 14 '17 at 21:04
Cye WaldmanCye Waldman
4,2112523
answered Jul 14 '17 at 21:04
Cye WaldmanCye Waldman
4,2112523
4,2112523
add a comment |
add a comment |
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My guess would be no, since $$(a)_{n}=(a)(a+1)cdots(a+n-1)$$ and $$(-a)_{n}=(-a)(-a+1)cdots(-a+n-1)=(-1)^{n}(a)(a-1)cdots(a-n+1)=(-1)^{n}a^{(n)}$$ are very different beasts. Here, $(a)_{n}$ is the rising factorial (a.k.a. Pochhammer symbol) and $a^{(n)}$ is the falling factorial.
$endgroup$
– parsiad
Jul 13 '17 at 21:23
$begingroup$
I guess no also. You asked about $F(-1/2, 1/2, 1, 1/2)$ which Mathematica evaluates to $sqrt{2}mathrm {EllipticE}[-1]/pi$ and I don't that can be expressed with $Gamma$ functions.
$endgroup$
– Somos
Jul 14 '17 at 1:10
$begingroup$
In your example, those $F$ values are $frac{2}{pi}K$ and $frac{2}{pi}E$ for $k^2=k'^2=frac{1}{2}$. Therefore $K=K'$, $E=E'$. Now use Legendre's $KE'+K'E-KK'=frac{pi}{2}$. Here that works, but that's a very special case of course.
$endgroup$
– ccorn
Jul 15 '17 at 13:02
2
$begingroup$
This is a particular case of your previous question. These hypergeometric series are essentially complete elliptic integrals and my answer (math.stackexchange.com/a/2354635/72031) shows that both $K, E$ can be expressed in terms of Gamma function and $pi$ for any singular modulus $k$. The hypergeometric function at the end of the answer is related to Ramanujan's theory of theta functions to alternative bases (Berndt calls this theories with signature $r$, classical case being $r=2$, and your hypergeometric geometric function belongs to $r=6$). The answer is yes for this case also.
$endgroup$
– Paramanand Singh
Jul 17 '17 at 7:49
$begingroup$
A minor correction in my previous comment : in the theory of signature $6$ the function ${}_{2}F_{1}(1/6,5/6,1,x)$ plays the role of complete elliptic integral $K$ and the role of $E$ is played by ${}_{2}F_{1}(-5/6,5/6,1,x)$ and it is this function which can be evaluated in terms of Gamma and $pi$ for singular moduli $k$ (or $x$). I doubt there is a connection between this function and $ F(-1/12,5/12,1,x)$ of your question.
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– Paramanand Singh
Jul 17 '17 at 17:27