Inverse of the asymptotic expansion of Gauss Hypergeometric function
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I am interested in obtaining the asymptotic expansion of $r(rho)$ (which is the inverse of $rho$ below). Basically I want to series expand $rho$ for large $r$ (i.e. as $rto infty$) and then invert the series to obtain $r(rho)$. I have tried some readily available asymptotic expansion of $_2F_1$ but I think I ruined my calculations and I am not really sure if I am doing the right thing. By the way, $b$ is just some positive constant while $-infty<q<1$.
$rho=frac{2b}{1-q}(1-(frac{b}{r})^{1-q})^{1/2}(_2F_1(frac{1}{2},1-frac{1}{q-1},frac{3}{2},1-(frac{b}{r})^{1-q}))$
asymptotics power-series inverse-function hypergeometric-function
asked Nov 23 at 3:32
user583893
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I am interested in obtaining the asymptotic expansion of $r(rho)$ (which is the inverse of $rho$ below). Basically I want to series expand $rho$ for large $r$ (i.e. as $rto infty$) and then invert the series to obtain $r(rho)$. I have tried some readily available asymptotic expansion of $_2F_1$ but I think I ruined my calculations and I am not really sure if I am doing the right thing. By the way, $b$ is just some positive constant while $-infty<q<1$.
$rho=frac{2b}{1-q}(1-(frac{b}{r})^{1-q})^{1/2}(_2F_1(frac{1}{2},1-frac{1}{q-1},frac{3}{2},1-(frac{b}{r})^{1-q}))$
asymptotics power-series inverse-function hypergeometric-function
asked Nov 23 at 3:32
user583893
264
Your function isn't analytic at $1/r=0$, it has a branch point with something like $r^{-1/2}$ and $r^{1-3/2-1+1/(q-1)}$ terms
– reuns
Nov 23 at 4:05
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up vote
0
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up vote
0
down vote
favorite
I am interested in obtaining the asymptotic expansion of $r(rho)$ (which is the inverse of $rho$ below). Basically I want to series expand $rho$ for large $r$ (i.e. as $rto infty$) and then invert the series to obtain $r(rho)$. I have tried some readily available asymptotic expansion of $_2F_1$ but I think I ruined my calculations and I am not really sure if I am doing the right thing. By the way, $b$ is just some positive constant while $-infty<q<1$.
$rho=frac{2b}{1-q}(1-(frac{b}{r})^{1-q})^{1/2}(_2F_1(frac{1}{2},1-frac{1}{q-1},frac{3}{2},1-(frac{b}{r})^{1-q}))$
asymptotics power-series inverse-function hypergeometric-function
asked Nov 23 at 3:32
user583893
264
I am interested in obtaining the asymptotic expansion of $r(rho)$ (which is the inverse of $rho$ below). Basically I want to series expand $rho$ for large $r$ (i.e. as $rto infty$) and then invert the series to obtain $r(rho)$. I have tried some readily available asymptotic expansion of $_2F_1$ but I think I ruined my calculations and I am not really sure if I am doing the right thing. By the way, $b$ is just some positive constant while $-infty<q<1$.
$rho=frac{2b}{1-q}(1-(frac{b}{r})^{1-q})^{1/2}(_2F_1(frac{1}{2},1-frac{1}{q-1},frac{3}{2},1-(frac{b}{r})^{1-q}))$
asymptotics power-series inverse-function hypergeometric-function
asymptotics power-series inverse-function hypergeometric-function
asked Nov 23 at 3:32
user583893
264
asked Nov 23 at 3:32
user583893
264
asked Nov 23 at 3:32
user583893
264
asked Nov 23 at 3:32
user583893
264
asked Nov 23 at 3:32
user583893
264
264
Your function isn't analytic at $1/r=0$, it has a branch point with something like $r^{-1/2}$ and $r^{1-3/2-1+1/(q-1)}$ terms
– reuns
Nov 23 at 4:05
add a comment |
Your function isn't analytic at $1/r=0$, it has a branch point with something like $r^{-1/2}$ and $r^{1-3/2-1+1/(q-1)}$ terms
– reuns
Nov 23 at 4:05
Your function isn't analytic at $1/r=0$, it has a branch point with something like $r^{-1/2}$ and $r^{1-3/2-1+1/(q-1)}$ terms
– reuns
Nov 23 at 4:05
Your function isn't analytic at $1/r=0$, it has a branch point with something like $r^{-1/2}$ and $r^{1-3/2-1+1/(q-1)}$ terms
– reuns
Nov 23 at 4:05
add a comment |
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We need an asymptotic for ${_2F_1}(a, b; c; z)$ with $z to 1^-$. Since $c - a - b = 1/(q - 1) < 0$, the leading term is
$$frac {Gamma(c) Gamma(a + b - c)} {Gamma(a) Gamma(b)} (1 - z)^{c - a - b}.$$
This gives $rho sim r sqrt{1 - (b/r)^{1 - q}}$, therefore $rho sim r$ and $r sim rho$.
answered Nov 23 at 8:10
Maxim
4,413219
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1 Answer
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We need an asymptotic for ${_2F_1}(a, b; c; z)$ with $z to 1^-$. Since $c - a - b = 1/(q - 1) < 0$, the leading term is
$$frac {Gamma(c) Gamma(a + b - c)} {Gamma(a) Gamma(b)} (1 - z)^{c - a - b}.$$
This gives $rho sim r sqrt{1 - (b/r)^{1 - q}}$, therefore $rho sim r$ and $r sim rho$.
answered Nov 23 at 8:10
Maxim
4,413219
add a comment |
up vote
0
down vote
We need an asymptotic for ${_2F_1}(a, b; c; z)$ with $z to 1^-$. Since $c - a - b = 1/(q - 1) < 0$, the leading term is
$$frac {Gamma(c) Gamma(a + b - c)} {Gamma(a) Gamma(b)} (1 - z)^{c - a - b}.$$
This gives $rho sim r sqrt{1 - (b/r)^{1 - q}}$, therefore $rho sim r$ and $r sim rho$.
answered Nov 23 at 8:10
Maxim
4,413219
add a comment |
up vote
0
down vote
up vote
0
down vote
We need an asymptotic for ${_2F_1}(a, b; c; z)$ with $z to 1^-$. Since $c - a - b = 1/(q - 1) < 0$, the leading term is
$$frac {Gamma(c) Gamma(a + b - c)} {Gamma(a) Gamma(b)} (1 - z)^{c - a - b}.$$
This gives $rho sim r sqrt{1 - (b/r)^{1 - q}}$, therefore $rho sim r$ and $r sim rho$.
answered Nov 23 at 8:10
Maxim
4,413219
We need an asymptotic for ${_2F_1}(a, b; c; z)$ with $z to 1^-$. Since $c - a - b = 1/(q - 1) < 0$, the leading term is
$$frac {Gamma(c) Gamma(a + b - c)} {Gamma(a) Gamma(b)} (1 - z)^{c - a - b}.$$
This gives $rho sim r sqrt{1 - (b/r)^{1 - q}}$, therefore $rho sim r$ and $r sim rho$.
answered Nov 23 at 8:10
Maxim
4,413219
answered Nov 23 at 8:10
Maxim
4,413219
answered Nov 23 at 8:10
Maxim
4,413219
answered Nov 23 at 8:10
Maxim
4,413219
4,413219
add a comment |
add a comment |
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Your function isn't analytic at $1/r=0$, it has a branch point with something like $r^{-1/2}$ and $r^{1-3/2-1+1/(q-1)}$ terms
– reuns
Nov 23 at 4:05