Inverse of the asymptotic expansion of Gauss Hypergeometric function
$begingroup$
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 $-infty<q<1$.
$$rho=frac{2b}{1-q}left(1-left(frac brright)^{1-q}right)^{1/2}left(_2F_1left(frac{1}{2},1-frac{1}{q-1},frac{3}{2},1-left(frac brright)^{1-q}right)right)$$
asymptotics power-series inverse-function hypergeometric-function
edited Jan 18 at 6:43
mrtaurho
6,19771641
asked Nov 23 '18 at 3:32
user583893user583893
336
$endgroup$
add a comment |
$begingroup$
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 $-infty<q<1$.
$$rho=frac{2b}{1-q}left(1-left(frac brright)^{1-q}right)^{1/2}left(_2F_1left(frac{1}{2},1-frac{1}{q-1},frac{3}{2},1-left(frac brright)^{1-q}right)right)$$
asymptotics power-series inverse-function hypergeometric-function
edited Jan 18 at 6:43
mrtaurho
6,19771641
asked Nov 23 '18 at 3:32
user583893user583893
336
$endgroup$
$begingroup$
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
$endgroup$
– reuns
Nov 23 '18 at 4:05
$begingroup$
What then can I do with the function? Thanks
$endgroup$
– user583893
Jan 26 at 5:25
add a comment |
$begingroup$
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 $-infty<q<1$.
$$rho=frac{2b}{1-q}left(1-left(frac brright)^{1-q}right)^{1/2}left(_2F_1left(frac{1}{2},1-frac{1}{q-1},frac{3}{2},1-left(frac brright)^{1-q}right)right)$$
asymptotics power-series inverse-function hypergeometric-function
edited Jan 18 at 6:43
mrtaurho
6,19771641
asked Nov 23 '18 at 3:32
user583893user583893
336
$endgroup$
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 $-infty<q<1$.
$$rho=frac{2b}{1-q}left(1-left(frac brright)^{1-q}right)^{1/2}left(_2F_1left(frac{1}{2},1-frac{1}{q-1},frac{3}{2},1-left(frac brright)^{1-q}right)right)$$
asymptotics power-series inverse-function hypergeometric-function
asymptotics power-series inverse-function hypergeometric-function
edited Jan 18 at 6:43
mrtaurho
6,19771641
asked Nov 23 '18 at 3:32
user583893user583893
336
edited Jan 18 at 6:43
mrtaurho
6,19771641
asked Nov 23 '18 at 3:32
user583893user583893
336
edited Jan 18 at 6:43
mrtaurho
6,19771641
edited Jan 18 at 6:43
mrtaurho
6,19771641
edited Jan 18 at 6:43
mrtaurho
6,19771641
6,19771641
asked Nov 23 '18 at 3:32
user583893user583893
336
asked Nov 23 '18 at 3:32
user583893user583893
336
asked Nov 23 '18 at 3:32
user583893user583893
336
336
$begingroup$
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
$endgroup$
– reuns
Nov 23 '18 at 4:05
$begingroup$
What then can I do with the function? Thanks
$endgroup$
– user583893
Jan 26 at 5:25
add a comment |
$begingroup$
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
$endgroup$
– reuns
Nov 23 '18 at 4:05
$begingroup$
What then can I do with the function? Thanks
$endgroup$
– user583893
Jan 26 at 5:25
$begingroup$
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
$endgroup$
– reuns
Nov 23 '18 at 4:05
$begingroup$
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
$endgroup$
– reuns
Nov 23 '18 at 4:05
$begingroup$
What then can I do with the function? Thanks
$endgroup$
– user583893
Jan 26 at 5:25
$begingroup$
What then can I do with the function? Thanks
$endgroup$
– user583893
Jan 26 at 5:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We need an asymptotic for ${_2hspace{-1px}F_1}(a, b; c; z)$ 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 $r sim rho$.
For $q < 0$, the next term in the expansion of ${_2hspace{-1px}F_1}$ is a constant, and we need only the first term from the $(1 - z)^{1/2}$ factor:
$${_2hspace{-1px}F_1}
{left( frac 1 2, p; frac 3 2; 1 - z right)} =
frac {z^{1 - p}} {2 (p - 1)} +
frac {sqrt pi ,Gamma(1 - p)}
{2 Gamma {left( frac 3 2 - p right)}} + dots, \
rho = r + frac {b sqrt pi ,Gamma(1 - p)}
{(1 - q) Gamma {left( frac 3 2 - p right)}} + dots, \
r = rho - frac {b sqrt pi ,Gamma(1 - p)}
{(1 - q) Gamma {left( frac 3 2 - p right)}} + dots,$$
where $p = (2 - q)/(1 - q)$. For $q > 0$ and excluding the logarithmic case $p in mathbb N$, the next term in the expansion of ${_2hspace{-1px}F_1}$ is of order $z^{2 - p}$, and we need two terms from $(1 - z)^{1/2}$:
$${_2hspace{-1px}F_1}
{left( frac 1 2, p; frac 3 2; 1 - z right)} =
frac {z^{1 - p}} {2 (p - 1)} +
frac {(2 p - 3) z^{2 - p}} {4 (p - 1) (p - 2)} + dots, \
rho = r + frac {b^{1 - q} r^q} {2 q} + dots, \
r = rho - frac {b^{1 - q} rho^q} {2 q} + dots,$$
the last step coming from the Lagrange reversion theorem.
answered Nov 23 '18 at 8:10
MaximMaxim
6,3231221
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$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 19 at 6:00
$begingroup$
@Maxim, why did you split the analysis into $q<0$ and $q>0$? In my numerical test of the magnitude of terms in the expansion, it seems that the constant in the expansion is practically zero (or has negative value) for all $q<0$ . And I guess all the correction terms come from the terms in parenthesis with the leading order.
$endgroup$
– user583893
Jan 20 at 10:31
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We need an asymptotic for ${_2hspace{-1px}F_1}(a, b; c; z)$ 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 $r sim rho$.
For $q < 0$, the next term in the expansion of ${_2hspace{-1px}F_1}$ is a constant, and we need only the first term from the $(1 - z)^{1/2}$ factor:
$${_2hspace{-1px}F_1}
{left( frac 1 2, p; frac 3 2; 1 - z right)} =
frac {z^{1 - p}} {2 (p - 1)} +
frac {sqrt pi ,Gamma(1 - p)}
{2 Gamma {left( frac 3 2 - p right)}} + dots, \
rho = r + frac {b sqrt pi ,Gamma(1 - p)}
{(1 - q) Gamma {left( frac 3 2 - p right)}} + dots, \
r = rho - frac {b sqrt pi ,Gamma(1 - p)}
{(1 - q) Gamma {left( frac 3 2 - p right)}} + dots,$$
where $p = (2 - q)/(1 - q)$. For $q > 0$ and excluding the logarithmic case $p in mathbb N$, the next term in the expansion of ${_2hspace{-1px}F_1}$ is of order $z^{2 - p}$, and we need two terms from $(1 - z)^{1/2}$:
$${_2hspace{-1px}F_1}
{left( frac 1 2, p; frac 3 2; 1 - z right)} =
frac {z^{1 - p}} {2 (p - 1)} +
frac {(2 p - 3) z^{2 - p}} {4 (p - 1) (p - 2)} + dots, \
rho = r + frac {b^{1 - q} r^q} {2 q} + dots, \
r = rho - frac {b^{1 - q} rho^q} {2 q} + dots,$$
the last step coming from the Lagrange reversion theorem.
answered Nov 23 '18 at 8:10
MaximMaxim
6,3231221
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 19 at 6:00
$begingroup$
@Maxim, why did you split the analysis into $q<0$ and $q>0$? In my numerical test of the magnitude of terms in the expansion, it seems that the constant in the expansion is practically zero (or has negative value) for all $q<0$ . And I guess all the correction terms come from the terms in parenthesis with the leading order.
$endgroup$
– user583893
Jan 20 at 10:31
add a comment |
$begingroup$
We need an asymptotic for ${_2hspace{-1px}F_1}(a, b; c; z)$ 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 $r sim rho$.
For $q < 0$, the next term in the expansion of ${_2hspace{-1px}F_1}$ is a constant, and we need only the first term from the $(1 - z)^{1/2}$ factor:
$${_2hspace{-1px}F_1}
{left( frac 1 2, p; frac 3 2; 1 - z right)} =
frac {z^{1 - p}} {2 (p - 1)} +
frac {sqrt pi ,Gamma(1 - p)}
{2 Gamma {left( frac 3 2 - p right)}} + dots, \
rho = r + frac {b sqrt pi ,Gamma(1 - p)}
{(1 - q) Gamma {left( frac 3 2 - p right)}} + dots, \
r = rho - frac {b sqrt pi ,Gamma(1 - p)}
{(1 - q) Gamma {left( frac 3 2 - p right)}} + dots,$$
where $p = (2 - q)/(1 - q)$. For $q > 0$ and excluding the logarithmic case $p in mathbb N$, the next term in the expansion of ${_2hspace{-1px}F_1}$ is of order $z^{2 - p}$, and we need two terms from $(1 - z)^{1/2}$:
$${_2hspace{-1px}F_1}
{left( frac 1 2, p; frac 3 2; 1 - z right)} =
frac {z^{1 - p}} {2 (p - 1)} +
frac {(2 p - 3) z^{2 - p}} {4 (p - 1) (p - 2)} + dots, \
rho = r + frac {b^{1 - q} r^q} {2 q} + dots, \
r = rho - frac {b^{1 - q} rho^q} {2 q} + dots,$$
the last step coming from the Lagrange reversion theorem.
answered Nov 23 '18 at 8:10
MaximMaxim
6,3231221
$endgroup$
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 19 at 6:00
$begingroup$
@Maxim, why did you split the analysis into $q<0$ and $q>0$? In my numerical test of the magnitude of terms in the expansion, it seems that the constant in the expansion is practically zero (or has negative value) for all $q<0$ . And I guess all the correction terms come from the terms in parenthesis with the leading order.
$endgroup$
– user583893
Jan 20 at 10:31
add a comment |
$begingroup$
We need an asymptotic for ${_2hspace{-1px}F_1}(a, b; c; z)$ 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 $r sim rho$.
For $q < 0$, the next term in the expansion of ${_2hspace{-1px}F_1}$ is a constant, and we need only the first term from the $(1 - z)^{1/2}$ factor:
$${_2hspace{-1px}F_1}
{left( frac 1 2, p; frac 3 2; 1 - z right)} =
frac {z^{1 - p}} {2 (p - 1)} +
frac {sqrt pi ,Gamma(1 - p)}
{2 Gamma {left( frac 3 2 - p right)}} + dots, \
rho = r + frac {b sqrt pi ,Gamma(1 - p)}
{(1 - q) Gamma {left( frac 3 2 - p right)}} + dots, \
r = rho - frac {b sqrt pi ,Gamma(1 - p)}
{(1 - q) Gamma {left( frac 3 2 - p right)}} + dots,$$
where $p = (2 - q)/(1 - q)$. For $q > 0$ and excluding the logarithmic case $p in mathbb N$, the next term in the expansion of ${_2hspace{-1px}F_1}$ is of order $z^{2 - p}$, and we need two terms from $(1 - z)^{1/2}$:
$${_2hspace{-1px}F_1}
{left( frac 1 2, p; frac 3 2; 1 - z right)} =
frac {z^{1 - p}} {2 (p - 1)} +
frac {(2 p - 3) z^{2 - p}} {4 (p - 1) (p - 2)} + dots, \
rho = r + frac {b^{1 - q} r^q} {2 q} + dots, \
r = rho - frac {b^{1 - q} rho^q} {2 q} + dots,$$
the last step coming from the Lagrange reversion theorem.
answered Nov 23 '18 at 8:10
MaximMaxim
6,3231221
$endgroup$
We need an asymptotic for ${_2hspace{-1px}F_1}(a, b; c; z)$ 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 $r sim rho$.
For $q < 0$, the next term in the expansion of ${_2hspace{-1px}F_1}$ is a constant, and we need only the first term from the $(1 - z)^{1/2}$ factor:
$${_2hspace{-1px}F_1}
{left( frac 1 2, p; frac 3 2; 1 - z right)} =
frac {z^{1 - p}} {2 (p - 1)} +
frac {sqrt pi ,Gamma(1 - p)}
{2 Gamma {left( frac 3 2 - p right)}} + dots, \
rho = r + frac {b sqrt pi ,Gamma(1 - p)}
{(1 - q) Gamma {left( frac 3 2 - p right)}} + dots, \
r = rho - frac {b sqrt pi ,Gamma(1 - p)}
{(1 - q) Gamma {left( frac 3 2 - p right)}} + dots,$$
where $p = (2 - q)/(1 - q)$. For $q > 0$ and excluding the logarithmic case $p in mathbb N$, the next term in the expansion of ${_2hspace{-1px}F_1}$ is of order $z^{2 - p}$, and we need two terms from $(1 - z)^{1/2}$:
$${_2hspace{-1px}F_1}
{left( frac 1 2, p; frac 3 2; 1 - z right)} =
frac {z^{1 - p}} {2 (p - 1)} +
frac {(2 p - 3) z^{2 - p}} {4 (p - 1) (p - 2)} + dots, \
rho = r + frac {b^{1 - q} r^q} {2 q} + dots, \
r = rho - frac {b^{1 - q} rho^q} {2 q} + dots,$$
the last step coming from the Lagrange reversion theorem.
answered Nov 23 '18 at 8:10
MaximMaxim
6,3231221
edited Jan 16 at 17:01
answered Nov 23 '18 at 8:10
MaximMaxim
6,3231221
answered Nov 23 '18 at 8:10
MaximMaxim
6,3231221
answered Nov 23 '18 at 8:10
MaximMaxim
6,3231221
6,3231221
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 19 at 6:00
$begingroup$
@Maxim, why did you split the analysis into $q<0$ and $q>0$? In my numerical test of the magnitude of terms in the expansion, it seems that the constant in the expansion is practically zero (or has negative value) for all $q<0$ . And I guess all the correction terms come from the terms in parenthesis with the leading order.
$endgroup$
– user583893
Jan 20 at 10:31
add a comment |
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 19 at 6:00
$begingroup$
@Maxim, why did you split the analysis into $q<0$ and $q>0$? In my numerical test of the magnitude of terms in the expansion, it seems that the constant in the expansion is practically zero (or has negative value) for all $q<0$ . And I guess all the correction terms come from the terms in parenthesis with the leading order.
$endgroup$
– user583893
Jan 20 at 10:31
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 19 at 6:00
$begingroup$
Comments are not for extended discussion; this conversation has been moved to chat.
$endgroup$
– Aloizio Macedo♦
Jan 19 at 6:00
$begingroup$
@Maxim, why did you split the analysis into $q<0$ and $q>0$? In my numerical test of the magnitude of terms in the expansion, it seems that the constant in the expansion is practically zero (or has negative value) for all $q<0$ . And I guess all the correction terms come from the terms in parenthesis with the leading order.
$endgroup$
– user583893
Jan 20 at 10:31
$begingroup$
@Maxim, why did you split the analysis into $q<0$ and $q>0$? In my numerical test of the magnitude of terms in the expansion, it seems that the constant in the expansion is practically zero (or has negative value) for all $q<0$ . And I guess all the correction terms come from the terms in parenthesis with the leading order.
$endgroup$
– user583893
Jan 20 at 10:31
add a comment |
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$begingroup$
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
$endgroup$
– reuns
Nov 23 '18 at 4:05
$begingroup$
What then can I do with the function? Thanks
$endgroup$
– user583893
Jan 26 at 5:25