Mixing
Asymptotics of Hypergeometric $_2F_1(a;b;c;z)$ for large $|z| to infty$?
$begingroup$
I found this list of asymptotics of the Gauss Hypergeometric function $_2F_1(a;b;c;z)$ here on Wolfram's site for large $|z| to infty$
In particular there is a general formula for $|z| to infty$
$$
_2F_1(a;b;c;z) approx frac{Gamma(b-a)Gamma(c)}{Gamma(b)Gamma(c-a)} (-z)^{-a} +frac{Gamma(a-b)Gamma(c)}{Gamma(a)Gamma(c-b)} (-z)^{-b}
$$
How is this derived? Also, is this always true (meaning, for all $a$, $b$, $c$)? There are no sources on the site I linked.
Is there also a way to determine the next-order terms?
complex-analysis asymptotics power-series hypergeometric-function
asked Jan 29 at 13:48
Greg.PaulGreg.Paul
787921
$endgroup$
add a comment |
$begingroup$
I found this list of asymptotics of the Gauss Hypergeometric function $_2F_1(a;b;c;z)$ here on Wolfram's site for large $|z| to infty$
In particular there is a general formula for $|z| to infty$
$$
_2F_1(a;b;c;z) approx frac{Gamma(b-a)Gamma(c)}{Gamma(b)Gamma(c-a)} (-z)^{-a} +frac{Gamma(a-b)Gamma(c)}{Gamma(a)Gamma(c-b)} (-z)^{-b}
$$
How is this derived? Also, is this always true (meaning, for all $a$, $b$, $c$)? There are no sources on the site I linked.
Is there also a way to determine the next-order terms?
complex-analysis asymptotics power-series hypergeometric-function
asked Jan 29 at 13:48
Greg.PaulGreg.Paul
787921
$endgroup$
add a comment |
$begingroup$
I found this list of asymptotics of the Gauss Hypergeometric function $_2F_1(a;b;c;z)$ here on Wolfram's site for large $|z| to infty$
In particular there is a general formula for $|z| to infty$
$$
_2F_1(a;b;c;z) approx frac{Gamma(b-a)Gamma(c)}{Gamma(b)Gamma(c-a)} (-z)^{-a} +frac{Gamma(a-b)Gamma(c)}{Gamma(a)Gamma(c-b)} (-z)^{-b}
$$
How is this derived? Also, is this always true (meaning, for all $a$, $b$, $c$)? There are no sources on the site I linked.
Is there also a way to determine the next-order terms?
complex-analysis asymptotics power-series hypergeometric-function
asked Jan 29 at 13:48
Greg.PaulGreg.Paul
787921
$endgroup$
I found this list of asymptotics of the Gauss Hypergeometric function $_2F_1(a;b;c;z)$ here on Wolfram's site for large $|z| to infty$
In particular there is a general formula for $|z| to infty$
$$
_2F_1(a;b;c;z) approx frac{Gamma(b-a)Gamma(c)}{Gamma(b)Gamma(c-a)} (-z)^{-a} +frac{Gamma(a-b)Gamma(c)}{Gamma(a)Gamma(c-b)} (-z)^{-b}
$$
How is this derived? Also, is this always true (meaning, for all $a$, $b$, $c$)? There are no sources on the site I linked.
Is there also a way to determine the next-order terms?
complex-analysis asymptotics power-series hypergeometric-function
complex-analysis asymptotics power-series hypergeometric-function
asked Jan 29 at 13:48
Greg.PaulGreg.Paul
787921
asked Jan 29 at 13:48
Greg.PaulGreg.Paul
787921
asked Jan 29 at 13:48
Greg.PaulGreg.Paul
787921
asked Jan 29 at 13:48
Greg.PaulGreg.Paul
787921
asked Jan 29 at 13:48
Greg.PaulGreg.Paul
787921
787921
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It follows from the "reciprocation" formula
$$
eqalign{
& {}_2F_1 (a,b;c;z)quad left| matrix{
;a - b notin Z hfill cr
;z notin left( {0,1} right) hfill cr} right.quad = cr
& {{Gamma (b - a)Gamma (c)} over {Gamma (b)Gamma (c - a)}}{1 over {left( { - z} right)^{,a} }}
{}_2F_1 left( {a,,a - c + 1;;a - b + 1;{1 over z}} right) + cr
& + {{Gamma (a - b)Gamma (c)} over {Gamma (a)Gamma (c - b)}}{1 over {left( { - z} right)^{,b} }}
{}_2F_1 left( {b,,b - c + 1;;b - a + 1;{1 over z}} right) cr}
$$
(re., e.g., to this link )
That, in turn, is derived from the solutions of the hypergeometric differential equation.
answered Jan 29 at 20:16
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
What do you do in the case that $a-b in mathbb{Z}$?
$endgroup$
– Greg.Paul
Jan 29 at 20:28
$begingroup$
@Greg.Paul: refer to this section of wikipedia, where it is said "Around $z=infty$, if $a-b$ is not an integer, one has two independent solutions (which are those combined above). Again, when the conditions of non-integrality are not met, there exist other solutions that are more complicated." However I am not a theorist, sorry.
$endgroup$
– G Cab
Jan 29 at 22:43
$begingroup$
@Greg.Paul: if you need a formula for the case $a-b$ integral, you may have a look at this book-pag.63
$endgroup$
– G Cab
Jan 29 at 23:22
add a comment |
$begingroup$
Converting ${_2hspace{-1px}F_1}$ to the Meijer G-function, we obtain
$${_2hspace{-1px}F_1}(a, b; c; z) =
frac {Gamma(c)} {Gamma(a) Gamma(b)}
G_{2, 2}^{1, 2} left( -z middle| {1 - a, 1 - b atop 0, 1 - c} right) = \
frac {Gamma(c)} {2 pi i Gamma(a) Gamma(b)} int_{mathcal L} frac
{Gamma(y + a) Gamma(y + b) Gamma(-y)}
{Gamma(y + c)} (-z)^y dy.$$
A left loop is a valid contour for $|z| > 1$, and the sum of the residues at $y = -a - k$ and $y = -b - k$ gives a complete asymptotic expansion for large $|z|$. Excluding the logarithmic cases,
$$operatorname*{Res}_{y = -a - k} frac
{Gamma(y + a) Gamma(y + b) Gamma(-y)}
{Gamma(y + c)} (-z)^y =
frac {(-1)^k Gamma(b - a - k) Gamma(a + k)} {Gamma(c - a - k)}
frac {(-z)^{-a - k}} {k!}.$$
answered Jan 29 at 18:32
MaximMaxim
6,1181221
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It follows from the "reciprocation" formula
$$
eqalign{
& {}_2F_1 (a,b;c;z)quad left| matrix{
;a - b notin Z hfill cr
;z notin left( {0,1} right) hfill cr} right.quad = cr
& {{Gamma (b - a)Gamma (c)} over {Gamma (b)Gamma (c - a)}}{1 over {left( { - z} right)^{,a} }}
{}_2F_1 left( {a,,a - c + 1;;a - b + 1;{1 over z}} right) + cr
& + {{Gamma (a - b)Gamma (c)} over {Gamma (a)Gamma (c - b)}}{1 over {left( { - z} right)^{,b} }}
{}_2F_1 left( {b,,b - c + 1;;b - a + 1;{1 over z}} right) cr}
$$
(re., e.g., to this link )
That, in turn, is derived from the solutions of the hypergeometric differential equation.
answered Jan 29 at 20:16
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
What do you do in the case that $a-b in mathbb{Z}$?
$endgroup$
– Greg.Paul
Jan 29 at 20:28
$begingroup$
@Greg.Paul: refer to this section of wikipedia, where it is said "Around $z=infty$, if $a-b$ is not an integer, one has two independent solutions (which are those combined above). Again, when the conditions of non-integrality are not met, there exist other solutions that are more complicated." However I am not a theorist, sorry.
$endgroup$
– G Cab
Jan 29 at 22:43
$begingroup$
@Greg.Paul: if you need a formula for the case $a-b$ integral, you may have a look at this book-pag.63
$endgroup$
– G Cab
Jan 29 at 23:22
add a comment |
$begingroup$
It follows from the "reciprocation" formula
$$
eqalign{
& {}_2F_1 (a,b;c;z)quad left| matrix{
;a - b notin Z hfill cr
;z notin left( {0,1} right) hfill cr} right.quad = cr
& {{Gamma (b - a)Gamma (c)} over {Gamma (b)Gamma (c - a)}}{1 over {left( { - z} right)^{,a} }}
{}_2F_1 left( {a,,a - c + 1;;a - b + 1;{1 over z}} right) + cr
& + {{Gamma (a - b)Gamma (c)} over {Gamma (a)Gamma (c - b)}}{1 over {left( { - z} right)^{,b} }}
{}_2F_1 left( {b,,b - c + 1;;b - a + 1;{1 over z}} right) cr}
$$
(re., e.g., to this link )
That, in turn, is derived from the solutions of the hypergeometric differential equation.
answered Jan 29 at 20:16
G CabG Cab
20.4k31341
$endgroup$
$begingroup$
What do you do in the case that $a-b in mathbb{Z}$?
$endgroup$
– Greg.Paul
Jan 29 at 20:28
$begingroup$
@Greg.Paul: refer to this section of wikipedia, where it is said "Around $z=infty$, if $a-b$ is not an integer, one has two independent solutions (which are those combined above). Again, when the conditions of non-integrality are not met, there exist other solutions that are more complicated." However I am not a theorist, sorry.
$endgroup$
– G Cab
Jan 29 at 22:43
$begingroup$
@Greg.Paul: if you need a formula for the case $a-b$ integral, you may have a look at this book-pag.63
$endgroup$
– G Cab
Jan 29 at 23:22
add a comment |
$begingroup$
It follows from the "reciprocation" formula
$$
eqalign{
& {}_2F_1 (a,b;c;z)quad left| matrix{
;a - b notin Z hfill cr
;z notin left( {0,1} right) hfill cr} right.quad = cr
& {{Gamma (b - a)Gamma (c)} over {Gamma (b)Gamma (c - a)}}{1 over {left( { - z} right)^{,a} }}
{}_2F_1 left( {a,,a - c + 1;;a - b + 1;{1 over z}} right) + cr
& + {{Gamma (a - b)Gamma (c)} over {Gamma (a)Gamma (c - b)}}{1 over {left( { - z} right)^{,b} }}
{}_2F_1 left( {b,,b - c + 1;;b - a + 1;{1 over z}} right) cr}
$$
(re., e.g., to this link )
That, in turn, is derived from the solutions of the hypergeometric differential equation.
answered Jan 29 at 20:16
G CabG Cab
20.4k31341
$endgroup$
It follows from the "reciprocation" formula
$$
eqalign{
& {}_2F_1 (a,b;c;z)quad left| matrix{
;a - b notin Z hfill cr
;z notin left( {0,1} right) hfill cr} right.quad = cr
& {{Gamma (b - a)Gamma (c)} over {Gamma (b)Gamma (c - a)}}{1 over {left( { - z} right)^{,a} }}
{}_2F_1 left( {a,,a - c + 1;;a - b + 1;{1 over z}} right) + cr
& + {{Gamma (a - b)Gamma (c)} over {Gamma (a)Gamma (c - b)}}{1 over {left( { - z} right)^{,b} }}
{}_2F_1 left( {b,,b - c + 1;;b - a + 1;{1 over z}} right) cr}
$$
(re., e.g., to this link )
That, in turn, is derived from the solutions of the hypergeometric differential equation.
answered Jan 29 at 20:16
G CabG Cab
20.4k31341
answered Jan 29 at 20:16
G CabG Cab
20.4k31341
answered Jan 29 at 20:16
G CabG Cab
20.4k31341
answered Jan 29 at 20:16
G CabG Cab
20.4k31341
20.4k31341
$begingroup$
What do you do in the case that $a-b in mathbb{Z}$?
$endgroup$
– Greg.Paul
Jan 29 at 20:28
$begingroup$
@Greg.Paul: refer to this section of wikipedia, where it is said "Around $z=infty$, if $a-b$ is not an integer, one has two independent solutions (which are those combined above). Again, when the conditions of non-integrality are not met, there exist other solutions that are more complicated." However I am not a theorist, sorry.
$endgroup$
– G Cab
Jan 29 at 22:43
$begingroup$
@Greg.Paul: if you need a formula for the case $a-b$ integral, you may have a look at this book-pag.63
$endgroup$
– G Cab
Jan 29 at 23:22
add a comment |
$begingroup$
What do you do in the case that $a-b in mathbb{Z}$?
$endgroup$
– Greg.Paul
Jan 29 at 20:28
$begingroup$
@Greg.Paul: refer to this section of wikipedia, where it is said "Around $z=infty$, if $a-b$ is not an integer, one has two independent solutions (which are those combined above). Again, when the conditions of non-integrality are not met, there exist other solutions that are more complicated." However I am not a theorist, sorry.
$endgroup$
– G Cab
Jan 29 at 22:43
$begingroup$
@Greg.Paul: if you need a formula for the case $a-b$ integral, you may have a look at this book-pag.63
$endgroup$
– G Cab
Jan 29 at 23:22
$begingroup$
What do you do in the case that $a-b in mathbb{Z}$?
$endgroup$
– Greg.Paul
Jan 29 at 20:28
$begingroup$
What do you do in the case that $a-b in mathbb{Z}$?
$endgroup$
– Greg.Paul
Jan 29 at 20:28
$begingroup$
@Greg.Paul: refer to this section of wikipedia, where it is said "Around $z=infty$, if $a-b$ is not an integer, one has two independent solutions (which are those combined above). Again, when the conditions of non-integrality are not met, there exist other solutions that are more complicated." However I am not a theorist, sorry.
$endgroup$
– G Cab
Jan 29 at 22:43
$begingroup$
@Greg.Paul: refer to this section of wikipedia, where it is said "Around $z=infty$, if $a-b$ is not an integer, one has two independent solutions (which are those combined above). Again, when the conditions of non-integrality are not met, there exist other solutions that are more complicated." However I am not a theorist, sorry.
$endgroup$
– G Cab
Jan 29 at 22:43
$begingroup$
@Greg.Paul: if you need a formula for the case $a-b$ integral, you may have a look at this book-pag.63
$endgroup$
– G Cab
Jan 29 at 23:22
$begingroup$
@Greg.Paul: if you need a formula for the case $a-b$ integral, you may have a look at this book-pag.63
$endgroup$
– G Cab
Jan 29 at 23:22
add a comment |
$begingroup$
Converting ${_2hspace{-1px}F_1}$ to the Meijer G-function, we obtain
$${_2hspace{-1px}F_1}(a, b; c; z) =
frac {Gamma(c)} {Gamma(a) Gamma(b)}
G_{2, 2}^{1, 2} left( -z middle| {1 - a, 1 - b atop 0, 1 - c} right) = \
frac {Gamma(c)} {2 pi i Gamma(a) Gamma(b)} int_{mathcal L} frac
{Gamma(y + a) Gamma(y + b) Gamma(-y)}
{Gamma(y + c)} (-z)^y dy.$$
A left loop is a valid contour for $|z| > 1$, and the sum of the residues at $y = -a - k$ and $y = -b - k$ gives a complete asymptotic expansion for large $|z|$. Excluding the logarithmic cases,
$$operatorname*{Res}_{y = -a - k} frac
{Gamma(y + a) Gamma(y + b) Gamma(-y)}
{Gamma(y + c)} (-z)^y =
frac {(-1)^k Gamma(b - a - k) Gamma(a + k)} {Gamma(c - a - k)}
frac {(-z)^{-a - k}} {k!}.$$
answered Jan 29 at 18:32
MaximMaxim
6,1181221
$endgroup$
add a comment |
$begingroup$
Converting ${_2hspace{-1px}F_1}$ to the Meijer G-function, we obtain
$${_2hspace{-1px}F_1}(a, b; c; z) =
frac {Gamma(c)} {Gamma(a) Gamma(b)}
G_{2, 2}^{1, 2} left( -z middle| {1 - a, 1 - b atop 0, 1 - c} right) = \
frac {Gamma(c)} {2 pi i Gamma(a) Gamma(b)} int_{mathcal L} frac
{Gamma(y + a) Gamma(y + b) Gamma(-y)}
{Gamma(y + c)} (-z)^y dy.$$
A left loop is a valid contour for $|z| > 1$, and the sum of the residues at $y = -a - k$ and $y = -b - k$ gives a complete asymptotic expansion for large $|z|$. Excluding the logarithmic cases,
$$operatorname*{Res}_{y = -a - k} frac
{Gamma(y + a) Gamma(y + b) Gamma(-y)}
{Gamma(y + c)} (-z)^y =
frac {(-1)^k Gamma(b - a - k) Gamma(a + k)} {Gamma(c - a - k)}
frac {(-z)^{-a - k}} {k!}.$$
answered Jan 29 at 18:32
MaximMaxim
6,1181221
$endgroup$
add a comment |
$begingroup$
Converting ${_2hspace{-1px}F_1}$ to the Meijer G-function, we obtain
$${_2hspace{-1px}F_1}(a, b; c; z) =
frac {Gamma(c)} {Gamma(a) Gamma(b)}
G_{2, 2}^{1, 2} left( -z middle| {1 - a, 1 - b atop 0, 1 - c} right) = \
frac {Gamma(c)} {2 pi i Gamma(a) Gamma(b)} int_{mathcal L} frac
{Gamma(y + a) Gamma(y + b) Gamma(-y)}
{Gamma(y + c)} (-z)^y dy.$$
A left loop is a valid contour for $|z| > 1$, and the sum of the residues at $y = -a - k$ and $y = -b - k$ gives a complete asymptotic expansion for large $|z|$. Excluding the logarithmic cases,
$$operatorname*{Res}_{y = -a - k} frac
{Gamma(y + a) Gamma(y + b) Gamma(-y)}
{Gamma(y + c)} (-z)^y =
frac {(-1)^k Gamma(b - a - k) Gamma(a + k)} {Gamma(c - a - k)}
frac {(-z)^{-a - k}} {k!}.$$
answered Jan 29 at 18:32
MaximMaxim
6,1181221
$endgroup$
Converting ${_2hspace{-1px}F_1}$ to the Meijer G-function, we obtain
$${_2hspace{-1px}F_1}(a, b; c; z) =
frac {Gamma(c)} {Gamma(a) Gamma(b)}
G_{2, 2}^{1, 2} left( -z middle| {1 - a, 1 - b atop 0, 1 - c} right) = \
frac {Gamma(c)} {2 pi i Gamma(a) Gamma(b)} int_{mathcal L} frac
{Gamma(y + a) Gamma(y + b) Gamma(-y)}
{Gamma(y + c)} (-z)^y dy.$$
A left loop is a valid contour for $|z| > 1$, and the sum of the residues at $y = -a - k$ and $y = -b - k$ gives a complete asymptotic expansion for large $|z|$. Excluding the logarithmic cases,
$$operatorname*{Res}_{y = -a - k} frac
{Gamma(y + a) Gamma(y + b) Gamma(-y)}
{Gamma(y + c)} (-z)^y =
frac {(-1)^k Gamma(b - a - k) Gamma(a + k)} {Gamma(c - a - k)}
frac {(-z)^{-a - k}} {k!}.$$
answered Jan 29 at 18:32
MaximMaxim
6,1181221
answered Jan 29 at 18:32
MaximMaxim
6,1181221
answered Jan 29 at 18:32
MaximMaxim
6,1181221
answered Jan 29 at 18:32
MaximMaxim
6,1181221
6,1181221
add a comment |
add a comment |
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