Is the hypergeometric function a simple resurgent function
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In the study of Airy functions you obtain, by Borel transforming components of a transseries, a hypergeometric function $_2F_1(a,b,c|z)$ where the constant are positive. Now I would like to know if this function is simple resurgent, by this I mean that it can be written in the form
$$
_2F_1(a,b,c|z)Big|_{z=A} sim frac{C}{2pi i(z-A)} + psi(z-A) frac{log{(z-A)}}{2pi i} + text{holomorphic}
$$
where $psi(z)$ is a meromorphic function and $Cin mathbb{C}$. My first guess would be start from the integral representation of the hypergeometric function
$$
_2F_1(a,b;c;z)=frac{Gamma(c)}{Gamma(b)Gamma(c-b)}int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}dt,
$$
however I do not know how I can start tackling this problem. From the integral representation it is clear that things go wrong when $zgeq 1$, and in this question 2656851 they determine the discontinuity. This argues in favour of being a simple resurgent formula since it has the required branch cut. Therefore I hope someone can help me rewriting the hypergeometric function in the above form of a simple resurgent function.
Thank you very much in advance.
complex-analysis hypergeometric-function branch-cuts
asked Nov 16 at 15:41
Dylan_VM
373112
add a comment |
up vote
0
down vote
favorite
In the study of Airy functions you obtain, by Borel transforming components of a transseries, a hypergeometric function $_2F_1(a,b,c|z)$ where the constant are positive. Now I would like to know if this function is simple resurgent, by this I mean that it can be written in the form
$$
_2F_1(a,b,c|z)Big|_{z=A} sim frac{C}{2pi i(z-A)} + psi(z-A) frac{log{(z-A)}}{2pi i} + text{holomorphic}
$$
where $psi(z)$ is a meromorphic function and $Cin mathbb{C}$. My first guess would be start from the integral representation of the hypergeometric function
$$
_2F_1(a,b;c;z)=frac{Gamma(c)}{Gamma(b)Gamma(c-b)}int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}dt,
$$
however I do not know how I can start tackling this problem. From the integral representation it is clear that things go wrong when $zgeq 1$, and in this question 2656851 they determine the discontinuity. This argues in favour of being a simple resurgent formula since it has the required branch cut. Therefore I hope someone can help me rewriting the hypergeometric function in the above form of a simple resurgent function.
Thank you very much in advance.
complex-analysis hypergeometric-function branch-cuts
asked Nov 16 at 15:41
Dylan_VM
373112
With $u_l,v_l$ the Taylor coefficients, for $M,N$ large enough and $z not in (1,infty)$ $int_0^1 (t^{b-1}-sum_{l=0}^M u_l(b) (t-1)^l) ((1-t)^{c-b-1}(1-zt)^{-a}-sum_{l=0}^N v_l(c-b-1,z,a) t^l)dt$ converges and is locally analytic in each variable. For $z in (1,infty)$ you need to subtract more $(t-1/z)^l$ terms
– reuns
Nov 16 at 19:51
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In the study of Airy functions you obtain, by Borel transforming components of a transseries, a hypergeometric function $_2F_1(a,b,c|z)$ where the constant are positive. Now I would like to know if this function is simple resurgent, by this I mean that it can be written in the form
$$
_2F_1(a,b,c|z)Big|_{z=A} sim frac{C}{2pi i(z-A)} + psi(z-A) frac{log{(z-A)}}{2pi i} + text{holomorphic}
$$
where $psi(z)$ is a meromorphic function and $Cin mathbb{C}$. My first guess would be start from the integral representation of the hypergeometric function
$$
_2F_1(a,b;c;z)=frac{Gamma(c)}{Gamma(b)Gamma(c-b)}int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}dt,
$$
however I do not know how I can start tackling this problem. From the integral representation it is clear that things go wrong when $zgeq 1$, and in this question 2656851 they determine the discontinuity. This argues in favour of being a simple resurgent formula since it has the required branch cut. Therefore I hope someone can help me rewriting the hypergeometric function in the above form of a simple resurgent function.
Thank you very much in advance.
complex-analysis hypergeometric-function branch-cuts
asked Nov 16 at 15:41
Dylan_VM
373112
In the study of Airy functions you obtain, by Borel transforming components of a transseries, a hypergeometric function $_2F_1(a,b,c|z)$ where the constant are positive. Now I would like to know if this function is simple resurgent, by this I mean that it can be written in the form
$$
_2F_1(a,b,c|z)Big|_{z=A} sim frac{C}{2pi i(z-A)} + psi(z-A) frac{log{(z-A)}}{2pi i} + text{holomorphic}
$$
where $psi(z)$ is a meromorphic function and $Cin mathbb{C}$. My first guess would be start from the integral representation of the hypergeometric function
$$
_2F_1(a,b;c;z)=frac{Gamma(c)}{Gamma(b)Gamma(c-b)}int_0^1t^{b-1}(1-t)^{c-b-1}(1-zt)^{-a}dt,
$$
however I do not know how I can start tackling this problem. From the integral representation it is clear that things go wrong when $zgeq 1$, and in this question 2656851 they determine the discontinuity. This argues in favour of being a simple resurgent formula since it has the required branch cut. Therefore I hope someone can help me rewriting the hypergeometric function in the above form of a simple resurgent function.
Thank you very much in advance.
complex-analysis hypergeometric-function branch-cuts
complex-analysis hypergeometric-function branch-cuts
asked Nov 16 at 15:41
Dylan_VM
373112
asked Nov 16 at 15:41
Dylan_VM
373112
asked Nov 16 at 15:41
Dylan_VM
373112
asked Nov 16 at 15:41
Dylan_VM
373112
asked Nov 16 at 15:41
Dylan_VM
373112
373112
With $u_l,v_l$ the Taylor coefficients, for $M,N$ large enough and $z not in (1,infty)$ $int_0^1 (t^{b-1}-sum_{l=0}^M u_l(b) (t-1)^l) ((1-t)^{c-b-1}(1-zt)^{-a}-sum_{l=0}^N v_l(c-b-1,z,a) t^l)dt$ converges and is locally analytic in each variable. For $z in (1,infty)$ you need to subtract more $(t-1/z)^l$ terms
– reuns
Nov 16 at 19:51
add a comment |
With $u_l,v_l$ the Taylor coefficients, for $M,N$ large enough and $z not in (1,infty)$ $int_0^1 (t^{b-1}-sum_{l=0}^M u_l(b) (t-1)^l) ((1-t)^{c-b-1}(1-zt)^{-a}-sum_{l=0}^N v_l(c-b-1,z,a) t^l)dt$ converges and is locally analytic in each variable. For $z in (1,infty)$ you need to subtract more $(t-1/z)^l$ terms
– reuns
Nov 16 at 19:51
With $u_l,v_l$ the Taylor coefficients, for $M,N$ large enough and $z not in (1,infty)$ $int_0^1 (t^{b-1}-sum_{l=0}^M u_l(b) (t-1)^l) ((1-t)^{c-b-1}(1-zt)^{-a}-sum_{l=0}^N v_l(c-b-1,z,a) t^l)dt$ converges and is locally analytic in each variable. For $z in (1,infty)$ you need to subtract more $(t-1/z)^l$ terms
– reuns
Nov 16 at 19:51
With $u_l,v_l$ the Taylor coefficients, for $M,N$ large enough and $z not in (1,infty)$ $int_0^1 (t^{b-1}-sum_{l=0}^M u_l(b) (t-1)^l) ((1-t)^{c-b-1}(1-zt)^{-a}-sum_{l=0}^N v_l(c-b-1,z,a) t^l)dt$ converges and is locally analytic in each variable. For $z in (1,infty)$ you need to subtract more $(t-1/z)^l$ terms
– reuns
Nov 16 at 19:51
add a comment |
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With $u_l,v_l$ the Taylor coefficients, for $M,N$ large enough and $z not in (1,infty)$ $int_0^1 (t^{b-1}-sum_{l=0}^M u_l(b) (t-1)^l) ((1-t)^{c-b-1}(1-zt)^{-a}-sum_{l=0}^N v_l(c-b-1,z,a) t^l)dt$ converges and is locally analytic in each variable. For $z in (1,infty)$ you need to subtract more $(t-1/z)^l$ terms
– reuns
Nov 16 at 19:51