Show that the complex integral is a solution of Gauss's hypergeometric equation and express it in terms of standard ones.Catagorising a Differential EquationConnections between the solution of simple ordinary equation, normal distribution and heat equationWhy are the standard form of non-homogeneous linear DVQ and the general solution equations true?What is $int_0^inftyfraccos^nx sin x ln xx dx$?Differential equations - Hypergeometric functionClosed-form Solution for series involving incomplete Gamma FunctionWriting the solution of a linear first order DE in the way that the null solution and the particular solutions are separatedUse of series and integral (Fourier or not) to write solution of three dimensional wave equation in different coordinates systemsCommon solution to Integral Equation and Differential Equationlinear differential equation with two regular, one irregular singular point
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Convert seconds to minutes
Show that the complex integral is a solution of Gauss's hypergeometric equation and express it in terms of standard ones.
Catagorising a Differential EquationConnections between the solution of simple ordinary equation, normal distribution and heat equationWhy are the standard form of non-homogeneous linear DVQ and the general solution equations true?What is $int_0^inftyfraccos^nx sin x ln xx dx$?Differential equations - Hypergeometric functionClosed-form Solution for series involving incomplete Gamma FunctionWriting the solution of a linear first order DE in the way that the null solution and the particular solutions are separatedUse of series and integral (Fourier or not) to write solution of three dimensional wave equation in different coordinates systemsCommon solution to Integral Equation and Differential Equationlinear differential equation with two regular, one irregular singular point
$begingroup$
There is Gauss's hypergeometric equation
$[z(1-z)fracd^2dz^2 + [c - (a+b+1)z]fracddz - ab]u(z) = 0$
Apparently, (see [1], example in 14.2), this equation can be rearranged in the form
$[(z fracddz + a) (z fracddz + b) + (fracddz+c)(fracddz)]u = 0$
Consider for $arg(1-z) < pi$, c is not integer, for simplicity.
$F(z) = frac12pi iint_-i infty^i inftyGamma(a+s)Gamma(b+s)Gamma(-s)Gamma(c-a-b-s)(1-z)^s ds$
Here the contour divides the "going to the left" and "going to the right" series of poles.
I want to show that $F(z)$ satisfies the equation above.
Step 2. Since this is a regular ODE, the equation has 2 non-proportional solutions, and the standard base is $F(a,b;c;z)$ and $z^1-cF(1+a-c,1+b-c; 2-c;z)$
I need to express F(z) as a linear combination of the standard solutions.
Some useful hints is in evaluation in [1]14.53.
They use Barnes' lemma ( [1] 14.52) , and the corollary from 14.53 that I will write down there.
$Gamma(-t)(1-z)^t = frac12 pi i int_-i infty^i inftyGamma(s-t)Gamma(-s)(-z)^s)ds$
But, actually, may be we could try evaluate the values of F(z) and standard solutions at 0 and 1. And then solve the linear equation system without those complex integrals.
If you know some more elementary literature about the question, or a hint, that could make the things easier, please, let me know.
[1] The book of Whittaker, Watson, A course of modern analysis.
UPD: continue thinking about the task.
ordinary-differential-equations improper-integrals transcendental-functions
asked Mar 28 at 14:55
Lada DudnikovaLada Dudnikova
34
New contributor
Lada Dudnikova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
There is Gauss's hypergeometric equation
$[z(1-z)fracd^2dz^2 + [c - (a+b+1)z]fracddz - ab]u(z) = 0$
Apparently, (see [1], example in 14.2), this equation can be rearranged in the form
$[(z fracddz + a) (z fracddz + b) + (fracddz+c)(fracddz)]u = 0$
Consider for $arg(1-z) < pi$, c is not integer, for simplicity.
$F(z) = frac12pi iint_-i infty^i inftyGamma(a+s)Gamma(b+s)Gamma(-s)Gamma(c-a-b-s)(1-z)^s ds$
Here the contour divides the "going to the left" and "going to the right" series of poles.
I want to show that $F(z)$ satisfies the equation above.
Step 2. Since this is a regular ODE, the equation has 2 non-proportional solutions, and the standard base is $F(a,b;c;z)$ and $z^1-cF(1+a-c,1+b-c; 2-c;z)$
I need to express F(z) as a linear combination of the standard solutions.
Some useful hints is in evaluation in [1]14.53.
They use Barnes' lemma ( [1] 14.52) , and the corollary from 14.53 that I will write down there.
$Gamma(-t)(1-z)^t = frac12 pi i int_-i infty^i inftyGamma(s-t)Gamma(-s)(-z)^s)ds$
But, actually, may be we could try evaluate the values of F(z) and standard solutions at 0 and 1. And then solve the linear equation system without those complex integrals.
If you know some more elementary literature about the question, or a hint, that could make the things easier, please, let me know.
[1] The book of Whittaker, Watson, A course of modern analysis.
UPD: continue thinking about the task.
ordinary-differential-equations improper-integrals transcendental-functions
asked Mar 28 at 14:55
Lada DudnikovaLada Dudnikova
34
New contributor
Lada Dudnikova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
I'm pretty sure that this integral DOES NOT satisfy the hypergeometric equation. Note that as you defined it $ F(z) = C G(z)$ where $G(z) = int_-iinfty^iinfty (1-z)^s ds$ and $C$ is a constant (does not depend on $z$). Since the hypergeometric equation is linear, for $F$ to be a solution you'd need $G$ to be a solution. But $G(z)$ does not depend on parameters $a$, $b$, $c$, so it couldn't possibly satisfy the hypergeometric equation for arbitrary parameters.
$endgroup$
– Adam Latosiński
Mar 28 at 15:47
$begingroup$
Thanks for your remark. Gammas can't be moved behind the integral, as you propose, because they depend on s, the integration variable.
$endgroup$
– Lada Dudnikova
Mar 28 at 16:08
$begingroup$
Right, my mistake. I'm so used to $z$ being the integration variable, I missed that.
$endgroup$
– Adam Latosiński
Mar 29 at 1:04
add a comment |
$begingroup$
There is Gauss's hypergeometric equation
$[z(1-z)fracd^2dz^2 + [c - (a+b+1)z]fracddz - ab]u(z) = 0$
Apparently, (see [1], example in 14.2), this equation can be rearranged in the form
$[(z fracddz + a) (z fracddz + b) + (fracddz+c)(fracddz)]u = 0$
Consider for $arg(1-z) < pi$, c is not integer, for simplicity.
$F(z) = frac12pi iint_-i infty^i inftyGamma(a+s)Gamma(b+s)Gamma(-s)Gamma(c-a-b-s)(1-z)^s ds$
Here the contour divides the "going to the left" and "going to the right" series of poles.
I want to show that $F(z)$ satisfies the equation above.
Step 2. Since this is a regular ODE, the equation has 2 non-proportional solutions, and the standard base is $F(a,b;c;z)$ and $z^1-cF(1+a-c,1+b-c; 2-c;z)$
I need to express F(z) as a linear combination of the standard solutions.
Some useful hints is in evaluation in [1]14.53.
They use Barnes' lemma ( [1] 14.52) , and the corollary from 14.53 that I will write down there.
$Gamma(-t)(1-z)^t = frac12 pi i int_-i infty^i inftyGamma(s-t)Gamma(-s)(-z)^s)ds$
But, actually, may be we could try evaluate the values of F(z) and standard solutions at 0 and 1. And then solve the linear equation system without those complex integrals.
If you know some more elementary literature about the question, or a hint, that could make the things easier, please, let me know.
[1] The book of Whittaker, Watson, A course of modern analysis.
UPD: continue thinking about the task.
ordinary-differential-equations improper-integrals transcendental-functions
asked Mar 28 at 14:55
Lada DudnikovaLada Dudnikova
34
New contributor
Lada Dudnikova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
There is Gauss's hypergeometric equation
$[z(1-z)fracd^2dz^2 + [c - (a+b+1)z]fracddz - ab]u(z) = 0$
Apparently, (see [1], example in 14.2), this equation can be rearranged in the form
$[(z fracddz + a) (z fracddz + b) + (fracddz+c)(fracddz)]u = 0$
Consider for $arg(1-z) < pi$, c is not integer, for simplicity.
$F(z) = frac12pi iint_-i infty^i inftyGamma(a+s)Gamma(b+s)Gamma(-s)Gamma(c-a-b-s)(1-z)^s ds$
Here the contour divides the "going to the left" and "going to the right" series of poles.
I want to show that $F(z)$ satisfies the equation above.
Step 2. Since this is a regular ODE, the equation has 2 non-proportional solutions, and the standard base is $F(a,b;c;z)$ and $z^1-cF(1+a-c,1+b-c; 2-c;z)$
I need to express F(z) as a linear combination of the standard solutions.
Some useful hints is in evaluation in [1]14.53.
They use Barnes' lemma ( [1] 14.52) , and the corollary from 14.53 that I will write down there.
$Gamma(-t)(1-z)^t = frac12 pi i int_-i infty^i inftyGamma(s-t)Gamma(-s)(-z)^s)ds$
But, actually, may be we could try evaluate the values of F(z) and standard solutions at 0 and 1. And then solve the linear equation system without those complex integrals.
If you know some more elementary literature about the question, or a hint, that could make the things easier, please, let me know.
[1] The book of Whittaker, Watson, A course of modern analysis.
UPD: continue thinking about the task.
ordinary-differential-equations improper-integrals transcendental-functions
ordinary-differential-equations improper-integrals transcendental-functions
asked Mar 28 at 14:55
Lada DudnikovaLada Dudnikova
34
New contributor
Lada Dudnikova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 14:55
Lada DudnikovaLada Dudnikova
34
New contributor
Lada Dudnikova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 30 at 20:39
Lada Dudnikova
asked Mar 28 at 14:55
Lada DudnikovaLada Dudnikova
34
New contributor
Lada Dudnikova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 14:55
Lada DudnikovaLada Dudnikova
34
asked Mar 28 at 14:55
Lada DudnikovaLada Dudnikova
34
34
New contributor
Lada Dudnikova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Lada Dudnikova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Lada Dudnikova is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I'm pretty sure that this integral DOES NOT satisfy the hypergeometric equation. Note that as you defined it $ F(z) = C G(z)$ where $G(z) = int_-iinfty^iinfty (1-z)^s ds$ and $C$ is a constant (does not depend on $z$). Since the hypergeometric equation is linear, for $F$ to be a solution you'd need $G$ to be a solution. But $G(z)$ does not depend on parameters $a$, $b$, $c$, so it couldn't possibly satisfy the hypergeometric equation for arbitrary parameters.
$endgroup$
– Adam Latosiński
Mar 28 at 15:47
$begingroup$
Thanks for your remark. Gammas can't be moved behind the integral, as you propose, because they depend on s, the integration variable.
$endgroup$
– Lada Dudnikova
Mar 28 at 16:08
$begingroup$
Right, my mistake. I'm so used to $z$ being the integration variable, I missed that.
$endgroup$
– Adam Latosiński
Mar 29 at 1:04
add a comment |
$begingroup$
I'm pretty sure that this integral DOES NOT satisfy the hypergeometric equation. Note that as you defined it $ F(z) = C G(z)$ where $G(z) = int_-iinfty^iinfty (1-z)^s ds$ and $C$ is a constant (does not depend on $z$). Since the hypergeometric equation is linear, for $F$ to be a solution you'd need $G$ to be a solution. But $G(z)$ does not depend on parameters $a$, $b$, $c$, so it couldn't possibly satisfy the hypergeometric equation for arbitrary parameters.
$endgroup$
– Adam Latosiński
Mar 28 at 15:47
$begingroup$
Thanks for your remark. Gammas can't be moved behind the integral, as you propose, because they depend on s, the integration variable.
$endgroup$
– Lada Dudnikova
Mar 28 at 16:08
$begingroup$
Right, my mistake. I'm so used to $z$ being the integration variable, I missed that.
$endgroup$
– Adam Latosiński
Mar 29 at 1:04
$begingroup$
I'm pretty sure that this integral DOES NOT satisfy the hypergeometric equation. Note that as you defined it $ F(z) = C G(z)$ where $G(z) = int_-iinfty^iinfty (1-z)^s ds$ and $C$ is a constant (does not depend on $z$). Since the hypergeometric equation is linear, for $F$ to be a solution you'd need $G$ to be a solution. But $G(z)$ does not depend on parameters $a$, $b$, $c$, so it couldn't possibly satisfy the hypergeometric equation for arbitrary parameters.
$endgroup$
– Adam Latosiński
Mar 28 at 15:47
$begingroup$
I'm pretty sure that this integral DOES NOT satisfy the hypergeometric equation. Note that as you defined it $ F(z) = C G(z)$ where $G(z) = int_-iinfty^iinfty (1-z)^s ds$ and $C$ is a constant (does not depend on $z$). Since the hypergeometric equation is linear, for $F$ to be a solution you'd need $G$ to be a solution. But $G(z)$ does not depend on parameters $a$, $b$, $c$, so it couldn't possibly satisfy the hypergeometric equation for arbitrary parameters.
$endgroup$
– Adam Latosiński
Mar 28 at 15:47
$begingroup$
Thanks for your remark. Gammas can't be moved behind the integral, as you propose, because they depend on s, the integration variable.
$endgroup$
– Lada Dudnikova
Mar 28 at 16:08
$begingroup$
Thanks for your remark. Gammas can't be moved behind the integral, as you propose, because they depend on s, the integration variable.
$endgroup$
– Lada Dudnikova
Mar 28 at 16:08
$begingroup$
Right, my mistake. I'm so used to $z$ being the integration variable, I missed that.
$endgroup$
– Adam Latosiński
Mar 29 at 1:04
$begingroup$
Right, my mistake. I'm so used to $z$ being the integration variable, I missed that.
$endgroup$
– Adam Latosiński
Mar 29 at 1:04
add a comment |
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$begingroup$
I'm pretty sure that this integral DOES NOT satisfy the hypergeometric equation. Note that as you defined it $ F(z) = C G(z)$ where $G(z) = int_-iinfty^iinfty (1-z)^s ds$ and $C$ is a constant (does not depend on $z$). Since the hypergeometric equation is linear, for $F$ to be a solution you'd need $G$ to be a solution. But $G(z)$ does not depend on parameters $a$, $b$, $c$, so it couldn't possibly satisfy the hypergeometric equation for arbitrary parameters.
$endgroup$
– Adam Latosiński
Mar 28 at 15:47
$begingroup$
Thanks for your remark. Gammas can't be moved behind the integral, as you propose, because they depend on s, the integration variable.
$endgroup$
– Lada Dudnikova
Mar 28 at 16:08
$begingroup$
Right, my mistake. I'm so used to $z$ being the integration variable, I missed that.
$endgroup$
– Adam Latosiński
Mar 29 at 1:04