Convert a polynomial to hypergeometric function

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Homework Help Overview

The discussion revolves around expressing a hypergeometric function, specifically (2F1(a,b;c,x)), in terms of a variable n to generate a series of polynomials. The polynomials are defined for specific values of n ranging from 0 to 4, with each corresponding to a specific polynomial expression involving the variable y and a parameter ω.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore how to derive the hypergeometric function from the given polynomial series. There are inquiries about the parameters a, b, c, and x in the hypergeometric function and the need for a method to analytically achieve this transformation.

Discussion Status

Some participants express a desire for a complete solution, while others suggest that verification of the results for specific values of n could be done independently. There is a recognition of the challenge in finding a general method for this conversion, with some skepticism about the feasibility of such a method.

Contextual Notes

Participants note the need for a deeper understanding of the relationship between the polynomial series and the hypergeometric function, indicating that assumptions about the parameters and the transformation process are under scrutiny.

azizianhra
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i want to write a hypergeometric function (2F1(a,b;c,x)) as function of n that generate polynomials below

n=0 → 1
n=1 → y
n=2 → 4(ω+1)y^2-1
n=3 → y(2(2ω+3)y^2-3)
n=4 → 8(ω+2)(2ω+3)y^4-6(6+4ω)y^2+3
... → ...


2F1(a,b;c,x)=1+(ab)/(c)x+(a(a+1)b(b+1))/(c(c+1))x^2/2!+...


the answer is 2F1(-n,n+4ω+2;2ω+3/2;(1-y)/2) but i want the solution completely.
 
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azizianhra said:
i want to write a hypergeometric function (2F1(a,b;c,x)) as function of n that generate polynomials below

n=0 → 1
n=1 → y
n=2 → 4(ω+1)y^2-1
n=3 → y(2(2ω+3)y^2-3)
n=4 → 8(ω+2)(2ω+3)y^4-6(6+4ω)y^2+3
... → ...


2F1(a,b;c,x)=1+(ab)/(c)x+(a(a+1)b(b+1))/(c(c+1))x^2/2!+...


the answer is 2F1(-n,n+4ω+2;2ω+3/2;(1-y)/2) but i want the solution completely.

What is preventing you from verifying the results for n = 0, 1, 2, 3, 4 for yourself? You can work out the solution just as easily as we can.
 
Ray Vickson said:
What is preventing you from verifying the results for n = 0, 1, 2, 3, 4 for yourself? You can work out the solution just as easily as we can.

i want to proof it. if i don't know the answer how can i yield a,b,c and x in hypergeometric function?
 
azizianhra said:
i want to proof it. if i don't know the answer how can i yield a,b,c and x in hypergeometric function?

You told us that the answer is 2F1(-n,n+4ω+2;2ω+3/2;(1-y)/2), so you are told what must be a, b and c. You prove it (not "proof" it) by straight verification.
 
I need a method of solution that starts form series of polynomials and convert it to hypergeometric function. i mean how to achieving answer that we already have it. I know that by substituting n in answer the polynomials generated, but this isn't the point that i need it. I'm looking for a analytical solution.
 
azizianhra said:
I need a method of solution that starts form series of polynomials and convert it to hypergeometric function. i mean how to achieving answer that we already have it. I know that by substituting n in answer the polynomials generated, but this isn't the point that i need it. I'm looking for a analytical solution.

Good luck with that; I don't think there is any reasonable and general method.
 

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