Convert a polynomial to hypergeometric function

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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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