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Convert a polynomial to hypergeometric function

  1. Sep 10, 2014 #1
    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.
     
  2. jcsd
  3. Sep 10, 2014 #2

    Ray Vickson

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    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.
     
  4. Sep 10, 2014 #3
    i want to proof it. if i dont know the answer how can i yield a,b,c and x in hypergeometric function?
     
  5. Sep 10, 2014 #4

    Ray Vickson

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    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.
     
  6. Sep 11, 2014 #5
    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.
     
  7. Sep 11, 2014 #6

    Ray Vickson

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