Writing this series as a hypergeometric series

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Homework Help Overview

The discussion revolves around expressing a series involving shifted factorials and powers of a variable in terms of the Gauss hypergeometric series. The original poster seeks to rewrite the series \(\sum_{k=0}^{\infty} \frac{1}{9^k (\frac{2}{3})_k} \frac{w^{3k}}{k!}\) in the form of the hypergeometric series \(_2 F_1(a,b;c;z)\).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the transformation of the series into the hypergeometric form, questioning how to incorporate the \(9^{-k}\) factor. The original poster considers substituting \(z = \frac{w^3}{9}\) and explores the implications of choosing parameters \(a\) and \(b\) in the hypergeometric series.

Discussion Status

There is an ongoing exploration of the correct parameters for the hypergeometric series. Some participants suggest verifying the proposed solution by checking it against the hypergeometric differential equation. The discussion reflects uncertainty about the conditions under which shifted factorials equal one.

Contextual Notes

Participants note that the shifted factorials \((0)_n\) do not equal one for all \(n\), raising questions about how to achieve the necessary conditions for the hypergeometric series representation.

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Homework Statement



Write \displaystyle \sum_{k=0}^{\infty} \frac{1}{9^k (\frac{2}{3})_k} \frac{w^{3k}}{k!} in terms of the Gauss hypergeometric series of the form _2 F_1(a,b;c;z).

Homework Equations



The Gauss hypergeometric series is http://img200.imageshack.us/img200/5992/gauss.png

The Attempt at a Solution



It's nearly a series of that form if I put z=w^3 and k=n but how do I get the 9^{-k} = 3^{-k}3^{-k} factors in terms of shifted factorials (that is if I need to)?
 
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That term does not go into the factorials, it goes into z^n.
 
clamtrox said:
That term does not go into the factorials, it goes into z^n.

Ah of course. So if I put z=\frac{w^3}{9} then the series can be written as _2 F_1 (a,b ; \frac{2}{3} ; \frac{w^3}{9}) for some a and b with (a)_n(b)_n = 1 for all n=0,1,2,... Can I just pick a=b=0?
 
I would do some extra checking to be sure that that's right. Can you plug the solution into the hypergeometric differential equation with a=b=0 and see if it solves it?
 
clamtrox said:
I would do some extra checking to be sure that that's right. Can you plug the solution into the hypergeometric differential equation with a=b=0 and see if it solves it?

Actually (0)_n (0)_n \neq 1 for n=0,1,2,... so how do I get 2 shifted factorials to equal 1?
 

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