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Particular identity with Hypergeometric Function.

I'm have found particular identity with Hypergeometric2F1.

enter image description here

{Abs[(2 (2 + x^2)^(5/4))/Sqrt[x^3 (2 + x^2)]] == 
5 Hypergeometric2F1[-(3/4), 1/2, 3/2, -(2/x^2)] - 3 Hypergeometric2F1[1/4, 1/2, 3/2, -(2/x^2)]} /. 
x -> {1/10, 1/5, 1.2, 1, 2, Pi} // N
(*True*)

Plot[{Abs[(2 (2 + x^2)^(5/4))/Sqrt[x^3 (2 + x^2)]], 5 Hypergeometric2F1[-(3/4), 1/2, 3/2, -(2/x^2)] - 
3 Hypergeometric2F1[1/4, 1/2, 3/2, -(2/x^2)]}, {x, -1, 1}, PlotStyle -> {Red, {Green, Dashed}}, PlotRange -> {{-1, 1}, {0, 222}}]

enter image description here

Are there many such identity's ? Maybe this has already been discovered?

POSTED BY: Mariusz Iwaniuk
2 Replies

Yes, there are many such cases known. Using the series representations, one can easily show it:

5 Pochhammer[-3/4, n] Pochhammer[1/2, n]/Pochhammer[3/2, n] z^n/n! - 
  3 Pochhammer[1/4, n] Pochhammer[1/2, n]/Pochhammer[3/2, n] z^n/n! // FullSimplify

$$\frac{2 z^n \Gamma \left(n-\frac{3}{4}\right)}{\Gamma \left(-\frac{3}{4}\right) \Gamma (n+1)}$$

Sum[%, {n, 0, \[Infinity]}]

$$2 (1-z)^{3/4}$$

The most comprehensive investigations are by Vidunas, e.g.

https://arxiv.org/abs/0807.4808

POSTED BY: Michael Trott

Thank you very much! You are such a great help! :)

Best regards Mariusz.

POSTED BY: Mariusz Iwaniuk
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