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Bug introduced in 13.0 or earlier and persisting through 13.2.0 or later.

If you were to evaluate these expressions, Mathematica returns the value shown.

HypergeometricPFQ[{-6/2, -(6 - 1)/2}, {-(6 - 1)}, 1]           = 1/32

HypergeometricPFQ[{-n/2, -(n - 1)/2}, {-(n - 1)}, 1] /. n -> 6 = 1/64

Obviously these shouldn't be different. The former should be correct. Any insight as to why the latter is off by a factor of 2?

Edit

This evaluation seems to be correct

FunctionExpand[ Hypergeometric2F1[a, a/2 + 1, a/2, z]]// Simplify

(1 - z)^(-1 - a) (1 + z)

With[{a = -2}, (1 - z)^(-1 - a) (1 + z)]

(1 - z) (1 + z)

However here the result is different

With[{a = -2}, Hypergeometric2F1[a, a/2 + 1, a/2, z]]

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    $\begingroup$ It's a bug. Please report it to support :) $\endgroup$ Commented May 11, 2022 at 22:55
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    $\begingroup$ While you're reporting the bug you might consider mentioning that Hypergeometric2F1 has the same issue: Hypergeometric2F1[-n/2, -(n - 1)/2, -(n - 1), 1] /. n -> 6 and Hypergeometric2F1[-6/2, -(6 - 1)/2, -(6 - 1), 1]. $\endgroup$ Commented May 11, 2022 at 23:13
  • $\begingroup$ The symbolic result prior to substitution is 2^(-n). This appears to agree with a [Wikipedia article]( en.wikipedia.org/wiki/…) (see "Special values at z = 1") and with the functions.wolfram.com page $\endgroup$ Commented May 11, 2022 at 23:25
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    $\begingroup$ I have emailed this issue to [email protected] and mentioned that Hypergeometric2F1 has the same behavior. Hopefully they'll get to the bottom of it. Thanks. $\endgroup$ Commented May 12, 2022 at 17:48
  • $\begingroup$ The former result is indeed correct, due to the Chu-Vandermonde identity. However, consider also the following evaluations: With[{n = 6, h = 1*^-11, prec = 35}, N[{HypergeometricPFQ[{-(n + h)/2, -((n + h) - 1)/2}, {-((n + h) - 1)}, 1], HypergeometricPFQ[{-(n - h)/2, -((n - h) - 1)/2}, {-((n - h) - 1)}, 1], HypergeometricPFQ[{-(n + I h)/2, -((n + I h) - 1)/2}, {-((n + I h) - 1)}, 1], HypergeometricPFQ[{-(n - I h)/2, -((n - I h) - 1)/2}, {-((n - I h) - 1)}, 1]}, prec]] (adjust h and prec as needed). "Obviously" isn't quite so. $\endgroup$ Commented May 17, 2022 at 22:27

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