Why this hypergeometric function equals a complex number

Question

According to the following Wolfram alpha calculator link, this hypergeomtric function is a complex number$${}_1F_0(\pi/4;2)=(-1)^{\pi/4}=-0.78121\ldots-i×0.62425\ldots$$

I dont get it. I know that raising (-1) to a real power can give a complex number, what i dont get is how adding an infinite amount of real numbers equals a complex number. How?

Yes, see here, why $(-1)^{\pi/4}$ is not a real number. And indeed, hypergeometric functions need not be real. — Dietrich Burde, Jun 01 '21 at 10:03
I checked your Wolfram Alpha link, and it shows ${}_1F_0(\pi/4,2)$ is an infinite series of real numbers ... times $(-2)^{-\pi/4}$, which is certainly not real. — anon, Jun 01 '21 at 10:05
@Dietrich Burde i know how raising -1 to a real power can give complex numbers. What i dont get is how adding an infinite amount of real numbers equals a complex number. — Robert Puscasu, Jun 01 '21 at 10:12
Robert, of course a real sequence cannot converge to a non-real number because the real numbers are a closed subset of the complex numbers. But your series is not a real sequence. — Dietrich Burde, Jun 01 '21 at 10:17
There's a typo: the answer given by WA is $(-1)^{-\pi/4}$, not $(-1)^{\pi/4}$. — Hans Lundmark, Jun 01 '21 at 10:52
@DietrichBurde What do you mean by "its not a real sequence"? It can be expressed as an infinite sum of real numbers. — Robert Puscasu, Jun 01 '21 at 13:26
I exactly mean what Hans says in his answer. It is not obtained by summing up only real numbers. — Dietrich Burde, Jun 01 '21 at 13:27

score 2 · Accepted Answer · answered Jun 01 '21 at 10:50

2

The radius of convergence of the power series is $1$, so the value at $z=2$ isn't obtained by summing a series of real numbers, but instead by analytic continuation. In this case, it's $$ {}_1 F_0(a;;z) = (1-z)^{-a} . $$ According to the Wolfram documentation: “For $p=q+1$, HypergeometricPFQ[alist,blist,z] has a branch cut discontinuity in the complex plane running from $1$ to $\infty$.”

answered Jun 01 '21 at 10:50

Hans Lundmark

53,395

Thank you for your time. – Robert Puscasu Jun 01 '21 at 13:27

Why this hypergeometric function equals a complex number

1 Answers1