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I am looking for a power-series expression of the form $\Gamma(z)=b+\sum_{k=0}^\infty a_kz^k$ where the $a_k$ can be calculated as some function of k.

Bumblebee
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JacksonFitzsimmons
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1 Answers1

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$\Gamma(z)=\dfrac1z+\displaystyle\sum_{k=0}^\infty a_k\cdot z^k,~$ where the terms $a_k$ form this “beautiful” sequence here. :-$)$

Lucian
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  • Exactly what I was looking for, thank you. – JacksonFitzsimmons Jul 08 '15 at 06:14
  • Is there a reasonable expression for the general term $a_k$? – Travis Willse Jul 08 '15 at 06:49
  • Also, it's worth noting that $\Gamma$ has an especially nice Laurent series about $z = 1$. – Travis Willse Jul 08 '15 at 06:50
  • @Travis: Could you please post its expression, or at least its few first terms ? – Lucian Jul 08 '15 at 06:55
  • @Lucian It's a bit involved, and involves a sum over partitions; Wikipedia has it here: https://en.wikipedia.org/wiki/Gamma_function#General (Search for the phrase "has the following Laurent expansion in 1".) One can in principle use expand the powers of $z - 1$ that appear in this series to answer my above question, though the resulting expression produced this way pushes the boundaries of "reasonable". – Travis Willse Jul 08 '15 at 07:06
  • @Travis: It's basically the same as the one I posted, only that it lacks the pole, and does not expand the expressions for $\Gamma^{(n)}(1),$ like I did in the link. – Lucian Jul 08 '15 at 07:15
  • @Lucian Yes, that's what I mean by "expand the powers...", but the expression for the general term (of the series around $z = 0$) becomes messy if one proceeds that way. My question is basically whether there's a better way to write the general term than that. – Travis Willse Jul 08 '15 at 07:37
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    @Travis: Not to my humble knowledge. – Lucian Jul 08 '15 at 07:56
  • I didn't know about the DLMF, this looks like a great resource! Anyway, perhaps this is another argument that the Gamma function should actually be defined as $\Gamma(x + 1)$ (i.e., so that it extends the factorial function). – Travis Willse Jul 08 '15 at 08:34
  • @TravisWillse i can't seem to find the info on the series about $z=1$ in that Wiki link. Do you have another link? Thanks. – pshmath0 Feb 04 '20 at 01:44
  • @Pixel Alas, it looks like it's been edited away sometime in the last five years. I don't recall the formula now, but there's a question here whose answer gives the first few terms: https://math.stackexchange.com/questions/1287555/how-to-obtain-the-laurent-expansion-of-gamma-function-around-z-0 See also: https://dlmf.nist.gov/5.7 – Travis Willse Feb 04 '20 at 02:55
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    @TravisWillse thank you. I also took Mathematica for a spin: $$\Gamma(s-1)=\frac{1}{s-1}-\gamma +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) (s-1)+\frac{1}{6} \left(-\gamma ^3-\frac{\gamma \pi ^2}{2}+\psi ^{(2)}(1)\right)(s-1)^2 +\frac{1}{24} \left(\gamma ^4+\frac{3 \pi ^4}{20}+\gamma ^2 \pi ^2-4 \gamma \psi ^{(2)}(1)\right)(s-1)^3+O\left((s-1)^4\right),$$ where I believe the polygamma's are some function of $\zeta(q)$, $q\in\mathbb{N}$. – pshmath0 Feb 04 '20 at 08:35
  • I would like to know how the "beautiful sequence" full of EulerGamma's is derived. If you like I can create a question for it. I am especially interested in references showing how it is done, which theories/techniques are used. Thank you. – nilo de roock Jan 25 '22 at 10:09
  • @niloderoock: As far as I can tell, it is simply Taylor's classical series expansion, coupled with certain special values of the polygamma function, which usually depend on the zeta. – Lucian Jan 25 '22 at 13:01
  • Thank you, @Lucian - I'll figure it out somehow, as I always do. :-) – nilo de roock Jan 25 '22 at 13:29