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Just a thought: In the Taylor expansion of an analytic function $f(x)$, the $\Gamma(n+1) = n!$ appears in the coefficient for $x^n$. So if we use a Puiseux series instead, would we get a $\Gamma(n/k)$ appearing in the coefficient for $x^{n/k}$?

I searched around and the only related result is that the Gamma function values appear in the general Newton binomial expansion theorem.

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    $\begingroup$ What's your question exactly? Isn't the Puisuex series expansion of an analytic function equal to its Taylor expansion? (i.e. the coefficients of fractional powers are all 0) $\endgroup$ – Dzoooks Jul 3 at 2:11
  • $\begingroup$ To make $1/\Gamma(n+1-a)$ appear look at the fractional derivative of $x^n 1_{x > 0}$ $\endgroup$ – reuns Jul 3 at 2:29
  • $\begingroup$ I meant that whether the Gamma function would appear in the general case of nonanalytic functions, with Puiseux series not being a Taylor series. $\endgroup$ – MaudPieTheRocktorate Jul 3 at 10:07
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If you take a function $f(z)$ with a converging Puiseux series in powers of $z^{1/k}$, say

$$ f(z) = \sum_{j=0}^\infty a_j z^{j/k}$$ then $$ f(z^k) = \sum_{j=0}^\infty a_j z^j$$ is analytic in a neighbourhood of $0$, with $$ a_n = \frac{1}{n!} \left. \dfrac{d^n}{dz^n} f(z^k) \right|_{z=0} $$

No, there's no reason for this to involve $\Gamma(n/k)$.

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