# The Puiseux series and gamma function

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.

• 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) – Dzoooks Jul 3 at 2:11
• To make $1/\Gamma(n+1-a)$ appear look at the fractional derivative of $x^n 1_{x > 0}$ – reuns Jul 3 at 2:29
• I meant that whether the Gamma function would appear in the general case of nonanalytic functions, with Puiseux series not being a Taylor series. – MaudPieTheRocktorate Jul 3 at 10:07

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 $$\Gamma(n/k)$$.