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.

  • 1
    $\begingroup$ 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

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 $\Gamma(n/k)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.