# Tag Archives: generating

## Stirling numbers of first kind generating function

Prove the following $$\sum_{n=k}^\infty(-1)^{n-k}\left[n\atop k\right] \frac{z^n}{n!} = \frac{\log^k(1+z)}{k!}$$ $$\textit{proof}$$ We start by the following $$(1+z)^u = \sum_{n=0}^\infty {u \choose n} z^n$$ Now use that $${u \choose n} = \frac{\Gamma(u+1)}{\Gamma(u-n+1)n!}$$ Now use that $$\frac{\Gamma(u+1)}{\Gamma(u-n+1)} = \frac{u(u-1)\cdots (u-n+1)\Gamma(u+1)}{\Gamma(u+1)} = (u)_n$$ This implies … Continue reading

Signed Stirling numbers of the first kind We define the following $$s(n,k) = (-1)^{n-k} \left[n\atop k\right]$$ Prove the following $$(x)_n = x(x-1)(x-2)\cdots (x-n+1) = \sum_{k=0}^n s(n,k)x^k$$ $$\textit{proof}$$ We already proved that $$x^{(n)} = \sum_{k=0}^n \left[n\atop k\right] x^k$$ Which can be … Continue reading
Prove that $$x^{(n)} = x(x+1)(x+2)\cdots(x+n-1) = \sum_{k=0}^n \left[n\atop k\right]x^k$$ Or $$\sum_{k=0}^n \left[n\atop k\right]x^k = \frac{\Gamma(x+n)}{\Gamma(x)}$$ $$\textit{proof}$$ By induction for $n=0$ we have $$\left[0\atop 0\right] = 1$$ Assume it is true for $n$ then for \( … Continue reading