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x_x [2016/04/05 09:33]
nikolaj
x_x [2020/01/10 23:08] (current)
nikolaj
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 Furthermore Furthermore
  
-^ $x^x = \sum_{n=0}^\infty \prod_{k=1}^n (1-x)\left(1-\frac{1+x}{k}\right) = \sum_{n=0}^\infty \frac{1}{n!}(1-x)_n(1-x)^n $ ^+^ $x^x = \sum_{n=0}^\infty \prod_{k=1}^n (1-x)\left(1-\frac{1+x}{k}\right) = \sum_{n=0}^\infty \frac{1}{n!}(1-x)_n(1-x)^n $ ^
  
 with the Pochhammer symbol $(1-x)_n:​=\prod_{k=0}^{n-1} (k-x)=\prod_{k=1}^n (k-(1+x))$. with the Pochhammer symbol $(1-x)_n:​=\prod_{k=0}^{n-1} (k-x)=\prod_{k=1}^n (k-(1+x))$.
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