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Both sides previous revision Previous revision Next revision | Previous revision | ||
x_x [2016/04/05 09:32] nikolaj |
x_x [2020/01/10 23:08] nikolaj |
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^ $x^x={\mathrm e}^{x\log(x)}=\left({\mathrm e}^x\right)^{\log(x)}$ ^ | ^ $x^x={\mathrm e}^{x\log(x)}=\left({\mathrm e}^x\right)^{\log(x)}$ ^ | ||
+ | |||
+ | >todo: write down the above with an expanded $\log$ to third order | ||
Because of this, the local minimum of $x^x$ is that of $x\log(x)$, namely $\frac{1}{\mathrm e}\approx 0.37$, and then see | Because of this, the local minimum of $x^x$ is that of $x\log(x)$, namely $\frac{1}{\mathrm e}\approx 0.37$, and then see | ||
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Furthermore | Furthermore | ||
- | ^ $x^x = x \sum_{n=0}^\infty \prod_{k=1}^n (1-x)\left(1-\frac{1+x}{k}\right) = x \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 $(1-x)_n:=\prod_{k=0}^{n-1} (k-x)=\prod_{k=1}^n (k-(1+x))$. |