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| Both sides previous revision Previous revision Next revision | Previous revision | ||
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x_x [2016/04/05 09:29] nikolaj |
x_x [2020/01/10 23:08] nikolaj |
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| == Representations == | == Representations == | ||
| - | ^ $x^x={\mathrm e}^{x\log(x)}$ ^ | + | ^ $x^x={\mathrm e}^{x\log(x)}=\left({\mathrm e}^x\right)^{\log(x)}$ ^ |
| + | |||
| + | >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 $(1-x)_n:=\prod_{k=0}^{n-1} (k-x)=\prod_{k=1}^n (k-(1+x))$. | ||
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| >this must come from the Expression at the beginning, but I don't know how I arrived there. | >this must come from the Expression at the beginning, but I don't know how I arrived there. | ||
| - | >Maybe I also switched from $\int_0^1$ to $\int_1^0$. | + | >Maybe there I switched from $x$ to $1-x$ or $\int_0^1$ to $\int_1^0$. |
| $\int_0^1 \, x^x \, dx = \sum_{n=0}^\infty \frac{1}{n!} \int_0^1 \, P_n(x) \, dx = \frac{1}{2} + \frac{1}{12} + \frac{1}{24} + \frac{131}{5040} + \frac{1093}{60480} + \dots$ | $\int_0^1 \, x^x \, dx = \sum_{n=0}^\infty \frac{1}{n!} \int_0^1 \, P_n(x) \, dx = \frac{1}{2} + \frac{1}{12} + \frac{1}{24} + \frac{131}{5040} + \frac{1093}{60480} + \dots$ | ||
