I am well aware of the expansion by using $e^{x*ln{x}}$ and am looking for a different way. Can somebody please tell me a different expansion for $x^x$?
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Please use [MathJax](https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – saulspatz Jul 23 '19 at 00:24
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2If you substitute $x\ln{x}$ into the Taylor series for $e^x$, you don't get a Taylor series. There are no $\ln{x}$ factors in a power series. Also, Taylor series about what point? Note that $x^x$ is not analytic at $x=0$, so it cannot be developed in a Taylor series about $x=0.$ – saulspatz Jul 23 '19 at 00:27
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1See: https://math.stackexchange.com/questions/2322635/taylor-expansion-of-xx-at-x-0 – NoChance Jul 23 '19 at 00:55
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You may want to consider using Pade Approximant: https://math.stackexchange.com/questions/860293/how-to-compute-the-pade-approximation – NoChance Jul 23 '19 at 01:00
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While there is no Taylor series about $x=0$, you could expand about any positive $x$. The series around $x=1$ is $$ 1+(x-1)+ \left( x-1 \right) ^{2}+{\frac{1}{2}} \left( x-1 \right) ^{3}+ {\frac{1}{3}} \left( x-1 \right) ^{4}+{\frac{1}{12}} \left( x-1 \right) ^{5}+{\frac{3}{40}} \left( x-1 \right) ^{6}-{\frac{1}{120}} \left( x-1 \right) ^{7}+{\frac{59}{2520}} \left( x-1 \right) ^{8}-{ \frac{71}{5040}} \left( x-1 \right) ^{9}+ \ldots $$
Robert Israel
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As suggested in comments, you could also consider Padé approximants built around $x=1$ (just as for the Taylor series given in Robert Israel's answer).
Choosing the $[5,5]$ approximant and using $t=x-1$ for an easier legibility, we should get $$x^x=\frac{1+\frac{129004 }{82639}t+\frac{166329 }{165278}t^2+\frac{250321 }{578473}t^3+\frac{386677 }{3470838}t^4+\frac{1341953 }{69416760}t^5 } {1+\frac{46365 }{82639}t-\frac{91679 }{165278}t^2-\frac{42594 }{578473}t^3+\frac{436889 }{3470838}t^4-\frac{599629 }{23138920}t^5 }$$ which is equivalent to a Taylor series to $O\left((x-1)^{11}\right)$.
Claude Leibovici
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