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Find the Taylor seriespolynomial form of:

$$x^x$$

My attempt:

I started calculating derivatives of $x^x$ for the series, $$f(x)=x^x$$ $$f'(x)=x^x(\ln x+1)$$ $$f''(x)=x^{x-1}+x^x(\ln x+1)$$ $$f'''(x)=\cdots$$ $$\vdots$$ But i could get only till certain terms and got frustrated,

Attempt no. 2:

$$f(x)=x^x$$ $$f(x)=e^{x\ln x}$$ $$f(x)=1+x\ln x+\frac{x^2\ln^2 x}{2!}+\cdots$$ $$f(x)=1+x({x-1}-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}+\cdots)+\frac{x^2}{2}({x-1}-\frac{(x-1)^2}{2}+\frac{(x-2)^3}{3}+\cdots)^2+\cdots$$ Now let's try to get $x^n$'s coefficient, $$C(x^0)=1$$ In the bracket multiplied to x, $$1+x(x-1-x^2+2x-1+x^3-3x+3x-1+\cdots)+\cdots$$ Following the lead $$1+x(g(x)-1-1-1-1\cdots)+\cdots$$ $$C(x^1)=-1-1-1-\cdots$$ So, $$C(x^1)=-\infty\cdots?$$ How do we write this equation?

Attempt 3:

$$f(x)=1+(x+1)\ln (x+1)+\frac{(x+1)^2}{2!}\ln^2(x+1)+\cdots$$ Now, $$f(x)=1+(x+1)(x+\frac{x^2}{2}+\frac{x^3}{3}\cdots)+\cdots$$ Now, $$C(x^0)=1$$ Considering $(1+x)^2,(1+x)^3\cdots$ contribution to$C(x^1)$ $$C(x^1)=1+1+1+1+\cdots$$ $$C(x^1)=\infty\cdots ?$$ Still.....stuck

Community
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neonpokharkar
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2 Answers2

2

Hint: We can find an elaboration of the $n$-th derivative of $x^x$ in an example (p. 139) of Advanced Combinatorics by L. Comtet. The idea is based upon a clever Taylor series expansion. Using the differential operator $D_x^j:=\frac{d^j}{dx^j}$ the following holds:

The $n$-th derivative of $x^x$ is

\begin{align*} \color{blue}{D_x^n x^x=x^x\sum_{i=0}^n\binom{n}{i}(\ln(x))^i\sum_{j=0}^{n-i}b_{n-i,n-i-j}x^{-j}}\tag{1} \end{align*} with $\color{blue}{b_{n,j}}$ the Lehmer-Comtet numbers.

These numbers follow the recurrence relation \begin{align*} b_{n+1,j}=(j-n)b_{n,j}+b_{n,j-1}+nb_{n-1,j-1}\qquad\qquad n,j\geq 1 \end{align*} and the first values, together with initial values are listed below.

\begin{array}{c|cccccc} n\setminus k&1&2&3&4&5&6\\ \hline 1&1\\ 2&1&1\\ 3&-1&3&1\\ 4&2&-1&6&1\\ 5&-6&0&5&10&1\\ 6&24&4&-15&25&15&1\\ \end{array}

The values can be found in OEIS as A008296. They are called Lehmer-Comtet numbers and were stored in the archive by N.J.A.Sloane by referring precisely to the example we can see here.

Example: $n=2$

Let's look at a small example. Letting $n=2$ we obtain from (1) and the table with $b_{n,j}$: \begin{align*} D_x^2x^x&=x^x\sum_{i=0}^2\binom{2}{i}(\ln(x))^i\sum_{j=0}^{2-i}b_{2-i,2-i-j}x^{-j}\\ &=x^x\left(\binom{2}{0}\sum_{j=0}^2b_{2,2-j}x^{-j}+\binom{2}{1}\ln(x)\sum_{j=0}^1b_{1,1-j}x^{-j}\right.\\ &\qquad\qquad\left.+\binom{2}{2}\left(\ln(x)\right)^2\sum_{j=0}^0b_{0,0-j}x^{-j}\right)\\ &=x^x\left(\left(b_{2,2}+b_{2,1}\frac{1}{x}+b_{2,0}\frac{1}{x^2}\right)+2\ln(x)\left(b_{1,1}+b_{1,0}\frac{1}{x}\right) +(\ln(x))^2b_{0,0}\right)\\ &=x^x\left(1+\frac{1}{x}+2\ln(x)+\left(\ln(x)\right)^2\right) \end{align*} in accordance with the result of Wolfram Alpha.

Note: A detailed answer is provided in this MSE post.

Markus Scheuer
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0

By the formula \begin{equation}\label{x-poer-x-thm3-driv-repl} \frac{\operatorname{d}^n}{\operatorname{d}x^n}\bigl(x^{tx}\bigr) =n!x^{tx-n}\sum_{k=0}^{n}t^kx^{k} \sum_{j=0}^{k} \Biggl[\sum_{q=0}^{n-k} \frac{s(q+j,j)}{(q+j)!} \binom{j}{n-k-q}\Biggr] \frac{(\ln x)^{k-j}}{(k-j)!}, \end{equation} one can derive Taylor series of the function $x^x$ around any point $x_0>0$.

References

  1. Jian Cao, Feng Qi, and Wei-Shih Du, Closed-form formulas for the $n$th derivative of the power-exponential function $x^x$, Symmetry 15 (2023), no. 2, Article 323, 13 pages; available online at https://doi.org/10.3390/sym15020323.
  2. https://mathoverflow.net/a/439188
qifeng618
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