Expanding $x^x$ to series with $o(x)$ polynomials

Question

I have some doubts in finding $x^x$ series.
I know that from Taylor theorem I have $$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k + o(x^k) $$

I want to have $o(x)$ polynomials so:
Let $$ f(x) = x^x $$ $$ f'(x) = (e^{x\cdot \ln x}) = (\ln x + 1)x^x$$ In use of taylor I should $$ f(x) = f(0) + \frac{f'(0)(x)}{1!} + o(x) $$ but $f(0)$ is not defined $0^0$. Also $f'(0)$ is not defined... But I found informations that it should be $$ 1+ \ln x \cdot x + o(x)$$ I don't know why :-(

Answer 1

As you have written $$x^x=e^{x\ln{x}}$$ and as the series expansion of $e^x$ converges for all values of $x\in\mathbb{C}$ one can write $$e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+...$$ $$e^{x\ln{x}}=\frac{(x\ln{x})^0}{0!}+\frac{(x\ln{x})^1}{1!}+\frac{(x\ln{x})^2}{2!}+...$$ $$\therefore x^x=1+x\ln{x}+\frac{x^2\ln^2{x}}{2}+\frac{x^3\ln^3{x}}{6}+...$$

Answer 2

Although $f(0)$ and $f'(0)$ are not defined, one can write$$\lim_{x\to 0^+} f(x)=\lim_{x\to 0^+} x^x=e^{\lim_{x\to 0^+} x\ln x}=e^0=1$$In that case $${f(x)=x^x\sim 1\\f'(x)=x^x(\ln x+1)\sim \ln x+1}$$both when $x\to0^+.$ Here is a plot of the function: