Taylor coefficients of the function $x^{1/x}$

Question

I'm interested in the following function:

$$\begin{aligned} f(x)&=x^{\frac{1}{x}}\\ &=e^{\frac{1}{x}\ln x}\end{aligned} $$

I'd like to find its Taylor coefficients (specifically around the point at $x=1$), because they form an interesting progression numerically that I can't quite find a pattern to.

I've been able to find the first few derivatives, but they don't seem to form a useful pattern.

$$ \mathrm{define} \ f_1(x)\equiv1-\ln x$$ $$\begin{align}f'(x)&=\frac{1}{x^2}f_1(x)f(x)\\ f''(x)&=\frac{1}{x^3}\left(\frac{1}{x}f_1(x)^2-2 f_1(x)-1\right)f(x)\end{align} $$

While this is sort of a recursive formula, it's just getting more complicated and doesn't seem particularly useful. It's clear that it will always be some function multiplied by $f(x)$, but what that function is for each derivative isn't clear. What are some strategies for finding Taylor coefficients in general, and for this function in particular? I'd like to find an expression in terms of a sum or something, or at least a recursion relation.

Note that I suspect (but have not proven) that this function is smooth but non-analytic at the origin, considering its relation to the famous function $e^{-\frac{1}{x}}$. I don't know if that's relevant in terms of how difficult its Taylor series is to find.

Answer 1

We have, for $|x|<1$, \begin{align*} f(x) &= \exp\! \bigg( {\frac{{\log (1 + (x - 1))}}{{1 + (x - 1)}}} \bigg) = \exp\! \bigg( {\sum\limits_{n = 1}^\infty {\bigg( {\sum\limits_{k = 1}^n {\frac{{( - 1)^{k + 1} }}{k}( - 1)^{n - k} } } \bigg)(x - 1)^n } } \bigg) \\ & = \exp\! \bigg( {\sum\limits_{n = 1}^\infty {( - 1)^{n + 1} H_n (x - 1)^n } } \bigg) = \sum\limits_{n = 0}^\infty {B_n (1, \ldots ,( - 1)^{n + 1} n!H_n )\frac{{(x - 1)^n }}{{n!}}} , \end{align*} where $H_n$ are the harmonic numbers and $B_n$ are the complete exponential Bell polynomials.

Alternatively, if we set $$ \exp\! \bigg( {\sum\limits_{n = 1}^\infty {( - 1)^{n + 1} H_n (x - 1)^n } } \bigg) = \sum\limits_{n = 0}^\infty {a_n (x - 1)^n } $$ then $a_0=1$ and $$ a_n = \frac{1}{n}\sum\limits_{k = 1}^n {( - 1)^{k + 1} kH_k a_{n - k} } $$ for $n\ge 1$. The OEIS entry $\text{A}008405$ provides the following formula in terms of the Stirling numbers of the first kind: $$ a_n = \frac{1}{n!}\sum\limits_{k = 1}^n {s(n,k)\sum\limits_{j = 0}^k {( - j)^{k - j} \binom{k}{j}} } . $$

Answer 2

Since Taylor series of $e^x$ is uniformly convergent we have that Taylor series of $e^{\log(x)/x} = \sum_i \frac{\log(x)^i}{x^i i !}$

Now apply the Taylor series of $\log(x)$ around $1$ in the above series and you get the Taylor series of $x^{1/x}$ around $x=1$.

Or the coefficients can be obtained from using the formula in Understanding an aplication of Faà di Bruno's formula