What is the series expansion of $ \sqrt[x] x $?

Given $\,f(x) := x^{1/x},\,$ assuming $\,x>0.\,$ Let $\,t>0,\,$ let $\,u := -\log(t),\,$ and $\, z := (x-t)/t^2.\,$ Expand $\,f(x)\,$ into a power series at $\,x=t\,$ getting $$ \frac{f(x)}{f(t)} = \sum_{n=0}^\infty a_n \frac{z^n}{n!} = 1 + (1+u)z + (u^2 + 2(1-t)u +(1-3t))\frac{z^2}{2!} +\dots$$ where $\,a_n\,$ is a polynomial in $\,t,u.\,$ For example, $$ a_3 = u^3 + 3(1-2t)u^2 +3(1-5t+2t^2)u +(1-9t+11t^2).$$

Note that the choice of $\,t\,$ is arbitrary and you can let it be $\,t=1,\,$ for example, but I have not yet determined the radius of convergence which is an important detail since it determines where the expansion is valid. It is known that all derivatives of $f(x)$ are $0$ at $x=0$ and so the Taylor series sums to $0$ and is not valid for $x>0.$ The plot of function $\,f(x)\,$ is:

enter image description here

Note also that the terms in the power series expansion of the function $\,f(x) = \exp(\log(x)/x)\,$ all had a factor of $\,f(t)\,$ which is why I factored it out in my expansion. Check this by finding succesive derivatives of $\,f(x).$ as follows.

Let $\,f(x) = \exp(g(x))\,$ where $\,g(x) = \log(x)/x.$ Let $\,d_n := \left(\frac{d}{dx}\right)^n f(x)\,$ be the $n$-th derivative of $\,f(x)\,$ so $\,d_0 = f(x)\,$ by definition. Next $\,d_1 = f(x)\,g'(x)\,$ where $\,g'(x) = (1 - \log(x))/x^2.\,$ Next $\,d_2 = f(x)(g'(x)^2 + g''(x)),\,$ and so on. By induction $\,d_n\,$ is a polynomial in $\,x\,$ and $\,\log(x)\,$ all divided by $\,x^{2n}.\,$


Easier than I thought.

We can write $ x ^ \frac {1}{x} $ as:

$$ x ^ \frac {1}{x}= e ^ { \frac {ln x} {x}} $$ The expansion of $e^x$ is: $$ e^x=\sum_{n=0}^\infty \frac {x^n}{n!}$$ Similarly the expansion of $e ^ { \frac {ln x} {x}}$ is: $$ e ^ { \frac {ln x} {x}}=\sum_{n=0}^\infty \frac {(\ln x)^n}{x^n.n!} $$

Tags:

Integration

Calculus

Power Series

Indefinite Integrals

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