Power series raised to a non integer power

I would like to know whether there is a general formula to raise a power series to a non integer exponent:

$$\left(\sum_{k = 0}^{+\infty} a_k x^k\right)^\alpha, \text{ where } \alpha \in \mathbb{R}.$$

I have seen previous posts related to an integer power: integer power 1, integer power 2, integer power 3. I have also come accross to the following couple of posts that might be closely related to my question: power series^power series and fractional exponent.

It seems that an answer to this problem could involve a generalisation of the multinomial theorem, maybe in the like of this.

EDIT : I am also interested in the corresponding questions of convergence!

There is a slight ambiguity in your question since you do not explain what kind of formula you would like to see: however, since it seems, from the references you cited, that you mean a power series formula, you can build it by using Newton's generalization of the binomial theorem $$t\mapsto (1+t)^\alpha = \sum_{k=0}^\infty \binom{\alpha}{k} t^k.\label{1}\tag{1}$$ By using it, you have that $$\left(\sum_{k=0}^\infty a_kx^k\right)^{\alpha}= \begin{cases} \displaystyle \sum_{n=0}^\infty \binom{\alpha}{n} \left[\sum_{k=1}^\infty a_kx^k -1\right]^n & a_0=0\\ \\ \displaystyle a_0^\alpha\sum_{n=0}^\infty \binom{\alpha}{n} \left[\sum_{k=1}^\infty a_kx^k\right]^n & a_0\neq0\\ \end{cases}\label{2}\tag{2}$$ and you could further expand this formula by using the generalization of Cauchy Product formula. However, the computational work involved in the whole procedure seems daunting.

The convergence properties of \eqref{2} can be inferred from the convergence properties of \eqref{1}, listed in the related wikipedia entry, and from the properties of the power series $$s(x) = \sum_{k=1}^\infty a_k x^k$$: call $$D(R)$$ the convergence disk (or interval, if $$x, a_k\in\Bbb R$$ for all $$n\in\Bbb N$$) of $$s(x)$$, and $$C\big(s^\alpha(x)\big)$$ the convergence region of \eqref{2}, i.e the largest open set in $$D$$ where \eqref{2} converges. Since \eqref{1} converges absolutely for any real $$\alpha$$ if $$|t|<1$$ then $$C\big(s^\alpha(x)\big) = \begin{cases} \big\{x\in D(R) : |s(x)-1|<1\big\} & a_0=0 \\ \\ \big\{x\in D(R) : |s(x)-a_0|<{a_0}\big\} & a_0\neq0 \\ \end{cases}$$ In a similar fashion, you can analyze the boundary of $$C\big(s^\alpha(x)\big)$$ by the convergence properties of \eqref{1} on the boundary $$|t|=1$$ of its convergence disk.

1. Following a comment of the Asker, let's prove the following formula $$s^n(x)=\left(\sum_{k=1}^{\infty} a_k x^k\right)^n = \sum_{k=1}^{\infty} \sum_{\substack{0< k_1, \ldots, k_n\le k \\ k_1+\ldots+k_n=k}}a_{k_1} \cdots a_{k_n} x^k \quad\forall n\in\Bbb N,\, n\ge 2\label{3}\tag{3}$$ Let's proceed by induction

• For $$n=2$$ equation \eqref{3} is simply the Cauchy Product of the power series $$s(x)$$ by itself.
• Assuming that \eqref{3} holds true for a general $$n\in\Bbb N$$ with $$n\ge 2$$, lets see if it holds for $$n+1$$. By using Cauchy's product formula, $$\begin{split} \left(\sum_{k=1}^{\infty} a_k x^k\right)^{n+1} & = \left(\sum_{k=1}^{\infty} a_k x^k\right)\cdot\left(\sum_{k=1}^{\infty} a_k x^k\right)^n \\ &= \Bigg( \sum_{k=1}^{\infty} a_k x^k\Bigg)\Bigg(\sum_{k=1}^{\infty} \sum_{\substack{0< k_1, \ldots, k_n\le k \\ k_1+\ldots+k_n=k}}a_{k_1} \cdots a_{k_n} x^k\Bigg)\\ &= \sum_{k=1}^{\infty} \Bigg(\sum_{m=1}^{k} a_m \sum_{\substack{0< k_1, \ldots, k_n\le k-m \\ k_1+\ldots+k_n=k-m}}a_{k_1} \cdots a_{k_n} \Bigg)x^k\\ &= \sum_{k=1}^{\infty} \Bigg(\sum_{\substack{0< k_1, \ldots, k_n, k_{n+1}\le k \\ k_1+\ldots+k_n + k_{n+1}=k}}a_{k_1} \cdots a_{k_n} a_{k_{n+1}}\Bigg)x^k \end{split}$$ and the validity of the formula for $$n$$ implies its validity for $$n+1$$.

Thus formula \eqref{3} is proven true for every $$n\in\Bbb N$$ by induction.

2. A final note. There are several ways of representing the composition of functions $$\left(\sum_{k=0}^\infty a_kx^k\right)^{\alpha}$$ for a real $$\alpha$$: for example it is possible to express it by using Faa di Bruno's formula in its various flavors. As stated in the comments, If you know the derivative of the composition of two analytic functions you know the Taylor series of their composition since this is likewise an analytic function. In our specific case we have $$s^\alpha(x)=\sum_{k=0}^\infty \frac{1}{k!}\frac{\mathrm{d}^k}{\mathrm{d}x^k}s^\alpha(x)\big|_{x=0} x^k$$ and, as remarked also in the Wikipedia entry on the said formula, you can calculate the Taylor expansion coefficients directly.
What is the "most convenient" way of proceeding? It depends heavily on the structure of the power series $$s(x)$$ and on on your purpose: in a word it depends on the context of your application.

• Thanks a lot, that's quite helpful !! Can we say that $\left(\sum_{k = 1}^{+\infty} a_k x^{k}\right)^{n} = \sum_{k = 1}^{+\infty} \left(\sum_{\substack{0 < r_1,\ldots,r_l \leq k \\ r_1 + \cdots + r_l = k}}a_{r_1} \cdots a_{r_l}\right)x^{k}$? Then we could go on with interchanging the sums. – maxbo Dec 5 '20 at 23:59
• @maxbo I've added a section which gives a formula for $s^n(x)$ for $n\ge 2$: interchanging the summations can be done provided the absolute convergence of the power series, which surely holds if $x\in C\big(s^n(x)\big)$. – Daniele Tampieri Dec 6 '20 at 13:27
• @maxbo the composition of analytic functions is analytic, therefore if you know an expression of the derivative of their composition, you know the Taylor series of their composition: $$f(g(x))=\sum_{k=0}^\infty \frac{1}{k!}\frac{\mathrm{d}^k}{\mathrm{d}x^k}f(g(x))\big|_{x=0} x^k$$ I'll add something about it in the "final note" above. – Daniele Tampieri Dec 8 '20 at 12:17
• Thanks again! My point is that $f(x) = x^\alpha$ is not analytic at $x = 0$ if $\alpha \notin \mathbb{N}$. But I think in this case we may do the following: $(\sum_{k = 0}^\infty a_k x^k)^\alpha = a_0^\alpha (1 + \sum_{k = 1}^\infty b_k x^k)^\alpha$ where we define $b_k = a_k/a_0$ provided $k \geqslant 1$ and $a_0 \neq 0$. Finally, Faa di Bruno's formula can be applied with $f(x) = (1 + x)^\alpha$. – maxbo Dec 8 '20 at 19:53
• Sorry if I am repeating exactly what you had in mind ! Thanks for the great answer – maxbo Dec 8 '20 at 20:06