How can I raise a Taylor Series to a power?

Question

I have recently been undertaking the challenge of finding the antiderivative of $x^x$. In doing so, I have come across the idea of raising a Taylor series to a variable exponent. I came to the following conclusion:

$$(\sum_{n=0}^\infty c_nx^n)^p = \sum_{{n_1}=0}^\infty \sum_{{n_2}=0}^\infty \cdots \sum_{{n_p}=0}^\infty c_{n_1}x^{n_1} c_{n_2}x^{n_2} \cdots c_{n_p}x^{n_p}.$$

Now, for multiplying two different Taylor series, this was as far as I could get, but assuming that it is a single Taylor series (as is the case for raising one to an exponent), I believe that one has:

$$c_{n_1}x^{n_1} = c_{n_2}x^{n_2} = \cdots = c_{n_p}x^{n_p}.$$

Following from this, one would have the result:

$$\left(\sum_{n=0}^\infty c_nx^n\right)^p = \sum_{n=0}^\infty \left(c_nx^n\right)^p.$$

Is my logic mathematically sound, and is this a proper result? I have tried Googling this, but no results have come up. Thank you for your help.

Answer 1

Recently, I had a similar problem.

The generating function for Bell polynomials seems to give the answer immediately:

$$ \left( \sum_{i=1}^\infty a_i x^i \right)^n = \sum_{m=n}^\infty B_{m,n}(a_1,a_2,\ldots, a_{m-n+1})x^m, $$

where $B_{m,n}$ are the partial Bell polynomials.

Answer 2

Suppose that you have known series $$f=\sum_{i=0}^\infty a_i x^i$$ and that you want to express $$f^2=\sum_{i=0}^\infty b_i x^i$$ and you want to express the $b_i$'s as functions of the $a_i$'s, it is slightly more complex since $$\sum_{i=0}^\infty b_i x^i=\Big(\sum_{i=0}^\infty a_i x^i\Big)\times \Big(\sum_{j=0}^\infty a_j x^j\Big)$$ For a given power $k$ in the lhs, there are many terms to be used in the rhs (the sums of powers of which being equal to $k$).

This will make $$b_k=\sum_{i=0}^k a_i\, a_{k-i}$$ For example $$b_0=a_0^2$$ $$b_1=2\,a_0\,a_1$$ $$b_2=2\,a_0\, a_2+a_1^2$$ $$b_3=2\,a_0\, a_3+2\,a_1\, a_2$$ $$b_4=2\,a_0\, a_4+2\,a_1\, a_3+a_2^2$$ $$b_5=2\,a_0\, a_5+2\,a_1\, a_4+2\,a_2\, a_3$$