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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.

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    $\begingroup$ Try looking up Cauchy-products. For example this article explains multiplication of taylor series $\endgroup$
    – Eoin
    Jul 4 '15 at 23:36
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    $\begingroup$ It's always good to do sanity checks. Try this out with c_n = 0 for n > 1, c_0 = c_1 = 1, and p = 2. What does your equality tell you in this case? $\endgroup$
    – Rikimaru
    Jul 4 '15 at 23:36
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    $\begingroup$ Unfortunately the power series $1+x$ squared is $1+2x + x^2$, not $1+x^2$, so this result is not correct. $\endgroup$ Jul 4 '15 at 23:38
  • $\begingroup$ Does this imply that $$ \sum_{n_1=0}^\infty \sum_{n_2=0}^\infty \cdots \sum_{n_p=0}^\infty \left(\prod_{p=1}^\infty c_{n_p}x^{n_p}\right) $$ is as far as one can go? $\endgroup$ Jul 4 '15 at 23:51
  • $\begingroup$ To answer the title: With difficulty. $\endgroup$ Jul 5 '15 at 11:28
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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.

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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 $$b_0=a_0^2$$ $$b_1=2\,a_0\,a_1$$ $$b_2=2\,a_0\, a_2+a_1^2$$ $$b_3=2\,a_0\, a_3+2\,a_1\, a_2$$ $$b_4=2\,a_0\, a_4+2\,a_1\, a_3+a_2^2$$ $$b_5=2\,a_0\, a_5+2\,a_1\, a_4+2\,a_2\, a_3$$

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  • $\begingroup$ (1) Feng Qi, Chao-Ping Chen, and Dongkyu Lim, Several identities containing central binomial coefficients and derived from series expansions of powers of the arcsine function, Results in Nonlinear Analysis 4 (2021), no. 1, 57--64; available online at doi.org/10.53006/rna.867047. (2) Bai-Ni Guo, Dongkyu Lim, and Feng Qi, Series expansions of powers of arcsine, closed forms for special values of Bell polynomials, and series representations of generalized logsine functions, AIMS Mathematics 6 (2021), no. 7, 7494--7517; available online at doi.org/10.3934/math.2021438. $\endgroup$
    – qifeng618
    Sep 28 at 10:58

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