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By an odd coincidence, I stumbled upon a second question in as many weeks about power series, and once again the only way I know how to prove the result is by complex methods; once again, I am leaving it here as a challenge to any interested readers, and I would be particularly interested in knowing of a proof that was not based on complex analysis (or thinly disguised versions thereof), or for a reference to previous literature where something like this identity has occured. (I suspect for instance that something like this may have shown up before in free probability, based on the answer to part (ii) of the problem.)

Here is a purely algebraic form of the problem:

Problem 1 Let ${F = F(z)}$ be a formal function of one variable ${z}$. Suppose that ${G = G(z)}$ is the formal function defined by

$\displaystyle G := \sum_{n=1}^\infty \left( \frac{F^n}{n!} \right)^{(n-1)}$

$\displaystyle = F + \left(\frac{F^2}{2}\right)' + \left(\frac{F^3}{6}\right)'' + \dots$

$\displaystyle = F + FF' + (F (F')^2 + \frac{1}{2} F^2 F'') + \dots,$

where we use ${f^{(k)}}$ to denote the ${k}$-fold derivative of ${f}$ with respect to the variable ${z}$.

• (i) Show that ${F}$ can be formally recovered from ${G}$ by the formula

$\displaystyle F = \sum_{n=1}^\infty (-1)^{n-1} \left( \frac{G^n}{n!} \right)^{(n-1)}$

$\displaystyle = G - \left(\frac{G^2}{2}\right)' + \left(\frac{G^3}{6}\right)'' - \dots$

$\displaystyle = G - GG' + (G (G')^2 + \frac{1}{2} G^2 G'') - \dots.$

• (ii) There is a remarkable further formal identity relating ${F(z)}$ with ${G(z)}$ that does not explicitly involve any infinite summation. What is this identity?

To rigorously formulate part (i) of this problem, one could work in the commutative differential ring of formal infinite series generated by polynomial combinations of ${F}$ and its derivatives (with no constant term). Part (ii) is a bit trickier to formulate in this abstract ring; the identity in question is easier to state if ${F, G}$ are formal power series, or (even better) convergent power series, as it involves operations such as composition or inversion that can be more easily defined in those latter settings.

To illustrate Problem 1(i), let us compute up to third order in ${F}$, using ${{\mathcal O}(F^4)}$ to denote any quantity involving four or more factors of ${F}$ and its derivatives, and similarly for other exponents than ${4}$. Then we have

$\displaystyle G = F + FF' + (F (F')^2 + \frac{1}{2} F^2 F'') + {\mathcal O}(F^4)$

and hence

$\displaystyle G' = F' + (F')^2 + FF'' + {\mathcal O}(F^3)$

$\displaystyle G'' = F'' + {\mathcal O}(F^2);$

multiplying, we have

$\displaystyle GG' = FF' + F (F')^2 + F^2 F'' + F (F')^2 + {\mathcal O}(F^4)$

and

$\displaystyle G (G')^2 + \frac{1}{2} G^2 G'' = F (F')^2 + \frac{1}{2} F^2 F'' + {\mathcal O}(F^4)$

and hence after a lot of canceling

$\displaystyle G - GG' + (G (G')^2 + \frac{1}{2} G^2 G'') = F + {\mathcal O}(F^4).$

Thus Problem 1(i) holds up to errors of ${{\mathcal O}(F^4)}$ at least. In principle one can continue verifying Problem 1(i) to increasingly high order in ${F}$, but the computations rapidly become quite lengthy, and I do not know of a direct way to ensure that one always obtains the required cancellation at the end of the computation.

Problem 1(i) can also be posed in formal power series: if

$\displaystyle F(z) = a_1 z + a_2 z^2 + a_3 z^3 + \dots$

is a formal power series with no constant term with complex coefficients ${a_1, a_2, \dots}$ with ${|a_1|<1}$, then one can verify that the series

$\displaystyle G := \sum_{n=1}^\infty \left( \frac{F^n}{n!} \right)^{(n-1)}$

makes sense as a formal power series with no constant term, thus

$\displaystyle G(z) = b_1 z + b_2 z^2 + b_3 z^3 + \dots.$

For instance it is not difficult to show that ${b_1 = \frac{a_1}{1-a_1}}$. If one further has ${|b_1| < 1}$, then it turns out that

$\displaystyle F = \sum_{n=1}^\infty (-1)^{n-1} \left( \frac{G^n}{n!} \right)^{(n-1)}$

as formal power series. Currently the only way I know how to show this is by first proving the claim for power series with a positive radius of convergence using the Cauchy integral formula, but even this is a bit tricky unless one has managed to guess the identity in (ii) first. (In fact, the way I discovered this problem was by first trying to solve (a variant of) the identity in (ii) by Taylor expansion in the course of attacking another problem, and obtaining the transform in Problem 1 as a consequence.)

The transform that takes ${F}$ to ${G}$ resembles both the exponential function

$\displaystyle \exp(F) = \sum_{n=0}^\infty \frac{F^n}{n!}$

and Taylor’s formula

$\displaystyle F(z) = \sum_{n=0}^\infty \frac{F^{(n)}(0)}{n!} z^n$

but does not seem to be directly connected to either (this is more apparent once one knows the identity in (ii)).