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Suppose we have a formal power series $h(t)=t+\sum_{k=2}^\infty h_k(t^k/k!)$. In principle, this can be inverted to obtain $g(x)=x+\sum_{k=2}^\infty g_k(x^k/k!)$ such that $h(g(x))=x$. The specific expressions are known in terms of Bell polynomials and can be found on Wikipedia's page on the Lagrange inversion theorem.

Now, suppose we have another formal power series $y(t)=t+\sum_{k=1}^\infty y_k (t^k/k!)$ with $y_1=1$. Then we can eliminate the $t$-dependence via the above reversion of series, yielding another formal power series $f(x)=(y\circ g)(x)=x+\sum_{k=2}^\infty f_k (x^k/k!)$. Are the expressions for the coefficients $\{f_k\}$, in terms of $\{y_k\}$ and $\{x_k\}$, known and documented somewhere?

Semiclassical
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  • Your notation has confused me somewhat, but it seems that the fact that $x$ and $g$ are inverses (reversions) of each other does not seem to matter. Rather, you seem to be asking about the coefficients of $y\circ g$ in terms of the coefficients of $y$ and of $g$. If this is not the case, could you please clarify? – Lubin Jun 28 '16 at 00:16
  • You are going to have to use the Bell polynomials twice: first in the Lagrangian inversion, and then in your use of Faà di Bruno to get the coefficients of the composition. Nothing simpler comes to mind at the moment. – J. M. ain't a mathematician Jun 28 '16 at 01:55
  • Is your question: given a power series $,y(t) = t + y_2 t^2+\cdots,,$ how to expand a power series $,f(t),$ in terms of $y$, i.e. how to compute $,f_i,$ in $,f(t) = f_0 + f_1 y(t) + f_2 y(t)^2 +, \cdots, ?\ $ – Bill Dubuque Jun 28 '16 at 01:58
  • @Lubin Noted. I've clarified the notation, I hope. – Semiclassical Jun 28 '16 at 02:03
  • @BillDubuque Correct, with the assumption that $f(0)=0$ (and hence $f'(0)=1$ by comparing the two series.) I've edited the question somewhat, so hopefully that's a bit clearer now. – Semiclassical Jun 28 '16 at 02:05
  • @Semiclassical Ok, thanks for clarifying. One approach to such expansions is via the Umbral Calculus - which leads to nice closed forms in many particular cases - see my answer. – Bill Dubuque Jun 28 '16 at 02:10
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    Yes, you'd need to read the first few pages of any exposition of the Umbral Calculus to grok it. It would be too long to reproduce in an answer. Iiirc there are some good expositions online. – Bill Dubuque Jun 28 '16 at 02:12

2 Answers2

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An elegant method is to use the Expansion Theorem in Umbral Calculus. Below is a typical statement, from p. 18 of Roman's book The Umbral Calculus.

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Bill Dubuque
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  • Thanks for this answer, though I think I'll have to stare at it quite a bit (and do some extra reading) before I entirely appreciate it. – Semiclassical Jun 28 '16 at 02:10
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Roman, St. M.; Rota, G.-C.: The Umbral Calculus. Adv. Math. 27 (1978) 95-188:
"$f(t)=\sum_{k=0}\frac{a_k}{k!}t^k$
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When $a_0=0$ and $a_1\neq 0$ we call $f(t)$ a delta series.
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Given two delta series $f(t)$ and $g(t)$, the nth coefficient in $g(f^{-1}(t))$, multiplied by $n$, equals the $(n-1)$st coefficient in $g'(t)\left(\frac{f(t)}{t}\right)^{-n}$."

IV_
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