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It is easy to get a expression for the nth-derivative of an inverse function ; starting from $(f^{-1})'=\frac{1}{f'\circ f^{-1}}$, one gets things like $(f^{-1})^{(n)}=\frac{\sum a_k\prod (f^{(n_j)}\circ f^{-1})^j}{(f'\circ f^{-1})^{2n-1}}$, with reasonably easy constraints on the $n_j$. But what are the values of the $a_k$? I believe I read somewhere this was an application of umbral calculus, but I don't see how, and inverting Faa di Bruno's formula on the identity $f\circ f^{-1}=id$ don't seem to get anywhere.

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  • $\begingroup$ Go to the Online Encyclopedia of Integer Sequences (OEIS) on the Net and search under Lagrange inversion and also series reversion and you will find many examples. $\endgroup$ – Tom Copeland Jun 1 '12 at 7:48
  • $\begingroup$ The survey “Lagrange inversion” by Ira Gessel is probably worth mentioning in the context of this question (especially as the author posted one of the answers to this question, but before the survey in question was written). $\endgroup$ – Gro-Tsen Dec 22 '19 at 22:34
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Riordan's Combinatorial identities has a chapter on partition polynomials that may be helpful. It specifically covers the question you are asking, but is in umbral calculus.

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See Warren P. Johnson, Combinatorics of Higher Derivatives of Inverses, American Mathematical Monthly, Vol. 109, No. 3 (Mar., 2002), pp. 273-277, http://www.jstor.org/stable/2695356

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To precise my question, I was asking for the exact values of the $a_k$. Thanks to Tom Copeland, I could find the sequence A176740 of OEIS, giving a complete answer (with useful links) to this problem.

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You should be able to get a formula, first by reducing to the case where f(0)=0 and the evaluation of the derivatives (for both f and its inverse) is at 0. Then, work formally by replacing f by its Taylor-MacLaurin series at 0. The problem then becomes that of the reversion of power series. It has been done in many places and typically involves summing over trees.

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This is sometime called the Lagrange inversion formula.

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My answer contains partial exponential Bell polynomials.

Because the problem concerns higher derivatives of compositions of functions, it can be answered by using partial exponential Bell polynomials.

Set $f^{-1}=\phi$.
Let $B$ denote partial exponential Bell polynomials.

Applying Faà di Bruno's formula (higher chain rule of differentiation) to the inverse rule of differentiation $\phi'=\frac{1}{f'\circ \phi}$, we get

$$\phi^{(n)}=\sum_{k=0}^{n-1}(-1)^kk!(f'\circ\phi)^{-(k+1)}B_{n-1,k}(f'\circ\phi).$$

That means, we still need an expression for $B_{n-1,k}(f'\circ\phi)$.
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The usual way is to use Lagrange inversion theorem instead. Lagrange inversion theorem gives the Taylor series expansion of the inverse function of an analytic function.

The inverse of an analytic function $f$ in a neighbourhood of a point $a$ can be written as a power series iff $f'(a)\neq 0$.

For $f(x)=\sum_{n=0}^\infty\frac{f_n}{n!}x^n$, $\phi(x)=\sum_{n=0}^\infty\frac{\phi_n}{n!}x^n$ and $f_0=0$, Lagrange inversion theorem gives

$$\phi_n=(n-1)!\left[x^{n-1}\right]\left(\frac{x}{f(x)}\right)^n.$$

This is sequence A176740 in the Online Encyclopedia of Integer Sequences (OEIS). See the other answers.

Clearly you can derive this also with umbral algebra.
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Let's look at the representation with Bell polynomials.

Applying Faà di Bruno's formula to $\left(\frac{x}{f(x)}\right)^n$ gives for $n\ge 2$

$$\phi_n=(n-1)!\frac{1}{f_1^n}\sum_{k=1}^{n-1}(-1)^k(n)_kB_{n-1,k}\left(\frac{f_2}{2f_1},\frac{f_3}{3f_1},...,\frac{f_{n-k+1}}{(n-k+1)f_1}\right)$$

with $(n)_k$ the rising factorial (Pochhammer symbol).

This is already written in the Wikipedia article for Lagrange inversion theorem.

$$B_{n,k}(f_1,f_2,...,f_n)=\sum_{1k_1+2k_2+...+(n-k+1)k_{n-k+1}=n\atop k_1+k_2+...+k_{n-k+1}=k}\frac{n!}{1!^{k_1}k_1!2!^{k_2}k_2!\cdot ...\cdot (n-k+1)!^{k_{n-k+1}}k_{n-k+1}!}f_1^{k_1}f_2^{k_2}\cdot ...\cdot f_{n-k+1}^{k_{n-k+1}}$$

$$B_{n-1,k}\left(\frac{f_2}{2f_1},\frac{f_3}{3f_1},...,\frac{f_{n-k}}{(n-k)f_1}\right)$$ $$=\sum_{1k_1+2k_2+...+(n-k)k_{n-k}=n-1\atop k_1+k_2+...+k_{n-k}=k}\frac{(n-1)!}{1!^{k_1}k_1!2!^{k_2}k_{2}!\cdot ...\cdot (n-k)!^{k_{n-k}}k_{n-k}!}\left(\frac{f_2}{2f_1}\right)^{k_1}\left(\frac{f_3}{3f_1}\right)^{k_2}\cdot ...\cdot \left(\frac{f_{n-k}}{(n-k)f_1}\right)^{k_{n-k}}$$ $$=\frac{1}{f_1^k}\sum_{1k_1+2k_2+...+(n-k)k_{n-k}=n-1\atop k_1+k_2+...+k_{n-k}=k}\frac{(n-1)!}{1!^{k_1}k_1!2!^{k_2}k_2!\cdot ...\cdot (n-k)!^{k_{n-k}}k_{n-k}!}\frac{1}{2^{k_1}3^{k_2}(n-k)^{k_{n-k}}}f_2^{k_1}f_3^{k_2}\cdot ...\cdot f_{n-k}^{k_{n-k}}$$

Now we have an explicit formula for $\phi_n$.

We could also explore $B_{n-1,k}\left(\frac{f_2}{2f_1},\frac{f_3}{3f_1},...,\frac{f_{n-k}}{(n-k)f_1}\right)$ to try to get some recurrence formulas.

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I. G. Macdonald gives a very explicit formula for the coefficients in Example 24, p35 of the 2nd Edition of Symmetric Functions and Hall Polynomials. The example begins `Another involution on the ring $\Lambda$ may be defined as follows...'

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