I have been wrestling with the following problem for some time now. If possible, I would prefer a hint rather than a full solution, because I would like to "solve" it myself.


Let $f(z)$ be a power-series and $[z^n]\{-\}$ denote the $n$'th coefficient. Show that the following holds, whenever $[z^0]{f(z)}=1$:


$$
\exp\left[\sum\_{n,m>0}\sum\_{j>0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\} \frac{q^{n+m}}{(nm)}\right] = \sum\_{n\geq 0}[z^n]\{f(z)^{n-1}\}q^n\,.
$$


**Remark:** This identity appears in comparing two generating series of the same geometric invariants computed using different methods.


**Edit 1:** Alternatively, if one wishes to get rid of the exponential, one can take a logarithm of the equation and derivative with respect to $q$. Then we are left to solve the following:


$$
\left[\sum\_{n,m>0}\sum\_{j>0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\} \frac{(n+m)}{(nm)}q^{n+m}\right]\sum\_{l\geq 0}[z^{l}]\{f(z)^{l-1}\}q^l = \sum\_{n\geq 0}n[z^{n}]\{f(z)^{n-1}\}q^n\,.
$$
**Edit 2:** To avoid it being pointed out again (also see Timothy's answer). It is easy to rewrite the identity using Lagrange inversion (this is where this identity comes from in the first place). The issue appears when one is only allowed to sum over $n\in \mathbb{Z}\_{>0}$. Timothy used for this the notation $[q^{>0}]\{w(q)^{-j}\}$. I have been unable to do anything with this expression, hence the more complicated form above.



Comparing two power-series
I decided to summarize the two main proofs that I liked. The first one was motivated by the answer by esg. The second one can be found in the answer of Alex Gavrilov and is made more explicit. I am very grateful for their help.


**1. proof:**
Using Gessel [(2.4.4)](https://arxiv.org/abs/1609.05988) and the unique solutions $H(q)=qf(q)$, we can write


\begin{align\*}
&\sum\_{j>0}\sum\_{n,m>0}[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\}\frac{p^n}{n}\frac{q^m}{m}= -\sum\_{j>0}\frac{1}{j}H^j(p)H^{-j}(q) \\
&-\sum\_{j>0}\sum\_{-j\leq n<0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\}\frac{p^n}{n}\frac{q^m}{m}-\sum\_{j>0}[t^{-1}]\big\{t^{-j-1}\log\Big(R(t)\Big)\big\}H^j \\
&=\log\Big(1-\frac{H(p)}{H(q)}\Big)+ \sum\_{n>0}\sum\_{j\geq n}[z^{j-n}]\big\{R^{-n}(z)\big\}H^jp^{-n}+\log\Big(R(H)\Big)
\end{align\*}
Therefore we obtain after taking exponential and limit $p\to q$:
$$
\lim\_{p\to q}\frac{H(p)-H(q)}{p-q}\frac{1}{R\big(H(q)\big)}\frac{p}{H(p)} =\Big(\frac{dq}{dH}\Big)^{-1}\Big(\frac{q}{H(q)}\Big)^2=\frac{1}{f\big(H(q)\big)}\Big(1-qf'(H)\Big)\,.
$$
The last term is clearly the right hand side by the Lagrange inversion formula.


**2. proof**


Without loss of generality assume that $f(q)$ is a polynomial in $q$. We will use the following identity
\begin{align\*}
[q^{>0}]H^{-j}(q)&=\sum\_{n>0}[z^n]\big\{H^{-j}(z)\big\}q^n = \frac{1}{2\pi i} \oint\_{|z|=R}\sum\_{n>0}\Big(\frac{q}{z}\Big)^n\frac{1}{z}H^{-j}(z)dz\\
 &= \frac{1}{2\pi i} \oint\_{|z|=R}\Big(\frac{1}{z-q}-\frac{1}{z}\Big)H^{-j}(z)dz\,.
\end{align\*}
Then we get
$$
\sum\_{j>0}\frac{1}{j}H^j(q)[q^{>0}]H^{-j}(q)=\frac{1}{2\pi i}\oint\_{|z|=R}\Big(\frac{1}{z-q}-\frac{1}{z}\Big)\log\Big(\frac{H(z)}{H(z)-H(q)}\Big)dz
$$
Note that
\begin{align\*}
&\frac{1}{2\pi i}\oint\_{|z|=R}\Big(\frac{1}{z-q}-\frac{1}{z}\Big)\log\Big(\frac{z}{z-q}\Big)dz\\
&= -\int\_{0}^{2\pi}\Big(\frac{\frac{r}{R}e^{2\pi i(\theta-\tau)}}{1-\frac{r}{R}e^{2\pi i(\theta-\tau)}}\Big)\log\Big(1-\frac{r}{R}e^{2\pi i(\theta-\tau)}\Big)d\tau=0\,,
\end{align\*}
where the integral vanishes, because it is proportional to the integral of a total derivative of a $2\pi$ periodic function.


Therefore, we may work instead with
$$
\frac{1}{2\pi i}\oint\_{|z|=R}\Big(\frac{1}{z-q}-\frac{1}{z}\Big)\log\Big(\frac{H(z)\big(z-q\big)}{z\big(H(z)-H(q)\big)}\Big)dz\,.
$$
Taking the residues of this integrand at $z=q$ and $z=0$, we obtain again
$$
-2\log\Big(\frac{H(q)}{q}\Big)-\log\big(H'(q)\big)\,.
$$


