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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 $[z^0]{f(z)}=1$:

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

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:

$$ \Big[\sum_{n,m>0}\sum_{j>0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\} \frac{(n+m)}{(nm)}q^{n+m}\Big]\sum_{l\geq 0}[z^{l}]\{f(z)^{l-1}\}q^l = \sum_{n\geq 0}n[z^{n}]\{f(z)^{n-1}\}q^n\,. $$ 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.

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.

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 $[z^0]{f(z)}=1$:

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

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:

$$ \Big[\sum_{n,m>0}\sum_{j>0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\} \frac{(n+m)}{(nm)}q^{n+m}\Big]\sum_{l\geq 0}[z^{l}]\{f(z)^{l-1}\}q^l = \sum_{n\geq 0}n[z^{n}]\{f(z)^{n-1}\}q^n\,. $$ 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 identity $[q^{>0}]\{w(q)^{-j}\}$. I have been unable to do anything with this expression, hence the more complicated form above.

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 $[z^0]{f(z)}=1$:

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

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:

$$ \Big[\sum_{n,m>0}\sum_{j>0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\} \frac{(n+m)}{(nm)}q^{n+m}\Big]\sum_{l\geq 0}[z^{l}]\{f(z)^{l-1}\}q^l = \sum_{n\geq 0}n[z^{n}]\{f(z)^{n-1}\}q^n\,. $$ 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.

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 $[z^0]{f(z)}=1$:

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

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:

$$ \Big[\sum_{n,m>0}\sum_{j>0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\} \frac{(n+m)}{(nm)}q^{n+m}\Big]\sum_{l\geq 0}[z^{l}]\{f(z)^{l-1}\}q^l = \sum_{n\geq 0}n[z^{n}]\{f(z)^{n-1}\}q^n\,. $$ 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 identity $[q^{>0}]\{w(q)\}$. I have been unable to do anything with this expression, hence the more complicated form above.

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 $[z^0]{f(z)}=1$:

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

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:

$$ \Big[\sum_{n,m>0}\sum_{j>0}j[z^{n+j}]\{f(z)^n\}[z^{m-j}]\{f(z)^m\} \frac{(n+m)}{(nm)}q^{n+m}\Big]\sum_{l\geq 0}[z^{l}]\{f(z)^{l-1}\}q^l = \sum_{n\geq 0}n[z^{n}]\{f(z)^{n-1}\}q^n\,. $$ 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 identity $[q^{>0}]\{w(q)^{-j}\}$. I have been unable to do anything with this expression, hence the more complicated form above.