Return to Answer

As in Timothy Budd's answer let $w=w(q)$ denote the (formal) solution of $q=\frac{w}{f(w)}$.

Let $p$ be another variable and consider the sum \begin{align*} S(p,q):=\sum_{n,m>0} \sum_{j>0} j[z^{n+j}]\{f(z)^n\} [z^{m-j}]\{f(z)^m\} \frac{p^n q^m}{nm} \end{align*}

Write it as \begin{align*} \sum_{n>0} \frac{p^n}{n}\Big(\sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j\Big) \end{align*}

and rewrite the inner sums as

\begin{align*} \sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j &=\frac{1}{w(q)^n}\Big(f(w(q))^n -\sum_{j=0}^{n-1}[z^j]\{f(z)^n\}w(q)^j\Big) -[z^n] \{f(z)^n\}\\ &=\frac{1}{q^n} -\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j} -[z^n] \{f(z)^n\} \end{align*}

Note that \begin{align*} \sum_{n>0} \frac{p^n}{n}[z^n]\{f(z)^n\}=-\log\big(f(w(p))\big) \end{align*} and that \begin{align*} \sum_{n>0} \frac{p^n}{nq^n}=-\log\big(1-\frac{p}{q}\big) \end{align*} The remaining sum can be written as \begin{align*} \sum_{n>0} \frac{p^n}{n}\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j}&=\sum_{j>0}w(q)^{-j}\sum_{n\geq j}\frac{1}{n}[z^{n-j}]\{f(z)^n\}p^n\\ &=\sum_{j>0}w(q)^{-j}\frac{w(p)^j}{j}\\ &=-\log\big(1-\frac{w(p)}{w(q)}\big) \end{align*} where we have used that for $n\geq j$ (by Lagrange-Bürmann) \begin{align*}\frac{1}{n}[z^{n-j}]\{f(z)^n\}=\frac{1}{n}[z^{n-1}]\{z^{j-1}f(z)^n\}=[q^n]\{\frac{w(q)^j}{j}\}\end{align*}

Thus $S(p,q)=-\log\big(1-\frac{p}{q}\big)+\log\big(1-\frac{w(p)}{w(q)}\big)-\log\big(f(w(p))$ and \begin{align*} \exp(S(p,q))=\frac{1}{f(w(p))}\,\frac{1-\frac{w(p)}{w(q)}}{1-\frac{p}{q}}\end{align*} Now \begin{align*} \frac{q-q\frac{w(p)}{w(q)}}{q-p}&=1+\frac{p}{f(w(q)}\frac{f(w(q))-f(w(p)}{q-p}\\ &=1 +\frac{p f^\prime(w(p))w^\prime(p)}{f(w(q))}+O(q-p)\\ &=1+\frac{f(w(p))}{f(w(q))}\frac{p f^\prime(w(p)}{1-pf^\prime(w(p))} +O(q-p) \end{align*}

so that $\exp(S(q,q)$ reduces to $$\exp(S(q,q))=\frac{1}{f(w(q))}\frac{1}{1-qf^\prime(w(q))}\;\;,$$ as desired.

As in Timothy Budd's answer let $w=w(q)$ denote the (formal) solution of $q=\frac{w}{f(w)}$.

Let $p$ be another variable and consider the sum \begin{align*} S(p,q):=\sum_{n,m>0} \sum_{j>0} j[z^{n+j}]\{f(z)^n\} [z^{m-j}]\{f(z)^m\} \frac{p^n q^m}{nm} \end{align*}

Write it as \begin{align*} \sum_{n>0} \frac{p^n}{n}\Big(\sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j\Big) \end{align*}

and rewrite the inner sums as

\begin{align*} \sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j &=\frac{1}{w(q)^n}\Big(f(w(q))^n -\sum_{j=0}^{n-1}[z^j]\{f(z)^n\}w(q)^j\Big) -[z^n] \{f(z)^n\}\\ &=\frac{1}{q^n} -\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j} -[z^n] \{f(z)^n\} \end{align*}

Note that \begin{align*} \sum_{n>0} \frac{p^n}{n}[z^n]\{f(z)^n\}=-\log\big(f(w(p))\big) \end{align*} and that \begin{align*} \sum_{n>0} \frac{p^n}{nq^n}=-\log\big(1-\frac{p}{q}\big) \end{align*} The remaining sum can be written as \begin{align*} \sum_{n>0} \frac{p^n}{n}\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j}&=\sum_{j>0}w(q)^{-j}\sum_{n\geq j}\frac{1}{n}[z^{n-j}]\{f(z)^n\}p^n\\ &=\sum_{j>0}w(q)^{-j}\frac{w(p)^j}{j}\\ &=-\log\big(1-\frac{w(p)}{w(q)}\big) \end{align*} where we have used that for $n\geq j$ (by Lagrange-Bürmann) \begin{align*}\frac{1}{n}[z^{n-j}]\{f(z)^n\}=\frac{1}{n}[z^{n-1}]\{z^{j-1}f(z)^n\}=[q^n]\{\frac{w(q)^j}{j}\}\end{align*}

Thus $S(p,q)=-\log\big(1-\frac{p}{q}\big)+\log\big(1-\frac{w(p)}{w(q)}\big)-\log\big(f(w(p))$ and \begin{align*} \exp(S(p,q))=\frac{1}{f(w(p))}\,\frac{1-\frac{w(p)}{w(q)}}{1-\frac{p}{q}}\end{align*} Now \begin{align*} \frac{q-q\frac{w(p)}{w(q)}}{q-p}&=1+\frac{p}{f(w(q))}\frac{f(w(q))-f(w(p)}{q-p}\\ &=1 +\frac{p f^\prime(w(p))w^\prime(p)}{f(w(q))}+O(q-p)\\ &=1+\frac{f(w(p))}{f(w(q))}\frac{p f^\prime(w(p))}{1-pf^\prime(w(p))} +O(q-p) \end{align*}

so that $\exp(S(q,q))$ reduces to $$\exp(S(q,q))=\frac{1}{f(w(q))}\frac{1}{1-qf^\prime(w(q))}\;\;,$$ as desired.

As in Timothy Budd's answer let $w=w(q)$ denote the (formal) solution of $q=\frac{w}{f(w)}$.

Let $p$ be another variable and consider the sum \begin{align*} S(p,q):=\sum_{n,m>0} \sum_{j>0} j[z^{n+j}]\{f(z)^n\} [z^{m-j}]\{f(z)^m\} \frac{p^n q^m}{nm} \end{align*}

Write it as \begin{align*} \sum_{n>0} \frac{p^n}{n}\Big(\sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j\Big) \end{align*}

and rewrite the inner sums as

\begin{align*} \sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j &=\frac{1}{w(q)^n}\Big(f(w(q))^n -\sum_{j=0}^{n-1}[z^j]\{f(z)^n\}w(q)^j\Big) -[z^n] \{f(z)^n\}\\ &=\frac{1}{q^n} -\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j} -[z^n] \{f(z)^n\} \end{align*}

Note that \begin{align*} \sum_{n>0} \frac{p^n}{n}[z^n]\{f(z)^n\}=-\log\big(f(w(p))\big) \end{align*} and that \begin{align*} \sum_{n>0} \frac{p^n}{nq^n}=-\log\big(1-\frac{p}{q}\big) \end{align*} The remaining sum can be written as \begin{align*} \sum_{n>0} \frac{p^n}{n}\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j}&=\sum_{j>0}w(q)^{-j}\sum_{n\geq j}\frac{1}{n}[z^{n-j}]\{f(z)^n\}p^n\\ &=\sum_{j>0}w(q)^{-j}\frac{w(p)^j}{j}\\ &=-\log\big(1-\frac{w(p)}{w(q)}\big) \end{align*} where we have used that for $n\geq j$ (by Lagrange-Bürmann) \begin{align*}\frac{1}{n}[z^{n-j}]\{f(z)^n\}=\frac{1}{n}[z^{n-1}]\{z^{j-1}f(z)^n\}=[q^n]\{\frac{w(q)^j}{j}\}\end{align*}

Thus $S(p,q)=-\log\big(1-\frac{p}{q}\big)+\log\big(1-\frac{w(p)}{w(q)}\big)-\log\big(f(w(p))$ and \begin{align*} \exp(S(p,q))=\frac{1}{f(w(p))}\,\frac{1-\frac{w(p)}{w(q)}}{1-\frac{p}{q}}\end{align*} Now \begin{align*} \frac{q-q\frac{w(p)}{w(q)}}{q-p}&=1+\frac{p}{f(w(q)}\frac{f(w(q))-f(w(p)}{q-p}\\ &=1 +\frac{p f^\prime(w(p))w^\prime(p)}{f(w(q))}+O(q-p)\\ &=1+\frac{f(w(p))}{f(w(q))}\frac{p f^\prime(w(p)}{1+pf^\prime(w(p))} +O(q-p) \end{align*}

so that $\exp(S(q,q)$ reduces to $$\exp(S(q,q))=\frac{1}{f(w(q))}\frac{1}{1-qf^\prime(w(q)}\;\;,$$ as desired.

As in Timothy Budd's answer let $w=w(q)$ denote the (formal) solution of $q=\frac{w}{f(w)}$.

Let $p$ be another variable and consider the sum \begin{align*} S(p,q):=\sum_{n,m>0} \sum_{j>0} j[z^{n+j}]\{f(z)^n\} [z^{m-j}]\{f(z)^m\} \frac{p^n q^m}{nm} \end{align*}

Write it as \begin{align*} \sum_{n>0} \frac{p^n}{n}\Big(\sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j\Big) \end{align*}

and rewrite the inner sums as

\begin{align*} \sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j &=\frac{1}{w(q)^n}\Big(f(w(q))^n -\sum_{j=0}^{n-1}[z^j]\{f(z)^n\}w(q)^j\Big) -[z^n] \{f(z)^n\}\\ &=\frac{1}{q^n} -\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j} -[z^n] \{f(z)^n\} \end{align*}

Note that \begin{align*} \sum_{n>0} \frac{p^n}{n}[z^n]\{f(z)^n\}=-\log\big(f(w(p))\big) \end{align*} and that \begin{align*} \sum_{n>0} \frac{p^n}{nq^n}=-\log\big(1-\frac{p}{q}\big) \end{align*} The remaining sum can be written as \begin{align*} \sum_{n>0} \frac{p^n}{n}\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j}&=\sum_{j>0}w(q)^{-j}\sum_{n\geq j}\frac{1}{n}[z^{n-j}]\{f(z)^n\}p^n\\ &=\sum_{j>0}w(q)^{-j}\frac{w(p)^j}{j}\\ &=-\log\big(1-\frac{w(p)}{w(q)}\big) \end{align*} where we have used that for $n\geq j$ (by Lagrange-Bürmann) \begin{align*}\frac{1}{n}[z^{n-j}]\{f(z)^n\}=\frac{1}{n}[z^{n-1}]\{z^{j-1}f(z)^n\}=[q^n]\{\frac{w(q)^j}{j}\}\end{align*}

Thus $S(p,q)=-\log\big(1-\frac{p}{q}\big)+\log\big(1-\frac{w(p)}{w(q)}\big)-\log\big(f(w(p))$ and \begin{align*} \exp(S(p,q))=\frac{1}{f(w(p))}\,\frac{1-\frac{w(p)}{w(q)}}{1-\frac{p}{q}}\end{align*} Now \begin{align*} \frac{q-q\frac{w(p)}{w(q)}}{q-p}&=1+\frac{p}{f(w(q)}\frac{f(w(q))-f(w(p)}{q-p}\\ &=1 +\frac{p f^\prime(w(p))w^\prime(p)}{f(w(q))}+O(q-p)\\ &=1+\frac{f(w(p))}{f(w(q))}\frac{p f^\prime(w(p)}{1-pf^\prime(w(p))} +O(q-p) \end{align*}

so that $\exp(S(q,q)$ reduces to $$\exp(S(q,q))=\frac{1}{f(w(q))}\frac{1}{1-qf^\prime(w(q))}\;\;,$$ as desired.

Here is an extended hint for the first conjectured formula.

As in Timothy Budd's answer let $w=w(q)$ denote the (formal) solution of $q=\frac{w}{f(w)}$.

Write the sum in the exponential above as

\begin{align*} \sum_{n>0} \frac{q^n}{n}\Big(\sum_{j>0}[z^{n+j}]\{f(z)^n\} w^j\Big) \end{align*}

and rewrite the inner sums as

\begin{align*} \sum_{j>0}[z^{n+j}]\{f(z)^n\} w^j =\frac{1}{w^n}\Big(f(w)^n -\sum_{j=0}^{n-1}[z^j]\{f(z)^n\}w^j\Big) -[z^n] \{f(z)^n\} \end{align*}

Note that \begin{align*} \sum_{n>0} \frac{q^n}{n}[z^n]\{f(z)^n\}=\log\big(f(w)\big) \end{align*} and that the remaining sum can be written as \begin{align*} \sum_{n>0} \frac{1}{n} [u^{n-1}]\Big\{\frac{1}{1-u} \big(1-\frac{f(uw)^n}{f(w)^n}\big)\Big\} \end{align*}

Summing up and evaluating the residue now gives the value $-\log\big(1-q\,f^\prime(w)\big)$
for this sum, as follows:

\begin{align*} &\sum_{n>0} \frac{1}{n} [u^{n-1}]\Big\{\frac{1}{1-u} \big(1-\frac{f(uw)^n}{f(w)^n}\big)\Big\}\\ &= [u^{-1}]\Big\{\frac{1}{1-u}\big(-\log(1-\frac{1}{u}) +\log(1-\frac{f(uw)}{uf(w)})\big)\Big\}\\ &= [u^{-1}]\Big\{\frac{1}{1-u}\Big(\log\big(\frac{1-\frac{f(uw)}{uf(w)}}{1-\frac{1}{u}}\big)\Big)\Big\}\\ \end{align*} Now substitute $u=z+1$ and use the residue composition rule (for formal Laurent series, see e.g. Goulden/Jackson, Combinatorial Enumeration, Thm 1.2.2). \begin{align*} &=[z^{-1}]\Big\{-\frac{1}{z}\log\Big(\frac{1+z-\frac{f((1+z)w)}{f(w)}}{z}\Big)\Big\}\\ &=[z^{0}]\Big\{-\log\Big(\frac{1+z-\frac{f((1+z)w)}{f(w)}}{z}\Big)\Big\}\\ &=-\log\big(1-qf^\prime(w)\big) \end{align*}

Finally, note that \begin{align*} \frac{1}{f(w)}\frac{1}{1-q\,f^\prime(w)}= \frac{q^2 w^\prime}{w^2}\;\;. \end{align*}

As in Timothy Budd's answer let $w=w(q)$ denote the (formal) solution of $q=\frac{w}{f(w)}$.

Let $p$ be another variable and consider the sum \begin{align*} S(p,q):=\sum_{n,m>0} \sum_{j>0} j[z^{n+j}]\{f(z)^n\} [z^{m-j}]\{f(z)^m\} \frac{p^n q^m}{nm} \end{align*}

Write it as \begin{align*} \sum_{n>0} \frac{p^n}{n}\Big(\sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j\Big) \end{align*}

and rewrite the inner sums as

\begin{align*} \sum_{j>0}[z^{n+j}]\{f(z)^n\} w(q)^j &=\frac{1}{w(q)^n}\Big(f(w(q))^n -\sum_{j=0}^{n-1}[z^j]\{f(z)^n\}w(q)^j\Big) -[z^n] \{f(z)^n\}\\ &=\frac{1}{q^n} -\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j} -[z^n] \{f(z)^n\} \end{align*}

Note that \begin{align*} \sum_{n>0} \frac{p^n}{n}[z^n]\{f(z)^n\}=-\log\big(f(w(p))\big) \end{align*} and that \begin{align*} \sum_{n>0} \frac{p^n}{nq^n}=-\log\big(1-\frac{p}{q}\big) \end{align*} The remaining sum can be written as \begin{align*} \sum_{n>0} \frac{p^n}{n}\sum_{j=1}^{n}[z^{n-j}]\{f(z)^n\}w(q)^{-j}&=\sum_{j>0}w(q)^{-j}\sum_{n\geq j}\frac{1}{n}[z^{n-j}]\{f(z)^n\}p^n\\ &=\sum_{j>0}w(q)^{-j}\frac{w(p)^j}{j}\\ &=-\log\big(1-\frac{w(p)}{w(q)}\big) \end{align*} where we have used that for $n\geq j$ (by Lagrange-Bürmann) \begin{align*}\frac{1}{n}[z^{n-j}]\{f(z)^n\}=\frac{1}{n}[z^{n-1}]\{z^{j-1}f(z)^n\}=[q^n]\{\frac{w(q)^j}{j}\}\end{align*}

Thus $S(p,q)=-\log\big(1-\frac{p}{q}\big)+\log\big(1-\frac{w(p)}{w(q)}\big)-\log\big(f(w(p))$ and \begin{align*} \exp(S(p,q))=\frac{1}{f(w(p))}\,\frac{1-\frac{w(p)}{w(q)}}{1-\frac{p}{q}}\end{align*} Now \begin{align*} \frac{q-q\frac{w(p)}{w(q)}}{q-p}&=1+\frac{p}{f(w(q)}\frac{f(w(q))-f(w(p)}{q-p}\\ &=1 +\frac{p f^\prime(w(p))w^\prime(p)}{f(w(q))}+O(q-p)\\ &=1+\frac{f(w(p))}{f(w(q))}\frac{p f^\prime(w(p)}{1+pf^\prime(w(p))} +O(q-p) \end{align*}

so that $\exp(S(q,q)$ reduces to $$\exp(S(q,q))=\frac{1}{f(w(q))}\frac{1}{1-qf^\prime(w(q)}\;\;,$$ as desired.