# Comparing two power-series

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.

• What is $q$? As of now the left hand side is independent of $q$. Jul 15 at 17:43
• Corrected. It was just another variable. Jul 15 at 17:44
• did you try to express the coefficients of powers via Lagrange–Bürmann formula? Jul 16 at 21:21
• Yes, I have attempted this in many different ways. However I always run into the issue that I am only summing over positive $n$. Jul 17 at 6:40
• Just a note, I have tried working with convolution polynomials/matrices, LB formulae, Bell polynomials, but nothing has led to me understanding the problem better. Jul 17 at 6:42

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.

• Could you perhaps say something about the step where you obtain $-\log (1-qf'(w))$? I have been unsuccessful with it. Aug 2 at 17:49
• Yes, of course. I've added more details. Sorry for being too succinct. It's not easy to give the exactly right amount of hint.
– esg
Aug 3 at 18:00
• There are multiple issues with the suggestion. Most importantly, the residue composition rule works only for variable change with no constant coefficient. If one does it in this general case one needs to take residues at 1. Aug 4 at 16:40
• You're right, the substitution step is false. Apologies for posting nonsense. Abobe is a new try to deal with the convergence issue, hoppefully correct to the end this time.
– esg
Aug 5 at 12:55

Here is a proof of the formula $$\sum_{j>0} \frac{1}{j} w(q)^j P_qw(q)^{-j}=2\log \frac{w(q)}{q}-\log w'(q).$$ (The notation is the same as in the Timothy's answer except I prefer $$P_qf(q)$$ to $$[q^{>0}]f(q)$$.) Consider $$q$$ fixed. We may assume without loss of generality that $$w(z)$$ is analytic in some circle $$|z|\le R$$ where $$R>|q|$$. We assume additionally that it is univalent in this circle and that $$|w(z)|>|w(q)|$$ when $$|z|=R$$.

We have a formula $$P_q f(q)=\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)f(z)\,dz$$ which is valid when $$f$$ is analytic in the punctured circle $$0<|z|\le R$$. A straightforward calculation then gives $$\sum_{j>0} \frac{1}{j} w(q)^j P_qw(q)^{-j}=\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{w(z)}{w(z)-w(q)}\,dz.$$ (Importantly, the series converges uniformly so there are no analytic issues.) This integral is problematic because of branching points of the logarithm but there is a trick to circumvent this obstacle, $$\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{w(z)}{w(z)-w(q)}\,dz=\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{w(z)(z-q)}{z(w(z)-w(q))}\,dz+\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{z}{z-q}\,dz.$$ Under close examination, $$\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{z}{z-q}\,dz=0.$$ (I found it out using the dilogarithm but there may be better proofs.) The logarithm in the first integral is now analytic in the whole circle so we can easily compute it with residues, $$\frac{1}{2\pi i}\int_{|z|=R} \left(\frac{1}{z-q}-\frac{1}{z}\right)\log\frac{w(z)(z-q)}{z(w(z)-w(q))}\,dz=2\log\frac{w(q)}{q}-\log w'(q).$$ This is it.

• Is there a way to remove the requirements on analyticity and that $|w(z)|>|w(q)|$ when $|z|=R$? Aug 5 at 5:44
• Definitely no way (for the calculation to make sense), but this is not a problem. The equality in question is a system of polynomial identities in the Taylor coefficients of $w$, so it is automatically extended to the case when $w$ is just a formal series. (A polynomial which is zero on an open set is zero everywhere.) Aug 5 at 16:52
• I would love to accept your question also, because I like your proof. But the combinatorial proof of esg is closer to what I imagined. Aug 5 at 17:12

You may have already tried this, but let me record it here anyway. Setting $$w(q) = q + \cdots$$ to be the power series solution to $$w(q) = q\,f(w(q))$$, Lagrange inversion implies that \begin{align} \sum_{n\geq 0}[z^n]\{f(z)^{n-1}\}q^n &= q^2 \frac{w'(q)}{w(q)^2} \\ \sum_{m>0} \frac{q^m}{m}[z^{m-j}]f(z)^m &= \frac{w(q)^j}{j} \\ \sum_{n>0} \frac{q^n}{n}[z^{n+j}]f(z)^n &= - [q^{>0}]\frac{w(q)^{-j}}{j}, \end{align} where by $$[q^{>0}]$$ I mean only keeping the positive terms in the Laurent series expansion. Your identity then becomes equivalent to \begin{align} \exp\left[-\sum_{j>0}\frac{1}{j} w(q)^j [q^{>0}] w(q)^{-j}\right] = q^2 \frac{w'(q)}{w(q)^2} \end{align} or if you prefer \begin{align} \sum_{j>0}\frac{1}{j} w(q)^j [q^{>0}] w(q)^{-j} = 2\log \frac{w(q)}{q} - \log w'(q). \end{align} This certainly looks more tractable.

• Thanks Timothy, I have tried this. The main difficulty for me here was precisely [q>0]w(q)^{-j}, which I was unable rephrase in any reasonable way. This is indeed the only difficult part for me. Jul 23 at 8:56
• Introduce new variable $z$ and think that $1\gg |z|\gg |q|$. Then for Laurent series $h(q)$ we have $[q^{>0}]h(q)={\rm CT}_z \frac{z}{1-z} h(q/z)$, where ${\rm CT}_z$ stands for the constant term w.r.t. $z$. Did you try this? Jul 26 at 19:17

Since you've asked for a hint rather than a complete solution I'm giving a hint. I think this will lead to complete solution, but I haven't worked out the details, which I'm leaving to you. Of course it's possible that this doesn't work.

This is based on Timothy Budd's formulation.

Let us look at a slightly different but closely related problem. Given a power series $$G(t)$$ with constant term 1, find $$\sum_{j>0}\frac{1}{j} x^j [q^{>0}] \left(\frac{G(q)}{q}\right)^{j}\tag{1}$$

If we can solve this problem then we can just replace $$G(q)$$ with $$q/w(q)$$. and then replace $$x$$ with $$w(q)$$.

The sum in $$(1)$$ is equal to $$[q^{>0}]\log\left(\frac{1}{1-xG(q)/q}\right)$$ (Here we are working in $$\mathbb{C}((q))[[x]]$$, or perhaps $$\mathbb{C}[[q,x/q]]$$—we allow negative powers of $$q$$ but not of $$x$$.)

Now let $$g(x)$$ be the formal power series solution of $$g(x) = x G(g(x))$$. Then have the formulas $$[q^{<0}]\log\left(\frac{1}{1-xG(q)/q}\right)=\log\left(\frac{1}{1-g(x)/q}\right) \tag{2}$$ and $$[q^{=0}]\log\left(\frac{1}{1-xG(q)/q}\right)=\log\left(\frac{g(x)}{x}\right) \tag{3}$$

Equation (2) is easily seen to be equivalent to ordinary Lagrange inversion. Equation (3) is not so well known, but it's easily derived from well-known forms of Lagrange inversion. It can be found, for example, in my paper on Lagrange inversion [I. M. Gessel, Lagrange inversion, J. Combin. Theory Ser. A 144 (2016), 212–249], equation (2.2.9).

Therefore \begin{aligned} \ [q^{>0}]\log\left(\frac{1}{1-xG(q)/q}\right) &= \log\left(\frac{1}{1-xG(q)/q}\right) -\log\left(\frac{1}{1-g(x)/q}\right)\\ &\qquad\qquad-\log\left(\frac{g(x)}{x}\right). \end{aligned}

• Thank you for this answer. It was a lot of fun going through this and filling in the details. However, when you plug in $w(q)=x$ at the end and take exponential, you end up with $0\cdot\infty\cdot(q/w(q))$. In fact, I was a bit proud at the end that I have done a very similar calculation separating the two diverging terms myself before. In both cases I don't see how to get around the convergence issue. Jul 25 at 17:14

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) 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 $$-2\log\Big(\frac{H(q)}{q}\Big)-\log\big(H'(q)\big)\,.$$