# Solving a transcendental equation, in closed form

There is a change of variable between two varibles $q$ and $Q$ as the following: $$q=Q\exp(2f(Q))\quad\quad \quad (*)$$ where $f(Q)$ is given by $$f(Q)=\sum_{d=1}^\infty \frac{(2d-1)!}{(d!)^2}Q^d$$ The problem is to find the inverse change of varible, which amounts to solving the transendental equation $(*)$. The formula is surprisingly simple: $$Q=\frac{q}{(1+q)^2}$$ Admittedly, it can be verified that the two formulas are inverse to each other, such as checking the differential equation with initial conditions. This example actually appears in one of Givental's famous papers. My question is, how can we find such an inverse by direct computational methods? I would appreciate any ideas.

• Starting from $Q=q(1+q)^{-2}$, the expression for the inverse, $f(Q)$, is given almost immediately by the Lagrange inversion formula. – Pietro Majer Apr 9 '17 at 6:07

First notice that $$f'(Q)=\frac{1}{2}\sum_{d=1}^\infty \binom{2d}{d}Q^d = \frac{(1-4Q)^{-1/2}-1}{2Q}.$$ It follows that differentiation of $q=Q\exp(2f(Q))$ with respect to $Q$ gives $$q'=\frac{\exp(2f(Q))}{\sqrt{1-4Q}}.$$ Then $$\frac{q'}{q}=\frac{1}{Q\sqrt{1-4Q}}.$$ Solution to this differential equation is $$q = C\frac{\sqrt{1-4Q}- 1}{\sqrt{1-4Q}+ 1}.$$ Since $q'(0)=1$, we get $C=-1$, and further the following rational relation between $Q$ and $q$: $$1-4Q = \left(\frac{1-q}{1+q}\right)^2$$ or $$Q=\frac{q}{(1+q)^2}.$$