Formal Lagrange inversion identities for period coordinates

Put $w(x)=x\mathrm{/}f(x)=x+O(x^2)$. Writing $u(z)={\sum }_{n\ge 1}{u}_n{z}^n$, the coefficient of $z^n$ in $w(u(z))$ is $u_n$ plus a polynomial in $u_1,\dots ,u_{n-1}$. Hence the equations $w(u(z))=z$ determine $u_n$ successively and uniquely, starting with $u_1=1$. The same coefficient recursion gives $u(w(x))=x$. Since $w(u)=u\mathrm{/}f(u)=z$, this inverse satisfies $u=zf(u)$; conversely that equation implies $w(u)=z$.

For a Laurent series $G(x)$ with finitely many negative powers, define ${\mathrm{Res}}_{x}G(x)dx=[x^{-1}]G(x)$. If $w(x)=x+O(x^2)$, then ${\mathrm{Res}}_{z}H(z)dz={\mathrm{Res}}_{x}H(w(x))w^{\mathrm{\prime }}(x)dx$: it suffices to check $H(z)=z^j$; for $j\mathrm{\ne }-1$ the substituted expression is a formal derivative, and for $j=-1$, $w^{\mathrm{\prime }}\mathrm{/}w=x^{-1}+(\log (w\mathrm{/}x){)}^{\mathrm{\prime }}$ has residue $1$. Applying this change of variable and $w=x\mathrm{/}f(x)$ gives, for any $F(x)\in A\mathrm{⟦}x\mathrm{⟧}$ and $n\ge 1$, $[z^n]F(u(z))={\mathrm{Res}}_{x}F(x)x^{-n-1}f(x{)}^{n-1}(f(x)-x{f}^{\mathrm{\prime }}(x))dx\mathrm{.}$ The residue of $(F(x)f(x{)}^n{x}^{-n}{)}^{\mathrm{\prime }}$ is zero. Expanding that derivative and rearranging the three resulting terms yields ${\mathrm{Res}}_{x}F(x)x^{-n-1}f(x{)}^{n-1}(f(x)-x{f}^{\mathrm{\prime }}(x))dx=\frac{1}{n}[x^{n-1}]F^{\mathrm{\prime }}(x)f(x{)}^n\mathrm{.}$ Therefore $[z^n]F(u(z))=\frac{1}{n}[x^{n-1}]F^{\mathrm{\prime }}(x)f(x{)}^n$.

Taking $F(x)=x$ in step 2.1 gives the first coefficient formula and hence $u(z)\mathrm{/}z={\sum }_{n\ge 1}\frac{1}{n}[x^{n-1}]f(x{)}^n{z}^{n-1}$. Taking $F(x)=\log f(x)$, whose derivative is $f^{\mathrm{\prime }}(x)\mathrm{/}f(x)$ by [A1], gives $[z^n]\log f(u(z))=\frac{1}{n}[x^{n-1}]f(x{)}^{n-1}{f}^{\mathrm{\prime }}(x)=\frac{1}{n}[x^n]f(x{)}^n,$ where the last equality follows by differentiating $f(x{)}^n$. Since $u\mathrm{/}z=f(u)$, exponentiating the coefficient identity proves the remaining equality.