Lagrange reversion theorem

We start by writing:

${\displaystyle g(v)=\int \delta (yf(z)-z+x)g(z)(1-yf'(z))\,dz}$

Writing the delta-function as an integral we have:

${\displaystyle {\begin{aligned}g(v)&=\iint \exp(ik[yf(z)-z+x])g(z)(1-yf'(z))\,{\frac {dk}{2\pi }}\,dz\\[10pt]&=\sum _{n=0}^{\infty }\iint {\frac {(ikyf(z))^{n}}{n!}}g(z)(1-yf'(z))e^{ik(x-z)}\,{\frac {dk}{2\pi }}\,dz\\[10pt]&=\sum _{n=0}^{\infty }\left({\frac {\partial }{\partial x}}\right)^{n}\iint {\frac {(yf(z))^{n}}{n!}}g(z)(1-yf'(z))e^{ik(x-z)}\,{\frac {dk}{2\pi }}\,dz\end{aligned}}}$

The integral over k then gives ${\displaystyle \delta (x-z)}$ and we have:

${\displaystyle {\begin{aligned}g(v)&=\sum _{n=0}^{\infty }\left({\frac {\partial }{\partial x}}\right)^{n}\left[{\frac {(yf(x))^{n}}{n!}}g(x)(1-yf'(x))\right]\\[10pt]&=\sum _{n=0}^{\infty }\left({\frac {\partial }{\partial x}}\right)^{n}\left[{\frac {y^{n}f(x)^{n}g(x)}{n!}}-{\frac {y^{n+1}}{(n+1)!}}\left\{(g(x)f(x)^{n+1})'-g'(x)f(x)^{n+1}\right\}\right]\end{aligned}}}$

Rearranging the sum and cancelling then gives the result:

${\displaystyle g(v)=g(x)+\sum _{k=1}^{\infty }{\frac {y^{k}}{k!}}\left({\frac {\partial }{\partial x}}\right)^{k-1}\left(f(x)^{k}g'(x)\right)}$

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