Lagrange reversion theorem

In mathematics, the Lagrange reversion theorem gives series or formal power series expansions of certain implicitly defined functions; indeed, of compositions with such functions.

Let ${\displaystyle v}${\displaystyle v} be a function of ${\displaystyle x}${\displaystyle x} and ${\displaystyle y}${\displaystyle y} in terms of another function ${\displaystyle f}${\displaystyle f} such that

${\displaystyle v=x+yf(v)}${\displaystyle v=x+yf(v)}

Then for any function ${\displaystyle g}${\displaystyle g}, for small enough ${\displaystyle y}${\displaystyle y}:

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

In particular, if ${\displaystyle g}${\displaystyle g} is the identity function ${\displaystyle g(x)=x}${\displaystyle g(x)=x}, this reduces to

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

in which case the equation can be derived using perturbation theory.

In 1770, Joseph Louis Lagrange (1736–1813) published his power series solution of the implicit equation for ${\displaystyle v}${\displaystyle v} mentioned above. However, his solution used cumbersome series expansions of logarithms.^{[1] [2]} In 1780, Pierre-Simon Laplace (1749–1827) published a simpler proof of the theorem, which was based on relations between partial derivatives with respect to the variable ${\displaystyle x}${\displaystyle x} and the parameter ${\displaystyle y}${\displaystyle y}.^{[3] [4] [5]} Charles Hermite (1822–1901) presented the most straightforward proof of the theorem by using contour integration.^{[6] [7] [8]}

Lagrange's reversion theorem is used to obtain numerical solutions to Kepler's equation.

We start by writing

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

Writing the delta-function as an integral, we have:

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

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

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

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

• Lagrange Inversion [Reversion] Theorem on MathWorld
• Cornish–Fisher expansion, an application of the theorem
• Article on equation of time contains an application to Kepler's equation.

• Theorems in mathematical analysis
• Inverse functions

• Articles with short description
• Short description is different from Wikidata