Lagrange inversion theorem: Difference between revisions

Revision as of 11:26, 6 December 2023

Revision as of 21:33, 7 December 2023

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By convergence tests, this series is in fact convergent for <math>|z| \leq (p-1)p^{-p/(p-1)},</math> which is also the largest disk in which a local inverse to {{mvar|f}} can be defined.

==Derivation==

==Sketch of the proof==

We can use Cauchy Integral theorem:

For simplicity suppose <math>z=0=f(w=0)</math>. We can then compute

:<math>

f^{-1}(z)

\oint_{w=0} \frac{d w}{2\pi i} \frac{1}{f(w) -z}

=

\oint_{w=0} \frac{d w1}{2\pi i} \fracoint_{1f(C)} \frac{f'^{-1}(g(z)\xi)}{\xi w- + O(w^2)z}d\xi

</math>

and substitute:

<math>

\xi=f(\omega)

</math>

<math>

d\xi=f'(\omega)d\omega

</math>

<math>

f(C)\rightarrow C

</math>

<math>

f^{-1}(z)

=

\frac{1}{2\pi i} \oint_{C} \frac{\omega}{f(\omega) - z} f'(\omega) d\omega

\frac{1}{f'(g(z))}

</math>

using geometric series:

<math>

\frac{1}{f(\omega) - z}

=

\frac{1}{f(\omega) - f(a) - z + f(a) }

g'(f(w))

=

\frac{1}{f(\omega) - f(a) }\frac{1}{1 - \frac{z - f(a) }{f(\omega) - f(a)} }

g'(z)

=

.

\frac{1}{f(\omega) - f(a) }\sum_{n=0}^\infty

\left(\frac{z - f(a) }{f(\omega) - f(a)}\right)^{n}

</math>

<math>

If we expand the integrand using the geometric series we get

f^{-1}(z)

:<math>

\oint_{w=0} \frac{d w}{2\pi i} \frac{1}{f(w) -z}

=

\frac{1}{2\pi i}\sum_{n=0}^\infty

\left({z - f(a) }\right)^{n}

z^n

\oint_{w=0C} \frac{d w}{2\pi i} omega f'(\frac{1omega)}{(f(w\omega) - f(a))^{n+1} } d\omega

</math>

now by integration by parts: <math>

u

=

\omega

</math> and <math>

dv

=

\sum_{n=0}^\infty

\frac{f'(\omega)}{(f(\omega) - f(a))^{n+1} }

z^n

\oint_{w=0} \frac{d w}{2\pi i} \frac{1}{w^{n+1}} \left(\frac{w}{f(w)}\right)^{n+1}

</math> where <math>

uv

=

\frac{-1 }{n} \frac{e^{i\theta} }{(f(e^{i\theta} ) - f(a))^{n}}\Biggl|_{0}^{2\pi }

=

0

</math> we get:

<math>

f^{-1}(z)

=

\frac{1}{2\pi i}\sum_{n=0}^\infty

\frac{({z - f(a) })^{n}}{n}

\oint_{C} \frac{1}{(f(\omega) - f(a))^{n} } d\omega

</math>

by residue theorem:

<math>

f^{-1}(z)

=

\sum_{n=0}^\infty

\frac{ ({z - f(a) })^{n }}{n!}

\left. \fracoperatorname{d^nRes}{ d w^n}\left(\frac{w1}{(f(w)}\rightomega) - f(a))^{n+1} \right|_{}, w=0}a)

,

</math>

where in the last step we used the fact that <math>f(w)</math> has one simple zero.

finally:

Finally we can integrate over <math>z</math> taking into account <math>g(0)=0</math>

:<math>

<math>

g(z) = \sum_{n=0}^\infty

f^{-1}(z)

\frac{ z^n }{n!}

=

\left. \frac{d^n}{ d w^n}\left(\frac{w}{f(w)}\right)^{n+1} \right|_{w=0}

\sum_{n=0}^\infty

~~\Longrightarrow~~

g\frac{({z) =- \sum_f(a) })^{n=0}^\infty}{n!}

\lim_{ \omega \to a} \frac{d^{n-1}}{d\omega ^{n-1}} \left(\frac{\omega - a }{f(\omega) - f(a)}\right)^{n}

\frac{ z^{n+1} }{(n+1)!}

\left. \frac{d^n}{ d w^n}\left(\frac{w}{f(w)}\right)^{n+1} \right|_{w=0}

.

</math>

Upon a redefiniton of the summation index we get the stated formula.

==Applications==