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→Derivation: Per talk: "here I only generalized something that a certain individual shared on Instagram" i.e. this is unambiguous WP:OR Tag: Reverted |
Undid revision 1231010840 by JayBeeEll (talk) |
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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==
We can use [[Cauchy's integral formula]]:
:<math>
f^{-1}(z)
=
\frac{1}{2\pi i} \oint_{f(C)} \frac{f^{-1}(\xi)}{\xi - 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
</math>
using [[geometric series]]:
:<math>
\frac{1}{f(\omega) - z}
=
\frac{1}{f(\omega) - f(a) - z + f(a) }
=
\frac{1}{f(\omega) - f(a) }\frac{1}{1 - \frac{z - f(a) }{f(\omega) - f(a)} }
=
\frac{1}{f(\omega) - f(a) }\sum_{n=0}^\infty
\left(\frac{z - f(a) }{f(\omega) - f(a)}\right)^{n}
</math>
:<math>
f^{-1}(z)
=
\frac{1}{2\pi i}\sum_{n=0}^\infty
\left({z - f(a) }\right)^{n}
\oint_{C} \frac{ \omega f'(\omega)}{(f(\omega) - f(a))^{n+1} } d\omega
</math>
now by [[integration by parts]]: <math>
u
=
\omega
</math> and <math>
dv
=
\frac{f'(\omega) d\omega}{(f(\omega) - f(a))^{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 the [[residue theorem]]:
:<math>
f^{-1}(z)
=
\sum_{n=0}^\infty
\frac{({z - f(a) })^{n}}{n}
\operatorname{Res}(\frac{1}{(f(\omega) - f(a))^{n} }, w=a)
</math>
finally:
:<math>
f^{-1}(z)
=
\sum_{n=0}^\infty
\frac{({z - f(a) })^{n}}{n!}
\lim_{ \omega \to a} \frac{d^{n-1}}{d\omega ^{n-1}} \left(\frac{\omega - a }{f(\omega) - f(a)}\right)^{n}
</math>
<em>In fact, integration by parts doesn't work for n=0, because in the formula after integrating by parts we would have to divide by zero which does not make any sense, so instead we can do the following:</em>
:<math>
n=0\rightarrow
\frac{1}{2\pi i}
\oint_{C} \frac{ \omega f'(\omega)}{ f(\omega) - f(a) } d\omega =
\frac{1}{2\pi i}
\oint_{f(C)} \frac{ f^{-1}(u)}{ u - f(a) } du=
f^{-1}(f(a))=a
</math>
<em>Fortunately, writing the integral as a residue in the last line fixes the problem.</em>
==Applications==
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