Lagrange inversion theorem: Difference between revisions

Revision as of 20:40, 3 June 2024

Revision as of 15:42, 15 June 2024

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We can use Cauchy Integral theorem:

:<math>

f^{-1}(z)

=

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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)

=

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using geometric series:

:<math>

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

=

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</math>

:<math>

f^{-1}(z)

=

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</math> we get:

:<math>

f^{-1}(z)

=

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by residue theorem:

:<math>

f^{-1}(z)

=

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finally:

:<math>

f^{-1}(z)

=

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''<em>In fact, integration by parts doesn't work for:''</em>

:<math>

n=0\rightarrow

\frac{1}{2\pi i}

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</math>

''<em>Fortunately, writing the integral as a residue in the last line fixes the problem.''</em>

==Applications==