Modding out functional relationships; An introduction to congruent integration.
#4
I thought I'd add a bit about Taylor Series too.

If I write,

\(
F_k(w,z) = \int_\gamma \frac{(w-\zeta)^k}{(s-\zeta)^{k+1}} f(s)z\,ds\bullet z = z e^{\displaystyle 2 \pi i \frac{f^{(k)}(\zeta)}{k!} (w-\zeta)^k}\\
\)

Then,

\(
\Omega_{k=1}^\infty F_k(w,z)\bullet z = z e^{\displaystyle 2 \pi i \sum_{k=1}^\infty\frac{f^{(k)}(\zeta)}{k!} (w-\zeta)^k} = z e^{2\pi i f(w)}= \int_\gamma \frac{f(s)z}{s-w}\,ds\bullet z\\
\)

Or,

\(
G_k(w,z) = \int_\gamma \frac{(w-\zeta)^k}{(s-\zeta)^{k+1}} f(s)z^2\,ds\bullet z = \frac{1}{\displaystyle 1/z + 2\pi i\frac{f^{(k)}(\zeta)}{k!}(w-\zeta)^k}\\
\)

Then, similarly,

\(
\Omega_{j=1}^\infty G_k(w,z)\bullet z = \int_\gamma \frac{f(s)z^2}{s-w}\,ds\bullet z\\
\)



This always extends to separable functions, in which,

\(
\Omega_{k=1}^\infty \int_\gamma \frac{(w-\zeta)^k}{(s-\zeta)^{k+1}} f(s)\phi(z)\,ds\bullet z=\int_\gamma \frac{f(s)\phi(z)}{s-w}\,ds\bullet z
\)

Using the congruent integral we can show,

\(
\Omega_{k=1}^\infty \oint_\gamma \frac{(w-\zeta)^k}{(s-\zeta)^{k+1}}\phi(s,z)\,ds\bullet z=\oint_\gamma \frac{\phi(s,z)}{s-w}\,ds\bullet z
\)
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Messages In This Thread
RE: Modding out functional relationships; An introduction to congruent integration. - by JmsNxn - 06/23/2021, 07:07 AM

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