Cauchy's Integral Formula/General Result/Corollary

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Theorem

Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.

Let the coefficient of $z^n$ extracted from $\map G z$ be denoted:

$\sqbrk {z^n} \map G z := a_n$

Let $\map G z$ be convergent for $z = z_0$ and $0 < r < \cmod {z_0}$.


Then:

$\sqbrk {z^n} \map G z = \displaystyle \frac 1 {2 \pi i} \oint_{\cmod z \mathop = r} \dfrac {\map G z \d z} {z^{n + 1} }$


Proof

\(\displaystyle \sqbrk {z^n} \map G z\) \(=\) \(\displaystyle \dfrac 1 {n!} \map {G^{\paren n} } 0\) Derivative of Generating Function/General Result/Corollary
\(\displaystyle \) \(=\) \(\displaystyle \dfrac 1 {n!} \paren {\dfrac {n!} {2 \pi i} \int_{\partial D} \frac {\map G z} {z^{n + 1} } \d z}\) Cauchy's Integral Formula/General Result
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 \pi i} \oint_{\cmod z \mathop = r} \dfrac {\map G z \d z} {z^{n + 1} }\)

$\blacksquare$


Sources