Cauchy's Integral Formula/General Result
Theorem
Let $D = \set {z \in \C: \cmod z \le r}$ be the closed disk of radius $r$ in $\C$.
Let $f: U \to \C$ be holomorphic on some open set $U$ such that $D \subseteq U$.
Let $n \in \N$ be a natural number.
Then for each $a$ in the interior of $D$:
- $\displaystyle f^{\paren n} \paren a = \dfrac {n!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.
Corollary
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
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
- $\displaystyle \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
Basis for the Induction
$\map P 0$ holds, as this is:
- $\displaystyle \map f a = \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}} \rd z$
which is Cauchy's Integral Formula.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\displaystyle \map {f^{\paren k} } a = \frac {k!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 1} } \rd z$
Then we need to show:
- $\displaystyle \map {f^{\paren {k + 1} } } a = \frac {\paren {k + 1}!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 2} } \rd z$
Induction Step
This is our induction step:
\(\ds \frac {\rd} {\rd a} \map {f^{\paren k} } a\) | \(=\) | \(\ds \frac {k!} {2 \pi i} \int_{\partial D} \frac {\rd} {\rd a} \frac {\map f z} {\paren {z - a}^{k + 1} } \rd z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {k!} {2 \pi i}\int_{\partial D} \frac {\paren {k + 1} \map f z} {\paren {z - a}^{k + 2} } \rd z\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {k + 1}!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{k + 2} } \rd z\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\displaystyle \forall n \in \N: \map {f^{\paren n} } a = \frac {n!} {2 \pi i} \int_{\partial D} \frac {\map f z} {\paren {z - a}^{n + 1} } \rd z$
$\blacksquare$
Also known as
This result can also be referred to as Cauchy's Integral Formula for Derivatives.
Source of Name
This entry was named for Augustin Louis Cauchy.