Cauchy's Integral Formula/General Result

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Let $D = \left\{ {z \in \C : \left|{z}\right| \le r}\right\}$ 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 $D$:

$\displaystyle f^{\left({n}\right)} \left({a}\right) = \frac {n!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{n+1} } \ \mathrm d z$

where $\partial D$ is the boundary of $D$, and is traversed anticlockwise.


Proof by induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle f^{\left({n}\right)} \left({a}\right) = \frac {n!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{n+1} } \ \mathrm d z$

Basis for the Induction

$P \left({0}\right)$ holds, as this is:

$\displaystyle f \left({a}\right) = \frac 1 {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)} \ \mathrm d z$

which is Cauchy's Integral Formula.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\displaystyle f^{\left({k}\right)} \left({a}\right) = \frac {k!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{k+1} } \ \mathrm d z$

Then we need to show:

$\displaystyle f^{\left({k + 1}\right)} \left({a}\right) = \frac {\left({k + 1}\right)!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{k+2} } \ \mathrm d z$

Induction Step

This is our induction step:

\(\displaystyle \frac {\mathrm d} {\mathrm d a} f^{\left({k}\right)} \left({a}\right)\) \(=\) \(\displaystyle \frac {k!} {2 \pi i} \int_{\partial D} \frac {\mathrm d} {\mathrm d a} \frac {f \left({z}\right)} {\left({z - a}\right)^{k + 1} }\ \mathrm d z\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {k!} {2 \pi i}\int_{\partial D} \frac {\left({k + 1}\right) f \left({z}\right)}{\left({z - a}\right)^{k + 2} }\ \mathrm d z\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\left({k+1}\right)!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{k + 2} }\ \mathrm d z\) $\quad$ $\quad$

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.


$\displaystyle \forall n \in \N: f^{\left({n}\right)} \left({a}\right) = \frac {n!} {2 \pi i} \int_{\partial D} \frac {f \left({z}\right)} {\left({z - a}\right)^{n+1} } \ \mathrm d z$


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.