📂Complex Anaylsis

Derivation of the Taylor Series Using Complex Analysis

Theorem ¹

If a function $f: A \subseteq \mathbb{C} \to \mathbb{C}$ is analytic on the disk $|z - \alpha| < r$, then $$ f(z) = \sum_{n = 0} ^{\infty} {{f^{(n)} (\alpha)} \over {n!}} (z - \alpha)^n $$

Explanation

One of the joys of mathematics is generalization. Even Taylor’s theorem can be regarded as a generalization of the mean value theorem, and this time let us extend the reals to the complex numbers. The interesting fact is that, rather than being extended piece by piece, it is essentially built up from scratch, yet the proof becomes even simpler. It is a proof method that is both novel and clean in form, giving one a sense of reward for studying.

Derivation

First, consider a circle $\mathscr{C}: |z| = r$ and a point $w$ inside it. By the Cauchy integral formula, $$ \begin{align*} f(w) =& {{1} \over {2 \pi i}} \int_{ \mathscr{C} } {{f(z)} \over {z - w}} dz \\ =& {{1} \over {2 \pi i}} \int_{ \mathscr{C} } {{f(z)} \over {z}} { {1} \over {1 - {{w} \over {z}} } } dz \end{align*} $$ Expressing it as an infinite geometric series, we have $\displaystyle {{1} \over {1 - { {w} \over {z} } }} = \sum_{n=0}^{\infty} \left( {{w} \over {z}} \right) ^{n}$, and substituting this back into the integral above gives $$ \begin{align*} f(w) =& {{1} \over {2 \pi i}} \int_{ \mathscr{C} } {{f(z)} \over {z}} \sum_{n=0}^{\infty} \left( {{w} \over {z}} \right) ^{n} dz \\ =& {{1} \over {2 \pi i}} \sum_{n=0}^{\infty} w^{n} \int_{\mathscr{C}} {{f(z)} \over {z^{n+1}}} dz \end{align*} $$

Cauchy integral formula generalized to derivatives: Suppose a function $f: A \subseteq \mathbb{C} \to \mathbb{C}$ is analytic on a simply connected region $\mathscr{R}$.

If a simple closed contour $\mathscr{C}$ inside $\mathscr{R}$ encloses some point $\alpha$, then for a natural number $n$

$$f^{(n)} (\alpha) = {{n!} \over {2 \pi i }} \int_{\mathscr{C}} {{f(z)} \over { (z - \alpha)^{n+1} }} dz$$

Once again, by the Cauchy integral formula, $\displaystyle {{1} \over {2 \pi i}} \int_{\mathscr{C}} {{f(z)} \over {z^{n+1}}} dz = {{f^{(n)} (0)} \over {n!}}$, so $$ f(w) = \sum_{n = 0} ^{\infty} {{f^{(n)} (0)} \over {n!}} w^n $$ Now, to generalize to the circle $|z- \alpha| = r$, setting $z-\alpha = w$ gives $$ f(z) = f(w+\alpha) = \sum_{n=0}^{\infty} {{f^{(n)}(\alpha) (z-\alpha)^{n}} \over {n!}} $$

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