📂Complex Anaylsis

Derivation of the Taylor Series Using Complex Analysis

Theorem ¹

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

Explanation

One of the joys of mathematics is generalization. The Taylor’s Theorem is considered a generalization of the Mean Value Theorem, and now we extend from real numbers to complex numbers. Interestingly, despite the process of extending it bit by bit, the proof becomes simpler. The proof method is fascinating and neat, providing a sense of accomplishment in studying.

Derivation

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

Generalized Cauchy Integral Formula for Differentiation: Let’s say function $f: A \subseteq \mathbb{C} \to \mathbb{C}$ is analytic in a simply connected region $\mathscr{R}$.

For a simple closed path $\mathscr{C}$ in $\mathscr{R}$ surrounding a point $\alpha$, for any 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 $\displaystyle {{1} \over {2 \pi i}} \int_{\mathscr{C}} {{f(z)} \over {z^{n+1}}} dz = {{f^{(n)} (0)} \over {n!}}$ it follows that $$f(w) = \sum_{n = 0} ^{\infty} {{f^{(n)} (0)} \over {n!}} w^n$$ Now, to generalize this for a circle $|z- \alpha| = r$, if we set $z-\alpha = w$, $$f(z) = f(w+\alpha) = \sum_{n=0}^{\infty} {{f^{(n)}(\alpha) (z-\alpha)^{n}} \over {n!}}$$

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