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8.2 Taylor's Theorem

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Taylor's theorem gives us a power series expansion for every holomorphic function. Since power series can be differentiated infinitely often, we will be able to deduce that holomorphic functions can be differentiated infinitely often!

Theorem (Taylor's Theorem)

Fix $f : D \to \C$ holomorphic on a domain $D$. On every ball $\ball(b,R) \subset D$ the function $f$ can be represented as a power series \[f(z) = \sum_{n=0}^\infty a_n (z-b)^n\] and moreoever all higher derivatives of $f$ exists and \[a_n = \dfrac{f^{(n)}(b)}{n!}\] for all $n \in \N$.

Proof:

Fix a ball $\ball(b,R) \subset D$. For $0 < r < R$ let $\gamma_r$ be the contour $\gamma_r(t) = b + re^{it}$ on $[0,2\pi]$. By Cauchy's integral formula \[f(b+h) = \frac{1}{2\pi i} \int\limits_{\gamma_r} \dfrac{f(z)}{z - (b+h)} \intd z\] whenever $|h| < r$. We have \[\dfrac{1}{z - (b+h)} - \dfrac{1}{z - b} = \dfrac{h}{(z-b)(z-(b+h))}\] and \[\dfrac{1}{z - (b+h)} - \dfrac{1}{z - b} - \dfrac{h}{(z - b)^2} = \dfrac{h^2}{(z-b)^2(z-(b+h))}\] while an inductive argument gives the more general \[ \begin{align*} & \dfrac{1}{z - (b+h)} - \dfrac{1}{z - b} - \dfrac{h}{(z - b)^2} - \cdots - \dfrac{h^{m-1}}{(z-b)^m} \\ = & \dfrac{h^m}{(z-b)^m(z-(b+h))} \end{align*}\] for all $m \in \N$. Plugging into Cauchy's integral formula gives \[ \begin{align*} f(b+h) & = \dfrac{1}{2 \pi i} \int\limits_{\gamma_r} \dfrac{f(z)}{z-b} \intd z + \dfrac{h}{2 \pi i} \int\limits_{\gamma_r} \dfrac{f(z)}{(z-b)^2} \intd z + \cdots \\ & \qquad + \dfrac{h^{m-1}}{2 \pi i} \int\limits_{\gamma_r} \dfrac{f(z)}{(z-b)^m} \intd z \\ & \qquad + \dfrac{h^m}{2 \pi i} \int\limits_{\gamma_r} \dfrac{f(z)}{(z-b)^m(z-(b+h))} \intd z \end{align*}\] so we will define \[a_n = \dfrac{1}{2 \pi i} \int\limits_{\gamma_r} \dfrac{f(z)}{(z-b)^{n+1}} \intd z\] for all $n \in \N$. It remains to prove that \[\lim_{m \to \infty} \dfrac{h^m}{2 \pi i} \int\limits_{\gamma_r} \dfrac{f(z)}{(z-b)^m(z-(b+h))} \intd z = 0\] holds.

We have \[ \begin{align*} & \dfrac{h^m}{2 \pi i} \int\limits_{\gamma_r} \dfrac{f(z)}{(z-b)^m(z-(b+h))} \intd z \\ ={} & \dfrac{h^m}{2 \pi i} \int\limits_0^{2\pi} \dfrac{f(b+re^{it})}{(re^{it})^m (re^{it} - h)} ire^{it} \intd t \end{align*}\] and since $f$ is holomorphic it is continuous and therefore bounded on $\{ \gamma(t) : 0 \le t \le 2\pi \}$. Fix $M > 0$ with $|f(b + re^{it})| \le M$ for all $0 \le t \le 2\pi$. Since $|r e^{it} - h| > |r - |h||$ for all $0 \le t \le 2 \pi$ the estimation lemma gives us \[ \begin{align*} & \left| \dfrac{h^m}{2 \pi i} \int\limits_0^{2\pi} \dfrac{f(b+re^{it})}{(re^{it})^m (re^{it} - h)} \intd t \right| \\ \le{} & \dfrac{|h|^m}{2\pi} \cdot 2 \pi \cdot \dfrac{Mr}{r^m (r - |h|)} = \dfrac{Mr}{r - |h|} \left( \dfrac{|h|}{r} \right)^m \end{align*}\] and this converges to zero as $m \to \infty$ because $|h| < r$. We conclude that \[f(b+h) = \sum_{n=0}^\infty a_n h^n\] whenever $|h| < r$. Taking $w = b+h$ gives \[f(w) = \sum_{n=0}^\infty a_n (w-b)^n\] whenever $|w-b| < r$. This is the desired power series representation of $f$ centered at $b$. $\square$

Corollary

If \[f(z) = \sum_{n=0}^\infty a_n (z-b)^n\] on $\ball(b,R)$ then \[a_n = \dfrac{f^{(n)}(b)}{n!} = \dfrac{1}{2 \pi i} \displaystyle\int\limits_{\gamma_r} \dfrac{f(z)}{(z-b)^{n+1}} \intd z\] for any $0 < r < R$ where $\gamma_r(t) = b + re^{it}$ on $[0,2\pi]$. In particular \[f^{(n)}(b) = \dfrac{n!}{2 \pi i} \int\limits_{\gamma_r} \dfrac{f(z)}{(z-b)^{n+1}} \intd z\] giving an integral formula for the higher derivatives of $f$ at $b$.

Although the integral expressions for the power series coefficients of $f$ are of theoretical interest, they are not always practical. We will usually determine the Taylor series of a function from known power series e.g. trigonometric functions and the geometric series.

Definition (Taylor series)

The power series representing $f$ on $\ball(b,R)$ developed in the proof is called the Taylor series of $f$ on $\ball(b,R)$.

Lemma

If a power series \[\sum_{n=0}^\infty a_n (z-b)^n\] is equal to 0 on $\ball(b,R)$ then all coefficients $a_n$ are zero.

Proof:

Plug in $z = b$ to get $a_0 = 0$. Differentiating gives another power series equal to zero, whose constant coefficient is $a_1$. Thus $a_1 = 0$ as well. Repeating this argument gives, by induction, that all coefficients are equal to 0. $\square$

Corollary

If \[f(z) = \sum_{n=0}^\infty b_n (z-b)^n\] on $\ball(b,R)$ then this is the Taylor series of $f$.

Example

The function \[f(z) = \dfrac{1}{1-z}\] is holomorphic on $\ball(0,1)$. We know \[\dfrac{1}{1-z} = \sum_{n=0}^\infty z^n = 1 + z + z^2 + z^3 + z^4 + \cdots\] so this is the Taylor series of $f$ on $\ball(0,1)$. We cannot have a Taylor series on a larger ball because $f$ is not defiend at 1.

Example

The function \[f(z) = \begin{cases} \dfrac{\sin(z)}{z} & z \ne 0 \\ 1 & z = 0 \end{cases}\] is holomorphic on $\C$ and \[f(z) = \dfrac{1}{z} \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} z^{2n+1} =\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!} z^{2n}\] on all of $\C$. This must be the Taylor series by uniqueness.

Definition (Entire Function)

An entire function is any function $f : \C \to \C$ that is holomorphic.

Taylor's theorem tells us entire functions are nothing but power series with infinite radii of convergence.