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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!
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 $n \in \N$.
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 $b$. $\square$
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.
The power series representing $f$ on $\ball(b,R)$ developed in the proof is called the Taylor series of $f$ on $\ball(b,R)$.
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.
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. $\square$
If \[f(z) = \sum_{n=0}^\infty b_n (z-b)^n\] on $\ball(b,R)$ then this is the Taylor series of $f$.
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.
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.
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.