# MATH3401

Lecture 30

### Taylor's Theorem in $$\C$$

Let $$f$$ be analytic on $$B_R(z_0)$$. Then $$f$$ has a power series representation on $$B_R(z_0)$$: $f(z) = \sum_{n=0}^{\infty} a_n\left(z-z_0\right)^n \text{ for } |z-z_0|\lt R,$ where $$a_n=\dfrac{f^{(n)}(z_0)}{n!}$$, $$n\in \N_0$$.

For $$z_0=0$$: Maclaurin series

### Taylor's Theorem in $$\C$$

Example: $$f(z) = e^z$$, find Maclaurin series.

Answer: Since $$f$$ is entire, $$\implies R=\infty$$.

$$f^{(n)}(z)=e^z$$ $$\forall n$$ $$\implies f^{(0)}(z)=e^0 = 1$$ $$\forall n$$.

Previous theorem $$\implies e^z = \ds \sum_{n=0}^{\infty} \frac{1}{n!} z^n$$.

### Proof of Taylor's Theorem

Take $$z_0=0$$, otherwise translate.

Choose $$z\in B_R$$, $$|z|=r$$. Fix some $$r_0 \in (r, R)$$ and set $$C = C_{r_0}$$ positively oriented.

 Cauchy $$\implies$$ $f(z) = \frac{1}{2 \pi i} \int_C \frac{f(\zeta) d\zeta}{\zeta - z} \quad (1)$

### Proof of Taylor's Theorem

Note

$$\dfrac{1}{\zeta - z} = \dfrac{1}{\zeta}\left( \dfrac{1}{1-\frac{z}{\zeta}}\right)$$ $$= \dfrac{1}{\zeta}\left( \ds \sum_{n=0}^{N-1}\left( \frac{z}{\zeta} \right)^n + \frac{\left(\frac{z}{\zeta}\right)^N}{1 - \frac{z}{\zeta}} \right)$$

$$\;\;\qquad = \ds \sum_{n=0}^{N-1} \frac{1}{\zeta ^{n+1}} z^n + \frac{z^N}{(\zeta-z)\zeta^N}$$

Multiply by $$\dfrac{f\left(\zeta\right)}{2 \pi i}$$ and then integrate over $$C$$.

### Proof of Taylor's Theorem

$$\implies \ds \frac{1}{2 \pi i } \int_C \frac{f(\zeta)}{\zeta-z}d\zeta$$

$$\;\;\qquad = \ds \sum_{n=0}^{N-1} \frac{1}{2 \pi i } \int_C \frac{f(\zeta) z^n}{\zeta^{n+1}}d\zeta + \underbrace{ \frac{z^N}{2 \pi i } \int_C \frac{f(\zeta)}{(\zeta -z)\zeta^N}d\zeta }_{ {\small \text{Call this } \rho_{{\tiny N-1}}(z)} }$$

$$(1)$$ and the extended Cauchy integral formula $$\implies$$ $f(z) = \sum_{n=0}^{N-1} \frac{f^{(n)}(0)z^n}{n!} + \rho_{{\tiny N-1}}(z).$

### Proof of Taylor's Theorem

If we can show that $$\ds \lim_{N\ra \infty } \rho_{{\tiny N-1}} (z) = 0$$ we are done.

Take $$\zeta \in C \implies |\zeta|=r_0$$. Suppose there exists $$M_N$$ such that $(H) \quad \left| \frac{f(\zeta)}{(\zeta-z)\zeta^N}\right|\leq M_N \text{ on } C.$

Then   $$\left| \rho_{{\tiny N-1}} (z) \right|\leq \dfrac{r^N}{2 \pi} M_N \cdot \ell(C)$$ $$= r_0 r^N M_N$$. $$\quad(**)$$

### Proof of Taylor's Theorem

To get $$M_N$$: $$f$$ is analytic $$\implies |\,f|$$ is continuous, $$C$$ is closed and bounded, so by the extreme value theorem there exists $$\mu$$ such that $|\,f|\leq \mu \text{ on } C.$

Further $$|\zeta|^N = r_0^N$$ and

$$\left| \zeta - z \right|\geq \left| \left| \zeta \right| - \left| z \right|\right|$$ $$= r_0-r$$.

$$\implies M_N = \dfrac{\mu}{r_0^N(r_0 - r)}$$ which suffices for $$(H)$$.

### Proof of Taylor's Theorem

Hence $$(**)$$ $$\implies$$

$$\left| \rho_{{\tiny N-1}} (z) \right| \leq \ds \frac{\mu r_0 r^N}{r_0^N (r_0 - r)}$$ $$= \ds \frac{\mu r_0}{r_0 - r}\left( \frac{r}{r_0} \right)^N$$

$$\ra 0$$   as   $$N\ra \infty. \;\square$$

### Taylor series in $$\C$$

In $$\R$$, Taylor series may converge, but fail to converge to the function: c.f. Lecture 16.

Now consider the series expansion $$\sum a_n (z- z_0)^n$$.

We can calculate the Radius of convergence $$R$$:

Set $$\Lambda = \ds \lim_{n\ra \infty} \left| \frac{a_{n+1}}{a_n} \right|$$, then set $$R= \dfrac{1}{\Lambda}$$.

$$\Lambda =0 \iff R =\infty$$ and $$\Lambda =\infty \iff R=0$$.

### Taylor series in $$\C$$

Example 1: For $$f(z) = e^z$$, we have $$e^z = \ds \sum_{n=0}^{\infty} \frac{1}{n!}z^n$$

Then  $$\Lambda = \ds \lim_{n\ra \infty} \left( \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}} \right)$$ $$= \ds \lim_{n\ra \infty} \frac{1}{n+1}$$ $$= 0$$.

Thus $$R=\infty$$ and $e^z = \sum_{n=0}^{\infty} \frac{1}{n!}z^n \text{ on } \C. \quad (*)$

### Taylor series in $$\C$$

Example 2: Find Maclaurin series for $$f(z) = z^2 e^{3z}$$.

Answer: Since $$f$$ is entire, $$(*)$$ implies $e^{3z} = \ds \sum_{n=0}^{\infty}\frac{3^nz^n}{n!}.$

Therefore $$z^2 e^{3z} = \ds \sum_{n=0}^{\infty}\frac{3^nz^{n+2}}{n!}$$ $$=\ds \sum_{n=2}^{\infty}\frac{3^{n-2}z^{n}}{(n-2)!}$$.