Taylor Series.

Theorem 8.1.1   Let $f$ be a function analytic at a point $z_0$ . Denote by $C$ the largest circle centered at $z_0$ such that $f$ is analytic at all the points interior to $C$ , and let $R$ be its radius. Then there exists a power series $\underset{k \geq 0}{\sum} a_k (z-z_0)^k$ which converges to $f(z)$ in $C$ .

This series is unique; it is called the Taylor series of $f$ at $z_0$ . The cofficients of this series are determined by the following formula:

$$a_n = \frac {f^{(n)}(z_0)}{n!}.$$

The Taylor series of a function about 0 is called the Maclaurin series of $f$ .

Example 8.1.2 (Maclaurin series)       

1. $e^z = 1 + z + \frac {z^2}{2!} + \frac {z^3}{3!} + \frac {z^4}{4!} + \dots = \underset{n=0}{\overset{+ \infty}{\sum}} \frac {z^n}{n!}$ .
2. $\cos z = 1 - \frac {z^2}{2!} + \frac {z^4}{4!} - \dots =\underset{n=0}{\overset{+ \infty}{\sum}} (-1)^n\frac {z^{2n}}{(2n)!}$ .
3. $\sin z = z - \frac {z^3}{3!} + \frac {z^5}{5!} - \dots = \underset{n=0}{\overset{+ \infty}{\sum}} (-1)^n\frac {z^{2n+1}}{(2n+1)!}$ .
4. $\cosh z = 1 + \frac {z^2}{2!} + \frac {z^4}{4!} + \dots = \underset{n=0}{\overset{+ \infty}{\sum}}\frac {z^{2n}}{(2n)!}$ .
5. $\sinh z = z + \frac {z^3}{3!} + \frac {z^5}{5!} + \dots =\underset{n=0}{\overset{+ \infty}{\sum}} \frac {z^{2n+1}}{(2n+1)!}$ .
6. $\frac {1}{1-z} = 1 + z + z^2 + z^3 + z^4 + \dots = \underset{n=0}{\overset{+ \infty}{\sum}} z^n$ .
7. $\log (1+z) = z -\frac {z^2}{2} + \frac {z^3}{3} - \frac {z^4}{4} + \dots = \underset{n=1}{\overset{+ \infty}{\sum}} (-1)^{n+1} \frac {z^n}{n}$ .
8. $(1+z)^{\alpha} = 1 + \alpha z + \frac {\alpha (\alpha -1)}{2!} z^2 + \frac {\alpha (\alpha -1)(\alpha -2}{3!} z^3 + \dots$ .

How to compute quickly a Taylor series expansion?

1. Derive term-by-term the Taylor series of $f(z)$ about $z_0$ to get the Taylor series of $f'(z)$ about the same point. The radius of convergence of both series is the same.
2. The Taylor series of $f(z)+g(z)$ is the sum of the Taylor series of $f(z)$ and of $g(z)$ .
3. The Taylor series of $f(z)g(z)$ is the product of the Taylor series of $f(z)$ and of $g(z)$ .
4. If $g(z_0) \neq 0$ , the Taylor series of $f(z)/g(z)$ is the quotient of the Taylor series of $f(z)$ by the Taylor series of $g(z)$ , according to increasing power order.

Example 8.1.3   Compute the Maclaurin series of $f(z)=\sin z \cdot e^z$ . Using 1.2, we have:

$$\sin z \cdot e^z = \left( z - \frac {z^3}{3!} + \frac {z^5}{5!} - \dots \right) \left( 1 + z + \frac {z^2}{2!} + \frac {z^3}{3!} + \frac {z^4}{4!} + \dots \right)$$

$$\quad = z +z^2+\frac {z^3}{3}-\frac {z^5}{30}-\frac{z^7}{630}+\dots$$

We can use Taylor series in order to find limits:

Example 8.1.4   Let $f(z)= \frac {e^z-1}{\sin z}$ . We wish to compute $\underset{z \rightarrow 0}{\lim} f(z)$ .

Using 1.2, we get:

$$f(z) = \frac {z + \frac {z^2}{2!} + \frac {z^3}{3!} + \frac {z^4}{4!} + \dots} {z + \frac {z^3}{3!} + \frac {z^5}{5!} + \dots}$$

$$\quad = \frac {1 + \frac {z}{2!} + \frac {z^2}{3!} + \frac {z^3}{4!} + \dots} {1 + \frac {z^2}{3!} + \frac {z^4}{5!} + \dots}$$

Therefore $\underset{z \rightarrow 0}{\lim} f(z)=1$ .

The following result is a consequence of Thm 3.5 and Thm 3.7.

Proposition 8.1.5   Let $f$ be a function analytic on a neighborhood of $z_0$ .

1. We get the Taylor series expansion of $f'$ by differentiating term-by-term the Taylor series expansion of $f$ .
2. If the Taylor series expansion of $f'$ is known, we get the expansion of $f$ by integrating term-by-term the Taylor series expansion of $f'$ (take care of the additive constant of integration!)

Example 8.1.6   In 1.2, we saw that

$$\sin z = z + \frac {z^3}{3!} + \frac {z^5}{5!} + \dots$$

By term-by-term differentiation, we have:

$$\cos z = 1 - \frac {z^2}{2!} + \frac {z^4}{4!} - \dots$$

and this fits 1.2.

Example 8.1.7   Let $f(z)=\frac {1}{1+z^2}$ . The MacLaurin series of $f(z)$ is:

$$f(z)=\frac {1}{1+z^2} = 1-z^2+z^4-z^6 + \dots + (-1)^nz^{2n} + \dots$$

By integration term-by-term, we have:

$$\arctan z = z - \frac 13 z^3 + \frac 15 z^5 - \frac 17 x^7 + \dots + \frac {(-1)^n}{2n+1} z^{2n+1} + \dots$$

Important remark: When we studied power series over the reals we had a surprise: the convergence domain of a power series is not always obvious.

Take $f(x)=1/(1-x)$ . This function is defined over $\mathbb{R}-\{ 1 \}$ . Its first MacLaurin expansions are given by:

$$P_1(x) = 1+x$$

$$P_2(x) = 1+ x + x^2$$

$$P_3 (x) = 1+ x + x^2+x^3$$

$$P_4 (x) = 1+ x + x^2+x^3+x^4.$$

The graphs of $f$ and of these approximations are displayed in Figure 1.

Figure 1: First approximations of a given function.

It seems that the visualization shows that the successive approximations tend to the original function only for $-1<x<1$ . The condition for a geometric sequence to be convergent supports this impression. First of all, the function is not defined at 1 (where it has a singular point) and this point acts as a "barrier". But, does the power series make sense out of the interval $(-1,1)$ ? Actually not in our frame of study. Maybe in other frames.

Now take $g(x)=1/(1+x^2)$ . It is obtained by the substitution of $-x^2$ instead of $x$ . The first MacLaurin expansions are given by:

$$P_1(x) = 1-x^2$$

$$P_2(x) = 1-x^2 + x^4$$

$$P_3 (x) = 1-x^2 + x^4-x^6$$

$$P_4 (x) = 1-x^2 + x^4-x^6+x^8.$$

The graphs of $f$ and of these approximations are displayed in Figure 2.

Figure 2: First approximations of a given function.

We have here the same visual impression: the MacLaurin series tends to the given function $g$ for $-1<x<1$ . But for the function $g$ , -1 and 1 are not points of discontinuity. So, what happens?

Passing to the complex setting, consider the function of the complex variable $z$ given by $G(z)=1/(1+z^2)$ . It is defined over $\mathbb{C} - \{ -i, i \}$ . The corresponding MacLaurin series is given by $S(z)=1-z^2+z^4-z^6+...$ and is convergent for $z$ in the open unit ball centered at the origin, i.e. on the largest ball centered at 0 at not touching the two points where $G$ fails to be defined (see Figure 3).

Figure 3: The largest ball for the function to be defined.

This example shows the importance of working in a complex setting. Without exaggeration, we could say that the complex setting is "more natural" than the real one.