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:

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

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:

$\displaystyle \sin z \cdot e^z$ $\displaystyle = \left( z - \frac {z^3}{3!} + \frac {z^5}{5!} - \dots \right) \left( 1 + z + \frac {z^2}{2!} + \frac {z^3}{3!} + \frac {z^4}{4!} + \dots \right)$    
$\displaystyle \quad$ $\displaystyle = z +z^2+\frac {z^3}{3}-\frac {z^5}{30}-\frac{z^7}{630}+\dots$    

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:

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

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

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

By term-by-term differentiation, we have:

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

and this fits 1.2.

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

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

By integration term-by-term, we have:

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

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:

$\displaystyle P_1(x)$ $\displaystyle = 1+x$    
$\displaystyle P_2(x)$ $\displaystyle = 1+ x + x^2$    
$\displaystyle P_3 (x)$ $\displaystyle = 1+ x + x^2+x^3$    
$\displaystyle P_4 (x)$ $\displaystyle = 1+ x + x^2+x^3+x^4.$    

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

Figure 1: First approximations of a given function.
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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:

$\displaystyle P_1(x)$ $\displaystyle = 1-x^2$    
$\displaystyle P_2(x)$ $\displaystyle = 1-x^2 + x^4$    
$\displaystyle P_3 (x)$ $\displaystyle = 1-x^2 + x^4-x^6$    
$\displaystyle P_4 (x)$ $\displaystyle = 1-x^2 + x^4-x^6+x^8.$    

The graphs of $ f$ and of these approximations are displayed in Figure 2.
Figure 2: First approximations of a given function.
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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.
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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.

Noah Dana-Picard 2007-12-24