4.4 Taylor Series

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We determine the Taylor series for a given analytic function.

Suppose $f$ is analytic at $z_0$ . Then the complex power series $\sum _{k=0}^\infty \frac {f^{(k)}(z_0)}{k!}(z-z_0)^k$ is called the Taylor series representation for $f$ centered at $z_0$ .

Suppose $f$ is analytic at $z_0$ . Then there exists $\rho >0$ such that for all $z \in D(z_0, \rho ), \,f(z)$ is equal to its Taylor series representation centered at $z_0$ .

example 1 Find the Taylor series representation for $f(z) = \sin (z)$ centered at the origin.
The derivatives of $\sin z$ evaluated at the origin form the sequence $\{0, 1, 0, -1,...\}$ , where this cycle of length four repeats indefinitely. Hence the series is $\sin z = z - \frac {z^3}{3!} + \frac {z^5}{5!} - \dots$ which can be written as $\sin z = \sum _{k=0}^\infty (-1)^k \frac {z^{2k+1}}{(2k+1)!}$ The ratio test shows that this series converges for all $z$ . When the center of the Taylor series is the origin, it is referred to as the Maclaurin series for the function.

(problem 1a) Find the Maclaurin series for $f(z) = e^z$ and find its radius of convergence.

(problem 1b) Find the Maclaurin series for $f(z) = \cos (z)$ and find its radius of convergence.

(problem 1c) Find the Taylor series for $f(z) = e^z$ centered at $\pi i$ and find its radius of convergence.

(problem 1d) Find the Taylor series for $f(z) = \Log (z)$ centered at $i$ and find its radius of convergence.

Here are video solutions to problems 1c and 1d:

example 2 Find the Maclaurin series for $f(z) = z\sin (\pi z)$ .
First note that $f(z)$ is an entire function, so it has a Maclaurin series representation. Using the Maclaurin series for $\sin z$ gives $z\sin (\pi z) = z\sum _{k=0}^\infty (-1)^k \frac {(\pi z)^{2k+1}}{(2k+1)!} = \sum _{k=0}^\infty (-1)^k \frac {\pi ^{2k+1} z^{2k+2}}{(2k+1)!}$ The ratio test shows that this series converges for all $z$ .

(problem 2a) Find the Maclaurin series for $f(z) = e^{-z^2}$ and find its radius of convergence.

(problem 2b) Find the Maclaurin series for $f(z) = z\cos (\pi z)$ and find its radius of convergence.

Here are video solutions to problems 2a and 2b:

example 3 Find the first few terms of the Maclaurin series for $f(z) = \tan z$ .
Since $\tan z = \frac {\sin z}{\cos z}$ and $\cos 0 \neq 0$ , $\tan z$ is analytic at the origin and has a Maclaurin series expansion: $\tan z = \sum _{k=0}^\infty c_k z^k = c_0 + c_1 z + c_2 z^2 + \cdots$ Using the Maclaurin series of $\sin z$ and $\cos z$ and writing $\sin z = \tan z \cdot \cos z$ we have $z-\frac {z^3}{3!} + \frac {z^5}{5!} - \cdots = \left (c_0 + c_1 z + c_2 z^2 + \cdots \right ) \cdot \left (1-\frac {z^2}{2!} + \frac {z^4}{4!} - \cdots \right )$ Multiplying on the RHS and comparing coefficients leads to a system of equations: \begin{align*} 0 & = c_0 \\ 1 & = c_1 \\ 0 & = c_2 -\frac{c_0}{2!} \\ -\frac{1}{3!} & = c_3- \frac{c_1}{2!} \\ 0 & = c_4 - \frac{c_2}{2!} + \frac{c_0}{4!} \\ \frac{1}{5!} & = c_5- \frac{c_3}{2!} + \frac{c_1}{4!}\\ & \text{etc.} \end{align*}

Solving this system yields $c_0 = c_2 = c_4 = 0$ , etc., and $c_1 = 1, c_3 = \frac 13, c_5 = \frac {2}{15}$ etc. Hence $\tan z = z + \frac 13 z + \frac {2}{15}z^3 + \cdots$

(problem 3) Find the first few terms of the Maclaurin series for $f(z) = \sec z$ .

Here is a video solution to problems 3:

2024-09-27 14:06:46