Taylor and Maclaurin Series

Definition: Power Series @power-series A power series in powers of $z - z_0$ is a series of the form $$ \sum_{n=0}^{\infty} a_n (z - z_0)^n = a_0 + a_1(z - z_0) + a_2(z - z_0)^2 + \cdots, $$ where $z$ is a complex variable, $a_0, a_1, \dots,$ are complex constants, called the coefficients of the series, and $z_0$ is a complex constant, called the center of the series. Referenced by (4 direct)Direct references:Taylor and Maclaurin SeriesTaylor Expansion Theoremtheorem-7A theorem about the uniqueness of Taylor series as @power-series expansions

Theorem: Taylor Expansion Theorem @taylor-expansion-theorem Let $f$ be analytic in a domain $D$ and $z_0$ be a point in $D$. Then $f$ can be expanded in a power series $$ f(z) = f(z_0) + f'(z_0)(z - z_0) + \frac{f''(z_0)}{2!}(z - z_0)^2 + \cdots, $$ valid in all circles $\|z - z_0\| < r$ containing only points of $D$.

Definition: Taylor series @taylor-series The expansion $$ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!} (z - z_0)^n $$ is called the Taylor series of $f$ about $z_0$. Referenced by (5 direct, 3 transitive)Direct references:Laurent Seriestheorem-1-intuitionMaclaurin seriesCircle of ConvergenceA theorem about the uniqueness of Taylor series as @power-series expansionsTransitive (depth 1):Radius of ConvergenceTransitive (depth 2):A theorem about radius of convergencetheorem-7

Definition: Maclaurin series @maclaurin-series The special case of the taylor series in which $z_0$ = 0 $$ f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} z^n $$ is called the Maclaurin series of f.

Definition: Circle of Convergence @circle-of-convergence The circle $\|z - z_0\| < $ in which the taylor series converges to the function is called the circle of convergence for the taylor series. Referenced by (2 direct, 3 transitive)Direct references:Taylor and Maclaurin SeriesRadius of ConvergenceTransitive (depth 1):A theorem about radius of convergencetheorem-7A theorem about the uniqueness of Taylor series as @power-series expansions

Definition: Radius of Convergence @radius-of-convergence The radius of the circle of convergence is called the radius of convergence. Referenced by (4 direct) Direct references: Taylor and Maclaurin Series theorem-7 A theorem about the uniqueness of Taylor series as @power-series expansions A theorem about radius of convergence

Theorem @theorem-7 Every power series representation of (or, Taylor series for) an entire-function has an infinite radius of convergence .

Theorem: A theorem about the uniqueness of Taylor series as @power-series expansions @theorem-8 If $f$ has a power series expansion about a point $z_0$ with nonzero radius of convergence, it must be the taylor series about $z_0$.

Theorem: A theorem about radius of convergence @theorem-9 The radius of convergence of the Taylor series for a function $f(z)$ about a point $z_0$ is the distance from $z_0$ to the nearest singularity of $f(z)$.

Here are some important and useful Maclaurin series that we can often use to find those of other functions:

$$ e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{1}{n!}z^n, |z| < \infty, $$

$$ \sin{z} = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}z^{2n+1}, |z| < \infty, $$

$$ \cos{z} = z - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}z^{2n}, |z| < \infty, $$

$$ \frac{1}{1 - z} = 1 + z + z^2 + z^3 + \cdots = \sum_{n = 0}^{\infty} z^n, |z| < 1. $$

Finding the Center and Radius of Convergence

Given a power series, we often want to know the center and radius of its circle of convergence. Finding its center is easy, from the form

$$ \sum_{n=0}^{\infty} a_n (z - z_0)^n, $$

the center is simply $z_0.$ When $z = z_0,$ all terms in the series become $0$ and so the power series always converges there.

To find the radius of convergence, we can shift the power series to be centered at $0,$ which makes it easier to manipulate.

To do this, let $w = z - z_0$ (or, if we have something like $\sum_{n=0}^{\infty} a_n (z - z_0)^{2n},$ then let $w = (z-z_0)^2.$

Now, the series becomes

$$ \sum_{n=0}^{\infty} a_n w^n. $$

Now, we can proceed with either to the ratio test or the root test. First, the root test.

$$ \left | \frac{a_{n+1}w^{a_{n+1}}}{a_n w^n} \right | = \left | \frac{a_{n+1}}{a_n} \right | |w|. $$

Here we can see that $w$ contributes only a factor of $|w|,$ which doesn't depend on $n.$ So, when we take the limit as $n \to \infty,$ all of the $n$-dependence is in the coefficient ratio $|a_{n+1}/a_n|.$ Thus,

$$ L = \lim_{n \to \infty} \left | \frac{a_{n+1}}{a_n} \right | \implies \text{series converges if} |w| < 1/L. $$

So, our radius of convergence for our $w$ series here is $1/L.$ We have to bring $z$ back, so for example, if we had something like $w = (z - z_0)^{mn},$ then our radius of convergence would be $(1/L)^{1/m}.$ We get that by

$$ |w| < R \implies |z^m| < R \implies |z| < R^{1/m}. $$

Similarly with the root test, we end up with something like

$$ |w| < \frac{1}{\limsup{|a_n|^{1/n}}}. $$

Binomial Series

Theorem: Binomial Series @binomial-series For complex $z$ and complex $\alpha,$ we have $$ (1 + z)^{\alpha} = \sum_{n=0}^{\infty} \binom{\alpha}{n} z^n. $$ If $|z| < 1,$ the series converges absolutely for any any complex $\alpha,$ including negative integers. Other convergence conditions are listed here. Referenced by (1 direct)Direct references:Laurent Series

Remark @remark-11 This can be viewed as an extension of the Binomial Theorem .