Taylor Series | Sahithyan's Notes

If $f(z)$ is analytic at $z = \alpha$ , the series

\sum_{n=0}^{\infty} \frac{f^{(n)}(\alpha)}{n!}(z-\alpha)^n = f(\alpha) + f'(\alpha)(z-\alpha) + \frac{f''(\alpha)}{2!}(z-\alpha)^2 + \cdots

is called the Taylor series of $f(z)$ centered at $\alpha$ . When $\alpha = 0$ , it is called the Maclaurin series of $f(z)$ .

Taylor’s Theorem

Let $f$ be analytic in a domain $G$ , and let the disk $D_R(\alpha)$ be contained in $G$ . Then the Taylor series of $f$ centered at $\alpha$ converges to $f(z)$ for all $z \in D_R(\alpha)$ .

f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(\alpha)}{n!}(z-\alpha)^n, \quad z \in D_R(\alpha)

The convergence is uniform on every closed disk $\{z : |z-\alpha| \le r\}$ with $0 < r < R$ .

Radius from Singular Points

Let $f$ be analytic in a domain $G$ containing $\alpha$ . Let $z_0$ be a nearest singular point of $f$ to $\alpha$ , and let $R = |z_0 - \alpha|$ .

The Taylor series of $f$ centered at $\alpha$ converges to $f(z)$ for each $z \in D_R(\alpha)$ .
No disk centered at $\alpha$ with a radius larger than $R$ can be a disk on which the series represents $f$ .

Uniqueness

Suppose

f(z) = \sum_{n=0}^{\infty} a_n (z-\alpha)^n = \sum_{n=0}^{\infty} b_n (z-\alpha)^n

in some disk $D_r(\alpha)$ with $r > 0$ . Then $a_n = b_n$ for each $n \in \mathbb{N} \cup \{0\}$ .

Any power series representation of $f(z)$ in a disk centered at $\alpha$ is automatically the Taylor series of $f$ centered at $\alpha$ .

Examples:

$\frac{1}{(1-z)^2} = \sum_{n=0}^{\infty} (n+1) z^n$ is valid for $z \in D_1(0)$
$\sin z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}$

Worked Examples

Maclaurin Series of Sine

Find the Maclaurin series of $\sin z$ .

The derivatives at $0$ cycle through $\sin z, \cos z, -\sin z, -\cos z$ , so

f(0) = 0, \quad f'(0) = 1, \quad f''(0) = 0, \quad f'''(0) = -1, \quad \ldots

Only odd powers survive.

\sin z = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+1}}{(2n+1)!}

$\sin z$ is entire, so the series is valid for all $z$ .

Series by Differentiation

Show that $\frac{1}{(1-z)^2} = \sum_{n=0}^{\infty} (n+1) z^n$ for $z \in D_1(0)$ .

Start with the geometric series.

\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n, \quad |z| < 1

Differentiate both sides term by term.

\frac{1}{(1-z)^2} = \sum_{n=1}^{\infty} n z^{n-1} = \sum_{n=0}^{\infty} (n+1) z^n

The only singular point is $z = 1$ , so $R = 1$ .

Series Centered at a Nonzero Point

Find the Taylor series of $\frac{1}{z}$ centered at $\alpha = 1$ and its disk of validity.

Write $z = 1 + (z-1)$ .

\frac{1}{z} = \frac{1}{1 + (z-1)} = \sum_{n=0}^{\infty} (-1)^n (z-1)^n

The geometric series converges for $|z-1| < 1$ . The nearest singular point is $0$ , so $R = |0 - 1| = 1$ and the disk is $D_1(1)$ .

Series Valid in a Disk

Find the Taylor series of $\frac{1}{z+2}$ centered at $0$ and its disk of validity.

\frac{1}{z+2} = \frac{1}{2} \cdot \frac{1}{1 + \frac{z}{2}} = \sum_{n=0}^{\infty} \frac{(-1)^n z^n}{2^{n+1}}

The nearest singular point is $-2$ , so $R = 2$ and the disk is $D_2(0)$ .

Coefficients from Uniqueness

Find $a_0, \ldots, a_4$ if $\sum_{n=0}^{\infty} a_n z^n = e^z + z^2$ near $0$ .

By uniqueness, the $a_n$ are the Taylor coefficients of $e^z + z^2$ .

e^z + z^2 = 1 + z + \left(\frac{1}{2} + 1\right) z^2 + \frac{z^3}{6} + \frac{z^4}{24} + \cdots

a_0 = 1, \quad a_1 = 1, \quad a_2 = \frac{3}{2}, \quad a_3 = \frac{1}{6}, \quad a_4 = \frac{1}{24}

Polynomial Times Cosine

Find the Maclaurin series of $z^2 \cos z$ and its radius of convergence.

z^2 \cos z = z^2 \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n}}{(2n)!} = \sum_{n=0}^{\infty} \frac{(-1)^n z^{2n+2}}{(2n)!}

$z^2 \cos z$ is entire, so $R = \infty$ .