8.3: Taylor Series

The previous section showed that a power series converges to an analytic function inside its disk of convergence. Taylor’s theorem completes the story by giving the converse: around each point of analyticity an analytic function equals a convergent power series.

Theorem \(\PageIndex{1}\): Taylor's Theorem

Suppose \(f(z)\) is an analytic function in a region \(A\). Let \(z_0 \in A\). Then,

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

where the series converges on any disk \(|z - z_0| < r\) contained in \(A\). Furthermore, we have formulas for the coefficients

\[a_n = \dfrac{f^{(n)} (z_0)}{n!} = \dfrac{1}{2\pi i} \int_{\gamma} \dfrac{f(z)}{(z - z_0)^{n + 1}} \ dz.\]

(Where \(\gamma\) is any simple closed curve in \(A\) around \(z_0\), with its interior entirely in \(A\).)

We call the series the power series representing \(f\) around \(z_0\).

Proof

The proof will be given below. First we look at some consequences of Taylor’s theorem.

Corollary

The power series representing an analytic function around a point \(z_0\) is unique. That is, the coefficients are uniquely determined by the function \(f(z)\).

Proof

Taylor’s theorem gives a formula for the coefficients.

Order of a Zero

Theorem \(\PageIndex{2}\)

Suppose \(f(z)\) is analytic on the disk \(|z - z_0| < r\) and \(f\) is not identically 0. Then there is an integer \(k \ge 0\) such that \(a_k \ne 0\) and \(f\) has Taylor series around \(z_0\) given by

\[\begin{align} f(z) &= (z - z_0)^k (a_k + a_{k + 1} (z - z_0) + ...) \\[4pt] &= (z - z_0)^k \sum_{n = k}^{\infty} a_n (z - z_0)^{n - k}. \label{8.4.3} \end{align}\]

Proof

Since \(f(z)\) is not identically 0, not all the Taylor coefficients are zero. So, we take \(k\) to be the index of the first nonzero coefficient.

Theorem \(\PageIndex{3}\): Zeros are Isolated

If \(f(z)\) is analytic and not identically zero then the zeros of \(f\) are isolated. (By isolated we mean that we can draw a small disk around any zeros that doesn’t contain any other zeros.)

Figure \(\PageIndex{1}\): Isolated zero at \(z_0\): \(f(z_0) = 0\), \(f(z) \ne 0\) elsewhere in the disk. (CC BY-NC; Ümit Kaya)

Proof

Suppose \(f(z_0) = 0\). Write \(f\) as in Equation 8.4.3. There are two factors:

\[(z - z_0)^k \nonumber\]

and

\[g(z) = a_k + a_{k + 1} (z - z_0) + ... \nonumber\]

Clearly \((z - z_0)^k \ne 0\) if \(z \ne z_0\). We have \(g(z_0) = a_k \ne 0\), so \(g(z)\) is not 0 on some small neighborhood of \(z_0\). We conclude that on this neighborhood the product is only zero when \(z = z_0\), i.e. \(z_0\) is an isolated 0.

Definition: Order of the Zero

The integer \(k\) in Equation \ref{8.4.3} is called the order of the zero of \(f\) at \(z_0\).

Note, if \(f(z_0) \ne 0\) then \(z_0\) is a zero of order 0.