8.3: Taylor Series
- Page ID
- 6518
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.
Suppose \(f(z)\) is an analytic function in a region \(A\). Let \(z_0 \in A\). Then,
\[f(z) = \sum_{n = 0}^{\infty} a_n (z - z_0)^n, \nonumber \]
where the series converges on any disk \(|z - z_0| < r\) contained in \(A\). Furthermore, we have formulas for the coefficients
\[a_n = \dfrac{f^{(n)} (z_0)}{n!} = \dfrac{1}{2\pi i} \int_{\gamma} \dfrac{f(z)}{(z - z_0)^{n + 1}} \ dz. \nonumber \]
(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
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The proof will be given below. First we look at some consequences of Taylor’s theorem.
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
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Taylor’s theorem gives a formula for the coefficients.
Order of a Zero
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
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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.
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.)
- Proof
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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.
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.