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# 8.3: Taylor Series

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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.)

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.

8.3: Taylor Series is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Jeremy Orloff (MIT OpenCourseWare) via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.