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# 8.4: Taylor Series Examples

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The uniqueness of Taylor series along with the fact that they converge on any disk around $$z_0$$ where the function is analytic allows us to use lots of computational tricks to find the series and be sure that it converges.

##### Example $$\PageIndex{1}$$

Use the formula for the coefficients in terms of derivatives to give the Taylor series of $$f(z) = e^z$$ around $$z = 0$$.

Solution

Since $$f'(z) = e^z$$, we have $$f^{(n)} (0) = e^0 = 1$$. So,

$e^z = 1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!} + \ ... = \sum_{n = 0}^{\infty} \dfrac{z^n}{n!} \nonumber$

##### Example $$\PageIndex{2}$$

Expand $$f(z) = z^8 e^{3z}$$ in a Taylor series around $$z = 0$$.

Solution

Let $$w = 3z$$. So,

$e^{3z} = e^w = \sum_{n = 0}^{\infty} \dfrac{w^n}{n!} = \sum_{k = 0}^{\infty} \dfrac{3^n}{n!} z^n \nonumber$

Thus,

$f(z) = \sum_{n = 0}^{\infty} \dfrac{3^n}{n!} z^{n + 8}. \nonumber$

##### Example $$\PageIndex{3}$$

Find the Taylor series of $$\sin (z)$$ around $$z = 0$$ (Sometimes the Taylor series around 0 is called the Maclaurin series.)

Solution

We give two methods for doing this.

Method 1.

$f^{(n)} (0) = \dfrac{d^n \sin (z)}{dz^n} = \begin{cases} (-1)^m & \text{ for } n = 2m + 1 = \text{ odd}, m = 0, 1, 2,\ ... \\ 0 & \text{ for } n \text{ even} \end{cases} \nonumber$

Method 2. Using

$\sin (z) = \dfrac{e^{iz} - e^{-iz}}{2i}, \nonumber$

we have

\begin{align*} \sin (z) &= \dfrac{1}{2i} \left[\sum_{n = 0}^{\infty} \dfrac{(iz)^n}{n!} - \sum_{n = 0}^{\infty} \dfrac{(-iz)^n}{n!}\right] \\[4pt] &= \dfrac{1}{2i} \sum_{n = 0}^{\infty} [(1 - (-1)^n)] \dfrac{i^n z^n}{n!}\end{align*}

(We need absolute convergence to add series like this.)

Conclusion:

$\sin (z) = \sum_{n = 0}^{\infty} (-1)^n \dfrac{z^{2n + 1}}{(2n + 1)!},\nonumber$

which converges for $$|z| < \infty$$.

##### Example $$\PageIndex{4}$$

Expand the rational function

$f(z) = \dfrac{1 + 2z^2}{z^3 + z^5}\nonumber$

around $$z = 0$$.

Solution

Note that $$f$$ has a singularity at 0, so we can’t expect a convergent Taylor series expansion. We’ll aim for the next best thing using the following shortcut.

$f(z) = \dfrac{1}{z^3} \dfrac{2(1 + z^2) - 1}{1 + z^2} = \dfrac{1}{z^3} [ 2 - \dfrac{1}{1 + z^2}].\nonumber$

Using the geometric series we have

$\dfrac{1}{1 + z^2} = \dfrac{1}{1 - (-z^2)} = \sum_{n = 0}^{\infty} (-z^2)^n = 1 - z^2 + z^4 - z^6 +...\nonumber$

Putting it all together

$f(z) = \dfrac{1}{z^3} (2 - 1 + z^2 - z^4 + ...) = \left(\dfrac{1}{z^3} + \dfrac{1}{z}\right) - \sum_{n = 0}^{\infty} (-1)^n z^{2n + 1}\nonumber$

Note: The first terms are called the singular part, i.e. those with negative powers of $$z$$. he summation is called the regular or analytic part. Since the geometric series for $$1/(1 + z^2)$$ converges for $$|z| < 1$$, the entire series is valid in $$0 < |z| < 1$$.

##### Example $$\PageIndex{5}$$

Find the Taylor series for

$f(z) = \dfrac{e^z}{1 - z}\nonumber$

around $$z = 0$$. Give the radius of convergence.

Solution

We start by writing the Taylor series for each of the factors and then multiply them out.

$\begin{array} {rcl} {f(z)} & = & {\left(1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!} + \ ...\right) (1 + z + z^2 + z^3 + \ ...)} \\ {} & = & {1 + (1 + 1) z + \left(1 + 1 + \dfrac{1}{2!}\right) z^2 + \left(1 + 1 + \dfrac{1}{2!} + \dfrac{1}{3!}\right) z^3 + \ ...} \end{array}\nonumber$

The biggest disk around $$z = 0$$ where $$f$$ is analytic is $$|z| < 1$$. Therefore, by Taylor’s theorem, the radius of convergence is $$R = 1$$. $$f(z)$$ is analytic on $$|z| < 1$$ and has a singularity at $$z = 1$$. (CC BY-NC; Ümit Kaya)
##### Example $$\PageIndex{6}$$

Find the Taylor series for

$f(z) = \dfrac{1}{1 - z}\nonumber$

around $$z = 5$$. Give the radius of convergence.

Solution

We have to manipulate this into standard geometric series form.

$f(z) = \dfrac{1}{-4(1 + (z - 5)/4)} = -\dfrac{1}{4} \left[1 - \left(\dfrac{z - 5}{4}\right) + \left(\dfrac{z - 5}{4}\right)^2 - \left(\dfrac{z - 5}{4}\right)^3 + \ ...\right]\nonumber$

Since $$f(z)$$ has a singularity at $$z = 1$$ the radius of convergence is $$R = 4$$. We can also see this by considering the geometric series. The geometric series ratio is $$(z - 5)/4$$. So the series converges when $$|z - 5|/4 < 1$$, i.e. when $$|z - 5| < 4$$, i.e. $$R = 4$$. Disk of convergence stops at the singularity at $$z = 1$$. (CC BY-NC; Ümit Kaya)
##### Example $$\PageIndex{7}$$

Find the Taylor series for

$f(z) = \log (1 + z)\nonumber$

around $$z = 0$$. Give the radius of convergence.

Solution

We know that $$f$$ is analytic for $$|z| < 1$$ and not analytic at $$z = -1$$. So, the radius of convergence is $$R = 1$$. To find the series representation we take the derivative and use the geometric series.

$f'(z) = \dfrac{1}{1 + z} = 1 - z + z^2 - z^3 + z^4 - \ ...\nonumber$

Integrating term by term (allowed by Theorem 8.3.1) we have

$f(z) = a_0 + z - \dfrac{z^2}{2} + \dfrac{z^3}{3} - \dfrac{z^4}{4} + \ ... = a_0 + \sum_{n = 1}^{\infty} (-1)^{n - 1} \dfrac{z^n}{n}\nonumber$

Here $$a_0$$ is the constant of integration. We find it by evalating at $$z = 0$$.

$f(0) = a_0 = \log (1) = 0.\nonumber$ Disk of convergence for $$\log (1 + z)$$ around $$z = 0$$. (CC BY-NC; Ümit Kaya)
##### Example $$\PageIndex{8}$$

Can the series

$\sum a_n (z - 2)^n\nonumber$

converge at $$z = 0$$ and diverge at $$z = 3$$.

Solution

No! We have $$z_0 = 2$$. We know the series diverges everywhere outside its radius of convergence. So, if the series converges at $$z = 0$$, then the radius of convergence is at least 2. Since $$|3 - z_0| < 2$$ we would also have that $$z = 3$$ is inside the disk of convergence.

## Proof of Taylor’s Theorem

For convenience we restate Taylor’s Theorem $$\PageIndex{1}$$.

##### Theorem $$\PageIndex{1}$$: Taylor’s Theorem (Taylor Series)

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

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

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

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

Proof

In order to handle convergence issues we fix $$0 < r_1 < r_2 < r$$. We let $$\gamma$$ be the circle $$|w - z_0| = r_2$$ (traversed counterclockise). Disk of convergence extends to the boundary of $$A$$ with $$r_1 < r_2 < r$$, but $$r_1$$ and $$r_2$$ can be arbitrarily close to $$r$$. (CC BY-NC; Ümit Kaya)

Take $$z$$ inside the disk $$|z - z_0| < r_1$$. We want to express $$f(z)$$ as a power series around $$z_0$$. To do this we start with the Cauchy integral formula and then use the geometric series.

As preparation we note that for $$w$$ on $$\gamma$$ and $$|z - z_0| < r_1$$ we have

$|z - z_0| < r_1 < r_2 = |w - z_0| \nonumber,$

so

$\dfrac{|z - z_0|}{|w - z_0|} < 1. \nonumber$

Therefore,

$\dfrac{1}{w - z} = \dfrac{1}{w - z_0} \cdot \dfrac{1}{1 - \dfrac{z - z_0}{w - z_0}} = \dfrac{1}{w - z_0} \sum_{n = 0}^{\infty} (\dfrac{z - z_0}{w - z_0})^n = \sum_{n = 0}^{\infty} \dfrac{(z - z_0)^n}{(w - z_0)^{n + 1}} \nonumber$

Using this and the Cauchy formula gives

$\begin{array} {rcl} {f(z)} & = & {\dfrac{1}{2\pi i} \int_{\gamma} \dfrac{f(w)}{w - z}\ dw} \\ {} & = & {\dfrac{1}{2\pi i} \int_{\gamma} \sum_{n = 0}^{\infty} \dfrac{f(w)}{(w - z_0)^{n + 1}} (z - z_0)^n\ dw} \\ {} & = & {\sum_{n = 0}^{\infty} (\dfrac{1}{2\pi i} \int_{\gamma} \dfrac{f(w)}{(w - z_0)^{n + 1}} \ dw) (z - z_0)^n} \\ {} & = & {\sum_{n = 0}^{\infty} \dfrac{f^{(n)} (z_0)}{n!} (z - z_0)^n} \end{array} \nonumber$

The last equality follows from Cauchy’s formula for derivatives. Taken together the last two equalities give Taylor’s formula. QED

8.4: Taylor Series Examples 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.