8.4: Taylor Series Examples

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,

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

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

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

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

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

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