8.4: Taylor Series Examples
- Page ID
- 6519
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.
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,
\[e^z = 1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!} + \ ... = \sum_{n = 0}^{\infty} \dfrac{z^n}{n!} \nonumber \]
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,
\[f(z) = \sum_{n = 0}^{\infty} \dfrac{3^n}{n!} z^{n + 8}. \nonumber \]
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 \(|z| < \infty\).
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 \(0 < |z| < 1\).
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 \(R = 1\).
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. \(R = 4\).
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 \(z = 0\).
\[f(0) = a_0 = \log (1) = 0.\nonumber \]
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 \(\PageIndex{1}\).
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