Taylor and Maclaurin Series: Learn It 4

Giselle Miller

Taylor and Maclaurin Series: Learn It 4

Convergence of Taylor Series

We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.

Find the Taylor series for [latex]f\left(x\right)=\frac{1}{x}[/latex] at [latex]x=1[/latex]. Determine the interval of convergence.

For [latex]f\left(x\right)=\frac{1}{x}[/latex], the values of the function and its first four derivatives at [latex]x=1[/latex] are

[latex]\begin{array}{cccccccc}\hfill f\left(x\right)& =\hfill & \frac{1}{x}\hfill & & & \hfill f\left(1\right)& =\hfill & 1\hfill \\ \hfill {f}^{\prime }\left(x\right)& =\hfill & -\frac{1}{{x}^{2}}\hfill & & & \hfill {f}^{\prime }\left(1\right)& =\hfill & -1\hfill \\ \hfill f^{\prime\prime}\left(x\right)& =\hfill & \frac{2}{{x}^{3}}\hfill & & & \hfill f^{\prime\prime}\left(1\right)& =\hfill & 2\text{!}\hfill \\ \hfill f^{\prime\prime\prime}\left(x\right)& =\hfill & -\frac{3\cdot 2}{{x}^{4}}\hfill & & & \hfill f^{\prime\prime\prime}\left(1\right)& =\hfill & -3\text{!}\hfill \\ \hfill {f}^{\left(4\right)}\left(x\right)& =\hfill & \frac{4\cdot 3\cdot 2}{{x}^{5}}\hfill & & & \hfill {f}^{\left(4\right)}\left(1\right)& =\hfill & 4\text{!.}\hfill \end{array}[/latex]

That is, we have [latex]{f}^{\left(n\right)}\left(1\right)={\left(-1\right)}^{n}n\text{!}[/latex] for all [latex]n\ge 0[/latex]. Therefore, the Taylor series for [latex]f[/latex] at [latex]x=1[/latex] is given by

[latex]\displaystyle\sum _{n=0}^{\infty }\frac{{f}^{\left(n\right)}\left(1\right)}{n\text{!}}{\left(x - 1\right)}^{n}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{\left(x - 1\right)}^{n}[/latex].

To find the interval of convergence, we use the ratio test. We find that

[latex]\frac{|{a}_{n+1}|}{|{a}_{n}|}=\frac{|{\left(-1\right)}^{n+1}{\left(x - 1\right)}^{n+1}|}{|{\left(-1\right)}^{n}{\left(x - 1\right)}^{n}|}=|x - 1|[/latex].

Thus, the series converges if [latex]|x - 1|<1[/latex]. That is, the series converges for [latex]0

[latex]\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{\left(2 - 1\right)}^{n}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}[/latex]

diverges by the divergence test. Similarly, at [latex]x=0[/latex],

[latex]\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{\left(0 - 1\right)}^{n}=\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{2n}=\displaystyle\sum _{n=0}^{\infty }1[/latex]

diverges. Therefore, the interval of convergence is [latex]\left(0,2\right)[/latex].

Watch the following video to see the worked solution to the example above.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “6.3 Taylor and Maclaurin Series” here (opens in new window).

We know that the Taylor series found in this example converges on the interval [latex]\left(0,2\right)[/latex], but how do we know it actually converges to [latex]f?[/latex] Let’s explore this by rewriting our function:

[latex]f\left(x\right)=\frac{1}{x}=\frac{1}{1-\left(1-x\right)}[/latex].

This representation shows that [latex]f[/latex] can be expressed as the geometric series [latex]\displaystyle\sum_{n=0}^{\infty}(1-x)^n[/latex]. Since geometric series converge to [latex]\frac{1}{x}[/latex] when [latex]|1-x| < 1[/latex], we’ve confirmed that our Taylor series does indeed converge to [latex]f(x) = \frac{1}{x}[/latex] on [latex]\left(0,2\right)[/latex].

Now for the broader question: If a Taylor series for a function [latex]f[/latex] converges on some interval, how do we know it converges to [latex]f[/latex] itself?

To answer this, remember that a series converges to a value if and only if its sequence of partial sums converges to that value. For a Taylor series centered at [latex]a[/latex], the [latex]n[/latex]th partial sum is the [latex]n[/latex]th Taylor polynomial [latex]p_n[/latex].

Therefore, the Taylor series converges to [latex]f[/latex] when:

[latex]\underset{n\to \infty }{\text{lim}}{p}_{n}\left(x\right)=f\left(x\right)[/latex].

Since the remainder [latex]{R}_{n}\left(x\right)=f\left(x\right)-{p}_{n}\left(x\right)[/latex], the Taylor series converges to [latex]f[/latex] if and only if

[latex]\underset{n\to \infty }{\text{lim}}{R}_{n}\left(x\right)=0[/latex].

We now state this theorem formally.

theorem: convergence of Taylor series

Suppose that [latex]f[/latex] has derivatives of all orders on an interval [latex]I[/latex] containing [latex]a[/latex]. Then the Taylor series

[latex]\displaystyle\sum _{n=0}^{\infty }\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}[/latex]

converges to [latex]f\left(x\right)[/latex] for all [latex]x[/latex] in [latex]I[/latex] if and only if

[latex]\underset{n\to\infty}\lim {R}_{n}\left(x\right)=0[/latex]

for all [latex]x[/latex] in [latex]I[/latex].

With this theorem, we can prove that a Taylor series for [latex]f[/latex] at [latex]a[/latex] converges to [latex]f[/latex] if we can prove that the remainder [latex]{R}_{n}\left(x\right)\to 0[/latex]. To prove that [latex]{R}_{n}\left(x\right)\to 0[/latex], we typically use the bound from Taylor’s theorem:

[latex]|{R}_{n}\left(x\right)|\le \frac{M}{\left(n+1\right)\text{!}}{|x-a|}^{n+1}[/latex]

where [latex]M[/latex] is an upper bound for the [latex](n+1)[/latex]th derivative on the interval.

In our next example, we’ll find the Maclaurin series for [latex]e^x[/latex] and [latex]\sin x[/latex], then prove these series converge to their respective functions for all real numbers by showing [latex]R_n(x) \to 0[/latex].

For each of the following functions, find the Maclaurin series and its interval of convergence. Use Taylor’s Theorem with Remainder to prove that the Maclaurin series for [latex]f[/latex] converges to [latex]f[/latex] on that interval.

[latex]e^{x}[/latex]
[latex]\sin{x}[/latex]

Using the nth Maclaurin polynomial for [latex]e^{x}[/latex] found in the example: Finding Maclaurin Polynomials (a), we find that the Maclaurin series for [latex]e^{x}[/latex] is given by

[latex]\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}[/latex].

To determine the interval of convergence, we use the ratio test. Since

[latex]\frac{|{a}_{n+1}|}{|{a}_{n}|}=\frac{{|x|}^{n+1}}{\left(n+1\right)\text{!}}\cdot \frac{n\text{!}}{{|x|}^{n}}=\frac{|x|}{n+1}[/latex],

we have

[latex]\underset{n\to \infty }{\text{lim}}\frac{|{a}_{n+1}|}{|{a}_{n}|}=\underset{n\to \infty }{\text{lim}}\frac{|x|}{n+1}=0[/latex]

for all x. Therefore, the series converges absolutely for all x, and thus, the interval of convergence is [latex]\left(-\infty ,\infty \right)[/latex]. To show that the series converges to [latex]e^{x}[/latex] for all [latex]x[/latex], we use the fact that [latex]{f}^{\left(n\right)}\left(x\right)={e}^{x}[/latex] for all [latex]n\ge 0[/latex] and [latex]e^{x}[/latex] is an increasing function on [latex]\left(-\infty ,\infty \right)[/latex]. Therefore, for any real number [latex]b[/latex], the maximum value of [latex]e^{x}[/latex] for all [latex]|x|\le b[/latex] is [latex]e^{b}[/latex]. Thus,

[latex]|{R}_{n}\left(x\right)|\le \frac{{e}^{b}}{\left(n+1\right)\text{!}}{|x|}^{n+1}[/latex].

Since we just showed that

[latex]\displaystyle\sum _{n=0}^{\infty }\frac{|x{|}^{n}}{n\text{!}}[/latex]

converges for all x, by the divergence test, we know that

[latex]\underset{n\to \infty }{\text{lim}}\frac{{|x|}^{n+1}}{\left(n+1\right)\text{!}}=0[/latex]

for any real number [latex]x[/latex]. By combining this fact with the squeeze theorem, the result is [latex]\underset{n\to \infty }{\text{lim}}{R}_{n}\left(x\right)=0[/latex].
Using the nth Maclaurin polynomial for [latex]\sin{x}[/latex] found in the example: Finding Maclaurin Polynomials (b), we find that the Maclaurin series for [latex]\sin{x}[/latex] is given by

[latex]\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{\left(2n+1\right)\text{!}}[/latex].

In order to apply the ratio test, consider

[latex]\frac{|{a}_{n+1}|}{|{a}_{n}|}=\frac{{|x|}^{2n+3}}{\left(2n+3\right)\text{!}}\cdot \frac{\left(2n+1\right)\text{!}}{{|x|}^{2n+1}}=\frac{{|x|}^{2}}{\left(2n+3\right)\left(2n+2\right)}[/latex].

Since

[latex]\underset{n\to \infty }{\text{lim}}\frac{{|x|}^{2}}{\left(2n+3\right)\left(2n+2\right)}=0[/latex]

for all [latex]x[/latex], we obtain the interval of convergence as [latex]\left(-\infty ,\infty \right)[/latex]. To show that the Maclaurin series converges to [latex]\sin{x}[/latex], look at [latex]{R}_{n}\left(x\right)[/latex]. For each [latex]x[/latex] there exists a real number [latex]c[/latex] between 0 and [latex]x[/latex] such that

[latex]{R}_{n}\left(x\right)=\frac{{f}^{\left(n+1\right)}\left(c\right)}{\left(n+1\right)\text{!}}{x}^{n+1}[/latex].

Since [latex]|{f}^{\left(n+1\right)}\left(c\right)|\le 1[/latex] for all integers [latex]n[/latex] and all real numbers [latex]c[/latex], we have

[latex]|{R}_{n}\left(x\right)|\le \frac{{|x|}^{n+1}}{\left(n+1\right)\text{!}}[/latex]

for all real numbers [latex]x[/latex]. Using the same idea as in part a., the result is [latex]\underset{n\to \infty }{\text{lim}}{R}_{n}\left(x\right)=0[/latex] for all [latex]x[/latex], and therefore, the Maclaurin series for [latex]\sin{x}[/latex] converges to [latex]\sin{x}[/latex] for all real [latex]x[/latex].

Watch the following video to see the worked solution to the example above.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “6.3 Taylor and Maclaurin Series” here (opens in new window).