Taylor and Maclaurin Series: Learn It 2

Giselle Miller

Taylor and Maclaurin Series: Learn It 2

Taylor Polynomials

The [latex]n[/latex]th partial sum of the Taylor series for a function [latex]f[/latex] at [latex]a[/latex] is called the [latex]n[/latex]th Taylor polynomial.

The first few Taylor polynomials are:

[latex]p_0(x) = f(a)[/latex]
[latex]p_1(x) = f(a) + f'(a)(x-a)[/latex]
[latex]p_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2[/latex]
[latex]p_3(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3[/latex]

These are the [latex]0[/latex]th, [latex]1[/latex]st, [latex]2[/latex]nd, and [latex]3[/latex]rd Taylor polynomials of [latex]f[/latex] at [latex]a[/latex], respectively. When [latex]a = 0[/latex], these are called Maclaurin polynomials. We now provide a formal definition of Taylor and Maclaurin polynomials for a function [latex]f[/latex].

Taylor polynomial

If [latex]f[/latex] has [latex]n[/latex] derivatives at [latex]x = a[/latex], then the [latex]n[/latex]th Taylor polynomial for [latex]f[/latex] at [latex]a[/latex] is:

[latex]{p}_{n}\left(x\right)=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f^{\prime\prime}\left(a\right)}{2\text{!}}{\left(x-a\right)}^{2}+\frac{f^{\prime\prime\prime}\left(a\right)}{3\text{!}}{\left(x-a\right)}^{3}+\cdots +\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}[/latex].

The [latex]n[/latex]th Taylor polynomial for [latex]f[/latex] at [latex]0[/latex] is known as the [latex]n[/latex]th Maclaurin polynomial for [latex]f[/latex].

Let’s see how to use this definition by finding several Taylor polynomials for [latex]f(x) = \ln x[/latex] at [latex]x = 1[/latex].

Find the Taylor polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[/latex] and [latex]{p}_{3}[/latex] for [latex]f\left(x\right)=\text{ln}x[/latex] at [latex]x=1[/latex]. Use a graphing utility to compare the graph of [latex]f[/latex] with the graphs of [latex]{p}_{0},{p}_{1},{p}_{2}[/latex] and [latex]{p}_{3}[/latex].

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

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 now show how to find Maclaurin polynomials for [latex]e^{x}[/latex], [latex]\sin{x}[/latex], and [latex]\cos{x}[/latex]. As stated above, Maclaurin polynomials are Taylor polynomials centered at zero.

For each of the following functions, find formulas for the Maclaurin polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[/latex] and [latex]{p}_{3}[/latex]. Find a formula for the nth Maclaurin polynomial and write it using sigma notation. Use a graphing utilty to compare the graphs of [latex]{p}_{0},{p}_{1},{p}_{2}[/latex] and [latex]{p}_{3}[/latex] with [latex]f[/latex].

[latex]f\left(x\right)={e}^{x}[/latex]
[latex]f\left(x\right)=\sin{x}[/latex]
[latex]f\left(x\right)=\cos{x}[/latex]

Since [latex]f\left(x\right)={e}^{x}[/latex], we know that [latex]f\left(x\right)={f}^{\prime }\left(x\right)=f^{\prime\prime}\left(x\right)=\cdots ={f}^{\left(n\right)}\left(x\right)={e}^{x}[/latex] for all positive integers n. Therefore,

[latex]f\left(0\right)={f}^{\prime }\left(0\right)=f^{\prime\prime}\left(0\right)=\cdots ={f}^{\left(n\right)}\left(0\right)=1[/latex]

for all positive integers n. Therefore, we have

[latex]\begin{array}{ccc}\hfill {p}_{0}\left(x\right)& =\hfill & f\left(0\right)=1,\hfill \\ \hfill {p}_{1}\left(x\right)& =\hfill & f\left(0\right)+{f}^{\prime }\left(0\right)x=1+x,\hfill \\ \hfill {p}_{2}\left(x\right)& =\hfill & f\left(0\right)+{f}^{\prime }\left(0\right)x+\frac{f^{\prime\prime}\left(0\right)}{2\text{!}}{x}^{2}=1+x+\frac{1}{2}{x}^{2},\hfill \\ \hfill {p}_{3}\left(x\right)& =\hfill & f\left(0\right)+{f}^{\prime }\left(0\right)x+\frac{f^{\prime\prime}\left(0\right)}{2}{x}^{2}+\frac{f^{\prime\prime\prime}\left(0\right)}{3\text{!}}{x}^{3}\hfill \\ & =\hfill & 1+x+\frac{1}{2}{x}^{2}+\frac{1}{3\text{!}}{x}^{3},\hfill \\ \hfill {p}_{n}\left(x\right)& =\hfill & f\left(0\right)+{f}^{\prime }\left(0\right)x+\frac{f^{\prime\prime}\left(0\right)}{2}{x}^{2}+\frac{f^{\prime\prime\prime}\left(0\right)}{3\text{!}}{x}^{3}+\cdots +\frac{{f}^{\left(n\right)}\left(0\right)}{n\text{!}}{x}^{n}\hfill \\ & =\hfill & 1+x+\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{3}}{3\text{!}}+\cdots +\frac{{x}^{n}}{n\text{!}}\hfill \\ & =\hfill & {\displaystyle\sum _{k=0}^{n}}\frac{{x}^{k}}{k\text{!}}.\hfill \end{array}[/latex]

The function and the first three Maclaurin polynomials are shown in Figure 2.

This graph has four curves. The first is the function f(x)=e^x. The second function is psub0(x)=1. The third is psub1(x) which is an increasing line passing through y=1. The fourth function is psub3(x) which is a curve passing through y=1. The curves are very close around y= 1. — Figure 2. The graph shows the function [latex]y={e}^{x}[/latex] and the Maclaurin polynomials [latex]{p}_{0},{p}_{1},{p}_{2}[/latex] and [latex]{p}_{3}[/latex].

For [latex]f\left(x\right)=\sin{x}[/latex], the values of the function and its first four derivatives at [latex]x=0[/latex] are given as follows:

[latex]\begin{array}{cccccccc}\hfill f\left(x\right)& =\hfill & \sin{x}\hfill & & & \hfill f\left(0\right)& =\hfill & 0\hfill \\ \hfill {f}^{\prime }\left(x\right)& =\hfill & \cos{x}\hfill & & & \hfill {f}^{\prime }\left(0\right)& =\hfill & 1\hfill \\ \hfill f^{\prime\prime}\left(x\right)& =\hfill & -\sin{x}\hfill & & & \hfill f^{\prime\prime}\left(0\right)& =\hfill & 0\hfill \\ \hfill f^{\prime\prime\prime}\left(x\right)& =\hfill & -\cos{x}\hfill & & & \hfill f^{\prime\prime\prime}\left(0\right)& =\hfill & -1\hfill \\ \hfill {f}^{\left(4\right)}\left(x\right)& =\hfill & \sin{x}\hfill & & & \hfill {f}^{\left(4\right)}\left(0\right)& =\hfill & 0.\hfill \end{array}[/latex]

Since the fourth derivative is [latex]\sin{x}[/latex], the pattern repeats. That is, [latex]{f}^{\left(2m\right)}\left(0\right)=0[/latex] and [latex]{f}^{\left(2m+1\right)}\left(0\right)={\left(-1\right)}^{m}[/latex] for [latex]m\ge 0[/latex]. Thus, we have

[latex]\begin{array}{c}{p}_{0}\left(x\right)=0,\hfill \\ {p}_{1}\left(x\right)=0+x=x,\hfill \\ {p}_{2}\left(x\right)=0+x+0=x,\hfill \\ {p}_{3}\left(x\right)=0+x+0-\frac{1}{3\text{!}}{x}^{3}=x-\frac{{x}^{3}}{3\text{!}},\hfill \\ {p}_{4}\left(x\right)=0+x+0-\frac{1}{3\text{!}}{x}^{3}+0=x-\frac{{x}^{3}}{3\text{!}},\hfill \\ {p}_{5}\left(x\right)=0+x+0-\frac{1}{3\text{!}}{x}^{3}+0+\frac{1}{5\text{!}}{x}^{5}=x-\frac{{x}^{3}}{3\text{!}}+\frac{{x}^{5}}{5\text{!}},\hfill \end{array}[/latex]

and for [latex]m\ge 0[/latex],

[latex]\begin{array}{cc}\hfill {p}_{2m+1}\left(x\right)& ={p}_{2m+2}\left(x\right)\hfill \\ & =x-\frac{{x}^{3}}{3\text{!}}+\frac{{x}^{5}}{5\text{!}}-\cdots +{\left(-1\right)}^{m}\frac{{x}^{2m+1}}{\left(2m+1\right)\text{!}}\hfill \\ & ={\displaystyle\sum _{k=0}^{m}}{\left(-1\right)}^{k}\frac{{x}^{2k+1}}{\left(2k+1\right)\text{!}}.\hfill \end{array}[/latex]

Graphs of the function and its Maclaurin polynomials are shown in Figure 3.

Figure 3. The graph shows the function [latex]y=\sin{x}[/latex] and the Maclaurin polynomials [latex]{p}_{1},{p}_{3}[/latex] and [latex]{p}_{5}[/latex].
For [latex]f\left(x\right)=\cos{x}[/latex], the values of the function and its first four derivatives at [latex]x=0[/latex] are given as follows:

[latex]\begin{array}{cccccccc}\hfill f\left(x\right)& =\hfill & \cos{x}\hfill & & & \hfill f\left(0\right)& =\hfill & 1\hfill \\ \hfill {f}^{\prime }\left(x\right)& =\hfill & -\sin{x}\hfill & & & \hfill {f}^{\prime }\left(0\right)& =\hfill & 0\hfill \\ \hfill f^{\prime\prime}\left(x\right)& =\hfill & -\cos{x}\hfill & & & \hfill f^{\prime\prime}\left(0\right)& =\hfill & -1\hfill \\ \hfill f^{\prime\prime\prime}\left(x\right)& =\hfill & \sin{x}\hfill & & & \hfill f^{\prime\prime\prime}\left(0\right)& =\hfill & 0\hfill \\ \hfill {f}^{\left(4\right)}\left(x\right)& =\hfill & \cos{x}\hfill & & & \hfill {f}^{\left(4\right)}\left(0\right)& =\hfill & 1.\hfill \end{array}[/latex]

Since the fourth derivative is [latex]\sin{x}[/latex], the pattern repeats. In other words, [latex]{f}^{\left(2m\right)}\left(0\right)={\left(-1\right)}^{m}[/latex] and [latex]{f}^{\left(2m+1\right)}=0[/latex] for [latex]m\ge 0[/latex]. Therefore,

[latex]\begin{array}{c}{p}_{0}\left(x\right)=1,\hfill \\ {p}_{1}\left(x\right)=1+0=1,\hfill \\ {p}_{2}\left(x\right)=1+0-\frac{1}{2\text{!}}{x}^{2}=1-\frac{{x}^{2}}{2\text{!}},\hfill \\ {p}_{3}\left(x\right)=1+0-\frac{1}{2\text{!}}{x}^{2}+0=1-\frac{{x}^{2}}{2\text{!}},\hfill \\ {p}_{4}\left(x\right)=1+0-\frac{1}{2\text{!}}{x}^{2}+0+\frac{1}{4\text{!}}{x}^{4}=1-\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{4}}{4\text{!}},\hfill \\ {p}_{5}\left(x\right)=1+0-\frac{1}{2\text{!}}{x}^{2}+0+\frac{1}{4\text{!}}{x}^{4}+0=1-\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{4}}{4\text{!}},\hfill \end{array}[/latex]

and for [latex]n\ge 0[/latex],

[latex]\begin{array}{cc}\hfill {p}_{2m}\left(x\right)& ={p}_{2m+1}\left(x\right)\hfill \\ & =1-\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{4}}{4\text{!}}-\cdots +{\left(-1\right)}^{m}\frac{{x}^{2m}}{\left(2m\right)\text{!}}\hfill \\ & ={\displaystyle\sum _{k=0}^{m}}{\left(-1\right)}^{k}\frac{{x}^{2k}}{\left(2k\right)\text{!}}.\hfill \end{array}[/latex]

Graphs of the function and the Maclaurin polynomials appear in Figure 4.

Figure 4. The function [latex]y=\cos{x}[/latex] and the Maclaurin polynomials [latex]{p}_{0},{p}_{2}[/latex] and [latex]{p}_{4}[/latex] are plotted on this graph.

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

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).