Taylor and Maclaurin Series: Learn It 1

  • Learn how to find Taylor polynomials of a given order for a function
  • Estimate the remainder when using a Taylor series to approximate a function
  • Determine when a Taylor series converges to the original function

Overview of Taylor/Maclaurin Series

In the previous sections, you learned to find power series representations for functions related to geometric series. Now we’ll tackle a broader question: Which functions can be represented by power series, and how do we find these representations?

We’ll also address a crucial follow-up: If we find a power series for a function [latex]f[/latex] that converges on some interval, how do we prove the series actually converges to [latex]f[/latex]?

Finding the Right Coefficients

Consider a function [latex]f[/latex] that has a power series representation at [latex]x = a[/latex]. The series takes the form:

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

The key question is: What should the coefficients be? For now, we’ll focus on determining the coefficients and address convergence later.

If this series represents [latex]f[/latex] at [latex]x = a[/latex], we want the series to equal [latex]f(a)[/latex] when [latex]x = a[/latex]. Evaluating the series at [latex]x = a[/latex]:

[latex]\begin{array}{cc}\hfill \displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}& ={c}_{0}+{c}_{1}\left(a-a\right)+{c}_{2}{\left(a-a\right)}^{2}+\cdots \hfill \\ & ={c}_{0}.\hfill \end{array}[/latex]

Therefore, for the series to equal [latex]f(a)[/latex], we need [latex]c_0 = f(a)[/latex].

Matching Derivatives

We also want the first derivative of our power series to equal [latex]f'(a)[/latex] at [latex]x = a[/latex]. Differentiating term-by-term:

[latex]\frac{d}{dx}\left(\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)={c}_{1}+2{c}_{2}\left(x-a\right)+3{c}_{3}{\left(x-a\right)}^{2}+\cdots[/latex].

At [latex]x = a[/latex], this becomes:

[latex]\begin{array}{}\\ \\ \hfill \frac{d}{dx}\left(\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& ={c}_{1}+2{c}_{2}\left(a-a\right)+3{c}_{3}{\left(a-a\right)}^{2}+\cdots \hfill \\ & ={c}_{1}.\hfill \end{array}[/latex]

So for the derivative to match, we need [latex]c_1 = f'(a)[/latex].

Continuing this pattern, the second and third derivatives are:

[latex]\frac{{d}^{2}}{d{x}^{2}}\left(\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)=2{c}_{2}+3\cdot 2{c}_{3}\left(x-a\right)+4\cdot 3{c}_{4}{\left(x-a\right)}^{2}+\cdots[/latex]

and

[latex]\frac{{d}^{3}}{d{x}^{3}}\left(\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)=3\cdot 2{c}_{3}+4\cdot 3\cdot 2{c}_{4}\left(x-a\right)+5\cdot 4\cdot 3{c}_{5}{\left(x-a\right)}^{2}+\cdots[/latex].

At [latex]x = a[/latex], these derivatives evaluate to:

[latex]\begin{array}{cc}\hfill \frac{{d}^{2}}{d{x}^{2}}\left(\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& =2{c}_{2}+3\cdot 2{c}_{3}\left(a-a\right)+4\cdot 3{c}_{4}{\left(a-a\right)}^{2}+\cdots \hfill \\ & =2{c}_{2}\hfill \end{array}[/latex]

and

[latex]\begin{array}{cc}\hfill \frac{{d}^{3}}{d{x}^{3}}\left(\displaystyle\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& =3\cdot 2{c}_{3}+4\cdot 3\cdot 2{c}_{4}\left(a-a\right)+5\cdot 4\cdot 3{c}_{5}{\left(a-a\right)}^{2}+\cdots \hfill \\ & =3\cdot 2{c}_{3}\hfill \end{array}[/latex]

For these to equal [latex]f''(a)[/latex] and [latex]f'''(a)[/latex] respectively, we need:

  • [latex]c_2 = \frac{f''(a)}{2}[/latex]
  • [latex]c_3 = \frac{f'''(a)}{3 \cdot 2} = \frac{f'''(a)}{3!}[/latex]

The General Pattern

Following this pattern, if [latex]f[/latex] has a power series representation at [latex]x = a[/latex], the coefficients must be:

[latex]{c}_{n}=\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}[/latex]

This gives us the complete series:

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

This power series for [latex]f[/latex] is known as the Taylor series for [latex]f[/latex] at [latex]a[/latex]. If [latex]a=0[/latex], then this series is known as the Maclaurin series for [latex]f[/latex].

Taylor and Maclaurin series

If [latex]f[/latex] has derivatives of all orders at [latex]x=a[/latex], then the Taylor series for the function [latex]f[/latex] at [latex]a[/latex] is

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

 

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

Uniqueness of Taylor Series

Power series representations are unique, which leads to an important result: if a function [latex]f[/latex] has a power series at [latex]a[/latex], then it must be the Taylor series for [latex]f[/latex] at [latex]a[/latex].

theorem: uniqueness of Taylor series

If a function [latex]f[/latex] has a power series at [latex]a[/latex] that converges to [latex]f[/latex] on some open interval containing [latex]a[/latex], then that power series must be the Taylor series for [latex]f[/latex] at [latex]a[/latex].

To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as Taylor polynomials.