Applications of Series: Learn It 1

  • Write out the terms in a binomial series
  • Find the Taylor series for different functions
  • Use Taylor series to solve differential equations
  • Use Taylor series to evaluate integrals that don’t have elementary antiderivatives

The Binomial Series

In the preceding section, we defined Taylor series and showed how to find the Taylor series for several common functions by explicitly calculating the coefficients of the Taylor polynomials. In this section we show how to use those Taylor series to derive Taylor series for other functions.

Our first goal is to determine the Maclaurin series for the function [latex]f\left(x\right)={\left(1+x\right)}^{r}[/latex] for all real numbers [latex]r[/latex]. The Maclaurin series for this function is known as the binomial series.

We begin by considering the simplest case: [latex]r[/latex] is a nonnegative integer. Let’s look at what happens when we expand [latex]f\left(x\right)={\left(1+x\right)}^{r}[/latex] for the first few values of [latex]r[/latex]:

[latex]\begin{array}{}\\ f\left(x\right)={\left(1+x\right)}^{0}=1,\hfill \\ f\left(x\right)={\left(1+x\right)}^{1}=1+x,\hfill \\ f\left(x\right)={\left(1+x\right)}^{2}=1+2x+{x}^{2},\hfill \\ f\left(x\right)={\left(1+x\right)}^{3}=1+3x+3{x}^{2}+{x}^{3},\hfill \\ f\left(x\right)={\left(1+x\right)}^{4}=1+4x+6{x}^{2}+4{x}^{3}+{x}^{4}.\hfill \end{array}[/latex]

The expressions on the right-hand side are known as binomial expansions and the coefficients are known as binomial coefficients.

binomial coefficients

For any nonnegative integer [latex]r[/latex], the binomial coefficient of [latex]{x}^{n}[/latex] in the binomial expansion of [latex]{\left(1+x\right)}^{r}[/latex] is given by:

[latex]\left(\begin{array}{c}r\ n\end{array}\right)=\frac{r!}{n!\left(r-n\right)!}[/latex]

Using this formula, we can write the general binomial expansion as:

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

For example, using this formula for [latex]r=5[/latex], we see that:

[latex]\begin{array}{cc}\hfill f\left(x\right)& ={\left(1+x\right)}^{5}\hfill \\ & =\left(\begin{array}{c}5\hfill \\ 0\hfill \end{array}\right)1+\left(\begin{array}{c}5\hfill \\ 1\hfill \end{array}\right)x+\left(\begin{array}{c}5\hfill \\ 2\hfill \end{array}\right){x}^{2}+\left(\begin{array}{c}5\hfill \\ 3\hfill \end{array}\right){x}^{3}+\left(\begin{array}{c}5\hfill \\ 4\hfill \end{array}\right){x}^{4}+\left(\begin{array}{c}5\hfill \\ 5\hfill \end{array}\right){x}^{5}\hfill \\ & =\frac{5\text{!}}{0\text{!}5\text{!}}1+\frac{5\text{!}}{1\text{!}4\text{!}}x+\frac{5\text{!}}{2\text{!}3\text{!}}{x}^{2}+\frac{5\text{!}}{3\text{!}2\text{!}}{x}^{3}+\frac{5\text{!}}{4\text{!}1\text{!}}{x}^{4}+\frac{5\text{!}}{5\text{!}0\text{!}}{x}^{5}\hfill \\ & =1+5x+10{x}^{2}+10{x}^{3}+5{x}^{4}+{x}^{5}.\hfill \end{array}[/latex]

We now consider the case when the exponent [latex]r[/latex] is any real number, not necessarily a nonnegative integer. If [latex]r[/latex] is not a nonnegative integer, then [latex]f\left(x\right)={\left(1+x\right)}^{r}[/latex] cannot be written as a finite polynomial. However, we can find a power series for [latex]f[/latex].

Specifically, we look for the Maclaurin series for [latex]f[/latex]. To do this, we find the derivatives of [latex]f[/latex] and evaluate them at [latex]x=0[/latex]:

[latex]\begin{array}{cccccccc}\hfill f\left(x\right)& =\hfill & {\left(1+x\right)}^{r}\hfill & & & \hfill f\left(0\right)& =\hfill & 1\hfill \\ \hfill {f}^{\prime }\left(x\right)& =\hfill & r{\left(1+x\right)}^{r - 1}\hfill & & & \hfill f\prime \left(0\right)& =\hfill & r\hfill \\ \hfill f^{\prime\prime}\left(x\right)& =\hfill & r\left(r - 1\right){\left(1+x\right)}^{r - 2}\hfill & & & \hfill f^{\prime\prime}\left(0\right)& =\hfill & r\left(r - 1\right)\hfill \\ \hfill f^{\prime\prime\prime}\left(x\right)& =\hfill & r\left(r - 1\right)\left(r - 2\right){\left(1+x\right)}^{r - 3}\hfill & & & \hfill f^{\prime\prime\prime}\left(0\right)& =\hfill & r\left(r - 1\right)\left(r - 2\right)\hfill \\ \hfill {f}^{\left(n\right)}\left(x\right)& =\hfill & r\left(r - 1\right)\left(r - 2\right)\cdots \left(r-n+1\right){\left(1+x\right)}^{r-n}\hfill & & & \hfill {f}^{\left(n\right)}\left(0\right)& =\hfill & r\left(r - 1\right)\left(r - 2\right)\cdots \left(r-n+1\right)\hfill \end{array}[/latex]

We conclude that the coefficients in the binomial series are given by:

[latex]\dfrac{{f}^{\left(n\right)}\left(0\right)}{n\text{!}}=\dfrac{r\left(r - 1\right)\left(r - 2\right)\cdots \left(r-n+1\right)}{n\text{!}}[/latex].
When [latex]r[/latex] is a nonnegative integer, the series automatically terminates (since higher derivatives become zero) and our generalized binomial coefficient formula reduces to the familiar [latex]\left(\begin{array}{c}r\ n\end{array}\right)=\frac{r!}{n!\left(r-n\right)!}[/latex]. This confirms our approach works for both finite and infinite cases.

More generally, to denote the binomial coefficients for any real number [latex]r[/latex], we define:

[latex]\left(\begin{array}{c}r\hfill \\ n\hfill \end{array}\right)=\frac{r\left(r - 1\right)\left(r - 2\right)\cdots \left(r-n+1\right)}{n\text{!}}[/latex].

With this notation, we can write the binomial series for [latex]{\left(1+x\right)}^{r}[/latex] as:

[latex]\displaystyle\sum _{n=0}^{\infty }\left(\begin{array}{c}r\hfill \\ n\hfill \end{array}\right){x}^{n}=1+rx+\frac{r\left(r - 1\right)}{2\text{!}}{x}^{2}+\cdots +\frac{r\left(r - 1\right)\cdots \left(r-n+1\right)}{n\text{!}}{x}^{n}+\cdots[/latex].

We now determine the interval of convergence for the binomial series using the ratio test:

[latex]\begin{array}{cc}\hfill \dfrac{|{a}_{n+1}|}{|{a}_{n}|}& =\dfrac{{|r\left(r - 1\right)\left(r - 2\right)\cdots \left(r-n\right)|x||}^{n+1}}{\left(n+1\right)\text{!}}\cdot \dfrac{n\text{!}}{|r\left(r - 1\right)\left(r - 2\right)\cdots \left(r-n+1\right)|{|x|}^{n}}\hfill \\ & =\dfrac{|r-n||x|}{|n+1|}.\hfill \end{array}[/latex]

 

Since [latex]\underset{n\to\infty}{\text{lim}}\frac{|{a}{n+1}|}{|{a}{n}|}=|x|[/latex], we have convergence when [latex]|x|<1[/latex].

The interval of convergence for the binomial series is [latex]\left(-1,1\right)[/latex]. The behavior at the endpoints depends on [latex]r[/latex]:

  • For [latex]r\geq 0[/latex]: the series converges at both endpoints
  • For [latex]-1 < r < 0[/latex]: the series converges at [latex]x=1[/latex] and diverges at [latex]x=-1[/latex]
  • For [latex]r<-1[/latex]: the series diverges at both endpoints

The binomial series does converge to [latex]{\left(1+x\right)}^{r}[/latex] in [latex]\left(-1,1\right)[/latex] for all real numbers [latex]r[/latex].

the binomial series

For any real number [latex]r[/latex], the Maclaurin series for [latex]f\left(x\right)={\left(1+x\right)}^{r}[/latex] is the binomial series:

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

for [latex]|x|<1[/latex].

We can use this definition to find the binomial series for [latex]f\left(x\right)=\sqrt{1+x}[/latex] and use the series to approximate [latex]\sqrt{1.5}[/latex].

  1. Find the binomial series for [latex]f\left(x\right)=\sqrt{1+x}[/latex].
  2. Use the third-order Maclaurin polynomial [latex]{p}_{3}\left(x\right)[/latex] to estimate [latex]\sqrt{1.5}[/latex]. Use Taylor’s theorem to bound the error. Use a graphing utility to compare the graphs of [latex]f[/latex] and [latex]{p}_{3}[/latex].

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

You can view the transcript for “6.4.1” here (opens in new window).