Common Functions Expressed as Taylor Series
At this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form [latex]f\left(x\right)={\left(1+x\right)}^{r}[/latex]. The table below summarizes these important series.
| Function | Maclaurin Series | Interval of Convergence |
|---|---|---|
| [latex]f\left(x\right)=\frac{1}{1-x}[/latex] | [latex]\displaystyle\sum _{n=0}^{\infty }{x}^{n}[/latex] | [latex]-1 < x < 1[/latex] |
| [latex]f\left(x\right)={e}^{x}[/latex] | [latex]\displaystyle\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}[/latex] | [latex]\text{-}\infty < x < \infty[/latex] |
| [latex]f\left(x\right)=\sin{x}[/latex] | [latex]\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{\left(2n+1\right)\text{!}}[/latex] | [latex]\text{-}\infty < x < \infty[/latex] |
| [latex]f\left(x\right)=\cos{x}[/latex] | [latex]\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{2n}}{\left(2n\right)\text{!}}[/latex] | [latex]\text{-}\infty < x < \infty[/latex] |
| [latex]f\left(x\right)=\text{ln}\left(1+x\right)[/latex] | [latex]\displaystyle\sum _{n=1}^{\infty }{\left(-1\right)}^{n+1}\frac{{x}^{n}}{n}[/latex] | [latex]-1 < x\le 1[/latex] |
| [latex]f\left(x\right)={\tan}^{-1}x[/latex] | [latex]\displaystyle\sum _{n=0}^{\infty }{\left(-1\right)}^{n}\frac{{x}^{2n+1}}{2n+1}[/latex] | [latex]-1 < x\le 1[/latex] |
| [latex]f\left(x\right)={\left(1+x\right)}^{r}[/latex] | [latex]\displaystyle\sum _{n=0}^{\infty }\left(\begin{array}{c}r\hfill \\ n\hfill \end{array}\right){x}^{n}[/latex] | [latex]-1 < x < 1[/latex] |
Earlier in the chapter, we showed how you could combine power series to create new power series. Here we use these properties, combined with the Maclaurin series above, to create Maclaurin series for other functions.
Find the Maclaurin series of each of the following functions by using one of the series listed in the table.
- [latex]f\left(x\right)=\cos\sqrt{x}[/latex]
- [latex]f\left(x\right)=\text{sinh}x[/latex]
You can view the transcript for “6.4.3” here (opens in new window).
We also showed previously in this chapter how power series can be differentiated term by term to create a new power series. In the next example, we differentiate the binomial series for [latex]\sqrt{1+x}[/latex] term by term to find the binomial series for [latex]\frac{1}{\sqrt{1+x}}[/latex]. Note that we could construct the binomial series for [latex]\frac{1}{\sqrt{1+x}}[/latex] directly from the definition, but differentiating the binomial series for [latex]\sqrt{1+x}[/latex] is an easier calculation.
Use the binomial series for [latex]\sqrt{1+x}[/latex] to find the binomial series for [latex]\frac{1}{\sqrt{1+x}}[/latex].
The ability to differentiate power series term by term makes them a powerful tool for solving differential equations. Lets see how this is accomplished.