Skip to main content

Section 2.1 Generating New Series from Known Series

In Table 2.1.1 below we have listed the Maclaurin series for a small set of basic functions of science and engineering. These series can all be relatively easily derived from first principals as discussed in previous lectures. However, for some more complex functions it is easier to find their Maclaurin series by starting from known series rather than trying to find the series from scratch. We will look at two commonly used techniques.

  • Manipulating/Substituting into known series.

  • Differentiating/Integrating known series.

Table 2.1.1. Some Important Maclaurin Series:
\(\dfrac{1}{1-x}=\sum_{k=0}^{\infty}x^{k}=1+x+x^{2}+x^{3}+\cdots\) \(\lvert x\rvert < 1\)
\(e^{x}=\sum_{k=0}^{\infty}\dfrac{x^{k}}{k!}=1+x+\dfrac{x^{2}}{2!} +\dfrac{x^{3}}{3!}+\cdots\) \(\mathbb{R}\)
\(\sin(x)=\sum_{k=0}^{\infty}(-1)^{k}\dfrac{x^{2k+1}}{(2k+1)!}=x-\dfrac{x^{3}}{3!} +\dfrac{x^{5}}{5!}-\cdots\) \(\mathbb{R}\)
\(\cos(x)=\sum_{k=0}^{\infty}(-1)^{k}\dfrac{x^{2k}}{(2k)!}=1-\dfrac{x^{2}}{2!} +\dfrac{x^{4}}{4!}-\cdots\) \(\mathbb{R}\)
\(\ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\dfrac{x^{k}}{k}=x-\dfrac{x^{2}}{2} +\dfrac{x^{3}}{3}-\cdots\) \(\lvert x\rvert < 1\)
\(\tan^{-1}(1+x)=\sum_{k=0}^{\infty}(-1)^{k}\dfrac{x^{2k+1}}{2k+1}=x-\dfrac{x^{3}}{3} +\dfrac{x^{5}}{5}-\cdots\) \(\lvert x\rvert < 1\)
\(\sinh(x)=\sum_{k=0}^{\infty}\dfrac{x^{2k+1}}{(2k+1)!}=x+\dfrac{x^{3}}{3!} +\dfrac{x^{5}}{5!}+\cdots\) \(\mathbb{R}\)
\(\cosh(x)=\sum_{k=0}^{\infty}\dfrac{x^{2k}}{(2k)!}=1+\dfrac{x^{2}}{2!} +\dfrac{x^{4}}{4!}+\cdots\) \(\mathbb{R}\)

Manipulating/Substituting into Known Series.

Find the Maclaurin series for \(f(x)=\dfrac{1}{1+x^{2}}. \)

Answer.

\(\dfrac{1}{1+x^{2}}=\displaystyle\sum_{k=0}^{\infty}(-1)^{k}x^{2k}\)

Solution.

We know that the Maclaurin series for\(f(x)=\dfrac{1}{1-x}\) on \(\lvert x\rvert < 1 \) is

\begin{equation*} \dfrac{1}{1-x}=\sum_{k=0}^{\infty}x^{k}=1+x+x^{2}+x^{3}+\cdots \cdot \end{equation*}

We can use this series by replacing \(x \) with \(-x^{2} \) to obtain

\begin{align*} \dfrac{1}{1+x^{2}} \amp =\dfrac{1}{1-(-x^{2})}\\ \amp = 1+(-x^{2})+(-x^{2})^{2}+(-x^{2})^{3}+\cdots\\ \amp = 1-x^{2} +x^{4}-x^6+\cdots\\ \amp = \sum_{k=0}^{\infty}(-1)^{k}x^{2k} \end{align*}

which will converge on the interval \(\lvert -x^{2}\rvert < 1, \) i.e., \(-1< x < 1. \)

Find the Maclaurin series for \(f(x)=\dfrac{x^2}{2-5x}. \)

Answer.

\(\dfrac{x^2}{2-5x}=\dfrac{x^2}{2}\displaystyle\sum_{k=0}^{\infty}\left(\dfrac{5x}{2}\right)^{k}\)

Solution.

Once again we will use the Maclaurin series for \(f(x)=\dfrac{1}{1-x}\) on \(\lvert x\rvert < 1. \) Now

\begin{align*} \frac{x^2}{2-5x} \amp =\dfrac{x^{2}}{2}\left(\dfrac{1}{1-(\frac{5x}{2})}\right)\\ \amp = \dfrac{x^2}{2}\left(1+\left(\frac{5x}{2}\right)+\left(\frac{5x}{2}\right)^{2}+\left(\frac{5x}{2}\right)^{3}+\cdots\right)\\ \amp = \dfrac{x^2}{2}\sum_{k=0}^{\infty}\left(\dfrac{5x}{2}\right)^{k} \end{align*}

which will converge on the interval \(\lvert \frac{5x}{2}\rvert < 1, \) i.e., \(-\frac{2}{5}< x < \frac{2}{5}. \)

Find the Maclaurin series for \(f(x)=e^{-x^2}. \)

Answer.

\(e^{-x^2}=\displaystyle\sum_{k=0}^{\infty}(-1)^{k}\dfrac{x^{2k}}{k!}\)

Solution.

Substituting into the Maclaurin series for \(f(x)=e^{x}\text{,}\) we obtain

\begin{align*} e^{-x^2} \amp =\sum_{k=0}^{\infty}\dfrac{(-x^2)^{k}}{k!}=\sum_{k=0}^{\infty}(-1)^{k}\dfrac{x^{2k}}{k!}\\ \amp = 1- x^{2}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+\cdots \end{align*}

which will converge for all real values of \(x. \)

Differentiating/Integrating Known Series.

As the theorem below says, we can also use differentiation and integration to work out new Maclaurin series expansions from ones that we already know. The proof of this theorem is beyond the scope of this course.

Note that the behaviour of the series (i.e. whether it converges or diverges) at the endpoints may change when it is differentiated or integrated.

Find the Maclaurin series for \(f(x)=\dfrac{1}{(1-x)^{2}}.\)

Answer.

\(\dfrac{1}{(1-x)^{2}}=\displaystyle\sum_{k=1}^{\infty} k x^{k-1}\)

Solution.

Notice that \(\int \frac{1}{(1-x)^{2}} dx=\frac{1}{1-x}, \) or put the other way, \(\frac{1}{(1-x)^{2}}=\frac{d}{dx} \left(\frac{1}{1-x}\right). \) Since we know the Maclaurin series for \(\frac{1}{1-x} \) we can differentiate this series to obtain the Maclaurin series for \(\frac{1}{(1-x)^{2}}. \) So, on \(\lvert x\rvert < 1, \)

\begin{align*} \frac{1}{(1-x)^{2}} \amp =\frac{d}{dx}\left(\sum_{k=0}^{\infty} x^{k}\right)\\ \amp = \sum_{k=1}^{\infty} k x^{k-1}\\ \amp = 1+2x+3x^2+4x^3+\dots \end{align*}

Find the Maclaurin series for \(f(x)=\ln(1+x).\)

Answer.

\(\ln(1+x)=\displaystyle\sum_{k=1}^{\infty}(-1)^{k-1}\dfrac{x^{k}}{k}\)

Solution.

Notice that \(\frac{d}{dx}(\ln(1+x)), \) or put the other way, \(\ln(1+x)=K+\int \frac{1}{1+x} dx\text{.}\) Since we know the Maclaurin series for \(\frac{1}{1+x} \) we can integrate this series to obtain the Maclaurin series for \(\ln(1+x).\) So, on \(\lvert x\rvert < 1, \)

\begin{align*} \ln(1+x) \amp = K+\int \left(\sum_{k=0}^{\infty} (-x)^{k}\right)dx\\ \amp = K+\sum_{k=0}^{\infty} (-1)^{k}\frac{x^{k+1}}{k+1}\\ \amp = K + x - \frac{1}{2}x^{2} +\frac{1}{3}x^{3} - \dots \end{align*}

Since \(\ln(1+x)\) is \(0 \) when \(x=0 \) we have \(K=0. \) Thus on \(\lvert x\rvert < 1 \)

\begin{equation*} \ln(1+x)=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^{k}}{k}=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots \cdot \end{equation*}

Exercises Example Tasks

1.

Find the Maclaurin series for \(f(x)=\dfrac{1}{3+2x}. \)

2.

Find the Maclaurin series for \(f(x)=\dfrac{1-x^2}{5-3x^2}. \)

3.

Find the Maclaurin series for \(f(x)=\dfrac{1}{(1-2x)^2}. \)

4.

Find the Maclaurin series for \(f(x)=\tan^{-1}(x). \)

5.

Use a sixth degree Maclaurin polynomial to estimate \(f(x)=\displaystyle\int_{0}^{1}e^{-x^{2}}dx. \)