Representing Functions as Power Series (College Board AP® Calculus BC): Revision Note

Find the Maclaurin series for the function $f\left(x\right)=\frac{1}{1+4x^2}$, up to and including the term in $x^8$. Be sure to include the interval of convergence.

Answer:

Start with the standard result for the 'outside function', $\frac{1}{1-x}$

$\frac{1}{1-x}=\sum _{n=0}^{\infty }{x}^n=1+x+x^2+x^3+x^4+...$

$\frac{1}{1+4x^2}=\frac{1}{1-\left(-4x^2\right)}$, so substitute the 'inside function' in brackets, $\left(-4x^2\right)$ everywhere $x$ appears in the original series and simplify

$\begin{array}{rcl}\frac{1}{1+4x^2}& =& 1+\left(-4x^2\right)+{\left(-4x^2\right)}^2+{\left(-4x^2\right)}^3+{\left(-4x^2\right)}^4+...\\ & =& 1-4x^2+16x^4-64x^6+256x^8-...\end{array}$

Now consider the interval of convergence

The series for $\frac{1}{1-x}$ converges for $\left|x\right|<1$, so substitute $-4x^2$ as $x$ into that

$\left|-4x^2\right|<1\iff 4x^2<1\iff x^2<\frac{1}{4}\iff -\frac{1}{2}<x<\frac{1}{2}$

$\begin{array}{rcl}& & 1-4x^2+16x^4-64x^6+256x^8\end{array}$, with interval of convergence $-\frac{1}{2}<x<\frac{1}{2}$

For composite functions and products you could work out the power series 'from scratch' using the Taylor series formula $\sum _{n=0}^{\infty }\frac{f^{\left(n\right)}\left(a\right)}{n!}{\left(x-a\right)}^n$. However using standard results and the methods here is usually much quicker.

For power series of products, it will save you work to use the minimum number of terms necessary for each of the series involved.

But if in doubt, always use more terms than you think you might need. You can always remove any unneeded extra terms at the end after carrying out the multiplication.

Find the Maclaurin series for the function $g\left(x\right)=e^x\sin \left(2x\right)$, up to and including the term in $x^4$.

Answer:

First use substitution to find the Maclaurin series for $\sin \left(2x\right)$

The 'outside function' is $\sin x=\sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{x^{2n+1}}{\left(2n+1\right)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...$ and the 'inside function' is $2x$ so substitute $\left(2x\right)$ everywhere $x$ appears and simplify

$\begin{array}{rcl}\sin \left(2x\right)& =& \left(2x\right)-\frac{{\left(2x\right)}^3}{3!}+\frac{{\left(2x\right)}^5}{5!}-...\\ & =& 2x-\frac{4x^3}{3}+\frac{4x^5}{15}-...\end{array}$

The Maclaurin series for $e^x$ is also a standard result

$\begin{array}{rcl}{e}^x& =& 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\\ & =& 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+...\end{array}$

Before multiplying those together, think about which terms you actually need

The question only asks for terms up to $x^4$

• That means you don't need the $x^5$ term (or any higher-order terms) from the series for $\sin \left(2x\right)$

• Also the smallest power in the $\sin \left(2x\right)$ series is the $2x$ term

  • So you can stop the $e^x$ series with the $\frac{x^3}{6}$ term

  • Because any higher-order term will multiply with $2x$ to give a term in $x^5$ or higher

$\begin{array}{rcl}{e}^x\sin \left(2x\right)& \approx & \left(1+x+\frac{x^2}{2}+\frac{x^3}{6}\right)\left(2x-\frac{4x^3}{3}\right)\\ & =& 1\left(2x-\frac{4x^3}{3}\right)+x\left(2x-\frac{4x^3}{3}\right)+\frac{x^2}{2}\left(2x-\frac{4x^3}{3}\right)+\frac{x^3}{6}\left(2x-\frac{4x^3}{3}\right)\\ & =& 2x-\frac{4x^3}{3}+2x^2-\frac{4x^4}{3}+x^3-\frac{2x^5}{3}+\frac{x^4}{3}-\frac{2x^6}{9}\\ & =& 2x+2x^2-\frac{x^3}{3}-x^4-\frac{2x^5}{3}-\frac{2x^6}{9}\end{array}$

For the answer, we only want terms up to $x^4$

Note that the terms in $x^5$ and $x^6$ there are not actually the correct terms for the power series of $e^x\sin \left(2x\right)$

• To find those terms properly, we would need to use more terms from the two original series in the product

$\begin{array}{rcl}& & 2x+2x^2-\frac{x^3}{3}-x^4\end{array}$

If the 'basic' series you are using converge for all values of $x$, then series found by using substitution or taking products will also converge for all values of $x$.

Confirm that differentiating the Maclaurin series for $\cos x$ term by term yields the Maclaurin series for $-\sin x$.

Answer:

The Maclaurin series for $\cos x$ is a standard result, $\cos x=\sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{x^{2n}}{\left(2n\right)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-...$

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-...$

Differentiate that term by term

$\begin{array}{rcl}\frac{d}{dx}\left(\cos x\right)& =& \frac{d}{dx}\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-...\right)\\ & =& 0-\frac{2x}{2!}+\frac{4x^3}{4!}-\frac{6x^5}{6!}+\frac{8x^7}{8!}-...\\ & =& -x+\frac{x^3}{3!}-\frac{x^5}{5}+\frac{x^7}{7}-...\end{array}$

The series for $\sin x$ is also a standard result, $\sin x=\sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{x^{2n+1}}{\left(2n+1\right)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$

$\begin{array}{rcl}& \Rightarrow & -\sin x=-\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...\right)\\ & =& -x+\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}-...\end{array}$

Indeed those two series match

Differentiating the Maclaurin series for $\cos x$ term by term gives the Maclaurin series for $-\sin x$

The radius of convergence of a power series found using term-by-term differentiation or term-by-term integration is the same as the radius of convergence of the original power series.

However convergence or divergence at the endpoints is not always the same, and needs to be checked separately.

Find the Maclaurin series for $\frac{1}{1+x^2}$, and use that result to determine the Maclaurin series for $\arctan x$ up to and including the term in $x^7$. Be sure to include the radius of convergence.

Answer:

Start with the standard Maclaurin series result for $\frac{1}{1-x}$, and substitute $-x^2$ everywhere

$\frac{1}{1-x}=1+x+x^2+x^3+x^4+...$

$\begin{array}{rcl}& \Rightarrow & \frac{1}{1+x^2}=1+\left(-x^2\right)+{\left(-x^2\right)}^2+{\left(-x^2\right)}^3+{\left(-x^2\right)}^4+...\\ & =& 1-x^2+x^4-x^6+x^8-...\end{array}$

Now use the standard integral result, $\int \frac{1}{1+x^2}dx=\arctan x+C$

That means you can integrate the series for $\frac{1}{1+x^2}$ term by term to get the series for $\arctan x$

Just remember to include a constant of integration, which it is easiest to write at the front of the series

$\begin{array}{rcl}\arctan x& =& \int \left(1-x^2+x^4-x^6+x^8-...\right)dx\\ & =& C+x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-...\end{array}$

To find the value of $C$, substitute in $x=0$; remember that $\arctan \left(0\right)=0$

$\begin{array}{rcl}\arctan \left(0\right)& =& C+\left(0\right)-\frac{{\left(0\right)}^3}{3}+\frac{{\left(0\right)}^5}{5}-\frac{{\left(0\right)}^7}{7}+\frac{{\left(0\right)}^9}{9}-\\ 0& =& C\end{array}$

Now consider the radius of convergence

The series for $\frac{1}{1-x}$ converges for $\left|x\right|<1$, so substitute $-x^2$ into that

$\left|-x^2\right|<1\iff x^2<1\iff -1<x<1\iff \left|x\right|<1$

Thus the radius of convergence for the $\frac{1}{1+x^2}$ series is 1 (i.e. the radius of convergence is 1, with the series centered on $x=0$);

The series we found by integrating that will have the same radius of convergence

And the question only wants terms up through $x^7$, so

$\begin{array}{rcl}& & x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\end{array}$, with the radius of convergence equal to 1