Learning Outcomes
- Compute power series by manipulating known exponential/trigonometric/binomial power series.
Section 9.3 Manipulation of Power Series (PS3)
Subsection 9.3.1 Activities
Activity 9.3.1.
How might we use the known geometric power series
\begin{equation*}
\frac{1}{1-x}=\sum_{n=0}^\infty x^n=1+x+x^2+x^3+x^4+\dots
\end{equation*}
to find the value of
\begin{equation*}
\unknown =\sum_{n=0}^\infty nx^{n-1}=0+1+2x+3x^2+4x^3+\dots\text{?}
\end{equation*}
(a)
Which operation describes the relationship between these two series?
- Bifurcation
- Composition
- Differentiation
- Multiplication
(b)
What is the result of applying this operation to \(\frac{1}{1-x}\text{?}\)
- \(\displaystyle 0\)
- \(\displaystyle \frac{1}{(1-x)^2}\)
- \(\displaystyle 1-\frac{1}{x}\)
- \(\displaystyle \frac{x}{1-x^2}\)
Fact 9.3.2.
Whenever a function is defined as a power series:
\begin{equation*}
f(x)=\sum_{n=0}^\infty a_n(x-c)^n
\end{equation*}
then its derivative and general antiderivative are also defined as power series with the same domain of convergence as \(f(x)\text{,}\) found by differentiating or integrating term-by-term:
\begin{align*}
\frac{d}{dx}[f(x)] \amp=\sum_{n=0}^\infty\frac{d}{dx}\left[a_n(x-c)^n\right]\\
\amp=\sum_{n=0}^\infty na_n(x-c)^{n-1}\\
\int f(x)\,dx\amp=C+\sum_{n=0}^\infty\left[\int a_n(x-c)^n\,dx\right]\\
\amp=C+\sum_{n=0}^\infty\frac{(x-c)^{n+1}}{n+1}
\end{align*}
Activity 9.3.3.
Let’s investigate the power series
\begin{equation*}
\exp(x)=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\dots\text{.}
\end{equation*}
(a)
What is the value of \(\exp(0)\text{?}\)
- \(0\text{.}\)
- \(1\text{.}\)
- \(2\text{.}\)
- \(\infty\text{.}\)
(b)
What is the value of \(\exp'(x)\text{?}\)
- \(0+1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots\text{.}\)
- \(1+x+\frac{x^2}{6}+\frac{x^3}{24}+\frac{x^4}{120}+\dots\text{.}\)
- \(0+1+x+\frac{x^2}{3}+\frac{x^3}{12}+\frac{x^4}{60}+\dots\text{.}\)
- \(1+x+\frac{x^2}{3}+\frac{x^3}{12}+\frac{x^4}{60}+\dots\text{.}\)
(c)
What can we conclude from our calculation of \(f'(x)\text{?}\)
- \(\exp'(x)=[\exp(x)]^2\text{.}\)
- \(\exp'(x)=\exp(x^2)\text{.}\)
- \(\exp'(x)=2\exp(x)\text{.}\)
- \(\exp'(x)=\exp(x)\text{.}\)
(d)
What function do we know of that shares each of these properites?
- \(\displaystyle \exp(x)=\frac{1}{1+x}\)
- \(\displaystyle \exp(x)=\cos(x)\)
- \(\displaystyle \exp(x)=e^x\)
- \(\displaystyle \exp(x)=0\)
Fact 9.3.4.
We have that
\begin{equation*}
\exp(x)=e^x=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n=\sum_{n=0}^\infty \frac{x^n}{n!}.
\end{equation*}
That is, for any real number \(x\text{,}\) the series \(\exp(x)=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}x^n\) will converge to \(e^x\text{.}\)
Fact 9.3.5.
We may similarly determine that
\begin{equation*}
\cos(x)=\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}=\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}
\end{equation*}
and
\begin{equation*}
\sin(x)=\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}
\end{equation*}
for all real numbers \(x\text{.}\) However, we will delay until Fact 9.4.6 to prove this fact another way.
Activity 9.3.6.
Suppose we wish to find the power series for the function \(f(x)=e^{2x}\) by modifying the power series \(\exp(z)=e^z=\displaystyle\sum_{n=0}^\infty \frac{z^n}{n!}.\)
(a)
Substituting \(z=2x\text{,}\) what is the power series for \(\exp(2x)\text{?}\)
- \(\exp(2x)=\displaystyle\sum_{n=0}^\infty \frac{2x^n}{n!}=2+2x+x^2+\frac{1}{3}x^3+\dots\text{.}\)
- \(\exp(2x)=\displaystyle\sum_{n=0}^\infty \frac{2x^{n+1}}{n!}=2x+2x^2+x^3+\frac{1}{3}x^4+\dots\text{.}\)
- \(\exp(2x)=\displaystyle\sum_{n=0}^\infty \frac{(2x)^n}{n!}=1+2x+2x^2+\frac{4}{3}x^3+\dots\text{.}\)
- \(\exp(2x)=\displaystyle\sum_{n=0}^\infty \frac{x^n}{(2n)!}=1+\frac{x}{2}+\frac{x^2}{4}+\frac{x^3}{720}+\dots\text{.}\)
(b)
What is the interval of convergence for \(x\) for this series?
- \(\left(-\infty,\infty\right)\text{.}\)
- \(\displaystyle \left(-\frac{1}{2},\frac{1}{2}\right)\text{.}\)
- \(\displaystyle \left(0,\frac{1}{2}\right)\text{.}\)
- \(\displaystyle \left(-\frac{1}{2},\frac{1}{2}\right]\text{.}\)
Fact 9.3.7.
If a power series
\begin{equation*}
f(x)=\sum_{n=0}^\infty a_n(x-c)^n
\end{equation*}
is known, then for any polynomial \(g(x)\) the composition \(f\circ g\) has a power series given by
\begin{equation*}
(f\circ g)(x)=f(g(x))=\sum_{n=0}^\infty a_n(g(x)-c)^n
\end{equation*}
where the domain of convergence is transformed based upon the transformation given by \(g(x)\text{.}\)
For example, if \(f(x)\) has the domain of convergence \(-2\leq x < 2\text{,}\) then \(f(2x+4)\) has the domain of convergence:
\begin{equation*}
-2\leq 2x+4 < 2
\end{equation*}
\begin{equation*}
-6\leq 2x < -2
\end{equation*}
\begin{equation*}
-3\leq x < -1
\end{equation*}
Activity 9.3.8.
Suppose we wish to find the power series for the function \(f(x)=\frac{1}{x}\text{.}\)
(a)
Which of the following represents the power series for \(g(r)=\frac{1}{1-r}\text{?}\)
- \(g(r)=\displaystyle\sum_{n=0}^\infty rx^n\text{.}\)
- \(g(r)=\displaystyle\sum_{n=0}^\infty (rx)^n\text{.}\)
- \(g(r)=\displaystyle\sum_{n=0}^\infty r^n\text{.}\)
- \(g(r)=\displaystyle\sum_{r=0}^\infty x^r\text{.}\)
(b)
For what value of \(r\) is \(\frac{1}{1-r}=\frac{1}{x}\text{?}\)
- \(r=x-1\text{.}\)
- \(r=1-x\text{.}\)
- \(r=x+1\text{.}\)
- \(r=-x\text{.}\)
(c)
Substituting \(r\) with this value, which of the following is a power series for \(f(x)=\frac{1}{x}\text{?}\)
- \(f(x)=\displaystyle\sum_{n=0}^\infty (-x)^n\text{.}\)
- \(f(x)=\displaystyle\sum_{n=0}^\infty (1-x)^n\text{.}\)
- \(f(x)=\displaystyle\sum_{n=0}^\infty (x-1)^n\text{.}\)
- \(f(x)=\displaystyle\sum_{n=0}^\infty (1+x)^n\text{.}\)
(d)
Given that the domain of convergence for \(r\) in \(f(r)\) is \(-1 < r < 1\text{,}\) what should be the domain of convergence for \(x\) in \(f(x)\text{?}\)
- \(-1 < x < 1\text{.}\)
- \(-2 < x < 0\text{.}\)
- \(-2 < x < 2\text{.}\)
- \(0 < x < 2\text{.}\)
Activity 9.3.9.
Suppose we wish to find the power series for the function \(f(x)=\frac{1}{3-2x}\text{.}\) Recall that \(g(x)=\frac{1}{1-r}=\displaystyle\sum_{n=0}^\infty r^n.\)
(a)
For what value of \(r\) is \(\frac{1}{1-r}=\frac{1}{3-2x}\text{?}\)
- \(r=2x-2\text{.}\)
- \(r=2-2x\text{.}\)
- \(r=2x-3\text{.}\)
- \(r=3-2x\text{.}\)
(b)
Evaluating \(r\) at the previously found value, which of the following is the power series of \(f(x)=\frac{1}{3-2x}\text{?}\)
- \(f(x)=\displaystyle\sum_{n=0}^\infty (3-2x)^n\text{.}\)
- \(f(x)=\displaystyle\sum_{n=0}^\infty (2x-3)^n\text{.}\)
- \(f(x)=\displaystyle\sum_{n=0}^\infty (2-2x)^n\text{.}\)
- \(f(x)=\displaystyle\sum_{n=0}^\infty (2x-2)^n\text{.}\)
(c)
Given that the interval of convergence for \(r\) is \(-1 < r < 1\text{,}\) what is the interval of convergence for \(x\text{?}\)
- \(-1 < x < \frac{3}{2}\text{.}\)
- \(-\frac{1}{2} < x < 1\text{.}\)
- \(\frac{1}{2} < x < \frac{3}{2}\text{.}\)
- \(-\frac{1}{2} < x < \frac{3}{2}\text{.}\)
Activity 9.3.10.
Suppose we wish to find the power series for the function \(f(x)=\frac{1}{1+x^2}\text{.}\) Recall that \(g(x)=\frac{1}{1-r}=\displaystyle\sum_{n=0}^\infty r^n.\)
(a)
For what value of \(r\) is \(\frac{1}{1-r}=\frac{1}{1+x^2}\text{?}\)
- \(r=x^2\text{.}\)
- \(r=-x^2\text{.}\)
- \(r=1-x^2\text{.}\)
- \(r=x^2-1\text{.}\)
(b)
Evaluating \(r\) at the previously found value, which of the following is the power series of \(f(x)=\frac{1}{1+x^2}\text{?}\)
- \(\displaystyle\frac{1}{1+x^2}=\sum_{n=0}^\infty (-1)^nx^{2n}\text{.}\)
- \(\displaystyle\frac{1}{1+x^2}=\sum_{n=0}^\infty (1-x^2)^n\text{.}\)
- \(\displaystyle\frac{1}{1+x^2}=\sum_{n=0}^\infty x^{2n}\text{.}\)
- \(\displaystyle\frac{1}{1+x^2}=\sum_{n=0}^\infty (x^2-1)^n\text{.}\)
(c)
Given that the interval of convergence for \(r\) is \(-1 < r < 1\text{,}\) what is the interval of convergence for \(x\text{?}\)
- \(-1 < x < 1\text{.}\)
- \(-1 < x < 0\text{.}\)
- \(0 < x < 1\text{.}\)
- \(0 < x < 4\text{.}\)
(d)
How can the power series for \(\frac{1}{1+x^2}\) be manipulated to obtain a power series for \(\arctan(x)\text{?}\)
- Differentiate each term.
- Integrate each term.
- Replace \(x\) with \(x^2\) in each term.
- Replace \(x\) with \(1/x\) in each term.
(e)
Which of these power series is the result of this manipulation?
- \(\displaystyle\arctan(x)=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}\text{.}\)
- \(\displaystyle\arctan(x)=\sum_{n=0}^\infty (-1)^n\frac{x^{2n-1}}{2n-1}\text{.}\)
- \(\displaystyle\arctan(x)=\sum_{n=0}^\infty (-1)^n(2n)x^{2n-1}\text{.}\)
- \(\displaystyle\arctan(x)=\sum_{n=0}^\infty (-1)^n(2n+1)x^{2n}\text{.}\)
Activity 9.3.11.
What function \(f(x)\) has power series \(f(x)=\displaystyle \sum_{n=0}^\infty \frac{(-1)^nx^n}{n!}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\cdots\text{?}\)
- \(f(x)=(-1)^ne^x\text{.}\)
- \(f(x)=-e^x\text{.}\)
- \(f(x)=e^{-x}\text{.}\)
- \(f(x)=-e^{-x}\text{.}\)
Activity 9.3.12.
What function \(f(x)\) has power series \(f(x)=\displaystyle \sum_{n=0}^\infty \frac{x^{n+3}}{n!}=x^3+x^4+\frac{x^5}{2}+\frac{x^6}{6}+\cdots\text{?}\)
- \(f(x)=e^{x+3}\text{.}\)
- \(f(x)=e^{x^3}\text{.}\)
- \(f(x)=e^{3x}\text{.}\)
- \(f(x)=x^3e^{x}\text{.}\)
Fact 9.3.13.
If a power series
\begin{equation*}
f(x)=\sum_{n=0}^\infty a_n(x-c)^n
\end{equation*}
is known, then for any polynomial \(g(x)\) the product \(fg\) has a power series given by
\begin{equation*}
(fg)(x)=f(x)g(x)=\sum_{n=0}^\infty a_ng(x)(x-c)^n
\end{equation*}
where the domain of convergence is the same as \(f(x)\text{.}\)
Activity 9.3.14.
What function \(f(x)\) has power series \(f(x)=\displaystyle \sum_{n=3}^\infty x^n=x^3+x^4+\cdots\text{?}\)
- \(f(x)=\frac{1}{1-3x}\text{.}\)
- \(f(x)=\frac{3}{1-x}\text{.}\)
- \(f(x)=\frac{1}{1-x}-x^2-x-1\text{.}\)
- \(f(x)=\frac{x^3}{1-x}\text{.}\)
Activity 9.3.15.
The function \(n(x)=e^{-x^2}\) is one whose integrals are very important for statistics. However, it does not admit an elementary antiderivative.
(a)
Which of the following best represents the power series for \(n(x)=e^{-x^2}\text{?}\)
- \(n(x)=\displaystyle -x^2\sum_{n=0}^\infty \frac{1}{n!}x^n=\displaystyle \sum_{n=0}^\infty -\frac{1}{n!}x^{n+2}\text{.}\)
- \(n(x)=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}(-x^2)^n=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}(-1)^{n}x^{2n}\text{.}\)
- \(n(x)=\displaystyle x^{-2}\sum_{n=0}^\infty \frac{1}{n!}(-x)^n=\displaystyle \sum_{n=0}^\infty \frac{1}{n!}(-1)^{n+2}x^{n+2}\text{.}\)
(b)
Which of the following best represents a degree 10 polynomial that approximates \(n(x)\text{?}\)
- \(n_{10}(x)=\displaystyle -x^2-x^3-\frac{1}{2}x^4-\frac{1}{6}x^5-\frac{1}{24}x^6-\frac{1}{120}x^7-\frac{1}{720}x^8-\frac{1}{5040}x^9-\frac{1}{40320}x^{10}\text{.}\)
- \(n_{10}(x)=\displaystyle x^2-x^3+\frac{1}{2}x^4-\frac{1}{6}x^5+\frac{1}{24}x^6-\frac{1}{120}x^7+\frac{1}{720}x^8-\frac{1}{5040}x^9+\frac{1}{40320}x^{10}\text{.}\)
- \(n_{10}(x)=1-x^2+\frac{1}{2}x^4-\frac{1}{6}x^6+\frac{1}{24}x^8-\frac{1}{120}x^{10}\text{.}\)
(c)
Use your choice of \(n_{10}(x)\) to estimate \(\displaystyle \int_0^1 n(x)dx\) by computing \(\displaystyle \int_0^1 n_{10}(x)dx\text{.}\)
Activity 9.3.16.
Recall that
\begin{equation*}
g(x)=\displaystyle\sum_{n=0}^\infty x^n=\frac{1}{1-x}
\end{equation*}
for \(-1< x< 1\text{.}\)
(a)
Which of the following represents an antiderivative of \(g(x)=\displaystyle \frac{1}{1-x}\text{?}\)
- \(G(x)=C+\displaystyle\sum_{n=0}^\infty x^{n+1}\text{.}\)
- \(G(x)=C+\displaystyle\sum_{n=1}^\infty \frac{1}{n}x^{n+1}\text{.}\)
- \(G(x)=C+\displaystyle\sum_{n=0}^\infty \frac{1}{n+1}x^{n+1}\text{.}\)
- \(G(x)=C+\displaystyle\sum_{n=1}^\infty \frac{1}{n+1}x^{n}\text{.}\)
(b)
Find the interval of convergence for this series.
(c)
Recall that \(\tilde{G}(x)=\ln|1-x|\) is an antiderivative of \(g(x)=\displaystyle \frac{1}{1-x}\text{.}\) For which \(C\) is your chosen \(G(x)=\ln|1-x|\text{?}\)
(d)
Use \(G_4(x)\) to estimate \(\displaystyle \int_2^4 \ln|1-x|dx\text{.}\)
Activity 9.3.17.
Recall that the power series for \(f(x)=\sin\left(x\right)\) is:
\begin{equation*}
\sin\left(x\right)=\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n} x^{2 \, n + 1}}{\left(2 \, n + 1\right)!}.
\end{equation*}
(a)
Find a power series for \(\sin\left(-5 \, x^{2}\right)\text{.}\)
(b)
Find a power series for \(x^{4} \sin\left(x\right)\text{.}\)
(c)
Find a power series for \(F(x)\text{,}\) an antiderivative of \(f(x)\) such that \(F(0)=4\text{.}\)
Activity 9.3.18.
Recall that the power series for \(f(x)=-\frac{1}{x - 1}\) is:
\begin{equation*}
-\frac{1}{x - 1}=\sum_{n=0}^{\infty} x^{n}.
\end{equation*}
(a)
Find a power series for \(\frac{1}{x^{4} + 1}\text{.}\)
(b)
Find a power series for \(-\frac{x^{5}}{x - 1}\text{.}\)
(c)
Find a power series for \(f'(x)\text{.}\)
Activity 9.3.19.
Recall that
\begin{equation*}
g(x)=\displaystyle\sum_{n=0}^\infty x^n=\frac{1}{1-x}
\end{equation*}
for \(-1< x< 1\) and \(\frac{d}{dx}[\arctan(x)]=\frac{1}{1+x^2}=g(-x^2)\text{.}\) We computed the power series for \(g(-x^2)\) in Activity 9.3.10.
(a)
Integrate this power series and find \(C\) to find a power series for \(H(x)=\arctan(x)\text{.}\) Recall that \(\arctan(0)=0\text{.}\)
(b)
Find the interval of convergence for this series.
Activity 9.3.20.
(a)
Find the power series for \(\alpha(x)=\ln|x|\text{.}\)
(b)
Find the interval of convergence for this series.
Activity 9.3.21.
(a)
Find the power series for \(\beta(x)=\arctan(-3x^2)\text{.}\)
(b)
Find the interval of convergence for this series.
Subsection 9.3.2 Videos
Subsection 9.3.3 Exercises
Exercises available at
https://tbil.org/calculus/preview/exercises/#/bank/PS3/
