Learning Outcomes
- Approximate functions defined as power series.
Section 9.1 Power Series (PS1)
Subsection 9.1.1 Activities
Activity 9.1.1.
Suppose we could define a function as an “infinite-length polynomial”:
\begin{equation*}
f(x)=1+x+x^2+x^3+x^4+\cdots\text{.}
\end{equation*}
(a)
Would \(f(1)\) be well-defined as a finite real number?
- No, the sum would diverge towards \(\infty\text{.}\)
- No, the sum would oscillate between \(0\) and \(1\text{.}\)
- Yes, the sum would be \(0\text{.}\)
- Yes, the sum would be \(1\text{.}\)
(b)
Would \(f(-1)\) be well-defined as a finite real number?
- No, the sum would diverge towards \(\infty\text{.}\)
- No, the sum would oscillate between \(0\) and \(1\text{.}\)
- Yes, the sum would be \(0\text{.}\)
- Yes, the sum would be \(1\text{.}\)
(c)
Would \(f(1/2)\) be well-defined as a finite real number?
- No, the sum would diverge towards \(\infty\text{.}\)
- Yes, the sum would be approximately \(1\text{.}\)
- Yes, the sum would be approximately \(2\text{.}\)
- Yes, the sum would be exactly \(2\text{.}\)
(d)
When is \(f(x)\) well-defined as a finite real number?
- Its value is \(\frac{x}{1-x}\) when \(|x|<1\text{.}\)
- Its value is \(\frac{x}{1-x}\) when \(x<1\text{.}\)
- Its value is \(\frac{1}{1-x}\) when \(|x|<1\text{.}\)
- Its value is \(\frac{1}{1-x}\) when \(x<1\text{.}\)
Definition 9.1.2.
Given a sequence of numbers \(a_n\) and a number \(c\text{,}\) we may define a function \(f(x)\) as a power series:
\begin{equation*}
f(x)=\sum_{n=0}^\infty a_n(x-c)^n = a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\cdots\text{.}
\end{equation*}
The above power series is said to be centered at \(c\). Often power series are centered at \(0\text{;}\) in this case, they may be written as:
\begin{equation*}
f(x)=\sum_{n=0}^\infty a_n x^n = a_0+a_1x+a_2x^2+a_3x^3+\cdots\text{.}
\end{equation*}
The domain of this function (often referred to as the domain of convergence or interval of convergence) is exactly the set of \(x\)-values for which the series converges.
Activity 9.1.3.
In Section 9.2 we will learn how to prove that \(\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}\) converges for each real value \(x\text{.}\) Thus the function
\begin{equation*}
f(x)=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}+\cdots
\end{equation*}
has the domain of all real numbers.
(a)
To estimate \(f(2)\text{,}\) use technology to compute the first few terms as follows:
\begin{align*}
f(2)=\sum_{n=0}^\infty \frac{2^n}{n!} \amp = 1+2+\frac{2^2}{2}+\frac{2^3}{6}+\frac{2^4}{24}+\frac{2^5}{120}+\cdots \\
\amp = \unknown +\cdots \\
\amp \approx \unknown
\end{align*}
Which of these choices is the closest to this value?
- \(\sqrt{2}\approx 1.414\text{.}\)
- \(e^2\approx 7.389\text{.}\)
- \(\sin(2)\approx 0.909\text{.}\)
- \(\cos(2)\approx -0.416\text{.}\)
(b)
Estimate \(f(-1)\) in a similar fashion:
\begin{align*}
f(-1)=\sum_{n=0}^\infty \frac{\unknown}{n!} \amp = \unknown+\unknown+\unknown+\unknown+\unknown+\unknown+\cdots \\
\amp = \unknown +\cdots \\
\amp \approx \unknown
\end{align*}
Which of these choices is the closest to this value?
- \(\frac{1}{\sqrt{1}}\approx 1.000\text{.}\)
- \(\frac{1}{e^1}\approx 0.369\text{.}\)
- \(\frac{1}{\sin(1)}\approx 1.188\text{.}\)
- \(\frac{1}{\cos(1)}\approx 1.851\text{.}\)
Activity 9.1.4.
The function
\begin{equation*}
f(x)=\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}=\sum_{n=0}^\infty \frac{1}{n!}(x-0)^n
\end{equation*}
is centered at \(0\text{.}\) Likewise, graphing the polynomial that uses the first six terms
\begin{equation*}
f_5(x)=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}
\end{equation*}
alongside the graph of \(e^x\) reveals the illustration given in the following figure.
Plots of \(y=f_5(x), y=e^x\text{.}\)
What might we conclude?
- \(e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\) near \(x=0\text{.}\)
- \(e^x= 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\) near \(x=0\text{.}\)
- \(e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\) for all \(x\text{.}\)
- \(e^x= 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\) for all \(x\text{.}\)
Definition 9.1.5.
Given a power series
\begin{equation*}
f(x)=\sum_{n=0}^\infty a_n(x-c)^n = a_0+a_1(x-c)+a_2(x-c)^2+a_3(x-c)^3+\cdots\text{,}
\end{equation*}
let
\begin{equation*}
f_N(x)=\sum_{n=0}^N a_n(x-c)^n = a_0+a_1(x-c)+a_2(x-c)^2+\cdots+a_N(x-c)^N
\end{equation*}
be its degree \(N\) polynomial approximation for \(x\) nearby \(c\text{.}\)
For example,
\begin{align*}
g_3(x)=\sum_{n=0}^3 n^2 (x-1)^n &= 0+(x-1)+4(x-1)^2+9(x-1)^3\\
&= -6+20x-23x^3+9x^3
\end{align*}
is a degree \(3\) approximation of \(g(x)=\sum_{n=0}^\infty n^2 (x-1)^n\) valid for \(x\) values nearby \(1\text{.}\)
Activity 9.1.6.
Consider a function \(p(x)\) defined by \(\displaystyle p(x)=\sum_{n=0}^\infty \frac{2^n}{(2n)!}x^n.\)
(a)
Find \(p_3(x)\text{,}\) the degree 3 polynomial approximation for \(p(x)\text{.}\)
(b)
Use \(p_3(x)\) to estimate \(p(-1)\text{.}\)
