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Section   9.1   Power Series (PS1) 
 
Learning Outcomes 
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 \(y=f_5(x), y=e^x\text{.}\) 
 
Figure   187.    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 \(p(-1)\text{.}\) 
Subsection   9.1.2   Videos 
 
Figure   188.    Video: Approximate functions defined as power series
Subsection   9.1.3   Exercises