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Calculus for Team-Based Inquiry Learning PREVIEW Edition

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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*}
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(a)
Would \(f(1)\) be well-defined as a finite real number?
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  1. No, the sum would diverge towards \(\infty\text{.}\)
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  2. No, the sum would oscillate between \(0\) and \(1\text{.}\)
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  3. Yes, the sum would be \(0\text{.}\)
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  4. Yes, the sum would be \(1\text{.}\)
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(b)
Would \(f(-1)\) be well-defined as a finite real number?
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  1. No, the sum would diverge towards \(\infty\text{.}\)
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  2. No, the sum would oscillate between \(0\) and \(1\text{.}\)
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  3. Yes, the sum would be \(0\text{.}\)
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  4. Yes, the sum would be \(1\text{.}\)
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(c)
Would \(f(1/2)\) be well-defined as a finite real number?
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  1. No, the sum would diverge towards \(\infty\text{.}\)
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  2. Yes, the sum would be approximately \(1\text{.}\)
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  3. Yes, the sum would be approximately \(2\text{.}\)
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  4. Yes, the sum would be exactly \(2\text{.}\)
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(d)
When is \(f(x)\) well-defined as a finite real number?
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  1. Its value is \(\frac{x}{1-x}\) when \(|x|<1\text{.}\)
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  2. Its value is \(\frac{x}{1-x}\) when \(x<1\text{.}\)
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  3. Its value is \(\frac{1}{1-x}\) when \(|x|<1\text{.}\)
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  4. Its value is \(\frac{1}{1-x}\) when \(x<1\text{.}\)
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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*}
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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*}
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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.
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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.
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(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*}
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Which of these choices is the closest to this value?
  1. \(\sqrt{2}\approx 1.414\text{.}\)
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  2. \(e^2\approx 7.389\text{.}\)
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  3. \(\sin(2)\approx 0.909\text{.}\)
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  4. \(\cos(2)\approx -0.416\text{.}\)
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(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*}
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Which of these choices is the closest to this value?
  1. \(\frac{1}{\sqrt{1}}\approx 1.000\text{.}\)
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  2. \(\frac{1}{e^1}\approx 0.369\text{.}\)
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  3. \(\frac{1}{\sin(1)}\approx 1.188\text{.}\)
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  4. \(\frac{1}{\cos(1)}\approx 1.851\text{.}\)
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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.
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What might we conclude?
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  1. \(e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\) near \(x=0\text{.}\)
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  2. \(e^x= 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\) near \(x=0\text{.}\)
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  3. \(e^x\approx 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\) for all \(x\text{.}\)
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  4. \(e^x= 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\) for all \(x\text{.}\)
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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 \(c\text{.}\)
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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 \(1\text{.}\)
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Activity 9.1.6.

Consider a function \(p(x)\) defined by \(\displaystyle p(x)=\sum_{n=0}^\infty \frac{2^n}{(2n)!}x^n.\)
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(a)
Find \(p_3(x)\text{,}\) the degree 3 polynomial approximation for \(p(x)\text{.}\)
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(b)
Use \(p_3(x)\) to estimate \(p(-1)\text{.}\)
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Subsection 9.1.2 Videos

Figure 120. Video: Approximate functions defined as power series
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Subsection 9.1.3 Exercises

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