Skip to main content Contents
Prev Up Next \(\newcommand{\markedPivot}[1]{\boxed{#1}}
\newcommand{\IR}{\mathbb{R}}
\newcommand{\IC}{\mathbb{C}}
\renewcommand{\P}{\mathcal{P}}
\renewcommand{\Im}{\operatorname{Im}}
\newcommand{\RREF}{\operatorname{RREF}}
\newcommand{\vspan}{\operatorname{span}}
\newcommand{\setList}[1]{\left\{#1\right\}}
\newcommand{\setBuilder}[2]{\left\{#1\,\middle|\,#2\right\}}
\newcommand{\unknown}{\,{\color{gray}?}\,}
\newcommand{\drawtruss}[2][1]{
\begin{tikzpicture}[scale=#1, every node/.style={scale=#1}]
\draw (0,0) node[left,magenta]{C} --
(1,1.71) node[left,magenta]{A} --
(2,0) node[above,magenta]{D} -- cycle;
\draw (2,0) --
(3,1.71) node[right,magenta]{B} --
(1,1.71) -- cycle;
\draw (3,1.71) -- (4,0) node[right,magenta]{E} -- (2,0) -- cycle;
\draw[blue] (0,0) -- (0.25,-0.425) -- (-0.25,-0.425) -- cycle;
\draw[blue] (4,0) -- (4.25,-0.425) -- (3.75,-0.425) -- cycle;
\draw[thick,red,->] (2,0) -- (2,-0.75);
#2
\end{tikzpicture}
}
\newcommand{\trussNormalForces}{
\draw [thick, blue,->] (0,0) -- (0.5,0.5);
\draw [thick, blue,->] (4,0) -- (3.5,0.5);
}
\newcommand{\trussCompletion}{
\trussNormalForces
\draw [thick, magenta,<->] (0.4,0.684) -- (0.6,1.026);
\draw [thick, magenta,<->] (3.4,1.026) -- (3.6,0.684);
\draw [thick, magenta,<->] (1.8,1.71) -- (2.2,1.71);
\draw [thick, magenta,->] (1.6,0.684) -- (1.5,0.855);
\draw [thick, magenta,<-] (1.5,0.855) -- (1.4,1.026);
\draw [thick, magenta,->] (2.4,0.684) -- (2.5,0.855);
\draw [thick, magenta,<-] (2.5,0.855) -- (2.6,1.026);
}
\newcommand{\trussCForces}{
\draw [thick, blue,->] (0,0) -- (0.5,0.5);
\draw [thick, magenta,->] (0,0) -- (0.4,0.684);
\draw [thick, magenta,->] (0,0) -- (0.5,0);
}
\newcommand{\trussStrutVariables}{
\node[above] at (2,1.71) {\(x_1\)};
\node[left] at (0.5,0.866) {\(x_2\)};
\node[left] at (1.5,0.866) {\(x_3\)};
\node[right] at (2.5,0.866) {\(x_4\)};
\node[right] at (3.5,0.866) {\(x_5\)};
\node[below] at (1,0) {\(x_6\)};
\node[below] at (3,0) {\(x_7\)};
}
\newcommand{\N}{\mathbb N}
\newcommand{\Z}{\mathbb Z}
\newcommand{\Q}{\mathbb Q}
\newcommand{\R}{\mathbb R}
\DeclareMathOperator{\arcsec}{arcsec}
\DeclareMathOperator{\arccot}{arccot}
\DeclareMathOperator{\arccsc}{arccsc}
\newcommand{\tuple}[1]{\left\langle#1\right\rangle}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Section 9.5 Taylor’s Theorem (PS5)
Learning Outcomes
Subsection 9.5.1 Activities
Activity 9.5.1 .
Recall that we can use a Taylor series for a function to approximate that function by using an \(k\) th degree Taylor polynomial.
(a)
Which of the following is the 3rd degree Taylor polynomial for \(f(x)=\sin x\) centered at 0.
\(\displaystyle 1-\dfrac{x^2}{2}\)
\(\displaystyle x-\dfrac{x^3}{3!}\)
\(\displaystyle x+\dfrac{x^3}{3!}\)
\(\displaystyle x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}\)
(b)
Use the 3rd degree Taylor polynomial for \(f(x)=\sin x\) to approximate \(\sin(1)\text{.}\)
(c)
Use technology to approximate \(\sin(1)\text{.}\)
Definition 9.5.2 .
Given a infinitely differentiable function
\begin{equation*}
f(x)=\displaystyle\sum_{n=0}^\infty \dfrac{f^{(n)}(c)}{n!}(x-c)^n\text{,}
\end{equation*}
we define the remainder , denoted \(R_k(x)\text{,}\) to be the difference between the function \(f(x)\) and its \(n\) th degree Taylor polynomial \(T_k(x)\text{.}\) That is,
\begin{equation*}
R_k(x)=f(x)-T_k(x).
\end{equation*}
Activity 9.5.3 .
We saw in
Fact 9.4.6 , the Maclaurin series for
\(f(x)=e^x\) is
\begin{equation*}
e^x=\displaystyle\sum_{n=0}^\infty \dfrac{1}{n!}x^n.
\end{equation*}
(a)
Compute \(R_2(4)\) using technology.
(b)
Compute \(R_3(4)\) using technology.
(c)
What do you expect from \(R_4(4)\text{?}\)
Not enough information.
It will be greater than both \(R_3(4)\) and \(R_4(4)\text{.}\)
It will be between \(R_3(4)\) and \(R_4(4)\text{.}\)
It will be less than both \(R_3(4)\) and \(R_4(4)\text{.}\)
Fact 9.5.4 .
Let \(f(x)\) be a function represented by a power series centered at \(x=c\)
\begin{equation*}
f(x)=\displaystyle\sum_{n=0}^\infty a_n(x-c)^n
\end{equation*}
with an interval of convegence \(I\text{.}\) Then for all \(x\in I\text{,}\)
\begin{equation*}
\lim_{n\rightarrow\infty} R_n(x)=0.
\end{equation*}
Subsection 9.5.2 Sample Problem
Here you are tasked with approximating the value of \(\cos(1)\text{.}\)
(a)
Calculate the 4th degree Taylor polynomial for \(f(x)=\cos x\) centered at \(\pi\text{,}\) then use it to approximate the value of \(\cos(1)\) to three decimal places.
(b)
Apply Taylor’s Theorem to find an upper bound for the error in this approximation.
(c)
Use technology to calculate \(|R_4(1)|\text{.}\) Is the error within the upper bound found in part (b)?
(d)
Explain whether the approximation error \(|R_{k}(1)|\) increases or decreases as \(k\rightarrow\infty\text{.}\)