Program: English — B30 · Bahasa Indonesia — B30 · Mathematics Program home · Beranda Program Matematika · Original source — Original authors’ website — CLP-2 textbook, problems and PDFs · Open related page · Open related page Mathematics program — English Program — Bahasa Indonesia Original authors’ website ↗ Book contents
Skip to main content
Contents
Dark Mode Prev Up Next
\(\require{cancel}\newcommand{\dee}[1]{\mathrm{d}#1}
\newcommand{\half}{ \frac{1}{2} }
\newcommand{\ds}{\displaystyle}
\newcommand{\ts}{\textstyle}
\newcommand{\es}{ {\varnothing}}
\newcommand{\st}{ {\mbox{ s.t. }} }
\newcommand{\pow}[1]{ \mathcal{P}\left(#1\right) }
\newcommand{\set}[1]{ \left\{#1\right\} }
\newcommand{\lin}{{\text{LIN}}}
\newcommand{\quot}{{\text{QR}}}
\newcommand{\simp}{{\text{SMP}}}
\newcommand{\diff}[2]{ \frac{\mathrm{d}#1}{\mathrm{d}#2}}
\newcommand{\bdiff}[2]{ \frac{\mathrm{d}}{\mathrm{d}#2} \left( #1 \right)}
\newcommand{\ddiff}[3]{ \frac{\mathrm{d}^#1#2}{\mathrm{d}{#3}^#1}}
\renewcommand{\neg}{ {\sim} }
\newcommand{\limp}{ {\;\Rightarrow\;} }
\newcommand{\nimp}{ {\;\not\Rightarrow\;} }
\newcommand{\liff}{ {\;\Leftrightarrow\;} }
\newcommand{\niff}{ {\;\not\Leftrightarrow\;} }
\newcommand{\De}{\Delta}
\newcommand{\bbbn}{\mathbb{N}}
\newcommand{\bbbr}{\mathbb{R}}
\newcommand{\bbbp}{\mathbb{P}}
\newcommand{\cA}{\mathcal{A}}
\newcommand{\cI}{\mathcal{I}}
\newcommand{\cR}{\mathcal{R}}
\newcommand{\cV}{\mathcal{V}}
\newcommand{\Si}{\Sigma}
\newcommand{\arccsc}{\mathop{\mathrm{arccsc}}}
\newcommand{\arcsec}{\mathop{\mathrm{arcsec}}}
\newcommand{\arccot}{\mathop{\mathrm{arccot}}}
\newcommand{\erf}{\mathop{\mathrm{erf}}}
\newcommand{\smsum}{\mathop{{\ts \sum}}}
\newcommand{\atp}[2]{ \genfrac{}{}{0in}{}{#1}{#2} }
\newcommand{\ave}{\mathrm{ave}}
\newcommand{\llt}{\left \lt }
\newcommand{\rgt}{\right \gt }
\newcommand{\YEaxis}[2]{\draw[help lines] (-#1,0)--(#1,0) node[right]{$x$};\draw[help lines] (0,-#2)--(0,#2) node[above]{$y$};}
\newcommand{\YEaaxis}[4]{\draw[help lines] (-#1,0)--(#2,0) node[right]{$x$};\draw[help lines] (0,-#3)--(0,#4) node[above]{$y$};}
\newcommand{\YEtaxis}[4]{\draw[help lines] (-#1,0)--(#2,0) node[right]{$t$};\draw[help lines] (0,-#3)--(0,#4) node[above]{$y$};}
\newcommand{\YEtaaxis}[4]{\draw[help lines] (-#1,0)--(#2,0) node[right]{$t$}; \draw[help lines] (0,-#3)--(0,#4) node[above]{$y$};}
\newcommand{\YExcoord}[2]{\draw (#1,.2)--(#1,-.2) node[below]{$#2$};}
\newcommand{\YEycoord}[2]{\draw (.2,#1)--(-.2,#1) node[left]{$#2$};}
\newcommand{\YEnxcoord}[2]{\draw (#1,-.2)--(#1,.2) node[above]{$#2$};}
\newcommand{\YEnycoord}[2]{\draw (-.2,#1)--(.2,#1) node[right]{$#2$};}
\newcommand{\YEstickfig}[3]{
\draw (#1,#2) arc(-90:270:2mm);
\draw (#1,#2)--(#1,#2-.5) (#1-.25,#2-.75)--(#1,#2-.5)--(#1+.25,#2-.75) (#1-.2,#2-.2)--(#1+.2,#2-.2);}
\newcommand{\IBP}[7]{
\begin{array}{|c | l | l |}
\hline
\color{red}{\text{Option 1:}}
& u=#2
&\color{red}{\dee{u}=#3 ~ \dee{#1}}
\\
& \dee{v}=#5~\dee{#1}
&\color{red}{v=#7}
\\
\hline
\color{blue}{\text{Option 2:}}
& u=#5
&\color{blue}{\dee{u}=#6 ~ \dee{#1}}
\\
&\dee{v}=#2 \dee{#1}
&\color{blue}{v=#4}
\\
\hline
\end{array}
}
\renewcommand{\textcolor}[2]{{\color{#1}{#2}}}
\newcommand{\trigtri}[4]{
\begin{tikzpicture}
\draw (-.5,0)--(2,0)--(2,1.5)--cycle;
\draw (1.8,0) |- (2,.2);
\draw[double] (0,0) arc(0:30:.5cm);
\draw (0,.2) node[right]{$#1$};
\draw (1,-.5) node{$#2$};
\draw (2,.75) node[right]{$#3$};
\draw (.6,1.1) node[rotate=30]{$#4$};
\end{tikzpicture}}
\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\,}$}}}
\)
Chapter 3 Sequences and series
You have probably learned about Taylor polynomials and, in particular, that
\begin{align*}
e^x &= 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} +E_n(x)
\end{align*}
where \(E_n(x)\) is the error introduced when you approximate \(e^x\) by its Taylor polynomial of degree \(n\text{.}\) You may have even seen a formula for \(E_n(x)\text{.}\) We are now going to ask what happens as \(n\) goes to infinity? Does the error go to zero, giving an exact formula for \(e^x\text{?}\) We shall later see that it does and that
\begin{align*}
e^x &=1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^\infty\frac{x^n}{n!}
\end{align*}
At this point we haven’t defined, or developed any understanding of, this infinite sum. How do we compute the sum of an infinite number of terms? Indeed, when does a sum of an infinite number of terms even make sense? Clearly we need to build up foundations to deal with these ideas. Along the way we shall also see other functions for which the corresponding error obeys \(\lim\limits_{n\rightarrow\infty}E_n(x)=0\) for some values of \(x\) and not for other values of \(x\text{.}\)
To motivate the next section, consider using the above formula with \(x=1\) to compute the number \(e\text{:}\)
\begin{align*}
e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots = \sum_{n=0}^\infty\frac{1}{n!}
\end{align*}
As we stated above, we don’t yet understand what to make of this infinite number of terms, but we might try to sneak up on it by thinking about what happens as we take more and more terms.
\begin{align*}
\text{1 term}\phantom{s} && 1&=1\\
\text{2 terms} && 1+1&=2\\
\text{3 terms} && 1+1+\frac{1}{2}&=2.5\\
\text{4 terms} && 1+1+\frac{1}{2}+\frac{1}{6}&=2.666666\dots\\
\text{5 terms} && 1+1+\frac{1}{2}+\frac{1}{6} + \frac{1}{24}&=2.708333\dots\\
\text{6 terms} && 1+1+\frac{1}{2}+\frac{1}{6} + \frac{1}{24} + \frac{1}{120}&=2.716666\dots
\end{align*}
By looking at the infinite sum in this way, we naturally obtain a sequence of numbers
\begin{gather*}
\{\ 1\,,\,2\,,\,2.5\,,\,2.666666\,,\cdots,\,2.708333\,,\cdots,\,
2.716666\,,\cdots,\,\cdots\ \}.
\end{gather*}
The key to understanding the original infinite sum is to understand the behaviour of this sequence of numbers — in particularly, what do the numbers do as we go further and further? Does it settle down to a given limit?
Mathematics program — English Program — Bahasa Indonesia Original authors’ website ↗ Book contents Program: English — B30 · Bahasa Indonesia — B30 · Mathematics Program home · Beranda Program Matematika · Original source — Original authors’ website — CLP-2 textbook, problems and PDFs · Open related page · Open related page