Revision as of 00:01, 3 November 2024 by Bot (Created page with "<div class="d-none"><math> \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow}...")
BBy Bot
Nov 03'24
Exercise
[math]
\newcommand{\ex}[1]{\item }
\newcommand{\sx}{\item}
\newcommand{\x}{\sx}
\newcommand{\sxlab}[1]{}
\newcommand{\xlab}{\sxlab}
\newcommand{\prov}[1] {\quad #1}
\newcommand{\provx}[1] {\quad \mbox{#1}}
\newcommand{\intext}[1]{\quad \mbox{#1} \quad}
\newcommand{\R}{\mathrm{\bf R}}
\newcommand{\Q}{\mathrm{\bf Q}}
\newcommand{\Z}{\mathrm{\bf Z}}
\newcommand{\C}{\mathrm{\bf C}}
\newcommand{\dt}{\textbf}
\newcommand{\goesto}{\rightarrow}
\newcommand{\ddxof}[1]{\frac{d #1}{d x}}
\newcommand{\ddx}{\frac{d}{dx}}
\newcommand{\ddt}{\frac{d}{dt}}
\newcommand{\dydx}{\ddxof y}
\newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}}
\newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}}
\newcommand{\dist}{\mathrm{distance}}
\newcommand{\arccot}{\mathrm{arccot\:}}
\newcommand{\arccsc}{\mathrm{arccsc\:}}
\newcommand{\arcsec}{\mathrm{arcsec\:}}
\newcommand{\arctanh}{\mathrm{arctanh\:}}
\newcommand{\arcsinh}{\mathrm{arcsinh\:}}
\newcommand{\arccosh}{\mathrm{arccosh\:}}
\newcommand{\sech}{\mathrm{sech\:}}
\newcommand{\csch}{\mathrm{csch\:}}
\newcommand{\conj}[1]{\overline{#1}}
\newcommand{\mathds}{\mathbb}
[/math]
The first of the following examples comes from the formula for a geometric series, and the last two follow from the theory developed later in this chapter:
- [math]\frac23 = \frac1{1+\frac12} = \sum_{i=0}^\infty (-\frac12)^i = 1 - \frac12 + \frac14 - \cdots[/math].
- [math]\ln 2 = \sum_{i=1}^\infty (-1)^{i+1} \frac1i = 1 - \frac12 + \frac13 - \frac14 + \cdots[/math].
- [math]\pi = 4 \arctan 1 = \sum_{i=0}^\infty (-1)^i \frac4{2i+1}= 4 - \frac43 + \frac45 - \frac47 + \cdots[/math].
If the value of each of these series is approximated by a partial sum [math]\sum_{i=m}^\infty a_i[/math], how large must [math]n[/math] be taken to ensure an error no greater than [math]0.1[/math], [math]0.01[/math], [math]0.001[/math], [math]10^{-6}[/math]?