#LyX 2.4 created this file. For more info see https://www.lyx.org/ \lyxformat 620 \begin_document \begin_header \save_transient_properties true \origin unavailable \textclass book \begin_preamble \input defs \end_preamble \use_default_options true \maintain_unincluded_children no \language english \language_package default \inputencoding utf8 \fontencoding auto \font_roman "default" "default" \font_sans "default" "default" \font_typewriter "default" "default" \font_math "auto" "auto" \font_default_family default \use_non_tex_fonts false \font_sc false \font_roman_osf false \font_sans_osf false \font_typewriter_osf false \font_sf_scale 100 100 \font_tt_scale 100 100 \use_microtype false \use_dash_ligatures true \graphics default \default_output_format default \output_sync 0 \bibtex_command default \index_command default \float_placement class \float_alignment class \paperfontsize default \spacing single \use_hyperref false \papersize default \use_geometry false \use_package amsmath 1 \use_package amssymb 1 \use_package cancel 1 \use_package esint 1 \use_package mathdots 1 \use_package mathtools 1 \use_package mhchem 1 \use_package stackrel 1 \use_package stmaryrd 1 \use_package undertilde 1 \cite_engine basic \cite_engine_type default \biblio_style plain \use_bibtopic false \use_indices false \paperorientation portrait \suppress_date false \justification true \use_refstyle 1 \use_formatted_ref 0 \use_minted 0 \use_lineno 0 \index Index \shortcut idx \color #008000 \end_index \secnumdepth 3 \tocdepth 3 \paragraph_separation indent \paragraph_indentation default \is_math_indent 0 \math_numbering_side default \quotes_style english \dynamic_quotes 0 \papercolumns 1 \papersides 1 \paperpagestyle default \tablestyle default \tracking_changes false \output_changes false \change_bars false \postpone_fragile_content false \html_math_output 0 \html_css_as_file 0 \html_be_strict false \docbook_table_output 0 \docbook_mathml_prefix 1 \end_header \begin_body \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash exerc1[00] \end_layout \end_inset What is the smallest positive rational number? \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset There isn't. \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash exerc2[00] \end_layout \end_inset Is \begin_inset Formula $1+0.239999999\dots$ \end_inset a decimal expansion? \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset Yes, as long as it means a number \begin_inset Formula $x$ \end_inset with \begin_inset Formula $1.239999999\leq x<1.24$ \end_inset , that is, as long as it doesn't end with an infinite number of nines. \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash exerc3[02] \end_layout \end_inset What is \begin_inset Formula $(-3)^{-3}$ \end_inset ? \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset \begin_inset Formula $(-3)^{-3}=\frac{1}{(-3)^{3}}=\frac{1}{-27}=-\frac{1}{27}$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash rexerc4[05] \end_layout \end_inset What is \begin_inset Formula $(0.125)^{-2/3}$ \end_inset ? \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset \begin_inset Formula $(0.125)^{-2/3}=(\frac{1}{8})^{-2/3}=\sqrt[3]{(\frac{1}{8})^{-2}}=\sqrt[3]{8^{2}}=\sqrt[3]{64}=4$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash exerc5[05] \end_layout \end_inset We defined real numbers in terms of a decimal expansion. Discuss how we could have defined them in terms of a binary expansion instead, and give a definition to replace Eq. (2). \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset A decimal expansion would have the form \begin_inset Formula $x=n+0.b_{1}b_{2}b_{3}\dots$ \end_inset , where each \begin_inset Formula $b_{i}$ \end_inset would be a binary digit, 0 or 1, and this would mean that \begin_inset Formula \[ n+\frac{b_{1}}{2}+\frac{b_{2}}{4}+\dots+\frac{b_{k}}{2^{k}}\leq xy$ \end_inset , based on the decimal representation. \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset We say that \begin_inset Formula $x=y$ \end_inset if \begin_inset Formula $m=n$ \end_inset and \begin_inset Formula $d_{k}=e_{k}$ \end_inset for all \begin_inset Formula $k$ \end_inset , and that \begin_inset Formula $x0$ \end_inset , then \begin_inset Formula $f(x)\leq0$ \end_inset . We have \begin_inset Formula \[ f'(x)=\frac{1}{n\sqrt[n]{1+x}}-\frac{1}{n}=\frac{1}{n}\left(\frac{1}{\sqrt[n]{1+x}-1}\right), \] \end_inset but since \begin_inset Formula $\sqrt[n]{1+x}>1$ \end_inset for \begin_inset Formula $x>0$ \end_inset , we have \begin_inset Formula $f'(x)<0$ \end_inset and \begin_inset Formula $f$ \end_inset is strictly decreasing on \begin_inset Formula $(0,+\infty)$ \end_inset . Since \begin_inset Formula $f(0)=0$ \end_inset , this proves the result. \end_layout \begin_layout Enumerate It's clear that \begin_inset Formula $b^{n+d_{1}/10+\dots+d_{k}/10^{k}}\leq b^{n+1}$ \end_inset , and then, taken \begin_inset Formula $x=b-1>0$ \end_inset and \begin_inset Formula $n=10^{k}$ \end_inset , we have \begin_inset Formula $\sqrt[10^{k}]{1+b-1}=b^{1/10^{k}}\leq(b-1)/10^{k}$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash exerc15[10] \end_layout \end_inset Prove or disprove: \begin_inset Formula \begin{align*} \log_{b}x/y & =\log_{b}x-\log_{b}y, & \text{if }x,y & >0. \end{align*} \end_inset \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset Let \begin_inset Formula $\alpha:=\log_{b}x$ \end_inset and \begin_inset Formula $\beta:=\log_{b}y$ \end_inset , we have \begin_inset Formula \[ \frac{b^{\alpha}}{b^{\beta}}=b^{\alpha}b^{-\beta}=b^{\alpha-\beta}, \] \end_inset and taking logarithms, we get \begin_inset Formula $\log_{b}\frac{b^{\alpha}}{b^{\beta}}=\log_{b}\frac{x}{y}=\log_{b}b^{\alpha-\beta}=\alpha-\beta=\log_{b}x-\log_{b}y.$ \end_inset \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash exerc16[00] \end_layout \end_inset How can \begin_inset Formula $\log_{10}x$ \end_inset be expressed in terms of \begin_inset Formula $\ln x$ \end_inset and \begin_inset Formula $\ln10$ \end_inset ? \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset \begin_inset Formula $\log_{10}x=\frac{\log_{e}x}{\log_{e}10}=\frac{\ln x}{\ln10}$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash rexerc17[05] \end_layout \end_inset What is \begin_inset Formula $\lg32$ \end_inset ? \begin_inset Formula $\log_{\pi}\pi$ \end_inset ? \begin_inset Formula $\ln e$ \end_inset ? \begin_inset Formula $\log_{b}1$ \end_inset ? \begin_inset Formula $\log_{b}(-1)$ \end_inset ? \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset \begin_inset Formula $\lg32=5$ \end_inset , \begin_inset Formula $\log_{\pi}\pi=1$ \end_inset , \begin_inset Formula $\ln e=1$ \end_inset , \begin_inset Formula $\log_{b}1=0$ \end_inset . As for \begin_inset Formula $\log_{b}(-1)$ \end_inset , it would be an \begin_inset Formula $x$ \end_inset such that \begin_inset Formula $b^{x}=-1$ \end_inset , which is not a real number most of the time. As a complex number, it would be \begin_inset Formula $\log_{b}(-1)=\frac{\ln(-1)}{\ln b}=\frac{i\pi}{\ln b}$ \end_inset , since \begin_inset Formula $e^{i\pi}=-1$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash exerc18[10] \end_layout \end_inset Prove or disprove: \begin_inset Formula $\log_{8}x=\frac{1}{2}\lg x$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset \begin_inset Formula $\log_{8}x=\frac{\lg x}{\lg8}$ \end_inset , but \begin_inset Formula $\lg8=3\neq2$ \end_inset , so this is false in the general case (this is only true when \begin_inset Formula $x=1$ \end_inset ). \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash rexerc19[20] \end_layout \end_inset If \begin_inset Formula $n$ \end_inset is an integer whose decimal representation is 14 digits long, will the value of \begin_inset Formula $n$ \end_inset fit in a computer word with a capacity of 47 bits and a sign bit? \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset The number of possibilities for a number with up to 14 digits is \begin_inset Formula $10^{14}$ \end_inset , and the number for 47 bits is \begin_inset Formula $2^{47}\approx1.3\cdot2^{14}$ \end_inset , so it would fit. \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash exerc20[10] \end_layout \end_inset Is there any simple relation between \begin_inset Formula $\log_{10}2$ \end_inset and \begin_inset Formula $\log_{2}10$ \end_inset ? \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset \begin_inset Formula $\log_{10}2=\frac{\log_{2}2}{\log_{2}10}=\frac{1}{\log_{2}10}$ \end_inset , so \begin_inset Formula $\log_{10}2\log_{2}10=1$ \end_inset . \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash rexerc22[20] \end_layout \end_inset (R. W. Hamming.) Prove that \begin_inset Formula \[ \lg x\approx\ln x+\log_{10}x, \] \end_inset with less than \begin_inset Formula $\unit[1]{\%}$ \end_inset error! (Thus a table of natural logarithms and of common logarithms can be used to get approximate values of binary logarithms as well.) \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset \begin_inset Formula \[ \ln x+\log_{10}x=\frac{\lg x}{\lg e}+\frac{\lg x}{\lg10}=\lg x\left(\frac{1}{\lg e}+\frac{1}{\lg10}\right)\cong0.994\lg x\approx\lg x, \] \end_inset and this gives us about \begin_inset Formula $\unit[0.6]{\%}$ \end_inset error. \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash rexerc27[M25] \end_layout \end_inset Consider the method for calculating \begin_inset Formula $\log_{10}x$ \end_inset discussed in the text. Let \begin_inset Formula $x'_{k}$ \end_inset denote the computed approximation to \begin_inset Formula $x_{k}$ \end_inset , determined as follows: \begin_inset Formula $x(1-\delta)\leq10^{n}x'_{0}\leq x(1+\epsilon)$ \end_inset ; and in the determination of \begin_inset Formula $x'_{k}$ \end_inset by Eqs. (18), the quantity \begin_inset Formula $y_{k}$ \end_inset is used in place of \begin_inset Formula $(x'_{k-1})^{2}$ \end_inset , where \begin_inset Formula $(x'_{k-1})^{2}(1-\delta)\leq y_{k}\leq(x'_{k-1})^{2}(1+\epsilon)$ \end_inset and \begin_inset Formula $1\leq y_{k}<100$ \end_inset . Here \begin_inset Formula $\delta$ \end_inset and \begin_inset Formula $\epsilon$ \end_inset are small constants that reflect the upper and lower errors due to rounding or truncation. If \begin_inset Formula $\log^{\prime}x$ \end_inset denotes the result of the calculations, show that after \begin_inset Formula $k$ \end_inset steps we have \begin_inset Formula \[ \log_{10}x+2\log_{10}(1-\delta)-1/2^{k}<\log^{\prime}x\leq\log_{10}x+2\log_{10}(1+\epsilon). \] \end_inset \end_layout \begin_layout Standard \begin_inset ERT status open \begin_layout Plain Layout \backslash answer \end_layout \end_inset We first prove by induction that \begin_inset Formula \[ x^{2^{k}}(1-\delta)^{2^{k+1}-1}\leq10^{2^{k}(n+b_{1}/2+\dots+b_{k}/2^{k})}x'_{k}\leq x^{2^{k}}(1+\epsilon)^{2^{k+1}-1}. \] \end_inset For \begin_inset Formula $k=0$ \end_inset , we have \begin_inset Formula $x(1-\delta)\leq10^{n}x'_{0}\leq x(1+\epsilon)$ \end_inset , which is given. If this holds up to a certain \begin_inset Formula $k$ \end_inset , then if \begin_inset Formula $(x'_{k})^{2}<10$ \end_inset , we have \begin_inset Formula $b_{k+1}=0$ \end_inset , \begin_inset Formula $(1-\delta)(x'_{k})^{2}\leq x'_{k+1}\leq(1+\epsilon)(x'_{k})^{2}$ \end_inset and \begin_inset Formula \begin{eqnarray*} x^{2^{k+1}}(1-\delta)^{2^{k+2}-1} & = & (x^{2^{k+1}}(1-\delta)^{2^{k+1}-1})^{2}(1-\delta)\\ & \leq & (10^{2^{k}(n+b_{1}/2+\dots+b_{k}/2^{k})}x'_{k})^{2}(1-\delta)\\ & \leq & 10^{2^{k+1}(n+b_{1}/2+\dots+b_{k}/2^{k}+b_{k+1}/2^{k+1})}x'_{k+1}\\ & \leq & (10^{2^{k}(n+b_{1}/2+\dots+b_{k}/2^{k})}x'_{k})^{2}(1+\epsilon)\\ & \leq & (x^{2^{k}}(1+\epsilon)^{2^{k+1}-1})^{2}(1+\epsilon)=x^{2^{k+1}}(1+\epsilon)^{2^{k+2}-1}. \end{eqnarray*} \end_inset If \begin_inset Formula $(x'_{k})^{2}\geq10$ \end_inset , we have \begin_inset Formula $b_{k+1}=1$ \end_inset , \begin_inset Formula $(1-\delta)(x'_{k})^{2}\leq10x'_{k+1}\leq(1+\epsilon)(x'_{k})^{2}$ \end_inset and \begin_inset Formula \begin{eqnarray*} x^{2^{k+1}}(1-\delta)^{2^{k+2}-1} & \leq & (10^{2^{k}(n+b_{1}/2+\dots+b_{k}/2^{k})}x'_{k})^{2}(1-\delta)\\ & \leq & 10^{2^{k+1}(n+b_{1}/2+\dots+b_{k}/2^{k})}10x'_{k+1}\\ & = & 10^{2^{k+1}(n+b_{1}/2+\dots+b_{k}/2^{k}+b_{k+1}/2^{k+1})}x'_{k+1}\\ & \leq & (10^{2^{k}(n+b_{1}/2+\dots+b_{k}/2^{k})}x'_{k})^{2}(1+\epsilon)\\ & \leq & x^{2^{k+1}}(1+\epsilon)^{2^{k+2}-1}. \end{eqnarray*} \end_inset Then, by taking logarithms on the expression, we get \begin_inset Formula \[ 2^{k}\log x+(2^{k+1}-1)\log(1-\delta)\leq2^{k}\log^{\prime}x+\log x'_{k}\leq2^{k}\log x+(2^{k+1}-1)\log(1+\epsilon). \] \end_inset Finally, subtracting \begin_inset Formula $\log x'_{k}$ \end_inset from all sides, using that \begin_inset Formula $-1<-\log(1-\delta)-\log x'_{k}=-\log(x'_{k}(1-\delta))$ \end_inset , because \begin_inset Formula $x'_{k}(1-\delta)<10$ \end_inset , using that \begin_inset Formula $-\log(1+\epsilon)-\log x'_{k}=-\log(x'_{k}(1+\epsilon))\leq0$ \end_inset , because \begin_inset Formula $x'_{k}(1+\epsilon)\geq1$ \end_inset , and dividing everything by \begin_inset Formula $2^{k}$ \end_inset , we get \begin_inset Formula \[ \log x+2\log(1-\delta)-\frac{1}{2^{k}}<\log^{\prime}x\leq\log x+2\log(1+\epsilon), \] \end_inset which is precisely what we wanted to prove. \end_layout \end_body \end_document