Commit 7ac475c2 authored by Jim Hefferon's avatar Jim Hefferon

new cover

parent 2f5b2f80
......@@ -124,4 +124,7 @@ General:
Jad Nohra, Tom Lahore, Lopatin Vladimir Many corrections, large and small.
2014-Jan-08
Mana Borwornpadungkitti Correction to exercise answer.
\ No newline at end of file
Mana Borwornpadungkitti Correction to exercise answer.
2014-Feb-21
Mana Borwornpadungkitti Number of corrections to homogeom.
\ No newline at end of file
// axes.asy
// Axes for book cover
import three;
import bsp;
import settings;
settings.outformat="pdf";
settings.render=0; // just draw it, not three-d shape
size(72*3); // 72 pts per inch
currentprojection=perspective(4,6.5,5);
currentlight=light(diffuse=gray(.6), ambient=yellow, specular=paleyellow,
specularfactor=0.1, viewport=false,(4,6.5,10));
material m_xz=
// diffusepen, ambientpen, emissivepen, specularpen
material( grey, yellow, gray(0.3), orange);
material m_xy=
// diffusepen, ambientpen, emissivepen, specularpen
material( gray(0.9), yellow, black, orange);
material m_yz=
// diffusepen, ambientpen, emissivepen, specularpen
material( grey, yellow, black, gray(0.1));
pen p=linewidth(0.4)+squarecap+miterjoin+black+opacity(0.2);
defaultpen(p);
pen slateblue = rgb(2/255, 101/255, 116/255);
pen puffblue = rgb(0/255, 194/255, 196/255);
pen beige = rgb(216/255, 195/255, 148/255);
pen burgundy = rgb(166/255, 5/255, 3/255);
real XLIMIT_POS=1;
real XLIMIT_NEG=-1*XLIMIT_POS;
real YLIMIT_POS=2;
real YLIMIT_NEG=-1*YLIMIT_POS;
real ZLIMIT_POS=4;
real ZLIMIT_NEG=-1*ZLIMIT_POS;
pen INSIDE_SHADE = gray(0.7);
pen OUTSIDE_SHADE = gray(0.99);
// boundary of quandrants to be shaded
path3 floor_q1 = (0,0,ZLIMIT_NEG)--(XLIMIT_POS,0,ZLIMIT_NEG)
--(XLIMIT_POS,YLIMIT_POS,ZLIMIT_NEG)--(0,YLIMIT_POS,ZLIMIT_NEG)--cycle;
path3 floor_q2 = (0,0,ZLIMIT_NEG)--(XLIMIT_NEG,0,ZLIMIT_NEG)
--(XLIMIT_NEG,YLIMIT_POS,ZLIMIT_NEG)--(0,YLIMIT_POS,ZLIMIT_NEG)--cycle;
path3 floor_q3 = (0,0,ZLIMIT_NEG)--(XLIMIT_NEG,0,ZLIMIT_NEG)
--(XLIMIT_NEG,YLIMIT_NEG,ZLIMIT_NEG)--(0,YLIMIT_NEG,ZLIMIT_NEG)--cycle;
path3 floor_q4 = (0,0,ZLIMIT_NEG)--(XLIMIT_POS,0,ZLIMIT_NEG)
--(XLIMIT_POS,YLIMIT_NEG,ZLIMIT_NEG)--(0,YLIMIT_NEG,ZLIMIT_NEG)--cycle;
// quadrants
surface q1 =surface(floor_q1,new pen[] {INSIDE_SHADE,OUTSIDE_SHADE,OUTSIDE_SHADE,OUTSIDE_SHADE});
surface q2 =surface(floor_q2,new pen[] {INSIDE_SHADE,OUTSIDE_SHADE,OUTSIDE_SHADE,OUTSIDE_SHADE});
surface q3 =surface(floor_q3,new pen[] {INSIDE_SHADE,OUTSIDE_SHADE,OUTSIDE_SHADE,OUTSIDE_SHADE});
surface q4 =surface(floor_q4,new pen[] {INSIDE_SHADE,OUTSIDE_SHADE,OUTSIDE_SHADE,OUTSIDE_SHADE});
picture shadow_pic;
size(shadow_pic,72*2);
draw(shadow_pic,q1,nolight);
draw(shadow_pic,q2,nolight);
draw(shadow_pic,q3,nolight);
draw(shadow_pic,q4,nolight);
shipout("shadow",shadow_pic);
// layer();
// // shadow
// import graph;
// // Find the projection to make this under the axes
// path3 floorplane_path=(0,0,ZLIMIT_NEG)--(1,0,ZLIMIT_NEG)
// --(1,1,ZLIMIT_NEG)--(0,1,ZLIMIT_NEG)--cycle;
// transform3 shade_transform=planeproject(floorplane_path,dir=normal(floorplane_path));
// pen[] pens = new pen[] {gray(.9), gray(.94), gray(.97), gray(.98),gray(.99),gray(.998),red};
// path outerpath = circle((0,0), 2);
// path innerpath = (1,0.8);
// path midpath(path p1, path p2, real t) {
// pair f(real s) {
// return (1-t)*relpoint(p1, s) + t*relpoint(p2, s);
// }
// path p = graph(f, 0, 1, operator ..) -- cycle;
// return p;
// }
// int lastpath = pens.length - 1;
// path[] paths = new path[lastpath + 1];
// surface[] grays = new surface[lastpath + 1];
// for (int i = 0; i <= lastpath; ++i) {
// paths[i] = midpath(innerpath, outerpath, i/lastpath);
// grays[i] = xscale3(5)*surface(paths[i]); // in XYPlane
// }
// draw(paths, pens);
path3 xz=(XLIMIT_POS,0,ZLIMIT_POS)
--(XLIMIT_NEG,0,ZLIMIT_POS)
--(XLIMIT_NEG,0,ZLIMIT_NEG)
--(XLIMIT_POS,0,ZLIMIT_NEG)
--cycle;
path3 xy=(XLIMIT_POS,YLIMIT_POS,0)
--(XLIMIT_NEG,YLIMIT_POS,0)
--(XLIMIT_NEG,YLIMIT_NEG,0)
--(XLIMIT_POS,YLIMIT_NEG,0)
--cycle;
path3 yz=(0,YLIMIT_POS,ZLIMIT_POS)
--(0,YLIMIT_NEG,ZLIMIT_POS)
--(0,YLIMIT_NEG,ZLIMIT_NEG)
--(0,YLIMIT_POS,ZLIMIT_NEG)
--cycle;
// surface xyplane = surface(xy);
// surface xzplane = surface(xz);
// surface yzplane = surface(yz);
// draw(xyplane,m_xy);
// draw(xzplane,m_xz);
// draw(yzplane,m_yz);
// draw(xy);
// draw(xz);
// draw(yz);
picture pic;
size(pic,72*3);
// Use asymptote "limited" hidden surface removal
face[] faces;
// pen[] p={red,green,blue,black};
// latticeshade(faces.push(floor_q1),floor_q1,new pen[][] {{INSIDE_SHADE,OUTSIDE_SHADE},{OUTSIDE_SHADE,OUTSIDE_SHADE}});
filldraw(faces.push(xy),project(xy),fillpen=beige,drawpen=p);
filldraw(faces.push(xz),project(xz),fillpen=puffblue,drawpen=p);
filldraw(faces.push(yz),project(yz),fillpen=slateblue,drawpen=p);
add(pic,faces);
// add(xscale3(4.0)*pic);
// add(currentpicture,shadow_pic);
add(currentpicture,pic);
shipout("axes",currentpicture);
......@@ -20513,12 +20513,12 @@ sage: p.save('bridges.pdf')
\includegraphics{ch3.96}
\end{center}
For instance, the effect of $H$ on the unit vector whose angle with
the $x$-axis is $\pi/3$ is this.
the $x$-axis is $\pi/6$ is this.
\begin{center}
\includegraphics{ch3.97}
\end{center}
Verifying that the resulting vector has unit length and forms an
angle of $-\pi/6$ with the $x$-axis is routine.
angle of $-\pi/12$ with the $x$-axis is routine.
\end{exparts}
\end{ans}
......@@ -63,6 +63,12 @@
\definecolor{lightcolor}{HTML}{DAE4FC} % {799FFC} %
\definecolor{bgcolor}{HTML}{FCFFDC} %
\definecolor{flourishcolor}{HTML}{F80018} %
% From https://kuler.adobe.com/Heisenberg-color-theme-3442140/
\definecolor{coverdarkcolor}{HTML}{026574} %
\definecolor{coverboldcolor}{HTML}{A60503} %
\definecolor{coverlightcolor}{HTML}{00C2C4} %
\definecolor{coverbgcolor}{HTML}{D8C394} %
\definecolor{coverflourishcolor}{HTML}{5A0000} %
% \usepackage[\dvidrv]{graphicx} %
\usepackage{graphicx} %
......@@ -899,34 +905,50 @@
% \usepackage[capitalise]{cleveref} % nameinlink option not in my LaTeX yet?
% ==================== Cover
\newcommand{\covergraphic}{%
\par\noindent\makebox[0pt][c]{\contourlength{1.4pt}
\rule{4.65in}{0em}
% \fontsize{260pt}{8pt}
\fontsize{235pt}{8pt}
{\fontfamily{qzc}\itshape\selectfont \contour{darkcolor}{\textcolor{lightcolor}{Linear}}}
} \\[0.05in]
% } \\[-2.15in]
% \noindent\makebox[0pt][c]{\rule{8.5in}{0pt}%
% \makebox[0pt][l]{\contourlength{0.6pt} %
% \fontsize{65pt}{10pt}
% \newcommand{\covergraphic}{%
% \par\noindent\makebox[0pt][c]{\contourlength{1.4pt}
% \rule{4.65in}{0em}
% % \fontsize{260pt}{8pt}
% \fontsize{235pt}{8pt}
% {\fontfamily{qzc}\itshape\selectfont \contour{darkcolor}{\textcolor{lightcolor}{Linear}}}
% } \\[0.05in]
% % } \\[-2.15in]
% % \noindent\makebox[0pt][c]{\rule{8.5in}{0pt}%
% % \makebox[0pt][l]{\contourlength{0.6pt} %
% % \fontsize{65pt}{10pt}
% % {\fontfamily{qzc}\itshape\selectfont \contour{flourishcolor}{\textcolor{white}{Algebra}}}
% % }
% % }
% \noindent\makebox[0pt][c]{\rule{7.4in}{0pt} % \rule{8.65in}{0pt}%
% \makebox[0pt][l]{\contourlength{0.9pt} %
% \fontsize{55pt}{6pt} % \fontsize{65pt}{10pt}
% {\fontfamily{qzc}\itshape\selectfont \contour{flourishcolor}{\textcolor{white}{Algebra}}}
% }
% }
\noindent\makebox[0pt][c]{\rule{7.4in}{0pt} % \rule{8.65in}{0pt}%
\makebox[0pt][l]{\contourlength{0.9pt} %
\fontsize{55pt}{6pt} % \fontsize{65pt}{10pt}
{\fontfamily{qzc}\itshape\selectfont \contour{flourishcolor}{\textcolor{white}{Algebra}}}
}
% }
% qzc
\newcommand{\covergraphic}{%
\setlength{\unitlength}{1in}
\begin{picture}(0,0)
% \put(0,0){\line(1,1){2}}
\put(-0.3,-2.5){\fontsize{50pt}{8pt}{\fontfamily{ugq}\selectfont LINEAR ALGEBRA}}
\thicklines
\put(-0.26,-2.75){\color{coverboldcolor}\rule{5.78in}{4pt}}
\put(3.75,-3.025){\fontsize{20pt}{8pt}{\fontfamily{ugq}\selectfont Jim Hef{}feron}}
\put(-0.0,-6.9){\includegraphics{shadow.pdf}}
\put(0,-6.5){\includegraphics{axes.pdf}}
\put(2.3,-6.4){{\color{coverdarkcolor}\large \texttt{http://joshua.smcvt.edu/linearalgebra}}}
\end{picture}
}
}
\newcommand{\coverauthor}{%
\noindent\makebox[0em][l]{\makebox[8.5in]{%
\begin{tabular}{r}
{\Large Jim Hef{}feron} \\[1ex]
{\large \url{http://joshua.smcvt.edu/linearalgebra}}
\end{tabular}}}
% \noindent\makebox[0em][l]{\makebox[8.5in]{%
% \begin{tabular}{r}
% {\Large Jim Hef{}feron} \\[1ex]
% {\large \url{http://joshua.smcvt.edu/linearalgebra}}
% \end{tabular}}}
\relax
}
\usepackage{conc}
......
......@@ -12,9 +12,10 @@
% {\Large Jim Hef{}feron} \\[1ex]
% {\large \url{http://joshua.smcvt.edu/linearalgebra}}
% \end{tabular}}}
\vspace*{.3in}
\coverauthor{}
\vspace*{3.6in}
% next three lines prior cover
% \vspace*{.3in}
% \coverauthor{}
% \vspace*{3.6in}
\covergraphic{}
% \noindent\makebox[0pt][l]{\contourlength{1.4pt}
% \rule{-1.9in}{0em}
......
......@@ -3561,19 +3561,23 @@ solution set's description.
\begin{lemma} \label{th:GenEqPartHomo}
%<*th:GenEqPartHomo>
For a linear system, where $\vec{p}\/$ is any particular solution,
the solution set equals this set.
\begin{equation*}
For a linear system and for any particular solution $\vec{p}\/$,
the solution set equals
% \begin{equation*}
$
\set{\vec{p}+\vec{h} \suchthat \text{ \( \vec{h} \) satisfies the
associated homogeneous system} }
\end{equation*}
associated homogeneous system} }$.
%\end{equation*}
%</th:GenEqPartHomo>
\end{lemma}
% So fixing any particular solution gives the above
% description of the solution set.
\begin{proof}
We will show mutual set inclusion, that any solution to the system is in
the above set and that anything in the set is a solution of the
system.\appendrefs{equality of sets}
system.\appendrefs{set equality}
%<*pf:GenEqPartHomo0>
For set inclusion the first way, that if a vector solves the system
......
......@@ -35,8 +35,8 @@ than is a domain interval near $x=0$.
The linear maps are nicer, more regular,
in that for each map all of the domain
spreads by the same factor.
The map~$f_1$ on the left spreads all intervals apart to be twice as wide
while on the right~$f_2$ keeps intervals the same length but reverses
The map~$h_1$ on the left spreads all intervals apart to be twice as wide
while on the right~$h_2$ keeps intervals the same length but reverses
their orientation, as with the rising interval from $1$ to $2$
being transformed
to the falling interval from $-1$ to~$-2$.
......@@ -169,7 +169,7 @@ The resulting shape has the same base and height as the square
\end{center}
For contrast, the next picture shows the effect of the map represented by
$C_{2,1}(1)$.
$C_{2,1}(2)$.
Here vectors are affected according to their
second component:
$\binom{x}{y}$ slides horizontally by twice $y$.
......@@ -431,12 +431,12 @@ The Chain Rule multiplies the matrices.
\includegraphics{ch3.96}
\end{center}
For instance, the effect of $H$ on the unit vector whose angle with
the $x$-axis is $\pi/3$ is this.
the $x$-axis is $\pi/6$ is this.
\begin{center}
\includegraphics{ch3.97}
\end{center}
Verifying that the resulting vector has unit length and forms an
angle of $-\pi/6$ with the $x$-axis is routine.
angle of $-\pi/12$ with the $x$-axis is routine.
\end{exparts}
\end{answer}
\item
......
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