Commit 2f5303c0 by Jim Hefferon

### homogeom

parent 65bd3df9
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 ... ... @@ -23,7 +23,7 @@ pref,% %dummy_chapter,% for printing the answers to Topic exercises gr1,gr2,gr3,cas,leontif,ppivot,network,% vs1,vs2,vs3,fields,crystal,voting,dimen,% map1,map2,map3,map4,map5,map6,lstsqs,erlang,markov,homogeom,% map1,map2,map3,map4,map5,map6,lstsqs,homogeom,markov,erlang,% det1,det2,det3,cramer,detspeed,projplane,% jc1,jc2,jc3,jc4,powers,pops,recur,%eigengeom,prinaxis,% appen,% ... ...
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bridges.sage 0 → 100644
 # bridges.sage # data on bridges and tolls # http://costoftolls.com/Tolls_in_New_York.html # http://maps.google.com/ # 2012-Jan-07 JH # Bring up sage in a terminal, then call # load "bridges.sage" # # Holland Tunnel 12.00 one way, 7 # Lincoln Tunnel 12.00 one way, 2 # George Washington Bridge 12.00 one way, 8 # Bear Mountain Bridge 1.00, 47 # Mid Hudson Bridge, Highland, NY, 1.00, 82 # Gov. Thomas E. Dewey Thruway bridge over hudson toll part of larger road toll # Kingston-Rhinecliff Bridge, 1.00 , 102 # Newburgh-Beacon Bridge, 1.00, 67 # Rip Van Winkle Bridge, 1.00, 120 # Tappan Zee Bridge, 5.00 one way, 27 # Verrazano-Narrows Bridge, 13.00 one way, 16 data=[ [2,6], # Lincoln Tunnel [7,6], # Holland Tunnel [8,6], # George Washington Bridge [16,6.5], # Verrazano-Narrows Bridge [27,2.5], # Tappan Zee Bridge [47,1], # Bear Mountain Bridge [67,1], # Newburgh-Beacon Bridge [82,1], # Mid Hudson Bridge [102,1], # Kingston-Rhinecliff Bridge [120,1], # Rip Van Winkle Bridge ] var('slope,intercept') model(x) = slope*x+intercept # use this to find the slope and intercept: find_fit(data,model) g=points(data)+plot(model(intercept=find_fit(data,model)[0].rhs(),slope=find_fit(data,model)[1].rhs()),(x,0,140),color='red',figsize=3) g.save("bridges.png") \ No newline at end of file
 ... ... @@ -2435,7 +2435,7 @@ beginfig(50) % rotation ccw numeric w; %horizontal scaling factor u:=.15in; w:=u; v:=u; save codomain_shift; pair codomain_shift; codomain_shift=(12.5w,0v); save codomain_shift; pair codomain_shift; codomain_shift=(15.5w,0v); % the axes save xmin, xmax, ymin, ymax; numeric xmin, xmax, ymin, ymax; xmin = -.5w; xmax = 3.5w; ymin = -.5v; ymax = 3.5v; ... ... @@ -2449,7 +2449,7 @@ beginfig(50) % rotation ccw xpart(point .5 of (xaxis shifted codomain_shift))]; y1 = .5[ypart(point .5 of yaxis), ypart(point .5 of (yaxis shifted codomain_shift))]; label(btex {\tiny $\mapsunder{\colvec{x \\ y} \mapsto \colvec{x\cos\theta-y\sin\theta \\ x\sin\theta+y\cos\theta}}$} etex,z1); label(btex $\xrightarrow{\scriptsize \colvec{x \\ y} \mapsto \colvec{x\cos\theta-y\sin\theta \\ x\sin\theta+y\cos\theta}}$ etex,z1); % the vectors z0 = (0w,0v); ... ... @@ -2531,7 +2531,7 @@ beginfig(51) % projection to xz-axis %z30 = .25*(z10+z11+z12+z13)+.5*codomain_shift; z30 = (point .5 of zaxis) shifted (.5*codomain_shift); label(btex {\tiny $\mapsunder{\colvec{x \\ y \\ z} \mapsto \colvec{x \\ 0 \\ z}}$} etex,z30); label(btex $\xrightarrow{\scriptsize \colvec{x \\ y \\ z} \mapsto \colvec{x \\ 0 \\ z}}$ etex,z30); % clean up the observer free_vect(lpn); ... ... @@ -2566,7 +2566,7 @@ beginfig(52) % tripling the x's xpart(point .5 of (xaxis shifted codomain_shift))]; y1 = .5[ypart(point .5 of yaxis), ypart(point .5 of (yaxis shifted codomain_shift))]; label(btex {\tiny $\mapsunder{\colvec{x \\ y} \mapsto \colvec{3x \\ y}}$} etex,z1); label(btex $\xrightarrow{\scriptsize \colvec{x \\ y} \mapsto \colvec{3x \\ y}}$ etex,z1); % the vectors z0 = (0w,0v); ... ... @@ -2605,7 +2605,7 @@ beginfig(53) % negative doubling the x's xpart(point .5 of (xaxis shifted codomain_shift))]; y1 = .5[ypart(point .5 of yaxis), ypart(point .5 of (yaxis shifted codomain_shift))]; label(btex {\tiny $\mapsunder{\colvec{x \\ y} \mapsto \colvec{-2x \\ y}}$} etex,z1); label(btex $\xrightarrow{\scriptsize \colvec{x \\ y} \mapsto \colvec{-2x \\ y}}$ etex,z1); % the vectors z0 = (0w,0v); ... ... @@ -2644,7 +2644,7 @@ beginfig(54) % negative doubling the x's xpart(point .5 of (xaxis shifted codomain_shift))]; y1 = .5[ypart(point .5 of yaxis), ypart(point .5 of (yaxis shifted codomain_shift))]; label(btex {\tiny $\mapsunder{\colvec{x \\ y} \mapsto \colvec{y \\ x}}$} etex,z1); label(btex $\xrightarrow{\scriptsize \colvec{x \\ y} \mapsto \colvec{y \\ x}}$ etex,z1); % the vectors z0 = (0w,0v); ... ... @@ -2688,7 +2688,7 @@ beginfig(55) % skew's action xpart(point .5 of (xaxis shifted codomain_shift))]; y1 = .5[ypart(point .5 of yaxis), ypart(point .5 of (yaxis shifted codomain_shift))]; label(btex {\tiny $\mapsunder{\colvec{x \\ y} \mapsto \colvec{x \\ 2x+y}}$} etex,z1); label(btex $\xrightarrow{\scriptsize \colvec{x \\ y} \mapsto \colvec{x \\ 2x+y}}$ etex,z1); % the vectors z0 = (0w,0v); ... ... @@ -2735,7 +2735,7 @@ beginfig(56) % skew on unit square xpart(point .5 of (xaxis shifted codomain_shift))]; y1 = .5[ypart(point .5 of yaxis), ypart(point .5 of (yaxis shifted codomain_shift))]; label(btex {\tiny $\mapsunder{\colvec{x \\ y} \mapsto \colvec{x \\ 2x+y}}$} etex,z1); label(btex $\xrightarrow{\scriptsize \colvec{x \\ y} \mapsto \colvec{x \\ 2x+y}}$ etex,z1); % the vectors z0 = (0w,0v); ... ... @@ -2783,7 +2783,7 @@ beginfig(57) % second skew on unit square xpart(point .5 of (xaxis shifted codomain_shift))]; y1 = .5[ypart(point .5 of yaxis), ypart(point .5 of (yaxis shifted codomain_shift))]; label(btex {\tiny $\mapsunder{\colvec{x \\ y} \mapsto \colvec{x+2y \\ y}}$} etex,z1); label(btex $\xrightarrow{\scriptsize \colvec{x \\ y} \mapsto \colvec{x+2y \\ y}}$ etex,z1); % the vectors z0 = (0w,0v); ... ...
 ... ... @@ -76,7 +76,7 @@ The obvious example is this \emph{translation}.\index{translation} \begin{equation*} \colvec{x \\ y} \quad\mapsto\quad \colvec{x \\ y}+\colvec{1 \\ 0}=\colvec{x+1 \\ y} \colvec{x \\ y}+\colvec[r]{1 \\ 0}=\colvec{x+1 \\ y} \end{equation*} However, this example turns out to be the only example, in the ... ... @@ -211,20 +211,20 @@ from $\vec{e}_1$ \begin{tabular}{@{}c@{}}\includegraphics{ch3.62}\end{tabular} \qquad $\rep{t}{\stdbasis_2,\stdbasis_2}= \begin{pmatrix} \begin{mat} a &-b \\ b &a \end{pmatrix}$ \end{mat}$\end{center} and one where is is mapped a quarter circle counterclockwise. \begin{center} \begin{tabular}{@{}c@{}}\includegraphics{ch3.63}\end{tabular} \qquad$\rep{t}{\stdbasis_2,\stdbasis_2}= \begin{pmatrix} \begin{mat} a &b \\ b &-a \end{pmatrix}$\end{mat}$ \end{center} We can geometrically describe these two cases. ... ... @@ -327,18 +327,18 @@ More on Klein and the Erlanger Program is in \cite{Yaglom}. \item Decide if each of these is an orthonormal matrix. \begin{exparts} \partsitem $\begin{pmatrix} \partsitem$\begin{mat}[r] 1/\sqrt{2} &-1/\sqrt{2} \\ -1/\sqrt{2} &-1/\sqrt{2} \end{pmatrix}$\partsitem$\begin{pmatrix} \end{mat}$\partsitem$\begin{mat}[r] 1/\sqrt{3} &-1/\sqrt{3} \\ -1/\sqrt{3} &-1/\sqrt{3} \end{pmatrix}$\partsitem$\begin{pmatrix} \end{mat}$\partsitem$\begin{mat}[r] 1/\sqrt{3} &-\sqrt{2}/\sqrt{3} \\ -\sqrt{2}/\sqrt{3} &-1/\sqrt{3} \end{pmatrix}$\end{mat}$ \end{exparts} \begin{answer} \begin{exparts} ... ... @@ -363,38 +363,38 @@ More on Klein and the Erlanger Program is in \cite{Yaglom}. \partsitem $\colvec{x \\ y} \mapsto \begin{pmatrix} \begin{mat} x\cdot\cos(\pi/6)-y\cdot\sin(\pi/6) \\ x\cdot\sin(\pi/6)+y\cdot\cos(\pi/6) \end{pmatrix} +\colvec{0 \\ 1} =\begin{pmatrix} \end{mat} +\colvec[r]{0 \\ 1} =\begin{mat} x\cdot(\sqrt{3}/2)-y\cdot(1/2)+0 \\ x\cdot(1/2)+y\cdot\cos(\sqrt{3}/2)+1 \end{pmatrix}$ \end{mat}$\partsitem The line$y=2x$makes an angle of$\arctan(2/1)$with the$x$-axis. Thus$\sin\theta=2/\sqrt{5}$and$\cos\theta=1/\sqrt{5}$. \begin{equation*} \colvec{x \\ y} \mapsto \begin{pmatrix} \begin{mat} x\cdot(1/\sqrt{5})-y\cdot(2/\sqrt{5}) \\ x\cdot(2/\sqrt{5})+y\cdot(1/\sqrt{5}) \end{pmatrix} \end{mat} \end{equation*} \partsitem$\colvec{x \\ y} \mapsto \begin{pmatrix} \begin{mat} x\cdot(1/\sqrt{5})-y\cdot(-2/\sqrt{5}) \\ x\cdot(-2/\sqrt{5})+y\cdot(1/\sqrt{5}) \end{pmatrix} +\colvec{1 \\ 1} =\begin{pmatrix} \end{mat} +\colvec[r]{1 \\ 1} =\begin{mat} x/\sqrt{5}+2y/\sqrt{5}+1 \\ -2x/\sqrt{5}+y/\sqrt{5}+1 \end{pmatrix}$\end{mat}$ \end{exparts} \end{answer} \item \label{exer:IsometryFacts} ... ... @@ -437,18 +437,18 @@ More on Klein and the Erlanger Program is in \cite{Yaglom}. Check that these two computations yield the same first two components. \begin{equation*} \begin{pmatrix} \begin{mat} a &c \\ b &d \end{pmatrix} \end{mat} \colvec{x \\ y} +\colvec{e \\ f} \qquad \begin{pmatrix} \begin{mat} a &c &e \\ b &d &f \\ 0 &0 &1 \end{pmatrix} \end{mat} \colvec{x \\ y \\ 1} \end{equation*} (These are ... ...
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 ... ... @@ -3261,10 +3261,10 @@ Note, as a check, that this result is indeed in $P$. See the third item of \nearbyexercise{exer:AlgOfPerps}.) \partsitem Generalize that to apply to any $\map{f}{\Re^n}{\Re^m}$. \end{exparts} This, and related results, is the In \cite{Strang93} this is called the \definend{Fundamental Theorem of Linear Algebra}% \index{Fundamental Theorem!of Linear Algebra} in \cite{Strang93}. \begin{answer} \begin{exparts} \partsitem The representation of ... ...
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