Commit 16e057b4 authored by Jim Hefferon's avatar Jim Hefferon

map6

parent 88c5cf63
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......@@ -17965,7 +17965,7 @@ sage: n(a_by_a)
\end{ans}
\section{Projection}
\subsection{Three.VI.1: Orthogonal Projection Into a Line}
\begin{ans}{Three.VI.1.7}
\begin{ans}{Three.VI.1.6}
Each is a straightforward application of the formula from
\nearbydefinition{def:ProjIntoLine}.
\begin{exparts*}
......@@ -18002,7 +18002,7 @@ sage: n(a_by_a)
\end{exparts*}
\end{ans}
\begin{ans}{Three.VI.1.8}
\begin{ans}{Three.VI.1.7}
\begin{exparts}
\partsitem
$\displaystyle
......@@ -18024,7 +18024,7 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.VI.1.9}
\begin{ans}{Three.VI.1.8}
$\displaystyle
\frac{\colvec[r]{1 \\ 2 \\ 1 \\ 3}\dotprod\colvec[r]{-1 \\ 1 \\ -1 \\ 1}}{%
\colvec[r]{-1 \\ 1 \\ -1 \\ 1}\dotprod\colvec[r]{-1 \\ 1 \\ -1 \\ 1}}
......@@ -18034,7 +18034,7 @@ sage: n(a_by_a)
=\colvec[r]{-3/4 \\ 3/4 \\ -3/4 \\ 3/4}$
\end{ans}
\begin{ans}{Three.VI.1.10}
\begin{ans}{Three.VI.1.9}
\begin{exparts*}
\partsitem $\displaystyle
\frac{\colvec[r]{1 \\ 2}\dotprod\colvec[r]{3 \\ 1}}{%
......@@ -18066,7 +18066,7 @@ sage: n(a_by_a)
\end{equation*}
\end{ans}
\begin{ans}{Three.VI.1.11}
\begin{ans}{Three.VI.1.10}
Suppose that $\vec{v}_1$ and $\vec{v}_2$ are nonzero and orthogonal.
Consider the linear relationship
$c_1\vec{v}_1+c_2\vec{v}_2=\zero$.
......@@ -18085,7 +18085,7 @@ sage: n(a_by_a)
Showing that $c_2$ is zero is similar.
\end{ans}
\begin{ans}{Three.VI.1.12}
\begin{ans}{Three.VI.1.11}
\begin{exparts}
\partsitem If the vector $\vec{v}\,$ is in the line then the
orthogonal projection is \( \vec{v} \).
......@@ -18131,7 +18131,7 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.VI.1.13}
\begin{ans}{Three.VI.1.12}
If $\vec{s}\,$ is the zero vector then the expression
\begin{equation*}
\proj{\vec{v}}{\spanof{\vec{s}\,}}=
......@@ -18144,7 +18144,7 @@ sage: n(a_by_a)
must be defined to be \( \zero \).
\end{ans}
\begin{ans}{Three.VI.1.14}
\begin{ans}{Three.VI.1.13}
Any vector in \( \Re^n \) is the projection of some other into a
line, provided that the dimension \( n \) is greater than one.
(Clearly, any vector is the projection of itself
......@@ -18188,7 +18188,7 @@ sage: n(a_by_a)
of itself.
\end{ans}
\begin{ans}{Three.VI.1.15}
\begin{ans}{Three.VI.1.14}
The proof is simply a calculation.
\begin{equation*}
\norm{\frac{\vec{v}\dotprod\vec{s}}{\vec{s}\dotprod\vec{s}}
......@@ -18204,7 +18204,7 @@ sage: n(a_by_a)
\end{equation*}
\end{ans}
\begin{ans}{Three.VI.1.16}
\begin{ans}{Three.VI.1.15}
Because the projection of \( \vec{v} \) into the line spanned by
\( \vec{s} \) is
\begin{equation*}
......@@ -18241,7 +18241,7 @@ sage: n(a_by_a)
\end{align*}
\end{ans}
\begin{ans}{Three.VI.1.17}
\begin{ans}{Three.VI.1.16}
Because square root is a strictly increasing function, we can
minimize \( d(c)=(cs_1-v_1)^2+(cs_2-v_2)^2 \) instead of the square root
of \( d \).
......@@ -18268,7 +18268,7 @@ sage: n(a_by_a)
etc.
\end{ans}
\begin{ans}{Three.VI.1.18}
\begin{ans}{Three.VI.1.17}
The Cauchy-Schwartz inequality
$\absval{\vec{v}\dotprod\vec{s}\,}
\leq\norm{\vec{v}\,}\cdot\norm{\vec{s}\,}$
......@@ -18289,14 +18289,14 @@ sage: n(a_by_a)
That is, \( \norm{\vec{v}\,} \) is larger than or equal to the fraction.
\end{ans}
\begin{ans}{Three.VI.1.19}
\begin{ans}{Three.VI.1.18}
Write \( c\vec{s} \) for \( \vec{q} \), and
calculate:
\( (\vec{v}\dotprod c\vec{s}/c\vec{s}\dotprod c\vec{s}\,)\cdot c\vec{s}=
(\vec{v}\dotprod \vec{s}/\vec{s}\dotprod \vec{s}\,)\cdot \vec{s} \).
\end{ans}
\begin{ans}{Three.VI.1.20}
\begin{ans}{Three.VI.1.19}
\begin{exparts}
\partsitem Fixing
\begin{equation*}
......@@ -18325,7 +18325,7 @@ sage: n(a_by_a)
\end{exparts}
\end{ans}
\begin{ans}{Three.VI.1.21}
\begin{ans}{Three.VI.1.20}
The sequence need not settle down.
With
\begin{equation*}
......@@ -19622,7 +19622,7 @@ sage: n(a_by_a)
The proof of \nearbylemma{le:OrthoProjWellDefd} shows that
each vector $\vec{v}\in\Re^n$
is the sum of its orthogonal projections
onto the lines spanned by the basis vectors.
into the lines spanned by the basis vectors.
\begin{equation*}
\vec{v}=\proj{\vec{v}\,}{\spanof{\vec{\kappa}_1}}
+\dots+\proj{\vec{v}\,}{\spanof{\vec{\kappa}_n}}
......@@ -19698,7 +19698,7 @@ sage: n(a_by_a)
$M$ and the second half is a basis for $M^\perp$.
The proof also checks that
each vector in the space is the sum of its orthogonal projections
onto the lines spanned by these basis vectors.
into the lines spanned by these basis vectors.
\begin{equation*}
\vec{v}=\proj{\vec{v}\,}{\spanof{\vec{\kappa}_1}}
+\dots+\proj{\vec{v}\,}{\spanof{\vec{\kappa}_n}}
......@@ -1020,10 +1020,10 @@ beginfig(25) % (2,2,1) and its projection
pickup pencircle scaled (line_width_light);
drawarrow z0--z1;
label.rt(btex \raisebox{0ex}[0pt][0pt]{\scriptsize $\colvec[r]{1 \\ 2 \\ 2}$} etex,z1);
label.rt(btex \raisebox{0ex}[0pt][0pt]{\tiny $\colvec[r]{1 \\ 2 \\ 2}$} etex,z1);
pickup pencircle scaled (line_width_dark);
drawarrow z0--z2;
%label.lrt(btex {\scriptsize $\colvec[r]{1 \\ 2 \\ 0}$} etex,z2);
%label.lrt(btex {\tiny $\colvec[r]{1 \\ 2 \\ 0}$} etex,z2);
draw_action_arrow((z1+(1pt,0))--(z2+(1pt,0)));
% shift so y-axis is on TeX's baseline (to line this pic with #26)
show ypart(llcorner currentpicture);
......@@ -1065,7 +1065,7 @@ beginfig(26) % (2,2,-1) and its projection
pickup pencircle scaled (line_width_light);
drawarrow z0--z1;
label.rt(btex \raisebox{0ex}[0pt][0pt]{\scriptsize $\colvec[r]{1 \\ 2 \\ -1}$} etex,z1);
label.rt(btex \raisebox{0ex}[0pt][0pt]{\tiny $\colvec[r]{1 \\ 2 \\ -1}$} etex,z1);
pickup pencircle scaled (line_width_dark);
drawarrow z0--z2;
%label.lrt(btex {\scriptsize $\colvec[r]{1 \\ 2 \\ 0}$} etex,z2);
......
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