Add footnote, a bit more clarification in 4.1, and some minor other

clarifications

CHANGES

@@ -47,6 +47,13 @@
   3 defective eigenvalues, and improve exposition on the higher
   multiplicity.
 
+* 4.1: Expand slightly more on the linear algebra connection for
+  eigenvalues/eigenvectors
+* 4.1: Add footnote with the definition of sinh, cosh, as some students
+  might be coming to this sections not having seen these (or not having
+  seen them recently), and may have seen the solution in terms of
+  exponentials.
+
 * 7.1: Add root test and an example for it.
 
 * 8.1: Add a short paragraph on what happens near noncritical points to

ch-first-order-ode.tex

@@ -464,7 +464,7 @@ \subsection{Slope fields}
 By looking at the slope field we get a lot of information
 about the behavior of solutions without having to solve
 the equation.  For
-example, in \figureref{1.3:fig2} we see what the solutions do when the initial conditions
+example, in \figurevref{1.3:fig2} we see what the solutions do when the initial conditions
 are $y(0) > 0$, $y(0) = 0$ and $y(0) < 0$.
 A small change in the
 initial condition causes quite different behavior.
@@ -1001,7 +1001,7 @@ \subsection{Examples}
 $t = - \frac{\ln \frac{60-22}{67}}{0.0616} \approx 9.21$ minutes.  So Bob can
 begin to drink the coffee at just over 9 minutes from the time Bob made
 it.  That is probably about the amount of time it took us to calculate how long
-it would take.  See \figureref{sintro:coffeefig}.
+it would take.  See \figurevref{sintro:coffeefig}.
 
 \begin{myfig}
 \capstart

ch-fourier-and-pde.tex

@@ -91,11 +91,17 @@ \subsection{Eigenvalue problems}
 
 Note the similarity to eigenvalues and eigenvectors of matrices.  The
 similarity is not just coincidental.  If we think of the equations as
-differential operators, then we are doing the same exact thing.  For example, 
-let $L = -\frac{d^2}{{dt}^2}$.  We are looking for nonzero functions $f$
+differential operators, then we are doing the same exact thing.
+Think of a function $x(t)$
+as a vector with infinitely many components (one for each $t$).
+Let $L = -\frac{d^2}{{dt}^2}$ be the linear operator.
+Then the eigenvalue/eigenfunction pair should be $\lambda$ and
+nonzero $x$ such that $Lx = \lambda x$.
+In other words,
+we are looking for nonzero functions $x$
 satisfying certain endpoint conditions that solve
-$(L- \lambda)f = 0$.  A lot of the formalism from linear algebra can still
-apply here, though we will not pursue this line of reasoning too far.
+$(L- \lambda)x = 0$.  A lot of the formalism from linear algebra still
+applies here, though we will not pursue this line of reasoning too far.
 
 \begin{example} \label{bvp:eig1ex}
 Let us find the eigenvalues and eigenfunctions of
@@ -122,15 +128,23 @@ \subsection{Eigenvalue problems}
  $\sqrt{\lambda} = k$ for a positive integer $k$.
 Hence the positive eigenvalues are
 $k^2$ for all integers $k \geq 1$.  Corresponding eigenfunctions
-can be taken as $x=\sin (k t)$.  Just like for eigenvectors, we get all the
-multiples of an eigenfunction, so we only need to pick one.
+can be taken as $x=\sin (k t)$.  Just like for eigenvectors, constant
+multiples of an eigenfunction are also eigenfunctions,
+so we only need to pick one.
 
 Now suppose that $\lambda = 0$.  In this case the equation is $x'' = 0$,
 and its general solution is $x = At + B$.  The condition $x(0) = 0$ implies
 that $B=0$, and $x(\pi) = 0$ implies that $A = 0$.  This means that $\lambda
 = 0$ is \emph{not} an eigenvalue.
 
-Finally, suppose that $\lambda < 0$.  In this case we have the general solution
+Finally, suppose that $\lambda < 0$.  In this case we have the general
+solution\footnote{Recall that
+$\cosh s = \frac{1}{2}(e^s+e^{-s})$
+and
+$\sinh s = \frac{1}{2}(e^s-e^{-s})$.  As an exercise
+try the computation with the general solution written as
+$x = A e^{\sqrt{-\lambda}\, t} + B e^{-\sqrt{-\lambda}\, t}$ (for
+different $A$ and $B$ of course).}
 \begin{equation*}
 x = A \cosh ( \sqrt{-\lambda}\, t) + B \sinh ( \sqrt{-\lambda}\, t ) .
 \end{equation*}
@@ -367,7 +381,7 @@ \subsection{Orthogonality of eigenfunctions}
 \quad
 \text{when } m \not = n.
 \end{equation*}
-Similarly
+Similarly,
 \begin{equation*}
 \int_{0}^\pi \cos (mt) \cos (nt) ~dt = 0 ,
 \quad
@@ -496,7 +510,7 @@ \subsection{Application}
 Let $L$ be the length of the string (in meters) and the string
 is fixed at the beginning and end
 points.  Hence, $y(0) = 0$ and $y(L) = 0$.  See
-\figureref{bvp:whirstringfig}.
+\figurevref{bvp:whirstringfig}.
 
 \begin{myfig}
 \capstart

ch-laplace.tex

@@ -1478,7 +1478,7 @@ \subsection{Rectangular pulse}
 0 & \text{if } \; b \leq t .
 \end{cases}
 \end{equation*}
-See \figureref{lt:sqpulse} for a graph.
+See \figurevref{lt:sqpulse} for a graph.
 Notice that
 \begin{equation*}
 \varphi(t) = M \bigl( u(t-a) - u(t-b) \bigr) ,

ch-nonlin-systems.tex

@@ -1924,7 +1924,7 @@ \subsection{Exercises}
 Van der Pol oscillator (for $\mu > 0$) must not lie completely in the set
 where 
 $-1 < x < 1$.
-Compare with \figureref{fig:nlin-van-der-fig}.
+Compare with \figurevref{fig:nlin-van-der-fig}.
 \end{exercise}
 \exsol{%
 $f(x,y) = y$, $g(x,y) = \mu(1-x^2)y-x$.  So

ch-power-ser.tex

@@ -888,7 +888,7 @@ \section{Series solutions of linear second order ODEs}
 \end{myfig}
 \end{example}
 The functions $y_1$ and $y_2$ cannot be written in terms of the elementary
-functions that you know.  See \figureref{ps:airyfig} for the plot of
+functions that you know.  See \figurevref{ps:airyfig} for the plot of
 the solutions $y_1$ and $y_2$.  These functions have many interesting
 properties.  For example, they are oscillatory for negative $x$
 (like solutions to $y''+y=0$) and

ch-systems.tex

@@ -3499,7 +3499,7 @@ \subsection{Examples}
 \end{bmatrix} .
 \end{equation*}
 The graphs of the two displacements, $x_1$ and $x_2$ of the two carts is in
-\figureref{sosa:superposfig}.
+\figurevref{sosa:superposfig}.
 \begin{myfig}
 \capstart
 \diffyincludegraphics{width=3in}{width=4.5in}{sosa-superpos}
@@ -4405,12 +4405,13 @@ \subsection{Defective eigenvalues}
 \right)
 = (2-\lambda)^2(1-\lambda) .
 \end{equation*}
-So the eigenvalues are 1 and 2, where 2 has multiplicity 2.
-For the eigenvalue $\lambda = 1$ we leave it to the reader to find that
+The eigenvalues are 1 and 2, where 2 has multiplicity 2.
+We leave it to the reader to find that
 $\left[ \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \right]$
-is an eigenvector.
+is an eigenvector
+for the eigenvalue $\lambda = 1$.
 
-Let's focus on $\lambda = 2$.  Let's compute eigenvectors,
+Let's focus on $\lambda = 2$.  We compute eigenvectors:
 \begin{equation*}
 \vec{0} =
 (A - 2 I) \vec{v}
@@ -4432,11 +4433,11 @@ \subsection{Defective eigenvalues}
 Problem is that setting $v_3$ to anything else just gets multiples
 of this vector and so we have a defect of 1.
 Let $\vec{v}_1$ be the eigenvector and let's look for
-a generalized eigenvector $\vec{v}_2$.
+a generalized eigenvector $\vec{v}_2$:
 \begin{equation*}
 (A - 2 I) \vec{v}_2 = \vec{v}_1 . 
 \end{equation*}
-or
+Or
 \begin{equation*}
 \begin{bmatrix}
 0 & -5 & 0 \\
@@ -4454,9 +4455,9 @@ \subsection{Defective eigenvalues}
 where we used $a$, $b$, $c$ as components of $\vec{v}_2$ for simplicity.
 The first equation says $-5b = 1$ so $b = \nicefrac{-1}{5}$.  The
 second equation says nothing.
-The last equation is $-a + 4b - c = -1$ or
-$a + \nicefrac{4}{5} + c = 1$ or
-$a + c = \nicefrac{1}{5}$.  We can let $c$ be the free variable, so let's
+The last equation is $-a + 4b - c = -1$, or
+$a + \nicefrac{4}{5} + c = 1$, or
+$a + c = \nicefrac{1}{5}$.  We let $c$ be the free variable and we
 choose $c=0$.  We find
 $\vec{v}_2 = \left[ \begin{smallmatrix} \nicefrac{1}{5} \\ \nicefrac{-1}{5}
 \\ 0 \end{smallmatrix} \right]$.