diff --git a/ca.tex b/ca.tex
index 10d7946..ee98279 100644
--- a/ca.tex
+++ b/ca.tex
@@ -15106,13 +15106,13 @@ \subsection{Schwarz reflection principle}
=
0 .
\end{equation*}
-In other words, the mean value equals $F(z) = 0$,
+That is, the mean value equals $F(z) = 0$,
and the mean-value property is satisfied for all small enough
$r$ at every $z \in L$.
We proved above that $F$ is harmonic on $U \setminus L$ and
so the mean-value property
-is satisfied for all small enough
-$r$ around any of $U \setminus L$ as well.
+is also satisfied for all small enough circles
+around any $z \in U \setminus L$.
By the mean-value property (\thmref{thm:meanprop}), $F$ is harmonic in $U$.
\end{proof}
@@ -15811,9 +15811,9 @@ \section{Infinite products}
\label{sec:prod}
If a
-function has zeros at $0$, $1$, and $i$ we
+function has zeros at $0$, $1$, and $i$, we
can write $f$ as
-$f(z) = g(z) z(z-1)(z-i)$. And if those are the only zeros, then
+$f(z) = g(z) z(z-1)(z-i)$. If those are the only zeros, then
$g$ is never zero. If there are infinitely many zeros, however,
things become difficult.
Can we factor out the zeros out of $\sin z$? Can we
@@ -15825,7 +15825,7 @@ \section{Infinite products}
we take care of convergence.\footnote{Beware of
formal expressions bearing gifts.
Especially ones with infinite sets in them.
-For example, $\prod_{n \in \Z} (z-\pi n)$ does not make sense.}
+For example, $\prod_{n \in \Z} (z-\pi n)$ does not actually make any sense.}
Of course, we need to first
figure out what we mean by convergence. Once we figure that out,
we will show that convergence happens the way we want to
@@ -15900,7 +15900,7 @@ \section{Infinite products}
For the other direction, suppose that the sequence of partial products
converges. For all but finitely many $n$, $\sabs{a_n} < 1$.
-Otherwise the product would double infinitely often and go to
+Otherwise the partial products would double infinitely often and go to
infinity. So suppose that $\sabs{a_n} < 1$ for all $n \geq N$.
If $\sabs{a_n} < 1$, then $\sabs{a_n} \leq 2 \log \bigl(1+\sabs{a_n}\bigr)$,
and
@@ -15917,7 +15917,7 @@ \section{Infinite products}
An immediate consequence is that if
$\prod_{n=1}^\infty (1+a_n)$ converges absolutely, then
-$\{ a_n \}$ converges to $0$. Actually much more than that, they have to
+$\{ a_n \}$ converges to $0$. Not only that, they
go to zero fast enough to be absolutely summable.
As for series, we need to know that absolute convergence really is convergence.
@@ -15959,11 +15959,11 @@ \section{Infinite products}
\\
& \leq
\sum_{n=k+1}^m
-\left(
+\Bigl(
\log \sabs{1+a_n}
+
\abs{ \Arg (1+a_n)}
-\right) .
+\Bigr) .
\end{split}
\end{equation}
As before,
@@ -15975,7 +15975,8 @@ \section{Infinite products}
\end{equation*}
and the series on the right converges via \propref{prop:infprodabsconvinfsum}.
Next, $\Arg(1+a_n)$ is between $\nicefrac{-\pi}{2}$ and
-$\nicefrac{\pi}{2}$, and $a_n$ is going to zero. So for all $n$ large
+$\nicefrac{\pi}{2}$, and $a_n$ goes to zero. A calculus exercise
+shows that for all $n$ large
enough, $\babs{\Arg(1+a_n)} \leq 2\sabs{a_n}$. As $\sum
\sabs{a_n}$ converges,
$\sum \babs{\Arg (1+a_n)}$ converges.
@@ -16109,9 +16110,8 @@ \section{Infinite products}
\begin{exbox}
\begin{exercise}
-Suppose $\sabs{1+a_n} \leq r < 1$ for all
-$n$,
-prove that $\prod_{n=1}^\infty (1+a_n)$ converges to $0$, but that the
+Suppose $\sabs{1+a_n} \leq r < 1$ for all $n$.
+Prove that $\prod_{n=1}^\infty (1+a_n)$ converges to $0$, but that the
convergence is not absolute.
\end{exercise}
@@ -17013,7 +17013,7 @@ \section{Runge's theorem}
\end{exercise}
\end{exbox}
-Let us now approximate simple poles of the form $\frac{1}{z-p}$ by rational
+We now approximate simple poles of the form $\frac{1}{z-p}$ by rational
functions with poles in a given set. This procedure is called
\emph{\myindex{pole pushing}} as we are going to \myquote{push} the poles along a
path to where we need them to be.
diff --git a/changes-draft.html b/changes-draft.html
index 8a42bc7..0b78945 100644
--- a/changes-draft.html
+++ b/changes-draft.html
@@ -83,4 +83,5 @@
Simplify Exercise 7.2.26 a little by assuming that \(U\) is
connected to having to think about the technicality of countably many
components which is not really important.
+In Figure 7.6 stop marking \(U_-,\) which we never defined.
Clarify the proof of Rado's theorem.
diff --git a/figures/schwarzreflection.fig b/figures/schwarzreflection.fig
index ee0277f..faa3234 100644
--- a/figures/schwarzreflection.fig
+++ b/figures/schwarzreflection.fig
@@ -1,4 +1,4 @@
-#FIG 3.2 Produced by xfig version 3.2.7a
+#FIG 3.2 Produced by xfig version 3.2.8b
Landscape
Center
Inches
@@ -9,8 +9,8 @@ Single
1200 2
0 32 #b9b9b9
0 33 #e6e6e6
-1 3 0 1 0 0 50 -1 20 0.000 1 0.0000 900 375 18 18 900 375 918 375
1 3 0 1 0 0 50 -1 20 0.000 1 0.0000 900 975 18 18 900 975 918 975
+1 3 0 1 0 0 50 -1 20 0.000 1 0.0000 900 375 18 18 900 375 918 375
2 1 0 1 0 7 50 -1 -1 0.000 0 0 -1 0 0 2
225 675 2925 675
2 1 0 3 0 7 50 -1 -1 0.000 0 0 -1 0 0 2
@@ -23,7 +23,6 @@ Single
0.000 -0.500 -0.500 -0.500 -0.500 0.000
4 1 0 50 -1 0 12 0.0000 2 150 330 2250 876 $L$\001
4 1 0 50 -1 0 12 0.0000 2 150 405 2775 876 $\\R$\001
-4 1 0 50 -1 0 12 0.0000 2 150 540 1500 300 $U^+$\001
-4 1 0 50 -1 0 12 0.0000 2 150 495 1500 1200 $U^-$\001
+4 1 0 50 -1 0 12 0.0000 2 165 555 1500 300 $U_+$\001
4 0 0 50 -1 0 12 0.0000 2 150 300 975 450 $z$\001
4 0 0 50 -1 0 12 0.0000 2 165 795 975 1050 $\\bar{z}$\001
diff --git a/figures/schwarzreflection.pdf b/figures/schwarzreflection.pdf
index 02adaac..f08f89f 100644
Binary files a/figures/schwarzreflection.pdf and b/figures/schwarzreflection.pdf differ
diff --git a/figures/schwarzreflection.pdf_t b/figures/schwarzreflection.pdf_t
index 911ffbf..83ab275 100644
--- a/figures/schwarzreflection.pdf_t
+++ b/figures/schwarzreflection.pdf_t
@@ -14,9 +14,7 @@
}}}}
\put(2776,-37){\makebox(0,0)[b]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$\R$}%
}}}}
-\put(1501,539){\makebox(0,0)[b]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$U^+$}%
-}}}}
-\put(1501,-361){\makebox(0,0)[b]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$U^-$}%
+\put(1501,539){\makebox(0,0)[b]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$U_+$}%
}}}}
\put(976,389){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\familydefault}{\mddefault}{\updefault}{\color[rgb]{0,0,0}$z$}%
}}}}