Fix erratum, minor fix

ca.tex

@@ -16278,7 +16278,7 @@ \subsection{Factorization of sine}
 The $\pi$ out front can be guessed by thinking what would we get if we
 differentiate $\sin(\pi z)$ and evaluate at $0$.
 Differentiating the product (using product rule) would give you $\pi \cdot 1
-+ 0=1$, as any time the derivative falls on some factor other that $z$, when
++ 0=\pi$, as any time the derivative falls on some factor other that $z$, when
 you evaluate at $0$, you get $0$.  We can do this formal computation on the
 finite products as the product converges uniformly on compact subsets and
 therefore so does the derivative.  See also \exerciseref{exercise:proddiff}.
@@ -16293,11 +16293,11 @@ \subsection{Factorization of sine}
 \pi z \prod_{n=1}^\infty 
 \left(1-\frac{z^2}{n^2}\right) .
 \end{equation*}
-That's a rather nice factorization.  Of course, we still do not know if $f(z)$ is
+That's a rather nice factorization.  We still do not know if $f(z)$ is
 $\sin(\pi z)$.
 All we know is that the two have the same zeros,
 and the derivative at $0$ is $\pi$ as it should be.
-As $f$ captures the zeros of $\sin(\pi z)$, we write
+Because $f$ captures the zeros of $\sin(\pi z)$, we write
 (as in the factorization theorem)
 \begin{equation*}
 \sin(\pi z)