From d7dd23fb05aec7aa1fc0561f931c9783e23ef24d Mon Sep 17 00:00:00 2001 From: "Jiri (George) Lebl" Date: Tue, 18 Jul 2023 23:58:09 -0500 Subject: [PATCH] Fix erratum, minor fix --- ca.tex | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/ca.tex b/ca.tex index 2ad26bf..6c7e79f 100644 --- a/ca.tex +++ b/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)