# [texhax] Math notation

Michael Doob mdoob at ccu.umanitoba.ca
Wed May 10 15:30:44 CEST 2006

On Tuesday 09 May 2006 9:46 pm, E. Krishnan wrote:

unnatural.
>
> Sure it does, since we've been seeing the whole "dx" in italics for all
> these years. But from a strictly notational point of view, it is
> equivalent to the product of the variables "d" and "x". So, a striaght "d"
> and a slanted "x"  seems to be a more consistent notation, emphasizing the
> fact that "d" is an operator and "x" is a variable. Remains to be seen if
> this finds acceptance among mathematicians. In the history of mathematical
> symbols, we've quite a few instances of innovations losing out.

However the "d" is not really an operator in the usual sense of the word
(I'm really trying to avoid the word infinitesimal here). Clearer cases are
the use of e and i, which are usually typeset in mathematical journals in
(math) italics even if they are being used as constants. According to ISO,
this is a no-no. In any case, I guess I support their case with the following
proof that $\pi=0$. It's a true typographical error :-).

\documentclass{article}
\begin{document}
\begin{center}
\textbf{A typographic proof that $\pi=0$}
\end{center}
We show that $\pi=0$ by considering
$$P=\left(\prod_{i=1}^n e^{i\pi}\right)^2$$
On the one hand we use the fact that $e^{i\pi}=-1$ to see that
$$P=(-1)^{2n}=1$$
On the other hand, we use the fact that $1+2+\cdots+n=\frac{n(n+1)}2$
to see that
$$\left(\prod_{i=1}^n e^{i\pi}\right)^2 =\left(e^{\left(\sum_{i=1}^ni\right)\pi}\right)^2 =e^{n(n+1)\pi}$$
Taking the log of the two values of $P$, and remembering that $n>0$,
we conlude that $n(n+1)\pi=0$ and so $\pi=0$.
\end{document}

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