[OS X TeX] Grammar checker

Roberto Avanzi roberto.avanzi at gmail.com
Tue Dec 2 11:51:23 EST 2008


On Dec 2, 2008, at 2:39 PM, David B. Thompson, Ph.D., P.E., D.WRE, CFM  
wrote:

>
> On Dec 1, 2008, at 23:32 , Roberto Avanzi wrote:
>
>> You make an example, that of "which" and "that", which is one of  
>> the most wicked aspects of English language for foreigners, with  
>> perhaps the exception of the correct usage of some phrasal verbs  
>> and some fine aspects of punctuation (reading "Eats, shoots, and  
>> leaves" may have helped me, though) ... I am pretty sure I got it  
>> wrong in this message as well!
>
>
> Actually, I think your English is fine. Your sentence is a bit of a  
> run-on, but not too much of one. Your reference to "Eats, Shoots,  
> and Leaves" is an excellent suggestion for some very nasty parts of  
> English usage. The reference to Strunk and White is also a good  
> idea. I keep a copy of the Chicago Manual of Style in my writing  
> space and a copy of the U.S. Government Style Guide for Authors as  
> well.

I have been lucky to have had a good english teacher at high school.   
This is a rare beast in Italy.  Then, my job requires me to write in  
English, and with about 30 peer-reviewed publications, one has to  
learn...

Grammar checking is a nearly hopeless task.  The "that and which"  
problem is paramount.

To return IT: it would probably almost meaningless to have an english  
grammar checker that must work with mathematical english.  How should  
it handle things like the following example?

The expansion is done as follows: \textit{Write}
\begin{align*}
  m &\equiv \textstyle  n_0+{\frac{m-n_0}{p}}
      ({-a_1} s- s^2) \enspace .\\
\intertext{\textit{Now put $\kappa_1:=\frac{m-n_0}{p}$ and let
   $n_1\equiv -a_1\kappa_1 \bmod p$, where we choose the remainder of
   smallest absolute value}}
%
m &\equiv \textstyle n_0+n_1 s
+ \frac{-a_1\kappa_1-n_1}{p}(-a_1 s^2- s^3)-\kappa_1 s^2 \enspace ,\\
\intertext{\textit{similarly, $\kappa_2:=\frac{-a_1\kappa_1-n_1}{p}$ and
   $n_2 \equiv -a_1\kappa_2-\kappa_1 \bmod p$} m &\equiv \textstyle
n_0+n_1 s+n_2 s^2+{ \frac{-a_1\kappa_2-\kappa_1-n_2}{p}}}
(-a_1 s^3- s^4)-\kappa_2 s^3\\
\intertext{\textit{and we continue putting
   $\kappa_{i+1}:=\frac{-a_1\kappa_i-\kappa_{i-1}-n_i}{p}$, $n_{i+1}
   \equiv -a_1\kappa_{i+1}-\kappa_i \bmod p$ for
$2\leq i <5$,
%
until, finally, we write}} m &\equiv \textstyle \sum_{i=0}^{5} n_i s^i +
    \kappa_6 (-a_1 s^6- s^7)-\kappa_5 s^6
  \bmod \ell \enspace .
\end{align*}


This *IS* from a paper of mine.

Oh,the humanity.

  Roberto





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