[Blindmath] LaTeX vs Nemeth Question

[Jared Wright]

Nemeth would be preferable, but "a timely manner' means instantaneously, 
because that's how fast I can get LaTeX documents. I don't think the 
answering is so much in reaidng raw LaTeX as much as having a lot of 
tools to convert the LaTeX material to whatever format is most easily 
consumed by the end user. I understand there are some good LaTeX to 
Nemeth translators now. I would love to see, but have no Braille display 
to even work with at present.

JW

On 11/6/2009 4:43 PM, Susan Jolly wrote:
> I'm curious whether any of you would actually prefer LaTeX to Nemeth 
> if you
> could get quality Nemeth in a timely manner. I've chosen an example of
> something a bit harder than algebra to contrast these two 
> representations.
>
> First is a LaTeX example from p. 263 of Sewell's "Weaving a Program."
>
> $$\eqalign{\sin[(2n+1)\theta&=
> \Im(e^{(2n+1)i\theta})\cr
> &=\Im\{[\cos(\theta)
> + i \sin(\theta)]^{2n+1}\}\cr
> &=\sum_{k=0}^n(-1)^k
> {{2n+1}\choose{2k+1}}\sin^{2k+1}
> (\theta)\cos^{2(n-k)}(\theta)\cr
> &=\Bigl[\sum_{k=0}^n (-1)^k
> {{2n+1}\choose{2k+1}}\cot^{2(n-k)}(\theta)
> \Bigr]\Bigl[\sin^{2n+1}(\theta)\Bigr]\)\cr$$
>
> Here's the same expression in Nemeth which I entered directly. Please 
> let me
> know if I've made any errors.
>
> sin (2n+1).t .k im e^(2n+1)i.t
> .k im (cos .t + isin .t)^2n+1
> .k ".,s<k .k #0%n]-#1^k(2n+1%2k+1)
> sin^2k+1 .t cos^2(n-k) .t
> .k @(".,s<n .k #0%n]-#1^k(2n+1%2k+1)
> cot^2(n-k) .t@) sin^2n+1 .t
>
>
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