\htmllevel
.
Hyperlatex understands the $
sign delimiting math mode as well as
\(
and \)
. Subscripts and superscripts produced using _
and
^
are understood. If the \htmllevel
is html2+
or
html3
, then HTML tags for subscripts and superscripts are created.
Otherwise, Hyperlatex creates a textual approximation of the form
a^2 = x^{2n}.
Hyperlatex now implements many common math mode commands, so simple
formulas in your text should be converted to some textual
representation. If you are not satisfied with that representation, you
can use the new \math
command:
In LaTeX, this command typesets the LaTeX-version, which is read in math mode (with all special characters enabled, if you have disabled some using\math[
HTML -version]{LaTeX-version}
\NotSpecial
).
Hyperlatex typesets the optional argument if it is present, or
otherwise the LaTeX-version.
If, for instance, you want to typeset the ith element
(the \math{i}th element
) of an array as ai in LaTeX,
but as a[i]
in HTML, you can use
\math[\code{a[i]}]{a_{i}}It takes a bit of care to find the best representation for your formula. This is an example of where any mechanical LaTeX-to-HTML converter must fail--I hope that Hyperlatex's
\math
command will
help you produce a good-looking and functional representation.
You could create a bitmap for a complicated expression, but you should
be aware that bitmaps eat transmission time, and they only look good
when the resolution of the browser is nearly the same as the
resolution at which the bitmap has been created, which is not a
realistic assumption. In many situations, there are easier solutions:
If xi is the ith element of an array, then I would rather write
it as x[i]
in HTML. If it's a variable in a program, I'd
probably write xi
. In another context, I might want to write
x_i. To write Pythagoras's theorem, I might simply use
a^2 + b^2 = c^2
, or maybe a*a + b*b = c*c. To express
"For any eps> 0 there is a delta> 0 such that for |x
- x0| < delta we have |f(x) - f(x0)| < eps" in HTML, I
would write "For any eps > 0 there is a
delta > 0 such that for
|x-x0| <
delta we have
|f(x)-f(x0)| <
eps."
More on math later.