It supports both MathJax and TeX the World, which is what /r/math uses. However, that is no longer being developed, so for sites other than /r/math I think people should stick to MathJax.
The great thing about the bookmarklet approach is that it also works great on mobile devices. The previous solution for adding math to sites where the site owner won't support it was to use a greasemonkey plug-in or something similar, which generally left out mobile.
Here's some math if anyone wants something to try out with the bookmarklet.
The solutions of \(a x^2+ b x + c = 0\) are: $$x = {{-b\pm\sqrt{b^2-4ac}}\over{2a}}$$
Another example. Consider the continued fraction expansion of an arbitrary real number \(\alpha\):$$\alpha=a_0+{1\over{a_1+{1\over{a_2+{1\over{a_3+...}}}}}}$$.
Let \(\left\{p_n\over{q_n}\right\}\) be the sequence of convergents for this continued fraction.
For almost all real \(\alpha\), the limit as \(n\to\infty\) of \(q_n^{1/n}\)exists and $$\lim_{n\to\infty}{q_n^{1/n}}=e^{\pi^2\over{12 \log 2}}$$
Here was the Reddit discussion: http://www.reddit.com/r/math/comments/o6a2u/math_bookmarklet...
It supports both MathJax and TeX the World, which is what /r/math uses. However, that is no longer being developed, so for sites other than /r/math I think people should stick to MathJax.
The great thing about the bookmarklet approach is that it also works great on mobile devices. The previous solution for adding math to sites where the site owner won't support it was to use a greasemonkey plug-in or something similar, which generally left out mobile.
Here's some math if anyone wants something to try out with the bookmarklet.
The solutions of \(a x^2+ b x + c = 0\) are: $$x = {{-b\pm\sqrt{b^2-4ac}}\over{2a}}$$
Another example. Consider the continued fraction expansion of an arbitrary real number \(\alpha\):$$\alpha=a_0+{1\over{a_1+{1\over{a_2+{1\over{a_3+...}}}}}}$$.
Let \(\left\{p_n\over{q_n}\right\}\) be the sequence of convergents for this continued fraction.
For almost all real \(\alpha\), the limit as \(n\to\infty\) of \(q_n^{1/n}\)exists and $$\lim_{n\to\infty}{q_n^{1/n}}=e^{\pi^2\over{12 \log 2}}$$
For the curious, more information on that interesting limit can be found here: http://mathworld.wolfram.com/Khinchin-LevyConstant.html