The polymath blog

How to use LaTeX in comments

WordPress has the ability to insert LaTeX math displays (e.g. \int_{-\infty}^\infty e^{-\pi x^2}\ dx = 1) into both posts and comments. The format for this is “$latex [Your LaTeX code]$”. See this announcement for details.

There used to be a number of quirks with the WordPress LaTeX plugin, but they have now largely been fixed. If you find any problems, please report them at this page.

WordPress also supports a certain amount of HTML. As a consequence, be careful with using the < and > signs in a comment, they may be misinterpreted as HTML tags! You can use &lt; and &gt; instead. (Inside of a LaTeX environment, you can use \lt and \gt.)

In case a comment really gets mangled up by formatting errors, contact one of the moderators of the polymath project, so that he or she can manually correct it.

The comments to this post will serve as the LaTeX help forum and LaTeX sandbox for this blog.  If you want to test out some LaTeX code in the comments below, you may wish to first describe the code without the “latex” symbol in order to show other readers what you are doing.  For instance: “Here is a LaTeX test: $a^n+b^n=c^n$ becomes a^n+b^n=c^n“. (Note that one can also mouse over a compiled LaTeX image to recover the original LaTeX source.)

14 Comments »

  1. [...] to have an environment where I could discuss Math. As documented by WordPress and used by the polymath blog Ah! I believe it now works. Alas, I should figure out something worthwhile to do with this [...]

    Pingback by math and latex test « Josh's Trial Blog — August 26, 2009 @ 4:11 pm | Reply

  2. Let’s see if this works, I’m going to try entering $x^2 + x – 1 = 0$

    x^2 + x - 1 = 0

    Comment by Asdquefty — November 20, 2009 @ 7:12 pm | Reply

  3. Awesome, so it works. Does this come with WordPress by default?

    Comment by Asdquefty — November 20, 2009 @ 7:13 pm | Reply

  4. Simply testing some latex: $e^{i\pi}-1=0$ e^{i\pi}-1=0

    Comment by Corey Brady — February 7, 2010 @ 1:19 am | Reply

  5. i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H\left|\Psi(t)\right>
    \LaTeX

    Comment by jhgf — April 30, 2010 @ 11:46 am | Reply

  6. \int x\,dx

    Comment by meme — June 9, 2010 @ 4:27 pm | Reply

  7. \displaystyle \int x\,dx

    Comment by meme — June 9, 2010 @ 4:28 pm | Reply

  8. 2^2

    Comment by Anonymous — July 8, 2010 @ 4:55 pm | Reply

  9. for test,

    $
    \int f \prime (x) dx
    $

    Comment by Anonymous — July 10, 2010 @ 6:35 pm | Reply

  10. sorry for test
    $ \int f \prime (x) dx $

    Comment by Anonymous — July 10, 2010 @ 6:35 pm | Reply

  11. from helpless,
    $ a^n – ln(x) = 0$

    Comment by Anonymous — July 10, 2010 @ 6:36 pm | Reply

  12. helpless
    \int f \prime (x)dx

    Comment by Anonymous — July 10, 2010 @ 6:37 pm | Reply

  13. “[\sum_1^n k^3]”.

    Comment by wb — August 14, 2010 @ 7:56 am | Reply

  14. Hi Terry,

    Another beautiful example of duality comes from optimization theory, in the form of the Fenchel dual to a function. The Fenchel dual has the following physical interpretation, which nicely illustrates your theme of “dual intrinsic/extrinsic” descriptions of an object:

    “A particle in a convex potential well F(z) can be pushed to any desired equilibrium point x by applying an appropriate force p. There is a bijective map relating x and p: the forward map is \nabla F, while the inverse is \nabla F^*, where F^* is the Fenchel dual to F.”

    To be more formal, given a nice convex function F : \mathbb{R} \rightarrow \mathbb{R}, we can interpret it as a potential function that a particle rolls around in.

    Let (a,b) denote the inner product. Then the Fenchel dual to F is defined as

    F^*(p) = \max_z (p,z) - F(z).

    Let x be the point where the max is achieved. The point x can be interpreted as the equilibrium position of a particle in potential F when this particle is subjected to a constant “applied force” p. The function

    F_p(z) = F(z) - (p,z)

    can be interpreted as an effective potential induced by the applied force, and the Fenchel dual is implicitly finding x, the equilibrium point where this potential is minimized. With a little calc and algebra we find

    \nabla F(x) = p
    F^*(p) = -F_p(x).

    (The negative is taken simply because the duality is clearer if $F^*$ is convex rather than concave.) A little calculus shows that

    (*) \nabla F^*(p) = x

    completing the justification of the physical interpretation at the top of this post.

    The derivation of this fact is intriguing in itself; by applying the multivariate chain rule to differentiate

    F^*(p) = g(x,p) = (x,p) - F(x)

    with respect to p, we find

    \nabla F^*(p) = (p - \nabla F(x)) + x = 0 + x = x.

    The fact p - \nabla F(x) = 0 is known from above, and arises from the fact that the particle is at equilibrium; thus the only change in the effective potential arises from the change in the “applied potential” with respect to p alone, holding x constant.

    Another interesting fact: a nonrigorous, geometric derivation of the Fenchel dual is possible by

    1. draw a contour plot of some nice F.
    2. draw a vector from 0 to a given equilibrium point x.
    3. draw the applied force p as a vector pointing from point x to x + p.

    Note that the two vectors drawn are now two sides of a parallelogram; the Fenchel dual is a function for which one can follow the exact same steps (1)-(3) again, but one uses the other two sides of the parallelogram and reverses the roles of p and x. After drawing this diagram, the involutive property of Fenchel duality becomes immediately obvious: taking the dual just means “flipping” the parallelogram over! :)

    Comment by Anonymous — August 18, 2010 @ 1:54 pm | Reply


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