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 “$latex [Your LaTeX code]$” (but without the brackets, of course). 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.)

12 Comments »

  1. the line \mathbf{R} v

    Comment by Anonymous — October 17, 2011 @ 2:15 pm | Reply

  2. this is a test
    $x^2 + x – 1 = 0$

    Comment by M Wall — June 20, 2013 @ 8:44 am | Reply

  3. This might be an obvious question, but to have a sunflower with three petals, is to find three sets A,B,C, so that A \cap B \subseteq C, and C \cap (A \triangle B)=\emptyset.

    What happens if instead of both properties, just one of them is forbidden? For example, if we forbid three sets A,B,C in our family, so that C \cap (A \triangle B)=\emptyset?

    Comment by ILan — November 8, 2015 @ 2:59 pm | Reply

  4. The \gt and \lt don’t seem to be working: I’m trying to compile $a \gt b$: a \gt b". But a>b$ works, see a>b

    Comment by Anonymous — March 7, 2016 @ 3:01 pm | Reply

  5. I will try to use Latex to express the process but there is no telling how it will turn out

    Z_t \: is \:IID \: N(0,1)\\\\     X_t=Z_t  \: if\: t \:is\: even\\\\      X_t=\frac{Z_{t-1}^2 -1}{\sqrt{2}}  \: if\: t \:is\: odd

    Comment by PJKar — May 24, 2016 @ 8:21 pm | Reply

  6. \displaystyle \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}

    Comment by Anonymous — December 11, 2016 @ 6:21 pm | Reply

  7. a_2^3

    Comment by junkmailer — February 10, 2017 @ 7:21 pm | Reply

  8. This is a test:
    $latex[x^2 + \phi]$

    Comment by Anonymous — March 7, 2017 @ 4:35 am | Reply

  9. I think I understand now. For the operator norm question, R has eigenvalues \frac{1}{\lambda_i - (a+bi) which are bounded above by 1/|b|.

    For the positive definite imaginary part, I am still a bit confused. The imaginary part of R does not even seem to be Hermitian. But I think I see how to get the conclusion that follows anyway: The eigenvalues above have positive imaginary part \frac{b}{(\lambda_i - a)^2 + b^2}. Using the fact that R has an orthogonal eigenbasis, you can now calculate that the X^* R X has a nonnegative imaginary part for any X.

    Is this what it means for the imaginary part of $\latex R$ to be positive definite?

    Comment by hhh — March 19, 2017 @ 6:11 pm | Reply

  10. I think I understand now. For the operator norm question, R has eigenvalues \frac{1}{\lambda_i - (a+bi)} which are bounded above by 1/|b|.

    For the positive definite imaginary part, I am still a bit confused. The imaginary part of R does not even seem to be Hermitian. But I think I see how to get the conclusion that follows anyway: The eigenvalues above have positive imaginary part \frac{b}{(\lambda_i - a)^2 + b^2}. Using the fact that R has an orthogonal eigenbasis, you can now calculate that the X^* R X has a nonnegative imaginary part for any X.

    Is this what it means for the imaginary part of $\latex R$ to be positive definite?

    Comment by hhh — March 19, 2017 @ 6:12 pm | Reply

  11. This proposition is true, though you’re quite right that I should have mentioned it. Here’s a proof sketch: If \mathcal{E} is immodest, then it generates what I called in the proof of Proposition 2 an ‘immodest S5’ Kripke frame—i.e., the relation R partitions \mathcal{W} into equivalence classes, where each equivalence class [w]_R = \langle \mathcal{E} = \mathcal{E}_w \rangle. Then, for each pair of worlds w, w' \in \mathcal{W}, either \mathcal{E}_w(\{ w' \}) = \mathcal{E}_{w'}(\{ w' \}) (if w' \in [w]_R) or \mathcal{E}_w(\{ w' \}) = 0 (if w' \notin [w]_R). So, for each w \in \mathcal{W},

    \langle \mathcal{E}(\{w\}) = \mathcal{E}_w(\{w\}) \rangle = \langle \mathcal{E} = \mathcal{E}_w \rangle

    And thus, if you treat \mathcal{E} as a local expert,

    Comment by jdmitrig — June 19, 2017 @ 10:04 pm | Reply

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