WordPress has the ability to insert LaTeX math displays (e.g. ) 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 < and > 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 “. (Note that one can also mouse over a compiled LaTeX image to recover the original LaTeX source.)

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[...] 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 |

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

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

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

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

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

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

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

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

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

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

for test,

$

\int f \prime (x) dx

$

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

sorry for test

$ \int f \prime (x) dx $

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

from helpless,

$ a^n – ln(x) = 0$

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

helpless

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

“”.

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

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 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 , while the inverse is , where is the Fenchel dual to .”

To be more formal, given a nice convex function , we can interpret it as a potential function that a particle rolls around in.

Let denote the inner product. Then the Fenchel dual to F is defined as

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

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

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

(*)

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

with respect to p, we find

The fact 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 |

$\sum_{t=1}^{+\infty} \frac{1}{n^2}$

Comment by Anonymous — August 7, 2011 @ 2:25 pm |

Comment by Anonymous — August 7, 2011 @ 2:28 pm |

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Pingback by using LaTeX in Wordpress « Golbing — August 16, 2011 @ 4:54 pm |

$a^n+b^n=c^n$

Comment by Anonymous — September 16, 2011 @ 2:06 pm |

Comment by Anonymous — September 16, 2011 @ 2:07 pm |

$\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}$

Comment by Anonymous — September 30, 2011 @ 1:18 am |

Comment by Anonymous — September 30, 2011 @ 1:18 am |

\[\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} > \sqrt {{b^2} - 4ac} \]

Comment by Anonymous — September 30, 2011 @ 1:19 am |

Maxwell’s equations:

$

\vec{\nabla} \cdot \vec{\mathbf{E}} = \frac{1}{\epsilon_0}\,\rho \\

\vec{\nabla} \times \vec{\mathbf{E}} = -\frac{\partial \vec{\mathbf{B}}{\partial t} \\

\vec{\nabla} \cdot \vec{\mathbf{B}} = 0 \\

\vec{\nabla} \times \vec{\mathbf{B}} = \mu_0\vec{mathbf{J}} + \epsilon_0\frac{\partial \vec{\mathbf{E}}{\partial t} \\

$

$latex[

\vec{\nabla} \cdot \vec{\mathbf{E}} = \frac{1}{\epsilon_0}\,\rho \\

\vec{\nabla} \times \vec{\mathbf{E}} = -\frac{\partial \vec{\mathbf{B}}{\partial t} \\

\vec{\nabla} \cdot \vec{\mathbf{B}} = 0 \\

\vec{\nabla} \times \vec{\mathbf{B}} = \mu_0\vec{mathbf{J}} + \epsilon_0\frac{\partial \vec{\mathbf{E}}{\partial t} \\

]$

Comment by Mike — September 30, 2011 @ 8:19 pm |

OK, I guess that newlines are to be added with HTML tags (<br>)

Maxwell’s equations:

$latex[\vec{\nabla} \cdot \vec{\mathbf{E}} = \frac{1}{\epsilon_0}\,\rho ]$ <br>

$latex[\vec{\nabla} \times \vec{\mathbf{E}} = -\frac{\partial \vec{\mathbf{B}}{\partial t} ]$ <br>

$latex[\vec{\nabla} \cdot \vec{\mathbf{B}} = 0 ]$ <br>

$latex[\vec{\nabla} \times \vec{\mathbf{B}} = \mu_0\vec{mathbf{J}} + \epsilon_0\frac{\partial \vec{\mathbf{E}}{\partial t} ]$ <br>

$latex[\vec{\nabla} \cdot \vec{\mathbf{E}} = \frac{1}{\epsilon_0}\,\rho ]$

$latex[\vec{\nabla} \times \vec{\mathbf{E}} = -\frac{\partial \vec{\mathbf{B}}{\partial t} ]$

$latex[\vec{\nabla} \cdot \vec{\mathbf{B}} = 0 ]$

$latex[\vec{\nabla} \times \vec{\mathbf{B}} = \mu_0\vec{mathbf{J}} + \epsilon_0\frac{\partial \vec{\mathbf{E}}{\partial t} ]$

Comment by Mike — September 30, 2011 @ 8:25 pm |

Another try

Maxwell’s equations:

<br>

<br>

<br>

<br>

Comment by Mike — September 30, 2011 @ 8:32 pm |

Me again. Could not handle the 2nd and 4th equations. Comparing those with the equations that worked, the only differences are with \times and \partial. Let’s test:

$\vec{A} \times \vec{B} = \vec{C}$ and $\frac{\partial \phi}{\partial t} = \nabla \cdot \vec{A}$

and $\frac{\partial \phi}{\partial t} = \nabla \cdot \vec{A}$

Comment by Mike — September 30, 2011 @ 8:40 pm |

oops…

$\frac{\partial \phi}{\partial t} = \nabla \cdot \vec{A}$

Comment by Mike — September 30, 2011 @ 8:43 pm |

$f(x)^2$

Comment by Anonymous — October 17, 2011 @ 2:13 pm |

Comment by Anonymous — October 17, 2011 @ 2:14 pm |

the line

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

the line

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

Comment by Anonymous — October 17, 2011 @ 2:18 pm |

The task is to find a vector which is not parallel to , since then is orthogonal to .

Clearly, any parameter-family which is not contained in the line will contain such a vector (this is just rephrasing the condition of being not parallel). Therefore, if we manage to find a one-parameter family which is not contained in any line through the origin then this construction will work for

arbitraryvectors .The simplest such construction is to simply fix two of the three components of your vector with at least one of them being non-zero (i.e., to choose as your one-parameter family a one-dimensional affine space which does not contain the origin), e.g.,

.

Then given a specific vector we only need to choose in such a way that the last two components of are not parallel to those of . Suppose that we had failed to do so and the two -vectors are in fact parallel. Then

.

It follows that we can simply choose

to ensure that the two vectors and are non-parallel (one quickly verifies that this also works for as long as one defines ).

Comment by Anonymous — October 17, 2011 @ 2:41 pm |

Heh, fail. value , is

The constraints are that the probabilities must sum to one,

and that the expected value is 10,

Using Lagrange multipliers, we solve for the probability distribution extremizing

The variation of this is

which vanishes for

We can solve for the constants by satisfying the constraints, which demand and . And so we have the probability distribution consistent with our knowledge,

which we can use to answer questions. What’s the probability of pulling a jeweled coin valued at least $100?

Comment by Anonymous — March 8, 2012 @ 1:53 am |

functional of the probability of pulling a jewel of value , is

The constraints are that the probabilities must sum to one,

and that the expected value is 10,

Using Lagrange multipliers, we solve for the probability distribution extremizing

The variation of this is

which vanishes for

We can solve for the constants by satisfying the constraints, which demand and . And so we have the precise probability distribution consistent with our knowledge:

What’s the probability of pulling a jeweled coin valued at least $100?

Comment by Anonymous — March 8, 2012 @ 1:57 am |

parallel and perp does not work

Comment by sporepigfish — March 30, 2012 @ 3:11 pm |

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Pingback by Polymath7 discussion thread « The polymath blog — June 9, 2012 @ 5:51 am |

The integral of $x$ can be written as $\int x \ dx$Comment by Anonymous — June 9, 2012 @ 6:35 pm |

\int x\ dxComment by Anonymous — June 9, 2012 @ 6:36 pm |

Comment by Anonymous — June 9, 2012 @ 6:37 pm |

G_nComment by paultupper — June 10, 2012 @ 3:51 pm |

$G_n$

Comment by paultupper — June 10, 2012 @ 3:52 pm |

$G_n$Comment by paultupper — June 10, 2012 @ 3:53 pm |

Comment by paultupper — June 10, 2012 @ 3:56 pm |

$E=Mc^2$

Comment by Anonymous — June 11, 2012 @ 2:40 am |

E=mc^2Comment by Anonymous — June 11, 2012 @ 2:41 am |

Comment by Anonymous — June 11, 2012 @ 2:46 am |

Theorem.In a right triangle with sides and , where is the hypothenuse, holds.Comment by Anonymous — June 11, 2012 @ 4:56 pm |

Comment by Anonymous — June 14, 2012 @ 3:17 pm |

Test

(1)

$ latex R=f(B) $

Comment by Anonymous — June 14, 2012 @ 4:01 pm |

Comment by Anonymous — June 14, 2012 @ 4:03 pm |

Comment by Anonymous — June 14, 2012 @ 4:11 pm |

Comment by Anonymous — June 14, 2012 @ 4:12 pm |

Comment by Anonymous — June 18, 2012 @ 9:10 pm |

Comment by Anonymous — June 18, 2012 @ 9:10 pm |

test 1: denoted respectively by and

Comment by meditationatae — June 19, 2012 @ 4:42 am |

It’s clear from the presentation in that report that is a non-zero constant function, whatever the domain. They state that the nodal set of the eigenfunction divides the region or domain into at most sign domains.

Comment by meditationatae — June 19, 2012 @ 5:03 am |

For the unit interval, I believe that eigenfunction is given by the formula . I find that has roots at and .

With respect to your second thought, a proof by contradiction is indeed what I have in mind. The laplacian of the composition of functions represented by seems to me to depend not only on , but also on . So an affine transformation map , where is everywhere constant, and allowing mappings from a domain to a possible different domain for illustration, transforms the laplacian in a simple way (noting that is then identically zero.) In case is not affine, I think the relation between the laplacian of composed with and the laplacian of isn’t easy to understand on an intuitive level …

Comment by meditationatae — June 19, 2012 @ 5:55 am |

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Comment by Jasmine Liveonea — August 17, 2012 @ 11:45 pm |

Comment by John Baez — August 19, 2012 @ 6:22 am |

Comment by Anonymous — September 15, 2012 @ 12:59 pm |

Comment by Anonymous — October 1, 2012 @ 5:13 am |

Comment by Pytha — October 26, 2012 @ 7:55 pm |

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Pingback by Problem(s) of the Week « Reddit Math — January 9, 2013 @ 12:50 am |

Test:

$\frac{a}{b}$

Comment by Anonymous — January 9, 2013 @ 6:06 am |

Comment by Anonymous — January 9, 2013 @ 7:32 am |

oops.

Comment by Anonymous — January 9, 2013 @ 7:35 am |

Comment by Anonymous — January 9, 2013 @ 4:50 pm |

Comment by Anonymous — January 11, 2013 @ 4:17 am |

Comment by Anonymous — January 12, 2013 @ 3:55 pm |

Comment by marnixklooster — March 28, 2013 @ 9:02 pm |

$\vec{E}\cdot\hat n\;\mathrm{d}A$

Comment by Anonymous — April 4, 2013 @ 7:43 pm |

Comment by Anonymous — April 4, 2013 @ 7:44 pm |

Comment by dsf — April 26, 2013 @ 4:17 am |

Comment by Anonymous — April 29, 2013 @ 7:31 pm |

Comment by Bob Saggit — May 19, 2013 @ 2:40 am |

$\int_0^1 x^2 dx$ or ?

Comment by Anonymous — June 5, 2013 @ 2:05 am |

$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=\frac{8 \pi G}{c^4}T_{\mu\nu}$

Comment by Anonymous — June 6, 2013 @ 12:41 pm |

this is a test

$x^2 + x – 1 = 0$

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

Lemat Hensela pozwala na podobne sztuczki w sytuacji gdy rozważamy równanie wielomianowe postaci . Metoda jest taka, że startujemy od rozwiązania a następnie wykonujemy “podnoszenie” do rozwiązania modulo . Uzyskane w ten sposób rozwiązania, zbiegają w naturalny sposób do rozwiązania tego równania w zbiorze liczb p-adycznych (czyli w pewnym sensie modulo . Ta teoria działa bardzo dobrze dla wielomianów, ponieważ są to obiekty czysto algebraiczne. Natomiast w tych zadaniach jest trudniej, bo mamy do czynienia z równaniem wykładniczym względem x. Dlaczego mimo wszystko ta technika działa? Według mnie, dlatego, że szczęśliwie oraz też bardzo przypomina potęgę dziesiątki. Ogólnie niestety nie zawsze wygląda jak (dokładniej: ma nowe czynniki pierwsze w stosunku do ).

Comment by damianstraszak — July 6, 2013 @ 8:08 pm |

I have to test < x^2 >

Comment by monsieurcactus — July 24, 2013 @ 12:14 pm |

Comment by zhengtianyu — December 4, 2013 @ 2:43 pm |

$ax^2+bx+c$

Comment by Anonymous — December 12, 2013 @ 2:34 am |

Comment by Anonymous — December 12, 2013 @ 2:35 am |

$x^2+x=1$

Comment by Anonymous — January 26, 2014 @ 2:17 pm |

Comment by Anonymous — January 26, 2014 @ 2:20 pm |

Comment by Anonymous — February 12, 2014 @ 5:00 am |

#1.73: Using the result from #1.70 state 1 is recurrent iff . But which converges. So state 1 isn’t recurrent and then by Thm. 1.29 the chain can’t have a stationary distribution. What do you guys think?

Comment by Maksim Levental — February 25, 2014 @ 11:00 pm |

Inline test.

Comment by Anonymous — March 12, 2014 @ 5:39 am |

Display

test.

Comment by Anonymous — March 12, 2014 @ 5:42 am |

Testing.

Comment by Anonymous — March 14, 2014 @ 11:06 am |

Comment by Anon — March 14, 2014 @ 11:09 am |

In the Dirac eq. we will pass to the covariant derivative …

Comment by Anon — March 14, 2014 @ 11:10 am |

$LATEX_{1} T_{E}ST $

Comment by Anonymous — April 25, 2014 @ 12:37 am |

TEST

$a^n+b^n=c^n$

Comment by Anonymous — April 25, 2014 @ 12:39 am |

2. Proof

We observe (), () and () are (commutative) groups, where + is the normal complex number addition, and and are respectively the term-wise (Hadamard) product and the inner product of infinite dimension vectors of the form where for complex variable .

Let be defined as , then is homomorphism (where is normal complex number multiplication):

Similarly, let be defined as , then is homomorphism:

and note:

We observe that , for integer set , is kernel for , therefore is isomorphism. Furthermore, since is on principal value, if we define , then .

Suppose and are two roots of , then:

(Q.E.D.)

Comment by tcne — May 2, 2014 @ 1:35 pm |

Summary of the Proof.

RH: all nontrivial roots are on real critical line x = 1/2.

Since we know all nontrivial roots are symmetric about the real line x = 1/2, the nontrivial roots are of the form . Therefore, RH is nothing but , which is equivalent to the statement: .

RH, in this interpretation, is to prove . It is no more fancier than that.

We know that both and are roots of , so .

If we can show implies , we will have RH settled.

To show the above, we only need to demonstrate that the equality in the range of is preserved to its domain which is nothing but (C, +), the group of complex addition.

We observe that equality preservation happens in group isomorphism. Therefore, our proof boils down to showing the isomorphism.

The follow-up post is the full proof.

Comment by tcne — May 2, 2014 @ 2:29 pm |

Comment by ME — June 8, 2014 @ 10:30 pm |

solve for ?

To solve this use,

multiply from right of equation both side.

We have:

we use the

So, we have

then multiply frm the left both side of equation above:

\\

changing j to i and i to j, we get:

Comment by Anonymous — June 20, 2014 @ 2:15 pm |

$a^n+b^n=c^n$

Comment by Aidia — July 5, 2014 @ 7:10 pm |

$ /int f\prime(x) DX $

Comment by wrfggg — August 4, 2014 @ 6:29 pm |

$x^2 + y^2$

Comment by jlyates — August 12, 2014 @ 6:08 pm |

$2+2=4$

Comment by Anonymous — August 26, 2014 @ 12:08 pm |

$e^{i\pi}_{h}-1=0$

Comment by Karl — September 9, 2014 @ 10:52 pm |

$x^4$

Comment by Anonymous — October 6, 2014 @ 7:13 am |

To achieve this dream of stress-free karaoke, you need

to follow a couple steps. One key for the event is to get each player to offer

a number of pledges. This kid’s karaoke machine

is inspired by popular pop star Hannah Montana.

Comment by basi karaoke gratis — October 20, 2014 @ 2:36 pm |