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.)

107 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

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

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

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

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

  17. [...] learned I could add in at the Polymath blog but the WordPress announcement can be found here. GA_googleAddAttr("AdOpt", "1"); [...]

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

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

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

  19. a^n+b^n=c^n

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

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

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

  21. \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} > \sqrt {{b^2} - 4ac}

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

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

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

  23. 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 | Reply

  24. 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 | Reply

  25. Another try

    Maxwell’s equations:

    [\vec{\nabla} \cdot \vec{\mathbf{E}} = \frac{1}{\epsilon_0}\,\rho ] <br>
    [\vec{\nabla} \times \vec{\mathbf{E}} = -\frac{\partial \vec{\mathbf{B}}{\partial t} ] <br>
    [\vec{\nabla} \cdot \vec{\mathbf{B}} = 0 ] <br>
    [\vec{\nabla} \times \vec{\mathbf{B}} = \mu_0\vec{mathbf{J}} + \epsilon_0\frac{\partial \vec{\mathbf{E}}{\partial t} ] <br>

    \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 @ 8:32 pm | Reply

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

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

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

  27. oops…

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

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

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

  28. $f(x)^2$

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

  29. f(x)^2

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

  30. the line \mathbf{R} v

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

  31. the line \mathbb{R} v

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

  32. \mathbb{R} v

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

  33. The task is to find a vector \mathbf{w} which is not parallel to v, since then \mathbf{w} \times \mathbf{v} is orthogonal to \mathbf{v}.

    Clearly, any parameter-family which is not contained in the line \mathbb{R} \mathbf{v} 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 arbitrary vectors \mathbf{v}.

    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.,

    \mathbf{w} \in \begin{pmatrix}0\\w\\1\end{pmatrix}.

    Then given a specific vector \mathbf{v} we only need to choose w in such a way that the last two components of \mathbf{w} are not parallel to those of \mathbf{v}. Suppose that we had failed to do so and the two 2-vectors are in fact parallel. Then

    \mathrm{sign}(yz) = \mathrm{sign}(w).

    It follows that we can simply choose

    w = -\mathrm{sign}(yz)

    to ensure that the two vectors \mathbf{v} and \mathbf{w} are non-parallel (one quickly verifies that this also works for yz = 0 as long as one defines \mathrm{sign}(0) = 1).

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

  34. Heh, fail. value v, is

    I = \int_0^\infty dv p(v) ln p(v)

    The constraints are that the probabilities must sum to one,

    1 = \int_0^\infty dv p(v)

    and that the expected value is 10,

    10 = \int_0^\infty dv v p(v)

    Using Lagrange multipliers, we solve for the probability distribution extremizing

    \int_0^\infty dv \{ p ln p + \lambda p + \mu v p \}

    The variation of this is

    \int_0^\infty dv \delta p \{ ln p + 1 + \lambda + \mu v \}

    which vanishes for

    p(v) = p_0 e^{-mu v}

    We can solve for the constants by satisfying the constraints, which demand p_0 = \mu and \mu = 1/10. And so we have the probability distribution consistent with our knowledge,

    p(v) = \frac{1}{10} e^{- \frac{v}{10}}

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

    P(v \geq 100) = \int_100^\infty dv p(v) = \int_100^\infty dv \frac{1}{10} e^{- \frac{v}{10}} = e^{-10} \simeq 0.000045

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

  35. functional of the probability of pulling a jewel of value v, is

    I = \int_0^\infty dv \, p(v) \, ln \, p(v)

    The constraints are that the probabilities must sum to one,

    1 = \int_0^\infty dv \, p(v)

    and that the expected value is 10,

    10 = \int_0^\infty dv \, v \, p(v)

    Using Lagrange multipliers, we solve for the probability distribution extremizing

    \int_0^\infty dv \{ p \, ln \, p + \lambda p + \mu v p \}

    The variation of this is

    \int_0^\infty dv \, \delta p \{ ln \, p + 1 + \lambda + \mu v \}

    which vanishes for

    p(v) = p_0 e^{-\mu v}

    We can solve for the constants by satisfying the constraints, which demand p_0 = \mu and \mu = 1/10. And so we have the precise probability distribution consistent with our knowledge:

    p(v) = \frac{1}{10} e^{- \frac{v}{10}}

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

    P(v \geq 100) = \int_{100}^\infty dv p(v) = \int_{100}^\infty dv \frac{1}{10} e^{- \frac{v}{10}} = e^{-10} \simeq 0.000045

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

  36. parallel and perp does not work

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

  37. [...] One thing I see is that not everybody who has participated knows how to make latex formatting such as appear in their comments.  The instructions for that (as well as a “sandbox” to try out the code) are at this link. [...]

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

  38. The integral of $x$ can be written as $\int x \ dx$

    Comment by Anonymous — June 9, 2012 @ 6:35 pm | Reply

  39. \int x\ dx

    Comment by Anonymous — June 9, 2012 @ 6:36 pm | Reply

  40. \int x\ dx

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

  41. G_n

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

  42. $G_n$

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

  43. $G_n$

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

  44. G_n

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

  45. $E=Mc^2$

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

  46. E=mc^2

    Comment by Anonymous — June 11, 2012 @ 2:41 am | Reply

  47. E=mc^2

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

  48. Theorem. In a right triangle with sides a,b and c, where c is the hypothenuse, a^2+b^2 = c^2 holds.

    Comment by Anonymous — June 11, 2012 @ 4:56 pm | Reply

  49. \sum_{y\sim y_0}p_t(x,y)

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

  50. Test

    (1) f(x)= | x − 2\lfloor (x + 1)/2\rfloor |

    $ latex R=f(B) $

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

  51. f(x)= | x − 2&lfloor (x + 1)/2&rfloor |

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

  52. f(x)= | x − 2h( (x + 1)/2) |

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

  53. f(x)= \| x − 2h( (x + 1)/2) \|

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

  54. \int_0^1 dx

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

  55. S_{\varepsilon(t+1)}

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

  56. test 1: denoted respectively by \psi_{ 0} and \psi_{1}

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

  57. It’s clear from the presentation in that report that \psi_{ 0} is a non-zero constant function, whatever the domain. They state that the nodal set of the ith eigenfunction \psi_{ i-1} divides the region or domain into at most i sign domains.

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

  58. For the unit interval, I believe that ith eigenfunction \psi_{ i-1} is given by the formula \psi_{i-1} \left( x\right) = \cos \left(\left(i-1\right)\pi x\right). I find that \cos \left(2\pi x\right) has roots at { 1 \over 4} and { 3 \over 4} .

    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 x \mapsto u_{2}\left(\phi \left( x \right) \right) seems to me to depend not only on \phi \prime \left( x\right), but also on \phi \prime \prime \left( x\right). So an affine transformation map \psi , where \phi \prime is everywhere constant, and allowing mappings from a domain D to a possible different domain D \prime for illustration, transforms the laplacian in a simple way (noting that \phi \prime \prime is then identically zero.) In case \phi is not affine, I think the relation between the laplacian of u_{2} composed with \phi and the laplacian of u_{2} isn’t easy to understand on an intuitive level …

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

  59. Love your website! Thanks for taking the time to share with everyone.

    Comment by Collin Weeks — August 16, 2012 @ 8:00 am | Reply

  60. Great site, Posted this on FB for some friends to checkout.

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

  61. E = mc^2

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

  62. \left({{\sqrt{11}}\over{3^{{{3}\over{2}}}}}+{{19}\over{27}}\right)  ^{{{1}\over{3}}}+{{4}\over{9\,\left({{\sqrt{11}}\over{3^{{{3}\over{2  }}}}}+{{19}\over{27}}\right)^{{{1}\over{3}}}}}+{{1}\over{3}}

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

  63. x^{2^2}

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

  64. [\lambda]

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

  65. [...] a guide to using LaTeX in [...]

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

  66. Test:

    $\frac{a}{b}$

    \frac{a}{b}

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

  67. 1+\frac{1}{\phi} = \phi

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

  68. oops. 1 + \frac{1}{\varphi} = \varphi

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

  69. \frac{1}{x}

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

  70. e^{i\pi}+1=0

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

  71. \sum_{k=0}^{n}binom{n}{k}x^{n-k}h^{k}

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

  72. \langle \forall x :: 2 * x \lt 2 * x + 1 \rangle

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

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

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

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

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

  75. \frac{\int_a^b f(x)dx} {b-a}

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

  76. \cal O(\cal M)

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

  77. x

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

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

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

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

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

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

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

  80. Lemat Hensela pozwala na podobne sztuczki w sytuacji gdy rozważamy równanie wielomianowe postaci q(x)=0 \pmod{p^k}. Metoda jest taka, że startujemy od rozwiązania q(x)=0 \pmod{p} a następnie wykonujemy “podnoszenie” do rozwiązania modulo p^2,p^3,...,p^k,p^{k+1},.... Uzyskane w ten sposób rozwiązania, zbiegają w naturalny sposób do rozwiązania tego równania w zbiorze liczb p-adycznych (czyli w pewnym sensie modulo p^\infty. 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 \phi(2^k)=2^{k-1} oraz \phi(10^k) też bardzo przypomina potęgę dziesiątki. Ogólnie niestety nie zawsze \phi(n) wygląda jak n (dokładniej: ma nowe czynniki pierwsze w stosunku do n).

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

  81. I have to test < x^2 >

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

  82. x+y=0

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

  83. $ax^2+bx+c$

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

  84. ax^2+bx+c

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

  85. $x^2+x=1$

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

  86. x^2+x=1

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

  87. e^{i \pi}+1=0

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

  88. #1.73: Using the result from #1.70 state 1 is recurrent iff \sum_{y=1}^\infty \prod_{x=1}^{y-1} \frac{q_x}{p_x} = \infty . But \sum_{y=1}^\infty \prod_{x=1}^{y-1} \frac{q_x}{p_x} = 0 + \frac{1}{2} +\frac{1}{2}\frac{1}{3} + \frac{1}{2}\frac{1}{3}\frac{1}{4} + \cdots 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 | Reply

  89. Inline 1+1 test.

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

  90. Display

    1 \times 1

    test.

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

  91. Testing. \int2x

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

  92. -i\hbar\gamma^\mu\theta^\mu\psi+mc\psi

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

  93. In the Dirac eq. -i\hbar\gamma^\mu\theta^\mu\psi+mc\psi = 0 we will pass to the covariant derivative D_\mu

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

  94. $LATEX_{1} T_{E}ST $

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

  95. TEST
    $a^n+b^n=c^n$

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

  96. 2. Proof

    We observe (\mathbb{C}, +), (V \backslash \{\mathbf{0}\}, \circ) and (V \backslash \{\mathbf{0}\}, \bullet) are (commutative) groups, where + is the normal complex number addition, and \circ and \bullet are respectively the term-wise (Hadamard) product and the inner product of infinite dimension vectors of the form V(c) = (v_{1}(c), v_{2}(c), v_{3}(c), ...) where v_{n}(c) = n^{-c} for complex variable c.

    Let f: (\mathbb{C}, +) \longrightarrow (V\backslash\{\mathbf{0}\}, \circ) be defined as f(c) = V(c), then f is homomorphism (where \times is normal complex number multiplication):

    f(c_{_{1}}+c_{_{2}}) = V(c_{_{1}}+c_{_{2}})
    = (v_{_{1}}(c_{_{1}}+c_{_{2}}), v_{_{2}}(c_{_{1}}+c_{_{2}}), v_{_{3}}(c_{_{1}}+c_{_{2}}), ...)
    = (1^{-(c_{_{1}}+c_{_{2}})}, 2^{-(c_{_{1}}+c_{_{2}})}, 3^{-(c_{_{1}}+c_{_{2}})}, ...)
    = (1^{-c_{_{1}}}\times1^{-c_{_{2}}}, 2^{-c_{_{1}}}\times2^{-c_{_{2}}}, 3^{-c_{_{1}}}\times3^{-c_{_{2}}}, ...)
    = (v_{_{1}}(c_{_{1}})\times v_{_{1}}(c_{_{2}}), v_{_{2}}(c_{_{1}})\times v_{_{2}}(c_{_{2}}), v_{_{3}}(c_{_{1}})\times v_{_{3}}(c_{_{2}}), ...)
    = V(c_{_{1}})\circ V(c_{_{2}})
    = f(c_{_{1}})\circ f(c_{_{2}})

    Similarly, let g: (\mathbb{C}, +) \longrightarrow (V\backslash\{\mathbf{0}\}, \bullet) be defined as g(c) = g(c+0) = V(c+0), then g is homomorphism:

    g(c_{_{1}}+c_{_{2}}) = V(c_{_{1}}+c_{_{2}})
    = (v_{_{1}}(c_{_{1}}+c_{_{2}}) + v_{_{2}}(c_{_{1}}+c_{_{2}}) + v_{_{3}}(c_{_{1}}+c_{_{2}}) + ...)
    = (1^{-(c_{_{1}}+c_{_{2}})} + 2^{-(c_{_{1}}+c_{_{2}})} + 3^{-(c_{_{1}}+c_{_{2}})} + ...) = \sum_{n=1}^{\infty}\frac{1}{n^{c_{_{1}}+c_{_{2}}}}
    = (1^{-c_{_{1}}}\times1^{-c_{_{2}}} + 2^{-c_{_{1}}}\times2^{-c_{_{2}}} + 3^{-c_{_{1}}}\times3^{-c_{_{2}}} + ...)
    = (v_{_{1}}(c_{_{1}})\times v_{_{1}}(c_{_{2}}) + v_{_{2}}(c_{_{1}})\times v_{_{2}}(c_{_{2}}) + v_{_{3}}(c_{_{1}})\times v_{_{3}}(c_{_{2}}) + ...)
    = V(c_{_{1}})\bullet V(c_{_{2}})
    = g(c_{_{1}})\bullet g(c_{_{2}})
    and note:
    g(c+0) = V(c+0) = V(c) \bullet V(0)

    We observe that 2{\pi}i\mathbb{Z}, for integer set \mathbb{Z}, is kernel for f, therefore f': (\mathbb{C}/2{\pi}i\mathbb{Z}, +) \longrightarrow (V\backslash\{\mathbf{0}\}, \circ) is isomorphism. Furthermore, since \zeta is on principal value, if we define g' : (\mathbb{C}/2{\pi}i\mathbb{Z}, +) \longrightarrow (V\backslash\{\mathbf{0}\}, \bullet), then g' = \zeta.

    Suppose c_{_{1}} = ((\frac{1}{2} + \delta) \pm it) and c_{_{2}} = ((\frac{1}{2} - \delta) \pm it) are two roots of \zeta, then:

    \zeta(c_{_{1}}) = 0 = \zeta(c_{_{2}})
    \Longrightarrow\ g'(c_{_{1}}) = g'(c_{_{2}})
    \Longrightarrow\ g'((\frac{1}{2} + \delta) \pm it)) = g'((\frac{1}{2} - \delta) \pm it))
    \Longrightarrow\ g'((\frac{1}{2} \pm it) + \delta) = g'((\frac{1}{2} \pm it) - \delta)
    \Longrightarrow\ g'(\frac{1}{2} \pm it) \bullet g'(\delta)) = g'(\frac{1}{2} \pm it) \bullet g'(-\delta)
    \Longrightarrow\ g'(\delta) = g'(-\delta)
    \Longrightarrow\ V(\delta) = V(-\delta)
    \Longrightarrow\ f'(\delta) = f'(-\delta)
    \Longrightarrow\ \delta = -\delta
    \Longrightarrow\ \delta = 0

    (Q.E.D.)

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

  97. 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 (1/2 \pm \delta) \pm it. Therefore, RH is nothing but (1/2 \pm \delta) \pm it = 1/2 \pm it, which is equivalent to the statement: \delta = 0.

    RH, in this interpretation, is to prove (1/2 + \delta) \pm it = (1/2 - \delta) \pm it. It is no more fancier than that.

    We know that both (1/2 + \delta) \pm it and (1/2 + \delta) \pm it are roots of \zeta, so \zeta((1/2 + \delta) \pm it) = \zeta((1/2 - \delta) \pm it) = 0.

    If we can show \zeta((1/2 + \delta) \pm it) = \zeta((1/2 - \delta) \pm it) implies (1/2 + \delta) \pm it = (1/2 - \delta) \pm it, we will have RH settled.

    To show the above, we only need to demonstrate that the equality in the range of \zeta 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 | Reply

  98. z^n = y^n + x^n

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

  99. e_{i}^{'}=A_{i}^{j}e_{j}
    solve for \sigma^{i} ?

    To solve this use,

    e^{k}\cdot e_{i}= \delta^{k}_{i}\\e^{'k}\cdot \sigma^{'j}= \delta^{'j}_{i}

    multiply \sigma^{'j} from right of equation e_{i}^{'}=A_{i}^{j}e_{j} both side.
    We have:

    e_{i}^{'} \sigma^{'j}=A_{i}^{j}e_{j}\sigma^{'j}\\ \delta^{'j}_{i}=A_{i}^{j}e_{j}\sigma^{'j}
    we use the \delta^{'i}=\delta^{j}_{i}
    So, we have
    e_{i}\cdot \sigma^{j}=A^{j}_{i}e_{j}\cdot \sigma^{'j}
    then multiply e^{i} frm the left both side of equation above:
    e^{i}\cdot e_{i}\cdot \sigma^{j}=A^{j}_{i}e^{i}\cdot e_{j}\cdot \sigma^{'j}\\ \delta^{i}_{i}\sigma^{j}=A^{j}_{i}\delta^{i}_{j} \sigma^{'j}\\ \sigma^{j}=A^{j}_{i}\sigma^{'i}\\
    changing j to i and i to j, we get:

    \sigma^{i}=A^{i}_{j}\sigma^{'i}

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

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

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

  101. $ /int f\prime(x) DX $

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

  102. $x^2 + y^2$
    x^2 + y^2

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

  103. $2+2=4$

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

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

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

  105. $x^4$

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

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