MikeMirzayanov's blog

Codeforces: MathJax in Blog Posts and Comments
By MikeMirzayanov, 6 years ago, translation, In English

Hello!

We supported the rendering of LaTeX-formulas with MathJax in new posts and comments. Now the formulas will be as beautiful as in problem statements. Drawback: they are displayed not immediately, but redrawn after the page is displayed. Old posts and comments are displayed in the old way (already a lot of old content, backward compatibility is very important). Note that if you edit old post, it will still be displayed in the old style.

Here are some example: $$$1 \le n \le 10^{12}$$$,  $$$c = \sqrt{a^2+b^2}$$$,  $$$i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H\left|\Psi(t)\right>$$$.

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6 years ago, # |
  Vote: I like it +9 Vote: I do not like it

Can you write something here to see how it looks like?

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    6 years ago, # ^ |
    ← Rev. 2   Vote: I like it +9 Vote: I do not like it
    $$$\sum_{i=1}^{n}i=\frac{n \cdot (n+1)}{2}$$$
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      6 years ago, # ^ |
      ← Rev. 2   Vote: I like it +5 Vote: I do not like it

      I understand. It need like this. yor need use double dollar sign to include your code

      $$$ \frac{a}{b} $$$
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      6 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thanks a lot.

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    6 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it
    $$$\forall \varepsilon\!>\!0 \ \exists k(\varepsilon)\!\in\!N: \forall n>k \Rightarrow |x_{n}-a|<\varepsilon$$$

    You have to use $ or $$ like in LateX

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    6 years ago, # ^ |
      Vote: I like it +45 Vote: I do not like it

    $a^2$
    $$$a^2$$$
    $$a^2$$

    $$$a^2$$$
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6 years ago, # |
  Vote: I like it +27 Vote: I do not like it
$$$G_{ij}=\sum_k F_{ik}F_{jk}$$$
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    6 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Sound matrix multiplications am I right? :D

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      6 years ago, # ^ |
        Vote: I like it +13 Vote: I do not like it

      Maybe Sound signal can use the function too? I use the Gamma Matrix to calculate the loss function with higher CNN-layer similarity in Picture Style Transfer.

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        6 years ago, # ^ |
          Vote: I like it +13 Vote: I do not like it

        Whoa, slow down senpai, I need a few more time to be familiar with ML models. :D

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6 years ago, # |
  Vote: I like it +244 Vote: I do not like it

Ah nice

The old style: previous style

The new one: $$$\lfloor \sqrt{x} \rfloor - \lfloor \sqrt{y} \rfloor \le \frac{x-y}{\sqrt{x}+\sqrt{y}} + 1 \le \frac{1}{\sqrt{x-y}+1}(x-y) + 1$$$

Looks so much better!

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6 years ago, # |
← Rev. 2   Vote: I like it -17 Vote: I do not like it

Good job

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6 years ago, # |
  Vote: I like it +23 Vote: I do not like it
$$$LOL$$$
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6 years ago, # |
← Rev. 2   Vote: I like it +130 Vote: I do not like it
$$$\text{C}_{\text{ode}}\text{F}_{\text{orces}}\text{ Round}_{547} \text{ Please}$$$
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    6 years ago, # ^ |
    ← Rev. 3   Vote: I like it +35 Vote: I do not like it

    It seems there is an error on displaying when I change the revision of comment to see. Check the revision 1 and move back to revision 2 of my comment then you will see the bug. MikeMirzayanov

    • It's fixed
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6 years ago, # |
  Vote: I like it +22 Vote: I do not like it

$$$\mathbb{CODEFORCES}$$$

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6 years ago, # |
  Vote: I like it +156 Vote: I do not like it
$$$\displaystyle \sum_{i=1}^{\infty} i = -\frac{1}{12}$$$
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6 years ago, # |
  Vote: I like it +55 Vote: I do not like it

What was the LaTeX renderer before? I always thought it was some kind of Russian bootleg MathJax.

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6 years ago, # |
  Vote: I like it +36 Vote: I do not like it
$$$\begin{matrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{matrix}$$$
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6 years ago, # |
← Rev. 2   Vote: I like it +13 Vote: I do not like it
$$$\int_G f(g) \, \text{d}g = \frac{1}{|W|} \int_G \int_T \mathopen| \Delta(t) \mathclose|^2 f(gtg^{-1}) \, \text{d}t \, \text{d}g $$$

Looks very pretty :)

$$$1+2=3<4$$$
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6 years ago, # |
  Vote: I like it +57 Vote: I do not like it

$$$ \huge \mathcal{O}( N ) $$$

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6 years ago, # |
  Vote: I like it +71 Vote: I do not like it

Testing

$$$\mathrm{f}[u]=b_u+\min\limits_{\substack{v\in anc_u\\\mathrm{dis}[u]-\mathrm{dis}[v]\le l_u}}(\mathrm{f}[v]-\mathrm{dis}[v]\times p_u)$$$

If $$$f(x)=\frac{1}{1-x^k}$$$, then $$$g(x)=(\ln f)(x)=\sum_{i=1}^{\infty}\frac{1}{i}x^{ik}$$$. Proof:

$$$\begin{aligned}g(x)&=\ln f(x)\\&=\int(\frac{\mathrm{d}}{\mathrm{d}x}\ln f)(x)\mathrm{d}x\\&=\int(\frac{f'(x)}{f(x)})\mathrm{d}x\\&=\int((1-x^k)f'(x))\mathrm{d}x\\&=\int((1-x^k)\sum_{i=1}^{\infty}k\cdot i\cdot x^{ki-1})\mathrm{d}x\\&=\int(\sum_{i=1}^{\infty}k\cdot i\cdot x^{ki-1}-\sum_{i=1}^{\infty}k\cdot i\cdot x^{ki-1}\cdot x^k)\mathrm{d}x\\&=\int(\sum_{i=1}^{\infty}k\cdot i\cdot x^{ki-1}-\sum_{i=1}^{\infty}k\cdot (i-1)\cdot x^{ki-1})\mathrm{d}x\\&=\int(\sum_{i=1}^{\infty}k\cdot x^{ki-1})\mathrm{d}x\\&=\sum_{i=1}^{\infty}\frac{1}{i}x^{ki}\end{aligned}$$$
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    6 years ago, # ^ |
    ← Rev. 2   Vote: I like it +73 Vote: I do not like it

    I think, that you forgot to try \left( instead ( and \right) instead ). This allows to auto detect correct height of parentheses under summation, like this:

    $$$\prod_{k=1}^{n-1} \left(\exp{\left(\dfrac{2 \cdot \pi \cdot i \cdot k}{n}\right)}-1\right) = \left(-1\right)^{n+1} \cdot n$$$
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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can explain me why backward compatibility is very important?

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6 years ago, # |
← Rev. 4   Vote: I like it 0 Vote: I do not like it

Deleted and screw uplaoding a photo

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6 years ago, # |
  Vote: I like it +8 Vote: I do not like it
$$$\text{summitwei} \in \text{geniosity}$$$

Proof left as an exercise to the reader.

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6 years ago, # |
  Vote: I like it +7 Vote: I do not like it

$$$a^2$$$

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6 years ago, # |
  Vote: I like it +227 Vote: I do not like it

Nice, Now I can draw beautiful owls in my posts. Thanks Mike!

$$$\underline{\widehat{\dbinom{\odot_\vee\odot}{{\raise-8pt"}\wr{\raise-8pt"}}}}$$$
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    6 years ago, # ^ |
    ← Rev. 2   Vote: I like it +36 Vote: I do not like it

    LOL

    $$$ \widetilde{ \dbinom{\odot_\vee\odot} {\underline{\ \smile\ }}} $$$
    $$$ \widetilde{ \dbinom{\odot_\vee\odot} {\underline{\ -\ }}} $$$
    $$$ \widetilde{ \dbinom{\odot_\vee\odot} {\underline{\ \frown\ }}} $$$
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6 years ago, # |
← Rev. 4   Vote: I like it +5 Vote: I do not like it
$$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$$
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6 years ago, # |
← Rev. 3   Vote: I like it +117 Vote: I do not like it

$$$\require{AMScd}$$$ \begin{CD} T @>h>> A\\ @. @VV n V\\ M @<<s< K\\ @VV iV @.\\ K @>>e> !\\ \end{CD}

$$$ \frac{\color{purple}{4}}{\color{blue}{\Pi}} = \color{green}{1} + \cfrac{\color{red}{1}}{\color{orange}{2} + \cfrac{\color{red}{9}}{\color{orange}{2} + \cfrac{\color{red}{25}}{\color{orange}{2} + \cfrac{\color{red}{49}}{\color{orange}{2} + \dots}}}} $$$
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6 years ago, # |
← Rev. 6   Vote: I like it +24 Vote: I do not like it
$$$ \displaystyle \sum_{i=1}^n \sum_{j=1}^m d_1(\gcd(i, j)) \\ \displaystyle\sum_{d=1}^n \sum_{i=1}^n \sum_{j=1}^m d_1(d)[\gcd(i,j)=d])\\ \displaystyle\sum_{d=1}^n d_1(d) \sum_{i=1}^n \sum_{j=1}^m [\gcd(i,j)=d])\\ \displaystyle\sum_{d=1}^n d_1(d) \sum_{k=1}^{\lfloor \frac nd \rfloor} \mu(k) \lfloor \frac n{kd} \rfloor \displaystyle\lfloor \frac m{kd} \rfloor \\ $$$

Let $$$T=kd$$$,

$$$ \displaystyle\sum_{T=1}^n \lfloor \frac nT \rfloor \lfloor \frac mT \rfloor \sum_{d|T}d_1(d) \mu(\frac Td) $$$
$$$ \displaystyle dp(i)=\min\limits_{i < j \leq n, dp(j)\leq \operatorname{sum}(j - 1) - \operatorname{sum}(i - 1)}(\operatorname{sum}(j - 1) - \operatorname{sum}(i - 1)) $$$

Why the \sum became \prod?....

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    6 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Maybe you can use..

    $$$\Sigma_{i=1}^n \Sigma_{j=1}^m d_1(\gcd(i, j))$$$

    \Sigma_{i=1}^n \Sigma_{j=1}^m d_1(\gcd(i, j))

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      6 years ago, # ^ |
      ← Rev. 2   Vote: I like it +6 Vote: I do not like it

      It seems that it was fixed now, also thank you...

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        6 years ago, # ^ |
          Vote: I like it +6 Vote: I do not like it

        I'm just crack a joke,if you use the \Sigma,it will be too small...

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6 years ago, # |
  Vote: I like it +40 Vote: I do not like it
$$$a^{n}\equiv a^{\varphi(m)+(n \bmod \varphi(m))}\pmod{m}$$$
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    6 years ago, # ^ |
      Vote: I like it +62 Vote: I do not like it

    It is not always true when $$$n<\varphi(m)$$$. For example, $$$6^{2}\not\equiv 6^{4+2}\pmod8$$$.

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      6 years ago, # ^ |
        Vote: I like it +45 Vote: I do not like it

      It's true for $$$n>\log_2 m$$$

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        6 years ago, # ^ |
          Vote: I like it +27 Vote: I do not like it

        More accurately, it's true for $$$n\ge\max_{i=1}^k\alpha_i$$$, where $$$m=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$$$.

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    6 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    what's the theory behind the fomula plz

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      6 years ago, # ^ |
        Vote: I like it +21 Vote: I do not like it

      Let $$$m=p_1^{d_1} \dots p_n^{d_n}$$$. Consider the periodicity of $$$a^n$$$ on each $$$p_k^{d_k}$$$ separately. By Chinese Remainder Theorem, pre-period of $$$a^n$$$ modulo $$$m$$$ will be equal to the maximum of pre-periods modulo $$$p_k^{d_k}$$$ and period itself will be least common multiple of periods modulo $$$p_k^{d_k}$$$.

      On each $$$p_k^{d_k}$$$ pre-period will be either $$$0$$$ if $$$\gcd(a,p)=1$$$ or at most $$$d_k \leq \log_2 m$$$ otherwise. And period will always be the divisor of $$$\varphi(p_k^{d_k})$$$ which is in turn the divisor of $$$\varphi(m)$$$. Thus their least common multiple will be the divisor of $$$\varphi(m)$$$. This gives you the result above, given that $$$\varphi(m) \geq \log_2 m$$$.

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        6 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        thank you vvery much. seems that I have a lot of things to catch up with

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6 years ago, # |
  Vote: I like it -10 Vote: I do not like it

I don't care. I want another codeforces round asap.

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    6 years ago, # ^ |
      Vote: I like it +7 Vote: I do not like it

    You can have hundreds of virtual contests.

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6 years ago, # |
← Rev. 2   Vote: I like it +22 Vote: I do not like it

WoW, it's cool! Thank you~

It was finally realized. A question about using LaTeX in blogs

$$$ \begin{bmatrix} F_i \\ F_{i-1} \\ (i+1)^3 \\ (i+1)^2 \\ i+1 \\ 1 \end{bmatrix} \Longleftarrow \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 3 & 3 & 1 \\ 0 & 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix} \times \begin{bmatrix} F_{i-1} \\ F_{i-2} \\ i^3 \\ i^2 \\ i \\ 1 \end{bmatrix} $$$
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6 years ago, # |
  Vote: I like it +44 Vote: I do not like it

The most convenient thing is that you can copy the formula by $$$\text{Right Click}\rightarrow\text{Show Math As}\rightarrow\text{TeX Commands}$$$.

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6 years ago, # |
  Vote: I like it +8 Vote: I do not like it

WAIT! Why all \sums became \prods? $$$\underset{\text{actually \sum}}{\underline{\sum}}\neq\underset{\text{real \prod}}{\underline{\prod}}$$$

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6 years ago, # |
← Rev. 2   Vote: I like it +38 Vote: I do not like it

Looks like I found the following feature: If you use \newcommand or \renewcommand, it will affect all comments below yours. Apologies to all the people between my two comments that got confused by this.

Stackexchange had the same issue, but there some people could at least edit the answers/comments and remove the malicious code. Their fix (mentioned in an answer to the linked post) was to wrap posts in \begingroup and \endgroup. Another fix mentioned there was to use \resetstack. MikeMirzayanov could you look into that?

Edit: It looks like \newcommand and \renewcommand now get stripped from mathjax parts of comments. This should deal with people replacing $$$\pi$$$ with 40 owls for now. Nevertheless, I would love to see support for a \newcommand whose scope is just the comment it was written in. (But I see that implementing and testing that might take some time.)

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6 years ago, # |
← Rev. 3   Vote: I like it +12 Vote: I do not like it
$$$\begin{align*}\sum_{i=1}^n\sum_{j=1}^nij\gcd\left(i,j\right)&=\sum_{d=1}^nd\sum_{i=1}^n\sum_{j=1}^nij[\gcd\left(i,j\right)==d]\\&=\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor\dfrac nd\rfloor}\sum_{j=1}^{\lfloor\dfrac nd\rfloor}ij[\gcd\left(i,j\right)==1]\\&=\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor\dfrac nd \rfloor}\sum_{j=1}^{\lfloor\dfrac nd\rfloor}ij\sum_{t|\gcd\left(i,j\right)}\mu\left(t\right)\\&=\sum_{d=1}^nd^3\sum_{t=1}^{\lfloor\dfrac nd \rfloor}\mu\left(t\right)t^2\sum_{i=1}^{\lfloor\dfrac n{td}\rfloor}\sum_{j=1}^{\lfloor\dfrac n{td}\rfloor}ij\\&=\sum_{d=1}^nd^3\sum_{t=1}^{\lfloor\dfrac nd \rfloor}\mu\left(t\right)t^2\left(1+2+3+\cdots+\lfloor\dfrac n{td}\rfloor\right)^2\\&=\sum_{d=1}^nd^3\sum_{d|T}^n\mu\left(\dfrac Td\right)\left(\dfrac Td\right)^2sum^2\left(\lfloor\dfrac nT\rfloor\right)\\&=\sum_{T=1}^nsum^2\lfloor\dfrac nT\rfloor T^2\sum_{d|T}d\mu\left(\dfrac Td\right)\\&=\sum_{T=1}^nsum^2\lfloor\dfrac nT\rfloor T^2\phi\left(T\right)\end{align*}$$$

It is a great joy to solve mathematical problems using $$$LaTeX$$$ !

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6 years ago, # |
  Vote: I like it +20 Vote: I do not like it
$$$gcd(Fib_{a},Fib_{b})=Fib_{gcd(a,b)}$$$
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6 years ago, # |
← Rev. 2   Vote: I like it +21 Vote: I do not like it

\begin{equation*} e^{\pi i} + 1 = 0 \end{equation*}

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6 years ago, # |
  Vote: I like it +5 Vote: I do not like it

Well, Let's see some tables

$$$ \begin{array}{c|lcr} n & \text{Left} & \text{Center} & \text{Right} \\ \hline 1 & 0.24 & 1 & 125 \\ 2 & -1 & 189 & -8 \\ 3 & -20 & 2000 & 1+10i \end{array} $$$
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6 years ago, # |
  Vote: I like it -14 Vote: I do not like it
      1. it doesnt work
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6 years ago, # |
← Rev. 6   Vote: I like it -9 Vote: I do not like it

Testing

$$$\sin \pi = 0$$$
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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it
$$$2019 \ good \ luck \ and \ high \ rating $$$
$$$2019 \ become \ master:D$$$
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    6 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Wow, I find a person use the same picture as me! orz.

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6 years ago, # |
  Vote: I like it +7 Vote: I do not like it

$$$-\dfrac{\hslash^2}{2m} \, \dfrac{\mathrm{d}^2 \psi}{\mathrm{d} x^2}$$$

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    6 years ago, # ^ |
      Vote: I like it +9 Vote: I do not like it

    Oh boi we should forbid posting Hamiltonians on codeforces...

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      6 years ago, # ^ |
        Vote: I like it +2 Vote: I do not like it
      $$$\begin{pmatrix}1&0\\0&-1\\\end{pmatrix}$$$
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      6 years ago, # ^ |
        Vote: I like it +14 Vote: I do not like it

      strange suggestion, coming from a string theorist...

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6 years ago, # |
  Vote: I like it -27 Vote: I do not like it

and what does this have to do on a programming website? this website is for competitive programming, not math.

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    6 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Relax competitive programming is just maths + coding

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      6 years ago, # ^ |
        Vote: I like it -14 Vote: I do not like it

      do you realise what you're saying? just because some random website called codeforces decides to mess around with us by giving us terrible, useless problems full of math doesn't mean the whole thing called COMPETITIVE PROGRAMMING is like this.

      read before you comment because people who don't know better will take what you say as being true.

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6 years ago, # |
  Vote: I like it +28 Vote: I do not like it

\begin{CD} Me\\ @VV V\\ @>>> \\ solving\ problems @VVV @.\\ @<<< \\ @VVV\\ playing\ with\ LaTeX \end{CD}

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6 years ago, # |
  Vote: I like it +6 Vote: I do not like it

It's fantastic! $$$f(x)=g(y)+1$$$

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6 years ago, # |
  Vote: I like it +6 Vote: I do not like it

Great!It will make codeforces better. :D

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

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6 years ago, # |
  Vote: I like it +30 Vote: I do not like it
$$$F(\omega_n^m)=F_0(\omega_{\frac{n}{2}}^m)+\omega_n^mF_1(\omega_{\frac{n}{2}}^m)$$$
$$$F(\omega_n^{m+\frac{n}{2}})=F_0(\omega_{\frac{n}{2}}^m)+\omega_n^{m+\frac{n}{2}}F_1(\omega_{\frac{n}{2}}^m)=F_0(\omega_{\frac{n}{2}}^m)-\omega_n^{m}F_1(\omega_{\frac{n}{2}}^m)$$$
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6 years ago, # |
  Vote: I like it +6 Vote: I do not like it
$$$ c_i = \sum_{\gcd(j,k)=i} a_j \cdot b_k $$$
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6 years ago, # |
  Vote: I like it +8 Vote: I do not like it
$$$2\sum_{n=1}^n n^3$$$
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6 years ago, # |
← Rev. 2   Vote: I like it +5 Vote: I do not like it
$$$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$$
$$$ \LaTeX $$$
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6 years ago, # |
  Vote: I like it +3 Vote: I do not like it
$$$\begin{aligned}\left[ mx\right] =\sum ^{m}_{j=1}\left[ x+\dfrac {j-1}{m}\right] \left( m\in \mathbb{N} ^{+}\right) \end{aligned}​$$$
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6 years ago, # |
  Vote: I like it +8 Vote: I do not like it

That's what you love :D AhmadLoiy

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6 years ago, # |
← Rev. 4   Vote: I like it +8 Vote: I do not like it
$$$a_n=\sum_{k=0}^{n}\binom{n}{k}b_k\iff b_n=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}a_k$$$
$$$\int_a^bf(x)\text{d}x\approx\frac{b-a}{6}(f(a)+4f(\frac{a+b}{2})+f(b))$$$
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6 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Is it rated ?

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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Does it support Markdown and MathJax in the same post/comment ?

Here I'm trying to test both of this:

Markdown & Latex Example

Codeforces is awesome!! Now let's try to write a matrix: Inline $$$\begin{bmatrix}a & b\ c & d\ e & f \end{bmatrix}$$$ and

Now in new line matrix:`

$$$\begin{bmatrix}a & b\\ c & d\\ e & f \end{bmatrix}$$$

Seems it doesn't work in inline form.

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    6 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    Use $$$\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}$$$, I succeeded.

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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Great!

$$$ \boldsymbol{f}(n) = \sum_{d \mid n} \boldsymbol{\mu} \left( \frac{n}{d} \right) \boldsymbol{g}(d) $$$
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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

(2 + 3) / 55

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6 years ago, # |
  Vote: I like it +5 Vote: I do not like it
$$$ \bbox[5px,border:2px solid #B050FF] { \begin{array}{c|ccc} a \oplus b & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 0 & 3 & 2 & 5 & 4 \\ 2 & 3 & 0 & 1 & 6 & 7 \\ 3 & 2 & 1 & 0 & 7 & 6 \\ 4 & 5 & 6 & 7 & 0 & 1 \\ 5 & 4 & 7 & 6 & 1 & 0 \\ \end{array} } $$$
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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it
$$$ \frac 1x = \frac 1a + \frac1b \\ \Downarrow \\ 1 = \frac xa + \frac xb \\ \Downarrow \\ 1 = \frac{xa+xb}{ab} \\ \Downarrow \\ xa+xb = ab \\ \Downarrow \\ xa = b\left(a - x\right) \Rightarrow xa - x^2 + x^2 = x\left(a - x\right) + x^2 \\ \Downarrow \\ x^2 = b(a-x) - x(a-x) = (a-x)(b-x) $$$
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6 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it
$$$ \begin{aligned} F(n) &= \sum_{d \mid n} f(d) \\ f(n) &= \sum_{d \mid n} F(d) \mu\!\left(\frac nd\right) \end{aligned} $$$
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6 years ago, # |
  Vote: I like it 0 Vote: I do not like it

new LaTex looks much better

<p>Unable to parse markup [type=CF_MATHJAX]

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    6 years ago, # ^ |
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    F(n)f(n)=∑d∣nf(d)=∑d∣nF(d)μ(nd)

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    6 years ago, # ^ |
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    umm…… seem to be unable to use LaTex on my computer……

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6 years ago, # |
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$$$ \begin{pmatrix} 1&2&3\\ 4&5&6\\ \end{pmatrix} $$$
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6 years ago, # |
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It's so cool, I want to comment too :

$$$ \frac{d}{dx} \int_{a}^{x} f(t) \hspace{0.2cm} dt = f(x) $$$
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5 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Am I the only one who feels that it doesn't work now?

Inline: $$$\frac{x}{y}\in\mathbb{R}$$$

Block:

$$$\frac{x}{y}\in\mathbb{R}$$$

Displaystyle:

$$$\displaystyle \frac{x}{y}\in\mathbb{R}$$$

UPD: never mind, I've changed my browser recently, and now I seem to have to change my mathjax math renderer. Sorry for necroposting

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