MikeMirzayanov's blog

By MikeMirzayanov, 11 days 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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11 days ago, # |
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Can you write something here to see how it looks like?

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    11 days ago, # ^ |
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    How to use it in comment? It dosen't work. if i type. \frac{1}{2}

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    11 days ago, # ^ |
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    $$$\sum_{i=1}^{n}i=\frac{n \cdot (n+1)}{2}$$$
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      11 days ago, # ^ |
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      I understand. It need like this. yor need use double dollar sign to include your code

      $$$ \frac{a}{b} $$$
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      4 days ago, # ^ |
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      Thanks a lot.

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    11 days ago, # ^ |
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    $$$\forall \varepsilon\!>\!0 \ \exists k(\varepsilon)\!\in\!N: \forall n>k \Rightarrow |x_{n}-a|<\varepsilon$$$

    You have to use $ or $$ like in LateX

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    11 days ago, # ^ |
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    $a^2$
    $$$a^2$$$
    $$a^2$$

    $$$a^2$$$
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11 days ago, # |
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$$$G_{ij}=\sum_k F_{ik}F_{jk}$$$
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    10 days ago, # ^ |
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    Sound matrix multiplications am I right? :D

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      10 days ago, # ^ |
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      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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        10 days ago, # ^ |
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        Whoa, slow down senpai, I need a few more time to be familiar with ML models. :D

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11 days ago, # |
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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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11 days ago, # |
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Good job

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11 days ago, # |
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$$$LOL$$$
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11 days ago, # |
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$$$\text{C}_{\text{ode}}\text{F}_{\text{orces}}\text{ Round}_{547} \text{ Please}$$$
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    11 days ago, # ^ |
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    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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11 days ago, # |
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$$$\mathbb{CODEFORCES}$$$

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11 days ago, # |
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$$$Cool$$$

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11 days ago, # |
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$$$\displaystyle \sum_{i=1}^{\infty} i = -\frac{1}{12}$$$
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11 days ago, # |
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What was the LaTeX renderer before? I always thought it was some kind of Russian bootleg MathJax.

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11 days ago, # |
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$$$\begin{matrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{matrix}$$$
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$$$\int_G f(g) \, \text{d}g = \frac{1}{|W|} \int_G \int_T \mathopen| \Delta(t) \mathclose|^2 f(gtg^{-1}) \, \text{d}t \, \text{d}g $$$

Looks very pretty :)

$$$1+2=3<4$$$
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11 days ago, # |
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$$$ \huge \mathcal{O}( N ) $$$

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10 days ago, # |
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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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    10 days ago, # ^ |
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    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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      10 days ago, # ^ |
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      Yeah, you're right. But sometimes it's too complicated.

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10 days ago, # |
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Can explain me why backward compatibility is very important?

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10 days ago, # |
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Deleted and screw uplaoding a photo

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10 days ago, # |
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$$$\text{summitwei} \in \text{geniosity}$$$

Proof left as an exercise to the reader.

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    10 days ago, # ^ |
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    $$$\text{fishy} \not\in \text{geniosity}$$$

    I have already proved this B).

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10 days ago, # |
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$$$a^2$$$

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10 days ago, # |
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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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    10 days ago, # ^ |
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    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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10 days ago, # |
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$$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$$
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10 days ago, # |
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$$$\require{AMScd}$$$ \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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10 days ago, # |
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$$$ \displaystyle \sum_{i=1}^n \sum_{j=1}^m d_1(\gcd(i, j)) \\ \displaystyle\sum_{d=1}^n \sum_{i=1}^n \sum_{j=1}^m d_1(d)[\gcd(i,j)=d])\\ \displaystyle\sum_{d=1}^n d_1(d) \sum_{i=1}^n \sum_{j=1}^m [\gcd(i,j)=d])\\ \displaystyle\sum_{d=1}^n d_1(d) \sum_{k=1}^{\lfloor \frac nd \rfloor} \mu(k) \lfloor \frac n{kd} \rfloor \displaystyle\lfloor \frac m{kd} \rfloor \\ $$$

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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    10 days ago, # ^ |
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    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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      10 days ago, # ^ |
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      It seems that it was fixed now, also thank you...

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        9 days ago, # ^ |
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        I'm just crack a joke,if you use the \Sigma,it will be too small...

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10 days ago, # |
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$$$a^{n}\equiv a^{\varphi(m)+(n \bmod \varphi(m))}\pmod{m}$$$
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    10 days ago, # ^ |
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    It is not always true when $$$n<\varphi(m)$$$. For example, $$$6^{2}\not\equiv 6^{4+2}\pmod8$$$.

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      10 days ago, # ^ |
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      It's true for $$$n>\log_2 m$$$

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        9 days ago, # ^ |
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        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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    10 days ago, # ^ |
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    what's the theory behind the fomula plz

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      9 days ago, # ^ |
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      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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        9 days ago, # ^ |
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        thank you vvery much. seems that I have a lot of things to catch up with

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10 days ago, # |
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I don't care. I want another codeforces round asap.

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    10 days ago, # ^ |
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    You can have hundreds of virtual contests.

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10 days ago, # |
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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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10 days ago, # |
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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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10 days ago, # |
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WAIT! Why all \sums became \prods? $$$\underset{\text{actually \sum}}{\underline{\sum}}\neq\underset{\text{real \prod}}{\underline{\prod}}$$$

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10 days 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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10 days 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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10 days ago, # |
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$$$gcd(Fib_{a},Fib_{b})=Fib_{gcd(a,b)}$$$
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10 days ago, # |
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\begin{equation*} e^{\pi i} + 1 = 0 \end{equation*}

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10 days ago, # |
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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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10 days ago, # |
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      1. it doesnt work
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10 days ago, # |
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$$$ \rm Test $$$

$$$ a ^ 2 + b ^ 2 = c ^ 2 $$$

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10 days ago, # |
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Testing

$$$\sin \pi = 0$$$
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10 days ago, # |
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$$$2019 \ good \ luck \ and \ high \ rating $$$
$$$2019 \ become \ master:D$$$
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    7 days ago, # ^ |
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    Wow, I find a person use the same picture as me! orz.

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9 days ago, # |
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$$$-\dfrac{\hslash^2}{2m} \, \dfrac{\mathrm{d}^2 \psi}{\mathrm{d} x^2}$$$

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    9 days ago, # ^ |
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    Oh boi we should forbid posting Hamiltonians on codeforces...

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      9 days ago, # ^ |
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      $$$\begin{pmatrix}1&0\\0&-1\\\end{pmatrix}$$$
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      8 days ago, # ^ |
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      strange suggestion, coming from a string theorist...

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9 days ago, # |
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and what does this have to do on a programming website? this website is for competitive programming, not math.

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    9 days ago, # ^ |
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    Relax competitive programming is just maths + coding

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      8 days ago, # ^ |
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      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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9 days ago, # |
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Looks pretty good! Gl for Codeforces in further developments.

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8 days ago, # |
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\begin{CD} Me\\ @VV V\\ @>>> \\ solving\ problems @VVV @.\\ @<<< \\ @VVV\\ playing\ with\ LaTeX \end{CD}

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8 days ago, # |
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$$$LATEX$$$
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8 days ago, # |
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It's fantastic! $$$f(x)=g(y)+1$$$

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8 days ago, # |
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Great!It will make codeforces better. :D

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

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8 days ago, # |
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$$$F(\omega_n^m)=F_0(\omega_{\frac{n}{2}}^m)+\omega_n^mF_1(\omega_{\frac{n}{2}}^m)$$$
$$$F(\omega_n^{m+\frac{n}{2}})=F_0(\omega_{\frac{n}{2}}^m)+\omega_n^{m+\frac{n}{2}}F_1(\omega_{\frac{n}{2}}^m)=F_0(\omega_{\frac{n}{2}}^m)-\omega_n^{m}F_1(\omega_{\frac{n}{2}}^m)$$$
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8 days ago, # |
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$$$ c_i = \sum_{\gcd(j,k)=i} a_j \cdot b_k $$$
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8 days ago, # |
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$$$2\sum_{n=1}^n n^3$$$
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8 days ago, # |
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$$$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$$
$$$ \LaTeX $$$
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8 days ago, # |
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$$$\begin{aligned}\left[ mx\right] =\sum ^{m}_{j=1}\left[ x+\dfrac {j-1}{m}\right] \left( m\in \mathbb{N} ^{+}\right) \end{aligned}​$$$
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8 days ago, # |
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That's what you love :D AhmadLoiy

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8 days ago, # |
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$$$a_n=\sum_{k=0}^{n}\binom{n}{k}b_k\iff b_n=\sum_{k=0}^{n}(-1)^{n-k}\binom{n}{k}a_k$$$
$$$\int_a^bf(x)\text{d}x\approx\frac{b-a}{6}(f(a)+4f(\frac{a+b}{2})+f(b))$$$
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8 days ago, # |
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Is it rated ?

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7 days ago, # |
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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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    3 days ago, # ^ |
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    Use $$$\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}$$$, I succeeded.

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7 days ago, # |
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$$$fb(n)=fb(n-1)+fb(n-2)$$$
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7 days ago, # |
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Great!

$$$ \boldsymbol{f}(n) = \sum_{d \mid n} \boldsymbol{\mu} \left( \frac{n}{d} \right) \boldsymbol{g}(d) $$$
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6 days ago, # |
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(2 + 3) / 55

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4 days ago, # |
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$$$ \bbox[5px,border:2px solid #B050FF] { \begin{array}{c|ccc} a \oplus b & 1 & 2 & 3 & 4 & 5 \\ \hline 1 & 0 & 3 & 2 & 5 & 4 \\ 2 & 3 & 0 & 1 & 6 & 7 \\ 3 & 2 & 1 & 0 & 7 & 6 \\ 4 & 5 & 6 & 7 & 0 & 1 \\ 5 & 4 & 7 & 6 & 1 & 0 \\ \end{array} } $$$
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4 days ago, # |
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$$$ \frac 1x = \frac 1a + \frac1b \\ \Downarrow \\ 1 = \frac xa + \frac xb \\ \Downarrow \\ 1 = \frac{xa+xb}{ab} \\ \Downarrow \\ xa+xb = ab \\ \Downarrow \\ xa = b\left(a - x\right) \Rightarrow xa - x^2 + x^2 = x\left(a - x\right) + x^2 \\ \Downarrow \\ x^2 = b(a-x) - x(a-x) = (a-x)(b-x) $$$
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4 days ago, # |
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$$$ \begin{aligned} F(n) &= \sum_{d \mid n} f(d) \\ f(n) &= \sum_{d \mid n} F(d) \mu\!\left(\frac nd\right) \end{aligned} $$$
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27 hours ago, # |
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new LaTex looks much better

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

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

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