MikeMirzayanov's blog

By MikeMirzayanov, 7 months 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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7 months 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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    7 months 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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    7 months ago, # ^ |
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    $$$\sum_{i=1}^{n}i=\frac{n \cdot (n+1)}{2}$$$
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      7 months 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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      7 months ago, # ^ |
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      Thanks a lot.

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    7 months 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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    7 months ago, # ^ |
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    $a^2$
    $$$a^2$$$
    $$a^2$$

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

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      7 months 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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        7 months 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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7 months 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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7 months ago, # |
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Good job

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7 months ago, # |
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$$$LOL$$$
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7 months ago, # |
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$$$\text{C}_{\text{ode}}\text{F}_{\text{orces}}\text{ Round}_{547} \text{ Please}$$$
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    7 months 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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7 months ago, # |
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$$$\mathbb{CODEFORCES}$$$

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7 months ago, # |
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$$$Cool$$$

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7 months ago, # |
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$$$\displaystyle \sum_{i=1}^{\infty} i = -\frac{1}{12}$$$
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7 months 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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7 months 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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7 months ago, # |
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$$$ \huge \mathcal{O}( N ) $$$

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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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    7 months 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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      7 months ago, # ^ |
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      Yeah, you're right. But sometimes it's too complicated.

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7 months ago, # |
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Can explain me why backward compatibility is very important?

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7 months ago, # |
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Deleted and screw uplaoding a photo

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7 months ago, # |
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$$$\text{summitwei} \in \text{geniosity}$$$

Proof left as an exercise to the reader.

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7 months ago, # |
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$$$a^2$$$

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7 months 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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    7 months 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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7 months ago, # |
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$$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$$
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7 months 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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7 months 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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    7 months 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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      7 months ago, # ^ |
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      It seems that it was fixed now, also thank you...

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        7 months 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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7 months ago, # |
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$$$a^{n}\equiv a^{\varphi(m)+(n \bmod \varphi(m))}\pmod{m}$$$
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    7 months 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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      7 months ago, # ^ |
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      It's true for $$$n>\log_2 m$$$

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        7 months 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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    7 months ago, # ^ |
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    what's the theory behind the fomula plz

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      7 months 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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        7 months 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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7 months ago, # |
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I don't care. I want another codeforces round asap.

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    7 months ago, # ^ |
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    You can have hundreds of virtual contests.

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7 months 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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7 months 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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7 months 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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7 months ago, # |
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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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7 months ago, # |
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$$$\begin{align*}\sum_{i=1}^n\sum_{j=1}^nij\gcd\left(i,j\right)&=\sum_{d=1}^nd\sum_{i=1}^n\sum_{j=1}^nij[\gcd\left(i,j\right)==d]\\&=\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor\dfrac nd\rfloor}\sum_{j=1}^{\lfloor\dfrac nd\rfloor}ij[\gcd\left(i,j\right)==1]\\&=\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor\dfrac nd \rfloor}\sum_{j=1}^{\lfloor\dfrac nd\rfloor}ij\sum_{t|\gcd\left(i,j\right)}\mu\left(t\right)\\&=\sum_{d=1}^nd^3\sum_{t=1}^{\lfloor\dfrac nd \rfloor}\mu\left(t\right)t^2\sum_{i=1}^{\lfloor\dfrac n{td}\rfloor}\sum_{j=1}^{\lfloor\dfrac n{td}\rfloor}ij\\&=\sum_{d=1}^nd^3\sum_{t=1}^{\lfloor\dfrac nd \rfloor}\mu\left(t\right)t^2\left(1+2+3+\cdots+\lfloor\dfrac n{td}\rfloor\right)^2\\&=\sum_{d=1}^nd^3\sum_{d|T}^n\mu\left(\dfrac Td\right)\left(\dfrac Td\right)^2sum^2\left(\lfloor\dfrac nT\rfloor\right)\\&=\sum_{T=1}^nsum^2\lfloor\dfrac nT\rfloor T^2\sum_{d|T}d\mu\left(\dfrac Td\right)\\&=\sum_{T=1}^nsum^2\lfloor\dfrac nT\rfloor T^2\phi\left(T\right)\end{align*}$$$

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

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

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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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7 months ago, # |
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      1. it doesnt work
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$$$ \rm Test $$$

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

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Testing

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

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

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    7 months ago, # ^ |
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    Oh boi we should forbid posting Hamiltonians on codeforces...

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

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7 months 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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    7 months ago, # ^ |
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    Relax competitive programming is just maths + coding

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      7 months 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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7 months ago, # |
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Looks pretty good! Gl for Codeforces in further developments.

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

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

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Great!It will make codeforces better. :D

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

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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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$$$ c_i = \sum_{\gcd(j,k)=i} a_j \cdot b_k $$$
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$$$2\sum_{n=1}^n n^3$$$
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$$$ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} $$$
$$$ \LaTeX $$$
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7 months 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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7 months ago, # |
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That's what you love :D AhmadLoiy

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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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7 months ago, # |
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Is it rated ?

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

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      4 months ago, # ^ |
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      Great!! You did it. Got it. Example: $$$\begin{bmatrix}a&b\\c&d\\e&f\end{bmatrix}$$$ inline matrix. Just need 3 backslashes (\) instead of 2 backslashes (\) . Thanks :)

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$$$fb(n)=fb(n-1)+fb(n-2)$$$
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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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7 months ago, # |
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(2 + 3) / 55

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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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$$$ \frac 1x = \frac 1a + \frac1b \\ \Downarrow \\ 1 = \frac xa + \frac xb \\ \Downarrow \\ 1 = \frac{xa+xb}{ab} \\ \Downarrow \\ xa+xb = ab \\ \Downarrow \\ xa = b\left(a - x\right) \Rightarrow xa - x^2 + x^2 = x\left(a - x\right) + x^2 \\ \Downarrow \\ x^2 = b(a-x) - x(a-x) = (a-x)(b-x) $$$
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$$$ \begin{aligned} F(n) &= \sum_{d \mid n} f(d) \\ f(n) &= \sum_{d \mid n} F(d) \mu\!\left(\frac nd\right) \end{aligned} $$$
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7 months ago, # |
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new LaTex looks much better

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

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

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

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4 months ago, # |
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$$$ \begin{pmatrix} 1&2&3\\ 4&5&6\\ \end{pmatrix} $$$
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4 months 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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3 months 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