Awesome work!! $\frac{-b \pm \sqrt{b^2+4ac}}{2a}$
Thanks!!!

$x^2$

That is great! Let me test:
$\sqrt{3x-1}+(1+x)^2$
It would be cool to integrate with Aloha Editor in some way. The problem is how to define the workflow, but looks very cool…

\lim_{n\rightarrow \infty}\sum_{i=0}^n \frac{1}{p_i}

$ \lim_{n\rightarrow \infty}\sum_{i=0}^n \frac{1}{p_i} $

Not bad if I can write Einstein’s equation!

$E=mc^2$

Hi, great works.
May I have the link to the code where you integrate writemaths with wysiwyg editor? I cannot get it to work. Thanks for your help

Do you mean in WordPress? I haven’t got it to work in WordPress’s wysiwyg editor.

Can you give me a link of demo to use it in Tinymce?

$\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{m^2\,n}
{3^m\left(m\,3^n+n\,3^m\right)}$

Great work !

Are you able to get it work on any wysiwyg editor?

$iint _{ 2 }^{ 4 }{ x{ y }^{ 2 } } .dz$

\iint _{ 2 }^{ 4 }{ x{ y }^{ 2 } } .dz

\[e^{\pi i} + 1 = 0\]

This is cool!

x^2

Again with dollar signs

$x^2$

$x^2$

$x^3+x^2+sqrt(4)$

\[ 1 + \frac{q^2}{(1-q)}+\frac{q^6}{(1-q)(1-q^2)}+\cdots =
\prod_{j=0}^{\infty}\frac{1}{(1-q^{5j+2})(1-q^{5j+3})},
\quad\quad \text{for $|q|<1$}. \]

cool man..

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

$I=\frac{1^3}{2}mR^2=\frac{1}{2}.1.(0,2)^2=0,02 kg.m^2$

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

Let $a = n_1p_j$ where $p_j \in \{2, 3, 5, 7, \ldots\},$ so $a \in {P_n}_1.$ Let $b = n_2p_k$ where $p_k \in \{2, 3, 5, 7, \ldots\},$ so $b \in {P_n}_2.$

Now assume that ${P_n}_1 \cap {P_n}_2$ is nonempty. Then there exists some $n_1$ and $n_2$ such that for some $p_j$ and $p_k$ such that $a = b$. Thus we have $n_1p_j = n_2p_k$.

Note that $p_kp_j = p_kp_j$ is clearly true, and multiplying both sides of the equation by some integer $m$ gives us $mp_kp_j = mp_kp_j$.

So let $n_1 = mp_k$ and $n_2 = mp_j$, which implies $n_1p_j = n_2p_k$, and thus $a = b$ as desired. This means that ${P_n}_1 \cap {P_n}_2$ is guaranteed to be nonempty if we can divide any common factors from $n_1$ and $n_2$ and be left with only prime numbers.

(A) is out as $1$ is not prime. (B) is is out because if we divide $7$ and $21$ by their common factor $7$, we are left with $1$ and $7$, and $1$ is not prime. (D) is out because $24/4 = 6$ which is not prime. (E) is out because $5/5 =1$ is not prime. (C) is correct as $12/4 = 3$ and $20/4 = 5$, which are both prime.

$$ sqrt(4) $$

$ sqrt(4) $

\(J_\alpha(x) = \sum\limits_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m + \alpha + 1)}{\left({\frac{x}{2}}\right)}^{2 m + \alpha}\)

Amazing !

$$ \frac{\partial f}{\partial t} = \frac{\partial ^{2} f}{\partial x^{2}} $$

I will try to install it on my own wordpress. Thanks a lot !

$a^2+b^4

$\log_2 \left( \frac{1}{x+2} \right)$

a $x+1$
\begin{align}
x &= 1 \\
y &= 2 \\
z &= 3
\end{align}

$\left [ – \frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V \right ] \Psi
= i \hbar \frac{\partial}{\partial t} \Psi
$\

test

$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$

$$ x^2 $$

גכשדכשד $$ 5x+3^2 = 3 $$

I wish music was as easy to work with as math… $$E\flat = D\sharp$$

$$ x = \sqrt \frac {2x + 3y^2 + 2} {2x +y} + a^3 + b^2 + 4$$

$\frac{1}{\Bigl(\sqrt{\phi \sqrt{5}}-\phi\Bigr) e^{\frac25 \pi}} = 1+\frac{e^{-2\pi}} {1+\frac{e^{-4\pi}} {1+\frac{e^{-6\pi}} {1+\frac{e^{-8\pi}} {1+\cdots} } } }$

$$y=a cos(\frac{t}{p}~qx)$$

\texttt{dst} (I) = \fork{\texttt{src}(I)^power}{if \texttt{power} is integer}{|\texttt{src}(I)|^power}{otherwise}

y = $\sqrt(x^2+2x^3)$

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