Testing latex

[John2020]

\[\frac{\Delta y}{\Delta x} = \frac{f(x+h)- f(x)}{h}\]

\[\Delta y \]

\[ y = \int f(x) dx \]

 

Edited September 24, 2020 by John2020

[John2020]

Some further tests:

\[\sum \vec{F}_{T} = \frac{\sum \vec{\tau}_{T}}{|\left(\vec{r}_{A} - \vec{r}_{R} \right)|}\]

Edited September 27, 2020 by John2020

[John2020]

Some further tests:

 

Edited September 27, 2020 by John2020

[Eise]

Testing relativity formulas for the thread which name we do not mention...

\[ t = \frac{{t}' + \frac{v{x}'}{c^{2}}}{\sqrt{1 - \frac{v^{2}}{c^{2}}}} \]

 

\[t = \frac{{0} + {0.8} . {0.6} }{\sqrt{1 - {0.8 ^{2}}}}\]

Edited October 9, 2020 by Eise

[joigus]

Testing more stuff.

\[ f\left(x\right)=\begin{cases} 1, & x\in\mathbb{Q}\\ -1, & x\in\mathbb{R}-\mathbb{Q} \end{cases} \]

Testing more stuff.

\[ \textrm{int}C\neq\textrm{Ø}\Rightarrow\textrm{if }x\in C\Rightarrow\in B_{\epsilon}\left(x\right)\subseteq C\Rightarrow\mu\left(C\right)\geq\mu\left(B_{\epsilon}\left(x\right)\right)>0 \]

 

 

Edited April 8, 2021 by joigus

[joigus]

Rather:

\[ \textrm{int}C\neq\textrm{Ø}\Rightarrow\textrm{if }x\in C\Rightarrow B_{\epsilon}\left(x\right)\subseteq C\Rightarrow\mu\left(C\right)\geq\mu\left(B_{\epsilon}\left(x\right)\right)>0 \]

[joigus]

Total energy:

\[E=\frac{mc^{2}}{\sqrt{1-v^{2}/c^{2}}}\]

Rest energy:

\[E_{0}=mc^{2}\]

Kinetic energy:

\[\textrm{K.E.}=\frac{mc^{2}}{\sqrt{1-v^{2}/c^{2}}}-mc^{2}\]

Edited June 4, 2021 by joigus

[Orion1]

[math]\begin{align} \text{text} \end{align}[/math]

[math]\color{blue}{\text{text}}[/math]

text

[math]F = ma[/math]

 

 

 

Edited July 4, 2021 by Orion1

[Orion1]

...

 

 

Edited July 4, 2021 by Orion1

[joigus]

\[\int (f(x)+ dy/2)dx= \int f(x)dx+ \frac{1}{2}\int dydx\]

\[ \frac{1}{2}\int dydx \]

 

Inline: \( dy\left( x \right) = y'\left( x \right) dx \).

Edited July 4, 2021 by joigus

[Orion1]

17 hours ago, joigus said:

 

∫(f(x)+dy/2)dx=∫f(x)dx+12∫dydx

 

 

12∫dydx

 

 

Inline: dy(x)=y′(x)dx .

@joigus, how did you center your equations?

I cannot get \begin{center} command to work.

 

 

Edited July 5, 2021 by Orion1

[joigus]

1 hour ago, Orion1 said:

@joigus, how did you center your equations?

I cannot get \begin{center} command to work.

 

 

I just use \[ (and the closing square brace) for centred (display) equations and \( (and its closing round brace) for inline maths. It seems to interact well with the MathJax "engine".

[Orion1]

 

y=∫f(x)dx

Thanks joigus!

\[ \color{blue}{\text{test}} \]

 

Edited July 5, 2021 by Orion1

[Orion1]

...

 

Edited July 5, 2021 by Orion1

[joigus]

\[ {\not}R \]

[joigus]

Overlapping symbols test:

\mathclap{\fullmoon}\fullmoon 

\[ \mathclap{\fullmoon}\fullmoon \]

Pitty.

[Ghideon]

Source:

[math]\begin{displaymath}
  T=\sum_{
    \mathclap{\substack{i_1,\ldots,i_n=1 \\ j_1,\ldots,j_m=1}}}^d
    T_{j_1,\ldots,j_m}^{i_1,\ldots,i_n}
    E_{i_1,\ldots,i_n}^{j_1,\ldots,j_m}
\end{displaymath}[/math]

[math]\begin{displaymath} T=\sum_{ i_1,\ldots,i_n,j_1,\ldots,j_m=1}^d \; T_{j_1,\ldots,j_m}^{i_1,\ldots,i_n} E_{i_1,\ldots,i_n}^{j_1,\ldots,j_m} \end{displaymath[/math]

Source:

\begin{displaymath}
  T=\sum_{
    i_1,\ldots,i_n,j_1,\ldots,j_m=1}^d \;
    T_{j_1,\ldots,j_m}^{i_1,\ldots,i_n}
    E_{i_1,\ldots,i_n}^{j_1,\ldots,j_m}
\end{displaymath}

Test result:
[math]\begin{displaymath}
  T=\sum_{
    i_1,\ldots,i_n,j_1,\ldots,j_m=1}^d \;
    T_{j_1,\ldots,j_m}^{i_1,\ldots,i_n}
    E_{i_1,\ldots,i_n}^{j_1,\ldots,j_m}
\end{displaymath}[/math]

Expected result:

 

(I got curious about @joigus test... wanted to learn)

 

Edited November 5, 2022 by Ghideon
trial & error

[joigus]

Maybe this one works for "male" and "female"?

\male

 

Undefined control sequence \male

 

 

\female

 

Undefined control sequence \female

DDo

 

On 11/5/2022 at 2:35 PM, Ghideon said:

Source:

[math]\begin{displaymath}
  T=\sum_{
    \mathclap{\substack{i_1,\ldots,i_n=1 \\ j_1,\ldots,j_m=1}}}^d
    T_{j_1,\ldots,j_m}^{i_1,\ldots,i_n}
    E_{i_1,\ldots,i_n}^{j_1,\ldots,j_m}
\end{displaymath}[/math]

Unknown environment 'displaymath'

Source:

\begin{displaymath}
  T=\sum_{
    i_1,\ldots,i_n,j_1,\ldots,j_m=1}^d \;
    T_{j_1,\ldots,j_m}^{i_1,\ldots,i_n}
    E_{i_1,\ldots,i_n}^{j_1,\ldots,j_m}
\end{displaymath}

Test result:
Unknown environment 'displaymath'

Expected result:

 

(I got curious about @joigus test... wanted to learn)

 

I think you don't need environment "displaymath" for that one. Simply:

\[ 
  T=\sum_{\begin{array}{c}
	i_{1},\cdots,i_{n}=1\\
	j_{1},\cdots,j_{m}=1
\end{array}}^{d}T_{i_{1}\cdots i_{n}}^{j_{1}\cdots j_{m}}E_{j_{1}\cdots j_{m}}^{i_{1}\cdots i_{n}}
\]

should work. Let's see:

\[ T=\sum_{\begin{array}{c} i_{1},\cdots,i_{n}=1\\ j_{1},\cdots,j_{m}=1 \end{array}}^{d}T_{i_{1}\cdots i_{n}}^{j_{1}\cdots j_{m}}E_{j_{1}\cdots j_{m}}^{i_{1}\cdots i_{n}} \]

[Mordred]

\[\frac{1}{2}\]

Edited December 9, 2022 by Mordred

[Mordred]

\[\array{ \mathfrak{g} \times X && \overset{R}{\longrightarrow} && T X \\ & {\llap{pr_2}}\searrow && \swarrow_{\rlap{p}} \\ && X }\]

\[{\small\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline Particle& Spin & g & Q &B&L_e &L_\mu&M (Mev)&\tau\\\hline \gamma&1&2&0&0&0&0&<3*10^{-33}&stable\\\hline e^-&1/2&2&-1&0&1&0&0.511&>2*10^{22}yrs\\\hline e+&1/2&2&1&0&-1&0&0.511&>2*10^{22}yrs\\\hline v_e&1/2&1&0&0&1&0&,5*10^{-5}&stable\\\hline \overline{v}_e&1/2&1&0&0&-1&0&<5*10^{-5}&stable\\\hline \mu^-&1/2&2&-1&0&0&1&105.7&2.2*10^{-6}sec\\\hline \mu^+&1/2&2&1&0&0&-1&105.7&2.2*10^{-5}sec\\\hline v_\mu&1/2&1&0&0&0&1&<0.25&>10^{32}yrs\\\hline \overline{v}_\mu&1/2&1&0&0&0&-1&<0.25&>10^{32}yrs \\\hline p&1/2&2&1&1&0&0&938.3&.10^{32}yrs\\\hline \overline{p}&1/2&2&-1&-1&0&0&938.3&>10^{32}yrs\\\hline n&1/2&2&0&1&0&0&939.6&898 sec\\\hline \overline{n}&1/2&2&&*-1&0&0&939.6&898 sec\\\hline \pi^0+&0&1&1&0&0&0&139.6&1.39*10^{-8}sec\\\hline \pi^-&0&1&0&0&0&0&135.0&8.7*10^{-17}sec\\\hline \pi^+&0&1&-1&0&0&0&139.6&2.6*10^{-8}sec\\\hline\end{array}}\]

 

Edited January 1 by Mordred

[joigus]

\[ \overset{m^{2}c^{4}}{\overbrace{E^{2}-c^{2}\boldsymbol{p}^{2}}}+\overset{0}{\overbrace{\hbar^{2}\omega^{2}-c^{2}\hbar^{2}\boldsymbol{k}^{2}}}+2\hbar E\omega-2c^{2}\hbar\boldsymbol{p}\cdot\boldsymbol{k}=m^{2}c^{4} \]

OK then.