Jump to content

Testing latex


John2020

Recommended Posts

  • 2 weeks later...

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 by Eise
Link to comment
Share on other sites

  • 5 months later...

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 by joigus
Link to comment
Share on other sites

Rather:

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

Link to comment
Share on other sites

  • 1 month later...

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 by joigus
Link to comment
Share on other sites

  • 5 weeks later...

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

\[ \frac{1}{2}\int dydx \]

 

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

Edited by joigus
Link to comment
Share on other sites

17 hours ago, joigus said:

 

(f(x)+dy/2)dx=f(x)dx+12dydx

 

 

12dydx

 

 

Inline: dy(x)=y(x)dx .

@joigus, how did you center your equations?

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

 

 

Edited by Orion1
Link to comment
Share on other sites

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".

Link to comment
Share on other sites

  • 1 year later...
  • 2 months later...

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:

image.png.a3f5117a2d531dc34a0a0485bfab952a.png

 

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

 

Edited by Ghideon
trial & error
Link to comment
Share on other sites

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:

image.png.a3f5117a2d531dc34a0a0485bfab952a.png

 

(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:

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

Link to comment
Share on other sites

  • 4 weeks later...

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

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

 

Edited by Mordred
Link to comment
Share on other sites

  • 2 months later...

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

OK then.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.