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Math Help - Testing LaTeX 2

  1. #31
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Jones View Post
    Hehe, how would i get a left bracket first to indicate an equation system, (or congruence system if you like)
    Try this :

    \begin{cases}<br />
\phi(n) \equiv q \pmod{42} \\<br />
p \equiv n \pmod{234^{432}} \\<br />
\end{cases}

    If you want to have the symbols \equiv aligned you can use the command & :

    \begin{cases}<br />
\phi(n) & \equiv q \pmod{42} \\<br />
p & \equiv n \pmod{234^{432}} \\<br />
\end{cases}

    If you don't want q and \phi(n) to be aligned on the left, you'll have to use an array instead of the cases environment :

    <br />
\left\{<br />
\begin{array}{ccl}<br />
\phi(n) & \equiv & q \pmod{42} \\<br />
p & \equiv & n \pmod{234^{432}} \\<br />
\end{array}<br />
\right.<br />

    <br />
\left\{<br />
\begin{array}{rcl}<br />
\phi(n) & \equiv & q \pmod{42} \\<br />
p & \equiv & n \pmod{234^{432}} \\<br />
\end{array}<br />
\right.<br />
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  2. #32
    A riddle wrapped in an enigma
    masters's Avatar
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    Arc symbol

    What do you use for the arc symbol? I've been using

    \widehat{ABC} to get \widehat{ABC}, but that's not quite it.

    I tried using: \stackrel{\frown}{ABC}

    \stackrel{\frown}{ABC}

    That just doesn't look right, either. It's too small.
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  3. #33
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    Opalg's Avatar
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    Quote Originally Posted by masters View Post
    What do you use for the arc symbol? I've been using

    \widehat{ABC} to get \widehat{ABC}, but that's not quite it.

    I tried using: \stackrel{\frown}{ABC}

    \stackrel{\frown}{ABC}

    That just doesn't look right, either. It's too small.
    You can make it slightly bigger by \stackrel{\textstyle\frown}{ABC}:

    \stackrel{\textstyle\frown}{ABC}
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  4. #34
    A riddle wrapped in an enigma
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    Quote Originally Posted by Opalg View Post
    You can make it slightly bigger by \stackrel{\textstyle\frown}{ABC}:

    \stackrel{\textstyle\frown}{ABC}
    Yeah, that'd be ok for minor arcs like \stackrel{\textstyle\frown}{AB}
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  5. #35
    MHF Contributor Mathstud28's Avatar
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    Let X be a metric space, and let E\subset{X}. Now define S\equiv\left\{d(p,q):q,p\in{E}\right\}, now if \sup\left(S\right)=\gamma we may equivalently write \diam\left(E\right)=\gamma...Is ther a code that gives \text{diam}\left(E\right)?
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  6. #36
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Let X be a metric space, and let E\subset{X}. Now define S\equiv\left\{d(p,q):q,p\in{E}\right\}, now if \sup\left(S\right)=\gamma we may equivalently write \diam\left(E\right)=\gamma...Is ther a code that gives \text{diam}\left(E\right)?
    From what I know, no. One can define this command...but you can't do so here on MHF. If you had a LaTeX editor [like TeXnic Center], you can post the following into the preamble of the document:

    Code:
    \newcommand{\diam}[1]{\text{diam}\left(#1\right)}
    And then when you type \diam E, it gives you \text{diam}\left(E\right)

    --Chris
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  7. #37
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    From what I know, no. One can define this command...but you can't do so here on MHF. If you had a LaTeX editor [like TeXnic Center], you can post the following into the preamble of the document:

    Code:
    \newcommand{\diam}[1]{\text{diam}\left(#1\right)}
    And then when you type \diam E, it gives you \text{diam}\left(E\right)

    --Chris
    Haha, that is a lot of work. I think I will just stick with \text{diam}\left(E\right)

    Thanks though!
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  8. #38
    Super Member flyingsquirrel's Avatar
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    \begin{pmatrix}<br />
a_{1,1} & a_{2,1} & \ldots & a_{n,1}\\<br />
a_{2,1} & a_{2,2} & \ldots & a_{n,2}\\<br />
\hdotsfor[1.7]{4}\\<br />
a_{n,1} & a_{2,n} & \ldots & a_{n,n}<br />
\end{pmatrix}

    \begin{pmatrix}<br />
a_{1,1} & a_{2,1} & \ldots & a_{n,1}\\<br />
a_{2,1} & a_{2,2} & \ldots & a_{n,2}\\<br />
\vdots & \vdots & \ddots & \vdots \\<br />
a_{n,1} & a_{2,n} & \ldots & a_{n,n}<br />
\end{pmatrix}
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  9. #39
    Rhymes with Orange Chris L T521's Avatar
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    <br />
\begin{matrix}0&1\\1&0\end{matrix} \hspace{1.5pc}<br />
\begin{pmatrix}0&1\\1&0\end{pmatrix} \hspace{1.5pc}<br />
\begin{bmatrix}0&1\\1&0\end{bmatrix} \hspace{1.5pc}<br />
\begin{Bmatrix}0&1\\1&0\end{Bmatrix} \hspace{1.5pc}<br />
\begin{vmatrix}0&1\\1&0\end{vmatrix} \hspace{1.5pc}<br />
\begin{Vmatrix}0&1\\1&0\end{Vmatrix}<br /> <br />
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  10. #40
    Member Greengoblin's Avatar
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    \sum_{k=0}^n

    \sum_{k=0}^n a^k

    \sum_{r=0}^n a^r

    <br />
\sum_{r=0}^n\ ^nC_r a^r

    \sum_{r=0}^n ^nC_r a^rb^{n-r}

    (x+y)^n

    \(x+y\)^n
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  11. #41
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    <br />
-e^{-x^2}<br />

    equals 0
    Last edited by iLikeMaths; December 14th 2008 at 06:23 AM.
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  12. #42
    MHF Contributor Mathstud28's Avatar
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    \int\limits_-\cdots \int\limits_\text{ }^-\cdots\int\limits_a^b

    Math \int\limits_- f~d\alpha=\sup_{P}L\left(P,f,\alpha\right)
    Last edited by Mathstud28; December 31st 2008 at 12:52 AM.
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  13. #43
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    . . . \frac{1{\color{red}\rlap{/}}6}{{\color{red}\rlap{/}}64} \;=\;\frac{1}{4} \qquad\qquad \frac{1{\color{red}\rlap{/}}9}{{\color{red}\rlap{/}}95} \;=\;\frac{1}{5} . . . . . \frac{2{\color{red}\rlap{/}}6}{{\color{red}\rlap{/}}65} \;=\;\frac{2}{5}\qquad\qquad \frac{4{\color{red}\rlap{/}}9}{{\color{red}\rlap{/}}98} \;=\;\frac{4}{8}\;=\;\frac{1}{2}



    . . . . . \frac{1-x^{{\color{red}\rlap{/}}2}}{(1+x)^{{\color{red}\rlap{/}} 2} }\;=\;\frac{1-x}{1+x}



    All that work for a bad joke . . .


    (Good luck trying to read my LaTeX code.)
    .
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  14. #44
    Member Last_Singularity's Avatar
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    A fairly comprehensive guide to LaTeX here:

    http://tobi.oetiker.ch/lshort/lshort.pdf

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  15. #45
    Eater of Worlds
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    Just trying something fancy in LaTex. This came from that tutorial.

    Thanks for that link.


    \text{Heron's formula}\setlength{\unitlength}{1cm}<br />
\begin{picture}(6,5)<br />
\thicklines<br />
\put(1,0.5){\line(2,1){3}}<br />
\put(4,2){\line(-2,1){2}}<br />
\put(2,3){\line(-2,-5){1}}<br />
\put(0.7,0.3){$A$}<br />
\put(4.05,1.9){$B$}<br />
\put(1.7,2.95){$C$}<br />
\put(3.1,2.5){$a$}<br />
\put(1.3,1.7){$b$}<br />
\put(2.5,1.05){$c$}<br />
\put(0.3,4){$F=<br />
\sqrt{s(s-a)(s-b)(s-c)}$}<br />
\put(3.5,0.4){$\displaystyle<br />
s:=\frac{a+b+c}{2}$}<br />
\end{picture}
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