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Math Help - Testing LaTeX

  1. #91
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    Quote Originally Posted by Mathstud28 View Post
    Hmm, this is interesting. I have never seen this identity before. I assume it is derived by converting the inverse sine function to the complex logarithmic form. But anyways, very interesting.

    Also note that

    \leq is gotten by \leq
    Or \le : \le

    The other one is \ge \ge (or \geq \geq)


    (he has posted a topic for this)
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  2. #92
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    How would you do qunitic integration?

    \iiiint\int{dxdydzdtdu}
    You can add "negative space" with \!. Try this:

    Code:
    \iiiint\!\!\!\int dx\,dy\,dz\,dt\,du
    which gives

    \iiiint\!\!\!\int dx\,dy\,dz\,dt\,du

    Also, note the use of the \, spacer. I don't think I have ever seen you use it, but it will make your integrals a bit nicer by putting space between the integrand and the variable of integration. e.g.

    Instead of \int xdx, you'd get \int x\,dx.
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  3. #93
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Reckoner View Post
    You can add "negative space" with \!. Try this:

    Code:
    \iiiint\!\!\!\int dx\,dy\,dz\,dt\,du
    which gives

    \iiiint\!\!\!\int dx\,dy\,dz\,dt\,du

    Also, note the use of the \, spacer. I don't think I have ever seen you use it, but it will make your integrals a bit nicer by putting space between the integrand and the variable of integration. e.g.

    Instead of \int xdx, you'd get \int x\,dx.
    I didnt think about using the classic \!..!

    and when I space I usually just do either ~ or \quad
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  4. #94
    Rhymes with Orange Chris L T521's Avatar
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    \mathop{\int\mkern-20.8mu \circlearrowleft}\nolimits_C <br />
 {{\mathbf{F}} \cdot d{\mathbf{r}}}  = \iint\limits_R {{\text{curl}}\,{\mathbf{F}} \cdot {\mathbf{k}}\,dA}<br />
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  5. #95
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    \mathop{\int\mkern-20.8mu \circlearrowleft}\nolimits_C <br />
{{\mathbf{F}} \cdot d{\mathbf{r}}} = \iint\limits_R {{\text{curl}}\,{\mathbf{F}} \cdot {\mathbf{k}}\,dA}<br />
    Wow thats a complicated for \oint_{C}\mathbf{F}\cdot{d\mathbf{r}}

    Or even less letters

    \oint_{C}\bold{F}\cdot{d\bold{r}}
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  6. #96
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Wow thats a complicated for \oint_{C}\mathbf{F}\cdot{d\mathbf{r}}

    Or even less letters

    \oint_{C}\bold{F}\cdot{d\bold{r}}
    Actually, I believe he was trying to express a counterclockwise contour integral (look closely and notice the arrow).

    I found this really old post on comp.text.tex by Donald Arseneau that provides a method to get it to look right, but I do not think these forums have the right packages.
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  7. #97
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Reckoner View Post
    Actually, I believe he was trying to express a counterclockwise contour integral (look closely and notice the arrow).

    I found this really old post on comp.text.tex by Donald Arseneau that provides a method to get it to look right, but I do not think these forums have the right packages.
    Hmm..I still do not see an arrow indicating direction.

    <br /> <br />
\def\Xint#1{\mathchoice <br />
{\XXint\displaystyle\textstyle{#1}}% <br />
{\XXint\textstyle\scriptstyle{#1}}% <br />
{\XXint\scriptstyle\scriptscriptstyle{#1}}% <br />
{\XXint\scriptscriptstyle\scriptscriptstyle{#1}}% <br />
\!\int} <br />
\def\XXint#1#2#3{{\setbox0=\hbox{$#1{#2#3}{\int}$} <br />
\vcenter{\hbox{$#2#3$}}\kern-.5\wd0}} <br /> <br />
\def\dashint{\Xint-} <br />
\def\ddashint{\Xint=} <br />
\def\clockint{\Xint\circlearrowright} <br />
\def\counterint{\Xint\rotcirclearrowleft} <br />
\def\rotcirclearrowleft{\mathpalette{\RotSymbol{-30}}\circlearrowleft} <br />
\def\RotSymbol#1#2#3{\rotatebox[origin=c]{#1}{$#2#3$}}
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  8. #98
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Or even less letters

    \oint_{C}\bold{F}\cdot{d\bold{r}}
    One can also get the same result without using any bracket :
    Code:
    \oint_C \bold F \cdot d \bold r
    gives \oint_C \bold F \cdot d \bold r
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  9. #99
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    Quote Originally Posted by flyingsquirrel View Post
    One can also get the same result without using any bracket :
    Code:
    \oint_C \bold F \cdot d \bold r
    gives \oint_C \bold F \cdot d \bold r
    Even smaller without the spaces:

    Code:
    \oint_C\bold F\cdot d\bold r
    gives \oint_C\bold F\cdot d\bold r
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  10. #100
    Rhymes with Orange Chris L T521's Avatar
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    Code:
    \mathop{\int\mkern-20.8mu \circlearrowleft}\nolimits_C 
     {{\mathbf{F}} \cdot d{\mathbf{r}}}  
    = \iint\limits_R {{\text{curl}}
    \,{\mathbf{F}} \cdot {\mathbf{k}}\,dA}
    Quote Originally Posted by Mathstud28 View Post
    Wow thats a complicated for \oint_{C}\mathbf{F}\cdot{d\mathbf{r}}

    Or even less letters

    \oint_{C}\bold{F}\cdot{d\bold{r}}
    Note the part in red. The intention was for a counterclockwise line integral...

    You probably don't see the arrow because the arrow is at the top of the circle: \circlearrowleft

    ...The integral sign blocks it...
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  11. #101
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    Thread Closed.
    Getting too long, start a Part II continuation.
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