# Testing LaTeX

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• Jul 1st 2008, 08:51 AM
Moo
Quote:

Originally Posted by Mathstud28
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

$\displaystyle \leq$ is gotten by \leq

Or \le : $\displaystyle \le$

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

(he has posted a topic for this)
• Jul 1st 2008, 12:39 PM
Reckoner
Quote:

Originally Posted by Mathstud28
How would you do qunitic integration?

$\displaystyle \iiiint\int{dxdydzdtdu}$

You can add "negative space" with \!. Try this:

Code:

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

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

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 $\displaystyle \int xdx$, you'd get $\displaystyle \int x\,dx$.
• Jul 1st 2008, 01:31 PM
Mathstud28
Quote:

Originally Posted by Reckoner
You can add "negative space" with \!. Try this:

Code:

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

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

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 $\displaystyle \int xdx$, you'd get $\displaystyle \int x\,dx$.

I didnt think about using the classic \!..!

and when I space I usually just do either ~ or \quad
• Jul 1st 2008, 09:38 PM
Chris L T521
$\displaystyle \mathop{\int\mkern-20.8mu \circlearrowleft}\nolimits_C {{\mathbf{F}} \cdot d{\mathbf{r}}} = \iint\limits_R {{\text{curl}}\,{\mathbf{F}} \cdot {\mathbf{k}}\,dA}$
• Jul 1st 2008, 09:54 PM
Mathstud28
Quote:

Originally Posted by Chris L T521
$\displaystyle \mathop{\int\mkern-20.8mu \circlearrowleft}\nolimits_C {{\mathbf{F}} \cdot d{\mathbf{r}}} = \iint\limits_R {{\text{curl}}\,{\mathbf{F}} \cdot {\mathbf{k}}\,dA}$

Wow thats a complicated for $\displaystyle \oint_{C}\mathbf{F}\cdot{d\mathbf{r}}$ :p

Or even less letters

$\displaystyle \oint_{C}\bold{F}\cdot{d\bold{r}}$
• Jul 1st 2008, 10:40 PM
Reckoner
Quote:

Originally Posted by Mathstud28
Wow thats a complicated for $\displaystyle \oint_{C}\mathbf{F}\cdot{d\mathbf{r}}$ :p

Or even less letters

$\displaystyle \oint_{C}\bold{F}\cdot{d\bold{r}}$

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.
• Jul 1st 2008, 10:49 PM
Mathstud28
Quote:

Originally Posted by Reckoner
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.

$\displaystyle \def\Xint#1{\mathchoice {\XXint\displaystyle\textstyle{#1}}% {\XXint\textstyle\scriptstyle{#1}}% {\XXint\scriptstyle\scriptscriptstyle{#1}}% {\XXint\scriptscriptstyle\scriptscriptstyle{#1}}% \!\int} \def\XXint#1#2#3{{\setbox0=\hbox{$#1{#2#3}{\int}$} \vcenter{\hbox{$#2#3$}}\kern-.5\wd0}} \def\dashint{\Xint-} \def\ddashint{\Xint=} \def\clockint{\Xint\circlearrowright} \def\counterint{\Xint\rotcirclearrowleft} \def\rotcirclearrowleft{\mathpalette{\RotSymbol{-30}}\circlearrowleft} \def\RotSymbol#1#2#3{\rotatebox[origin=c]{#1}{$#2#3$}}$
• Jul 1st 2008, 10:57 PM
flyingsquirrel
Quote:

Originally Posted by Mathstud28
Or even less letters

$\displaystyle \oint_{C}\bold{F}\cdot{d\bold{r}}$

One can also get the same result without using any bracket :
Code:

\oint_C \bold F \cdot d \bold r
gives $\displaystyle \oint_C \bold F \cdot d \bold r$
• Jul 1st 2008, 11:01 PM
Reckoner
Quote:

Originally Posted by flyingsquirrel
One can also get the same result without using any bracket :
Code:

\oint_C \bold F \cdot d \bold r
gives $\displaystyle \oint_C \bold F \cdot d \bold r$

Even smaller without the spaces:

Code:

\oint_C\bold F\cdot d\bold r
gives $\displaystyle \oint_C\bold F\cdot d\bold r$
• Jul 1st 2008, 11:57 PM
Chris L T521
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
Wow thats a complicated for $\displaystyle \oint_{C}\mathbf{F}\cdot{d\mathbf{r}}$ :p

Or even less letters

$\displaystyle \oint_{C}\bold{F}\cdot{d\bold{r}}$

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: $\displaystyle \circlearrowleft$

...The integral sign blocks it...
• Jul 2nd 2008, 10:08 AM
ThePerfectHacker