trying out LATEX

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March 21st 2009, 05:44 PM
manyarrows

trying out LATEX

$\int$ x dx = $\frac{x^2}{2}$ +C
March 23rd 2009, 06:03 PM
lmk866

This is what must be typed: [tex]\int x ~dx = \frac{x^2} {2} + C[/tex]

the "~" adds a space. alternatively, you can use "\," or "\;"
March 23rd 2009, 06:40 PM
Reckoner

It is probably best to include the whole expression in one set of [tex][/tex] tags. lmk866 has it.

Quote:

Originally Posted by lmk866

the "~" adds a space. alternatively, you can use "\," or "\;"

\, is the one that is typically used for this purpose. The other two give a bit more space:

$a\,a$ : [tex]a\,a[/tex]
$a~a$ : [tex]a~a[/tex]:
$a\;a$ : [tex]a\;a[/tex]
April 27th 2009, 10:21 PM
fardeen_gen

Test

\sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}\left[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)\right]8}}

This isn't working! Why?

EDIT:
Thanks Opalg! Also, eight shouldn't be there.
April 28th 2009, 02:58 AM
Opalg

Quote:

Originally Posted by fardeen_gen

\sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}\left[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)\right]8}}

This isn't working! Why?

I don't know why, but for some reason the compiler doesn't like the unnecessary braces that you have used to wrap around some sections of the formula. Remove the red ones in this code
\sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}\left[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)\right]8}}
and you get $\sum_{n = 0}^{ \infty } \frac{1}{n!} \left[\sum_{k = 0}^{n}(k + 1)\left(\int_0^1 2^{-(k + 1)x}\, dx\right)\right]8$

(And it looks as though the 8 probably shouldn't be there at all?)

Also note that you don't need to use \limits with \sum. They are supplied automatically.
April 29th 2009, 04:56 PM
lmk866

i am not exactly sure about the equation you are trying to write but here are the syntax problems

correct form
\sum_{n = 0}^{\infty} {\frac{1}{n!} \left[ \sum_{k = 0}^{n} {(k+1)\left( \int_{0}^{1} 2^{-(k + 1)x} dx \right)} \right]}

problems
1 - \dx
2 - the location of closed }

note: you can eliminate the \limits. \sum_{lowerbound}^{upperbound} works

i hope this helps..
April 29th 2009, 05:00 PM
chella182

The .pdf file in this thread is really helpful for LaTex codes I find :)
August 31st 2009, 10:01 AM
edutabacman

how about this $\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ here, or

$\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

like this?
August 31st 2009, 10:02 AM
edutabacman

well, \n $\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ \n
eh?
August 31st 2009, 10:05 AM
edutabacman

or, better
$\frac{{n!}}{{r!\left( {n - r} \right)!}}$
like $\frac{{n!}}{{r!\left( {n - r} \right)!}}$ this?
August 31st 2009, 10:08 AM
edutabacman

but
$\begin{array}{*{20}{c}} {\sqrt {{b^2} - 4ac} } & {\frac{1}{2}} \\ {\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0} & {\sum\limits_{i = 0}^\infty {{x_i}} } \\ \end{array}$
August 31st 2009, 10:09 AM
edutabacman

$\left[ {\begin{array}{*{20}{c}} {\sqrt {{b^2} - 4ac} } & {\frac{1}{2}} \\ {\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0} & {\sum\limits_{i = 0}^\infty {{x_i}} } \\ \end{array}} \right]$
August 31st 2009, 10:40 AM
edutabacman

but <latex>$\left[ {\begin{array}{*{20}{c}}
{\sum\limits_{i = 1}^n {{X_i}{Y_i}} } & {\frac{{x - \mu }}{\sigma }} \\
{\frac{{{\partial ^2}\Omega }}{{\partial u\partial v}}} & {\sqrt {{b^2} - 4ac} } \\
\end{array}} \right]$</latex>
August 31st 2009, 10:41 AM
edutabacman

I know
$\left[ {\begin{array}{*{20}{c}} {\sum\limits_{i = 1}^n {{X_i}{Y_i}} } & {\frac{{x - \mu }}{\sigma }} \\ {\frac{{{\partial ^2}\Omega }}{{\partial u\partial v}}} & {\sqrt {{b^2} - 4ac} } \\ \end{array}} \right]$
right?