# Math Help - trying out LATEX

1. ## trying out LATEX

$\int$ x dx = $\frac{x^2}{2}$+C

2. This is what must be typed: $$\int x ~dx = \frac{x^2} {2} + C$$

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

3. It is probably best to include the whole expression in one set of  tags. lmk866 has it.

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:

$a\,a$: $$a\,a$$
$a~a$: $$a~a$$:
$a\;a$: $$a\;a$$

4. ## 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.

5. 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.

6. 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..

7. The .pdf file in this thread is really helpful for LaTex codes I find

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

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

like this?

9. well, \n $\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$\n
eh?

10. or, better
$\frac{{n!}}{{r!\left( {n - r} \right)!}}$
like $\frac{{n!}}{{r!\left( {n - r} \right)!}}$this?

11. but
$\begin{array}{*{20}{c}}
{\sqrt {{b^2} - 4ac} } & {\frac{1}{2}} \\
{\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0} & {\sum\limits_{i = 0}^\infty {{x_i}} } \\
\end{array}$

12. $\left[ {\begin{array}{*{20}{c}}
{\sqrt {{b^2} - 4ac} } & {\frac{1}{2}} \\
{\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0} & {\sum\limits_{i = 0}^\infty {{x_i}} } \\
\end{array}} \right]$

13. 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>

14. 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?