x dx = +C

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- Mar 21st 2009, 06:44 PMmanyarrowstrying out LATEX
x dx = +C

- Mar 23rd 2009, 07:03 PMlmk866
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 "\;" - Mar 23rd 2009, 07:40 PMReckoner
- Apr 27th 2009, 11:21 PMfardeen_genTest
\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. - Apr 28th 2009, 03:58 AMOpalg
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

(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. - Apr 29th 2009, 05:56 PMlmk866
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.. - Apr 29th 2009, 06:00 PMchella182
The .pdf file in this thread is really helpful for LaTex codes I find :)

- Aug 31st 2009, 11:01 AMedutabacman
how about this here, or

like this? - Aug 31st 2009, 11:02 AMedutabacman
well, \n \n

eh? - Aug 31st 2009, 11:05 AMedutabacman
or, better

like this? - Aug 31st 2009, 11:08 AMedutabacman
but

- Aug 31st 2009, 11:09 AMedutabacman
- Aug 31st 2009, 11:40 AMedutabacman
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> - Aug 31st 2009, 11:41 AMedutabacman
I know

right?