Results 1 to 14 of 14

Math Help - trying out LATEX

  1. #1
    Junior Member
    Joined
    Mar 2009
    From
    Kentucky
    Posts
    72

    trying out LATEX

    \int x dx = \frac{x^2}{2}+C
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Mar 2009
    Posts
    2
    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 "\;"
    Last edited by Jhevon; March 23rd 2009 at 06:12 PM. Reason: clarified
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

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

    Quote Originally Posted by lmk866 View Post
    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]
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member fardeen_gen's Avatar
    Joined
    Jun 2008
    Posts
    539

    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.
    Last edited by fardeen_gen; April 28th 2009 at 05:14 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by fardeen_gen View Post
    \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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2009
    Posts
    2
    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..
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member chella182's Avatar
    Joined
    Jan 2008
    Posts
    267
    The .pdf file in this thread is really helpful for LaTex codes I find
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Aug 2009
    Posts
    7
    how about this \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} here, or

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

    like this?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Aug 2009
    Posts
    7
    well, \n \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\n
    eh?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Aug 2009
    Posts
    7
    or, better
    \frac{{n!}}{{r!\left( {n - r} \right)!}}
    like \frac{{n!}}{{r!\left( {n - r} \right)!}}this?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Aug 2009
    Posts
    7
    but
    \begin{array}{*{20}{c}}<br />
   {\sqrt {{b^2} - 4ac} } & {\frac{1}{2}}  \\<br />
   {\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0} & {\sum\limits_{i = 0}^\infty  {{x_i}} }  \\<br />
\end{array}
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    Aug 2009
    Posts
    7
    \left[ {\begin{array}{*{20}{c}}<br />
   {\sqrt {{b^2} - 4ac} } & {\frac{1}{2}}  \\<br />
   {\mathop {\lim }\limits_{x \to \infty } \frac{1}{x} = 0} & {\sum\limits_{i = 0}^\infty  {{x_i}} }  \\<br />
\end{array}} \right]
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Aug 2009
    Posts
    7
    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>
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Newbie
    Joined
    Aug 2009
    Posts
    7
    I know
    \left[ {\begin{array}{*{20}{c}}<br />
   {\sum\limits_{i = 1}^n {{X_i}{Y_i}} } & {\frac{{x - \mu }}{\sigma }}  \\<br />
   {\frac{{{\partial ^2}\Omega }}{{\partial u\partial v}}} & {\sqrt {{b^2} - 4ac} }  \\<br />
\end{array}} \right]
    right?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 10th 2011, 06:40 PM
  2. LaTex inside of LaTex
    Posted in the LaTeX Help Forum
    Replies: 2
    Last Post: November 12th 2009, 11:07 AM
  3. Some little help more with Latex, please
    Posted in the LaTeX Help Forum
    Replies: 3
    Last Post: October 29th 2009, 09:30 AM
  4. Latex Help
    Posted in the LaTeX Help Forum
    Replies: 1
    Last Post: February 16th 2009, 09:48 AM
  5. Latex help
    Posted in the LaTeX Help Forum
    Replies: 1
    Last Post: February 8th 2008, 10:57 AM

Search Tags


/mathhelpforum @mathhelpforum