This is my first question on tex.stackexchange, so the formatting will be terrible.

I've got a problem with three lengthy equations - the closing brackets, in all three formulas, are not displayed.

Technical info - MikTex x64 (2.9), latest TeXnicCenter (Beta 1 I think it is), Windows 8 x64. In TeXnicCenter I have selected XeLaTeX -> PDF, as it is the optimal way for display of my language.

My document settings are following:

\usepackage[top=2.5cm, left=3cm, bottom=2.5cm, right=2.5cm]{geometry}
\setmainfont{Times New Roman}
\setotherlanguages{english, russian}

And the equations are:

F'=\delta m\left\{f\frac{M_1l}{r^3}\left[3\cos^2\phi\cos^2(\lambda-D)-1\right]+f\frac{M_2l}{R^3_1}\left(3\cos^2\phi\cos^2\lambda-1\right)-\\-\frac{1}{2}f\frac{M_2r_1}{R^3_1}\cos\phi\cos\left(\lambda-D\right)\left(1+3\cos2D\right)-\frac{3}{2}f\frac{M_2r_1}{R^3_1}\cos\phi\sin\left(\lambda-D\right)\sin2D\right\}
F^n=\delta m\left\{-\frac{3}{2}f\frac{M_1l}{r^3}\sin 2\phi\cos^2\left(\lambda-D\right)-\frac{3}{2}f\frac{M_2l}{R^3_1}\sin 2\phi\cos^2\lambda-\\-\frac{1}{2}f\frac{M_2r_1}{R^3_1}\sin\phi\left[3\sin\left(\lambda-D\right)\sind 2D-\cos\left(\lambda-D\right)\left(1_3\cos 2D\right)\right]\right\}
F^m=\sigma m\left[-\frac{3}{2}f\frac{M_1l}{r^3}\cos\phi\sin 2\left(\lambda-D\right)-\frac{3}{2}f\frac{M_2l}{R^3_1}\cos\phi\sin 2\lambda+\\+\frac{1}{2}f\frac{M_2}{R^3_1}r_1\sin\left(\lambda-D\right)\left(1+3\cos 2D\right)+\frac{3}{2}f\frac{M_2}{R^3_1}r_1\cos\left(\lambda-D\right)\sin 2D\right]

I've made a screenshot from TeXworks (PDF mode) with the problem -

enter image description here

And I've also posted my compile log to the pastebin - http://pastebin.com/CEZ2EUjU

And what I have noticed - each equation has added me, approximately, 12 new errors.

  • Welcome to TeX.SX! Usually, we don't put a greeting or a “thank you” in our posts. While this might seem strange at first, it is not a sign of lack of politeness, but rather part of our trying to keep everything very concise. Accepting and upvoting answers is the preferred way here to say “thank you” to users who helped you. – Marco Daniel May 14 '13 at 16:11
  • 1
    You can't have \left in one equation and \right in a different one. – egreg May 14 '13 at 16:13

You can't use \left in one line of a split and \right in another one. You should also use an align* environment (and don't use redundant \left and \right); the big delimiters must be set by hand.

F'&=\delta m\biggl\{f\frac{M_1l}{r^3}\bigl[3\cos^2\phi\cos^2(\lambda-D)-1\bigr]
    +f\frac{M_2l}{R^3_1}(3\cos^2\phi\cos^2\lambda-1)-{} \\
   -\frac{3}{2}f\frac{M_2r_1}{R^3_1}\cos\phi\sin(\lambda-D)\sin2D\biggr\} \\[2ex]
F^n&=\delta m\biggl\{-\frac{3}{2}f\frac{M_1l}{r^3}\sin 2\phi\cos^2(\lambda-D)
    -\frac{3}{2}f\frac{M_2l}{R^3_1}\sin 2\phi\cos^2\lambda-{} \\
    \cos(\lambda-D)(1_3\cos 2D)\bigl]\biggl\} \\[2ex]
F^m&=\sigma m\biggl[-\frac{3}{2}f\frac{M_1l}{r^3}\cos\phi\sin 2(\lambda-D)
    -\frac{3}{2}f\frac{M_2l}{R^3_1}\cos\phi\sin 2\lambda+{} \\
   &\qquad{}+\frac{1}{2}f\frac{M_2}{R^3_1}r_1\sin(\lambda-D)(1+3\cos 2D)
    +\frac{3}{2}f\frac{M_2}{R^3_1}r_1\cos(\lambda-D)\sin 2D\biggr]

enter image description here

  • First of all I did an align*, so the three equals sign can be aligned to each other. It's not good to stack \[...\] formulas.

  • I removed all the inner \left and \right that do nothing except adding unwanted spaces. However, I increased the size of a [...] pair to make clearer their correspondence (it's in the second equation; there's no need for doing this in the first formula)

  • In order to make clear that each second line is a continuation, I added a \qquad of space to push it to the right of the alignment point.

  • Before or after the "isolated" minus or plus signs, I put {} in order to get correct spacing, otherwise they would not work as binary operation because of how TeX determines the difference between $-1$ and $2-1$.

  • Most important, I set by hand the size of the main delimiters, because so you have full control over them even if they are in different lines.

As a side note, I wouldn't repeat the operation sign at the break point; it's a bad habit of Russian typography, that's not used much in Western countries. I find it distracting and ambiguous: in the first equation is it "minus minus" that makes "plus"? It isn't, I know, but why repeating it? The reader finds the break, goes on the next line where it's clear that the formula continues.

  • Interesting, I'll have to look into it. Your result looks way better than mine, albeit I do not understand sever parts of it. – user30709 May 14 '13 at 16:33
  • Thanks, I've read the edited answer and now it is clear on how do I improve my equations. I have a question on \[...\] - what do they do, in fact. I just put the there because the TeXnicCenter does so, if I use 'Formula' button. As for the operation sign at break point - it is USSR legacy that I have to stick to, as it is how professors here write it as well. – user30709 May 14 '13 at 17:32

The -\\- parts are main source of the problem. You should add closing \right's, in particular \right., before \\, and opening \left's after it.

  • 1
    This works because, by chance, the second part of the formula has the same size as the first one. It wouldn't give the same size of the delimiters in other cases. Plus it's wrong to use three consecutive \[...\] environments. – egreg May 14 '13 at 16:33
  • @egreg Certainly. And, additionally, instead of (1_3\cos 2D) should be (1-3\cos 2D). (And I think your solution should be accepted). – Przemysław Scherwentke May 14 '13 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy