# Does MathJaX have a command to stretch or reduce vertical space?

The equations below: \begin{align} \sum_{i=1}^n F_{2i-1}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\ &=1+2+5+\cdots+F_{2n-1}\\ &=F_{2n}\\ \end{align} have a problem with too much space between the first and second lines—at least that's how it displays in Opera. The code is below:

\begin{align} \sum_{i=1}^n F_{2i-1}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\ &=1+2+5+\cdots+F_{2n-1}\\ &=F_{2n}\\ \end{align}


Is there a way to ajdust the vertical space in MathJaX?

Another possibility is to use \smash to stop MathJAX from adjusting the line spacing to accomodate mathematical expressions:

\begin{align} \smash{\sum_{i=1}^n}F_{2i-1}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\ &=1+2+5+\cdots+F_{2n-1}\\ &=F_{2n}\\ \end{align}

\begin{align} \smash{\sum_{i=1}^n}F_{2i-1}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\ &=1+2+5+\cdots+F_{2n-1}\\ &=F_{2n}\\ \end{align}

• \smash worked perfectly! Thank you! May 9, 2014 at 20:54
• I'm not sure why, but smash in your screenshot has adjusted the spacing within the sum (as compared to the other screenshots) in that the F is closer to the summation symbol. It doesn't do so in latex...might be a mathjax thing? May 9, 2014 at 20:59
• Also, just FYI (not necessarily for your information, but for others reading that might not know) I'm not sure if they're implemented in mathjax, but the latex smash has optional arguments: \smash[t]{stuff} has zero height above the baseline, while \smash[b]{stuff} has zero depth below the baseline. In some cases you might want to retain the height or depth while "smashing" the other. May 9, 2014 at 21:07
• @Scott, that solves the other issue I had when I first implemented it&mdash;namely the collapsing of the line feed above the first line. Also helpful. Can we use \smash[r] or \smash[l] as well? May 9, 2014 at 21:16
• @BrianJ.Fink In latex there are \llap{stuff} and \rlap{stuff} but I'm not sure they work in mathmode (or if there are corresponding mathmode versions). The mathtools package provides \mathrlap etc. so you could try loading that package in mathjax to see if it works. May 9, 2014 at 21:23
• @Scott I meant "it--namely," but the dash didn't translate. May 9, 2014 at 21:23
• @Scott MathJaX does have \llap and \rlap, but they make the text completely overlap the text next to it. May 9, 2014 at 21:26
• @BrianJ.Fink That's the intent! You're "smashing" the width (?) of the text from the left/right in the same way that smash does vertically. The difference being that with vertical smashing you have the natural line height to provide some buffer space whereas horizontally you don't. May 9, 2014 at 21:32
• @Scott Good eye with that $F$ in the summation. I discovered that if I enclose the $F_{2i-1}$ along with the $\sum$, the issue is resolved. May 9, 2014 at 22:44

You can adjust the space in a line break with \\[10pt] \begin{align} \sum_{i=1}^n F_{2i-1}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\ &=1+2+5+\cdots+F_{2n-1}\\[10pt] &=F_{2n}\\ \end{align} Right-click on the rendered MathJax and choose "Show Math As > TeX Commands"

• This is very helpful. Do you mind if I essentially copy and paste the first line of this answer into the end of the answer about the align environment on the big Mathjax formatting thread? I think I searched the web only to be directed right back here. Jan 9, 2017 at 21:22
• Sure, however, David K has added a comment with a bit more detail. You might want to include some of that detail as well.
– robjohn Mod
Apr 11, 2017 at 15:29

The reason why the space is different is because of the $i=1$ under the sigma symbol. Let's write this without the $i=1$ and see how it would look. \begin{align} \sum^n F_{2i-1}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\ &=1+2+5+\cdots+F_{2n-1}\\ &=F_{2n}\\ \end{align} See, now the spacing is the same. Now I will show you how to make the spacing the same if it is ever different. \begin{align} \sum_{i=1}^n F_{2i-1}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\ &=1+2+5+\cdots+F_{2n-1}\\[10pt] &=F_{2n}\\ \end{align} Simply add \\[10pt] between the code for the second line and the third line to space out the lines. Note that it will not always be \\[10pt]. For something like $\lim\limits_{x\to 1}$, then it will be something else.