4
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In this formula:

$$\frac{1}{\binom{n}{1}}+\frac{1}{\binom{n}{2}}+\frac{1}{\binom{n}{3}}+\cdots+\frac{1}{\binom{n}{n}}=?$$

The $\binom{n}{3}$ renders much larger than the other \binom expressions. If I change the 3 to a 2 it looks correct.

Screenshot:

enter image description here

Here's my browser version information:

  • Google Chrome: 19.0.1084.46 (Official Build 135956)
  • OS: Linux
  • WebKit: 536.5 (@116441)
  • JavaScript: V8 3.9.24.21
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  • $\begingroup$ I get the same effect (OS X 10.6.8, FF 12.0) when I right click on the math and choose SVG under Math Settings -> Math Renderer. With HTML-CSS and MathML it looks fine. It has likely to do with the fact that the glyph of 3 larger than the ones of 1,2 and n. $\endgroup$ – t.b. May 30 '12 at 13:26
  • $\begingroup$ I didn't know that one could right-click a formula, thanks! I was using HTML_CSS; switching my renderer to SVG does not fix the problem. Screenshot on request. $\endgroup$ – MJD May 30 '12 at 13:31
  • $\begingroup$ I can reproduce on Debian linux, IceWeasel version 10.0.4. The chosen renderer is already HTML-CSS. $\endgroup$ – Willie Wong May 30 '12 at 13:38
  • $\begingroup$ @t.b. While the MathML corrects the problem it does not look fine! :-) $\endgroup$ – Asaf Karagila May 30 '12 at 13:38
3
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The problem is with the digit; neither 1 nor 2 cause the problem, but switch it to 3, 5, 6, 7, 8, 9, or 0 (in any denominator) and it looks similar; e.g., $$\frac{1}{\binom{n}{1}}+\frac{1}{\binom{n}{3}}+\frac{1}{\binom{n}{2}}+\cdots+\frac{1}{\binom{n}{8}}=?$$

A workaround is to use \smash on the 3 (or any items that cause the issue), so force $\LaTeX$ to think the box with the 3 is "standard" size. Thus,

\frac{1}{\binom{n}{1}}+\frac{1}{\binom{n}{2}}+\frac{1}{\binom{n}{\smash{3}}}+\cdots+\frac{1}{\binom{n}{n}}=?

produces $$\frac{1}{\binom{n}{1}}+\frac{1}{\binom{n}{2}}+\frac{1}{\binom{n}{\smash{3}}}+\cdots+\frac{1}{\binom{n}{n}}=?$$

Note that the problem is not related to the denominators, but rather to the rendering of the binomial coefficient when not in displaymode. For example, $\binom{n}{1}\binom{n}{2}\binom{n}{3}\binom{n}{4}\binom{n}{5}\binom{n}{6}\binom{n}{7}\binom{n}{8}\binom{n}{9}\binom{n}{0}$ gives

$\binom{n}{1}\binom{n}{2}\binom{n}{3}\binom{n}{4}\binom{n}{5}\binom{n}{6}\binom{n}{7}\binom{n}{8}\binom{n}{9}\binom{n}{0}$

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  • $\begingroup$ Thanks. I seem to remember that DEK guaranteed that all digit symbols would be the same height, to avoid problems like this. I don't know how MathJax chooses box heights. Do you suppose it ought to enforce a similar rule? $\endgroup$ – MJD May 30 '12 at 20:00
  • $\begingroup$ @Mark What's DEK? Just curious. $\endgroup$ – Dylan Moreland May 30 '12 at 23:57
  • $\begingroup$ Donald E. Knuth, inventor of $\TeX$. $\endgroup$ – MJD May 31 '12 at 0:38
  • $\begingroup$ I was remembering wrong anyway; the digits usually not always) have the same width. Of course they often don't have the same height; for example. $\endgroup$ – MJD May 31 '12 at 0:49

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