# Some Cute Math

Back before my now-ended blogging hiatus, the server machinery that keeps ScienceBlogs running was not so snazzy as it is now. Now it’s running a WordPress implementation that includes LaTeX support. LaTeX is a free environment for (among other things) typesetting mathematics. Let’s give it a test run:

$3 = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$

I actually came across the above expression in a Mathematica book, which used it as an example of how Mathematica could do typesetting. It didn’t include anything other than the expression – no proof or anything else. We might as well test it out to see if it’s true. First, take a pocket calculator and start with the number 6. Take the square root and get (2.44949…). Add 6, and get (8.44949…). Take the square root and get (2.9068…). Add 6 and take the square root and get 2.98443…

Keep going and it sure looks like you’re closing in on 3, which is a good sign. But that’s not a proof. We could make our train of thought more systematic. We’re talking about a sequence of numbers $S_n$, where each $S_n$ in the sequence is given by the definition:

$S_{1} = \sqrt{6}$
$S_{n+1} = \sqrt{6 + S_n}$

Well, if this sequence converges, we’re just saying that each successive $S_{n}$ starts to close in on the final result $S$.

$\displaystyle \lim_{n \to \infty}S_{n} = S$

So the defining relationship becomes:

$S = \sqrt{6 + S}$

Which we can solve by inspection (or by algebra, if you prefer): $S = 3$.

Now notice that I said this logic was valid if and only if the sequence converges. Does it? Yes, but I’ll leave proving it as a challenge for people who’re looking for Calc 2 practice problems.

Hey, that wasn’t bad. This post took me massively less time than it would have back when I had to generate each equation image separately and upload it to the website – especially if I discovered a typo. I’m not sure if LaTeX works in comments, but if you’d like to give it a shot the WordPress-specific syntax is here. (Also for new readers: I put math posts in the Physical Science category because ScienceBlogs doesn’t have a dedicated math category.)

1. #1 david
September 14, 2012

nice.

by induction, S(n+1) is < 3, and so the sequence is bounded above. it's non-decreasing. therefore…

2. #2 Eric Lund
September 14, 2012

You can generate a similar series for any positive integer greater than 1: if the desired limit is n, then the number in the radical is n(n-1). Which means, for instance, that (LaTeX experiment follows) $2 = \sqrt{2 + \sqrt{2 + \dots}}$. (In case that didn’t work: for n = 2 the number in the radical is 2.)

It clearly fails for n = 1, because every term in the sequence is 0, but it works for n = 0 for the same reason.

Generalizing to the real numbers, it works for all n > 1, and fails for all n < 0 (if you restrict yourself to positive square roots). The range 0 < n < 1 is tricky because you are dealing with square roots of complex numbers.

3. #3 Matt Springer
September 14, 2012

Hey, LaTeX does work in the comments!

Another version of Eric’s argument is this: we know these expressions satisfy $S = \sqrt{n + S}$, and so we can generate the appropriate n by picking an S and calculating n via $S^2 - S = n$.

4. #4 Jon Wharf
September 14, 2012

or vice versa: we can solve for S, given n(>0):
$S = \frac{1+\sqrt{1+4n}}{2}$

For the better-behaved sequence case arising when taking $S_1=1$, we can allow $n \geq -0.25$ as expected from the quadratic solution (which takes us to the minimum of $S^2-S$) – even though that doesn’t look pretty in the original formulation.

5. #5 Tualha
October 5, 2012

I wonder if $\LaTeX$ works in the comments? Or $\LaTeXe$?