Some Cute Math


$latex 3 = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}} &s=2$

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 $latex S_n$, where each $latex S_n$ in the sequence is given by the definition:

$latex S_{1} = \sqrt{6} &s=2$
$latex S_{n+1} = \sqrt{6 + S_n} &s=2$

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

$latex \displaystyle \lim_{n \to \infty}S_{n} = S &s=2$

So the defining relationship becomes:

$latex S = \sqrt{6 + S} &s=2$

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

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.)

More like this


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

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) $latex 2 = \sqrt{2 + \sqrt{2 + \dots}}$. (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.

By Eric Lund (not verified) on 14 Sep 2012 #permalink

Hey, LaTeX does work in the comments!

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

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

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

By Jon Wharf (not verified) on 14 Sep 2012 #permalink

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