Built on Facts

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.)

Comments

  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?

The site is currently under maintenance and will be back shortly. New comments have been disabled during this time, please check back soon.