# 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:

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

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.)

Tags

### More like this

##### Quick, hit the brakes!
A reader emailed me a fun question from a physics exam he took, along these lines: A car driver going at some speed v suddenly finds a wide wall at a distance r. Should he apply brakes or turn the car in a circle of radius r to avoid hitting the wall? My first thought was that surely the question…
##### How often does the sun emit 1 TeV photons?
I had an interesting question posed to me recently: how frequently does the sun emit photons with an energy greater than 1 TeV? All of you know about the experiments going on at the LHC, where particles are accelerated to an energy which is equivalent to an electron being accelerated through a…
##### Rearrangements of Series
Blake Stacey directed me towards a terrific tool for embedding TeX code into a web page. So how about we do ourselves a math post! Remember the harmonic series? No doubt you encountered it in some calculus class or other. It's the one that goes like this:  1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4…
##### The Mathematics of Reddit Rankings, or, How Upvotes Are Time Travel
Ok, so this isn't really physics as such, but it's pretty fascinating. There's a very large online community called Reddit in which users submit links which interest them. These links come with two little arrows beside them, and the users can vote the link up or down. Here's a screenshot of how the…

nice.

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 $latex \LaTeXe$?