Comments Off on A neat trick (warning: contains maths)

A neat trick (warning: contains maths)

\sqrt{1}

This is a radical.

\sqrt{1 + \sqrt{2}}

This is a nested radical.

\sqrt{1+\sqrt{1+{\sqrt{1+...}}}}

This is an infinitely nested radical; it goes on, and on, repeating itself.

You may now be asking yourself two obvious questions: (a) what has this got to do with me, and (b) is this going to end up in a joke about political radicals saying the same things over and over again. The answers, respectively, are they present an interesting example of how redefining a problem can make it easier to solve, and that it probably would have done if I’d thought of it prior to your suggestion.

Let’s take \sqrt{1+\sqrt{1+{\sqrt{1+...}}}}.

Let X = \sqrt{1+\sqrt{1+{\sqrt{1+...}}}}

Then X =\sqrt{1+X}

Therefore X^2 = 1+X, X^2 -X = 1, X = \frac{1}{2} + \frac{\sqrt{5}}{2}. You might recognise this last as the golden ratio; when the ratio of X to Y is the same as the ratio of Y to X+Y. This number pops upĀ everywhere; the patterns of spirals in leaves follow it to maximise density, designers use it in the layout of books and playing cards, architects employ it in the design of their buildings.

Here’s a more complicated infinitely nested radical.

\sqrt{1+2\sqrt{1+3\sqrt{1+4{\sqrt{1+...}}}

Finding the pattern here isn’t quite so easy, so we’re going to steal Ramanujan’s solution (this is fair as it’s also his question). And if you want the answer before we start, mouse over this footnote 1.

Let’s have another look at the shape of that problem. What we’d like to do is something similar to what we did with the first one; turning it into a recursive problem.

Ramanujan started by pointing out that

n(n+2) = n\sqrt{1+(n+1)(n+3)} (2)

If we then label f(n) = n(n+2), we can write the right hand side as f(n) = n\sqrt{1+f(n+1)}

Or as f(n) = n\sqrt{1+(n+1)\sqrt{1+f(n+2)}}

Or as… and so on down the line.

What we end up with is

f(n) = n\sqrt{1+(n+1)\sqrt{1+(n+2)\sqrt{1+n+3\sqrt{...}}}}

And in our original problem, we had n=1.

So the right hand side collapses back to \sqrt{1+2\sqrt{1+3\sqrt{1+4{\sqrt{1+...}}}, and the left hand side collapses to… 3.

There isn’t much of a point to this post other than presenting a problem that I found neat, and a lovely little solution to it, so I’ll leave off here.

  1. It’s 3. Isn’t that an anticlimax?
  2. as \sqrt{1+n^2 +4n + 3} = \sqrt{n^2 +4n +4} =\sqrt{(n+2)^2}