# Recursive Square Root Futility Closet

This post on Futility Closet the other day: http://www.futilitycloset.com/2013/12/05/emptied-nest/ asked for the solution to this equation:

\begin{equation}\sqrt{x+\sqrt{x+\sqrt{x...}}} = 2\end{equation}

The problem can be described recursively as

\begin{equation} \sqrt{x + 2} = 2 \end{equation} and then solved for x = 2. This is very similar to the solution on Futility Closet.

It can be extended into the more general \begin{equation} \sqrt{x + y} = y \end{equation}

and then \begin{equation} x = y^2 - y \end{equation} which works, as far as I'm aware for all x>1.

When 1 is plugged in for y, however, the output becomes 0. This is not altogether unfathomable (though it is quite strange) in and of itself, but it is the same result as when 0 is plugged in. That means that either

\begin{equation}\sqrt{0+\sqrt{0+\sqrt{0...}}} = \sqrt{1+\sqrt{1+\sqrt{1...}}} \end{equation}

or one of these values is undefined. I expect that either imaginary numbers or dividing by zero is involved, but I can't quite figure out how. Why does the reasoning fall apart here?

## 1 Answer

Some steps for further investigation.

• If $x > 0$ is fixed, show that the nested radical $$\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$$ converges to a positive number.

• Define the function $$f(x) = \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}.$$ Show that $f(x) > 1$ for all $x > 0$.

• Calculate $$\lim_{x\to 0^+} f(x).$$