If $x=-1/4$ does $S=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=1/2$?

Question

If $x=-1/4$ does $S=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=1/2$?

My attempt: As $S^2=x+S$, it holds that $S=\frac{1\pm \sqrt{(-1)^2+4x}}{2}$, which has a real solution for $x\ge -1/4$.

Therefore, if $x=-1/4$, $S=1/2$, meaning that this infinite nested quadratic radical of a negative number, gives a positive answer.

Does it make sense or am I missing something?

Answer 1

Try considering this problem in $\mathbb{C}$: Let $x_0=x=-\frac{1}{4}$,so $x_1=\frac{i}{2}$. And $x_{n+1}=\sqrt{x_0+x_n}$

Answer 2

One way to make sense of the dots is to start with any number $S_0$ and calculate $$S_{n+1}=\sqrt{x+S_n}$$ For example, if $S_0=1$, then the sequence of $S_n$ is $$1,0.866,0.785, 0.731,...$$ and the limit turns out to be 0.5.

If you start with $S_0=-0.25$ then you get complex numbers. One of the square-roots of a complex number always has positive-or-zero real part and so long as you pick that one, the limit of $S_n$ is still $0.5$.

Answer 3

This is not an answer, but is a little more than a comment.

The quadratic for $S$ apparently appears because there are two choices for the sign of the square root, and if these are taken as consistently positive (the presumptive value as given) or consistently negative, you can get a solution which makes sense. Squaring the expression brings in a second, parasitic solution.

But that is not the whole story. Take $x=0$. It is "obvious" that $S=0$ whatever choice of signs is made. But the quadratic also has the solution $S=1$ which seems to make no sense at all (it is a limit, though - see below). The correct value $0$ requires the negative square root in solving the quadratic.

But if $x=2$ we have the solution $S=2$, which requires the positive root in solving the quadratic, with alternative $S=-1$, which makes sense with all negative square roots.

If $S$ is positive we have $S=1+\frac xS$ so that for $x\gt 0$ we have $S\gt 1$. Although everything looks smooth, neat and well-defined, this gives us a discontinuity at $x=0$ where we switch to the "other" solution to the quadratic.

This tells us we have to take care about exactly what our terms mean and whether they are well-defined. This doesn't quite answer your question, but does show that everything is not quite what it may seem.