$\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}$

Question

Let $\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}=x$. Therefore, $$n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}=x^2$$ $$x^2-x-n=0$$ $$x=\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}=\frac{1 \pm \sqrt{1+4n}}{2}$$

It seems counter-intuitive at first that $f(n)$ doesn't have an effect on the final answer. If $f(n)$ were $\frac{1}{n!}$ or $e^{2.3n^2}$, or anything else for that matter, we still get the same formula for the result.

My question regarding this is:

1. Is this applicable for any $f(n)$ at the end of the infinite nested radical (I haven't studied much about properties of functions but have heard of terms like 'odd function', 'even function', '$f: \mathbb{R} \to \mathbb{R}$' etc)?
2. Is this result because of the infinitesimally small effect of $f(n)$ on the result?

EDIT : Some people have commented that there is '1 less $n$ post squaring'. There are infinite '$\sqrt{n+\sqrt{}}$' before I added $f(n)$ at the end.

Answer 1

Firstly, note that you actually don't need the function $f$. You only need its value at point $n$ and we can mark it $x_0=f(n)$.

You actually invented the fixed point iteration. You are iterating the equation $x_{m+1}=\sqrt{n+x_m}$. You noticed that this may have two solutions. Let's mark them: $$ \bar{x}_1=\frac{1+\sqrt{1+4n}}{2}\\ \bar{x}_2=\frac{1-\sqrt{1+4n}}{2} $$ The solution $\bar{x}_1$ will be always valid, while $\bar{x}_2$ will be valid only for $n \in \left[-\frac14;0\right]$. For small $n$, it will be imaginary, and for big $n$, it will be negative and square root can't be negative in real numbers.

The iterated function $x_{m+1}=\sqrt{n+x_m}$ is contractive (i.e. its first derivation is in range $(-1;1)$) in $\bar{x}_1$ except one case but never in $\bar{x}_2$. The only case, in which the function isn't contractive in $\bar{x}_1$ is when $n=-\frac14$ and $\bar{x}_1=\bar{x}_2$. So there are a few cases which may happen:

1. For $n<-\frac14$, there is no real solution and $x_m$ will diverge.
2. For $n>0$, the solution $\bar{x}_2$ doesn't make sense and the iterated function is contractive at $\bar{x}_1$. So $x_m$ converges to $\bar{x}_1$ for any $x_0$ for which the function makes sense.
3. For $n \in \left(-\frac14;1\right]$, the iterated function is contractive only at $\bar{x}_1$. If $x_0>\bar{x}_2$, $x_m$ will converge to $\bar{x}_1$. If $x_0=\bar{x}_2$, $x_m$ will stay $\bar{x}_2$ forever. And if $x_0<\bar{x}_2$, $x_m$ will diverge.
4. For $n = -\frac14$, it holds that $\bar{x}_1=\bar{x}_2=\bar{x}$. If $x_0 \geq\bar{x}$, $x_m$ will slowly converge to $\bar{x}$. Otherwise, it will diverge.

As you see, in cases 3 and 4, it actually depends on the choice of $x_0$.

Answer 2

If you set $$x=\sqrt{n+\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}}$$ when you square $$x^2=n+\left[\sqrt{n+\sqrt{n+...+\sqrt{n+f(n)}}}\right]$$ The part in the square brackets no longer has the same number of square roots, therefore it is not $x$. It is something else.

Maybe you saw something like this with infinite square roots. In that case, under some conditions, the trick of squaring actually works.

Not in the finite case.