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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.

sato
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  • Your computation is not correct...post squaring the radical term has one fewer $n$ in it. – lulu Feb 05 '21 at 11:36
  • The illusion is that the $f(n)$ comes at the end of an apparently infinite process. The expression itself is not well-defined. It's like writing the "number" $0.00...0001$. Such a number doesn't exist. – David P Feb 05 '21 at 11:41
  • There are infinite n's @lulu – sato Feb 05 '21 at 11:44
  • That doesn't make sense. – lulu Feb 05 '21 at 11:45
  • @DavidP I just realized my point 2 in the list is obvious. – sato Feb 05 '21 at 11:51
  • To be clear, my point was that your question doesn't make sense. Has nothing to do with $f(n)$. – David P Feb 05 '21 at 12:28
  • This notation with "an infinite number of $n$s" is not a standard mathematical notion, so you must define what you mean by it. Your contraction here shows that, whatever it means, it does not have the properties you used. – GEdgar Feb 05 '21 at 12:29
  • You can consider the sequence ${x_k}{k=1}^\infty$ which obeys $x_k^2=n+x{k-1}$ with $x_1=\sqrt{n+f(n)}$ and then prove that there is a limit to this sequence. – am301 Feb 05 '21 at 12:37
  • 'This notation with "an infinite number of ns" is not a standard mathematical notion' What is the standard notation in this case then @GEdgar? – sato Feb 05 '21 at 13:16
  • I do not know the "standard notation" because I do not understand what it means. – GEdgar Feb 05 '21 at 19:34

2 Answers2

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 $x_0$.

Hume2
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0

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.

Raffaele
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