# Iterative square root? sqrt(2+sqrt(2+sqrt(...

by Damidami
Tags: iterative, root, sqrt, sqrt2, square
 P: 94 The other day I was playing with my calculator and noticed that $$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}} \approx 2$$ But, what is that kind of expression called? How does one justify that limit? And, to what number exactly does converge, for example: $$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}} \approx 1.6161$$ $$\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+...}}}} \approx 2.3027$$ Any references where I could read about these subjects? Another question. Considering real $$x>1$$, we have: $$\Gamma(x) - x^1 = 0$$ then $$x \approx 2$$ But how does one justify that? And what are the exact values of these functions: $$\Gamma(x) - x^2 = 0$$ then $$x \approx 3.562382285390898$$ $$\Gamma(x) - x^3 = 0$$ then $$x \approx 5.036722570588711$$ $$\Gamma(x) - x^4 = 0$$ then $$x \approx 6.464468490129385$$ Thanks, Damián.
 P: 898 They are called Nested Radicals. There are references in the link. I am unfortunately not familiar with their theory.
P: 94
 Quote by slider142 They are called Nested Radicals. There are references in the link. I am unfortunately not familiar with their theory.
Thank you slider142!
 P: 810 Iterative square root? sqrt(2+sqrt(2+sqrt(... $$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}} = x$$ For something like this, you can rewrite the equation as $$\sqrt{2+x} = x$$ And then the infinite equation is captured in a finite form. From there, you simply square both sides. $$2+x = x^2$$ And solve for x. But I don't know much more than that! Don't forget that square-roots are non-negative.
 P: 286 For nested radicals of the form $$\sqrt{a + \sqrt{a + \sqrt{a + \dots}}}$$ using the trick $$\sqrt{a + x} = x$$ works very well. Two roots will emerge, but only one is positive (the other is extraneous). In the case when a = 2, x = 2. When a = 1, then x = $\phi = \frac{1 + \sqrt{5}}{2} \approx 1.618034$ a.k.a. the golden ratio. As to the second question regarding the Gamma function, I'm not sure much theory is available. --Elucidus
 Emeritus Sci Advisor PF Gold P: 16,091 For a dose of rigor -- we have to be sure the limit really exists before we can compute it with such tricks! In this case, it's easy: the value is the limit of an increasing sequence, and limits of increasing, extended real number-valued sequences always exist. It's important to notice that extended real numbers come into play here! The equation2 + x = x²has three relevant solutions: -1, 2, and $+\infty$. We know the limit exists, so it has to have one of those three values. It's easy to rule out -1, but more work is needed to decide between 2 and $+\infty$.
 P: 286 If we examine the sequence $\{a_n\}_{n=0}^{\infty}$ when $a_0 = \sqrt{2}$ and $$a_{n+1}=\sqrt{2+a_n}$$ Then it is possible to show by induction that $a_n \leq 2$ for all n so the $+\infty$ case is impossible. But you are correct, this possibility does need to be ruled out, Hurkyl. --Elucidus