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

1. ### Damidami

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:

$$\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}} \approx 1.6161$$

$$\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+...}}}} \approx 2.3027$$

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.

2. ### slider142

956
They are called Nested Radicals. There are references in the link. I am unfortunately not familiar with their theory.

3. ### Damidami

94
Thank you slider142!
Does anyone know about my second question? Or any further references?
Thanks,
Damián.

4. ### Tac-Tics

810
$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}} = x$$

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.

5. ### Elucidus

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

6. ### Hurkyl

16,090
Staff Emeritus
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 equation
2 + x = x²​
has 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 $+\infty$.

7. ### Elucidus

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

Last edited: Sep 8, 2009