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

[Damidami]

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.

 

[slider142]

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

 

[Damidami]

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

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

 

[Tac-Tics]

\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.

 

[Elucidus]

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

 

[Hurkyl]

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.

 

[Elucidus]

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