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

by Damidami
Tags: iterative, root, sqrt, sqrt2, square
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Damidami
#1
Sep7-09, 08:51 PM
P: 94
The other day I was playing with my calculator and noticed that

[tex]\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}} \approx 2[/tex]

But, what is that kind of expression called? How does one justify that limit?
And, to what number exactly does converge, for example:

[tex]\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}} \approx 1.6161[/tex]

[tex]\sqrt{3+\sqrt{3+\sqrt{3+\sqrt{3+...}}}} \approx 2.3027[/tex]

Any references where I could read about these subjects?

Another question. Considering real [tex]x>1[/tex], we have:
[tex]\Gamma(x) - x^1 = 0[/tex] then [tex] x \approx 2[/tex]

But how does one justify that? And what are the exact values of these functions:

[tex]\Gamma(x) - x^2 = 0[/tex] then [tex] x \approx 3.562382285390898[/tex]
[tex]\Gamma(x) - x^3 = 0[/tex] then [tex] x \approx 5.036722570588711[/tex]
[tex]\Gamma(x) - x^4 = 0[/tex] then [tex] x \approx 6.464468490129385[/tex]

Thanks,
Damián.
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slider142
#2
Sep7-09, 08:56 PM
P: 896
They are called Nested Radicals. There are references in the link. I am unfortunately not familiar with their theory.
Damidami
#3
Sep7-09, 09:37 PM
P: 94
Quote Quote by slider142 View Post
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
#4
Sep7-09, 11:14 PM
P: 810
Iterative square root? sqrt(2+sqrt(2+sqrt(...

[tex]\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}} = x[/tex]

For something like this, you can rewrite the equation as

[tex]\sqrt{2+x} = x[/tex]

And then the infinite equation is captured in a finite form. From there, you simply square both sides.

[tex]2+x = x^2[/tex]

And solve for x.

But I don't know much more than that! Don't forget that square-roots are non-negative.
Elucidus
#5
Sep7-09, 11:43 PM
P: 286
For nested radicals of the form

[tex]\sqrt{a + \sqrt{a + \sqrt{a + \dots}}}[/tex]

using the trick

[tex]\sqrt{a + x} = x[/tex]

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 = [itex]\phi = \frac{1 + \sqrt{5}}{2} \approx 1.618034[/itex] 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
#6
Sep7-09, 11:53 PM
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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 [itex]+\infty[/itex]. 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 [itex]+\infty[/itex].
Elucidus
#7
Sep8-09, 12:30 AM
P: 286
If we examine the sequence [itex]\{a_n\}_{n=0}^{\infty}[/itex] when [itex]a_0 = \sqrt{2}[/itex] and

[tex]a_{n+1}=\sqrt{2+a_n}[/tex]

Then it is possible to show by induction that [itex]a_n \leq 2 [/itex] for all n so the [itex]+\infty[/itex] case is impossible.


But you are correct, this possibility does need to be ruled out, Hurkyl.

--Elucidus


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