Register to reply 
Iterative square root? sqrt(2+sqrt(2+sqrt(... 
Share this thread: 
#1
Sep709, 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. 


#2
Sep709, 08:56 PM

P: 896

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



#3
Sep709, 09:37 PM

P: 94

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


#4
Sep709, 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 squareroots are nonnegative. 


#5
Sep709, 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 


#6
Sep709, 11:53 PM

Emeritus
Sci Advisor
PF Gold
P: 16,092

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 numbervalued sequences always exist. It's important to notice that extended real numbers come into play here! The equation 2 + x = x²has 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]. 


#7
Sep809, 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 


Register to reply 
Related Discussions  
[itex]\lim_{x\rightarrow\infty} \frac{\sqrt{x^{2}+5}  x}{\sqrt{x^{2}+2}  x}[/itex]  Calculus & Beyond Homework  6  
Is there a way to show that sqrt(2)+sqrt(3)+sqrt(5) is algebraic over Q?  General Math  4  
Simplify $ln(x(\sqrt{1+e^x}\sqrt{e^x})) + ln(\sqrt{1+e^{x}}+1)$  Calculus & Beyond Homework  4  
Does this proof for irrationality of sqrt(2)+sqrt(3) work?  Calculus & Beyond Homework  5  
Proof that sqrt(6)sqrt(2)sqrt(3) is irrational  General Math  10 