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.
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.
[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.
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
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 = 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].
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