Dismiss Notice
Join Physics Forums Today!


  1. Sep 7, 2009 #1
    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]

  2. jcsd
  3. Sep 7, 2009 #2
    They are called Nested Radicals. There are references in the link. I am unfortunately not familiar with their theory.
  4. Sep 7, 2009 #3
    Thank you slider142!
    That answered my first question.
    Does anyone know about my second question? Or any further references?
  5. Sep 7, 2009 #4
    [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.
  6. Sep 7, 2009 #5
    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.

  7. Sep 7, 2009 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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].
  8. Sep 8, 2009 #7
    If we examine the sequence [itex]\{a_n\}_{n=0}^{\infty}[/itex] when [itex]a_0 = \sqrt{2}[/itex] and


    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.

    Last edited: Sep 8, 2009

Have something to add?
Similar Discussions: Iterative square root? sqrt(2+sqrt(2+sqrt(
  1. Graphing sqrt(x^2 + 1) (Replies: 12)

  2. Sqrt(x^2) and (Sqrt x)^2 (Replies: 14)