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Homework Help: Closed form of an infinitely nested radical.

  1. Dec 9, 2012 #1


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    Does anybody happen to know a closed form of this infinitely nested radical?
    http://imageshack.us/a/img268/6544/radicals.jpg [Broken]
    By any chance, maybe you even saw it somewhere?

    I haven't had too much success so far. At the moment I am so desperate that I'm even willing to try and guess the solution, then prove that it is equal to my nested radical. For any real positive x the limit indeed exists (various criteria can be found for that very reason). Numerical limits can be seen in the plot:
    http://imageshack.us/a/img842/876/plote.jpg [Broken]

    Also here is a convergence plot:
    http://img100.imageshack.us/img100/64/convergence.jpg [Broken]
    It is made in a sense that using double precision variables computer sees no difference between a_{k}(x) and a_{k+1}(x) which in turn means that ~16 decimal digits have already been found. In fact it's so nasty that a{6}(50000) - a{5}(50000) < 10^(-24).
    Notably the bigger my argument, the faster it converges (although I'm not sure what useful conclusions I can draw from that).

    Pretty much the only known elegant cases: a(1) is equal to golden ratio, a(4)=2.

    What would you suggest?

    Best regards,
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 9, 2012 #2

    I like Serena

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    Hi Pzi! :smile:

    You can write:
    $$\sqrt{x^1 + \sqrt{x^2 + ...}} = \sqrt x \cdot \sqrt{x^0 + \sqrt{x^1 + ...}} = \sqrt x \cdot a(x)$$

    If you substitute that in the expression for a(x), you get an equation that you can solve.......
  4. Dec 9, 2012 #3


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    Not really.
    Your idea requires powers like 1, 2, 4, 8, 16, 32... whereas we actually have 1, 2, 3, 4, 5, 6...
  5. Dec 11, 2012 #4


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    Still looking for any insightful ideas!
  6. Dec 12, 2012 #5


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    I get a good fit to 1.19 x1/4
  7. Dec 12, 2012 #6


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    You say you are desperate for a solution. Do you have any reason to believe there is an analytically exact solution?
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