Results 1 to 6 of 6

Math Help - Closed form of an infinitely nested radical.

  1. #1
    Member Pranas's Avatar
    Joined
    Oct 2010
    From
    Europe. Lithuania.
    Posts
    81

    Closed form of an infinitely nested radical.

    Hello.

    Does anybody happen to know a closed form of this infinitely nested radical?

    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:


    Also here is a convergence plot:

    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,
    Pranas.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Pranas's Avatar
    Joined
    Oct 2010
    From
    Europe. Lithuania.
    Posts
    81

    Re: Closed form of an infinitely nested radical.

    Still looking for any insightful ideas!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: Closed form of an infinitely nested radical.

    I'm not sure your equations are correct, but I think you mean that a_{n+1}(x)=\sqrt{x+a_n(x)}, and you would like to find a(x)=\lim_{n\rightarrow\infty}a_n(x).

    If you can prove that the limit exists, then it's easy to determine what that limit is. Take the limit of the recurrence relation a_{n+1}(x)=\sqrt{x+a_n(x)}, giving a(x)=\sqrt{x_0+a(x)}. Then you can solve for a(x).

    How do you prove that the limit exists? You can probably prove that a_n(x)\ge{0}, so it's bounded below. Then you would need to prove that it is decreasing: a_{n+1}(x)<{a_n(x)}. And there's a theorem that a decreasing sequence that is bounded below always converges.

    You might have to vary your approach for different ranges of x and make some assumptions about x_0.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Pranas's Avatar
    Joined
    Oct 2010
    From
    Europe. Lithuania.
    Posts
    81

    Re: Closed form of an infinitely nested radical.

    Quote Originally Posted by hollywood View Post
    ... a_{n+1}(x)=\sqrt{x+a_n(x)} ...
    Thanks for your reply, but unfortunately this is not the same thing at all.
    Essentially that's the main problem - there is no obvious convenient way to express a_{n+1}(x) in terms of a_{n}(x).

    Example:
    {a_3}(x) = \sqrt {{x^0} + \sqrt {{x^1} + \sqrt {{x^2} + \sqrt {{x^3}} } } }
    {a_4}(x) = \sqrt {{x^0} + \sqrt {{x^1} + \sqrt {{x^2} + \sqrt {{x^3} + \sqrt {{x^4}} } } } }

    So a_{4}(x) in terms of a_{3}(x) is... ???
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,853
    Thanks
    321
    Awards
    1

    Re: Closed form of an infinitely nested radical.

    Quote Originally Posted by Pranas View Post
    Still looking for any insightful ideas!
    Please do not bump your thread. It is rude.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Mar 2010
    Posts
    980
    Thanks
    236

    Re: Closed form of an infinitely nested radical.

    So for example,

    a_0(2) = \sqrt{2^0} = \sqrt{1} = 1

    a_1(2) = \sqrt{2^0+\sqrt{2^1}} = \sqrt{1+\sqrt{2}} \approx 1.554

    a_2(2) = \sqrt{2^0+\sqrt{2^1+\sqrt{2^2}}} = \sqrt{1+\sqrt{2+\sqrt{4}}} = \sqrt{3} \approx 1.732

    a_3(2) = \sqrt{2^0+\sqrt{2^1+\sqrt{2^2+\sqrt{2^3}}}}=\sqrt{  1+\sqrt{2+\sqrt{4+\sqrt{8}}}} \approx 1.774

    I think we can still prove the sequence is bounded, and it's obviously increasing, but you're right - there seems to be no obvious recurrence relation.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Prove that there are infinitely many primes in the form 6k+5
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: June 13th 2011, 10:43 PM
  2. radical to exponential form
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 1st 2010, 06:22 PM
  3. infinitely many primes of the form 6k + 5 and 6K + 1
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 22nd 2009, 05:25 AM
  4. answer in radical form
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: November 25th 2008, 08:15 AM
  5. [SOLVED] [SOLVED] Nested Radical
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 27th 2007, 01:45 PM

Search Tags


/mathhelpforum @mathhelpforum