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Problem -

  1. Mar 19, 2009 #1
    Problem --

    1. The problem statement, all variables and given/known data

    Give me a hint
    2. Relevant equations

    [tex] \sqrt{1 + \sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + 4\sqrt{1 + \cdots}}}}} = ? [/tex]


    3. The attempt at a solution

    Can someone help get in the right direction, give a little hint?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Mar 19, 2009
  2. jcsd
  3. Mar 19, 2009 #2

    Dick

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    Re: Problem --

    Cute. VERRY cute. Contemplate the following series of equations:
    2=sqrt(1+1*3)
    3=sqrt(1+2*4)
    4=sqrt(1+3*5)
    5=sqrt(1+4*6)
    ...
     
  4. Mar 20, 2009 #3
    Re: Problem --

    seems likely to fit for 2

    Then

    [tex]

    2 = \sqrt{1 + \sqrt{1 + 2\sqrt{1 + 3\sqrt{...}}}}
    [/tex]

    Then n must be such that

    [tex]

    n = \sqrt{n + ((n-1)(n +1))}

    [/tex]

    Which fits for all n.. Is this correct resoning? Can someone give me a more formal explanation if it is (or isn't)..
     
  5. Mar 20, 2009 #4

    Dick

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    Re: Problem --

    2 doesn't solve n=sqrt(n+(n-1)(n+1)). But yes, I would say the whole thing is equal to 2. If you want to get formal, then you have to define what the whole iterated square root means to begin with. How to truncate it and define it as a convergent series?
     
  6. Mar 22, 2009 #5
    Re: Problem --

    actually ment

    [tex] n = \sqrt{1 + (n + 1)(n - 1)} [/tex]



    [/tex]
    an so on. But that doesn't get me any further. Don't really know how to handle the rootexpressions. However, it seems to have som similarities with the root form of the golden ratio
    [tex]
    \phi^2 = \phi + 1
    [/tex]

    [tex]
    \phi = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \cdots}}}}

    [/tex]
     
    Last edited: Mar 22, 2009
  7. Mar 22, 2009 #6

    Dick

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    Re: Problem --

    I don't think you are seeing it. Substitute 3=sqrt(1+2*4) for the '3' in 2=sqrt(1+1*3). Now substitute 4=sqrt(1+3*5) into that. Etc etc.
     
  8. Mar 22, 2009 #7
    Re: Problem --

    Ok, I see know. Thanks a lot
     
  9. Mar 27, 2009 #8
    Re: Problem --

    But inserting it just gives av pinpoint of the solution. tried using mathematical induction to prove it formula, but i did'nt seem correct. are there better methods to prove this, eventually define this as a series in a way?
     
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