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Root sequence question

  1. Apr 15, 2005 #1
    Short root sequence question

    For the recursive sequence
    [tex] R_n = x + \sqrt {x - \sqrt {R_{n - 2} } } [/tex]
    [tex] R_0 = x = k^2 - k + 1 [/tex]

    [tex] \forall k \in \mathbb{N},\;k > 1 [/tex] why does [tex] \mathop {\lim }\limits_{n \to \infty } R_n = k^2 [/tex] ??
     
    Last edited: Apr 15, 2005
  2. jcsd
  3. Apr 16, 2005 #2
    Perhaps if I redefine the whole problem and open it to the reals ([tex] \mathbb{R} [/tex]),
    it will become clearer and easier to solve :smile: .

    First off, the sequence shall be defined recursively as
    [tex] R_n = \sqrt {x + \sqrt {x - \sqrt {R_{n - 2} } } } [/tex]
    (just takin' the root of what's in my original post! (It's a root sequence anyway))

    So:
    If we start with [tex] R_0 [/tex], we will follow with [tex] \left\{ {R_2 ,R_4 , \ldots ,R_n } \right\} [/tex]
    If we start with [tex] R_1 [/tex], we will follow with [tex] \left\{ {R_3 ,R_5 , \ldots ,R_n } \right\} [/tex]

    Such that:
    [tex] R_0 \in \left\{ {x|x > \sqrt {x + \sqrt {x - \sqrt x } } > 1 ,x \in \mathbb{R}} \right\} [/tex] (numerically, x > 1.311 for [tex] R_0 [/tex])
    [tex] R_1 \in \left\{ {x|x > \sqrt {x + \sqrt x } > 1 ,x \in \mathbb{R}} \right\} [/tex] (numerically, x > 1.755 for [tex] R_1 [/tex])

    We will find that:
    [tex] \mathop {\lim }\limits_{n \to \infty } R_n = \frac{{1 + \sqrt {4x - 3} }}{2} [/tex]
    (doesn't matter which [tex] R_0 [/tex] or [tex] R_1 [/tex] you choose...
    just as long as it's inside the appropriate interval!:biggrin:)
    ------------------------------------------------------------------
    Now, why exactly does:
    [tex] \mathop {\lim }\limits_{n \to \infty } R_n = \frac{{1 + \sqrt {4x - 3} }}{2} [/tex] ?
     
    Last edited: Apr 17, 2005
  4. Apr 17, 2005 #3

    CRGreathouse

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    If the limit exists, you should just be able to set [tex]R_n=R_{n-2}[/tex] and solve. This gives

    [tex]((R-k^2+k-1)^2-k^2+k-1)^2-R=0[/tex]

    which has solutions

    [tex](R-k^2)(R-(k^2-2k+1))(R^2+(-2k^2+2k-3)R+(k^4-2k^3+3k^2-2k+1))[/tex]

    Why this is defined for all [tex]k\in\mathbb{Z}^+[/tex] and why it has the one particular solution out of the four possible ones I don't know offhand.
     
  5. Apr 17, 2005 #4

    Hurkyl

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    [tex]
    \lim_{n \rightarrow \infty} R_n =
    \lim_{n \rightarrow \infty} \sqrt{x + \sqrt{x - \sqrt{R_{n-2}}} }
    = \sqrt{ \lim_{n \rightarrow \infty}x + \sqrt{x - \sqrt{R_{n-2}}} }
    = \cdots
    [/tex]
     
  6. Apr 17, 2005 #5
    But how does that simplify [tex] \mathop {\lim }\limits_{n \to \infty } R_n [/tex] down to '"[tex] \frac{{1 + \sqrt {4x - 3} }}{2} [/tex]"' ?
     
    Last edited: Apr 17, 2005
  7. Apr 17, 2005 #6

    Hurkyl

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    Continue through the dots. :tongue2: It might help you visualize to use L to denote [itex]\lim_{x \rightarrow \infty} R_n[/itex].
     
  8. Apr 17, 2005 #7
    Oh--ic now
    !
     
  9. Apr 17, 2005 #8

    saltydog

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    You mind explaining this a little further?

    I realize the limit of an infinitely nested radical:

    [tex]\sqrt{x+\sqrt{x+\sqrt{x+...}}}[/tex]

    is:

    [tex]\frac{1+\sqrt{1+4x}}{2}[/tex]

    but don't see how to get the 3 in there.
     
    Last edited: Apr 17, 2005
  10. Apr 17, 2005 #9

    Hurkyl

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    I think your problem is that you're considering a different sequence. :smile:
     
  11. Apr 18, 2005 #10
    My sequence alternates the signs of the square roots as I embed them, as
    [tex]\sqrt{x+\sqrt{x-\sqrt{x+...}}}[/tex]
     
  12. Apr 19, 2005 #11

    saltydog

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    Hello Bomba. Thanks for responding, you too Hurkyl. I know you're just telling me the minimum for my own good but you know what, I couldn't even get past the definition and I'd like to understand it. This is what I have:

    If I start from [itex]R_0[/itex] and you say calculate R2, R4, R6,. . ., and the recursion relation is always in reference to R(n-2), then it's just like doing it for each N and set up the relation for R(n-1). Is that not correct? That is, if I start with [itex]R_0[/itex], then it seems to me, just ignore the odd members in the sequence. Perhaps that is not correct though. When I experiment with it though in Mathematica, say for R0=1.9, it doesn't seem to converge to the value you specify. This is how I set it up in Mathematica. Perhaps you can correct my interpretation:

    [tex]f[x,xm1]=\sqrt{x+\sqrt{x-\sqrt{xm1}}}[/tex];

    [tex]xstart=1.9[/tex]

    [tex]ntotal=alargenumber[/tex]

    [tex]valist=Table[{0},{ntotal}][/tex]

    [tex]valist[[1]]=f[xstart,xstart][/tex]

    [tex]For[n=2,n<ntotal,n++,[/tex]

    [tex]valist[[n]]=f[xstart,valist[[n-1]]];[/tex]

    When I do that, it seems to converge to 1.639 . . .

    But:

    [tex]\frac{1+\sqrt{4xstart-3}}{2}=1.572...[/tex]

    Would you (or Hurkyl or anyone else) tell me where my problem is?

    Thanks,
    Salty

    Edit: corrected to reflect relation of even members in terms of R0
     
    Last edited: Apr 19, 2005
  13. Apr 19, 2005 #12

    Hurkyl

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    Hrm. The problem is that I had assumed that what was posted as "the answer" was correct. 1.639... is, in fact, the correct limit. (Assuming that my calculuator is sufficiently smart and would tell me if there were additional real solutions to the fixed point equation)
     
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