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Sequence Proof

  1. Dec 11, 2007 #1
    Given the following sequence:

    [tex]x_0 = 1, \quad x_1 = \sqrt{3+1}, \quad x_2 = \sqrt{3+\sqrt{4}}, \quad x_3 = \sqrt{3+\sqrt{5}},[/tex]
    [tex]x_4 = \sqrt{3+\sqrt{3+\sqrt{5}}}, \quad x_5 = \sqrt{3+\sqrt{3+\sqrt{3+\sqrt{5}}}} \ldots[/tex]

    prove the above sequence converges and determine the limit.

    ...................

    So from [tex]n=3[/tex] onwards, I notice that the sequence is recursively defined:

    [tex]n \geq 4,\;x_n = \sqrt {3 + x_{n - 1} }[/tex]

    To prove convergence, I'd stimply have to show sequence is bounded above and that it's increasing.

    I'm not quite sure how to do this with a recursive function.

    To find the limit, I realized that

    [tex]x_n = \sqrt {3 + x_{n - 1} }[/tex] is at it's "equilibrium point" when [tex]x = 3 + \sqrt{x}[/tex]. I solved for [tex]\sqrt{x}[/tex] and found the limit to be:

    [tex]\frac{1\pm\sqrt{13}}{2}[/tex]. I then see that we go towards:

    [tex]\frac{1+\sqrt{13}}{2}[/tex]

    So I know the limit will be the above if the sequence converges.

    Any help on the proof?
     
  2. jcsd
  3. Dec 11, 2007 #2
    what you call "equilibrium point"?
     
  4. Dec 11, 2007 #3
    Stable point..
     
  5. Dec 11, 2007 #4
    i'm not sure what a stable or equilibrium point is. You can show the sequence is bounded and monotonic using induction, show that it's bounded first, this will help you show it's monotonic.

    This means lim x_{n-1} = lim x_n exists. The recursive definition and limit laws show that the limit, call it, x must satisfy the equation:

    x=squareroot(3+x) or the quadratic
    x^2-x-3, solve and deduce which solution is x.
     
  6. Dec 11, 2007 #5
    I already know the limit. I'm having a hard time with the proof.
     
  7. Dec 11, 2007 #6
    I gave you the outline of a rigorous proof.
     
  8. Dec 11, 2007 #7
    [tex]\begin{array}{l}
    x_2 < x_3 < 9 \\
    \sqrt {x_2 } < \sqrt {x_3 } < 3 \\
    3 + \sqrt {x_2 } < 3 + \sqrt {x_3 } < 6 \\
    \underbrace {\sqrt {3 + \sqrt {x_2 } } }_{x_3 } < \underbrace {\sqrt {3 + \sqrt {x_3 } } }_{x_4 } < \sqrt 6 \\
    x_3 < x_4 < 9 \\
    \end{array}[/tex]

    Now I show it from the [tex]k+1[/tex] term? Not exactly sure how. But does that look right?
     
  9. Dec 11, 2007 #8
    show x_3 is less than 3, then assume x_n is less than 3. what can you say then?

    x_{n+1}=root(3+x_n)<root(3+3)<root(3+6)=3

    done, so know x_n is bounded, now show it is monotonic.
     
  10. Dec 12, 2007 #9
    x_n is positive for all n = {0,1,2,3,...}

    [tex]x_{n + 1} = \sqrt {3 + x_n}[/tex] ==> [tex]x_{n + 1} > x_n[/tex] for all n
     
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