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Math Help - Convergence of sequence

  1. #1
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    Convergence of sequence

    A) How do you prove that if 0(</=)x(</=)10, then 0(</=)sqrt(x+1)(</=)10?

    B) So once that is found, then how can you prove that if 0(</=)u(</=)v(</=)10, then 0(</=)sqrt(u+1)(</=)sqrt(v+1)(</=)10?

    C) They give a recursively defined sequence: a_1=0.3; a_(n+1)=sqrt((a_n)+1)for n>1
    How do you find out the first five terms for it (I know what they are). then prove that this sequence converges. What is a specific theorem that will guarantee convergence, along with the algebraic results of parts A and B?

    D) How do you find out the exact limit of the sequence defined in part C? Are you supposed to square the recursive equation and take limits using limit theorems? If so, then which are these theorems?
    Last edited by Brotha; April 4th 2010 at 12:11 PM. Reason: addition
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  2. #2
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    A) Graph it or use a table. Also remember that sqrt (|x|) is going to be less than |x| for |x| > 1 and if |x| < 1 than sqrt (|x|) < 1 though greater than x
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  3. #3
    MHF Contributor chisigma's Avatar
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    The sequence can be expressed as...

    \Delta_{n} = a_{n+1} - a_{n} = f(a_{n}) = 1 + \sqrt{a_{n}} - a_{n} (1)

    The function that generates the sequence...

    f(x) = 1 + \sqrt{x} - x (1)

    ... is represented in figure...



    ... and, because is has only one fixed point at x_{0} = \frac{3 + \sqrt{5}}{2} = 2,6180339887\dots and that is an attractive fixed point, any a_{0} \ge 0 will produce a sequence convergent at x_{0} without oscillations, because the slope of f(x) in x=x_{0} is in absolute value less than 1...

    Kind regards

    \chi \sigma
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  4. #4
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    Quote Originally Posted by chisigma View Post
    The sequence can be expressed as...

    \Delta_{n} = a_{n+1} - a_{n} = f(a_{n}) = 1 + \sqrt{a_{n}} - a_{n} (1)

    The function that generates the sequence...

    f(x) = 1 + \sqrt{x} - x (1)

    ... is represented in figure...



    ... and, because is has only one fixed point at x_{0} = \frac{3 + \sqrt{5}}{2} = 2,6180339887\dots and that is an attractive fixed point, any a_{0} \ge 0 will produce a sequence convergent at x_{0} without oscillations, because the slope of f(x) in x=x_{0} is in absolute value less than 1...

    Kind regards

    \chi \sigma
    How did you find out delta_n?
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  5. #5
    MHF Contributor chisigma's Avatar
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    The sequence is defined as...

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

    ... so that is...

    \Delta_{n} = a_{n+1} - a_{n} = 1 + \sqrt{a_{n}} - a_{n} (2)

    Kind regards

    \chi \sigma
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  6. #6
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    If 0\le x\le 10 then 1\le x+1\le 11 so 1\le \sqrt{x+1}\le \sqrt{10}< 10
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  7. #7
    MHF Contributor chisigma's Avatar
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    After a more carefull reading it seems to Me that the sequence is defined as...

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

    If that is true I apologize for my mistake ... fortunately the solving procedure is almost the same we have described. The (1) can be alternatively written as...

    \Delta_{n} = a_{n+1} - a_{n} = f(a_{n}) = \sqrt{1 + a_{n}} - a_{n} (2)

    ... the 'generating function' of which is...

    f(x)= \sqrt{1 + x} - x (3)

    ... that is represented in figure...




    As in the previous case there is only one attractive fixed point in x_{0} = \frac{1 + \sqrt{5}}{2} = 1,6180339887\dots and, since the slope of f(x) at x=x_{0} is in absolute value less than 1, all a_{0} \ge -1 will produce a sequence converging at x_{0} without oscillations...

    Kind regards

    \chi \sigma
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  8. #8
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    Yes the correction is true. Thank you very much.
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  9. #9
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    So, what would be the exact limit of the sequence defined in part c? This is just for myself. But would you square the recursive equation and take limits using some limit theorems?
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  10. #10
    MHF Contributor chisigma's Avatar
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    Let suppose that a sequence is defined by an initial term a_{0} an the recursive relation...

    \Delta_{n} = a_{n+1} - a_{n} = f(a_{n}) (1)

    ... where f(x) is a continous function in a \le x \le b. From (1) it follows immediately that ...

    \lim_{n \rightarrow \infty} a_{n} = a_{0} + \sum_{n=0}^{\infty} \Delta_{n} (2)

    The limit (2) will exist only if the series converges and a necessary condition for that is...

    \lim_{n \rightarrow \infty} \Delta_{n} =0 (3)

    In order to find a sufficient condition let suppose now that f(x) has in [a,b] a single zero in x=x_{0} and that in that interval is |f(x)|\le |r(x)| where r(x) is a linear function crossing the x-axis in x_{0} with slope equal to -1. Such situation is illustrated in figure...



    ... where r(x) is the red line. Under these condition we can use the root test to verify the convergence of the series in (2) and we find that \forall n is...

    \frac{\Delta_{n+1}}{\Delta_{n}} < 1 (4)

    ... so that the series converges and for (3) the sequence converges at x_{0}...

    Kind regards

    \chi \sigma
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