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Math Help - Convergence of Sequence

  1. #1
    MHF Contributor harish21's Avatar
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    Convergence of Sequence

    For c>0, consider the following quadratic equation:

    x^2- x- c=0 , x>0

    Define the sequence { x_n} recursively fixing x_1 > 0 and then, if n is an index for which { x_n} has been defined, defining

    x_{n+1} = sqrt{c+x_n}

    Prove that the sequence { x_n} converges monotonically to the solution of the above equation!

    --------------------

    I tried to attempt this problem by setting x_1 = 1, and creating a sequence. It turns out to be a monotonic incereasing sequence. But how do we show that the sequence converges to the solution of the quadratic equation?

    Also I am unclear about the starting value of x_1. Is it ok to start with x_1 = 1, or does the starting value have to be another number?
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  2. #2
    MHF Contributor
    Prove It's Avatar
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    Quote Originally Posted by harish21 View Post
    For c>0, consider the following quadratic equation:

    x^2- x- c=0 , x>0

    Define the sequence { x_n} recursively fixing x_1 > 0 and then, if n is an index for which { x_n} has been defined, defining

    x_{n+1} = sqrt{c+x_n}

    Prove that the sequence { x_n} converges monotonically to the solution of the above equation!

    --------------------

    I tried to attempt this problem by setting x_1 = 1, and creating a sequence. It turns out to be a monotonic incereasing sequence. But how do we show that the sequence converges to the solution of the quadratic equation?

    Also I am unclear about the starting value of x_1. Is it ok to start with x_1 = 1, or does the starting value have to be another number?
    Isn't it because for a monotonically convergent sequence, then as n \to \infty, x_{n + 1} \approx x_n?

    So if x_{n + 1} = \sqrt{c + x_n}

    x_{n} = \sqrt{c + x_n} for large n

    x_n^2 = c + x_n

    x_n^2 - x_n - c = 0, the same Quadratic Equation given above...
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  3. #3
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by Prove It View Post
    Isn't it because for a monotonically convergent sequence, then as n \to \infty, x_{n + 1} \approx x_n?

    So if x_{n + 1} = \sqrt{c + x_n}

    x_{n} = \sqrt{c + x_n} for large n

    x_n^2 = c + x_n

    x_n^2 - x_n - c = 0, the same Quadratic Equation given above...
    got it. thanks
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  4. #4
    MHF Contributor chisigma's Avatar
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    The difference equation that generates the sequence is...

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

    ... and its solution depends from c and x_{0}. The 'fixed point' [if any] are the solution of the equation...

    f(x) = \sqrt{x + c} - x = 0 (2)

    ... that is, in the case we consider the positive value of the 'square root' ...

    x_{+} = \frac{1 + \sqrt{1+4c}}{2} (3)

    From (3) we deduce that x_{+} exists only if is  c > -\frac{1}{4} and from (2) that in that case any x_{0} > -c will produce a sequence converging at x_{+}. The function f(x) is represented in the figure for c=2\rightarrow x_{+}=2 ...



    The point x_{+} is an 'actratting fixed point' and, because the 'slope' of f(x) is negative and in absolute value less than 1 [see red line...] for any x_{0}> -2 the generated sequence will converge to x_{+} without oscillations...

    Kind regards

    \chi \sigma
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