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Square Root

  1. Jan 30, 2005 #1

    dduardo

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    Can someone get me going in the right direction.

    For the given function:

    y[n] = (1/2)(y[n-1] + x[n]/y[n-1])

    where x[n] = a * u[n] (u[n] is the unit step function)

    and y[-1] = 1

    prove that y[n] as n -> infinity is equal to sqrt(a)

    -----------

    I know that this is the Newton-Raphson Method, but how do I go about analytically proving the above.

    I've tried writing out a few terms and seeing if there is a pattern, but couldn't find anything.

    The inside looks like a accumulator and tried to do a subsitution, but that didn't work.

    Any help would be appreciated.
     
    Last edited: Jan 30, 2005
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  3. Jan 30, 2005 #2

    Gokul43201

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    When you see limiting behavior, it means that the differencebetween successive terms gets smaller and smaller. In the infinite limit, the difference between successive terms should be 0.

    So, set y[n] = y[n-1] and see what happens.
     
  4. Jan 30, 2005 #3

    dduardo

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    I think I need a little more information.

    So you are saying lim n-> inf (y[n] - y[n-1]) = 0 ? But since i'm dealing with a circular function how would I go about taking the limit?
     
  5. Jan 30, 2005 #4

    Gokul43201

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    Yes, does it seem reasonable to you that this should be true ?

    There really isn't much to do. Let's call [itex]\lim_{n \rightarrow \infty} y[n] \equiv y [/itex]

    Then you have [tex] y = \frac {y}{2} + \frac {x}{2y} [/tex]

    This gives you a quadratic in y.
     
  6. Jan 30, 2005 #5

    dduardo

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    Ok, now I understand. Thanks
     
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