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Approximation for Newton-Raphson Inverse Algorithm

  1. Jun 19, 2009 #1
    I am attempting to make an initial approximation for the inverse algorithm (1/x)

    Code (Text):

    n = NUMBER TO INVERSE
    a = APPROXIMATION

    a = a*(2-(n*a))
     
    'a' gets closer to the actual result each time the algorithm is preformed

    The problem is finding the initial approximation. An exponential equation seems to fit the best

    Code (Text):

    a = .5^n
     
    The equation gets more accurate as n increases
    http://www09.wolframalpha.com/input/?i=%281%2Fx%29-%28.5^x%29

    I chose .5, because in binary, dividing by two is as simple as shifting to the right.

    Is there any other way to make a close approximation that is better than .5^n?
     
  2. jcsd
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