Approximation for Newton-Raphson Inverse Algorithm


by calebh
Tags: algorithm, approximation, inverse, newtonraphson
calebh
calebh is offline
#1
Jun19-09, 05:05 PM
P: 1
I am attempting to make an initial approximation for the inverse algorithm (1/x)

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

a = .5^n
The equation gets more accurate as n increases
http://www09.wolframalpha.com/input/...%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?
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