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Useful Computational Methods


My square-root iteration formula

Let Q be a positive number.
Then the successive approximations to √Q are given by
xn+1 = (xn + Q)/(xn +1) with x0 = 1.
I first discovered this simple formula using the formulas given in the Bhaskara-Brouncker Algorithm. Then a second proof was found.

Example:
For Q = 5, the first few approximations are x0 = 1
x1 = 3
x2 = 2
x3 = 7/3 = 2.333...
x4 = 11/5 = 2.2
x5 = 9/4 = 2.25
x6 = 29/13 = 2.230769231...
x8 = 38/17 = 2.23529...
x9 = 123/55 = 2.23636...
x10 = 199/89 = 2.235955056...
x11 = 161/72 = 2.236111...
x12 = 521/233 = 2.236051502...
etc.
Compare the values to √5 = 2.236067978...
So this method is not fast, but it does slowly converge to √5.
The beauty of the algorithm is that we may substitute any positive number for the starting value x0. The algorithm is not sensitive at all to the starting value. For instance, try x0 = 1000000. Then x1 = 1.0000039... so that we are almost as well off as assuming x0 = 1.

Note that the formuala xn+1 = (xn + Q)/(xn +1) will appear less surprising if you replace xn by √Q.
Then xn+1 = (√Q + Q)/(√Q +1) = √Q(1 + √Q)/(1 + √Q) = √Q.

-- George R. Thomas

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Last updated - December 20, 2003