Is this a fix? - Solution

To find a suitable fix point function to find the square root of 5 using a Fix point method, we can begin with:

t^2= 5

subtract the square of an integer from both sides (in this case 4) gives:

t^2-4=1

use the difference of two squares factorization on the left-hand-side to get

(t-2)(t+2)=1

divide by one of the factors on the left-hand-side giving (t-2)=1/(t+2)

and finally add 2 to both sides giving

t=2+1/(t+2)

i.e. the fix point function to use is fix(t)=2+1/(t+2).






To find a suitable fix point function to find the cubed root of 5, using a Fix point method, we can begin with:

t^3=5

In order to use the well-known factorisation of the difference of two cubes, we subtract an integer cube from both sides, in this case 1, giving

t^3-1=4

Factorise to give:
(t-1)(t^2+t+1)=4

Divide both sides by (t^2+t+1)
t-1 = 4/(t^2+t+1)
Finally add 1 to both sides to give

t =1+4/(t^2+t+1)

So fix(t)=1+4/(t^2+t+1)

To use this fix point function we increase the potential number of iterations to the maximum of 100, giving e.g. the result below.

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