# Is this a fix?

To find a fix-point function suitable for finding a square root, for instance of 2, we can begin with

t^2= 2

subtract the square of an integer from both sides (in this case 1) gives:

t^2-1=1

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

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

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

and finally add 1 to both sides giving

t=1+1/(t+1)

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

Plot the graph of y(t)=t^2-2 and perform zero finding using fix(t). The value of t found should be the square root of 2.

Use the same method to find the square root of 5.

Can a similar method of rearrangement work for cube roots?

#sequences #equationsolving #numericalmethods

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