x = 2.5, a better approximation to the square root of 4.0.
Now using x = 2.5 in the formula yields:
x' = (1/2)(2.5 + 4/2.5)
= (1/2)(2.5 + 1.6)
= (1/2)(4.1)
= 2.05
This is a better yet approximation, and could be fed back into the formula to get an even better approximation.
## newton.asm -- compute sqrt(n)
## given an approximation x to sqrt(n),
## an improved approximation is:
## x' = (1/2)(x + n/x)
## $f0 --- n
## $f1 --- 1.0
## $f2 --- 2.0
## $f3 --- x : current approx.
## $f4 --- x' : next approx.
## $f8 --- temp
## $f10 --- accuracy limit, a small value
.text
.globl main
main:
l.s $f0,n # get n
li.s $f1,1.0 # constant 1.0
li.s $f2,2.0 # constant 2.0
li.s $f3,1.0 # x == first approx.
li.s $f10,1.0e-5 # accuracy limit
loop:
. . . . # to be continued
## Data Segment
##
.data
n: .float 3.0
This example program implements this procedure. It repeatedly uses the formula to calculate increasingly accurate approximations to the square root.
The program will stop looping when no further improvement is possible. We can't expect to get the answer exactly correct.
The program uses a hard-coded value for n,
but could easily be improved by asking the user for it.
It should also check that the value is greater than zero.
(Memory test: ) How many decimal places of precision can be expected with floating point?