I am trying to find the cube root of a number using Newton's method. I wrote scheme procedures as follows:
(define (cbrt x)
(cbrt-iter 1.0 x))
(define (cbrt-iter guess x)
(if (good-enough? guess x) guess (cbrt-iter (improve guess x) x)))
(define (good-enough? guess x)
(< (- guess (improve guess x)) 0.00001))
(define (improve guess x)
(/ (+ (/ x (* guess guess)) (* 2 guess)) 3))
(cbrt 27)
(cbrt 8)
(cbrt 64)
Actually I am working on Exercise 1.8 of the famous (or may be infamous) book SICP. then I run scheme < cuberoot.scm and got the following result:
MIT/GNU Scheme running under GNU/Linux
Type `^C' (control-C) followed by `H' to obtain information about interrupts.
Copyright (C) 2019 Massachusetts Institute of Technology
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
Image saved on Thursday September 5, 2019 at 11:51:46 AM
Release 10.1.10 || Microcode 15.3 || Runtime 15.7 || SF 4.41 || LIAR/x86-64 4.118
1 ]=> (define (cbrt x)
(cbrt-iter 1.0 x))
;Value: cbrt
1 ]=> (define (cbrt-iter guess x)
(if (good-enough? guess x) guess (cbrt-iter (improve guess x) x)))
;Value: cbrt-iter
1 ]=> (define (good-enough? guess x)
(< (- guess (improve guess x)) 0.00001))
;Value: good-enough?
1 ]=> (define (improve guess x)
(/ (+ (/ x (* guess guess)) (* 2 guess)) 3))
;Value: improve
1 ]=> (cbrt 27)
;Value: 1.
1 ]=> (cbrt 8)
;Value: 1.
1 ]=> (cbrt 64)
;Value: 1.
1 ]=>
End of input stream reached.
Post proelium, praemium.
The program is always producing 1. as a result. I also tried adjusting the threshold value in good-enough? procedure from 0.00001 to 0.0001 and so on but that didn't worked.
Please explain what went wrong and how to fix that.
You need to include an (abs ...) in good-enough?, otherwise you don't just see if two values are close enough, but just if one value is greater than another (approximately).
(define (good-enough? guess x)
(< (abs (- guess (improve guess x))) 0.00001))