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# SICP exercise 1.40

From Drewiki

## Problem

Define a procedure `cubic` that can be used together with the `newtons-method`
procedure from the text in expressions of the form `(newtons-method (cubic a b c) 1)`
to approximate zeros of the cubic <math>x^3 + ax^2 + bx + c</math>.

## Solution

Here's `cubic`:

(define (cube x) (* x x x)) (define (square x) (* x x)) (define (cubic a b c) (lambda (x) (+ (cube x) (* a (square x)) (* b x) c)))

And here's `newtons-method` and friends from the text:

(define tolerance 0.00001) (define (fixed-point f first-guess) (define (close-enough? v1 v2) (< (abs (- v1 v2)) tolerance)) (define (try guess) (let ((next (f guess))) (if (close-enough? guess next) next (try next)))) (try first-guess)) (define dx 0.00001) (define (deriv g) (lambda (x) (/ (- (g (+ x dx)) (g x)) dx))) (define (newton-transform g) (lambda (x) (- x (/ (g x) ((deriv g) x))))) (define (newtons-method g guess) (fixed-point (newton-transform g) guess))

Tests performed using Chicken Scheme 3.1 on a MacBook Pro running Mac OS X 10.5.

(newtons-method (cubic 1 1 1) 1)

*Output:*

`
`

-0.99999999999978

(newtons-method (cubic 0 0 0) 1)

*Output:*

`
`

2.65319902917972e-05

(newtons-method (cubic 0 0 -1) 1)

*Output:*

`
`

1.0

(newtons-method (cubic 2 1 2) 1)

*Output:*

`
`

-1.99999999998065