Using Scheme to calculate Newton-Raphson's

Question

According to Newton-Raphson's:

X_n+1 = X_n - f(X_n) / f'(X_n)

(newtonRhap x f fx)

(newtonRhap 0.1 sin cos) => 0

(newtonRhap 2.0 
            (lambda (x) (- (* x x) x 6))             ; f
            (lambda (x) (- (* 2 x) 1  )) ) => 3      ; f'

How can I implement this function?

The routine will stop iteration when the change in solution is less than pre-set tolerance.

I use a global variable for it, (define TOL 1e-6):

#lang racket
(define TOL 1e-6)
(define (func f x) (f x))
(define (f x) (- (* x x) x 6))
(define (fx x) (- (* 2 x) 1))
(define x 2.0)
(define (newtonRhap x ff ffx)  
    ( (> (- x (/ ff ffx)) TOL)
      (newtonRhap (- x (/ ff ffx)) ff ffx)  
      (list x) ) )

(display (newtonRhap x (f x) (fx x)) )

If you just want to copy some code, this looks good: https://gist.github.com/steshaw/ac0f6e5c697cf75bd3ab (I haven't tried it myself.) — Ben Kovitz, Apr 08 '19 at 18:14
FYI the code in [this answer](https://stackoverflow.com/a/55852661/849891) is wrong. (to a moderator: this comment is not offensive. it is factual.) — Will Ness, May 28 '19 at 06:20

score 0 · Answer 1 · answered Apr 25 '19 at 15:17

0

Good attempt, you need to modify newtonRaph function to use IF though, like this

(define (newtonRhap x ff ffx)
  (if (> (- x (/ ff ffx)) TOL)
      (newtonRhap (- x (/ ff ffx)) ff ffx))
      (list x)))

answered Apr 25 '19 at 15:17

river

1,028
6
16

score 0 · Answer 2 · answered May 27 '19 at 15:45

First, you're missing an if there:

(define (newtonRhap x ff ffx)
    ( if (> (- x (/ ff ffx)) TOL)
      (newtonRhap (- x (/ ff ffx)) ff ffx)
      (list x) ) )

Without it you would be calling the result of the Boolean test as a function with the following two forms values as arguments (it could work in a lazy functional language with Church-encoded conditionals, but that's besides the point, and Scheme / Racket is not lazy).

Second, and even more crucially, the last two arguments to newtonRhap are functions, but you're treating them as if they were numeric values. Correcting this gives us

(define (nl x)     ; for debugging insight;
   (newline)       ; for production redefine it as
   (display x)     ;    (define (nl x) x)
   x)  

(define (newtonRhap x ff ffx)
  (let ((new-x (nl (- x (/ (ff x) (ffx x))))))  ; Follow The Formula
    ( if (> (abs (- x new-x)) TOL)             ; use abs
      (newtonRhap new-x ff ffx)          ; make next step
      (list new-x) ) ) )        ; return the new x, it's closer to the true result

Now we can run

(display (newtonRhap x f fx))

3.333333333333333
3.019607843137255
3.0000762951094835
3.000000001164153
3.0(3.0)
>

Using Scheme to calculate Newton-Raphson's

2 Answers2