my attempt to do the exercises in sicp.

Monday, June 30, 2008

sicp exercise 1.36

;  Exercise 1.36. Modify fixed-point so that it prints the sequence of approximations it generates, using
;  the newline and display primitives shown in exercise 1.22. Then find a solution to x^x = 1000 by finding
;  a fixed point of x --> log(1000)/log(x). (Use Scheme's primitive log procedure, which computes natural
;  logarithms.) Compare the number of steps this takes with and without average damping. (Note that you
;  cannot start fixed-point with a guess of 1, as this would cause division by log(1) = 0.)

(define tolerance 0.00001)
(define (fixed-point f first-guess)
   (define (close-enough? v1 v2)
      (< (abs (- v1 v2)) tolerance))
   (define (try guess)
      (display (rationalize guess 0.00001))(newline)
      (let ((next (f guess)))
        (if (close-enough? guess next)
            (try next))))
   (try first-guess))

(define (average x y) (/ (+ x y) 2))
(define (func x) (/ (log 1000) (log x)))
(define (func-avg-damp x) (average x (/ (log 1000) (log x))))

;(fixed-point func 2)(newline)
(fixed-point func-avg-damp 2)(newline)

; Answer:
; 4.55554089709763  in 34 steps without average damping
; 4.55555555555556  in 9 steps with average damping

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