1 day ago

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)

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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