연습문제 1.36

변경된 fixed-point는 아래와 같다.

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

이제 주어진 식 x->log(1000)/log(x)의 고정점을 찾아보자.

> (fixed-point (lambda (x) (/ (log 1000) (log x))) 2.0)

2.0
9.965784284662087
3.004472209841214
6.279195757507157
3.759850702401539
5.215843784925895
4.182207192401397
4.8277650983445906
4.387593384662677
4.671250085763899
4.481403616895052
4.6053657460929
4.5230849678718865
4.577114682047341
4.541382480151454
4.564903245230833
4.549372679303342
4.559606491913287
4.552853875788271
4.557305529748263
4.554369064436181
4.556305311532999
4.555028263573554
4.555870396702851
4.555315001192079
4.5556812635433275
4.555439715736846
4.555599009998291
4.555493957531389
4.555563237292884
4.555517548417651
4.555547679306398
4.555527808516254
4.555540912917957

4.555532270803653

35단계를 거친다.

이제 평균을 구해 그 값을 통해 구해보면..

> (fixed-point (lambda (x) (average x (/ (log 1000) (log x)))) 2.0)

2.0
5.9828921423310435
4.922168721308343
4.628224318195455
4.568346513136242
4.5577305909237005
4.555909809045131
4.555599411610624
4.5555465521473675
4.555537551999825

10단계를 거치게된다.