問題1-36 - 理系学生日記

近似値を印字するようにfixed-pointを修正してやった。

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

で、 $x^x=1000$ の解を求めてやるお！
まずは、average dampingを使わないで、 $x\mapsto \log (1000)/\log(x)$ の不動点をさがす。

gosh> (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

なげーーー。
次はaverage damping使って、 $x\mapsto \frac 12 \left(\frac{\log 1000}{\log x}+x\right)$ の不動点をさがす。

gosh> (fixed-point (lambda (x)
	       (/ (+ (/ (log 1000) (log x)) x) 2))
	     2.0)

2.0
5.9828921423310435
4.922168721308343
4.628224318195455
4.568346513136242
4.5577305909237005
4.555909809045131
4.555599411610624
4.5555465521473675
4.555537551999825

というわけで、average dampingはすごい！というおはなし。