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 xx = 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.)
如下来求解不动点，初始值只要不为 1.0 就可以：
(define (fixed-point f first-guess)
(define tolerance 0.00001)
(define (close-enough? v1 v2)
(< (abs (- v1 v2)) tolerance))
(define (try guess)
(let ((next (f guess)))
(newline)
(display guess)
(if (close-enough? guess next)
next
(try next))))
(try first-guess))
; 1
(fixed-point (lambda (x)
(/ (log 1000) (log x)))
2.0)
结果为：
2.
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
然后使用 average damping 来求不动点：
(fixed-point (lambda (x)
(/ (+ x (/ (log 1000) (log x)))
2))
2.0)打印的结果为：
2.
5.9828921423310435
4.922168721308343
4.628224318195455
4.568346513136242
4.5577305909237005
4.555909809045131
4.555599411610624
4.5555465521473675
average damping 因为每次都会求 x 与现在的值的平均值，会比直接求值收敛得更快。