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 guess) (newline)
    (let ((next (f guess)))
      (if (close-enough? guess next)
          next
          (try next))))
  (try first-guess))

1 ]=> (fixed-point (lambda (x) (/ (log 1000) (log x))) 3.0)
3.
6.287709822868153
3.757079790200296
5.218748919675315
4.180797746063314
4.828902657081293
4.386936895811029
4.671722808746095
4.481109436117821
4.605567315585735
4.522955348093164
4.577201597629606
4.541325786357399
4.564940905198754
4.549347961475409
4.5596228442307565
4.552843114094703
4.55731263660315
4.554364381825887
4.556308401465587
4.555026226620339
4.55587174038325
4.555314115211184
4.555681847896976
4.555439330395129
4.555599264136406
4.555493789937456
4.555563347820309
4.555517475527901
4.555547727376273
4.555527776815261
4.555540933824255
;Value: 4.555532257016376

1 ]=> (fixed-point (lambda (x) (/ (+ x (/ (log 1000) (log x))) 2)) 3.0)
3.
4.643854911434076
4.571212264484558
4.558225323866829
4.555994244552759
4.555613793442989
4.5555490009596555
4.5555379689379265
;Value: 4.55553609061889