1 (define tolerance 0.0000001)
2 (define (fixed-point f first-guess)
3   (define (close-enough? v1 v2)
4     (< (abs (- v1 v2)) tolerance))
5   (define (try guess)
6     (let ((next (f guess)))
7       (if (close-enough? guess next)
8 	  next
9 	  (try next))))
10   (try first-guess))
11 
12 (define (average x y)
13   (/ (+ x y) 2.0))
14 (define (average-damp f)
15   (lambda (x) (average x (f x))))
16 (define (fixed-point-of-transform g transform guess)
17   (fixed-point (transform g) guess))
18 (define (sqrt x)
19   (fixed-point-of-transform (lambda (y) (/ x y))
20 			    average-damp
21 			    1.0))
22 (define (cube-root x)
23   (fixed-point-of-transform (lambda (y) (/ x (square y)))
24 			    average-damp
25 			    1.0))
26 
27 (define (compose f g)
28   (lambda (x)
29     (f (g x))))
30 
31 (define (test-case actual expected)
32   (load-option 'format)
33   (newline)
34   (format #t "Actual: ~A Expected: ~A" actual expected))
35 
36 (define (square x) (* x x))
37 (define (inc x) (1+ x))
38 ;; (test-case ((compose square inc) 6) 49)
39 
40 (define (repeated f n)
41   (if (= n 0)
42       (lambda (x) x)
43       (compose f (repeated f (- n 1)))))
44 
45 ;; (test-case ((repeated square 2) 5) 625)
46 ;; (test-case (cube-root 5) 1.70997594668)
47 
48 
49 ;; Exercise 1.44.  The idea of smoothing a function is an important concept in signal processing. If f is a function and dx is some small number, then the smoothed version of f is the function whose value at a point x is the average of f(x - dx), f(x), and f(x + dx). Write a procedure smooth that takes as input a procedure that computes f and returns a procedure that computes the smoothed f. It is sometimes valuable to repeatedly smooth a function (that is, smooth the smoothed function, and so on) to obtained the n-fold smoothed function. Show how to generate the n-fold smoothed function of any given function using smooth and repeated from exercise 1.43. 
50 
51 (define dx 0.01)
52 
53 (define (smooth f)
54   (lambda (x)
55     (/ (+ (f x) 
56 	  (f (+ x dx))
57 	  (f (- x dx)))
58        3.0)))
59 (define (n-fold-smooth f n)
60   ((repeated smooth n) f))
61 
62 ;; Exercise 1.45.  We saw in section 1.3.3 that attempting to compute square roots by naively finding a fixed point of y x/y does not converge, and that this can be fixed by average damping. The same method works for finding cube roots as fixed points of the average-damped y x/y2. Unfortunately, the process does not work for fourth roots -- a single average damp is not enough to make a fixed-point search for y x/y3 converge. On the other hand, if we average damp twice (i.e., use the average damp of the average damp of y x/y3) the fixed-point search does converge. Do some experiments to determine how many average damps are required to compute nth roots as a fixed-point search based upon repeated average damping of y x/yn-1. Use this to implement a simple procedure for computing nth roots using fixed-point, average-damp, and the repeated procedure of exercise 1.43. Assume that any arithmetic operations you need are available as primitives. 
63 
64 (define (fast-expt b n)
65   (cond ((= n 0) 1)
66 	((even? n) (square (fast-expt b (/ n 2))))
67 	(else (* b (fast-expt b (- n 1))))))
68 
69 
70 (define (quartic-root x)
71   (fixed-point-of-transform (lambda (y) (/ x (cube y)))
72 			    (repeated average-damp 2)
73 			    1.0))
74 (define (nth-root-test x n d)
75   (fixed-point-of-transform (lambda (y) (/ x (expt y (- n 1))))
76 			    (repeated average-damp d)
77 			    1.0))
78 
79 ;; (test-case (nth-root-test 19 1 0) 19)
80 ;; (test-case (nth-root-test 19 2 1) 4.35889894)
81 ;; (test-case (nth-root-test 19 3 1) 2.66840165)
82 ;; (test-case (nth-root-test 19 4 2) 2.08779763)
83 ;; (test-case (nth-root-test 19 5 2) 1.80198313)
84 ;; (test-case (nth-root-test 19 6 2) 1.6335243)
85 ;; (test-case (nth-root-test 199 7 2) 2.13013723)
86 ;; (test-case (nth-root-test 199 8 3) 1.93801277)
87 ;; (test-case (nth-root-test 199 9 3) 1.80064508)
88 ;; (test-case (nth-root-test 199 10 3) 1.69779522)
89 ;; (test-case (nth-root-test 19999 11 3) 2.46037187)
90 ;; (test-case (nth-root-test 19999 12 3) 2.28253453)
91 ;; (test-case (nth-root-test 19999 13 3) 2.14213477)
92 ;; (test-case (nth-root-test 19999 14 3) 2.02868621)
93 ;; (test-case (nth-root-test 19999 15 3) 1.93523578)
94 ;; (test-case (nth-root-test 1999999 16 4) 2.47636331)
95 ;; (test-case (nth-root-test 1999999 17 4) 2.34773357)
96 
97 
98 ;; At first, my conclusion: average damp = sqrt(n), rounded down -- EXCEPT for 8...why is that?
99 ;; Maybe: 2^average damp = n
100 
101 (define (nth-root x n)
102   (fixed-point-of-transform (lambda (y) (/ x (expt y (- n 1))))
103 			    (repeated average-damp (truncate (/ (log n) (log 2))))
104 			    1.0))
105 
106 (test-case (nth-root 19 1) 19)
107 (test-case (nth-root 19 2) 4.35889894)
108 (test-case (nth-root 19 3) 2.66840165)
109 (test-case (nth-root 19 4) 2.08779763)
110 (test-case (nth-root 19 5) 1.80198313)
111 (test-case (nth-root 19 6) 1.6335243)
112 (test-case (nth-root 199 7) 2.13013723)
113 (test-case (nth-root 199 8) 1.93801277)
114 (test-case (nth-root 199 9) 1.80064508)
115 (test-case (nth-root 199 10) 1.69779522)
116 (test-case (nth-root 19999 11) 2.46037187)
117 (test-case (nth-root 19999 12) 2.28253453)
118 (test-case (nth-root 19999 13) 2.14213477)
119 (test-case (nth-root 19999 14) 2.02868621)
120 (test-case (nth-root 19999 15) 1.93523578)
121 (test-case (nth-root 1999999 16) 2.47636331)
122 (test-case (nth-root 1999999 17) 2.34773357)