本文共 2700 字，大约阅读时间需要 9 分钟。
练习 1.35
φ^2 = φ+1 ==> φ = 1 + (1/φ) (define tolerance 0.00001)(define (fixed-point f first-guess) (define (close-enough? v1 v2) (< (abs (- v1 v2)) tolerance)) (define (try guess) (let ((next (f guess))) (if (close-enough? guess next) next (try next)))) (try first-guess)) > (fixed-point (lambda (x) (+ 1 (/ 1 x))) 1.0)1.6180327868852458
练习 1.36
(define tolerance 0.00001)(define (fixed-point f first-guess) (define (close-enough? v1 v2) (and (report v1) (< (abs (- v1 v2)) tolerance))) (define (report v) (display " *** ") (display v) (newline)) (define (try guess) (let ((next (f guess))) (if (close-enough? guess next) next (try next)))) (try first-guess));; 不使用平均阻尼需要34次 > (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 *** 4.5555409129179574.555532270803653;; 使用平均阻尼需要9次 > (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 *** 4.55554655214736754.555537551999825
练习 1.37
;; 线性递归(define (cont-frac n d k) (if (= k 1) 1 (/ (n k) (+ (d k) (cont-frac n d (- k 1)))))) ;; 线性迭代(define (cont-frac n d k) (cont-frac-iter n d (/ (n 1) (d 1)) 1 k))(define (cont-frac-iter n d v c k) (if (= c k) v (cont-frac-iter n d (/ (n c) (+ (d c) v)) (+ c 1) k))) ;; k取值11时即可达到4位精度(取8位精度为1.61803399)1 ]=> (cont-frac (lambda (i) 1.0) (lambda (i) 1.0) 11)0.6180555555555556
练习 1.38
;; 欧拉连分式的Di过程define (d-euler i) (if (= (remainder (+ i 1) 3) 0) (* 2.0 (/ (+ i 1) 3.0)) 1.0)) ;; 求值1 ]=> (+ 2 (cont-frac (lambda (i) 1.0) d-euler 100));Value: 2.5037311291101405
练习 1.39
;; 过程定义应该如下, 但是并没有得到正确结果(define (tan-cf x k) (define (n-tan i) (if (= i 1) x (- (* x x)))) (cont-frac n-tan (lambda (i) (- (* 2.0 i) 1.0)) k))1 ]=> (tan-cf (/ 3.1415926 4) 100);Value: -3.1312698448737733e-3
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