問題1.6

#lang racket

;;;;Newton法による平方根

;;;2乗
(define (square x)
  (* x x))

;;;平均
(define (average x y)
  (/ (+ x y) 2))

;;;予測値の2乗が近似値に近いか
(define (good-enough? guess x)
  (< (abs (- (square guess) x)) 0.001))

;;;予想値の商と平均で次の予想値を求める
(define (improve guess x)
  (average guess (/ x guess)))

;;;ifの新版を定義
(define (new-if predicate then-clause else-clause)
  (cond (predicate then-clause)
        (else else-clause)))

(define (sqrt-iter guess x)
  (new-if (good-enough? guess x)
      guess
      (sqrt-iter (improve guess x)
                 x)))

;;;平方根の近似値を求める
#|
(define (sqrt-iter guess x)
  (if (good-enough? guess x)
      guess
      (sqrt-iter (improve guess x)
                 x)))
|#
;;;開始
(define (sqrt x)
  (sqrt-iter 1.0 x))

(sqrt 2)
(sqrt 3)
(sqrt (+ 100 37))

;;;ifの新版を定義
(define (new-if predicate then-clause else-clause)
  (cond (predicate then-clause)
        (else else-clause)))

(define (sqrt-iter guess x)
  (new-if (good-enough? guess x)
      guess
      (sqrt-iter (improve guess x)
                 x)))

;;;平方根の近似値を求める
#|
(define (sqrt-iter guess x)
  (if (good-enough? guess x)
      guess
      (sqrt-iter (improve guess x)
                 x)))
|#