SICP 1.6-1.8

1.6
由于scheme应用序求值的性质,该函数会陷入循环(一直计算 improve guess x)

1.7

值过小,0.001这个精度远远不够...

值过大,由于浮点数精度有限,(比如使用IEEE754浮点数标准, 32位浮点数表示123456789这样的数会有严重的精度丢失), 所以无法得出正确求两个较大的数的差.

 (define (abs x)
   (if (>= x )
       x
       (- x)))

 (define (average a b)
   (/ (+ a b) ))

 (define (improve guess x)
   (average guess
            (/ x guess)))

 (define (good-enough? guess x)
   (< (/ (abs (- (improve guess x)
                 guess))
         guess)
      0.01))

 (define (sqrt-iter guess x)
   (cond ((good-enough? guess x) guess)
         (else (sqrt-iter (improve guess x) x))
         ))

 (define (sqrt x)
   (sqrt-iter 1.0 x))

---> 原则上可以的,但是其实精度损失是肯定的...

1.8

 (define (improve2 guess x)
   (/ (+ (/ x
            (* guess guess))
         (*  guess))
      ))

 (define (good-enough2? guess x)
   (< (/ (abs (- (improve2 guess x)
                 guess))
         guess)
      0.01))

 (define (cbrt-iter guess x)
   (cond ((good-enough2? guess x) guess)
         (else (cbrt-iter (improve2 guess x)
                          x))))

 (define (cbrt x)
   (cbrt-iter 1.0 x))