2.13

练习 2.13 请证明，在误差为很小的百分数的条件下，存在着一个简单公式，利用它可以从两个被乘区间的误差算出乘积的百分数误差值。你可以假定所有的数为正，以简化这一问题。

在前面的练习2.11 中，已经讨论了区间相乘的 9 种情况。在所有的数都为正的情况下，有：

$$[x_1, x_2] \cdot [y_1, y_2] = [x_1\cdot y_1, x_2 \cdot y_2]$$

一个区间的百分数误差值和宽度以及中心点的关系是：

$$center \cdot percent = width$$

(define (make-interval a b) (cons a b))
(define upper-bound cdr)
(define lower-bound car)
(define (add-interval x y) 
    (make-interval (+ (lower-bound x) (lower-bound y))
                    (+ (upper-bound x) (upper-bound y))))
(define (sub-interval x y) (add-interval x (make-interval (- 0 (upper-bound y)) (- 0 (lower-bound y)))))
(define (subtract-interval x y)
    (define p1 (- (lower-bound x) (lower-bound y)))
    (define p2 (- (lower-bound x) (upper-bound y)))
    (define p3 (- (upper-bound x) (lower-bound y)))
    (define p4 (- (upper-bound x) (upper-bound y)))

    (make-interval (min p1 p2 p3 p4) (max p1 p2 p3 p4)))

(define (mul-interval x y)
    (define p1 (* (lower-bound x) (lower-bound y)))
    (define p2 (* (lower-bound x) (upper-bound y)))
    (define p3 (* (upper-bound x) (lower-bound y)))
    (define p4 (* (upper-bound x) (upper-bound y)))

    (make-interval (min p1 p2 p3 p4) (max p1 p2 p3 p4))
)

(define int1 (make-interval 2 4))
(define int2 (make-interval 3 6))
(mul-interval int1 int2)

(define (make-center-percent center percent)
  (make-interval (- center (* center percent))
                 (+ center (* center percent))))

(define (center x) (/ (+ (lower-bound x) (upper-bound x)) 2))

(define (percent x)
  (/ (width x) (center x)))

(define int (make-center-percent 2 0.5))
int

(center int)

(define (width x) (/ (- (upper-bound x) (lower-bound x)) 2))
(percent int)

先用几个例子找一下规律：

(define i1 (make-center-percent 2 0.5))
(define i2 (make-center-percent 2 0.25))

i1

i2

(define i3 (mul-interval i1 i2))
i3

(percent i3)

看起来是 (1/2) * (1/4) = (1/3) = (1+1) / (2+4)

不过，这个直观的规律和并不能和以上的定义产生什么联系。还是老老实实地计算吧。

证明

令 $$int1 = [x_1, x2], int2 = [y_1, y_2]$$ 以及 $$[x_1, x_2] \cdot [y_1, y_2] = [x_1\cdot y_1, x_2 \cdot y_2] = int3$$

int3 的百分数误差值是

$$p_3 = \frac{\frac{x_2 y_2 - x_1 y_1}{2}}{\frac{x_1 y_1 + x_2 y_2}{2}} = \frac{x_2 y_2 - x_1 y_1}{x_1 y_1 + x_2 y_2}$$

又

$$p_1 = \frac{x_2 - x1}{x_1 + x_2} p_2 = \frac{y_2 - y1}{y_1 + y_2}$$

引入中心点的关系，

$$c_1 = \frac{x_1 + x_2}{2}, c_2 = \frac{y_1 + y_2}{2}$$

即：

$$x_1 = c_1 - c_1 p_1, x_2 = c_1 + c_1 p_1, y_1 = c_2 - c_2 p_2, y_2 = c_2 + c_2 p_2$$

代入上面的公式，得到

$$p_3 = \frac{x_2 y_2 - x_1 y_1}{x_1 y_1 + x_2 y_2}$$ $$= \frac{(c_1 + c_1 p_1) (c_2 + c_2 p_2) - (c_1 - c_1 p_1) (c_2 - c_2 p_2)}{(c_1 - c_1 p_1) (c_2 - c_2 p_2) + (c_1 + c_1 p_1) (c_2 + c_2 p_2)}$$ $$= \frac{(c_1 c_2 + c_1 c_2 p_2 + c_1 p_1 c_2 + c_1 p_1 c_2 p_2) - (c_1 c_2 - c_1 c_2 p_2 - c_1 p_1 c_2 - c_1 p_1 c_2 p_2)}{(c_1 - c_1 p_1) (c_2 - c_2 p_2) + (c_1 + c_1 p_1) (c_2 + c_2 p_2)}$$ $$= \frac{p_1+p_2}{1+p_1 p_2}$$

由于 $p1, p2$ 都很小，所以相乘后的结果几乎为0，从而以上公式可以得到

$$p_3 \approx p_1 + p_2$$