Exercise 2.13
Show that under the assumption of small percentage tolerances there is a simple formula for the approximate percentage tolerance of the product of two intervals in terms of the tolerances of the factors. You may simplify the problem by assumingthat all numbers are positive.
After considerable work, Alyssa P. Hacker delivers her finished system. Several years later, after she has forgotten all about it, she gets a frenzied call from an irate user, Lem E. Tweakit. It seems that Lem has noticed that the formula for parallel resistors can be written in two algebraically equivalent ways:
He has written the following two programs, each of which computes the parallel-resistors formula differently:
(define (par1 r1 r2)
(div-interval (mul-interval r1 r2)
(add-interval r1 r2)))
(define (par2 r1 r2)
(let ((one (make-interval 1 1)))
(div-interval
one (add-interval (div-interval one r1)
(div-interval one r2)))))
Lem complains that Alyssa's program gives different answers for the two ways of computing. This is a serious complaint.
這道題一是要修改乘法函式，二是要演示兩種計算並聯電阻的方法計算的結果不一致。
(define (mul-interval x y)
(let ((cx (center x))
(cy (center y))
(px (/ (percent x) 100))
(py (/ (percent y) 100)))
(make-interval (* (- cx (* cx px)) (- cy (* cy py)))
(* (+ cx (* cx px)) (+ cy (* cy py))))))
(define r1 (make-interval 6.12 7.48))
(define r2 (make-interval 4.465 4.935))
(display-interval-tolerance (par1 r1 r2))
(display-interval-tolerance (par2 r1 r2))
; 執行結果
2.844199964577264 ± 22.613352145193332%
2.777440701636504 ± 7.05260392723452%
Exercise 2.14
Demonstrate that Lem is right. Investigate the behavior of the system on a variety of arithmetic expressions. Make some intervals A and B, and use them in computing the expressions A/A and A/B. You will get the most insight by using intervals whose width is a small percentage of the center value. Examine the results of the computation in center-percent form (see Exercise 2.12).
(define r1 (make-interval 6.12 7.48))
(define r2 (make-interval 4.465 4.935))
(display-interval (div-interval r1 r1))
(display-interval (div-interval r1 r2))
(newline)
(display-interval-tolerance (div-interval r1 r1))
(display-interval-tolerance (div-interval r1 r2))
; 執行結果
[0.8181818181818182, 1.222222222222222]
[1.2401215805471126, 1.6752519596864504]
1.02020202020202 ± 19.801980198019795%
1.4576867701167815 ± 14.925373134328357%
根據執行結果來看，Lem顯然是對的，對於 A/A，答案顯然應該是 1 才對，但是程式計算出來的卻是一個相對很大的範圍，這是因為計算區間除法時，我們並沒有定義“相等”這個概念，所以計算 A/A 時，它也按照通用的方法去計算了。
Exercise 2.15
Eva Lu Ator, another user, has also noticed the different intervals computed by different but algebraically equivalent expressions. She says that a formula to compute with intervals using Alyssa's system will produce tighter error bounds if it can be written in such a form that no variable that represents an uncertain number is repeated. Thus, she says, par2 is a "better" program for parallel resistances than par1. Is she right? Why?
Eva 說的是對的，對於練習 2.13 計算出的兩個答案，par2 更準確，跟書上給出的答案是一致的，至於原因我覺得應該是區間之間的運算每次都會增加不確定性，par2 其實只有相加那一步是兩個區間進行運算，其他的都可以看作確定數字與區間的運算，沒有增加額外的不確定性；而 par1 進行了3次區間之間的運算————乘法、加法和除法，乘法和除法都會使結果變得不確定。
Exercise 2.16
Explain, in general, why equivalent algebraic expressions may lead to different answers. Can you devise an interval-arithmetic package that does not have this shortcoming, or is this task impossible? (Warning: This problem is very difficult.)
(define a (make-interval 2 4))
(define b (make-interval -2 0))
(define c (make-interval 3 8))
(define x (mul-interval a (add-interval b c)))
(define y (add-interval (mul-interval a b)
(mul-interval a c)))
(display-interval x)
(display-interval y)
; 執行結果
[2, 32]
[-2, 32]
根據上面的結果可以看到，區間運算連乘法分配律都無法保證。至於有沒有無缺陷的區間運算規則，我大概想了一下，沒有什麼頭緒，就沒繼續想了。。