Exercise 2.13 of SICP

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 assuming that all numbers are positive.

(a-\epsilon_1,a+\epsilon_1) \cdot (b-\epsilon_2,b+\epsilon_2)
(a+\epsilon_1)\cdot(b+\epsilon_2)=ab+a\epsilon_2+b\epsilon_1+\epsilon_1\epsilon_2
(a+\epsilon_1)\cdot(b-\epsilon_2)=ab-a\epsilon_2+b\epsilon_1-\epsilon_1\epsilon_2
(a-\epsilon_1)\cdot(b+\epsilon_2)=ab+a\epsilon_2-b\epsilon_1-\epsilon_1\epsilon_2
(a-\epsilon_1)\cdot(b-\epsilon_2)=ab-a\epsilon_2-b\epsilon_1+\epsilon_1\epsilon_2

Since a>0 and b>0:
(a+\epsilon_1)\cdot(b+\epsilon_2) \approx ab+ a\epsilon_2+b\epsilon_1+\epsilon_1\epsilon_2
(a-\epsilon_1)\cdot(b-\epsilon_2) \approx ab -a\epsilon_2-b\epsilon_1+\epsilon_1\epsilon_2
Assuming \epsilon_1 and \epsilon_2 are small, the quantity \epsilon_1\epsilon_2 can be discarded.
(a+\epsilon_1)\cdot(b+\epsilon_2) \approx ab+a\epsilon_2+b\epsilon_1
(a-\epsilon_1)\cdot(b-\epsilon_2) \approx ab-a\epsilon_2-b\epsilon_1

Since we are interested in the tolerance terms:
(a+\epsilon_1)\cdot(b+\epsilon_2) \approx a\epsilon_2+b\epsilon_1
(a-\epsilon_1)\cdot(b-\epsilon_2) \approx -a\epsilon_2-b\epsilon_1

The new product interval in terms of small tolerance terms looks like this:
[ab-(a\epsilon_2+b\epsilon_1),ab+(a\epsilon_2+b\epsilon_1)]

Leave a Reply