Exercise 2.13 of SICP

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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−ϵ1,a+ϵ1)⋅(b−ϵ2,b+ϵ2) (a-\epsilon_1,a+\epsilon_1) \cdot (b-\epsilon_2,b+\epsilon_2)

(a+ϵ1)⋅(b+ϵ2)=ab+aϵ2+bϵ1+ϵ1ϵ2 (a+\epsilon_1)\cdot(b+\epsilon_2)=ab+a\epsilon_2+b\epsilon_1+\epsilon_1\epsilon_2

(a+ϵ1)⋅(b−ϵ2)=ab−aϵ2+bϵ1−ϵ1ϵ2 (a+\epsilon_1)\cdot(b-\epsilon_2)=ab-a\epsilon_2+b\epsilon_1-\epsilon_1\epsilon_2

(a−ϵ1)⋅(b+ϵ2)=ab+aϵ2−bϵ1−ϵ1ϵ2 (a-\epsilon_1)\cdot(b+\epsilon_2)=ab+a\epsilon_2-b\epsilon_1-\epsilon_1\epsilon_2

(a−ϵ1)⋅(b−ϵ2)=ab−aϵ2−bϵ1+ϵ1ϵ2 (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+ϵ1)⋅(b+ϵ2)≈ab+aϵ2+bϵ1+ϵ1ϵ2 (a+\epsilon_1)\cdot(b+\epsilon_2) \approx ab+ a\epsilon_2+b\epsilon_1+\epsilon_1\epsilon_2

(a−ϵ1)⋅(b−ϵ2)≈ab−aϵ2−bϵ1+ϵ1ϵ2 (a-\epsilon_1)\cdot(b-\epsilon_2) \approx ab -a\epsilon_2-b\epsilon_1+\epsilon_1\epsilon_2

Assuming $$ \epsilon_1$% and %$latex \epsilon_2$% are small, the quantity %$latex \epsilon_1\epsilon_2 $$ can be discarded.

(a+ϵ1)⋅(b+ϵ2)≈ab+aϵ2+bϵ1 (a+\epsilon_1)\cdot(b+\epsilon_2) \approx ab+a\epsilon_2+b\epsilon_1

(a−ϵ1)⋅(b−ϵ2)≈ab−aϵ2−bϵ1 (a-\epsilon_1)\cdot(b-\epsilon_2) \approx ab-a\epsilon_2-b\epsilon_1

Since we are interested in the tolerance terms:

(a+ϵ1)⋅(b+ϵ2)≈aϵ2+bϵ1 (a+\epsilon_1)\cdot(b+\epsilon_2) \approx a\epsilon_2+b\epsilon_1

(a−ϵ1)⋅(b−ϵ2)≈−aϵ2−bϵ1 (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−(aϵ2+bϵ1),ab+(aϵ2+bϵ1)] [ab-(a\epsilon_2+b\epsilon_1),ab+(a\epsilon_2+b\epsilon_1)]

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