Unformatted text preview: , b1 = 11π , b−1 = − π , b2 = 17π and
b−2 = − 73 . Hence, in terms of the a’s and b’s, our domain is
. . . (a−2 , b−2 ) ∪ (b−2 , a−1 ) ∪ (a−1 , b−1 ) ∪ (b−1 , a0 ) ∪ (a0 , b0 ) ∪ (b0 , a1 ) ∪ (a1 , b1 ) ∪ . . .
If we group these intervals in pairs, (a−2 , b−2 ) ∪ (b−2 , a−1 ), (a−1 , b−1 ) ∪ (b−1 , a0 ), (a0 , b0 ) ∪ (b0 , a1 )
and so forth, we see a pattern emerge of the form (ak , bk ) ∪ (bk , ak + 1 ) for integers k so that
our domain can be written as
∞ ∞ (ak , bk ) ∪ (bk , ak + 1 ) =
k=−∞ k=−∞ (6k + 1)π (6k + 5)π
3 ∪ (6k + 5)π (6k + 7)π
3 A second approach to the problem exploits the periodic nature of f . It is based on the
same premise on which our equation solving technique is based – for a periodic function,
if we understand what happens on a fundamental period, we can know what is happening
everywhere by adding integer multiples of the period. Since cos(x) and sin(x) have period
2π , it’s not too diﬃcult to show the function f is perio...
View Full Document