Adv Alegbra HW Solutions 18

Adv Alegbra HW Solutions 18 - n = 0 is also f ne; the only...

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18 Write n = 2 m + 1. Only odd powers of i are imaginary, so that, if r = 2 k + 1, sin nx = X 0 k m ³ n 2 k + 1 ´ ( 1 ) k sin 2 k + 1 x cos 2 ( m k ) x . But cos 2 ( m k ) x = ( cos 2 x ) m k = ( 1 sin 2 x ) m k , and so we have expressed sin nx as a polynomial in sin x . 1.38 (i) What is the coef f cient of x 16 in ( 1 + x ) 20 ? Solution. Pascal s formula gives ( 20 16 ) = 4845. (ii) How many ways are there to choose 4 colors from a palette con- taining paints of 20 different colors? Solution. Pascal s formula gives ( 20 4 ) = 4845. One could also have used part (i) and Exercise 1.30. 1.39 Give at least two different proofs that a set X with n elements has exactly 2 n subsets. Solution. There are many proofs of this. We offer only three. Algebraic . Let X ={ a 1 , a 2 ,..., a n } . We may describe each subset S of X by a bitstring ; that is, by an n -tuple 1 2 ,...,² n ), where ² i = 0i f a i is not in S 1i f a i is in S . (after all, a set is determined by the elements comprising it). But there are exactly 2 n such n -tuples, for there are two choices for each coordinate. Combinatorial . Induction on n 1 (taking base step
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Unformatted text preview: n = 0 is also f ne; the only set with 0 elements is X = , which has exactly one subset, itself). If X has just one element, then there are two subsets: and X . For the inductive step, assume that X has n + 1 elements, of which one is colored red, the other n being blue. There are two types of subsets S : those that are solid blue; those that contain the red. By induction, there are 2 n solid blue subsets; denote them by B . But, there are as many subsets R containing the red as there are solid blue subsets: each R arises by adjoining the red element to a solid blue subset, namely, B = R { red } (even the singleton subset consisting of the red element alone arises in this way, by adjoining the red element to ). Hence, there are 2 n + 2 n = 2 n + 1 subsets....
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