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# psol09 - Solutions to Homework 9 Math 55 Section 6.4 2 The...

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Unformatted text preview: Solutions to Homework 9, Math 55 Section 6.4 2. The generating function is 1 + 4 x + 16 x 2 + 64 x 3 + 256 x 4 = ∑ 4 n =0 (4 x ) n = 1- (4 x ) 5 1- 4 x = 1- 1024 x 5 1- 4 x . 8. (a) Since ( x 2 + 1) 3 = (1 + x 2 ) 3 = ∑ 3 i =0 ( 3 i ) x 2 i , we have a n = ( 3 n/ 2 ) if 0 ≤ n ≤ 6 and n is even; otherwise, a n = 0. (b) Similarly, (3 x- 1) 3 = ∑ 3 n =0 ( 3 n ) (3 x ) n (- 1) 3- n ; thus, if 0 ≤ n ≤ 3, then a n = (- 1) 3- n 3 n ( 3 n ) , and otherwise a n = 0. (c) We have 1 1- 2 x 2 = ∑ ∞ k =0 (2 x 2 ) k = ∑ ∞ k =0 2 k x 2 k . Thus, if n is even, then a n = 2 n/ 2 , while if n is odd, then a n = 0. (d) The given function is equal to x 2 (1- x )- 3 . By the binomial theorem, this is equal to x 2 ∑ ∞ k =0 (- 3 k ) (- 1) k x k = ∑ ∞ k =0 (- 3 k ) (- 1) k x k +2 . Thus, to get a term x n we set k = n- 2; if n < 2, this gives a n = 0, while if n ≥ 2, this gives a n = (- 3 n- 2 ) (- 1) n . By the result of example 8 on page 438, we can write this as a n = ( n n- 2 ) = ( n 2 ) . (e) The given function is equal to x- 1 + ∑ ∞ n =0 3 n x n . Thus, a = 3- 1 = 0; a 1 = 3 1 + 1 = 4; and if n ≥ 2, then a n = 3 n . (f) We rewrite 1+ x 3 (1+ x ) 3 = 1- 3 x +3 x 2 (1+ x ) 3 = 1- 3 x (1+ x ) 2 . This is equal to 1- 3 x ∑ ∞ k =0 (- 2 k ) x k = 1- ∑ ∞ k =0 3 (- 2 k ) x k +1 . Thus, we have a = 1, while for n ≥ 1, we set k = n- 1 to get the x n term, so a n =- 3 (- 2 n- 1 ) . Again using example 8, we can rewrite this as a n = 3 n (- 1) n . (g) If we multiply the top and bottom by 1- x , then this is equal to x- x 2 1- x 3 = ( x- x 2 ) ∑ ∞ k =0 x 3 k = ∑ ∞ k =0 ( x 3 k +1- x 3 k +2 ). Thus, we get a n = , if n ≡ (mod 3); 1 , if n ≡ 1 (mod 3);- 1 , if n ≡ 2 (mod 3) . Alternately, let ω =- 1+ i √ 3 2 ; then we can factor 1 + x + x 2 = (1- ωx )(1- ωx ), where ω =- 1- i √ 3 2 is the complex conjugate. We can now use partial fractions to write x 1 + x + x 2 = 1 i √ 3 1 1- ωx- 1 1- ωx = 1 i √ 3 ∞ X n =0 ( ω n- ω n ) x n . Thus, a n = 1 i √ 3 ( ω n- ω n ). Since ω = e 2 πi/ 3 and ω = e- 2 πi/ 3 , this can be rewritten in terms of real numbers as a n = 2 √ 3 sin( 2 πn 3 ). (h) The given function expands to- 1 + ∑ ∞ k =0 (3 x 2 ) k /k ! =- 1 + ∑ ∞ k =0 (3 k /k !) x 2 k . Thus, if n is odd, then a n = 0; a =- 1 + 1 = 0; and if n ≥ 2 is even, then a n = 3 n/ 2 / ( n/ 2)!....
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psol09 - Solutions to Homework 9 Math 55 Section 6.4 2 The...

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