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HWSet-Chapter4

# HWSet-Chapter4 - Math 326 Fall 2010 Homework Set Chapter 4...

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Math 326 Fall 2010 Homework Set Chapter 4 Solutions 1

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4.1-7 Let a k = 2 k + 1 adn b k = ( k - 1) 3 + k + 2 for all integers k 0. Show that the first three terms of these sequences are identical but that their 4th terms differ. a 0 = 2 · 0 + 1 = 1 a 1 = 2 · 1 + 1 = 3 a 2 = 2 · 2 + 1 = 5 a 3 = 2 · 3 + 1 = 7 . . . b 0 = (0 - 1) 3 + 0 + 2 = ( - 1) 3 + 2 = 1 b 1 = (1 - 1) 3 + 1 + 2 = (0) 3 + 3 = 3 b 2 = (2 - 1) 3 + 2 + 2 = (1) 3 + 4 = 5 b 3 = (3 - 1) 3 + 3 + 2 = (2) 3 + 5 = 13 . . . 4.1-13 Find an explicit formula for the sequence: 1 - 1 2 , 1 2 - 1 3 , 1 3 - 1 4 , 1 4 - 1 5 , 1 5 - 1 6 , 1 6 - 1 7 , . . . a k = 1 k - 1 k + 1 , k = 1 , 2 , . . . 4.1-16 Find an explicit formula for the sequence: 3 , 6 , 12 , 24 , 48 , 96 , . . . a k = 3 · 2 k , k = 0 , 1 , 2 , . . . 4.1-22 Compute the summation 4 Y j =0 ( - 1) j 4 Y j =0 ( - 1) j = ( - 1) 0 ( - 1) 1 ( - 1) 2 ( - 1) 3 ( - 1) 4 = 1 4.1-33 (1 3 - 1) - (2 3 - 1) + (3 3 - 1) - (4 3 - 1) + (5 3 - 1) = 5 X k =1 ( - 1) k +1 ( k 3 - 1) 4.1-34 (2 2 - 1)(3 2 - 1)(4 2 - 1) = 4 Y k =2 ( k 2 - 1) 4.1-37 (1 - t )(1 - t 2 )(1 - t 3 )(1 - t 4 ) = 4 Y k =1 (1 - t k ) 4.1-39 1 2! + 2 3! + · · · + n ( n + 1)! = n X i =2 k ( k + 1)! 4.1-53 Change variables i = k + 1 in n Y k =1 k k 2 + 4 n Y k =1 k k 2 + 4 = n +1 Y i =1+1 i - 1 ( i - 1) 2 + 4 2
4.1-57 Change the variables using j = i - 1 in 2 n Y i = n n - i + 1 n + i 2 n Y i = n n - i + 1 n + i = 2 n - 1 Y j = n - 1 n - ( j + 1) + 1 n + j + 1 = 2 n - 1 Y j = n - 1 n - j n + j + 1 4.1-59 Write as a single sum: 2 n X k =1 (3 k 2 + 4) + 5 n X k =1 (2 k 2 - 1) 2 n X k =1 (3 k 2 + 4) + 5 n X k =1 (2 k 2 - 1) = n X k =1 2(3 k 2 + 4) + 5(2 k 2 - 1) = n X k =1 6 k 2 + 8 + 10 k 2 - 5 = n X k =1 16 k 2 + 3 4.1-62 Suppose a[1],a[2],...,a[m] is a one dimensional array and consider the following algorithm segment: sum = 0; for k=1 to m sum = sum + a[k] next k Fill in the blanks below so that each algorithm segment performs the same job as the one given above. (a) sum = 0; for i=0 to ____ sum = _____ next i answer: m - 1; sum + a[i+1] (b) sum = 0; for j=2 to ____ sum = _____ next j answer: m + 1; sum + a[j-1] 4.1-69 Write an informal description of an algorithm using repeated division by 16 to convert a non-negative integer from decimal notation to hexadecimal. Let a non-negative integer a be given. Divide a by 16 using the quotient remainder theorem to obtain a quotient q[0] and a re- mainder r[0] . If the quotient is non-zero, divide by 16 to obtain a quotient q[1] and a remainder r[1] . 3

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Continue this process until a quotient of 0 is obtained. The sequence of remainders is the list of hexadecimal digits of a . 4.1-73 Write a formal version of the algorithm you developed in exercise 69. Input: a Algorithm Body: q=a; i=0; while i=0 or q != 0, r[i]= q mod 16 q = q div 16 i = i + 1 end while Output: r[0], r[1], r[2], ..., r[i-1] End Algorithm 4
4.2-2 Use mathematical induction to show that any postage of at least 12 cents can be obtained using the 3 cent and 7 cent stamps. Proof that the Property holds for n = 12 . The property is true for n = 12 because we can use four 3-cent stamps. Show that for all integers k 12 , if the property is true for n = k then it is true for n = k + 1 . Suppose that k cents can be obtained using some combination of 3 cent and 7 cent stamps for some integer n 12. We must show that this implies that that k +1 cents can be obtained using some combination of the 3 cent and 7 cent stamps.

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