handouts4_seq_ind_rec

handouts4_seq_ind_rec - Sequences and Recursion Mariusz...

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Unformatted text preview: Sequences and Recursion Mariusz Bajger COMP2781/8781 School of Computer Science, Engineering and Mathematics April 4, 2011 1/ 35 Reading and Exercises Reading Epp, Chapter 5 Exercises Start with the blue ones in Sec. 5.1 to 5.8 Good learning strategy: BE ACTIVE! I Regularly revise lectures I Solve the suggested exercises I Be critical when reading textbook/lectures I Ask your colleagues, ask the lecturer, don’t be shy! I It is OK to ask for help any time 2/ 35 Sequences Definition A sequence is a function whose domain is either all the integers between two given integers or all the integers greater than or equal to a given integer. A sequence can be finite a m , a m +1 , a m +2 ,..., a n n − m terms or infinite a m , a m +1 , a m +2 ,... infinite number of terms EXAMPLE (explicitly defined sequence) Let s n = 2 n + 4 · 3 n , n ∈ Z , n ≥ 2. Then s n − 1 = 2 n − 1 + 4 · 3 n − 1 . Find s n − 2 ? 3/ 35 Summation notation Summation from k ( summation index ) equals m ( lower limit ) to n ( upper limit ) of a-sub- k : ∑ n k = m a k = a m + a m +1 + a m +2 + ··· + a n For example, 5 ∑ i =0 ( − 1) i i + 1 = 0 − 1 2 + 2 3 − 3 4 + 4 5 − 5 6 EXAMPLE (Telescoping sum) Use the formula 1 k ( k +1) = 1 k − 1 k +1 to show that n ∑ k =1 1 k ( k + 1) = 1 − 1 n + 1 4/ 35 Product notation Product from k equals m to n of a-sub- k : ∏ n k = m a k = a m · a m +1 · a m +2 ··· a n For example, ∏ 5 i =1 i i +1 = 1 2 · 2 3 · 3 4 · 4 5 · 5 6 = 1 6 Theorem (properties of sums and products) n ∑ k = m a n + n ∑ k = m b k = n ∑ k = m ( a k + b k ) c · n ∑ k = m a k = n ∑ k = m c · a k ( n ∏ k = m a k ) · ( n ∏ k = m b k ) = n ∏ k = m ( a k · b k ) 5/ 35 Change of variable in sums Similar to changing variable in integrals. For example, replace the index i by k in the sum ∑ n i =1 i 2 r n − i where i = k + 1. Steps: I find the lower and upper limits of the new summation when i = 1 , k = 0 and when i = n , k = n − 1 I find the general term of the new summation (replace i by the new index k ) i 2 r n − i = ( k + 1) 2 r n − k − 1 I put it together n ∑ i =1 i 2 r n − i = n − 1 ∑ k =0 ( k + 1) 2 r n − k − 1 6/ 35 ’n Choose r’ notation ( n r ) Reminder: (factorial definition), for n ∈ Z + n ! = n · ( n − 1) ··· 3 · 2 · 1 = n ∏ k =1 k and 0! = 1 Definition For n , r ∈ Z : 0 ≤ r ≤ n we define ’ n choose r ’ ( n r ) which represents the number of subsets of size r that can be selected from a set of n elements. Later, we will prove that ( n r ) = n ! r !( n − r )! 7/ 35 Sequences in computer programming One-dimensional arrays a [0] , a [1] , a [2] ,..., a [10] are finite sequences....
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handouts4_seq_ind_rec - Sequences and Recursion Mariusz...

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