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Unformatted text preview: A LGORITHMS — S AMPLE S OLUTIONS — M ATHEMATICAL P RELIMINARIES These solutions are contributed by several teaching assistants including Grigore Rosu, and Diem Vu. 1 Formulae you should know You should remember these properties from previous calculus classes: • log a ( b · c ) = log a ( b ) + log a ( c ) , • log a ( b/c ) = log a ( b ) log a ( c ) , • log a ( b c ) = c · log a ( b ) , • c log a ( b ) = b log a ( c ) , • 1 + x + x 2 + ··· + x n = x n +1 1 x 1 . 2 Summation Problem 1: (Summation) Let S k n be the sum 1 k + 2 k + 3 k + ··· + n k . Then 1. Calculate S 1 n and S 2 n , and 2. Show that S k n = O ( n k +1 ) . Proof: 1. We recall the formulas ( x + 1) 2 = x 2 + 2 x + 1 and ( x + 1) 3 = x 3 + 3 x 2 + 3 x + 1 . Then S 2 n = 1 2 + 2 2 + 3 2 + ··· + n 2 = 1 + (1 + 1) 2 + (2 + 1) 2 + ··· + (( n 1) + 1) 2 = 1 + (1 2 + 2 · 1 + 1) + (2 2 + 2 · 2 + 1) + ··· + (( n 1) 2 + 2( n 1) + 1) = (1 2 + 2 2 + ··· + ( n 1) 2 ) + 2(1 + 2 + ··· + ( n 1)) + n = ( S 2 n n 2 ) + 2( S 1 n n ) + n = S 2 n + 2 S 1 n n ( n + 1) . Simplifying by S 2 n , one gets S 1 n = n ( n +1) 2 . 1 Similarly, S 3 n = 1 3 + 2 3 + 3 3 + ··· + n 3 = 1 + (1 + 1) 3 + (2 + 1) 3 + ··· + (( n 1) + 1) 3 = 1 + (1 3 + 3 · 1 2 + 3 · 1 + 1) + (2 3 + 3 · 2 2 + 3 · 2 + 1)+ ··· + (( n 1) 3 + 3( n 1) 2 + 3( n 1) + 1) = (1 3 + 2 3 + ··· + ( n 1) 3 ) + 3(1 2 + 2 2 + ··· + ( n 1) 2 )+ +3(1 + 2 + ··· + ( n 1)) + n = ( S 3 n n 3 ) + 3( S 2 n n 2 ) + 3 S 1 n 1 + n = S 2 n + 3 S 2 n n 3 3 n 2 + n + 3 · n ( n 1) 2 = S 2 n + 3 S 2 n n (2( n 2 +3 n 1) 3( n 1)) 2 = S 2 n + 3 S 2 n n (2 n 2 +3 n +1) 2 = S 2 n + 3 S 2 n n ( n +1)(2 n +1) 2 . Simplifying by S 3 n , one gets S 2 n = n ( n +1)(2 n +1) 6 . 2. We recall that ( n + 1) k +1 = n k +1 + ( k + 1) n k + P k 1 ( n ) , where P k 1 is a polynomial of degree k 1 . Then S k +1 n = 1 k +1 + 2 k +1 + 3 k +1 + ··· + n k +1 = 1 + (1 + 1) k +1 + (2 + 1) k +1 + ··· + (( n 1) + 1) k +1 = 1 + (1 k +1 + ( k + 1) · 1 k + P k 1 (1)) + (2 k +1 + ( k + 1) · 2 k + P k 1 (2))+ ··· + (( n 1) k +1 + ( k + 1) · ( n 1) k + P k 1 ( n 1)) = (1 k +1 + 2 k +1 + ··· + ( n 1) k +1 ) + ( k + 1)(1 k + 2 k + ··· + ( n 1) k )+ +(1 + P k 1 (1) + P k 1 (2) + ··· + P k 1 ( n 1)) = ( S k +1 n n k +1 ) + ( k + 1)( S k n n k ) + O ( n k ) = S k +1 n + ( k + 1) S k n n k +1 + O ( n k ) Simplifying S k n , one gets S k n = n k +1 k +1 O ( n k ) , that is, S k n = O ( n k +1 ) . Problem 2: (Summation) Calculate the sums n X k =0 k 2 k and n X k =0 k 2 2 k . Proof: Recall that n X k =0 x k = x n +1 1 x 1 . By differentiation, one obtains: n X k =0 kx k 1 = ( n + 1) x n ( x 1) x n +1 + 1 ( x 1) 2 = nx n +1 ( n + 1) x n + 1 ( x 1) 2 ....
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 Winter '08
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 LG, Natural logarithm, Logarithm, Recurrence relation, Fibonacci number

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