new_hand1 - n + 1) 6 n s k =1 k 3 = 1 + 8 + + n 3 = n 2 ( n...

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CMSC 351:Spring 2011 Michelle Hugue CMSC351 Test 1 Reference Asymptotic Notations. Θ( g ( n )) = { f ( n ): there exist positive constants c 1 , c 2 , and n 0 such that 0 c 1 g ( n ) f ( n ) c 2 g ( n ) for all n n 0 } . O ( g ( n )) = { f ( n ): there exist positive constants c and n 0 such that 0 f ( n ) cg ( n ) for all n n 0 } . Ω( g ( n )) = { f ( n ): there exist positive constants c and n 0 such that 0 cg ( n ) f ( n ) for all n n 0 } . f ( n ) = o ( g ( n )) if lim n →∞ f ( n ) g ( n ) = 0. f ( n ) = ω ( g ( n )) if lim n →∞ f ( n ) g ( n ) = . f ( n ) g ( n ) if f ( n ) = g ( n ) + o ( g ( n )). Logarithms. a = b log b a log c ( ab ) = log c a + log c b log b a n = n log b a log b a = log c a log c b log b (1 /a ) = log b a log b a = 1 log a b a log b n = n log b a a f ( n ) = e f ( n ) ln a (log a f ( n )) = f ( n ) f ( n ) ln a ( a f ( n ) ) = ln af ( n ) a f ( n ) Quadratic Formula. ax 2 + bx + c = 0 x = b ± b 2 4 ac 2 a 1
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Summations. Simple Arithmetic Series: n s k =1 k = 1 + 2 + ··· + n = n ( n + 1) 2 n s k =1 k 2 = 1 + 4 + ··· + n 2 = n ( n + 1)(2
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Unformatted text preview: n + 1) 6 n s k =1 k 3 = 1 + 8 + + n 3 = n 2 ( n + 1) 2 4 General Arithmetic Series: n s k = m k = 1 + 2 + + n = ( n m + 1)( n + m ) 2 Geometric series: n s k =0 x k = 1 + x + + x 2 + x n = x n +1 1 x 1 x n = 1 s k =0 x k = 1 1 x | x | < 1 Recurrences SMT: Simplifed Master Theorem: T ( n ) = b aT ( n/b ) + n k n > 1 1 n = 1 implies T ( n ) = ( n log b a ) a > b k ( n k ) a < b d ( n k log b n ) a = b k . 2...
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This note was uploaded on 01/13/2012 for the course CMSC 351 taught by Professor Staff during the Fall '11 term at University of Louisville.

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new_hand1 - n + 1) 6 n s k =1 k 3 = 1 + 8 + + n 3 = n 2 ( n...

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