hand3 - f ( x ): Z b a-1 f ( s ) ds b X i = a f ( i ) Z b...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
CMSC 351:Spring 2011 Michelle Hugue CMSC351 Final Exam 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 )) 0 = f 0 ( n ) f ( n )ln a ( a f ( n ) ) 0 = ln af 0 ( n ) a f ( n ) Quadratic Formula. ax 2 + bx + c = 0 x = - b ± b 2 - 4 ac 2 a 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Summations. Simple Arithmetic Series: n X k =1 k = 1 + 2 + ··· + n = n ( n + 1) 2 n X k =1 k 2 = 1 + 4 + ··· + n 2 = n ( n + 1)(2 n + 1) 6 n X k =1 k 3 = 1 + 8 + ··· + n 3 = n 2 ( n + 1) 2 4 General Arithmetic Series: n X k = m k = 1 + 2 + ··· + n = ( n - m + 1)( n + m ) 2 Geometric series: n X k =0 x k = 1 + x + + x 2 ··· + x n = x n +1 - 1 x - 1 x 6 = 1 X k =0 x k = 1 1 - x | x | < 1 Integration rules - for an increasing function
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: f ( x ): Z b a-1 f ( s ) ds b X i = a f ( i ) Z b +1 a f ( s ) ds for a decreasing function g ( x ): Z b +1 a g ( s ) ds b X i = a g ( i ) Z b a-1 g ( s ) ds Recurrences. AMT: Ambitious Master Theorem: T ( n ) = aT ( n/b ) + cn d n &amp;gt; 1 f n = 1 implies T ( n ) = f + c ab-d-1 n log b a-( cn d ab-d-1 ) = ( n log b a ) a &amp;gt; b d ( n d ) a &amp;lt; b d n d ( f + c log b n ) = ( n d log b n ) a = b d . Miscelleneous. Stirlings Approximation: n ! 2 n n e n 2...
View Full Document

Page1 / 2

hand3 - f ( x ): Z b a-1 f ( s ) ds b X i = a f ( i ) Z b...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online