Asymptotic Dominance Problems. is not constant but for n 0, f ( n) 11. 0, so that for n N f

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1 Asymptotc Domce Prolems Dsply ucto : N R tht s Ο( ) ut s ot costt 0 = 0 The ucto ( ) = > 0 s ot costt ut or 0, ( ) Dee the relto " " o uctos rom N to R y g d oly = Ο( g) Prove tht s relexve d trstve (Recll: to e relexve, you must To prove relexvty, otce tht or y : N R d ll 0, ( ) ( ) To prove trstvty, suppose = Ο( g) d g = Ο( h), the y deto, there exst N 0, M 0, Ng 0, Mg 0, so tht or N, ( ) M g( ) d or N g, g ( ) Mg h ( ) Thus or mx{ N, Ng}, ( ) M Mg h( ) We my coclude tht =Ο( h) 3 Suppose = Ο(g) d g = Ο(h), prove or dsprove (wth smple couter-exmple) tht = Ο(h) Suppose =Ο( g) d g =Ο( h), the y deto, there exst N 0, M 0, Ng 0, Mg 0, so tht or N, ( ) M g( ) d or N g, g ( ) Mg h ( ) Thus or mx{ N, Ng}, ( ) M Mg h( ) We my coclude tht =Ο( h) 4 Suppose = ο( g) d g =Ο( h) Prove tht = ο( h) Sce g = Ο(h), there exst M d N so tht N g( ) M h( ) Gve ε > 0, let ε = ε / M Sce = ο(g), there exst N such tht N ( ) ε g( ) = ε / M g( ) Thus lettg N = mx{ N, N }, or N we hve ( ) ε / M g( ) ε h( ) so = ο(h) 5 Suppose =Ο( g) d g =Ο( h) I h = Ο( ), prove tht h = Ο( g) By deto, there exst N 0, M 0, Nh 0, Mh 0, so tht or N, ( ) M g( ) d or N h, h ( ) Mh ( ) Thus or mx{ N, Nh}, h ( ) M M g ( ) We my coclude tht h = Ο( g) h

2 6 Usg Theorem d ducto prove tht or =,,,, =Ο( g = Ο( g ) ), the For =, we hve = = Ο( g ) = Ο( g ) Now ssume = Ο( g ) d cosder + Sce = g Ο + Ο( ) d = Ο( g ), Theorem gurtees tht = + = ( g + g ) = Ο( g ) 7 Employg ducto d Theorem 3, prove tht or =,,,, =Ο( g), the = Ο( g) For =, we hve = = Ο( g) y hypothess Now ssume = Ο( g) d cosder + + Sce = Ο( g) d + = Ο( g), Theorem 3 gurtees tht = + + = Ο(mx{ g, g }) = Ο( g ) = Ο( g) 8 Show tht ( )= + 3 d g ( )=, the = O(g) Let N = 3 d M = 3 For N : ( ) = + 3 = = 3 3 = 3 = M Thus = O(g) g( ) 00 0 or = 7 9 Dee : N R y ( ) = or 7 For 8, ( ) =, so = Ο( ) Prove tht = Ο( )

3 0 Cosder the uctos d g deed o N y ( ) = g ( ) = Show tht =Ο( g) ut tht ο( g) d g Ο( ) or eve or odd d =Ο( g): Sce or 0, ; we hve tht d, so ( ) g( ) Thus = Ο( g) ο( g): Suppose = ο( g), the or ε = / there s o-egtve N so tht or ll N, ( ) ε g( ) But lettg = N = 0 d = N or N+ (whchever s eve) N s postve, we hve ( ) = > = ε g( ) Ths s cotrdcto, so ο( g) g Ο( ): Suppose g =Ο( ), the there exst oegtve M d N so tht or ll N, g ( ) M ( ) But lettg e odd d greter th N d M, the we hve g ( ) = = > M= M = M ( ) Ths s cotrdcto, so g Ο( ) Show tht =Ο(!) For d = 3,,,, we hve, thus = = = =! d we hve =Ο(!) Thereore,!, thus Show tht or y rel vlue o, =Ο(!) (Ht: e creul to cosder egtve vlues o ) Dee K = (e K s the rst teger greter th or equl to ) For K d = K, K +,,, we hve, thus Thereore, K K K K K K K K K d N = K, we hve M! or ll N = = =! So wth M = Thus =Ο(!)

4 3 Show tht or y >, log =ο ( ) Cosder y postve ε, d choose N = + ε ( ) The, > N, we hve > + ε, thus ( ) ε ( ) >, d ( ) ( ε ) > But usg the ( ) oml theorem, we hve ε ε ε ε j ε = = + = j j > ( ) ( ( )) ( ) ( ) > = 0 By tg se logrthms, we hve ε ε = ε = log > log = log 4 Prove tht 0 <, the = ο ( ) I = 0, the or ll ε > 0 d ll, we hve = 0 ε Assume ow tht > 0 Te N = l( ε ) / l( / ) d (ssumg ε < ), or N, l( / ) l( ε ) d = ε = ε (I ε the = ε = ε or 0 ) Thus = ο ( ) 5 Prove tht 0 <, the = ο ( ) Gve y ε > 0, let N /ε, d = ο ( ) ( ) = ( / ) /( ) 6 Prove tht 0 < <, the Ο( ) Gve M 0 d N 0, let M ε Notce the or N = ( / ε ) /( ), ε So = ( ) = ( ) ε Thereore, = mx{ M, } thus M M d l( M) 0 Notce tht l( ) > 0 d choose N l( M) M = mx{, } + For ths we hve > l( ), l( ) l( ) thus l( ) > l( M) d ( ) M M > But the M M = > = so Ο( )

5 7 Prove tht =Ο( ) Let M = d N = For, 3/ N Thus = 3/ = =, so = Ο( ) ( ) 8 Prove tht e ο ( e ) Let ε =, cosder d N, d choose mx{n,} Sce, > e ( ) > e = ε so e ο( e ) d Usg oly Deto, prove tht 3 = Ο( ) Let M = 3 d N = For N =, we hve 4 4, so = Thus 3 = Ο( ) 0 Usg oly Deto, prove tht 5 ο ( 4 ) Let ε = / 4 d suppose there exsts N so tht or ll N, 5 ε 4 But or 5 = mx{, N }, we hve N d, so ( ) > d 5 > 4, thus 4 5 = 5 > 4 = / 4 = ε 4 d 5 ο ( 4 ) Show tht ( )= d g( )=, the o(g) Let ε = d cosder y postve N Let = N + so d N ( ) = = > ε = ε g( ) Thus o(g) Show tht log! =Ο ( log ) d log = Ο (log!) For, we hve log! = log ( ) = log log = We hve: log Thus log! log d log! = Ο ( log ) To show log =Ο (log!) let N = 8 d M = 3 Notce tht 8,, so 3 ( ) = Also otce tht 8 8, so + = Flly 3 3( ) / 3 / 3 / = ( ) (( ) ) = ( ) ( ) = (!) = = = =

6 By tg logs, we hve or 8, log = log 3log! = 3 log!