Solutions to Exercises Chapter 4: Recurrence relations and generating functions


 Catherine Brown
 4 years ago
 Views:
Transcription
1 Solutios to Exercises Chapter 4: Recurrece relatios ad geeratig fuctios 1 (a) There are seatig positios arraged i a lie. Prove that the umber of ways of choosig a subset of these positios, with o two chose positios cosecutive, is F +1. (b) If the positios are arraged aroud a circle, show that the umber of choices is F + F 2 for 2. (a) Proof by iductio. If g() deotes this umber, the we have g(1) = 2 = F 2, g(2) = 3 = F 3. (For = 2, we caot occupy both positios; but all other choices are possible.) For > 2, we separate the seatig selectios ito those i which the last positio is uoccupied ad those i which it is occupied. There are g( 1) of the first kid. If the last positio is take, the the oe before it must be free, ad we have a arbitrary seatig pla o the first 2 positios; so there are g( 2) of these. Hece ad the iductive step is proved. g() = g( 1) + g( 2) = F + F 1 = F +1, (b) Cosider a particular positio o the circle. If it is uoccupied, we ca break the circle at that poit, ad obtai a lie with 1 positios, which ca be filled i g( 1) = F ways. If the positio is occupied, the its eighbours o either side are uoccupied, ad (if 3) we ca remove the positio ad its two eighbours, obtaiig a lie of 3 positios which ca be filled i g( 3) = F 2 ways. The same holds if = 2, sice there is just oe seatig pla with the give positio occupied, ad F 0 = 1. So we have F + F 2 seatig plas i all. 1
2 2 Prove the followig idetities: (a) F 2 F +1 F 1 = ( 1) for 1. (b) F i = F (c) F F 2 = F 2, F 1 F + F F +1 = F 2+1. (d) F = /2 ( i i ). (a) Iductio: the result holds for = 1. For 2, we have F 2 F +1 F 1 = F 2 (F +F 1 )F 1 = (F F 1 )F F 2 1 = (F 2 1 F F 2 ), so, if F 2 1 F F 2 = ( 1) 1 the F 2 F +1 F 1 = ( 1). (b) Iductio. The result is true for = 0 (the empty sum is zero). Assumig it for 1, we have F i = (F +1 1) + F = F (c) Agai, iductio. The result holds for = 1 by ispectio. Assume it for ; that is, F F 2 1 = F 2 2, F 2 F 1 + F 1 F = F 2 1. Addig these equatios ad usig the Fiboacci recurrece, we get F 2 F + F 1 F +1 = F 2. Usig (a) twice, this implies that F F2 = F 2. Now add this equatio to the secod displayed equatio usig the Fiboacci recurrece to get F 1 F + F F +1 = F 2+1. (d) Most easily, this follows from our origial iterpretatio of F as the umber of expressios for as a ordered sum of 1s ad 2s. Such a expressio with i 2s will have 2i 1s, hece i summads altogether; there are ( i) i ways to choose the positios of the 2s i the sum. Now summig over i gives the result. 2
3 3 Show that F is composite for all odd > 3. By 2(c), F 2+1 = F (F 1 + F +1 ); ad if > 1, the both factors are greater tha 1. 4 Show that for 1. ( 1)/2 F 2i = F +1 1 The proof is by a iductio which goes from 2 to, so the iitial cases = 1 ad = 2 must both be checked. Assumig the result for 2, we have ( 1)/2 F 2i = (F 1 1) + F = F i=1 5 Prove that every oegative iteger x less tha F +1 ca be expressed i a uique way i the form F i1 + F i F ir, where i 1,i 2,...,i r {1,...,}, i 1 > i 2 + 1, i 2 > i 3 + 1,... (i other words, i 1,...,i r are all distict ad o two are cosecutive). Deduce Exercise 1(a). Follow the hit. If we had ay expressio of this form usig Fiboacci umbers below F, the we could if ecessary replace the summads by larger oes ad add ew summads to obtai F 1 + F = F 1 (by Questio 4). So the sum of the origial expressio was at most F 1. Hece ay expressio summig to x, with F x < F +1, must iclude F. Now x F < F +1 F = F 1, so by iductio there is a uique expressio for x F, ad hece for x (sice the expressio for x F caot ivolve F 1, so does ot cotai cosecutive Fiboacci umbers). Hece, the umber of expressios of this form (that is, the umber of ways of choosig a subset of the idices 1,2,..., with o two cosecutive) is equal to the umber of possible sums 0,1,...,F +1 1, that is, F +1, as asserted i 1(a). 6 Fiboacci umbers are traditioally associated with the breedig of rabbits. Assume that a pair of rabbits does ot breed i its first moth, ad that it produces a pair of offsprig i each subsequet moth. Assume also that rabbits live forever. Show that, startig with oe ewbor pair of rabbits, the umber of pairs alive i the th moth is F. 3
4 I the 0th ad 1st moths, oe pair is alive, so the result is true for = 0,1. Assume it holds for 1. The, i the ( 1)st moth, F 1 pairs of rabbits are alive, of whom F 2 were alive i the precedig moth (ad hece old eough to breed), providig F 2 ewbor pairs i the th moth. Thus, the total umber of pairs i the th moth is F 1 + F 2 = F. 7 Prove that the umber of additios required to compute the Fiboacci umber F accordig to the iefficiet algorithm described i the text is F 1. Iductio: check the result for small. Now F 1 takes F 1 additios, ad F 2 takes F 1 1 additios; oe further additio is required to combie them, givig i all (F 1) + (F 1 1) + 1 = F +1 1 additios. 8 (a) Prove that F m+ = F m F +F m 1 F 1 for m, 0 (with the covetio that F 1 = 0). (b) Use this to derive a algorithm for calculatig F usig oly clog arithmetic operatios. (c) Give that multiplicatio is slower tha additio, is this algorithm really better tha oe ivolvig 1 additios? (a) Iductio o m. For m = 0, the result is a tautology, ad for m = 1 it is the Fiboacci recurrece. For m > 1 we have F (m+1)+ = F m+ + F (m 1)+ = F m F + F m 1 F 1 + F m 1 F + F m 2 F 1 = F m+1 F + F m F 1, usig the covetio that F 1 = 0 ad the fact that, the, the relatio F m = F m 1 + F m 2 holds also for m = 1. (b) The trick is to calculate pairs (F 1,F ) of Fiboacci umbers, observig that we ca go efficietly from the pairs (F m 1,F m ) ad (F 1,F ) to the pair (F m+ 1,F m+ ) usig (a). A cleaer way to express this is i terms of matrices. We have (F F +1 )A = (F +1 F +2 ), where A is the 2 2 matrix ( ). Hece (F F +1 ) = (1,1)A. By the aalogue for powers of Russia peasat multiplicatio (Chapter 2, Exercise 12(iii)), we ca fid A i about 2log 2 matrix multiplicatios, each requirig eight iteger multiplicatios ad four additios (though this ca be improved). (c) Sice F is expoetially large, the umber of its digits is proportioal to. Hece a log multiplicatio ivolves a umber of additios proportioal to, ad we have ot really made much savig over a method ivolvig additios. 4
5 9 (a) Solve the followig recurrece relatios. (i) f ( + 1) = f () 2, f (0) = 2. (ii) f ( + 1) = f () + f ( 1) + f ( 2), f (0) = f (1) = f (2) = 1. 1 (iii) f ( + 1) = 1 + f (i), f (0) = 1. (b) Show that the umber of ways of writig as a sum of positive itegers, where the order of the summads is sigificat, is 2 1 for 1. (a) (i) Let g() = log 2 f (). The g( + 1) = 2g(), g(0) = 1; so g() = 2 (Sectio 3.1), ad f () = 2 2. (ii) The characteristic equatio (see Sectio 4.3) is x 3 = x 2 + x + 1. This cubic has three distict real roots α, β, γ, ad so the geeral solutio is f () = aα + bβ + cγ, where a,b,c are determied by the three equatios a + b + c = 1, aα + bβ + cγ = 1, aα 2 + bβ 2 + cγ 2 = 1. (iii) We have f (1) = 1, sice the empty sum is zero. For 1, we have ( ) 1 2 f ( + 1) = 1 + f (i) = 1 + f (i) + f ( 1) = f () + f ( 1), ad so by iductio f () is the Fiboacci umber F. (b) There is oe expressio cosistig of the sigle umber. Ay other expressio eds with some positive umber j <, preceded by a expressio with sum j; so the umber f () of expressios satisfies 1 1 f () = 1 + f ( j) = 1 + f (i). j=1 i=1 Just as i (a)(iii), we fid that f ( + 1) = f () + f () = 2 f (). Sice f (1) = 1, we have f () = 2 1, as claimed. 5
6 10 The umber f () of steps required to solve the Chiese rigs puzzle with rigs satisfies f (1) = 1 ad { 2 f (), odd, f ( + 1) = 2 f () + 1, eve. Prove that f ( + 2) = f ( + 1) + 2 f () + 1. Hece or otherwise fid a formula for f (). If is odd, the f ( + 2) = 2 f ( + 1) + 1 ad f ( + 1) = 2 f (); if is eve, the f ( + 2) = 2 f ( + 1) ad f ( + 1) = 2 f () + 1. So the result holds i either case. This is ot a pure recurrece relatio because of the added 1. But if we set g() = f () + 2 1, the g( + 1) + 2g() = f ( + 1) + 2 f () = g( + 2). The characteristic equatio is x 2 = x + 2, with solutios 2 ad 1, so g() = a2 + b( 1). The iitial values g(1) = 3 2 ad g(2) = 5 2 yield a = 2 3 ad b = 1 6. So f () = ( 1) (a) Let s() be the umber of sequeces (x 1,...,x k ) of itegers satisfyig 1 x i for all i ad x i+1 2x i for i = 1,...,k 1. (The legth of the sequece is ot specified; i particular, the empty sequece is icluded.) Prove the recurrece s() = s( 1) + s( /2 ) for 1, with s(0) = 1. Calculate a few values of s. Show that the geeratig fuctio S(t) satisfies (1 t)s(t) = (1 +t)s(t 2 ). (b) Let u() be the umber of sequeces (x 1,...,x k ) of itegers satisfyig 1 x i for all i ad x i+1 > i j=1 x j for i = 1,...,k 1. Calculate a few values of u. Ca you discover a relatioship betwee s ad u? Ca you prove it? (a) We have s(0) = 1, sice oly the empty sequece occurs. For > 0, divide the sequeces couted by s() ito two classes: those ot cotaiig, ad those cotaiig. There are s( 1) of the first class. For the secod, all terms other tha are at most /2, so such a sequece is obtaied from a sequece of umbers with 1 x i /2 ad x i+1 2x i by adjoiig ; so there are s( /2 ) of these. The recurrece relatio follows. If S(t) = s()t, the we have S(t) = ts(t) + (1 +t)s(t 2 ); 6
7 for the coefficiet of t o the right is s( 1) + s( /2 ), ad the costat term o both sides is equal to 1. (b) The relatioship is u() u( 1) = s()/2 for 2. The proof is quite itricate, ad depeds o defiig aother sequece v() as follows: v() is the umber of sequeces (x 1,...,x k ) of positive itegers which satisfy x i > i 1 j=1 x j for all i k ad k i=1 x i =. Now such a sequece must have last term x k > /2, ad is obtaied by takig a sequece with sum x k ( 1)/2 ad appedig x k to it. So we have the recurrece v() = ( 1)/2 v(i). i=1 It follows from this that v(2) = v(2 1) = v(2 2) + v( 1) for all > 0. Now we claim that v(2) = s()/2 for all 1. This is proved by iductio, beig true for = 1. Assumig the result for 1, we have v(2) = v(2 2) + v( 1) = s( 1)/2 + s( /2 )/2 = s()/2, ad the iductio goes through. Now u() u( 1) is the umber of sequeces satisfyig the specificatio for u which have last term. These are obtaied by appedig to a sequece with sum strictly smaller tha ; so we have u() u( 1) = 1 v(i) = v(2) = s()/2. Remark: Sequeces satisfyig this coditio are called supericreasig; they arose i the first example of a publickey cryptosystem, the Merkle Hellma kapsack system. The two sequeces s ad u occur as umbers M1011 ad M1053 i the Ecyclopedia of Iteger Sequeces, where further refereces ca be foud, or you ca fid them i the olie versio here (for s) ad here (for u). 12 Let F(t) be a formal power series with costat term 1. By fidig a recurrece relatio for its coefficiets, show that there is a multiplicative iverse G(t) of F(t). Moreover, if the coefficiets of F are itegers, so are those of G. 7
8 Let F(t) = f t ad G(t) = g t, with f 0 = 1. The relatio F(t)G(t) = 1 is equivalet to g 0 = 1 ad the recurrece relatio g = i=1 f i g i for g i terms of earlier values. So the values g are determied, ad are itegers if the values f are. 13 A permutatio π of the set {1,...,} is called coected if there does ot exist a umber k with 1 k < such that π maps the subset {1,2,...,k} ito itself. Let c be the umber of coected permutatios. Prove that i=1 c i ( i)! =! Deduce that, if F(t) = 1!t ad G(t) = 1 c t are the geeratig fuctios of the sequeces (!) ad (c ) respectively, the 1 G(t) = (1 + F(t)) 1. For a permutatio π, let i(π) be the least positive iteger k such that π maps the set {1,...,k} ito itself. (Such a k exists, sice certaily π maps {1,...,} to itself.) The π is the composite of a coected permutatio o {1,...,i(π)} ad a arbitrary permutatio o {i(π)+1,...,} (this last set beig empty if i(π) = ). Summig over i, we obtai the stated recurrece. As i the precedig questio, this recurrece is equivalet to (1 + F(t))(1 G(t)) = Let (1 +t ) = a t. 1 0 Prove that a is the umber of ways of writig as the sum of distict positive itegers. (For example, a 6 = 4, sice 6 = = = ) A term i the ifiite product (1 + t ) is obtaied by selectig t m from the mth factor for a fiite umber of values of, ad 1 from all the other factors. So there is a cotributio of t for every expressio of as a sum of distict positive itegers. 8
9 15 (a) I a electio, there are two cadidates, A ad B; the umber of votes cast is 2. Each cadidate receives exactly votes; but, at every itermediate poit durig the cout, A has received more votes tha B. Show that the umber of ways this ca happe is the Catala umber C. Ca you costruct a bijectio betwee the bracketed expressios ad the votig patters i (a)? (b) I the above electio, assume oly that, at ay itermediate stage, A has received at least as may votes as B. Prove that the umber of possibilities is ow C +1. (a) As i the Hit, let f () be the umber of ways of coutig. Now A leads by just oe vote after the first vote is couted. Suppose that this ext occurs after 2i + 1 votes have bee couted. (It must happe agai, sice B evetually catches A.) The there are f (i) choices for the cout betwee these poits, sice each cadidate receives i votes ad A is always strictly ahead. Also, there are f ( i) choices for the rest of the cout; for if we preted at this stage that A has just oe vote ad B oe, we are ruig the cout satisfyig the same coditios with just 2( i) votes altogether. So we have f () = 1 i=1 f (i) f ( i), which is the Catala recurrece. (b) Agai follow the hit: I the modified electio where A gets a extra vote at the start ad B a extra vote at the ed, there are + 1 votes for each cadidate, ad A is alway strictly ahead. So there are C +1 ways of doig the cout uder this coditio. 16 A clow stads o the edge of a swimmig pool, holdig a bag cotaiig red ad blue balls. He draws the balls out oe at a time ad discards them. If he draws a blue ball, he takes oe step back; if a red ball, oe step forward. (All steps have the same size.) Show that the probability that the clow remais dry is 1/( + 1). Imagie that the clow wears a divig suit, ad cotiues to draw balls eve after he gets wet. There are ( 2) ways i which the balls could be draw. The clow stays dry if ad oly if the umber of red balls ever exceeds the umber of blue oes; accordig to the votig iterpretatio, this is C +1. The ratio is 1/( + 1). For a harder exercise, fid a direct proof of this exercise, ad reverse the above argumet to deduce the formula for the Catala umbers. 17 Prove that lim ( ) 1/ B = 0.! 9
10 of The radius of covergece of the power series 0 B t /! is the reciprocal lim sup ( ) 1/ B.! But the radius of covergece is ifiite, sice the series coverges everywhere. (I geeral, the radius of covergece is the distace from the origi to the earest sigularity i the complex plae; if the sum fuctio has o sigularities, the radius of covergece is ifiite.) So the limsup is zero, which meas (sice all the values are positive) that the limit is zero. 18 (a) Prove that the expoetial geeratig fuctio for the umber s() of ivolutios o {1,...,} (Sectio 4.4) is exp(t t2 ). (b) Prove that the expoetial geeratig fuctio for the umber d() of deragemets of {1,...,} is 1/((1 t)exp(t)). (a) The fuctio S(t) = exp(t t2 ) is the uique solutio of the differetial equatio S (t) = (1 +t)s(t) with the iitial coditio S(0) = 1. So it suffices to check that the e.g.f. of the umbers s() satisfies this differetial equatio. (Clearly it satisfies the iitial coditio.) Now, if we put S(t) = 0 s()t /!, the the coefficiet of t 1 /( 1)! i (1 + t)s(t) is s( 1) + ( 1)s( 2), whereas the coefficiet i S (t) is s() (sice differetiatio of the e.g.f. correspods to a left shift of the sequece). By the recurrece relatio i Sectio 4.4, these two expressios are equal. (b) We have by (4.4.1). Hece d()t 0! d() =! = = ( ) ( 1) i. i! ( ) ( ) ( t) i i 0 i! t k k 0 1 e t (1 t). (I the first lie, we put k = i; the i ad k ru idepedetly over the oegative itegers.) 10
11 19 The Beroulli umbers B() (ot to be cofused with the Bell umbers B ) are defied by the recurrece B(0) = 1 ad k=0 ( + 1 k ) B(k) = 0 for 1. Prove that the expoetial geeratig fuctio B()t f (t) = 0! is give by f (t) = t/(exp(t) 1). Show that f (t) t is a eve fuctio of t, ad deduce that B() = 0 for all odd 3. What is the solutio of the similarlookig recurrece b(0) = 1 ad for 1? k=0 The recurrece ca be writte as ( k=1 k ( ) b(k) = 0 k ) B(k) = B( + 1) for 2. Multiplyig by t +1 /(+1)! ad summig over 2, the two sides are f (t)exp(t) ad f (t) with the costat ad liear terms omitted. Thus f (t)exp(t) 1 t t = f (t) t, whece f (t) = t/(exp(t) 1), as required. Now f (t)+ 2 1t = 1 2 t(exp( 1 2 t)+exp( 2 1t))/(exp( 1 2 t) exp( 1 2 t)) = 1 2 t coth 1 2 t, a eve fuctio. The last recurrece obviously has the solutio b(k) = ( 1) k. 11
12 20 For eve, let e be the umber of permutatios of {1,...,} with all cycles eve; o, the umber of permutatios with all cycles odd; ad p =! the total umber of permutatios. Let E(t), O(t) ad P(t) be the expoetial geeratig fuctios of these sequeces. Show that (a) P(t) = (1 t 2 ) 1 ; (b) E(t) = (1 t 2 ) 1/2 ; [HINT: Exercise 15 of Chapter 3] (c) E(t).O(t) = P(t); (d) e = o for all eve. Fid a bijective proof of the last equality. (a) Sice we are oly cosiderig eve vales of, we have ( ) (2m)! P(t) = t 2m = 1 m 0 (2m)! 1 t 2. (b) By Chapter 3, Exercise 16, the umber e 2m of permutatios of a (2m)set with all cycles eve is ((2m)!!) 2. So we have ( ((2m)!!) 2 ) E(t) = t 2m = (1 t 2 ) 1/2, m 0 (2m)! the secod equality beig verified by expadig the last expressio usig the Biomial Theorem. (c) This is a istace of a very geeral coutig priciple. If F(t) ad G(t) are the e.g.f.s for labelled structures i two classes F ad G, with F(t) = f t /! ad G(t) = g t /!, the F(t)G(t) is the e.g.f. of structures cosistig of a partitio of the poit set with a F structure o oe part ad a Gstructure o the complemet. For, if H is the class of such composite structures, the the umber of H structures o a set is give by h = k=0 ( k ) f k g k, from which a simple calculatio shows that H(t) = h t /! = F(t)G(t). Now (c) follows o applyig this result, where F ad G deote the classes of permutatios with all cycles eve, resp., all cycles odd: give a arbitrary 12
13 permutatio π o {1,...,2m}, there is a uique partitio of the set ito two parts o which the iduced permutatios satisfy these coditios; ad both parts have eve cardiality. (d) From (a), (b) ad (c), we deduce that O(t) = (1 t 2 ) 1/2 = E(t); so o = e for all. Here is a outlie of the costructio of a bijectio: you should fill i the details. Give the cycle decompositio of a permutatio of {1,...,}, we ca rotate each cycle so that the least elemet comes first, ad the order the cycles accordig to their least elemets. Call this the caoical form of the cycle decompositio. We ow defie the bijectios: (a) Give a caoical form C 1,...,C 2r, where each cycle C i has odd legth, for i = 1,...,r remove the last elemet of the cycle C 2i ad add it at the ed of the cycle C 2i 1. (Some cycles may disappear i this process.) The resultig cycles all have eve legth, ad the expressio is i the caoical form. (b) Give a caoical form C 1,...,C m, where each cycle has eve legth, proceed recursively as follows. Let x be the last elemet of C 1, ad y the first elemet of C 2. (Take y = if C 2 does t exist, that is, if there is oly oe cycle.) If x > y, remove x from C 1 ad add it at the ed of C 2 ; the process C 3,...,C m. If x < y, remove x from C 1 ad add it as a ew sigleto cycle after C 1 ; the process C 2,...,C m. The resultig cycles all have odd legth ad the expressio is i caoical form. It remais to show that these two maps are mutually iverse. See R. P. Lewis ad S. P. Norto, Discrete Mathematics 138 (1995), , for further details. 21 Show that QUICKSORT sometimes requires all ( 2) comparisos to sort a list. For how may orderigs does this occur? Oe such orderig is the case whe the list is already sorted is this a serious defect of QUICKSORT? Suppose that we start with a sorted list. The QUICKSORT selects the first elemet (which is the smallest), ad partitios the remaider ito the empty list ad all the other elemets (with 1 comparisos). Oe brach of the recursio is ( trivial, but i the other we agai have a sorted list, requirig (by iductio) 1 ) ( 2 comparisos. So ) 2 comparisos are eeded altogether. The sorted list is ot the oly permutatio requirig so may comparisos. As log as each elemet is either smaller tha or greater tha all its successors, the same ueve split will occur at eac stage. So there are 2 1 bad permutatios, sice there are two choices for each elemet except the last. 13
14 This is a problem, but i practice ot too serious. If the lists to be sorted are truly radom, the the proportio of bad permutatios decreases faster tha expoetially. If we are likely to meet sorted or partially sorted lists quite ofte, we ca modify the algorithm by choosig a to be the middle item of the list, rather tha the first. 22 Let m be the miimum umber of comparisos required by QUICKSORT to sort a list of legth. Prove that, for each iteger k > 1, m is a liear fuctio of o the iterval from 2 k 1 1 to 2 k 1, with m 2 k 1 = (k 2)2k + 2. If = 2 k 1, what ca you say about the umber of orderigs requirig m comparisos? The miimum umber of comparisos will occur whe the sublists are as early equal as possible: both of legth ( 1)/2 if is odd, or of legths ( 2)/2 ad /2 if is eve. So we have the recurrece { 1 + 2m( 1)/2, if is odd, m = 1 + m ( 2)/2 + m /2, if is eve. Suppose that m = a + b for c d. The, for 2c + 1 2d + 1, we have m = 1 + 2(a( 1)/2 + b) = (a + 1) + (2b a 1). But m is liear (i fact idetically zero) for 0 1; by iductio o k, it is liear o each iterval 2 k k 1. If the value o this iterval is give by m = a k +b k, the we have a 0 = b 0 = 0, a k+1 = a k + 1, ad b k+1 = 2b k a k 1. By iductio, we fid that a k = k ad b k = 2 k+1 + k + 2. Settig = 2 k 1, we have as required. m 2 k 1 = k (2k 1) 2 k+1 + k + 2 = (k 2)2 k + 2, To cout the umber of orders which require the miimum umber of comparisos, ote that the first step i the algorithm splits the list ito sublists of legth i ad i 1, say; ow we require that each of these sublists ca be sorted with the miimum umber of comparisos, ad also that both i ad 1 i lie i a iterval i which m j is a liear fuctio of j, that is, oe of the form [2 d 1 1,2 d 1]. However, the two sublists ca be merged arbitrarily. So the umber f () of orders requirig the miimum umber of comparisos satisfies the recurrece ( ) 1 f () = f (i) f ( i 1), i 14
15 where the sum is over all i such that i ad i 1 lie i the same iterval of the form [2 d 1 1,2 d 1]. For example, if = 5 ad the list cotais 1,...,5 i some order, the it requires the miimum umber 6 of comparisos to sort if ad oly if oe of the followig holds: the first elemet is 3; the first elemet is 2, ad 4 precedes 3 ad 5; the first elemet is 4, ad 2 precedes 1 ad 3. There are 40 such lists, out of 120. As i the previous questio, the maximum umber of comparisos is 10, ad 16 lists require this umber. Check that the umbers requirig 7, 8, 9 comparisos are 32, 24, 8 respectively. Now check that the average agrees with the value calculated i the recurrece of Sectio
16 23 This exercise justifies the Twety Questios priciple. We are give N objects ad required to distiguish them by askig questios, each of which has two possible aswers. The aim of this exercise is to show that, o matter what scheme of questioig is adopted, o average the umber of questios required is at least log 2 N. (For some schemes, the average may be much larger. If we ask Is it a 1?, Is it a 2?, etc., the o average (N + 1)/2 questios are eeded!) A biary tree is a graph (see Chapter 2) with the followig properties: there is a vertex (the root) lyig o just two edges; every other vertex lies o oe or three edges (ad is called a leaf or a iteral vertex accordigly); there are o circuits (closed paths of distict vertices), ad every vertex ca be reached by a path from the root. It is coveiet to arrage the vertices of the tree o successive levels, with the root o level 0. The ay oleaf is joied to two successors o the ext level, ad every vertex except the root has oe predecessor. The height of a vertex is the umber of the level o which it lies. I our situatio, a vertex is ay set of objects which ca be distiguished by some sequece of questios. The root correspods to the whole set (before ay questios are asked), ad leaves are sigleto sets. The two successors of a vertex are the sets distiguished by the two possible aswers to the ext questio. The height of a leaf is the umber of questios required to idetify that object uiquely. STEP 1. Show that there are two leaves of maximal height (h, say) with the same predecessor. Deduce that, if there is a leaf of height less tha h 1, we ca fid aother biary tree with N leaves havig smaller average height. Hece coclude that, i a tree with miimum average height, every leaf has height m or m + 1, for some m. STEP 2. Sice there are o leaves at height less tha m, there are altogether 2 m vertices o level m. STEP 3. If there are p iteral vertices o level m, show that there are 2p leaves of height m + 1, ad N 2p = 2 m p of height m; so N = 2 m + p, where 0 p < 2 m. STEP 4. Prove that log 2 (2 m + p) m+2p/(2 m + p), ad deduce that the average height of leaves is at least log 2 N. 16
17 Follow the suggested proof. Step 1: If v is a vertex of maximum height ad w its predecessor, the the other successor v of w would also be a leaf (else its ow successors would be higher tha v). If there is a leaf x with height less tha h 1, the remove the two leaves v,v of height h ad add two ew successors y,y of x with height less tha h, reducig the average height without chagig the umber of leaves. So, i a tree of miimum height, o leaf has height less tha h 1 (where h is the maximum height). Put m = h 1. Step 2: If there are o leaves o level k, the each vertex o this level has two successors, ad so the umber of vertices o level k + 1 is twice the umber o level k. I our case, this holds for k = 0,1,...,m 1, so there are 2 m vertices o level m. Step 3: Suppose that there are p iteral vertices o level m. The there are 2p vertices o level m + 1, all of them leaves; ad there are 2 m p leaves o level m, givig 2 m + p altogether. Thus, if N is the umber of leaves, the m ad p are determied by N: 2 m is the largest power of 2 ot exceedig N, ad p = N = 2 m. (If N is a power of 2, we ca take it to be 2 m rather tha 2 m+1, to simplify thigs.) Step 4: The average height of the leaves is (2 m p)m + 2p(m + 1) 2 m + p = m + 2p 2 m + p. By calculus, show that log 2 (1 x) 2x for 0 x 2 1. (The two expressios are equal whe x = 0 ad whe x = 1 2 ; ad their differece has a uique statioary value i the iterval, at x = 1 1/(2log2), which is a miimum.) Apply this iequality with x = p/(2 m + p), so that m log 2 (1 x) = m + log 2 (1 + p/2 m ) = log 2 (2 m + p). Thus, log 2 N m + 2p/(2 m + p), ad we are doe. 24 Suppose that the two successors of each oleaf ode i a biary tree are distiguished as left ad right. Show that, with this covetio, the umber of biary trees with leaves is the Catala umber C. Let T be the umber of biary trees with leaves, with the leftright distictio. The T 1 = 1 (the root is the uique leaf); ad, for > 1, a tree with leaves is specified by a left subtree with k leaves ad a right subtree with k leaves, where 1 k 1. Hece T = 1 k=1 T kt k for > 1. By (4.5.1) ad iductio, T = C for all. 17
Soving Recurrence Relations
Sovig Recurrece Relatios Part 1. Homogeeous liear 2d degree relatios with costat coefficiets. Cosider the recurrece relatio ( ) T () + at ( 1) + bt ( 2) = 0 This is called a homogeeous liear 2d degree
More informationIn nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008
I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces
More informationCS103X: Discrete Structures Homework 4 Solutions
CS103X: Discrete Structures Homewor 4 Solutios Due February 22, 2008 Exercise 1 10 poits. Silico Valley questios: a How may possible sixfigure salaries i whole dollar amouts are there that cotai at least
More informationDepartment of Computer Science, University of Otago
Departmet of Computer Sciece, Uiversity of Otago Techical Report OUCS200609 Permutatios Cotaiig May Patters Authors: M.H. Albert Departmet of Computer Sciece, Uiversity of Otago Micah Colema, Rya Fly
More informationInfinite Sequences and Series
CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...
More informationNotes on exponential generating functions and structures.
Notes o expoetial geeratig fuctios ad structures. 1. The cocept of a structure. Cosider the followig coutig problems: (1) to fid for each the umber of partitios of a elemet set, (2) to fid for each the
More informationAsymptotic Growth of Functions
CMPS Itroductio to Aalysis of Algorithms Fall 3 Asymptotic Growth of Fuctios We itroduce several types of asymptotic otatio which are used to compare the performace ad efficiecy of algorithms As we ll
More informationSequences and Series
CHAPTER 9 Sequeces ad Series 9.. Covergece: Defiitio ad Examples Sequeces The purpose of this chapter is to itroduce a particular way of geeratig algorithms for fidig the values of fuctios defied by their
More informationSAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx
SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval
More informationCS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations
CS3A Hadout 3 Witer 00 February, 00 Solvig Recurrece Relatios Itroductio A wide variety of recurrece problems occur i models. Some of these recurrece relatios ca be solved usig iteratio or some other ad
More information1. MATHEMATICAL INDUCTION
1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1
More information5.3. Generalized Permutations and Combinations
53 GENERALIZED PERMUTATIONS AND COMBINATIONS 73 53 Geeralized Permutatios ad Combiatios 53 Permutatios with Repeated Elemets Assume that we have a alphabet with letters ad we wat to write all possible
More information5 Boolean Decision Trees (February 11)
5 Boolea Decisio Trees (February 11) 5.1 Graph Coectivity Suppose we are give a udirected graph G, represeted as a boolea adjacecy matrix = (a ij ), where a ij = 1 if ad oly if vertices i ad j are coected
More informationSection 11.3: The Integral Test
Sectio.3: The Itegral Test Most of the series we have looked at have either diverged or have coverged ad we have bee able to fid what they coverge to. I geeral however, the problem is much more difficult
More informationTrigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is
0_0605.qxd /5/05 0:45 AM Page 470 470 Chapter 6 Additioal Topics i Trigoometry 6.5 Trigoometric Form of a Complex Number What you should lear Plot complex umbers i the complex plae ad fid absolute values
More informationWeek 3 Conditional probabilities, Bayes formula, WEEK 3 page 1 Expected value of a random variable
Week 3 Coditioal probabilities, Bayes formula, WEEK 3 page 1 Expected value of a radom variable We recall our discussio of 5 card poker hads. Example 13 : a) What is the probability of evet A that a 5
More informationDiscrete Mathematics and Probability Theory Spring 2014 Anant Sahai Note 13
EECS 70 Discrete Mathematics ad Probability Theory Sprig 2014 Aat Sahai Note 13 Itroductio At this poit, we have see eough examples that it is worth just takig stock of our model of probability ad may
More informationSECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES
SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces,
More informationBasic Elements of Arithmetic Sequences and Series
MA40S PRECALCULUS UNIT G GEOMETRIC SEQUENCES CLASS NOTES (COMPLETED NO NEED TO COPY NOTES FROM OVERHEAD) Basic Elemets of Arithmetic Sequeces ad Series Objective: To establish basic elemets of arithmetic
More information.04. This means $1000 is multiplied by 1.02 five times, once for each of the remaining sixmonth
Questio 1: What is a ordiary auity? Let s look at a ordiary auity that is certai ad simple. By this, we mea a auity over a fixed term whose paymet period matches the iterest coversio period. Additioally,
More information23 The Remainder and Factor Theorems
 The Remaider ad Factor Theorems Factor each polyomial completely usig the give factor ad log divisio 1 x + x x 60; x + So, x + x x 60 = (x + )(x x 15) Factorig the quadratic expressio yields x + x x
More informationFIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. 1. Powers of a matrix
FIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. Powers of a matrix We begi with a propositio which illustrates the usefuless of the diagoalizatio. Recall that a square matrix A is diogaalizable if
More informationChapter 7 Methods of Finding Estimators
Chapter 7 for BST 695: Special Topics i Statistical Theory. Kui Zhag, 011 Chapter 7 Methods of Fidig Estimators Sectio 7.1 Itroductio Defiitio 7.1.1 A poit estimator is ay fuctio W( X) W( X1, X,, X ) of
More informationHere are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.
This documet was writte ad copyrighted by Paul Dawkis. Use of this documet ad its olie versio is govered by the Terms ad Coditios of Use located at http://tutorial.math.lamar.edu/terms.asp. The olie versio
More information4.3. The Integral and Comparison Tests
4.3. THE INTEGRAL AND COMPARISON TESTS 9 4.3. The Itegral ad Compariso Tests 4.3.. The Itegral Test. Suppose f is a cotiuous, positive, decreasig fuctio o [, ), ad let a = f(). The the covergece or divergece
More informationLecture 4: Cauchy sequences, BolzanoWeierstrass, and the Squeeze theorem
Lecture 4: Cauchy sequeces, BolzaoWeierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits
More informationIncremental calculation of weighted mean and variance
Icremetal calculatio of weighted mea ad variace Toy Fich faf@cam.ac.uk dot@dotat.at Uiversity of Cambridge Computig Service February 009 Abstract I these otes I eplai how to derive formulae for umerically
More informationTHE ARITHMETIC OF INTEGERS.  multiplication, exponentiation, division, addition, and subtraction
THE ARITHMETIC OF INTEGERS  multiplicatio, expoetiatio, divisio, additio, ad subtractio What to do ad what ot to do. THE INTEGERS Recall that a iteger is oe of the whole umbers, which may be either positive,
More informationLecture 13. Lecturer: Jonathan Kelner Scribe: Jonathan Pines (2009)
18.409 A Algorithmist s Toolkit October 27, 2009 Lecture 13 Lecturer: Joatha Keler Scribe: Joatha Pies (2009) 1 Outlie Last time, we proved the BruMikowski iequality for boxes. Today we ll go over the
More informationI. Chisquared Distributions
1 M 358K Supplemet to Chapter 23: CHISQUARED DISTRIBUTIONS, TDISTRIBUTIONS, AND DEGREES OF FREEDOM To uderstad tdistributios, we first eed to look at aother family of distributios, the chisquared distributios.
More informationProperties of MLE: consistency, asymptotic normality. Fisher information.
Lecture 3 Properties of MLE: cosistecy, asymptotic ormality. Fisher iformatio. I this sectio we will try to uderstad why MLEs are good. Let us recall two facts from probability that we be used ofte throughout
More informationINFINITE SERIES KEITH CONRAD
INFINITE SERIES KEITH CONRAD. Itroductio The two basic cocepts of calculus, differetiatio ad itegratio, are defied i terms of limits (Newto quotiets ad Riema sums). I additio to these is a third fudametal
More information4. Trees. 4.1 Basics. Definition: A graph having no cycles is said to be acyclic. A forest is an acyclic graph.
4. Trees Oe of the importat classes of graphs is the trees. The importace of trees is evidet from their applicatios i various areas, especially theoretical computer sciece ad molecular evolutio. 4.1 Basics
More informationCHAPTER 3 DIGITAL CODING OF SIGNALS
CHAPTER 3 DIGITAL CODING OF SIGNALS Computers are ofte used to automate the recordig of measuremets. The trasducers ad sigal coditioig circuits produce a voltage sigal that is proportioal to a quatity
More informationSEQUENCES AND SERIES
Chapter 9 SEQUENCES AND SERIES Natural umbers are the product of huma spirit. DEDEKIND 9.1 Itroductio I mathematics, the word, sequece is used i much the same way as it is i ordiary Eglish. Whe we say
More informationTheorems About Power Series
Physics 6A Witer 20 Theorems About Power Series Cosider a power series, f(x) = a x, () where the a are real coefficiets ad x is a real variable. There exists a real oegative umber R, called the radius
More informationExample 2 Find the square root of 0. The only square root of 0 is 0 (since 0 is not positive or negative, so those choices don t exist here).
BEGINNING ALGEBRA Roots ad Radicals (revised summer, 00 Olso) Packet to Supplemet the Curret Textbook  Part Review of Square Roots & Irratioals (This portio ca be ay time before Part ad should mostly
More informationChapter 5 O A Cojecture Of Erdíos Proceedigs NCUR VIII è1994è, Vol II, pp 794í798 Jeærey F Gold Departmet of Mathematics, Departmet of Physics Uiversity of Utah Do H Tucker Departmet of Mathematics Uiversity
More informationChapter 6: Variance, the law of large numbers and the MonteCarlo method
Chapter 6: Variace, the law of large umbers ad the MoteCarlo method Expected value, variace, ad Chebyshev iequality. If X is a radom variable recall that the expected value of X, E[X] is the average value
More informationA probabilistic proof of a binomial identity
A probabilistic proof of a biomial idetity Joatho Peterso Abstract We give a elemetary probabilistic proof of a biomial idetity. The proof is obtaied by computig the probability of a certai evet i two
More informationNATIONAL SENIOR CERTIFICATE GRADE 12
NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P EXEMPLAR 04 MARKS: 50 TIME: 3 hours This questio paper cosists of 8 pages ad iformatio sheet. Please tur over Mathematics/P DBE/04 NSC Grade Eemplar INSTRUCTIONS
More informationTHE ABRACADABRA PROBLEM
THE ABRACADABRA PROBLEM FRANCESCO CARAVENNA Abstract. We preset a detailed solutio of Exercise E0.6 i [Wil9]: i a radom sequece of letters, draw idepedetly ad uiformly from the Eglish alphabet, the expected
More information1.3. VERTEX DEGREES & COUNTING
35 Chapter 1: Fudametal Cocepts Sectio 1.3: Vertex Degrees ad Coutig 36 its eighbor o P. Note that P has at least three vertices. If G x v is coected, let y = v. Otherwise, a compoet cut off from P x v
More informationFactors of sums of powers of binomial coefficients
ACTA ARITHMETICA LXXXVI.1 (1998) Factors of sums of powers of biomial coefficiets by Neil J. Cali (Clemso, S.C.) Dedicated to the memory of Paul Erdős 1. Itroductio. It is well ow that if ( ) a f,a = the
More information3. Greatest Common Divisor  Least Common Multiple
3 Greatest Commo Divisor  Least Commo Multiple Defiitio 31: The greatest commo divisor of two atural umbers a ad b is the largest atural umber c which divides both a ad b We deote the greatest commo gcd
More informationEngineering 323 Beautiful Homework Set 3 1 of 7 Kuszmar Problem 2.51
Egieerig 33 eautiful Homewor et 3 of 7 Kuszmar roblem.5.5 large departmet store sells sport shirts i three sizes small, medium, ad large, three patters plaid, prit, ad stripe, ad two sleeve legths log
More informationTHE HEIGHT OF qbinary SEARCH TREES
THE HEIGHT OF qbinary SEARCH TREES MICHAEL DRMOTA AND HELMUT PRODINGER Abstract. q biary search trees are obtaied from words, equipped with the geometric distributio istead of permutatios. The average
More informationModified Line Search Method for Global Optimization
Modified Lie Search Method for Global Optimizatio Cria Grosa ad Ajith Abraham Ceter of Excellece for Quatifiable Quality of Service Norwegia Uiversity of Sciece ad Techology Trodheim, Norway {cria, ajith}@q2s.tu.o
More informationCooleyTukey. Tukey FFT Algorithms. FFT Algorithms. Cooley
Cooley CooleyTuey Tuey FFT Algorithms FFT Algorithms Cosider a legth sequece x[ with a poit DFT X[ where Represet the idices ad as +, +, Cooley CooleyTuey Tuey FFT Algorithms FFT Algorithms Usig these
More informationBuilding Blocks Problem Related to Harmonic Series
TMME, vol3, o, p.76 Buildig Blocks Problem Related to Harmoic Series Yutaka Nishiyama Osaka Uiversity of Ecoomics, Japa Abstract: I this discussio I give a eplaatio of the divergece ad covergece of ifiite
More informationConvexity, Inequalities, and Norms
Covexity, Iequalities, ad Norms Covex Fuctios You are probably familiar with the otio of cocavity of fuctios. Give a twicedifferetiable fuctio ϕ: R R, We say that ϕ is covex (or cocave up) if ϕ (x) 0 for
More informationA Recursive Formula for Moments of a Binomial Distribution
A Recursive Formula for Momets of a Biomial Distributio Árpád Béyi beyi@mathumassedu, Uiversity of Massachusetts, Amherst, MA 01003 ad Saverio M Maago smmaago@psavymil Naval Postgraduate School, Moterey,
More informationTaking DCOP to the Real World: Efficient Complete Solutions for Distributed MultiEvent Scheduling
Taig DCOP to the Real World: Efficiet Complete Solutios for Distributed MultiEvet Schedulig Rajiv T. Maheswara, Milid Tambe, Emma Bowrig, Joatha P. Pearce, ad Pradeep araatham Uiversity of Souther Califoria
More informationS. Tanny MAT 344 Spring 1999. be the minimum number of moves required.
S. Tay MAT 344 Sprig 999 Recurrece Relatios Tower of Haoi Let T be the miimum umber of moves required. T 0 = 0, T = 7 Iitial Coditios * T = T + $ T is a sequece (f. o itegers). Solve for T? * is a recurrece,
More informationChapter 5: Inner Product Spaces
Chapter 5: Ier Product Spaces Chapter 5: Ier Product Spaces SECION A Itroductio to Ier Product Spaces By the ed of this sectio you will be able to uderstad what is meat by a ier product space give examples
More informationCHAPTER 3 THE TIME VALUE OF MONEY
CHAPTER 3 THE TIME VALUE OF MONEY OVERVIEW A dollar i the had today is worth more tha a dollar to be received i the future because, if you had it ow, you could ivest that dollar ad ear iterest. Of all
More informationOur aim is to show that under reasonable assumptions a given 2πperiodic function f can be represented as convergent series
8 Fourier Series Our aim is to show that uder reasoable assumptios a give periodic fuctio f ca be represeted as coverget series f(x) = a + (a cos x + b si x). (8.) By defiitio, the covergece of the series
More informationConfidence Intervals for One Mean
Chapter 420 Cofidece Itervals for Oe Mea Itroductio This routie calculates the sample size ecessary to achieve a specified distace from the mea to the cofidece limit(s) at a stated cofidece level for a
More informationMARTINGALES AND A BASIC APPLICATION
MARTINGALES AND A BASIC APPLICATION TURNER SMITH Abstract. This paper will develop the measuretheoretic approach to probability i order to preset the defiitio of martigales. From there we will apply this
More informationApproximating Area under a curve with rectangles. To find the area under a curve we approximate the area using rectangles and then use limits to find
1.8 Approximatig Area uder a curve with rectagles 1.6 To fid the area uder a curve we approximate the area usig rectagles ad the use limits to fid 1.4 the area. Example 1 Suppose we wat to estimate 1.
More informationUC Berkeley Department of Electrical Engineering and Computer Science. EE 126: Probablity and Random Processes. Solutions 9 Spring 2006
Exam format UC Bereley Departmet of Electrical Egieerig ad Computer Sciece EE 6: Probablity ad Radom Processes Solutios 9 Sprig 006 The secod midterm will be held o Wedesday May 7; CHECK the fial exam
More informationFast Fourier Transform
18.310 lecture otes November 18, 2013 Fast Fourier Trasform Lecturer: Michel Goemas I these otes we defie the Discrete Fourier Trasform, ad give a method for computig it fast: the Fast Fourier Trasform.
More information2. Degree Sequences. 2.1 Degree Sequences
2. Degree Sequeces The cocept of degrees i graphs has provided a framewor for the study of various structural properties of graphs ad has therefore attracted the attetio of may graph theorists. Here we
More informationNotes on Combinatorics. Peter J. Cameron
Notes o Combiatorics Peter J Camero ii Preface: What is Combiatorics? Combiatorics, the mathematics of patters,, helps us desig computer etwors, crac security codes, or solve sudous Ursula Marti, VicePricipal
More information0.7 0.6 0.2 0 0 96 96.5 97 97.5 98 98.5 99 99.5 100 100.5 96.5 97 97.5 98 98.5 99 99.5 100 100.5
Sectio 13 KolmogorovSmirov test. Suppose that we have a i.i.d. sample X 1,..., X with some ukow distributio P ad we would like to test the hypothesis that P is equal to a particular distributio P 0, i.e.
More informationLecture 2: Karger s Min Cut Algorithm
priceto uiv. F 3 cos 5: Advaced Algorithm Desig Lecture : Karger s Mi Cut Algorithm Lecturer: Sajeev Arora Scribe:Sajeev Today s topic is simple but gorgeous: Karger s mi cut algorithm ad its extesio.
More informationSolutions to Selected Problems In: Pattern Classification by Duda, Hart, Stork
Solutios to Selected Problems I: Patter Classificatio by Duda, Hart, Stork Joh L. Weatherwax February 4, 008 Problem Solutios Chapter Bayesia Decisio Theory Problem radomized rules Part a: Let Rx be the
More informationClass Meeting # 16: The Fourier Transform on R n
MATH 18.152 COUSE NOTES  CLASS MEETING # 16 18.152 Itroductio to PDEs, Fall 2011 Professor: Jared Speck Class Meetig # 16: The Fourier Trasform o 1. Itroductio to the Fourier Trasform Earlier i the course,
More informationOutput Analysis (2, Chapters 10 &11 Law)
B. Maddah ENMG 6 Simulatio 05/0/07 Output Aalysis (, Chapters 10 &11 Law) Comparig alterative system cofiguratio Sice the output of a simulatio is radom, the comparig differet systems via simulatio should
More informationTHE REGRESSION MODEL IN MATRIX FORM. For simple linear regression, meaning one predictor, the model is. for i = 1, 2, 3,, n
We will cosider the liear regressio model i matrix form. For simple liear regressio, meaig oe predictor, the model is i = + x i + ε i for i =,,,, This model icludes the assumptio that the ε i s are a sample
More information5: Introduction to Estimation
5: Itroductio to Estimatio Cotets Acroyms ad symbols... 1 Statistical iferece... Estimatig µ with cofidece... 3 Samplig distributio of the mea... 3 Cofidece Iterval for μ whe σ is kow before had... 4 Sample
More informationCME 302: NUMERICAL LINEAR ALGEBRA FALL 2005/06 LECTURE 8
CME 30: NUMERICAL LINEAR ALGEBRA FALL 005/06 LECTURE 8 GENE H GOLUB 1 Positive Defiite Matrices A matrix A is positive defiite if x Ax > 0 for all ozero x A positive defiite matrix has real ad positive
More informationCHAPTER 7: Central Limit Theorem: CLT for Averages (Means)
CHAPTER 7: Cetral Limit Theorem: CLT for Averages (Meas) X = the umber obtaied whe rollig oe six sided die oce. If we roll a six sided die oce, the mea of the probability distributio is X P(X = x) Simulatio:
More informationPermutations, the Parity Theorem, and Determinants
1 Permutatios, the Parity Theorem, ad Determiats Joh A. Guber Departmet of Electrical ad Computer Egieerig Uiversity of Wiscosi Madiso Cotets 1 What is a Permutatio 1 2 Cycles 2 2.1 Traspositios 4 3 Orbits
More informationChapter 7  Sampling Distributions. 1 Introduction. What is statistics? It consist of three major areas:
Chapter 7  Samplig Distributios 1 Itroductio What is statistics? It cosist of three major areas: Data Collectio: samplig plas ad experimetal desigs Descriptive Statistics: umerical ad graphical summaries
More informationGCE Further Mathematics (6360) Further Pure Unit 2 (MFP2) Textbook. Version: 1.4
GCE Further Mathematics (660) Further Pure Uit (MFP) Tetbook Versio: 4 MFP Tetbook Alevel Further Mathematics 660 Further Pure : Cotets Chapter : Comple umbers 4 Itroductio 5 The geeral comple umber 5
More informationAnalysis Notes (only a draft, and the first one!)
Aalysis Notes (oly a draft, ad the first oe!) Ali Nesi Mathematics Departmet Istabul Bilgi Uiversity Kuştepe Şişli Istabul Turkey aesi@bilgi.edu.tr Jue 22, 2004 2 Cotets 1 Prelimiaries 9 1.1 Biary Operatio...........................
More informationBINOMIAL EXPANSIONS 12.5. In this section. Some Examples. Obtaining the Coefficients
652 (1226) Chapter 12 Sequeces ad Series 12.5 BINOMIAL EXPANSIONS I this sectio Some Examples Otaiig the Coefficiets The Biomial Theorem I Chapter 5 you leared how to square a iomial. I this sectio you
More informationSEQUENCES AND SERIES CHAPTER
CHAPTER SEQUENCES AND SERIES Whe the Grat family purchased a computer for $,200 o a istallmet pla, they agreed to pay $00 each moth util the cost of the computer plus iterest had bee paid The iterest each
More informationDefinition. A variable X that takes on values X 1, X 2, X 3,...X k with respective frequencies f 1, f 2, f 3,...f k has mean
1 Social Studies 201 October 13, 2004 Note: The examples i these otes may be differet tha used i class. However, the examples are similar ad the methods used are idetical to what was preseted i class.
More information1. C. The formula for the confidence interval for a population mean is: x t, which was
s 1. C. The formula for the cofidece iterval for a populatio mea is: x t, which was based o the sample Mea. So, x is guarateed to be i the iterval you form.. D. Use the rule : pvalue
More informationThe Stable Marriage Problem
The Stable Marriage Problem William Hut Lae Departmet of Computer Sciece ad Electrical Egieerig, West Virgiia Uiversity, Morgatow, WV William.Hut@mail.wvu.edu 1 Itroductio Imagie you are a matchmaker,
More informationOverview of some probability distributions.
Lecture Overview of some probability distributios. I this lecture we will review several commo distributios that will be used ofte throughtout the class. Each distributio is usually described by its probability
More informationYour organization has a Class B IP address of 166.144.0.0 Before you implement subnetting, the Network ID and Host ID are divided as follows:
Subettig Subettig is used to subdivide a sigle class of etwork i to multiple smaller etworks. Example: Your orgaizatio has a Class B IP address of 166.144.0.0 Before you implemet subettig, the Network
More informationInteger Factorization Algorithms
Iteger Factorizatio Algorithms Coelly Bares Departmet of Physics, Orego State Uiversity December 7, 004 This documet has bee placed i the public domai. Cotets I. Itroductio 3 1. Termiology 3. Fudametal
More informationConfidence Intervals. CI for a population mean (σ is known and n > 30 or the variable is normally distributed in the.
Cofidece Itervals A cofidece iterval is a iterval whose purpose is to estimate a parameter (a umber that could, i theory, be calculated from the populatio, if measuremets were available for the whole populatio).
More informationLecture 5: Span, linear independence, bases, and dimension
Lecture 5: Spa, liear idepedece, bases, ad dimesio Travis Schedler Thurs, Sep 23, 2010 (versio: 9/21 9:55 PM) 1 Motivatio Motivatio To uderstad what it meas that R has dimesio oe, R 2 dimesio 2, etc.;
More information5.4 Amortization. Question 1: How do you find the present value of an annuity? Question 2: How is a loan amortized?
5.4 Amortizatio Questio 1: How do you fid the preset value of a auity? Questio 2: How is a loa amortized? Questio 3: How do you make a amortizatio table? Oe of the most commo fiacial istrumets a perso
More informationHypergeometric Distributions
7.4 Hypergeometric Distributios Whe choosig the startig lieup for a game, a coach obviously has to choose a differet player for each positio. Similarly, whe a uio elects delegates for a covetio or you
More information*The most important feature of MRP as compared with ordinary inventory control analysis is its time phasing feature.
Itegrated Productio ad Ivetory Cotrol System MRP ad MRP II Framework of Maufacturig System Ivetory cotrol, productio schedulig, capacity plaig ad fiacial ad busiess decisios i a productio system are iterrelated.
More informationFOUNDATIONS OF MATHEMATICS AND PRECALCULUS GRADE 10
FOUNDATIONS OF MATHEMATICS AND PRECALCULUS GRADE 10 [C] Commuicatio Measuremet A1. Solve problems that ivolve liear measuremet, usig: SI ad imperial uits of measure estimatio strategies measuremet strategies.
More informationIrreducible polynomials with consecutive zero coefficients
Irreducible polyomials with cosecutive zero coefficiets Theodoulos Garefalakis Departmet of Mathematics, Uiversity of Crete, 71409 Heraklio, Greece Abstract Let q be a prime power. We cosider the problem
More informationPerfect Packing Theorems and the AverageCase Behavior of Optimal and Online Bin Packing
SIAM REVIEW Vol. 44, No. 1, pp. 95 108 c 2002 Society for Idustrial ad Applied Mathematics Perfect Packig Theorems ad the AverageCase Behavior of Optimal ad Olie Bi Packig E. G. Coffma, Jr. C. Courcoubetis
More informationHypothesis testing. Null and alternative hypotheses
Hypothesis testig Aother importat use of samplig distributios is to test hypotheses about populatio parameters, e.g. mea, proportio, regressio coefficiets, etc. For example, it is possible to stipulate
More information. P. 4.3 Basic feasible solutions and vertices of polyhedra. x 1. x 2
4. Basic feasible solutios ad vertices of polyhedra Due to the fudametal theorem of Liear Programmig, to solve ay LP it suffices to cosider the vertices (fiitely may) of the polyhedro P of the feasible
More informationwhere: T = number of years of cash flow in investment's life n = the year in which the cash flow X n i = IRR = the internal rate of return
EVALUATING ALTERNATIVE CAPITAL INVESTMENT PROGRAMS By Ke D. Duft, Extesio Ecoomist I the March 98 issue of this publicatio we reviewed the procedure by which a capital ivestmet project was assessed. The
More informationAP Calculus BC 2003 Scoring Guidelines Form B
AP Calculus BC Scorig Guidelies Form B The materials icluded i these files are iteded for use by AP teachers for course ad exam preparatio; permissio for ay other use must be sought from the Advaced Placemet
More information1 Correlation and Regression Analysis
1 Correlatio ad Regressio Aalysis I this sectio we will be ivestigatig the relatioship betwee two cotiuous variable, such as height ad weight, the cocetratio of a ijected drug ad heart rate, or the cosumptio
More informationChair for Network Architectures and Services Institute of Informatics TU München Prof. Carle. Network Security. Chapter 2 Basics
Chair for Network Architectures ad Services Istitute of Iformatics TU Müche Prof. Carle Network Security Chapter 2 Basics 2.4 Radom Number Geeratio for Cryptographic Protocols Motivatio It is crucial to
More informationProject Deliverables. CS 361, Lecture 28. Outline. Project Deliverables. Administrative. Project Comments
Project Deliverables CS 361, Lecture 28 Jared Saia Uiversity of New Mexico Each Group should tur i oe group project cosistig of: About 612 pages of text (ca be loger with appedix) 612 figures (please
More information