Applications of Regular Closure
|
|
- Chad Palmer
- 5 years ago
- Views:
Transcription
1 Applictios of Regulr Closure 1
2 The itersectio of cotext-free lguge d regulr lguge is cotext-free lguge L1 L2 cotext free regulr Regulr Closure L1 L 2 cotext-free 2
3 Liz 6 th, sectio 8.2, exple 8.7, pge 227 L={^ ^ 0, 100} is cotext free 3
4 A Applictio of Regulr Closure Prove tht: L { : 100} is cotext-free 4
5 We kow: { } is cotext-free 5
6 We lso kow: 100 L1 { 100 } is regulr L1 {( ) } { * } is regulr 6
7 { } cotext-free 100 L1 {( ) } { * regulr 100 } (regulr closure) { } L 1 is cotext-free { } L1 { : 100} L is cotext-free 7
8 Liz 6 th, sectio 8.2, exple 8.8, pge 227 L={w # (w) = # (w) = # c (w)} is ot cotext free 8
9 Aother Applictio of Regulr Closure Prove tht: L { w: c } is ot cotext-free 9
10 If L { w: c } is cotext-free (regulr closure) The L { * * c*} { c } cotext-free regulr cotext-free Ipossile!!! Therefore, L is ot cotext free 10
11 Decidle Properties of Cotext-Free Lguges 11
12 Meership Questio: for cotext-free grr fid if strig w L(G) G Meership Algoriths: Prsers Exhustive serch prser CYK prsig lgorith 12
13 Epty Lguge Questio: for cotext-free grr fid if L(G) G Algorith: 1. Reove useless vriles 2. Check if strt vrile is useless S 13
14 Ifiite Lguge Questio: for cotext-free grr fid if L(G) is ifiite G Algorith: 1. Reove useless vriles 2. Reove uit d productios 3. Crete depedecy grph for vriles 4. If there is loop i the depedecy grph the the lguge is ifiite 14
15 Exple: S A A C C cs Depedecy grph Ifiite lguge A C S 15
16 S A C A C cs S => A => => => i => i 16
17 17 cs C C A A S cs cs C A S i i S c S c cs S ) ( ) ( ) ( ) ( 2 2
18 There is o lgorith to deterie whether two cotext-free grrs geerte the se lguge. For the oet we do ot hve the techicl chiery for defiig the eig of there is o lgorith. 18
19 The Pupig Le for Cotext-Free Lguges 19
20 Tke ifiite cotext-free lguge Geertes ifiite uer of differet strigs Exple: S A A S 20
21 S A A S A derivtio: Vriles re repeted S A S A 21
22 Derivtio tree S strig A S A 22
23 Derivtio tree S strig A S A repeted 23
24 S A S A 24
25 Repeted Prt S A 25
26 Aother possile derivtio S A S A 26
27 S A S A S 27
28 S A S A S A S 28
29 S A S A S A S 29
30 S Therefore, the strig is lso geerted y the grr 30
31 We kow: S We lso kow this strig is geerted: S 31
32 We kow: S Therefore, this strig is lso geerted: S 32
33 We kow: S Therefore, this strig is lso geerted: S ( ) ( ) ( ) ( ) 2 2 ( ) ( )
34 We kow: S Therefore, this strig is lso geerted: S ( ) ( ) i i ( ) ( ) i i 34
35 Therefore, kowig tht is geerted y grr, we lso kow tht G ( ) i ( ) i is geerted y G 35
36 I geerl: We re give ifiite cotext-free grr G Assue G hs o uit-productios o -productios 36
37 Tke strig w L(G) with legth igger th > (Nuer of productios) x (Lrgest right side of productio) Cosequece: Soe vrile ust e repeted i the derivtio of w 37
38 Strig u, v, x, y, z w uvxyz : strigs of S terils u Lst repeted vrile A z v repeted A y x 38
39 S Possile derivtios: u z S uaz A vay v A A y A x x 39
40 We kow: S uaz AvAy A x This strig is lso geerted: S uaz * uxz uv 0 xy 0 z 40
41 We kow: S uaz AvAy A x This strig is lso geerted: S * uazuvayzuvxyz * The origil w uv 1 xy 1 z 41
42 We kow: S uaz AvAy A x This strig is lso geerted: S * * uazuvayzuvvayyzuvvxyyz * uv 2 xy 2 z 42
43 We kow: S uaz AvAy A x This strig is lso geerted: * S uaz uvayz * uvvayyz uvvvayyyzuvvvxyyyz uv * 3 xy 3 z * 43
44 We kow: S uaz AvAy This strig is lso geerted: S * * * uaz * uvvvayyyz uvayz * * uvvv vay yyyz * uvvv vxy yyyz uvvayyz * * A x uv i xy i z 44
45 Therefore, y strig of the for uv i xy i z i 0 is geerted y the grr G 45
46 Therefore, kowig tht uvxyz L(G) i i we lso kow tht uv xy z L(G) 46
47 S u A z v A y x Oservtio: Sice A vxy is the lst repeted vrile 47
48 S u A z v A y Oservtio: x vy 1 Sice there re o uit or productios 48
49 The Pupig Le: For ifiite cotext-free lguge L there exists iteger such tht for y strig we c write w L, w w uvxyz with legths vxy d vy 1 d it ust e: uv i xy i z L, for ll i 0 49
50 Applictios of The Pupig Le 50
51 No-cotext free lguges { c : 0} Cotext-free lguges { : 0} 51
52 Liz 6 th, sectio 8.1, exple 8.1, pge 216 { c 0 } 52
53 Theore: The lguge L { c : is ot cotext free 0} Proof: Use the Pupig Le for cotext-free lguges 53
54 L { c : 0} Assue for cotrdictio tht is cotext-free L Sice L is cotext-free d ifiite we c pply the pupig le 54
55 L { c : 0} Pupig Le gives gic uer such tht: Pick y strig w L with legth w We pick: w c 55
56 L { c : 0} w c We c write: w uvxyz with legths vxy vy 1 d 56
57 L { c : 0} c w uvxyz w vxy vy 1 Pupig Le sys: uv i xy i z L for ll i 0 57
58 L { c : 0} c w uvxyz w vxy vy 1 We exie ll the possile loctios of strig vxy i w 58
59 L { c : 0} c w uvxyz w vxy vy 1 Cse 1: vxy is withi ccc... ccc u vxy z 59
60 L { c : 0} c w uvxyz w vxy vy 1 Cse 1: v d y cosist fro oly ccc... ccc u vxy z 60
61 L { c : 0} c w uvxyz w vxy vy 1 Cse 1: Repetig v d y k 1 k ccc... ccc u 2 xy 2 v z 61
62 L { c : 0} c w uvxyz w vxy vy 1 Cse 1: Fro Pupig Le: uv 2 xy 2 z L k 1 k ccc... ccc u 2 xy 2 v z 62
63 L { c : 0} c w uvxyz w vxy vy 1 Cse 1: Fro Pupig Le: uv 2 xy 2 z L k 1 However: uv 2 xy 2 z k c L Cotrdictio!!! 63
64 L { c : 0} c w uvxyz w vxy vy 1 Cse 2: vxy is withi ccc... ccc u vxy z 64
65 L { c : 0} c w uvxyz w vxy vy 1 Cse 2: Siilr lysis with cse ccc... ccc u vxy z 65
66 L { c : 0} c w uvxyz w vxy vy 1 Cse 3: vxy is withi c ccc... ccc u vxy z 66
67 L { c : 0} c w uvxyz w vxy vy 1 Cse 3: Siilr lysis with cse ccc... ccc u vxy z 67
68 L { c : 0} c w uvxyz w vxy vy 1 Cse 4: vxy overlps d ccc... ccc u vxy z 68
69 L { c : 0} c w uvxyz w vxy vy 1 Cse 4: Possiility 1: v y cotis oly cotis oly ccc... ccc u vxy z 69
70 L { c : 0} c w uvxyz w vxy vy 1 Cse 4: Possiility 1: k1 k2 1 v y cotis oly cotis oly k 1 k ccc... ccc u 2 xy 2 v z 70
71 L { c : 0} c w uvxyz w vxy vy 1 Cse 4: Fro Pupig Le: uv 2 xy 2 z L k1 k2 1 k 1 k ccc... ccc u 2 xy 2 v z 71
72 L { c : 0} c w uvxyz w vxy vy 1 Cse 4: Fro Pupig Le: uv 2 xy 2 z L k1 k2 1 However: uv 2 xy 2 z k1 k2 c L Cotrdictio!!! 72
73 L { c : 0} c w uvxyz w vxy vy 1 Cse 4: Possiility 2: v cotis d y cotis oly ccc... ccc u vxy z 73
74 L { c : 0} c w uvxyz w vxy vy 1 Cse 4: k1 k2 k v y Possiility 2: cotis d cotis oly 1 k1 k2 k ccc... ccc u 2 xy 2 v z 74
75 L { c : 0} c w uvxyz w vxy vy 1 Cse 4: Fro Pupig Le: uv 2 xy 2 z L k1 k2 k 1 k1 k2 k ccc... ccc u 2 xy 2 v z 75
76 L { c : 0} c w uvxyz w vxy vy 1 Cse 4: Fro Pupig Le: uv 2 xy 2 z L However: k1 k2 k 1 uv 2 xy 2 z k 1 k 2 k c L Cotrdictio!!! 76
77 L { c : 0} c w uvxyz w vxy vy 1 Cse 4: Possiility 3: v y cotis oly cotis d ccc... ccc u vxy z 77
78 L { c : 0} c w uvxyz w vxy vy 1 Cse 4: Possiility 3: v y cotis oly cotis d Siilr lysis with Possiility 2 78
79 L { c : 0} c w uvxyz w vxy vy 1 Cse 5: vxy overlps d c ccc... ccc u vxy z 79
80 L { c : 0} c w uvxyz w vxy vy 1 Cse 5: Siilr lysis with cse ccc... ccc u vxy z 80
81 There re o other cses to cosider vxy (sice, strig cot vxy overlp, d c t the se tie) 81
82 I ll cses we otied cotrdictio Therefore: The origil ssuptio tht L { c : 0} is cotext-free ust e wrog Coclusio: L is ot cotext-free 82
Formal Languages The Pumping Lemma for CFLs
Forl Lguges The Pupig Le for CFLs Review: pupig le for regulr lguges Tke ifiite cotext-free lguge Geertes ifiite uer of differet strigs Exple: 3 I derivtio of log strig, vriles re repeted derivtio: 4 Derivtio
More informationCD5080 AUBER RECAPITULATION. Context-Free Languages. Models of Computation, Languages and Automata. S asb. S bsb
CD5080 AUBR RCAPIULAION Models of Coputtio, Lguges d Autot Leture 1 Cotext-ree Lguges, CL Pushdow Autot, PDA Pupig Le for CL eleted CL Proles Cotext-ree Lguges Mälrdle Uiversity 00 1 3 Cotext-ree Lguges
More informationReview of CFGs and Parsing I Context-free Languages and Grammars. Winter 2014 Costas Busch - RPI 1
Review of CFGs d Prsig I Cotext-free Lguges d Grmmrs Witer 2014 Costs Busch - RPI 1 Cotext-Free Lguges { b : { ww } 0} R Regulr Lguges *b* ( b) * Witer 2014 Costs Busch - RPI 2 Cotext-Free Lguges Cotext-Free
More informationNamely, for regular languages L1. Concatenation. Regular Languages. Star operation L 1. Complement. For regular language L the complement L is regular
CD5560 FABER For guges, Autot d Modes of Coputtio eture 5 Märde Uiversit 2003 Cotet More Properties of Regur guges (R) Stdrd Represettios of R Eeetr Questios out R No-Regur guges The Pigeohoe Priipe The
More informationThe limit comparison test
Roerto s Notes o Ifiite Series Chpter : Covergece tests Sectio 4 The limit compriso test Wht you eed to kow lredy: Bsics of series d direct compriso test. Wht you c ler here: Aother compriso test tht does
More informationPositive Properties of Context-Free languages
CS 30 - Leture 8 Properties of Cotext Free Graars Fall 008 Review Laguages ad Graars Alphabets, strigs, laguages Regular Laguages Deteriisti Fiite ad Nodeteriisti Autoata Equivalee of NFA ad DFA Regular
More informationContent. Languages, Alphabets and Strings. Operations on Strings. a ab abba baba. aaabbbaaba b 5. Languages. A language is a set of strings
CD5560 FABER Forml guges, Automt d Models of Computtio ecture Mälrdle Uiversity 006 Cotet guges, Alphets d Strigs Strigs & Strig Opertios guges & guge Opertios Regulr Expressios Fiite Automt, FA Determiistic
More informationCourse Material. CS Lecture 1 Deterministic Finite Automata. Grading and Policies. Workload. Website:http://www.cs.colostate.
Course Mteril CS 301 - Lecture 1 Determiistic Fiite Automt Fll 2008 Wesite:http://www.cs.colostte.edu/~cs301 Syllus, Outlie, Grdig Policies Homework d Slides Istructor: D Mssey Office hours: 2-3pm Tues
More informationMATH 118 HW 7 KELLY DOUGAN, ANDREW KOMAR, MARIA SIMBIRSKY, BRANDEN LASKE
MATH 118 HW 7 KELLY DOUGAN, ANDREW KOMAR, MARIA SIMBIRSKY, BRANDEN LASKE Prt 1. Let be odd rime d let Z such tht gcd(, 1. Show tht if is qudrtic residue mod, the is qudrtic residue mod for y ositive iteger.
More informationSection IV.6: The Master Method and Applications
Sectio IV.6: The Mster Method d Applictios Defiitio IV.6.1: A fuctio f is symptoticlly positive if d oly if there exists rel umer such tht f(x) > for ll x >. A cosequece of this defiitio is tht fuctio
More informationGeneral properties of definite integrals
Roerto s Notes o Itegrl Clculus Chpter 4: Defiite itegrls d the FTC Sectio Geerl properties of defiite itegrls Wht you eed to kow lredy: Wht defiite Riem itegrl is. Wht you c ler here: Some key properties
More informationALGEBRA II CHAPTER 7 NOTES. Name
ALGEBRA II CHAPTER 7 NOTES Ne Algebr II 7. th Roots d Rtiol Expoets Tody I evlutig th roots of rel ubers usig both rdicl d rtiol expoet ottio. I successful tody whe I c evlute th roots. It is iportt for
More informationM098 Carson Elementary and Intermediate Algebra 3e Section 10.2
M09 Crso Eleetry d Iteredite Alger e Sectio 0. Ojectives. Evlute rtiol epoets.. Write rdicls s epressios rised to rtiol epoets.. Siplify epressios with rtiol uer epoets usig the rules of epoets.. Use rtiol
More informationChapter 7 Infinite Series
MA Ifiite Series Asst.Prof.Dr.Supree Liswdi Chpter 7 Ifiite Series Sectio 7. Sequece A sequece c be thought of s list of umbers writte i defiite order:,,...,,... 2 The umber is clled the first term, 2
More informationDiscrete Mathematics I Tutorial 12
Discrete Mthemtics I Tutoril Refer to Chpter 4., 4., 4.4. For ech of these sequeces fid recurrece reltio stisfied by this sequece. (The swers re ot uique becuse there re ifiitely my differet recurrece
More information10.5 Power Series. In this section, we are going to start talking about power series. A power series is a series of the form
0.5 Power Series I the lst three sectios, we ve spet most of tht time tlkig bout how to determie if series is coverget or ot. Now it is time to strt lookig t some specific kids of series d we will evetully
More informationWeek 13 Notes: 1) Riemann Sum. Aim: Compute Area Under a Graph. Suppose we want to find out the area of a graph, like the one on the right:
Week 1 Notes: 1) Riem Sum Aim: Compute Are Uder Grph Suppose we wt to fid out the re of grph, like the oe o the right: We wt to kow the re of the red re. Here re some wys to pproximte the re: We cut the
More information, we would have a series, designated as + j 1
Clculus sectio 9. Ifiite Series otes by Ti Pilchowski A sequece { } cosists of ordered set of ubers. If we were to begi ddig the ubers of sequece together s we would hve series desigted s. Ech iteredite
More informationSequence and Series of Functions
6 Sequece d Series of Fuctios 6. Sequece of Fuctios 6.. Poitwise Covergece d Uiform Covergece Let J be itervl i R. Defiitio 6. For ech N, suppose fuctio f : J R is give. The we sy tht sequece (f ) of fuctios
More informationLAWS OF INDICES M.K. HOME TUITION. Mathematics Revision Guides Level: GCSE Higher Tier
Mthetics Revisio Guides Lws of Idices Pge of 7 Author: Mrk Kudlowski M.K. HOME TUITION Mthetics Revisio Guides Level: GCSE Higher Tier LAWS OF INDICES Versio:. Dte: 0--0 Mthetics Revisio Guides Lws of
More informationSummer MA Lesson 4 Section P.3. such that =, denoted by =, is the principal square root
Suer MA 00 Lesso Sectio P. I Squre Roots If b, the b is squre root of. If is oegtive rel uber, the oegtive uber b b b such tht, deoted by, is the pricipl squre root of. rdicl sig rdicl expressio rdicd
More information0 otherwise. sin( nx)sin( kx) 0 otherwise. cos( nx) sin( kx) dx 0 for all integers n, k.
. Computtio of Fourier Series I this sectio, we compute the Fourier coefficiets, f ( x) cos( x) b si( x) d b, i the Fourier series To do this, we eed the followig result o the orthogolity of the trigoometric
More informationFourier Series and Applications
9/7/9 Fourier Series d Applictios Fuctios epsio is doe to uderstd the better i powers o etc. My iportt probles ivolvig prtil dieretil equtios c be solved provided give uctio c be epressed s iiite su o
More informationIn an algebraic expression of the form (1), like terms are terms with the same power of the variables (in this case
Chpter : Algebr: A. Bckgroud lgebr: A. Like ters: I lgebric expressio of the for: () x b y c z x y o z d x... p x.. we cosider x, y, z to be vribles d, b, c, d,,, o,.. to be costts. I lgebric expressio
More informationMTH 146 Class 16 Notes
MTH 46 Clss 6 Notes 0.4- Cotiued Motivtio: We ow cosider the rc legth of polr curve. Suppose we wish to fid the legth of polr curve curve i terms of prmetric equtios s: r f where b. We c view the cos si
More informationFirst Midterm Examination
24-25 Fll Semester First Midterm Exmintion ) Give the stte digrm of DFA tht recognizes the lnguge A over lphet Σ = {, } where A = {w w contins or } 2) The following DFA recognizes the lnguge B over lphet
More information( a n ) converges or diverges.
Chpter Ifiite Series Pge of Sectio E Rtio Test Chpter : Ifiite Series By the ed of this sectio you will be ble to uderstd the proof of the rtio test test series for covergece by pplyig the rtio test pprecite
More informationdegree non-homogeneous Diophantine equation in six unknowns represented by x y 2z
Scholrs Jourl of Egieerig d Techology (SJET Sch. J. Eg. Tech., ; (A:97- Scholrs Acdeic d Scietific Pulisher (A Itertiol Pulisher for Acdeic d Scietific Resources www.sspulisher.co ISSN -X (Olie ISSN 7-9
More informationPresentation for use with the textbook, Algorithm Design and Applications, by M. T. Goodrich and R. Tamassia, Wiley, Divide-and-Conquer
Presettio for use with the textook, Algorithm Desig d Applictios, y M. T. Goodrich d R. Tmssi, Wiley, 25 Divide-d-Coquer Divide-d-Coquer Divide-d coquer is geerl lgorithm desig prdigm: Divide: divide the
More informationMATH 104 FINAL SOLUTIONS. 1. (2 points each) Mark each of the following as True or False. No justification is required. y n = x 1 + x x n n
MATH 04 FINAL SOLUTIONS. ( poits ech) Mrk ech of the followig s True or Flse. No justifictio is required. ) A ubouded sequece c hve o Cuchy subsequece. Flse b) A ifiite uio of Dedekid cuts is Dedekid cut.
More informationf(bx) dx = f dx = dx l dx f(0) log b x a + l log b a 2ɛ log b a.
Eercise 5 For y < A < B, we hve B A f fb B d = = A B A f d f d For y ɛ >, there re N > δ >, such tht d The for y < A < δ d B > N, we hve ba f d f A bb f d l By ba A A B A bb ba fb d f d = ba < m{, b}δ
More informationALGEBRA. Set of Equations. have no solution 1 b1. Dependent system has infinitely many solutions
Qudrtic Equtios ALGEBRA Remider theorem: If f() is divided b( ), the remider is f(). Fctor theorem: If ( ) is fctor of f(), the f() = 0. Ivolutio d Evlutio ( + b) = + b + b ( b) = + b b ( + b) 3 = 3 +
More informationELEG 3143 Probability & Stochastic Process Ch. 5 Elements of Statistics
Deprtet of Electricl Egieerig Uiversity of Arkss ELEG 3143 Probbility & Stochstic Process Ch. 5 Eleets of Sttistics Dr. Jigxi Wu wuj@urk.edu OUTLINE Itroductio: wht is sttistics? Sple e d sple vrice Cofidece
More informationLecture 38 (Trapped Particles) Physics Spring 2018 Douglas Fields
Lecture 38 (Trpped Prticles) Physics 6-01 Sprig 018 Dougls Fields Free Prticle Solutio Schrödiger s Wve Equtio i 1D If motio is restricted to oe-dimesio, the del opertor just becomes the prtil derivtive
More informationThe total number of permutations of S is n!. We denote the set of all permutations of S by
DETERMINNTS. DEFINITIONS Def: Let S {,,, } e the set of itegers from to, rrged i scedig order. rerrgemet jjj j of the elemets of S is clled permuttio of S. S. The totl umer of permuttios of S is!. We deote
More informationApproximate Integration
Study Sheet (7.7) Approimte Itegrtio I this sectio, we will ler: How to fid pproimte vlues of defiite itegrls. There re two situtios i which it is impossile to fid the ect vlue of defiite itegrl. Situtio:
More informationSummer Math Requirement Algebra II Review For students entering Pre- Calculus Theory or Pre- Calculus Honors
Suer Mth Requireet Algebr II Review For studets eterig Pre- Clculus Theory or Pre- Clculus Hoors The purpose of this pcket is to esure tht studets re prepred for the quick pce of Pre- Clculus. The Topics
More informationSUTCLIFFE S NOTES: CALCULUS 2 SWOKOWSKI S CHAPTER 11
SUTCLIFFE S NOTES: CALCULUS SWOKOWSKI S CHAPTER Ifiite Series.5 Altertig Series d Absolute Covergece Next, let us cosider series with both positive d egtive terms. The simplest d most useful is ltertig
More informationFirst Midterm Examination
Çnky University Deprtment of Computer Engineering 203-204 Fll Semester First Midterm Exmintion ) Design DFA for ll strings over the lphet Σ = {,, c} in which there is no, no nd no cc. 2) Wht lnguge does
More informationTopic 4 Fourier Series. Today
Topic 4 Fourier Series Toy Wves with repetig uctios Sigl geertor Clssicl guitr Pio Ech istrumet is plyig sigle ote mile C 6Hz) st hrmoic hrmoic 3 r hrmoic 4 th hrmoic 6Hz 5Hz 783Hz 44Hz A sigle ote will
More informationRULES FOR MANIPULATING SURDS b. This is the addition law of surds with the same radicals. (ii)
SURDS Defiitio : Ay umer which c e expressed s quotiet m of two itegers ( 0 ), is clled rtiol umer. Ay rel umer which is ot rtiol is clled irrtiol. Irrtiol umers which re i the form of roots re clled surds.
More informationThe Reimann Integral is a formal limit definition of a definite integral
MATH 136 The Reim Itegrl The Reim Itegrl is forml limit defiitio of defiite itegrl cotiuous fuctio f. The costructio is s follows: f ( x) dx for Reim Itegrl: Prtitio [, ] ito suitervls ech hvig the equl
More informationThe Elementary Arithmetic Operators of Continued Fraction
Americ-Eursi Jourl of Scietific Reserch 0 (5: 5-63, 05 ISSN 88-6785 IDOSI Pulictios, 05 DOI: 0.589/idosi.ejsr.05.0.5.697 The Elemetry Arithmetic Opertors of Cotiued Frctio S. Mugssi d F. Mistiri Deprtmet
More information n. A Very Interesting Example + + = d. + x3. + 5x4. math 131 power series, part ii 7. One of the first power series we examined was. 2!
mth power series, prt ii 7 A Very Iterestig Emple Oe of the first power series we emied ws! + +! + + +!! + I Emple 58 we used the rtio test to show tht the itervl of covergece ws (, ) Sice the series coverges
More informationModule 9: String Matching
Module 9: Strig Mtchig CS 240 Dt Structures d Dt Mgemet T. Biedl K. Lctot M. Sepehri S. Wild Bsed o lecture otes y my previous cs240 istructors Dvid R. Cherito School of Computer Sciece, Uiversity of Wterloo
More informationUSA Mathematical Talent Search PROBLEMS / SOLUTIONS / COMMENTS Round 4 - Year 13 - Academic Year solutions edited by Erin Schram
USA Mthemticl Tlet Serch PROBLEMS / SOLUTIONS / COMMENTS Roud - Yer 13 - Acdemic Yer 2001-2002 solutios edited y Eri Schrm 1//13. I strge lguge there re oly two letters, d, d it is postulted tht the letter
More informationUNIVERSITY OF BRISTOL. Examination for the Degrees of B.Sc. and M.Sci. (Level C/4) ANALYSIS 1B, SOLUTIONS MATH (Paper Code MATH-10006)
UNIVERSITY OF BRISTOL Exmitio for the Degrees of B.Sc. d M.Sci. (Level C/4) ANALYSIS B, SOLUTIONS MATH 6 (Pper Code MATH-6) My/Jue 25, hours 3 miutes This pper cotis two sectios, A d B. Plese use seprte
More informationAdvanced Algorithmic Problem Solving Le 6 Math and Search
Advced Algorithmic Prolem Solvig Le Mth d Serch Fredrik Heitz Dept of Computer d Iformtio Sciece Liköpig Uiversity Outlie Arithmetic (l. d.) Solvig lier equtio systems (l. d.) Chiese remider theorem (l.5
More informationNumbers (Part I) -- Solutions
Ley College -- For AMATYC SML Mth Competitio Cochig Sessios v.., [/7/00] sme s /6/009 versio, with presettio improvemets Numbers Prt I) -- Solutios. The equtio b c 008 hs solutio i which, b, c re distict
More informationFig. 1. I a. V ag I c. I n. V cg. Z n Z Y. I b. V bg
ymmetricl Compoets equece impedces Although the followig focuses o lods, the results pply eqully well to lies, or lies d lods. Red these otes together with sectios.6 d.9 of text. Cosider the -coected lced
More informationQuiz No. 1. ln n n. 1. Define: an infinite sequence A function whose domain is N 2. Define: a convergent sequence A sequence that has a limit
Quiz No.. Defie: a ifiite sequece A fuctio whose domai is N 2. Defie: a coverget sequece A sequece that has a limit 3. Is this sequece coverget? Why or why ot? l Yes, it is coverget sice L=0 by LHR. INFINITE
More informationTaylor Polynomials. The Tangent Line. (a, f (a)) and has the same slope as the curve y = f (x) at that point. It is the best
Tylor Polyomils Let f () = e d let p() = 1 + + 1 + 1 6 3 Without usig clcultor, evlute f (1) d p(1) Ok, I m still witig With little effort it is possible to evlute p(1) = 1 + 1 + 1 (144) + 6 1 (178) =
More informationTheorem 5.3 (Continued) The Fundamental Theorem of Calculus, Part 2: ab,, then. f x dx F x F b F a. a a. f x dx= f x x
Chpter 6 Applictios Itegrtio Sectio 6. Regio Betwee Curves Recll: Theorem 5.3 (Cotiued) The Fudmetl Theorem of Clculus, Prt :,,, the If f is cotiuous t ever poit of [ ] d F is tiderivtive of f o [ ] (
More informationRepeated Root and Common Root
Repeted Root d Commo Root 1 (Method 1) Let α, β, γ e the roots of p(x) x + x + 0 (1) The α + β + γ 0, αβ + βγ + γα, αβγ - () (α - β) (α + β) - αβ (α + β) [ (βγ + γα)] + [(α + β) + γ (α + β)] +γ (α + β)
More informationIntermediate Arithmetic
Git Lerig Guides Iteredite Arithetic Nuer Syste, Surds d Idices Author: Rghu M.D. NUMBER SYSTEM Nuer syste: Nuer systes re clssified s Nturl, Whole, Itegers, Rtiol d Irrtiol uers. The syste hs ee digrticlly
More informationSimilar idea to multiplication in N, C. Divide and conquer approach provides unexpected improvements. Naïve matrix multiplication
Next. Covered bsics of simple desig techique (Divided-coquer) Ch. of the text.. Next, Strsse s lgorithm. Lter: more desig d coquer lgorithms: MergeSort. Solvig recurreces d the Mster Theorem. Similr ide
More informationSummary: Binomial Expansion...! r. where
Summy: Biomil Epsio 009 M Teo www.techmejcmth-sg.wes.com ) Re-cp of Additiol Mthemtics Biomil Theoem... whee )!!(! () The fomul is ville i MF so studets do ot eed to memoise it. () The fomul pplies oly
More informationGRAPHING LINEAR EQUATIONS. Linear Equations. x l ( 3,1 ) _x-axis. Origin ( 0, 0 ) Slope = change in y change in x. Equation for l 1.
GRAPHING LINEAR EQUATIONS Qudrt II Qudrt I ORDERED PAIR: The first umer i the ordered pir is the -coordite d the secod umer i the ordered pir is the y-coordite. (, ) Origi ( 0, 0 ) _-is Lier Equtios Qudrt
More informationGraphing Review Part 3: Polynomials
Grphig Review Prt : Polomils Prbols Recll, tht the grph of f ( ) is prbol. It is eve fuctio, hece it is smmetric bout the bout the -is. This mes tht f ( ) f ( ). Its grph is show below. The poit ( 0,0)
More informationSUTCLIFFE S NOTES: CALCULUS 2 SWOKOWSKI S CHAPTER 11
UTCLIFFE NOTE: CALCULU WOKOWKI CHAPTER Ifiite eries Coverget or Diverget eries Cosider the sequece If we form the ifiite sum 0, 00, 000, 0 00 000, we hve wht is clled ifiite series We wt to fid the sum
More informationAPPLICATION OF DIFFERENCE EQUATIONS TO CERTAIN TRIDIAGONAL MATRICES
Scietific Reserch of the Istitute of Mthetics d Coputer Sciece 3() 0, 5-0 APPLICATION OF DIFFERENCE EQUATIONS TO CERTAIN TRIDIAGONAL MATRICES Jolt Borows, Le Łcińs, Jowit Rychlews Istitute of Mthetics,
More informationIf a is any non zero real or imaginary number and m is the positive integer, then a...
Idices d Surds.. Defiitio of Idices. If is o ero re or igir uer d is the positive iteger the...... ties. Here is ced the se d the ide power or epoet... Lws of Idices. 0 0 0. where d re rtio uers where
More informationUnit 1. Extending the Number System. 2 Jordan School District
Uit Etedig the Number System Jord School District Uit Cluster (N.RN. & N.RN.): Etedig Properties of Epoets Cluster : Etedig properties of epoets.. Defie rtiol epoets d eted the properties of iteger epoets
More informationMAT136H1F - Calculus I (B) Long Quiz 1. T0101 (M3) Time: 20 minutes. The quiz consists of four questions. Each question is worth 2 points. Good Luck!
MAT36HF - Calculus I (B) Log Quiz. T (M3) Time: 2 miutes Last Name: Studet ID: First Name: Please mark your tutorial sectio: T (M3) T2 (R4) T3 (T4) T5 (T5) T52 (R5) The quiz cosists of four questios. Each
More informationg as the function in which for every element x be the set of polynomials of a degree less than or equal to n with , for each i from 0 to n )
Vector Spce Observtio: I this book, we will distiguish vectors d sclr eleets. For sclr eleets we will use Greek letters,,,, etc. d for the vectors i V we will use letters x,y,u,v,w,s,, etc., fro our ow
More informationLinford 1. Kyle Linford. Math 211. Honors Project. Theorems to Analyze: Theorem 2.4 The Limit of a Function Involving a Radical (A4)
Liford 1 Kyle Liford Mth 211 Hoors Project Theorems to Alyze: Theorem 2.4 The Limit of Fuctio Ivolvig Rdicl (A4) Theorem 2.8 The Squeeze Theorem (A5) Theorem 2.9 The Limit of Si(x)/x = 1 (p. 85) Theorem
More informationMathematical Induction (selected questions)
Mtheticl Iductio (selected questios). () Let P() e the propositio : For P(), L.H.S. R.H.S., P() is true. Assue P() is true for soe turl uer, tht is, () For P( ),, y () By the Priciple of Mtheticl Iductio,
More informationSection 6.3: Geometric Sequences
40 Chpter 6 Sectio 6.: Geometric Sequeces My jobs offer ul cost-of-livig icrese to keep slries cosistet with ifltio. Suppose, for exmple, recet college grdute fids positio s sles mger erig ul slry of $6,000.
More informationApproximations of Definite Integrals
Approximtios of Defiite Itegrls So fr we hve relied o tiderivtives to evlute res uder curves, work doe by vrible force, volumes of revolutio, etc. More precisely, wheever we hve hd to evlute defiite itegrl
More informationA general theory of minimal increments for Hirsch-type indices and applications to the mathematical characterization of Kosmulski-indices
Mlysi Jourl of Librry & Iformtio Sciece, Vol. 9, o. 3, 04: 4-49 A geerl theory of miiml icremets for Hirsch-type idices d pplictios to the mthemticl chrcteriztio of Kosmulski-idices L. Egghe Uiversiteit
More informationReview of the Riemann Integral
Chpter 1 Review of the Riem Itegrl This chpter provides quick review of the bsic properties of the Riem itegrl. 1.0 Itegrls d Riem Sums Defiitio 1.0.1. Let [, b] be fiite, closed itervl. A prtitio P of
More informationNational Quali cations AHEXEMPLAR PAPER ONLY
Ntiol Quli ctios AHEXEMPLAR PAPER ONLY EP/AH/0 Mthemtics Dte Not pplicble Durtio hours Totl mrks 00 Attempt ALL questios. You my use clcultor. Full credit will be give oly to solutios which coti pproprite
More informationBasic Limit Theorems
Bsic Limit Theorems The very "cle" proof of L9 usig L8 ws provided to me by Joh Gci d it ws this result which ispired me to write up these otes. Absolute Vlue Properties: For rel umbers x, d y x x if x
More informationFACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING. Lectures
FACULTY OF MATHEMATICAL STUDIES MATHEMATICS FOR PART I ENGINEERING Lectures MODULE 0 FURTHER CALCULUS II. Sequeces d series. Rolle s theorem d me vlue theorems 3. Tlor s d Mcluri s theorems 4. L Hopitl
More informationEXERCISE a a a 5. + a 15 NEETIIT.COM
- Dowlod our droid App. Sigle choice Type Questios EXERCISE -. The first term of A.P. of cosecutive iteger is p +. The sum of (p + ) terms of this series c be expressed s () (p + ) () (p + ) (p + ) ()
More informationConvergence rates of approximate sums of Riemann integrals
Covergece rtes of pproximte sums of Riem itegrls Hiroyuki Tski Grdute School of Pure d Applied Sciece, Uiversity of Tsuku Tsuku Irki 5-857 Jp tski@mth.tsuku.c.jp Keywords : covergece rte; Riem sum; Riem
More information1. (25 points) Use the limit definition of the definite integral and the sum formulas to compute. [1 x + x2
Mth 3, Clculus II Fil Exm Solutios. (5 poits) Use the limit defiitio of the defiite itegrl d the sum formuls to compute 3 x + x. Check your swer by usig the Fudmetl Theorem of Clculus. Solutio: The limit
More informationHomework 3 Solutions
CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.
More informationRiemann Integral and Bounded function. Ng Tze Beng
Riem Itegrl d Bouded fuctio. Ng Tze Beg I geerlistio of re uder grph of fuctio, it is ormlly ssumed tht the fuctio uder cosidertio e ouded. For ouded fuctio, the rge of the fuctio is ouded d hece y suset
More informationMath 153: Lecture Notes For Chapter 1
Mth : Lecture Notes For Chpter Sectio.: Rel Nubers Additio d subtrctios : Se Sigs: Add Eples: = - - = - Diff. Sigs: Subtrct d put the sig of the uber with lrger bsolute vlue Eples: - = - = - Multiplictio
More information1.3 Continuous Functions and Riemann Sums
mth riem sums, prt 0 Cotiuous Fuctios d Riem Sums I Exmple we sw tht lim Lower() = lim Upper() for the fuctio!! f (x) = + x o [0, ] This is o ccidet It is exmple of the followig theorem THEOREM Let f be
More informationFOURIER SERIES PART I: DEFINITIONS AND EXAMPLES. To a 2π-periodic function f(x) we will associate a trigonometric series. a n cos(nx) + b n sin(nx),
FOURIER SERIES PART I: DEFINITIONS AND EXAMPLES To -periodic fuctio f() we will ssocite trigoometric series + cos() + b si(), or i terms of the epoetil e i, series of the form c e i. Z For most of the
More informationFourier Series. Topic 4 Fourier Series. sin. sin. Fourier Series. Fourier Series. Fourier Series. sin. b n. a n. sin
Topic Fourier Series si Fourier Series Music is more th just pitch mplitue it s lso out timre. The richess o sou or ote prouce y musicl istrumet is escrie i terms o sum o umer o istict requecies clle hrmoics.
More informationApplied Databases. Sebastian Maneth. Lecture 16 Suffix Array, Burrows-Wheeler Transform. University of Edinburgh - March 10th, 2016
Applied Dtbses Lecture 16 Suffix Arry, Burrows-Wheeler Trsform Sebsti Meth Uiversity of Ediburgh - Mrch 10th, 2016 2 Outlie 1. Suffix Arry 2. Burrows-Wheeler Trsform 3 Olie Strig-Mtchig how c we do Horspool
More informationPOWER SERIES R. E. SHOWALTER
POWER SERIES R. E. SHOWALTER. sequeces We deote by lim = tht the limit of the sequece { } is the umber. By this we me tht for y ε > 0 there is iteger N such tht < ε for ll itegers N. This mkes precise
More information10. 3 The Integral and Comparison Test, Estimating Sums
0. The Itegrl d Comriso Test, Estimtig Sums I geerl, it is hrd to fid the ect sum of series. We were le to ccomlish this for geometric series d for telescoig series sice i ech of those cses we could fid
More informationis infinite. The converse is proved similarly, and the last statement of the theorem is clear too.
12. No-stdrd lysis October 2, 2011 I this sectio we give brief itroductio to o-stdrd lysis. This is firly well-developed field of mthemtics bsed o model theory. It dels ot just with the rels, fuctios o
More informationINTEGRATION TECHNIQUES (TRIG, LOG, EXP FUNCTIONS)
Mthemtics Revisio Guides Itegrtig Trig, Log d Ep Fuctios Pge of MK HOME TUITION Mthemtics Revisio Guides Level: AS / A Level AQA : C Edecel: C OCR: C OCR MEI: C INTEGRATION TECHNIQUES (TRIG, LOG, EXP FUNCTIONS)
More informationProject 3: Using Identities to Rewrite Expressions
MAT 5 Projet 3: Usig Idetities to Rewrite Expressios Wldis I lger, equtios tht desrie properties or ptters re ofte lled idetities. Idetities desrie expressio e repled with equl or equivlet expressio tht
More informationChapter 2 Infinite Series Page 1 of 9
Chpter Ifiite eries Pge of 9 Chpter : Ifiite eries ectio A Itroductio to Ifiite eries By the ed of this sectio you will be ble to uderstd wht is met by covergece d divergece of ifiite series recogise geometric
More informationINFINITE SERIES. ,... having infinite number of terms is called infinite sequence and its indicated sum, i.e., a 1
Appedix A.. Itroductio As discussed i the Chpter 9 o Sequeces d Series, sequece,,...,,... hvig ifiite umber of terms is clled ifiite sequece d its idicted sum, i.e., + + +... + +... is clled ifite series
More informationInfinite Sequences and Series. Sequences. Sequences { } { } A sequence is a list of number in a definite order: a 1, a 2, a 3,, a n, or {a n } or
Mth 0 Clculus II Ifiite Sequeces d Series -- Chpter Ifiite Sequeces d Series Mth 0 Clculus II Ifiite Sequeces d Series: Sequeces -- Chpter. Sequeces Mth 0 Clculus II Ifiite Sequeces d Series: Sequeces
More informationCOMP 2804 Solutions Assignment 1
COMP 2804 Solutios Assiget 1 Questio 1: O the first page of your assiget, write your ae ad studet uber Solutio: Nae: Jaes Bod Studet uber: 007 Questio 2: I Tic-Tac-Toe, we are give a 3 3 grid, cosistig
More informationChapter Real Numbers
Chpter. - Rel Numbers Itegers: coutig umbers, zero, d the egtive of the coutig umbers. ex: {,-3, -, -,,,, 3, } Rtiol Numbers: quotiets of two itegers with ozero deomitor; termitig or repetig decimls. ex:
More informationSM2H. Unit 2 Polynomials, Exponents, Radicals & Complex Numbers Notes. 3.1 Number Theory
SMH Uit Polyomils, Epoets, Rdicls & Comple Numbers Notes.1 Number Theory .1 Addig, Subtrctig, d Multiplyig Polyomils Notes Moomil: A epressio tht is umber, vrible, or umbers d vribles multiplied together.
More informationInfinite Series Sequences: terms nth term Listing Terms of a Sequence 2 n recursively defined n+1 Pattern Recognition for Sequences Ex:
Ifiite Series Sequeces: A sequece i defied s fuctio whose domi is the set of positive itegers. Usully it s esier to deote sequece i subscript form rther th fuctio ottio.,, 3, re the terms of the sequece
More informationLEVEL I. ,... if it is known that a 1
LEVEL I Fid the sum of first terms of the AP, if it is kow tht + 5 + 0 + 5 + 0 + = 5 The iterior gles of polygo re i rithmetic progressio The smllest gle is 0 d the commo differece is 5 Fid the umber of
More informationUNIT 4 EXTENDING THE NUMBER SYSTEM Lesson 1: Working with the Number System Instruction
Lesso : Workig with the Nuber Syste Istructio Prerequisite Skills This lesso requires the use of the followig skills: evlutig expressios usig the order of opertios evlutig expoetil expressios ivolvig iteger
More informationSect Simplifying Radical Expressions. We can use our properties of exponents to establish two properties of radicals: and
128 Sect 10.3 - Simplifyig Rdicl Expressios Cocept #1 Multiplictio d Divisio Properties of Rdicls We c use our properties of expoets to estlish two properties of rdicls: () 1/ 1/ 1/ & ( Multiplictio d
More informationMATH 104: INTRODUCTORY ANALYSIS SPRING 2009/10 PROBLEM SET 8 SOLUTIONS. and x i = a + i. i + n(n + 1)(2n + 1) + 2a. (b a)3 6n 2
MATH 104: INTRODUCTORY ANALYSIS SPRING 2009/10 PROBLEM SET 8 SOLUTIONS 6.9: Let f(x) { x 2 if x Q [, b], 0 if x (R \ Q) [, b], where > 0. Prove tht b. Solutio. Let P { x 0 < x 1 < < x b} be regulr prtitio
More information