4 Linear Homogeneous Recurrence Relation 4-1 Fibonacci Rabbits. 组合数学 Combinatorics

Transcription

1 4 Ler Homogeeous Recurrece Relto 4- bocc Rbbts 组合数学 ombtorcs

2 The delt of the th moth d - th moth s gve brth by the rbbts - moth. o = Moth Moth Moth Moth 4 I the frst moth there s pr of ewly-bor rbbts; If pr of rbbts could gve brth to ew pr every moth oe mle, oe femle; New rbbts could strt gvg brth sce the thrd moth; The rbbts ever de; How my rbbts would there be the th moth? Moth

3 bocc umber 8 4 OEI: 4 Recurrece Relto: =-+- Itl vlues: =, = I, Id mthemtcs reserched the umber of rrgemets to pcge tems wth legth d wdth to boes. d they descrbed ths sequece for the frst tme. I the wester world, bocc metoed problem bout the reproducto of rbbts Lber bbc. bocc,leordo 7- Member of the Bocc fmly. Trvelled to s d frc t wth hs fther d lered to clculte wth Id dgts; Plyed mportt role the recovery of Wester Mthemtcs. d coected Wester d Oretl mthemtcs. G.rdo: We could ssume tht ll mthemtcs we ow ecept the cet Gree oes re gotte by bocc. Leordo of Ps bocc,bocc s so Bocc: good, turl, smple

4 bocc umber 8 4

5 bocc Numbers The bocc Qurterly fouded 96 especlly publsh the ewest reserches o ths sequece. Whch cludes: The lst dgt loops every 6 umbers; the lst dgts loops every umbers; the lst dgts loops every umbers; the lst 4 dgts loops every umbers; the lst dgts loops every umbers. Every rd umber could be dvded by. Every 4 th umber could be dvded by. Every th umber could be dvded by. Every 6 th umber could be dvded by 8, etc. These dvsors c lso costruct bocc equece.

6 bocc prme equece 478 OEI I the bocc equece, there re prmes:,,,, 89,, 97, 867, 49, , 977, , Ecept = 4, the dees of ll bocc Prmes re prmes. However, ot ll prme de bocc Numbers re prmes. ojecture: re there fte prmes mog bocc Numbers? The lrgest ow prme s the 889 th bocc Number, whch hs 7 dgts.

7 re of the rectgle = um of multple qudrtes Proof wthout words vs Logc deducto

8 Proof: 4 4 =, =, = = Recurrece Relto Prove the detty: = +

9 Recurrece Relto =, =, = = Proof: = = +

10 Recurrece Relto =, =, = = Proof: Detled Epressos?

11 4 Ler Homogeeous Recurrece Relto 4- Epressos of bocc Numbers 组合数学 ombtorcs

12 Mgc There s 8cm 8cm qudrte tblecloth. How to covert t to.m cm oe?,,,,,, 8,,,. * -+ = - =,, Lrger tblecloths? =? 8 8 Drect epressos?

13 G : : 4 4 G G G G ssume bocc Recurrece B G =, =, = - + -

14 B B { { B B, B [ ] G, ] [ 68. bocc Recurrece

15 bocc equece =

16 pplctos Optmzto Methods ssume tht fucto f reches ts mmum t = ξ.,desg optmzto lgorthm to fd the etreme pot to cert etet wth fte tertos. The smplest wy s to trsect the tervl,b. b, b 如下图 : y y f f f b 6

17 .4 pplctos Optmzto Method Dscuss ccordg to the szes of f, f Whe f > f, the mmum ξ must be,, tervl, b could be removed. y y f y f f b f f 7

18 .4 pplctos Optmzto Method Whe f < f, the mmum ξ must be, b, the rge, could be removed. y y f y f f f f b b 8

19 .4 pplcto Optmzto Method Whe f = f, the mmum ξ must be,, so both, d, b could be removed. y y f f f y b f f 9

20 y y f f f o wth tests, t lest we could reduce the rge to / of the org dom. or emple f > f d we fd the mmum,. If cotue usg threefold dvso method, t s fct tht s ot used. o we mge to hve symmetrcl pot, l, to test.

21 If eep,, the we eep testg t pots,,. If The the test t could be used g. o we sve oe test. We hve: ,.68.6,.8.46,.6

22 pplcto Optmzto Method Ths s the.68 optmzto method. Whe fdg umodl mmums,, we could test t:.68,,68.8 pots. or emple eep,.68, s.68.8, we oly eed to test oce t

23 pplctos Optmzto Method We could use bocc equece Optmzto Method. Its dfferece from.68 method s to decde the umber of tests before testg. We troduce stutos. The possble testg umber s some

24 t ths pot testg pots re dvso pots - d -. If - s better, remove the prt smller th - ; The remed prt cots - - = - dvso pots, I whch test pts - d - re correspodet to former de = - d = -. Just rght, - hs bee tested the prevous roud = = -

25 If - s better, remove the prt lrger th -. I the remed prt there re - dvso pots, the et step mog test pots - d -, - hs bee tested. o mog the possble tests, we could fd the etreme wth t most - tests. Oe dfferece betwee the bocc Method d.68 method s tht bocc Method could be used whe the prmeters re ll tegers. If the umber of possble tests re smller th but lrger th, we could dd severl mge pots to me t pots. We could ssume these mge pots to be worse th y other pots wthout ctully compre them.

26 Ellott wve complete loop cludes 8 wves crese, decrese complete perod cludes 8 wves, whch re cresg, re decresg. They re ll bocc Numbers. I detls we could get 4 wves d 44 wves, they re lso bocc Numbers. ommo retrcemet rto re.8. d.68. It mly reflects the psychology of vestors. -

27 y y - - If lwys vest uformly from to, d the t decreses by The sustetto =? bocc retrcemet Ellott Wve crese prt, - decrese prt. y Proft= Loss Proft:y / Loss: +y/ +y= =.8

28 bocc retrcemet

29

30 4 Ler Homogeeous Recurrece Relto 4- Ler Homogeeous Recurrece Relto 组合数学 ombtorcs

31 = h h-+h-= ummry Ler summto RH = oeffcets re costts Def If sequece { } stsfes: d, d,, d,,,, d d, d,, d re costts,, so ths epresso s clled th -order ler homogeeous recurrece relto of { }., h h, h = = = =

32 bocc Recurrece = =, = ssume G G G ctorg?... B - -

33 ctor Theorem If s root of ler polyoml f, whch mes f =, the polyoml f hs fctor. We eed fctor, If s root of ler polyoml f, whch mes f =, the polyoml f hs fctor - -=-/ = = m -m- Let m= - m = m -m-=m-m- ubsttute m= -, get = - + = = ,

34 Ler Homogeeous Recurrece Relto The recurrece epresso of bocc equece =, =, = - + G Deomtor becomes = = m -m- Let m= - m = m -m-=m-m- ubsttute m= -,, get = =- - B G Recurrece epresso of Ho Tower h h H h h ubtrct d get = -+ h h h Roots of = re d

35 Ler Homegeeous Recurrece Relto ssume G s geertg fucto of { }: G dds up both sdes of these equtos, get G G G,

36 Ler Homegeeous Recurrece Relto j j j j G Let, the order of polyoml P. j j j j P P G P m m m m m m m m..., m= -

37 Ler Homogeeous Recurrece Relto = --= h h-+h-= Def f sequece { } stsfes: d, d,, d,,, d d, d, d re costts,, the ths epresso s clled th -order ler homogeeous recurrece relto of { }. -+= hrcterstc Polyoml,

38 = h h-+h-= Def f sequece { } stsfes:,,,,, d d d,, d d, d, d re costts,, the ths epresso s clled th -order ler homogeeous recurrece relto of { }. hrcterstc Polyoml

39 Ler Homegeeous Recurrece Relto Now we dscuss the clculto by stutos hrcterstc Polyoml hs dstct rel roots ssume G could be smplfed s l l l l G l... l l l,,, could be solved by whch d,d d - re tl vlues of... d l l l d l l l d l l l

40 Ler Homegeeous Recurrece Relto bocc equece =, =, = - + = --=-- hrcterstc Polyoml B B { ; B Def If sequece { } stsfes:,,,,, d d d hrcterstc Polyoml hrcterstc polyoml hs dstct rel roots, dfferet rel roots l l l Ler Homogeeous Recurrece Relto,, d d, d, d re costts. =, =

41 Geertg ucto l l l Def - or sequece,,, costruct fucto G= + + +, G s the geertg fucto of,, Euler 764- D Berell 7- D eems to be fuctos but t s ctully mppg Lple 8 D Iteger egmetto, l l l Recurrece Geertg fucto G s brdge

42 Ler Homogeeous Recurrece Relto Def If sequece { } stsfes:,,,, d d d,, d d, d, d re costts hrcterstc Polyoml hrcterstc polyoml hs dstct rel roots l l l I whch l, l, l re udetermed coeffcets.,

43 ...! R hrcterstc Polyoml hs multple roots Eg 4, 4 Geertg ucto Method 4 4 4, : 4 4 : , hrcterstc Equto Method hrcterstc Equto: -4+4=- Geertg ucto orm: Prtl rctos: b B B ' B ', B 4 ', B

44 hrcterstc Polyoml hs multple roots ssume β s -multple root of, t could be smplfed s j j j 4 t j j j G j j j s j--order polyoml of. o s the product of β d --order polyoml of. The term relted to the soluto of recurrece relto s: whch re udetermed coeffcets. s coeffcets = j= j j+ β, whch,,, R!...

45 Eg 4 4,, 4 hrcterstc Equto s : 4 4 4,

46 Dstct rel roots Multple rel rootsojugte comple roots?

47 ojugte omple Roots Qudrtc ormul: Whe b -4c<, there s o rel root, two comple roots. cos s, cos s Trgoometrcl form of comple umber z = + b: z = ρcosθ+ sθ b

48 . Ler Homogeeous Recurrece Relto hrcterstc Polyoml hs cojugte comple roots ssume tht, re pr of cojugte comple roots of. s cos, s cos I the coeffcet of s: s cos s cos s cos s cos s cos

49 I whch, B Whe clcultg relty, we could solve the cojugte comple roots t frst, the clculte udetermed coeffcets, B to vod the termedte comple umber clcultos. B s cos s cos

50 Eg,, hrcterstc equto: e s cos s cos ; s cos B s cos

51 ummry of Ler Recurrece Relto ccordg to the o-zero roots of dstct o- rel roots I whch l l, l,, l l l, re udetermed coeffcets. pr of cojugte comple root e d e : cos B s I whch,b re udetermed coeffcets. Hs root α wth multplcty of. I whch,,, re udetermed coeffcets.

52 4 Ler Homogeeous Recurrece Relto 4-4 pplctos 组合数学 ombtorcs

53 Ler Homogeeous Recurrece Relto Eg:olve mlrly ubtrct, get mlrly,,

54 orrespodg hrcterstc Equto s m m m m,, m s -multple root B B, B, 4 B B,, o Ths proves

55

56 Ler Homogeeous Recurrece Relto Eg:lculte mlrly ubtrct, get mlrly 4,,, ubtrct, get mlrly 4

57 4,,, orrespodet chrcterstc equto s: r r r r r s 4-multple root r D B s we hve equto group bout B D: 4,,, D B D B D B

58 B D B D B D B D. j j D = =

59 pplctos of geertg fucto d recurrece relto Eg:There s pot P o the ple. It s the cross of felds D, D, D. olor these felds wth colors. We requre the color of two djcet res to be dfferet. lculte the umber of rrgemets. Let be the umber of rrgemet to color these res. There re stutos: D D D P D D 9

60 pplctos of geertg fucto d recurrece relto D d D - hve the sme color; D hs - choces, whch s ll colors ecept the oe used by D d D - ; the rrgemets for D - to D 的 re oeto-oe correspodet to the rrgemets for - res. D d D - hve dfferet colors. D hs - choces; the rrgemets from D to D - re oe-to-oe correspodet to the rrgemets for - res.,,. D D D D P D D D P D D D D 6

61 6 pplctos of geertg fucto d recurrece relto.,,.,..,, B., B B..,., B

62 Def - or sequece,,, costruct fucto G= + + +, The G s clled the geertg fucto of,, Lplce 8 D Geertg fuctos re hger to hg serres of umbers. Herbert 4 4