Diakritˆ Majhmatikˆ I. Leutèrhc KuroÔshc (EÔh Papaðwˆnnou)

Diakritˆ Majhmatikˆ I Leutèrhc KuroÔshc (EÔh Papaðwˆnnou)

PlhroforÐec... Tetˆrth, 09.00-11.00, Paraskeu, 18.00-20.00 SÔggramma 1: Λ. Κυρούσης, Χ. Μπούρας, Π. Σπυράκης. Διακριτά Μαθηματικά: Τα Μαθηματικά της Επιστήμης των Υπολογιστών. Gutenberg, 1992 Γ. Βουτσαδάκης, Λ. Κυρούσης, Χ. Μπούρας και Π. Σπυράκης, Διακριτά Μαθηματικά: Προβλήματα και Λύσεις. Gutenberg, 1994. SÔggramma 2: C.L. Liu. Introduction to Combinatorial Mathematics. McGraw-Hill, 1969. IstoselÐda maj matoc: http://lca.ceid.upatras.gr/courses/diakrita/index.html EpikoinwnÐa: papaioan@ceid.upatras.gr Diafˆneiec: http://www.ceid.upatras.gr/papaioan/dm10-dd.mm.pdf

H Ôlh sunoptikˆ... Stoiqei dhc sunduastik Genn triec sunart seic JewrÐa Polyá Egkleismìc - Apokleismìc

Stoiqei dhc sunduastik Eisagwg Omˆdec Ðdiwn (mh diakekrimènwn) antikeimènwn SunduasmoÐ kai diatˆxeic me epanˆlhyh UposÔnola Dianomèc antikeimènwn se upodoqèc DiwnumikoÐ suntelestèc

Kanìnec ginomènou kai ajroðsmatoc Kanìnac ajroðsmatoc: An èna gegonìc mporeð na sumbeð katˆ m trìpouc kai èna ˆllo gegonìc mporeð na sumbeð katˆ n trìpouc, tìte upˆrqoun m + n trìpoi, katˆ touc opoðouc èna apì ta duo gegonìta mporeð na sumbeð. Kanìnac ginomènou: An èna gegonìc mporeð na sumbeð katˆ m trìpouc kai èna ˆllo gegonìc mporeð na sumbeð katˆ n trìpouc, tìte upˆrqoun m n trìpoi, katˆ touc opoðouc kai ta duo gegonìta mporeð na sumboôn.

Kanìnec ginomènou kai ajroðsmatoc Sto Tm ma diatðjentai 7 prwinˆ maj mata kai 5 apogeumatinˆ. Kanìnac ajroðsmatoc: Pìsec epilogèc èqei ènac foitht c pou endiafèretai na pˆrei 1 mìno mˆjhma; 7 + 5 = 12 Kanìnac ginomènou: Pìsec epilogèc èqei ènac foitht c gia na pˆrei 1 prwinì kai 1 apogeumatinì mˆjhma; 7 5 = 35

Diatˆxeic (Permutations) r antikeimènwn epilegmènwn apì n antikeðmena qwrðc epanatopojèthsh: P(n, r) = n(n 1)...(n r + 1) = } {{ } r 'Eqei shmasða h seirˆ Antimetajèseic r antikeimènwn: P(r, r) = r! n! (n r)!, n, r N

Diatˆxeic (Permutations) Parˆdeigma: Me pìsouc trìpouc mporoôme na programmatðsoume exètash 3 majhmˆtwn se 5 hmèrec ste na mhn pèftoun dôo maj mata thn Ðdia mèra; Apˆnthsh: Mac endiafèrei poiec ja eðnai oi 3 mèrec; Nai. 'Ara jèloume metajèseic. P(5, 3) = 5 4 3 = 60

SunduasmoÐ (Combinations) r antikeimènwn epilegmènwn apì n antikeðmena qwrðc epanatopojèthsh: C(n, r) = P(n, r) P(r, r) = n(n 1)...(n r + 1) r! = n! (n r)!r! = ( ) n, r n, r N Den èqei shmasða h seirˆ

SunduasmoÐ (Combinations) Parˆdeigma: Me pìsouc trìpouc mporoôme na programmatðsoume katˆ th diˆrkeia miac ebdomˆdac 3 geômata me krèac; Apˆnthsh: Mac endiafèrei poiec ja eðnai oi 3 mèrec; 'Oqi. Aplˆ dialègoume... 'Ara jèloume sunduasmoôc. P(7, 3) C(7, 3) = P(3, 3) = 7 6 5 3 2 1 = 35

Pìsoi akèraioi upˆrqoun metaxô tou 100 kai tou 199 oi opoðoi èqoun diaforetikˆ yhfða; Pìsoi apì touc akèraiouc autoôc eðnai perittoð; Oi zhtoômenoi arijmoð apoteloôntai apì 3 jèseic stic opoðec to pr to yhfðo eðnai 1 kai gia ta ˆlla 2 prokôptoun apì tic diatˆxeic 2 yhfðwn apì ta 9 diajèsima (de sumperilambˆnoume to yhfðo 1 pou èqei dh qrhsimopoihjeð): P(9, 2) = 9 8 = 72 Oi perittoð arijmoð ja katal goun se 3,5,7 9 (afoô èqoun diaforetikˆ yhfða kai to 1 apokleðetai) Gia kˆje mða apì autèc tic epilogèc upˆrqoun 8 epilogèc gia to mesaðo yhfðo Epomènwc, sunolikˆ, upˆrqoun 4 8 = 32 perittoð akèraioi me diaforetikˆ yhfða metaxô 100 kai 199

Pìsoi apì touc pr touc 10.000 jetikoôc akèraiouc èqoun diaforetikˆ yhfða; H apˆnthsh prokôptei apì to ˆjroisma twn apant sewn stic ex c erwt seic: πόσοι είναι οι 4-ψήϕιοι ακέραιοι με διαϕορετικά ψηϕία; 9 9 8 7 = 4536 (όχι 0 στην πρώτη θέση, όχι ό,τι επιλέχθηκε για την πρώτη θέση και το 0 στη δεύτερη, όχι ό,τι επιλέχθηκε για τις δύο πρώτες θέσεις στην τρίτη, όχι ό,τι επιλέχθηκε για τις τρεις πρώτες θέσεις στην τέταρτη) πόσοι είναι οι 3-ψήϕιοι ακέραιοι με διαϕορετικά ψηϕία; 9 9 8 = 648 (όχι 0 στην πρώτη θέση, όχι ό,τι επιλέχθηκε για την πρώτη θέση και το 0 στη δεύτερη, όχι ό,τι επιλέχθηκε για τις δύο πρώτες θέσεις στην τρίτη) πόσοι είναι οι 2-ψήϕιοι ακέραιοι με διαϕορετικά ψηϕία; 9 9 = 81 (όχι 0 στην πρώτη θέση, όχι ό,τι επιλέχθηκε για την πρώτη θέση και το 0 στη δεύτερη) πόσοι είναι οι 1-ψήϕιοι ακέραιοι με διαϕορετικά ψηϕία; 10 Επομένως, συνολικά: 5275

Pìsouc perittoôc akèraiouc mporoôme na sqhmatðsoume me ta yhfða 1,2,3,4,5 oi opoðoi èqoun 4 yhfða kai ta yhfða autˆ eðnai diaforetikˆ metaxô touc; Oi zhtoômenoi 4-y fioi akèraioi prèpei na èqoun 1 3 5 sth dexiìterh jèsh. 4-y fioi me 1 sth dexiìterh jèsh: 4 3 2 = 24 4-y fioi me 3 sth dexiìterh jèsh: 4 3 2 = 24 4-y fioi me 5 sth dexiìterh jèsh: 4 3 2 = 24 Epomènwc, sunolikˆ 3 24 = 72 arijmoð

Pìsoi pentay fioi akèraioi upˆrqoun oi opoðoi eðnai megalôteroi tou 53000 kai èqoun tautìqrona tic ex c idiìthtec: a) ta yhfða touc eðnai diaforetikˆ kai b) den emfanðzontai se autoôc ta yhfða 0 kai 9; ArijmoÐ pou xekinˆne apì 53 kai akoloujoôn 3 yhfða pou de mporoôn na eðnai 0 kai 9, oôte 5 kai 3 kai prèpei na eðnai kai diaforetikˆ metaxô touc: 6 5 4 = 120 ArijmoÐ pou xekinˆne apì 5 kai akoloujoôn 4 yhfða pou de mporoôn na eðnai 0 kai 9, oôte 5, prèpei na eðnai kai diaforetikˆ metaxô touc kai ta yhfða thc aristerìterhc jèshc prèpei na eðnai megalôtera tou 3: 4 6 5 4 = 480 5-y fioi arijmoð oi opoðoi sthn aristerìterh jèsh èqoun yhfðo megalôtero tou 5 kai ìqi 0 9, sthn epìmenh jèsh ìqi ì,ti sthn prohgoômenh oôte 0 9, sthn epìmenh jèsh ìqi ì,ti stic dôo prohgoômenec oôte 0 9, kok: 3 7 6 5 4 = 2520 Epomènwc, sunolikˆ 120 + 480 + 2520 = 3120 arijmoð

Pìsec lèxeic me 0 kai 1 upˆrqoun pou èqoun 8 grˆmmata kai a) xekinˆne me 1100 b) perièqoun akrib c dôo 1; Gia to a) 2 4 lèxeic Gia to b) C(8, 2) = P(8, 2) P(2, 2) = 8 7 = 28 lèxeic 1 2

Ta grˆmmata A,B,G,D qrhsimopoioôntai gia na sqhmatðsoume lèxeic m kouc 3. a) Pìsec lèxeic upˆrqoun pou perièqoun to grˆmma A epitrepomènwn twn epanal yewn; b) Pìsec lèxeic upˆrqoun pou arqðzoun me A epitrepomènwn twn epanal yewn; 'Olec oi pijanèc lèxeic me 3 grˆmmata apì ta A,B,G,D eðnai 4 3. Autèc pou den perièqoun kanèna A eðnai 3 3. Epomènwc, oi zhtoômenec eðnai h diaforˆ touc: 4 3 3 3 = 64 27 = 37 lèxeic. To aristerìtero grˆmma eðnai A. Opìte zhtˆme lèxeic 2 grammˆtwn pou sqhmatðzontai apì ta 4 dosmèna grˆmmata: 4 2 = 16 lèxeic.

To agglikì alfˆbhto perièqei 26 grˆmmata ek twn opoðwn ta 5 eðnai fwn enta. a) Pìsec lèxeic m kouc 5 mporoôme na sqhmatðsoume qrhsimopoi ntac 3 diaforetikˆ sômfwna kai 2 diaforetikˆ fwn enta; b) Pìsec apì autèc perièqoun to grˆmma b; Gia to a): Zhtˆw lèxeic twn 5 qarakt rwn apì touc opoðouc 3 prèpei na eðnai sômfwna kai 2 fwn enta - ìla diaforetikˆ metaxô touc. Με πόσους τρόπους μπορώ να διαλέξω ποια από τα 5 γράμματα θα είναι τα σύμϕωνα; Με C(5, 3) = 5 4 3 1 2 3 = 10 τρόπους. Προϕανώς, για κάθε μία τέτοια περίπτωση, τα εναπομείναντα γράμματα θα είναι τα ϕωνήεντα. Πόσες διαϕορετικές 3άδες διαϕορετικών συμϕώνων μπορώ να έχω; P(21, 3) = 21 20 19 = 7980 Πόσες διαϕορετικές 2άδες διαϕορετικών ϕωνηέντων μπορώ να έχω; P(5, 2) = 5 4 = 20 'Ara, o zhtoômenoc arijmìc lèxewn eðnai 10 7980 20 = 1596000

To agglikì alfˆbhto perièqei 26 grˆmmata ek twn opoðwn ta 5 eðnai fwn enta. a) Pìsec lèxeic m kouc 5 mporoôme na sqhmatðsoume qrhsimopoi ntac 3 diaforetikˆ sômfwna kai 2 diaforetikˆ fwn enta; b) Pìsec apì autèc perièqoun to grˆmma b; Gia to b): upologðzw pìsec lèxeic den perièqoun to b kai tic afair apì autèc thc er thshc a) Με πόσους τρόπους μπορώ να διαλέξω ποια από τα 5 γράμματα θα είναι τα σύμϕωνα; Με C(5, 3) = 5 4 3 1 2 3 = 10 τρόπους. Προϕανώς, για κάθε μία τέτοια περίπτωση, τα εναπομείναντα γράμματα θα είναι τα ϕωνήεντα. Πόσες διαϕορετικές 3άδες διαϕορετικών συμϕώνων που δεν είναι b μπορώ να έχω; P(20, 3) = 20 19 18 = 6840 Πόσες διαϕορετικές 2άδες διαϕορετικών ϕωνηέντων μπορώ να έχω; P(5, 2) = 5 4 = 20 'Ara, to pl joc twn lèxewn pìsec lèxeic den perièqoun to b eðnai 10 6840 20 = 1368000 Opìte oi zhtoômenec lèxeic eðnai: 1596000 1368000 = 228000

Pìsoi akèraioi metaxô tou 1 kai tou 10000000000 perièqoun to yhfðo 1 kai pìsoi den to perièqoun; MetaxÔ tou 0 kai tou 9999999999, 9 10 arijmoð den perièqoun to 1 Αυτός είναι ο αριθμός των διατάξεων 9 στοιχείων (0, 2, 3,..., 9) σε 10 θέσεις με επαναλήψεις Epomènwc, metaxô tou 1 kai tou 10000000000: 10 10 9 10 + 1 περιέχουν το 1

Pìsoi tetray fioi arijmoð tou dekadikoô sust matoc den èqoun dôo yhfða Ðdia; Gia na eðnai tetray fioc kˆpoioc arijmìc den prèpei na èqei 0 sthn aristerìterh jèsh, sthn opoða mporeð na brðsketai èna apì ta enapomeðnanta 9 yhfða (1,...9). 'Ara, to pl joc twn zhtoômenwn arijm n eðnai: 9 9 8 7 = 4.536

Pìsec eðnai oi sumboloseirèc thc morf c ww R m kouc 10 me kefalaða grˆmmata tou ellhnikoô alfab tou qwrðc tìnouc; Ta pènte pr ta grˆmmata kajorðzoun kai ta pènte epìmena. Epomènwc, asqoloômai mìno me ta pènte pr ta kai upologðzw pìsouc diaforetikoôc trìpouc mpor na sunjèsw pentˆdec. H epilog tou kˆje grˆmmatoc eðnai anexˆrthth kai kajèna mporeð na pˆrei 24 diaforetikèc timèc. 'Ara sunolikˆ, mporoôme na ftiˆxoume 24 } 24 24 {{ 24 24} = 24 5 5 forèc diaforetikèc lèxeic twn pènte grammˆtwn. (Kanìnac ginomènou) Tìsec eðnai kai oi zhtoômenec sumboloseirèc, afoô ta pènte pr ta grˆmmata kajorðzoun kai ta pènte epìmena.

'Eqw 24 arijmhmènec (diaforetikèc) prˆsinec mpˆlec kai 24 arijmhmènec kìkkinec mpˆlec. Me pìsouc diaforetikoôc trìpouc mpor na dialèxw mða prˆsinh kai mða kìkkinh mpˆla; Prˆsinh mpˆla mpor na dialèxw me 24 trìpouc. Kìkkinh mpˆla mpor na dialèxw me 24 trìpouc. Gia na sumbaðnoun kai ta dôo mazð upˆrqoun 24 24 = 576 diaforetikoð trìpoi. (Kanìnac ginomènou)

'Eqw 24 arijmhmènec (diaforetikèc) prˆsinec mpˆlec kai 24 arijmhmènec kìkkinec mpˆlec. Me pìsouc diaforetikoôc trìpouc mpor na dialèxw dôo mpˆlec qwrðc periorismì qr matoc; Mpor na dialèxw dôo mpˆlec me 3 diaforetikoôc trìpouc: Na dialèxw mða kìkkinh kai mða prˆsinh: autì mporeð na gðnei me 24 24 = 24 2 diaforetikoôc trìpouc (Kanìnac ginomènou) Na dialèxw dôo kìkkinec: autì mporeð na gðnei me 24 23 diaforetikoôc trìpouc, thn pr th th dialègw apì 24 mpˆlec kai th deôterh apì tic upìloipec 23 Na dialèxw dôo prˆsinec: ìmoia me prin (Kanìnac ginomènou) Epeid mpor na dialèxw 2 mpˆlec me ènan apì touc parapˆnw trìpouc (kanìnac ajroðsmatoc), autì shmaðnei ìti upˆrqoun sunolikˆ: 24 2 + 24 23 + 24 23 = 1680 diaforetikoð trìpoi.

'Eqw 5 Ellhnikˆ, 7 Agglikˆ kai 10 Ispanikˆ biblða (sunolikˆ 22). (a) Me pìsouc diaforetikoôc trìpouc mpor na dialèxw dôo biblða; (b) Me pìsouc diaforetikoôc trìpouc mpor na dialèxw dôo biblða diaforetik c gl ssac; Gia to (a): Το πρώτο το διαλέγω από 22 διαθέσιμα και το δεύτερο από τα υπόλοιπα 21 διαθέσιμα. 'Ara sunolikˆ me 22 21 = 462 diaforetikoôc trìpouc. Gia to (b): Mpor na dialèxw: Ελληνικό και Αγγλικό με 5 7 = 35 τρόπους Ελληνικό και Ισπανικό με 5 10 = 50 τρόπους Αγγλικό και Ισπανικό με 7 10 = 70 τρόπους 'Ara sunolikˆ me 35 + 50 + 70 = 155 diaforetikoôc trìpouc.

Me pìsouc diaforetikoôc trìpouc mpor na dialèxw kai na bˆlw se seirˆ 5 apì 20 majhtèc; Me 20 19 18 17 16 = P(20, 5) = 20! 15! diaforetikoôc trìpouc.

Me pìsouc trìpouc mporoôn na kaj soun 5 ˆtoma se mia seirˆ apì 12 kajðsmata; Diˆlexe poia 5 apì ta 12 diajèsima kajðsmata ja qrhsimopoihjoôn. Autì mporeð na gðnei me C(12, 5) = 12! 5! 7! trìpouc. Me pìsouc trìpouc mporeðc na diatˆxeic ta 5 ˆtoma pou prìkeitai na kˆtsoun sta kajðsmata autˆ; Me P(5, 5) = 5! trìpouc 'Ara, 5 ˆtoma mporoôn na kaj soun se mia seirˆ apì 12 12! kajðsmata me 5! = 95.040 trìpouc 5! 7!

Me pìsouc trìpouc mporoôn na kaj soun 5 ˆtoma se mia seirˆ apì 12 kajðsmata; Enallaktik lôsh Me pìsouc diaforetikoôc trìpouc mpor na diatˆxw ta 12 kajðsmata; Pìsec metajèseic twn 12 kajismˆtwn upˆrqoun; 12! Jewr ìti pˆnta topojet ta 5 ˆtoma sta pr ta 5 kajðsmata kˆje metˆjeshc Gia tic 7-ˆdec pou mènoun ˆdeiec, de me endiafèrei h seirˆ gia kˆje metˆjesh kratˆw 1 apì tic 7! diaforetikèc 'Ara, èqw sunolikˆ: 12! = 95.040 trìpouc 7!

Ta prohgoômena aforoôn sta maj mata 6 kai 8/10/2010

Me pìsouc trìpouc mporoôn na kaj soun 5 ˆtoma se mia seirˆ apì 12 kajðsmata; Enallaktik lôsh pou protˆjhke apì mða sunˆdelfo sto prohgoômeno mˆjhma Upojètoume ìti ta ˆtoma eðnai stajerˆ kai ta kajðsmata pˆne proc ta ˆtoma... Ta kajðsmata den eðnai plèon Ðdia allˆ èqei to kajèna èna diaforetikì tampelˆki... Gia to pr to ˆtomo upˆrqoun 12 epilogèc kajismˆtwn, gia to epìmeno 11, k.o.k. Me pìsouc diaforetikoôc trìpouc mpor na diatˆxw ta 12 kajðsmata se 5-adec; Me P(12, 5) = 12 11 10 9 8 = 12! 7! diaforetikoôc trìpouc.

Me pìsouc trìpouc mporoôn na epilegoôn 3 arijmoð apì to 1 èwc to 300 ètsi ste to ˆjroismˆ touc na eðnai diairetì me 3; QwrÐzoume touc arijmoôc apì 1 èwc 300, S = {1,..., 300}, se trða uposônola A, B kai G ètsi ste: A B Γ = S A B =, A Γ = kai B Γ = A = {x S : x mod 3 = 0}, A = {3, 6, 9,..., 300} B = {y S : y mod 3 = 1}, B = {1, 4, 7,..., 298} Γ = {z S : z mod 3 = 2}, Γ = {2, 5, 8,..., 299} A = B = Γ = 100 Gia na eðnai to ˆjroisma tri n arijm n diairetì me 3, prèpei είτε και οι τρεις να ανήκουν στο A, είτε και οι τρεις στο B, είτε και οι τρεις στο Γ, δηλ. ( ) ( 100 3 + 100 ) ( 3 + 100 ) 3 είτε ένας αριθμός να ανήκει σε καθένα από τα τρία αυτά υποσύνολα, δηλ. ( )( 100 100 )( 100 ) 1 1 1 'Ara sunolikˆ: 3 (100 ) + 100 3 = 1.485.100 trìpoi 3

Poioc eðnai o arijmìc twn diairet n tou 180; 180 = 2 90 = 2 2 45 = 2 2 3 15 = 2 2 3 3 5 = 2 2 3 2 5 Apì JewrÐa Arijm n, diairètec tou 180 eðnai oi arijmoð thc morf c 2 k 3 n 5 m me 0 k 2, 0 n 2 kai 0 m 1 Mpor na epilèxw to k me 3 trìpouc, to n me 3 trìpouc kai to m me 2 trìpouc kai jèlw ìla autˆ na isqôoun mazð 'Epomènwc, èqw sunolikˆ, 3 3 2 = 18 trìpouc

Pìsouc diairètec èqei o arijmìc 1400; 1400 = 2 700 = 2 2 350 = 2 2 2 175 = 2 2 2 5 35 = 2 2 2 5 5 7 = 2 3 5 2 7 Apì JewrÐa Arijm n, diairètec tou 1400 eðnai oi arijmoð thc morf c 2 k 5 n 7 m me 0 k 3, 0 n 2 kai 0 m 1 Mpor na epilèxw to k me 4 trìpouc, to n me 3 trìpouc kai to m me 2 trìpouc kai jèlw ìla autˆ na isqôoun mazð 'Epomènwc, èqw sunolikˆ, 4 3 2 = 24 trìpouc

Stoiqei dhc sunduastik SunduasmoÐ kai diatˆxeic me epanˆlhyh

Diatˆxeic me epanˆlhyh: Me pìsouc trìpouc mpor na diatˆxw r apì n antikeðmena ìtan epitrèpontai epanalhptikèc emfanðseic twn antikeimènwn; Sthn arq èqw n antikeðmena apì ta opoða na dialèxw èna. Metˆ thn epilog mou aut, topojet to antikeðmeno pou diˆlexa mazð me ta upìloipa. 'Ara thn deôterh forˆ èqw pˆli n antikeðmena sthn diˆjes mou gia na dialèxw. 'Omoia gia kˆje mia apì tic r epilogèc mou. Upˆrqoun n epilogèc gia to pr to antikeðmeno, n epilogèc gia to deôtero antikeðmeno,..., n epilogèc gia to r -tì antikeðmeno n } n... {{ n} = n r r forèc

Pìsoi tetray fioi arijmoð upˆrqoun sto dekadikì sôsthma; Kajèna apì ta 4 yhfða mpor na to dialèxw me 10 diaforetikoôc trìpouc. 'Ara sunolikˆ 10 4 teray fioi arijmoð.

Pìsa diaforetikˆ uposônola enìc sunìlou S me n stoiqeða upˆrqoun; Kajèna apì ta n stoiqeða tou dosmènou sunìlou S mpor na to dialèxw me 2 diaforetikoôc trìpouc: na to epilèxw na mhn to epilèxw na an kei se kˆpoio uposônolo tou S. 'Ara sunolikˆ 2 n uposônola.

SunduasmoÐ me epanˆlhyh: Me pìsouc trìpouc mpor na dialèxw r apì n antikeðmena ìtan epitrèpontai epanalhptikèc emfanðseic twn antikeimènwn; To sônolo twn antikeimènwn apì to opoðo gðnetai h epilog eðnai to {1,..., n} 'Enac sunduasmìc r antikeimènwn apì n me epanˆlhyh eðnai mia akoloujða x 1, x 2,..., x r me 1 x i n kai x i x j ìtan 1 i j r alli c ta antikeðmena kˆje akoloujðac pou prokôptei ta diatˆssw se aôxousa seirˆ afoô ta antikeðmena me endiafèroun kai ìqi h seirˆ touc Allˆ epeid epitrèpetai na dialèxw se diaforetikèc jèseic to Ðdio antikeðmeno prèpei na brw ènan trìpo na orðsw touc ìrouc kˆje akoloujðac monos manta Qrhsimopoi thn 1-1 kai epð antistoiqða (x 1, x 2,..., x n ) (x 1 + 0, x 2 + 1,..., x n + r 1) Opìte plèon dialègw r antikeðmena apì n + r 1 qwrðc epanˆlhyh, dhl. C(n + r 1, r) (pou xèrw na to upologðzw)

SunduasmoÐ me epanˆlhyh: Me pìsouc trìpouc mpor na dialèxw r apì n antikeðmena ìtan epitrèpontai epanalhptikèc emfanðseic twn antikeimènwn; JewroÔme ìti ta n dosmèna antikeðmena prosdiorðzontai apì touc akèraiouc 1, 2, 3,..., n JewroÔme ìti ta r antikeðmena pou epilègoume prosdiorðzontai apì mia lðsta akeraðwn {j, k, l,..., m} se aôxousa diˆtaxh Π.χ., η λίστα {1, 1, 1, 3, 4, 5, 5,...} σημαίνει ότι το 1ο αντικείμενο έχει επιλεχθεί 3 ϕορές, το 2ο καμμία, το 3ο μία, το 4ο μία, το 5ο δύο κ.ο.κ. TropopoioÔme thn akoloujða aut twn r antikeimènwn wc ex c: j + 0, k + 1, l + 2,..., m + (r 1) se aôxousa diˆtaxh Π.χ., η προηγούμενη λίστα θα γινόταν {1 + 0, 1 + 1, 1 + 2, 3 + 3, 4 + 4, 5 + 5, 5 + 6,...} = {1, 2, 3, 6, 8, 10, 11,...} Kai plèon kˆje epilog r antikeimènwn prosdiorðzetai monadikˆ sa mia epilog r diaforetik n akeraðwn apì touc 1, 2,..., n + (r 1)

SunduasmoÐ me epanˆlhyh: Me pìsouc trìpouc mpor na dialèxw r apì n antikeðmena ìtan epitrèpontai epanalhptikèc emfanðseic twn antikeimènwn; DÐnontai ta (n =) 4 antikeðmena {1, 4, 7, 11} kai mˆc zhtˆne na epilèxoume (r =) 3 apì autˆ ìtan epitrèpontai epanal yeic Ta dosmèna antikeðmena prosdiorðzontai apì th lðsta twn akeraðwn 1, 2, 3, 4 'Estw ìti h epilog mac eðnai h lðsta {2, 4, 4} (me ta stoiqeða thc topojethmèna se aôxousa diˆtaxh) pou shmaðnei ìti sth lðsta aut to 1o dosmèno antikeðmeno den emfanðzetai, to 2o emfanðzetai mða forˆ, to 3o den emfanðzetai kai to 4o emfanðzetai 2 forèc Tropopoi th lðsta {2, 4, 4}: {2 + 0, 4 + 1, 4 + 2} = {2, 5, 6} Dhl., h lðsta twn epilegmènwn antikeimènwn proèkuye epilègontac 3 apì touc arijmoôc {1, 2, 3, 4, 5, 6} qwrðc epanal yeic antð 3 apì touc {1, 2, 3, 4} me epanal yeic

SunduasmoÐ me epanˆlhyh: Me pìsouc trìpouc mpor na dialèxw r apì n antikeðmena ìtan epitrèpontai epanalhptikèc emfanðseic twn antikeimènwn; AntÐstrofa, an epèlexa th lðsta {4, 5, 6} shmaðnei ìti epèlexa th {4 0, 5 1, 6 2} = {4, 4, 4} dhl., to antikeðmeno epèlexa 3 forèc to antikeðmeno 4, dhl., epèlexa 3 forèc to 11 Dhlad : den èqei shmasða poia antikeðmena eðnai sto dosmèno sônolo allˆ pìsa kai epeid oi epanal yeic èqoun plèon shmasða allˆ eg xèrw na metrˆw sunduasmoôc qwrðc epanal yeic na tropopoi sw anˆloga to sônolo twn dosmènwn antikeimènwn

SunduasmoÐ me epanˆlhyh r antikeimènwn epilegmènwn apì n antikeðmena ( ) n + r 1 r

Pìsec zarièc upˆrqoun sto tˆbli; Kˆje zˆri mporeð na pˆrei n = 6 diaforetikèc timèc, èqw r = 2 zˆria kai metrˆw tic epanal yeic Dialègw r = 2 apì n = 6 antikeðmena me epanˆlhyh Yˆqnw touc sunduasmoôc r = 2 apì n = 6 antikeimènwn me epanˆlhyh ( ) ( n+r 1 'Ara, èqw sunolikˆ, r = 6+2 1 ) ( 2 = 7 ) 2 = 7! 2! 5! = 21 zarièc

Apì ènan megˆlo arijmì apì nomðsmata twn 0,05E, 0,10E, 0,20E kai 0,50E, me pìsouc trìpouc mporoôme na epilèxoume 6 kèrmata; Dialègw 6 (= r ) kèrmata apì èna sônolo me 4 (= n) antikeðmena kai epitrèpontai oi epanal yeic C(n + r 1, r) = C(4 + 6 1, 6) = C(9, 6) = 9! 6! 3! = 84

Stoiqei dhc sunduastik Omˆdec mh diakekrimènwn (dhl. Ðdiwn) antikeimènwn

Metr seic antikeimènwn se omˆdec n antikeðmena diaqwrismèna se t omˆdec Ðdiwn (mh diakekrimènwn) antikeimènwn, me pl joc stoiqeðwn q 1, q 2,..., q t antðstoiqa. n! Diatˆxeic n tètoiwn antikeimènwn: q 1!q 2!...q t! Upˆrqoun n! trìpoi na diatˆxw n diaforetikˆ (diakekrimèna) antikeðmena. Epeid ìmwc ta antikeðmena kˆje omˆdac eðnai Ðdia (den eðnai diakekrimèna) metaxô touc, diair me ton arijmì twn dunat n diatˆxewn kˆje omˆdac (q 1!...q t!). SunduasmoÐ n tètoiwn antikeimènwn: (q 1 + 1)(q 2 + 1)...(q t + 1) 1 Mpor na epilèxw to pr to antikeðmeno twn n me (q 1 + 1) trìpouc: eðte na mhn to epilèxw, eðte na to epilèxw mia forˆ, eðte duo forèc,..., eðte q 1 forèc. OmoÐwc gia ta upìloipa. O ìroc 1 mpaðnei sto tèloc gia na mhn metr sw to endeqìmeno na mhn epilèxw kanèna antikeðmeno.

'Eqw 7 a, 8 b, 5 g kai 4 d. Pìsec diaforetikèc sumboloseirèc mpor na ftiˆxw me autˆ; 24! 7!8!5!4!

Me pìsouc trìpouc mporoôn na qrwmatistoôn 12 mpˆlec ètsi ste 3 na eðnai kìkkinec, 2 roz, 2 ˆsprec kai oi upìloipec prˆsinec; Me pìsouc diaforetikoôc trìpouc mpor na diatˆxw tic 12 mpˆlec; Pìsec metajèseic twn 12 mpal n upˆrqoun; 12! Jewr ìti pˆnta qrwmatðzw kìkkinec tic 3 pr tec, roz tic 2 epìmenec, ˆsprec tic 2 epìmenec kai prˆsinec tic upìloipec, qwrðc na me endiafèrei h seirˆ mèsa stic omˆdec autèc Metrˆw 1 apì tic 3! arqikèc (kìkkinec) triˆdec, 1 apì tic 2! epìmenec (roz) duˆdec, mða apì tic 2! epìmenec (ˆsprec) duˆdec kai 1 apì tic epìmenec 5! (prˆsinec) pentˆdec 12! 'Ara, èqw sunolikˆ = 166.320 trìpouc 3! 2! 2! 5!

Stoiqei dhc sunduastik UposÔnola

UposÔnola Den epitrèpontai epanal yeic kai de metrˆei h seirˆ. Me pìsouc trìpouc mpor na epilèxw èna perissìtera apì n antikeðmena; Για κάθε ένα από τα n αντικείμενα υπάρχουν 2 τρόποι: να το επιλέξω ή να μην το επιλέξω και θέλω να επιλέξω τουλάχιστον 1 αντικείμενο 2 } 2 {{... 2 } 1 = 2 n 1 τρόποι n ϕορές (αϕαιρώ 1 για να εξαιρέσω την περίπτωση που δεν επιλέγω κανένα αντικείμενο) Upˆrqoun C(n, k) trìpoi na dialèxw k antikeðmena apì n 2 n 1 = n C(n, k) 2 n = k=1 n C(n, k)+1 2 n = k=1 n C(n, k) k=0

UposÔnola 'Eqoume t omˆdec pou perièqoun antðstoiqa q 1, q 2,..., q t Ðdia (mh diakekrimèna) antikeðmena. De metrˆei h diˆtaxh. Me pìsouc trìpouc mpor na epilèxw èna perissìtera antikeðmena; Για κάθε ομάδα t i έχω q ti επιλογές + 1 (δεν επιλέγω κανένα στοιχείο από την ομάδα) Άρα συνολικά έχω (q 1 + 1)(q 2 + 1)...(q t + 1) 1 τρόπους

Poioc eðnai o arijmìc twn diairet n tou 180; (Enallaktik lôsh) 180 = 2 90 = 2 2 45 = 2 2 3 15 = 2 2 3 3 5 Oi diairètec sqhmatðzontai apì sunduasmoôc apì tic ex c treic omˆdec: A = {2, 2}, B = {3, 3} kai Γ = {5} Από την ομάδα A μπορώ να επιλέξω 1, 2 ή κανένα στοιχείο (3 επιλογές) από την ομάδα B μπορώ να επιλέξω 1, 2 ή κανένα στοιχείο (3 επιλογές) από την ομάδα Γ μπορώ να επιλέξω 1 ή κανένα στοιχείο (2 επιλογές) 'Epomènwc, o arijmìc twn diairet n tou 180 eðnai (2 + 1)(2 + 1)(1 + 1) = 18. Den afair 1 gia thn perðptwsh 2 0 3 0 5 0 = 1 giatð dðnei diairèth tou 180

Poioc eðnai o arijmìc twn diairet n tou 1400; (Enallaktik lôsh) 1400 = 2 700 = 2 2 350 = 2 2 2 175 = 2 2 2 5 35 = 2 2 2 5 5 7 Oi diairètec sqhmatðzontai apì sunduasmoôc apì tic ex c treic omˆdec: A = {2, 2, 2}, B = {5, 5} kai Γ = {7} Από την ομάδα A μπορώ να επιλέξω 1, 2, 3 ή κανένα στοιχείο (4 επιλογές) από την ομάδα B μπορώ να επιλέξω 1, 2 ή κανένα στοιχείο (3 επιλογές) από την ομάδα Γ μπορώ να επιλέξω 1 ή κανένα στοιχείο (2 επιλογές) 'Epomènwc, o arijmìc twn diairet n tou 1400 eðnai (3 + 1)(2 + 1)(1 + 1) = 24. Den afair 1 gia thn perðptwsh 2 0 5 0 7 0 = 1 giatð dðnei diairèth tou 1400

SÔntomh epanˆlhyh... 'Ena gegonìc mporeð na sumbeð katˆ m trìpouc kai èna ˆllo katˆ n trìpouc tìte èna apì ta dôo mporeð na sumbeð me m + n trìpouc en kai ta dôo mporoôn na sumboôn katˆ mn trìpouc. QwrÐc epanal yeic stoiqeðwn P(n, r) = n(n 1)...(n r + 1) } {{ } r όροι ( ) n C(n, r) = = r Me epanal yeic stoiqeðwn P(n, r) = n } n... {{ n } = n r r ( ϕορές ) n + r 1 C(n, r) = r = n! (n r)! P(n, r) P(r, r) = n! r!(n r)!

SÔntomh epanˆlhyh... DÐnontai n antikeðmena diaqwrismèna se t omˆdec, me pl joc stoiqeðwn q 1, q 2,..., q t antðstoiqa, kai ta antikeðmena kˆje omˆdac eðnai Ðdia (den eðnai diakekrimèna) n! Diatˆxeic: q 1!q 2!...q t! SunduasmoÐ: (q 1 + 1)(q 2 + 1)...(q t + 1) 1 Den epitrèpontai epanal yeic kai de metrˆei h seirˆ. Me pìsouc trìpouc mpor na epilèxw èna perissìtera apì n antikeðmena; 2 } 2 {{... 2} 1 = 2 n 1 trìpoi n forèc Upˆrqoun C(n, k) trìpoi na dialèxw k antikeðmena apì n 2 n 1 = n C(n, k) 2 n = k=1 n C(n, k)+1 2 n = k=1 n C(n, k) k=0

Na apodeiqjeð ìti oi arijmoð 1) (3n)! akèraioi. (n 2 )! 2 n 3 n kai 2) (n!) n+1 eðnai Gia to 1): 'Eqoume 3n stoiqeða qwrismèna se n omˆdec. Kˆje mða apì tic n omˆdec perièqei 3 Ðdia stoiqeða. O arijmìc twn metajèsewn twn 3n stoiqeðwn eðnai: (3n)! 3!3!...3! = (3n)! (3!) n = (3n)! (2 3) n = (3n)! 2 n 3 n pou eðnai akèraioc giatð anaparistˆ arijmì metajèsewn.

Na apodeiqjeð ìti oi arijmoð 1) (3n)! akèraioi. Gia to 2): (n 2 )! 2 n 3 n kai 2) (n!) 'Eqoume n 2 stoiqeða qwrismèna se n omˆdec. Kˆje mða apì tic n omˆdec perièqei n Ðdia stoiqeða. O arijmìc twn metajèsewn twn n 2 stoiqeðwn eðnai: (n 2 )! n!n!...n! = (n2 )! (n!) n pou eðnai arijmìc diairetìc me n! afoô gia kˆje mia apì tic metajèseic twn n stoiqeðwn èqoume Ðso arijmì metajèsewn twn n 2 stoiqeðwn. Epomènwc kai o arijmìc (n 2 )! eðnai akèraioc. (n!) n+1 n+1 eðnai

(a) Pìsec apì tic 2 r akoloujðec r duadik n yhfðwn perièqoun zugì arijmì monˆdwn; (b) Pìsec apì tic 5 r pentadikèc akoloujðec r yhfðwn perièqoun zugì arijmì monˆdwn; Gia to (a): QwrÐzw tic 2 r duadikèc akoloujðec m kouc r se 2 r 1 zeugˆria ètsi ste oi akoloujðec se kˆje zeugˆri na diafèroun mìno sto dexiìtero yhfðo touc (dhl. sto stoiqeðo thc r - c jèshc touc). Se kˆje èna apì ta zeôgh autˆ mìno mða apì tic akoloujðec èqei ˆrtio arijmì monˆdwn. Epomènwc, upˆrqoun 2 r 1 akoloujðec r duadik n yhfðwn me zugì arijmì monˆdwn.

(a) Pìsec apì tic 2 r akoloujðec r duadik n yhfðwn perièqoun zugì arijmì monˆdwn; (b) Pìsec apì tic 5 r pentadikèc akoloujðec r yhfðwn perièqoun zugì arijmì monˆdwn; Gia to (b): Apì tic 5 r pentadikèc akoloujðec 3 r perièqoun mìno ta yhfða 2,3 kai 4. Autèc sumperilambˆnontai se ekeðnec pou èqoun ˆrtio arijmì monˆdwn. Tic upìloipec 5 r 3 r akoloujðec tic qwrðzoume se omˆdec anˆloga me to sqhmatismì twn yhfðwn 2,3 kai 4 pou perièqoun. Se kˆje mia apì autèc tic omˆdec oi misèc akoloujðec ja perièqoun ˆrtio arijmì monˆdwn. Epomènwc, o sunolikìc arijmìc twn pentadik n akolouji n r yhfðwn me ˆrtio pl joc monˆdwn eðnai: 3 r + 1 2 (5r 3 r )

MetaxÔ 2n antikeimènwn ta n eðnai Ðdia. Me pìsouc trìpouc mpor na epilèxw n apì ta dosmèna 2n antikeðmena; MporoÔme na dialèxoume n antikeðmena apì ta 2n wc ex c: dialègw me 1 trìpo n ( n Ðdia antikeðmena kai me 0) trìpouc 0 apì ta diaforetikˆ n antikeðmena, dialègw me 1 trìpo n 1 ( ) n Ðdia antikeðmena kai me trìpouc 1 apì ta diaforetikˆ n antikeðmena, dialègw me 1 trìpo n 2 ( ) n Ðdia antikeðmena kai me trìpouc 2 apì ta diaforetikˆ n antikeðmena, k.o.k. dialègw me 1 trìpo 0 ( n Ðdia antikeðmena kai me n) trìpouc n apì ta diaforetikˆ n antikeðmena. 'Ara sunolikˆ: ( n 0) + ( n 1) +... + ( n n ) = n i=0 ( ) n = 2 n trìpouc. i 1 2

MetaxÔ 2n antikeimènwn ta n eðnai Ðdia. Me pìsouc trìpouc mpor na epilèxw n apì ta dosmèna 2n antikeðmena; Enallaktik lôsh: JewroÔme n diadoqikèc jèseic ìpou kˆje mða antiproswpeôei èna apì ta n diaforetikˆ antikeðmena. Gia kˆje mða apì autèc tic jèseic èqoume 2 epilogèc: na epilèxoume to antikeðmenì thc na mhn to epilèxoume kai na bˆloume sth jèsh tou èna apì ta n Ðdia antikeðmena. 'Ara sunolikˆ: 2 n trìpoi.

MetaxÔ 3n + 1 antikeimènwn ta n eðnai Ðdia. Me pìsouc trìpouc mpor na epilèxw n apì ta dosmèna 3n + 1 antikeðmena; MporoÔme na dialèxoume n antikeðmena apì ta 3n + 1 wc ex c: dialègw me 1 trìpo n ( ) 2n+1 Ðdia antikeðmena kai me trìpouc 0 apì ta diaforetikˆ 2n + 1 antikeðmena, dialègw me 1 trìpo n 1 ( ) 2n+1 Ðdia antikeðmena kai trìpouc 1 apì ta diaforetikˆ 2n + 1 antikeðmena, dialègw me 1 trìpo n 2 ( ) 2n+1 Ðdia antikeðmena kai trìpouc 2 apì ta diaforetikˆ 2n + 1 antikeðmena, kìk dialègw me 1 trìpo 0 ( ) 2n+1 Ðdia antikeðmena kai trìpouc n apì ta diaforetikˆ 2n + 1 antikeðmena. 'Ara sunolikˆ: ( 2n+1 ) ( 0 + 2n+1 1 ) +... + ( 2n+1 n ) n ( ) 2n + 1 = = 2 2n trìpouc. i i=0 n 0 1 2

MetaxÔ 3n + 1 antikeimènwn ta n eðnai Ðdia. Me pìsouc trìpouc mpor na epilèxw n apì ta dosmèna 3n + 1 antikeðmena; 2 2n+1 = ( ) ( 2n+1 0 + 2n+1 ) ( 1 +... + 2n+1 ) ( n + 2n+1 ) ( n+1 +... + 2n+1) 2 2n+1 = ( ) ( 2n+1 0 + 2n+1 ) ( 1 +... + 2n+1 ) ( n + 2n+1 ) ( n +... + 2n+1 ) [ 0 (2n+1 2 2n+1 ) ( = 2 0 + 2n+1 ) ( 1 +... + 2n+1 ) ] n [ (2n+1 ) ( 0 + 2n+1 ) ( 1 +... + 2n+1 ) ] n = 2 2n ) ( = n n k) GiatÐ; 'Eqw k Ðdia kapèla kai jèlw na ta d sw se n ˆtoma. EÐte dialèxw ta k ˆtoma sta opoða ja d sw kapèlo eðte ta n k sta opoða de ja d sw eðnai to Ðdio. ( n k

Ta prohgoômena aforoôn sta maj mata 12 kai 15/10/2010

Stoiqei dhc sunduastik Dianom antikeimènwn se upodoqèc

Dianom Antikeimènwn se Upodoqèc Me pìsouc trìpouc mporoôme na dianeðmoume r antikeðmena (diakekrimèna ìqi) se n upodoqèc. DiakrÐnoume peript seic: 1. Τα αντικείμενα είναι διαϕορετικά (διακεκριμένα) και η σειρά στις υποδοχές δε μετράει. } n n {{... n } = n r r Για το πρώτο αντικείμενο έχουμε n επιλογές ως προς το που θα το τοποθετήσουμε. Ομοια για το δεύτερο, αϕού δεν απαγορεύεται να ρίξουμε δυο αντικείμενα στην ίδια υποδοχή, κοκ. Άρα συνολικά μπορώ να τοποθετήσω τα r αντικείμενα στις n υποδοχές με n r τρόπους.

Dianom Antikeimènwn se Upodoqèc 2. Ta antikeðmena eðnai diaforetikˆ (diakekrimèna) kai h seirˆ se kˆje upodoq metrˆei. (n + r 1)! (n 1)! Arqikˆ diatˆssw ta r antikeðmena. Ja prosomoi sw tic n upodoqèc me (n + 1) kˆjetec grammèc pou ja topojet sw anˆmesa sta diatetagmèna antikeðmena. AntikeÐmena pou brðskontai metaxô thc i-st c kai thc (i + 1)-st c gramm c ja jewroôme ìti brðskontai sthn i-st upodoq. α 1 α 3 α 5 α 2 α 6... } {{ } n upodoqèc

Dianom Antikeimènwn se Upodoqèc 2. Ta antikeðmena eðnai diaforetikˆ (diakekrimèna) kai h seirˆ se kˆje upodoq metrˆei. (n + r 1)! (n 1)! α 1 α 3 α 5 α 2 α 6... } {{ } n upodoqèc Apì tic n + 1 grammèc pou topojèthsa, mìno oi n 1 orðzoun tic upodoqèc (mpor na agno sw tic akraðec). O sunolikìc arijmìc antikeimènwn pou èqw eðnai r + (n 1) (r arqikˆ antikeðmena kai n 1 grammèc), kai mpor na ta diatˆxw me (n + r 1)! trìpouc. 'Omwc ta n 1 antikeðmena (oi grammèc) eðnai Ðdia (den eðnai diakekrimèna), ˆra èqw mìno (n+r 1)! (n 1)! diaforetikoôc trìpouc.

Pìsoi eðnai oi diaforetikoð trìpoi na perˆsoun k (diaforetikˆ) autokðnhta apì n diaforetikoôc upall louc diodðwn ìtan paðzei rìlo h seirˆ me thn opoða kˆje upˆllhloc exuphreteð ta autokðnhta; To pr to autokðnhto èqei n epilogèc ìsoi kai oi upˆllhloi. To deôtero autokðnhto èqei n + 1 epilogèc: an pˆei ston upˆllhlo pou p ge kai to pr to autokðnhto na pˆei prðn metˆ apì autì. To trðto autokðnhto èqei n + 2 epilogèc: an pˆei ston upˆllhlo pou p ge kai to pr to autokðnhto na pˆei prðn metˆ apì autì, an pˆei ston upˆllhlo pou p ge kai to deôtero autokðnhto na pˆei prðn metˆ apì autì. To k autokðnhto èqei n + k 1 epilogèc: ìmoia ìpwc prohgoumènwc. 'Ara sunolikˆ (apì kanìna ginomènou) upˆrqoun (n + k 1)! n(n + 1)...(n + k 1) = diaforetikoð trìpoi. (n 1)!

Dianom Antikeimènwn se Upodoqèc 3. Ta antikeðmena eðnai Ðdia (den eðnai diakekrimèna). ( ) n + r 1 'Opwc sthn prohgoômenh perðptwsh, mìno pou aut th forˆ kai ta r antikeðmena eðnai Ðdia (den eðnai diakekrimèna), epomènwc èqoume mìno (n+r 1)! r!(n 1)! diaforetikoôc trìpouc na ta diatˆxoume. r = ( ) n+r 1 r

Dianom antikeimènwn se upodoqèc: me pìsouc trìpouc mporoôme na topojet soume r diaforetikˆ (diakekrimèna) antikeðmena se n diaforetikèc (diakekrimènec) upodoqèc ìtan de metrˆei h seirˆ Gia to antikeðmeno 1 èqoume n trìpouc (dhl. mporoôme na to topojet soume se kˆje mða apì tic n diajèsimec upodoqèc) Gia to antikeðmeno 2 èqoume n trìpouc (dhl. mporoôme na to topojet soume se kˆje mða apì tic n diajèsimec upodoqèc)... Gia to antikeðmeno r èqoume n trìpouc (dhl. mporoôme na to topojet soume se kˆje mða apì tic n diajèsimec upodoqèc) 'Ara, sunolikˆ n } {{... n} = n r r forèc

Dianom antikeimènwn se upodoqèc: me pìsouc trìpouc mporoôme na topojet soume r diaforetikˆ (diakekrimèna) antikeðmena se n diaforetikèc (diakekrimènec) upodoqèc ìtan metrˆei h seirˆ 1oc trìpoc Οι n υποδοχές είναι τα διαστήματα που ορίζονται από n + 1 σημεία πάνω σε μια ευθεία. Το πρώτο από αυτά τα σημεία μπαίνει πάντα στην αρχή και το τελευταίο πάντα στο τέλος, επομένως υπάρχουν n 1 ίδια (μη διακεκριμένα) σημεία που πρέπει να διαχωρίσουν r διαϕορετικά (διακεκριμένα) αντικείμενα Επομένως, ζητάμε τις διατάξεις r + n 1 αντικειμένων από τα οποία τα n 1 είναι ίδια (μη διακεκριμένα). Αυτός είναι: (r+n 1)! (n 1)!

Dianom antikeimènwn se upodoqèc: me pìsouc trìpouc mporoôme na topojet soume r diaforetikˆ (diakekrimèna) antikeðmena se n diaforetikèc (diakekrimènec) upodoqèc ìtan metrˆei h seirˆ 2oc trìpoc Για το 1ο από τα r αντικείμενα, υπάρχουν n τρόποι τοποθέτησης σε κάθε μία από τις n διαθέσιμες υποδοχές. Αϕού επιλεγεί μία, αυτή χωρίζεται στα δύο. Για το 2ο από τα r αντικείμενα, υπάρχουν n + 1 τρόποι τοποθέτησης σε κάθε μία από τις n διαθέσιμες υποδοχές και τη 1 νέα. Αϕού επιλεγεί μία, αυτή χωρίζεται στα δύο. Για το 3ο από τα r αντικείμενα, υπάρχουν n + 2 τρόποι τοποθέτησης σε κάθε μία από τις n διαθέσιμες υποδοχές και τις 2 νέες. Αϕού επιλεγεί μία, αυτή χωρίζεται στα δύο, κοκ Για το r-ό από τα r αντικείμενα, υπάρχουν n + r 1 τρόποι τοποθέτησης σε κάθε μία από τις n διαθέσιμες υποδοχές και τις r 1 νέες. Άρα, συνολικά: n (n + 1) (n + 2)... (n + r 1) τρόποι

Dianom antikeimènwn se upodoqèc: me pìsouc trìpouc mporoôme na topojet soume r Ðdia (mh diakekrimèna) antikeðmena se n diaforetikèc (diakekrimènec) upodoqèc ìtan metrˆei h seirˆ 1oc trìpoc Με πόσους τρόπους μπορούμε να τοποθετήσουμε r διαϕορετικά (διακεκριμένα) αντικείμενα σε n διαϕορετικές (διακεκριμένες) υποδοχές όταν μετράει η σειρά; (r+n 1)! (n 1)! Αλλά τώρα τα r αντικείμενα είναι ίδια (μη διακεκριμένα) οι ζητούμενοι τρόποι είναι (r+n 1)! (n 1)!r!

Dianom antikeimènwn se upodoqèc: me pìsouc trìpouc mporoôme na topojet soume r Ðdia (mh diakekrimèna) antikeðmena se n diaforetikèc (diakekrimènec) upodoqèc ìtan metrˆei h seirˆ 2oc trìpoc Οπως πριν με τα διαστήματα στην ευθεία, μόνο που τώρα τόσο η ομάδα που περιέχει τα n 1 σημεία όσο και αυτή που περιέχει τα r αντικείμενα περιέχουν ίδια (μη διακεκριμένα) στοιχεία: (r+n 1)! (n 1)!r!

'Estw ìti topojetoôme me kˆpoion trìpo tic Ðdiec mpˆlec sta diaforetikˆ koutiˆ. Den allˆzei kˆti an eujugrammðsoume tic mpˆlec. Den allˆzei kˆti an sb soume kˆpoia ìria apì ta koutiˆ. To p c topojet same tic mpˆlec sta koutiˆ mporeð na parastajeð me mia akoloujða apì 0 kai 1: 0001001100

An upˆrqoun r mpˆlec kai n koutiˆ tìte prèpei na topojet soume r 0 kai n 1 1 se n + r 1 jèseic. Autì gðnetai me C(r + n 1, r) trìpouc.

SÔntomh epanˆlhyh: Me pìsouc trìpouc mporoôme na dianeðmoume r antikeðmena se n upodoqèc; Diaforetikˆ antikeðmena: Δε μετράει η σειρά: } n n {{... n } = n r r (n + r 1)! Μετράει η σειρά: (n 1)! ( ) (n + r 1)! n + r 1 Idia antikeðmena: = r!(n 1)! r

ParadeÐgmata Me pìsouc trìpouc mporoôme na topojet soume r Ðdiec mpˆlec se n diaforetikˆ koutiˆ an kˆje koutð qwrˆei mða mìno mpˆla; C(n, r) Pìsec duadikèc akoloujðec 32 yhfðwn èqoun akrib c 7 ˆssouc; C(32, 7)

Me pìsouc trìpouc r Ðdiec mpˆlec mporoôn na topojethjoôn se n diaforetikˆ koutiˆ ètsi ste kˆje koutð na perièqei toulˆqiston 9 mpˆlec; toulˆqiston 9 mpˆlec se kˆje koutð: 9n mpˆlec. Mènoun r 9n Ðdiec mpˆlec tic opoðec mporoôme na topojet soume sta n diaforetikˆ koutiˆ me ìlouc touc dunatoôc trìpouc (r 9n+n 1)! (n 1)!(r 9n)! = ( ) r 9n+n 1 r 9n

Me pìsouc trìpouc r Ðdia antikeðmena mporoôn na topojethjoôn se n upodoqèc ètsi ste ìlec oi upodoqèc na pˆroun toulˆqiston 1 antikeðmeno; TopojetoÔme 1 antikeðmeno se kˆje mða apì tic n upodoqèc gia na mh meðnei kamða ˆdeia. Mènoun r n antikeðmena. 'Eqoume r n Ðdia antkeðmena ta opoða mporoôme na topojet soume stic n ( ) n+r n 1 upodoqèc me r n trìpouc

'Eqw 7 a, 8 b, 5 g kai 4d. Pìsec sumboloseirèc mpor na ftiˆxw an den prèpei na emfanðzetai to ga se kamða apì autèc; Mìno me ta 7 a, ta 8 b kai ta 4d mpor na ftiˆxw 19! 7!8!4! diatˆxeic qwrðc kanènan periorismì. Gia kˆje mða apì tic parapˆnw diatˆxeic, jewr ìti ta 5 g eðnai 5 Ðdia antikeðmena pou topojetoôntai se diaforetikèc upodoqèc pou sqhmatðzontai apì thn upˆrqousa diˆtaxh. 'Eqoume sunolikˆ 19 grˆmmata pou sqhmatðzoun 20 diaforetikèc upodoqèc: mða prin apì kˆje grˆmma kai mða sto tèloc. To g de mporeð na topojethjeð prin apì a giatð ja sqhmatisteð ga. Opìte mènoun 13 upodoqèc gia na topojethjeð to g: prin apì kˆpoio b, prin apì kˆpoio d sto tèloc. ( Oi diaforetikoð ) trìpoi na topojethjoôn ta 5 g 13 + 5 1 eðnai: = 17! 5 5!12!. 19! 'Ara sunolikˆ: 7!8!4! + 17! 5!12!

'Eqw n jèseic sth seirˆ kai jèlw na topojet sw k foithtèc gia na grˆyoun exetˆseic ste metaxô kˆje dôo foitht n na upˆrqei mða ken jèsh (n 2k 1). Me pìsouc diaforetikoôc trìpouc mpor na to kˆnw; Bgˆzw foithtèc kai jranða apì thn aðjousa. DÐnw èna jranðo se kˆje foitht (1 trìpoc upˆrqei afoô ta jranða eðnai Ðdia, desmeôw k). Diatˆssw touc foithtèc me ta jranða touc: upˆrqoun k! diaforetikoð trìpoi afoô oi foithtèc eðnai diaforetikoð. Topojet èna ˆdeio jranðo anˆmesa se kˆje zeugˆri foitht n (1 trìpoc afoô ta jranða eðnai Ðdia, desmeôw k 1 jranða). Moirˆzw ta n 2k + 1 Ðdia jranða pou perðsseyan se ( k + 1 diaforetikèc upodoqèc. ) ( Autì gðnetai ) me k + 1 + n 2k + 1 1 n k + 1 = n 2k + 1 n 2k + 1 ( ) n k + 1 'Ara sunolikˆ upˆrqoun k! diaforetikoð trìpoi. n 2k + 1

SÔntomh epanˆlhyh: Me pìsouc trìpouc mporoôme na dianeðmoume r antikeðmena se n upodoqèc; Diaforetikˆ antikeðmena: Δε μετράει η σειρά: } n n {{... n } = n r r (n + r 1)! Μετράει η σειρά: (n 1)! ( ) (n + r 1)! n + r 1 Idia antikeðmena: = r!(n 1)! r

Jèloume na steðloume èna m numa me 12 diaforetikˆ sômbola pˆnw apì èna thlepikoinwniakì kanˆli. Epiplèon twn sumbìlwn jèloume na steðloume kai 45 kenˆ metaxô twn sumbìlwn, me toulˆqiston 3 kenˆ metaxô diaforetik n sumbìlwn. Me pìsouc diaforetikoôc trìpouc mporoôme na metad soume to m numa; DiaforetikoÐ trìpoi gia na metad sw mìno ta 12 diaforetikˆ sômbola: 12! Gia kajènan apì autoôc touc trìpouc, sqhmatðzontai 11 jèseic metaxô twn 12 sumbìlwn, se kˆje mða apì tic opoðec prèpei na bˆloume toulˆqiston 3 kenˆ: ˆra qrhsimopoioôme ˆllouc 33 apì touc 45 kenoôc qarakt rec. Mènoun 12 kenˆ. 12 kenˆ pou mènoun 12 Ðdia antikeðmena pou jèloume na topojet soume se 11 diaforetikˆ koutiˆ ( tic jèseic pou ) ( 11+12 1 eðpame prin. Autì mporeð na gðnei me = 22 ) diaforetikoôc trìpouc. 12 12

Jèloume na steðloume èna m numa me 12 diaforetikˆ sômbola pˆnw apì èna thlepikoinwniakì kanˆli. Epiplèon twn sumbìlwn jèloume na steðloume kai 45 kenˆ metaxô twn sumbìlwn, me toulˆqiston 3 kenˆ metaxô diaforetik n sumbìlwn. Me pìsouc diaforetikoôc trìpouc mporoôme na metad soume to m numa; Epomènwc ( ) mporoôme na metad soume to m numa me 22 12! = 3.097 10 14 diaforetikoôc trìpouc. 12

Me pìsouc trìpouc mporoôme na topojet soume r diaforetikèc shmaðec se n diaforetikoôc istoôc me dedomèna ìti (a) èqei shmasða h seirˆ me thn opoða emfanðzontai oi shmaðec stouc istoôc kai (b) kanènac istìc den prèpei na meðnei ˆdeioc (r n); Gia na isqôei to (b): r! P(r, n) = r(r 1)...(r n + 1) = (r n)! trìpoi Gia na isqôei kai to (a): Epomènwc, sunolikˆ: (r n + n 1)! (n 1)! = r! (r 1)! (r n)! (n 1)! trìpoi. (r 1)! (n 1)! trìpoi

Ta prohgoômena aforoôn sto mˆjhma 22/10/2010

Me pìsouc trìpouc mporoôme na topojet soume 4 Ðdia portokˆlia kai 6 diaforetikˆ m la se 5 diaforetikˆ koutiˆ; Se poio posostì aut n twn trìpwn topojetoôntai 2 akrib c froôta se kˆje koutð; Ta 4 Ðdia portokˆlia mporoôme na ta topojet soume sta (4 + 5 1)! 5 koutiˆ me = 8! = 70 diaforetikoôc 4!(5 1)! 4!4! trìpouc. Ta 6 diaforetikˆ m la mporoôme na ta topojet soume sta 5 koutiˆ (qwrðc na metrˆei h seirˆ) me 5 6 = 15.625 diaforetikoôc trìpouc. Epomènwc, sunolikˆ èqoume 70 15.625 = 1.093.750 diaforetikoôc trìpouc.

Me pìsouc trìpouc mporoôme na topojet soume 4 Ðdia portokˆlia kai 6 diaforetikˆ m la se 5 diaforetikˆ koutiˆ; Se poio posostì aut n twn trìpwn topojetoôntai 2 akrib c froôta se kˆje koutð; Gia na èqoume 2 akrib c froôta se kˆje koutð, asqoloômaste pr ta me ta 4 portokˆlia pou eðnai Ðdia: Bˆzoume apì 2 portokˆlia se 2 apì ta pènte koutiˆ. Pìsoi diaforetikoð trìpoi na dialèxw ta koutiˆ autˆ upˆrqoun; C(5, 2) = 5! 2!3! = 10. Ta 6 diaforetikˆ m la mpor na ta qwrðsw se 3 dimeleðc omˆdec (mða gia kajèna apì ta upìloipa 3 koutiˆ) me 6! 2!2!2! = 90 trìpouc. 'Ara sunolikˆ 10 90 = 900 trìpoi.

Me pìsouc trìpouc mporoôme na topojet soume 4 Ðdia portokˆlia kai 6 diaforetikˆ m la se 5 diaforetikˆ koutiˆ; Se poio posostì aut n twn trìpwn topojetoôntai 2 akrib c froôta se kˆje koutð; Bˆzoume 2 portokˆlia se 1 koutð me C(5, 1) = 5! 1!4! = 5 trìpouc, kai apì èna portokˆli se 2 apì ta pènte koutiˆ me C(4, 2) = 4! 2!2! = 6 trìpouc. Ta 6 diaforetikˆ m la mpor na ta qwrðsw se 2 dimeleðc omˆdec (mða gia kajèna apì ta upìloipa 2 koutiˆ) me 6! 2!2! = 180 trìpouc. 'Ara sunolikˆ 5 6 180 = 5.400 trìpoi.

Me pìsouc trìpouc mporoôme na topojet soume 4 Ðdia portokˆlia kai 6 diaforetikˆ m la se 5 diaforetikˆ koutiˆ; Se poio posostì aut n twn trìpwn topojetoôntai 2 akrib c froôta se kˆje koutð; Bˆzoume apì 1 portokˆlia se 4 koutiˆ me C(5, 4) = 5! 4!1! = 5 trìpouc. Ta 6 diaforetikˆ m la mpor na ta qwrðsw se zeugˆria (gia to koutð pou mènei) me 6! 2! = 360 trìpouc. 'Ara sunolikˆ 5 360 = 1.800 trìpoi. Epomènwc, sunolikˆ 900 + 5.400 + 1.800 = 8.100 trìpoi. Opìte to zhtoômeno posostì eðnai 8.100 100 1.093.750 % = 7.4%

(a) Pìsec akèraiec lôseic thc exðswshc x 1 + x 2 + x 3 + x 4 = 12 me x i 0 upˆrqoun; (b) Pìsec me x i > 0; (g) Pìsec me x 1 2, x 2 2, x 3 4, x 4 0; (a) To prìblhma eðnai Ðdio me thn eôresh twn diaforetik n trìpwn topojèthshc 12 Ðdiwn antikeimènwn (twn monˆdwn) se 4 diaforetikˆ koutiˆ (touc ìrouc thc exðswshc). (12 + 4 1)! Epomènwc upˆrqoun = 455 trìpoi, dhl., 12!(4 1)! diaforetikèc lôseic.

(a) Pìsec akèraiec lôseic thc exðswshc x 1 + x 2 + x 3 + x 4 = 12 me x i 0 upˆrqoun; (b) Pìsec me x i > 0; (g) Pìsec me x 1 2, x 2 2, x 3 4, x 4 0; (b) To prìblhma eðnai Ðdio me thn eôresh twn diaforetik n trìpwn topojèthshc 12 Ðdiwn antikeimènwn (twn monˆdwn) se 4 diaforetikˆ koutiˆ (touc ìrouc thc exðswshc) allˆ kˆnena koutð na mhn eðnai ˆdeio. Epomènwc upˆrqoun (8 + 4 1)! = 165 trìpoi, dhl., diaforetikèc lôseic. 8!(4 1)!

(a) Pìsec akèraiec lôseic thc exðswshc x 1 + x 2 + x 3 + x 4 = 12 me x i 0 upˆrqoun; (b) Pìsec me x i > 0; (g) Pìsec me x 1 2, x 2 2, x 3 4, x 4 0; (g) To prìblhma eðnai Ðdio me thn eôresh twn diaforetik n trìpwn topojèthshc 12 Ðdiwn antikeimènwn (twn monˆdwn) se 4 diaforetikˆ koutiˆ (touc ìrouc thc exðswshc) me touc dosmènouc periorismoôc. Epomènwc upˆrqoun (4 + 4 1)! = 35 trìpoi, dhl., diaforetikèc lôseic. 4!(4 1)!

(a) Pìsec mh arnhtikèc akèraiec lôseic èqei h exðswsh x 1 + x 2 +... + x 6 = 10; (b) Pìsec mh arnhtikèc akèraiec lôseic èqei h anðswsh x 1 + x 2 +... + x 6 < 10; (a) To prìblhma eðnai Ðdio me thn eôresh twn diaforetik n trìpwn topojèthshc 10 Ðdiwn antikeimènwn (twn monˆdwn) se 6 diaforetikˆ koutiˆ (touc ìrouc thc exðswshc). Epomènwc upˆrqoun (6+10 1)! 10!(6 1)! = 3003 trìpoi, dhl., diaforetikèc lôseic. (b) AntÐ na asqolhj me thn anðswsh x 1 + x 2 +... + x 6 < 10 yˆqnw isodônama tic lôseic thc exðswshc: x 1 + x 2 +... + x 6 + x 7 = 10 me touc periorismoôc: x i 0, 0 i 6 kai x 7 > 0. To prìblhma eðnai Ðdio me thn eôresh twn diaforetik n trìpwn topojèthshc 10 Ðdiwn antikeimènwn (twn monˆdwn) se 7 diaforetikˆ koutiˆ, allˆ 1 koutð prèpei pˆnta na mhn eðnai ˆdeio. Epomènwc = 5005 trìpoi, dhl., diaforetikèc lôseic. upˆrqoun (7+9 1)! 9!(7 1)!

Se èna magazð ta antikeðmena A, B kai G kostðzoun apì 5 eur kai to antikeðmeno D 20 eur. An jèlw na xodèyw sunolikˆ 100 eur pìsec diaforetikèc agorèc mpor na kˆnw; To mikrìtero posì eðnai ta 5 eur kai ta ˆlla eðnai pollaplˆsiˆ tou. Opìte jewr autì wc monˆda kai mpor na diatup sw thn parapˆnw er thsh wc: A + B + Γ + 4 = 20 me A, B, Γ, 0. Epeid to D èqei suntelest 4, prèpei na poôme rhtˆ pìsa antikeðmena tôpou D agorˆzoume. 'Estw i to pl joc touc. Opìte h exðswsh gðnetai: A + B + Γ + 4i = 20 A + B + Γ = 20 4i me A, B, Γ, 0 kai 0 i 5.

Se èna magazð ta antikeðmena A, B kai G kostðzoun apì 5 eur kai to antikeðmeno D 20 eur. An jèlw na xodèyw sunolikˆ 100 eur pìsec diaforetikèc agorèc mpor na kˆnw; To prìblhma eðnai Ðdio me thn eôresh twn diaforetik n trìpwn topojèthshc 20 4i Ðdiwn antikeimènwn (twn monˆdwn) se 3 diaforetikˆ koutiˆ, gia kˆje tim 0 i 5. 5 (20 4i + 3 1)! Epomènwc upˆrqoun (20 4i)!(3 1)! = 5 (20 4i + 3 1)! (20 4i)!2! i=0 diaforetikèc lôseic. i=0 = 5 i=0 (11 2i)! = 536 trìpoi, dhl., (20 4i)!

Me pìsouc trìpouc mporoôn na dianemhjoôn 2n + 1 jèseic enìc sunedriakoô kèntrou se 3 omˆdec, ste o sunaspismìc opoiwnd pote dôo omˆdwn na touc exasfalðsei pleioyhfða; Tìso oi jèseic ìso kai oi sônedroi thc kˆje omˆdac de diakrðnontai metaxô touc DÐnetai ìti: tìso oi jèseic ìso kai oi sônedroi thc kˆje omˆdac de diakrðnontai metaxô touc. Epomènwc, dianomèc twn jèsewn pou prokôptoun h mða apì thn ˆllh me antallag jèsewn sunèdrwn metaxô touc, pros pwn sunèdrwn, qwrðc allag tou arijmoô twn sunèdrwn thc kˆje omˆdac, de metrˆne wc diaforetikèc. 'Etsi, to prìblhma metatrèpetai se prìblhma topojèthshc mh diakekrimènwn (dhl., Ðdiwn) antikeimènwn se diakekrimènec upodoqèc (tic diaforetikèc omadec).

Me pìsouc trìpouc mporoôn na dianemhjoôn 2n + 1 jèseic enìc sunedriakoô kèntrou se 3 omˆdec, ste o sunaspismìc opoiwnd pote dôo omˆdwn na touc exasfalðsei pleioyhfða; Tìso oi jèseic ìso kai oi sônedroi thc kˆje omˆdac de diakrðnontai metaxô touc Idèa: An mða omˆda èqei n + 1 jèseic, tìte oi dôo ˆllec de mporoôn potè na èqoun pleioyhfða. 'Ara gia na apant soume sthn er thsh prèpei 1. να βρούμε με πόσους τρόπους μπορούν να διανεμηθούν 2n + 1 θέσεις σε 3 ομάδες χωρίς περιορισμό, και Trìpoi topojèthshc ( ) 2n ( + 1 Ðdiwn antikeimènwn se 3 2n+1+3 1 upodoqèc: 2n+1 = 2n+3 ) 2n+1 2. να αϕαιρέσουμε από αυτούς εκείνους τους τρόπους που δίνουν τουλάχιστον n + 1 θέσεις σε μία μόνο ομάδα DÐnoume n + 1 jèseic se mða apì tic omˆdec opìte mènoun gia dianom 2n + 1 n 1 = n jèseic ( Trìpoi topojèthshc n Ðdiwn antikeimènwn se 3 upodoqèc: n+3 1 ) ( n = n+2 ) n Sunolikˆ, afoô upˆrqoun 3 omˆdec: 3 (n+2 ) n

Me pìsouc trìpouc mporoôn na dianemhjoôn 2n + 1 jèseic enìc sunedriakoô kèntrou se 3 omˆdec, ste o sunaspismìc opoiwnd pote dôo omˆdwn na touc exasfalðsei pleioyhfða; Tìso oi jèseic ìso kai oi sônedroi thc kˆje omˆdac de diakrðnontai metaxô touc Opìte oi zhtoômenoi trìpoi eðnai: ( 2n+3 ) ( 2 3 n+2 2 (n + 1) ) = n 2

SÔnoyh-Praktikˆ (1) 'Eqw 10 yhfða (diakekrimèna antikeðmena). Mpor na ftiˆxw 10 9 8 7 diaforetikèc 4-ˆdec (diatˆxeic 4 apì 10 antikeimènwn = P(10, 4)). PARATHRHSH: h tetrˆda 1,2,3,4 eðnai diaforetik apì thn tetrˆda 4,3,2,1 Mpor na dialèxw 10 9 8 7 diaforetikèc 4-ˆdec 4! (sunduasmoð 4 apì 10 antikeimènwn = C(10, 4) = P(10,4) P(4,4) ). PARATHRHSH: h tetrˆda 1,2,3,4 metrˆei 1 forˆ kai den eðnai diaforetik apì thn tetrˆda 4,3,2,1

SÔnoyh-Praktikˆ (2) 'Eqw 100 mpˆlec (mh diakekrimèna antikeðmena). Tic qwrðzw se omˆdec apì prˆsinec, kìkkinec kai mple mpˆlec. Oi prˆsinec mpˆlec eðnai 60, oi kìkkinec 30 kai oi mple 10. 100! Upˆrqoun trìpoi na diatˆxw tic 100 mpˆlec. 60!30!10! 'Eqw 2 diaforetikˆ duadikˆ yhfða (diakekrimèna antikeðmena): 0 kai 1. 'Eqw ìsa jèlw apì autˆ (epitrèpontai oi epanal yeic). Mpor na ftiˆxw 2 5 diaforetikèc 5-ˆdec. 'Eqw 2 zˆria allˆ de me endiafèrei h seirˆ pou pèftoun (sunduasmoð). Kˆje zˆri èqei 6 pijanèc timèc kai mporeð na fèrnei thn Ðdia tim suneq c (epitrèpontai epanal yeic). Upˆrqoun sunolikˆ C(6 + 2 1, 2) zarièc.

SÔnoyh-Praktikˆ (3) 'Eqw n antikeðmena, èna apì to kajèna kai de me noiˆzei h seirˆ. Upˆrqoun 2 n 1 trìpoi gia na epilèxw èna perissìtera apì autˆ. 2: επιλέγω/δεν επιλέγω κάποιο αντικείμενο -1: πρέπει να επιλέξω τουλάχιστον ένα 'Eqw t omˆdec mh diakekrimènwn antikeimènwn, kˆje mða me mègejoc q 1, q 2,..., q t, antðstoiqa. Μπορώ να διαλέξω ένα ή περισσότερα αντικείμενα με (q 1 + 1)(q 2 + 1)...(q t + 1) 1 τρόπους.

SÔnoyh-Praktikˆ (4) n diakekrimènec upodoqèc, r diakekrimèna antikeðmena, de metrˆei h seirˆ stic upodoqèc: n r trìpoi (mporeð na upˆrqoun kai ˆdeiec upodoqèc) n diakekrimènec upodoqèc, r diakekrimèna antikeðmena, (n + r 1)! metrˆei h seirˆ stic upodoqèc: trìpoi (mporeð (n 1)! na upˆrqoun kai ˆdeiec upodoqèc) ( n diakekrimènec ) upodoqèc, r mh diakekrimèna antikeðmena: n + r 1 (n + r 1)! = trìpoi (mporeð na upˆrqoun r (n 1)!r! kai ˆdeiec upodoqèc)

Stoiqei dhc sunduastik DiwnumikoÐ Suntelestèc

DiwnumikoÐ Suntelestèc (1 + x) n = di numo { }} { (1 + x) (1 + x)... (1 + x) = } {{ } n n k=0 ( ) n x k k Gia ton suntelest tou x k : Apì kˆje di numo mporoôme na pˆroume èna x gia na sqhmatðsoume to x k, mporeð kai ìqi. Epomènwc èqoume n eukairðec (ìsa kai ta di numa) na pˆroume k antikeðmena x. O arijmìc twn trìpwn pou mporoôme na pˆroume k antikeðmena apì n qwrðc na mac endiafèrei h seirˆ, eðnai ( n ) k. Genikˆ: (x + y) n = n k=0 ( ) n x k y n k k

Idiìthtec Diwnumik n Suntelest n ( ) ( ) n n 1. = k n k Oi trìpoi pou mpor na dialèxw k antikeðmena apì n eðnai Ðsoi me touc trìpouc pou mpor na dialèxw ta (upìloipa) n k antikeðmena apì ta n. ( ) ( ) ( ) n + 1 n n 2. = + k + 1 k + 1 k 'Estw ìti xeqwrðzw èna antikeðmeno apì ta n + 1. An pˆrw k + 1 antikeðmena sunolikˆ, mpor na sumperilˆbw kai autì pou xeq risa ìqi. Sthn pr th perðptwsh prèpei na pˆrw ta upìloipa k antikeðmena apì ta (upìloipa) n, en sthn deôterh, prèpei na pˆrw kai ta k + 1 antikeðmena apì ta n. 3. ( ) n + r + 1 = r r ( ) n + k k=0 k

Idiìthtec Diwnumik n Suntelest n 4. ( ) n + 1 = r + 1 ( ) n n k=r ( ) k r ) 5. = n ( n 1 r r r 1 Gia na dialèxw r antikeðmena apì n arkeð na dialèxw pr ta èna antikeðmeno, kai ta upìloipa r 1 antikeðmena na ta dialèxw apì ta upìloipa n 1. To pr to antikeðmeno mpor na to epilèxw me n trìpouc, kai ta upìloipa r 1 ( n 1 me r 1) trìpouc. Epeid ìmwc den èqei shmasða poio apì ta r antikeðmena eðnai pr to (dhlad ja mporouse na eðnai pr to opoiod pote apì ta upìloipa r 1 antikeðmena pou epelèghsan ston sunduasmì, qwrðc na prokôptei diaforetikìc sunduasmìc), diair me r.

Idiìthtec C(n, r) = C(n 1, r) + C(n 1, r 1) 'Enac sunduasmìc mporeð eðte na perièqei to n-stì stoiqeðo eðte ìqi. Upˆrqoun C(n 1, r) sunduasmoð pou den perièqoun to n-stì stoiqeðo kai C(n 1, r 1) sunduasmoð pou to perièqoun. C(n, r) = n r C(n 1, r 1) Kˆje stoiqeðo metèqei se C(n 1, r 1) sunduasmoôc twn n anˆ r. Sunolikèc summetoqèc gia ta n stoiqeða: n C(n 1, r 1) Oi sunolikèc summetoqèc stoiqeðwn eðnai r C(n, r).

DiwnumikoÐ suntelestèc (1 + x): di numo (1 + x) n : polu numo bajmoô n O suntelest c tou x r sto polu numo isoôtai me to pl joc twn trìpwn na epilèxoume r apì touc n ìrouc x pou emfanðzontai sto ginìmeno (1 + x)...(1 + x) } {{ } n forèc Epomènwc, suntelest c tou x r sto polu numo (1 + x) n = C(n, r) anˆptugma diwnômou tou NeÔtwna: (1 + x) n = Genikˆ: (s + t) n = n C(n, r)s r t n r r=0 n C(n, r)x r r=0

Poio eðnai o stajerìc ìroc sto anˆptugma tou (x 2 + 1 x )12 ; IsqÔei (s + t) n = n C(n, r)s r t n r r=0 Gia s = x 2 kai t = 1 x = x 1 èqoume: 12 (x 2 + 1 x )12 = (x 2 + x 1 ) 12 = C(n, r)x 2r x 12+r = 12 r=0 C(12, r)x 3r 12 r=0 O stajerìc ìroc eðnai o sunetelest c tou x 0. Dhlad, prèpei 3r 12 = 0 r = 4. Epomènwc, o zhtoômenoc suntelest c eðnai o C(12, 4) = 495

Poioc eðnai o suntelest c tou x 40 sthn parˆstash (1 + x 4 + x 5 ) 100 ; Parathr ste ìti sthn exðswsh 4i + 5j = 40 oi akèraiec lôseic eðnai oi (i, j) {(0, 8), (5, 4), (10, 0)} IsqÔei (s + t) n = n C(n, r)s r t n r r=0 Gia s = x 5 kai t = 1 + x 4 èqoume: 100 ( ) 100 (1 + x 4 + x 5 ) 100 = x 5 r (1 + x 4 ) 100 r r r=0 100 r ( ) 100 r (1 + x 4 ) 100 j = x 4 k k k=0

100 ( 100 (1 + x 4 + x 5 ) 100 = r r=0 )x 5 r 100 r k=0 ( 100 r k ) x 4 k Gia na upologðsoume to suntelest tou x 40 prèpei na epilèxoume katˆllhla zeôgh tim n gia k kai r tètoia ste 4k + 5r = 40, me k kai r jetikoôc akèraiouc mikrìterouc tou 100. Autˆ dðnontai apì thn ekf nhsh: {(0, 8), (5, 4), (10, 0)} Opìte o suntelest c tou x 40 eðnai: ( 100 )( 92 ) ( 8 0 + 100 )( 96 ) ( 4 5 + 100 )( 100 ) 0 10

Gia kˆje fusikì arijmì n isqôei: n i=0 ( ) n = 2 n i n ( ) n Gia kˆje x isqôei: x i = (1 + x) n i i=0 n ( ) n Gia x = 1 h parapˆnw sqèsh dðnei: = (1 + 1) n = 2 n i i=0

Gia kˆje fusikì arijmì n isqôei: n i=0 ( ) n 2 i = 3 n i Gia kˆje x isqôei: i=0 n i=0 ( ) n x i = (1 + x) n i Gia x = 2 h parapˆnw sqèsh dðnei: n ( ) n 2 i = (1 + 2) n = 3 n i

Gia kˆje fusikì arijmì n isqôei: ( n ) ( 0 + n ) ( 2 + n ) ( 4 +... = n ) ( 1 + n 3 ) ( + n ) 5 +... Gia kˆje x isqôei: n i=0 ( ) n x i = (1 + x) n i Gia x = 1 h parapˆnw sqèsh dðnei: n i=0 ( n 0) ( n 1) + ( n 2) ( n 3) + ( n 4) ( n 5) +... = 0 ( n 0) + ( n 2) + ( n 4) +... = ( n 1) + ( n 3) + ( n 5) +... ( ) n ( 1) i = 0 i

Gia kˆje fusikì arijmì n isqôei: n i i=0 ( ) n = n2 n 1 i Gia kˆje x isqôei: n ( ) n x i = (1 + x) n i i=0 ParagwgÐzoume kai ta dôo mèlh thc isìthtac: n i i=0 ( n i ) x i 1 = n(1 + x) n 1 Jètoume x = 1 kai h parapˆnw sqèsh dðnei: n ( ) n i = n(1 + 1) n 1 = n2 n 1 i i=0

Gia kˆje fusikì arijmì n isqôei: ( ) ( n Epeid i = n n i i=0 ), eðnai n i=0 n i=0 ( ) n 2 = i ( ) 2 n = i n i=0 ( ) 2n n ( )( n n ) i n i ( ) 2n n shmaðnei na dialèxoume n apì 2n antikeðmena. Autì mporeð na gðnei wc ex c: Χωρίζουμε τα 2n αντικείμενα σε 2 ομάδες ώστε κάθε μία να έχει n αντικείμενα. Για i = 0, 1, 2, 3,..., n, διαλέγουμε i αντικείμενα από την πρώτη ομάδα με ( ) n i τρόπους και n i αντικείμενα από τη δεύτερη ομάδα με ( n n i) τρόπους. Για i = 0, 1, 2, 3,..., n, υπάρχουν ( )( n n i n i) τρόποι επιλογής. Τα ενδεχόμενα για διαϕορετικές τιμές του i είναι αμοιβαία αποκλειόμενα, οπότε με κανόνα αθροίσματος έχουμε: ( ) 2n n ( )( ) n n = n i n i

ApodeÐxte ìti: ( ) m + n = r r ( )( ) m n r k k ( ) k=0 m + n : me pìsouc trìpouc mpor na dialèxw r apì èna r sônolo me m arijmhmènec prˆsinec mpˆlec kai n arijmhmènec kìkkinec mpˆlec. EÐnai san na dialègw (1) r k apì tic m arijmhmènec prˆsinec mpˆlec kai (2) tic upìloipec k apì tic n arijmhmènec kìkkinec mpˆlec. ( ) m To (1) mporeð na gðnei me trìpouc kai to (2) me ( ) r k ( )( ) n m n trìpouc. 'Ara sunolikˆ upˆrqoun k r k k trìpoi gia kˆje tim tou k. Ta gegonìta eðnai amoibaða apokleiìmena gia diaforetikèc timèc tou k. Opìte apì kanìna ajroðsmatoc, upˆrqoun r ( )( ) m n sunolikˆ trìpoi. r k k k=0

ApodeÐxte ìti: ( ) m + n = r r k=0 ( )( ) m n r k k 'Amesh sunèpeia: ( ) 2n = n n k=0 ( )( ) n n = n k k n k=0 ( ) n 2 k

TrÐgwno tou Pascal Anadromikìc trìpoc upologismoô diwnumik n suntelest n. ( ) ( ) ( ) n n 1 n 1 +, an 0 < k < n = k 1 k k 1, diaforetikˆ.

H Ôlh sunoptikˆ... Stoiqei dhc sunduastik Genn triec sunart seic Egkleismìc - Apokleismìc JewrÐa Polyá

H Ôlh sunoptikˆ... Genn triec sunart seic

Ti eðnai h genn tria Sthn akoloujða twn arijm n α 0, α 1, α 2, α 3,... antistoiqeð h genn tria sunˆrthsh: A(z) = α 0 + α 1 z + α 2 z 2 + α 3 z 3 +... pou grˆfetai pio sôntoma: α n z n n=0 An h akoloujða eðnai peperasmènh (stamatˆei p.q., ston ìro α 10 ) tìte kai h genn tria èqei peperasmèno m koc: 10 α n z n n=0

Ti eðnai h genn tria 'Estw h akoloujða α 0 = 1, α 1 = 0, α 2 = 2, α 3 = 5 en gia n 4 eðnai α n = 0. H antðstoiqh genn tria sunˆrthsh eðnai h: 3 α n z n = 1 + 2z 2 + 5z 3 n=0

Qrhsimìthta gennhtri n sunart sewn Se orismèna polôploka probl mata eðnai dôskolo na doulèyoume me akoloujðec. Metatrèpoume tic akoloujðec se genn triec sunart seic (tic opoðec mporoôme na diaqeiristoôme eukolìtera giatð moiˆzoun me polu numa), katal goume se mia telik genn tria sunˆrthsh kai epistrèfoume sthn telik akoloujða pou dðnei lôsh sto prìblhmˆ mac.

Qr simec sqèseic Apì gewmetrikèc proìdouc ja qreiastoôme touc tôpouc: 1 + λ + λ 2 + λ 3 +... + λ n = 1 λn+1 1 λ en ìtan èqoume ˆpeirouc ìrouc: 1 + λ + λ 2 + λ 3 +... = 1, isqôei mìnon ìtan λ < 1 1 λ H ekjetik sunˆrthsh e x mporeð na grafeð kai wc ˆpeirh seirˆ wc: e x = 1 + x + x 2 2! + x 3 3! +...

Parˆdeigma Poia eðnai h genn tria sunˆrthsh thc stajer c akoloujðac 5, 5,...; Prìkeitai gia thn akoloujða α n = 5, gia kˆje n. Epomènwc h antðstoiqh genn tria sunˆrthsh eðnai: 5z n = 5 + 5z + 5z 2 + 5z 3 +... = 5(1 + z + z 2 + z 3 +...) = n=0 5 1 1 z = 5 1 z

Parˆdeigma Poia eðnai h genn tria sunˆrthsh thc stajer c akoloujðac α n = n + 1; Prìkeitai gia thn akoloujða α 0 = 1, α 1 = 2, α 2 = 3, α 3 = 4,... Epomènwc h antðstoiqh genn tria sunˆrthsh eðnai: 1 + 2z + 3z 2 + 4z 3 +... Gia na broôme kleistì tôpo, parathroôme ìti oi ìroi eðnai parˆgwgoi twn dunˆmewn tou z: z + (z 2 ) + (z 3 ) + (z 4 ) +... = (z + z 2 + z 3 + z 4 +...) = [ z(1 + z + z 2 + z 3 +...) ] [ ] 1 = z = 1 z z ( 1 z ) = 1 (1 z) 2

Genn triec sunart seic se probl mata sunduastik c Me pìsouc trìpouc mporoôme na dialèxoume n antikeðmena ìtan...; AfoÔ kˆnoume diergasðec (ja doôme sth sunèqeia p c) katal goume se mða genn tria sunˆrthsh α 0 + α 1 z + α 2 z 2 + α 3 z 3 +... kai h apˆnthsh brðsketai sto suntelest tou z n, dhl. mporoôme na dialèxoume n antikeðmena me α n trìpouc: 3 antikeðmena me α 3 trìpouc 100 antikeðmena me α 100 trìpouc... Dhl. de qreiˆzetai na lôsoume to prìblhma qwristˆ gia n = 3, gia n = 100 ktl: to lônoume mia kai kal kai oi suntelestèc α n dðnoun thn apˆnthsh gia kˆje n

Praktikˆ stic ask seic 'Otan epilègoume antikeðmena enìc sugkekrimènou eðdouc, antistoiqeð ènac parˆgontac pou eðnai kommˆti tou: z 0 + z 1 + z 2 + z 3 + z 4 +... Oi ekjètec deðqnoun tic epilogèc pou mporoôme na kˆnoume: An jèloume na epilèxoume apì 3 mèqri 6 antikeðmena: (z 3 + z 4 + z 5 + z 6 ) An jèloume na epilèxoume apì 0 eðte 1 eðte 5 eðte 7 antikeðmena: (z 0 + z 1 + z 5 + z 7 ) = (1 + z + z 5 + z 7 ) An jèloume na epilèxoume to polô 2 antikeðmena: (z 0 + z 1 + z 2 ) = (1 + z + z 2 ) An jèloume na epilèxoume toulˆqiston 5 antikeðmena: (z 5 + z 6 + z 7 +...)

Praktikˆ stic ask seic 'Otan epilègoume antikeðmena enìc sugkekrimènou eðdouc, antistoiqeð ènac parˆgontac pou eðnai kommˆti tou: z 0 + z 1 + z 2 + z 3 + z 4 +... Oi ekjètec deðqnoun tic epilogèc pou mporoôme na kˆnoume: An jèloume na epilèxoume perittì pl joc antikeimènwn, to polô 5: (z 1 + z 3 + z 5 ) = (z + z 3 + z 5 ) An den upˆrqei periorismìc, tìte antistoiqeð olìklhrh h parapˆnw parˆstash. Epanalambˆnoume th diadikasða gia kˆje eðdoc Ðdiwn antikeimènwn xeqwristˆ, kai sto tèloc pollaplasiˆzoume touc parˆgontec pou br kame.

'Eqoume 2 ˆsprec kai 3 maôrec mpˆlec. Jèloume na dialèxoume merikèc mpˆlec (de mˆc endiafèrei h seirˆ) allˆ me ton ex c periorismì: na dialèxoume upoqrewtikˆ toulˆqiston 2 maôrec mpˆlec Me pìsouc trìpouc mpor na dialèxw 0 mpˆlec; Me kanènan afoô prèpei na èqw toulˆqiston 2 maôrec mpˆlec. Me pìsouc trìpouc mpor na dialèxw 1 mpˆla; Me kanènan afoô prèpei na èqw toulˆqiston 2 maôrec mpˆlec. Me pìsouc trìpouc mpor na dialèxw 2 mpˆlec; Me 1: 2 maôrec - afoô prèpei na èqw toulˆqiston 2 maôrec mpˆlec.

'Eqoume 2 ˆsprec kai 3 maôrec mpˆlec. Jèloume na dialèxoume merikèc mpˆlec (de mˆc endiafèrei h seirˆ) allˆ me ton ex c periorismì: na dialèxoume upoqrewtikˆ toulˆqiston 2 maôrec mpˆlec Me pìsouc trìpouc mpor na dialèxw 3 mpˆlec; Me 2: 3 maôrec 1 ˆsprh kai 2 maôrec mpˆlec. Me pìsouc trìpouc mpor na dialèxw 4 mpˆlec; Me 2: 3 maôrec kai 1 ˆsprh 2 maôrec kai 2 ˆsprec. Me pìsouc trìpouc mpor na dialèxw 5 mpˆlec; Me 1: na tic epilèxw ìlec - 3 maôrec kai 2 ˆsprec. P c mporoôme na pˆroume ta Ðdia apotelèsmata me mia genn tria sunˆrthsh;

'Eqoume 2 ˆsprec kai 3 maôrec mpˆlec. Jèloume na dialèxoume merikèc mpˆlec (de mˆc endiafèrei h seirˆ) allˆ me ton ex c periorismì: na dialèxoume upoqrewtikˆ toulˆqiston 2 maôrec mpˆlec 'Eqoume 2 ˆsprec mpˆlec, ˆra sthn epilog twn ˆsprwn mpal n antistoiqeð o parˆgontac: (z 0 + z 1 + z 2 ) = (1 + z + z 2 ) 'Eqoume 3 maôrec mpˆlec kai prèpei upoqrewtikˆ na epilèxoume 2 maôrec mpˆlec, ˆra sthn epilog twn maôrwn mpal n antistoiqeð o parˆgontac: (z 2 + z 3 ) Epomènwc, h genn tria sunˆrthsh eðnai: A(x) = (1 + z + z 2 )(z 2 + z 3 ) = z 2 + 2z 3 + 2z 4 + z 5 = 0z 0 + 0z 1 + 1z 2 + 2z 3 + 2z 4 + 1z 5

Parat rhsh An se mia omˆda èqoume ˆpeira antikeðmena apì ta opoða mporoôme na dialèxoume, o antðstoiqoc parˆgontac eðnai kommˆti tou (1 + z + z 2 + z 3 +...) to opoðo an to doôme san ˆjroisma ìrwn gewmetrik c proìdou isoôtai me: 1 = (1 z) 1 1 z

'Eqoume 2 ˆsprec kai ˆpeirec maôrec mpˆlec. Me pìsouc trìpouc mporoôme na dialèxoume n mpˆlec ìtan dialègoume upoqrewtikˆ toulˆqiston 2 maôrec mpˆlec; Sthn epilog twn ˆsprwn mpal n antistoiqeð o parˆgontac (1 + z + z 2 ) Sthn epilog twn maôrwn mpal n, epeid jèloume toulˆqiston 2 maôrec mpˆlec, antistoiqeð o parˆgontac (z 2 + z 3 + z 4 +...) Epomènwc, h genn tria sunˆrthsh eðnai: A(x) = (1 + z + z 2 )(z 2 + z 3 + z 4 +...) = (z 2 + z 3 + z 4 +...) + (z 3 + z 4 + z 5 +...) + (z 4 + z 5 + z 6 +...) = z 2 + 2z 3 + 3z 4 + 3z 5 + 3z 6 +... Oi sunetelestèc deðqoun tic antðstoiqec apant seic.

'Eqoume 2 ˆsprec kai ˆpeirec maôrec mpˆlec. Me pìsouc trìpouc mporoôme na dialèxoume n mpˆlec ìtan dialègoume upoqrewtikˆ toulˆqiston 2 maôrec mpˆlec; z 2 + 2z 3 + 3z 4 + 3z 5 + 3z 6 +... Parˆdeigma: Me pìsouc trìpouc mpor na epilèxw 5 mpˆlec; H apˆnthsh brðsketai sto suntelesth tou z 5 : me 3 trìpouc. Prˆgmati, oi dunatèc epilogèc eðnai: 5 maôrec mpˆlec 4 maôrec kai 1 ˆsprh mpˆla 3 maôrec mpˆlec kai 2 ˆsprec mpˆlec

'Otan ta antikeðmena èqoun kˆpoio mègejoc (p.q., bˆroc, nomismatik axða, ktl) Parˆdeigma: 'Eqoume 2 ˆsprec twn 5 kil n kai ˆpeirec maôrec mpˆlec twn 10 kil n. Allˆzei kˆti sto er thma Me pìsouc trìpouc mpor na dialèxw 5 mpˆlec; OQI, afoô to bˆroc den paðzei kanèna rìlo, h apˆnthsh kai pˆli eðnai 3. 'Allo er thma Me pìsouc trìpouc mpor na dialèxw mpˆlec ste na sugkentr sw bˆroc 40 kil n; Praktikˆ, oi trìpoi eðnai 2: 5 + 5 + 10 + 10 + 10 kai 10 + 10 + 10 + 10 P c ja pèrname aut thn apˆnthsh me genn triec;

Me pìsouc trìpouc mpor na dialèxw mpˆlec ste na sugkentr sw bˆroc 40 kil n; Se kˆje parˆgonta thc genn triac sunˆrthshc oi dunˆmeic tou z ja deðqnoun pìsa kilˆ mporoôme na sugkentr soume: Sthn epilog twn 5-kilwn (ˆsprwn) mpal n antistoiqeð o parˆgontac (z 0 + z 5 + z 10 ) afoô me dôo 5-kilec mpˆlec mporoôme na sugkentr soume 0 5 10 kilˆ. Sthn epilog twn 10-kilwn (maôrwn) mpal n, epeid jèloume toulˆqiston 2 maôrec mpˆlec, antistoiqeð o parˆgontac (z 20 + z 30 + z 40 +...) Epomènwc, h genn tria sunˆrthsh eðnai: A(x) = (1 + z 5 + z 10 )(z 20 + z 30 + z 40 +...) = (z 20 +z 30 +z 40 +...)+(z 25 +z 35 +z 45 +...)+(z 30 +z 40 +z 50 +...) = z 20 + 2z 30 + z 35 + 2z 40 + z 45 + 2z 50 +... Prˆgmati, h sunetelest c tou z 40 pou mˆc endiafèrei eðnai 2.

Qr simec sqèseic gia upologismì gennhtri n sunart sewn 1. 1 + z + z 2 + z 3 +... + z n = 1 zn+1 1 z n ( ) n 2. (1 + z) n = z k k k=0 n ( ) n 3. (1 + z) n = z k ìpou k ( ) k=0( ) n k + n 1 = ( 1) k k k ( ) n = k n! k!(n k)! kai An stic parapˆnw sqèseic sth jèsh tou z bˆloume z èqoume:

Qr simec sqèseic gia upologismì gennhtri n sunart sewn n ( ) n n ( ) n 4. (1 z) n = ( z) k = ( 1) k z k k k k=0 k=0 n ( ) n 5. (1 z) n = ( z) k = k k=0 n ( ) k + n 1 n ( ) k + n 1 ( 1) k ( 1) k z k = z k k k k=0 k=0

ParadeÐgmata AkoloujÐa: 1, 0, 0, 1 Genn tria sunˆrthsh: 1 z 0 + 0 z 1 + 0 z 2 + 1 z 3 = 1 + z 3 AkoloujÐa: 1, 2, 1, 0, 0, 0,... Genn tria sunˆrthsh: 1 z 0 + 2 z 1 + 1 z 2 + 0 z 3 + 0 z 4 +... = 1 + 2z + z 2 = (1 + z) 2 AkoloujÐa: 1, 1, 1,..., 1 } {{ } n+1 ìroi Genn tria sunˆrthsh: 1 + z + z 2 + z 3 +... + z n = 1 zn+1 1 z AkoloujÐa: 1, 1, 1, 1,... Genn tria sunˆrthsh: 1 z + z 2 z 3 +... = 1 ( ( 1 + z n AkoloujÐa: 0), n ( 1), n ( 2),..., n ) n Genn tria sunˆrthsh: ( n 0) z 0 + ( ) n 1 z 1 + ( ) n 2 z 2 + ( ) n 3 z 3 +... + ( ) n n z n = (1 + z) n

Ta prohgoômena aforoôn sta maj mata 27 kai 29/10/2010

'Eqoume 10 ˆsprec mpˆlec kai dialègoume kˆpoiec apì autèc. Me pìsouc trìpouc mporeð na gðnei autì; MporoÔme na dialèxoume 0,1,2,...,10 ˆsprec mpˆlec me 1 trìpo kˆje forˆ, en 10,11,... me kanènan trìpo. 'Ara, h antðstoiqh genn tria sunˆrthsh eðnai: 1 + z + z 2 + z 3 +... + z 10 + 0 + 0 +... = 1 z11 1 z

'Eqoume ˆsprec mpˆlec kai dialègoume kˆpoiec apì autèc. Me pìsouc trìpouc mporeð na gðnei autì; MporoÔme na dialèxoume 0,1,2,... ˆsprec mpˆlec me 1 trìpo kˆje forˆ. 'Ara, h antðstoiqh genn tria sunˆrthsh eðnai: 1 + z + z 2 + z 3 +... = 1 1 z

'Eqoume ˆsprec, prˆsinec kai kìkkinec mpˆlec. Me pìsouc trìpouc mporoôme na dialèxoume r apì autèc; Apì tic ˆsprec mpˆlec mporoôme na dialèxoume me 1 trìpo kˆje forˆ 0,1,2,3,... apì autèc Genn tria sunˆrthsh gia tic ˆsprec mpˆlec eðnai: 1 + z + z 2 + z 3 +... = 1 1 z Apì tic prˆsinec mpˆlec mporoôme na dialèxoume me 1 trìpo kˆje forˆ 0,1,2,3,... apì autèc Genn tria sunˆrthsh gia tic prˆsinec mpˆlec: 1 + z + z 2 + z 3 +... = 1 1 z Apì tic kìkkinec mpˆlec mporoôme na dialèxoume me 1 trìpo kˆje forˆ 0,1,2,3,... apì autèc Genn tria sunˆrthsh gia tic kìkkinec mpˆlec: 1 + z + z 2 + z 3 +... = 1 1 z

'Eqoume ˆsprec, prˆsinec kai kìkkinec mpˆlec. Me pìsouc trìpouc mporoôme na dialèxoume r apì autèc; 'Ara, h telik genn tria sunˆrthsh gia to dialègw kˆpoiec apì ˆsprec, prˆsinec kai kìkkinec mpˆlec eðnai 1 h: ( 1 z )3 To pl joc twn trìpwn me touc opoðouc mpor na dialèxw r apì autèc dðnetai apì to suntelest tou z r 1 sto ( 1 z )3 1 ( ) 2 + k ( 1 z )3 = (1 z) 3 = z k k k=0 ( ) Opìte o suntelest c tou z r 1 2 + r sto ( 1 z )3 eðnai:, r pou eðnai to pl joc twn zhtoômenwn trìpwn.

'Eqoume ˆsprec, prˆsinec kai kìkkinec mpˆlec. Me pìsouc trìpouc mporoôme na dialèxoume r apì autèc; Me apl sunduastik : jèlw na dialèxw r apì 3 antikeðmena me epanal yeic: autì mporeð na gðnei me ( ) ( ) 3 + r 1 2 + r = r r trìpouc.

'Eqoume ˆsprec, prˆsinec kai kìkkinec mpˆlec. Me pìsouc trìpouc mporoôme na dialèxoume r apì autèc ìtan prèpei na pˆroume perittì arijmì apì ˆsprec kai ˆrtio arijmì apì kìkkinec mpˆlec; Gia tic ˆsprec mpˆlec o aparijmht c eðnai: z + z 3 + z 5... = z(1 + z 2 + z 4...) Gia tic kìkkinec mpˆlec o aparijmht c eðnai: 1 + z 2 + z 4 +... Gia tic prˆsinec mpˆlec o aparijmht c eðnai: 1 + z + z 2 + z 3 +... = 1 1 z To pl joc twn trìpwn pou zhtˆw dðnetai apì to suntelest tou z r sto z(1 + z 2 + z 4...) 2 1 1 z = z 1 z ( 1 1 z 2 )2

Me pìsouc trìpouc mporoôme na topojet soume 2n + 1 Ðdiec mpˆlec se 3 diaforetikˆ koutiˆ ste kˆje koutð na èqei to polô n mpˆlec; Gia kˆje koutð h genn tria sunˆrthsh eðnai: 1 + z + z 2 + z 3 +... + z n = 1 zn+1 1 z Epomènwc, h telik genn tria sunˆrthsh eðnai: ( ) 1 z n+1 3 A(z) = 1 z To zhtoômeno pl joc trìpwn dðnetai apì to suntelest tou z 2n+1 sthn parapˆnw parˆstash. A(z) = (1 3z n+1 + 3z 2n+2 ( z 3n+3 )(1 z) 3 = ( ) k + 2 (1 3z n+1 + 3z 2n+2 z 3n+3 ) 1 + )z k 2 k=1 O suntelest c tou z 2n+1 eðnai: ( 2n+3 2 ) 3 ( n+2 2 ) = n(n + 1) 2

Poioc eðnai o arijmìc lôsewn thc exðswshc z 1 + z 2 + z 3 + z 4 = 30 stouc fusikoôc an z 1 ˆrtioc 10, z 2 perittìc 11, 3 z 3 10, 0 z 4 15; Oi genn triec sunart seic eðnai: Gia th metablht z 1 : 1 + x 2 + x 4 +... + x 10 Gia th metablht z 2 : x + x 3 + x 5 +... + x 11 Gia th metablht z 3 : x 3 + x 4 + x 5 +... + x 10 Gia th metablht z 4 : 1 + x + x 2 + x 3 + x 4 +... + x 15 Dhl. telikˆ: (1 + x 2 + x 4 +... + x 10 )(x + x 3 + x 5 +... + x 11 )(x 3 + x 4 + x 5 +... + x 10 )(1 + x + x 2 + x 3 + x 4 +... + x 15 ) To zhtoômeno pl joc trìpwn dðnetai apì to suntelest tou x 30 sthn parapˆnw parˆstash pou eðnai 185. PROSOQH: EÐnai diaforetikìc apì to suntelest tou x 30 sthn parˆstash: (1 + x 2 + x 4 +...)(x + x 3 + x 5 +...)(x 3 + x 4 + x 5 +...)(1 + x + x 2 + x 3 + x 4 +...)

'Eqoume kèrmata twn 20 lept n, 50 lept n, 1 eur kai 2 eur. Me pìsouc trìpouc mpor na dialèxw kèrmata sunolik c axðac n eur, dialègontac upoqrewtikˆ toulˆqiston 1 kèrma apì kˆje eðdoc; KwdikopoioÔme ston ekjèth thn axða twn kermˆtwn se leptˆ. Oi genn triec sunart seic eðnai: Gia ta 20-lepta: z 20 + z 40 + z 60 +... Gia ta 50-lepta: z 50 + z 100 + z 150 +... Gia ta 1-eura: z 100 + z 200 + z 300 +... Gia ta 2-eura: z 200 + z 400 + z 600 +... Dhl. telikˆ: (z 20 + z 40 + z 60 +...)(z 50 + z 100 + z 150 +...)(z 100 + z 200 + z 300 +...)(z 200 + z 400 + z 600 +...) To zhtoômeno pl joc trìpwn dðnetai apì to suntelest tou z 100n sthn parapˆnw parˆstash.

'Eqoume 20 markadìrouc, 6 maôrouc, 10 prˆsinouc kai 4 kìkkinouc. Me pìsouc trìpouc mporoôme na touc moirˆsoume se 2 ˆtoma ste kajèna na pˆrei 10 markadìrouc kai toulˆqiston 1 apì kˆje qr ma; UpologÐzoume me pìsouc trìpouc mporoôme na d soume markadìrouc sto 1o ˆtomo sômfwna me touc periorismoôc afoô autì kajorðzei monadikˆ autoôc pou ja pˆrei to 2o ˆtomo. Oi genn triec sunart seic eðnai: Gia touc maôrouc markadìrouc: z + z 2 + z 3 +... + z 5 Gia touc prˆsinouc markadìrouc: z + z 2 + z 3 +... + z 9 Gia touc kìkkinouc markadìrouc: z + z 2 + z 3 Dhl. telikˆ: (z + z 2 + z 3 +... + z 5 )(z + z 2 + z 3 +... + z 9 )(z + z 2 + z 3 ) To zhtoômeno pl joc trìpwn dðnetai apì to suntelest tou z 10 sthn parapˆnw parˆstash pou eðnai 15.

Me pìsouc trìpouc 100 Ðdioi epibˆtec mporoôn na kateboôn se 4 diaforetikèc stˆseic; AnazhtoÔme tic akèraiec lôseic thc exðswshc x 1 + x 2 + x 3 + x 4 = 100 me x 1, x 2, x 3, x 4 0 H genn tria sunˆrthsh gia kˆje stˆsh eðnai: 1 + z + z 2 + z 3 +... - de stamatˆw sto 100 giatð mporeð na upˆrqoun ki ˆlloi epibˆtec H telik genn tria sunˆrthsh eðnai: ( ) 1 4 (1 + z + z 2 + z 3 +...) 4 = = 1 z 1 (1 z) 4 To zhtoômeno pl joc trìpwn dðnetai apì to suntelest tou z 100 sthn parapˆnw parˆstash pou eðnai C(103, 100).

Me pìsouc trìpouc 100 Ðdioi epibˆtec mporoôn na kateboôn se 4 diaforetikèc stˆseic ìtan prèpei pl joc epibat n sthn 3h stˆsh pl joc epibat n sthn 2h stˆsh pl joc epibat n sthn 1h stˆsh; Prèpei x 2 = x 1 + k, k 0, x 3 = x 2 + λ = x 1 + k + λ, λ 0 AnazhtoÔme tic akèraiec lôseic thc exðswshc 3x 1 + 2k + λ + x 4 = 100 me x 1, k, λ, x 4 0 H genn tria sunˆrthsh eðnai: (1 + z 3 + z 6 +... + z 99 )(1 + z 2 + z 4 +... + z 100 )(1 + z + z 2 + z 3 +... + z 100 )(1 + z + z 2 + z 3 +... + z 100 ) = (1 + z 3 + z 6 +... + z 99 )(1 + z 2 + z 4 +... + z 100 )(1 + z + z 2 + z 3 +... + z 100 ) 2 To zhtoômeno pl joc trìpwn dðnetai apì to suntelest tou z 100 sthn parapˆnw parˆstash pou eðnai 30.787.

Ekjetikèc genn triec sunart seic 'Otan endiaferìmaste gia DIATAXEIS kai ìqi gia SUNDUASMOUS qrhsimopoioôme ekjetik genn tria sunˆrthsh. Skeftìmaste akrib c ìpwc prin mìno pou t ra paðrnoume kommˆtia thc seirˆc: 1 + z + z2 2! + z3 +... = ex 3! GIATI; Se ì,ti èqoume dei èwc t ra, o suntelest c tou z r deðqnei to pl joc twn sunduasm n r antikeimènwn apì n antikeðmena. P(n, r) 'Omwc C(n, k) = P(r, r). Opìte gia na deðqnei o suntelest c tou z r diatˆxeic prèpei na diairoôme me P(r, r). 'Otan upologðsoume thn telik ekjetik genn tria sunˆrthsh, h apˆnthsh sto er thma me pìsouc trìpouc mporoôme na diatˆxoume n antikeðmena ìtan... ja brðsketai sto suntelest tou zn n!

Me pìsouc trìpouc mporoôme na ektup soume 25 diaforetikˆ arqeða se 3 diaforetikoôc ektupwtèc me ton periorismì ìti kˆje ektupwt c prèpei na ektup sei toulˆqiston èna arqeðo; Ousiastikˆ yˆqnoume to pl joc twn trìpwn me touc opoðouc mporoôme na diatˆxoume 25 antikeðmena epilegmèna apì 3 antikeðmena ìtan epitrèpontai epanal yeic twn 3 antikeimènwn. Gia ton 1o ektupwt - afoô prèpei opwsd pote na lˆbei èna toulˆqiston arqeðo - h genn tria sunˆrthsh eðnai: z + z2 2! + z3 z25 +... + 3! 25! = ez 1. 'Omoia gia to 2o kai 3o ektupwt. H telik genn tria sunˆrthsh gia ìlouc touc ektupwtèc eðnai: (e z 1) 3 kai to pl joc twn zhtoômenwn trìpwn dðnetai apì to suntelest tou z25 25! sto (ez 1) 3.

Me pìsouc trìpouc mporoôme na ektup soume 25 diaforetikˆ arqeða se 3 diaforetikoôc ektupwtèc me ton periorismì ìti kˆje ektupwt c prèpei na ektup sei toulˆqiston èna arqeðo; Kˆnoume prˆxeic: (e z 1) 3 = 3 3z 3e 2z + 3e z 1 = r=0 3 r zr r! 3 r=0 2 r zr r! + 3 z r r! 1 = r=0 (3 r 3 2 r + 3) zr r! 1 r=0 Epomènwc, o suntelest c tou z25 25! eðnai 325 3 2 25 + 3.

Poio eðnai to pl joc twn pentadik n sumboloseir n m kouc n me ˆrtio pl joc 1 kai perittì pl joc 0 stic opoðec ta yhfða 2,3,4 emfanðzontai toulˆqiston 1 forˆ; Ekjetikìc aparijmht c gia ta yhfða 2,3,4: (e x 1) 3 Ekjetikìc aparijmht c gia to yhfðo 1: x 0 + x 2 2! + x 4 4! + x 6 6! +... = (ex + e x ) 2 Ekjetikìc aparijmht c gia to yhfðo 0: x + x 3 3! + x 5 5! + x 7 7! +... = (ex e x ) 2 H telik genn tria sunˆrthsh eðnai: (e x 1) 3 (ex + e x ) (ex e x ) 2 2 O suntelest c tou x n pou deðqnei to zhtoômeno pl joc trìpwn n! eðnai: 5 n 3 4 n + 3 n+1 2 n + ( 2) n 3 ( 1) n 1

Qr simoi tôpoi 'Ajroisma n arqik n ìrwn gewmetrik c proìdou: 1 + x + x 2 + x 3 +... + x n = 1 x n+1 1 x 'Ajroisma ˆpeirwn ìrwn gewmetrik c proìdou (ìtan x 1): 1 + x + x 2 + x 3 +... = 1 1 x Diwnumikì anˆptugma: (1 + x) n = Gia n: (1 + x) n = 'Otan n < 0: ( ) n k ( ) n k ( ) n x k k k=0 = ( 1) k ( k + n 1 k k=0 ( ) k n 1 = ( 1) k ) k ( ) n x k k

Qr simoi tôpoi Gia x: ( ) n ( ) n (1 x) n = (1 + ( x)) n = ( x) k = ( 1) k x k k k k=0 k=0 ( ) n (1 x) n = (1 + ( x)) n = ( x) k = k k=0 ( ) n ( ) k + n 1 ( 1) k x k = ( 1) k ( 1) k x k = k k k=0 k=0 ( ) k + n 1 ( ) k + n 1 ( 1) 2 k x k = x k k k k=0 k=0

Qr simoi tôpoi 1 + x 1 1! + x 2 2! + x 3 3! + x 4 e αx = r=0 α r x r r! 4! +... = 1 + x 2 2! + x 4 4! +... = ex + e x 2 x + x 3 3! + x 5 5! +... = ex e x 2 r=0 x r r! = ex

Poio eðnai to pl joc twn pentadik n sumboloseir n m kouc n me ˆrtio pl joc 1; Ekjetikìc aparijmht c gia ta yhfða 0,2,3,4: (1 + x 1 1! + x 2 2! + x 3 3! + x 4 4! +...)4 = (e x ) 4 Ekjetikìc aparijmht c gia to yhfðo 1: x 0 + x 2 2! + x 4 4! + x 6 6! +... = (ex + e x ) 2 H telik genn tria sunˆrthsh eðnai: e 4x (ex + e x ) = 2 (e 5x + e 3x ) = 1 2 2 ( 5 k x k k! + 3 k x k k! ) = 1 (5 k + 3 k ) x k 2 k! k=0 k=0 k=0 O suntelest c tou x n pou deðqnei to zhtoômeno pl joc trìpwn n! eðnai: 1 2 (5n + 3 n )

Me pìsouc trìpouc èna sônolo apì r 6 Ðdia antikeðmena mporeð na diaqwristeð se 3 diaforetikˆ uposônola, anˆ dôo xèna metaxô touc, ste kˆje uposônolo na èqei toulˆqiston 2 antikeðmena (h ènwsh twn 3 uposunìlwn ja perièqei ìla ta r antikeðmena); Gia na èqei kˆje èna apì ta 3 uposônola toulˆqiston dôo antikeðmena, paðrnoume 6 antikeðmena kai ta topojetoôme apì 2 se kˆje uposônolo. AfoÔ ta antikeðmena eðnai Ðdia, de mˆc endiafèrei poia antikeðmena ja topojet soume se kˆje uposônolo, arkeð na eðnai 2. O sunolikìc arijmìc twn trìpwn na moirˆsoume ta upìloipa r 6 Ðdia antikeðmena prokôptei wc ex c: Σε κάθε αντικείμενο από τα r 6 αναθέτουμε ένα υποσύνολο από τα 3. Αυτό γίνεται με τους τρόπους που μπορούμε να διαλέξουμε ((r 6) αντικείμενα ) ( από 3) με επανάληψη: r 6 + 3 1 r 4 (r 4)(r 5) = = r 6 r 6 2

Me pìsouc trìpouc èna sônolo apì r 6 Ðdia antikeðmena mporeð na diaqwristeð se 3 diaforetikˆ uposônola, anˆ dôo xèna metaxô touc, ste kˆje uposônolo na èqei toulˆqiston 2 antikeðmena (h ènwsh twn 3 uposunìlwn ja perièqei ìla ta r antikeðmena); H genn tria sunˆrthsh èqei sa suntelestèc ton arijmì twn trìpwn pou mporoôme na topojet soume ta r antikeðmena sta 3 uposônola, me ton periorismì ìti kˆje uposônolo èqei toulˆqiston 2 antikeðmena. A(x) = (x 2 +x 3 +x 4 +...)(x 2 +x 3 +x 4 +...)(x 2 +x 3 +x 4 +...) ( ) 1 A(x) = x 6 (1+x+x 2 +...)(1+x+x 2 +...)(1+x+x 2 +...) = x 6 1 x ( ) 3 ( ) 3 + k 1 x 6 (1 x) 3 = x 6 ( 1) k x k = x 6 x k k k k=0 k=0

Me pìsouc trìpouc èna sônolo apì r 6 Ðdia antikeðmena mporeð na diaqwristeð se 3 diaforetikˆ uposônola, anˆ dôo xèna metaxô touc, ste kˆje uposônolo na èqei toulˆqiston 2 antikeðmena (h ènwsh twn 3 uposunìlwn ja perièqei ìla ta r antikeðmena); ( ) 3 + k 1 A(x) = x 6 x k k k=0 Gia na broôme to suntelest tou x r pou eðnai h apˆnthsh sthn ( er thsh jètoume ) ( k = ) r 6 kai èqoume r 6 + 3 1 r 4 (r 4)(r 5) = = r 6 r 6 2

Ta prohgoômena aforoôn sto mˆjhma 3/11/2010

DeÐxte ìti h genn tria sunˆrthsh thc akoloujðac α r = ( 2r r ) eðnai h A(x) = (1 4x) 1/2. Apì to diwnumikì anˆptugma èqoume: (1 + x) n = n(n 1)(n 2)...(n r + 1) 1 + x r (1 4x) 1/2 = r! r=1 ( 1 2 1 + )( 1 2 1)( 1 2 2)...( 1 2 r + 1) ( 4x) r = r! r=1 4 r ( 1 2 1 + )( 3 2 )( 5 2r 1 2 )...( 2 ) x r = r! r=1 2 r (1 3 5... (2r 1)) 1 + x r = r! r=1 2 r r!(1 3 5... (2r 1)) 1 + x r = r!r! r=1

DeÐxte ìti h genn tria sunˆrthsh thc akoloujðac α r = ( 2r r ) eðnai h A(x) = (1 4x) 1/2. (2 4 6... 2r)(1 3 5... (2r 1)) 1 + x r = r!r! r=1 (2r)! 1 + r!r! x r = r=1 ( ) 2r 1 + x r = r r=1 ( ) 2r x r r r=0 'Ara h akoloujða α r = ( 2r r A(x) = (1 4x) 1/2 ) èqei genn tria sunˆrthsh th

Poia akoloujða antistoiqeð sth genn tria sunˆrthsh 1 1 + x ; Apì to diwnumikì anˆptugma èqoume: (1 + x) n n(n 1)(n 2)...(n r + 1) = 1 + x r r! r=1 (1 + x) 1 = ( 1)( 1 1)( 1 2)...( 1 r + 1) 1 + x r r! r=1 (1 + x) 1 ( 1)( 2)( 3)...( r) = 1 + x r r! r=1 (1 + x) 1 = 1 + ( 1) r 1 2 3... r x r r! r=1 (1 + x) 1 = ( 1) r x r r=0 'Ara, h zhtoômenh akoloujða eðnai: 1, 1, 1, 1,..., ( 1) n,...

'Eqoume aperiìristo arijmì kermˆtwn twn 50, 20 kai 10 lept n kai jèloume na dialèxoume 10 kèrmata. D ste th genn tria sunˆrthsh kai to pl joc twn trìpwn gia aut thn epilog. 'Estw α k o arijmìc twn diaforetik n trìpwn na dialèxoume k kèrmata. Gia kˆje eðdoc kermˆtwn, mporoôme na dialèxoume kanèna, 1,2,... Autì kwdikopoieðtai sth GS 1 + x + x 2 + x 3 +... = 1 1 x. AfoÔ èqoume 3 diaforetikˆ eðdh kermˆtwn, h GS gia thn α k eðnai (1 x) 3. To zhtoômeno dðnetai apì to α 10 pou eðnai o suntelest c tou x 10 sto anˆptugma thc GS. Efarmìzoume to diwnumikì anˆptugma ( kai brðskoume ìti o ) ( 3+10 1 zhtoômenoc suntelest c eðnai: 10 = 12 ) 2 = 12 11 2 = 66 H lôsh prokôptei kai me sunduasmoôc antikeimènwn me epanˆlhyh.

Ta prohgoômena aforoôn sto mˆjhma 12/11/2010

'Eqoume aperiìristo arijmì kermˆtwn twn 50, 20 kai 10 lept n kai jèloume na dialèxoume 10 kèrmata ste na èqoume toulˆqiston 1 kèrma kai to polô 8 kèrmata twn 50 lept n, ˆrtio arijmì kermˆtwn twn 10 lept n kai o arijmìc twn kermˆtwn twn 20 lept n na eðnai perittìc kai na mhn xepernˆ to 5. D ste th genn tria sunˆrthsh kai to pl joc twn trìpwn gia aut thn epilog. 'Estw α k o arijmìc twn diaforetik n trìpwn na dialèxoume k kèrmata me touc parapˆnw periorismoôc. Gia ta 50-lepta, h GS eðnai: x + x 2 + x 3 +... + x 8. Gia ta 10-lepta, h GS eðnai: 1 + x 2 + x 4 + x 6 +... Gia ta 20-lepta, h GS eðnai: x + x 3 + x 5.

'Eqoume aperiìristo arijmì kermˆtwn twn 50, 20 kai 10 lept n kai jèloume na dialèxoume 10 kèrmata ste na èqoume toulˆqiston 1 kèrma kai to polô 8 kèrmata twn 50 lept n, ˆrtio arijmì kermˆtwn twn 10 lept n kai o arijmìc twn kermˆtwn twn 20 lept n na eðnai perittìc kai na mhn xepernˆ to 5. D ste th genn tria sunˆrthsh kai to pl joc twn trìpwn gia aut thn epilog. Pollaplasiˆzontac tic GS gia kˆje eðdoc kermˆtwn èqoume th GS gia thn akoloujða α k (x + x 2 + x 3 +... + x 8 )(1 + x 2 + x 4 + x 6 +...)(x + x 3 + x 5 ) To zhtoômeno dðnetai apì to suntelest tou x 10 sto anˆptugma thc GS pou (kˆnontac prˆxeic) eðnai 11.

'Eqoume aperiìristo arijmì kermˆtwn twn 50, 20 kai 10 lept n. D ste th genn tria sunˆrthsh kai to pl joc twn trìpwn gia na sqhmatðsoume to posì twn 2 eur. 'Estw α k o arijmìc twn diaforetik n trìpwn na sqhmatðsoume to posì twn k lept n dialègontac apì ta parapˆnw kèrmata. Gia ta 50-lepta, h GS eðnai: 1 + x 50 + x 100 +... = 1 1 x 50. Gia ta 10-lepta, h GS eðnai: 1 + x 20 + x 40 +... = 1 1 x 20. Gia ta 20-lepta, h GS eðnai: 1 + x 10 + x 20 +... = 1 1 x 10. Pollaplasiˆzontac tic GS gia kˆje eðdoc kermˆtwn èqoume th GS gia thn akoloujða α k 1 ( 1 x 50 )( 1 1 x 20 )( 1 1 x 10 ) AfoÔ jèloume na sqhmatðsoume to posì twn 200 lept n (=2 eur ), to zhtoômeno dðnetai apì to suntelest tou x 200 sto anˆptugma thc GS pou (kˆnontac prˆxeic) eðnai 29. GiatÐ;

1 ( 1 x 50 )( 1 1 x 20 )( 1 1 x 10 ) = ( ) 1 ( ) 1 ( 1) k x 50k + ( 1) l x 20l + k l k=0 l=0 ( ) 1 ( 1) r x 10r = x 50k + x 20l + x 10r = r r=0 k=0 l=0 r=0 k,l,r=0 x 50k+20l+10r Ousiastikˆ anazht to suntelest tou x 200 sto parapˆnw anˆptugma, dhl., tic akèraiec lôseic thc exðswshc: 50k + 20l + 10r = 200 5k + 2l + r = 20 me 0 k 4, 0 l 10, 0 r 20. Gia k = 0, prokôptoun 11 lôseic Gia k = 1, prokôptoun 8 lôseic Gia k = 2, prokôptoun 6 lôseic Gia k = 3, prokôptoun 3 lôseic Gia k = 4, prokôptoun 1 lôsh 'Ara, sunolikˆ: 11 + 8 + 6 + 3 + 1 = 29 lôseic.

Mia omˆda dðnei 30 ag nec. Ta dunatˆ apotelèsmata gia kˆje ag na eðnai nðkh, tta, isopalða. Me qr sh gennhtri n sunart sewn na upologisteð o arijmìc twn diaforetik n apotelesmˆtwn an o sunolikìc arijmìc twn nik n eðnai perittìc, o sunolikìc arijmìc twn htt n eðnai ˆrtioc, en oi isopalðec eðnai toulˆqiston 2. 'Estw α k o arijmìc twn diaforetik n apotelesmˆtwn me touc parapˆnw periorismoôc gia k ag nec sunolikˆ. Gia tic nðkec h GS eðnai: z + z 3 + z 5 +...z 29. Gia tic ttec h GS eðnai: 1 + z 2 + z 4 +...z 30. Gia tic isopalðec h GS eðnai: z 2 + z 3 + z 4 +...z 30. Pollaplasiˆzontac tic GS gia kˆje eðdoc kermˆtwn èqoume th GS gia thn akoloujða α k (z + z 3 + z 5 +...z 29 )(1 + z 2 + z 4 +...z 30 )(z 2 + z 3 + z 4 +...z 30 ) To zhtoômeno prokôptei apì to suntelest tou x 30 sto anˆptugma thc parapˆnw GS.

Mia omˆda dðnei 30 ag nec. Ta dunatˆ apotelèsmata gia kˆje ag na eðnai nðkh, tta, isopalða. Me qr sh gennhtri n sunart sewn na upologisteð o arijmìc twn diaforetik n apotelesmˆtwn an o sunolikìc arijmìc twn nik n eðnai perittìc, o sunolikìc arijmìc twn htt n eðnai ˆrtioc, en oi isopalðec eðnai toulˆqiston 2. Gia na upologðsoume to suntelest, epekteðnoume to ˆjroisma mèqri to ˆpeiro (kai èqoume ˆpeirouc ìrouc gewmetrik c proìdou). Autì den ephreˆzei to suntelest tou z 30 epeid oi ìroi pou prostðjentai èqoun ekjèth megalôtero tou 30. 'Etsi h GS gðnetai: A(z) = (z +z 3 +z 5 +...)(1+z 2 +z 4 +...)(z 2 +z 3 +z 4 +...) = z 1 z 2 z 2 1 1 z 2 = z 3 (1 z) 3 (1 + z) 2 = 1 z z 3 (1 + z) (1 z) 3 (1 + z) 3 = z3 + z 4 (1 z 2 ) 3 = z 3 (1 z 2 ) 3 + z 4 (1 z 2 ) 3

Mia omˆda dðnei 30 ag nec. Ta dunatˆ apotelèsmata gia kˆje ag na eðnai nðkh, tta, isopalða. Me qr sh gennhtri n sunart sewn na upologisteð o arijmìc twn diaforetik n apotelesmˆtwn an o sunolikìc arijmìc twn nik n eðnai perittìc, o sunolikìc arijmìc twn htt n eðnai ˆrtioc, en oi isopalðec eðnai toulˆqiston 2. Efarmìzoume to diwnumikì anˆptugma kai èqoume: z 3 (1 z 2 ) 3 = ( ) k + 2 z 2k+3 kai 2 k=0 z 4 (1 z 2 ) 3 = ( k + 2 2 ) z 2k+4 k=0 Sto pr to anˆptugma den upˆrqei to z 30, en sto deôtero to z 30 emfanðzetai ( gia k = 13 kai o antðstoiqoc ) 15 suntelest c eðnai 2 = 15 14 2 = 105 dðnei to sunolikì arijmì apotelesmˆtwn gia 30 ag nec me touc dosmènouc periorismoôc.

Jèloume na moirˆsoume 24 karamèlec se 4 paidiˆ ste kˆje paidð na pˆrei toulˆqiston 3 kai to polô 8 karamèlec. D ste th genn tria sunˆrthsh kai to pl joc twn trìpwn gia na gðnei autì. H GS gia kajèna apì ta 4 paidiˆ eðnai: z 3 + z 4 +... + z 8. H telik GS eðnai: ( ) 1 z (z 3 +z 4 +...+z 8 ) 4 = z 12 (1+z+z 2 +...+z 5 ) 4 = z 12 6 4. 1 z To zhtoômeno pl joc trìpwn dðnetai apì to suntelest tou z 24 sto anˆptugma thc parapˆnw GS o opoðoc eðnai Ðdioc me to suntelest tou z 12 sto anˆptugma thc ( ) 1 z 6 4 sunˆrthshc. 1 z PROSOQH: De mporoôme na epekteðnoume to ˆjroisma sto ˆpeiro giatð ta z 9,..., z 24 suneisfèroun sto suntelest tou z 24.

Jèloume na moirˆsoume 24 karamèlec se 4 paidiˆ ste kˆje paidð na pˆrei toulˆqiston 3 kai to polô 8 karamèlec. D ste th genn tria sunˆrthsh kai to pl joc twn trìpwn gia na gðnei autì. Efarmìzoume to diwnumikì anˆptugma gia th sunˆrthsh: ( 1 z 6 1 z ) 4. (1 z 6 ) 4 (1 z) 4 = (1 4z 6 + 6z 12 +...) ( ) k + 3 z k 3 k=0 To z 12 sqhmatðzetai apì to z 0 ston pr to ìro kai apì to z 12 sto deôtero ìro (suntelest c ( 15 3 ) ), apì to z 6 apì kˆje ìro (suntelest c 4 ( 9 3) ) kai to z 12 ston pr to ìro kai to z 0 sto deôtero ìro (suntelest c 6). 'Ara o zhtoômenoc suntelest c eðnai: ( 15 ) ( 3 + ( 4) 9 ) 3 + 6 = 125.

Me pìsouc trìpouc mporeð kˆpoioc na moirˆsei 24 (Ðdia) antikeðmena se 4 ˆtoma ètsi ste kˆje ˆtomo na pˆrei toulˆqiston 3 kai ìqi parapˆnw apì 8 antikeðmena; Gia kˆje ˆtomo h GS eðnai: x 3 + x 4 +... + x 8, opìte kai gia ta 4 ˆtoma h GS eðnai: (x 3 + x 4 +... + x 8 ) 4 To pl joc twn zhtoômenwn trìpwn dðnetai apì to suntelest tou x 24 sto anˆptugma thc parapˆnw GS. (x 3 +x 4 +...+x 8 ) 4 = x 12 (1+x +x 2 +...+x 5 ) 4 = x 12 ( 1 x 6 1 x 'Ara anazhtoôme to suntelest tou x 24 sto anˆptugma ( ) tou 1 x 6 4 1 x = (1 x 6 ) 4 (1 x) 4 = [1 ( ) 4 1 x 6 + ( ) 4 2 x 12...+x 24 ][ ( ) ( 4 0 + 4 ) ( 1 ( x)+ 4 ) 2 ( x) 2 +...] O suntelest c tou x 12 eðnai: ( ) ( )( ) ( )( ) 4 4 4 4 4 [ ( 1) 12 ( 1) 6 + ] = 12 1 6 2 0 ( ) ( )( ) ( ) 15 4 9 4 [ + ] = 125 12 1 6 2 ) 4

Na upologisteð h akoloujða me genn tria sunˆrthsh 1 + 2x x 2 x 3 1 + x x 2 x 3 A(x) = 1 + x x 2 x 3 1 + x 2 x 3 + x 1 + x x 2 x 3 = x 1 + 1 + x x 2 (1 + x) = 1 + x (1 + x)(1 x 2 ) = x 1 + (1 + x) 2 (1 x) x (1+x) 2 se ˆjroisma apl n (1 x) klasmˆtwn (to p c sthn epìmenh diafˆneia). 'Eqoume: AnalÔoume to klˆsma A(x) = 1 + 1 4 (1 x) 1 + 1 4 (1 + x) 1 1 2 (1 + x) 2 = 1 + 1 ( ) 1 ( 1) k x k + 1 ( ) 1 x k 1 ( ) 2 x k = 4 k 4 k 2 k k=0 k=0 k=0 1 + 1 x k + 1 ( 1) k x k 1 ( 1) k (k + 1)x k = 4 4 2 k=0 k=0 k=0

Na upologisteð h akoloujða me genn tria sunˆrthsh 1 + 2x x 2 x 3 1 + x x 2 x 3 1 + k=0 [ 1 4 + 1 4 ( 1)k 1 ] 2 ( 1)k (1 + k) x k = 1 + x x 2 + 2x 3 2x 4 + 3x 5 3x 6 +...

Na upologisteð h akoloujða me genn tria sunˆrthsh 2 + 3x 6x 2 1 + 2x 2 + 3x 6x 2 3 1 + 6x 3x 6x 2 A(x) = = = 1 + 2x 1 + 2x 3x + 3 1 1 + 2x = ( ) 1 3 3x (1 + 2x) 1 = 3 3x 2 k x k = k k=0 3 3x ( 1) k 2 k x k = 2 x 4x 2 + 8x 3 16x 4 +... k=0

Na upologisteð h akoloujða me genn tria sunˆrthsh 2 1 4x 2 A(x) = 2 2 1 4x 2 = 2(1 4x 2 ) 1 = 2 ( 1 k k=0 4 k x 2k = 2 + 0 x + 8x 2 + 0 x 3 + 32x 4 +... k=0 ) ( 1) k (4x 2 ) k =

Na upologisteð h genn tria sunˆrthsh thc akoloujðac 2, 5, 13, 35,..., 2 n + 3 n,... A(x) = 2 0 + 3 0 + (2 1 + 3 1 )x + (2 2 + 3 2 )x +... = (2 0 + 2x + 2 2 x 2 +...) + (3 0 + 3x + 2 3 x 2 +...) [(2x) 0 + (2x) 1 + (2x) 2 +...] + [(3x) 0 + (3x) 1 + (3x) 2 +...] = 1 1 2x + 1 1 3x + 1 2x 2 5x = 1 3x 1 5x + 6x 2 = 1 5x + 6x 2

Na upologisteð h genn tria sunˆrthsh thc akoloujðac 1, 2, 3,..., r,... A(x) = 1 + 2x + 3x 2 +... + rx r 1 +... = (x+x 2 +x 3 +...+x r +...) = [(1+x+x 2 +x 3 +...+x r +...) 1] = ( ) 1 1 x 1 1 = (1 x) 2

Na apodeiqjeð ( me qr sh gennhtri n sunart sewn h ) ( n sqèsh k = n 1 ) ( k + n 1 k 1). ( n r) eðnai oi suntelestèc tou z r sto anˆptugma tou (1 + z) n. 'Omwc (1 + z) n = (1 + z)(1 + z) n 1 = (1 + z) n 1 + z(1 + z) n 1 Sto parapˆnw ˆjroisma, oi suntelestèc tou z r eðnai: ( n 1 ) ( n 1 k apì ton pr to ìro kai k 1) apì to deôtero ìro, dhl. sunolikˆ (n 1 ) ( k + n 1 ) k 1

Na brejeð h genn tria sunˆrthsh thc akoloujðac α 0, α 1,..., ìpou α r eðnai o arijmìc twn trìpwn na epilèxoume (me epanal yeic) r grˆmmata apì to alfˆbhto 0, 1, 2 me ton periorismì ìti to grˆmma 0 ja epilegeð ˆrtio arijmì for n. Me bˆsh th genn tria sunˆrthsh upologðste to α r. H GS eðnai: A(x) = (1 + x 2 + x 4 +...) (1 + x + x 2 +...) (1 + x + x 2 +...) = } {{ } } {{ } } {{ } gia to 0 gia to 1 gia to 2 1 1 1 x 2 (1 x) 2 = 1 1 1 + x (1 x) 3 AnalÔontac se aplˆ klˆsmata (to p c sthn epìmenh diafˆneia): A(x) = 1 8 (1 + x) 1 + 1 8 (1 x) 1 + 1 4 (1 x) 2 + 1 2 (1 x) 3 =

1 8 1 8 ( ) 1 x k + 1 k 8 k=0 ( 1) k x k + 1 8 k=0 ( 1 k k=0 + 1 2 ) ( 1) k x k + 1 4 ( ) 3 ( 1) k x k = k k=0 x k + 1 4 k=0 (k+1)x k + 1 2 k=0 α r = 1 8 ( 1)r + 1 8 + r + 1 4 + ( ) 2 ( 1) k x k k k=0 k=0 (r + 1)(r + 2) 4 1 8 [1 + (r + 1)(r + 3) ( 1)r ] + 4 (k + 1)(k + 2) x k 2 =

Na brejeð h genn tria sunˆrthsh thc akoloujðac α 0, α 1,..., ìpou o ìroc α r isoôtai me ton arijmì twn trìpwn na epilèxoume (me epanal yeic) r apì 10 sunolikˆ antikeðmena, metaxô twn opoðwn to antikeðmeno Q mporeð na epilegeð to polô 2 forèc, to antikeðmeno U to polô 3 forèc kai ta upìloipa antikeðmena apì mða forˆ to polô. GS gia to Q: z 0 + z 1 + z 2 = 1 + z + z 2 GS gia to U: z 0 + z 1 + z 2 + z 3 = 1 + z + z 2 + z 3 GS gia kajèna apì ta upìloipa 8 antikeðmena: z 0 + z 1 = 1 + z H telik GS eðnai: A(z) = (1 + z + z 2 )(1 + z + z 2 + z 3 )(1 + z) 8 To pl joc twn trìpwn na epilèxoume r apì ta antikeðmena autˆ dðnetai apì to suntelest tou z r sto anˆptugma thc A(z).

Na brejeð me qr sh aparijmht n o arijmìc twn trìpwn na epilèxoume r antikeðmena me aperiìristec epanal yeic apì n antikeðmena. GS gia kajèna apì ta n antikeðmena: 1 + x + x 2 + x 3 +... H telik GS eðnai: A(x) = (1 + x + x 2 + x 3 +...) n A(x) = (1 + x + x 2 +...) n = ( 1 1 x )n = (1 x) n = ( n ) k=0 k ( 1) k x k = ( n)( n 1)...( n k+1) k=0 k! ( 1) k x k = (n)(n+1)...(n+k 1) k=0 k! x k = (n+k 1)! k=0 k!(n 1)! x k = ( n+k 1 ) k=0 k x k To pl joc twn trìpwn na epilèxoume r apì ta antikeðmena autˆ dðnetai apì to suntelest tou x r sto anˆptugma thc A(x) ( ) n+r 1 pou eðnai r.

Na brejeð o aparijmht c gia tic epilogèc r antikeimènwn apì n antikeðmena (r n) me aperiìristec epanal yeic ìtan kˆje antikeðmeno epilègetai toulˆqiston mða forˆ. H GS gia kajèna apì ta n antikeðmena eðnai: x + x 2 + x 3 +... H telik GS eðnai: A(x) = (x + x 2 +...) n = ( 1 1 x 1)n = ( ) n ( x n 1 x = x n (1 x) n = x n k k=0 ( ) n + k 1 x n+k k=0 k ( 1 1+x 1 x ) n = ) ( 1) k x k = To pl joc twn trìpwn na epilèxoume r apì ta antikeðmena autˆ dðnetai apì to suntelest tou x r sto ( anˆptugma ) thc r 1 A(x) pou prokôptei gia k = r n kai eðnai. r n

Ta prohgoômena aforoôn sto mˆjhma 19/11/2010

Na prosdioristeð o suntelest c tou x 15 sto anˆptugma thc f (x) = (x 2 + x 3 + x 4 +...) 4. (x 2 + x 3 + x 4 +...) 4 = [ x 2 (1 + x + x 2 +...) ] 4 = [ ] 4 x 2 ( 1 1 x ) = x 8 ( 1 1 x )4 = x 8 (1 x) 4 = ) x k = x 8 ) x k ( 3+k k x 8 k=0 ( 1)k( 4 k k=0 Ousiastikˆ anazht to suntelest tou x 7 ston ìro k=0 ( 3+k k ) ( x k 10 ) pou eðnai: 7 = 120

Me pìsouc trìpouc mporoôme na topojet soume 25 (Ðdia) antikeðmena se 7 diaforetikˆ koutiˆ me ton periorismì ìti to pr to koutð den epitrèpetai na èqei pˆnw apì 10 antikeðmena; H GS gia to 1o koutð eðnai: 1 + x + x 2 +... + x 10 H GS gia kajèna apì ta upìloipa 6 koutiˆ eðnai: 1 + x + x 2 +... Opìte, h telik GS gia ta 7 koutiˆ eðnai: (1 + x + x 2 +... + x 10 )(1 + x + x 2 +...) 6 = ( ) ( ) 1 x 11 1 6 = (1 x) 7 (1 x 11 ) = [ 1 x 1 x ( ] [ 7 ( ] 6 + k ( 1) )x k k (1 x 11 ) = )x k (1 x 11 ) k k k=0 k=0

Me pìsouc trìpouc mporoôme na topojet soume 25 (Ðdia) antikeðmena se 7 diaforetikˆ koutiˆ me ton periorismì ìti to pr to koutð den epitrèpetai na èqei pˆnw apì 10 antikeðmena; To zhtoômeno pl joc trìpwn[ dðnetai apì to suntelest ( ] 6 + k tou x 25 sto anˆptugma tou )x k (1 x 11 ) k k=0 Autìc mporeð ( na prokôyei eðte gia k = 25 eðte gia ) ( k = 14 6+25 kai eðnai: 25 6+14 ) ( 14 = 31 ) ( 25 20 14) = 697521

Me pìsouc trìpouc mporoôme na epilèxoume 25 paiqnðdia apì 7 diaforetikˆ pou upˆrqoun sunolikˆ ìtan mporoôme na epilèxoume apì 2 èwc 6 kommˆtia apì kˆje paiqnðdi; H GS gia kˆje paiqnðdi eðnai: x 2 + x 3 + x 4 + x 5 + x 6 H GS gia ìla ta paiqnðdia eðnai: (x 2 + x 3 + x 4 + x 5 + x 6 ) 7 = ( x 2 (1 + x + x 2 + x 3 + x 4 ) ) 7 = ( ) 1 x x 14 (1 + x + x 2 + x 3 + x 4 ) 7 = x 14 5 7 = x 14 (1 1 x x) 7 (1 x 5 ) 7 = x 14 k=0 ( 1)k( ) 7 k x k ( ) r=0 ( 1)r 7 r x 5r Ousiastikˆ anazht to suntelest tou x 11 ston ìro k=0 ( 1)k( ) 7 k x k ( r=0 ( 1)r 7 r) x 5r = ( 6+k ) k=0 k x k ( ) r=0 ( 1)r 7 r x 5r Autìc prokôptei eðte gia k = 11, r = 0, eðte gia k = 6, r = 1, eðte gia k = 1, r = 2. 'Ara: ( 17 ) ( 11 + 12 ) ( 6 ( 1) 7 ( 1) + 7 7 1)( 2) = 12.376 6.468 + 147 = 6.055

Na brejeð genn tria sunˆrthsh gia thn akoloujða α r pou èqei wc ìrouc thc touc arijmoôc twn trìpwn èkfrashc tou r san ˆjroisma diaforetik n akeraðwn. Gia kˆje akèraio k, to ˆjroisma 1 + x k ekfrˆzei to ìti mporoôme na qrhsimopoi soume ton akèraio k kamða mða forˆ sto ˆjroisma gia to sqhmatismì tou akeraðou r. Opìte, gia ìlouc touc akèraiouc h GS eðnai: (1 + x)(1 + x 2 )(1 + x 3 )(1 + x 4 )...(1 + x k )... H GS deðqnei ìti o r mporeð na sqhmatisteð me qr sh kˆje akeraðou to polô mða forˆ.

Na brejeð h ekjetik genn tria sunˆrthsh gia ton arijmì α r twn diaforetik n metajèsewn r antikeimènwn pou epilègontai apì 4 diaforetikoôc tôpouc antikeimènwn, ek twn opoðwn ta antikeðmena kˆje tôpou epilègontai toulˆqiston 2 kai ìqi perissìterec apì 5 forèc. H GS gia kˆje antikeðmeno kˆje tôpou eðnai: x 2 2! + x 3 3! + x 4 4! + x 5 5! H GS gia touc 4 tôpouc antikeimènwn eðnai: ( x 2 2! + x 3 3! + x 4 4! + x 5 ) 4 5! O zhtoômenoc arijmìc α r dðnetai apì to suntelest tou x r sto anˆptugma thc parapˆnw sunˆrthshc. r!

Na brejeð h ekjetik genn tria sunˆrthsh gia ton arijmì α r twn trìpwn topojèthshc r (diaforetik n) anjr pwn se 3 diaforetikèc aðjousec me 1 toulˆqiston ˆnjrwpo se kˆje aðjousa. Poia eðnai h genn tria sunˆrthsh ìtan tejeð o periorismìc ìti prèpei na topojethjeð ˆrtioc arijmìc anjr pwn se kˆje aðjousa; Kai oi ˆnjrwpoi kai oi aðjousec eðnai diaforetikèc, opìte anazhtoôme diatˆxeic me ton periorismì ìti 1 toulˆqiston ˆtomo ja pˆei se kˆje aðjousa. H GS gia kˆje aðjousa eðnai: x + x2 2! + x3 x2 3! +... = 1 + x + 2! + x3 3! +... 1 = ex 1 Epomènwc, h GS gia tic 3 aðjousec eðnai: (e x 1) 3 = e 3x 3e 2x + 3e x 1 = 3 k x k k! 3 2 k x k k! +3 x k k! 1 = (3 k 3 2 k +3) x k k! 1 k=0 k=0 k=0 k=0

Na brejeð h ekjetik genn tria sunˆrthsh gia ton arijmì α r twn trìpwn topojèthshc r (diaforetik n) anjr pwn se 3 diaforetikèc aðjousec me 1 toulˆqiston ˆnjrwpo se kˆje aðjousa. Poia eðnai h genn tria sunˆrthsh ìtan tejeð o periorismìc ìti prèpei na topojethjeð ˆrtioc arijmìc anjr pwn se kˆje aðjousa; Kai oi ˆnjrwpoi kai oi aðjousec eðnai diaforetikèc, opìte anazhtoôme diatˆxeic me ton periorismì ìti 1 toulˆqiston ˆtomo ja pˆei se kˆje aðjousa. H GS gia kˆje aðjousa eðnai: x + x2 2! + x3 x2 3! +... = 1 + x + 2! + x3 3! +... 1 = ex 1 Epomènwc, h GS gia tic 3 aðjousec eðnai: (e x 1) 3 = e 3x 3e 2x + 3e x 1 = 3 k x k k! 3 2 k x k k! + k=0 k=0 x k 3 k! 1 = (3 k 3 2 k x k + 3) } {{ } k! 1 k=0 k=0 α r gia k=r

Na brejeð h ekjetik genn tria sunˆrthsh gia ton arijmì α r twn trìpwn topojèthshc r (diaforetik n) anjr pwn se 3 diaforetikèc aðjousec me 1 toulˆqiston ˆnjrwpo se kˆje aðjousa. Poia eðnai h genn tria sunˆrthsh ìtan tejeð o periorismìc ìti prèpei na topojethjeð ˆrtioc arijmìc anjr pwn se kˆje aðjousa; Kai oi ˆnjrwpoi kai oi aðjousec eðnai diaforetikèc, opìte anazhtoôme diatˆxeic me ton periorismì ìti ˆrtio pl joc atìmwn ja pˆei se kˆje aðjousa. H GS gia kˆje aðjousa eðnai: 1 + x2 2! + x4 4! +... = ex +e x 2 = 1 2 (ex + e x ) Epomènwc, h GS gia tic 3 aðjousec eðnai: [ 1 2 (ex + e x )] 3 = 1 8 (e3x 3e x + 3e x e 3x ) = 1 8 ( 3 k x k k! 3 x k k! + 3 ( 1) k x k k! 3 k ( 1) k x k k! ) k=0 k=0 k=0 k=0

Na brejeð h ekjetik genn tria sunˆrthsh gia ton arijmì α r twn trìpwn topojèthshc r (diaforetik n) anjr pwn se 3 diaforetikèc aðjousec me 1 toulˆqiston ˆnjrwpo se kˆje aðjousa. Poia eðnai h genn tria sunˆrthsh ìtan tejeð o periorismìc ìti prèpei na topojethjeð ˆrtioc arijmìc anjr pwn se kˆje aðjousa; 1 8 ( 3 k x k k! 3 x k k! + 3 ( 1) k x k k! 3 k ( 1) k x k k! ) = ( k=0 k=0 k=0 k=0 ) 1 [3 k 3 + 3( 1) k 3 k ( 1) k ] x k 8 k! k=0

Na brejeð h ekjetik genn tria sunˆrthsh gia ton arijmì α r twn trìpwn topojèthshc r (diaforetik n) anjr pwn se 3 diaforetikèc aðjousec me 1 toulˆqiston ˆnjrwpo se kˆje aðjousa. Poia eðnai h genn tria sunˆrthsh ìtan tejeð o periorismìc ìti prèpei na topojethjeð ˆrtioc arijmìc anjr pwn se kˆje aðjousa; 1 8 ( 3 k x k k! 3 x k k! + 3 ( 1) k x k k! 3 k ( 1) k x k k! ) = k=0 k=0 k=0 k=0 1 [3 k 3 + 3( 1) k 3 k ( 1) k x k ] 8 } {{ } k! k=0 α r gia k=r

BreÐte mia apl èkfrash gia th genn tria sunˆrthsh thc arijmhtik c akoloujðac 1, 1, 2, 2, 3, 3, 4, 4,... (1 z 2 )A(z) = 1 1 z 1 A(z) = (1 z 2 )(1 z) = 1 (1 + z)(1 z) 2 'Ara h genn tria sunˆrthsh thc akoloujðac dðnetai apì th sqèsh: A(z) = 1 (1 + z)(1 z) 2

Me pìsouc trìpouc mporoôme na diatˆxoume r diaforetikˆ antikeðmena pou epilègontai apì aperiìristo arijmì antikeimènwn n diaforetik n eid n; H GS gia kˆje eðdoc antikeimènwn eðnai: 1 + x + x2 2! + x3 3! +... Epomènwc, ( h GS gia ìla ) ta n eðdh antikeimènwn eðnai: n 1 + x + x2 2! + x3 3! +... = (e x ) n = e xn IsqÔei ìti: e αx = e xn = r=0 n r x r r! r=0 α r x r r!, opìte: To zhtoômeno pl joc diatˆxewn dðnetai apì to suntelest tou xr r! sto parapˆnw anˆptugma pou eðnai n r.

'Ena ploðo èqei 48 shmaðec, apì tic opoðec 12 kìkkinec, 12 ˆsprec, 12 mple kai 12 maôrec. 12 apì autèc tic shmaðec topojetoôntai se ènan katakìrufo istì gia na antallˆssontai mhnômata me ˆlla ploða. Pìsa apì autˆ ta mhnômata qrhsimopoioôn ˆrtio arijmì mple shmai n kai perittì arijmì maôrwn shmai n; GS gia ˆrtio arijmì mple shmai n: 1 + x2 2! + x4 4! +... GS gia perittì arijmì maôrwn shmai n: x + x3 3! + x5 5! +... GS gia kìkkinec kai ˆsprec (den upˆrqei kˆpoioc periorismìc): (1 + x + x2 2! + x3 3! + x4 4! + x5 Epomènwc, h telik GS eðnai: 5! +...)2 f (x) = (e x ) 2 ( ex + e x )( ex e x ) = 1 2 2 4 e2x (e 2x e 2x ) ( = 1 4 (e4x 1) = 1 ) (4x) i 1 = 1 (4x) i 4 i! 4 i! i=0 i=1

'Ena ploðo èqei 48 shmaðec, apì tic opoðec 12 kìkkinec, 12 ˆsprec, 12 mple kai 12 maôrec. 12 apì autèc tic shmaðec topojetoôntai se ènan katakìrufo istì gia na antallˆssontai mhnômata me ˆlla ploða. Pìsa apì autˆ ta mhnômata qrhsimopoioôn ˆrtio arijmì mple shmai n kai perittì arijmì maôrwn shmai n; Epomènwc, o suntelest c tou x12 12! sth sunˆrthsh pou dðnei to zhtoômeno arijmo mhnumˆtwn eðnai: 1 4 412 = 4 11 = 4.194.304 i=1 (4x) i i!

Mia etaireða proslambˆnei 11 nèouc upall louc. Kˆje upˆllhloc topojeteðtai se èna apì ta 4 upokatast mata thc etaireðac kai kˆje upokatˆsthma enisqôetai me ènan toulˆqiston nèo upˆllhlo. Me pìsouc trìpouc mporoôn na gðnoun oi topojet seic twn neoproslhfjèntwn upall lwn; Onomˆzoume A, B, G kai D ta 4 upokatast mata. Oi zhtoômenoi trìpoi eðnai ousiastikˆ apant seic sthn er thsh: kajènac apì touc upall louc se poio katˆsthma phgaðnei (upì touc dosmènouc periorismoôc). 'Ara anazhtoôme to pl joc twn diaforetik n 11-ˆdwn pou mporoôme na pˆroume qrhsimopoi ntac ta grˆmmata A, B, G kai D, me ton periorismì ìti to kajèna prèpei na emfanðzetai opwsd pote mða forˆ sthn 11-ˆda. Profan c, h seirˆ èqei shmasða, ˆra anazhtoôme diatˆxeic kai epomènwc ekjetik genn tria sunˆrthsh.

Mia etaireða proslambˆnei 11 nèouc upall louc. Kˆje upˆllhloc topojeteðtai se èna apì ta 4 upokatast mata thc etaireðac kai kˆje upokatˆsthma enisqôetai me ènan toulˆqiston nèo upˆllhlo. Me pìsouc trìpouc mporoôn na gðnoun oi topojet seic twn neoproslhfjèntwn upall lwn; Gia to grˆmma A (ìmoia kai gia ta upìloipa) h GS eðnai: x + x2 2! + x3 'Ara telikˆ h GS eðnai: 3! +... (x + x 2 2! + x 3 ) 4 3! +... = (e x 1) 4 = e 4x 4e 3x +6e 2x 4e x +1 O zhtoômenoc arijmìc topojet sewn dðnetai apì to suntelest tou x11 11! sto anˆptugma thc GS. 4 r x r r! 4 3 r x r r! + 6 2 r x r r! 4 1 r x r r! + 1 = r=0 r=0 r=0 r=0

Mia etaireða proslambˆnei 11 nèouc upall louc. Kˆje upˆllhloc topojeteðtai se èna apì ta 4 upokatast mata thc etaireðac kai kˆje upokatˆsthma enisqôetai me ènan toulˆqiston nèo upˆllhlo. Me pìsouc trìpouc mporoôn na gðnoun oi topojet seic twn neoproslhfjèntwn upall lwn; (4 r 4 3 r + 6 2 r 4r) x r r! + 1 r=0 UpologÐzoume to suntelest tou x11 11! jètontac r = 11 sthn parˆstash: (4 r 4 3 r + 6 2 r 4r) 4 r 4 3 r + 6 2 r 4r = 4 11 4 3 11 + 6 2 11 4 11

Na brejeð o arijmìc twn trìpwn na rðxoume k diakekrimèna antikeðmena se n diakekrimènec upodoqèc ètsi ste kˆje upodoq na deqjeð toulˆqiston èna antikeðmeno. Se kˆje upodoq prèpei na upˆrqei toulˆqiston èna antikeðmeno kai metrˆei h seirˆ. H GS gia kˆje upodoq eðnai: x + x2 2! + x3 3! +... = ex 1 H GS gia ìlec tic n upodoqèc eðnai: (e x 1) n = n ( n k ( n k=0 r=0 n k=0 ( n k ) ( 1) k e x(n k) = k=0 ( n k ) ( 1) k (e x ) n k = n ( n k k=0) )( 1) k (n k) r ) ( 1) k x r r! r=0 (n k) r x r r! =

Na brejeð o arijmìc twn trìpwn na diamerðsw èna sônolo apì k diakekrimèna antikeðmena se n uposônola mh kenˆ kai anˆ duo xèna. An jewr soume ìti ta n uposônola antistoiqoôn se n upodoqèc, tìte oi apait seic gia ta uposônola na eðnai anˆ duo xèna kai mh kenˆ metatrèpontai stouc periorismoôc gia tic upodoqèc na eðnai diakekrimènec kai na èqoun toulˆqiston èna antikeðmeno antðstoiqa. Epomènwc, o arijmìc pou yˆqnoume eðnai to apotèlesma tou prohgoômenou paradeðgmatoc, me th diaforˆ ìti afoô den mac endiafèrei h seirˆ twn stoiqeðwn mèsa sta uposônola ( twn antikeimènwn stic upodoqèc antðstoiqa), ja ( prèpei na diairèsoume me n!. 'Ara: n ( ) n )( 1) k (n k) r 1 k n! k=0

Na brejeð o arijmìc twn trìpwn na diamerðsw èna sônolo apì k diakekrimèna antikeðmena se n uposônola mh kenˆ kai anˆ duo xèna. O prohgoômenoc arijmìc lègetai kai arijmìc Stirling deutèrou eðdouc kai( sumbolðzetai me: n S(k, n) = { n ( ) n )(n k } = 1) k ( 1) i 1 i n! i=1 Gia touc arijmoôc Stirling deutèrou eðdouc isqôei o parakˆtw anadromikìc tôpoc: S(k + 1, n + 1) = S(k, n) + (n + 1)S(k, n + 1)

Idiìthtec gennhtri n sunart sewn JewroÔme akoloujðec α = (α 0, α 1, α 2,..., α n,...) kai β = (β 0, β 1, β 2,..., β n,... me genn triec sunart seic A(x) kai B(x) antðstoiqa. IsqÔoun oi akìloujec idiìthtec: Grammik idiìthta Idiìthta klðmakac Idiìthta olðsjhshc Idiìthta merik n ajroismˆtwn Idiìthta sumplhrwmatik n merik n ajroismˆtwn Idiìthta parag gou Idiìthta olokl rwshc Idiìthta sunèlixhc

Grammik idiìthta H akoloujða α = (α 0, α 1, α 2,..., α n,...) èqei genn tria sunˆrthsh thn A(x) = α r x r. r=0 H akoloujða β = (β 0, β 1, β 2,..., β n,... èqei genn tria sunˆrthsh th B(x) = b r x r. r=0 'Estw c, d stajerèc. H GS thc akoloujðac c α + d β eðnai h c A(x) + d B(x). GiatÐ; H GS thc c α + d β ja eðnai h Γ(x) = (c α r + d β r )x r = (c α r x r + d β r x r ) = r=0 r=0 c α r x r + d β r x r = c α r x r +d β r x r = ca(x)+db(x) r=0 r=0 r=0 r=0

Grammik idiìthta H GS thc akoloujðac 4 n + 9 2 n eðnai h 1 1 4x + 9 10 38x = 1 2x 1 6x + 8x 2 9 47x AntÐstrofa, h akoloujða me GS 1 10x + 21x 2 prokôptei an analôsoume th GS se merikˆ klˆsmata 5 1 3x + 4 1 7x. H akoloujða eðnai h 5 3n + 4 7 n.

Idiìthta klðmakac H akoloujða α = (α 0, α 1, α 2,..., α n,...) èqei genn tria sunˆrthsh thn A(x) = α r x r. r=0 H GS thc akoloujðac b r = λ r α r eðnai h A(λx). GiatÐ; B(x) = λ 0 α 0 x 0 + λ 1 α 1 x 1 + λ 2 α 2 x 2 +... + λ r α r x r +... = α 0 λ 0 x 0 + α 1 λ 1 x 1 + α 2 λ 2 x 2 +... + α r λ r x r +... = α 0 (λx) 0 + α 1 (λx) 1 + α 2 (λx) 2 +... + α r (λx) r +... = α r (λx) r = A(λx) r=0

Idiìthta olðsjhshc H akoloujða α = (α 0, α 1, α 2,..., α n,...) èqei genn tria sunˆrthsh thn A(x) = α r x r. r=0 r=0 r=0 H GS thc akoloujðac: b r = 0 gia r = 0,..., n 1 kai b r = α r n gia r = n, n + 1,... eðnai h B(x) = x n A(x). GiatÐ; n 1 B(x) = b r x r = b r x r + b r x r = 0 + α r n x r r=n Jètoume r n = k, opìte: B(x) = α k x n+k = α k x k x n = x n k=0 k=0 k=0 r=n α k x k = x n A(x)

Idiìthta olðsjhshc H akoloujða 0,0,0,0,1,1,1,1,... prokôptei apì thn 1,1,1,1,... an thn olisj soume proc ta dexiˆ katˆ 4 jèseic: 0,0,0,0,1,1,1,1,... Opìte, h GS thc akoloujðac 0,0,0,0,1,1,1,1,... eðnai x 4 A(x) = x 4 1 1 x = x4 1 x H akoloujða 0,0,1,2,4,8,...,2 n,... prokôptei apì thn 1,2,4,8,...,2 n,... an thn olisj soume proc ta dexiˆ katˆ 2 jèseic: 0,0,1,2,4,8,...,2 n,... Opìte, h GS thc akoloujðac 0,0,1,2,4,8,...,2 n,... eðnai x 2 A(x) = x 2 1 1 2x = x2 1 2x

BreÐte th diakrit arijmhtik sunˆrthsh pou antistoiqeð sth genn tria sunˆrthsh A(z) = z 5 5 6z + z 2. H genn tria sunˆrthsh A(z) isoôtai me z 5 B(z) ìpou 1 B(z) = 5 6z + z 2. An broôme thn arijmhtik sunˆrthsh b n pou antistoiqeð sth genn tria sunˆrthsh B(z), h arijmhtik sunˆrthsh α n pou antistoiqeð sth genn tria sunˆrthsh A(z) ja eðnai h α n = S 5 b n. 1 B(z) = 5 6z + z 2 = 1 (5 z)(1 + z) = 1 4 1 5 z + 1 4 1 1 z = 1 20 1 1 1 5 z + 1 4 1 1 z

BreÐte th diakrit arijmhtik sunˆrthsh pou antistoiqeð sth genn tria sunˆrthsh A(z) = 1 B(z) = 5 6z + z 2 = 1 (5 z)(1 + z) = 1 4 1 5 z + 1 4 1 1 z = 1 20 H genn tria sunˆrthsh sunˆrthsh ( 1 5) n, n 0 H genn tria sunˆrthsh 1 z 5 5 6z + z 2. 1 1 1 5 z + 1 4 1 1 z 1 1 1 antistoiqeð sthn arijmhtik z 5 1 z antistoiqeð sthn (stajer ) arijmhtik sunˆrthsh 1 'Ara: b n = 1 ( 1 ) n 20 5 + 1 4 {, n 0 0, 0 n 4, Epomènwc: α n = 1 ( 1 ) n 5 20 5 + 1 4, n 5.

Idiìthta merik n ajroismˆtwn H akoloujða α = (α 0, α 1, α 2,..., α n,...) èqei genn tria sunˆrthsh thn A(x) = α r x r. r=0 H GS thc akoloujðac b k = k α r, k = 0, 1, 2,... eðnai r=0 B(x) = A(x) 1 x. GiatÐ; α k = b k b k 1 = grammik idiìthta α k x k = k=0 b k x k k=0 k=0 A(x) = B(x) xb(x) B(x) = A(x) 1 x b k 1 x k = idiìthta olðsjhshc

Idiìthta merik n ajroismˆtwn H akoloujða γ n = n + 1 mporeð na jewrhjeð san h akoloujða twn merik n ajroismˆtwn thc α n = 1: γ n = 1, 1 + 1, 1 + 1 + 1,... H GS thc α n = 1 eðnai h A(x) = 1 1 x Opìte h GS thc γ n = n + 1 ja eðnai h A(x) 1 x = 1 (1 x) 2 Qrhsimopoi ntac thn idiìthta thc olðsjhshc, èqoume ìti h GS thc akoloujðac γ n 1 = n eðnai h xγ(x) = x (1 x) 2 Qrhsimopoi ntac xanˆ thn idiìthta twn merik n ajroismˆtwn, èqoume ìti h akoloujða δ n = n i=0 i èqei sa GS th (x) = x (1 x) 3

D ste kleistì tôpo gia thn akoloujða (x) = x (1 x) 3 (ìpwc deðxame se prohgoômenh ˆskhsh) H (x) = eðnai GS thc akoloujðac δ n = n i=0 i Efarmìzoume to diwnumikì anˆptugma kai èqoume: (1 x) 3 = 1 + ( ) 3 + k 1 x k = 1 + k k=1 Apì thn idiìthta olðsjhshc: x(1 x) 3 = 0x 0 + 1x + = 0 + x + k=1 ( k ) + 1 x k = 2 k =2 ( k + 2 k=1 ( ) k + 2 x k+1 2 ( k + 1 k =1 2 2 ) x k O ìroc δ n eðnai o ( suntelest c tou x n sto parapˆnw ) n+1 anˆptugma, dhl. 2. x (1 x) 3 ) x k

Na upologisteð kleistìc tôpoc gia to ˆjroisma 1 2 + 2 2 + 3 2 +... + n 2 me qr sh gennhtri n sunart sewn. Upìdeixh: BreÐte pr ta th genn tria sunˆrthsh thc akoloujðac α k = k 2. Idèa: BrÐskw pr ta th genn tria sunˆrthsh A(x) thc akoloujðac α k = k 2 Parathr ìti to zhtoômeno ˆjroisma eðnai o n-stìc ìroc k thc akoloujðac b k = α r, k = 0, 1, 2,... twn merik n r=0 ajroismˆtwn thc α k = k 2 H genn tria sunˆrthsh thc b k eðnai B(x) = A(x) 1 x (èqei apodeiqjeð h sqetik idiìthta) Opìte to zhtoômeno ˆjroisma isoôtai me to suntelest tou x n sth genn tria sunˆrthsh B(x)

Na upologisteð kleistìc tôpoc gia to ˆjroisma 1 2 + 2 2 + 3 2 +... + n 2 me qr sh gennhtri n sunart sewn. Upìdeixh: BreÐte pr ta th genn tria sunˆrthsh thc akoloujðac α k = k 2. Gia th genn tria sunˆrthsh thc akoloujðac α k = k 2 èqoume: A(x) = 0 + 1 2 x + 2 2 x 2 + 3 2 x 3 + 4 2 x 4 +... = x(1 2 + 2 2 x + 3 2 x 2 + 4 2 x 3 +...) = x(x + 2x 2 + 3x 3 + 4x 4 +...) = x [ x(1 + 2x + 3x 2 + 4x 3 +...) ] [ = x x(x + x 2 + x 3 + x 4...) ] = [ ( ) 1 ] [ ] x x 1 x 1 x x(1 + x) = x x (1 x) 2 = (1 x) 3

Na upologisteð kleistìc tôpoc gia to ˆjroisma 1 2 + 2 2 + 3 2 +... + n 2 me qr sh gennhtri n sunart sewn. Upìdeixh: BreÐte pr ta th genn tria sunˆrthsh thc akoloujðac α k = k 2. Apì thn idiìthta twn merik n ajroismˆtwn prokôptei ìti h genn tria sunˆrthsh thc akoloujðac b = 0 2, 0 2 + 1 2, 0 2 + 1 2 + 2 2,..., 0 2 + 1 2 +... + r 2,... gia thn opoða eðnai b k = k r=0 α r eðnai h S(x) = A(x) 1 x = x(1 + x) (1 x) 4 Opìte gia na upologðsoume to zhtoômeno ˆjroisma, prèpei na upologðsoume to suntelest tou x r sthn parapˆnw sunˆrthsh. Apì to diwnumikì anˆptugma èqoume gia to suntelest tou x r sto (1 x) 4 : ( 4)( 4 1)...( 4 r + 1) ( 1) r 4 5... (r + 3) = r! r! (r + 1)(r + 2)(r + 3) = 1 2 3

Na upologisteð kleistìc tôpoc gia to ˆjroisma 1 2 + 2 2 + 3 2 +... + n 2 me qr sh gennhtri n sunart sewn. Upìdeixh: BreÐte pr ta th genn tria sunˆrthsh thc akoloujðac α k = k 2. P c mporoôme na èqoume x r sth dik mac parˆstash; EÐte san x r = x 1 x r 1 eðte san x r = x 2 x r 2 Opìte o suntelest c tou x r x(1 + x) sth sunˆrthsh S(x) = (1 x) 4 eðnai o: r(r + 1)(r + 2) 1 2 3 + 'Ara: 1 2 + 2 2 + 3 2 +... + r 2 = (r 1)r(r + 1) 1 2 3 = r(r + 1)(2r + 1) 6 r(r + 1)(2r + 1) 6

Idiìthta sumplhrwmatik n merik n ajroismˆtwn H akoloujða α = (α 0, α 1, α 2,..., α n,...) èqei genn tria sunˆrthsh thn A(x) = α r x r. r=0 H GS thc akoloujðac b k = α r, k = 0, 1, 2,... eðnai r=k A(1) xa(x) B(x) =. GiatÐ; 1 x k 1 b k = α r = α r α r r=k r=0 k 1 b k = A(1) α r k=0 r=0 r=0 k=0 k 1 b k x k = A(1) x k x ( α r )x k 1 k=0 r=0 1 B(x) = A(1) 1 x xa(x) 1 x k 1 A(1)x k ( α r )x k = k=0 = A(1) xa(x) 1 x k=0 r=0

Idiìthta parag gou H akoloujða γ n = nα n èqei GS th Γ(x) = xa (x), ìpou A (x) eðnai h pr th parˆgwgoc thc sunˆrthshc A(x). GIATI; Γ(x) = xa (x) = x (α n x n ) = x nα n x n 1 = n=0 n=0 (nα n )x n n=0

Idiìthta oloklhr matoc H akoloujða δ n = α n n+1 èqei GS th (x) = 1 x x 0 A(z)dz. H parˆgousa tou z n eðnai zn+1 n+1, opìte èqoume: (x) = 1 x A(z)dz = 1 x α n z n dz = x 0 x n=0 0 1 α n x n + 1 x n+1 α n = n + 1 x n n=0 n=0

Na upologisteð h GS thc akoloujðac α n = n(n + 1). H akoloujða β n = n èqei GS th B(x) = deðxame se prohgoômenh ˆskhsh). x (1 x) 2 (ìpwc Apì thn idiìthta thc parag gou, h γ n = n 2 = nb n èqei GS Γ(x) = xb (x) = x(x+1) (1 x) 3. Apì th grammik idiìthta, h akoloujða α n = n(n + 1) = n 2 + n èqei GS th Γ(x) + B(x) = x(x + 1) (1 x) 3 + x (1 x) 2 = 2x (1 x) 3

Idiìthta sunèlixhc 'Estw akoloujða α me ìrouc: α 0, α 1, α 2, α 3, α 4,... 'Estw akoloujða β me ìrouc: b 0, b 1, b 2, b 3, b 4,... H sunèlix touc eðnai h akoloujða me ìrouc: γ 0 = α 0 b 0 γ 1 = α 0 b 1 + α 1 b 0 γ 2 = α 0 b 2 + α 1 b 1 + α 2 b 0 γ 3 = α 0 b 3 + α 1 b 2 + α 2 b 1 + α 3 b 0... DHLADH: Sunèlixh twn akolouji n α kai β eðnai h akoloujða d k = k r=0 α r β k r, k = 0, 1, 2,...

Idiìthta sunèlixhc H akoloujða d k = k r=0 α r β k r, k = 0, 1, 2,... onomˆzetai sunèlixh twn akolouji n α kai β kai sumbolðzetai α β. H GS thc akoloujðac d k eðnai h D(x) = A(x)B(x), ìpou A(x) eðnai h GS thc akoloujðac α r kai B(x) eðnai h GS thc akoloujðac b r. GiatÐ; A(x)B(x) = (α 0 + α 1 x + α 2 x 2 +...)(b 0 + b 1 x + b 2 x 2 +...) = (α 0 b 0 ) + (α 0 b 1 )x + (α 0 b 2 )x 2 +...(α 1 b 0 )x + (α 1 b 1 )x 2 + (α 1 b 2 )x 3 +... + (α 2 b 0 )x 2 + (α 2 b 1 )x 3 + (α 2 b 2 )x 4 +... = d k x k = D(x) k=0

Idiìthta sunèlixhc H akoloujða γ n = n i=1 α iβ n 1 onomˆzetai sunèlixh twn akolouji n α kai β kai sumbolðzetai α β. 'Estw Γ(x) h GS thc akoloujðac γ. EÐnai Γ(x) = A(x)B(x) pio aplˆ h GS thc sunèlixhc dôo akolouji n dðnetai apì to ginìmeno twn GS touc. Autì prokôptei eôkola apì ton orismì tou ginomènou poluwnômwn: o sunetelest c tou x n sto ginìmeno A(x)B(x) n isoôtai me i=0 α iβ n i epeid ìloi oi dunatoð trìpoi na pˆroume to x n sto ginìmeno prokôptoun pollaplasiˆzontac to x i sto A(x) me to x n i sto B(x), gia ìla ta i = 0,..., n.

ApodeÐxte ìti h prˆxh thc sunèlixhc eðnai prˆxh antimetajetik, dhl., ìti gia opoiesd pote akoloujðec α kai β isqôei ìti α β = β α. To ginìmeno poluwnômwn eðnai antimetajetik prˆxh. Apì thn idiìthta thc sunèlixhc, oi akoloujðec α β kai β α èqoun thn Ðdia GS. 'Ara prìkeitai gia tic Ðdiec akoloujðec.

ApodeÐxte thn idiìthta twn merik n ajroismˆtwn qrhsimopoi ntac thn idiìthta thc sunèlixhc. H sunèlixh thc akoloujðac α me thn akoloujða β n = 1 eðnai n i=0 α i, dhl., h akoloujða twn merik n ajroismˆtwn thc α. 'Estw A(x) h GS thc α. H GS thc β n = 1 eðnai B(x) = (1 x) 1. Apì thn idiìthta thc sunèlixhc, h GS thc akoloujðac twn merik n ajroismˆtwn thc α eðnai A(x) 1 x.

UpologÐste to ˆjroisma n i=0 3i 2 n i qrhsimopoi ntac genn triec sunart seic. To zhtoômeno ˆjroisma eðnai o n-ostìc ìroc thc sunèlixhc twn akolouji n α n = 3 n kai β n = 2 n. H α n èqei GS 1 1 3x kai h β n èqei GS B(x) = 1 1 2x. H GS thc sunèlix c touc eðnai 1 (1 3x)(1 2x) = 3 1 3x 2 1 2x Autì prokôptei me merik klasmatik anˆlush. Apì th grammik idiìthta, h akoloujða pou antistoiqeð se aut th GS èqei n-ostì ìro 3 n+1 2 n+1. Epomènwc, n i=0 3i 2 n i = 3 n+1 2 n+1

Anˆlush se klˆsmata kai antistrof gennhtri n sunart sewn Jèloume na upologðsoume thn akoloujða me genn tria sunˆrthsh P(x), ìpou P(x) kai D(x) eðnai polu numa wc D(x) proc x. Efarmìzoume merik klasmatik anˆlush eðte apeujeðac sth sunˆrthsh P(x) D(x) eðte sth sunˆrthsh 1 D(x). Apì to apotèlesma thc anˆlushc se merikˆ klˆsmata upologðzoume thn akoloujða me thn antðstoiqh genn tria sunˆrthsh. 1 An èqoume thn qrhsimopoioôme thn idiìthta thc D(x) olðsjhshc kai th grammik idiìthta.

Na upologisteð h akoloujða me genn tria sunˆrthsh thn F (x) = 4x 2 (1 8x) (1 4x)(1 2x) 2 BrÐskoume pr ta th GS tou parˆgonta: 1 8x G(x) = (1 4x)(1 2x) 2 Me klasmatik anˆlush: 1 8x G(x) = (1 4x)(1 2x) 2 = 4 (1 4x) + 3 (1 2x) 2 + 2 (1 2x) H akoloujða pou antistoiqeð sth genn tria sunˆrthsh G(x) eðnai h: α r = 4 4 r + 3 (r + 1)2 r + 2 2 r, r = 0, 1, 2,... Epomènwc, h F (x) = 4x 2 G(x) (me bˆsh idiìthta olðsjhshc) eðnai h GS{ thc akoloujðac 4α r 2, (3r 1)2 dhl.:α r = r 4 r, an r = 2, 3, 4,..., 0, an r < 2.

Na upologisteð h akoloujða me genn tria sunˆrthsh thn F (x) = 4x 2 (1 8x) (1 4x)(1 2x) 2 1 GiatÐ h eðnai genn tria sunˆrthsh thc (1 2x) 2 akoloujðac (r + 1)2 r ; 1 AnaptÔssw to kai anazht to suntelest tou (1 2x) 2 x k pou ja d sei to genikì ìro thc zhtoômenhc akoloujðac 1 ( ) 2 (1 2x) 2 = (1 2x) 2 = ( 2x) k = k k=0 ( ) k + 2 1 ( ) k + 1 ( 1) k ( 1) k 2 k x k = 2 k x k = k k k=0 k=0 (k + 1)2 k x k k=0

Na upologisteð h akoloujða me genn tria sunˆrthsh 2 1 4x 2 2 Me klasmatik anˆlush: 1 4x 2 = 1 1 2x + 1 1 + 2x 1 Gia th genn tria sunˆrthsh h akoloujða eðnai h 1 2x β n = 2 n. Gia th genn tria sunˆrthsh γ n = ( 2) n. 1 1 + 2x h akoloujða eðnai h Apì th { grammik idiìthta h zhtoômenh akoloujða eðnai: 2 α n = n+1, an n ˆrtioc, 0, an n perittìc.

Na upologisteð h akoloujða me genn tria sunˆrthsh 22x 3 9x 2 14x 1 (1 + x)(1 + 3x)(1 2x) 2 1 Me klasmatik anˆlush: (1 + x)(1 + 3x)(1 2x) 2 = 1/18 1 + x + 27/50 1 + 3x + 56/225 1 2x + 4/15 (1 2x) 2 H akoloujða me GS (1 + x) 1 eðnai h ( 1) n. Me grammik idiìthta kai idiìthta olðsjhshc: akoloujða me GS ( 1/18)(22x 3 9x 2 14x 1) eðnai h: (1 + x) 1 [ 18 22( 1) n 3 9( 1) n 2 14( 1) n 1 ( 1) n] = 1 18 [ 22 9 + 14 1] ( 1)n = ( 1) n

Na upologisteð h akoloujða me genn tria sunˆrthsh 22x 3 9x 2 14x 1 (1 + x)(1 + 3x)(1 2x) 2 H akoloujða me GS (1 + 3x) 1 eðnai h ( 3) n. Me grammik idiìthta kai idiìthta olðsjhshc: akoloujða me GS (27/50)(22x 3 9x 2 14x 1) eðnai h: [ (1 + 3x) 27 50 22( 3) n 3 9( 3) n 2 14( 3) n 1 ( 3) n] = 27 50 [ 22/27 9/9 + 14/3 1] ( 3)n = ( 3) n H akoloujða me GS (1 2x) 1 eðnai h 2 n. Me grammik idiìthta kai idiìthta olðsjhshc: akoloujða me GS (56/225)(22x 3 9x 2 14x 1) eðnai h: [ (1 + 3x) 56 225 22 2 n 3 9 2 n 2 14 2 n 1 2 n] = 56 225 [ 22/8 9/4 14/2 1] 2n = 28 15 2n

Na upologisteð h akoloujða me genn tria sunˆrthsh 22x 3 9x 2 14x 1 (1 + x)(1 + 3x)(1 2x) 2 H akoloujða me GS (1 2x) 2 eðnai h (n + 1)2 n. Me grammik idiìthta kai idiìthta olðsjhshc: akoloujða me GS (4/15)(22x 3 9x 2 14x 1) (1 2x) 2 eðnai h: [ 4 15 22(n 2)2 n 3 9(n 1)2 n 2 14n2 n 1 (n + 1)2 n] = 4 15 [22/8 9/4 14/2 1] n2n + 4 15 [( 2)22/8 + 9/4 1] 2n = 2n2 n 17 15 2n Apì ìla ta parapˆnw, katal goume ìti h zhtoômenh akoloujða eðnai h: α n = ( 1) n + ( 3) n 2n2 n 28 + 17 2 n = 15 ( 1) n + ( 3) n n2 n+1 3 2 n

Na upologisteð h akoloujða me genn tria sunˆrthsh 22x 3 9x 2 14x 1 (1 + x)(1 + 3x)(1 2x) 2 H anˆlush eðnai eukolìterh an qrhsimopoi soume th 22x 3 9x 2 14x 1 diˆspash: (1 + x)(1 + 3x)(1 2x) 2 = 1 1 + x + 1 1 + 3x 1 1 2x 2 (1 2x) 2

Ta prohgoômena aforoôn sta maj mata 8 kai 10/12/2010

H Ôlh sunoptikˆ... Stoiqei dhc sunduastik Genn triec sunart seic Egkleismìc - Apokleismìc JewrÐa Polyá

Pìsec apì autèc tic skakièrec eðnai al jeia diaforetikèc;

Autèc oi skakièrec eðnai sthn ousða Ðdiec... dhl. isodônamec!

H kˆje mða prokôptei apì thn prohgoômenh me dexiìstrofh peristrof katˆ 90 o

Autèc oi skakièrec eðnai sthn ousða Ðdiec... dhl. isodônamec!

H kˆje mða prokôptei apì thn prohgoômenh me dexiìstrofh peristrof katˆ 90 o

Autèc oi skakièrec eðnai sthn ousða Ðdiec... dhl. isodônamec!

H kˆje mða prokôptei apì thn prohgoômenh me dexiìstrofh peristrof katˆ 90 o

Autèc oi skakièrec eðnai sthn ousða Ðdiec... dhl. isodônamec!

H kˆje mða prokôptei apì thn prohgoômenh me dexiìstrofh peristrof katˆ 90 o

Autèc oi skakièrec eðnai isodônamec wc proc thn antðjesh qrwmˆtwn...!

Pìsouc trìpouc na anadiatˆxw th skakièra èqw; 6 Να τα αϕήσω όλα ως έχουν Να τα κάνω όλα πράσινα Να κάνω 1 πράσινο Να κάνω 2 διπλανά πράσινα Να κάνω 2 μη διπλανά διπλανά πράσινα Να κάνω 3 πράσινα

Profan c, den eðnai apodotikì gia kˆje parapl sia er thsh na kˆnoume thn exantlhtik anaz thsh pou kˆname prin na basizìmaste aplˆ sthn parat rhsh gia na metr soume to pl joc twn diaforetik n antikeimènwn ìtan emfanðzontai omoiìthtec lìgw kˆpoiac morf c summetrðac sth dom twn antikeimènwn... Upˆrqei kˆpoia idiìthta pou na èqoun ta diaforetikˆ ( ta Ðdia) antikeðmena bˆsei thc opoðac na èqoume èna susthmatikì trìpo prosdiorismoô twn diaforetik n antikeimènwn; Sth sunèqeia ja doôme th jewrða mètrhshc diaforetik n antikeimènwn ìtan lambˆnontai upìyh jèmata summetrðac pou anaptôqjhke apì ton Pólya to 1938.

Basikˆ stoiqeða jewrðac sunìlwn SÔnolo, uposônolo, gn sio uposônolo 'Enwsh, Tom, Diaforˆ Diamèrish Diatetagmèno zeôgoc Kartesianì ginìmeno Duadikèc sqèseic

SÔnolo, uposônolo, gn sio uposônolo SÔnolo: sullog diaforetik n antikeimènwn pou kaloôntai stoiqeða tou sunìlou, p.q., S = {a, b, c, x, z} Δε μετράει η σειρά των στοιχείων: {a, b, c} = {c, b, a} Δεν έχουν νόημα οι επαναλήψεις ίδιων στοιχείων: {a, a, b, c} = {a, b, c} {} = : σύνολο χωρίς στοιχεία T S: To T eðnai uposônolo tou S dhl., kˆje stoiqeðo tou T eðnai kai stoiqeðo tou S, p.q., {a, b, x} {a, b, c, x, z}, {a, b, y} {a, b, c, x, z} Kˆje sônolo eðnai uposônolo tou eautoô tou. T S: To T eðnai gn sio uposônolo tou S dhl., to T eðnai uposônolo tou S kai upˆrqei èna toulˆqiston stoiqeðo sto S pou den eðnai stoiqeðo tou T. α S: to α eðnai stoiqeðo tou sunìlou S S : pl joc stoiqeðwn tou sunìlou S To S eðnai èna k sônolo an perièqei k stoiqeða.

'Enwsh, Tom, Diaforˆ, Diamèrish 'Estw dôo sônola A kai B. A B: 'Enwsh twn sunìlwn A kai B, perièqei ìla ta stoiqeða twn sunìlwn A kai B. {a, b, c, d} {a, d, e, j} = {a, b, c, d, e, j} A B: Tom twn sunìlwn A kai B, perièqei ta koinˆ stoiqeða twn sunìlwn A kai B. {a, b, c, d} {a, d, e, j} = {a, d} A B: Diaforˆ twn sunìlwn A kai B, perièqei ta stoiqeða tou sunìlou A pou den an koun sto B. {a, b, c, d} {a, d, e, j} = {b, c} Diamèrish enìc sunìlou eðnai h upodiaðresh twn stoiqeðwn tou se xèna metaxô touc mh kenˆ uposônola ALLIWS Diamèrish enìc sunìlou eðnai mia sullog uposunìlwn tou tètoia ste kˆje stoiqeðo tou sunìlou na an kei se akrib c èna uposônolo, p.q., to sônolo {{a, b, x}, {d}, {c, z}} eðnai mia diamèrish tou sunìlou {a, b, c, d, x, z}.

Diatetagmèno zeôgoc, Kartesianì ginìmeno, Duadik sqèsh Diatetagmèno zeôgoc (a, b) eðnai mia diˆtaxh dôo - ìqi aparaðthta diaforetik n - stoiqeðwn a kai b. Ta (a, b) kai (b, a) eðnai dôo diaforetikˆ diatetagmèna zeôgh. Kartesianì ginìmeno dôo sunìlwn S kai T - S T - eðnai to sônolo ìlwn twn diatetagmènwn zeug n (x, y) sta opoða x S kai y T, p.q., {a, b, c} {1, 2} = {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)} Mia Duadik sqèsh metaxô twn sunìlwn S kai T eðnai èna uposônolo diatetagmènwn zeug n apì to kartesianì ginìmeno S T, p.q., {(a, 1), (a, 2), (c, 2)} eðnai mia duadik sqèsh metaxô twn sunìlwn {a, b, c} kai {1, 2}.

Diatetagmèno zeôgoc, Kartesianì ginìmeno, Duadik sqèsh Mia Duadik sqèsh metaxô dôo sunìlwn anaparðstatai me ènan pðnaka, p.q., h duadik sqèsh {(a, 1), (a, 3), (b, 4), (d, 2), (d, 4)} metaxô twn sunìlwn {a, b, c, d} kai {1, 2, 3, 4, 5} faðnetai ston parakˆtw pðnaka : Mia Duadik sqèsh se èna sônolo S eðnai mia duadik sqèsh metaxô tou S kai tou eautoô tou, p.q., {(a, a), (a, c), (b, a), (b, c), (c, b)} eðnai mia duadik sqèsh sto sônolo {a, b, c}

Sqèsh isodunamðac Mia Duadik sqèsh se èna sônolo S kaleðtai Sqèsh isodunamðac an ikanopoioôntai oi parakˆtw sunj kec: 1. Κάθε στοιχείο στο σύνολο σχετίζεται με τον εαυτό του (ανακλαστική ιδιότητα) 2. Για οποιαδήποτε στοιχεία a, b του συνόλου, αν το a σχετίζεται με το b τότε και το b σχετίζεται με το a (συμμετρική ιδιότητα) 3. Για οποιαδήποτε στοιχεία a, b, c του συνόλου, αν το a σχετίζεται με το b και b σχετίζεται με το c το τότε και το a σχετίζεται με το c (μεταβατική ιδιότητα) H duadik sqèsh aristerˆ eðnai sqèsh isodunamðac en aut sta dexiˆ den eðnai.

Klˆseic isodunamðac kai diamerðseic An èqoume mia sqèsh isodunamðac se èna sônolo S, mporoôme na qwrðsoume ta stoiqeða tou S se klˆseic - pou kaloôntai klˆseic isodunamðac - ètsi ste dôo stoiqeða na an koun sthn Ðdia klˆsh mìno an sqetðzontai metaxô touc. Κάθε στοιχείο ανήκει σε κάποια κλάση ισοδυναμίας αϕού μπορεί να είναι σε τουλάχιστον μία κλάση από μόνο του (λόγω της ανακλαστικής ιδιότητας). Δεν υπάρχει ασάϕεια σχετικά με το αν κάποιο στοιχείο ανήκει σε κάποια κλάση ισοδυναμίας (λόγω της συμμετρικής ιδιότητας) Κάθε στοιχείο δε μπορεί να ανήκει σε παραπάνω από μία κλάσεις ισοδυναμίας (λόγω της μεταβατικής ιδιότητας)

Klˆseic isodunamðac kai diamerðseic Mia sqèsh isodunamðac se èna sônolo orðzei mia diamèrish tou sunìlou sthn opoða ta xèna metaxô touc uposônola eðnai oi klˆseic isodunamðac, p.q., h diamèrish pou orðzetai apì th sqèsh isodunamðac sto sônolo {a, b, c, d, e} eðnai h {{a, b}, {c, d, e}}. DÔo stoiqeða eðnai isodônama an an koun sthn Ðdia klˆsh isodunamðac.

Sunart seic

Metajèseic Metˆjesh, SÔnolo metajèsewn G H metˆjesh pou antistoiqeð kˆje stoiqeðo ston eautì tou, af nei dhl. ta stoiqeða wc èqoun, lègetai tautotik. Duadik sqèsh epagìmenh apì sônolo metajèsewn G eðnai sqèsh isodunamðac

Metajèseic DÐnetai sônolo S = {a, b,...} kai èna sônolo metajèsewn G gia ta stoiqeða tou S Mia Duadik sqèsh sto S eðnai duadik sqèsh epagìmenh apì to G ìtan èna stoiqeðo a sqetðzetai me èna stoiqeðo b an kai mìnon an upˆrqei metˆjesh sto G pou apeikonðzei to a sto b. ( ) ( ) ( ) ( ) abcd abcd abcd abcd 'Estw G = {,,, }. abcd bacd abdc badc H Duadik sqèsh pou epˆgetai apì to G faðnetai ston pðnaka: H Duadik sqèsh se èna sônolo pou epˆgetai apì sônolo metajèsewn G eðnai sqèsh isodunamðac.

Je rhma Burnside ZhtoÔmeno: Pìsec diaforetikèc morfèc, p.q., qrwmatismoð, stoiqeðwn enìc sunìlou upˆrqoun ìtan emfanðzontai isodônamoi sqhmatismoð lìgw summetrðac; Idèa: Πάρε τις ενδεχόμενες μεταθέσεις των στοιχείων του συνόλου. Κάποιες από αυτές είναι ισοδύναμες και ϕτιάχνουν κλάσεις ισοδυναμίας. Οσες είναι οι κλάσεις ισοδυναμίας τόσοι είναι και οι διαϕορετικοί σχηματισμοί που ψάχνεις. Πόσες είναι οι κλάσεις ισοδυναμίας που δημιουργούνται από τις μεταθέσεις; Για να το βρεις μέτρα τα στοιχεία που μένουν ίδια από τις μεταθέσεις, δηλ. στοιχεία που η μετάθεση τα απεικονίζει στον εαυτό τους, άθροισέ τα και διαίρεσε το άθροισμα με το πλήθος των μεταθέσεων.

Je rhma Burnside ZhtoÔmeno: na metr sw diaforetikèc morfèc enìc sunìlou (antikeimènou) ìtan anadiatˆssontai ta stoiqeða (mèrh) tou. Parat rhsh: pl joc klˆsewn isodunamðac = pl joc diaforetik n metajèsewn DiatÔpwsh: To pl joc twn klˆsewn isodunamðac stic opoðec diamerðzetai èna sônolo S apì th sqèsh isodunamðac pou epˆgetai apì èna sônolo metajèsewn G 1 tou S eðnai: ψ(π) G π G Efarmog : DÐnetai èna sônolo S. 1. BrÐskw (ektìc an dðnetai) to sônolo metajèsewn G. 2. Se kˆje metˆjesh sto G brðskw to pl joc twn stoiqeðwn pou den allˆzoun. 3. Ta ajroðzw gia ìlec tic metajèseic kai diair to ˆjroisma me to pl joc twn metajèsewn G kai èqw to zhtoômeno.

Je rhma Burnside: Apìdeixh 'Estw sônolo S kai sônolo metajèsewn G stoiqeðwn tou S. Pìsa stoiqeða tou S mènoun Ðdia sunolikˆ se ìlec tic metajèseic tou G; Upˆrqoun dôo trìpoi na ta metr soume: eðte anˆ stoiqeðo eðte anˆ metˆjesh: Profan c: n(s) = y(π) π G s S

Je rhma Burnside: Apìdeixh 'Estw stoiqeða α kai b tou S pou eðnai isodônama upˆrqei metˆjesh sto G me α b. Pìse tètoiec metajèseic upˆrqoun; 'Osec - n(α) - af noun to stoiqeðo α Ðdio, dhl., perièqoun α α afoô gia kˆje mða mporoôme na èqoume α b wc ex c: α α b. Gia ton Ðdio lìgo n(α) eðnai kai oi metajèseic me α c...α h, gia kˆje ˆllo stoiqeðo sthn Ðdia klˆsh isodunamðac me to α.

Je rhma Burnside: Apìdeixh Oi metajèseic tou G mporoôn na omadopoihjoôn se autèc pou af noun to α Ðdio, autèc me α b, autèc me α c... gia kˆje stoiqeðo thc klˆshc isodunamðac pou perièqei to α. Kˆje mða apì tic omˆdec autèc perièqei n(α) metajèseic, ìpwc deðxame prin, dhl., n(α) = n(b) =... = n(h). 'Ara n(a) # stoiqeðwn sthn kl. isodunamðac pou perièqei to a = G G n(a) = # stoiqeðwn sthn klˆsh isodunamðac pou perièqei to a

Je rhma Burnside: Apìdeixh Opìte gia ìla ta stoiqeða sthn klˆsh isodunamðac eðnai: n(s) = G ìla ta s sthn klˆsh isodunamðac Kai gia ìlec tic klˆseic isodunamðac, dhl., ìla ta stoiqeða tou S: n(s) #klˆsewn isodunamðac stic opoðec qwrðzetai to S = G y(π) π G #klˆsewn isodunamðac stic opoðec qwrðzetai to S = G afoô n(s) = y(π) s S π G s S

Je rhma Burnside

Je rhma Burnside

Je rhma Burnside: Apìdeixh To na metr sw ta sunolikˆ stoiqeða pou mènoun Ðdia gia ìlec tic metajèseic eðnai san na metrˆw gia kˆje stoiqeðo tou dosmènou sunìlou to pl joc twn metajèsewn pou den to allˆzoun (dhl., to apeikonðzoun ston eautì tou). An èqoume dôo stoiqeða a kai b sthn Ðdia klˆsh isodunamðac, upˆrqoun tìsec metajèseic pou apeikonðzoun to a sto b ìsec eðnai oi metajèseic pou af noun to a Ðdio (dhl., to apeikonðzoun ston eautì tou). GiatÐ;

Je rhma Burnside

Je rhma Burnside

Je rhma Burnside: Apìdeixh An èqoume dôo stoiqeða a kai b sthn Ðdia klˆsh isodunamðac, upˆrqoun tìsec metajèseic pou apeikonðzoun to a sto b ìsec eðnai oi metajèseic pou af noun to a Ðdio (dhl., to apeikonðzoun ston eautì tou). GiatÐ; Αϕού τα a και b είναι ισοδύναμα, υπάρχει τουλάχιστον μία μετάθεση που απεικονίζει το a στο b. Κοίτα τώρα συνολικά τις μεταθέσεις που αϕήνουν το a ως έχει. Μπορώ να ϕτιάξω ένα σύνολο μεταθέσεων που απεικονίζουν το a στο b ως εξής: κάνω πρώτα κάποια από τις μεταθέσεις που αϕήνουν το a ως έχει και μετά τη μετάθεση που απεικονίζει το a στο b. Οι μεταθέσεις στο σύνολο αυτό: είναι μεταθέσεις που απεικονίζουν το a στο b, είναι όλες διαϕορετικές μεταξύ τους, γιατί αν δεν ήταν και τις έκανα αντίστροϕα θα έδιναν δυο διαϕορετικές μορϕές για το ίδιο στοιχείο a αυτές είναι οι μόνες δυνατές μεταθέσεις που απεικονίζουν το a στο b

Je rhma Burnside

Je rhma Burnside: Apìdeixh To na metr sw ta sunolikˆ stoiqeða pou mènoun Ðdia gia ìlec tic metajèseic eðnai san na metrˆw gia kˆje stoiqeðo tou dosmènou sunìlou to pl joc twn metajèsewn pou den to allˆzoun (dhl., to apeikonðzoun ston eautì tou). An èqoume dôo stoiqeða a kai b sthn Ðdia klˆsh isodunamðac, upˆrqoun tìsec metajèseic pou apeikonðzoun to a sto b ìsec eðnai sunolikˆ oi metajèseic pou af noun to a Ðdio (dhl., to apeikonðzoun ston eautì tou). GiatÐ; Koitˆw t ra ta stoiqeða tou S pou eðnai se mða klˆsh isodunamðac: {a, b, c,..., h}. Omadopoi tic metajèseic tou G se autèc pou apeikonðzoun to a sto a, to a sto b, to a sto c,..., to a sto h. 'Opwc dh deðxame se kˆje mia apì autèc tic omˆdec upˆrqoun tìsec metajèseic ìsec eðnai sunolikˆ oi metajèseic pou af noun to a Ðdio (dhl., to apeikonðzoun ston eautì tou), tic opoðec sumbolðzw n(a).

Je rhma Burnside

Je rhma Burnside: Apìdeixh Oi metajèseic tou G mporoôn na omadopoihjoôn se autèc pou af noun to α Ðdio, autèc me α b, autèc me α c... gia kˆje stoiqeðo thc klˆshc isodunamðac pou perièqei to α. Kˆje mða apì tic omˆdec autèc perièqei n(α) metajèseic, ìpwc deðxame prin. 'Ara n(a) # stoiqeðwn sthn kl. isodunamðac pou perièqei to a = G G n(a) = # stoiqeðwn sthn klˆsh isodunamðac pou perièqei to a

Je rhma Burnside: Apìdeixh Dhl., n(a) = G # stoiqeðwn sthn klˆsh isodunamðac pou perièqei to a Kˆnw to parapˆnw gia kˆje stoiqeðo b, c,..., h, opìte: n(a) + n(b) +... + n(h) = G. Epomènwc gia kˆje klˆsh isodunamðac sto S eðnai: n(s) = G ìla ta s sthn klˆsh isodunamðac 'Ara gia ìlo to S isqôei: n(s) = #klˆsewn isodunamðac stic opoðec qwrðzetai to S G s S

Je rhma Burnside

Je rhma Burnside

Parˆdeigma DÐnetai sônolo S = {a, b, c, d} ( kai sônolo ) metajèsewn ( ) abcd abcd G = {π 1, π 2, π 3, π 4 }, me π 1 =, π abcd 2 =, bacd ( ) ( ) abcd abcd π 3 =, π abdc 4 =. Pìsoi diaforetikoð badc sqhmatismoð upˆrqoun sto S lìgw tou G; Perimènoume na broôme dôo diaforetikoôc sqhmatismoôc afoô 2 eðnai kai oi klˆseic isodunamðac sto S lìgw tou G : Sth metˆjesh π 1 mènoun Ðdia 4 stoiqeða, sthn π 2 2, sthn π 3 2 kai sthn π 4 0 stoiqeða. Sunolikˆ, upˆrqoun 4 metajèseic sto G. Opìte to pl joc klˆsewn isodunamðac eðnai 1 4 (4 + 2 + 2 + 0) = 2.

Pìsa diaforetikˆ braqiolˆkia pou na èqoun 2 qˆntrec, mple kai kðtrinec, mporoôme na ftiˆxoume ìtan oi qˆntrec mporoôn na peristrèfontai; DÔo braqiolˆkia eðnai ta Ðdia an peristrèfontac to èna prokôptei to ˆllo. Oi pijanèc duˆdec apì qˆntrec eðnai oi: bb, by, yb, yy. Oi dunatèc metajèseic - na meðnoun wc èqoun na enallagoôn ( ta ˆkra touc ) - eðnai ( 2: bb, by, yb, yy bb, by, yb, yy π 1 =, π bb, by, yb, yy 2 = bb, yb, by, yy Sthn π 1 upˆrqoun 4 stoiqeða analloðwta. ) Sthn π 2 upˆrqoun 2 stoiqeða analloðwta. 'Ara oi klˆseic isodunamðac eðnai: 1 2 (4 + 2) = 3

Pìsa diaforetikˆ braqiolˆkia pou na èqoun 3 qˆntrec, mple kai kðtrinec, mporoôme na ftiˆxoume ìtan oi qˆntrec mporoôn na peristrèfontai; DÔo braqiolˆkia eðnai ta Ðdia an peristrèfontac to èna prokôptei to ˆllo. Oi pijanèc triˆdec apì qˆntrec eðnai oi: bbb, bby, byb, byy, ybb, yby, yyb, yyy. Oi dunatèc metajèseic - na meðnoun wc èqoun na enallagoôn ( ta ˆkra touc - eðnai 2: ) bbb, bby, byb, byy, ybb, yby, yyb, yyy π 1 =, bbb, bby, byb, byy, ybb, yby, yyb, yyy ( ) bbb, bby, byb, byy, ybb, yby, yyb, yyy π 2 = bbb, ybb, byb, yyb, bby, yby, byy, yyy Sthn π 1 upˆrqoun 8 stoiqeða analloðwta. Sthn π 2 upˆrqoun 4 stoiqeða analloðwta. 'Ara oi klˆseic isodunamðac eðnai: 1 2 (8 + 4) = 6

Pìsa diaforetikˆ braqiolˆkia pou na èqoun 5 qˆntrec, mple, kðtrinec kai ˆsprec, mporoôme na ftiˆxoume ìtan oi qˆntrec mporoôn na peristrèfontai; 'Estw S to sônolo apì ta 3 5 = 243 braqiolˆkia (dhl. ìlec tic dunatèc 5-ˆdec qwrðc na lambˆnoume upìyh summetrðec). Oi dunatèc metajèseic eðnai: Sthn π 1 mènoun Ðdia 243 stoiqeða. Gia kˆje mia apì tic upìloipec metajèseic, 3 5-ˆdec ja dðnoun Ðdia braqiolˆkia: ìtan ìlec oi qˆntrec èqoun to Ðdio qr ma, giatð mìno tìte h peristrof af nei analloðwto to braqiolˆki. 'Ara, upˆrqoun 1 5 (243 + 3 + 3 + 3 + 3) = 51 klˆseic isodunamðac (= diaforetikˆ braqiolˆkia).

Me pìsouc trìpouc mpor na bˆlw n ˆtoma na kaj soun se kôklo; Upˆrqoun n! metajèseic gia ta n ˆtoma an de lˆboume upìyh tic summetrðec. Upˆrqoun G = {π 1,...π n } metajèseic: π 1, h tautotik kai kˆje mða apì tic epìmenec prokôptei apì thn prohgoômenh me metatìpis thc katˆ mða jèsh katˆ th forˆ tou rologioô. Sthn π 1 mènoun Ðdiec ìlec oi jèseic, dhl., n! stoiqeða. Se kˆje mia apì tic upìloipec metajèseic kamða jèsh de mènei h Ðdia, ˆra analloðwta mènoun 0 stoiqeða. 'Ara, o sunolikìc arjmìc twn klˆsewn isodunamðac eðnai 1 n (n! + 0 +... + 0) = (n 1)!.

Jèlw na tup sw 5-y fia noômera se qartˆkia, èna se kˆje qartˆki. Pìsa diaforetikˆ qartˆkia prèpei na tup sw gia na èqw ìla ta noômera; Upˆrqoun 10 5 diaforetikˆ 5-y fia noômera an de lˆboume upìyh tic summetrðec. Pìsec metajèseic upˆrqoun sto sônolo G; Τα ψηϕία 0,1,6,8,9 είναι ίδια αν τα διαβάσω πάνω κάτω και δεξιά και πάνω (π.χ., 89166-99168) Opìte eðnai G = {π 1, π 2 } π 1 : tautotik metˆjesh - af nei ìla ta noômera ìpwc eðnai - mènoun Ðdia 10 5 stoiqeða π 2 : af nei ton arijmì Ðdio ìtan de mporeð na diabasteð pˆnw kˆtw (13765 13765) kai kˆnei ton arijmì Ðdio me ton antðstoiqo ìtan mporeð na diabasteð anˆpoda (p.q., 89166 99168). Ed mènoun Ðdia (10 5 5 5 ) + 3 5 2 stoiqeða. GiatÐ; 'Ara, o sunolikìc arjmìc twn klˆsewn isodunamðac eðnai 1 2 (105 + 10 5 5 5 + 3 5 2 ).

Jèlw na tup sw 5-y fia noômera se qartˆkia, èna se kˆje qartˆki. Pìsa diaforetikˆ qartˆkia prèpei na tup sw gia na èqw ìla ta noômera; Ta yhfða 2,3,4,5,7 de mporoôn na diabastoôn anˆpoda kai mporoôn na d soun 5 5 pentay fiouc. Gia na diabˆzontai pentay fioi Ðdia eðte pˆnw kˆtw eðte dexiˆ kai pˆnw prèpei to mesaðo yhfðo na eðnai 0,1,8 (3 epilogèc) to pr to yhfðo prèpei na eðnai to teleutaðo gurismèno anˆpoda: 0,1,8,6,9 (5 epilogèc) to deôtero yhfðo prèpei na eðnai to tètarto gurismèno anˆpoda: 0,1,8,6,9 (5 epilogèc)

Ta prohgoômena aforoôn sta maj mata 15 kai 17/12/2010

Mpor kalôtera...;;; Je rhma Burnside bohjˆei polô afoô antð na metrˆw klˆseic isodunamðac koitˆw tic metajèseic kai metrˆw mìno ta stoiqeða pou oi metajèseic af noun analloðwta... ALLA Πάλι μετράω πολλά στοιχεία και η διαδικασία παραμένει πολύπλοκη... Εκτός από το πλήθος των κλάσεων ισοδυναμίας δε μπορώ να έχω πληροϕορία και για άλλες ιδιότητες ισοδύναμων στοιχείων π.χ., αν θέλαμε να βρούμε τον αριθμό από διαϕορετικές σκακιέρες που να έχουν 2 πράσινα και 2 άσπρα τετραγωνάκια; Το θεώρημα Burnside δε βοηθάει...

Idèa... f : trìpoc na rðxw D antikeðmena se R koutiˆ pl joc f apì D R: pl joc trìpwn na rðxw D antikeðmena se R koutiˆ

Klˆseic isodunamðac sunart sewn 'Estw D kai R sônola kai G sônolo metajèsewn twn stoiqeðwn tou D. OrÐzoume thn ex c dimel sqèsh sto sônolo twn sunart sewn apì to D sto R: f 1, f 2 sqetðzontai an f 1 (d) = f 2 (π(d)), d D pou eðnai sqèsh isodunamðac. Epomènwc, oi sunart seic apì D R qwrðzontai se klˆseic isodunamðac pou kaloôntai prìtupa (patterns): antistoiqoôn se diaforetikoôc trìpouc na moirˆsw D antikeðmena se R koutiˆ ìtan h isodunamða metaxô twn moirasmˆtwn kajorðzetai apì to G.

Parˆdeigma 'Estw D = {a, b, c, d}, R = {x, ( y} kai ) sônolo metajèsewn ( ) abcd abcd G = {π 1, π 2, π 3, π 4 }, me π 1 =, π bcda 2 =, cdab ( ) ( ) abcd abcd π 3 =, π dabc 4 =. abcd Oi 16 sunart seic f : D R eðnai oi ex c:

Parˆdeigma 'Estw D = {a, b, c, d}, R = {x, ( y} kai ) sônolo metajèsewn ( ) abcd abcd G = {π 1, π 2, π 3, π 4 }, me π 1 =, π bcda 2 =, cdab ( ) ( ) abcd abcd π 3 =, π dabc 4 =. abcd Oi 16 sunart seic f : D R eðnai oi ex c:

Xanˆ to jèma me tic skakièrec... Onomˆzw ta koutˆkia thc skakièrac. 'Eqw dôo qr mata: ˆspro (x)-prˆsino (y ) Opìte D = {a, b, c, d}, R = {x, y} f : D R deðqnei th skakièra To sônolo metajèsewn pou prokôptoun me peristrof twn cells eðnai ( ) ( ) ( ) ( ) abcd abcd abcd abcd G = {,,, }. bcda cdab dabc abcd Akrib c to prohgoômeno parˆdeigma. Upˆrqoun 6 klˆseic isodunamðac, dhl., 6 diaforetikoð sqhmatismoð 2 qrwmˆtwn.