{"id":9023,"date":"2004-07-01T00:25:50","date_gmt":"2004-06-30T21:25:50","guid":{"rendered":"http:\/\/109.232.216.219\/~bilimvegelecek\/?p=9023"},"modified":"2017-05-16T12:33:27","modified_gmt":"2017-05-16T09:33:27","slug":"cahit-arf-matematik-gunleri-2004","status":"publish","type":"post","link":"https:\/\/bilimvegelecek.com.tr\/index.php\/2004\/07\/01\/cahit-arf-matematik-gunleri-2004","title":{"rendered":"Cahit Arf Matematik G\u00fcnleri 2004"},"content":{"rendered":"

\u0130kinci G\u00fcn \/ 21 Nisan 2004<\/strong>
\n\u0130stanbul Bilgi \u00dcniversitesi Matematik B\u00f6l\u00fcm\u00fc<\/strong><\/p>\n

Problem 1. <\/strong>D\u00fczlemdeki \u2124x\u2124 \u0131zgaras\u0131nda (yandaki \u015fekil) (0, 0) \"\"noktas\u0131ndan herhangi bir (k<\/em>, n<\/em>) noktas\u0131na hep kuzeye ve do\u011fuya gitmek ko\u015fuluyla ve \u0131zgaray\u0131 takip ederek ka\u00e7 de\u011fi\u015fik bi\u00e7imde gidilir?<\/p>\n

Problem 2.<\/strong> n<\/em> ve k<\/em> birer do\u011fal say\u0131 olsunlar.<\/p>\n

A<\/em>(n<\/em>, k<\/em>) = {(a<\/em>0<\/sub>, …, ak<\/sub><\/em>-1<\/sub>) : aj<\/sub><\/em> \u00ce \u2115 ve a<\/em>0<\/sub> + a<\/em>1<\/sub> + … + ak<\/sub><\/em>-1<\/sub> = n<\/em>}<\/p>\n

k\u00fcmesinin eleman say\u0131s\u0131n\u0131 bulun.<\/p>\n

Problem 3. <\/strong>n<\/em> elma, adlar\u0131 A<\/em>0<\/sub>, …, Ak<\/sub><\/em>-1<\/sub> olan k<\/em> ki\u015fi aras\u0131nda ka\u00e7 t\u00fcrl\u00fc da\u011f\u0131t\u0131l\u0131r? (Dikkat:<\/strong> Ki\u015filer aras\u0131nda ayr\u0131m yap\u0131yoruz, ama elmalar aras\u0131nda yapm\u0131yoruz.)<\/p>\n

Problem 4.<\/strong> n<\/em> \u00b3 1 ve k<\/em> \u00b3 1 birer do\u011fal say\u0131 olsunlar.<\/p>\n

B<\/em>(n<\/em>, k<\/em>) = {(b<\/em>1<\/sub>, …, bn<\/sub><\/em>) : bi<\/sub><\/em> \u00ce \u2115 ve 0 \u00a3 b<\/em>1<\/sub> \u00a3 b<\/em>2<\/sub> \u00a3 … \u00a3 bn<\/sub><\/em> \u00a3 k <\/em>– 1}<\/p>\n

k\u00fcmesinin eleman say\u0131s\u0131n\u0131 bulun.<\/p>\n

Problem 5. <\/strong>Birbirinden ay\u0131rt edilemeyen n<\/em> top ve bu toplar\u0131 boyayabilece\u011fimiz k<\/em> de\u011fi\u015fik renkte boyam\u0131z var. Bu n<\/em> topu bu k<\/em> de\u011fi\u015fik renge ka\u00e7 t\u00fcrl\u00fc boyayabiliriz? Dikkat: <\/strong>Toplar ay\u0131rt edilmiyor ama renkler ay\u0131rt ediliyor. \u00d6rnek: <\/strong>\u00dc\u00e7 top ve k\u0131rm\u0131z\u0131 ve mavi olmak \u00fczere iki rengimiz varsa (\u00fc\u00e7\u00fc mavi), (iki mavi, bir k\u0131rm\u0131z\u0131), (bir mavi, iki k\u0131rm\u0131z\u0131), (\u00fc\u00e7\u00fc k\u0131rm\u0131z\u0131) olmak \u00fczere d\u00f6rt t\u00fcrl\u00fc boyayabiliriz. E\u011fer \u00fc\u00e7 top ve \u00fc\u00e7 rengimiz varsa boyamay\u0131 10 t\u00fcrl\u00fc yapabiliriz.<\/p>\n

Problem 6. <\/strong>n<\/em> \u00b3 1 bir do\u011fal say\u0131 olsun. n<\/em>\u2019yi ka\u00e7 de\u011fi\u015fik bi\u00e7imde pozitif do\u011fal say\u0131lar\u0131n toplam\u0131 olarak yazabiliriz? (De\u011fi\u015fik s\u0131ralamalar ayr\u0131 ayr\u0131 say\u0131lacak, \u00f6rne\u011fin, n<\/em> = 14 ise, 1 + 3 + 5 + 5 ve 5 + 3 + 5 + 1 toplamlar\u0131 ayr\u0131 ayr\u0131 say\u0131lacak.)<\/p>\n

Problem 7.<\/strong> 1 + t<\/em> – 2t<\/em>2<\/sup> + 6t<\/em>3<\/sup> – 3t<\/em>6<\/sup> + … gibi sonsuza dek uzayabilen (polinoma benzeyen ancak sonsuza dek uzayabildiklerinden illa polinom olmayan) terimlere g\u00fc\u00e7 serisi<\/strong> ad\u0131 verilir. Dikkat:<\/strong> Buradaki t<\/em> bir say\u0131 de\u011fildir, de\u011fi\u015fken ad\u0131 verilen bir \u201c\u015fey\u201ddir. G\u00fc\u00e7 serileri de aynen polinomlar gibi toplan\u0131p \u00e7arp\u0131labilirler:<\/p>\n

(a<\/em>0<\/sub> + a<\/em>1<\/sub>t <\/em>+ a<\/em>2<\/sub>t<\/em>2<\/sup> + … )(b<\/em>0<\/sub> + b<\/em>1<\/sub>t <\/em>+ b<\/em>2<\/sub>t<\/em>2<\/sup> + … ) =<\/p>\n

a<\/em>0<\/sub>b<\/em>0<\/sub> + (a<\/em>0<\/sub>b<\/em>1<\/sub>+a<\/em>1<\/sub>b<\/em>0<\/sub>)t<\/em> + (a<\/em>0<\/sub>b<\/em>2<\/sub>+a<\/em>1<\/sub>b<\/em>1<\/sub>+a<\/em>2<\/sub>b<\/em>0<\/sub>)t<\/em>2<\/sup> + (a<\/em>0<\/sub>b<\/em>3<\/sub>+a<\/em>1<\/sub>b<\/em>2<\/sub>+a<\/em>2<\/sub>b<\/em>1<\/sub>+a<\/em>3<\/sub>b<\/em>0<\/sub>)t<\/em>3<\/sup> + …<\/p>\n

Bir g\u00fc\u00e7 serisi k\u0131saca\"\" bi\u00e7iminde yaz\u0131labilir. Burada, am<\/sub><\/em> katsay\u0131lar\u0131n\u0131 ger\u00e7el say\u0131lar olarak alaca\u011f\u0131z. E\u011fer am<\/sub><\/em> katsay\u0131lar\u0131 belli bir a\u015famadan sonra hep 0\u2019sa, o zaman g\u00fc\u00e7 serisi bir polinomdur, yoksa de\u011fildir. E\u011fer t\u00fcm am<\/sub><\/em>\u2019ler 0\u2019sa o zaman g\u00fc\u00e7 serisi de 0\u2019d\u0131r.<\/p>\n

Kolayca g\u00f6r\u00fclece\u011fi \u00fczere g\u00fc\u00e7 serilerini toplama ve \u00e7arpma kurallar\u0131 \u015f\u00f6yledir:
\n\"\"<\/p>\n

7.1.i. <\/strong>(1 – t<\/em>)(1+ t<\/em> + t<\/em>2<\/sup> + t<\/em>3<\/sup> + t<\/em>4<\/sup> + … ) = 1 e\u015fitli\u011fini g\u00f6sterin. (Not: Her polinom gibi, 1-t<\/em> polinomu da bir g\u00fc\u00e7 serisidir.)
\n7.2.ii. <\/strong>(1 + t<\/em>)g<\/em>(t<\/em>) = 1 e\u015fitli\u011fini sa\u011flayan bir g<\/em>(t<\/em>) g\u00fc\u00e7 serisi bulun.
\n7.2.iii. <\/strong>(1 – 2t<\/em>)g<\/em>(t<\/em>) = 1 e\u015fitli\u011fini sa\u011flayan bir g<\/em>(t<\/em>) g\u00fc\u00e7 serisi bulun.
\n7.3.<\/strong> \"\"\u00a0<\/strong>e\u015fitli\u011fini g\u00f6sterin.
\n7.4.<\/strong> \u0130ki g\u00fc\u00e7 serisini \u00e7arpabildi\u011fimiz gibi, \u00fc\u00e7 g\u00fc\u00e7 serisini de \u00e7arpabiliriz:
\n\"\"<\/p>\n

Anla\u015f\u0131laca\u011f\u0131 \u00fczere sonlu tane g\u00fc\u00e7 serisini \u00e7arpabiliriz.
\n\"\"e\u015fitli\u011fini g\u00f6sterin.<\/p>\n

7.5. <\/strong>A\u015fa\u011f\u0131daki \u00e7arp\u0131mlar\u0131 bulun:
\n(1+t<\/em>)(1+t<\/em>2<\/sup>)
\n(1+t<\/em>)(1+t<\/em>2<\/sup>)(1+t<\/em>4<\/sup>)
\n(1+t<\/em>)(1+t<\/em>2<\/sup>)(1+t<\/em>4<\/sup>)(1+t<\/em>8<\/sup>)
\n(1+t<\/em>)(1+t<\/em>2<\/sup>)(1+t<\/em>4<\/sup>)(1+t<\/em>8<\/sup>)(1+t<\/em>16<\/sup>)
\nBundan kimi zaman sonsuz tane g\u00fc\u00e7 serisini \u00e7arpabildi\u011fimiz \u00e7\u0131kar.
\n\"\"
\n\u00e7arp\u0131m\u0131n\u0131n ne olmas\u0131 gerekti\u011fini s\u00f6yleyin.
\n7.6. <\/strong>(1 + t<\/em>)(1 + t<\/em>)(1 + t<\/em>)(1 + t<\/em>)(1 + t<\/em>) …. sonsuz \u00e7arp\u0131m\u0131 bir g\u00fc\u00e7 serisi olarak yaz\u0131lamaz. Neden?<\/p>\n

Problem 8. <\/strong>A\u015fa\u011f\u0131daki \u00e7arp\u0131mlar\u0131 bulun:
\n(1+t<\/em>)(1+t<\/em>2<\/sup>)
\n(1+t<\/em>)(1+t<\/em>2<\/sup>)(1+t<\/em>3<\/sup>)
\n(1+t<\/em>)(1+t<\/em>2<\/sup>)(1+t<\/em>3<\/sup>)(1+t<\/em>4<\/sup>)
\n(1+t<\/em>)(1+t<\/em>2<\/sup>)(1+t<\/em>3<\/sup>)(1+t<\/em>4<\/sup>)(1+t<\/em>5<\/sup>)
\nBundan,
\n\"\"
\nsonsuz \u00e7arp\u0131m\u0131n\u0131n var oldu\u011fu anla\u015f\u0131lmal\u0131.
\nE\u011fer n<\/em> pozitif bir tamsay\u0131ysa, q<\/em>(n<\/em>), n<\/em>\u2019yi birbirinden de\u011fi\u015fik pozitif tamsay\u0131lar\u0131n toplam\u0131 olarak yazma bi\u00e7imi olsun, yani q<\/em>(n<\/em>),
\nA<\/em>(n<\/em>) = {(a<\/em>1<\/sub>, …, ak<\/sub><\/em>) : k<\/em> \u00ce \u2115, k<\/em> \u00b3 1, a<\/em>1<\/sub>, …, ak<\/sub><\/em> \u00ce \u2115, 1 \u00a3 a<\/em>1<\/sub> < … < ak<\/sub><\/em>\u00a0 ve a<\/em>1<\/sub> + a<\/em>2<\/sub> + … + ak<\/sub><\/em> = n<\/em>}
\nk\u00fcmesinin eleman say\u0131s\u0131 olsun. Ayr\u0131ca q<\/em>(0) = 1 olsun. \u00d6rne\u011fin q<\/em>(0) = 1, q<\/em>(1) = 1, q<\/em>(2) = 1, q<\/em>(3) = 2, q<\/em>(4) = 2, q<\/em>(5) = 3, q<\/em>(6) = 4, q<\/em>(7) = 5, q<\/em>(8) = 6.
\n\"\"
\ne\u015fitli\u011fini kan\u0131tlay\u0131n.
\n<\/strong>Problem 9. \"\"<\/strong>\u00a0sonsuz \u00e7arp\u0131m\u0131, yani,
\n\"\"
\nya da
\n(1 + t<\/em> + t<\/em>2<\/sup> + t<\/em>3<\/sup> + … )(1 + t<\/em>2<\/sup> + t<\/em>4<\/sup> + t<\/em>6<\/sup> + … )(1 + t<\/em>3<\/sup> + t<\/em>6<\/sup> + t<\/em>9<\/sup> + … )(1 + t<\/em>4<\/sup> + t<\/em>8<\/sup> + t<\/em>12<\/sup> + … )…
\nsonsuz \u00e7arp\u0131m\u0131 bir g\u00fc\u00e7 serisidir. Bu sonsuz \u00e7arp\u0131m\u0131
\n\"\"
\nolarak yazal\u0131m. Buradaki p<\/em>(n<\/em>) say\u0131lar\u0131n\u0131n matemati\u011fin daha g\u00fcncel ve daha anla\u015f\u0131l\u0131r bir dilinde (yukardaki sorulardan esinlenerek belki) neyi ifade etti\u011fini bulun.<\/p>\n","protected":false},"excerpt":{"rendered":"

\u0130kinci G\u00fcn \/ 21 Nisan 2004 \u0130stanbul Bilgi \u00dcniversitesi Matematik B\u00f6l\u00fcm\u00fc Problem 1. D\u00fczlemdeki \u2124x\u2124 \u0131zgaras\u0131nda (yandaki \u015fekil) (0, 0) noktas\u0131ndan herhangi bir (k, n) noktas\u0131na hep kuzeye ve do\u011fuya gitmek ko\u015fuluyla ve \u0131zgaray\u0131 takip ederek ka\u00e7 de\u011fi\u015fik bi\u00e7imde gidilir? Problem 2. n ve k birer do\u011fal say\u0131 olsunlar. A(n, k) = {(a0, …, ak-1) […]<\/p>\n","protected":false},"author":22,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[43,25,514],"tags":[208],"class_list":["post-9023","post","type-post","status-publish","format-standard","hentry","category-5-sayi","category-matematik","category-matematik-sohbetleri","tag-matematik"],"acf":[],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t