{"id":73,"date":"2007-12-22T00:47:28","date_gmt":"2007-12-21T22:47:28","guid":{"rendered":"http:\/\/www.umitcanli.com\/school\/index.php\/2007\/12\/22\/ev-yapimi-mobius-seritleri\/"},"modified":"2025-05-31T14:09:46","modified_gmt":"2025-05-31T12:09:46","slug":"ev-yapimi-mobius-seritleri","status":"publish","type":"post","link":"http:\/\/www.umitcanli.com\/school\/index.php\/2007\/12\/22\/ev-yapimi-mobius-seritleri\/","title":{"rendered":"M\u00f6bius \u015eeridi I"},"content":{"rendered":"

M\u00f6bius \u015feridine, internetteki \u00e7e\u015fitli pop\u00fcler matematik yaz\u0131lar\u0131nda s\u0131k\u00e7a rastlan\u0131r. Bir \u00f6nceki yaz\u0131da,<\/a> \u00e7ok yerde okuyabilmemize kar\u015f\u0131n, g\u00f6z\u00fcm\u00fczde canland\u0131rmam\u0131z\u0131n zor oldu\u011fu bir d\u00f6n\u00fc\u015f\u00fcm\u00fcn videosu vard\u0131 . Tabii ki b\u00f6yle bir d\u00f6n\u00fc\u015f\u00fcm i\u00e7in, “Nas\u0131l oluyor da oluyor?” demek veya daha \u00f6ncesi bunu merak etmek i\u00e7in, \u00f6ncelikle onu tan\u0131mak gerekir. Bu y\u00fczden amac\u0131m\u0131z, m\u00f6bius \u015feridini ve belki ba\u015fka bir yaz\u0131da da Klein \u015fi\u015fesini tan\u0131tmak. \u00d6ncelikle, M\u00f6bius \u015feridi ne imi\u015f bir g\u00f6relim. \u0130\u015fte size Wikipedia’dan al\u0131nan, estetik bir m\u00f6bius \u015feridinin resmi.<\/p>\n

<\/p>\n

Yapmak ve g\u00f6rmek, anlaman\u0131n en iyi yoludur (Bug\u00fcnk\u00fc e\u011fitimin de en b\u00fcy\u00fck eksikli\u011fi) . Malzemelerimiz \u015funlar: K\u00e2\u011f\u0131t,bant ve makas. K\u0131rtasiyeci amcadan ald\u0131\u011f\u0131n\u0131z ka\u011f\u0131d\u0131, a\u015fa\u011f\u0131da g\u00f6r\u00fcld\u00fc\u011f\u00fc \u00fczere bir \u015ferit olacak \u015fekilde kesin.<\/p>\n

<\/p>\n

\u015eimdi \u015feridi al\u0131p, A ile C, ve B ile D \u00e7ak\u0131\u015facak \u015fekilde bantlay\u0131p birle\u015ftirin. Elinizde tabii ki gayet normal bir \u015ferit var \u015fu an. [AC] ve [BD] (do\u011fru par\u00e7alar\u0131na) “kenar”, A,B,C ve D’nin s\u0131n\u0131rlad\u0131\u011f\u0131 k\u0131sma da “y\u00fczey” diyelim. \u015eimdi \u015feridin i\u00e7erde kalan y\u00fczeyine parma\u011f\u0131n\u0131z\u0131 koyun ve \u015ferit boyunca dola\u015ft\u0131r\u0131n. Ayn\u0131 \u015feyi d\u0131\u015ftaki y\u00fczey i\u00e7in de yapt\u0131\u011f\u0131n\u0131zda, ka\u011f\u0131d\u0131n \u00e7ok \u00e7ok ince olan kenar\u0131n\u0131 a\u015fmadan bir y\u00fczeyden di\u011ferine ge\u00e7menin m\u00fcmk\u00fcn olmad\u0131\u011f\u0131n\u0131 g\u00f6rmemiz tabii ki ileri bir anlay\u0131\u015f\u0131n g\u00f6stergesi de\u011fil. Bu s\u0131k\u0131c\u0131 bir deneydi ama haz\u0131r parma\u011f\u0131n\u0131z \u015feridin \u00fczerindeyken, bir de ayn\u0131 deneyi ka\u011f\u0131d\u0131n kenar\u0131 \u00fczerinde yap\u0131n. Burada da g\u00f6zlemleyece\u011finiz \u00fczere parma\u011f\u0131n\u0131z\u0131 ka\u011f\u0131ttan kald\u0131rmadan, bir kenardan di\u011ferine ge\u00e7menin tek yolu y\u00fczeyi kullanmakt\u0131r. \u015eimdi M\u00f6bius \u015feridini enteresan k\u0131lan \u00f6zelli\u011fe gelelim. M\u00f6bius \u015feridini yapmak \u00fczere, \u015fekilde g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi, C ve D nin oldu\u011fu ucu kendi etraf\u0131nda d\u00f6nd\u00fcr\u00fcyoruz. D\u00f6nd\u00fcrd\u00fc\u011f\u00fcn\u00fczde olmas\u0131 gereken durum a\u015fa\u011f\u0131da g\u00f6r\u00fcld\u00fc\u011f\u00fc gibi.<\/p>\n

<\/p>\n

\u015eimdi A ile D, ve B ile C \u00e7ak\u0131\u015facak \u015fekilde, k\u00e2\u011f\u0131d\u0131n iki ucunu bantlayarak birle\u015ftirin ve ev yap\u0131m\u0131 ilk M\u00f6bius \u015feridinize bak\u0131n. \u0130lk \u015feritte yapt\u0131\u011f\u0131m\u0131z deneyleri M\u00f6bius \u015feridi \u00fczerinde tekrarlay\u0131n. Ka\u011f\u0131d\u0131n y\u00fczeyine parma\u011f\u0131n\u0131z\u0131 koyun ve az \u00f6nce yapt\u0131\u011f\u0131n\u0131z gibi dola\u015ft\u0131rmaya ba\u015flay\u0131n. Az \u00f6ncekinin aksine, t\u00fcm y\u00fczeyi dola\u015fabildi\u011finizi farkedeceksiniz. Bu da \u015fu anlama gelir: M\u00f6bius \u015feridinin herhangi bir y\u00fczeyini boyamaya ba\u015flarsan\u0131z, \u015feridin t\u00fcm y\u00fczeyini boyaman\u0131z gerekecek. Yani elinizde tuttu\u011funuz \u015feridin, az \u00f6nce oldu\u011fu gibi i\u00e7 ve d\u0131\u015f olarak ay\u0131rabilece\u011finiz iki y\u00fczeyi de\u011fil, tek bir y\u00fczeyi vard\u0131r. Bunu daha da somutla\u015ft\u0131rman\u0131n yolu, (e\u011fer estetik kayg\u0131n\u0131z varsa \u015f\u00f6yle g\u00fczelinden) bir kalem al\u0131p az \u00f6nce parma\u011f\u0131n\u0131z\u0131n izledi\u011fi yolu y\u00fczey boyunca \u00e7izmektir. G\u00f6rece\u011finiz \u00fczere, \u00e7izdi\u011finiz yol t\u00fcm \u015feridi dola\u015f\u0131yor.<\/p>\n

\u015eimdiye kadar yapt\u0131klar\u0131m\u0131z\u0131 \u00f6zetlersek: K\u00e2\u011f\u0131t bir \u015ferit al\u0131p herhangi bir k\u0131v\u0131rma yapmadan u\u00e7lar\u0131 birle\u015ftirdik. \u0130lk yapt\u0131\u011f\u0131m\u0131z bu \u015feride K\u00b0 diyelim. K\u00b0 \u0131n 2 y\u00fczeyi ve 2 kenar\u0131 oldu\u011funu g\u00f6rd\u00fck. Ard\u0131ndan \u015feridin ucunu bir kere k\u0131v\u0131rd\u0131k ve M\u00f6bius \u015feridini elde ettik. M\u00f6bius \u015feridini de K\u00b9 olarak adland\u0131r\u0131rsak, K\u00b9 in tek bir y\u00fczeyi ve tek bir kenar\u0131 oldu\u011funu anlad\u0131k. Burada 0 ve 1, tabii ki \u015feridin birle\u015ftirilmeden \u00f6nceki k\u0131vr\u0131lma say\u0131s\u0131n\u0131 g\u00f6steriyor. Daha ileri gitmeden sorulabilecek \u00e7e\u015fitli sorular var. Sorulardan biri \u015fu olabilir:<\/p>\n

n’nin hangi de\u011ferleri i\u00e7in, Kn <\/sup>tek y\u00fczeye (ve tek kenara), hangi de\u011ferleri i\u00e7in iki y\u00fczeye (ve iki kenara) sahiptir ? (Verece\u011finiz yan\u0131t\u0131n do\u011frulu\u011fundan emin de\u011filseniz veya bir yan\u0131t\u0131n\u0131z yoksa en uygun yol, bilimum K2<\/sup><\/span>, K3<\/sup>, K4<\/sup>,<\/span>… \u015feritlerini yap\u0131p incelemek olacakt\u0131r)<\/p><\/blockquote>\n

Bunun yan\u0131t\u0131n\u0131 verenler de \u015feritleri daha estetik hale getirip \u00fczerlerine K\u00b0, K\u00b9 yazarak “m\u00fchim \u015feyler yap\u0131yorum” hissine kap\u0131lma yolunda \u00f6nemli bir ad\u0131m atabilir, \u015feritleri mat bir zemin \u00fczerine koyup seyrederken, bir yandan da elektronik m\u00fczik dinleyebilirler. \u0130ste\u011fe ba\u011fl\u0131 olarak yap\u0131labileceklerden baz\u0131lar\u0131, renkli kalemler kullan\u0131p estetik bir ok bi\u00e7imiyle y\u00fczeyleri i\u015faretlemek veya kenarlar\u0131 boyamak olabilir.<\/p>\n

Ve i\u015fler kar\u0131\u015fmaya ba\u015flar…<\/strong><\/p>\n

\u015eimdi yapaca\u011f\u0131m\u0131z ise, \u015feritleri belli oranlarda kesmek ve bu yap\u0131lan i\u015flem ile elde etti\u011fimiz par\u00e7ay\u0131 veya par\u00e7alar\u0131 incelemek. \u00d6ncelikle bir M\u00f6bius \u015feridini a\u015fa\u011f\u0131daki \u015fekilde oldu\u011fu gibi g\u00f6rmeye \u00e7al\u0131\u015fal\u0131m, yani boylamas\u0131na b\u00f6lgelerden yap\u0131land\u0131r\u0131lm\u0131\u015f gibi. Az \u00f6nce yapt\u0131klar\u0131m\u0131z g\u00f6z \u00f6n\u00fcne al\u0131nd\u0131\u011f\u0131nda, her bir b\u00f6lgenin i\u00e7indeki bir okun, \u015feridin t\u00fcm y\u00fczeyini tamamen dola\u015ft\u0131\u011f\u0131n\u0131 anlamak g\u00fc\u00e7 olmayacakt\u0131r. \u015eekildeki M\u00f6bius \u015feridi, (birbirine paralel olarak) 4 b\u00f6lgeye ayr\u0131lm\u0131\u015f. Bu \u015feridi, K1 <\/sup>B4 <\/sup>olarak adland\u0131rabiliriz.<\/p>\n

<\/p>\n

A\u00e7\u0131kt\u0131r ki, K0<\/sup><\/span> \u0131 ortadan bir \u00e7izgiyle boylamas\u0131na ay\u0131r\u0131rsak, yine iki adet K0 <\/sup><\/span> elde ederiz. Bu i\u015flemi :<\/p>\n

K0<\/sup><\/span> B2 <\/sup><\/span>\u2192 K0<\/sup><\/span> + <\/span>K0<\/sup><\/span><\/p><\/blockquote>\n

olarak g\u00f6sterebiliriz. Elde edilen her bir \u015ferit kal\u0131nl\u0131k olarak tabii ki yar\u0131ya inmi\u015ftir ama uzunluklar\u0131 ayn\u0131d\u0131r. \u015eimdi ayn\u0131 kesme i\u015flemini K1 <\/sup><\/span>e yapal\u0131m. \u00d6nerilen, mevcut \u015feritleri al\u0131p kesmek de\u011fil, kesilmeden \u00f6nceki hallerini kar\u015f\u0131la\u015ft\u0131rmak \u00fczere yeni \u015feritler yap\u0131p onlar\u0131 kesmenizdir. K1<\/sup> i al\u0131p kesti\u011finizde, \u015fa\u015f\u0131rt\u0131c\u0131 ve sezgi g\u00fcc\u00fcn\u00fcz\u00fc biraz y\u0131k\u0131ma u\u011fratan (u\u011framad\u0131 deyip kendinizi kand\u0131rmay\u0131n) bir sonu\u00e7la kar\u015f\u0131la\u015fmaya haz\u0131r olun. Bu kez ne yaz\u0131k ki iki \u015ferit de\u011fil, tek bir \u015ferit elde ediyoruz. A\u015fa\u011f\u0131daki \u015fekilde amac\u0131na uygun olarak yap\u0131lm\u0131\u015f bir M\u00f6bius \u015feridi, yani K1 <\/sup>g\u00f6r\u00fclmekte. Yap\u0131lan i\u015flem:<\/p>\n

K1<\/sup><\/span> B2 <\/sup><\/span>\u2192 K2<\/sup><\/span><\/p><\/blockquote>\n

Before<\/strong><\/p>\n

<\/p>\n

Ve ortadan kesilince \u015fu hale geliyor.<\/p>\n

After<\/strong><\/p>\n

<\/p>\n

\u015eimdi \u015feridi biraz evirip \u00e7evirelim. Keserek elde etti\u011fimiz yeni \u015feridin incelenecek bir \u00e7ok \u00f6zelli\u011fi var. \u00d6rne\u011fin bu \u015ferit, art\u0131k bir K1 <\/sup> de\u011fildir. Arka k\u0131sma do\u011fru bakt\u0131\u011f\u0131n\u0131zda, bu \u015feridin iki defa katlanm\u0131\u015f oldu\u011funu, yani art\u0131k bir K2<\/sup><\/span> haline geldi\u011fini g\u00f6rebilirsiniz. Art\u0131k bu \u015feridin tek bir y\u00fczeyi de\u011fil, K0<\/sup><\/span> da oldu\u011fu gibi iki y\u00fczeyi vard\u0131r. Ayn\u0131 sonuca, \u015feridin y\u00fczeyindeki oklar\u0131 takip ederek de ula\u015fabiliriz. Oklar sadece tek bir y\u00fczeyde bulunmaktad\u0131r ama kesme i\u015flemini yapmadan \u00f6nce, \u015feridin tamam\u0131n\u0131 dola\u015f\u0131yorlard\u0131. Yap\u0131labilecek ba\u015fka bir g\u00f6zlem, \u015feridin uzunlu\u011funun iki kat\u0131na \u00e7\u0131km\u0131\u015f olmas\u0131d\u0131r (\u00e7\u00fcnk\u00fc tek par\u00e7a olarak kald\u0131) . Uzunlu\u011fu, az \u00f6nceki sembolizasyona adapte ederek U ile g\u00f6sterelim ve yukar\u0131daki d\u00f6n\u00fc\u015f\u00fcm\u00fc bir daha yazal\u0131m.<\/p>\n

K1<\/sup><\/span> B2 <\/sup><\/span>\u2192 K2<\/sup><\/span>U2<\/sup><\/span><\/p><\/blockquote>\n

\u0130\u015flerin kar\u0131\u015fmaya ba\u015flad\u0131\u011f\u0131 nokta da tam burada, \u015feritlerin kesilme oran\u0131nda ortaya \u00e7\u0131kmaktad\u0131r. Yani, \u015feridi yanyana ka\u00e7 b\u00f6lgeye ay\u0131rd\u0131\u011f\u0131m\u0131za ba\u011fl\u0131 olarak \u015fa\u015f\u0131rt\u0131c\u0131 sonu\u00e7larla kar\u015f\u0131la\u015f\u0131yoruz. Bu kesme i\u015flemiyle elde edilen \u015feritler ya da halkalar , ingilizce bir ka\u00e7 kaynakta “paradromik halka” olarak ge\u00e7iyor (paradromic ring<\/a>). TDK’n\u0131n 2000 bas\u0131m\u0131 “Matematik Terimleri S\u00f6zl\u00fc\u011f\u00fc” nde “paradromik” terimi yer alm\u0131yor, yani bu terimin hen\u00fcz bir t\u00fcrk\u00e7e kar\u015f\u0131l\u0131\u011f\u0131 var m\u0131 bunu bilmiyorum. Asl\u0131nda, yap\u0131lan mevcut aramalarla, “paradromic ring” teriminin ge\u00e7ti\u011fi yeg\u00e2ne kaynak Wolfram olarak g\u00f6r\u00fcnmekte. Bloglarda okuyaca\u011f\u0131n\u0131z M\u00f6bius \u015feridi ile ilgili bir \u00e7ok i\u00e7eri\u011fin metni de direkt olarak Wolfram’dan al\u0131nma. Wolfram’da paradromik halkalarla ilgili a\u015fa\u011f\u0131daki tablo var.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
half-twists<\/td>\ncuts<\/td>\ndivs.<\/td>\nresult<\/td>\n<\/tr>\n
1<\/td>\n1<\/td>\n2<\/td>\n1 band, length 2<\/td>\n<\/tr>\n
1<\/td>\n1<\/td>\n3<\/td>\n1 band, length 2<\/td>\n<\/tr>\n
<\/td>\n<\/td>\n<\/td>\n1 M\u00f6bius strip, length 1<\/td>\n<\/tr>\n
1<\/td>\n2<\/td>\n4<\/td>\n2 bands, length 2<\/td>\n<\/tr>\n
1<\/td>\n2<\/td>\n5<\/td>\n2 bands, length 2<\/td>\n<\/tr>\n
<\/td>\n<\/td>\n<\/td>\n1 M\u00f6bius strip, length 1<\/td>\n<\/tr>\n
1<\/td>\n3<\/td>\n6<\/td>\n3 bands, length 2<\/td>\n<\/tr>\n
1<\/td>\n3<\/td>\n7<\/td>\n3 bands, length 2<\/td>\n<\/tr>\n
<\/td>\n<\/td>\n<\/td>\n1 M\u00f6bius strip, length 1<\/td>\n<\/tr>\n
2<\/td>\n1<\/td>\n2<\/td>\n2 bands, length 1<\/td>\n<\/tr>\n
2<\/td>\n2<\/td>\n3<\/td>\n3 bands, length 1<\/td>\n<\/tr>\n
2<\/td>\n3<\/td>\n4<\/td>\n4 bands, length 1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Merakl\u0131s\u0131 tabloyu inceleyebilir. Az \u00f6nce bahsedilen ve sonucu foto\u011frafta g\u00f6r\u00fcl\u00fcyor olan i\u015flem, tablonun ilk sat\u0131r\u0131nda yer al\u0131yor. Bu yaz\u0131da, “\u015feridi bir kere k\u0131v\u0131rma” olarak anlat\u0131lan i\u015flem, tabloda “half-twist” olarak ge\u00e7iyor. “Divs” s\u00fctunu, \u015feridin yanyana ka\u00e7 b\u00f6lgeye ayr\u0131ld\u0131\u011f\u0131n\u0131 g\u00f6steriyor. A\u015fa\u011f\u0131da, hevesle boyanan ve kesilen di\u011fer \u015feritleri inceleyebilirsiniz.<\/p>\n

K1<\/sup><\/span> B3 <\/sup><\/span>\u2192 <\/span>K1<\/sup><\/span> <\/strong>U1<\/sup><\/span><\/strong> + K2 <\/sup><\/span>U2<\/sup><\/span><\/strong><\/p>\n

K1<\/sup><\/span> B3<\/sup><\/span><\/p>\n

<\/p>\n

K1<\/sup><\/span> U1<\/sup><\/span> + K2 <\/sup><\/span>U2<\/sup><\/span><\/p>\n

<\/p>\n

Yukar\u0131daki \u015feridin ayn\u0131s\u0131 (K1<\/sup><\/span> U1<\/sup><\/span> sola do\u011fru kayd\u0131r\u0131lm\u0131\u015f)<\/p>\n

<\/p>\n

K1<\/sup><\/span> B5 <\/sup><\/span>\u2192 <\/span>K1<\/sup><\/span> <\/strong>U1<\/sup><\/span><\/strong> + 2 K2 <\/sup><\/span>U2<\/sup><\/span><\/strong><\/p>\n

K1<\/sup><\/span> B5<\/sup><\/span> de\u011fil, d\u00f6n\u00fc\u015f\u00fcm sonucu elde edilen K1<\/sup><\/span>U 1<\/sup><\/span> + 2 K2 <\/sup><\/span>U2<\/sup><\/span> g\u00f6sterilmi\u015ftir.<\/p>\n

<\/p>\n

Paradromik halkalar ve M\u00f6bius \u015feridiyle ilgili yaz\u0131l\u0131p \u00e7izilecek olduk\u00e7a fazla \u00f6zellik mevcut. Burada, sadece \u00f6nemli g\u00f6r\u00fclen ve keyifle yap\u0131lan bir ka\u00e7 \u00e7al\u0131\u015fma ve bunlar\u0131n g\u00f6zlemleri yer almakta. \u0130lgililenenler daha fazlas\u0131n\u0131 yaz\u0131 i\u00e7inde ge\u00e7en ba\u011flant\u0131lardan inceleyebilirler. Y\u00f6nlendirilebilir y\u00fczey, bir torusun (k\u0131saca simidin) \u00e7e\u015fitli \u015feritlere b\u00f6l\u00fcnebilmesi veya tekrar yap\u0131\u015ft\u0131r\u0131larak bir torus elde edilebilmesi, Klein \u015fi\u015fesiyle ili\u015fkileri gibi bir \u00e7ok konu, ara\u015ft\u0131rmaya ve merak etmeye de\u011fer.<\/p>\n

M\u00f6bius \u015feridine, “sanatsal \u00e7al\u0131\u015fma” olarak adland\u0131rd\u0131\u011f\u0131m\u0131z bir \u00e7ok i\u015fte de rastlamak m\u00fcmk\u00fcn. Akla gelmesi gereken ilk ki\u015fi Escher olmal\u0131. Bunun yan\u0131nda \u0130lhan Koman,<\/a> “3-D M\u00f6bius t\u00fcrevleri ve piramitler<\/a>” ad\u0131 alt\u0131ndaki serisinde M\u00f6bius formunu kullanm\u0131\u015ft\u0131r. Koman’\u0131n M\u00f6bius uygulamalar\u0131ndan biri<\/a> Stockholm \u015eehir Terminalinde yer al\u0131yor. Ayr\u0131ca “paradromik halka” terimi arat\u0131l\u0131rken, bir de \u015funa<\/a> rastland\u0131.<\/p>\n

\u015eeritlerle u\u011fra\u015f\u0131p, biraz da unutal\u0131 yakla\u015f\u0131k bir ay kadar oldu. O y\u00fczden bu yaz\u0131 da biraz da\u011f\u0131n\u0131k ve b\u00f6l\u00fck p\u00f6r\u00e7\u00fck oldu. Bug\u00fcnl\u00fck M\u00f6bius \u015feridi yetsin, yeniden ilgimizi \u00e7ekti\u011fi ve vakit bulabildi\u011fimiz bir zamanda daha fazlas\u0131n\u0131 yazar\u0131z…<\/p>\n\"1\"2\"3\"4\"5 (21<\/strong> ki\u015finin oyu ile 5<\/strong> \u00fczerinden ortalama 4,90<\/strong> verilmi\u015f)<\/span>Loading...<\/span>\n

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M\u00f6bius \u015feridine, internetteki \u00e7e\u015fitli pop\u00fcler matematik yaz\u0131lar\u0131nda s\u0131k\u00e7a rastlan\u0131r. Bir \u00f6nceki yaz\u0131da, \u00e7ok yerde okuyabilmemize kar\u015f\u0131n, g\u00f6z\u00fcm\u00fczde canland\u0131rmam\u0131z\u0131n zor oldu\u011fu bir d\u00f6n\u00fc\u015f\u00fcm\u00fcn videosu vard\u0131 . Tabii ki b\u00f6yle bir d\u00f6n\u00fc\u015f\u00fcm i\u00e7in, “Nas\u0131l oluyor da oluyor?” demek veya daha \u00f6ncesi bunu … Okumaya devam et →<\/span><\/a><\/p>\n

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