{"id":20518,"date":"2018-03-01T12:16:16","date_gmt":"2018-03-01T09:16:16","guid":{"rendered":"https:\/\/bilimvegelecek.com.tr\/?p=20518"},"modified":"2018-03-01T12:18:28","modified_gmt":"2018-03-01T09:18:28","slug":"20518","status":"publish","type":"post","link":"https:\/\/bilimvegelecek.com.tr\/index.php\/2018\/03\/01\/20518","title":{"rendered":"Sonsuz k\u00fc\u00e7\u00fckler krizi!"},"content":{"rendered":"<p>\u00dcnl\u00fc Alman filozof Nietzsche\u2019nin \u201cTanr\u0131 \u00f6ld\u00fc\u201d s\u00f6z\u00fcyle bilimsel geli\u015fmeler sonras\u0131nda Tanr\u0131ya ihtiya\u00e7 kalmad\u0131\u011f\u0131n\u0131 anlatmak istedi\u011fi s\u00f6ylenebilir. Nietzsche\u2019nin b\u00f6ylesi bir sonuca varmas\u0131ndaki en \u00f6nemli iki geli\u015fme, Newton\u2019un matematiksel fizi\u011fi, Darwin\u2019in canl\u0131 t\u00fcrlerinin k\u00f6kenini ke\u015ffetmesi olarak say\u0131labilir.<\/p>\n<p>Do\u011fal d\u00fcnyay\u0131 anlamada en etkili yol olan matematiksel fizik b\u00fct\u00fcn\u00fcyle t\u00fcrev ve integralleri konu eden diferansiyel hesaba dayan\u0131r. Diferansiyel hesab\u0131n \u00f6nc\u00fcleri addedilen Newton ve Leibniz\u2019in ya\u015fad\u0131\u011f\u0131 17\u2019inci y\u00fczy\u0131l\u0131 izleyen y\u00fcz elli y\u0131l i\u00e7inde bu alanda \u00e7ok h\u0131zl\u0131 ilerlemeler ya\u015fand\u0131. Do\u011fan\u0131n bir\u00e7ok yasas\u0131 t\u00fcrev ve integral hesab\u0131n diliyle yaz\u0131ld\u0131. \u00a0T\u00fcrev ve integraller, kab\u0131na s\u0131\u011fmaz bir a\u015fk\u0131nl\u0131kla fizik, astronomi, m\u00fchendislik gibi alanlarda uygulanarak yeni bulu\u015f ve y\u00f6ntemlerin ortaya \u00e7\u0131kmas\u0131n\u0131 sa\u011flad\u0131, ama bu \u00fcretken s\u00fcre\u00e7 beraberinde baz\u0131 sorunlar\u0131n, belirsizliklerin ve zay\u0131fl\u0131klar\u0131n olu\u015fmas\u0131na da neden oldu. Diferansiyel hesab\u0131n yap\u0131s\u0131ndaki \u00e7eli\u015fkilere kar\u015f\u0131n do\u011fadaki uygulamalar\u0131nda elde edilen ola\u011fan\u00fcst\u00fc ba\u015far\u0131lar kan\u0131t ve kavramlardaki bulan\u0131kl\u0131\u011f\u0131 gizliyordu.<\/p>\n<p>Diferansiyel hesab\u0131n in\u015fas\u0131nda yer alan \u201csonsuz k\u00fc\u00e7\u00fck\u201d kavram\u0131na ilk ele\u015ftiri, daha do\u011frusu ilk sald\u0131r\u0131 1734\u2019de \u0130rlandal\u0131 piskopos George Berkeley taraf\u0131ndan yap\u0131ld\u0131. Berkeley, <em>The Analyst(Analizci)<\/em> isimli kitab\u0131nda sonsuz k\u00fc\u00e7\u00fckleri \u201cyok olmu\u015f niceliklerin hayaletleri\u201d olarak niteledi. \u00a0Asl\u0131nda Berkeley\u2019in bu kitab\u0131yla Newton\u2019un arkada\u015f\u0131 ve Halley kuyruklu y\u0131ld\u0131z\u0131n\u0131n isim babas\u0131 olan astronom Edmund Halley\u2019i hedef ald\u0131\u011f\u0131 tahmin edilir. \u00a0Halley ateisttir ve Berkeley\u2019in bir arkada\u015f\u0131n\u0131 dinin imkans\u0131zl\u0131\u011f\u0131 konusunda ikna etmi\u015ftir. Belki Berkeley Halley\u2019den intikam al\u0131yordu ve belki sonras\u0131nda Nietzsche\u2019nin dedi\u011fi gibi \u201cTanr\u0131 \u00f6ld\u00fc\u201d denilmesinden korkuyordu, ama \u00f6ne s\u00fcrd\u00fc\u011f\u00fc matematiksel ele\u015ftiriler son derece do\u011fruydu.<\/p>\n<p>Sonsuz k\u00fc\u00e7\u00fck kavram\u0131ndaki sorun neydi? Newton ve Leibniz\u2019in kulland\u0131\u011f\u0131 sonsuz k\u00fc\u00e7\u00fck kavram\u0131n\u0131 basite indirgeyerek \u015f\u00f6yle a\u00e7\u0131klayabiliriz: B\u00fct\u00fcn pozitif say\u0131lardan k\u00fc\u00e7\u00fck ama s\u0131f\u0131rdan b\u00fcy\u00fck bir say\u0131! B\u00f6yle bir say\u0131 bulabilir miyiz? Elbette bulamay\u0131z, \u00e7\u00fcnk\u00fc bir say\u0131 s\u0131f\u0131r olmaks\u0131z\u0131n ne kadar k\u00fc\u00e7\u00fck olabilir ki? \u00a0\u201cSonsuz k\u00fc\u00e7\u00fck\u201d ad\u0131n\u0131 verdi\u011fimiz say\u0131yla s\u0131f\u0131r aras\u0131nda hi\u00e7bir say\u0131n\u0131n olmamas\u0131 gerekir ki bu m\u00fcmk\u00fcn de\u011fildir, \u00e7\u00fcnk\u00fc \u201csonsuz k\u00fc\u00e7\u00fc\u011f\u00fcn\u201d \u00a0yar\u0131s\u0131n\u0131 alarak daha k\u00fc\u00e7\u00fck bir say\u0131 ve bu i\u015fleme devam ederek daha da k\u00fc\u00e7\u00fck say\u0131lar elde edebiliriz. Bu k\u00fc\u00e7\u00fcltme i\u015flemlerini yaparak s\u0131f\u0131ra \u00e7ok yak\u0131n say\u0131lar bulabiliriz, ama \u201cen k\u00fc\u00e7\u00fck say\u0131 \u015fu\u201d veya \u201csonsuz k\u00fc\u00e7\u00fck say\u0131 \u015fu\u201d diyemeyiz.<\/p>\n<p><strong>Yolum yanl\u0131\u015f ama sonu\u00e7 do\u011fru\u2026<\/strong><\/p>\n<p>\u00d6\u011frencili\u011fimizde bir matematik problemini \u00e7\u00f6zerken \u00e7ok s\u0131k kar\u015f\u0131la\u015ft\u0131\u011f\u0131m\u0131z bir durumdur, \u015f\u00f6yle s\u00f6yleriz: \u201cYolum yanl\u0131\u015f ama sonu\u00e7 do\u011fru\u201d. \u00a0Matematik tarihinde sonsuz k\u00fc\u00e7\u00fcklerle yap\u0131lan i\u015flemlerde de tam b\u00f6yle olmamakla birlikte benzer bir s\u00fcre\u00e7 ya\u015fand\u0131. Sonsuz k\u00fc\u00e7\u00fckleri kullanarak do\u011fru sonu\u00e7lara ula\u015f\u0131l\u0131yordu ama gidilen yol belirsizdi. Bu durumu geometriden bir \u00f6rnekle a\u00e7\u0131klayal\u0131m, sonra t\u00fcrev ve integraldeki sonsuz k\u00fc\u00e7\u00fck kullan\u0131m\u0131n\u0131 ele alal\u0131m.<\/p>\n<p>Bir dairenin alan\u0131n say\u0131sal de\u011ferinin \u00e7evresinin say\u0131sal de\u011ferine oran\u0131n\u0131n yar\u0131\u00e7ap uzunlu\u011funun yar\u0131s\u0131na e\u015fit oldu\u011funu biliyoruz. (\u03c0r2\/2\u03c0r = r\/2). Bu e\u015fitli\u011fi \u015fu y\u00f6ntemle kan\u0131tlamaya \u00e7al\u0131\u015fal\u0131m. Dairenin i\u00e7ine sonsuz say\u0131da yar\u0131\u00e7ap \u00e7izdi\u011fimizi varsayal\u0131m ve yar\u0131\u00e7aplar\u0131 y\u00fcksekli\u011fi r olan sonsuz k\u00fc\u00e7\u00fck \u00fc\u00e7genler gibi d\u00fc\u015f\u00fcnelim. A\u015fa\u011f\u0131daki \u015fekilde \u00e7izilmi\u015f olan yar\u0131\u00e7aplar\u0131n sonsuz say\u0131da oldu\u011funu ve daireyi kaplad\u0131\u011f\u0131n\u0131 hayal edelim.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-20543 alignright\" src=\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2018\/03\/daire.jpg\" alt=\"\" width=\"228\" height=\"214\" \/>B\u00f6ylece daire her birinin y\u00fckseklik uzunlu\u011fu r birim olan sonsuz k\u00fc\u00e7\u00fck \u00fc\u00e7genden olu\u015fmaktad\u0131r. Bir \u00fc\u00e7genin alan\u0131 taban uzunlu\u011fuyla o tabana ait y\u00fckseklik uzunlu\u011funun \u00e7arp\u0131m\u0131n\u0131n yar\u0131s\u0131na e\u015fit oldu\u011fundan \u00fc\u00e7genlerin alanlar\u0131n\u0131n toplam\u0131 taban uzunluklar\u0131n\u0131n toplam\u0131yla r\u2019nin \u00e7arp\u0131m\u0131n yar\u0131s\u0131na e\u015fit olacakt\u0131r. Taban uzunluklar\u0131n\u0131n toplam\u0131 da dairenin \u00e7evre uzunlu\u011funa e\u015fit oldu\u011fundan \u00fc\u00e7genlerin alanlar\u0131 toplam\u0131 yani dairenin alan\u0131, \u00e7evresinin r\/2 kat\u0131na e\u015fit olur.<\/p>\n<p>Bu sonu\u00e7 elbette do\u011frudur, ama sonuca giderken at\u0131lan ad\u0131mlar olduk\u00e7a sorunludur. Bir \u00fc\u00e7genin sonsuz k\u00fc\u00e7\u00fck taban\u0131 olabilir mi? S\u00f6z\u00fc edilen \u00fc\u00e7genlerin taban uzunluklar\u0131 s\u0131f\u0131r de\u011filse (ki s\u0131f\u0131r olamaz \u00e7\u00fcnk\u00fc o zaman \u00fc\u00e7genden s\u00f6z edemeyiz) ne kadar k\u00fc\u00e7\u00fck olurlarsa olsunlar taban uzunluklar\u0131ndan olu\u015fan sonsuz say\u0131da terimi toplad\u0131\u011f\u0131m\u0131zda sonsuz b\u00fcy\u00fckl\u00fckte bir toplam buluruz. Bu itiraz matematikte <em>Ar\u015fimet \u00f6zelli\u011fi<\/em> olarak bilinen, s\u0131f\u0131rdan b\u00fcy\u00fck \u00e7ok k\u00fc\u00e7\u00fck bir say\u0131n\u0131n bile kendisiyle \u201cdefalarca\u201d toplanmas\u0131 halinde sonlu bir b\u00fcy\u00fckl\u00fc\u011fe ula\u015f\u0131laca\u011f\u0131 \u00f6nermesine dayan\u0131r ki, burada sonsuz k\u00fc\u00e7\u00fckler sonsuz say\u0131da topland\u0131\u011f\u0131ndan sonsuz b\u00fcy\u00fckl\u00fckte bir toplama ula\u015fmam\u0131z gerekirdi.<\/p>\n<p>Newton, t\u00fcrevi ke\u015ffederken bir e\u011fri boyunca hareket eden ve h\u0131z\u0131 s\u00fcrekli de\u011fi\u015fen bir cismin anl\u0131k h\u0131z\u0131n\u0131n ne olaca\u011f\u0131 sorusuna yan\u0131t aram\u0131\u015ft\u0131r. Bu hesab\u0131 basit\u00e7e a\u00e7\u0131klayabilmek i\u00e7in bir \u00f6rnek verelim. D\u00fc\u015fen bir ta\u015f\u0131n s = t<sup>2<\/sup> parabol\u00fc \u00fczerinde hareket etti\u011fini varsayal\u0131m ve verili bir andaki, mesela\u00a0t = 1 an\u0131ndaki h\u0131z\u0131n\u0131 hesaplamaya \u00e7al\u0131\u015fal\u0131m. Burada s, ta\u015f\u0131n kat etti\u011fi yolu, t de ta\u015f\u0131n b\u0131rak\u0131lmas\u0131ndan itibaren ge\u00e7en zaman\u0131 g\u00f6steriyor.<\/p>\n<p>Sonlu bir zamanda ortalama h\u0131z\u0131 hesaplarken kulland\u0131\u011f\u0131m\u0131z, ortaokuldan bu yana bildi\u011fimiz bir form\u00fcl var: H\u0131z = Yol \u00a0Zaman. Acaba\u00a0t = 1 an\u0131ndaki h\u0131z\u0131 bu form\u00fclle bulabilir miyiz? \u015e\u00f6yle yapal\u0131m: t = 1\u00a0an\u0131ndan sonra ge\u00e7en sonsuz k\u00fc\u00e7\u00fck bir zaman oldu\u011funu varsayal\u0131m ve bu sonsuz k\u00fc\u00e7\u00fck zaman art\u0131\u015f\u0131n\u0131 da <em>dt<\/em> simgesiyle g\u00f6sterelim. Bu durumda cismin t = 1 ile t = 1+<em>dt<\/em> zamanlar\u0131 aras\u0131nda alaca\u011f\u0131 yolu \u00a0e\u015fitli\u011fiyle hesaplayabiliriz:\u00a0(1+<em>dt<\/em>)<sup>2<\/sup>\u20131<sup>2<\/sup> = 2<em>dt<\/em>+<em>dt<\/em><sup>2<\/sup>.<\/p>\n<p>Zamandaki sonsuz k\u00fc\u00e7\u00fck art\u0131\u015f, yolda da sonsuz k\u00fc\u00e7\u00fck bir art\u0131\u015fa kar\u015f\u0131l\u0131k gelecektir. Yukar\u0131da hesaplanan yoldaki sonsuz k\u00fc\u00e7\u00fck art\u0131\u015f\u0131 da \u00a0simgesiyle g\u00f6sterelim. \u015eimdi, H\u0131z = Yol \/ Zaman form\u00fcl\u00fcyle sonlu olmas\u0131 gereken <em>ds\/dt<\/em>\u00a0oran\u0131n\u0131 bulal\u0131m.<\/p>\n<p><em>ds\/dt =\u00a02dt+dt<sup>2\u00a0<\/sup>\/\u00a0dt = 2+dt<sup>\u00a0<\/sup><\/em><\/p>\n<p>Bu oran sonlu olmas\u0131 gerekti\u011finden sonsuz k\u00fc\u00e7\u00fck bir terim olan <em>dt&#8217;<\/em>yi atar ve t = 1\u00a0an\u0131ndaki anl\u0131k h\u0131z\u0131n de\u011ferini 2 olarak buluruz.<\/p>\n<figure id=\"attachment_20549\" aria-describedby=\"caption-attachment-20549\" style=\"width: 225px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-20549\" src=\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2018\/03\/berkeley.jpg\" alt=\"\" width=\"225\" height=\"300\" \/><figcaption id=\"caption-attachment-20549\" class=\"wp-caption-text\">Berkeley.<\/figcaption><\/figure>\n<p><strong>\u0130tiraz f\u0131rt\u0131nas\u0131\u2026<\/strong><\/p>\n<p>\u0130\u015fte, tam burada Berkeley\u2019in itiraz f\u0131rt\u0131nas\u0131 \u015f\u00f6yle ba\u015flar: \u201cNe kadar k\u00fc\u00e7\u00fck olursa olsun bir \u015fey ihmal ediliyorsa, h\u0131z\u0131n kesin de\u011fere de\u011fil ancak yakla\u015f\u0131k bir de\u011fere sahip oldu\u011funu s\u00f6yleyebiliriz. (\u2026) Art\u0131\u015flar\u0131n yok oldu\u011funu varsay\u0131yorsak,\u00a0 art\u0131\u015flar\u0131n var oldu\u011fu y\u00f6n\u00fcndeki ilk varsay\u0131m terk edilmi\u015f demektir, bu durumda o ilk varsay\u0131m\u0131n sonu\u00e7lar\u0131ndan biri sayesinde elde edilmi\u015f bir sonu\u00e7la kar\u015f\u0131la\u015f\u0131r\u0131z ki bu da yanl\u0131\u015f bir ak\u0131l y\u00fcr\u00fctmedir. Peki, bu anl\u0131k sonsuz k\u00fc\u00e7\u00fck art\u0131\u015flar nedir? Bunlar ne sonlu niceliklerdir, ne de hi\u00e7liktir. Bunlara \u00f6lm\u00fc\u015f niceliklerin hayaletleri dememiz gerekmez mi?\u201d<\/p>\n<p>Berkeley\u2019in tepkisini yukar\u0131da verdi\u011fimiz \u00f6rne\u011fe g\u00f6re \u015f\u00f6yle \u00f6zetleyebiliriz: Eninde sonunda, <em>dt<\/em>\u00a0ya s\u0131f\u0131ra e\u015fittir ya da s\u0131f\u0131ra e\u015fit de\u011fildir. E\u011fer <em>dt<\/em>\u00a0s\u0131f\u0131ra e\u015fit de\u011filse 2+<em>dt<\/em>\u00a0de 2\u2019ye e\u015fit de\u011fildir, e\u011fer <em>dt<\/em>\u00a0s\u0131f\u0131ra e\u015fitse mesafedeki <em>ds<\/em>\u00a0art\u0131\u015f\u0131 da s\u0131f\u0131ra e\u015fit olur ki bu durumda <em>dt\/ds\u00a0<\/em>oran\u0131 da 2\u2019ye e\u015fit de\u011fil, 0\/0 gibi anlams\u0131z bir ifadeye e\u015fit olur.<\/p>\n<p>Berkeley\u2019in t\u00fcrev ve integralin mant\u0131ksal temellerine y\u00f6nelik ele\u015ftirileri do\u011fruydu. Acaba bu g\u00f6rkemli yap\u0131 \u00e7\u00f6kecek miydi? Hi\u00e7 de beklendi\u011fi gibi olmad\u0131; matematik\u00e7iler, fizik\u00e7iler, m\u00fchendisler bir y\u00fczy\u0131l daha \u00fcstelik b\u00fcy\u00fck bir ba\u015far\u0131yla sonsuz k\u00fc\u00e7\u00fcklere dayanan t\u00fcrev ve integrali kullanmaya devam ettiler. Sonsuz k\u00fc\u00e7\u00fckler kavram\u0131 \u015f\u00fcphe ve tepki \u00e7ekse de m\u00fckemmel ve \u201cdo\u011fru sonu\u00e7lar\u201d elde edildi\u011finden t\u00fcrev ve integral hesaba olan g\u00fcvenin sars\u0131lmad\u0131\u011f\u0131n\u0131 s\u00f6yleyebiliriz.\u00a0 Newton ve Leibniz\u2019in y\u00f6ntemlerinin tarihsel \u00f6nemine b\u00fcy\u00fck bir de\u011fer bi\u00e7en Karl Marx, sonsuz k\u00fc\u00e7\u00fckler krizi \u00fczerine <em>Matematiksel Elyazmalar\u0131\u2019<\/em>nda \u015fu notu d\u00fc\u015fm\u00fc\u015ft\u00fcr: \u201cMatematik\u00e7iler, do\u011fru sonuca yanl\u0131\u015f bir matematiksel i\u015flemle varan yeni bulunmu\u015f hesaplama arac\u0131n\u0131n gizemli karakterine ger\u00e7ekten g\u00fcvendiler\u201d.<\/p>\n<p>Bu krizden \u00e7\u0131kmak i\u00e7in yakla\u015f\u0131k y\u00fcz y\u0131l gibi bir s\u00fcrenin ge\u00e7mesi gerekiyordu. Soyut matematik ayr\u0131 bir disiplin olarak ortaya \u00e7\u0131kt\u0131\u011f\u0131nda matematik\u00e7iler t\u00fcrev ve integralin temellerinde hi\u00e7bir \u00e7eli\u015fki olmad\u0131\u011f\u0131na emin oldular. Bug\u00fcnk\u00fc limit kavram\u0131n\u0131 1916\u2019da Bolzano, 1821\u2019de Cauchy birbirlerinden ba\u011f\u0131ms\u0131z olarak ke\u015ffederek sonsuz k\u00fc\u00e7\u00fck kavram\u0131n\u0131n t\u00fcrev ve integral hesab\u0131n\u0131n i\u00e7inden at\u0131lmas\u0131n\u0131 sa\u011flad\u0131lar. 1870\u2019lerde Alman matematik\u00e7i Weierstrass, limitin \u201cdelta-epsilon\u201d tan\u0131m\u0131n\u0131 yaparak t\u00fcrev ve integral hesab\u0131n sars\u0131lmaz bi\u00e7imde temellerini atm\u0131\u015f oldu.<\/p>\n<p>Weierstrass\u2019\u0131n limit tekni\u011fini kullanarak anl\u0131k h\u0131z\u0131 bulurken, h\u0131z\u0131 bir oran olarak de\u011fil bir limit, sonlu art\u0131\u015flar\u0131n oranlar\u0131yla yakla\u015ft\u0131\u011f\u0131m\u0131z bir limit olarak ele al\u0131r\u0131z. Diyelim ki, \u0394t\u00a0simgesi de\u011fi\u015fken bir sonlu zaman art\u0131\u015f\u0131n\u0131, \u0394s\u00a0simgesi de de\u011fi\u015fken mesafe art\u0131\u015f\u0131n\u0131 g\u00f6stersin. Bu durumda \u0394s\/\u0394t, 2+\u0394t de\u011fi\u015fken niceli\u011fine e\u015fit olacakt\u0131r. \u0394t\u2019yi limit durumunda s\u0131f\u0131ra yakala\u015f\u0131rken \u0394s\/\u0394t\u2019nin de 2 de\u011ferine yakla\u015ft\u0131\u011f\u0131n\u0131 ve tan\u0131m itibariyle\u00a0 t = 1\u2019deki h\u0131z\u0131n tam olarak 2 oldu\u011fu sonucuna ula\u015f\u0131r\u0131z. Bu att\u0131\u011f\u0131m\u0131z ad\u0131mlar sadece bir a\u00e7\u0131klama olup, merakl\u0131 okura daha derinlemesine bilgi i\u00e7in Kaynak 3\u2019\u00fc incelemesini \u00f6neririm.<\/p>\n<p>Sonu\u00e7 olarak, sonsuz k\u00fc\u00e7\u00fckler krizi 19\u2019uncu y\u00fczy\u0131l\u0131n ikinci yar\u0131s\u0131nda tamamen a\u015f\u0131lmakla birlikte, her matematiksel krizin sonras\u0131nda oldu\u011fu gibi matematik d\u00fcnyas\u0131nda parlak ke\u015fiflerin yap\u0131lmas\u0131na \u00f6n ayak oldu ve tarih Newton\u2019la Leibniz\u2019i hakl\u0131 \u00e7\u0131kard\u0131. Bu s\u00fcreci matematik tarih\u00e7isi Judith Grabiner\u2019in \u015fu s\u00f6zleri \u00e7ok iyi a\u00e7\u0131klar: \u201cT\u00fcrev ilk \u00f6nce kullan\u0131ld\u0131, sonra ke\u015ffedildi, daha sonra ara\u015ft\u0131r\u0131l\u0131p geli\u015ftirildi ve en sonunda tan\u0131mland\u0131\u201d.<\/p>\n<p><strong>Kaynaklar<\/strong><\/p>\n<p>1) Ian Stewart, <em>Matemati\u011fin K\u0131sa Tarihi<\/em>, \u00c7ev. Sibel Sevin\u00e7, Alfa yay\u0131nevi, 2016.<\/p>\n<p>2) Philip J. Davis, Reuben Hersh, <em>Matemati\u011fin Seyir Defteri<\/em>, \u00c7ev. Ender Abado\u011flu, Doruk yay\u0131nevi, 2009.<\/p>\n<p>3) Ali Nesin, <em>Analiz I-II<\/em>, NMK E-K\u00fct\u00fcphane- Nesin Matematik K\u00f6y\u00fc.<\/p>\n<p>4) Karl Marx, <em>Matematiksel Elyazmalar\u0131<\/em>, \u00c7ev. \u00d6ner \u00dcnalan, Ba\u015fak yay\u0131nevi, 1990.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>\u00dcnl\u00fc Alman filozof Nietzsche\u2019nin \u201cTanr\u0131 \u00f6ld\u00fc\u201d s\u00f6z\u00fcyle bilimsel geli\u015fmeler sonras\u0131nda Tanr\u0131ya ihtiya\u00e7 kalmad\u0131\u011f\u0131n\u0131 anlatmak istedi\u011fi s\u00f6ylenebilir. Nietzsche\u2019nin b\u00f6ylesi bir sonuca varmas\u0131ndaki en \u00f6nemli iki geli\u015fme, Newton\u2019un matematiksel fizi\u011fi, Darwin\u2019in canl\u0131 t\u00fcrlerinin k\u00f6kenini ke\u015ffetmesi olarak say\u0131labilir. Do\u011fal d\u00fcnyay\u0131 anlamada en etkili yol olan matematiksel fizik b\u00fct\u00fcn\u00fcyle t\u00fcrev ve integralleri konu eden diferansiyel hesaba dayan\u0131r. Diferansiyel hesab\u0131n [&hellip;]<\/p>\n","protected":false},"author":375,"featured_media":20544,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[2434,25,514],"tags":[2479,1699,208,1530],"class_list":["post-20518","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-169-sayi","category-matematik","category-matematik-sohbetleri","tag-daire","tag-hiz","tag-matematik","tag-sayilar"],"acf":[],"aioseo_notices":[],"aioseo_head":"\n\t\t<!-- All in One SEO 4.9.10 - aioseo.com -->\n\t<meta name=\"robots\" content=\"max-image-preview:large\" \/>\n\t<meta name=\"author\" content=\"Ali T\u00f6r\u00fcn\"\/>\n\t<link rel=\"canonical\" href=\"https:\/\/bilimvegelecek.com.tr\/index.php\/2018\/03\/01\/20518\" \/>\n\t<meta name=\"generator\" content=\"All in One SEO (AIOSEO) 4.9.10\" \/>\n\t\t<meta property=\"og:locale\" content=\"tr_TR\" \/>\n\t\t<meta property=\"og:site_name\" content=\"Bilim ve Gelecek\" \/>\n\t\t<meta property=\"og:type\" content=\"article\" \/>\n\t\t<meta property=\"og:title\" content=\"Sonsuz k\u00fc\u00e7\u00fckler krizi! | Bilim ve Gelecek\" \/>\n\t\t<meta property=\"og:url\" content=\"https:\/\/bilimvegelecek.com.tr\/index.php\/2018\/03\/01\/20518\" \/>\n\t\t<meta property=\"fb:app_id\" content=\"2104805563100892\" \/>\n\t\t<meta property=\"fb:admins\" content=\"1250955469\" \/>\n\t\t<meta property=\"og:image\" content=\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2018\/03\/daireler.jpg\" \/>\n\t\t<meta property=\"og:image:secure_url\" content=\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2018\/03\/daireler.jpg\" \/>\n\t\t<meta property=\"og:image:width\" content=\"800\" \/>\n\t\t<meta property=\"og:image:height\" content=\"451\" \/>\n\t\t<meta property=\"article:published_time\" content=\"2018-03-01T09:16:16+00:00\" \/>\n\t\t<meta property=\"article:modified_time\" content=\"2018-03-01T09:18:28+00:00\" \/>\n\t\t<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/bilimvegelecekdergisi\/\" \/>\n\t\t<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n\t\t<meta name=\"twitter:site\" content=\"@bilimvegelecek\" \/>\n\t\t<meta name=\"twitter:title\" content=\"Sonsuz k\u00fc\u00e7\u00fckler krizi! | Bilim ve Gelecek\" \/>\n\t\t<meta name=\"twitter:image\" content=\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2018\/03\/daireler.jpg\" \/>\n\t\t<script type=\"application\/ld+json\" class=\"aioseo-schema\">\n\t\t\t{\"@context\":\"https:\\\/\\\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\\\/\\\/bilimvegelecek.com.tr\\\/index.php\\\/2018\\\/03\\\/01\\\/20518#article\",\"name\":\"Sonsuz k\\u00fc\\u00e7\\u00fckler krizi! | Bilim ve Gelecek\",\"headline\":\"Sonsuz k\\u00fc\\u00e7\\u00fckler krizi!\",\"author\":{\"@id\":\"https:\\\/\\\/bilimvegelecek.com.tr\\\/index.php\\\/author\\\/atorun#author\"},\"publisher\":{\"@id\":\"https:\\\/\\\/bilimvegelecek.com.tr\\\/#organization\"},\"image\":{\"@type\":\"ImageObject\",\"url\":\"https:\\\/\\\/bilimvegelecek.com.tr\\\/wp-content\\\/uploads\\\/2018\\\/03\\\/daireler.jpg\",\"width\":800,\"height\":451},\"datePublished\":\"2018-03-01T12:16:16+03:00\",\"dateModified\":\"2018-03-01T12:18:28+03:00\",\"inLanguage\":\"tr-TR\",\"mainEntityOfPage\":{\"@id\":\"https:\\\/\\\/bilimvegelecek.com.tr\\\/index.php\\\/2018\\\/03\\\/01\\\/20518#webpage\"},\"isPartOf\":{\"@id\":\"https:\\\/\\\/bilimvegelecek.com.tr\\\/index.php\\\/2018\\\/03\\\/01\\\/20518#webpage\"},\"articleSection\":\"169. 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