{"id":1206,"date":"2018-02-21T16:07:52","date_gmt":"2018-02-21T13:07:52","guid":{"rendered":"http:\/\/www.akademikidea.org\/a-kitap\/?p=1206"},"modified":"2023-02-02T09:15:43","modified_gmt":"2023-02-02T06:15:43","slug":"euler-ozdesligi","status":"publish","type":"post","link":"https:\/\/www.akademikidea.org\/a-kitap\/euler-ozdesligi\/","title":{"rendered":"Euler \u00d6zde\u015fli\u011fi"},"content":{"rendered":"
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Matemati\u011fin 5 temel sabit say\u0131s\u0131 olan 0, 1, \u03c0=3,141592653…., e=2,718281828…., <\/em>ve
\n<\/em>\\(i=\\sqrt { – 1} \\)\u00a0say\u0131lar\u0131n\u0131n bir araya getirildi\u011fi ve\u00a0 toplama, \u00e7arpma ve \u00fcs alma i\u015flemlerinin yaln\u0131zca bir kere kullan\u0131ld\u0131\u011f\u0131 \u0130svi\u00e7reli matematik\u00e7i Leonhard Euler(1707-1783)’\u0131n<\/p>\n<\/dd>\n

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$$e^{i\\pi}+1=0…………………………………………………(1)$$ \u00f6zde\u015fli\u011fi, matemati\u011fin g\u00fczelli\u011fini sergileyen en g\u00fczel \u00f6rneklerden biridir.<\/p>\n

Euler \u00f6zde\u015fli\u011fini elde etmek i\u00e7in \u00f6nce \\(e\\) say\u0131s\u0131 ve \\(e^z\\) \u00fcstel i\u015flevini tan\u0131mlayacak ve ard\u0131ndan i\u015flevlerin seri a\u00e7\u0131l\u0131mlar\u0131ndan yararlanaca\u011f\u0131z.<\/p>\n

Euler(1748)’den sonra \\(e\\)\u00a0ile g\u00f6sterilen ve Jacob Bernoulli(1683)’yi\u00a0 \u00a0$$e=\\lim _{n\\to \\infty }\\left(1+{\\frac {1}{n}}\\right)^{n}………………………………………….(2)$$ say\u0131s\u0131na g\u00f6t\u00fcren s\u00fcrekli bile\u015fik faiz <\/em>sorununda oldu\u011fu gibi\u00a0bir b\u00fcy\u00fckl\u00fc\u011f\u00fcn mevcut de\u011ferine orant\u0131l\u0131 bir h\u0131zda artmas\u0131 ya da azalmas\u0131, $$e = x = 2,718281828….$$ de\u011feri,<\/p>\n<\/dd>\n

$$\\int\\limits_1^x {\\frac{1}{t}dt} $$<\/p>\n

denkleminin \u00e7\u00f6z\u00fcm\u00fc olmak \u00fczere,<\/p>\n

$$e^z………………………………………………………..(3)$$<\/p>\n

\u00fcstel i\u015flevi ile g\u00f6sterilir. \u00dcstel i\u015flev hesab\u0131 \u00fczerine \u00e7al\u0131\u015fmalar daha sonra 1697’da Johann Bernoulli ile devam etti.<\/p>\n

Jacob Bernoulli’nin \\(e\\) say\u0131s\u0131na nas\u0131l ula\u015ft\u0131\u011f\u0131n\u0131 g\u00f6rmek i\u00e7in, y\u0131ll\u0131k enflasyonun %100’den y\u00fcksek oldu\u011fu ancak vadeli hesaplara uygulanabilecek y\u0131ll\u0131k faiz oran\u0131n\u0131n da %100’le s\u0131n\u0131rland\u0131r\u0131ld\u0131\u011f\u0131 bir ortamda rekabet eden bankalar\u0131n m\u00fc\u015fteri \u00e7ekmek i\u00e7in vade s\u00fcrelerini k\u0131saltmaktan ba\u015fka se\u00e7eneklerinin bulunmad\u0131\u011f\u0131n\u0131 d\u00fc\u015f\u00fcnelim. Bu durumda 1\u20ba’l\u0131k vadeli hesap a\u00e7an birisinin y\u0131l sonuna kadar hesab\u0131ndan hi\u00e7 para \u00e7ekmemesi durumunda se\u00e7mi\u015f oldu\u011fu vade s\u00fcrelerine g\u00f6re y\u0131l sonunda hesab\u0131ndaki miktarlar a\u015fa\u011f\u0131daki \u00e7izelgede g\u00f6sterildi\u011fi gibi olurdu.<\/p>\n<\/dd>\n

<\/dd>\n<\/dl>\n\n\n\n\n\n\n\n\n
Vade<\/th>\nVade Sonlar\u0131nda<\/th>\nY\u0131l Sonunda<\/th>\n<\/tr>\n
12 ay<\/td>\n\\((1+\\frac {1}{1})\\times 1\u20ba =2\u20ba\\)<\/td>\n\\(2\u20ba\\)<\/td>\n<\/tr>\n
6 ay<\/td>\n\\((1+\\frac {1}{2})\\times 1\u20ba =1,5\u20ba\\)<\/td>\n\\((1+\\frac {1}{2})^2\\times 1\u20ba =2,25\u20ba\\)<\/td>\n<\/tr>\n
3 ay<\/td>\n\\((1+\\frac {1}{4})\\times 1\u20ba =1,25\u20ba\\)<\/td>\n\\((1+\\frac {1}{4})^4\\times 1\u20ba =2,44\u20ba\\)<\/td>\n<\/tr>\n
1 ay<\/td>\n\\((1+\\frac {1}{12})\\times 1\u20ba =1,0833\u20ba\\)<\/td>\n\\((1+\\frac {1}{12})^{12}\\times 1\u20ba =2,613\u20ba\\)<\/td>\n<\/tr>\n
1 g\u00fcn<\/td>\n\\((1+\\frac {1}{365})\\times 1\u20ba =1,00027\u20ba\\)<\/td>\n\\((1+\\frac {1}{365})^{365}\\times 1\u20ba =2,715\u20ba\\)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
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Sonsuz k\u0131sa vade s\u00fcresinde yukar\u0131daki \u00f6rnek i\u00e7in bile\u015fik faiz uygulanmas\u0131 durumunda 1\u20ba s\u0131n\u0131n y\u0131l sonunda ula\u015fabilece\u011fi \u00fcst s\u0131n\u0131r\u0131n matematiksel t\u00fcmevar\u0131m yoluyla (1)’de tan\u0131mlanan \\(e\\) say\u0131s\u0131 olaca\u011f\u0131 kolayca g\u00f6r\u00fclecektir.<\/dd>\n
\\(\\pi\\) ve \\(\\sqrt 2\\) gibi \\(e\\) say\u0131s\u0131 da irrasyonel say\u0131lard\u0131r ve hesaplama yolu ile haneleri t\u00fcketilemez. \\(e\\) say\u0131s\u0131 i\u00e7in yakla\u015f\u0131k bir de\u011fer elde etmek i\u00e7in \\(e^z\\) i\u015flevinin,\u00a0 $${e^z} = \\sum\\limits_{n = 0}^\\infty {\\frac{{{z^n}}}{{n!}}} ……………………………………………….(4)$$ bi\u00e7imindeki Maclaurin serisine a\u00e7\u0131l\u0131m\u0131ndan yararlanabiliriz. \u00d6rne\u011fin (4) deki sonsuz serinin ilk 5 terimine g\u00f6re \\(e\\)’nin yakla\u015f\u0131k de\u011feri, $$e = {e^1} \\approx \\frac{{{1^0}}}{{0!}} + \\frac{{{1^1}}}{{1!}} + \\frac{{{1^2}}}{{2!}} + \\frac{{{1^3}}}{{3!}} + \\frac{{{1^4}}}{{4!}} = \\frac{1}{1} + \\frac{1}{1} + \\frac{1}{2} + \\frac{1}{6} + \\frac{1}{{24}} = {\\rm{2}}{\\rm{,708\\bar 3}}$$olarak elde edilir.<\/dd>\n
Euler \u00f6zde\u015fli\u011findeki \\({\\rm{sin}}(z)\\) ve \\({\\rm{cos}}(z)\\) i\u015flevlerinin, s\u0131ras\u0131yla$${\\rm{sin}}(z) = \\sum\\limits_{n = 1}^\\infty {\\frac{{{{( – 1)}^{n – 1}}}}{{(2n – 1)!}}{z^{2n – 1}}} ……………………………………(5)$$ve$${\\rm{cos}}(z) = \\sum\\limits_{n = 0}^\\infty {\\frac{{{{( – 1)}^n}}}{{(2n)!}}{z^{2n}}} ………………………………………(6) $$bi\u00e7iminde tan\u0131mlanan sonsuz seri a\u00e7\u0131l\u0131mlar\u0131n\u0131n, $${\\rm{sin}}(z) = \\frac{{{z}}}{{1!}} – \\frac{{{z^3}}}{{3!}} + \\frac{{{z^5}}}{{5!}} – \\frac{{{z^7}}}{{7!}} + \\cdots …………………………….(7)$$ve $${\\rm{cos}}(z) = \\frac{{{1}}}{{0!}} – \\frac{{{z^2}}}{{2!}} + \\frac{{{z^4}}}{{4!}} – \\frac{{{z^6}}}{{6!}} + \\cdots…………………………….(8)$$bi\u00e7imindeki ilk 4 terimini toplarsak,$${\\rm{sin}}(z) + {\\rm{cos}}(z) = 1 + z -\\frac{{{z^2}}}{{2!}}- \\frac{{{z^3}}}{{3!}} + \\frac{{{z^4}}}{{4!}} + \\frac{{{z^5}}}{{5!}} – \\frac{{{z^6}}}{{6!}} – \\frac{{{z^7}}}{{7!}} +\\cdots……(9)$$elde edilir.<\/dd>\n
\u00dcstel i\u015flevin (4) deki sonsuz seriye a\u00e7\u0131l\u0131m\u0131ndaki ilk 8 terimi i\u00e7eren$${e^z} = 1 + \\frac{z}{1!} + \\frac{{{z^2}}}{{2!}} + \\frac{{{z^3}}}{{3!}} + \\frac{{{z^4}}}{{4!}} + \\frac{{{z^5}}}{{5!}} + \\frac{{{z^6}}}{{6!}} + \\frac{{{z^7}}}{{7!}} + \\cdots …………..(10)$$ile sin\u00fcs ve cosin\u00fcs i\u015flevlerinin yine 8 terimli (9) daki toplam\u0131 aras\u0131nda terimlerin ayn\u0131 olmas\u0131na dayal\u0131 bir benzerlik hemen g\u00f6ze \u00e7arp\u0131yor.<\/dd>\n
(9) ve (10) aras\u0131ndaki bu benzerli\u011fi bir denkli\u011fe d\u00f6n\u00fc\u015ft\u00fcrmek (10)’daki kimi terimlerin i\u015faretini (9)’daki kar\u015f\u0131l\u0131klar\u0131n\u0131nki ile \u00f6rt\u00fc\u015fecek bi\u00e7imde eksilere d\u00f6n\u00fc\u015ft\u00fcrecek matemati\u011fin sihirli de\u011fneklerinden birini kullanmam\u0131z gerekecek. Bu sihirli de\u011fnek, karesi -1 olan ve $$i=\\sqrt {-1}$$bi\u00e7iminde tan\u0131mlanan sanal say\u0131.<\/dd>\n
\u015eimdi bu sihirli de\u011fne\u011fi, (10)’a de\u011fdirerek \u00f6nce t\u00fcm \\(z\\)’leri $$e^{ix} = 1 + \\frac{{{ix}}}{{1!}} + \\frac{{{(ix)^2}}}{{2!}} + \\frac{{{(ix)^3}}}{{3!}} + \\frac{{{(ix)^4}}}{{4!}} + \\frac{{{(ix)^5}}}{{5!}} + \\frac{{{(ix)^6}}}{{6!}} + \\frac{{{(ix)^7}}}{{7!}} + \\cdots(11)$$bi\u00e7iminde\u00a0\\(ix\\)’lere d\u00f6n\u00fc\u015ft\u00fcrelim ve sonra da \\(i\\)’ye g\u00f6re, $$e^{ix} = 1 + \\frac{{{ix}}}{{1!}} – \\frac{{{x^2}}}{{2!}} – \\frac{{{ix^3}}}{{3!}} + \\frac{{{x^4}}}{{4!}} + \\frac{{{ix^5}}}{{5!}} – \\frac{{{x^6}}}{{6!}} – \\frac{{{ix^7}}}{{7!}} + \\cdots……………….(12)$$ bi\u00e7iminde sadele\u015ftirelim.<\/dd>\n
E\u011fer (12)’deki sanal say\u0131 i\u00e7eren terimler \\(i\\) ortak \u00e7arpan\u0131 ile,$$e^{ix} = 1 – \\frac{{{x^2}}}{{2!}} +\u00a0\\frac{{{x^4}}}{{4!}} – \\frac{{{x^6}}}{{6!}} +\u00a0\\cdots + i\\left( {\\frac{{{x}}}{{1!}}\u00a0 – \\frac{{{x^3}}}{{3!}}\u00a0 + \\frac{{{x^5}}}{{5!}} – \\frac{{{x^7}}}{{7!}} + \\cdots}\\right)…………(13)$$bi\u00e7iminde bir araya getirilirse, (7) ve (8)’den, $$e^{ix} = {\\rm{cos}}(x) + i{\\rm{sin}}(x)………………………………………….(14)$$bi\u00e7imindeki Euler denklemi elde edilir ve (1)’deki Euler \u00f6zde\u015fli\u011finin \\(x=\\pi\\) i\u00e7in (14)’\u00fcn \u00f6zel bir durumu oldu\u011fu kolayca g\u00f6r\u00fcl\u00fcr.<\/dd>\n
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Kaynaklar<\/h2>\n

Boyer, Carl B. – Merzbach, Uta C.(2011) A History of Mathematics<\/a>,<\/em> Third Edition, Wiley. s:419.<\/h6>\n
\u00a0Coolman, Robert(2015)\u00a0Euler\u2019s Identity: ‘The Most Beautiful Equation’<\/a>, LiveScience.<\/h6>\n
Jacob Bernoulli(1683) Qu\u00e6stiones nonnull\u00e6 de usuris, cum solutione problematis de sorte alearum<\/a><\/em>, propositi in Ephem. Gall. A. s:222.<\/h6>\n
Leonard Euler(1748)\u00a0Introductio in analysin infinitorum<\/i><\/a>, Tomi Primi, Caput VII. s:128.<\/h6>\n
\u00a0e (Euler’s Number)<\/a>,\u00a0Euler’s Identity<\/a>,\u00a0e to the pi i for dummies<\/a><\/h6>\n
Intuitive Understanding Of Euler\u2019s Formula<\/a><\/h6>\n","protected":false},"excerpt":{"rendered":"

Matemati\u011fin 5 temel sabit say\u0131s\u0131 olan 0, 1, \u03c0=3,141592653…., e=2,718281828…., ve \\(i=\\sqrt { – 1} \\)\u00a0say\u0131lar\u0131n\u0131n bir araya getirildi\u011fi ve\u00a0 toplama, \u00e7arpma ve \u00fcs alma i\u015flemlerinin yaln\u0131zca bir kere kullan\u0131ld\u0131\u011f\u0131 \u0130svi\u00e7reli matematik\u00e7i Leonhard Euler(1707-1783)’\u0131n $$e^{i\\pi}+1=0…………………………………………………(1)$$ \u00f6zde\u015fli\u011fi, matemati\u011fin g\u00fczelli\u011fini sergileyen en g\u00fczel \u00f6rneklerden biridir. Euler \u00f6zde\u015fli\u011fini elde etmek i\u00e7in \u00f6nce \\(e\\) say\u0131s\u0131 ve \\(e^z\\) \u00fcstel i\u015flevini…<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[26],"tags":[],"class_list":["post-1206","post","type-post","status-publish","format-standard","hentry","category-matematik"],"_links":{"self":[{"href":"https:\/\/www.akademikidea.org\/a-kitap\/wp-json\/wp\/v2\/posts\/1206","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.akademikidea.org\/a-kitap\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.akademikidea.org\/a-kitap\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.akademikidea.org\/a-kitap\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/www.akademikidea.org\/a-kitap\/wp-json\/wp\/v2\/comments?post=1206"}],"version-history":[{"count":116,"href":"https:\/\/www.akademikidea.org\/a-kitap\/wp-json\/wp\/v2\/posts\/1206\/revisions"}],"predecessor-version":[{"id":1469,"href":"https:\/\/www.akademikidea.org\/a-kitap\/wp-json\/wp\/v2\/posts\/1206\/revisions\/1469"}],"wp:attachment":[{"href":"https:\/\/www.akademikidea.org\/a-kitap\/wp-json\/wp\/v2\/media?parent=1206"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.akademikidea.org\/a-kitap\/wp-json\/wp\/v2\/categories?post=1206"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.akademikidea.org\/a-kitap\/wp-json\/wp\/v2\/tags?post=1206"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}