6.2-bo'lim: Teylor seriyasi
Tekshiruvdan chiqqanimizdan so'ng biz Teylor seriyalari bilan ko'p ish qilmoqchi emasmiz, ammo ular bizni kuch seriyalari bilan ishlashga qaytarishning yaxshi usuli. Ko'pchilik talabalar matematik karerasining ushbu bosqichiga etib kelganlarida, ular kamida bir yoki ikki semestr davomida kuch seriyalari bilan shug'ullanishlari shart emas. Teylor seriyasining qanday ishlashini eslab qolish, differensial tenglamalarni ko'rib chiqishni boshlashdan oldin kuch seriyalari bilan qulay bo'lishning juda qulay usuli bo'ladi.
Teylor seriyasi
Agar \(f(x)\) cheksiz differensiallanuvchi funksiya bo'lsa, \(x = {x_0}\) haqidagi \(f(x)\) ning Teylor seriyasi,
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( {{x_0}} \right)}}{{n!}}{{\left( {x - {x_0}} \right)}^n}} \]Shuni eslang
\[{f^{\left( 0 \right)}}\left( x \right) = f\left( x \right)\hspace{0.25in}{f^{\left( n \right)}}\left( x \right) = n^{\text{th}}{\mbox{ derivative of }}f\left( x \right)\]Keling, bir misolni ko'rib chiqaylik.
Yechim
Bu, ehtimol, Teylor seriyasini topishning eng oson funktsiyalaridan biridir. Faqat shuni eslashimiz kerak,
\[{f^{\left( n \right)}}\left( x \right) = {{\bf{e}}^x}\hspace{0.25in}n = 0,1,2, \ldots \]va shuning uchun biz olamiz,
\[{f^{\left( n \right)}}\left( 0 \right) = 1\hspace{0.25in}n = 0,1,2, \ldots \]Teylor bu misol uchun qator, keyin
\[{{\bf{e}}^x} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \]Albatta, \(x=0\) haqidagi Teylor seriyasini topish ko'pincha osonroq, lekin biz buni har doim ham qilmaymiz.
Yechim
Bu muammo oldingi muammo bilan deyarli bir xil. Bunday holda biz shuni ta'kidlashimiz kerakki,
\[{f^{\left( n \right)}}\left( { - 4} \right) = {{\bf{e}}^{ - 4}}\hspace{0.25in}n = 0,1,2, \ldots \]Teylor bu misol uchun qator, keyin
\[{{\bf{e}}^x} = \sum\limits_{n = 0}^\infty {\frac{{{{\bf{e}}^{ - 4}}}}{{n!}}{{\left( {x + 4} \right)}^n}} \]Keling, biroz ko'proq mehnat talab qiladigan Teylor seriyasini yarataylik.
Yechim
Bu safar har bir \(n\) uchun hosila beradigan formula yo'q, keling, hosilalarni olib, \(x = 0\) ni ulashni boshlaylik.
\[\begin{align*}{f^{\left( 0 \right)}}\left( x \right) & = \cos \left( x \right) & \hspace{0.25in}{f^{\left( 0 \right)}}\left( 0 \right) & = 1\\ {f^{\left( 1 \right)}}\left( x \right) & = - \sin \left( x \right) & \hspace{0.25in}{f^{\left( 1 \right)}}\left( 0 \right) & = 0\\ {f^{\left( 2 \right)}}\left( x \right) & = - \cos \left( x \right) & \hspace{0.25in}{f^{\left( 2 \right)}}\left( 0 \right) & = - 1\\ {f^{\left( 3 \right)}}\left( x \right) & = \sin \left( x \right) & \hspace{0.25in}{f^{\left( 3 \right)}}\left( 0 \right) & = 0\\ {f^{\left( 4 \right)}}\left( x \right) & = \cos \left( x \right) & \hspace{0.25in}{f^{\left( 4 \right)}}\left( 0 \right) & = 1\\ & \hspace{0.1in} \vdots & \hspace{0.25in} & \hspace{0.1in} \vdots \end{align*}\]Bu nuqtaga yetganimizdan so'ng, bu erda bir naqsh paydo bo'lishi aniq bo'ladi. Bu naqsh nima ekanligini hali aniqlab bo'lmaydi, ammo bu naqsh mavjudligi aniq ko'rinadi.
Keling, Teylor seriyasi formulasiga nima borligini kiritamiz va biz nimani olganimizni ko'ramiz.
\[\begin{align*}\cos \left( x \right) & = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}} \\ & = \frac{{{f^{\left( 0 \right)}}\left( 0 \right)}}{{0!}} + \frac{{{f^{\left( 1 \right)}}\left( 0 \right)}}{{1!}}x + \frac{{{f^{\left( 2 \right)}}\left( 0 \right)}}{{2!}}{x^2} + \frac{{{f^{\left( 3 \right)}}\left( 0 \right)}}{{3!}}{x^3} + \cdots \\ & = \frac{1}{{0!}} + 0 - \frac{{{x^2}}}{{2!}} + 0 + \frac{{{x^4}}}{{4!}} + 0 - \frac{{{x^6}}}{{6!}} + 0 + \frac{{{x^8}}}{{8!}} + \cdots \end{align*}\]Demak, har bir boshqa atama nolga teng.
Biz buni ketma-ketlikda yozmoqchimiz, ammo har bir atamada nolga teng bo'lgan va nolga teng bo'lmaganlar uchun to'g'ri javob beradigan formulani topish keraksiz darajada murakkab bo'ladi. Shunday qilib, keling, yuqoridagi narsalarni qayta yozamiz va shu vaqt ichida shartlarni quyidagicha qayta nomlaymiz:
\[\cos \left( x \right) = \underbrace {\,\frac{1}{{0!}}\,}_{n = 0} - \underbrace {\frac{{{x^2}}}{{2!}}}_{n = 1} + \underbrace {\frac{{{x^4}}}{{4!}}}_{n = 2} - \underbrace {\frac{{{x^6}}}{{6!}}}_{n = 3} + \underbrace {\frac{{{x^8}}}{{8!}}}_{n = 4} + \cdots \]Ushbu "qayta raqamlash" yordamida biz \(x = 0\) haqidagi kosinus funktsiyasining Teylor seriyasi uchun formulani osongina olishimiz mumkin.
\[\cos \left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}} \]Amaliyot uchun siz \(x = 0\) ning sinus funksiyasi uchun Teylor qatori ekanligini tekshirishingiz mumkinmi?
\[\sin \left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} \]Biz Teylor seriyasining yana bir misolini ko'rib chiqishimiz kerak. Bu misol ham qiyin, ham juda oson.
Yechim
Bu erda ko'p narsa qilishning hojati yo'q, ba'zi hosilalarni olish va shu nuqtada baholash.
\[\begin{align*}f\left( x \right) & = 3{x^2} - 8x + 2 & \hspace{0.25in}f\left( 2 \right) & = - 2\\ f'\left( x \right) & = 6x - 8 & f'\left( 2 \right) & = 4\\ f''\left( x \right) & = 6 & \hspace{0.25in}f''\left( 2 \right) & = 6\\ {f^{\left( n \right)}}\left( x \right) & = 0,\,\,n \ge 3 & \hspace{0.25in}{f^{\left( n \right)}}\left( 2 \right) & = 0,\,\,n \ge 3\end{align*}\]Shunday qilib, bu holda hosilalar ma'lum bir tartibdan keyin nolga teng bo'ladi. Bu vaqti-vaqti bilan sodir bo'ladi va bizning ishimizni osonlashtiradi. Teylorni sozlash seriya keyin beradi,
\[\begin{align*}3{x^2} - 8x + 2 & = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( 2 \right)}}{{n!}}{{\left( {x - 2} \right)}^n}} \\ & = \frac{{{f^{\left( 0 \right)}}\left( 2 \right)}}{{0!}} + \frac{{{f^{\left( 1 \right)}}\left( 2 \right)}}{{1!}}\left( {x - 2} \right) + \frac{{{f^{\left( 2 \right)}}\left( 2 \right)}}{{2!}}{\left( {x - 2} \right)^2} + \frac{{{f^{\left( 3 \right)}}\left( 2 \right)}}{{3!}}{\left( {x - 2} \right)^3} + \cdots \\ & = - 2 + 4\left( {x - 2} \right) + \frac{6}{2}{\left( {x - 2} \right)^2} + 0\\ & = - 2 + 4\left( {x - 2} \right) + 3{\left( {x - 2} \right)^2}\end{align*}\]Bu holda Teylor seriyasi tugaydi va faqat uchta shartga ega edi. E'tibor bering, chunki biz Teylordan keyinmiz qatorni ikkinchi hadda 4 ga ko'paytirmaymiz yoki uchinchi hadni kvadratga chiqarmaymiz. Konstantadan tashqari barcha shartlar \(x - 2\) ni o'z ichiga olishi kerak.
Ushbu oxirgi misolga e'tibor bering, agar biz Teylor qatorini ko'paytirsak, biz asl polinomimizni olamiz. Bu juda ajablanarli bo'lmasligi kerak, chunki ikkalasi ham polinom va ular teng bo'lishi kerak.
Bizga hozir Teylor seriyasi bilan bog'liq bo'lganligi sababli, bizga kerak bo'lgan keyingi bo'limda emas, balki bu erda berish mantiqiyroq bo'lgan tezkor ta'rifga muhtojmiz.
Analitik funktsiyaning ta'rifi
\(f(x)\) funksiyasi chaqiriladi analitik haqida \(f(x)\) uchun Teylor qator bo'lsa \(x = a\) da x=a musbat yaqinlashuv radiusiga ega va \(f(x)\) ga yaqinlashadi.
Keyingiga o'tishdan oldin biz bitta yakuniy eslatma berishimiz kerak. Biz ushbu bo'limni ushbu bo'limdan keyin Teylor seriyasi bilan ko'p ish qilmasligimizdan boshladik. Bu to'g'ri bo'lsa-da, bu faqat to'g'ri, chunki biz muammolarni juda oddiy tutamiz. Murakkabroq muammolar uchun biz bir nechta Teylordan foydalanamiz seriya.