10.16-bo'lim: Teylor seriyasi
Oldingida biz funktsiyaning darajali seriyali tasvirini yozishni ko'rib chiqdik. Ushbu bo'limdagi yondashuv bilan bog'liq muammo shundaki, hamma narsa funktsiyani qandaydir tarzda bog'lash zarurati tug'ildi.
\[\frac{1}{{1 - x}}\]va bu funktsiya bilan bog'liq bo'lishi mumkin bo'lgan ko'plab funktsiyalar mavjud bo'lsa-da, bunga aloqador bo'lmagan yana ko'p funktsiyalar mavjud.
Shunday qilib, biz oldingi bo'limda ko'rib chiqqan jarayondan hech narsani olib tashlamasdan, biz qilishimiz kerak bo'lgan narsa, funktsiya uchun quvvat seriyasining tasvirini yozishning umumiy usulini topishdir.
Shunday qilib, hozircha ikkita taxmin qilaylik. Birinchidan, \(f\left( x \right)\) funksiyasi aslida \(x = a\) haqida quvvat seriyali ko‘rinishga ega deb faraz qilaylik,
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {{c_n}{{\left( {x - a} \right)}^n}} = {c_0} + {c_1}\left( {x - a} \right) + {c_2}{\left( {x - a} \right)^2} + {c_3}{\left( {x - a} \right)^3} + {c_4}{\left( {x - a} \right)^4} + \cdots \]Keyinchalik, \(f\left( x \right)\) funksiyasi har bir tartibning hosilalariga ega va biz ularning barchasini topa olamiz deb taxmin qilishimiz kerak.
Endi biz kuch seriyasining ko'rinishi mavjud deb taxmin qilganimizdan so'ng, \({c_n}\) koeffitsientlari nima ekanligini aniqlashimiz kerak. Bu birinchi qarashda ko'rinadiganidan osonroq. Keling, avval hamma narsani \(x = a\) da baholaylik. Bu beradi,
\[f\left( a \right) = {c_0}\]Shunday qilib, birinchisidan tashqari barcha shartlar nolga teng va endi biz \({c_0}\) nima ekanligini bilamiz. Afsuski, \(x\) ning boshqa koeffitsientlarni tezda topishga imkon beradigan funktsiyaga ulashimiz mumkin bo'lgan boshqa qiymati yo'q. Ammo, agar biz funktsiyaning hosilasini (va uning quvvat seriyasini) olsak, \(x = a\) ni ulaymiz, biz quyidagilarni olamiz:
\[\begin{align*}f'\left( x \right) & = {c_1} + 2{c_2}\left( {x - a} \right) + 3{c_3}{\left( {x - a} \right)^2} + 4{c_4}{\left( {x - a} \right)^3} + \cdots \\ f'\left( a \right) & = {c_1}\end{align*}\]va endi biz \({c_1}\) ni bilamiz.
Keling, bu fikrni davom ettiramiz va ikkinchi hosilani topamiz.
\[\begin{align*}f''\left( x \right) & = 2{c_2} + 3\left( 2 \right){c_3}\left( {x - a} \right) + 4\left( 3 \right){c_4}{\left( {x - a} \right)^2} + \cdots \\ f''\left( a \right) & = 2{c_2}\end{align*}\]Shunday qilib, shunday ko'rinadi,
\[{c_2} = \frac{{f''\left( a \right)}}{2}\]Uchinchi hosiladan foydalanish,
\[\begin{align*}f'''\left( x \right) & = 3\left( 2 \right){c_3} + 4\left( 3 \right)\left( 2 \right){c_4}\left( {x - a} \right) + \cdots \\ f'''\left( a \right) & = 3\left( 2 \right){c_3}\hspace{0.75in} \Rightarrow \hspace{0.75in}{c_3} = \frac{{f'''\left( a \right)}}{{3\left( 2 \right)}}\end{align*}\]To'rtinchi hosiladan foydalanish,
\[\begin{align*}{f^{\left( 4 \right)}}\left( x \right) & = 4\left( 3 \right)\left( 2 \right){c_4} + 5\left( 4 \right)\left( 3 \right)\left( 2 \right){c_5}\left( {x - a} \right) \cdots \\ {f^{\left( 4 \right)}}\left( a \right) & = 4\left( 3 \right)\left( 2 \right){c_4}\hspace{0.75in} \Rightarrow \hspace{0.75in}{c_4} = \frac{{{f^{\left( 4 \right)}}\left( a \right)}}{{4\left( 3 \right)\left( 2 \right)}}\end{align*}\]Umid qilamanki, bu vaqtga kelib siz bu erda naqshni ko'rgansiz. Umuman olganda, biz koeffitsientlar uchun quyidagi formulani oldik.
\[{c_n} = \frac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}\]Agar siz \(0! = 1\) ni eslab, \({f^{\left( 0 \right)}}\left( x \right) = f\left( x \right)\) ni aniqlasangiz, bu hatto \(n = 0\) uchun ham ishlaydi.
Shunday qilib, \(x = a\) haqida \(f\left( x \right)\) funktsiyasi uchun quvvat seriyasi taqdimoti mavjud. Teylor seriyasi uchun \(f\left( x \right)\) haqida \(x = a\) is,
Teylor seriyasi
Agar biz \(a = 0\) dan foydalansak, shuning uchun biz \(x = 0\) haqidagi Teylor seriyasi haqida gapiramiz, biz seriyani a deb ataymiz. Maclaurin seriyasi \(f\left( x \right)\) uchun yoki,
Maclaurin seriyasi
Teylor seriyasining har qanday misollarini ko'rib chiqishdan oldin, biz birinchi navbatda Teylor seriyasi ma'lum bir funktsiya uchun mavjud bo'ladi degan taxminni ko'rib chiqishimiz kerak. Keling, bizga kerak bo'lgan ba'zi belgilar va ta'riflar bilan boshlaylik.
Funktsiya uchun Teylor qatori mavjud bo'lishi uchun to'g'ri bo'lishi kerak bo'lgan shartni aniqlash uchun, avvalo, buni aniqlaymiz nth darajali Teylor polinomi \(f\left( x \right)\) as,
\[{T_n}\left( x \right) = \sum\limits_{i = 0}^n {\frac{{{f^{\left( i \right)}}\left( a \right)}}{{i!}}{{\left( {x - a} \right)}^i}} \]E'tibor bering, bu haqiqatan ham \(n\) darajasidagi polinomdir. Agar yig'indini yig'indi belgisisiz yozadigan bo'lsak, bu aniq n bo'ladith darajali polinom. Biz Teylorning yaxshi ilovasini ko'ramiz keyingi bo'limda polinomlar.
E'tibor bering, To'liq Teylor seriyasi uchun,
\[\sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}{{\left( {x - a} \right)}^n}} \]\(n\)th Teylor darajali polinomi qator uchun faqat qisman yig'indidir.
Keyingi, qolgan deb belgilangan,
\[{R_n}\left( x \right) = f\left( x \right) - {T_n}\left( x \right)\]Demak, qolganlari haqiqatan ham xato \(f\left( x \right)\) va \(n\) funksiyalari o'rtasidath Berilgan \(n\) uchun Teylor darajasidagi polinom.
Ushbu ta'rif bilan biz funktsiyani quyidagicha yozishimiz mumkinligini unutmang:
\[f\left( x \right) = {T_n}\left( x \right) + {R_n}\left( x \right)\]Endi bizda quyidagi teorema mavjud.
Teorema
Aytaylik, \(f\left( x \right) = {T_n}\left( x \right) + {R_n}\left( x \right)\). Keyin agar,
\[\mathop {\lim }\limits_{n \to \infty } {R_n}\left( x \right) = 0\]keyin \(\left| {x - a} \right| < R\) uchun,
\[f\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( a \right)}}{{n!}}{{\left( {x - a} \right)}^n}} \]on \(\left| {x - a} \right| < R\).
Umuman olganda, \[\mathop {\lim }\limits_{n \to \infty } {R_n}\left( x \right) = 0\] biroz qiyin jarayon ekanligini ko'rsatish va shuning uchun biz ko'rib chiqadigan barcha misollarda buni ba'zi \(R\) uchun qilish mumkin deb taxmin qilamiz.
Endi ba'zi misollarni ko'rib chiqaylik.
Yechim
Bu aslida bizdan hisoblashimiz so'raladigan Teylor seriyalaridan biri. Funktsiya uchun Teylor seriyasini topish uchun biz \({f^{\left( n \right)}}\left( a \right)\) uchun umumiy formulani aniqlashimiz kerak. Bu boshidanoq bajarish oson bo'lgan bir nechta funktsiyalardan biridir.
\({f^{\left( n \right)}}\left( 0 \right)\) formulasini olish uchun shuni tan olishimiz kerak,
\[{f^{\left( n \right)}}\left( x \right) = {{\bf{e}}^x}\hspace{0.5in}n = 0,1,2,3, \ldots \]va shuning uchun,
\[{f^{\left( n \right)}}\left( 0 \right) = {{\bf{e}}^0} = 1\hspace{0.5in}n = 0,1,2,3, \ldots \]Shuning uchun, \(x = 0\) haqidagi \(f\left( x \right) = {{\bf{e}}^x}\) uchun Teylor seriyasi,
\[{{\bf{e}}^x} = \sum\limits_{n = 0}^\infty {\frac{1}{{n!}}{x^n}} = \sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}} \]Yechim
Ushbu muammoni hal qilishning ikki yo'li mavjud. Ikkalasi ham juda oddiy, ammo ulardan biri sezilarli darajada kamroq ish talab qiladi. Biz ikkala yechimni ham ishlab chiqamiz, chunki uzoqroq bo'lsa, boshqa misollarda ko'rib chiqamiz.
Yechim 1
Birinchi misolda bo'lgani kabi biz \({f^{\left( n \right)}}\left( 0 \right)\) uchun formulani olishimiz kerak. Biroq, birinchisidan farqli o'laroq, bizda biroz ko'proq ish bor. Keling, avval ba'zi hosilalarni olamiz va ularni \(x = 0\) da baholaymiz.
Bir nechta hisob-kitoblardan so'ng biz \({f^{\left( n \right)}}\left( x \right)\) va \({f^{\left( n \right)}}\left( 0 \right)\) uchun umumiy formulalarni olishga muvaffaq bo'ldik. Biz ko'pincha \({f^{\left( n \right)}}\left( x \right)\) uchun umumiy formulani ololmaymiz, shuning uchun formulani olishdan juda hayajonlanmang. Bundan tashqari, biz ko'rib turganimizdek, \({f^{\left( n \right)}}\left( a \right)\) uchun umumiy formulani olish har doim ham oson bo'lmaydi.
Shunday qilib, bu holda bizda umumiy formulalar bor, shuning uchun biz qilishimiz kerak bo'lgan narsa ularni Teylor seriyasi formulasiga ulash va muammoni hal qilishdir.
\[{{\bf{e}}^{ - x}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^n}}}{{n!}}} \]Yechim 2
Oldingi yechim unchalik yomon emas edi va biz ko'pincha shunday qilishimiz kerak. Biroq, bu holatda ancha qisqaroq yechim usuli mavjud. Oldingida biz yangi seriyani topishga yordam berish uchun allaqachon topgan seriyalardan foydalanganmiz. Keling, bu bilan ham xuddi shunday qilaylik. Biz \({{\bf{e}}^x}\) uchun \(x = 0\) haqida Teylor seriyasini allaqachon bilamiz va bu holatda yagona farq shundaki, bizda faqat \(x\) o‘rniga eksponentda “-\(x\)” mavjud.
Shunday qilib, biz qilishimiz kerak bo'lgan yagona narsa, biz birinchi misolda topilgan Teylor seriyasidagi \(x\) ni "-\(x\)" bilan almashtirishdir.
\[{{\bf{e}}^{ - x}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - x} \right)}^n}}}{{n!}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^n}}}{{n!}}} \]Bu bir xil javobga erishishning ancha qisqa usuli, shuning uchun iloji bo'lsa (va, albatta, ruxsat etilgan) oldindan hisoblangan seriyalardan foydalanishni unutmang.
Yechim
Ushbu misol uchun bizda \(x = 0\) haqida \({{\bf{e}}^x}\) uchun Teylor seriyasi mavjudligidan foydalanamiz. Ushbu misolda, oldingi misoldan farqli o'laroq, buni to'g'ridan-to'g'ri bajarish ancha uzoqroq va qiyinroq bo'ladi.
\[\begin{align*}{x^4}{{\bf{e}}^{ - 3{x^2}}} & = {x^4}\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 3{x^2}} \right)}^n}}}{{n!}}} \\ & = {x^4}\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 3} \right)}^n}{x^{2n}}}}{{n!}}} \\ & = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 3} \right)}^n}{x^{2n + 4}}}}{{n!}}} \end{align*}\]Shu paytgacha biz faqat \(x = 0\) (Maklaurin seriyasi sifatida ham tanilgan) haqidagi Teylor seriyasini ko'rib chiqdik, shuning uchun keling, \(x = 0\) haqida bo'lmagan Teylor seriyasini ko'rib chiqaylik. Bundan tashqari, biz eksponensial funktsiyani yana bir bor tanlaymiz, chunki u ba'zi ishlarni osonlashtiradi. Bu ushbu bo'limdagi eksponentlar uchun oxirgi Teylor seriyasi bo'ladi.
Yechim
\({f^{\left( n \right)}}\left( { - 4} \right)\) uchun umumiy formulani topish juda oddiy.
\[{f^{\left( n \right)}}\left( x \right) = {\left( { - 1} \right)^n}{{\bf{e}}^{ - x}}\hspace{0.5in}{f^{\left( n \right)}}\left( { - 4} \right) = {\left( { - 1} \right)^n}{{\bf{e}}^4}\]Teylor seriyasi o'sha paytda,
\[{{\bf{e}}^{ - x}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{{\bf{e}}^4}}}{{n!}}} {\left( {x + 4} \right)^n}\]OK, endi biz eksponensial funktsiyani o'z ichiga olmaydigan ba'zi misollar bilan ishlashimiz kerak, chunki ular biroz ko'proq ishlashni talab qiladi.
Yechim
Birinchidan, biz funktsiyaning ba'zi hosilalarini olishimiz va ularni \(x = 0\) da baholashimiz kerak.
\[\begin{align*}{f^{\left( 0 \right)}}\left( x \right) & = \cos x & \hspace{0.5in} & {f^{\left( 0 \right)}}\left( 0 \right) = 1\\ {f^{\left( 1 \right)}}\left( x \right) & = - \sin x & \hspace{0.5in} & {f^{\left( 1 \right)}}\left( 0 \right) = 0\\ {f^{\left( 2 \right)}}\left( x \right) & = - \cos x & \hspace{0.5in} & {f^{\left( 2 \right)}}\left( 0 \right) = - 1\\ {f^{\left( 3 \right)}}\left( x \right) & = \sin x & \hspace{0.5in} & {f^{\left( 3 \right)}}\left( 0 \right) = 0\\ {f^{\left( 4 \right)}}\left( x \right) & = \cos x & \hspace{0.5in} & {f^{\left( 4 \right)}}\left( 0 \right) = 1\\ {f^{\left( 5 \right)}}\left( x \right) & = - \sin x & \hspace{0.5in} & {f^{\left( 5 \right)}}\left( 0 \right) = 0\\ {f^{\left( 6 \right)}}\left( x \right) & = - \cos x & \hspace{0.5in} & {f^{\left( 6 \right)}}\left( 0 \right) = - 1\\ \vdots \,\,\,\,\,\,\,\,\,\, & & \hspace{0.5in} & \,\,\,\,\,\,\, \vdots \end{align*}\]Bu misolda, avvalgilaridan farqli o'laroq, umumiy hosila uchun ham, hosilani baholash uchun ham oson formula mavjud emas. Biroq, baholashning aniq namunasi mavjud. Shunday qilib, keling, Teylorga nima borligini kiritamiz seriyali va biz nimaga erishganimizni ko'ring,
\[\begin{align*}\cos x & = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}} \\ & = f\left( 0 \right) + f'\left( 0 \right)x + \frac{{f''\left( 0 \right)}}{{2!}}{x^2} + \frac{{f'''\left( 0 \right)}}{{3!}}{x^3} + \frac{{{f^{\left( 4 \right)}}\left( 0 \right)}}{{4!}}{x^4} + \frac{{{f^{\left( 5 \right)}}\left( 0 \right)}}{{5!}}{x^5} + \cdots \\ & = \underbrace {\,\,1\,\,}_{n = 0} + \underbrace {\,0\,}_{n = 1} - \underbrace {\frac{1}{{2!}}{x^2}}_{n = 2} + \underbrace {\,0\,}_{n = 3} + \underbrace {\frac{1}{{4!}}{x^4}}_{n = 4} + \underbrace {\,0\,}_{n = 5} - \underbrace {\frac{1}{{6!}}{x^6}}_{n = 6} + \cdots \end{align*}\]Shunday qilib, biz faqat \(x\)-da teng kuchga ega shartlarni olamiz. Bu bizga Teylor seriyasi uchun umumiy formulani olishga yordam bermaydi. Biroq, keling, nollarni tashlab, nima olishimiz mumkinligini ko'rish uchun atamalarni quyidagi tarzda "qayta raqamlang".
\[\cos x = \underbrace {\,\,1\,\,}_{n = 0} - \underbrace {\frac{1}{{2!}}{x^2}}_{n = 1} + \underbrace {\frac{1}{{4!}}{x^4}}_{n = 2} - \underbrace {\frac{1}{{6!}}{x^6}}_{n = 3} + \cdots \]Biz atamalarni qayta raqamlash orqali biz aslida Teylor seriyasi uchun umumiy formulani ishlab chiqishimiz mumkin va mana bu,
\[\cos x = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n}}}}{{\left( {2n} \right)!}}} \]Oldingi misolda bo'lgani kabi ketma-ket atamalarni qayta raqamlash g'oyasi unchalik tez-tez ishlatilmaydi, lekin vaqti-vaqti bilan juda foydali. Biz buni qilishimiz kerak bo'lgan yana bir qator bor, shuning uchun seriya atamalarini qayta raqamlashning yana bir misolini ko'rib chiqamiz.
Yechim
Oxirgi misolda bo'lgani kabi, biz ham xuddi shu tarzda boshlaymiz.
\[\begin{align*}{f^{\left( 0 \right)}}\left( x \right) & = \sin x & \hspace{0.5in} & {f^{\left( 0 \right)}}\left( 0 \right) = 0\\ {f^{\left( 1 \right)}}\left( x \right) & = \cos x & \hspace{0.5in} & {f^{\left( 1 \right)}}\left( 0 \right) = 1\\ {f^{\left( 2 \right)}}\left( x \right) & = - \sin x & \hspace{0.5in} & {f^{\left( 2 \right)}}\left( 0 \right) = 0\\ {f^{\left( 3 \right)}}\left( x \right) & = - \cos x & \hspace{0.5in} & {f^{\left( 3 \right)}}\left( 0 \right) = - 1\\ {f^{\left( 4 \right)}}\left( x \right) & = \sin x & \hspace{0.5in} & {f^{\left( 4 \right)}}\left( 0 \right) = 0\\ {f^{\left( 5 \right)}}\left( x \right) & = \cos x & \hspace{0.5in} & {f^{\left( 5 \right)}}\left( 0 \right) = 1\\ {f^{\left( 6 \right)}}\left( x \right) & = - \sin x & \hspace{0.5in} & {f^{\left( 6 \right)}}\left( 0 \right) = 0\\ \vdots \,\,\,\,\,\,\,\,\,\, & & \hspace{0.5in} & \,\,\,\,\,\,\, \vdots \end{align*}\]Shunday qilib, biz shunga o'xshash naqshni olamiz. Raqamlarni Teylor seriyasiga kiritamiz.
\[\begin{align*}\sin x & = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}} \\ & = \frac{1}{{1!}}x - \frac{1}{{3!}}{x^3} + \frac{1}{{5!}}{x^5} - \frac{1}{{7!}}{x^7} + \cdots \end{align*}\]Bunday holda, biz faqat \(x\) da toq ko'rsatkichga ega bo'lgan atamalarni olamiz va oxirgi muammoda bo'lgani kabi, nol atamalarni e'tiborsiz qoldirganimizda aniq naqsh va formula mavjud. Shunday qilib, oldingi misolda bo'lgani kabi atamalarni qayta raqamlash orqali biz quyidagi Teylor seriyasini olamiz.
\[\sin x = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{x^{2n + 1}}}}{{\left( {2n + 1} \right)!}}} \]Biz haqiqatan ham \(f\left( x \right)\) \(x = 0\) haqida bo'lmagan yana bir yoki ikkita misolni ishlashimiz kerak.
Yechim
Bu erda dastlabki bir nechta hosilalar va baholashlar.
\[\begin{align*}{f^{\left( 0 \right)}}\left( x \right) & = \ln \left( x \right) & \hspace{0.5in} & {f^{\left( 0 \right)}}\left( 2 \right) = \ln 2\\ {f^{\left( 1 \right)}}\left( x \right) & = \frac{1}{x} & \hspace{0.5in} & {f^{\left( 1 \right)}}\left( 2 \right) = \frac{1}{2}\\ {f^{\left( 2 \right)}}\left( x \right) & = - \frac{1}{{{x^2}}} & \hspace{0.5in} & {f^{\left( 2 \right)}}\left( 2 \right) = - \frac{1}{{{2^2}}}\\ {f^{\left( 3 \right)}}\left( x \right) & = \frac{2}{{{x^3}}} & \hspace{0.5in} & {f^{\left( 3 \right)}}\left( 2 \right) = \frac{2}{{{2^3}}}\\ {f^{\left( 4 \right)}}\left( x \right) & = - \frac{{2\left( 3 \right)}}{{{x^4}}} & \hspace{0.5in} & {f^{\left( 4 \right)}}\left( 2 \right) = - \frac{{2\left( 3 \right)}}{{{2^4}}}\\ {f^{\left( 5 \right)}}\left( x \right) & = \frac{{2\left( 3 \right)\left( 4 \right)}}{{{x^5}}} & \hspace{0.5in} & {f^{\left( 5 \right)}}\left( 2 \right) = \frac{{2\left( 3 \right)\left( 4 \right)}}{{{2^5}}}\\ \vdots \,\,\,\,\,\,\,\,\,\, & & \hspace{0.5in} & \,\,\,\,\,\,\, \vdots \\ {f^{\left( n \right)}}\left( x \right) & = \frac{{{{\left( { - 1} \right)}^{n + 1}}\left( {n - 1} \right)!}}{{{x^n}}} & \hspace{0.5in} & {f^{\left( n \right)}}\left( 2 \right) = \frac{{{{\left( { - 1} \right)}^{n + 1}}\left( {n - 1} \right)!}}{{{2^n}}}\hspace{0.25in}n = 1,2,3, \ldots \end{align*}\]E'tibor bering, bizda umumiy formula mavjud bo'lsa-da, u \(n = 0\) uchun ishlamaydi. Bu ba'zida sodir bo'ladi, shuning uchun bu sodir bo'lganda tashvishlanmang.
Buni Teylor seriyali formulasiga ulash uchun avval \(n = 0\) atamasini olib tashlashimiz kerak.
\[\begin{align*}\ln \left( x \right) & = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( 2 \right)}}{{n!}}{{\left( {x - 2} \right)}^n}} \\ & = f\left( 2 \right) + \sum\limits_{n = 1}^\infty {\frac{{{f^{\left( n \right)}}\left( 2 \right)}}{{n!}}{{\left( {x - 2} \right)}^n}} \\ & = \ln \left( 2 \right) + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}\left( {n - 1} \right)!}}{{n!\,\,\,\,{2^n}}}{{\left( {x - 2} \right)}^n}} \\ & = \ln \left( 2 \right) + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{n\,{2^n}}}{{\left( {x - 2} \right)}^n}} \end{align*}\]E'tibor bering, biz bu holatda faktoriallarni soddalashtirdik. Agar bir nechta bo'lsa va ularni soddalashtirish mumkin bo'lsa, ularni har doim soddalashtirishingiz kerak.
Bundan tashqari, serial oldida o'tirgan atama haqida hayajonlanmang. Ba'zan biz \(n\) ning barcha qiymatlariga mos keladigan umumiy formulani ololmasak, buni qilishimiz kerak.
Yechim
Shunga qaramay, bu erda hosilalar va baholashlar.
\[\begin{align*}{f^{\left( 0 \right)}}\left( x \right) & = \frac{1}{{{x^2}}} & \hspace{0.5in} & {f^{\left( 0 \right)}}\left( { - 1} \right) = \frac{1}{{{{\left( { - 1} \right)}^2}}} = 1\\ {f^{\left( 1 \right)}}\left( x \right) & = - \frac{2}{{{x^3}}} & \hspace{0.5in} & {f^{\left( 1 \right)}}\left( { - 1} \right) = - \frac{2}{{{{\left( { - 1} \right)}^3}}} = 2\\ {f^{\left( 2 \right)}}\left( x \right) & = \frac{{2\left( 3 \right)}}{{{x^4}}} & \hspace{0.5in} & {f^{\left( 2 \right)}}\left( { - 1} \right) = \frac{{2\left( 3 \right)}}{{{{\left( { - 1} \right)}^4}}} = 2\left( 3 \right)\\ {f^{\left( 3 \right)}}\left( x \right) & = - \frac{{2\left( 3 \right)\left( 4 \right)}}{{{x^5}}} & \hspace{0.5in} & {f^{\left( 3 \right)}}\left( { - 1} \right) = - \frac{{2\left( 3 \right)\left( 4 \right)}}{{{{\left( { - 1} \right)}^5}}} = 2\left( 3 \right)\left( 4 \right)\\ \vdots \,\,\,\,\,\,\,\,\,\, & & \hspace{0.5in} & \,\,\,\,\,\,\, \vdots \\ {f^{\left( n \right)}}\left( x \right) & = \frac{{{{\left( { - 1} \right)}^n}\left( {n + 1} \right)!}}{{{x^{n + 2}}}} & \hspace{0.5in} & {f^{\left( n \right)}}\left( { - 1} \right) = \frac{{{{\left( { - 1} \right)}^n}\left( {n + 1} \right)!}}{{{{\left( { - 1} \right)}^{n + 2}}}} = \left( {n + 1} \right)!\end{align*}\]E'tibor bering, barcha salbiy belgilar baholashda bekor qilinadi. Bundan tashqari, ushbu formula oldingi misoldan farqli o'laroq, barcha \(n\) uchun ishlaydi.
Bu funksiya uchun Teylor seriyasi.
\[\begin{align*}\frac{1}{{{x^2}}} & = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( { - 1} \right)}}{{n!}}{{\left( {x + 1} \right)}^n}} \\ & = \sum\limits_{n = 0}^\infty {\frac{{\left( {n + 1} \right)!}}{{n!}}{{\left( {x + 1} \right)}^n}} \\ & = \sum\limits_{n = 0}^\infty {\left( {n + 1} \right){{\left( {x + 1} \right)}^n}} \end{align*}\]Keling, ushbu bo'limdagi eng oson misollardan birini ko'rib chiqaylik. Aksariyat talabalar uchun muammo shundaki, bu birinchi qarashda unchalik oson bo'lmasligi (yoki juda oson bo'lib tuyulishi mumkin).
Yechim
Mana bu muammoning hosilalari.
\[\begin{align*}{f^{\left( 0 \right)}}\left( x \right) & = {x^3} - 10{x^2} + 6 & \hspace{0.5in} & {f^{\left( 0 \right)}}\left( 3 \right) = - 57\\ {f^{\left( 1 \right)}}\left( x \right) & = 3{x^2} - 20x & \hspace{0.5in} & {f^{\left( 1 \right)}}\left( 3 \right) = - 33\\ {f^{\left( 2 \right)}}\left( x \right) & = 6x - 20 & \hspace{0.5in} & {f^{\left( 2 \right)}}\left( 3 \right) = - 2\\ {f^{\left( 3 \right)}}\left( x \right) & = 6 & \hspace{0.5in} & {f^{\left( 3 \right)}}\left( 3 \right) = 6\\ {f^{\left( n \right)}}\left( x \right) & = 0 & \hspace{0.5in} & {f^{\left( 4 \right)}}\left( 3 \right) = 0\hspace{0.25in}n \ge 4\end{align*}\]Ushbu Teylor seriyasi \(n = 3\) dan keyin tugaydi. Bu har doim polinomning Teylor seriyasini topganimizda sodir bo'ladi. Mana bu uchun Teylor seriyasi.
\[\begin{align*}{x^3} - 10{x^2} + 6 & = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( 3 \right)}}{{n!}}{{\left( {x - 3} \right)}^n}} \\ & = f\left( 3 \right) + f'\left( 3 \right)\left( {x - 3} \right) + \frac{{f''\left( 3 \right)}}{{2!}}{\left( {x - 3} \right)^2} + \frac{{f'''\left( 3 \right)}}{{3!}}{\left( {x - 3} \right)^3} + 0\\ & = - 57 - 33\left( {x - 3} \right) - {\left( {x - 3} \right)^2} + {\left( {x - 3} \right)^3}\end{align*}\]Ko'phadning Teylor seriyasini topishda biz o'ng tomonni soddalashtirmaymiz. Biz uni xuddi shunday qoldiramiz. Aslida, agar biz hamma narsani ko'paytiradigan bo'lsak, biz faqat asl polinomga qaytamiz!
Ko'pnom uchun Teylor seriyasini yozish foydali ekanligi aniq bo'lmasa-da, buni qilish kerak bo'lgan holatlar mavjud. Muammo shundaki, ular ushbu kurs doirasidan tashqarida va shuning uchun bu erda yoritilgan emas. Masalan, Differensial tenglamalar sohasida ketma-ketliklarga bitta ilova mavjud bo'lib, buni vaqti-vaqti bilan qilish kerak.
Shunday qilib, biz Teylor seriyasining bir nechta misollarini ko'rdik va ularning barchasida biz seriya uchun umumiy formulalarni topa oldik. Bu har doim ham shunday bo'lmaydi. Umumiy formulaga ega bo'lmagan misolni ko'rish uchun keyingi bo'limdagi oxirgi misolni ko'rib chiqing.
Ushbu bo'limni tark etishdan oldin biz ushbu bo'limda olingan uchta muhim Teylor seriyasi mavjud bo'lib, ularni bir joyda umumlashtirishimiz kerak. Mening sinfimda siz ushbu formulalarni shu vaqtdan boshlab bilasiz deb taxmin qilaman.