10.16-bo'lim : Teylor seriyasi
5. \(x = - 3\) haqida \(\displaystyle f\left( x \right) = \frac{7}{{{x^4}}}\) uchun Teylor seriyasini toping.
Yechim
OK, biz ushbu muammoni funktsiyaning bir nechta hosilalarini olishdan boshlashimiz kerak.
\[\begin{align*}{n = 0:} & & f\left( x \right) & = \frac{7}{{{x^4}}} = 7{x^{ - 4}}\\{n = 1:} & & f'\left( x \right) & = - 7\left( 4 \right){x^{ - 5}}\\{n = 2:} & & f''\left( x \right) & = 7\left( 4 \right)\left( 5 \right){x^{ - 6}}\\{n = 3:} & & {f^{\left( 3 \right)}}\left( x \right) & = - 7\left( 4 \right)\left( 5 \right)\left( 6 \right){x^{ - 7}}\\{n = 4:} & & {f^{\left( 4 \right)}}\left( x \right) & = 7\left( 4 \right)\left( 5 \right)\left( 6 \right)\left( 7 \right){x^{ - 8}}\end{align*}\]Esda tutingki, umuman olganda, biz Teylor seriyasida umumiy atama formulasini olishimiz uchun ushbu muammolarning aksariyati uchun kamida \(n = 4\) ga borishimiz kerak.
Bundan tashqari, lotinlarni olishda narsalarni ko'paytirmaslikni unutmang! Bu umumiy formulani topish vaqti kelganda hayotingizni ancha qiyinlashtiradi. Bu holda biz qilgan yagona "soddalashtirish" hosilalarni bajarishda ko'rsatkichlardan kelib chiqadigan minus belgilarini ko'paytirish edi. Bu kabi muammolar bilan shug'ullanish juda keng tarqalgan narsa.
2-qadam
Teylor seriyasida umumiy atama uchun formulani olishimiz mumkinligini ko'rish vaqti keldi.
Umid qilamanki, yuqoridagi lotinlardagi naqshni ko'rishingiz mumkin. Umumiy atama tomonidan berilgan:
\[\begin{align*}{f^{\left( n \right)}}\left( x \right) & = 7{\left( { - 1} \right)^n}\frac{{\left( 2 \right)\left( 3 \right)}}{{\left( 2 \right)\left( 3 \right)}}\left( 4 \right)\left( 5 \right)\left( 6 \right) \cdots \left( {n + 3} \right){x^{ - (n+4)}}\\ & = 7{\left( { - 1} \right)^n}\frac{{\left( 2 \right)\left( 3 \right)\left( 4 \right)\left( 5 \right)\left( 6 \right) \cdots \left( {n + 3} \right)}}{6}{x^{ - (n+4)}}\\ & = \frac{7}{6}{\left( { - 1} \right)^n}\left( {n + 3} \right)!{x^{ - \left( {n + 4} \right)}} \hspace{0.5in} n = 0,1,2,3, \ldots \end{align*}\]Ushbu formulani olish biroz qiyinroq bo'lishi mumkin. Bizda hosilalarda faktorial deyarli bor edi, lekin ularning har birida faktorialni ko'rsatish uchun zarur bo'lgan \(\left( 2 \right)\left( 3 \right)\) qismi etishmayapti. Bu yetishmayotgan narsa va hosilalarning har birida yo'qligi sababli biz har bir hosilani \(\frac{{\left( 2 \right)\left( 3 \right)}}{{\left( 2 \right)\left( 3 \right)}}\) ga ko'paytirdik (i.e. yozishning chindan ham ajoyib usuli...). Keyin faktorialni bajarish uchun bu raqamdan foydalanishimiz mumkin va maxraj yolg'iz qoldi.
Bundan tashqari, ta'kidlanganidek, bu formula \(n = 0\) ga qadar ishlaydi. Qanchalik orqada ishlashini aniqlash uchun ushbu formulani tekshirganingizga ishonch hosil qilish muhimdir. Biz ba'zida \(n\) ning birinchi juftligi uchun ishlamaydigan formulalarni olamiz va Teylor seriyasini yozishni boshlashdan oldin buni bilishimiz kerak.
3-qadam
Endi eslang, biz hech qanday \(x\) da umumiy atamani xohlamaymiz. Biz \(x = - 3\) da umumiy atamani xohlaymiz. Bu,
\[\begin{align*}{f^{\left( n \right)}}\left( { - 3} \right) & = \frac{7}{6}{\left( { - 1} \right)^n}\left( {n + 3} \right)!{\left( { - 3} \right)^{ - \left( {n + 4} \right)}}\\ & = \frac{{7{{\left( { - 1} \right)}^n}\left( {n + 3} \right)!}}{{6{{\left( { - 3} \right)}^{n + 4}}}}\,\\ & = \frac{{7{{\left( { - 1} \right)}^n}\left( {n + 3} \right)!}}{{6{{\left( { - 1} \right)}^{n + 4}}{{\left( 3 \right)}^{n + 4}}}}\,\,\\ & = \frac{{7\left( {n + 3} \right)!}}{{6{{\left( { - 1} \right)}^4}{{\left( 3 \right)}^{n + 4}}}}\\ & = \frac{{7\left( {n + 3} \right)!}}{{6{{\left( 3 \right)}^{n + 4}}}} \hspace{0.5in} n = 1,2,3, \ldots \end{align*}\]Biz bu erda biroz soddalashtirdik, shuning uchun atamada mavjud bo'lgan barcha o'zgaruvchan belgilarni bekor qilishimiz mumkin.
4-qadam
Xo'sh, hozir biz ushbu muammo uchun Teylor seriyasini rasmiy ravishda yozishimiz mumkin.
\[\frac{7}{{{x^4}}} = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( { - 3} \right)}}{{n!}}{{\left( {x + 3} \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{{7\left( {n + 3} \right)!}}{{6{{\left( 3 \right)}^{n + 4}}n!}}{{\left( {x + 3} \right)}^n}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\sum\limits_{n = 0}^\infty {\frac{{7\left( {n + 3} \right)\left( {n + 2} \right)\left( {n + 1} \right)}}{{6{{\left( 3 \right)}^{n + 4}}}}{{\left( {x + 3} \right)}^n}} }}\]Yakuniy javobda qayerda bo'lishimiz mumkinligini soddalashtirishni/bekor qilishni unutmang. Bunday holda biz faktoriallar yordamida biroz soddalashtirishimiz mumkin.