10.16-bo'lim : Teylor seriyasi
4. \(x = 0\) haqida \(f\left( x \right) = \ln \left( {3 + 4x} \right)\) uchun Teylor seriyasini toping.
Yechim
OK, biz ushbu muammoni funktsiyaning bir nechta hosilalarini olishdan boshlashimiz kerak.
\[\begin{align*}{n = 0:} & & f\left( x \right) & = \ln \left( {3 + 4x} \right)\\{n = 1:} & &f'\left( x \right) & = \frac{4}{{3 + 4x}} = 4{{\left( {3 + 4x} \right)}^{ - 1}}\\{n = 2:} & & f''\left( x \right) & = - {4^2}{{\left( {3 + 4x} \right)}^{ - 2}}\\{n = 3:} & & {f^{\left( 3 \right)}}\left( x \right) & = {4^3}\left( 2 \right){{\left( {3 + 4x} \right)}^{ - 3}}\\{n = 4:} & & {f^{\left( 4 \right)}}\left( x \right) & = - {4^4}\left( 2 \right)\left( 3 \right){{\left( {3 + 4x} \right)}^{ - 4}}\\{n = 5:} & & {f^{\left( 5 \right)}}\left( x \right) & = {4^5}\left( 2 \right)\left( 3 \right)\left( 4 \right){{\left( {3 + 4x} \right)}^{ - 5}}\end{align*}\]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. Bunday holda, biz zanjir qoidasidan kelgan barcha 4 ni bitta atamaga "birlashtirdik", lekin haqiqiy qiymatni olish o'rniga uni eksponent sifatida qoldirdik. Bunday muammolar bilan bu odatiy emas. Ko'rsatkichlarni biz lotinlar uchun tushirdik, belgilar bilan ishlash bundan mustasno.
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:
\[\begin{align*}{f^{\left( 0 \right)}}\left( x \right) & = \ln \left( {3 + 4x} \right) & \hspace{0.25in} & n = 0\\ {f^{\left( n \right)}}\left( x \right) & = {\left( { - 1} \right)^{n + 1}}{4^n}\left( {n - 1} \right)!{\left( {3 + 4x} \right)^{ - n}} & \hspace{0.25in} & n = 1,2,3, \ldots \end{align*}\]Qayd etilganidek, bu formula \(n = 1\) ga qadar ishlaydi, lekin \(n = 0\) uchun aniq ishlamaydi. Aynan shunga o'xshash muammolar nima uchun biz taklif qilgan umumiy yechim formulamiz qanchalik orqada ishlashini ko'rish uchun doimo tekshirishimiz kerakligini aniq ko'rsatib beradi.
3-qadam
Endi eslang, biz hech qanday \(x\) da umumiy atamani xohlamaymiz. Biz \(x = 0\) da umumiy atamani istaymiz. Bu,
\[\begin{align*}{f^{\left( 0 \right)}}\left( 0 \right) & = \ln \left( 3 \right) & \hspace{0.25in} & n = 0\\ {f^{\left( n \right)}}\left( 0 \right) & = {\left( { - 1} \right)^{n + 1}}{4^n}\left( {n - 1} \right)!{\left( 3 \right)^{ - n}} & &\\ & = {\left( { - 1} \right)^{n + 1}}{4^n}\left( {n - 1} \right)!\frac{1}{{{3^n}}} & & \\ & = {\left( { - 1} \right)^{n + 1}}{\left( {\frac{4}{3}} \right)^n}\left( {n - 1} \right)! & \hspace{0.25in} & n = 1,2,3, \ldots \end{align*}\]Biz ikkinchisini biroz soddalashtirish uchun biroz soddalashtirdik.
4-qadam
Xo'sh, hozir biz ushbu muammo uchun Teylor seriyasini rasmiy ravishda yozishimiz mumkin. Biroq, buni amalga oshirishdan oldin, bizning umumiy atama formulamiz \(n = 0\) uchun ishlamaganligini esga oling va shuning uchun biz umumiy formulani kiritishdan oldin uni seriyadan olib tashlashimiz kerak.
\[\begin{align*}\ln \left( {3 + 4x} \right) & = \sum\limits_{n = 0}^\infty {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}} \\ & = f\left( 0 \right) + \sum\limits_{n = 1}^\infty {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n}} \\ & = \ln \left( 3 \right) + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}{{\left( {\frac{4}{3}} \right)}^n}\left( {n - 1} \right)!}}{{n!}}{x^n}} \\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\ln \left( 3 \right) + \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}{{\left( {\frac{4}{3}} \right)}^n}}}{n}{x^n}} }}\end{align*}\]Yakuniy javobda qayerda bo'lishimiz mumkinligini soddalashtirishni/bekor qilishni unutmang. Bunday holda biz faktoriallar yordamida biroz soddalashtirishimiz mumkin.