10.18-bo'lim: Binom seriyasi
Ushbu bobning ushbu yakuniy qismida biz funktsiyaning boshqa seriyali tasvirini ko'rib chiqamiz. Buni amalga oshirishdan oldin, keling, quyidagi teoremani eslaylik.
Binom teoremasi
Agar \(n\) har qanday musbat son bo'lsa,
\[\begin{align*}{\left( {a + b} \right)^n} & = \sum\limits_{i = 0}^n {n \choose i} {a^{n - i}}\,{b^i} \,\\ & = {a^n} + n{a^{n - 1}}b + \frac{{n\left( {n - 1} \right)}}{{2!}}{a^{n - 2}}{b^2} + \cdots + na{b^{n - 1}} + {b^n}\end{align*}\]qayerda,
\[\begin{align*}{n \choose i} & = \frac{{n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - i + 1} \right)}}{{i!}}\hspace{0.25in}i = 1,2,3, \ldots n\\ {n \choose 0} & = 1\end{align*}\]Bu to'g'ridan-to'g'ri ko'paytirish oson bo'lmaganda, katta \(n\) uchun \({\left( {a + b} \right)^n}\)-ni kengaytirish uchun foydalidir. Keling, bir misolni tez ko'rib chiqaylik.
Yechim
Haqiqatan ham teoremaga ulanishdan boshqa qiladigan ko'p narsa yo'q.
\[\begin{align*}{\left( {2x - 3} \right)^4} & = \sum\limits_{i = 0}^4 { {4 \choose i} \,{{\left( {2x} \right)}^{4 - i}}\,{{\left( { - 3} \right)}^i}} \\ & = {4 \choose 0}{\left( {2x} \right)^4} + {4 \choose 1}{\left( {2x} \right)^3}\left( { - 3} \right) + {4 \choose 2}{\left( {2x} \right)^2}{\left( { - 3} \right)^2} + {4 \choose 3}\left( {2x} \right){\left( { - 3} \right)^3} + {4 \choose 4}{\left( { - 3} \right)^4}\\ & = {\left( {2x} \right)^4} + 4{\left( {2x} \right)^3}\left( { - 3} \right) + \frac{{4\left( 3 \right)}}{2}{\left( {2x} \right)^2}{\left( { - 3} \right)^2} + 4\left( {2x} \right){\left( { - 3} \right)^3} + {\left( { - 3} \right)^4}\\ & = 16{x^4} - 96{x^3} + 216{x^2} - 216x + 81\end{align*}\]Endi binom teoremasi \(n\) musbat butun son boʻlishini talab qildi. Buning kengaytmasi mavjud, ammo u har qanday raqamga ruxsat beradi.
Binom seriyasi
Agar \(k\) har qanday raqam va \(\left| x \right| < 1\) bo'lsa,
\[\begin{align*}{\left( {1 + x} \right)^k} & = \sum\limits_{n = 0}^\infty { {k \choose n} {x^n}} \,\\ & = 1 + kx + \frac{{k\left( {k - 1} \right)}}{{2!}}{x^2} + \frac{{k\left( {k - 1} \right)\left( {k - 2} \right)}}{{3!}}{x^3} + \cdots \end{align*}\]qayerda,
\[\begin{align*}{k \choose n} & = \frac{{k\left( {k - 1} \right)\left( {k - 2} \right) \cdots \left( {k - n + 1} \right)}}{{n!}}\hspace{0.25in}n = 1,2,3, \ldots \\ {k \choose 0} & = 1\end{align*}\]Shunday qilib, binomial teoremaga o'xshash, bundan tashqari bu cheksiz qator va konvergentsiyani olish uchun bizda \(\left| x \right| < 1\) bo'lishi kerak.
Keling, bunga misolni ko'rib chiqaylik.
Yechim
Shunday qilib, bu holda \(k = \frac{1}{2}\) va biz atamani kerakli shaklga kiritish uchun biroz qayta yozishimiz kerak bo'ladi.
\[\sqrt {9 - x} = 3{\left( {1 - \frac{x}{9}} \right)^{\frac{1}{2}}} = 3{\left( {1 + \left( { - \frac{x}{9}} \right)} \right)^{\frac{1}{2}}}\]Binom qatoridagi dastlabki to'rtta atama shunday bo'ladi:
\[\begin{align*}\sqrt {9 - x} & = 3{\left( {1 + \left( { - \frac{x}{9}} \right)} \right)^{\frac{1}{2}}}\\ & = 3\sum\limits_{n = 0}^\infty { {\frac{1}{2} \choose n} {{\left( { - \frac{x}{9}} \right)}^n}} \,\\ & = 3\left[ {1 + \left( {\frac{1}{2}} \right)\left( { - \frac{x}{9}} \right) + \frac{{\frac{1}{2}\left( { - \frac{1}{2}} \right)}}{2}{{\left( { - \frac{x}{9}} \right)}^2} + \frac{{\frac{1}{2}\left( { - \frac{1}{2}} \right)\left( { - \frac{3}{2}} \right)}}{6}{{\left( { - \frac{x}{9}} \right)}^3} + \cdots } \right]\\ & = 3 - \frac{x}{6} - \frac{{{x^2}}}{{216}} - \frac{{{x^3}}}{{3888}} - \cdots \end{align*}\]