Appendix A.8: Summation notation
Ushbu bo'limda biz yig'indi belgisi yoki sigma belgisini qisqacha ko'rib chiqishimiz kerak. Biz ikkita butun son, \(n\) va \(m\), \(n < m\) va quyidagi tarzda belgilangan raqamlar ro'yxati bilan boshlaymiz,
\[{a_n},\,\,{a_{n + 1}},\,\,{a_{n + 2}},\,\, \ldots ,\,\,{a_{m - 2}},\,\,{a_{m - 1}},\,\,{a_m}\]We want to add them up, in other words we want,
\[{a_n} + \,{a_{n + 1}} + \,\,{a_{n + 2}} + \,\, \ldots + \,\,{a_{m - 2}} + \,\,{a_{m - 1}} + \,\,{a_m}\]Katta ro'yxatlar uchun bu juda og'ir belgi bo'lishi mumkin, shuning uchun biz bu turdagi summalarni belgilash uchun yig'indi belgisini kiritamiz. The case above is denoted as follows.
\[\sum\limits_{i = n}^m {{a_i}} = {a_n} + \,{a_{n + 1}} + \,\,{a_{n + 2}} + \,\, \ldots + \,\,{a_{m - 2}} + \,\,{a_{m - 1}} + \,\,{a_m}\]\(i\) yig'indisi indeksi deb ataladi. Bu belgi bizga \(n\) dan boshlab \(m\) da tugaydigan barcha tamsayılar uchun barcha \({a_i}\) ni qo‘shishni bildiradi.
Masalan,
\[\begin{align*}\sum\limits_{i = 0}^4 {\frac{i}{{i + 1}}} & = \frac{0}{{0 + 1}} + \frac{1}{{1 + 1}} + \frac{2}{{2 + 1}} + \frac{3}{{3 + 1}} + \frac{4}{{4 + 1}} = \frac{{163}}{{60}} = 2.7166\overline{6}\\ \sum\limits_{i = 4}^6 {{2^i}{x^{2i + 1}}} & = {2^4}{x^9} + {2^5}{x^{11}} + {2^6}{x^{13}} = 16{x^9} + 32{x^{11}} + 64{x^{13}}\\ \sum\limits_{i = 1}^4 {f\left( {x_i^*} \right)} & = f\left( {x_1^*} \right) + f\left( {x_2^*} \right) + f\left( {x_3^*} \right) + f\left( {x_4^*} \right)\end{align*}\]Properties of Summation Notation
Bu erda yig'indini belgilash uchun bir nechta formulalar mavjud.
- \(\sum\limits_{i\, = \,{i_{\,0}}}^n {c{a_i}} = c\sum\limits_{i\, = \,{i_{\,0}}}^n {{a_i}} \) where \(c\) is any number. So, we can factor constants out of a summation.
- \(\sum\limits_{i\, = \,{i_{\,0}}}^n {\left( {{a_i} \pm {b_i}} \right)} = \sum\limits_{i\, = \,{i_{\,0}}}^n {{a_i}} \pm \sum\limits_{i\, = \,{i_{\,0}}}^n {{b_i}} \) Shunday qilib, biz yig'indi yoki farq bo'yicha yig'indini ajratishimiz mumkin.
Esda tutingki, biz seriyani \({i_{\,0}}\) da, ular \(i\) ning bizga kerak bo‘lgan istalgan qiymatidan boshlanishi mumkinligini bildirish uchun boshladik. Shuni ham yodda tutingki, biz avvalgidek yig'indi va farqlarni ajratishimiz mumkin 2 yuqorida biz mahsulotlar va ko'rsatkichlar uchun xuddi shunday qila olmaymiz. Boshqacha aytganda,
\[\sum\limits_{i\, = \,{i_{\,0}}}^n {\left( {{a_i}{b_i}} \right)} \ne \left( {\sum\limits_{i\, = \,{i_{\,0}}}^n {{a_i}} } \right)\left( {\sum\limits_{i\, = \,{i_{\,0}}}^n {{b_i}} } \right)\hspace{0.75in}\sum\limits_{i\, = \,{i_{\,0}}}^n {\frac{{{a_i}}}{{{b_i}}}} \ne \frac{{\sum\limits_{i\, = \,{i_{\,0}}}^n {{a_i}} }}{{\sum\limits_{i\, = \,{i_{\,0}}}^n {{b_i}} }}\]Formulalar
Mana bir necha bo'limda foydali bo'ladigan bir nechta chiroyli formulalar. E'tibor bering, bu formulalar faqat \(i = 1\) dan boshlanganda to'g'ri bo'ladi. Agar kerak bo'lsa, siz, albatta, turli boshlang'ich nuqtalar uchun boshqa formulalarni olishingiz mumkin.
Xulosa formulalari
- \(\displaystyle \sum\limits_{i = 1}^n c = cn\)
- \(\displaystyle \sum\limits_{i = 1}^n i = \frac{{n\left( {n + 1} \right)}}{2}\)
- \(\displaystyle \sum\limits_{i = 1}^n {{i^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\)
- \(\displaystyle \sum\limits_{i = 1}^n {{i^3}} = {\left[ {\frac{{n\left( {n + 1} \right)}}{2}} \right]^2}\)
Qo'lda qilish oson bo'lmagan summani tezda baholash uchun ushbu xususiyatlardan qanday foydalanish haqida qisqacha misol.
Yechim
Biz qilishimiz kerak bo'lgan birinchi narsa - yig'ilgan narsalarni kvadratga aylantirish va keyin quyidagi xususiyatlardan foydalanib yig'indini ajratish:
\[\begin{align*}\sum\limits_{i = 1}^{100} {{{\left( {3 - 2i} \right)}^2}} & = \sum\limits_{i = 1}^{100} {9 - 12i + 4{i^2}} \\ & = \sum\limits_{i = 1}^{100} 9 - \sum\limits_{i = 1}^{100} {12i} + \sum\limits_{i = 1}^{100} {4{i^2}} \\ & = \sum\limits_{i = 1}^{100} 9 - 12\sum\limits_{i = 1}^{100} i + 4\sum\limits_{i = 1}^{100} {{i^2}} \end{align*}\]Endi formulalar yordamida buni hisoblash oson,
\[\begin{align*}\sum\limits_{i = 1}^{100} {{{\left( {3 - 2i} \right)}^2}} & = 9\left( {100} \right) - 12\left( {\frac{{100\left( {101} \right)}}{2}} \right) + 4\left( {\frac{{100\left( {101} \right)\left( {201} \right)}}{6}} \right)\\ & = 1293700\end{align*}\]Buni qo'lda qilish, albatta, biroz vaqt talab etadi va biz biron bir joyda kichik xatoga yo'l qo'ygan bo'lishimiz uchun yaxshi imkoniyat bor.