"Sevgilim" Taylor

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Men Taylorni yaxshi ko’raman
Men ham blyat

Matematik muhabbatim Taylorni qalbiga qanday yo’l topganim haqida.

Avval sizlarni tanishtirsam. Taylor bu - murakkab funksiyani hosilalar yordamida oddiy ifodaga aylantirish. Sevgilim yordamida sin(x) yoki $e^x$ kabi funksiyalarni tez va taxminiy hisoblash mumkin.

Oxirgi kunlarda miyyamni itargan mavzu.

Boshlaymiz:

Binomial qator dastlab

Bu nima? bu $(1+x)r$ kabi ifodani qator (cheksiz yoki chekli) ko‘rinishda yozish usuli. Bu ayniqsa $r$ har qanday haqiqiy son bo‘lishi mumkin bo‘lganda foydali.

Oddiy binomial formulasi (butun sonlar uchun):

Agar $r$ noldan katta butun son bo‘lsa, u holda biz maktabdan yaxshi biladigan binomial formulani olamiz(maktabda xuy o’rgatadi buni)

(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k = 1 + nx + \frac{n(n-1)}{2!}x^2 + \cdots + x^n

Bu yerda $(n k)$ — binomial koeffitsiyentlar (kombinatsiyalar soni).

Binomial qatori (har qanday r uchun):

Agar $r$ har qanday haqiqiy son bo‘lsa (masalan, $r=2/1$ yoki $r=−3.7$ ), u holda $r(1+x)^r$ ni cheksiz qator shaklida yozishga to’g’ri keladi. Bu yerda:

\binom{r}{k} = \frac{r(r - 1)(r - 2) \cdots (r - k + 1)}{k!}

va bu qator faqat $|x| < 1$ bo‘lganda yaqinlashadi (konvergent bo‘ladi). Bu rostan maktabda o’tilishi kerak edi mayli endi asosiy mavzuga qaytamiz.

Quyidagi funksiyalar:

\begin{aligned} f(x) &= (1 + x)^0 = 1 \\ f(x) &= (1 + x)^1 = 1 + x \\ f(x) &= (1 + x)^2 = 1 + 2x + x^2 \\ f(x) &= (1 + x)^3 = 1 + 3x + 3x^2 + x^3 \\ f(x) &= (1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4 \end{aligned}

yuqoridagi o‘ng tarafdagi ifodalar binomial kengaytmalar (binomial expansions) deb ataladi. Ushbu kengaytmadagi sonlar esa binomial koeffitsiyentlar (binomial coefficients) deb yuritiladi.Bir biridan ybancha nomlar.

Umuman olganda, har qanday musbat butun son $r$ uchun, $(1+x)^r$ kengaytmasidagi $x^n$ ning binomial koeffitsiyenti quyidagicha aniqlanadi:

\binom{r}{n} = \frac{r!}{n!(r - n)!}

va yana

f(x) = (1 + x)^r = \sum_{n=0}^{r} \binom{r}{n} x^n = \binom{r}{0} + \binom{r}{1}x + \binom{r}{2}x^2 + \cdots + \binom{r}{r}x^r

Misol uchun: $r=5$ bo‘lsa:

(1 + x)^5 = \binom{5}{0} + \binom{5}{1}x + \binom{5}{2}x^2 + \binom{5}{3}x^3 + \binom{5}{4}x^4 + \binom{5}{5}x^5

= 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5

Hop faqat faqat musbat butun emas haqiqiy son bo‘lgandachi?

Agar $r$ musbat butun son bo‘lmasa, unda: $f(x)=(1+x)^r$

ifoda chekli darajali polinom sifatida yozib bo‘lmaydi. Biroq, biz uning uchun quvvatlar qatori(o’zi power-series bilaman tarjima gavno) topishimiz mumkin.Biz bu funksiyaning Makloren qatorini topmoqchimiz.

Buning uchun:

Funksiya hosilalarini olamiz.
Ularni $x=0$ nuqtada qo’yamiz

Hosilalar ketma-ketligi:

\begin{aligned} f(x) &= (1 + x)^r \\ f'(x) &= r(1 + x)^{r - 1} \\ f''(x) &= r(r - 1)(1 + x)^{r - 2} \\ f^{(3)}(x) &= r(r - 1)(r - 2)(1 + x)^{r - 3} \\ f^{(n)}(x) &= r(r - 1)(r - 2) \cdots (r - n + 1)(1 + x)^{r - n} \end{aligned}

Hosilalarni $x=0$ da hisoblash:

\begin{aligned} f(0) &= 1 \\ f'(0) &= r \\ f''(0) &= r(r - 1) \\ f^{(3)}(0) &= r(r - 1)(r - 2) \\ f^{(n)}(0) &= r(r - 1)(r - 2) \cdots (r - n + 1) \end{aligned}

Binomial qatorning koeffitsiyentlari:

\binom{r}{n} = \frac{r(r - 1)(r - 2) \cdots (r - n + 1)}{n!}

Cheksiz binomial qatori (umumiy ko‘rinish):

(1 + x)^r = \sum_{n=0}^{\infty} \binom{r}{n} x^n = 1 + rx + \frac{r(r - 1)}{2!}x^2 + \frac{r(r - 1)(r - 2)}{3!}x^3 + \cdots

Agar rr musbat butun son bo‘lsa, u holda f(r+1)f(r+1) — ya’ni (r+1)-darajali hosila — nol funksiyaga aylanadi, ya’ni hosilalar tugaydi. Shuning uchun qator cheklanadi, ya’ni tugaydi.

Shuningdek, agar $r$ musbat butun son bo‘lsa, u holda tenglama haqiqiy $r$ uchun va butun $r$ uchun mos keladi, va binomial qator formulasi ham formula bilan bir xil bo‘ladi.

Har qanday haqiqiy son $r$ uchun binomial koeffitsiyent quyidagicha aniqlanadi:

\binom{r}{n} = \frac{r(r - 1)(r - 2) \cdots (r - n + 1)}{n!}

*Eslatma: Ba’zida yuqoridagi ifoda faqat $r−n+1$ gacha bo‘lgan $n$ ta ko‘paytuvchi bilan yoziladi, chunki bu umumiylashtirilgan faktorialning (Pochhammer symbol) bir ko‘rinishi hisoblanadi

binomial qator har qanday $r∈R$ uchun quyidagicha yoziladi:

(1 + x)^r = \sum_{n=0}^{\infty} \binom{r}{n} x^n = 1 + rx + \frac{r(r - 1)}{2!}x^2 + \frac{r(r - 1)(r - 2)}{3!}x^3 + \cdots

yana masalan:

yaqinlashuv oraliğini (interval of convergence) aniqlashimiz kerak:

\sum_{n=0}^{\infty} \binom{r}{n} x^n = 1 + rx + \frac{r(r - 1)}{2!}x^2 + \cdots + \frac{r(r - 1) \cdots (r - n + 1)}{n!}x^n + \cdots

Shunda nisbat:

\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{r - n}{n + 1} \cdot x \right|

Limit Test:

\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{r - n}{n + 1} \cdot x \right| = |x|

Yakuniy:

\text{Binomial qator } \sum_{n=0}^{\infty} \binom{r}{n} x^n \text{ quyidagi oraliqda yaqinlashadi: } (-1, 1)

Mashhur funksiyalarning Makloren qatorlari:

Funksiya	Makloren qatori	Yaqinlashuv oraliği
$f(x) = \frac{1}{1 - x}$	$\sum_{n=0}^{\infty} x^n$	$-1 < x < 1$
$f(x) = e^x$	$\sum_{n=0}^{\infty} \frac{x^n}{n!}$	$-\infty < x < \infty$
$f(x) = \sin x$	$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$	$-\infty < x < \infty$
$f(x) = \cos x$	$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}$	$-\infty < x < \infty$
$f(x) = \ln(1 + x)$	$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}$	$-1 < x \leq 1$
$f(x) = \tan^{-1}(x)$	$\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$	$-1 \leq x \leq 1$
$f(x) = (1 + x)^r$	$\sum_{n=0}^{\infty} \binom{r}{n} x^n$	$-1 < x < 1$

Misol yechamiz

Ma’lum qatorlardan Maclaurin qatorini hosil qilish:

a misol) $f(x) = \sqrt{\cos x}$

$\cos x$ ning Maclaurin qatori quyidagicha:

\cos x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}

Endi $f(x) = \sqrt{\cos x}$ funksiyasining Maclaurin qatorini topamiz.

$\sqrt{\cos x}$ ning Maclaurin qatori quyidagicha ifodalanadi:

f(x) = \sqrt{\cos x} = \sum_{n=0}^{\infty} (-1)^n \frac{(\sqrt{x})^{2n}}{(2n)!} = \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{(2n)!}

Yoki ochilgan shaklda:

f(x) = 1 - \frac{x}{2!} + \frac{x^2}{4!} - \frac{x^3}{6!} + \frac{x^4}{8!} - \cdots

Ushbu qator $x \geq 0$ da aniqlangan va $\sqrt{\cos x}$ funksiyasiga yaqinlashadi.

b misol) $f(x) = \sinh x$

sinhx funksiyasini quyidagicha ifodalaymiz:

\sinh x = \frac{e^x - e^{-x}}{2}

Eksponentaning Maclaurin qatoridan foydalanib, $sinh⁡x$ ning $n$ chi hadini quyidagicha yozamiz:

f(x) = \sqrt{\cos x} = \sum_{n=0}^{\infty} (-1)^n \frac{(\sqrt{x})^{2n}}{(2n)!} = \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{(2n)!}

Agar $n$ juft bo‘lsa, ushbu had nolga teng. Agar $n$ toq bo‘lsa, bu had $2x^n$ / $n!$ ga teng bo‘ladi. Shuning uchun, $sinh⁡x$ ning Maclaurin qatori faqat toq tartibli hadlardan iborat va quyidagicha ifodalanadi:

f(x) = 1 - \frac{x}{2!} + \frac{x^2}{4!} - \frac{x^3}{6!} + \frac{x^4}{8!} - \cdots

Umumiy bilish kerak bo’lgan narsalar shular,qolgani Taylorni boshqa funksiyalar bilan bog’lash va ishlatish. So’ngi xulosa qilib, Taylor qatori funksiyaning tanlangan nuqtadagi qiymati va uning birinchi, ikkinchi, uchinchi va boshqa hosilalari (derivativlari) orqali quriladi desak bo’ladi.