Taylorov razvoj iracionalne funkcije

Razvijte u Taylorov red oko točke funkciju zadanu s

$\displaystyle f(x)=\frac{1}{\sqrt[3]{(5x-4)^7}}$

i odredite područje konvergencije dobivenog reda.

Rješenje. Iz

$\displaystyle f{'}(x)$	$\displaystyle =-\left(\frac{7}{3}\right) (5x-4)^{-\frac{10}{3}} \cdot 5,$
$\displaystyle f{''}(x)$	$\displaystyle =\left(-\frac{7}{3}\right) \left(-\frac{10}{3}\right) (5x-4)^{-\frac{13}{3}}\cdot 5^2,$
$\displaystyle f{'''}(x)$	$\displaystyle =\left(-\frac{7}{3}\right) \left(-\frac{10}{3}\right) \left(-\frac{13}{3}\right) (5x-4)^{-\frac{16}{3}}\cdot 5^3,$

slijedi

$\displaystyle f^{(n)}(x)$	$\displaystyle =\left(-\frac{7}{3}\right) \left(-\frac{10}{3}\right)\cdots \left(-\frac{3n+4}{3}\right) (5x-4)^{-\frac{3n+7}{3}} \cdot 5^n$
	$\displaystyle =(-1)^n \frac{7\cdot 10\cdots(3n+4)}{3^n}\cdot (5x-4)^{-\frac{3n+7}{3}}\cdot 5^n,$

pa je

$\displaystyle f^{(n)}(1)=(-1)^n\,\, 7\cdot 10\cdots(3n+4)\cdot\left(\frac{5}{3}\right)^n.$

Prema

[M1, teorem 6.18], Taylorovog razvoj funkcije

oko točke

glasi

$\displaystyle f(x) = f(1)+\sum_{n=1}^\infty \frac{f^{(n)}(1)}{n!} (x-1)^n,$

odakle uvrštavanjem

i $f^{(n)}(1)$ slijedi

$\displaystyle f(x) = 1+\sum_{n=1}^\infty (-1)^n \frac{7\cdot 10\cdots(3n+4)}{n!} \cdot\left(\frac{5}{3}\right)^n (x-1)^n,$

Budući je

$\displaystyle \lim_{n\to \infty} \left\vert\frac{a_{n+1}}{a_n}\right\vert$	$\displaystyle = \lim_{n\to \infty} \left\vert\frac{(-1)^{n+1} \displaystyle\fra... ...{7\cdot 10\cdots(3n+4)}{n!} \cdot\left(\frac{5}{3}\right)^n (x-1)^n}\right\vert$
	$\displaystyle =\lim_{n\to \infty} \frac{3n+7}{n+1}\cdot\frac{5}{3}\cdot\vert x-1\vert=5\vert x-1\vert,$

D'Alembertov kriterij povlači da red konvergira za $x\in \left\langle\frac{4}{5},\frac{6}{5} \right\rangle$ .