Question

Cum se rezolva exercitiul din imagine?

Answer 1

$f:\mathbb{R} \to \mathbb{R}, f(x)=x^n-2x+2,~~n\geq 3 \text{ cu radacinile:} \\ x_1,x_2,...~,x_n.\\ \\ S = \dfrac{1}{(1-x_{1})^2} + \dfrac{1}{(1-x_2)^2} +...+\dfrac{1}{(1-x_n)^2} \\ \\ \text{f(x) se mai poate scrie in forma generala ca:}\\ f(x) = (x-x_1)(x-x_2)(x-x_3)\cdot ...\cdot (x-x_{n})\\ \\ \text{Daca diferentiem si aplicam regula produsului obtinem:}\\ f'(x) = \sum\limits_{k=1}^{n}\dfrac{f(x)}{x-x_{k}}$

$\text{Iar a doua derivata este:}\\ \\ f''(x) = \sum\limits_{k=1}^n\sum\limits_{l\neq k}\dfrac{f(x)}{(x-x_k)(x-x_l)}\\ \\ \text{Formulele astea se pastreaza in general,}\\ \text{pana acum nu am avut nevoie de forma specifica a polinomului }\\ f(x) = x^n-2x+2.\\ \\ \text{Dar de acum punem x = 1 in formulele de sus:}$

$\dfrac{1}{(1-x_1)^2} + \dfrac{1}{(1-x_2)^2} +...+\dfrac{1}{(1-x_n)^2} = \\ \\ = \dfrac{\left(\sum\limits_{k=1}^n \dfrac{f(1)}{1-x_k}\right)^2 - f(1)\cdot \sum\limits_{k=1}^n\sum\limits_{l\neq k}\dfrac{f(1)}{(1-x_k)(1-x_l)}}{f^2(1)}$

$\Rightarrow S = \dfrac{\big[f'(1)\big]^2 - f(1)\cdot f''(1)}{f^2(1)} \\ \\ \\ \text{In cazul nostru: }\\\\f(x)=x^n-2x+2\\f'(x) = nx^{n-1}-2\\ f''(x) = n(n-1)x^{n-2}\\ \\ f'(1) = n-2 \text{ si } f''(1) = n(n-1) \\ \\ \Rightarrow S = (n-2)^2 - n(n-1) = n^2-4n+4-n^2+n = 4-3n \\ \\ \Rightarrow \text{raspuns corect: }~~ b) \quad 4-3n$