Question

Salut, aveti o idee la problema numarul 15 ???

Answer 1

$\displaystyle 1.~\underline{Parte~teoretica} \\ \\Ideea~de~baza~a~acestei~probleme~este~urmatoarea~proprietate: \\ \\ Daca~x_1,x_2,...,x_n~sunt~(toate)~radacinile~unui~polinom~P,~atunci \\ \\ \sum\limits_{k=1}^n \frac{1}{x_k}=- \frac{P'(0)}{P(0)}. ~(in~ipoteza~x_k \neq 0,~k= \overline{1,n}) \\ \\ O~sa~demonstrez~acest~rezultat~pentru~polinoamele~de~gradul~2~si~3, \\ \\ apoi~pentru~n.$

$\displaystyle \bullet~P(x)=ax^2+bx+c \\ \\ \frac{1}{x_1}+ \frac{1}{x_2}= \frac{x_1+x_2}{x_1x_2}= \frac{- \frac{b}{a}}{\frac{c}{a}}=- \frac{b}{c}= -\frac{P'(0)}{P(0)}. \\ \\ \bullet~ P(x)=ax^3+bx^2+cx+d \\ \\ \frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}= \frac{x_2x_3+x_3x_1+x_1x_2}{x_1x_2x_3}= \frac{\frac{c}{a}}{-\frac{d}{a}}= -\frac{c}{d}=- \frac{P'(0)}{P(0)}.$

$\displaystyle Am~inceput~cu~aceste~cazuri~particulare~pentru~a~fixa~un~"pattern". \\ \\  \bullet P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0. \\ \\ Dupa~amplificare~suma~devine~un~raport~de~polinoame~simetrice~si \\ \\ omogene~pe~care~le~putem~calcula~cu~relatiile~lui~Viete. \\ \\ Numitorul~este~produsul~radacinilor,~egal~cu~(-1)^n \frac{a_0}{a_n}. \\ \\ Numaratorul~va~corespunde~atunci~coeficientului~a_1,~fiind~egal~cu~ \\ \\ (-1)^{n-1}\frac{a_1}{a_n}.$

$\displaystyle Atunci~suma~este~\frac{(-1)^{n-1} \frac{a_1}{a_n}}{(-1)^n \frac{a_0}{a_n}}=- \frac{a_1}{a_0}=- \frac{P'(0)}{P(0)}.$

$\displaystyle 2. ~\underline{Rezolvarea~problemei} \\ \\ Avem~ \sum\limits_{k=1}^{20} \frac{1}{x_k(1-x_k)}= \sum\limits_{k=1}^n \left( \frac{1}{x_k} + \frac{1}{1-x_k } \right) = \sum\limits_{k=1}^n \frac{1}{x_k} + \sum\limits_{k=1}^n \frac{1}{1-x_k }. \\ \\ \sum\limits_{k=1}^n \frac{1}{x_k}=- \frac{P'(0)}{P(0)} \\ \\ Pentru~calculul~celei~de-a~doua~sume,~vom~forma~polinomul~al~carui \\ \\ radacini~sunt~y_k=1-x_k.~Acesta~este~P(1-y). \\ \\ (Pentru~ca~y=1-x~implica~x=1-y.) \\ \\ Fie~Q(x)=P(1-x).$

$\displaystyle  \sum\limits_{k=1}^n \frac{1}{1-x_k }= -\frac{Q'(0)}{Q(0)}. \\ \\ Mai~trebuie~sa~calculam~ -\frac{P'(0)}{P(0)}- \frac{Q'(0)}{Q(0)},~iar~acesta~va~fi~raspunsul. \\ \\ P'(0)=0. \\ \\ Q(0)=P(1)=5. \\ \\ Q'(x)= \Big( P(1-x) \Big)' =-P'(1-x) \\ \\ Q'(0)=-35 \\ \\ -\frac{P'(0)}{P(0)}- \frac{Q'(0)}{Q(0)}=7.$