Question

Rădăcinile ecuaţiei polinomiale P(s) = 0 sunt: s1=1; s2=2; s3=3; s4=4
Se cere:
a) Exprimaţi forma polinomială P(s) pe baza rădăcinilor precizate prin enunţ;
b) Descompuneţi expresia F(s)=1/P(s) într-o sumă de patru rapoarte;

Answer 1

$P(s) = 0\\ s_1 = 1, s_2 = 2, s_3 = 3, s_4 = 4 \\ \\ a)\\ \\P(s) = s^4-(s_1+s_2+s_3+s_4)s^3+(s_1s_2+s_1s_3+s_1s_4+s_2s_3+s_2s_4+s_3s_4)s^2- \\ -(s_1s_2s_3+s_1s_2s_4+s_1s_3s_4+s_2s_3s_4)s+s_1s_2s_3s_4 \\ \\ \dfrac{1}{a}P(s) = s^4 - (1+2+3+4)s^3+(1\cdot 2+1\cdot 3+1\cdot 4+2\cdot 3+2\cdot 4+3\cdot 4)s^2-\\ -(1\cdot 2\cdot 3+1\cdot 2\cdot 4+1\cdot 3\cdot 4+2\cdot 3\cdot 4)s+1\cdot 2\cdot 3\cdot 4$

$\dfrac{1}{a}P(s) = s^4-10s^2+(2+3+4+6+8+12)s^2- (6+8+12+24)s+24 \\ \\ P(s) = a(s^4-10s^3+35s^2-50s+24), \quad a\in \mathbb{R}\\ \\b)\\ \\ F(s) = \dfrac{1}{P(s)} = \dfrac{1}{a(s-1)(s-2)(s-3)(s-4)} \\ \\ aF(s) = \dfrac{1}{(s-1)(s-2)(s-3)(s-4)} =\dfrac{1}{(s-1)(s-4)(s-2)(s-3)} = \\ \\ = \dfrac{1}{(s^2-5s+4)(s^2-5s+6)} \\ \\ s^2-5s+4 = t \\ \\ aF(s) =\dfrac{1}{t(t+2)} = \dfrac{1}{2}\cdot \dfrac{t+2-t}{t(t+2)} = \dfrac{1}{2}\cdot \Big(\dfrac{1}{t}- \dfrac{1}{t+2}\Big)$

$2aF(s) = \dfrac{1}{t} - \dfrac{1}{t+2} = \dfrac{1}{s^2-5s+4}- \dfrac{1}{s^2-5s+6} = \\ \\ = \dfrac{1}{(s-1)(s-4)}- \dfrac{1}{(s-2)(s-3)} = \dfrac{1}{3}\cdot \dfrac{s-1-(s-4)}{(s-1)(s-4)}- \dfrac{s-2-(s-3)}{(s-2)(s-3)} = \\ \\ =\dfrac{1}{3}\cdot \Big(\dfrac{1}{s-4}-\dfrac{1}{s-1}\Big) - \Big(\dfrac{1}{s-3}- \dfrac{1}{s-2}\Big) \\ \\ \Rightarrow aF(s) = \dfrac{1}{6}\cdot \Big(\dfrac{1}{s-4}-\dfrac{1}{s-1}\Big) - \dfrac{1}{2}\cdot \Big(\dfrac{1}{s-3}- \dfrac{1}{s-2}\Big)$

$aF(s) = \dfrac{-1}{6(s-1)} +\dfrac{1}{2(s-2)}+\dfrac{-1}{2(s-3)}+\dfrac{1}{6(s-4)} \\ \\ \Rightarrow F(s) = \dfrac{-1}{6a(s-1)} +\dfrac{1}{2a(s-2)}+\dfrac{-1}{2a(s-3)}+\dfrac{1}{6a(s-4)},\quad a\in \mathbb{R}$