if-a-1-a-2-a-14-are-roots-of-the-polynomial-p-x-x-14-x-8-2x-1-calculate-i-1-14-1-a-i-1-2-

Question Number 28544 by abdo imad last updated on 26/Jan/18

i f a_{1}, a_{2}, \dots a_{14} a r e r o o t s o f t h e p o l y n o m i a l

p (x) = x^{14} + x^{8} 2 x + 1 c a l c u l a t e \sum_{i = 1}^{14} \frac{1}{{(a_{i} - 1)}^{2}} .

Commented by abdo imad last updated on 26/Jan/18

p (x) = x^{14} + x^{8} + 2 x + 1 .

Commented by abdo imad last updated on 28/Jan/18

w e k n o w t h a t \frac{p^{^{'}} (x)}{p (x)} = \sum_{i = 1}^{14} \frac{1}{x - x_{i}}

\frac{p^{^{″}} p (x) - p^{' 2} (x)}{p^{2} (x)} = - \sum_{i = 1}^{14} \frac{1}{{(x - x_{i})}^{2}} b u t w e h a v e

p^{^{'}} (x) = 14 x^{13} + 8 x^{7} + 2 a n d p^{(2)} (x) = 14 \times 13 x^{12} + 56 x^{6}

\sum_{i = 2}^{14} \frac{1}{(x - a_{i}) 2} = - \frac{(14 \times 13 x^{12} + 56 x^{6}) (x^{14} + x^{8} + 2 x + 1) - {(14 x^{13} + 8 x^{7} + 2)}^{2}}{{(x^{14} + x^{8} + 2 x + 1)}^{2}}

a n d f o r x = 1 ? w e f i n d

\sum_{i = 1}^{14} \frac{1}{{(a_{i} - 1)}^{2}} = - \frac{(14 \times 13 + 56) \times 5 - (24)}{25} \dots