Given the following polynomial: $$P(x)=(x^2+x+1)^{100}$$ How do I find : $$\sum_{k=1}^{200} \frac{1}{1+x_k} $$ Is there a general solution for this type of problem cause I saw they tend to ask the same thing for $\sum_{k=1}^{200} \frac{1}{x_k}$? Also how do I find the coefficient of $a_1$ and the remainder for $$P(x)/(x^2+x)$$ (/=divided) (I found the coef of a1 is 100 adn the remainder is 1 but im not sure)
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Does $x_k$ denote the roots of $P$? – florence May 19 '17 at 14:55
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yes,xk are the roots of p which i found to be complex roots – Lola May 19 '17 at 14:56
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Note that the roots of $x^2+x+1$ are $$\zeta_3 =e^{2\pi i/3} = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$$ and $$\zeta_3^2 =e^{4\pi i/3} = -\frac{1}{2}-i\frac{\sqrt{3}}{2}$$ Both of these are roots of $P(x)$ with multiplicity $100$. Therefore, $$\sum_{k=1}^{200}\frac{1}{1+x_k} = 100\left( \frac{1}{1+\zeta_3}+\frac{1}{1+\zeta_3^2}\right)$$ $$ = 100\left(\frac{1+\zeta_3+1+\zeta_3^2}{1+\zeta_3+\zeta_3^2+\zeta_3^3} \right) = 100$$
As for the polynomial division, suppose $$P(x) = g(x)(x^2+x)+r(x)$$ where $r$ is linear. We have $P(0) = 1$, which implies that $r(0) = 1$. Further, $P(-1) = 1$, and so $r(-1)=1$. Consequently, $r(x)=1$ for all $x$, so $a_1=0$.
florence
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Is there a general solution for this type of problem
Consider for example $P(x)=(x^2+x+1)^n\,$.
$\sum_{k=1}^{2n} \frac{1}{x_k}$
Let $y_k = \frac{1}{x_k} \ne 0\,$ then $P(\frac{1}{y_k})=P(x_k)=0\,$, so $y_k$ are the roots of: $$Q(y)=y^{2n} P\left(\frac{1}{y}\right)=y^{2n}\left(\frac{1}{y^2}+\frac{1}{y}+1\right)^n=(1+y+y^2)^n=y^{2n}+\binom{n}{0,1,n-1}y^{2n-1}+\cdots\,$$
Then by Vieta's relations $\sum_{k=1}^{2n} y_k=-\binom{n}{0,1,n-1}=-n\,$.
$\sum_{k=1}^{2n} \frac{1}{1+x_k}$
Let $z_k=x_k+1\,$, then $P(z_k-1)=P(x_k)=0\,$, so $z_k$ are the roots of:
$$\require{cancel} R(z) = P(z-1)=\left((z-1)^2+(z-\bcancel{1})+\bcancel{1}\right)^n=(z^2-z+1)^n $$
Then, by an argument entirely similar to the above, it follows that $\sum_{k=1}^{2n} \frac{1}{z_k}=\binom{n}{0,1,n-1}=n\,$.
find the coefficient of $a_1$ and remainder for $P(x)/(x^2+x)$
$$ P(x)=\big((x^2+x)+1\big)^n=(x^2+x)^n+\binom{n}{1}(x^2+x)^{n-1}+\cdots+\binom{n}{n-1}(x^2+x) + 1 $$
All terms are multiples of $x^2+x$ except the very last one, so the remainder of the division is $1\,$ which is a constant polynomial so $a_1=0\,$.
dxiv
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