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Question :

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My approach :

Now as I had to obtain a remainder of $\frac{(x-1)^{2017}}{x^2 - x +1}$ So, I could write this as $\frac{(x-1)^{2017}}{(x - 1)^2 + x}$

now I substitute $t = (x-1)$, so $\frac{(x-1)^{2017}}{(x - 1)^2 + x}$ could be written as $\frac{t^{2017}}{t^{2} + t + 1}$

From here we can easily obtain the remainder by plain division itself, so $P^{1}(x) = -(t+1)t^{2015} = -x(x-1)^{2015}$

Since we need $P^{2017}(2016)$, using the above obtained $P^{1}(x)$, I first obtained $P^{1}(2016) = -2016(2015)^{2015}$.

Similarly $P^{2}(x) = -P^{1}(x)(P^{1}(x) - 1)^{2015}$ by question definition, but there's no further way to simplify it, and also $P^{n}(x)$ gets bigger and bigger as $n$ increases.

So I genuinely feel this approach is wrong way, so could you help me out upon this question ?

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  • $\begingroup$ But.. Doesn't it satisfy $a = bq + r$ ?? $\endgroup$ – Arnav Das Nov 30 '16 at 16:49
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Hint:

You mean remainder right, not quotient.

$t^3\equiv1\pmod{t^2+t+1}$

As $3|2016, t^{2016}\equiv1$

$\implies t^{2017}\equiv t$

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$p(x) = q(x) d(x) + P^1(x)$

$P^{1}(x)$ is at most a 1 degree polynomial

If $\omega$ is a root of $d(x).$ $p(\omega) = P^1(\omega)$

From this we find $P^{1}(\omega_1), P^{1}(\omega_2)$ and the line between them.

$\omega_1, \omega_2 = \frac 12 + i \frac {\sqrt 3}{2}, \frac 12 - i \frac {\sqrt 3}{2}$

$p(\omega_1) = $$(-\frac 12 + i \frac {\sqrt 3}{2})^{2017}\\ e^{\frac {2017\cdot2 \pi}{3}}\\ -\frac 12 + i \frac {\sqrt 3}{2}\\ \omega_1 - 1$

$p(\omega_2) = \omega_2 - 1$

$P^1(x) = x-1$

$P^2(x) =$$ P^1(P^1(x))\\ x-2$

$P^{k}(x) =x-k$

$P^{2017}(2016)=-1$

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