Question

Explain the proof :

Answer 1

Help me understand the corollary

Answer 2

@satellite73

Answer 3

consider if we have \(n+1\) values \(a_0,\dots,a_n\). if the functions match at all points, we have: $$P(a_0)=Q(a_0)\\P(a_1)=Q(a_1)\\\dots\\P(a_n)=Q(a_n)$$
for a particular \(P(a_i)=Q(a_i)\), we can equivalently write $$P(a_i)-Q(a_i)=0$$
define the polynomial function \(R(x)=P(x)-Q(x)\), so the earlier list of equations tells us:$$R(a_0)=0\\R(a_1)=0\\\dots\\R(a_n)=0$$
so it follows that \(R\) has \(n+1\) zeros. but since \(R(x)=P(x)-Q(x)\), it follows that \(R\) is at most dimension \(n\)

Answer 4

at most degree* n

Answer 5

so it follows that we must have that \(R(x)=0\) everywhere (in other words, in the representation \(R(x)=c(x-x_1)\dots(x-x_n)\), we must have \(c=0\))

Answer 6

and if \(R(x)=0\), well \(R(x)=P(x)-Q(x)\) so that means \(P(x)-Q(x)=0\implies P(x)=Q(x)\)

Answer 7

Okay for the $$P(a_0)=Q(a_0)$$ what does it basically explain the value of the function being equal or the term $$a_0$$

Answer 8

it means the two different polynomials \(P,Q\) match at \(x=a_0\)

Answer 9

in the first part why is it taken that 6|K please explain :)

Answer 10

I want to understand the approach in the question. Please help

Answer 11

\(\epsilon_{1,2}\) are roots of unity -- recall that: $$\cos(\pm\pi/3)=\pm\frac12\\\sin(\pm\pi/3)=\frac{\sqrt3}2\\$$

Answer 12

so we have: $$\epsilon_1=\cos(\pi/3)+i\sin(\pi/3)$$now by de Moivre's theorem we see $$\epsilon_1^3=\cos(3\cdot\pi/3)+i\sin(3\cdot \pi/3)=\cos(\pi)+i\sin(\pi)=-1$$ and so \(\epsilon_1^6=(\epsilon_1^3)^2=(-1)^2=1\)

Answer 13

similarly, $$\epsilon_2=\cos(-\pi/3)+i\sin(-\pi/3)\\\epsilon_2^3=\cos(-\pi)+i\sin(-\pi)=-1\\\epsilon_2^6=(\epsilon_2^3)^2=(-1)^2=1$$

Answer 14

equivalently, you could've just done de Moivre's theorem directly for the sixth power, so: $$\epsilon_1^6=\cos(6\cdot \pi/3)+i\sin(6\cdot \pi/3)=\cos(2\pi)+i\sin(2\pi)=1\\\epsilon_2^6=\cos(6\cdot-\pi/3)+i\sin(6\cdot-\pi/3)=\cos(-2\pi)+i\sin(-2\pi)=1$$

Answer 15

and yes @ParthKohli I meant to say: $$\cos\left(\pm\frac{\pi}3\right)=\frac12\\\sin\left(\pm\frac\pi3\right)=\pm\frac{\sqrt3}2$$

Answer 16

anyways, they factor \(x^2-x+1=(x-\epsilon_1)(x-\epsilon_2)\) because that way we can check if \(x^2-x+1\) divides \(x^n-x+1\) by merely checking if the individual factors \(x-\epsilon_1,x-\epsilon_2\) divide \(x^n-x+1\). by the factor theorem, we know that \(x-\epsilon\) is a factor of \(f(x)=x^n-x+1\) if and only if \(f(\epsilon)=0\); in other words, $$\epsilon^n-\epsilon+1=0$$ so in this case we're checking both factors \(x-\epsilon_1,x-\epsilon_2\) so we have: $$\epsilon_1^n-\epsilon_1+1=0\\\epsilon_2^n-\epsilon_2+1=0$$

Answer 17

anyways, since \(\epsilon_1,\epsilon_2\) are roots of \(x^2-x+1=0\), consider: $$x^2-x+1=0\\x(x-1)=-1\\x(1-x)=1$$so it follows that \(x(1-x)=1\) or \(1-x=\frac1x=x^{-1}\) for \(x\in\{\epsilon_1,\epsilon_2\}\) -- in other words, $$1-\epsilon_1=1/\epsilon_1\\1-\epsilon_2=1/\epsilon_2$$

Answer 18

also i had the \(n\)-th degree polynomial wrong, I meant to write \(f(x)=x^n+x-1\) so $$\epsilon_1^n+\epsilon_1-1=0\\\epsilon_2^n+\epsilon_2-1=0$$

Answer 19

in which case we can rewrite the equations like: $$\epsilon_1^n=1-\epsilon_1\\\implies \epsilon_1^n=1/\epsilon_1\\\epsilon_2^n=1-\epsilon_2\\\implies \epsilon_2^n=1/\epsilon_2$$

Answer 20

in other words, \(\epsilon_1^{n+1}=1\) and \(\epsilon_2^{n+1}=1\) by multiplying both sides by \(\epsilon_1,\epsilon_2\) respectively

Answer 21

and yet we saw that: $$\epsilon_1^6=\epsilon_2^6=1$$so taking both sides to any integer power \(i\) we have $$\epsilon_1^{6i}=\epsilon_2^{6i}=1$$ so it follows that we must have \(n+1=6i\implies n=6i-1\)

Answer 22

thank u :)

Answer 23

the reason I showed earlier that \(\epsilon_1^3=\epsilon_2^3=-1\) earlier is to show that \(k=6\) is the *smallest* positive integer such that \(\epsilon_1^k=\epsilon_2^k=1\), otherwise we could've had other solutions we missed

Answer 24

oops, I said show and earlier too many times

Answer 25

so that \(i\) in \(6i-1\) has nothing to do with the imaginary unit \(i\) haha!