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praxer
 one year ago
Explain the proof :
praxer
 one year ago
Explain the proof :

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praxer
 one year ago
Best ResponseYou've already chosen the best response.0Help me understand the corollary

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider 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 \(n\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so it follows that we must have that \(R(x)=0\) everywhere (in other words, in the representation \(R(x)=c(xx_1)\dots(xx_n)\), we must have \(c=0\))

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and if \(R(x)=0\), well \(R(x)=P(x)Q(x)\) so that means \(P(x)Q(x)=0\implies P(x)=Q(x)\)

praxer
 one year ago
Best ResponseYou've already chosen the best response.0Okay for the $$P(a_0)=Q(a_0)$$ what does it basically explain the value of the function being equal or the term $$a_0$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it means the two different polynomials \(P,Q\) match at \(x=a_0\)

praxer
 one year ago
Best ResponseYou've already chosen the best response.0in the first part why is it taken that 6K please explain :)

praxer
 one year ago
Best ResponseYou've already chosen the best response.0I want to understand the approach in the question. Please help

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\epsilon_{1,2}\) are roots of unity  recall that: $$\cos(\pm\pi/3)=\pm\frac12\\\sin(\pm\pi/3)=\frac{\sqrt3}2\\$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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 \(\epsilon_1^6=(\epsilon_1^3)^2=(1)^2=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0similarly, $$\epsilon_2=\cos(\pi/3)+i\sin(\pi/3)\\\epsilon_2^3=\cos(\pi)+i\sin(\pi)=1\\\epsilon_2^6=(\epsilon_2^3)^2=(1)^2=1$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0equivalently, you could've just done de Moivre's theorem directly for the sixth power, so: $$\epsilon_1^6=\cos(6\cdot \pi/3)+i\sin(6\cdot \pi/3)=\cos(2\pi)+i\sin(2\pi)=1\\\epsilon_2^6=\cos(6\cdot\pi/3)+i\sin(6\cdot\pi/3)=\cos(2\pi)+i\sin(2\pi)=1$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and yes @ParthKohli I meant to say: $$\cos\left(\pm\frac{\pi}3\right)=\frac12\\\sin\left(\pm\frac\pi3\right)=\pm\frac{\sqrt3}2$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anyways, they factor \(x^2x+1=(x\epsilon_1)(x\epsilon_2)\) because that way we can check if \(x^2x+1\) divides \(x^nx+1\) by merely checking if the individual factors \(x\epsilon_1,x\epsilon_2\) divide \(x^nx+1\). by the factor theorem, we know that \(x\epsilon\) is a factor of \(f(x)=x^nx+1\) 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: $$\epsilon_1^n\epsilon_1+1=0\\\epsilon_2^n\epsilon_2+1=0$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0anyways, since \(\epsilon_1,\epsilon_2\) are roots of \(x^2x+1=0\), consider: $$x^2x+1=0\\x(x1)=1\\x(1x)=1$$so it follows that \(x(1x)=1\) or \(1x=\frac1x=x^{1}\) for \(x\in\{\epsilon_1,\epsilon_2\}\)  in other words, $$1\epsilon_1=1/\epsilon_1\\1\epsilon_2=1/\epsilon_2$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also i had the \(n\)th degree polynomial wrong, I meant to write \(f(x)=x^n+x1\) so $$\epsilon_1^n+\epsilon_11=0\\\epsilon_2^n+\epsilon_21=0$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in which case we can rewrite the equations like: $$\epsilon_1^n=1\epsilon_1\\\implies \epsilon_1^n=1/\epsilon_1\\\epsilon_2^n=1\epsilon_2\\\implies \epsilon_2^n=1/\epsilon_2$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in other words, \(\epsilon_1^{n+1}=1\) and \(\epsilon_2^{n+1}=1\) by multiplying both sides by \(\epsilon_1,\epsilon_2\) respectively

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and 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 \(n+1=6i\implies n=6i1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, I said show and earlier too many times

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so that \(i\) in \(6i1\) has nothing to do with the imaginary unit \(i\) haha!
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