A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

praxer

  • one year ago

Explain the proof :

  • This Question is Closed
  1. praxer
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Help me understand the corollary

    1 Attachment
  2. praxer
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @satellite73

  3. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \(n\)

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    at most degree* n

  5. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \(c=0\))

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  7. praxer
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  9. praxer
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

    1 Attachment
  10. praxer
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  12. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \(\epsilon_1^6=(\epsilon_1^3)^2=(-1)^2=1\)

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  14. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  16. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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: $$\epsilon_1^n-\epsilon_1+1=0\\\epsilon_2^n-\epsilon_2+1=0$$

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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, $$1-\epsilon_1=1/\epsilon_1\\1-\epsilon_2=1/\epsilon_2$$

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \(n+1=6i\implies n=6i-1\)

  22. praxer
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thank u :)

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oops, I said show and earlier too many times

  25. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  26. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.