Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Im trying to show that the binary form $x^{n−1}+x^{n−2}yα+x^{n−3}y^2α^2+...+y^{n−1}α^{n−1}$ is bounded below by $ c*y^{n−1}$ where c is some explicit constant.

For the case n=3 this is fairly easy, using the discrimiant of the binary quadratic form, $c=3/4\alpha^2 $

Yes this equation is from the factorization of $x^n-(\alpha y)^n$ and it is of degree n-1

share|cite|improve this question
How is it that the power of $y$ is always one less than that of $\alpha$ on the left of the $\dots$, but on the right we have $y^{n-1}\alpha^{n-1}$? Is it true that $x^{n-1}$ does not have an $\alpha^1$, nor $x^{n-n}$ an $\alpha^n$ coefficient yet $x^{n-k}$ has an $\alpha^k$ for every $k$ in between? This is degree $n-1$ isn't it? Also posted MO over 2 hours ago. I think you should clarify so your question can be understood. – anon Apr 30 '12 at 23:41
Ha - I asked the same question about the powers of $\alpha$ at the MO posting. – Gerry Myerson May 1 '12 at 0:07
Based on the last two lines I guess that the form should be $(x^n-(\alpha y)^n)/(x-\alpha y)$ so I edited the initial presentation. – Zander May 1 '12 at 3:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.