Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

How can I show that $a_0x^n+...+a_n$ and $a_nx^n+...a_0$ have the same discriminant?

You can use two different definition of the discriminant of the polynomial $f(x)=a_nx^n+...a_0$.

The first is $$D(f)=a_n^{2n-2}\prod_{i<k}(\alpha_i-\alpha_j),$$ where $\{\alpha_i\}_{i=1,...n}$ are the roots of the polynomial.

The second is $$D(f)=(-1)^{\frac{n(n-a)}{2}}\frac{1}{a_n}R(f,f')$$ where $R(f,f')$ is the resultant of $f$ and $f'$.

share|cite|improve this question
Do you want to show it with the two definitions or only with one? – Davide Giraudo Oct 15 '12 at 21:35
Is $f'$ the formal derivative of $f$? – Alexander Gruber Oct 15 '12 at 21:37
either can be used... and $f'$ is the usual derivate – Elmo goya Oct 15 '12 at 21:44
up vote 3 down vote accepted

The roots of $a_nx^n+\cdots +a_0$ are the reciprocals of the roots of $a_0x^n+\cdots +a_n$. Therefore $\prod (\alpha_i-\alpha_j)$ is replaced with $\prod (\frac1{\alpha_i}-\frac1{\alpha_j})=\prod \frac{\alpha_j-\alpha_i}{\alpha_i\alpha_j}$. The numerators yield the original discriminant (check that there is no sign change!). The denominators produce $(\prod\alpha_i)^{2(n-1)}=(\frac{a_0}{a_n})^{2n-2}$, so that everything sorts out.

share|cite|improve this answer
Well, but... I can't see clearly why $\prod \alpha_i^{2(n-1)}=(\frac{a_0}{a_n})^{2n-2}$ @HagenvonEitzen – Elmo goya Oct 15 '12 at 22:08
Camilo, are you familiar with $\prod\alpha_i=a_0/a_n$ in this context? Then raise to the power $2n-2$. – Gerry Myerson Oct 15 '12 at 22:57

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.