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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 $f'$.

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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

1 Answer 1

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.

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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

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