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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 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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    either can be used... and $f'$ is the usual derivate2012-10-15

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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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    Camilo, are you familiar with $\prod\alpha_i=a_0/a_n$ in this context? Then raise to the power $2n-2$.2012-10-15