Let $p(t)=(t^2-a_1^2)\ldots(t^2-a_n^2)$ be an even polynomial with distinct real non-zero roots. Can the roots of its derivative $p'(t)$ be expressed nicely (e.g. as rational symmetric functions) in terms of the roots of $p(t)$? How?
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Rational symmetric functions of the roots of the derivative can be expressed as rational symmetric functions of the roots of the polynomial, because they are rational functions of the coefficients of the derivative, thus rational functions of the coefficients.
But the individual roots cannot be:
$\frac{d}{dt}(t^2-a_1^2)(t^2-a_2^2) = 4t^3 - 2 (a_1^2 + a_2^2)t$, so the nonzero roots are $\pm \sqrt{\frac{a_1^2 + a_2^2}{2} }$.
For higher degrees, we will get more complicated algebraic functions. For instance:
$\frac{d}{dt}(t^2-a_1^2)(t^2-a_2^2)(t^2-a_3^2) = 6 t^5 - 4 (a_1^2+a_2^2+a_3^2)t^3 + 2 (a_1^2a_2^2 + a_2^2a_3^2 + a_1^2a_3^2)t $ so the nonzero roots are:
$$\pm \sqrt{ \frac{a_1^2+a_2^2+a_3^2\pm \sqrt{ \left(a_1^2+a_2^2+a_3^2\right)^2 - 3\left(a_1^2a_2^2+a_1^2a_3^2+a_2^2a_3^2\right)}}{3} }$$
and it just gets worse as the degree increases.
answered Oct 16, 2013 at 4:00
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$\begingroup$ What happens when $n\ge6$? We know there's no nice formula in terms of the coefficients of the polynomial but in this case we have that the coefficients are those symmetric functions evaluated at the $a_i$'s which we know, maybe that extra information could help to produce an explicit formula? Or perhaps it's possible to prove there will be no such nice formula (radicals and symmetric functions on the $a_i$'s) similar to the last one above? $\endgroup$ Commented Oct 18, 2013 at 1:59
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$\begingroup$ The fact that the coefficients are symmetric functions evaluated at the $a_i$s is not very helpful - for a generic set of symmetric functions, you will not be able to do any better than the case of completely general functions. Clearly there will be no formula in terms of the radicals of the coefficients, like these ones, for $n\geq 6$, because then for appropriate coefficients it would work for every polynomial. $\endgroup$ Commented Oct 18, 2013 at 5:55
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$\begingroup$ It seems really unlikely that going to all the symmetric functions, instead of just symmetric functions of squares, and getting the last symmetric function, would help at all. There's just no relation of that to the roots of the polynomial. I don't currently have a rigorous proof of that - would you like to see one? $\endgroup$ Commented Oct 18, 2013 at 5:57