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Robert Israel <israel@xxxxxxxxxxx> wrote:
>eugene <jane1806@xxxxxxx> wrote:
>>
>> Prove that polynomial p(x) = x^100 + x^98 + a_3 x^97 +...+ a_n
>> cannot have all real zeros.
>
> Use Descartes' Rule of Signs. The number of changes of sign in the
> coefficients of p(x) plus the number of changes of sign in the
> coefficients of p(-x) is at most 98. In fact, if a
> polynomial P(x) = sum_j c_j x^j has all real zeros and one of its
> coefficients c_j = 0 (where there is some nonzero coefficient
> c_k with k < j), then c_{j-1} c_{j+1} < 0.
It's more elementary via root symmetric functions:
ri roots of X^n + a X^(n-1) + b X^(n-2) + ...
-> -a = sum ri
b = sum ri rj, i<j
-> sum ri^2 = a^2 - 2b
a=0 & ri real -> 0 <= sum ri^2 = -2b -> b <= 0
so, indeed, c{n-1} = 0 -> c_{n} c{n-2} = b <= 0
Polya often posed practice problems similar to this.
--Bill Dubuque
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