2
$\begingroup$

Given the quartic polynomial:

$$q(x) = 16 x^4 - 40 a x^3 + (15 a^2 + 24 b) x^2 - 18 a b x + 3 b^2,$$ where $a < 0$ and $b \in \mathbb{R}$. By making use of Mathematica, it can be verified that the above quartic has four real roots $x_1 \leq x_2 \leq x_3 \leq x_4$ if and only if $$0 < b < \frac{a^2}{32}\left(27 - \sqrt{111+46 \sqrt{6}} - 3 \sqrt{6}\right).$$

I wish to prove that the closer $b$ gets to $0$, the closer $x_4$ will be to zero or in symbols: $b \to 0$ implies $x_4 \to 0$. In the extreme case $b = 0$ this is indeed true. Doing some numerical tests the statement seems true. However, I wish to prove it mathematically. One way could be of trying to write down an analytic expression for $x_4$. However, I wish to prove it in a more elegant way. Any hints would be helpful.

Thanks in advance.

  • 1
    Hint : Show that the roots depend on $b$ continously. – 2017-02-02
  • 0
    Thanks for the quick response. If I recall correctly it is true that the roots of a polynomial are continuous functions of its coefficients if one works over the complex numbers. Are you referring to this? Furthermore how would this help me to prove the fact that only the fourth root goes to 0 if b does so? – 2017-02-02
  • 0
    Can we assume, that $b$ remains positive ? In this case (considering $a<0$) it is easy to see that $f(x)$ has no positive roots. In this case, $x4\rightarrow 0$ follows immediately if we have proven that $x_4$ depends continously on $b$. – 2017-02-02
  • 0
    Yes, we can assume that. To prove that $x_4$ depends continuously on $b$ I think I should have some kind of analytical expression for $x_4$. – 2017-02-02
  • 0
    Seems to be a promising approach. And it would be the last step for the proof ... – 2017-02-02

0 Answers 0