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Consider the following polynomial: $p(x)=x^{3}-(k-1)x^{2}-(2k-1)x+(k-1)^{2}$, where $k \geq 5$ is a fixed parameter. I am trying to find a strong lower bound on the largest root $x_{\max}$ of the polynomial of the form $x_{\max} \geq f(k)$.

So far I was able to show that $x_{\max} \geq k$ which is quite close to the actual value of the root but I need some help going the extra mile.

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  • $\begingroup$ Newton's method starting near $k$ should be good enough? $\endgroup$
    – Lucia
    Commented Nov 14, 2013 at 0:21
  • $\begingroup$ @Lucia But how do you ensure that Newton's iterates are lower approximations to $x_{\max}$? (From some point, of course) $\endgroup$
    – Felix Goldberg
    Commented Nov 14, 2013 at 0:37
  • $\begingroup$ Eventually the polynomial is positive. If you plug in a value given by a Newton approximation and find a negative value then the root lies above it. I think your polynomial is simple enough that this will work without too much trouble. $\endgroup$
    – Lucia
    Commented Nov 14, 2013 at 1:18
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    $\begingroup$ Actually $p(x)$ is increasing and convex in $[k,\infty)$, therefore the Newton iteration starting at $x_0$ is decreasing provided $x_0 \ge k$ and $f(x_0)\ge0$. $\endgroup$
    – Pietro Majer
    Commented Nov 14, 2013 at 17:18
  • $\begingroup$ @PietroMajer: Note that $f(k)$ is negative. So starting Newton approximations from $k$ gives exactly what the problem wants. $\endgroup$
    – Lucia
    Commented Nov 14, 2013 at 19:11

2 Answers 2

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I would write this as a comment, but as I'm new here, it doesn't allow me to do so.

Anyway, one can make your bound better by replacing $x_{max}\ge k$ by $x_{max}\ge k+\frac{1}{2k}$. This can be seen by considering $q(y)=p(y+k)=y^3+(2k+1)y^2+(k^2+1)y-k+1=0$. Definitely $q(\frac{1}{k})>0$ and actually, it's not difficult to see $q(\frac{1}{2k})<0$ for $k>3$.

I guess you can keep doing this by adding negative degrees of $k$ to your expression and make your bound better and better. Hope this will help you.

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With a little help from Maple, one can derive that asymptotically (in $k$) the largest root satisfies

$$ k + \frac{1}{k} - \frac{1}{k^2} - \frac{3}{k^3} + \frac{4}{k^4} + \frac{14}{k^5} +O(\frac{1}{k^6}) $$

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    $\begingroup$ If you want something non-asymptotic that's easily certifiable, you can expand $k^{15}f(k+\frac{1}{k}-\frac{1}{k^2}-\frac{3}{k^3}+\frac{4}{k^4}+\frac{7}{k^5})$ in powers of $(k-5)$. Every coefficient is negative, so $f$ must be less than $0$ there. $\endgroup$
    – Kevin P. Costello
    Commented Nov 14, 2013 at 23:38

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