# Comparing largest root of degree three polynomial

Given $n\ge 12$, consider the three polynomials: $$f(x)=x^2-\left(\frac{(n-4)^2}{n-3}+\frac{n-2}{3(n-3)}+\frac{4}{3}\right)x+\frac{4(n-4)^2}{3(n-3)},$$ $$g(x)=x^3-\left(2+\frac{(n-5)^2}{n-4}\right)x^2+\left(\frac{3}{4}+\frac{3(n-5)^2}{2(n-4)}\right)x-\frac{(n-5)^2}{4(n-4)},$$ $$h(x)=x^3-\left(2+\frac{(n-5)(n-4)}{n-3}\right)x^2+\left(\frac{3}{4}+\frac{(n-5)(n-4)}{n-3}\right)x-\frac{(n-5)(n-4)}{4(n-3)},$$

Suppose $f_\lambda, g_\lambda,h_\lambda$ are the largest root of $f(x),g(x),h(x)$ respectively. In order to prove one result I need to show that $f_\lambda >g_\lambda >h_\lambda.$ For some specific values I have verified it. Is it true for all $n\ge 12$ ?

Partial Answer: As mathlove mentioned in the below answer, the result does not hold. But still I feel at least $f_{\lambda}>h_{\lambda}$ is true. Any idea for this one?

• Where did these three polynomials come from? – J. M. isn't a mathematician Mar 5 '18 at 13:54
• It can be shown that the largest roots behave like $$x_f = n - \frac{22}{3} + O\!\left(\frac{1}{n}\right), \\ x_g = n - \frac{9}{2} + O\!\left(\frac{1}{n}\right), \\ x_h = n - 5 + O\!\left(\frac{1}{n}\right).$$ So for $n$ large enough we have $x_f < x_h < x_g$. – Antonio Vargas Mar 5 '18 at 18:29
• They arose while calculating the spectral radii of three symmetric matrices of some particular pattern. – rationalize Mar 6 '18 at 7:28

No, it is not true for all $n\ge 12$.

For $n=12$, we have $g_{\lambda}\lt h_{\lambda}$.

Proof :

For $n=12$, we have $$g(0)=-\frac{49}{32},g(1)=\frac{41}{32},g(2)=-\frac{197}{32},g(7)=\frac{413}{32}\implies g_{\lambda}\lt 7$$ and $$h(x)=\frac{1}{36}(2x-1)(18x^2-139x+56)\implies h_{\lambda}=\frac{139+\sqrt{15289}}{36}\gt \frac{139+122}{36}=7.25$$ from which $$g_{\lambda}\lt 7\lt 7.25\lt h_{\lambda}$$ follows.

Next, let us prove that $f_{\lambda}\gt h_{\lambda}$ for all $n\ge 12$.

Proof :

We have $$f_{\lambda}=\frac{3(3 n^2 - 19 n + 34)+\sqrt{(9n^2-81n+182)^2+64(9n-31)}}{18(n-3)}\tag1$$

Also, noting that $x=\frac 12$ is a root of $h(x)$, we have $$h(x)=\frac{(2x-1)(2(n-3)x^2+(15n-2n^2-31)x+n^2-9n+20)}{4(n-3)}$$ So, $$h(x)=0\iff x=\frac 12,\quad \frac{2n^2-15n+31\pm\sqrt{(2n^2-17n+39)^2+20(n-4)}}{4(n-3)}$$ Now, for $n\ge 12$, we have \begin{align}&\frac 12\lt \frac{2n^2-15n+31+\sqrt{(2n^2-17n+39)^2+20(n-4)}}{4(n-3)} \\\\&\iff 2(n-3)-(2n^2-15n+31)\lt \sqrt{(2n^2-17n+39)^2+20(n-4)} \\\\&\iff -\frac 18(4n-17)^2-\frac 78\lt \sqrt{(2n^2-17n+39)^2+20(n-4)} \end{align} which holds for all $n\ge 12$.

It follows that $$h_{\lambda}=\frac{2n^2-15n+31+\sqrt{(2n^2-17n+39)^2+20(n-4)}}{4(n-3)}\tag2$$

From $(1)(2)$, \small\begin{align}&f_{\lambda}\gt h_{\lambda} \\\\&\iff \frac{3(3 n^2 - 19 n + 34)+\sqrt{(9n^2-81n+182)^2+64(9n-31)}}{18(n-3)}\\\\&\qquad\qquad \gt \frac{2n^2-15n+31+\sqrt{(2n^2-17n+39)^2+20(n-4)}}{4(n-3)} \\\\&\iff 6(3 n^2 - 19 n + 34)+2\sqrt{(9n^2-81n+182)^2+64(9n-31)} \\\\&\qquad\qquad \gt 9(2n^2-15n+31)+9\sqrt{(2n^2-17n+39)^2+20(n-4)} \\\\&\iff 3(7n-25)+2\sqrt{(9n^2-81n+182)^2+64(9n-31)} \\\\&\qquad\qquad\gt 9\sqrt{(2n^2-17n+39)^2+20(n-4)} \\\\&\iff 9(7n-25)^2+6(7n-25)\sqrt{(9n^2-81n+182)^2+64(9n-31)} \\\\&\qquad\qquad +4((9n^2-81n+182)^2+64(9n-31))\gt 81((2n^2-17n+39)^2+20(n-4))\\\\&\iff 6(7n-25)\sqrt{(9n^2-81n+182)^2+64(9n-31)}\gt 36((9n-104)n^2+(361n-374)) \\\\&\iff (7n-25)\sqrt{(9n^2-81n+182)^2+64(9n-31)}\gt 6((9n-104)n^2+(361n-374)) \\\\&\iff (7n-25)^2((9n^2-81n+182)^2+64(9n-31))\gt 36((9n-104)n^2+(361n-374))^2 \\\\&\iff 39 n^6 - 1200 n^5 + 15542 n^4 - 104636 n^3 + 381471 n^2 - 712796 n + 534332\gt 0\end{align} which does hold for $n\ge 12$ according to WolframAlpha.

Therefore, we can say that $f_{\lambda}\gt h_{\lambda}$ for all $n\ge 12$.

• I am extremely sorry. While typing, I missed a $3$ in the denominator of last term of $f(x).$ I have edited it now. Inconvenience is deeply regreted. – rationalize Mar 6 '18 at 7:27
• Thank you ! It helped but now I am struck at $f_{\lambda}>h_{\lambda}$. I think it is true for $n\ge12$. Any help? – rationalize Mar 6 '18 at 13:28
• @rationalize: I've just added a proof that $f_{\lambda}\gt h_{\lambda}$ for all $n\ge 12$. – mathlove Mar 6 '18 at 15:50
• Thank you very much. Appreciations!! – rationalize Mar 7 '18 at 14:12