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I am stuck on the following problem:

Given \begin{align*} f(x) &= x^4 + (-m - 2n + 6)x^3 + (mn - 5m - 8n + 10)x^2\\ &\quad + (3mn - 7m - 8n)x + mn-m-n-7, \end{align*} if $\alpha_1,\alpha_2$ are two negative roots of $f(x)=0$, show that $\alpha_1+\alpha_2< -3$ for $m,n\ge 4$.

I took some values of $m,n$ and found the following:

$m=n=4$, $\alpha_1=-2.476, \alpha_2=-0.621$;

$m=5,n=4$, $\alpha_1=-2.49, \alpha_2=-0.59$

I tried some other values of $m,n$, In all the above cases, I am getting $\alpha_1+\alpha_2<-3$.

But I am not able to prove it. Can someone please help me out?

NOTE:

I found the following. I obtained $-1<\alpha_1<-k$ and $-3<\alpha_2<-3+k$ where $k$ is some function of $m$ and $n$.

I took $m=4,n=5$ and I got $\alpha_1=-2.492<-3+0.55, \alpha_2=-0.592<-0.55$. I took some other values of $m,n$ and obtained similar stuff. So I think my guess is correct.

Is there any way to find $k$ such that $k$ is a function of $m,n$?

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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Feb 13, 2021 at 14:36

3 Answers 3

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We have \begin{align} f(-3) &= mn + 2m + 5n + 2 > 0, \\ f(-3/2) &= \frac{-20mn + 26m - 4n + 5}{16} < 0,\\ f(0) &= mn - m - n - 7 > 0, \\ f(m) &= -m^3 n+m^3-5 m^2 n+3 m^2-7 m n-m-n-7 < 0, \\ f(2m + 2n) &= 8 m^4+28 m^3 n+32 m^2 n^2+12 m n^3+28 m^3+78 m^2 n+66 m n^2\\ &\quad +16 n^3+26 m^2+51 m n+24 n^2-m-n-7\\ & > 0. \end{align} (Note: Simply letting $m = 4 + s, n = 4 + t$ for $s, t \ge 0$, all the inequalities are obvious.)

Thus, $f(x) = 0$ has four real roots $x_1, x_2, x_3, x_4$ located in the following intervals respectively $$x_1 \in (-3, -3/2), \quad x_2 \in (-3/2, 0), \quad x_3 \in (0, m), \quad x_4 \in (m, 2m+2n).$$

We need to prove that $x_1 + x_2 < -3$ or $x_2 < -3 - x_1$.

Since $-3 - x_1 \in (-3/2, 0)$, it suffices to prove that $f(-3-x_1) > 0$.

Since $f(x_1) = 0$, it suffices to prove that $f(-3-x_1) - f(x_1) > 0$ that is $$(2x_1 + 3)\Big((m+2n)x_1^2 + 3(m+2n)x_1 + m+2n + 3\Big) > 0.$$

It suffices to prove that $$x_1 > - \frac{3}{2} - \sqrt{\frac{5m + 10n - 12}{4m + 8n}}.$$

It suffices to prove that $$f\left(- \frac{3}{2} - \sqrt{\frac{5m + 10n - 12}{4m + 8n}}\right) > 0$$ that is $$\frac{m^3+2 m^2 n+2 m n^2+4 n^3-m^2-10 m n-16 n^2+3 m+6 n+9}{(m+2 n)^2} > 0$$ which is true. (Note: Simply letting $m = 4 + s, n = 4 + t$ for $s, t\ge 0$, this inequality is obvious.)

We are done.

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For @Math_Freak: The Maple code for the last two equations is given by

f := x^4+(-m-2*n+6)*x^3+(m*n-5*m-8*n+10)*x^2+(3*m*n-7*m-8*n)*x+m*n-m-n-7
x := -3/2+Q
f1 := collect(expand(f), Q)
f2 := subs({Q^2 = (5*m+10*n-12)/(4*m+8*n), Q^3 = (5*m+10*n-12)*Q/(4*m+8*n), Q^4 = ((5*m+10*n-12)/(4*m+8*n))^2}, f1)
factor(f2)
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  • $\begingroup$ Nice ,but how did you even think of checking the sign of $g(y)$ for the ertain values ,is it jsut random guessing or motivated by computer ,and also why did you think of subbing $x=-3/2+\sqrt{5}/2y$ $\endgroup$ Feb 14, 2021 at 15:43
  • $\begingroup$ @AlbusDumbledore Actually, if $f(a) f(b) < 0$, then there is at least one real root on $(a, b)$. We need to prove that $x_1 + x_2 < -3$ that is $(x_1 + 3/2) + (x_2 + 3/2) < 0$. Let $x_1 + 3/2 = y_1, x_2 + 3/2 = y_2$. We need to prove that $y_1 + y_2 < 0$. $\endgroup$
    – River Li
    Feb 14, 2021 at 16:10
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    $\begingroup$ LOL that was not my question I know that, I am asking How you thought of g(2m+2n+3/2) and checking its sign $\endgroup$ Feb 14, 2021 at 16:11
  • $\begingroup$ @Albus Dumbledore I edited the answer since the substitution is not necessary. For your question, we may use bisection to find the points. For example, first, $f(0) > 0$; second, since $f(\infty) = \infty$, we may find a large $c$ such that $f(c) > 0$, e.g., $c = 10m + 10n$. Then we check the middle point of $0$ and $10m + 10n$, i.e., $c_1 = 5m + 5n$, if $f(c_1) < 0$, OK, otherwise, check the middle point of $0$ and $c_1$ and so on. Finally, we find $c_0 = \frac{5}{8}(m+n)$ such that $f(c_0) < 0$. Thus, $x_3 \in (0, \frac{5}{8}(m+n))$ and $x_4 \in (\frac{5}{8}(m+n), 10m + 10n)$. $\endgroup$
    – River Li
    Feb 15, 2021 at 2:14
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    $\begingroup$ @AlbusDumbledore Yes, the last one is quite complicated. $\endgroup$
    – River Li
    Feb 15, 2021 at 16:11
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Let $x_1 < x_2 < 0$ be two negative solutions and I will obtain a bound: $$x_2 <-\dfrac{3}{10}(5+\sqrt{10})\approx-2.4487.$$

First, rewrite the polynomial as: \begin{align} f(x,m,n) = (x^2+3x+1)mn \\-(x^3+5x^2+7x+1)m \\-(2x^3+8x^2+8x+1)n \\+(x^4+6x^3+10x^2-7) \end{align} As my previous update and @RiverLi's comment said, we can obtain the simple bound: \begin{equation} \lambda_1 < x_1 < x_2 < \lambda_2 < 0, \end{equation} where $\lambda_{1,2} = \dfrac{-3+\sqrt{5}}{2}$ are the roots of the equation $x^2+3x+1 = 0,$ by simply considering $f(\lambda_i,m,n).$ But this is not strong enough because we want an upper bound on $x_1.$ Therefore, for a small $\varepsilon > 0$ to be chosen later, consider the equation: $$x^2+3x+1 = -\varepsilon\,\,(1)$$ and call its roots $x_{1,2}^{\varepsilon} = \dfrac{-3\pm\sqrt{5-4\varepsilon}}{2},\,x_1^{\varepsilon}<x_2^{\varepsilon} .$ The nice thing about this is that if $x < 0$ is root of $(1),$ then we have a nice form for: \begin{align} f(x,m,n) = -\varepsilon mn \\+(x\varepsilon+2\varepsilon +1)m \\+(2x\varepsilon+2\varepsilon+1)n \\+(\varepsilon^2+\varepsilon-7-3x) \end{align} which we will call $(2).$ We want to choose $\varepsilon$ so that $f(x_1^{\varepsilon})$ is negative, which will give $x_1 < x_1^{\varepsilon}.$

One can immediately see that: $$2x_1^{\varepsilon}\varepsilon+2\varepsilon+1 = 1-\varepsilon -\varepsilon\sqrt{5-4\varepsilon}<0\iff 0.344446\approx\varepsilon_0 < \varepsilon < 1.25$$ where $\varepsilon_0$ is the unique positive solution of $1-\varepsilon -\varepsilon\sqrt{5-4\varepsilon}=0.$ Choosing this $\varepsilon = \varepsilon_0$ will give a slightly better bound, but when I was doing it by hand I found $0.35$ works, which gives the quadratic equation: $$20x^2+60x+27 = 0\implies x_1^{\varepsilon} = -\dfrac{3}{10}(5+\sqrt{10})\approx-2.4487.$$

For this chosen value, the only thing left to show is that $f(x_1^\varepsilon) < 0.$ We automatically know that the coefficient of $n$ is negative by our choice and the free coefficient is trivially negative as well. For the remaining part: $$-\varepsilon mn+(x\varepsilon+2\varepsilon +1)m \leq m\left(1+x\varepsilon-2\varepsilon\right)=m\cdot\dfrac{2-7\varepsilon - \varepsilon\sqrt{5-4\varepsilon}}{2}<0,$$ which can be verified by hand for $\varepsilon = 0.35.$

To conclude, we then have: $$x_2+x_1 < -\dfrac{3}{10}(5+\sqrt{10}) +\dfrac{-3+\sqrt{5}}{2}\approx -2.83.$$ But all of this was done to obtain an explicit upper bound on $x_1$, which we previously do not have. So now by considering $x^2+3x+1 = \varepsilon$ for small, suitable $\varepsilon >0,$ I hope to better the not-so-tight upper bound $x_2:$ $$x_2<\lambda_2 = \dfrac{-3+\sqrt{5}}{2}\approx -0.381,$$ next time I got some time on my hand.

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    $\begingroup$ Let $x_1 < x_2 < 0$ be the two negative real roots. We have $f(-3)>0, f(-1) < 0, f(0) > 0$ and $f(\frac{-3 - \sqrt{5}}{2}) > 0$, $f(\frac{-3 + \sqrt{5}}{2}) > 0$. Thus $\frac{-3 - \sqrt{5}}{2} < x_1$ and $x_2 < \frac{-3 + \sqrt{5}}{2}$. $\endgroup$
    – River Li
    Feb 13, 2021 at 16:05
  • $\begingroup$ @RiverLi, yea that's exactly what I did in the edit. The difference is we don't have to explicitly plug in $\dfrac{-3\pm\sqrt{5}}{2}$ to see their signs. I am thinking that by considering the roots of $x^2+3x+1 = \pm\varepsilon$, we can probably get the reverse bound on $x_1.$ $\endgroup$
    – dezdichado
    Feb 13, 2021 at 17:52
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Let $\alpha_{i,m,n}$ be the $i^\text{th}$ root of $f(x;m,n)$ in ascending order. It is shown that for $m>4,n>4$

$$ \begin{align} &\alpha_{1,\infty,\infty}<\alpha_{1,m,n}<\alpha_{1,4,4}\\ &\alpha_{2,4,4}<\alpha_{2,m,n}<\alpha_{2,\infty,\infty} \end{align} $$

Numerically,

$$ \begin{align} &-2.618\ldots<\alpha_{1,m,n}<-2.476\ldots\\ &-0.621\ldots<\alpha_{2,m,n}<-0.3819\ldots \end{align} $$ leading to the bound $$ \alpha_{1,m,n}+\alpha_{2,m,n}<-2.858\ldots $$

The proof is via observing the change of sign of $f$ at the boundaries of the intervals $(\alpha_{1,\infty,\infty}~,\alpha_{1,4,4})$ and $(\alpha_{2,4,4}~,\alpha_{2,\infty,\infty})$, as tabulated below:

$x$ $sgn(f(x,m,n))$
$\alpha_{1,\infty,\infty}$ $+1$
$\alpha_{1,4,4}$ $-1$
$\alpha_{2,4,4}$ $-1$
$\alpha_{2,\infty,\infty}$ $+1$

The infinity roots are $(-3\pm\sqrt{5})/2$.

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