Proving that extrema of cubic with 3 distinct roots always happen to fall between the roots

functionspolynomialsreal-analysis

By Rolle's Theorem, it is possible to prove that between points $a$ and $b$ there is a point $c$ at which the value of $f'(c)=0$.

Now, consider a cubic polynomial function with 3 distinct real roots,

$f(x)=A(x-a)(x-b)(x-c)$

It is now necessary to prove that $x$-coordinates of extrema of $f(x)$ fall between $a, b$ and $c$. Or otherwise stated,
$a<x_1<b<x_2<c$, where $x_1$ and $x_2$ are $x$-coordinates of extrema.

The problem is, however, in the fact that this should be proven without the usage of Rolle's Theorem. I can prove this fact for quadratic, I get $x=(a+b)/2$, which suggests that extremum is halfway between the roots. With cubic, however, it is much more difficult. Any ideas?

Best Answer

I'm taking capital $A=1$ and then $a < b < c$ as the roots. we get the cubic $$ x^3 - \sigma_1 x^2 + \sigma_2 x - \sigma_3 \; , $$ where $$ \sigma_1 = a + b + c, $$ $$ \sigma_2 = bc + ca + ab, $$ $$ \sigma_3 = abc. $$ The first derivative is $3 x^2 - 2 \sigma_1 x + \sigma_2,$ with roots $$ \frac{\sigma_1 \pm \sqrt{\sigma_1^2 - 3 \sigma_2}}{3} \; .$$

Worth emphasizing that, with $a,b,c$ distinct, we get $$ \sigma_1^2 - 3 \sigma_2 = a^2 + b^2 + c^2 - bc - ca - ab = \frac{1}{2} \left( (b-c)^2 + (c-a)^2 + (a-b)^2 \right) $$ being strictly positive, so the two roots of $3 x^2 - 2 \sigma_1 x + \sigma_2$ are real and distinct.

The claim that $c > \frac{\sigma_1 + \sqrt{\sigma_1^2 - 3 \sigma_2}}{3}$ comes to combining the observation that $c > \frac{\sigma_1}{3}$ along with $$ \left( c - \frac{\sigma_1}{3} \right)^2 -\left( \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-a)(c-b) > 0 $$

The claim that $a < \frac{\sigma_1 - \sqrt{\sigma_1^2 - 3 \sigma_2}}{3}$ comes to combining the observation that $a < \frac{\sigma_1}{3}$ along with $$ \left( a - \frac{\sigma_1}{3} \right)^2 -\left( \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-a)(b-a) > 0 $$

The final claim is that $b$ lies between the critical points, in that the distance between $b$ and $\sigma_1/3$ is smaller than $\sqrt{\sigma_1^2 - 3 \sigma_2}/3.$ Indeed, $$ \left( b - \frac{\sigma_1}{3} \right)^2 -\left( \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-b)(a-b) < 0 $$

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[Math] Generalised formula for fitting a cubic between two points with specified slopes at each point

Your polynomial is some $p(x)=ax^3+bx^2+cx+d$, $p'(x)=3ax^2+2bx+c$, with four unknown parameters $a,b,c,d$.

Two points give two equations $p(x_i)=y_i$, two slopes two more: $p'(x_i) = m_i$.

In matrix form: $$ A= \left( \begin{matrix} x_1^3 & x_1^2 & x_1 & 1 \\ x_2^3 & x_2^2 & x_2 & 1 \\ 3x_1^2 & 2x_1 & 1 & 0 \\ 3x_2^2 & 2x_2 & 1 & 0 \end{matrix} \right) \quad x= \left( \begin{matrix} a \\ b \\ c \\ d \end{matrix} \right) \quad b= \left( \begin{matrix} y_1 \\ y_2 \\ m_1 \\ m_2 \end{matrix} \right) \\ Ax=b \\ x = A^{-1} b $$ Then for $x_1 \ne x_2$: $$ \left[ \begin{array}{cccc|c} x_1^3 & x_1^2 & x_1 & 1 & y_1 \\ x_2^3 & x_2^2 & x_2 & 1 & y_2 \\ 3x_1^2 & 2x_1 & 1 & 0 & m_1 \\ 3x_2^2 & 2x_2 & 1 & 0 & m_2 \end{array} \right] \to \left[ \begin{array}{cccc|c} x_1^3 & x_1^2 & x_1 & 1 & y_1 \\ 3x_1^2 & 2x_1 & 1 & 0 & m_1 \\ 3(x_2^2 - x_1^2) & 2(x_2 - x_1) & 0 & 0 & m_2 - m_1 \\ x_2^3 & x_2^2 & x_2 & 1 & y_2 \\ \end{array} \right] \to \left[ \begin{array}{cccc|c} x_1^3 & x_1^2 & x_1 & 1 & y_1 \\ 3x_1^2 & 2x_1 & 1 & 0 & m_1 \\ 3(x_2^2 - x_1^2) & 2(x_2 - x_1) & 0 & 0 & m_2 - m_1 \\ x_2^3 - x_1^3 & x_2^2 - x_1^2 & x_2 - x_1 & 0 & y_2 - y_1 \\ \end{array} \right] \to \left[ \begin{array}{cccc|c} x_1^3 & x_1^2 & x_1 & 1 & y_1 \\ 3x_1^2 & 2x_1 & 1 & 0 & m_1 \\ 3(x_2^2 - x_1^2) & 2(x_2 - x_1) & 0 & 0 & m_2 - m_1 \\ x_1^2 + x_1 x_2 + x_2^2 & x_1 + x_2 & 1 & 0 & \frac{y_2 - y_1}{x_2 - x_1} \\ \end{array} \right] \to \left[ \begin{array}{cccc|c} x_1^3 & x_1^2 & x_1 & 1 & y_1 \\ 3x_1^2 & 2x_1 & 1 & 0 & m_1 \\ 3(x_2^2 - x_1^2) & 2(x_2 - x_1) & 0 & 0 & m_2 - m_1 \\ -2 x_1^2 + x_1 x_2 + x_2^2 & x_2 - x_1 & 0 & 0 & \frac{y_2 - y_1}{x_2 - x_1} - m_1 \\ \end{array} \right] \to \left[ \begin{array}{cccc|c} x_1^3 & x_1^2 & x_1 & 1 & y_1 \\ 3x_1^2 & 2x_1 & 1 & 0 & m_1 \\ 3(x_2^2 - x_1^2) & 2(x_2 - x_1) & 0 & 0 & m_2 - m_1 \\ -4 x_1^2 + 2 x_1 x_2 + 2 x_2^2 & 2(x_2 - x_1) & 0 & 0 & 2\frac{y_2 - y_1}{x_2 - x_1} - 2m_1 \\ \end{array} \right] \to \left[ \begin{array}{cccc|c} x_1^3 & x_1^2 & x_1 & 1 & y_1 \\ 3x_1^2 & 2x_1 & 1 & 0 & m_1 \\ x_1^2 - 2 x_1 x_2 + x_2^2 & 0 & 0 & 0 & m_1 + m_2 - 2\frac{y_2 - y_1}{x_2 - x_1}\\ -4 x_1^2 + 2 x_1 x_2 + 2 x_2^2 & 2(x_2 - x_1) & 0 & 0 & 2\frac{y_2 - y_1}{x_2 - x_1} - 2m_1 \\ \end{array} \right] \to \left[ \begin{array}{cccc|c} x_1^3 & x_1^2 & x_1 & 1 & y_1 \\ 3x_1^2 & 2x_1 & 1 & 0 & m_1 \\ 3 (x_2^2-x_1^2) & 2(x_2 - x_1) & 0 & 0 & m_2 - m_1 \\ x_1^2 - 2 x_1 x_2 + x_2^2 & 0 & 0 & 0 & m_1 + m_2 - 2\frac{y_2 - y_1}{x_2 - x_1}\\ \end{array} \right] \to \left[ \begin{array}{cccc|c} x_1^3 & x_1^2 & x_1 & 1 & y_1 \\ 3x_1^2 & 2x_1 & 1 & 0 & m_1 \\ \frac{3}{2} (x_1+x_2) & 1 & 0 & 0 & \frac{m_2 - m_1}{2(x_2-x_1)} \\ (x_1 - x_2)^2 & 0 & 0 & 0 & m_1 + m_2 - 2\frac{y_2 - y_1}{x_2 - x_1}\\ \end{array} \right] \to \left[ \begin{array}{cccc|c} x_1^3 & x_1^2 & x_1 & 1 & y_1 \\ 3x_1^2 & 2x_1 & 1 & 0 & m_1 \\ \frac{3}{2} (x_1+x_2) & 1 & 0 & 0 & \frac{m_2 - m_1}{2(x_2-x_1)} \\ 1 & 0 & 0 & 0 & \frac{m_1 + m_2 - 2\frac{y_2 - y_1}{x_2 - x_1}}{(x_1 - x_2)^2} \\ \end{array} \right] $$ This gives: \begin{align} a &= \frac{m_1 + m_2 - 2\frac{y_2 - y_1}{x_2 - x_1}}{(x_1 - x_2)^2} \\ b &= \frac{m_2 - m_1}{2(x_2-x_1)} - \frac{3}{2} (x_1+x_2) a \quad (*) \\ c &= m_1 - 3x_1^2 a - 2x_1 b \\ d &= y_1 - x_1^3 a - x_1^2 b - x_1 c \end{align}

The last two equations seem obvious in hind sight.

Test:

octave:1> x1 = 2
x1 =  2
octave:2> x2 = 5
x2 =  5
octave:3> m1 = 3
m1 =  3
octave:4> m2 = -1
m2 = -1
octave:6> A = [x1**3, x1**2, x1, 1; x2**3, x2**2, x2**1, 1;
3*x1**2,2*x1,1,0; 3*x2**2,2*x2,1,0]
A =
     8     4     2     1
   125    25     5     1
    12     4     1     0
    75    10     1     0

octave:17> y1=-2
y1 = -2
octave:18> y2=5
y2 =  5
octave:19> b = [y1;y2;m1;m2]
b =
  -2
   5
   3
  -1

octave:20> x = inv(A)*b
x =
  -0.29630
   2.44444
  -3.22222
  -2.96296

Now we compare this with equation $(*)$

octave:22> a = (m1+m2-2*(y2-y1)/(x2-x1))/(x1-x2)**2
a = -0.29630
octave:23> b = (m2-m1)/(2*(x2-x1))-(3/2)*(x1+x2)*a
b =  2.4444
octave:24> c = m1-3*(x1**2)*a -2*x1*b
c = -3.2222
octave:25> d = y1-(x1**3)*a -(x1**2)*b-x1*c
d = -2.9630

[Math] Relation between roots of function and roots of its derivative

If $p(x)=x^2+1$, then $p(x)$ has zero roots. However, $p'(x)=2x$, so $p'(x)$ has one root $x=0$.

That argument from your Calculus textbook proves that between any two roots of the original polynomial $p(x)$ there is at least one root of $p'(x)$ between them, but there may be other roots.

Best Answer

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[Math] Generalised formula for fitting a cubic between two points with specified slopes at each point

[Math] Relation between roots of function and roots of its derivative

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