Number of real roots in polynomial equation

[juantheron]

Evaluate number of real roots of the equation $$x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5} = 0$$

 

[MarkFL]

Evaluate number of real roots of the equation $$x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5} = 0$$

My solution:

Spoiler

Let \(\displaystyle f(x)=x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5}\)

And then:

\(\displaystyle f'(x)=6x^5-5x^4+4x^3-3x^2+2x-1\)

\(\displaystyle f''(x)=30x^4-20x^3+12x^2-6x+2\)

\(\displaystyle f'''(x)=120x^3-60x^2+24x-6\)

\(\displaystyle f^{(4)}(x)=360x^2-120x+24\)

The 4th derivative has a negative discriminant, therefore as an upward opening parabolic function, is positive for all $x$. This means the third derivative is strictly increasing and can have only 1 real root. Using a numeric root-finding technique, we find this root is approximated by:

\(\displaystyle x\approx0.342384094858369\)

We then look at:

\(\displaystyle f''(0.342384094858369)=0.961949707437654530>0\)

This means we may conclude that the 2nd derivative is positive for all $x$, and so the first derivative is strictly increasing with only 1 real root, which we find at about:

\(\displaystyle x\approx0.67033204760309682774\)

We then look at:

\(\displaystyle f(0.67033204760309682774)=0.03509389397174151671752\)

And so we conclude that for all real $x$, we have $f(x)>0$, and thus $f$ has no real roots.

 

[Fernando Revilla]

Another way:

Spoiler

If $p(x)=x^6-x^5+x^4-x^3+x^2-x+\dfrac{2}{5}$ then $p(-x)=x^6+x^5+x^4+x^3+x^2+x+\dfrac{2}{5}$ and by the Descartes' Rule of Signs there are no negative roots for $p(x)$.

On the other hand, if $f(x)=a_nx^n+\ldots +a_1x+a_0\in\mathbb{R}[x]$ with $a_n\neq 0$ and $c$ a root of de $f(x)$ then $|c|\leq M$ where $$M=\max\left \{\left(n\left| \frac{a_{i-1}}{a_n}\right|\right)^{1/i}:i=1,\ldots,n\right\}.$$ Now, we can use the Sturm's Theorem on the closed interval $\left[0,\lfloor M\rfloor +1\right].$

 

[juantheron]

My Solution::

Spoiler

$\bullet\; $ If $x\leq 0\;,$ Then $\displaystyle f(x) = x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5}>0$

$\bullet\; $ If $x\geq 1\;,$ Then $\displaystyle f(x) = x^5(x-1)+x^3(x-1)+x(x-1)+\frac{2}{5}>0$

$\bullet \; 0<x<1\;,$ Let $\displaystyle f(x)-\frac{2}{5} = x^6-x^5+x^4-x^3+x^2-x=x(x-1)(x^4+x^2+1)$

So $\displaystyle f(x)-\frac{2}{5} = -x(1-x)(1+x+x^2)(1-x+x^2) = -(1-x^3)(x-x^2+x^3)$

So $\displaystyle -f(x)+\frac{2}{5} = (1-x^3)(x-x^2+x^3) \leq \left(\frac{1-x^3+x-x^2+x^3}{2}\right)^2$

So $\displaystyle -f(x)+\frac{2}{5}\leq \left(\frac{x-x^2+1}{2}\right)^2 = \frac{1}{64}\left(4x-4x^2+4\right)^2$

So $\displaystyle -f(x)+\frac{2}{5} \leq \frac{1}{64}\left(5-(2x-1)^2\right) \leq \frac{25}{64}$

So $\displaystyle -f(x)+\frac{2}{5}\leq \frac{25}{64}\Rightarrow f(x)-\frac{2}{5}\geq \frac{25}{64}$

So $\displaystyle f(x)\geq \frac{2}{5}-\frac{25}{64} = \frac{3}{320}>0$

So $\displaystyle f(x) = x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5}>0\;\forall x \in \mathbb{R}$