MHB Number of real roots in polynomial equation

  • Thread starter Thread starter juantheron
  • Start date Start date
  • Tags Tags
    Polynomial Roots
juantheron
Messages
243
Reaction score
1
Evaluate number of real roots of the equation $$x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5} = 0$$
 
Mathematics news on Phys.org
jacks said:
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:

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

And then:

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

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

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

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

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:

$$x\approx0.342384094858369$$

We then look at:

$$f''(0.342384094858369)=0.961949707437654530>0$$

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:

$$x\approx0.67033204760309682774$$

We then look at:

$$f(0.67033204760309682774)=0.03509389397174151671752$$

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

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 $\left[0,\lfloor M\rfloor +1\right].$
 
My Solution::

$\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 $\displaystyle f(x) = x^6-x^5+x^4-x^3+x^2-x+\frac{2}{5}>0\;\forall x \in \mathbb{R}$
 
Last edited by a moderator:
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top