3
$\begingroup$

I tried to solve below equation $$x^{10}-2x^9+3x^8-...-10x+11=0$$ I plot the graph and see there is no real root , But I get stuck how to show analytical .Can some one help me or give an idea ?

Thanks in advance.
This is graph of the function $$f(x)=x^{10}-2x^9+3x^8-...-10x+11$$ https://www.desmos.com/calculator/6k1lj498ra

$\endgroup$
  • 3
    $\begingroup$ Looks like coefficients are alternating, so why $...+10x+11$ at the end? $\endgroup$ – Ennar Dec 2 '17 at 18:14
  • $\begingroup$ Your link has $-10x+11$ $\endgroup$ – Mark Bennet Dec 2 '17 at 18:21
  • $\begingroup$ And what exactly did you try? $\endgroup$ – Professor Vector Dec 2 '17 at 18:21
10
$\begingroup$

There is some confusion over the sign of the coefficients. But it might prove useful to consider $$x^8(x-1)^2+2x^6(x-1)^2+3x^4(x-1)^2+4x^2(x-1)^2 =$$$$=x^{10}-2x^9+x^8+2x^8-4x^7+2x^6+3x^6-6x^5+3x^4+4x^4-8x^3+4x^2 =$$$$=x^{10}-2x^9+3x^8-4x^7+5x^6-6x^5+7x^4-8x^3+4x^2$$which is clearly non-negative, or some variant of this. You should be able to conclude from there.

$\endgroup$
  • $\begingroup$ Nice solution. Perhaps one can add a term $5(x-1)^{2} + 6$ to both members ? $\endgroup$ – Panurge Dec 2 '17 at 18:35
  • $\begingroup$ Well, I guess the intended, final form of this is $(x^8 + 2x^6 + 3x^4 + 4x^2 + 5)(x-1)^2+6$, right? $\endgroup$ – Professor Vector Dec 2 '17 at 19:36
  • $\begingroup$ @Panurge Indeed that can be done, but I left some work for OP, and the signs at the end had been left ambiguous $\endgroup$ – Mark Bennet Dec 2 '17 at 19:49
  • 1
    $\begingroup$ @ProfessorVector You could do that, or just leave it as a rather more explicit sum of squares. When I put this the signs of the terms had not been clarified and I thought to leave something for OP to do. $\endgroup$ – Mark Bennet Dec 2 '17 at 19:51
9
$\begingroup$

If you multiply the given polynomial by $(1+x)^2$ you get $q(x)=x^{12}+12x+11$.
Of course we have a double root at $x=-1$ and $q(x)$ is a convex function since $q''(x)\geq 0$.
Since $x=-1$ is the only real root of $q'$, $q(-1)=0$ is an absolute minimum for $q(x)$ and $q(x)$ does not vanish over $\mathbb{R}\setminus\{-1\}$. It follows that the original polynomial has no real zeroes.

$\endgroup$
  • $\begingroup$ This works for general n. $\endgroup$ – marty cohen Dec 2 '17 at 19:54
7
$\begingroup$

Let's consider the polynomial from the link, $p(x)=x^{10}-2x^9+3x^8-4x^7+5x^6-6x^5+7x^4-8x^3+9x^2-10x+11$. Then, $q(x)=(x+1)^2\,p(x)=x^{12}+12x+11$. This is certainly positive for large enough $x$. The only zero of the derivative is $x=-1$, where $q(x)=0$. This must be the global minimum of $q$, so it takes only non-negative values, and is $0$ only for $x=-1$. This means $p(x)>0$ for all $x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.