2
$\begingroup$

Show that the equation $x^n=f(x)$ where $f(x)$ is a polynomial with positive coefficients of degree $n-1$, has only one positive root.

I found this problem but I'm having trouble solving it and I would really like some help.

I thought proof by contradiction by assuming that we have at least two positive roots that satisfy the equation but I don't really know where to go from there.

Sorry for any mistakes in my English. It's not my native language

$\endgroup$
  • 1
    $\begingroup$ @NoChance $f(x)$ is supposed to have +ve coeffs $\endgroup$ – Anvit Jun 18 '19 at 11:08
  • $\begingroup$ @OK, thanks for the remark. $\endgroup$ – NoChance Jun 18 '19 at 11:09
2
$\begingroup$

This is (a particular case of) Descartes' rule of signs

Since $x^n-f(x)$ has exactly one sign change, the number of positive real roots is either 1 or an odd number less than 1. This means it has exactly 1 positive root.

You can find a proof for example here, or many other places

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think that the sign of the constant term must be considered. For example $x^3-2x^2+1=0$ has 2 positive roots, (1 and 1.618) whereas $x^3-2x^2-1=0$ has only 1 positive root of value 2.206 (assuming no complex roots). $\endgroup$ – NoChance Jun 18 '19 at 11:21
  • $\begingroup$ @NoChance What do you mean? In $$x^3=2x^2-1=:f(x)$$ $f(x)$ does not have positive coefficients. The constant term is the coefficient of $x^0$, it is still called a coefficient. $\endgroup$ – N. S. Jun 18 '19 at 11:24
  • $\begingroup$ You are correct, thank you for your reply. $\endgroup$ – NoChance Jun 18 '19 at 11:24
  • 1
    $\begingroup$ @NoChance No problem... $\endgroup$ – N. S. Jun 18 '19 at 11:26
0
$\begingroup$

Let $g(x)=x^n -f(x)$. Let $M$ be the number of positive roots of $g(x)$. Observe that $g(0)=-k$ where $k>0$ is constant term in $f(x)$ . Also $lim_{x\rightarrow\infty}g(x)\rightarrow\infty$ so there is at least one root in $(0,\infty)$ which sets the lower bound for $M$ i.e. $M\ge 1$.

Descarte's rule of sign changes in $g(x)$ sets the upper bound for $M$ i.e. $M\le 1$. So $M=1$.

Furthermore, if $f(x)$ doesn't have a constant term then $g(x)=x^r.h(x)$ where $g(x)$ has a root at $x=0$ with multiplicity $r$. You can work along the same lines on $h(x)$ now.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

HINT:

It's enough to show that $$1=\frac{f(x)}{x^n}$$ has one root in $(0,\infty)$. Note that the RHS is decreasing.

| cite | improve this answer | |
$\endgroup$

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.