6
$\begingroup$

In Chapter 1 of Polynomials by Victor Prasolov, Springer, 2001, the following theorem is proved. (p.3)

Theorem 1.1.4 (Ostrovsky). Let $f(x)=x^{n}-b_{1}x^{n-1}-\cdots -b_{n}$, where all the numbers $b_{i}$ are non-negative and at least one of them is nonzero. If the greatest common divisor of the indices of the positive coefficients $b_{i}$ is equal to $1$, then $f$ has a unique positive root $p$ and the absolute value of the other roots are $<$ p.

The following is one of the Problems to Chapter 1 (p.41).

Problem 1.5 - Find the number of real roots of the following polynomials

a) ...

b) $nx^{n}-x^{n-1}-\cdots -1$

Question: How to solve this Problem?


Added: $nx^{n}-x^{n-1}-\cdots -1=0$ $\Leftrightarrow x^{n}-\dfrac{1}{n}x^{n-1}-\cdots -\dfrac{1}{n}=0$

Added 2: Sturm's Theorem.

  • 0
    Anyway, for the others: [here](http://books.google.com/books?id=qIJPxdwSqlcC&pg=PA41) is the problem in the book, and [here](http://books.google.com/books?id=qIJPxdwSqlcC&pg=PA3) is where Ostrovsky's theorem is stated.2010-11-19

1 Answers 1

14

I think there is some value here in knowing how to do such problems "by hand." The proof in this case is quite simple. If $|x| > 1$, then $|x^n| > |x^k|$ for $k < n$, hence

$n |x|^n > |x|^{n-1} + |x|^{n-2} + ... + |1| \ge |x^{n-1} + x^{n-2} + ... + 1|$

by the triangle inequality, so this polynomial $f(x)$ has no roots of absolute value greater than $1$. It follows that any real roots lie in $[-1, 1]$. By inspection $x = 1$ is a root and $x = 0, -1$ are not, so any remaining roots lie in $(0, 1)$ or $(-1, 0)$. If $x \in (0, 1)$, then

$x^{n-1} + x^{n-2} + ... + 1 > nx^n$

so there are no roots in $(0, 1)$. To find any remaining roots in $(-1, 0)$, let

$g(x) = f(x) (x - 1) = nx^n(x - 1) - x^n + 1 = nx^{n+1} - (n+1) x^n + 1.$

Then g'(x) = n(n+1) x^n - n(n+1) x^{n-1} = n(n+1) x^{n-1}(x - 1) has roots $x = 0, 1$, hence $g$ is monotonic on $(-1, 0)$, so to determine if there are roots on this interval it suffices to compute $g(-1)$ and $g(0)$. We have $g(0) = 1$ and $g(-1) = -2n$ if $n$ is even and $g(-1) = 2n+2$ if $n$ is odd. In the first case there is one real root in $(-1, 0)$ by the IVT and in the second case there are none.

  • 2
    I suppose part of the argument constitutes a sketch of the proof of the quoted theorem. But my point is that one doesn't need to use the result as a black box when basic principles of one-variable calculus are enough to solve the problem directly.2010-11-19