# Problem from Victor Prasolov's Polynomials — Finding the number of real roots of $nx^{n}-x^{n-1}-\cdots -1$

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

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A sketch: monicize your polynomial so that it fits the conditions of Ostrovsky's theorem, and see what you can make of it. I'd have suggested Sturm, but I'm getting the impression that you're not supposed to use it just yet at that stage in the book. –  Ｊ. Ｍ. Nov 19 '10 at 10:49
Otherwise, note that all the members of the family have $x-1$ as a factor. –  Ｊ. Ｍ. Nov 19 '10 at 10:52
@J.M. Thanks for the suggestion! –  Américo Tavares Nov 19 '10 at 11:02
Anyway, for the others: here is the problem in the book, and here is where Ostrovsky's theorem is stated. –  Ｊ. Ｍ. Nov 19 '10 at 11:17

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.

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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. –  Qiaochu Yuan Nov 19 '10 at 14:14