Given a polynomial $x^n=m \cdot \sum_{k=0}^{n-1} x^k$ (for all $m,n \in \mathbb{N}, m \geq 2,n\geq 2$), the numerical computation of roots for different $n$ and $m$ shows, that the absolute values of roots are strictly less than $m + 1$. The Cauchy's bound gives a non-strict bound for the given polynomial: the absolute values of roots are less or equal to $m+1$.
I have come up with the approach for showing the strict inequality, and would like to ask, if my approach is valid?
Below is a description of my approach.
If we show that the roots don't belong to the complex circle with radius $m+1$, then in combination with the Cauchy's bound we will obtain a strict inequality.
As far as $x=1$ is not a root of this polynomial, we can transform the polynomial as follows:
$$ x^n - m \cdot \sum_{k=0}^{n-1} x^k = x^n - m \cdot {x^n-1 \over x - 1} = 0 $$
So, equivalently, we need to consider the roots of the polynomial: $x^n \cdot ((m+1) - x) - m = 0$.
For the sake of contradiction let's assume, that $x =(m+1) \cdot e^{i \phi}$ (for some $\phi \in \mathbb{R}$) is a root of the polynomial $x^n \cdot ((m+1) - x) - m = 0$.
After doing the substitution we obtain:
$$ (m+1)^n \cdot e^{i n \phi} \cdot ((m+1) - (m+1) \cdot e^{i \phi}) = m \iff \\ \iff (m+1)^{n+1} \cdot e^{i n \phi} \cdot (1 - e^{i \phi}) = m \iff \\ \iff e^{i n \phi} \cdot (1 - e^{i \phi}) = {m \over (m+1)^{n+1}} \iff \\ \iff e^{i n \phi} - e^{i (n+1) \phi} = {m \over (m+1)^{n+1}} $$
Using the trigonometric form of complex numbers we can rewrite the latter equality as follows:
$$ \left( cos(n\phi) - cos((n+1)\phi) \right) - i \cdot \left( sin(n\phi) - sin((n + 1)\phi) \right) = {m \over (m+1)^{n+1}} $$
As far as the imaginary part on the left hand side is $0$ we have a following system:
$$ \left\{ \begin{aligned} sin(n\phi) - sin((n+1)\phi) &= 0 \\ cos(n\phi) - cos((n+1)\phi) &= {m \over (m+1)^{n+1}} \end{aligned}\right. $$
Using the sum-to-product trigonometric identities we can rewrite the system as follows: $$ \left\{ \begin{aligned} 2 \cdot cos \left({2n + 1 \over 2} \phi \right) \cdot sin \left( - {\phi \over 2} \right) &= 0 \\ -2 \cdot sin \left({2n + 1 \over 2} \phi \right) \cdot sin \left( - {\phi \over 2} \right) &= {m \over (m+1)^{n+1}} \end{aligned}\right. $$
Looking at the first equation: $2 \cdot cos \left({2n + 1 \over 2} \phi \right) \cdot sin \left( - {\phi \over 2} \right)$ we conclude that some of its multipliers should be $0$. So, we have two cases:
- Case 1: $sin \left( - {\phi \over 2} \right) = 0$. In this case the second equation shows a contradiction:
$$ -2 \cdot sin \left({2n + 1 \over 2} \phi \right) \cdot sin \left( - {\phi \over 2} \right) = 0 \neq {m \over (m+1)^{n+1}} $$
- Case 2: $cos \left({2n + 1 \over 2} \phi \right) = 0$. In this case we can see that: $\phi = {2k + 1 \over 2n + 1} \pi$ for $k \in \mathbb{Z}$. Let's substitute $\phi$ into the second equation:
$$ -2 \cdot sin \left({2n + 1 \over 2} \cdot {2k + 1 \over 2n + 1} \pi \right) \cdot sin \left( - {1 \over 2} \cdot {2k + 1 \over 2n + 1} \pi \right) = {m \over (m+1)^{n+1}} \iff \\ \iff sin \left({2k + 1 \over 2 \cdot (2n + 1)} \pi \right) = {m \over 2 \cdot (m+1)^{n+1}} $$
So, from the second equation we have obtained the equation $sin(a \cdot \pi) = b$, where $a$ and $b$ are rational numbers ($a={2k + 1 \over 2 \cdot (2n + 1)}$ and $b={m \over (m+1)^{n+1}}$). Furthermore, in case if $m \geq 2, n \geq 2$ we see that $b \not \in \{ 0, \pm 1, \pm {1 \over 2}\}$. In this case, we have a contradiction with the Niven's Theorem (that states, that if $sin(a \cdot \pi) = b$ and $a, b \in \mathbb{Q}$, then the sine takes only the values $0, \pm 1, \pm {1 \over 2}$).
So, we see that the original assumption (that $(m+1) \cdot e^{i \phi}$ is a root of the polynomial) leads to the contradiction. Consequently, $(m+1) \cdot e^{i \phi}$ can't be a root of the polynomial.
In combination with the Cauchy's bound we have a strict bound on the absolute values of roots: the roots are strictly less than $m + 1$.
So, I would like to know if my approach is valid?
And, the second question: if there are any simpler approaches that allow to obtain a strict bound in this case?
Your proof seems to be good, I could not find an error. It should work even for $m \ge 1$ and $n \ge 1$.
Yes: If $x$ with $|x| > 1$ is a root of that polynomial then $$ |x|^n \le m\left(1 + |x| + \ldots + |x|^{n-1} \right) = m \frac{|x|^n-1}{|x|-1} \, . $$ Multiplication with the (positive) number $(|x|-1)/|x|^n$ gives $$ |x| -1 \le m \frac{|x|^n-1}{|x|^n} < m $$ and therefore the strict inequality $|x| < m +1$.
This also works for all positive real numbers $m$, not only for integers.
More generally, an inspection of the proof of Cauchy's bound (e.g. the one here) shows that if $$ h = \max\{ |a_0|, |a_1|, \ldots, |a_{n-1}| \} > 0 $$ then all roots of $$ x^n + a_{n-1}x^{n-1} + a_1 x + x_0 = 0 $$ satisfy the strict inequality $|x| < 1 + h$.