Compare largest root of two polynomials

Question

For $1\leq a, a+1<b$ with $a,b\in\mathbb{N}$ let

• $f:=f(a,b):=\sum_{i=2}^{b-a}i,$
• $g:=g(a,b):=f(a,b)+a+b+1$
• $h:=h(a,b):=f(a,b)-(b-a-1)$

and consider the polynomial $$ p(t):=t^g-\left(\sum_{i=h}^f t^i\right). $$

I would like to show that the largest (in absolute value) root of $p(t)$ is smaller than the largest (in absolute value) root of the polynomial $$ q(t):=t^{a+b+1}-t^{a+b}-1. $$

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We can factor out $t^h$, i.e. $$ p(t)=t^h\cdot\left(t^{g-h}-\sum_{i=1}^{b-a-1}t^i -1\right), $$ so $$ p(t)=0\Leftrightarrow t^h=0\quad\text{ or }\quad t^{g-h}-\sum_{i=1}^{b-a-1}t^i-1=0 $$

Answer 1

In this proof, we only need that $a,b\in\mathbb{Z}_{\geq 0}$ with $a<b$. We have $$p(t)=t^{\frac{(b-a-1)(b-a-2)}{2}}\left(t^{2b}-\sum_{k=0}^{b-a-1}\,t^k\right)\,.$$ Observe that each of the two polynomials $p(t)$ and $q(t)$ has a unique positive real root (using Descartes's Rule of Signs, or something similar---there are a multitude of ways to show this simple result).

Write $\alpha$ and $\beta$ for the positive real roots of $p(t)$ and $q(t)$, respectively. Using Rouché's Theorem, $\alpha$ and $\beta$ are roots of $p(t)$ and $q(t)$, respectively, with the largest moduli amongst all (real or complex) roots. (Rouché's Theorem is not necessary here; this result can be proven via the Triangle Inequality for complex numbers.)

Let $B(r,z) \subseteq \mathbb{C}$ denote the open ball of radius $r>0$ centered at $z\in\mathbb{C}$. To use Rouché's Theorem for $p$, consider the disc $B(\alpha+\epsilon,0)$, where $\epsilon>0$ is arbitrary. At the boundary, $$\big|z^{2b}\big|>\left|-\sum_{k=0}^{b-a-1}\,z^k\right|$$ for all $z\in \partial B(\alpha+\epsilon,0)$, whence the polynomial $p(t)$ has the same number of roots in $B(\alpha+\epsilon,0)$ as the polynomial $t^{2b}$, which is $2b$. That is, all roots of $p(t)$ are in $\bigcap\limits_{\epsilon>0}\,B(\alpha+\epsilon,0)$, which is the same as the closed ball $\bar{B}(\alpha,0)$.
For $q$, you can do something similar, but I am using the Triangle Inequality to deal with it now. Note that $\big|q(z)\big|\geq q\big(|z|\big)$ for all $z\in\mathbb{C}$ by the Triangle Inequality. If $z$ is a complex number such that $|z|>\beta$, we get $$\big|q(z)\big|\geq q\big(|z|\big)>0\,.$$ Thus, all roots of $q$ are in the closed ball $\bar{B}(\beta,0)$.

If $b=a+1$, then clearly $\alpha=1$ and $\beta>1$ (because $q(1)<0$). From now on, suppose that $b>a+1$, which means $\alpha>1$ and $\beta>1$ (noting that $p(1)<0$ and $q(1)<0$).

Suppose contrary that $\alpha\geq \beta$. Since $\beta^{a+b}(\beta-1)=q(\beta)+1=1$, we obtain $$\alpha^{a+b}\left(\alpha-1\right)\geq \beta^{a+b}\left(\beta-1\right)=1\,.$$ Now, since $p(\alpha)=0$ and $\alpha\neq 0$, we get that $$\alpha^{2b}=\sum_{k=0}^{b-a-1}\,\alpha^k=\frac{\alpha^{b-a}-1}{\alpha-1}\,.$$ That is, $$1\leq \alpha^{a+b}(\alpha-1)=\frac{\alpha^{b-a}-1}{\alpha^{b-a}}<1\,,$$ which is a contradiction.

Answer 2

Part of the claim (namely about the maximal positive root of $p$) follows by straightforward computation:

Note that $$\tag1 q(t)=(t-1)t^{a+b}-1$$ As $q(1)=-1$ and $q(t)\to+\infty$ as $t\to+\infty$, there is clearly a root of $q$ outside $[-1,1]$, i.e., we need only consider $|t|>1$.

Let $t_\min$, $t_\max$ be the smallest and largest root of $q$, respectively. As seen above, $t_\max>1$ and $q(t)>0$ for $t>t_\max$. Using the formula for geometric series, we verify $$\tag2\frac{ p(t)}{t^{f+1}}=\frac{q(t)}{t-1}+\frac{t^{b-1}}{t-1}$$ Then using $q(t)\ge 0$ for $t\ge t_\max>1$, it follows from $(2)$ that $p(t)>0$ for all $t\ge t_\max$. In particular,

The maximal root of $p$ is strictly smaller than the maximal root of $q$.