# Polynomials with well separated roots

The following proposition shows how roots of polynomials and their derivatives are somewhat attracted to each other in a rough sense. They exhibit some sort of clustering at the ends of intervals of reasonable logarithmic length.

Proposition 1 Let ${p(t)}$ be a real polynomial of degree ${d}$ s.t. ${p(0)=0}$, and call ${\zeta_i}$, for ${i=1,...,d}$ its ${d}$ complex roots – which we assume are all distinct – ordered by increasing modulus (${\zeta_1=0}$ thus). Let ${A>C_d>0}$ be big (with ${C_d}$ constant depending only on the degree of ${p}$) and suppose there exists a index ${m for which

$\displaystyle A|\zeta_m|

holds. Then

• when ${A|\zeta_m| < |t| < A^{-1}|\zeta_{m+1}|}$ one has

$\displaystyle \left(1-A^{-1}\right)^{d-1}\left(\prod_{i=m+1}^{d}|\zeta_i|\right)|t^m|\leq |p(t)| \leq \left(1+A^{-1}\right)^{d-1}\left(\prod_{i=m+1}^{d}|\zeta_i|\right)|t^m|;$

• there exists a constant ${c_d>0}$ depending only on the degree of ${p}$ such that on the same interval as in the previous point it holds

$\displaystyle \left|t\frac{p^\prime (t)}{p(t)}\right|>c_d.$

The first conclusion of the theorem is telling us that in between – but far enough from – well separated roots (${\zeta_m}$ and ${\zeta_{m+1}}$ are separated by a factor of at least ${C_d^2}$), ${|p(t)|\sim |t|^m}$, i.e. the polynomial behaves essentially like a monomial ${t^m}$, whose degree can be less that the degree of ${p}$. The second conclusion is telling us that ${|p'(t)/p(t)|>0}$ on the interval ${A|\zeta_m| < |t| < A^{-1}|\zeta_{m+1}|}$, that is ${|\left(\log p(t)\right)'|>0}$, so that ${\log p(t)}$, and therefore ${p(t)}$, is either strictly increasing or strictly decreasing. In particular ${|p(t)|}$ is strictly increasing on ${[A|\zeta_m|,A^{-1}|\zeta_{m+1}|]}$ and strictly decreasing on ${[-A|\zeta_m|,-A^{-1}|\zeta_{m+1}|]}$.

Proof: As for the first point, write ${p(t)=\prod_{i=1}^{d}{(t-\zeta_i)}}$ and observe that when ${i\leq m}$ the term ${|t-\zeta_i|}$ is essentially ${|t|}$ because the roots are smaller than that by a huge factor, and when ${i>m}$ the opposite happens, i.e. ${|t-\zeta_i|\sim |\zeta_i|}$. To be more precise, when ${2\leq i\leq m}$ one has ${|\zeta_i|\leq A^{-1}|t|}$ and thus

$\displaystyle \left(1-A^{-1}\right)|t|\leq ||t|-|\zeta_i||\leq |t-\zeta_i| \leq |t|+|\zeta_i|\leq \left(1+A^{-1}\right)|t|,$

while if ${i>m}$ it is ${A|t|\leq |\zeta_i|}$ and thus

$\displaystyle \left(1-A^{-1}\right)|\zeta_i|\leq ||t|-|\zeta_i||\leq |t-\zeta_i| \leq |t|+|\zeta_i|\leq \left(1+A^{-1}\right)|\zeta_i|.$

Taking the product of all factors one recognizes the upper and lower terms above.

Now, since ${p'(t)/p(t)=(\log p(t))'}$ one has

$\displaystyle \left|\frac{p^\prime (t)}{p(t)}\right|=\left|\sum_{i=1}^{d}{\frac{1}{t-\zeta_i}}\right|.$

Before going further, a little heuristics about the sum. The terms in the sum for ${i\leq m}$ are now of order of magnitude ${\sim \frac{1}{|t|}}$, while the terms for ${i>m}$ have order of magnitude ${\frac{1}{|\zeta_i|}\lesssim \frac{1}{A|t|}}$. Since ${A}$ is a huge constant (and we can set how big it is) one should expect that the first ${m}$ terms dominate on the last ${d-m}$, which of course it is. So, assume wlog that ${t>0}$,

$\displaystyle \left|\sum_{i=1}^{d}{\frac{1}{t-\zeta_i}}\right|\geq \left|\sum_{i=1}^{m}{\frac{1}{t-\zeta_i}}\right|-\left|\sum_{i=m+1}^{d}{\frac{1}{t-\zeta_i}}\right|$

and each term of the second sum contributes with at most

$\displaystyle \frac{1}{|t-\zeta_i|}\leq \frac{1}{|\zeta_i|-|t|}\leq \frac{1}{A|t|-|t|}=\frac{1}{(A-1)|t|},$

while the first term being a modulus is certainly bigger or equal that its real part, and then for each term

$\displaystyle \Re\frac{1}{t-\zeta_i}=\frac{t-\Re\zeta_i}{|t-\zeta_i|^2}\geq \frac{\left(1-A^{-1}\right)t}{\left(1+A^{-1}\right)^2|t|^2},$

so in the end the logarithmic derivative is bounded from below by

$\displaystyle \left(m\frac{\left(1-A^{-1}\right)}{\left(1+A^{-1}\right)^2}-(d-m)\frac{1}{A-1}\right)\frac{1}{|t|},$

and the constant can be made positive – bigger than a constant ${c_d>0}$ – for every ${m}$ by taking ${A}$ big enough (the first term in ${A}$ inside the brackets is ${\gtrsim 1}$ and the second one is ${\sim A^{-1}}$). $\Box$
Notice that if ${0<|t|\leq A^{-1}|\zeta_2|}$ then the logarithmic derivative is still bounded from below, and that tells us ${p}$ is monotonic on ${[-A^{-1}|\zeta_2|,A^{-1}|\zeta_2|]}$, since ${\zeta_1=0}$ is a simple root (thus ${p}$ is monotonic through ${0}$). On the other hand if ${|t|>A|\zeta_d|}$ then ${|p(t)|\sim |t|^d}$, as it is to be expected, and moreover the logarithmic derivative is again bounded from below, so that ${|p(t)|}$ is increasing on ${[A|\zeta_d|,+\infty[}$ and decreasing on ${]-\infty, A|\zeta_d|]}$.

One last comment about the structure that the proposition yields us. Between intervals of the form ${[A|\zeta_i|,A^{-1}|\zeta_{i+1}|]}$ there are intervals of the form ${[A^{-1}|\zeta_i|,A|\zeta_i|]}$ that contain at least a root and where the polynomial doesn’t behave so nicely. We can call them Type 1 intervals, and call Type 2 intervals the others. This leads to a partition of the positive reals (that extends to all reals by taking the symmetric intervals, reflected with respect to 0) into Type 1 and Type 2 intervals, given by

$\displaystyle [0, A^{-1}|\zeta_2|]\cup[A^{-1}|\zeta_2|,A|\zeta_2|]\cup[A|\zeta_2|,A^{-1}|\zeta_3|]\cup[A^{-1}|\zeta_3|,A|\zeta_3|]\cup \ldots,$

where some (or all, if the roots are concentrated around ${0}$) of the intervals ${[A|\zeta_i|,A^{-1}|\zeta_{i+1}|]}$ might be empty (we adopt the convention that if ${a>b}$ then ${[b,a]}$ is a proper interval but ${[a,b]=\emptyset}$). They are not empty when

$\displaystyle \frac{|\zeta_{i+1}|}{|\zeta_i|}> A^2\geq C_d^2,$

but this doesn’t prevent ${[A|\zeta_i|,A^{-1}|\zeta_{i+1}|]}$ from having a very small measure, so in some applications it might be better to assume that Type 2 intervals have the property

$\displaystyle \frac{|\zeta_{i+1}|}{|\zeta_i|}>(2A)^2\geq 4 C_d^2,$

so that the measure of ${[A|\zeta_i|,A^{-1}|\zeta_{i+1}|]}$ is bounded from below by ${3A|\zeta_i|}$, and all the other intervals are of Type 1 instead.