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<d} for which

\displaystyle A|\zeta_m| <A^{-1}|\zeta_{m+1}|

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.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s