I am currently facing the following problem:

Given a polynomial $f(x) = \sum_{s \in S_f} u_s x^s$, $f(0)\neq 0$, $\lvert S_f \rvert \leq t$ (i.e. $f$ is $t$-sparse) with $u_s$ coming as samples from i.i.d. $\mathcal{N}(0,1)$-distributed variables, bound

$$ \int_0^1 \bigg(\frac{f'(x)}{f(x)}\bigg)^2 dx = \int_0^1 \bigg(\frac{d}{dx} \log(\lvert f(x) \rvert)\bigg)^2 dx $$

in terms of the coefficients $u_s$ and the sparsity $t$, but not in terms of $\deg(f)$. It is not too difficult to bound $\int_0^1 \frac{f'(x)}{f(x)} dx = \log(\lvert f(x) \rvert) \rvert_0^1 = \log(\lvert f(1) \rvert) - \log(\lvert f(0) \rvert)$ if $f(0) \neq 0$, as we can plug in upper and lower bounds for $\log$. This makes me hopeful a bound of the squared integrand should exist too. In the worst case, a bound of the expectation of the integral with respect to the $u_s$ would also suffice, i.e. a bound for

$$ \mathbb{E}_{u_s, s\in S_f} \bigg[\int_0^1 \bigg(\frac{f'(x)}{f(x)}\bigg)^2 dx \bigg]. $$

It would take too long to explain where this comes from - I arrived at this problem looking at zero distributions of certain polynomials.

Thank you for all your ideas!


1 Answer 1


If the cardinality of the set $S_f$ is $\ge1$, then the value of the integral $$ I:=\int_0^1\Big(\frac{f'(x)}{f(x)}\bigg)^2 dx$$ will be $\infty$ with nonzero probability, because with nonzero probability the polynomial $f$ will have a non-multiple root in the interval $[0,1]$. This will then also imply that the expectation of $I$ is $\infty$.

  • $\begingroup$ Well that is slightly disappointing but I see your point. Thanks very much. I'll have to be more careful with my bounds getting there. I am a little confused as to why without the square I can bound it from above then, do you have an idea why that is? $\endgroup$
    – Azad Tasan
    Commented Apr 14, 2021 at 19:51
  • 1
    $\begingroup$ @AzadTasan : Without squaring $f'/f$, the integral will in general exist only in the principal value en.wikipedia.org/wiki/Cauchy_principal_value sense; consider e.g. $f(x)\equiv x-1/2$. When you do square $f'/f$, even the principal value may be infinite. $\endgroup$ Commented Apr 14, 2021 at 20:04

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