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Here is an animation of the zeros of the first $100$ Bernoulli polynomials, produced using Maple. For the number of real roots, see OEIS sequence A094937 and references there. EDIT: As requested by Wolfgang, here is a plot of the real roots for even $n$ up to $200$.


It's false in general: consider $a=\sqrt[4]{2},b=-\sqrt[4]{2}$. Then $p_4(a,b)=p_6(a,b)=0,p_5(a,b)=2$ yet $p_3(a,b)=\sqrt{2}$.


For exponential generating function, $$ f_m(z) = \sum_{j=0}^\infty \frac{q_m(j)}{j!}\;z^j $$ I get $$ f_m(z) = \frac{1}{(1-z)^{m+1/2}(1+z)^{1/2}} $$


In case anyone is interested, here's a run-down of the history of this result and a comparison of available proofs, as far as I could uncover in a rainy evening. To be clear, Kronecker's original article, cited in P.A. Damianou's article, actually proves the following: given a monic polynomial with integer coefficients all of whose roots have norm 1, the ...


Write the equation in the form $1=x^{-1}+x^{-2}+x^{-3}+x^{-4}+cx^{-5}:=f(x) $. If $|\beta|\geqslant \alpha$, the RHS has absolute value at most $f(\alpha) =1$ with equality if and only if $|\beta|=\alpha$ and all five summands $\beta^{-1} $ etc are positive reals. That is, $\beta=\alpha$.


Everything you wanted to know (and a little more) is in this paper by John Abbott Abbott, John, Bounds on factors in $\Bbb Z[x]$, J. Symb. Comput. 50, 532-563 (2013). ZBL1295.12010.:


Of course we must assume some $a_j \ne 0$. Say $a_j$ is the one with least index. Then you want $\varepsilon$ such that $p(x) = a_n x^{n-j} + \ldots + a_j \ne 0$ for $|x| < \varepsilon$. You may use inequalities such as $|p(x)| \ge |a_j| - \sum_{k=j+1}^{n} |a_k| |x|^{k-j} \ge |a_j| - m \sum_{k=j+1}^n |a_k|$ where $m = \max(|x|^{n-j}, |x|)$. Thus $p(x)...


Slightly generalized from Ex. 1: $$ \frac{1}{1-r x} = \prod_{j=0}^\infty \sum_{i=0}^{m-1} (r x)^{i m^j} $$ for integers $m \ge 2$.


Try $T f(x) = f(-x)$, and $S = \{-1, 0, 1\}$ (or any finite subset of $\mathbb C$ that is invariant under multiplication by $-1$). EDIT: Of course, the vector space is polynomials of degree $\le n$, not $=n$. Try $Tf(x) = x^n f(1/x)$, with $S$ the union of $\{0\}$ and the $m$'th roots of unity. Still more generally, let $g(z) = (a z + b)/(c z + d)$ be a ...


$f(x)$ is the characteristic polynomial of its companion matrix $$ A = \pmatrix{0 & 0 & 0 & 0 & c\cr 1 & 0 & 0 & 0 & 1\cr 0 & 1 & 0 & 0 & 1\cr 0 & 0 & 1 & 0 & 1\cr 0 & 0 & 0 & 1 & 1\cr}$$ and $A^5$ has all entries $> 0$. Therefore by the Perron-Frobenius theorem there is a ...


$$\mathbb{E}[\text{#|roots of P|}]=\mathbb{E}(\sum_{k\in \mathbb{Z}}1_{k \text{ is a root of P}})=\sum_{k\in\mathbb{Z}}\mathbb{P}(P(k)=0)$$ For $k=0$, $\mathbb{P}(P(0)=0)=\frac{1}{(2B+1)}$. For $k\neq 0$, because the coefficients are independent (if $B$ and $d$ large) $P(k)$ should behave like a Gaussian if $d$ is large of variance $\sigma^2=\sum_{j=0}^d \...

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