Stack Exchange Network

Stack Exchange network consists of 174 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.

Visit Stack Exchange
12

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 $200$.


7

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 $p_3(a,b)=\sqrt{2}$.


6

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


5

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 ...


4

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, $\beta=\alpha$.


3

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.:


3

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)...


2

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 $m \ge 2$.


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 ...


2

$f(x)$ 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 ...


2

$$\mathbb{E}[\text{#|roots of P|}]=\mathbb{E}(\sum_{k\in \mathbb{Z}}1_{k \text{ is a root of P}})=\sum_{k\in\mathbb{Z}}\mathbb{P}(P(k)=0)$$ 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 \...


Only top voted, non community-wiki answers of a minimum length are eligible