New answers tagged polynomials
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 ...
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)...
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 ...
6
For exponential generating function,
$$
f_m(z) = \sum_{j=0}^\infty \frac{q_m(j)}{j!}\;z^j
$$
I get
$$
f_m(z) = \frac{1}{(1-z)^{m+1/2}(1+z)^{1/2}}
$$
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 ...
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, $\beta=\alpha$.
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 $200$.
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 \...
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