25
votes
Accepted
Irreducibility of the polynomial $x^n+5x+3$ over $\mathbb{Q}$
This should be provable using the results of On the irreducibility of the non-reciprocal part of polynomials of the form $f(x) x^n+g(x)$ by Filaseta-Li-Patane-Skabelund Acta Arithmetica 196 (2020), ...
10
votes
Accepted
Polynomials such that $|p(z)|\leq p(|z|)$
The claim holds with $p$ replaced by any positive real analytic function on $(0,+\infty)$ that is not a monomial function $z \mapsto C z^\alpha$ (which also satisfies $|p(z)| \leq p(|z|)$, but for ...
9
votes
What relationship is there between repeated roots of discriminants and orders of roots of the original polynomials?
(1) There is the following indirect explanation:
For a generic curve neither of these phenomena happen - the discriminant has no repeated roots and the branch points all have ramification index two. ...
9
votes
Accepted
What relationship is there between repeated roots of discriminants and orders of roots of the original polynomials?
$\def\P{{\mathbb{P}}}
\def\A{{\mathbb{A}}}
\newcommand{\O}{\mathcal{O}}
\DeclareMathOperator{\Disc}{Disc}$I think the story goes like this. The multiplicity of a zero of the discriminant counts ...
8
votes
Polynomials such that $|p(z)|\leq p(|z|)$
The result proved in the answer of @Terry Tao is actually due to Teichmuller, see, for example
L. Ahlfors, Conformal invariants, section 3-4.
For an interesting related result, see
MR0344465
Boĭčuk, V....
7
votes
Accepted
On the derivative of the Bernstein polynomial
It's equation (2.1.9) of Generalized Bernstein Polynomials by Ceren Ustaoğlu (2014). Note that the stepsize is chosen as $1/n$, so
Ustaoğlu cites a source: George M. Phillips, Interpolation and ...
5
votes
On the derivative of the Bernstein polynomial
I may be wrong, but it seems to me that $(B_n g)$ is a 1D Bézier curve. If you write $P_k$ instead of $g(k)$, Wikipedia show its derivative.
Update in response to comment: Bézier curves is a standard ...
4
votes
Rational functions of order $3$
Since you ask, there has been much research on the analogous question of elements of finite order in the Nottingham group, which is the group of power series under composition:
$$ N(\mathbb K) = \left\...
4
votes
Accepted
Solving a recursion for polynomials defined by a matrix product
Your polynomial is precisely
$$
\sum_{k_1+2k_2+\cdots+nk_n=n}\binom{k_1+\cdots+k_n}{k_1,\ldots,k_n}X_1^{k_1}\cdots X_n^{k_n}.
$$
The proof is straightforward by induction: you have
$$
p_n(X)=\sum_{i=...
3
votes
Accepted
Bound for the $n$-th derivative of a proper rational function with no poles on the right half-plane
I will address the second, more general question (the answer implies the answer to the first one). Let us write the rational function as $p/q$, and assume for simplicity that zeros of $q$ are simple, ...
3
votes
Number of real roots of 0,1 polynomial
Using a different approach outlined below, I found the two dual polynomials of order $\nu=41$,
\begin{align}
P_7(x)&=x+x^{2}+x^{4}+x^{7}+x^{11}+x^{12}+x^{15}+x^{17}+x^{18}\\
&\quad{}+x^{...
2
votes
Why is this polynomial factorizable?
Here is an idea that should lead to a human way to verify this.
Take a triple $(a_1,a_2,a_3)$ for which $a_1+a_2+a_3=0$. For example, $a_1=a=-a_2$, and $a_3=0$ or $a_i=a\cdot\omega^i$ where $\omega^3=...
1
vote
Accepted
Where to find or how to prove that the ratio of two Bernoulli polynomials is increasing?
This answer is a slight revision of the proof of Proposition 1 in Section 4 of the forthcoming paper:
Gui-Zhi Zhang, Zhen-Hang Yang, and Feng Qi, On normalized
tails of series expansion of generating ...
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