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), ...
Will Sawin's user avatar
  • 135k
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 ...
Terry Tao's user avatar
  • 108k
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. ...
Will Sawin's user avatar
  • 135k
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 ...
Jack Huizenga's user avatar
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....
Alexandre Eremenko's user avatar
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 ...
Carlo Beenakker's user avatar
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 ...
Pablo H's user avatar
  • 153
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\...
Joe Silverman's user avatar
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=...
Vladimir Dotsenko's user avatar
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, ...
Alexandre Eremenko's user avatar
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^{...
Fred Hucht's user avatar
  • 2,705
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=...
Sándor Kovács's user avatar
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 ...
qifeng618's user avatar
  • 836

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