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Questions in which polynomials (single or several variables) play a key role. It is typically important that this tag is combined with other tags; polynomials appear in very different contexts. Please, use at least one of the top-level tags, such as nt.number-theory, co.combinatorics, ac.commutative-algebra, in addition to it. Also, note the more specific tags for some special types of polynomials, e.g., orthogonal-polynomials, symmetric-polynomials.

5 votes
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Guess the next polynoms in the sequence (MO vs. AI :), count anticommuting $F_p$-matrices, P...

It's not entirely clear to me how much data your guesses are based on, so I present a table with calculated data and guessed polynomials based on that data and the assumption that $f(1) = f(-1) = 1$. …
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2 votes
Accepted

Simple question about 0,1-polynomials

But if there are carries then $P_a P_b$ is not a $0,1-$polynomial, and furthur multiplication by polynomials with no negative coefficients cannot restore the property of being a $0,1-$polynomial. …
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1 vote

An identity for the ratio of two partial Bell polynomials

Counterexample: consider $\ell = 3$, $m = 1$. The LHS is $$\frac{B_{7,3}({\color{red} 1}, {\color{green} 0}, {\color{blue} 1}, 0, 9)} {B_{5,3}({\color{red} 1}, {\color{green} 0}, {\color{blue} 1})} = …
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1 vote

Getting a bound on the coefficients of the factor polynomial

The review Bounds on Factors of $\mathbb{Z}[x]$ by John Abbott gives various such bounds and shows that none of them is strictly better than the others.
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1 vote

How to interpret this result modulo $(y-1)^{n+1}$?

$$\frac{\partial}{\partial x} \frac{(xy-1)^a}{(1-x)^b} = ay \frac{(xy-1)^{a-1}}{(1-x)^b} + b \frac{(xy-1)^a}{(1-x)^{b+1}}$$ from which it's easy to see either by induction or combinatorially that $$\f …
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4 votes

Vanishing of a product of cyclotomic polynomials in characteristic 2

The question title talks about cyclotomic polynomials, so consider instead $$q_{n,j}(x)=\frac{\prod_{i=1}^{n+1} \Psi_i(x)}{\prod_{i=1}^j \Psi_i(x) \prod_{i=1}^{n-j+1} \Psi_i(x)}$$ where $\Psi_k(x) = \frac …
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1 vote
Accepted

A Newton identity and the primes--the Faber partition polynomials and modular arithmetic

From $$\ln[A(x)] = \ln[1 + a_1 x + a_2 x^2 + \cdots ] = \sum_{n \geq 1} -F_n(a_1,...,a_n)\; \frac{x^n}{n}$$ and letting $A'(x) = A(x) - 1$ we have $$\begin{eqnarray*} F_n(a_1,...,a_n) &=& n [x^n] \sum …
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15 votes
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Multiple roots of polynomials with coefficients $\pm 1$

The following four Littlewood polynomials: $z^{27} + z^{26} + z^{25} + z^{24} + z^{23} - z^{22} - z^{21} + z^{20} + z^{19} + z^{18} - z^{17} - z^{16} - z^{15} - z^{14} - z^{13} - z^{12} - z^{11} - z^{ …
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2 votes
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Analytic expression for the coefficient of a multivariate polynomial

I was coming to the same conclusion that Brendan McKay posted in the comments at about the same time: the efficient way to calculate this is the direct approach $$\sum_{r,s,t,u} \binom{k}{r} \binom{k} …
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0 votes

Method to solve modular quadratic polynomial

If $q$ is prime then first solve $f(x) \equiv 0 \pmod q$ using the standard expression $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$; there are three cases for $\sqrt{y} \pmod q$: If $y$ is not a quadratic resi …
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25 votes
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How can you find an integer coefficient polynomial knowing its values only at a few points (...

This Python code implements the idea and finds two polynomials with sum of absolute values of 29: $$-2x^3 + 4x^4 + 3x^5 + 11x^6 + x^7 + 5x^8 + 3x^{10} \\ -2x^3 + 4x^4 + 3x^5 - 4x^6 + 9x^7 + 4x^8 + 3x^{ …
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14 votes
Accepted

Are there exotic polynomial bijections from $\mathbb N^d$ onto $\mathbb N$?

Lew, Morales, and Sánchez proved (Diagonal polynomials for small dimension, Math. Sys. … Applied Math. 34 (2005) 316-334 that the six polynomials which Morales and Sánchez describe are all the diagonal polynomials of dimension 4 (up to permutation of variables). …
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3 votes

Best way to introduce B-splines?

The undergrad course I took which included B-splines spent a lot of time first on Bézier curves. You might not necessarily want to spend much time on them, but I think they can motivate a variant on m …
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4 votes
Accepted

Can a polynomial be evaluated from evaluations of partial interpolations? (Or: can the uniqu...

No. The interpolating polynomial is a weighted sum $a(x) = \sum_{x_i} a(x_i) P_i(x)$ and the independence of the $a(x_i)$ from each other imposes the independence of the $P_i$ from each other, which i …
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