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Results tagged with polynomials
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user 290
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.
50
votes
Accepted
Given a polynomial f, can there be more than one constant c such that every root of f(x)-c i...
This is impossible by the Mason-Stothers theorem (which holds over any algebraically closed field of characteristic zero).
We want to find $f, g, h$ such that $f + g = h$ where $g$ is a constant and …
- 118k
30
votes
1
answer
2k
views
Which of the proofs of the fundamental theorem of algebra can actually produce bounds on whe...
One of the old classic MO questions is a big-list of proofs of the fundamental theorem of algebra. Here is a second big-list question about this big list:
Which of the FTA proofs can, even in prin …
- 118k
30
votes
7
answers
4k
views
When is a monic integer polynomial the characteristic polynomial of a non-negative integer m...
Suppose $P(x)$ is a monic integer polynomial with roots $r_1, ... r_n$ such that $p_k = r_1^k + ... + r_n^k$ is a non-negative integer for all positive integers $k$. Is $P(x)$ necessarily the charact …
- 118k
26
votes
Accepted
What's an example of a transcendental power series?
If $k$ has characteristic zero, then $\displaystyle e^t = \sum_{n \ge 0} \frac{t^n}{n!}$ is certainly transcendental over $k[t]$; the proof is essentially by repeated formal differentiation of any pur …
- 118k
23
votes
Accepted
Is there an integer a such that f(X)+a is irreducible in Z[X]?
Yes, and you don't need $f$ irreducible. The following irreducibility criterion suffices and shows that infinitely many $a$ work.
Lemma: Let $g(x) = a_n x^n + ... + a_0 \in \mathbb{Z}[x]$ be such th …
- 118k
23
votes
Irreducibility of polynomials in two variables
If $k$ is algebraically closed, then any two components of the projective closure of $\text{Spec } k[x, y]/(f(x, y))$ intersect by Bezout's theorem, and one can check for the existence of such points …
- 118k
22
votes
Accepted
Integer valued polynomial through some points with rational coordinates
The integer-valued polynomials have a basis (over $\mathbb{Z}$) given by the Newton polynomials
$$\displaystyle {x \choose n} = \frac{x (x - 1)\dots(x - (n-1))}{n!} …
- 118k
16
votes
Accepted
Finding all roots of a polynomial
There are also lots of specialized algorithms for finding roots of polynomials at the Wikipedia article. …
- 118k
15
votes
Surprising behaviour of polynomial that generates the series 1,2,4,8,...2^(k-1)
The second observation is true of all polynomials which interpolate an integer sequence. …
- 118k
14
votes
Accepted
Todd polynomials
.$$
As a sanity check, expanding the terms in WolframAlpha up to $t^4$ in terms of elementary symmetric polynomials / Chern classes gives
$$T_1 = \frac{c_1}{2}$$
$$T_2 = \frac{c_1^2 + c_2}{12}$$
$$T_3 …
- 118k
13
votes
Accepted
Integer polynomial (of degree >1) all of whose values are square-free
No. WLOG $A$ is irreducible. Pick a sufficiently large prime $p$ dividing $A(k)$ for some $k$ (there are infinitely many such primes, for example by the argument here). In particular pick $p$ large …
- 118k
10
votes
Accepted
No simple duplication formula for factorials?
It's equivalent to show that there is no polynomial relationship f({2n choose n}, n!) = 0. On the other hand, we know that {2n choose n} ~ 4^n/sqrt{n} asymptotically and n! grows much faster.
Tere …
- 118k
9
votes
Polynomial representing all nonnegative integers
But if $f$ is quadratic, it is a constant plus the sum of squares of two polynomials with rational coefficients and there are many integers not representable as the sum of squares of two rational numbers …
- 118k
9
votes
Factorizing polynomials of several variables (in a different perespective)
As David indicates, the truth is much more interesting: the zero set of polynomials in two or more variables generically describe interesting geometric structures called algebraic varieties, and the case … in which the polynomials factor into linear factors corresponds to the most boring structure possible: a bunch of straight lines. …
- 118k
8
votes
If a polynomial f is irreducible then (f) is radical, without unique factorization?
No, in the sense that this statement is false in a ring without unique factorization. For example, the element $2 + \sqrt{-5}$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$, and $9 \in (2 + \sqrt{-5})$ bu …
- 118k