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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.
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
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
2
votes
Examples of nice families of irreducible polynomials over Z
But I am not really sure what you want, since already Eisenstein's criterion lets you write down large parameterized families of irreducible polynomials. Can you be more specific? …
- 118k
4
votes
Decomposition result for multivariate polynomial
Suppose there exist integer polynomials $f_1, ... f_n$ with the desired property. … Since the image of an odd degree polynomial contains negative numbers, the polynomials $f_i$ must have degree at least $2$. …
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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
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 …
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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!} …
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6
votes
Which positive definite symmetric matrices have solvable characteristic polynomial?
As a subspace of the space $\mathbb{Q}^n$ of monic polynomials of degree $n$ with rational coefficients, the solvable polynomials are dense (and so in particular are not contained in an algebraic or even … To see this it suffices to observe that any such polynomial is a product of real linear or quadratic polynomials and that we can approximate these by rational linear or quadratic polynomials. …
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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 …
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5
votes
How to factorize X^n - 1 in Z/pZ?
However, in this particular problem you are probably supposed to use the fact that the divisors of $x^{p^n} - x$ over $\mathbb{F}_p$ are precisely the irreducible polynomials of degree dividing $n$. …
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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 …
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5
votes
Are plethories a theory of basis-free polynomials?
Some naive remarks. It seems to me that the simplest reason to choose the standard basis is because it exhibits the universal property of a polynomial ring.
One way to exhibit a partially basis-free …
- 118k
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 …
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8
votes
Are quotients of polynomial rings almost UFDs?
So here the failure of unique factorization is quite simple: certain polynomials in $t$ are being treated as prime which "shouldn't be." … as a rational function in $t$, avoids these anomalous primes, will have the usual prime factorization properties as a polynomial in $t$, but these prime factors will not necessarily always come from polynomials …
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1
vote
When is a monic integer polynomial the characteristic polynomial of a non-negative integer m...
For monic quadratic polynomials it's necessary and sufficient that both roots be real and one be positive with absolute value at least the other. … On the other hand polynomials such as the polynomial with roots $5, 5, 3 + 4i, 3 - 4i$ don't have this property even though they satisfy the non-negativity condition. …
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