40 votes

Is the notion "If a polynomial has small coefficients (relative to the exponent), then it has small roots" true?

There exist estimates for the size of the largest root. The most general go back to the idea that $z$ is not a root if $|z|>R>0$ and $$ |a_n|R^n\ge |a_{n-1}|R^{n-1}+...|a_1|R+|a_0| $$ This ...
Lutz Lehmann's user avatar
20 votes

Is the notion "If a polynomial has small coefficients (relative to the exponent), then it has small roots" true?

No. Take a polynomial with all roots "very large" (according to the context), and scale it down by dividing by an even larger number. To employ the counterexample in the comments, take $p(x) ...
D S's user avatar
  • 4,288
6 votes

proving that a polynomial has at least two roots

It is interesting that you listed infinitesimals as one of the tags for this question, even though you did not mention them in the body of the question. Infinitesimals are indeed relevant here, as ...
Mikhail Katz's user avatar
  • 41.6k
5 votes

Strange behaviour of $x^2+5x+7$ under iteration

Oscar Lanzi has already given quite a nice overview over a general method to attack a certain set of polynomials trying to disprove eventual divisibility. Let's explore this idea a bit further and ...
TheOutZ's user avatar
  • 1,176
2 votes

A question about derivative and polynomial

I assume that, by saying "has a double root", you mean $\alpha$ is a root of multiplicity. Then we can assume $$f(x) = (x-\alpha)^2(x-\beta)$$ because $\deg(f) = 3$. So we get $$f(x) = (x-\...
ZYX's user avatar
  • 778
2 votes

Show that if $x=-1$ is a solution of $x^{3}-2bx^{2}-a^{2}x+b^{2}=0$, then $1-\sqrt{2}\le b\le1+\sqrt{2}$

You seem to have made a mistake during simplification and/or completing the square. Substituting $x = -1$ into the given equation, we get $$\begin{align*} -1 - 2b + a^2 + b^2 &= 0 \\[0.3cm] \...
Haris's user avatar
  • 2,898
1 vote

matrix with univariate entries: rank deficit of specialization ≤ vanishing order of determinant.

A matrix over a commutative ring is invertible whenever its determinant is, so multiplying $M$ by invertible matrices would change neither quantity. $F[X]$ is a PID, so we have the Smith normal form, ...
Amateur_Algebraist's user avatar
1 vote

proving that a polynomial has at least two roots

Let p(x) be a polynomial of exact even degree n >= 2. Assume that the highest power has a coefficient of 1, otherwise we divide by the highest coefficient. Now if we calculate the limit if p(x) for ...
gnasher729's user avatar
  • 10.1k
1 vote

Does a constant $C>0$ exist such that for $\forall\ p\in\mathbb{C}[z_1,z_2]$ we have: $\sup_{z\in rD^2}|p(z_1,z_2)|\le C\sup_{z\in D^2}|p(z_1,z_2)|$?

(I think this works. Maybe I'm tireder than I think.) Let $p_n(z) = n z_1^{2n}$. (This is holomorphic in the first coordinate and oblivious to the second coordinate. So we know that the supremum is ...
Eric Towers's user avatar
  • 66.6k
1 vote

Can I find a non-trivial cubic polynomial with rational zeroes *and* rational turning points?

$a,b,c \in \mathbb Q.$ $(x-a)(x-b)(x-c)=0$ $x^3-(a+b+c)x^2+(ab+ac+bc)x-abc=0$ $3x^2-2(a+b+c)x+(ab+ac+bc)=0$ $x=\frac{(a+b+c)\pm \sqrt{(a+b+c)^2-3(ab+ac+bc)}}{3}$ $w=a+b+c$ $z=ab+ac+bc$ $x=\frac{w \pm \...
TurlocTheRed's user avatar
  • 5,485
1 vote

Prove that if polynomial with integral coefficients has a solution in integers, then the congruence is solvable for any value of modulus $m$.

Can't you just say $F(x_1,\ldots, x_n) = 0$ $\implies$ $F(x_1, \ldots, x_n) \equiv_m 0$ $\forall m \in \mathbb{N}$, as $0 \equiv_m 0$ for all such $m$?
Mike's user avatar
  • 20.2k
1 vote

$\ \forall x_1,x_2,...,x_n \in \mathbb{R} (x_i\not=x_j)$ in the range of $[-1,1]$ prove:$\sum_{i=1}^{n}\frac{1}{\Pi_{k\not=i}|x_k-x_i|}\ge2^{n-2}$

One may follow a proof of Chebyshev's $\|\cdot\|_\infty$-minimality. Suppose $j<k\implies x_j<x_k$ (w.l.o.g.). Denote our sum $S=\sum_{k=1}^n 1/\prod_{j\neq k}|x_k-x_j|$. Assume $S<2^{n-2}$, ...
metamorphy's user avatar
  • 38.9k

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