# Tagged Questions

96 views

### Roots of modified polynomials

Consider the following two polynomials: $$g=x^3 - x^2 - (c + 2)x + c$$ and $$h=x^3 - x^2 - cx + c$$ The roots of $h$ are $1$ and $\pm \sqrt{c}$. I am interested in obtaining the roots of $g$, ...
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### Polynomials are dense in $A_{B(0,1)}$

Let $D(0,1)$ be the disk of center 0 and radius 1 and call $A_{D(0,1)}= \{ f:\overline{D(0,1)} \rightarrow \mathbb{C} : f \text{ is continuous and } f|_{D(0,1)} \text{ is holomorphic} \}$. Can ...
543 views

### Do approximately the same polynomials have approximately the same roots? [closed]

"If $U$ is an open subset of the complex plane, then matrices $X\in\textrm{M}(n,\mathbb C)$ all of whose eigenvalues belong to $U$ make up an open subset of $\textrm{M}(n,\mathbb C)$." Trying to prove ...
300 views

### Uniform convergence of a series to exponent [closed]

I'm trying to prove that in the complex plane $\left(1+\frac{z}{n}\right)^n$ converges uniformly to $e^z$ in every closed disc $|z|\leq c$. I thought about showing the sequence as a logarithm of ...
541 views

### distinct zero points for polynomial

I met an interesting phenomenon. Suppose $f(z)=\frac{1}{p(z)}$ where p(z) is a polynomial in $\mathbb{C}[z]$. If there exists a $k \in \mathbb{N}$ and $k>1$ such that after you take $k$-th ...
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### Are there polynomials (almost) all of whose intersection numbers are divisible by some integer?

I've been playing around with some basic intersection theory, and I've wondered the following: For every two integers $n$ and $m$, and complex numbers $a_1,...,a_n$, are there polynomials ...
826 views

### When are complex polynomial maps almost surjective?

Consider a complex polynomial map $f: \mathbb{C}^n \rightarrow \mathbb{C}^n$. For $n = 1$, the fundamental theorem of algebra says that, for any $y \in \mathbb{C}$ there exists $x \in \mathbb{C}$ ...
The inverse of the Weierstrass transform expands a function as a series of Hermite polynomials $H_{n}$. There are several ways to invert the Weierstrass transform which led me to the following ...