## Computing epsilon-closure. What does $E(n) \leftarrow \{n\};$ and $E(p)$ mean?

I’m currently reading “Engineering a Compiler” book. In the chapter that explanes computing epsilon-closure there is listed the following algorithm:

But I couldn’t understand what does $$E(n)$$ and $$E(p)$$ mean. I know that it has something to do with sets.

## f(x) is invertible polynomial function of degree ‘n’ {n≥3} then f”(x) = 0 has exactly ‘n – 2’ distinct real roots if

A)f′(x)=0 has n−1/2 distinct real roots B)f′(x)=0 has n−1 distinct real roots C)all the roots of f′(x)=0 are distinct D)none of these

## A finite group of order $n$ has exponent $n$.

Definitions: The order $$|G|$$ of a group $$G$$ is the number of elements of $$G$$. The exponent of a group is an integer $$n$$ such that $$x^n = e$$ for all $$x\in G$$ ($$e$$ is the neutral element).

Particularly, the definition doesn’t say it’s the least integer with the property. So in this definition the 4-group $$V$$ has exponent $$4$$ as well as exponent $$2$$.

But I’m having a hard time proving the following statement:

Lagrange’s Theorem shows that a finite group $$G$$ of order $$n$$ has exponent $$n$$.

I know it must be very simple because it’s not even stated as a corollary of Lagrange’s.

## biggest common divisor of the polynomials $(x+1)^{4n+3}+x^{2n},n\in \mathbb N^*$ and $x^3-1$

How can you find the biggest common divisor of the polynomials $$(x+1)^{4n+3}+x^{2n},n\in \mathbb N^*$$ and $$x^3-1$$? The answers are $$x^3-1,x-1,x^2+x+1,prime,(x+1)^{4n+3}+x^{2n}$$. Only one is correct.

## If $(N,\|.\|)$ is a non trivial vector space with norm, then the diameter of the unit sphere is $\ge2$

It is easy to see that $$\delta(B_1(0))\le 2$$ because $$\|x-y\|\le\|x\|+\|-y\|=\|x\|+|-1|\|y\|=\|x\|+\|y\|<1+1=2$$ so $$\sup\{\|x-y\|:x,y\in B_1(0)\}\le 2$$

But to show that $$\delta(B_1(0))\ge 2$$ we chose $$a\in B_1(0)\backslash\{0\}$$ and $$s\in[0,1)$$ and observe that $$\|s\frac{a}{\|a\|}\|=s$$ $$\implies s\frac{a}{\|a\|},-s\frac{a}{\|a\|}\in B_1(0)$$ and $$\|s\frac{a}{\|a\|}-(-s\frac{a}{\|a\|})\|=2s$$

But why does that imply that $$\delta(B_1(0))\ge2s$$?

This is how I think I understand it: the diameter of $$B_1(0)$$ is greater than that of $$B_s(0)$$ for all $$s\in[0,1)$$ because we it contains it. And the diameter of $$B_s(0)$$ is $$2s$$ for each $$s$$, which is true because there exists a sequence $$(\epsilon_n\frac{a}{\|a\|},-\epsilon_n\frac{a}{\|a\|})$$ of pairs of points, with $$0\le\epsilon_n and $$\lim\limits_{n\rightarrow\infty}\epsilon_n=s$$ such that it is always contained in $$B_s(0)$$ and $$\lim\limits_{n\rightarrow\infty}d(\epsilon_n\frac{a}{\|a\|},-\epsilon_n\frac{a}{\|a\|})=2s$$

I’m not sure if I got it right…

## Simple way to generate (or characterize) real $N \times N$ matrices $K$ satisfying $(-1)^{|A|}\det(K-I_A) \ge 0,\;\forall A \subseteq [\![N]\!]$

Let $$N$$ be a large positive integer.

# Question

What is a simple way to generate (or characterize) real $$N \times N$$ matrices $$K$$ such that $$(-1)^{|A|}\det(K-I_A) \ge 0,\;\forall A \subseteq [\![N]\!],$$ where $$A$$ is masked version of the $$N$$-by-$$N$$ identity matrix defined by $$(I_A)_{i,j}=1$$ if $$i=j \in A$$ and $$0$$ otherwise.