I am studying over my notes, and there is something I don’t understand about $$e_m$$. We represent the floating point numbers as $$1.d_1d_2…d_t \times \beta^e$$. Now, my professor defines $$\epsilon_m$$ as the smallest $$x$$ such that $$fl(1+x) > 1$$. But then she writes

$$\epsilon_m = \beta^{1-t} \hspace{1cm} \text{ (for “chopping”)}$$ and

$$e_m = \dfrac 12 \beta^{1-t} \hspace{1cm} \text{ (for rounding)}$$ However, I think these should be $$\beta^{-t}$$ and $$\dfrac 12 \beta^{-t}$$, respectively.

The number $$1$$, and the next number right after, are

$$1.00 \cdots 00 \times \beta^0$$ and $$1.00 \cdots 01 \times \beta^0$$ where the $$1$$ is in the $$t^{th}$$ decimal place. Therefore their difference is $$\beta^{-t}$$ so if $$x = \beta^{-t}$$, then $$1+x$$ is itself a floating point number, so $$fl(1+x) = 1+x>1$$.

Am I wrong, or are the notes wrong?

Thank you very much.

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## Let M be an $\epsilon$-NFA and let $S\subseteq Q$. Prove $\epsilon (S) = \epsilon (\epsilon (S))$

Let M be an $$\epsilon$$-NFA and let $$S\subseteq Q$$. Prove $$\epsilon (S)= \epsilon (\epsilon (S))$$.

I would like to prove this by contradiction but I don’t know if my idea is correct.

Definition of $$\epsilon -closure$$: $$\epsilon : 2^Q \rightarrow 2^Q$$

a) $$S \subseteq \epsilon (S)$$ Base case

b) If $$q \in \epsilon (S)$$ then $$\delta(q,\epsilon )\subseteq \epsilon (S)$$ Recursive case

c) and nothing else is in $$\epsilon (S)$$

Prove i) That $$\epsilon (S) \subseteq \epsilon (\epsilon (S))$$.

By contradiction, suppose $$\exists x \in \epsilon (S)$$ such that $$x \notin \epsilon (\epsilon (S))$$.

Using definitions of $$\epsilon (S)$$ , we know $$S \in \epsilon (S)$$. So $$x \in S \in \epsilon (S)$$, and $$\delta(x,\epsilon ) \subseteq \epsilon (S)$$.

Using definitions again, we know $$S \in \epsilon (S)$$, so $$x \notin \epsilon (\epsilon (S))\notin \epsilon (S)$$. Also $$\delta (x, \epsilon) \nsubseteq \epsilon (\epsilon (S))$$

We have a contradiction because $$x \notin \epsilon (S)$$ and we said $$x \in \epsilon (S)$$.

Therefore i) is true.

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## Show that $L:=\{(a^{k}b)^{i}|i,k \epsilon \mathbb{N}_{+} \}$ is context-sensitve. (With context-sensitive/noncontracting grammar)

I am studying for an upcoming exam and this is an old exam question from two years ago (all exams were made available through our lecturer):

Show that $$L:=\{(a^{k}b)^{i}|i,k \epsilon \mathbb{N}_{+} \}$$ is context-sensitve.

I could easily construct a LBA for this language. But since the notation/construction of LBAs weren’t really explained, I would have to define it in the exam.

So I assume this task was expected to be done by constructing a context-sensitive grammar.

NOTE: In our lecture the definition of a context-sensitive grammar was in fact the definition of a noncontracting grammar. So any rule like this x -> y is allowed if |x| <= |y|

So this is allowed:

aAb -> bXaa 

My best idea goes like this:

S  -> AB B  -> bB | b A  -> CA | C Cb -> abC Ca -> aC 

so to generate aaabaaabaaab I do this:

S AB AbB AbbB Abbb CAbbb CCAbbb CCCbbb (let all C-Variables run through the word and leave an 'a' before every 'b') CCabCbb CCababCb CCabababC ... aaabaaabaaabCCC 

But I can’t make all the ‘C’-Variables disappear.

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## How do I use the epsilon N definition to prove this?

I must use the $$\epsilon-N$$ definition to prove $$\lim_{n\to\infty}\dfrac{a_1+a_2 \hspace{3px}+\hspace{3px}…\hspace{3px}+\hspace{3px}a_2}{n} = L$$ when $$\displaystyle \lim_{n\to \infty} a_n = L$$. Does anybody have any pointers as to how to start this problem? I’m stumped.

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## Handling epsilon productions in recursive descent parsing

I am working on a recursive descent parser for lambda calculus. In my grammar, after removing left-recursion, I am left with the following two productions:

# APPLICATION  -> ATOM APPLICATION' # APPLICATION' -> ATOM APPLICATION' | ε        

Both APPLICATION and APPLICATION’ correspond to the same type of node in the AST. How should I handle parsing the second production? I know that choosing between two alternatives for a production is performed by looking up the next token in buffer, but the epsilon production confuses me.

EDIT:

Here is current implementation (Ruby) that fails to left-associate applications.

require_relative 'token.rb' require_relative 'ast.rb'  class Parser     def self.call(tokens)         new(tokens).send(:parse_expr)     end      def self.to_proc         method(:call).to_proc     end      private         def initialize(tokens)             @tokens = tokens             @l = -1         end          def parse_expr             if lookahead.type == :tklambda                 return ExprNode.new([parse_abstraction], @l)             else                 return ExprNode.new([parse_application], @l)             end         end          def parse_abstraction             match_token(:tklambda)             id = parse_identifier             match_token(:tkdot)             expr = parse_expr             return AbstrNode.new([id, expr], @l)         end          def parse_identifier             id_token = match_token(:tkid)             val = id_token.value             return IdNode.new(nil, @l, val)         end          def parse_application             left_child = parse_atom             first_atom = [:tklparen, :tkid] # FIRST(ATOM)             # application -> atom application'             while !lookahead.nil? and first_atom.include?(lookahead.type)                 return AppNode.new([left_child, parse_application], @l)             end             # application' -> atom application' | ε             return left_child         end          def parse_atom             if lookahead.type == :tklparen                 match_token(:tklparen)                 expr = parse_expr                 match_token(:tkrparen)                 return AtomNode.new([expr], @l)             else                 return AtomNode.new([parse_identifier], @l)             end         end          def match_token(type = nil)             if !type.nil? and lookahead.type != type                 raise "Unexpected token at #{@l + 1}. Expected: #{type}, "\                       "got: #{lookahead.type}."             end             @l += 1             return @tokens[@l]         end          def lookahead             return @tokens[@l + 1]         end end 
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## Asymptotic Bound on Minimum Epsilon Cover of Arbitrary Manifolds

Let $$M \subset \mathbb{R}^d$$ be a compact smooth $$k$$-dimensional manifold embedded in $$\mathbb{R}^d$$. Let $$\mathcal{N}(\epsilon)$$ denote the size of the minimum $$\epsilon$$ cover $$P$$ of $$M$$; that is for every point $$x \in M$$ there exists a $$p \in P$$ such that $$\| x – p\|_{2}$$.

Is it the case that $$\mathcal{N}(\epsilon) \in \Theta\left(\frac{1}{\epsilon^k}\right)$$? If so, is there a reference?

## Proving equivalence between $\epsilon$ based & $lub$ definition of supremum.

Based on $$\epsilon$$ have a new definition of supremum:

Let there be a nonempty set $$X$$ with supremum $$s$$, then $$X\cap(s – \epsilon, s]\ne \emptyset, \,\, \forall \epsilon\gt 0$$.

The conventional definition is given by:

Let $$X$$ be a nonempty set of real numbers. The number $$s$$ is called the supremum of $$X$$ if $$s$$ is an upper bound of $$X$$ and $$s \le y$$ for every upper bound of $$X$$.

Let, the conventional definition be denoted by ‘Def. 1’, while the new definition by ‘Def. 2’.

Have two questions below. I need help in attempting them.

Q. 1 : Need show that the two definitions are equivalent by proving the following two conditional statements:

(i) If $$s = sup(X)$$, as given by Defn. 1, then $$s$$ is the supremum, as given by Defn. 2. Here, assume that Defn. 1 holds, and use this assumption to prove that Defn. 2 holds.

Let $$s’$$ is supremum as per Defn. 2. Also, the relation between the magnitudes of $$s,s’$$ is unknown, & need be established.

$$s$$ will have set $$X$$ elements in the range $$(s-\epsilon, s]$$ if $$s-s’ \lt \epsilon$$, by the below proof:

Let us assume that $$s-s’ \ne 0$$, let $$s-s’=k.\epsilon, k\lt 1$$, then $$s = s’+k.\epsilon \implies s -\epsilon = s’+(k-1).\epsilon \implies s -\epsilon \lt s’$$.

$$s-\epsilon\lt s’\implies \exists x \in X: X\cap (s – \epsilon, s]\ne \emptyset$$.
But, Def. 2 can take any $$\epsilon\gt 0$$ to ensure $$\exists x \in X: X\cap (s’ – \epsilon, s’]\ne \emptyset$$.
So, if Def. 1 is to have ability to take any $$\epsilon\gt 0$$, need the lower bound of $$(s – \epsilon, s]$$ to equal at least to $$s’ – \epsilon$$.
But, $$s – \epsilon= s’+(k-1)\epsilon \ge s- \epsilon, \forall k, 0\lt k\lt 1$$.
So, the only possible value is $$k=0$$ to have the lower bound of $$(s – \epsilon, s]$$ equal to $$s’ – \epsilon$$.

But, by this cannot impose any restriction on the upper bound $$s$$ (of Def. 1) to equal $$s’$$ (of Def. 2).

(ii) If $$s = sup(X)$$, as given by Defn. 2, then $$s$$ is the supremum, as given by Defn. 1. Here, assume that Defn. 2 holds, and use this assumption to prove that Defn. 1 holds.

Let us modify for consistency with part (i) sake, $$s$$ replaced by $$s’$$.

If Defn. 2 holds, then the upper bound of the interval is bounded by $$s’$$, which is also the last element that can possibly be (if, $$s’\in X$$) in $$X$$. For Defn. 1 to hold, the upper bound must then be the same as the upper bound of Defn. 2, i.e. $$s’$$.

Q. 2: What is the practical significance of showing that these two definitions are logically equivalent?

The step (i) of showing that if Defn. 1 holds, then Defn. 2 holds, leads to having the lower bound of $$(s – \epsilon, s]=s’ – \epsilon$$.

The step (ii) of showing that if Defn. 2 holds, then Defn. 1 holds, leads to having the upper bound of $$(s – \epsilon, s]=s’$$per bound implies the max. value is $$s=1$$ at $$n= \infty$$. All possible values of $$n$$ are covered in the interval $$(s-\epsilon,s]$$ with $$n=\infty$$ at the upper bound.

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## Proving there is an eigenvalue $\lambda$ for which $|\lambda – b_{jj}| < \epsilon \sqrt{n}$

Let $$A$$ be an $$n\times n$$ real symmetric matrix. By applying Jacobi’s method, suppose we have generated an orthogonal matrix $$R$$ and a symmetric matrix $$B$$ such that the equality

$$B = R^{T}AR$$

holds. Moreover, suppose the inequality $$|b_{ij}| < \epsilon$$ holds for all $$i \neq j$$.

Show that for each $$j = 1, 2, \ldots, n$$, there is at least one eigenvalue $$\lambda$$ of $$A$$ such that $$|\lambda – b_{jj}| < \epsilon \sqrt{n}$$ holds.

This is an exercise that I am doing to study for my final exam. So, I’ve just recently learned Jacobi’s method, and I know that the eigenvalues and eigenvectors are related to the matrices $$B$$ and $$R$$; however, I have no idea how to use those results to prove an inequality. I also have no idea how to get the $$\sqrt{n}$$ term in there. I would greatly appreciate any help in this exercise.

Thanks

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## kPDA handling multiple epsilon transtions

I’m assigned to build a kPDA with 2 stacks that handles {w#w, where w is a string of (0,1)*}. I understand the # delineates the two strings, but I’m unsure of the logic when popping off stacks with epsilon. I asked my professor, verbally, if this works and he told me no, but I can’t think of any other way. Posted on Categories proxies

## Solving $u_{\epsilon \eta} = \frac{ u_\epsilon – u_eta}{4 (\epsilon – \eta) } .$

I’m trying to solve the following PDE

$$y^2 u_{xx} – u_{yy} = 0.$$

I’ve found its canonical form as $$u_{\epsilon \eta} = \frac{ u_\epsilon – u_eta}{4 (\epsilon – \eta) } .$$

However, I’m having trouble in solving this PDE. I’ve tried solving with mathematica, and that gave me

$$u = \frac{ x + y^3}{ 3} ,$$ but, I need to know how to obtain such a result from the canonical form of the given PDE.

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