I must use the $ \epsilonN$ 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.
Tag: \{\epsilon
Handling epsilon productions in recursive descent parsing
I am working on a recursive descent parser for lambda calculus. In my grammar, after removing leftrecursion, 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 leftassociate 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
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 $ ss’ \lt \epsilon$ , by the below proof:
Let us assume that $ ss’ \ne 0 $ , let $ ss’=k.\epsilon, k\lt 1$ , then $ s = s’+k.\epsilon \implies s \epsilon = s’+(k1).\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’+(k1)\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.
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
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.
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.
Use of the reverse triangle inequality in epsilon proof
I’m new to analysis and trying to prove something about a converging series.
Now I want to get from $ x_{n}\bar{x} < \frac{\bar{x}}{2}$ to the following statement $ x_{n} > \frac{\bar{x}}{2}$ using the reverse triangle inequality, but I just don’t seem to get it right.
As for as my knowledge goes, the reverse inequality states that $ ba \leq ba$ . Any suggestions on how to apply this?
PS: it is a bout a converging sequence $ x_{n}$ with limit $ \bar{x}$ .
Given least upper bound $\alpha$ for $\{\ f(x) : x \in [a,b] \ \}$, $\forall \epsilon > 0 \ \exists x$ s.t. $\alpha – f(x) < \epsilon$
I can’t figure out how all of this follows. Taken from Ch.8 of Spivak’s Calculus.
If $ \alpha$ is the least upper bound of $ \{\ f(x) : x \in [a,b] \ \}$ then, $ $ \forall \epsilon > 0 \ \exists x\in [a,b] \ \ \ \ \ \ \ \alpha – f(x) < \epsilon$ $ This, in turn, means that $ $ \frac{1}{\epsilon} < \frac{1}{\alpha – f(x)}$ $
Anticoncentration: upper bound for $P(\sup_{a \in \mathbb S_{n1}}\sum_{i=1}^na_i^2Z_i^2 \ge \epsilon)$
Let $ \mathbb S_{n1}$ be the unit sphere in $ \mathbb R^n$ and $ z_1,\ldots,z_n$ be a i.i.d sample from $ \mathcal N(0, 1)$ .
Question
Given $ \epsilon > 0$ (may be assumed to be very small), what is a reasonable upper bound for the tail probability $ P(\sup_{a \in \mathbb S_{n1}}\sum_{i=1}^na_i^2z_i^2 \ge \epsilon)$ ?
Observations

Using ideas from this other answer (MO link), one can establish the nonuniform anticoncentration bound: $ P(\sum_{i=1}^na_i^2z_i^2 \le \epsilon) \le \sqrt{e\epsilon}$ for all $ a \in \mathbb S_{n1}$ .

The uniform analogue is another story. May be one can use covering numbers ?