Search found 375 matches

by Papapetros Vaggelis
Wed Oct 05, 2016 9:33 am
Forum: Real Analysis
Topic: Sequence of functions
Replies: 2
Views: 2976

Re: Sequence of functions

If \(\displaystyle{\left(H,\langle{,\rangle}\right)}\) is a Hilbert space and \(\displaystyle{x_1,...,x_n\in H}\)

such that \(\displaystyle{\langle{x_i,x_j\rangle}=0\,,\forall\,i\,,j\in\left\{1,...,n\right\}\,,i\neq j}\), then

\(\displaystyle{||x_1+...+x_n||^2=||x_1||^2+...+||x_n||^2}\).
by Papapetros Vaggelis
Sun Oct 02, 2016 12:21 pm
Forum: Number theory
Topic: Subset of natural numbers
Replies: 2
Views: 5231

Subset of natural numbers

Let \(\displaystyle{S\in\mathcal{P}(\mathbb{N})\setminus \left\{\varnothing\right\}}\) such that the set \(\displaystyle{\mathbb{N}\setminus S}\) is infinite. Suppose that \(\displaystyle{x\,,y\in S\implies x+y\in S}\). Prove that there exists \(\displaystyle{n\in\mathbb{N}\,,n\geq 2}\) which divide...
by Papapetros Vaggelis
Tue Sep 27, 2016 8:24 pm
Forum: General Topology
Topic: Question on a metric
Replies: 1
Views: 6119

Re: Question on a metric

Let \(\displaystyle{x\,,y\in S^{m-1}}\) such that \(\displaystyle{d(x,y)=\theta}\). Consider the perpendicular to \(\displaystyle{xy}\) - line from the origin. Then, \(\displaystyle{\sin\,\dfrac{\theta}{2}=\dfrac{||x-y||_{2}}{2}\iff ||x-y||_{2}=2\,\sin\,\dfrac{\theta}{2}}\). Yes, the function \(\dis...
by Papapetros Vaggelis
Thu Sep 08, 2016 1:37 pm
Forum: Complex Analysis
Topic: \( \int_{0}^{\infty}\frac{1-\cos x}{x^2}\,dx \)
Replies: 2
Views: 3624

Re: \( \int_{0}^{\infty}\frac{1-\cos x}{x^2}\,dx \)

Here is another solution using Laplace trasformations. \(\displaystyle{\begin{aligned} \int_{0}^{\infty}\dfrac{1-\cos\,x}{x^2}\,\mathrm{d}x&=\int_{0}^{\infty}\dfrac{\sin\,x}{x}\,\mathrm{d}x\\&=\int_{0}^{\infty}e^{-x}\,\dfrac{e^x\,\sin\,x}{x}\,\mathrm{d}x\\&=\int_{1}^{\infty}\mathcal{L}(e...
by Papapetros Vaggelis
Thu Sep 08, 2016 12:19 pm
Forum: Calculus
Topic: Double integral and sum
Replies: 2
Views: 3025

Re: Double integral and sum

If \(\displaystyle{y\in\left[0,1\right]}\), then \(\displaystyle{\begin{aligned}\int_{0}^{1}|x-y|^n\,\mathrm{d}x&=\int_{0}^{y}(y-x)^n\,\mathrm{d}x+\int_{y}^{1}(x-y)^n\,\mathrm{d}x\\&=\left[-\dfrac{(y-x)^{n+1}}{n+1}\right]_{0}^{y}+\left[\dfrac{(x-y)^{n+1}}{n+1}\right]_{y}^{1}\\&=\dfrac{y^...
by Papapetros Vaggelis
Wed Sep 07, 2016 12:17 pm
Forum: Algebraic Structures
Topic: On Solvable Sylow groups
Replies: 1
Views: 2689

Re: On Solvable Sylow groups

Hello. i. Firstly, there exist \(\displaystyle{q}\) - Sylow subgroups of \(\displaystyle{G}\). If \(\displaystyle{n_{q}}\) measures the \(\displaystyle{q}\) - Sylow subgroups of \(\displaystyle{G}\), then, \(\displaystyle{n_{q}\equiv 1 mod(q)}\) and \(\displaystyle{n_q\mid p}\), so \(\displaystyle{n...
by Papapetros Vaggelis
Wed Aug 10, 2016 6:00 pm
Forum: Calculus
Topic: Integration
Replies: 2
Views: 2884

Integration

Calculate, if it exists, the integral

\(\displaystyle{\int_{0}^{\infty}\dfrac{x^3}{e^x-1}\,\mathrm{d}x}\)
by Papapetros Vaggelis
Tue Aug 02, 2016 3:48 pm
Forum: Algebraic Structures
Topic: Finite Boolean Ring
Replies: 1
Views: 2601

Re: Finite Boolean Ring

Hi Nickos. Firstly, if \(\displaystyle{X}\) is a non-empty set, then, the power set \(\displaystyle{\mathbb{P}(X)}\) is a \(\displaystyle{\rm{Boolean}}\) ring with addition \(\displaystyle{A+B:=(A-B)\cup(B-A)\,,\forall\,A\,,B\in\mathbb{P}(X)}\) and multiplication \(\displaystyle{A\cdot B:=A\cap B\,,...
by Papapetros Vaggelis
Thu Jul 14, 2016 7:43 pm
Forum: Real Analysis
Topic: A limit from derivative
Replies: 1
Views: 2139

A limit from derivative

Let \(\displaystyle{f:\mathbb{R}\longrightarrow \mathbb{R}}\) be a differentiable function at \(\displaystyle{x_0=0}\) and \(\displaystyle{f(0)=0}\) . If \(\displaystyle{k\in\mathbb{N}}\), then evaluate the limit : $$\lim_{x\to 0}\dfrac{1}{x}\,\left[f(x)+f\,\left(\dfrac{x}{2}\right)+...+f\,\left(\df...
by Papapetros Vaggelis
Thu Jul 14, 2016 7:26 pm
Forum: Calculus
Topic: Series and Integral
Replies: 3
Views: 3214

Re: Series and Integral

2. Here is another solution. Firstly, \(\displaystyle{0<a_{n}=\dfrac{n}{(n+1)!}\leq \dfrac{n!}{(n+1)!}=\dfrac{1}{n+1}\,,\forall\,n\in\mathbb{N}}\) . Since, \(\displaystyle{\dfrac{1}{n+1}\longrightarrow 0\,,n\longrightarrow +\infty}\), we get : \(\displaystyle{\dfrac{n}{(n+1)!}\longrightarrow 0\,,n\...