## Prove that if a_{n} \to Aa_{n} \to A for some real AA then a_{n}^{2} \to A^{2}a_{n}^{2} \to A^{2}

Let A \in \mathbb{R} and let (a_{n}) be a sequence of reals convergent to A. We have |a_{n}^{2} – A^{2}| = |a_{n} – A||a_{n} + A|. I am stuck at majorizing |a_{n} + A|. I know that convergent sequences are bounded, but, that kind of bounds, which takes the form \max \{ |a_{1}|, \dots, |a_{N-1}|, … Read more

## x+ϵf(x)x + \epsilon f(x) is injective when ϵ\epsilon is small

If f:R→R is differentiable on R and if |f′(t)|≤M for all t∈R, then there exists ϵ0>0 such that for all 0<ϵ≤ϵ0 , the function g(x)=x+ϵf(x) is injective. Answer Hint: Find an ϵ such that g′(x)>0. AttributionSource : Link , Question Author : bimal maiti , Answer Author : Ben Grossmann

## Polygamma reflection formula

How does one prove the polygamma reflection formula: ψ(n)(1−z)+(−1)n+1ψ(n)(z)=(−1)nπdndzncotπz Do we have to invoke the power of contour integration and kernels? I searched the web but the proof was to be found nowhere. Answer You just need to prove the reflection formula: ψ(1−z)−ψ(z)=πcot(πz) then differentiate it multiple times. In order to prove (1), let’s start … Read more

## Understanding L’hospital

I have some trouble understanding L’Hospital’s rule. Let’s say I am given the function f(x)=2x5x+6×2 I am interested in lim. (1) \lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{2}{5+12x} = \frac{2}{5} (2) \lim\limits_{x \to 0} f(x) = \lim\limits_{x \to 0} \frac{2}{5+12x} = \lim\limits_{x \to 0} \frac{0}{12} = 0 Where’s the mistake? Answer The second … Read more

## Existence of a metric space where each open ball is closed and has a limit point

Show that there exists a metric space in which every open ball is closed and contains a limit point. I think that the space {1n∣n∈N,n>0}∪{0} with the standard Euclidean metric is an answer, but it is not true open ball with center 1 is closed. Answer Z with the p-adic metric d(m,n)=p−νp(m−n) has the property … Read more

## Do we need to have a subsequence such that limk→∞‖\lim_{k\to\infty}\left\|x_{n_k}\right\|=\liminf_{n\to\infty}\left\|x_n\right\|?

Let (X,\left\|\;\cdot\;\right\|) be a normed space and (x_n)_{n\in\mathbb{N}}\subseteq X. Can we prove that there is a subsequence \left(x_{n_k}\right)_{k\in\mathbb{N}}\subseteq(x_n)_{n\in\mathbb{N}} such that \lim_{k\to\infty}\left\|x_{n_k}\right\|=\liminf_{n\to\infty}\left\|x_n\right\|\;? Answer Yes, and this holds more generally for real sequences. In fact, the limit inferior of a sequence can be defined as the least limit point of any subsequence of the sequence. To prove … Read more

## Is this a Norm?

How is the following formula calculated, assuming $a$ and $b$ are $n$-dimensional vectors? $\parallel \overrightarrow{a} – \overrightarrow{b}\parallel^2$ Answer as $(\vec{a}-\vec{b}).(\vec{a}-\vec{b})$ AttributionSource : Link , Question Author : Scholle , Answer Author : John McGee

## Convergence of improper integral ∫π60x√1−2sinxdx\int_{0}^{\frac{\pi}{6}}\dfrac{x}{\sqrt{1-2\sin x}}dx

I’m trying to determine whether the following improper integral is convergent or divergent. ∫π/60x√1−2sinxdx At first, I substituted t=π2−x and then I used 1−12×2≤cosx. But I couldn’t determine. 🙁 Second attempt, I used sinx≤x on [0,π6]. But I couldn’t determine. :-[ Could you give me some advice? Thanks in advance. Answer Since (√1−2sinx)′=−cosx√1−2sinx, using integration … Read more

## Double integral on a compact subset

Calculate: $$\int \int _D \left(6x+2y^2 \right) dxdy$$ where D is a compact subset of $\mathbb{R}^2$ enclosed by a parabola $y=x^2$ and a line $x+y=2$. How to find that, how to find the limits of integration in this case? I think then i will manage to calculate the whole integral. I think the area is the … Read more

## What is the expansion in power series of {x \over \sin x}{x \over \sin x}

How can I expand in power series the following function: {x \over \sin x} ? I know that: \sin x = x – {x^3 \over 3!} + {x^5 \over 5!} – \ldots, but a direct substitution does not give me a hint about how to continue. Answer A simple way is to manipulate the generating … Read more