Hot answers tagged

3

I will evaluate this integral indirectly by recognizing it for what it is: an integral over the surface of a unit sphere. First of all, for simplicity sake, I am going to replace your notation so it looks like we are integrating over solid angle: $$\int_0^{2 \pi} d\phi \, \int_0^{\pi} d\theta \, \sin{\theta} \, e^{\sin{\theta} (\cos{\phi}-\sin{\phi})} $$ ...


2

If $(X,d)$ denotes a metric space then we can construct a metric $d'$ on $X$ by stating: $$d'(x,y)=\min(d(x,y),1)$$ This function can be shown to be a metric and induces the same topology as the original $d$. However every subset of $X$ is bounded with respect to the constructed metric.


2

For completeness sake, the following are useful. For $|k|<1$ we have: $\sum\limits_{n=0}^\infty k^n = \frac{1}{1-k}$ Deriving each side with respect to $k$, we get: $\sum\limits_{n=0}^\infty nk^{n-1} = \frac{1}{(1-k)^2}$ Multiplying by $k$ yields $\sum\limits_{n=0}^\infty nk^n = \frac{k}{(1-k)^2}$ Deriving the expression by $k$ yields ...


2

The easiest way is to use the Weirestrass M-test. The $n$-th term is uniformly bounded by $e^{-n}$, which is summable. Thus, the sum converges uniformly. Each term of the sum is continuous, so you have a uniform limit of continuous functions, and therefore the limit function is continuous.


2

If $f$ is continuous on a compact set $K$, then there is a constant $M$ such that $|f(x)|\leq M$ for all $x\in K$. Therefore $$ \int_K|f(x)|\;dx\leq Mm(K)<\infty $$ so $f$ is integrable on $K$.


2

The definition you give is correct, but rather confusingly written. What it means is the set of values $\sum_{i=1}^n \lambda_i x_i$ where $\lambda_1,\ldots,\lambda_n$ are positive real numbers with sum $1$ and $x_1,\ldots,x_n$ are taken amongĀ $M$. Now in fact we can replace "positive" by "nonnegative" in this definition, because we can always remove those ...


2

Take $l_\infty$ and $S = \{e_k\}_{k \in \mathbb{N}}$. The $S$ is closed, bounded, but not totally bounded since $\|e_i - e_j\| = 1$ for all $i \neq j$ (hence there can be no finite $\epsilon$-net for $\epsilon <1$).


1

Essentially the same example as given by copper.hat, but perhaps written in a simpler way. Let $X=\mathbb{N}$ and let $d$ be the discrete metric given by $d(m,n)=1$ if $m\neq n$ and $d(m,n)=0$ if $m=n$. Every subset is closed and bounded by $1$. But only finite subsets admit an $1/4$-net.


1

Let $\lambda$ denote the measure on $\mathbb R$ which is being used (I assume it is the Lebesgue measure, but this argument works for whatever measure you wish to use). We see \begin{align*} \lambda(f^{-1}((a,\infty))) = \lambda(\{x \in \mathbb R : f(x) > a \}) &= \int_{\{x \in \mathbb R : f(x) > a \}} d\lambda(x) \\ &\le \int_{\{x \in \mathbb ...


1

In order to correctly construct confidence interval, first of all, you need a "pivot" T such that T is a function of all data $X_1,X_2,...,X_n$ T is a function of $\mu$ The distribution of T is know, and it is not a function of $\mu$ For example, $T=\frac{\bar{x}-\mu}{\sigma/\sqrt{n}}$ is a pivot, and $T$~$N(0,1)$ regardless of the value of $\mu$ (if the ...


1

Perhaps the simplest counterexample is the discrete metric [ $d(p,q)=1$ when $p\ne q$ ] on any infinite set $X.$ Then every subset of $X$ is bounded and open and closed, but only the finite subsets of $X$ are compact.


1

It is not true in general, the closed unit ball of an infinite dimensional Banach space is not compact


1

In vol. 1, Chap.II, Sec.3.3 of the book Generalized functions by Gelfand-Shilov you will find the formula $$ F\bigl[\Vert x\Vert^\lambda\bigr](\omega)=2^{\lambda+d}\pi^{\frac{n}{2}}\frac{\Gamma\left(\frac{\lambda+d}{2}\right)}{\Gamma\left(-\frac{\lambda}{2}\right)} \Vert\omega\Vert^{-\lambda-d}, $$ or all $\lambda\neq -d,-d-2,\dotsc $, where, for ...



Only top voted, non community-wiki answers of a minimum length are eligible