Hot answers tagged

5

We have that $\sin(x)-x\cos(x)$ behaves like $x^3$ in a right neighbourhood of the origin, hence integrability over there is ensured by $\color{red}{\alpha < 4}$. By Dirichlet's test, $$ \int_{1}^{+\infty}\frac{\sin x}{x^\beta}\,dx,\qquad \int_{1}^{+\infty}\frac{\cos x}{x^\beta}\,dx $$ are convergent as soon as $\color{red}{\beta>0}$, hence the ...


4

Most probably though I can obviously not guarantee it, it is just $\frac{5}{6}$ times $\sum\frac{1}{n(n+1)}$. $$\sum_{n=1}^{30}\frac{1}{n(n+1)}$$. $$\sum_{n=1}^{30}\frac{1}{n}-\frac{1}{n+1} = 30/31.$$ Therefore, one probable answer is 25/31.


4

Your "kinda think not" is correct, but so is your intuition that the unit interval should have the points $0$ and $1$ on its boundary. The structure you want to make this idea work is a manifold. It's a metric space plus some other stuff that makes it resemble a part of $\mathbb{R}^n$ without necessarily being embedded there. The "manifold with boundary" ...


4

I think that the derivative of $f$ is indeed bounded on $[0,\infty),$ which implies $f$ is uniformly continuous there. I'll give an outline: Let's write $$f(x) = x^2 \sum_{n=1}^{\infty} \frac{n}{n^3 + x^3}.$$ This will give $$\tag 1 f'(x) = 2x \sum_{n=1}^{\infty} \frac{n}{n^3 + x^3} + x^2 \sum_{n=1}^{\infty} \frac{-3x^2n}{(n^3 + x^3)^2}.$$ (You verify ...


3

Clearly if you regard $E$ as embedded in $\mathbb{R}$, then it has a boundary. But from your comments, I assume you are regarding $E$ as not embedded. In that case, it is trivial that $E$ is open and closed and hence has no boundary (given the usual definitions - there is a good Wikipedia article on the detail). But if you are interested in the intuitive ...


3

Note that $$ \lim_{n \to \infty} \frac{2^n + 3^{m+1}}{2^n + 3^m} = \lim_{n \to \infty} \frac{1 + \frac{3^{m+1}}{2^n}}{1 + \frac{3^m}{2^n}} = 1 $$ Thus, we may conclude that each element $a_k$ is part of a subsequence $\{a_{k_1},a_{k_2}, \dots\}$ in which $$ \frac{a_{k_{j + 1}}}{a_{k_j}} \to 1 $$ From this, we may conclude that $a_{k+1}/a_k \to 1$ as ...


3

$$\frac{1}{(k+1)\sqrt{k}+k\sqrt{k+1}}$$ $$=\frac{1}{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt k)}$$ $$=\frac{\sqrt{k+1}-\sqrt k}{\sqrt{k(k+1)}({k+1}- k)}=?$$ Can you recognize the Telescoping series?


2

The answer is that we cannot find such subsets $A$ and $B$. Of course, $A$ and $B$ need to be Lebesgue measurable so that $\mu (A)$ and $\mu(B)$ are well-defined. Let us defined $\nu (C) = \mu (A \cap C)$ for all Lebesgue measurable subsets $C$. Then: $\nu$ is a measure on Lebesgue-measurable sets, as it satisfies all axioms; $\nu(C) = \mu(C)/2$ for all ...


2

Suppose, for the moment, that the family $\{f_n\}_n \cup \{f\} \subseteq L^1$ is uniformly integrable, i.e. that for any $\epsilon>0$ there exists $\delta>0$ such that $$\sup_{n \in \mathbb{N}} \int_A |f_n| \, d\mu + \int_A |f| \, d\mu < \epsilon \tag{1}$$ for any Borel set $A$ with $\mu(A)\leq\delta$ (here, and in what follows, $\mu$ denotes a ...


2

Here's another solution — this one doesn't require knowing about the exponential function. We're given that $f$ is differentiable on $\mathbb{R}$, $f'(x)=f(x)$ for all $x \in \mathbb{R}$, and $f(0)=1.$ Assume that there is some $a \in \mathbb{R}$ such that $f(a) = 0$. We must have $a \ne 0$ since $f(0)=1.$ Note that, since $f$ is differentiable, it ...


2

No for $p=1$, yes for $1<p<\infty$. If $\phi\in \Delta C^\infty_c$ then $\int\phi=0$; this shows that $\Delta C^\infty_c$ is not dense in $L^1$. One might think at first that this shows the same thing for other $p$, but it doesn't, because the integral is not a bounded linear functional. Suppose from now on that $1<p<\infty$. Suppose that $K\...


2

I believe your approach is correct. I wrote something up before realizing you had already provided a proof (more-or-less what you have, just by proving the contrapositive): Let $E_{n,\epsilon}=\{x\colon|f_{n}(x)-f(x)|\geq\epsilon\}$. Suppose $f_{n}$ does not converge to $f$ in measure so that there exists an $\epsilon>0$ and $\delta>0$ with $\mu(E_{n,\...


2

This is called differentiating under the integral sign. It is valid if the function in the integrand, as well as the partial you are interested in is continuous in the argument with respect to which you are integrating. Since that was a mouthful, with your notation: $$\int{\dfrac{\partial}{\partial t}f(x,t)\,\mathrm dx} = \dfrac{d}{dt}\int{f(x,t)\,\mathrm ...


2

Here's one of my favorite tricks. If a sequence $x_n$ has a limit, you can define a new sequence $\bar{x}$ on $\mathbb{N} \cup \{ \infty \}$ by setting $$ \bar{x}_n = \begin{cases} x_n & n \in \mathbb{N} \\ \lim_{m \to \infty} x_m & n = \infty \end{cases} $$ $\bar{x}_n$ is a continuous function on the compact set $\mathbb{N} \cup \{ \infty \}$, and ...


1

$m(n) =\inf\{m: \frac{1}{2^m}\le \frac{1}{n} m^{3/2}\} $ If $\frac{1}{2^m} \le \frac{1}{n} m^{3/2} $, taking the reciprocal gives us $2^m \ge n/ m^{3/2} $, and taking logs gives $m \log 2 \ge \log n-\frac32 \log m $ or $m \log 2+\frac32 \log m \ge \log n $. So we want to see what this tells us about $m$ in terms of $n$. If $m \ge \dfrac{\log n}{\log 2} =\...


1

Your heuristics are in fact correct. One heuristic is to isolate the main term involving the unknown variable, which in this case is $2^m$. Solving, we get $2^m \ge \frac{n}{m^{3/2}}$. To solve for $m$, we take logarithms, which gives $m \ge \log_2 n - \frac32 \log_2 m$. Since $\log_2 m$ is insignificant compared to $m$, we can ignore the $\log_2 m$ term,...


1

...But since $\delta < \epsilon$ , $\forall \epsilon > 0, \epsilon \in \mathbb Q $... This part is not right. Make sure to use the right argument with the right inequality!


1

The problem is not true as stated. If $(x_0,y_0)$ is fixed (as it is in the current statement), your implication is true as long as $g(x_0)\neq 0$ and $h(y_0) \neq 0$. A counterexample is easily constructed by picking $g$ a function such that $g(x_0)=0=\lim_{x \to x_0} g(x)$ and $h$ any discontinuous function which is bounded near $y_0$. To do this, use ...


1

It is not closed since there is a sequence of elements in this set which converge to $5$, yet $5$ is not in this set. Notice that it is also not open set. If $S \subset \mathbb{R}$ is open, then for every point in $x\in S$, there exists a ball $B_{\epsilon}(x)$ such that $B_\epsilon(x)\subset S$. This is clearly not true by taking $x = 4$.


1

One way to show that $A\cup \{b\}$ is countably infinite is by defining a bijection $f:\Bbb{N}\to A\cup \{b\}$. Now, since $A$ is countably infinite, there exist $g:\Bbb{N}\to A$ which is bijective. Let's call $g(1)=a_1, g(2)=a_2,...,g(n)=a_{n},..$ So lets define $f$ in the following way: $$f(1)=b, f(2)=a_1,...,f(n)=g(n+1)=a_{n+1},...$$ and let's see that $...


1

Re-reading the textbook problem description one last time, in fact we see the additional information $x>0$. Oops.


1

Showing the limits at infinity is straightforward. Simply consider the related rational function $$g(x) = \frac{3-x}{1+x^2}$$ which, by definition of the floor function, satisfies $$f(x) \le g(x) < f(x) + 1$$ for all $x$, and for which $$\lim_{x \to -\infty} g(x) = \lim_{x \to -\infty} \frac{\frac{3}{x^2} - \frac{1}{x}}{\frac{1}{x^2} + 1} = 0,$$ and ...


1

See, how many numbers $a_k$ are there below some huge $N$? At least this includes all of those for which $2^n\le{N\over2}$ and $3^m\le{N\over2}$ (and probably more), which is already $O(\log^2N)$, which pretty much forbids the existence of a limit > 1.


1

Hint: $$f(\zeta+h)-f(\zeta)=[f(\zeta+h)-f(\zeta+(0,h_2)]+[f(\zeta+(0,h_2))-f(\zeta)].$$The continuity of $\partial f/\partial x$ lets you handle $f(\zeta+h)-f(\zeta+(0,h_2)$, and then you can use existence of $\partial f(\zeta)/\partial y$ for $f(\zeta+(0,h_2))-f(\zeta)$. Slightly more detailed hint: Say $\nabla f(0)=(\alpha,\beta)$. The triangle inequality ...


1

May be useful \begin{align} & \int_{0}^{\infty }{\frac{\sin x}{{{x}^{\alpha }}}}dx=\frac{\pi }{2\Gamma (\alpha )\sin \left( \frac{\alpha \pi }{2} \right)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\,\,\,\,\,\,\,0<\alpha <2 \\ \\ & \int_{0}^{\infty }{\frac{\cos x}{{{x}^{\alpha -1}}}}dx=\frac{\pi }{2\Gamma (\alpha-1 )\sin \left( \frac{\alpha\pi }{2} \...


1

a) Suppose $A\subset B$. Now suppose that $a \in \bar{A}$. Then $a$ is the limit of a sequence of terms in $A$ and thus of terms in $B$ and so $a \in \bar{B}$. b) Note that $A\subset \bar{A}$ and $B\subset \bar{B}$ and so $A\cup B \subset \bar{A}\cup \bar{B}$. But the closure of a set, say $X$, is the smallest closed set containing $X$. So $\overline{A\cup ...


1

Your work is almost done. Now $a_k=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}$. Now let $V_k=\frac{1}{\sqrt{k}}$, then $a_k=V_k-V_{k+1}$. Now start putting the values of $k$ and add $a_k$. Finally you will get $$S=\sum_{k=1}^{n} a_k=V_1-V_{n+1}=1-\frac{1}{\sqrt{n+1}}$$ Now if $n$->$\infty$, then $S=1$ which is the final answer. Hope this will be helpful !


1

The answer is already given, but in case you don't see how to interchange the order clearly: \begin{align} \int_{0}^{+\infty}\int_{\{t\in \mathbb{R}:g(t)>x\}}2h(t)x\,dt\,dx &= \int_{\{(x,t)\in \mathbb{R}^2:0\leq x<g(t)\}} 2h(t)x\,dt\,dx \\ &= \int_{\mathbb{R}}h(t)\int_0^{g(t)}2x\,dx\,dt \\ &= \int_{\mathbb{R}}g^2h\,dm. \end{align}


1

As the integrands are non-negative, Tonelli's theorem tells us order of integration is interchangable. $$\mathrm{ \int_0^\infty\int_{\{t:g(t)>x\}}2h(t)x\,dt\,dx=\int_{-\infty}^\infty\int_{\{x:\,0<x<g(t)\}}2h(t)x\,dx\,dt\\ =\int_{-\infty}^\infty\int^{g(t)}_{0}2h(t)x\,dx\,dt=\int_{-\infty}^\infty g(t)^2h(t)\,dt }$$


1

I think you're making this a bit harder than it needs to be. You don't need an exact formula for $\text{vol}(C_{\delta_n})$, just upper and lower bounds. Furthermore, these can be quite crude - within a constant multiple. To make this precise and to set notation, define an $\varepsilon$-mesh interval to be one of the form $(j\varepsilon,(j+1)\varepsilon)$ ...



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