The terms of the ordered pairs must be separated by a comma. The right answer is $\{(2,a),(3,a),(4,a),(2,b),(3,b),(4,b)\}$.

Let $R \equiv P \lor Q$ and $S \equiv \neg P \lor \neg Q$. When you say "when both $R$ and $S$ are true", you are looking for $R \land S$. If $R \land S$ is always false no matter who is $P$ or $Q$, ...

The definition of the square root, $\sqrt{\,\cdot\,}$, in real numbers is a function that given a non-negative number $y$, we take the non-negative number $x$ such that $x^2 = y$. That's why we write $... View answer 2 votes No, there is not. Not in the standard set theory, which is the ZFC Set Theory. The reason is the power set (the set of all subsets of a given set), as explained by the other answers here, we have$|\...

Ok, finally got it. Define, for every $k\in \{0,\dots, n\}^N$, \begin{align*} r_k: \mathbb{R}^N& \to (\mathrm{Lip}(\mathbb{R}^N))^\ast\\ x & \mapsto \begin{aligned}[t] ...

Hint: A function $f:U\subset\mathbb{R}^m\rightarrow \mathbb{R}^n$ is differentiable at $a \in U$ if there exists a linear transformation $T_a:\mathbb{R}^m\rightarrow \mathbb{R}^n$ such that $$f(a+h) =... View answer 1 votes Since \lim\limits_{n\rightarrow\infty} a_n = a, then \forall\varepsilon>0, \exists n_0\in \mathbb{N} such that n>n_0\Rightarrow |a_n-a|<\varepsilon/2. For n>n_0, we have \begin{... View answer 1 votes Hint: Given z = r e^{i\theta}, we have$$\log z = \log r +i\theta,$$and, of course,$$\log r = \int_1^r \frac{1}{t} \mathrm{d}t.$$View answer 1 votes \forall \varepsilon>0, \exists n_0\in \mathbb{N} such that n>n_0\Rightarrow 1/n<\varepsilon. By assumption,$$|X_n|<\frac{1}{n}<\varepsilon,\quad \forall n>n_0,$$which is ... View answer 1 votes Using the Change of Variables Theorem: Let f:[a,b]\rightarrow \mathbb{R} a continuous function and g:[c,d]\rightarrow \mathbb{R} with continuous derivative and g([c,d])\subseteq [a,b]. Then ... View answer Accepted answer 1 votes No. If you want a combination, then "Apple-Orange-Boat" is the same as "Orange-Boat-Apple". The order does not matter and therefore 2000^3 is counting too much. For combination with repetitions, ... View answer Accepted answer 0 votes Using a diagonal argument, we can consider (L_n)_{n\in\mathbb{N}} such that \|L_n\|_{\mathrm{Lip}} < 1 + 1/n for every n\in \mathbb{N}. The secret to guarantee the Lipschitz constant for the ... View answer 0 votes As @lulu said in his comment, the linear approximation or a first order (order n=1) Taylor approximation of a function f around x_0 is$$f(x)\approx f(x_0)+(x-x_0)f'(x_0),$$since the Taylor ... View answer 0 votes Hint: If you think about it, this comes down to prove that$$\left|\int_a^b f(x) \mathrm{d}x\right|\leq \int_a^b |f(x)|\mathrm{d}x.$$To help you, let us prove that |f|:x\mapsto |f(x)| is ... View answer 0 votes It does not matter if it is decreasing. Given the set A=\{1/2,...,1/n\}, define your partition P=\{x_0,x_1...,x_{2n+2}\} such that for every k\in\{1,...,n\},$$\frac{1}{k}-\frac{\varepsilon}{2n}\...

For the function $f$, you are right. For $g$, the result comes from this theorem: Let $f:X\rightarrow \mathbb{R}$ be a continuous function. If $X$ is compact, then $f(X)$ is also compact. ...
You can easily see that the tangent of a angle $\phi$ of a parametrized curve $(y(t),x(t))$, at $t=\theta$, with respect to the $x$-axis is \begin{equation}\tan{\phi} = \lim_{t\rightarrow \theta} \...