# Tag Info

3

No, that makes no sense, I'm afraid. The set $$U-W=\{u-w:u\in U,w\in W\}$$ is exactly the same as $U+W$, because whenever $w\in W$, also $-w\in W$. There's no way, in general, to recover $W$ from $V=U+W$ and $U$. Just to make an example, let $V=\mathbb{R}^2$ and $U$ be the subspace spanned by $e_1=(1,0)$. Then $$V=U+W$$ for $W$ the subspace spanned by ...

3

Like Geoff suggests in the comments, you can use $\text{Ker}$ for the object and $\text{ker}$ for the morphism. But I don't think this convention is universal so you should probably say that you're using it. I sometimes use $\text{ker}$ for both.

3

In my experience, no. The choice of whether to use cal or scr is generally just stylistic. I don't think I've ever seen $\mathcal{F}$ and $\mathscr{F}$ both appearing in the same paper; authors (or their typesetters) choose one or the other. They don't use both to refer to two different objects.

2

I've often seen $a\,\#\,b$ or $a\star b$ for custom-defined or arbitrary operations.

2

Binary operations also are represented as $a \circ b$

2

\begin{align} 5e2 & = 5\times 10^2 = 500 \\ 5e{-2} & = 5\times 10^{-2} = 0.05 \\ -5e2 & = -5\times 10^2 = -500 \\ -5e{-2} & = -5 \times 10^{-2} = -0.05 \end{align}

2

A possibly related concept is that of an Iterated Function System, and the Hutchinson operator: $$H(S) = \bigcup_{i=1}^n f_i(S)$$. In this setting, the collection of all iterates of your collection of functions $\{f_i\}$ applied to $x$ could be described as $$\bigcup_{j=0}^\infty H^{j}(\{x\})$$ where the exponent represents repeated applications of the ...

2

One usually writes "$y_n\nearrow x$" (resp. "$y_n\searrow x$"), or "$y_n\uparrow y$" (resp. "$y_n\downarrow x$"), to indicate an ascending (resp. descending) sequence $(y_n)_{n\in\Bbb N}$ such that $\lim\limits_{n\to\infty} y_n=x$. I think "$y\nearrow x$" should be a placeholder for the more common "$y\to x^-$" ($y$ approaches $x$ from the left), due to the ...

2

Let's say $\succ$ is a total ordering on $S$. Therefore, either $a \succ b$ or $b \succ a$ for all $a, b \in S$. In this case, to do what you want to do, you need to create a function from $S^2$ to $S^2$. It will take as input $(a, b)$ for $a, b$ in $S$ and then output the ordered tuple. Here is the formal definition: $$f(a, b)=\begin{cases}(a, b) \ \ \ ... 1 No, that's not union notation. I will create some variables to highlight the difference. Let N be the number of sixes in four rolls. Let R_i = 0 if the ith roll is not a six, and R_i = 1 if it is a six. Notice that the complement of \{N\geq 1\} is \{N = 0\}. Finally notice that$$\{N =0\}\iff\{R_1=0\text{ and }R_2 = 0\text{ and }R_3 = 0\text{ ...

1

Here is a link to a textbook that adopts the convention $\mathfrak{p}^0 = R$ explicitly (see the bottom line).

1

Notice that, in general, $\sin x \le 1 = 1 \cdot 1$, so it is trivial to understand that $\sin x$ is $O(1)$ where $c=1$. For the second, assume that there exist $c>0$ such that $\frac {x^3 + x} {x+1} \le cx$; usually these problems are given for $x>0$, so your assumption would be equivalent to $x^3+x \le cx^2 + c$, or $x^3 - cx^2 + x - c \le 0$, but ...

1

The correct statement is that $$U_k \cap \sum_{i \in I, \; I \neq k} U_i = \{0_V\}$$ for all $k$, where as you note the summation just means the set of things that can be written as finite linear sums of elements in the subspaces you're summing over. The key point is that this allows you to show that anything in $\sum_{i \in I} U_i$ has a unique expression ...

1

You could write $$\sum_{i=1}^nA_j^{ki}=X$$ with the point being that this equation is true for all $j,k$. You have $mn$ equations here, each one summing over one of the indices on $A$.

1

Here it is a recursive definition of $\Gamma(x)$, the orbit of $x$: Let $\Gamma_1(x)=\{f_1(x),\ldots,f_n(x)\}$. Let $\Gamma_{m+1}(x)=\{ f_1(y),\ldots,f_n(y) / y\in \Gamma_{m}(x)\}$. Finally, $$\Gamma(x)=\bigcup_{m\geq 1}\Gamma_m(x).$$

1

What you could write is just $$\{(g_m \circ g_{m-1} \circ \ldots \circ g_1)(x) | m\leq k; g_i \in F\, \forall i=1,\ldots,m\}$$ This does cover repetitions and allows any sequence of $f_i$'s up to length $k$.

1

Brian M. Scott provided the answer in a comment: The default interpretation is the limit as $n \rightarrow \infty$

1

The sequent calculus is based on the notation $\Gamma \Rightarrow \Delta$ (or $\Gamma \vdash \Delta$), with $\Gamma, \Delta$ finite (possibly empty) sequences of formulas, called a sequent. The intuitionistic sequent calculus is obtained with the restriction that $\Delta$ consists of at most one formula. For the semantics for sequents, see Gaisi ...

1

Just a suggestion.. Usually $\sigma(A)$ denotes the set of all possible permutations of elements of $A$ To simplify you could write $\sigma(n) = \sigma(\{1, \dots, n\})$ and write then $f_{\sigma(n)}$ the set of all functions you describe Or if you mean also subsequence (i.e. Permutations shorter than $n$) you could define $\sigma(n)= \cup_{i = 1}^n ... 1 To follow your line of reasoning, we would actually have $$E\left[g(X)\right]=\sum_{x}g(x)P(g(X)=g(x)).$$ Assume that$g$is injective. Then,$g(X)=g(x)$if and only if$X=x$, and the above becomes $$E\left[g(X)\right]=\sum_{x}g(x)P(X=x).$$ You can make a similar argument in the case that$g$is not injective. 1 Generally speaking, we start with the expectations of sums of indicator random variables, which are defined as:$\Sigma_x x P(\{\omega: X(\omega)=x\})$as you correctly noted. Then, for such a simple random variable, if we take$g(X)$, we have:$P(\{\omega: g(X)(\omega)=y\}=P(\bigcup_{x:g(x)=y} \{\omega: X(\omega)=x\})=\sum_{x:g(x)=y} P(\{\omega: ...

1

I created a small illustration with the expectation (no pun intended) that it might help some learners. HTH

1

This is a common notation: Suppose you have a set $\mathscr{C}$, all whose elements are also sets. The you might want to consider the intersection of all elements of $\mathscr{C}$. We might denote this as $\bigcap\mathscr{C}$. Example 1: $\mathscr{C}=\left\{\left\{1,2,3\right\},\left\{1,4\right\}\right\}$. Then ...

1

We can write ${\bf A}\cdot {\bf e}_i = A_i$, so $\nabla({\bf A}\cdot {\bf e}_i)$ is the gradient of $A_i$, To pick the $i-$th component, we dot that with ${\bf e}_i$ again to obtain $\frac{\partial A_i}{\partial x_i} = \nabla({\bf A}\cdot {\bf e}_i)\cdot {\bf e}_i$, and the desired field is: \sum_{i=1}^n \frac{\partial A_i}{\partial x_i}\,{\bf e}_i = ...

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