# Tag Info

-1

I've seen it used before, not sure where. Doesn't seem to have caught on. It's $\ni$ = "\ni". See this tex question.

2

I've searched over 900 symbols using this tool and I wasn't able to find such one, nor have I ever seen it used. My advice would be to just ask your professor if he invented that symbol, and if he did, for what reason.

0

Uhm.. In such a case you can just write as following. Let's say that you want to write down the $i-$ row and the $j-$ column of matrix $A$. Then you can write: $e_i^T \cdot A,$ where $e_i^T$ is the $i-$ row of the identity matrix. $A\cdot e_j$, where $e_j$ is the $j-$ column of the identity matrix.

1

$$a \mid b \iff b=ak \text{ for some } k\in \Bbb Z$$ This is a true/ false kind of statement. As in, either $a \mid b$ is true, or it's not (in which case we write $a\not\mid b$). $$a/b = \frac ab$$ This is just a number. And not even necessarily an integer. Here's a cool example of when we can have $a\mid b$ that wouldn't even make sense as a ...

0

$a|b$ means $a$ divides $b$.And $a/b$ represents a fraction $\frac ab$.

0

Yes you are correct, $(a_1,a_2)$ is what we call a strategy profile: in this profile, player 1 plays $a_1$ while player 2 plays $a_2$. Remember the definition of a strategic game in Osborne: players, strategy sets for each one of the players and payoff functions. The payoff functions are maps from the space of strategy profiles (also referred as outcomes) to ...

0

The pair $(x,y)$ means that Player $1$ choose $x$ and Player $2$ chooses $y$. Thus, $(a_1,a_2)$ means that Players $1$ and $2$ choose $a_1$ and $a_2$, respectively. The question asks you to prove that there is a Nash equilibrium $(a_1,a_1)$, which means that there is some choice $a_1$ that both players will make.

0

Einstein summation is a notational convention for simplifying expressions including summations of vectors, matrices, and general tensors. There are essentially three rules of Einstein summation notation, namely: Repeated indices are implicitly summed over. Each index can appear at most twice in any term. Each term must contain identical non-repeated ...

3

A very famous textbook using the notation $[n]=\{1,2,\dots,n\}$ is Enumerative Combinatorics (Volume 1) by Richard Stanley. It's in the list of notation at the beginning of the book, and is first used in Example 1.1.16.

1

Statement: "All cats either love or hate dogs." I am sure there are different ways to interpret this statement, but how about: $\forall x:[ Cat(x)\implies \neg Dog(x)]$ $\space\space\space\land \forall x:[Cat(x) \implies \forall y:[Dog(y) \implies Loves(x,y) \land \neg Hates(x,y)]$ $\space\space\space\space\space\space\lor \space \forall y:[Dog(y) ... 5 Neither of your suggestions captures the meaning I get from the original sentence. However, the second one is the most wrong. It will be true as soon as there exists anything that is not a dog. Namely, no matter what$x$is, you can choose the non-dog to be$y$, and then${\rm cat}(x)\land{\rm dog}(y)$will be false, which automatically makes the entire ... 0 I have always seen the notation $$\begin{cases} &x_1 = 2r+s-t \\ &x_2 = r \\ &x_3 = -2s + 2t \\ &x_4 = s \\ &x_5 = t \\ \end{cases}$$ I'm fairly sure this is standard notation, but you can of course adapt your notation as long as the reader understands (clarification might be needed). 0 Let$f:A\to B$be a map. Prop1 Suppose the fibers of$f$are nonempty. Then$f$is surjective. Proof: Let$b\in B$. We need to show that there exists$a\in A$such that$f(a)=b$. Well, the fibers are nonempty, so there exists$a\in f^{-1}(b)=\{x\in A\mid f(x)=b\}$. Therefore,$f(a)=b$as required. Prop2 Suppose every fiber of$f$contains at most one ... 1 The former. The latter makes as much sense as, for example,$x \mapsto x\sin$, which I hope you wouldn't write in a similar context! 0 As the indicator function takes$t$as an argument,$t\mathbf 1_B(t)$is the better notation. 2 There is no "normative answer," there is no "normative reference." There is no Holy Bible of Mathematics, no Supreme Court of Mathematical Notations. There are only people who find different conventions useful at different times. If you want to use Excel, you have to do it the way Excel does it; if you want to use Wolfram Alpha, you have to do it the way ... 0 This is covered by the standard "Pedmas" ("Please excuse my dear aunt Sally") or "Bedmas" to our British friends. First parentheses (Brackets) then exponentials then division then multiplication then subtraction. In "-7^2", the "negative" is multiplying by -1 so the exponent, 2, takes precedence. 1 You can not assume anything. All you can use is what is explicitly stated and what you can deduce from that. If coordinates are stated for the points, and$c$is equal to$(a+b)/2$, then$c$is the midpoint of$ab$and you can use that. 2 No. Even if it looks like the midpoint, it should be explicitly stated in the problem description, or you should be able to prove it from other information given. If neither of these cases hold, don't assume you have a midpoint. 2 The ISO has built such a list, although naturally there are disagreements since math uses a lot of domain-specific notation. See ISO 31-11, superseded by ISO 80000-2. 1$A \not \subset B$means$A$is not a subset of$B$. So does$A\not\subseteq B$.$A\subsetneq B$means$A$is a proper subset of$B$. I don't know a single binary relation symbol that means$A$is not a proper subset of$B$. One could write$A \not\subset B$on the theory that$A\subset B$means$A$is a proper subset of$B$, but that could easily be ... 0 For simulation we have to specify the values of parameters; in such a context, the notation$f(x \mid \theta)$suggests that$\theta$(may be a vecotr) is "known". For statistical inference the values of parameters are to be investigated via data at hand; in this context, the notation$f(\theta \mid x)$is preferable, which suggests that the "data"$x$are ... 0 It appears that this is notation meant to convey specific submatrices through indexing rows and columns. For example, if we let$A=\left[\begin{matrix}1&2&3\\4&5&6\\7&8&9\\ \end{matrix}\right]$,$A[\{1,3\},\{1,2,3\}]=\left[\begin{matrix}1&2&3\\7&8&9\\ \end{matrix}\right]$as we took the first and third row and the ... 3 Yes,$\rightharpoonup$is usually used to indicate weak convergence. 1 One interpretation that may clear things up is that$K[S]$is not only a ring, but an algebra, specifically, a$K$-algebra, specifically, the$K$-algebra generated by$S$. That clarifies the asymmetry:$\mathbb{Q}[\sqrt 3]$is an algebra over the rational numbers, not over the square root of 3! But as a ring, it's perfectly fine to say$\mathbb{Q}[\sqrt ...

1

There is an elementary proof. Write $$M=\left[\begin{array}{cc} a & b\\ c& d \end{array}\right].$$ Since $M^{3}=0$, taking the determinant of both sides and using the determinant of a product is the product of the determinants, gives $\text{det}\,M=0$. Hence, $$ad-bc=0.$$ By matrix multiplication, the following is found. ...

0

$M^2 (M + I) = M^3 + M^2$ $M^2 (M+I)=M^2$ since $M^3 = 0$. For this last equation to hold, either $M^2 = 0$ or $M + I = I \Rightarrow M = 0$. In both cases, $M^2 = 0$.

2

Cases should be the environment you're looking for $$\begin{cases} a + b = c &\hbox{when } a > 0 \\ a + b = 2c & \hbox{when } a\leq 0 \end{cases}$$ begin{cases} a + b = c &\hbox{when } a > 0 \ a + b = 2c & \hbox{when } a\leq 0 \end{cases} with \ before the begin.

0

Before calculators existed, logarithms were computed from tables. those tables listed the (base 10) logarithms of numbers $1 \le x \lt 10$. Thus, for example the log of $2$ is $0.3010$. So log $2 \times 10^5 = 5.3010$.

1

$F[x]$ represents the ring of polynomials over the field F. Formally, this ring can be defined as the set of functions with finite support (taking only finitely many nonzero values) from the natural numbers into the field. The operations are defined as follows: $$(f+g)(i):= f(i) + g(i) \text{ }\forall f,g \in F[x] \text{ and } i \in \mathbb{N} ... 0$$ F[x]=\left\{\sum_{i=0}^n a_ix^i\colon a_i\in F,n\in\{0,1,\cdots\}\right\}  F[[x]]=\left\{\sum_{i=0}^\infty a_ix^i\colon a_i\in F\right\} $$1 By now you may have noticed that, if put to a vote, "Yes, you can use negative numbers too." would win but not by a unanimous vote. There was a time, when people used slide rulers, that scientific notation was needed to perform calculations. It still is now, but not nearly as much. Now it's just a simple way to represent really large and really small ... 4 I what you are asking about in fact is an indicator function. Consider a set A\subset \mathbb{R}^n, then the indicator function for the set A as a function of x\in\mathbb{R}^n, is defined as$$ \mathbb{1}_{A}\left(x\right) := \begin{cases} 1,\qquad \forall x\in A,\\ 0,\qquad \forall x\not\in A. \end{cases} $$This is very useful in many areas of ... 0 Let z=a+bi. Then we define \bar z = a-bi. So, for example, if z = 5+3i, then z\bar z = (5+3i)(5-3i), which you can expand and simplify. 2 There is no problem having a negative number shown in scientific notation, computers and calculators have been doing this for ages. The quote from the textbook shouldn't be taken too literally as they were probably thinking of the magnitude of the number and forgot about this trivia. In any case, the problem collapses if you read -4\cdot10^{50} as ... 2 Yes, it means the hom spaces have all been tensored with \overline{R}. (Note that this notation is consistent with the case that C has one object, in which case it is a commutative ring.) 0 If I had seen the notation you suggest, I would have been quite confused about it. If there is a clear explanation in text, it might have helped, but unfortunately not everyone provides textual explanation, and it's not always clear. I'd prefer to use the following:$$\mathcal G=\left\{\bigcup_{i\in I} F_i\mathrel{}\middle|\mathrel{} ...

2

The simplest solution is $$\mathscr{G}=\left\{\bigcup\mathscr{A}:\mathscr{A}\subseteq\mathscr{F}\right\}\;.$$

1

I think it's just part of a more general notation for cosets of a subgroup, which has been popular for most of the 20th century. If $G$ is a group, and $H$ is a normal subgroup, then $G/H$ is the set of cosets $xH : x \in G$. With this notation, an integer modulo $n$ is literally a subset of $\mathbb{Z}$. For example $1$ modulo $\mathbb{Z}$ is the ...

0

I would write $S = \{\bar u_i: i ∈ I\}$. And then $∑_{i ∈ I} a_i \bar u_i$. But note that this is an actual linear combination only if $\{i ∈ I: a_i ≠ 0\}$ is finite. If you want the index set to be well-ordered, you may write $S = \{\bar u_α: α < |S|\}$ or $\{\bar u_α: α < κ\}$ and then $∑_{α < κ} a_α \bar u_α$. If $S$ is finite, you may use $n$ ...

2

Non-seriously: $$f(x) = \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!} f^{(k)} (a) \;\;\approx\;\; \text{true}$$

0

The short answer is YES, $p$ can take on the "value" NULL for the function you have provided. For this function, you have allowed $p$ to contribute to the value if and only if it is not NULL. This implies that it has a value in your domain. If $p$ is NULL, it should not contribute to the value of the function, because it wouldn't make any sense. Consider ...

1

The inputs to a function can come from any set you like. There is nothing wrong with saying $p$ belongs to say, the set $$\{\text{NULL}, 5,🌊\}$$ But of course, care is needed to make sure you have a well-defined function.  I see from the comments that you don't just want it to be NULL, you want to completely omit the parameter. You can do this if ...

0

What about: $$\mathrm{P} \left( f(x) \approx \sum_{k=0}^{\infty} \frac{(x-a)^k}{k!} f^{(k)} (a) \right) \approx 1$$

0

For example: Let $\mathcal M_{mn}$ be the set of all $m\times n$ matrices with real (complex) entries. Define $F:\mathcal M_{wh}\to \mathcal M_{rs}$ by the formula $$F(A)_{ij} = \text{something with i,j,A_{kl}}$$

0

The question is little bit soft. I find following variants and variations useful. Worth noting, I am not a native speaker. Naming the roots: A number $x^*$ is a root of $x^2 -1$ if (and only if) ... $x^* = -1$ or $x^* = 1$ $x^* =-1, 1$ $x^*\in \{-1, 1\}$ Not naming: The roots of $x^2 - 1$ are (exactly) $-1,1$. The set of roots of $x^2 - 1$ is $\{-1, ... 2 I would say there is no need to make nonessential distinction, unless confusion may arise. The comma appearing in a phrase such as$x = 1,2$is usually understood as "or". In this sense to write$x = 1,2$or to write$x \in \{ 1,2 \}$are equally clear. (And the latter one may look "spurious"...) But if instead what you are working with already employs ... 2 It is probably the ball centered at$f$with radius$a$: $$B(f,a)= \left\{ g \in L^p(\mathbb{R}) \mid \|f-g\|<a \right\}.$$ 0 The commend by Did was an answer: For every signed bounded measure$\rho$and every$A$in$\mathcal A$, $$\rho_+(A)=\sup\{\rho(B)\mid B\in\mathcal A,B\subseteq A\}.$$ Likewise, $$\rho_-=(-\rho)_+.$$ Then, for every$A$in$\mathcal A$, $$\rho(A)=\rho_+(A)-\rho_-(A),$$ and one defines $$|\rho|=\rho_+ +\rho_-.$$ Related topic: Jordan measure ... 32 When I’m writing up tentative results that remain yet to prove (perhaps subject to some additional regularity conditions or assumptions), I usually do this: $$x\,\overset{?}{{=}}\,y+z.$$ On a less serious note: Let$f:\mathbb R\to\mathbb R$be a function of class$\mathcal C^{\infty}$. Then, one has, for any$x\in\mathbb R$and$a\in\mathbb R\$, that ...

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