New answers tagged

1

If you are going to talk about a "last" element, you have to put an order on the set. So instead of thinking of $S$ as a set, you want to think of it as a tuple $\langle 23,45,32,56\rangle$, which is really a function $\sigma:3\to \mathbb R$ by $\sigma(0)=23$, $\sigma(1)=45$, $\sigma(2)=32$, and $\sigma(3)=56$. If you want everything but the last element, ...


-1

Actually, all of these are the subsets of $S=\{23,45,32,56\}$:$\{23\},\{45\},\{32\}$,$\{23,45\},\{23,32\}$,$\{23,45,32\}$


1

$$f:\mathcal{P}(\mathbb{K})\to \mathcal{P}(\mathbb{K})$$ Where $\mathcal{P}(\mathbb{K})$ denotes the power set of $\mathbb{K}$.


0

Yes. There's nothing preventing functions being in all three at the same time, and it's really easy to verify that your function indeed is.


5

$(0:x)$ is the set of elements $r$ of $R$, such that $rx=0$; called annihilator of $x$. Note that In the complex $K(x): 0\to R\xrightarrow{x} R$, the homomorphism $\xrightarrow{x}$ is multiplication by $x$. So $ker (\xrightarrow{x}) = \{r\in R| rx=0\}=(0:x)$ (and this is the homology). The complex is not an exact sequence unless the homology is ...


0

Let $\mathbb{Z}_n$ be the set of integers $\{0,1,\ldots,n-1\}$ equipped with the operations of addition mod $n$ and multiplication mod $n$. It can be shown this structure is a ring. $\mathbb{Z}_n[x]$ is defined as the set of polynomials of the form $a_n x^n + \cdots + a_1 x + a_0$, where $a_i \in \mathbb{Z}_n$ equipped with the usual operations of addition ...


1

$p$ and $p'$ are both arbitrary elements of $P$. They are possibly equal, but need not be. Furthermore, $P$ need not have the concept of multiplication and therefore doesn't need to have the concept of a multiplicative identity. Multiplicative identities only make sense if we have a multiplication to talk about. For example: $P=\{red,blue,yellow\}$ and ...


1

A skew diagram is not a partition, so you cannot write it as a tuple. I would just write "$\lambda/\mu$ where $\lambda = 4444$ and $\mu = 421$." You need to keep both $\lambda$ and $\mu$ around to describe $\lambda/\mu$. Of course, if you're interested in some special situation you could make up your own notation.


1

Never thought I'd give handwriting advice here, but I suppose this does fall under notation. I agree, the dagger can look like a plus sign. My rendition adds a guard to the dagger's pommel. Additional advantage: it's a single stroke.


3

Lots of people use $A^*$ to represent the Conjugate Transpose. To actually draw the dagger, draw a really long vertical line, dashed with a small horizontal. I personally dash the horizontal at an angle when I'm doing this so I know that it's a dagger. I personally use the third style like I mentioned.


1

Converted to an answer by popular request: If you want such a notation, use $\Bbb B$; just explain the notation at the start. The notations $\Bbb Z_2$ or $\Bbb Z/2\Bbb Z$ might be regarded as referring to a set with those same two elements, but they have sort of a different flavor, referring to a field (in the abstract-algebra sense), not a Boolean algebra. ...


1

Considering that we have that the characteristic is $n$, the writer probably has addapted the notation $$\mathbb Z_n:=\mathbb Z/n\mathbb Z.$$


0

It's the same thing as $(\mathbb{Z} / n\mathbb{Z})[x]$, the ring of polynomials with coefficients being the integers $mod$ $n$.


0

Let $C$ be the [infinite] set of all colors, and $P$ be the [finite] set of all people. For all $p \in P$, $f_p: C \rightarrow \mathbb{R}$ is a color-liking function where a higher value means a greater satisfaction with that color. Assumptions: The total color satisfaction for a person is the sum of the satisfactions for each color and the total color ...


0

TLDR: It is personal preference and largely doesn't matter Notice that $\frac{11}{3}+\frac{5}{3}n = \frac{5}{3}n+\frac{11}{3}$ due to the commutativity of addition. These two answers are not fundamentally different. What order you write things in is largely personal preference, and can change depending on context. One of the most common ordering of ...


0

If you find yourself repeatedly using expressions of this form you could define your own notation. For example (uploading image since MathJax cannot handle the $\LaTeX$) \left.\sqrt{(a_k+(-1)^{p_k}(}\right\vert_{k=1}^n=\sqrt{(a_1+(-1)^{p_1}\sqrt{(a_2+(-1)^{p_2}\sqrt{(\cdots+\sqrt{a_n}}}}


1

This is not a standard notation, though I have seen notation like this used occasionally (not specifically in group theory) when you need to repeatedly refer to lots of restrictions in the course of a single argument. I would never use a notation like this without defining it for my readers. The standard notation is $\phi|_H$, not $\phi_H$.


0

I think the restriction $\phi_H$ to $H$ indicated is not only restriction of application, but it seems more, this is a group homomorphism


1

For $r>0$ take $M>1$ such that $1/M<r.$ For $y>M$ we have $$0<\ln y=\int_1^y (1/z)\;dz= \int_1^M(1/z)\;dz+\int_M^y(1/z)\;dz=$$ $$=\ln M+\int_M^y(1/z)\;dz<\ln M+\int_M^y(1/M)\;dz=\ln M+(y-M)/M.$$ $$\text { So, }\quad 0<(\ln y)/y<(\ln M)/y+(1-M/y)/M.$$ Therefore $$0\leq \sup_{x\geq y}\;(\ln x)/x\leq \sup_{x\geq y}(\;(\ln ...


0

The usual notation is that $f^0(y)=y$ and $f^{n+1}(y)=f(f^n(y))$. That is, $f^n(y)$ is the $n$-th iterate of $f$ applied to $y,$ and that $f(t)\cdot f(t)$ is written $f(x)^2.$ There are some customary exceptions to this, especially with trigonometric functions, where, for example $\cos^2 y$ means $(\cos y)^2$, probably because it appears so often and someone ...


3

The first point is: when you use asymptotic notations, you need to specify with regard to what point the asymptotics is taken. Here, it looks like this is when $x\to 0^+$; note that this could have equally been $x\to\infty$, so specifying it is required. Now, you have that for any fixed $\alpha > 0$, $$ \ln x = O(x^{-\alpha}) $$ when $x\to 0^+$, since ...


3

The symbol $\mid$ means "divides" (in this context, at least). That is, if $a$ and $b$ are integers then $a\mid b$ (read "$a$ divides $b$") means that there exists an integer $c$ such that $b=ac$. Except in the case $a=b=0$, this just means that $b/a$ is an integer. (More generally, you can make the same definition for elements of any ring.)


1

Working from the linked blog, $X_j \approx \mathcal X$ means that $X_j$ is homeomorphic to $\mathcal X$. This makes sense, as he then immediately gives a name to a homeomorphism. Classes are confusing. But there is a lot written about them already on MSE. I direct you to the question Difference between a class and a set as a good jumping point here. ...


0

[Note: of course for one integer $n$, there is no such decomposition.] The most common way to say this is indeed some variation of "Write $n$ as $2^r m$ with $m$ odd." If you really want alternatives (there might be contexts where these read better due to placing emphasis on $r$ or on $m$), you might also consider: "Let $2^r$ be the largest power of $2$ ...


2

In many common programming languages, such as java, c, c++, the percent sign is used as a modulo operator. see this page for example. In practice, one has $q\% p =q-p\lfloor q/p\rfloor$ for $p\neq 0$ and $0$ otherwise. In other words, the modulo operator outputs the remainder of the division operation $q/p$. For examples, $23\% 5 = 3$ and $15\% 3 = 0$ ...


1

I can't totally parse the post, but A Gröbner basis is a special generating set for an ideal in a ring $R$ considered as an $R$-submodule of $R$. and A basis for a vector space is a special generating set for the vector space as an $\Bbb F$-module for some field (or even division ring) $\Bbb F$. Yes, if $1$ were part of a Gröbner basis for an ...


1

There are several typos which make this question confusing: Not $\prod_{\alpha} U_\alpha$ but $\coprod_{\alpha} U_\alpha$, the disjoint union of the sets $U_\alpha$. Secondly, it is $f|U_\alpha$, not $f|\prod_{\alpha} U_\alpha$. Thirdly, an important assumption is missing: Each $U_\alpha$ is an open subset of $R^n$. With this in mind, $f|U_\alpha$ is just ...


33

To answer the question, yes. $$ a \nless b\\ a \ngtr b\\ a \nleq b\qquad a \nleqq b\qquad a \nleqslant b\\ a \ngeq b\qquad a \ngeqq b\qquad a \ngeqslant b $$ and so on for many other mathematical relations $$ a \nleftarrow b\\ a \nLeftarrow b\\ A \nsupseteqq B\\ A \nvdash \phi\qquad A \nVdash \phi\\ \nexists x $$


11

I would probably use $850 \le 950$, as order is defined for integers.


9

Equality is special in that there are two ways that two real numbers $a$ and $b$ can be not equal: $$a>b,b>a$$ So, instead of saying $a>b \;\textrm {or}\;b>a$, we write $b\neq a$. For the others, each negation has an existing symbol, so: $$a\not>b \iff a\leq b,\;\,a\nleq b\iff a>b$$ etc. But like the comments say, either is OK.


1

I post here the corrected example. Let $(\Omega,\mathcal{A},P)$ be a probability space and $(E,\mathcal{E})$, $(E_1,\mathcal{E}_1),\dotsc,(E_n,\mathcal{E}_n)$ be measurable spaces. The random variable $X\colon \Omega\to E$ is distributed according to $P_X$. Let $X_i\colon \Omega\to E_i$ also be random variables and define a sample as ...


4

See Restricted quantifiers: $(\forall x \in D)P(x)$ is equivalent to: $\forall x (x \in D \to P(x))$. Thus, your formula: $∀x \in U : f(x) \in V$ is equivalent to: $\forall x (x \in U \to f(x) \in V)$.


1

$\sum_{ij}$ is just a shorthand for $\sum_i\sum_j$. They both mean "sum over index $i$ and index $j$". In some cases the order counts, in that case the notation with two sums is preferable.


4

The way I interpret your question: If $a_{ij}$ are scalars indexed by some subset of $\Bbb N^2$, then are the two summation signs the same thing? I'll assume you know some basics of sequence and series theory. (If not, then this answer is not helpful to you.) In the absence of absolute convergence, $\sum_{ij}$ itself makes little sense because it's ...


1

As explained by DylanSp and Paolo the right relation would be inclusion, not pertenence but in addition the way you expressed those intervals is wrong, they should be: $$(2,\infty) \ \subset \ (0,\infty) $$ or $$\lbrace x|x>2 \rbrace \subset \lbrace x|x>0 \rbrace$$ Added: I'm using the convention that the symbol $\subset$ denotes the non-strict ...


0

No, it wouldn't; $x \in y$ denotes that $x$ is an element of the set $y$. What you want is to denote that an interval $i$ is a subset of another interval $j$. The notation for this is $i \subseteq j$. ($i \subset j$ denotes that $i$ is a "proper" subset of $j$, meaning that $i \neq j$.)


1

In this case $\hbar$ is just a parameter. Frequently, one also denotes $U_{\hbar}(g)$ as $U_q(g)$. $U(g)[[\hbar]]$ denotes the space of formal power series over $U(g)$, i.e. if $f\in U(g)[[\hbar]]$, then $$f=\sum_{n\in\mathbb{Z}}a_n \hbar^n$$ with $a_n\in U(g)$ for every $n$. (Sometimes people reindex so that the sum is over $\mathbb{N}$ or some other ...


-1

Usually $\hbar$ is simply the reduced Planck constant $h/(2\pi)$. I have sadly no idea why it is featured such prominently in your context.


1

I FINALLY found an answer to what I was imagining. Called a hybrid set, it can contain negative multiplicity, which would allow the single hybrid set to represent both of the "opposing multisets."


0

Using these operations on multisets, you could say that a pair $(A,B)$ of multisets is reduced to the canonical form $$ (A-(A\cap B),B-(A\cap B)) $$ and all pairs that reduce to the same canonical form are equivalent. Equivalently, you could say that two pairs $(A,B)$ and $(C,D)$ of multisets are equivalent exactly if there are multisets $E$ and $F$ with ...


1

I am not sure what you mean by one-sided derivatives especially when you say that they exist. According to wikipedia, the notation for left and right derivative of $f(x)$ at $x=a$ is $$\partial_+f(a)=\lim_{x\to a^+}\frac{f(x)-f(a)}{x-a}$$ $$\partial_-f(a)=\lim_{x\to a^-}\frac{f(x)-f(a)}{x-a}$$ According to that definition, for your function ...


1

Seems the question is about left/right derivative. The notation is normally: $f'(x^+)$ and $f'(x^-)$. In your case it would be: $f'(0^+)$ and $f'(0^-)$.


1

I think the most simple notation would be to write $$ 0_{M_{0,0}}$$ for a matrix of size $0\times 0$, and $$ 0_{M_{3,0}}$$ for a matrix of size $3\times 0$. It's very similar when you want to write the zero of an unknown field $F$, when you write $0_F$.


0

The third one is the better because it shows the dependency of $F$, and it shows clearly that $F$ is a function from $\mathbb{R}^3$ to $\mathbb{R}^3$, i.e. it associates a vector to each vector.


0

The first uses unit vectors. This is called 'unit vector notation'. The second and third are the same notation, but the third is more explicit/unambiguous. The notation $(2x,3y,5z)$ or $\left( \begin{array}{c}2x\\3y\\5z\end{array}\right)$ is called 'matrix notation', because it's a matrix with respectively 1 row and 1 column. Source: ...


0

Such a notation often means a vector that is a solution of minimization problem in linear programming. I googled for another article on MDS problem, it uses just ordinary vector notation (see p.4)


0

In Hermann Sohr's The Navier-Stokes Equations --- An Elementary Functional Analytic Approach (p.23), $C^k(\overline\Omega)$ means the space of all restrictions $u|_{\overline\Omega}$ to $\overline\Omega$ of functions $u\in C^k(\mathbb{R}^n)$ such that $$ \sup_{|\alpha|\leq k,x\in\mathbb{R}^n}|D^\alpha u(x)|<\infty. $$ Here $|\alpha|\leq k$ is replaced by ...


-1

One of my teacher used that for vectors, instead of the usual notation with an arrow over the variable. But it was a physic teacher. Never trust physicists!


0

It's not an often-used convention, but in physics, matrices are sometimes appended with a double line underneath and vectors a single line underneath. This somewhat unifies the matrix/vector notation without the clumsiness of vector notation (and how to extend that to matrices).


1

Note that since $n \mid (9^n-1)$ and $n \mid (9^{\varphi(n)} - 1)$, it divides any common divisors of $9^n - 1$ and $9^{\varphi(n)} - 1$. However, we note that for any integers $a,b$, we have $$ 9^{ab} - 1 = (9^{a})^b - 1 = (9^a - 1)(1 + 9^a + 9^{2a} + \cdots + 9^{a(b-1)}) $$ Conclude that $9^d - 1$ is a common divisor of $(9^{n} - 1)$ and $(9^{\varphi(n)} ...



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