New answers tagged

0

There is also an ambiguity between a decimal fraction with a dot, as in $3.5^2$, and multiplication with a centre dot, as in $3\cdot5^2$, particularly if the latter doesn't have spacing around the dot to give context, as in $3\!\cdot\!5^2$. In fact, I seem to remember some textbooks using a centre dot for decimal fractions.


0

I have not seen any special symbol for this. I would write $S=\{\{a,b\}:a\in A\land b\in B\}.$ Of course if $x\in A\cap B$ then $\{x\}$ belongs $S.$ If you wish to exclude one-member sets you can write $T=\{\{a,b\}:a\in A\land b\in B\land a\ne b\}. $.... However there is are symbols $[A]^2$ for the set of all subsets of $A$ that have exactly $2$ members ...


4

The lowercase letter $x$ and the multiplication cross $\times$ (\times in TeX, × in HTML) are very different symbols. One can be used to represent a variable, as you have already halfway surmised, but it shouldn't be used to denote any kind of multiplication, whereas the other can be used to denote multiplication but should not be used to represent ...


3

It's because . stands for any binary operation which might look like 'multiplication' in some particular setting, or might be a substitute for multiplication. Hence, it is more general in nature.


11

There are multiple reasons. Perhaps the most important is a desire to make the notation as concise as possible. The change is not really from $a\times b$ to $a\cdot b$. It is from $a\times b$ to $ab$. In most undergraduate algebra books the "multiplication" operation is just denoted by juxtaposition. This is also linked to a desire for speed, or maybe one ...


2

This is primarily done to emphasize different multiplication operations in terms of vector and multidimensional calculus. In particular, this is to emphasize that the dot product $\cdot$ is mechanically different from the cross product $\times$, although in operations on objects of one dimension, they are virtually the same.


0

It's unclear the question but i think i can give an answer: You can think "the set of all different sets consisting on exactly one element of each subset of a a set $S$" as the cartesian product: \begin{equation} \prod\mathcal{P}(X)\setminus\{\emptyset\}=\{f\mid f:I\to\bigcup_{i\in I}X_i, \text{ }f(i)\in X_i\} \end{equation} where I is an index set for $\...


2

In $\mathbb R^n$, it is common to use $S^{n-1}$ to denote $\{v \in \mathbb R^n : ||v|| = 1\}$.


0

That "big caret" is "and". This says "Argmax(f) is the set of all values of x such that f(x) is larger than equal to f(y) for all y in the set S." Equivalently, "Argmax(f) is the set of all x that give the maximum value of f."


0

The caret (the symbol is $\wedge$, written as \wedge in LaTeX) means "and". The definition says that the argmax is the set of all $x$ which are in $S$ and satisfy $f(y) \leq f(x)$ whenever $y$ is also in $S$. We need this funny definition mainly because the argmax might have more than one element, for example this occurs with $f(x)=-x^4/4+x^2/2$. It also ...


0

Given a set $S$ and a real valued function $f$ then $\operatorname{argmax}_S f = \{ s \in S | f(s) = \sup_{x \in S} f(x) \}$.


1

To avoid clutter, let $$f(i_1,i_2,\ldots,i_u)=\frac{\alpha_1}{r^{i_1}}\cdot\frac{\alpha_2}{r^{i_2}}\cdot\ldots\cdot\frac{\alpha_u}{r^{i_u}}\;.\tag{1}$$ (It could just as well be any function of $u$ variables.) Then $$\sum_{\substack{1\le i_j\le m\\i_1,i_2,\ldots,i_u}}f(i_1,\ldots,i_u)\tag{2}$$ is the sum of the values $f(i_1,\ldots,i_u)$ as $\langle i_1,...


2

I think as others are getting at, the confusion probably arises from the (great) question: what is $dx$ anyway? I will try and give an intuitive answer that should help a little with the confusion. The rigorous answer is that $dx$ is a differential 1-form, which you will learn about if you do real analysis, but I think a perfectly good intuitive response is ...


1

An operator is an object that acts on another object, typically to the right, though proper definition of the space you're working in will clarify that. We require that operators and the objects they act on fulfill certain requirements. Integrals and derivatives acting on functions fulfill the requirements for a vector space of functions. When an integral is ...


1

Would this help? \begin{align} \left( \sum_{j=1}^2 \dfrac{\alpha_j}{r^j} \right)^3 &= \left( \dfrac{\alpha_1}{r}+\dfrac{\alpha_2}{r^2} \right)^3 \\ &= \left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right) \left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \right) \left( \dfrac{\alpha_1}{r^1}+\dfrac{\alpha_2}{r^2} \...


2

The hyperreal view helps clarify the picture with regard to the question whether $f(x)dx$ in the expression $\int_a^b f(x)dx$ is a product or not, as well as the issue of commutativity. From the hyperreal viewpoint, the integral is defined as the standard part of an infinite Riemann sum $\sum_i f(x_i)\Delta x$. In the Riemann sum, the term $f(x_i)\Delta x$ ...


0

in the set theory we have $\{x_1,...,x_n\}=\{x_i\mid 1\leq i\leq n\}=\cup_{1\leq i\leq n}\{x_i\}$ so is the set of $n$ elements, and the set $\{x_i\}^{n}_{i=1}=\{x_1\}\{x_2\}\cdot\cdot\cdot\{x_n\}=\{(x_1,\cdot\cdot\cdot x_n)\}$ is a set of one element.


0

What you described is not a set, you probably mean $\{{x_1,x_2...,x_n}\}$ This is usually denoted as ${A=\{{x_i|i\in I}\}}$ Where $I$ is your index set $I=\{1,2,...n\}$


0

For the set $\{x_1,\ldots,x_n\}$ you can write $$\begin{cases}\{x_i\}_{i=1}^n \\ \\ \{x_i , 1\leq i\leq n\} \\ \\ \{x_i\}_{1\leq i\leq n}. \end{cases}$$


3

There is an index of notations in the book on page 573. If you look up "$\Delta$" in it, you see that this is the symmetric difference of sets: $$X \mathbin\Delta Y = (X \setminus Y) \cup (Y \setminus X).$$ It does not appear to be an operation on graphs, just on sets.


3

In plain language, you can interpret the $\int$ and $\mathrm{d}$ as a verb or a command: "integrate" or "differentiate" (the following) and the $\mathrm{d}x$ as a complement to the verb: "with respect to variable $x$". With $\dfrac{\mathrm{d}}{\mathrm{d}x}$, the verb and its complement are generally close by, and the fractional separation makes it quite ...


13

Others have explained the syntax and semantics of this notation, but no one seems to have addressed why it is a good thing to do... The notation "$\dfrac{\mathrm{d}}{\mathrm{d}u}$" is used for the differentiation operator (a gadget which takes functions as input and produces functions as output). This is the "differentiate with respect to $u$ operator". ...


5

The meaning is the same. Placing dx at the beginning of the integral has advantages when you have nested integrals: $$\int_{x_0}^{x_1} dx \int_{y_0}^{y_1} dy \int_{z_0}^{z_1} dz \; f(x,y,z)$$ Otherwise, you need lots of parentheses in order to know which variable is integrated in which interval, and becomes less legible: $$\int_{x_0}^{x_1} \left( \int_{...


8

While, as other answers have noted, some authors write the differential (the $du$) next to the integral sign, I would recommend avoiding it. It's uncommon, meaning many people who see it will have the same confusion you did. But also, having the differential at the end of an integral is a notational convenience, as it shows the end of the integrand ...


0

If there is "co-exponential" $f:B\to C\times A$ as you suggest, let $f_1:B\to C$ and $f_2:B\to A$ be the compositions of $f$ with the projections. The universal property of $f$ says that given any $g=(g_1,g_2):B\to D\times A$, there is a unique $h:C\to D$ such that $g=(h\times 1)f$. Composing with the projections, $g=(h\times 1)f$ just says that $g_1=hf_1$ ...


31

The notation $\int f(x) \, dx $ and $\int dx \; f(x)$ mean the exact same thing.


6

This is just another way of writing $\int \frac{1}{(u-1)u^2} \, du$.


13

First, note that $$ \frac{1}{u-1}-\frac{1}{u}-\frac{1}{u^{2}}=\frac{1}{u-1}-\frac{u+1}{u^{2}}=\frac{1}{\left(u-1\right)u^{2}} $$ from which the equality follows: $$ \int du \left(\frac{1}{\left(u-1\right)u^2}\right)=\int du\left(\frac{1}{u-1}-\frac{1}{u}-\frac{1}{u^{2}}\right). $$ Note that $\int fdu$ and $\int du f$ mean the same thing.


1

I'm not aware of any conventional method of writing this. That said, if the application is such that it is important that this be able to be written in one expression, I'd suggest doing it in two steps: First, let $S' \subset \{0,1\}\times S$. Define $\hat{y}_{S'}$ as the vector whose $i$th component is $y_i$ if $i \not \in S$, and $\pi_i(S')$ if $i\in S$ (...


1

If I understand your question correctly, you could write $$\sum_{s \in S}\sum_{y_s\in\{0,1\}}.$$


1

Usually, if $\mu $ is a measure and $f $ is a function, then $f\mu $ is the measure given by $$ (f\mu)(A) =\int_A f \, d\mu. $$ This is indeed a measure as long as $f \in L^1 (\mu)$ or if $f : X \to [0,\infty] $.


-2

To a category theorist, this is called functoriality. In particular, observe the similarities: If $f : X \rightarrow Y$ is a function between sets and $x,x'$ are elements of $X$, then we get an implication like so: $$x = x' \rightarrow f(x) = f(x')$$ If $f : X \rightarrow Y$ is an order-preserving mapping between posets and $x,x'$ are elements of $X$, ...


2

You can be less formal by saying something like this, and everyone will understand you: Let $\cup_{i \in I} R_i$ denote a disjoint union of copies $R_i$ of $\mathbb{R}$...


1

In all rigor, you should use distinct names for the dummy variables $$ \int_{\nu=0}^{v} d\nu = \int_{\tau=0}^{t} -4 d\tau ,$$ then $$ \nu\bigg|_{\nu=0}^{v} = -4\tau\bigg|_{\tau=0}^{t}.$$


2

You understand that the usual implementation of disjoint union is as tagged elements (ordered pairs): $$ \bigsqcup_{i \in I} A_i = \bigcup_{i \in I} \{(i,x) \mid x \in A_i\} $$ So your $x$ and $y$ are pairs indicating, respectively, the index of their member of the disjoint union and element from that member. Note that it is very difficult to ...


4

I don't think there's any notation that is more standard than $\coprod_{i\in I}\mathbb{R}$ that makes it clear you are talking about a disjoint union of copies of $\mathbb{R}$. But if you want a nice way to be able to refer to the individual copies, you could just write it as $I\times\mathbb{R}$, with the $i$th copy being $\{i\}\times\mathbb{R}$. The ...


2

It means that it is a function that maps the numbers in the interval $[a,b]$ to numbers in the interval $[c,d]$. $[a,b]$ is the domain and $[c,d]$ is the codomain. $f: X \rightarrow Y$ is fairly standard notation. Example if $f(x) = x^2 - 7$ for $-1 \le x \le 5$ then $f:[-1,5] \rightarrow [-7,18]$ because $f$ maps the numbers in the interval $[-1,5]$ to ...


1

We read $f: [a,b] \rightarrow [c,d]$ as "$f$ is a function from the closed interval $[a,b]$ to the closed interval $[c,d]$". Alternatively, "$f$ maps the closed interval $[a,b]$ to the closed interval $[c,d]$". In general, $f: A \rightarrow B$ means that $f$ is a function that maps elements from its domain $A$ to elements in its range $B$.


2

It means that $f$ is a function whose domain is the closed interval from $a$ to $b$ and whose codomain is the closed interval from $c$ to $d$.


3

Where did this notation come from? In lattice theory we have join and meet [see: Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), page 34] : the least upper bound of $a, b \in A$ will be denoted by $a \cup b$ and called the join of elements $a, b$, and the greatest lower bound of $a, b \in A$ will be denoted by $a \cap b$ ...


1

How to remember them? Long ago someone showed me his method. I still use it sometimes. Read the three corners like this:


0

Normally the notation $g:Y\to X$ is used to mean that $g$ is a function from $Y$ to $X$ (though it can have other meanings in certain contexts). So by writing $g:Y\to X$, you would be asserting that $g$ is a function with domain $Y$, which is true only when $f$ is a bijection. The notation $g:Y\looparrowright X$ is sometimes used to denote "$g$ is a ...


1

Writing $(-\infty,-1]\cup\{-1\}$ is redundant since $-1\in(-\infty,-1]$. When you write cl in LaTeX, write \operatorname{cl} instead; it will give it the right styling. Lastly, $\operatorname{cl}\mathbb{Q}=\mathbb{R}$, since each sequence in $\mathbb{Q}$ converges to a real number and any real number can be written as a convergent sequence of rationals.


1

I would like to say a few things here...are you studying analysis or topology? If you are studying analysis then you can talk about closure which directly means closure in standard topology but if you are studying topology you need to ask first is it standard topology or discrete topology or lower limit topology or K topology and stuff like that. Assuming ...


1

This notation is borrowed from Computer science and means (in Computer science) ‘takes the value…’. I prefer the more explicit old style $$f'(a) \stackrel{\text{def}}{=}\mkern1mu \lim_{h\to 0} .\frac{f(a+h)-f(a)}{h}$$


4

It is common to use $:=$ when you are defining something. That way you communicate that it isn't a formula that is derived, but something that is defined. For example, I might say that the velocity of a particle is $$ v := \frac{d}{dt} s(t) $$ where $s(t)$ is the position. By using $:=$ I have told the reader that $v$ is defined as the derivative of the ...


2

The "=" is equality. The case you give is definition. It is also sometimes used as redefinition, e.g. x := x + 1 for a programming language.


0

Its fine if it dont lead to some ambiguity. An alternative notation can be $x,y\in\Bbb R_{>0}$. Or just $x,y\in[0,\infty)$.


5

It's commonly used notation, so yes. And even if it wasn't, you can always define notation in the beginning, and then use it. In this case, you'd see something like "For brevity, we will use $x,y,\ldots>0$ to mean $x>0$, $y>0$, etc. in the rest of the article."



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