New answers tagged

0

The bar notation is used to indicate the determinant of a matrix. It converts a matrix into a scalar. For example, $$\begin{align}\begin{vmatrix} a & b\\c & d \end{vmatrix}=ad - bc \end{align}$$ Other common notations are $\det A$ and $\det (A)$. In particular, for your case, the mentioned determinant becomes $$\begin{align}\begin{vmatrix} \ln (2x+1) ...


1

The bar notation means taking the determinant of the matrix enclosed. This is a shorthand for $\det M$.


3

The most common use of this symbol is as logical operator "or", which connects two statements. So for two statements $A$ and $B$ the expression $A\vee B$ would read "A or B". As many other symbols this has other uses too, so it depends on the context. You linked a set-theory related topic. The other symbol "$\wedge$" is the ...


1

I guess I would choose something along the line of $[a]_m \mapsto [a]_n$ for the first one, if I really wanted to specify that with symbols. Otherwise there is nothing wrong with describing the map with words, which is not uncommon at all. You are reducing mod $n$. For the second one I would always use the standard notation $$\overline{a_0} + \overline{a_1}t ...


1

You already wrote down a Euclidean space in your question: $\mathbb{R}$. The only other thing I can think of that you might want to include is your metric. Say $(\mathbb{R},d)$ is a metric space and define d, which is the distance of any two points. There are some axioms to remember for metrics: $d(x,x)=0$ $d(x,y)>0$ $d(x,y)=d(y,x)$ $d(x,z)\leq d(x,y)+...


0

I'm 8 years late, but I've seen that this answer is not on the list. In Romania we use a symbol similar to a pair of scissors, that I have learned from both my teachers I've studied with. Unfortunately, I was unable to find such a symbol in MathJax, but I will try reproducing it: $\require{HTML} \style{display: inline-block; transform: rotate(30deg) scaleX(-...


0

The reason that people use alternative notations is because there is no/little consensus in some areas of mathematics. Note, that mathematics is the science that deals with logic and arrangements. Moreover, it is all around us. Whether or not you believe that mankind had a single origin, by the time humans started writing, in particular writing down ...


0

If you want understandability and not formality here's what I'd go for: $B(M) = \begin{cases} 1: & E_M \neq \emptyset \\ 0: & E_M = \emptyset \end{cases}$ Where $E_M$ is the set of events that happen between time $(t - M)$ and $t$: $E_M = \{ e_p: e_p \in E \text{ and } t-m \leq p \leq t\}$ Where $e_p$ is an event that occurs at time $p$, and $E$ is ...


0

Let $R$ be a ring (in your example $(\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$) ) equipped with addition $+$ and multiplication $\times$. In $\mathbb{Z}$ or $\mathbb{R}$ or $\mathbb{Q}$ (or choose familiar favourite ring) you'll be familiar that if $n \geq 0$ is an integer, we write $nx$ to mean $x +... + x$ ($n$ times) and $x^n$ to mean $x \...


2

$X \ll_{n,\epsilon} Y$ typically means that there's a "constant" $C$ which depends on the parameters $n$ and $\epsilon$ such that $X \leq C \cdot Y$. This is meaningful when you consider $X$ and $Y$ as functions of some other variable besides $n$ and $\epsilon$, and treat $n$ and $\epsilon$ as parameters.


1

$\mathbf{Z}$ is an $n \times k$ matrix, and the first $\mathbf{1}$ is a $k \times 1$ matrix (column vector) of 1s (often written as $\mathbf{e}$). The second $\mathbf{1}$ is an $n \times 1$ matrix (column vector) of 1s, and $\mathbf{Z 1} = \mathbf{1}$ means $$\sum_{j=1}^k Z_{i,j} = 1$$ for $i \in \{1,\dots,n\}$. Because $Z_{i,j} \in \{0,1\}$, this implies ...


0

There's notation and there's notation! If you are using notation in the sense of shorthand symbolism helpfully added to mathematicians' English, with "$\forall$" shorthand for "for all", then "$\forall x, y \in \mathbb{R}$" is perfectly acceptable shorthand "for all $x$ and $y$ in the real numbers". There are no strict ...


2

This is entirely standard shorthand notation. It could be argued that it is slightly unclear whether $x\in\Bbb R$, but mostly this is apparent from context. If you want to remedy this minor issue without being too verbose, you could, for instance, say $$\forall (x,y)\in\Bbb R^2$$and there is absolutely no more room for any reasonable ambiguity.


2

$\mathbb I$ is an indicator function. It returns $1$ if the predicate inside is true and $0$ if false. Sometimes this is written $[P]$, the Iverson bracket.


1

This 1909 trig textbook mentions $\mathrm{inv}\sin{x}$ as an alternate form of $\arcsin{x}$ and $\sin^{-1}{x}$ and says it may be better than those because of its "general applicability". Applying this to $f(x)$ would presumably give either $\mathrm{inv\ }f(x)$ or $\mathrm{inv}f(x)$. Wikipedia also cites a 2009 textbook when it talks about putting $...


1

Trigonometric Functions (And Functions In General) Aren't Considered As Operators, But As Functions Instead, So, Technically Speaking, They Can't Belong In The Order Of Operation (A.K.A. BODMAS/PEDMAS/BIDMAS). BUT, If We Would Count Them As Operators, Then They Would Be Computed In The Very Beginning Of A Equation. That Is, For Example, If We Have A Equation ...


4

I think the convention is just to remember which is which. I can see how your example may be confusing. I will give you two strategies for making it easier to distinguish which is which: Name your variables consistently and intuitively! Your example is en excellent illustration of a poor choice of variable names. Since $f$ and $g$ are functions, $h$ is a ...


1

A good resource for all things Sobolev-space related is Sobolev Spaces by R. A. Adams and J. J. F. Fournier, $2^\mathrm{nd}$ edition. Page $60$ there provides the definition that you're looking for: $$H^m_q = H^{m,q}(\Omega) \equiv \mbox{the completion of } \{u\in C^m(\Omega) \mid \|u\|_{m,q} \lt \infty \} $$ where the completion is with respect to the ...


0

There is no single correct notation. Often images are defined as a function on a discrete domain $D \in \mathbb{Z}^2$. But when talking about binary images as sets, only the true (foreground) pixels are considered part of the set. So the cardinality of the binary image set is the number of true (foreground) pixels. This explains the Dice score definition.


0

You mentioned in a comment that this came up in an algorithmic context. In programming terms, the general expression for this is reduce(operation, iterable) (of course, this implies that you're dealing with an associative binary operator; reduce(mean, iterable) isn't going to get the mean of the iterable), e.g. reduce(or, (f(_) for _ in range(k))). There's ...


0

This is the inverse limit, also known as projective limit.


1

The blue path is the boundary of $S$, denoted $\partial S$. We cannot define an inverse operator $\partial^{-1}$ to apply to this boundary so as to recover $S$, because $S$'s complement $S^\complement$ has the same boundary, i.e. $\partial S=\partial S^\complement$. However, you can disambiguate which set with the given boundary is intended. An important ...


1

AFAIK there is no standard notation for this. Best to say it in words.


2

We tend to introduce $\log(x) = \log_{10}(x)$ because before you know anything about logarithms it's most convenient to think about things in terms of base $10$. Once you learn enough about logs, you know that base $e$ is really the most convenient to work with, but at that point you've probably been introduced to the notation $\ln(x) = \log_{e}(x)$. For ...


1

In pure mathematics, we hardly ever need to use $\log_{10}$. It's true that using $\ln$ would probably be more clear, but we tend to read out "$\ln$" as "log" anyway (out loud), so there is an argument to be made to just use $\log$ since it's the most natural notation which reflects how we communicate.


0

It really depends. Mostly in physics $\log(x)$ is assumed as base e but to be safe you should probably us $\ln(x)$. It highly depends on the context. In elementary math, $\log(x)$ is assumed as base $10$.


1

It means composition, for Instance $g(x)=x^2$ and $f(x)=-1/x$. So $g(f(x)) = (-1/x)^2 = \frac{1}{x^2}$ and $f(g(x))= \frac{-1}{x^2}$


0

Tuples (in general points in $R^n$) are points in space. Vectors have a length and direction. A one to one correspondence between vectors and points in space can be made by fixing the tail of a vector to the origin and the head of the vector at that point in space. Note that vectors themselves are free floating - allowing addition, etc.


2

Guide: The $4$ cases are $1 \in B$, $0 \in B$. $1 \in B$, $0 \notin B$. $1 \notin B$, $0 \in B$. $1 \notin B$, $0 \notin B$. Evaluate the set for each case. It is possible that $B$ contain elements other than $0$ and $1$, of which these elements will be ignored in this particular computation. For example $f^{-1}(\{0,1,2\})=f^{-1}(\{0,1\})$ for our $f$. $$f^...


1

The thing to keep in mind is that partial differentials, as they are commonly notated, are not distinct. That is, $\partial \phi$ can refer to any number of values. This is an unfortunate development in the history of mathematics. One way (but not the only way!) to do it is to notate on the differential which other variables were free (i.e., independent) ...


2

Interestingly, you can quite easily use a statement like $$(\forall i\in I)(\exists a_i\in A_i)$$ to get repeated "there exist" statements and similarly you could use $$(\forall i\in I)(\forall a_i \in A_i)$$ to get repeated "for all" statements. You would just need to define the set $I$ and the sets $A_i$. Note that each $a_i$ is not ...


9

Use \bigvee and \bigwedge , $$ \bigvee_{i=1}^{100} X_i $$ and $$ \bigwedge_{i \in I} X_i \text{,} $$ respectively. See also What is the meaning of $\bigvee$ (bigvee) operator Be aware that these are also used for meets and joins (lattice theory). Depending on your context, it could be a good idea to explicitly introduce this notation.


3

If you plan on specifying the parameter $a$, it might be best to choose beforehand a notation that reflects that intention. For example, either of the notational choices $y(x,a)=3x^2+ax+b$ or $y_a(x)=3x^2+ax+b$ makes it easy to show what you want to show. (And similar notational choices can be used for $b$, with or without $a$.)


0

The notation $dx^2=2x$ is wrong and might confuses you. The abstract derivative as a limit of differential quotient comes first. The following notation is all the same: $$ \lim \limits_{x\to a} \frac{f(x)-f(a)}{x-a}\equiv f'(a)\equiv \left.\frac{dy}{dx}\right\vert _{a}.$$ However, since a derivative is computed by the quotient, we might design a new notation ...


0

To provide a proof of $$t\frac{\partial f}{\partial x}(tx,ty)=t^r\frac{\partial f}{\partial x}(x,y)$$ it is sufficient to show $\frac{\partial f}{\partial x}$ is homogeneous of degree $r-1$. By definition $$\frac{\partial f}{\partial x}(tx,ty)=\lim_{h\to 0}\frac{f(tx+h,ty)-f(tx,ty)}{h}.$$ Using homogeneity, we can rewrite this as $$t^r\lim_{h\to 0}\frac{f(x+\...


0

allow me to take one more tiny little step forward based on your excellent calculations. $$ \sqrt{1+x}=\sum _{n=0}^{\infty } \frac{ 1}{(1-2 n)} \frac{(2 n)!}{n! n!}\left(-\frac{x}{4}\right)^n $$ or $$ \sqrt{1+x}=\sum _{n=0}^{\infty } \frac{ C_{2n}^n}{(1-2 n)} \left(-\frac{x}{4}\right)^n $$


0

It is true that 5 is a divisor of 0, and yet it is true that there are no divisors of 0.


3

The notation working out nicely for Lagrange's theorem is one reason. There's a couple others: $\mathbb{Z}/n\mathbb{Z}$ is our prototypical case for quotient groups, and here it really is very closely related to division; the equivalence class of $x$ mod $n$ is the (equivalence class of the) remainder of $x$ divided by $n$. The notation also works out very ...


4

Your set is $ (\mathbb Q + \mathbb R) \cup (\mathbb Q + i\mathbb R) = \mathbb R \cup (\mathbb Q + i\mathbb R) = \{ z \in \mathbb C : \operatorname{Im}(z)=0 \text{ or } \operatorname{Re}(z) \in \mathbb Q \} $. Choose the expression that makes your intention clear to the reader. I guess it is the first one.


2

$\mathbb Z[x]$ is the ring of all polynomials over the variable $x$ with integer coefficients. So this includes $a+bx$ (for all $a,b \in \mathbb Z$), but also things like $x^2$, and the other higher order polynomials. Whereas $\mathbb Z[i]$ is what we get when we evaluate the polynomials at $i$. That is, we take an integer polynomial $f(x)$ and plug in $i$ ...


2

$(2)$ is just a rewrite of $(1)$, using simple logical transformations and the definition of $\cup$ and of the comprehension set notation: $$\begin{align}t\in \{\,x\in A\mid \phi(x)\,\}\cup \{\,y\in A\mid \psi(y)\,\} &\iff t\in\{\,x\in A\mid \phi(x)\,\}\lor t\in \{\,y\in A\mid \phi(y)\,\}\\ &\iff (t\in A\land \phi(t))\lor (t\in A\land \psi(t))\\ &...


0

$$(1)\space X \cup Y = \{x \in \Bbb Z \mid \exists_{a}[a \in \Bbb Z: x = 2a +1]\} \cup \{y \in \Bbb Z\mid \exists_{b}[b \in \Bbb Z: y = 5b]\} $$ The above statement is a union of X and Y. Here X is a set of integers, z, for which there exists an integer, $a$, such that z = 2a + 1. we see that X is a set of odd numbers. Y is a set of integers, y, for which ...


1

Yes indeed this is just the notation for a 'part' of the Jacobian. In general, if you have a function $f:\mathbb{R}^{d}\to\mathbb{R}^{k}$, then you define $\partial_{p}f$ with $p=(x_{1},\dots,x_{l})$ for some $l\in\{1,\dots,d\}$ as $$\mathcal{J}_{f}(x)=:(\partial_{p}f(x),\partial_{q}f(x))$$ where $\mathcal{J}_{f}(x)$ denotes the Jacobian and $q=(x_{l+1},\...


3

The third one, $\lim_{x \to a^+} f'(x)$, is different! If that limit exists and $f$ is continuous at $a$, then it equals the right-hand derivative $f'_+(a)$, as can be shown using the mean value theorem for derivatives. But it may happen that it doesn't exist even if $f'_+(a)$ exists, as shown by the standard examples of differentiable but not continuously ...


-2

I think it's due to history. Groups were discovered/invented as a complete unit and sub groups were found later(Starting with Lagrange).First field existed intuitively as rational numbers,Later Galois extended them with roots looking for 5th degree equation solutions.


1

The expression $$ x^n=1\land \forall k<n\colon x^k\ne 1 $$ is a logical conjunction of the two terms $$ x^n = 1$$ and $$ \forall k<n\colon x^k\ne 1 .$$ You might write it like this to make it clearer how the symbols are grouped into expressions: $$ (x^n=1)\land (\forall k<n\colon x^k\ne 1). $$ This conjunction is true if and only if $x^n=1$ is ...


1

In this context, $\land$ is the logical AND operator \land.


1

Hint: You want to bound the denominator from below. The denominator equals zero when $x=\pm\frac{\sqrt{2}}{2}$. Set some $c<\frac{\sqrt{2}}{2}$. Then when $|x|\leq c$, it follows that $|1-2x^2|\geq 1-2c^2>0.$


0

Very often it works to just think of $\mathrm{d}f$ as an alternative notation to $f'$ or the derivative of $f$ with respect to some unnamed variable. For example, assume that $x$ and $y$ are functions of some common variable. Then we have $$ y\,x' - x\,y' = -x^2 (y/x)' $$ Just using the notation $\mathrm{d}f$ instead of $f'$ gives $$ y\,\mathrm{d}x - x\,\...


4

You should know that the differential at a point $\mathbf x_0$ of a function $\;\mathbf R^m\longrightarrow \mathbf R^n$ is the linear map $\:\ell:\mathbf R^m\longrightarrow \mathbf R^n$, that yields the best linear approximation of $f(\mathbf x_0)$ in a neighbourhood of $\mathbf x_0$, in the sense that we have $$f(\mathbf x_0+\mathbf h)=f(\mathbf x_0)+\ell(\...


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