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0

It is not conventional and will not be broadly understood without an explanation of what it means, but there's nothing incorrect about it if you explain it before using it. Here's another example of its use: Suppose $H_1,\ldots,H_n$ are mutually exclusive hypotheses one of which must be true. Their probabilities given some new data $D$ are desired. Then $$ ...


4

This notation is not standard. One can define an equivalence relation on $\mathbb{R}^{n}$ s.t $v\sim u\iff u=\alpha v$ for $0\neq\alpha\in\mathbb{R}$ and then we can write for example $$ \begin{pmatrix}6\\ 2\\ 2 \end{pmatrix}\sim\begin{pmatrix}3\\ 1\\ 1 \end{pmatrix} $$ Regarding the use of the hat symbol and vectors - it is standard that if $0\neq ...


1

I don't think the particular one that you mention is valid. See here for a full list. He or she may have meant to write $ \cong $ or $ \equiv $; both of them make sense, since they both imply that the two vectors are equivalent and of the same form and meaning.


0

If you represent the complex no. as: $$z=a+bi=\sqrt{a^2+b^2}e^{i\arctan(b/a)}$$ Then the flipped number can be written as: $$z'=b+ai=\sqrt{a^2+b^2}e^{i\cdot(\Pi/2-arctan(b/a))}$$


6

As per the comments, if $$z = a + bi$$ then $$i\cdot\bar z = b+ai$$ where $\bar z$ is the conjugate of $z$.


3

There is no difference in meaning between "class" and "set" in this particular usage. He is correct on the right way to formalize these notations, but the shorthand is so much easier to read, and, as long as everybody knows what you mean, it is acceptable.


1

I don't know such notation, but note that in general, $$(a+bi)\times \left(\frac{2ab}{a^2+b^2}+\frac{a^2-b^2}{a^2+b^2}i\right)=b+ai$$ where $(a,b)\not=(0,0).$


0

a) Hint: $4^{\log_4 n} = n$. Note that these are all asymptotic statements, so you can pick a number, say $N=100000000$, and note that for $n \geq N$, $g(n)/10 \leq f(n) \leq 10 g(n)$. Thus, $f(n)$ is $\Theta(g(n))$. b) Note $g(n) = (2^{1/2})^{\log_2 n} = (2^{\log_2 n})^{1/2} = n^{1/2} = \sqrt{n}$. It should be obvious that $\sqrt{n} \leq n^2$ for $n ...


2

$\approx$ is used as the mathematical "approximately" symbol $-$ this means $ \Delta x $ has approximately the same value as $ \frac {\lambda}{\sin \alpha} $. However, $\sim$ is used a proportionality symbol, so there is some factor tacked onto $\frac{h}{mc\sin \alpha}$, which may be $1000$ or $2$ or something else.


0

If you're familiar with the indexed sum notation that looks like: $$\sum_{i=0}^n x^2 + 1$$ There's a similar union notation. $$\bigcup_{i=0}^n S_i$$


0

In France too, we use $\geqslant$ and $\leqslant$, at least in high school teaching. The $\geq$ and $\leq$ signs are understandable though, and used by pocket calculators.


1

It is standard to use $\oplus$ for the biproduct of modules. (aside: for infinite products of modules, $\oplus$ is interpreted as the coproduct) If the modules $M$ and $N$ have an algebra structure, then they induce a canonical algebra structure on their direct sum $M \oplus N$. It is standard, but strange, notation to write $R \oplus S$ for the algebra ...


1

Every element of $S$ can be written additively as $r_1+\cdots+r_n$ where $r_i\in R_i$. Every element of $U(S)$ can be written multiplicatively as $u_1\cdots u_n$ where $u_i\in U(R_i)$. We view the $R_i$s as subrings of $R$, so the sums/products are "internal."


1

Well, we could use $\mu(n)\ne 0$. Here $\mu$ is the Möbius function. I prefer (by a lot) to say square-free.


1

$n=ab$ implies $gcd(a,b)=1$ for all $a,b$


1

$\bf{Question\ 1:}$ The plus or minus sign does not distribute like that. To see this, consider the equation $x^{2}-4=0$. It should be obvious that the roots of this equation are $\pm 2$. We can rewrite that as $\pm (10-8)$ since $10-8=2$. However, Distributing the $\pm$ sign through $(10-8)$ would imply that $18$ and $-18$ are also roots of $x^{2}-4=0$. ...


5

If the $\pm$ sign is confusing you, get rid of it. If you have $$15 = \pm(a+x)$$ you can turn that into two equations: $$15 = a+x\\15 = -(a+x)$$ and then deal with the two equations separately, one at a time. That is exactly the meaning of the $\pm$ sign. The reason you're confused is because the notation is confusing! The expression $\pm a \pm b$ is ...


-1

1: Yes, you can distribute it. Well, it works for all cases. $$\pm(x + y) = \pm x \pm y \quad \forall x,y \in \mathbb R$$ The basic properties are: $$(-)\times (\pm) = (\mp)\\ (\pm)\times(\pm) = (+)\\ (\mp)\times(\mp) = (+)\\ (\pm)\times(\mp) = (-) $$ 2: As you said, $\pm$ is used to represent $+$ and $-$ in separate equations. It's a bit more ...


6

Should mean that $f$ is continuous on $\mathbb R$. This is also often written as $C^0(\mathbb R)$ where $C^n(\mathbb R)$ means that the functions in this set are $n$ times continuously differentiable.


0

It represents an open ball. In a normed space, $B(\epsilon)$ represents the set of all $x$ that are $\epsilon$-distant (under the norm $|| \cdot ||$ or generally under some metric) from a fixed point $x^*$. $B$ saysthat it is a ball. $\epsilon $ denotes the distance. $()$ and $[]$ generally stand for open and closed balls which contain $\lt $ and $\le $ ...


1

$x\in (-\infty,-4)\cup(4,+\infty)$ is a an option. It is a union of two disjoint intervals. There is no way of writing it as one interval.


1

There are two nested suprema, and as written, they are evaluated consecutively. So if $f\in C^\infty$ and $x\in\mathbb R^n$, you have $$ \sup_{x\in\mathbb R^n}\sup_{|\alpha|<k}(1+\|x\|^2)^k \lvert D^\alpha f(x) \rvert = \sup_{x\in\mathbb R^n}a(x), $$ where $a(x)=\sup_{|\alpha|<k}(1+\|x\|^2)^k \lvert D^\alpha f(x) \rvert$. But, fortunately enough, the ...


1

If you have a set $S$ and you want the set of values that you can get by applying $f$ to elements of $S$, you can write $$\{f(s) \mid s\in S\}.$$ This is the set of all values of the form $f(s)$, for some $s$ in $S$. This notation is standard. If you say up front “We will use the notation $f^*(S)$ to abbreviate the set $\{f(s) \mid s\in S\}$”, nobody will ...


1

No, that is not standard and is not likely to be understood as intended. As Sam points out in a comment, $C(M)$ is usually understood as $C(M,\mathbb R)$. So for your purpose a better notation should be $C(C(\mathbb R),C(\mathbb R))$. Furthermore, echoing PVAL's comment, it is not obvious what topology you have in mind on $C(\mathbb R)$, and this ...


3

It is the slice category (a special case of comma categories), sometimes also denoted by $\,1\!\downarrow {\bf Set}$, its objects are the arrows $1\to A$ of ${\bf Set}$ and its morphisms between $a:1\to A$ and $b:1\to B$ are arrows $f:A\to B$ that makes the triangle commutative, i.e. satisfying $\ f\circ a=b$.


1

The $\LaTeX$ code \restriction for $\restriction$ itself provides a clue to the meaning of $\preceq_Y\upharpoonright(X\times X)$ and it is $$\preceq_Y\upharpoonright(X\times X):=\preceq_Y\cap \,(X\times X).$$ Do not forget that binary relations, in particular partial orders, are just sets of ordered pairs. The obvious choice for an upper bound of $\mathscr ...


1

My goal here is to extend the idea proposed by Liu Gang. You could write $$S:=\{x \in X: x\text{ satisfy property } P\},$$ where $P$ is a property that characterizes local minima. Liu Gang chose $P$ to be the mathematical definition of a local minimum. You could also make it even easier by putting $P$ to be the property "is a local minimum", which would ...


0

Instead of specifying $A$ and then working with its inverse $A^{-1}$, do the opposite. Say calculate $B^{-1} = ..$ and the index with $B_{ij}$. I have done this often when I include the $\mbox{ }^{-1}$ in my matrix definition/calculation. Example: $$\Lambda^{-1} = s^\top T I^{-1}$$ $$\Lambda_{11} = m_1+m_2$$


0

Means the Power set. i.e. if $W = \{w_1, w_2, w_3\}$ then $2^W = \{\emptyset,\{w_1\},\{w_2\},\{w_3\},\{w_1,w_2\},\{w_1,w_3\},\{w_2,w_3\},\{w_1,w_2,w_3\}\}$


2

$$1\frac12=1+\frac12$$ and not $$1\cdot\frac12$$


2

I would write as $$\{x\in X:x=\mathop{\arg\min}_{y\in U^\circ,\ U\subset X}f(y)\}$$ where $U^\circ$ is the interior of $U$, hence open.


0

Like wrote Liu Gang in the comment, using a compact notation for neighborhood such as for example: $$ f(\bar{x}) \le f(x) \quad \forall x \in U(\bar{x}) $$ Or also: $$ \bar{x} = \mathop{\mathrm{arg\, min}}_{x \in U(\bar{x})} f(x) $$ where $U(\bar{x}) \in \mathcal{J}(\bar{x})$, a neighborhood of $\bar{x}$. I never saw more compact notation


1

Most likely it is the direct sum; this symbol is still alive in Banach space theory.


0

Fulton denotes $\mbox{ord}_{P}(C)$ to be the order function on $k(C)$ defined by $\mathcal{O}_{P}(C)$. Where $C$ is an arbitrary curve and $P$ is a point on that curve. This is found on the first two paragraphs of chapter 7, Resolution of Singularities, section 7.1, page 81. Hopefully this can give you some insight on how this applies to ...


6

Besides the complex-valued case, I suspect it also has to do with the existence of other Lp spaces; Wikipedia gives the general definition as $$ \|f\|_p = \left(\int_S |f|^p d\mu\right)^{1/p} $$ Since the absolute value symbols are redundant only for even integral $p$, omitting them disrupts the uniformity of the notation without buying a whole lot.


1

Now, thank to @GEdgar, I can see all my options but 4 are wrong. Just writing $(0,+\infty)\times \Bbb{R}$ or, using the definition of Cartesian Product, $\{ (x,y) : x,y\in\Bbb{R}, x>0 \}$ is correct. Thanks.


2

The quater-imaginary base $2i$ is quite amusing, being able to express every complex number using only digits in $\{0, 1, 2, 3\}$. Being the only base yet proposed here that includes the elements of this extended system, it is clearly the best one for this purpose :)


22

Because complex-valued functions are used. The square of a complex number need not be non-negative.


0

As I see in books and papers physicist often use d notation, mathematician $d$. Maybe because some paper think about d as an operator, and often divide with it, and calculate with it as a number, as an infinitesimal small difference of $x$. I think it is not always mathematically correct, but this is just my opinion. People who are not doing this prefer $d$ ...


1

I have seen the following notation, for the $i,j$th entry of $A$ use $A_{i,j}$, then for the $i,j$th entry of $A^{-1}$ use $A^{i,j}$.


1

There are two standard ways to index an element in a matrix. First. If you note the matrix with uppercase letters from the beginning of the english alphabet, then you can use the lowercase version of the letter while indexing. For example the matrix is $A$ and the element in the $i$-th row and $j$-th column is $a_{ij}$. Or for $B$ you use $b_{ij}$. Etc. ...


2

The notation $o(h)$ is not a specific function. In a limit settings, it is usually used to denote a small quantity compared to the quantity $h$. In this given settings, one is working with $\lim\limits_{n\to\infty}$. Then one should take $o(1)$ as some quantity that $$\lim_{n\to\infty}\frac{o(\color{red}{1})}{\color{red}{1}}=0,$$ and $o(\frac1n)$ some ...


0

Found an another (and arguably better) reference, which uses different notation. In Section 8.1.3 of Handbook of Discrete and Combinatorial Mathematics by Rosen, this graph is called the dipole graph (of size $m$) and denoted by $D_m$.


5

According to Ian Stewart, the symbol "!" was introduced because of printability. Before 1808 $\underline{n\big|} = n \cdot (n-1) \cdots 3 \cdot 2$ was [widely?] used to denote the factorial. Because it was hard to print [in non-computer ages], the French mathematician Christian Kramp chose "!". Source: Professor Stewart's Hoard of Mathematical Treasures


0

You should probably check this wiki page $n=O(n^2)$ is mathematically correct because $\frac{n}{n^2}=\frac{1}{n} < 1$. It is also true that $n=o(n^2)$ as $\lim_{n\rightarrow +\infty} |\frac{n}{n^2}|=0$ I don't have any reference on the subject, but the books I've seen usually use the notations properly, even if they don't go into much details about what ...


7

As pointed out in another answer, the notation $\int \ldots\mathrm dx$ is consistent with the typesetting of other mathematical symbols, since $\mathrm d$ is the name of a specific operator. There is also an ISO standard governing these things, which purportedly specifies $\int \ldots\mathrm dx$ as the correct notation, but a copy of the latest standard, ...


4

The underlying rule (which is often violated) is that variables should be in italic, but names should not. In ${\rm d}x$, $x$ is a variable which could be exchanged with any other letter, but ${\rm d}$ is the name of the differential operator and cannot be exchanged with any other letter. For the same reason, a general function $f$ is in italic, but the ...


11

$$\int f(x) dx$$ is just fine, though some people, as a matter of preference, write $$\int f(x) \mathrm{d}x$$ (perhaps to indicate that we are not taking the product of $d$ and $x$.) Just as there are folks, like me, who like to insert space between the function and $dx$: E.g. $$\int f(x)\,dx$$ But rest assured that the appearance of the integral sign makes ...


0

The radical operator may suit your need. If $m$ is an integer such that $m = \Pi_{1}^{r}p_{j}^{m_j}$ for some integer $r > 0$ and some integers $m_{1}, \dots, m_{r} \geq 0$ and some primes $p_{1}, \dots, p_{r}$, the radical of $m$ is defined to be $\Pi_{1}^{r}p_{j}$ and is denoted by $rad(m)$. Then two integers $m, n$ have the same prime divisors if and ...


0

Considering only the local minima does not seem particularly natural and it is unlikely that there is a standard piece of notation for it. On the other hand, the set of all extrema can be written simply as the level set of the derivative: $\{f'=0\}$.



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