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1

Yeah $N$ is a natural number. It's just the standard $N$ dimensional real vector space.


0

There is no privileged or standardized notation. It is a matter of definition, which is dependent on what is convenient for you. Sometimes you want to have a function that outputs an $n$-th root of the input, in which case you would define: $a^b = \exp(b\ln_π(a))$ where $\ln_π$ is the principal branch of the natural logarithm. This gives a well-defined ...


4

In my experience, the expression $x\bmod y$ is a binary operation which returns the remainder when you divide $y$ into $x$. By contrast, the parenthesized $\pmod y$ is used to qualify an equivalence. Thus one might write "$x_1\equiv x_2 \pmod y$". Usually the parenthesized $\pmod y$ goes at the right of the line, right-justified. So it is a qualifier which ...


1

You are right. There is no difference between the two.


0

Yes, both are preferred. There are various considerations. For example, some people use $\vert$ for set notation like $\{x\in\mathbb{R}\,\vert\,x<0\}$. Then $\Vert$ helps to avoid ambiguity. Sometimes $\vert$ is for an absolute value of an element of a simple algebraic structure such as a field, whereas $\Vert$ is then used for higher structures, like ...


0

It is a reimander calculation In the standard format mod(8,4) equals 0 because there is no reimander in 8/4 but for mod(8,6) equals 2 because there is a reimander. It's just the first number divided by the second and the reimander that comes from it.


0

Modular Arithmetic is a topic in number theory that deals with division of numbers and their remainders (but not restricted to this). I encourage you to read more about it online here. As an example $20 \equiv0 \ (mod\ 2) $ because 20 does not leave a remainder when divided by 2. It has very interesting applications in the field of cryptography and internet ...


2

Let $a,b$ be integers and let $m \neq 0$ be an integer. The integer $a$ is said to be congruent to $b$ modulo $m$ if there is some integer $q$ such that $a-b = qm$. Thus that $a$ is congruent to $b$ modulo $m$ means that $m$ divides $a-b$. In this case, we write $a \equiv b \mod m$. This notation is due to K. F. Gauss. You can find it in Gauss's book ...


2

It depends on the context. The usual one is with an "equivalent to" sign $\equiv$: for example, $26\equiv5\;(\!\!\!\!\mod3)$, which is read "$26$ is congruent (or equivalent) to $5$ modulo $3$", and means that $26$ and $5$ differ by a multiple of $3$. Generally, $m\equiv n\;(\!\!\!\!\mod k)$ means that $m$ and $n$ differ by an integer multiple of $k$. Here ...


0

Read $\forall$ as "for all" or "for each" and $\in$ as "in" or "belongs to".


0

For details a reference is the book of Hartshorne "Algebraic Geometry", page 116. If $S$ is a graded ring, then a graded $S$-module $M$ gives a sheaf $\tilde{M}$ on $Proj(S)$. On the basic open, $\tilde{M}(D(f)) \cong \tilde{M_f}$. Now, it's only pure algebra. There is a general construction with $S$ a graded ring, and $M$ a graded $S-$module. We can ...


1

Check this. Search the best (efficient and/or clear) way to construct a list of the 10 biggest values from the set of frequencies and name it, by example as $F_{10}$. The concepts of max and min maybe essential, and relations of order between elements trough their index. By example you can define the set of frequencies in some way, it depends of your ...


0

$$ \int_0^1 f(x, t) \sin (m \pi x)dx = \int_0^1 [\sum_{k = 0}^n f_k(t) \sin (k \pi x)] \sin (m \pi x)dx = \int_0^1 \sum_{k = 0}^n f_k(t) \sin (k \pi x) \sin (m \pi x)dx = \sum_{k = 0}^n \int_0^1 f_k(t) \sin (k \pi x) \sin (m \pi x)dx = \sum_{k = 0}^n f_k(t) \int_0^1 \sin (k \pi x) \sin (m \pi x)dx = \sum_{k=1}^n f_k(t) \frac{1}{2} \delta_k(m) = ...


1

If $e_i$ is the $i$th standard basis vector (all zero but one at the $i$th position), then $e_i^\top X$ is the $i$th row of $X$.


0

For $A^Ti,∗$, it's unclear whether you take the row and then transpose it into a column, or whether you take the transpose of the matrix and then take the row. It can only be the first. If it were the second there would be a dimension mismatch if $n \neq m$ And $Y$ also cannot be a matrix if you are really doing what you are writing. If you ...


1

According to Change/Keisler, "Model theory", 3rd edition, page 209, the double vertical bar turnstile means the "finite forcing relation". The detailed explanation is somewhat technical. PS. Much the same definition for the double vertical turnstile is given by Smullyan/Fitting, "Set theory and the continuum problem", page 210. They say that $p\Vdash X$ ...


3

That symbol usually denotes the set of all functions $B\to A$. I suspect the notation is suggested by the fact that the size $|A^B|$ of this set is $|A|^{|B|}$. Addendum: To see this, consider some simple examples. Suppose $A$ consists of a single point, say $A=\{a\}$. Then the only function $f:B\to \{a\}$ is the constant function with $f(b)\equiv a$ for ...


2

For any sets, finite or not, $$A^B = \{ f : B \to A \}$$ the set of functions from $B$ to $A$. For example, in that sense $\mathbb R^2$ is the set of functions $f : 2 = \{ 0, 1\} \to \mathbb R$.


0

Any notation in mathematics is dependent on context, which you did not provide much of. (Where are you seeing these symbols - in your textbook? If so, which textbook? What is the surrounding sentence, or paragraph, where these notations show up?) However, given that you've said this is group theory, what seems likely to me is that you are learning about the ...


0

I'd recommend $f^I$ for the elementwise version of $f$, following standard conventions in category theory. In particular, observe that given a set $I$, the function $\mathbf{Set} \rightarrow \mathbf{Set}$ given by $X \mapsto X^I$ becomes a functor as follows. For all functions $f : X \rightarrow Y$, the corresponding function $f^I : X^I \rightarrow Y^I$ is ...


1

There is. Let $\pi : \mathbb R^4\to \mathbb R^2$ be the projection to the $y_1, y_2$-plane and $i : \mathbb R^2 \to \mathbb R^4$ be $(y_1, y_2) \mapsto (0,0, y_1, y_2)$. Then your $P_{[dy_1,dy_2]}$ is $\pi^* \circ i^* = (i\circ\pi)^*$.


1

$$z^n=c\implies z=\omega^k\sqrt[n]c$$ Where $\omega$ is a primitive $n^{th}$ root of unity, and $0\le k\in\Bbb{Z}\le n-1$


1

For a non-negative real number $x$ there is always a unique choice of non-negative real $n$-th root, which is usually denoted by $\sqrt[n]{x}$. Furthermore, if $x$ is negative there is a unique choice of $n$-th root if $n$ is odd and none if $n$ is even. In short, if $x \geq 0$ is real and $n$ is even, then the only real $n$-th roots of $x$ are $\pm ...


0

Writing instead of symbolizing is a very bad idea but some rare cases. The reasons are: Doesnt exist an universal language more than one that is symbolic Symbols have visual meaning, words not... it is slow reading (and writing) instead of just knowing Symbols carry more meanings than just words, this is exactly the meaning of "symbol" For this I think ...


1

In complex analysis, $\sqrt[n]{x}$ is regarded as a multivalued function. Or you can write it as $$\sqrt[n]{x}=\exp{\frac{\operatorname{Log}(x)}{n}},\space x\ne0.$$ $\operatorname{Log}(x)$ is the inverse function of $\exp(x)$, see here.


2

I'd say: $$r=z^{\frac{1}{n}}e^{\frac{2i\pi k}{n}}$$ It is a multivalued function with $k=0,\dots,n-1$


1

For 2, if the collection is finite, then you can consider its interval graph, which is a graph representation of the intersections among the intervals in the collection. Now I quote Wikipedia: The interval graphs that have an interval representation in which every two intervals are either disjoint or nested are the trivially perfect graphs.


0

For question 1, this article of mine gives the answer. For question 2, you could disambiguate the situation by factoring. Remember that the plus-or-minus symbol is just a convenience, and, like most conveniences, involves some loss of something, clarity in this case. Nothing is free, and the cost of convenience is loss of clarity, or safety (as in trying to ...


0

It's hard to say whether your expression is valid. However, it is certainly unclear since you include more than one $\mid$ in your set definition. I would write something like this: $$ \newcommand{\ip}[1]{\langle#1 \rangle} (U^\perp)^\perp = \{w \in V: v \in U^\perp \implies \ip{v,w} = 0\}\\ = \{w \in V: (\forall u \in U, \ip{v,u} = 0) \implies \ip{v,w} = ...


1

As Jonathan Hebert noted, $\Bbb R / 2\pi$ is probably the set of real numbers modulo $2\pi$. Are you familiar with $\Bbb Z_{n}$ from Abstract Algebra? It's the set of integers modulo the natural number $n$, and the concept here is the same. Basically, you look at $\Bbb R$ and for each $x \in \Bbb R$, you start forming its equivalence class. You say $y ...


1

No, $\partial x$ cannot be understood as a product. If it could be, then you would get $$\frac{\partial y}{\partial x} = \frac{y}{x}$$ which obviously is not true in general. The expression $\frac{\partial}{\partial x}$ is also known as derivation operator. This can be understood as follows: Given a function of several variables, say $f(x,y)$, partial ...


3

Here is 'physical' derivation for Gauss's linking number formula. In contrast with David H's remark above, I will talk in terms of magnetic circulation rather than the magnetic force between two current-carrying wires. Suppose I have two closed curves, with paths $C_1,C_2$ respectively. We can generate a magnetic field by running a uniform current $I$ ...


2

It is a subscript, indicating that $q_1$ is not necessarily the same as $q$. -- hardmath The author has established that $p^2$ is divisible by $4$. This makes it possible to write $p^2=4q_1$ where $q_1$ is some integer. One could use another letter, say, "$p^2=4r$ where $r$ is an integer". But if this process continues, one runs out of letters very ...


0

Your blurb is pointing out the difference between $\infty$ and practically every other symbol used in math to denote a number, such as in this case with the variable $a$. However large you define the value of $a$ to be, it is understood to be a finite, unique number. It follows that for any $a$, there is a finite, unique number $a+1$. This is true in the ...


0

Start with the set $\mathbb{R}$. And adjoin two new elements to it, $\mathbb{R}\cup \{ +\infty,-\infty\}$. Denote this resulting set as $\overline{ \mathbb{R}}$. On this new set we do not define any algebra, i.e. we do not define what it means to add/multiply across infinities (because it is not possible to preserve all the rules we would like). However, we ...


1

In the sense that Spivak discusses (i.e. the way it is generally used in calculus), "$\infty$" is really a sort of abbreviation to allow you to write certain limits and sets in a way consistent with other sets. Here are a couple of examples: The interval $(a,\infty)$ is the set of real numbers that are larger than $a$. This is by analogy with the interval ...


3

You could put it this way: If we write $$\begin{align}\lim_{x\to a}f_1(x)&=b\\ \lim_{x\to c}f_2(x)&=\infty\\ \lim_{x\to \infty}f_3(x)&=d\\ \end{align}$$ then there is a qualitative difference between these three lines. That is, the second is not just the same as the first with $c$ in place of $a$ and $f_2$ in place of $f_1$ and $\infty$ in ...


4

In short, the symbol has separate but analogous meanings in different contexts. While we often use it in place of a real number in notation, careful usage will not treat it as a real number---consider, for example, the notation $\lim_{x \to \infty} f(x)$. To expand orthogonally to Simon S' excellent answer (and perhaps besides the cursory paragraph above, ...


-4

You may have heard the adage, "infinity is not a number, it's a concept" (or at least you have now). This is in fact true, since $\infty$ is not a real number (if it were, terrible things would happen). Infinity $-$ $\infty$ $-$ is the object such that no number is greater than it: $$ \forall x\in \mathbb R : |\infty| > |x|$$ We use this object to talk ...


7

$\infty$ and $-\infty$ are not real numbers. So when ever we use them in real analysis we need to define their use carefully, as he has done in the extract you quote. For instance the interval $(a,b)$ is defined as $\{ x \in \mathbb R \ : \ a < x < b \}$ whenever $a, b$ are both real numbers. This definition is in trouble if the symbols $a$ or $b$ ...


2

For any set $S$, we have the following two notations: $$|S|=\text{the cardinality of }S\qquad\qquad S^2=S\times S=\{(a,b):a,b\in S\}$$ Thus, $|X^2|$ is the cardinality of the set of ordered pairs of elements of $X$. which, incidentally, is equal to $|X|^2$. For example, if $X$ were a set with $2$ elements, then $|X^2|=4$.


4

If the goal is to be readable and understandable by others: use the standard notations. So that people don't have to put too much effort in order to understand your equations. Attention $$\text{compact notation $\neq $ easy to read}$$ Sometimes it helps, sometimes it just makes things worst even more if the reader is not familiar with the notation used in ...


1

It's a Neumann boundary condition. Here $n$ stands for the outward-pointing unit normal vector on the boundary of your domain and $\nabla u \cdot n = \sum_{i=1}^2 \partial_{x_i} u n_i$ is the dot product of the gradient, $\nabla u$, and the vector $n$.


0

I would go for something like this (assuming we have a set of actors $A$, a set of states $S$, and a set of levels of security $L$): $$\forall a \in A \forall s \in S( aDs \to \neg \exists s'\in S (c(a,s') > c(a,s))$$ where $D \subseteq A \times S$ is the "desires" relation, $c: A \times S \to L$ is the "level of security function" and $\mathord < ...


0

$(a,b)$ can be interpreted as an element of a product of vector spaces. In your case $(a,b)\in\Bbb R^m\times \Bbb R^n \cong \Bbb R^{m+n}$. So it is ok and commonly understandable to write it like that. If you want to emphasisze that $m,n\neq 1$ you could also write $(\mathbf{a},\mathbf{b})$ for $\mathbf{a}\in\Bbb R^m$ and $\mathbf{b}\in\Bbb R^n$. Actually ...


7

This is Gauss' Linking Number Formula, for two space curves $\vec{A}, \vec{B}: S^1 \to \mathbb{R}^3$ $$ \textrm{link}(A,B) = \oint_A \oint_B \frac{\vec{A}-\vec{B}}{|\vec{A}-\vec{B}|^3} \cdot (d\vec{A} \times d\vec{B})$$ In our case, $\vec{A}(t) = (\cos t, \sin t, 0)$ and $\vec{B}(t) = ( 1+ \cos 2t, \frac{1}{2}\sin t, \sin 2t)$ . How to picture these two ...


5

Here I am interpreting it as the scalar triple product. We have vector functions $\vec A, \vec B:[0,2\pi] \to \mathbb R^3$, given by $$ \vec A(\alpha) = (\cos\alpha,\sin\alpha,0) \\ \vec B(\beta) = (2\cos^2\beta,\tfrac12 \sin\beta,\sin(2\beta)) .$$ I believe you are being asked to calculate $$ \int_{\alpha=0}^{2\pi} \int_{\beta=0}^{2\pi} ...


1

In the first case, what you have is that $a_{n}=b_{n}+o(1)b_{n}$. In the limit, any term that is $o(1)$ goes to zero, so you can think of the $b_{n}o(1)$ term as "lower order terms." This essentially says that $$a_{n}=b_{n}+\text{lower order terms}.$$ You can also think of this as saying that $a_{n}$ and $b_{n}$ are asymptotically the same. The latter is ...


3

The equal sign $=$ means that the LHS and the RHS indicate the same number so, if we write $\theta=\sin^{-1} \left(\frac {\sqrt{3}-1}{2} \right)$, this means that $\theta$ is exactly the real number indicated by the symbols in the RHS. The symbol of approximation $\approx$ means that the LHS and the RHS can differ by a quantity that is less than a value, ...



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