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3

Versine is $\displaystyle1 - \cos(\theta) $ or $ \displaystyle2\sin^2\left(\frac\theta2\right)$ 'versin' is the Versine here, 'ver' by itself is nothing. It is short for 'versed sine'. See here.


3

As Robert Soupe says, either $\mathbf{R}$ or $\Bbb{R}$ is unlikely to confuse a reader without explanation. Until the last decade or two of the 20th Century, $\mathbf{R}$ was common notation for the real numbers in print journals. As Raclette notes, $\Bbb{R}$ is called blackboard boldface because lecturers would put a double spine on the "R" to make it look ...


2

The proper notation for the set of all real numbers is either $\mathbb{R}$ or $\textbf{R}$. It really comes down to your choice, and whichever you choose you can back with plenty of precedent. But most of the time people will understand what you mean without you having to explain it. The problem with $R$ is that in the context of algebraic number theory it ...


0

The presented notation (what is written under the $\sum$) is not standard, not self-explaining, and does not reflect your intention. What you mean is the following (it cannot be done without text): One has $$s(t_1,t_2)=\sum_{k=1}^N|x(\tau_k)-x(\tau_{k-1})|$$ for any partition $${\cal P}:\quad t_1=\tau_0<\tau_1<\ldots<\tau_N=t_2$$ with $x'$ ...


6

I'm going to say that it's also because of the way we read. $2^3$ is "two to the cube" or "two to the third" whereas $^23$ would be "to two, the three?". Of course, this could be very bad reason since you can argue that the operation $3^2$ existed before we decided to read it... (I wanted to write this as a comment better than a proper answer, but I ...


7

Wikipedia says "The modern notation for exponentiation was introduced by René Descartes in his Géométrie of 1637", and has a link to a page from Descartes.


1

You have a function $f(x,y,z)$ presumably, and then you take a composition $h(x,y) = f(x,y,g(x,y))$. The chain rule here is $$ \frac{\partial h}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial z}\frac{\partial g}{\partial x} $$ and similarly for $y$. You should verify that on your own, and check a couple examples to convince ...


1

If you connect two logical statements, you must use the according logical operators. Usually, you do not write a $\&$, but rather a $\wedge$. Although one may use a separation by commas I prefer also using the logical conjunction $\wedge$. Another optional thing is writing the main Set the elements are contained in before the bar. But a major issue is ...


5

This is a general way to talk about structures that consist of several "parts". For example a field is a set $F$ together with two operations ($+$, $\cdot$) and two special elements ($0$ and $1$). So to unambiguosly specify a field, we could denote it with a pentuple $(F,+,\cdot,0,1)$. Then for example a field homomoprhism to another field ...


2

As mentioned above, $\ll$ and $\lll$ can refer to much less than, or very much less than respectively. There do exist other uses for the symbols as well. In particular, in measure theory (seemingly unrelated to your specific example), for two measures $\lambda$ and $\mu$, you have that $\lambda \ll \mu$ (i.e., $\lambda$ is absolutely continuous with ...


4

The symbols $\ll$ and $\gg$ denote "asymptotically very much less than" and "asymptotically very much greater than," respectively.


1

The standard abbreviation for "there are at least $482$ objects $x$ satisfying $\phi(x)$" is $$ (\exists^{482} x)\phi(x) $$ The extra quantifiers of the form $(\exists^{n} x)$ can either be defined directly in the metatheory, or viewed as abbreviations for longer formulas. So you could write $$ (\exists^{482} x)\phi(x) \land \lnot(\exists^{483} x)\phi(x). ...


0

You are quite right. The second bullet list is full of muddled notation, which shows that the author has vague and confused ideas about probability. Rather than trying to interpret and learn from this sort of stuff, you would be much better off sticking to materials by authors whose notation makes sense.


1

$$\exists x_1, \ldots, x_{482}\left(\bigwedge_{i \neq j \leq 482}x_i \neq x_j \wedge \forall y \left(\phi(y) \leftrightarrow (\bigvee_{i \leq 482} y = x_i)\right)\right)$$ Or, since it's not in doubt that there is such a first-order formula, create some ad-hoc shorthand and write "$\exists^{482} x\, \phi(x)$ where $\exists^{482}x$ is a quantifier denoting ...


1

Let $S_P$ be the set of individuals characterized by $P$ and $S_Q$ the set of individuals characterized by $Q$, $\mid S_P \cap S_Q\mid=482$. Where $\mid S \mid$ is the cardinality of the set $S$. You can possibly define these sets as $S_P = \{x \mid P(x)\}$.


0

It means "repeat the terms I've written, but with these indices swapped". So in your case, $$i\hbar\Big(g^{\mu\rho}M^{\nu\sigma}-(\mu\leftrightarrow\nu)\Big)-(\rho\leftrightarrow\sigma)$$ is short for $$i\hbar\Big(g^{\mu\rho}M^{\nu\sigma}-g^{\nu\rho}M^{\mu\sigma}\Big)-i\hbar\Big(g^{\mu\sigma}M^{\nu\rho}-g^{\nu\sigma}M^{\mu\rho}\Big).$$


1

Notation was introduced by Leibniz..Earlier he used $\overline{omn} l$ where omn stands for sum and $l$ for differences.Later he used $\int$ for $\overline{omn}$. It was elongated S from sum. Before Newton and Leibniz mathematicians like Roberval has arrived at relation between area under the curve and anti-derivative[Fundamental theorem]. But they cannot ...


2

To supplement MPW's answer and the comments, the other important detail about this notation is absolutely nothing is implied about the cardinality (size) of the index set $I$. It could be a finite collection, countably infinite collection, uncountably infinite....we use the term "an arbitrary collection" to indicate any particular size.


3

It denotes a collection $\mathscr A$ of objects, and it means there is surjection $f: I\rightarrow \mathscr A$. One often has that $f$ is injective, so that objects $f(i) = A_i\in \mathscr A$ and $f(j)= A_j\in\mathscr A$ corresponding to distinct indices $i\in I$ and $j\in I$ are themselves distinct, but this need not be the case. It simply means that ...


7

The comments have addressed the notational issues, but I will discuss a comment you made. Isn't $3$ just a 1D vector? No. $3$ is a scalar unless we define a 1-D vector space over the reals. There are some issues with considering every scalar a de facto 1D vector. For instance, scalar-matrix multiplication is well defined, e.g. $3\begin{pmatrix} 1 ...


1

It is convenient to immediately make it clear that whatever structures set you define, is a set together with this and that extra things. You immediately know how many things to expect. Also, for pedagogical reasons, when first learning about, e.g., vector spaces, it is important to drive the fact that a vector space is a set together with extra structure. ...


1

This notation allows for unambiguous reference to the object. For instance $\mathbb{R}$ could refer to the group $(\mathbb{R},+)$ or to the ring $(\mathbb{R},+,\cdot)$. Usually either the distinction is clear from context, or it doesn't matter, and so the -tuple notation is dispensed with. There can be situations where a set has, for instance, two ...


0

$\alpha$ is used to denote the density function in a Riemann-Stjeltjes integral. The Riemann-Stjeltjes integral is a generalization of the ordinary Riemann Integral. Given a function $\alpha$, we sum $f(x_i) (\alpha(x_{i+1}) - \alpha(x_i))$ over our partition instead of $f(x_i)$. To prove Riemann Integrability, start with given $\epsilon > 0$. Then ...


2

The notations $\lim_{x \to a +0}$ and $\lim_{x \to a - 0}$ can be used to signify one-sided limits; the meaning is exactly the same as that of $\lim_{x \to a^+}$ and $\lim_{x \to a^-}$, respectively. This is a so-called "abuse of notation"; one assigns to the string $a+0$ a meaning other than the usual one, "$a$ plus $0$." The rational is to avoid a ...


2

$3-0=3+0=3$, so each of the corresponding limits are two sided limits. The other limits, $x\to 3^-$ and $x\to 3^+$, are one sided limits. In general, therefore, they are not the same.


0

After a long Facebook conversation, it turned out that $\overline{f}(x)$ is the inverse function of $f(x)$ in this context. I prefer using $f^{-1}(x)$, but at least I've learnt something new today.


0

It is correct. However, it can be written easier. We have $$ \mathbb{Z}\left[1+\sqrt{-5}\right]=\{ a+b(1+\sqrt{-5})\mid a,b\in \mathbb{Z}\}=\mathbb{Z}[\sqrt{-5}]. $$ This is exactly the ring of integers in the quadratic number field $\mathbb{Q}(\sqrt{-5})$. For integers not congruent $1$ modulo $4$ this is different. So, for example, the ring of interegs ...


1

Yes. In general, $\mathbb Z[\alpha]$ denotes the set of polynomial expressions in $\alpha$ (with integer coefficients), i.e., $$ \mathbb Z[\alpha]=\Bigl\{\,\sum_{k=0}^na_k\alpha^k\Bigm| n\in\mathbb N_0; a_0,\ldots,a_n\in\mathbb Z\,\Bigr\}.$$ In your case, $\alpha^2$ can be expressed with lower degrees, hence it is not necessary to go beyond elements of the ...


0

The primary reason is that $\mathbb{R}_+^*$ has a star in the superscript, and it would look silly to put a plus sign there as well. As to why we put the star there in the first place, it is because $\mathbb{R}^*$ is common notation for the set $\{x\in\mathbb{R}:x \neq 0\}$ equipped with the operation of multiplication of real numbers. The reason to ...


0

We can think $s$ as a derivative operator in time domain, e.g. $(s^2 + s + 1)u(t) = \ddot{u}(t) + \dot{u}(t) + u(t)$. But it is not appropriate to use it like this because $s$ is a variable in frequency domain. I think that's why the authors chose the notation $\left( \left( \frac{d}{dt} \right)^2 + \frac{d}{dt} + 1 \right) u(t)$.


0

I think that you're going to have some cluttered-looking formulas no matter what. However, instead of $[X, Y]$, one could write, say, $\mathcal{L}_X(Y)$. You could obviously even shorten this to $X(Y)$, like function notation. If you then switch to exponent notation for functions (as, $f(x) = x^f$), and especially if you're willing to mix these notations, ...


3

Nothing particular standard, but $\rm Card$ has been used before, I have seen $\Gamma(M)$ denoting the cardinals of a model $M$, although more in the context of failure of choice where this is a partial order. Cantor originally used Tav (ת) the last letter of the Hebrew alphabet to denote the class of cardinals. But that didn't stick.


17

This is the character $$\Huge \mathfrak{X} $$ or \mathfrak{X}, which is just a capital X in the Fraktur typeface.


2

The cartesian product is your answer. \begin{equation} \mathcal{A}^{k} = \mathcal{A} \times \cdots \times \mathcal{A} = \{ (t_{1},\ldots, t_{k}) \, | \, t_{j} \in \mathcal{A} \textrm{ for } 1 \leq j \leq k \}. \end{equation} In the case that $\mathcal{A} = \mathbb{Z}_{\geq 0}$, such $k$-tuples are often called multi-indices,since they are used in the ...


2

$\mathbb{Z}^+$ means "positive integers." $\mathbb{N}$ means "natural numbers," which includes zero, no, wait, it doesn't, yeah, it does... Look at this page: http://oeis.org/wiki/Latin_alphabet On the rightmost column of the table, it says: $\mathbb{N}$ The set of natural numbers, including 0[31], though sometimes also used to mean the same thing but ...


2

There has always been some ambiguity as to whether $\Bbb N$, the natural numbers, includes zero or not due to inconsistent usage by past authors. Where as using $\Bbb Z^+$, the positive integers, or $\Bbb Z^\ast$, the nonnegative integers, is unambiguous about the matter.


3

I really think it is just a notation thing. For sure humans considered the natural numbers before we considered the negatives and even zero. Why I think it is a handy notation. There is no universal agreement on what we call the natural numbers, some people consider zero to be a natural number, while I have never had a book that did not start the ...


7

The answer is different for different symbols. It wasn't invented all at once, but accreted gradually. Even closely-related symbols like $\land$ and $\lor$ or $\forall $ and $\exists $ were introduced separately. (Strange but true.) $\lor $ dates to the 19th century and is an abbreviation for Latin vel. $\land$ didn't come until the following century by ...


0

Base 10 is a positional fixed base (The value of the digit depends on its position in the number as well as on its value) and it is neither to large nor to small which is best suited for most purposes. This has many advantages and the andvantage this has over factorial base is, that we need only 10 symbols to represent every possible number. Factorial ...


1

You can indeed start by $i$ or by $j$ as you prefer. However, if you say $\beta_{i,i}$ is not defined, probably you should include the condition $j\neq i$, so like $$ \sum_{\substack{j=1\\j\neq i}}^{m} \gamma_i \cdot \beta_{i,j} \cdot \alpha_j $$ using \sum_{\substack{j=1\\j\neq i}}^{m} \gamma_i \cdot \beta_{i,j} \cdot \alpha_j


0

Literally, the sum reads: Take the sum over the expression: $$(-1)^{|J|-1}\left|\bigcap_{i\in J}A_i\right|$$ for all non-empty subsets $J$ of the integers $\{1,2,\ldots,n\}$. The expression is the size of the intersection of the subsets $A_i$ where $i\in J$, negated when $|J|$ is odd. More intutively, you can read this as: The size of the union of the ...


2

The sum runs over every nonempty subset $J$ of $\{1, 2, 3, ..., n\}$. There will be exactly $2^n-1$ of these subsets, so there will be $2^n-1$ terms in the summation. I think this is most easily seen with an example. Say $n=2$. Then the sum would be: $\left| \cup_{i=1}^n A_i \right| = (-1)^{|{\{1\}|-1}}|A_1| + (-1)^{|{\{2\}|-1}}|A_2| + (-1)^{|{\{1, ...


0

From the context, it seems $[n]$ is the power set of $\{1,2,3,....,n\}$, so its summing over all nonempty subsets $J$ of the power set. So, for subsets of length 1 (that's the $|J|$), you ADD the order of each $A_i$, then for subsets of length 2, it's saying to subtract each pairwise intersection, then add each triple-wise intersection, etc.


2

You can read the symbols $\mid$ and $,$ as follows: $"\mid"=$ given that. So the LHS can be read as: Probability that $X_1$ is less than $X_2$ given that the minimum of $X_1, X_2$ equals $t$. $","=$ and. So the RHS (numerator) can be read as: Probability that $X_1$ is less than $X_2$ and that the minimum of $X_1, X_2$ equals $t$. In the first case, it ...


1

In your case, $| I |$ stands for the number of rows in the bicluster and similarly, $| J | = $ number of columns. More generally, the notation $|A|$ (usually) means the number of elements in a set $A$.


4

Notation is a matter of convention. In your case the notation $$\log_b a$$ is much (much) more common, but $${}^b \log a$$ is also a notation and you may or may not like and/or use it. If your teacher says that you must use his notation, you better do. Keep in mind that the one you prefer is more common for others, though. A related example of convention is ...


2

The circle $\circ$ is the symbol for composition of functions. In General, if you have to functions $g\colon X\rightarrow Y$ and $f\colon Y\rightarrow Z$, then $f\circ g$ is a function from $X$ to $Z$. For $x\in X$ one has $(f\circ g)(x) = f(g(x))$. In your case one has: $f(x) = 2x-1$, $g(x) = 3x+2$ and $$ (f\circ g)(x) = f(g(x) = 2(g(x))-1 = 2(3x+2) -1 = ...


3

This notation means that you take the output of $h$ and use it as the input of $f$. When we are working with a specific $x$ value, we can suggestively write $f(h(x))$ instead. For instance if $f(z)=1/z$ and $h(x)=2+3x$ then $$(f\circ h)(x) = f\big(h(x)\big) = f(2+3x) = \frac{1}{2+3x}.$$ (Note: I only used $z$ as the variable for $f$ to avoid confusion; in ...


0

http://www.extension.harvard.edu/open-learning-initiative/abstract-algebra https://www.youtube.com/playlist?list=PLA58AC5CABC1321A3 This course was taught using the Artin book so i think this one is the best one to follow Dr. Gross explanation is very good you may need to pause and rewind and digest some material at times but i am sure these lectures ...


0

Building on @Henning Makholm's approach, here's what I did: $$\left(\begin{bmatrix}1&0\\0&1\end{bmatrix}\begin{bmatrix}2\\3\end{bmatrix}\right)^{-1} \begin{bmatrix}1&1&0&0\\1&0&1&1\end{bmatrix} =\begin{bmatrix}1/2&1/2&0&0\\1/3&0&1/3&1/3\end{bmatrix}$$



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