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0

I believe the value you are asking about is the $\textbf{separable degree}$. Read the comments and answers at this previous question for more information. This is a very important concept in field theory.


1

So, the answer seems to be, $$\sum_{i\le r,\ j\le s,\ i+j\le\max(r,s)}p_{ij}x^iy^j$$


0

The notation is indicating that we are thinking of the expression $\phi_t(x)$ as a function of $t$ (with $x$ fixed; or rather that there is a family of functions of $t$ parameterized by $x$). The notation is read "$t$ maps to $\phi_t(x)$." One may also see something such as "Define $g_x:\mathbb{R}\to\mathbb{R}^n$ by $t\mapsto \phi_t(x)$ to be the solution ...


0

I haven't tested it yet but http://www.georgehart.com/virtual-polyhedra/conway_notation.html?notation=sA4 might work.


1

It means that $N_S \subset M_S$ where $ N_S$ and $M_S$ are both modules.


5

It means submodule. In fact when there is an inclusion map $K \to L$, then we consider $K$ as submodule of $L$. Here we have $0\to N\to M$ which gives $0\to N_S\to M_S$. So one can consider $N_S$ as submodule of $M_S$.


0

There is no symbol to denote the angle that a line defined by $p$ and $q$ makes with the x-axis because that's quite an arbitrary thing. We could measure the angle relative to any other line we want. Instead, you could specify $X = (1,0)$ then say that $\angle q p X = 45^\circ$ and $\angle p q X = 135^\circ$.


0

Two points do not determine an angle. You implied x-axis as a standard reference line. Label or name the three points say $A,B,C$. Then $\measuredangle ABC $ is angle at the middle point B.


0

Long comment You have to note that Peano's "inverted-C" symbols plays a double role. Regarding propositions : $a \supset b$, with $a,b \in \mathsf P$ [where $\mathsf P$ stands for propositio] has to be read as : "$a$ deducitur $b$", i.e. (more or less) our : $a \to b$ [see : page VIII]. Regarding classes : $a \supset b$, with $a,b \in \mathsf K$ ...


3

There is no special notation for powers which indicates specifically that composition is the operation. You will have to depend on context. For example, the set of continuous functions from $\Bbb R\to \Bbb R$ can be made into a ringmonoid with composition or pointwise multiplication, and in both cases you would use ordinary power notation for its elements. ...


1

$\dot H^{-1}$ is dual homogeneous Sobolev space http://sma.epfl.ch/~wwywong/papers/sobolevnotes.pdf


1

The trick I used to memorize them actually stemmed from formal logic (which you may or may not have had any exposure to): The symbol $\land$ is a way to symbolize the binary connective "and". Notice it looks like a "pointy" $\cap$. Similarly $\lor$ (or) looks similar to $\cup$. Now, $$x\in A\cap B$$ can be read as $$x\text{ is in } A \textbf{ and } B$$ and ...


3

For general sets, $A^B$ denotes the set of functions from $B$ into $A$. In your case, $\Bbb R^{\Bbb R}$ denotes the set of functions from $\Bbb R$ to $\Bbb R$. A way to see that this makes sense is that $\Bbb R^3$ is usually viewed as a vector space with 3 basis vectors. We write a general element as $(x_1,x_2,x_3)$. We could instead view this as a function ...


0

That would work (provided all the points have different distances to $p_k$, so the maximizing argument is unique). However, there is a much simpler solution: The value of a function at it’s arg max is it’s maximum, so for each $p \in P$ the maximum distance to any element of $Q$ is simply$$\max_{q \in Q} d(p, q).$$


1

In some italian textbooks we use two different symbols: as $x \to x_0$ $f\sim g$ to mean that $\lim_{x \to x_0} \frac{f(x)}{g(x)}=\ell \in \mathbb{R} \setminus \{0\}$, and $f \asymp g$ to mean that $\lim_{x \to x_0} \frac{f(x)}{g(x)}=1$. This notation is not really popular, but it helps you to understand where a multiplicative constant is not enough.


0

There is no contradiction between the two. Your difficulty stems from the circumstance that the symbol $x$ is used for two different purposes: (a) as variable for points on ${\Bbb R}$, and (b) as name for a certain function, namely the coordinate function $$\iota:\quad{\Bbb R}\to{\Bbb R},\qquad x\mapsto\iota(x):=x\ .$$ From $\iota(x+h)-\iota(x)=h$ it ...


2

I'm not 100% sure the following will answer your answer, but by definition, $f \sim g$ (implicitly, at $+\infty$) if $|f(x) - g(x)| = o(|f(x)|)$ (this is the little o notation). In other words, $$\forall \epsilon > 0, \exists x_0 \text{ s.t. } \forall x \ge x_0, |f(x) - g(x)| < \epsilon |f(x)|.$$ When $g(x)$ does not vanish, it is equivalent to saying ...


1

The first is a good book. The second is a poor book: it "defines" $du$ by means of an undefined symbol $dx$. The author could have written $$ du = u'(x) \spadesuit, $$ and call this a forma differential. The differential is precisely the linear map $T$, and introducing $dx$ as something that we don't want to define is rather old-fashioned. Finally, ...


5

Sometimes you see $f^{\leftarrow}$ as an alternate notation for an inverse function $f^{-1}$. This was proposed to reserve the $-1$ exponent for $1/f$ only. But this has not caught on. So let's see if this works. What is $V=(-1/\log F)^\leftarrow$? $$ V(\theta)=(-1/\log F)^\leftarrow(\theta)\qquad\Longleftrightarrow\qquad \theta = ...


0

A note to explain (hopefully) why you may have had this empty set in mind. Let's count all subsets of a set with $n$ elements $S=\{a_1,a_2,\dots,a_n\}$. As you have suggested, there are $n \choose k$ subsets with $k$ elements. It is easy to prove for $k>0$, and then remains the case $k=0$, which is conventionally $1$, since there is only one subset with ...


0

Here's a bit of a speculative idea. I'm making this community wiki; everyone is free to modify it as they wish, you can work on little bits of it, etc. Maybe this will eventually evolve into a genuinely workable solution. Write $\langle x:X,y:Y\rangle$ to mean $X \times Y$, or perhaps $Y \times X$, but instead of referring to the ordinates using natural ...


2

If you are looking for the proper three element subsets of $A$ then the answer is given $\binom{7}{3}$. Your intuition about the empty set is incorrect. When constructing the power set of a set, then we care about the empty set. However, there is no such element $\lbrace \rbrace$ in your set $A$. Just as a side note, you should stray away from using your ...


2

I agree that it's bad notation to write it that way. I would read what you wrote as asserting that all three statements below hold, so it's somewhat ambiguous and nonstandard at best. Putting the word "or" in the equation as pjs36 suggests is better, but it's kind of strange to have a word in the middle of an expression like that. You could write $$(x ...


2

The acceptability of any abuse of notation depends on whether the meaning is clear. For instance if you are asked to solve the equation $x^2 - 3x + 2=0$ and write $x=1,2$ I think most everyone will know what you mean, although to be precise you should write $x = 1$ or $x=2$, or possibly even $x \in \{1,2\}$. At first glance the meaning of $x \equiv a,b,c ...


1

There is a certain ambiguity here. In one context, as when one writes $S(X_1,\ldots,X_n)$, one regards $S$ as simply a function of $n$ variables, $$S:\mathcal X^n \to \mathbb R, \tag 0$$ but in another context, as when one inquires about the probability distribution of $S$, one regards it as a function whose domain is $\Omega$, thus \begin{align} S : ...


0

In addition to the answers given, I would like to add: If you cannot easily find the notation for what you are doing, it is definitely not standardized (the Schur/Hadamard product comes to mind). You can also expect the reader not to be familiar with any notation, and thus choose what's most convenient. Just be sure to define the notation clearly in ...


1

It seems that strict ceil is equal to $ 1 - \lceil - x \rceil $, and srtict floor is $ - 1 - \lfloor - x \rfloor $. Hope that helps.


0

There is no such notation (to the best of my knowledge) but you can define such a function yourself. Consider the function you need as $\mathcal{P_{c,f}(x)}$ for strict ceiling and strict floor respectively. Then, for strict ceiling, $$\mathcal{P_c}(x)=\begin{cases}\lceil x\rceil~,~x\not\in\Bbb{Z}\\ \lceil x\rceil +1~,~x\in\Bbb{Z}\end{cases}$$ And for ...


1

It tells you things like $(Au)\cdot v = u \cdot (A^Tv)$


2

Am I right in saying that the absolute value symbols act like a function such that if $x$, for example, is $x<0$ then $x=-x$ Yes, if you would add bars: $\lvert x \rvert = -x$ for $x < 0$. The definition is $$ \lvert x \rvert = \left\{ \begin{array}{cccl} x & \mbox{ for } & x > 0 & \Rightarrow \mbox{positive sign} \\ 0 & ...


1

Yes, the absolute value of a number is always non-negative (Zero if the input is zero, positive otherwise). This is the point of the absolute value function; it is in some sense a measure of the size of the number. To be more precise, the absolute value of a number is the distance from the number to the origin.


1

That depends. For instance, one would usually argue that $\cos^5 x$ is simpler than $$\frac1{16}(\cos 5x+5\cos 3x+10\cos x)$$ However, in the context of integration, the latter is actually simpler (they are always equal, in case you wonder). It's really a matter of context, and of experience you'll acquire. Then there are also habits. I have had ...


2

A general rule is that you want the fewest number of special functions ($\sin, \cos, \exp, \dots$). It's not less "simple" to nest basic math like multiplication and radicals, but nesting special functions should be avoided. Off course there are no real rules, so you'll have to decide for yourself what's easier to read.


1

The star means nothing in itself, it's like the prime (') sign, it means $\mathrm{SO}^*(n)$ something analogous but different from $\mathrm{SO}(n)$. Actually, $\mathbf{H}$ denoting the quaternions, $\mathrm{SO}^*(n)$ (or $\mathrm{SO}^*(2n)$, I'm not sure of conventions) is the group of $\mathbf{H}$-linear automorphisms of $\mathbf{H}^n$ preserving a ...


-1

The notation should be interpreted as magnitude, not absolute value signs. If we let $z$, the complex number, be represented as $z = a + bi$, we can then say that: $$\left|a+1 + i(b+ 3)\right| = \sqrt{(a+1)^2 + (b+3)^2}$$ and that: $$\left|a-5 + i(b-7)\right| = \sqrt{(a-5)^2 + (b-7)^2}$$ Setting them equal to one another, we get that: $$\sqrt{(a+1)^2 + ...


1

It has nothing to see with $\lvert z+1+3\mathrm i\rvert=0$. Geometrically, the equation means the image of $z$ is at the same distance from the image $A$ of $-1-3\mathrm i$ as from the image $B$ of $5+7\mathrm i$. The locus of such points is the midperpendicular of $[AB]$; the midpoint of $[AB]$ is the point with affix $\frac12(4+4\mathrm i)$. Now the ...


-1

There is an ambiguity in the notation $$\sum_{k=1}^n ak+b $$ does it means $$\left(\sum_{k=1}^n ak\right)+b $$ or $$\sum_{k=1}^n \left( ak+b \right)$$


4

You can see the correct answer by writing the sums explicitly: $$\begin{align} \sum_{r=1}^n(ar+b) &= (a1+b) + (a2+b) + \dots + (an+b)\\ &=(a1 + a2 + \dots + an) + (b + b + \dots + b)\\ &=\sum_{r=1}^nar+\sum_{r=1}^nb \end{align}$$


1

This seems to be based on the meaning of the parentheses, in terms of notation. Yes, when you remove the parentheses, you should distribute the "sigma" to both the term ar AND the term b.


2

One paragraph above: Let $\binom{n}{r_1, r_2}$ denote a "trinomial" quantity, i.e. the number of ways of choosing $r_1$ positions of one kind and $r_2$ positions of another kind in a vector of length $n$. It is simple to confirm the following relationship between trinomial and binomial quantities: $\binom{n}{r_1, r_2} = \binom{n}{r_1}\binom{n-r_1}{r_2}$. ...


3

$\mathbb R^\times$ is the set of invertible real numbers, that is, $$\mathbb R^\times = \{x\in\mathbb R: x^{-1}\text{ exists}\} = \{x\in\mathbb R: x\ne 0\}$$ However, $(\mathbb R^\times,+)$ is not a group, therefore I strongly suspect that the equation in your text actually reads $f(x+y) = f(x)f(y)$, with multiplication instead of addition on the right hand ...


2

If it is a multinomial coefficient, then $d_f+d_f=N$, $c+c=2N-y_2$, and $d_f-c+d_f-c=y_2-N$. These three equations are consistent, since in the order I wrote them, the first minus the second gives the third. So that's supporting evidence that you have multinomial coefficients. What feels strange is that usually with multinomial coefficients that involve ...


1

As per request, I post my comment as an answer: I'd use the notation $\mathcal D\to M_{|R|\times |\mathcal D|}(\{0,1\})$ to insist that we are talking about matrices ($M$) of certain dimensions ($|R|\times |\mathcal D|$) with elements in a certain set ($\{0,1\}$).


2

As 5xum mentioned in the comments, union starts with a u and has a symbol $\cup$ that looks very much like a $u$. Then intersection is simply the same symbol flipped $\cap$. As for what they mean, you can think of union $A \cup B$ as a cup (indeed, the LaTeX command for it is \cup) in which you pour all of the elements of both $A$ and $B$, whereas $A \cap ...


0

The comments provided the requested notation accurately, namely, that if we have $$n=p_0^{a_0}\cdot p_1^{a_1}\cdots p_r^{a_r}\tag 1$$ then we use $\Omega(n)=\sum_{i=0}^ra_i$ to denote the count of prime factors of $n$ including multiplicity. I believe I have even found a reasonable arithmetic function composition that can be used to "calculate" ...


1

A group $G$ is a set endowed with a binary operation, the map $\cdot :G\times G \to G$ that obeys some properties. The notation is then $\cdot(g,h) := g\cdot h$. Of course, nothing stops us from using the plus symbol for the map $+ : G\times G \to G$. But mathematicians are lazy, and a dot is much easier to write than a plus sign! Also from a categorical ...


2

I guess the main reason is representation theory: You can represent every group by linear operators on a vector space, where the group operation maps to the multiplication of operators. You cannot generally map the group operation to addition of operators.


0

For the $R^2$, 2 is plenty. If your data is simulated, then getting a value of $1$ isn't that weird. Leave it at $0.99$ if you don't like $1$ For the coefficients, it depends on the context. The key question to ask is: Does adding a digit provide any useful information to your audience? Personally, I don't like having more than 3. If more are needed, then ...


2

The $n$-th preimage of $A$ under $T$, that is $$ T^{-1}(A) = \{x \in X \mid Tx \in A \} $$ denotes the preimage of $A$ under $T$, and we interate this by $$ T^{-n}(A) := T^{-1}\bigl(T^{1-n}(A)\bigr), n \ge 2 $$ or what is the same, the preimage under $T^n = T \circ \cdots \circ T$ of $A$, that is $$ T^{-n}(A) = \{x \in X \mid T^nx \in A\} $$



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