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2

In several complex variables there is a standard term domain of holomorphy. Hardly anyone ever writes "domain of holomorphicity". However, outside of "domain of" combination, both holomorphicity and holomorphy are about equally used. My feeling is that people using complex analysis in relation to Banach spaces, Lie groups, etc are more likely to write ...


1

Arnold E. Ross and Alexander V. Oppenheim were students of Leonard Eugene Dickson. You mostly want to borrow two of Dickson's many books; Modern Elementary Theory of Numbers (1939), especially page 161 where he gives references about universal indefinite ternary forms; Studies in the Theory of Numbers (1929) which has more but is harder reading. For an ...


1

If speaking of an algorithm, saying a specific algorithm is "O of n squared" is commonly used, especially in the vicinity of computer scientists.


1

I've usually head big oh of x but as long as your audience know what you're talking about it doesn't really matter too much, unless you're working with a function O it's not likely that anyone will mistake you. Unless you're working with little o as well. Little o doesn't seem to be used that often though so the chances for confusion should be pretty small.


0

My attempt: a right triangle with unit hypothenuse and angle $x$ has base $\cos x$.


1

In my opinion (and I have to agree with Jp McCarthy: without more context, we cannot be certain what you mean), when someone says that maps $f,g$ commute whenever $f\circ g=g\circ f$. However, an alternative meaning might be that they make a particular diagram commute. For example, if $f_{ab}:A\to B$, $f_{bd}:B\to D$, $f_{ac}:A\to C$, and $f_{cd}:C\to D$, ...


2

Presumably (and it is difficult to be 100% sure without further context), maps $f:A\rightarrow A$ and $g:A\rightarrow A$ are said to be commute if $$f\circ g=g\circ f\Leftrightarrow f(g(a))=g(f(a)),$$ for all $a\in A$.


3

Yes, $1$ and $n$ are, trivially, factors or divisors of $n$. In particular, if $n$ is a prime number, then the prime factorization of $n$ is simply $n$ itself. That said, when speaking of the factors of a number, it is often useful to exclude these trivial factors. The usual term for that is "proper factor" or "proper divisor". Specifically, a proper ...


9

It is meaningless to talk about factors without agreeing in advance what "numbers" we are using. For example, $3$ is not a factor of $5$ if we use integers, but it is a factor of $5$ if we use fractions, because $3\cdot \frac{5}{3}=5$. With this understanding, we say that $a$ is a factor of $b$ if there is another "number" $c$ (of the agreed-upon ...


0

This is a common misuse of the word infinite. Anything that is infinite or approaching infinity, is not quantifiable, regardless of the context. There exists no such quantity that can ever get close to infinity. Therefore, it will never make sense to say that a quantity is "almost infinite".


0

Here is a practical example. Suppose that $A$ and $B$ are two genes, each of which is found in exactly $50\%$ of the population. Suppose also that we have established that there is zero correlation between the occurrence of one gene and another. So what percentage of the population has both genes? $25\%$, right? Well, no—in fact, it is impossible that ...


0

Saying that something massively big is 'almost infinite' is no different from saying that 1 is almost infinite, since the difference between infinity and 1 and between infinity and massively big is exactly the same - namely, infinite.


0

Again you can't say bounded in place of finiteness also. Consider $[2.3]$ which is contained in $\cup_{n=1}^\infty (\frac 1n,\infty) $. Then it is contained in a finite subcollection but namely $(1, \infty)$ which is not bounded, moreover any finite subcollection will not be bounded here.


2

No. A finite subcollection of the collection of open intervals means finitely many of the open intervals. This is not the same as a bounded subcollection. For instance, the set of open intervals $(1/n,1)$, for positive integers $n$, is bounded (all are contained in $(0,1)$) but not finite. And no finite subcollection will cover $(0,1)$.


-1

First, consider the possibility that you misunderstood your teacher or that your words above do not quite capture what (s)he meant. That said, the "percentage with respect to the score of the highest achiever" has nothing to do with percentiles, and I am not sure there is a single word to express "I got within 89% of the best". Percentiles (or, more ...


0

"It is an important and popular fact that things are not always what they seem." - Hitchhiker's Guide to The Galaxy Your teacher meant Percentile Rank. In the Wikipedia Article, it is verifiably stated that: "In test theory, the percentile rank of a raw score is interpreted as the percentages of examinees in the norm group who scored at or below the score ...


0

If your memory is correct, your teacher was wrong. The percentile tells you the value below which a certain percentage of the observations fall. Thus, if the median score on the test was $80$, a score of $80$ would put you at the fiftieth percentile regardless of what the high score was.


3

It is a perfectly legitimate Venn Diagram which shows two, disjoint sets. They don't have to overlap to be a Venn Diagram. This type of diagram could also be called a mapping diagram because it shows you how the elements of one set (the oval on the left) are mapped to elements of another set (the oval on the right).


3

It's called the Sign function, also called the Signum function: $$\large \text{sgn}(x) = \begin{cases} \frac{d}{dx} \left| x\right| = \frac{x}{\left| x\right|} , \text{when } x \neq0\\ 0\space, \text{when} \space x=0 \end{cases} $$ $\leftarrow$ It is clearly a piecewise step function.


3

That's that signum function. Actually: $$\text{signum}(x)=\begin{cases}\begin{align}1,\quad x>0\\0,\quad x=0\\-1,\quad x<0\end{align}\end{cases}$$ See it at wikipedia or WolframMathworld.


2

It means that if you restrict $\phi$ to the field $K^{un}$, you get the Frobenius automorphism.


4

If $x$ is a probability, then $1-x$ is the complementary probability.


0

The set $\tau$ of open subsets of a set $X$ is an algebraic structure with $\cup$ and $\cap$. The intersection of two sets in $\tau$ should be a set in $\tau$. And the union $\displaystyle\bigcup_{i\in I}\mathcal O_{i}$ should be in $\tau$ for any set of open sets $\{\mathcal O_{i}\}_{i\in I}$. Also $\emptyset,X\in\tau$. And that's all. The set $\sigma$ ...


1

This question has been answered in comments: Those are usually just called "distributions." Here is a good overview of the different varieties of distributions. Hope it helps. – icurays1 Nov 20 '12 at 6:10


8

Yes! It is the Sierpinski Space. You will find most answers in the wikipedia link. It is a connected two point set and is really useful for plenty counterexamples and/or constructions. From a categorical viewpoint the Sierpinski Space represents the functor $X \mapsto \tau (X)$, $f\mapsto f^{-1}$.


2

Consider the function $\frac{\sin x}{x}$ as x approaches 0 for a case where the limit does exist and is equal to 1. Reference if you want one for this case. If you want another example where the limit doesn't exist consider either $\sin x$ as x tends to infinity or $(-1)^n$ as n tends to infinity, consider the cases where n is a sequence of odd numbers ...


0

This link offers a few general differences between British and American terms. http://www.waldomaths.com/ukus.jsp


2

It's called a "spherical shell".


0

The idea is not that you can count the entire infinite collection. But that you can index the collection so that every element can be counted after a finite number of steps. Meaning there is an element counted the first step, $a_1$ then the second step, $a_2$ and so on. And each element has a unique index which only finitely many previous indices before we ...


0

The mathematical usage "$A$ is a countable set" doesn't imply there is an algorithm which counts each element, one after another. It instead asserts that there exists an (injective) function $f : A \rightarrow \mathbb N$; the distinction will become clearer when you study things like the Halting Problem. Finite sets are of course countable; and an infinite ...


0

When you count things, you set up a map to the integers. That is, you pick a first object, then a second one, then a third etc. A set is countable, then, if there is a way of assigning a nice order to the elements, so that for any given element you could count up to it. On the other hand, sets are uncountable if, no matter how you started counting, there ...


0

Well, the term is really self explanatory, since we are talking about sets X that either are finite, perfectly countable, or that there exists a bijection $f: \mathbb{N} \rightarrow X$ which is giving us a "natural" enumeration of something, when we write $$f(1) = x_{1}, f(2) = x_{},...,f(n)= x_{n},...$$ We should have then $X = \lbrace ...


6

An infinite set is countable if there's a way to start counting its elements such that, although you might never finish the count, each element will eventually be counted. What is counting? It is making a bijection between a set of numbers and a set of things. When you point towards a collection of objects and say "1, 2, 3, ..." you are assigning a natural ...


3

I think some authors require that "simply connected" only be applied to spaces that are already known to be path connected, and then the definition is that a path connected space is simply connected if its fundamental group is trivial (with some basepoint). Here is an article from 1955 which explicitly uses some version of this convention. The problem is ...


1

The most general ways to "pronounce" $\in$ certainly are "is an element of" or "is a member of". However, in a case like this one where the set is not only finite but also very small it might make sense to read "$y \in \{1,2,3\}$" as "$y$ is either $1$ or $2$ or $3$" or "$y$ is one of the values $1$, $2$, and $3$". This is in accordance with, say, reading ...


9

The relation $\in$ is typed \in (in TeX) and often pronounced "is in", or "is a member of", or "is an element of". But many variations occur occuring to the particular set, so you might pronounce $z\in\mathbf C$ as "$z$ is a complex number" rather than "$z$ is an element of the set of complex numbers". Your confusion is due to the two different roles of a ...


1

As for "$\in$" see the other answers. Concerning the idea of a variable: Each letter, say $y$, denoting a variable comes a priori with its domain $D_y$, a certain set. We are allowed to replace this $y$ in the formula by any element $a\in D_y$ and obtain a proposition about constants which is either true or false. See also here: High school math ...


3

"You belong to me", "I belong to you"... Possession always causes confusion. A set, by definition, is a collection of elements. A given element $x$ can either be in a given collection, or not in the collection. I read "$y\in\{0,1,2,3\}$" as "$y$ is one of the elements $0,1,2,3$". No belonging involved.


0

I'm not sure I understand the question "does the word 'ten' have a base?" but what your friend says is correct. You can think of 'ten' mapping to $||||||||||$, and then that maps to various representations like $10$, $1010$, and $\text{A}$ depending on the base. And if the base is known the mapping can be reversed from the representation to the word. If it ...


2

Your friend is right. "Ten" is indeed a baseless number, which exists whether you write it down in digits or not. It's worth mentioning, however, that the names we give to numbers are based on the significance of this specific one. I guess the reason for that is the number of fingers each one of us has on both their hands. Having said that, still, your ...


2

If $(A,\leq)$ is a partial order, then we define these two definitions for $a\in A$: $a$ is maximal if whenever $a\leq b$, then $a=b$. $a$ is maximum if for every $b\in A$, $b\leq a$. You can prove that every maximum is maximal, but a maximal element need not be a maximum. In particular there can be many maximal elements. So being maximal and maximum are ...


2

Here is the answer that I wrote a few months ago, which should pertain to this question as well: Both categories are denoted by $\mathrm{Rel}$. In most cases, however, the author either explicitly tells which $\mathrm{Rel}$ he/she is using, or reproduces the definition of $\mathrm{Rel}$. (Note: this applies to any definition/term in math which can be ...


2

Stochastic comes from Ancient Greek whereas random is an old French word. (fun fact: random has totally disappeared in modern French and was replaced by aléatoire which comes from... Latin) Otherwise there is no difference between them in the realm of Probability Theory.


10

Usually, the acronym $s.t.$ means such that. In the context of optimization, it means subject to. Also note that such that does not have the same meaning as so that. Such that, describes how something should be done. So that, describes why something should be done. For clarity, it's usually best to avoid $s.t.$ and simply write such that.


2

Maybe what you mean is that $1-x$ is a sort of additive inverse of $x$ with respect to $1$. Moreover, the function is an involution as long as $x\in[0, 1]$, so applying this "inverse" twice gets you $x$ back. This operation is useful in probability, for instance, and I don't think it has a name, although if you called it the complement people wouldn't think ...


0

If you're speaking of $x^{-1}$, then that's been answered in the comments. If you're talking about $x - 1$ and $1 - x$, then the correct answer is they are each other's additive inverses. The additive inverse of $x$ is the number $-x$ such that $x + (-x) = 0$. If we add $(x - 1) + (1 - x)$, we have $$(x - 1) + (1 - x) = x - 1 + 1 - x = x - x + 1 - 1 = ...


3

If you add $x-1$ and $1-x$, you get $$(1-x)+(x-1) = 0$$ Indeed, $(x-1)= -(1-x)$. Each is the additive inverse of the other.


5

To the best of my knowledge, there is no commonly used term for the relationship between $x$ and $1-x$.


1

When $d = 1$, this is often called a flat ($n$-)torus. We say this because the metric induced on it by the ambient Euclidean space $\mathbb{R}^n$ is (locally) flat---by contrast, this is not the case for any $2$-torus embedded in $\mathbb{R}^3$. Note, though, that this is a statement about the metric geometry on the space, not the topology, where there is no ...


3

Recall that a sequence is really a function from $\Bbb N$ into some set, $\Bbb R$ in this case (guessing by the choice of tags). If the sequence is $x\colon\Bbb N\to\Bbb R$ then we write, for convenience, $x(n)$ as $x_n$. But since $x$ is a function, it has a range $\{x_n\mid n\in\Bbb N\}$ (where we forget about the enumeration). The sequence $x_n=0$, for ...



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