# Tag Info

1

The term "identity" is vague without a specified domain. The above is not an identity for $x \in \mathbb{R}$ because it isn't true for $x = a$, but it is true for $x \in \mathbb{R} \setminus \{a\}$ because, by the definition, the statement is true for all $x$ in the domain. This answer takes into account the typo Emilio Novati noted.

1

I suppose that your question is about $$\dfrac{x^2-a^2}{x-a}=x+a$$ and this is an idenity for $x\ne a$ but is not an identity for $x=a$ because the Left side is not defined.

1

First of all, I suspect there's a typo: I think it should be $$"{x^2-a^2\over x-a}=x+a."$$ (The "$x^2$" is an "$x$" in the OP.) E.g., set $x=2, a=1$ - in the OP this would yield 1=3. Second, as the comments have pointed out, even with that fix, it's false if $x=a$. There's an important conceptual issue lurking here: there are some functions, which are ...

1

While sample set could also be worthwhile terminology, it is worth recalling that any space in mathematics is always a set with some sort of operations defined. For example, in the theory of stochastic processes, we could let $\Omega = \mathcal{D}(\mathbb{R}_+, \mathbb{R})$, where $\mathcal{D}(\mathbb{R}_+, \mathbb{R})$ is the Skorokhod space of real-valued ...

3

We say "$A$ injects into $B$", and may write $f: A \hookrightarrow B$.

7

Probably the most standard term of the sort you're looking for is "equipotent", though it isn't used particularly often (more often, people will just say two sets "have the same cardinality" or "are in bijection"). People who think about categories a lot sometimes say "isomorphic as sets".

8

You can call them equinumerous.

1

The $L_p$ norm is more general, but you need to specify a measure space for it to make sense. You're integrating over $\mathbb X$ after all! The $l_p$ norm can be seen as a particular case of the above, as @Stephen Montgomery-Smith noted, with the counting measure on positive integers. So I don't think there really is any source of ambiguity: either I ...

0

When taking a calculus course a few years back some of the students were deaf. At every lecture there were 2 translators. It seemed that some concepts like "The mean value theorem" were easy to translate, but others consisting of words rarely used in everyday speech demanded more effort. I might be wrong, but I think that it is possible to "type out" words ...

1

It is a polynomial in $y=x^{\frac 1{12}}$ namely $$p(y)=a_4y^3+a_3y^4+a_2y^6+a_1y^{12}+a_0$$and is thus an element of the polynomial ring obtained by adjoining a $12^{th}$ root of $x$ to the original polynomial ring.

4

There's no name in common use for the set of expressions in exactly the form you quote. One problem with the concept is that they are not closed under multiplication, like polynomials are. For example we have $$(a_2\sqrt x + a_1x)(b_2\sqrt x + b_1 x) = a_2b_2 x + (a_1b_2+a_2b_1) x^{3/2} + a_1b_1 x^2$$ where the middle term looks neither like $ax^n$ nor ...

2

You could call it a Puiseux polynomial.

1

I would write this as $$\nexists k: (2^k \equiv 12 \mod 14)$$

1

$\{k \in \mathbb Z \mid 2^k \equiv 12 \mod 14 \} = \emptyset$

0

The term codomain is mostly used in an unsound way. Indeed, the majority of mathematics texts define a function as a functional relation, that is: a relation (set of ordered pairs) in which no two pairs have the same first member. Given a function f, the set of the first elements of all pairs in f is uniformly called the domain of f; for the set of second ...

1

If $M$ is a vector space, then we refer $A$ as an endomorphism. There are differences when a function takes an element from one space to an element in the same space. Take for example: $f:C[0,1]_{\|\cdot\|_1}\to C[0,1]_{\|\cdot\|_2}$, where $\|\cdot\|_1 \to$ integral norm $\|\cdot\|_2 \to$ maximum norm Note that $C[0,1]$ is complete under maximum ...

2

I think the idea is that a particular player set contains all the nodes in which a given player makes a decision. There are $N+1$ of them because there are $N$ players and then we consider nature as an additional random player. In your diagram, if we call Nature player zero, then $g^0$ would include just the initial node. $g^1$ would include both of P1's ...

2

A mapping $f:X\to X$, for a set $X$, which satisfy $f\circ f=1$ is called involution.

2

As others have noted, the term that you want is anticompact. This is an example of the kind of property studied by Paul Bankston in The total negation of a topological property, Illinois J. of Math., Vol. 23, Nr. 2 (1979), 241-252. I quote: Let $K$ be a topological class. The spectrum $\operatorname{Spec}(K)$ of $K$ is the class of cardinal numbers ...

1

There is a subtle difference between the two: A proposition is a statement in either a natural or a formal language, for which it makes sense to ask whether it is either true or false. An assertion is a statement which one claims to be true. As such, assertions are more restrictive than propositions. For example: In a logic textbook, we read "It's ...

0

Yes, there is a graph called GRAIL GRAPH. I have found it first by trial-and-error searching technique among many papers of Network (branch of graph theory) and Information theory, a paper is published and written by C. Wang and N. B. Shroff. Moreover, I received a couple days ago an email from my friend telling me the reference for the name of the Grail ...

4

The term "one-to-one" is ambiguous. Some authors use it to mean "injective", while others use it to mean "bijective" (i.e. both injective and onto). "injective" means properties 1, 2, and 4 hold. (Note that properties 1 and 2 are just the definition of a function, so an "injective function" usually is defined as a function with property 4. The term ...

2

In general, the terminology "1-1" (not to be confused with 1-1 correspondence) means injective and onto means surjective. If a function $f$ is injective and surjective, it is said to be bijective, or sometimes 1-1 correspondence. There are many examples which $f$ is injective but not surjective. In particular, if you look for one on vector spaces, consider ...

1

It is sometimes referred to as Hölder exponent, see for example the abstract in this paper. This terminology is also used in Signal and Image Multiresolution Analysis. I would use this terminology with caution since on the Wikipedia page, for example, it appears that if $f$ is Hölder continuous with parameter $\alpha$, then $\alpha$ is a Hölder exponent ...

2

The way people make up words for things is far more haphazard than you seem to believe, both in math and in the broader world. There are many examples of terms that don't really make sense when interpreted too literally. For instance, an antlion bears little direct resemblance to either an ant or a lion. That said, I strongly disagree with your ...

3

The notion of a multi-set is not exact opposite of set. The fact is that in a multi-set, "the notion of importance to presence of multiplicity of same element" is exact opposite of the same notion in a set. Every set is a multi-set, with highest multiplicity 1. Here is a good example of importance of multiplicity in some natural ways.

2

There are two major notions in control theory: an open loop system and a closed loop one. The open loop means that you consider a system as an input-output function $$y=Gv.$$ One can plug in different input functions $v$ and see what happens with the output signal. The input is often split into the control input $u$ (the one you can choose) and the ...

0

Remember that the vertical asymptote is not the function or even part of the function. It's a line that a point on a curve (aka your function) approaches as its horizontal position becomes arbitrarily close to $2$. Just to reiterate (reiterate: is the redundant?): the vertical asymptote is not the function.

0

The answer depends on your definition of "vertical asymptote". Unfortunetaly, many authors (and teachers, and online encyclopedias) don't give a careful definition. If possible, it's a good idea to check with your instructor. Absent further information, if you have a continuous, real-valued function $f$ with domain $(2, \infty)$ and if $$\lim_{x \to 2^{+}} ... 1 Interesting question. When people say that, i don't think it is strictly mathematical. As far as i'm concerned what is implied is the following. 100\% faster charging = half charging time required = \frac{5}{2} = 2.5 minutes to charge (twice as fast). 125\% faster charging = 2.25 times less charging time = \frac{5}{2.25} = 2.22 min. to charge. ... 0 "350% faster" is promotional vocabulary for "3.5 times faster". A ratio of 3.5 between speeds or times. This is inconsistent with the usage for smaller percentages. 10% faster would mean subtraction; the action happens in 0.9 times the original duration. Where is the boundary between the two modes of speech? I'd guess the descriptions switch to the ... 13 "Faster" refers to speed (units of something per unit of time) or velocity (units of something per unit of time, with a directional component). In this case you are talking about charging a battery in 5 minutes. The speed of that operation is (1.0\,\text{battery})/(5.0\,\text{minutes}), or 0.2 batteries per minute. 100\% faster than 0.2 batteries ... 0 I think you're being too literal. 100% faster, in this context, means you cut the time in half. Definitely not the correct usage though. For instance, saying "That shirt is 100% off" when you mean "that shirt is half off" is really bad form. If you see that battery x is charging b% faster, which as a proportion is p=1+{{b} \over {100}}, what is ... 1 You're applying the percentage to the wrong magnitude. 100% faster means speed is increased by 100%, i.e., double speed. So, you don't reduce time by 100%, but by 50%. If it's 200% faster, you get 3x speed. In your example, you'd charge the battery in 2.5 and 1.6 minutes, respectively. 0 Percentage is nothing but a number: 100\% = 1. So, 200\% = 2 \cdot 100\% = 2. Saying that something is 200\% faster is same as saying it is 2 times faster, which essentially translates to your battery being charged in half the time. EDIT: It seems that "faster" means +200\% and not \times 200\%. Oh, well... luckily enough, it could just as well ... 2 In naive set theory, we usually have the operation$$A \setminus B := \{x: x \in A, x \notin B\}$$For ordinary sets, we have:$$A \setminus B = A - (A \cap B) = (A \cup B) - B$$although usually the latter two are also written using \setminus. Therefore, it would seem reasonable to retain this definition for multisets. Indeed, this definition also ... 2 These are strictly triangular matrices. A matrix A=(a_{ij})\in M_n(F) is strictly (upper) triangular matrix if for i\geq j, a_{ij}=0. 0 This can be one visual alternative. Concepts of hypergraphs won’t be addressed in this answer. In any case, the paper “Directed hypergraph and applications” of Giorgio Gallo, Giustino Longo, Sang Nguyen,and Stefano Pallottino is a good support. In order to explain the concept, i'll use a form of graph coloring applied to hypergraphs. Contact Coloring and ... 1 The error in the question is to assume that people say "linear" only in the context of algebra (where log can be interpreted as linear in the way you explained) when in fact the usage almost everywhere else is incompatible with this alternative interpretation. Examples include linear functions in geometry (and physics, science, economics, etc), as ones ... 14 it's worth noting that with your argument, any bijection is linear! We have a set X and a vector space Y. We have a bijection$$ f: X \to Y. $$We simply define the operations$$ x+y = f^{-1}(f(x) + f(y)), \;\;\; \lambda x = f^{-1}(\lambda(f(x)).  Now $f$ is linear. The issue is that when we say $f: V \to W$ is linear, we generally already have ...

9

I will work through this, trying to justify everything I do to the detail. What I'll end up with will probably be something like a summary of the other answers and the question statement, but it is done with the hope that nothing at all "sketchy" appears in arguments. The fundamental definition. A function $f:(V,+,\cdot)\to(W,\oplus,\odot)$ of vector ...

3

A subgroup is a subset of a group that is itself closed under the group operation. A semigroup is a set equipped with an operation that is merely associative, different from a group in that we assume the binary operation of a group is associative and invertible, i.e. each element has an inverse with respect to the operation.

3

It is a subgroup of $\mathbb{C}-\{0\}$ equiped with multiplication. So it's also a group (a subgroup is a group). A semigroup is just a set with an associative operation. So every group is also a semigroup, but the converse is false.

13

The first statement you have is: The logarithm is not the restriction of linear map $(\mathbb R,+)\rightarrow (\mathbb R,+)$ to $\mathbb R_{>0}$. The second one you have is: The logarithm is a linear map $(\mathbb R_{>0},\cdot)\rightarrow(\mathbb R,+)$. I don't see any contradiction there. Certainly $f(xy)=f(x)+f(y)$ does not imply ...

4

You are correct, $\ln$ is linear for this vector space structure on $\mathbb{R}_*^+$ (though not for the usual vector space structure on $\mathbb R$. The thing is that this vector space structure you defined has no real interest, so people don't use it. In fact probably most functions could be made into something linear for a weird, custom-made vector space ...

35

You are correct if we endow $\Bbb R_{> 0}$ with the strange vector space structure in which "addition" is given by the usual multiplication, and "scalar multiplication" is given by exponentiation. When people say that logarithms are not linear, they are usually thinking of giving $\Bbb R$ the usual vector space structure, and with this being understood, ...

2

In the 17th and 18th centuries, the name was commonly spelled l'Hospital, and, as far as I learned when I was a student, even himself used to spell his name that way. However, with time going, many French spellings have been altered: in particular, the silent s has been removed and replaced with the circumflex over the preceding o making l'Hôpital. The ...

1

I know you problem! I live in a city where English is popular, but people are not so good at it. I know for sure that it is pronounced without the 's', (L'Hôpital's Rule). But whenever I ask a question related to it to my teacher, instead of saying "lop-it-als" I say "eL-Hos-pi-tals", and then only he understands. So I don't think it will be a problem, just ...

0

It's better to think of each as its own type of object, but where some types can be naturally converted into others (similar to type conversion in computer programming). Some of the conversions faithfully translate all of the information, some are reversible, and others lose part of the information.

1

To summarize the comments: The concepts in their general meaning do not admit a linear hierarchy. Vectors are elements of a vector space. For example, $\mathbb R^n$, the set of $\mathbb R$ valued function over a non empty set $X$, $\mathbb R$ as $\mathbb Q$ vector space. Sequences are function form $\mathbb N$ into any non empty set $X$. If $X$ is a $K$ ...

Top 50 recent answers are included