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3

For any function $f : \mathbb R \rightarrow \mathbb R$, the set $f^{-1}( a, \infty)$ always exists, and it is defined by $$ \{ x: \in \mathbb R : f(x) > a \}, $$ although it is not always measurable.


4

$f^{-1}$ here does not mean the inverse of $f$. It is a horrifically bad overuse of notation and can be very misleading. $f^{-1}(A)$ where $A$ is a set is the pre-image, i.e. $$f^{-1}(A) = \{x\in X: f(x) \in A\}.$$ The pre-image always exists. If $f$ were invertible, then this would coincide with what you think; however the pre-image takes care of the ...


7

$f^{-1}(A)$ is the preimage of the set $A$, it exists even if $f$ is not invertible. For $f: X\to Y$, it is defined as $$f^{-1}(A) = \{x\in X: f(x)\in A\}.$$ The notation is slightly confusing, I would say, but one gets used to it. It isn't really that bad of a notation, since, if $f$ is actually invertible and its inverse is $g$, then $g(A) = f^{-1}(A)$ ...


1

Given a measureable $f:(X,\mathcal{M},\mu)\to(Y,\mathcal{N})$ (a measure space to a measurable space) one can define a measure $\nu=f_{*}\mu$ on $(Y,\mathcal{N})$ by $$ \nu(B)=\mu(f^{-1}(B)). $$ Both $f$ and $\mu$ are necessary to define it, so $f_{*}\mu$ seems like good notation to me.


2

Just so that everyone knows what we are talking about here, let me rephrase in more familiar notation. Suppose $(\Omega, \mathcal{F}, P)$ is a probability space, and $(M, \mathcal{M})$ is a measurable space. If $X : \Omega \to M$ is a random variable (i.e. a $(\mathcal{F}, \mathcal{M})$-measurable function), it induces a pushforward measure on $(M, ...


0

First: Yes, the square root of the determinant of the $i$-th covariance matrix. Second: No, the inverse of the $i$-th covariance matrix. Yes, raising $2\pi$ to the $D/2$-th power. If $D$ is odd, this is a half integer; you still need to use $D/2$, without the rounding suggested in your code. $x^{D/2}=\sqrt{x^D}$.


3

Opinions on this issue differ, but I strongly believe that a basis (particularly in finite-dimensional linear algebra) should be a list, not a set. Here I am using "list" to mean the same thing as "ordered set". Here are two reasons why using sets does not work well: It is often convenient to talk about the matrix of a linear map $T \colon V \to W$ with ...


1

You're on the right track, but you're having trouble with the endpoints. The points $(-3, -2)$ and $(4, -2)$ are on the graph in a), which means that your domain should actually be $x \in [-3, 4]$ in interval notation to indicate that $x = -3$ and $x = 4$ are included. This would be reflected in the set-builder notation by using inequalities with "or equal ...


0

I know that, $\sin\theta=\frac{y}{r}$, and, $\cos\theta=\frac{x}{r}$. Probably you mean the "trigonometric functions as quantities computed from a right triangle" definition. To make it clearer that everything is well defined it is better to use the more modern "point on a fixed circle of radius $r$" definition, but this answer will use the presumed ...


0

There are various conventions regarding the nomenclature for relations. Unfortunately, they are often conflicting. On the one hand, we have the "axiomatic-set-theoretic" names: $A, B$ have no name $\{1,3\} = \{x \mid \exists y: (x,y)\in R\}$ is called the domain of $R$ $\{20,40\} = \{y \mid \exists x: (x,y)\in R\}$ is called the codomain of $R$ On the ...


0

It means $p$ is a divisor of $a$, i.e $a$ leaves a remainder of $0$ when divided by $p$. You can also write $a \equiv 0 \text{ mod }p$.


2

If you wish to think about it from the perspective of "why isn't $\theta$ on the right hand side," it may be helpful to approach sin and cos backwards - start with $\sin^{-1}(x)$ and work our way backwards. Why? Because doing so makes it look more like a typical definition of a variable, with a variable on one side and some expression on the other side. ...


-1

Besides customary uses, I've seen it used to emphathize the differences of two types of addition, for example: $(\alpha \oplus \beta)A+B$. The $\oplus$ refers to regular scalar addition, and the $+$ refers to matrix addition.


10

If $A$ and $B$ are modules over a ring, their direct product $A \times B$ and their tensor product $A \otimes B$ are different things, so it would be unhelpful to use the same notation for them.


9

These symbols have different meanings in different contexts. For instance, if we are talking about vector spaces then saying $V=U+W$ is different from $V=U\oplus W$


1

Like others said, it does depend on the context—but here's the way I intuitively answer your questions: What are the trigonometric functions? $\sin$, $\cos$, and $\tan$ are the ratios of the length of the sides of a right triangle They are functions of $\theta$, but all $\theta$ is doing is telling you which sides to look at. As an example, let's ...


8

Looking at your profile, you post a lot on StackOverflow, so maybe a programming analogy will help. Say you want to make a Point class. There's a lot of ways you could do it. You could have member variables p.x and p.y, for the $x$ and $y$ coordinates (duh). That's the most common way. But in theory, you could also write it in terms of p.rad and p.theta. ...


5

It depends on how you want to define $\sin$ and $\cos$. I suspect you're looking for an explicit definition in terms of things you already know (such as polynomials), in which case @Karl's answer (the MacLaurin series definition) is what you're looking for. However, others like me find it more elegant to define them implicitly as bases for the the set of ...


17

I think the historical reason for the confusion stems from graphing trigonometric functions in polar form versus rectangular form. In rectangular form, the following statement below is true. $$ \theta \quad =\quad x $$ (Where the meaning of this equality is that we let the measure of the angle on the unit circle be representative of the rectangular distance ...


15

There is a $\theta$ on the right hand side. The definition of $\sin(\theta)$ is not just $y/r$; instead it is something like: $y/r$, after you've drawn a right triangle with $\theta$ as an angle, and where $y$ is the length of the side opposite $\theta$ and $r$ is the length of the hypotenuse. As you can see, that full definition does in fact contain a ...


11

I guess it depends on how you define $\sin\theta$. One possible definition is $f(\theta)\equiv \theta- \frac{\theta^3}{3!}+\frac{\theta^5}{5!}...$ This series converges for all $\theta$ and is called $\sin\theta$. For more info look up Maclaurin series. More generally I think you are confused about what a function is. The technical definition is daunting ...


1

$\sin(\theta) = \sin \theta$, without brackets is just a short hand notation. And yes, it means that $\sin$ depends on $\theta$ as its argument like a function $f(\theta)$. In this sense your expression $\theta = \sin^{-1}(\frac y r)$ means that you use the inverse of $\sin$ (actually the inverse of $\sin$ is $\arcsin$) to get the related $\theta$ to the ...


5

Yes $\sin$ is just a function on the real (or complex) numbers. People often write $\sin(\theta)$ or $\sin\theta$ because the argument of the $\sin$ function is often an angle in physical applications, and $\theta$ is often used to denote angles. For your follow-up: $\sin^{−1}$ is the inverse function of the $\sin$ function. Similar to how $\log$ is the ...


1

It means different things in different contexts. However, it is function notation. $\sin(\theta)$ is the function that takes an input angle $\theta$ and outputs the ratio of the opposite side and hypotenuse for a right triangle. $\sin^{-1}(y/r)=\theta$ is the inverse function. It takes as input a ratio of opposite to hypotenuse and outputs the angle to ...


1

P|a means P divides a.For P|a we can also write this as Pc=a where c is a constant.It simply means that $$P*c=a$$ or P is a factor of a.


2

It means $P$ is a divisor of $a$.


0

A thing that maps indices to objects is called a tuple or a sequence. It's richer than a set, because it is ordered, and it can include an object more than once. You typically use round parentheses to write the tuple and subscripts to write its elements: $$A=(1,2,3,4)$$ $$A_2=2$$


1

If you only use this type of set, then this is impossible, because sets have no order, as you said. But you can use other objects, which are often helpful: use tupels (or vectors, which are basically the same). also, you can instead use $A$ as a function: $A : \{1,2,3,4\} \to \mathbb R, \, A(k) = k$. then you can freely "access" the second element.


0

You (almost) can't. You are thinking of sets as subsets of some ordered set, like the integers or the reals. If $A=\{1,\text{ahdwhelrj},<\text{some picture file}>\}$, what is $A(2)?$ As there is no natural order, you can't distinguish this from $A=\{\text{ahdwhelrj},<\text{some picture file}>,1\}$ If you have a global axiom of choice, you can ...


3

The only difference it that it sometimes unclear if you consider $0$ as an element of $\mathbb{N}$ so that $$ \mathbb{N}=\{0,1,2,...\} $$ or that $$ \mathbb{N}=\{1,2,...\} $$ The first notation removes this ambiguity and makes things more clear. At the end - I would say that its a matter of preference and convention, I have seen both used many times ...


3

If you define $\Bbb N = \{1,2,3,\dots\}$, then yes: the two sets you've defined are identical, and describe the same infinite union. Note that some define $\Bbb N = \{0,1,2,3,\dots\}$


0

You are defining a partial order on the lists. You could use $\ge$ to show that, but people may confuse it with the usual greater than order. You might want to use $\succ$ or $\succeq$, which are often used with some other order relation.


1

a) $1/999 = 0.001001001\ldots = 0.\overline{001}$ Either of these is fairly standard notation. The overline format $0.\overline{001}$ is a little more explicit, so I think it would be preferred. b) $.001 + .000001 + .000000001 + \cdots$ This denotes an infinite sequence in the way that $1, 2, 3, \ldots$ indicates an infinite sequence: a little ...


1

In Hatcher's Algebraic Topology, these are called characteristic maps, and the restriction $\varphi\colon \partial\Delta^n\to \Delta$ to the boundary is referred to as the attaching map. In my experience, this usage is fairly standard.


0

For fixed $x$, call $p = P(Y=1|x)$. Then $1-p = P(Y=-1|x)$. Let $G(f) = E_{Y|x}\left( \exp(-Yf) \right) = p\exp(-f) + (1-p)\exp(f)$. Setting $\frac{dG}{df} = 0$ yields the result.


2

In Germany, there is the DIN 1338 standard, according to which the differntial operator d, as, e.g., e for the Euler number, should be typeset as an upright letter. According to Wikipedia, these letters are typeset in italic if AMS conventions are used.


11

Many excellent journals and books use $d$ in the italics form, such as the Journal of the American Mathematical Society (e.g., recent article by Terence Tao), London Mathematical Society Proceedings (e.g., equations 74 and 75 of this recent paper) and Spivak's Calculus. Given that reference quality publications use $d$--and that it is faster and cleaner to ...


18

This is more of an extended comment than an answer, but with regards to typesetting in $\LaTeX$, let me point out that typing \mathrm{d} should not take longer than typing d, as you shouldn't be doing either throughout your paper! The semantically correct thing to do is to define a macro representing your desired differential operator, for example, ...


2

By logic I think that is a good idea differentiate the symbol with the roman notation but there isnt a "standard", you can use any of them. In the same sense doesnt exist any kind of "standard" mathematical notation. I read a lot of books of many mathematical topics, everyone with different notations, not only just the infinitesimal symbol. The problem, ...


4

There is no difference between them. It merely comes as a result of a choice in LaTeX formatting; specifically, some people write "\text{d}" (or some equivalent) for the upright formatting, but many other people don't do this for the sake of speed, and instead just write "d".


8

Quick answer: there is a standard to follow. Longer answer: while physicists write differential operators in upright fonts (because they follow the standards), mathematicians tend to typeset differential operators as variables (because we are lazy). I am joking, but it should be clear that $dx$ is not $d \cdot x$, and that $d$ is essentially an operator: ...


0

OK, I think I got it. First $D_A(i)$ is defined (as a matrix) for $n$ particular values in your field $K=\Bbb Z/p\Bbb Z$. Then for every matrix position, there is a unique univariate polynomial $P\in K[x]$ of degree strictly less than $n$ (your book is off by one here) such that the corresponding entry of $D_A(i)$ equals the evaluation $P[x:=i]$ (this is a ...


1

The symbol for denoting similar triangles is ($\color{blue}{\sim}$) Notice, suppose $\triangle ABC$ & $\triangle PQR$ are similar then in LaTex it is written as $\text{"\triangle ABC \sim \triangle PQR"}$ surrounded in-between by 2 or 4 dollar signs which appears as follows $$\color{blue}{\triangle ABC \sim \triangle PQR}$$


-2

$f(x)=\exp(O(|x|^2))$ means that there is a $C$ such that $$\lim_{x \to \infty} \frac{f(x)}{\exp (C|x|^2)} \leq 1$$


4

$\triangle ABC \sim \triangle DEF$.


3

The following also works : $$\{n\mid n\in\mathbb N,n\not=1,n^2\equiv 1,19\pmod{30}\}$$


11

The other answers are correct, but here is an attempt at a compact approach: $$\{n\in\mathbb{N}_{>1}:\gcd(n,30)=1\}$$


5

You could use $(\mathbb{N}+6) \setminus (2\mathbb{Z} \cup 3\mathbb{Z} \cup 5\mathbb{Z})$.


9

These are the positive integers that are not 5-smooth. These numbers are also called 7-rough. Alternatively, $$S = \{n : n \in \mathbb N_{>1}, 2 \nmid n, 3 \nmid n, 5 \nmid n\}$$ works.


1

Lets define $\models\alpha$ as a shorthand of $\emptyset\models\alpha$ and we will proof that $\models\alpha$ if and only if $\alpha$ is a tautology. This implies taking $\models \alpha$ to mean "$\alpha$ is tautology" is an equivalent definition. Given an arbitrary valuation $v$ there are four cases: 1) $\forall x\in\emptyset([x]_v=0)$ 2) $\forall ...



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