New answers tagged

0

Let $R(x)=f(x)-f(0)-f'(0) x$ so that $f(x)=f(0)+f'(0)x + R(x)$. What they mean by $R(x)$ having complexity $O(x^2)$ (sometimes written $R(x)=O(x^2)$) as $x \to 0$ is that for all $x$ sufficiently close to $0$, we have $|R(x)| \le C x^2$ for some constant $C$. That is, the value of $R(x)$ as $x\to 0$ is asymptotically no bigger than some multiple of $x^2$. ...


1

Depending on the audience (if they're coming from a slightly more computer-scientific background), treating as the function foldl could perhaps be the clearest. See also How to write a functional fold in mathematics?


1

You could write e.g. write $x = f(\ldots f(f(i,a_k),a_{k-1})\ldots ,a_0)$ or alternatively define $g_n (y) := f(y,a_n)$ and write the equation as $$x = \left(\bigcirc_{n=0}^k g_n\right)(i)$$


2

I can think of two possibilities. One is $$ x = f(f(\cdots f(f(i, a_k), a_{k-1}), \ldots, a_1), a_0) $$ and the other is $$ x = b_0 $$ where $$ b_j = f(b_{j+1}, a_j) \qquad 0 \leq j < k $$ $$ b_k = f(i, a_k) $$


0

The following notation is used by R.P. Stanley in Topics in Algebraic Combinatorics. See section Basic Notation (p.6) \begin{array}{ccl} &2^S\qquad&\qquad \text{the set of all subsets of the set } S\\ &\binom{S}{k}\qquad&\qquad \text{the set of $k$-element subsets of } S\\ \end{array} We can find in section 1.2, p.23 in his classic ...


1

You want a notation for the set of all $k$-element subsets of some set $X$. There’s a standard notation for this: $[X]^k$. However, it’s most commonly found in set theory and some areas of combinatorics and for a more general audience is quite likely to be unfamiliar to some readers, so you’d do well to define it anyway. Added: I don’t care for the notation ...


1

In plain english: Any integer number is twice another integer number or its successive number is twice an integer number (, or both). Where I addes the or both to emphasize that the mathematical or is non-exclusive (that would be xor) Or equivalently, by definition of even and odd: Any integer number is even or odd (, or both).


1

I find it helps to read "$\exists$" as "we can find," and $\forall$ as "for each." While not technically true, as there may be no algorithm for finding the thing, it helps straighten out the difference between $\exists, \forall$ and $\forall, \exists$. The first says "We can find a thing, which for all other things, blah blah blah blah." The second says "...


0

A lot of the time, mathematical notation is decided by historical reasons rather than logical ones. This is true with function application. These all mean the same thing, unless it's obvious from context that they don't. For example, the notation $fx$ is common when $f$ is a linear transformation. Indeed, $Ax$ is common shorthand for $A(x)$; they mean the ...


0

Unfortunately I discovered that I've mixed the notation in a work on Lie Groups. It was unfortunated because I than realized that I could use $exp(X)$ for the application between the Lie Algebra and the Lie Group, while using $e^X$ for the proper matrix exponential and then I could conclude in a theorem that $exp(X)=e^X$. It would have been a great theorem....


1

My reading of this quote is that $\overline{\bf R}$ refers to $\Bbb{C}$, and that "impressive finiteness result" refers to the (not that obvious) fact that $\Bbb{C}$ is algebraically closed. Most of us hear about $\Bbb{C}$ being algebraically closed early in our studies, may be in the same course the complex numbers are first introduced? More often than ...


1

$\beta_{\frac{1}{2}}=\frac{1}{2}(\delta_0+\delta_1)$ is a probability measure on the real numbers: $\beta_{\frac{1}{2}}(A)=1,\frac{1}{2},0$ depending on whether the set $A$ contains both of $0,1$, exactly one of $0,1$, or neither $0$ nor $1$. The factor $\frac{1}{2}$ is needed to make $\beta_{\frac{1}{2}}$ a probability measure. Moreover, $\beta_{\frac{1}{2}...


1

The coefficients given by the formula are correct; for instance, ${2}\choose{1}$$=2$, so it gives you $x^2+2xy+y^2$. In general, ${n}\choose{k} $$=$ $$\frac{n!}{k!(n-k)!}$$ e.g. ${2}\choose{1}$$=$ $$\frac{2!}{1!\cdot1!} = \frac{2}{1} = 2$$


0

I don;t know exactly what this notation means, but I can tell you how homogeneous coordinates are used in things like computer graphics. (Indeed, shameless plug: I wrote a book describing this (among other things) in considerable detail, and if you want the longer explanation, that's one good place to get it.) Here's the idea: linear transformations of the ...


3

Let me expand on Dave L. Renfron's answer: While all notations listed thus far are commonly used and I don't think there is anything wrong with it, there is a reason why $$ A^{B} = \{ f \in \mathcal P(B \times A) \mid f \text{ is a function} \} $$ is preferable. At least until one is comfortable with these notations and there is no fear that these semi-...


3

Perhaps better still is $$A^B = \{f \in {\mathcal P}(B \times A) \mid f \text{ is a function}\},$$ because this fits the more formal way of defining sets using the Axiom schema of specification (i.e. set builder notation in school math).


7

The notation $f:B\to A$ is typically used to denote that $f$ is a function from $B$ into $A$. Thus, saying $$A^B = \{f :B \to A| f \text{ is a function}\}$$ is like saying $A^B$ is the set of all functions $f:B\to A$ such that $f$ is a function. So there is some redundancy. I, personally, would go with the first option, however your intentions seem ...


5

The expression "$n/p$" reprsentsthe number that you get when you divide $n$ by $p$, whereas the expression "$p\mid n$" represents the statement that $p$ divides evenly into $n$.


9

For nonzero integers $x$ and $y$, the following are equivalent statements: $x$ divides $y$ $x\mid y~~~~~~~~~~~~$(this is read aloud the same way as the previous line) $y$ is divisible by $x$ $y\equiv 0\pmod{x}$ $\frac{y}{x}$ is an integer There exists some integer $k$ for which $xk = y$ $y$ is a multiple of $x$ $x$ is a factor of $y$ On the other hand, $...


4

$p \mid n$ means "p divides n" or $\exists k\in\mathbb{Z}$ so $n=pk$


1

All you need to do is show $f,g$ continuous $\Rightarrow$ $f^g$ continuous, the general result follows by induction. The basic results follows from $f(x)^{g(x)}=e^{g(x)\ln f(x)}$.


2

The familiar a % b or mod(a, b) found in many programming languages is a function from the integers to the set $\{0, 1, \ldots, b-1\}$ (or something very similar). That function finds surprisingly little use in mathematics, however, so little that there's no standard notation for it that I know of. On the other hand, the notion of "equivalence mod $...


2

The notation $m\equiv n\pmod{k}$ means $$ k\mid (m-n) $$ In words, $k$ divides $m-n$. On the other hand, $m\bmod k$ denotes the unique number $n$ with $0\le n<k$ such that $$m\equiv n\pmod{k}$$ (under a very common convention). As an example, $42\bmod 8=2$, but we can also write $42\equiv 34\pmod{8}$.


3

Because $A\equiv C\pmod{B}$ and $A\bmod B=C$ mean very different things. The first just says that $A-C$ is a multiple of $B$, so for a given $A$ and $B$ there are infinitely many different values of $C$ making it true. The second says that $C$ is the unique integer in the set $\{0,1,\ldots,B-1\}$ for which $A-C$ is a multiple of $B$. To put it a bit ...


2

This is not correct: you should not have the $\exists a,b$. Indeed, reading your set $$\{(a,b)\mid \exists a,b:(a,b \mid n \ \land a \mid b)\}$$ aloud, it is $$\text{the set of $(a,b)$ such that there exist $a$ and $b$ such that $a$ and $b$ divide $n$ and $a$ divides $b$.}$$ This doesn't make sense! You can't say "there exist $a$ and $b$" because you ...


1

The Cartesian product will work, but you want to specify how the concatenated vector is composed. $$\left\{\vec z\in X{\times}Y ~\middle\vert~ \vec z=\vec x\Vert\vec y\,, \vec x\in X\,, \vec y\in Y\,,\mathbf 1^\top\vec x=1\,, \mathbf 1^\top\vec y=0\right\}$$ Or more elegantly: $$\left\{\vec x \Vert\vec y~\middle\vert~\vec x\in X\,, \vec y\in Y\,:\,\mathbf ...


0

Why not? Because $X\times Y=\{(x,y): x\in X,\, y\in Y\}$ so no matter what set is $X$ or what set is $Y$.


0

I'd assume the latter is a fixed constant "the expected value of random variable $X$ given the value of random variable $Y$". $\mathsf E(X\mid Y)$ is rather: "the expected value of random variable $X$ for any given value of random variable $Y$."   You don't know what value you might be given, so how can this be a fixed constant?†   The ...


2

The intended interpretation is that $K^*=K\setminus\{0\}$ (the group of multiplicative units of $K$), so an element of $K^*$ is just a nonzero element of $K$. Another common notation for this set is $K^\times$.


3

Yes, the notation is confusing: $E(X)$ is a constant, but $E(X\mid Y)$ is a random thing. If you haven't gotten to advanced (measure-theoretic) probability yet, then $E(X\mid Y)$ is usually introduced by considering $E(X\mid Y=y)$, the conditional expectation of $X$ given that $Y=y$. Now think of this as a function of $y$, i.e. it's a function $h(y)$ ...


1

This is a bit unnecessarily elaborate machinery, but if you want, you can view $E[X]$ as $E[X \mid \{ \emptyset,\Omega \}]$. Thus, $E[X]$ is measurable with respect to $\{ \emptyset,\Omega \}$. But the only such functions are constant (exercise). It is more convenient to view constant functions as just being literally equal to their value than to view them ...


3

I'm not familiar with the measure-theoretic approach to conditional expectation and would appreciate it if someone could correct my response here (or even better, post a dumbed-down explanation of how conditional expectation is defined). But I suppose one way you could look at this is by noticing that - for example, if $X$ and $Y$ are continuous - $$\mathbb{...


0

Shortly, from a formal perspective, introducing new notation amount to expanding the language of the formal system (theory) you are working on, adding some axioms that in some way provide a definition for the new symbols. The new symbols can be predicate or operation-symbols. For predicate-symbols the new axioms are logical equivalences between the ...


2

What you may be looking for in your formal system is variously called full abbreviation power or definitorial expansion. Basically, it comprises rules that allows you to create on the fly new symbols extending the original language. We need one type of rule for each kind of symbol: $\def\eq{\leftrightarrow}$ For each $k$-parameter sentence $φ$ over the ...


3

I think the main problem is not non-closedness but non-compactness. Consider $f(x) =-e^x$. Its supremum is zero, but the function is nowhere zero. However, as every supremum, you can approximate its supremum by function values, i. e. there is a sequence $x_n$ such that $f(x_n) $ converges to 0, but the sequence itself does not converge. This problem would ...


1

I thought it was all points $0 \le x \le1$. If it is, I have no idea what the heck the answer is. The notation $[0,1]$ means all the points $0 \le x \le 1.$ The notation $f : [0,1] \to [0,1]$ is different. It means $f$ is a function that takes its input values from $[0,1]$, and its output values are (also) part of $[0,1]$. More generally, and as ...


2

what about $\{y\in\Bbb{R}^n| \forall 1 \le i < j \le n, y_i\ne y_j\}$?


1

It means the domain and codomain of the function are the closed interval from $0$ to $1$. Given any $x$ such that $0 \le x \le 1$, you must have $0 \le f(x) \le 1$. This is true for $f(x)=x^2$. You don't have to have that every point in $[0,1]$ is the image of some point in the domain. If that is true, the function is surjective.


0

To convert from Polish notation to Reverse Polish notation, the variables and constants stay in the same order, but the operations go at the end instead of the front. So, let's say you have something like C(C(x, C(y, z)), C(C(x, y), C(x, z))). Each C corresponds to a left parenthesis and a right parenthesis. We just move each C from preceding it's left ...


-1

The choice of symbol, depends mainly on the mathematics, physics, or programming discipline the equation is used in. Merrian-Webster defines minimum as "the least quantity assignable, admissible, or possible." For ordering, minimum means 'less than or equal to', which is symbolized in some/many mathematics disciplines as ≤. List A = [1 0 1 0], List B = [...


1

Suppose I wrote something meaningless and terrible such as $\sqrt{\mathbb R} + 27 \ne 13 \implies $ $\sqrt{\mathbb R} \ne -14 \implies $ $\mathbb R \ne 196$ Your assumption ought to be that I clearly don't have the foggiest idea what I'm talking about and I am clearly ... lost. Then supposed I argued "But every statement is true! $\sqrt{\mathbb R}$ is ...


3

Firstly, to answer your question, if the limit $\lim\limits_{n\to\infty}a_n$ does not exist, it makes no sense to say that $\lim\limits_{n\to\infty}a_n=a$, where $a$ does not exist. We would say that the sequence $\{a_n\}_{n=1}^\infty$ is divergent. In short, no, it is not good notation to say that $a\neq 0$, in any of the cases you stated. Secondly, the ...


2

The standard way to write this is $f(x)^2$ (this always means $(f(x))^2$ rather than $f(x^2)$, so you don't have to worry about any ambiguity). You will occasionally see $f^2(x)$, but I would not recommend this--$f^2(x)$ more often means $f(f(x))$. The notation $f^2(x)$ for squaring $f(x)$ is generally used for only a few particular functions (typically ...


0

Square of a function $f(x)$ can be denoted as $f^2(x)$, similar to $sin^2x$ and $cos^2x$.


9

The intent ought to be what you said. So, it is $a ⋇ b$ is short for "$a \ b$ or $a \ b^{-1}$" that is the multiplicative analogue of $\pm$. Yet, a more common way to write $a ⋇ b$ is just $a \ b^{\pm 1}$. In addition, I never heard of this operator, all the early search hits are for Unicode-tables not math, and at least to me it looks quite similar to ...


4

No, it doesn't behave like $\pm$. An important part of how $\pm$ behaves is that it is likely to be understood without further explanation, and the $⋇$ sign does not have that property. You can use it for "divided or multiplied by" if you want, but you will have to explain this usage to your reader before you do so, because you can't expect it to be clear ...


1

I can spot the following mistake in your attempt: And by using the properties of limits we can say: $$\frac{\lim_{h \to 0}{f(x + h) - f(x)}}{\lim_{h \to 0}{h}} = D$$ You cannot actually do that, as smcc has said. You must note that $$\lim_{x\to 0} \frac{f(x)}{g(x)}=\frac{\lim_\limits{x\to 0} f(x)}{\lim_\limits{x\to 0} g(x)} \,\,\,\,\,\,\,\,\,\,...


2

Yes, $X/A$ is the space where you identify everything in $A$ to a single point, also known as the quotient space of $X$ by $A$.


1

There seems to be a confusion as how you read the question you link. It asks you to find the summation notation for $$ \frac{1}{n}+\frac{2}{n+1}+\frac{3}{n+2}+\cdots+\frac{n+1}{2n} \tag{1}. $$ $(1)$ is then called expanded form of the sum, because... well, it lists all the terms that are summed. Then, the answer says that the summation notation for the ...


4

As a spinoff of that notation that $n \choose k$ denotes the number of $k$-element subsets of a set of size $n$, we can define $S \choose k$ to denote the set of all $k$-element subsets of a set $S$. So, to say that you're thinking about one of the sets $\{A,B\},\, \{B,C\}$ or $\{A, C\}$, you might write something like "$\Delta \in {{\{A, B, C\}}\choose{2}}$"...



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