# Tag Info

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I imagine that $f'$ is the gradient, so it is a $n$-vector. $f''(x+t(y-x))(x-y)$ is the product of an $n\times n$ matrix and a $n$-vector, so it is again a $n$-vector.

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It means that $q$ divides $k$, or $k$ is divisible by $q$. Add a little slash and it negates that meaning: $q \nmid k$ means $q$ does not divide $k$. For example: $3 \mid 1728$, $3 \nmid 1729$. They are not relatively prime, unless $q = 1$ or $-1$. In fact, if $q \mid k$ then $\gcd(q, k) = |q|$.

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The notation means : $q$ divides $k$ ; There is an integer $m$ with $qm=k$.

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$x+x-2x \equiv 0$ but $x^2+1=0$ is for only $x= \pm 1$. The first represents an identity which holds for all $x$ while the other is conditional equal which may or may not have solutions.

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The sign you are referring to is the standard symbol for "identically equal to". For example, a more usual context in which to see that symbol (at least in pre-calculus classes) might be: $$(x-1)^{2} \equiv x^{2}-2x+1,$$ where the presence of the symbol indicates that the equation holds for all values of $x$. The reason for its use in this specific context ...

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The second notation means "decreasing to", the first just "going to". In the same way you can use the "increasing to" arrow: $\nearrow$. Both "increasing to" and "decreasing to" implies "going to" but the vice versa does not hold. Sometimes $\uparrow$ and $\downarrow$ are also used.

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$n!!$ often denotes the semi-factorial, i.e. the product of every other integer up to $n$.

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Well i can only assume what you are looking for. By the term in your question i think you might mean a superfactorial. See this Wikipedia Article on Factorials - it might help. If you mean a product of factorials it would be a notation like $$sf(n)=\prod^n_{k=1}k!$$

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A double factorial is something else. In your case, it would be something like $$1!2!3!\dotsm n! = \prod_{k=1}^n k!$$ That's about it.

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It is the boundary of the ball.

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It is one of the common notations for numeral systems in different bases. In this case it is the ternary system. See, $12_3=1\cdot3^1+2\cdot3^0=5$.

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$u$ is a two variable function $u(x,t)$, $$u_{xx}=\frac{\partial^2x}{\partial x^2}$$.

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subscript is a derivative with respect to this variable. $$u_{xx} = \frac{\partial^2u}{\partial x \partial x}$$

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The paper explains the notation in the footnote on page 2: It uses hollow square brackets $[\![\cdots]\!]$ as a notation for the Iverson bracket, which by definition means "$1$ if the claim inside the bracket is true; $0$ if it is false". Thus, $y=2[\![c_k=c_{\rm min}]\!]-1$ is just a terse way to write $$y = \begin{cases} 1 & \text{if }c_k=c_{\rm min} ... 0 You can hardly find it defined "alone" because it is part of the "complex" : function f from A to B (denoted : f : A \to B). See : Ethan Bloch, Proofs and Fundamentals : A First Course in Abstract Mathematics (2nd ed - 2011), page 131 : Definition 4.1.1 and Index, page 354, sub voce : function. 0 If we write y = f(x), then f:A\to B indicates that x \in A and y\in B. It is required that f(x) be defined for every x in A and that every value of f(x) should be in B. Some times the relationship between x and f(x) is written as x \mapsto y=f(x), or more commonly, as x \mapsto f(x). A is called the domain and B is called the ... 2 This is not a mathematical term but a figure of speech. One says that A is an honest B to assert that A satisfies the definition of B. Usually the reason to emphasize honest is that A was informally called "B" earlier in the text. This is associated with a somewhat conversational style of writing. Specifically for distance functions, this word would ... 0 I'm not sure there is a universal way of saying each equation, and I think it will sound clunky trying to say it all in one sentence anyway. Perhaps there is a way around it? But anyway, here is how I think they could be said:$$\bar{x_k}^C = \frac{\sum_{i=1}^{m_k}\bar{x_i}}{m_k}$$"x bar sub k super C equals the sum of x bar sub i from i equals one to m ... 2 U is some other normed vector space. In this case L^2([0,T];U), sometimes lazily written as L^2(0,T;U), consists of functions f from [0,T] to U such that \int_0^T \| f(t) \|^2 dt<\infty, where \| \cdot \| is the norm on U. This notation is used in, for instance, Partial Differential Equations by Evans. Most commonly the U in question ... 2 The colon : inside set brackets means "such that". Here, (x,y)\in$$R$iff$(x,y)\in$$A$$\times$$A and (f(x),f(y))\in$$S$. 4 Just a correction to the answer of Tien Kha Pham (since I don't have enough reputation to add a comment): it should be $$\frac{2}{\pi} = \prod_{k = 2}^{\infty} \cos \Big( \frac{\pi}{2^k} \Big),$$ of which the$\piappears in numerator, not denominator. To prove the previous result, you can follow these steps: Writing the double-angle formula $$... 0 This is an example of an equality$$2x^2 {-x}-3=0$$only certain values of x will satisfy it, in this case x=\frac{3}{2} or x=-1 This is an example of an equivalence or identity$$\frac{512}{(16-x^2)^\frac{3}{2}} ≡ \frac{x^2}{2(1-\frac{x^2}{16})^\frac{3}{2}}+\frac{8}{(1-\frac{x^2}{16})^\frac{1}{2}}$$this statement is true for all values of x. ... 0 In this context p(x_1\ldots x_n\mid y) is the same as p(x_1,\ldots,x_n\mid y). I would say that in the former notation the comma (,) is reserved to distinguish the random variables x_i (what is called the attributes in the video lecture) from the random variable y (the class) when you are conditioning over both x_i and y. This becomes more ... 1 Ask your professor, but I think you are right, and if so, the former notation is ambiguous. (Assuming we are right in #1) This is not the chain rule in calculus/analysis but rather the chain rule in probability: For events A_1, ..., A_n, = P(A_n | A_1, ..., A_{n-1}) P(A_{n-1} | A_1, ..., A_{n-2})P(A_2 | A_1)P(A_1) For random variables X_1, ..., ... 2 To search for such symbols, you can try Detexify or Shapecatcher. Both worked for \aleph when I tried. Also, usually you could search for the term used, in this case Cardinality. 2 That is the Hebrew letter "aleph" \aleph. The subscript zero means we read the letter \aleph_0 as "aleph null." It canonically denotes the cardinality of a countably infinite set, namely the natural numbers, \mathbb{N}. 0 Other suggestions: 1) Perhaps this notation means that f is C^1 and has a derivative that has finite norm in L^2. 2) It's also possible that this notation means that f is C^1 and has a derivative that is Lipschitz continuous with Lipschitz constant 2. 1 I've never seen this notation but I'd guess that it means functions that are C^1 as a function of the first variable and C^2 as a function of the second variable, in both cases the other variable being held fixed. 1 In section 2, the paper gives this definition: 2 (comment from 3 weeks ago, with an additional paragraph, given as an answer, as requested by @Jesse P Francis. I'll add more later if I find that I have anything more to say that might be of interest.) I think it's because the widespread use of the \ln notation is relatively recent (50 or 60 years?) and mostly restricted to "school mathematics", which ... 0$$\prod_{j=1}^n(1+a_j)$$The letter \pi is for product, of course. 0 The product series symbol is Pi: \quad\prod\limits_{i=1}^n (1+a_i) This and other mathematical symbols can be found here. 1 No. For a single absolutely continuous random variable, yes. Consider the first expectation:$$E(X)= \langle x_1+ x_2\rangle=\iint(x_1+x_2)\rho(x_1,x_2)\mathrm{d}x_1\mathrm{d}x_2$$This is supposed to be:$$E(X)= \langle x_1+ x_2\rangle=\iint_{\mathbb R^2}(x_1+x_2)\rho(x_1,x_2)\mathrm{d}x_1\mathrm{d}x_2$$or$$E(X)= \langle x_1+ ... 0 I think the rule you are looking for is that $$\left(x^a\right)^b = x^{ab}.$$ Applying it to your case: $$\left(6.3\times10^{2}\right)^{-6}=\left(6.3^{-6}\times10^{-12}\right)\approx 1.599398 \times 10^{-17}.$$ 3 If I understand the question, then what you wrote is correct. You can verify properties of exponentiation on corresponding Wikipedia page, for example. In particular, \begin{align} \big(\,b^m\,\big)^n &= b^{m\cdot n}& \text{ and }& & b^{-n} &= \dfrac{1}{b^n} \end{align} 2 You can use the second form: Letf\colon (X,\sigma)\to(X,\tau)$be continuous. You're right that in the universe of sets, taken literally this notation implies that$f$is one of the four functions from$\{\{X\}, \{X,\sigma\}\}$to$\{\{X\}, \{X,\tau\}\}$— hardly what you mean. The highlighted statement implicitly means that$f$is a morphism in the ... 2 The second and the third are both fine, and I consider them synomyms. It's quite common to "abuse notation" and say that the domain of the function is the whole structure instead of just the underlying set, if you're interested in structure preserving maps, like here. If the topologies are clear from context, they're often omitted, but here they should be ... 1 Think about it: it certainly makes sense. For a function$f$,$2*f$is the function$(2*f)(x) = 2*f(x)$. For example, in linear algebra, the linear transformations between two vector spaces form a linear algebra, and its vectors, which are functions, can be multiplied by scalars. By extensionality, functions are uniquely determined by the values they take on ... 3$f\circ g$refers to a function.$f\circ g(x)$is the result of evaluating the function$f\circ g$at a point$x$. In order to show that two functions are equal, it suffices to show that they are equal when evaluated at every point of their (common) domains. In your case, this means that you want to show that$(f\circ(g\circ h))(x) = ((f\circ g)\circ h)(x)$... 0 The comments tell us that The second notation is generally preferred, The line informs that it is an augmented matrix. 3 The formula $$\phi(n) = \prod_{p \mid n} (p-1)$$ is correct for squarefree$n$, but incorrect otherwise. Generally correct is $$\phi(n) = \prod_{p \mid n} (p-1)\cdot p^{v_p(n) - 1},$$ where$v_p(n)$is the exponent of$p$in the prime factorisation of$n$. But in comparison to that, $$\phi(n) = n \prod_{p \mid n} \bigl(1 - \tfrac{1}{p}\bigr)$$ is more ... 3 It should be the symmetric$(0,2)$-tensor given by $$dz^2 = dz\otimes dz.$$ 4$\bigcup_{i=1}^4 D_i = D_1 \cup D_2 \cup D_3 \cup D_4$. This is analogous to$\sum_{i=1}^4 a_i = a_1 + a_2 + a_3 + a_4$, e.g.$\bigcap_{i=1}^4 D_i = D_1 \cap D_2 \cap D_3 \cap D_4$, and the last one goes to$n$(which is fixed for the sum), the$i$is the running index that takes values$1$to$n$. 0 Probably not, notations differ from author to author. Not even$A\subset S$is unambiguous - it can mean proper subset or just subset. This means that if you're going to use a shorthand notation you would probably need to define the meaning of it. The mentioned$\mathcal P_{\ge 1}$for example is nothing I've encountered, and the same applies to ... 0 I would express the statement as: every function in$W^{m,p}(\Omega)$can be approximated, in the norm, by functions that are$C^\infty$smooth on all of$\mathbb{R}^n$, with compact support. The validity of this statement depends on the geometry of$\Omega$; I don't know what is meant by the segment property but it must be something that enables the ... 1 Often$ \langle f \rangle_\mathcal{F}$denotes the average or expected value of$f$over the family/region$\mathcal{F}$. In my experience, it's typically normalized (but not always) when that makes sense. So if$\mathcal{F}$is a discrete family, then$\langle f \rangle_\mathcal{F}$might typically be defined as $$\langle f \rangle_\mathcal{F} = ... 2 From John L. Kelley's General Topology (1955 edition, July 1957 printing, p. 65): A directed set is a pair (D,\ge) such that \ge directs D. [. . . .] A net \{S_n,n\in D,\ge\} is in a set A iff S_n\in A for all n; it is eventually in A iff there is an element m of D such that, if n\in D and n\ge m, then S_n\in A. The net is ... 0 I'm aware of only one way how to write this clearly -- in symbols:$$ \exists_\infty n\in\mathbb{N} : P(n).$$However, unless you need this a lot you shouldn't use this notation. If you need it a lot, you can introduce it: We suppose that$P(n)$holds for infinitely many$n$, i.e., that$\exists_\infty n\in\mathbb{N}:P(n)$. Some people use: "$P(n)$... 2 In some contexts (set theory, order theory, point set topology, though probably never in probability) you can say cofinally, or cofinally many, cofinally often. Given a preorder$(A,\preceq)$, a subset$X\subseteq A$is cofinal in$A \Leftrightarrow$for every$a\in A$there is$x\in X$with$a\preceq x$. A predicate$\varphi(x)\$ holds cofinally often, and ...

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