# Tag Info

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The order of a method refers to the speed with with it approaches the objective. If you have $x_{n + 1}$ computed from $x_n$, converging to $x^*$, the method is called order $\alpha$ if, approximately when near the objective: $$\lvert x_{n + 1} - x^* \rvert \approx c \lvert x_n - x^* \rvert^\alpha$$ Newton's root finding method (either on the function or ...

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For a ring to be a subring it does not need to have the unit element. It only has to have the zero element, closure under addition and multiplication and an additive inverse, meaning a+(some element)=(zero element). If k={0} it has the zero element of the integers, it is closed under Addition and multiplication and 0+(-0)=0.

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Isn't it just the rule to judge if a number is divided by 9?

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The word quartile refers to both the four partitions (or quarters) of the data set, and to the three points that mark these divisions. After all, we can't have one without the other. When citing a value for a quartile, though, we are specifically referring to the three dividing points, else it'd be meaningless. Thus, the first, second, and third quartiles ...

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If you look in a non-mathematical dictionary, you will often find both definitions. For example, http://www.oxforddictionaries.com/us/definition/american_english/quartile defines quartile as 1 Each of four equal groups into which a population can be divided according to the distribution of values of a particular variable. 1.1 Each of the three ...

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Note: This is really just a long comment - but maybe it's helpful. This seems to me to be purely a matter of context. I have never seen anything like this before, with only 3 quartiles - it's written into the word itself that there should be 4 (QUART-iles). That said, this kind of thing happens in mathematics relatively often - there will be multiple uses ...

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In the context of subspaces of a vector space (or more generally, subgroups of a group), it is common to abuse terminology and call such subspaces disjoint. This abuse is generally harmless since it is obviously impossible for them to actually be literally disjoint. You could also call them (linearly) independent, though personally I think "disjoint" ...

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I don't think there is a common term in set theory for that. But in the context of linear algebra, it is frequent to say that two subspaces are independent when their intersection is the zero space.

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Sets that have empty intersection are said to be disjoint. Subgroups of the same group, subspaces of the same vector space, and so on will never be disjoint, as they'll have at least one common element (the identity element of the group, the 0 vector, etc.) Sets with just one element are singletons, but it would be unusual to use that term about algebraic ...

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It's traditional mathematical English to say "$x$ vanishes" to mean $x = 0$.

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Your interpretation is correct. "The Wronskian vanishes" means the Wronskian is equal to $0$.

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After much thought, I've settled on the following terminology. Definition. An object of a category is mono-terminal $\;$ iff there is at most one arrow into it from each object. epi-initial $\;\;\;\;\;\;\;$iff there is at most one arrow out of it to each object. This terminology is good because irrespective of whether or not initial or ...

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(The question is basically answered in the comment) As a topological manifold is locally path connected, a connected manifold is automatically path connected, as a connected locally path connected topological space is path connected

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"Meta" means something like "mentally take a step back and look at what your doing". Compare https://en.wikipedia.org/wiki/Metaphilosophy. In my experience this is actually what you do when calculating a "hyperoperator". That's why I'd actually argue, "metaoperator" would be a more appropriate name for it.

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"Unit" from "unity", from Latin unus meaning "one" (see uno, un, etc. in Romance languages). Thus, "1-like thing". In fact, even 1 itself is sometimes referred to as "unity", as in the term roots of unity.

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I don't think $σ^2_{x,y}$ is good notation, since the covariances can be negative. Thinking of the "co-standard deviations" as being imaginary isn't very helpful either, because the covariance matrix is symmetric, so its eigenvalues are real. So in some sense everything that can be said about covariance "lives in the real numbers".

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It is valid only if the operator norm is induced by the Euclidean norm. If you use $||\cdot||_1$ or $||\cdot||_{\infty}$, you'll get different expressions of the operator norm. In order to show that this equality holds when the norm on $\mathbb{R}^n$ is the Euclidean norm, you can diagonalize $A^TA$ in an orthonormal basis : there exists an orthogonal ...

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It would be reasonable to call such things "coreflexive profunctors." Here's why. The notion of a profunctor comes from category theory. When we specialize to the $\mathrm{Bool}$-enriched case, the following definition is obtained. Definition 0. Let $P$ and $Q$ denote preordered sets. Then a relation $R : P \nrightarrow Q$ is said to be a ...

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The equation is a single thing. The plural of maximum is maxima. There are multiple places of maxima. So the proper sentence is: "$t=\frac{(2n+1)\pi}{2}$ represents the places of maxima". "represets" is singular, "places" and "maxima" plural. Better explanation of period of 5/2? The period isn't 5/2. There were 5/2 periods. I would say "the function ...

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To get the temperatur, you have to get the mean over all atoms. The Kinetic theory tells us $$k_B T = \frac{m \bar{v^2}}{2},$$ where $k_b$ is the Boltzmann constant, $m$ the mass of the particles (they have to be the same for this), $T$ is the temperature of the gas as one unit and $\bar{v^2}$ is the expected value of the square of the velocity.

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It is correct that temperature is proportional to the average energy of the particles. If all the particles have the same mass and the only energy of interest is kinetic energy, which you did not specify, the temperature is proportional to the average square of the velocity as you say.

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Note that "Leading zeros are never significant". When approximating/rounding-off to n significant figures, your n-th digit should be rounded-off. In this case, your answer would be 0.383. Please refer to the Wiki link above and this site for explanations and examples on working with significant digits.

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Let $f:A\to B$ be a map. Prop1 Suppose the fibers of $f$ are nonempty. Then $f$ is surjective. Proof: Let $b\in B$. We need to show that there exists $a\in A$ such that $f(a)=b$. Well, the fibers are nonempty, so there exists $a\in f^{-1}(b)=\{x\in A\mid f(x)=b\}$. Therefore, $f(a)=b$ as required. Prop2 Suppose every fiber of $f$ contains at most one ...

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The word "calculus" has different meanings. On the one hand it denotes the parts of analysis first met by the student, as well as the adopted way of handling these parts. On the other hand it denotes the set of computational rules valid in any particular branch of mathematics. In this sense we have the "calculus" of propositional logic, or of quantum ...

1

It depends on the expression at hand: I would say differential, integral, or integro-differential according to the (respective) appearance of derivatives, integrals, and both. Edit To address the question as modified: Probably the best modifier is just calculus itself; e.g., one writes calculus identity just as readily as algebraic identity. See this ...

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Such functors are precisely the pseudomonomorphisms in the 2-category of categories. (A pseudomonomorphism is a morphism $f : X \to Y$ in a bicategory such that $$\require{AMScd} \begin{CD} X @= X \\ @| @VV{f}V \\ X @>>{f}> Y \end{CD}$$ is a bicategorical pullback square.)

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Prime numbers are the building blocks of numbers; every integer has one unique representation as the product of prime numbers. Ex: 54 = 2 x 3 x 3 x 3. Here 2 and 3 are the prime number building blocks. 2 x 3 x 3 x 3 is the only way to represent 54 as the product of prime numbers. If one was a prime number, then there would be an infinite number of ...

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There is a metric space structure on a normed linear space $(X, \|\cdot \|)$: define the distance between $x, y\in X$ by $$d(x, y) = \|x-y\|.$$ On the other hand, whenever you have a metric on $X$, you can use that to define a topology on $X$: A set $U \subset X$ is called open if for all $x\in U$, there is $\epsilon >0$ so that $$d(x, y) < ... 3 Yes, \rightharpoonup is usually used to indicate weak convergence. 1 One interpretation that may clear things up is that K[S] is not only a ring, but an algebra, specifically, a K-algebra, specifically, the K-algebra generated by S. That clarifies the asymmetry: \mathbb{Q}[\sqrt 3] is an algebra over the rational numbers, not over the square root of 3! But as a ring, it's perfectly fine to say \mathbb{Q}[\sqrt ... 2 There is nothing wrong with using valley in this context. It is a visual metaphor which will be readily understood. An alternative word might be trough, which is a bit more common in this abstract sense; but is still a visual metaphor. 0 From Wikipedia A function can be defined by any mathematical condition relating each argument (input value) to the corresponding output value. If the domain is finite, a function f may be defined by simply tabulating all the arguments x and their corresponding function values f(x). More commonly, a function is defined by a formula, or (more generally) an ... 0 It is not true, in general, that there is a 1-to-1 correspondence between antichains and lower sets. It is true if the poset satisfies the ascending chain condition. Here's a counterexample. Take as the poset the unit interval with the usual order: ([0,1], \leq). Consider the set S = [0, 1). Then S is a lower set, because for any x\in S, any y\leq ... 1 The discriminant is closely related to the (possible) number of real roots of the polynomial. For degree 2 and 3, the situation is particularlay easy : If the discriminant is positive, we have two real roots for degree 2 and three real roots for degree 3. If it is negative, we have no real roots for degree 2 and 1 real root for degree 3. If ... 3 Another example of such a space: let \mathcal{F} be any ultrafilter on a set X. Define a topology \mathcal{T} = \mathcal{F} \cup \{\emptyset\}. One easily checks this is a topology, and it clearly has the required property (as for any subset A \subseteq X we have A \in \mathcal{F} or X \setminus A \in \mathcal{F}. If \mathcal{F} is free, the ... 9 IMO that is a needlessly confusing application of the word 'pointwise' in that article intro. I have always seen "pointwise multiplication" refer to the operation of multiplying two functions via the rule fg(x)=f(x)g(x). That is, you multiply the values as if they were parallel lists, and you multiply corresponding terms. Similarly, you could say that ... 5 It means that the composition of f and g is defined as$$(f \circ g)(x) = f\big(g(x)\big)$$where f is applied to the result of g separately for each x, rather than applying f globally to g itself like this:$$x \mapsto \big(f(g)\big)(x).$$To give you an example, let f(h) = x \mapsto \langle h(x), h(x) \rangle and define g as$$ g(x) ...

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As a start: any space having exactly one limit point has your property. In this case every set is either open or closed. For call the limit point $x$. Then every set $A$ not containing $x$ is open (indeed, $A$ is discrete) and any set containing $x$ is closed (its complement does not contain $x$ and thus is open). An example of such a space is $X_0 = ... 0 Injectivity is derived more in Latin. In French, injectivité is widely used. I wonder how to say "there is an injection from A to B" as easily as in French : "A s'injecte à B". Invectiveness sounds more Anglo-Saxon, but we rarely see its usage. So we can suppose the employment of such idea is not really English .... 1 It is not the usual terminology. "Canonical" has a well established use in linear algebra as a construction that is independent of a basis, or at least invariant with respect to different choices of basis. The thing in the lecture slides (around 25min in the video) is ordinarily called the "standard" inner product associated to a basis, or the "Euclidean" ... 0 Analytic in the generic math sense essentially means to solve using Algebra (properties, rules, or theorems, or use trig/functions properties), or in other words without the use of a calculator, graph, or by plugging in values (which is similar to a table of values). 1 I suggest you do the following: First, show that if$I$is a left ideal in a ring$R$, then$I$(the exact same set) is a right ideal in the opposite ring$R^{\mathrm{op}}$. Next, show that if$I$is a maximal left ideal in a ring$R$, then$I$is a maximal right ideal in the opposite ring$R^{\mathrm{op}}$. It follows from this that the set of maximal ... 1 A$\chi = 1 $"torus" may be. With a single hole what you are left with is homeomorphic to a torus. Depending on whether there is drilled one,two.. holes we have a doubly, triply connected surfaces/solids as 2-torus/ 3 -torus etc. They are topologically characterized by Euler characteristic$ V+F-E-2 =\chi$that is respectively as solid, Clifford torus,.. ... 0 One thing I thought about was the discrete case. Actually that works here too. One just has to be careful of dimensions. Notice that $$I = \int_{\mathbb R} (f(x))^2 \, d x$$ has dimensions of$[x]^{-1}$. Therefore, in order to interpret this as a (dimensionless) probability, it's necessary to multiply it by something else with dimensions of$[x]$. ... 0 When we say "proportional to" we often mean a linear variation.We are taught that way in the first place.$y$is proportional to$x^2 $appears as a contradiction in its own terms, I for one agree with you. A better way to say it is :$ y $varies as the$ square $of$x$. In this general sense even if we say$y$varies as the sine of square root of ... 0 Substitute$z = x^2$. Clearly in$y = z$, we can say "$y$is proportional to$z$". (It's the same as$y = kz$with$k = 1$). Then by back-substitution, we can say "$y$is proportional to$x^2$". 0 Let$f: \mathbb{R} \to \mathbb{R}$. Then$f$is called homothetic iff$f(ax) = af(x)$for all$a,x \in \mathbb{R}$. In this sense, we say that$f(x)$is proportional to$x$. If$f: x \mapsto x^{2}$and$a \in \mathbb{R}$, then$f(ax) = a^{2}x^{2},$which need not be$= af(x) = ax^{2}$, so$f(x)$is not proportional to$x$. 1 The definition of proportionality is that$A \propto B \iff A = kB$. So A is linearly related to B, and also$A=0$iff$B=0$. In your case,$y = (1) x^2$so y is actually proportional to$x^2$. I think you are confused by "y is linear to x squared". Actually linear means that the power is 1. But when I say that y is linear to$x^2\$ it means that if you ...

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I studied Calculus using the "early transcendentals" version of Anton [look this link of the book http://www.amazon.com/Calculus-Transcendentals-Combined-Howard-Anton/dp/0471472441] and I remember that the book explain natural logarithms without integrals, and after, explain again natural logarithms using the integral notation. I liked the "early ...

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