# Tag Info

9

"Angle" is a touchy subject. There are (at least) the following three interpretations, applicable depending on circumstances. (a) The simplest is the following: Given two nonzero vectors ${\bf x}$, ${\bf y}\in{\mathbb R}^n$, $\>n\geq2$, the (nonoriented) angle $\alpha$ between them is the nonnegative number $$\alpha:=\arccos{{\bf x}\cdot{\bf ... 7 Although the definition that Trevor quoted from Cantor's Attic fails, as he says, to match the usual meaning of "hereditarily", it is, as far as I know, fairly standard. I suspect the reason (or at least a reason) for that is that it works well in Lévy's cardinal boundedness theorem: Anything \Delta_1-definable from sets in H_\lambda is itself in ... 7 If T is a monad on a category \mathcal{E}, then one can define T-algebras (I prefer the terminology T-modules) and this generalizes all the usual notions of algebraic structures. They are objects A equipped with a morphism T(A) \to A such that the two obvious compatibility conditions are satisfied. For example, if \mathcal{E}=\mathsf{Set} and ... 6 In universal algebra and predicate logic, the signature is essentially defined as the non-logical symbols. The axioms are not part of this. Even "sort U" is not part of this, at least for the prevalent single-sorted logic. As long as your axioms are universal Horn sentences, you have a canonical notion for homomorphisms, and hence a direct connection to ... 4 The definition you propose is not equivalent to the usual definition. Consider the function f:\mathbb R \to \mathbb R given by f(x)=x if x is rational and f(x)=-x when x is irrational. With the usual definition of continuity, f(x) is continuous at x=0. However, under your definition condition 2 fails. You conditions also do not imply ... 4 You are correct. For the reasons you discuss, the appropriate definition of H_\lambda for singular \lambda is the set of x such that x and all sets in its transitive closure have size below \lambda. (Though I could not instantly find a reference, this is actually standard among those that consider the set at all, which are not too many, and its use ... 4 There are two wonderful resources: Foundations for Category Theory, by Daniel Murfet Joy of Cats 4 In this context, \mid is a relation: m \mid n means that m divides n, or equivalently n is a multiple of m. This means that there is an integer \ell such that$$ n = \ell m.$$If m and n are positive integers with m \mid n, then we must have m \le n. The symbol "\mid" is read divides and is produced by \mid in \LaTeX. Therefore ... 3 In the theorem you have x^n, but in the series you have x^{2n} and x^{2n+1}. In the theorem you drop the n=0 term because that's the constant term (whose derivative is zero); in the case of cosine, the n=0 term is again the constant term, so it works out; but in the case of sine it's the x^1 term. To apply the theorem very carefully for cosine: ... 3 It doesn't describe all the elements. All this definition says, is that all the points on the line segment from x to y also lie in C. EDIT (after the question was changed): You have two points x,y in C, and the equation for the linear space with direction y-x is t(y-x), for t\in\mathbb R. But what you want is the affine line parallel to ... 3 1*a=2a+1\not\equiv a and so 1 is not an identity. Instead, what you demonstrated is that 1 is a non-trivial center, which all elements other than 0 will be since this operation is abelian. It is not possible for a set to have two distinct left-right identities, since if e_1 and e_2 are identities, then e_1*e_2=e_1 and e_1*e_2=e_2 and so ... 3 So why these two different terms? Because the history of mathematical terms is long and complicated. At least we stopped talking about monogenic functions and regular functions, which are two more terms for the same concept (as far as complex analysis is concerned). Quoting HOMT site: In modern analysis the term ANALYTIC FUNCTION is used in two ... 3 Consider \alpha(t)=(t,t^2) with t\in\boldsymbol{R}_+ and \beta(t)=\bigl(\exp(t),\exp(2t)\bigr) for t\in\boldsymbol R. The traces are the same, but \alpha and \beta are different paths: the velocity vectors are different, for example. Edit (an even more simple example): the traces of t\mapsto t(1,0) and t\mapsto t(-1,0) for t\in\boldsymbol ... 3 Imagine you're on an infinitely large, flat piece of paper with a transparent beach ball of diameter 1m. Assume the ball is perfectly round. You have a lazer pen. The paper is the complex plane. Place the beach ball down so that it rests on 0, its centre is 0.5m off the ground and the point directly above 0 is 1m off the ground. Take your lazer ... 2 The notation must be used only when \dagger is satisfied.$$\lim_{N_1 \to \infty} \lim_{N_2 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k \text{ and }\lim_{N_2 \to \infty} \lim_{N_1 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k \text{ exists and }\lim_{N_1 \to \infty} \lim_{N_2 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k = \lim_{N_2 \to \infty} \lim_{N_1 \to \infty} ...

2

Since there is a natural Riemannian metric on $\mathbb R^n$ and $\mathbb R^m$, a notion of pull back for vector fields could be defined as follows: let $v$ be a vector field on $V$. Then $f^*v$ is a vector field on $U$ so that for all vector fields $w$ on $U$ we have $\langle f^* v,w\rangle = \langle v, f_* w \rangle$, where $\langle\cdot,\cdot\rangle$ is ...

2

It seems you should actually start with $\{(0,0),(0,1),(1,0),(1,1)\}$ as you have all four corners of the paper. Then you define what constructions are allowed. I imagine that folding on a line through any points you already have is one. Another would be bringing together two pairs of points that are the same distance apart. There may be more operations ...

2

There is a bijection between functions $h: X \rightarrow Y$ and functions $h': X'\rightarrow Y'$. Given $h: X\rightarrow Y$, we can compose with the isomorphisms $f^{-1}$ and $g$ to get $g\circ h\circ f^{-1}: X'\rightarrow Y'$. Conversely, given $h': X'\rightarrow Y'$, we can compose with the isomorphisms $f$ and $g^{-1}$ to get $g^{-1}\circ h'\circ f: X ... 2 The action of$f$does indeed define a partial function from$S$to itself. But that's intentional. Category actions are equivalent to diagrams. The "picture" you should have in mind is that$S$is the disjoint union of all of the objects in the diagram, so that you really do want the action of$f$to be a partial operation, defined only on those objects of ... 2 Since the variable$x$is referenced in neither the expression for$M$,$M = (a_{ij})$, nor in the defining equation for the a$a_{ij}$,$a_{ij} = 4i + 2j - 6$, it strikes me that the ocurrances of$1 \le x \le 3$are misprints for$1 \le i \le 3$and$1 \le j \le 3$; under this assumption, the whole thing makes sense, and we have for example$a_{23} = 4(2) ...

2

You misunderstood why you go from $\sum\limits_{n=0}^\infty$ to $\sum\limits_{n=1}^\infty$. It's not that you're getting rid of the lowest power term, as you have done. But rather that you're getting rid of the $x^0$ term, which is constant. So when you differentiate $\sin(x)$, you don't get rid of the $n=0$ term, because that's the $x^1$ term.

1

This is wrong: $$\sin'(x)= \left(\sum_{n=0}^{+ \infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \right)'=\sum_{n=1}^{+ \infty} \frac{(-1)^n(2n+1)x^{2n}}{(2n+1)!}$$ The point of discarding the derivative of the first term $a_0x^0$ is simply that the derivative of a constant term is zero, so it can be left out. Otherwise the first term would be $0 \times a_0x^{-1}$, ...

1

You are confusing homeomorphism, a bijection that is in both directions continuous, and homotopy, which is a deformation from one form to the other. In the case of knots, one needs ambient homotopies, and only the trivial knots are, by definition, ambient homotopic to the circle. Added: You can of course deform every simple knot into a circle. However, ...

1

From your recursive definition, it follows that $X$ is populated by elements of the form $-2^n3^m$ (for any non-negative integers $n,m$). Thus, to determine if a given number $y$ is in $X$, we need to see if there are integer solutions for $n$ and $m$ such that $$y = (-1)2^n3^m$$ Since $2$ and $3$ are both prime, one way of finding $n$ and $m$, if they ...

1

Let $\langle a_n:n\in\Bbb N\rangle$ be a sequence, and let $S$ be a set. The assertion that $S$ contains $a_n$ for all but finitely many $n\in\Bbb N$ means that the set $\{n\in\Bbb N:a_n\notin S\}$ is finite: there are only finitely many indices $n$ such that $a_n$ is not in $S$. Note that zero is finite: it’s quite possible that $a_n\in S$ for every ...

1

There are many different ways in which the number 'one' can be defined. A discussion of the text you reference can be found in Poincare's "Science in Method" (pp. 458-459) where the author, too, is unsure of Burali-Forti's notation. (Note I am using a translation by Halsted from 1982.) I understand Peanian [the language of Peano] too ill to dare risk a ...

1

Unfortunately we can't multiply a given form or a function globally by $dx_j$. For one, this is because such a form is nowhere zero: in local coordinates it would be equal to $dx_j$, which is most certainly nonzero. However, there are manifolds on which any differential $1$-form must be zero at some point, like the $2$-dimensional sphere (see the hairy ball ...

1

Given a smooth function $f$ on $\mathbf R^n$, the expression $f(x_1, \dots, x_n) dx_i$ is a $1$-form on $\mathbf R^n$. However, if $f$ is a function on an arbitrary manifold, the expression $f(x_1, \dots, x_n) dx_i$ makes no sense because it depends on a choice of coordinates. A manifold $M$ of dimension $n$ has $k$-forms of every degree $0 \leq k\leq n$. ...

1

The analogue of the quotient field for a ring with zero divisors is the total ring of fractions: basically, just invert everything that is not a zero divisor. Geometrically, an element of this ring can be viewed as a collection of rational functions, one on each irreducible component of $X$, such that they coincide on intersections. Rational functions are ...

1

Good question. Your worry about convergent vs. abosolutely convergent shows your attention to detail, which will serve you well in math. First of all, on the left hand side, you should have $|f_c- c|$, not simply $f_c$. There is a special fact about power series which guarantees that they will be absolutely convergent on the interior of their domain of ...

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