# Tag Info

6

What this gives is something remarkably powerful. Specifically, the result tells you that there's a way to isolate the various terms of a polynomial, and in fact a power series. Think of it this way: take $z_0=0$. We then have that \begin{equation*} \frac{1}{2\pi i}\int_{C_0}{z^{n-1}}\,dz=\begin{cases}1&n=0,\\0&\text{otherwise}.\end{cases} ...

4

Consider the function $f(z)=\sum_{-\infty}^\infty c_n z^n$. Your result tells us that if we want to determine the coefficient $c_{-1}$ we just need to integrate $f(z)$ over a closed contour about the origin. $$\int_C f(z) dz = 2\pi i c_{-1}$$ If we want some other coefficient we can multiply by some power of $z$ first and then integrate. For instance ...

4

A homomorphism $\phi:\mathbb Z_3\to G$ is uniquely determined by the image $\phi([1]_3)$ of the standard generator of $\mathbb Z_3$. Also, $g\in G$ can serve as such image $\phi([1]_3)$ if and only if $g^3=1$. Thus $g$ is either $1$ (which leads to the trivial homomorphism) or an element of order $3$. In other words, the number of nontrivial homomorphisms ...

4

I find this a natural question. Particularly because I recently heard that I would be teaching group theory for advanced undergrads next year (or the year after), so I want to test my motivational skills here :-) Indeed, one of the aspects of representation theory is to study the groups being represented. It is nice that we can take some abstract group, and ...

3

I don't think there is a reasonable general answer to this very genuine phenomenon: that's just the way the (mathematical) world is. In a related area, most of modern cryptography (and thus of modern economics: think banking) relies on the fact that it is very easy to multiply two huge prime numbers $p, q$ but extremely difficult, given just their product ...

3

You do not have to go through factor groups. $G'$ is a normal subgroup of $G$, so there are two possibilities: $G'=\{1\}$ or $G=G'$. The first case is being equivalent to $G$ being abelian (note that every commutator is the identitiy element!). Abelian and simple means isomorphic to a cyclic group of prime order. So your statement is false, unless you ...

3

You should go hurt whoever told you those things because they didn't talk about the inner product they'd silently fixed in the background. What is happening here is that fixing an inner product induces an isomorphism between $\bigwedge^k V^*$ and $\bigwedge^{n-k} V$, which permets us to identity a $k$-form with an $(n-k)$-vector. To fix ideas, let $V$ be an ...

3

(Too long for a comment:) I can offer an explanation showing that dividing by $n$ would give an underestimation of the variance. The sum of squares $\sum (X_i - \overline{X})^2$, where $\overline{X}$ is the sample mean, is smaller than the sum $\sum (X_i - \mu)^2$ where $\mu$ is the true mean. This is the case since $\overline{X}$ is expected to be ...

2

This answer is predicated on Did's, but is not a duplicate. To obviate litotes, I define "accomplish = not neglect". Based upon “If you neglect your homework, then you’ll fail the course,” could I not assert that its "intended meaning" is $\color{#C41E3A}{\text{$P∨¬Q$= "You neglect your homework or you won't fail the course"}}$? Answer: No you could ...

2

$1. (a_1, a_2, ···,a_n)= (a_1, a_n)(a_1, a_{n-1}) · · · (a_1, a_3)(a_1, a_2).$ You can visualize the LHS as moving every letter to its destination all at once, while the RHS moves each letter one at a time, keeping $a_1$ as a "working" slot until the end. For example, let's look at $\sigma \triangleq (1,2,3,\cdots, n)$. You can envision picking all ...

2

2– The absolute value is always $\geq 0$. This means that $x$ satisfies $|x-c|<\delta$ –which appears in $\color{red}{(I)}$– if and only if $\quad 0<|x-c|<\delta \,$ or $\, \color{dodgerblue}{|x-c|=0}$. You know that $|x-c|=0$ just means that $x=c$, so that the same $|x-c|<\delta$ is equivalent to $\quad0<|x-c|<\delta \,$ or $\, ... 2 We seem to be missing "...where none of the boxes are empty." Assume the boxes are labeled (we divide by$k!$at the end for unlabeled boxes). We can put the balls in the boxes in$k^n$ways: but this overcounts since it allows the possibility for some of the boxes to be empty. So we need to subtract the number of balls-in-box-arrangements in which there ... 2 Since$\log$is an increasing function, it's equivalent to minimize the logarithm of your expression, which is $$n\log k - \frac{\log v}{\log2}\log n + \big\{ \log\frac{\log 2^v}{\log v} + \frac{\log v}{\log2} \log kvc \big\};$$ and the last expression, being independend of$n$, is irrelevant. The derivative of this function of$n$is simply$\log k - ...

2

A simple point of view on the matter is that wedge products of vectors represent subspaces. Each subspace has a weight and an orientation. Think of vectors for a moment: a vector can be taken to represent a 1d subspace. Such a 1d subspace has two unit vectors--some unit vector $\hat v$ and its additive inverse $-\hat v$. A vector representing this ...

2

So I think this is what we may want to do. Suppose for contradiction that $\Gamma$ is not satisfiable. This means that $\Gamma$ has no models. Now, fix some sentence $P$ and let $S \equiv P \wedge \neg P$. Now, $\Gamma \models S$ will be vacuously true since there are no models of $\Gamma$, (i.e. any model of $\Gamma$ will satisfy $S$). So by the ...

2

$\Gamma \models S$ means 'Every model of $\Gamma$ satisfies wff $S$ ' So given statement can be written as follows: For all wff $S$ on the language of $\Gamma$, if every model of $\Gamma$ satisfies $S$ then $\Gamma$ proves $S$. Formally, this statement is written as $\forall S[ (\forall \mathcal{M} :\mathcal{M}\models \Gamma \implies \mathcal{M}\models ... 2 The symbol :$\Gamma \vDash S$means that the sentence$S$is a logical consequence of the set of sentences$\Gamma$. This means, as said in the above comments : every interpretation that satisfies all sentences$\varphi \in \Gamma$(i.e. is a model of$\Gamma$), satisfies also$S$. The symbol :$\Gamma \vdash S$means that the sentence ... 2 The idea behind this in the end is that a useful scalar product on$\mathbb C^n$is given by$\langle x, y\rangle=\sum_{k=1}^n \overline {x_k}y_k$(and not$\sum_{k=1}^n {x_k}y_k$as in the real case). Given any complex$n\times n$matrix$Awe can consider the bilinear map \begin{align}f_A\colon \mathbb C^n\times \mathbb C^n&\to\mathbb ... 2 Intuitively, it's because while functions have to be well-defined, they're allowed to be non-injective, so if you're placing arrows between the elements of two objects, you have a lot more freedom in choosing where to make your arrows point than where you can choose where to make them point from. To put it more succinctly, you can find more ways to send ... 2 I'm not sure if this is the type of thing you are looking for, and it is only a partial answer. I don't know about \operatorname{curl}, but for \operatorname{div} I've always thought that this came from expectations of what we are looking for. In a physical sense it seems sensible that, given a domain \Omega to study, we may want to know how much ... 2 The complete answer is given on pages 22-27 of my 2011 vector calculus notes. I think many good calculus text include these heuristic arguments, I found them in Thomas' calculus a few editions back. Long story short, what you should really do to understand is to prove Greene's and Stokes' Theorems, this will give you deeper insight into the nature of your ... 2 Mauro makes part of this point, but it's worth stressing: the two different ways of notating a set are used for instances of application of two different axioms (strictly, axiom schemas) in ZF: B = \{x \in A : \phi(x)\} $$is an application of the subset axiom (a.k.a. separation):$$\forall y_1, \ldots, y_n \forall A\, \exists B\, \forall x (x \in B ... 2 No parametrization is needed, just some careful uncovering of definitions. Maybe the following can help: LetS^n(r)=\{x\in\mathbb R^{n+1}|\|x\|=r\}\subset\mathbb R^{n+1}$(the sphere of radius$r>0$centered at the origin). There is a natural volume form$\mu_{r}$on$S^n(r)$, i.e. a nowhere-vanishing$n$-form, induced by the volume form ... 2 The idea is, as always when dealing with local notions on manifolds, to pass to a chart and see what happens. On a chart (i.e. Euclidean space) you have an obvious correspondence between vectors and directional derivatives, in the sense that derivation in direction$v$is given by$Df\cdot v$. If you write this down component-wise and lift it up to the ... 2 You've already said that you're using$r$to denote clockwise rotation by$90^\circ$. Let's agree that the flip$f$means "flip across its horizontal axis", so that $$\fbox{\begin{matrix} 1 & 2\\ 3 & 4 \end{matrix}}\;\; \xrightarrow{f}\;\;\fbox{\begin{matrix} 3 & 4\\ 1 & 2 \end{matrix}}$$ (Note that the symbols are there only to denote ... 2 Open mapping: were it invertible, its inverse would be continuous! :-) Take an open mapping $$f: A \to B.$$ If$b \in B$is in the image of$f$, then, if$b_{\lambda}$approaches$b$,$a_{\lambda}$such that$f(a_{\lambda}) = b_{\lambda}$"approaches"$f^{-1}(b)$. Technically, it works as follows... If$f(A) \subsetneq B$, then lets just trim$B$down ... 2 I realised after writing this that I had misread your question - I thought you wanted$i^*F$and$i^*G$to be isomorphic, but you actually said they should be equal. Still, here are my thoughts. == A formal neighborhood is definitely smaller than a Zariski-open neighborhood. It can be helpful to think of$\hat{X}$as the union or direct limit of all ... 2 In my experience, there is no plan of making a proof bidirectional. Really, you just try to prove$P\implies Q$and see what happens. When you have your proof that$P$implies$Q$, you take a good look at it and try to reverse every step of it. The usual case is that you can reverse some of the steps, but not all of them, but sometimes you get lucky and can ... 1 As said in the above comments, the cited statements are meaningful as "advices" about proof-strategy ... nothing is deterministic in this topic. What is sound is the observation : if you are trying to prove$P \equiv Q$, it is wrong to start your write-up of the proof with the unjustified statement$P \equiv Q$... He says the (obvious but useful) ... 1 An example of what Velleman refers to might be the following: Claim$\lambda $is an eigenvalue of the matrix$A$if and only if it solves the characteristic equation$det(A-\lambda I)=0$. Proof If$\lambda$is an eigenvalue of$A$then there is a$x\neq0$such that$Ax=\lambda x$, implying that$(A-\lambda I)x=0$some$x\neq 0\$, implying that ...

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