# Tag Info

6

This is not an open problem anymore. In 1980 Moti Gitik published a solution. Starting with a proper class of strongly compact cardinals, one can construct a model of $\sf ZF$ in which every limit ordinal has countable cofinality. Therefore there are no uncountable regular $\aleph$'s. M. Gitik, All uncountable cardinals can be singular, Israel J. Math. ...

5

In terms of intuitions, I think that the best way to think about an injection is that it is a function that preserves difference; if two elements are different before having $f$ applied to them, then they are still different after having $f$ applied to them. Symbolically: $$x_1\neq x_2 \Rightarrow f(x_1) \neq f(x_2).$$ Another way to think about this is ...

4

Think of indivisible bricks, and houses: $\,\,\,\,$ You can cut the first house into two equal parts. But you're not strong enough to cut the second house into two equal part, since you'd be left with one brick. For divisibility, build houses with bases containing $m$ bricks. Then the total number of full floors is the quotient of the total number of ...

3

Even numbers are just numbers that can be cut in half. Or, another way, if you have an even number of objects, you can sort them in pairs of two, or you can split them into two equally sized groups. The odd numbers are those where you can't do that. This is generalizable to other numbers, so $n$ is divisible by $k$ if you can sort $n$ objects into groups of ...

3

If I look in the complex plane at the eight numbers that when raised to the eighth power form $-1$, they are $\exp(\frac {2k+1}8\pi i)$ for $0 \le k \le 7, k$ integer. The seventh powers of these are the same set. I can certainly pick two that have a real part of their sum being $0$, for example $\exp(\frac {1}8\pi i)$ and $\exp(\frac {-1}8\pi ... 3 The rate of change at a single point (i.e. the derivative at this point) is the slope of the best linear approximation the function has at this point. This means, if you have a function$f:D\rightarrow \mathbb R$with the derivative$c$at the point$x_0$, than the best linear approximation of the function at the point$x_0$is given by$y\mapsto f(x_0) + ...

3

This is very good intuitive question to ask, since when we say "rate of change at a point", what we should really be saying is "rate of change around a point". As I'm sure you know, in one variable the derivative of a function $f(x)$ can be defined as follows: $f'(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$ But if we step back and consider ...

3

Let's say the function is $f(t)$ which gives distance, so that the derivative is $f'(t)$ or instantaneous speed. You can think of it as follows: what speed is the car (for example) travelling at that precise moment in time. Or: if the car was to continue at the same speed, what would it be going. $f'(t)$ gives the speed at time t exactly. That is the ...

2

The Yoneda Lemma in the stated form tells you about natural transformations between the functors $\text{Hom}(-,A)$ and $F$ from $\mathbf{Fin}$ to $\mathbf{Set}$. The crucial point is that $\text{Hom}(-,A)$ is contravariant, and this requires the functor $F$ to be contravariant, too. But the $F$ you chose, namely $\text{id}_\mathbf{Fin}$ is covariant. Let's ...

2

This is not a real answer but more an alternative proof. If it is useless in this context then please let me know. In that case I will delete the answer. If $g\circ f=id$ then $g$ is surjective since $g\left(f\left(a\right)\right)=a$ for each $a\in A$. So if $g$ is injective as well then it is bijective (i.e. isomorphic in the category of sets) and, as its ...

2

How to determine truth values? By finding a proof, or a counterexample. How does one do that? How does one approach a problem in mathematics to begin with? One finds some examples, plays with them, tests whether or not they can be modified to be counterexamples (in the case they are not counterexamples to begin with), and if one suspects that the answer is ...

2

I'm pretty sure that in just the plain old complex numbers, $Re((-1)^{1/8} + (-1)^{7/8})$ is $\cos(\frac 18 \pi) + \cos(\frac 78 \pi) = 0$. In fact, in general, if $a+b=1$, $(−1)^a+(−1)^b$ will have a real component of $0$. However, you seem to be focusing on modular arithmetic and number systems, so I'll try to give you an answer in modular arithmetic ...

2

The exponential function on the real line $e^{x}$ can be defined by the power series $$e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$ It extends as an analytic function to the complex plane in precisely one way $$e^{z} = \sum_{n=0}^{\infty} \frac{z^n}{n!}$$ This is the definition of the complex exponential, and all of the properties you are observing ...

2

First of all, you must understand that mathematicians like to abuse notation. So, in truth, $e^z$ with $z \in \mathbb{C}$ is just an analogue (or the generalization) of it real counterpart. In complex analysis one is often interested in the following complex power series $$f(z) = \sum_{n = 0}^\infty \frac{z^n}{n!}, \quad z \in \mathbb{C}.$$ Because $f(z)$ ...

2

This is not exactly your scenario, but it might help to gain intuition. Inside the interval $[0,1]$, consider the set of rationals of the form $m/2^n$ for $m,n \in \mathbb{N}$. These are the numbers that, expressed in binary, have finite fractional part. This set is dense in $[0,1]$. However, if we place over each one a ball of radius $r=2^{-(n+3)}$ we still ...

2

In this question you have to integrate over the whole disk. The best way there is to use the polar coordinates. The normal to the disk is on the direction $-j$ so we have to reverse the orientation as follows so we take $x=r\sin \theta, z=r\cos \theta, x^2+z^2=r^2$ and integrate from $r=0$ to $r=1$. So the integrand is ...

2

The homomorphism you mention is indeed important because, if $N\triangleleft G$ then the projection $$p:G\to G/N$$ of $G$ onto its quotient $G/N$ is "universal among morphisms killing $N$": Suppose $N\triangleleft G$ and let $\phi:G\to L$ be a group homomorphism with $\phi (N)=1$. Then there is a unique morphism $f:G/N\to L$ such that $$\phi=f\circ p$$ ...

2

The zero morphism $A \to B$ can be factored into $$A \to 0 \to B$$ where $0$ is a zero object. (i.e. it is a terminal object and an initial object) As an aside, when you wrote it as "$0$-morphism", my first reaction was that you were referring to the concept from higher category theory; e.g. in $\mathbf{Cat}$, categories are $0$-morphisms (i.e. ...

2

Note this is the map $(x,y)\mapsto (x-y)\cdot (x-y)$, that is $(x,y)\mapsto x\cdot x-2x\cdot y+y\cdot y$. It suffices we show each of the summands is differentiable. But the summands are bilinear, so it is not too hard to obtain their derivative. Claim Let $B:\Bbb R^n\times\Bbb R^m\to\Bbb R^p$ bilinear. Then $DB(x,y)(a,b)=B(x,b)+B(a,y)$. More generally, if ...

1

When solving this kind of problems, your eyes are often more important than your memory. I mean: look at the function and try to collect as much information as you can. For instance, your function is sum of squares, and this may be really helpful. Then you can try to apply the criterion of the directional derivative: if $$L=\lim_{\varepsilon \to 0} ... 1 Another way to put it. An abelian category \mathscr M is in particular a category enriched over the category \mathsf{Ab} of abelian groups. So for two objects A,B of \mathscr M, the hom-set \hom_{\mathscr M}(A,B) actually carries a structure of abelian group : in particular, it has a neutral, which is the zero map from A to B. 1 About 2. In the plane the 1-volume of the maximal hyperplane section, is nothing else than the diameter of the set. In fact since the set is convex, every section is a segment, and every diameter is a section. So the problem becomes the isodiametric problem, whose solution is known to be the disk. 1 My gut feeling is that the shape that would give the smallest hyperplane section would be the d-ball. If that is the case, the d-volume of a d-ball of radius r is \nu_dr^d where \nu_d=\pi^{d/2}/\Gamma(d/2+1). If r_d=\nu_d^{-1/d} is the radius that gives volume 1 and we plug this into the d-1-volume of the equatorial hyperplane section ... 1 Here's an attempt to provide intuition as to why we can work with \sin(\sqrt x) in a way that we can't with \sin(x): The equidistribution theorem tells us that 2 \pi n - \lfloor 2 \pi n \rfloor will stay "evenly" spread out between 0 and 1. On average, it will "miss the mark" of 0 by 1/2. Not particularly useful in producing an upper bound ... 1 The formula is a compact representation of the nine equations for the cross products of the basis vectors. I'll call the basis vectors e_1,e_2,e_3 in this answer, i.e. I'm not going to draw arrows over them. You might want to start by recognizing that "k" is a "dummy variable" and then summing on it. That gives you: e_i \times e_j = ... 1 \binom{n}{h} means possible numbers of choosing h objects out of n. So, possible numbers of choosing h out of n then choosing k objects out of n-h objects which are not chosen before would be \binom{n}{h}\binom{n-h}{k}. Similarly, choosing k objects first and h objects later out of n would be \binom{n}{k}\binom{n-k}{h} cases. Since ... 1 The first thing to note is that in an optimal solution, all coefficients \alpha_i are either 0 or 1. To see this, suppose you have an optimal solution with some \alpha_i satisfying 0 < \alpha_i < 1. Then this solution is better than setting \alpha_i = 0, and it can be shown that this implies that the objective function is strictly ... 1 just to give you a different and possibly more general point of view: let v_i= (sin\theta_i, \cos\theta_i) for i=1,\ldots,n, and V=[v_1,v_2...], then your problem is \max || v_0 ||^2 s.t.  v_0 = V\alpha ,  \alpha_i \in [0,1] or more compactly \max \frac{1}{2} \alpha^T V^T V \alpha  s.t.  \alpha_i \in [0,1] From the optimality conditions ... 1 You already know all coefficients are either 0 or 1 based on some answers and your own observations. So to get an O(n^2) time solution, it suffices to prove that the vectors in the optimal solution are contiguous when the original set of vectors are ordered according to angle circularly. For this first we prove a lemma. Lemma: The set of vectors in an ... 1 Suppose the root is$$a\pm bi$$Then the polynomial will have a factor of the form$$(x-a-bi)(x-a+bi)=(x-a)^2-(bi)^2=(x-a)^2+b^2$$So we will have for some polynomial g,$$f(x)=(x-a)^2g(x)+b^2g(x)$$Thus$$f'(x)=2(x-a)g(x)+\big((x-a)^2+b^2\big)\,g'(x)$$and$$f'(a)=b^2\,g'(a)$$If f is quadratic, then g will be constant, so we will get$$f'(a)=0 In other ...

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