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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

As Cameron Williams pointed out, it suffices to consider the non-squared version. But note that $$ \prod_{k=0}^{n} (2^{k} + 1) = \prod_{k=0}^{n} 2^{k}(1 + 2^{-k}) = \left( \prod_{k=0}^{n} 2^{k} \right)\left( \prod_{k=0}^{n} (1 + 2^{-k}) \right). $$ Taking logarithm, we find that $$ \log \prod_{k=0}^{n} 2^{k} = \sum_{k=0}^{n} k \log 2 = ...


4

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 ...


3

It's hard to tell you what it means "intuitively" because studying mathematics one has to break away from previous intuition, and work hard on constructing a new intuition. But if you just write it out into words, what does a choice function on $\mathcal P(X)\setminus\{\varnothing\}$ accept? It accepts non-empty subsets of $X$; what does it return? It ...


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

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

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 ...


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

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

Notice that you are not using the size of the derivative in discussing limits at $\infty.$ So, you don't emphasize the derivative in limits at $0.$ You just talk about which function gets close to $0$ earlier, which at $0^+$ means for larger $x.$ Also, at $\infty^+,$ the derivative information (in absolute value) agrees with the size information; note that ...


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

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 ...


1

May be an example clarify the situation: Consider $f(x)=e^{-\frac{1}{x^2}}$ $f'(x)=\frac{1}{2x^3}e^{-\frac{1}{x^2}}$ which converges to $0$ as $x\to 0$. Similarly you can check that $f$ is infintely many times differentiable and that all the derivatives of $f$ at $0$ vanishes. Therefore, the Taylor series fo $f$ is... the zero function! The explanation ...


1

My professor used to say: You might want to do calculus in $\Bbb{R}$, but the functions themselves naturally live in $\Bbb{C}$. Euler was the first to discover that if you don't look at what they do everywhere in the complex plane, you don't really understand their habits. This is as subjective as it gets, but it has always helped my intuition. In ...


1

The Babylonians solved problems, in many cases stated geometrically, essentially using quadratic equations. For example, the area of a square diminished by the side of the square is (a given number); find the side of the square. I don't think it's at all clear what were the practical problems (if any) that motivated the discovery of these methods: what we ...


1

Here is all that is known about this sequence on OEIS. (Or rather, the sequence before squaring it.) I see no formulas that are not recursive or do not involve $\sum$ or $\prod$.


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 ...


1

Ok, let's get through this step by step: The set $(\mathrm{position}\setminus\mathrm{error})\times\mathrm{tile}$ is the set of all tuples $(p,t)$ for which $p$ is a position, and $t$ is a tile. Such a tuple just says that the tile $t$ is currently at position $p$. A set $C$ is just a subset of $(\mathrm{position}\setminus\mathrm{error})\times\mathrm{tile}$ ...


1

I would recall/derive it simply by rewritting: Suppose $c | a_1, \dots, a_k$. Then $a_1 = ca_1', \dots, a_k = ca_k'$ where $a_i' = \frac{a_i}{c}$. Then, $$a_1u_1 + \dots + a_ku_k \\ = ca_1'u_1 + \dots + ca_k'u_k\\ = c(a_1'u_1 + \dots + a_k'u_k)$$ which is clearly divisible by $c$!


1

I tend to visualize functions from $\mathbb{C}$ to $\mathbb{C}$ as a vector field on $\mathbb{R}^2$. In other words, given an $f$, there is an arrow for every $x \in \mathbb{C}$ which starts at $x$ and ends at $x + f(x)$. For $z \to e^z$, for example, if you look at a line with $\textrm{Re}(z) = u$, the arrows all have the same length $e^u$, but keep ...


1

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 ...


1

This is one of those cases where the proof strategy is just to write down your conclusion and your premise, and at every step just write down the first thing you can do with the tools you are given. Always start by writing down the premise: Suppose that $g \circ f = id$ and $f$ is onto. We would like to show that $g$ is one-to-one, which means ...


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

You might be interested in Michael Bensimhoun's article: Historical account and ultra-simple proofs of Descartes's rule of signs I've just had a quick look at it and it seems very accessible.


1

Basically, at different values of $x$ different terms in the polynomial "dominate." So every the sign switches, there will be a change in the direction of the curve. Either (1) this will result in crossing the x-axis and a root or (2) there will have to be another change, meaning "losing roots" will always happen in pairs. So the roots are equal to or less ...


1

Here is another perspective, which perhaps shifts where the intrigue should be directed. An important question to consider: How do you define $\pi$? If you take the definition of $\pi$ to be the usual one we first encounter involving circles, then it is fairly remarkable that this geometric constant should have anything to do with the complex function ...



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