# Tag Info

## New answers tagged infinity

0

Let $$a_n = (n!)^{\frac2n}$$ $$\log(a_n)={\frac2n} \log(n!)$$and, as suggested by GPerez, use Stirling approximation, that is to say $$n!\approx n^n \sqrt{2\pi n}e^{-n}$$ or, even better, Stirling series $$\log(n!) \sim n\log(n)-n+\frac 12 \log(2\pi n)+\frac{1}{12n}+\cdots$$ So $$\log(a_n)\sim 2\frac{ n\log(n)-n+\frac 12 \log(2\pi n)+\cdots}{n}> ... 1 Let c_n=(n!)^2; then \displaystyle\frac{c_{n+1}}{c_n}=\frac{((n+1)!)^2}{(n!)^2}=(n+1)^2\to\infty, \;\;\;so (n!)^{\frac{2}{n}}=(c_n)^{\frac{1}{n}}\to\infty 2 Note that x(n+1-x)\ge (1)(n) for 1\le x\le n. It follows that$$(n!)^2=(1)(n)(2)(n-1)(3)(n-2)\cdots (n)(1)\ge n^n$$and therefore (n!)^{2/n}\ge n. 4 Pick K as large as you want, then n!\ge K^{n-K} (for n\ge K), so$$(n!)^{1/n}\ge K^{(n-K)/n}=K^{1-K/n}\to K\text{ as }n\to\infty.$$4 Hint:$$n!>\left(\frac{n}{2}\right )^n$$for large enough n. 0 The Cantor-Schroder-Bernstein theorem tells you that if there exists an injection from set A to set B and an injection from set B to set A, there exists a bijection between A and B and they have the same cardinality. An injection f:\mathbb{R}^+ \to \mathbb{R} is easy - just use the inclusion map (f(x)=x). An injection g:\mathbb{R} \to \mathbb{R}+ ... 0 Two sets A and B are said to be equicardinal if there exist at least one bijective function from A to B. What you have shown is that a certain f between \mathbb{R} and \mathbb{R}^+ is not bijective. But that doesn't prove that each function between \mathbb{R} and \mathbb{R}^+ isn't bijective. 4 In fact, infinite numbers are a part of the Hyperreal Numbers. Just check this out on Wikipedia. 4 I think the fundamental error here is a mistaken notion of what the Pigeonhole Principle is. Nowhere in the Wikipedia page cited in the question does it say there is any rule over which hole each pigeon may go into. As is evident from some comments on other answers the argument in the question is based on the misconception that the Pigeonhole Principle ... 2 This answer is specifically about the Pigeonhole Principle, unlike some other answers which have been about infinity. A correct use of the Pigeonhole Principle requires the following: I define what my pigeonholes and pigeons are. The cardinality of the pigeons must be strictly larger than the cardinality of the pigeonholes. My nemesis assigns pigeons to ... 1 The fault in the logic in your argument is that the Pigeonhole Principle only applies to finite numbers. From Wikipedia, the statement of the Pigeonhole Principle is: "if n items are put into m containers, with n>m, then at least one container must contain more than one item" Your argument attempts to use the Pigeonhole Principle on an ... 0 A typical beginning use of the distinction between infinities is the following theorem: Let f(x) be an increasing function on the real line. There can be at most countably many discontinuities of f(x). Indeed, we can find a function which is increasing and with discontinuities at every rational number, but we can't find an increasing function that ... 5 As mentioned by others as well, the statement is that A\sim B \Leftrightarrow \exists \phi:A\leftrightarrow B (i.e., A and B are of the same cardinality if there exists a bijection between A and B) That is not to say that all maps between them must be bijective, just that there must be at least one such map. Consider A=\mathbb{N} and ... 15 The definition of equicardinal is that there exists a bijection between the sets. You are trying to define "not equicardinal" as "there exists a bijection between one set and a strict subset of another". This definition is not a good one, as all Dedekind infinite sets (such as \mathbb{Z}, \mathbb{R}) have the property that they are bijective with strict ... 3 I think that the issue here is mostly linguistic. Saying that something is infinite simply says that it is not finite. Saying that I have more than two students in my class tells you nothing about whether there are three, or six, or 42 students in my class. It is true that both \Bbb N and \Bbb R are infinite. But it tells you nothing about comparing ... 3 The assertion that Cantor was the first to deal with infinities of different sizes is a bit misleading. This may be true in the context of set theory that he introduced, but there are different ways of formalizing infinity. Two centuries before Cantor, Leibniz dealt with a hierarchy of infinite numbers (inverses of his hierarchy of infinitesimals dx, ... 1 In the ancient India, Mahavira, the founder of Jainism became very close to the modern notion of "different sizes of infinities". In the mathematical part of Jain philosophy, there are several kinds of infinities. Quoted from Wikipedia: The Jain mathematical text Surya Prajnapti (c. 400 BC) classifies numbers into three sets: enumerable, innumerable, ... 2 In some cases, it is interesting (or simply makes notation simpler) to consider the extended real numbers: \overline{\mathbb{R}}=[-\infty,\infty]. Formally, pick your favorite objects which are not real numbers, and let's conveniently call them -\infty and \infty. Define the set \overline{\mathbb{R}}=\mathbb{R}\cup\left\{-\infty,\infty\right\}. ... 2 Hint:$$ \frac{1^{p}+2^{p}+\ldots+n^{p}}{n^{p+1}}\sim\frac{\int_{0}^{n}m^{p}dm}{n^{p+1}}=\frac{\frac{n^{p+1}}{p+1}}{n^{p+1}}=\frac{1}{p+1} $$2 Note that \infty \notin \mathbb{R}, so it doesn't make sense to say that [0, \infty] is a subset of \mathbb{R}. On the other hand, the set$$[0, \infty) = \{x \in \mathbb{R} \mid 0 \leq x\}$$completely makes sense and defines a subset of \mathbb{R}. However, one can consider the extended real numbers \overline{\mathbb{R}} = ... 1 Sometimes, we want to consider \infty as a value that can be attained. For example, we could define$$ f(x) = \begin{cases} 1/|x| & x \neq 0\\ \infty & x = 0 \end{cases} $$In this case, we would say that f:\Bbb R \to (0,\infty] is an onto function We can consider \Bbb R = (-\infty,\infty) as a subset of the larger space \overline{\Bbb R} ... 4 When you "close" the bracket, it implicitly means that you're working in the space \Bbb R\cup\{\pm\infty\} (and we omit the + from the positive infinity sign), and in that space [0,\infty) is not closed, since \infty is indeed a limit point of that set. [It might be the case that you are working in the 1-point compatification of \Bbb R, which is ... 2 Suppose |A|<|\mathbb{N}|. Then there is an injection A\to \mathbb{N} onto a proper subset that has smaller cardinality than \mathbb{N}. However, all infinite subsets of \mathbb{N} have the same cardinality as \mathbb{N}, so it follows that A is finite. 0 Transfinite numbers are numbers that are "infinite" in the sense that they are larger than all finite numbers, yet not necessarily absolutely infinite. The term transfinite was coined by Georg Cantor, who wished to avoid some of the implications of the word infinite in connection with these objects, which were nevertheless not finite. Few contemporary ... 2 In general, number systems exist because they usefully quantify things. We invent the extended real numbers, which includes two extra numbers, +\infty and -\infty, because they are fantastically useful for doing calculus! For example, it lets us understand$$ \lim_{x \to +\infty} \frac{x^2 + 1}{x + 1} = +\infty $$as saying that the function is ... 0 We can make it a convention that we would only use +\infty in that way. For example, the natural interpretation (via transfer) of that symbol in non-standard analysis is, in fact, the largest extended hyperreal. This isn't too different from the idea that we often use symbols 1 and 0 to denote the largest and smallest elements in a bounded lattice. ... 1$$\infty \cdot (0+1) = \infty $$is indeed correct, assuming by \infty you mean the element of the extended real numbers. The problem is that you can't say$$ \infty \cdot (0+1) = \infty \cdot 0 + \infty \cdot 1 $$The distributive law for extended real numbers is only valid where it is defined. 1 No you can't. It is still indeterminate form. The results may be different. Look at the examples: Let (a_n), (b_n), (c_n) be infinity sequence such that a_n= n^4, b_n = \frac{-1}{n}, c_n = n.$$\lim_{n\to\infty}a_n\cdot b_n + c_n = \lim_{n\to\infty}n^4 \cdot\frac{-1}{n}+n = \lim_{n\to\infty} n(1-n^2) = -\infty$$Or let (a_n), (b_n), (c_n) be ... 2 No, you can't do that. Such manipulations with \infty aren't mathematically permitted. Perhaps more importantly for usage of \infty, they don't cooperate with limits, either. For example, let f(x)=\frac{1}{x^2}, g(x)=x, and h(x)=\frac{1}{\sqrt{-x}}. Then$$\lim_{x \to 0^-} f(x) g(x) + h(x) = \lim_{x \to 0^-} \frac{1}{x} + \frac{1}{\sqrt{-x}} = ...

0

Change $x=\pi/2-t$: then the limit becomes $$\lim_{t\to0}(\cot t)^{\sin t}$$ Take the logarithm: $$\lim_{t\to0}\sin t\log\cot t=\lim_{t\to0}(\sin t\log\cos t-\sin t\log\sin t)$$ Can you go on by computing the limit of the two summands separately?

0

\begin{equation*} \lim_{x \rightarrow 0} (\cot x )^{\sin x} \end{equation*} is a more convenient expression for the same thing. Now \begin{equation*} \lim_{x \rightarrow 0} (\cos x )^{\sin x} = 1 \end{equation*} presents no problem. So take $u = \sin x$. We are left with \begin{equation*} \lim_{u \rightarrow 0} (1/u)^{u} \end{equation*} This is easily ...

2

We have: $\cos x\log(\tan x) = \dfrac{\log (\tan x)}{\sec x}$, and using L'hospital's rule: $\dfrac{\dfrac{\sec^2 x}{\tan x}}{\sec x\tan x} = \dfrac{\sec x}{\sec^2 x - 1} = \dfrac{\cos x}{1 - \cos^2 x} \to 0$ as $x \to \pi/2^{-}$. Thus $\tan x^{\cos x} \to e^{0} = 1$

1

This time the hotel has rooms that are labeled not with single numbers, but with sets of numbers, possibly infinite sets. One room corresponds to the empty set, and has the label $\{\}$ on the door. One room has the label $\{5,23,119\}$ on the door. The Prime Presidential Penthouse Suite has every prime number on its door. The hotel is full, with exactly ...

6

In standard set theory, they are the same set, but seen from different perspectives. The notation $\omega$ emphasizes that it is an ordinal number, whereas the notation $\aleph_0$ emphasizes its role as a cardinal number. Now whereas $\omega$ and $\aleph_0$ are two different notations for the same set, $+$ and $+$ is the same notation being used for two ...

6

You’re talking about two different operations. In $\omega+\omega\ne\omega$ you’re talking about ordinal addition; in $\aleph_0+\aleph_0=\aleph_0$ you’re talking about cardinal addition. When you consider $\omega$ as a cardinal number and perform cardinal arithmetic on it, $\omega+\omega=\omega$. Similarly, the ordinal sum $\omega_1+\omega$ is the ordinal ...

1

$\omega$ is a well ordered set, while $\aleph_0$ is the cardinality of that set. $|\omega|=\aleph_0$. As I've been informed, since $\omega$ is an initial ordinal, it is also considered a cardinal number, so indeed $\omega=\aleph_0$, and $|\omega|=\aleph_0$. However, when we use ordinal arithmetic $|\omega+\omega|=\aleph_0$ as well, even though ...

5

$$\frac{n!}{(n+1)!} = \frac{\require{cancel}\cancel{n!}}{(n+1)\cancel{n!}}=\frac 1{n+1}$$ Can you take it from here?

4

Hint: $$\frac{n!}{(n+1)!}=\frac{n!}{(n+1)(n!)}=\frac{1}{n+1}$$ Game Over!

0

@sonystarmap, I think I get the idea now: you denote $G'=G_1 - G_2$, a sphere $B_R$ with radius $R$ and functions $$\phi_R(\textbf{r}) = \{ {\begin{array}{cc} arbitrary & |\textbf{r}| < R \\ 0 & \textbf{r} \ge R \end{array}}$$ Then with $\langle \nabla G', \nabla \phi_R \rangle_{B_R}=\langle \nabla G_1, \nabla \phi_R ... 0 Maybe you could use the following argumentation: Let$B_R =\{x\in \mathbb{R}^n | ||x||\leq R \}$be a ball of size$R$. Let$\phi_R\in C_0^\infty(B_R)$with$\phi_R(0)\not = 0$be arbitrary. Then from$G(r)=\delta(r), we get \begin{align} &\langle \nabla^2 G,\phi_R\rangle_{B_R} = \langle \delta,\phi_R\rangle_{B_R}=\phi_R(0)\\ ... 1 If you denote byI_n$the set of all irrational numbers between$n$and$n+1$, then for any$n,m$the function$f\colon I_n\to I_m$given by$f(x)=x+m-n$is easily seen to be a bijection. Therefore, the cardinalities of$I_n$and$I_m$are the same. 0 Yes. There are many irrational numbers between$n$and$n+1$. In fact, It equals$\mathcal c = |\Bbb R|$. 3 The way we measure (possibly infinite) cardinality is by using bijections (one-to-one) mappings. Since the irrational numbers in the open interval$I=I_0=(0,1)$are in bijection with those in$I_n=(n,n+1)$by the map$f_n:x\mapsto x+n$, each set has the same cardinality. Since$n\in\mathbb{Z}$was arbitrary, they all have the same cardinality. Why does the ... 0 Here's a simple test of logic: Given that two numbers are equal when there are no possible intermediate values, and there are no possible intermediate values between 0.999999... and 1.000000..., the only logical conclusion is that 0.999999... is equal to 1.000000... 0 I don't know if this is quite what you're looking for, but I know I have used this one is class before and if'd you'd like to look I've also used it in a question that i posted on here. Hope it helps! $$\sum_{k=1}^nku^k={u\over (1-u)^2}\bigg[nu^{n+1}-(n+1)u^n+1\bigg]\forall u\ge 1$$ P.S.- I know it's not for$k=0$to$n$, but i thought it might help still. ... 5 Note that for$|x| < 1$, we have $$\sum_{n=0}^\infty nx^n = x \sum_{n=0}^\infty nx^{n-1} = x \sum_{n=0}^\infty \frac{d\ }{dx} x^n = x \frac{d\ }{dx} \sum_{n=0}^\infty x^n = x \frac{d\ }{dx} \frac{1}{1-x} = \frac{x}{(1-x)^2}$$ You can now write down an expression for your sum. 2 Differentiate the first expression with respect to$x$, then multiply by$x$, then add missing initial term(s)? 3 The game where one looses \$1 when the toss is heads and one wins \$2 when the toss is tails corresponds to a random walk with positive bias. These have positive probability to never hit \$0. The game where one looses \$1 when the toss is heads and one doubles one's fortune when the toss is tails, is at least as favorable as the first one at every step such ... 2 Begin by defining $$S(1;9) := \sum_{i=2}^9 \frac{1}{i}$$ Let$a = S(1;9)$. Then$a < 2$. Let$S(10;99)$be the sum over the fractions of all numbers between$10$and$99$that don't contain a$1$in their depiction. Prove that$S(10;99) < \frac{9}{10}a$. Similarly, define$S(100;999)$to be the sum over the fractions of all numbers between$100$... 0 I think I can give a proof of divergence at$ \infty \Rightarrow $divergence: (THe other case$-\infty$is similar. Assume that$(a_n)$diverges to$+\infty$. Then let$a\in \mathbb R$. Let$\epsilon = a+1 \in \mathbb R$. Then by definition there is$K\in \mathbb N$so that$a_n \geq a+1$for all$N\in \mathbb K$. In particular, for all$N\in \mathbb N\$, ...

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