# Tag Info

1

In the case of the circle, you take a polygon and say that it approximates a circle, but it is not precise. As you increase the number of points on the polygon, it approximates a circle more and more, but never quite reaches it. It is said that as the limit of points on the polygon approaches infinity, you have yourself a circle, which is why one ...

1

There are more than two points on a line segment. In fact, there are infinitely many points on the line segment. For example, there is a point between the two endpoints that is also in the line. And a point between the middle point and the start. And one between the middle and the end. And one at a distance of $\frac{\pi^2}{4}$ from the start and on and ...

4

Both the circumference and the segment have infinitely many points, but that's not the point (heh, heh). What matters is their length. The picture presumably is trying to show the ratio between the diameter and the circumference, which turns up to be $\pi$, or 3 and then something.

10

You are conflating end points and points. The circumference of a circle does not have any end points but infinitely many points. The line segment has two end points and still infinitely many points. Sidenote: The length of the segment is not as important as you may think. We cannot distinguish the number of points on a line segment of length $π$ from ...

-4

I have a beautifully simple argument which will help you to show many doubters the true situation. If you project the real number line on to a circle it is clear that the real numbers are dense everywhere, while the natural numbers are dense only at infinity. This creates a strong, visualisable argument that there are more real numbers than integers. The ...

1

Related to joriki's answer, if Inhabitant $n$ goes to a room between $1$ and $n\log n$, each room is also almost surely occupied, but if they go to a room between $1$ and $n^2$, the probability is less than $1$.

2

The limit you took doesn't correspond to any limiting distribution. For each $x$, each occupant is assigned one of the rooms with the same probability $\frac1x$. But infinitely many occupants cannot be assigned rooms with the same probability "$\frac1\infty$". Here's a distribution for the infinite case, though: Inhabitant $n$ takes one of the rooms $1$ ...

1

You can use the squeeze theorem: $$-|1/2|^{x}\leq (-1/2)^{x}\leq|1/2|^{x}.$$ And, as you noted, both terms in the left and right of the above inequality go to $0$ as $x\to\infty$.

0

Infinity can be a number, and then we can interpret the usual algebraic operations as applying to them. For example, to illustrate your first principle, for each positive infinite hyperreal $H$ and appreciable number $b<0$ we will have that $H \cdot b\;$ is an infinite negative hyperreal. The so-called "indeterminate forms" are also a lot less mysterious ...

1

But why then are the following results true : (...) Logic of these "results" can be explained the following way: 1) In addition to usual operations on $\mathbb{R}$ $+$ and $\cdot$ let's introduce symbols $+_L$ and $\cdot_L$, when $a +_L b = c$, by definition, is a short for statement "$\forall (a_n), (b_n), (a_n \to a, b_n \to b) \implies (\exists ... 3 Strictly speaking we can define the symbols to work however we want. We choose to define the symbols to work that way because it matches up with the behavior of limits. For example, if$a>0$and$f(x) \to \infty$as$x \to c$then$af(x) \to \infty$as$x \to c$while$-af(x) \to -\infty$as$x \to c$. (Note that$f(x) \to \infty$and similar expressions ... 2 I like your starting line "I'm having trouble with the limit approach to calculus ever since I heard about the infinitesimal definition." +1 for that. The short answer to your question lies in that line because of the following non-mathematical theorem: A non-mathematical theorem: There is no sound theory of infinitesimals available for a beginner in ... 1 To answer the query contained in the title of your question, the derivative is the shadow of the ratio of infinitesimals$\frac{\Delta y}{\Delta x}$. Sometimes the shadow is referred to as the standard part. Thus even in the infinitesimal approach the derivative is not exactly the ratio but only the ratio rounded off to the nearest real number. This ... 3 You're treating the notion of "tangent" as pre-existing and asking how we can say that the derivative is (the slope of) the tangent. But the situation is more complex than that – by defining the derivative, we're defining what we mean by tangent in many cases in which there was no such pre-existing concept. So the task is to clarify the pre-existing, ... 4 Because from my understanding, in order for it to be a tangent line, it intersects the curve at one point only, however Δx approaches zero, it never reaches it, so Δx must be greater than zero, however infinitesimally small, correct? You're right. We don't ever reach that point. We take a limit. The colloquialism, "reaching the point" is a good ... 0 The two are equivalent. If$\existsR \in \Bbb R$such that for all$x > R$,$|f(x) - L| < \epsilon$, then if$R >0$there's nothing to prove, and if$R \le 0$, take$R' = 1$, then$R' > 0$and if$x>R'$, then$x > R$and so$|f(x) - L| < \epsilon$. 2 Yes, there is a problem in your approximations. Recall that $$\lim_{n \to \infty} \biggl(1 + \frac{x}{n}\biggr)^n = e^x.$$ Effectively, you have replaced this limit with$1$four times, and since the various$x$s don't cancel, you got a wrong result. The square root does indeed tend to$1$, so we can ignore that. Assuming that$a_2$and$p$are constant ... 1$[-\infty,+\infty]$refers to the extended reals. In general, though,$\mathbb{R}=(-\infty,+\infty)$. 1 Some items have been dealt with in comments, so we look only at c). We want to show that for any$\epsilon\gt 0$, there is a$B$such that if$x\gt B$then $$|\sin(1/x)-0|\lt \epsilon.$$ Let$\epsilon\gt 0$. Since$\lim_{t\to 0}\sin t=0$(given), there is a$\delta\gt 0$such that if$0\lt |t-0|\lt \delta$, then$|\sin t-0|\lt \epsilon$. Let$B=1/\delta$. ... 1 Hmmm.... Couldn't we say$\frac {a+b}{ab} = \frac a {ab} +\frac b {ab}= \frac 1b +\frac 1a $? Thus if$a $is "infinite" then$\frac {infinite + b}{infinite * b} =\frac {infinite}{infinite*b} =\frac 1b$is consistent with$\frac 1b + \frac 1 {infinite}= \frac 1b +0 =\frac 1b $. I think your confusion is inf/inf can be anything-- not just 1. In the case ... 1 Hmm, after trying a squinty-eyes interpretation of the Dutch, it looks like the question is asking for both the maximal and minimal value that$\frac{a+b}{ab}$can take for any$a\in[0.15,\infty)$when$b=0.017$. In other words you're interested in the behavior of $$\frac{a+0.017}{a\cdot 0.017}$$ for large$a$. We can rewrite this as $$... 0 In modern mathematics we have several ways of formalizing infinity. The one that is most relevant to your question was provided by Abraham Robinson; see here. Following his framework, there are both infinitesimals and infinite numbers. Thus if H denotes an infinite number, one can indeed divide a unit interval into H parts. Each part would have ... 0 Calculate the probability that you lose every time, using an infinite product. For example, for starting at 1/2 probability and cutting in half every time, the probability that you have not won after playing n times is$$\prod_{i=1}^{n} \left(\frac{2^{i}-1}{2^{i}}\right) = \prod_{i=1}^{n}(1-2^{-i}).$$I'm not motivated work out the exact formula, but ... 0 On the first flip, you have a \frac12 chance of winning. Assuming you lost (with probability \frac12) you win with probability \frac14, which gives you an additional \frac18 probability of winning, for a total of \frac58. Assuming you lost again (probability \frac38) you get another shot at probability \frac18, for an additional \frac3{64}, ... 1 f:(\color\red{-1},\color\green{1})\rightarrow(\color\orange{0},\color\purple{4}) f(x)=(x-(\color\red{-1}))\cdot\frac{\color\purple{4}-\color\orange{0}}{\color\green{1}-(\color\red{-1})}+\color\orange{0}=2x+2 g:(\color\orange{0},\color\purple{4})\rightarrow(\color\red{-1},\color\green{1}) ... 3 From the things you say it seems clear that when you talk about splitting AB into infinitely many parts, you mean infinitely many parts of equal length. You simply can't do that. You seem to realize what the problem is, but you're not drawing the right conclusion: The problem is if the parts have positive length that would mean AB had infinite length, ... 1 Length is formalized by the mathematical concept of Lebesgue measure, which assumes real number values. A subset of the real number line cannot have infinitessimal measure, it is either zero or positive. Some subsets are nonmeasurable though. Also, there is no "value that comes immediately after zero." One is free to create their own partially ordered sets, ... 2 The conversation seems almost equivalent to this hypothetical one: ME: Suppose there are positive integers m, n such that m^2=2n^2. Then we can generate a list of all prime factors and read off the exponent of 2. This exponent should be even (because it is in the unique factorization of m^2) but also odd (because it is in the unique factorization ... -6 The key here is to let it go. If you let it bother you then you yourself risk becoming the "pro-Cantor crank". Brouwer, Poincare and Wittgenstein were all opponents of Cantor's work and many respectable, published, logicians have disputed Cantor's diagonal argument. Since it's a proof by contradiction it relies heavily upon the law of the excluded middle, ... 5 One natural use for smallish infinite ordinals, which makes clear that different infinite ordinals have distinct properties, is the length of games. I'm playing a chess-like (turn-taking, deterministic, complete information) board game of some kind against an opponent. My teammate is also playing a separate game against my opponent's teammate. Moves are ... 0 Summing over an uncountable index results in an infinite sum if uncountably many terms are non-zero. To see this, we prove the contrapositive: that if a sum over an uncountable index is finite implies that at most countably many terms are non-zero. Proof. Let \sum_{\alpha \in A} x_\alpha = L. Let S_n = \{\alpha \in A \mid x_\alpha > 1/n\}. Then$$L ... 1 Here's an intuitive way to think of it. Consider a more general case:$$y=\sin(\frac{1}{x})$$ When$x$gets closer and closer to$0$, there will be an infinite amount of times it crosses the x-axis. This is because$\frac{1}{x}$approaches$-\infty$from the left and$\infty$from the right. When$\frac{1}{x}$gets very close to the y-axis it will grow ... 1 Since$\sin x$is zero at$\pi,2\pi,\ldots$,$(\sin x)(\sin x^{-2}) $is zero at$\dfrac{1}{\sqrt \pi},\dfrac{1}{\sqrt {2\pi}},\ldots $, so the answer to your question is "yes".$\sin\left(\dfrac1x\right)$is a simpler example with the same property. 0 First, this function$f$can be defined at$0$by continuity, with value$0$. Now take$x_k = \frac{1}{\sqrt{2\pi k}}$,$k >0$. All$x_k$are in$[0,1]$, and$f(x_k)=0$, an infinity of times in the interval, and increasingly denser near$0$. A lot of them are quite easy to build, and motivate of lot of exercices for students to study continuity, ... 7 One issue not yet addressed is why we have both$\aleph$s and$\omega$s. These both exist because Cantor introduced two separate concepts about infinite sets - their sizes ($\aleph$) and their order-types ($\omega$). Everyone else explained sizes, so I won't go over that. For order-types, begin by considering the following set: { 1/2, 2/3, 3/4, ..., ... 4 It is an indeterminate form and as such cannot be assigned any value. It is better expressed as$\lim_\limits{{x\to \infty}\\,\\{y\to 0}}x^y$. As commented by Did, it is true that$x^y$has no limit when$x\to \infty$and$y\to 0$. And$\infty^0$has no definite meaning in mathematics. It is basically some kind of a meaningless statement where the ... 4 First of all: you cannot just substract infinity from infinity. Infinity is not a real number so you can't simply use the basic operations as you're used to do with (real) real numbers. However, in the context of limits, there are things you can do. As pointed out, the limit in your question is$0$because$x-x=0$for all$x$, so the 'calculation' would be ... 0 The answer is$0$as first you simplify limit expression and then plug in value to check the limit if it's not indeterminate. So the limit is$0$1 For a beginner, I would simply introduce the idea of one-to-one correspondence and not blow him away with Cantor and Hilbert. Thus, two sets A and B have the same number of elements if each element of A can be paired with exactly one element of B, and that pairing covers every element of B. Thus, 2 is paired with 3, 4 with 6, 6 with 9, 8 with 12, etc, as ... 2 Your question touches on issues both mathematical and philosophical; in fact, you might enjoy an introductory text in philosophy of mathematics and logic, or on the foundations of mathematics. Truth is, an infinity is not just an infinity. One reason for this peculiarity is due to the seminal work of Georg Cantor. Cantor proved that the set of real numbers ... 27 As an insight, think of the size of the usual sets of numbers. Think of the set of all positive integers, and the set of all positive multiples of 5. It's a bit strange to the "uninitiated", but it's ultimately not hard to wrap your head around the fact that there's the same amount of each, because we can list them side by side without any problem: ... 4 If it's any consolation, in practice you won't encounter more than two different types of infinity: that corresponding to the natural numbers, and that corresponding to the real numbers (cardinality of the continuum). The reason why it's necessary to differentiate between those two is that the "size" (cardinality) of the real numbers corresponds to the ... 5 The major motivation for classifying infinities (other then the intrinsic enjoyment of mathematics) is that different infinities permit different properties. Countable infinities can be reasoned about using inductive proofs. On the other hand many of the properties analysis makes use of requires uncountable sets. Having different infinities often makes ... 9 I won't comment on your more philosophical questions, but I will give what I think is one of the more important applications of different sizes of infinity. There is a rigorous mathematical way of thinking about a computer program, called a Turing machine. One can show that the cardinality of the set of Turing machines is$\aleph_0\$, however the set of all ...

2

I wish to propose another simple, perhaps, less mathematical but more intuitive understanding of 'infinities'. It also happens to be by an author that you may possibly be familiar with: “There are infinite numbers between 0 and 1. There's .1 and .12 and .112 and an infinite collection of others. Of course, there is a bigger infinite set of numbers between ...

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