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1

"Indeterminate forms" are a vague concept and it is better to keep them "vague" rather than define them properly. Limit evaluation are done on the basis of certain limit theorems which include the "algebra of limits" in particular. Theorems dealing with "algebra of limits" suffice to calculate limit of expressions which are composed of sub-expressions ...


6

That is not a standard notation for any ordinal or cardinal. I'm not sure how to answer any better.


1

Note: This answer uses Iverson brackets to cope with the floor function. Although this may need some more lines, the calculations are easy and can be performed this way more or less mechanically. The answer was inspired by example (1.14) of Two Notes on Notation by D. Knuth. For convenience only we set $N=2^k$. We obtain \begin{align*} ...


0

$\infty$ is not a real number (it's not in $\mathbb R$). It's not even an imaginary or complex number. Basically, $\infty$ isn't one of the things people usually call "number," and, as such, $+$, $-$, $\times$, and $\div$ are undefined for it. However, there are ways to get around this. Define the set $\overline{\mathbb R}$ to be $\mathbb ...


0

First: Can $1/\infty = 0$, or $1/\infty\ne0$? Neither is true, according to ordinary algebraic conventions, but there are two ways in which the former can be correct. Notational abuse. We can't use $\infty$ like a number in our expressions, because it isn't a number in any of the usual senses. But let's be honest, everyone knows what you mean when you say ...


0

If we are to understand "$=$" as a binary relation of identity and "$\neq$ as a binary relation of non-identity, then in most contexts, formally, $1/\infty$ is not identical to zero. Therefore it would make logical sense to write $1/\infty \neq 0$ as long as we "define what we mean by $1/\infty$". This is more in a strictly logical sense than in a ...


16

It is worth noting that since $\infty$ is not a number, the expression $\frac{1}{\infty}$ is not meaningful. That is, you cannot evaluate this expression. It is not a number. You can think of it in much the same way as $\frac{1}{0}$ --- an expression without any interpretation. The only means that we have to talk about expressions involving $\infty$ is ...


6

Your argument makes use of stochastics. If you want to do things the mathematically exact way, you would first need to define a probability measure on $(0,\infty)$. Assuming that you choose a continuous distribution, then yes, the Probability that you hit $5$ is $0$ - but there are infinitely many numbers, so there is no contradiction. (Keep in mind that $0 ...


0

In some contexts it makes sense to say $1/0=\infty$ and $1/\infty=0$. If you're working in a context in which there is just one $\infty$ that can be approached in either the positive or the negative direction, then it makes sense. This makes sense for the values of (but not the arguments to) the tangent and secant functions. For rational functions it ...


3

Operations in the real line are maps: \begin{align} +&\colon \Bbb R \times \Bbb R \to \Bbb R \\ -&\colon \Bbb R \times \Bbb R \to \Bbb R \\ \cdot &\colon \Bbb R \times \Bbb R \to \Bbb R \\ \div &\colon \Bbb R \times (\Bbb R \setminus \{0\})\to \Bbb R \end{align} But $\infty \not\in \Bbb R$, so there is no sense trying to do things like ...


2

No. Specifically, $\infty$ is not a real number. It is a limit. Over the real numbers, ${1 \over \infty}$ is meaningless because you're trying to divide a number by something that isn't a number. Now of course, using calculus, we can get around some of this. By observation, $\displaystyle \lim_{x\to\infty} {1 \over x} = 0$. We can intuitively see this ...


0

Unfortunately, infinity is not a number. When you get a bit older (calculus), you'll learn the concept of "limit." A limit allows us to look at what one side of an equation approaches while the other side is approaching some number or infinity. Moreover, a limit only exists if the left-hand limit and right-hand limit are the same. Imagine looking at a graph ...


0

Let $\Omega$ be any set and $p:\Omega\to[0,1]$. For any subset $A\subseteq \Omega$ define $$P(A)=\sup\left\{\sum_{x\in F}p(x): F\subseteq A, F\mbox{ is finite}\right\}.$$ Exercise: If $P(\Omega)=1$, then $P$ defines a probability measure on the $\sigma$-algebra of all subsets of $\Omega.$ Is this your question? If so, let me know where you get stuck ...


0

This fails for the infinite case. Consider the following $P$ $$P(A) = \begin{cases}0 & \text{If $A$ is finite}\\ 1 & \text{If $A$ is infinite}\end{cases} $$ then consider $P$ defined on the subsets of $\Bbb{N}$. you can see that $\Bbb{N} = \cup_{n} \{n\}$ but $$1 = P(\Bbb{N}) = P(\cup_{n} \{n\}) \neq \sum_n P(\{n\}) = 0 $$


0

Other answers show that there is a bijection between $\mathbb N$ and $\mathbb{N}\setminus\{2\}$, so $\mathbb{N}\setminus\{2\}$ is also infinite just like $\mathbb N$. If you don't have knowledge of cardinality you may still wonder why it is also infinite? Cardinality is, roughly, an equivalence relation between different sets. Formally, two sets $A$ and $B$ ...


3

Consider $f : \Bbb N \setminus \{ 2 \} \to \Bbb N$, $f(0)=0, f(1)=1, f(n) = n-1$ for $n \ge 3$. Check for yourself that this is bijective. Since $\Bbb N$ is infinite, so will then be $\Bbb N \setminus \{ 2 \}$. In general, removing a finite number of elements from an infinite set leaves it infinite.


3

Yes, it is. Here is a bijection from $\mathbf{N}$ to $\mathbf{N}\setminus\{2\}$: $$f(x) = \left\{\begin{array}{l l} x & \text{if $x < 2$} \\ x+1 & \text{otherwise} \end{array}\right..$$ For an arbitrary infinite set, it is most easily seen by contraposition. Namely, for a set $A$ (which we can assume to be non-empty since the empty set is ...


4

Just observe that $$\tan\alpha=\cot\alpha-2\cot 2\alpha,$$ and you'll get $$S_k=\sum_{n=0}^k \frac{1}{2^n}\tan\frac{\alpha}{2^n}=\tan\alpha+\sum_{n=1}^k\frac{1}{2^n}\cot\frac{\alpha}{2^n}-\sum_{n=1}^k\frac{1}{2^{n-1}}\cot\frac{\alpha}{2^{n-1}}=$$ $$=\tan\alpha-\cot\alpha+\frac{1}{2^k}\cot\frac{\alpha}{2^k}.$$ Since $x\cot\alpha x\to \frac{1}{\alpha}$ when ...


2

I assuming that is $\lim_{k\rightarrow\infty}f_{k}\left(x\right)=\sum_{n\geq0}\frac{\tan\left(x/2^{n}\right)}{2^{n}}. $ Using the Taylor series of tangent and cotangent we have ...


3

The left hand limit will not be defined for $\ln$, be careful. Then note that \begin{align*} \lim_{x \rightarrow 0^{+}} \ln\bigg(\frac{1}{x}\bigg) &= \lim_{x \rightarrow 0^{+}} \ln(1) - \ln(x) \\ &= \lim_{x \rightarrow 0^{+}} - \ln(x) \\ &= \infty \end{align*}


3

The domain of the (real) logarithm function is the set $(0,\infty)$. In particular, $\displaystyle \lim_{\alpha \to 0^-} \ln \left( \frac 1 \alpha \right)$ is undefined. On the other hand, $\displaystyle \lim_{\alpha \to 0^+} \ln \left( \frac 1 \alpha \right) = \infty$ via the limit laws.


1

Thank you everyone for your answers. They were all informative, but I wasn't able to intuitively understand them until I saw this visual: http://tube.geogebra.org/student/m107691 The key for me was to see that there are two kinds of revolution happening: Revolutions of Circle A with respect to Circle B, and Revolutions of Circle A with respect to the ...


6

I was bored, so I read the paper in order to figure out what's wrong with it. The paper purports to create an injection from $\mathbb{X}_+$ (the positive irrationals) to $\mathbb{Q}_+$ (the positive rationals) by repeatedly removing one element from each set. The glaring flaw I found was this (emphasis preserved): If the set $\mathbb{Q}_+$ were exhausted ...


10

This author is notorious for his claims about set theory, and mainstream mathematics does not take them seriously. For a detailed criticism of his style of argument, see this review (in German) by Franz Lemmermeyer of a book he wrote. Regarding the specific text you link to, arguments like this: The rational numbers are countable whereas the ...


3

The natural numbers are, almost by definition: 0 (sometimes) the number after 0 the number after that the number after that the number after that etc. Nowhere in that list is $\infty$. (Note: Sometimes, 0 isn't considered a natural number. Then just replace 0 by 1.)


0

EDIT: I agree with many of the comments. I guess I am being informal with my language here, and that example was not so great, but I still hold to the notion that we treat infinity as something other than a number. Sorry for confusion.


10

No, it is not. ${}{}{}{}{}{}{}{}$


3

Note that the non-transcendental irrational numbers form a subset of $\Bbb A = \overline {\Bbb Q} \subset \Bbb C$, the set of complex algebraic numbers (i.e. the algebraic closure of $\Bbb Q$). From a purely set-theoretic perspective, let us show that $\Bbb A$ is countable. For $a_0, \dots, a_n \in \Bbb Q$, let $R(a_0, \dots, a_n)$ be the set of the roots ...


24

The non-transcendental numbers (otherwise known as the algebraic numbers – Wikipedia link) comprise a countably infinite set, whereas the transcendental numbers are uncountably infinite. (Why are there only countably many algebraic numbers? Because we can group them according to what polynomial in $\mathbb{Q}[x]$ they are a root of, and any such ...


12

The set of algebraic numbers $\mathbb A$ is countable, so $\mathbb A\cap (\mathbb R\setminus \mathbb Q)$ is also countable. On the other hand, the set of transcendental numbers $\mathbb R\setminus\mathbb A$ must be uncountable, so $$|\mathbb R\setminus\mathbb A|>|\mathbb A\cap(\mathbb R\setminus\mathbb Q)|.$$


3

Hint: the set of algebraic numbers is Countable. For proving this, you can show that the polynomials with integer coefficients form a countable set, and that the set of their roots is also countable. The latter is just the Set of Algebraic Numbers (over $\mathbb{C}$). As the Algebraic Numbers over $\mathbb{R}$ form a subset of the former set, it must also ...


1

You are right in assuming that iff $c=0$, we define $s_\alpha(\infty) = \infty$. As in the case $c = 0$, we know that $\alpha$ (considered as a linear map on $F^2$) maps $F \times \{0\}$ onto itself. If we consider $P(F)$ as $F^2-\{0\} / \sim$, where $\sim$ is defined by $x \sim y$ iff for some $\lambda \in F^\times$ we have $\lambda x = y$, then $F^\times ...


5

For any real number $a>1$, the sequence $x_n=\sqrt[n]{a}$ has the property that $x_n\to 1^+$, and $$\lim_{n\to\infty}(x_n)^n=\lim_{n\to\infty}a=a$$ So the limit you're asking about doesn't exist.


7

It depends on how $x$ approaches zero. Let $(x_n)$ be a sequence converging to $1$, write $y_n := n(x_n - 1)$, then $$ x_n^n = \left(1 + \frac{y_n}n\right)^n $$ which converges iff $y_n$ converges to some $y \in \mathbf R$, namely to $e^y$. So $x_n^n \to e^{\lim_n n(x_n - 1)}$ if $\lim n(x_n - 1)$ exists. So, for example $$\left(1 + \frac 1{\sqrt ...


0

A humble engineer's thought: think about what $(-1)^t$ for $t\in\mathbb{R}$ would be if you let $t\rightarrow\infty$. When you substitute $(-1)=\mathrm{e}^{-\mathrm{j}\pi}$ you see that the expression is actually just a complex phasor, which keeps spinning around the origin. The limit is thus undefined. Maths experts, please don't kill me :)


1

For $\pi$ the answer is not known...as far as I am aware, nothing out there would suggest that the digits are distributed non-uniformly. Here is an article which, among other things, gives a histogram of the occurrence digits in the first trillion digits of $\pi$: http://www.ams.org/samplings/math-history/hap-6-pi.pdf Still, a trillion is finite and ...


2

The distinction here is really philosophical rather than mathematical, but I wouldn't say it's hand-wavy at all. In any case, here's my best attempt at an answer. Suppose we want to make a mathematical distinction between an actual infinitude and a potential infinitude, where by 'mathematical distinction' I mean a negation of an equation between two ...


1

I don't know what you mean by "the points map to each other". To "map" one set to another implies a function that maps each point in one set to a unique point in the other; to "map to each other" implies that the function is one-to-one. There is no one-to-one mapping between points: each point on the small circle meets three different points on the large ...


3

The mapping of points from one circle to the other is no different than when the small circle is rolling on a straight line; the extra revolution is purely due to bending. It might be easier to see this in the case of gliding, when there is a single point $A$ on the small circle which is in contact with the line or the large circle throughout the process. ...


2

As it is depicted on the figure below the smaller circle has to turn around $3$ times if it travels along a straight distance equaling the perimeter of the larger circle. In this case the center of the smaller circle takes the same distance. Now imagine that the straight line of length $6\pi r_B$ takes a complete rotation around its center while the small ...


1

I'm assuming no finite equivalence classes are allowed. A model $\mathcal{M}$ of this theory is basically just a family $\{E_\eta: \eta<\lambda\}$ of sets (the equivalence classes), where each $E_\eta$ has cardinality $\lambda_\eta\ge\aleph_0$. EXERCISE: two models $\mathcal{M}=\{E_\eta: \eta<\lambda_0\}$ and $\mathcal{N}=\{F_\eta: ...


0

A small alteration to the working provided in the "Example" section of the Partial fractions in complex analysis Wikipedia page (linked to by Jyrki Lahtonen above) leads to a similar representation for $\sec x$. The residues of $\tan z$ are calculated as follows (using L'Hopital for the limit): $$ Res\left[\tan z ; (n+\tfrac{1}{2})\pi\right]=\lim_{z \to ...


0

Other than that, you can create any number of your own operattions that grow faster than exponential and produce continuous curves with infinite sums. You need not be thrown off by the fact that it's a discrete summation, any certified mathematician will confirm that an operator is an operator is an operator is an operator is an operator is an operator, and ...


5

Recall that the infinite product representation of the cosine function is given by $$\cos z=\prod_{n=1}^{\infty}\left(1-\frac{z^2}{\pi^2(n-1/2)^2}\right) \tag 1$$ Now, just take the logarithmic derivative of both sides of $(1)$ and multiply by $-1$ to expose that $$\bbox[5px,border:2px solid #C0A000]{\tan ...


2

If the first row is infinte, then it doesn't matter how much time she has on her hands, she will never finish laying it down, and thus never get to row number two. However, in the comic, while she says that the desert expands seemingly infinitely, she doesn't say anything about her computer being infinitely large. Just that there's more than enough room for ...


-1

Assuming the wiring is 100% efficient and the speed of the electrical current is instantaneous (which is just as illogical as the idea of unreal numbers such as infinite), there would be no final flip, as the time it takes to switch it is always divided by half. However, in a real-world application, where wiring has latency and electrical current has a set ...


0

It depends entirely on how division is defined on the real projective line (which I will denote $S^1$). We could say that $S^1$ inherits the same operations as the reals (keeping in mind that $\infty$ is not a real number, and it's role in algebraic operations can only be defined in terms of limits, which inevitably leads to ambiguity) then the answer to ...


1

To answer that question, you must first answer the question Is Infinity Even or Odd? Assume the lamp is initially on. After one minute, if you flip the switch an even number of times, the lamp will be on. However, if after that minute, you had flipped the switch an odd number of times, the lamp will be off. So if you flip the switch an infinite number of ...


4

Some infinite series converge, and some do not. In this case, because of the halving of the interval between each switching event, the total time for an infinite number of such events does converge, and converges to a small number (1 minute if you start with a 30 second gap). However the state of the lamp is represented by a different infinite series which ...


0

The lamp is ill-defined. You've given us a very thorough description of what happens before the minute is up. But you've not told us what happens next, so the best we can do is guess. We could imagine a lamp which is on after the minute ends, and we could equally well imagine a lamp which is off. Both possibilities are compatible with all of the ...



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