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4

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 ... 1 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 ... 5 This is known as Thomson's lamp. The Wikipedia article is probably less enlightening than the answers given here, and I don't think that there's much more to be said about it, but the search term will lead you to more articles if you want them. Added: Your reasoning is correct. By definition, the lamp must be in one of two states: on or off. This state is ... 2 Consider the sequence a_n=1-2^{-n}. The situation that you describe could be mathematically described this way: For each x\in[0,1) there exists some unique n\in\Bbb N such that x\in[a_n,a_{n+1}). Define f(x)=1 if this n is even and f(x)=0 otherwise. As you can see, the value f(1) remains undefined. So, assuming that the job of toggling ... 34 The apparent paradox in this question arises from the conflict between the following two ideas: Our intuition about physical lamps. We expect them to have certain properties, such as being on or off, but not both or neither. A complicated function that oscillates between two values at an increasing pace, whose oscillations become arbitrarily rapid. It ... 1 You have to be really careful when making an argument that has a process continuing to infinity. You have to decide exactly what that means, and how to interpret the result. For example, this question shows a "proof" that \pi=4 based performing this type of process. The mistake is that the argument does not rigorously prove that what they are showing is ... 3 The problem is that any argument involving infinity needs to be incredibly precise. (Well, all math needs to be precise, that's sort of the point; but in particular, when talking about infinity, naive intuition is dangerous.) Unfortunately, everything you've written is incredibly vague. Basically, limits are complicated - it took a long time for calculus, ... 0 In the world of natural numbers it is known that 2 ^ a \neq 3 ^ b for any pair of positive integers a and b. This is true for any pair of primes. So if we believe there is only one infinitely large natural number (\infty) From the above statement: 2 ^ \infty \neq 3 ^ \infty Let 2 ^ \infty  be \infty_{2} Let 3 ^ \infty  be \infty_{3} Then ... 0 I think it is praiseworthy that you are interested in these infinite sums and are thinking of ways to combine them in order derive new results. However there is a problem: all these sums are not strictly convergent, but exist only if one extends a correct result to a case where they are not really properly defined. Unfortunately this means that the ... -1 There is no official way to add these divergent alternating series, but \sum (-1)^n is known as Grandi's series:$$ \sum_{n=0}^\infty (-1)^n = \begin{cases} 1 & \text{ if }n\text{ is odd} \\ 0 & \text{ if }n\text{ is even} \\ \end{cases} \approx \frac{1}{2}$$And for your second series (1 - 1 + 1 - 1 + \dots)^2 = 1 - 2 + 3 - 4 + \dots = ... 1 As the other answers have said, your reasoning is very off: these are divergent sequences, and don't behave the way you think they do. On the other hand, a reasonable question to ask is: "Can we make sense of these divergent series?" The answer is yes, although it is not easy: there are several methods for summing (some) classically divergent series. See ... 3 Every claim you started with is wrong. Your S_{1}, S_{2}, S_{3} are all divergent series. It looks like you declared these series to be equal to their Ramanujan summation assignment, (which is akin to setting a matrix equal to its own determinant), and then proceeded as if you actually meant it converged to that value. 1 You start with wrong conditions. For example, the infinite sum \sum_n^{\infty} (-1)^n  is not equal to 1/2. 0 Your blurb is pointing out the difference between \infty and practically every other symbol used in math to denote a number, such as in this case with the variable a. However large you define the value of a to be, it is understood to be a finite, unique number. It follows that for any a, there is a finite, unique number a+1. This is true in the ... 0 Start with the set \mathbb{R}. And adjoin two new elements to it, \mathbb{R}\cup \{ +\infty,-\infty\}. Denote this resulting set as \overline{ \mathbb{R}}. On this new set we do not define any algebra, i.e. we do not define what it means to add/multiply across infinities (because it is not possible to preserve all the rules we would like). However, we ... 1 In the sense that Spivak discusses (i.e. the way it is generally used in calculus), "\infty" is really a sort of abbreviation to allow you to write certain limits and sets in a way consistent with other sets. Here are a couple of examples: The interval (a,\infty) is the set of real numbers that are larger than a. This is by analogy with the interval ... 3 You could put it this way: If we write$$\begin{align}\lim_{x\to a}f_1(x)&=b\\ \lim_{x\to c}f_2(x)&=\infty\\ \lim_{x\to \infty}f_3(x)&=d\\ \end{align}$$then there is a qualitative difference between these three lines. That is, the second is not just the same as the first with c in place of a and f_2 in place of f_1 and \infty in ... 4 In short, the symbol has separate but analogous meanings in different contexts. While we often use it in place of a real number in notation, careful usage will not treat it as a real number---consider, for example, the notation \lim_{x \to \infty} f(x). To expand orthogonally to Simon S' excellent answer (and perhaps besides the cursory paragraph above, ... -4 You may have heard the adage, "infinity is not a number, it's a concept" (or at least you have now). This is in fact true, since \infty is not a real number (if it were, terrible things would happen). Infinity - \infty - is the object such that no number is greater than it:$$ \forall x\in \mathbb R : |\infty| > |x|$$We use this object to talk ... 7 \infty and -\infty are not real numbers. So when ever we use them in real analysis we need to define their use carefully, as he has done in the extract you quote. For instance the interval (a,b) is defined as \{ x \in \mathbb R \ : \ a < x < b \} whenever a, b are both real numbers. This definition is in trouble if the symbols a or b ... 0 We write P(-1^+) as$$\begin{align} P(-1^+)&=\arctan(1/\sqrt{2})-\pi/2+\sum_{n=2}^{\infty}\left(\arctan\left(\frac{1}{\sqrt{n+1}}\right)-\arctan\left(\frac{1}{\sqrt{n-1}}\right)\right)\\\\ &=\sum_{n=1}^{\infty}\left(\arctan \sqrt{n-1}-\arctan \sqrt{n+1}\right)\\\\ &=\lim_{N\to \infty}\sum_{n=1}^{N}\left(\arctan \sqrt{n-1}-\arctan ...

2

$$\sum_{n\geq 1}\left(\arctan\frac{1}{\sqrt{n+1}}-\arctan\frac{1}{\sqrt{n-1}}\right)=-\frac{\pi}{2}-\frac{\pi}{4}=\color{red}{-\frac{3\pi}{4}}$$ since the LHS is a telescopic series. Here I assumed $\arctan\frac{1}{\sqrt{0}}=\lim_{x\to 0^+}\arctan\frac{1}{\sqrt{x}}=\frac{\pi}{2}$. We are allowed to switch the limit and the integral since for any fixed $x$ ...

1

A (positive) real number is rational if it is the ratio of two whole numbers. When a number is presented as decimal expansion, a theorem states that, a number is rational if and only if, after some initial hiccups, the decimals start repeating in a cyclic manner. So this theorem gives handle: The square roots, namely $\sqrt2,\sqrt3, \sqrt5,\ldots$ ...

0

An easy way to "Generate" long strings of numbers is to just roll a 10 sided dice (with numbers 0-9) and put the number which comes up in the current possition. So if you roll four times you may get a $3, 7, 6$ and a $3$ in which case you will get the number 3.763 . Assymptotically i.e. when this process goes to infinity, you will have probability one of ...

22

The Champernowne constant $0.123456789101112131415\dots$ is guaranteed to have every finite string of digits represented, and asymptotically in the proper proportion. We do not know if $\pi$ satisfies that. It is also easy to compute any given digit. See this post for how.

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$0.123456789101112131415...$ is the one such number. Given any string of $n$ digits, the density of occurrences of the string is the ideal $1/10^n$, I believe.

3

Don't use L'Hospital's rule. It won't work here, and when it works, it is equivalent with Taylor's polynomial at order $1$, which is much less error-prone. It is a problem of Asymptotic analysis: set $u=\dfrac1x$. Then $$\lim_{x\to 0^+}\frac{\mathrm e^{-\tfrac1x}}{x^2}=\lim_{u\to+\infty}\frac{u^2}{e^u}=0$$ since a basic result in Asymptotic analysis is ...

6

You can do without the application of the L'Hospitals rule. Hint : $\frac{e^{-\frac{1}{x}}}{x^{2}}=\frac{1}{x^{2}e^{\frac{1}{x}}}$

6

Let $y = \dfrac{1}{x}$, then $x = \dfrac{1}{y} \Rightarrow L = \displaystyle \lim_{y \to +\infty} \dfrac{y^2}{e^y}= \displaystyle \lim_{y \to +\infty} \dfrac{2y}{e^y}=\displaystyle \lim_{y \to +\infty} \dfrac{2}{e^y}= 0$

1

I don't think this is true. For example, take $u_{k}=\frac{(-1)^{k}}{k^{3/4}}$. Of course, you can show that the series converges uniformly on subsets of $R^{+}$ of the form $[a,\infty )$ if you write $f(t) = \sum_{k=0}^\infty ke^{-t\sqrt{k}}u_k=f(t) = \sum_{k=0}^\infty \sqrt{k}u_k(\sqrt{k}e^{-t\sqrt{k}})$ and use the Weierstrass M-test.

1

Your intuition is correct but your calculation of entropy is not, the differential entropy of the uniform distribution on $(0,\frac{1}{2^n})$ is $$-2^n\int_0^{\frac{1}{2^n}}\log_22^ndx=-n$$ The concept of entropy for continuous random variables (differential entropy) is not well defined, since a real number requires a theoretically infinite number of bits ...

0

$\def\pp{\mathbb{P}}$The usual definition of conditional entropy of $A$ given $B$ works. It is defined as the expected conditional entropy, which is the expected weighted negative log conditional probability. This corresponds to your correct intuition since in this case the weight function is uniform since every value in $[0,1)$ is equally likely. You can ...

0

Assuming n is a continuous variable and you wish to take the limit $n \to \infty$ (please clarify your question), the behavior might be more easily seen: let $z \equiv (-1)^n = (e^{i \pi})^n = e^{i n \pi}$. This is not "divergent" in the sense that it goes to (complex) infinity, since $|z|$ remains unity, but does not converge to a single value. It could ...

2

The definition of a divergent sequence $a_n$ is that there exists $\varepsilon>0$ such that for all $x\in \mathbb R$, there is no $N$ such that $|a_n-x|<\varepsilon$ for all $n>N$. This is a formal way of saying that there is no point that the sequence stays close to after a long time. If $$a_n=(-1)^n=-1,1,-1,1,...,$$ then the only sensible point ...

2

The Mandelbrot set can't have infinite area, since the entire set is contained in the disk of radius $2$ centered at the origin.

3

The essential mistake here is the assumption that if you can establish a One-To-Many mapping from A to B, then you have proven that Cardinality(A) < Cardinality(B), or just in general, that they are not equal. While this happens to be true for finite sets, this most definitely is not true for sets of transfinite cardinality. Rather, you must rely on the ...

1

This doesn't explicitly answer your question, but may contribute to your understanding. What you have observed (essentially) is that there is a proper subset of the real numbers with the same cardinality of the whole set. Here "cardinality" means "matches using a one to one and onto function". The next step is to realize that's not a contradiction - it's ...

5

Your argument does not show a larger cardinality for $(1,\infty)$. You have shown that $\mathfrak {c=c}\cdot \aleph_0$, which is true

34

Your argument shows that the cardinality of $(1,\infty)$ equals the cardinality of $(0,1)$ times the cardinality of the positive integers. In other words, $$\mathfrak c=\mathfrak c\cdot\aleph_0$$ That is also known in other ways. The equation comes from the definition of the product of cardinal numbers. The source of the "paradox" is that an infinite set ...

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The problem is you do not have one mapping, you have a family of maps indexed by $\mathbb{N}$. In order to deduce something formally about cardinality, one must talk about a single map, not a bunch of maps. Also, each of your maps embeds $(0,1)$ into $(1,\infty)$, which only shows that the cardinality of $(0,1)$ is less than or equal to the cardinality of ...

0

Hint: Let $f(x)=\dfrac1x$ and $g(x)=\dfrac1{x+1}$ . Apply the former on $[1,\infty)$, and the latter on $[0,\infty)$.

0

Let $y \in W_F-\{0\}$, we have the embedding: $\phi \in Hom(V_F,F) \mapsto g_{\phi} \in Hom(V_F,W_F)$ where $g_{\phi}(x)=\phi(x)y$ for all $x \in V_F$. It's an embedding because $g_{\phi}=g_{\psi}$ implies $\forall x, \phi(x)y=\psi(x)y$. So, $\forall x \in V_F, (\phi(x)-\psi(x))y=0$. And $y \neq 0 \implies \phi(x)=\psi(x)$ for all $x$. So $\phi=\psi$. So, ...

1

The accepted answer is correct and answers your question perfectly. You might read later in your education the sense in which $A:=[0,1]$ is almost finite while $B:=[1,\infty)$ is not. This almost finiteness is known as compactness and sets that are compact are in a different way smaller than non-compact sets. For example, given any $\varepsilon>0$, you ...

0

Another alternative: There is an injection $[0,1] \to [1,\infty)$ given by $x \mapsto x + 1$. Hence the cardinalities $\|\,[0,1]\, \| \leq \|\,[1,\infty)\,\|$. And there is an injection $[1,\infty) \to [0,1]$ given by $x \mapsto 1/x$. Hence $\|\, [1,\infty)\, \| \leq \|\,[0,1]\,\|$. Therefore $$\| [0,1] \| = \|[1,\infty)\|$$

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