# Tag Info

## New answers tagged infinity

0

observe that $|B_n|=2^n$ so $|B_n|\to 2^\omega=c$ as $n \to \infty$ but this is not true about $A$. and observe that $B$ is not union of $B_n$.

2

$$\int^∞_{-∞}\cos(\pi t) \,dt= \int^∞_{0}\cos(\pi t)\,dt + \int^0_{-∞}\cos(\pi t)\,dt$$ Indeed, you are indeed correct that $\int^0_{-\infty}cos(\pi t)\,dt$ diverges. And as you stated at the start, the integral is convergent if and only if split integrals are both convergent. Since you showed one of the two integrals in divergent, the entire ...

0

A function isn't just an expression, but you can think whether a single expression can be applied to an argument. The expression $0^{-1}$ is rather meaningless, so you don't know how to get the behavior of the function $f(x)=0\cdot x^{-1}$ at $x=0$ from the expression. Limits are just a way to describe the behavior (if it looks consistent enough that the ...

4

Note that writing $f(x) = \frac0x$ results in $f(0) = \frac00$ wich is undefined. However, the singularity of $f$ is nice in the way that is can be continuously defined by $f(0) := 0$ (note the colon for defining the value). A limit is exactly this concept: What is the value of $f(x)$ when $x$ comes arbitrarily close to $0$, but not equal to $0$. The ...

2

You are in fact considering function $f: \mathbb R \setminus \{0\} \mapsto \mathbb R$ that is defined $f(x) = \frac 0x$ so it i equal $0$ on all its domain. Lets look at limit $\lim \limits_{x \rightarrow 0^+} f(x)$. Function is identically equal $0$ on every open interval $(0,\varepsilon)$ for $\varepsilon>0$. Hence right side limit is equal $0$. By ...

1

The reason is if you look at the function $f(x,y)=y/x$ then when you approach $(0,0)$ this should be invariant to the path you take in your case you walk along the path where $y=0$ (constant $0$) but when approaching over the path $y=x$ then it is constant $1$. So we say it is not defined.

7

The limits are not about a value at a point, but about the values approaching that point. I.e. why is the function $\frac0x = f(x)$, not undefined at $x=0$? It is undefined at that point. However, its "neighbourhood" is defined, and that's what the limit gives you. Thus $\lim_{x\to0}\frac0x=0$ means that $x$ approaching $0$ from both sides results in ...

0

When we deal with limits we first see the result of the part after the sign $\lim$. In this case we have $\frac{0}{x}=0$. Than we have $\lim_{x\to0}{\frac{0}{x}}=\lim_{x\to0}0=0$.

3

The probability of getting all tails is $$\prod_{k=1}^\infty\left(1-\frac1{2^k}\right)\tag{1}$$ For $0\le x\le\frac12$, we have that $$\log(1-x)\ge-2\log(2)\,x\tag{2}$$ Therefore, since $$\sum_{k=1}^\infty\frac1{2^k}=1\lt\infty\tag{3}$$ we have $$\prod_{k=1}^\infty\left(1-\frac1{2^k}\right)\ge\frac14\tag{4}$$ Thus, the probability of getting a head ...

4

The probability of not getting a head on the first toss is $\frac{1}{2}$, the probability of not getting a head on the second toss is $\frac{3}{4}$, the probability of not getting a head on the third toss is $\frac{7}{8}$, and so on. So the probability of not getting a head on the first $n$ tosses is $$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{7}{8}\cdots \cdot ... 1 We can show that the expected number of heads is 1 by saying that each flip is an independent Bernoulli trial and we can write that the PDF is p(x)=\frac{1}{2^k}, therefore the Expected Value E(x) (or number of heads you expect to see) in n flips is \sum\limits_{k=1}^n \frac{1}{2^k}, which means after an infinite number of flips we can say E(x) = ... 3 The law of large numbers does not apply, because the tosses are not identically distributed. The law only applies to many repetitions of the same experiment, or samples from the same distribution – but you sample from a different distribution every time. The total probability that you ever toss a head is just the complement of tossing tails forever. ... 0 Let R be a local ring. If M is faithfully flat over R, then \text{depth}_R(M) = \text{depth}(R), since for any R-ideal I and any x \in R,$$0 \to R/I \xrightarrow{x} R/I \to R/(I, x) \to 0$$is exact iff$$0 \to M/IM \xrightarrow{x} M/IM \to M/(I, x)M \to 0$$is exact, so a sequence of elements x_1, \ldots, x_n is regular on R iff it ... 1 Between two points A and B on the circle one can draw an arc, and at the midpoint of the arc there is a point C. Then there is a point at the midpoint between A and C, and another at the midpoint between B and C, and so on . . . 1 This may be very counterintuitive, but if you accept the axiom of choice, it is possible to design a method that predicts the future with probability 1. See this amazing paper. However, it's non-constructive, so although mathematically one may do this, you can't program a computer to do so. -1 I don't know why cardinality even exists but here is how it goes. Positive integers are the smallest of all the sets. Add 0 and you get a slightly larger infinite set. Add negative integer and you get an infinitely larger infinite set. These are all countable. There are however infinite rational numbers between rationals and so on and so forth. They are ... 1 Note that for any n,$$ \frac{f(y)}{g(y)} = \frac{y^n}{y^{n+1}} = \frac{1}{y} $$Now fix \epsilon > 0. Choose N = 1. For n > N, we have$$ \left| \frac{f_{(n)}(y)}{g_{(n)}(y)} - \frac{1}{y} \right| = \left| \frac{1}{y} - \frac{1}{y} \right| = 0 < \epsilon. $$Q. E. D. 0 As \frac{f(y)}{g(y)} = \frac{1}{y} and does not depend on n, then this limit equals exactly \frac{1}{y}. Using the epsilon-delta definition of limit you can write \forall \varepsilon > 0 \; \exists n_0 = 1: \forall n > n_0 \; \left|\frac{f(y)}{g(y)} - \frac{1}{y}\right| < \varepsilon, because \left|\frac{f(y)}{g(y)} - \frac{1}{y}\right| = ... 0 For 0<y<1, it must be L=1/y 2$$\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x}=\frac{(5^x-1)(2^x-1)}{x\tan x}=\frac{\dfrac{5^x-1}x\cdot\dfrac{2^x-1}x\cdot\cos x}{\dfrac{\sin x}x}$$Now as \displaystyle a^h=\left(e^{\ln a}\right)^h=e^{h\cdot\ln a}, \displaystyle\lim_{h\to0}\frac{a^h-1}h=\ln a\cdot\lim_{h\to0}\frac{e^{h\ln a}-1}{h\ln a}=\ln a 1 \infty, as you said in the last paragraph of your question, is not a number (natural or real), so the usual arithmetical operations on numbers are not defined on it. This is why \frac{\infty}{\infty} is not equal to 1. 5 Through your thought experiments, you and your friend have discovered the reason that \infty/\infty cannot be assigned a consistent meaning. The "answer" you get, if any, will depend completely on the relative rate at which the numerator and denominator grow, as you have seen. 9 The following limits all have the indeterminate form of \frac{\infty}{\infty}, but they are not all 1.$$\lim \limits_{x \to \infty} \frac{x^2}{x}\lim \limits_{x \to \infty} \frac{x}{x^2}\lim \limits_{x \to \infty} \frac{x}{x}$$However, if you are given \frac{\infty}{\infty} without context, the value is indeterminate. Furthermore, note ... 3 If we replace in the expression -1 by 1 we have in this case the indeterminate form \frac 00 and we calculate the limit using the Taylor series:$$\frac{10^x - 2^x - 5 ^ x + 1 } {x\tan x}=\frac{\exp(x\log10) -\exp(x\log2) - \exp(x\log5) + 1 } {x\tan ...

4

By Faulhaber's formula the sum of squares up to $k$ is $\frac 16k(k+1)(2k+1)$ and you can write a $30th$ degree polynomial for the sum of $29th$ powers. You can also work modulo 10 to see when the zeros appear, or modulo 1000 to see when there are three zeros at the end. Note that $i^{29} \equiv i \pmod {10}$ so to find single zeros at the end you want ...

2

Hint: Let $f: \Bbb N \rightarrow \Bbb N \cup \{a\}$ be defined thus. $$f(1) = a \; \text{and } \; f(x) = x - 1 \; \text{for every member in \Bbb N \setminus \{1\}}$$ You can show that this is a bijection.

3

Presumably the monkey only chooses countably many numbers. You can adapt the proof that the measure of $\Bbb Q$ is zero to show that the chance any given real is chosen is still $0$. Choosing from an unbounded interval doesn't change anything. To show that, take your favorite bijection between $(0,1)$ and $\Bbb R$ Specifically for your limit, you need ...

0

Hint All the arguments of the trigonometric functions are very small when $n$ is large. So, taking into account what you are supposed to use, rewrite you expression replacing $\csc(x)$ by $1/ \sin(x)$ and $\cot(x)$ by $1/ \tan(x)$ and, taking into account the limits you know, replace $\sin(x)$ by $x$ and $\tan(x)$ by $x$ with the proper corresponding ...

0

No, for all positive n, case 2 is the constant 1/y. Why the doubt that you cannot "manipulate the exponents in the same way"? For all n, $\frac{y^n}{y^{n+1}} = \frac{1}{y}$. To convince yourself of this, plug in say, 2 for n, then plug in 3, then plug in 4 etc. But you could just as easily write $\frac{y^n}{y^{n+1}} = \frac{1}{y^{(n + 1) - n}} = ... 1 Looks good. For case 1, the denominator grows faster and thus the limit goes to 0. The second case is wrong however. It can be simplified like this: $$\lim_{n \to \infty} \frac{y^n}{y^{n+1}} = \lim_{n \to \infty} \frac{1}{y} = \frac{1}{y}$$ In other words, you can use the same method as for case 1. 1 Hint: Let$A_n$be the number of$n-$permutations with an odd number of$a$. Let$B_n$be the number of$n-$permuations with an even number of$ba$. Hint: Establish some recurrence equations with a combinatorial argument: Roll over if you want to see what these are Hint: Hence show that$A_{n+1} = 6A_n - 8A_{n-1}$. Solve it as a linear recurrence. with ... 1 One negative point is that the use of$\pm\infty$with$+$and$\cdot$is awkward to define. In textbooks, a list of relations like$\infty+\infty=\infty$or$1/\infty=0$is given, but usually this list is not exhaustive and it's up to the reader to define the rest himself. When definitions are left to the reader, it always leaves a nagging uneasiness (at ... 3 I'm not sure if explaining why the extended reals are natural is an appropriate answer to this question, but just in case: As Baby Dragon and Bill Dubuque have pointed out, various notions of "compactifications" obtained by adding "points at infinity" are truly ubiquitous in mathematics, especially in geometry. The entire notion of projective geometry ... -2 The extended reals are not a particularly natural way to think about infinity in the context of calculus. Historically, calculus was developed as the study of infinitesimals, and in some approaches (such as non-standard analysis, NSA), these infinitesimals are invertible, and their inverses are infinite. Other reasons to dislike the extended reals are that ... 4 You have the special set$\mathbb{R}$. Addition and multiplication are defined and you have a total order on it. So, for all$a, b \in \mathbb{R}\,(a+b)$,$ab$in$\mathbb{R}$too, and it is true that$a\leq b$or$b\leq a$is true. Now you define$\overline{\mathbb{R}}=\mathbb{R}\cup\{-\infty\}\cup\{+\infty\}$(I will distinguish$+\infty$and ... 3 One can pinpoint the reason for such an antagonism to the notion of an infinite number (as opposed to cardinality) rather precisely due to the able work of the historian Joseph Dauben. Dauben wrote as follows: Cantor devoted some of his most vituperative correspondence, as well as a portion of the Beiträge, to attacking what he described at one point ... 20 The extended reals are a way of thinking, but hardly the only way. It is normal to think of a line as having no endpoints, rather than as ends that meet. The reason you see antagonism is that this site is frequented by learners of mathematics, that often have very muddled ideas about infinity. Introducing the extended reals and allowing$\infty$to be a ... 0 It is unknown whether the series:$\sum_n \frac{(-1)^n n}{p_n}$converges. Here,$p_n$is the$n$-th prime number. This problem is posed in Guy's book on unsolved problems in number theory and I am pretty sure that it originated from Erdos. 1 The former does imply the latter: inversion is continuous everywhere except$0$. Thus, the limit of the inverse is the inverse of the limit: $$\lim f(x) = \frac{1}{\lim 1/f(x)} = \frac{1}{+\infty} = 0$$ Inversion is not defined at$0$, however, so we cannot hope to argue in the other direction: $$\lim \frac{1}{f(x)} ?= \frac{1}{\lim f(x)} ?= ... 1 You can define a number like this one if you consider hyperreal numbers: and it would be written n + \epsilon, which is greater than n but less than any other number greater than n. Indeed, if you could define n.00000...[infinite zeros]...1 , then you should be able to define n.00000...[infinite zeros]...2 and many other numbers. This 0.000[infinite ... -1 This is not possible because the set of |R is uncountable, which means that there is no defined next element. This has nothing to do with ininity itself , e.g the set of |N (all positive Integers) is infinite set of /Z (all Integers) is infinite, too. But its clear it must have more elements. The set of all rational numbers is clearly infinite and must ... 2 It looks as though you're trying to extend an algebraic structure. Would you like Clippy to help you with that? ;-) why is c necessarily divisible? There are special exceptions for other numbers. There is one special exception in the real numbers, which is that you can't divide by 0. Whether you define the reals axiomatically or via a construction from ... -1 Just look at this in the first place: the reason why probably, .9999... = 1, no rounding! {Perhaps one will argue about '0.9999...' being just a teensy amount less than 1. Well, but isn’t '.9999…' just 3 times '.3333…'? Everyone will agree with this one. And isn’t '.3333…' the same as 1/3? Yes, of course. And isn’t 3 times 1/3 equal to 1?} Back to our ... 1 Hint As you correctly found, the antiderivative of \sec ^2(x) is \tan (x). If the bounds of integration are 0 and a, the value of the integral is \tan (a) which means that the results approached infinity when a approached \pi/2. I am sure that you can take from here. 0 sec^{2}(x) = \frac{1}{cos^2(x)} As x goes from 0 to \frac{\pi}{2} what happends? Well, think about this: cos(\frac{\pi}{2})=0. As we approach \frac{\pi}{2} from either side, we have \frac{1}{cos^2(x)} \rightarrow \infty. Then, if you think of the integral as measuring the area under the curve, you see why this integral goes to \infty. . 0 You can't really take the integral on [0,\pi/2] since \sec^2x=1/\cos^2 x is discontinuous at \pi/2. So what we really want is$$ \lim_{\theta\to\pi/2^-}\int_0^{\theta}\sec^2x\,dx+\lim_{\psi\to\pi/2^+}\int_{\psi}^\pi\sec^2x\,dx. $$We have to split the integral up around the singularity. In this case, we have$$ ... 3 It isn't necessary that$0. \overline 01$describes a process for constructing a real number in countable steps. The real numbers that can be so defined are a subset with measure zero. If we allow$0. \overline 01$to be notation for a nonzero real number, there is indeed a nonzero real number exactly half that, which might be denoted$0. \overline 005$... 1$\dfrac1{x^n}\to0$as$x\to\infty$, for positive n. Hence, if the primitive were positive, then, by subtracting something positive from$0$, you'd really get a negative number, and then things would really make no sense, since then you'd really have a sum of positive quantities yielding a negative result! But, by being negative, we have$-(-a)=+a>0$for ... 0$ \lim_{z\to\infty}1/f(z)=\infty $means that, for any$M>0$, there exists$N\in\mathbb N$so that, when$z>N$,$1/f(z)>M$. Therefore$f(z)<1/M$for all$z>N$. That means that, as$z\to\infty$, we can always shrink$f(z)$as small as possible, so yes,$\lim_{z\to\infty}f(z)=0$. I'm assuming here that$z\in\mathbb R\$. Something needs tweaking ...

0

A simple proof that may or may not work: .999... = 1 .999... + c = 1 c = 0 n + c = n + 0 n + c = n Just an idea that is, at the very least, easy to understand.

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