# Tag Info

-1

Partial solution: If there are only a finite number of hats of a certain color, then everyone would see that and should say the opposite color.

0

The $\mathcal{H}^{\infty}$ norm is specified from one vector input to one vector output. Right now, you have written things with two inputs. So you need to reorganize your $A,B,C,D$ matrices.

3

No, not necessarily. Consider $p=2$ and $$A=\begin{pmatrix} n & n+1 \cr n-1 & n \end{pmatrix},$$ for $n\to \infty$. Then we have $$A^{-1}=\begin{pmatrix} n & -n-1 \cr -n+1 & n \end{pmatrix},$$ but still the main diagonals go to $\infty$. Edit: OK, the question was changed. $A$ now has to be a symmetric, positive definite $p \times p$ ...

-1

Infinity is not a "real" number. It is not even a number. At best it is an abstract of numbers. It is a state. The equality sign, equating an entity to infinity, is merely idiomatic. A state is an abstract, that comprises various components of a state-machine. To fully describe a state, one must describe the events or processes that led to that state. ...

7

If that seems puzzling to you, this will surely blow your mind: $$\lim_{x \to \infty} \frac{1}{x} = \lim_{x \to \infty} \frac{1}{x^2} = 0$$ even though $\dfrac{1}{x} > \dfrac{1}{x^2}$ for all $x > 1$. Are there different sizes of $0$, too? Actually, there's nothing special about $\infty$ or $0$ in this respect. It's perfectly possible for two ...

1

There were various theories extending the real numbers to include infinitesimals throughout the period from 1870 until 1960, but Robinson was the first to introduce a system that can be used in analysis. Earlier systems were critized by Klein, Fraenkel, and others on the grounds that they were not proven to satisfy the mean value theorem for, e.g., ...

-2

The OP asked: Why don't we explicitly define infinity so that we can show differences in sizes? The answer is, indeed we do. Ever since the pioneering work of E. Hewitt (1948), J. Los (1955), A. Robinson (1961) and others, today we have a number system including both infinitesimals and infinite numbers (of course, of different orders) suitable for ...

-3

You are losing information about a function if you take a limit or even a normal functional value* and only look at the resulting value. You can have 2 functions approaching the number 5 from 2 different sides. That doesn't make the functions equal. One function will always be bigger than 5 and the other smaller. saying $\underset{x->\infty}{lim} f(x) = ... 1 Functions may grow in quite complicated manner – it is not that simple as discerning between$x$and$x^2$. For example $$f(x) = \frac{x(x+1)}2 + \frac{x(x-1)}2\cos x$$ oscillates between$x$and$x^2$: for$x=2n\pi$with natural$n$, that is whenever$\cos x = 1$, we get$f(x)=x^2$, but for each$x=(2n+1)\pi$we have$\cos x=-1$and$f(x)=x$. The function ... -3 Im just going to add to current answers, not referring to the limits but to infinity size. There is a definition for infinity "size" - at least in set theory. For example, there are more real numbers than integers, that is obvious. However the set of positive integers and all integers is of the same size, as every integer can be mapped through a ... 1 Infinite limits and limits at infinity have standard ε-δ definitions using real numbers, and$\infty$is not a real number. It is possible to use the affinely-extended real line and extend our definitions for limits to cater for these, but they will still not satisfy the properties of limits, such as: ?$\lim_{x \to a} ( f(x) + g(x) ) = \lim_{x \to a} f(...

1

Big $O$ notation is a way of comparing asymptotic behaviours no matter the expression is convergent or not. Prime counting function $$\pi(x) = O \left(\frac{x}{\ln x} \right)$$ $n$-th prime number $$p_{n}=n \left[ \ln n+\ln \ln n-1+O\left( \frac{\ln \ln n}{\ln n} \right) \right]$$ Stirling Series $$n! = \sqrt{2\pi n}\left( \frac{n}{e}\right)^n \left[1 +\... 7 You misremember how limits interact with inequalities: Theorem: If f(x) < g(x) for all x sufficiently near a \lim_{x \to a} f(x) exists \lim_{x \to a} g(x) exists Then \lim_{x \to a} f(x) \leq \lim_{x \to a} g(x) Note that "less than" becomes "less than or equal to" after taking the limits; the point being ... 8 The notation \lim _{x\to \infty}f(x)=\infty is just an abbreviation for$$\forall y\in \mathbb R\;\exists z\in \mathbb R\;\forall x\in \mathbb R\; (x>z\implies f(x)>y).$$Literally.Nothing more. It does not assume the existence of a "number" \infty. It is useful, in pure math, as well as in applied math, to intuit x as an object in motion, ... 49 The notation \displaystyle \lim_{x\rightarrow\infty}f(x)=\infty where f is any real-valued function of a real variable means exactly that as x gets arbitrarily large, so does f(x). That's all it means. This usage has no relation to any metaphysical ideas about infinity. It's also the case that in the mathematical field of set theory, there is an ... 31 I like this question a lot. The pedantic answer (and the right answer on an exam) is, of course, "They are the same. As limits, they are both equal to the formal symbol "+\infty"." That's pretty boring, though. There is a clear intuition that the behavior of x^2 at infinity is "bigger" than the behavior of x; the challenge is to try to figure out how ... 2 The fact that x^2 > x eventually should not necessarily mean that we could have defined different kind of infinities i.e..$$\lim _{x\rightarrow \infty }x^2> \lim _{x\rightarrow \infty} x$$This is intuitive since when restricting taking limit whose limits point are real numbers does not suggest such a property. That is even if f(x)>g(x), x<... 2 Infinity is not quite a number. If you must make it one, you have to allow for \bar {\Bbb R}, the extended real line. But even in the sense of \bar{\Bbb R}, the "number" +\infty is a single point, so the second infinity is no different than the first one because they are just the same point. The gist here, however, is not to compare the two resulting ... 1 It seems to me that you have more of a misunderstanding regarding the limit concept rather than anything having to do with infinity. One can have two functions f,g such that f(x)>g(x) for all x but \lim_{x \to \infty} f(x) = \lim_{x \to \infty} g(x) = L. That is, both f and g tend to the same limit. Take, for example f,g:\mathbb{R}_{\geq ... 2 When we say "infinite" we don't mean some specific number. It is more along the lines of, "without bound." Or, more formally, "pick a number, bigger than that." x= \infty \implies \forall M>0, x>M 2 Well first of all it is because \infty is not a number, you can't do what you usually do with numbers but let's suppose here that it is, let's suppose \infty is a number, the bigest number. Then you can imagine a finite constant k added to \infty which has to be \infty also i.e. k+\infty=\infty. That then implies that$$\infty-\infty=k+\infty-\...

0

the problem is that when we define point we consider point to be dimensionless but line is comprised of points , so it should be dimensionless as well but that's not the case . To go around the problem consider an $\epsilon$ value that is the minimum distance between two points , now you will realize for that $\epsilon$ value in $DE$ the distance between ...

0

Take this example. Find any point between $(1,\infty)$ , invert it and you will get a point in $(0,1)$. Since there are infinitely many points you can choose , according to you it must mean $\infty=\infty$ i.e. $(0,1)=(1,\infty)$?

1

Infinity was a number to both Kepler and Leibniz who spoke of a circle as an infinite-sided polygon. This point of view is useful in analyzing the properties of the circle, as well as more general curves, in infinitesimal calculus. Today we have rigorous mathematical frameworks incorporating both infinite and infinitesimal numbers, vindicating Leibniz's ...

0

The formula for a n sided regular polygon that inscribes a circle of radius 1 is 2n*sin(pi/2n). We can see this by drawing lines that connect the center of the polygon to the corners. This creates n isosceles triangles with side lengths 1 and the central angle is pi/n. Breaking each of those triangles in half and we see that half of each side length is sin(...

0

There are multiple interpretations to this question. One scenario that I can think of is that the question asks to prove 'unit' distance between one black dot and one white dot. Note the term 'one unit'. Now a unit is not defined as such. You can define a metre, a centimetre, etc. But a unit can be anything. It may be equal to one fermi-metre, or 56 ...

-2

$$e^{-k} = \dfrac{1}{e^{k}}$$ Also $e^{\infty} = \infty$ and $e^{-k}= \dfrac{1}{e^k} \Rightarrow e^{-\infty} = \dfrac{1}{\infty}= 0$ (anything divided by infinity is zero)

2

The topology $\mathcal{T}_{\overline{\mathbb{R}}}$ you are looking for is defined in the following way: $A\in\mathcal{T}_{\overline{\mathbb{R}}}$ iff $A \cap \mathbb{R}$ is open in the standard topology of $\mathbb{R}$; if $-\infty \in A$ then there is $r\in \mathbb{R}$ such that $[-\infty, r]\subseteq A$, and if $+\infty \in A$ then there is $s\in \... 0 Your definition of "limit point" is wrong. The expression "a point whose neighbourbood" makes no sense, since a point generally has more than one neighbourhood. Also, the point contained in the neighbourhood may not be the point itself. Rather, a limit point of a set is a point$x$each of whose neighbourhoods contains a point in the set other than$x$. It'... 1 The function you have is$f:\mathbb{R} \setminus \{-1\} \to \mathbb{R}$,$x \mapsto x-1$. Every$c \in \mathbb{R} \setminus \{-1\}$can be enclosed inside an open interval where$f$does not have a hole (and is hence locally a polynomial) and so$f'(c), f''(c), f^{(n)}(c)$exist and are continuous for all$n$. 1 Note that$x^2-1=(x-1)(x+1)$and hence the point$x=-1$is a removable discontinuity of$f$. Therefore, if you define the derivative with, say, the central difference $$f^{\prime}(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x-h)}{2h},$$$f^\prime(-1)$is well-defined (and so too are all subsequent derivatives). Having said that, this approach is messy, and it ... 2 The way this is handled in calculus courses is that you have to show that the sequence is, say monotone increasing and bounded from above. Then applying the least upper bound property of the real numbers you can conclude that the sequence converges. Once you know it converges you can apply the argument you gave. However, checking boundedness and ... 1 The existing comments and answer don't seem to mention the pertinent fact that in complex analysis it is indeed correct to assert that$\frac{1}{0}=\infty$because the complex line$\mathbb{C}$is completed to the Riemann sphere$\mathbb{C}\cup \{ \infty \}$by means of adding a single point at infinity, which is the reciprocal of$0$. -1 The simplest answer to your question, without either sequences or limits, is that there exist distinct infinite numbers. The most commonly used extension of the real numbers to an ordered field is the field of hyperreals. If$H$is an infinite hyperreal, then so is$2H$or for example$H^2$or$\sqrt{H}$, not to speak of$H+1$:-) 0 Let$g: \Bbb N \to X$be a bijection. If$\sum_{x \in X} f(x)$is absolutely convergent, then$\sum_{n \in \Bbb N} |f(g(n))|$is convergent, hence: $$\infty > \sum_{n=0}^{\infty} |f(g(n))| = : L$$ Given$A \subset X$a finite set, we have: $$\sum_{x \in A} |f(x)| = \sum_{n \in g^{-1}(A)} |f(g(n))| \le \sum_{n \in \Bbb N} |f(g(n))| = L$$ Hence $$\sup\... 1 Infinite often is not an intuitive concept, but when a function goes to infinity its important to consider how it quickly grow. For example if you take f(x)=3^x and g(x)=x when x \to +\infty, \space both functions go to infinity but f is "faster" than g so in this case 3^x-x=f_1(x)\to +\infty when x\to +\infty. 2 if we have$$\lim_{n\to \infty}n^3=\infty$$and$$\lim_{n \to \infty}n^2=\infty$$then the difference goes to$$\infty$$4 Hint. Two examples. Observe that, as x \to \infty,$$ (x^2+x)-x^2=x \to \infty $$whereas, as x \to \infty,$$ \left(x+\frac1x\right)-x=\frac1x \to 0. $$In each case you have an indeterminate form \infty-\infty. 1 The form \infty-\infty is what is known as an indeterminate form. This means as limits are concerned, the limit \infty-\infty does not have a value. 1 I guess it's an exercise meant to show how things might go wrong without being careful doing algebraic manipulations. Since they're evaluating the limit at -\infty, it's not restrictive to assume x<0. Divide numerator and denominator by -x:$$ \frac{x+1}{\sqrt{x^2}}= \frac{-1-\dfrac{1}{x}}{\dfrac{\sqrt{x^2}}{-x}} $$Since x<0, we have -x>0... 3 The solution in the book multiplies (x+1)/\sqrt{x^2} by (-1/x)/(-1/x)\ (=1). Since x\lt 0, the denominator will be$$\sqrt{x^2}\times\left(-\frac 1x\right)=\sqrt{x^2}\times\frac{1}{-x}=\sqrt{x^2}\times \frac{1}{\sqrt{(-x)^2}}=\sqrt{\frac{x^2}{x^2}}$$0 The reasoning seems to be, for x<0,$$ \sqrt{x^2}\cdot(-1/x)=\sqrt{x^2}\cdot(1/|x|)=\sqrt{x^2}\cdot(1/\sqrt{x^2})=\sqrt{x^2/x^2}. $$1 The comment made by YoTengoUnLCD is not exactly correct. From the definition of limits,$$ \lim_{x \to \inf} \frac{1}{x}=0 $$means that, for each \epsilon >0, there exists N such that$$ \lvert\frac{1}{x}\rvert<\epsilon $$whenever x>N. (Although, in practical terms, it would be wise to treat infinity vaguely as a really big number in many ... 0 I don't think we learn anything by trying to compare the \infty of the extended real line to the cardinal number \aleph_0. But if you're interested in extending the real numbers into the realm of the transfinite, you may be interested in the surreal numbers. In the world of the surreals, weird-looking formulae like$$\frac{1}{1+\infty^2}$\$ actually make ...

Top 50 recent answers are included