qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
780018 | bounding the sum of squares of lengths of a quadrilateral inscribed in a unit square
Consider this nice little problem: if [imath]ABCD[/imath] is a quadrilateral inscribed in a unit square, then [imath]2\leq AB^2+BC^2+CD^2+DA^2\leq4[/imath] (Evidently this is problem 1 on paper 1 of the 1989 Irish Mathematical Olympiad. I also found it in another nice collection of problems, a recent admissions exam for the Indian Statistical Institute.) Here's the straightforward algebraic proof I used: let [imath]x_1[/imath], [imath]x_2[/imath], [imath]x_3[/imath], [imath]x_4[/imath] denote legs of the right triangles with hypotenuses [imath]AB[/imath], [imath]BC[/imath], [imath]CD[/imath], [imath]DA[/imath], with no two of the [imath]x_i[/imath] lying on the same side of the square. Then the Pythagorean theorem gives [imath]AB^2+BC^2+CD^2+DA^2=\sum_{i=1}^4 x_i^2+(1-x_i)^2=\sum 1-2x_i(1-x_i)=4-2\sum x_i(1-x_i)[/imath] But [imath]0\leq x_i(1-x_i)\leq\frac{1}{4}[/imath] and the claim follows. (The lower bound of 2 is achieved when the vertices of [imath]ABCD[/imath] are the midpoints of the sides of the square, and the upper bound of 4 is achieved when [imath]ABCD[/imath] is the square itself.) I'm curious about other approaches to the problem. Perhaps the result falls out of a more high-powered theorem in geometry? Or does the problem essentially boil down to a nice application of clever quadratic algebra? | 2996633 | The [imath]1997[/imath] IIT JEE problem
Let [imath]S[/imath] be a square of unit area. Consider any quadrilateral whose [imath]4[/imath] vertices lie on each side square [imath]S[/imath]. Let the length of the sides of this quadrilateral be [imath]a,b,c,d[/imath]. Then prove that [imath]2 \leq a^2+b^2+c^2+d^2 \leq 4[/imath] This problem appeared in IIT JEE [imath]1997[/imath] (re-exam).I really do not know know how to approach this problem properly.I have managed to prove [imath]a^2 + b^2 + c^2 +d^2 \leq 4[/imath] but not the initial inequality. |
2655249 | What is the use of considering non-reduced schemes?
According to Liu's "Algebraic Geometry and Arithmetic Curves", Proposition 2.4.2, if [imath]X[/imath] is any scheme, there is a unique closed subscheme [imath]X_{red} \to X[/imath] with the same underlying topological space, and this satisfies the property that each morphism [imath]Y \to X[/imath] from a reduced scheme [imath]Y[/imath] factors uniquely through [imath]X_{red}[/imath]. Of course, it is also true that any morphism of schemes [imath]X \to Y[/imath] can be extended uniquely to a morphism [imath]X_{red} \to Y[/imath] by composition. Since "nice" schemes are usually required to be reduced, I wonder what the use of considering reduced schemes is. More in detail, I have the following questions: Is the purpose of having non-reduced schemes solely to make definitions clearer and more consistent? (For instance, if [imath]A[/imath] is a non-reduced ring, we would need to change the definition of [imath]\mathrm{Spec}A[/imath] to make it a reduced scheme) Is it true that one can, for all practical purposes, just assume that all schemes he's dealing with are reduced? This seems reasonable in view of the fact that morphisms to and from [imath]X[/imath] can be seen as morphisms to and from [imath]X_{red}[/imath]. But is it possible to have two different morphisms from [imath]X[/imath] to a reduced scheme [imath]Y[/imath] that become the same when we compose them with the "inclusion" [imath]X_{red} \to X[/imath]? Is this a major problem that hinders us from assuming that all schemes are reduced, or not? | 2311905 | What is the significance of allowing nilpotents in scheme theory?
I'm trying to learn the theory of algebraic groups using "modern" algebraic geometry, going past the traditional, very standard books by Borel, Springer, and Humphreys, all of which try to avoid the use of group schemes. One of the motivations for using schemes is that in the classical setting, "nilpotents are not allowed," a sentence I don't understand. For example, on page 18 in Milne's Algebraic Groups (which is still in draft form), he writes, "Since we allow nilpotents in the structure sheaf, the points of an algebraic group with coordinates in a field, even algebraically closed, do not convey much information about the group. Thus, it is natural to consider its points in a [imath]k[/imath]-algebra. Once we do that, the points capture all information about the algebraic group." This is clearly not meant to be totally precise. But I still wonder what he means. I'm thinking of the curves [imath]x^2=0[/imath] and [imath]x=0[/imath] as one example which is sometimes used to motivate the definition of scheme (the behavior at 0 is different in those two examples). Here are my questions, presumably related: By "nilpotents in the structure sheaf," does he mean nilpotents in the coordinate ring? What does "allowing nilpotents" mean, and what is wrong with not allowing nilpotents? Why doesn't allowing nilpotents force the points to "convey much information" about the group? What information does thinking of algebraic groups as functors capture? I should add that I'm mostly used to thinking of algebraic groups as functors. For example, [imath]SL_n[/imath] is thought of as a functor that takes a ring [imath]R[/imath] to the matrix group [imath]SL_n(R)[/imath]. (I can post this as more than one question if it would be better, but I don't know enough about the answer I'm looking for to know whether or not that is a good idea.) |
2996783 | Single point function is not a continuous function according to the intermediate value theorem
According to the intermediate value theorem, we can infer that a single point function, for example [imath]f(x) = \sqrt x + \sqrt(-x)[/imath], which has a range of only [imath]0[/imath] and a domain of only [imath]0[/imath] is not a continuous function. Is my conclusion right or there is actually exception when it is continuous? or outside of the calculus realm of thinking single point function CAN BE continuous? | 1095887 | Is a function defined at a single point continuous?
Is a function defined at a single point continuous? For example [imath]f:\{0\}\to\{0\}[/imath] defined by [imath]f(x)=\sqrt{x}+\sqrt{-x}[/imath] is a sum of two continuous functions and is therefore continuous, however for [imath]f[/imath] to be continuous, it has to have the limit [imath]\lim\limits_{x_0\to x}f(x_0)=f(x)[/imath] but it dosen't seem to have that limit. |
2996379 | For an infinite dimensional Banach space, [imath]X^*[/imath] when given the weak* topology is of the first category in itself
Let [imath]X[/imath] be an infinite dimensional Banach space. Why is [imath]X^*[/imath] of the first category in itself when given the weak* topology. Very closely related to [imath]X^*[/imath] with its weak*-topology is of the first category in itself, but I can't follow all the steps. The first step (hopefully) is to show that [imath]B_n = \{x^* : \lVert x^* \rVert \leq n\}[/imath] is nowhere dense, i.e. its closure is has an empty interior. Then [imath]X = \bigcup_n B_n[/imath], so [imath]X[/imath] is meagre. The answer claims that "It suffices to prove that [imath]\text{int}_{w∗}B_n=\emptyset[/imath]", but don't we have to show that [imath]\text{int}_{w∗}\overline{B_n}=\emptyset[/imath]? So I'm getting stuck trying to say anything about the closure of [imath]B_n[/imath]. | 153822 | [imath]X^*[/imath] with its weak*-topology is of the first category in itself
Let [imath]X[/imath] be an infinite-dimensional Fréchet space. Prove that [imath]X^*[/imath],with its weak*-topology is of the first category in itself. |
2997424 | Round table seating probability with 8 people
Mr. A, Mr. B and [imath]6[/imath] more people wants to sit around round table. We need to find out the probability that Mr. A and Mr. B will sit near each other. First of all we need to describe [imath](Ω,F,P)[/imath] I think n=(8−7)!=5040 than Mr. A and Mr. B we can keep as one because we want them to sit near each other. Tham m=(7−1)!∗2=1440 Than P=m/n=1440/5040=2/7 Is ir right? And how I should describe (Ω,F,P)? | 2997366 | Probability. About round table
Mr. A, Mr. B and 6 more people wants to sit around round table. We need to find out the probability that Mr. A and Mr. B will sit near each other. First of all we need to describe [imath](\Omega,F,P)[/imath] I think [imath]n= (8-7)!=5040[/imath] than Mr. A and Mr. B we can keep as one because we want them to sit near each other. Tham [imath]m=(7-1)!*2=1440[/imath] Than [imath]P=m/n=1440/5040=2/7[/imath] Is ir right? And how I should describe [imath](\Omega,F,P)[/imath]? |
2996903 | Proving [imath](1-a)(1-b)(1-c)(1-d) > 1-a-b-c-d[/imath]
Prove the following inequality for [imath]a,b,c,d \in (0,1)[/imath]: [imath](1-a)(1-b)(1-c)(1-d) > 1-a-b-c-d[/imath] I have a problem. I don't know if my idea is good [imath]a=b=c=d [/imath] [imath](1-a)^4 > 1- 4a [/imath] So, this is true.. | 536048 | Polynomial inequality proof
Prove [imath](1-a)(1-b)(1-c)(1-d)>1-a-b-c-d[/imath] and [imath] a,b,c,d[/imath] are real and between 0 and 1. I can do this with [imath](1-a)(1-b)>1-a-b \\ 1-a-b+ab>1-a-b \\ ab>0 [/imath] But with [imath]c[/imath] and [imath]d[/imath], this becomes much more difficult. Can someone nudge me in the right direction? |
2998406 | Example of E(E(X|F)|G) \neq E(E(X|G)|F)
Can you find an example where E(E(X|F)|G) [imath]\neq[/imath] E(E(X|G)|F) (F and G is [imath]\sigma[/imath]-field in probability theory) | 1565688 | An example s.t. [imath]E(E(X|\mathcal F_1)|\mathcal F_2)\neq E(E(X|\mathcal F_2)|\mathcal F_1)[/imath]
If [imath]\Omega=\{a,b,c\}[/imath] give an example that [imath]E(E(X|\mathcal F_1)|\mathcal F_2)\neq E(E(X|\mathcal F_2)|\mathcal F_1)[/imath] If I choose the [imath]\sigma[/imath]-algebras disjoint (or better said: one is not fully contained in the other) is the inequality always satisfied ? |
2997184 | Computing joint density
Suppose [imath]X[/imath] and [imath]Y[/imath] are independent and identically distributed uniform r.v on [imath](0,1)[/imath]. Compute the joint density of [imath]U=X+Y[/imath] and [imath]V = \dfrac{X}{X+Y}[/imath]. How about if [imath]X[/imath] and [imath]Y[/imath] are iid exponential with [imath]\lambda=1[/imath]? Try Notice that [imath]f_{X,Y}(x,y)=1[/imath] and [imath]X= V(X+Y)= UV [/imath] and so [imath]Y = U -X = U - UV = U(1-V)[/imath]. We compute the Jacobian of map [imath](U,V)[/imath] and obtain [imath] J = 1 \cdot \left( - \frac{X}{(X+Y)^2} \right) - 1 \cdot \left( \frac{Y}{(X+Y)^2} \right) = - \frac{1}{X+Y} = \frac{-1}{U} [/imath] And since [imath] f_{X,Y}(X,Y) = 1 [/imath] then [imath] f_{U,V}(u,v) = f_{X,Y} (x,y) \frac{1}{|J|} = u [/imath] Since [imath]0 \leq UV \leq 1[/imath] and [imath]0 \leq U(1-V) \leq 1[/imath], we have [imath]0 \leq U \leq \frac{1}{V}[/imath] and [imath]1 - V \leq \frac{1}{U}[/imath] and so [imath]1 - \frac{1}{U} \leq V \leq 1[/imath] is this correct? | 780990 | Joint distribution of [imath]X+Y[/imath] and [imath]\frac{X}{X+Y}[/imath]
Let [imath]X[/imath] and [imath]Y[/imath] be two random variables i.i.d [imath]U(0,1)[/imath]. Find the joint pdf of [imath]T = X+Y[/imath] and [imath]U = \frac{X}{X+Y}[/imath] and the marginal densities of [imath]T[/imath] and [imath]U[/imath] My attempt: We will have the following transformation: [imath]X = TU[/imath] and [imath]Y = T - TU[/imath]. The jacobian is [imath]J = -ut -t(1-u) = -t[/imath] and the joint pdf: [imath]f_{T,U}(t,u) = I_{(0,1)}(tu)I_{(0,1)}(t-tu) |t|[/imath] where [imath]I[/imath] is the indicatr function Note that [imath]U[/imath] and [imath]T[/imath] will be jointly defined in the following region: 1) [imath]0 < t < 2[/imath] 2) [imath]0 < u < \infty[/imath] 3) [imath]0 < tu < 1 \Rightarrow t<\frac{1}{u} (u>0)[/imath] 4)[imath] 0 < t -tu < 1[/imath] 4.1) [imath]u < 1 (t>0)[/imath] 4.2) [imath]t < \frac{1}{1-u}[/imath] Integrating in respect of [imath]t[/imath], I could obtain th right pdf of [imath]U[/imath] because I know the answer. But in respect of [imath]u[/imath] I couldn't. Is my region wrong? P.S: I know how to obtain the pdf of [imath]X+Y[/imath] using other ways, I want it using this joint distribution. Thanks! |
2996736 | A uniformly continuous function on a bounded set whose image is not bounded?
I'm trying to find an example of a uniformly continuous function [imath]f:X\to Y[/imath] such that [imath]f(X)[/imath] is not bounded, for some bounded metric space [imath]X[/imath] and a metric space [imath]Y[/imath]. The classic exercise is to show that if [imath]X\subseteq\mathbb{R}[/imath] is bounded then any uniformly continuous [imath]f[/imath] sends bounded sets onto bounded sets, this is because such sets are totally bounded in this case. So what is an example of a bounded metric space [imath]X[/imath], not totally bounded, such that a uniformly continuous function sends it to a not bounded set? | 125292 | Uniformly Continuous Function sending Bounded Set to Unbounded One
Let [imath](X,d)[/imath] and [imath](Y,\rho)[/imath] be metric spaces, and let [imath]f: X \to Y[/imath] be a uniformly continuous function. If [imath]A \subset X[/imath] is bounded, must [imath]f(A) \subset Y[/imath] be bounded? It is clear to me that in metric spaces that satisfy the Heine-Borel property, such as [imath]\mathbb{R}^{n}[/imath], the answer to this question is yes. However, I can see no reason why this should hold for arbitrary metric spaces. Any ideas? Thanks! |
2999055 | Show that Riemann integrable function [imath]f[/imath] on [imath][a,b][/imath] must be a bounded function.
I see these two: How to show that a Riemann integrable function is bounded If a function [imath]f(x)[/imath] is Riemann integrable on [imath][a,b][/imath], is [imath]f(x)[/imath] bounded on [imath][a,b][/imath]? The first uses a very different definition of Riemann integrable functions. The second post offers casual intuition, not a formal proof. The definitions I'm working with: A Riemann Sum is defined for a partition [imath]\mathcal{P}[/imath] of [imath][a,b][/imath] as: [imath]\begin{align*} \mathcal{R}(f, \mathcal{P}) &= \sum\limits_{j=1}^k f(s_j) \Delta_j \\ \end{align*}[/imath] The function is Riemann integrable if Riemann sums converge to a number [imath]\ell[/imath] as the mesh sizes of the partitions approach zero. A function [imath]f[/imath] is Riemann integrable if for any [imath]\epsilon > 0[/imath], there must exist some [imath]\delta > 0[/imath] and some partition [imath]\mathcal{P}[/imath] such that: [imath]\begin{align*} m(\mathcal{P}) < \delta &\implies |\mathcal{R}(f, \mathcal{P}) - \ell| < \epsilon \\ \end{align*}[/imath] Intuitively, if [imath]f[/imath] is unbounded, it looks like that Riemann sum will not converge, but I can't see how to formally demonstrate that. | 2250722 | Proving a function is bounded
Hi i was wondering if anyone could help me with my revision This is a question of a past paper i'm stuck on. Let [imath]f:[a,b]\to \Re[/imath] be Riemann intergrable. Part (a) is to prove that f is bounded and part (b) is to Give an example of a bounded function f that is not Riemann integrable my attempt for part (a) is since f is Riemann intergrable then it is continuous on [a,b] which means that it must also be bounded? or is this incorrect i find this topic rather difficult For part (b) i dont really have a clue i cant recall any function like the one they ask for |
2999377 | Prove convergence from [imath]|\sum_{k = 1}^n a_k |\leq \sqrt{n}[/imath]
Given [imath]|\sum_{k = 1}^n a_k |\leq \sqrt{n}[/imath], how to show that [imath]\sum_{k = 1}^\infty a_k/k [/imath] converges?. I've not encountered a convergence problem where I don't know the convergence of [imath](a_n)[/imath]. So the usual test are out of option. | 636130 | Does [imath]\big|\sum_{k=1}^na_k\big|\leq\sqrt{n}[/imath] imply the convergence of [imath]\sum_{1}^\infty \frac{a_k}{k}[/imath]?
Suppose [imath]\big|\sum_{k=1}^na_k\,\big|\leq\sqrt{n}[/imath] for all [imath]n\geq 1[/imath]. Show that [imath]\sum_{1}^\infty \frac{a_k}{k}[/imath] converges. Summation by part doesn't help. ([Added: Thanks to the comments, I should have made my thought clear so that it would not cause confusion. It is not that "Summation by part doesn't help", but that "I don't see how 'summation by part' might work".]) I tried several convergence tests without any progress. The assumption gives an estimate for the Cesaro mean: [imath] \left|\frac{\sum_{k=1}^n a_k}{n}\right|\leq \frac{1}{\sqrt n}. [/imath] It seems that if I can get [imath] a_n\sim \left|\frac{\sum_{k=1}^n a_k}{n}\right| [/imath] then things will be done since [imath] \sum\frac{1}{\sqrt k \cdot k} [/imath] converges. How shall I go on? |
2998791 | Sum of algebraic number of prime degree.
I'm looking to prove the following: Let [imath]K[/imath] be a field and suppose that [imath][K(\alpha):K]=p[/imath] and [imath][K(\beta):K]=q[/imath] for [imath]p,q[/imath] distinct primes. Then [imath]K(\alpha+\beta):K[/imath] has degree [imath]pq[/imath]. Of course I know that [imath]K(\alpha+\beta)[/imath] has either degree [imath]1,p,q[/imath] or [imath]pq[/imath] over [imath]K[/imath]. But I do not know how to rule out the cases where the degree is [imath]p[/imath] or [imath]q[/imath]. Is there another way? | 2623906 | If [imath][K(\alpha):K]=p\neq q=[K(\beta):K][/imath] then [imath][K(\alpha+\beta):K]=pq[/imath]
I am having some trouble with the following problem: Let [imath]K\subseteq L[/imath] be fields. Suppose [imath]\alpha,\beta\in L[/imath] are algebraic elements over [imath]K[/imath] of degrees [imath]p,q[/imath] respectively, where [imath]p[/imath] and [imath]q[/imath] are distinct primes. Show that [imath]\alpha+\beta[/imath] is algebraic over [imath]K[/imath] with degree [imath]pq[/imath]. I have shown that [imath]\alpha+\beta[/imath] is algebraic over [imath]K[/imath], but I am having trouble showing that the degree is [imath]pq[/imath]. I have previously shown that [imath][K(\alpha,\beta):K]=pq[/imath], so I have been trying to use this result. I thought that if I could show that [imath]K(\alpha,\beta)=K(\alpha+\beta)[/imath] then I'd be done. To show this I first noted that clearly [imath]K(\alpha+\beta)\subseteq K(\alpha,\beta)[/imath]. Then we have [imath]pq=[K(\alpha,\beta):K]=[K(\alpha,\beta):K(\alpha+\beta)][K(\alpha+\beta):K][/imath] so that [imath][K(\alpha+\beta):K]=1,p,q,pq[/imath]. I tried looking at the cases when it is equal to [imath]1,p,q[/imath] and deriving a contradiction, but have been unsuccessful. Another way I thought of solving this is to show that [imath]\alpha,\beta\in K(\alpha+\beta)[/imath], which would allow me to conclude that [imath]K(\alpha+\beta)=K(\alpha,\beta)[/imath], but I am not sure how to complete this either. I am looking for some assistance to show that [imath][K(\alpha+\beta):K]\neq 1,p,q[/imath], or that [imath]\alpha,\beta\in K(\alpha,\beta)[/imath]. If you have any other solutions to this problem I would like to see those too. |
3000184 | How to evaluate [imath]\int \sin^2 x \ dx[/imath]
The fact that sin is squared is really throwing me off, can't seem to relate it to any standard integrals. | 892725 | About integrating [imath]\sin^2 x[/imath] by parts
This is about that old chestnut, [imath]\newcommand{\d}{\mathrm{d}} \int \sin^2 x\,\d x[/imath]. OK, I know that ordinarily you're supposed to use the identity [imath]\sin^2 x = (1 - \cos 2x)/2[/imath] and integrating that is easy. But just for the heck of it, I tried using the [imath]u[/imath]-[imath]v[/imath] substitution method (otherwise known as integration by parts). [imath] \int \sin^2 x\,\d x = \int \sin x \sin x\,\d x [/imath] We can say [imath]u=\sin x[/imath] and [imath]\d u=\cos x\,\d x[/imath] while [imath]\d v = \sin x\,\d x[/imath] and [imath]v = -\cos x[/imath]. When we put it all together: [imath]\int \sin^2 x\,\d x = u v - \int v\,\d u = -\sin x \cos x - \int -\cos^2 x\,\d x[/imath] and doing the same routine with [imath]\cos^2 x[/imath] we get [imath]u = \cos x[/imath], [imath]\d u = -\sin x\,\d x[/imath], [imath]\d v = \cos x\,\d x [/imath] and [imath]v = -\sin x[/imath], leading to: [imath]\begin{align} \int \sin^2 x\,\d x = uv - \int v\,\d u &= -\sin x \cos x - (-\cos x \sin x - \int -\sin^2 x\,\d x) \\ &= -\sin x \cos x + \sin x \cos x - \int \sin^2 x\,\d x \end{align}[/imath] which eventually works out to [imath]2\int \sin^2 x\,\d x = 0[/imath] So I wanted to get an idea why this didn't work. Maybe it's higher math and the why will be beyond me (I would think that might be the case), or maybe it's one of those proofs that looks absurdly simple when shown that I am just unaware of. |
2999995 | Axioms that characterize the notion of dimension
Let [imath]\mathcal{C}[/imath] be a class of sets/spaces/structures among which we have a dimension. Namely a map [imath]d:\mathcal{C}\rightarrow \mathbb{N}[/imath] defined in a certain manner that motivated the appellative dimension. What properties would you expect [imath]d[/imath] to have? E.g. if [imath]Y, X\in\mathcal{C}[/imath] and [imath]Y\subseteq X[/imath] then [imath]\dim Y\leq \dim X[/imath]. Another way of phrasing this question is, what properties are shared by every standard notion of dimension in mathematics? | 1358723 | When is a function a dimension?
The concept of dimension is used in many different contexts. Generally a dimension is a function that has as domain some family of sets ad has value on a set that, in the most common situations, is [imath]\mathbb{N}[/imath] or [imath]\mathbb{R}[/imath]. As an example of the first case we can think to the dimension of vector spaces (but this can also be infinite) or to the Krull dimension of commutative rings. As an example of the second case we can think to the Hausdorff dimension for metric spaces ( and its variants). But we can also define dimensions that have as range a family of ordered sets and as range an interval of [imath]\mathbb{R}[/imath], as in the case of continuous geometry, and it seems that we can also define a dimension functions on super vector space that can have negative values, and a dimension with complex values for self-similar sets (Has the notion of having a complex amount of dimensions ever been described? And what about negative dimensionality?). All these dimensions are different in their definitions and properties and, if I well understand there is not an axiomatic definition of dimension that can be used to identify a function as a dimension function (see:https://mathoverflow.net/questions/80708/is-there-an-axiomatic-approach-of-the-notion-of-dimension). So my question is why mathematicians call all those different function with the same name? I understand that the name come from our common intuition of dimension but I don't understand how such intuition apply to such sophisticated notions called dimension. More precisely: I'm curious to know what is the inspiration that guide a mathematician to recognize that a particular function can be called a dimension. I see that this is not a question that can have a unique and well defined answer, but I suppose (or I hope) that there is some common mathematical meaning about this so used word. |
2999220 | Uniformly Distributed Marginal Density
A point [imath](X,Y)[/imath] in the Cartesian plane is uniformly distributed within the unit circle if [imath]X[/imath] and [imath]Y[/imath] have joint density: [imath]f(x,y) = \begin{cases} \frac{1}{π} & \mathrm{x^2+y^2} \le 1 \\ 0 & \mathrm{otherwise} \end{cases} [/imath] Find the marginal densities [imath]f_X[/imath] and [imath]f_Y[/imath] and state whether [imath]X[/imath] and [imath]Y[/imath] are independent or not. Provide a mathematical justification for your answer. I am really confused, I don't know how to set up the two integrals, please help :) | 1092669 | Uniform distribution on unit disk
Let [imath](X, Y)[/imath] be a random point chosen according to the uniform distribution in the disk of radius 1 centered at the origin. Compute the densities of [imath]X[/imath] and of [imath]Y[/imath]. I know that the joint density of [imath]X[/imath] and [imath]Y[/imath] is [imath]\frac{1}{\pi}[/imath] since when we integrate [imath]\frac{1}{\pi}[/imath] over the unit circle, we get [imath]1[/imath]. So if I wanted to find the density of [imath]X[/imath], I was thinking of finding the cumulative distribution of [imath]X[/imath] and the differentiate it to get its density. In order to get its cumulative distribution function, I was going to use the fact that [imath]P(X<x)=P(X<x, -\infty < Y < \infty)[/imath], but this integral doesn't seem nice to work with. Am I on the right track or is there a better way? |
3000075 | find a number [imath]aabb[/imath] such that is full square
I know that I can write [imath]aabb=1000a+100a+10b+b[/imath], [imath]a,b\in \{0,1,2,3,4,5,6,7,8,9\}, a\neq 0[/imath], so [imath]aabb=1100a+11b=m^2[/imath], where [imath]m\in \mathbb N[/imath], I notice that [imath]m^2=11(100a+b)[/imath] so I need to find some number that [imath]11\mid m^2[/imath] and [imath]11\mid m[/imath],[imath]1100<m^2<9999[/imath] or [imath]33<m<99[/imath] so I just put number in calculator and 88 is answer, so [imath]aabb=7744[/imath], but is there any other better solution? | 2026973 | Find all digits [imath]a,b[/imath] such that the number [imath]\overline{aabb}[/imath] is a perfect square
What is the method for finding numbers [imath]a[/imath] and [imath]b[/imath]? |
3000329 | Give two specific examples of a non-zero matrix X such that: AX = XA.
If matrix A = [imath]\begin{bmatrix}3&1\\-2&-1\end{bmatrix}[/imath] I am trying to give two specific examples of a non-zero matrix X such that: AX = XA. so far I let X = [imath]\begin{bmatrix}a&b\\c&d\end{bmatrix}[/imath] And so AX = XA, and I multiplied X and A respectivley to give the linear equations: 3a + c = 3a - 2b 3b + d = a - b -2a + c = 3c - 2d -2b - d = c - d I believe i am supposed to show this in matrix form and solve the system to get a 4x4 matrix containing expressions with two variables where any number can be substituted so that AX = XA. However I am unsure on how to do this and need some help. | 2522260 | Finding two non-zero matrices where [imath]AX = BA[/imath]
I have a question where I have a matrix [imath]A=\begin{bmatrix}2&-1\\-1&1\end{bmatrix}[/imath] and I need to find two examples of a non-zero matrix [imath]X[/imath] such that [imath]AX = XA[/imath]. My first matrix is \begin{bmatrix}1&0\\0&1\end{bmatrix} the identity matrix but I cant think of another one. Any ideas? |
3000968 | on such group whose inner automorphisms group isomorphic to S3.
Let [imath]\frac{G}{Z(G)}≅S_3[/imath], such that [imath]S_3[/imath] is permutation group on 3 letters and [imath]Z(G)[/imath] is non trivial central subgroup of [imath]G[/imath]. What are the possibility of group [imath]G[/imath] and does there exist always an non-inner automorphism group [imath]G[/imath] ? or if inner automorphism group is given then what we can say about [imath]G[/imath] ? Thanks. | 2875892 | Is there a way to describe all finite groups [imath]G[/imath] such that [imath]\operatorname{Aut}(G) \cong S_3[/imath]?
Is there a way to describe all finite groups [imath]G[/imath] such that [imath]\operatorname{Aut}(G) \cong S_3[/imath]? Two groups that definitely satisfy that condition are [imath]S_3[/imath] itself (as it is a complete group) and [imath]\mathbb{Z}_2 \times \mathbb{Z}_2[/imath] (as [imath]S_3[/imath] is isomorphic to [imath]GL(2, 2)[/imath]). I have read somewhere that those two groups are the only two groups that satisfy that condition. There was no proof of this statement given, however, so I do not know whether it is true or false (and if it is true, it would be interesting to know the proof). Any help will be appreciated. |
3001243 | Random "start afresh" of Brownian Motion
In the book "Brownian Motion and Stochastic Calculus" by Karatzas and Shreve I encountered Problem 6.1 which the authors refer to as "not be hard to show": Let [imath]\{B_t, \cal{F}_t;t\geq 0\}[/imath] be a standard, one-dimensional Brownian motion. Give an example of a random time [imath]S[/imath] with [imath]P[0\leq S<\infty]=1[/imath], such that with [imath]W_t:=B_{S+t}-B_S[/imath], the process [imath]W=\{W_t, \cal{F}^W_t;t\geq 0\}[/imath] is not a Brownian motion. Surely, if S being a stopping time [imath](W_t)_t[/imath] would still be a Brownian motion. I tried out some distributions for [imath]S[/imath] but none of them gave a sufficient answer. Most likely I am missing a crucial point here and appreciate any kind of help. | 2713265 | An example of almost sure finite random time S such that [imath]W_t=B_{S+t}-B_S[/imath] is not a brownian motion
Let [imath]\{B_t,\mathcal{F}_t,t \geq 0\}[/imath] be a standard one dimensional Brownian motion. Give an example of a random time [imath]S[/imath] with [imath]P[0\leq S< \infty]=1[/imath] such that with [imath]W_t:=B_{S+t}-B_S[/imath], the process [imath]W=\{W_t, \mathcal{F}^w_t;t\geq0\}[/imath] is not a brownian motion. How could I go about constructing such an example. I have no idea of how to proceed. I tried to invent an almost sure finite random time using the law of large numbers for Brownian motion [imath]T_{\epsilon}=\{\sup_{s \geq0}: B_s\geq s\epsilon\}[/imath] but I am not sure on how to handle the object [imath]B_{T_{\epsilon}+t}-B_{T_{\epsilon}}[/imath]. Any help would be appreciated |
3001070 | Behavior of [imath]\sum_{n=1}^\infty \frac{1}{n} z^{n!}[/imath] on the unit circle
I'm trying to understand the behavior of [imath]\sum_{n=1}^\infty \frac{1}{n} z^{n!}[/imath] on the unit circle. Since for each [imath]m[/imath]th root of unity [imath]\zeta_m[/imath] [imath]\sum_{n=1}^\infty \frac{1}{n} \zeta_m^{n!} = C + \sum_{n=m}^\infty \frac{1}{n} = \infty[/imath] holds for some [imath]C \in \mathbb{C}[/imath], the series diverges for all [imath]e^{\varphi \pi i}[/imath] with [imath]\varphi \in \mathbb{Q}[/imath]. But what happens for [imath]\varphi \in \mathbb{R} \setminus \mathbb{Q}[/imath]? Does the series diverge everywhere, or are there points where it is convergent? | 2176288 | Series with radius of convergence 1 that diverges on roots of unity, converges elsewhere on the circle.
I recall from complex analysis a while ago that the series [imath]\sum_{n=1}^\infty \frac{z^{n!}}{n}[/imath] has radius of convergence 1, diverges on roots of unity, and converges elsewhere on the circle. It's simple enough to see that it has to diverge on roots of unity: consider [imath]z[/imath] such that [imath]z^k = 1[/imath] for some [imath]k \in \mathbb{N}.[/imath] Plug this into the series and you see that past the [imath]k[/imath]'th term, [imath]z^{n!} = (z^k)^m = 1[/imath] for some [imath]m \in \mathbb{N}[/imath]; hence, the series becomes harmonic and must diverge. But I don't recall how to show that the series converges for other values on the circle. If this is simple I'd like just a hint, if it is involved I would prefer a sparse outline of the proof. Thank you! |
3001700 | How to solve the equation [imath]15x- 16y= 10[/imath]
I am trying to find an [imath]x[/imath] and [imath]y[/imath] that solve the equation [imath]15x - 16y = 10[/imath], usually in this type of question I would use Euclidean Algorithm to find an [imath]x[/imath] and [imath]y[/imath] but it doesn't seem to work for this one. Computing the GCD just gives me [imath]16 = 15 + 1[/imath] and then [imath]1 = 16 - 15[/imath] which doesn't really help me. I can do this question with trial and error but was wondering if there was a method to it. Thank you | 3000011 | Find the value of [imath]x,y[/imath] for the equation [imath]12x+5y=7[/imath] using number theory .
I tried the solution using other ways like: [imath]12x+5y=7[/imath] [imath]12x=7-5y[/imath] [imath]x=7-5y/12[/imath] putting the value of [imath]x[/imath] [imath]12(7-5y/12)+5y=7[/imath] both [imath]12[/imath] will cancel out [imath]7-5y+5y=7[/imath] here comes the problem please help me ? |
3002109 | Can a sequence be undefined at a point?
A sequence which is a mapping from [imath]\mathbb{N} \to \mathbb{R}[/imath]. For example can the sequence [imath]\{a_n\} = 1/(3-n)[/imath]. This would be undefined at [imath]3[/imath]. Is it a sequence? | 143780 | Are undefined terms allowed in a sequence?
Is this a valid sequence: [imath]\{\frac1{(n-3)}\}[/imath]? I.e., can a sequence have individual terms that are undefined? And if so, does this mean that the above sequence is unbounded (since the third term is not smaller than any real number)? [What I do know is that if the above sequence were valid, it would be convergent (by definition), which would in turn mean that it is bounded (by the Boundedness Theorem).] p.s. Subsequent to the main discussion here, I have just come across an old exam question that asks us to show that [imath]$\langle\tan(\frac{\sqrt n\pi}4):n∈N\rangle$[/imath] is divergent. But this sequence is undefined at say, n=4, so like the one above, it isn't even a valid sequence, is it? (I've used a different notation for sequences on Brian M Scott's advice.) |
3001330 | Statistical Integral in Financial Mathematics
I need to show that [imath]\frac{1}{\sqrt {2\pi}} \int_{-\infty}^{+\infty} x^{2n} e^{\frac{-x^2}{2}}dx = (2n-1)!![/imath] Integration by parts seems to be the best apporach but I cannot seem to figure my way through it. | 1796806 | Calculate [imath]E(X^{2n})[/imath] where [imath]X[/imath] is normal (0,1)
I need help proving the following: Let [imath]X[/imath] be normally distributed with parameters [imath]\sigma=0[/imath] and [imath]\mu=1[/imath]. Let [imath]n[/imath] be a positive integer. Show that: [imath]E(X^{2n})=\frac{(2n)!}{2^nn!}=:(2n-1)!![/imath] I've tried the change of variables [imath]Y=X^2[/imath] and then calculating the integral, but got nowhere. Any hints? Thanks in advance! |
3002147 | Finding an Inverse Function and Composition of Functions? I add for all y ∈ R
The functions of each pair are inverse to each other. For each pair, check that both compositions give the identity function. [imath]F: \mathbb{R} \to \mathbb{R}[/imath] and [imath]F^{−1}:\mathbb{R} \to \mathbb{R}[/imath] are defined by [imath]F(x)=3x+2[/imath] and [imath]F^{−1}(y)=\dfrac{y−2}{3}[/imath], for all [imath]y \in \Bbb{R}.[/imath] My attempt: Inverse Function For each particular but arbitrarily chosen [imath]y \in \mathbb{R}[/imath], according to the definition of [imath]f^{-1}[/imath], [imath]f^{-1}(y) = \dfrac{y-2}{3}[/imath] is a unique real number [imath]x[/imath] such that [imath]f(x) = y[/imath]. [imath]\begin{align*} F(x) & = y\\ 3x + 2 & = y\\ x & = \frac{y-2}{3} \end{align*}[/imath] Therefore, [imath]f^{-1}(y) = \frac{y-2}{3}[/imath]. Compositions of Functions. The functions [imath]g \circ f[/imath] and [imath]f \circ g[/imath] are defined as follows: [imath](g \circ f)(x) = g(f(x)) = g(3x + 2) = 3x + 2[/imath] for all [imath]x \in \mathbb{Z}[/imath]. | 2986479 | Finding an Inverse Function and Composition of Functions?
The functions of each pair are inverse to each other. For each pair, check that both compositions give the identity function. [imath]F: \mathbb{R} \to \mathbb{R}[/imath] and [imath]F^{−1}:\mathbb{R} \to \mathbb{R}[/imath] are defined by [imath]F(x)=3x+2[/imath] and [imath]F^{−1}(y)=\dfrac{y−2}{3}[/imath]. for all y ∈ R My attempt: Inverse Function For each particular but arbitrarily chosen [imath]y \in \mathbb{R}[/imath], according to the definition of [imath]f^{-1}[/imath], [imath]f^{-1}(y) = \dfrac{y-2}{3}[/imath] is a unique real number [imath]x[/imath] such that [imath]f(x) = y[/imath]. [imath]\begin{align*} F(x) & = y\\ 3x + 2 & = y\\ x & = \frac{y-2}{3} \end{align*}[/imath] Therefore, [imath]f^{-1}(y) = \frac{y-2}{3}[/imath]. Compositions of Functions The functions [imath]g \circ f[/imath] and [imath]f \circ g[/imath] are defined as follows: [imath](g \circ f)(x) = g(f(x)) = g(3x + 2) = 3x + 2[/imath] for all [imath]x \in \mathbb{Z}[/imath]. |
3002760 | Lagrange Multipliers of distance from point to a function with respect to tangent and normal spaces.
I have a question with regards to lagrange multiplers. The question is as follows: Find the points on the sphere x^2+y^2+z^2=4 that are closest to and farthest from the point (3,1,-1). Now, it is clear to me that the distance from any point to (3,1,-1) is simply using the R^3 metric, d=[imath]\sqrt((x-3)^2+(y-1)^2+(z+1)^2[/imath] and I can make the algebra easier by considering [imath]d^2[/imath]. However, my question is as follows: How can I be sure that the maximum or minimum occurs when [imath]\nabla(d^2(x,y,z))=\lambda\nabla(g(x,y,z))[/imath] where [imath]g(x,y,z)=x^2+y^2+z^2[/imath]? I would like an explanation in terms of tangent and normal spaces please? As i'm unable to find one. Furthermore, it is to my knowledge that the lagrange multiplier works when the contours of the function f(x,y) just touches the constraint g(x,y)=c. However, is it possible to have contours of f(x,y) where the contours increase and decrease and wouldn't then the lagrange multiplier not work? | 1290487 | Why do Lagrange Multipliers work?
I know that the Lagrange multiplier method helps us evaluate critical points of [imath]f[/imath] on the closed boundary of the restriction. In other words we solve:[imath]\nabla f=\lambda \nabla g[/imath] But why does this method actually work? Can someone please give me an explanation. Thank you! |
3002736 | Some integer solutions for [imath]a^2+b^2+c^2=d^2[/imath]
I'm looking for integer solutions to the equation [imath]a^2+b^2+c^2=d^2[/imath] such that [imath]\gcd(a,b,c,d) = 1[/imath]. | 2088960 | Solutions for [imath]a^2+b^2+c^2=d^2[/imath]
I have to find an infinite set for [imath]a,b,c,d[/imath] such that [imath]a^2+b^2+c^2=d^2[/imath] and [imath]a,b,c,d\in\mathbb N[/imath] I have found one possible set which is [imath]x,2x,2x,3x[/imath] and [imath]x\in\mathbb N[/imath] But I want another infinite sets and how to reach them because I have found my solution by hit and trial . |
2746421 | Why is : [imath] \mathrm{dim} \ ( S_X )_d = \begin{pmatrix} n+d \\ n \end{pmatrix} - \begin{pmatrix} n+d-s \\ n \end{pmatrix} [/imath]?
In page : [imath]206[/imath] of the following electronic textbook : https://scholar.harvard.edu/files/joeharris/files/000-final-3264.pdf , the author says that the Hilbert function of [imath] S_X [/imath] is : [imath] \mathrm{dim} \ ( S_X )_d = \begin{pmatrix} n+d \\ n \end{pmatrix} - \begin{pmatrix} n+d-s \\ n \end{pmatrix} [/imath] with, [imath] X [/imath] is any hypersurface of degree [imath]s[/imath] in [imath] \mathbb{P}^n [/imath]. Can you explain to me please, what is [imath] S_X [/imath], and how do we obtain this formula above ? Thanks in advance for your help. | 573817 | Hilbert polynomial of an hypersurface in projective space
Let [imath]X[/imath] an hypersurface in [imath]\mathbb{P}^{n}[/imath] of degree [imath]d[/imath]. I would like to prove that the Hilbert polynomial of [imath]X[/imath] is [imath]\qquad \qquad \qquad \qquad \qquad \qquad p(n)= \begin{pmatrix} n+r \\ n \end{pmatrix} - \begin{pmatrix} n+d-r \\ n \end{pmatrix} [/imath] In the book "Geometry of Algebraic Curves Vol. II", Arbarello, Cornalba, Griffith, p. 7, I read that this result can be proved by taking the cohomology of the following exact sequence [imath] \qquad \qquad \qquad \qquad \qquad 0 \longrightarrow \mathcal{O}_{\mathbb{P}^{r}}(n-d) \longrightarrow \mathcal{O}_{\mathbb{P}^{r}}(n) \longrightarrow \mathcal{O}_{\mathbb{P}^{r}}(d) \longrightarrow 0 [/imath] Nevertheless, I am not able to compute [imath]p(n)[/imath]. |
3003237 | How to solve this limit (with factorial)?
How to solve this limit?? [imath]\lim_{k \to \infty} \frac{(2k)!}{2^{2k} (k!)^2}[/imath] It's a limit, not a series | 2738541 | Does series [imath]4^{n}×\frac{(n!)^2}{(2n)!}[/imath] converge or diverge?
I know [imath]4^{n}[/imath] diverges, and [imath]\frac{(n!)^2}{(2n)!}[/imath] converges. I also think that their product diverges since the term of this series is increasing but I don't know how to prove it. I tried ratio test which is inconclusive, and some other tests. |
3003012 | Ideal in polynomial ring extension
Let [imath]K \subset L[/imath] be a field extension, [imath]K[X][/imath] and [imath]L[X][/imath] the corresponding polynomial rings (in one variable) and [imath]I \subset K[X][/imath] an ideal. I want to show that [imath]I=K[X] \cap IL[X][/imath], where [imath]IL[X][/imath] denotes the ideal generated by [imath]I[/imath] in [imath]L[X][/imath]. I was told that while there are many ways to show this abstractly, there is supposed to be a very simple proof only involving Linear Algebra. I don't really know where to start here. The inclusion from left to right is trivial, but I haven't got much more. Any help - even just a hint - would be appreciated. | 764488 | Divisor in [imath]\mathbb{C}[X][/imath] [imath]\implies[/imath] divisor in [imath]\mathbb{R}[X][/imath]?
let [imath]P \in \mathbb{R}[X][/imath] be a real polynomial divisible by a polynomial [imath]Q \in \mathbb{R}[X][/imath] in [imath]\mathbb{C}[X][/imath]. How can I easily show that [imath]P[/imath] is also divisible by [imath]Q[/imath] in [imath]\mathbb{R}[X][/imath]? A simple argument without using higher algebraic theorems is desirable. If I could use instruments of higher algebra, the exercise I have to do in whole would be done in two lines. But I'm not allowed to use. I think there would be an easy argument which I can't see yet because of my mental fogginess that I have sometimes. Thank you beforehand. |
3003102 | Zeros of partial sums of the exponential
I am trying to show that if [imath]f_n(z)=1+z+\frac{z^2}{2!}+...+\frac{z^n}{n!}[/imath] Then [imath]f_n(z)[/imath] don’t have zeros inside the unitary disk. I have tryied to use Rouche’s theorem or use that in the limit the polinomial converges to the exponential, but i dont get hoy to do this. | 1029199 | How prove this [imath]|z|>1[/imath] with [imath]1+z+\frac{z^2}{2!}+\cdots+\frac{z^n}{n!}=0[/imath]
For give the postive integer [imath]n[/imath],and [imath]z\in C[/imath] such this [imath]1+z+\dfrac{z^2}{2!}+\cdots+\dfrac{z^n}{n!}=0[/imath] show that [imath]|z|> 1[/imath] maybe we Assmue that exst [imath]z[/imath] such [imath]|z|\le 1[/imath] then we have [imath]z(1+\dfrac{z}{2}+\cdots+\dfrac{z^{n-1}}{n!})=-1[/imath] so [imath]|z|\cdot\left|1+\dfrac{z}{2}+\cdots+\dfrac{z^{n-1}}{n!}\right|=1[/imath] then we have [imath]\left|\dfrac{z}{2}+\cdots+\dfrac{z^{n-1}}{n!}\right|\ge 1[/imath] then I can't have idea |
279725 | Evaluation of the double integral [imath]\int_{[0,1]×[0,1]} \max\{x, y\} dxdy[/imath]
Evaluate: [imath]\int_{[0,1]×[0,1]} \max\{x, y\} dxdy[/imath] I am totally stuck on it. How can I solve this? | 646387 | Evaluate the integral [imath]\int\int _{[0,1]^2} \max \left\{x,y\right\} dx\, dy[/imath]
Find the integral [imath]\int\int _{[0,1] \times[0,1]} \max \left\{x,y\right\} dx \,dy[/imath] |
3004198 | Prove [imath]\sum\limits_{n=0}^\infty\frac{x^n}{n!}=\lim\limits_{n\rightarrow\infty}\left(1+\frac{x}{n}\right)^n[/imath]
[imath]x\in\mathbb{R}[/imath]. I can prove that both sides converge, and maybe we should show [imath]\limsup\limits_{n\rightarrow\infty}\left|\sum\limits_{k=0}^{n}\frac{x^k}{n!}-\left(1+\frac{x}{n}\right)^n \right|<\varepsilon[/imath], but the construction of the right hand side is a little tricky for me. | 2348229 | Validity and Equivalence of two definitions of the real exponential function
The Problem : We state the following two definitions of the real exponential function from the Pr[imath]\infty[/imath]fWiki page. We're interested in showing that the two definitions are valid [imath]([/imath]i.e. the defining sequence/series does converge to a unique real number[imath])[/imath] and that the two definitions are equivalent. I'm stuck at a couple of points [imath]([/imath]which are described in highlighted lines[imath])[/imath]. Any help would be much appreciated. Thank you! Definition [imath]1[/imath]. The exponential function can be defined as the following limit of a sequence [imath]\exp x := \lim_{n \to \infty} \left({1 + \frac x n}\right)^n[/imath] Definition [imath]2[/imath]. The exponential function can be defined as a power series [imath]\exp x := \sum_{n = 0}^\infty \frac {x^n} {n!}[/imath] My Progress and two places where I'm stuck : Essentially the solution consists of three parts, namely validity of definition [imath]1[/imath], validity of definition [imath]2[/imath] and equivalence of the two definitions. Validity of Definition [imath]1[/imath]. I'm stuck here! Can we show that the sequence [imath](a_n)[/imath] given by [imath]a_n = \left(1+\frac{x}{n}\right)^n[/imath] converges for every [imath]x \in \mathbb{R} ??[/imath] Is it eventually monotone and bounded [imath]??[/imath] Validity of Definition [imath]2[/imath]. The radius of convergence of the power series is [imath]r=\lim_{n \to \infty}\left| \frac{\frac{1}{n!}}{\frac{1}{(n+1)!}} \right|=\lim_{n \to \infty}\left| \frac{(n+1)!}{n!} \right|=\lim_{n \to \infty}(n+1)=+\infty[/imath] Thus the infinite series in the right-hand side of Definition [imath]2[/imath] converges to a unique real number for all [imath]x \in \mathbb{R}[/imath]. Hence the definition is well-defined. Equivalence of Definition [imath]1[/imath] and Definition [imath]2[/imath]. There's this proof of Definition [imath]1[/imath] [imath]\implies[/imath] Definition [imath]2[/imath] in the Pr[imath]\infty[/imath]fWiki page, but it's kind of under construction and I'm not really convinced by it. So I decided to try to take my own shot at it. For all [imath]n \in \mathbb{N} \cup \{0\},[/imath] let [imath]T_n=\left(1+\frac{x}{n} \right)^n, ~S_n=\sum_{k=0}^n \frac{x^k}{k!}[/imath] We have to show that [imath]\lim_{n \to \infty} T_n = \lim_{n \to \infty} S_n[/imath] Now, \begin{align} T_n &= \left(1+\frac{x}{n}\right)^n\\ &= 1+n\cdot\frac{x}{n}+\frac{n(n-1)}{2!}\cdot\frac{x^2}{n^2}+\cdots +\frac{n(n-1)\cdots 1}{n!}\cdot\frac{x^n}{n^n}\\ &= 1+x+\left(1-\frac{1}{n}\right)\cdot \frac{x^2}{2!}+\cdots +\left(1-\frac{1}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)\cdot \frac{x^n}{n!} \end{align} Clearly, [imath]S_n-T_n=\left\{1-\left(1-\frac{1}{n}\right)\right\}\frac{x^2}{2!}+\cdots +\left\{1-\left(1-\frac{1}{n}\right)\cdots \left(1-\frac{n-1}{n}\right)\right\}\cdot \frac{x^n}{n!}\geq 0[/imath] I'm stuck at this point. Can we show that [imath]S_n-T_n \leq B_n[/imath] such that [imath]B_n \to 0[/imath] as [imath]n \to \infty ??[/imath] |
1691456 | [imath]f_\ast\mathcal{O}_X[/imath] is locally free of rank [imath]r=\deg f[/imath] for a finite [imath]r[/imath]-to-one morphism [imath]f[/imath].
Let [imath]f\colon X\to Y[/imath] be a finite [imath]r[/imath]-to-one morphism between smooth projective varieties. How does one show that [imath]f_\ast\mathcal{O}_X[/imath] is a locally free sheaf of rank [imath]r[/imath] on [imath]Y[/imath] ? Any insights for [imath]f_\ast\mathcal{O}_X[/imath] would be very much appreciated. A local section over [imath]U\subset Y[/imath] for [imath]f_\ast\mathcal{O}_X[/imath] is just a regular function over [imath]f^{-1}(U)[/imath], so it seems as if one can write this uniquely as a sum of [imath]r[/imath] "pieces"...? An aside second question: is there a naturally defined map [imath]f^{\ast}f_\ast\mathcal{O}_X(D)\to\mathcal{O}_X(D)[/imath] where [imath]D[/imath] is a divisor on [imath]X[/imath] ? | 909013 | Push forward of the structure sheaf along covering
Let [imath]f: X \to Y[/imath] be a covering (proper, surjective, finite regular map) of smooth projective varieties of degree [imath]d[/imath]. How one can show that in this case [imath]f_* \mathcal{O}_X[/imath] is a locally free sheaf of rank [imath]d[/imath]? |
3005363 | Function with countably infinite number of discontinuities is integrable
Suppose that [imath]g:[a,b]\to\mathbb{R}[/imath] is bounded and continuous on [imath][a,b][/imath] except at [imath]\{x_n\}_{n=1}^\infty\subset[a,b][/imath] and [imath]\{x_n\}[/imath] converges to [imath]x_0[/imath]. Prove that [imath]g[/imath] is Riemann integrable on [imath][a,b][/imath]. So I know that this should be true because you can pick a partition such that you can just isolate the discontinuities and shrink them to less than some [imath]\epsilon[/imath]. However, I'm not really sure how to formalize that as a proof. I'm not very good at this so I'd really appreciate a thorough explanation. | 263189 | Proof that a function with a countable set of discontinuities is Riemann integrable without the notion of measure
Let [imath]f:[a,b]\to \mathbb{R}[/imath] be a bounded function and [imath]A[/imath] be the set of its discontinuities. I am asking for a (direct) proof that if [imath]A[/imath] is countable then [imath]f[/imath] is Riemann integrable in [imath][a,b][/imath] that doesn't explicitely, or implicitly, require the notion of sets of measure [imath]0[/imath] ( and of course without the use of the Lebesgue Criterion). One could take a typical proof of the Lebesgue Criterion, make the neccessary adjustments and give me the proof of what I am asking. I don't want that however, but rather a simpler and more direct proof that heavily relies on the fact that [imath]A[/imath] is countable. A proof that can't be trivially altered so that it holds even if [imath]\lambda(A)=0[/imath] EDIT: Here is the proof of WimC with all the details: Let [imath]\epsilon>0[/imath] and [imath]D=\left\{d_1,d_2,...\right\}\subseteq A[/imath] be the countable set of discontinuities of [imath]f[/imath]. Define: [imath]I=\left\{x\in [a,b]:\exists \delta>0: \omega f((x-\delta,x+\delta)\cap [a,b])<\epsilon\right\}[/imath] Now [imath]x\in I\iff \exists \delta>0: \omega f((x-\delta,x+\delta)\cap [a,b])<\epsilon\iff [\left|y-x\right|<\delta\implies \omega f(y)<\epsilon][/imath] and because [imath]\epsilon[/imath] is in fact arbitrary, [imath]x\in I\iff \text{ $f$ is continuous at $x$}[/imath] In addition, if [imath]x\in I[/imath], [imath]\exists \delta>0[/imath] so that [imath]\omega f((x-\delta,x+\delta)\cap [a,b])<\epsilon[/imath] If [imath]y\in B(x,\delta)\cap [a,b][/imath]. Then [imath]y\in I[/imath] and so [imath]I[/imath] is open relative to [imath][a,b][/imath]. Because [imath]I=[a,b]\setminus D[/imath], [imath][a,b]=I\cup D[/imath]. For [imath]k\in \mathbb{N}[/imath] define [imath]D_k=\left(d_k-\frac{\epsilon}{M2^{k+1}},d_k+\frac{\epsilon}{M2^{k+1}}\right)\cap [a,b][/imath] Obviously [imath]D\subset \bigcup_{k=1}^{\infty}D_k[/imath] and [imath][a,b]=I\cup \bigcup_{k=1}^{\infty}D_k[/imath] (since [imath]D_k\subseteq [a,b][/imath]). The compactness of [imath][a,b][/imath] implies [imath][a,b]=I\cup \bigcup_{k=1}^{N}D_k[/imath]. Now [imath][a,b]\setminus \bigcup_{k=1}^{N}D_k[/imath] is compact (closed and bounded) and included in [imath]I[/imath]. As such it can be covered by [imath]F_x=(x-\delta_x,x+\delta_x)[/imath] where [imath] \delta_x>0:[/imath] is chosen so that [imath]\omega f((x-\delta_x,x+\delta_x)\cap [a,b])<\epsilon[/imath]. Compactness implies the existence of a finite subcover, [imath][a,b]\setminus \bigcup_{k=1}^{N}D_k\subseteq \bigcup_{i=1}^{M}(x_i-\delta_i,x+\delta_i)[/imath] As we can replace the intervals that intersect we can suppose [imath] \bigcap_{i=1}^{M}[x_i-\delta_i,x+\delta_i]=\emptyset[/imath] Therefore, [imath][a,b]= \bigcup_{k=1}^{N}\overline{D}_k\cup \bigcup_{i=1}^{M}[x_i-\delta_i,x+\delta_i]=\bigcup_{i=0}^{n}[t_{i-1},t_i][/imath] where for [imath]i\le n[/imath], [imath][t_{i-1},t_i]= [x_k-\delta_k,x+\delta_k][/imath] or [imath][t_{i-1},t_i]= \overline{D}_k[/imath] because [imath]\bigcup_{k=1}^{N}\overline{D}_k\cap \bigcup_{i=1}^{M}[x_i-\delta_i,x+\delta_i]=\emptyset[/imath] Considering all the endpoints of the above (that are pairwise different) we can create a partition [imath]\mathcal{P}=\left\{a=t_0<...<t_n=b\right\}[/imath] of [imath][a,b][/imath]. We separate the indices: [imath]A=\left\{i:[t_{i-1},t_i]= [x_k-\delta_k,x+\delta_k]\right\}[/imath] and [imath]B=\left\{i:[t_{i-1},t_i]= \overline{D}_k \right\}[/imath]. Therefore, \begin{gather}U_{f,\mathcal{P}}-L_{f,\mathcal{P}}=\sum_{i=1}^n\omega f([t_{i-1},t_i])(t_i-t_{i-1})=\sum_{i\in A}\omega f([t_{i-1},t_i])\ell([t_{i-1},t_i])+\sum_{i\in B}\omega f([t_{i-1},t_i])\ell([t_{i-1},t_i])\\ \le \sum_{i\in A}2\left\|f\right\|\ell([t_{i-1},t_i])+\sum_{i\in B}\epsilon\ell(\overline{D}_k)\le 2\left\|f\right\|\frac{\epsilon}{M}+\epsilon(b-a)=2\epsilon+\epsilon(b-a)\end{gather} My questions are: Is this proof correct? ( I doubt the point: "where for [imath]i\le n[/imath], [imath][t_{i-1},t_i]= [x_k-\delta_k,x+\delta_k][/imath] or [imath][t_{i-1},t_i]= \overline{D}_k[/imath] because [imath]\bigcup_{k=1}^{N}\overline{D}_k\cap \bigcup_{i=1}^{M}[x_i-\delta_i,x+\delta_i]=\emptyset[/imath]") Second, can it be simplified? |
3004804 | Is [imath] f \colon (1, +\infty) \to \mathbb{R} , f(x) = \sin \frac{1}{x} [/imath] uniformly continuous?
Is [imath] f \colon (1, +\infty) \to \mathbb{R} , f(x) = \sin \frac{1}{x} [/imath] uniformly continuous? I think it could be but can't prove it. Would appreciate the help. Edit: I'm looking for an answer in which mean value theorem and boundedness of the derivative isn't used. So, I'm left with the definition and Lipschitz | 1488432 | [imath]\sin \frac{1}{x}[/imath] is not uniformly continuous on [imath][1, \infty][/imath]
[imath]\sin \frac{1}{x}[/imath] is not uniformly continuous on [imath][1, \infty][/imath] I was trying to come up with a counter example using sequence, but I could not find one. |
3004963 | Probability of algebraic solutions for polynomials with integer coefficients
Polynomials of degree larger than 4 with integer coefficients may have non-algebraic solutions, that is, zeros that do not have an algebraic expression (over the integers). How likely is this to happen? For example, of the [imath](2M+1)^{n}[/imath] polynomials of degree less than or equal to [imath]n\geq 5[/imath] with coefficients in [imath]\{-M,\dots,M\}[/imath], what is the asymptotic behavior, as [imath]M\to\infty[/imath], of the proportion of those polynomials with no/at least one/only algebraic solutions? | 1019449 | Natural density of solvable quintics
A recent question asked about the topological density of solvable monic quintics with rational coefficients in the space of all monic quintics with rational coefficients. Robert Israel gave a nice proof that both solvable and unsolvable quintics are dense in [imath]\Bbb Q^5[/imath]. A natural (non-topological) way to ask about the relative density of solvable quintics in all quintics is to ask about their natural density, as follows: Write our quintics as [imath]x^5+a_0x^4 + \dots + a_4[/imath], with [imath]a_i \in \Bbb Z[/imath]. Write [imath]f(N) := \frac{\text{# of solvable quintics with } |a_i| < N}{(2N+1)^5}.[/imath] Robert Israel's answer to the linked question gives some data that supports our intuition that yes, the solvable quintics are in fact extremely rare. Do we indeed have [imath]\lim_{N \to \infty} f(N) = 0[/imath]? Are there known nice asymptotics for [imath]f(N)[/imath]? |
3006087 | How to divide numbers [imath]1,\ldots,n[/imath] into [imath]3[/imath] equal groups?
The sum of numbers [imath]1,\ldots,n[/imath] is equal [imath]n(n+1)/2[/imath] so it will be divisible by [imath]3[/imath] only for [imath]n=3k[/imath] and [imath]n=3k+2[/imath]. I am interested is that partition always possible for that cases. Let's exclude [imath]n=2[/imath] and [imath]n=3[/imath] because clearly it is not possible to do partition them. For [imath]n=8[/imath] we can do [imath]\{4,8\}[/imath], [imath]\{1,5,6\}[/imath] and [imath]\{2,3,7\}[/imath]. | 1559650 | Partitioning [imath]\{1,\cdots,k\}[/imath] into [imath]p[/imath] subsets with equal sums
Let [imath]p[/imath] be a prime. For which [imath]k[/imath] can the set [imath]\{1,\cdots,k\}[/imath] be partitioned into [imath]p[/imath] subsets with equal sums of elements ? Obviously, [imath]p\mid k(k+1)[/imath]. Hence, [imath]p\mid k[/imath] or [imath]p\mid k+1[/imath]. All we have to do now is to show a construction. But I can't find one. I have tried partitioning the set and choose one element from each set but that hasn't yielded anything. Any hint will be appreciated. |
3006078 | Example of subgroup of Direct Product which is not in usual form
I wanted to find group [imath]A\times B[/imath] such that it has subgroup which is not of form [imath]C\times D[/imath] where [imath]C < A[/imath] and [imath]D <B[/imath]. Actually I think this is wrong I tried to get proof. But my friend told this is true . that it possible to have group with such prpoerty ANy suggestion is appreciated | 313224 | subgroup of direct product of two groups
Its seems true but I did not see in any group theory book. Let [imath]G\times_{c} H[/imath] be product of two abelian groups (both are infinite). Then [imath]I[/imath] is subgroup of product iff [imath]I = A\times_{c} B[/imath] where [imath]A[/imath] is a subgroup of [imath]G[/imath] and [imath]B[/imath] is a subgroup of [imath]H[/imath]. Thanks |
3006319 | Prove [imath]\exists \xi \in (-2,2):f(\xi)+f''(\xi)=0.[/imath]
Problem Let [imath]f(x)[/imath] be twice-differentiable over [imath][-2,2][/imath], [imath]|f(x)|\leq 1[/imath] and [imath]f^2(0)+[f'(0)]^2=4.[/imath] Prove that [imath]\exists \xi \in (-2,2):f(\xi)+f''(\xi)=0.[/imath] My Proof First，by Lagrange' Mean Value Theorem，we have [imath]\exists \xi_1 \in (-2,0),\xi_2 \in (0,2):f'(\xi_1)=\frac{f(0)-f(-2)}{2},f'(\xi_2)=\frac{f(2)-f(0)}{2}.[/imath] Thus [imath]|f'(\xi)|\leq \frac{|f(0)|+|f(-2)|}{2}\leq \frac{1+1}{2}=1,[/imath] and likewise [imath]|f'(\xi_2)|\leq 1.[/imath] Denote[imath]F(x):=f^2(x)+[f'(x)]^2.[/imath] It's easy to verify that [imath]F(x)[/imath] is continuous and differentiable over [imath][\xi_1,\xi_2][/imath]. Therefore, [imath]F(x)[/imath] can reach its maximum value over the interval above. But [imath]F(\xi_1)\leq 1+f'^2(\xi_1)\leq 2<F(0),[/imath] and similarily [imath]F(\xi_2)\leq 2<F(0),[/imath] the maximum value can not come out at the end points of the interval.Hence, it must be taken over [imath](\xi_1,\xi_2)[/imath]. Let the point be [imath]x=\xi\in(\xi_1,\xi_2)\subset (-2,2)[/imath]. Then [imath]F'(\xi)=0,[/imath] which implies that [imath]2f'(\xi)[f(\xi)+f''(\xi)]=0.[/imath] Now, Notice that [imath]4\leq F(\xi)=f^2(\xi)+f'^2(\xi)\leq 1+f'^2(\xi),[/imath] which implies [imath]f'(\xi)\geq 3[/imath]，and hence[imath]f'(\xi) \neq 0[/imath]. It follows that [imath]f(\xi)+f''(\xi)=0,[/imath] which is desired. Please correct if I'm wrong. Hope to see other solutions. THX. | 1685616 | Prove that there exists at least one [imath]x_0\in\mathbb{R}[/imath], such that [imath]f(x_0)+f''(x_0)=0[/imath]
Let [imath]f:\mathbb{R}\to\mathbb{R}[/imath] be a function, two times differentiable with [imath]\left|f(x)\right|\leq1,\forall x\in\mathbb{R}[/imath] and [imath]f^2(0)+\left(f'(0)\right)^2=4[/imath]. Prove that there exists at least one [imath]x_0\in\mathbb{R}[/imath], such that [imath]f(x_0)+f''(x_0)=0[/imath]. I have tried the usual Rolle method by [imath]e^x[/imath], but didn't go far... Any help available? |
3005344 | Prove that in finite steps you can know if the solution set of linear programming problem is unbounded
Given is [imath]\max\left\{c^T \cdot x | Ax \leq b, x \geq 0\right\}[/imath] which is a linear programming problem and its solution set [imath]M[/imath]. Prove that you can find out in finite steps if [imath]M[/imath] is unbounded. It will surely work using the Simplex algorithm but I'm not sure how to prove this correctly? I think the same proof to this is: Prove that the Simplex algorithm will stop (in finite steps). It will stop in finite steps because there are finite number of feasible bases. In addition, every feasible basis determines a basic feasible solution with an objective value. To make sure that a feasible basis is never used more than once, one can strictly increase the objective value in every step. For this reason, the Simplex algorithm will come to an end in finite steps. But I'm not sure if this is fine like that or if it's really a proof at all? :S | 2992493 | If the solution set of linear programming problem is unbounded, can you find that out in finite steps?
Let [imath](P) \max\left\{c^T \cdot x \mid A \cdot x \leq b, x \geq 0\right\}[/imath] be an arbitrary linear programming problem and [imath]M[/imath] its solution set. Is it possible to find out if [imath]M[/imath] is unbounded (in finite steps)? I think the answer is yes because we have linear program here and that's why it's possible to tell in polynomial time if it has feasible solution or infeasible. So there must be an algorithm to do this in polynomial time e.g. in finite steps. But I don't know any algorithm to do this. So far I've read about the simplex method in a book and if I understood correctly, if you run simplex method on the LP above, you will see in the table that the problem is unbounded because there will appear one or more variables where you can pivot these to [imath]\infty[/imath]. But this is only thought is it correct like that? Maybe it's possible with another way too? |
3006010 | Counting 2-partitions of [imath]\{1, 2, \cdots, n\}[/imath] of equal size and sum
The following question solves the existence of such a partition: Partitioning [imath]\{1,\cdots,k\}[/imath] into [imath]p[/imath] subsets with equal sums. However, I'm trying to count how many such 2-partitions exist. Specifically, Let [imath]X = \{1, \cdots, 2n\}[/imath]. How many unique [imath]S \subset X[/imath] has cardinality [imath]n[/imath] and [imath]\sum S = 2n^2+n[/imath]? I'm trying to count the number of exact sequences on finite dimensional vector spaces [imath]\mathbb R^1, \cdots, \mathbb R^{2n}[/imath]. | 186530 | The number of ways to represent [imath](n^2+n)/4[/imath] as a sum of [imath]n/2[/imath] distinct integers in [imath]1,\dots,n[/imath]
For any positive integer [imath]n[/imath] (using integer division only), let [imath]P(n)[/imath] denote the number of ways in which the integer [imath](n^2+n)/4[/imath] can be expressed as a sum of exactly [imath]n /2[/imath] distinct elements of the set [imath]\{1,2,3,\dots, n\}[/imath]. What is [imath]P(n)[/imath] in terms of n? Specifically, how exponential is it? Is this less than [imath]2^{n/2}[/imath]? |
3006841 | Does [imath]\sum_{n=2}^\infty \frac{1}{n\log(n)}[/imath] converge or diverge?
So I know that [imath]\sum_{n\in\mathbb{N}}1/n[/imath] diverges and [imath]\sum_{n\in\mathbb{N}}1/n^2[/imath] converges. What about the series [imath]\sum_{n=2}^\infty1/n(\log(n))[/imath]? I'm pretty confident that it diverges but is there a quick justification? | 574503 | Infinite series [imath]\sum _{n=2}^{\infty } \frac{1}{n \log (n)}[/imath]
Recently, I encountered a problem about infinite series. So my question is how to know whether the infinite series [imath]\sum _{n=2}^{\infty } \frac{1}{n \log (n)}[/imath] is convergent? |
3007162 | In the Category of [imath]\mathbf{ Set } [/imath] is "the product of an empty set of sets a one-element set"?
I was reading these notes on Category Theory and it said (paraphrased to add context): Exercise 4: Explain why in Set (the Category of Sets), the product of an empty set of sets is a one-element set. which I think is incorrect. The product of two empty sets (or any number) is empty because we are considering: [imath] \emptyset \times \emptyset = \{ (a,b) : a \in \emptyset, b \in \emptyset \} = \emptyset[/imath] where [imath]a \in \emptyset , b \in \emptyset[/imath] are false, so the above is the [imath]\emptyset[/imath] which is NOT a one element set (its a zero element set). This should be trivial so I am assuming I am somewhere mis reading the natural language of the exercise. Someone help me catch where is it? i.e whats being asked and what the answer is? | 3007160 | In the category [imath]\mathbf{Set}[/imath] is "the product of an empty set of sets a one-element set"?
I was reading these notes on Category Theory and it said (paraphrased to add context): Exercise 4: Explain why in [imath]\textbf{Set}[/imath] (the Category of Sets), the product of an empty set of sets is a one-element set. which I think is incorrect. The product of two empty sets (or any number) is empty because we are considering: [imath] \emptyset \times \emptyset = \{ (a,b) : a \in \emptyset, b \in \emptyset \} = \emptyset[/imath] where [imath]a \in \emptyset , b \in \emptyset[/imath] are false, so the above is the [imath]\emptyset[/imath] which is NOT a one element set (its a zero element set). This should be trivial so I am assuming I am somewhere mis reading the natural language of the exercise. Someone help me catch where is it and what the answer should be? i.e. whats being asked and the answer? |
3006574 | Finding probability distributions associated with moment generating functions
I think the answer to my question is pretty simple, but I'm not able to figure it out. The question is: Find the distribution which corresponds to the moment generating function [imath]\frac{2e^t}{3-e^t}[/imath]. I've checked a few tables that list common MGFs and their associated distributions, but couldn't find any that matched this MGF. Do I need to modify the function before using a table? | 3005546 | Find the Distribution that corresponds to the given MGF
I am working on a problem and am a little bit confused. I need to find the distribution that corresponds to the MGF: [imath]2e^t\over3-e^t[/imath] Do we need to separate this into something like: 2e[imath]^t[/imath] and [imath]1\over3-e^t[/imath] It seems like we might be able to separate this into X ~ Exp() or Poi() |
1011498 | How big can the deck get while still allowing this puzzle to be solvable?
Here's a classic puzzle (I think Martin Gardner talks about it somewhere, though I'm not sure exactly where): Alice and Bob are co-conspirators. Alice is dealt five random cards from a standard deck; she looks at the cards, and then shows four of them to Bob in some chosen order. Using only the four cards and the order they are shown in, Bob must determine the identity of the fifth card. The solution that I know of is: Alice and Bob agree on an ordering of the [imath]52[/imath] cards, and a bijection between [imath]S_3[/imath] to [imath]\{1,\dots,6\}[/imath]. With this information, an ordering of any three cards gives an integer between [imath]1[/imath] and [imath]6[/imath] inclusive. Now, Alice uses the pigeonhole principle to find two cards of the same suit. Without loss of generality, their ranks are [imath]r_1[/imath] and [imath]r_2[/imath] where [imath]r_1+a \equiv r_2 \pmod{13}[/imath] and [imath]1 \leq a \leq 6[/imath] (if they weren't originally, swap [imath]r_1[/imath] and [imath]r_2[/imath]). Alice puts the card of rank [imath]r_1[/imath] first (naming the suit) and then orders the other three cards in a way that tells Bob [imath]a[/imath] (naming the rank). Notice that [imath]52[/imath] cards is the maximum possible number that this solution works for, since it depends heavily on there being only four suits and thirteen cards in each suit. However, it seems like it wastes a fair amount of information whenever the suit distribution isn't 2-1-1-1. Is there a method to do this which works on a deck containing more than [imath]52[/imath] cards? The obvious upper bound on deck size is that Bob sees a list of four ordered cards and can deduce from it the original five cards Alice was dealt, so it it works for a deck of size [imath]n[/imath] we must have [imath] 4!\binom{n}{4} \geq \binom {n}{5} [/imath] or [imath]n \leq 124[/imath]. But that still leaves quite a gap... | 800351 | 52-card trick for a larger deck?
Long ago someone demonstrated the following card trick with a standard 52-card deck: (1) A volunteer selects 5 cards from a shuffled deck, which the performer does not see. (2) The assistant puts exactly 4 of those in a pile and passes it to the performer. (3) The performer can then identify the 5th card. There are a few ways to execute this trick, one of which goes as follows. (1) The assistant finds a suit common to 2 of the 5 cards, with values [imath]x[/imath] and [imath]y[/imath] modulo [imath]13[/imath]. (2) He then removes one of them, say [imath]x[/imath], such that [imath]x-y \in [1..6][/imath] modulo [imath]13[/imath]. (3) He then puts the other one at the top of the pile. (4) And he puts the 3 other cards after that in a permutation corresponding to [imath]x-y[/imath] modulo [imath]13[/imath]. Since (1) works by pigeonhole, (2) works because [imath]\frac{13-1}{2} = 6[/imath], and (4) works because there are 6 possible permutations of 3 cards, the trick succeeds. My question is whether the same trick can work with more than 52 cards. It is not obvious to me how to do so, nor compute the maximum number of (pairwise distinct) cards possible. Also, it seems to me that even if the trick can be made to work for [imath]n[/imath] cards, it may not work for [imath]n-1[/imath] cards! Any ideas? |
3007367 | Proving that two real functions are equal
Let [imath]f : \mathbb{R} \to \mathbb{R}[/imath] and [imath]g : \mathbb{R} \to \mathbb{R}[/imath] be continuous functions. Suppose that [imath]D ⊆ \mathbb{R}[/imath], and that [imath]D[/imath] is dense in [imath]\mathbb{R}[/imath]. Suppose that [imath]f(x) = g(x)[/imath] for every [imath]x ∈ D[/imath]. Prove that [imath]f(x) = g(x)[/imath] for every [imath]x ∈ \mathbb{R}[/imath]. Any tips for where I can start here? I'm pulling up blanks for this. EDIT: there is a very similar question asked, but I have no idea what a metric space is so the whole thing didn't mean a ton to me unfortunately. | 1146931 | Continuous function and dense set
Let [imath]f: X \rightarrow Y [/imath] and [imath]g:X \rightarrow Y[/imath] continuous functions and [imath](X,d_X),(Y,d_Y)[/imath] metric space. Let [imath]E[/imath] by a dense subset in [imath]X[/imath]. Prove that [imath]f(E)[/imath] is a dense subset in [imath]f(X)[/imath]. If [imath]f(p)=g(p)[/imath] for all [imath]p\in E[/imath], prove that [imath]f(x)=g(x)[/imath] for all [imath]x\in X[/imath]. My attempt: since [imath]E \subset X \Rightarrow f(E)\subset f(X)[/imath] and from the dense property we have [imath]f(X) \subset f(\overline{E})[/imath] and since [imath]f[/imath] is continuous, [imath]f(X) \subset f(\overline{E}) \subset \overline{f(E)}[/imath]. Then, [imath]f(E)[/imath] is dense in [imath]f(X)[/imath]. Suppose that [imath]f(p)=g(p), \; \forall p \in E[/imath]. Hence, we have [imath]d_{Y}(f(x),g(x)) \leq d_{Y}(f(x),f(p)) + d_{Y}(f(p),g(x)) = d_{Y}(f(x),f(p)) + d_{Y}(g(p),g(x)).[/imath] By the fact that [imath]f,g[/imath] are continuous, there exist a [imath]\delta_1[/imath] and [imath]\delta_2[/imath] such that [imath]d_X(x,p) < \delta_1 \qquad d_X(x,p) < \delta_2[/imath] implies [imath]d_{Y}(f(x),f(p))<\epsilon /2 \qquad d_{Y}(g(p),g(x)) < \epsilon /2.[/imath] for all [imath]\epsilon >0.[/imath] Is this correct? |
3007453 | find [imath]\limsup[/imath] and [imath]\liminf[/imath] of [imath]a_n=\{\sqrt{n} - \lfloor\sqrt{n}\rfloor[/imath] [imath]n\in\mathbb{N}\}[/imath]
I have to find [imath]\limsup[/imath] and [imath]\liminf[/imath] of [imath]a_n=\{\sqrt{n} - \lfloor\sqrt{n}\rfloor : n \in\mathbb{N}\}[/imath] I suppose that I have to relate to subsequences . First one is [imath]a_{n_k}[/imath] for all n that maintains [imath]\sqrt n\in\mathbb {N}[/imath] , so the limit is [imath]0[/imath]. Second one [imath]a_{n_j}[/imath] includes the rest of the elements.I suppose it converges to 1, but dont know how to prove it. | 2015590 | Supremum of [imath]\sqrt{n} -\left\lfloor\sqrt{n}\right\rfloor [/imath]
I've got the following question: [imath]A = \{\sqrt{n} - \left\lfloor\sqrt{n}\right\rfloor : n \in \mathbb{N}\}[/imath] Find inf A, sup A inf A is no big deal, but I have a hard time proving sup A is 1 I've tried to prove that: [imath]\sqrt{n^2 - 1} -\left\lfloor\sqrt{n^2-1}\right\rfloor > \sqrt{n} -\left\lfloor\sqrt{n}\right\rfloor [/imath] but couldn't find a way to do that |
3007449 | Bolzano-Weistrass theorem true in any vector space?
Let’s say I have a vector space [imath]V[/imath] endowed with an order relation : [imath]\leq[/imath]. Then is Bolzano-Weirstrass theorem true in [imath]V[/imath] ? In class we have seen Bolzano-Weirstrass theorem only in [imath]\mathbb{R}[/imath], that’s why I am wondering if this theorem is true in arbitrary ordered structure like for example a vector space. In the case where it’s not true for every vector space, is it possible to classify the vector space in which Bolzano-Weirstrass theorem holds ? | 202012 | In what spaces does the Bolzano-Weierstrass theorem hold?
The Bolzano-Weierstrass theorem says that every bounded sequence in [imath]\Bbb R^n[/imath] contains a convergent subsequence. The proof in Wikipedia evidently doesn't go through for an infinite-dimensional space, and it seems to me that the theorem ought not to be true in general: there should be some metric in which [imath]\langle1,0,0,0,\ldots\rangle, \langle0,1,0,0,\ldots\rangle, \langle0,0,1,0,\ldots\rangle, \ldots [/imath] is bounded but fails to contain a convergent subsequence. Let [imath]M[/imath] be a general metric space. What conditions on [imath]M[/imath] are necessary and sufficient for every bounded sequence of elements of [imath]M[/imath] to contain a convergent subsequence? |
3006488 | [imath]p[/imath] and [imath]q[/imath] orthogonal projections in [imath]B(H)[/imath], then the orthogonal projection onto the intersection space is the strong operator limit of [imath](pq)^n[/imath]
[imath]p[/imath] and [imath]q[/imath] orthogonal projections why is it true that the orthogonal projection onto [imath]p(H) \cap q(H)[/imath] is the strong operator limit of [imath](pq)^n[/imath] ? | 3006463 | Let [imath]H[/imath] be a Hilbert space with projections [imath]P,Q[/imath], then [imath]\lim_{n\to\infty}(PQ)^nx= Rx[/imath] for all [imath]x\in H[/imath], where [imath]R[/imath] is the projection to [imath]PH\cap QH[/imath]
Suppose we have a Hilbert space [imath]\mathcal H[/imath], with projections [imath]P,Q[/imath], and let [imath]R[/imath] be the projection onto [imath]P\mathcal H\cap Q\mathcal H[/imath], then I would like to show that [imath](PQ)^n\to R[/imath] strongly. We can clearly reduce to the case where [imath]P\mathcal H\cap Q\mathcal H=0[/imath] and [imath]P\mathcal H+Q\mathcal H=\mathcal H[/imath], in which case we must show that [imath](PQ)^n\to0[/imath] strongly. Now, if the limit [imath]y=\lim_{n\to\infty}(PQ)^n x[/imath] exists, then if it is nonzero, then [imath]\|y\| = \|PQ y\|<\|Q y\|\le\| y\|[/imath] which is a contradiction, so once we have convergence, we're done. However, I don't see quite how to show convergence in a general infinite dimensional case. |
3007428 | All real number for which [imath]n[/imath] in [imath]5^n+7^n+11^n=6^n+8^n+9^n[/imath]
Finding all real number [imath]n[/imath] in [imath]5^n+7^n+11^n=6^n+8^n+9^n[/imath] Try: From given equation [imath]n=0,1[/imath] are the solution But i did not understand any other solution exists or not Although i have tried like this way [imath]\bigg(\frac{5}{9}\bigg)^n+\bigg(\frac{7}{9}\bigg)^n+\bigg(\frac{11}{9}\bigg)^n = \bigg(\frac{6}{9}\bigg)^n+\bigg(\frac{8}{9}\bigg)^n+1[/imath] Right side is strictly increasing function. but i have a confusion whether left side is strictly increasing or not could some help me how to solve it, thanks | 2840394 | Find the number of natural solutions of [imath]5^x+7^x+11^x=6^x+8^x+9^x[/imath]
Find the number of natural solutions of [imath]5^x+7^x+11^x=6^x+8^x+9^x[/imath] It's easy to see that [imath]x=0[/imath] and [imath]x=1[/imath] are solutions but are these the only one? How do I demonstrate that? I've tried to write them either: [imath]5^x+7^x+11^x=2^x*3^x+2^{3x}+3^{2x}[/imath] or [imath]5^x+7^x+11^x=(5+1)^x+(7+1)^x+(11-2)^x[/imath] and tried to think of some AM-GM mean inequality or to divide everything by [imath]11^x[/imath], but those don't seem like the way to go. Any hints? |
3007240 | If [imath]M/N\cong M[/imath], can we conclude that [imath]N=0[/imath]?
Let [imath]M[/imath] be an [imath]R[/imath]-module, where we may assume that [imath]R[/imath] is an integral domain. Let [imath]N[/imath] be a submodule of [imath]M[/imath]. Suppose that [imath]M/N\cong M[/imath]. Can we conclude that [imath]N=0[/imath]? (If no, what are some sufficient conditions that make it true?) Update: I learn that for "infinite dimensional" cases it can fail. How about when [imath]M[/imath] is finitely generated, does it work? Thanks a lot. | 708633 | Is there a (f.g., free) module isomorphic to a quotient of itself?
My question is as in the title: is there an example of a (unital but not necessarily commutative) ring [imath]R[/imath] and a left [imath]R[/imath]-module [imath]M[/imath] with nonzero submodule [imath]N[/imath], such that [imath]M \simeq M/N[/imath]? What if [imath]M[/imath] and [imath]N[/imath] are finitely-generated? What if [imath]M[/imath] is free? My intuition is that if [imath]N[/imath] is a submodule of [imath]R^n[/imath], then [imath]R^n/N \simeq R^n[/imath] implies [imath]N=0[/imath]. It seems like [imath]N\neq 0[/imath] implies [imath]R^n/N[/imath] has nontrivial relations, so [imath]R^n/N[/imath] can't be free. If [imath]R^n/N \simeq R^n[/imath], we'd have an exact sequence [imath]0 \rightarrow N \hookrightarrow R^n \twoheadrightarrow R^n/N \simeq R^n \rightarrow 0[/imath] which splits since [imath]R^n[/imath] is free, so [imath]R^n \simeq R^n \oplus N[/imath]. Does this imply [imath]N=0[/imath]? What if we assume [imath]R[/imath] is commutative, or even local? Maybe Nakayama can come in handy. I'm interested in noncommutative examples too. Thanks! |
3007937 | If X is a binomial random variable, find [imath]E(X(X - 1) (X - 2))[/imath]
If X is a binomial random variable, how do you find [imath]E(X(X - 1)(X - 2))[/imath]? Here's my approach: [imath]E(X(X - 1)(X - 2)) = E(X^3 - 3X^2 + 2X)[/imath] [imath]E(X) = np, Var(X) = np(1 - p)[/imath] [imath]Var(X) = E(X^2) - (E(X))^2[/imath] [imath]E(X^2) = np(1 - p) + (np)^2[/imath] So, [imath]E(X^3 - 3X^2 + 2X) = E(X^3) - 3E(X^2) + 2E(X)[/imath] How should I deal with [imath]E(X^3)[/imath]? | 3007315 | Given that [imath]X \sim \operatorname{Binomial}(n,p)[/imath], Find [imath]\mathbb{E}[X(X-1)(X-2)(X-3)][/imath]
Given that [imath]X \sim \operatorname{Binomial}(n,p)[/imath], Find [imath]\mathbb{E}[X(X-1)(X-2)(X-3)][/imath]. It is suggested that I can transform it into [imath]\begin{align} \mathbb{E}[X(X-1)(X-2)(X-3)] &=\sum_{k=0}^n k(k-1)(k-2)\mathbb{P}\{X=k\}\\ &=\sum_{k=3}^{n+3} (k-3)(k-4)(k-5)\mathbb{P}\{X=k-3\}\\ &=\sum_{k=0}^n i(i-1)(i-2)\mathbb{P}\{X=i\} \end{align}[/imath] But then I just have no idea about how can i do it. I suspect that it needs something similar to this post but the steps are quite different from this one. Please help. |
3007835 | Elementary probability theory
I would like to ask a simple question but I'm unable to understand the logic behind it. From an urn containing [imath]M[/imath] balls [imath]n[/imath] balls are drawn with replacement. What is the probability that at least one ball is drawn more than once? Thank you in advance. | 489772 | Probability of sampling with and without replacement
In sampling without replacement the probability of any fixed element in the population to be included in a random sample of size [imath]r[/imath] is [imath]\frac{r}{n}[/imath]. In sampling with replacement the corresponding probability is [imath]\left[1- \left(\frac{1}{1-n}\right)^r\right][/imath]. Please help me show how this is proved. |
3008068 | Use the [imath]\epsilon - \delta[/imath] definition to verify that [imath]\lim_{(x,y)\to(0,0)}\frac{x^3-y^3}{x^2+y^2} = 0[/imath]?
I know that I have to find a [imath]\epsilon > 0[/imath] s.t. |[imath]\frac{x^3-y^3}{x^2+y^2}[/imath]|<[imath]\epsilon[/imath] whenever [imath]0<\sqrt{x^2+y^2}<\delta[/imath] but what are the steps to proving this? I'm looking for an answer which shows the steps as well as explains the intuition. Thanks | 23199 | Finding the limit of a 2-dimensional function
How can I prove that [imath]\displaystyle{\lim_{(x, y) \to (0, 0)} \frac{x^3 - y^3}{x^2 + y^2}}=0[/imath]? The method I've been taught is the pinching one, where you compare the absolute value of the function to greater limits that are known to equal zero, but I haven't managed to find a comparison that works. Could someone point me in the right direction? |
3007953 | Prove that number [imath]4p^2+1[/imath] can show as sum of squares of three different numbers
Let [imath]p>3[/imath] is prime number. Prove that number [imath]4p^2+1[/imath] can show as sum of squares of three different numbers. Only what I know that every prime number [imath]p>3[/imath] can show as [imath]p=6k+1[/imath] or [imath]p=6k-1[/imath], such that [imath]k \in \mathbb Z[/imath]. If I put [imath]p=6k+1[/imath], then [imath]4(6k+1)^2+1=(12k)^2+(24k+3)^2-(24k+2)^2[/imath], here I did not show what they want in task. For [imath]p=6k-1[/imath] things do not change, do you have some idea? | 2230884 | Prove that for every prime number [imath]p>3[/imath], [imath]4p^2+1[/imath] can be written as the sum of three square numbers
Given that [imath]p>3[/imath] prove that [imath]4p^2+1[/imath] can be written as the sum of three distinct positive square numbers. Plugging in [imath]5[/imath] I get [imath]101=49+36+16=7^2+6^2+4^2[/imath] I also know that all primes greater than [imath]3[/imath] can be written in the form [imath]3k+1[/imath] and [imath]3k+2[/imath] but plugging those values in I get: [imath]36k^2+24k+5[/imath], [imath]36k^2+48k+17[/imath] and the solution probably lies in arranging these numbers in such a way that we get the desired squares, but I can't come up with a combination ,or is my "idea" not even in the right direction? |
3000650 | Number Theory Euler totient function
Prove that [imath]\sum_{d\mid n}(-1)^{n/d}\varphi(d)=\begin{cases}-n&2\nmid n\\0&2\mid n\end{cases}[/imath] I have came across the above question. I have done the following: [imath]n[/imath] is odd then so is [imath]n/d[/imath] which would result [imath](-1)^{n/d}[/imath] [imath]=-1[/imath] I am assuming you would have to use Gauss: [imath]\sum_{d/n} φ(d) = n[/imath] [imath]===>[/imath] [imath]\sum_{d/n} (-1)^{n/d}φ(d) = -\sum_{d/n}φ(d)=-n[/imath] [imath]n[/imath] is even then [imath]n=2^{a}m[/imath], where we can say [imath]m[/imath] is odd. I am not sure what else to do from here. Any help would be appreciated. | 2079155 | Proof on Euler totient function
For a positive integer [imath]n[/imath], prove that: [imath]\sum_{d\mid n} (-1)^{n/d} \cdot \varphi(d) = \begin{cases} 0 & \text{if $n$ is even} \\ -n & \text{if $n$ is odd} \end{cases}[/imath] I can't even think, how, to begin with, this problem, please provide easy proof of this. |
3008520 | Proving that [imath]\mathbb{Z}_{n}[/imath] is cyclic
It should be very basic, but I'm struggling to understand the solution. I'm trying to prove that to group [imath]\mathbb{Z}_{n}[/imath] is cyclic. Everywhere I tried to find a solution I came across of the saying [imath]\mathbb{Z}_{n}[/imath] is cyclic because it has a generator [imath]1[/imath] (meaning [imath]\mathbb{Z}_{n} = \langle 1 \rangle[/imath]). I understand the meaning of generator, but I don't understand why the generator is [imath]1[/imath]. Probably I'll feel stupid when I'll finally understand. Why [imath]\mathbb{Z}_{n} = \langle 1 \rangle[/imath] when [imath]\mathbb{Z}_{n}[/imath] is the Sum group of integers modulo [imath]n[/imath]. | 494811 | Is [imath]\mathbb{Z}/\langle n\rangle[/imath] cyclic and of order [imath]n[/imath]?
1) Is [imath]\mathbb{Z}/\langle n\rangle[/imath] cyclic and of order [imath]n[/imath]? Why or why not? ([imath]\mathbb{Z}/\langle n\rangle[/imath] is defined to be the factor group of [imath]\mathbb{Z}[/imath] determined by [imath]\langle n\rangle[/imath].) I guess I'm having trouble understanding factor groups (quotient groups make much more sense) and cyclic groups. My textbook isn't the best at explaining what's going on. I just chose some past homework problems that looked related to my problems. This is my first abstract algebra course and I'm floundering a lot. |
3005789 | Showing when two functors are naturally isomorphic, if one is faithful, then the other also is.
Supposing we have a natural isomorphism [imath]\tau : S \rightarrow T[/imath] between functors [imath]S,T : \mathscr{C} \rightarrow \mathscr{D}[/imath], how exactly do we show that if [imath]S[/imath] is faithful, then so is [imath]T[/imath]? If [imath]S[/imath] is faithful, then it is injective on the hom-sets, i.e. if [imath]f = g[/imath], then [imath]S(f) = S(g)[/imath]. I also know that the components of the natural transformation form a commutative square defined by [imath]\tau \circ S(f) = T(f) \circ \tau[/imath], and that the ultimate implication I want to show is (probably along the lines of?) [imath]f = g \;\;\; \Rightarrow \;\;\; S(f)=S(g) \;\;\; \Rightarrow \;\;\; T(f)=T(g)[/imath] (with the first implication being given as stated by [imath]S[/imath] being faithful). I've tried playing with around with some of the compositions but the closest I've gotten would be [imath]T(\tau(f)) = T(\tau(g))[/imath] (which came from [imath]S(f)=S(g)[/imath], composing on the right by [imath]\tau[/imath] which is injective, and then applying the definition from the commuting square) which I don't think is sufficient. Really, I'm starting to think that the composition idea isn't going to work, and that proving such things from a categorical perspective is a little bit more nuanced than we'd see in, say, a set theory class, and maybe I'm just not thinking "categorically", so to speak? I'm not sure. Anyhow, sorry for the dumb question, but any ideas? | 581555 | What properties do natural isomorphisms between functors preserve?
If [imath]F,G: C \rightarrow D[/imath] are functors between regular categories and [imath]F \Rightarrow G[/imath] is a natural isomorphism, is it true that if [imath]F[/imath] is faithful {resp. full) then [imath]G[/imath] is full? I believe this is true but would like some confirmation. What about if [imath]F[/imath] is essentially surjective? Is it necessarily true that [imath]G[/imath] is essentially surjective? Thanks. |
3008614 | Does [imath]\aleph_0!=\omega[/imath]?
More generally, is the order type of some cardinal [imath]\alpha[/imath] equal to [imath]\alpha![/imath]? Related What is [imath]\aleph_0![/imath]? Factorial of Infinite Cardinal factorial of infinite Cardinals | 2551471 | What is [imath]\aleph_0![/imath]?
What is [imath]\aleph_0![/imath] ? I know that in the original definition the factorial is defined for natural numbers but, what if we extend this concept to cardinal numbers? This concept has been extended to the real numbers by the [imath]\Gamma[/imath] function but I never see this kind of extension before. This is a proof that I made by myself and can be incorrect but still interesting for me. [imath]\aleph_0\times(\aleph_0 - 1)\times(\aleph_0 - 2)\times ...[/imath] We can rewrite this as [imath]\aleph_0! = \prod_{i = 1}^{\infty}(\aleph_0 - i) = \prod_{i = 1}^{\infty}(\aleph_0)[/imath] But, is this equal to: [imath]\aleph_0^{\aleph_0}[/imath] Also, if we assume the continumm hypothesis [imath]2^{\aleph_0} = \mathfrak{c} \leq \aleph_0^{\aleph_0} \leq \mathfrak{c}[/imath] Hence, [imath]\aleph_0! = \mathfrak{c}[/imath] |
2686575 | Find integer roots for quadratic
Find all positive integers [imath]a[/imath], [imath]b\neq1[/imath] such that each of the equations:- [imath]x^2-ax+b=0[/imath] and [imath]x^2-bx+a=0[/imath] has distinct positive integer roots. I have taken x and y as roots for first equation and p and q for the second equation and got two equations from the sum of roots and product of roots but I am not able to progress further.Any idea will be appreciated | 2451636 | Find all positive integers [imath]a,b[/imath] such that each of the equations [imath]x^2-ax+b=0[/imath] and [imath]x^2-bx+a=0[/imath] has distinct positive integral roots.
Find all positive integers [imath]a,b[/imath] such that each of the equations [imath]x^2-ax+b=0[/imath] and [imath]x^2-bx+a=0[/imath] has distinct positive integral roots. Suppose [imath]\alpha,\beta[/imath] are roots of the first equation and [imath]\gamma,\delta[/imath] are of the second, then we have these two sets of equations [imath]a=\alpha+\beta[/imath] [imath]b=\alpha\beta[/imath] and [imath]b=\gamma+\delta[/imath] [imath]a=\gamma\delta[/imath] How to proceed? |
2095843 | Find the maximum value of the expression [imath]{\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}[/imath].
Find the maximum value of the expression : [imath]{\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}[/imath] where [imath]x,y,z[/imath] are real numbers satisfying the condition that [imath]x+y+z=1[/imath]. Taking [imath]x=y=z=\frac {1}{3}[/imath], I get the expression as [imath]\frac {3x}{1+x^2}[/imath], which is equal to [imath]\frac {1}{1+{\frac{1}{9}}}[/imath] or [imath]\frac {9}{10}[/imath]. How can I actually solve the problem without making unnecessary assumptions ? | 266893 | Prove that [imath]\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \leq \frac{9}{10}[/imath] if [imath]a+b+c=1[/imath].
Let [imath]a,b,c[/imath] are real number such that [imath]a+b+c=1[/imath]. Prove that: [imath]\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \leq \frac{9}{10}.[/imath] |
3009283 | Probability that the man is the first to die
If p is the probability that a man aged x will die in a year then the probability that out of n men [imath]A_1, A_2 ,A_3 ...A_n[/imath], each aged x, [imath]A_1[/imath] will die in an year and be the first to die is: ? a) [imath]1- (1- p)^{n-1}[/imath] b)[imath](1-p)^n[/imath] c) [imath]\dfrac 1n [1- (1-p)^n][/imath] d) [imath]\dfrac 1n (1-p)^n[/imath] Attempt: Order of dying matters. Choose [imath]r[/imath] out of [imath]n-1[/imath] and arrange them (i.e. their deaths) in [imath]r![/imath] ways. So [imath]P(E) = p\times{^{n-1}C_0} (1- p)^{n-1}+ p \times p\times {^{n-1}C_1} (1-p)^{n-1} + p\times p^2 \times 2!\times {^{n-1}C_2} (1- p)^{n-2}+...[/imath] [imath]\implies P(E) = p \sum_{r=0}^{n-1} {^{n-1}}C_r (1-p)^{n-1-r}p^r \times r![/imath] But this doesn't match any of the options. Could someone please tell me where I have gone wrong? Edit: My prime focus is the mistake that I have made in solving which that question doesn't cover. therefore, please don't mark it as a duplicate | 1453169 | Then what is the probability that out of [imath]n[/imath] men [imath]A_1,A_2,....,A_n[/imath] each aged [imath]x[/imath] years,[imath]A_1[/imath] will die and will be the first to die
Let [imath]p[/imath] be the probability that a man aged [imath]x[/imath] years will die in a year time.Then what is the probability that out of [imath]n[/imath] men [imath]A_1,A_2,....,A_n[/imath] each aged [imath]x[/imath] years,[imath]A_1[/imath] will die and will be the first to die? The probability that a man aged [imath]x[/imath] years will die in a year time[imath]=p[/imath] The probability that a man aged [imath]x[/imath] years will not die in a year time[imath]=1-p[/imath] But from here on,i got stuck.Some hints and suggestions are welcome. |
3009498 | Is [imath]F[/imath] continuously differentiable at [imath]x=0[/imath]?
Do you have any tips? Especially for the second part? IS it enough to say since [imath]f(x)[/imath] is continuous at [imath]x=0[/imath] and [imath]f[/imath] is differentiable at [imath]x=0[/imath]. Is [imath]f[/imath] continuously differentiable? Here is the question. Let [imath] f(x)= \begin{cases} x^2\sin\big(\frac{1}{x}\big) & \text{ if }x\neq 0\\ 0 & \text{ if }x= 0 \end{cases} [/imath] Show that [imath]f[/imath] is differentiable at [imath]x=0[/imath] and compute [imath]f^\prime(0)[/imath]. Is [imath]F[/imath] continuously differentiable at [imath]x=0[/imath]? Edit: For the second part, I used the fundamental theorem of calculus part 2. f is continuous and according to that theorem, so F is continuously differentiable. P.S: I don't think this question is a duplicate of another question. In that question, it is asking for the derivative of f. But, in this question, it is asking for the integral of f as the capital F is a symbol for the integral of f. I don't understand why you keep insisting that this question is a duplicate. | 1391544 | Differentiable but not continuously differentiable.
Given [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] defined as [imath]f(x)=\left\{\begin{array}{cc}x^2\sin\left(\frac{1}{x}\right)&,x\neq 0\\ 0&,x=0\end{array}\right\}.[/imath] I am trying to prove [imath]f[/imath] is differentiable at [imath]x=0[/imath] but not continuously differentiable there. |
2092957 | Morphisms [imath]X\to\text{Spec }A[/imath] are in natural bijection with ring morphisms [imath]A\to\Gamma(X,\mathscr O_X)[/imath].
[imath]\newcommand{\Spec}{\operatorname{Spec}}[/imath]I'm working on the following problem from Vakil's notes: Show that (scheme) morphisms [imath]X\to\Spec A[/imath] are in natural bijection with ring morphisms [imath]A\to\Gamma(X,\mathscr O_X)[/imath]. Of course, any map [imath]X\to\Spec A[/imath] induces a map of global sections [imath]A\to\Gamma(X,\mathscr O_X)[/imath]. I am working out the other direction. It is certainly true if [imath]X[/imath] is affine, because affine scheme morphisms [imath]\Spec B\to\Spec A[/imath] are in bijection with ring maps [imath]A\to B[/imath]. In the general case, we cover [imath]X[/imath] with affine open sets [imath]\Spec B_i[/imath]. Then for each [imath]i[/imath] we get a map [imath]A\to\Gamma(X,\mathcal O_X)\overset{\text{res}}{\to}\Gamma(\Spec B_i,\mathcal O_X)=B_i[/imath] which in turn gives a map [imath]\pi_i:\Spec B_i\to\Spec A[/imath] (in particular, if we let [imath]\pi_i^{\#}[/imath] be the map [imath]A\to B_i[/imath], then the map is given by [imath]\pi_i(p)=(\pi_i^{\#})^{-1}(p)[/imath]). How can I show that these glue correctly to give a morphism of schemes [imath]\pi:X\to\Spec A[/imath]? | 1523392 | Prove that the natural map [imath]\alpha : \text{Hom}(X,\text{Spec} A) \rightarrow \text{Hom}(A,\Gamma(X,\mathcal{O}_X))[/imath] is an isomorphism
This is question 2.4 in Hartshorne. Let [imath]A[/imath] be a ring and [imath](X,\mathcal{O}_X)[/imath] a scheme. We have the associated map of sheaves [imath]f^\#: \mathcal{O}_{\text{Spec } A} \rightarrow f_* \mathcal{O}_X[/imath]. Taking global sections we obtain a homomorphism [imath]A \rightarrow \Gamma(X,\mathcal{O}_X)[/imath]. Thus there is a natural map [imath]\alpha : \text{Hom}(X,\text{Spec} A) \rightarrow \text{Hom}(A,\Gamma(X,\mathcal{O}_X))[/imath]. Show [imath]\alpha[/imath] is bijective. I figure we need to start off with the fact that we can cover [imath]X[/imath] with affine open [imath]U_i[/imath], and that a homomorphism [imath]A \rightarrow \Gamma(X,\mathcal{O}_X)[/imath] induces a morphism of schemes from each [imath]U_i[/imath] to [imath]\text{Spec} A[/imath] and some how glue them together. But I have no idea how to show that the induced morphisms agree on intersections. How does this work? |
3010146 | Question on Infinitely Differentiable Functions
Let [imath]f:(-1,1)\rightarrow\mathbb R[/imath] be infinitely differentiable on [imath](-1,1)[/imath] with [imath]f(0)=1[/imath] and has the following properties: (1) [imath]\vert f^{(n)}(x)\vert \le n![/imath] for every [imath]x\in (-1, 1)[/imath] and for every [imath]n\in \mathbb N[/imath] (2) [imath]f'(\frac{1}{m+1})=0[/imath] for every [imath]m\in \mathbb N[/imath] I need to: Find the value of [imath]f^{(n)}(0)[/imath] for each [imath]n\in \mathbb N[/imath] Determine the value of [imath]f(x)[/imath] for every [imath]x\in(-1,1)[/imath] I claim that [imath]f^{(n)}(0)=0[/imath] for all [imath]n\in \mathbb N[/imath] and I wanted to prove this by induction. When [imath]n=1[/imath], we note that by (2), [imath]f'(0)=0[/imath] by the sequential criterion for limits with [imath]m\rightarrow \infty[/imath]. However, I am not sure on how to prove the inductive step. I note that from (1), we have [imath]f(x)\le\frac{1}{1-x}[/imath] by Taylor's Theorem. Any help is greatly appreciated! | 2013147 | Let [imath]f:(-1,1) \rightarrow \mathbb{R}[/imath] be smooth on [imath](-1,1)[/imath], [imath]f(0) = 1[/imath]. Find [imath]f(x)[/imath].
Let [imath]f:(-1,1) \rightarrow \mathbb{R}[/imath] be infinitely differentiable on [imath](-1,1)[/imath], [imath]f(0)=1[/imath] and [imath]|f^{(n)}(x)| \le n![/imath] for every [imath]x\in(-1,1)[/imath] and for every [imath]n\in \mathbb{N} [/imath] [imath]f'(\frac{1}{m+1})=0[/imath] for every [imath]m \in \mathbb{N}[/imath] Determine the value of [imath]f(x)[/imath] for every [imath]x\in(-1,1)[/imath]. I guess [imath]f(x)=1[/imath] for every [imath]x\in(-1,1)[/imath]. |
3010424 | Evaluate [imath]\int_{2}^{4}\frac{\sqrt{\ln(9-x)} dx}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}[/imath]
Help with the evaluation of this definite integral. I don't know where to start. So we have :[imath]\frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}}=\frac{\ln(9-x)-\sqrt{\ln(9-x)}\sqrt{\ln(x+3)} }{\ln(9-x)-\ln(x+3)}[/imath] What's now? | 728164 | Integrating [imath] \int_2^4 \frac{ \sqrt{\ln(9-x)} }{ \sqrt{\ln(9-x)}+\sqrt{\ln(x+3)} } dx. [/imath]
Compute [imath] \int_2^4 \frac{ \sqrt{\ln(9-x)} }{ \sqrt{\ln(9-x)}+\sqrt{\ln(x+3)} } dx. [/imath] I am not sure how to start this one...I am thinking of a substitution to get started. |
3010477 | Proving that [imath]\sqrt 3\not\in Q(\sqrt[4]2)[/imath]
I came across this problem while solving another one. I will show how far I could get on my own: Suppose that [imath]\sqrt 3 \in Q(\sqrt[4]2)[/imath]. Since [imath]Q(\sqrt3)[/imath] is the smallest field containing both [imath]Q[/imath] and [imath]\sqrt3[/imath], thus [imath]Q(\sqrt3)\subset Q(\sqrt[4]2)[/imath]. Now [imath][Q(\sqrt[4]2):Q] =4 [/imath] and [imath][Q(\sqrt3):Q]=2[/imath] (I've already proven both), so[imath][Q(\sqrt[4]2):Q(\sqrt 3)] =2[/imath]. Therefore by the definition of degree extension, there exist [imath]p(x)\in Q(\sqrt 3)[X][/imath] of degree 2 realizing [imath]\sqrt[4]2[/imath], that is, exists: [imath]a,b,c\in Q(\sqrt 3)[/imath] such that: [imath]a(\sqrt[4]2)^2+b \sqrt[4]2+c =0.[/imath] How to proceed now? I should use the fact that those extensions lie in [imath]\mathbb R[/imath] to get a contradiction? I only know that [imath]a\neq 0[/imath], but what about the other coefficients? EDIT: I could find an argument below, check the answers. Check it out if you agree! The argument is just a continuation of the above one. | 1159217 | Show that [imath]\sqrt{2}\notin \mathbb{Q}(\sqrt[4]{3})[/imath]
I want to show that [imath]\sqrt{2}\notin \mathbb{Q}(\sqrt[4]{3})[/imath]. I think it would be easier to prove it using the following: [imath]\mathbb{Q}\subset\mathbb{Q}(\sqrt{3})\subset\mathbb{Q}(\sqrt[4]{3})[/imath]. Then [imath]\sqrt{2}\notin\mathbb{Q}(\sqrt{3})[/imath] so [imath]\sqrt{2}\in \mathbb{Q}(\sqrt[4]{3})\Longleftrightarrow \sqrt{2}=a+b\sqrt[4]{3}[/imath], with [imath]a,b\in\mathbb{Q}(\sqrt{3})[/imath]. I tried simplifying the above equation, but I did not get anything. |
3008777 | Question on conditional variance in the book Introduction to Probability 2nd edition
I read the book of Introduction to Probability SECOND EDITION by John N. Tsitsiklis et al. and in Sec. 4.3 Conditional Expectation and Variance Revisited, I think the [imath]var(X|Y)[/imath] is not [imath]nY(1-Y)[/imath] but [imath]nY(1-nY)[/imath]. My understanding is that since [imath]var(X|Y)=E(X^2|Y)-[E[X|Y]]^2[/imath] and [imath]E[X|Y]=nY[/imath] and [imath]E(X^2|Y)=nY[/imath], so [imath]var(X|Y)=nY(1-nY)[/imath] rather than [imath]nY(1-Y)[/imath]. How do you think about it? | 322232 | Variance of Binomial Distribution [imath]E[X^2][/imath]
so I'm trying to use the equation: [imath]Var(X) = E[X^2] - (E[x])^2[/imath], And for the [imath]E[X^2][/imath] part, I'm trying to use the method of indicators... However, when I do that, I get the same value as with [imath]E[X][/imath]... Is it wrong to try to use the method of indicators for this case? Basically, I end up with [imath]Var(X) = np(1-np)[/imath], when it should be [imath]Var(X) = np(1-p)[/imath]. |
3010444 | Is [imath]g[/imath] differentiable in [imath]0[/imath]?
Let [imath]g[/imath] be a continuous function on [imath]\mathbb{R}[/imath] such that: [imath]\displaystyle\lim_{x \to 0} \:\dfrac{g(2x)-g(x)}{x}=l \in \mathbb{R}.[/imath] I am supposed to prove that [imath]g[/imath] is differentiable at [imath]0[/imath] and [imath]g'(0)=l[/imath] Here is what I did: [imath]\displaystyle\begin{align} \lim_{x \to 0} \:\dfrac{g(2x)-g(x)}{x}&=\lim_{x \to 0}\:\frac{2(g(2x)-g(0))}{2x}-\dfrac{g(x)-g(0)}{x}\\&=\lim_{X \to 0}\:2\:\dfrac{g(X)-g(0)}{X} - \lim_{x \to 0}\:\dfrac{g(x)-g(0)}{x}\\ &=\lim_{x \to 0}\:\dfrac{g(x)-g(0)}{x}\\ &=l \end{align}[/imath] Since [imath]\displaystyle\lim_{x \to 0} \:\dfrac{g(x)-g(0)}{x} = l[/imath], then g is differentiable in [imath]0[/imath] and [imath]g'(0)=l[/imath] Is this valid? I'd appreciate any suggestion thanks. | 1864058 | Prove that [imath]f'(0)[/imath] exists and [imath]f'(0) = b/(a - 1)[/imath]
Problem: If [imath]f(x)[/imath] is continous at [imath]x=0[/imath], and [imath]\lim\limits_{x\to 0} \dfrac{f(ax)-f(x)}{x}=b[/imath], [imath]a, b[/imath] are constants and [imath]|a|>1[/imath], prove that [imath]f'(0)[/imath] exists and [imath]f'(0)=\dfrac{b}{a-1}[/imath]. This approach is definitely wrong: \begin{align} b&=\lim_{x\to 0} \frac{f(ax)-f(x)}{x}\\ &=\lim_{x\to 0} \frac{f(ax)-f(0)-(f(x)-f(0))}{x}\\ &=af'(0)-f'(0)\\ &=(a-1)f'(0) \end{align} I will show you a case why this approach is wrong: \[f(x)= \begin{cases} 1,&x\neq0\\ 0,&x=0 \end{cases}\] [imath]\lim_{x\to0}\dfrac{f(3x)-f(x)}{x}=\lim_{x\to0} \dfrac{1-1}{x}=0[/imath] but [imath]\lim_{x\to0}\dfrac{f(3x)}{x}=\infty[/imath],[imath]\lim_{x\to0}\dfrac{f(x)}{x}=\infty[/imath] Does anyone know how to prove it? Thanks in advance! |
3010998 | Why do similar matrices represent the same linear transformation?
I don’t understand the theorem: [imath]A[/imath] and [imath]B[/imath] are similar if and only if they represent the same linear transformation. I know one direction "If [imath]A[/imath] and [imath]B[/imath] represent the same linear transformation then they are similar", but why is the other direction also true? (I did only basic linear algebra so could you please give an easy proof if possible! Or any intuitions for this theorem! Thank you!) | 813785 | Similar Matrices and Linear Transformations
There is a theorem in the course notes but the lecturer didn't given the proof: If [imath]A,B\in M_n(K)[/imath], where [imath]K[/imath] is a field. Then [imath]A\sim B[/imath] if and only if they represent the same linear transformation [imath]T[/imath] of an [imath]n[/imath]-dimensional vector space [imath]V[/imath] over [imath]K[/imath] with respect to bases [imath]\mathcal{B}[/imath] and [imath]\mathcal{C}[/imath]. i.e., [imath] A\sim B\iff A=[T]_\mathcal{B},\ B=[T]_{\mathcal{C}}. [/imath] I know how to prove the '[imath]\Leftarrow[/imath]' but I dont think I can prove the other direction '[imath]\Rightarrow[/imath]'. Could please somebody help me? |
3011491 | [imath]\tan^2 10^\circ+\tan^2 50^\circ+\tan^2 70^\circ=9[/imath]
Strangest thing...*: [imath]\tan^2 10^\circ+\tan^2 50^\circ+\tan^2 70^\circ=9\tag{1}[/imath] The trick, as always, is how to prove it. My idea was to add a "missing" tangent and analyze a similar expression: [imath]\tan^2 10^\circ+\tan^2 30^\circ+\tan^2 50^\circ+\tan^2 70^\circ[/imath] ...and then to attack this sum pairwise (first and the last term, second and third). Despite the fact that I got the same angle ([imath]80^\circ[/imath]) here and there, I got pretty much nowhere with this approach. The other interesting fact is that (1) can be rewritten as: [imath]\cot^2 20^\circ+\cot^2 40^\circ+\cot^2 80^\circ\tag{1}[/imath] ...and now the angles are in nice geometric progression. That's the vector of attack that I'm trying to exploit now, but maybe you can entertain youself a little bit too. *Borrowed from "Usual suspects" | 1198248 | Trigonomnetric equality involving tg
I need help proving the following identity. [imath]\tan^210^\circ+\tan^250^\circ+\tan^270^\circ=9[/imath]. I am not sure if it is even true. |
3011750 | Trigonometric Limit (No L'Hôpital)
I came across with this limit: [imath]\lim_{x \to 0}\frac{\tan x - \sin x}{x^3}[/imath] I started working it out this way: [imath]\lim_{x \to 0}(\frac{\tan x}{x}\times \frac{1}{x^2}) - (\frac{sin x }{x}\times \frac{1}{x^2})[/imath] = [imath]\lim_{x \to 0}(1\times \frac{1}{x^2}) - (1\times \frac{1}{x^2})[/imath] = [imath]\lim_{x \to 0}\frac{1}{x^2} - \frac{1}{x^2}[/imath] = [imath]0[/imath] This is wrong since the solution is [imath]\frac{1}{2}[/imath]. So my question is why can't I do this? | 3008071 | [imath]\lim\limits_{x\to 0} \frac{\tan x - \sin x}{x^3}[/imath]?
[imath]\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}[/imath] Solution [imath]\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\ &= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\sin x}{x}\lim_{x\to 0} \frac{1}{x^2}\\&= \lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{1}{x^2}\\ &= \lim_{x\to 0} \frac{1}{x^2} -\frac{1}{x^2}\\&=0 \end{align}[/imath] But the answer is [imath]\dfrac{1}{2}[/imath] by L'Hopital's Rule. |
3011939 | [imath]G[/imath] finite group and [imath]\mathbb{C}[G][/imath] its group ring, show: [imath]K_i = \sum_{a_k\in cl_G(a_i)} a_k [/imath], then [imath]Z(\mathbb{C}[G])=span_\mathbb{C}(K_1,...,K_n)[/imath]
Given a finite group [imath]G = \{a_1 ,a_2,...,a_n\}[/imath] and [imath]\mathbb{C}[G][/imath] the respective group ring, [imath]\mathbb{C}[G] = \{\sum z_ia_i : z_i \in\mathbb{C} , a_i\in G\}[/imath]. Defining [imath]K_i := \sum_{a_k\in cl_G(a_i)} a_k[/imath] , when [imath]cl_G(a_i)[/imath] the conjugacy class, means [imath]K_i =a_1a_ia_1^{-1} + a_2a_ia_2^{-1} +...+a_na_ia_n^{-1}[/imath]. I wish to show that [imath](K_1 ,K_2 ,...,K_n)[/imath] spans Z([imath]\mathbb{C}[G])[/imath]. I had already shown that for [imath]a_i \in Z(\mathbb{C}[G])[/imath], [imath]K_i = \{a_i\}[/imath] thus [imath]a_i = 1\cdot K_i[/imath], as needed. But I failed to show the other direction, means that for [imath]w\in span_\mathbb{C}(K_1,K_2,...,K_n)[/imath] then [imath]w[/imath] commutes with any element of [imath]\mathbb{C}[G][/imath], means [imath]w\in Z(\mathbb{C}[G])[/imath]. | 2806252 | Basis for Center of Group Ring
There is already a similar post, that has an answer to my question, but it is quite short and I don't get it... So I am trying to show that \begin{align} \{e_K |~ K \subset G~ conjugacy~ class \} \end{align} is a basis for [imath]Z(\mathbb{Z}[G])[/imath]. Serre even writes "one immediately checks that the [imath]e_K[/imath] form a basis", but even reading the other post I don't see that... My approach to this is: Be [imath]x \in Z(\mathbb{Z}[G])[/imath], which is equivalent to: [imath]x = gxg^{-1}[/imath] for all [imath]g \in G[/imath]. Writing [imath]x = \sum\limits_{h \in G} \lambda_h h[/imath], we get [imath]x = \sum\limits_{h \in G} \lambda_h ghg^{-1}[/imath]. I don't really know how to go on at this point. I mean, I know that when [imath]a,b \in K[/imath] (same conjugacy class) you get [imath]\sum\limits_{s \in K} \mu_s s = \mu\sum\limits_{s \in K} s[/imath], since \begin{align} a = hbh^{-1} \Rightarrow \mu_a = \mu_b ~\forall a,b\in K \end{align} So all elements in the same conjugacy class have the same scalar. But from here I don't really know how I can generate the [imath]x = \sum\limits_{h \in G} \lambda_h ghg^{-1}[/imath] with the [imath]e_K[/imath] ... I would be really thankful, if someone could give me quite a basic proof that even I understand... |
2461252 | How many elements are there in [imath]\Bbb Z_5[i]/\langle 1+i\rangle[/imath]
How many elements are there in [imath]\Bbb Z_5[i]/\langle 1+i\rangle[/imath]? Any element of [imath]\Bbb Z_5[i]/\langle 1+i\rangle[/imath] is of the form [imath]a+bi+ \langle 1+i\rangle[/imath] Now [imath]1+i=0\implies i=-1\implies 2=0[/imath] [imath]a+bi+ \langle 1+i\rangle=a-b+ \langle 1+i\rangle[/imath]. Since [imath]2=0[/imath] so we can take [imath]a-b[/imath] to be odd (say [imath]1[/imath])and hence the only elements are [imath] \langle 1+i\rangle,1+ \langle 1+i\rangle[/imath] and hence [imath]\Bbb Z_5[i]/\langle 1+i\rangle\cong \Bbb Z_2[/imath] Is the solution correct? Please help | 2113667 | Cardinality of the quotient ring [imath]\mathbb Z_5[i]/\langle 1+i\rangle[/imath]
Find the cardinality of the quotient ring, [imath]\mathbb Z_5[i]/\langle 1+i\rangle[/imath] My Attempt: Since, [imath]\mathbb Z_5[i]\cong Z[x]/\langle x^2+1,5\rangle[/imath] Hence, [imath]\mathbb Z_5[x]/\langle 1+i\rangle\cong \mathbb Z[x]/\langle x^2+1,x+1,5\rangle[/imath] In the quotient ring we have the following, [imath]x^2+1=0,x+1=0[/imath] and [imath]5=0[/imath], i.e., [imath]x=-1\;\; \implies 2=(-1)^2+1=0[/imath] And, [imath]5=0 \implies 1=0[/imath]. Thus, [imath]\mathbb Z_5[i]/\langle 1+i\rangle\;\;=\{0+\langle 1+i\rangle\}[/imath]. Is the above reasoning correct? Initially, I had come across the following argument,[imath]1=(3+2i)(1+i)\implies 1\in \langle 1+i\rangle[/imath] And hence the quotient ring has just one element namely the zero element. This argument is quite good but it is not always easy to make such guesses. So I wanted to use the traditional method to find the cardinality. Moreover, is there any different approach to the problem? |
3012151 | Prove the sequence [imath]a_n =\frac{1\cdot 3\cdot…\cdot(2n-1)}{2\cdot4\cdot…\cdot2n}[/imath] has a limit
I have several questions to ask: 1) Show increasing, find the upper bound if you can of [imath]\sqrt{(n^2-1)}/n[/imath]. [imath]\sqrt{(n^2-1)}/n= |n|\sqrt{1-1/n^2}/n[/imath] if [imath]n[/imath] is positive than [imath]\sqrt1[/imath] else [imath]-\sqrt1[/imath]; bound: [imath]\sqrt{(n^2-1)}/n \le \sqrt {n^2}/n = |n|/n=1[/imath] To show if it is increasing should I do [imath]\frac{\sqrt{(n+1)^2-1}}{n+1} \ge \frac{\sqrt{(n)^2-1}}{n}[/imath]? 2)Prove the sequence [imath]a_n = \frac{1\cdot 3\cdot…\cdot(2n-1)}{2\cdot4\cdot…\cdot2n}[/imath] has a limit [imath]a_n[/imath] is a decreasing sequence, so to have a limit it must be bounded from below. [imath]a_n = (1-1/2)(1-1/4)…(1-1/2n)[/imath] | 1693377 | What is the limit of [imath]\frac{\prod\mathrm{Odd}}{\prod\mathrm{Even}}?[/imath]
What is this limit [imath] \frac{1\times3\times5\times\cdots}{2\times4\times6\times8\times\cdots} = \lim_{n \rightarrow \infty}\prod_{i=1}^{n}\frac{(2i-1)}{2i} [/imath] I remember that it was something involving [imath]\pi[/imath]. How can I compute it? In addition; how can I compute it's sum series limit? |
3012414 | Show that, for every integer [imath]a[/imath] such that [imath]\gcd(a,100)=1[/imath], we have [imath]a^{20}≡1\pmod{100}[/imath].
Show that, for every integer [imath]a[/imath] such that [imath]\gcd(a,100)\equiv 1[/imath], we have [imath]a^{20}\equiv 1\pmod{100}[/imath]. Based on Euler's theorem, I have [imath]a^{40}\equiv 1\pmod{100}[/imath], but I have problem getting to [imath]a^{20}\equiv 1\pmod{100}[/imath]. And I am wondering if it has to do with the fact that [imath]20\mid 40[/imath]? Any hints would be appreciated! | 1799351 | [imath]x^{20}=1[/imath] for all [imath]x\in U(100)[/imath]
Show that [imath]x^{20}=1[/imath] for all [imath]x\in U(100)[/imath]. In the book that I'm reading, it says "Since [imath]U(100)\thickapprox Z_2\oplus\,Z_{20}[/imath] we see that [imath]x^{20}=1[/imath] for all [imath]x[/imath] in [imath]U(100)[/imath]".I think maybe that remark is due to this theorem which states "The order of an element of a direct product of a finite groups is the least common multiple of the orders of the components of the elements". i.e [imath]|(g_1,g_2,.\,.\,.,g_n)|=lcm(|g_1|,|g_2|,.\,.\,.,|g_n|).[/imath] We define [imath]U(n)[/imath] to be the set of all positive integers less than [imath]n[/imath] and relatively prime to [imath]n[/imath] with the operation multiplication modulo [imath]n[/imath] |
3012645 | Prove That [imath]T(B)=AB-BA[/imath] has [imath]\dim \operatorname{Im}(T)\leq n^{2}-n[/imath].
Let [imath]A[/imath] be an arbitrary [imath]n\times n[/imath] matrix over [imath]\mathbb{C}[/imath] and define the linear transformation [imath]T:F^{n\times n}\rightarrow F^{n\times n}\text{,}\quad T(B)=AB-BA[/imath]. I want to prove that [imath]\dim \operatorname{Im}(T)\leq n^{2}-n[/imath] The rank nullity theorem gives [imath]\dim\operatorname{Im}(T)=n^{2}-\dim \operatorname{Ker}(T)[/imath], so this is equivalent to proving that [imath]\dim \operatorname{Ker}(T)\geq n[/imath], i.e. the subspace of matrices commuting with [imath]A[/imath] is at least [imath]n[/imath] dimensional. In the case that [imath]A[/imath] is diagonalizable, it isn't too hard to show that [imath]n\leq \dim\operatorname{Ker}(T)\leq n^{2}[/imath], the minimal case occuring for [imath]A[/imath] having [imath]n[/imath] distinct eigenvalues (in this caase the kernel is the subspace of diagonal matrices), and the maximal case provided by [imath]A=I[/imath]. It is the general/non diagonalisable case I'm not sure on, and any hints or pointers would be appreciated. I'm not sure how relevant it is, but by noting that [imath]T(A)=T(0)=0[/imath], while [imath]A\neq 0[/imath] in general we see that [imath]T[/imath] is not injective, so [imath]\det T(B)=0[/imath]. | 122503 | Dimension of set of commutable matrices
Let [imath]A[/imath] be an [imath]n \times n[/imath] complex matrix and [imath]$V = \{B\mid AB=BA\}$[/imath]. I've proved that [imath]V[/imath] is a vector space. How can I prove that [imath]\dim V \ge n[/imath] for any [imath]A[/imath]? |
3012860 | Making sense of a bijection between proper classes?
I am trying to solve the following problem in ZF: "Show that the collection [imath]\{\kappa \ | \ \kappa=\aleph_\kappa\}[/imath] is a proper class. My thinking is that in the construction any element of this collection, one takes an arbitrary ordinal [imath]\alpha[/imath], and that therefore one might be able to show it is a proper class by showing that there is a bijection between the class of all ordinals and the class given above, since the class of all ordinals is a proper class. However, I'm not sure how to fully make sense of the concept of a function between proper classes, since a function is normally defined as being between sets. Is this a valid line of reasoning, and if so, how does one formalise this using the axioms of ZF? | 337047 | Can proper classes also have cardinality?
In some set theories such as ZF+GAC, in which GAC is global axiom of choice, the Von Neumann universe [imath]V[/imath] bijects to [imath]Ord[/imath], the class of ordinals. It suggests us that proper classes may also have cardinality，in the example is [imath]|V|=|Ord|[/imath]. In addition, if we are in ZF+GAC+ALS, it seems [imath]|V|[/imath] is the only cardinality which is not a cardinal number. Moreover, it seems some properties such as Cantor–Bernstein–Schroeder theorem also holds for cardinality of proper classes, but I'm not sure if it is well-defined and won't cause any paradox... |
3013091 | If [imath]f_m[/imath] is a sequence of continuous functions which converges uniformly to a function. then that function is continuous as well.
I was looking for the proof is this theorem, but I couldn't find it anywhere. the theorem is stated formally: If [imath]f_m[/imath] is a sequence of continuous functions defined on [imath]D[/imath] (subset of [imath]R[/imath]) such that [imath]f_m$$\to$$f[/imath] uniformly on [imath]D[/imath] then [imath]f[/imath] is continuous. can someone give the stepwise proof? | 2164642 | Proof of uniform limit of Continuous Functions
The uniform limit of Continuous Functions is continuous. Proof Let [imath]\varepsilon > 0[/imath]. There exists [imath]N[/imath] in natural numbers such that [imath]n>N[/imath] implies [imath]|f_n(x)-f(x)| < \frac{\varepsilon}{3}[/imath] for all [imath]x[/imath] in [imath]S[/imath]. In particular [imath]|f_{N+1}(x)-f(x)| < \frac{\varepsilon}{3}[/imath] for all [imath]x[/imath] in [imath]S[/imath]. Since [imath]f_{N+1}[/imath] is continuous at [imath]x_0[/imath] there is a [imath]\delta>0[/imath] such that, [imath]x[/imath] in [imath]S[/imath] and [imath]|x-x_0| < \delta[/imath] imply [imath]|f_{N+1}(x)-f_{N+1}(x_0)|< \frac{\varepsilon}{3}[/imath] Now we conclude [imath]x[/imath] in [imath]S[/imath] and [imath]|x-x_0| < \delta[/imath] imply [imath]|f(x)-f(x_0)|< 3 \cdot \frac{\varepsilon}{3} = \varepsilon[/imath]. Can someone explain what is happening here? I don't follow it all. |
3013965 | Is a "local enough" class number always equal to one?
Let [imath]F[/imath] be a number field, and [imath]\mathcal{O}[/imath] its ring of integers. Is there always a finite set [imath]S[/imath] of places of [imath]F[/imath] such that [imath]\mathcal{O}_S[/imath] has class number one? Is it a consequence of standard results on class field theory or is it something else? | 2934348 | Standard argument for making the class group of a number field trivial
Let [imath]K[/imath] be a number field and let [imath]R=\mathcal{O}_K[/imath] be its ring of integers. Let [imath]a_1, \ldots, a_h[/imath] be ideals of [imath]R[/imath] generating the class group [imath]Cl(K)[/imath]. Let [imath]S[/imath] denote the set of valuations corresponding to prime ideals dividing [imath]a_1\cdots a_h[/imath]. Let [imath]R_S = \{a \in K \mid v(a)\geq 0,\ v \text{ non-Archimedean valuation on } K, \ v \notin S \}[/imath] Silverman (p.213 Arithmetic of Elliptic Curves) claims that [imath]R_S[/imath] has trivial class group. I’m sure this is a standard and simple fact, but I’m having trouble proving it. We have an inclusion [imath]R \to R_S[/imath] and a map [imath]Cl(R) \to Cl(R_S)[/imath]. Let [imath]I[/imath] be an ideal of [imath]R_S[/imath], we want to show it is principal. We have [imath]I \cap R \equiv \prod a_i^{e_i}[/imath] in the class group so there exists principal ideal [imath]P,P’[/imath] of [imath]R[/imath] such that [imath]P(I\cap R) = P’ \prod a_i^{e_i}[/imath] [imath]P(I\cap R)R_S = P’ \prod a_i^{e_i}R_S[/imath] Got stuck. Somehow the [imath]a_iR_S[/imath] become principal in [imath]R_S[/imath]. I know [imath]R=\bigcap\limits_\mathfrak{p} R_\mathfrak{p}[/imath] [imath]R_S=\bigcap\limits_{\mathfrak{p} \notin S} R_\mathfrak{p}[/imath] Where [imath]\mathfrak{p}[/imath] ranges over all maximal ideals of [imath]R[/imath] |
3014130 | Let (X, d) be a metric space and let A be a subset of X and O be an open subset of X. Prove that [imath]O \cap \overline{A} \subseteq \overline{O\cap A}[/imath]
I don't know how to use the fact that O is open. Its interior must be itself, so I guess I should take the complement of the sets in question. But I'm not sure how to proceed. | 2502937 | The intersection of a closure with an open set
I have been stuck on this problem : Prove :If [imath]A[/imath] is Open , then [imath]A\cap \overline B \subset \overline {A \cap B}[/imath].I have tried to use the previously proven fact that [imath]\overline {A \cap B}\subset \overline{A} \cap \overline{B}[/imath] , but I cannot seem to get anywhere ...In particular, I do not see how A being open is relevant here.......... |
1924177 | How to show that [imath]\Bbb R^m[/imath] is not diffeomorphic to [imath]\Bbb R^n[/imath] when [imath]n \neq m[/imath]
Suppose that there is indeed an diffeomorphism [imath]f: \Bbb R^m \to \Bbb R^n[/imath], and [imath]f(a) = 0[/imath]. Then [imath]f \circ f^{-1} = identity[/imath]. In class my teacher asserts that [imath]Df^{-1}_a \cdot Df_a = identity[/imath] and [imath]Df_a \cdot Df_a^{-1} = identity[/imath]. Why this is true and how does it imply that [imath]f[/imath] cannot exist? | 2137954 | Diffeomorphism [imath]f:U\to V[/imath] then [imath]Df:\mathbb{R^n}\to\mathbb{R^m}[/imath] will be linear isomorphism hence [imath]n=m[/imath]
[imath]{\large let \ U\subset \mathbb{R^n,}V\subset \mathbb{R^m}}[/imath] be open subsets. map [imath]f:{\large U }\to{\large V }[/imath] is called a diffeomorphism if is bijective and both [imath]f[/imath] and [imath]f^{-1}[/imath] are continuously differentiable. once there is diffeomorphism [imath]f:{\large U }\to{\large V }[/imath] then [imath]Df:\mathbb{R^n}\to\mathbb{R^m}[/imath] will be [imath]{\large linear \ isomorphism }[/imath] hence [imath]n=m[/imath]. how is this a linear isomorphism??. i know [imath]Df(a)[/imath] is a linear map , but why this is one one and on to ?? |
1069387 | Show that [imath]\mathbb Q(\sqrt p) \not\simeq\mathbb Q(\sqrt q)[/imath]
I'd like to show that for [imath]p,q[/imath] distinct primes, the extensions [imath]\mathbb Q(\sqrt p),\mathbb Q(\sqrt q)[/imath] are not isomorphic. I don't really have knowledge of the "high-level language" of algebraic number theory, so my approach is going to be very bit-wise, if you will. Here's my try: If there were such an isomorphism, call it [imath]\varphi[/imath]. I know that we must have: [imath]\left\{\begin{align}&\varphi\mid_{\mathbb Q} = \text{id} \\ &\varphi(\sqrt p) =\pm\sqrt p \end{align}\right.[/imath] So essentially the existence of [imath]\varphi[/imath] means that for any [imath]a+b'\sqrt p[/imath] there exist [imath]c,d[/imath] such that [imath]\varphi(a+b'\sqrt p) = a+b\sqrt p = c+d\sqrt q[/imath] (where [imath]b[/imath] might be a sign change of [imath]b'[/imath]). Take, to simplify [imath]a=0, b=1[/imath]. Let [imath]\alpha = \sqrt p - d\sqrt q = c[/imath], from above. We can rearrange to obtain [imath]\begin{align}&(\alpha-\sqrt p)^2 = d^2q \Rightarrow \\ \Rightarrow& \ \alpha^2 - 2\alpha\sqrt p + p - d^2q = 0 \\ \Rightarrow& \ \sqrt p -\frac{c^2+p-d^2q}{2c} = \sqrt p - \lambda = 0\end{align}[/imath] But then it has to be that [imath](x^2-p) \mid (x-\lambda)[/imath], which cannot be. This proof however doesn't satisfy me for two reasons. First of all, I'd like to have stopped at the polynomial I found in [imath]\alpha[/imath], shown that it was irreducible in [imath]\mathbb Q(\sqrt2)[/imath], and argued in terms of extension degrees. However I don't know how to show that. The second reason is that I'm not using the primality of [imath]p,q[/imath] other than having it imply the irreducibility of [imath]x^2-p[/imath]. It seems to me from the way the exercise is worded that if they were compound and [imath]x^2-p,x^2-q[/imath] were still irreducible, we might have a homomorphism. Can someone clarify? | 3028343 | [imath]\Bbb Q(\sqrt 2)[/imath] and [imath]\Bbb Q(\sqrt 3)[/imath] are not isomorphic
How to prove that [imath]\Bbb Q(\sqrt 2)[/imath] and [imath]\Bbb Q(\sqrt 3)[/imath] are not isomorphic. I thought that they are but I got this problem in Dummit Foote Section 14.1. Question no 4. As they extension over [imath]\Bbb Q[/imath] by the polynomials [imath]x^2-2[/imath] and [imath]x^2-3[/imath] resp. |
3013803 | Not continuity of a multivariable function given by $\frac{2x^2y}{x^2+y^2}$
Let [imath]f(x,y)= \begin{cases} \frac{2x^2y}{x^2+y^2} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y)=(0,0) . \end{cases}[/imath] (i) Prove the directional derivatives of [imath]f[/imath] exist in any direction at the point [imath](0,0)[/imath]. (ii)¿Is [imath]f[/imath] continuous on [imath](0,0)[/imath]? For (i) I took [imath]u=(u_{1},u_{2}) \in \mathbb{R^{2}}[/imath] such [imath]$\|u\|=1$[/imath] and [imath]0=(0,0)[/imath]. So [imath]\lim_{t \to 0} \frac{f(0+ t u)-f(0)}{t}=\lim_{t \to 0}\frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=\lim_{t \to 0}\frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}[/imath]. But I cannot find the limit when [imath]t[/imath] approaches to [imath]0[/imath]. Which by finding them I would prove the directional derivative exist at every direction right? For (ii) I got the intuition [imath]f[/imath] is not continuous at [imath](0,0)[/imath] so I took [imath]\lbrace (\frac{1}{k},\frac{1}{k}) \rbrace_{k \in \mathbb{N}}[/imath] and [imath]\lbrace (\frac{1}{k},0) \rbrace_{k \in \mathbb{N}}[/imath] which are two different sequences in [imath]\mathbb{R}^{2}[/imath] converging to [imath](0,0)[/imath]. However, [imath]\lbrace f(\frac{1}{k},\frac{1}{k}) \rbrace_{k \in \mathbb{N}}=\lbrace \frac{1}{k^{2}} \rbrace_{n \in \mathbb{N}} \to 0[/imath] and [imath]\lbrace f(\frac{1}{k},0) \rbrace_{k \in \mathbb{N}} \to 0[/imath]. So maybe my intuition was not right? Also I've have approach [imath](0,0)[/imath] through [imath](x,mx)[/imath] and when this and gives me 0 as this value approaches to zero bit I cannot find another trajectory approaching through [imath](0,0)[/imath] which gives me another value distinct to [imath]0[/imath]. :( Can anyone help me end the proof of continuity or not continuity, please? | 3012478 | Continuity and existence of directional derivatives at all directions of a [imath]f:\mathbb{R}^{2} \to \mathbb{R}[/imath] given by $f(x,y)=\frac{2x^2y}{x^2+y^2}$
Let [imath]f(x,y)= \begin{cases} \frac{2x^2y}{x^2+y^2} & if & (x,y) \neq (0,0) \\ 0 & if & (x,y)=(0,0) . \end{cases}[/imath] (i) Prove the directional derivatives of [imath]f[/imath] exist in any direction at the point [imath](0,0)[/imath]. (ii)¿Is [imath]f[/imath] continuous on (0,0)? For (i) I took [imath]u=(u_{1},u_{2}) \in \mathbb{R^{2}}[/imath] such [imath]$\|u\|=1$[/imath] and [imath]0=(0,0)[/imath]. So [imath]$$\lim_{t \to 0} \frac{f(0+ t u)-f(0)}{t}=\lim_{t \to 0}\frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=\lim_{t \to 0}\frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}.$$[/imath] But I cannot find the limit when [imath]t[/imath] aproaches to [imath]0[/imath]. Which by finding them I would prove the directional derivative exist at every direction right? For (ii) I got the intuition [imath]f[/imath] is not continuous at [imath](0,0)[/imath] so I took [imath]\lbrace (\frac{1}{k},\frac{1}{k}) \rbrace_{k \in \mathbb{N}}[/imath] and [imath]\lbrace (\frac{1}{k},0) \rbrace_{k \in \mathbb{N}}[/imath] which are two different sequences in [imath]\mathbb{R}^{2}[/imath] converging to [imath](0,0)[/imath]. However, [imath]\lbrace f(\frac{1}{k},\frac{1}{k}) \rbrace_{k \in \mathbb{N}}=\lbrace \frac{1}{k^{2}} \rbrace_{n \in \mathbb{N}} \to 0[/imath] and [imath]\lbrace f(\frac{1}{k},0) \rbrace_{k \in \mathbb{N}} \to 0[/imath]. So maybe my intuition was not right? Can anyone help me end the proof of continuity or not continuity , please? |
2713310 | Calculate the Pontragin dual [imath]\text{Hom}_{\mathbb{Z}}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})[/imath].
I'm calculating the [imath]\text{Ext}^1_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Z}, \mathbb{Z})[/imath]. In particular, [imath]\text{Ext}^1_{\mathbb{Z}}(\mathbb{Q}, \mathbb{Z}, \mathbb{Z}) \cong \text{Hom}_{\mathbb{Z}}(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})[/imath] the Pontragin dual. However how should I see what this group is? I only know that for finite abelian groups on the first position, this dual is isomorphic to it self. But no idea for infinitely generated groups. | 795543 | What is [imath]\operatorname{Hom}_\mathbb{Z}(\mathbb{Q}/\mathbb{Z},\mathbb{Q}/\mathbb{Z})[/imath]?
Is [imath]\operatorname{Hom}_\mathbb{Z}(\mathbb{Q}/\mathbb{Z},\mathbb{Q}/\mathbb{Z})[/imath] isomorphic to any "known" group? I suppose what I mean is, is it isomorphic to a group that isn't a Hom group? If such an isomorphism is known, is there a reference which sketches it out? I know [imath]\mathbb{Q}/\mathbb{Z}=\bigoplus_p\mathbb{Z}_{p^\infty}[/imath], but using the various properties to pull direct sums out doesn't seem fruitful. My motivation is that I know [imath]\operatorname{Ext}^0_\mathbb{Z}(A,B)[/imath] is naturally isomorphic to [imath]\operatorname{Hom}_\mathbb{Z}(A,B)[/imath], and pairs like [imath]\operatorname{Hom}(\mathbb{Z},\mathbb{Z})[/imath], [imath]\operatorname{Hom}(\mathbb{Q},\mathbb{Q})[/imath], etc., come up frequently in algebraic topology/basic examples in homological algebra. Since [imath]\mathbb{Q}/\mathbb{Z}[/imath] shows up commonly in topology, I'm curious about that case as well. |
3013980 | Alternative definition of positive definite matrix
In a text book I am following, there is a definition of positive definite matrices, which I did not see before: [imath]x^TAx \geq \alpha x^Tx[/imath] where [imath]\alpha[/imath] is a positive scalar and [imath]x \in \mathbb{R^n}[/imath]. The usual definition I know is for every non-zero vector [imath]x \in \mathbb{R^n}[/imath], it is [imath]x^TAx > 0[/imath]. These definitions must be equivalent. It is trivial to show that the first definition implies [imath]x^TAx > 0[/imath]. But I failed to see a way to show the other way around of the equivalence. I thought of applying SVD to the matrix [imath]A[/imath] and tried to find a way to show that [imath]x^TAx[/imath] is always lower bounded by a [imath]\alpha x^Tx[/imath] with [imath]\alpha[/imath] being positive, using singular values and vectors of [imath]A[/imath] but couldn't move forward. What is the correct way of approach here? (The other question in the board assumes symmetric matrices. This one does not). | 2030774 | Is [imath]{\bf x^TAx} \ge { a \bf x^Tx}[/imath] for some [imath]a>0[/imath] not dependent on [imath]\bf x[/imath] if [imath]\bf A[/imath] is positive definite matrix?
Is there a theorem saying [imath]{\bf x^TAx} \ge { a \bf x^Tx}[/imath] for some [imath]a>0[/imath] not dependent on [imath]\bf x[/imath] if [imath]\bf A[/imath] is positive definite matrix? If there is, can some help provide a reference or some explanation? Thank you! |
3013195 | Can the empty set be the image of a function on [imath]\mathbb{N}[/imath]?
I cannot find any example of function [imath]f:\mathbb{N}\rightarrow\mathbb{N}[/imath] of which we can say that [imath] f(\mathbb{N})=\emptyset [/imath] Does there exists any? | 789123 | Why is there no function with a nonempty domain and an empty range?
Let [imath]A[/imath] to be a nonempty set and [imath]B= \emptyset[/imath]; then [imath] A \times B[/imath] is a set. And let [imath]F[/imath] be a function [imath]A \to B[/imath]. Then [imath]F \subseteq A \times B[/imath]. By the axiom of specification, [imath]F[/imath] must exists (if I didn't mess up something). But the book I'm reading, Elements of Set Theory by Enderton, says that no function could have a nonempty domain and an empty range, and no more detail is given. So my confusion arises. His statement against my proof. The only axiom as far as I know to prove such a set does not exist is the axiom of regularity. But I can't give such a proof. So I need help. I hope someone could clearly explain why such a function doesn't exist, and why this doesn't contradict the axiom of specification. Let me explain my confusion in more detail: First of all, I know that [imath]A \times \emptyset[/imath] is empty. But [imath] \emptyset \times A[/imath] is also an empty set, yet there is no problem with functions with an empty domain. The book I'm reading defines a function as: "A function is a relation such that for each [imath]x[/imath] in [imath]\operatorname{dom} F[/imath] there is only one [imath]y[/imath] such that [imath]x \mathop F y[/imath]." and a relation as: "A relation is a set of ordered pairs." Now, let me define the function [imath]F[/imath] as: [imath]F = \{\langle x,y \rangle \mid x \in A \text{ and } y \in B \text{ and ... other conditions} \}[/imath] so it's equal to: [imath]F = \{ \langle x,y \rangle \in A \times B \mid \text{... other conditions} \}.[/imath] Doesn't that mean that [imath]F[/imath] meets the conditions of the axiom of specification, and therefore exists? What I would specifically like to ask is: How do you define a function [imath]F[/imath] (as precisely as possible)? Why does the argument above not imply that a function [imath]F: A \to B[/imath] always exists? Does [imath]F \subseteq A \times B[/imath] still hold when [imath]B = \emptyset[/imath]? (The answer to #2 cannot simply be that [imath]A \times B = \emptyset[/imath], because [imath] B \times A = \emptyset[/imath] as well, but a function [imath]B \to A[/imath] does exist.) |
2119691 | Proving Order of Double Coset
Problem: Prove that if [imath]G[/imath] is a finite group, then the number of elements in the double coset [imath]AxB[/imath] is [imath]\frac{|A| * |B|}{|A \cap xBx^{-1}|}[/imath]. So here's what I know, and I don't know if these pieces of information will be useful. [imath]|AB| = \frac{|A| * |B|}{|A \cap B|}[/imath] and [imath][G:B] = [G:A][A:B][/imath]. I am especially caught up on the [imath]xBx^{-1}[/imath] part. I understand that [imath]xBx^{-1} = B[/imath] if the left and right cosets are equal (that's how the professor explained it anyway) and I know that the left and right cosets are equal if the subgroup is normal. I don't even know if that is useful for this problem though. I really don't have any level of understanding when it comes to cosets beyond the definitions as my professor has really not taught this and the textbook does not seem very clear. There was a preceding problem and the link will be provided in case that is necessary to prove this problem. Equivalence class of double coset | 533639 | Order of Double Coset
I am working on a homework problem (so don't just give me the answer) from Herstein's Topics in Algebra, which goes as follows: If [imath]G[/imath] is a finite group, show that the number of elements in the double coset [imath]AxB[/imath] is [imath]\dfrac{o(A)o(B)}{o(A\cap xBx^{-1})}[/imath] It makes sense to me, but my attempts at a proof seem to fall short of undeniability. I have been trying to show that [imath]o(A \cap x B x^{-1})[/imath] equals the number of duplicate terms in the list of all possible products of an element from [imath]A[/imath] and an element from [imath]xB[/imath]. Suppose [imath]a_0 \in A \cap xBx^{-1}[/imath]. Then [imath]\exists b_0 \in B : a_0 = x b_0 x^{-1}[/imath]. In the list of products of the form [imath]axb, a \in A, b \in B[/imath], any term involving [imath]a_0[/imath] will be of the form [imath]a_0 x b = x b_0 x^{-1} x b = x b_0 b[/imath] for some [imath]b \in B[/imath]. But then this list is just the left coset [imath]xB[/imath], which is already accounted for in the product list of [imath]AxB[/imath] by setting [imath]a=e[/imath]. This is where I run into trouble. I can't seem to crystallize this argument. Am I going about this the right way? And if you see how I should extend this argument, can you nudge me in the right direction? Thanks Math.SE. |
3014875 | How to find the sum of this Convergent series
I am given the following convergent series asked to find the value of: [imath]\sum_{n=1}^{\infty} \left(\frac{4n}{(n^4+n^2 +1)}\right)[/imath] According to Sum calculators, I have found the value to be 2. Though, I don't know what approach to take to prove this. To aid us in the question, we've been given the following hint: [imath]n^4 + n^2 + 1 = (n^2 + 1)^2 − n^2[/imath] Can someone point me in the right direction? | 1645776 | Evaluate [imath]\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}[/imath]
I am trying to re-learn some basic math and I realize I have forgotten most of it. Evaluate [imath]\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}[/imath] Call the terms [imath]S_n[/imath] and the total sum [imath]S[/imath]. [imath]S_n < \frac{1}{n^3} \Rightarrow \sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1} = S < \infty[/imath] [imath]S_n = \frac{n}{n^4+n^2+1} = \frac{n}{(n^2+1)^2-1}[/imath] It has been more than a few years since I did these things. I would like a hint about what method I should try to look for? Thanks. |
3014886 | suppose n-by-n matrix A only has zero eigenvalues, and [imath]A^{n-1} \neq 0[/imath] show there exists x such that x, Ax, ...[imath]A^{n-1}[/imath]x are linearly independent.
I know [imath]p_a(A) = 0[/imath] so [imath]A^n = 0[/imath] and since [imath]A^{n-1} \neq 0[/imath] so there exist y such that [imath]A^{n-1} y \neq 0[/imath] . The hint says dimension of nullspace of A is 1 and up to a constant multiple there is only one eigenvector for A, which I do not understand how to prove. Thanks | 2242327 | Prove that [imath]v, Tv, T^2v, ... , T^{m-1}v[/imath] is linearly independent
Suppose [imath]T[/imath] is in [imath]L(V)[/imath], [imath]m[/imath] is a positive integer, and [imath]v[/imath] in vector space [imath]V[/imath] is such that [imath](T^{m-1})v \neq 0[/imath], and [imath](T^m)v = 0[/imath]. Prove that [imath][v, Tv, T^2v, ... , T^{m-1}v][/imath] is linearly independent I get that [imath](T^j)v \neq 0\ \forall\ j < m[/imath]. Additionally, since [imath]T[/imath] is nilpotent, [imath]V[/imath] has a basis with respect to which the matrix of [imath]T[/imath] has [imath]0[/imath]s on and below the diagonal. However, I'm not sure if these can be used to show linear independence or if they're even relevant to the problem at all. Any help is appreciated! |
2630506 | Show that [imath]\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right][/imath] is a Euclidean Domain.
Let [imath]F = \mathbb{Q}(\sqrt{-3})[/imath] be a quadratic field with associated integer ring [imath]\mathcal{O} = \mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right] = \left\{a+b\frac{1+\sqrt{-3}}{2} : a,b \in \mathbb{Z} \right\}[/imath] and field norm [imath]N[/imath] defined by [imath]N(a+b\sqrt{-3}) = a^2 + 3b^2[/imath]. Prove that [imath]\mathcal{O}[/imath] is a Euclidean Domain with respect to [imath]N[/imath]. Prove that every element of [imath]F[/imath] differs from an element of [imath]\mathcal{O}[/imath] by an element whose norm is at most [imath]1/3[/imath] (this is what I'm stuck on). Following almost the exact steps to this proof (Proving that [imath]\mathbb{Z}[\sqrt{2}][/imath] is a Euclidean domain) I write [imath]a = \beta q + r[/imath] and get that [imath]N(r) < \left[\left(\frac{1}{2}\right)^2 + 3 \left(\frac{1}{2}\right)^2\right] *N(\beta) = 1*N(\beta)[/imath], but I'm not sure how I'm supposed to get a bound of [imath]1/3[/imath] instead of [imath]1[/imath]. | 23086 | Prove that the class number of [imath]\mathbb{Z}[\zeta_3][/imath] is [imath]1[/imath]
How does one prove that the class number of [imath]\mathbb{Z}[\zeta_3][/imath] is [imath]1[/imath]? |
1370857 | Proof that [imath]\sqrt6 - \sqrt2 - \sqrt3[/imath] is irrational.
I want to prove that: [imath]\sqrt6 - \sqrt2 - \sqrt3[/imath] is irrational. I have tried using squares, the [imath]p/q[/imath] definition of rationality and the facts that 1)rational[imath]\times[/imath] irrational=irrational (unless rational=0), 2)rational[imath]+[/imath]irrational=irrational. However, I haven't been able to reach some conclusion. Things seem harder than when you have two square roots. Any help would be appreciated! | 1126255 | [imath]\sqrt{m_1}+\sqrt{m_2}+ \cdots + \sqrt{m_n}[/imath] is Irrational
If [imath]m_1 , m_2, \cdots m_n[/imath] are natural numbers where at least one of them is not a perfect square, then how do I prove that the sum [imath]\sqrt{m_1}+\sqrt{m_2}+ \cdots + \sqrt{m_n}[/imath] is irrational? I'm hoping to prove this using methods from pre-calculus level algebra. |
3015290 | The closed form for [imath]\sum_{n\geq1}\frac1{n(n+1)(n+2)\cdots(n+m)}[/imath]
Just for fun, I defined the following series. For [imath]m\in\Bbb N[/imath], [imath]g_m=\sum_{n\geq1}\prod_{i=0}^{m}\frac1{n+i}[/imath] I am seeking a closed form for [imath]g_m[/imath]. I found [imath]g_1,\ g_2[/imath], and a general expression involving a lot of integrals. [imath] \begin{align} g_1=&\sum_{n\geq1}\frac1{n(n+1)}\\ =&\sum_{n\geq1}\frac1n\int_0^1 t_0^n\mathrm{d}t_0\\ =&\sum_{n\geq1}\int_0^1\int_0^{t_0} t_1^{n-1}\mathrm{d}t_1\mathrm{d}t_0 \\ =&\int_0^1\int_0^{t_0}\sum_{n\geq0}t_1^{n}\mathrm{d}t_1\mathrm{d}t_0 \\ =&\int_0^1\int_0^{t_0}\frac{\mathrm{d}t_1}{1-t_1}\mathrm{d}t_0 \\ =&1 \\ \end{align} [/imath] Through the same process, it can be shown that [imath]g_2=\frac14[/imath]. In fact, through the same process, it can be shown that [imath]g_m=\int_0^1\int_0^{t_0}\int_0^{t_1}\cdots\int_0^{t_{m-2}}\frac{\mathrm{d}t_{m-1}}{1-t_{m-1}}\mathrm{d}t_{m-2}\cdots\mathrm{d}t_2\mathrm{d}t_1\mathrm{d}t_0[/imath] Which I don't even know how to approach. Any ideas? Thanks. This isn't a duplicate question, because I would like to know how to evaluate those nested integrals, if possible. | 2592581 | Sum of infinite series [imath]\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}[/imath]
Lets define the following series [imath]S_n =\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}.[/imath] I know that [imath]S_0[/imath] does not converge so let's suppose [imath]n \in N[/imath] and define [imath]S_n[/imath] for some [imath]n[/imath]. We have [imath]S_1=1[/imath] , [imath]S_2=\frac1 4[/imath] , [imath]S_3=\frac 1 {18}[/imath] , [imath]S_4=\frac 1{96}[/imath] , [imath]S_5=\frac 1{600}[/imath] etc.. the numerator in all results is 1the pattern in denominator is [imath][1,4,1,96,600...][/imath] and can be found here that is equal to [imath]n*n![/imath] Finally I want to prove the general equality: [imath]\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}[/imath] |
1499831 | When do we have equality in the third axiom of valuations?
Let [imath]R[/imath] be a ring and [imath]v: R \to \mathbb Z \cup \{+\infty\}[/imath] a map that meets following axioms: [imath]v(a) = +\infty \iff a=0[/imath] [imath]v(ab) = v(a)+v(b)[/imath] [imath]v(a+b) \geq \min\{v(a),v(b)\}[/imath] I have to show that [imath]v(a) \neq v(b)[/imath] implies [imath]v(a+b) = \min\{v(a),v(b)\}[/imath], but I have no idea where to begin, can anyone give me some hints? | 102044 | How do I compute the discrete valuation of the sum of two elements
Let [imath]O[/imath] be a discrete valuation ring with valuation [imath]v[/imath]. We normalize by [imath]v(\pi) =1[/imath], with [imath]\pi[/imath] a prime element in [imath]O[/imath]. By definition, for all [imath]x,y[/imath] in [imath]O[/imath], we have [imath]v(x+y) \geq \min (v(x),v(y))[/imath]. For some reason, the texts (Neukirch and Serre) suggest that one has equality when [imath]v(x) \neq v(y)[/imath]. Where does this come from? Does one also have an upper bound for [imath]v(x+y)[/imath]? I actually have elements [imath]x_1,\ldots,x_n[/imath] and I want to show that [imath]v(x_1+\ldots+x_n) = \min(v(x_i))[/imath]. Does it suffice to show that [imath]v(x_i) \neq v(x_j)[/imath] for all [imath]i\neq j[/imath]? |
3014660 | How to find [imath]\partial\chi^2/\partial b[/imath] when $\chi^2=\sum_{i=1}^N\frac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} $?
How do I find How to find [imath]\partial\chi^2/\partial b[/imath] when [imath]\chi^2=\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} [/imath]? My attempt: [imath]\begin{align*} \dfrac{\partial}{\partial b}\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} &= \dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-a}{\sigma_i^2}-\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= \dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-a}{\sigma_i^2}\Bigg]-\dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= 0-\dfrac{\partial}{\partial b}\Bigg[\sum_{i=1}^N\dfrac{b(x_i)^2}{\sigma_i^2}\Bigg]\\ &= -\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_1)^2}{\sigma_1^2}+\dfrac{b(x_2)^2}{\sigma_2^2}+\cdots+\dfrac{b(x_N)^2}{\sigma_N^2}\Bigg]\\ &= -\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_1)^2}{\sigma_1^2}\Bigg]+\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_2)^2}{\sigma_2^2}\Bigg]+\cdots+\dfrac{\partial}{\partial b}\Bigg[\dfrac{b(x_N)^2}{\sigma_N^2}\Bigg]\\ &= -\dfrac{1}{\sigma_1^2}\dfrac{\partial}{\partial b}\Bigg[b(x_1)^2\Bigg]-\dfrac{1}{\sigma_2^2}\dfrac{\partial}{\partial b}\Bigg[b(x_2)^2\Bigg]-\cdots-\dfrac{1}{\sigma_N^2}\dfrac{\partial}{\partial b}\Bigg[b(x_N)^2\Bigg]\\ &= -\dfrac{1}{\sigma_1^2}\cdot2\cdot b(x_1)-\dfrac{1}{\sigma_2^2}\cdot2\cdot b(x_2)-\cdots-\dfrac{1}{\sigma_N^2}\cdot2\cdot b(x_N)\\ &= -\Bigg[\dfrac{2}{\sigma_1^2}\cdot b(x_1)+\dfrac{2}{\sigma_2^2}\cdot b(x_2)+\cdots+\dfrac{2}{\sigma_N^2}\cdot b(x_N)\Bigg]\\ &= -\sum_{i=1}^N\dfrac{2b(x_i)}{\sigma_1^2} \end{align*}[/imath] Is this correct? EDIT: I previously asked this question. But the fact that [imath]b[/imath] is a function and changes with each different [imath]x_i[/imath] confuses me. | 3014606 | How to find [imath]\partial\chi^2/\partial a[/imath] where $\chi^2=\sum_{i=1}^N\frac{D(x_i)-a-b(x_i)^2}{\sigma_i^2}$?
How to find [imath]\partial\chi^2/\partial a[/imath] where [imath]\chi^2=\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2}[/imath]? My attempt: [imath]\begin{align} \dfrac{\partial}{\partial a}\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} &= \dfrac{\partial}{\partial a}\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\sum_{i=1}^N\dfrac{-a}{\sigma_i^2}\\ &=\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\dfrac{\partial}{\partial a}\sum_{i=1}^N\dfrac{-a}{\sigma_i^2}\\ &=\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\dfrac{1}{\sigma_i^2}\cdot\dfrac{\partial}{\partial a}\sum_{i=1}^N-a\\ &=\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\dfrac{1}{\sigma_i^2}\cdot\dfrac{\partial}{\partial a}a^N\\ &=\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\dfrac{1}{\sigma_i^2}\cdot N\cdot a^{N-1}\\ \end{align}[/imath] Is this correct? EDIT: As per the accepted answer, this is how I got my final partial derivative. [imath]\begin{align} \dfrac{\partial}{\partial a}\sum_{i=1}^N\dfrac{D(x_i)-a-b(x_i)^2}{\sigma_i^2} &= \dfrac{\partial}{\partial a}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}-\sum_{i=1}^N\dfrac{-a}{\sigma_i^2}\Bigg]\\ &=\dfrac{\partial}{\partial a}\Bigg[\sum_{i=1}^N\dfrac{D(x_i)-b(x_i)^2}{\sigma_i^2}\Bigg]-\dfrac{\partial}{\partial a}\Bigg[\sum_{i=1}^N\dfrac{-a}{\sigma_i^2}\Bigg]\\ &= 0-\dfrac{\partial}{\partial a}\Bigg[\dfrac{-a}{\sigma_1^2}+\dfrac{-a}{\sigma_2^2}+\cdots+\dfrac{-a}{\sigma_N^2}\Bigg]\\ &= -\Bigg[\dfrac{\partial}{\partial a}\Bigg[\dfrac{-a}{\sigma_1^2}\Bigg]+\dfrac{\partial}{\partial a}\Bigg[\dfrac{-a}{\sigma_2^2}\Bigg]+\cdots+\dfrac{\partial}{\partial a}\Bigg[\dfrac{-a}{\sigma_N^2}\Bigg]\Bigg]\\ &= -\Bigg[\dfrac{-1}{\sigma_1^2}\cdot\dfrac{\partial}{\partial a}[a]+\dfrac{-1}{\sigma_1^2}\cdot\dfrac{\partial}{\partial a}[a]+\cdots+\dfrac{-1}{\sigma_N^2}\cdot\dfrac{\partial}{\partial a}[a]\Bigg]\\ &= -\Bigg[\dfrac{-1}{\sigma_1^2}+\dfrac{-1}{\sigma_2^2}+\cdots+\dfrac{-1}{\sigma_N^2}\Bigg]\\ &= \sum_{i=1}^{N}\dfrac{1}{\sigma_i^2}\\ \end{align}[/imath] |
299652 | Let the matrix [imath]A=[a_{ij}]_{n×n}[/imath] be defined by [imath]a_{ij}=\gcd(i,j )[/imath]. How prove that [imath]A[/imath] is invertible, and compute [imath]\det(A)[/imath]?
Let [imath]A=[a_{ij}]_{n×n}[/imath] be the matrix defined by letting [imath]a_{ij}[/imath] be the rational number such that [imath]a_{ij}=\gcd(i,j ).[/imath] How prove that [imath]A[/imath] is invertible, and compute [imath]\det(A)[/imath]? thanks in advance | 1199581 | Surprising result between determinant of a matrix and product of Euler totient function [imath]\prod_{i=1}^n\phi(i)[/imath]
Let [imath]A[/imath] be the [imath]n[/imath] square matrix with as entries the greatest common divisors of the respective indices: [imath](A)_{i,j}=\gcd(i,j)[/imath], [imath]A=\begin{Bmatrix}\gcd(1,1) & \gcd(1,2) & ... & \gcd(1,n) \\ \gcd(2,1) & \gcd(2,2) & ... & \gcd(2,n) \\ \vdots & \vdots & \ddots & \vdots \\\gcd(n-1,1) & \gcd(n-1,2) & ... & \gcd(n-1,n)\\\gcd(n,1) & \gcd(n,2) & ... & \gcd(n,n)\end{Bmatrix}[/imath] ok so [imath]\phi(n)[/imath] denotes Euler's totient function, then [imath]\det(A)=\phi(1)\phi(2)...\phi(n)=\prod_{i=1}^n\phi(i)\quad\quad(1)[/imath] this can permit us to determine [imath]\phi(n)[/imath] if we know the values of [imath]\phi[/imath] for the numbers less than [imath]n[/imath] because [imath]\underbrace{\det(A) = \sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma_i}}_{\text{Leibniz formula}}\to\dfrac{\det(A)}{\phi(1)...\phi(n-1)}=\dfrac{\sum_{\sigma \in S_n} \mathrm{sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma_i}}{\phi(1)...\phi(n-1)}=\phi(n)[/imath] but idk if this is useful so (1) is a surprising result, but how to prove it? are there similar examples where [imath](A)_{i,j}=f(i,j)[/imath] and [imath]\det(A)=\prod_{i=1}^n g(i)?[/imath] |
3006385 | How to show [imath]D_3\oplus D_4[/imath] is not isomorphic to [imath]D_{24}[/imath]?
How to show [imath]D_3\oplus D_4[/imath] is not isomorphic to [imath]D_{24}[/imath]? Here [imath]D_n[/imath] is the dihedral group of order [imath]2n[/imath]. I am not sure how to prove this. I am not very good with the dihedreal groups. | 1437850 | [imath]D_3\oplus D_4[/imath] not isomorphic to [imath]D_{24}[/imath]
We need to prove that [imath]D_{3} \oplus D_{4}[/imath] is not isomorphic to [imath]D_{24}[/imath] . The way in which I approach such type of questions is to count the number of elements of order [imath]x[/imath] in one group and then in the other group, and then conclude that they aren't equal and hence there can't be any isomorphism between them. But this approach here doesn't look easy, with [imath]D_{24}[/imath] involved. Could anyone suggest another way of solving this ? |
3016116 | Proving [imath]a_{0} = 1[/imath], [imath]a_{n + 1} = \sqrt{1 + a_{n}}[/imath] converges to the golden ratio
Define [imath]a_{0} = 1[/imath] and [imath]a_{n + 1} = \sqrt{1 + a_{0}}[/imath]. Then we have [imath]\lim_{n\to\infty} a_{n} = \sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \ldots }}}} [/imath] So I let [imath]x = \sqrt{1 + \sqrt{ 1 + \ldots }}[/imath]. Then we have [imath]x = \sqrt{1 + x} \Longleftrightarrow x = \frac{1 \pm \sqrt{5}}{2}.[/imath] So I think it converges to the golden ratio. But how can I prove this? Clearly, it's monotone increasing. Now I can just show the golden ratio is a supremum for the sequence. But how can I do that? (Monotone Convergence Theorem) | 930038 | Prove that the limit of this sequence is [imath]\frac{1}{2}(1 + \sqrt{5})[/imath]
I have the following exercise: Let [imath]S_1[/imath] = 1 and for [imath]n \geq 1[/imath] let [imath]S_{n+1} = \sqrt{S_n + 1}[/imath]. Prove that the limit of this sequence is [imath]\frac{1}{2}(1 + \sqrt{5})[/imath] |
3016761 | Solving [imath](\cos x-\sin x)\left(2 \tan x+\frac{1}{\cos x}\right)+2=0[/imath]
Please help me to solve this question [imath](\cos x-\sin x)\left(2 \tan x+\frac{1}{\cos x}\right)+2=0[/imath] I tried to open the whole LHS expression and the tried to bring it in square format. Something like this [imath]\left(\cos x +\frac32\right)^2 - \left(\sin x+\frac12\right)^2-(\sin x-\cos x)^2 =2[/imath] But after this I am not able to proceed forward. | 3015467 | Real solution of [imath](\cos x -\sin x)\cdot \bigg(2\tan x+\frac{1}{\cos x}\bigg)+2=0.[/imath]
Real solution of equation [imath](\cos x -\sin x)\cdot \bigg(2\tan x+\frac{1}{\cos x}\bigg)+2=0.[/imath] Try: Using Half angle formula [imath]\displaystyle \cos x=\frac{1-\tan^2x/2}{1+\tan^2 x/2}[/imath] and [imath]\displaystyle \sin x=\frac{2\tan^2 x/2}{1+\tan^2 x/2}[/imath] Substuting These values in equation we have an polynomial equation in terms of [imath]t=\tan x/2[/imath] So our equation [imath]3t^{4}+6t^{3}+8t^{2}-2t-3=0[/imath] Could Some Help me how to Factorise it. OR is there is any easiest way How to solve it, Thanks |
3015499 | Convergence of [imath]a_n[/imath] given [imath]a_{\lfloor{x^n}\rfloor}[/imath] converges to [imath]0[/imath]
Sequence [imath]a_n[/imath] has a property that for every real [imath]x > 1[/imath], sequence [imath]a_{\lfloor{x^n}\rfloor}[/imath] converges to [imath]0[/imath]. Does that mean that [imath]a_n[/imath] converges to [imath]0[/imath]? I have tried to find a counterexample, but failed at it, so I think it might be true, but I don't know how to prove it. | 1107604 | Convergence of a sequence with assumption that exponential subsequences converge?
Problem One of my best friends asked me to think about the following problem: Suppose a sequence [imath]\{a_n\}_{n=1}^\infty[/imath] satisfies [imath]\lim_{n\to\infty}a_{\lfloor\alpha^n\rfloor}=0[/imath] for each [imath]\alpha>1[/imath]. Is it true that [imath]\lim_{n\to\infty}a_n=0[/imath]? He told me that the preceding proposition (if true) implies Kolmogorov's strong law of large numbers in probability theory. I don't know how, but it's irrelevant. Thoughts Suppose [imath]E\subseteq(1,+\infty)[/imath] is countable (for example, [imath]E=(1,+\infty)\cap\mathbb Q[/imath]) and [imath]\lim_{n\to\infty}a_{\lfloor\alpha^n\rfloor}=0[/imath] for each [imath]\alpha\in E[/imath], we cannot conclude that [imath]\lim_{n\to\infty}a_n=0[/imath]. In fact, we can choose an infinite set [imath]S\subseteq\mathbb Z_{>0}[/imath] such that [imath]S\cap\{\lfloor\alpha^n\rfloor\colon n\in\mathbb Z_{>0}\}[/imath] is finite for each [imath]\alpha\in E[/imath] as follows: Suppose [imath]E=\{\alpha_1,\alpha_2,\dotsc\}[/imath]. We choose [imath]S[/imath] inductively. Suppose [imath]T_0=\mathbb Z_{>0}[/imath]. Given [imath]T_{n-1}[/imath], we set [imath]s_n=\min T_{n-1}[/imath] and [imath]T_n=(T_{n-1}\setminus\{\lfloor\alpha_n^k\rfloor\colon k\in\mathbb Z_{>0}\})\setminus\{s_n\}[/imath]. By a density argument, it's easy to see that [imath]T_n[/imath] are infinite therefore the process doesn't terminate. Let [imath]S=\{s_n\colon n\in\mathbb Z_{>0}\}[/imath]. It's apparent that [imath]\#(S\cap\{\lfloor\alpha_n^m\rfloor\colon m\in\mathbb Z_{>0}\})\le n[/imath]. Given [imath]S[/imath], we set [imath]a_n=1/n[/imath] if [imath]n\not\in S[/imath], and [imath]a_n=1[/imath] if [imath]n\in S[/imath], then [imath]\lim_{n\to\infty}a_{\lfloor\alpha^n\rfloor}=0[/imath] for each [imath]\alpha\in E[/imath], but [imath]\lim_{n\to\infty}a_n[/imath] doesn't exist. Here we choose [imath]S[/imath] by a diagonal process, therefore we cannot mimic the construction when [imath]E[/imath] is uncountable. In fact, the falsehood of the original statement is equivalent to the existence of [imath]S[/imath], therefore it's essential combinatorial, or related to some topological structure of [imath]\mathbb Z_{>0}[/imath] (say, compactness or Baire category, etc). I have no idea on the general case. Any idea? Thanks! |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.