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2000950
|
Discrete martingales in [imath]L^p[/imath]
Let [imath](\Omega,\mathcal{F},(\mathcal{F_n})_n,P)[/imath] be a filtered probability space, [imath]p \in (1,\infty)[/imath], and [imath]M_n[/imath] an [imath]\mathcal{F_n}[/imath] martingale uniformly bounded in [imath]L^p[/imath]. I used this fact to show that [imath](M_n)_n[/imath] is uniformly integrable and that it converges in [imath]L^1[/imath] to some integrable [imath]M_{\infty}[/imath] and [imath](M_n)_{\mathbb{N} \cup \infty}[/imath] is a martingale. How can I show that [imath]|M_n-M_\infty|^p[/imath] is uniformly integrable which will imply [imath]M_n \to M_\infty[/imath] in [imath]L^p[/imath]?
|
1435903
|
[imath]L^p[/imath] submartingale convergence theorem
In class today, we learned about the familiar [imath]L^p[/imath] martingale convergence theorem: For [imath]p >1,[/imath] if [imath]X_n[/imath] is a martingale with [imath]\sup \mathbb{E}|X_n|^p <\infty[/imath], then [imath]X_n \rightarrow X[/imath] a.s. and in [imath]L^p[/imath]. Comparing this result with many other convergence theorems, we find that is requires X to be a martingale instead of just a submartingale. I assume this must have some reason, i.e. this result does not hold for submartingales. Intuitively, I would assume that this is caused by something related to a restriction to the nonnegativity of the submartingale, i.e. the result will not hold when [imath]X_n[/imath] is a negative submartingale. Is this true? Or does the theorem also hold for submartingales? If it does not hold, could you give me a counterexample?
|
2014077
|
why aren't the integers open?
In principals of mathematical analysis on page 32, Rudin defines (definition 2.18e-f): a point [imath]p[/imath] is an interior point of [imath]E[/imath] of there is a neighborhood [imath]N[/imath] of [imath]p[/imath] such that [imath]N\subset E[/imath], [imath]E[/imath] is open if every point of [imath]E[/imath] is an interior point of [imath]E[/imath] By this definition, shouldn't [imath]\mathbb{Z}[/imath] be open? Because [imath]\forall n \in \mathbb{Z} \exists m \in \mathbb{Z}s.t. |n-m|>0[/imath] and so [imath]m[/imath] is in a neighborhood of [imath]n[/imath] and in [imath]\mathbb{Z}[/imath]
|
1653234
|
The set of integers is not open or is open
Baby Rudin gives the example of the set of all integers being not open if it is a subset of [imath]\mathbb{R}^2[/imath]. If we consider the set of integers in [imath]\mathbb{R}[/imath], is this set also not open? I can find a neighbourhood which will contain any point, [imath]p[/imath], however is it a requirement that a neighbourhood contains more than one point? I'm trying to understand this fully and have searched through the various posts that have a slight relation and can not find out specifically how these take interior and isolated points into account and how these relate to openess.
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2014026
|
Lattice of congruences of a lattice is distributive
Let [imath]\mathcal{L}=(L;\land,\lor)[/imath] be a lattice. Show that the lattice [imath](\operatorname{Con}\mathcal{L},\subseteq)[/imath] is distributive. I understand that I only need to show that for any [imath]a,b \in L[/imath] and congruences [imath]A,B,C[/imath] that [imath](a,b ) \in(A\land(B\ \lor C)) \Rightarrow (a,b) \in ((A\land B) \lor (A\land C)).[/imath] If we take [imath](a,b) \in (A\land(B\ \lor C))[/imath] then [imath](a,b) \in A[/imath] and [imath](a,b) \in (B \lor C)[/imath] and we only need to show that [imath](a,b) \in B[/imath] or [imath](a,b) \in C.[/imath] It should be easy, but for some reason I can't prove this. Any help would be greatly appreciated.
|
142371
|
Lattices are congruence-distributive
[imath]\newcommand{\r}[1]{\mathrel{#1}}[/imath] First, a few definitions. Given a lattice [imath]L[/imath], a congruence on [imath]L[/imath] is an equivalence relation [imath]\theta[/imath], compatible with the lattice operations, i.e. if [imath]x_1\r{\theta}x_2[/imath] and [imath]y_1\r{\theta}y_2[/imath], then [imath]x_1\wedge x_2\r{\theta}y_1\wedge y_2[/imath] and [imath]x_1\vee x_2\r{\theta}y_1\vee y_2[/imath]. The congruences on [imath]L[/imath] form a lattice, with meet being intersection and join being the equivalence envelope. I'm trying to prove that this lattice of congruences is distributive. So let's take three congruences [imath]\theta_1,\theta_2,\theta_3[/imath] and attempt to prove that [imath]\theta_1\cap(\theta_2\vee\theta_3)\subseteq (\theta_1\cap\theta_2)\vee(\theta_1\cap\theta_3)[/imath] For a start, let's try to prove something easier which should lead me to the general idea. Take elements [imath]x,y,z[/imath] such that [imath]x\r{\theta_1}y,x\r{\theta_2}z[/imath] and [imath]z\r{\theta_3}y[/imath]. We want to prove that [imath]x(\r{(\theta_1\cap\theta_2)\vee(\theta_1\cap\theta_3)})y[/imath]. Start with [imath](x\wedge z)\r{\theta_1}(y\wedge z)[/imath], which gives [imath]x\r{\theta_1}((y\wedge z)\vee x)[/imath]. Similarly, [imath](x\wedge y)\r{\theta_2}(z\wedge y)[/imath] and [imath]x\r{\theta_2}((y\wedge z)\vee x)[/imath]. This gives [imath]x\r{(\theta_1\cap\theta_2)}((y\wedge z)\vee x)[/imath] I expect a similar manipulation should now give [imath]((y\wedge z)\vee x)\r{\theta_1\cap\theta_3} y[/imath] but I can't see how to manage this. Is this even in the right direction? I suppose something other than [imath]((y\wedge z)\vee x[/imath] could be the middle link, but nothing obviously better comes to mind.
|
2014841
|
Prove that [imath]\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}\binom{n}{k} = 1+\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}[/imath]
For the past 30 minutes I've tried to solve this identity but with no luck. I've tried with math induction,perturbation method and to write the sum and find some simetry but non of them worked. Could someone please help me? Thank you. [imath]\sum_{k=1}^{n}(-1)^{k+1}\frac{1}{k}\binom{n}{k} = 1+\frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}[/imath]
|
1842969
|
Proving [imath]\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}[/imath]
I've been trying to prove [imath]\sum_{k=1}^{n}{(-1)^{k+1} {{n}\choose{k}}\frac{1}{k}=H_n}[/imath] I've tried perturbation and inversion but still nothing. I've even tried expanding the sum to try and find some pattern that could relate this to the harmonic series but this just seems surreal. I can't find any logical reasoning behind this identity. I don't understand Wikipedia's proof. Is there any proof that doesn't require the integral transform?
|
2011985
|
[imath]f[/imath] real-valued, continuously differentiable, then [imath](\int |f|^2\,dx)^2 \le 4(\int |xf(x)|^2\,dx)(\int |f'|^2\,dx)[/imath]?
If [imath]f[/imath] is real-valued and continuously differentiable on [imath]\mathbb{R}[/imath], do we have the following inequality:[imath]\left(\int |f|^2\,dx\right)^2 \le 4\left(\int |xf(x)|^2\,dx\right)\left(\int |f'|^2\,dx\right)?[/imath]
|
1526513
|
Prove an integral inequality: [imath] \left(\int|f|^2dx\right)^2\le 4\left(\int|xf(x)|^2dx\right)\left(\int|f'|^2dx\right) [/imath]
If [imath]f[/imath] is real-valued and continuously differentiable on [imath]\mathbb{R}[/imath], prove that [imath] \left(\int|f|^2dx\right)^2\le 4\left(\int|xf(x)|^2dx\right)\left(\int|f'|^2dx\right) [/imath] Attempt: I tried the Cauchy-Schwarz inequality as well as the Plancherel theorem, but none of them seems to work.
|
2015618
|
[imath]|\bar{F}:F|=\infty[/imath] implies that the degree of the irreducible polynomials in [imath]F[x][/imath] is unbounded?
Let [imath]F[/imath] be a field and [imath]\bar{F}[/imath] be its algebraic closure. Is it true that [imath]|\bar{F}:F|=\infty[/imath] implies that the degree of the irreducible polynomials in [imath]F[x][/imath] is unbounded? I believe it is, and in this case an hint in the right direction for a proof would be enough, but I don't know where to begin
|
292133
|
[imath]R[/imath] with an upper bound for degrees of irreducibles in [imath]R[x][/imath]
One very convenient property of [imath]\mathbb{R}[/imath] as a ring is that there is an upper bound for the degree of irreducible polynomials in [imath]\mathbb{R}[x][/imath], as If [imath]f\in\mathbb{R}[x][/imath] has degree larger than [imath]2[/imath], then [imath]f[/imath] is reducible. However, the proof as I know depends highly on the fact that [imath]\mathbb{C}[/imath] is algebraically closed, and the very nice property: [imath]z\in\mathbb{C}[/imath] is in [imath]\mathbb{R}[/imath] if and only if [imath]z=\bar{z}[/imath]. This makes a generalization to other integral domains rather difficult. So the problem is For what kind of integral domain [imath]R[/imath], we have a finite upper bound on the degree of irreducible elements in [imath]R[x][/imath]? Some most familiar examples are ruled out: in [imath]\mathbb{Z}[x][/imath], [imath]\mathbb{Q}[x][/imath] and [imath]\mathbb{F}_{p}[x][/imath], there is not such a bound. Unfortunately these exhaust all integral domains about which I have a working knowledge. Another result might help is Eisenstein's criterion and its generalized form, which says if we can find a prime ideal [imath]\mathfrak{p}[/imath] in [imath]R[/imath] such that [imath]\mathfrak{p}^2\neq\mathfrak{p}[/imath], then by picking [imath]a\in\mathfrak{p}^2\backslash\mathfrak{p}[/imath] we have an irreducible [imath]a+x^d[/imath], where [imath]d[/imath] can be arbitrary, and hence the upper bound is not possible. So we only need to focus on domains where [imath]\mathfrak{p}^2=\mathfrak{p}[/imath] for all prime ideals. This seems to be a quite strong restriction but I am not sure what to make of it. Can someone give a hint? Thanks!
|
1999737
|
If we let [imath]T : R^2 \to R^2[/imath] be a linear transformation. Consider two bases [imath]B[/imath] and [imath]D[/imath] of the vector space [imath]R^2[/imath]
Having trouble with this problem and not only looking for a answer but some explanation of what is going on in the problem. If we let [imath]T : R^2 \to R^2[/imath] be a linear transformation. Consdider two bases [imath]B[/imath] and [imath]D[/imath] of the vector space [imath]R^2[/imath], where [imath]B=\{v_1,v_2\}[/imath] with [imath] v_1 = \left(\begin{bmatrix} -2 \\1 \end{bmatrix}\right)[/imath] and [imath] v_2 = \left(\begin{bmatrix} 1 \\1 \end{bmatrix}\right)[/imath] and where [imath]D = \{w_1,w_2\}[/imath] with [imath] w_1 = \left(\begin{bmatrix} -1 \\4 \end{bmatrix}\right)[/imath] and [imath] w_2 = \left(\begin{bmatrix} 0 \\1 \end{bmatrix}\right)[/imath]. Suppose that [imath] A = \left(\begin{bmatrix} 1&2 \\-1 &4 \end{bmatrix}\right) = M_{D_<-D(T)}[/imath]. I need to find the matrix [imath]B = M_{B<-B}(T)[/imath]
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2014642
|
Computing matrices of linear transformation under different basis
Came across a problem in my textbook and I am having trouble with it, the book provides an answer but no workings. Looking to get an explanation of what is going on help I prepare for an exam. If we let [imath]T : \mathbb R^2 \to \mathbb R^2[/imath] be a linear transformation. Consdider two bases [imath]V[/imath] and [imath]W[/imath] of the vector space [imath]\mathbb R^2[/imath], where [imath]V=\{v_1,v_2\}[/imath] with [imath] v_1 = \begin{bmatrix} -2 \\1 \end{bmatrix}[/imath] and [imath] v_2 = \begin{bmatrix} 1 \\1 \end{bmatrix}[/imath] and where [imath]W = \{w_1,w_2\}[/imath] with [imath] w_1 = \begin{bmatrix} -1 \\4 \end{bmatrix}[/imath] and [imath] w_2 = \begin{bmatrix} 0 \\1 \end{bmatrix}[/imath]. Suppose that [imath] A = \begin{bmatrix} 1&2 \\-1 &4 \end{bmatrix}[/imath] is the matrix of [imath]T[/imath] under basis [imath]W[/imath]. I need to find the matrix [imath]B[/imath] the matrix of [imath]T[/imath] under basis [imath]V[/imath]. The solution is: [imath] B = P^{-1}AP[/imath] = [imath] \begin{bmatrix} -30&22 \\-48 & 35 \end{bmatrix} [/imath] But I don't know why.
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2016240
|
How do I find [imath]x[/imath] in the following figure
Can anyone help me out in finding the value of x in the following figure?? While drawing i missed out a value. [imath]\angle DCA = 20^{\circ}[/imath]. Not sure how to go about this.
|
63819
|
Determine angle [imath]x[/imath] using only elementary geometry
Using only elementary geometry, determine angle x. You may not use trigonometry, such as sines and cosines, the law of sines, the law of cosines, etc.
|
2015999
|
What is the set of functions from [imath]X[/imath] into [imath]\emptyset[/imath].
Let [imath]X[/imath] be a set. What is the set of functions from [imath]\emptyset[/imath] into [imath]X[/imath] . What is the set of functions from [imath]X[/imath] into [imath]\emptyset[/imath]. My answers: It is not a function because there is not a any element in set of domain. It is empty set because for all [imath]x\in X[/imath] there is no any element in the set of range. Can check you my answer?
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475613
|
Nonempty set mapped to [imath]\emptyset[/imath] and vice-versa
[imath](a)[/imath] How many functions are there from a nonempty set [imath]S[/imath] into [imath]\emptyset[/imath]? [imath](b)[/imath] How many functions are there from [imath]\emptyset[/imath] into an arbitrary set [imath]S[/imath]? This question seems very simplistic but I don't know the answer. I think for [imath](a)[/imath] that there isn't a function that maps a set [imath]S[/imath] into a empty set? For [imath](b)[/imath] I assume it to be all function that map the empty set to an arbitrary set since all sets contain the empty set?
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2017106
|
Algebraic independence of exponentials
Friends, Do you know of a neat proof of the linear independence (over [imath]\mathbb{C}[/imath]) of the functions [imath]f(t) = e^{at}[/imath] and [imath]g(t)=e^{bt}[/imath] when [imath]a[/imath] and [imath]b[/imath] are linearly independent over [imath]\mathbb{Q}[/imath]?
|
1451281
|
Showing that [imath]n[/imath] exponential functions are linearly independent.
I have [imath]n[/imath] lambdas, which are all different real and positive numbers, where: [imath]\lambda_1 < \lambda_2 < \cdots < \lambda_n[/imath]. I then have to show that these functions are linearly independent: [imath]e^{\lambda_1 t}, e^{\lambda_2 t}, \ldots, e^{\lambda_n t}[/imath] So I guess what I want to show is that the only solution to this equations: [imath]c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t}+ \ldots +c_n e^{\lambda_n t}=0, \quad \text{for all } t[/imath] is that all constants are zero. I am not entirely sure how to do this when it is for [imath]n[/imath] lambdas - I did it earlier for just 3 lambdas, where I differentiated the function, so I have tried to use the same approach. I wanted to get [imath]n[/imath] equations with [imath]n[/imath] unknown, so I differentiated the function [imath]n-1[/imath] times, and then I chose to look at the case where [imath]t = 0[/imath], so I have something looking like this: [imath]c_1+c_2+ \ldots +c_n = 0[/imath] [imath]c_1 \lambda_1+c_2 \lambda_2+ \ldots +c_n \lambda_n = 0[/imath] [imath] \cdots [/imath] [imath]c_1 \lambda_1^{n-1}+c_2 \lambda_2^{n-1}+ \ldots +\lambda_n^{n-1} = 0[/imath] Then I put this into a matrix [imath]Ac = 0[/imath], and now I want to show that the [imath]\det(A)[/imath] doesn't equal zero, so that the only solution is c = 0, but I'm not quite sure how to do this, or whether there is an easier way to do it ?
|
1009835
|
Algebraic closed field has infinite many elements
Show that an algebraic closed field must have infinite many elements. Let's suppose that an algebraic closed field [imath]K[/imath] has finite many elements. But how could I get a contradiction??
|
416764
|
Show that an algebraically closed field must be infinite.
Show that an algebraically closed field must be infinite. Answer If F is a finite field with elements [imath]a_1, ... , a_n[/imath] the polynomial [imath]f(X)=1 + \prod_{i=1}^n (X - a_i)[/imath] has no root in F, so F cannot be algebraically closed. My Question Could we not use the same argument if F was countably infinite? Couldn't we say that if F was a field with elements [imath]a_1, a_2, ... [/imath] then the polynomial [imath]f(X) = 1 + \prod_{i=1}^{\infty} (X - a_i)[/imath] does not split over F? Thank you in advance
|
2016900
|
Matrix representation of [imath]\operatorname{Aut}(\mathbb{Z}_2\times\mathbb{Z}_2)\cong\mathbb{S}_3[/imath]
We can show that [imath]\operatorname{Aut}(\mathbb{Z}_{2}\times\mathbb{Z}_2)\cong\mathbb{S}_3[/imath] in two ways: (a). One way is to show that any automorphism is determined by an invertible 2-by-2 matrix with entries in [imath]\mathbb{Z}_2[/imath], that there are six such matrices, and that they form a group isomorphic to [imath]S_3[/imath]. I don't quite understand how to show that any automorphism is determined by such 2-by-2 matrix, e.g. what matrix are we looking for and how do they determine the automorphism. Can someone please enlighten me with, maybe, some concrete examples? Thanks! Edit: This is not a duplicate as I'm not looking for a proof but rather seeking examples of this particular way of looking at the problem.
|
693646
|
Isomorphism from [imath]\mathrm{Aut}(\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z})[/imath] to [imath]S_3[/imath]
I know that [imath]\mathrm{Aut}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} ) \cong S_3[/imath], where [imath]S_3[/imath] is the symmetric group. I do not know how to prove that they are isomorphic, however. What I tried was finding a specific [imath]\phi:\mathrm{Aut}(\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z})\rightarrow S_3[/imath], but the part that confuses me is that the automorphism group is a group of isomorphisms itself, so I don't know how to send an isomorphism to [imath]S_3[/imath].
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2017776
|
Prove that 9 | [imath]a_k +... +a_1+a_0[/imath] implies that 9 | [imath]a_k 10^k +... + a_1 10^1 + a_o 10^0[/imath]
How to prove that : 9 | [imath]a_k+... +a_1+a_0[/imath] implies that 9 | [imath]a_k 10^k +... +a_1 10^1 + a_o 10^0[/imath] where | denotes the divisibility sign (that is, I need to prove that if [imath]a_k+... a_1+a_0[/imath] is divisible by 9 then so is [imath]a_k 10^k +... a_1 10^1 + a_o 10^0[/imath]).
|
1217594
|
Divisibility Rule for 9
I'm working through an elementary number theory course right now and I think I've come up with a proof here but wanted some feedback on my logic. Question: If the sum of the digits in base 10 is divisible by 9, then the number itself is divisible by 9. Proof: Suppose that [imath]9|d_1+d_2+...+d_n[/imath] then [imath]d_1+d_2+...+d_n=0\mod9[/imath] Now consider [imath]d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)[/imath] Each power of [imath]10[/imath] is equivalent to [imath]1\mod9[/imath] therefor [imath]d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)=(1\mod9)(d_1+d_2+...d_n)[/imath] [imath]9|(d_1+d_2+...d_n)[/imath] by our assumption, thus [imath]9|(1\mod9)(d_1+d_2+...d_n)[/imath] Thus we have shown that if 9 divides the sum of the digits in base 10, 9 divides the number itself. The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of 10 being [imath]1\mod 9[/imath]. I think this is fair game here but not 100% confident. Thanks.
|
2017716
|
Show that every compact subspace of a metric space is bounded in that metric and is closed
For the closed part, I just noted that it's a compact subspace of a Hausdorff space and therefore it's closed. For the bounded part, I know intuitively that since every open cover has a finite subcover, I just have to take the largest ball that includes all these covers, but I don't know how to write it rigorously. I know they'd all fit a large ball... I also must find a metric space in which not every closed subspace is compact. Which is an example of a metric space in which not every closed is compact? Because once I know that, I could just take the metric [imath]0,1[/imath]
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1529481
|
Are compact subsets of metric spaces closed and bounded?
Rubin's Theorem 2.34 says that "Compact subsets of metric spaces are closed." But I'm wondering should it also be true that, compact subsets of metric spaces are closed and bounded at the same time? The closeed-ness comes from Rudin's proof, and the bounded-ness comes from the fact that for a compact subset [imath]E[/imath], we can take the finite open cover and than this open cover is what bounds [imath]E[/imath]? Discussions are welcomed! Thank you!!!
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2018544
|
if [imath]a\mid x[/imath] and [imath]b\mid x[/imath] and [imath]\gcd(a,b) = 1[/imath], prove [imath]a\cdot b\mid x[/imath]
if [imath]a\mid x[/imath] and [imath]b\mid x[/imath] and [imath]\gcd(a,b) = 1[/imath], prove that [imath]a\cdot b\mid x[/imath] Well, I've started by saying that [imath]x = q_1a[/imath] and [imath]x = q_2b[/imath], and I know that [imath]a,b[/imath] are prime numbers. And, I'm not sure how to proceed from here. Edit: The answer here is better than in the posts listed here
|
194961
|
Prove: If [imath]a\mid m[/imath] and [imath]b\mid m[/imath] and [imath]\gcd(a,b)=1[/imath] then [imath]ab\mid m[/imath]
Prove: If [imath]a\mid m[/imath] and [imath]b\mid m[/imath] and [imath]\gcd(a,b)=1[/imath] then [imath]ab\mid m[/imath] I thought that [imath]m=ab[/imath] but I was given a counterexample in a comment below. So all I really know is [imath]m=ax[/imath] and [imath]m=by[/imath] for some [imath]x,y \in \mathbb Z[/imath]. Also, [imath]a[/imath] and [imath]b[/imath] are relatively prime since [imath]\gcd(a,b)=1[/imath]. One of the comments suggests to use Bézout's identity, i.e., [imath]aq+br=1[/imath] for some [imath]q,r\in\mathbb{Z}[/imath]. Any more hints? New to this divisibility/gcd stuff. Thanks in advance!
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2018958
|
A question about Stieltjes integral
While studying the Riemann - Stieltjes integral, I read this exercise in my book: "Let [imath]f:[a, b]\to \mathbb{R}[/imath] be a bounded function. Denoted [imath]L(P, f)[/imath] the lower Darboux sum and [imath]|P|[/imath] the mesh of the partition [imath]P[/imath]. Prove that for every [imath]\epsilon>0[/imath], there exists [imath]\delta>0[/imath] such that for every partition [imath]P[/imath] with [imath]|P|<\delta[/imath], we have: [imath]\underline{\int_{a}^{b}}f(x)dx-L(P, f)<\epsilon[/imath]." I have a solution for this exercise, but my question is: Can we generalize this problem for Stieltjes integral, with [imath]\underline{\int_{a}^{b}}f(x)d\alpha[/imath] instead of [imath]\underline{\int_{a}^{b}}f(x)dx[/imath]? Particularly, can we have this theorem? Let [imath]f:[a, b]\to \mathbb{R}[/imath] be a bounded function and [imath]\alpha[/imath] is a monotonic function on [imath][a, b][/imath] . Denoted [imath]L(P, f, \alpha)[/imath] the lower Darboux sum and [imath]|P|[/imath] the mesh of the partition [imath]P[/imath]. Prove that for every [imath]\epsilon>0[/imath], there exists [imath]\delta>0[/imath] such that [imath]\underline{\int_{a}^{b}}f(x)d\alpha-L(P, f, \alpha)<\epsilon[/imath] for every partition [imath]P[/imath] with [imath]|P|<\delta[/imath]?
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2018109
|
A result of Stieltjes Integral
While studying the Riemann - Stieltjes integral, I read this exercise in my book: "Let [imath]f:[a, b]\to \mathbb{R}[/imath] be a bounded function. Denoted [imath]L(P, f)[/imath] the lower Darboux sum and [imath]|P|[/imath] the mesh of the partition [imath]P[/imath]. Prove that for every [imath]\epsilon>0[/imath], there exists [imath]\delta>0[/imath] such that for every partition [imath]P[/imath] with [imath]|P|<\delta[/imath], we have: [imath]\underline{\int_{a}^{b}}f(x)dx-L(P, f)<\epsilon[/imath]." My question is: Can we generalize this problem for Stieltjes integral, with [imath]\underline{\int_{a}^{b}}f(x)d\alpha[/imath] instead of [imath]\underline{\int_{a}^{b}}f(x)dx[/imath]? Particularly, can we have this theorem? Let [imath]f:[a, b]\to \mathbb{R}[/imath] be a bounded function and [imath]\alpha[/imath] is a monotonic function on [imath][a, b][/imath] . Denoted [imath]L(P, f, \alpha)[/imath] the lower Darboux sum and [imath]|P|[/imath] the mesh of the partition [imath]P[/imath]. Prove that for every [imath]\epsilon>0[/imath], there exists [imath]\delta>0[/imath] such that [imath]\underline{\int_{a}^{b}}f(x)d\alpha-L(P, f, \alpha)<\epsilon[/imath] for every partition [imath]P[/imath] with [imath]|P|<\delta[/imath]? I don't know how the mesh of the partition relate to the function [imath]\alpha[/imath].
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2018640
|
Does [imath]f_n(x) = x^{1/n}[/imath] converge?
How would you prove that [imath]f_n(x) = x^{1/n}[/imath] converges pointwise to [imath]1[/imath] for [imath]x \in (0, \infty)[/imath]? It's obvious if you take the limit as [imath]n \to \infty[/imath] but how would you prove it using epsilon?
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488014
|
How can one prove that $\lim_{n \to \infty}a^{1/n}=1$ for every $a>0$?
Prove that [imath]\lim_{n \to \infty} a^{\frac{1}{n}} = 1[/imath] if [imath]a >0[/imath]. In my textbook, we are given a suggestion to let [imath]a^{\frac{1}{n}} = (1+h_n)[/imath] and then show that the [imath]h_n[/imath] term goes to zero using a Theorem that states the following conditions: [imath]\lim_{n \to \infty} = \infty,[/imath] if [imath]a >1, 1,[/imath] if [imath]a = 1,[/imath] and 0, if [imath]\vert a \vert < 1[/imath]. I apologize for not formatting the above cases appropriately; I could not figure out how to use a giant left brace to group them altogether. If I rewrite it as [imath]a^{n^{-1}}[/imath], then I thought that this might help, but I don't think the Binomial Theorem would then apply since the exponent is negative. Furthermore, the problem is only concerned with the positive, real [imath]n[/imath]-th roots. I am quite stuck on this problem and any suggestions or advice would be greatly appreciated. I am using the textbook Introduction to Analysis by Arthur Mattuck.
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2019054
|
Invertibility of polynomials
The following statement popped up on a review sheet for an exam I have coming up. I understand the statement but am unsure how to prove it. I'm also a little unsure about how invertibility works in polynomials. Any sources about the proof or inverting polynomials would be great, thanks! All non-zero elements of [imath]\mathbb{F}_p[x]/(h(x))[/imath] are invertible iff [imath]h(x)[/imath] is irreducible
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816364
|
Polynomial irreducible - maximal ideal
I have a couple of ideals which I wonder if I correctly classify as maximal/prime ideal. [imath]I_1 = \langle 2x^2 + 9x -3\rangle[/imath], [imath]I_2 = \langle x - 1\rangle[/imath] [imath]\mathbf 1)[/imath] Is [imath]I_1[/imath] a maximal ideal in [imath]\mathbb{Q}[x][/imath]? Yes, since [imath]I_1[/imath] is irreducible with [imath]p=3[/imath] using Eisenstein's criterion, thus maximal ideal. [imath]\mathbf 2)[/imath] Is [imath]I_2[/imath] a prime ideal in [imath]\mathbb{Q}[x][/imath]? Yes, since [imath]I_2[/imath] is obviously irreducible, and thus a maximal ideal, and every maximal ideal is a prime ideal. [imath]\mathbf 3)[/imath] Is [imath]I_2[/imath] a maximal ideal in [imath]\mathbb{Z}[x][/imath]? Yes, [imath]I_2[/imath] is obviously irreducible, and thus a maximal ideal. [imath]\mathbf{Edit:}[/imath] No, as it is not a field. [imath] [/imath] Am I right in my conclusions? Appreciate any help.
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2018942
|
Finding rational solutions to this system of equations
For a Linear Algebra problem I need to find [imath]3[/imath] distinct solutions, [imath](x,y,z)[/imath] to the following system of equations: [imath]x^2+y^2+z^2=1 \\ x+y+z=\pm 1[/imath] such that [imath]x,y,z\in \mathbb Z[/imath] I have found [imath](x,y,z)=(\pm 1, 0, 0) , ({-1\over 3}, {2\over 3}, {2\over 3})[/imath] But I'm having trouble finding a third solution. I'm not sure if it even exists. Just to be clear, by "distinct" I mean that that [imath](0, \pm 1, 0)[/imath] and [imath](0, 0, \pm 1)[/imath] count the same as the first solution.
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2017818
|
Find three distinct triples (a, b, c) consisting of rational numbers that satisfy [imath]a^2+b^2+c^2 =1[/imath] and [imath]a+b+c= \pm 1[/imath].
Find three distinct triples (a, b, c) consisting of rational numbers that satisfy [imath]a^2+b^2+c^2 =1[/imath] and [imath]a+b+c= \pm 1[/imath]. By distinct it means that [imath](1, 0, 0)[/imath] is a solution, but [imath](0, \pm 1, 0)[/imath] counts as the same solution. I can only seem to find two; namely [imath](1, 0, 0)[/imath] and [imath]( \frac{-1}{3}, \frac{2}{3}, \frac{2}{3})[/imath]. Is there a method to finding a third or is it still just trial and error?
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2019155
|
Non-trivial definition of logarithmic function
Prove that [imath]\log(t):=\lim_{n\to\infty}n(t^{1/n}-1),[/imath] for any [imath]t>0[/imath].
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1769256
|
Trying to show that [imath]\ln(x) = \lim_{n\to\infty} n(x^{1/n} -1)[/imath]
How do I show that [imath]\ln(x) = \lim_{n\to\infty} n (x^{1/n} - 1)[/imath]? I ran into this identity on this stackoverflow question. I haven't been able to find any proof online and my efforts to get from [imath]\ln(x) := \int_1^x \frac{\mathrm dt}t[/imath] to that limit have been a failure.
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2016920
|
Find limit of [imath]\sqrt[3]{n+1}-\sqrt[3]{n}[/imath]
I have to find the limit of: [imath]\sqrt[3]{n+1}-\sqrt[3]{n}[/imath] What I know is that multiplying can be the key here, so I simplified it to [imath]\frac{1}{(n+1)^{\tfrac{2}{3}}+n^{\tfrac{1}{3}}(n+1)^{\tfrac{1}{3}}+n^{\tfrac{2}{3}}}.[/imath] Now what I do not understand: I know that this converges to zero, since all the terms of the denominator grow as [imath]n[/imath] increases. But how do I formally finish this proof? Wouldn't I have to show for each term that it grows continually? I am not quite sure how to do this. Any hints would be very kindly appreciated, thanks!
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1200544
|
Finding the limit [imath]\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}[/imath]
I am trying to find this limit [imath]\lim_{x\rightarrow \infty} \sqrt[3]{x+1}-\sqrt[3]{x}[/imath] My so far method is this [imath]f(x)>0.[/imath] [imath]f^{\prime}(x)=\frac{1}{3\sqrt[3]{(x+1)^2}}-\frac{1}{3\sqrt[3]{x^2}}<0.[/imath] For every [imath]0<y<1[/imath] the equation [imath]f(x)=y[/imath] has a unique solution(found in maple): [imath]x=\frac{1}{3}\cdot \frac{\frac{1}{3}y^2(3y^2+\sqrt{-3y^4+12y})+\frac{1}{6}\frac{3y^2+\sqrt{-3y^4+12y}}{y}-y^4-2y}{y}[/imath] The previous facts implies that the limit is [imath]0[/imath]. I am wondering if there is any way easier than this to find the limite. thanks in advance.
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2018891
|
Proving continuity in case of unifrom convergence
I got stuck a little bit and I could use some help. Thanks in advance for your time! Let [imath]f_n[/imath] be a sequence of continuous functions and [imath]f_n[/imath] uniformly converges to [imath]f[/imath]. Prove that f is also a continuous function.
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1378563
|
Uniform Continuity implies Continuity
Let [imath]f[/imath] be a function from a metric space [imath]X[/imath] to a metric space [imath]Y[/imath]. Show that if [imath]f[/imath] is uniformly continuous on [imath]X[/imath] then [imath]f[/imath] is continuous on [imath]X[/imath]. Show that the converse is not true. Uniform continuity definition: [imath]∀ε>0:∃δ>0:∀p,q∈X:d_{X}(p,q)<δ⟹d_{Y}(f(p),f(q))<ε[/imath] Continuity definition: [imath]∀ε>0:∀p∈X:∃δ>0:∀q∈X:d_{X}(p,q)<δ⟹d_{Y}(f(p),f(q))<ε[/imath] I wonder if these proofs are correct and if they are formal enough? Uniform continuity ⟹ Continuity Let [imath]f[/imath] be uniformly continuous. Fix [imath]ε_{0}[/imath], obtain [imath]δ_{0}(ε_{0})[/imath] (as a function of [imath]ε_{0}[/imath]), fix any [imath]p_{0}[/imath] and [imath]q_{0}[/imath] and we know that: [imath]d_{X}(p_{0},q_{0})<δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}[/imath] If we want to prove that [imath]f[/imath] is continuous at [imath]p_{0}[/imath], we fix [imath]ε_{0}[/imath] and we pick the same [imath]δ_{0}[/imath] as above and fix any [imath]q_{0}[/imath] and we are assured by 1. that [imath]d_{X}(p_{0},q_{0})<δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}[/imath] Since [imath]p_{0}[/imath] was arbitrary, [imath]f[/imath] is continuous at [imath]X[/imath]. Continuity [imath]\color{red}{\implies}[/imath] Uniform continuity Let [imath]f[/imath] be continuous. Fix [imath]ε_{0}[/imath], fix any obtain [imath]p_{0}[/imath] and [imath]p_{1}[/imath], obtain both [imath]δ_{0}(ε_{0},p_{0})[/imath] and [imath]δ_{1}(ε_{0},p_{1})[/imath] (not necessarily equal), fix any [imath]q_{0}[/imath] and [imath]q_{1}[/imath] and we know that: [imath]d_{X}(p_{0},q_{0})<δ_{0}⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}[/imath] [imath]d_{X}(p_{1},q_{1})<δ_{1}⟹d_{Y}(f(p_{1}),f(q_{1}))<ε_{0}[/imath] For [imath]f[/imath] to be uniformly continuous, if we fix [imath]ε_{0}[/imath], obtain [imath]δ[/imath], fix any [imath]p_{0}[/imath], [imath]p_{1}[/imath], [imath]q_{0}[/imath] and [imath]q_{0}[/imath] we must have that: [imath]d_{X}(p_{0},q_{0})<δ⟹d_{Y}(f(p_{0}),f(q_{0}))<ε_{0}[/imath] [imath]d_{X}(p_{1},q_{1})<δ⟹d_{Y}(f(p_{1}),f(q_{1}))<ε_{0}[/imath] But those two implications need not hold for the same [imath]δ[/imath] as shown above. Suppose we pick [imath]δ_{0}[/imath] and w.l.o.g. we suppose [imath]δ_{0}>δ_{1}[/imath]. Then it could be the case that [imath]δ_{1}<d_{X}(p_{1},q_{1})<δ_{0}[/imath] but we cannot conclude ⟹[imath]d_{Y}(f(p_{1}),f(q_{1}))<ε_{0}[/imath]. Therefore, continuity does not imply uniform continuity.
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2019341
|
Show that [imath]6^{(k+1)+1} + 7^{2(k+1)-1}[/imath] is a multiple of [imath]43[/imath]
Show that [imath]6^{(k+1)+1} + 7^{2(k+1)-1}[/imath] is a multiple of [imath]43[/imath] knowing that [imath]6^{k+1} + 7^{2k-1}[/imath] is a multiple of 43. I have written it as [imath]6^{k+1} 6^1 + 7^{2k-1} 7^2[/imath] but idk what to do after.
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794536
|
[imath]6^{(n+2)} + 7^{(2n+1)}[/imath] is divisible by [imath]43[/imath] for [imath]n \ge 1[/imath]
Use mathematical induction to prove that 6(n+2) + 7(2n+1) is divisible by 43 for n >= 1. So start with n = 1: 6(1+2) + 7(2(1)+1) = 63 + 73 = 559 -> 559/43 = 13. So n=1 is divisible Let P(k): 6(k+2)+7(2k+1) , where k>=1 Show that P(k+1): 6((k+1)+2) + 7(2(k+1)+1) is true = 6(k+1+2) + 7(2k+2+1) = 6(k+3) + 7(2k+3) I'm unsure where to go from here, I've tried several directions after this but have got nowhere. I don't know how to get the 43 or 559 out the front. Any help would be great
|
1977422
|
Intuition behind the sigma algebra generated by a series of random variables?
I am reading David Williams' "Probability with Martingales" and in this book he supposes that you are given the following information about some 'outcome' [imath]\omega \in \Omega[/imath]: [imath]X_n(\omega) \ \text{s.t.} \ n\in \mathbb{N}[/imath] for a series of random variables [imath]X_n:\Omega \rightarrow \mathbb{R}[/imath]. He then says, based on this information, [imath]\sigma(X_n : n \in \mathbb{N})[/imath] consists of 'events' [imath]F[/imath] such that for each point [imath]\omega[/imath], you can tell whether the event [imath]F[/imath] has occurred or not (that [imath]\omega \in F[/imath]). However, no matter how hard I try to wrap my head around this vague idea, I can't seem to understand it. I don't understand how the 'information' provided helps to understand whether some event has occurred or not and what this means in terms of a sigma algebra generated by random variables (which also confuses me). Is there any easier way to understand what a sigma algebra generated by these functions actually is?
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1273287
|
Intuition behind measurable random variables and [imath]\sigma[/imath]-algebra
I've been trying to understand [imath]\sigma[/imath]-algebras and how it encodes information in context of filtration. While certain parts seem clear and logical, I can't say I get the whole picture. I'll try to explain the counter-intuition I get with the classical example of the coin tossing: the probability space [imath]\Omega = \{ HH, HT, TH, TT \}[/imath] and a r.v. [imath]X(\omega)[/imath] equal to the number of heads. At times [imath]0[/imath], [imath]1[/imath] and [imath]2[/imath] the available information is represented using [imath]\sigma[/imath]-algebras [imath]\mathcal{F}_0=\{\emptyset,\Omega\}[/imath], [imath]\mathcal{F}_1=\{\emptyset, \Omega, \{HH,HT\},\{TH,TT\}\}[/imath] and [imath]\mathcal{F}_2=\{\emptyset, \Omega,\{HH,HT\},\{TH,TT\},\{HH\},\{HT\},\{TH\},\{TT\}\}[/imath]. One can notice that [imath]X(\omega)[/imath] is not measurable with respect to [imath]\mathcal{F}_0[/imath] and [imath]\mathcal{F}_1[/imath], because [imath]X^{-1}((\frac{3}{2}; +\infty))=\{HH\}[/imath]. To me it is quite surprising: intuitively [imath]X[/imath] makes perfect sense at all times. In particular it has an expected value at time [imath]0[/imath], which I interpret as that the probability and value of all outcomes [imath]\{\omega\}[/imath] can be computed. How do you think of a non-measurable function? Here's another way of expressing the same confusion. The most natural choice of [imath]\sigma[/imath]-algebra in a finite discrete case is [imath]\mathcal{F}=2^\Omega[/imath], and it is implicitly used in all elementary probability problems. However, this choice of [imath]\mathcal{F}[/imath] does not reflect the fact that some information is known or unknown, conditional probability does. Does it mean that the statement "[imath]\sigma[/imath]-algebra is known information" make sense only in conditioning? Why is it convenient then?
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2012468
|
Prove limit of [imath]\sum_{n=1}^\infty n/(2^n)[/imath]
How do you prove the following limit? [imath]\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{k}{2^k}\right)=2[/imath] Do you need any theorems to prove it?
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1325254
|
What does [imath]\sum_{k=0}^\infty \frac{k}{2^k}[/imath] converge to?
This problem comes from another equation on another question (this one). I tried to split it in half but I found out that [imath]\sum_{k=0}^\infty \frac{k}{2^k}[/imath] can't be divided. Knowing that [imath]\sum_{k=0}^\infty x^k=\frac{1}{1-x}[/imath] I wrote that [imath]\sum_{k=0}^\infty \frac{k}{2^k}=\sum_{k=0}^\infty \left(\frac{\sqrt[k] k}{2}\right)^k=\frac{1}{1-\frac{\sqrt k}{2}}=\frac{2}{2-\sqrt[k] k}[/imath] But that's not what I wanted. Could anyone help me?
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2019118
|
Let [imath]f : [a,b] \rightarrow \mathbb{R}[/imath] be continuous, such that [imath]f(x) ≥ 0[/imath]. Show that if [imath]\int^b_a{f(x)dx} = 0[/imath] then [imath]f(x) =0[/imath]
Let [imath]f : [a,b] \rightarrow \mathbb{R}[/imath] be continuous, such that [imath]f(x) ≥ 0[/imath], for all [imath]x \in \mathbb{R}[/imath]. Show that: if [imath]\int^b_a{f(x)dx} = 0[/imath] then [imath]f(x) =0[/imath] for all [imath]x \in \mathbb{R}[/imath] Is the idea, schematically, that as x > [imath]0[/imath], we find the graph of x equaling the line y = [imath]0[/imath]?
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1729092
|
Show that if a function is not negative and its integral is [imath]0[/imath] than the function is [imath]0[/imath]
Suppose [imath]f[/imath] is continuous, [imath]f(x) \ge 0[/imath] on [imath][a,b][/imath] and [imath]\int_a^b f(x) dx=0[/imath]. Show that [imath]f(x)=0[/imath] on [imath][a,b][/imath]. I know that if [imath]f[/imath] in continuous at [imath]x_0[/imath] and [imath]f(x_0) > 0[/imath], then there exists [imath]\delta > 0[/imath] s.t. [imath]f(x) \ge \frac{1}{2} f(x_0)[/imath] for [imath]|x-x_0| < \delta[/imath] and I think it should help me but I'm stuck at this point.
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2012704
|
Show that [imath]\lim\limits_{x \to \infty} \frac{\lfloor x \rfloor}{x} = 1[/imath]
I know that [imath]x-1 < \lfloor x \rfloor \leq x[/imath] hence [imath]1-\dfrac{1}{x}<\dfrac{\lfloor x \rfloor}{x} \leq 1[/imath] for [imath]x>0[/imath]. I think it is easy to see from this that the limit should be 1 but I don't know how to formally prove this.
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1503018
|
How to find [imath]\lim_{x \to \infty} [x]/x[/imath]?
Find the limit [imath]\lim_{x\to \infty } \ \frac {[x]}{x}.[/imath] Does [imath][x][/imath] means greatest integer in this case?
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2019250
|
An application of the intermediate value theorem
Show that between any two real solutions of [imath]e^x \sin x = 1 [/imath], there is at least one real solution of [imath]e^x \cos x = -1 [/imath] Try: Let [imath]f(x) = e^x \sin x - 1 [/imath]. Consider [imath]I = [0, \pi/2][/imath], Then, [imath]f( 0) = -1 [/imath] and [imath]f( \pi/2) = e^{\pi/2} - 1 >0 [/imath]. Thus, one can find a solution [imath]c_1 \in I [/imath] such that [imath]f(c_1) = 0[/imath]. Similarly, if [imath]J= [ \pi/2, \pi ] [/imath], one can find [imath]c_2 \in J[/imath] such that [imath]f(c_2) = 0 [/imath]. Consider the interval [imath][c_1,c_2][/imath]. Let [imath]F(x) = e^x \cos x + 1 [/imath]. We have [imath]F(c_1 ) = e^{c_1} \cos (c_1) + 1 < e^{c_1} \cos 0 +1 = e^{c_1} + 1 >0 [/imath] and [imath] F(c_2) = e^{c_2} \cos (c_2) - 1 < e^{c_2} \cos \pi - 1 = - e^{c_2} - 1 < 0 [/imath] so, [imath]\mathbf{there \; exists}[/imath] some [imath]\alpha \in [c_1,c_2][/imath] such that [imath]F(\alpha) = 0[/imath] as we wanted to prove. Is this a correct proof?
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1578696
|
Alternating roots of [imath]f(x) = \exp(x) \sin(x) -1[/imath] and [imath]\exp(x)\cos(x) +1[/imath]
The following is a homework problem: Prove that between two roots of the function [imath]f(x) = \exp(x) \sin(x) -1[/imath] there must be at least one root of the function [imath]\exp(x)\cos(x) +1[/imath]. I think it is possible to prove it by simply computing the signs of the functions at [imath]\frac{\pi}{2}k, \pi k, \frac{3\pi}{2}k, 2\pi k[/imath]. However, I'm looking for a more elegant way. It is intuitively clear that the roots of [imath]f[/imath] are ˋˋconverging'' to the roots of [imath]sin[/imath] and the ones of [imath]g[/imath] to the ones of [imath]cos[/imath]. Do you have any hint how to construct a nice argument based on this observation?
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2019420
|
Is this going to converge? [imath]\sum_{n=1}^{\infty}\frac{1}{n^{n}}[/imath]
I have this series [imath]\sum_{n=1}^{\infty}\frac{1}{n^{n}}[/imath] is it converging, or diverging, what test do you use?
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773469
|
test for convergence [imath]\sum_{n=1}^{\infty} \frac{1}{n^n}[/imath]
Test for convergence [imath]\sum_{n=1}^{\infty} \frac{1}{n^n}[/imath] I'm at a loss on what to do, is this a geometric series [imath]\frac{1^n}{n^n}[/imath]?
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2008774
|
Suppose a matrix [imath]A[/imath] has two distinct eigenvalues
Suppose [imath]A\in M_n(F)[/imath] has two distinct eigenvalues [imath]\lambda_1[/imath] and [imath]\lambda_2[/imath] and that [imath]dim(E_{\lambda_1})=n-1[/imath] Prove [imath]A[/imath] is diagonalizable. I'm wondering if there is some sort of theorem that can help as I really have no idea where to start. I believe [imath]dim(E_{\lambda_1})=n-1[/imath] has to mean [imath]dim(E_{\lambda_2})=1[/imath]? But am still not sure how to proceed. Hints appreciated.
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596838
|
If [imath]A[/imath] has two eigenvalues [imath]\lambda _1, \lambda_2[/imath] and [imath]\dim (E_{\lambda_1})=n-1[/imath], then [imath]A[/imath] is diagonalizable
Suppose that [imath]A \in M_{n\times n}(\Bbb F)[/imath] has two distinct eigenvalues [imath]\lambda_{1}[/imath] and [imath]\lambda_{2}[/imath] and that [imath]\dim (E_{\lambda_1})=n-1[/imath] show that [imath]A[/imath] is diagnolizable.
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2001839
|
Prove this relation for nested radicals
For [imath]a_1=\sqrt a , a_2=\sqrt{a-\sqrt{a}}, a_3=\sqrt{a-\sqrt{a+\sqrt {a}}} ,\ldots [/imath] , We have [imath]a_n[/imath] as [imath]\lim_{n\rightarrow\infty}a_n=\frac {A-1}6+\frac 23\sqrt{4a+A}\sin\left(\frac 13\arctan\frac {2A+1}{3\sqrt3}\right)\tag1[/imath] where [imath]A=\sqrt{4a-7}[/imath] Note: We have [imath]- , + , +[/imath] every three terms. How we can prove this relation? (I want to prove it in general not only [imath]a=2[/imath] and period is [imath]3[/imath] not [imath]4[/imath])
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1969087
|
Formula for [imath]\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}[/imath]
Supposedly, the infinitely nested radical [imath]\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}\tag1\label{1}[/imath] converges to [imath]\frac {A-1}{6}+\frac 23\sqrt{4a+A}\sin\left(\frac 13\arctan\frac {2A+1}{3\sqrt 3}\right)\tag2[/imath] where [imath]A=\sqrt{4a-7}[/imath]. In fact, that's how Ramanujan arrived at the famous nested radical [imath]\sqrt{2-\sqrt{2+\sqrt{2+\ldots}}}=2\sin\left(\frac {\pi}{18}\right)\tag3[/imath] So I'm wondering how you would prove [imath]\ref{1}[/imath]. My best try was to set [imath](1)[/imath] equal to [imath]x[/imath] and substitute to get [imath]x=\sqrt{a-\sqrt{a+\sqrt{a+x}}}[/imath]to get an octic polynomial. But I'm not too sure how the trigonometry made its way into the generalization. And I also wonder if there is a way to generalize this even further to possible [imath]\sqrt{a-\sqrt{a-\sqrt{a+\sqrt{a+\ldots}}}}[/imath] which has period [imath]4[/imath] instead of [imath]3[/imath].
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2019470
|
Solving [imath]\lim \limits_{x \to 2}{\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}}[/imath]
I'm trying to solve the following limit. [imath]\lim \limits_{x \to 2}{\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}}[/imath] Unfortunately, I haven't made any headway. I tried multiplying by the conjugates of both the numerator and denominator (in 2 different attempts, of course), but ended up with a monster of an expression either time.
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195532
|
Limit of quotients with square roots: [imath]\lim_{x\to2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}[/imath]
You can't use L'Hospital's rule :S [imath]\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}[/imath] I've tried to multiply by conjugates but ended up with a so complex equation, please help, anyone? :S
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2005670
|
[imath]x^2 \equiv 1 \pmod{p^\alpha}[/imath] - number of solutions
I have the following problem: Let [imath]p[/imath] be prime, [imath]\alpha \in \mathbb{N}[/imath] and [imath]x \in \mathbb{Z}_{p^\alpha}[/imath]. What is the number of solutions to the equation [imath]x^2 \equiv 1 \pmod{p^\alpha}[/imath]? What I've got so far is [imath]p^\alpha|(x - 1)(x + 1)[/imath] but I don't know what to do next.
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366800
|
solution set for congruence [imath]x^2 \equiv 1 \mod m[/imath]
if [imath]m[/imath] is an integer greater than 2, and a primitive root modulo [imath]m[/imath] exists, prove that the only incongruent solutions of [imath]x^2 \equiv 1 \mod m[/imath] are [imath]x \equiv \pm 1 \mod m[/imath]. I know that if a primitive root mod [imath]m[/imath] exists, then [imath]m = 1, 2, 4, p^m,[/imath] or [imath]2p^m[/imath], where p is an odd prime and [imath]m[/imath] is a positive integer. Obviously I can check the cases where [imath]m = 1, 2, 4[/imath] by hand but I am having trouble proving the other two cases.
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2019823
|
Extensions of f to a continuous linear functional
I am working on the following exercise, but I really don't know how to start this: Let [imath]X[/imath] be a normed space, and let [imath]W[/imath] be a linear subspace. Fix a [imath]f_W \in W'[/imath]. Prove that the following are equivalent: a) [imath]W[/imath] is dense in [imath]X[/imath] b) there exists a unique extension of [imath]f_W[/imath] to a continuous linear functional on [imath]X[/imath]. I know that a linear functional [imath]f_X[/imath] is an extension of [imath]f_W[/imath] if [imath]f_X(u)=f_W(u)[/imath] for all [imath]u \in W[/imath]. And [imath]W[/imath] is dense if every point in [imath]X[/imath] is contained in [imath]W[/imath], or a limit point of [imath]W[/imath]. I am very thankful for help!
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332154
|
Functional Analysis. Continuous extensions of linear functionals.
I would like to show that a continuous functional on a subspace [imath]W[/imath] of a normed space [imath](V,\|\cdot \|)[/imath] has a unique continuous extension to [imath]V[/imath] iff [imath]W[/imath] is dense in [imath]V[/imath]. I have proved [imath](\Leftarrow[/imath]). But the converse is currently eluding me! Help would be much appreciated!
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2021374
|
Prove that [imath]e^x>x+1 \forall x\ne 0[/imath]
I need to prove that [imath]e^x>x+1 \forall x\ne 0[/imath]. Any ideas of hints about how to begin? I don't have any idea except the graphical way.
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1714172
|
Proof that [imath]e^{-x} \ge 1-x[/imath]
My aim is to prove that [imath]e^{-x} \geq 1-x[/imath] for any [imath]x \geq 0[/imath]. What I found so far is Bernoulli's inequality, which states that [imath]1+x\leq\left(1+\frac{x}{n}\right)^n\xrightarrow [n\to\infty]{} e^x[/imath] Is there a way to utilize this proven fact in order to prove my problem?
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2021653
|
To what extent is [imath]\textbf{Fld}[/imath] not a filtered category?
EDIT: Basically, I want to see if one can construct examples of fields [imath]F,K[/imath] such that they don't have a common extension field. Apologies in advance if this is obvious, but for example, is there a field containing isomorphic copies of both [imath]\mathbb{C}[/imath] and [imath]\mathbb{Q}_p[/imath] for any prime [imath]p[/imath]? If not, what's a good field invariant that could tackle a lot of these kinds of questions? I was thinking of using Galois groups somehow, but not sure where to begin. This comes from the observation that the underlying set of the geometric realisation (as a locally ringed space) [imath]|F|[/imath] of a functor [imath]F: \textbf{CRing} \to \textbf{Set}[/imath] can be given by [imath]\text{colim}_\textbf{Fld} F(k)[/imath] (limit taken over the functor sending a field [imath]k[/imath] to [imath]F(k)[/imath] and sending [imath]f: k \to K[/imath] to [imath]F(f): F(k) \to F(K)[/imath]), where [imath]\textbf{Fld}[/imath] is the full subcategory of [imath]\textbf{CRing}[/imath] where the objects are fields. If you want to see what this actually looks like, you can consider [imath]\bigsqcup_{\textbf{Fld}} F(k)/\sim[/imath], where [imath]\sim[/imath] is the equivalence relation generated by the relation [imath]x \sim y[/imath] for [imath]x \in F(k)[/imath] and [imath]y \in F(K)[/imath] if there exists [imath]L[/imath] and [imath]f: k \to L[/imath] and [imath]g: K \to L[/imath] such that [imath]F(f)(x) = F(g)(y)[/imath]. Now, if such an [imath]L[/imath] existed always, then this would be an equivalence relation, but if not, then you have to basically force transitivity. So I'm wondering to what extent you really are doing this.
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1098660
|
Field extensions with(out) a common extension
Let [imath]K[/imath] be a field having two field extensions [imath]L\supseteq K[/imath] and [imath]M\supseteq K[/imath]. Does there exist a field [imath]N[/imath] along with embeddings [imath]L\to N[/imath] and [imath]M\to N[/imath], such that the diagram [imath] \require{AMScd} \begin{CD} K @>>> L\\ @V V V @VV V\\ M @>>> N \end{CD} [/imath] is commutative? To put it less formally, do [imath]L[/imath] and [imath]M[/imath] have a common field extension [imath]N[/imath] (with [imath]K[/imath] lying in the intersection)? If yes, consider this side question (but do leave an answer even if you can only answer the main question!): Does the above property of fields generalise to the following, stronger property? Let [imath]p[/imath] be either [imath]0[/imath] or a prime number. Does there exist a sequence of fields [imath] \mathbb F_p = L_0\subseteq L_1\subseteq L_2\subseteq L_3\subseteq L_4\subseteq\cdots[/imath] (where I use the convention [imath]\mathbb F_0 = \mathbb Q[/imath]) such that any field [imath]K[/imath] of characteristic [imath]p[/imath] has an extension field among the [imath]L_\alpha[/imath]? This sequence should be understood as enumerated by ordinal numbers. In other words, what I need is a function that assigns to each ordinal number [imath]\alpha[/imath] a field [imath]L_\alpha[/imath] containing all [imath]L_\beta[/imath] with [imath]\beta < \alpha[/imath]. (Intuitively, I suspect that this might depend on the Axiom of Choice.) If the second property holds, it shows that you can essentially only extend a field in one "direction." It also shows that the class (it is obviously not a set) [imath]\mathbb M_p = \bigcup_\alpha L_\alpha[/imath] is a field (class). We can define all the usual field operations (addition, multiplication, division) here since all pairs of elements lie in [imath]L_\alpha[/imath] for a sufficiently large [imath]\alpha[/imath]. This "monster field" of characteristic [imath]p[/imath] then contains all other set fields of that characteristic.
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2021740
|
If [imath]m=a_1x+b_1y[/imath], [imath]n=a_2x+b_2y[/imath] and [imath]a_1b_2-a_2b_1=1[/imath] then prove that [imath]GCD(m,n)=GCD(x,y)[/imath]
If [imath]m=a_1x+b_1y[/imath], [imath]n=a_2x+b_2y[/imath] and [imath]a_1b_2-a_2b_1=1[/imath] then prove that [imath]GCD(m,n)=GCD(x,y)[/imath] Using a little bit of algebra, I arrived at: [imath]x=b_2m-b_1n\tag{1}[/imath] and [imath]y=a_1n-a_2m\tag{2}[/imath] But then I'm at a loss. What should I consider next? I did plug in a couple of values and it holds.
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1911382
|
Proving that if [imath]ad-bc = \pm 1[/imath], then [imath]\gcd(x,y) = \gcd(ax +by, cx + dy)[/imath]
I'm having issues figuring out how to approach this problem: Conclude that if [imath]ad-bc = \pm 1[/imath], then [imath]\gcd(x,y) = \gcd(ax +by, cx + dy)[/imath] The fact that [imath]\gcd(x,y) = \gcd(x+ky, y)[/imath] is a very special case of this exercise I believe there is a property that if [imath]a = bq_1 + 0[/imath] where 0 is r, then b divides a, and [imath]b=\gcd(a,b)[/imath] Is that at all relevant in this case?
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2021318
|
Show that for every [imath]A\in M_{n}(\mathbb{R})[/imath] with rank(A)=1 that [imath]||A||_{op}=||A||_{HS}[/imath]. What happens in the general case?
Show that for every [imath]A\in M_{n}(\mathbb{R})[/imath] with rank(A)=1 that [imath]||A||_{op}=||A||_{HS}[/imath]. What happens in the general case? Just to clarify the notations: [imath]||A||_{op}= \sup\left\{\frac{\|Av\|}{\|v\|} : v\in V \mbox{ with > }v\ne 0\right\}.[/imath] [imath]\|A\|^2_{HS}={\rm Tr} (A^{T}A) := \sum_{i \in I} \|Ae_i\|^2[/imath] So using this inequality that I proved before: [imath]\|AB\|_{HS} \leq \|A\|_{\mathrm{op}} \|B\|_{HS} [/imath] If I Substitute B for I I get the weak inequality, but I can't see how rank=1 helps me here.. Any help?
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115020
|
Hilbert Schmidt Norm-Rank-inequality
Problem: Let [imath]A_{n.n}[/imath] be square complex matrix. Prove the following: [imath]\left \| A \right \|=\left \| A \right \|_{HS}\Leftrightarrow rank(A)\leqslant 1[/imath]. Where [imath]\left \| . \right \|_{HS} [/imath] is the Hilbert Schmidt Norm. Please read my solution and tell me whether it is correct. If not, let me know where the mistake is. Proof of the implication [imath]\Leftarrow [/imath]: If [imath]rank(A)=0[/imath], then in this case [imath]A=0[/imath]. It follows that [imath]\left \| A \right \|=\left \| A \right \|_{HS}=0[/imath] If [imath]rank(A)=1[/imath], than: [imath] A=\begin{pmatrix} A_{1}\\ \alpha _{2}A_{1}\\ ...\\ \alpha _{n}A_{1} \end{pmatrix}[/imath] where [imath]A_{1}[/imath] is the first row of [imath]A[/imath] and in this case: [imath]\left \| A \right \|=\left \| A \right \|_{HS}=\left \| A_{1} \right \|\sqrt{1+\alpha _{1}^{2}+...+\alpha _{n}^{2}}[/imath] Proof of the implication [imath]\Rightarrow [/imath] We know that [imath]\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|[/imath] On the other hand: [imath] A=\begin{pmatrix} A_{1}\\ A_{2}\\ ... \\ A_{n} \end{pmatrix}[/imath]. So, [imath]\left \| Ax \right \|=\left \| \begin{pmatrix} \left \langle A_{1},x \right \rangle\\ \left \langle A_{2},x \right \rangle\\ ...\\ \left \langle A_{n},x \right \rangle\end{pmatrix} \right \|=\sqrt{\left \langle A_{1},x \right \rangle^{2}+\left \langle A_{2},x \right \rangle^{2}+...+\left \langle A_{n},x \right \rangle^{2}} [/imath] Where [imath] \left \langle A_{i},x \right \rangle[/imath] is the inner product of [imath]A_{i}[/imath] and [imath]x[/imath] Using the Cauchy-Schwarz inequality: [imath]\left \langle A_{i},x \right \rangle^{2}\leq \left \| A_{i} \right \|^{2}\left \| x \right \|^{2}[/imath], we get: [imath]\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|\leq max_{\left \| x \right \|=1}\left \| x \right \|\left \| A \right \|_{HS}=\left \| A \right \|_{HS}[/imath]. In order to have the equality: [imath]\left \| A \right \|=\left \| A \right \|_{HS}[/imath], we should have: [imath]\left \langle A_{i},x \right \rangle^{2}=\left \| A_{i} \right \|^{2}\left \| x \right \|^{2} [/imath] which occurs only if [imath]A_{i}=\lambda _{i} x[/imath]. So, the rows of A are dependent, which implies that [imath]rank(A)=1[/imath]. Note that the case [imath]rank(A)=0[/imath] happens when [imath]A=0[/imath] Please let me know if my solution makes sense.
|
2021352
|
Prove that the function [imath]x\mapsto\sin(x^{2})[/imath] is not periodic
Prove that the function [imath]f:\Bbb R \rightarrow \Bbb R ;x\mapsto\sin(x^{2})[/imath] is not periodic. Let's assume the opposite, i.e that [imath]f[/imath] is periodic. Then for all [imath]\tau \in\Bbb R[/imath], [imath]\sin(\tau +x)^{2}=\sin(x^{2})[/imath]. How do I continue from here?
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282644
|
Is [imath]f(x)=\sin(x^2)[/imath] periodic?
Is the function [imath]f:\Bbb R \rightarrow \Bbb R[/imath] defined as [imath]f(x)=\sin(x^2)[/imath], for all [imath]x\in\Bbb R[/imath], periodic? Here's my attempt to solve this: Let's assume that it is periodic. For a function to be periodic, it must satisfy [imath]f(x)=f(T+x)[/imath] for all [imath]x\in\Bbb R[/imath], so it must satisfy the relation for [imath]x=0[/imath] as well. So we get that [imath]T^2=k\pi \iff T=\sqrt{k\pi}[/imath], [imath]k\in\Bbb N[/imath] (since [imath]T[/imath] must be positive, we remove the [imath]-\sqrt{k\pi}[/imath] solution). So what now? I tried taking [imath]x=\sqrt\pi[/imath] and using the [imath]T[/imath] I found, and I get this: [imath] \sin\pi=\sin(T+\sqrt\pi)\iff-1=\sin(\pi(\sqrt k+1)^2)\iff k+2\sqrt k+1=3/2+l [/imath] Is this enough for contradiction? The left side of equation is sometimes irrational and gets rational only when [imath]k[/imath] is perfect square, which doesn't happen periodic, while the right hand side is always rational. Or I'm still missing some steps? Thanks.
|
2021727
|
Finding a stationary matrix
I am having trouble understanding how to answer this question. I have a transition matrix P: [imath] \begin{matrix} .7 & 0 & .3 \\ 0 & 1 & 0 \\ .2 & 0 & .8 \\ \end{matrix} [/imath] I've figured out the first part of the question which asked me to prove that two 1x3 matrices were stationary which I did with SP=S. But for the second part of the question it asks me to find another stationary matrix for P and it provides me with a hint. The hint is: [imath] \begin{matrix} \\T=aR+(1-a)S, \\ 0<a<1 \end{matrix} [/imath] I'm not sure what to do with that equation. Am I supposed to turn it into another 1x3 matrix and prove it's stationary with TP=T? Do I multiply the equation as it is now with P and then figure out something with that result? Or do I substitute a number in for 'a' and solve for that and then multiply that result with P?
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2004383
|
(Markov Chains)Help finding another stationary matrix using the hint provided in the book
This is my first post here so I apologize if I format something incorrectly. So I have a transition matrix P, [imath] \begin{matrix} .7 & 0 & .3 \\ 0 & 1 & 0 \\ .2 & 0 & .8 \\ \end{matrix} [/imath] and I have two stationary matrices R and S respectively, [imath] \begin{matrix} .4 & 0 & .6 \\ \end{matrix} [/imath] and [imath] \begin{matrix} 0 & 1 & 0 \\ \end{matrix} [/imath] I was able to prove that they are both stationary matrices by showing SP=S and RP=R but the next question asks me to find [imath]\mathbf another[/imath] stationary matrix for P using a provided hint. The hint is [imath] \begin{matrix} \\T=aR+(1-a)S, \\ 0<a<1 \end{matrix} [/imath] I am unsure how to get a stationary matrix from that. Do I just multiply the R matrix by a and then multiply (1-a)S and then add the two resulting matrices together? If I do that then I end up with 1-a in the matrix and that just confuses me more. Sorry if this is supposed to be basic stuff, I always struggle with math :/
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2018933
|
Is there any difference between the free homotopy classes and homology classes of oriented loops in a closed oriented surface
Is there any difference between the free homotopy classes and homology classes of oriented loops in a closed oriented surface [imath]S[/imath]? I know that the conjugacy classes of the fundamental group [imath]\pi_1(S)[/imath] are in one to one correspondence with the free homotopy classes of oriented loops in [imath]S[/imath]. But the conjugacy classes of the fundamental group [imath]\pi_1(S)[/imath] are identified with the homology classes of loops in [imath]S[/imath]. Therefore, free homotopy classes and homology classes of loops in [imath]S[/imath] are in one to one correspondence. So there is no difference if we say homology class or free homotopy class. Am I right?
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604395
|
Two homologous but not homotopic loops on a closed surface of genus greater than one
Suppose [imath]S[/imath] is a closed surface with genus greater than [imath]1[/imath]. Give an example of two loops on [imath]S[/imath] which are not homotopic, but are homologous to each other. I need help solving this question. I have read the Wikipedia page on it and I realize that it does not contain the necessary knowledge to visualize this problem. And I was also wondering the relation between the fundamental group [imath]\pi_1(S)[/imath] and the first homology group [imath]H_1(S)[/imath]? Some restrictions is that [imath]S[/imath] is a closed surface and the genus is greater than [imath]1[/imath].
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2022260
|
Atomic domain that doesn't satisfy ACCP
I recently just completed an exercise which states an atomic domain [imath]A[/imath] is a UFD [imath]\iff[/imath] it satisfies the ascending chain condition on principal ideals. This is all well and good, but because this is an if and only if statement, that seems to suggest there are atomic domains that DON'T satisfy the ACCP. Does anyone know of such an example? I would imagine you need something like the algebraic numbers but I am unsure.
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1941913
|
Example of FD not satisfying ACCP
Let [imath]A[/imath] be an integral domain. We know that if [imath]A[/imath] satisfies ACC on principal ideals then it is a Factorisation Domain. I want to know about the converse, i.e., Does there exist a FD not satisfying ACC on principal ideals? Edit. A Factorisation Domain is an integral domain whose every non-zero non-unit element is a finite product of irreducible elements. Thank You.
|
2010489
|
how to calulate the number of powers in finite field
For any finite field [imath]F_q[/imath] where [imath]q[/imath] is a prime power, I want to calulate the number of powers. Specifically, for 3[imath]\le k \le q-1[/imath], how to determine the size of the set: [imath]\{x^k \mid x \in F_q\}[/imath]. Since the order of [imath]F_q^{\times}[/imath] is [imath]q-1[/imath], I only need to consider the case 2[imath] \le k \le q-1[/imath]. Also I have known the case [imath]k=2[/imath]. Thanks for any advise. addition:I read the post:Number of elements which are cubes/higher powers in a finite field. . Maybe this problem is to determine the kernel of the group endomorphism: [imath]\phi:F_q^\times \rightarrow F_q^\times \text{ sending } x\mapsto x^k[/imath]
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627996
|
Number of elements which are cubes/higher powers in a finite field.
This question is a slight generalization of This Question. How many elements are there in a finite field of order [imath]q[/imath] which are : Squares. Cubes. Higher powers. I mean : How many elements are there which are squares... How many elements are there which are cubes... how many elements are there which are higher powers... What all I could see is : For finding elements which are cubes I would consider : [imath]\eta : F^*\rightarrow F^*[/imath] sending [imath]x\rightarrow x^3[/imath] For this If I know the kernel of [imath]\eta[/imath] then I would have [imath]F^*/Ker(\eta) \cong \{x^3 : x\in F^*\}[/imath] Which says that : No of elements which are cubes are [imath]\dfrac{|q-1|}{|Ker (\eta)|}[/imath]. Now, Kernel of [imath]\eta[/imath] would be [imath]\{x\in F^* : x^3=1\}[/imath] All I know is if there is some subgroup (I am mentioning [imath]\{1,x,x^2\}[/imath] for [imath]\{x\in F^* : x^3=1\}[/imath]) Then It would be unique as a finite cyclic group can not have two subgroups of same order. So, Now the problem is how to see for existence as we are through with uniqueness. Suppose I prove uniqueness then I would say : Number of elements of [imath]F^*[/imath] which are squares are : [imath]\dfrac{q-1}{3}[/imath] If [imath]F^*[/imath] have an element of order [imath]3[/imath] [imath]q-1[/imath] If [imath]F^*[/imath] have no element of order [imath]3[/imath] (I believe this would be totally dependent of nature of [imath]q[/imath]) Now the question is how do i make sure of existence with given nature of [imath]q[/imath]. I would be Thankful If some one can help me to see this and I would be happy to see further generalization. How many elements of order [imath]n[/imath] are there in a Finite field of cardinality [imath]q[/imath] Thank you :)
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2022058
|
Prove that [imath]S_3\times\mathbb{Z}_2[/imath] is isomorphic to [imath]D_6[/imath]
Prove that [imath]S_3\times \mathbb{Z}_2[/imath] is isomorphic to [imath]D_6[/imath]. My attempt: I tried to decompose [imath]D_6[/imath] as the internal direct product of subgroups [imath]H[/imath] and [imath]K[/imath] isomorphic to [imath]S_3[/imath] and [imath]\mathbb{Z}_2[/imath] respectively: Let [imath]H=\{1, r^2, r^4, s, r^2s, r^4s\}[/imath] Let [imath]K=\{1, r^3s\}[/imath] [imath]HK=D_6[/imath] [imath]H\cap K=\{1\}[/imath] However, taking [imath]h=r^2[/imath] and [imath]k=r^3s[/imath], \begin{equation*} \begin{split} hk&=r^2\cdot r^3s \\ &=r^5s \end{split} \end{equation*} \begin{equation*} \begin{split} kh&=r^3s\cdot r^2 \\ &=r^3sr^2ss \\ &=r^3r^{-2}s \\ &= rs \end{split} \end{equation*} [imath]hk\neq kh[/imath] So, this doesn't work. How do I solve this problem?
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2749304
|
How do I prove that [imath]S_3 \times \mathbb{Z}_2[/imath] is isomorphic to [imath]D_6[/imath]?
I was able to find an element of order 6 within [imath]S_3 \times \mathbb{Z}_2[/imath]. I was also able to define a function mapping all 6 elements of [imath]S_3 \times \mathbb{Z}_2[/imath] to [imath]D_6[/imath] I the homomorphism portion would be too exhaustive to prove all 36 cases. How do I prove this.
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2023006
|
Number Theory equality
Suppose that the equation [imath]x^4+y^4=z^2[/imath] has solutions [imath](x,y,z)\in\mathbb{Z}^{3+}[/imath], let [imath]c[/imath] be the smaller [imath]z[/imath] for which [imath](x,y,c)[/imath] is a solution. (1) If [imath]x[/imath] is even, prove [imath]x^2=m^2-n^2, y^2=2mn, c=m^2+n^2[/imath], [imath]m,n[/imath] relatively prime. (2) Prove [imath]x=r^2 - s^2, n=2rs, m=r^2+s^2[/imath], [imath]r,s[/imath] relatively prime. (1) I have the following theorem: if [imath]x^2+y^2=z^2[/imath] then [imath]x^2=2ab, y^2=b^2-a^2, c=b^2+a^2[/imath]. Taking the [imath]x^2[/imath] to be [imath]y[/imath] in the latter, [imath]y^2[/imath] to be [imath]x[/imath] and [imath]c[/imath] to be [imath]z[/imath], then there is [imath]m,n[/imath] such that [imath]x^2=m^2-n^2, y^2=2mn, c=m^2+n^2[/imath]. (2) [imath]x^2+y^2=m^2-n^2+2mn=(m-n)^2[/imath]. By means of the same theorem, there is [imath]r,s[/imath] such that [imath]x=r^2-s^2, y=2rs, m+n=r^2+s^2[/imath]. This gives the equation for [imath]x[/imath], but the other two don't seem two follow, because it needs too be [imath]n=y[/imath] and if [imath]m+n=r^2+s^2[/imath] then [imath]m=(r-s)^2[/imath] instead of [imath]m=r^2+s^2[/imath].
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276515
|
[imath]x^4 + y^4 = z^2[/imath]
[imath]x, y, z \in \mathbb{N}[/imath], [imath]\gcd(x, y) = 1[/imath] prove that [imath]x^4 + y^4 = z^2[/imath] has no solutions. It is true even without [imath]\gcd(x, y) = 1[/imath], but it is easy to see that [imath]\gcd(x, y)[/imath] must be [imath]1[/imath]
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412714
|
Trouble with proving [imath]A[/imath] is an integrally closed domain [imath]\Rightarrow[/imath] [imath]A[t][/imath] is integrally closed domain
This problem has been bugging me for a while. As was stated in the title, I wish to prove: [imath]A[/imath] is an integrally closed domain [imath]\Rightarrow[/imath] [imath]A[t][/imath] is integrally closed domain Here's what I have so far... Suppose [imath]f \in k(t) = Frac(A[t])[/imath] is integral over [imath]A[t][/imath]. Then, trivially, it is integral over [imath]k[t][/imath]. However, [imath]k[t][/imath] is a PID, hence a UFD, and is thus integrally closed. So [imath]f \in k[t][/imath]. So the problem is reduced to: [imath]A[/imath] is an integrally closed domain [imath]\Rightarrow[/imath] [imath]A[t][/imath] is integrally closed in [imath]k[t][/imath]. Now, [imath]f[/imath] is integral. We may (with no loss of generality) assume [imath]f[/imath] is monic by adding a high power of [imath]t[/imath]. (Since [imath]t^n[/imath] is integral, [imath]t^n + f[/imath] is integral if and only if [imath]f[/imath] is.) We write [imath]f^n + a_{n-1}f^{n-1} + \ldots + a_1 f + a_0 = 0[/imath] [imath]f (f^{n-1} + a_{n-1} f^{n-2} + \ldots + a_2 f + a_1 ) = -a_0[/imath] Now I'd like to be able to use the Gauss lemma and win, but every time I try to do that it beats me. Attempt 1: [imath]f[/imath] is monic, so I'd like to make [imath]f^{n-1} + a_{n-1} f^{n-2} + \ldots + a_2 f + a_1[/imath] monic as well. However, if I divide by the leading coefficient [imath]q[/imath], I have to divide [imath]-a_0[/imath] by [imath]q[/imath]. Unfortunately, there's no guarantee (that I know of) keeping [imath]-a_0/q[/imath] in [imath]A[t][/imath]. Attempt 2: Add a really high power of [imath]t[/imath] to [imath]f[/imath]. But there is no bound on the degree of the coefficients, so again there's no guarantee this will be monic. I think I'm in the right direction, but this problem is getting into my head. Any thoughts are welcome! Thanks (For those who are interested, this is a question from Neukirch's Algebraic Number Theory Chapter 1, Section 2, Question 2)
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2208543
|
If an integral domain [imath]A[/imath] is integrally closed, then so is [imath]A[T][/imath]
Show that if an integral domain [imath]A[/imath] is integrally closed in its field of fractions [imath]K[/imath], then so is [imath]A[T][/imath] in its ring of fractions, [imath]K(T) := \mathrm{Frac}(A[T])[/imath]. What I have done so far (albeit not much) is the following. Pick any [imath]p(T) \in K(T)[/imath] that is integral over [imath]A[T][/imath], that is, the following is true. [imath]p(T)^n + a_{n-1}p(T)^{n-1} + \cdots+ a_1p(T) + a_0 = 0[/imath] with each [imath]a_i \in A[T][/imath]. One hint I got was that I could try eliminating the denominator of [imath]p(T) = \frac{f(T)}{g(T)}[/imath] by multiplying out by [imath]g(T)^n[/imath]. But that is about it, I'm stuck quite some time trying to do the next step. So any help or insights regarding this is deeply appreciated.
|
1753938
|
Is there an easy way to see that this simple recurrence is 9-periodic?
In a colloquium talk yesterday, Robert Bryant pointed out that for all initial values [imath]a_0, a_1 \in \mathbb{R}[/imath], the sequence generated by the recurrence relation [imath] a_{n+1} = |a_n| - a_{n-1} [/imath] turns out to be periodic with period 9, i.e., [imath] a_n = a_{n+9} = a_{n+18} = a_{n+27} = \cdots [/imath] This is an interesting and unexpected fact, especially when you consider that to directly prove the periodicity (by writing down expressions for [imath]a_2, \dots, a_9[/imath] in terms of [imath]a_0, a_1[/imath]), you would have to prove the following monstrous identity: [imath] a_0 = -\left| \left| a_1\right| -a_0\right| -\left| -\left| \left| \left| a_1\right| -a_0\right| -a_1\right| +\left| a_1\right| +\left| -\left| \left| a_1\right| -a_0\right| +\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| -\left| a_1\right| +a_0\right| +a_1\right| -a_0\right| +\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| -\left| a_1\right| +a_0\right| +\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| +\left| \left| \left| a_1\right| -a_0\right| +\left| -\left| \left| \left| a_1\right| -a_0\right| -a_1\right| +\left| a_1\right| +\left| -\left| \left| a_1\right| -a_0\right| +\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| -\left| a_1\right| +a_0\right| +a_1\right| -a_0\right| -\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| -\left| a_1\right| +a_0\right| -a_1\right| -\left| a_1\right| -\left| -\left| \left| a_1\right| -a_0\right| +\left| \left| \left| \left| a_1\right| -a_0\right| -a_1\right| -\left| a_1\right| +a_0\right| +a_1\right| +a_0\right| +a_1 [/imath] Of course, a straightforward "brute-force" approach this problem would be to consider sequentially the cases [imath]a_1 > a_0 > 0[/imath], [imath]a_0 > 0 > a_1[/imath], etc. and show that the RHS whittles down to the LHS in each case. But due to the simplicity of the original recurrence relation, I wonder if there isn't a more elegant way (hopefully involving less casework) to see that this recurrence must be 9-periodic.
|
890745
|
[imath]a_{n+1}=|a_n|-a_{n-1} \implies a_n \; \text{is periodic}[/imath]
Prove that any sequence of real numbers satisfying [imath]a_{n+1}=|a_n|-a_{n-1}[/imath] is periodic. Although it looks simple, I can't prove this statement... I tried rewriting the first few terms of the sequence, but nothing interesting showed up... I'd be mostly interested in hints for this one.
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2024027
|
Banach Spaces and Closure of Direct Sums
Let [imath]A_1, A_2[/imath] be closed subspaces of a Banach space [imath]B[/imath] such that [imath]A_1 \cap A_2 = \{0\}[/imath] and: [imath]\inf\{\|x-y\|:x\in A_{1}, y\in A_{2}, \|x\|=\|y\|=1\}>0[/imath] Show that [imath]A_{1}+A_{2}[/imath] is closed. I'm completely lost on how to use the infimum assumption given. I know of a similar result for metric spaces, in where if we have a compact set [imath]A[/imath] and a closed set [imath]B[/imath] with trivial intersection, then we must have that [imath]d(A,B)>0[/imath]. As well, I'm not sure on how to use the fact that the space is Banach. I was thinking of taking a sequence [imath]x_n+y_n \to z[/imath] and then showing [imath]z \in A_{1}+A_{2}[/imath], first by showing that the component sequences [imath]x_n[/imath] and [imath]y_n[/imath] are Cauchy, and then using completeness (i.e. closed subsets of a complete vector space) to find that [imath]z=x+y[/imath] for some [imath]x\in A_1, y\in A_2[/imath]. What would be the best/simpler way to solve this problem, preferably avoiding quotient spaces? Thanks for the help.
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747611
|
Showing the set [imath]A+B[/imath] is closed.
Let [imath]X[/imath] be a banach space, and let [imath]A[/imath], and [imath]B[/imath] be closed linear subspaces. Assume that [imath]\inf\{\|x-y\|\mid x\in A, y\in B, \|x\|=\|y\|=1\}>0[/imath] I want to show that [imath]A+B[/imath] is closed. I was thinking of doing something like, let [imath]z[/imath] be a limit point of [imath]A+B[/imath], then there exist [imath]z_n\in A+B[/imath] such that [imath]z_n\rightarrow z[/imath], each [imath]z_n[/imath] can be written as [imath]a_n+b_n[/imath]. Then I wanted to do something along the lines of determining whether [imath]a_n[/imath] and [imath]b_n[/imath] have a limit (if they do, then call them [imath]a[/imath] and [imath]b[/imath] and then [imath]a\in A[/imath], [imath]b\in B[/imath], and [imath]z=a+b\in A+B[/imath]), but I can't seem to be able to use the condition given. I am studying for a qual, so you can go ahead and either give a solution or sketch it.
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2023797
|
Find the limit of the function [imath]f(x,y)=\frac{x^2 y^2}{x^2 y^2 + (x-y)^2}[/imath] whenever [imath]x^2 y^2 + (x-y)^2 \not= 0[/imath] as [imath](x,y) \to (0,0) [/imath]
Find the limit of the function [imath]f(x,y)=\frac{x^2 y^2}{x^2 y^2 + > (x-y)^2}[/imath] whenever [imath]x^2 y^2 + (x-y)^2 \not= 0[/imath] as [imath](x,y) \to (0,0) [/imath] I thought I can split this question into 4 limit question like [imath]\lim_{(x,y) \to (0^+, 0^+)} f(x,y)[/imath] [imath]\lim_{(x,y) \to (0^+, 0^-)} f(x,y)[/imath] [imath]\lim_{(x,y) \to (0^-, 0^+)} f(x,y)[/imath] [imath]\lim_{(x,y) \to (0^-, 0^-)} f(x,y)[/imath] Logically, this way of looking the question should not be a problem, but it is to a efficient way to do it, especially if f(x,y) is more complicated than the one in this question. So, how can I solve this question more efficiently ? and what is the correct solution ?
|
1862235
|
Find the limit of [imath]\lim\limits_{(x,y)\to (0,0)} \frac{x^2y^2}{x^2y^2+(x-y)^2}[/imath]
Find the limit of [imath]\lim\limits_{(x,y)\to (0,0)} \frac{x^2y^2}{x^2y^2+(x-y)^2}[/imath] So, I know that [imath]\lim\limits_{x \to x_0} f(x)=c \Leftrightarrow \forall (x_n)\subseteq D\setminus\{x_0\}, x_n\to x_0: f(x_n)\to c \, (n\to \infty)[/imath] Let [imath]x_n=\left(\frac{1}{n},\frac{1}{n}\right), y_n=\left(\frac{1}{n},0\right):[/imath] [imath]f(x_n)=\frac{\frac{1}{n}^4}{\frac{1}{n}^4}=1\\ f(y_n)=\frac{0}{...}=0\\ \Rightarrow f(x_n)\neq f(y_n)\, (n\to \infty)[/imath] So the limit doesn't exist. Correct?
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2023540
|
Prime and Maximal Ideals of [imath]\mathbb{Z}_2 \times \mathbb{Z}_4[/imath]
I'm having some trouble finding ideals in general. The problem I'm stuck on is: Find all prime and maximal ideals of [imath]\mathbb{Z}_2 \times \mathbb{Z}_4[/imath]. I know that a finite integral domain is a field, which means the prime and maximal ideals are the same. That means I just have to find one or the other, right? But I don't really know how to go about finding ideals for [imath]\mathbb{Z}_n \times \mathbb{Z}_m[/imath]. Any help would be great. Thank you!
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1216203
|
Prime ideals in a finite direct product of rings
Let [imath]S=\prod_{i=1}^{n}{R_i}[/imath] where each [imath]R_i[/imath] is a commutative ring with identity. The prime ideals of [imath]S[/imath] are of the form [imath]\prod_{i=1}^{n}{P_i}[/imath] where for some [imath]j[/imath], [imath]P_j[/imath] is a prime ideal of [imath]R_j[/imath] and for [imath]i\neq j[/imath], [imath]P_i=R_i[/imath].
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2024277
|
R is a quasi-semi-local ring, I is ideal of R. Prove Jac(R/I)=(JacR + I)/I
R is a quasi-semi-local ring (R has finitely many maximal ideals), I is ideal of R. Prove Jac(R/I)=(JacR + I)/I, in which Jac is Jacobson radical. It's a problem in my teacher's book. I think we should divide maximal ideals of R into [imath]M_1,...,M_k[/imath] and [imath]M_{k+1},...,M_n[/imath], the first set consists of maximal ideals containing I. But I don't know what to do next
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1424906
|
Image of Jacobson Radical is the Jacobson Radical
I'm stuck on the following problem: Let [imath]R[/imath] be a commutative ring with [imath]1[/imath], and suppose [imath]R[/imath] is semilocal with maximal ideals [imath]\mathfrak m_1, ... , \mathfrak m_s[/imath]. Let [imath]\mathfrak a[/imath] be an ideal of [imath]R[/imath]. The problem is to show that under the usual surjection [imath]\pi:R \rightarrow R/\mathfrak a[/imath], the Jacobson radical of [imath]R[/imath] is sent to the Jacobson radical of [imath]R/\mathfrak a[/imath]. My attempt: suppose that [imath]\mathfrak m_1, ... , \mathfrak m_t[/imath] contain [imath]\mathfrak a[/imath], but [imath]\mathfrak m_{t+1}, ... , \mathfrak m_s[/imath] do not. Then [imath]Jac(R/\mathfrak a) = (\mathfrak m_1/ \mathfrak a) \cap \cdots \cap (\mathfrak m_t/ \mathfrak a)[/imath]. Obviously [imath]\pi Jac(R) \subseteq Jac(R/ \mathfrak a)[/imath], and the converse inclusion is equivalent to the following claim: if [imath]x \in \mathfrak m_1 \cap \cdots \cap \mathfrak m_t[/imath], then there exists a [imath]y \in \mathfrak m_1 \cap \cdots \cap \mathfrak m_s[/imath] such that [imath]x - y \in \mathfrak a[/imath]. This should be some application of the Chinese remainder theorem, but I'm not seeing it. All I have to go on is that there exist [imath]m_i \in \mathfrak m_i[/imath] ([imath]t+1 \leq i \leq s[/imath]) with [imath]m_i \not\in I[/imath]. I also haven't found out a way to use the fact that [imath]\mathfrak m_1 ,... , \mathfrak m_s[/imath] are all the maximal ideals of [imath]R[/imath]. I was also thinking about looking at some explicit maps. If we let [imath]\Gamma[/imath] be the composition [imath]\frac{(R/\mathfrak a)}{Jac(R/ \mathfrak a)} \xrightarrow{\cong} \frac{(R/\mathfrak a)}{(\mathfrak m_1/\mathfrak a)} \oplus \cdots \oplus \frac{(R/\mathfrak a)}{(\mathfrak m_t/\mathfrak a)} \xrightarrow{\cong} R/\mathfrak m_1 \oplus \cdots \oplus R/\mathfrak m_t[/imath] then we have a commutative diagram [imath]\begin{matrix} R/Jac(R) & \rightarrow & \frac{(R/\mathfrak a)}{Jac(R/\mathfrak a)} \\ \downarrow & & \downarrow \Gamma \\ R/\mathfrak m_1 \oplus \cdots R/\mathfrak m_s & \rightarrow & R/\mathfrak m_1 \oplus \cdots \oplus R/\mathfrak m_t \end{matrix}[/imath] where the vertical arrows are isomorphisms, and the horizontal arrows are surjections.
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2023853
|
two definitions of isometry
Let [imath]V[/imath] be an inner product space, and [imath]F :V\to V[/imath] a linear map. Some say that [imath]F[/imath] is an isometry when [imath]\langle F(u),F(v)\rangle = \langle u,v\rangle \forall u,v\in V[/imath]. Others say that [imath]F[/imath] is an isometry when [imath]||F(u)||=||u||[/imath] [imath]\forall u\in V[/imath]. Show that these two meanings are equivalent.
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684843
|
Equivalent definitions of isometry
Consider a map [imath]T:\mathbb{R}^2\to\mathbb{R}^2[/imath] such that [imath]\lVert T(x)\rVert=\lVert x\rVert[/imath]. Is this equivalent to stating that [imath]\langle x, y\rangle=\langle T(x), T(y)\rangle[/imath] for all [imath]x,y\in\mathbb{R}^2[/imath], where [imath]\langle\cdot,\cdot\rangle[/imath] denotes the usual inner product?
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2024880
|
Prove that if [imath]f(d) = g(d)[/imath] for every [imath]d\in D[/imath] then [imath]f=g[/imath].
Where [imath]f,g:\mathbb{R}\to\mathbb{R}[/imath] be continuous functions and [imath]D[/imath] is a [imath]\textit{dense}[/imath] set. I can only use the definition of a continuous function and definition of density (since that is all I learned) But I can seem to connect the two here, any guidance is appreciated.
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1202585
|
Real analysis: density and continuity
I am supposed to prove that if two continuous functions [imath]f:\mathbb R\rightarrow\mathbb R[/imath] and [imath]g:\mathbb R\rightarrow\mathbb R[/imath] coincide on a dense subset [imath]\\D[/imath] of [imath]\mathbb R[/imath], then they coincide everywhere. I know that since [imath]\\f[/imath] and [imath]\\g[/imath] are continuous at every point in [imath]\mathbb R[/imath], then the preimage of any open neighborhood of any real number [imath]\\a[/imath] must also be an open neighborhood of [imath]\\f^{-1}(a)[/imath]. But I'm lost on how to use that to get to the conclusion. Do I have to use the concept of connected sets?
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2004856
|
Is [imath]\sum\limits_{n = 1}^\infty (-1)^n\frac{H_n}{n}=\frac{6\ln^2(2)-\pi^2}{12}[/imath] a valid identity?
A while ago I asked a question about an identity that I found while playing with series involving harmonic numbers. However, since the methods I used were not the focus of my question, I never explained how I got this result. So here it goes... NOTE: It may be of help to work backwards through this problem instead. This is just how I did it. We start by defining [imath]H_n[/imath] as the [imath]n^{th}[/imath] harmonic number. We also take note of the identities: [imath] \sum_{n=1}^\infty \frac{(-1)^n}{n} = -\ln(2)[/imath] [imath] \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}[/imath] Now we note that: [imath] \tag{1}H_n = \sum_{k=1}^\infty \frac{1}{k} - \frac{1}{k+n} = \sum_{k=1}^\infty \frac{n}{k(k+n)}[/imath] Since [imath]\sum_{k=1}^\infty \frac{1}{k} - \frac{1}{k+n}[/imath] is a telescoping series. Dividing by [imath]n[/imath] we get: [imath]\tag{2} \frac{H_n}{n} = \sum_{k=1}^\infty \frac{1}{k(k+n)}[/imath] Since [imath] \sum_{n=1}^\infty\frac{H_n}{n}[/imath] is obviously divergent, I took a look at the alternating series: [imath] \tag{3}\sum_{n=1}^\infty(-1)^n\frac{H_n}{n} = \sum_{n=1}^\infty(-1)^n\left(\sum_{k=1}^\infty \frac{1}{k(k+n)}\right)[/imath] We must now multiply this series by 2 and add [imath]\sum_{n=1}^\infty\frac{1}{n^2} [/imath] which is necessary for the next step. So we have: [imath]\begin{align}\tag{4}\sum_{n=1}^\infty\frac{1}{n^2} + 2 \sum_{n=1}^\infty(-1)^n\left(\sum_{k=1}^\infty \frac{1}{k(k+n)}\right) & = \left(\frac{1}{1}\cdot\frac{1}{1} + \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{3}\cdot\frac{1}{3}+\cdots\right) \\ & -2 \left(\frac{1}{1}\cdot\frac{1}{2} + \frac{1}{2}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{4}+\cdots\right) \\ & +2 \left(\frac{1}{1}\cdot\frac{1}{3} + \frac{1}{2}\cdot\frac{1}{4} + \frac{1}{3}\cdot\frac{1}{5}+\cdots\right) \\ & -2 \left(\frac{1}{1}\cdot\frac{1}{4} + \frac{1}{2}\cdot\frac{1}{5} + \frac{1}{3}\cdot\frac{1}{6}+\cdots\right) \\ & + \cdots \end{align}[/imath] With a very careful (and possibly illegal) rearrangement of this expanded sum, we can finally obtain: [imath]\\[/imath] (NOTE: It may be easier to work backwards through this step) [imath] \begin{align} \tag{5}\sum_{n=1}^\infty\frac{1}{n^2} + 2 \sum_{n=1}^\infty(-1)^n\left(\sum_{k=1}^\infty \frac{1}{k(k+n)}\right) &= \left(-\frac{1}{1} + \frac{1}{2} - \frac{1}{3} + \cdots\right)\left(-\frac{1}{1} + \frac{1}{2} - \frac{1}{3} + \cdots\right) \\ & = \ln^2(2)\end{align}[/imath] Thus, we have: [imath]\tag{6}\sum_{n=1}^\infty(-1)^n\frac{H_n}{n} = \frac{6\ln^2(2) - \pi^2}{12} = -0.5822...[/imath] Here are 3 graphs from computing the actual value of the sum. The first graph is of the difference between the actual sum and [imath]\frac{6*ln^2(2)-\pi^2}{12}[/imath] for k up to 14000. As you can see it seems to converge on 0. You can ignore the 2nd and 3rd graphs. QUESTIONS: Is the method used from step 4-5 (or vise versa) legal even though it involves conditionally convergent series? If not, is there any restrictions and/or rearrangements I can use to make it legal? Is there any known generalizations for [imath]\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^q}[/imath] that extend this result? Recommended readings that cover this topic would be greatly appreciated as well. Thanks, Dom
|
275643
|
Proving an alternating Euler sum: [imath]\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2[/imath]
Let [imath]A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},[/imath] where [imath]H^{(p)}_n = \sum_{i=1}^n i^{-p}[/imath], the [imath]n[/imath]th [imath]p[/imath]-harmonic number. The [imath]A(p,q)[/imath]'s are known as alternating Euler sums. Can someone provide a nice proof that [imath]A(1,1) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2?[/imath] I worked for a while on this today but was unsuccessful. Summation by parts, swapping the order of summation, and approximating [imath]H_k[/imath] by [imath]\log k[/imath] were my best ideas, but I could not get any of them to work. (Perhaps someone else can?) I would like a nice proof in order to complete my answer here. Bonus points for proving [imath]A(1,2) = \frac{5}{8} \zeta(3)[/imath] and [imath]A(2,1) = \zeta(3) - \frac{1}{2}\zeta(2) \log 2[/imath], as those are the other two alternating Euler sums needed to complete my answer. Added: I'm going to change the accepted answer to robjohn's [imath]A(1,1)[/imath] calculation as a proxy for the three answers he gave here. Notwithstanding the other great answers (especially the currently most-upvoted one, the one I first accepted), robjohn's approach is the one I was originally trying. I am pleased to see that it can be used to do the [imath]A(1,1)[/imath], [imath]A(1,2)[/imath], and [imath]A(2,1)[/imath] derivations.
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2025118
|
Linear Algebra, determinants
Let [imath]A_{n}[/imath] be the matrix with zeros on the diagonal and ones everywhere else. Show that the [imath]\det(A_n) = 1-n [/imath] . This is the question that my professor asked but I think it must be a typo. Is it not [imath](-1)^n(1-n)[/imath]? For the proof I was thinking of doing it by induction but I'm a little confused about the inductive step.
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1312849
|
Matrix with zeros on diagonal and ones in other places is invertible
Hi guys I am working with this and I am trying to prove to myself that n by n matrices of the type zero on the diagonal and 1 everywhere else are invertible. I ran some cases and looked at the determinant and came to the conclusion that we can easily find the determinant by using the following [imath]\det(A)=(-1)^{n+1}(n-1)[/imath]. To prove this I do induction n=2 we have the [imath]A=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}[/imath] [imath]\det(A)=-1[/imath] and my formula gives me the same thing (-1)(2-1)=-1 Now assume if for [imath]n \times n[/imath] and [imath]\det(A)=(-1)^{n+1}(n-1)[/imath] Now to show for a matrix B of size [imath]n+1 \times n+1[/imath]. I am not sure I was thinking to take the determinant of the [imath]n \times n[/imath] minors but I am maybe someone can help me. Also is there an easier way to see this is invertible other than the determinant? I am curious.
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170936
|
Verifying some trigonometric identities: [imath]\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta[/imath]
Prove the following: 46. [imath]\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta[/imath] I got as far as Right Side: [imath]\tan\theta\sin\theta[/imath] to [imath]\dfrac{\sin\theta}{\cos\theta}\dfrac{\sin\theta}{1}[/imath] and then; [imath]\dfrac{\sin^2\theta}{\cos\theta}[/imath] Left Side: [imath]\begin{align*} \dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta} &= \dfrac{\frac{1}{\sin^2\theta}-{\frac{\cos^2\theta}{\sin^2\theta}}}{\frac{\cos\theta}{\sin\theta}-{\frac{1}{\sin^2\theta}}}\\ &= \dfrac{\frac{1-\cos^2\theta}{\sin^2\theta}}{\frac{\cos\theta}{\sin^2\theta}}\\ &= \dfrac{\frac{\sin^2\theta}{\sin^2\theta}}{\frac{\cos\theta}{\sin^2\theta}}\\ &= \frac{1}{\frac{\cos\theta}{\sin^2\theta}}\\ &= \frac{\sin^2\theta}{\cos\theta} \end{align*}[/imath] Thanks a lot!
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185205
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Proving that [imath]\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta[/imath]
Prove [imath]\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta[/imath] So, LS= [imath]\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}[/imath] [imath]\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\right)-\left(\dfrac{1}{\tan\theta}\cdot \dfrac{\sin\theta}{1}\right)[/imath] [imath]\dfrac{\tan\theta}{\sin\theta}-\dfrac{\sin\theta}{\tan\theta}[/imath] Now, considering the fact that I must have a common denominator to subtract, would this be correct: [imath]\dfrac{\tan^2\theta}{\tan\theta\sin\theta}-\dfrac{\sin^2\theta}{\tan\theta\sin\theta}\Rightarrow \dfrac{\tan^2\theta-\sin^2\theta}{\tan\theta\sin\theta}[/imath] I feel like I'm close to the answer because the denominator is the RS of the OP. Please help. Do not give me the answer.
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2025402
|
Prove that if [imath]a[/imath] is odd, then [imath]a^2\equiv 1\pmod 8[/imath]
Prove that if [imath]a[/imath] is odd, then [imath]a^2\equiv 1\pmod 8[/imath]I got this question in my discrete mathematics class, can anyone help me?Thanks.
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2020095
|
Proof by Cases involving divisibility
To prove that if [imath]n[/imath] is odd, then [imath]8 \mid n^2−1[/imath], observe that as [imath]n[/imath] is odd either [imath]n = 4k+1[/imath] for some integer [imath]k[/imath] or [imath]n=4k+3[/imath] for some integer [imath]k[/imath]. In the first case [imath]n^2 -1=8(2k^2+k)[/imath] and in the second case [imath]n^2-1 = 8(2k^2 + 3k + 1)[/imath]. Given an odd integer [imath]a[/imath], establish that [imath]a^2 +(a+2)^2 +(a+4)^2 + 1[/imath] is divisible by [imath]12[/imath]. Solve by consideration of six cases.
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2001532
|
When are the 2 fields isomorphic?
Let [imath]F[/imath] be a field. Let [imath]F[x][/imath] be the polynomial ring and [imath]f(x)[/imath] and [imath]g(x)[/imath] be 2 irreducible polynomials of same degree in [imath]F[x][/imath]. When do we have the 2 fields [imath]F[x]/<f(x)> \cong F[x]/<g(x)>[/imath] Well, I can think of [imath]F[x]/<f(x)>[/imath] as the field [imath]F(\alpha)[/imath] where [imath]\alpha[/imath] is the root of [imath]f(x)[/imath] and check if any algebraic combination of [imath]\alpha[/imath] with elements of [imath]F[/imath] is a root of [imath]g(x)[/imath]. If it is then both the fields are the same/isomorphic. But is there any general condition on the 2 polynomials such that the isomorphism holds?
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1595094
|
Easy criteria to determine isomorphism of fields?
Let [imath]K[/imath] be a field and [imath]f,g[/imath] irreducible polynomials in [imath]K[X][/imath], is there a nice iff condition for [imath]K[X]/(f)\cong K[X]/(g)[/imath]? ([imath]\cong[/imath] denotes an isomorphism that is the identity on restriction to [imath]K[/imath]). Thoughts: It is sufficient that they are [imath]K^\times[/imath] multiples of each other. I'd hoped this was necessary but it isn't as [imath]\mathbb{Q}[X]/(X^2-2)\cong\mathbb{Q}(\sqrt2)=\mathbb{Q}(\sqrt2+1)\cong\mathbb{Q}[X]/(X^2-2X-1)[/imath] with the middle two fields viewed as subfields of [imath]\mathbb{C}[/imath]. It is necessary that they have the same degree. Please let me know if there are any other simple necessary conditions. Thanks!
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2026365
|
Smallest n>0 such that [imath]2001^n-1\:%[/imath] is divisible by [imath]\:2^{2002}[/imath]
Smallest n>0 such that [imath]2001^n-1\:%[/imath] is divisible by [imath]\:2^{2002}[/imath], can you give me some hints as to what theorem/algorithm to use to approach this problem?
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2021531
|
Find min natural number [imath]n[/imath] so that [imath]2^{2002}[/imath] divides [imath]2001^{n}-1[/imath]
Can someone explain me how this type of examples are being solved? I know that I can watch this example like this [imath]2001^{n}\equiv 1\ ({\rm mod}\ 2^{2002})[/imath] but I don't know what to do when the divisor is such a large number like [imath]2^{2002}[/imath] [imath]2^{2002}|2001^{n}-1[/imath]
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2025904
|
Prove [imath]\frac{k^7}7+\frac{k^5}5+\frac{2k^3}3-\frac k{105}[/imath] is a integer by mathematical induction
This is what I got: If this must be possible i.e.if it must be an integer then [imath](\frac{15k^7}{7} + \frac{21k^3}{3}+ \frac{70k^2}{2}-k)[/imath] must be divisible by 105. [imath]k(\frac{15k^6}{7} + \frac{21k^2}{3} + \frac{70k}{2}-1)[/imath] must be divisible by 105. For the purpose of using math induction I found out htks, but then I am stuck!! Therefore it works for 1 but how do I further prove it for [imath]k+1[/imath]?
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1916947
|
proof using the mathematical induction
Prove [imath]\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}[/imath] is an integer for every positive integer k. In proving for (n+1) integer,the expression is integer,I found [imath](n+1)^7[/imath] term.I use binomial theorem to expand but finally it won't work(some term become integer While some other remaining as rational) Any hint to solve the problem is appreciated.Thanks.
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2026724
|
Composition of decreasing functions is increasing
I want to prove the following: If [imath]f[/imath] and [imath]g[/imath] are decreasing functions on [imath]I[/imath] and [imath]f \circ g[/imath] is defined on [imath]I[/imath], then [imath]f \circ g[/imath] is increasing on [imath]I[/imath]. My attempt: Let [imath]x_1,x_2[/imath] be in [imath]I[/imath] such that [imath]x_1\lt x_2[/imath] and let [imath]f[/imath] and [imath]g[/imath] be decreasing. Then [imath]f(x_1)\gt f(x_2)[/imath] and [imath]g(x_1)\gt g(x_2)[/imath]. Then [imath](f \circ g)(x_1)= f(g(x_1))[/imath] and [imath](f \circ g)(x_2)=f(g(x_2))[/imath]. Consequently, we have [imath]f(g(x_1)) < f(g(x_2))[/imath] so [imath]f \circ g[/imath] is increasing on [imath]I[/imath].
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1554956
|
if [imath]f: (0,\infty) \to (0,\infty)[/imath] is a strictly decreasing then [imath]f \circ f[/imath] is decreasing?
I need to find the truth value of the statement "for each strictly decreasing function [imath]f :(0,\infty)\to(0,\infty)[/imath],the composition function function [imath]f\circ f[/imath] is decreasing". Can I find a function to disprove this statement? Can I use the function [imath]f :(0,\infty)\to(0,\infty)[/imath], where [imath]f(x)=\dfrac1x[/imath]?
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2027354
|
[imath]n[/imath] men and [imath]n[/imath] women form a queue, how many ways are there so that for every man, there are more women in front of him?
[imath]n[/imath] men and [imath]n[/imath] women form a queue. How many ways are there to line them up so that for every man, the number of women in front of him exceeds the number of men in front of him? I don't know how to generalize. [imath]n=1[/imath]: BA [imath]n=2[/imath]: BABA, BBAA [imath]n=3[/imath]: BBBAAA, BBAABA, BABBAA, BABABA, BBABAA
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1727593
|
Catalan number - combinatoric
Alice and bob are playing cards. for each round of the game, player can win 1 card if he or she won the round, otherwise, he or she loses [imath]1[/imath]. It's a [imath]30[/imath] rounds game, and player can hold negative amount of cards which he or she owes to the other player. the game result will be presented by series of [imath]30[/imath] characters, built from [imath]A,B[/imath]. A- if Alice won the round. B- if Bob won the round. What is the number of possible sequences (game results) under the following restriction: Bob started with 2 cards, finished with no cards and during the competition Bob never owed Alice cards. if I wasn't clear enough, please tell me and I'll fix it.
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2027452
|
[imath][(-2)^\frac{3}{8}]^\frac{8}{3}[/imath]
If we have to evaluate [imath][(-2)^\frac{3}{8}]^\frac{8}{3}[/imath] would the answer be -2 or 2? My teacher says that in this case the exponents can't be multiplied as they are not natural numbers. Then how do we simplify it solving one bracket in a step?
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317528
|
How do you compute negative numbers to fractional powers?
My teachers have gone over rules for dealing with fractional exponents. I was just wondering how someone would compute say: [imath](-5)^{2/3}[/imath] I have tried a couple ways to simplify this and I am not sure if the number stays negative or turns into a positive. I know that if a negative number is raised to an odd power it is negative, but fractional powers are neither odd or even. Is there a general rule for dealing with these types of problems?
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2026935
|
Example of ideal which is not primary, but its radical is prime.
I want to find an example of ideal [imath]Q[/imath] such that [imath]\sqrt{Q}[/imath] is prime, but [imath]Q[/imath] is not primary. It is clear that our domain would not be PID because [imath]\sqrt{Q}[/imath] should not be maximal ideal.
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733735
|
Is it true that an ideal is primary iff its radical is prime?
Is it true that an ideal [imath]I[/imath] in a commutative ring is primary iff [imath]Rad(I)[/imath] is prime? If not, what are some nice counterexamples?
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2027879
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Relation between divisions [imath]\Bbb Z[/imath] and [imath]K[t][/imath]
Let K be a field and a,b [imath]\in \Bbb N[/imath] Show that [imath]a\text{ divides } b \text{ in } \Bbb Z \iff t^a-1\text{ divides } t^b-1 \text{ in } K[t] [/imath] My attempt at this: I found a formula (speculation) for the polynomial division. It states as follows [imath](t^b-1):(t^a-1)=\sum_{i=1}^{\frac{b}{a}}t^{b-ia}[/imath] of course this is only defined as long as [imath]\frac{b}{a}[/imath] is a natural number, which is only the case when a divides b. So (I think) it's sufficient to show that this formula is true. I try to show this by induction over [imath]\frac{b}{a}[/imath] as follows. [imath]\frac{b}{a}=1[/imath] [imath]\Rightarrow a = b[/imath], so [imath](t^a-1):(t^a-1) = 1[/imath] and [imath]\sum_{1}^{1}t^{b-a} = 1[/imath] [imath]\frac{b}{a} \to \frac{b}{a}+1[/imath] We have [imath]\frac{b}{a}+1=\frac{b+a}{a}[/imath] So [imath](t^{b+a}-1):(t^a-1) = 1[/imath] and [imath]\sum_{i=1}^{\frac{b+a}{a}}t^{b-ia+a}[/imath] (we have to prove this equality holds) [imath]\iff t^b(t^a-1):(t^a-1)=\sum_{i=1}^{\frac{b}{a}}t^{b-ia+a} + 1 [/imath] Now susing the induction hypothesis we get [imath]\iff t^b\sum_{i=1}^{\frac{b}{a}}t^{b-ia} =\sum_{i=1}^{\frac{b}{a}}t^{b-ia}t^a + 1 [/imath] I am stuck here. Thanks in advance
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609900
|
How to prove [imath]\,x^a-1 \mid x^b-1 \iff a\mid b[/imath]
How to prove [imath]x^a-1\mid x^b-1 \iff a \mid b[/imath], where [imath]x \ge 2[/imath] and [imath]a,b,x \in \Bbb Z[/imath]. I've tried the following in attempting to solve this: [imath]a\mid b \Rightarrow aq=b \Rightarrow x^{aq}=x^b \Rightarrow x^ax^q=x^b[/imath] Because [imath]x^q \in \Bbb Z[/imath], it follows that [imath]x^a\mid x^b[/imath]. This is as far as I have gotten; any help getting further is appreciated. Note: It may be that this identity is not true at all?
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1948422
|
The sigma function (sum of divisors) multiplicative proof
I am trying to prove that [imath]\sigma(p_1^a\cdot p_2^b) =\sigma(p_1^a)\cdot\sigma(p_2^b)[/imath] where [imath]p_1[/imath] and [imath]p_2[/imath] are prime numbers. We know that [imath]\sigma(p_1^a) = \frac{p_1^{a+1}-1}{p_1-1}[/imath] and [imath]\sigma(p_2^b) = \frac{p_2^{b+1}-1}{p_2-1}[/imath]. Now I am trying to find the divisors of [imath]p_1^a\cdot p_2^b[/imath] and add them: I found that the divisors are [imath]1[/imath], [imath]p_1[/imath], [imath]p_1^{2},\dotsc,p_1^{a}[/imath], [imath]p_2[/imath], [imath]p_2^{2},\dotsc,p_2^{b}[/imath], [imath]p_1\cdot p_2[/imath], [imath]p_1\cdot p_2^2,\dotsc,p_1\cdot p_2^{b},\dotsc,p_1^{a}\cdot p_2,\dotsc,p_1^{a}\cdot p_2^{b}[/imath]. Now when we do their summation we get [imath]\sum_{k=0}^a[/imath] [imath]p_1^k[/imath] + [imath]\sum_{k=1}^b[/imath] [imath]p_2^k[/imath] + ([imath]\sum_{k=1}^{a}[/imath] [imath]p_1^k\cdot\sum_{k=1}^b[/imath] [imath]p_2^k[/imath]), is this right? If yes I can't reach [imath]\frac{p_1^{a+1}-1}{p_1-1}\cdot\frac{p_2^{b+1}-1}{p_2-1}[/imath].
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1500089
|
Sum of Divisors Multiplicative
I wish to prove that the sum of the divisors function, [imath]\sigma(x)[/imath], is multiplicative, i.e. [imath]\sigma(m\cdot n)=\sigma(m)\cdot\sigma(n)[/imath] if [imath]gcd(m,n)=1[/imath]. I start by claiming that since [imath]m[/imath] and [imath]n[/imath] can be written uniquely as a product of prime factors, let [imath]m=\prod_{i=1}^k \, p_i = p_1 \cdot p_2 \cdots p_k[/imath] [imath]n=\prod_{j=1}^l \, q_j = q_1 \cdot q_2 \cdots q_l[/imath] such that [imath]p_i \neq q_j[/imath] for any [imath]i,j \in \mathbb{N}[/imath]. Then the sum of the divisors of [imath]m[/imath] is \begin{align} \sigma(m) = \,\,& p_1 +p_1 \cdot p_2 + p_1 \cdot p_2 \cdot p_3 + \cdots + m + \\ &p_2+p_2 \cdot p_3 + p_2 \cdot p_3 \cdot p_4 + \cdots + \prod_{i=2}^k \,p_i + \\ &p_3+p_3 \cdot p_4 + p_3 \cdot p_4 \cdot p_5 + \cdots + \prod_{i=3}^k \,p_i + \\ &\vdots \\ &+ p_k +1 \end{align} Or, [imath]\sigma(m)=\sum_{z=1}^k \left(\sum_{h=z}^{k}\left(\prod_{i=z}^h \,p_i\right)\right)+1.[/imath] Similarly, [imath]\sigma(n)=\sum_{y=1}^l \left(\sum_{r=y}^{l}\left(\prod_{j=y}^r \,q_j\right)\right)+1.[/imath] Then the product [imath]m \cdot n[/imath] will have for divisors every divisor of [imath]m[/imath] excluding [imath]1[/imath], every divisor of [imath]n[/imath] excluding [imath]1[/imath], [imath]1[/imath], and every possible combination of [imath]m[/imath]'s divisors with [imath]n[/imath]'s divisors. The combinations are given by the product of the two expressions for [imath]\sigma(m)[/imath] and [imath]\sigma(n)[/imath] less one. Then, \begin{align} \sigma(m \cdot n)=&\sum_{z=1}^k \left(\sum_{h=z}^{k}\left(\prod_{i=z}^h \,p_i\right)\right) \cdot \sum_{y=1}^l \left(\sum_{r=y}^{l}\left(\prod_{j=y}^r \,q_j\right)\right) + \\[2ex] &\sum_{z=1}^k \left(\sum_{h=z}^{k}\left(\prod_{i=z}^h \,p_i\right)\right) + \sum_{y=1}^l \left(\sum_{r=y}^{l}\left(\prod_{j=y}^r \,q_j\right)\right) + 1 \end{align} Thus, \begin{align} \sigma(m \cdot n) &= \left[\sum_{z=1}^k \left(\sum_{h=z}^{k}\left(\prod_{i=z}^h \,p_i\right)\right)+1\right] \cdot \left[\sum_{y=1}^l \left(\sum_{r=y}^{l}\left(\prod_{j=y}^r \,q_j\right)\right)+1 \right] \\[2ex] \sigma(m \cdot n) &=\sigma(m) \cdot \sigma(n) \end{align} [imath]\tag*{$\blacksquare$}[/imath] My question is two-fold. [imath](a)[/imath] Is this a valid proof? [imath](b)[/imath] Is the Sigma-Sigma-Pi notation for [imath]\sigma(m)[/imath], [imath]\sigma(n)[/imath] an appropriate notation, i.e. is there a more concise way to express them?
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2028633
|
Prove that [imath] \mathbb{R}[/imath] and the interval [0,1 ] have the same cardinality.
I'm having troubles with a question. Prove that [imath]\mathbb{R}[/imath] and the interval [0,1] have the same cardinality. Can you help me, please? Thank you in advance!
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660997
|
Proving that cardinality of the reals = cardinality of [imath][0,1][/imath]
Homework problem, intro to topology. Here's what I've done so far. Am I on the right track? And, how would you advise me to proceed from here? I have already established that [imath]\left |[0,1] \right | = \left |[0,1) \right|[/imath]. Now I wish to show that [imath]\left |[0,1) \right| = \left |\mathbb{R} \right |[/imath]. I will try to do this by constructing a bijection [imath]f: [0,1) \rightarrow \mathbb{R}[/imath]. Given [imath]x \in [0,1)[/imath], consider the following countably infinite subset of [imath]\mathbb{R}[/imath]: [imath]x^c = \{x+1, x-1, x+2,x-2,x+3,x-3,\ldots \}[/imath]. Then we have that [imath] \bigcup_{x \in [0,1)} x^c = \mathbb{R}, [/imath] and if [imath]x_1, x_2 \in [0,1)[/imath] with [imath]x_1 \neq x_2[/imath], then [imath]x_1^c \cap x_2^c = \varnothing[/imath].
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175736
|
Evaluate [imath]\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})[/imath]
Evaluate [imath]\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})[/imath]. Can anyone help me with this? Thank You!
|
2862068
|
Proving that [imath]\cot^220^\circ + \cot^240^\circ + \cot^280^\circ = 9[/imath].
Prove that [imath]\cot^220^\circ + \cot^240^\circ + \cot^280^\circ = 9[/imath]. I tried bringing them all to [imath]\cot^220^\circ[/imath] but it didn't work. How do I proceed?
|
2028708
|
Commutative matrices - approximation by commutative simultaneously diagonalizable matrices.
It is well known that the set [imath]\mathcal{D}_n(\mathbb{C})[/imath] of all complex, diagonalizable, [imath]n \times n[/imath] matrices is dense in [imath]\mathcal{M}_n(\mathbb{C})[/imath], the set of all complex [imath]n \times n[/imath] matrices. And that two diagonalizable matrices which commute are simultaneously diagonalizable. The exercise, which I am having difficulty to solve, is the following. Let [imath]A,B \in \mathcal{M}_n(\mathbb{C})[/imath]. Suppose [imath]AB = BA[/imath] and let [imath]\varepsilon >0[/imath]. Show that there exists two simultaneously diagonalizable complex, [imath]n \times n [/imath] matrices, [imath]C[/imath] and [imath]D[/imath] such that [imath]\| A - C \| \leq \varepsilon [/imath] and [imath]\| B - D \| \leq \varepsilon[/imath]. I have managed to find a solution when [imath]A[/imath] (or [imath]B[/imath]) is diagonalizable... But I really cannot find the answer in the general case. Anyone able to help me out ?
|
227842
|
Approximating commuting matrices by commuting diagonalizable matrices
Suppose the matrices [imath]A[/imath] and [imath]B[/imath] commute. Do there exists sequences [imath]A_n[/imath] and [imath]B_n[/imath] of matrices such that [imath]A_n \rightarrow A[/imath], [imath]B_n \rightarrow B[/imath]. Each [imath]A_n[/imath] is diagonalizable and the same for each [imath]B_n[/imath]. For every [imath]n[/imath], [imath]A_n[/imath] commutes with [imath]B_n[/imath]. Moreover, it would be nice if the following property was additionally satisfied: if [imath]A,B[/imath] are real, then [imath]A_n,B_n[/imath] can be chosen to be real as well.
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2029443
|
Are the numbers [imath]0[/imath] and [imath]13[/imath] coprime with each other?
I couldn't find an answer on the internet but my script fails solving this. I have to know this because this is required to prove something else (it has to do with random numbers and their period). I cannot say if they are coprime because I don't know if they are coprime because of the fact that division by [imath]0[/imath] doesn't work. On the other hand I say they are not coprime because we don't get to [imath]\text{gcd}(0,13)=1[/imath]. But this is very confusing for me, please help!
|
27719
|
What is [imath]\gcd(0,a)[/imath], where [imath]a[/imath] is a positive integer?
I have tried [imath]\gcd(0,8)[/imath] in a lot of online gcd (or hcf) calculators, but some say [imath]\gcd(0,8)=0[/imath], some other gives [imath]\gcd(0,8)=8[/imath] and some others give [imath]\gcd(0,8)=1[/imath]. So really which one of these is correct and why there are different conventions?
|
2028260
|
[imath]\lim a^{b_n}=a^b[/imath] if [imath]\lim b_n = b[/imath]
Let [imath]a \in \mathbb{R}[/imath] and [imath]\lim b_n = b[/imath] (for a real sequence [imath]b_n[/imath]). I am trying to show that [imath]\lim a^{b_n}=a^b[/imath] by using the definition of the convergence of a sequence, but I cannot see how to estimate [imath]|a^{b_n} - a^b|[/imath] against [imath]|b_n - b|[/imath] properly.
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1382367
|
how to prove that [imath] \lim x_n^{y_n}=\lim x_n^{\lim y_n}[/imath]?
Let [imath](x_n)[/imath] and [imath](y_n)[/imath] be two sequences of real numbers, such that: [imath]\lim x_n = a > 0[/imath] and [imath]\lim y_n = b \in \mathbb R[/imath] I need to prove that: [imath]\lim \left( x_n ^{y_n} \right) = a^b[/imath] I tried using Cauchy's criterion: [imath]|x_n-a|<\epsilon[/imath] and [imath]|y_n-b|<\epsilon[/imath], and then we need to prove that [imath]\left| x_n ^{y_n} -a^b\right|<\epsilon[/imath], but I didn't find a way of proving it.
|
2027389
|
A vector field orthogonal to it's curl at every point
Is there a nice geometric interpretation of what a vector field [imath]V[/imath] in [imath]\mathbf{R}^3[/imath], that satisfies [imath]\epsilon_{ijk} V_i \partial_j V_k=0[/imath] is? I.e. if I know a vector field at every point in some region in [imath]\mathbf{R}^3[/imath] is orthogonal to it's curl, can I say anything nice about it geometrically? I know this is somewhat of a vague question, but I can't make it any more concrete at the moment. Context: This is a necessary and sufficient condition that the velocity vector field of a steady flow in fluid dynamics has to satisfy, in order for Bernoulli's theorem to apply not only on a streamline, but across the streamlines. In textbooks it is usually not treated however, as they usually just state that if the curl is zero, the theorem applies.
|
1054887
|
geometric interpretation of [imath]\vec v \cdot \operatorname{curl} \vec v = 0 [/imath]
There is a family of surfaces orthogonal to the vector field [imath]\vec v \in \mathbb R^3[/imath] iff [imath]\vec v \cdot \operatorname{curl} \vec v = 0 [/imath]. Now the necessity part is trivial, but the proof of sufficiency I have seen in physics textbooks, e.g. Kestin: Thermodynamics, is kind of mindless integration similar to the one usually offered to prove Poincare's lemma as in Flanders, or just asserted as in Born&Wolf: Optics. Is there an intuitive and geometric interpretation of this condition that would make it obvious why its sufficiency must be true?
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2030273
|
Number Theory Greatest Common Divisor proof
Prove the following: [imath]\gcd(a^2,b^2)=\gcd(a,b)^2[/imath]
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1199945
|
How to prove [imath]\gcd(a^2,b^2) = (\gcd(a,b))^2[/imath]?
How to prove [imath]\gcd(a^2, b^2) = (\gcd(a, b))^2[/imath]? My attempt: Let [imath]\gcd(a, b) = d[/imath]. Then [imath]d|a[/imath] and [imath]d|b[/imath] then [imath]d^2|a^2[/imath] and [imath]d^2|b^2[/imath]. i.e [imath]d^2[/imath] divides [imath]a^2 ~~\&~~ b^2[/imath].
|
2030204
|
Evaluation of [imath]\int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx[/imath]
Evaluation of [imath]\int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx[/imath] Let [imath]I = \int^{1}_{0}\bigg(\frac{1}{1-x}+\frac{1}{\ln x}\bigg)dx = \int^{1}_{0}\frac{(1-x)+\ln x}{(1-x)\ln x}dx[/imath] Now How can i solve after that , Help required, Thanks
|
980593
|
Some integral representations of the Euler–Mascheroni constant
What kind of substitution should I use to obtain the following integrals? [imath]\begin{align} \int_0^1 \ln \ln \left(\frac{1}{x}\right)\,dx &=\int_0^\infty e^{-x} \ln x\,dx\tag1\\ &=\int_0^\infty \left(\frac{1}{xe^x} - \frac{1}{e^x-1} \right)\,dx\tag2\\ &=-\int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx\tag3\\ &=\int_0^\infty \left( e^{-x} - \frac{1}{1+x^k} \right)\,\frac{dx}{x},\qquad k>0\tag4\\ \end{align}[/imath] This is not homework problems and I know that the above integrals equal to negative of the Euler–Mascheroni constant. I got these integrals while reading this Wikipedia page: The Euler–Mascheroni constant. According to Wikipedia, the Euler–Mascheroni constant is defined as the limiting difference between the harmonic series and the natural logarithm: [imath]\gamma=\lim_{N\to\infty} \left(\sum_{k=1}^N \frac{1}{k} - \ln N\right)[/imath] but I don't know why can this definition be associated to the above integrals? I can obtain the equation [imath](1)[/imath] using substitution [imath]t=\ln \left(\frac{1}{x}\right)\rightarrow x=e^{-t} \rightarrow dx=-e^{-t}\,dt[/imath] and I know that [imath]\int_0^\infty e^{-x} \ln x\,dx=\Gamma'(1)=\Gamma(1)\psi(1)=-\gamma[/imath] but I can't obtain the rest. Any idea? Any help would be appreciated. Thanks in advance.
|
2030179
|
Solution of this summation: [imath]\sum_{i=1}^{10} {\frac{i}{i^4+i^2+1}}[/imath]
The summation in question: [imath]\sum_{i=1}^{10} {\frac{i}{i^4+i^2+1}}[/imath] I have been able to factorize [imath]i^4+i^2+1[/imath] as [imath](i^2+i+1)(i^2-i+1)[/imath] but I doubt this will help. What is the solution?
|
2029773
|
Find [imath]S_n [/imath] if [imath]t_n=\frac{n}{1+n^2+n^4}[/imath]
If [imath]S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}[/imath], find the value of [imath]14S[/imath]. The question can ve simplified to: Find [imath]S [/imath] if [imath]t_n=\frac{n}{1+n^2+n^4}[/imath] [imath]t_n=\frac{n}{1+n^2+n^4}[/imath] As [imath]1+n^2+n^4[/imath] forms a GP, [imath]t_n=\frac{n(n^2-1)}{n^6-1}[/imath] But I can't figure out how to solve further. It would be great if someone could help.
|
2030634
|
How can we factorize [imath]x^4 - 2x^2 + 49[/imath] with coefficients in [imath]\mathbb{R}[/imath]?
How can we factorize [imath]{x^4} - 2{x^2} + 49[/imath] with coefficients in [imath]\mathbb{R}[/imath]? A problem would be easier if it was a quadratic equation - we could simply find the roots and get the linear factors. Moreover, the polynomial [imath]x^4 - 2x^2 + 49[/imath] does not have a real root which would be easy to guess. (If we have one root, we could divide by linear factor determined by this root.) WolframAlpha says that this can be factorized as [imath]x^4 - 2x^2 + 49 = (x^2-4x+7)(x^2+4x+7).[/imath] But how can we get to this factorization? Factorization of this polynomial also appears as an example in an answer tp another question: Does the Rational Root Theorem ever guarantee that a polynomial is irreducible?
|
876244
|
How to factorize [imath]x^4+2x^2+4[/imath] to a product of polynomials with real coefficients?
How do you factor [imath]x^4+2x^2+4 [/imath] so it can be written as [imath] (x^2+2x+2)(x^2-2x+2) [/imath]
|
2030771
|
Why should be accept this [imath]{\displaystyle {n \choose k}}=0 [/imath] for [imath]k> n[/imath]?
I'm confused how [imath]{\displaystyle {n \choose k}}=0 [/imath] for [imath]k> n[/imath] however the ratio :[imath]{\displaystyle\frac{n!}{(n-k)!k!}}[/imath] is not defined for [imath]k>n[/imath] ? My question here is: Why should be accept this [imath]{\displaystyle {n \choose k}}=\frac{n!}{(n-k)!k!}=0 , for , k> n[/imath] ? Note: I don't accept this if it were by convention because it's not defined !!!! Thank you for any help
|
267644
|
[imath]n[/imath] choose [imath]k[/imath] where [imath]n[/imath] is less than [imath]k[/imath]
I am working on parameter estimation and one of the estimators involves a summation of [imath]_nC_k[/imath] ([imath]n[/imath] choose [imath]k[/imath]) expressions. For some iterations, I need to compute expressions like [imath]_0C_1[/imath], [imath]_0C_2[/imath], etc. In general how do we compute [imath]_nC_k[/imath] when [imath]n[/imath] is less than [imath]k[/imath]? Do we still use the formula [imath]\frac{n!}{(n-k)!k!}[/imath] and use the gamma function to compute the negative factorial? Thanks!
|
1473743
|
Why not generalize the Intermediate Value Theorem
So I know the IVT says that given a continuous function on a closed interval [a,b] then if [imath]f(a) < \gamma <f(b)[/imath] then there is [imath]c\in(a,b)[/imath] such that [imath]f(c) = \gamma[/imath]. Is there any reason for not replacing the image of the end-points with the image of the maximum and minimum of [imath]f[/imath] on [imath][a,b][/imath]. In other words, why isn't the mean value theorem stated as above but with [imath]f(a)[/imath] replaced by the minimum value of [imath]f[/imath] taken on [imath][a,b][/imath] and similarly [imath]f(b)[/imath] replaced with the maximum value of [imath]f[/imath] taken on [imath][a,b][/imath]?
|
2021838
|
Why does Intermediate Value Theorem only consider end-points?
The IVT says that, given a closed interval of real numbers, [imath][a, b][/imath], and a real-valued function [imath]f[/imath] continuous on [imath][a,b][/imath], for any point [imath]z\in [f(a), f(b)][/imath] (WLOG, [imath]f(a) < f(b)[/imath]), there exist a point [imath]c\in [a,b][/imath] such that [imath]f(c) = z[/imath]. Now, I'm wondering, why is the theorem only applicable to [imath]z\in[f(a), f(b)][/imath] and not to [imath]z\in [\min\limits_{x\in [a,b]}\{f(x)\},\max\limits_{x\in [a,b]}\{f(x)\}][/imath]?
|
2031166
|
How do I find a function that is proportional to the definite integral of itself and a trigonometric function in a limited space?
[imath] f(x) = x - \int_0^{\pi/2} f(x)\sin(x)dx [/imath] How do you solve this equation for the function [imath]f(x)[/imath]? The question had a hint to use integration by parts.
|
2029538
|
If [imath]f(x)=x-\int^{\frac{\pi}{2}}_0f(x)\sin(x)dx[/imath] find the explicit function for [imath]f(x)[/imath] in its simplest form
A function [imath]f(x)[/imath], where [imath]x[/imath] is a real number , is defined implicitly by the following formula: [imath]f(x)=x-\int^{\frac{\pi}{2}}_0f(x)\sin(x)dx[/imath] Find the explicit function for [imath]f(x)[/imath] in its simplest form. This question appeared in the recent New Zealand Qualifications Authority 2016 Scholarship Calculus examination. What I have done Let [imath]f(x)=y[/imath] [imath]f(x)=x-\int^{\frac{\pi}{2}}_0f(x)\sin(x)dx \Rightarrow y=x-\int^{\frac{\pi}{2}}_0y\sin(x)dx[/imath] [imath] y=x-\int^{\frac{\pi}{2}}_0y\sin(x)dx \Leftrightarrow \int^{\frac{\pi}{2}}_0y\sin(x)dx=x-y[/imath] Consider the integral [imath] \int^{\frac{\pi}{2}}_0y\sin(x)dx[/imath] Let [imath]u=y\Rightarrow du=dy[/imath] and [imath]dv= \sin(x) \Rightarrow v=-\cos(x)[/imath] [imath] \int^{\frac{\pi}{2}}_0y\sin(x)dx = \left[y\sin(x) \right]^{\frac{\pi}{2}}_0+\int^{\frac{\pi}{2}}_0\cos(x) dydx[/imath] How can I continue?
|
2031187
|
Can an analytic function on the open unit disk blow up near the boundary?
Can we have a map [imath]f[/imath], holomorphic on the open unit disk, such that [imath]|f(z)|\rightarrow \infty[/imath] as [imath]|z|\rightarrow 1[/imath]? I think not, (at least I can't think of any such map), but I'd like to be able to prove this.
|
1829735
|
There is not an holomorphic function on a bounded domain such that [imath]\lim_{z \to w} f(z)= \infty[/imath] for every [imath]w \in Fr(\Omega)[/imath]
I had to solve the following question in an exam: Prove that there is not a holomorphic function on a bounded domain [imath]\Omega[/imath], [imath]f:\Omega \to \mathbb{C}[/imath] such that [imath]\lim_{z \to w} f(z)= \infty[/imath] for every [imath]w \in Fr(\Omega)[/imath] where Fr denotes the boundary of [imath]\Omega[/imath]. The exam suggested that one should proof that in this situation [imath]f[/imath] could not have zeros in the domain. I did the exercise in this case, but how can I proof that [imath]f[/imath] has no zeros in [imath]\Omega[/imath]?
|
2029354
|
Bijective map [imath][0,1][/imath] to [imath]\mathbb{R}[/imath]
If I wanted a bijective map [imath][0,1][/imath] to [imath]\mathbb{R}[/imath] and I would define [imath]f[/imath] as: \begin{equation} f(x)=\begin{cases} 0&x=0\\{}\\ 1&x=1\;\\{}\\ -cot\pi x,&\text{otherwise}\\ \end{cases} \end{equation} Would it be a legitimate bijection?
|
2029021
|
Is there a bijection from [0,1] to R?
I'm looking for a bijection from the closed interval [0,1] to the real line. I have already thought of [imath]\tan(x-\frac{\pi}{2})[/imath] and [imath]-\cot\pi x[/imath], but these functions aren't defined on 0 and 1. Does anyone know how to find such a function and/or if it even exists? Thanks in advance!
|
2029299
|
[imath]\lim\limits_{x\to\infty}xf(x) = 0 [/imath]?
I've got this task to prove that, given [imath]\forall x \geq a[/imath] : f is monotonically decreasing, [imath] f \geq {0}[/imath] [imath]\int_a^\infty f(x)dx [/imath] converges. Then [imath]\lim\limits_{x\to\infty}xf(x) = 0 [/imath] I proved successfully that [imath]\lim\limits_{x\to\infty}f(x) = 0[/imath] but I don't know if it helps me anyhow.
|
1398810
|
Proving [imath]\lim\limits_{x \to \infty} xf(x)=0[/imath] if [imath]\int_{0}^{\infty}f(x) dx[/imath] converges.
Let [imath]f(x)[/imath] be a monotone non-increasing function such that [imath]\int_{0}^{\infty}f(x) dx[/imath] converges. Prove: [imath]\lim\limits_{x \to \infty} xf(x)=0[/imath]. My question is, why can't I simply contradict any other possibility by using the Integral Limit Comparison Test with [imath]1\over x[/imath]? I am, after all, to show that [imath]f(x)=o(x)[/imath] as [imath]x\to \infty[/imath]. I really don't understand why monotony is crucial here. I could use some help.
|
2029631
|
let [imath]x[/imath] be a eigenvector for [imath]Q[/imath]
Searching for some help with this exam review proof. If we let [imath]x[/imath] be a eigenvector for [imath]Q[/imath], that is, a non zero vector satisfying [imath]Qx=cx[/imath] for some scalar [imath]c[/imath]. How do I show that [imath]c= \pm1.[/imath] EDIT: [imath]Q[/imath] is a matrix with orthonormal columns
|
653133
|
Eigenvalues in orthogonal matrices
Let [imath]A \in M_n(\Bbb R)[/imath]. How can I prove, that 1) if [imath] \forall {b \in \Bbb R^n}, b^{t}Ab>0[/imath], then all eigenvalues [imath]>0[/imath]. 2) if [imath]A[/imath] is orthogonal, then all eigenvalues are equal to [imath]-1[/imath] or [imath]1[/imath]
|
1226127
|
Is a discrete subgroup of a Hausdorff group closed?
Let [imath]G[/imath] be a Hausdorff topological group. Let [imath]H[/imath] be a subgroup of [imath]G[/imath] such that [imath]H[/imath] is a discrete subspace of [imath]G[/imath]. Is [imath]H[/imath] a closed subgroup of [imath]G[/imath]? I thought this is obviously true, but I failed to prove it.
|
29515
|
Why is every discrete subgroup of a Hausdorff group closed?
I have just begun to learn about topological group recently and is still not familiar with combining topology and group theory together. I have read a useful property of discrete group on the wikipedia: every discrete subgroup of a Hausdorff group is closed But I have no idea how to prove it. I find that it cannot be proved only considering the topological structure, since [imath]\left\{\frac{1}{n}: n=1,2,3,...\right\}[/imath] is a discrete subspace of [imath]\Bbb R[/imath], which is not closed. I don't know how to use the group structure here. Can you please help? Thanks.
|
2031677
|
Is [imath](x^3)/x[/imath] really the same as [imath]x^2[/imath]?
Is [imath]f(x)=\dfrac{x^3}{x}[/imath] really the same as [imath]x^2[/imath]? At [imath]x=0[/imath], [imath]f(x)=\dfrac{x^3}x[/imath] is undefined. So then why can't I just say [imath]f(x) = \dfrac{x^3}{x} = x^2[/imath] therefore [imath]f(x)=\dfrac{x^3}{x} [/imath] at [imath]x=0[/imath] is [imath]0[/imath] and not undefined!
|
2027810
|
How can [imath]\frac{x^3-4x^2+4x}{x^2-4}[/imath] be both [imath]0[/imath] and "undefined" when [imath]x = 2[/imath]?
Suppose I have a function defined as [imath]F(x)= \frac{x^3-4x^2+4x}{x^2-4}[/imath] Now I want to find the value of [imath]F(2)[/imath]. I can do it in 2 ways: Put [imath]x=2[/imath] and solve the function. It will give: [imath]F(2)=\frac{0}{0}[/imath] which is not defined. Solve [imath]F(x)[/imath] first and then put [imath]x=2[/imath]. [imath]F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}[/imath] It will give [imath]{F(2)=\frac{0}{4}}[/imath] which is zero. How can zero equal not defined?
|
2031899
|
Is the following ~ transitive?
I was given the following problem in my assignment: Define [imath]a[/imath]~[imath]b[/imath] on the rationals by [imath]a[/imath]~[imath]b[/imath] iff [imath]b=ak^2[/imath] for some rational number [imath]k[/imath]. Is ~ transitive?. Please somebody explain to me how to do this problem. Thanks!
|
2031013
|
Relations of a~b iff b = ak^2
Just need some verification for this question or another way to approach it. Given: a~b is defined on the rationals by a~b iff [imath]b = ak^2[/imath] for [imath]some[/imath] rational number [imath]k[/imath]. Is it reflexive? Symmetric? Transitive?(Show the Equivalence relation) Reflexive: Must show that a~a [imath]a=ak^2[/imath] [imath]1=k^2[/imath] which checks out. Its Reflexive Symmetric: Must show that a~b and b~a a~b: [imath]b=ak^2[/imath], [imath]\frac ba[/imath]=[imath]k^2[/imath] b~a: [imath]a=bk^2[/imath], [imath]\frac ab[/imath]=[imath]k^2[/imath] Not symmetric. Transitive: a~b and b~c than we must show that a~c a~b: [imath]b=ak^2[/imath], b~c: [imath]c=bk^2[/imath] a~c: [imath]c=ak^2[/imath] This is were I am sorta stuck not sure if I am allowed to do the following: since [imath]c=ak^2[/imath] and [imath]c=bk^2[/imath] I can find that a=b?not sure any help would be appreciated. Noticing that if zero is excluded then we would have an equivalence relation. However, when zero is included would symmetric fail and both reflexive and transitive still hold?
|
2032220
|
What is the role of inequality in this problem?
Let [imath]\{e_1,e_2,\ldots,e_n\}[/imath] be orthonormal basis of [imath]V[/imath] and let [imath]v_1,v_2,\ldots,v_n[/imath] are vectors in [imath]V[/imath]. If [imath]\|e_j-v_j\|<1/ \sqrt n[/imath] for [imath]j=1,2,\ldots,n[/imath] then how can we prove that [imath]\{v_1,v_2,\ldots,v_n\}[/imath] is basis for [imath]V[/imath]?
|
1732753
|
Proving a basis for inner product space V when [imath]||e_j-v_j||< \frac{1}{\sqrt{n}}[/imath].
Suppose [imath](e_1,e_2,...,e_n)[/imath] is an orthonormal basis of the inner product space [imath]V[/imath] and [imath]v_1,v_2,...,v_n[/imath] are vectors of [imath]V[/imath] such that [imath]||e_j-v_j||< \frac{1}{\sqrt{n}}[/imath] for each [imath]j \in \left\{1,2,...,n \right \}[/imath]. Prove that [imath](v_1,v_2,...,v_n)[/imath] is a basis of [imath]V[/imath]. I am completely lost and just starting to learn about inner product spaces. Could someone provide a proof with the explanation of how you got there?
|
2032677
|
Evaluation of [imath]\int^{1}_{0}\frac{\ln^2(1+x)}{x}dx[/imath]
Evaluation of [imath]\int^{1}_{0}\frac{\ln^2(1+x)}{x}dx[/imath] [imath]\bf{My\; Try::}[/imath] Let [imath]I = \int^{1}_{0}\frac{\ln^2(1+x)}{x}dx = \int^{1}_{0}\ln^2(1+x)\cdot\frac{1}{x}dx[/imath] Using By parts, We get [imath]I = \left[\ln^2(1+x)\cdot \ln(x)\right]^{1}_{0}-2\int^{1}_{0}\frac{\ln(1+x)\cdot \ln x}{1+x}dx[/imath] So we get [imath]I = -2\int^{1}_{0}\frac{\ln(1+x)\cdot \ln x}{1+x}dx[/imath] Now how can i solve above Integral , Help required, Thanks
|
316745
|
How to evaluate [imath]\int_0^1\frac{\log^2(1+x)}x\mathrm dx[/imath]?
The definite integral [imath]\int_0^1\frac{\log^2(1+x)}x\mathrm dx=\frac{\zeta(3)}4[/imath] arose in my answer to this question. I couldn't find it treated anywhere online. I eventually found two ways to evaluate the integral, and I'm posting them as answers, but they both seem like a complicated detour for a simple result, so I'm posting this question not only to record my answers but also to ask whether there's a more elegant derivation of the result. Note that either using the method described in this blog post or substituting the power series for [imath]\log(1+x)[/imath] and using [imath]\frac1k\frac1{s-k}=\frac1s\left(\frac1k+\frac1{s-k}\right)\;[/imath] yields [imath] \int_0^1\frac{\log^2(1+x)}x\mathrm dx=2\sum_{n=1}^\infty\frac{(-1)^{n+1}H_n}{(n+1)^2}\;. [/imath] However, since the corresponding identity without the alternating sign is used to obtain the sum by evaluating the integral and not vice versa, I'm not sure that this constitutes progress.
|
2018451
|
How can we prove that we've discovered a strange attractor?
So, this webpage has a strange attractor of the form: [imath]x_{n+1} = a x_n + y_n[/imath] [imath]y_{n+1} = b + (x_n)^2[/imath] ...it precedes the graphic of a strange attractor itself (the initial conditions are also given for the strange attractor). My question is, how can we prove that this system defines a strange attractor? Also, is there any way to get more exact values for the system - for example, more accurate numbers? EDIT I asked this question for a general reference on chaos and strange attractors. For this question (the current question), I'm wondering what exactly we need to do to prove that we have a strange attractor. In other words, the last question was a reference request, and this question is the mathematics itself - how do we go about proving that we've found a strange attractor?
|
2017231
|
Where can I learn "everything" about strange attractors?
I have a (possibly very) limited math background, but I would like to find a book, website, or other type of work that will teach almost everything about strange attractors. As a motivating example, I found a recurrence of the following form: [imath]x_{n+1} = a x_n + y_n[/imath] [imath]y_{n+1} = b + (x_n)^2[/imath] ...at this Mathematica website. I'm very interested in finding more exact values (for [imath]a[/imath] and [imath]b[/imath]) for the recurrence, and knowing how to do this. I'd also like to be able to prove that this defines a strange attractor. So anything that would allow me to do this would be great, but I'd be especially interested in finding a "Bible" on chaos and strange attractors, kind of like Artin's Algebra, Herbert Wilf's Generatingfunctionology, or some other work that can be considered to be the guide to the subject.
|
2031390
|
Finding limit of a trigonometric expression
Question: Find the value of the limit of the following expression such that [imath]x[/imath] approaches to zero [imath]\frac{\cos(\sin x)-\cos x}{x^4}.[/imath] My attempt:
|
582275
|
Calculating limit of function
To find limit of [imath]\lim_{x\to 0}\frac {\cos(\sin x) - \cos x}{x^4} [/imath]. I differentiated it using L Hospital's rule. I got [imath]\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}\text{.}[/imath] I divided and multiplied by [imath]\sin x[/imath]. Since [imath]\lim_{x\to 0}\frac{\sin x}{x} = 1[/imath], thus I got [imath]\frac{1-\cos x}{4x^2}[/imath].On applying standard limits, I get answer [imath]\frac18[/imath]. But correct answer is [imath]\frac16[/imath]. Please help.
|
2033095
|
Sum of Two Triangular Numbers?
What numbers can be written as the sum of two triangular numbers? i.e., [imath]\frac{x(x+1)}{2}+\frac{y(y+1)}{2}=z [/imath] [imath]4z+1[/imath] is either the sum of two squares or itself a square. Why?
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1164862
|
Determine if a number is the sum of two triangular numbers.
Is it possible to figure out if a number [imath]z[/imath] is the addition of two triangular number without recursion or finding the values to [imath]x[/imath] and [imath]y[/imath]? [imath]\frac{x(x+1)}{2} + \frac{y(y+1)}{2} = z[/imath] An example of this would be following: [imath]\frac{345(345+1)}{2} + \frac{234(234+1)}{2} = 87180[/imath] Is it possible to determine if [imath]87180[/imath] originated from the addition of two triangular numbers without recursively going back and plugging every possibility for [imath]x[/imath] and [imath]y[/imath]?
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2002917
|
[imath]TT^*=T^2[/imath], show that [imath]T[/imath] is self-adjoint
Let [imath]V[/imath] be an inner product space, finitely generated over [imath]\mathbb{C}[/imath], [imath]T\in \operatorname{End}(V)[/imath] that satisfies [imath]TT^*=T^2[/imath], show that [imath]T[/imath] is self-adjoint. I know that [imath]TT^*[/imath] is positive so has positive square-root, thus the square root is positive definite. But I don't feel like this is a proof. Can anyone prove it without considering square root?
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2011189
|
Show [imath]\alpha[/imath] is selfadjoint.
Question: Let [imath]V[/imath] be an inner product space finitely generated over [imath]\Bbb C[/imath] and let [imath]\alpha[/imath] be an endomorphism of [imath]V[/imath] satisfying [imath]\alpha \alpha^* = \alpha^2[/imath]. Show that [imath]\alpha[/imath] is selfadjoint. **Edited after more thought: I know that in order to be selfadjoint [imath]\alpha^*=\alpha[/imath]. Becuase this inner product space is over [imath]\Bbb C[/imath] does it change the proof? Or, can I show selfadjoint for [imath]\Bbb R[/imath] and then state the proposition that for an inner product space over [imath]\Bbb C[/imath] and an End[imath](V)[/imath], If [imath]\langle{\alpha(v), v}\rangle \in \Bbb R[/imath] for all [imath]v \in V[/imath], then [imath]\alpha[/imath] is selfadjoint.
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2032622
|
Why is [imath]0\cdot x[/imath] defined in fields even though [imath]F\setminus\{0\}[/imath] is the set for multiplication
I am somewhat confused about a definition thing about fields. It says that [imath]\otimes[/imath] is defined on [imath]F \setminus \{0\}[/imath], but in all tables I find, the zero row and column do have entries. From [imath]\otimes[/imath] beig defined for [imath]F \setminus \{0\}[/imath] I would expect the [imath]0[/imath] element to not occur in multiplication tables. Consider these tables: https://en.wikipedia.org/wiki/Finite_field#Field_with_four_elements Why is [imath]0[/imath] defined in the multiplication table or conversely if [imath]0[/imath] is defined for multiplication why does it say in the definition that [imath]\otimes[/imath] is defined on [imath]F \setminus \{0\}[/imath]?
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1597535
|
Multiplication by [imath]0[/imath] in a field
Background A field [imath]\left( F,+,\cdot,0,1 \right)[/imath] is an algebraic structure that is composed of the two Abelian groups [imath]\left( F,+,0 \right)[/imath] and [imath]\left( F \setminus \{ 0 \}, \cdot, 1 \right)[/imath]. The two distributive laws, which must hold, are irrelevant for what follows. I rather want to advert to the fact that multplication is a map [imath] F \setminus \{ 0 \} \times F \setminus \{ 0 \} \longrightarrow F \setminus \{ 0 \}. [/imath] Question Obviously, neither [imath](0,x)[/imath] nor [imath](x,0)[/imath] are in [imath]F \setminus \{ 0 \} \times F \setminus \{ 0 \}[/imath] and, therefore, elements of the domain of the multiplication. But doesn't this imply that expressions such as [imath] 0 \cdot x = x \cdot 0 = 0 [/imath] cannot be written down, because multiplication by [imath]0[/imath] is acually impermissible? Update As the comments and answers clarify, the multiplication of a given field is, in fact, a map [imath] \cdot : F \times F \longrightarrow F, [/imath] and the multiplication of the underlying multiplicative group is the restiction of [imath]\cdot[/imath] to [imath]F \setminus \{ 0 \} \times F \setminus \{ 0 \}[/imath] and would actually deserve another symbol. My false reasoning was the other way round: Given two groups as above, a field is constructed by simply inheriting the addition and the multiplication.
|
1993923
|
Prove that random variable is Poisson Distributed.
Assume in a certain year the number of credit obligors of a bank who are at risk of default is Poisson distributed with intensity [imath]\lambda = 10[/imath]. Each individual obligor may actually default in that year with probability [imath]p = 0.2[/imath]. Assume mutual independence of the obligors. Let [imath]X[/imath] be the number of obligors that default in that year. Show that [imath]X[/imath] is Poisson distributed and determine its intensity. I denote [imath]N[/imath] the number of obligors at risk. And by using the identity [imath]P(X = k) = \sum_E P(X = k\mid N = n)P(N = n)[/imath] to find the pmf I think I am supposed to arrive at the Poisson distribution. [imath]P(N=n)[/imath] is already Poisson. I am having trouble figuring out the [imath]P(X = k\mid N = n)[/imath] and coming up with the probabilities of the pmf which need to be summed up to yield the final Poisson.
|
1665672
|
Marginal probability derived from joint probability distribution of a poisson process and binomial process
I thought I figured out this problem, but my answer is different than the solutions manual. Hoping someone here can stear me in the correct direction: Assume the number of automobile collisions that occur on a given stretch of highway per year is a Poisson random variable [imath]X[/imath] with [imath]\mu = 20[/imath]. The probability is [imath]p=0.05[/imath] that there will be one or more fatalities in each accident; occurrences of fatalities are independent from one collision to the next. If [imath]Y[/imath] is the number of collisions with one or more fatalities on this stretch of road in one year, find the probability law for [imath]Y[/imath]. So as said, X is poisson, so: [imath]p_X(x) = \frac{20^x}{x!}e^{-20}[/imath] The conditional probability of [imath]Y[/imath] given [imath]X=x[/imath] is a Binomial random variable: [imath]p_{Y\mid X}(y\mid x) = \binom{x}{y}p^yq^{x-y}[/imath] where [imath]Y =[/imath] number of fatal traffic accidents and [imath]x =[/imath] given number of accidents (from poisson distribution) Therefore the joint probability for [imath]X[/imath] and [imath]Y[/imath] is: [imath]p_{X,Y}(x,y)= p_{Y\mid X}\cdot p_X(x) = \binom{x}{y}p^yq^{x-y}\frac{20^x}{x!} e^{-20}[/imath] to find the marginal probability of [imath]Y[/imath], I sum over [imath]P_{X,Y}[/imath] for all [imath]x[/imath]: \begin{align} p_Y(y) & = \sum_{x=y}^\infty{\binom{x}{y}p^yq^{x-y}\frac{20^x}{x!}e^{-20}} = \frac{(20p)^ye^{-20}}{y!} \sum_{x=y}^\infty {20^{x-y}\frac{q^{x-y}}{(x-y)!}} \\[10pt] & = \frac{(20p)^ye^{-20}}{y!} \sum_{x-y=0}^\infty {20^{x-y}\frac{q^{x-y}}{(x-y)!}} = \frac{(20p)^ye^{-20}}{y!}e^{20q} \end{align} So: [imath]p_Y(y) = \frac{(20p)^ye^{-20p}}{y!}[/imath] however, the correct answer is apparently [imath]p_Y(y)=\dfrac{e^{-1}}{y!}[/imath]
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2033929
|
If [imath]G[/imath] is abelian, [imath]|G|=p^k[/imath] for [imath]k\in\mathbb{Z}[/imath], and [imath]G[/imath] has a unique subgroup of order [imath]p[/imath], then [imath]G[/imath] is cyclic.
If [imath]G[/imath] is abelian, [imath]|G|=p^k[/imath] for some [imath]k\in\mathbb{Z}[/imath], and [imath]G[/imath] has a unique subgroup of order [imath]p[/imath], then [imath]G[/imath] is cyclic. I'm having difficulty getting anywhere on this problem. The supposition that there is a unique subgroup leads me to believe Sylow's theorem should be used. In particular, if there is a unique Sylow p-subgroup P, then [imath]P\triangleleft G[/imath]. So for all [imath]a\in P, g\in G[/imath], [imath]gag^{-1}\in P[/imath]. Moreover, by Cauchy's theorem, [imath]P[/imath] is cyclic. How can I relate these facts to show that [imath]G[/imath] is cyclic?
|
81607
|
Finite abelian [imath]p[/imath]-group with only one subgroup size [imath]p[/imath] is cyclic
My goal is to prove this: If [imath]G[/imath] is a finite abelian [imath]p[/imath]-group with a unique subgroup of size [imath]p[/imath], then [imath]G[/imath] is cyclic. I tried to prove this by induction on [imath]n[/imath], where [imath]|G| = p^n[/imath] but was not able to get very far with it at all (look at the edit history of this post to see the dead ends). Does anyone have any ideas for a reasonably elementary proof of this theorem?
|
2033746
|
Basic Infinite Series Question: [imath]\sum_{n=0}^\infty\frac{2^n+4^n}{6^n}=?[/imath]
The question: [imath]\sum_{n=0}^\infty\frac{2^n+4^n}{6^n}=\;?[/imath] I can't figure out how to approach this question as I can't find a constant value for [imath]r[/imath].
|
1753313
|
Finding the sum of series [imath]\sum_{n=0}^∞ \frac{2^n + 3^n}{6^n}[/imath]
I am being asked to find the sum of the following convergent series : [imath]\sum_{n=0}^∞ \frac{2^n}{6^n} + \frac{3^n}{6^n}[/imath] Attempting to generalize from partial sums yields nothing of interest: [imath]s_1 = \frac{5}{6}[/imath] [imath]s_2 = \frac{5}{6} + \frac{13}{36} = \frac{43}{36}[/imath] [imath]s_3 = \frac{43}{36} + \frac{35}{216} = \frac{293}{216}[/imath] [imath]s_4 = \frac{293}{216} + \frac{97}{1296} = \frac{1855}{1296} [/imath] I do not see a pattern here... How must I proceed to find the sum of this series?
|
2034314
|
Prove that if [imath]a>0\ then\ 1+a^9\le \frac1a+a^{10}[/imath]
Prove that if [imath]a>0\ then\ 1+a^9\le \frac1a+a^{10}[/imath] I started off by moving every term to the left side of the inequality and multiplied every term by a since a>0 but not sure where to go after that. The hint given is to complete the square. [imath]a+a^{10}-a^{11}\le0[/imath]
|
2032507
|
Algebraic proof that if [imath]a>0[/imath] then [imath]1+a^9 \leq \frac{1}{a}+a^{10} [/imath]
Prove that if [imath]a>0[/imath] then [imath]1+a^9 \leq \frac{1}{a}+a^{10} [/imath]. Using the fact that [imath]a \gt 0[/imath], multiply by [imath]a[/imath] on both sides and get everything to one side we have; [imath]a^{11}-a^{10}-a+1 \geq 0[/imath]. By factoring [imath](a^{10}-1)(a-1) \geq 0 [/imath]. I am not sure how to proceed any further.
|
2034556
|
Prove that n choose k is even for 0<k<n if and only if n is a power of 2
Prove that [imath]\binom{n} k[/imath] is even for all [imath]k[/imath] between [imath]1[/imath] and [imath]n-1[/imath], if and only if [imath]n[/imath] is a power of [imath]2[/imath]. I have tried strong induction on Pascal's triangle but I've got nowhere. I think that there is a generating function approach to this question that I'd like to find.
|
1836342
|
[imath]\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}[/imath] are all even numbers.
Consider the binomial coefficients [imath]\binom{n}{k}=\frac{n!}{k!(n-k)!}\ (k=1,2,\ldots n-1)[/imath]. Determine all positive integers [imath]n[/imath] for which [imath]\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}[/imath] are all even numbers. The answer is [imath]n\in\{2^k|k\in\mathbb{N}\}[/imath], but we have to show two things. First, if [imath]n = 2^k[/imath] then [imath]\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}[/imath] are all even numbers and second if [imath]n \neq 2^k[/imath] then [imath]\binom{n}{1},\binom{n}{2},\ldots ,\binom{n}{n-1}[/imath] are not all even numbers. I think we can use this for the first part but what about the second part?
|
2034580
|
Field homomorphism in both directions
If [imath]E[/imath] and [imath]F[/imath] are two fields and [imath]f : E \to F[/imath] and [imath]g : F \to E[/imath] be field homomorphisms. Is it true that [imath]E[/imath] and [imath]F[/imath] are isomorphic? We know that both [imath]f[/imath] and [imath]g[/imath] are injective which prompts the question. EDIT: As pointed out in comments and other posts, this is not true in general. Are there any conditions on [imath]E[/imath], [imath]F[/imath] which makes this result true? Algebraically closed, characteristic zero, positive etc.?
|
803332
|
Embeddings [imath]A → B → A[/imath], but [imath]A \not\cong B[/imath]?
Are there any nice examples of structures (groups, modules, rings, fields) [imath]A[/imath] and [imath]B[/imath] such that there are embeddings [imath]A → B → A[/imath] while [imath]A \not\cong B[/imath]? I would especially like to see an example for modules [imath]A[/imath], [imath]B[/imath]. Or is it even true that the existence of such embeddings implies [imath]A \cong B[/imath]? Background: I’m correcting exercises and I wanted to give a counterexample to a failing argument. (Well, I’m not certain it fails, but I’m pretty sure it does and it’s not sufficiently justified at least.)
|
2034785
|
For any smooth manifold [imath]M[/imath], does there exist some coordinate chart [imath](U,(x^i))[/imath] such that [imath]M\setminus U[/imath] has measure zero?
Assume that [imath]M[/imath] is connected. Ideally, I would like [imath]U[/imath] to be connected, though that's not strictly necessary. If this were true, it would instantly make integration on compact oriented manifolds much easier, since we can simply pass to Euclidean space [imath]\mathbb{R}^n[/imath] in order to evaluate the integral. Intuitively, it seems like it must be true. Consider the following examples: [imath]\mathbb{S}^n[/imath] – we use the stereographic projection [imath]\mathbb{T}^n \cong (\mathbb{R}/\mathbb{Z})^n[/imath] – we consider [imath]U=(0,1)^n[/imath] [imath]\mathbb{RP}^n[/imath] – we consider [imath]U_0=\{[1:x^1:\cdots:x^n]\, \big\vert\, x^1,\ldots,x^n\in\mathbb{R}^n\}[/imath] If [imath]M[/imath] and [imath]N[/imath] have smooth charts [imath](U,(x^i)[/imath] and [imath](V,(y^j))[/imath] defined almost everywhere, respectively, then [imath]M\times N[/imath] has smooth chart [imath](U\times V, (x^1,\ldots,x^n,y^1,\ldots,y^m))[/imath] defined almost everywhere. However, I have no clue about how to prove a statement like this, and I suspect that any counterexample would be very difficult to find.
|
322027
|
How much of an [imath]n[/imath]-dimensional manifold can we embed into [imath]\mathbb{R}^n[/imath]?
I observed some naive examples. Spheres, for example, when we cut out one point, can be embedded into [imath]\mathbb{R}^n[/imath]. And if we cut out a measure zero set of a projective space, it can be embedded into the Euclidean space of the same dimension. So I wonder if all manifolds can be embedded into a same dimensional Euclidean space when we cut out a measure zero set? Can anyone prove it or disprove it by giving me some counterexamples?
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