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1218788 | Let [imath]G[/imath] be a finite abelian group. Show that there is a Galois extension [imath]K/\Bbb Q[/imath] with [imath]\text{Gal}(K/\Bbb Q) \cong G[/imath].
Let [imath]G[/imath] be a finite abelian group. Show that there is a Galois extension [imath]K/\Bbb Q[/imath] with [imath]\text{Gal}(K/\Bbb Q) \cong G[/imath]. I have seen one proof using For a fixed positive integer [imath]n[/imath], there are infinitely many prime numbers p such that [imath]p ≡ 1[/imath] (mod [imath]n\Bbb Z)[/imath]. But is there any easy way to prove this? Please reply | 131376 | Every finite abelian group is the Galois group of some finite extension of the rationals
I'm trying to prove that every finite abelian group is the Galois group of of some finite extension of the rationals. I think I'm almost there. Given a finite abelian group [imath]G[/imath], I have constructed field extensions whose Galois groups are the cyclic groups occurring in the direct product of [imath]G[/imath]. How do I show that the compositum of these fields has Galois group [imath]G[/imath]. Cheers |
1219014 | Summation of binomial coefficients
Is there a closed formula for: [imath]\sum_{i=1}^{N}{\binom{i+k}{i}}[/imath] ( k is a constant whole number ) | 1218470 | Sum of combinations with varying [imath]n[/imath]
What is the sum of number of ways of choosing [imath]n[/imath] elements from [imath](n+r)[/imath] elements where [imath]r[/imath] is fixed and [imath]n[/imath] varies from [imath]1[/imath] to [imath]m[/imath] ? Can this be reduced to a formula ? [imath] \sum ^m _{n=1} \binom{n + r}n [/imath] |
1218896 | Fractional Laplacian on the torus
Consider the Laplacian on the [imath]n[/imath] dimensional torus [imath]T[/imath], given by [imath]-\Delta : L^2 \rightarrow L^2[/imath]. Let the domain of [imath]-\Delta[/imath] be all [imath]C^\infty[/imath] functions initially. Now consider the Friedrichs extension of [imath]-\Delta[/imath], still called [imath]-\Delta[/imath]. I want to find explicitly what the domain of [imath](-\Delta)^{1/2}[/imath] is. Intuitively it seems to me that it should be all [imath]L^2[/imath] functions [imath]u[/imath] such that [imath]\frac{\partial}{\partial x_k} u \in L^2, k = 1, 2,...,n[/imath], but cannot prove it. Any help would be appreciated. Thanks! Edit: After Michael Renardy's comment: Instead of [imath]-\Delta[/imath], let's say, we consider the following operator [imath]-A = \sum^n_{i = 1} X_i^2[/imath], where [imath]X_i[/imath] are vector fields generating the tangent space at each point, and [imath]A \geq 0[/imath], symmetric and densely defined on [imath]L^2[/imath], so that we can use the Friedrich's extension on [imath]A[/imath]. Can we still say that the domain of [imath]A^{1/2}[/imath] consists exactly of all [imath]L^2[/imath] functions [imath]u[/imath] such that [imath]X_i u \in L^2[/imath] for all [imath]i[/imath]? Thanks for your time. | 1218704 | Fractional Laplacian on the torus
Consider the Laplacian on the [imath]n[/imath] dimensional torus [imath]T[/imath], given by [imath]-\Delta : L^2 \rightarrow L^2[/imath]. Let the domain of [imath]-\Delta[/imath] be all [imath]C^\infty[/imath] functions initially. Now consider the Friedrichs extension of [imath]-\Delta[/imath], still called [imath]-\Delta[/imath]. I want to find explicitly what the domain of [imath](-\Delta)^{1/2}[/imath] is. Intuitively it seems to me that it should be all [imath]L^2[/imath] functions [imath]u[/imath] such that [imath]\frac{\partial}{\partial x_k} u \in L^2, k = 1, 2,...,n[/imath], but cannot prove it. Any help would be appreciated. Thanks! Edit after Neal's comment: Instead of [imath]-\Delta[/imath], let's say, we consider the following operator [imath]-A = \sum^n_{i = 1} X_i^2[/imath], where [imath]X_i[/imath] are vector fields generating the tangent space at each point, and [imath]A \geq 0[/imath], symmetric and densely defined on [imath]L^2[/imath], so that we can use the Friedrich's extension on [imath]A[/imath]. Can we still say that the domain of [imath]A^{1/2}[/imath] consists exactly of all [imath]L^2[/imath] functions [imath]u[/imath] such that [imath]X_i u \in L^2[/imath] for all [imath]i[/imath]? Thanks for your time. |
1215776 | Computing the limit [imath]\lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{n^4+k}\sin \frac{2k \pi}{n}[/imath]
I'm pretty sure this is a Riemann sum, but I can't solve it. I tried to write it as the integral of a function which equals the product of [imath]\sin(2 \pi x)[/imath] and something, but so far it didn't work. [imath]\lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{n^4+k}\sin \frac{2k \pi}{n}[/imath] | 920949 | The limit of a sum (a Riemann sum?)
[imath] \lim_{n \rightarrow \infty} \sum_{k=1}^n \left( \sqrt{n^4+k}\ \sin\frac{2k\pi}{n} \right) = ? [/imath] I tried to transform it into a Riemann sum, to use Taylor-series of the sine function, to estimate, but nothing. Any help would be great. |
1218879 | Test To Determine If Two Circles Intersect In A 2-D Plane
Given two circles described by the 3-tuples (x-coordinates, y-coordinates, radius) I read on StackOverflow that if the following inequality holds, then the two circles intersect: [imath](r_0 - r_1)^2 \leq (x_0 - x_1)^2 + (y_0 - y_1)^2 \leq (r_0 + r_1)^2[/imath] Intersection is defined as two circles having at least one common point. I wrote a computer program to apply this result to a list of circles to see which pairs of circles intersect and I am getting the wrong results. For example, the solution says that the following circles [imath](0,0,1)[/imath] and [imath](0,1,5)[/imath] do have at least one common point and so are defined as intersecting, but the above inequality gives : [imath]16 \leq 1 \leq 36[/imath] which is not true and so the circles are not marked as intersecting. My question is whether the above inequality is a correct test to determine if two circles intersect? If not, then what is the correct geometric way to determine this. | 275514 | Two circles overlap?
If we have two circles in the plane described by [imath](x_1, y_1, r_1)[/imath] and [imath](x_2, y_2, r_2)[/imath] we can determine if they are completely disjoint by simply: [imath](x_1 - x_2)^2 + (y_1 - y_2)^2 < (r_1 + r_2)^2[/imath] Assume this is not the case, we now want to know if one completely overlaps the other. (That is: if the second circles interior is a subset of the first circles interior.) If [imath](x_1, y_1) = (x_2, y_2)[/imath] than we can trivially compare radii, so lets assume their centers are distinct. The way I have imagined is to create a parametric equation of the line that connects the two centers: \begin{align} x_p(t) &= x_1 + t(x_2 - x_1) \\ y_p(t) &= y_1 + t(y_2 - y_1) \end{align} Then we calculate the two line segments (represented as two pairs of t values) where the circles intersect this line. One is a subset of the other if and only if the corresponding circles are subsets of each other. Is there a simpler approach I am overlooking? |
1218325 | Why is F[x] a UFD?
When reading the proof for if [imath]R[/imath] is a UFD, then [imath]R[x][/imath] is a UFD, the author uses a fact that [imath]F[x][/imath] is a UFD. I don't quite understand this. Why [imath]F[x][/imath] is a UFD? ([imath]F[/imath] is the fraction field of [imath]R[/imath]). Can someone explain to me why? Thanks. | 1218261 | Prove that any polynomial in [imath]F[x][/imath] can be written in a unique manner as a product of irreducible polynomials in F[x].
Prove that any polynomial in [imath]F[x][/imath] can be written in a unique manner as a product of irreducible polynomials in [imath]F[x][/imath]. [imath]F[/imath] is a field of fractions of a UFD. Can someone tell me how I can solve this problem. It is one of my lecture note result, but my lecturer did not provide a proof. |
1219901 | Steepest descent method and how to find the weighting factor
I am trying to understand the numerical method of finding a minimum, the steepest descent method. \begin{equation} x_1 = x_0 - \alpha \frac{df}{dx} \end{equation} I understand the idea behind it and how to set up the equation, but I do not get: How to find the "right" weighting factor [imath]\alpha[/imath]. Why the next search has to occur orthogonal to the previous search direction. Thanks in advance for the answers. | 373868 | Optimal step size in gradient descent
Suppose a differentiable, convex function [imath]F(x)[/imath] exists. Then [imath]b = a - \gamma\nabla F(a)[/imath] implies that [imath]F(b) \leq F(a)[/imath] given [imath]\gamma[/imath] is chosen properly. The goal is to find the optimal [imath]\gamma[/imath] at each step. In my book, in order to do this, one should minimize [imath]G(\gamma)=[/imath] [imath]F(x-\gamma\nabla F(x))[/imath] for [imath]\gamma[/imath]. It also says that it should be minimized via a line search. Why can't this function be minimized by simple calculus? I am unsure what it means to perform a line search on this function. I am also unsure why we want to minimize this function to find the optimal step size. |
1219965 | Is it true that every two Hamel Basis for a Vector Space have the same cardinality?
Let [imath]X[/imath] be a vector space over any field [imath]F[/imath], and let [imath]\mathcal{B}[/imath] and [imath]\mathcal{C}[/imath] Hamel basis for [imath]X[/imath]. My question is Is there a bijection [imath]\phi : \mathcal{B}\to \mathcal{C}\ ?[/imath] i.e. do [imath]\mathcal{B}[/imath] and [imath]\mathcal{C}[/imath] have the same cardinality? We know this fact for finite dimensional vector spaces but, does this hold in general? (Please ignore any concept of norm: I know that if [imath]X[/imath] is a Bannach space, then its dimension is at least [imath]2^{\aleph_0}[/imath], however, this doesn't prove my question, even in this case) | 52667 | Proof that two basis of a vector space have the same cardinality in the infinite-dimensional case
I am having a difficulty setting up the proof of the fact that two basis of a vector space have the same cardinality for the infinite-dimentional case. In particular, let [imath]V[/imath] be a vector space over a field [imath]K[/imath] and let [imath]\left\{v_i\right\}_{i \in I}[/imath] be a basis where [imath]I[/imath] is infinite countable. Let [imath]\left\{u_j\right\}_{j \in J}[/imath] be another basis. Then [imath]J[/imath] must be infinite countable as well. Any ideas on how to approach the proof? |
1219806 | Prove that [imath]\{ ax+by\mid x,y\in\mathbb Z\} = \{ n(a,b) \mid n\in\mathbb Z\}[/imath]
Prove the following proposition: Suppose [imath]a,b[/imath] are fixed integers. Then [imath]\{ ax+by\mid x,y\in\mathbb Z\} = \{ n(a,b) \mid n\in\mathbb Z\}[/imath]. | 717771 | Show [imath]a\Bbb Z+b\Bbb Z = \gcd(a,b)\Bbb Z[/imath]
I have the following problem: Let [imath]a, b \in\mathbb{Z}[/imath]. Show that [imath]\,\{ ax + by\ :\ x, y \in \mathbb{Z}\} = \{ n \gcd(a,b)\ :\ n\in \mathbb{Z} \}[/imath] I understand that the Bezout's lemma says that [imath]gcd(a,b) = ax +by[/imath], so Im not really how you would go about proving the above, it doesn't really make sense to me. Any help is appreciated1 |
1219120 | Equivalence of norms in [imath]C^1[0,1][/imath].
Consider the set of all mappings of class [imath]C^1[/imath] defined from [imath][-1,1][/imath] into [imath]\mathbb{R}[/imath]. Consider the two norms [imath]N_1=\sup_{x\in[-1,1]}|f(x)|[/imath] and [imath]N_2=|f(0)|+\sup_{x\in[-1,1]}|f'(x)|[/imath]. Are they equivalent? | 595047 | Deciding whether two metrics are topologically equivalent in the space [imath]C^1([0,1])[/imath]
Consider the space [imath]C^1([0,1])[/imath] and the function [imath]d:C([0,1])\times C([0,1]) \to \mathbb R[/imath] defined as [imath]d(f,g)=|f(0)-g(0)|+sup_{x \in [0,1]}|f'(x)-g'(x)|[/imath]. Decide whether the metrics [imath]d[/imath] and [imath]d_{\infty}[/imath] are topologically equivalent in [imath]C^1([0,1])[/imath] (where [imath]d_{\infty}=sup_{x \in [0,1]}|f(x)-g(x)|)[/imath] My attempt at a solution: If two metrics are topologically equivalent, then they have the same convergent sequences. Honestly, I couldn't do anything. I am trying to define a sequence of functions [imath]\{f_n\}_{n \in \mathbb N}[/imath] such that [imath]f_n \to f[/imath] in, for instance, [imath](C^1([0,1]),d)[/imath] but [imath]f_n \not \to f[/imath] in [imath](C^1([0,1]),d_{\infty})[/imath]. Could it be this two metrics are topologically equivalent? If this is the case, how could I prove it? If not, I would appreciate any hint to find an adequate sequence of functions that works for what I am trying to prove. |
1220090 | Understanding algebraic closure
I am trying to understand what it means to have an extension that is an algebraic closure of the base field. I'm looking for someone who can help conceptually. I understand how [imath]C/R[/imath] looks. The basis of the extension is [imath]\{1,i\}[/imath], so the degree must be [imath]2[/imath]. Moreover, the extension is Galois, and [imath]\mathrm{Gal}(C/R)[/imath] is isomorphic to C2. However, I am unclear about how this works when we are working with finite fields. If we let [imath]K[/imath] be the algebraic closure of [imath]F_p[/imath], then I know [imath]K[/imath] as equivalent to [imath]F_p[/imath] adjoin every single root of every single polynomial in [imath]F_p[X][/imath]. I know, given [imath]\alpha \in K[/imath] implies [imath]\alpha[/imath] is the root of some polynomial, [imath]p(x)[/imath] of deg [imath]n[/imath] in [imath]F_p[X] \Rightarrow [F_p(\alpha):F_p]=n[/imath], and [imath][K/F_p] = |N|[/imath], as was kindly pointed out below. This extension is Galois (finite fields are always separable, and [imath]K[/imath] is clearly the splitting field of each xˆ(p)ˆn - 1, for every [imath]n \in N[/imath]). So [imath]\mathrm{Gal} (K/F_p) = |N|[/imath]. I want to clarify: what does [imath]\mathrm{Gal} (K/F_p)[/imath] actually look like? Am I correct to write [imath]\mathrm{Gal} (K/F_p) = \{\sigma: a \mapsto aˆpˆn | n \in N\}[/imath] ? | 27119 | The algebraic closure of a finite field and its Galois group
[imath]F[/imath] is an extension field of a field [imath]K[/imath]. Let [imath]F[/imath] be an algebraic closure of [imath]\mathbb{Z}_p [/imath] ([imath]p[/imath] prime). Show that [imath](i)[/imath] [imath]F[/imath] is algebraic Galois over [imath]\mathbb{Z}_p[/imath] [imath](ii)[/imath] The map [imath]\alpha:F\rightarrow F[/imath] given by [imath]u\mapsto u^p[/imath] is a nonidentiy [imath]\mathbb{Z}_p[/imath]-automorphism of [imath]F[/imath]. [imath](iii)[/imath] The subgroup [imath]H=\langle \alpha \rangle[/imath] is a proper subgroup of Aut[imath](F/\mathbb{Z}_p)[/imath] where the fixed field is [imath]\mathbb{Z}_p[/imath], which is also the fixed field of Aut[imath](F/\mathbb{Z}_p)[/imath] by [imath](i).[/imath] So, here is my attempt for (i). Let [imath]S \subset \mathbb{Z}_p[x])[/imath] of monic polynomials of the form [imath]x^{p^n}-x[/imath]. Then for all [imath]f\in S[/imath], gcd[imath](f,f^{\prime})=1[/imath](i.e, the polynomials are separble) and [imath]F=\mathbb{Z}_p(a\in F:f(a)=0)[/imath]. So [imath](F/\mathbb{Z}_p)[/imath] is Galois. I'd be glad if I could get assistance for (ii) and (iii) as well. Thanks. ADDED: Attempt at (iii). but I know that since [imath]F/\mathbb{Z}_p[/imath] is a finite Galois extension, its fixed field is [imath]\mathbb{Z}_p[/imath] hints... The field [imath]\mathbb{Z}_p[/imath] must be contained in [imath]F[/imath]. For [imath]a\in \mathbb{Z}_p[/imath], [imath]\alpha(a)=a^p=a[/imath]. Thus the polynomial [imath]x^p-x[/imath] has [imath]p[/imath] zeros in [imath]F[/imath], namely, the elements of [imath]\mathbb{Z}_p[/imath]. But the elements fixed under [imath]\alpha[/imath] are precisely the zeros in [imath]F[/imath] of [imath]x^p-x[/imath]. Hence the fixed field of [imath]\alpha[/imath] is [imath]\mathbb{Z}_p[/imath] which is also the fixed field of Aut[imath](F/\mathbb{Z}_p)[/imath]. Left to show that [imath]H[/imath] is a proper subgroup... If it helps I know that the order of [imath]\langle \alpha \rangle [/imath] is [imath]n[/imath]... |
1220463 | Let [imath]n[/imath] be a positive integer. Show that the smallest integer greater than [imath](\sqrt{3} + 1)^{2n}[/imath] is divisible by [imath]2^{n+1}.[/imath]
Let [imath]n[/imath] be a positive integer. Show that the smallest integer greater than [imath](\sqrt{3} + 1)^{2n}[/imath] is divisible by [imath]2^{n+1}.[/imath] We know that [imath](\sqrt 3-1)^{2n}<1[/imath] | 1113099 | How to show [imath]\lceil ( \sqrt3 +1)^{2n}\rceil[/imath] where [imath]n \in \mathbb{ N}[/imath] is divisible by [imath]2^{n+1}[/imath]
Show that [imath] \left\lceil( \sqrt3 +1)^{2n}\right\rceil[/imath] where [imath]n \in \mathbb{N}[/imath] is divisible by [imath]2^{n+1}[/imath]. I wrote the binomial expansion of [imath] ( \sqrt3 +1)^{2n}[/imath] and [imath]( \sqrt3 -1)^{2n}[/imath] and then added them to confirm that the next integer is even. Afterwards I applied [imath]AM \ge GM[/imath] on the two terms to get [imath] ( \sqrt3 +1)^{2n} + ( \sqrt3 -1)^{2n} \ge (2^{n+1})[/imath]. Now I'm unable to figure out the next step. Any help would be appreciated. :) |
1220459 | Proving Trig. Identities
[imath]\cot x=\sin x \sin(\pi/2 -x) + \cos^2x \cot x[/imath] I'm having difficulty with figuring out how to prove trigonometric identities. I know that in order to do these you need to use the trig ratios reciprocal, quotient identities and compound angle formulas. If someone could guide me through these kind of questions. That would be really appreciated! | 1220323 | Prove: [imath]\cot x=\sin x\,\sin\left(\frac{\pi}{2}-x\right)+\cos^2x\,\cot x[/imath]
Prove: [imath]\cot x=\sin x\,\sin\left(\frac{\pi}{2}-x\right)+\cos^2x\,\cot x[/imath] Hi there! So this problem asks to prove this trigonometric identity. I am not sure how to approach these problems other than needing to know the quotient,p ythagorean, and reciprocal identities. From here I can see that [imath]\cot x[/imath] can be changed to [imath]1/\tan x[/imath], but is it really necessary? If someone could help with this, it'd be very appreciated! |
1221039 | Which is more preferable to write [imath]\log(x)[/imath] or [imath]\ln(x)[/imath]
Which one is more preferable to write when you are writing an exam. Is it [imath]\log(x)[/imath] which denotes the natural logarithm or is it [imath]\ln(x).[/imath] | 1694 | How did the notation "ln" for "log base e" become so pervasive?
Wikipedia sez: The natural logarithm of [imath]x[/imath] is often written "[imath]\ln(x)[/imath]", instead of [imath]\log_e(x)[/imath] especially in disciplines where it isn't written "[imath]\log(x)[/imath]". However, some mathematicians disapprove of this notation. In his 1985 autobiography, Paul Halmos criticized what he considered the "childish [imath]\ln[/imath] notation," which he said no mathematician had ever used. In fact, the notation was invented by a mathematician, Irving Stringham, professor of mathematics at University of California, Berkeley, in 1893. Apparently the notation "[imath]\ln[/imath]" first appears in Stringham's book Uniplanar algebra: being part I of a propædeutic to the higher mathematical analysis. But this doesn't explain why "[imath]\ln[/imath]" has become so pervasive. I'm pretty sure that most high schools in the US at least still use the notation "[imath]\ln[/imath]" today, since all of the calculus students I come into contact with at Berkeley seem to universally use "[imath]\ln[/imath]". How did this happen? |
1220376 | For two path-connected homeomorphic spaces, is there a homeomorphism with a prescribed value at a point?
Let [imath]X[/imath], [imath]Y[/imath] be homeomorphic path-connected topological spaces. Is is true that for any pair [imath]x\in X[/imath], [imath]y\in Y[/imath] a homeomorphism can be chosen such that [imath]h(x)=y[/imath]? | 25326 | Homogeneous topological spaces
Let [imath]X[/imath] be a topological space. Call [imath]x,y\in X[/imath] swappable if there is a homeomorphism [imath]\phi\colon X\to X[/imath] with [imath]\phi(x)=y[/imath]. This defines an equivalence relation on [imath]X[/imath]. One might call [imath]X[/imath] homogeneous if all pairs of points in [imath]X[/imath] are swappable. Then, for instance, topological groups are homogeneous, as well as discrete spaces. Also any open ball in [imath]\mathbb R^n[/imath] is homogeneous. On the other hand, I think, the closed ball in any dimension is not homogeneous. I assume that these notions have already been defined elsewhere. Could you please point me to that? Are there any interesting properties that follow for [imath]X[/imath] from homogeneity? I think for these spaces the group of homeomorphisms of [imath]X[/imath] will contain a lot of information about [imath]X[/imath]. |
1221092 | Prove there does not exist any epimorphism of [imath](\mathbb{Q}, +)[/imath] onto [imath](\mathbb{Z}, +)[/imath].
Prove there does not exist any epimorphism of [imath](\mathbb{Q}, +)[/imath] onto [imath](\mathbb{Z}, +)[/imath]. How do I proceed on this? | 963870 | Show that there is no epimorphism from ([imath]\mathbb{Q}[/imath], +) to ([imath]\mathbb{Z}[/imath], +)
I can't figure out why such an epimorphism cannot exist. I think the fact that a bijection exists between the [imath]\mathbb{Q}[/imath] and [imath]\mathbb{Z}[/imath] throws me off a little. What is it about having a homomorphic surjection that causes problems? |
1219326 | Derivative: [imath]e^x[/imath].
How do you differentiate [imath]e^x[/imath]? I looked on many sites, including similar questions here but most answers seemed circular. The only known definition of [imath]e[/imath] to be used in this proof is [imath] e=\lim_{n \to\infty} \left(1+\frac{1}n \right)^n [/imath] What I did is: [imath] \begin{align*} (e^x)' &=\lim_{h\to0}\frac{e^{x+h}-e^x}h \\ &= e^x\lim_{h\to0}\frac{e^{h}-1}h \end{align*} [/imath] But I don't know how to go on, I know [imath]\lim_{h\to0}\frac{e^{h}-1}h=1[/imath] but I don't know how to prove it, I can't use the [imath]e^x[/imath] taylor expansion as that would imply diferentiating [imath]e^x[/imath]. Edit: I also can't use the derivative of [imath]\ln(x)[/imath]. | 359023 | Using the Limit definition to find the derivative of [imath]e^x[/imath]
I was wondering how we could use the limit definition [imath] \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}[/imath] to find the derivative of [imath]e^x[/imath], I get to a point where I do not know how to simplify the indeterminate [imath]\frac{0}{0}[/imath]. Below is what I have already done [imath]\begin{align} &\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} \\ &\lim_{h \rightarrow 0} \frac{e^x (e^h-1)}{h} \\ &e^x \cdot \lim_{h \rightarrow 0} \frac{e^h-1}{h} \end{align}[/imath] Where can I go from here? Because, the [imath]\lim[/imath] portion reduces to indeterminate when [imath]0[/imath] is subbed into [imath]h[/imath]. |
1220960 | What is the sum of reciprocals of Natural Numbers?
I want to calculate the sum of first [imath]n[/imath] natural numbers. I used the following C program to compute the first '[imath]n[/imath]' digits : #include <stdio.h> int main() { int n; double m; double sum=0; scanf("%d", &n); while (n>0) { m=1.0/n; sum+=m; n--; } printf("%lf",sum); return(0); } What I get is that the sum is very slowly diverging. For [imath]10^{1}[/imath], I get the value : [imath]2.92...[/imath] For [imath]10^{14}[/imath], I get the value : [imath]18.807..[/imath] For [imath]10^{16}[/imath], I get the value : [imath]21.92..[/imath] So it appears to diverge. How can we prove it? Actually sum of reciprocals is divergent, even after being a subseries of this series, so it naturally should be divergent. How can we show this? | 1130923 | Proof that the harmonic series diverges (without improper integrals)
Show that [imath]\sum\limits_{k=1}^n \frac{1}{k} \geq \log(n)[/imath] Use this to deduce that the series [imath]\sum\limits_{k=1}^\infty \frac{1}{k}[/imath] diverges. Hint: Use the estimate [imath]\frac{1}{k} \geq \int_k^{k+1} \frac{1}{x}\,\mathrm{d}x[/imath] Note that all instances of "[imath]\log[/imath]" mean the natural logarithm, the inverse of the exponential. After doing the integration, I've concluded that [imath]\frac{1}{k} \geq \log(1+\frac{1}{k})>-\log(k)[/imath] I suspect I'm supposed to use the hint to arrive at the first inequality, and then since [imath]\log(n)[/imath] can become arbitrarily large as [imath]n[/imath] increases, so must the harmonic series (essentially the comparison test). I'm having trouble making this jump. Hints are appreciated, but please no solutions. |
644930 | If [imath]p,q[/imath] are prime, solve [imath]p^3-q^5=(p+q)^2[/imath].
If [imath]p,q[/imath] are prime, solve [imath]p^3-q^5=(p+q)^2[/imath] I can't think of a nice idea for the solution. Since there's a solution [imath](7;3)[/imath], consisting of two distinct numbers, I really doubt modular arithmetic would help here in any way. I wonder if there's an interesting and simple approach. Thanks. | 474768 | Diophantine equation involving prime numbers : [imath]p^3 - q^5 = (p+q)^2[/imath]
Find all pairs of prime nummbers [imath]p,q[/imath] such that [imath]p^3 - q^5 = (p+q)^2[/imath]. It's obvious that [imath]p>q[/imath] and [imath]q=2[/imath] doesn't work, then both [imath]p,q[/imath] are odd. Assuming [imath]p = q + 2k[/imath] we conclude, by the equation, that [imath]k|q^3 - q - 4[/imath] because [imath]\gcd(k,q)=1[/imath] (else [imath]p[/imath] is not prime) and [imath]k=1[/imath] has no solution. I also tried to use some modules, but I couldn't. |
1221618 | [imath]\sigma[/imath]-algebra with cardinality [imath]\aleph_0[/imath]
Can a [imath]\sigma[/imath]-algebra in a set [imath]X[/imath] have cardinality [imath]\aleph_0[/imath], the cardinality of the naturals? I do not have a clue on how to start with this? Can someone please give me a hint? | 986248 | Is there an infinite countable [imath]\sigma[/imath]-algebra on an uncountable set
Let [imath]\Omega[/imath] be a set. If [imath]\Omega[/imath] is finite, then any [imath]\sigma[/imath]-algebra on [imath]\Omega[/imath] is finite. If [imath]\Omega[/imath] is infinite and countable, a [imath]\sigma[/imath]-algebra on [imath]\Omega[/imath] cannot be infinite and countable. What about if [imath]\Omega[/imath] is not countable ? Is it possible to find an uncountable [imath]\Omega[/imath] with a [imath]\sigma[/imath]-algebra that is infinite and countable ? |
1221490 | Floor and Ceiling Proofs
I wish to prove that if [imath]x[/imath] is a real number then [imath]\lfloor−x\rfloor=−\lceil x\rceil[/imath] and [imath]\lceil−x\rceil=−\lfloor x\rfloor[/imath]. Would I be able to prove it by contradiction? [imath]\lfloor−x\rfloor \neq −\lceil x\rceil[/imath] and [imath]\lceil−x\rceil \neq −\lfloor x\rfloor[/imath], then [imath]x[/imath] is not a real number. Or would I just be able to simply plug in numbers and solve it that way? Any help would be great. | 1216615 | Floor and Ceiling function
prove that x is a real number then [imath]\lfloor −x \rfloor = −\lceil x\rceil[/imath] and [imath]\lceil −x \rceil = −\lfloor x \rfloor[/imath]. If I were to put a real number say like [imath]1[/imath], wouldn't it be right? But if I put a value like [imath]1.5[/imath] it would be incorrect? I don't think this is the way I should be proving it though so could anyone help me out. |
437350 | The sum of square roots of non-perfect squares is never integer
My question looks quite obvious, but I'm looking for a strict proof for this: Why can't the sum of two square roots of non-perfect squares be an integer? For example: [imath]\sqrt8+\sqrt{15}[/imath] isn't an integer. Well, I know this looks obvious, but I can't prove it... | 2028856 | Sum of radicals is rational only for perfect squares
How can we prove that if [imath]a_1\sqrt{n_1}+a_2\sqrt{n_2}+\dots+a_k\sqrt{n_k}\in\mathbb{Q}[/imath] with [imath]a_1,a_2\dots, a_k\in\mathbb{Q}^{*}_{+}[/imath], then the natural numbers [imath]n_1,n_2,\dots, n_k[/imath] are perfect squares? I have proved it for [imath]k=1,2,3[/imath] but for [imath]k\geq 4[/imath] I don't find anything. I couldn't show even that [imath]\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7}\notin\mathbb{Q}[/imath]. |
1222200 | Entropy for three random variables
I'm just working through some information theory and entropy, and I've come into a bit of a problem. In many texts, it's easy to find the "chain rule" for entropy in two variables, and the "conditional chain rule" for three variables, respectively; [imath]H(Y|X) = H(X,Y) - H(X)[/imath] [imath]H(X,Y|Z) = H(Y|Z) + H(X|Y,Z) = H(X|Z) + H(Y|X,Z)[/imath] However, I'm trying to determine the entropy of three random variables: [imath]H(X,Y,Z)[/imath]. I haven't done a lot of probability/statistics before, and googling hasn't really turned up anything too fruitful. Can anyone help me derive this result?? | 168316 | Calculating conditional entropy given two random variables
I have been reading a bit about conditional entropy, joint entropy, etc but I found this: [imath]H(X|Y,Z)[/imath] which seems to imply the entropy associated to [imath]X[/imath] given [imath]Y[/imath] and [imath]Z[/imath] (although I'm not sure how to describe it). Is it the amount of uncertainty of [imath]X[/imath] given that I know [imath]Y[/imath] and [imath]Z[/imath]? Anyway, I'd like to know how to calculate it. I thought this expression means the following: [imath]H(X|Y,Z) = -\sum p(x,y,z)log_{2}p(x|y,z)[/imath] and assuming that [imath]p(x|y,z)[/imath] means [imath]\displaystyle \frac{p(x,y,z)}{p(y)p(z)}[/imath], then \begin{align} p(x|y,z)&=\displaystyle \frac{p(x,y,z)}{p(x,y)p(z)}\frac{p(x,y)}{p(y)}\\&=\displaystyle \frac{p(x,y,z)}{p(x,y)p(z)}p(x|y) \\&=\displaystyle \frac{p(x,y,z)}{p(x,y)p(x,z)}\frac{p(x,z)}{p(z)}p(x|y)\\&=\displaystyle \frac{p(x,y,z)}{p(x,y)p(x,z)}p(x|z)p(x|y) \end{align} but that doesn't really help. Basically I wanted to get a nice identity such as [imath]H(X|Y)=H(X,Y)-H(Y)[/imath] for the case of two random variables. Any help? Thanks |
1220205 | Finding density function, plus showing [imath]X \sim F[/imath] where F is cdf of X, [imath]X = F^{-1}(U)[/imath], [imath]U\sim unif(0,1)[/imath]
Suppose [imath]X[/imath] has a continuous, strictly increasing cdf [imath]F[/imath]. Let [imath]Y = F(X)[/imath]. What is the density of [imath]Y[/imath]? Then let [imath]U \sim unif(0,1)[/imath] and let [imath]X = F^{-1}(U)[/imath]. Show that [imath]X\sim F[/imath]. The first part seems like the density of [imath]Y[/imath] is just the pdf [imath]f(X)[/imath]. The wording of the problem is confusing me though. | 868400 | Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X
Let [imath]X[/imath] be a random variable with a continuous and strictly increasing c.d.f. function [imath]F[/imath] (so that the quantile function [imath]F^{−1}[/imath] is well-defined). Define a new random variable [imath]Y[/imath] by [imath]Y = F(X)[/imath]. Show that [imath]Y[/imath] has a uniform distribution on the interval [imath][0, 1][/imath]. My initial thought is that [imath]Y[/imath] is distributed on the interval [imath][0,1][/imath] because this is the range of [imath]F[/imath]. But how do you show that it is uniform? |
1222668 | Problem about finite group whose proper subgroups are abelian.
Let [imath]G[/imath] be a finite group with the property that all of its proper subgroups are abelian. Let [imath]N[/imath] be a proper normal subgroup of [imath]G[/imath]. Show that either [imath]N[/imath] is contained in the center of [imath]G[/imath] or else [imath]G[/imath] has a normal subgroup [imath]H[/imath] which is a maximal subgroup of [imath]G[/imath]. Here is my attempt at trying to rationalize the problem. Now, I believe that since all proper subgroups are abelian, then [imath]G[/imath] must be normal. Also, if [imath]N[/imath] is contained in the center of [imath]G[/imath], we know that the [imath]p-[/imath]groups of [imath]G[/imath] contain more than one element. But if [imath]N[/imath] has a normal subgroup [imath]H[/imath] that is maximal, then it must be proper, so [imath]H[/imath] cannot be the identity or [imath]G[/imath] itself. Is this on the right track? If so, how can I prove this? Or if not correct, what should I be thinking instead? | 625623 | A finite group with the property that all of its proper subgroups are abelian
Let [imath]G[/imath] be a finite group with the property that all of its proper subgroups are abelian. Let [imath]N[/imath] be a normal subgroup of [imath]G[/imath]. Prove that either [imath]N[/imath] is contained in the center of [imath]G[/imath] or else [imath]G[/imath] has a normal abelian subgroup of prime index. I think [imath]G[/imath] is solvable. http://crazyproject.wordpress.com/2010/06/08/every-finite-group-whose-every-proper-subgroup-is-abelian-is-solvable/. I hope that idea maybe usefull. Help me some hints. Thanks a lot. P/s: This is a question comes from a qualifying exam in Algebra ( Wisconsin August [imath]1979[/imath] ) |
1222700 | Showing [imath]s^2=\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{x})^2=\frac{1}{n-1}\left [\sum_{i=1}^n x_i^2-\frac{1}{n}\left ( \sum_1^nx_i\right )^2 \right ][/imath]
I've got as far as this [imath]s^2=\frac{1}{n-1}\sum_{i=1}^n(x_i-\bar{x})^2=\frac{1}{n-1}\sum_{i=1}^n\left ( x_i^2-2x_i\bar{x}+\bar{x}^2\right )[/imath] [imath]=\frac{1}{n-1}\left [\sum_{i=1}^nx_i^2+\bar{x}^2-2\bar{x}\sum_{i=1}^n x_i \right ][/imath] [imath]\frac{1}{n-1}\left [\sum_{i=1}^nx_i^2 +\left ( \bar{x}-\sum_{i=1}^n x_i\right )^2 -\left (\sum_{i=1}^n x_i \right )^2\right ][/imath] And now I'm stuck. I can use [imath]\sum_{i=1}^n\frac{x_i}{n}=\bar{x}[/imath] However can't get the result. If someone could point me in the right direction I'd be grateful. | 152707 | Sample variance derivation
I have quite a simple question but I can't for the life of me figure it out. For a set of iid samples [imath]\,\,X_1, X_2, \ldots, X_n\,\,[/imath] from distribution with mean [imath]\,\mu[/imath]. If you are given the sample variance as [imath] S^2 = \frac{1}{n-1}\sum\limits_{i=1}^n \left(X_i - \bar{X}\right)^2 [/imath] How can you write the following? [imath] S^2 = \frac{1}{n-1}\left[\sum\limits_{i=1}^n \left(X_i - \mu\right)^2 - n\left(\mu - \bar{X}\right)^2\right] [/imath] All texts that cover this just skip the details but I can't work it out myself. I get stuck after expanding like so [imath] S^2 = \frac{1}{n-1}\sum\limits_{i=1}^n \left(X_i^2 -2X_i\bar{X} + \bar{X}^2\right) [/imath] What am I missing? Edit: A similarly equivalent expression is often given that I also can't derive but which may be more obvious is [imath] S^2 = \frac{1}{n-1}\left[\sum\limits_{i=1}^n X_i^2 - n\bar{X}^2\right] [/imath] |
1222861 | Definite integral with trigonometric functions
I have problem finding, how to solve this integral [imath] \int _0 ^{\frac{\pi}{4}} \frac{3 \sin x + 2 \cos x}{2 \sin x + 3 \cos x}dx [/imath] This I can rewrite as [imath] \int _0 ^{\frac{\pi}{4}} \frac{12 \sin ^2 x - 5 \cos x \sin x - 6}{4-13 \cos ^2 x}dx = [/imath] [imath] = \int _0 ^{\frac{\pi}{4}} \frac{12 \sin ^2 x}{4-13 \cos ^2 x}dx - \int _0 ^{\frac{\pi}{4}} \frac{5 \cos x \sin x}{4-13 \cos ^2 x}dx - \int _0 ^{\frac{\pi}{4}} \frac{6}{4-13 \cos ^2 x}dx [/imath] But from this point, I can't get any closer to the answer. Can you, please, help me? | 1219016 | Indefinite Integral with "sin" and "cos": [imath]\int\frac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx [/imath]
Indefinite Integral with sin/cos I can't find a good way to integrate: [imath]\int\dfrac{3\sin(x) + 2\cos(x)}{2\sin(x) + 3\cos(x)} \; dx [/imath] |
1222393 | Find all the positive natural solutions of [imath]x^2+y^2=3z^2[/imath].
Find all the positive natural solutions of [imath]x^2+y^2=3z^2[/imath]. I guess it has something to do with Pythagorean triples, but I don't know how to relate it properly. Suggestions, hints, or any sort of assisting is welcome. | 1217070 | The diophantine equation [imath]x^2+y^2=3z^2[/imath]
I tried to solve this question but without success: Find all the integer solutions of the equation: [imath]x^2+y^2=3z^2[/imath] I know that if the sum of two squares is divided by [imath]3[/imath] then the two numbers are divided by [imath]3[/imath], hence if [imath](x,y,z)[/imath] is a solution then [imath]x=3a,y=3b[/imath]. I have [imath]3a^2+3b^2=z^2[/imath] and that implies [imath] \left(\frac{a}{z}\right)^2+\left(\frac{b}{z}\right)^2=\frac{1}{3} [/imath] so I need to find the rational solutions of the equation [imath]u^2+v^2=\frac{1}{3}[/imath] and I think that there are no solutions for that because [imath]\frac{1}{3}[/imath] doesn't have a rational root, but I dont know how to explain it. Thanks |
1221506 | The function [imath]f:G \rightarrow G[/imath] defined by [imath]f(x)=x^2[/imath] is a homomorphism iff G is abelian.
Let G be a group. The function [imath]f:G \rightarrow G[/imath] defined by [imath]f(x)=x^2[/imath] is a homomorphism iff G is abelian. I am having trouble with the 2nd part of the proof. Proof: Assume the function [imath]f:G \rightarrow G[/imath] is defined by [imath]f(x)=x^2[/imath] then: Assume [imath]f[/imath] is a homomorphism then [imath]f[/imath] satisfies the property: [imath] f(xy)=f(x)f(y) [/imath] where [imath]x,y \in G[/imath] Notice how: [imath]f(xy)=(xy)^2=xyxy[/imath] [imath]f(x)f(y)=x^2y^2=xxyy [/imath] Since [imath]f(xy)=f(x)f(y)[/imath] then by substitution: [imath]xyxy=xxyy[/imath] Multiplying the right by [imath]y^{-1}[/imath] and the left by [imath]x^{-1}[/imath] then: [imath]yx=xy[/imath] which means G is abelian Assume G is abelian, we want to show [imath]f[/imath] is a homomorphism For the 2nd part of the proof, would I just work backwards from the first part of the proof? Here is an outline: then for [imath]x,y \in G[/imath] [imath]yx=xy[/imath] by definition of abelian groups. Multiple the right by y and the left by x, we have: [imath]xyxy=xxyy[/imath] which is equal to: [imath](xy)^2=(x)^2(y)^2[/imath] since we are given [imath]f(x)=x^2[/imath] then translating this to the above statement: [imath]f(xy)=f(x)f(y)[/imath] which satisfies the property of homomorphism. Does that look right? | 1065396 | Prove that [imath]G[/imath] is abelian iff [imath]\varphi(g) = g^2[/imath] is a homomorphism
I'm working on the following problem: Let [imath]G[/imath] be a group. Prove that [imath]G[/imath] is abelian if and only if [imath]\varphi(g) = g^2[/imath] is a homomorphism. My solution: First assume that [imath]G[/imath] is an abelian group and let [imath]g, h \in G[/imath]. Observe that [imath]\varphi(gh) = (gh)^2 = (gh)(gh) = g^2h^2 = \varphi(g)\varphi(h)[/imath]. Thus, [imath]\varphi[/imath] is a homomorphism. I'm having trouble completing the proof in the reverse direction. Assume that [imath]\varphi[/imath] is a homomorphism. We then know that [imath]\varphi(gh) = \varphi(g)\varphi(h)[/imath] and [imath]\varphi(hg) = \varphi(h)\varphi(g)[/imath]. However, I don't see a way to use this to show that [imath]gh = hg[/imath]. Could anyone lend a helping hand? |
1220967 | How to simplify [imath]\sum_{i=1}^{k}\binom{n + i - 1}{i}[/imath]?
How to simplify [imath]\sum_{i=1}^{k}\binom{n + i - 1}{i}[/imath]? I tried reducing the sum to [imath]\binom{n}{1}, \binom{n}{2}, \binom{n}{3}[/imath] and so on but couldn't get a pattern. | 1223627 | Combinatorics Summation Formula
Can anybody please tell me the summation formula for: [imath]\sum_{i=1}^NC(k+i-1, i) = ?[/imath] Where ' k ' is some constant. Here C(x, y) stands for Combinatorics formula e.g. C(12, 6) = 924. |
1222423 | Solve this combination with a summation. Edited and maybe Solved
I have been quite stuck on this problem and with the help of others, I may have solved it. This is what I have to prove. [imath]\binom{N}{n+1}=\sum^{N-1}_{k=n} \binom{k}{n}.[/imath] So far I have this [imath]{N \choose {n+1}}={{N-1}\choose{n+1}}+{{N-1}\choose n}.[/imath] by Pascals triangle. If I repeat this idea, I believe I end up with: [imath]={{N-x}\choose{n+1}}+{{N-x}\choose{n}}...+{{{N-1}\choose n}}[/imath]For some [imath]x[/imath] such that [imath]x=n+1[/imath]. But lets go one further. [imath]={{N-x-1}\choose{n+1}}+{{N-x-1}\choose{n}}...+{{{N-1}\choose n}}.[/imath] But this reduces to [imath]{{n}\choose{n+1}}+{{n}\choose{n}}...+{{{N-1}\choose n}}.[/imath] However [imath]{{n}\choose{n+1}}=0[/imath]. Therefore we are just left with: [imath]\sum^{N-1}_{k=n} \binom{k}{n}.[/imath] as desired. | 833451 | Prove [imath]\sum\limits_{i=0}^n\binom{i+k-1}{k-1}=\binom{n+k}{k}[/imath] (a.k.a. Hockey-Stick Identity)
Let [imath]n[/imath] be a nonnegative integer, and [imath]k[/imath] a positive integer. Could someone explain to me why the identity [imath] \sum_{i=0}^n\binom{i+k-1}{k-1}=\binom{n+k}{k} [/imath] holds? |
880082 | Searching two matrix A and B, such that exp(A+B)=exp(A)exp(B) but AB is not equal to BA.
We know that if two matrix [imath]A[/imath] and [imath]B[/imath] commutes then [imath]\exp(A+B)=\exp(A)\exp(B)[/imath]. I am trying to find two matrix that does not commute but [imath]\exp(A+B)=\exp(A)\exp(B)[/imath] is true for them. Can anybody give exact example. Thanks | 638627 | [imath]\exp(A+B)[/imath] and Baker-Campbell-Hausdorff
A few years ago, I did research in quantum mechanics, specifically dealing with generalized displacement operators. In such musings, BCH lights (or gets in, depending on your viewpoint) the way. A question that struck me today was: does [imath]\exp(A+B) = \exp(A)\exp(B)[/imath] hold if and only if [imath]A[/imath] and [imath]B[/imath] commute? Clearly, if they do commute this is true but I have not seen anything detailing the opposite direction. Given the complexity of BCH, I would be inclined to think that it's not simple to prove if it is true. I've thought about it on my own but haven't been able to come to any sort of conclusion one way or the other. |
1222285 | Eigen values of the operator [imath]T : V \rightarrow V : T(p(t)) = p(t+1)[/imath]
Let [imath]V[/imath] be the linear space of all real polynomials [imath]p(x)[/imath] of degree [imath]\le n[/imath]. If [imath]p \in V,[/imath] define [imath]q = T(p)[/imath] to mean that [imath]q(t) = p(t+1) ~\forall ~t \in \mathbb R.[/imath] Prove that [imath]T[/imath] has only the eigenvalue [imath]1[/imath]. Attempt: Given that [imath]T(p(t)) = p(t+1).[/imath] Suppose [imath]\lambda[/imath] is an eigenvalue of [imath]T[/imath]. Then : [imath]p(t+1) = \lambda p(t) ~~~~......................(1)[/imath]. Could anyone please tell me how do I move from here? Thank you for your help. | 737932 | Eigenvalue of a linear transformation substituting [imath]t+1[/imath] for [imath]t[/imath] in polynomials.
Let [imath]V[/imath] be the linear space of all real polynomial [imath]p(x)[/imath] of degree [imath]\leq n[/imath]. If [imath]p \in V[/imath], define [imath]q=T(p)[/imath] to mean that [imath]q(t)=p(t+1)[/imath] for all real [imath]t[/imath]. Prove that [imath]T[/imath] has only the eigenvalue [imath]1[/imath]. What are the eigenfunctions belonging to this eigenvalue? It is obvious that constant polynomials are eigenfunctions with eigenvalues 1. But how to prove that this is the only eigenvalue. |
1223354 | find laurent series of [imath]\frac{1}{1 - cos z} [/imath]
So it is not to solve everything with regards to the series. I was talking to my professor today and he mentioned something about since my denominator is an analytic function itself I could bring it into the numerator in some fashion. Sort of in the same way it is done when finding the laurent of [imath]\frac{1}{e^z - 1}[/imath] where in this one after factoring out I'm left with an expression of the form: [imath]\frac{1}{z} \frac{1}{(1+ \frac{z}{2!} + \frac{z^2}{3!}+...)}[/imath] which could then inverted be said along these lines: [imath]\frac{1}{z} (1+ \frac{z}{2!} + \frac{z^2}{3!}+...) [/imath] where the coefficients ended up being for the residue and terms after: [imath] 1, \frac{1}{2!} , \frac{1}{3!} [/imath] how can I apply this idea to my question? | 1195528 | Bernoulli Number analog using Cosine
I know that Bernoulli Numbers can be found with the generating function [imath]\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n[/imath] I was wondering if any work has been done using a similar equation [imath]\frac{x^2}{\cos{x}-1}=\sum_{n=0}^{\infty}\frac{C_n}{n!}x^{2n}[/imath] I'm particularly interested in the [imath]C_n[/imath]. Can anyone help me with a reference to work with this particular generating function or to the coefficients [imath]C_n[/imath]? |
1223599 | How to prove Poincare-like inequality?
Suppose [imath]u\in W^{1,1}[/imath] and [imath]\partial u[/imath] is [imath]C^1[/imath]. I want to prove the following: [imath]\int_{\partial\Omega}|u-\bar u|\leq A\int_{\Omega}|\nabla u|[/imath], where [imath]\bar u=\dfrac{1}{|\Omega|}\int_{\Omega}u[/imath] and [imath]A>0[/imath]. Note that unlike Poincare inequality, the left integral is over [imath]\partial\Omega[/imath] and not [imath]\Omega[/imath]. How can I do that? | 1220342 | How to prove Poincaré-like inequality for the integral over the boundary?
Suppose [imath]u\in W^{1,1}[/imath] and [imath]\partial u[/imath] is [imath]C^1[/imath]. I want to prove the following: [imath]\int_{\partial\Omega}|u-\bar u|\leq A\int_{\Omega}|\nabla u|[/imath] where [imath]\bar u=\frac{1}{|\Omega|}\int_{\Omega}u[/imath] and [imath]A>0[/imath]. Note that unlike Poincaré inequality, the left integral is over [imath]\partial\Omega[/imath] and not [imath]\Omega[/imath]. How can I do that? I tried to prove in a similar way to Poincare inequality proof (arguing by contradiction), but with no success. I think I am missing something because of the [imath]\partial\Omega[/imath] issue. |
1224186 | Is Spec R compact?
And if so, why? I'm having some trouble with this. I know that the [imath]D_{f}[/imath] (set of primes not containing [imath]f[/imath]) are the open sets and form a basis for the Zariski topology; but I do not know how to go about even constructing an open cover. Thanks. | 104248 | Compactness of [imath]\operatorname{Spec}(A)[/imath]
In an exercise in Atiyah-Macdonald it asks to prove that the prime spectrum [imath]\operatorname{Spec}(A)[/imath] of a commutative ring [imath]A[/imath] as a topological space [imath]X[/imath] (with the Zariski Topology) is compact. Now because the basic open sets [imath]X_f = \{\mathfrak{p} \in \operatorname{Spec} (A) : \{f\} \not\subseteq \mathfrak{p} \}[/imath] form a basis for the Zariski Topology it suffices to consider the case when [imath]X = \bigcup_{i \in I} X_{f_i}[/imath] where [imath]I[/imath] is some index set. Then taking the complement on both sides we get that [imath]\emptyset = \bigcap_{i \in I} X_{f_i}^c[/imath] so there is no prime ideal [imath]\mathfrak{p}[/imath] of [imath]A[/imath] such that all the [imath]f_i[/imath]'s are in [imath]\mathfrak{p}[/imath]. Now from here I am able to show that the ideal generated by the [imath]f_i[/imath]'s is the whole ring as follows. Since there is no prime ideal [imath]\mathfrak{p}[/imath] such that all the [imath]f_i \in \mathfrak{p}[/imath], it is clear that there is no [imath]\mathfrak{p}[/imath] such that [imath](f_i) \subseteq \mathfrak{p}[/imath] for all [imath]i \in I.[/imath] Taking a sum over all the [imath]i[/imath] then gives [imath]\sum_{i \in I} (f_i) = (1).[/imath] Now here's the problem: How do I show from here that there is an equation of the form [imath]1 = \sum_{i \in J} f_ig_i,[/imath] where [imath]g_i \in A[/imath] and [imath]J[/imath] some finite subset of [imath]I[/imath]? This part has been giving me a headache. I am not sure if the finiteness bit has to do with algebra, topology or the fact that we are dealing with prime ideals. This is not a homework problem but rather for self-study. [imath]\textbf{Edit:}[/imath] I have posted my answer below after the discussion with Dylan and Pierre. |
1224338 | Intuitive explanation for why the group [imath]S_n[/imath] has [imath]n![/imath] elements
It's not entirely obvious to me that the group of permutations of a set A under permutation multiplication / function composition has [imath]n![/imath] functions in it. Would someone be able to provide an intuitive explanation? I am taking an undergraduate course on abstract algebra at the moment. | 156285 | Proving that [imath]S_n[/imath] has order [imath]n![/imath]
I have been working on this exercise for a while now. It's in B.L. van der Waerden's Algebra (Volume I), page [imath]19[/imath]. The exercise is as follows: The order of the symmetric group [imath]S_n[/imath] is [imath]n!=\prod_{1}^{n}\nu[/imath]. (Mathematical induction on [imath]n[/imath].) I don't comprehend how we can logically use induction here. It seems that the first step would be proving [imath]S_1[/imath] has [imath]1!=1[/imath] elements. This is simply justified: There is only one permutation of [imath]1[/imath], the permutation of [imath]1[/imath] to itself. The next step would be assuming that [imath]S_n[/imath] has order [imath]n![/imath]. Now here is where I get stuck. How do I use this to show that [imath]S_{n+1}[/imath] has order [imath](n+1)![/imath]? Here is my attempt: I am thinking this is because all [imath]n![/imath] permutations of [imath]S_n[/imath] now have a new element to permutate. For example, if we take one single permutation [imath] p(1,\dots,n) = \begin{pmatrix} 1 & 2 & 3 & \dots & n\\ 1 & 2 & 3 & \dots & n \end{pmatrix} [/imath] We now have [imath]n[/imath] modifications of this single permutation by adding the symbol [imath](n+1)[/imath]: \begin{align} p(1,2,\dots,n,(n+1))&= \begin{pmatrix} 1 & 2 & \dots & n & (n+1)\\ 1 & 2 & \dots & n & (n+1) \end{pmatrix}\\ p(2,1,\dots,n,(n+1))&= \begin{pmatrix} 1 & 2 & \dots & n & (n+1)\\ 2 & 1 & \dots & n & (n+1) \end{pmatrix}\\ \vdots\\ p(n,2,\dots,1,(n+1))&= \begin{pmatrix} 1 & 2 & \dots & n & (n+1)\\ n & 2 & \dots & 1 & (n+1) \end{pmatrix}\\ p((n+1),2,\dots,n,1)&= \begin{pmatrix} 1 & 2 & \dots & n & (n+1)\\ (n+1) & 2 & \dots & n & 1 \end{pmatrix} \end{align} There are actually [imath](n+1)[/imath] permutations of that specific form, but we take [imath]p(1,\dots,n)=p(1,\dots,n,(n+1))[/imath] in order to illustrate and prove our original statement. We can make this general equality for all [imath]n![/imath] permutations: [imath]p(x_1,x_2,\dots,x_n)=p(x_1,x_2,\dots,x_n,x_{n+1})[/imath] where [imath]x_i[/imath] is any symbol of our finite set of [imath]n[/imath] symbols and [imath]x_{n+1}[/imath] is strictly defined as the symbol [imath](n+1)[/imath]. We can repeat this process for all [imath]n![/imath] permutations in [imath]S_n[/imath]. This gives us [imath]n!n[/imath] permutations. Then, adding in the original [imath]n![/imath] permutations, we have [imath]n!n+n!=(n+1)n!=(n+1)![/imath]. Consequently, [imath]S_n[/imath] has order [imath]n![/imath]. How is my reasoning here? Furthermore, is there a more elegant argument? I do not really see my argument here as incorrect, it just seems to lack elegance. My reasoning may well be very incorrect, however. If so, please point it out to me. |
1224747 | Formal proof that [imath]N[/imath] is the smallest infinite set
I wish to write a formal proof of the following statement: For any infinite set [imath]X[/imath], there exists an injection [imath]f:\mathbb{N}\to X[/imath]. I'd like the proof to explicitly use the full axiom of choice (for every family of sets [imath]\{S_\alpha\}[/imath] there exists a family of elements [imath]\{x_\alpha\}[/imath] such that each [imath]x_\alpha\in S_\alpha[/imath]). When this was asked before, none of the answers were explicit about where choice is invoked. Motivation: I'm TAing a course in discrete math and was embarrassed to find that I can't prove this homework question. | 250119 | Proving Dedekind finite implies finite assuming countable choice
I'd like to show that if a set [imath]X[/imath] is Dedekind finite then is is finite if we assume [imath](AC)_{\aleph_0}[/imath]. As set [imath]X[/imath] is called Dedekind finite if the following equivalent conditions are satisfied: (a) there is no injection [imath]\omega \hookrightarrow X[/imath] (b) every injection [imath]X \to X[/imath] is also a surjection. Countable choice [imath](AC)_{\aleph_0}[/imath] says that every contable family of non-empty, pairwise disjoint sets has a choice function. There is the following theorem: from which I can prove what I want as follows: Pick an [imath]x_0 \in X[/imath]. Define [imath]G(F(0), \dots, F(n-1)) = \{x_0\}[/imath] if [imath]x_0 \notin \bigcup F(k)[/imath] and [imath]G(F(0), \dots , F(n-1)) = X \setminus \bigcup F(k)[/imath] otherwise. Also, [imath]G(\varnothing) = \{x_0\}[/imath]. Let [imath]F: \omega \to X[/imath] be as in the theorem. Then [imath]F[/imath] is injective by construction. The problem with that is that I suspect that the proof of theorem 24 needs countable choice. So what I am after is the following: consider the generalisation of theorem 24: (note the typo in [imath](R^\ast)[/imath], it should be [imath]F(z) \in G^\ast (F \mid I(z), z)[/imath]), and its proof (assuming AC): I want to modify this proof to prove the countable version of the theorem. But I can't seem to manage. I need a countable set [imath]\{G^\ast \mid \{\langle f,z \rangle \} : \langle f,z \rangle \in dom(G^\ast) \}[/imath]. Ideas I had were along the lines of picking [imath]f_0(x) = x_0[/imath] the constant function and then to consider [imath]\{G^\ast \mid \{\langle f_0,n \rangle \} : \langle f_0,n \rangle \in dom(G^\ast) \}[/imath] but what then? Thanks for your help. |
1224780 | I need help with a proof in Probability
I could use help with the following problem: Show that if [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] are three events such that [imath]P(A \cap B \cap C) \neq 0[/imath] and [imath]P(C \mid A \cap B)[/imath] = [imath]P(A \mid B)[/imath], then [imath]P(C) = P(A \mid B)[/imath]. Here is my attempt to solve the problem: [imath]P(C \mid A \cap B) = { {P(A \cap B \cap C)} \over {P(A \cap B)}} = P(C \mid B) = {{P(C \cap B)} \over {P(B)}}[/imath] However, at this point I do not know what to do. Please help. Note: This is not a homework problem Bob | 1224826 | Help with a Probability Proof
How do I prove the following: Show that if [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] are three events such that [imath]P(A \cap B \cap C) \neq 0[/imath] and [imath]P(C | A \cap B)[/imath] = [imath]P(C|B)[/imath], then [imath]P(A| B \cap C) = P(A | B)[/imath]. Here is my attempt [imath]P(C|A \cap B) = { {P(A \cap B \cap C)} \over {P(A \cap B)}} = P(C|B) = {{P(C \cap B)} \over {P(B)}}[/imath] Now, I do not know what to do. I am hoping somebody can help me. This is not a homework problem. Bob |
1216775 | How do we know Euclid's sequence always generates a new prime?
How do we know that [imath](p_1 \cdot \ldots \cdot p_k)+1[/imath] is always prime, for [imath]p_1 = 2[/imath], [imath]p_2 = 3[/imath], [imath]p_3 = 5[/imath], [imath]\ldots[/imath] (i.e., the first [imath]k[/imath] prime numbers)? Euclid's proof that there is no maximum prime number seems to assume this is true. | 631977 | Why is Euclid's proof on the infinitude of primes considered a proof?
I've expressed Euclid's proof on the infinitude of primes on Mathematica: f[x_] := Product[Prime[n], {n, 1, x}] + 1 TableForm[Table[{f[x], PrimeQ[f[x]]}, {x, 1, 20}]] Which results in: [imath]\begin{array}{ll} 3 & \text{True} \\ 7 & \text{True} \\ 31 & \text{True} \\ 211 & \text{True} \\ 2311 & \text{True} \\ 30031 & \text{False} \\ 510511 & \text{False} \\ 9699691 & \text{False} \\ 223092871 & \text{False} \\ 6469693231 & \text{False} \\ 200560490131 & \text{True} \\ 7420738134811 & \text{False} \\ 304250263527211 & \text{False} \\ 13082761331670031 & \text{False} \\ 614889782588491411 & \text{False} \\ 32589158477190044731 & \text{False} \\ 1922760350154212639071 & \text{False} \\ 117288381359406970983271 & \text{False} \\ 7858321551080267055879091 & \text{False} \\ 557940830126698960967415391 & \text{False} \\ \end{array}[/imath] The proof flaws for all those values, how is it considered a proof then? I guess that there might be infinite prime numbers according to the proof, but what is the guarantee that at some point it won't fail indefinitely? |
1224945 | Continuity of convex function
Let [imath]f[/imath] be a proper convex lower semi-continuous function on [imath]\mathbb R^n[/imath], how can we prove [imath]f[/imath] is continuous in the interior of [imath]Dom(f)=\{f<\infty\}[/imath] ? | 216778 | Continuity of a convex function
I'm trying to solve the following problem: Let [imath]f:K\rightarrow \mathbb{R} [/imath], [imath]f[/imath] convex and [imath]K \subseteq \mathbb{R}^n[/imath] convex. Then [imath]f[/imath] is continuous on [imath]K[/imath]. I have proved the only case [imath]n=1[/imath], but for an arbitrary [imath]n[/imath]?? |
1225203 | Prove that [imath]b_n = \frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \dotsb + \frac{1}{\sqrt{n^2 + n}} \to 1[/imath] as [imath]n \to \infty[/imath]
I need to prove that [imath]b_n = \frac{1}{\sqrt{n^2 + 1}} + \frac{1}{\sqrt{n^2 + 2}} + \dotsb + \frac{1}{\sqrt{n^2 + n}} \to 1[/imath] as [imath]n \to \infty[/imath] My initial thoughts are to sandwich this (seems sensible to me). Finding the above inequality [imath]b_n < 1[/imath] was easy, since clearly [imath]\frac{1}{\sqrt{n^2 + k}} < \frac{1}{n}[/imath] for integer k. But I'm stuck on how to compare it with something from below. A hint I've been given is that if [imath]a_n \to a[/imath], then [imath]\sqrt{a_n} \to \sqrt{a}[/imath] as [imath]n \to \infty[/imath], though I'm struggling to see a way of applying this. | 948049 | How to prove that [imath] \lim_{n\to\infty}\left( \sum_{r=1}^n \dfrac 1 {\sqrt{n^2 + r}} \right) = 1[/imath]
We have to show that [imath] \lim_{n\to\infty} \left( \dfrac 1 {\sqrt{n^2 +1}} +\dfrac 1 {\sqrt{n^2 +2}} + \dfrac 1 {\sqrt{n^2 +3}} + \ldots + \dfrac 1 {\sqrt{n^2 +n}} \right) = 1[/imath] Now, I thought of doing it as a definite integral, however the terms don't translate into a function of [imath]\left(\dfrac r n\right)[/imath]. I tried searching wolfram-alpha, but it used Hurwitz-zeta function, of which I have no clue. (This is an assignment problem) |
877689 | Solve the given differential equation by using Green's function method
I am really struggling with the concept and handling of the Green's function. I have to solve the given differential equation using Green's function method [imath]\frac{d^{2}y}{dx^{2}}+k^{2}y=\delta (x-x');\qquad y(0)=y(L)=0[/imath] | 877470 | Solve the given differential equation by using Green's function method
I am really struggling with the concept and handling of the green's function. I have to solve the given differential equation using Green's function method [imath]\frac{d^{2}y}{dx^{2}}+k^{2}y=\delta (x-x');y(0)=y(L)=0[/imath] |
1224956 | Showing that a divisor of zero in a commutative ring with unity can have no multiplicative inverse"
A divisor of zero in a commutative ring with unity can have no multiplicative inverse. I don't understand why this statement is true. So for [imath]a,b[/imath] in the ring, [imath]ab=0[/imath] by zero divisor. How can we guarantee that there does not exist [imath]c[/imath] such that [imath]ac=1[/imath]? | 493660 | If [imath]a[/imath] is a zero divisor of [imath]\mathbb{F}[/imath], then does [imath]a[/imath] fail to have a multiplicative inverse?
Let F be a ring. Show that if [imath]a[/imath] is a zero divisor of F, then [imath]a[/imath] fails to have a multiplicative inverse. All I understand is that [imath]ab = 1[/imath] doesn't hold and that [imath]ac = 0[/imath] for some [imath]c \in F[/imath] holds. Basically, only the definitions, but I don't know what to do with them. |
458420 | Subadditivity inequality and power functions
Is it true that if [imath]a,b\in\mathbb{R}[/imath] with [imath]a,b\geq 0[/imath] and [imath]0<r<1[/imath], then [imath](a+b)^r\leq a^r+b^r[/imath]? | 1225144 | Proving the inequality [imath](x+y)^\alpha \leq x^\alpha + y^\alpha[/imath]
Why does this inequality hold? [imath](x+y)^\alpha \leq x^\alpha + y^\alpha[/imath] where [imath]x[/imath], [imath]y\geq 0[/imath] and [imath]\alpha \in (0,1)[/imath]. Thanks in advance! |
1027974 | Why is this limit [imath]\lim_{n \to \infty}(1 + \frac{1}{n})^n[/imath] equal to [imath]e[/imath]?
I dont get it, if I look at [imath]\lim \limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n = e[/imath] It seems like it should be [imath]1[/imath] because [imath]1 \over n[/imath] is almost zero and [imath]1^\infty[/imath] is still [imath]1[/imath] | 136784 | Why [imath]\lim\limits_{n\to \infty}\left(1+\frac{1}{n}\right)^n[/imath] doesn't evaluate to 1?
I am trying to identify what the flaw is exactly when reasoning about a limit such as the definition of [imath]\mathbf e[/imath]: [imath] \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n={e} [/imath] Now, I know there are ways of proving this limit, such as by considering the binomial expansion of [imath](1+\frac{1}{n})^n[/imath] and comparing that to the Maclaurin series of [imath]e[/imath]. So to make this clear, I am not looking for a proof of the limit definition of [imath]e[/imath]. I tried searching for "limit laws/rules" but none of the rules I found described the above case. Hence, I am looking for a specific rule (or perhaps a certain perspective) that will help me realize why the above does not in fact evaluate to 1. My train of thought is as follows: at a first glance, if I wasn't already familiar with what the limit evaluates to, I probably would evaluate the expression inside the brackets first, and then apply the limit to the power. So, [imath]\displaystyle\lim_{n\to\infty}\frac{1}{n}=0\quad\text{and then}[/imath] [imath]\displaystyle\lim_{n\rightarrow \infty}\left(1+0\right)^n=\lim_{n\to\infty}1^n=1[/imath] Why is my reasoning flawed? |
1225244 | System of first order differential equations
Regard the diff equation that includes [imath]sin(\phi)[/imath]: [imath]mϕ′′+aϕ′+(mg/L)sin(ϕ)=0[/imath] [imath]ϕ(0)=0.1[/imath] [imath]ϕ′(0)=0[/imath] where [imath]m=0.1,L=1,a=2,[/imath] How can i rewrite the second order diff equation as a system of first order linear equations? | 1222542 | Find the first order system of linear equations
Regard the diff equation: [imath]mϕ′′+aϕ′+(mg/L)ϕ=0[/imath] [imath]ϕ(0)=0.1[/imath] [imath]ϕ′(0)=0[/imath] where [imath]m=0.1,L=1,a=2,[/imath] 1) Rewrite the second order diff equation as a system of first order linear equations. 2) What is the matrix that comes from the rewritten equations? |
1225643 | Show that [imath]L^{\infty}[/imath] space does not have a countable dense set.
I was able to show that when [imath]p ≥ 1[/imath], the [imath]L^p[/imath] space on the interval [imath][0,1][/imath] has a countable dense set. However, when [imath]p[/imath] is infinite, how to prove that [imath]L^p[/imath] space on the interval [imath][0,1][/imath] does not have a countable dense set? I can't find some way to approach. | 97648 | Why is [imath]L^{\infty}[/imath] not separable?
[imath]l^p (1≤p<{\infty})[/imath] and [imath]L^p (1≤p<∞)[/imath] are separable spaces. What on earth has changed when the value of [imath]p[/imath] turns from a finite number to [imath]{\infty}[/imath]? Our teacher gave us some hints that there exists an uncountable subset such that the distance of any two elements in it is no less than some [imath]\delta>0[/imath]. Actually I don't understand the question very well, but I hope I have made the question clear enough. Thank you in advance. |
1225629 | Find the last three digits of [imath]17^{256}[/imath]
Find the last three digits of [imath] 17 ^{256} [/imath] We have to check mod [imath]1000[/imath] I tried to check some patterns but in vain.! | 619810 | What is the shortest way to compute the last 3 digits of [imath]17^{256}[/imath]?
What is the shortest way to compute the last 3 digits of [imath]17^{256}[/imath] ? My solution: \begin{align} 17^{256} &=289^{128} \\ &=(290 - 1)^{128}\\ &=\binom{128}{0}290^{128} - ... +\binom{128}{126}290^2 - \binom{128}{127}290 + \binom{128}{128} \end{align} Computed the last 3 terms whose last 3 digits gave the last 3 digits of [imath]17^{256}[/imath]. Is there any shorter method to do this(which requires much less computation) ? |
1225241 | Prob. 4 (a), Sec. 20 in Munkres' TOPOLOGY, 2nd edition: Which of these functions are continuous in these topologies?
Let [imath]\mathbb{R}^\omega[/imath] denote the set of all infinite sequences of real numbers. Let the uniform metric [imath]\tilde{\rho}[/imath] on [imath]\mathbb{R}^\omega[/imath] be defined as follows: [imath]\tilde{\rho}(x,y) \colon= \sup \left\{ \ \min \left( \ \vert x_i - y_i \vert \ , 1 \ \right) \ \colon \ i = 1, 2, 3, \ldots \ \right\} \ \ \ \forall x = (x_i)_{i \in \mathbb{N}}, \ y=(y_i)_{i \in \mathbb{N}} \in \mathbb{R}^\omega.[/imath] Then the topology induced by this metric is called the uniform topology on [imath]\mathbb{R}^\omega[/imath]. And, the box topology on [imath]\mathbb{R}^\omega[/imath] is the one havaing as a basis all sets of the form [imath] (a_1, b_1) \times (a_2, b_2) \times (a_3, b_3) \times \cdots, [/imath] where [imath]\left(a_i \right)_{i \in \mathbb{N}} \in \mathbb{R}^\omega[/imath] and [imath]\left(b_i \right)_{i \in \mathbb{N}} \in \mathbb{R}^\omega[/imath] are such that [imath]a_i < b_i[/imath] for each [imath]i = 1, 2, 3, \ldots[/imath], and [imath](a_i, b_i)[/imath] denotes the segment (i.e. open interval) with [imath]a_i[/imath] as the left endpoint and [imath]b_i[/imath] as the right endpoint. Now with respect to which of the above two topologies on [imath]\mathbb{R}^\omega[/imath] are the following functions from [imath]\mathbb{R}[/imath] to [imath]\mathbb{R}^\omega[/imath] continuous? [imath]f(t) \colon= (t, 2t, 3t, \ldots),[/imath] [imath]g(t) \colon= (t, t, t, \ldots), [/imath] [imath]h(t) \colon= (t, \frac{t}{2}, \frac{t}{3}, \ldots).[/imath] My effort: The functions [imath]f[/imath], [imath]g[/imath], and [imath]h[/imath] are not continuous when [imath]\mathbb{R}^\omega[/imath] is given the box topology. The inverse image under each of [imath]f[/imath], [imath]g[/imath], and [imath]h[/imath] of the basis element [imath]B \colon= \left( -1, 1 \right) \times \left(-\frac{1}{2^2}, \frac{1}{2^2} \right) \times \left( -\frac{1}{3^2}, \frac{1}{3^2} \right) \times \cdots, [/imath] for example, contains the point [imath]t = 0[/imath] since [imath]f(0) = g(0) = h(0) = (0, 0, 0, \ldots) \in B.[/imath] So, in order for this inverse image to be open in [imath]\mathbb{R}[/imath] with the usual topology, there must be an open interval [imath]\left(-\delta_f, \delta_f \right)[/imath], [imath]\left( -\delta_g, \delta_g \right)[/imath], and [imath]\left( -\delta_h, \delta_h \right)[/imath], for some positive real numbers [imath]\delta_f[/imath], [imath]\delta_g[/imath], and [imath]\delta_h[/imath], respectively, such that [imath] \left(-\delta_f, \delta_f \right) \subset f^{-1}(B),[/imath] [imath]\left( -\delta_g, \delta_g \right) \subset g^{-1}(B), [/imath] [imath]\left( -\delta_h, \delta_h \right) \subset h^{-1}(B). [/imath] In particular, we must have [imath]f\left( \frac{\delta_f}{2} \right) \in B,[/imath] [imath]g\left( \frac{\delta_g}{2} \right) \in B,[/imath] [imath]h\left( \frac{\delta_h}{2} \right) \in B.[/imath] So, for each [imath]n \in \mathbb{N}[/imath], we have [imath]\frac{n \delta_f}{2} \in \left( - \frac{1}{n^2}, \frac{1}{n^2}\right), \ \frac{\delta_g}{2} \in \left( - \frac{1}{n^2}, \frac{1}{n^2}\right), \ \frac{\delta_h}{2n} \in \left( - \frac{1}{n^2}, \frac{1}{n^2}\right), [/imath] and hence [imath]n^3 < \frac{2}{ \delta_f } \ \mbox{ for all} \ n \in \mathbb{N},[/imath] [imath]n^2 < \frac{2}{ \delta_g } \ \mbox{ for all} \ n \in \mathbb{N},[/imath] [imath]n < \frac{2}{ \delta_h } \ \mbox{ for all} \ n \in \mathbb{N},[/imath] each of which is impossible. Am I right? That [imath]g[/imath] is not continuous was also shown by Munkres himself in Example 2, Sec. 19 on page 117. What is the situation for the uniform topology? That is, which of these three functions is continuous relative to the uniform topology? | 297303 | What about the continuity of these functions in the uniform topology?
Let [imath]f[/imath], [imath]g[/imath], [imath]h \colon \mathbf{R} \to \mathbf{R}^\omega[/imath] be defined by [imath]\begin{align*} f(t)&:=(t,2t,3t,\ldots),\\\\ g(t)&:=(t,t,t,\ldots),\\\\ h(t)&:=\left(t,\frac{t}{2},\frac{t}{3},\ldots\right) \end{align*}[/imath] for all [imath]t \in \mathbf{R}[/imath]. How to determine whether these functions are continuous relative to the uniform topology on [imath]\mathbf{R}^\omega[/imath]? We of course assume that [imath]\mathbf{R}[/imath] is given the usual topology. |
1226171 | Greatest Common Divisor Integer
I need help with the following problem: Let [imath]n[/imath] and [imath]m[/imath] be positive integers. Prove that [imath]\frac{\gcd(n,m)}{n}{n \choose m}[/imath] is an integer. | 1165229 | [imath]{\gcd(n,m)\over n}{n\choose m}[/imath] is an integer
Prove that for every [imath]n\geq m \geq1[/imath] natural numbers, the following number is an integer: [imath]{\gcd(n,m)\over n}\cdot{n\choose m}[/imath] Where [imath]\gcd[/imath] is the greatest common divisor. I tried to make it simpler by cancelling the [imath]n[/imath] from the left side, and making it [imath](n-1)![/imath] on the right: [imath]\gcd(n, m) \cdot \frac{(n-1)!}{m!(n-m)!}[/imath], but can't really go further. This was problem B-2 on the 2000 Putnam exam. |
1225702 | Prove that there is not a retraction of the Möbius band to its boundary
I would like a hint to solve the following problem: Let [imath]X[/imath] be the Möbius band and [imath]A[/imath] its boundary (which is a circumference). Prove that there is not a retraction [imath]r:X \rightarrow A[/imath]. Each try I did gave no solution. Any hint to solve this problem? Thanks! | 202447 | Retraction of the Möbius strip to its boundary
Prove that there is no retraction (i.e. continuous function constant on the codomain) [imath]r: M \rightarrow S^1 = \partial M[/imath] where [imath]M[/imath] is the Möbius strip. I've tried to find a contradiction using [imath]r_*[/imath] homomorphism between the fundamental groups, but they are both [imath]\mathbb{Z}[/imath] and nothing seems to go wrong... |
1226317 | Converge the sequence [imath]\left(\left(1+\frac{1}{n}\right) \left(1+\frac{2}{n}\right)\cdots\left(1+\frac{n}{n}\right)\right)^{1/n}[/imath]
Does this series converge or diverge? [imath] \left[\prod_{k=1}^{n}\left(1+\frac{k}{n}\right)\right]^{\frac{1}{n}} [/imath] | 1071053 | To evaluate limit of sequence [imath]\left(\left( 1 + \frac1n \right) \left( 1 + \frac2n \right)\cdots\left( 1 + \frac nn \right) \right)^{1/n}[/imath]
How do I evaluate the limit of the following sequence [imath]a_n = \left(\left( 1 + \frac1n \right) \left( 1 + \frac2n \right)\cdots\left( 1 + \frac nn \right) \right)^{1/n}[/imath] I tried to take log and then using cauchy [imath]1^{st}[/imath] theorem of limits but I coudn't do it. Could someone kindly help? |
1168033 | Proving measurable function: Real versus rational number
I am working on this problem[imath]^{(1)}[/imath] on measurable function like this: Show that [imath]f[/imath] is measurable if [imath](X, \mathcal A)[/imath] is measurable space, [imath]f[/imath] is real-value function and [imath]\{x \mid f(x) > r \} \in \mathcal A[/imath] for each rational number [imath]r.[/imath] Earlier on the text gives this definition[imath]^{{2}}[/imath] of measurable function: A function [imath]f : X \to \mathbb R[/imath] is measurable or [imath]\mathcal A[/imath]-measurable if [imath]\{x \mid f(x) > a \} \in \mathcal A[/imath] for all [imath]a \in \mathbb R.[/imath] To an untrained eyes, the problem with this problem is that there seems to be no problem at all, except, perhaps, that the exercise says "[imath]\forall r \in \mathbb Q[/imath]", whereas the definition says "[imath]\forall a \in \mathbb R.[/imath]" Am I on the right track here? Please help me going forward if this is the right issue to tackle. Thank you for your time and help. Footnotes: (1) Richard F. Bass' Real Analysis, 2nd. edition, chapter 5: Measurable Functions, Exercise 5.1, page 44. (2) Definition 5.1, page 37. | 228605 | Let [imath]E[/imath] be measurable and define [imath]f:E\rightarrow\mathbb{R}[/imath] such that [imath]\{x\in E : f(x)>c\}[/imath] is measurable for all [imath]c\in\mathbb{Q}[/imath], is [imath]f[/imath] measurable?
Let [imath]E[/imath] be measurable and define [imath]f:E\rightarrow\mathbb{R}[/imath] such that [imath]\{x\in E : f(x)>c\}[/imath] is measurable for all [imath]c\in\mathbb{Q}[/imath], is [imath]f[/imath] measurable? There are a number of equivalent definitions for the measurability of a function and the most obvious one would be to show that [imath]\{x\in E : f(x)>c\}[/imath] is measurable for all [imath]c\in\mathbb{R}[/imath]. Thus my strategy has been to consider an arbitrary irrational [imath]y[/imath] and use the density of the rationals in the reals to show that there exists some open set [imath]O[/imath] such that [imath]m^*(O-\{x\in E : f(x)>y\})< \varepsilon[/imath]. I would do this by choosing some rational [imath]q<y[/imath], close enough to [imath]y[/imath], such that the set of values in [imath]E[/imath] which get sent under [imath]f[/imath] to values strictly greater than [imath]q[/imath] and less than or equal to [imath]y[/imath] is small enough. But that doesn't in general seem possible since I could keep sending intervals of some fixed positive length to smaller and smaller regions just beneath [imath]y[/imath]. However conversely, such a function would not be enough for a counter example, since finding such a function only rules out this strategy, but doesn't necessarily entail non-measurability. Is there a better definition of the measurability of a set I should use? or does the statement actually not imply measurability? Thanks. |
1179981 | If [imath]\phi(g)=g^3[/imath] is a homomorphism and [imath]3 \nmid |G|[/imath], [imath]G[/imath] is abelian.
As the title suggests. Let [imath]G[/imath] be a group, and suppose the function [imath]\phi: G \to G[/imath] with [imath]\phi(g)=g^3[/imath] for [imath]g \in G[/imath] is a homomorphism. Show that if [imath]3 \nmid |G|[/imath], [imath]G[/imath] must be abelian. By considering [imath]\ker(\phi)[/imath] and Lagrange's Theorem, we have [imath]\phi[/imath] must be an isomorphism (right?), but I'm not really sure where to go after that. This is a problem from Alperin and Bell, and it is not for homework. | 2261710 | Problem in working out a question related to group theory.
The question is : Let [imath]G[/imath] be a finite group whose order is not divisible by [imath]3[/imath].Suppose that [imath](ab)^{3} = a^{3} b^{3}[/imath] for all [imath]a,\ b \in G[/imath]. Prove that [imath]G[/imath] must be abelian. By the given condition it can be deduced that [imath](ba)^{2} = a^{2} b^{2}[/imath] for all [imath]a,\ b \in G[/imath].But now how can proceed? Actually I think I have failed to do it because I have failed to use the condition that order of [imath]G[/imath] is not divisible by [imath]3[/imath]. Please help me. Thank you in advance. |
1226985 | Can a cyclic group have more than two generators?
Can a cyclic group have more than two generators? for example the group [imath]\mathrm{Z}[/imath] has two generators [imath]-1[/imath] and [imath]1[/imath], but can a group have more than two generators? | 964784 | Cyclic group generators
My question is: Can you find a cyclic group with n generators? I know that zero (or any other identity element for that matter) is included, so there would be for [imath]Z_n[/imath] at most n-1 generators. However, is it possible to say that [imath]Z_{n+1}[/imath] could provide n generators? Any help would be much appreciated! If you wouldn't mind, I would appreciate it if you provide enough detail to really help me understand so I can learn. Thanks in advance! |
1226046 | Showing that [imath]\{x\mid f(x)\leq g(x)\}[/imath] is closed.
Problem statement: Let [imath]Y[/imath] be an ordered set in the order topology and [imath]X[/imath] a topological space. [imath]f,g:\, X\to Y[/imath] are continuous. Show that [imath]\{x\mid f(x)\leq g(x)\}[/imath] is closed. My thoughts: In case [imath]Y=\mathbb{R}[/imath] I would define [imath]h=g-f[/imath] and say that the above set is [imath]h^{-1}([0,\infty))[/imath] which is closed by continuity of [imath]h[/imath]. But in an ordered set [imath]Y[/imath] there is no meaning of [imath]g-f[/imath]. Can someone give me a hint ? | 165432 | Let [imath]Y[/imath] be an ordered set in the order topology with [imath]f,g:X\rightarrow Y[/imath] continuous, show that [imath]\{x:f(x)\leq g(x)\}[/imath] is closed in [imath]X[/imath]
Let [imath]Y[/imath] be an ordered set in the order topology with [imath]f,g:X\rightarrow Y[/imath] continuous. Show that the set [imath]A = \{x:f(x)\leq g(x)\}[/imath] is closed in [imath]X[/imath]. I am completely stumped on this problem. As far as I can tell I've either got the task of proving [imath]A[/imath] gathers its limit points, or showing that there is a closed set [imath]V \subset Y[/imath] such that [imath]f^{-1}(V)[/imath] or [imath]g^{-1}(V)[/imath] is equal to A, which would prove [imath]A[/imath] closed since both [imath]f[/imath] and [imath]g[/imath] are continuous. The latter strategy seems like the more likely to succeed. Unfortunately I can't find any way of constructing this set [imath]V[/imath]. The fact that I don't know if [imath]f[/imath] or [imath]g[/imath] are injective means I keep running into the problem of having [imath]f^{-1}(V)[/imath] or [imath]g^{-1}(V)[/imath] give me extra points not in [imath]A[/imath]. And even if I were able to construct [imath]V[/imath] this wouldn't guarantee it was closed. I suspect I may need to use the continuity of both [imath]f[/imath] and [imath]g[/imath] together in some way but I can't see how. Can anyone give me some guidance on this problem? Thanks. |
1227052 | Calculation of [imath]\prod_{k=1}^\infty \left( 1 + \frac{a}{k^2} \right)[/imath]?
I am curious how to calculate the infinite product [imath] \prod_{k=1}^\infty \left( 1 + \frac{a}{k^2} \right). [/imath] WolframAlpha reports that it is equal to approximately [imath] \prod_{k=1}^\infty \left( 1 + \frac{a}{k^2} \right) \approx \frac{0.31831 \sinh (3.14159 \sqrt{a})}{\sqrt{a}} [/imath] which I suspect might be given by [imath] \frac{\sinh (\pi \sqrt{a})}{\pi \sqrt{a}}. [/imath] This in turn suggests that [imath] \sinh a = a \prod_{k=1}^\infty \left( 1 + \frac{a^2}{\pi^2 k^2} \right), [/imath] which is confirmed by WolframAlpha. However, I don't know how this is derived. | 359638 | Infinite Product [imath]\prod_{n=1}^\infty\left(1+\frac1{\pi^2n^2}\right)[/imath]
How do I find: [imath]\prod_{n=1}^\infty\; \left(1+ \frac{1}{\pi ^2n^2}\right) \quad[/imath] I am pretty sure that the infinite product converges, but if it doesn't please let me know if I have made an error. Also, could I have a nice explanation as to how would someone arrive to the answer. Thanks alot. |
1226975 | Can find the determinant of a matrix A of size [imath]n[/imath] in terms of the traces of [imath]A^m[/imath]
We can find the determinant of a matrix A of size [imath]n[/imath] in terms of the traces of [imath]A^m[/imath], for [imath]m=1,…,n[/imath] ? It's det of a matrix with term are traces, but i saw but i can't remember Expressing the determinant in terms of the trace of a matrix and the trace of its square when [imath]n=2[/imath]. | 668374 | Determinant of a finite-dimensional matrix in terms of trace
I have noticed that for the case of [imath]1 \times 1[/imath], [imath]2 \times 2[/imath] and [imath]3 \times 3[/imath] matrices [imath]A[/imath], [imath]B[/imath], I can write the determinant of their commutator [imath]C=[A,B][/imath] in terms of traces: [imath]1 \times 1[/imath] matrices [imath]A[/imath], [imath]B[/imath]: [imath]\det(C)=\text{tr}(C)[/imath] [imath]2 \times 2[/imath] matrices [imath]A[/imath], [imath]B[/imath]: [imath]\det(C)=-\frac{1}{2}\text{tr}(C^2)[/imath] [imath]3 \times 3[/imath] matrices [imath]A[/imath], [imath]B[/imath]: [imath]\det(C)=\frac{1}{3}\text{tr}(C^3)[/imath] But I can't find a simple formula for [imath]4 \times 4[/imath] matrices--I have no idea how to generalize this general finite-dimensional matrices: I can't quite understand the origin of this pattern for low-dimensional matrices. So my question is can this be generalized to: Larger [imath]n\times n[/imath] matrices, [imath]n>3[/imath]? Anticommutators [imath]C=\{A,B\} = AB+BA[/imath]? |
1227664 | Is this statement about measurable function true?
If [imath]f[/imath] is a measurable function on [imath][0,1][/imath], then there exists a measurable set [imath]D\subset [0,1][/imath] such that [imath]mD>0.9[/imath], and a continuous function [imath]g:[0,1]\to R[/imath] such that [imath]f=g[/imath] on [imath]D[/imath]. My question is that how can we approximate a measurable function by continuous function and how good could it be? Is there any theorem for it? | 1217495 | Proof of a theorem saying that we can make a measurable function continuous by altering it by a set of arbitarily small measure
I found the following problem in the book of kolmogorov fomin Introductory real analysis (p.293 problem 10) which I have no idea how to show. Prove that a function [imath]f[/imath] defined on a closed interval [imath][a,b][/imath] is [imath]m[/imath] measurable if and only if given any [imath]\epsilon>0[/imath] there is a continuous function [imath]g[/imath] on [imath][a,b][/imath] such that [imath]m[/imath]{[imath]x:f(x)\neq g(x)[/imath]}[imath]<\epsilon[/imath] Hint use Egorov's theorem Does anyone have any idea how to prove this Thanks in advance |
1227333 | Prove that [imath]F[x]/(f(x))[/imath] has [imath]q^n[/imath] elements
let [imath]F[/imath] be a finite field of order [imath]q[/imath] and let [imath]f(x)[/imath] be a polynomial in [imath]F[x][/imath] of degree [imath]n \geq 1[/imath]. Prove that [imath]F[x]/(f(x))[/imath] has [imath]q^n[/imath] elements. I know that if a field is finite then the order is prime so that means q is prime, but what does that says about [imath]F[x]/(f(x))[/imath]. Can anyone try to help me understand this problem because i just recently learn about field extension. thanks | 166741 | Number of elements in a finite field extension for finite fields
Given an arbitrary finite field [imath]K[/imath] (not necessarily [imath]\mathbb{F}_p[/imath] with [imath]p \in \mathbb{P}[/imath]) with [imath]|K| = q[/imath] and an irreducible polynomial [imath]f[/imath] with [imath]\alpha[/imath] as root and degree of [imath]n[/imath]. Is [imath]|K(\alpha)| = q^n[/imath] and why? Its clear to me for [imath]K[/imath] isomorphic to [imath]\mathbb{Z}_p[/imath] with [imath]p[/imath] is a prime number. |
1228098 | Invertible Matrices as a Manifold
I've been trying to prove that the set of all invertible [imath]n \times n[/imath] matrices is a differentiable manifold. My attempt is as follows: Define a map [imath]\alpha : X \to XX^{-1} - I[/imath] I take the inverse: [imath]\alpha^{-1}(0)[/imath] which would map to the set of invertible [imath]n \times n[/imath] matrices. I believe this suffices to show it is a manifold, but I am stumped in finding it's dimension. How can I approach this? | 175443 | Proving that the general linear group is a differentiable manifold
We know that the the general linear group is defined as the set [imath]\{A\in M_n(R): \det A \neq 0\}[/imath]. I have a homework on how to prove that it is a smooth manifold. So far my only idea is that we can think of each matrix, say [imath]A[/imath], in that group as an [imath]n^2-[/imath]dimensional vector. So i guess that every neighborhood of [imath]A[/imath] is homeomorphic to an open ball in [imath]\mathbb{R}^{n^2}[/imath] (However, i don't know how to prove this.) Now, I'm asking for help if anyone could give me a hint on how to prove that the general linear group is a smooth manifold since I really don't have an idea on how to do this. (By the way, honestly, I don't really understand what a [imath]C^{\infty}-[/imath]smooth structure means which is essential to the definition of a smooth manifold.) Your help will be greatly appreciated. :) |
1228243 | Show that if [imath]f,g:\Bbb{N}\to \Bbb{C}[/imath] are multiplicative, then so is [imath]f*g(n)=\sum_{d|n}{f(d)g({n\over d})}[/imath].
Show that if [imath]f,g:\Bbb{N}\to \Bbb{C}[/imath] are multiplicative, then so is [imath]f*g(n)=\sum_{d|n}{f(d)g({n\over d})}[/imath]. What I did is: Let [imath]m[/imath] and [imath]n[/imath] be co-primes. [imath]m's[/imath] divisors=[imath]\{1,m_1,...,m_s,m\}[/imath], [imath]n's[/imath] divisors=[imath]\{1,n_1,...,n_r,n\}[/imath], therefore: [imath]f*g(mn)=f(1)g(mn)+f(m_1)g({mn\over m_1})+...+f(m_s)g({mn\over m_s})+f(m)g({n})+f(n_1)g({mn\over n_1})+...+f(n_r)g({mn\over n_r})+f(n)g({m})[/imath]. Denoting [imath]{m\over m_i}=m_i'[/imath] and given [imath]f,g[/imath] are multiplicative: [imath]f*g(mn)=f(1)g(m)g(n)+f(m_1)g(m_1')g(n)+...+f(m_1)g(m_s')g(n)+f(m)g({n})+f(n_1)g(n_1')g(m)+...+f(n_r)g(n_r')g(m)+f(n)g({m})[/imath]. Now, the problem is when I try to calculate [imath](f*g(m))\cdot (f*g(n))[/imath]: there are many factors of which I can't get rid! I would appreciate your help a lot. | 692784 | Multiplicative Functions
Prove that if [imath]f[/imath] and [imath]g[/imath] are multiplicative, then so is [imath]F(n) = \sum_{d|n} f(d)g\left(\frac nd\right) [/imath] I have an understanding of how to prove it for a single function, but 2 functions and a summation added in confused me. |
1228236 | Bound on the eigenvalues of PSD matrix
Given That A and B are two PSD (positive semi-definite) real matrices and the following holds [imath] A \leq B [/imath] (meaning that [imath] B-A [/imath] is also PSD) can I bound the eigenvalues of A using the eigenvalues of B? Thanks | 1225138 | Comparing Eigenvalues of Positive Semidefinite Matrices
If [imath]B\succeq A[/imath], show that [imath]\lambda_n(B)\ge\lambda_n(A)[/imath], where [imath]\lambda_n[/imath] is the [imath]n[/imath]th largest eigenvalue. It is Theorem 6 in this paper but the proof is only given as "by characterization." Do they mean the min/max characterization of eigenvalues, [imath]\lambda_k=\min\max\frac{x^TAx}{x^Tx}[/imath] and if so, how is that applied? Thanks. |
1228287 | What is the origin of the use of [imath]\Pi[/imath] and [imath]\Sigma[/imath] for dependent function and dependent product types in type theory?
In the type theory I have read (e.g. homotopy type theory) I have seen the following notion used for dependent function types: [imath]\prod_{x : A} B(x)[/imath] and the following for dependent product types: [imath]\sum_{x : A} B(x)[/imath] I am curious as to the origin of this notation; given the use of [imath]\prod[/imath] for indexed products in algebra, I would have guessed that it would be used for the product type. It is also strange because I generally see independent products expressed as [imath]A \times B[/imath] (which is consistent with cartesian products in set theory), and coproducts expressed as [imath]A + B[/imath] (which is consistent with the [imath]\oplus[/imath] notation for categorical coproducts). Based on this, you would expect [imath]\sum_{x : A} B(x) = A + B[/imath] when [imath]B[/imath] is independent of [imath]x[/imath], not [imath]A \times B[/imath]. Does anyone know where these notations come from? | 458991 | Confusion about Homotopy Type Theory terminology
I've picked up the Homotopy Type Theory book for leisure. I'm comfortable with strongly typed languages and familiar with dependently typed languages and I enjoy topology, so I thought that the HoTT book was a good opportunity to learn some of the math underlying the type systems (and, by HoTT's reputation, a new way of looking at many other things I know.) I'd dabbled a bit in formal type theory previous to HoTT. I've long thought of (independent) types like this: Unit data types are types with only one value, isomorphic to: data Unity = Unity You can sum types, creating a type inhabited by as many values as A + B data Sum a b = Left a | Right b more commonly known as Either a b in programming circles. Then there are product types, which house as many types as A × B data Product a b = Both a b more commonly known as (a, b). Finally there are exponential types inhabited by as many values as there are mappings A → B (which is [imath]B^A[/imath]). These are more commonly known as functions, and defining them is my day job. The little algebraic type theory I know is founded in those assumptions. (I'm curious what's beyond exponential types in this progression, but that's besides the point.) The HoTT book starts out by defining the exponential type [imath]\rightarrow[/imath], the dependent function type [imath]\Pi[/imath], the product type [imath]\times[/imath], and the dependent pair type [imath]\Sigma[/imath]. The definitions all make sense, but I find the naming confusing. Function types seem pretty exponential to me: why are dependent function types (a generalization of [imath]\rightarrow[/imath]) denoted [imath]\Pi[/imath] ('product')? This doesn't seem like just a silly mistake: the product type [imath]\times[/imath] is generalized to a dependent product type [imath]\Sigma[/imath] ('summation'). Both types are "demoted" when they're generalized (exponential to product, product to sum). There's obviously some reasoning behind this naming, but it's going right over my head. Why does Homotopy Type Theory use this naming scheme? What's the isomorphism that I'm missing? |
1228667 | Showing that [imath]f_n\to f[/imath] a.e. [imath]\implies f_n\to f[/imath] almost uniformly where [imath]|f_n|\leq g\in L_1[/imath]
The following problem is from Carothers' Real Analysis: Suppose [imath]f_n[/imath] is a measurable sequence of functions such that [imath]|f_n|\leq g\in L_1[/imath] for all [imath]n[/imath]. Prove that [imath]f_n\to f[/imath] almost everywhere implies [imath]f_n\to f[/imath] almost uniformly. Thoughts/ Attempt: At first glance, this looks like a mix of the Dominated Convergence Theorem and Egorov's Theorem. A sequence of functions [imath](f_n)[/imath] is said to converge almost uniformly on [imath]D[/imath] if: [imath]\forall\epsilon>0\exists N\in\mathbb{N}:\forall n\geq N\implies \sup_{x\in D\setminus E}|f_n(x)-f(x)|<\epsilon \\\text{ (where $E\subset D$ is measurable and $m(E)<\epsilon$) }[/imath] It is clear that [imath]f\in L_1[/imath]. I tried a few things that didn't lead anywhere such as: [imath]f_n\to f \text{ a.e.} \implies \int f_n\to \int f \text{ a.e. }\implies |\int f_n-\int f|<\epsilon \text{ a.e. }\implies \int |f_n-f|<\epsilon \text{ a.e. }[/imath] But doesn't lead anywhere about showing almost uniform convergence. I think I have an idea on how to start now by somewhat imitating the proof of Egorov's Theorem, which defines a measurable set [imath]E(n,k)=\bigcup_{m=n}^{\infty}\left\{x\in D: |f_m(x)-f(x)|\geq \frac {1}{k}\right\}[/imath] and shows the measure of this set decreases to [imath]0[/imath] and then a subsequence [imath](n_k)[/imath] is chosen such that [imath]m(E(n_k,k))<\epsilon/2^k[/imath]. Then it is shown that [imath]f_n\to f[/imath] uniformly on [imath]D\setminus E[/imath], where [imath]E=\cup_{k=1}^{\infty}E(n_k,k)[/imath]. However I'm having trouble deciding how to define my set [imath]E(n,k)[/imath] given the hypotheses of the Dominated Convergence Theorem. Any ideas on how to proceed would be appreciated. Hints are preferred over full solutions. Thanks! | 83424 | Omitting the hypotheses of finiteness of the measure in Egorov theorem
I want to prove that if I omit the fact that [imath]\mu (X) < \infty[/imath] in Egorov theorem and place instead that our functions [imath]|f_n| <g[/imath] and [imath]g[/imath] is integrable, we still get the result of Egorov's theorem. Fix [imath]m[/imath] a natural number. I took [imath] w_{n} = |f_n-f|[/imath] and thus by DCT [imath]\int |f_{n} - f|[/imath] goes to zero. Then I took [imath]\bigcup_n {( w_{n} \geq 1/m)}[/imath]. I need its measure to be finite. Its measure is less than the sum of the measures of each [imath] w_n\geq 1/m)[/imath] varying [imath]n[/imath], and by Tchebychev, this is less than [imath]m\int|f_{n} - f|[/imath]. But I got stuck here. Any help is appreciated. Thanks! |
1228576 | Vector spaces with complex field as scalar.
Sorry for stating the question informally. If we have a vector space whose scalars are the field [imath]\mathbb{R}[/imath], if we change the field to be [imath]\mathbb{C}[/imath] and "adapt" the addition and scalar multiplication operations, does it necessarily yield a new vector space? Are there theorems developed in this manner or similar tests? | 634104 | Vector space of real vectors over field complex scalars.
Let [imath]V = \{(a_1,a_2,\ldots,a_n): a_i \in \mathbb{R}\ \text{ for } i = 1,2,\ldots,n\};[/imath] So [imath]V[/imath] is a vector space over [imath]\mathbb{R}[/imath]. Is [imath]V[/imath] a vector space over the field of complex numbers with the operations of coordinatewise addition and multiplication? I don't need an actual answer for this, as I just want to confirm that I understand what the question is asking. Paraphrasing: If [imath]V = \mathbb{R}^n[/imath] and [imath]\mathbb{F} = \mathbb{R}[/imath], then we know that [imath]V[/imath] is a vector space. If [imath]\mathbb{F} = \mathbb{C}[/imath], then is [imath]V[/imath] still a vector space (Is the set of real vectors [imath]v \in \mathbb{R}^n[/imath] over the field of complex scalars a vector space)? |
1229449 | Proof on sets [imath](A\cap B)\cup(A\cap \bar{B}) = A[/imath]
Original question : To Prove : [imath](A\cap B)\cup(A\cap \bar{B}) = A[/imath] My Response to it : We have, [imath](A\cap B)\cup(A\cap \bar{B}) = A[/imath] [imath]\Rightarrow (A \cap B) + (A \cap \bar{B}) - (A \cap B \cap A \cap \bar{B})[/imath] [imath]\Rightarrow (A \cap B) + (A \cap \bar {B}) - \emptyset[/imath] [imath]\Rightarrow (A \cap B) + (A \cap \bar {B})[/imath] [imath]\Rightarrow (A \cap B) + A \cap (U-B)[/imath] [imath]\Rightarrow (A \cap B) + (A \cap U) - (A \cap B)[/imath] [imath]\Rightarrow A[/imath] The identity I used above was: [imath](X \cup Y) = X + Y - (X \cap Y)[/imath] However, my professor says the identity I used is flawed and is wrong thus making my solution wrong too. Please provide your insight on what you think. | 1227555 | Show that if [imath]A[/imath] and [imath]B[/imath] are sets, then [imath](A\cap B) \cup (A\cap \overline{B})=A[/imath].
Show that if [imath]A[/imath] and [imath]B[/imath] are sets, then [imath](A\cap B) \cup (A\cap \overline{B})=A[/imath]. So I have to show that [imath](A\cap B) \cup (A\cap \overline{B})\subseteq A[/imath] and that [imath]A \subseteq(A\cap B) \cup (A\cap \overline{B})[/imath]. Lets begin with the first one: If [imath]x \in (A \cap B)[/imath] it means [imath]x \in A \wedge x \in B[/imath]. If [imath]x \in (A \cap \overline{B})[/imath] it means [imath]x \in A \wedge x \in \overline{B}[/imath]. And the second one: If [imath]x \in A[/imath] it means [imath]x \in (A \cap B)[/imath]. But here after I am confused. |
1229616 | Isomorphisms between finite fields of same characteristic
Here I am taking the definition of isomorphism to be an injective homomorphism. Suppose we have two finite fields of the same characteristic, [imath]\mathbb{F}_{p^n}[/imath] and [imath]\mathbb{F}_{p^m}[/imath] with [imath]m<n[/imath]. Is there always an isomorphism [imath]\phi:\mathbb{F}_{p^m}\rightarrow \mathbb{F}_{p^n}[/imath]? If [imath]m|n[/imath] then we could use the same construction [imath]\mathbb{F}_{p^m}[X]/(P)[/imath] where [imath]P[/imath] is an irreducible polynomial in [imath]\mathbb{F}_{p^m}[X][/imath] such that [imath]\deg(P)m=n[/imath]. Since this would produce a field of the same characteristic and cardinality, it would provide an isomorphism. Does anyone have any references on this topic? Thanks in advance. | 894667 | Inclusion of Fields whose order is a prime power
Blue was correct, I need to fix my understanding of this: Finite fields have cardinality of a prime order because they have a prime subfield that has finite characteristic. I do not completely understand why finite field of each characteristic is unique. Perhaps we look at a polynomial and express the finite field as a splitting field? And we use uniqueness of splitting fields? Then, why is it that [imath]\mathbb F_{p^m} \subseteq \mathbb F_{p^n}[/imath] if and only if [imath]m[/imath] divides [imath]n[/imath]? |
692588 | How can you show that [imath]\binom {n}{7}=\sum_{k=7}^n \binom {k-1} {6}[/imath]?
How can you show that [imath]\binom {n}{7}=\sum_{k=7}^n \binom {k-1} {6}[/imath]? This counts the number of subsets from [imath]\{1,2,3,\dots,n\}[/imath] having size [imath]7[/imath]. To me, the summation part counts subsets of size [imath]6[/imath]. Can someone please tell me how this is done? P.S. I don't know how to edit in the proper way. | 1229396 | Number of subsets of length 7
I have the following summation: [imath]\sum\limits_{k=7}^{n} {k-1\choose 6} [/imath] and apparently it counts the number of subsets of {1, 2, . . . , n} having size 7. Why is this? |
1230119 | What is the shortest/longest distance from [imath]9x^2 + 4y^2 = 36[/imath] to [imath](5,5)[/imath]?
What is the shortest/longest distance from [imath]9x^2 + 4y^2 = 36[/imath] to [imath](5,5)[/imath]? Using Langrange Multipliers, I've set up the standard equation with [imath]g(x,y) = (x/2)^2 + (y/3)^2 = 1[/imath] [imath]f(x,y) = (x-5)^2 + (y-5)^2[/imath] and [imath] \nabla f = -\lambda \nabla g.[/imath] This gives us [imath]2(x-5) = - \lambda x / 2[/imath] [imath](y-5) = - \lambda y / 9.[/imath] Solving for [imath]y[/imath] and [imath]x[/imath], I have [imath] y = 9x / (x + 4) [/imath] [imath] x = 4y / (9-y) .[/imath] But if I plug this value for [imath]y[/imath] into the original ellipse, I get [imath] x^4 + 8x^3 + 48x^2 - 32x - 64 = 0.[/imath] Somehow this doesn't seem quite right as it's now cumbersome to solve for [imath]x[/imath]. Where am I going wrong and is there a better approach? | 828803 | How to find the minimum/maximum distance of a point from elipse
I have the point [imath](1,-1)[/imath] and the ellipse [imath]x^2/9 + y^2/5 = 1 [/imath] How to find the minimum and maximum distance of the point from the ellipse ? from exploring the ellipse I know that [imath]a = 3[/imath] , [imath]b =\sqrt{5}[/imath] [imath] c = \sqrt{a^2-b^2} =\sqrt{9-5} = \sqrt{4}=2[/imath] the eccentricity of the ellipse is [imath]e=c/a = 2/3 [/imath] the center is [imath](0,0)[/imath] and the guides are [imath]x=3,~~x=-3,~~y=\sqrt{5},~~y=-\sqrt{5}[/imath] the focus points are : [imath](2,0)[/imath] and [imath](-2,0)[/imath] how from all of that do I find the requested in the question? |
1230675 | for which odd number n does 503 divide [imath]n^2+1[/imath]
The question is already mentioned in the title: For which odd number n does 503 divide [imath]n^2+1[/imath] [imath]503\cdot q=n^2+1[/imath] | 1228448 | Maybe is right [imath]\frac{n^2 + 1}{4k + 3} \notin \mathbb{Z}, n, k \in \mathbb{N}^{+}[/imath]
Prove or disprove [imath]\dfrac{n^2 + 1}{4k + 3} \notin \mathbb{Z}, n, k \in \mathbb{N}^{+}[/imath] I know if [imath]n^2 + 1[/imath] is prime if and only if [imath]n^2 + 1 \equiv 1 \pmod 4[/imath]. |
1230736 | Prove that the isomorphism between vector spaces and their duals is not natural
In preparation for an introductory talk on category theory, I recently spent some time thinking about natural transformations. The first example, or maybe the second, that everyone gives to motivate the concept of a natural transformation is the double dual: a vector space is naturally isomorphic to its double dual, and category theory makes this notion precise by saying that there is a natural isomorphism between the identity functor and the double dual functor [imath]\text{Vec}_k\to\text{Vec}_k[/imath]. At this point, whoever is giving the example cautions that the dual functor [imath]\text{Vec}_k^{\text{op}}\to\text{Vec}_k[/imath] is not naturally isomorphic to the identity functor, and this is because making an isomorphism between a vector space and its dual requires choosing a basis. But no one ever proves it! Implicitly, there is a "theorem" here to the following effect: "Theorem": There is no natural isomorphism between the identity functor [imath]\text{Vec}_k\to\text{Vec}_k[/imath] and the dual functor [imath]\text{Vec}_k^{\text{op}}\to\text{Vec}_k[/imath]. The problem with this "theorem" is that, to my knowledge, it doesn't make sense to talk about a natural transformation between a covariant and a contravariant functor. The obvious commutative diagram to write down, something like [imath] \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\la}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{ll} V & \ra{f} & W \\ \da{\eta_V} & & \da{\eta_W} \\ V^* & \la{f^*} & W^* \\ \end{array}, [/imath] will almost certainly not commute -- take, for instance, [imath]f=0[/imath]. This failure is caused by the contravariance of the dual functor, not its unnaturality. My question, then, is this: make precise the claim that there does not exist a natural isomorphism between the identity functor and the dual functor, and prove it. | 622589 | In categorical terms, why is there no canonical isomorphism from a finite dimensional vector space to its dual?
I've read in several places that one motivation for category theory was to be able to give precise meaning to statements like, "finite dimensional vector spaces are canonically isomorphic to their double duals; they are isomorphic to their duals as well, but not canonically." I've finally sat down to work through this, and - Okay, yes, it is easy to see that the "canonical isomorphism" from [imath]V[/imath] to [imath]V^{**}[/imath] is a functor that has a natural isomorphism (in the sense of category theory) to the identity functor. Also, I see that there is no way that the functor [imath]V\mapsto V^*[/imath] could have a natural isomorphism to the identity functor, because it is contravariant whereas the identity functor is covariant. My question amounts to: Is contravariance the whole problem? To elaborate: I was initially disappointed by the realization that the definition of natural isomorphism doesn't apply to a pair of functors one of which is covariant and the other contravariant, because I was hoping that the lack of a canonical isomorphism [imath]V\rightarrow V^*[/imath] would feel more like a theorem as opposed to an artifact of the inapplicability of a definition. Then I tried to create a definition of a natural transformation from a covariant functor [imath]F:\mathscr{A}\rightarrow\mathscr{B}[/imath] to a contravariant functor [imath]G:\mathscr{A}\rightarrow\mathscr{B}[/imath]. It seems to me that this definition should be that all objects [imath]A\in\mathscr{A}[/imath] get a morphism [imath]m_A:F(A)\rightarrow G(A)[/imath] such that for all morphisms [imath]f:A\rightarrow A'[/imath] of [imath]\mathscr{A}[/imath], the following diagram (in [imath]\mathscr{B}[/imath]) commutes: [imath]\require{AMScd}\begin{CD} F(A) @>m_A>> G(A)\\ @VF(f)VV @AAG(f)A\\ F(A') @>>m_{A'}> G(A') \end{CD}[/imath] This is much more stringent a demand on the [imath]m_A[/imath] than the typical definition of a natural transformation. Indeed, it is asking that [imath]m_A=G(f)\circ m_{A'}\circ F(f)[/imath], regardless of how [imath]f[/imath] or [imath]A'[/imath] may vary. Taking [imath]\mathscr{A}=\mathscr{B}=\text{f.d.Vec}_k[/imath], [imath]F[/imath] the identity functor and [imath]G[/imath] the dualizing functor, it is clear that this definition can never be satisfied unless [imath]m_V[/imath] is the zero map for all [imath]V\in\text{f.d.Vec}_k[/imath] (because take [imath]f[/imath] to be the zero map). In particular, it cannot be satisfied if [imath]m_V[/imath] is required to be an isomorphism. Is this the right way to understand (categorically) why there is no natural isomorphism [imath]V\rightarrow V^*[/imath]? As an aside, are there any interesting cases of some kind of analog (the above definition or another) of natural transformations from covariant to contravariant functors? Note: I have read a number of math.SE answers regarding why [imath]V^*[/imath] is not naturally isomorphic to [imath]V[/imath]. None that I have found are addressed to what I'm asking here, which is about how categories make the question and answer precise. (This one was closest.) Hence my question here. |
1230500 | Let [imath]p[/imath] be a prime. Why is [imath]{p^mn \choose p^m}[/imath], where [imath]p \nmid n[/imath], not divisible by [imath]p[/imath]?
Let [imath]p[/imath] be a prime. Why is [imath]{p^mn \choose p^m}[/imath], where [imath]p \nmid n[/imath], not divisible by [imath]p[/imath]? [imath]{p^mn \choose p^m} = \frac{(p^mn)!}{p^m!(p^mn-p^m)!} = \frac{p^mn(p^mn-1)(p^mn-2)\cdots(p^mn-p^m+1)}{p^m(p^m-1)(p^m-2)\cdots(p^m-p^m+1)}[/imath] I can see [imath]p^m[/imath] cancel out in the first term, but why is this not divisible by [imath]p[/imath]? | 480059 | A problem regarding the proof of [imath]{p^nk\choose p^n}\equiv k\mod p[/imath], where [imath]p\nmid k[/imath].
In this proof, there is a statement where: [imath](a+b)^{p^nk}\equiv (a^{p^n}+b^{p^n})^k\mod p[/imath] I understand this part. But then it expands both sides binomially, and compares coefficients of [imath]\displaystyle{a^{p^n(k-1)}b^{p^n}}[/imath] to conclude [imath]\displaystyle{{p^nk\choose p^n}}\equiv k\mod p[/imath]. If [imath]k<p[/imath], then none of the coefficients in the binomial expansion of [imath](a^{p^n}+b^{p^n})^k[/imath] are divisible by [imath]p[/imath]. I don't understand why comparing coefficients would give us the answer here. Thanks in advance! |
1230686 | Evaluating arithmetic sum using prime factorization
Please help! I have no idea how to start this problem / what to do to evaluate this. For m>0 , let f(m) = [imath]\sum_{r=1}^{m} \frac{m}{gcd(m,r)}[/imath] . Evaluate f(m) in terms of the prime factorization of m. | 210005 | How can I compute the sum of [imath] {m\over\gcd(m,n)}[/imath]?
[imath] \sum_{m =1}^n {m\over\gcd(m,n)}[/imath] For example, for 1 it is [imath]{1\over\gcd(1,1)} =1;[/imath] for 5 it is [imath]{1\over \gcd(1,5)}+{2\over \gcd(2,5)}+{3\over \gcd(3,5)}+{4\over \gcd(4,5)}+{5\over \gcd(5,5)}=\\ \frac11+\frac21+\frac31+\frac41+1 = \\ 1+2+3+4+1= \\ 11[/imath] |
1227207 | Find the minimum value of [imath]P=\frac{\sqrt{3(2x^2+2x+1)}}{3}+\frac{1}{\sqrt{2x^2+(3-\sqrt{3})x+3}}+\frac{1}{\sqrt{2x^2+(3+\sqrt{3})x+3}}[/imath]
Let [imath]x[/imath] be a real number. Find the minimum value of [imath]P=\frac{\sqrt{3(2x^2+2x+1)}}{3}+\frac{1}{\sqrt{2x^2+(3-\sqrt{3})x+3}}+\frac{1}{\sqrt{2x^2+(3+\sqrt{3})x+3}}[/imath] This is a problem from 2015 Vietnamese University Entrance Exam. I tried plotting and finding all local minima but that seems to be impossible as the derivative is too complicated. I guess this problem can only be solved by using inequalities. | 1215784 | Find minimum of [imath]P=\frac{\sqrt{3(2x^2+2x+1)}}{3}+\frac{1}{\sqrt{2x^2+(3-\sqrt{3})x +3}}+\frac{1}{\sqrt{2x^2+(3+\sqrt{3})x +3}}[/imath]
For [imath]x\in\mathbb{R}[/imath] find minimum of [imath]P[/imath]. [imath]P=\dfrac{\sqrt{3(2x^2+2x+1)}}{3}+\dfrac{1}{\sqrt{2x^2+(3-\sqrt{3})x +3}}+\dfrac{1}{\sqrt{2x^2+(3+\sqrt{3})x +3}}[/imath] Source : Viet Nam national test for high school student. I can't slove this problem. |
116350 | Continuous injective map [imath]f:\mathbb{R}^3 \to \mathbb{R}[/imath]?
How would you show that there is no continuous injective map [imath]f:\mathbb{R}^3 \to \mathbb{R}[/imath]? I tried the approach where if [imath]a \in \mathbb{R}[/imath] then [imath]\mathbb{R} \backslash \{a\} [/imath] is not clearly connected so there can't exist a continuous function otherwise [imath]\mathbb{R} \backslash \{a\} [/imath] would be connected? | 1808749 | Nonexistence of a continuous injection from [imath]\mathbb{R}^n[/imath] to [imath]\mathbb{R}[/imath], for all [imath]n \geq 2[/imath].
I'm trying to do the following exercise from my lecture notes: There does not exist a continuous injection from [imath]\mathbb{R}^n[/imath] to [imath]\mathbb{R}[/imath], for all [imath]n \geq 2[/imath]. I don't really know where to start, but I think it helps that the exercise is marked as harder than the following exercises before it: [imath]\mathbb{R}^n\setminus\{0\}[/imath] is connected, for all [imath]n \geq 2[/imath]. [imath]\mathbb{R}^n[/imath] is nonhomeomorphic to [imath]\mathbb{R}[/imath], for all [imath]n \geq 2[/imath]. |
1232325 | Is trace of regular representation in Lie group a delta function?
My major is physics. I need to use some tools in group theory, but I am really confused by the trace in compact infinite groups. The following is my question: In discrete group theory, the irreducible representation of identity group element [imath]e[/imath] is always an identity matrix. So the trace of [imath]e[/imath] under regular representation is the order of the group : [imath]\chi(e) = |G|[/imath]. I hope to get similar result for Lie group. For example, SO(2) has infinite number of irreducible representations, [imath]D^{(m)}[/imath], where [imath]m=0,\pm 1, \pm 2,\cdots [/imath]. All of them are 1-dimensional [imath]D^{(m)} = e^{im\phi}[/imath]. Here [imath]\phi[/imath] is the rotation angle. For element [imath]R(\phi)[/imath] in SO(2), the trace in [imath]m[/imath]th irreducible representation should be [imath]\chi(R(\phi))=e^{im\phi}[/imath], if we sum all these traces up, we get the trace in regular representation. So what is the trace of [imath]e[/imath] in the regular representation ? It seems to be infinite. | 1232328 | Is trace of regular representation in Lie group a delta function?
My major is physics. I need to use some tools in group theory, but I am really confused by the trace in compact infinite groups. The following is my question: In discrete group theory, the irreducible representation of identity group element [imath]e[/imath] is always an identity matrix. So the trace of [imath]e[/imath] under regular representation is the order of the group : [imath]\chi(e) = |G|[/imath]. I hope to get similar result for Lie group. For example, SO(2) has infinite number of irreducible representations, [imath]D^{(m)}[/imath], where [imath]m=0,\pm 1, \pm 2,\cdots [/imath]. All of them are 1-dimensional [imath]D^{(m)} = e^{im\phi}[/imath]. Here [imath]\phi[/imath] is the rotation angle. For element [imath]R(\phi)[/imath] in SO(2), the trace in [imath]m[/imath]th irreducible representation should be [imath]\chi(R(\phi))=e^{im\phi}[/imath], if we sum all these traces up, we get the trace in regular representation. So what is the trace of [imath]e[/imath] in the regular representation ? It seems to be infinite. |
1232388 | [imath]f(x)[/imath] is continuously differentiable on [imath][0,1][/imath], prove that [imath] (\int_0^1 f(x)dx)^2 \leq \frac {1} {12} \int_0^1 (f'(x))^2 dx [/imath]
[imath]f(x)[/imath] is a real function continuously differentiable on [imath][0,1][/imath]. [imath]f(0) = f(1) = 0[/imath] Prove that [imath] \left(\int_0^1 f(x)\mathrm{dx}\right)^2 \leq \frac {1} {12} \int_0^1 (f'(x))^2 \mathrm{dx} [/imath] | 1018659 | How prove this inequality [imath]\left(\int_{0}^{1}f(x)dx\right)^2\le\frac{1}{12}\int_{0}^{1}|f'(x)|^2dx[/imath]
Let [imath]f\in C^{1}[0,1][/imath] such that [imath]f(0)=f(1)=0[/imath]. Show that [imath]\left(\int_{0}^{1}f(x)dx\right)^2\le\dfrac{1}{12}\int_{0}^{1}|f'(x)|^2dx.[/imath] I think we must use Cauchy-Schwarz inequality [imath]\int_{0}^{1}|f'(x)|^2dx\ge \left(\int_{0}^{1}f(x)dx\right)^2[/imath] but this maybe is not useful to problem. The coefficient [imath]\dfrac{1}{12}[/imath] is strange. How find it ? |
1231189 | Is this module injective?
Here's the problem: (This problem is from Hungerford's Algebra Chapter 5 exercise 6.7 ) Let R be a principal ideal domain and [imath]p[/imath] a prime in [imath]R[/imath] and [imath]n[/imath] a positive integer. Then [imath]R/(p^{n})[/imath] is an unitary [imath]R/(p^{n})[/imath]-module with [imath](r+(p^{n}))(s+(p^{n}))=rs+(p^{n})[/imath]. I attempt to demonstrate that the [imath]R/(p^{n})[/imath]-module [imath]R/(p^{n})[/imath] is injective as following steps: (1) I notice that the ring [imath]R/(p^{n})[/imath] is a principal ideal ring since it is a quotient ring of [imath]R[/imath] which is a PID. And by this fact, I prove that each proper ideal of [imath]R/(p^{n})[/imath] is generated by [imath]p^{i}+(p^{n})[/imath] with [imath]0<i<n[/imath]. (2) Now, in order to prove that module [imath]R/(p^{n})[/imath] is injective, it is sufficient to prove that for each ideal [imath]I=(p^{i}+(p^{n}))[/imath] of the ring [imath]R/(p^{n})[/imath] the module homomorphism [imath]f: I\rightarrow R/(p^{n})[/imath] can be extended to a homomorphism [imath]R/(p^{n})\rightarrow R/(p^{n})[/imath]. The module [imath]R/(p^{n})[/imath] is injective? how to extend this module homomorphism [imath]f: I\rightarrow R/(p^{n})[/imath]? Anyway, any and all help is appreciated. Thanks. | 947940 | Any artinian chain ring is self-injective.
Let [imath]R[/imath] be an artinian chain (uniserial) ring. I want to prove that [imath]R[/imath] is self-injective, that is, any [imath]\alpha:I\rightarrow R[/imath] right [imath]R[/imath]-module homomorphism can be extended to a homomorphism [imath]\tilde{\alpha}:R\rightarrow R[/imath] where [imath]I[/imath] is a right ideal of [imath]R[/imath]. Let [imath]J[/imath] be the jacobson radical of [imath]R[/imath]. We knov that all right ideals of [imath]R[/imath] form a chain [imath]R\supset J\supset J^2\supset ...\supset J^{m-1}\supset J^m=0[/imath] and [imath]J^k=p^kR[/imath] for some [imath]p\in J\setminus J^2[/imath] and any [imath]k=1,2,...[/imath] since [imath]R[/imath] is an artinian chain ring. So I need to show that any [imath]\alpha:p^k\rightarrow R[/imath] right [imath]R[/imath]-module homomorphism can be extended to a homomorphism [imath]\tilde{\alpha}:R\rightarrow R[/imath]. How can i find the extension. I need some help. Thanks a lot... |
1233263 | Evaluate the following series using elementary methods.
[imath]\frac{1} {\sqrt 1} + \frac{1} {\sqrt2} + \dots + \frac{1} {\sqrt{ 100}}[/imath] Just bear in mind that I'm going to say the solution to a person of low education. So please provide creative hints or solutions. I don't want to use derivatives or other limit-related stuff. Thanks in advance. EDIT:Sorry for posting a duplicate question but I have another question now.Can you prove that the series above does not have rational answer? | 516846 | How to find [imath]\lfloor 1/\sqrt{1}+1/\sqrt{2}+\dots+1/\sqrt{100}\rfloor [/imath] without a calculator?
[imath] \left\lfloor\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +\dots+ \frac{1}{\sqrt{100}}\right\rfloor =\,? [/imath] I rationalized the denominator and then I think I should somehow group the numbers, but i don't know how. Thanks in advance! |
291349 | Looking for a simple proof that groups of order [imath]2p[/imath] are up to isomorphism [imath]\mathbb{Z}_{2p}[/imath] and [imath]D_{p}[/imath] .
I'm looking for a simple proof that up to isomorphism every group of order 2p (p prime) is either [imath]\mathbb{Z}_{2p}[/imath] or [imath]D_{p}[/imath] (The Dihedral group of order 2p). I should note that by simple I mean short and elegant and not necessarily elementary. So feel free to use tools like Sylow Theorems, Cauchy Theorem and similar stuff. Thanks a lot! | 2572450 | Are all groups of order [imath]2\cdot p[/imath] abelian?
Are all groups of order [imath]2\cdot p[/imath] abelian? I would like to prove that Dihedral group [imath]D(p)[/imath] is the only non-abelian group of that order. [imath]p[/imath] is any prime [imath]>2[/imath]. Thanks for advice. |
1233904 | unknown permutation
Can someone help on this. I'm stuck on this part. I am trying to find a permutation [imath]\sigma[/imath] such that [imath]\sigma(1,2)(3,4)\sigma^{-1} = (5,6)(3,1)[/imath]. By a particular theorem, I know I can have this one [imath] \sigma (1,2)\sigma^{-1}\cdot \sigma(3,4)\sigma^{-1} =(5,6)(3,1)[/imath]Then, [imath](\sigma(1) , \sigma(2)) = (5,6) [/imath] and [imath](\sigma(3) , \sigma(4)) = (3,1) [/imath] Am I right on doing this? [imath]\sigma(1) = 5 \qquad \mbox{and} \qquad \sigma(2)=6[/imath] and [imath]\sigma(3) = 3 \qquad \mbox{and} \qquad \sigma(4)=1[/imath] I don't really know what to do next. I thought of making them as one permutation but it doesn't really makes sense to me. Please help. | 1232082 | Finding a permutation to satisfy given condition
In [imath]S_n[/imath] with n=10, find a permutation [imath]a[/imath] such that [imath]axa^{-1} = y[/imath] if [imath]x=(1,2)(3,4)[/imath] and [imath]y=(5,6)(3,1)[/imath] I don't know how to start doing this. I read something like I need to get, for this case, [imath](a(1),a(2)) \qquad \mbox{ and} \qquad (a(3),a(4))[/imath] but I am not sure where will I get their equivalents. Also, will those be my permutation [imath]a[/imath]? Please help. I have problems very similar to this. |
1233623 | Consider the following system of differential equations
Consider the following model. Find the general solution [imath]\frac{dx}{dt}=-2y[/imath], [imath]\frac{dy}{dt}=8x[/imath] So far this is what I have: [imath]\frac{dx}{dt}=-2y[/imath] [imath]x'=-2y[/imath] [imath]y=-\frac{x'}{2}[/imath] [imath]y=-\frac{x''}{2}[/imath] and then [imath]\frac{dy}{dt}=8x[/imath] [imath]y'=8x[/imath] [imath]-\frac{x''}{2}=8x[/imath] [imath]-x''=16x[/imath] [imath]x''+16x=0[/imath] Is this correct so far? What step should I take next? | 1233783 | Differential Equations: solve the system
Solve the following system: [imath]dx/dt=-.2(y-2)[/imath] [imath]dy/dt=.8(x-2)[/imath] This is what I have so far, but I got stuck.. [imath]\begin{eqnarray} dx/dt&=&-.2y-.4\\ x'&=&-.2y-.4\\ x'+.4&=&-.2y\\ y&=&-x/2'-2\\ y&=&-x''/2-2\\ \\ dy/dt&=&.8(x-2)\\ dy/dt&=&.8x-1.6\\ y'&=&.8x-1.6\\ -x''-2&=&.8x-1.6\\ -x''&=&.8x+.4\\ x''+16x+.8&=&0 \end{eqnarray}[/imath] when I used the quadratic formula I got stuck. Is this correct so far? Can someone finish it off for me? |
1224865 | For a cyclotomic polynomial with p = 5, prove that [imath]G(Q(\alpha)/Q)[/imath] is isomorphic to [imath]C_4[/imath]
I'm currently self-studying and would like to understand automorphisms in relation to splitting fields. Could someone please help me with this question? I'm trying explicitly to write down each automorphism of the splitting field [imath]Q(\alpha)[/imath] (with [imath]Q[/imath] fixed) in terms of a single generator for the group. I believe that the automorphism - [imath]\psi_{\alpha, \alpha^2}[/imath] will be of use here. [imath]\alpha[/imath] is a root. Thanks in advance. | 894406 | Showing Galois Group is Abelian
I'm having trouble showing that [imath]\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})[/imath] is Abelian. First I want to be able to show that [imath]\mathbb{Q}(\zeta_n)/\mathbb{Q}[/imath] is Galois, but I'm also not sure how to do this. Any help is appreciated! |
1233408 | Field extension in rational functions
I'm facing the following problem: Let [imath] F [/imath] be a field, and let [imath] F(x) [/imath] denote all rational functions over [imath] F [/imath] (functions of form [imath]\frac{P(x)}{Q(x)}[/imath], where [imath] P,Q[/imath] are polynomials over [imath] F [/imath]). Find the degree [imath] \left[ F(x): F \left( x^2 + \frac{1}{x^2}\right)\right] [/imath] Actually it's a homework exercise and I'm not sure what [imath] F \left( x^2 + \frac{1}{x^2}\right) [/imath] is supposed to mean. Is it the smallest field contained in [imath] F(x) [/imath] containing [imath] F \cup \{x^2 + \frac{1}{x^2}\} [/imath]? If so, how do I find the degree of the extension? I'm not even sure where to begin. | 1230287 | Degree of field extension [imath]F(x) / F(x^2 + 1 / x^2)[/imath]
Let [imath]y=\frac{x^4+1}{x^2} \in F(x)[/imath]. Then [imath]g(x)=0[/imath] for the polynomial [imath]g(s) = (s^4+1)-ys^2[/imath]. How to show that it is the minimal polynomial over the field [imath]F(y)[/imath]? |
1213089 | Let [imath]\gcd(m,n)=1[/imath], prove that [imath]m\mathbb Z/mn\mathbb Z[/imath] is isomorphic to [imath]\mathbb Z/n\mathbb Z[/imath]
Let [imath]\gcd(m,n)=1[/imath], prove that [imath]m\mathbb Z/mn\mathbb Z[/imath] is isomorphic to [imath]\mathbb Z/n\mathbb Z[/imath] | 547826 | If [imath]\gcd(m,n) =1, [/imath] then [imath]m\mathbb{Z}/ mn\mathbb{Z} \cong \mathbb{Z}/(n)[/imath]
If [imath]\gcd(m,n) =1, [/imath] then [imath]m\mathbb{Z}/ mn\mathbb{Z} \cong \mathbb{Z}/(n)[/imath]. Here is my proof to the a/m problem. But I did not make use of [imath]\gcd(m,n) =1.[/imath] Could anyone hence point out and explain the mistakes in my proof? Thank you. [imath]\text{Define}\ \phi: m\mathbb{Z}/ mn\mathbb{Z} \to \mathbb{Z}/(n)[/imath] by [imath]\phi(ms + mn\mathbb{Z})= [s']_n, \text{where} \ s \in [s']_n \ \text{and} \ s' \in \{0,...,n-1\}[/imath] [imath]\phi[/imath] is well-defined: [imath]mx + mn\mathbb{Z} = my + mn\mathbb{Z}[/imath] [imath]\implies mx = my + mr,[/imath] for some [imath]r\in \mathbb{Z}[/imath] [imath]\implies n | x-y[/imath] [imath]\implies [x]_n = [x']_n= \phi(mx + mn\mathbb{Z}) = [y]_n = [y']_n= \phi(my+ mn\mathbb{Z})[/imath] [imath]\phi[/imath] is injective: It suffices to prove that [imath]\ker(\phi) =\{mn\mathbb{Z}\}.[/imath] Since [imath]\phi(0+mn\mathbb{Z}) = [0]_n, \ker(\phi) \neq \emptyset.[/imath] Given [imath]x = ms+mn\mathbb{Z} \in \ker(\phi),\ [/imath] [imath] \phi(x)= [s]_n = [0]_n[/imath] [imath]\text{hence}\ s = nk, \text{for some}\ k \in \mathbb{Z}.[/imath] Clearly, [imath]mnk + mn\mathbb{Z}\subseteq mn\mathbb{Z}.[/imath] Given [imath]y \in mn\mathbb{Z},[/imath] let [imath]y = mnk + (mnc -mnk),[/imath] for some [imath]c \in \mathbb{Z}.[/imath] Then [imath]y \in mnk + mn\mathbb{Z},[/imath] i.e. [imath]mnk + mn\mathbb{Z} = mn\mathbb{Z} [/imath] and hence [imath]\ker(\phi) =\{mn\mathbb{Z}\}.[/imath] [imath]\phi[/imath] is homomorphism: Let [imath]a = ms + mn\mathbb{Z}, a' = ms'+ mn\mathbb{Z} \in m\mathbb{Z}/mn\mathbb{Z}. [/imath] Then, [imath]\phi(a+a')= \phi(m(s+s') + mn\mathbb{Z}) [/imath] [imath]\exists r \in \{1,...,n-1\}[/imath] such that [imath][r]_n = [s+s']_n[/imath] and so [imath]\phi(m(s+s') + mn\mathbb{Z}) = \phi(mr + mn\mathbb{Z}) = [r]_n = \phi(a) +_n \phi(a').[/imath] [imath]\phi[/imath] is surjective: [imath]\text{Given} \ [r]_n \in \mathbb{Z}/(n),\ \text{let} \ x= mr + mn\mathbb{Z}. \ \text{Then} \ \phi(x) =[r]_n[/imath] |
105877 | Scaling property of Fourier series and Fourier Transform
This question about the intuition behind the scaling property of the Fourier transform made me wonder about the corresponding notion for a Fourier series. The Fourier transform of [imath]f(ax)[/imath] is [imath]\frac{1}{|a|} \mathcal{F(\frac{u}{a})}[/imath]. If [imath]a>1[/imath] then the graph of [imath]f(ax)[/imath] is [imath]f[/imath] compressed and so its Fourier transform has frequencies that are higher. However they are scaled down in magnitude. On the other hand take [imath]f(x)= \cos x[/imath]. Its fourier series is trivially itself, with the coefficient of [imath]\cos x[/imath] being [imath]1[/imath]. Scaling it to [imath]f(ax) = \cos ax[/imath] where [imath]a>1[/imath] is an integer still has a trivial fourier series, and the coefficient of [imath]\cos ax[/imath] is [imath]1[/imath]. This is unlike the Fourier transform where the "coefficients" of each frequency get scaled as the formula shows: [imath]\frac{1}{|a|} \mathcal{F(\frac{u}{a})}[/imath]. Is there a conceptual way to explain this discrepancy? | 1457131 | On a property of the Fourier transform.
There's a simple property of the Fourier transform on [imath]\mathbb{R}^n[/imath] that I'm having trouble establishing. Let [imath]\lambda > 0[/imath]. I would like to show that [imath](D_\lambda f)^\hat{}(\xi) = \lambda^{-n}(D_{\lambda^{-1}}\hat{f})(\xi)[/imath], where [imath]D_{\lambda}f(x) := f(\lambda x)[/imath]. The substitution [imath]u = \lambda^{-1}x[/imath] gives me the following: [imath]\lambda^{-n}(D_{\lambda^{-1}}\hat{f})(\xi)[/imath] = [imath]\lambda^{-n} \int_{\mathbb{R}^n} f(x)e^{-2\pi i x \cdot \lambda^{-1}\xi}dx[/imath] = [imath]\lambda^{1-n}\int_{\mathbb{R}^n} f(\lambda u)e^{-2\pi i u \cdot \xi}du[/imath] = [imath]\lambda^{1-n} (D_\lambda f)^\hat{}(\xi)[/imath], which isn't quite what I want. What am I doing wrong here? |
1234222 | A∩B∩C=∅, then A∩B=∅,A∩C=∅ or B∩C=∅
Sorry everyone, I made a mistake in my question itself for my previous posting: Show that [imath]A∩B∩C= ∅[/imath] is only true when [imath]A∩B = ∅, A∩C = ∅[/imath] or [imath]B∩C = ∅[/imath] or show a counterexample. It should have been: If A∩B∩C=∅, then A∩B=∅,A∩C=∅ or B∩C=∅. Prove or show counterexample. So false, if I take A={1,2}, B={2,3}, and C={1,3}. then A∩B∩C=∅, but A∩B={2},A∩C={1},B∩C={3}. Thus leads to contradiction. Is this correct? By the way, can someone explain to me the difference in the prove when the question is or and and. | 1234178 | Show that [imath]A∩B∩C= ∅[/imath] is only true when [imath]A∩B = ∅, A∩C = ∅[/imath] or [imath]B∩C = ∅[/imath] or show a counterexample.
Show that [imath]A∩B∩C= ∅[/imath] is only true when [imath]A∩B = ∅, A∩C = ∅[/imath] or [imath]B∩C = ∅[/imath] or show a counterexample. My answer: True, Let set A={a,b,c,...}, set B={1,2,3,...} and set C={-1,-2,-3...}. Then there is not common elements in A, B or C. Thus, A∩B∩C= ∅. A∩B = ∅, A∩C = ∅ or B∩C = ∅. Is my prove right? |
1232040 | Determine the units of the ring [imath]A= \mathbb Z[X]/(X^3)[/imath] and the structure of the group [imath]A^*[/imath]
Determine the units of the ring [imath]A= \mathbb Z[X]/(X^3)[/imath] and the structure of the group [imath]A^*[/imath] I've only managed to show that the free coefficient of any unit in [imath]A[/imath] is a unit in [imath]\mathbb Z[/imath]. | 74782 | Units in quotient ring of [imath]\mathbb Z[X][/imath]
An exercise from Dummit & Foote: Determine the units of the ring [imath]A = \mathbb{Z}[X]/(X^{3})[/imath] and the structure of the unit group [imath]A^{\times}[/imath]. Help would be great. Thanks! |
1233756 | a question about the evaluation of integral
Let [imath]\alpha:[0,1] \to R[/imath] be the Cantor function. Evaluate [imath]\int_{0}^{1}xd\alpha [/imath]and [imath]\int_{0}^{1}x^2d\alpha.[/imath] I know that the Cantor function is continuous and monotone increasing, how can I evaluate the integral above using the properties of [imath]\alpha[/imath]? Can someone help me solve this question? | 1139217 | Calculate integral with cantor measure
Calculate the integral [imath]\int_{[0,1]}x^2d\mu_F[/imath] where F is the cantor function. Use the following hints about the cantor function: [imath]F(1-x)=1-F(x)[/imath] [imath]F(\frac x 3)=\frac{F(x)}{2}\quad\forall x\in[0,1][/imath] [imath]F(0)=0[/imath] I thought that [imath]\int_{[0,1]}x^2d\mu_F=\int_{[1,0]}(1-x)^2d\mu_{F}=\int_{[0,1]}x^2d\mu_{1-F(x)}[/imath] but here I'm stuck and I don't know how to continue calculating this integral. Furthermore, how do we use the second and third properties when given the cantor function above? |
1234593 | Continued fraction for [imath][1,2,3,4,5,6,\dots][/imath]
Any continued fraction that does not terminate or repeat can't be rational or a quadratic irrational. It is not hard to write something that does not fit these two categories. Can we still get a closed form for something like: [imath] [1,2,3,4,5, \dots ] = 1 + \cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{\dots}}}[/imath] | 69519 | Closed form for a pair of continued fractions
What is [imath]1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cdots}}}[/imath] ? What is [imath]1+\cfrac{2}{1+\cfrac{3}{1+\cdots}}[/imath] ? It does bear some resemblance to the continued fraction for [imath]e[/imath], which is [imath]2+\cfrac{2}{2+\cfrac{3}{3+\cfrac{4}{4+\cdots}}}[/imath]. Another thing I was wondering: can all transcendental numbers be expressed as infinite continued fractions containing only rational numbers? Of course for almost all transcendental numbers there does not exist any method to determine all the numerators and denominators. |
1235200 | How to integrate [imath]\frac{1}{(1+a\cos x)}[/imath] from [imath]-\pi[/imath] to [imath]\pi[/imath]
How to solve the following integration?[imath]\int_{-\pi}^\pi\frac{1}{1+a \cos x}[/imath] | 263397 | How to evaluate [imath]\int_0^{2\pi} \frac{d\theta}{A+B\cos\theta}[/imath]?
I'm having a trouble with this integral expression: [imath]\int_0^{2\pi} \frac{d\theta}{A+B \cos\theta}[/imath] I've done this substitution: [imath]t= \tan(\theta/2)[/imath] and get: [imath]\displaystyle \cos\theta= \frac{1-t^2}{1+t^2}[/imath] and [imath]\displaystyle d\theta=\frac{2}{1+t^2}dt[/imath] where [imath]\displaystyle \cos^2\theta/2=\frac{1}{1+t^2}[/imath] then the integral becomes: [imath]\int\frac{2 \, dt}{(A-B)t^2+(A+B)}= \sqrt\frac{A+B}{A-B} \arctan \left(\left(\sqrt\frac{A+B}{A-B} \right) t\right)[/imath] However, I'm not sure about the new limits since [imath]\tan[/imath] has period [imath]\pi[/imath] so what I have to do at this point to decide the new limits? And of course if you find some mistake in what I've done before, please let me know! |
1235456 | A problem about an algebraic number
Show that [imath]2^{\frac{1}{2}}+5^{\frac{1}{3}}[/imath] is algebraic over [imath]\mathbb{Q}[/imath] of degree [imath]6[/imath]. Can I just construct [imath]x=2^{\frac{1}{2}}+5^{\frac{1}{3}}[/imath], [imath]x-2^{\frac{1}{2}}=5^{\frac{1}{3}}[/imath], [imath]x^3-3*2^{\frac{1}{2}}x^2+6x-2\frac{1}{2}=5[/imath], [imath]x^3+6x-5=\sqrt{2}(3x^2+2)[/imath], [imath](x^3+6x-5)^2=2(3x^2+2)^2[/imath], [imath]x^6+12x^4-10x^3+36x^2-60x+25=18x^4+24x^2+8[/imath], [imath]x^6-6x^4-10x^3+12x^2-60x+17=0[/imath]. Does it show what is required? Then, [imath]2^{\frac{1}{2}}+5^{\frac{1}{3}}[/imath] is algebraic over [imath]\mathbb{Q}[/imath] of degree 6. | 1131350 | The degree of [imath]\sqrt{2} + \sqrt[3]{5}[/imath] over [imath]\mathbb Q[/imath]
I know that the degree is at most [imath]6[/imath], since [imath]\sqrt{2} + \sqrt[3]{5} \in \mathbb Q(\sqrt{2}, \sqrt[3]{5})[/imath], which has degree [imath]6[/imath] over [imath]\mathbb Q[/imath]. I'm trying to construct a polynomial with root [imath]\sqrt{2} + \sqrt[3]{5}[/imath] and coefficients in [imath]\mathbb Q[/imath] of degree [imath]6[/imath], and then show that it is irreducible over [imath]\mathbb Q[/imath]. I managed to find that it is a root of the polynomial [imath]x^6 - 6x^4 - 10x^3 +12x^2 - 60x +17[/imath]. This is where I run in to some problems. I don't know how to show that this is irreducible over [imath]\mathbb Q[/imath] (or [imath]\mathbb Z[/imath]). The only criteria we have learned is Eisenstein's criteria, which clearly does not apply here. How else can I show that the degree of this number over [imath]\mathbb Q[/imath] is [imath]6[/imath]? |
1232956 | Property of free submodules for a module over a PID
This question was asked here and remains without solution. It's possible to produce an example of an integral domain [imath]R[/imath] and a free [imath]R[/imath]-module [imath]M[/imath] with free submodules [imath]L, L'[/imath] such that [imath]L+L'[/imath] is not free. We can take [imath]R=M=K[x,y][/imath] , [imath]L=<x>[/imath] , [imath]L'=<y>[/imath]. If [imath]R[/imath] is a PID and [imath]M[/imath] is free [imath]R[/imath]-module, then for every pair of submodules [imath]L, L'[/imath] of [imath]M[/imath] we know that [imath]L+L'[/imath] is free. My question is the following. If [imath]R[/imath] is a PID and [imath]M[/imath] is an [imath]R[/imath]-module, is it true that [imath]L+L'[/imath] is free whenever [imath]L[/imath] and [imath]L'[/imath] are free submodules of [imath]M[/imath]? | 1219987 | Sum of free submodules of a module over a PID
It's possible to produce an example of an integral domain [imath]R[/imath] and a free [imath]R[/imath]-module [imath]M[/imath] with free submodules [imath]L, L'[/imath] such that [imath]L+L'[/imath] is not free. We can take [imath]R=M=K[x,y][/imath] , [imath]L=\left<x\right>[/imath] , [imath]L'=\left<y\right>[/imath]. If [imath]R[/imath] is a PID and [imath]M[/imath] is free [imath]R[/imath]-module, then for every pair of submodules [imath]L, L'[/imath] of [imath]M[/imath] we know that [imath]L+L'[/imath] is free. My question is the following. If [imath]R[/imath] is a PID and [imath]M[/imath] is an [imath]R[/imath]-module, is it true that [imath]L+L'[/imath] is free whenever [imath]L[/imath] and [imath]L'[/imath] are free submodules of [imath]M[/imath]? Thanks! |
1235650 | Find all ring homomorphisms from [imath]\Bbb Q[/imath] to [imath]\Bbb R[/imath]
My question is find all homomorphism [imath] f: \Bbb Q \to \Bbb R[/imath]. I think I should use ring isomorphism theorem to do this problem, but I just don't know how to do this. | 924060 | Ring homomorphisms from [imath]\Bbb Q[/imath] into a ring
Let [imath]A[/imath] be a ring. I'm trying to prove that there is only one ring homomorphism (different from the zero one) from [imath]\Bbb Q[/imath] into [imath]A[/imath] or there are no ring homomorphisms between [imath]\Bbb Q[/imath] and [imath]A[/imath]. I have proven that if [imath]A=\Bbb Z[/imath] or [imath]A=\Bbb R[/imath] then there are no ring homomorphisms between [imath]A[/imath] and [imath]\Bbb R[/imath] (not taking into account the zero one) but I don't know how to prove it for any [imath]A[/imath]. |
1235569 | Prove by strong induction that [imath]2^n[/imath] divides [imath]p_n[/imath] for all integers n ≥ 1
Let [imath]p_1 = 4[/imath], [imath]p_2 = 8[/imath], and [imath]p_n = 6p_{n−1} − 4p_{n−2}[/imath] for each integer [imath]n ≥ 3[/imath]. Prove by strong induction that [imath]2^n[/imath] divides [imath]p_n[/imath] for all integers [imath]n ≥ 1[/imath] I got up to the base step where by you prove for [imath]p_3[/imath] but unsure about the strong induction step | 1229554 | Proving by strong induction for a sequence of integers, [imath]2^n[/imath] divides term [imath]n[/imath]
Provided the following sequence of integers [imath]t_1, t_2, t_3[/imath],... is defined as: [imath]t_1 =4, t_2 =8[/imath] and [imath]t_n= [/imath] $ 6t_n[imath]_-[/imath]_1$ - $4t_n[imath]_-[/imath]_2$ for all integers [imath]n \geq 3[/imath] How do we prove that [imath]2^n[/imath] divides [imath]t_n[/imath] for all integers [imath]n \geq 1[/imath] by strong induction? So far, I am having problems with the induction step as I am still trying to grasp the concept of strong induction, but this is what I came up with so far: Base step: [imath]t_3 = 32[/imath], which is divisible by 8; and [imath]t_4 = 160[/imath], which is divisible by 16. [imath]\therefore[/imath] Statement is true when [imath]n=3[/imath] and [imath]n=4[/imath]. Induction step: Assuming [imath]k=n[/imath] and the statement is true for [imath]3,4,...k[/imath]. Prove that [imath]2^k[/imath] divides [imath]t_k[/imath] for [imath]k+1[/imath]: [imath]t_3 = ((2^3 - 2)(2^3) - (2^2)(2^2)) = 2^6 - 2^4 - 2^4 [/imath]; dividing this by [imath]2^3[/imath] gives [imath]2^3 - 2^2[/imath] which would be an integer result. [imath]t_4 = ((2^3 - 2)(2^5) - (2^2)(2^3)) = 2^8 - 2^6 - 2^5 [/imath]; dividing this by [imath]2^4[/imath] gives [imath]2^4 - 2^2 - 2[/imath] which would also be an integer result. [imath]\therefore[/imath] By the logic above, the statement also holds for [imath]k-1[/imath], [imath]k[/imath] as well as [imath]k+1[/imath], thus as required to prove by strong induction. I believe this is a fundamentally flawed way of trying to prove by strong induction, but this is the only way I can possibly think of. Can someone show a correct method of solving this? |
1236256 | Integrating [imath]\int x \sin \sqrt x dx[/imath] and substituting [imath]u=\sqrt x[/imath]
Is it wrong to substitute [imath]u=\sqrt x[/imath] when integrating? Here's what I mean, I have to integrate: [imath]\int x \sin \sqrt x dx[/imath]. I defined [imath]u=\sqrt x[/imath] so it's: [imath]\int u^2\sin u du=2u{\cdot}\sin\left(u\right)+\left(2-{u}^{2}\right){\cdot}\cos\left(u\right)+{C}[/imath] But when placing back [imath]\sqrt x[/imath] I get a totally different result from the integration of the original expression: http://www.integral-calculator.com/#expr=xsinsqrtx [imath]2{\cdot}\left(\cos\left(\sqrt{x}\right){\cdot}\left(6{\cdot}\sqrt{x}-{x}^{\frac{3}{2}}\right)+\sin\left(\sqrt{x}\right){\cdot}\left(3x-6\right)\right)+{C}[/imath] Shouldn't both expressions be equal and why aren't they? | 428415 | How to integrate [imath]\int x\sin {(\sqrt{x})}\, dx[/imath]
I tried using integration by parts twice, the same way we do for [imath]\int \sin {(\sqrt{x})}[/imath] but in the second integral, I'm not getting an expression that is equal to [imath]\int x\sin {(\sqrt{x})}[/imath]. I let [imath]\sqrt x = t[/imath] thus, [imath]\int t^2 \cdot \sin({t})\cdot 2t dt = 2\int t^3\sin(t)dt = 2[(-\cos(t)\cdot t^3 + \int 3t^2\cos(t))] = 2[-\cos(t)\cdot t^3+(\sin(t)\cdot 3t^3 - \int 6t \cdot \sin(t))]][/imath] which I can't find useful. |
1236570 | Solve congruence using fermat's theorem
Hi I am given this problem and I am supposed to use fermat's theorem. Here is it is: Prove that [imath]24^{31} \equiv 23^{32} \pmod{19}[/imath] We are supposed to solve it by setting up the congruence like this: [imath]24^{31} \equiv x \pmod{19}[/imath] [imath]23^{32} \equiv x \pmod{19}[/imath] I am just confused how to break it down. I was hoping someone can guide me through this question | 1224268 | Prove congruence using fermat's thm
I was given this problem (I have to prove) and not sure if I use fermat's theorem [imath]24^{31} ≡ 23^{32} (mod 19)[/imath] If I do use fermat's is this right: I would do the LHS first: [imath]24^{18}·24^{13} ≡ 1·24·24^{12}[/imath] RHS: [imath]23^{18}·23^{14} ≡ 1·23^{14}[/imath] I am not sure where to from here or if it is correct thus far. I have looked at some other posts but they have not helped with this problem. |
1236510 | Constructive Expectation
For any subset [imath]S\subseteq\{1,2,\ldots,15\}[/imath], call a number [imath]n[/imath] an anchor for [imath]S[/imath] if [imath]n[/imath] and [imath]n+\#(S)[/imath] are both elements of [imath]S[/imath]. For example, [imath]4[/imath] is an anchor of the set [imath]S=\{4,7,14\}[/imath], since [imath]4\in S[/imath] and [imath]4+\#(S) = 4+3 = 7\in S[/imath]. Given that [imath]S[/imath] is randomly chosen from all [imath]2^{15}[/imath] subsets of [imath]\{1,2,\ldots,15\}[/imath] (with each subset being equally likely), what is the expected value of the number of anchors of [imath]S[/imath]? Mila has four ropes. She chooses two of the eight loose ends at random (possibly from the same rope) and ties them together, leaving six loose ends. She again chooses two of these six ends at random and joins them, and so on, until there are no loose ends. At this point she has somewhere between one and four loops of rope (inclusive). Find the expected value of the number of loops Mila ends up with. I don't really know how to start, but thanks for helping. | 905663 | What is the expected value of the number of anchors of [imath]S[/imath]?
For any subset [imath]S\subseteq\{1,2,\ldots,15\}[/imath], call a number [imath]n[/imath] an anchor for [imath]S[/imath] if [imath]n[/imath] and [imath]n+ |S|[/imath] are both elements of [imath]S[/imath]. For example, [imath]4[/imath] is an anchor of the set [imath]S=\{4,7,14\}[/imath], since [imath]4\in S[/imath] and [imath]4+ |S| = 4+3 = 7\in S[/imath]. Given that [imath]S[/imath] is randomly chosen from all [imath]2^{15}[/imath] subsets of [imath]\{1,2,\ldots,15\}[/imath] (with each subset being equally likely), what is the expected value of the number of anchors of [imath]S[/imath]? |
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