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2244318 | If the number of functions from [imath]A \to B[/imath] is equal to [imath]A[/imath], then the cardinality of A and B is [imath]1[/imath].
If the number of functions from [imath]A \to B[/imath] is equal to [imath]A[/imath], then the cardinality of A and B is [imath]1[/imath]. In set theory notation: If [imath]|B|^{|A|}=|A|[/imath] then [imath]|A|=|B|=1[/imath]. Any ideas? I was think of assuming [imath]|A|[/imath] or [imath]|B| \neq 1[/imath] but can't think of anything I can do after/ | 2226353 | ZF Set Theory - If [imath]|A|^{|B|} = |B|[/imath], then [imath]|A| = 1 = |B|.[/imath]
Show (within ZF Set Theory without choice) that for any sets [imath]A,B,C:[/imath] If [imath]|A|^{|B|} = |B|[/imath], then [imath]|A| = 1 = |B|.[/imath] That is, if there is a bijection from [imath]|A|^{|B|}[/imath] to [imath]|B|[/imath], then [imath]|A| = 1 = |B|.[/imath] Please help me.. |
2244592 | A garage has [imath]20[/imath] places for parking ; if [imath]2[/imath] cars are to be parked ; What is the probability that they are sit next to each other?
A garage has [imath]20[/imath] places for parking (one single row ); if [imath]2[/imath] cars are to be parked ; What is the probability that they are sit next to each other ? My try follows: - Number of adjacent places is [imath]19[/imath] Number of ways for chossing [imath]2[/imath] places out of [imath]20[/imath] is [imath]20C2=190[/imath] Probability =[imath]19/190=0.1[/imath] Is my answer right ? Thank you for your help | 1475031 | We align [imath]N[/imath] persons including persons A and B, what is the probability that they sit next to each other?
We align [imath]N[/imath] persons including person A and B in a row, what is the probability that A and B sit next to each other? I can't find the right equation for this one, so if anyone could point me in the right direction it would help me a lot. Thank you |
2240231 | Continuously bijective but not homotopy equivalent.
Give an example of topological spaces [imath]X[/imath],[imath]Y[/imath] with bijective continuous maps between them(so 2 maps in total!), such that [imath]X[/imath] is not homotopy equivalent to [imath]Y[/imath].Be aware of the theorem stating that bijective continuous map from compact space to Hausdorff space must be a homeomorphism. There are also examples such that [imath]f[/imath] and [imath]g[/imath] don't provide a homotopy equivalence, but still the spaces are. P.S. The problem has already been asked 2 years ago, but there is no satisfactory answer: Example of topological spaces with continuous bijections that are not homotopy equivalent | 1624375 | Finding an example of nonhomeomorphic closed connected sets
Question: Find two closed, connected subsets in [imath]\mathbb{R}^2[/imath], [imath]A[/imath] and [imath]B[/imath], such that [imath]A[/imath] is not homeomorphic to [imath]B[/imath], but there is a continuous bijection [imath]f:A \rightarrow B[/imath] and a continuous bijection [imath]g:B \rightarrow A[/imath]. This is a homework question, so please only very small hints. I realize that both [imath]A[/imath] and [imath]B[/imath] must not be compact. Since they both must be closed, then they must be unbounded. However, I am having a hard time getting started on this. It is very easy to find two closed, unbounded, connected subsets of the plane that are not homeomorphic to each other, but it is hard to find the continuous bijections required. I know the classic example of a continuous bijection with a discontinuous inverse is a map [imath]f: [0,2\pi) \rightarrow \mathbb{S}^1[/imath] given by [imath]f(x) = (\cos x, \sin x)[/imath]. I am trying to use this map as a template to come up with my sets but I am having no success. |
2245102 | Finding the closed form [imath]\phi(n) / n^x[/imath]
[imath]\sum_{1}^{\inf}{\phi(n)/n^x}[/imath] Get an closed form expression for the value in terms of x. Also, is there a closed form for the function [imath]f(x) = \sum_{1}^{\inf}{\pi(n)/n^x}[/imath]. If you use PNT you can get some expressions in terms of li integrals but that's probably a dead end and since PNT is only an approximation its not the actual function. | 2244218 | totient function series diverges?
Earlier today I thought I proved that the following series diverged: [imath]\sum_{n=2}^\infty\frac{\phi(n)}{n^2}[/imath] as a result of a misapplication of the prime number theorem. I mistook [imath]\phi(n)[/imath] for [imath]\pi(n)[/imath] in the statement. Is this salvageable? I've been trying to find lower bounds for [imath]\phi(n)[/imath] that I may pass to, but the literature on this is a bit dense for me. Thanks |
2245496 | Prove that an entire function with modulus bounded from below by a polynomial is a polynomial itself.
Suppose [imath]f[/imath] is an entire function and that [imath]\exists k > 0, R > 0[/imath] and [imath]n \in \mathbb{N}[/imath] such that [imath]|f(z)| > k|z|^n[/imath] whenever [imath]|z| > R[/imath]. Prove that [imath]f[/imath] is a polynomial. I am not sure how to go about this and I would greatly appreciate some hints. | 2241860 | If a complex function [imath]f[/imath] is Entire, and there exist [imath]k,R > 0[/imath], [imath]n \in N[/imath] such that [imath]|f(z)| > k|z|^n[/imath] for all [imath]|z| > R[/imath], then f is a polynomial.
As in the title, I must prove that: If a complex function [imath]f[/imath] is Entire, and there exist [imath]k,R > 0[/imath], [imath]n \in N[/imath] such that [imath]|f(z)| > k|z|^n[/imath] for all [imath]|z| > R[/imath]: Then f is a polynomial. We've already had a similar question with the inequality reversed, and found that it had to be a polynomial with degree less than n. I was able to show that if it was a polynomial, then it had to have degree less than n, but not that it had to be a polynomial in the first place. Once again, I'm face with the same problem. I assume this time, the degree of the polynomial will have degree greater than or equal to n, but how do I show that such [imath]f[/imath] must be a polynomial? It's not like I can go through every other possible function there is, can I? I can show it doesn't work for exponentials, for example, but I just can't think of how to show that this ONLY works for a polynomial... |
2238680 | Let [imath]f[/imath] be a differentiable function with [imath]|f'(x)|\leq1[/imath] and [imath]f(-3)=-3, f(3)=3[/imath]. Then find [imath]f(0)[/imath].
Let [imath]f[/imath] be a differentiable function with [imath]|f'(x)|\leq1[/imath] and [imath]f(-3)=-3, f(3)=3[/imath]. Then find [imath]f(0)[/imath]. I think this is mean value theorem problem. But I can't solve... help me please. | 1108314 | Find a function value given 2 points
Given [imath]f(x)[/imath], which is differentiable at every point such that: [imath]f'(x) \ge -5[/imath] for every [imath] x \in R[/imath] [imath]f(2) = -13[/imath], [imath]f(9) = -48[/imath] Prove that:[imath] f(3) = -18[/imath] Now it's quite obvious that [imath] f(3) = -18[/imath] since that if we know that the lowest value of the derivative is -5, the difference between [imath]f(2)[/imath] and [imath]f(9)[/imath] is [imath]-48-(-13)=-35[/imath] that means that that in the area [imath][2,9][/imath], [imath]f'(x)=-5[/imath] right? that's why I told that [imath]f(3) = -18[/imath], but I feel that my proof is insufficient, probably because I'm missing some theorems or such. can someone help me structure this proof into something RIGHT :)? Thanks a lot! |
2245890 | Prove [imath]\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y\rfloor+\lfloor x+y\rfloor[/imath]
Prove that [imath]\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor[/imath] for all real [imath]x[/imath] and [imath]y[/imath]. If anybody could post a simple solution (no complicated abstract theories or calc :D) to the above question, I would greatly appreciate it. Thanks! Also, if possible, the solution shouldn't wander too much away from floor, ceiling, fraction functions, and other related functions. | 1082211 | Show that [imath][2x]+[2y] \geq [x]+[y]+[x+y][/imath]
Prove that [imath][2x]+[2y] \geq [x]+[y]+[x+y][/imath] whenever [imath]x[/imath] and [imath]y[/imath] are real numbers. The [imath][][/imath] symbol is the greatest integer or floor function. I have proved this fact by cases, but I stumbled upon what I believe to be another way to prove the above inequality, and I was wondering if my sequence of statements are legitimate. I make use of two lemmas that I have proved Lemma 1. If [imath]x[/imath] is a real number and m is an integer, then [imath][x+m] = [x]+m[/imath]. Lemma 2. [imath]\displaystyle [x]+\left[x+\frac{1}{2} \right] = [2x][/imath]. My proof begins with an "obvious" statement [imath]x+y \leq x+y+1[/imath] I then take the floor of the inequality to get [imath][x+y] \leq [x+y]+1[/imath] (1) which is true in virtue of lemma 1. Furthermore, if I add the following statements [imath][x+1/2] \leq x + 1/2[/imath] [imath][y+1/2] \leq y + 1/2[/imath] I procure [imath]\left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2}\right] \leq x+y+1[/imath] which by definition of the floor function renders the equation [imath][x+y]+1 = \left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2} \right][/imath] (2) Substituting (2) for (1), I have [imath][x+y] \leq \left[x+\frac{1}{2} \right]+\left[y+\frac{1}{2} \right][/imath] I then add [imath][x][/imath] and [imath][y][/imath] to the above inequality to produce [imath][x]+[y]+[x+y] \leq [x]+\left[x+\frac{1}{2} \right]+[y]+\left[y+\frac{1}{2} \right][/imath] And in virtue of Lemma 2, the right hand side of the inequality becomes [imath][x]+[y]+[x+y] \leq [2x]+[2y].[/imath] I personally don't see anything wrong, except maybe for the implication made to establish (2). Solving these kinds of problems is solely for personal gratification, so I will greatly appreciate feedback. Thanx. |
2245536 | Find the interval of convergence for [imath]\sum_{k=0}^{\infty}\frac{x^k}{k!}[/imath].
Find the interval of convergence for [imath]\sum_{k=0}^{\infty}\frac{x^k}{k!}[/imath]. I tried using the ratio test to find the radius of convergence. \begin{align} \lim_{k\to\infty}\dfrac{\left|\frac{x^{k+1}}{(k+1)!}\right|}{\left|\frac{x^k}{k!}\right|} &= \lim_{k\to\infty}\left|\dfrac{x}{k+1}\right| =0 \end{align} Doesn't this tell us that we can choose any [imath]x\in\mathbb{R}[/imath] because [imath]\lim_{k\to\infty}\left|\dfrac{x}{k+1}\right| =0[/imath] for any [imath]x[/imath]? | 2054653 | Proof of radius of convergence exponential function
Suppose one is trying to find the radius of convergence of [imath] \exp(x)=\sum\limits_{k=0}^{\infty} \frac{x^k}{k!} [/imath] This "proof" was given in the lecture: [imath] \begin{align} r =& \frac{1}{\limsup\limits_{n \rightarrow \infty}\sqrt [n]{|a_n|}} \\ =& \frac{1}{\limsup\limits_{n \rightarrow \infty}\sqrt [n]{\frac{1}{n!}}} \\ =&\lim\limits_{n \rightarrow \infty} \sqrt [n]{n!}\\ r=& \infty \end{align}[/imath] Other proofs involving the ratio test also seem to inspect the convergence of [imath]\frac{1}{k!}[/imath] instead of [imath]\frac{x^k}{k!}[/imath]. Why is one allowed to substitute [imath]\frac{1}{k!}[/imath]? |
2244500 | Finding eigenvector of linear map defined by matrix whose columns each add to [imath]k[/imath]
Let [imath]M[/imath] be an [imath]n \times n[/imath] matrix over [imath]\mathbb F[/imath], where [imath]\mathbb F\in \{\mathbb C, \mathbb R\}[/imath]. Define the linear map [imath]T: \mathbb F^n\rightarrow \mathbb F^n[/imath] by [imath]Tv= Av[/imath]. Suppose the sum of the entries in each column of [imath]M[/imath] is equal to [imath]k[/imath]. I want to determine that [imath]k[/imath] is an eigenvalue and find a corresponding eigenvector. I am able to show that it's an eigenvalue indirectly, namely by showing that the map [imath]T-kI[/imath] is not invertible. But I think that I should show that [imath]k[/imath] is an eigenvalue by obtaining an eigenvector. This way I find the eigenvector I want, too. However, I cannot come up with one. How could I find such an eigenvector? Note that I am following Axler's book, which avoids using determinants when dealing with eigenvalues. So I would appreciate an explanation that does the same. Edit: I was linked to a possible duplicate. The question is similar to mine, but the only direct answers involved this fact: [imath][1, \ldots, 1]A= [k, \ldots, k]=k[1, \ldots, 1][/imath]. The problem here is that the operator [imath]T[/imath] I defined above has the form [imath]Av[/imath]. So I would want [imath]Av =kv[/imath] for some [imath]v[/imath]. This means that [imath]v[/imath] must be a column vector, not a row vector as described in the possible duplicate. I am specifically looking for a nonzero vector [imath]v[/imath] such that [imath]Tv=kv[/imath], were [imath]v[/imath] is considered as a column vector. | 682922 | Algebra-sum of entries in each column of a sqaure matrix = constant
This is a question from an algebra homework and I am just looking for some tips. The question is: We have: [imath]M[/imath]: an [imath]n\times n[/imath] matrix with real entries [imath]c[/imath]: a real constant the sum of the entries of each column of [imath]M[/imath] is equal to [imath]c[/imath] We have to prove that [imath]c[/imath] is an eigenvalue of [imath]M[/imath]. I started out by writing out the form of the matrix and see if anything was popping up without much success. I found a similar question except that they were working with a [imath]2\times 2[/imath] matrix and that the sum of entries in each column [imath]= c[/imath] as well as the sum of entries in each row. Also, in this question they are looking for eigenvectors. In this case the question is pretty easy as we can notice that each column or row is composed of the same entries (in different order or not). link (p4): http://www.math.upenn.edu/~deturck/m260/hw3sols.pdf Hence I looked at the sum in the columns to find some kind of pattern or hint that could help me solve this question without any success. It is obvious that the entries in each column might not be the same. Anyhow I am stuck there. Again I am not asking for the answer, just a hint. Thank you for your time and good luck. |
2228027 | How to parametrise a circle in a plane
I've been studying for an exam and in one of the exams it asks me the parametrise the unit circle in [imath]R^3[/imath] which lines in the plane [imath]x+y=0[/imath]. I know how to parametrise a normal circle, but I don't really have any idea how to parametrise it in a plane like that. Any help would be appreciated, thanks | 492607 | For a general plane, what is the parametric equation for a circle laying in the plane
Given a general equation for a plane through the origin [imath]\vec{n}\cdot\vec{r}=0[/imath] With no assumptions made on [imath]\vec{n}[/imath] except having unit modulus, real [imath]3\times1[/imath] vector. How can you describe a unit circle, centred at the origin, laying in this plane? I can only seem to find parametric equations that rely on knowing two vectors in the plane, but with no knowledge of the vector [imath]\vec{n}[/imath] you can't generally create two such vectors, as some component(s) of [imath]\vec{n}[/imath] may be zero. All the information you need to define such a circle is contained within the normal to the plane, so I am confused as to why there is not a form defined only with reference to this vector. EDIT#1: With reference to this matrix. Can we start with in the [imath]xy[/imath] plane [imath](x,y,z)=(\cos(\theta),\sin(\theta),0)[/imath] Then rotate this about the axis ([imath]\vec{u}[/imath] in the link) [imath]\vec{u}=(-n_2,n_1,0)[/imath] about an angle [imath]\phi[/imath] that satisfies [imath]\tan(\phi)=\frac{n_3}{\sqrt{n_1^2+n_2^2}}.[/imath] I claim that [imath]\vec{u}[/imath] is the axis of rotation as this vector is perpendicular to the normal of the plane [imath]\vec{n}[/imath] and lies in the [imath]xy[/imath] plane. Also that [imath]\phi[/imath] is the angle which the [imath]xy[/imath] plane is rotate about [imath]\vec{u}[/imath] by. Therefore by substituting into the matrix linked to at the beginning of this edit, transforming [imath](x,y,z)=(\cos(\theta),\sin(\theta),0)[/imath] by said matrix will give parametric coordinates for the tilted circle in terms of [imath]\vec{n}[/imath]? EDIT #2: I find this for the rotation matrix from the [imath]xy[/imath] plane to the plane with normal [imath]\vec{n}[/imath], from the method described above. [imath]Q=\small{\left(\begin{array}{ccc} {\mathrm{n_2}}^2 - {\mathrm{n_2}}^2\, \sqrt{1 - {\mathrm{n_3}}^2} + \sqrt{1 - {\mathrm{n_3}}^2} & \mathrm{n_1}\, \mathrm{n_2}\, \left(\sqrt{1 - {\mathrm{n_3}}^2} - 1\right) & \mathrm{n_1}\, \mathrm{n_3}\\ \mathrm{n_1}\, \mathrm{n_2}\, \left(\sqrt{1 - {\mathrm{n_3}}^2} - 1\right) & {\mathrm{n_2}}^2\, \sqrt{1 - {\mathrm{n_3}}^2} + {\mathrm{n_3}}^2\, \sqrt{1 - {\mathrm{n_3}}^2} - {\mathrm{n_2}}^2 - {\mathrm{n_3}}^2 + 1 & \mathrm{n_2}\, \mathrm{n_3}\\ - \mathrm{n_1}\, \mathrm{n_3} & - \mathrm{n_2}\, \mathrm{n_3} & \sqrt{1 - {\mathrm{n_3}}^2} \end{array}\right)}[/imath] This is found from this MATLAB code. EDIT #3: Using [imath]\vec{n}=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})[/imath] I find this parametrically plots |
2246305 | Show that [imath]f[/imath] is constant. Liouville Thm
Let [imath]f[/imath] : [imath]\mathbb C \to \mathbb C[/imath] be analytic such that [imath]|Re(f(z)) Im(f(z))|[/imath] [imath]\le[/imath] [imath]1[/imath] for every [imath]z \in \mathbb C[/imath]. Show that [imath]f[/imath] is constant. I know the set is bounded hence I should be able to apply Liouville Thm. Other than this information I do not know how to approach this question. Any help will be appreciated. | 2220778 | Liouvilles theorem question - show [imath]f[/imath] is constant
Let [imath]f: \mathbb C \to \mathbb C[/imath] be an analytic function, such that for all [imath]z\in\mathbb C[/imath], [imath]| \operatorname{Re}(f(z))\operatorname{Im}(f(z)) | \le 1.[/imath] I have to show that [imath]f[/imath] is constant. I don't know how to apply Liouville's theorem to the question, but I can tell it is bounded, thus the theorem should apply. Any help would be appreciated. |
2246651 | Show that if [imath]p[/imath] and [imath]q[/imath] are primes [imath]\equiv 3[/imath](mod [imath]4[/imath]) then at least one of the equations [imath]px^{2}-qy^{2} = \pm 1[/imath] is soluble in integers [imath]x, y[/imath].
So far we have only talked about equations in the form [imath]x^{2} - dy^{2} = \pm 1[/imath] and I'm unsure how to handle the specific coefficients. We have normally found [imath]\sqrt{d}[/imath] and looked at the convergents of the continued fraction to find solutions. I was just wondering how to get started on this type of problem. | 2245758 | How to deal with a Pell's equation type problem with two primes
I have some understanding of Pell's equation [imath]x^2-dy^2=1[/imath] where [imath]d[/imath] is a prime. I know that you can take the continued fraction of [imath]\sqrt{d}[/imath] and use information about the period and convergents to find things out. Now I need to show that if [imath]p,q[/imath] are primes [imath]\equiv 3 \pmod{4}[/imath] then at least one of the equations [imath]px^2-qy^2=\pm 1[/imath] is soluble in integers [imath]x,y[/imath]. I have not learned what to do when there is a prime in front of the [imath]x^2[/imath]. How should I begin doing this? |
2245392 | Prove a binomial coefficient identity.
[imath] \sum_{k=0}^{n}{\binom{p}{k}\binom{q}{k}\binom{n+k}{p+q}} = \binom{n}{p}\binom{n}{q} [/imath] My try was to use multivariable generating functions(I don't know, I just thought). I hope that there is a solution using combinatorical analogy. | 2088564 | Sum of binomial coefficients involving [imath]n,p,q,r[/imath]
Sum of binomial product [imath]\displaystyle \sum^{n}_{r=0}\binom{p}{r}\binom{q}{r}\binom{n+r}{p+q}[/imath] Simplifying [imath]\displaystyle \frac{p!}{r!\cdot (p-r)!} \cdot \frac{q!}{r!\cdot (q-r)!}\cdot \frac{(n+r)!}{(p+q)! \cdot (n+r-p-q)!}[/imath]. Could some help me with this, thanks |
2246905 | Order of the elements in [imath]Z_n[/imath]
What are the orders of the elements in [imath]\mathbb{Z}_n[/imath]? If [imath]n[/imath] is prime then 0 has order 1 and the rest order [imath]n[/imath]. If [imath]a|n[/imath] then [imath]a[/imath] has order [imath]n/a[/imath], and if [imath]a[/imath] is prime then it has order [imath]n[/imath]. So are the orders of the elements divisors of n? Say, we take [imath]\mathbb{Z}_{120}[/imath]. Are the orders appearing the divisors of 120? | 347186 | Order of elements in [imath]Z_n[/imath]
I have this question: Let [imath]x, n[/imath] be integers with [imath]n \geq 2[/imath] and [imath]n[/imath] not dividing [imath]x[/imath]. Show that the order o([imath]\bar{x}[/imath]) of [imath]x \in Z_n[/imath] is [imath]o(\bar{x})= \frac{n}{HCF(x, n)}[/imath] I've been thinking about it for ages but I still don't get why. A hint would be appreciated. |
2245815 | compute [imath]\displaystyle \lim_{n \to \infty}\frac{\sum_{i=1}^{n-2} \sum_{j = 1} ^{n-i-1} ( n-(i + j))}{n^3}[/imath]?
I was trying to solve the following problem If we pick k numbers from the interval (0, 1), what is the probability the sum of those numbers is <1 when ? (k = 6) for k = 2, got [imath]\displaystyle\lim_{n \to \infty}\frac{\sum_{i = 1}^{n-1} i}{n^2}=0.5[/imath] for k = 3, I got the expression [imath]\displaystyle \lim_{n \to \infty}\frac{\sum_{i=1}^{n-2} \sum_{j = 1} ^{n-i-1} ( n-(i + j))}{n^3}[/imath] how to compute it and is there a way to generalize it for picking k numbers from the interval (0,1)? | 1448908 | Probability of the sum of three random variables being less than 1
Given three Uniform random variables between 0 and 1, [imath](x,y,z)[/imath], that are i.i.d., what is the probability [imath]x+y+z < 1[/imath]? |
2247051 | [imath]M[/imath] is path connected if and only if [imath]M[/imath] is connected
Let [imath]M[/imath] be a manifold. Prove that [imath]M[/imath] is path connected if and only if [imath]M[/imath] is connected. My attempt: ([imath]\Rightarrow[/imath]) Let [imath]M \subset \mathbb{R}^n[/imath] be path connected. Every path connected subset of [imath]\mathbb{R}^n[/imath] is connected. ([imath] \Leftarrow[/imath]) Let [imath]M \subset \mathbb{R}^n[/imath] be connected. I think that the goal here is to show that for any [imath]x,y \in M[/imath], there exists [imath]a,b \in \mathbb{R}[/imath] and a path [imath]\Phi:[a,b] \rightarrow M[/imath] such that [imath]\Phi(a)=x[/imath] and [imath]\Phi(b)=y[/imath]. I know that since [imath]M[/imath] is a manifold, there exists an open neighborhood [imath]U[/imath] around [imath]x,y \in M[/imath]. I also know that [imath]M[/imath] contains a diffeomorphism. Not sure how to put this all together in my proof.. | 1145293 | connected manifolds are path connected
prove every connected manifold is path connected manifold . my thought: connected space : Let [imath] X[/imath] be a topological space. A separation of [imath] X [/imath] is a pair [imath]U, V[/imath] of disjoint nonempty open subsets of [imath] X [/imath] whose union is [imath]X[/imath]. The space [imath] X [/imath] is said to be connected if there does not exist a separation of [imath]X[/imath] . Components: Given [imath]X[/imath] , define an equivalence relation on [imath]X[/imath] by setting x~y if there is a connected subspace of [imath]X[/imath] containing both [imath] x[/imath] and [imath] y[/imath] . The equivalence classes are called the components (or the "connected components") of [imath] X[/imath]. Path Component: I define another equivalence relation on the space [imath] X[/imath] by defining x ~ у if there is a path in [imath] X[/imath] from [imath] x[/imath] to [imath] y[/imath] . The equivalence classes are called the path components of [imath] X[/imath] . Theorem : The path components of [imath] X[/imath] are path connected disjoint subspaces of [imath]X[/imath] whose union is [imath]X[/imath] , such that each nonempty path connected subspace of [imath]X[/imath] intersects only one of them. thank you so much |
2245443 | Let [imath]A[/imath] be a square matrix that commutes with its transpose. Show that the nullspaces of [imath]A[/imath] and [imath]A^T[/imath] coincide.
Am completely stuck on this, any help would be appreciated. | 1129712 | Range of A and null space of the transpose of A
So I'm complete stuck with something. I know it the following statements are true (or at least the seem to be from the results that I got from messing around with it a bit on MATLAB), but I don't understand why they are true or how to show so. Let [imath]A[/imath] be and [imath]m[/imath]X[imath]n[/imath] matrix. Show that: a) if [imath]x \in N(A^TA)[/imath] then [imath]Ax[/imath] is in both [imath]R(A)[/imath] and [imath]N(A^T)[/imath]. For this one I messed around with it with my own examples and I got [imath]Ax=0[/imath], therefore satisfing the statement, but I don't understand what's actually going on. b) [imath]N(A^TA)=N(A)[/imath] again, makes sense when I see the results in MATLAB, but don't undestand why it works. c) [imath]A[/imath] and [imath]A^TA[/imath] have the same rank d) If [imath]A[/imath] has linearly independent columns, the [imath]A^TA[/imath] is nonsingular. For the last two I have no idea on how to even start showing the relationship. I feel like I'm missing some crucial relationship between [imath]A[/imath] and [imath]A^TA[/imath], but I'm just not seeing it. I would greatly appreciate any help of sugestions on how to show that these statements are true. Thank you very much. |
2248204 | What is [imath](1+x+\cdots+x^{n-1})^2[/imath]?
This may seem like a pretty basic question, but I am struggling to find a nice way of expressing [imath](1+x+\cdots+x^{n-1})^2[/imath] where [imath]n\in\mathbb{Z}_{\geq 2}[/imath]. Plugging in some numbers yields a clear pattern, [imath](1+x+\cdots+x^{n-1})^2=1+2x+\cdots+nx^n+\cdots+2x^{2n-3}+x^{2n-2}.[/imath] However, I can't seem to find a formal expression of this, or a proof. Can anyone point me in the right direction? | 2129326 | On an expansion of [imath](1+a+a^2+\cdots+a^n)^2[/imath]
Question: What is an easy or efficient way to see or prove that [imath] 1+2a+3a^2+\cdots+na^{n-1}+(n+1)a^n+na^{n+1}+\cdots+3a^{2n-2}+2a^{2n-1}+a^{2n}\tag{1} [/imath] is equal to [imath] (1+a+a^2+\cdots+a^n)^2\tag{2} [/imath] Maybe this is a particular case of a more general, well-known result? Context: This is used with [imath]a:=e^{it}[/imath] to get an expression in terms of [imath]\sin[/imath] for the Fejér kernel. Thoughts: I thought about calculating the coefficient [imath]c_k[/imath] of [imath]a^k[/imath]. But my method is not so obvious that we can get from [imath](1)[/imath] to [imath](2)[/imath] in the blink of an eye. [imath]\mathbf{k=0}[/imath] : clearly [imath]c_0=1[/imath]. [imath]\mathbf{1\leq k\leq n}[/imath] : [imath]c_k[/imath] is the number of integer solutions of [imath]x_1+x_2=k[/imath] with [imath]0\leq x_1,x_2\leq k[/imath], which in turn is the number of ways we can choose a bar [imath]|[/imath] in [imath] \underbrace{|\star|\star|\cdots|\star|}_{k\text{ stars}} [/imath] So [imath]c_k=k+1[/imath]. [imath]\mathbf{k=n+i\quad(1\leq i\leq n)}[/imath] : [imath]c_k[/imath] is the number of integer solutions to [imath]x_1+x_k=n+i[/imath] with [imath]0\leq x_1,x_2\leq n[/imath], which in turn is the number of ways we can choose a bar [imath]|[/imath] in [imath] \underbrace{|\star|\star|\cdots|\star|}_{n+i\text{ stars}} [/imath] different from the [imath]i[/imath]-th one from each side. So [imath]c_k=(n+i)+1-2i=n-i+1[/imath]. |
2248306 | Comparison test of series with ln function
How to test convergence of series with comparison test in the next examples: [imath] \sum_{n=1}^\infty \ln\left(1+\frac{1}{2^n} \right) [/imath] I know that: (incorrect) [imath] \ln\left(1+\frac{1}{2^n}\right) < 1 + \frac{1}{2^n} [/imath] Correct: [imath] \ln\left(1+\frac{1}{2^n}\right) \leq \frac{1}{2^n} [/imath] Therefore (incorrect) [imath] \sum_{n=1}^\infty 1+\frac{1}{2^n}\ [/imath] Therefore: [imath] \sum_{n=1}^\infty \frac{1}{2^n}\ [/imath] Is divergent, but that doesn't tell me anything because it is larger than the starting series. So how to solve this using comparison test? Is convergent which implies that starting series is convergent as of rule: Suppose that [imath] 0 \leq a_n \leq b_n [/imath] for sufficiently large n, then: If [imath]\sum_{n=1}^{\infty} a_n[/imath] diverges, then [imath]\sum_{n=1}^{\infty} b_n[/imath] diverges. If [imath]\sum_{n=1}^{\infty} b_n[/imath] converges, then [imath]\sum_{n=1}^{\infty} a_n[/imath] converges. | 1172424 | Convergence of [imath]\sum^\infty_{n=1} \ln(1+\frac 1 {2^n})[/imath]
Does [imath]\displaystyle\sum^\infty_{n=1} \ln(\frac {2^n+1}{2^n})=\sum^\infty_{n=1} \ln(1+\frac 1 {2^n})[/imath] converge? Condensation, root, ratio and limit comparison tests don't help, I don't know what to do... Any hints please? Note: no integral test. |
191077 | Evaluate [imath]\int\frac{dx}{\sin(x+a)\sin(x+b)}[/imath]
Please help me evaluate: [imath] \int\frac{dx}{\sin(x+a)\sin(x+b)} [/imath] | 778479 | Evaluate [imath]\int\frac{1}{\sin(x-a)\sin(x-b)}\,dx[/imath]
I'm stuck in solving the integral of [imath]\dfrac{1}{\sin(x-a)\sin(x-b)}[/imath]. I "developed" the sin at denominator and then I divided it by [imath]\cos^2x[/imath] obtaining [imath]\int\frac{1}{\cos(a)\cos(b)\operatorname{tan}^2x-\cos(a)\sin(b)\operatorname{tan}x-\sin(a)\cos(b)\operatorname{tan}x+\sin(a)\sin(b)}\frac{1}{\cos^2x}dx[/imath] Then I made a substitution by [imath]t=\operatorname{tan}x[/imath] arriving to this [imath]\int\frac{1}{\cos(a)\cos(b)t^2-(\cos(a)\sin(b)+\sin(a)\cos(b))t+\sin(a)\sin(b)}dt[/imath] How can I solve it now? (probably I forgot something, it easy to make mistakes here) Thank you in advance! |
2248467 | Geometric interpretation of contour integral?
If [imath]C[/imath] is a contour in the complex plane, is there a geometric interpretation of the contour integral [imath]\int_Cf(z)dz[/imath]? What does the value of this integral mean/tell us about anything? Why does the sign of the integral change when the contour is traversed in the opposite direction? | 111368 | What is a geometric explanation of complex integration in plain English?
Im trying to get my head around complex integration/complex line integrals. Real integration can be thought of as the area under a curve or the opposite of differentiation. Thinking of it geometrically as the area under a curve or the volume under a surface in 3 dimension is very intuitive. So is there a geometric way of thinking about complex integration? Or should I just be viewing it as process that reverses differentiation? Or has integration other meanings in complex analysis? Here is an example, could someone explain this to me - Here's the definition of the integral along a curve gamma in C, parameterized by [imath]w:[a, b] ->C[/imath] \begin{equation} \int_\gamma f(z)dz = \int_a^b f(w(t)).w{(t)}'dt \end{equation} So I have - [imath]\gamma[/imath] is the unit circle with anti-clockwise orientation parameterized by [imath]w:[0, 2\pi]\to C[/imath] [imath]w(t) = e ^{it} = Cos(t) + iSin(t)[/imath] So if use the definition of the integral, [imath]\int_\gamma f(z)dz = \int^b_a f(w(t)).w{(t)}'dt[/imath], and work this out it comes to [imath]\int^{2\pi}_0 i dt = 2{\pi}i[/imath] So what does this [imath]2{\pi}i[/imath] represent? Does it mean anything geometrically, like if a regular integral works out to be 10 that means the area under the curve between 2 points is 10... |
2246222 | Unbiased estimator of a uniform distribution
For a random sample [imath]X_1,X_2,\ldots,X_n[/imath] from a [imath]\operatorname{Uniform}[0,\theta][/imath] distribution, with probability density function [imath] f(x;\theta) = \begin{cases} 1/\theta, & 0 \le x \le \theta \\ 0, & \text{otherwise} \end{cases} [/imath] Let [imath]X_{\max} = \max(X_1,X_2,\ldots,X_n).[/imath] What is the value of k such that [imath]\hat \theta = kX_{\max}[/imath] is an unbiased estimator of [imath]\theta[/imath] ? I'm not sure if there is more to this question, because my intuitive answer answer is just [imath]k=1[/imath]. This is because if you order the sample like [imath]x_{(1)} \le x_{(2)} \le \cdots \le x_{(n)}[/imath] such that [imath]x_{(n)} = E[X_{\max}][/imath]. and the fact that the distribution is uniform, the estimator of [imath]\theta[/imath] should just be [imath]X_{\max}[/imath]. Unbiased estimator -> [imath]E\left[\widehat{\theta\,}\right] = kE[X_{\max}] = \theta[/imath] Is my logic wrong here? | 2497654 | What makes the cdf curve of a uniform distribution a constant positive slope?
Assignment asks to explain why max of X will have the highest probability.. What is the reason why the probabilities of each random variable create a constant positive slope? (max[imath]_k[/imath]≤)=(1 ≤ ∩ 2 ≤ ∩ ⋯ n ≤ ) explain why OR =(1≤)(2≤)⋯ ( ≤) explain why is this because of linearity and independence ? This is the very first statistics class I've taken. I'm still learning. I am suppose to find an unbiased estimator for a given pmf for a uniform distribution. pmf: [imath]() = \frac{1}\theta{} \ \ \ \ (0 < < )[/imath] |
2230151 | Characterization of (Lebesgue) measurable sets
I am stuck trying to show a characterization of (Lebesgue) measurable sets from Caratheodory's definition of measurable. This means we do not have the notion of an inner measure, as it is not required to define measure via Caretheodory's criterion. Show that a bounded subset [imath]E\subset \mathbb{R}[/imath] is (Lebesgue) measurable if there is some bounded interval [imath]I\supset E[/imath] such that [imath]\lambda^*(I)=\lambda^*(E)+\lambda^*(I\setminus E)[/imath] I am trying to prove that [imath]E[/imath] is measurable by showing that if such an [imath]I[/imath] exists, then for any interval [imath]J[/imath] [imath]\lambda^*(J)=\lambda^*(J\cap E)+\lambda^*(J\setminus E)[/imath] I am trying to prove this when [imath]J\subset I[/imath] (it is clearly true when [imath]J\cap E=\emptyset[/imath], and then the result will follow). So I require [imath]\lambda^*(J)\ge\lambda^*(J\cap E)+\lambda^*(J\setminus E)[/imath] for [imath]J\subset I[/imath]. Manipulating the set relations doesn't seem to get me anywhere. | 2008508 | Proving Caratheodory measurability if and only if the measure of a set summed with the measure of its complement is the measure of the whole space.
Suppose we have a premeasure [imath]\mu[/imath] on a space [imath]X[/imath] such that [imath]\mu(X) < \infty[/imath]. Prove that [imath]E \subset X[/imath] is Caratheodory measurable iff [imath] \mu^*(E)+ \mu^*(E^C) = \mu(X)[/imath]. Going in the forward direction is fairly easy. Assuming that E is Caratheodory measurable, we can just substitute X into [imath]\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^C) [/imath], and then we note that the outer measure and the premeasure of X themselves would have the same value. The other direction is more difficult however. My primary idea of how to solve this part is to show that [imath]\mu^*(A)[/imath] and [imath]\mu^*(A \cap E) + \mu^*(A \cap E^C) [/imath] are both greater than and less than each other to show equality. However, I am not completely sure how to proceed with this. Can anybody provide any pointers as to how to prove the equality between these two expressions? |
2246474 | If [imath]f[/imath] is differntiable at [imath]x_0[/imath], then [imath]f[/imath] is continuous at [imath]x_0[/imath]. How does this work?
There's a similar post here. but I haven't got the answer: If [imath]f[/imath] is differentiable at [imath]x = x_0[/imath] then [imath]f[/imath] is continuous at [imath]x = x_0[/imath]. I just started learning calculus for physics, so I don't understand well yet. I was watching an MIT OpencourseWare Lecture (link below), and I just couldn't understand the last section of the video where the proof of a theorem was being explained. I did search online for answers but didn't understand their explanations, and it feels like I'm missing something really obvious. The theorem states: If [imath]f[/imath] is differentiable at [imath]x_0[/imath], then [imath]f[/imath] is continuous. The proof goes like this: 1) [imath] \lim_{x\to x_0} f(x)-f(x_0) = 0 [/imath] 2) [imath] \lim_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}\cdot(x-x_0)=f'(x_0)\cdot0=0 [/imath] What did I understand is that the first equation is just the definition of continuity: [imath]\lim\limits_{x\to x_0} f(x) = f(x_0)[/imath] rearranged. But I do not understand the second part at all. Can someone please explain what is happening in the second equation? Also: 1) Where does the [imath]f'(x_0)[/imath] suddenly come from? 2) Why is it being multiplied with [imath]\dfrac{x-x_0}{x-x_0}[/imath]? It does look vaguely familar to the difference quotient: [imath]\lim\limits_{\Delta x\to 0} f'(x_0) = \dfrac{f(x_0+\Delta x) - f(x_0)}{\Delta x}[/imath] I am suspecting that it has something to do with [imath]\Delta x[/imath] being implied somewhere. Here are the links: Video (starts at 46:10) PDF version | 395827 | If [imath]f[/imath] is differentiable at [imath]x = x_0[/imath] then [imath]f[/imath] is continuous at [imath]x = x_0[/imath].
Claim: if [imath]f[/imath] is differentiable at [imath]x = x_0[/imath] then [imath]f[/imath] is continuous at [imath]x = x_0[/imath]. Please, see if I made some mistake in the proof below. I mention some theorems in the proof: The condition to [imath]f(x)[/imath] be continuous at [imath]x=x_0[/imath] is [imath]\lim\limits_{x\to x_0} f(x)=f(x_0)[/imath]. (1) If [imath]f(x)[/imath] is differentiable at [imath]x-x_0[/imath], then [imath]f'(x)=\lim\limits_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}[/imath] exists and the function is defined at [imath]x=x_0[/imath]. (2) Therefore, by the Limit Linearity Theorem, [imath]\lim\limits_{x\to x_0} f(x)[/imath] exists and we'll show it is equals [imath]f(x_0)[/imath]. (3) We'll do this by the Precise Limit Definion: given [imath] \epsilon>0, \exists\delta|0<|x-x_0|<\delta[/imath], then [imath]0<|f(x)-f(x_0)|<\epsilon[/imath]. As this limit exists by (2), we can make [imath]f(x)[/imath] as close to [imath]f(x_0)[/imath] as one wishes, therefore [imath]\lim\limits_{x\to x_0} f(x)=f(x_0)[/imath], what satisfies the condition for [imath]f(x)[/imath] be differentiable at [imath]x=x_0[/imath]. The end. |
2249467 | injective ring homomorphism inducing a non-injective homomorphism?
I am looking for an example of an injective ring homomorphism [imath]\phi\colon A\to B[/imath] such that the following holds: there exists a prime ideal [imath]\frak q[/imath] of [imath]B[/imath] such that [imath]\phi^{-1}(\frak q) = \frak p[/imath], and the induced map [imath]A_{\frak p} \to B_{\frak q}[/imath] is not injective. The algebro-geometric picture should be clear: if we put [imath]X = \mathrm{Spec\, A}[/imath] and [imath]Y = \mathrm{Spec\, B}[/imath], then [imath]\phi[/imath] induces a morphism [imath](\psi,\theta)\colon Y\to X[/imath] of affine schemes with [imath]\psi(\frak q) = \frak p[/imath]. The map [imath]A_{\frak p} \to B_{\frak q}[/imath] corresponds to the induced map on the level of stalks at [imath]\frak q[/imath] of the morphism of sheaves [imath]\theta^\sharp \colon \psi^* \mathcal O_X \to \mathcal O_Y[/imath] [I have used the notation from EGA I]. | 1786346 | Does such localization of integral extension preserve inclusion?
Let [imath]R\subset T[/imath] be two commutative rings, and [imath]T[/imath] is integral over [imath]R[/imath]. Let [imath]\mathfrak m\in \operatorname{Max} R,\mathfrak n\in\operatorname{Max}T[/imath] such that [imath]\mathfrak m=\mathfrak n\cap R[/imath]. Show that [imath]R_\mathfrak m\subset T_\mathfrak n[/imath], i.e. the canonical map [imath]\{\frac ru\mid r\in R,u\in R\setminus\mathfrak m\}\to\{\frac tv\mid t\in T,v\in T\setminus\mathfrak n\}[/imath] must be an injection. I feel confused because the exercise 4 from Chapter 5 in the book of Atiyah and Macdonald just assume that the localization is a ring extension. And I don't know if the condition "integral over" is necessary. |
2249494 | Standard Graded Algebra over a field
My question is What is standard graded Algebra over a field [imath]K[/imath]? I know this question is already posted here Standard graded algebra , But still I'm repeating because the concept is not clear to me. What I know a graded ring [imath]R[/imath] with gradation [imath]{\{R_n\}}[/imath], where [imath]R_n[/imath] is additive subgroup satisfies [imath]1.[/imath] [imath]R=\oplus[/imath][imath]R_n[/imath] [imath]2.[/imath] [imath]R_mR_n \subseteq R_{m+n}, \forall m,n\ge0[/imath] and similarly we can define graded [imath]R[/imath] module [imath]M[/imath] with gradation [imath]{\{M_n\}}[/imath] satisfying [imath]1.[/imath] [imath]M=\oplus M_n[/imath] [imath]2.[/imath] [imath]R_m M_n \subseteq M_{m+n}, \forall m,n \ge0[/imath] But the definition of standard graded algebra [imath]R[/imath] over a field [imath]K[/imath] is according to the article http://www.dima.unige.it/~conca/Articoli%20Conca%20PDF/PDF%20da%20rivista/(2014)%20Conca%20(LNM%20Levico)%20Koszul%20Algebras%20and%20Their%20Syzygies.pdf is [imath]R=\oplus R_i[/imath] such that [imath]R_0 =K[/imath], the vector space [imath]R_1[/imath] has finite dimension and [imath]R_iR_j=R_{i+j}, \forall i,j \ge 0[/imath] Can anyone please explain this definition in a simpler way? Thanks in advance. | 133484 | Standard graded algebra
I am so sorry if you feel this kind of question is not appropriate for MS. But I hope you can sympathize with me, I tried to find the answer in all my books and even Google but I found nothing. My question is : What is a standard graded algebra over a ring? Please help me. Thanks. Edit: In the following paper On the asymtotic linearity of Castelnuovo-Mumford regularity, there is a definition of standard graded algebra. Suppose that [imath]A[/imath] is a ring, [imath]R[/imath] is a standard graded [imath]A[/imath] algebra if [imath]R_{0}=A[/imath] and [imath]R[/imath] is generated by the element of [imath]R_1[/imath]. I did not fully understand this definition. Can anyone give here an example? |
2249747 | Show that J is a maximal ideal
Let [imath]R[/imath] be the ring of all functions [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] with the operations [imath](f + g)(x) = f(x) + g(x)[/imath] [imath](fg)(x) = f(x)g(x)[/imath] Let [imath]c ∈ \mathbb{R}[/imath]. Show that [imath]J = \{f ∈ R | f(c) = 0\}[/imath] is a maximal ideal.Can you give me some hints ? | 2230974 | [imath]I:=\{f(x)\in R\mid f(1)=0\}[/imath] is a maximal ideal?
Let [imath]R[/imath] be a ring of the real functions with "+" and "x" and let [imath]I_1:=\{f(x)\in R\mid f(1)=0\}[/imath], [imath]I_1[/imath] is a maximal ideal? I think yes, but don't know how to prove it. |
2247683 | real numbers of the form [imath]\frac{m}{10^n} [/imath] with [imath]m,n \in \mathbb{Z} [/imath] and [imath]n \geq 0[/imath] is dense in [imath]\mathbb{R}[/imath] .
Problem : Verify if the statement if true of false - The set [imath]S[/imath] of all real numbers of the form [imath]\frac{m}{10^n} [/imath] with [imath]m,n \in \mathbb{Z} [/imath] and [imath]n \geq 0[/imath] is dense in [imath]\mathbb{R}[/imath] . I think this true . Reason : [imath]S[/imath] is actually a subgroup of [imath]\mathbb{R}[/imath] with respect to addition . Now i know the theorem that any subgroup of [imath]\mathbb{R}[/imath] is either cyclic or dense . If [imath]S[/imath] were cyclic then there non-zero [imath]m[/imath] and an [imath]n\geq 0[/imath] such that [imath]\frac{m}{10^n} [/imath] generates [imath]S[/imath] . So there is a non-zero [imath]i\in \mathbb{Z} [/imath] such that [imath]0= \frac{m}{10^n}+..\frac{m}{10^n}[/imath] ( [imath]i[/imath] times adiition )[imath] = \frac {mi}{10^n} [/imath] which is not possible because neither [imath]m[/imath] or [imath]i[/imath] is [imath]0[/imath] .So [imath]S[/imath] can be cyclic . Hence it is dense . | 1646096 | Is [imath]\{\frac{m}{10^n}\mid m,n\in\mathbb Z,\ n\geq 0\}[/imath] dense in [imath]\mathbb R[/imath]?
The set [imath]S[/imath] of real numbers of the form [imath]m/10^n[/imath], [imath]m,n[/imath] integers and [imath]n[/imath] greater than or equal to [imath]0[/imath], is a dense subset of [imath]\mathbb R[/imath] or not?? I know dense means closure of [imath]S[/imath] in [imath]\mathbb R[/imath] is [imath]\mathbb R[/imath] then it will be dense. How to prove or disprove I have no idea. |
184760 | Brute force method of solving the cube: How many moves would it take?
Given that Rubik's cube has finitely many positions, one possible "brute force" method to solve it would be to determine once a sequence of moves which eventually reaches every possible position of the cube, and then whenever you want to solve a cube, you just mindlessly follow the sequence until you eventually reach the solved cube. This strategy is absolutely failsafe, provided you have an extraordinary memory and enough time (so, it's not a strategy for humans, but maybe for bored gods waiting for their just created universe to finally develop intelligent life ;-)). However, the question is: How many moves does the shortest possible sequence which visits every reachable position have? Of course, it's easy to give a lower bound: At least as many moves as there are reachable positions, which according to Wikipedia is [imath]43\,252\,003\,274\,489\,856\,000 \approx 4{.}3 \cdot 10^{19}[/imath]. However, I doubt that there's a sequence which gives a not yet visited position at every move, so the actual number of moves is likely even higher. Is there maybe even a constructive method to create this "bored god's algorithm" (so that even gods with less than stellar memory can apply it)? | 1694536 | Running through all permutations of a Rubik's cube
According to Wikipedia a [imath]3 \times3\times3[/imath] Rubik's cube has [imath]43252003274489856000[/imath] permutations. I never tried solving one myself (too tedious), however I wondered, if one could miraculously "solve" the cube in color blind mode (all faces of the cube have the same color to him or the person is just blind folded). (Yes, it is tedious, but I guess sooner or later he will just hit the right permutation and hopefully someone tells him, that it's good and he may stop.) I wondered, if there were some kind of "best" approach to run all through those permutations without too many repetitions, for example the algorithm makes sure one gets any permutation at most [imath]n[/imath] times. I don't know the answer yet and probably won't have it anytime soon. However I would be quite interested in seeing this algorithm and thus I hope the question has been asked before and been solved. Any constructive comment/answer is appreciated. |
2249611 | Statistics - Proving -uniform random variables
Suppose [imath]U_1[/imath] and [imath]U_2[/imath] are independent uniform random variables on [imath][0,1][/imath]. Show that [imath]\frac{\log(U_1)}{\log(U_1)+\log(U_2)}[/imath] is also uniform random variable. Appreciate your help. | 2246756 | Statistics-Uniform random variable
Suppose [imath]U_1[/imath] and [imath]U_2[/imath] are independent uniform random variables on [imath][0,1][/imath]. Show that [imath]\frac{\log(U_1)}{\log(U_1)+\log(U_2)}[/imath] is also uniform random variable. Appreciate your help. |
2249313 | There exist elements of finite order [imath]n[/imath] in [imath]GL_2(\Bbb Z)[/imath] for every positive [imath]n[/imath].
I am trying to prove the title statement. So far, I have tried to do this algebraically. For example, for [imath]n = 2[/imath], and a matrix [imath]\begin{pmatrix} a & b \\ c & d \end{pmatrix},[/imath] I tried to find such [imath]a, b, c, d[/imath] where [imath]ab - cd = 1[/imath] or [imath]-1[/imath] and where [imath]\begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & cd + d^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.[/imath] I found that [imath]\begin{pmatrix} 4 & 3 \\ -5 & -4 \end{pmatrix}[/imath] works. However, is there a way to do this without algebra for larger [imath]n?[/imath] Any hints as to how I would go about this? | 939953 | Elements of [imath]GL_{2}(\mathbb{Z})[/imath] of finite order
Prove that any element of [imath]GL_{2}(\mathbb{Z})[/imath] of finite order has order [imath]1,2,3,4,6[/imath] using FIELD THEORY. My idea is to reduce such a finite order matrix say [imath]A[/imath] with order [imath]n[/imath] to modulo a prime [imath]p[/imath]. [imath]Det A=\pm1[/imath] so [imath]A[/imath] will land inside [imath]G=\{M\in GL_{2}(\mathbb{F}_{p}):Det M=\pm 1\}[/imath]. [imath]|G|=2p(p^{2}-1)[/imath]. So [imath]n|2p(p^{2}-1)[/imath] for all prime [imath]p[/imath]. I don't know how to proceed after this. |
1574317 | The probability that the product of 100 die rolls [imath]\le[/imath] [imath]a^{100}[/imath] for [imath]1 < a < 6[/imath]
Suppose that a fair [imath]6[/imath] sided die is rolled 100 times. Let [imath]X_i[/imath] be the value obtained on the [imath]i[/imath]th roll. Compute an approximation for [imath] P\left(\prod X_i \le a^{100}\right)[/imath] for [imath]1 < a < 6[/imath]. I'm thinking about using Central Limit Theorem but not sure how. Any help is appreciated! | 2248193 | Rolling a die 100 times and probability of product of results
Ello, my teacher gave me a simple exercise but I'm stuck. Rolling a die 100 times, denote the outcome of roll [imath]i[/imath] by [imath]X_i[/imath]. Estimate the probability [imath] \Pr\{\prod_{i=1}^{100}X_i\leq a^{100}\}[/imath] for real [imath]1<a<6[/imath] I think that [imath] \Pr\{\prod_{i=1}^{100}X_i\leq a^{100}\}=(\Pr\{X_1\leq a\})^{100}[/imath] by trying to find a solution numerically. Is it correct and why ? Thanks, Herosix |
2063696 | How do I prove this limit does not exist: [imath]\lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} [/imath] i ve a doubt
How do I prove that this limit does not exist? [imath]\lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} [/imath] My attempt: When you approach from from left towards zero , say i take -0.00000000000001 . i substitute in expression i get (-1) . But if i take 0.000000000001 and substitue i get (=1) (by applying L'Hop) . But if i dont do this and apply L'Hop straighaway i get 1 . | 1032039 | How do I prove this limit does not exist: [imath]\lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} [/imath]
How do I prove that this limit does not exist? [imath]\lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} [/imath] My attempt: When you approach from from left towards zero , say i take -0.00000000000001 . i substitute in expression i get (-1) . But if i take 0.000000000001 and substitue i get (=1) (by applying L'Hop) . But if i dont do this and apply L'Hop straighaway i get 1 . |
624984 | How prove this [imath]H_{2n}-H_{n}+\frac{1}{4n}>\ln{2}[/imath]
Show that, for every positive integer [imath]n[/imath], [imath]\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}+\dfrac{1}{4n}>\ln{2}[/imath] I know this [imath]\lim_{n\to\infty}\dfrac{1}{n+1}+\dfrac{1}{n+2}+\cdots+\dfrac{1}{2n}=\ln2[/imath] and use this [imath]\ln{(1+\dfrac{1}{n})}<\dfrac{1}{n}[/imath] is not useful But for this inequality I can't. Thank you | 549686 | How to prove this inequality?[imath]\frac{1}{n+1} + \frac{1}{n+2} + \cdots+\frac{1}{n+n} + \frac{1}{4n} > \ln 2[/imath]
[imath]\frac{1}{n+1} + \frac{1}{n+2} + \cdots +\frac{1}{n+n} + \frac{1}{4n} > \ln 2[/imath] [imath]n[/imath] is positive integer. Thank you ! |
2252248 | Infinite Product of 1-1/n^4
When I look at wolframalpha, I get [imath]\prod_{n=2}^\infty \left(1-\frac{1}{n^4}\right) = \frac{\sinh(4\pi)}{4\pi}.[/imath] My only guess where this comes from would be the euler's sine product formula[imath]\sin(z) = z\prod_{n=1}^\infty \left(1-\frac{z^2}{n^2\pi^{2}}\right).[/imath] But, this gives [imath]\sinh(π)=-i\sin(i\pi)= i\pi\prod\left(1+\frac{1}{n^2}\right).[/imath] I've fiddled around and cannot manage to get an [imath]n^{4}[/imath] showing up anywhere. | 1781117 | Prove that [imath]\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}[/imath]
The question Prove that: [imath]\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}[/imath] What I've tried Knowing that: [imath]\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)[/imath] evaluating at [imath]z=i[/imath] gives [imath] \frac{e^π - e^{-π}}{2i} = \sin(πi) = πi \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right)[/imath] so: [imath] \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right) = \frac{e^π - e^{-π}}{2π}[/imath] I'm stucked up and don't know how to continue, any help? |
2251664 | Complex integration of [imath]\int_{-\infty}^\infty \frac{dx}{(x^2+1)^2(x^2+16)}[/imath]
Complex integration of [imath]\int_{-\infty}^\infty \frac{dx}{(x^2+1)^2(x^2+16)}[/imath]. I see that the poles are at [imath]i,-i,4i,-4i[/imath] and only [imath]i,4i[/imath] are in the upper half plane. Then [imath]\int_{-\infty}^\infty \frac{dx}{(x^2+1)^2(x^2+16)}= 2\pi i[ Res(f(x),i)+Res(f(x),4i)][/imath]. [imath]Res(f(x),4i)=\frac{1}{(x^2+1)^2} |_{4i} = 1/225[/imath] since 4i is of order 1. [imath]Res(f(x),4i)=\frac{d}{dx}[\frac{1}{x^2+16}]|_i=\frac{-2x}{(x^2+16)^2}|_i=-2i/225[/imath] since i is a pole of order 2. This gives that the integral is [imath]2\pi i(\frac{1}{225}+\frac{2i}{225})[/imath] but the answer isn't right. It should be [imath]\frac{3\pi}{100}[/imath] according to wolfram. | 1725288 | Evaluate the integral using the theory of residues: [imath]\int_{-\infty}^{\infty} \frac{dx}{(x^2+1)^2(x^2+16)}[/imath]
[imath]\int_{-\infty}^{\infty} \frac{dx}{(x^2+1)^2(x^2+16)}[/imath] My attempt: We are integrating over the real axis, which is the real part of a set of complex numbers, so [imath]=\int_{-\infty}^{\infty} \frac{dz}{(z^2+1)^2(z^2+16)}[/imath] [imath]=\int_{-\infty}^{\infty} \frac{dz}{((z+i)(z-i))^2(z+4i)(z-4i)}[/imath] But I notice that all of the singularities are imaginary! Should this give me pause? What should I do now? |
2252457 | Solving a problem on limits
[imath]\lim_{x\to 1}\left(\frac{p}{1-x^p}-\frac{q}{1-x^q}\right) , p,q\in N[/imath] I tried fitting the formula [imath]\lim_{x\to a}\frac{x^n - a^n}{x-a} = na^{n-1}[/imath] but could not proceed further. Also I don't think that we should directly apply L'Hôpital's rule as it will be too lengthy. | 1945523 | Showing that [imath]\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)=6[/imath]
How does one evaluate the following limit? [imath]\lim_{x \to 1} \left(\frac{23}{1-x^{23}}-\frac{11}{1-x^{11}} \right)[/imath] The answer is [imath]6[/imath]. How does one justify this answer? Edit: So it really was just combine the fraction and use L'hopital's rule twice (because function and its first derivative are of indeterminate form at [imath]x=1[/imath]). This problem is more straightforward than it seems at first. |
2252602 | polynomial such that p([imath]\sqrt2[/imath]+[imath]\sqrt3[/imath])=0
Are these following exercises equivalent? Let [imath]\mathbb R[/imath] be an extension of [imath]\mathbb Q[/imath].Find a polynomial p(x) in [imath]\mathbb Q[/imath][x]-{[imath]0[/imath]} such that p([imath]\sqrt2[/imath]+[imath]\sqrt3[/imath])=[imath]0[/imath]. and 2.Find the minimal polynomial of [imath]\sqrt2[/imath]+[imath]\sqrt3[/imath] over[imath]\mathbb Q[/imath][x] | 1662080 | Find the minimal polynomial of [imath]\sqrt2 + \sqrt3 [/imath] over [imath]\mathbb Q[/imath]
I have no idea how to do this. To find the minimal polynomial of say [imath]\sqrt2 + \sqrt3[/imath], we need to find the monic polynomial [imath]p \in \mathbb Q[/imath] (correct if I am wrong but monic polynomial is when the coefficient of the highest degree term is [imath]1[/imath]) of the smallest possible degree such that [imath]\sqrt2 + \sqrt3[/imath] is a root of [imath]p[/imath]. If we let [imath]u=\sqrt2 + \sqrt3[/imath] then [imath]u ^2 = 5+ 2 \sqrt6 \iff u^2 - 5 = 2 \sqrt6 [/imath], then [imath](u^2 - 5)^2=24 \iff u^4 -10u^2 +1=0[/imath] All I did was keep squaring until all of the irrational terms go away. But what next? Am I doing this correctly and what do we do next if I am? |
2252868 | Monstrous integral [imath]\int_{0}^{1}\frac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}dx[/imath]
How to show the following [imath]\displaystyle \int_{0}^{1}\dfrac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}dx \\ = \pi\sqrt{\frac{1+\sqrt{13}}{78}} \log \left(\frac{1+\sqrt{13}+\sqrt{2 \sqrt{13}-2}}{4} \right)+\pi\sqrt{\frac{\sqrt{13}-1}{78}}\tan ^{-1}\left(\frac{\sqrt{5+2 \sqrt{13}}}{3} \right)[/imath] It seems that it can be transformed into an elliptic integral. Maybe we can define [imath]I(a) = \int_{0}^{1}\dfrac{\arcsin{a\sqrt{x}}}{x^4-2x^3+2x^2-x+1}dx [/imath] Then by differentiation [imath]I'(a) = \int_{0}^{1}\dfrac{\sqrt{x}}{(\sqrt{1-a^2x})(x^4-2x^3+2x^2-x+1)}dx [/imath] Another approach might involve series expansion of [imath]\arcsin(x)[/imath] [imath]\arcsin(x) = \sum_{n\geq 0}\frac{{2n \choose n} }{4^n(2n+1)}x^{2n+1}[/imath] To be honest it seems helpless. Reference http://integralsandseries.prophpbb.com/topic808.html | 683454 | Evaluating [imath]\int_{0}^{1}\frac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}\operatorname d\!x[/imath]
Find this integral [imath]\operatorname I=\int\limits_{0}^{1}\dfrac{\arcsin{\sqrt{x}}}{x^4-2x^3+2x^2-x+1}\operatorname d\!x[/imath] My try: let [imath]f(x)=x^4-2x^3+2x^2-x+1[/imath] I found [imath]f(1-x)=(1-x)^4-2(1-x)^3+2(1-x)^2-x+1=x^4-2x^3+2x^2-x+1=f(x)[/imath] so [imath]I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx[/imath] so [imath]2I=\int_{0}^{1}\dfrac{\arcsin{\sqrt{x}}+\arcsin{\sqrt{(1-x)}}}{x^4-2x^3+2x^2-x+1}dx[/imath] then I can't,Thank you very much |
2253224 | Show that for any reals [imath]a; b; x; y \ge 0, (ax + by)^2 \le (a^2 + b^2)(x^2 + y^2)[/imath]. (Hint: Use Cauchy-Schwarz inequality in [imath]\mathbb{R}^{2}[/imath].)
Having a little trouble with this. The Cauchy-Schwatrz inequality for vectors [imath]u[/imath] and [imath]v[/imath] is generally [imath]|(u,v)^2|\le (u,u)\cdot(v,v).[/imath] Then for a two dimensional space its [imath]|(u,v)^2|= (\|u\| \|v\|\cos(\theta))^2\le \|u\|^2\|v\|^2.[/imath] What I did was set two vectors [imath]u[/imath] and [imath]v[/imath] with the values [imath]u=(a,b), v=(x,y).[/imath] Then I used the inequality to plug in the values. [imath]||u||=(a^2+b^2),\qquad ||v||=(x^2+y^2),[/imath] \begin{align*} \cos(\theta)&= (u\cdot v)/(||u||\cdot||v||)\\ &= (ax+by)/((a^2+b^2)\cdot(x^2+y^2)) \end{align*} [imath]((a^2+b^2)\cdot (x^2+y^2)\cdot(ax+by)/((a^2+b^2)\cdot(x^2+y^2)))^2\le (a^2+b^2)^2\cdot(x^2+y^2)^2[/imath] after canceling becomes [imath](ax+by)^2\le(a^2+b^2)^2\cdot(x^2+y^2)^2.[/imath] How could I get rid of the square on the terms on the right side of the inequality? Edited: Looks like a duplicate but when I searched it did not come up on SE as the inequality was rearranged in the previous question. Thanks DcMcMor for the editing. | 1925766 | Prove [imath](a^2+b^2)(c^2+d^2)\ge (ac+bd)^2[/imath] for all [imath]a,b,c,d\in\mathbb{R}[/imath].
Prove [imath](a^2+b^2)(c^2+d^2)\ge (ac+bd)^2[/imath] for all [imath]a,b,c,d\in\mathbb{R}[/imath]. So [imath](a^2+b^2)(c^2+d^2) = a^2c^2+a^2d^2+b^2c^2+b^2d^2[/imath] and [imath](ac+bd)^2 = a^2c^2+2acbd+b^2d^2[/imath] So the problem is reduced to proving that [imath]a^2d^2+b^2c^2\ge2acbd[/imath] but I am not sure how to show that |
2252862 | If [imath]f:[0,1]\rightarrow [0,1][/imath] is increasing function. Suppose [imath]0. Show f has fixed point.[/imath]
A Fixed point of [imath]f[/imath] is a point [imath]r[/imath] such that [imath]f(r)=r[/imath]. I am not really convinced by this theorem holds for any function (even discontinuous functions). For example, I sketched this function: This function has [imath]f(0)>0, f(1)<1[/imath], it is increasing and has no value [imath]r[/imath] where [imath]f(r)=r[/imath]. Am I missing something or did the question miss a condition? | 791411 | Every increasing function from a certain set to itself has at least one fixed point
I need a hint for the following question: Let [imath]S[/imath] be a nonempty ordered set such that every nonempty subset [imath]E\subseteq S[/imath] has both a least upper bound and a greatest lower bound. Suppose [imath]f:S \rightarrow S[/imath] is a monotonically increasing function. Show that there exists an [imath]x\in S[/imath] so that [imath]f(x)=x[/imath]. My reasoning so far has been as follows: I wanted to prove this by contradiction. If we assume the statement is false, this implies that for each [imath]x\in S[/imath], [imath]f(x) > x[/imath] or [imath]f(x) < x[/imath]. Let [imath]A:=\{x:f(x)<x\}[/imath], [imath]B:=\{x:f(x)>x\}[/imath]. Let's assume [imath]A[/imath] is empty. Then it is easy to show a contradiction. We know that [imath]\sup(B)\in B[/imath]. But that means that [imath]f(\sup(B))>\sup(B)[/imath] which is impossible. The same idea applies if we assume [imath]B[/imath] is empty. What I am trying to do is prove the case where both [imath]A[/imath] and [imath]B[/imath] are nonempty. To prove the first part I haven't even used the monoticity of the function [imath]f[/imath], so I know that I have to use it at this point. That means trying to find a contradiction by finding [imath]x,y\in S[/imath], [imath]x\le y[/imath], [imath]f(x)>f(y)[/imath]. I was trying to look at the supremum and infimum of [imath]A[/imath] and [imath]B[/imath] but that didn't seem to lead me anywhere. I would appreciate hints! |
2253004 | When can we say two matrices have the same Jordan form?
When can we say two matrices have the same Jordan form? Well, it's obvious that if they're similar then they have the same Jordan form, but can we look further? What if they have the same eigenvalues? That's not enough either because [imath]\begin{bmatrix}0 & 1\\0&0\end{bmatrix}[/imath] and [imath]\begin{bmatrix}0 & 0\\0&0\end{bmatrix}[/imath] have the same values but they are not similar and therefore don't have the same Jordan form. I can't find an example where two matrices have the same characteristic equation and minimal polynomial yet they are not similar but I am unable to prove it. Is it enough to have the same algebraic multiplicity and geometric multiplicity? I am not looking for the minimal condition where two matrices have the same Jordan form | 56673 | Does equality of characteristic polynomials guarantee equivalence of matrices?
I have a qualifying exam coming up in a couple days and I am just trying to understand some pathological examples I have in my notes. I will list a similar problem which I know the solution to and then the question. True or False: let [imath]A,B \in \mathbb{Q}^{n \times n }[/imath] Suppose [imath]xI-A[/imath] and [imath]xI-B[/imath] are equivalent then [imath]\det(xI-A) = \det(xI-B)[/imath]. I wrote in my notes this is true because any two matrices are equivalent in [imath]\mathbb{Q[x]}^{n \times n}[/imath] if and only if they have the same invariant factors. Since the characteristic polynomial is a product of the invariant factors it follows that [imath]\det(xI-A) = \det(xI-B)[/imath]. True or False: let [imath]A,B \in \mathbb{Q}^{n \times n }[/imath]. If [imath]\det(xI-A) = \det(xI-B)[/imath] (in [imath]\mathbb{Q}[x][/imath]) then [imath]xI-A[/imath] and [imath]xI-B[/imath] are equivalent in [imath]\mathbb{Q}[x]^{n \times n}[/imath] I think this is false because of the fact that equality of determinants is not strong enough to guarantee all invariant factors are equal but I do not have an example. |
1195825 | Anybody know a proof of [imath]\prod_{n=1}^\infty\cos(x/2^n)=\sin x/x[/imath].
This is actually an exercise from Apostol's Mathematical Analysis. Ch. 8 Ex 42. which asks to find all real values [imath]x[/imath] for which [imath]\prod_{n=1}^\infty \cos\left(\large\frac{x}{2^n}\right)[/imath] converges. I've shown that the product converges for all [imath]x[/imath]. The problem then asks to find what values the product converges to. By playing around with Wolfram Alpha, I found that [imath]\large\prod_{n=1}^\infty\cos\left(\frac{x}{2^n}\right)=\frac{\sin (x)}{x}.[/imath] I can't figure out how to prove this. | 2791593 | To prove [imath] (\sin x)/x=\cos(x/2)×\cos(x/4)×\cos(x/8)......[/imath]
I inverted the LHS such that first term is [imath]\cos(x/2^n)[/imath] then and multiplied it by [imath]2\sin(x/2^n)/2\sin(x/2^n)[/imath] and evaluated it by [imath]\sin2A=2\sin A\cos A[/imath] but now I am stuck.Are any other ways to prove it? Please give suggestions.what should I do? |
2253771 | Number of partitions of an integer, if the partitions are no larger than [imath]2[/imath]
Find an explicit formula for the number of partitions of a positive integer [imath]k[/imath] into parts no larger than [imath]2[/imath]? I was thinking about it and worked on it, I've come up with [imath]1^k+2^k= 3^n[/imath]. | 1269615 | How many ways can [imath]133[/imath] be written as sum of only [imath]1s[/imath] and [imath]2s[/imath]
Since last week I have been working on a way, how to sum [imath]1[/imath] and [imath]2[/imath] to have [imath]133[/imath]. So for instance we can have [imath]133[/imath] [imath]1s[/imath] or [imath]61[/imath] [imath]s[/imath]2 and one and so on. Looking back to the example: if we sum: [imath]1 + 1... + 1 = 133[/imath] there is only one way. But for the second one there will be [imath]131[/imath] possible ways. And I have to do this for every possible combination. I am stuck with it and I have no idea whatsoever on how to begin. Any ideas or methods I could use guys? |
2254304 | Groups with order [imath]p^nq[/imath] with [imath]p[/imath] and [imath]q[/imath] primes.
Show that if [imath]G[/imath] is a group with [imath]|G| = p^nq[/imath], with [imath]p[/imath] and [imath]q[/imath] are primes and [imath]p > q[/imath], then [imath]G[/imath] contains a unique normal subgroup of index [imath]q[/imath]. | 379349 | If [imath]|G|=p^nq[/imath], then [imath]G[/imath] contains a unique normal subgroup of index [imath]q[/imath]
I'm trying to prove if [imath]|G|=p^nq[/imath] with [imath]p\gt q[/imath], primes, then [imath]G[/imath] contains a unique normal subgroup of index [imath]q[/imath]. I know by the first Sylow theorem that G has a Sylow p-subgroup [imath]P[/imath] with [imath][G:P]=q[/imath]. My problem is prove that it's unique. I need help Thanks in advance |
839350 | Find the remainder when [imath]x^{100}[/imath] is divided by [imath]x^2-3x+2[/imath]
We have to find the remainder when [imath]x^{100}[/imath] is divided by [imath]x^2-3x+2[/imath].I tried to use the remainder theorem but am not just able to solve it.please help. | 772326 | Give the remainder of [imath]x^{100}[/imath] divided by [imath](x-2)(x-1)[/imath].
What will be the remainder obtained when the polynomial [imath]x^{100}[/imath] is divided by the polynomial [imath](x-2)(x-1)[/imath]. I used remainder theorem but it had no impact in the solution. |
2253911 | Mth position of the Nth Ordered permutation of an ordrered set all elements taken at once
Since there is no ordering in sets I am not sure what would one call a set that is strictly ordered, aka an array in programming e.g. [imath]S={a_1,a_2,\cdots,a_n}[/imath] where [imath]a_k< a_{k+1}[/imath] the first ordered permutation will be [imath][a_1,a_2,\cdots,a_n][/imath] the second will be [imath][a_2,a_1,\cdots,a_n][/imath] and so on Is there a non recursive formula to calculate the Mth position of the Nth Ordered permutation? so that one will be able to calculate all the elements of the Nth entry without having to calculate all the ordered permutations before it? For the above two example Let [imath]P[N,M,S][/imath] be the Mth position in the Nth ordered permutaion of S then [imath]P[1,1,S] = a_1,\quad P[1,2,S] = a_2,\quad P[1,n,S] = a_n[/imath] [imath]P[2,1,S] = a_2,\quad P[2,2,S] = a_1,\quad P[2,3,S] = a_3,\quad P[2,n,S] = a_n[/imath] | 84112 | Elementary formula for permutations?
Suppose I fix [imath]n[/imath] and let [imath]\sigma_k[/imath] represent the [imath]k[/imath]th permutation of [imath]S_n[/imath] with respect to some ordering (whatever ordering might serve my purpose). Is there an elementary formula for [imath]\sigma_k(i)[/imath] which requires only [imath]i, k,[/imath] and [imath]n[/imath]? Is one known for small [imath]n[/imath], perhaps even as small as 4? |
1438443 | A question on the group algebra [imath]k[G][/imath] of a finite abelian group
Let [imath]G[/imath] be a finite abelian group and [imath]k[/imath] a field in which the group order is invertible. Then the group algebra [imath]k[G][/imath] is a semisimple ring by Maschke's theorem. As [imath]k[G][/imath] is also abelian, the Artin-Wedderburn theorem implies, that [imath] k[G]=K_1\oplus\ldots\oplus K_n [/imath] for fields [imath]K_i[/imath]. I have two questions and I have to admit that they are both a little awkward: Is [imath]K_i[/imath] an algebraic field extension of [imath]k[/imath]? I am sure that this is true but I can't find an argument. If (1) is true and [imath]k[/imath] is algebraically closed, I can deduce, that each irreducible representation of [imath]G[/imath] is one-dimensional. However, this is quite easy to prove directly and the structure of [imath]k[G][/imath] (obtained by non-trivial theorems) seems to contain more information. Can I deduce something about the number of irreducible representations? What can I say for [imath]k[/imath] not algebraically closed? | 1233403 | Corollary to Maschke's Theorem.
If in Maschke's Theorem, for group ring KG where K is any field s.t. char [imath]K \nmid |G|[/imath], I take G to be finite , then I know Maschke implies that KG will be semisimple so it is isomorphic to a direct sum of Matrices over division rings, but how does adding that G is also abelian implies that KG will be direct sum of fields? |
2251032 | Image of first quadrant under [imath]f(z)=\frac{z-i}{z+i}[/imath]
help with this excercise Find the image of region [imath]K=\{z=x+iy:x>0,y>0\}[/imath] under the function [imath]f(z)=\frac{z-i}{z+i}[/imath] [imath]w=\frac{z-i}{z+i}=\frac{(x^2+y^2-1)+((y-1)x-(y+1)x)i}{x^2+(y+1)^2}[/imath] I have. [imath]u(x,y)=\frac{x^2+y^2-1}{x^2+(y+1)^2}, v(x,y)=\frac{(y-1)x-(y+1)x)}{x^2+(y+1)^2}.[/imath] ok? then? | 1158885 | Describe the image of the set [imath]\{z=x+iy:x>0,y>0\}[/imath] under the mapping [imath]w=\frac{z-i}{z+i}[/imath]
Describe the image of the set [imath]\{z=x+iy:x>0,y>0\}[/imath] under the mapping [imath]w=\frac{z-i}{z+i}[/imath] So from this mapping , I can see that [imath]a=1, b=-i, c=1, d=i[/imath] thus [imath]ad-bc=i+i=2i \not =0[/imath] so this is a Mobius transformation. Solving for [imath]z[/imath] I got [imath]z=\frac{i+iw}{1-w}[/imath] for [imath]w=u+iv[/imath], we have [imath]z=\frac{-2v+i(1-u^2-v^2)}{(1-u^2)+v^2}[/imath] so [imath]x=\frac{-2v}{(1-u^2)+v^2}[/imath] and [imath]y=\frac{1-u^2-v^2}{(1-u^2)+v^2}[/imath] Since [imath]x>0[/imath], [imath]v<0[/imath] and since [imath]y>0[/imath], [imath]1-u^2 -v^2 >0[/imath], thus [imath]u^2 +v^2 <1[/imath], this implies that the image is the interior of a unit circle center at the origin, but since [imath]v<0[/imath], we only take the negative part. |
2256272 | Convergence of [imath]\int^\infty_0\frac{1}{1+x^2\sin^2(5x)}\,dx[/imath]
I need to find out whether the following improper integral converges: [imath]\int^\infty_0\frac{1}{1+x^2\sin^2(5x)}\,dx[/imath] I tried two comparison tests that failed, any ideas? | 86837 | Does [imath]\int_{0}^{\infty}\frac{dx}{1+(x\sin5x)^2}[/imath] converge?
I would like your help with deciding whether the following integral converges or not: [imath]\int_{0}^{\infty}\frac{dx}{1+(x\sin5x)^2}.[/imath] I tried to compare it to other functions and to change the variables, but it didn't work for me. Thanks a lot! |
2256167 | What's the meaning of factors 'collapsing' in quotients like [imath]\mathbb{Z}_4\times\mathbb{Z}_6/\langle(0,1)\rangle[/imath]?
My book says that in [imath]\mathbb{Z}_4\times\mathbb{Z}_6/\langle(0,1)\rangle[/imath] since we're setting everything in [imath]\langle(0,1)\rangle[/imath] to [imath]0[/imath], it's like. That is, the whole second factor [imath]\mathbb{Z}_6[/imath] of [imath]\mathbb{Z}_4\times\mathbb{Z}_6[/imath] is collapsed, leaving just the factor [imath]\mathbb{Z}_4[/imath]. Then, for [imath]\mathbb{Z}_4\times\mathbb{Z}_6/\langle(0,2)\rangle[/imath] It says that the same thing happens, but now [imath]\mathbb{Z}_6[/imath] is collapsed by a subgroup of order [imath]3[/imath], therefore givinf a group in the second factor of order [imath]2[/imath], so it's isomorphic to [imath]\mathbb{Z}_4\times\mathbb{Z}_2[/imath] I tried to visualize the cosets for both groups, here they are: [imath]\mathbb{Z}_4\times\mathbb{Z}_6/\langle(0,1)\rangle[/imath] [imath]H = \{(0,0),(0,1),(0,2),(0,3),(0,4),(0,5)\}[/imath] [imath]H_1 = \{(1,0),(1,1),(1,2),(1,3),(1,4),(1,5)\}[/imath] [imath]H_2 = \{(2,0),(2,1),(2,2),(2,3),(2,4),(2,5)\}[/imath] [imath]H_3 = \{(3,0),(3,1),(3,2),(3,3),(3,4),(3,5)\}[/imath] And [imath]\mathbb{Z}_4\times\mathbb{Z}_6/\langle(0,2)\rangle[/imath] [imath]H = \{(0,0),(0,2),(0,4)\}[/imath] [imath]H_1 = \{(1,0),(1,2),(1,4)\}[/imath] [imath]H_2 = \{(2,0),(2,2),(2,4)\}[/imath] [imath]H_3 = \{(3,0),(3,2),(3,4)\}[/imath] But I cannot visualize this 'collapse' thing | 518572 | Calculate the Factor Group [imath](\mathbb{Z}_4 \times \mathbb{Z}_6)/\langle(0,2)\rangle[/imath]
I am attempting to understand and compute: [imath](\mathbb{Z}_4 \times \mathbb{Z}_6)/\langle(0,2)\rangle[/imath] I know [imath](0,2)[/imath] generates [imath]H = \{(0,0),(0,2),(0,4)\}[/imath], which has order 3 because there are 3 elements. Now, I must find all the cosets which there should be 8. The reason there should be 8 is because [imath]$4\cdot 6/3 = 8$[/imath]. This is where I run into the issue. I have a hard time generating the cosets, but furthermore, once I actually do get the cosets, I have a hard time analyzing them and knowing what it means. If someone could walk me through a step-by-step solution that would be great. The answer is that the factor group is isomorphic to [imath]\mathbb{Z}_4 \times \mathbb{Z}_2[/imath]. Reference: Fraleigh, A First Course in Abstract Algebra, p. 147, Example 15.10. |
2257000 | Prove or disprove there always exists a set of four numbers of which product is a perfect square out of arbitrarily chosen [imath]48[/imath] positive integers ...
Question: Assume you choose a set of arbitrary [imath]48[/imath] positive integers so that the number of distinct prime factors of all those numbers is [imath]10[/imath]. Prove or disprove there always exists a set of four numbers out of the [imath]48[/imath] positive integers of which product is a perfect square. I think that it uses pigeon hole, but I couldn't make a progress. | 2256988 | Prove or disprove there exists a set of four numbers out of the [imath]48[/imath] positive integers of which product is a perfect square
Question: Assume you choose any [imath]48[/imath] positive integers so that the number of different prime factors of all those numbers is [imath]10[/imath]. Prove or disprove there exists a set of four numbers out of the [imath]48[/imath] positive integers of which product is a perfect square. I think that it uses pigeon hole, but I couldn't make a progress. |
2257433 | Solving integral 1/(1+x^2)
Is it true that [imath]\int_{- \infty }^{\infty} \frac{1}{1+x^2}\, dx=[\tan^{-1} x]_{- \infty }^{\infty} = \frac{\pi}{2}-(-\frac{\pi}{2})=\pi[/imath]? I'm integrating after a long period of time of inactivity, and sometimes I have doubts. | 1122837 | How [imath]\int_{-\infty}^{\infty}\frac{dx}{1+x^2}[/imath] exists?
How [imath]\int_{-\infty}^{\infty}\frac{dx}{1+x^2}[/imath] exists? It is difficult question to me. i have tried to evaluate by using fact that [imath]\int_{-\infty}^{\infty} f(x) \ dx =\int_{-\infty}^{0} f(x)\, dx + \int_{0}^{\infty} f(x)\, dx[/imath] but i have failed in this one. any hints on this one? |
2257393 | Using the defenition of a limit, show this: [imath]\lim_{x\rightarrow a} f(x)= \lim_{h\rightarrow 0}f(a+h)[/imath] when at least one of the two limits exist.
Using the defenition of a limit, show that this is true: [imath]\lim_{x\rightarrow a} f(x)= \lim_{h\rightarrow 0}f(a+h)[/imath] when at least one of the two limits exist (thus also proof that the other one exists). The definition of limits for which I got is [imath]\lim_{x\rightarrow a} f(x) = b[/imath] this is true if with [imath]x\in dom(f)[/imath] and [imath]d(x,a)<\delta[/imath] then [imath]d(f(x),b)<\epsilon[/imath] where [imath]\epsilon, \delta[/imath] are positive real numbers. I have no idea where to start here. | 2251458 | Given [imath]f: \Bbb R\rightarrow\Bbb R[/imath] and a point [imath]a\in\Bbb R[/imath]. Prove [imath]lim_{x\rightarrow a} f(x)=\lim_{h\rightarrow 0} f(a+h)[/imath] if 1 of the limits exists.
We are given a function [imath]f: \Bbb R \rightarrow \Bbb R[/imath] and a point [imath]a \in \Bbb R[/imath]. Prove that [imath]lim_{x \rightarrow a} f(x)=\lim_{h \rightarrow 0} f(a+h)[/imath] if one of the two limits exists. So I thought I had to prove : 1) If [imath]lim_{x \rightarrow a} f(x)[/imath] exists, does [imath]\lim_{h \rightarrow 0} f(a+h)[/imath] exist, and are they equal? 2) If [imath]\lim_{h \rightarrow 0} f(a+h)[/imath] exists, does [imath]lim_{x \rightarrow a} f(x)[/imath] exist and are they equal? I was thinking about introducing a function [imath]g(x)[/imath] and letting [imath]lim_{x \rightarrow a} f(x)=b[/imath], and [imath]lim_{x \rightarrow a} g(x)=c[/imath], and use that [imath]lim_{x \rightarrow a} (f(x)+g(x))=b+c[/imath], but I'm not sure if you need that here or how to continue. |
2257904 | Minimum distance between [imath]e^x[/imath] and [imath]ln x [/imath]
We have to find minimum distance between [imath]e^x[/imath] and [imath]ln x[/imath] I thought they are mirror image along [imath]y=x[/imath] . So the point which would be at minimum distance would have slope -1 . from that I got the answer as [imath]\sqrt 2[/imath] . which is correct . But I want to kno2 ifbthere is some other good method . | 1644512 | Minimum distance between the curves [imath]f(x) =e^x[/imath] and [imath]g(x) =\ln x[/imath]
What is the minimum distance between the curves [imath]f(x) =e^x[/imath] and [imath]g(x) = \ln x[/imath]? I didn't understand how to solve the problem. Please help me. |
2258072 | Direct Product of a Graded Ring.
Let [imath]R[/imath] be a graded ring, for example with nonnegative grading such that [imath]R=R_0 \oplus R_1 \oplus \dots[/imath] Is then [imath]R^n= R \times \cdots \times R[/imath] a graded ring (not with a trivial grading) ? If so, with what grading? | 666906 | Grading on the graded direct product
This question is related to this one. Probably it's obvious but could you tell me what is the grading on the graded direct product? I was thinking about [imath]^*\Pi M^i=\oplus_j(\Pi_i M^i_j)[/imath] where [imath]M^i_j[/imath] is the degree [imath]j[/imath]-part of [imath]M^i[/imath], but I couldn't prove it and I'm not sure I'm correct, especially because if [imath]R=M^i=k[x][/imath], then [imath](1,x,x^2,x^3,\ldots)[/imath] seems to belong to the graded direct product but I don't think it belongs to [imath]\oplus_j(\Pi_i M^i_j)[/imath]. |
2254097 | How does greatest common divisor effected by field extension
Let [imath]p(x)\in K[x][/imath] and [imath]q(x)\in K[x][/imath], [imath]F[/imath] is an extension of field [imath]K[/imath]. Will [imath]gcd(p(x),q(x))[/imath] over [imath]K[x][/imath] equal to [imath]gcd(p(x),q(x))[/imath] over [imath]F[/imath]. My guess is yes. I couldn't find an elegant way to prove it. | 84551 | GCD in polynomial rings with coefficients in a field extension
Let [imath]E/F [/imath] be a field extension and [imath]f,g[/imath] [imath]\in[/imath] [imath]F[x][/imath] (the polynomial ring with coefficients in [imath]F[/imath] ). Let's denote with [imath]\gcd_F(f,g)[/imath] the greatest common divisor of [imath]f[/imath] and [imath]g[/imath] in [imath]F[x][/imath]. Is it true that [imath]\gcd_F(f,g)[/imath] [imath]=[/imath] [imath]\gcd_E(f,g)[/imath] ? |
2257406 | Maximum value of [imath]4x-9y[/imath]
Suppose xand y are real numbers and that [imath]x^2 +9y^2 -4x +6y+4=0[/imath] then we have to find the maximum vale of 4x-9y I found that the equation given is that of ellipse . I can consider a line [imath]4x-9y=0[/imath] which cut the ellipe . bt by doing this its getting very comlicated . | 618106 | Finding the maximum value of a function on an ellipse
Let [imath]x[/imath] and [imath]y[/imath] be real numbers such that [imath]x^2 + 9 y^2-4 x+6 y+4=0[/imath]. Find the maximum value of [imath]\displaystyle \frac{4x-9y}{2}[/imath]. My solution: the given function represents an ellipse. Rewriting it, we get [imath]\displaystyle (x-2)^2 + 9(y+\frac{1}{3})^2=1[/imath]. To find the maximum of [imath]\displaystyle \frac{4x-9y}{2}[/imath], [imath]x[/imath] should be at its maximum and [imath]y[/imath] at its minimum. Solving the equations, I get that [imath]x = 3[/imath] and [imath]\displaystyle y = -\frac{2}{3}[/imath], but the answer i get is wrong. What am I doing wrong? |
2258521 | For positive integers [imath], [/imath] and [imath]$ with [/imath] = ^2[imath] and [/imath]\gcd(a,b)=1[imath], show that [/imath]a[imath] and [/imath]b$ are squares
Let [imath], [/imath] and [imath] be positive integers such that $ = ^2$. If $gcd(, ) = 1$, prove that there exist positive integers [/imath] and [imath]$ such that [/imath] = ^2[imath] and [/imath] = ^2[imath]. [/imath] I'm a bit lost. I know that 1 = ac + bd[imath], and I can simplify [/imath]ab = n^2[imath] to [/imath]a = \frac{n^2}{b}[imath] and [/imath]b = \frac{n^2}{a}[imath], but I don't know where to go from there. I try substituting [/imath]a[imath] and [/imath]b[imath] but cannot figure out how to get that [/imath]a = c^2[imath] and [/imath]b = d^2$. | 675917 | If [imath]\gcd(a, b) = 1[/imath] and if [imath]ab = x^2[/imath], prove that [imath]a, b[/imath] must also be perfect squares; where [imath]a,b,x[/imath] are in the set of natural numbers
Problem: If [imath]\gcd(a, b) = 1[/imath] and If [imath]ab = x^2[/imath] ,prove that [imath]a[/imath], [imath]b[/imath] must also be perfect squares; where [imath]a[/imath],[imath]b[/imath],[imath]x[/imath] are in the set of natural numbers I've come to the conclusion that [imath]a \ne b[/imath] and [imath]a \ne x[/imath] and [imath]b \ne x[/imath] but I guess that won't really help me.. I understand that if the [imath]\gcd[/imath] between two numbers if [imath]1[/imath] then they obviously have no common divisors but where do I go from this point? Any tips at tackling this would be great. It looks quite easy though I'm still trying to get my hand around these proofs! Any pointers in the right direction would be great. Thank you in advance, |
2257442 | Limits of functions proof help [imath]\lim_{x\to 2} \left(x^3\right) = 8[/imath]
Can anyone help me finish this proof? Prove that [imath]\lim_{x\to 2} \left(x^3\right) = 8[/imath] [imath]|x^3-8|=|(x-2)(x^2+2x+4)|=|x-2|<1[/imath] [imath]1<x<3[/imath] | 1870711 | Please verify my epsilon - delta proof [imath]\lim_{x\to 2}(x^3 )=8,[/imath] and [imath]0 < x < 4[/imath]
I am concern about the delta calculation given [imath]0 < x < 4[/imath]. I believe it works, but I am not 100 percent sure as I am new to proofs. Consider the function [imath]f(x) = x^3[/imath] for [imath]x ∈ ℝ[/imath] and [imath]0 < x < 4[/imath]. Prove that [imath]\lim _{ x\rightarrow 2 }{ { x }^{ 3 }=8 } [/imath] Proof: Let [imath]\varepsilon > 0[/imath] be given, then let [imath]\delta = \min {(\varepsilon/19,1)}[/imath]. Now suppose [imath]|x -2| < \delta = \min {(\varepsilon /19,1)}[/imath]. Then [imath]|x^3 - 8| = |x - 2||x^2 + 2x + 4|< |x - 2|<19[/imath] [imath](|x^2 + 2x + 4| < 19~\text{if}~|x - 2| < 1)< \varepsilon /19 [/imath] [imath](|x - 2| < \varepsilon /19 )= \varepsilon [/imath]. Thus this [imath]\delta[/imath] makes [imath]|x^3 - 8| < \varepsilon [/imath] whenever [imath]0 < |x - 2| < \delta[/imath]. Therefore, it follows that [imath]\lim _{ x\rightarrow 2 } x^3 = 8[/imath]. |
2258326 | M is a square martix, and [imath]φ^*[/imath] : [imath]V^* \to V^*[/imath], then why [imath]M^T=M_{φ^*_M}[/imath]?
Let M be a matrix, transpose [imath]M^T[/imath] is the matrix who's i,j component is the j,i component of M. And let [imath] \phi\in Hom(V,V)[/imath], we define [imath]\phi^*: V^* \to V^*[/imath] by [imath]\phi^*(f)(v)=f(\phi(v))[/imath] where [imath]V^*[/imath] is the dual space of [imath]V[/imath]. [imath]M_{φ^*_M}[/imath] is the matrix representation of [imath]φ^*_M[/imath] Then how to prove [imath]M^T=M_{φ^*_M}[/imath] | 1138462 | Transpose matrix dual map
how do I see that the representing matrix of the dual map [imath]f^*[/imath] between finite-dimensional dual spaces is given by the transpose of the representing matrix of [imath]f[/imath]? Here I want to assume that the matrix [imath]f^*[/imath] is represented with respect to the dual basis. Apparently this result is very well-known but I would like to see a proof of this. |
2259401 | Question with matrices, determinant and adjugate
Let [imath]A[/imath] be a [imath]n \times n[/imath] matrix with complex elements, [imath]L[/imath] a [imath]1 \times n[/imath] matrix with complex elements and [imath]C[/imath] a [imath]n \times 1[/imath] matrix with complex elements. Show that [imath]\det (A-CL)=\det(A)-LBC[/imath], where [imath]B=\textrm{adj}(A)[/imath] is the adjugate matrix of [imath]A[/imath]. | 1514408 | Matrix determinant lemma with adjugate matrix
I would like a proof of the following result, given on wikipedia. For all square matrices [imath]\mathbf{A}[/imath] and column vectors [imath]\mathbf{u}[/imath] and [imath]\mathbf{v}[/imath] over some field [imath]\mathbb{F}[/imath], [imath] \det(\mathbf{A}+\mathbf{uv}^\mathrm{T}) = \det(\mathbf{A}) + \mathbf{v}^\mathrm{T}\mathrm{adj}(\mathbf{A})\mathbf{u}, [/imath] where [imath]\mathrm{adj}(\mathbf{A})[/imath] is the adjugate matrix of [imath]\mathbf{A}[/imath]. Note that [imath]\mathbf{A}[/imath] may be singular. However, the proof given on wikipedia requires that [imath]\mathbf{A}[/imath] is nonsingular. |
776413 | Limit of a function with log
Find the value of [imath]\lim_{t\to 0}\left(\frac{1}{\ln(1 + t)}+\frac{1}{\ln(1-t)}\right).[/imath] I tried L'Hospital's Rule but can't get it to work. I can't seem to find the right algebra tricks to apply before attempting L'Hospitals' Rule. | 132329 | Evaluate [imath]\lim_{x \to 0} \left(\frac{1}{\ln(1+x)} + \frac{1}{\ln(1-x)}\right)[/imath]
Evaluate [imath]\displaystyle \lim_{x \to 0} \left(\frac{1}{\ln(1+x)} + \frac{1}{\ln(1-x)}\right)[/imath] I am having trouble starting this one. I couldn't see any log laws that I'm familiar with to rearrange the formula. I also tried combining the fractions and using L'Hospital's, but it only seemed to make things worse. What direction should I take with this? |
2259283 | prove that [imath]\lim_{n\to\infty}\int_{-\infty}^\infty \frac n {\sqrt {\pi}}e^{-(nx)^2} f(x) dx = f(0)[/imath]
prove that for any integrable function [imath]f[/imath] defined on [imath](-\infty,\infty)[/imath] (or maybe square integrable function.. my professor didn't gave exact imformation about [imath]f[/imath].) [imath]\lim_{n\to\infty}\int_{-\infty}^\infty \frac n {\sqrt {\pi}}e^{-(nx)^2} f(x) dx = f(0)[/imath] I think I have to get some [imath]\delta>0[/imath] and divide integral into three terms. [imath]\int_{-\infty}^{-\delta} f*g_n +\int_{-\delta}^{\delta}f*g_n + \int_\delta ^\infty f*g_n[/imath] ([imath]g_n(x)=\frac n {\sqrt {\pi}}e^{-(nx)^2}[/imath]) since [imath]f*g_n[/imath] converges uniformly to [imath]0[/imath] if [imath]x\neq0[/imath], first and third term vanish. so what I have to do is to take [imath]N[/imath] and [imath]\delta[/imath] (from the given value [imath]\epsilon>0[/imath]) so that if [imath]n>N[/imath] [imath]\left|\int_{-\delta}^{\delta}\frac n {\sqrt {\pi}}e^{-(nx)^2} f(x)dx - f(0) \right|<\epsilon[/imath] I can't go any further. | 1715781 | How to show this sequence is a delta sequence?
Consider the sequence [imath](\phi_n)_{n\in \mathbb{N}}[/imath] of test-functions [imath]\phi_n\in \mathcal{D}(\mathbb{R})[/imath] defined by [imath]\phi_n(x) = \dfrac{n}{\sqrt{\pi}}e^{-n^2x^2}.[/imath] I want to show that this is a delta sequence, in the sense that: [imath]\lim_{n\to\infty}(\phi_n,\psi)=(\delta,\psi)=\psi(0), \quad \forall \psi\in \mathcal{D}(\mathbb{R}).[/imath] In that case we know that we have [imath](\phi_n,\psi)=\dfrac{n}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-n^2x^2}\psi(x)dx.[/imath] The problem here is to compute this integral. The function [imath]x\mapsto e^{-n^2x^2}\psi(x)[/imath] has no singularities, hence contour integration wouldn't help. Integration by parts or substitution also doesn't seem to be of great help. The major problem is that there is a function [imath]\psi[/imath] inside the integral about which we don't know really anything apart from the fact that it is [imath]C^\infty[/imath] and with compact support. How can we compute this integral and in the end show that the sequence [imath](\phi_n)[/imath] is a delta sequence? Just to make clear, I'm here doing all of this in the context of the Riemann integral. |
2259549 | Floors Complicated Proof
Problem: Prove that [imath]\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor[/imath] for all real [imath]x[/imath] and [imath]y[/imath]. I have my proof down below. Its a little complicated though. This is not a duplicate of the same question on another page on math.stackexchange.com because I provide my answer. | 1920924 | Prove that [imath]\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor[/imath] for all real [imath]x[/imath] and [imath]y[/imath].
Prove that [imath]\lfloor 2x \rfloor + \lfloor 2y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor + \lfloor x+y \rfloor[/imath] for all real [imath]x[/imath] and [imath]y[/imath]. How should I solve this? I can't think of a way with casework and I can't really simplify it more. Thanks in advance for posting a proof! |
2259313 | How to sum up the series of x + 1/x + 1/(x + 1/x) + 1/(x + 1/x + 1/(x + 1/x)) for some finite number of terms?
I'm having trouble in coming up with a formula for this series: [imath]S = x + \frac1x + \cfrac1{x + \frac1x} + \cfrac1{x + \frac1x + \cfrac1{x + \frac1x}} +\cfrac1{x + \frac1x + \cfrac1{x + \frac1x} + \cfrac1{x + \frac1x + \cfrac1{x + \frac1x}} } + \cdots + a_n [/imath] How can I find [imath]S[/imath] ? Thanks :) | 2259263 | How to simplify the series of [imath]x + x^{-1} + (x + x^{-1})^{-1} + (x + x^{-1} + (x + x^{-1})^{-1})^{-1}[/imath]?
[imath]x + \frac1x + \cfrac1{x + \frac1x} + \cfrac1{x + \frac1x + \cfrac1{x + \frac1x}} +\cfrac1{x + \frac1x + \cfrac1{x + \frac1x} + \cfrac1{x + \frac1x + \cfrac1{x + \frac1x}} } + \cdots [/imath] How can this series be simplified? Consider me as a high school graduate. |
2259878 | Is [imath]L^p[/imath] isomorphic to [imath]l^p[/imath]?
From the Fourier transform we see that [imath]L^2([0,2\pi])[/imath] is isometrically isomorphic to [imath]l^2(\mathbb{Z})[/imath]. Is this true for other spaces and values of [imath]p[/imath] besides [imath]p=2[/imath]? What about if we ask about isomorphisms that are not necessarily isometric? | 97126 | If [imath]1\leq p < \infty[/imath] then show that [imath]L^p([0,1])[/imath] and [imath]\ell_p[/imath] are not topologically isomorphic
If [imath]1\leq p < \infty[/imath] then show that [imath]L^p([0,1])[/imath] and [imath]\ell_p[/imath] are not topologically isomorphic unless [imath]p=2[/imath]. Maybe I would have to use the Rademacher's functions. |
2258984 | What's wrong with my "proof" that the Lebesgue measure of [imath][0,1][/imath] is [imath]0[/imath]?
Proposition. [imath]\lambda [0,1] = 0[/imath] Proof. Let [imath]\varepsilon>0[/imath] be arbitrary. We will prove that [imath]\lambda[0,1] <\varepsilon[/imath]. Let [imath]q : \mathbb{N} \rightarrow \mathbb{R}[/imath] denote an injection with image equal to [imath]\mathbb{Q} \cap [0,1][/imath]. Let [imath]p : \mathbb{N} \rightarrow \mathbb{R}_{>0}[/imath] denote a sequence with [imath]\sum_{i \in \mathbb{N}} p_i < \varepsilon[/imath]. Then [imath]\lambda [0,1] = \lambda \bigcup_{i \in \mathbb{N}} (q_i+[-p_i/2,p_i/2]) \leq \sum_{i \in \mathbb{N}} \lambda (q_i+[-p_i/2,p_i/2]) = \sum_{i\in \mathbb{N}} p_i < \varepsilon[/imath] So [imath]\lambda[0,1] < \varepsilon[/imath]. Since this is true for all [imath]\varepsilon>0[/imath], we deduce that [imath]\lambda [0,1] = 0.[/imath] Question. What gives? | 1780580 | What is wrong in this proof: That [imath]\mathbb{R}[/imath] has measure zero
Consider [imath]\mathbb{Q}[/imath] which is countable, we may enumerate [imath]\mathbb{Q}=\{q_1, q_2, \dots\}[/imath]. For each rational number [imath]q_k[/imath], cover it by an open interval [imath]I_k[/imath] centered at [imath]q_k[/imath] with radius [imath]\epsilon/2^k[/imath]. The total length of the intervals is a geometric progression that sums up to [imath]\epsilon[/imath]. Each real number is arbitrarily close to a rational number since [imath]\mathbb{Q}[/imath] is dense in [imath]\mathbb{R}[/imath]. Thus, each real number is in one of the open intervals. Thus the entire real line is covered by the union of the [imath]I_k[/imath], thus [imath]\mathbb{R}[/imath] is a null set with measure zero. Clearly there is something wrong in the above proof, however I am not sure where is it? Thanks for any help. |
2259761 | Galois Theory Quadratic Subfield
Let [imath]ζ_7=e^{i2\pi/7}[/imath] be a 7th root of unity. The field [imath]\Bbb{Q}(ζ_7)[/imath] contains a quadratic subfield that can be expressed in the form of [imath]\Bbb{Q}(\sqrt{D})[/imath] where D is an integer. What is D? I understand that there is a field extension of order 6 and therefore there will be a quadratic subfield, but how do we find out what D is? | 2254683 | Find the quadratic sub field of [imath]\mathbb Q(\zeta_7)[/imath] which can be expressed in the form [imath]\mathbb Q(\sqrt D)[/imath],where [imath]D[/imath] is an integer.
Let [imath]\zeta_7= e^{i2\pi/7}[/imath] be the [imath]7[/imath]th root of unity.Find the quadratic sub field of [imath]\mathbb Q(\zeta_7)[/imath] which can be expressed in the form [imath]\mathbb Q(\sqrt D)[/imath],where [imath]D[/imath] is an integer. My try:let [imath]\omega=\zeta_7+\zeta_7^{-1}\in\mathbb{Q}(\zeta_7)[/imath] is real which is [imath]\omega=2\cos(2\pi/7)[/imath]. Thus [imath]K=\Bbb{Q}(\omega)[/imath] is a proper subfield of [imath]\Bbb{Q}(\zeta_7)[/imath]. And the polynomial [imath] p(x)=(x-\zeta_7)(x-\zeta_7^{-1})=x^2-\omega x+1 [/imath] has its coefficients in the subfield [imath]K[/imath]. Thus it has to be the minimal polynomial of [imath]\zeta_7[/imath], and [imath][K(\zeta_7):K]=2[/imath]. But the problem is that what is [imath]D?[/imath] and how can i find that?Thank you |
2254811 | Intersection of lines [imath]y = mx[/imath] and [imath]y = mx+c[/imath] in projective geometry.
I'm trying to solve the following problem: Embed [imath]\mathbb{R^2}[/imath] in the projective plane [imath]\mathbb{RP^2}[/imath] by the map [imath](x,y)\rightarrow [1,x,y][/imath]. Find the point of intersection in [imath]\mathbb{RP^2}[/imath] of the projective lines corresponding to the parallel lines [imath]y = mx[/imath] and [imath]y = mx+c[/imath] in [imath]\mathbb{R^2}[/imath]. So in the projective plane the two lines correspond to [imath][1,x,mx][/imath] and [imath][1,x,mx+c][/imath]I get that somehow the point of intersection is the point at infinity, but that point would have coordinates [imath][0,1,m][/imath] and that doesn't fall on our line. So I'm not sure I understand how this works | 2259422 | Projective geometry general question
Can someone please help me solve this problem? I am a bit confused. Embed [imath]R^2[/imath] in the projective plane [imath]RP^2[/imath] by the map [imath](x, y) → [1, x, y][/imath]. Find the point of intersection in [imath]RP^2[/imath] of the projective lines corresponding to the parallel lines [imath]y = mx[/imath] and [imath]y = mx + c[/imath] in [imath]R^2[/imath] (where [imath]c \neq 0[/imath]). My attempt: So this transformation sends each point in [imath]R^2[/imath] to the line through the origin which is generated by the vector [imath](1,x,y)[/imath]. Then the line [imath]y=mx[/imath] will go to the set [imath][1,x,mx], x \in \Re, x \neq 0 [/imath]. By the same reasoning, [imath]y=mx+c[/imath] will go to the set {[imath][1,x,mx+c], x \in \Re, x \neq 0 [/imath]} The problem is I can't quite understand how to define the points at infinity.I don't want to just do the question, I'd like to understand the basics and will be very grateful if someone could help me. |
2259897 | How do I prove that powers of 5 are the sum of two squares using mathematical induction
Using mathematical induction I need to prove [imath]5^n=a_n^2 +b_n^2[/imath] What I've tried P(1): [imath]5^1 = 1^2 +2^2[/imath] Which is true P(2): [imath]5^2 = 3^2 +4^2[/imath] Which is also true Now for the induction hypothesis P(k): [imath]5^k=a_k^2+b_k^2[/imath] Then the question suggests to prove for (k+2) which I guess still makes sense. P(k+2): [imath]5^{k+2} = 5^2 . 5^k[/imath] [imath]=(3^2 +4^2)(a_k^2+b_k^2)[/imath] This is where I get stuck. Apologies for formatting I'm fairly new to the site | 472192 | [imath] \exists a, b \in \mathbb{Z} [/imath] such that [imath] a^2 + b^2 = 5^k [/imath]
I saw this problem recently and found an elegant solution to it, and was curious to see if anybody would think of something else. Nice solutions to nice problems are fun to see! Problem: Prove that, for all non-negative integers [imath]k[/imath], there exist integers [imath]a[/imath], [imath]b[/imath] such that [imath]a^2+b^2=5^k[/imath]. Bonus: Prove that there exist such [imath]$a$[/imath] and [imath]$b$[/imath] that furthermore satisfy [imath](a,b)=1[/imath]. |
2260381 | What is the connection between the standard basis and some other basis in [imath]\mathbb{R}^n[/imath]?
For example if i have a linear operator [imath]\ T:\mathbb{R}^n\to \mathbb{R}^n[/imath] and some basis [imath]\ B=\{v_1,v_2,..v_n\}[/imath]. Let [imath]\ [T]_B[/imath] be the representation matrix of [imath]\ T [/imath] with respect to the basis [imath]\ B[/imath]. for example if I do [imath]\ [T]_B*v_2 [/imath] do I get the coordinates of [imath]\ v_2[/imath] in the standard basis? | 721222 | How to change this matrix with respect to standard basis?
Given the basis [imath]\beta = \{ (1, 1, 0),\ (1, 0, -1),\ (2, 1, 0)\}[/imath] and the matrix: [imath] A = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} [/imath] with respect to [imath]\beta[/imath]. How to change this matrix with respect to standard basis? |
2087229 | A question regarding existence and uniqueness in IVP
Consider the IVP [imath]y'(t)=f(y(t)), \ \ \ \ y(0)=a \in \mathbb{R}[/imath] [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath] Which of the following is/are true [imath](A)[/imath] There exists a continuous function [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath] and [imath]a \in \mathbb{R}[/imath] such that the above problem does not have a solution in any nbd of [imath]0[/imath]. [imath](B)[/imath] The problem has unique solution for every [imath]a \in \mathbb{R}[/imath] when [imath]f[/imath] is Lipschitz continuous [imath](C)[/imath] When [imath]f[/imath] is twice continuously differentiable the maximal interval of existence for the above IVP is [imath]\mathbb{R}[/imath] [imath](D)[/imath] The maximum interval of existence for the IVP is [imath]\mathbb{R}[/imath] when [imath]f[/imath] is bounded and continuously differentiable. It is obvious to me that [imath]A \ \& \ B[/imath] are false from traditional existance and uniqueness theorems (viz Picards Theorem). I am not sure about [imath]C \ \& \ D[/imath] and this is really bugging me. Please could anyone shed light on this. PS: Multiple correct options are allowed. | 2285428 | Ordinary differential equation objective question.
Consider the ODE [imath]y'=f(y(t)),y(0)=a\in\mathbb{R}[/imath] Where [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] Which of the following statements are necessarily true? [imath]A.[/imath] There exist a continuous function [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] and [imath]a\in\mathbb{R}[/imath] such that the above problem does not have a solution in any neighbourhood of [imath]0.[/imath] [imath]B.[/imath] The problem has a unique solution for every [imath]a\in\mathbb{R}[/imath] where [imath]f[/imath] is lipscitz continuous. [imath]C.[/imath] When [imath]f[/imath] is twice continuously differentiable, the maximal interval of existance for the above I.V.P. is [imath]\mathbb{R}.[/imath] [imath]D.[/imath] The maximal interval of existance for the above problem is [imath]\mathbb{R}[/imath] when [imath]f[/imath] is bounded and continuously differentiable. According to me [imath]A[/imath] is false by Picard existance theorem, and [imath]B[/imath] is true and for [imath]C[/imath] [imath]y'=y^{2}[/imath] is counterexample. I have no idea about option [imath]D.[/imath] Please help me to handle option [imath]D.[/imath] In answer key option [imath]D[/imath] is given correct. Thanks in advance. |
2260990 | How to find ma and nb that multiply to c?
I'm trying to figure out the solution to the following problem: [imath]4585a+6734b=7[/imath], where [imath]a[/imath] and [imath]b[/imath] are integers. I am familiar with how to calculate the [imath]\gcd[/imath], but how can [imath]a[/imath] and [imath]b[/imath] be directly calculated? | 626374 | Euclidean Algorithm - find [imath]\gcd(172, 20)[/imath] and solve [imath]172a + 20b = 1000[/imath].
I am revising for an exam and have just realised that the euclidean algorithm questions in past exams are much harder than in homeworks! So i need some help please. I have a question here, i already have the solution to it but i dont understand it that well, so i would like someone to please explain what is going on ect, and dont worry about avoiding giving me the full solution as i already have it so a full solution would be good! Here is the question: Use euclids Algorithm to find [imath]gcd(172,20)[/imath]. Hence solve [imath]172a + 20b = 1000[/imath] giving all solutions in terms of parameter t. List any solutions for which [imath]a,b>0[/imath]. So it wants the answers in terms of parameter t, which is something i am not familiar with and cannot find in my books. I have the solution in the solutions paper as [imath]172= 8\times20+12 [/imath] [imath]20=1\times 12 + 8[/imath] [imath]12=1 \times 8 +4[/imath] [imath]8=2\times 4+0[/imath] thus [imath]gcd(172,20)=4[/imath] then [imath]4= 12-8=12-(20-12)=2\times 12 - 20[/imath] [imath]=2\times(172-8\times20)-20[/imath] [imath]=2 \times 172 -17 \times20[/imath] multiplying by 250 we get [imath]1000=500 \times 172 -4250 \times 20[/imath] so one solution is [imath]a=500, b=-4250[/imath] So, I understand everything up until here!! Now it says, Generally, [imath]a=500-t \frac{20}{4}=500-t[/imath] [imath]b=-4250+t \frac{172}{4}=-4250+43t[/imath] To have [imath]a,b>0[/imath], we need [imath]500-5t>0, (100>t)[/imath] to have [imath]a>b[/imath] we need [imath]-4250+43t>0,(t>98.8)[/imath][imath][/imath] Need t=99[imath], [/imath]a=5[imath], [/imath]b=7$ Could someone please help explain what is going on in this last bit of working out? I am pretty good at euclidean algorithm usually but havent seen this type before an am panicking bit as so close to exam! Any help appreciated how to find t and the other positive solutions would be great! Please could you show me exactly how it is done here as my lecturer is quite strict with us using his exact methods. Many thanks |
1033206 | If [imath]A = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots+\frac{1}{\sqrt{999}}+\frac{1}{\sqrt{1000}}.[/imath] Then [imath]\lfloor A \rfloor[/imath] is,
If [imath]\displaystyle A = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots\cdots\cdots+\frac{1}{\sqrt{999}}+\frac{1}{\sqrt{1000}}.[/imath] Then [imath]\lfloor A \rfloor[/imath] is, where [imath]\lfloor A\rfloor = A-\{A\}.[/imath] [imath]\bf{My\; Try::}[/imath] Using [imath]\left(\sqrt{k}+\sqrt{k-1}\right)<2\sqrt{k}<\left(\sqrt{k+1}+\sqrt{k}\right)\;,[/imath] where [imath]k\in N[/imath] and [imath]k\geq 2[/imath] Then [imath]\displaystyle \left(\sqrt{k+1}-\sqrt{k}\right)<\frac{1}{2\sqrt{k}}<\left(\sqrt{k}-\sqrt{k-1}\right)[/imath] So [imath]\displaystyle 2\sum_{k=2}^{1000}\left(\sqrt{k+1}-\sqrt{k}\right)<\sum_{k=2}^{1000}\frac{1}{\sqrt{k}}<2\sum_{k=2}^{1000}\left(\sqrt{k}-\sqrt{k-1}\right)[/imath] So we get [imath]\displaystyle 2(10\sqrt{10}-\sqrt{2}) < A < 2(10\sqrt{10}-1)[/imath] So we get [imath]60.41\approx <A<\approx 61.24[/imath]. So [imath]\lfloor A \rfloor = 61[/imath] My Question is how can we solve using Concept of definite Integral, plz explain me Thanks | 2168520 | Find the value of [imath]\left[\frac{1}{\sqrt 2}+\frac{1}{ \sqrt 3}+......+\frac{1}{\sqrt {1000}}\right][/imath]
Find the value of [imath]\left[\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+.....+\frac{1}{\sqrt {1000}}\right][/imath].Where [•] denote the greatest integer function. I am very confused about this problem. I tried to find the upper and lower bound of the function. But I can't find any formulas to find the bounds. Somebody please help me. |
2260808 | dense in Topology
Let [imath]M[/imath] be any subset of a Topological space [imath]X[/imath]. We say [imath]M[/imath] is dense in [imath]X[/imath] if [imath]\overline{M}=X[/imath]. Now I have to prove the question: [imath]M[/imath] dense in [imath]X[/imath] [imath]\Leftrightarrow[/imath] for any subset [imath]V \subseteq X : V \cap M \neq \emptyset [/imath] where [imath]V[/imath] is open and not empty. I tried to prove that by contradiction but didnt get far i hope you can help me out here. | 1168969 | Topology proof: dense sets and no trivial intersection
I was wondering if this proof of this basic topological result concerning the closure works. Proposition: Let [imath]A \subseteq (X,\tau)[/imath]. Then, [imath]A[/imath] is dense in [imath]X[/imath] if and only if every non-empty open subset of [imath]X[/imath] intersects [imath]A[/imath] non trivially. Proof: [only if] Assume that [imath]A[/imath] is dense in [imath]X[/imath]. This means that [imath]A \cup A' = X[/imath], where [imath]A'[/imath] denotes the set of all the limit points of [imath]A[/imath]. Let [imath]G \in \tau[/imath] be non-empty. Thus, [imath]G \subseteq (A \cup A') = X[/imath]. Hence, either [imath]G \cap A \neq \varnothing[/imath], which establishes the result, or [imath]G \cap A' \neq \varnothing[/imath], which leads to a contradiction. [if] Assume that [imath]A[/imath] is not dense in [imath]X[/imath]. Thus, there is a [imath]x^* \in X \setminus (A \cup A')[/imath]. Notice, that [imath]X \setminus (A \cup A')[/imath] is open, because [imath]A \cup A'[/imath] is closed, and it is not empty. Hence, we can conclude by realizing that [imath]X \setminus (A \cup A') \cap A = \varnothing[/imath], which completes the proof. [imath]\square[/imath] It is one of the first proofs I attempt in topology, and I am not planning to see how it is proved in the book I am reading (unless you answer me that my proof is completely wrong). Thus, I am really looking forward to any feedback, also contentwise. Thank you for you time. |
2261756 | Discrete math(Pascal triangle)
What is the coefficient of [imath]x^{10}[/imath] in the expansion of [imath](5x^2 +3)^{14}[/imath] Leave answer as an expression rather than number. Do I have to expand using the pascal triangle 10 times or is there a formula cos they said leave it as a mathematical expression? | 2261462 | Computing the coefficient in a large polynomial
What is the coefficient of [imath]x^{10}[/imath] in the expansion of [imath](5 x^2 + 3)^{14}[/imath]? (I would prefer to know the answer as a mathematical expression rather than a number.) May I know how to approach this question and what will be the answer? Thanks |
1870377 | The pigeonhole principle - how to solve questions like that?
We have two sequences , [imath](a_i)_{i=1}^{2n}[/imath] and [imath](b_i)_{i=1}^{2n}[/imath] such that [imath]1\leq a_i, b_i\leq n[/imath] for every [imath]i[/imath]. Show that there are two sets of indexes [imath]I, J \subseteq \left \{ 1,2, ... 2n \right \}[/imath] such what [imath]\sum_{i\in I}a_i=\sum_{j\in J}b_j[/imath]. Well, the question didn't say anything about those sets being empty but I believe that's not what they meant. I don't know that to do with questions like these. There are obviously much more subsets that possible sums ([imath]2^{2n}-1[/imath] compared to [imath]2n^2[/imath]) but it doesn't really help. I'd be glad to hear ideas, hints or solutions. | 2277019 | Equality of sums of sequences
We have two sequences [imath](a_n)[/imath] and [imath](b_n)[/imath], of positive integers, such that, [imath]1\leq a_1,...,a_m\leq n[/imath], [imath]1\leq b_1,...,b_n\leq m[/imath] prove that there exist [imath]p\leq q[/imath], [imath]r\leq s[/imath] such that, [imath]\sum_{i=p}^{q} a_i =\sum_{i=r}^{s}b_i[/imath] I think Pigeon Hole Principle can be applied, but I'm not sure how. Any help will be appreciated. |
2262167 | [imath]\mathbb{Z}[G][/imath] isomorphic to [imath]\mathbb{Z}[H][/imath] then what can be said about [imath]G[/imath] and [imath]H[/imath]?
The question in title has been considered for finite groups [imath]G[/imath] and [imath]H[/imath], but I do not know its situation, how far it is known whether [imath]G[/imath] and [imath]H[/imath] could be isomorphic. I have two simple questions regarding it. Q.0 If [imath]\mathbb{Z}[G]\cong \mathbb{Z}[H][/imath] then [imath]|G|[/imath] should be [imath]|H|[/imath]; because, [imath]G[/imath] is a free basis for the free additive abelian group [imath]\mathbb{Z}[G][/imath], am I right? (I am asking this because in Isaac's character theory, I saw something different argument, not too lengthy, but I was thinking for the above natural arguments.) Q.1 Are there known examples of finite groups [imath]G\ncong H[/imath] with [imath]\mathbb{Z}[G]\cong \mathbb{Z}[H][/imath]? (In the book of character theory by Isaacs, he stated for metablelian groups [imath]G,H[/imath], [imath]\mathbb{Z}[G]\cong \mathbb{Z}[H][/imath] implies [imath]G\cong H[/imath]; it was proved by Whitcomb, in [imath]1970[/imath]; but book has not been further revised, I don't known what is done after [imath]1970[/imath]). | 632372 | Minimal counterexamples of the isomorphism problem for integral group rings
The isomorphism problem for integral group rings asks if two finite groups [imath]G,H[/imath] are isomorphic when their integral group rings [imath]\mathbb{Z}[G][/imath], [imath]\mathbb{Z}[H][/imath] are isomorphic. Quite a lot has been done towards solving this and related problems. See for example Chapter 9 in Milies, Sehgal, "An Introduction to Group Rings". Now it seems that the first (?!) counterexample has been found by Martin Hertweck in his 2001 paper "A Counterexample to the Isomorphism Problem for Integral Group Rings". He has constructed two counterexamples, the one group has order [imath]2^{25} \cdot 97^2[/imath] and the other group has [imath]2^{21} \cdot 97^{28}[/imath]. Do we really have to consider such huge groups? The thesis by Geoffrey Janssens discusses Hertweck's construction in detail, and claims that this is the only known counterexample. Is this still correct? Question. What is known about the minimal counterexamples of the isomorphism problem for integral group rings? |
2262464 | Given [imath]A\sim \operatorname{Bin}(n,p), B\sim \operatorname{Bin}(n,q), p\geq q[/imath] show [imath]\Pr[A\geq k] \geq\Pr[B\geq k][/imath] for all [imath]k[/imath].
My intuition as to why the statement is true is that the binomial distribution with the higher probability will always have more weight 'further right' if you were to plot it, however I can't turn this notion into a proof. An equivalent statement is that [imath]\Pr[A = 0] + \cdots + \Pr[A = k - 1] \leq \Pr[B = 0] + \cdots + \Pr[B = k - 1].[/imath] All this is saying is that the binomial distribution with the lower probability as more weight 'further left'. Inductive arguments seem to go nowhere... Any help would be much appreciated. | 2139315 | How to show a binomial random variable dominates another binomial random variable with a smaller success value?
Let [imath]X\sim B(n,p_h)[/imath] and [imath]Y\sim B(n,p_\ell)[/imath] be two random variables following a respective binomial distribution, where [imath]p_h>p_\ell[/imath]. I want to show that [imath]P(X\ge\alpha)\ge P(Y\ge\alpha),[/imath] for any [imath]\alpha\in\{0,1,\dots,n\}[/imath]. In other words, I want to show [imath]P(X\ge\alpha)=\sum_{i=\alpha}^{n}\binom{n}{i}p_h^i(1-p_h)^{n-i}\ge P(Y\ge\alpha)=\sum_{i=\alpha}^{n}\binom{n}{i}p_\ell^i(1-p_\ell)^{n-i}.[/imath] The statement to be proven is obviously true, but proving it is extremely difficult (at least for me). Can anyone give me some guidance on proving it? It would be nice if someone can give an algebraic proof too, that is, show that [imath]\sum_{i=\alpha}^{n}\binom{n}{i}p_h^i(1-p_h)^{n-i}\ge \sum_{i=\alpha}^{n}\binom{n}{i}p_\ell^i(1-p_\ell)^{n-i}.[/imath] Thank you very much! |
2262720 | Is a continuous function defined on a closed interval always riemann integrable
Is a continuous function defined on a closed interval on [imath]R[/imath] always Riemann integrable or are there counter examples? I'm thinking it is true because it sounds like it would always be possible to find step functions to bound it? | 890953 | Riemann integrability of continuous function defined on closed interval
Theorem says that if [imath]f:[a,b]\longrightarrow \mathbb{R}[/imath] is a continuous function, then it's integrable in Riemann's sense. Here's my proof: Proof Let [imath]\mathfrak{P}=\{x_0,x_1,...x_n\}[/imath] be a subdivision of [imath][a,b][/imath]. Let denote [imath]m_k=\inf_{x\in[x_{k-1},x_k]}f(x)[/imath], [imath]M_k=\sup_{x\in[x_{k-1},x_k]}f(x)[/imath] and a diameter as [imath]\max_{k=0,1...,n}(x_{k+1}-x_k)[/imath]. Then, the lower Darboux sum is [imath]s(f,\mathfrak{P})=\sum_{k=1}^n m_k(x_k-x_{k-1})[/imath] and upper sum is [imath]S(f,\mathfrak{P})=\sum_{k=1}^n M_k(x_k-x_{k-1})[/imath]. We know that [imath]f\in\mathfrak{R}([a,b])\iff \forall_{\epsilon >0}\exists_{\mathfrak{P}}:S(f,\mathfrak{P})-s(f,\mathfrak{P})<\epsilon[/imath] Every funtion continuous on closed interval is also uniformly continuous, so [imath]\forall_{\epsilon >0}\exists_{\delta >0}\forall_{x_1,x_2\in [a,b]}:|x_1-x_2|<\delta \Rightarrow |f(x_1)-f(x_2)|<\frac{\epsilon}{b-a}[/imath] This means that for every subdivision [imath]\mathfrak{P}[/imath] of a diameter smaller than [imath]\delta[/imath] we get [imath]S(f,\mathfrak{P})-s(f,\mathfrak{P})=\sum_{k=1}^n(M_k-m_k)(x_k-x_{k-1})\leq \frac{\epsilon}{b-a}(b-a)=\epsilon[/imath] which ends the proof. Is it understandable and, what's more important, correct? |
2262799 | Let f be differentiable, prove that there exists [imath]c \in (0,1)[/imath] such that [imath]f(c) + f'(c) = 0 [/imath]
Let [imath]f[/imath] be differentiable on [imath]R[/imath] and [imath]f(0) = f(1) = 0[/imath]. Prove that there exists [imath]c \in (0,1)[/imath] such that [imath]f(c) + f'(c) = 0 [/imath] Can someone help me with this proof? | 1424999 | Prove [imath]kf(x)+f'(x)=0 [/imath] when conditions of Rolle's theorem are satisfied .
Prove that if [imath]f[/imath] is differentiable on [imath] [a,b][/imath] and if [imath] f(a)=f(b)=0[/imath] then for any real [imath]k[/imath] there is an [imath] x \in (a,b) [/imath]such that [imath]kf(x)+f'(x)=0 [/imath] As all the conditions of Rolle's theorem are satisfied one can say that there is at least one [imath]c \in (a,b) [/imath] such that [imath]f'(c) =0[/imath] How should I proceed furthur ? How can I use this to get to the required equation ? |
2262558 | Computing the quotient [imath]\mathbb{Q}_p[x]/(x^2 + 1)[/imath]
I am trying to compute the quotient [imath]\mathbb{Q}_p[x]/(x^2 + 1)[/imath] where [imath]\mathbb{Q}_p[/imath] represents the [imath]p[/imath]-adic numbers. I already proved that if [imath]p \equiv 2,3\mod4[/imath], then [imath]\mathbb{Q}_p[x]/(x^2 + 1) \cong \mathbb{Q}_p(i)[/imath]. Moreover, I also proved that if [imath]x \equiv 1 \mod 4[/imath], then [imath]x^2 + 1[/imath] has a root in [imath]Q_p[/imath] using Hensel's lemma. Online, I found that in this case [imath]\mathbb{Q}_p[x]/(x^2 + 1) \cong \mathbb{Q}_p \times \mathbb{Q}_p[/imath]. Why? A similar problem was posted here:Decomposition of the tensor product [imath]\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}[i][/imath] into a product of fields. However, they do not address my question. Thanks in advance! | 2259645 | Decomposition of the tensor product [imath]\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}[i][/imath] into a product of fields
I am trying to solve the following problem: For each rational prime [imath]p[/imath], describe the decomposition of the tensor product [imath]\mathbb{Q}_p \otimes_{\mathbb{Q}} \mathbb{Q}[i][/imath] into a product of fields, where [imath]\mathbb{Q}_p[/imath] is the field of [imath]p[/imath]-adic numbers. I know that the tensor product of [imath]2[/imath] extensions of a field one of which is finite is Artinian and is therefore a product of Artinian local rings, but I do not know how to compute these Artinian local rings. Please if you can help me with this, I'll really appreciate it. |
1252232 | Evaluate the limit [imath]\displaystyle\lim_{x \to 0}\frac{(e-\left(1 + x\right)^{1/x})}{\tan x}[/imath].
How to evaluate the following limit [imath]\displaystyle\lim_{x\to 0} \dfrac{e-\left(1 + x\right)^{1/x}}{\tan x}[/imath] I have tried to solve it using L-Hospital's Rule, but it creates utter mess. Thanks for your generous help in advance. | 1248815 | Limit [imath](0÷0)[/imath] and use L'hopital Rule
find value of the [imath]\lim_{x\to0}\frac{e-(1+x)^{\frac{1}{x}}}{x}[/imath]I use hospital law and can't find answer |
2262015 | How do i solve this limit: [imath]\lim_{x\to 0}{x-\sin(\sin(...(\sin x)))\over x^{3}}[/imath]
I have the next limit : [imath]\large \lim_{x\to 0}{x-\sin(\sin(\overbrace {\cdot \ \cdot \ \cdot }^n(\sin(x))\overbrace {\cdot \ \cdot \ \cdot }^n))\over x^{3}}[/imath] [imath]\sin(\sin(...(\sin(x))...))[/imath]-is n times. I have no idea. Someone can help me? Thank you! | 1345313 | Evaluating limit (iterated sine function)
The limit is [imath]\lim_{x\rightarrow0} \frac{x-\sin_n(x)}{x^3},[/imath] where [imath]\sin_n(x)[/imath] is the [imath]\sin(x)[/imath] function composed with itself [imath]n[/imath] times: [imath]\sin_n(x) = \sin(\sin(\dots \sin(x)))[/imath] For [imath]n=1[/imath] the limit is [imath]\frac{1}{6}[/imath], [imath]n=2[/imath], the limit is [imath]\frac{1}{3}[/imath] and so on. Can we define a recurrent relation upon that given hypothesis? Also, how do I involve [imath]n[/imath] into calculation, because the final limit will depend on it? Any suggestions on how to tackle this? Thank you! |
2262462 | Show that is not possible to form a unity ring structure on [imath](\Bbb Q /\Bbb Z , + )[/imath]
I'm trying to show exactly what the title says. I tried to construct a multiplication from [imath]\mathbb{Q}/\mathbb{Z}\times \mathbb{Q}/\mathbb{Z}\to \mathbb{Q}/\mathbb{Z}[/imath] defined as [imath]\hat{x} \ast\hat{y} =\widehat{xy}[/imath] and find a contradiction. But that didn't get me anywhere. Any help please ? Thank you ! | 106933 | Showing there is no ring whose additive group is isomorphic to [imath]\mathbb{Q}/\mathbb{Z}[/imath]
Show that there is no commutative ring with the identity whose additive group is isomorphic to [imath]\mathbb{Q}/\mathbb{Z}[/imath]. |
2263866 | [imath]\int_{1}^{\infty} \frac{p(x)}{e^x} dx[/imath] converges?
How to decide whether the improper integral [imath]\int_{1}^{\infty} \frac{p(x)}{e^x} dx[/imath] converges or diverges, when [imath]p(x) \in \mathbb R[x][/imath]? | 2261008 | does [imath]\int _{1}^{\infty}\frac{p(x)}{e^x}[/imath] converge?
We're learning Riemann integrals, and I'm trying to see whteher the following indefinite integral [imath]\int _{1}^{\infty}\frac{p(x)}{e^x}[/imath] converges, when p(x) is a polynomial. I don't know where to start this question - does it depends on p(x)? What theorem should I use? |
2242184 | Extension of Arzela-Ascoli Theorem in [imath]\mathbb R\;[/imath]
One version of Arzela-Ascoli theorem can be stated as follows: Consider a sequence of real-valued continuous functions [imath]\;\{ f_n \}_{n \in \mathbb N}\;[/imath] defined on a closed and bounded interval [imath]\;[a, b]\;[/imath] of the real line. If this sequence is uniformly bounded and equicontinuous, then there exists a subsequence [imath]\; \{ {f_n}_k \}_{k \in \mathbb N} \;[/imath] that converges uniformly. I am trying to extend the theorem for sequences of real-valued functions defined on [imath]\; \mathbb R\;[/imath]. One hint I've got , is to use a diagonal argument...but I'm not very familiar with this. EDIT: I need to derive that a sequence of real-valued functions defined on [imath]\; \mathbb R\;[/imath] which satisfy the conditions of boundedness and equicontinuity on a closed interval of [imath]\; \mathbb R\;[/imath], has a uniformly convergent subsequence on compact intervals. How do I proceed? I would appreciate if somebody could enlighten me about this. Any help would be valuable! Thanks in advance!!! | 2118703 | Applying Arzela-Ascoli to show pointwise convergence on [imath]\mathbb{R}[/imath].
The statement I am trying to prove: Let [imath]\{ f_n \}[/imath] be a sequence of equicontinuous, real valued, uniformly bounded continuous functions on [imath]\mathbb{R}[/imath]. Show that [imath]\{ f_n \}[/imath] has a convergent subsequence which converges uniformly on any bounded subset of [imath]\mathbb{R}[/imath] and pointwise on all of [imath]\mathbb{R}[/imath] to a continuous function. (Royden pg 210, problem 9). I believe the first portion follows directly by Arzela Ascoli, once it is noted that if we are given some bounded subset [imath]E[/imath], we can look at [imath]\overline{E}[/imath] a compact set on which the family is equicontinuous and uniformly bounded. For pointwise convergence on all of [imath]\mathbb{R}[/imath], the question seems to require some sort of diagonalization argument: Take some dense sequence of [imath]\{x_i\}[/imath] (the reals are separable) and write columns of [imath]f_n[/imath] applied to each individual point. There will be a convergent subsequence by Bolzano-Weirstrass and since the outputs are real and bounded by the uniform boundedness of the family. I am a bit fuzzy on this part: Repeating this literally pointwise and moving far enough diagonally, we can correct for different convergence "speeds" pluck out a single convergent subsequence for any given epsilon and have [imath]\sup|f_n(x)-f(x)|<\epsilon[/imath] for any given epsilon. Finally, continuity should follow from an [imath]\epsilon/3[/imath] argument and equicontinuity of the family. However, in formalizing the above, I am getting caught up in the notation of that sort of argument and am hoping for a more elegant solution. Is there a better way? Am I just generally confused and that's why I'm having trouble writing things down formally? Thanks! |
2264043 | [imath]x^6+x+a=0[/imath] has a zero or two distinct solutions in [imath]\mathbb{F}_{2^m}[/imath]
Let [imath]m \geq 3[/imath] be an odd integer. Prove that for any [imath]a \in \mathbb{F}_{2^m}[/imath], the equation [imath]x^6+x+a=0[/imath] has a zero or two distinct solutions in [imath]\mathbb{F}_{2^m}[/imath]. I have proven the following facts: Let [imath]a>1[/imath] be an integer. For any positive [imath]n,d\in \mathbb{Z}[/imath], [imath]\,[/imath] [imath]d[/imath] divides [imath]n[/imath] if and only if [imath]a^d-1[/imath] divides [imath]a^n-1[/imath]. In particular, [imath]\mathbb{F}_{p^d}\subseteq \mathbb{F}_{p^n}[/imath]. Let [imath]\phi[/imath] denote the Frobenius map [imath]x\mapsto x^p[/imath] on the finite field [imath]\mathbb{F}_{p^n}[/imath]. Then [imath]\phi[/imath] gives an isomorphism of [imath]\mathbb{F}_{p^n}[/imath] to itself. Also, [imath]\phi^n[/imath] is the identity map and no lower power of [imath]\phi[/imath] is the identity. For any prime [imath]p[/imath] and any nonzero [imath]a \in \mathbb{F}_p[/imath] prove that [imath]x^p-x+a[/imath] is irreducible and separable over [imath]\mathbb{F}_p[/imath]. I don't know if any of these facts will help me prove this. I am currently studying Chapter 13 from Dummit and Foote. Any help on how to prove the statement is greatly appreciated, thank you. ***My question is not exactly the same as the post in Number of solns of [imath]x^6+x=a[/imath] in [imath]\mathbb{F}_{2^m}[/imath], where [imath]m\geq 3[/imath] is odd is same as number of solns of [imath]x^2+ax+1=0[/imath] because I found a counterexample for [imath]x^2+ax+1=0[/imath] having the same number of solutions as [imath]x^6+ax+1=0[/imath] in [imath]\mathbb{F}_{2^m}[/imath] whenever [imath]m \geq 3[/imath] is odd. | 2258556 | Number of solns of [imath]x^6+x=a[/imath] in [imath]\mathbb{F}_{2^m}[/imath], where [imath]m\geq 3[/imath] is odd is same as number of solns of [imath]x^2+ax+1=0[/imath]
I am reading a proof that narrows down to the following statement: It is easy to see that the number of solutions of [imath]x^6+x=a[/imath] in [imath]\mathbb{F}_{2^m}[/imath], where [imath]a\in \mathbb{F}_{2^m}[/imath] and [imath]m\geq 3[/imath] is odd, is same as number of solutions of [imath]x^2+ax+1=0[/imath] in [imath]\mathbb{F}_{2^m}[/imath]. Why is this true? They say that [imath]x^7=1[/imath] for [imath]x \neq 0[/imath], which I also don't see. Thank you! |
2263698 | What does a linear transformation do to a vector in [imath]\mathbb{R}^2[/imath] 1
As in the last question i had, I'm doing some linear algebra. This question has me stumped however. I can't seem to visualize it. [imath] T_{B}: R^2 \to R^2,\qquad B = 1/2 \left[\begin{array} {rr}1 & 1 \\ 1 & 1 \\ \end{array}\right]. [/imath] What, geometrically, does the linear transformation [imath]T_{B}[/imath] do to a vector in [imath]R^2[/imath]? Thanks | 2263614 | What does a linear transformation do to a vector in [imath]\mathbb{R}^2[/imath]
[imath]\newcommand{\Reals}{\mathbb{R}}[/imath]Doing some linear algebra. This question has me intrigued. I imagine it causes the vector to rotates the [imath]x[/imath] axis. But I mean, is there more to it? [imath] T_{A(\theta)}: \Reals^2 \to \Reals^2,\qquad A(\theta) = \left[\begin{array} {rr}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{array}\right]. [/imath] What, geometrically, does the linear transformation [imath]T_{A(\theta)}[/imath] do to a vector in [imath]\Reals^2[/imath]? |
2264782 | Canonical rotation to take a unit vector to another
In 3 (and 2 for that matter) dimensions, there is a "canonical" (independent of arbitrary choices) rotation that takes a given unit vector [imath]u[/imath] to another unit vector [imath]v[/imath] (edit: when [imath]u,v[/imath] are in general position): just rotate around the direction orthogonal to both vectors. However, is there such a canonical rotation in higher dimensions? | 197772 | Generalized rotation matrix in N dimensional space around N-2 unit vector
There is a 2d rotation matrix around point [imath](0, 0)[/imath] with angle [imath]\theta[/imath]. [imath] \left[ \begin{array}{ccc} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array} \right] [/imath] Next, there is a 3d rotation matrix around point [imath](0, 0, 0)[/imath] and unit axis [imath](u_x, u_y, u_z)[/imath] with angle [imath]\theta[/imath] (Rodrigues' Rotation Formula). \begin{bmatrix} \cos \theta +u_x^2 \left(1-\cos \theta\right) & u_x u_y \left(1-\cos \theta\right) - u_z \sin \theta & u_x u_z \left(1-\cos \theta\right) + u_y \sin \theta \\ u_y u_x \left(1-\cos \theta\right) + u_z \sin \theta & \cos \theta + u_y^2\left(1-\cos \theta\right) & u_y u_z \left(1-\cos \theta\right) - u_x \sin \theta \\ u_z u_x \left(1-\cos \theta\right) - u_y \sin \theta & u_z u_y \left(1-\cos \theta\right) + u_x \sin \theta & \cos \theta + u_z^2\left(1-\cos \theta\right) \end{bmatrix} How it is possible to generalize rotation matrix on [imath]N[/imath] dimension around zero point and [imath]N-2[/imath] dimensional unit axis with angle [imath]\theta[/imath]? |
2264639 | Define [imath]S_n=f(S_{n-1})[/imath] for [imath]n\geq 1[/imath]. Prove [imath]S_n[/imath] is a convergent sequence.
Let [imath]f[/imath] be differentiable on [imath]\mathbb R[/imath] with [imath]a=\sup \{|f'(x)| :x\in \mathbb R\}<1[/imath] Select [imath]S_0\in \mathbb R[/imath] and define [imath]S_n=f(S_{n-1})[/imath] for [imath]n \geq 1.[/imath] Prove that [imath]\{S_n\}[/imath] is a convergent sequence. My attempt I know that I might have to show that [imath]S_n[/imath] is cauchy since it's convergent. By definition [imath]S_n[/imath] is cauchy if there exists a natural number [imath]N[/imath] such that [imath]m,n>N[/imath] implies [imath]|X_m - X_n| < \epsilon[/imath]. I was given a hint that we have to use the IVT to show that it is cauchy. The professor wrote on the board: By IVT, [imath]|S_{n+1}-S_n|=|f(S_n)-f(S_{n-1})|[/imath] I understand this step but I don't understand the next few steps [imath] = f'(c)|S_n - S_{n-1}|<a|S_n-S_{n-1}| \\ < a^2 |S_{n-1}-S_{n-2}| \\ a^n|S_1 - S_0|[/imath] | 2225451 | Contraction mapping in the context of [imath]f(x_n)=x_{n+1}[/imath].
I'm interested in convergence of [imath]f(x_n)=x_{n+1}[/imath] and often hear this term referenced. What does it mean to be a contraction mapping in the context of the sequence of real numbers given by [imath]f(x_n)=x_{n+1}[/imath]? And what does it tell us about such sequence? A search online gives the answer given by Wikipedia, that it is a function [imath]f[/imath] defined on a metric space [imath](M,d)[/imath] from [imath]M[/imath] to itself with the property that for some real number [imath]k \in [0,1)[/imath], [imath]d(f(x),f(y)) \leq k d(x,y)[/imath] This definition is quite unhelpful because I barely know what a metric space is. I feel like this definition is to generalized, and I'm interested in in the specific case of recursively defined real sequences [imath]f(x_n)=x_{n+1}[/imath] that there should be a more specific answer to my question. |
2263569 | Does [imath]\int_{1}^{\infty} g(x)\ dx[/imath] imply [imath]\lim\limits_{x\to \infty}g(x)=0[/imath]?
Let [imath]g:[1, \infty)\rightarrow \mathbb R[/imath] a continuous non-negative function, such that [imath]\int_{1}^{\infty} g(x)\ dx[/imath] converges. Is it true that [imath]\lim\limits_{x\to \infty}g(x)=0[/imath] ? I tried to find a counter-example but I can't figure a trivial one. I also tried to prove it by the definition of the convergence of [imath]g(x)[/imath] but couldn't show that limit is really [imath]0[/imath]. | 2354009 | Is it true that if [imath]\int_{1}^\infty f(x)dx[/imath] converges, then [imath]\lim_{x\to\infty}f(x)=0[/imath]?
Please help me decide if the following statement is true or false: If [imath]\int_{1}^\infty f(x)dx[/imath] converges then [imath]\lim_{x\to\infty}f(x)=0[/imath] I tried many counter examples with no luck so I tried to prove it but couldn't pull it off either... |
1458361 | Product of nilpotent matrices is nilpotent
Let [imath]A[/imath] and [imath]B[/imath] be two nilpotent [imath]n\times n[/imath] matrices that commute (so [imath]AB=BA[/imath]), how do I show that [imath]AB[/imath] is nilpotent as well? I have frankly no idea how to start this proof, so excuse me for not showing what I have done so far. | 1140141 | Let [imath]A[/imath] and [imath]B[/imath] be square matrices of the same size such that [imath]AB = BA[/imath] and [imath]A[/imath] is nilpotent. Show that [imath]AB[/imath] is nilpotent.
A square matrix [imath]A[/imath] is called nilpotent is [imath]A^k=0[/imath] for some positive integer [imath]k[/imath]. Let [imath]A[/imath] and [imath]B[/imath] be square matrices of the same size such that [imath]AB = BA[/imath] and [imath]A[/imath] is nilpotent. Show that [imath]AB[/imath] is nilpotent. Following that is [imath]AB[/imath] not equal to [imath]BA[/imath] in part (b), must [imath]AB[/imath] be nilpotent? I don't get how to prove this. Since [imath]A[/imath] can represent any matrix. I have no idea how to show it. |
2260332 | Using the [imath]\epsilon-\delta[/imath] definition of the limit, evaluate, [imath]\lim\limits_{x\to a}f(x)=x^2[/imath]
Using the [imath]\epsilon-\delta[/imath] definition of the limit, evaluate [imath]\lim\limits_ {x\to a}f(x)=x^2[/imath] I am stuck on [imath]|x−a||x+a|<\epsilon[/imath] and can't eliminate [imath]|x+a|[/imath]. | 1568784 | [imath] \lim x^2 = a^2[/imath] as [imath]x[/imath] goes to [imath]a[/imath]
Prove that [imath]\displaystyle \lim_{x \to a} x^2 = a^2[/imath] Let [imath]\varepsilon > 0[/imath], and let [imath]\delta = \min(\frac{\varepsilon}{2|a|+1}, 1)[/imath]. Suppose [imath]x \in\mathbb{R} - \left\{a\right\} [/imath] and [imath]|x-a| < \delta.[/imath] Then [imath]|x-a| < 1[/imath] hence [imath] -1 < x-a <1 [/imath] hence [imath] a-1 < x < a+1[/imath] therefore [imath]2a-1< x+a <2a+1.[/imath] Thus [imath] |x+a| < 2|a|+1.[/imath] So [imath]|x^2-a^2| = |x-a||x+a|[/imath] [imath] \displaystyle < (2|a|+1)\delta =(2|a|+1)\frac{\varepsilon}{(2|a|+1)} = \varepsilon. [/imath] Is this correct? Is there another choice of delta we could have chosen? |
360469 | The cardinality of Lebesgue sets
Suppose [imath]A=\{S\;|\;S \subset \mathbb R^n, S\text{ is Lebesgue measurable}\}[/imath]. What is the cardinality of [imath]A[/imath]? Is it the same as the cardinality of all of the real numbers? | 2544561 | Show that Lebesgue [imath]\sigma[/imath]-algebra has the same cardinality as [imath]\mathcal{P}(\Bbb R)[/imath]
We recently started with measure theory in my Analysis class and I got stuck trying to solve the following exercise: Show that Lebesgue [imath]\sigma[/imath]-algebra on [imath]\Bbb R[/imath] has the same cardinality as [imath]\mathcal{P}(\Bbb R)[/imath]. There is also a tip, suggesting that I should have a look at subsets of [imath]\mathcal{C}[/imath] (standard Cantor set on unit interval). Using the tip I believe I should look for bijection between [imath]\mathcal{C}[/imath] and [imath]\mathcal{P}(\Bbb R)[/imath]. Since both of those sets are uncountable, how can I come up with such bijection? Also to be honest, I do not fully get the argument why is it sufficient to search for bijection between [imath]\mathcal{C}[/imath] and [imath]\mathcal{P}(\Bbb R)[/imath] instead of bijection between Lebesgue [imath]\sigma[/imath]-algebra on [imath]\Bbb R[/imath] and [imath]\mathcal{P}(\Bbb R)[/imath]. Any tip\solution would be greatly appreciated. Thanks for the help in advance! |
2265600 | [imath]S_k \triangleq \sum_{i=1}^{p-1}i^k[/imath] implies [imath]p | S_k[/imath]
Here is a small cute lemma, that I have encountered while solving a problem. Hope you will enjoy it. Let [imath]p[/imath] be an odd prime. Define [imath]S_k[/imath] via [imath] S_k \triangleq \sum_{i=1}^{p-1}i^k. [/imath] Show that [imath]p \mid S_k[/imath] for all [imath]1 \leq k < p-1[/imath]. | 433678 | Sums of powers below a prime
Given a prime [imath]p[/imath] and a natural number [imath]k[/imath], such that [imath]k[/imath] is not divisible by [imath]p - 1[/imath], prove that [imath]\sum_{i = 1}^{p - 1}i^k \equiv 0 \pmod p[/imath]. I split the proof into two cases: one where [imath]k[/imath] is odd and another where it is even. The case where [imath]k[/imath] is odd can be proven as follows: [imath]\sum_{i = 1}^{p - 1}i^k \equiv \sum_{i = 1}^{\frac{p - 1}{2}}i^k + \sum_{i = 1}^{\frac{p - 1}{2}}(-i)^k \pmod p[/imath] Since [imath]k[/imath] is odd each [imath](-i)^k = -i^k[/imath], and will cancel with the positive [imath]i[/imath] leaving a zero. Therefore, [imath]\sum_{i = 1}^{p - 1}i^k \equiv 0 \pmod p \text{ if $k$ is odd.} [/imath] I approached the case where [imath]k[/imath] is even in a similar fashion. [imath]\sum_{i = 1}^{p - 1}i^k \equiv \sum_{i = 1}^{\frac{p - 1}{2}}i^k + \sum_{i = 1}^{\frac{p - 1}{2}}(-i)^k \equiv 2(\sum_{i = 1}^{\frac{p - 1}{2}}i^k) \space \pmod p[/imath] Since [imath]p[/imath] is prime, we just need to prove that [imath]\displaystyle\sum_{i = 1}^{\frac{p - 1}{2}}i^k \equiv 0 \pmod p[/imath]. I was stumped after this, so I considered a counterexample, one where [imath]k[/imath] is divisible by [imath]p - 1[/imath]. In that I considered a case where [imath]p = 5, k = 8[/imath]: [imath]1^8 + 2^8 + 3^8 + 4^8 \equiv 1 + 1 + 1 + 1 \space \pmod p[/imath] This and a few other cases led to the conjecture that all the numbers below [imath]p[/imath] have a mod cycle [imath](mod \space p)[/imath] that is divisible by [imath]p - 1[/imath]. I tested it with [imath]p = 7[/imath]. [imath]2: 2, 4, {\color{red}1}[/imath] [imath]3: 3, 2, 6, 4, 5, {\color{red}1}[/imath] [imath]4: 4, 2, {\color{red}1}[/imath] [imath]5: 5, 4, 6, 2, 3, {\color{red}1}[/imath] After this I was stuck. I couldn't see how this would help me solve the problem. And, more importantly, I couldn't prove my conjecture. |
2265404 | Factorize [imath]X^p+1[/imath] in a field of characteristic [imath]p[/imath].
How can I factorize [imath]X^p+1[/imath] in a field of characteristic [imath]p[/imath] ? If the field is [imath]\mathbb Z/p\mathbb Z[/imath], then [imath]X^p+1=X^p+1^p=(X+1)^p,[/imath] but if the field is [imath]\mathbb F_{p^n}[/imath] or if the characteristic is [imath]q\neq p[/imath], I don't know how to do it. It's not duplicate, but it solve my problem for [imath]\mathbb F_{p^n}[/imath]. But what happen for a field of characteristic [imath]q\neq p[/imath] ? | 1238531 | A freshman's dream
If [imath]p[/imath] is prime, then [imath](x+y)^p=x^p+y^p[/imath] holds in any field of characteristic [imath]p[/imath]. However all the proofs I have seen use induction and some relatively nasty algebra despite how fundamental this fact seems. What is the nicest, "highest level proof" you know? |
579817 | How many transitive relations on a set of [imath]n[/imath] elements?
If a set has [imath]n[/imath] elements, how many transitive relations are there on it? For example if set [imath]A[/imath] has [imath]2[/imath] elements then how many transitive relations. I know the total number of relations is [imath]16[/imath] but how to find only the transitive relations? Is there a formula for finding this or is it a counting problem? Also how to find this for any number of elements [imath]n[/imath]? | 243773 | Amount of transitive relations on a finite set
In counting the amount of relations on finite sets, we can quite easily count the amount of reflexive and symmetric relations on a finite set. We just consider (in accordance with the definition of a relation on a set as a subset of the cartesian product of the set with itself) a grid with on the vertical axis all the elements of the set and on the horizontal axis all elements of the set. We can then mark the points on the grid which are elements of the relation. Consider a set [imath]S[/imath] with [imath]|S|=n[/imath] for some [imath]n\in\mathbb{N}[/imath]. The amount of relations on this set is simply [imath]|\mathcal{P}(S^2)|=2^{|S^2|}=2^{n^2}[/imath]. The amount of reflexive relations [imath]S[/imath] can be found by considering that these relations on the grid are all the relations in which in any case the diagonal is in the relation. We see that [imath]n[/imath] pairs are already "chosen", so we can still choose for [imath]n^2-n[/imath] pairs whether or not they are in the relation while keeping the relation reflexive. So there are [imath]2^{n^2-n}[/imath] reflexive relations on [imath]S[/imath]. For the amount of symmetric relations we know that the pairs under the diagonal are in the relation if the corresponding ones above the diagonal are also in the relation. We have the freedom to choose any of the pairs above or on the diagonal of the grid, the amount of which is equal to [imath]T_n=\frac{n(n+1)}{2}[/imath]. So there are [imath]2^{\frac{n(n+1)}{2}}[/imath] symmetric relations on [imath]S[/imath]. Relations both symmetric and reflexive are relations for which we can freely choose any pair above the diagonal of the grid. There are [imath]T_{n-1}=\frac{n(n-1)}{2}[/imath] pairs above the diagonal, so there are [imath]2^{\frac{n(n-1)}{2}}[/imath] relations both symmetric and reflexive on [imath]S[/imath]. Now I'm sorry for not getting the point sooner and please correct me if I am wrong about any of them, but I am wondering if there are similar simple expressions for the amount of antisymmetric and transitive relations on this set [imath]S[/imath]. I have been told before it is very difficult to visualise transitive relations with the grid I suggested. For antisymmetric relations I think it means the diagonal is definitely in the relation and either a pair above the diagonal is in the relation or the corresponding pair below the diagonal, but not both. I don't really have any ideas as to how I could use these things to come up with a formula for the amount of relations. Any help and/or comments are appreciated. Thank you |
2257647 | Add a point to the Moore plane to get a normal space
I've got the following question, and I'm having trouble with it. I was hoping that someone here could help me. Show that by adding one point to the Moore plane [imath]\mathbf{M}[/imath] one can obtain a normal space of which [imath]\mathbf{M}[/imath] is a subspace. My definition of the Moore plane is just like on Wikipedia. That is, as a set [imath]\mathbf{M} = \{ (x,y) \in \mathbb{R}^2 : y \geq 0 \}[/imath], the points [imath](x,y)[/imath] for [imath]y > 0[/imath] have as basic open neighborhoods the usual open discs centered at [imath](x,y)[/imath], and the points [imath](x,0)[/imath] have as basic open neighborhoods open discs of radius [imath]r>0[/imath] centered at [imath](x,r)[/imath] together with the point [imath](x,0)[/imath]. I know that [imath]\mathbf{M}[/imath] is not normal (and for us normal implies Hausdorff), since, for example, the closed sets [imath]A = \{ (x,0) : x \in \mathbb{Q} \}[/imath] and [imath]B = \{ (x,0) : x \in \mathbb{R} - \mathbb{Q} \}[/imath] cannot be separated by disjoint open sets. I know that [imath]A[/imath] and [imath]B[/imath] cannot both be closed in this new space. But I'm having trouble figuring out how to add a point to [imath]\mathbb{M}[/imath] to get a normal space. Any help would be appreciated! | 770140 | Moore plane / Niemytzki plane (Topology)
I am supposed to show that by adding a point to the Moore plane, that the subspace of a [imath]T_4[/imath]-space is not necessarily a [imath]T_4[/imath]-space. I do know that the Moore plane is NOT a [imath]T_4-[/imath]space. Does anybody here know which point could be meant? |
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