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582787 | Calculating [imath]\int_{0}^{\infty} x^{a-1} \cos(x) \ \mathrm dx = \Gamma(a) \cos (\pi a/2)[/imath]
My goal is to calculate the integral [imath]\int_{0}^{\infty} x^{a-1} \cos(x) dx = \Gamma(a) \cos (\pi a/2)[/imath], where [imath]0<a<1[/imath], and my textbook provides the hint: integrate [imath]z^{a-1} e^{iz}[/imath] around the boundary of a quarter disk. However, I couldn't figure out how to control the integral over the quarter arc. Any hints? | 479586 | evaluate [imath]\int_{0}^{\infty}\cos(t) t^{z-1}dt=\Gamma(z)\cos(\frac{\pi z}{2})[/imath]
I need to prove [imath]\int_{0}^{\infty}\cos(t) t^{z-1}dt=\Gamma(z)\cos(\frac{\pi z}{2})[/imath] for [imath]0<Re(z)<1[/imath]. I tried [imath]\cos t=\frac{e^{it}+e^{-it}}{2}[/imath] and to integrate it along the contour from [imath]\epsilon [/imath] to R, then to [imath]Ri[/imath] via a quarter of a circle, then downward to [imath]\epsilon i[/imath], finally back to [imath]\epsilon[/imath] via a small quarter of a circle. However, I find that [imath]e^{-it}t^{z-1} [/imath] does not converge when [imath]R\to \infty[/imath], can anyone give a correction of my integration or offer another method to evaluate the integral?(would be better if using complex contours) thanks in advance. |
1926972 | Question about evaluating the integral [imath]\int_0^\infty \frac{1-\cos(ax)}{x^2}dx[/imath] using the residue theorem
Question: Evaluate the integral [imath]\int_0^\infty \frac{1-\cos(ax)}{x^2}dx[/imath] where [imath]a \in R[/imath]. My attempt: So, I actually have a worked out solution for this, but I'm confused on a few points and I'm hoping to get some help. The solution considers the function [imath]f(z) = \frac{1-e^{i|a|z}}{z^2},[/imath] and evaluates it around the contour [imath]\Gamma = C_R \cup [-R,-\epsilon] \cup C_\epsilon \cup [\epsilon,R][/imath], for [imath]R[/imath] large, [imath]\epsilon[/imath] small, [imath]C_R[/imath] the semicircle in the upper half plane of radius [imath]R[/imath], and [imath]C_\epsilon[/imath] the semicircle in the upper half plane of radius [imath]\epsilon[/imath] but with negative orientation. The next step they do is show that [imath]|f(z)|[/imath] on [imath]C_R[/imath] goes to [imath]0[/imath] as [imath]R \to \infty[/imath]. I think that I understand this part, but I'm not 100% sure, so I'd just like to check: [imath]\left|\frac{1-e^{i|a|z}}{z^2}\right| \le \frac{|1-e^{i|a|R\cos\theta} - e^{i|a|R\sin\theta}|}{R^2}\le \frac{2+|e^{-|a|R\sin\theta}}{R^2},[/imath] and the RHS of the above goes to [imath]0[/imath] as [imath]R \to \infty[/imath]. So, first question: 1) is my above reasoning correct? In the solution, they have a [imath]1 + |e^{-|a|R\sin\theta}[/imath] as the numerator in the last term, but I'm not sure how to get that stronger bound, so I'm unsure if my weaker one is correct. Ok, so after this they write: [imath]\int_0^\infty \frac{1-\cos(ax)}{x^2}dx = \frac{1}{2}\Re\left(\pi i \cdot \text{res}_{0}f(z)\right)[/imath] $=\frac{\pi |a|}{2}.[imath][/imath] This is the step that is driving me a little insane. I guess I must be missing something. So, I have a few questions here: 2) First of all, what I think they are saying is that \int_\Gamma f(z)dz = \int_{C_R}f(z)dz + 2\int_{\epsilon}^{R}f(z)dz + \int_{C_\epsilon}f(z)dz,$[imath][/imath] and then taking limits as \epsilon \to 0[imath], and [/imath]R \to \infty[imath]. But what happens to the integral over [/imath]C_\epsilon[imath]??? [/imath] 3) I guess maybe they are saying that as \epsilon \to 0[imath], the integral over [/imath]C_\epsilon[imath] approaches the residue at [/imath]0$. But I don't understand why that's valid - why are we allowed to use the residue theorem on this integral? It's not even a contour, it doesn't have an interior, and I guess I'm just very confused here. Thanks! | 2118704 | Detail in the use of the Residue Theorem for [imath]\int\limits_{-\infty}^{\infty}\frac{\cos(2x)-1}{x^2}dx[/imath]
As usual, I have a ridiculously specific question about a tiny detail of a tiny calculation. There is a tremendous amount of context, so fair warning. I don't need a solution to this whole problem. I need a solution to one specific part, and it is a minor detail. I'll show what I've got so far, and point out my concern at the end. Our problem is to use residue calculus to compute: [imath]\int\limits_{-\infty}^{\infty}\frac{\cos(2x)-1}{x^2}dx.[/imath] We note that this is equal to the real part of the easier-to-work-with integral: [imath]\int\limits_{-\infty}^{\infty}\frac{e^{2ix}-1}{x^2}dx,[/imath] and we use a "hippo's back" contour, formed by the positively-oriented half-circle [imath]|z|\leq R[/imath] in the upper half-plane and the line along the real axis from [imath]-R[/imath] to [imath]R[/imath], interrupted from [imath]-\varepsilon[/imath] to [imath]\varepsilon[/imath] by the negatively oriented half-circle in the upper half-plane of radius [imath]\varepsilon[/imath], so as to 'jump' over the simple pole at [imath]z=0[/imath]. Since this contour contains no poles or singularities, etc., the integral of our integrand around it is zero by Cauchy's Theorem. It is also a simple matter to show that the large half-circle's contribution is zero in the limit [imath]R\to\infty[/imath]. So now we are left with the fact that in the limit [imath]\varepsilon\to 0[/imath], we obtain: [imath]\int\limits_{-\infty}^{\infty}\frac{e^{2ix}-1}{x^2}dx=\lim\limits_{\varepsilon\to 0}\int\limits_{C_\varepsilon}\frac{e^{2iz}-1}{z^2}dz,[/imath] where [imath]C_\varepsilon[/imath] is the little negatively-oriented half-circle mentioned above. Now, I know the trick in which the value of the right-hand side happens to give exactly [imath]\pi i \text{Res}(f,0)[/imath] (that is, half of what the full circle would give), and can use it to show that the result is [imath]-2\pi[/imath]. Okay, thanks for reading this far. Here is my actual question: Without resorting to tricks like the half-residue thing, how can we get this result from the right-hand side? That is, how do we prove rigorously that: [imath]\lim\limits_{\varepsilon\to 0}\int\limits_{C_\varepsilon}\frac{e^{2iz}-1}{z^2}dz=-2\pi[/imath] I have been using the parametrization [imath]z=\varepsilon e^{-it}[/imath] with [imath]t\in[0,\pi][/imath], where the negative in the exponential deals with the negative orientation, which almost always works in these situations, but not here, since the [imath]\varepsilon[/imath] does not entirely cancel from the denominator then as it usually does in these types of problems. I have confirmed the result with Wolfram-Alpha using the original problem statement, and also noted a few other things, like that the original integrand can be expressed as [imath]\frac{\sin^2(x)}{x^2}[/imath], but gotten nowhere with that. In a pinch, I'll use the 'half-residue' trick, but being able to do this simply from basic principles would be nice. |
493309 | How find this limit [imath]\lim \limits_{x\to+\infty}e^{-x}\left(1+\frac{1}{x}\right)^{x^2}[/imath]
find this limit [imath]\lim \limits_{x\to+\infty}e^{-x}\left(1+\dfrac{1}{x}\right)^{x^2}[/imath] my idea: [imath]\lim \limits_{x\to+\infty}e^{-x}\left(1+\dfrac{1}{x}\right)^{x^2}=\lim \limits_{x\to+\infty}e^{-x}\cdot e^x=1[/imath] But book is answer is not 1? and How about find it? Thank you | 480003 | finding the limit [imath]\lim\limits_{x \to \infty }(\frac{1}{e}(1+\frac{1}{x})^x)^x[/imath]
Can someone show me how to calculate the limit: [imath]\lim_{x \to \infty }\left(\frac{1}{e}\left(1+\frac{1}{x}\right)^x\right)^x [/imath] I tried to use taylor series but failed. Thanks |
2027566 | How is the Riemann zeta function zero at the negative even integers?
I'm not sure if I'm misunderstanding this in any way, but surely evaluating the zeta function at, say, -2, would give [imath]1 + 2^2 + 3^2 + 4^2 + 5^2 + ...[/imath] which seems to diverge to infinity. What am I missing? | 726506 | Trivial zeros of the Riemann Zeta function
A question that has been puzzling me for quite some time now: Why is the value of the Riemann Zeta function equal to [imath]0[/imath] for every even negative number? I assume that even negative refers to the real part of the number, while its imaginary part is [imath]0[/imath]. So consider [imath]-2[/imath] for example: [imath]f(-2) = \sum_{n=1}^{\infty}\frac{1}{n^{-2}} = \frac{1}{1^{-2}}+\frac{1}{2^{-2}}+\frac{1}{3^{-2}}+\dots = 1^2+2^2+3^2+\dots = \infty[/imath] What am I missing here? |
2512403 | Eigenvalues and -vectors of product matrix
Let [imath]A,B\in \mathbf{R}^{3\times 3}[/imath]. Let [imath]A[/imath] have 3 distinct eigenvalues [imath]\lambda_1[/imath], [imath]\lambda_2[/imath], [imath]\lambda_3[/imath] with corresponding eigenspaces [imath]E_{\lambda_1}[/imath], [imath]E_{\lambda_2}[/imath], [imath]E_{\lambda_3}[/imath]. Let [imath]B[/imath] have 2 distinct eigenvalues [imath]\mu_1[/imath], [imath]\mu_2[/imath] with corresponding eigenspaces [imath]E_{\mu_1}=\operatorname{span}(E_{\lambda_1}+E_{\lambda_2})[/imath] and [imath]E_{\mu_2}=E_{\lambda_3}[/imath]. (a) Determine the eigenvalues and eigenvectors of [imath]AB[/imath]. (b) Prove that [imath]AB=BA[/imath]. I don't understand what my book means by "[imath]\operatorname{span}(E_{\lambda_1}+E_{\lambda_2})[/imath]", is this simply the sum [imath]E_{\lambda_1}+E_{\lambda_2}[/imath] or something more complicated? For part (a). I took a vector [imath]v\in E_{\mu_2}[/imath], then [imath]v\in E_{\lambda_3}[/imath] so [imath](AB)v=A(Bv)=A(\mu_2 v)=\mu_2 AV=\mu_2 \lambda_3 v[/imath] so [imath]\lambda_3 \mu_2[/imath] is an eigenvalue (corresponding to eigenvector [imath]v\in E_{\mu_2}=E_{\lambda_3}[/imath]?). Similarly, take a vector [imath]E_{\lambda_1}[/imath], then (about this step I am not sure) [imath]w\in E_{\mu_1}[/imath]. Therefore [imath](AB)w=A(Bw)=A(\mu_1 w)=\mu_1 (Aw)=\mu_1\lambda_1 w[/imath] so [imath]\lambda_1 \mu_1[/imath] is an eigenvalue (corresponding to what eigenvector? Since [imath]w[/imath] must lie in [imath]E_{\lambda_1}[/imath] and [imath]E_{\mu_1}[/imath] I would say [imath]E_{\lambda_1} \cap E_{\mu_1}[/imath], but is this correct? For part (b), I know that simultaneously diagonalizable matrices commute, so I need to show that [imath]A[/imath] and [imath]B[/imath] have the same basis of eigenvectors, which again is trivial if I know what my book means by the [imath]\operatorname{span}[/imath]-thing. | 1364925 | Eigenvalues and eigenspaces of AB
Problem: Consider two matrices [imath]A, B \in \mathbb{R}^{3 \times 3}[/imath]. Suppose [imath]A[/imath] has three distinct real eigenvalues [imath]\lambda_1, \lambda_2[/imath] and [imath]\lambda_3[/imath] with respective eigenspaces [imath]E_{\lambda_1}, E_{\lambda_2}[/imath] and [imath]E_{\lambda_3}[/imath]. Suppose furthermore that [imath]B[/imath] has two distinct real eigenvalues [imath]\mu_1[/imath] and [imath]\mu_2[/imath] with respective eigenspaces [imath]E_{\mu_1} = \text{span}(E_{\lambda_1}, E_{\lambda_2})[/imath] (the space spanned by [imath]E_{\lambda_1}[/imath] and [imath]E_{\lambda_2}[/imath]) and [imath]E_{\mu_2} = E_{\lambda_3}[/imath]. 1) Determine the eigenvalues and corresponding eigenspaces of [imath]AB[/imath]. 2) Show that [imath]AB = BA[/imath]. Attempt at solution: I have no idea how to do this. I tried writing [imath]\det(A - x \mathbb{I}_3) = (-1)^3 (x- \lambda_1) (x- \lambda_2) (x- \lambda_3)[/imath] and then using the fact that [imath]\det(AB) = \det(A) \det(B)[/imath]. But then I figured that the characteristic equation doesn't necessarily have to split like that ? |
2514393 | Uniqueness of IVP
I want to show that [imath]x''=x|x|, \quad[/imath] [imath]x(0)=a, \quad[/imath] [imath]x'(0)=b[/imath] has a unique solution for all [imath]a,b \in \mathbb{R}[/imath]. I tried to show that the function is globally lipschitz, but did not succeed. And since it is continous, if it were globally lipschitz that would have worked to show uniqueness for all [imath]a,b \in \mathbb{R}[/imath]. In the solution sheet, they just mention that '[imath]f(t,x)=x|x|[/imath] has a continous first derivative'. I cant really grasp this, how is it possible to show uniqueness for all [imath]a,b \in \mathbb{R}[/imath] by just arguing that 'it has continuous first derivative' ? | 1451282 | Show that the Cauchy problem [imath]x''=x|x|,x(0)=a,x^{\prime}(0)=b[/imath] ,has a unique solution for all [imath]a,b\in \Bbb R[/imath].
Show that the Cauchy problem [imath]x''=x|x|,x(0)=a,x^{\prime}(0)=b[/imath],has a unique solution for all [imath]a,b\in \Bbb R[/imath]. (The Problem is from A text book on ordinary differential equation by Shair Ahmad and Antonio Ambrosetti,exercise 4.3,pg:70) My attempt: I am using this theorem(given in book,page 69): [imath]\begin{cases} x^{(n)}=f(t,x',x'',...,x^{(n-1)})\\ x(t_0)=\alpha_1,x'(t_0)=\alpha_2,...,x^{(n-1)}(t_0)=\alpha_n \end{cases}[/imath] (1) If [imath]f:\Omega \to \Bbb R[/imath] is continuous and locally lipschitzian with respect to [imath](x_1,x_2,...,x_n),[/imath] then (1) has a unique solution,defined in a suitable nbd of [imath]t_0[/imath].([imath]\Omega[/imath] is an interval containing point [imath](t_0,\alpha_1,...,\alpha_n)[/imath]) (The function [imath]f[/imath] is called locally lipschitzian at [imath](t_0,\alpha_1,...,\alpha_n) \in \Omega[/imath] with respect to [imath](x_1,x_2,...,x_n)[/imath] if [imath]\exists[/imath] a neighborhood [imath]U \subset \Omega[/imath] of [imath](t_0,\alpha_1,...,\alpha_n)[/imath] and a number [imath]L>0[/imath] such that, denoting by [imath]|.|[/imath] the euclidean norm in [imath]\Bbb R^n[/imath], one has [imath]|f(t,x_1,x_2,...,x_n)-f(t,z_1,z_2,...,z_n)|\leq L |(x_1,x_2,...,x_n)-(z_1,z_2,...,z_n)|[/imath] for every[imath](x_1,x_2,...,x_n),(z_1,z_2,...,z_n) \in U[/imath]. ) clearly [imath]f[/imath] is continuous.therefore we will show that [imath]f[/imath]is lipschitzian at point [imath](0,a,b)[/imath]. To prove lipschitzian at point [imath](0,a,b)[/imath] we have to show that [imath]\exists[/imath] a nbd U of [imath]\Bbb R^3[/imath] containing point[imath](0,a,b)[/imath] and a number [imath]L>0[/imath] such that [imath]|f(t,x,x_1)-f(t,z,z_2)| \leq L||(x,x_1)-(z,z_1)||[/imath] [imath]\forall (t,x,x_1),(t,z,z_1) \in U[/imath] here [imath]||(x,x_1)-(z,z_1)||=\sqrt{(x-z)^2+(x_1-z_1)^2}[/imath] Now [imath]x|x|[/imath] is differentiable everywhere on [imath]\Bbb R[/imath].therefore it is locally lipschitz on any finite interval of [imath]\Bbb R[/imath]. therefore we have (for finite interval) [imath]|x|x|-z|z||\leq L|x-z|[/imath] for some number [imath]L[/imath]. Now [imath]|x|x|-z|z||\leq L|x-z| \leq L \sqrt{(x-z)^2+(x_1-z_1)^2}[/imath] [imath]\therefore[/imath][imath]|f(t,x,x_1)-f(t,z,z_2)| \leq L||(x,x_1)-(z,z_1)||[/imath] [imath]\forall (t,x,x_1),(t,z,z_1) \in U[/imath] where [imath]U[/imath] is any nbd of finite radius of [imath]\Bbb R^3[/imath] which contains the point [imath](0,a,b)[/imath] Hence given problem has a unique solution for all [imath]a,b\in \Bbb R[/imath]. I am confusing whether am I correct or not?If it is not correct then where am I making mistake?and how to solve this problem?Thank you in advance. |
2462557 | taking coefficient of highest order term zero in a polynomial and solving it
The solutions of the quadratic equation [imath]ax^2+bx+c=0[/imath] are [imath]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/imath]. If we take the coefficient of x i.e [imath]b[/imath] as zero in the equation and also in solution, then we get the same value of x in both cases but If we take [imath]a[/imath] as zero then we don't get same value of x. Is there any way to get the result from the solution when the coefficient of highest order term is zero? | 866331 | Numerically stable algorithm for solving the quadratic equation when [imath]a[/imath] is very small or [imath]0[/imath]
Solving [imath]a x^2 + bx +c=0[/imath] for [imath]x[/imath] gives [imath]x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \text{, for } a \ne 0[/imath] But for [imath]a = 0[/imath] we get [imath]x=-\frac{c}{b}[/imath] How to implement a numerically stable algorithm for computing [imath]x[/imath] from [imath]a,b,c[/imath] that allows [imath]a[/imath] to be close to zero or zero? |
2513995 | How to evaluate [imath]\lim_{n\to\infty}\frac{\ln n}{n}[/imath] without the exponent?
I’d like to calculate [imath] \lim_{n\to\infty}\frac{\ln n}{n}[/imath] by using the squeeze theorem. So I’m looking for a sequence with members that are bigger than [imath]\frac{\ln n}{n}[/imath], but which also converges to zero. How is it possible to evaluate this sequence without the exponent? | 2377412 | How do I solve [imath]\lim\limits_{x\to\infty} \ \frac{\ln \ x}{x} = 0[/imath]?
[imath]\lim\limits_{x\to\infty} \ \frac{\ln \ x}{x} = 0[/imath] How can I know this without looking at the graph? What's the easiest way to equate that expression to 0? I understand that an integer over x approaches 0 as x approaches 0, but since ln x is an increasing function as x approaches infinity, how do we know which effect outweighs the other? |
2042755 | Number of [imath]3 \times 3[/imath] matrices with integer entries such that [imath]AA^t[/imath] is the identity matrix
Let [imath]S[/imath] be the set of all [imath]3 \times 3[/imath] matrices [imath]A[/imath] with integer entries such that the product [imath]AA^t[/imath] is the identity matrix. Then [imath]|S|[/imath] is : 23. 24 48 60 Answer is [imath]48[/imath] but I don't know how to solve it. | 2038752 | How many [imath]3 \times 3[/imath] integer matrices are orthogonal?
Let [imath]S[/imath] be the set of [imath]3 \times 3[/imath] matrices [imath]\rm A[/imath] with integer entries such that [imath]\rm AA^{\top} = I_3[/imath] What is [imath]|S|[/imath] (cardinality of [imath]S[/imath])? The answer is supposed to be 48. Here is my proof and I wish to know if it is correct. So, I am going to exploit the fact that the matrix A in a set will be orthognal, so if the matrix is of the form \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix} Then each column and row will have exactly one non-zero element which will be +1 or -1. Thus, I have split possibilities for the first column into three cases and counted the possibilities in each case as follows :- [imath]a_{11} \neq 0[/imath] or [imath] a_{21} \neq 0[/imath] or [imath] a_{31} \neq 0[/imath] In case 1), we obviously have two possibilities(+1 or -1) so we consider the one where the entry is +1. Now, notice that the moment we choose the next non-zero entry, all the places for non-zero entries will be decided because of the rule 'each column and row will have exactly one non-zero element'. Meaning, if b and c are remaining two non-zero entries, we only have two possibilities left \begin{bmatrix} 1 & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \\ \end{bmatrix} or \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & c \\ 0 & b & 0 \\ \end{bmatrix} Using the fact that b and c are simply [imath]\pm1[/imath] In each of the above matrices, we get 4 possibilities for each of the matricies. Thus, 8 possibilities in totality. Basically, we are getting 8 possibilities on the assumption that [imath]a_{11} = 1[/imath] Thus, we get 16 possibilities on the case that [imath]a_{11} \neq 0[/imath] Following, the second and third cases analogously, we get a total of 16 possibilities in each of them and 48 possibilities in total. Source :- Tata Institute of Fundamental Research Graduate School Admissions 2016 |
1580683 | [imath]f:\mathbb R\rightarrow [0,\infty )[/imath] is continuous such that [imath]g(x)={(f(x))}^2[/imath] is uniformly continuous .
[imath]f:\mathbb R\rightarrow [0,\infty )[/imath] is continuous such that [imath]g(x)={(f(x))}^2[/imath] is uniformly continuous . Then which of the following is always true [imath]?[/imath] [imath]A.[/imath] [imath]f[/imath] is bounded. [imath]B.[/imath] [imath]f[/imath] may not be uniformly continuous. [imath]C.[/imath] [imath]f[/imath] is uniformly continuous. [imath]D.[/imath] [imath]f[/imath] is unbounded. I ticked option [imath]C.[/imath] Following is my logic : [imath]f:\mathbb R\rightarrow [0,\infty)[/imath] [imath]h:[0,\infty)\rightarrow [0,\infty) ; h(x)=x^2[/imath] Then [imath]g=h\circ f[/imath] Now the function [imath]h[/imath] is bijective(no [imath]?[/imath]) and [imath]h^{-1}(x)=\sqrt x[/imath] So composing both sides with [imath]h^{-1}[/imath] we get [imath]f=h^{-1}\circ g[/imath] Also both [imath]g[/imath] and [imath]h^{-1}[/imath] being uniformly continuous , [imath]f[/imath] is uniformly continuous. Was I correct [imath]?[/imath] Also , if my proof was correct I could really use some help to find counterexamples for the rest or say , why they cannot be true . Thank you . | 1576885 | Let [imath]f: \Bbb R \to [0,\infty)[/imath] be a continuous function such that [imath]g(x)=(f(x))^2[/imath] is uniformly continuous. Which of the following is always true?
A. [imath]f[/imath] is bounded B. [imath]f[/imath] may not be uniformly continuous C. [imath]f[/imath] is uniformly continuous D. [imath]f[/imath] is unbounded. Let [imath]f(x)=\sqrt x[/imath]. Then [imath]g(x)=x[/imath] is uniformly continuous and unbounded. Hence option A can not be true. Let [imath]f(x)=1[/imath]. Then [imath]g(x)=1[/imath] is uniformly continuous and bounded. Hence the option D can not be true. How should I check uniform continuity? EDIT: Since [imath]g[/imath] is uniformly continuous, hence for a given [imath]\epsilon \gt 0[/imath], [imath]\exists \delta \gt 0[/imath] depending only upon [imath]\epsilon[/imath] such that [imath]|g(x)-g(y)| \lt \epsilon[/imath] whenever [imath]|x-y| \lt \delta[/imath] [imath]\forall x,y[/imath]. [imath]\Rightarrow |\sqrt g(x) -\sqrt g(y)||\sqrt g(x)+\sqrt g(y)|\lt \epsilon[/imath] whenever [imath]|x-y| \lt \delta[/imath]. [imath]\Rightarrow |f(x)-f(y)||f(x)+f(y)|\lt \epsilon[/imath] whenever [imath]|x-y| \lt \delta[/imath]. What should be my next step now? How can I use the continuity of [imath]f[/imath]? |
2515116 | Proof of sum equals to [imath]0[/imath]
Proof that [imath]\sum_{k=1}^n (-1)^kk^m{n \choose k} = 0[/imath] for every [imath]n>m[/imath] and [imath]m>1[/imath]. I tried to use properties of [imath]{n \choose k}.\;[/imath] I tried to use [imath]{n \choose k} = {n-1 \choose k} + {n-1 \choose k-1}[/imath] and using induction got stuck here [imath]\sum_{k=1}^t (-1)^kk^m{t-1 \choose k-1} = 0[/imath], where [imath]t = n - 1[/imath]. | 1955339 | A property of every real polynomial.
Question: For every polynomial [imath]P(x)[/imath] of degree at n we have [imath]\sum_{i=0}^{n+1}(-1)^i\binom{n+1}{i}P(i)=0[/imath] Well I was inducting on degree of the polynomial, for [imath]n=0[/imath] its true and I got a bit confused in the inductive step, actually I cant figure the inductive step. Is this property (if true) a famous one? Thank you for your help. |
2515252 | Find [imath]\lim_{n\to \infty} \left[\frac{1}{n^k}(1^k+2^k+...+n^k)-\frac{n}{k+1}\right].[/imath]
Problem: Find [imath]\lim_{n\to \infty} \left[\frac{1}{n^k}(1^k+2^k+...+n^k)-\frac{n}{k+1}\right].[/imath] We solve this problem by taking [imath]x_n=(k+1)(1^k+2^k+...+n^k)-n^{k+1}[/imath] and [imath]y_n=n^k(k+1)[/imath] then by Stolz's theorem we have that [imath]\lim_{n\to \infty}\frac{x_{n}-x_{n-1}}{y_{n}-y_{n-1}}=\frac{(k+1)n^{k}-(n)^{k+1}+(n-1)^{k+1}}{(k+1)[n^{k}-(n-1)^k]}[/imath] [imath]=\frac{(k+1)/n-1+(1-1/n)^{k+1}}{\frac{k+1}{n}[1-(1-1/n)^k]}[/imath] [imath]=\frac{(k+1)/n-1+(1-1/n)^{k+1}}{\frac{k+1}{n^2}[(1-1/n)^{k-1}+(1-1/n)^{k-2}+...+1]}[/imath] [imath]=\frac{nk+n-n^2+n^2(1-1/n)^{k+1}}{(k+1)[(1-1/n)^{k-1}+(1-1/n)^{k-2}+...+1]}[/imath] I am not sure how to proceed from here and thus any hints will be much appreciated. | 2513406 | A familiar limit, but in general form .
Suppose [imath]a \in \mathbb{R} ,a>1[/imath] Is there an idea to compute the limit below ? [imath]\lim_{n \to \infty}\left( \frac{1^a+2^a+3^a+...+n^a}{n^a}-\frac{n}{a+1} \right)[/imath] I tried it for [imath]a=1,2,3[/imath] but I get stuck in general form . Can someone help me ? Thanks in advance. |
2514449 | Strong law of large numbers --> convergence in L^{1}?!
Let [imath]X_{1}, X_{2}, X_{3}, ...[/imath] be a sequence of iid and integrable random variables with [imath]E[X_{1}] = \mu[/imath]. Show that [imath] \frac{1}{n} (X_{1}+X_{2}+X_{3}+...+X_{n}) \to \mu [/imath] in [imath]L^{1}[/imath]. My thoughts so far: Strong law of large numbers implies almost surely convergence. So I still need to get from a.s. convergence to [imath]L^{1}[/imath] convergence. In my lectrure notes I have found two possible ways of doing so: (a) If [imath]\quad[/imath] [imath]X_{n} \to X \; a.s.[/imath] [imath]\quad[/imath] and [imath]\quad[/imath] [imath]E(|X_{n}|^{p}) \to E(|X|^{p})[/imath] [imath]\quad[/imath] then $\quad[imath]X_{n} \to X$ in $L^{p}$.[/imath] (b) If $\quad$ $X_{n} \to X \;$ in prob. $\quad$ and $\quad$ $\{X_{n} : n \in \mathbb{N}\}$ is UI $\quad$ then $\quadX_{n} \to X$ in [imath]L^{1}[/imath]. Which is more promising? I tried both but...it didnt really work out... Many thanks in advance. | 78445 | Does law of large numbers converge in [imath]L^1[/imath]?
I've seen the law of large numbers stated mainly in two (or three) forms: [imath]S_n/n[/imath] converges in probability (weak law) and converges almost surely (strong law). Also, there is convergence in the [imath]L^2[/imath]-norm for uncorrelated random variables ([imath]L^2[/imath] weak law). However there is a backwards martingale proof of the strong law of large numbers (any graduate level probability theory book, for example Durrett, should have it). The important thing is that [imath]M_{-n}:=S_n/n[/imath] is a backward martingale, and backward martingales converge both a.e. and in the [imath]L^1[/imath]-norm. Then, in particular, [imath]S_n/n[/imath] converges in the [imath]L^1[/imath]-norm. Does [imath]S_n/n[/imath] really converge in the [imath]L^1[/imath]-norm? If yes, why is this never mentioned? If no, what is wrong with my above proof? |
2513458 | setting of this induction proof
I would like to see if this is a correct induction proof and whether or not this is a good setting out of it A sequence is defined by [imath]a_n = a_{n-1} + a_{n-2} + a_{n-3}[/imath] for [imath]n\geq 3, a_0 = 1, a_1 = 2, a_2 = 4[/imath]. Prove that [imath]a_n \leq 4^n[/imath] for all [imath]n\in\mathbb{N}[/imath]. Let [imath]P(n)[/imath] be the proposition that [imath]''a_n\leq 4^n{''}.[/imath] Now since we have [imath]a_0 = 1 = 4^0 \leq 4^0[/imath] then [imath]a_0 \leq 4^0[/imath]. Also, [imath]a_1 = 2 < 4 \leq 4 = 4^1[/imath] then [imath]a_1 \leq 4^1[/imath]. Also, [imath]a_2 = 4<16 = 4^2 \leq 4^2[/imath] so [imath]a_2 \leq 4^2[/imath]. Hence, [imath]P(0),P(1),P(2)[/imath] are true. Now, let [imath]k-3\in\mathbb{N}[/imath] and suppose [imath]P(k-3),P(k-2),P(k-1)[/imath] is true. We must show that [imath]P(k)[/imath] is true. Now by definition [imath]\begin{align}a_n &= a_{n-1} + a_{n-2} + a_{n-3} \\ &\leq 4^{n-1} + 4^{n-2} + 4^{n-3} \qquad \text{by the inductive hypothesis}\\ &= 21\times 4^{k-3} \\ &\leq 64\times 4^{k-3} \\ &= 4^{k}.\end{align}[/imath] Hence, [imath]P(k)[/imath] is true if [imath]P(k-3),P(k-2),P(k-1)[/imath] is true for [imath]k\geq 3[/imath]. So by induction, [imath]P(n)[/imath] is true for all [imath]n\in\mathbb{N}[/imath]. Would appreciate any advice. | 947850 | A proof by strong induction that [imath]a_n\le3^n[/imath] where [imath]a_n=a_{n-1}+a_{n-2}+a_{n-3}[/imath]
I am not sure whether this is right. Can anyone verify, whether this proof is valid? Thanks! Define a sequence [imath]\{a_n\}_{n\ge0}[/imath] as follows: [imath]a_0=1,\qquad,a_1=3,\qquad,a_2=9,\qquad,a_n=a_{n-1}+a_{n-2}+a_{n-3}\text{ for }n\ge3.[/imath] Prove that for any positive integer [imath]n[/imath], [imath]a_n\le3^n[/imath]. Proof. Let [imath]P(n)[/imath]: [imath]a_n\le 3^n[/imath], [imath]n[/imath] is a non-negative integer. (i) Base case: Consider when [imath]n=0[/imath]. LHS[imath]=a_0=1[/imath], RHS=[imath]3^0=1[/imath] [imath]\therefore[/imath] LHS[imath]\le[/imath]RHS, then [imath]P(0)[/imath] holds. Consider when [imath]n=1[/imath]. LHS[imath]=a_1=3[/imath], RHS=[imath]3^1=3[/imath] [imath]\therefore[/imath] LHS[imath]\le[/imath]RHS, then [imath]P(1)[/imath] holds. Consider when [imath]n=1[/imath]. LHS[imath]=a_2=9[/imath], RHS=[imath]3^2=9[/imath] [imath]\therefore[/imath] LHS[imath]\le[/imath]RHS, then [imath]P(2)[/imath] holds. (ii) Inductive case: Assume [imath]P(i)[/imath] is true for [imath]0 \le i \le k[/imath], [imath]k\ge2[/imath]. (iii) Inductive conclusion: Consider [imath]n=k+1[/imath]. [imath] \begin{align*} \mathrm{LHS}=a_{k+1}&=a_k+a_{k-1}+a_{k-2} \qquad\text{(by definition, since $k\ge 2$)}\\ &\le 3^k+3^{k-1}+3^{k-2} \qquad\text{(by Induction Hypothesis)}\\ &=3^k(3^{-1}+3^{-2}+3^{-3})\\ &=3^k\left(\frac13+\frac19+\frac1{27}\right)\\ &=\frac{13}{27} 3^k\\ &\le 3^{k+1} = \mathrm{RHS} \end{align*} [/imath] Therefore, by the Principle of Strong Induction, [imath]P(n)[/imath] is true for all non-negative positive integers [imath]n[/imath]. |
2201232 | Using of Newton's method to find a complex root of a polynomial
I have studied the Newton's method for real functions, and these are the conditions to find a function real root on [imath][a,b][/imath]: [imath]f(a) \ f(b)<0[/imath] [imath]f'(x)>0 \quad (\text{or} \quad f'(x)<0) \qquad \forall \in [a,b] [/imath] Convergence theorem: [imath]\text{sign}(f' f'') = \text{sign}(x_0 -x^*)[/imath] Iterations: [imath] x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \qquad x_n\in \mathbb{R}[/imath] How can I use Newton's method to find the complex roots of a polynomial (e.g. [imath]z^2+1[/imath]) ? I know a real coefficients polynomial of degree [imath]p[/imath] has [imath]p[/imath] different complex roots (for Fundamental theorem of algebra) But, how can I verify the convergence of the Newton succession with an initial value [imath]z_0[/imath]? [imath]z_{n+1}=z_n-\frac{f(z_n)}{f'(z_n)} \qquad z_n\in \mathbb{C}[/imath] Thanks! | 1958956 | How to find a starting point for root finding in complex plane?
The function [imath]f(x) = x^2 +1[/imath] has zeros in the complex plane at [imath]x = +i[/imath] and [imath]x = -i[/imath]. Is there a real starting point for complex Newton's method such that the iterates converge to either of these zeros? Is there a complex starting point too? I know that Newton's method for finding roots is [imath]x_{n+1} = x_n - \frac {f(x)}{f'(x)}[/imath]. But I don't know how to find the proper starting point. |
2100018 | Evaluate [imath]S_n(n)[/imath] for [imath]S_r(n)=\sum_{m=0}^n(-1)^m m^r\binom{n}{m}[/imath]
Let [imath]S_r(n)=\sum_{m=0}^n(-1)^m m^r\binom{n}{m}[/imath] Evaluate [imath]S_n(n)[/imath] For this question, the first part I did is to prove that for integer r, [imath]0<r<n[/imath], [imath]S_r(n)=0[/imath] I did this by induction, given the binomial coefficient [imath](1+z)^n=\sum_{m=0}^nz^m \binom{n}{m}[/imath] and differentiate it [imath]r[/imath] times. The result can be obtained for [imath]S_1(n)=S_2(n)=\dots=S_k(n)=0 \implies S_{k+1}=0, [/imath] given [imath]0<k<n[/imath] I want to show that [imath]S_{n}(n)[/imath] is a linear combination of all [imath]S_k(n)[/imath] for [imath]0<k<n[/imath], but is seems not the case. How should I proceed? | 1852540 | Find [imath]\sum_{m=0}^n\ (-1)^m m^n {n \choose m}[/imath]
I'm going to university in October and thought I'd have a go at a few questions from one of their past papers. I have completed the majority of this question but I'm stuck on the very last part. In honesty I've been working on this paper a while now and I'm a bit tired so I'm probably giving up earlier than I usually would. I won't write out the full question, only the last part: Let [imath]S_r(n) = \sum_{m=0}^n\ (-1)^m m^r {n \choose m}[/imath] where r is a non-negative integer . Show that [imath]S_r(n)=0[/imath] for [imath]r<n[/imath]. Evaluate [imath]S_n(n)[/imath]. I have shown that [imath]S_r(n)=0[/imath] for [imath]r<n[/imath] by taking [imath](1+z)^n= \sum_{m=0}^n\ z^m {n \choose m}[/imath], letting [imath]D_r(f(z))=z\frac d{dz}(z\frac d{dz}...(\frac d{dz}(f(z)))...)[/imath] where [imath]z\frac d{dz}[/imath] is applied [imath]r[/imath] times and applied it to both sides. The left hand side gives a polynomial, degree n which has factor [imath](1+z)[/imath] for all [imath]r<n[/imath] and the right hand side gives [imath]\sum_{m=0}^n\ z^m (m)^r {n \choose m}[/imath]. Setting [imath]z=-1[/imath] yields the required result. There is some build up to this, so I'm fairly certain that this was the intended method. I'm stuck however on the very last part. I have tried finding a form for [imath]D_r((1+z)^n)[/imath], but I'm fairly sure that this isn't the correct approach, as the wording of the question implies that [imath]S_n(n)[/imath] needs to be considered separately. I'm surprised that I didn't find that this question had already been asked, so apologies if it has been. Thank you. |
2479799 | Compute [imath]\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}.[/imath]
NOTE 1: L'Hospitals and Taylor expansions are not allowed. NOTE 2: I really appreciate if someone would correct my attempt, however any other easier method only involving single variable calculus (excluding the concepts in NOTE 1) are welcome. PROBLEM: Compute [imath]\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}.[/imath] I'll just manipulate without writing out the limit, for now. I have \begin{array}{lcl} \left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n} & = & \left( 1+\frac{1}{n}\right)^{n^2}\cdot e^{-n} \\ & = & \exp\left( n^2\ln\left(1+\frac{1}{n}\right)-n\right) \\ & = & \exp((n\ln(1+\frac{1}{n})-1)n) \\ \end{array} And proceeding: \begin{array}{lcl} \exp((n\ln(1+\frac{1}{n})-1)n) & = & e^{((n\ln(1+\frac{1}{n})-1)n} \\ & = & (e^{((n\ln(1+\frac{1}{n})-1)})^n \\ & = & \left(\frac{(e^{((n\ln(1+\frac{1}{n})-1)}-1+1)}{(n\ln(1+\frac{1}{n})-1}\cdot{((n\ln(1+\frac{1}{n})-1})\right)^n\\ \end{array} It gets quite ugly very quickly as you can see. I'm trying to rewrite it so I can apply standard limits like [imath]\lim_{x\rightarrow\infty}\frac{e^x-1}{x}=\infty \quad \text{and} \quad \lim_{x\rightarrow\infty}\frac{\ln{(1+x)}}{x}=0.[/imath] | 553763 | Evaluate of [imath]\lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n}[/imath]
Evaluate the limit [imath] \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} [/imath] My Attempt: [imath] \lim_{n\rightarrow \infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot \frac{1}{e^n} = \lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^{n^2}\cdot \frac{1}{e^n} [/imath] How can I solve the problem from this point? |
2515279 | Reunion of disjoint stable sets in a graph
I need to prove that given a graph G=(V,E), |V| = n and for any v in V with deg(v) at most k then there are up to [imath]2k+1[/imath] disjoint stable sets (let's say [imath]S_1[/imath] ... [imath]S_{2k+1}[/imath]) and their reunion is V. Now, I suppose that in a coloring the vertices from a stable set could be the same color because the vertices are non-adjacent. And since the sets are disjoint we could say that each set needs a different color. So then the chromatic number of the original graph is [imath]2k+1[/imath] (or at least [imath]2k+1[/imath]). Now, I know that this is one way to look at the problem, but I have no idea how to connect that to the original hypothesis or how to start a demonstration from the hypothesis to the conclusion or from the conclusion to the hypothesis. Thanks in advance! | 2514016 | Prove that at most [imath]2k+1[/imath] independent sets form the set of vertices of a graph, where [imath]k[/imath] is the maximum degree.
I have a graph [imath]G(V,E)[/imath] where each vertex has degree at most [imath]k[/imath]. How can I prove that I need at most [imath]2k + 1[/imath] disjoint sets of vertices in which there is no connection between two vertices from the same set to create the set of vertices [imath]V[/imath]. |
2508200 | Convergence of Series in Hilbert Spaces
Let [imath]\{e_j\}_{j\in\mathbb N}[/imath] be some subset of a Hilbert space [imath]H[/imath]. If [imath]\|f\|^2=\sum|\langle f,e_j\rangle|^2[/imath] for every [imath]f\in H[/imath], how may I show that [imath]\sum\langle f,e_j\rangle e_j[/imath] converges for every [imath]f\in H[/imath]? | 166803 | a representation condition of Hilbert space
Let [imath]H[/imath] be a Hilbert space over [imath]\mathbb{C}[/imath]. Let [imath]\phi_{j} \in H[/imath] and let [imath]c_j[/imath] be scalars. Prove that there exists a representation [imath]f=\sum_{j}c_{j}\langle f,\phi_{j}\rangle\phi_{j}, \forall f\in H[/imath] if and only if [imath]\|f\|^2=\sum_{j}c_{j}|\langle f,\phi_{j}\rangle|^2, \forall f\in H[/imath] My queries: For [imath]\implies[/imath]: [imath]\|f\|^2= \langle f,f \rangle = \sum_{i,j} c_{i} \overline{c_{j}} \langle f,\phi_{i} \rangle \overline{\langle f,\phi_{j} \rangle}\langle \phi_{i}, \phi_{j}\rangle[/imath]. It can not be simplified further due to [imath]\phi_{j}[/imath] are may not be orthogonal. Even if one assumes orthogonality it would become [imath]\|f\|^2=\sum_{j}|c_{j}|^2|\langle f,\phi_{j}\rangle|^2\|\phi_{j}\|^2[/imath] which is still not the desired form. Is there any way to apply the fact that the representation for [imath]f[/imath] is valid for "all" [imath]f[/imath] in some way to deduce the desired equality? For sufficiency I intend to apply the polarization identity to [imath]\langle f,\phi_{j}\rangle[/imath] then substitute it into the desired formula for [imath]f[/imath], but the polarization identity implied [imath]\langle f,\phi_{j} \rangle =0[/imath] which is not true in general, is there any point I have missed? |
2516005 | Prove that [imath] \ A \ [/imath] is a [imath] \ G_\delta \ [/imath] subset of [imath] \ X \ [/imath].
Let [imath] \ (X, d) \ [/imath] be a metric space and let [imath] \ A ⊆ X \ [/imath]. Suppose that [imath] \ (A, d) \ [/imath] is homeomorphic to a complete metric space [imath] \ (Y, d_Y ) \ [/imath]. Prove that [imath] \ A \ [/imath] is a [imath] \ G_\delta \ [/imath] subset of [imath] \ X \ [/imath]. Answer: Since [imath] \ (A,d ) \ [/imath] is homeomorphic to a complete space , then [imath] (Y,d_Y) \ [/imath] can be embedded into [imath] \ X \ [/imath] . Now since [imath] A \ [/imath] is complete , we can write [imath] A=\cup_{n \in \mathbb{N}} \ C_n \ [/imath] , where [imath] C_n \ [/imath] are closed set . But now can not conclude the proof . please help me out | 1151207 | Proof of Mazurkiewicz theorem
Wikipedia states the following theorem: Theorem (Mazurkiewicz): Let [imath](\mathcal{X},\rho)[/imath] be a complete metric space and [imath]A\subset\mathcal{X}[/imath]. Then the following are equivalent: [imath]A[/imath] is a [imath]G_\delta[/imath] subset of [imath]\mathcal{X}[/imath]; There is a metric [imath]\sigma[/imath] on [imath]A[/imath] which is equivalent to [imath]\rho|A[/imath] such that [imath](A,\sigma)[/imath] is a complete metric space. I have tried Googling for proof of Mazurkiewicz theorem, but the result was I found two (seemingly) different theorems: The Hahn-Mazurkiewicz theorem, about which I found a Math SX question, a Wikipedia article and a pdf from a random website, plus a whole lot of other links; The Knaster-Kuratowski-Mazurkiewicz (or in short KKM) theorem, about which I found a pdf with proof and two previews, [1] and [2], which may well be the same pdf on two different links, but out of both of which the proof has been cut. Now the first theorem is stated at the Math SX question and is totally different. It seems to be related to "space-filling curves", which I know nothing at all about. The other theorem, from what I read, is also rather different, since it has to do with a map's fixed points. So here I am. Can you post a proof of the Mazurkiewicz theorem above or a reference to one? Thx. PS Feeling like only looking at the first page of Google results might not be judged enough research, I went on to page three, and found nothing about my theorem, seemingly. I did find what seems to be yet another couple of Mazurkiewicz theorems, none of which are mine. Then I got tired, thought I had searched well enough, posted the question and went to bed :). |
2468053 | Evalutaion [imath]\int _{-\infty }^{\infty }\frac{\sin\left(wz\right)}{w}\:dw[/imath]
Let [imath]f(z)=\int _{-\infty }^{\infty }\frac{\sin\left(wz\right)}{w}\:dw[/imath] [imath]f'(z)=\int _{-\infty }^{\infty }\sin\left(wz\right)\:dw=\lim _{x\to \infty }\left(\frac{\cos\left(wz\right)}{z}\right)-\lim _{x\to -\infty }\left(\frac{\cos\left(wz\right)}{z}\right)= \cos\left(\infty \right)-\cos\left(-\infty \right)=\cos\left(\infty \right)-\cos\left(\infty \right)=0[/imath] [imath]f(z)= 0 + C[/imath]; [imath]f(0)=\int _{-\infty }^{\infty }\frac{\sin\left(w0\right)}{w}\:dw \Leftrightarrow 0 = 0 + C \Leftrightarrow C=0 [/imath] I should have obtained [imath]\pi [/imath], since [imath]\frac{\sin\left(w\right)}{w}[/imath] is an even function and [imath]\int _{0\:}^{\infty \:}\frac{\sin\left(w\right)}{w}\:dw[/imath] = [imath]\frac{\pi }{2}[/imath] | 257148 | Evaluation of [imath]\int_{-\infty}^\infty \frac{\sin(at)}t\ dt[/imath]
I just need to know if this integral converges. If it does, what is the value? If not can anyone tell me what is the value of the [imath] \int_{-\infty}^\infty 2\,\frac{\cos(w)\sin(w)}{w}\ dw ? [/imath] |
2515872 | Finding order of convergence using Taylor Series
The fixed point method [imath]g(x)= x - \frac{f(x)}{2f ^\prime (x)}[/imath] (can also be written as [imath]x_{k+1} = x_k - \frac{f(x_k)}{2f ^\prime (x_k)}[/imath]) is similar to Newton's method except for the 1/2 coefficient. My question is how do I show the order of convergence for this method using Taylor series? EDIT I believe the goal is to get it into the form [imath]|x_{k+1} - a| = C|x_{k} - a|^n[/imath] and that [imath]n[/imath] is the order of convergence. I tried getting it into the form [imath]g(y) = g(x) + g^\prime(x)(y-x) + \frac{g ^{\prime\prime}(x)}{2!}(y-x)^2 + ...[/imath] Since [imath]g^\prime (x) = \frac{1}{2} + \frac{1}{2} \cdot \frac{f ^{\prime \prime} (x) f(x)}{[f^\prime (x)]^2}[/imath] Substituting in I get [imath]g(x_k) = g(a) + \frac{1}{2}(x_k -a) + \frac{g^{\prime \prime}(a)}{2}(x_k-a) [/imath] But I'm not sure what to do from here | 389368 | Convergence rate of Newton's method
Let [imath]f(x)[/imath] be a polynomial in one variable [imath]x[/imath] and let [imath]\alpha[/imath] be its [imath]\delta[/imath]-multiple root ([imath]\delta\ge2[/imath]). Show that in the Newton's [imath]x_{k+1}=x_k-f(x_k)/f'(x_k)[/imath], the rate of convergence to [imath]\alpha[/imath] is not quadratic. My solution: Suppose that [imath]\alpha[/imath] is one regular root of equation.Then [imath]x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}=\phi(x_k)[/imath] convergence rate to [imath]\alpha[/imath]: [imath]x_k-\alpha=\phi(x_k)-\phi(\alpha)=(x_k-\frac{f(x_k)}{f'(x_k)})-(x-\frac{f(x)}{f'(x)})[/imath] [imath](x_k-\alpha)=(f'(x_k))^{-1}(f(x_k)-f(x))=(x_k-\alpha)+(f'(x_k))^{-1}\{ f'(x_k)(x_k-\alpha)+O((x_k-\alpha)^2)\}=f'(x_k)^{-1}O((x_k-\alpha)^2)) \tag{1}[/imath] so the convergence rate to [imath]\alpha[/imath] is quadratic. But if [imath]\alpha[/imath] is not regular root, then [imath](f'(x))^{-1}[/imath] has no meaning. We need do slightly change in [imath](1)[/imath], [imath](x_k-f'(x_k)^{-1}f(x_k))-(x-f^{(\delta)}(x)\delta!f(x))=(x_k-x)-f'(x_k)(f(x_k)-f(x))[/imath] [imath]=(x_k-x)-f'(x_k)^{-1}(\frac{f^{(\delta)}(x_k) (x_k-x)^{\delta}}{\delta!}+O((x_k-x)^{\delta+1}))[/imath] convergence rate to non regular root [imath]\alpha[/imath] is one. Is my solution correct? Then I do same thing to next nonlinear equation: [imath]f(x)=x^2(x-1) [/imath], m is integer. so the newton's formula is above, and how about convergence rate to [imath]0,1[/imath]? I think convergence to 1 is one, absolutely convergence to 0 is quadratic. when convergence rate is 1, the how about the convergence rate? |
2517484 | Modular arithmetic with a gigantic exponent (ie. [imath]2^{2^{403}} \equiv x[/imath] (mod [imath]23[/imath]))
Suppose I want to find [imath]2^{2^{403}} \equiv x[/imath] (mod [imath]23[/imath]). Now I know by Fermat's Little Theorem that [imath]2^{22} \equiv 1 \space (\mod 23)[/imath] and I know how to show that [imath]2^{403} \equiv 8 \space (\mod 22)[/imath]. But I don't know how to put these two together to solve the congruence, or if I'm missing some crucial simplification step or theorem. I can't even use WolframAlpha because the number is too big. My question is: How do you solve [imath]2^{2^{403}} \equiv x[/imath] (mod [imath]23[/imath])? | 2515915 | Solve [imath][2^{(2^{403})}] = [a][/imath] in [imath]\mathbb Z_{23}[/imath] where [imath]0 \le a < 23[/imath].
Solve [imath][2^{(2^{403})}] = [a][/imath] in [imath]\mathbb Z_{23}[/imath] where [imath]0 \le a < 23[/imath]. I've tried to combine corollaries of Fermat's little theorem and various other methods to solve this, but have always been stuck. |
2517239 | Computing limit of [imath]\frac{x-\sin x}{x-\tan x}[/imath] without L'Hôpital
I want to compute following: [imath] \lim_{x \rightarrow 0} \frac{x-\sin(x)}{x-\tan(x)}[/imath] I have tried to calculate this with l’Hôpital's rule. l’Hôpital's rule states that: [imath] \lim_{x\rightarrow a} \frac{f(x)}{g(x)}=\lim_{x \rightarrow a } \frac{f'(x)}{g'(x)} [/imath] Now i can get to the right result with l’Hôpital's rule but it took a little over two pages on paper. Had to use l’Hôpital's rule 4 times. How do you solve this without l’Hôpital's rule ? The result appears to be(with l’Hôpital's rule): [imath] \lim_{x \rightarrow 0} \frac{x-\sin(x)}{x-\tan(x)}=-\frac{1}{2} [/imath] If someone can provide alternative solution to this problem that would be highly appreciated. | 508733 | [imath]\lim_{x\to0} \frac{x-\sin x}{x-\tan x}[/imath] without using L'Hopital
[imath]\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?[/imath] I tried using [imath]\lim_{x\to0} \frac{\sin x}{x}=1[/imath]. But it doesn't work :/ |
2517954 | Asking a variant of a question
I have a question which is based on the question here. The question reads: Show that if A is compact, show that [imath]d(x,A)= d(x,a)[/imath] for some [imath]a \in A[/imath]. What if I change the question to Show that if X is complete and [imath]A \subset X[/imath] is closed, show that [imath]d(x,A)= d(x,a)[/imath] for some [imath]a \in A[/imath]. I'm at a lost here. Can I get a hint? I can't seem to use to hints at the original post. My initial thoughts are to construct a sequence such that it converges to [imath]inf\{d(x,a) | a \in A\}[/imath]. However, I need to show that the set [imath]\{d(x,a)|a \in A\}[/imath] contains its [imath]inf[/imath]. I'm not sure how to proceed. | 508937 | If [imath](X,d)[/imath] is a complete metric space and [imath]A[/imath] is closed then show that for [imath]x \in X[/imath] there exists an element [imath]a_0 \in A[/imath] such that [imath]d(x,A)=d(x,a_0)[/imath]
If [imath](X,d)[/imath] is a complete metric space and [imath]A[/imath] is closed in [imath](X,d)[/imath] then show that for each [imath]x \in X[/imath] there exists an element [imath]a_0 \in A[/imath] such that [imath]d(x,A)=d(x,a_0).[/imath] I tried this problem several times but always got stuck at some point or another. For instance, I initially thought about showing the existence of a Cauchy sequence [imath](a_n)[/imath] in A such that [imath]d(x,a_n) \to \operatorname{inf}\{d(x,a):a \in A\}=d(x,A)[/imath] and then by completeness of [imath]A[/imath] and by continuity of the function [imath]f(a)=d(x,a)[/imath], to prove the existence of an element [imath]a_0[/imath] in [imath]A[/imath] such that [imath]a_n \to a_0[/imath] which would imply [imath]d(x,a_n)=f(a_n) \to f(a_0)=d(x,a_0)=d(x,A)[/imath]. But the problem was that I could not construct such a Cauchy sequence in [imath]A[/imath]. All my other attempts have similarly gone in vain. If anybody has any answer, please reply. Thank you. |
2040774 | Which continuous function satisfies [imath]\int\limits_0^x f(t) dt \ge f(x)[/imath] for all [imath]x[/imath] in [imath][0,1] [/imath]?
Written with StackEdit. Let [imath]f:[0,1] \to [0,\infty)[/imath] be continuous. Suppose [imath] \int\limits_0^x f(t) dt \ge f(x), \text{ for all } x \in [0,1][/imath] Then which of the following is true A. No such function exists B. There are infinitely many such functions C. There is only one such function D. There are exactly two such functions See the end of the question for correct answer. I have been able to figure out that, perhaps [imath]e^x[/imath] is one such function so A is NOT the answer. Correct answer : - C Source - Tata Institute of Fundamental Research, Graduate School Admissions 2014 | 1423234 | [imath]\int_{0}^{x} f(t) dt \ge {f(x)}[/imath] holds or not on [imath][0,1][/imath]
Let [imath]f\colon [0,1]\rightarrow [0, \infty )[/imath] be continuous. Suppose that [imath]\int_{0}^{x} f(t)\,\mathrm dt \ge {f(x)} \quad\text{for all }x\in[0,1].[/imath] Then [imath]A.[/imath] No such function exists. [imath]B.[/imath] There are infinitely many such functions. [imath]C.[/imath] There is only one such function [imath]D.[/imath] There are exactly two functions. Now I was thinking, since [imath]\int_{0}^{x} f(t) dt[/imath] is the area under the graph of [imath]f(t)[/imath] from [imath]0[/imath] to [imath]x[/imath] so it can be easily made greater than the value of [imath]f(t)[/imath] at one point, namely [imath]x[/imath]. So there will be infinitely many such functions satisfying this condition. Is my reasoning correct or not? Thanks. |
2208487 | Simplified method for symmetric matrices
I am trying to find the determinant of the following [imath]4\times 4[/imath] symmetric matrix [imath]A[/imath] given as [imath]A= \begin{pmatrix} a & b & c & d \\ b & c & d & a \\ c & d & a & b \\ d & a & b & c \end{pmatrix}[/imath] Any solution and help would be appreciated. Thanks. | 770117 | Determinant of circulant matrix
Find the determinant of the following matrix in the terms of [imath]a_1,a_2,\cdots,a_n[/imath] explicitly, [imath] \begin{bmatrix} a_1 & a_2 & a_3 & \cdots & a_n\\ a_2 & a_3 & a_4 & \cdots & a_1\\ a_3 & a_4 & a_5 & \cdots & a_2\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_n & a_1 & a_2 & \cdots & a_{n-1}\\ \end{bmatrix} [/imath] When the determinant is zero? |
2513771 | Determining the countability of a set with a particular property
Let [imath]E \subset \mathbb{C}[/imath] be a set with the following property: for any sequence of elements [imath](e_n)_{n \in \mathbb{N}}[/imath] with [imath]e_n \neq e_m[/imath] for [imath]m \neq n[/imath], [imath]e_n \to 0[/imath]. Is [imath]E[/imath] necessarily countable? (Also convergence is in norm, of course). It seems like this problem should yield to contrapositive, that is, given an uncountable set, I can always find at least one nonrepeating sequence that does not converge to [imath]0[/imath]. My idea is that for some [imath]\epsilon >0[/imath], if [imath]E[/imath] is uncountable, there must exist infinitely many distinct elements in the complement of [imath]B_\epsilon (0)[/imath]. Then we can construct a sequence not converging to [imath]0[/imath] from that (so the statement is true). Is there a direct proof for the above? | 468746 | Uncountable set with exactly one limit point
Is there any uncountable subset of [imath]\mathbb{C}[/imath] with exactly one limit point? |
2517663 | Find the surface area of S which is is the part of [imath]x^2 + y^2 + z^2 = a^2[/imath] which lies in the cylinder [imath]x^2 + y^2 = ay[/imath]
How do I set up the integral to find the surface area if I only have the two shapes a sphere and a cylinder ? | 893918 | How to calculate this area? (portion of a sphere inside a cylinder )
The area of the portion of the sphere [imath] x^{2} + y^{2} +z^{2} = 1[/imath] located inside of the cylinder [imath]x = x^{2} + y^{2}[/imath], and above the plane [imath]z = 0[/imath]. I'm stuck, so any tip will be helpful Thanks in advance! |
2518492 | Number of ways of distributing identical items among people.
The number of ways of distributing [imath]n[/imath] identical items among [imath]r[/imath] people such that each one of them receives at least one is [imath]\binom{n-1}{r-1}[/imath]. Can you please give a simple proof to explain how is this theorem derived ? Thanks . | 2352032 | Stars and Bars Derivation
I am trying to understand the derivation of the stars and bars formula. I understand how a problem can be converted to the stars and bars format, but I am confused about how the problem of choosing all the different ways to put k stars in n sections can be written as [imath]{n+k-1 \choose k}[/imath]. Generally "n choose k" is for counting the number of combinations of k objects we can get given n distinct objects right? That I understand since we're doing permutation divided by the number of orderings for each group since order doesn't matter. I need help figuring out how stars and bars fits into this. |
2517555 | Consider the set of all [imath]n×n[/imath] invertible real matrices. Is that set connected?
I thought of creating a path from a matrix [imath]A[/imath] to a matrix [imath]B[/imath] using their traces, but got nowhere. | 898688 | invertible matrices connected or not
The question asks "Is the set of all [imath]3\times3[/imath] real invertible matrices connected?" My intuitive idea is that we can establish a separation consisting of matrices with positive and negative determinant respectively, whose union will be the whole set but with no intersections. However I am not sure how to show intersection of one set with the closure of the other set is empty, according to the definition of being disconnected. (My guess is that the closure for real matrices with negative determinant is just itself union matrices with 0 determinant) So what will be a rigorous proof of this, using only tools in point-set topology and knowledge in linear algebra? I know that [imath]3\times3[/imath] matrices can be viewed as homeomorphic to [imath]\mathbb{R}^9[/imath], but how do we define the topology on this subspace (i.e. set of all [imath]3\times3[/imath] matrices)? What will an open set look like in this topology? |
2518747 | Proof of [imath](ab)^n=a^nb^n[/imath] is fulfilled when it is true for [imath]n = 2[/imath]
Prove that for given elements [imath]a,b[/imath] from group [imath]G[/imath], equation [imath](ab)^n = a^nb^n[/imath] is fulfilled for all [imath]n∈ {\displaystyle \mathbb {Z} }, [/imath] if and only if equation is true for [imath] n=2 [/imath]. I have no idea how to prove it. I would be gratefulfor any help. | 492332 | Let [imath]a,b[/imath] be in a group [imath]G[/imath]. Show [imath](ab)^n=a^nb^n[/imath] [imath]\forall n\in\mathbb{Z}[/imath] if and only if [imath]ab=ba[/imath].
Let [imath]G[/imath] be a group and [imath]a,b\in G[/imath]. Show [imath](ab)^n=a^nb^n[/imath] for all [imath]n\in\mathbb{Z}[/imath] if and only if [imath]ab=ba[/imath]. I don't known where to start. It seems trivially. |
908947 | Sum of an irreducible character over all [imath]G[/imath]
Let [imath]\phi: G\to GL_n(\mathbb C)[/imath] be an irreducible representation of a finite group [imath]G[/imath]. Let [imath]\chi: G\to \mathbb C[/imath] be the character of [imath]G[/imath]. Prove that: [imath]\displaystyle \sum_{g\in G}{\chi(g)}=0 [/imath] I know that if [imath]e[/imath] is the identity then [imath]\chi(e)=n[/imath], if [imath]g\in G[/imath] then [imath]\chi(g^{-1})=\overline{\chi(g)}[/imath]. Thus if [imath]g=g^{-1}[/imath] then [imath]\chi(g)=\chi(g^{-1})=\overline{\chi(g)}[/imath] then [imath]\chi(g)\in \mathbb R[/imath]. If [imath]g \ne g^{-1}[/imath] then [imath]\chi(g)+\chi(g^{-1})=2Re(\chi(g))[/imath]. Thus we can conclude at least that the sum: [imath]\displaystyle \sum_{g\in G}{\chi(g)}=n+\sum_{g\ne g^{-1}}{\chi(g)}+\sum_{g=g^{-1}}{\chi(g)} \in \mathbb R [/imath] But I don't know how to prove that it's in fact [imath]0[/imath]. Thanks! | 768051 | Sum of irreducible character values in a row of the character table
If [imath]\chi[/imath] is a nontrivial irreducible character of [imath]G[/imath] (a finite group), define [imath]S_{\chi}:= \sum_{x \in G} \chi(x)[/imath]. In terms of conjugacy classes [imath]\mathcal{C}[/imath], this is [imath]\sum_{\mathcal{C}} |\mathcal{C}| \chi(\mathcal{C})[/imath]. Is there a nice condition that guarantees [imath]S_{\chi}=0[/imath]? I've noticed that this occurs, for instance, with [imath]S_5[/imath]. I'd love a description of this phenomenon and a proof, if possible. |
2519605 | Normal extension of a group
Let [imath]\alpha=\sqrt{2+\sqrt{3}}[/imath]. Let [imath]L=\mathbb{Q}(\alpha)[/imath]. Show that [imath]L[/imath] is a normal extension of [imath]\mathbb{Q}[/imath]. I know that we need to prove that the Galois group for this extension is isomorphic to [imath]\mathbb{Z}_2 \times \mathbb{Z_2}[/imath] (which i dont know how to prove ). An extension is normal if and only if the subgroup of automorphisms which fix each intermediate field is a normal subgroup of the Galois group. When this group is abelian, every subgroup is normal and thus every extension is Galois. Thanks | 1367383 | Determine the Galois group of [imath]\mathbb{Q}(\sqrt{a+b\sqrt{d}})[/imath]
Suppose [imath]L=\mathbb{Q}(\sqrt{a+b\sqrt{d}})[/imath],([imath]d[/imath] and [imath]a+b\sqrt{d}[/imath] are square free algebraic integers) when is [imath]L/\mathbb{Q}[/imath] a normal extension? When does [imath]Aut(L/\mathbb{Q})=\mathbb{Z}/4\mathbb{Z}[/imath] or [imath]Aut(L/\mathbb{Q})=\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}[/imath]? |
2519523 | Does anybody known the relationship between the [imath]\Gamma (a)[/imath] and [imath]\Gamma (a+\frac{1}{2})[/imath]
Does anybody known the relationship between the [imath]\Gamma (a)[/imath] and [imath]\Gamma (a+\frac{1}{2})[/imath], is there a equation or approximate equality between this two? | 846626 | Duplication formula for gamma function
Using the Weierstrass definition for [imath]\Gamma(x)[/imath] and [imath]\Gamma\Big(x + \frac12\Big)[/imath], how can I prove the duplication formula? This is problem [imath]10.7.3[/imath] in the book Irresistible Integrals, by Boros and Moll. Any help is highly appreciated. |
2518426 | From [imath](1+z)^p(1+z)^q=(1+z)^{p+q}[/imath] show that [imath]\sum\limits_{n=0}^r\binom{p}{n}\binom{q}{r-n}=\binom{p+q}{r}[/imath] without using the Cauchy product
I'm stuck on what was an easy exercise at first sight. It's exercise 18, PSet 15.3, Kreyszig "Advanced Engineering Mathematics" 10th ed. It asks to prove that given [imath](1+z)^p(1+z)^q=(1+z)^{p+q}[/imath] then [imath]\sum\limits_{n=0}^r\binom{p}{n}\binom{q}{r-n}=\binom{p+q}{r}[/imath]. Taking for granted that [imath](1+z)^p=\sum\limits_{k=0}^p\binom{p}{k} z^k[/imath] and from [imath]\sum\limits_{n=0}^p\binom{p}{n} z^n \sum\limits_{m=0}^q\binom{q}{m} z^m=\sum\limits_{r=0}^{p+q}\binom{p+q}{r} z^r[/imath] I should be able to manipulate the indices to get [imath]\sum\limits_{r=0}^{p+q}\sum\limits_{n=0}^{r}\binom{p}{n} \binom{q}{r-n}z^r[/imath] but I can't. My attempt was noticing that there is a constraint for which [imath]n+m=r[/imath] and [imath]m,n>0[/imath]: [imath] \sum\limits_{n=0}^p\binom{p}{n} z^n \sum\limits_{m=0}^q\binom{q}{m} z^m=\sum\limits_{n=0}^p\sum\limits_{m=0}^q\binom{p}{n}\binom{q}{m} z^{n+m}=\sum\limits_{n=0}^p\sum\limits_{r=n}^{q+n}\binom{p}{n}\binom{q}{r-n} z^r=\\ =\sum\limits_{r=n}^{q+n}\sum\limits_{n=0}^p\binom{p}{n}\binom{q}{r-n} z^r[/imath] This is the closest I can get to [imath]\sum\limits_{r=0}^{p+q}\sum\limits_{n=0}^{r}\binom{p}{n} \binom{q}{r-n}z^r[/imath]. Can anyone help me? Thanks, Luca | 2516782 | Binomial-Theorem proof
I have the following problem: [imath]n,m \in \mathbb{N_0}[/imath]. Write [imath](1+x)^{n+m}=(1+x)^{n}(1+x)^{m}[/imath] with the help of binomial formulas, multiply the right side, and deduce the following Identity between binomial coefficients: [imath]\begin{pmatrix} n+m\\k\\ \end{pmatrix}= \sum_{l=0}^{k}{\begin{pmatrix} n\\l\\ \end{pmatrix}}{\begin{pmatrix} m\\k-l\\ \end{pmatrix}} \forall k \in \mathbb{N_0}[/imath] My idea: I know that: [imath](a+b)^n=\sum_{k=0}^{n}{\begin{pmatrix} n\\k\\ \end{pmatrix}a^{n-k}b^k=\sum_{k=0}^{n}{\begin{pmatrix} n\\k\\ \end{pmatrix}a^{k}b^{n-k}}}[/imath] So:[imath](1+x)^{n+m}=\sum_{k=0}^{n}{{\begin{pmatrix} n\\k\\ \end{pmatrix}x^k}*\sum_{k=0}^{n}{{\begin{pmatrix} m\\k\\ \end{pmatrix}x^k}}} =\sum_{k=0}^{n}{{\begin{pmatrix} n+m\\k\\ \end{pmatrix}x^k}}=(1+x)^{n+m}[/imath] But this doens't seem right.. Can anyone help me here? |
2519005 | If the [imath]Aut(C)[/imath] is trivial for a chain [imath]C[/imath] then [imath]C[/imath] or [imath]C^{op}[/imath] is a well order?
If you have a chain that is a well order (all well orders are chains), then [imath]Aut(C)[/imath] is trivial. I'm asking if the converse is also true. | 437826 | Is this a characterization of well-orders?
While grading some papers and thinking about a question related to well-orders (in particular, pointing a mistake in a solution), I came to think of a reasonable characterization for well-orders. I can immediately see it's true for countable orders, but not for uncountable orders. Definition. Let [imath]\cal L[/imath] be a first-order language, and [imath]\cal M[/imath] an [imath]\cal L[/imath]-structure. We say that [imath]\cal M[/imath] is rigid if [imath]\text{Aut}(\cal M)=\{\rm id\}[/imath]. Conjecture. Let [imath]\cal L=\{<\}[/imath]. Then [imath]\mathcal M=\langle M,<\rangle[/imath] is rigid if and only if it is a well-order or its reverse order is a well-order. One direction is trivial. Well-orders are rigid (and therefore their reversed orders are rigid as well). In the other direction, if [imath]\cal M[/imath] is countable, then it is easy. Suppose it's not a well-order. If [imath]\cal M[/imath] contains a convex copy of [imath]\Bbb Z[/imath] then fix everything else and shift that copy by [imath]1[/imath]. Otherwise [imath]\cal M[/imath] contains a convex copy of [imath]\Bbb Q[/imath] and then we can shift that copy by [imath]1[/imath] (or multiply it by [imath]\frac12[/imath], whatever floats your boat). The problem is that for uncountable order types the plot thickens and they might be [imath]\kappa[/imath]-dense, so neither of the arguments would work (and there is no additive structure - that I know of - that we can exploit like in the countable case). Question: Does the conjecture hold, or is there some intricate counterexample? |
2519807 | Numerical evaluation of a sine integral
Considering the integral [imath] \int_{-\pi}^{\pi} \left( \frac{\sin x}{x} \right)^{300} \text{d}x [/imath] Is it possible to find for it a reasonable numerical approximation, accurate up to two significant digits? Obviously by carrying out computations by hand... I don't know how to approach this problem, so thanks a lot to anybody willing to help! :) | 2479265 | How to estimate [imath]\int^{1}_{-1} \left(\frac{\sin{x}}{x}\right)^{300} dx[/imath] to 1 significant figure?
I would like to estimate [imath]\int^{1}_{-1} \left(\frac{\sin{x}}{x}\right)^{300} dx[/imath] to [imath]1[/imath] significant figure. (This question is taken from a quant exam). My (vague) idea is to use Taylor series expansion and to estimate the remainder term. But then I run into problems immediately as I don't see a straightforward way to compute the first few terms of Taylor series for [imath]\left(\frac{\sin{x}}{x}\right)^{300}[/imath]... Any ideas? |
521906 | Proof by contradiction that [imath]\sqrt{2} + \sqrt{6} < \sqrt{15}[/imath]
Claim: [imath]\sqrt{2} + \sqrt{6} < \sqrt{15}[/imath] a) State the negation of the claim. b) Use proof by contradiction to prove the claim. Can anyone explain to me how I would do this question? A computation suggests that these numbers are relatively close to each other; [imath]\sqrt{15}-\sqrt2-\sqrt6 \approx 0.01[/imath]. | 2520000 | Prove by contradiction (not using a calculator) that [imath]\sqrt6 + \sqrt2 < \sqrt{15}[/imath]?
Prove by contradiction (not using a calculator) that [imath]\sqrt6 + \sqrt2 < \sqrt{15}[/imath]. How do you approach such a problem? I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. I tried to square it as I read online but I'm stuck with having [imath]\sqrt3[/imath] and not knowing how to get rid of it to have the answer as clear as the sun. squaring them yielded: [imath](\sqrt2 + \sqrt6)^2 = 8 + 4\sqrt3 < 15[/imath] which indeed holds but the problem is that I am not allowed to use calculator to check the little difference. My final answer was [imath]4\sqrt3 < 7[/imath]. |
1984432 | multiplicity of an irreducible representation
Here's a quite basic question, and I'd be very glad for an answer! The multiplicity [imath]m_V[/imath] of an irreducible representation [imath]V[/imath] in the representation [imath]W[/imath] of the group [imath]G[/imath] is given by [imath] m_V = \dim \left( Hom(V,W)^G \right) [/imath] where [imath]Hom(V,W)^G[/imath] is the space of [imath]G[/imath]-module homomorphisms from [imath]V[/imath] to [imath]W[/imath]. How to see this? | 858852 | Multiplicity of G-module
I am currently working on Bruce Sagan's The Symmetric Group. The following proposition is given without proof: Let [imath]V[/imath] and [imath]W[/imath] be [imath]G[/imath]-modules with [imath]V[/imath] irreducible. Then dim Hom([imath]V[/imath],[imath]W[/imath]) is the multiplicity of [imath]V[/imath] in [imath]W[/imath]. This proposition is a strengthening of the following corollary of Schur's Lemma: Let [imath]V[/imath] and [imath]W[/imath] be two [imath]G[/imath]-modules with [imath]V[/imath] being irreducible. Then dim Hom([imath]V[/imath],[imath]W[/imath]) = [imath]0[/imath] if and only if [imath]W[/imath] contains no submodule isomorphic to [imath]V[/imath]. I understand why the corollary is true: Let us suppose [imath]W[/imath] contains a submodule isomorphic to [imath]V[/imath]. Then clearly, dim Hom([imath]V[/imath],[imath]W[/imath]) > [imath]0[/imath] because of the isomorphism from [imath]V[/imath] to the submodule. Let us suppose im Hom([imath]V[/imath],[imath]W[/imath]) > [imath]0[/imath]. Then there must be some non-zero homomorphism from [imath]V[/imath] to some submodule of [imath]W[/imath]. As the homomorphism is non-zero, its kernel is not [imath]V[/imath], so the kernel is zero, by Schur's Lemma. Hence it is injective. As it is surjective to its image, it is an isomorphism. How could one prove this proposition? Is the proof similar to the proof of the corollary? Thank you very much for your help! |
2520116 | Prove that [imath]\lim_{x\to 0}\frac{\ln (1+x)}{x}=1[/imath] by Taylor series.
I want to prove that [imath]\lim_{x\to 0}\frac{\ln (1+x)}{x}=1[/imath] by Taylor series. I have get to the step: [imath]\lim_{x\to 0}\frac{\ln (1+x)}{x}=\lim_{x\to 0}\frac{x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots}{x}[/imath] Next, when [imath]x\neq0[/imath], [imath]\frac{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots}{x}=\frac{1}{x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\right)=1-\frac{x}{2}+\frac{x^2}{3}-\cdots[/imath]. This is valid because [imath]\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}{n}[/imath] converges for every [imath]x\neq 0[/imath] near [imath]0[/imath], and then we can multiply a constant [imath]\frac{1}{x}[/imath] inside it. However, how can I rigorously and easily say that [imath]\lim_{x\to 0}(1-\frac{x}{2}+\frac{x^2}{3}-\cdots)=1[/imath]? | 985058 | Use the definition of derivative to prove [imath]\lim_{x\to 0}\ln(x+1)/x =1[/imath]
How would I use the definition of derivative to prove [imath]\lim_{x\to 0} \frac{\ln(1+x)}{x} = 1[/imath] I got to [imath]\frac{\frac{\ln(1+x+h)}{(x+h)} - \frac{\ln(1+x)}{x}}{h}[/imath] but have no idea where to go from here. On another site I found someones answer where they stated the following: [imath] \lim_{x\to 0} \frac{\ln(1+x) - \ln(1+0)}{x-0} = [\ln(1+x)]'\rvert_x = 0 [/imath] but I am unsure why the [imath]x[/imath] in the [imath]x-0[/imath] is removed. Can someone please explain? |
2519822 | Graph Colouring proof
Prove that any (simple, undirected) graph [imath]G = (V, E)[/imath], with [imath]m = |E|[/imath] edges has chromatic number [imath]χ(G) ≤\sqrt {2m} + 1[/imath]. Could any please give me any hint about this question? I don’t know how to start it. Maybe induction? Edit: I'd say they seem similar but it's hardly the same question | 645339 | Prove, that graph [imath]G[/imath] has at least [imath]\chi(G)(\chi(G)-1)/2[/imath] edges.
Can anybody give me any hints about how to prove that for any graph [imath]G[/imath] the number of edges in it is at least [imath]\chi(G)(\chi(G)-1)/2[/imath]? [imath]\chi(G)[/imath] is the minimal number of colors we need to use to color the graph (it's chromatic index in other words). I tried induction but I'm stuck at an induction step. I do the induction on the number of edges. When [imath]|E(G)|=0[/imath] then obviously we can color the graph using only one color, so the theorem holds. Now induction step from [imath]n \Rightarrow n+1[/imath] edges. Let's erase any edge [imath]e=\{v,u\}[/imath] from [imath]G[/imath]. Then [imath]\chi(G-e)[/imath] is either the same, or [imath]\chi(G-e)<\chi(G)[/imath], because we relaxed the graph an [imath]v[/imath] and [imath]u[/imath] no longer need to have different colors. From the first case the thesis is clear, because [imath]G-e[/imath] has more then [imath]\chi(G)(\chi(G)-1)/2[/imath] edges so with the one we erased it's still more. But from the second case we only know that [imath]G \geq \chi'(G)(\chi'(G)-1)/2[/imath] and [imath]\chi'(G)(\chi'(G)-1)/2 \leq \chi(G)(\chi(G)-1)/2[/imath] so that gives me nothing... |
2520399 | [imath]a_1 = 2, a_2 =7[/imath] and [imath]a_{n+1} = \frac{1}{2}\cdot(a_{n} + a_{n-1})[/imath], Find [imath]a_n[/imath]
For this sequence: [imath]a_1 = 2~~~a_2 =7[/imath] and [imath]a_{n+1} = \frac{1}{2}\cdot(a_{n} + a_{n-1})[/imath] The questions asks to find the exact formula for the [imath]n[/imath]th term of the sequence and hence find the limit as [imath]n \to \infty[/imath] They also gave us this formulae: http://puu.sh/ylNZ1/5e595930e0.jpg, I am not particularly sure how to use them though. Anyways I tried listing out more terms and I found that [imath]a_3 = \frac{9}{2} ~~~ a_4 = \frac{23}{4}[/imath] And I found out the difference between these terms is [imath]5,\frac{5}{2},\frac{5}{4}[/imath] thus we can write [imath]a_{n+1}=a_{n} + \frac{5}{2^{n-1}}[/imath] I am not sure how to continue though. | 2150607 | Recursive sequence convergence.: [imath]s_{n+1}=\frac{1}{2} (s_n+s_{n-1})[/imath] for [imath]n\geq 2[/imath], where [imath]s_1>s_2>0[/imath]
The problem is the following: suppose [imath]s_1>s_2>0[/imath] and let [imath]s_{n+1}=\frac{1}{2} (s_n+s_{n-1})[/imath] for [imath]n\geq 2[/imath]. Show that ([imath]s_n[/imath]) converges. Now, here is what I figured out: [imath]s_2<s_4[/imath]: Base Case for induction that [imath]s_{2n}[/imath] is an increasing sequence. Assume [imath]s_{2n-2}<s_{2n}[/imath]. Induction step: [imath]s_{2n}<s_{2n+2}[/imath] [imath]s_1>s_3[/imath]: Base Case for induction that [imath]s_{2n-1}[/imath] is a decreasing sequence. Assume [imath]s_{2n-1}<s_{2n-3}[/imath]. Induction step: [imath]s_{2n+1}<s_{2n-1}[/imath]. I have proved those two. However arguing in favor of convergence has me going around in circles. Since [imath]s_1>s_2[/imath] and (as discovered during the formulation of Base Cases) [imath]s_3>s_4[/imath], I figured it might be a good idea ot assume that if every odd member of the original sequence ([imath]s_n[/imath]) is greater than the following even member, then the limit would be somewhere in between, the two (odd and even) sequences won't cross. Hence the upper and lower bounds would be [imath]s_1[/imath] and [imath]s_2[/imath] respectively. Here is how I approach this: Assume [imath]s_{2n-1}>s_{2n}[/imath] Show that [imath]s_{2n+1}>s_{2n+2}[/imath] The proof as I mentioned has me running in circles. Any assistance? |
2520111 | Elementary set theory equivalent statements
Let [imath]A, B\subset M[/imath]. Prove the following statement: [imath]A\subset B \iff A^c\cup B = M[/imath] M is universal set and [imath]A^c[/imath] is complement of A. I can kind a solve this using bunch of reasoning but in no real way using set laws or using arbitrary elements to prove that [imath]A^c\cup B \subset M[/imath] and [imath]M\subset A^c\cup B[/imath]. I'm pretty sure its very trivial but I'm just looking for a very formal way to prove. | 2141704 | Proving set theory question
I need to prove the following letting A and B be subsets of a universal set U. [imath]A\cup B = U \iff A^c\subset B[/imath] It just isn't straight forward to me. |
2521048 | Determining the limit of [imath]\lim_{h \to 0} \frac{cos(x+h) - cos(x)}{\sqrt{x+h}-\sqrt{x}}[/imath] when x > 0
I'm pretty sure that this uses L'Hopital's rule but when I try to apply this I end up going in circles and am unable to ever find a limit. Taking derivatives of numerator and denominator: [imath]\lim_{h \to 0} \frac{-sin(x+h) +sin(x)}{\frac{1}{2}({x+h})^{-1/2}-\frac{1}{2}{x}^{-1/2}}[/imath] This still results in [imath]\frac{0}{0}[/imath] though. | 2520657 | Determining the limit [imath]\lim_{h\to0} \frac{\cos(x+h) - \cos(x)}{(x+h)^{1/2} - x^{1/2}}[/imath]
Determine the limit: [imath]\lim_{h\to0} \frac{\cos(x+h) - \cos(x)}{(x+h)^{1/2} - x^{1/2}}[/imath] After taking the conjugate, I got: [imath]\lim_{h\to 0} \frac{\big(\cos(x+h) - \cos(x)\big)\big((x+h)^{1/2} + x^{1/2})\big)} h[/imath] I took the conjugate of this, but I don't see how I can cancel out the [imath]h[/imath]. Any tips? |
2519639 | [imath]\frac{\mathbb{C}[x,y,z]}{x^5+y^3+z^2}[/imath] is UFD
I wish to show [imath]A=\dfrac{\mathbb{C}[x,y,z]}{x^5+y^3+z^2}[/imath] is UFD What I have tried so far: I was thinking of applying Nagata's criterion. Indeed, [imath]-(x^5+y^3)[/imath] cannot be a square in [imath]\mathbb{C}[x,y][/imath]. Hence [imath]A[/imath] is a domain. Note that [imath]\frac{A}{xA}=\frac{\mathbb{C}[x,y,z]}{(x,x^5+y^3+z^2)} \cong \frac{\mathbb{C}[y,z]}{y^3+z^2}[/imath] which is a domain as [imath]-y^3[/imath] is not a square in [imath]\mathbb{C}[y][/imath]. Hence [imath]x[/imath] is a prime in [imath]A[/imath]. Thus enough to check [imath]A[\frac1x][/imath] is a UFD. Now I am stuck here. I cannot simplify the ring [imath]A[\frac1x][/imath]. Any help/ suggestions to tackle this problem. | 1424190 | [imath](K[x,y,z]/(x^2+y^3+z^7))_{(x,y,z)}[/imath] is a UFD.
How can I show given a field [imath]K[/imath] that [imath](K[x,y,z]/(x^2+y^3+z^7))_{(x,y,z)}[/imath] is a UFD? I found that statement in a Wikipedia page so i'm not 100% sure it's true, maybe it's true for some field only? |
2519480 | Solve [imath]AX^2 + BX + C = 0[/imath]
How does one solve equations of the form [imath]AX^2+BX+C=0[/imath] where [imath]A,B,C[/imath] are square matrices and [imath]X[/imath] is a matrix to be solved for? More generally how does one solve equations of the form [imath]AX^2B+CXD+E=0[/imath]? Even more generally how does one solve higher order equations of this form such as [imath]AX^3B+CX^2D+EXF+G=0[/imath] In all cases, I could simply express the entries of the matrix [imath]X[/imath] as variables [imath]x_1,x_2,\dots[/imath] and multiply out the matrices and derive equations corresponding to each entry of the matrices however this would result in (for the first and most simple case at least) a system of 4 quadratic equations in 4 variables which I have no experience with. Is this the best approach or can the fact that we are dealing with matrices help to simplify the problem? Thanks | 95981 | Is there a unique solution for this quadratic matrix equation?
Here is the quadratic matrix equation I've been looking at lately: [imath] Q_{r,r}=A_{r,r}X_{r,r}^2+B_{r,r}X_{r,r}+C_{r,r}=0_{r,r} [/imath] Note that [imath]A, B, C,[/imath] and [imath]X[/imath] are [imath]r \times r[/imath] matrices. [imath]A[/imath] contains known elements, [imath]B[/imath] contains known elements, [imath]C[/imath] contains known elements, and [imath]X[/imath] contains the unknown elements that you are solving for. [imath] 0_{r,r} [/imath] is just the [imath]r \times r[/imath] null matrix. Is there any solution for [imath]X[/imath] in terms of [imath]A, B,[/imath] and [imath]C[/imath] (making no easy assumptions)? (e.g. [imath]X[/imath] is a diagonal matrix, [imath]A=B=C[/imath], or anything of that sort.) I have tried to solve this and nothing has worked out. I attempted solving it generally by manipulating the matrices in variable form (i.e. actually writing out the matrices [imath]A, B, C,[/imath] and [imath]X[/imath] in variables) and finding a unique solution for all of the elements of [imath]X[/imath] in terms of the elements of [imath]A, B,[/imath] and [imath]C[/imath]. That didn't work out beyond the case of [imath]r=1[/imath]. Trying to solve it by looking at [imath]r[/imath] at different values did not work out either; I ended up with very abysmal equations at just [imath]r=2[/imath]. I don't know exactly how to make this appealing to the denizens of math.stackexchange, but it (as far as I know) isn't a heavily studied problem. There is a very high possibility that I've just been doing elementary techniques and nothing of note, so I hope someone or a group of people could shed light on this. |
2521157 | Where am I going wrong in derivation of the following trig formula?
I was reading a trig book in which the author had mentioned a trick to jump from [imath](\theta+90^\circ)[/imath] to [imath](\theta-90^\circ)[/imath] formulas. I understood it for Sine and Cosine but couldn't for Tangent. Here is the [imath](\theta+90^\circ)[/imath] formula… [imath]\tan(\theta+90^\circ)=-\cot\theta[/imath] He mentioned to replace [imath]\theta[/imath] in the above formula by [imath](\theta-90^\circ[/imath] to obtain the desired [imath](\theta-90^\circ)[/imath] formula. I did the same and ended up getting this. [imath]\tan(\theta)=-\cot(\theta-90^\circ)[/imath] which is wrong according to the book. It says the answer should be… [imath]\tan(\theta-90^\circ)=-\cot(\theta)[/imath] I do not where am I going wrong. What is incorrect in my approach? What is the flaw? Please help me track it down. | 2521110 | How have the "following trig formulas" been modified to obtain the new trig formulas?
I was reading a trig book in which the author had deduced the following formulas from a drawing. I am mentioning those as follows... By looking at the above drawing the author deduced these formulas for [imath](\theta+90^\circ)[/imath]... 1. [imath]\sin (\theta + 90^\circ)=\cos\theta[/imath] 2. [imath]\cos (\theta + 90^\circ)=-\sin\theta[/imath] 3. [imath]\tan (\theta + 90^\circ)=-\cot\theta[/imath] I understood how did he conclude above formulas from the drawing. As you can see the drawing depicts author's approach to derive the [imath](\theta+90^\circ)[/imath] formulas by taking a similar triangle in the second quadrant, in the same way, to derive [imath](\theta-90^\circ)[/imath] formulas, he advised to take a similar triangle (as in the drawing) in quadrant 4 as well. But, to the contrary he mentioned a trick to get those formulas which I do not understand. The author told to replace [imath]\theta[/imath] in formulas 1,2 and 3 by [imath](\theta-90^\circ)[/imath]. So what I got by doing that is... 1. [imath]\sin ((\theta-90^\circ) + 90^\circ)=\cos(\theta-90^\circ)[/imath] 2. [imath]\cos ((\theta-90^\circ) + 90^\circ)=-\sin(\theta-90^\circ)[/imath] 3. [imath]\tan ((\theta-90^\circ) + 90^\circ)=-\cot(\theta-90^\circ)[/imath] By evaluating them further I get... 1. [imath]\sin (\theta)=\cos(\theta-90^\circ)[/imath] (I got this one correct) 2. [imath]\cos (\theta)=-\sin(\theta-90^\circ)[/imath] (wrong) 3. [imath]\tan (\theta)=-\cot(\theta-90^\circ)[/imath] (wrong) But the author says that the outcome of the formulas should be... 1.[imath]\sin (\theta)=\cos(\theta-90^\circ)[/imath] (correct.) 2. [imath]-\cos (\theta)=\sin(\theta-90^\circ)[/imath] 3. [imath]\tan (\theta-90^\circ) =-\cot(\theta)[/imath] I don't know where am I going wrong. The author told to replace [imath]\theta[/imath] by [imath](\theta-90^\circ)[/imath] then why am I getting my answers incorrect? Where is the flaw? Please help me to track it down. |
2519227 | If [imath]Z[/imath] is a standard normal random variable and [imath]g(z)[/imath] is differentiable, show that [imath]\Bbb{E}[g'(z)]=\Bbb{E}[Zg(z)][/imath]
I am looking for help proving this proposition: "If [imath]Z[/imath] is a standard normal random variable and [imath]g(z)[/imath] is differentiable, then [imath]\mathbb{E}[g'(Z)]=\mathbb{E}[Zg(Z)][/imath]" I have tried evaluating [imath]\mathbb{E}[g'(Z)][/imath] using integration by parts as follows. [imath]\mathbb{E}[g'(z)]=\int_{-\infty}^\infty{g'(z)}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}dz=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty{g'(z)}e^{-\frac{z^2}{2}}dz[/imath] Let [imath]u=e^{-\frac{z^2}{2}}[/imath] and [imath]dv=g'(z)dz[/imath]. Now, [imath]\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}g(z)\Big|_{-\infty}^{\infty}+\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}zg(z)e^{-\frac{z^2}{2}}dz[/imath] Interestingly the second half of this equation is equal to [imath]\mathbb{E}[Zg(Z)][/imath]. I am not sure how to show that the first half of the above equation equals zero and is thus equal to [imath]\mathbb{E}[Zg(Z)][/imath]. I am also not sure how to evaluate [imath]\mathbb{E}[Zg(Z)][/imath] as to yield the same equation, so it seems I have hit a dead end. I would like some help finishing this proof. Preferably, I want to know the simplest method. | 280297 | Stein's lemma condition
(Apologies if I break some conventions, this is my first time posting!) I am working on proving Stein's characterization of the Normal distribution: for Z [imath]\sim N(0,1)[/imath] and some differentiable function [imath]f[/imath] with [imath]E[|f'(Z)|] < \infty[/imath], [imath]E[Zf(Z)] = E[f'(Z)][/imath] Writing the LHS expression in integral form and integrating by parts, I eventually obtain: [imath]E[Zf(Z)] = \frac{1}{\sqrt{2\pi}} \left[ -f(z) \cdot \exp \left\{ \frac{-z^2}{2} \right\} \right] \Bigg|_{-\infty}^{\infty} + E[f'(Z)][/imath] Now I need to show that the first expression on the right hand size is zero. Intuitively, this seems clear because of the exponential term, but I am having trouble explicitly applying the condition on [imath]f'[/imath] to prove this rigorously. Any ideas? |
2521204 | Given [imath]f(x+y)=f(x)+f(y)+x^2y+xy^2[/imath] ,find [imath]f'(x)[/imath]
This question has already been asked but the solution was not satisfactory Suppose [imath]f[/imath] is a function satisfying the equation [imath]f(x+y)=f(x)+f(y)+x^2y+xy^2[/imath] for all real numbers [imath]x[/imath] and [imath]y[/imath] Suppose also that [imath]\lim \limits_{x \to 0}\frac{f(x)}{x}=1[/imath] Find [imath]f'(x)[/imath] By using the definition of derivative,I obtained [imath]f(0)=0[/imath] and [imath]f'(0)=1[/imath] How will I get [imath]f'(x)[/imath]? | 857229 | How do I find [imath]f(0)[/imath], [imath]f'(0)[/imath], and [imath]f'(x)[/imath] given [imath]f(x+y)=f(x)+f(y)+x^2y+xy^2[/imath] and [imath]\lim_{x\to0}\frac{f(x)}{x}=1[/imath]?
How can I find [imath]f(0)[/imath], [imath]f'(0)[/imath], and [imath]f'(x)[/imath] given that [imath]f(x+y)=f(x)+f(y)+x^2y+xy^2[/imath] and [imath]\lim_{x\to0}\frac{f(x)}{x}=1[/imath]. |
2521224 | Are [imath]\mathbb{Q}[/imath] and [imath]\mathbb{Z}[/imath] homeomorphic?
How can I check whether [imath]\mathbb{Q}[/imath] and [imath]\mathbb{Z}[/imath], with their usual topologies inherited from [imath]\mathbb{R}[/imath], are homeomorphic? | 1522889 | Proving [imath]\mathbb{Q}[/imath] isn't homeomorphic to [imath]\mathbb{Z}[/imath]
Prove that [imath]\mathbb{Q}[/imath] isn't homeomorphic to [imath]\mathbb{Z}[/imath] I know that a homeomorphism is a bijection function from two sets where both the function and its inverse are continuos. So I figure that if I show that there is no continuos invective mapping from [imath]\mathbb{Q}\rightarrow\mathbb{Z}[/imath] then I have done the proof. But I do not know how to do this (if his is the right way). How can I start to show this? |
1523230 | Primitive Pythagorean triple divisible by 3
Prove that for any primitive Pythagorean triple (a, b, c), exactly one of a and b must be a multiple of 3, and c cannot be a multiple of 3. My attempt: Let a and b be relatively prime positive integers. If [imath]a\equiv \pm1 \pmod{3}[/imath] and [imath]b\equiv \pm1 \pmod{3}[/imath], [imath]c^2=a^2+b^2\equiv 1+1\equiv 2 \pmod{3}[/imath] This is impossible as the only quadratic residues modulo 3 are 0 and 1. So far, so good. If one of a, b is [imath]\equiv 0 \pmod{3}[/imath] and the other is [imath]\equiv \pm1 \pmod{3}[/imath], [imath]c^2=a^2+b^2\equiv 0+1\equiv 1 \pmod{3}[/imath] This is the part I don't understand. Just because [imath]c^2\equiv 1\pmod{3}[/imath] doesn't mean that [imath]c^2[/imath] must be a perfect square. For example, [imath]a=12[/imath] and [imath]b=13[/imath] satisfy the above conditions but [imath]c^2=a^2+b^2=313[/imath], which isn't a perfect square. | 2603394 | Show that [imath](x,y,z)[/imath] is a primitive Pythagorean triple then either [imath]x[/imath] or [imath]y[/imath] is divisible by [imath]3[/imath].
Show that [imath](x,y,z)[/imath] is a primitive Pythagorean triple then either [imath]x[/imath] or [imath]y[/imath] is divisible by [imath]3[/imath]. The solution is given by 13.1.2: But how do you become the 2 congruences? |
257634 | Let T be a linear transformation on the real vector space [imath]\mathbb{R^n}[/imath] over [imath]\mathbb R[/imath] such that [imath]T ^2 =μT[/imath] for same [imath]μ∈\mathbb R[/imath]
Let [imath]T[/imath] be a linear transformation on the real vector space [imath]\mathbb R^n[/imath] over [imath]\mathbb R[/imath] such that [imath]T^2 =\mu T[/imath] for some [imath]\mu\in\mathbb R[/imath] . Then which of the following is/are true? [imath]\|Tx\| = |\mu| \|x\|[/imath] for all [imath]x \in\mathbb {R^n}[/imath] If [imath]\|Tx\| = \| x\| [/imath]for some non zero vector [imath]x \in\mathbb R^n[/imath], then [imath]\mu=\pm1[/imath] [imath]T= \mu I[/imath] where [imath]I[/imath] is the identity transformation on [imath]\mathbb R^n[/imath] If [imath]\|Tx \|>\|x\|[/imath] for a non zero vector [imath]x \in \mathbb R^n[/imath], then [imath]T[/imath] is necessarily singular. I am completely stuck on it. Can anybody help me please? | 1895376 | Linear transformation [imath]T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}[/imath] such that [imath]T^{2}=\lambda T.[/imath]
Let [imath]T[/imath] be a linear transformation [imath]T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}[/imath] such that [imath]T^{2}=\lambda T[/imath] for some [imath]\lambda\in\mathbb{R}.[/imath] Then which of the following is/are true? [imath]1.\|T(x)\|=|\lambda|\|x\|[/imath] for all [imath]x\in\mathbb{R}^n.[/imath] [imath]2.[/imath] If [imath]\|T(x)\|=\|x\|[/imath] for some non-zero [imath]x\in\mathbb{R}^{n}[/imath], then [imath]\lambda=\pm 1.[/imath] [imath]3.T=\lambda I[/imath] where [imath]I[/imath] is the identity transformation on [imath]\mathbb{R}^{n}.[/imath] [imath]4.[/imath] If [imath]\|T(x)\|>\|x\|[/imath] for a nonzero vector [imath]x\in\mathbb{R}^{n},[/imath] then [imath]T[/imath] is singular. According to me since [imath]T^{2}=\lambda T[/imath] we have [imath]T[/imath] satisfies the polynomial [imath]x^{2}-\lambda x.[/imath] So choices for its minimal polynomial are [imath]x,x-\lambda,x(x-\lambda).[/imath] So [imath]3[/imath]rd option is clearly incorrect and [imath]4[/imath]th option is not satisfied if [imath]T=I[/imath] and hence correct one. I am confused in first two options . Please help me. Thanks in advance. |
149822 | If [imath]T\colon \mathbb R^n \to \mathbb R^n [/imath] linear and [imath]T^2 = kT[/imath]
It is given that [imath]T[/imath] is a linear transformation from [imath]\mathbb R^n[/imath] to [imath]\mathbb R^n[/imath] such that [imath]T^2 = k T [/imath] for some [imath]k\in \mathbb R[/imath]. Then, one or more of the options are true [imath]\|T(x)\| = |k| \|x\|[/imath] for all [imath]x\in \mathbb R^n[/imath]. If [imath]\|T(x)\| = \|x\|[/imath] , for some non- zero vector [imath]x\in \mathbb R^n[/imath] then [imath] k =\pm 1 [/imath]. [imath]\|T(x)\| > \|x\|[/imath] for some non- zero vector [imath]x\in \mathbb R^n[/imath], then T is singular. [imath] T = k I [/imath] where I is an identity transformation. Please suggest how to proceed. | 2651295 | [imath]T^2 = \lambda T[/imath]
Let [imath]T[/imath] be a linear transformation on the real vector space [imath]\mathbb R^{n}[/imath] over [imath]\mathbb R[/imath] such that [imath]T^2 = \lambda T[/imath] for some [imath]\lambda \in R[/imath]. Then which of the following options are true [imath]||Tx|| = |\lambda| ||x||[/imath] for all [imath]x \in R^n.[/imath] If [imath]\|Tx\| = \| x\| [/imath]for some non zero vector [imath]x \in\mathbb R^n[/imath], then [imath]\lambda=\pm1[/imath] [imath]T= \lambda I[/imath] where [imath]I[/imath] is the identity transformation on [imath]\mathbb R^n[/imath] If [imath]\|Tx \|>\|x\|[/imath] for a non zero vector [imath]x \in \mathbb R^n[/imath], then [imath]T[/imath] is necessarily singular. I tried to solve by trying to find [imath]T(T(x)) = \lambda T(x)[/imath]. Let [imath]y= T(x)[/imath] for all [imath]x \in \mathbb R^n[/imath]. Then [imath]T(y) = \lambda y[/imath]. implies that [imath]\lambda[/imath] is an eigen value of [imath]T[/imath] and entire image of [imath]T[/imath] is an eigen vector corresponding to [imath]\lambda[/imath]. I dont know whether I am right or not? and i cannot connect with the conclusions with this logic... I have searched this question in mathstack, I found one and i am not convinced with the answer given there....So before marking this question duplicate, please tell me how can i solve this question... |
2521132 | What justify the division of power series?
It is sometimes seen that people divide power series of two functions [imath]f[/imath] and [imath]g[/imath] to get the power series representation of [imath]\frac{f}{g}[/imath]. For example, [imath]\tan x=\frac{\sin x}{\cos x}[/imath], so compute [imath]\frac{x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\cdots}{1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\cdots}[/imath] as if it were two finite polynomials by long division, and get the quotient of first few terms, and see the pattern of what this whole power series are. However, I haven't read any analysis books which mentioned the division of power series. What justifies doing so? Why is it valid? Is there something we should be careful when doing this? Why we can do the traditional long divison on infinite terms polynomials? | 1606336 | Division between power series that converge at least for |x| < r, only valid for |x| sufficiently small?
I have this book Calculus, Ninth Edition by Varberg, Purcell, and Rigdon; there's a particular point of a theorem (and another line after that) about Infinite Series that I really don't understand. I quote from Section 9.7, Operations on Power Series: Let [imath]f(x) = \sum_{n=0}^{\infty} a_n \cdot x^n[/imath] and [imath]g(x) =\sum_{n=0}^{\infty} b_n x^n[/imath], with both of these series converging at least for [imath]|x| < r[/imath]. If the operations of addition, subtraction, and multiplication are performed on these series as if they were polynomials, the resulting series will converge for [imath]|x| < r[/imath] and represent [imath]f(x) + g(x)[/imath], [imath]f(x) - g(x)[/imath], and [imath]f(x) \cdot g(x)[/imath], respectively. If [imath]b_0 \neq 0[/imath] the corresponding result holds for division, but we can guarantee its validity only for [imath]|x|[/imath] sufficiently small....We mention that the operation of substituting one power series into another is also legitimate for [imath]|x|[/imath] sufficiently small, provided that the constant term of the substituted series is zero. What does it mean by "sufficiently small"? How do I find the limit of how large [imath]|x|[/imath] can be before divisions and substitutions between power series become invalid? My book doesn't go into detail on the exact nature of this restriction and doesn't include a proof of the theorem. If I were dividing one power series by another I don't really know what I should be looking for. I wasn't able to find information about convergence sets for a series generated by two power series divided by each other. |
2521443 | If [imath](G,*)[/imath] is a group and [imath]a^{-1}=a, \forall a \in G[/imath] then [imath]G[/imath] is abelian .
If [imath](G,*)[/imath] is a group and [imath]a^{-1}=a, \forall a \in G[/imath] then [imath]G[/imath] is abelian . Is it true or false and why? | 924742 | Are groups satisfying [imath]g=g^{-1}[/imath] for all [imath]g \in G[/imath] abelian?
Is the following statement true or false: If [imath]G[/imath] is a group with the property that [imath]g=g^{-1}[/imath] for all [imath]g \in G[/imath], then [imath]G[/imath] is abelian. I believe it is false since I know that abelian or commutative property implies that every element in [imath]G[/imath] must have an inverse. Thus [imath]g\cdot g^{-1}= g^{1}\cdot g = e[/imath]. I need someone to check my attempt. |
2521355 | Find all continuous functions [imath]f:\mathbb{R}\longrightarrow \mathbb{R}[/imath] so that [imath]f(x) = f(x^2+{1\over 4})[/imath] is valid for all [imath]x[/imath].
Find all continuous functions [imath]f:\mathbb{R}\longrightarrow \mathbb{R}[/imath] so that [imath]f(x) = f(x^2+{1\over 4})[/imath] is valid for all [imath]x[/imath]. Well, I thought that I solved this one, but then one of my student noticed obvious mistake. Namely I defined the sequence [imath]x_{n+1}= x^2_n +1/4[/imath] and [imath]x_0=a[/imath] where [imath]a[/imath] is arbitrary real number. It is easy to see that it is increasing and it has lover bound. I was trying to arguing that it converge to [imath]1/2[/imath] and since [imath]f(x_{n+1})=f(x_n)[/imath] it has constant value on this sequence and so [imath]f(a)=1/2[/imath] is for all [imath]a[/imath]. But this sequence is not convergent for all [imath]a[/imath]. Any ideas how to repair this ''prove''. | 487455 | [imath]f(x)=f(x^2+ 1/4)[/imath] , [imath]f[/imath] is continuous from [imath]\mathbb{R}[/imath] to [imath]\mathbb{R}[/imath]
Find all continous functions [imath]f\colon\mathbb{R}\rightarrow\mathbb{R}[/imath] satisfying [imath]f(x)=f(x^2+ 1/4)[/imath] What I've tried so far: suppose that [imath]f[/imath] is one-one thus [imath]x=x^2+1/4[/imath] ... [imath]x=1/2[/imath] then [imath]f(x)=f(1/2)[/imath] is constant That's it, I'm stuck here. |
2521499 | Limit of [imath](1+\frac{1}{2})(1+\frac{1}{4})...(1+\frac{1}{2^n})[/imath] as [imath]n \to \infty[/imath] and the rationality of the answer
What is the Limit of [imath](1+\frac{1}{2})(1+\frac{1}{4})...(1+\frac{1}{2^n})[/imath] at n = [imath]\infty[/imath]? At first I think it is divergent but it is surely convergent to something between [imath]2[/imath] and [imath]3[/imath]. More exactly it is between [imath]2.3[/imath] to [imath]2.4[/imath]. something about [imath]2.38[/imath]. but I don't know how to find its limit. What can we say about rationality of the answer? (Is it rational or irrational?) My question is made of two parts. first finding the limit and second, discussing about its rationality. | 1924882 | What is the product [imath]\prod\limits_{n=1}^{\infty}\left(1+\frac{1}{2^n}\right)[/imath]?
What is the value of this product? [imath]\prod_{n=1}^{\infty}\left(1+\frac{1}{2^n}\right)[/imath] Me and my friend came across this product. Is the product till infinity equal to [imath]1[/imath]? If no, what is the answer? |
2521624 | Cardinality without Axiom of Choice
Let [imath]A^A[/imath] be the set of all functions from [imath]A[/imath] to [imath]A[/imath]. Prove [imath]\mathcal{P}(A)[/imath] and [imath]A^A[/imath] have the same cardinality for infinite sets [imath]A[/imath]. ([imath]\mathcal{P}(A)[/imath] is the power set of [imath]A[/imath].) Is it possible to show this without invoking axiom of choice? It can be proven with [imath]A \times A \sim A[/imath] for infinite sets [imath]A[/imath] but that is equivalent to the axiom of choice. | 1660457 | How does equality [imath]\mathfrak{m}^\mathfrak{m} = 2^\mathfrak{m}[/imath] in [imath]\sf{ZF}[/imath] relate to the axiom of choice?
Usually, the equality [imath]\mathfrak{m}^\mathfrak{m} = 2^\mathfrak{m}[/imath] for infinite cardinal [imath]\mathfrak{m}[/imath] is proved like that: [imath] 2^\mathfrak{m} \leqslant \mathfrak{m}^\mathfrak{m} \leqslant (2^\mathfrak{m})^\mathfrak{m} = 2^{\mathfrak{m} \cdot \mathfrak{m}} = 2^\mathfrak{m} [/imath] The last step relies much on the axiom of choice, because the fact [imath]\mathfrak{m}\cdot\mathfrak{m} = \mathfrak{m}[/imath] does. Moreover, Tarski's theorem states that in [imath]\sf{ZF}[/imath] this fact is equivalent to [imath]\sf{AC}[/imath]. Note, by a cardinal I mean an equipollence class of an arbitrary set, not necessarily admitting a well-ordering. That has motivated my question: what can we say about relationship of the initial equality and the axiom of choice in pure [imath]\sf{ZF}[/imath]? To avoid question about definition of cardinals without [imath]\sf{AC}[/imath], we may consider the following statement instead: For any infinite set [imath]X[/imath] exists a bijection [imath]\; F \, \colon X^X \rightarrow 2^X[/imath] where [imath]X^X[/imath] is the set of all functions from [imath]X[/imath] to [imath]X[/imath], [imath]\; 2^X = \mathcal P(X)[/imath] is the power set, and an infinite set is such that cannot be bijectively mapped onto a finite ordinal. |
2521662 | Can functions satisfying these criteria exist?
I've been wondering whether or not such functions can exist: (1) [imath]f: \emptyset \rightarrow \emptyset[/imath] I can think of only one relation that satisfies this criterion - the empty relation defined on the empty set, namely: [imath]R = \{\}[/imath] but, my question is - is this even a function? The definition of a function says that [imath]f[/imath] is a function iff if two elements have the same first entry, they must have the same second entry - but I think this is not a problem here. (2) [imath]X[/imath] is non-empty: [imath]g: \emptyset \rightarrow X[/imath] As for this function, there are no possible first entries, and so it is impossible to map anything into set [imath]X[/imath] (3) [imath]X[/imath] is non-empty [imath]h: X \rightarrow \emptyset[/imath] I cannot think of anything that would map something into the empty set. Any suggestions? | 45625 | Why is an empty function considered a function?
A function by definition is a set of ordered pairs, and also according the Kuratowski, an ordered pair [imath](x,y)[/imath] is defined to be [imath]\{\{x\}, \{x,y\}\}.[/imath] Given [imath]A\neq \varnothing[/imath], and [imath]\varnothing\colon \varnothing \rightarrow A[/imath]. I know [imath]\varnothing \subseteq \varnothing \times A[/imath], but still an empty set is not an ordered pair. How do you explain that an empty function is a function? |
2519177 | prove that there exists a perfect cube that divides 135 consecutive integers
how would you prove that for 135 consecutive integers [imath]a_1,a_2,...,a_{135}[/imath] such that for all integers [imath]k[/imath], [imath]1 \leq k \geq 135[/imath] there exists a perfect cube [imath]z_k > 1[/imath] such that [imath]z_k[/imath] divides [imath]a_k[/imath]? Im stuck on this question and don't even know how to start. All I know is that if [imath]z_k[/imath] divides [imath]a_k[/imath] then [imath]a_k = mod (z_k)[/imath] | 2515945 | Prove there exist 135 consecutive integers that each have a perfect cube [imath]>1[/imath] that divides it.
As of 22:13, the answer below does not satisfy the constraints of my question, so I'm still accepting answers. Prove there exist 135 consecutive integers, [imath]x_1, x_2, ... , x_{135}[/imath], such that for each of these integers, [imath]x_i[/imath], there exists a number [imath]k_i[/imath] such that [imath]k_i>1[/imath] is a perfect cube where [imath]k_i | x_i[/imath]. I was told that I could use GCRT to help with this. I got stuck after rewriting the given statement as a system of linear congruences. |
2521727 | Proof that we are allowed to make substitutions to evaluate limits?
Basically, if we want [imath]\lim_{x\to x_0}f(x)[/imath], why are we 'allowed' to instead find the limit [imath]\lim_{u\to u_0}f(u)[/imath] where [imath]x = k(u)[/imath], where [imath]k(u)[/imath] is a function in [imath]u[/imath] such that [imath]\lim_{u\to u_0}k(u)= x_0[/imath]. Sorry if this isn't precise enough.. essentially, I want to know why, for example, when we find the limit [imath]\lim_{x\to 0}f(x)[/imath] we're allowed to instead find the limit [imath]\lim_{2t\to 0}f(t)[/imath] with [imath]x = 2t[/imath]. I think it's trivial because it's basically 'the same limit', but I'd like to verify that intuition. | 167926 | Formal basis for variable substitution in limits
So we do a lot of variable substitution in limits at school. Stuff like [imath]\lim\limits_{x\to5}\ (x+1)^2\ =\ \lim\limits_{y\to6}\ y^2[/imath], where we define the substitution [imath]y = x + 1[/imath]. But I've never been clear on what exactly the theoretical basis for this is. What is the formula that you're actually applying when you do variable substitution? What are the formal conditions under which it is possible? My conjecture would be the following: For all continuous [imath]f[/imath], and all real [imath]a[/imath]: [imath]\lim\limits_{x\to a}\ (f\circ g)(x)\ =\lim\limits_{x\to g(a)}\ f(x)[/imath], where [imath]g[/imath] is a continuous function So to take my first example, [imath]f[/imath] would be [imath]x^2[/imath], [imath]g[/imath] would be [imath]x + 1[/imath], and [imath]a[/imath] would be [imath]5[/imath]. Am I in the right area? If this is correct, can it be proven using [imath]\epsilon[/imath]-[imath]\delta[/imath]? I had a half-hearted shot at it the other night and didn't get anywhere. |
2522504 | Is [imath]t\sin\left(\frac1t\right)[/imath] uniformly continuous on [imath](0,1)[/imath]?
Let [imath]f:\mathbb{R}\to (0,1)[/imath], [imath]f=t\sin\left(\frac1t\right)[/imath]. I think this function is uniformly continuous, but I don't see how to prove this formally. I can observe that if [imath]s,t\in (0,1)[/imath] and [imath]|s-t|<\delta[/imath], then the smaller [imath]\delta[/imath] we take, the smaller the quantity [imath]\left|t\sin\left(\frac1t\right)-s\sin\left(\frac1s\right)\right|\mbox{ (*)}[/imath] becomes. So we can take any [imath]\varepsilon>0[/imath] and any [imath]s,t[/imath], such that (*) is less than [imath]\varepsilon[/imath], and there will necessarily be a [imath]\delta>0[/imath] such that [imath]|s-t|<\delta[/imath]. I would appreciate some suggestions on what approach to take. | 2190124 | show that [imath]x\sin\left(\frac{1}{x}\right)[/imath] is uniformly continuous on [imath](0,1)[/imath]
A student showed me the following exercise: Is [imath]h(x) = x \sin\left(\frac{1}{x}\right)[/imath] uniformly continuous on [imath](0,1)[/imath]? Admittedly, it has been a while since I have looked at problems involving uniform continuity and so I am not sure how to go about this problem. My guess is that [imath]h(x)[/imath] is uniformly continuous on this interval but I am not sure how to go about picking an appropriate [imath]\delta > 0[/imath] such that [imath]|x-c|< \delta[/imath] implies [imath]|h(x) - h(c)| < \epsilon[/imath] for [imath]x,c \in (0,1)[/imath]. The student and I looked at [imath]|x \sin(x) - c \sin(c)| \leq |x \sin(x)| + |c \sin(c)| \leq |x| + |c|[/imath] although this feel like the wrong path to go down. I appreciate any help! Thanks in advance! |
2522992 | If [imath]p[/imath] a prime number [imath]\mathbb{Z}^{\times}_p[/imath] is always isomorphic to [imath](\mathbb{Z}_{p-1},+)[/imath]
Here https://math.stackexchange.com/a/1127555/283262 someone stated: "If [imath]p[/imath] a prime number [imath]\mathbb{Z}^{\times}_p[/imath] is always isomorphic to [imath](\mathbb{Z}_{p-1},+)[/imath]" What is this Theorem called? Where can i find it? How to proove it? Thank you in advance. | 2349414 | Show that the multiplicative group [imath](\mathbb{Z}/p\mathbb{Z})^*[/imath] is cyclic
I'm looking for a proof of this that does not rely on the Chinese remainder theorem, since the exercise is from a book that has not reached such a point (Aluffi, Algebra). Show that for [imath]p[/imath] prime, the multiplicative group [imath]G=(\mathbb{Z}/p\mathbb{Z})^*[/imath] is cyclic. You may use that for [imath]p[/imath] prime, the equation [imath]x^d=1[/imath] has at most [imath]d[/imath] solutions in [imath]\mathbb{Z}/p\mathbb{Z}[/imath]. (hint: for [imath]g\in G[/imath] of maximal order, [imath]\forall h\in G, |h|\;|\;|g|[/imath], and in particular, [imath]h^{|g|}=1[/imath].) My only idea for now is constructing an element [imath]a\in G[/imath] such that [imath]|a|=|G|[/imath], using the fact that a group of order [imath]n[/imath] is cyclic [imath]\iff\exists[/imath] an element of order [imath]n[/imath]. |
2451722 | Calculate the area of center square in the following figure
Calculate the area of center square in the following figure:(the big square has a side length of 1 and each vertex of big square has been connected to the midpoint of opposite side) answer choices : [imath]\color{blue}{\frac14 , \frac15 , \frac16 , \frac17 , \frac18}[/imath] Please review and correct my solution: [imath]S_{BigSquare}=2S_{BigTriangle}+S_{Paralleogram} \Rightarrow[/imath] [imath]S_{Paralleogram}=\frac12 [/imath] Now let the side lentgh of center square be [imath]x[/imath] , then we have: [imath]S_{Paralleogram}=\frac12 =2\times Base\times x =2\times\frac{\sqrt5}{2}\times x \Rightarrow [/imath] [imath]x=\frac{\sqrt5}{10} \Rightarrow [/imath] [imath]S_{CenterSquare}=\frac{1}{20}[/imath] ???!!! Which is not among the choice!! What's wrong with my solution??!!! | 341754 | Finding the area of the shaded square inside a square created by connecting point-opposite midpoint
If the lines intersect the vertices of the square. The area of the square is [imath]1[/imath] and the lines also intersect the midpoints of the square lines. How to find the area of shaded region? |
2520101 | Differentiation of the cosine and sine power series
I know that if [imath]\sum_{n=0}^{\infty} c_n (x-a)^n[/imath] is a power series with radius of convergence [imath]R[/imath] and [imath]f\colon (a-R,a+R)\to\mathbb{R}[/imath] is the function such that [imath]f(x):=\sum_{n=0}^{\infty} c_n (x-a)^n[/imath] then [imath]f[/imath] is differentiable on [imath](a-R,a+R)[/imath] and for any [imath]0<r<R[/imath], the series [imath]\sum_{n=1}^{\infty} nc_n (x-a)^{n-1}[/imath] converges uniformly to [imath]f'[/imath] on the interval [imath][a-r,a+r][/imath]. Now, since [imath]\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}[/imath] and [imath]\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}[/imath] we have [imath]\cos'(x)=\sum_{n=1}^{\infty}(2n)\frac{(-1)^n}{(2n)!}x^{2n-1}=\sum_{n=0}^{\infty}(2(n+1))\frac{(-1)^{n+1}}{(2(n+1))!}x^{2(n+1)-1}=-\sum_{n=0}^{\infty} \frac{2n+2}{2n+2}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=-\sin(x)[/imath] and [imath]\sin'(x)=\sum_{n=1}^{\infty}(2n+1)\frac{(-1)^n}{(2n+1)!}x^{2n+1-1}=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}[/imath] and now I'm stuck: I don't know how to recover from here the series of [imath]\cos(x)[/imath]. So, where I'm going wrong? | 2521541 | Understanding ProofWiki's proof of differentiation of sine power series
At ProofWiki's page on the derivative of the sine power series I've read that being [imath]\sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}[/imath] by using the Power Series Differentiable on Interval of Convergence result we get [imath]\sin'(x)=\sum_{n=0}^{\infty}(2n+1) \frac{(-1)^n}{(2n+1)!}x^{2n}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=\cos(x)[/imath] but the page Power Series Differentiable on Interval of Convergence result says that if [imath]f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n[/imath] then inside the radius of convergence of the power series we have [imath]f'(x)=\sum_{n=1}^{\infty}nc_n(x-a)^{n-1}[/imath] so shouldn't it be [imath]\sin'(x)=\sum_{n=1}^{\infty}(2n+1)\frac{(-1)^n}{(2n+1)!}x^{2n+1-1}=\sum_{n=1}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}[/imath]? Now I understand that this last result has a missing term due to the index of the series being [imath]1[/imath] and not [imath]0[/imath] but being the [imath]1[/imath] mandatory because of the above theorem and since trying to shift index doesn't help (we get [imath]\sin'(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{(2n+2)!}x^{2n+2}[/imath]) how can the result that ProofWiki shows be rigorously justified? (I mean, I think we can't just say "oh, there's a missing term!" after having applied the theorem and magically shift the index of the series to obtain the desired result) |
2521606 | Compactness of three dimensional torus
The three dimensional torus (the periodic domain in Navier-Stokes equations [imath]\mathbb{T}^3=(\mathbb{R}/\mathbb{Z})^3[/imath]) is compact. Can anyone here give us a detailed proof. | 537209 | show that torus is compact
I am having difficuties in showing a torus is compact. Initially I wanted to use Heine-Borel theorem, but after that I realise we are not working in [imath]\mathbb{R}^n[/imath] space. So a simple way to show torus is compact is by definition. But after defining an open cover for torus, I don't know how to proceed. Can anyone guide me? |
1999762 | Proof that [imath] f: X \times \mathbb{R} \to X: (x,\lambda) \mapsto \lambda x[/imath] is continuous
Let [imath](X, \|\cdot\|)[/imath] be a normed space, and [imath]\|\cdot\|_M : X \times \mathbb{R} \to \mathbb{R}^+: (x,y) \mapsto \max(\|x\|,|y|)[/imath]. Proof that [imath] f: X \times \mathbb{R} \to X: (x,\lambda) \mapsto \lambda x[/imath] is continuous by using the [imath]\epsilon[/imath] - [imath]\delta[/imath] defintion of continuity. This is what I've got so far: Take [imath](a,b) \in X \times \mathbb{R}[/imath], [imath]\forall \epsilon >0[/imath], [imath]\exists \delta >0[/imath], [imath]\forall (x,y) \in X \times \mathbb{R}[/imath]: [imath]\|(x,y)-(a,b)\|_M=||(x-a,y-b)\|_M=\max(\|x-a\|,|y-b|)<\delta[/imath] [imath]\|f(x,y)-f(a,b)\|=\|yx-ab\|=\|y(x-a)+a(y-b)\| \le |y|\|x-a\| + \|a\||y-b|[/imath] I don't know what to do now, because I don't have any information about [imath]|y|[/imath] and [imath]\|a\|[/imath]... | 167890 | Proof that every normed vector space is a topological vector space
The topology induced by the norm of a normed vector space is such that the space is a topological vector space. Can you tell me if my proof is correct? Of course we have to show that addition and scalar multiplication are continuous with respect to the product topology (induced by the norm). (1) To show that [imath](x,y) \mapsto x + y[/imath] is continuous let [imath]\varepsilon > 0[/imath]. I can show that the norm [imath]\|\cdot\|_{V \times V}: V \times V \to \mathbb R[/imath] defined as [imath]\|(x,y) - (x_0, y_0) \| = \|x-x_0\| + \|y_0 - y\|[/imath] induces the same topology as the product topology on [imath]V \times V[/imath]. Hence we can choose [imath]\delta = \varepsilon[/imath] to get [imath] \| (x+y) - (x_0+y_0)\| \leq \|x-x_0\| + \|y-y_0\| < \delta = \varepsilon[/imath] (2) To show that [imath]V \times K \to V[/imath], [imath](v, \alpha) \mapsto \alpha v[/imath] is continuous at [imath](v,\alpha)[/imath], observe that [imath]\| \alpha v - \beta w\| = \| \alpha v - \beta w + \alpha w - \alpha w\| = \|\alpha(v-w) + (\alpha - \beta) w\| \leq |\alpha| \|v-w\| + |\alpha - \beta| \|w\|[/imath] Hence [imath]\| \alpha v - \beta w\| < \varepsilon[/imath] if [imath]\|v-w\| < \frac{\varepsilon}{2 |\alpha|}[/imath] and [imath]|\alpha - \beta| < \frac{\varepsilon}{2 \|w\|}[/imath]. Unfortunately, the second inequality depends on [imath]w[/imath]. How do I make it independent of [imath]w[/imath]? Thanks. |
2523326 | Holder's inequality and Minkowski's inequality
Can somebody explain me how to use Holder's inequality and Minkowski's inequality. Can these inequalities be used to solve the problem that if [imath]x, y, k[/imath] are non negative reals then prove that [imath]\left(1+\frac{x}{y}\right)^k +\left(1+ \frac{y}{x}\right)^k \ge 2^{k+1}[/imath] Edit 1- My main question is about what Holder's inequality and minkowski's inequality are and how do we use them | 753593 | Prove the inequality [imath]({1+\frac{a}b})^n[/imath] + [imath](1+\frac{b}a)^n[/imath] [imath]\geq[/imath] [imath]2^{n+1}[/imath]
Let [imath]a[/imath] and [imath]b[/imath] be positive real numbers and let [imath]n[/imath] be a natural number prove that [imath]\left({1+\frac ab}\right)^n+\left(1+\frac ba\right)^n\ge2^{n+1}.[/imath] |
2519156 | Partition Generating Functions
Prove the following identity by counting two sets of partitions in two different ways. \begin{equation}\prod_{i\geq 0}(1+x^{2i+1})=1 + \sum_{n\gt 0}x^{n^2}\prod_{j=1}^n\frac{1}{1-x^{2j}}.\end{equation} I know the partition on the left side, it's the number of partitions with [imath]i[/imath] distinct odd parts. I just need help doing the right side. The second part of the second term , the product is the number of partitions of [imath]n[/imath] and for each [imath]i[/imath], the parts of size [imath]i[/imath] occur an even number of times. Now it's being multiplied by the summations of [imath]x^{n^2}[/imath] times. How would that effect the partition? | 2502299 | Counting set for generating functions
Give a proof of the following identity by counting two sets of partitions in two different ways. \begin{equation*} \prod_{i\geq 0}(1+x^{2i+1})=1 + \sum_{n\geq 1}x^{n^2}\prod_{j=1}^n\frac{1}{1-x^{2j}} \end{equation*} To prove this inductively, i only successfully complete the base cases, however the counting set for the left side i concluded that would be the number of partitions of n with distinct odd parts and for RS the product part would be the number of partitions of n with even parts, how would the summation of [imath]x^{n^2}[/imath] come into play with counting set. How would i initiate my proof? |
2524890 | Equal traces and similar matrix
I know that if matrix [imath]a[/imath] is similar to matrix [imath]b[/imath] then [imath]\operatorname{trace} a=\operatorname{trace} b[/imath]. Does it go to the other side? Thanks. | 1114792 | If two matrices have the same characteristic polynomials, determinant and trace, are they similar?
If two [imath]n \times n[/imath] matrices have the same characteristic polynomials, determinant and trace, are they similar, EVEN if ([imath] \lnot \#Spec= 0[/imath])? |
2524200 | Fact about Integrally closed domains.
Let [imath]A[/imath] be an integral domain. Then [imath]\forall m[/imath] where [imath]m[/imath] is maximal ideal in [imath]A[/imath],[imath]\; A_{m}[/imath] is integrally closed[imath]\implies[/imath] [imath]A[/imath] is integrally closed. This fact is proved in Atiyah-MacDonald in a different way. I am trying to prove it in the following way and hoping to complete it: The field of fractions of [imath]A_{m}[/imath] is same as the field of fractions of [imath]A, K_{A}.\;[/imath]Then we want to prove that [imath]a/b \in K_{A}[/imath] is integral over [imath]A\implies a/b\in A.\;[/imath] Now, [imath]a/b[/imath] satisfies some monic polynomial over [imath]A[/imath] means it satisfies some monic polynomial over [imath]A_{m}, \forall m.\;[/imath] Hence, [imath]a/b[/imath] is integral over [imath]A_{m},\forall m.\;[/imath] Now I just have to prove that [imath]b\notin m,\forall m.[/imath] Suppose, [imath]b\in m[/imath] for some maximal ideal [imath]m[/imath]. Then [imath]a\in m.[/imath] After that I'm stuck. Thanks in advance! | 119566 | [imath]A[/imath] is integrally closed if and only if [imath]A_P[/imath] is integrally closed for all maximal ideals [imath]P[/imath] of [imath]A[/imath].
I've managed [with help from the wonderful people lurking on this site :)] to prove that for an integral domain [imath]A[/imath], if [imath]A[/imath] is integrally closed, then [imath]S^{-1}A[/imath] is integrally closed for all multiplicatively closed subsets [imath]S[/imath] of [imath]A[/imath]. The problem that I want to apply this to is: [imath]A[/imath] is integrally closed if and only if [imath]A_{P}[/imath] is integrally closed for all maximal ideals [imath]P[/imath] of [imath]A[/imath]. So far: [imath](\Rightarrow)[/imath] If [imath]P[/imath] is a maximal ideal of [imath]A[/imath], then [imath]A\setminus P[/imath] is multiplicatively closed, because maximal ideals are prime. So by the result mentioned above, [imath]A_{P} = (A \setminus P)^{-1}A[/imath] is integrally closed. Update: Attempting to prove the contrapositive. Assume there is some [imath]\frac{r}{s}[/imath] in [imath]F[/imath] (the field of fractions of [imath]A[/imath]) such that [imath]\frac{r}{s}[/imath] is integral over [imath]A[/imath] but not in [imath]A[/imath]. I feel like I haven't done hardly anything at all with your hint, but I suppose that I would want to somehow get a maximal ideal of [imath]A[/imath] from this [imath]\frac{r}{s}[/imath] such that [imath]A_{P}[/imath] is not maximal, thus proving the converse. |
2524607 | Exercise 4 from Chapter 2 in Real Analysis by Elias M. Stein
Suppose [imath]f[/imath] is integrable on [imath][0,b][/imath], and [imath]g(x) = \int_x^b \frac{f(t)}{t}dt[/imath] for [imath]0 \lt x \le b[/imath] prove that [imath]g[/imath] is integrable on [imath][0,b][/imath], and [imath]\int_0^b g(x)dx= \int_0^bf(t)dt.[/imath] I know that the key point is how to prove slow increase of [imath]\frac{f(t)}{t}[/imath]. I prepare to find a function sequence that has compact support and converge to [imath]g[/imath], and use the monotone convergence theorem. but I can't do that. Can anybody help me? | 532912 | A Problem on Improper Integrals
Let [imath]f(x)[/imath] be continuous except at [imath]x = 0[/imath] and let [imath]a > 0[/imath]. Assume that the improper integral [imath]\int_{0}^{a}f(x)\,dx = \lim_{\epsilon \to 0+}\int_{\epsilon}^{a}f(x)\,dx[/imath] exists and let [imath]g(x) = \int_{x}^{a}\frac{f(t)}{t}\,dt[/imath] Show that [imath]\int_{0}^{a}g(x)\,dx = \int_{0}^{a}f(x)\,dx[/imath] I tried integration by parts noting that [imath]g'(x) = -f(x)/x[/imath] and obtained for [imath]0 < \epsilon < a[/imath] the following [imath]\int_{\epsilon}^{a}g(x)\,dx = [xg(x)]_{x = \epsilon}^{x = a} - \int_{\epsilon}^{a}xg'(x)\,dx[/imath] or [imath]\int_{\epsilon}^{a}g(x)\,dx = -\epsilon g(\epsilon) + \int_{\epsilon}^{a}f(x)\,dx[/imath] The problem is solved if we can somehow show that [imath]\lim_{\epsilon \to 0+}\epsilon g(\epsilon) = 0[/imath]. Looking at the definition [imath]g(x)[/imath] we see that we have no information of the behavior of [imath]f(t)[/imath] at [imath]t = 0[/imath] and the [imath]t[/imath] in denominator complicates the analysis of [imath]g(\epsilon)[/imath]. Please suggest some hints which can lead to the solution. Note: This problem is taken from G. H. Hardy's "A Course of Pure Mathematics" 10th ed. Page 397. |
2518059 | Count triangularizations of convex polygons such that each triangle created by triangularizations have atleast one side as polygon side.
Let [imath]n[/imath] be an integer greater than four, and let [imath]P_1,P_2,\dots, P_n[/imath] be a convex [imath]n[/imath]-sided polygon. Zach wants to draw [imath]n-3[/imath] diagonals that partition the region enclosed by the polygon into [imath]n-2[/imath] triangular regions and that may intersect only at the vertices of the polygon. In addition, he wants each triangular region to have at least one side that is also a side of the polygon. In how many different ways can Zach do this? | 2017996 | Path to Combinatorics Problem [[imath]1.11[/imath]]
Problem: Let n be a positive integer greater than four and let [imath]P_1P_2P_3..P_n[/imath] be a convex n-sided polygon. Zachary wants to draw [imath]n-3[/imath] diagonals that partition the region enclosed by the polygon into [imath]n-2[/imath] triangular regions and that may intersect only at vertices of the polygon. In addition, he wants each triangular region to have at least one side that is a side of the polygon. In how many different ways can Zachary do this? My attempt: I began with the case [imath]n=5.[/imath] I observed that the number of ways in which [imath]3[/imath] triangular regions can be constructed is [imath]5[/imath]. I then went on to consider a hexagon and observed that the number of ways to partition was [imath](6-1)*5.[/imath] Furthermore, I deduced for [imath]n[/imath] sided polygons we must have [imath](n-1)(n-2)...5*5[/imath] ways to partition the polygon. I would like to know whether this answer is correct and if not, then a proof or a sketch would be much appreciated. |
2442897 | Epsilon-Delta Proof [imath] \lim\limits_{n\to\infty}\frac{n+\sin(n)}{n+1} = 1[/imath]
How To Construct a Epsilon-Delta proof [imath]\displaystyle \lim\limits_{n\to\infty}\frac{n+\sin(n)}{n+1} = 1[/imath] ? The beginning: Fix [imath]\epsilon[/imath] > 0. Is there an [imath]N\in \mathbb{N},[/imath] such that [imath]n\ge N \implies\left|\frac{n+\sin(n)}{n+1}-1 \right|<\epsilon?[/imath] [imath]\left|\frac{n+\sin(n)}{n+1}-1 \right|<\epsilon \iff[/imath] [imath]\left|\frac{\sin(n)-1}{n+1}\right|<\epsilon \iff[/imath] [imath]\ldots[/imath] I got that [imath]n > \frac{2}{\epsilon}-1[/imath], is that right? N=(max or min ?) {0, [imath]\lfloor\frac{2}{\epsilon}-1\rfloor[/imath]} I hope that the task is understandable. Thank you for answering. | 2442833 | How can I prove that [imath]\lim\limits_{n\to\infty}\frac{n+\sin n}{n+1} = 1[/imath]
How to prove [imath]\displaystyle \lim\limits_{n\to\infty}\frac{n+\sin(n)}{n+1} = 1[/imath] ? The beginning: Fix [imath]\epsilon[/imath] > 0. Is there an [imath]N\in \mathbb{N},[/imath] such that [imath]n\ge N \implies\left|\frac{n+\sin(n)}{n+1}-1 \right|<\epsilon?[/imath] [imath]\left|\frac{n+\sin(n)}{n+1}-1 \right|<\epsilon \iff[/imath] [imath]\left|\frac{\sin(n)-1}{n+1}\right|<\epsilon \iff[/imath] [imath]\ldots[/imath] I got that [imath]n > \frac{2}{\epsilon}-1[/imath], is that right? N=(max or min ?) {0, [imath]\lfloor\frac{2}{\epsilon}-1\rfloor[/imath]} I hope that the task is understandable. Thank you for answering. |
2526124 | Find all positive integers n such that [imath]n + 3[/imath] divides [imath]n^2 + 27[/imath]
Find all positive integers n such that [imath]n + 3[/imath] divides [imath]n^2 + 27[/imath]. I am quite stuck on how to conclusively show all of the solutions. I tried look at the intersections of [imath]y = n^2 + 27[/imath] and [imath]f = x(n + 3)[/imath] where is [imath]x[/imath] is some number. But it still involves setting [imath]x[/imath] values. Any help would be much appreciated. | 2509155 | Division in [imath]1[/imath] variable
Let the sum of all integers [imath]n[/imath] such that [imath](2n^2+9/n+3)[/imath] is an integer be [imath]A[/imath], what is [imath]|A|[/imath]? I tried expressing it to: [imath]k[/imath] is an integer, [imath](n+3)k + 0 = 2n^2 + 9[/imath], [imath]n(2n-k) + 3(3-k) = 0[/imath], from which I couldn't move on, I also tried [imath](n+3)k + 0 = 2n^2 + 9[/imath], [imath]k = 2n[/imath], [imath]2n^2 - 2n^2 + 9 - 6n[/imath]. [imath]9-6n/n+3[/imath] is an integer, [imath]9 - 6n = 3(3-n)[/imath] though now I also do not know how to move on. |
2526275 | Show that [imath]\int_0^\pi f(\sin\theta) \, d\theta = 2\int_0^{\pi/2}f(\sin\theta) \, d\theta[/imath]
The full question states: Show that [imath]\int_0^\pi f(\sin\theta)\,d\theta = 2\int_0^{\pi/2}f(\sin\theta)\,d\theta[/imath] for an arbitrary function [imath]f[/imath] defined in the interval [imath][-1, 1].[/imath] Having tried out different possible functions for [imath]f[/imath] (rational, radical, exponential, etc.) I know that the equation indeed holds true, but I have no clue as to how to go about proving it. It has been suggested to me that the proof might involve the use of multivalued functions since a particular interval was defined but I don't see how this could be the case since neither the sine nor complex sine functions are multivalued as far as I know. Any help would be greatly appreciated. | 2523420 | Show that [imath]\int_0^\pi f(\sin x)\,\mathrm{d}x = 2\int_0^{\pi/2}f(\sin x) \, \mathrm{d}x[/imath]
Any tips on how to show that [imath]\int_0^\pi f(\sin x) \, \mathrm{d}x = 2\int_0^{\pi/2} f(\sin x) \, \mathrm{d}x \text{ ?}[/imath] |
2525439 | Proof that [imath]d^2 \phi = 2\pi \delta^{(2)}(r)dx\wedge dy[/imath] in polar coordinates around a puncture?
In the literature on point particles in 2+1 gravity, I often see the identity [imath]d^2 \phi = 2\pi \delta^{(2)}(r)dx\wedge dy[/imath] being used. Here, the particle is a puncture located at the origin and [imath](r,\phi)[/imath] are polar coordinates around this puncture. One example is in this paper (unfortunately it's behind a paywall), where in equation 3.4 the author uses the above identity to show that the torsion and curvature are both described by delta distributions. For example, for the curvature, since the connection is given in equation 3.3b by [imath]\omega=M d\phi[/imath], the curvature is [imath]F=d\omega=Md^2\phi=2\pi M \delta^{(2)}(r)dx\wedge dy.[/imath] Other papers seem to be using this result as well, and cite the above paper. One example is this paper, equations 4.6-4.7. I know that [imath]d^2[/imath] is supposed to be always zero because the partial derivatives commute, so this has to fail somehow for the identity to be satisfied, but I don't know how that would happen in this case. | 2473275 | How to use Stokes theorem in the presence of holes?
So I know Stokes theorem for integrating exact differential forms over a chain: [imath]\int_cd\omega=\int_{\partial c}\omega.[/imath] But what if there are holes? More precisely, consider the following famous example: Consider the punctured plane [imath]U=\Bbb R^2\setminus\{(0,0)\}[/imath] and the differential form [imath]\omega=\frac{1}{x^2+y^2}(-ydx+xdy)[/imath] on [imath]U[/imath]. It is known that line integral of [imath]\omega[/imath] along a circle [imath]c:[0,1]\to U, c(t)=(\cos 2\pi t,\sin 2\pi t)[/imath], is [imath]2\pi[/imath]. Stokes theorem does not really apply here because neither is [imath]\omega[/imath] exact nor is [imath]c[/imath] the boundary of another chain. But it is possible to "define" the following (I saw similar things from physics textbook and notes):[imath]d\omega=2\pi\delta(x,y)dx\wedge dy[/imath]where [imath]\delta[/imath] is the Dirac delta function, giving infinity when [imath]x=y=0[/imath], and [imath]0[/imath] otherwise, such that the integral of [imath]\delta(x,y)dx\wedge dy[/imath] over a subset [imath]S[/imath] of [imath]\Bbb R^2[/imath] is [imath]1[/imath] if the origin is an interior point of [imath]S[/imath]. Regard [imath]c[/imath] as the boundary of the unit disk [imath]D[/imath] at the origin and Stokes theorem can be "applied" just fine, by[imath]\int_Dd\omega=\int_{c}\omega.[/imath] Is there a way to formalise these ideas? If possible, I would like to see some reference to formal definitions of such generalised exterior derivative and proofs of such generalised Stokes theorem. I know some complex analysis (Cauchy's integral formula), the definition of a distribution, and the definition of de Rham cohomology if it will help. |
2525137 | How to prove that the closed unit ball in [imath]\mathbb{R}^n[/imath], with the subspace topology, is not a manifold?
Our definition of a manifold [imath]M[/imath] is a Hausdorff topological space such that for every [imath]x \in M[/imath], there exists a neighborhood [imath]U_x[/imath] that is homemorphic to [imath]\mathbb{R}^m[/imath] for some [imath]m[/imath]. We define the closed unit ball in [imath]\mathbb{R}^n[/imath] to be the set [imath]\{x \in \mathbb{R}^n \colon \|x\| \leq 1\}[/imath]. The claim is that the closed unit ball is not a manifold. The open unit ball is clearly a manifold, so I assume that for every point [imath]x[/imath] on the boundary of the closed unit ball, none of the neighborhoods of [imath]x[/imath] are homemorphic to [imath]\mathbb{R}[/imath]. However, I am having trouble doing so. I have tried proving this by contradiction by supposing such a homemorphism exists and showing there exists a topological property of [imath]\mathbb{R}^n[/imath] that the closed unit ball doesn't have. | 339767 | Closed ball not a manifold.
My book on differential geometry claims that a closed ball in [imath]\Bbb R^n[/imath] can never be a differentiable manifold because of the boundary points. The book doesn't really give an explanation for why this is true. Could someone provide more details please? thanks |
2526583 | find the limits : [imath]\lim_{n}( \frac{1\cdot3\cdot5\cdot \cdot \cdot(2n-1)}{2\cdot4\cdot6 \cdot \cdot \cdot (2n)} ) =?[/imath]
Find the limits (without Stirling's approximation ) : [imath]\lim_{n}( \frac{1\cdot3\cdot5\cdot \cdot \cdot(2n-1)}{2\cdot4\cdot6 \cdot \cdot \cdot (2n)} ) =?[/imath] My Try : [imath]\frac{1\times 3\times 5\cdots \times(2n-1)}{2\times 4\times 6\cdots \times(2n)}=\frac{(2n)!}{4^n(n!)^2}[/imath] now what ? | 248881 | limit of [imath]\frac{(2n)!}{4^n(n!)^2}[/imath]
I'd love to understand the behaviour of the sequence [imath] \frac{(2n)!}{4^n(n!)^2} \text{as } n \to \infty [/imath] the first step would be to simplify this to [imath] \frac{(2n)(2n-1)(2n-2)\cdots(n+1)}{4^n \cdot n(n-1)(n-2)\cdots 2} [/imath] and then factor out [imath]2[/imath] to get [imath] \frac{1}{2^n}\cdot\frac{(n)(n-1/2)(n-1)\cdots(n - (n-1)/2)}{n(n-1)(n-2)\cdots 2} [/imath] if I can now get the second term to be strictly larger than [imath]2^n[/imath] then I would be done - but how can I do this ? thanks so much for help!! P.S. this not a HW question - though it grew out of one where I had to find the radius of convergence for a power series - this is the series evaluated at the end points. If I can show that the above sequence does not converge to [imath]0[/imath] then I know that the power series diverges at the endpoints, this is what I'd love to find out! |
2526632 | Property of convergent series.
If [imath]\sum a(n)[/imath] and [imath]\sum b(n)[/imath] are convergent series with positive terms,can I say that [imath]\sum a(n)b(n)[/imath] is also convergent? | 2075511 | Convergence of product of convergent series
Given [imath]a_n, b_n[/imath] such that [imath]\exists N \forall n>N a_n, b_n \geq 0[/imath] I want to show that if [imath]\sum_{n=0}^{\infty}a_n[/imath] and [imath]\sum_{n=0}^{\infty}b_n[/imath] converge then [imath]\sum_{n=0}^{\infty}a_n b_n[/imath] also converges. I tried to show it in the following way, if [imath]\sum_{n=0}^{\infty}b_n[/imath] converges, then [imath]\lim_{n\rightarrow\infty}b_n=0[/imath], so there exists [imath]n_0[/imath] such that [imath]b_{n_0}\geq b_n[/imath] for all [imath]n\gt n_0[/imath], therefore [imath]a_n b_{n_0}\geq a_n b_n[/imath], so that [imath]\sum_{n=0}^{\infty}a_n b_n\leq \sum_{n=0}^{n_0}a_n b_n+b_{n_0}\sum_{n=n_0+1}^\infty a_n[/imath], and as my series is bounded by sum of finitely many terms and tail of convergent series multiplied by a constant it's also convergent. Is my proof correct, or are there any flaws in my reasoning? |
2526683 | Find the necessity and sufficiency condition for [imath]Z_m[/imath] to be a Field
If we have a group like [imath]Z_m = \{0,1,2,...,m-1 \}[/imath] and define an addition and multiplication on this group as the remainder of ordinary addition and multiplication division by [imath]m[/imath] (modulo m), prove that the necessity and sufficiency condition for [imath]Z_m[/imath] to be a Field is that the [imath]m[/imath] be a prime number! e.g.: [imath]m=5[/imath] [imath]2\oplus4=1[/imath] [imath]2\otimes4=3[/imath] for [imath]m=2[/imath] I guess it's like Boolean algebra! | 610383 | Alternative methods to prove that [imath]Z_m[/imath] is a field under addition and multiplication [imath]\bmod\ m[/imath] iff [imath]m[/imath] is a prime
I am looking for ways to prove that [imath]\mathbb{Z}_m=\{0,1,2,\dots,m-1\}[/imath] is a field under addition and multiplication [imath]\bmod\ m[/imath] iff [imath]m[/imath] is a prime. I tried it this way: If [imath]m[/imath] is a prime [imath]p[/imath],it is obvious that most importantly, we need to prove that every non-zero element in [imath]\mathbb{Z_p}[/imath] has an inverse in [imath]\mathbb{Z_p}[/imath].We note that if [imath]a,b\in \mathbb{Z_p}[/imath], then [imath]a\cdot b\in \mathbb{Z_p} \dots \boxed{*}[/imath]. Next, [imath]a^p\equiv a \bmod p[/imath] for prime [imath]p[/imath](the proof follows by induction on [imath]a[/imath]).As [imath]\displaystyle 0\leq a\leq p-1[/imath],[imath]a\not =p[/imath]. Thus [imath]a^{p-1}\equiv 1 \bmod p[/imath] for prime [imath]p[/imath] which,together with [imath]\boxed{*}[/imath] guarantees the existence of an inverse in [imath]\mathbb{Z_p}[/imath].Proving the other properties is simple. Conversely,if [imath]\mathbb{Z_m}[/imath] is a field,then every non-zero element of [imath]\mathbb{Z_m}[/imath] has an inverse.Note that inverse of an element is unique.For if [imath]a\cdot b\equiv a\cdot c\equiv 1 \bmod m[/imath],then [imath](a\cdot b)\cdot c\equiv c\equiv (a\cdot c)\cdot b\equiv b \bmod m[/imath] i.e [imath]m|(b-c)\implies b-c=0(0\leq a\leq m-1,a\in \mathbb{Z_m})[/imath].So, inverses must be unique and every element is either an inverse of itself or has in inverse in the field.That allows us to conclude that [imath][(m-1)!]^2\equiv 1 \bmod m[/imath] which can only mean that either [imath](m-1)!\equiv 1 \bmod m[/imath] or [imath](m-1)!\equiv -1 \bmod m[/imath]. If [imath]m[/imath] is not a prime, it has a prime factor, say [imath]p[/imath]. [imath] 0\leq p\leq m-1\implies p|(m-1)!\implies p|1[/imath]or [imath]-1[/imath] which is absurd, a contradiction.Thus [imath]m[/imath] is prime. I feel that this method is too involved and complicated. Can anyone please suggest some other method?(Some context: This is a problem from Finite Dimensional Vector Spaces by Paul Halmos. I am learning linear algebra for the first time.It seems some of the comments have been posted on the assumption that I know rings and fields; I don't. ) |
1567525 | The set [imath]S[/imath] contains......
Let [imath]S[/imath] be the set of 3[imath]\times[/imath]3 real matrix [imath]A[/imath] with [imath]A^TA= \left( \begin{array}{} 1 & 0 & 0\\ 0 & 0 & 0\\ 0& 0 & 0 \end{array}\right).[/imath] Then the set [imath]S[/imath] contains a nilpotent matrix a matrix of rank one a matrix of rank two a non-zero skew-symmetric matrix Here [imath]S[/imath] is contains a matrix of rank 1, Take [imath]\left( \begin{array}{} 1 & 0 & 0\\ 0 & 0 & 0\\ 0& 0 & 0 \end{array}\right)[/imath] itself. I am confusing with other options, help me. | 1423224 | [imath]S[/imath] is a [imath]3\times 3[/imath] real matrices. Then [imath]S[/imath] contains which matrices
Let , [imath]S[/imath] be the set of [imath]3\times 3[/imath] real matrices [imath]A[/imath] with [imath]AA^T=\left(\begin{matrix}1&0&0\\0&0&0\\0&0&0\end{matrix}\right).[/imath]Then the set [imath]S[/imath] contains : (A) a nilpotent matrix. (B) a matrix of rank [imath]1[/imath]. (C) a matrix of rank [imath]2[/imath]. (D) a non-zero skew-symmetric matrix. Attempt : Clearly , [imath]\left(\begin{matrix}1&0&0\\0&0&0\\0&0&0\end{matrix}\right)\in S[/imath]. So, (B) is TRUE. Again [imath]rank(AA^T)=rank(A)[/imath]. As , [imath]rank(AA^T)=1[/imath] , so [imath]S[/imath] does not contain a matrix of rank [imath]2[/imath]. So (C) is FALSE. Again we know the rank of a non-zero skew-symmetric matrix can never be [imath]1[/imath]. So , [imath]S[/imath] does not contain a non-zero skew-symmetric matrix. So (D) is FALSE. But I am unable to understand the option (A) whether it is correct or NOT ? Any hint ? |
2525013 | How many ways to form 3 unordered partitions from n-element set?
Let's say a set [imath]S[/imath] has [imath]n[/imath] elements, and it needs to be partitioned into [imath]3[/imath] different, unordered partitions. How do I obtain a general formula for this? I think I can calculate it if I know the value of [imath]n[/imath]. For example, if [imath]n=3[/imath] then it's partitioned as [imath]1,1,1[/imath] so there's only [imath]1[/imath] way. If [imath]n=4[/imath] then it's [imath]2,1,1[/imath] so there's [imath]{4\choose2}=6 [/imath] ways. If [imath]n=5[/imath] then it's [imath]2,2,1[/imath] or [imath]3,1,1[/imath] and there's [imath]15+10=25[/imath] ways. I can't figure out [imath]n[/imath] in general though. My latest attempt was summation: [imath]\sum_{a=1}^{n-2} \quad \sum_{b=1}^{n-2-a} \quad {n\choose a}{{n-a}\choose b} [/imath] but the partitions won't be unordered (and the summation is probably wrong too). I have a hint to regard the first [imath]x[/imath] elements in the first partition, and the [imath]x+1[/imath]-th element in the second, but I'm not sure where to proceed from this. | 650791 | Number of partitions of an [imath]n[/imath]-element set into [imath]k[/imath] classes
A partition of a set [imath]S[/imath] is formed by disjoint, nonempty subsets of [imath]S[/imath] whose union is [imath]S[/imath]. For example, [imath]\{\{1,3,5\},\{2\},\{4,6\}\}[/imath] is a partition of the set [imath]T=\{1,2,3,4,5,6\}[/imath] consisting of subsets [imath]\{1,3,5\},\{2\}[/imath] and [imath]\{4,6\}[/imath]. However, [imath]\{\{1,2,3,5\},\{3,4,6\}\}[/imath] is not partition of [imath]T[/imath]. If there are [imath]k[/imath] nonempty subsets in a partition, then it is called a partition into [imath]k[/imath] classes. Let [imath]S_k^n[/imath] stand for the number of different partitions of a set with [imath]n[/imath] elements into [imath]k[/imath] classes. Find [imath]S_2^n[/imath]. Show that [imath]S_k^{n+1}=S_{k-1}^n+kS_k^n[/imath]. -- My work: From the definition of [imath]S[/imath], [imath]S_2^n=2^n[/imath]. I think I am wrong somewhere, because when I put this formula into the second part to prove, I get, [imath]k^{n+1}=(k-1)^n+k \cdot k^n.[/imath] Please tell me where I am wrong. I think this problem cannot be solved by star-and-bar method as that method finds value for [imath]k[/imath] but does not prove it. Please help! |
2526913 | How to calculate [imath]276^{246} \mod 323[/imath] without a calculator?
How do I simplify this expression? [imath]276^{246} \mod 323[/imath] Thank you so much. | 2531407 | Figuring out a high power modulo
Could you walk me through the most efficient way to solve this modulo? [imath]276^{247} \mod 323[/imath] |
2527133 | Show that [imath]\text{sinc}(x)\not\in L^1(\mathbb{R}^+)[/imath].
I have to show that [imath]\int_{0}^{\infty} |\frac{\sin x}{x}|dx[/imath] diverges. A previous question proves that [imath]\int_{0}^{\infty} \frac{\sin x}{x}dx[/imath] converges. Maybe something can be done about the series [imath]|\frac{\sin(n)}{n}|[/imath] though I'm not sure what. | 2148525 | Show that [imath]\int_{0}^{\infty}|\frac{\sin x}{x}|\,dx=\infty [/imath]
Problem: Show that [imath]\int_{0}^{\infty}|\frac{ \sin x}{x}|=\infty [/imath] . Here is what i have tried : I found out that for all [imath]x>0[/imath] , [imath]x-\frac{x^3}{6} < \sin x <x [/imath] . So [imath]1-\frac{x^2}{6}<\frac{ \sin x}{x}<1[/imath] . So applying integration to both side from [imath]0[/imath] to [imath]M[/imath] we have [imath]M-\frac{M^3}{18}<\int_{0}^{M}|\frac{ \sin x}{x}|< M[/imath] . Now we send [imath]M \rightarrow \infty[/imath] and observe that left-hand side goes to infinity . Hence proved . Please point out if there is anything wrong with my approach . I will also be delighted if could provide alternative solutions . Thank you . |
2526572 | Proving a limit using the formal definition of a limit
Prove by using the formal definition of limit that: [imath]\displaystyle\lim_{x\to \infty }\dfrac{1}{x^4 + x^2 + 5} = 0[/imath] Can anyone please lend a hand here? | 1690758 | Prove using the formal definition of a limit that
How would I go about proving this limit? [imath]\lim_{x\to\infty}\frac{1}{x^4+x^2+5}=0[/imath] so far I have: [imath]|f(x) -L|< ϵ[/imath] wherever [imath]x > N[/imath] [imath]|\frac{1}{x^4+x^2+5} - 0| < ϵ [/imath] wherever [imath]x > N[/imath] [imath]|\frac{1}{x^4+x^2+5}| < ϵ ,\; x > ∞[/imath], assuming, [imath]x > 0[/imath] taking the absolute value [imath]\frac{1}{x^4+x^2+5} < ϵ[/imath] [imath]{x^4+x^2+5}[/imath] > [imath]\frac{1}{ϵ}[/imath] i am not sure whether this is correct, but what would be the next step? |
2527435 | there is no entire function f(z) which verifies that |f(z)| > |z| for every z ∈ C.
Hi I came across this question, prove there is no entire function f(z) which verifies that |f(z)| > |z| for every z ∈ C. I was gonna try use the fact that [imath]g∈C[/imath] such that [imath]f(g)=0[/imath] and maybe that function [imath]1/|f(z)|[/imath] is bounded. Thank you | 2526781 | Function [imath]f(z)[/imath] such that [imath]|f(z)| > |z|[/imath] for every [imath]z\in\mathbb{C}, 1/f(z)[/imath]
[imath]f(z)[/imath] is a function such that [imath]|f(z)| > |z|[/imath] for all [imath]z\in\mathbb{C}[/imath] How would you prove the function [imath]\frac{1}{f(z)}[/imath] is bounded. I think you have to use the [imath]|f(z)|\leq M[/imath] but I'm not sure. |
2527053 | Rudin's proof that rationals have no GLB (or LUB)
In Rudin's 3e of Principles of Mathematical analysis, he associates [imath]q = (2p+1)/(p+1)[/imath]. How does he create this number? What is the method for discovering this answer? | 1932371 | How to prove that the set [imath]A = \left\{ {p:{p^2} < 2,p \in {\Bbb Q^+}} \right\}[/imath] has no greatest element?
More specifically, I have to find a [imath]q[/imath] in [imath]A[/imath] such that [imath]q[/imath] is larger than any [imath]p[/imath] in [imath]A[/imath]. The only thing I can think of is using ([imath]{2 - {p^2}}[/imath]) somehow, or [imath]{(p+x)^2}<2[/imath], but other than that I don't know how to proceed. This exercise is taken from the very first chapter of Walter Rudin's Principles of Mathematical Analysis. Quoting the exercise: To do this, we associate with each rational [imath]p > 0[/imath] the number: [imath]q = p - \frac{{{p^2} - 2}}{{p + 2}} = \frac{{2p + 2}}{{p + 2}}[/imath] But he didn't explain how he came up with that number. The only thing I've thought of is the following procedure: [imath]\eqalign{ & {p^2} + (2 - {p^2}) = 2 \Rightarrow {p^2} + (\frac{{2 - {p^2}}}{2}) < 2,{\text{ and since }}2 - {p^2}{\text{ is always positive,}} \cr & {p^2} < \left[ {{p^2} + (\frac{{2 - {p^2}}}{2})} \right] < 2 \cr & {\text{Therefore, let }}q = \sqrt {{p^2} + (\frac{{2 - {p^2}}}{2})} ,{\text{ then:}} \cr & p < q < \sqrt 2 \cr} [/imath] However, my result doesn't necessarily belong to [imath]{\Bbb Q}[/imath]. I wonder how did he obtain the value of [imath]q[/imath] in the book. EDIT: Feel free to point out if there are more appropriate tags for this question. |
2526446 | Find all functions [imath]f:\Bbb N\rightarrow \Bbb R [/imath] such that: [imath]f(m+k)=f(mk-n)[/imath]
Find all functions [imath]f:\Bbb N\rightarrow \Bbb R [/imath] such that for a given value [imath]n\in \Bbb N[/imath] , the following identity holds: [imath]f(m+k)=f(mk-n) ,m,k \in \Bbb N , mk>n[/imath] This problem has already been asked here . I tried to get more deep solutions using my reputation but failed,so I repeat it here. | 2385733 | find functions that [imath]f(m+k)=f(mk-n)[/imath]
Find all functions from naturals excluding [imath]0[/imath] to reals that for a particular amount of [imath]n[/imath] we have: [imath]f(m+k)=f(mk-n):(mk>n , m,k,n \in \Bbb Z^+)[/imath] by putting [imath]k=1[/imath] and [imath]m>n[/imath] we can see that the function is periodic with cycle of [imath]n+1[/imath].But I can't continue. |
2526926 | How do I prove that [imath]r(3,t+1) ≥ 3t[/imath]
For each [imath]t ≥ 2[/imath], show that a [imath]t[/imath]-regular Ramsey graph has [imath]r(3,t+1) ≥ 3t[/imath] I'm stuck on where to go with this question. I've tried using what few theorems I know, but I've gotten caught up with inequalities that are not helping with anything, such as [imath]r(3,t+1)≤r(3,t)+r(2,t+1)=r(3,t)+t+1[/imath]. | 1596100 | Lower bound for Ramsey numbers: [imath]R( n + 2 , 3 )>3n[/imath]
I need to prove the following inequality: [imath]R( n + 2 , 3 )>3n[/imath] where [imath]n>1[/imath] and [imath]R(s,t)[/imath] is a Ramsey number. The most general way to prove such inequalities is to paint a graph with [imath]3n[/imath] nodes then show it could not have a sub graph with [imath]n+2[/imath] in red or [imath]3[/imath] in blue, I've also used induction but no answers yet. |
2528580 | Proving existence of points in a twice differentiable function
Let [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] be a twice differentiable function. Prove that if [imath]x_1<x_2<x_3[/imath] and [imath]f(x_1)=f(x_2)=f(x_3)=0,[/imath] then there exists a point [imath]q\in(x_1,x_3)[/imath] such that [imath]f''(q)=0[/imath]. I think it's based on Rolle's theorem, but I can't prove it rigorously. Would anybody have a rigorous way to prove this? | 2525548 | Existence of points in a differentiable function
Let [imath]g:\mathbf{R\rightarrow R}[/imath] be a twice differentiable function. a) Prove that if [imath]a_1 < a_2[/imath] and [imath]g(a_1)=g(a_2)=0[/imath], then there exists a point [imath]m \in (a_1,a_2)[/imath] such that [imath]g'(m)=0[/imath]. b) Prove that if [imath]a_1 < a_2 < a_3[/imath] and [imath]g(a_1)=g(a_2)=g(a_3)=0[/imath], then there exists a point [imath]n \in (a_1,a_3)[/imath] such that [imath]g''(m)=0[/imath]. I am just curious about how to prove this very rigorously. My intuition tells me that both parts are true, but I have a hard time putting it down in writing. Any general tips about how I can learn to do this better are also much appreciated! |
2520903 | Probably that an [imath]80\%[/imath]-truthful person actually rolled a [imath]6[/imath]
A person, [imath]A[/imath], speaks the truth [imath]4[/imath] out of [imath]5[/imath] times. The person throws a die and reports that he obtained a [imath]6[/imath]. What is the probability that he actually rolled a [imath]6[/imath]? I know there is a similar question like this but my doubts are different from it and also I want to identify and solve total probability theorem questions so I posted a side doubt also. In my attempt, I defined the events \begin{align*} E_1&: \text{The person tells the truth.} \\ E_2&: \text{The person lies.} \\ E_3&: \text{The person reports that the die landed on a 6.} \end{align*} I noted that [imath]P(E_1)=\frac{4}{5}[/imath], [imath]P(E_2)=\frac{1}{5}[/imath], [imath]P(E_3|E_1)=6^{-1}[/imath] and [imath]P(E_3|E_2)=0[/imath] and obtained \begin{align*} P(E_3) = \frac{4}{5} \cdot \frac{1}{6} + \frac{1}{5} \cdot 0 = \frac{2}{15}. \end{align*} However, the correct answer is, [imath]\frac{4}{9}[/imath]. What did I do wrong? Side doubt: Even though the first experiment (truth and lying) is different from the second experiment, can we still apply total probability theorem? In my book the dependent experiment lies inside the sample space associated with the mutually and exhaustive events. | 1084785 | A man who lies a fourth of the time throws a die and says it is a six. What is the probability it is actually a six?
First, I apologise for the vague title. I couldn't think of a short way to represent the problem. Also, I am aware that a similar question exists, but I have a little bit more insight. The problem is: A man is known to speak the truth three out of four times. He throws a die and reports that it is a six. Find the probability that it actually is a six. The answer in the book is [imath] \frac{3}{8} [/imath], which sounds completely off to me, given the numbers, on first glance. I decided to try it on my own. After some contemplation, it indeed is true that [imath] \frac{3}{8} [/imath] is correct. But, the question's vagueness makes it interesting. It's sort of a naive answer! First, here's the book's solution: Let [imath] E [/imath] be the event that the man reports that a six occurs, [imath] S_1 [/imath] be the event that six occurs and [imath] S_2 [/imath] be the event that six does not occur. [imath] P(S_1) [/imath] = Probability that six occurs = [imath] \frac{1}{6} [/imath] [imath] P(S_2) [/imath] = Probability that six does not occur = [imath] \frac{5}{6} [/imath] [imath] P(E~|S_1) [/imath] = Probability that the man reports that a six occurred when it has actually occurred (truth) = [imath] \frac{3}{4} [/imath] [imath] P(E~|S_2) [/imath] = Probability that the man reports that a six occurred when it hasn't actually occurred = [imath] \frac{1}{4} [/imath] [imath] P(S_1|E) [/imath] = Probability that a six has actually occurred when the man claims so = [imath] \frac{P(S_1)~P(E~|S1)}{P(S_1)~P(E~|S_1)+P(S_2)~P(E~|S_2)} = \frac{3}{8} [/imath] Link to original: Page 25 of this (link updated 22/11/2015; tends to break often). Here's the first of my two solutions: The question is a little vague and it doesn't tell us what the man says when he doesn't get a six. Assuming he lies every time he doesn't get a six (which is not quite what the question says), would bring up [imath] P(E~|S_2) [/imath] to [imath] 1 [/imath], which skyrockets [imath] P(lie) [/imath] to [imath] \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6}.\frac{1}{4} = \frac{7}{8} [/imath]. Also, [imath] P(E~|S_1) = \frac{3}{4} [/imath] and [imath] P(E) = \frac{5}{6} + \frac{1}{6}.\frac{3}{4} = \frac{23}{24} [/imath]. This brings us to, [imath] P(S_1|E) = \frac{P(S_1)P(E~|S_1)}{P(E)} = \frac{3}{23} = 13.04\% [/imath] Please note that I don't care about the sixes rolled when he lies about it being a six, because that condition is not applicable to the question. I'm very skeptical about my calculations! I decided to whip up a simple program and see what the probability distribution is for [imath] \frac{Number~of~sixes~rolled~while ~saying~it's~a~six}{Number~of~dice~rolls} [/imath] Here's the result: Considering the book says [imath] P(E~|S_2) = \frac{1}{4} [/imath], which is in essence is contradictory to the question, let's also consider the other possibility: It's even more vague in this scenario! If the man lies after any given roll, there is a [imath] \frac{1}{5} [/imath] probability that he will lie about it being any other number. This creates a new branch at each lie, which splits into five. In this case, [imath] P(E) = \frac{1}{6}.\frac{1}{4}.\frac{1}{5}.\frac{5}{1}+\frac{1}{6}.\frac{3}{4} = \frac{1}{6} [/imath] [imath] P(E~|S_1) [/imath] still remains [imath] \frac{3}{4} [/imath], but [imath] P(E~|S_2) = \frac{1}{4}.\frac{1}{5} = \frac{1}{20} [/imath]. [imath] P(S_1|E) = \frac{P(S_1)P(E~|S_1)}{P(E)} = \frac{3}{4} = 75\% [/imath] This seems like the more intuitive answer, and it is why I doubted the original answer in the first place. Given a truth probability of [imath] \frac{3}{4} [/imath], there was no way that the probability of a six was [imath] 37.5\% [/imath]! The ambiguity of the question is what makes [imath] \frac{3}{8} [/imath] wrong to me. Entirely out of curiosity, which answer fits best with the question and are these calculations correct or baloney? |
2528770 | Direct Product of Group: Isomorphic Subgroups
Show that a Direct product of groups, [imath]G \times H[/imath], contains subgroups isomorphic to both [imath]G[/imath] and [imath]H[/imath]. This seems like a simple questions but I can't wrap my head around it. I've already proved that [imath]G \times H[/imath] is a group | 1861875 | isomorphic groups related in direct product.
I want to prove that the direct product [imath]G\times H[/imath] of two groups has a subgroup isomorphic to [imath]G[/imath] and a subgroup isomorphic to [imath]H[/imath]. How I thought to prove is that taking a pair [imath](g,h)[/imath] from [imath]G\times H[/imath] and then to show that for some [imath]g[/imath] in [imath]G[/imath] there is an isomorphism with the pair [imath](g,h)[/imath] of [imath]G\times H[/imath]. and the same for [imath]h[/imath] in [imath]H[/imath] also. Is this method correct? Or should I try to show that there is a bijection between the groups and there exists an equivalence relation between the groups? Please explain how should I proceed. |
2526255 | [imath]H\otimes l^{2}(\Gamma)\cong\bigoplus\limits_{g\in \Gamma}H[/imath]
I have no idea of proving [imath]H\otimes l^{2}(\Gamma)\cong\bigoplus\limits_{g\in \Gamma}H[/imath],where H is a hilbert space,[imath]\Gamma[/imath] is a discrete group, [imath]l^{2}(\Gamma)[/imath] is the set of all complex valued square summable fuctions on [imath]\Gamma[/imath]. | 791150 | How to verify [imath]H\otimes K \cong \bigoplus\limits_{i\in I}H[/imath]
Let [imath]H,~K[/imath] be the Hilbert space. if [imath]\{v_{j}\}_{j\in J}\subset H[/imath] and [imath]\{w_{i}\}_{i\in I}\subset K[/imath] are the orthonormal bases, then how to construct the isomorphic mapping: [imath]H\otimes K \rightarrow \bigoplus\limits_{i\in I}H[/imath]. Here, the [imath]H \otimes K[/imath] is the tensor product of two Hilbert space. |
2528602 | Factor ring of the Gaussian integers
Let [imath]R=\mathbb{Z}[i]/\langle2-2i\rangle[/imath]. I want to know the number of elements in [imath]R[/imath]. The function [imath]N[/imath] given by [imath]N(a+bi)= a^2+b^2[/imath] is the norm, so [imath]N(a+bi) < N(2-2i)[/imath]. This leaves us with representatives [imath] \{0,1,-1,i,-i,1+i,1-i,-1+i,-1-i,2,-2,2i,-2i\} \, . [/imath] So in this particular factor ring, [imath]2 = 2i[/imath]. Then squaring both sides we see [imath]1 = i[/imath], so [imath]R = \{I,1+I,-1-I, 2+I,-2+I\}[/imath], where [imath]I=\langle2-2i\rangle[/imath]. So I find that the number of elements of [imath]R[/imath] is five. Is this solution correct? | 1868345 | Quotient ring of Gaussian integers [imath]\mathbb{Z}[i]/(a+bi)[/imath] when [imath]a[/imath] and [imath]b[/imath] are NOT coprime
The isomorphism [imath]\mathbb{Z}[i]/(a+bi) \cong \Bbb Z/(a^2+b^2)\Bbb Z[/imath] is well-known, when the integers [imath]a[/imath] and [imath]b[/imath] are coprime. But what happens when they are not coprime, say [imath](a,b)=d>1[/imath]? — For instance if [imath]p[/imath] is prime (which is not coprime with [imath]0[/imath]) then [imath]\mathbb{Z}[i]/(p) \cong \mathbb{F}_p[X]/(X^2+1) \cong \begin{cases} \mathbb{F}_{p^2} &\text{if } p \equiv 3 \pmod 4\\ \mathbb{F}_{p} \times \mathbb{F}_{p} &\text{if } p \equiv 1 \pmod 4 \end{cases}[/imath] (because [imath]-1[/imath] is a square mod [imath]p[/imath] iff [imath](-1)^{(p-1)/2}=1[/imath]). — More generally, if [imath]n=p_1^{r_1} \cdots p_m^{r_m} \in \Bbb N[/imath], then each pair of integers [imath]p_j^{r_j}[/imath] are coprime, so that by CRT we get [imath]\mathbb{Z}[i]/(n) \cong \mathbb{Z}[i]/(p_1^{r_1}) \times \cdots \times \mathbb{Z}[i]/(p_m^{r_m})[/imath] I was not sure how to find the structure of [imath]\mathbb{Z}[i]/(p^{r}) \cong (\Bbb Z/p^r \Bbb Z)[X] \,/\, (X^2+1)[/imath] when [imath]p[/imath] is prime and [imath]r>1[/imath]. — Even more generally, in order to determine the structure of [imath]\mathbb{Z}[i]/(a+bi)[/imath] with [imath]a+bi=d(x+iy)[/imath] and [imath](x,y)=1[/imath], we could try to use the CRT, provided that [imath]d[/imath] is coprime with [imath]x+iy[/imath] in [imath]\Bbb Z[i][/imath]. But this is not always true: for [imath]d=13[/imath] and [imath]x+iy=2+3i[/imath], we can't find Gauss integers [imath]u[/imath] and [imath]v[/imath] such that [imath]du + (x+iy)v=1[/imath], because this would mean that [imath](2+3i)[(2-3i)u+v]=1[/imath], i.e. [imath]2+3i[/imath] is a unit in [imath]\Bbb Z[i][/imath] which is not because its norm is [imath]13 \neq ±1[/imath]. — I was not able to go further. I recall that my general question is to known what [imath]\mathbb{Z}[i]/(a+bi)[/imath] is isomorphic to, when [imath]a[/imath] and [imath]b[/imath] are integers which are not coprime (for instance [imath]a=p^r,b=0[/imath] or [imath]d=(a,b) = a^2+b^2>1[/imath]). Thank you for your help! |
2529194 | Proof that [imath]\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq a+b+c[/imath]
I've tried getting everything on the left side and transforming it into something squared so that I can prove it's bigger or equals to 0 but I've been unsuccessful. | 2072018 | AM-GM inequality: [imath]\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d[/imath]
Let [imath]a, b, c, d > 0[/imath]. Prove that [imath]\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d[/imath]. I'm supposed to prove this by AM-GM, but I can't see how. Any help would be appreciated. |
1363460 | One particular permutation [imath]\sigma [/imath] of [imath]5[/imath] elements such that [imath]\sigma ^{-1} (j) \le \sigma (j)[/imath]
As for options [imath]1.[/imath] and [imath]2.[/imath] I have taken a few examples randomly and found them to be correct but could not generalize and I am clueless about options [imath]3.[/imath] and [imath]4.[/imath] The answer keys that came with it says all the options are correct. | 1821248 | Let [imath]\sigma :\{1,2,3,4,5\} \rightarrow \{1,2,3,4,5\}[/imath] be permutations such that [imath] \sigma^{-1}(j) \leq \sigma(j)~\forall j, 1 \leq j \leq 5[/imath].
Which of the following are true? [imath]\sigma \circ \sigma(j)=j~\forall j, 1 \leq j \leq 5[/imath]. [imath]\sigma^{-1}(j)=\sigma(j)~\forall j, 1 \leq j \leq 5[/imath]. The set [imath] \{k: \sigma(k) \neq k \}[/imath] has even number of elements. The set [imath]\{k:\sigma(k) =k \}[/imath] has an odd number of elements . Can someone tell me to solve it in [imath]1[/imath] or [imath]2[/imath] min....trick..or some concept behind it? |
2529554 | For non-algebraically closed field [imath]k[/imath] and integer [imath]n>1[/imath], there is a polynomial in [imath]n[/imath]-variables over [imath]k[/imath] having only one zero in [imath]k^n[/imath]
Let [imath]k[/imath] be a field which is not algebraically closed and [imath]n>1[/imath] be an integer . Then does there exist [imath]f\in k[X_1,...,X_n][/imath] such that [imath]Z(f) (:=\{(a_1,...,a_n)\in k^n : f(a_1,...,a_n)=0\}) =\{0\}[/imath] ? My Work : I was thinking like this : Considering [imath]k[/imath] as a sub-field of its algebraic closure [imath]\bar k[/imath], we have [imath]k \subsetneq \bar k[/imath] . Also, there is an irreducible polynomial [imath]g(X) \in k[X][/imath] such that deg [imath]g(X)>1[/imath] . May be we could somehow apply Hilbert Nullstelensatz to [imath]\bar k[/imath], but I am not sure . I can easily do it for a finite field [imath]k[/imath]; in that case just take [imath]f(X_1,...,X_n)=\sum_{(y_1,...,y_n)\in k^n \setminus \{0\} } \prod_{i=1}^n (1-(X_i-y_i)^{q-1})[/imath] , where [imath]|k|=q[/imath] . I can also do it if [imath][\bar k : k] [/imath] is finite and [imath]n=2[/imath] . By Artin-Schreier theorem (http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/artinschreier.pdf) , we will have [imath][\bar k : k]=2[/imath] . Let [imath]u\in \bar k \setminus k[/imath], then the minimal polynomial of [imath]u[/imath] over [imath]k[/imath] has degree 2, and let [imath]\bar u[/imath] be another root of the minimal polynomial of [imath]u[/imath] over [imath]k[/imath] . Then [imath]f(X_1,X_2)=(X_1+uX_2)(X_1+\bar u X_2)[/imath] satisfies our required condition . Even in this case when [imath]\bar k[/imath] is a finite extension over [imath]k[/imath], I don't know what happens if [imath]n>2[/imath] . Please help . Thanks in advance | 1642593 | Affine variety over a field which is not algebraically closed can be written as the zero set of a single polynomial
I am trying to prove the following statement. If a field [imath]K[/imath] is not algebraically closed, then any [imath]K[/imath]-variety [imath]V\subset\mathbb{A}^n[/imath] can be written as the zero set of a single polynomial in [imath]K[X_1,X_2,...,X_n].[/imath] Suppose [imath]V=V(f_1,f_2,...,f_n)[/imath], and I would want to find a polynomial [imath]g\in K[X_1,X_2,...,X_n][/imath] such that [imath]V=V(g)[/imath]. I know that if I can find a polynomial [imath]\phi\in K[X_1,X_2,...,X_n][/imath] whose only zero is [imath](0,0,...,0)[/imath], then it is done, since we could take [imath]g=\phi(f_1,f_2,...,f_n)[/imath]. For [imath]n=1[/imath], then I would just choose [imath]\phi=X[/imath], but I do not even know how to proceed to the case [imath]n=2[/imath]. Maybe we could choose an irreducible polynomial of degree [imath]>1[/imath]. Any help will be appreciated. |
2529316 | Method to find the differential equation given the solution
I don't know how to proceed with this problem: Find the homogeneous differential equation with constant coefficients and minimum order that has as a (one of them) solution: [imath]t\cdot sin(t)[/imath]. I don't know how to start. How we can inverse the process to find the DE that has this solution without even know the order of it? Thanks. | 1127703 | [imath]x\sin x[/imath] is a solution of [imath]n[/imath]th order linear differential equation. Find minimum [imath]n[/imath]
Let [imath]y(x)=x\sin x[/imath] be one of solution of [imath]n^{th}[/imath] order linear differential equation with constant coefficients. Find the minimum value of [imath]n[/imath]. I have no idea where how to approach this problem. In book the answer is [imath]4[/imath]. |
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