qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
2323024 | If [imath]\mathbb{H}[/imath] is a finite field and [imath]a\neq0, b\neq0[/imath], are elements of [imath]\mathbb{H}[/imath] then exists [imath]u,v\in\mathbb{H}[/imath] such that [imath]1+au^{2}+bv^{2}=0 [/imath]
i need help with this exercise. If [imath]\mathbb{H}[/imath] is a finite field and [imath]a\neq0, b\neq0[/imath], are elements of [imath]\mathbb{H}[/imath] then exists [imath]u,v\in\mathbb{H}[/imath] such that [imath]1+au^{2}+bv^{2}=0 [/imath] I don't have idea of how attack this exercise. Can someone help me? | 572975 | Weighted sum of squares, in a finite field
Let [imath]\mathbb{F}[/imath] be a finite field. Let [imath]a_1,\dots,a_n \in \mathbb{F}[/imath] be given. I want to know whether there exists [imath]x_1,\dots,x_n \in \mathbb{F}[/imath] such that [imath]a_1 x_1^2 + a_2 x_2^2 + \dots + a_n x_n^2 = 1.[/imath] My question: Can we characterize when this equation has a solution, i.e., by some criterion on [imath]a_1,\dots,a_n[/imath]? Is there a clean characterization, in terms of [imath]a_1,\dots,a_n[/imath] and the properties of [imath]\mathbb{F}[/imath]? Here's what I can see immediately. First, if [imath]n=1[/imath], there is an easy answer (the equation has a solution iff [imath]a_1[/imath] is a square in [imath]\mathbb{F}[/imath]). Second, if any [imath]a_i[/imath] is a square in [imath]\mathbb{F}[/imath], the equation has a solution (set all other [imath]x_j[/imath]'s to [imath]0[/imath]), so we can assume that all the [imath]a_i[/imath]'s are non-squares and that [imath]n \ge 2[/imath]. Third, if [imath]-1[/imath] is a square in [imath]\mathbb{F}[/imath], then the equation has a solution: we can take [imath]x_2 = \sqrt{-a_1/a_2} \cdot x_1[/imath]. But what about the case where [imath]-1[/imath] is a non-square in [imath]\mathbb{F}[/imath]? Is it always guaranteed to have a solution in this case? |
2323350 | Volume of the parallelepiped
Let vectors [imath]u_1,\ u_2,\ \cdots u_n[/imath] and [imath]\mathcal{P}[/imath] the parallelepiped determined by these vectors : [imath]\mathcal{P}=\left\{\lambda_1 u_1+\cdots +\lambda_nu_n/\ \lambda_k\in [0,1]\right\}[/imath] How we can prove that : [imath]V=\int_{\mathcal{P}}dx_1\cdots dx_n=\det\left(u_1, u_2, \cdots, u_n\right)[/imath] Thanks you very much. | 427528 | why determinant is volume of parallelepiped in any dimensions
for [imath]n = 2,[/imath] I can visualize that the determinant [imath]n \times n[/imath] matrix is the area of the parallelograms by actually calculate the area by coordinates. But how can one easily realize that it is true for any dimensions? |
2322547 | Improper integral of sin(x)/x from zero to infinity
I was having trouble with the following integral: [imath]\int_{0}^\infty \frac{\sin(x)}{x}dx[/imath]. My question is, how does one go about evaluating this, since its existence seems fairly intuitive, while its solution, at least to me, does not seem particularly obvious. | 192685 | Are there any ways to evaluate [imath]\int^\infty_0\frac{\sin x}{x}dx[/imath] without using double integral?
Are there any ways to evaluate [imath]\int^\infty_0\frac{\sin x}{x}dx[/imath] without using double integral? I can't find any this kind of solution. Can anyone please help me? Thank you. |
2324774 | A trigonometry equation
I am trying to solve this trigonometric equation: [imath]\sin 84^\circ \sin 18^\circ \sin x = \sin 12^\circ \sin 48^\circ \sin (18^\circ -x)[/imath] I used calculator to get an answer [imath]x=6[/imath] but I don't know how to solve it. Please help. Thanks! Best Regards, Michael | 2324425 | Trigonometric Equation : [imath]\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)[/imath]
Please help solving this equation: [imath]\sin 96^\circ \sin 12^\circ \sin x = \sin 18^\circ \sin 42^\circ \sin (12^\circ -x)[/imath] I used numerical method to solve it and got [imath]x=6^\circ[/imath] but I am not able to solve it by trigonmetry. Thank! Best Regards, Michael. |
2325059 | If [imath]G[/imath] is a group of order [imath]p^2[/imath] with [imath]p[/imath] a prime number then [imath]G \simeq C_{p^2}[/imath] or [imath]G \simeq C_p \times C_p[/imath]
If [imath]G[/imath] is a group of order [imath]p^2[/imath] with [imath]p[/imath] a prime number then [imath]G \simeq C_{p^2}[/imath] or [imath]G \simeq C_p \times C_p[/imath] Here's my attempt We have [imath]|G| = p^2[/imath], Therefore, [imath]o(g)= 1[/imath] or [imath]p[/imath] or [imath]p^2 \forall g[/imath] where [imath]o(g)[/imath] denotes the order of the element of [imath]g \in G[/imath] Let's differentiate the [imath]3[/imath] cases. If [imath]o(g)= 1[/imath] then [imath]g=e[/imath] by unicity of the identity of G If [imath]o(g) = p^2[/imath] then [imath]\langle g\rangle = G \implies G \simeq C_p^2[/imath] Now it's obvious that I have to show that if [imath]o(g)= p[/imath] then [imath]G \simeq C_p \times C_p[/imath] But I'm not sure how to do that. | 1170842 | If order of group is [imath]p^2[/imath], where [imath]p[/imath] is prime, how can you deduce [imath]G[/imath] is isomorphic to [imath]C_{p^2}[/imath] or [imath]C_p \times C_p[/imath]?
Given [imath]|G|=p^2[/imath] then how can you deduce [imath]G\cong C_{p^2}[/imath] or [imath]G \cong C_p \times C_p[/imath] I have shown that G is abelian, not sure what to do next |
2324843 | Find [imath]S=\sum_{n=1}^{\infty}\frac{1}{n^3 2^n}[/imath]
Mathematica gives: [imath]S= -\frac{1}{12}\pi^2\log(2)+\frac{\log(2)^3}{6}+ \frac{7}{8}\zeta(3)[/imath] How can I prove it? | 412413 | The value of the trilogarithm at [imath]\frac{1}{2}[/imath]
From the functional equation [imath]\text{Li}_{3}(z) + \text{Li}_{3}(1-z)+ \text{Li}_{3} \Big( 1 - \frac{1}{z} \Big) = \zeta(3) + \frac{\ln^{3} (z)}{6}+ \frac{\pi^{2} \ln (z) }{6}- \frac{\ln^{2} (z) \ln(1-z)}{2}[/imath] one can determine the value of [imath]\text{Li}_{3} (\frac{1}{2} )[/imath] since [imath] \text{Li}_{3}(-1) = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{3}} = -\Big(1 - 2^{-2} \Big) \zeta(3) = - \frac{3}{4} \zeta(3) [/imath] But how does one derive that functional equation? A similar functional equation for the dilogarithm can be derived using the integral representation [imath] \text{Li}_{2}(z) = - \int_{0}^{z} \frac{\ln(1-t)}{t} \ dt [/imath] Is there a better way to find the value of [imath]\text{Li}_{3} (\frac{1}{2} )[/imath]? |
2324772 | Mistake in naming quadratic equation?
cubic equation: [imath]ax^3+bx^2+cx+d=0[/imath] Cube means 3. Quadratic equation: [imath]ax^2+bx+c=0[/imath] Quad means 4. But it is a power of two, should it be called bi... Equation. My question is: Is this a mistake in naming the quadratic equation? | 2259595 | Why is it called a quadratic if it only contains [imath]x^2[/imath], not [imath]x^4[/imath]?
Why is it called a quadratic if it only contains [imath]x^2[/imath], not [imath]x^4[/imath]? Should it not be bidratic or didratic etc? |
571538 | [imath]G[/imath] is finite, [imath]A \leq G[/imath] and all double cosets [imath]AxA[/imath] have the same cardinality, show that [imath]A \triangleleft G[/imath]
If [imath]G[/imath] is a finite group and [imath]A[/imath] is a subgroup of [imath]G[/imath] such that all double cosets [imath]AxA[/imath] have the same number of elements, show that [imath]gAg^{-1}=A[/imath] for all [imath]g \in G[/imath]. Here is my attempt, I guess it's correct but please verify it. I looked for this problem on the internet but I found it nowhere, so I thought it might be a good idea to have an answer for it on MSE: We know that for [imath]x,y \in G[/imath] we have [imath]|AxA|=|AyA|[/imath] by hypothesis of the problem. In particular for any [imath]g \in G[/imath] we have [imath]|AgA|=|AeA|=|AA|=|A|[/imath]. In other words, all double cosets [imath]AgA[/imath] have the same number of elements as [imath]A[/imath] for any [imath]g \in G[/imath]. Now, we use the following counting formula: [imath]|AxB|=\frac{|A||B|}{|A \cap xBx^{-1}|}[/imath] with [imath]A=B[/imath] and [imath] \forall g \in G:|AgA|=|A|[/imath] we obtain: [imath] |A| = \frac{|A||A|}{|A \cap gAg^{-1}|} \implies |A \cap gAg^{-1}|=|A| [/imath] But [imath]A \cap gAg^{-1} \subseteq A[/imath] and since [imath]G[/imath] is finite and [imath]|A \cap gAg^{-1}|=|A|[/imath] it forces [imath]A \cap gAg^{-1} = A[/imath]. That implies [imath]A \subseteq gAg^{-1}[/imath] for any [imath]g \in G[/imath], which is the same as [imath]g^{-1}Ag \subseteq A[/imath] for any [imath]g \in G[/imath] and this proves the normality of [imath]A[/imath] in [imath]G[/imath]. Q.E.D. | 2150826 | [imath]|AxA|[/imath] is a constant implies normal subgroup
I have an algebra problem: Let [imath]G[/imath] be a finite group, and [imath]A \le G[/imath]. Show that if [imath]|AxA|[/imath] is a constant for all [imath]x \in G[/imath], then [imath]A[/imath] is a normal subgroup of [imath]G[/imath]. My first question is about the meaning of the notation [imath]AxA[/imath]. Is this: [imath] \{a_1xa_2 | a_1, a_2 \in A\}?[/imath] Then what would the proof strategy be? |
2325390 | Continunuity of the generalized primitive
Let [imath]f:I\subset \mathbb R\to \mathbb R[/imath] be a Riemann locally integrable function on the interval [imath]I[/imath]. Show that, for a fixed [imath]a\in I[/imath] [imath] x\mapsto F(x) =\int_a^x f(t)dt ~~~~ [/imath] is continuous on [imath]I[/imath]. This a claim in Brezis book's page 205 lemma 8.2 | 2215698 | Why the primitive of a Riemann functions must be continuous?
Intuitively why must the primitive [imath]F[/imath] of a function [imath]f[/imath] that is Riemann integrable in [imath][a,b][/imath], be continuous in [imath][a,b][/imath]? Thanks |
2325865 | Prove or disprove [imath]\frac{1}{5}(3-4i)[/imath] is a root of unity.
Prove or disprove [imath]\frac{1}{5}(3-4i)[/imath] is a root of unity. Here is the definition of root of unity: An nth root of unity, where [imath]n[/imath] is a positive integer [imath](i.e. n = 1, 2, 3, …)[/imath], is a number [imath]z[/imath] satisfying the equation [imath]z^n=1[/imath]. | 1721406 | How to prove that a complex number is not a root of unity?
[imath]\frac35+i\frac45[/imath] is not a root of unity though its absolute value is [imath]1[/imath]. Suppose I don't have a calculator to calculate out its argument then how do I prove it? Is there any approach from abstract algebra or can it be done simply using complex numbers? Any help will be truly appreciated. |
2326036 | Anyone help give a counterexample when [imath]\inf_{x∈X}(\sup_{y∈Y}F(x,y) )\le \sup_{y∈Y}(\inf_{x∈X}F(x,y) )[/imath] is false
Let [imath]F(x,y)[/imath] be a [imath]X\times Y \to \Bbb R[/imath] function and [imath]X,Y[/imath] are subsets in [imath]\Bbb R^n[/imath], anyone help give a counterexample when [imath]\inf_{x∈X}(\sup_{y∈Y}F(x,y) )\le \sup_{y∈Y}(\inf_{x∈X}F(x,y) )[/imath] is false? PS: The other direction is true because | 374316 | An example of a function which satisfies [imath]\sup_{y}\inf_{x}f(x,y)<\inf_{x}\sup_{y}f(x,y)[/imath]
Let [imath]f:X\times Y\to \mathbb{R}[/imath]. I could not come up with an example that satisfies [imath]\sup_{y\in Y}\inf_{x\in X}f(x,y)<\inf_{x\in X}\sup_{y\in Y}f(x,y).[/imath] Any help would be appreciated. Thanks! |
2326449 | How did the convention of 'multiplying' using [imath]4\times 3[/imath] become [imath]4\cdot 3[/imath]?
How did the convention of 'multiplying' using the cross symbol [imath]4\times 3[/imath] become [imath]4\cdot 3[/imath], using a dot in the middle? | 1844374 | Why does the symbol for the multiplication operation change shape?
Why does the "[imath]\times[/imath]" used in arithmetic change to a "[imath]\cdot[/imath]" as we progress through education? The symbol seems to only be ambiguous because of the variable [imath]x[/imath]; however, we wouldn't have chosen the variable [imath]x[/imath] unless we were already removing [imath]\times[/imath] as the symbol for multiplication. So why do we? I am very curious. It seems like [imath]\times[/imath] is already quite sufficient as a descriptive symbol. |
2326316 | Measure theory, product of [imath]L^2[/imath] functions, convergence in [imath]L^1[/imath]
How can I prove this? Let [imath]f_n \to f[/imath] in [imath]L^2[/imath] and [imath]g_n \to g[/imath] in [imath]L^2[/imath]. Then [imath]f_n \cdot g_n \to f \cdot g[/imath] in [imath]L^1[/imath]. | 2326424 | Convergence in quadratic mean and in mean
Let [imath]\mathrm{X, (X_n)_{n\in\mathbb{N}}, Y, (Y_n)_{n\in\mathbb{N}}}[/imath] [imath]\in[/imath] [imath]\mathcal{L^2}[/imath] be random variables. Prove that ([imath]\mathrm{X_n}[/imath] converge to [imath]\mathrm{X}[/imath] in [imath]\mathcal{L^2}[/imath] and [imath]\mathrm{Y_n}[/imath] converge to [imath]\mathrm{Y}[/imath] in [imath]\mathcal{L^2}[/imath]) [imath]\Rightarrow[/imath] [imath]\mathrm{X_nY_n}[/imath] converge to [imath]\mathrm{XY}[/imath] in [imath]\mathcal{L^1}[/imath]. My attempts: I think Jensen's inequality is needed here, with the function [imath]\phi[/imath]([imath]\mathcal{x}[/imath])=$\vert[imath]\mathcal{x}[/imath]\vert$, because the assumption is similar to a property stated here https://en.wikipedia.org/wiki/Convergence_of_random_variables#Convergence_in_mean, but I am unable to clearly formulate a strategy to solve this problem |
2326496 | Suppose f is a twice differentiable function with [imath]f(0)=0[/imath] and [imath]f(1)=1[/imath], [imath]f'(0)=f'(1)=0[/imath]. Then show that [imath]|f''(x)| > 4[/imath] for some [imath]x[/imath] in [imath](0,1)[/imath]
Any help would be appreciated. My initial attempts were using Lagrange's Mean Value Theorem first between [imath]f(0)[/imath] and [imath]f(1)[/imath] to show that [imath]f'(t)=1[/imath] for some [imath]t[/imath] in [imath](0,1)[/imath]. Now applying LMV between [imath]f'(t)[/imath] and [imath]f'(0)[/imath], and [imath]f'(t)[/imath] and [imath]f'(1)[/imath] I could prove it if t lies in either [imath](0,1/4][/imath] or in [imath][3/4,1)[/imath]. Don't really know where to go next. | 2327539 | Let f be a differentiable function with f(0)=0 and f(1)=1, f'(0)=f'(1)=0. Prove that |f''(x)| > 4 for some x in (0,1). (Without invoking integrals)
Let [imath]f[/imath] be a twice differentiable function such that [imath]f(0)=0[/imath] and [imath]f(1)=1[/imath]. Also, [imath]f'(0)=f'(1)=0[/imath]. Prove that [imath]f''(x)>4[/imath] for some [imath]x \in (0,1)[/imath]. Any help would be appreciated. My initial attempts were using Lagrange's Mean Value Theorem first between [imath]f(0)[/imath] and [imath]f(1)[/imath] to show that [imath]f'(t)=1[/imath] for some [imath]t[/imath] in [imath](0,1)[/imath]. Now applying LMV between [imath]f'(t)[/imath] and [imath]f'(0)[/imath], and [imath]f'(t)[/imath] and [imath]f'(1)[/imath] I could prove it if t lies in either [imath](0,1/4][/imath] or in [imath][3/4,1)[/imath]. Don't really know where to go next. |
2327049 | Is the ideal [imath]\langle x^2,x+1 \rangle[/imath] of [imath]\mathbb{Z}[X][/imath] a principal ideal ?
I computed the gcd of the 2 polynomials in the ideal and their gcd is equal to 1 ,using euclid algorithm. I want to know if the ideal in the title is a principal ideal and why ? I know a principal ideal is the ideal generated by only one element.So [imath]\langle x^2,x+1 \rangle = \langle 1 \rangle[/imath] but if that was a principal ideal then any ideal generated by 2 or more ideals would be principal because he could be generated by [imath]1[/imath] or some gcd of those polynomials. I think the same for non polynomial rings .. since the gcd can be 1 or any number , any ideal can be generated by one element so any ideal would be principal.What am I missing ? | 2326259 | Prove that ideal [imath]I = \langle X^2,X+1\rangle[/imath] is principal or not (in [imath]\mathbb{Z}[X][/imath] and [imath]\mathbb{Q}[X][/imath] )
I tried the following [imath]I = \langle X^2,X+1\rangle =\langle X^2,X+1,X^2+2(X+1)\rangle =\langle X^2,X+1,(X+1)^2+1 \rangle[/imath] Yet no matter how I arrange it, I cannot obtain [imath]1[/imath]. Can someone help me out? |
2327022 | Bounded linear integral operator and its norm
Let [imath] Tx(t) = \int_{0}^{1}K(t,s)x(s) ds, \quad x\in C[0, 1], [/imath] be an operator, where [imath]K(\cdot,\cdot) [/imath] is a continuous function on [imath][0,1]\times[0,1][/imath]. Prove that [imath]T \in \mathcal{L}(C[0,1])[/imath] (bounded linear) and calculate the norm of T. I would like to ask you for some help as I have no idea how I should approach the problem, the operator topics seem somewhat abstract to me. (There is no norm candidate provided as opposed to other problems you may find, I would like to know how to figure out one, and as far as the first part of the question goes, does it suffice to show either continuity or boundedness? Which one is easier, intuitive? I would appreciate some explanation.) | 759095 | norm of integral operator in [imath]C([0,1])[/imath]
If we define on [imath]C([0,1])[/imath] the operator [imath] Tf(x) = \int_{0}^{1} K(t,s) f(s) ds[/imath] where [imath]K[/imath] is a continous function on two variables. I want to show that: [imath]1)[/imath] [imath]||T|| = \displaystyle\max_{t} {\int_{0}^{1} |K(t,s)| ds} [/imath] [imath]2)[/imath] When [imath]K(t,s) = \displaystyle\min(t,s)[/imath], to prove that [imath]T[/imath] is compact Thanks. |
2327266 | Show the [imath]n[/imath]th prime is [imath]\le 2^{2^n}[/imath]
Let [imath]P_n[/imath] be the [imath]n[/imath]th prime. Ie [imath]P_1 = 2, P_2 = 3, P_3 = 5, ...[/imath] Show that [imath]P_n \le 2^{2^n}[/imath] Induction seems to be useless here. Is there a hint someone can provide? | 2138890 | n-th prime number is less than [imath]4^n[/imath]
Prove that [imath]\forall n \in \mathbb{N}^*, P_n \lt 4^n[/imath] where [imath]P_n[/imath] is the [imath]n[/imath]th prime number. I'm searching for a proof that doesn't use induction and uses only the elementary concepts of number theory. This theorem appears in an old french book on Arithmetics destined to high school level students (lycée). It is introduced right after the concept of decomposition into prime factors. The book includes a proof that I didn't understand. You can see below a screenshot taken from the book and a translation attempt. Theorem For all integers [imath]n \gt0,[/imath] the [imath]nth[/imath] prime number [imath]q[/imath] verifies the inequality [imath]q \lt 4^n[/imath]. Demonstration The [imath][1,q][/imath] interval is included in the set of integers [imath]k[/imath] that can be written, (after grouping all the squares that can be formed with the [imath]n[/imath] prime numbers [imath]p_i \le q [/imath] in a decomposition of [imath]k[/imath]), as [imath]k = m^2p_1^{e_1}p_2^{e_2}...p_n^{e_n}[/imath] with [imath]e_i \in {0,1}[/imath]. (Here, [imath]p_i[/imath] is the i-th prime number, and [imath]p_n = q[/imath].). There are no more than [imath]\sqrt{q}[/imath] integers like m, and at most [imath]2^n[/imath] choices for the exponents [imath]e_i[/imath], what shows that [imath]q[/imath], cardinal of this set, is strictly lower than the product [imath]2^n\sqrt{q}[/imath], then the result. |
2326137 | Cumulative distribution function of continuous random variables
Marginal density functions: [imath]f_{V}(v)=f_{W}(w)=\frac{1}{60}[/imath] Joint density function: [imath]f_{V,W}(v,w)=\frac{1}{3600}[/imath], for [imath]0\leq x\leq60[/imath] and [imath]0\leq y\leq60[/imath] I need to find the cumulative density function of [imath]T = W - V[/imath] I'm thinking it's something along the lines of this: [imath]F_{t}(t)=\left\{\begin{matrix} 0 & w-v<0\\ \frac{t}{60} & 0 \leq w-v \leq 60\\ 1 & w-v>60 \end{matrix}\right.[/imath] Any help appreciated. Thanks! | 203477 | Problem with Density of [imath]X-Y[/imath]
Assuming [imath]f(x,y) = 1[/imath], [imath]0<x<1[/imath], [imath]0<y<1[/imath], obtain the density of [imath]Z = X-Y[/imath]. |
107567 | the hyperbolic plane is complete
I'd like to prove that the hyperbolic plane is complete in a short, nice way. Here's my proof, but I'm not convinced of its correctness yet. Let [imath]\mathbb{H}=\left\{(x,y)\in\mathbb{R}^2\big\vert y>0\right\}[/imath]. Since this is a connected, Riemannian manifold with infinitesimal metric [imath]dh^2=ds^2/y^2)[/imath] (where [imath]ds[/imath] is the usual Euclidean metric) we can use the Hopf-Rinow theorem and say that completeness is equivalent to having geodesics defined from the whole real line. Hence, it suffices to show that every point [imath]p\in\mathbb{H}[/imath] is at infinite distance from the boundary. Take [imath]q\in\partial\mathbb{H}[/imath]. By moving the points via isometries, we can assume that [imath]p[/imath] lies in the positive, pure imaginary, open semi-line and have coordinates [imath]p=(0,\bar{y})[/imath] and that [imath]q[/imath] is the point [imath]\infty[/imath]. Then we may finish the proof just computing the integral [imath]\int_{\bar{y}}^\infty \sqrt{\frac{dy^2}{y^2}}=log(|y|)\Big\vert_{\bar{y}}^\infty=\infty.[/imath] Is there any wrong step? Do you know any other concise proof of the completeness of the hyperbolic plane? | 1870686 | Hyperbolic metric geodesically complete
Consider the upper half plane model of the hyperbolic space ([imath]\mathbb{H}[/imath] with the riemannian metric [imath]g=\frac{dx^2+dy^2}{y^2}[/imath]). It is known that [imath](\mathbb{H},g)[/imath] is geodesically complete, which means that no geodesic can reach the border [imath]\partial \mathbb{H}[/imath] in a finite time. Why is that? Of course if I consider particular geodesics such as [imath]t\mapsto (0,e^{-t})[/imath] this is true, but I can't figure out a general proof for this fact which doesn't rely on the particular form of the geodesic considered. My intuition is that it must depend on the fact that, as a geodesic approaches the border, the denominator of [imath]g[/imath] goes to [imath]0[/imath]. Can you point me out an explanatory proof which shows that the length of a hyperbolic geodesic which goes to the border can't be finite? EDIT: Thank you so much for the answers, I realized I didn't make myself clear. What I mean is: is there a way to prove directly for any hyperbolic geodesic (using the fact that [imath]g[/imath] has [imath]y^2[/imath] at the denominator) that its length is infinite? |
2327905 | Solve X=sqrt(A)^sqrt(A)^sqrt(A)^..............infinty?
If [imath]X= \newcommand{\W}{\operatorname{W}}\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{\sqrt{A}^{.{^{.^{\dots}}}}}}}}}}} [/imath] then what is the value of [imath]X^2-e^{1/X}[/imath] ? | 1948778 | What is the maximum convergent [imath]x[/imath] in the power tower [imath]x^{x^{x^{x\cdots}}}[/imath]?
In the power tower [imath]x^{x^{x^{x\cdots}}}[/imath] where there is an infinite stack of [imath]x[/imath]'s, what is the maximum convergent number? I know the answer by playing with the form [imath]x^y=y[/imath] and using Mathematica, but I don't know how to solve this by hand. |
2328829 | How to prove this inequality [imath]\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ge 1.5[/imath]?
Given [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath] are three positive real numbers. How to prove that sum of [imath]\frac{a}{b+c}[/imath], [imath]\frac{b}{c+a}[/imath] and [imath]\frac{c}{a+b}[/imath] is greater than or equal to [imath]1.5[/imath]? | 901012 | Nesbitt inequality symmetric proof
I was trying to prove Nesbitt inequality using symmetry. So I assumed that [imath]a+b=x , b+c =y ,c+a=z[/imath] and then I rewrite inequality as:[imath]\sum \limits_{cyc} \frac {\frac {x+z-y}{2}}{y}\geq \frac{3}{2}[/imath] multiply both sides by [imath]2[/imath]:[imath]\sum \limits_{cyc} \frac {x+z-y}{y}\geq 3[/imath] and adding [imath]3[/imath] to both sides:[imath]\sum \limits_{cyc} \frac {x+z}{y}\geq 6[/imath] And I stuck here.I searched on google and only source I found that solved this inequality like that was Wikipedia. And Wiki mentioned that [imath]\sum \limits_{cyc} \frac {x+z}{y}\geq 6[/imath] is true by am-gm, but I can't figure out how. |
2328917 | proving [imath]f[/imath] and [imath]g[/imath] are constant
Let [imath]f[/imath] and [imath]g[/imath] be entire functions such that [imath]e^f + e^ g=1[/imath] then how to show [imath]f[/imath] and [imath]g[/imath] are constant. I am having trouble in proving these. any help would be appreciated. Thanks in advance. | 300552 | Complex Analysis, Entire functions
Prove if [imath]f[/imath] and [imath]g[/imath] are entire and [imath]e^f+e^g=1[/imath], then [imath]f[/imath] and [imath]g[/imath] are constant. I believe the simplest way would be to use Louiville's theorem by using Pick's theorem but I am not sure on how to go about this. |
2329019 | Differential equation with matrix
How can we find [imath]x′ = Ax[/imath] for the matrix \begin{pmatrix}3 &1 &0\\ 0& 3 &0\\ 0& 0& 2\end{pmatrix} | 2328814 | system of differential equations with matrix
[imath]x′ = Ax[/imath] where [imath]A = \begin{pmatrix} 3&1&0\\ 0&3&0\\ 0&0&2 \end{pmatrix} [/imath] what is the logic to solve such type of equations? I found that [imath]det(A-λI) = (3-λ)^2(2-λ)=0[/imath], so the eigenvalues are 3 and 2 |
2328877 | Do exist an injective linear map from [imath]\mathbb{R}^5\rightarrow\mathbb{R}^4[/imath]
I already asked If there exists an injective function [imath]f:\mathbb{R}^5\rightarrow\mathbb{R}^4[/imath] and I found that it exists a such function: Do exist an injective function from [imath]\mathbb{R}^5 \rightarrow \mathbb{R}^4[/imath] Now, I would like to know If there exist such an injective linear map. I was trying to find out but I am unable. In think it does not exists because of the difference between dimmnsions but I am really unsure. Please help me. Thank you!!!! | 2168833 | If [imath]L:\mathbb{R}^n \rightarrow \mathbb{R}^m[/imath] and [imath]m linear trans. What does this tell about injectivity and surjectivity?[/imath]
Let [imath]L:\mathbb{R}^n \rightarrow \mathbb{R}^m[/imath] be a linear transformation such that [imath]m<n[/imath]. What does this tell you about injectivity and surjectivity? I can't really create a (counter) argument against any of these alternatives: [imath]L[/imath] is injective [imath]L[/imath] is surjective [imath]L[/imath] is not injective [imath]L[/imath] is not surjective My intuition says that one could find examples where [imath]L[/imath] is both injective and/or surjective, but not for all inputs outputs. |
789848 | Prove that [imath]g(x) > 0[/imath]
If [imath]f(x)[/imath] is a quadratic expression such that [imath]f(x) > 0,\ x\in\mathbb{R}[/imath] and if [imath]g(x)= f(x) + f'(x) + f''(x)[/imath], then prove that [imath]g(x) > 0, \ x\in\mathbb{R}[/imath]. | 942446 | If [imath]f[/imath] is a quadratic and [imath]f(x)>0\;\forall x[/imath], and [imath]g= f + f' + f''[/imath], prove [imath]g(x)>0\; \forall x[/imath]
If [imath]f(x)[/imath] is a quadratic expression such that [imath]f(x)>0\;\forall x\in\mathbb{R},[/imath] and if [imath]g(x)=f(x)+f'(x)+f''(x),[/imath] Then prove that [imath]g(x)>0\; \forall \; x\in \mathbb{R}[/imath]. [imath]\bf{My\; Trial \; Solution::}[/imath] If [imath]f(x)>0\;\forall x\in \mathbb{R}[/imath]. Then function [imath]f(x)[/imath] has Minimum value . Let Minimum occur at [imath]x=x_{0}[/imath]. Then [imath]f(x)_{Min} = f(x_{0})>0[/imath] Now Given [imath]g(x) = f(x)+f'(x)+f''(x)[/imath]. Then [imath]g'(x) = f'(x)+f''(x)+f'''(x).[/imath] Now at [imath]x=x_{0}\;\;,[/imath] Value of [imath]g(x_{0}) = f'(x_{0})+f''(x_{0})+f'''(x_{0}) = 0+f''(x_{0})+f'''(x_{0})[/imath] Now I did not understand How can i solve after that, Help me Thanks |
1941503 | Number of equivalence classes of matrices under switching rows and columns
I have been thinking about this problem: Suppose we have all [imath]R[/imath] by [imath]C[/imath] matrices, where the values are integers in [imath][1,n][/imath]. Two matrices are equivalent under interchanging rows and columns. For example, 1 5 0 0 would be equal to itself and to 0 0 1 5 and would be equal to 0 0 5 1 and would be equal to 5 1 0 0 How many unique such matrices are there? Any ideas how to go about it? Any help much appreciated! | 2056708 | Can this be counted with stars and bars method?
If I have a [imath]w \times h[/imath] matrix where each value is an integer [imath]1 \lt n \lt 20[/imath], how can I count the number of distinct configurations, where [imath]2[/imath] configurations are "distinct" if there is no way to reshuffle the rows and columns that would produce the same matrix? for example these are equal (we swapped a row, then a column) 0 0 0 2 0 4 0 2 4 0 0 0 but these are distinct (no way to swap rows or columns to produce the other) 0 0 0 2 0 0 0 2 4 0 4 0 it seems like there aught to be a way to count the rows or columns as "bins" and the values as balls. I realize that in this case there are [imath]18[/imath] different colored balls, but even if the only values possible were [imath]1[/imath] and [imath]0[/imath], (ball or no ball) I can't see how to represent it as stars and bars. |
2329211 | Group action with a single orbit has a [imath]g \in G[/imath] with no fixed points
Let [imath]G[/imath] be a group acting on a set [imath]S[/imath] where [imath]|S| \geq 2[/imath]. If this action has one orbit, show that there exists a [imath]g \in G[/imath] which has no fixed points. I feel I am having trouble proving this because I am not sure exactly what it means to have one orbit. I am taking the set [imath]Gs = \{g\cdot s : g \in G\}[/imath] to be the orbit of [imath]s \in S[/imath]. For every [imath]s \in S[/imath], we have [imath]e \cdot s = s[/imath]. In this case doesn't 'one orbit' mean that the orbit would be all of [imath]S[/imath]? Any help is appreciated. | 1820572 | G acts on X transitively, then there exists some element that does not have any fixed points
Let [imath]X[/imath] be a transitive [imath]G[/imath]-set. ([imath]G[/imath] acts on [imath]X[/imath] transitively.) If [imath]X[/imath] is finite and has at least two elements, show that there is some element [imath]g[/imath] [imath]\in[/imath] G which does not have any fixed points; that is, such that $g[imath]([/imath]x$[imath])[/imath] [imath]\ne[/imath] [imath]x[/imath] for all [imath]x[/imath] [imath]\in[/imath] [imath]X[/imath] I am trying to use contradiction, but it is not very clear to me why this is true. |
1019177 | How is [imath]\ln(-1) = i\pi[/imath]?
How do I derive: [imath]\ln(-1)=i\pi[/imath] and [imath]\ln(-x)=\ln(x)+i\pi[/imath] for [imath]x>0[/imath] and [imath]x \in\mathbb R[/imath] Thanks for any and all help! | 2821105 | The irrational of such constant
In this question, I needed to assume in my answer that [imath]e^{e^{e^{79}}}[/imath] is not an integer. Is there some standard result in number theory that applies to situations like this? After several years, it appears this is an open problem. As a non-number theorist, I had assumed there would be known results that would answer the question. I was aware of the difficulty in proving various constants to be transcendental -- such as [imath]e + \pi[/imath], which is not known to be transcendental at present. However, I was looking at a question that seems simpler, naively: whether a number is an integer, rather than whether it is transcendental. It seems that what appeared to be possibly simpler is actually not, with current techniques. The main motivation for asking about this particular number is that it is very large. It is certainly possible to find a pair of very large numbers, at least one of which is transcendental. But the current lack of knowledge about this particular number is even an integer shows just how much progress remains to be made, in my opinion. Any answers that describe techniques that would suffice to solve the problem (perhaps with other, unproven assumptions) would be very welcome. |
2329248 | To compute [imath]\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2|^2 dz[/imath] where [imath]\mathcal{C}[/imath] is the unit circle in [imath]\mathbb{C}[/imath]
Let [imath]\mathcal{C}[/imath] denote the unit circle in [imath]\mathbb{C}[/imath] centred at the origin taken anticlockwise. Compute the value of the integrtal [imath]\frac{1}{2\pi i}\int_\mathcal{C} |1+z+z^2|^2 dz[/imath] I don't recon we can use the residue theorem to compute this integral because the integrand is not analytic anywhere. I then tried substituting [imath]z=e^{i\theta}[/imath] and integrating it the traditional way but cant make much of the integrated [imath]|1+e^{i\theta}+e^{2i\theta}|^2[/imath]. Any help would be much appreciated. Thanks. | 2328658 | Complex integral
Let [imath]C[/imath] denote the unit circle centered at the origin in Complex Plane What is the value of [imath] \frac{1}{2\pi i}\int_C |1+z+z^2 |dz,[/imath] where the integral is taken anti-clockwise along [imath]C[/imath]? 0 1 2 3 What I have answered is 0 because it seems like [imath]f(z)[/imath] is analytic at 0 hence by Cauchy's Theorem. |
2329597 | If [imath]M[/imath] is finitely generated left module over a left noetherian ring [imath]R[/imath], then [imath]M[/imath] is a noetherian module.
If [imath]M[/imath] is finitely generated left module over a left noetherian ring [imath]R[/imath], then [imath]M[/imath] is a noetherian module. I need to show that an arbitrary ascending chain of submodules of [imath]M[/imath] must satisfy ACC, or equivalently, every submodule of [imath]M[/imath] is finitely generated. Since [imath]R[/imath] is a left noetherian ring, every left ideal in [imath]R[/imath] is finitely generated. I am trying to find a 1-1 correspondence between left ideals of [imath]R[/imath] and submodules of [imath]M[/imath]. Is it a correct way, or how can I show the assertion by a different way? Thanks. | 533873 | Finitely generated modules over a Noetherian ring are Noetherian
I'm trying to prove that if the ring [imath]R[/imath] is Noetherian then every finitely generated [imath]R[/imath]-module is Noetherian. First of all, it is known that every module is a homomorphic image of a free module, so we can take [imath]R[/imath] as a module over itself, and then we will have a homomorphism [imath]f:R \rightarrow M[/imath], where [imath]M[/imath] is some arbitrary module. Next, according to the well-known theorem, [imath]\operatorname{Im}(f) \cong \frac{R}{\ker(f)},[/imath] and [imath]\frac{R}{\ker(f)}[/imath] is Noetherian because [imath]R[/imath] is Noetherian (we can take finite set of generators for some submodule in [imath]R[/imath] and then its canonical image in [imath]\frac{R}{\ker(f)}[/imath] generates submodule there). Thus, [imath]\operatorname{Im}(f)[/imath] is Noetherian as well, but [imath]\operatorname{Im}(f) \subseteq M[/imath]. How then one could say the same about the whole module [imath]M[/imath]? |
2329833 | Proving the inequality [imath]2^n ≤2^{n+1}−2^{n−1}−1[/imath]
I've tried multiplying each side by [imath]2^n[/imath], factoring out a 2, but I find myself going in a circle. What's the best way to solve this? [imath]2^n ≤2^{n+1}−2^{n−1}−1[/imath]. For all integers n | 1208449 | Proving [imath]2^n\leq 2^{n+1}-2^{n-1}-1[/imath] for all [imath]n\geq 1[/imath] by induction
I am trying to prove that for every element of [imath]\mathbb{N}[/imath], that [imath]2^n \leq 2^{n+1} - 2^{n-1} - 1.[/imath] I started by showing that initial case, of [imath]n=1[/imath], is true. Then I proceed to the inductive step by assuming that for some [imath]k[/imath] in [imath]\mathbb{N}[/imath], [imath]2^k \leq 2^{k+1} - 2^{k-1} - 1[/imath]. Then, for [imath]n=k+1[/imath], [imath]2^{k+1} \leq 2^{k+2} - 2^{k} - 1[/imath]. Is this sufficient to prove the statement by induction or do I have to expand the exponentials to rearrange the inequality in a different way? |
2330065 | Proof by induction of inequality
Let p(n) be the mathematical statement: [imath]2^n \leq 2^{n+1} - 2^{n-1} - 1[/imath]. Base Case: When n = 1 we have [imath]2^1 \leq 2^{1+1} - 2^{1-1} - 1[/imath] which simplifies to [imath]2 \leq 4[/imath]. So P(1) is correct. Induction hypothesis: Assume that P(k) is correct for some integer k. Induction step: We will now show that P(k + 1) is correct. \begin{align} 2^{k+1} &= 2\cdot 2^{k}\tag{by definition}\\[0.5em] &\leq 2\cdot(2^{k+1}-2^{k-1}-1)\tag{by inductive hypothesis}\\[0.5em] &= 2^{k+2}-2^k-2\tag{expand}\\[0.5em] &\leq 2^{k+2}-2^k-1. \end{align} Is this proof sufficient? | 1550380 | Proving by induction that [imath]2^n \le 2^{n+1}-2^{n-1} - 1[/imath] . Does my proof make sense?
I'm not sure if this proof is valid or not and could use some feedback. Let P be the supposition that [imath]2^n \le 2^{n+1}-2^{n-1} - 1[/imath] for all positive integers n. Base Case: Let n=1 [imath]2^1 \le 2^2-2^0-1[/imath] [imath]2\le2[/imath] Suppose that P holds true for some positive integer n=k [imath]2^{k+1}=2*2^k[/imath] By inductive hypothesis: [imath]2*2^k \le 2(2^{k+1}-2^{k-1}-1)[/imath] [imath]2^{k+1} \le 2^{k+2}-2^{k}-2[/imath] Since we know [imath]2^{k+1} \le 2^{k+2}-2^{k}-2[/imath], we also know [imath]2^{k+1} \le 2^{k+2}-2^{k}-1[/imath] because the former is always smaller. This proves that: [imath]2^k \le 2^{k+1}-2^{k-1} - 1\Rightarrow 2^{k+1} \le2^{k+2}-2^{k}-1 [/imath] |
2330205 | Prove [imath]\lim_{n\to\infty} \frac{f(b_n)-f(a_n)}{b_n-a_n} =f'(c)[/imath]
Let [imath]f[/imath] be differentiable at [imath]c[/imath] and let [imath]\{a_n\}[/imath] and [imath]\{b_n\}[/imath] be sequences such that [imath]a_n<c<b_n[/imath] and [imath]a_n,b_n \rightarrow c[/imath]. Prove [imath]\displaystyle \lim_{n\to\infty} \frac{f(b_n)-f(a_n)}{b_n-a_n} = f'(c)[/imath] I know this problem is very similar to mean value theorem. However, it shows up before that chapter. I am wondering how to prove this one without using Rolle theorem and Mean Value theorem. I hope someone can give me some hint to think about. | 1820848 | Prove that [imath]f'(a)=\lim_{n\to\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}[/imath] under some conditions
Prove that [imath]f'(a)=\lim_{n\to\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}[/imath], with [imath]f: (k_1,k_2)\to\Bbb R[/imath] differentiable at [imath]a\in (k_1,k_2)[/imath], [imath]\lim b_n=\lim a_n=a[/imath], and [imath]a_n<a<b_n[/imath]. My attempt: Im trying to write the above equality as a particular case of this other equality that hold by any interior point [imath]a[/imath] [imath]f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a-h)}{2h}[/imath] Because [imath]a_n<a<b_n[/imath] then I can write [imath]f'(a)=\lim_{n\to\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}=\lim_{n\to\infty}\frac{f(h_n+a)-f(a-j_n)}{h_n+j_n}[/imath] Where [imath]h_n=b_n-a>0[/imath] and [imath]j_n=a-a_n>0[/imath], where these two new sequences tends monotonically to zero. Now my problem: my thought is that [imath]h_n[/imath] and [imath]j_n[/imath] are particular sequences then I can build a sequence [imath]k_n[/imath] such that [imath]j_n[/imath] and [imath]h_n[/imath] are subsequences so I can write: [imath]\{h_n:n\in\Bbb N\}\subset\{k_n:n\in\Bbb N\}\text{ and }\{j_n:n\in\Bbb N\}\subset\{k_n:n\in\Bbb N\}[/imath] (I can build this [imath]k_n[/imath] by example doing [imath]k_{2n}=h_n[/imath] and [imath]k_{2n-1}=j_n[/imath]) and then I have the desired expression [imath]f'(a)=\lim_{n\to\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}=\lim_{n\to\infty}\frac{f(k_n+a)-f(a-k_n)}{2k_n}[/imath] My question: Is this proof right? In particular Im unsure of the implicit statement "if the limit exist for [imath]k_n[/imath] then it must exist too for the particular case with [imath]h_n[/imath] and [imath]j_n[/imath]". The use of two sequences, [imath]h_n[/imath] and [imath]j_n[/imath] in the limit make me unsure about the validity of this manipulation and assumption. There is a simpler/different way to prove the same? Thank you in advance. |
2330281 | Prove that if [imath]\lim_{x\to0}{\frac{f(x)}{f'(x)}}[/imath] exists, then it must equal 0
Suppose that, in a deleted neighborhood of 0, f is differentiable with [imath]f'\neq 0[/imath] and that [imath]\lim_{x\to0}f(x)=0[/imath]. Prove that if [imath]\lim_{x\to0}{\frac{f(x)}{f'(x)}}[/imath] exists, then it must equal 0. This question is included in L'Hospital Rule's section, but I don't know how to apply the rule here. It seems to me that if f is differentiable with [imath]f'\neq 0[/imath], then [imath]\lim_{x\to0}{\frac{f(x)}{f'(x)}}[/imath] is not a form of [imath]\frac{0}{0}[/imath]. Then I don't know what to show now. Any hints? | 1619990 | Behavior of derivative near the zero of a function.
Suppose the function [imath]f:[0,\delta) \to \mathbb{R}[/imath] is continuous, differentiable in [imath](0,\delta)[/imath] and [imath]f(0)=0[/imath]. If the limit [imath]\displaystyle \lim_{x \to 0+}\frac{f(x)}{f'(x)}= L[/imath] exists, then is it always the case that [imath]L = 0[/imath]. This seems to be true both for functions with well-behaved derivatives such as [imath]f(x) = x,[/imath] [imath]\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x}{1}=0,[/imath] as well as functions with "bad" derivatives such as [imath]f(x) = \begin{cases}x \ln x &\mbox{if }x>0 ,\\0 &\mbox{if } x=0, \end{cases},[/imath] where [imath]\lim_{x \to 0+} \frac{f(x)}{f'(x)}=\lim_{x \to 0+} \frac{x \ln x }{1+ \ln x}=0.[/imath] Is there a simple proof or counterexample? |
2330609 | Trouble evaluating this limit
I have to evaluate [imath]\lim_{n \to \infty}n/(n^2+1) + n/(n^2+2)+n/(n^2+3)+....+n/(n^2+n)[/imath] Now, for the rth term: [imath]\lim_{n \to \infty}n/(n^2+r) = 1/(n+(r/n)) = 0.[/imath] Since this is true for any term, and limits are distributive over addition, the limit should be zero. But this is wrong. Why? | 2325093 | Can I break this limit into individual terms?
[imath]\lim_{x\to \infty} {\frac{x}{x^2+1} +\frac{x}{x^2+2} + ... + \frac{x}{x^2+x} }[/imath] It seems obvious that the result is zero for each term but in order to break the limit into its individual parts we must know that every term's limit exists . |
2330649 | Intermediate Rings that are not finitely generated
Let [imath]R_1[/imath] and [imath]R_2[/imath] are two rings such that [imath]R_1[/imath] [imath]\subseteq[/imath] [imath]R_2[/imath] and [imath]R_2[/imath] is a finitely generated ring over [imath]R_1[/imath]. We know that, if [imath]R_1[/imath] is Noetherian and [imath]S[/imath] is a ring such that [imath]R_1[/imath] [imath]\subseteq[/imath] [imath]S[/imath] [imath]\subseteq[/imath] [imath]R_2[/imath] and [imath]R_2[/imath] is a finitely generated module over [imath]S[/imath], then [imath]S[/imath] is a finitely generated ring over [imath]R_1[/imath]. I found out in the June edition of The American Mathematical monthly that if we remove the condition that "[imath]R_2[/imath] is a finitely generated module over [imath]S[/imath]", then there exists examples where [imath]S[/imath] might not be a finitely generated ring over [imath]R_1[/imath] and they construct an example using heavy machineries, I was thinking whether a 'simpler' example could be constructed. And Also if one could provide a general algorithm on constructing such rings, it will be very much appreciable. | 105164 | Are intermediate rings of finitely generated ring extensions also finitely generated?
It's well known that if [imath]K[/imath] is a finitely generated extension of some field [imath]E[/imath], then any intermediate field [imath]F[/imath], [imath]E\subseteq F\subseteq K[/imath], is also finitely generated over [imath]E[/imath]. I'm curious, does the same hold for rings? Say [imath]S[/imath] is a ring, and [imath]R\supset S[/imath] is a finitely generated extension of [imath]S[/imath]. If [imath]T[/imath] is any intermediate ring, is it necessarily true that [imath]T[/imath] is finitely generated over [imath]S[/imath] as a ring? Is it as simple as saying that for any [imath]t\in T[/imath], [imath]T[/imath] can be generated by the generators of [imath]R[/imath] over [imath]S[/imath]? I feel unsure about this statement, since it's not clear to me that the generators of [imath]R[/imath] over [imath]S[/imath] need be in [imath]T[/imath]. If not, what is an example what shows otherwise? Thanks. |
2330090 | Two nonabelian two dimensional Lie algebras are isomorphic
Let [imath]\mathfrak{g}[/imath] and [imath]\mathfrak{h}[/imath] be two-dimensional non abelian Lie-algebras. Show that they are isomorphic. Suppose that [imath]\{E_1,E_2\}[/imath] and [imath]\{F_1,F_2\}[/imath] be basis for [imath]\mathfrak{g}[/imath] and [imath]\mathfrak{h}[/imath] respectively. Since both [imath]\mathfrak{g}[/imath] and [imath]\mathfrak{h}[/imath] are nonabelian, [imath][E_1,E_2] \ne 0[/imath] and [imath][F_1,F_2] \ne 0[/imath]. Suppose that [imath][E_1,E_2]=aE_1+bE_2[/imath]. If [imath]a=0[/imath], then dividing by nonzero [imath]b[/imath] we have [imath][E_1,\frac{1}{b}E_2]=E_2[/imath]. If [imath]a \ne 0[/imath], then dividing by [imath]a[/imath] throughout we have [imath][E_1,\frac{1}{a}E_2]=E_1+\frac{b}{a}E_2[/imath]. Then [imath][E_1+\frac{b}{a}E_2-\frac{b}{a}E_2,\frac{1}{a}E_2]=[E_1+\frac{b}{a}E_2,\frac{1}{a}E_2]=E_1+\frac{b}{a}E_2[/imath]. So to start with, we can choose [imath]\{E_1,E_2\}[/imath] and [imath]\{F_1,F_2\}[/imath] such that [imath][E_1,E_2]=E_1[/imath] and [imath][F_1,F_2]=F_1[/imath]. Now define [imath]\phi: \mathfrak{g} \to \mathfrak{h}[/imath] by [imath]\phi(aE_1+bE_2)=aF_1+bF_2[/imath]. Then [imath]\phi[/imath] is linear, one-to-one and onto. Moreover, we have [imath]\phi([E_1,E_2])=\phi(E_1)=F_1=[\phi(E_1),\phi(E_2)][/imath]. Thus, [imath]\phi[/imath] is an isomorphism. I was wondering if I could do this more directy, without any assumption on the Lie brackets of both sides. If I do that and define [imath]\phi[/imath] as above, then I just have to show the nice behavior of [imath]\phi[/imath] with the Lie brackets. Now [imath]\phi([E_1,E_2])=aF_1+bF_2[/imath] but I don't know how to show that [imath][F_1,F_2]=aF_1+bF_2[/imath]. Thanks for the help!! | 24601 | Classsifying 1- and 2- dimensional Algebras, up to Isomorphism
I am trying to find all 1- or 2- dimensional Lie Algebras "a" up to isomorphism. This is what I have so far: If a is 1-dimensional, then every vector (and therefore every tangent vector field) is of the form [imath]cX[/imath]. Then , by anti-symmetry, and bilinearity: [imath][X,cX]=c[X,X]= -c[X,X]==0[/imath] I think this forces a unique Lie algebra because Lie algebra isomorphisms preserve the bracket. I also know Reals [imath]\mathbb{R}[/imath] are the only 1-dimensional Lie group, so its Lie algebra ([imath]\mathbb{R}[/imath] also) is also 1-dimensional. How can I show that every other 1-dimensional algebra is isomorphic to this one? Do I use preservation of bracket? For 2 dimensions, I am trying to use the fact that the dimension of the Lie algebra g of a group [imath]G[/imath] is the same as the dimension of the ambient group/manifold [imath]G[/imath]. I know that all surfaces (i.e., groups of dimension 2) can be classified as products of spheres and Tori, and I think the only 2-dimensional Lie group is [imath]S^1\times S^1[/imath], but I am not sure every Lie algebra can be realized as the Lie algebra of a Lie group ( I think this is true in the finite-dimensional case, but I am not sure). I know there is a result out there that I cannot yet prove that all 1- and 2-dimensional Lie algebras are isomorphic to Lie subalgebras of [imath]GL(2,\mathbb{R})[/imath] (using matrix multiplication, of course); would someone suggest how to show this last? Thanks. |
2331998 | Find expected value of balls sampling
I have a difficulty with understanding following example: We have [imath]N[/imath] balls amongs them there are [imath]b[/imath] white colored. We take from them [imath]n[/imath] balls(we take them without replecments). Let [imath]X[/imath] be a number of white balls we get. Calculate [imath]EX[/imath]. Now lets have [imath]X=\sum^{N} X_i[/imath] where [imath]X_i[/imath] is [imath]1[/imath] if we get white ball in [imath]i-th[/imath] sampling and [imath]0[/imath] otherwise. Then we calculte [imath]E X_i=P(X_i=1)=\frac{b}{N}[/imath] And so [imath]E X=\frac{n\cdot b}{N}[/imath]. I can agree that if we take one ball probability will be indeed [imath]\frac{b}{N}[/imath], but if we take another there will be one ball less, intuitively that should affect probability. | 1657367 | Intuitive reason why sampling without replacement doesnt change expectation?
Suppose we have [imath]10[/imath] red balls and [imath]10[/imath] blue balls. We pick [imath]5[/imath] balls. What is the expected number of red balls we have? Let [imath]X_i[/imath] be the event that the [imath]i[/imath]th ball is red. Then the answer is [imath]E[\sum _1 ^5 X_i]= \sum _1 ^5 E[x_i][/imath]. Buy why are all the [imath]E[X_i]=\frac12[/imath]? Obviously I can verify this for [imath]X_1[/imath] and the computation works out for [imath]X_2[/imath], but it is not intuitive to me at all why this is true for all [imath]X_i[/imath], especially when it seems to me like [imath]X_i[/imath] is dependent on all the [imath]X_j[/imath]s before it- what is the intuitive reason for why this computation works out so nicely? |
2332418 | [imath]f(x)=0[/imath] a.e. on [imath][a,b][/imath] implies [imath]f(x)=0[/imath] everywhere on [imath][a,b][/imath]
If [imath]f(x)[/imath] is a real continuous function that is zero almost everywhere on the interval [imath][a,b][/imath], how can I prove that it is zero everywhere in that interval? My apologies for the probably silly question. | 543648 | If two continuous functions are equal almost everywhere on [imath][a,b][/imath], then they are equal everywhere on [imath][a,b][/imath]
Suppose [imath]f[/imath] and [imath]g[/imath] are continuous functions on [imath][a,b][/imath]. Show that if [imath]f=g[/imath] almost everywhere on [imath][a,b][/imath], then, in fact, [imath]f=g[/imath] on [imath][a,b][/imath]. Is a similar assertion true if [imath][a,b][/imath] is replaced by a general measurable set [imath]E[/imath]? I have known that the set [imath]A=\{x \mid f(x) \neq g(x)\}[/imath] has measure zero and we want to show that A is empty. Now let's assume [imath]A[/imath] is not empty. I am stuck in getting the contradiction. Thanks for your hints and answers. |
2332986 | Pointwise convergence of [imath]h_{n}(x)=x^{1+\frac{1}{2n-1}}[/imath]
The following statement appeared in my textbook: Consider [imath]h_{n}(x)=x^{1+\frac{1}{2n-1}}[/imath] on the set [imath][-1, 1][/imath], for a fixed [imath]x \in [-1, 1][/imath], we have [imath]\lim_{n \rightarrow \infty} h_n(x) = |x|[/imath]. Can someone explain why [imath]\lim_{n \rightarrow \infty} h_{n}(x) = |x|[/imath]? Specifically, where did the absolute value sign come from? What limit laws/rules are used to obtain this result? I would have thought [imath]\lim_{n \rightarrow \infty} h_n(x) = x \lim_{n \rightarrow \infty} x^{\frac{1}{2n-1}} = x[/imath]? EDIT: I had a look at the other question suggested, but my question is different because I want to know the exact limit laws used in each step of the justification that [imath]\lim_{n \rightarrow \infty} h_n(x) = |x|[/imath]. | 802971 | Pointwise convergence of [imath]h_n(x) = x^{1+\frac{1}{2n-1}}[/imath]
Okay, maybe I'm missing something obvious but consider [imath]h_n(x) = x^{ 1+ \frac{1}{2n-1}}[/imath] on the set [imath][-1,1][/imath]. Now consider a fixed [imath]x \in [-1,1][/imath], we have: [imath]\lim_{n\rightarrow \infty} h_n(x) = x \lim_{n \rightarrow \infty} x^\frac{1}{2n-1} = x[/imath] But my book says it's [imath]|x|[/imath]? Why? |
2332500 | Examples for a K-module with K ring, but not an integral domain?
I am looking for an example for such a module, because I want to find a module for which the set of torsion elements is not a submodule of [imath]M[/imath]. | 1939173 | Torsion elements do not form a submodule.
Clearly if [imath]A[/imath] is a commutative domain then the torsion elements of an [imath]A[/imath]-module form a submodule. I'm having trouble finding an example of an [imath]A[/imath]-module such that [imath]A[/imath] is a noncommutative domain and the torsion elements of the module do not form a submodule. Is anyone aware of a good example of this? Thanks in advance |
2332502 | How to find the length between two points or a whole graph of a function?
You are given [imath]f(x)[/imath], where [imath]f(x)[/imath] is a function such as [imath]3x^4-6x^2-4x+2[/imath]. You are asked to find the length of the curve between two points [imath](a,f(a)) \text{ and } (b,f(b))[/imath]. How would I do this for: Polynomials [imath]3x^2, 4x^3-5x^2-x+2[/imath] Exponentials [imath]e^x, 5^x[/imath] Rationals [imath]\dfrac{x+1}{x},\dfrac{3x^2-5x+2}{x^2+2x+2}[/imath] Logarithmic [imath]\ln x, \log x^3[/imath] Trigonometric [imath]4\sin x, 5\cos x[/imath] | 922098 | Arc length of general polynomial
I am coding a computer program and I am in need of calculating the arc length of a curve described by a polynomial [imath]P(x)=ax^n+bx^{n-1}+\dots+cx+d[/imath] within [imath][x_0, x_1][/imath], where [imath]n\ge2[/imath]. Fast methods for [imath]P'(x)[/imath] and [imath]\int P(x) \, dx[/imath] calculations are available and can be used at will. I have tried expanding [imath]\displaystyle\int_{x_0}^{x_1}\sqrt{1+(P'(x))^2} \, dx[/imath] without success. However, it can be used any other formula or method. The big issue here is getting a result - or approximation of it - fast. |
2333095 | Can a 3x3 Matrix in the Integers with Determinant 2 have this product with a 3x1 Matrix?
I've been doing some old Linear Algebra papers that my university library had on file just for some practise and came across this question which caused many a debate among my friends. True or False: There exists a Matrix A [imath]\in \mathbb{M}_{33}(\mathbb{Z})[/imath] with determinant 2 such that: [imath] A \left[ {\begin{array}{cc} 2 \\ 1 \\ 4 \end{array} } \right] = \left[ {\begin{array}{cc} 4 \\ -8 \\ 16 \end{array} } \right][/imath] I started by writing out each of the [imath][a]_{ij}[/imath] entries and multiplying it out to look for a contradiction but that didn't get me very far. I then thought maybe since the determinant is 2 it may have something to do with the fact that [imath]A^{-1}[/imath] can be written as [imath]\frac{1}{det(A)}B[/imath] for some matrix B but a concrete answer still eludes me. A pointer in the right direction would be great! Thanks! | 2330869 | Condition on a matrix so that its determinant is [imath]2[/imath]
Is the following statement true or false? There exists [imath]A \in M_{3,3}(\mathbb{Z})[/imath] with determinant [imath]2[/imath] such that [imath]A\begin{pmatrix}2\\1\\4\end{pmatrix} = \begin{pmatrix}4\\-8\\16\end{pmatrix}[/imath] I first thought of eigenvalues but it doesn't look like the second vector is a multiple of the first. What are the conditions on [imath]A[/imath] so that it's "true"? Or is it always false? |
2333191 | If [imath]|f(x)-f(y)|<(x-y)^2[/imath] for all [imath]x,y \in \mathbb{R}[/imath]. Then [imath]f(x)[/imath] is constant.
If [imath]|f(x)-f(y)|<(x-y)^2[/imath] for all [imath]x,y \in \mathbb{R}[/imath]. Then [imath]f(x)[/imath] is constant. [imath]\bf{Attempt}[/imath] Put [imath]x=y+h[/imath] where [imath]h \rightarrow 0[/imath] Then [imath]\displaystyle |f(y+h)-f(y)|<(y+h-y)^2 = h^2[/imath] So [imath]\displaystyle \displaystyle \lim_{h\rightarrow 0}\bigg|\frac{f(y+h)-f(y)}{h}\bigg|<\lim_{h\rightarrow 0}h[/imath] So [imath]\displaystyle |f'(y)|<0[/imath] Now how can I calculate [imath]f(x)[/imath]? Could someone help me? Thanks | 164804 | Show [imath]f[/imath] is constant if [imath]|f(x)-f(y)|\leq (x-y)^2[/imath].
Problem: Let [imath]f[/imath] be defined for all real [imath]x[/imath], and suppose that [imath]|f(x)-f(y)|\le (x-y)^2[/imath] for all real [imath]x[/imath] and [imath]y[/imath]. Prove [imath]f[/imath] is constant. Source: W. Rudin, Principles of Mathematical Analysis, Chapter 5, exercise 1. |
2333341 | Find formula for solution of differential equation
When \begin{eqnarray*} y'[x] &=& \frac{r y[x] }{ x } \; \;\text{and} \;\; y[1] = a \\ y[x] &=& a x^r \end{eqnarray*} Use derivative formulas to explain the output. What I have so far: We know that \begin{eqnarray*} y'[x] = r y[x] = k e^{rx} \end{eqnarray*} So we can rewrite as \begin{eqnarray*} y[x] = \frac{k e^{rx} } { x} \end{eqnarray*} To get the value of k: \begin{eqnarray*} a = \frac{k e^{r} }{ 1} \\ k = \frac{a}{e^r} \end{eqnarray*} So now we have \begin{eqnarray*} \frac{(a/(e^r)) e^{rx} } { x} \end{eqnarray*} But when I plug this into a derivative calculator I get a solution much different than [imath]y[x] = a x^r[/imath]. Perhaps I am approaching this wrong. What is the correct way to approach this problem? | 2333372 | How does [imath]y[x] = a x^r[/imath] when [imath]y'[x] = \frac{(r y[x])}{x}[/imath] and [imath]y[1] = a[/imath]
When \begin{eqnarray*} y'[x] &=& \frac{r y[x] }{ x } \; \;\text{and} \;\; y[1] = a \\ y[x] &=& a x^r \end{eqnarray*} Use derivative formulas to explain the output. What I have so far: We know that \begin{eqnarray*} y'[x] = r y[x] = k e^{rx} \end{eqnarray*} So we can rewrite as \begin{eqnarray*} y[x] = \frac{k e^{rx} } { x} \end{eqnarray*} To get the value of k: \begin{eqnarray*} a = \frac{k e^{r} }{ 1} \\ k = \frac{a}{e^r} \end{eqnarray*} So now we have \begin{eqnarray*} \frac{(a/(e^r)) e^{rx} } { x} \end{eqnarray*} But when I plug this into a derivative calculator I get a solution much different than [imath]y[x] = a x^r[/imath]. Perhaps I am approaching this wrong. What is the correct way to approach this problem using derivative formulas? |
2333924 | Provide a [imath]2[/imath]-tensor which is not decomposable on [imath]\mathbb R^4[/imath]
Definitions: A function [imath]f:V^k\to \mathbb R[/imath] is called a [imath]k[/imath]-tensor if it takes [imath]k[/imath] members of a vector space [imath]V[/imath], as the input, and is linear with respect to each of the input elements. For example, [imath]f(\alpha v_1,\dots,v_k)=\alpha f(v_1,\dots,v_k)[/imath] and [imath]f(v_1+v',\dots,v_k)=f(v_1,\dots,v_k)+f(v',\dots,v_k)[/imath] [imath]L^k(V)[/imath] denotes the set of all [imath]k[/imath]-tensors on the vector space [imath]V[/imath]. Assume that [imath]V[/imath] is a vector space. We say [imath]T \in L^2(V)[/imath] is decomposable if there exist [imath]A,B\in L^1(V)[/imath] such that [imath]T=A \otimes B[/imath]. Question: On the vector space [imath]\mathbb R^4[/imath], Provide a [imath]2[/imath]-tensor which is not decomposable and prove your claim. Note: My problem is that i don't know a way of finding such a [imath]2[/imath]-tensor. If the question was about finding a decomposable [imath]2[/imath]-tensor, I could just find two tensors and the product of them would be the answer. But it's not the case... So, I'm stuck on this... Any idea? | 2314697 | Showing a tensor is not a simple tensor
Let [imath](e_1,e_2)[/imath] be the standard basis of [imath]\mathbb{R}^2[/imath]. Why can the element [imath]e_1\otimes e_2 + e_2\otimes e_1[/imath] in [imath] \mathbb{R}^2\otimes_{\mathbb{R}} \mathbb{R}^2 [/imath] not be written as a simple tensor [imath]v\otimes w[/imath] for [imath]v,w\in \mathbb{R}^2[/imath]? If we suppose [imath]e_1\otimes e_2 + e_2\otimes e_1=v\otimes w[/imath], then can we conclude that [imath]e_1=\lambda e_2[/imath] for some [imath]\lambda \in \mathbb{R}[/imath]? This would give a contradiction since [imath]e_1[/imath] and [imath]e_2[/imath] are linearly independent. However, I am not sure if the argument is valid. |
2334039 | Maximal and Prime Ideals in a finite Ring of Functions
I was trying to solve a question asked in some PhD entrance exam but I got stuck in some options. The question is: Let [imath]R[/imath] be the set of all functions from the set [imath]\{1,2,...,10\}[/imath] to [imath]\mathbb{Z}_2[/imath]. Then [imath]R[/imath] is a commutative ring with pointwise addition and pointwise multiplication of functions. Then which of the following statements are true? [imath]R[/imath] has a unique maximal ideal. Every prime ideal of [imath]R[/imath] is maximal. Number of Proper ideals of [imath]R[/imath] is [imath]511[/imath]. Every element of [imath]R[/imath] is idempotent. Now it is obvious that statement [imath]4[/imath] is correct. And I found that option [imath]1[/imath] is false because for each [imath]a \in X[/imath], the set of all functions which maps [imath]a[/imath] to [imath]0[/imath] forms a maximal ideal (consider the kernel of evaluation at [imath]a[/imath], i.e. the kernel of the map [imath]T[/imath] from [imath]R[/imath] to [imath]\mathbb{Z}_2[/imath] sending [imath]f[/imath] to [imath]f(a)[/imath]. From here we get that [imath]R/\ker T[/imath] is isomorphic to [imath]\mathbb{Z}_2[/imath] and hence the ideal will be maximal). So, we get at least [imath]10[/imath] maximal ideals. But I could not get any hint in proving or disproving options [imath]2[/imath] and [imath]3[/imath]. Any help would be appreciated. | 2328376 | Ring of [imath]\mathbb{Z}_2[/imath]-valued functions
Let [imath]R =\{f:\{1,2,3,4,\cdots ,10\}\to \mathbb{Z}_2 \}[/imath] be the set of all [imath]\mathbb{Z}_2[/imath] -valued functions on the set [imath]\{1,2,3......10\}[/imath] of the first ten positive integer. Then [imath]R[/imath] is a commutative ring with point-wise addition and multiplication of functions. Which of the following is true? 1) [imath]R[/imath] has a unique maximal ideal. 2) Every prime ideal of [imath]R[/imath] is also maximal. 3 ) The number of proper ideals of [imath]R[/imath] is [imath]511[/imath]. 4 ) Every element of [imath]R[/imath] is idempotent. R is finite commutative ring so every prime ideal is maximlal ideal ,so I think option 2 is true .But for the other options I am helpless . |
2333882 | [imath]3[/imath] does not divide [imath]|G| \implies \forall \ g \in G, g=h^3[/imath] , for some [imath]h[/imath] in [imath]G[/imath]
If [imath]G[/imath] be a finite group such that [imath]3[/imath] does not divide the order of [imath]G[/imath]. Then show that every element of [imath]G[/imath] can be expressed as [imath]h^3[/imath] , for some [imath]h \in G[/imath]. Thought: I have just figured out that if [imath]g[/imath] has order [imath]3[/imath] then [imath]g= h^3[/imath] is not possible. But nothing else. | 1891398 | [imath]|a|=m,\,\gcd(m,n)=1 \implies a[/imath] is an [imath]n[/imath]'th power
Let [imath]a[/imath] belong to a group and [imath]|a|=m[/imath]. If [imath]n[/imath] is relatively prime to [imath]m[/imath] show that [imath]a[/imath] can be written as the [imath]n^{th}[/imath] power of some element in the group. We need to show that if [imath]a\in G[/imath] and [imath]a^m=e\implies \exists \ b\in G [/imath] such that [imath]b^n=a[/imath] i.e.[imath]b^{nm}=e=({b^m})^n[/imath] i.e. [imath]\exists[/imath] an element of order [imath]mn[/imath] in [imath]G[/imath]. How to show this? |
2332690 | Finding [imath]f(x)[/imath] from integral equation
I have this problem: Find a contnuous function [imath]f[/imath] such that: [imath]\int_0^x f(t)\,dt = (x-1)e^{2x} + \int_0^x e^{-t} f(t)\,dt[/imath] and it's required to find [imath]f(x)[/imath] so [imath] \int_0^x f(t)\,dt - \int_0^x e^{-t} f(t)\,dt= (x-1)e^{2x} [/imath] here I differentiated both sides and factor [imath]f(x)[/imath] on the left [imath] f(x)\cdot (1-e^{-x})= 2(x-1)e^{2x}+e^{2x}[/imath] [imath]f(x)= \frac{2(x-1)e^{2x}+e^{2x}}{(1-e^{-x})}[/imath] but the answer appointed is [imath]f(x)= \frac{(1-2x)e^{2x}}{(e^{-x}-1)}[/imath] did I mess up somewhere or am I lacking some algebra work here? here is a similar question alredy answered Finding [imath]f(x)[/imath] from integral equations | 2332912 | Deduce function f from integral
I do not understand how to perform this exercise: If [imath]f[/imath] is a continuous function such that:[imath]\int_0^xf(t)\ dt=(x-1)e^{2x}+\int_0^xe^{-t}f(t) \ dt[/imath] For all [imath]x[/imath], what function [imath]f[/imath] results? |
2333618 | Circle inscribed in spherical triangle
Let [imath]u,v,w \in S^2[/imath] and let [imath][u,v,w][/imath] be a spherical triangle. I want to find its inscribed circle. However, I don't know how to approach this problem. Would someone please explain? In my understanding, one needs to somehow find the altitudes of the spherical triangle and then the point of its intersection (circumcentre)? Then what about the radius? | 2333613 | Finding circle circumscribed in spherical triangle
Given three vectors [imath]u,v,w \in S^2[/imath] and the triangle [imath][u,v,w][/imath] I want to find its circumscribed circle. However, I don't know how to approach this problem. Would some one please explain? In my understanding, one needs to somehow find the altitudes of the spherical triangle and then the point of its intersection (circumcentre)? Then what about the radius? |
2334305 | Prove [imath]F(x)=1-e^{- \lambda x}, 0 \lt x.[/imath]
Let [imath]F[/imath] be the distribution function of the continuous-type random variable [imath]X[/imath], and assume that [imath]F(x)=0[/imath] for [imath]x \le 0[/imath] and [imath]0 \lt F(x) \lt1[/imath] for [imath]0 \lt x[/imath]. Prove that if [imath]P(X \gt x+y |X \gt x)=P(X \gt y),[/imath] then [imath]F(x)=1-e^{- \lambda x}, 0 \lt x.[/imath] My attempt: Let [imath]F(x)=P(X \le x)[/imath]. Then [imath] \frac {P(X \gt x+y)}{P(X \gt x)}=P(X \gt y)[/imath] [imath]\Rightarrow \frac {1-P(X\le x+y)}{1-P(X \le x)}=1-p(X \le y)[/imath] [imath] \Rightarrow \frac {1-F(x+y)}{1-F(x)}=1-F(y)[/imath] Then I have no idea where to go from here. Any help is appreciated. | 107075 | Reliability function, proving exponential distribution
We are given [imath]R(t)[/imath] = [imath]P(X>t)[/imath] for all [imath]x > 0[/imath] and [imath]R(0) = 1 - Fx(0) = 1\text{ and }\lim\limits_{t \to \infty} R(t) = 0[/imath] The random variable [imath]X[/imath] also satisfies the memoryless property: [imath]P(X>s+t|X>t) = P(X>s)\text{ for }s>0\text{ and }t>0[/imath] Let [imath]R'(0) = - \lambda\ where \ \lambda\ ,[/imath] is some positive constant. I need to show that X must be exponentially distributed. Given that [imath]\dfrac{R(t + h) - R(t)}{h}[/imath] = [imath]R(t)\left[\dfrac{R(h) - 1}{h}\right][/imath] Show that by letting [imath]\lim\limits_{h\to \infty}[/imath] $\dfrac{dR(t)}{dt} = -\lambda[imath]R(t)$ (I think we should use Hopital's rule here I am not sure by differentiating $\left[\dfrac{R(h) - 1}{h}\right]$ and letting $h$ tend to $0$, we will get $-\lambda$ for this part but I got stuck afterwards).[/imath] Also argue that $X$ is an exponential random variable with rate parameter $\lambda$ by solving the differential equation above respecting the conditions: R(0) = 1 - Fx(0) = 1\text{ and }\lim\limits_{t \to \infty} R(t) = 0$$ |
2334414 | Proving [imath]\mathbb{Q}[/imath] not isomorphic to [imath]\mathbb{R}^{+}[/imath]
Question: Prove that [imath]\mathbb{Q}[/imath] under addition is not isomorphic to [imath]\mathbb{R^{+}}[/imath] under multiplication. I wonder what is a good way to approach this? Would constructing a map be more helpful? Thank in advance. Hints only* | 1203841 | Showing [imath](\mathbb{Q},+)[/imath] is not isomorphic to [imath](\mathbb{R},+)[/imath]
To show: [imath](\mathbb{Q},+)[/imath] is not isomorphic to [imath](\mathbb{R},+)[/imath] Now, the equation [imath]x^{2} =3[/imath] has a solution in [imath]\mathbb{R}[/imath], but not in [imath]\mathbb{Q}[/imath]. Hence they are not isomorphic to each other. Is that right, or do I need to prove something else? Thanks. |
2334980 | Special polynomials in [imath]\mathbb{Z}[X][/imath] have no roots in [imath]\mathbb{Z}[/imath] and [imath]\mathbb{Q}[/imath]
Given is a polynomial [imath]f \in \mathbb Z [X][/imath], such that [imath]2\nmid f(0)[/imath] and [imath]2\nmid f(1)[/imath]. Show: (I) [imath]f[/imath] has no roots in [imath]\mathbb Z[/imath]. (II) If the leading coefficient of [imath]f[/imath] is odd, [imath]f[/imath] has no roots in [imath]\mathbb Q[/imath]. My idea: [imath]f=a_nX^n+a_{n-1}X^{n-1}+...+a_1X+a_0 \in \mathbb Z[X][/imath]. I know that the constant term [imath]a_0[/imath] exists, because of [imath]2 \nmid f(0)[/imath]. (I) [imath]2\nmid f(0)[/imath] leads to [imath]2 \nmid a_0[/imath], which means, that [imath]a_0[/imath] is odd. [imath]2\nmid f(1)[/imath] leads to [imath]a_n+...+a_0[/imath] is odd. Now I can conclude, that [imath]a_n+...+a_1[/imath] is even, so [imath]2 \mid a_n+...+a_1[/imath]. But now I don't know how to go on. I thought about assuming, that f has at least one root, i.e. [imath]z[/imath]. Than [imath]f[/imath] factors into [imath]f(X)=g(X)\cdot(X-z)[/imath], where [imath]g \in \mathbb Z [X][/imath] with [imath]\mbox{deg}(g)+1= \mbox{deg}(f)[/imath]. Since [imath]a_0[/imath] is odd, [imath]z[/imath] and the constant term of [imath]g[/imath] are also odd. | 1280561 | How to show [imath]f(x)[/imath] has no root within [imath]\Bbb Q[/imath]
A polynomial problem from my old algebra textbook: [imath]f(x)\in\Bbb Z[x][/imath] with leading coefficient [imath]1[/imath], [imath]\deg f(x)\ge 1[/imath], and both [imath]f(0)[/imath] and [imath]f(1)[/imath] are odd numbers, prove: [imath]f(x)[/imath] has no root within [imath]\Bbb Q[/imath] Eisenstein's criterion seems to be of little help here because we know virtually nothing about the coefficients. So I tried some other approach. Let [imath]f(x)=a_0+a_1x+a_2x^2+\cdots+a_{n-1}x^{n-1}+x^n\in \Bbb{Z}[x][/imath] What we have known so far is [imath]f(0)=a_0\text{ is an odd number}[/imath] and [imath]f(1)=a_0+a_1+a_2+\cdots+a_{n-1}+1 \text{ is an odd number}[/imath] from which we can also conclude that [imath]f(1)-f(0)=a_1+a_2+\cdots+a_{n-1}+1\text{ is an even number}[/imath] To be honest I don't know how to proceed now. It isn't until just now that I realized "irreduicible" is not equivalent to "has no roots".. So all my previous attempt is fundamentally wrong.. Can anybody help or drop a hint? Best regards! |
2335024 | What is the difference between equality and congruence outside of geometry?
Yes, I have seen that this question has been asked and answered before in this same website, but answers given there were mostly in regards to geometry, or non-mathematical examples (e.g. "the 'e's in the word 'between' are congruent, but not equal", "two triangles with the same dimensions and points are equal"). I can understand this just fine, but I can't use this advice when it comes to pure numbers, like mods. For example: [imath]17 \equiv 5 (mod(6))[/imath]. How is this statement correct? If we solve [imath]5 (mod(6))[/imath] we get [imath]5[/imath], so that would mean [imath]17 \equiv 5[/imath]. Also, [imath]17 \equiv 4(mod(13))[/imath], which means [imath]17 \equiv 4[/imath], and [imath]17 \equiv 3(mod(7))[/imath], which is [imath]17 \equiv 3[/imath] So then, [imath]17 \equiv 5 \equiv 4 \equiv 3[/imath] is true? That doesn't seem right. Are all positive integers congruent to one another? I know this may seem like a simple matter to some, but I'm seriously stuck. | 328053 | Congruent Modulo [imath]n[/imath]: definition
In an Introduction to Abstract Algebra by Thomas Whitelaw, he gives examples of the congruence mod operation, such as [imath]13 \equiv5 \pmod4[/imath], and [imath]9 \equiv -1 \pmod 5[/imath]. But when I first learned about the modulo operation my junior year, I would have told you that [imath]13 \equiv 1 \pmod 4[/imath], and that [imath]9 \equiv 4 \pmod 5[/imath]. Is this just a difference in the definition of modulo? Or is this pretty typical (to not take it to the largest factor, or one over)? |
2334150 | Is an injective ring morphism always a monomorphism?
My question is about rings. I know there exists epimorphisms that are not surjective (e.g. the inclusion of [imath]\mathbb{Z}[/imath] in [imath]\mathbb{Q}[/imath]), but I want to know if there is an equivalence (on rings, more exactly, commutative rings with unity) between monomorphisms and injective homomorphisms. If there's that equivalence, would you provide an elementary proof? Or some book where I can find them? | 183337 | Are monomorphisms of rings injective?
Let [imath]R[/imath] and [imath]S[/imath] be rings and [imath]f:R\to S[/imath] a monomorphism. Is [imath]f[/imath] injective? |
2334528 | Prove a sequence [imath](a_n)_{n \geq 1}[/imath] is convergent
Given a sequence [imath](a_n)_{n \geq 1}[/imath] in [imath]\mathbb{R}[/imath], and given that [imath] a_1 > 0 \ \ \ \ \text{and} \ \ \ \ \ a_{n+1} = \sqrt{a_n + 6}\ \ \ (n\geq1)[/imath] Prove that [imath](a_n)_{n \geq 1}[/imath] converges. My thoughts for [imath]a_1 > 3[/imath], we know that [imath]3<a_{n+1} < a_n \ \ (n\in \mathbb{N})[/imath] (1) for [imath]a_1 < 3[/imath], we know that [imath]0<a_{n} < a_{n+1}<3 \ \ (n\in \mathbb{N})[/imath] (2) Now I now how to prove convergence. However, I'm having difficulty proving my two thoughts about the situation where [imath]a_1 >3[/imath] and [imath]a_1 < 3[/imath]. Any help on how I can show this correctly? | 1187550 | [imath]\{a_n\}[/imath] sequence [imath]a_1=\sqrt{6}[/imath] for [imath]n \geq 1[/imath] and [imath]a_{n+1}=\sqrt{6+a_n}[/imath] show it that convergence and also find [imath]\lim_{x \to \infty} \{a_n\}[/imath]
[imath]\{a_n\}[/imath] sequence [imath]a_1=\sqrt{6}[/imath] for [imath]n \geq 1[/imath] and [imath]a_{n+1}=\sqrt{6+a_n}[/imath] show that it convergence and as well find [imath]\lim \limits_{n \to \infty} a_n[/imath] In order to show that that sequence convergence I need to show that : [imath]\lim_{n \to \infty} a_n= L[/imath] While [imath]L[/imath] is finite. Using the calculator. I assume that L=3 because : [imath]\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6........}}}}=2.999 \cong 3[/imath] I really don't think that this method is good enough to established that [imath]\lim \limits_{n \to \infty} a_n= 3[/imath] since it based on intuition. I'll be glad to hear any ideas for an established method to show this.? Any help will be appreciated. |
2335271 | Given two vertices and incenter find third vertex
[imath]u=(3,1)[/imath], [imath]v=(3,6)[/imath], [imath]i=(2,3)[/imath], need find [imath]w\in \mathbb{R}^2[/imath] such that [imath]i[/imath] is the incenter of the triangle [imath][u,v,w][/imath]. I have looked at some other people's approaches to this problem, but I think those approaches are too complicated, and are more like for some advanced intuitive geometer. With the fact that I'm just taking a course in geometry, my approach is probably not so elegant, but it should work. The problem is that I'm still not getting the correct answer. Let [imath]\alpha, \beta, \gamma[/imath] be the three angles of [imath][u,v,w][/imath]. Then bisect the three angles to assume [imath]\alpha/2, \beta/2, \gamma/2[/imath]. We can find that [imath]a=(3,3)[/imath] is the point closest to [imath]i[/imath] on the edge [imath][u,v][/imath]. We can now set up the following equations: [imath]\alpha/2 = \arccos\left(\frac{(v-u)\cdot (v-a)}{\|v-u\|\|v-a\|}\right)=\arccos\left(\frac3{\sqrt{10}}\right)\implies \alpha = 2\arccos\left(\frac3{\sqrt{10}}\right)[/imath] [imath]\gamma/2 = \arccos\left(\frac{(v-w)\cdot (w-u)}{\|v-w\|\|w-u\|}\right)=\arccos\left(-\frac{9}{10}\right)\implies \gamma = 2\arccos\left(-\frac9{10}\right)[/imath] [imath]\beta = \pi-\alpha-\beta[/imath] [imath] \cos\alpha = \frac{(v-u)\cdot (w-u)}{\|v-u\|\|w-u\|} [/imath] Can someone please suggest what is wrong with my approach? Is there an additional equation missing? Also, I'm getting some "crazy" expression for [imath]\beta[/imath], which even WolframAlpha finds difficult to deal with. Is this geometric problem really that complicated? | 2299476 | Given vectors [imath]u,v, i[/imath], find vector [imath]w[/imath] such that [imath]i[/imath] is incenter of triangle [imath][u,v,w][/imath]
Given [imath]u=(3,1)[/imath], [imath]v=(3,6)[/imath], [imath]i=(2,3)[/imath], find [imath]w\in \mathbb{R}^2[/imath] such that [imath]i[/imath] is the incenter of the triangle [imath][u,v,w][/imath]. To my understanding, in order to come up with the coordinate pair [imath](w_1, w_2)[/imath], one needs to come up with six equations. I was only able to come up with four, so I'm asking to help me find the other two equations. The distances from the incenter [imath]i[/imath] to the three sides of [imath][u,v,w][/imath] are equal, so I find the point [imath]r[/imath] on the line [imath][u,v][/imath] as follows: [imath](v-u)\cdot (i-r)=0\iff (0,5)\cdot (2-r_1, 3-r_2)=0[/imath] [imath]\implies 5(3-r_2)=0\implies r_2=3[/imath] [imath]r_1 = 3[/imath], since the difference between the first coordinates of [imath]u[/imath] and [imath]v[/imath] is zero. So we have that [imath]r=(3,3)[/imath], so that [imath]\|i-r\|=1[/imath]. So, now we can come up with the equations: Let [imath]s=(s_1,s_2)[/imath] be the point on [imath][w,u][/imath] and [imath]t=(t_1,t_2)[/imath] be the point on [imath][v,w][/imath]. Now, [imath](w_1-3, w_2-1)\cdot(2-s_1,3-s_2)=0[/imath] Also, [imath](3-w_1)(2-t_1)+(6-w_2)(3-t_2)=0[/imath] (from the other dot product). [imath](2-s_1)^2+(3-s_2)^2=1[/imath] (from the fact that the incenter is equidistant from all sides of [imath][u,v,w][/imath]. [imath](2-t_1)^2+(3-t_2)^2=1[/imath] Now, to find [imath]w=(w_1, w_2)[/imath] precisely, I need two more equations. I would appreciate some hints. Moreover, I think this procedure is rather tedious, and I was wondering if there's a more elegant analytic (or maybe axiomatic) approach to this problem. |
2335710 | How to compute this limit: [imath]\lim_{x\to\infty}x\left(\frac 1e-\left(\frac x{x+1}\right)^x\right)[/imath]?
I tried to compute a limit: [imath]\lim_{x\to\infty}x\left(\frac 1e-\left(\frac x{x+1}\right)^x\right)[/imath] What I've done is that: [imath]\begin{align} &\lim_{x\to\infty}x\left(\frac 1e-\left(\frac x{x+1}\right)^x\right)\\ =&\lim_{x\to\infty}x\left(\frac 1e-\left(\frac x{x+1}\right)^{x+1}\left(\frac{x+1}{x}\right)\right)\\ =&\lim_{x\to\infty}\left(1-\frac 1{x+1}\right)^{x+1}x\left(1-\frac{x+1}{x}\right)\\ =&-\frac1e \end{align}[/imath] But Wolfram|Alpha seems not to agree with me, in which it gave [imath]-\frac1{2e}[/imath] So what actually does the error come from in my approach? And how should I compute the limit? Thanks in advance. | 2332509 | Finding limit when limit approches to infinity
Finding [imath]\displaystyle \lim_{y\rightarrow \infty}\bigg[\frac{y}{e}-y\bigg(\frac{y}{y+1}\bigg)^y\bigg][/imath] Attempt: Put [imath]\displaystyle y=\frac{1}{z}[/imath] . Then [imath]\displaystyle \lim_{z\rightarrow 0}\bigg[\frac{1}{ze}-\frac{1}{z}\bigg(\frac{z}{z+1}\bigg)^{\frac{1}{z}}\bigg][/imath] Could some help me to solve it, thanks |
2335797 | Is the formula [imath]|G|=|\ker \varphi|.|\text{im}\space \varphi|[/imath] connected to the formula [imath]ab = \gcd(a,b) \cdot \operatorname{lcm}(a,b)[/imath]?
I was reading the corollary; Let [imath]\varphi :G\rightarrow G'[/imath] be a homomorphism of finite groups. Then [imath]|G|=|\ker \varphi| \cdot |\text{im}\space \varphi|[/imath] Then I suddenly remember one of my 6-th or 7-th grade formula Let [imath]a,b\in \mathbb{N}[/imath], then [imath]a \cdot b=\gcd (a,b) \cdot\text{lcm} (a,b)[/imath] Are these two result really related!! If so then what is [imath]\varphi[/imath] here? | 144709 | Proof via Group Theory : [imath]\mathrm{lcm}(a,b) \cdot \gcd(a,b) = |ab|[/imath]
Recently, I was informed that we can verify the famous formula about [imath]\mathrm{lcm}(a,b)[/imath] and [imath]\gcd(a,b)[/imath] which is [imath]\mathrm{lcm}(a,b)=\frac{|ab|}{\gcd(a,b)} [/imath] via group theory. The least common multiple of two integers [imath]a[/imath] and [imath]b[/imath], usually denoted by [imath]\mathrm{lcm}(a,b)[/imath], is the smallest positive integer that is a multiple of both [imath]a[/imath] and [imath]b[/imath] and the greatest common divisor ([imath]\gcd[/imath]), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder. I do not know if we can prove this equation by using the groups or not, but if we can I am eager to know the way someone face it. Thanks. |
2336015 | Prove that : [imath] \varsigma (z)=\sum _{n=1}^\infty n^{-z} [/imath] converges in [imath] D=\left\{z:\operatorname{Re}(z) > 1\right\} \ [/imath]
Lets define the following function series (Riemann Zeta Function): [imath]\varsigma (z)=\sum _{n=1}^\infty n^{-z} [/imath] where the power is defined by the primary branch of the Logarithm. A) prove that the series converges in : [imath] D= \left\{ z:\operatorname{Re}(z) > 1 \right\}[/imath] and also converges absolutely and uniformly in [imath]D_\delta = \left\{z:\operatorname{Re}(z)>1+ \delta \right\} [/imath] B) find [imath]\varsigma '(z) [/imath]. Do guys have any hints? I thought about calculating the radius of convergence, but how do I do it in this case? and how do I prvoe that it converges with this minus power? also I didnt get the information about the power being defined that way, im not sure also that I translated this accurately from hebrew. | 1897433 | Convergence of [imath]\zeta(s)[/imath] on [imath]\Re(s)> 1[/imath]
I'm aware there are been several similar questions, and some great answers with hints on how to prove this, for example here and here. But I've never really seen a detailed proof of the absolute convergence of the Riemann zeta function in the half-plane [imath]\Re(s)> 1[/imath]. I hope there's interest on a detailed proof of this for reference. Here is my attempt. Please, fill in any detail that might be missing, or point out any mistake! On [imath]\Re(s)=\sigma> 1[/imath], we have [imath]\sum_{n=1}^{\infty}\bigg|\frac{1}{n^s}\bigg|=\sum_{n=1}^{\infty}\frac{1}{n^\sigma}[/imath] and [imath]\frac{1}{n^\sigma}\leq\frac{1}{n^{1+\epsilon}}[/imath] for any [imath]\epsilon >0[/imath], so by the direct comparison test, if [imath]\sum_{n=1}^{\infty}\frac{1}{n^{1+\epsilon}}[/imath] converges absolutely, then so do [imath]\sum_{n=1}^{\infty}|1/n^s|[/imath] and [imath]\zeta(s)[/imath]. By the integral test, the convergence of [imath]\sum_{n=1}^{\infty}1/n^{1+\epsilon}[/imath] is equivalente to the finitude of the integral [imath]\int_1^\infty \frac{1}{n^{1+\epsilon}}=\bigg[-\frac{1}{\epsilon x^\epsilon}\bigg]=\frac{1}{\epsilon}[/imath] which holds for any [imath]\epsilon < \infty[/imath]. Any correction or improvement is welcomed! Thanks in advance. |
2336278 | [imath]\mathbb{Q}(\sqrt{n}) \cong \mathbb{Q}(\sqrt{m})[/imath] iff [imath]n=m[/imath]
Page 105 of D. Burton's A First Course in Rings and Ideals reads It is not difficult to show that if (I'll call them [imath]n[/imath] and [imath]m[/imath] instead of [imath]n_1[/imath] and [imath]n_2[/imath]) [imath]n[/imath], [imath]m[/imath] are square free integers, then [imath]\mathbb{Q}(\sqrt{n}) \cong \mathbb{Q}(\sqrt{m})[/imath] if and only if [imath]n=m[/imath]. Well, it is getting difficult for me, any help or would be appreciated. My progress so far: Suppose [imath]\phi :\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})[/imath] is an isomorphism. Then [imath]\phi(u)=u[/imath] for every [imath]u \in \mathbb{Q}[/imath], so that [imath]\phi(\sqrt{n})^2=\phi(n)=n[/imath]. If [imath]\phi(\sqrt{n})=a+b\sqrt{m}[/imath] for some [imath]a,b\in\mathbb{Q}[/imath], then [imath](a+b\sqrt{m})^2=a^2+b^2m + 2ab\sqrt{m}=n[/imath] implies [imath]a^2+b^2m = n[/imath] and [imath]2ab=0.[/imath] Then [imath]b=0[/imath] cannot happen since that would imply that [imath]\sqrt{n}\in\mathbb{Q}[/imath], where [imath]n[/imath] is a square-free integer. Also [imath]a=b=0[/imath] cannot happen. Hence [imath]a=0[/imath] and we have [imath]b^2m=n.[/imath] If [imath]\psi:\mathbb{Q}(\sqrt{n})\to\mathbb{Q}(\sqrt{m})[/imath] were an isomorphism then there would exist some [imath]s\in\mathbb{Q}[/imath] such that [imath]s^2m=n,[/imath] then [imath]s=b[/imath] or [imath]s=-b[/imath]. Therefore the only isomorphisms from [imath]\mathbb{Q}(\sqrt{n})[/imath] to [imath]\mathbb{Q}(\sqrt{n})[/imath] are [imath]\phi(u+v\sqrt{n})=u+vs\sqrt{m}[/imath] and [imath]\psi(u+v\sqrt{n})=u-vs\sqrt{m}[/imath]. I don't know if there's an easier way and I don't know if I'm in the correct way here. Thanks beforehand :) | 3013930 | If [imath]p, q[/imath] are prime integers, then [imath]\mathbb{Q}(\sqrt{p})[/imath] is not isomorphic (as a field) to [imath]\mathbb{Q}(\sqrt{q})[/imath]
My strategy is something like this: suppose [imath]\phi: \mathbb{Q}(\sqrt{p}) \to \mathbb{Q}(\sqrt{q})[/imath] is a isomophism such that [imath]\phi(x) = x[/imath] for all [imath]x \in \mathbb{Q}[/imath] and let [imath]\phi(\sqrt{p}) = a + b \sqrt{q}[/imath]. Then [imath]p = \phi(p) = \phi(\sqrt{p}^2) = \phi(\sqrt{p})^2 = (a + b \sqrt{q})^2 = a^2 + 2ab \sqrt{q} + b^2 p[/imath]. I feel like '[imath]\phi(x) = x[/imath] for all [imath]x \in \mathbb{Q}[/imath]' can be proved from the supposition that [imath]\phi: \mathbb{Q}(\sqrt{p}) \to \mathbb{Q}(\sqrt{q})[/imath] is an isomorphism, but I am not sure how to prove it. Can anyone help me? |
2336784 | How many generator has a cyclic group of order n?
I need to find how many generators has a cyclic group [imath]G=<g>[/imath] of order [imath]n[/imath]. I know that I have to prove that if [imath]G[/imath] is a cyclic group with order [imath]n[/imath], then the number of generators of [imath]G[/imath] is [imath]\phi(n)[/imath]. But I don't know how can I prove that. I already know that [imath]<g^k>=<g^{gcd(k,n)}>[/imath], so the generators of [imath]G[/imath] will be [imath]g^k[/imath] where [imath]gcd(k,n)=1[/imath] | 2942205 | How many elements generate the cyclic group [imath]\mathbb{Z}/p^r\mathbb{Z}[/imath]?
I've been searching for an answer to this problem and I haven't found any duplicate, but excuse me if that's the case. I want to find how many elements generate the cyclic group [imath]\mathbb{Z}/p^r\mathbb{Z}[/imath]. I have already proved that for [imath]\mathbb{Z}/p\mathbb{Z}[/imath] there are [imath]\phi(p)=p-1[/imath] generators, because the order of the coprime numbers to [imath]p[/imath] is exactly [imath]p[/imath]. However I am struggling with the cases [imath]\mathbb{Z}/p^r\mathbb{Z}[/imath] and [imath]\mathbb{Z}/p^rq^s\mathbb{Z}[/imath]. I would appreciate a lot if you could help me with this proof. Thanks in advanced. |
2336918 | What is the difference between implication symbols: [imath]\rightarrow[/imath] and [imath]\Rightarrow[/imath]?
I do not understand the difference between [imath]\rightarrow[/imath] and [imath]\Rightarrow[/imath]. Sometimes I see implication truth tables labeled with the former, sometimes with the latter. Aren't they synonyms of logical implication or is there any difference? | 630391 | What does [imath]\rightarrow[/imath] mean in [imath]p \rightarrow q[/imath]
I was looking at an exercise where it asked the following: [imath]\begin{array}{ccc} p&q&p\rightarrow q \\ T&T&T \\ &\ldots \end{array}[/imath] So, for the third column, I just put [imath]T[/imath] which was correct but I didn't understand what [imath]\rightarrow[/imath] meant. I have seen [imath]\implies[/imath] but I haven't the arrow. Are they the same thing? Thanks a bunch! |
2337004 | Does one always have [imath]f_{xy}=f_{yx}[/imath]?
I would like to know if when we take a second partial derivative of a function do we always get: [imath]\frac{\partial^2{u}}{\partial{x} \, \partial{y}} = \frac{\partial{^2u}}{\partial{y} \, \partial x}[/imath] if not, what does it mean about the function if the condition happens or not? | 1075714 | Equality of mixed partial derivatives
Is the following statement [imath]\frac{\partial^2 f}{\partial x \, \partial y}=\frac{\partial^2 f}{\partial y \, \partial x}[/imath] always true? If not what are the conditions for this to be true? |
2337182 | Proof: if [imath]AB=BA[/imath], prove that [imath](AB)^2 = A^2B^2[/imath]
I got a little stuck with this proof. It is given that [imath]AB=BA[/imath]. Proof: [imath](AB)^2 = A^2 \cdot B^2[/imath] I have been thinking of several ways to solve it. I got to this point: [imath](AB)^2 = AB\cdot AB = BA\cdot AB = B\cdot A^2\cdot B[/imath] But I don't know how to proceed. I think I am missing some general rule. Can somebody help me? | 1282046 | If [imath]ab=ba[/imath], Prove [imath]a^2[/imath] commutes with [imath]b^2[/imath]
From Dr. Pinter's "A Book of Abstract Algebra": Given [imath]a[/imath] and [imath]b[/imath] are in [imath]G[/imath] and [imath]ab=ba[/imath], we say that [imath]a[/imath] and [imath]b[/imath] commute. Prove [imath]a^2[/imath] commutes with [imath]b^2[/imath] I tried: [imath] ab=ba [/imath] [imath] a^{2}b^{2}=b^{2}a^{2} \text{ // to prove that $a^2$ commutes with $b^2$} [/imath] [imath] aabb=bbaa [/imath] Then, knowing that [imath]a=b^{-1}ab,[/imath] I substituted that '[imath]a[/imath]' into '[imath]aabb=bbaa[/imath]'. But that simplified to '[imath]aabb=bbaa[/imath]', i.e. I got nowhere with it. How can I prove that [imath]aabb=bbaa[/imath] per this problem? |
1932735 | Constant sectional curvature and the Riemann tensor
I am trying to prove the theorem that: Let [imath](M,g)[/imath] be a Riemannian [imath]n[/imath]-manifold of constant sectional curvature [imath]C[/imath]. Then the curvature tensor is [imath]R(X,Y)Z=C( \langle Y,Z \rangle X - \langle X, Z \rangle Y)[/imath]. I can show by straight-forward calculation (using the definition of sectional curvature) that if the curvature tensor is as stated, then [imath]M[/imath] has constant sectional curvature. But the other way around is giving me some trouble. Does anybody have a hint? | 2222171 | Compute curvature tensor from constant sectional curvature
Given sectional curvature as a constant, i.e. [imath]\dfrac{R_m(X,Y,Y,X)}{|X|^2|Y|^2-<X,Y>^2} = C[/imath], I want to compute the curvature tensors [imath]R(X,Y)Z[/imath] and [imath]R_m(X,Y,Z,W)[/imath]. I believe I need to use the identity [imath]-6R_m(X,Y,Z,W) = \partial_t\partial_s\{R_m(X+sZ, Y+tW, Y+tW, X+sZ) - R_m(X+sW, Y+tZ, Y+tZ, X+sW)\}[/imath] I can almost see what to do for [imath]R_m(X,Y,Z,W)[/imath] and how this identity and the defintion of sectional curvature are related but can't really see how to finish it off. I don't really even know where to begin with [imath]R(X,Y)Z[/imath]. |
2337408 | Short exact sequence of sheaves induces long exact sequence on sections
Given a short exact sequence of sheaves on [imath]X[/imath]: [imath]0 \rightarrow \mathcal{A} \rightarrow \mathcal{B} \rightarrow \mathcal{C} \rightarrow 0[/imath] Then there is an induced exact sequence on the left on global sections: [imath]0 \rightarrow \mathcal{A}(X) \rightarrow \mathcal{B}(X) \rightarrow \mathcal{C}(X)[/imath] I'm studying on Bredon's book, and there the proof of this statement is said to be trivial because "look at stalks". But to me is not so trivial, because with stalks one loses all the global information about a sheaf. So, why is this true? | 318684 | Induced sequence of global sections
I'm reading Differential Analysis on Complex Manifolds by Raymond O. Wells. It states the following in the beginning of section 3 of chapter 2 on page 51: Consider a short exact sequence of sheaves: [imath]0 \to \mathcal{A} \to \mathcal{B} \to \mathcal{C} \to 0[/imath] Then it is easy to verify that the induced sequence [imath]0 \to \mathcal{A}(X) \to \mathcal{B}(X) \to \mathcal{C}(X) \to 0[/imath] is exact at [imath]\mathcal{A}(X)[/imath] and [imath]\mathcal{B}(X)[/imath] but not necessarily at [imath]\mathcal{C}(X)[/imath]. After that they give an example of failing exactness at [imath]\mathcal{C}(X)[/imath]. My problem is that Ii dont see why the induced sequence has to be exact at [imath]\mathcal{A}(X)[/imath] and [imath]\mathcal{B}(X)[/imath]. I tried to find an answer in other books. In the book Sheaf Theory by Bredon they make the same statement in proposition 2.2 but they also state it is easy to verify. I tried to come up with an argument of my own and got something like this (the argument is wrong i think): If we have the zero element [imath]0 \in \mathcal{B}(X)[/imath] this should induce a section [imath]\tilde{0} [/imath] of the stalks of [imath]\mathcal{B}[/imath] and should take the value of the zero element in every stalk. Then since the sequence of sheaves is exact (and thus exact in stalk level) we get that around every [imath]x \in X[/imath] there should be an open [imath]V[/imath] such that [imath]\tilde{0}|_{V}[/imath] is the image of the zero section in [imath]\mathcal{A}(V)[/imath] and since [imath]\mathcal{A}[/imath] is a sheaf we can patch these together to get a global zero section. Also while i get the example that shows inexactness at [imath]\mathcal{C}(X)[/imath] i dont get why this should generally be the case. Any help would be much appreciated |
1694472 | Airline binomial distribution overbooking question
An airline has a plane with 400 seats. The probability that a passenger fails to turn up for their flight is 0.04. The airline has an overbooking policy. Find the largest number of passengers this airline can book and still be at least 85% sure that everyone who shows up has a seat. Working: Need to find a value [imath]x[/imath] when [imath]1 - \text{cumulative probability}[/imath] is at least 0.85 or more, but as close to 0.85 as possible. If we call this [imath]x[/imath]-value [imath]Z[/imath] we need: [imath][1 – P(X = 0) + P(X = 1) +…+ P(X = Z)] [/imath] So [imath]n=400 p=0.04 q=0.96 x=0,1,2,...[/imath] Can't seem to add a picture of the answers but I got it as [imath]411[/imath]. Would anyone be able to verify that this method is correct ? | 2203271 | Binomial Distribution Problem - Airline Overbooking
Here's the problem: A Flight has 20 seats and enough demand to sell out all flights (not enough to justify buying bigger plane). All seats sell for [imath]200. [/imath] Probability a passenger with reservation shows up = p. The probability of “No show”=1-p. The occurrence of “Show/No show” is independent among passengers. Passengers who are turned away due to “overbooking” are given 240 (the purchase price plus a 20% penalty to airline) The number of reservations made for each flight is chosen by airline (not random): n The number of booked passengers who show up is a random variable: X What is the probability distribution of X? Write out the Revenues to the airline for a flight as a function of X Write out the expected Revenues to the airline What is the effect of increasing n on expected revenues for a given p? What is the effect of increasing p on expected revenues for a given n? When would it make no sense to overbook flights? My thoughts: I think the answer to the first part is that it is a binomial distribution (since we have two distinct outcomes and a constant and independent probability of success). However, I am facing problem in the remaining parts. Can someone please explain it to me? Thanks so much :) |
2337572 | Are there matrices such that [imath]AB=I[/imath] and [imath]BA \neq I[/imath]
Are there matrices such that [imath]AB=I[/imath] and [imath]BA \neq I[/imath] ? [imath]A[/imath] and [imath]B[/imath] are square matrices | 2524430 | Linear algebra, construction of counter example for two rectangular matrices such that [imath]AB=I[/imath] but [imath]BA\neq I[/imath]
We have a matrix [imath]A[/imath] of order [imath]m[/imath] by [imath]n[/imath] and a matrix [imath]B[/imath] of order [imath]n[/imath] by [imath]m[/imath] where entries are from real numbers. We are given that [imath]AB =I[/imath] and we need to check whether [imath]BA =I[/imath] or not. If we have square matrices then it is true but here are rectangular matrices. I know that it doesn't hold true for rectangular matrices and I am trying hard to find some counter example but unable to find such matrix. I want to learn about how to construct matrices as counter examples in linear algebra as in this case. Thanks |
2335309 | Show that there are infinitely many positive integers [imath]N[/imath] that cannot be written in the form [imath]a^n+b^n+c^n[/imath]
Show that there are infinitely many positive integers [imath]N[/imath] that cannot be written in the form [imath]a^n+b^n+c^n[/imath] where [imath]a,b,c[/imath] are positive integers and [imath]n \geq 2[/imath]. I thought about considering solutions such that [imath]a^n+b^n+c^n < N^n.[/imath] Thus [imath]a,b,c < N[/imath]. Thus there are less than [imath]N^3[/imath] positive integers expressible in the required form that satisfy the above inequality. Therefore, there are at least [imath]N^n-N^3[/imath] that aren't for a given [imath]n[/imath]. I didn't see how to continue from here since the exponent [imath]n[/imath] can vary. | 2337427 | Infinitely many numbers not expressible in the form [imath]a^n+b^n+c^n[/imath]
Show that there are infinitely many positive integers [imath]N[/imath] that cannot be written in the form [imath]a^n+b^n+c^n[/imath] where [imath]a,b,c[/imath] are positive integers and [imath]n \geq 2[/imath]. Note that in the question above, [imath]n[/imath] can vary and is not fixed. I thought about using a modular arithmetic argument by finding a set of values which are not expressible when the exponent is [imath]3n,3n+2,[/imath] or [imath]3n+4[/imath] and then use the Chinese Remainder Theorem. Note that for all expressions of the form [imath]a^{3n}+b^{3n}+c^{3n},[/imath] the numbers which are [imath]4,5,6[/imath] modulo [imath]9[/imath] are not expressible in these forms since each of the terms is [imath]-1,0,1 \pmod{9}[/imath]. For expressions of the form [imath]a^{3n+2}+b^{3n+2}+c^{3n+2} = a^{3n} \cdot a^2+b^{3n} \cdot b^2+c^{3n} \cdot c^2,[/imath] note that [imath]x^2 \equiv -1,0,1,4 \pmod{9}[/imath] and that [imath]x^3 \equiv -1,0,1 \pmod{9}[/imath]. But I couldn't find a set of numbers modulo [imath]9[/imath] which weren't expressible in this form. For expressions of the form [imath]a^{3n+4}+b^{3n+4}+c^{3n+4},[/imath] note that [imath]x^4 \equiv 0,1,3,9 \pmod{13}[/imath] and [imath]x^{3n} \equiv -1,0,1,8 \pmod{13}[/imath]. Therefore, [imath]x^{3n+4} \equiv 0,1,3,7,8,9,10,11,12[/imath], but I couldn't find a set of numbers modulo [imath]13[/imath] which weren't expressible in this form. Can we use a modular arithmetic argument similar to this to solve the cases for [imath]3n+2[/imath] and [imath]3n+4[/imath] one or do we need to use a higher exponent such as [imath]6n[/imath] instead of [imath]3n[/imath]? |
2338046 | Find [imath]x,y,z[/imath] such that [imath]6x+15y+20z=17[/imath]
Find [imath]x,y,z[/imath] such that: [imath] 6x+15y+20z=17 [/imath] I found this question in Hua Loo Keng's Number Theory under the Greatest Common Factor and Least Common Multiple section. | 620939 | Finding integer solutions for [imath]6x+15y+20z=1[/imath]
Problem: Find integers [imath]x[/imath], [imath]y[/imath], and [imath]z[/imath] that satisfy the equation [imath]6x+15y+20z=1[/imath]. I noticed that [imath]\gcd(6,15)=3[/imath], [imath]\gcd(15,20)=5[/imath], and that [imath]\gcd(6,20)=2[/imath]. And of course [imath]\gcd(6,15,2)=1[/imath]. Of course I know to set one of the variables (x, y, or z) to be zero. But I want to know how to get the more trivial answers. But where do I go from here to find the solutions? |
1475093 | Prove there is a polynomial [imath]P_0[/imath] with the property that [imath]\mathcal{I}[/imath] consists precisely of the multiples of [imath]P_0[/imath].
Let [imath]\mathcal{I}[/imath] be a proper ideal of [imath]\mathbb{F}[x][/imath] where [imath]\mathbb{F}[/imath] is a field. Show that there is a polynomial [imath]P_0(x)\in \mathbb{F}[x][/imath] with degree larger than [imath]0[/imath] with the property that [imath]\mathcal{I}=\{P_0(x)\cdot q(x):q(x)\in \mathbb{F}[x]\}.[/imath] (In other words, prove there is a polynomial [imath]P_0[/imath] with the property that [imath]\mathcal{I}[/imath] consists precisely of the multiples of [imath]P_0[/imath].) Previously, I have proved Let [imath]\mathbb{F}[/imath] denote the field described the set [imath]\mathbb{F}=\{r+si: r,s\in \mathbb{Q}\}[/imath] (Here [imath]i[/imath] is the infamous element of [imath]\mathbb{C}[/imath] with [imath]i^2=-1[/imath].) Consider the function [imath]\lambda: \mathbb{Q}[x] \to \mathbb{F}[/imath] defined by [imath]\lambda(p)=p(i) \qquad \text{ for all } p\in \mathbb{Q}[x][/imath] 1) Show that [imath]\lambda[/imath] is a ring homomorphism. 2) Show that the kernel of [imath]\lambda[/imath] is the set of all polynomials [imath]p\in \mathbb{Q}[x][/imath] that are divisible by [imath]x^2+1[/imath]. Proof of (i): For [imath]\lambda[/imath] to be a ring homomorphism, we need to show that > a) [imath]\lambda(f+g)=\lambda(f)+\lambda(g)[/imath] b) [imath]\lambda(fg)=\lambda(f)\lambda(g)[/imath] c) [imath]\lambda(1)=1[/imath] \textbf{\textit{Proof of a:}} Let [imath]f,g \in \mathbb{Q}[x][/imath]. Then we need to show that [imath]\lambda(f+g)=\lambda(f)+\lambda(g)[/imath]. So [imath]\lambda(f+g)=(f+g)(i)=f(i)+g(i)=\lambda(f)+\lambda(g).[/imath] \textbf{\textit{Proof of b:}} Let [imath]f,g \in \mathbb{Q}[x][/imath]. Then we need to show that [imath]\lambda(fg)=\lambda(f)\lambda(g)[/imath]. So [imath]\lambda(fg)=(fg)(i)=f(i)g(i)=\lambda(f)\lambda(g).[/imath] \textbf{\textit{Proof of c:}} Let [imath]f=1 \in \mathbb{Q}[x][/imath]. Then we need to show that [imath]\lambda(1)=1[/imath]. So [imath]\lambda(1)=1(i)=1.[/imath] Hence [imath]\lambda[/imath] is a ring homomorphism. Proof of (ii): We need to show that the kernel of [imath]\lambda[/imath] is the set of all polynomials [imath]p\in > \mathbb{Q}[x][/imath] that are divisible by [imath]x^2+1[/imath]. Suppose [imath]x^2+1|p(x)[/imath]. Then [imath]p(x)=(x^2+1)g(x)[/imath] for some [imath]g(x)\in > \mathbb{Q}[x][/imath]. Using part (i), we can show that [imath]p(x) \in > ker(\lambda)[/imath] (i.e. that [imath]p(x)=0[/imath]): [imath]\lambda(p(x))=(x^2+1)g(x)=(i^2+1)g(i)=0\cdot g(i)=0[/imath] Hence we have show that [imath]p(x)\in ker(\lambda)[/imath]. Suppose [imath]p(x) \in ker(\lambda)[/imath]. Using the Euclidean division over [imath]\mathbb{Q}[x][/imath] to divide [imath]p(x)[/imath] by [imath]x^2+1[/imath]. We obtain: \begin{equation*} \begin{aligned} 0 & =\lambda(p(x)) \\ & > =\lambda(q(x)(x^2+1)+r(x)) \\ & =\lambda(q(x))\lambda(x^2+1)+\lambda(r(x)) \\ & =p(i)(i^2+1)+r(i) \\ & =r(i) \\ \end{aligned} \end{equation*} Hence we have show that the remainder is [imath]0[/imath], which implies that [imath]p(x)=(x^2+1)g(x)[/imath]. Hence, the kernel of [imath]\lambda[/imath] is the set of all polynomials [imath]p\in \mathbb{Q}[x][/imath] that are divisible by [imath]x^2+1[/imath]. I know both proofs are related to one another. So what is needed to change in the previous proof for the new proof? | 1164195 | Ideal and minimal polynomial
I am looking at https://math.berkeley.edu/~ecarter/Summer08/110/notes/lec19.pdf and trying understand the proof for the theorem that states: Given a nonzero ideal I in [imath]P(F)[/imath], there is a monic polynomial [imath]p(t)[/imath] such that [imath]I=\left\{q(t)p(t) | q(t) \in P(F) \right\}[/imath] Proof goes like Since [imath]I[/imath] contains at least one nonzero element, we can let [imath]p(t)[/imath] be a nonzero element of [imath]I[/imath] of minimum degree. We can show that every element of [imath]I[/imath] is a multiple of [imath]p(t)[/imath]. Let [imath]f(t) \in I[/imath]. Then there exist [imath]q(t),r(t) \in P(F)[/imath] such that [imath]f(t) = q(t)p(t) + r(t)[/imath] and deg [imath]r(t)[/imath] < deg [imath]p(t)[/imath]. Since [imath]p(t) \in I[/imath], so is [imath]q(t)p(t)[/imath], and therefore so is [imath]r(t) = f(t) -q(t)p(t)[/imath]. By the choice of [imath]p(t),r(t) = 0[/imath], so [imath]f(t)[/imath] is a multiple of [imath]p(t)[/imath]. What I am having trouble understanding is why [imath]r(t)[/imath] must equal [imath]0[/imath]. If someone can explain why it is, I'd appreciate it. |
579729 | The quotient of a free group, of rank [imath]n[/imath], and its commutator subgroup is isomorphic to [imath]\mathbb{Z}^n[/imath]
Let [imath]F[/imath] be a free group of rank [imath]n[/imath]. Let [imath]G[/imath] be the commutator subgroup of [imath]F[/imath]. I need prove that [imath]F/G\cong\mathbb{Z}^n[/imath]. I have tried with the isomorphism theorem: With [imath]\varphi[/imath], I send the [imath]x_i[/imath] element in the basis of [imath]F[/imath] to the vector with [imath]1[/imath] in the i-th entree and [imath]0[/imath] in the others, in [imath]\mathbb{Z}^n[/imath]. But I can't prove that [imath]ker\varphi\subseteq G[/imath]. Any help. | 666155 | Abelianization of free group is the free abelian group
How does one prove that if [imath]X[/imath] is a set, then the abelianization of the free group [imath]FX[/imath] on [imath]X[/imath] is the free abelian group on [imath]X[/imath]? |
2338102 | Center of Lie algebra and Lie algebra of center
I am learning some Lie group and I met the following problem: Let [imath]G[/imath] be a Lie group and [imath]Z[/imath] its center. Then [imath]Z[/imath] is also a Lie group. Let [imath]\mathfrak{g}[/imath] be the Lie algebra of [imath]G[/imath] and [imath]\mathfrak{z}[/imath] be the center of [imath]\mathfrak{g}[/imath]. Is the Lie algebra of [imath]Z[/imath] equal to [imath]\mathfrak{z}[/imath]? It seems that this correspondence is natural, yet I cannot find any source containing this. Could anyone suggest a (sketch of) proof, or give a counter example? Thank you very much! | 1883131 | Center of Lie group and Lie algebra
Let [imath]G[/imath] be a Lie group and [imath]\mathfrak{g} = T_eG[/imath] its Lie algebra (where [imath]e \in G[/imath] is the neutral element). Denote [imath]Z(\mathfrak{g})[/imath] the center of [imath]\mathfrak{g}[/imath] and [imath]Z(g)[/imath] the center of [imath]G[/imath]. I've read the following statement but don't see how to prove it: [imath]Z(\mathfrak{g}) = 0 \Longleftrightarrow Z(G) \text{ is zero dimensional.}[/imath] I'm mainly interested in the "[imath]\Leftarrow[/imath]" direction, as this is used in a corollary of the Bonnet-Myers theorem in the text I'm reading. Now, my knowledge of Lie groups is very basic. For instance, I guess that the dimension of [imath]Z(G)[/imath] means its dimension as a manifold. So I would think that [imath]Z(G)[/imath] is a submanifold of [imath]G[/imath]. But how do I even see this in general? According to this question When is the Lie algebra of the center of Lie group the center of its Lie algebra, it seems to be difficult to find conditions when [imath]Z(\mathfrak{g}) = Z(G)[/imath] holds in general, however the statement above looks more simple. Would someone be able to prove the above equivalence? |
2338467 | getting negative logarithm in the answer
[imath]\int_{-3}^{-2} \frac{dx}{2x+1}[/imath] On solving this I'm getting [imath]\log(-3) - \log(-5)[/imath] and isn't negative log not possible? | 206032 | What is the integral of 1/x?
What is the integral of [imath]1/x[/imath]? Do you get [imath]\ln(x)[/imath] or [imath]\ln|x|[/imath]? In general, does integrating [imath]f'(x)/f(x)[/imath] give [imath]\ln(f(x))[/imath] or [imath]\ln|f(x)|[/imath]? Also, what is the derivative of [imath]|f(x)|[/imath]? Is it [imath]f'(x)[/imath] or [imath]|f'(x)|[/imath]? |
2338346 | End[imath]_A(A)\cong A[/imath]
I am wondering the following question. Let [imath]A[/imath] be a ring. Is it always true that End[imath]_A(A)\cong A[/imath]. Is true when [imath]A[/imath] is semisimple? I tried to give an isomorphism [imath]\phi[/imath] between End[imath]_A(A)[/imath] and [imath]A[/imath]. [imath]\phi(f)=f(1)[/imath], but I don't know if I am right. Thanks for your help. | 2856959 | Show that [imath]\text{End}_R(R_R, R_R) = R[/imath] as rings.
The standard way of showing that two objects are isomorphic is by constructing an invertible function between the two. How can I do that for these two objects? I think I need a hint to construct such an isomorphism. |
2338985 | Let [imath]G[/imath] be a group. Prove that [imath]o(xy) = o(yx)[/imath] for all [imath]x,y \in G[/imath]
Let [imath]G[/imath] be a group. Prove that [imath]o(xy) = o(yx)[/imath] for all [imath]x,y \in G[/imath]. Edit: I wanted to prove it with basic properties of groups and orders, not with isomorphisms. I will write the answer I worked on my notebook using the answer by the user that solved my problem (please stop downgrading my question / reputation, I've started learning this 2 days ago by myself, and I believe this site is about helping people to learn): Let [imath]m = o(xy)[/imath]. Then: [imath](xy)^m = e_G[/imath] As [imath]G[/imath] is a group, there exists inverse elements for [imath]x[/imath] and [imath]y[/imath]. Let [imath]y^{-1}[/imath] be the inverse of [imath]y[/imath]. As [imath]ye_G = y[/imath], [imath]e_Gy^{-1} = y^{-1}[/imath] and [imath]yy^{-1} = e_G[/imath] we can multiply by [imath]y[/imath] (by the left) and [imath]y^{-1}[/imath] (by the right) obtaining: [imath]e_G = ye_Gy^{-1}[/imath] Now, as we know that [imath](xy)^m = e_G[/imath], we can substitute [imath]e_G[/imath] in the previous formula: [imath]ye_Gy^{-1} = y(xy)^my^{-1}[/imath] Now, we know that [imath](xy)^m = \overbrace{(xy)(xy)\dots(xy)}^{\text{m times}}[/imath] So, using the previous equation: [imath]y(xy)^my^{-1} = y\overbrace{(xy)(xy)\dots(xy)}^{\text{m times}}y^{-1}[/imath] And because [imath]G[/imath] is a group, it's associative: [imath]y\overbrace{(xy)(xy)\dots(xy)}^{\text{m times}}y^{-1} = \overbrace{(yx)(yx)\dots(yx)}^{\text{m times}}yy^{-1}[/imath] And this is: [imath](yx)^myy^{-1} = (yx)^me_G = (yx)^m[/imath] So we obtained: [imath]e_G = (yx)^m[/imath] And as [imath](xy)^m = e_G = (yx)^m[/imath], [imath]o(yx) = m[/imath], and we conclude: [imath]o(xy) = o(yx)[/imath] for all [imath]x,y \in G[/imath] I find it interesting solving this kind of problems just by using properties. Using more math structures made the problem trivial. | 1081667 | Order of a product of elements in a generic group
I'm dealing with the following question: having a group [imath]G[/imath] (of which nothing else is specified, doesn't have to be abelian), prove that [imath]|ab|=|ba|[/imath] for any [imath]a[/imath], [imath]b[/imath]. Since [imath]G[/imath] is not abelian I'm not sure what to use here. I was thinking that except the case where [imath]b=a^{-1}[/imath], the order of those would be [imath]lcm(m,n)[/imath] where [imath]m[/imath] and [imath]n[/imath] are the orders of a and b. But looking it up it seems not to be so simple, even if all the info I could find was for abelian groups, so I would like some help with this specific case. |
2339264 | Show that a holomorphic function satisfying [imath]|f(z)|\leq c|z|^n[/imath] must be [imath]f(z)=a\cdot z^n[/imath] for some a.
Let [imath]f:\mathbb{C}\to\mathbb{C}[/imath] be a holomorphic function such there exists a [imath]c\in\mathbb C[/imath] with [imath]|f(z)|\leq c\cdot |z|^n[/imath] for all [imath]z\in\mathbb C[/imath]. Show that there exists an [imath]a\in\mathbb C[/imath] with [imath]f(z)=a\cdot z^n[/imath]. Proof. My general idea is to show that the taylor series only has one non-zero coefficient: [imath]\displaystyle |a_k|\leq\frac{k!}{2\pi}\oint_{|w|=r}\frac{|f(w)|}{|w|^{k+1}}dw\leq \frac{k!}{2\pi}\oint_{|w|=r}\frac{c\cdot|w|^n}{|w|^{k+1}}dw\leq \\\displaystyle\leq \frac{c\cdot k!}{2\pi}\oint_{|w|=r}|w|^{n-k-1}dw[/imath] Now using the standard estimate since [imath]\oint=2\pi r[/imath]: [imath]\displaystyle \leq c\cdot k!\cdot r^{n-k}\longrightarrow\begin{cases}0,\ k>n\\c\cdot n!,\ k=n\\ \infty,\ k<n\end{cases}[/imath] as [imath]r\to\infty[/imath] In conlusion the taylor series for now looks like [imath]f(z)=\sum_{k=0}^{n-1}a_k z^k+ a_n z^n[/imath] with [imath]|a_n|\leq n!\cdot c[/imath]. What do I do with the remaining summands? Edit: I would say that for [imath]k<n[/imath] we have [imath]|z|^k>|z|^n[/imath] for [imath]z\in D_1(0)[/imath]. So since the estimate has to hold for all [imath]z\in \mathbb C[/imath] we can't have that either. | 2338123 | Removable singularities and an entire function
Function [imath]f(z)[/imath] is an entire function such that [imath]|f(z)| \le |z^{n}|[/imath] for [imath]z \in \mathbb{C}[/imath] and some [imath]n \in \mathbb{N}[/imath]. Show that the singularities of the function [imath]\frac {f(z)}{z^{n}}[/imath] are removable. What can be implied about the function [imath]f(z)[/imath] if moreover [imath]f(1) = i[/imath]? Draw a far-reaching conclusion. My attempt: If the singularities of [imath]\frac {f(z)}{z^{n}}[/imath] are removable, it is entire (not sure, need help with the justification) and bounded, so constant from Liouville's theorem, the constant value of the funcion is [imath]i[/imath], hence [imath]f(z)=iz^{n}[/imath]. But what about the [imath]n[/imath] here, is it arbitrary? Could somebody help me prove the removability of the singularities and suggest if my attempt is going the right way? |
2339322 | MLE when the density is [imath]f(x;\theta )=\theta x^{\theta−1}[/imath]
Working through this given problem on maximum likelihood estimation (MLE). The density is given as [imath]f(x;\theta) = \theta x^{\theta -1} [/imath] transforming the above equation to MLE, we have [imath]L(x;\theta) = \prod_{i=1}^{n} \theta x_i^{\theta - 1} = {\theta}^n \prod_{i=1}^{n} x_i^{\theta - 1}. [/imath] Taking [imath]\ln[/imath] [imath] \begin{align} \ln L(x;\theta) &= \ln{\theta}^n +\ (\theta - 1) \sum_{i=1}^{n} \ln(x_i)\\ \end{align}[/imath] then we have [imath]L'(x;\theta) = {n \over \theta} - \sum_{i=1}^{n} \ln(x_i). [/imath] [imath]\theta = {n \over \sum_{i=1}^{n} \ln(x_i)}. [/imath] Is it correct? How can I apply the law of the strong numbers in this case :? Thanks for any help | 441135 | Maximum likelihood estimation when the density is [imath]f(x;\theta) = \theta x^{\theta -1} [/imath]
Working through this given problem on maximum likelihood estimation (MLE). The density is given as [imath]f(x;\theta) = \theta x^{\theta -1} [/imath] transforming the above equation to MLE, we have [imath]L(x;\theta) = \prod_{i=1}^{n} \theta x^{\theta - 1} [/imath] then we have [imath]L(x;\theta) = {\theta}^n \sum_{i=1}^{n} x^{\theta - 1}. [/imath] Taking [imath]\ln[/imath] [imath] \begin{align} \ln L(x;\theta) &= \ln{\theta}^n \ln\sum_{i=1}^{n} x^{\theta - 1}\\ &=n\ln{\theta} ({\theta - 1})\ln\sum_{i=1}^{n} x_{i}\\ &=n\ln{\theta} ({\theta - 1})\ln\sum_{i=1}^{n} x_{i}. \end{align} [/imath] Differentiating with respect to [imath]\theta[/imath] is what is turning me off here, or is there any thing i have done wrongly. Thanks for any help |
2339658 | Sum of Absolute Differences Inequality
Given 2 sequences: [imath]\quad a_1 \ge a_2 \ge \ldots \ge a_n[/imath] [imath]\quad b_1 \ge b_2 \ge \ldots \ge b_n[/imath] [imath]a_1b_1 + a_2b_2 + \cdots + a_nb_n \ge a_1b_{\pi(1)} + a_2b_{\pi(2)} + \cdots + a_nb_{\pi(n)}[/imath] where [imath]\pi(1), \pi(2), \ldots, \pi(n)[/imath] is any permutation of [imath]1, 2, \ldots, n[/imath] is the Rearrangement Inequality. Given 2 sequences: [imath]\quad a_1 \le a_2 \le \ldots \le a_n[/imath] [imath]\quad b_1 \le b_2 \le \ldots \le b_n[/imath] Does anyone know if there is a name for the following inequality? [imath]\left| a_1 - b_1 \right| + \left| a_2 - b_2 \right| + \cdots + \left| a_n - b_n \right| \le \left| a_1 - b_{\pi(1)} \right| + \left| a_2 - b_{\pi(2)} \right| + \cdots + \left| a_n - b_{\pi(n)} \right|[/imath] | 1888769 | Inequality involving rearrangement: [imath] \sum_{i=1}^n |x_i - y_{\sigma(i)}| \ge \sum_{i=1}^n |x_i - y_i|. [/imath]
If [imath]x_1 \ge x_2 \ge \cdots \ge x_n[/imath] and [imath]y_1 \ge y_2 \ge \cdots \ge y_n[/imath] are real numbers, and [imath]\sigma[/imath] is any permutation, then [imath] \sum_{i=1}^n |x_i - y_{\sigma(i)}| \ge \sum_{i=1}^n |x_i - y_i|. [/imath] This must be a known inequality. What is it called, and how is it proven? (Just a reference is OK.) The conditions are similar to rearrangement inequality. The inequality is a simple statement about minimizing the [imath]\ell^1[/imath] distance between a finite sequence and any rearrangement of another finite sequence. I searched around and clicked through various pages but couldn't find something relevant. If it is true, perhaps a proof could be constructed by decomposing the permutation into a sequence of transpositions. |
2339671 | Computing exponential object [imath]A^{A}[/imath] in category of irreflexive graphs
On p.333 of Conceptual Mathematics 2nd ed. we are asked to compute the exponential object [imath]A^{A}[/imath] in the category of irreflexive graphs, where [imath]A[/imath] is the graph of a single arrow s -> t. (For context there is a graph [imath]D[/imath] that is the graph of a single dot *. The book suggests using the properties of exponentials and graphs. For example to find the number of dots in [imath]A^{A}[/imath], find the number of maps [imath]D \rightarrow A^{A}[/imath], which is equivalent to the number of maps [imath]A \times D \rightarrow A[/imath]. This is the same as the number of maps from the dots of [imath]A[/imath] to [imath]A[/imath], of which there are 4. Similarly we can find the number of arrows by counting the maps of [imath]A \rightarrow A^{A}[/imath] which is the same as counting the maps of [imath]A \times A \rightarrow A[/imath] which there are 4, which are.. (s, s) and (t, t) must be mapped to s and t respectively, but (s, t) and (t, s) can be mapped to either s or t, hence 4. So I know there are 4 dots and 4 arrows.. but I don't know how to map the arrows onto the dots. Context: In the category of graphs an object is two sets [imath]X[/imath] (the edges of the graph) and [imath]P[/imath] (the nodes of the graph) equipped with two arrows [imath]s, t: X \rightarrow P[/imath] which correspond to the source and target of the edge. An arrow between two graphs is two arrows [imath]f: X \rightarrow X'[/imath] and [imath]g: P \rightarrow P'[/imath] which preserves source and target: [imath] g \circ s = s' \circ f [/imath] [imath] g \circ t = t' \circ f [/imath] | 1825765 | [imath]A^A[/imath] in category of graphs
(reference is Lawvere/Schanuel, Session 31, Ex. 1) I'm trying to calculate the exponential object [imath]A^A[/imath] and its evalution map [imath]e \colon A \times A^A \to A[/imath] in the category of graphs, where [imath]A[/imath] is the "arrow graph" (ie. one arrow and two dots). In the following, [imath]D[/imath] is the graph with one dot and no arrows, [imath]1[/imath] is the terminal object in this category (graph with one dot and one arrow, the loop). So far I have: The points of [imath]\mathbf{1}\to A^A[/imath] correspond to the maps [imath]A\to A[/imath] (via two standard isomorphisms), and since [imath]\mathbf{1}[/imath] is the loop, and there is one map of graphs [imath]A \to A[/imath], there is one loop in [imath]A^A[/imath]. The dots [imath]D\to A^A[/imath] correspond to the maps [imath]A \times D \to A[/imath], of which there are four, hence four dots in [imath]A^A[/imath]. The arrows [imath]A \to A^A[/imath] correspond to the maps [imath]A \times A \to A[/imath], of which there are four, hence four arrows in [imath]A^A[/imath]. But I'm stuck on how to put these together to constitute [imath]A^A[/imath] and its evaluation map. |
2340112 | If [imath]U\subset \mathbb{R^n}[/imath] is open then [imath]\dim U=n?[/imath]
If [imath]U\subset \mathbb{R^n}[/imath] is open then [imath]\dim U=n?[/imath] I know this is true from open submanifolds but not sure if it works for any open set. | 2225877 | Dimension of open subsets of [imath]R^n[/imath]
Does an open subset of [imath]R^n[/imath] exist that has dimension less than [imath]n[/imath] in the standard topology? All the less than [imath]n[/imath] dimensional subsets I can think of are not open. |
2339448 | A converse proposition to the Mean Value Theorem
This is just a question that popped up in my head while going through basic real analysis. The ordinary Mean Value Theorem (MVT) is given as follows. Let [imath]f:[a,b] \to \mathbb{R}[/imath] be a function satisfying the following conditions: [imath](i)[/imath] [imath]f[/imath] is continuous on [imath][a,b][/imath] and [imath](ii)[/imath] [imath]f[/imath] is differentiable on [imath](a,b)[/imath]. Then [imath]\exists c \in (a,b),[/imath] such that [imath]f(b)-f(a)=f'(c) \cdot (b-a)[/imath] Of course, we can restrict [imath]f[/imath] to any interval [imath][x,y] \subset [a,b][/imath] and apply the theorem on [imath]f|_{[c,d]}[/imath] to state that [imath]\exists z \in (x,y)[/imath] such that [imath]f(y)-f(x)=f'(z) \cdot (y-x)[/imath] Can I formulate a converse proposition as follows? Let [imath]f:[a,b] \to \mathbb{R}[/imath] be a function satisfying the following conditions: [imath](i)[/imath] [imath]f[/imath] is continuous on [imath][a,b][/imath] and [imath](ii)[/imath] [imath]f[/imath] is differentiable on [imath](a,b)[/imath]. Then [imath]\forall c \in (a,b), \exists x,y \in [a,b][/imath] such that [imath]f(y)-f(x)=f'(c) \cdot (y-x)[/imath] Does it hold? If yes, how do I prove it? (How does one find out the points [imath]x[/imath] and [imath]y[/imath], which need not be unique, that work for a given point [imath]c?)[/imath] Does it hold under weaker assumptions on the function? If no, can you provide me a counter example? Can I impose stronger conditions to make it work? Any help would be much appreciated. Thank you. EDIT : As pointed out by @Henrik, the proposition does not hold with it's current assumptions. On the other hand, @MANMAID claims that it does hold with the additional assumption that [imath]f[/imath] has no point of inflection. It seems very interesting. I require a formal proof, though. | 776693 | Converse of mean value theorem
I am wondering if the following converse (or modification) of the mean value theorem holds. Suppose [imath]f(\cdot)[/imath] is continuously differentiable on [imath][a,b][/imath]. Then for all [imath]c \in (a,b)[/imath] there exists [imath]x[/imath] and [imath]y[/imath] such that [imath] f'(c)=\frac{f(y)-f(x)}{y-x} [/imath] |
2339253 | Applications of Sylow theorems
Almost all the books of algebra or group theory give following types of applications of Sylow theorems: A group of order [imath]...[/imath] is not simple. A group of order ... has normal Sylow subgroup. Even in math.stackexchange, there are many questions with above title and they involve questions which are related to proving above type of statements. Are there any other type of applications of these theorems of Sylow? | 1561986 | Different Applications of Sylow Theorems
The theorems of Sylow are very well known and almost every mathematician learns in his undergraduate course. The applications of Sylow theorems given in books are of the kind "If [imath]|G|=....[/imath] then show that [imath]G[/imath] is not simple/ [imath]G[/imath] is solvable/ ..." I would like to know if there are some other, interesting, applications of this theorem. |
2339927 | how to solve SDE
How to solve this SDE: [imath]dX(t)=X^\alpha(t)dt+\sigma X(t)dW(t), for X(0)=x_0[/imath] [imath]dW(t)[/imath] is Wiener process. Also I have to use [imath]f(t)=X(t)exp(-\alpha W(t)+1/2\alpha^2t)[/imath] as integration factor. I tried to solve it looking at solutions of other SDE, but can't find the way for solving non-linear one as I started studying SDE just recently. The only idea I have is to multiply both sides by integration factor. But what I can do with non-linear part? | 468091 | how to do such stochastic integration [imath]dS = a S^b dt + c S dW[/imath]?
How to do stochastic integration [imath]dS = a S^b dt + c S dW[/imath], where [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath] are constant, [imath]b > 0[/imath], and [imath]W[/imath] is the Wiener process. I know how to do integration for [imath]dS = aS dt + cS dW[/imath], or [imath]dS / S = a dt + c dW[/imath]: According to Ito's lemma, for [imath]dx = \mu dt + \sigma dW[/imath], and [imath]f(t,x)[/imath], [imath]df(t,x) = (\frac{\partial f}{\partial t} + \mu \frac{\partial f}{\partial x} + \frac{1}{2} \sigma^2 \frac{\partial^2f}{\partial x^2} ) dt + \sigma \frac{\partial f}{\partial x} dW [/imath] Submitting in [imath]S\rightarrow x[/imath], [imath]aS \rightarrow \mu[/imath], [imath]cS \rightarrow \sigma[/imath], and [imath]ln(S)\rightarrow f(t,x)[/imath], will get: [imath]d[ln(S)] = (a - \frac{1}{2} c^2) dt + c dW[/imath] , then we'll have [imath]S(t) = exp[(a - \frac{1}{2} c^2) t + cW_t][/imath] But if [imath]S^b[/imath] is introduced, how could I solve it? |
2340409 | Show that there exist a continuous function [imath]f:X\to [0,1][/imath] such that [imath]f^{-1}(\{0\})=A[/imath] and [imath]f(B)=\{1\}[/imath].
Let [imath]A,B[/imath] disjoint closed subsets of a normal topological space [imath](X,\Gamma)[/imath]. Show that there exist a continuous function [imath]f:X\to [0,1][/imath] such that [imath]f^{-1}(\{0\})=A[/imath] and [imath]f(B)=\{1\}[/imath] if and only if [imath]A[/imath] is a [imath]G_{\delta}[/imath] on [imath](X,\Gamma)[/imath]. Recall that [imath]A[/imath] is a [imath]G_{\delta}[/imath] set in [imath]X[/imath], if [imath]A[/imath] is the intersection of a countable collection of open sets of [imath]X[/imath]. Let [imath]A,B[/imath] disjoint closed subsets of a normal topological space [imath](X,\Gamma)[/imath], and suppose there exist a continuous function [imath]f:X\to [0,1][/imath] such that [imath]f^{-1}(\{0\})=A[/imath] and [imath]f(B)=\{1\}[/imath], but note that [imath]\{0\}=\bigcap_{n=1}^{\infty}{\left[0,\dfrac{1}{n}\right)}[/imath] [imath]A=f^{-1}(\{0\})=f^{-1}\left(\bigcap_{n=1}^{\infty}{\left[0,\dfrac{1}{n}\right)}\right)=\bigcap_{n=1}^{\infty}{f^{-1}\left[0,\dfrac{1}{n}\right)}[/imath] Thus, [imath]f^{-1}\left[0,\dfrac{1}{n}\right)[/imath] is in [imath]\Gamma[/imath] for all [imath]n\in\mathbb{N}^{+}[/imath], so [imath]A[/imath] is a [imath]G_{\delta}[/imath] in [imath](X,\Gamma)[/imath]. But, I really need help with the other part of the proof. Thanks!! | 73798 | a construction of a continuous function from a normal space to [0,1]
Let [imath]A[/imath] be closed in [imath]X[/imath] , where [imath]X[/imath] is a normal space, and [imath]A[/imath] is also a countable intersection of open sets. Prove that there exist a continuous function such that [imath] \eqalign{ & f\left( x \right) = 0\,\,\text{ if }\,x\, \in A \cr & f\left( x \right) > 0\,\,\text{ if }\,x \notin A \cr} [/imath] EDITED: I mean intersection, sorry |
2339908 | If [imath]\lim_{x\to c} f'(x)[/imath] exists, does [imath]f'(c)[/imath] exist?
Here's a problem on differentiability at a point and continuity of the derivative function at the same point. I'm stating the problem and presenting my solution to it. I'm not entirely confident on my solution (especially about interchanging the limits). So I'd greatly appreciate if someone checks the solution and tell me if there's any gap in my arguments. Thank you. The Problem : Let [imath]f:(a,b) \to \mathbb{R}[/imath] is continuous on [imath](a,b)[/imath] and is differentiable except possibly at [imath]c \in (a,b)[/imath]. Assume that [imath]\lim_{x \to c} f'(x)[/imath] exists. Show that [imath]f'(c)[/imath] exists and [imath]f'[/imath] is continuous at [imath]c[/imath]. My Approach : Given that [imath]\lim_{x \to c}f'(x)[/imath] exists, we can write \begin{align} \lim_{x \to c}f'(x)&=\lim_{x \to c} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ &=\lim_{h \to 0} \lim_{x \to c} \frac{f(x+h)-f(x)}{h} ~~\text{[Since the double-limit exists]}\\ &=\lim_{h \to 0} \Big[\frac{1}{h} \cdot \lim_{x \to c}\big\{f(x+h)-f(x)\big\}\Big]\\ &=\lim_{h \to 0} \frac{f(c+h)-f(c)}{h} ~~\text{[Since [imath]f[/imath] is continuous]}\\ \end{align} Since [imath]\lim_{x \to c}f'(x)[/imath] exists, so does [imath]\lim_{x \to c} \frac{f(c+h)-f(c)}{h},[/imath] i.e. [imath]f'(c)[/imath]. Clearly, [imath]\lim_{x \to c}f'(x)=f'(c)[/imath]. Hence [imath]f'[/imath] is continuous at [imath]c[/imath]. [imath]~\blacksquare[/imath] P.S. The problem is from an exercise consisting of Mean Value Theorem (MVT)-related problems. So it is highly likely that MVT should play a role here, which makes me suspect this is a comparatively simple, probably too-simple-to-be-correct solution. | 221273 | Question on differentiability of a continuous function
Suppose that [imath]\mathcal f[/imath] is continuous on a closed interval and [imath]\mathcal v[/imath] is a point in its interior. [imath]\mathcal f[/imath] is differentiable when [imath]\mathcal x\not=v[/imath], and suppose that [imath]\,\displaystyle{\lim_{x\rightarrow v}f'(x)}\,[/imath] exists. Show that [imath]\mathcal f[/imath] is differentiable at [imath]\mathcal v[/imath]. Attempt at a solution I want to show that [imath] \lim_{h\rightarrow 0}\frac {f(v+h)-f(v)}{h}[/imath] exists. Based on the fact that [imath] \lim_{x\rightarrow v}f'(x)=\lim_{x\rightarrow v}\ \lim_{h\rightarrow 0}\frac {f(x+h)-f(x)}{h}[/imath] exists. Using the fact that [imath]\mathcal f[/imath] is continuous, I think if somehow I could interchange the limits [imath]\mathcal {\lim_{x\rightarrow v}\ \lim_{h\rightarrow 0}\frac {f(x+h)-f(x)}{h}}={\lim_{h\rightarrow 0}\ \lim_{x\rightarrow v}\frac {f(x+h)-f(x)}{h}}=\lim_{h\rightarrow 0}\frac {f(v+h)-f(v)}{h}[/imath] then the result comes out. But I don't think that I could do that, based on continuity. Other than that, I am not sure how to approach this. |
2340771 | Does the interval [imath][0,1][/imath] contains more real numbers than [imath](0,2)[/imath]?
It is a common saying that there are more real numbers than integers, since you can not uniquely map integers to real numbers - even after using up all integers, there will still be at least one real number left that you haven't mapped to any integer. So I applied the same argument to the two intervals in question. One obvious way to map number from [imath][0,1][/imath] to [imath][0,2][/imath] is [imath]\times 2[/imath] - [imath]0[/imath] is still [imath]0[/imath], [imath]0.25[/imath] becomes [imath]0.5[/imath], and [imath]1[/imath] becomes [imath]2[/imath]. Good enough. However, if we change the second interval to an open interval, namely [imath](0,2)[/imath], then the number [imath]0[/imath] and [imath]1[/imath] in the original mapping is now the "remaining" number, while all numbers in [imath](0,2)[/imath] have already been "used". Does that mean that [imath][0,1][/imath] have more real numbers than [imath](0,2)[/imath]? (even though [imath](0,2)[/imath] seems to be bigger than [imath][0,1][/imath]…) Does my argument even make sense? I did some Googling but didn't found anything useful. | 1327444 | More numbers between [imath][0,1][/imath] or [imath][1,\infty)[/imath]?
There are infinitely many real numbers between any two real numbers, therefore there are infinitely many real numbers in the range [imath][0,1][/imath] as there are in [imath][1, \infty)[/imath]. In a mathematical sense, are there more numbers in one range as opposed to the other? What about [imath][0,1][/imath] and [imath][0, \infty)[/imath] ? I'd assume this is akin to how limits work in that the limit as [imath]x\to\infty[/imath] of [imath]\frac{x^x}{\log(x)}[/imath] is [imath]\infty[/imath], even though both the numerator and denominator approach infinity. |
2340584 | What does it mean to define something as a curve?
What does it mean to define something as a curve? Does writing [imath]y=x[/imath] means I'm explicitly talking about a curve and writing [imath]f(x)=x[/imath] means I'm defining it as a function? | 1845169 | Is there a difference between [imath]y(x)[/imath] and [imath]f(x)[/imath]
Oftentimes functions described by [imath]f(x) = 2x+4[/imath], and when this is mapped to the Cartesian plane, [imath]f(x) = y[/imath]. This surely implies that [imath]y = 2x+4[/imath]. Is there a difference between this and [imath]y(x) = 2x+4[/imath]? |
277083 | How to show [imath]f'(0)[/imath] exist and is equal to [imath]1[/imath]?
Assume that [imath]f[/imath] be continuous on [imath]\mathbb{R}[/imath], [imath]f'(x)[/imath] exists for all [imath]x\neq 0[/imath], and [imath]\lim_{x\rightarrow 0} f'(x)=1[/imath]. We need to show [imath]f'(0)[/imath] exist and is equal to [imath]1[/imath]. [imath]f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}[/imath], [imath]\lim_{x\rightarrow 0}f'(x)=1\Rightarrow\lim_{x\rightarrow 0}\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=1[/imath]...am I going in the right direction? Please help. | 2227106 | Proving derivative is continuous using L'Hospital's Rule
Suppose that [imath]h[/imath] is continuous on [imath][a,b][/imath], differentiable on [imath](a,b)[/imath], and that [imath]c \in (a,b)[/imath]. Suppose also that [imath]\lim_{x \to c} h'(x)[/imath] exists. Prove that [imath]h'[/imath] is continuous at [imath]c[/imath]. I have found this answer here Prove derivative is continuous but can't follow the process. Any help/detailed explanations would be REALLY appreciated! I'd rather use L'Hospital Rule instead of the Mean Value Theorem, but I can't follow how to go from one step to the other... Thank you! |
2341128 | [imath]Null (A) \subseteq Null(A^T A)[/imath] if and only if [imath]Null(A^T A) \subseteq Null(A)[/imath]
Let [imath]A[/imath] be an [imath]m \times n[/imath] matrix. Show that [imath]Null (A) \subseteq Null(A^TA)[/imath] if and only if [imath]Null(A^TA) \subseteq Null(A)[/imath]. I can not really come up with the idea how to show that. Can someone help me? I would really appreciate that. | 66560 | Null space for [imath]AA^{T}[/imath] is the same as Null space for [imath]A^{T}[/imath]
[imath]A[/imath] is an [imath]n\times m[/imath] matrix and [imath]AA^{T}[/imath] is a symmetric real matrix. Also, we have: [imath]\operatorname{rank}(AA^{T})=r\stackrel{?}{=}\operatorname{rank}(A)[/imath]. Let [imath]Q= \begin{Bmatrix} q_1,...,q_{n-r} \end{Bmatrix}[/imath] be a basis for the Null space of [imath]AA^{T}[/imath]. i.e. [imath]AA^{T}q_i=0[/imath], show that [imath]A^{T}q_i=0[/imath]. I guess one proof can be that the Null space for [imath]A^{T}[/imath] is a subspace for Null space for [imath]AA^{T}[/imath], then the question would be why [imath]\operatorname{rank}(A)=\operatorname{rank}(AA^{T})[/imath]? |
2340939 | How many subgroups of order 8 of an abelian group of order 72?
In Fraleigh, there is a question asking if you can say how many subgroups of order 8 an abelian group [imath]G[/imath] of order 72 has, and also if you can say how many subgroups of order 4 that group has. The answer key says there is only one of order 8, but I don't understand why. I know that [imath]G[/imath] can be decomposed into a direct product of cyclic groups. But both [imath]Z_8 \times Z_9[/imath] and [imath] Z_2 \times Z_4 \times Z_9[/imath] are of order 72; while the first one has a subgroup [imath]Z_8[/imath], the second one has a subgroup of [imath]Z_2 \times Z_4[/imath], and these two groups are not isomorphic. What am I missing? Also, why couldn't we say how many groups of order 4 [imath]G[/imath] has? | 1013945 | How many subgroups or order 8 an abelian Group of order 72 can have
Let [imath]G[/imath] be an abelian group of order 72. How many subgroups of order 8 and 4 can it have? I have listed all possible abelian groups there are 6. Then i said that if I'm looking for an abelian group of order 8 I have 3 possible abelian groups and for each 1 of the 6 first abelian groups must check if 1 of the 3 of the later ones are subgroups. But that is taking forever and I don't have any easy way to check if they are subgroups. I use "If A is a subgroup of G and B is a subgroup of H, then the direct product A × B is a subgroup of G × H". So I found that for every possible G there is 1 subgroup of order 8. But is that correct? If I have a subgroup of a direct product say A × B of a G SxS1xS2x...xSn then A must be subgroup of S and B of S1? |
2342171 | How to evaluate [imath]\lim_{n \to 0} n\cot(n)[/imath]
I already know that the value of the limit [imath]\lim_{n \to 0} n\cot(n)[/imath] is equal to [imath]1[/imath], but I'm not quite sure how to evaluate it in an airtight way. I tried L'Hopital, but it just becomes circular. Does anybody have any hints? | 931212 | How do I find [imath]\lim\limits_{x \to 0} x\cot(6x) [/imath] without using L'hopitals rule?
How do I find [imath]\lim\limits_{x \to 0} x\cot(6x) [/imath] without using L'hopitals rule? I'm a freshman in college in the 3rd week of a calculus 1 class. I know the [imath]\lim\limits_{x \to 0} x\cot(6x) = \frac{1}{6}[/imath] by looking at the graph, but I'm not sure how to get here without using L'hopital's rule. Here is how I solved it (and got the wrong answer). Hopefully someone could tell me where I went wrong. [imath]\lim\limits_{x \to 0} x\cot(6x) = (\lim\limits_{x \to 0} x) (\lim\limits_{x \to 0}cot(6x)) = (0)((\lim\limits_{x \to 0}\frac{cos(6x)}{sin(6x)}) = (0)(\frac{1}{0}). [/imath] Therefore the limit does not exist. I am also unsure of how to solve [imath]\lim\limits_{x \to 0} \frac{\sin(5x)}{7x^2} [/imath] without using L'hopital's rule. |
2342329 | Probability proof with independent events
If events [imath]A_1,...A_n[/imath] are independent (in total), show that [imath]P\left(\bigcup\limits_{i=1}^nA_i\right)=1-\prod\limits_{i=1}^{n}\left(1-P(A_i)\right).[/imath] Could someone show how to prove this statement? | 2341408 | Proving an equality regarding fully independent events
I was asked to prove the following: If [imath]\{ E_1,E_2,\ldots,E_n\}[/imath] is a fully independent set of events, prove that: [imath] P\left(\bigcup_{i = 1}^n E_i \right) = 1 - \prod_{i = 1}^n \left[ 1 - P(E_i) \right] [/imath] (Hint: use complements) I got the following: Let [imath]\alpha = E_1\cup E_2 \cup \cdots \cup E_n = \{E^C_1 \cap E^C_2 \cap \cdots\cap E_n^C\}^C[/imath]. Then, [imath]\alpha^C = \{ E^C_1 \cap E^C_2 \cap \cdots \cap E^C\} [/imath]. Assuming that the set of complements of each [imath]{E_i}[/imath] is also fully independent, we have: \begin{align} & P(\alpha^C) = \prod_{i = 1}^n P(E_i^C) = \prod_{i = 1}^n \left[ 1 - P(E_i) \right] \\ & P(\alpha) = 1 - P(\alpha^C) = 1 - \prod_{i = 1}^n \left[ 1 - P(E_i) \right] \\ & P(\bigcup_{i = 1}^n E_i) = 1 - \prod_{i = 1}^n \left[ 1 - P(E_i) \right] \end{align} However, even though I know (by Googling it) that my assumption is true, I'm unable to prove it so I could validly use it as a lemma. Is there another way to prove this equality? |
2342478 | Shewing that a function attains every value.
Let [imath]f[/imath] be an entire non-constant function, such that [imath]f(1-z) = 1 - f(z)[/imath]. Shew that [imath]f({\bf C}) = {\bf C}[/imath] | 188139 | Showing that the image of a function is [imath]\mathbb{C}[/imath] if it satisfies a nice functional equation
Let [imath]f[/imath] be entire and non-constant. Assuming [imath]f[/imath] satisfies the functional equation [imath]f(1-z)=1-f(z)[/imath], can one show that the image of [imath]f[/imath] is [imath]\mathbb{C}[/imath]? The values [imath]f[/imath] takes on the unit disc seems to determine [imath]f[/imath]... Any ideas? |
2342592 | [imath]f(0)=g(0)[/imath] and [imath]f'(x) \leq g'(x) ~\forall~ x \in \mathbb{R} \implies f(x) \leq g(x) ~\forall~ x>0[/imath]
Here's a problem which I thought to be lacking an additional condition. So I imposed the condition and presented my solution. Please check whether it can be done without the additional condition. Thank you. The Problem : Let [imath]f,g:\mathbb{R} \to \mathbb{R}[/imath] be differentiable. Assume that [imath]f(0)=g(0)[/imath] and [imath]f'(x) \leq g'(x) ~\forall~ x \in \mathbb{R}[/imath]. To show that [imath]f(x) \leq g(x) ~\forall~ x \geq 0[/imath]. If we impose the condition that [imath]g'(x) \neq 0 ~\forall~ x>0,[/imath] then it is a quick consequence of Cauchy's Mean Value Theorem. For every [imath]x > 0, \exists c \in (0,x)[/imath] such that [imath]\frac{f(x)-f(0)}{g(x)-g(0)}=\frac{f'(c)}{g'(c)} \tag{1}[/imath] The right hand side of [imath](1)[/imath] is [imath]\leq 1[/imath] because of given assumption. Coupled with the fact that [imath]f(0)=g(0),[/imath] we have the desired result. [imath]\blacksquare[/imath] What can I do when [imath]g'(x)=0[/imath] at some points of [imath](0,\infty)??[/imath] | 2309532 | If [imath]f'(x)\le g'(x)[/imath], prove [imath]f(x)\le g(x)[/imath]
I have to do the following exercise: Let [imath]f[/imath] and [imath]g[/imath] two differentiable functions such that [imath]f(0)=g(0)[/imath] and [imath]f'(x)\leq g'(x)[/imath] for all [imath]x[/imath] in [imath]\mathbb{R}[/imath]. Prove that [imath]f(x)\leq g(x)[/imath] for any [imath]x\geq0[/imath]. Now, I know this is true because the first derivative of a function is the angular coefficient of the function in a point [imath]x[/imath]. So, [imath]f'(x)\leq g'(x)[/imath] means, in other words, that the function [imath]g(x)[/imath] grows faster than [imath]f(x)[/imath]. I think this is the base for a more formal proof, could someone help me to figure out a more formal proof? |
2328758 | Find Number Of Roots of Equation [imath]11^x + 13^x + 17^x =19^x [/imath]
The Equation [imath]11^x + 13^x + 17^x =19^x [/imath] Has No Real Roots Only One Real Roots Exactly Two Real Roots More than Two Real Roots What I have done is The function [imath]f(x)=11^x + 13^x + 17^x -19^x [/imath] is strictly increasing and being always positive it have no roots | 2408618 | What are the number of solutions of [imath]11^x+13^x+17^x-19^x[/imath]?
My attempt: [imath] 11^x+13^x+17^x-19^x=0\implies (\frac{11}{19})^x+(\frac{13}{19})^x+(\frac{17}{19})^x=1 [/imath]. Now taking limit as [imath]x\rightarrow\infty[/imath],we get [imath]0=1[/imath],which is absurd.Hence,the equation has no solution. Is it correct?? |
2343132 | Prove (ma, mb) =m(a,b)
I am trying to prove [imath](ma, mb) =m(a,b)[/imath] where [imath]m[/imath] is a positive integer. Should I start like this : [imath](ma, mb) =d\\ d \mid ma, \quad d \mid mb\\ d \mid max + may\\ d \mid m(ax+by)[/imath] Then what should I do? | 705862 | Prove that $(ma, mb) = |m|(a, b)\ $ [GCD Distributive Law]
I'm trying to prove that [imath](ma, mb) = [/imath]|[imath]m[/imath]|[imath](a, b)[/imath] , where [imath](ma, mb)[/imath] is the greatest common divisor between [imath]ma[/imath] and [imath]mb[/imath]. My thoughts: If [imath](ma, mb) = d[/imath] , then [imath]d[/imath]|[imath]ma[/imath] and [imath]d[/imath]|[imath]mb[/imath] → [imath]d[/imath]|[imath]max + mby[/imath] → [imath]d[/imath]|[imath]m(ax+by)[/imath]. This implies that [imath]d[/imath]|[imath]m[/imath] or [imath]d[/imath]|[imath](ax+by)[/imath]. This is the same as [imath]d[/imath]|[imath]m[/imath] or [imath]d[/imath]|[imath]a[/imath] and [imath]d[/imath]|[imath]b[/imath], so [imath]d[/imath]|[imath]m[/imath] or [imath]d[/imath]|[imath](a,b)[/imath]. This is the same as [imath]d[/imath]|[imath]m|[/imath] or [imath]d|(a,b)[/imath], so [imath]d[/imath]|[imath]|m|(a,b)[/imath]. I don't know what to do. Thanks |
2343269 | Finding a root of transcendental equation
I am having a problem with the following equation [imath]e^{2x}(2x−1)+1=0[/imath] By trial and error, you can see that the root is [imath]x=0[/imath], but how can I actually find it? [imath]e^{2x}=\frac{-1}{2x-1}[/imath] As we can see, [imath]e^{2x} > 0, \forall x \in \Re[/imath], but [imath]\frac{-1}{2x-1} > 0 , x \in (-\infty,\frac{1}{2})[/imath], and [imath]\frac{-1}{2x-1}<0, x\in(\frac{1}{2},\infty)[/imath]. When the [imath]x\to-\infty[/imath], LHS is decreasing, and RHS is increasing, but when [imath]x\to\infty[/imath] LHS is increasing, and RHS is decreasing. How can I find the root of this equation? It is clear that it has a root, because one can easily find two points where in one function is positive, and in the other one it is negative. | 2342089 | Examining the function
I am having trouble with this function [imath]f(x) = e^{2x}(2x-1)+1[/imath] I found the domain, domain is [imath]\mathbb R[/imath], it is neither even nor odd function. I found that [imath]f(x)=0 \Leftrightarrow x=0[/imath], because it is obvious. I am having trouble with finding the sign of this function. That "+1" is making my life miserable. I can easily find the sign of [imath]e^{2x}(2x-1)[/imath], but what do I do with the constant? Do I find the minimum of [imath]e^{2x}(2x-1)[/imath] and then add constant to it, and then check the sign? That seems like too much work, considering that this function is actually a numerator of the first derivative of function that appeared in other graphing assignment. There has to be a better way... The original graphing assignment was [imath]g(x)=\ln(e^{\frac{2x}{1-e^{2x}}})[/imath] You can ignore this one when writing the answer. I included it for the sake of context. EDIT: By finding the "sign" of a function I mean finding the intervals for which [imath]f(x)>0[/imath] and [imath]f(x)<0[/imath]. |
2314535 | Prove that xf(x) is continuous with a specific case.
We got [imath] f(x) = \begin{cases} \sin \frac1x,&x\neq 0 \\ 0,&x = 0 \end{cases} [/imath] I have to prove that g is continuous, where [imath] g: \mathbb R \rightarrow \mathbb R, x \rightarrow xf(x)[/imath]. How can I do this? | 1551257 | Derivative of piece-wise function given by [imath]x\sin\frac1x[/imath] at [imath]x=0[/imath]
Given the function: [imath]f(x) = \begin{cases} x\cdot\sin(\frac{1}{x}) & \text{if $x\ne0$} \\ 0 & \text{if $x=0$} \end{cases}[/imath] Question 1: Is [imath]f(x)[/imath] continuous at [imath]x=0[/imath]? Question 2: What is the derivative of [imath]f(x)[/imath] at [imath]x=0[/imath] and how do I calculate it? |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.