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1300320 | A question on [imath]k[/imath]-connected graphs
I'm looking for a proof of this theorem by Dirac: If a graph is [imath]k[/imath]-connected for [imath]k \ge 2[/imath], then for every set of [imath]k[/imath] vertices in the graph there is a cycle that passes through all the vertices in the set. I know a proof probably uses Menger's theorem but I haven't been able to find one or come up with one. | 156032 | Every [imath]k[/imath] vertices in an [imath]k[/imath] - connected graph are contained in a cycle.
Let [imath]G[/imath] be a [imath]k[/imath]-connected graph. Meaning, [imath]G[/imath] has no fewer than [imath]k[/imath] vertices, and for every set of [imath]k-1[/imath] or fewer vertices, if we remove them from [imath]G[/imath], the graph stays connected (Of course, [imath]G[/imath] itself is also connected). I want to prove that for any [imath]k>1[/imath], if [imath]G[/imath] is [imath]k[/imath]-connected, then every set of [imath]k[/imath] vertices is contained in a cycle. I have tried some ways - mainly using induction by removing one of the vertices of the set from the graph, and/or using Menger's theorem to construct the cycle. But I always encounter problems with making sure that the cycle I'm building deosn't have repeating edges etc. Help would be greatly appreciated :) Thanks! |
1300126 | If [imath] Y [/imath] is irreducible set so is [imath]cl(Y)[/imath].
If [imath] Y [/imath] is irreducible set so is [imath]cl(Y)[/imath]. If [imath]cl(Y)[/imath] is reducible then [imath]cl(Y)= A \cup B[/imath] where both [imath]A[/imath] and [imath]B[/imath] is closed in [imath]cl(Y)[/imath]. Now how do we proceed? | 314311 | The closure of an irreducible subset of an irreducible space is irreducible.
Start with an irreducible space [imath]X[/imath]. Take a subset [imath]Y[/imath] that is irreducible. Show that the closure of [imath]Y[/imath] is still irreducible. I imagine we are supposed to start with saying, assume we have a decomposition for [imath]\bar Y = S\cup T[/imath] and then somehow derive a contradiction to [imath]X[/imath] or [imath]Y[/imath]'s irreducibility, but I am struggling. |
605369 | If [imath]\int_a^b \left |{f^{\prime}}\right |=V\left. [ a,b \right ][/imath], [imath]f[/imath] of bounded variation, then [imath]f [/imath] absolutely continuous.
I need a little help with this. If [imath]f:\mathbb{\left. [ a,b \right ]}\longrightarrow{\mathbb{R}}[/imath] is a function of bounded variation on [imath]\left. [ a,b \right ][/imath], and if [imath]\int_a^b \left |{f^{\prime}}\right |=V\left. [ a,b \right ][/imath], where [imath]V\left. [ a,b \right ][/imath] is the variation of [imath]f[/imath] on [imath]\left. [ a,b \right ][/imath]. Show that [imath]f[/imath] is absolutely continuous on [imath]\left. [ a,b \right ][/imath] I think this must be easy, but I don't see the way of prove it. Thanks. | 418374 | Bounded variation and [imath]\int_a^b |F'(x)|dx=T_F([a,b])[/imath] implies absolutely continuous
If [imath]F[/imath] is of bounded variation defined on [imath][a,b][/imath], and [imath]F[/imath] satisfies [imath]\int_{a}^b |F'(x)|dx=T_F([a,b])[/imath] where [imath]T_F([a,b])[/imath] is the total variation. How to prove that [imath]F[/imath] is absolutely continuous? My Attempt: I used the inequality, that [imath]P_F([a,x])[/imath] and [imath]N_F([a,x])[/imath] (positive and negative variation )are both monotonic non-decreasing function, and thus their derivative exists a.e. Thus [imath]\int_{a}^b |F'(x)|dx\le \int_a^b |P_F'([a,x])|dx+\int_a^b|N_F'([a,x])|dx\le P_f([a,b])+N_F([a,b])=T_F([a,b])[/imath] Then by the condition, the middle inequality should all be replaced by equality. But I cannot derive any useful information from the equalities, since the annoying absolute value cannot be diminished. I tried to prove [imath]F(x)=F(a)+\int_{a}^x F'(t)dt[/imath] but this doesn't work. Applying the definition of absolute continuity also failed to give me a clearer view. Thanks for your attention! |
1299906 | How to prove that [imath]f_n(x)=\frac{nx}{1+n\sin(x)}[/imath] does not converge uniformly on [imath][0, \pi/2][/imath]?
If [imath]f_n[/imath] is a sequence of functions over [imath][0, \pi/2][/imath] given by [imath]f_n(x) = \frac {nx} {1+n\sin(x)},[/imath] then how would I go about proving that [imath]f_n[/imath] does not converge uniformly to a function [imath]f[/imath] on [imath][0, \pi/2][/imath]? | 1300277 | Does [imath]\frac{nx}{1+n \sin(x)}[/imath] converge uniformly on [imath][a,\pi/2][/imath] for all [imath]a \in (0,\pi/2][/imath]?
Edit: the question had some missing details. It should read as follows: Prove for all [imath]a \in (0,\frac{\pi}{2}][/imath], [imath]f_n \rightarrow f[/imath] uniformly on [imath][a,\frac{\pi}{2}][/imath]. Here [imath]f_n(x) = \frac{n x}{1 + n \sin(x)}.[/imath] Here is my attempt at the problem: If [imath]x \in (0, \frac{\pi}{2}][/imath] then [imath]| f_n(x) - f(x) |[/imath] [imath] =\left| \frac{nx}{1 + n\sin(x)} - \frac{x}{\sin(x)}\right| [/imath] [imath] = \left|\frac{nx \sin(x) - x(1 + n \sin(x))}{[1+ n \sin(x)]\sin(x)}\right|[/imath] [imath] = \frac{x}{\sin(x) + n \sin^2(x)} \leq \frac{1}{n} \space \space \text{(is this line correct)}[/imath] So [imath]\forall \epsilon > 0[/imath], we may choose [imath]N \geq \frac{1}{\epsilon}[/imath] such that when [imath]n \geq N \implies |f_n(x)-f(x)| \leq \epsilon \space \space \forall x \in (0, \frac{\pi}{2}][/imath] It's already been established that [imath]f_n(x)[/imath] converges pointwise to: [imath]f(x) = 0, x =0[/imath] and [imath]f(x) = \frac{x}{\sin(x)} ,x \in (0, \frac{\pi}{2}][/imath] |
1300307 | Girsanov theorem
I work on an exercice and I have to calculate: [imath]E(W_{t}^2e^{(\int_{0}^{T}\theta_{s}dW_{s}-\frac{1}{2}\int_{0}^{T}\theta_{s}^2ds)})[/imath] [imath]\theta[/imath] is deterministic function I don't know how to resolve this type of question neither how to use Girsanov Thanks in advance for your help | 1298699 | positive martingale process
I would to like to prove that the process: [imath]e^{\int_{0}^{T}\theta _{s}\,dW_{s}-\frac{1}{2}\int_{0}^{T}\theta _{s}^2\,ds}[/imath] is a martingale which is positive and has a mean=1, where [imath]\theta_s[/imath] is continuous deterministic function. Thank you |
1301095 | Evaluate [imath]\large \int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx [/imath] using elementary, high school techniques
Evaluate [imath]\large \int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx [/imath] [imath]$$ I was given this integral by a friend who saw this here on MSE. He asked me if I could solve it using the very basic tools I have. I was of course unsuccessful.[/imath][imath][/imath] My short and unsuccessful attempt: \large \int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx [imath][/imath] \int_0^1\left(\frac{1}{(\ln x)^2} + \frac{1}{(1-x)^2} +\frac{2}{\ln x(1-x)}\right) \mathrm dx [imath] [/imath]\int_0^1\frac{1}{(\ln x)^2}+\int_0^1\frac{1}{(1-x)^2}+\int_0^1\frac{2}{\ln x(1-x)}[imath][/imath] The problem (apart from the fact that I don't know as yet how to attempt the third integral) is that the second integral (I think) diverges. Could somebody please help me solve it using preferably only these: Integration by Parts, Integration by Substitution, Partial Fraction Decomposition or Differentiating under the Integral Sign?$$ Thanks so much in advance, and I'm truly sorry for the trouble I have caused. | 281406 | Evaluate [imath]\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx[/imath]
Evaluate [imath]\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx[/imath] |
1293559 | [imath]f(x,y)=4x^3y^2[/imath] Directional Derivative...
Let [imath]f(x,y)=4x^3y^2[/imath] How do I find the directional derivative of [imath]f[/imath] at [imath](2,1)[/imath] in the direction of the vector [imath]3i-4j[/imath]? What would be a unit vector in the direction in which [imath]f[/imath] decreases most rapidly from the point [imath](2,1)[/imath]? And, what is the rate of change of [imath]f[/imath] in the direction given in the second question? My Work (What I have done so far): For the first question: [imath]\nabla f(x,y)=12x^2y^2\mathbf i+8x^3y\mathbf j[/imath] [imath]\nabla f(2,1)=48\mathbf i+64\mathbf j[/imath] [imath]u=(3/5)\mathbf i-(4/5)\mathbf j[/imath] therefore [imath]D_u f=\nabla f\bullet u=0[/imath]. I'm not entirely sure if this is right? Can someone please verify? For the second question: [imath]u=(3/5)\mathbf i-(4/5)\mathbf j[/imath] therefore [imath]-\|\nabla f(2,1)\|=?[/imath] How do I find this? And am I doing this right? I have no idea what to do for the third question. | 1293447 | [imath]f(x,y)=4x^3y^2[/imath] Dealing with Directional Derivatives and Vectors
Let [imath]f(x,y)=4x^3y^2[/imath] How do I find the directional derivative of [imath]f[/imath] at [imath](2,1)[/imath] in the direction of the vector [imath]3i-4j[/imath]? What would be a unit vector in the direction in which [imath]f[/imath] decreases most rapidly from the point [imath](2,1)[/imath]? And, what is the rate of change of [imath]f[/imath] in the direction given in the second question? My Work (What I have done so far): For the first question: [imath]\nabla f(x,y)=12x^2y^2[/imath]i[imath]+8x^3y[/imath]j[imath]\nabla f(2,1)=48[/imath]i[imath]+64[/imath]j[imath]u=(3/5)[/imath]i[imath]-(4/5)[/imath]j [imath]\therefore D_u f=\nabla f\bullet u=0[/imath] I'm not entirely sure if this is right? Can someone please verify? For the second question: [imath]u=(3/5)[/imath]i[imath]-(4/5)[/imath]j [imath]\therefore -||\nabla f(2,1)||=[/imath]? How do I find this? And am I doing this right? I have no idea what to do for the third question. |
1301217 | zero drift brownian motions and barriers problem
Given two same brownian motion with no drift and different variances: [imath](dG_1/G_1)= \sigma_1dW_g [/imath] [imath](dG_2/G_2)= \sigma_2dW_g [/imath] and two barriers [imath]P_1 > P_2[/imath] assuming that [imath] \sigma_1 > \sigma_2 [/imath] is it possible to find the probability that a certain time [imath]\tau\ [/imath] "G1 will hit the barrier P1 faster than G2 hits P2"? | 1290818 | No drift brownian motion problem
Given two same brownian motion with no drift and different variances: [imath]dG_1= \sigma_1 G_1 dW [/imath] [imath]dG_2= \sigma_2 G_2 dW [/imath] and two barriers [imath]P_1 > P_2[/imath] assuming that [imath] \sigma_1 > \sigma_2 [/imath] and given [imath]τ_1=\inf [t:G_1(t)≥P_1] [/imath]; [imath]τ_2= \inf [t:G_2(t)≥P_2][/imath] is it possible to find [imath]P(τ_1<τ_2)[/imath]? (the probability that a certain time [imath]\tau\ [/imath] "G1 will hit the barrier P1 earlier than G2 hits P2"?) |
1300425 | Find all continuous real valued function such that [imath](f(x))^2+C=\int\limits_0^xf(t)dt[/imath]
Find all continuous real valued function such that [imath](f(x))^2+C=\int\limits_0^xf(t)dt[/imath] for some [imath]C\in\mathbb{R}[/imath] If I set [imath]F(x)=\int\limits_0^xf(t)dt[/imath] then [imath]F[/imath] is differentiable and [imath]F'(x)=f(x)[/imath], so [imath](f(x))^2+C[/imath] is diferentible as it does [imath]\sqrt{F(x)-C}=|f(x)|[/imath], and then [imath]F'(x)=(|f(x)|^2+C)'=2|f(x)||f(x)|'=f(x)[/imath]. and then I don't know how to continue. I just see the solution [imath]f(x)=0[/imath] whith [imath]C=0[/imath]. | 1135029 | Find all continuous functions satisfying [imath]\int_0^xf=(f(x))^2+C[/imath] for some constant [imath]C \neq 0[/imath].
Find all continuous functions [imath]f[/imath] satisfying [imath]\int_0^xf=(f(x))^2+C[/imath] for some constant [imath]C \neq 0[/imath], assuming that [imath]f[/imath] has at most one [imath]0[/imath]. I have a question about the solution to this problem. It says that Clearly [imath]f^2[/imath] is differentiable everywhere, its derivative at [imath]x[/imath] is [imath]f(x)[/imath] (I'm assuming that this is because the left hand side is differentiable, the right hand side must be differentiable as well). So [imath]f[/imath] is differentiable at [imath]x[/imath] whenever [imath]f(x) \neq 0[/imath], and [imath]f(x)=2f(x)f'(x),[/imath] so [imath]f'(x) = \frac{1}{2}[/imath] at such points. Thus, [imath]f= \frac{1}{2}x+b[/imath] for some [imath]b[/imath] on any interval where it is non-zero; if [imath]f[/imath] has a zero, this gives two possible solutions, with two possible values of [imath]b[/imath], but since [imath]f[/imath] is assumed continuous, they must be the same. So we need [imath]\int_0^x(\frac{1}{2}t+b)dt=(\frac{1}{2}x+b)^2+C[/imath] so we must have [imath]b=\sqrt{-C}[/imath] or [imath]-\sqrt{-C}[/imath], leading to two solutions for [imath]C \lt 0[/imath]. I don't understand the bolded statements. First, how does the differentiability of [imath]f^2[/imath] guarantee that [imath]f[/imath] is differentiable, and why only at [imath]x[/imath] where [imath]f(x) \neq 0[/imath]? Next, why are there two possible solutions if [imath]f[/imath] has a zero, but since [imath]f[/imath] is continuous they must be the same? Finally, does the last statement mean that if [imath]C \gt 0[/imath], then there is no solution, since we're restricted to real numbers here? I would appreciate it if anyone answers my questions. |
1301768 | Is the cartesian product associative
Does the cartesian product have an associative property such that [imath]M_1\times(M_2\times M_3) = (M_1\times M_2)\times M_3[/imath], or does the different order result in different ordered pairs? | 338319 | Associativity of Cartesian Product
I have a basic doubt about the associativity of the cartesian product. Well, first wikipedia says that the cartesian product isn't associative, and there's a good argument for it: if [imath]x\in E[/imath], [imath]y\in F[/imath] and [imath]z \in G[/imath] the identity [imath]((x,y), z)=(x,(y,z))[/imath] would imply that [imath](x,y) =x[/imath] and [imath](y,z) = z[/imath] so that [imath]((x,y),z)=(x,y,z)[/imath] would mean nothing. That's fine, I like this argument. However, in his book Calculus on Manifolds, Spivak says that the cartesian product is associative. He says: if [imath]A \subset \mathbb{R}^m[/imath], [imath]B\subset \mathbb{R}^n[/imath] and [imath]C \subset \mathbb{R}^p[/imath] then [imath](A\times B)\times C = A \times (B \times C)[/imath], and both of these are denoted simply by [imath]A \times B \times C[/imath]. Well, this confuses me because Spivak is always very rigorous, so that he wouldn't state something that's not true in such way. Is Spivak or Wikipedia right ? Or Spivak's statement only works for subsets of euclidean spaces ? Thanks in advance and sorry if this question is too basic. |
947971 | Is it possible to extend a commutative ring to have a unity?
Let [imath]R[/imath] be a commutative ring. Then, is it possible to extend this to have a unity? That is, is there a commutative ring with unity [imath]R'[/imath] such that [imath]R[/imath] is a subring of [imath]R'[/imath]? | 1301432 | For any rng [imath]R[/imath], can we attach a unity?
Let [imath]R[/imath] be an rng. (There may be no unity) Then, does there always exist a ring(with unity) [imath]A[/imath] such that [imath]R[/imath] is a subrng of [imath]A[/imath]? |
914621 | Sum of a series problem involving cubes
For any odd integer [imath]n[/imath], evaluate [imath]n^3 - (n-1)^3 + \cdots + (-1)^{n-1} \cdot 1^3 = \sum_{k=1}^n (-1)^{k+1}\cdot k^3[/imath] How would you go about solving such a problem? Any help would be appreciated. | 1301913 | Alternating Sum of Cubes
How is it possible to evaluate: [imath]\sum_{k=1}^n{((-1)^{n-k}\cdot k^3)}=n^3 - (n-1)^3 + (n - 2)^3 - \cdots \pm 1^3[/imath] The fact that there is the [imath]\pm[/imath] at the end makes it difficult. |
1302535 | Why Square Brackets for Expectation
I've often seen [imath]\mathbb{E}[X][/imath] instead of [imath]\mathbb{E}(X)[/imath], but it seems variance is almost always [imath]Var(X)[/imath]. E.g., Wikipedia for Expected Value and Variance. Is there a good mathematical reason for using square brackets for one and not the other? | 573148 | Why is the expected value (mean) of a variable written using square brackets?
My question is told in a few words: Why do you write [imath]E[X][/imath] in square brackets instead of something like [imath]E(X)[/imath]? Probably it is not a "function". How would you call it then? This question also applies for [imath]Var[X][/imath]. |
1302991 | ML-inequality: Why does this hold [imath]|e^{-3y+3ix}| = e^{-3y}[/imath] during numerator estimation of [imath]f(z) = \frac{e^{3iz}}{z^2 + 1}[/imath]
Given the following related to an ML-inequality for [imath]R > 1[/imath]: Estimation of the numerator from the function [imath]f(z)[/imath] is supposed to develop as follows: I'm wondering why and how exactly the power [imath]3ix[/imath] cancels out (why does it vanish/equal to zero) in the equality [imath]|e^{-3y+3ix}| = e^{-3y}[/imath] ? | 1297025 | Why is [imath]|e^{i \lambda z}| |e^{- \lambda y}|= |e^{- \lambda y}|[/imath] here?
Let [imath]z \in \Gamma (R)[/imath] where this is the upper semi circle centred at the origin with radius [imath]R>1[/imath]. Let [imath]z=x+iy[/imath] with [imath]x \in \mathbb{R}[/imath] and [imath]y \geq 0[/imath]. So [imath]|e^{i \lambda z}|=|e^{i \lambda z}| |e^{- \lambda y}|= |e^{- \lambda y}|[/imath] where [imath]\lambda \in \mathbb{R}^{+}[/imath]. How can the last equality possibly be true? |
1302659 | Prove by combinatorial method that [imath] \frac{(2m)! \cdot (2n)!}{(m)! \cdot (n)! \cdot (m+n)!} [/imath] is an integer
Prove that [imath] \dfrac{(2m)! \cdot (2n)!}{(m)! \cdot (n)! \cdot (m+n)!} [/imath] is a positive integer, where [imath](m,n) \in \mathbb{Z^{+}}[/imath] I have already solved it using Legendre's Formula which states that [imath]e_{p}(n)=\sum_{i=1}^{\infty} \bigg\lfloor \dfrac{n}{p^{i}} \bigg\rfloor[/imath] where [imath]e_{p}(n)[/imath] is the exponent of a prime [imath]p[/imath] in [imath]n![/imath]. For the problem it was sufficient to show that [imath] e_{p}(2m) + e_{p}(2n) \ge e_{p}(m) + e_{p}(n) + e_{p}(m+n) [/imath] which I can show using the properties of floor function. However, I'm seeking a combinatorial approach to this problem. For example, using basic combinatorics, I can show that the number of ways to divide [imath]A[/imath] objects into [imath]k[/imath] persons such that the [imath]i^{th}[/imath] person receives [imath]a_{i}[/imath] objects is [imath] \dfrac{A!}{\displaystyle\prod_{i=1}^{k}{(a_{i})!}} = \dfrac{\left(\displaystyle\sum_{i=1}^k (a_{i})\right)!}{\displaystyle\prod_{i=1}^{k}{(a_{i})!}} [/imath] here, the set [imath]\{a_{i}\}_{i=1}^k[/imath] is exhaustive, i.e, [imath] A = \displaystyle\sum_{i=1}^k a_{i} [/imath]. Using this, I can show the following numbers to be integer [imath] \dfrac{(2m)! \cdot (2n)!}{[(m)!]^{2} \cdot [(n)!]^{2} } [/imath] [imath] \dfrac{(2m)! \cdot (2n)!}{(m-n)! \cdot [(n)!]^2 \cdot (m+n)!} [/imath] ; if [imath]m \geq n[/imath] [imath] \dfrac{(2m)! \cdot (2n)!}{(n-m)! \cdot [(m)!]^2 \cdot (m+n)!} [/imath] ; if [imath]n \geq m[/imath] However, I can't seem to find a way to tackle this problem using my approach. Edit: I'm specifically asking for an answer using my combinatorics approach as I've already solved it using the answer given in the other question. Any help will be appreciated. Thanks. | 215355 | Prove that for all non-negative integers [imath]m,n[/imath], [imath]\frac{(2m)!(2n)!}{m!n!(m + n)!}[/imath] is an integer.
Prove that for all non-negative integers [imath]m,n[/imath], [imath]\frac{(2m)!(2n)!}{m!n!(m + n)!}[/imath] is an integer. I'm not familiar to factorial and I don't have much idea, can someone show me how to prove this? Thank you. |
1302554 | Inequality and Trigonometric Substitution
Prove that for all positive real [imath]a,b,c[/imath], we have [imath](a^2+2)(b^2+2)(c^2+2) \geq 9(ab+bc+ca).[/imath] Because of the term [imath]a^2+2[/imath], this motiveates me to substitute [imath]a=\sqrt{2}\tan A, b=\sqrt{2}\tan B, c=\sqrt{2}\tan C[/imath] and the fact that [imath]1+\tan^2x=\sec^2x[/imath], then the desired inequality becomes [imath]2^2\sec^2 A\sec^2 B \sec^2 C\geq 3^2(\tan A\tan C + \tan B\tan C + \tan C \tan A).[/imath] But then I was stuck and couldn't move further, please helps. | 834025 | Showing $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$
Let [imath]a[/imath], [imath]b[/imath], [imath]c[/imath] be nonnegative real numbers. Prove [imath](a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)[/imath]. |
1302470 | The Imaginary number
Sometimes one writes [imath]i^2=-1[/imath] to characterize the imaginary number and sometimes as the root of [imath]-1[/imath]. So when I resolve the first equation, I get for the imaginary number two roots of [imath]-1[/imath] and thus two complex numbers. I am confused. Is it a question of definition or are the both definitions the same? Thanks for your comment. | 177594 | How to tell [imath]i[/imath] from [imath]-i[/imath]?
Suppose now we are trying to explain to students who do not know complex numbers, how do we distinguish [imath]i[/imath] and [imath]-i[/imath] to them? They will object that they both squared to [imath]-1[/imath] and thus they are indistinguishable. Is there a way of explaining this in an elementary way without go into introductory things in complex analysis? |
1303378 | How is the first Sylow theorem a strenghtening of Cauchy's theorem?
Taken from Wikipedia: Theorem 1: For any prime factor p with multiplicity [imath]n[/imath] of the order of a finite group [imath]G[/imath], there exists a Sylow [imath]p[/imath]-subgroup of [imath]G[/imath], of order [imath]p^n[/imath]. The following weaker version of theorem 1 was first proved by Cauchy, and is known as Cauchy's theorem. Corollary: Given a finite group [imath]G[/imath] and a prime number [imath]p[/imath] dividing the order of G, then there exists an element (and hence a subgroup) of order [imath]p[/imath] in [imath]G.[/imath] My confusion arises when we consider a case where a prime [imath]p^n[/imath] appears with multiplicity [imath]n\geq 2;[/imath] according to the first Sylow theorem above, there exists a [imath]p[/imath]-subgroup of order [imath]p^n[/imath] in [imath]G,[/imath] but a priori we know nothing about the existence of [imath]p[/imath]-subgroups of order [imath]p^i[/imath] for [imath]i=1,\ldots,n-1.[/imath] | 635031 | How does Cauchy's theorem follow from Sylow's theorem?
Very quickly, Sylow's first theorem says a sylow p-subgroup of order [imath]p^rm[/imath] exists and Cauchy's theorem says if [imath]p \vert |G|[/imath] then there is an element of order [imath]p[/imath]. It's often said that Cauchys follows easily from Sylows, but I just don't see it! I don't see why a sylow p-subgroup must have an element of order [imath]p[/imath]; why couldn't they all be of order [imath]p^n,\ 2<n<r[/imath]? |
1303106 | what is the difference between [imath](\cos A)^2[/imath] and [imath]\cos^2 A[/imath]
[imath] (|\cos A|)^2 \qquad\text{and}\qquad \cos^2 A [/imath] For example if [imath]\cos A = 0.5[/imath], and [imath]0.5 \times 0.5 = 0.25[/imath], are here some difference in the notations or are they equal? | 1172760 | meaning of powers on trig functions
I always forget this, when a trig function has an exponent does that mean multiply itself or apply itself to the result recursivly? e.g. does [imath]\sin(x)^2=\sin(x)\sin(x)[/imath] or [imath]=\sin(\sin(x))[/imath]? What about [imath]\sin^2x[/imath]? |
1282331 | Localization of an integral A-algebra is not always integral.
Let [imath]A[/imath], [imath]B[/imath] rings with a morphism [imath]f : A \to B[/imath] and suppose that [imath]B[/imath] is integral over [imath]A[/imath]. Let [imath]\mathfrak{n} \subseteq B[/imath] a maximal ideal, and [imath]\mathfrak{m}[/imath] its preimage under [imath]f[/imath] (so [imath]\mathfrak{m}[/imath] is maximal in [imath]A[/imath]). The question is: Is the induced ring homomorphism [imath]A_{\mathfrak{m}} \to B_{\mathfrak{n}}[/imath] always integral? The answer will be "No", but of course my aim it to find a counterexample. Luckily, I have an hint: Consider [imath]A = k[X^2 - 1] \subseteq B = k[X][/imath] I see that in that case we have an integral A-algebra, but unfortunately I am finding some difficulties in proceeding. I need a maximal ideal of [imath]B[/imath] that does the job, so I tried with [imath](x)[/imath]. After having done a lot of computation, I am feeling a bit lost in my results...please, could you help me in any way? Also just saying "Yes, [imath](x)[/imath] works, try to think better." or "Instead of [imath](x)[/imath], try to consider the ideal [imath]\mathfrak{b}[/imath]" will be appreciated comments. Any suggestion is welcome :) Cheers | 1194636 | Atiyah-Macdonald, Exercise 5.4
I was having some trouble with the following exercise from Atiyah-Macdonald. Let [imath]A[/imath] be a subring of [imath]B[/imath] such that [imath]B[/imath] is integral over [imath]A[/imath]. Let [imath]\mathfrak{n}[/imath] be a maximal ideal of [imath]B[/imath] and let [imath]\mathfrak{m}=\mathfrak{n} \cap A[/imath] be the corresponding maximal ideal of [imath]A[/imath]. Is [imath]B_{\mathfrak{n}}[/imath] integral over [imath]A_{\mathfrak{m}}[/imath]? The book gives a hint which serves as a counter-example. Consider the subring [imath]k[x^{2}-1][/imath] of [imath]k[x][/imath] where [imath]k[/imath] is a field, and let [imath]\mathfrak{n}=(x-1)[/imath]. I am trying to show that [imath]1/(x+1)[/imath] could not be integral over [imath]k[x^{2}-1]_{\mathfrak{n}^{c}}[/imath]. I have understood why this situation serves as a counterexample. But I am essentially stuck at trying to draw a contradiction. A hint or any help would be great. |
1303758 | Why does [imath]rank(A) \ge \dfrac{{{{(trA)}^2}}}{{(tr{A^2})}}[/imath]?
Let [imath]A \in {M_n}[/imath] and Hermitian.Why does [imath]rank(A) \ge \dfrac{{{{(trA)}^2}}}{{(tr{A^2})}}[/imath]? | 72545 | Prove [imath]\text{rank}(A) \geq \frac{(\text{tr}(A))^2}{\text{tr}(A^2)}[/imath] when [imath]A[/imath] is Hermitian
If [imath]A \in \mathbb{C}^{n \times n}[/imath] is a non-zero Hermitian matrix, I need to show that [imath]\text{rank}(A) \geq \frac{(\text{tr}(A))^2}{\text{tr}(A^2)}[/imath] and reason out when the equality is attained? Can anyone help me in showing this result? |
1303395 | Is there a constructive discontinuous exponential function?
It is well-known that the only continuous functions [imath]f\colon\mathbb R\to\mathbb R^+[/imath] satisfying [imath]f(x+y)=f(x)f(y)[/imath] for all [imath]x,y\in\mathbb R[/imath] are the familiar exponential functions. (Prove [imath]f(x)=f(1)^x[/imath] successively for integers [imath]x[/imath], rationals [imath]x[/imath], and then use continuity to get all reals.) The usual example to show that the identity [imath]f(x+y)=f(x)f(y)[/imath] alone doesn't characterize the exponentials requires the axiom of choice. (Define [imath]f[/imath] arbitrarily on the elements of a Hamel basis for [imath]\mathbb R[/imath] over [imath]\mathbb Q[/imath], then extend to satisfy the identity.) Is there an explicit construction of a discontinuous function satisfying the identity? On the other hand, does the existence of such a function imply the axiom of choice or some relative? | 492751 | Is Zorn's lemma necessary to show discontinuous [imath]f\colon {\mathbb R} \to {\mathbb R}[/imath] satisfying [imath]f(x+y) = f(x) + f(y)[/imath]?
A UC Berkeley prelim exam problem asked whether an additive function [imath]f\colon {\mathbb R} \to {\mathbb R}[/imath], i.e. satisfying [imath]f(x + y) = f(x) + f(y)[/imath] must be continuous. The counterexample involved taking a positive-valued Hamel basis [imath]X[/imath] of [imath]\mathbb{R}[/imath] as a vector space over [imath]{\mathbb Q}[/imath], and then letting [imath]f(x_1) =1[/imath] and [imath]f(x_2)=-1[/imath] for two different [imath]x_1,x_2 \in X[/imath], and letting [imath]f(x)[/imath] be arbitrary for other [imath]x \in X[/imath], and then extending the function to all of [imath]{\mathbb R}[/imath] using the property of additivity. Then a sequence [imath]a_n = {p_nx_1 + q_nx_2}[/imath] could be found with rational [imath]p_n,q_n[/imath] such that [imath]a_n \to 0[/imath] but [imath]\lim \limits_{n\to \infty} f(a_n) \neq 0[/imath], showing discontinuity. But is Zorn's Lemma necessary to produce such an example? In other words, is Zorn's Lemma saying we can find a Hamel basis of [imath]{\mathbb R}[/imath] over [imath]{\mathbb Q}[/imath] equivalent to being able to construct a discontinuous additive function [imath]f\colon {\mathbb R} \to {\mathbb R}[/imath]? |
536664 | Orthogonal projection and two subspaces
Let [imath]\mathcal{S}[/imath] and [imath]\mathcal{T}[/imath] be two subspaces of [imath]\mathbb{R}^n[/imath], let [imath]P[/imath] be the orthogonal projection of [imath]\mathbb{R}^n[/imath] on [imath]\mathcal{S}[/imath] and let [imath]Q[/imath] be the orthogonal projection of [imath]\mathbb{R}^n[/imath] onto [imath]\mathcal{T}[/imath]. Show that if [imath]P[/imath] and [imath]Q[/imath] commute, then [imath]PQ[/imath] is a projection and [imath]PQ[/imath] is the projection onto [imath]\mathcal{S}\cap \mathcal{T}[/imath]. Is the converse assertion true? Suppose [imath]PQ[/imath] is the orthogonal projection of [imath]\mathbb{R}^n[/imath] onto the intersection [imath]\mathcal{S}\cap \mathcal{T}[/imath]. Must [imath]P[/imath] commute with [imath]Q[/imath]. Anybody has advice on how i should start proving this assertion? | 2408084 | when do orthogonal projection [imath]\ P_UP_V = P_VP_U[/imath]
U, V are subspaces of a finite dimensional vector space W, Let [imath]\ P_U[/imath] and [imath]\ P_V[/imath] be the orthogonal projections onto U and V respectively. When is it true that [imath]\ P_UP_V = P_VP_U[/imath]? |
455193 | For [imath]f[/imath] continuous, show [imath]\lim_{n\to\infty} n\int_0^1 f(x)x^n\,dx = f(1).[/imath]
Suppose [imath]f:[0,1]\to \mathbb{R}[/imath] is continuous. Show that [imath]\lim_{n\to\infty} n\int_0^1 f(x)x^n\,dx = f(1).[/imath] My answer so far: First I want to assume that [imath]f\in C^1[/imath]. Then [imath]n\int_0^1f(x)x^n\,dx = \left[\frac{n}{n+1}x^{n+1}f(x)\right]_0^1 - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\\ \frac{n}{n+1}f(1) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx,[/imath] which goes to [imath]f(1)[/imath] because the last integral goes to zero. But approximating [imath]f[/imath] by [imath]\phi\in C^1[/imath] won't necessarily work, because [imath]\phi(1)[/imath] may not equal [imath]f(1)[/imath]... how can we finish the argument? | 1392619 | Uniform Convergence and limit [imath](n+1)\int_0^1 x^nf(x) \; dx[/imath]
If [imath]f[/imath] is a continuous real-valued function, show that [imath] f(1)=\lim_{n\to \infty} \int_0^1 (n+1)\,x^n \,f(x) \; dx [/imath] I am looking for a general hint or steps to proceed but I want to fill them in. Looking at this I see the '[imath]n+1[/imath]' along with the [imath]\int x^n[/imath] makes me very suspicious. My first thought is that [imath](n+1)x^nf(x)[/imath] converges to [imath]f(1)[/imath] uniformly on [imath][0,1][/imath] but I feel as though this is a fools thought. I then sat and tried the case [imath]f(x)=x[/imath] and [imath]f(x)=x^{n_0}[/imath]. This was a bit illuminating. I then tried the case where [imath]f(x)[/imath] is a general polynomial and I can show that case. But of course [imath]f(x)[/imath] need not be a polynomial. To try to make this method work, I considered using Stone-Weierstrass. I can find a sequence of polynomials [imath]p_n(x)[/imath] such that [imath]p_n(x) \to f(x)[/imath] uniformly. But I am unsure of how to make it fit in with the integral. I assume it is trivial to show [imath]x^np_n(x) \to x^nf(x)[/imath]. I know that the above limit works with [imath]x^np_n(x)[/imath]. However, does this imply that this -- under the limit -- must be the same as [imath]x^nf(x)[/imath]? Or is this not at all how I should go about it? Is there a better or easier way? Or perhaps a more enlightening or 'quaint' way? |
1303212 | The Taylor series of the squared logarithm
Prove that the Taylor series about the origin of the function [imath][\log(1-z)]^2[/imath] is given by [imath]\sum_{n=1}^{\infty} \frac{2H_{n}}{n+1} z^{n+1}[/imath] where [imath]H_{n} = \sum_{j=1}^{n}\frac{1}{j}[/imath] is the [imath]n[/imath]-th partial sum of the harmonic series. (Hint: Write [imath]\log(1-z)[/imath] as a power series for [imath]|z| < 1[/imath] and use Cauchy products for [imath][\log(1-z)]^2[/imath]) I've been working on this question since last night and I've been trying to follow the hint and look at the derivative of [imath]\log(1-z)[/imath] and then integrate that, but all attempts have been futile. Really lost here... | 1303510 | Taylor series of [imath][\log(1-z)]^2 [/imath]
I'm having some trouble proving that the Taylor series about the origin of the function [imath][Log(1-z)]^2[/imath] to be [imath]\sum_{n=1}^\infty \frac{2H_n}{n+1}z^{n+1}[/imath] where [imath]H_n = \sum_{j=1}^n \frac{1}{j}[/imath] So far, I've been trying to use the definition of Log(1-z) = log|1-z| + iArg(1-z). I've also noted that the Taylor expansion for log|1-z| = [imath]\sum_{k=0}^\infty \frac{-z^{k+1}}{k+1}[/imath] from integrating the geometric series [imath]\sum_{k=0}^\infty z^k[/imath] given |z|<1. Because log|1-z| = [imath]\sum_{k=0}^\infty \frac{-z^{k+1}}{k+1}[/imath] converges for |z|<1, I've tried to use this for the Cauchy product of [Log(1-z)]^2 [imath][Log(1-z)]^2 = Log(1-z) Log(1-z)[/imath] [imath]=\left(\sum_{i=0}^\infty \frac{-z^{i+1}}{i+1}\right)\left(\sum_{j=0}^\infty \frac{-z^{j+1}}{j+1}\right)[/imath] [imath]=\sum_{n=0}^\infty \sum_{k=0}^n\frac{-z^{k+1}}{k+1} \frac{-z^{n-k+1}}{n-k+1}[/imath] [imath]= \sum_{n=0}^\infty \sum_{k=0}^n\frac{z^{n+2}}{(k+1)(n-k+1)}[/imath] Really lost here, so any clues would be helpful! Thank you in advance. |
1304409 | Suppose that [imath]f:(0,\infty)\rightarrow \mathbb{R}[/imath] is a function satisfying [imath]f'(x)=1/x[/imath] for all [imath]x\in (0,\infty)[/imath], show [imath]f(xy)=f(x)+f(y)[/imath].
I would like to ask you for some help with the following problem. Suppose that [imath]f:(0,\infty)\rightarrow \mathbb{R}[/imath] is a function satisfying [imath]f'(x)=1/x[/imath] for all [imath]x\in (0,\infty)[/imath], and [imath]f(1)=0[/imath]. Show, without using the fact that [imath]f[/imath] must be the natural logarithm, that for all [imath]x,y \in (0,\infty)[/imath], [imath]f(xy)=f(x)+f(y)[/imath]. Hint: fix [imath]y[/imath] and let [imath]g(x)=f(xy)[/imath]. Find [imath]g'(x)[/imath]. The furthest I managed to go is to find the [imath]g'(x)=f'(x)[/imath], but I have no idea how to proceed. Have you got any suggestions? | 501388 | [imath]g'(x) = \frac{1}{x}[/imath] for all [imath]x > 0[/imath] and [imath]g(1) = 0[/imath]. Prove that [imath]g(xy) = g(x) + g(y)[/imath] for all [imath]x, y > 0[/imath].
I'm trying to write a proof for this, but I don't know how to begin. Any help appreciated. Suppose [imath]g[/imath] is a function such that [imath]g'(x) = \frac{1}{x}[/imath] for all [imath]x > 0[/imath] and [imath]g(1) = 0[/imath]. Prove that [imath]g(xy) = g(x) + g(y)[/imath] for all [imath]x, y > 0[/imath]. Let [imath]h(x) = g(x,y)[/imath] |
1304493 | For which points of circle function [imath]\frac{1-y}{x}[/imath] is regular
Consider circle [imath]X = V(x^2+y^2-1)[/imath] and rational function [imath]\varphi = \frac{1-y}{x}[/imath] on [imath]X[/imath]. I want to know for which points [imath]\varphi[/imath] is regular(in sense of algebraic geometry). Intuitively i think it's [imath]X\setminus(-1,0)[/imath], but i can't prove that simple fact. | 727622 | when is rational function regular?
In general, how does one determine if a rational function is regular? I have the particular problem of determining in which points of the circle [imath]V(x^2+y^2-1) \subseteq A^2[/imath]is the rational function [imath]\alpha= \frac{y-1}{x}[/imath] regular? |
1304395 | Identity Element and Identity Properties
Learning more abstract algebra, really not the most enjoyable of subjects, as nothing seems all that clear cut, but here goes anyway. I have a set [imath]\mathbb Q = \{{p \over q} : p,q\in \mathbb Z \text{ and } q \neq 0\}[/imath] which is the set of rational numbers and for [imath]x,y\in \mathbb Q[/imath] defined the binary operation [imath]*[/imath] on [imath]\mathbb Q[/imath] by [imath]x*y = x + y + xy.[/imath] The pair [imath](\mathbb Q, * )[/imath] has an identity element. Find the identity element then verify the identity properties for [imath](\mathbb Q,*)[/imath]. Where do I begin? | 513045 | Need to prove that [imath](S,\cdot)[/imath] defined by the binary operation [imath]a\cdot b = a+b+ab[/imath] is an abelian group on [imath]S = \Bbb R \setminus \{-1\}[/imath].
So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity element, other than 0. Because if 0 is the identity element, then this group won't have inverses. The set explicitly excludes -1, which I found to be its identity element, which makes going about proving that this is a group mighty difficult. |
1304601 | how to prove [imath]2^n = {n \choose 0} +{n \choose 1} + \cdots {n \choose n}[/imath]
I have studying my maths book induction chapter and I found things to solve this but I am failed, somebody help me to solve this problem by simple method of mathematical induction. [imath]2^n = {n \choose 0} +{n \choose 1} + \cdots {n \choose n}[/imath] How to prove it by mathematical induction. | 519832 | Proving by induction that [imath] \sum_{k=0}^n{n \choose k} = 2^n[/imath]
Prove by induction that for all [imath]n \ge 0[/imath]: [imath]{n \choose 0} + {n \choose 1} + ... + {n \choose n} = 2^n.[/imath] In the inductive step, use Pascal’s identity, which is: [imath]{n+1 \choose k} = {n \choose k-1} + {n \choose k}.[/imath] I can only prove it using the binomial theorem, not induction. |
1200054 | Let [imath]G[/imath] be a group of order 35. Show that [imath]G \cong Z_{35}[/imath]
Let [imath]G[/imath] be a group of order 35. Show that [imath]G \cong Z_{35}[/imath] Now, I am going to assume that this is a LaGrange based question. Also, I know that in order to be isomorphic, it must be one to one and onto. | 76112 | If G is a group of order n=35, then it is cyclic
I've been asked to prove this. In class we proved this when [imath]n=15[/imath], but our approached seemed unnecessarily complicated to me. We invoked Sylow's theorems, normalizers, etc. I've looked online and found other examples of this approach. I wonder if it is actually unnecessary, or if there is something wrong with the following proof: If [imath]|G|=35=5\cdot7[/imath] , then by Cauchy's theorem, there exist [imath]x,y \in G[/imath] such that [imath]o(x)=5[/imath], [imath]o(y)=7[/imath]. The order of the product [imath]xy[/imath] is then [imath]\text{lcm}(5,7)=35[/imath]. Since we've found an element of [imath]G[/imath] of order 35, we conclude that [imath]G[/imath] is cyclic. Thanks. |
1304366 | Find the equation of the line perpenducular to the plane [imath]x-y+z = 4[/imath]
In geometry i am having difficulties in understanding how to find the equations.. e.g given [imath]x-y+z =4 [/imath]which passes through the point [imath](2,1,5)[/imath] when asked to find the equation of the line perpendicular to the plane. Any help is appreciated. | 646420 | Find the Vector Equation of a line perpendicular to the plane.
Question: Find the vector equation [imath]r(t)[/imath] for the line through the point [imath]P = (-1, -5, 2)[/imath] that is perpendicular to the plane [imath]1 x - 5 y + 1 z = 1[/imath]. Use [imath]t[/imath] as your variable, [imath]t = 0[/imath] should correspond to [imath]P[/imath], and the velocity vector of the line should be the same as the standard normal vector of the plane. This one is really giving me a hard time. I know that to find the plane perpendicular to the line I can use the vector n between two points on the line and and the plane. I cannot wrap my mind around how to reverse this process, particularly because the plane is equal to 1 and not zero. Any help would be greatly appreciated. |
1304736 | Combinatorial express of n^3
I know following expression [imath]n = {n\choose 1 } [/imath] [imath]n^2 = {n\choose 2 } + {n+1\choose 2 }[/imath] but how about [imath]n^3 = [/imath]? Are there simple expression? | 358165 | Expression for power of a natural number in terms of binomial coefficients
Is there a general expression for the pth power of a natural number k in terms of binomial coefficients? I found this identity in a high-school text, which was obtained by simply equating coefficients: [imath]k^3 = 6 \binom{k}{3} + 6 \binom{k}{2} + \binom{k}{1}[/imath] I would like to know the general expression for [imath]k^p[/imath] in terms of binomial coefficients. I would also appreciate it if someone could tell me the names of texts which might cover such expressions, and what they can be used for. |
165667 | [imath]X = \log_{12} 18[/imath] and [imath]Y= \log_{24} 54[/imath]. Find [imath]XY + 5(X - Y)[/imath]
[imath]X = \log_{12} 18[/imath] and [imath]Y= \log_{24} 54[/imath]. Find [imath]XY + 5(X - Y)[/imath] I changed the bases to 10, then performed manual addition/multiplication but it didn't yield me any result except for long terms. Please show me the way. All I'm getting is [imath]\frac{\lg 18\lg54 + 5 \lg18\lg24 - 5\lg54\lg12}{\lg12\lg24} [/imath] | 852407 | Proof the expession [imath]\log_{12}{18}\times\log_{24}{54} + 5(\log_{12}{18}-\log_{24}{54})=1[/imath]
I am trying to proof the following expression (without a calculator of course). [imath]\log_{12}{18}\times\log_{24}{54} + 5(\log_{12}{18}-\log_{24}{54})=1[/imath] I know this isn't a difficult task but it's just killing me. I have tried many things, among which was base transformation to 12 and expressing every logarithm in terms of [imath]\log_{12}{3}[/imath] and [imath]\log_{12}{2}[/imath] but every time I try to do it, I mess up something. I don't know if my concentration is terrible or I'm doing something wrong. Thanks ;) ( if there are more levels to this task, I'd like a hint, not a complete solution) |
1304592 | Singular matrix with entries in a ring.
Given a matrix [imath]M\in A^{n\times n}[/imath], where [imath]A[/imath] is a commutative ring different from [imath]\{0\}[/imath], then we know that if there exists a vector [imath]x\in A^n[/imath] such that [imath]Mx=0[/imath], then [imath]\det M[/imath] must be a zero divisor. The converse is true? That is, if [imath]\det M[/imath] is a divisor of zero, is it true that his kernel is non-trivial? And if [imath]\det M=0[/imath]? The only thing I obtained is that it's true for Principal Ideal Domains, thanks to the Smith Form. | 1189763 | Zero divisors in matrix rings
Let [imath]R[/imath] be a commutative ring, [imath]P \in M_n(R)[/imath] and [imath]\det(P)[/imath] is a zero divisor of [imath]R[/imath]. Must [imath]P[/imath] be a zero divisor of [imath]M_n(R)[/imath]? Here rings mean unital rings, [imath]M_n(R)[/imath] denotes the ring of square matrices over [imath]R[/imath] of order [imath]n[/imath], and zero divisor is understood to be nonzero. The difficulty lies in that the adjugate matrix of [imath]P[/imath] may well be [imath]0[/imath]. |
1305228 | Is there a math basis for what I observe in simulations?
I am stuck in my research with the following problem which occurs when proving a theorem. If [imath]x,y\in \mathbb{R}^+[/imath] and [imath]x>y[/imath], is it true that [imath](x-y)^{((x-y)/(2x-y))}\times (x+y)^{(x/(2x-y))}>x[/imath]? I plugged in a range of values for [imath]x[/imath] and [imath]y[/imath] and always found this holds. However I am not sure how one can prove a statement of this nature? Can someone help me with this? | 1305175 | Proving using AM-GM inequality
If [imath]x,y\in \mathbb{R}[/imath], and [imath]x>y[/imath], how to show [imath](x-y)^{((x-y)/(2x-y))}\times (x+y)^{((x)/(2x-y))}>x[/imath]? I know I have to use AM-GM inequality, but it is not clear how. |
1303173 | Can two perfect squares average to a third perfect square?
My question is does there exist a triple of integers, [imath]a<b<c[/imath] such that [imath]b^2 = \frac{a^2+c^2}{2}[/imath] I suspect that the answer to this is no but I have not been able to prove it yet. I realize this is very similar to the idea of Pythagorean triples but I am not versed enough in this subject to try and modify the theory for this case. One simple observation is that in order to have any hope of this working is that [imath]a[/imath] and [imath]c[/imath] must be of the same parity. Furthermore if such a triple exists we can build an infinite sequence since [imath](2b)^2 = \frac{(2a)^2+(2c)^2}{2}[/imath] if and only if [imath]b^2 = \frac{a^2+c^2}{2}[/imath] Any help on determining this to be either true or false would be much appreciated. I am hoping it does end up being impossible, so if someone does find a desired triple I would next move up to cubes instead of squares Edit: Thank you for the comments, I have foolishly overlooked a simple example and see that there are many solutions to this based on the theory of diophantine equations. However this is unfortunate for me because i was hoping NOT to be able to solve this. This question arose while studying a certain type of graph labeling. What I desire is to be able to create a sequence, [imath]S[/imath], of arbitrary length (since each member of the sequence is to be a label of a vertex in the graph) such that for every [imath]x \in S[/imath], [imath]|x-s_i| \neq |x-s_j|[/imath] for [imath] i \neq j[/imath]. I was naively hoping that the sequence of squares was sufficient to satisfy this condition. Further edit, I have found that the sequence [imath]2^n[/imath] works but it would be nice if I could find a polynomial sequence. | 1250912 | Diophantine Equations : Solving [imath]a^2+ b^2=2c^2[/imath]
I was working through some number theory problems , when I came across the following question : Find all solutions of [imath]a^2+b^2=2c^2[/imath] My Solution (Partial) : We can rewrite the above equation as : [imath]c^2 = (a^2 + b^2)/2 [/imath] So [imath]\Rightarrow[/imath] [imath]a^2 , c^2 , b^2[/imath] are in an Arithmetic Progression [imath]\Rightarrow[/imath] there exist infinite solutions WLOG , let [imath]a^2[/imath] be [imath]t[/imath] , [imath]b^2[/imath] be [imath]s[/imath] and [imath]c^2[/imath] be [imath]m[/imath] The equation can be re-written as [imath] t + s = 2m[/imath] Now , what I was thinking was that should I solve this Diophantine equation for [imath]t[/imath] & [imath]s[/imath] in terms of [imath]m[/imath] ; substitute their values back into the equation and find out [imath]m[/imath] ; and then finally put this value of [imath]m[/imath] back into the values of [imath]t[/imath] and [imath]s[/imath] to derive a general solution Can someone help me out ? Maybe a hint ... |
1305254 | Can someone please point out the flaw in my proof?
Let [imath]f:X \to Y[/imath] be a proper map.Show that [imath]f[/imath] takes discrete sets to discrete sets. Proof:Let [imath]A[/imath] be discrete in [imath]X[/imath] and let [imath]K[/imath] be compact in [imath]Y[/imath] then [imath]f(A) \cap K=f(A \cap f^{-1}(K))[/imath],is finite since [imath]A \cap f^{-1}(K)[/imath] is finite.Hence [imath]f(A)[/imath] is discrete But @Alex Ravsky here Does proper map [imath]f[/imath] take discrete sets to discrete sets? gave a counter example. So can Someone please point out the flaw in my proof? | 1296450 | Does proper map [imath]f[/imath] take discrete sets to discrete sets?
Suppose [imath]f:X \to Y[/imath] is a continuous proper map between locally compact Hausdorff spaces. Are the following results true? [imath]1[/imath]. The map [imath]f[/imath] takes discrete sets to discrete sets. [imath]2[/imath]. If [imath]f[/imath] is injective, then [imath]f[/imath] must be a homeomorphism onto its image. Edit1:Let [imath]A[/imath] be discrete in [imath]X[/imath] and let [imath]K[/imath] be compact in [imath]Y[/imath] then [imath]f(A) \cap K=f(A \cap f^{-1}(K))[/imath],is finite since [imath]A \cap f^{-1}(K)[/imath] is finite.Hence [imath]f(A)[/imath] is discrete. As there is a counterexample in answer below so can someone please point out the error in my proof? |
1287570 | An application of the uniform boundedness principle
Can someone provide me with a sketch of proof / hint for this exercise: Let K [imath]\subset[/imath] [imath]L^1([0,1]\,,\, \mu)[/imath] be a closed linear subspace. If [imath]\forall[/imath] [imath]f \in K[/imath], [imath]\exists \, p > 1[/imath] such that [imath]f \in L^p[/imath], then [imath]\exists \, p > 0 \Rightarrow\, K \subset L^p[/imath]. I understand that the uniform boundedness principle should be applied here, but I have no clue how. Thanks for any help in advance. | 167475 | Inclusion of [imath]L^p[/imath] spaces
Let [imath]X \subset L^1(\mathbb{R})[/imath] a closed linear subspace satisfying \begin{align} X\subset \bigcup_{p>1} L^p(\mathbb{R})\end{align} Show that [imath]X\subset L^{p_0}(\mathbb{R})[/imath] for some [imath]p_0>1.[/imath] I guess the problem is that in infinite measure spaces the inclusion [imath]L^p\subset L^q[/imath] only holds for [imath]p=q[/imath]. Is it maybe possbile to apply Baire's Theorem in some way? |
1277250 | Proof of Levy's theorem without Ottaviani inequality
Suppose [imath]X_1,⋯,X_n[/imath] are independent r.v., Let [imath]S_n=X_1+⋯+X_n[/imath], I am looking to show that convergence on [imath]S_n[/imath] in probability implies almost sure convergence by showing that [imath]P(\sup_{m\geq 1}|S_m|>4\epsilon)\leq 4\sup_{m\geq 1} P(|S_m|>\epsilon)[/imath]. I have been trying using Kolmogorov's maximal inequality but not much luck. Do I need to be looking at some martingale? Any hints? | 940147 | Maximal inequality for a sequence of partial sums of independent random variables
Let [imath]X_n[/imath], [imath]n=1,2,3,...[/imath] be a sequence of independent (not necessarily identically distributed) random variables, let [imath]S_n=\sum_{i=1}^nX_i[/imath]. Prove the following maximal inequality: For all [imath]t>0[/imath],[imath]\mathbb{P}\left(\max_{1\leq i\leq n}|S_i|\geq t \right)\leq 3\max_{1\leq i\leq n}\mathbb{P}\left(|S_i|\geq\frac{t}{3} \right)[/imath] |
1306011 | Annihilator for a tensor [imath]T\in\wedge V^{\ast}[/imath]
For [imath]T\in\wedge^{k} V^{\ast}[/imath] the annihilator is set [imath]an(T)= \{\phi\in V^{\ast}\mid \phi\wedge T=0\}[/imath] Then I need to prove that [imath]dim(an(T))\leq k[/imath] and is equal iff [imath]T[/imath] is decomposable ([imath]i.e.[/imath], [imath]T=\phi_1\wedge\cdots \wedge \phi_k[/imath] for [imath]\phi_j \in V^{\ast}[/imath]). I do not know how to attack this problem, suppose that has dimension greater that [imath]k[/imath] but I get no contradiction or at least do not know how to find it. | 341540 | Decomposable elements of [imath]\Lambda^k(V)[/imath]
I have a conjecture. I have a problem proving or disproving it. Let [imath]w \in \Lambda^k(V)[/imath] be a [imath]k[/imath]-vector. Then [imath]W_w=\{v\in V: v\wedge w = 0 \}[/imath] is a [imath]k[/imath]-dimensional vector space if and only if [imath]w[/imath] is decomposable. For example, for [imath]u=e_1\wedge e_2 + e_3 \wedge e_4[/imath] we have [imath]W_u = 0[/imath]. |
1306132 | Bisecting line segments in a tetrahedron.
Suppose that [imath]OABC[/imath] is a regular tetrahedron with base [imath]ABC[/imath]. Suppose further that [imath]T[/imath] is the mid-edge of [imath]AC[/imath], [imath]Q[/imath] is the mid-edge of [imath]OB[/imath], [imath]P[/imath] is the mid-edge of [imath]OA[/imath], and [imath]U[/imath] is the mid-edge of [imath]CB[/imath]. How can one show that [imath]QT[/imath] bisects [imath]PU[/imath]? | 1306237 | Vector proof for a tetrahedron.
Let [imath]OABC[/imath] be a regular tetrahedron. Using [imath]OA=a,~OB=b,~OC=c[/imath], prove that [imath]QT[/imath] bisects [imath]PU[/imath] where [imath]P,U,Q,T[/imath] are mid points on the edges of a regular tetrahedron. P is the midpoint of OA. T is the midpoint of AC. U is the midpoint of BC. Q is the midpoint of OB. Midpoints; P and Q are midpoints on the edges of the base of the tetrahedron. Going about this I thought that I would prove that [imath]PUQT[/imath] forms a a parallelogram and then prove that [imath]QT[/imath] bisects [imath]PU[/imath], however I may be wrong. |
1289572 | What are all the [imath]k[/imath]-dimensional unimodular subspaces of [imath]\mathbb{Z}^n[/imath]?
I am trying to prove the following assertion: The set of subgroups of [imath](S^1)^n[/imath] which are isomorphic by an element of [imath]\operatorname{Aut}((S^1)^n)[/imath] to the standard copy [imath](S^1)^k[/imath] is naturally parametrized by the set of [imath]k[/imath]-dimensional unimodular subspaces of [imath]\mathbb{Z}^n[/imath]. By a [imath]k[/imath]-dimensional unimodular subspace of [imath]\mathbb{Z}^n[/imath] we mean a rank [imath]k[/imath] submodule which is a direct summand. My guess is that the [imath]k[/imath]-dimensional unimodular subspaces of [imath]\mathbb{Z}^n[/imath] would be those that are isomorphic by an element of [imath]\operatorname{Aut}(\mathbb{Z}^n)[/imath] to the standard copy of [imath]\mathbb{Z}^k[/imath]. Is that correct? In that case, since [imath]\operatorname{Aut}(\mathbb{Z}^n)\cong GL(n,\mathbb{Z})\cong\operatorname{Aut}((S^1)^n)[/imath] I thought I could do the following: Given any subgroup [imath]H[/imath] of [imath](S^1)^n[/imath], as in the stated assertion, there is an [imath]f \in GL(n,\mathbb{Z})[/imath] such that [imath]f(\mathbb{(S^1)^k}) = H[/imath]. So I can send this subgroup to [imath]f(\mathbb{Z}^k)[/imath]. But this map need not be well defined. So what do I do? Thank you. | 1305479 | What are the direct summands in [imath]\mathbb{Z}^n[/imath]?
I am interested in knowing the rank [imath]k[/imath] submodules of the [imath]\mathbb{Z}[/imath]-module [imath]\mathbb{Z}^n[/imath] which are are also direct summands. I know that [imath]\mathbb{Z}^k[/imath] sitting in [imath]\mathbb{Z}^n[/imath] in the standard way, i.e., [imath](z_1, \dots, z_k) \mapsto (z_1, \dots, z_k, 0, \dots, 0)[/imath] is one such summand. So if I look at all the different ways in which [imath]\mathbb{Z}^k[/imath] is embedded in [imath]\mathbb{Z}^n[/imath] will that give me all the required summands or will there be more? If so, how do I characterize all the embeddings, i.e., injective [imath]\mathbb{Z}[/imath]-module homomorphisms from [imath]\mathbb{Z}^k[/imath] into [imath]\mathbb{Z}^n[/imath]? Thank you. |
1306334 | greatest common divisor of a cyclic group generator
I have been struggling with this for a very long time-in fact for a few days and it's preventing me from progressing-a severe hurdle. Mainly, 1) what is [imath]\left \langle a^{k} \right \rangle[/imath]? Is it read as the cyclic group generated by some element a to the power of k? 2) Can I have a step-by-step explanation as to how [imath]\left \langle a^{k} \right \rangle \sqsubseteq \left \langle a^{d} \right \rangle[/imath]? To be specific, I cannot make the connection involved in 'and so' I spent a couple of hours a day looking at this but it's not helping. Edit: The link provided to the duplicate was not at all clear. Topic in the other post clearly did not addressed the poster's question. It has also not answered my question on the implicit assumption made. | 1303868 | Proof involving Cyclic group, generator and GCD
Theorem: [imath]\left\langle a^k \right\rangle = \left\langle a^{\gcd(n,k)}\right\rangle[/imath] Let G be a group and [imath] a \in G[/imath] such that [imath]|a|=n[/imath] Then: [imath]\left\langle a^k \right\rangle = \left\langle a^{\gcd(n,k)}\right\rangle[/imath] The proof begins by letting d = gcd(n,k) such that d is a divisor of k so there exists an integer r such that k = dr. So, [imath]a^k=(a^d)^r[/imath]. [imath]\left\langle a^k \right\rangle \subseteq \left\langle a^{\gcd(n,k)}\right\rangle[/imath] I've spent a very long time on understanding this proof but found it to be obscure. I suspect there are some gaps in my understanding. If someone could show me the light I'll be really glad. I do not understand why the exponent r on a 'vanishes'. Secondly, how does [imath]a^k\in \left\langle a^d\right\rangle[/imath] follows? Thirdly, where does closure plays a role? |
1306817 | Finding the order of Z(G) in a non-Abelian group of order 8
Let [imath]G[/imath] be a non-Abelian group of order [imath]8[/imath]. Prove that [imath]|Z(G)|[/imath] is less or equal to [imath]2[/imath]. First I must say this is a question about normal subgroups. I haven't yet studied homomorphisms or more complex material. This is my try to solve the problem. Since [imath]Z(G)[/imath] is a subgroup of [imath]G[/imath], its order should be [imath]8, 4, 2[/imath] or [imath]1[/imath]. Because [imath]G[/imath] is non-Abelian, it can't be [imath]8[/imath]. Then I assumed the order is [imath]4[/imath] and tried to get to a contradiction. Since [imath]|Z(G)|=4[/imath], there are only two cosets of [imath]Z(G)[/imath] in [imath]G[/imath]. One of them is [imath]Z(G)[/imath] itself, and the other one is all the elements that are not in [imath]Z(G)[/imath]. Here I am stuck. Maybe it's not the right way to think about this problem.. Help would be greatly appreciated. | 1306671 | The center of a non-Abelian group of order 8
Let G be a non-Abelian group of order 8. Prove that [imath]|Z(G)|\leq2[/imath]. (The center [imath]Z(G)[/imath] is defined as [imath]Z(G)=\{ a\in G | ag=ga[/imath] for all [imath]g\in G \}[/imath]). I deduced from Lagrange's theorem that [imath]|Z(G)|\in\{1,2,4,8 \}[/imath] and ruled out [imath]8[/imath] because [imath]|Z(G)|=8\Rightarrow Z(G)=G[/imath] and [imath]Z(G)[/imath] is Abelian while [imath]G[/imath] is non-Abelian. I'm trying without success to rule out 4. Any ideas? |
1306798 | What does the "[imath]\cdot[/imath]" mean in an equation
I am trying to solve an equation for a project that I am undertaking. The equation is very long and its probably not necessary to show it all here. Most of the equation is fairly straightforward; i.e., [imath]1+(\frac{w}{W})(\frac{d}{t})[/imath], etc. but at the very end it reads [imath]\left(1+\frac{w}{W}\right)\left(\frac{D}{T}\right) \cdot \frac{1}{T}[/imath]. [imath]w[/imath], [imath]W[/imath], [imath]D[/imath], and [imath]T[/imath] are all either constants or variables that I have solved already. My question is, what is the "[imath]\cdot[/imath]" symbol asking me to do? In the equation in the book, the "[imath]\cdot[/imath]" is in the centre of the dividing line between [imath]1[/imath] and [imath]T[/imath], as opposed to down low like a period or full stop. I appreciate that I have phrased this question in an awkward manner, but as is obvious, I am no maths expert. Here is a photo of the equation in the book: | 1117833 | The meaning of dot centered vertically, as in [imath]3\cdot 5[/imath]
I just went to a site to practise some maths as I am studying some maths on my OU course in computing and IT and I ran into some symbols that I did not understand The questions were solve [imath]16+14/2+3\cdot 5-11[/imath] (the point was placed halfway between the top and bottom of the numbers) [imath](11-5)\cdot (63-59+6)/12[/imath], once again point is halfway between top and bottom Can someone explain what this means if it is different to a decimal point? Also this symbol was used in other questions [imath]|[/imath] and I was wondering what this meant as well In context it was the number [imath]-5[/imath] can be written as and one answer was [imath]|-5|[/imath] and another choice was [imath]-|-4|-1[/imath] |
1307259 | Proof of direct sum with span and basis
Let [imath]v_1, v_2,\dots, v_n[/imath] be a basis of the vector space [imath]V[/imath]. Let [imath]k[/imath] be an integer such that [imath]1 \leq k \leq n[/imath], and put [imath]U = \operatorname{Span}\{v_1, . . . , v_k\}[/imath] and [imath]W = \operatorname{Span}\{v_{k+1}, . . . , v_n\}[/imath]. I want to show that [imath]V = U\oplus W[/imath]. Would it be enough to prove that [imath]U[/imath] and [imath]W[/imath] are bases of [imath]V[/imath] using linear independence, and then show that there are unique [imath]v\in V[/imath] for every [imath]v = u + w[/imath]? | 761040 | Prove that if $W_1\subseteq V$ finite-dimensional, then there is $W_2\subseteq V$ such that $V=W_1\oplus W_2$
Prove that if [imath]W_1[/imath] is any subspace of a finite-dimensional vector space [imath]V[/imath], then there exists a subspace [imath]W_2[/imath] of [imath]V[/imath] such that [imath]V = W_1 \oplus W_2[/imath] What I have done so far is to note that since [imath]V[/imath] is finite and [imath]W_1[/imath] is a subspace of [imath]V[/imath], we have [imath]\dim(W_1) \leq \dim(V)[/imath]. If we have equality, then let [imath]W_2 = \{0\}[/imath], so that we have [imath]V= W_1 \oplus W_2[/imath]. So now I need to look at the case when [imath]\dim(W_1) \lt \dim(V)[/imath]. What I have tried for this case is to let [imath]\beta = \{v_1,v_2,..,v_n\}[/imath] be a basis for [imath]V[/imath] and [imath]\gamma=\{u_1,u_2,..,u_m\}[/imath] a basis for [imath]W_1[/imath]. My idea was to extend [imath]\gamma[/imath] to a basis for [imath]V[/imath], so let [imath]\alpha=\{u_1,u_2,..,u_m,w_1,w_2,..,w_{n-m}\}[/imath] be the extension of [imath]\gamma[/imath] to [imath]V[/imath], where [imath]w_1,w_2,..,w_{n-m}[/imath] are basis vectors for [imath]W_2[/imath]. If I'm doing this right then I would just have to show that [imath]W_1 \cap W_2 = \{0\}[/imath] and [imath]W_1 + W_2= V[/imath] Am I heading in the right direction? Any hints would be greatly appreciated. |
1307282 | Derivative of [imath](\ln x)^e[/imath]
In Randall Munroe's What If, he says that "if you want to be mean to first-year calculus students, you can ask them to take the derivative of [imath](lnx)^e[/imath]" He says, as I would expect, that the result "looks like it should be [imath]1[/imath] or something, but it's not." Why is this? And what's the actual answer? | 1179376 | "What if" math joke: the derivative of [imath]\ln(x)^e[/imath]
Randall Munroe, the creator of xkcd in his latest book What if writes (p. 175) that the mathematical analog of the phrase "knock me over with a feather" is seeing the expression [imath] \ln( x )^{e}[/imath]. And he writes regarding this expression: "it's not that, taken literally, it doesn't make sense - it's that you can't imagine a situation where this would apply." In the footer (same page) he also states that "if you want to be mean to first year calculus students, you can ask them to take the derivative of [imath] \ln( x )^{e}[/imath]. It looks like it should be "[imath]1[/imath]" or something but it's not." I don't get the joke. I think I am not understanding something correctly and I'm not appreciating the irony. Any help? |
1301692 | [imath]M_1 \to M_2 \to M_3 \to 0[/imath] exact iff [imath]0 \to \text{Hom}_A(M_3, N) \to \text{Hom}_A(M_2, N) \to \text{Hom}_A(M_1, N)[/imath] is exact.
Let [imath]M_1 \to M_2 \to M_3 \to 0[/imath] be a sequence of homomorphisms of [imath]A[/imath]-modules. Is this sequence exact if and only if the induced sequence of abelian groups[imath]0 \to \text{Hom}_A(M_3, N) \to \text{Hom}_A(M_2, N) \to \text{Hom}_A(M_1, N)[/imath]is exact for every [imath]A[/imath]-module [imath]N[/imath]? | 375763 | Characterization of short exact sequences
The following is the first part of Proposition 2.9 in "Introduction to Commutative Algebra" by Atiyah & Macdonald. Let [imath]A[/imath] be a commutative ring with [imath]1[/imath]. Let [imath]M' \overset{u}{\longrightarrow}M\overset{v}{\longrightarrow}M''\longrightarrow 0\tag{1} [/imath] be sequence of [imath]A[/imath]-modules and homomorphisms. Then the sequence (1) is exact if and only if for all [imath]A[/imath]-modules [imath]N[/imath], the sequence [imath]0\longrightarrow \operatorname{Hom}(M'', N) \overset{\overline{v}}{\longrightarrow}\operatorname{Hom}(M, N)\overset{\overline{u}}{\longrightarrow}\operatorname{Hom}(M', N) \tag{2} [/imath] is exact. Here, [imath]\overline{v}[/imath] is the map defined by [imath]\overline{v}(f)=f\circ v[/imath] for every [imath]f\in\operatorname{Hom}(M'', N)[/imath] and [imath]\overline{u}[/imath] is defined likewise. The proof one of direction, namely [imath](2)\Rightarrow (1)[/imath] is given in the book, which I am having some trouble understanding. So assuming (2) is exact sequence, the authors remark that "since [imath]\overline{v}[/imath] is injective for all [imath]N[/imath], it follows that [imath]v[/imath] is surjective". Could someone explain why this follows? Given that [imath]\overline{v}[/imath] is injective, we know that whenever [imath]f(v(x))=g(v(x))[/imath] for all [imath]x\in M[/imath], we have [imath]f=g[/imath]. I am not sure how we conclude from this surjectivity of [imath]v[/imath]. Thanks! |
1307297 | How to understand Weyl chambers?
Recall the definition of the Weyl Chambers: A Weyl Chamber is a region of [imath]V \setminus \bigcup_{\alpha \in \Phi} H_{\alpha}[/imath], where [imath]V[/imath] is underlying Euclidean space, and [imath]H_\alpha[/imath] the hyperplane that is perpendicular to [imath]\alpha[/imath] the elements of root system [imath]\Phi[/imath] in [imath]V[/imath]. I want to show that: The Weyl chambers are open, convex and connected. My attempts: A single hyperplane [imath]H_\alpha[/imath] separates the space [imath]V[/imath] into two half-spaces: positive half-space [imath]H_\alpha^+=\{x \in V \mid (\alpha,x)>0\}[/imath] and negative half-space [imath]H_\alpha^-=\{x \in V \mid (\alpha,x)<0\}[/imath]. It is clear that [imath]V \setminus H_\alpha = H_\alpha^+\cup H^-_\alpha.[/imath] As the result, the Weyl chambers can be regarded as components of [imath]V \setminus \bigcup_{\alpha \in \Phi} H_{\alpha}= \bigcap_{\alpha \in \Phi} \left( V \setminus H_{\alpha} \right)= \bigcap_{\alpha \in \Phi} \left( H_\alpha^+\cup H_\alpha^- \right).[/imath] By the discussion above, a Weyl chamber is simply a [imath]\textbf{intersection}[/imath] of finite half-spaces. It is the case since one may write [imath]\left( H_\alpha^+\cup H_\alpha^- \right) \cap \left( H_\beta^+\cup H_\beta^- \right) =\left( H_\alpha^+\cap H_\beta^+ \right) \cup\left( H_\alpha^+\cap H_\beta^- \right) \cup \left( H_\alpha^-\cap H_\beta^+ \right) \cup\left( H_\alpha^-\cap H_\beta^- \right).[/imath] Therefore openness and convexity of the Weyl chambers are straightforward since half-spaces are open and convex, so are their intersections. Moreover, we would replace [imath]V[/imath] by [imath]\Bbb R^{\dim V}[/imath] since they are isomorphic. Now we can prove that the Weyl chamber is connected by showing that every convex set in [imath]\Bbb R^{\dim V}[/imath] is connected. And I'm sure that proof of this statement existed. So am I done? | 1303440 | How to understand Weyl chambers?
Recall the definition of the Weyl Chambers: A Weyl Chamber is a region of [imath]V \setminus \bigcup_{\alpha \in \Phi} H_{\alpha}[/imath], where [imath]V[/imath] is underlying Euclidean space, and [imath]H_\alpha[/imath] the hyperplane that is perpendicular to [imath]\alpha[/imath] the elements of root system [imath]\Phi[/imath] in [imath]V[/imath]. I want to show that: The Weyl chambers are open, convex and connected. My attempts: A single hyperplane [imath]H_\alpha[/imath] separates the space [imath]V[/imath] into two half-spaces: positive half-space [imath]H_\alpha^+=\{x \in V \mid (\alpha,x)>0\}[/imath] and negative half-space [imath]H_\alpha^-=\{x \in V \mid (\alpha,x)<0\}[/imath]. It is clear that [imath]V \setminus H_\alpha = H_\alpha^+\cup H^-_\alpha.[/imath] As the result, the Weyl chambers can be regarded as components of [imath]V \setminus \bigcup_{\alpha \in \Phi} H_{\alpha}= \bigcap_{\alpha \in \Phi} \left( V \setminus H_{\alpha} \right)= \bigcap_{\alpha \in \Phi} \left( H_\alpha^+\cup H_\alpha^- \right).[/imath] By the discussion above, a Weyl chamber is simply a [imath]\textbf{intersection}[/imath] of finite half-spaces. It is the case since one may write [imath]\left( H_\alpha^+\cup H_\alpha^- \right) \cap \left( H_\beta^+\cup H_\beta^- \right) =\left( H_\alpha^+\cap H_\beta^+ \right) \cup\left( H_\alpha^+\cap H_\beta^- \right) \cup \left( H_\alpha^-\cap H_\beta^+ \right) \cup\left( H_\alpha^-\cap H_\beta^- \right).[/imath] Therefore openness and convexity of the Weyl chambers are straightforward since half-spaces are open and convex, so are their intersections. Moreover, we would replace [imath]V[/imath] by [imath]\Bbb R^{\dim V}[/imath] since they are isomorphic. Now we can prove that the Weyl chamber is connected by showing that every convex set in [imath]\Bbb R^{\dim V}[/imath] is connected. And I'm sure that proof of this statement existed. So am I done? |
1306676 | Reasoning in "Prove [imath]X[/imath] is a martingale" / Different approach using indicator functions
From here. The solution given is that [imath]E[X_{n+1}\mid X_n] = 1/2\times 2X_n + 1/2\times 0 = X_n[/imath] Why exactly? In retrospect, I'm not sure I really got it. I'm trying to think about it in terms of indicator functions over events we must define: Let [imath]A_n = \{ \omega\mid X_{n+1}(\omega) = 2 X_n(\omega) \}[/imath]. Note: [imath]A_n \in \mathscr{F}[/imath] and [imath]P(A_n) = 1/2 \ \forall \ n \in \mathbb{N}[/imath] It looks like [imath]X_{n+1} = 1_{A_n} (2X_n) + 1_{A_n^C}(0)[/imath] [imath]\to E[X_{n+1}\mid X_n] = E[1_{A_n} (2X_n) + 1_{A_n^C}(0)\mid X_n][/imath] [imath]= E[1_{A_n} (2X_n)\mid X_n][/imath] [imath]= (2X_n) E[1_{A_n}\mid X_n][/imath] [imath]= (2X_n) E[1_{A_n}][/imath] (*) [imath]= (2X_n) P({A_n}) = X_n[/imath] ? (*) Is [imath]\sigma(A_n)[/imath] independent of [imath]\sigma(X_n)[/imath] ? What is [imath]\sigma(X_n)[/imath] in terms of [imath]A_n[/imath] anyway? I guess that: [imath]\sigma(X_0) = \{ \emptyset, \Omega \}[/imath] [imath]\sigma(X_1) = \sigma(A_0) \because X_1 = 2 \times 1_{A_0}[/imath] [imath]\sigma(X_2) \cdots \subseteq \sigma(\sigma(A_0) \cup \sigma(A_1)) = \sigma(A_0, A_1)[/imath] | 974969 | Prove X is a martingale
Prove [imath]X = (X_n)_{n \geq 0}[/imath] is a martingale w/rt [imath]\mathscr{F}[/imath] where X is given by: [imath]X_0 = 1[/imath] and for [imath]n \geq 1[/imath] [imath]X_{n+1} = 2X_n[/imath] w/ prob 1/2 [imath]X_{n+1} = 0[/imath] w/ prob 1/2 and [imath]\mathscr{F_n} = \mathscr{F_n}^{X} \doteq \sigma(X_0, X_1, ..., X_n)[/imath]. I think that [imath]X_{n} = 2^{n} \prod_{i=1}^{n} 1_{A_i} \forall n \geq 0[/imath] where [imath]A_1 = \{ \omega \in \Omega | X_{2}(\omega) = 2 X_1(\omega) \} \in \mathscr{F}[/imath] as follows: [imath]X_{1} = 2X_0* 1_{A_1} + 0*1_{A_1^c}[/imath] [imath]X_{2} = 2X_1* 1_{A_2} + 0*1_{A_2^c}[/imath] [imath]=2(2X_0* 1_{A_1} + 0*1_{A_1^c})*1_{A_2} + 0*1_{A_2^c}[/imath] Is that right? If not, why? If so, here is my attempt: (leaving out adaptability and integrability stuff) We must show that [imath]E[X_{n+1}|\mathscr{F_n}] \equiv E[2^{n+1} \prod_{i=1}^{n+1} 1_{A_i}|\mathscr{F_n}] = 2^n \prod_{i=1}^{n} 1_{A_i}[/imath]? Is that right? help please? |
1305077 | How do I find [imath] Pr\{X_1 < k \} [/imath] and [imath] Pr\{X_1 > k \} [/imath]if [imath]X_1 : G(p_1)[/imath]- geometric distribution
I would think the song like [imath]1-Pr\{X_1 < k \} [/imath] but what is confusing to me is the fact that this is a discrete random variable, and these inequalities ussually apply to absolute continuous variables, so I just don;t know what to do? I have to prove: Geometric random variables [imath]X_1:G(p_1) X_2:G(p_2) X_3:G(p_3)[/imath] are independent, prove the following : [imath]Pr(X_1 < X_2 < X_3)= \frac{(1-p_1)(1-p_2)p_2p_3^2}{(1-p_2p_3)(1-p_1p_2p_3)}[/imath] I got started like: [imath]\begin{align*} \Pr[X_1 < X_2 < X_3] &= \sum_{k=0}^\infty \Pr[X_1 < k < X_3]\Pr[X_2 = k] \\ &\overset{\text{ind}}{=} \sum_{k=0}^\infty \Pr[X_1 < k]\Pr[k < X_3]\Pr[X_2 = k]. \end{align*}[/imath] | 1304563 | Geometric random variables [imath]X_1:G(p_1)[/imath] [imath]X_2:G(p_2)[/imath] [imath]X_3:G(p_3)[/imath] are independent, prove the following :
[imath]P(X_1 < X_2 < X_3)= \frac{(1-p_1)(1-p_2)p_2p_3^2}{(1-p_2p_3)(1-p_1p_2p_3)}[/imath] To be frank I do not know where to start with this question, I would like an idea to get me going, or better yet an answer. This was on an earlier exam, I would like to be ready if something like this were to come up again tomorrow. What makes this a little bit more difficult than usual is the fact that geometric distribution is a discrete variable.. |
1307454 | Maclaurin Series of [imath]\frac{2x}{e^{2x}-1}[/imath]
Calculate the Maclaurin series of [imath]\frac{2x}{e^{2x}-1} [/imath] I've tried to calculate it but the series [imath]\frac{1}{e^{2x}-1}[/imath] divides by 0 when x is equal to 0 | 311817 | Maclaurin series for [imath]\frac{x}{e^x-1}[/imath]
Maclaurin series for [imath]\frac{x}{e^x-1}[/imath] The answer is [imath]1-\frac x2 + \frac {x^2}{12} - \frac {x^4}{720} + \cdots[/imath] How can i get that answer? |
444887 | Solving the trigonometric equation [imath]A\cos x + B\sin x = C[/imath]
I have a simple equation which i cannot solve for [imath]x[/imath]: [imath]A\cos x + B\sin x = C[/imath] Could anyone show me how to solve this. Is this a quadratic equation? | 213545 | Solving trigonometric equations of the form [imath]a\sin x + b\cos x = c[/imath]
Suppose that there is a trigonometric equation of the form [imath]a\sin x + b\cos x = c[/imath], where [imath]a,b,c[/imath] are real and [imath]0 < x < 2\pi[/imath]. An example equation would go the following: [imath]\sqrt{3}\sin x + \cos x = 2[/imath] where [imath]0<x<2\pi[/imath]. How do you solve this equation without using the method that moves [imath]b\cos x[/imath] to the right side and squaring left and right sides of the equation? And how does solving [imath]\sqrt{3}\sin x + \cos x = 2[/imath] equal to solving [imath]\sin (x+ \frac{\pi}{6}) = 1[/imath] |
1307721 | Finding the smallest square inside a parabola.
I just thought of a problem earlier today, but wanted to know if there was an easier way of acquiring the answer. Say I have a standard parabola [imath]y=x^2[/imath] with 3 points on it [imath]P,Q,R[/imath] and another point [imath]S[/imath] such that [imath]PQRS[/imath] is a square. Now I can clearly see that if I have the origin [imath](1,1), (-1,1)[/imath] and [imath](0,2)[/imath], then I form a square with area 2. However, this may not necessarily be the smallest square. My idea for finding the square with the smallest area is to parameterise the parabola with 3 points [imath](t,t^2)[/imath] where [imath]t=p,q,r[/imath], then to find the area of that triangle (perhaps using the determinant) and then minimise that quantity with the constraint that the interval [imath]PQ=PR[/imath] and that [imath]PQ \perp PR[/imath]. However, this seems too tedious. Is there a better way perhaps? | 1297478 | The square of minimum area with three vertices on a parabola
I was given this question by a friend and after working tirelessly on it I have not come up with anything substantial. I was hoping someone in the community could provide a pointer or possibly a solution. The person who gave me the question says he has the answer to it, but I am more curious about the working. Here is the question: Three points [imath](B, C, D)[/imath] lie on parabola [imath]y = x^2[/imath] and a fourth point [imath]A[/imath] is placed such that [imath]ABCD[/imath] forms a square. Find the minimal area of [imath]ABCD[/imath]. So far my working out has involved a collection of distance/mid-point formula's and I haven't even reached the point where I differentiate to minimize the area. Any pointers would be welcome. Edit Apparently the correct answer is [imath]\sqrt{2}[/imath] units squared. |
1307580 | Calculate the ratio of the side length of a tetrahedron to the side length of the tetrahedron which its centroids form
[imath]OABC[/imath] is a regular tetrahedron. Let [imath]E[/imath], [imath]F[/imath], [imath]G[/imath], [imath]H[/imath] be the centroids of the triangles [imath]OBA[/imath], [imath]OCB[/imath], [imath]OAC[/imath], [imath]ABC[/imath], respectively. You are given that EFGH is also a regular tetrahedron. Using [imath]OA=a[/imath], [imath]OB=b[/imath], [imath]OC=c[/imath], find the ratio of the side length of tetrahedron [imath]EFGH[/imath] to the side length of tetrahedron [imath]OABC[/imath]. Show using vectors. Please give full proof to the answer. Link to the picture of the tetrahedron used in this question: http://imgur.com/SFzGA5E the correct answer is a 1:3 ratio, however i am unsure as to how this is calculated | 1305976 | Relationship between the side lengths of a tetrahedron and an inscribed tetrahedron with vertices at the centroids
Suppose that [imath]OABC[/imath] is a regular tetrahedron with sides having centroids [imath]\lbrace E,F,G,H\rbrace[/imath] also forming a regular tetrahedron. What is the relationship between the side lengths of [imath]OABC[/imath] and [imath]EFGH[/imath]? I suspect that tetrahedron [imath]EFGH[/imath] has side lengths [imath]1/3[/imath] of that of [imath]OABC[/imath] but I'm unsure as to how to prove it. |
1307992 | A question in matrix norm.
Let [imath]A \in {M_n}[/imath] and [imath]\left\| {\left| . \right|} \right\|[/imath] be a matrix norm on [imath]{M_n}[/imath].Why does [imath]{\left\| {\left| A \right|} \right\|_2} \le \left\| {\left| A \right|} \right\|_1^{\frac{1}{2}}\left\| {\left| A \right|} \right\|_\infty ^{\frac{1}{2}}[/imath]? | 1154483 | Show that [imath] \lVert A \rVert_2^2 \leq \lVert A \rVert _1 \lVert A \rVert _ \infty [/imath]
With the definition of [imath] \lVert A \rVert_2[/imath] and [imath]\lVert A \rVert_1[/imath] and [imath]\lVert A \rVert_ \infty[/imath] that is: \begin{gather} \lVert A\rVert_1 = \max_{j} \sum_{i=1}^m \lvert a_{ij}\rvert\\ \lVert A\rVert_2 = \sqrt{\rho(A^HA)}\\ \lVert A\rVert_\infty = \max_{i} \sum_{j=1}^n \lvert a_{ij}\rvert \end{gather} prove that: [imath]\lVert A \rVert_2^2 \leq \lVert A \rVert_1 \lVert A \rVert_\infty[/imath] |
661535 | [imath]\mathrm{Hom}[/imath] and [imath]^*\mathrm{Hom}[/imath] for graded modules: Exercise 1.5.19(f) of Bruns-Herzog
Assume [imath]R[/imath] is a graded ring and [imath]M[/imath] and [imath]N[/imath] graded modules. Denote by [imath]^*\mathrm{Hom}_R(M,N)[/imath] the set of all homogeneous [imath]R[/imath]-linear maps from [imath]M[/imath] to [imath]N[/imath]. How can I prove that if [imath]M[/imath] is finitely generated then [imath]^*\mathrm{Hom}_R(M,N)=\mathrm{Hom}_R(M,N)[/imath]? Do you have a counterexample of a not finitely generated module [imath]M[/imath] such that this equality doesn't hold? | 19976 | Homomorphisms of graded modules
Let [imath]M[/imath] and [imath]N[/imath] be graded [imath]R[/imath]-modules (with [imath]R[/imath] a graded ring). [imath]\varphi:M\rightarrow N[/imath] is a homogeneous homomorphism of degree [imath]i[/imath] if [imath]\varphi(M_n)\subset N_{n+i}[/imath]. Denote by [imath]\mathrm{Hom}_i(M,N)[/imath] the group of homogeneous homomorphisms of degree [imath]i[/imath]. We define [imath]^*\mathrm{Hom}_R(M,N)=\bigoplus_{i\in\mathbb{Z}}\mathrm{Hom}_i(M,N)[/imath]. This is a (graded) [imath]R[/imath]-submodule of [imath]\mathrm{Hom}_R(M,N)[/imath]. How can I prove that these two modules are equal if [imath]M[/imath] is finite? And do you know a counterexample if [imath]M[/imath] is not finite? |
1308749 | Let [imath]\left| {{a_{ii}}} \right| > \sum\limits_{i \ne j} {\left| {{a_{ij}}} \right|} [/imath].Why does [imath]A[/imath] is nonsingular? .
Let [imath]A \in {M_n}[/imath] and [imath]\left| {{a_{ii}}} \right| > \sum\limits_{j \ne i} {\left| {{a_{ij}}} \right|} [/imath].Why does [imath]A[/imath] is nonsingular? | 456722 | Strictly diagonally dominant matrices are non singular
I try to find a good proof for invertibility of strictly diagonally dominant matrices (defined by [imath]|m_{ii}|>\sum_{j\ne i}|m_{ij}|[/imath]). There is a proof of this in this paper but I'm wondering whether there are are better proof such as using determinant, etc to show that the matrix is non singular. |
1307782 | A Simple Fourier Transform
I am studying about the randomprocess thesedays. I am stuck on solving the discrete signal to show the fourier transform the formula is that [imath] w_b(k) = {N-|k|\over N} \quad \quad when\ \ |k| <= N [/imath] [imath] w_b(k) = 0 \ \ \ \quad \quad \quad else where [/imath] and the problem is the fourier transform of [imath]w_b(k), i.e \ W_B(f)[/imath]. I check the answer is [imath] W_B(f) = {1\over N}[sin(N*pie*f)/sin(pie*f)]^2[/imath] But I couldnt figure out the process of this fourier transform Please help me for solving this | 119302 | DTFT of a triangle function in closed form
I am sampling a continuous signal [imath]x_c(t)[/imath] that follows a triangle function in the time domain, meaning: [imath]x_c(t)=\left\{\begin{array}{rl}1-|t/a|,&|t|<|a|\\ 0,&\mbox{otherwise}\end{array}\right.[/imath] Parameter [imath]a[/imath] is an integer multiple of my sampling interval [imath]T_0[/imath] such that [imath]a=kT_0[/imath]. Thus: [imath]x_d[n]=\left\{\begin{array}{rl}1-|n/k|,&|n|<|k|\\ 0,&\mbox{otherwise}\end{array}\right.[/imath] I am wondering if a discrete-time Fourier transform [imath]x_d[n][/imath] exists in closed-form, like the continuous-time Fourier transform of [imath]x_c(t)[/imath]. I know that the continuous time triangle function is the convolution of two rectangular functions, and I know that the discrete-time Fourier transform exists in closed form for the rectangular function, however, I am having trouble writing down my sampled discrete version of the triangle function as a convolution of two discrete rectangles. It is interesting that the inverse DTFT for the triangle function (i.e. triangle function in the frequency domain) is listed in this table and it looks very much like the Fourier transform of the continuous-time version. If the other side of the DTFT doesn't exist (in closed form), I'd be interested in learning the reason why, as I am new to the signal processing and Fourier analysis world. |
241235 | proof of [imath]\sum\nolimits_{i = 1}^{n } {\prod\nolimits_{\substack{j = 1\\j \ne i}}^{n } {\frac{{x_i }}{{x_i - x_j }}} } = 1[/imath]
i found a equation that holds for any natural number of n and any [imath]x_i \ne x_j[/imath] as follows: [imath]\sum\limits_{i = 1}^{n } {\prod\limits_{\substack{j = 1\\j \ne i}}^{n } {\frac{{x_i }}{{x_i - x_j }}} } = 1[/imath] when n=2, it is given by [imath]\frac{x_1}{x_1-x_2}+\frac{x_2}{x_2-x_1}=\frac{x_1 - x_2}{x_1 - x_2} = 1[/imath] when n=3, it is given by [imath]\frac{x_1^2}{(x_1-x_2)(x_1-x_3)}+\frac{x_2^2}{(x_2-x_1)(x_2-x_3)}+\frac{x_3^2}{(x_3-x_1)(x_3-x_2)}=1[/imath] But, how can I prove for general [imath]n[/imath]? | 24535 | Is this algebraic identity obvious? [imath]\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1[/imath]
If [imath]\lambda_1,\dots,\lambda_n[/imath] are distinct positive real numbers, then [imath]\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1.[/imath] This identity follows from a probability calculation that you can find at the top of page 311 in the 10th edition of Introduction to Probability Models by Sheldon Ross. Is there a slick or obvious explanation for this identity? This question is sort of similar to my previous problem; clearly algebra is not my strong suit! |
1309126 | Are linear transformations [imath]T_1[/imath], [imath]T_2[/imath] invertible?
Let [imath]T_1, T_2[/imath] be two linear transformations from [imath]\mathbb{R}^n[/imath] to [imath]\mathbb{R}^n[/imath]. Let [imath]\{ x_1, x_2,....x_n\}[/imath] be a basis of [imath]\mathbb{R}^n[/imath]. Suppose that [imath]T_1 x_i \neq 0[/imath] for every [imath]i= 1,2,...,n[/imath] and that [imath]x_i \perp [/imath]Ker [imath] T_2 [/imath] for every [imath]i= 1,2,...,n.[/imath] Which of the following is/ are necessarily true? [imath]T_1[/imath] is invertible. [imath] T_2[/imath] is invertible. Both [imath]T_1, T_2[/imath] are invertible. Neither [imath] T_1[/imath] nor [imath] T_2[/imath] is invertible. | 618531 | Is Linear transformations [imath]T_1,T_2 : \mathbb{R}^n\rightarrow \mathbb{R}^n[/imath] Invertible?
Let [imath]T_1[/imath] and [imath]T_2[/imath] be two Linear transformations from [imath]\mathbb{R}^n[/imath] to [imath]\mathbb{R}^n[/imath]. Let [imath]\{x_1,x_2,\cdots,x_n\}[/imath] be a basis of [imath]\mathbb{R}^n[/imath]. Suppose that [imath]T_1(x_i)\neq 0[/imath] for every [imath]1\leq i\leq n[/imath] and that [imath]x_i\perp Ker (T_2)[/imath] for every [imath]1\leq i\leq n[/imath]. Which of the following is true? [imath]T_1[/imath] is invertible [imath]T_2[/imath] is invertible Both [imath]T_1[/imath] and [imath]T_2[/imath] are invertible Neither [imath]T_1[/imath] nor [imath]T_2[/imath] is invertible. As [imath]T_1(x_i)\neq 0[/imath] for each [imath]1\leq i\leq n[/imath] we do not have [imath]T_1(a_1x_1+a_2x_2+\dots+a_nx_n)=0 [/imath] unless each [imath]a_i=0[/imath] i.e.,[imath]a_1x_1+a_2x_2+\dots+a_nx_n=0[/imath] i.e., [imath]T_1[/imath] is one one thus invertible. I am not sure if [imath]T_2[/imath] is invertible or not. we have [imath]x_i\perp Ker (T_2)[/imath] for all [imath]x_i[/imath]. Would that be a good idea say something like [imath]\langle x_i :1\leq i\leq n\rangle \perp Ker(T_2)[/imath] and as they span whole space we would have [imath]\mathbb{R}^n\perp Ker(T_2)[/imath] and thus [imath]Ker(T_2)=0[/imath] so it is injective so it is invertible... Thus both [imath]T_1[/imath] and [imath]T_2[/imath] are invertible? I would be so thankful if someone can assure what it has been done here is sufficient/clear. THank you |
1309943 | Does [imath] G_1 / H \cong G_2 / H [/imath] imply [imath] G_1 \cong G_2[/imath]?
The following question is taken from page 105 of Aluffi's Algebra. If a group [imath] H [/imath] may be realized as a normal subgroup of two groups [imath] G_1 [/imath] and [imath] G_2 [/imath] and if [imath] G_1 / H \cong G_2 / H [/imath] does it follow that [imath] G_1 \cong G_2 [/imath]? Feel free to give just a hint and not a full solution. | 260761 | [imath]H_1\triangleleft G_1[/imath], [imath]H_2\triangleleft G_2[/imath], [imath]H_1\cong H_2[/imath] and [imath]G_1/H_1\cong G_2/H_2 \nRightarrow G_1\cong G_2[/imath]
Find a counterexample to show that if [imath] G_1 [/imath] and [imath]G_2[/imath] groups, [imath]H_1\triangleleft G_1[/imath], [imath]H_2\triangleleft G_2[/imath], [imath]H_1\cong H_2[/imath] and [imath]G_1/H_1\cong G_2/H_2 \nRightarrow G_1\cong G_2[/imath] I tried but I did not have success, I believe that these groups are infinite. |
1309686 | Show [imath]\sqrt{1+\sqrt{2}}[/imath] is algebraic over [imath]\mathbb{Q}[/imath] with degree [imath]4[/imath]
Show [imath]\sqrt{1+\sqrt{2}}[/imath] is algebraic over [imath]\mathbb{Q}[/imath] with degree [imath]4[/imath] Let [imath]\alpha=\sqrt{1+\sqrt{2}}[/imath], and it is a root of [imath]f(x)=x^4-2x^2+1\in \mathbb{Q}[X][/imath], so [imath]f(x)[/imath] is irreducible in [imath]\mathbb{Q}[X][/imath], thus [imath]\alpha[/imath] is algebraic over [imath]\mathbb{Q}[/imath]. Clearly, [imath]\alpha[/imath] is not a root of polynomial with degree one. How to show [imath]\sqrt{1+\sqrt{2}}[/imath] has degree [imath]4[/imath]? My thought is let [imath]g(x_1)=a_1x^2+b_1x+c_1, q(x)=a_2x^3+b_2x^2+c_2x+d[/imath] and let [imath]g(x_1)=0[/imath], [imath]q(x_2)=0[/imath], but is seems not right. Can anyone give a hit to do it? Thanks | 193317 | Degree of the extension [imath]\mathbb{Q}(\sqrt{a+\sqrt{b}})[/imath] over [imath]\mathbb{Q}[/imath]
Following my previous question the book then asks Use this to determine when the field extension [imath]\mathbb{Q}(\sqrt{a+\sqrt{b}})[/imath] over [imath]\mathbb{Q}[/imath] is biquadratic (where [imath]a,b\in\mathbb{Q}[/imath]) "this" means what proved in the linked post. My thoughts: First, I don't know what to say when [imath]a^2-b[/imath] is not a square. When [imath]a^2-b[/imath] is a square I think that the extension is of degree [imath]2[/imath] if [imath]a=b[/imath], this leaves the case [imath]a\neq b[/imath] which I am also having problems with and I don't know what to do (I have some cases like if [imath]m,n[/imath] are different primes then the degree of the extension is [imath]4[/imath], but I don't have something that is general) Can someone please help my do this exercice ? |
1309783 | Is f(A) Lebesgue measurable when A is lebesgue measurable and f is a function of the class C1?
Let A be a Lebesgue measurable set. Let f: [imath]\mathbb{R} \rightarrow \mathbb{R}[/imath] be a function of the class [imath]C^1[/imath]; Is this true that f(A) is lebesgue measurable? I know that this is true when f is injective. But I don't know that when f is not injective. Could you teach me it is true or not? Sorry for my poor English. | 59105 | Is the image of a null set under a differentiable map always null?
Let [imath]n[/imath] be a positive integer. [imath]\; [/imath] Let [imath]f : \mathbb{R}^n \to \mathbb{R}^n[/imath] be everywhere Frechet differentiable. Let [imath]S[/imath] be a subset of [imath]\mathbb{R}^n[/imath] with Lebesgue measure zero. Does it follow that [imath]\{f(s) : s\in S\}[/imath] has Lebesgue measure zero? (I know it would if [imath]D(f)[/imath] is continuous.) |
1310471 | Absolute Value Expressions
Are the following expressions are equal? [imath]\sqrt{x^2}=(\sqrt{x})^2=|x|[/imath] As there both are: [imath]x^\frac{2}{2}=x^1[/imath] | 887209 | Difference between [imath]\sqrt{x^2}[/imath] and [imath](\sqrt{x})^2[/imath]
According to my logic, [imath]\large\sqrt{x^2} = x^{2\times \frac{1}{2}} = x = x^{\frac{1}{2}\times 2}={(\sqrt{x})}^2[/imath] But when I look at the graphs of these guys, they're totally different. Edit: Complex answers are okay, if you know what I mean. |
1310634 | Determine all ring homomorphisms from [imath] \Bbb Z[/imath] [imath]\oplus[/imath] [imath]\Bbb Z [/imath] to [imath]\Bbb Z[/imath].
I got [imath](a,b) \to a[/imath], [imath](a,b) \to b[/imath] and [imath](a,b) \to 0[/imath] these mappings to be homomorphisms just by hit and trial. So when I looked for it's solution, these were the ONLY homomorphisms from [imath] \Bbb Z[/imath] [imath]\oplus[/imath] [imath]\Bbb Z [/imath] to [imath]\Bbb Z[/imath]. There was a hint, which is as follows: "Observe that an idempotent must map to an idempotent." Does this always happen? Why in this case it happens? (EDIT)How should I approach to construct those three homomorphisms by using the hint? This problem belongs to the book Contemporary abstract algebra by Gallian. | 380843 | Describe all ring homomorphisms
Describe all ring homomorphisms of: a) [imath]\mathbb{Z}[/imath] into [imath]\mathbb{Z}[/imath] b) [imath]\mathbb{Z}[/imath] into [imath]\mathbb{Z} \times \mathbb{Z}[/imath] c) [imath]\mathbb{Z} \times \mathbb{Z}[/imath] into [imath]\mathbb{Z}[/imath] d) How many homomorphisms are there of [imath]\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}[/imath] into [imath]\mathbb{Z}[/imath] Note: These were past homework questions and my professor already gave out answers. I just need someone to help me understand and approach this type of problem. Thank you. |
480177 | Let [imath]f:G\to G'[/imath] be a surjective homomorphism. Prove that if [imath]G[/imath] is cyclic, then [imath]G'[/imath] is cyclic.
Just would like to see if I'm right. Proof: Let [imath]G[/imath] be generated by [imath]x[/imath] (Let [imath]G=\langle x \rangle[/imath] for some [imath]x[/imath] in [imath]G[/imath].) We wish to show that there exists an [imath]x'[/imath] in [imath]G'[/imath] such that [imath]G'=\langle x' \rangle[/imath]. First, we will show that for all [imath]x^n[/imath] in [imath]G[/imath], [imath]f(x^n)=f^n(x)[/imath]. We proceed by induction. Base case: [imath]n=1[/imath]. [imath]f(x*e_G)=f(x)f(e_G)=f(x)[/imath]. Now, assume that [imath]f(x^k)=f^k(x)[/imath] for some arbitrary positive integer k. We show that [imath]f(x^{k+1})=f^{k+1}(x)[/imath]. Observe that [imath]f(x^{k+1})=f(x^kx)=f(x^k)f(x)=f^k(x)f(x)=f^{k+1}(x).[/imath] Next we will show that for all [imath]x^{-n}[/imath] in [imath]G[/imath], [imath]f(x^{-n})=f^{-n}(x)[/imath]. We proceed by induction. Base case: [imath]n=-1[/imath]. [imath]f(x^{-1}*e_G)=f(x^{-1})f(e_G)=f(x)[/imath]. Now, assume that [imath]f(x^{-k})=f^{-k}(x)[/imath] for some arbitrary positive integer k. We show that [imath]f(x^{-k-1})=f^{-k-1}(x)[/imath]. Observe that [imath]f(x^{-k-1})=f(x^{-k}x)=f(x^{-k})f(x)=f^{-k}(x)f(x)=f^{-k-1}(x).[/imath] Because [imath]f[/imath] is onto, [imath]Im(f)=G'[/imath] and therefore [imath]Im(f)=(...,f^{-2}(x),f^{-1}(x),e_G,f(x),f^2(x),...)=G'[/imath].Thus, there exists an [imath]x'[/imath] in [imath]G'[/imath] such that [imath]G'=\langle x' \rangle[/imath] and the proof is complete by the principle of mathematical induction. Q.E.D. I have a feeling the proof shouldn't be this long. Any suggestions for a shorter, cleaner proof? | 2207317 | Proving [imath]\phi(G)[/imath] is cyclic if [imath]\phi[/imath] : [imath]G \to H[/imath] is group homomorphism and [imath]G[/imath] is cyclic
Qusetion: Prove [imath]\phi(G)[/imath] is cyclic if [imath]\phi[/imath] : [imath]G \to H[/imath] is group homomorphism and [imath]G[/imath] is cyclicDo I do something like this to prove this: Let [imath]G[/imath] be a cyclic group and [imath]G[/imath]=[imath]\langle g \rangle[/imath]Then [imath]\phi[/imath] : [imath]\mathbb Z \to \langle g \rangle[/imath] by [imath]\phi(n)[/imath]= [imath]g^{n}[/imath], [imath]n\in \mathbb Z[/imath] I am not sure of my approach to this. Need some help |
1310824 | binomial expression of a powered term
One answer to a previous question of mine asserted that [imath]k^2=\binom k2+\binom {k+1}2.[/imath] I checked that the formula is true. However, it intrigued me. Is there a similar expression for [imath]k^3[/imath]? How would I find a binomial for [imath]k^n[/imath]? This is not a duplicate question to the best of my knowledge. | 1114651 | Curious Binomial Coefficient Identity
Consider the following set of identities: [imath]{m+1\choose 1}={m\choose 1}+1[/imath], [imath]{m+1\choose 2}=2\binom m 2 - {m-1\choose 2}+1[/imath], [imath]{m+1\choose 3}=3\binom m3-3{m-1\choose 3}+{m-2\choose 3}+1[/imath], ... This set of identities can be written as [imath]{m+1\choose k}=1+\sum_{i=0}^{k-1}(-1)^i{k\choose i+1}{m-i\choose k}\tag 1[/imath] or alternatively as [imath]\sum_{i=0}^k(-1)^i{k\choose i}{m-i\choose k}=1[/imath] Interestingly, the form in [imath](1)[/imath] suggests that every binomial coefficient can be written as a recurrence in only terms from its own line (ignoring the [imath]k\choose i+1[/imath] terms as "predetermined constants"): [imath]a_{n+1}=1+\sum_{i=0}^{k-1}(-1)^i{k\choose i+1}a_{n-i}[/imath] What is the best way to prove these identities hold in general? The proofs for [imath]k=1,2,3[/imath] were very straightforward with simple factorial expansion, but even induction doesn't seem to be working in the general case. Also, do these identities have a name? These identities appeared as I was considering the quantity [imath]\frac{x^n+y^n}{x+y}\pmod{(x+y)^k}[/imath] |
1262144 | Compute the following sum
I'm struggling with this problem Compute the following sum: [imath] \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times4} + \cdots +\frac{1}{99 \times 100} [/imath] Hint: Consider telescoping series, Please help me. | 194846 | Find the sum [imath]\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}[/imath]
Please help me calculate the following sum [imath]\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\dots+\frac{1}{99\cdot 100}[/imath] |
1308811 | Inequality with the variable in both the base and the exponent.
I realised that what I took as terseness of my question, actually made it look like a lazy attempt to get a homework answer. The following is the edited question, hopefully up to the standards of this great community. I am in process of self-studying analysis using Spivak's calculus. The book is amazing, but the answers for problems are very brief, sometimes not sufficient for my limited knowledge. In this problem, I am interested in the mechanics of obtaining the reduced inequality (i.e. x Find all values of x for which: [imath]\ 3^x+x < 4 [/imath] I managed to find the answer (x<1) using "brute-force", i.e. pondering what values of x would satisfy the equation, and obtaining the result without manipulating the formula, but my curiosity remains unsatisfied so as to the issue of how to manipulate this simple inequality to reduce it to solution. | 1273619 | How to solve the exponential inequality [imath]x+3^x<4[/imath]
How to solve the inequality [imath]x+3^x<4[/imath] This problem is found in Spivak's calculus, ch 1 - the highly praised work - which is supposed to be a gentle introduction for beginners in mathematics. Please do not tell me that it cannot be solved algebraically and one needs to use Lambert's function. I have heard of Lambert's function and understand that it requires some advanced knowledge of differential equations. Neither me nor the average student of Spivak's calculus is that advanced to know that stuff. Thanks in advance. |
1311023 | Showing [imath] \int_0^{2 \pi } \frac{dt}{a^2 \cos^2 t + b^2 \sin^2 t} = \frac{2 \pi}{ab}[/imath]
The question: Let [imath]\gamma[/imath] be a contour such that [imath]0 \in I(\gamma),[/imath] where [imath]I[/imath] is the interior of the contour. Show that [imath]\int_\gamma z^n \, \text{d}z = \begin{cases} 2\pi i & \text{if } n = -1 \\ 0 & \text{otherwise} \end{cases}[/imath] By taking [imath]\gamma[/imath] as the ellipse [imath]\{ (x,y) : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \},[/imath] show that [imath] \int_0^{2 \pi } \frac{dt}{a^2 \cos^2 t + b^2 \sin^2 t} = \frac{2 \pi}{ab}.[/imath] The question's answer uses the Deformation theorem and the fact that the first integral has the given value if the contour is a unit circle. However, the two contours must not overlap, so it seems like this should only be true for a contour that either always has magnitude less than one or greater than one. In Mathematica, the final integral was true for the case that [imath]a = 1.5[/imath] and [imath]b=0.4[/imath]. What am I missing? Edit: The radius of the circle cancels in the integral if [imath]n = -1[/imath], so then it does hold irrespective of the radius. | 518173 | Calculate the integral [imath]\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2t+b^{2}\sin^{2}t}dt[/imath], by deformation theorem.
I want to prove: [imath]\int_{0}^{2\pi}\frac{1}{a^{2}\cos^2t+b^{2}\sin^{2}t}dt=\frac{2\pi}{ab}[/imath] by the deformation theorem of complex variable. Then I consider a parameterization [imath]\gamma:[0,2\pi]\rightarrow A[/imath], traveled in the opposite direction of the clock hand, of the ellipse (in [imath]\mathbb{C}[/imath]): [imath]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/imath] I thought of searching for a function [imath]f:A\rightarrow \mathbb{C}[/imath] (analytic in [imath]A[/imath]) and a curve [imath]\lambda:[0,2\pi]\rightarrow A[/imath] (homotopic to [imath]\gamma[/imath]) and use the deformation theorem: [imath]\int_{\gamma}f = \int_{\lambda}f[/imath] But I can not find a function [imath]f(z)[/imath] that meets what I need, can you help me to find the f? P.D: Necessarily I have to use the theorem of deformation |
773568 | limit of p norm as p goes to 0!
Suppose we have a measure [imath]\mu[/imath] and a space [imath]X[/imath] such that [imath]\mu(X)=1[/imath], and a function [imath]f \in L^r[/imath] for some [imath]r > 0[/imath], where [imath]L^r[/imath] is defined in the usual way even for numbers less than [imath]1[/imath]. Show that [imath]lim_{p \to 0} ||f||_p =[/imath] exp[imath](\int [/imath]log[imath]|f|d\mu)[/imath], where we note that exp(-[imath]\infty)=0[/imath] by convention. Stuck on this one. Trying to use Holder's! Thoughts? | 115428 | Limit of [imath]L^p[/imath] norm when [imath]p\to0[/imath]
Let ([imath]\Omega[/imath], [imath]\cal{F}[/imath], [imath]\mu[/imath]) be a probability space and [imath]f\in L^1(\Omega)[/imath]. Prove that [imath]\displaystyle\lim_{p\to 0} \left[ \int_{\Omega}|f|^pd\mu \right]^{\frac{1}{p}}=\exp \left[ \int_{\Omega}\log|f| d\mu \right],[/imath] where [imath]\exp[-\infty]=0[/imath]. To simplify the problem, we may assume [imath]\log|f|\in L^1(\Omega).[/imath] |
1310283 | Show [imath][K : F] = [K : E][E : F][/imath].
Let [imath]E\subset F\subset K[/imath] be fields. Show that if [imath]K[/imath] is a field extension of finite degree over [imath]F[/imath] and [imath]F[/imath] is a field extension of finite degree over [imath]E[/imath] then [imath][K : E] = [K : F][F : E][/imath]. Since [imath]F[/imath] is a vector space over [imath]E[/imath] with degree [imath]m[/imath] and [imath]K[/imath] is a vector space over [imath]F[/imath] with degree [imath]n[/imath], then [imath]K[/imath] is a vector space over [imath]F[/imath]. So [imath][K : E] = [K : F][F : E]=mn[/imath]. I think I still miss some details need to show. Can anyone tell me what they are? Thanks! | 78517 | Given fields [imath]M/E/F[/imath], why does [imath][M:F] = [M:E][E:F][/imath]?
Let [imath]M[/imath] be a finite extension of [imath]E[/imath] and let [imath]E[/imath] be a finite extension of [imath]F[/imath]. Then [imath]M[/imath] is a finite extension of [imath]F[/imath] and [imath][M:F] = [M:E][E:F][/imath]. Is there an easy explanation and/or proof for this theorem? My instructor gave an incomplete proof in class but got stuck and I'm kind of confused of what's going on. |
1308088 | Norm of the quotient map for a normed space
Let [imath]X[/imath] be a normed space and [imath]F[/imath] a closed subspace. On [imath]X/F[/imath] let us take the quotient norm [imath]||[x]|| = \inf_{y \in F} ||x - y||[/imath]. Consider the quotient [imath]q : X \rightarrow X/F[/imath]. I can see that, if [imath]||x|| = 1[/imath], then [imath]||q(x)|| = \inf_{y \in F} ||x - y|| \leq ||x - 0|| = 1[/imath] since [imath]0 \in F[/imath]. This proves that [imath]q[/imath] is bounded and [imath]||q|| \leq 1[/imath]. How may I show that [imath]||q|| = 1[/imath]? | 704958 | Showing that the norm of the canonical projection [imath]X\to X/M[/imath] is [imath]1[/imath]
How do I show that given [imath]M[/imath] a closed subspace of a normed space [imath]X[/imath], and let [imath]\pi[/imath] be the canonical projection of X onto [imath]X/M[/imath]. Prove that [imath]\|\pi\| = 1[/imath]. I figure I could use Riesz' lemma and set [imath]\|\pi\| = 1[/imath], that's as far as I got. I could also use the fact that the canonical map is a contraction so I would have [imath]\|\pi(x)\| \leq \|x\| \Rightarrow \frac{\|\pi(x)\|}{\|x\|} \leq 1[/imath] and taking a supremum then we get the desired result. This doesn't seem as rigorous. |
1311315 | Prove [imath]\sin(kx) \rightharpoonup 0[/imath] as [imath]k \to \infty[/imath] in [imath]L^2(0,1)[/imath]
I want to show that [imath]u_k(x)= \sin(kx) \rightharpoonup 0[/imath] as [imath]k \to \infty[/imath] in [imath]L^2(0,1)[/imath]. We know trivially that [imath]0 \in L^2(0,1)[/imath]. I need to show that [imath]\langle u^*,\sin(kx) \rangle \to \langle u^*, 0 \rangle[/imath] for each bounded linear functional [imath]u^* \in L^2(U)[/imath], where [imath]L^2(U)[/imath] is a dual space of itself (since [imath]L^2[/imath] is a Hilbert space). I think I need to show that, as [imath]k \to \infty[/imath], [imath]\int_0^1 u^* \sin(kx) \, dx \to 0.[/imath] | 150645 | Why [imath]\sin(nx)[/imath] converges weakly in [imath]L^2(-\pi,\pi)[/imath]?
Can anybody tell me why [imath]\sin(nx)[/imath] converges weakly in [imath]L^2(-\pi,\pi)[/imath]. I can't see how [imath]\sin(nx)[/imath] can converge? Explanation with any other example will be nice as well. |
1311702 | Find all primes [imath]p>2[/imath] for which [imath]x^2+x+1[/imath] is irreducible in [imath]\mathbb{F}_p[x][/imath]
Find all primes [imath]p>2[/imath] for which [imath]x^2+x+1[/imath] is irreducible in [imath]\mathbb{F}_p[x][/imath] Attempt. Since [imath]x^2+x+1[/imath] is of degree 2, it is reducible iff it has a root in [imath]\mathbb{F}_p[/imath]. It has a root in [imath]\mathbb{F}_p[/imath] iff [imath]x^3-1=(x-1)(x^2+x+1)[/imath] has a root other than 1. The latter has a root other than [imath]1[/imath] iff [imath]\mathbb{F}_p[/imath] has elements of order [imath]3[/imath], which happens iff [imath]3\mid |\mathbb{F}_p^{\times}|=p-1[/imath], which happens iff [imath]p\equiv 1\bmod 3[/imath]. So [imath]x^2+x+1[/imath] is irreducible iff [imath]p\not \equiv 1\bmod3 [/imath]. I am getting a different answer than my professor, who has [imath]p\equiv 1\bmod 3[/imath]. If I am incorrect, were am I going wrong? Update. Now I am unsure about my own reasoning. What if [imath]1[/imath] is a root of [imath]x^2+x+1[/imath]? I don't know why I am considering roots of [imath]x^2+x+1[/imath] other than [imath]1[/imath] when [imath]1[/imath] can be a root. | 696417 | [imath]\mathbb{F}_p[X]/(X^2+X+1)[/imath] is a field iff [imath]p \equiv 2 \bmod 3[/imath]
Let [imath]p[/imath] be a prime. Prove that [imath]\mathbb{F}_p[X]/(X^2+X+1)[/imath] is a field iff [imath]p \equiv 2\bmod3[/imath]. So: If [imath]p \equiv 2 [/imath] mod [imath] 3[/imath], I have to show that every element of [imath]\mathbb{F}_p[X]/(X^2+X+1)[/imath] has an inverse. If [imath]\mathbb{F}_p[X]/(X^2+X+1)[/imath] is a field I have to show [imath]p \equiv 2 [/imath] mod [imath] 3[/imath]. |
1098954 | Integral [imath] I=\int_{-r}^r \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \sqrt{1 - \frac{x^2 + y^2}{x^2 + y^2 - r^2}} dy dx [/imath]
I have to solve this integral [imath] I=\int_{-r}^r \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \sqrt{1 - \frac{x^2 + y^2}{x^2 + y^2 - r^2}} dy dx [/imath] with substitution and then the trick that [imath]\dfrac{1}{\sqrt{1 - x^2}} = \dfrac{ d{\arcsin(x)}}{dx}[/imath] Can someone give me a suggestion on what I should substitute in order to continue? Thank you in advance. | 1098755 | Integrate via substitution and derivation rule
I have to solve this integral [imath]\int_{-r}^{+r}\int_{-\sqrt{r^2-x^2}}^{+\sqrt{r^2-x^2}} \sqrt{1-\frac{x^2+y^2}{x^2+y^2-r^2}} \operatorname d y \operatorname d x[/imath] with substitution and then the trick that [imath]\dfrac 1 {\sqrt{1-x^2}} = \dfrac{\mathsf d\;\arcsin(x)}{\mathsf d\;x\qquad\quad\;\,} [/imath] can someone give me a tip on what I should substitute in order to continue? |
1312926 | Find the limit of the following.
[imath]\lim_{x\to \ 0}\frac{{(1+x)}^\frac{1}{x}-e+ex}{x^2}[/imath] Using Taylor Expansion, [imath](1+x)^{1/x}=e(1- \frac{x}{2} + \frac{11x^2}{24} + ...)[/imath]. Finally, I found that [imath]\lim_{x\to \ 0}\frac{{(1+x)}^\frac{1}{x}-e+ex}{x^2}=\lim_{x\to \ 0}\frac{\frac{ex}{2}+\frac{11ex^2}{24}+...}{x^2}[/imath] [imath]=\lim_{x\to \ 0}\frac{\frac{e}{2}+\frac{11ex}{12}+...}{x}[/imath] Now, It seems limit doesn't exist, but options are given for this question as 1) [imath]\frac{11e}{12}[/imath] 2)[imath]-\frac{11e}{12}[/imath] 3)[imath]\frac{11e}{24}[/imath] 4)[imath]-\frac{11e}{24}[/imath]. So How to proceed further? Thank you. | 1284801 | [imath] \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} [/imath]
[imath] \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} [/imath] (can this be duplicate? I think not) I tried it using many methods [imath]1.[/imath] Solve this conventionally taking [imath]1^\infty[/imath] form in no luck [imath]2.[/imath] Did this, expand [imath] {(1+x)^{\frac1x}}[/imath] using binomial theorem got [imath]\frac13[/imath] then grouped coefficients of [imath]x^0[/imath] and it cancelled with [imath]e[/imath] then took coefficient of [imath]x[/imath] cancelled with [imath]\frac{ex}{2}[/imath] and so on very messy right ? at last I got [imath]\frac13[/imath] but that's not the expected answer! I must have went wrong somewhere can anyone help me with this. |
1313694 | Prove that [imath]L(s)[/imath] converges for [imath]s > 0,[/imath] or more generally for all complex [imath]s \in \mathbb{C}[/imath] with [imath]\Re(s) > 0.[/imath]
Let [imath]L(s) = \sum_{n=1}^{\infty} a_n/n^s[/imath] be a Dirichlet series. Suppose that the partial sums of the coefficients [imath]A_n = \sum_{i=1}^n a_i[/imath] are bounded, i.e. there exists a constant [imath]C[/imath] such that [imath]|A_n| \leqslant C[/imath] for all [imath]n.[/imath] Then prove that [imath]L(s)[/imath] converges for [imath]s > 0,[/imath] or more generally for all complex [imath]s \in \mathbb{C}[/imath] with [imath]\Re(s) > 0.[/imath] How do we prove this? I've only taken algebra courses so my analysis background seems to be lacking. | 743224 | how to show that this complex series converge?
If [imath]\sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}}[/imath] Converges( s is real) and [imath]\operatorname{Re}(z)>s[/imath]. Then [imath]\sum_{n=1}^{\infty} \frac{a_{n}}{n^{z}}[/imath] also converges. [imath]a_n[/imath] is complex sequence. |
1313855 | example for such a non-integrable function on [imath][0,1][/imath]
This is an exercise from the book "Real Analysis for Graduate Students": Find a non-negative function [imath]f[/imath] on [imath][0,1][/imath] such that [imath]\lim_{t\rightarrow \infty} t m(\{x: f(x)\geq t\})=0[/imath], but [imath]f[/imath] is not integrable, where [imath]m[/imath] is Lebesgue measure. So, it should be non-integrable with respect to Lebesgue integral. In one of the previous questions, I had proven that if it's integrable, then given limit is really zero. But, I couldn't come up with such a non-integrable function yet.Thanks! | 872067 | Find a non-negative function on [0,1] such that [imath]t\cdot m(\{x:f(x) \geq t\}) \to 0[/imath] that is not Lebesgue Integrable
Problem: Find a non-negative function [imath]f[/imath] on [imath][0,1][/imath] such that [imath]\lim_{t\to\infty} t\cdot m(\{x : f(x) \geq t\}) = 0,[/imath] but [imath]f[/imath] is not integrable, where [imath]m[/imath] is Lebesgue measure. My Attempt: Let [imath]f(x) = \frac{\chi_{(0,1]}}{\sqrt{x}}[/imath]. Then, \begin{align*} \lim_{t\to\infty} t\cdot m(\{x \in [0,1]: f(x) \geq t\}) &= \lim_{t\to\infty} t \cdot m(\{x\in (0,1]: 1/\sqrt{x} \geq t\})\\ &=\lim_{t\to\infty} t \cdot m(\{x\in (0,1]: x \leq (1/t^2)\})\\ &= \lim_{t\to\infty} t \cdot m((0, (1/t^2)))\\ &= \lim_{t\to\infty} \frac{t}{t^2} = 0. \end{align*} However, [imath]f(x)[/imath] is integrable over this interval. I have tried functions looking like [imath]f(x) = 1/x^p[/imath] but I cannot find any that will work here. There is a hint that says there is a monotonic function that fits this description. Also, does anyone know of a list of non-Lebesgue integrable functions on [imath][0,1][/imath]? I feel as though I could use this for many counterexamples if one were to exist. Thanks! |
1313989 | Uniformly continuous functions on the interval
Let [imath]f:[1,\infty)\to\mathbb R[/imath] be uniformly continuous. Prove [imath]\exists[/imath] [imath]M > 0[/imath] s.t [imath]\frac{\big|f(x)\big|}{x} \leq M, \hspace{11pt} \forall x\in[1,\infty)[/imath] | 305014 | Every uniformly continuous real function has at most linear growth at infinity
Assuming [imath]f:\mathbb R\to\mathbb R [/imath] be an uniform continuous function, how to prove [imath]\exists a,b\in \mathbb R^+ \quad \text{such that}\quad |f(x)|\le a|x|+b.[/imath] |
1314371 | Simplifying the product [imath]\prod\limits_{k=2}^n \left(1-\frac1{k^2}\right)[/imath]
Can we simplify the given product to a general law? [imath]\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{n^2}\right)[/imath] | 286798 | Find [imath]\lim\limits_{n \to \infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)[/imath]
Find the limit [imath]\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right][/imath] I take log and get [imath]\lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)[/imath] |
1314270 | Prob. 10, Sec. 3.10 in Kreyszig's functional analysis book: Every isometric linear operator on a finite-dimensional inner product space is unitary?
Let [imath]X[/imath] be an inner product space such that [imath]\dim X < \infty[/imath], and let [imath]T \colon X \to X[/imath] be an isometric linear operator. Since [imath]\dim X < \infty[/imath], [imath]X[/imath] is complete and thus a Hilbert space; since [imath]T[/imath] is isometric, [imath]T[/imath] is also injective and hence also surjective and thus bijective, because [imath]\dim X < \infty[/imath]. So [imath]T^{-1}[/imath] exists. How to show that [imath]T[/imath] is unitary. That is, how to show that the Hilbert adjoint operator [imath]T^*[/imath] of [imath]T[/imath] equals [imath]T^{-1}[/imath]? Since [imath]X[/imath] is finite-dimensional, we can choose an orthonormal basis for [imath]X[/imath]; let [imath]n \colon= \dim X[/imath], and let [imath]\{e_1, \ldots, e_n \}[/imath] be an orthonormal basis for [imath]X[/imath]. Then, for each [imath]i, j = 1, \ldots, n[/imath], we have [imath]\langle Te_i , e_j \rangle = \langle e_i, T^* e_j \rangle,[/imath] and, [imath]\langle T^* T e_i, e_j \rangle = \langle T e_i , T e_j \rangle = \langle e_i , e_j \rangle = \begin{cases} 1 \ & \mbox{ if } \ i = j \\ 0 \ & \mbox{ if } \ i \neq j. \end{cases} [/imath] What next? | 690337 | Linear surjective isometry then unitary
Basically what I'm trying to show is [imath]\forall h_1, \ h_2 \in \mathscr{H}[/imath] and [imath]U: \mathscr{H} \rightarrow \mathscr{K}[/imath] then [imath]\langle Uh_1, \ Uh_2\rangle_\mathscr{K} = \langle h_1, \ h_2 \rangle_\mathscr{H}[/imath] (So your standard definition). Now I figure the easiest way to do so is by polarization. Since we have an isometry then [imath]\begin{eqnarray*} \langle h_1, \ h_2 \rangle_\mathscr{H} &=& \frac{1}{4} \bigg( \|h_1 + h_2\|_\mathscr{H}- \|h_1-h_2\|_\mathscr{H} +i\|h_1+ih_2\|_\mathscr{H} - i\|h_1 - ih_2\|_\mathscr{H} \bigg) \\&=&\frac{1}{4} \bigg( \|Uh_1 + Uh_2\|_\mathscr{K}- \|Uh_1-Uh_2\|_\mathscr{K} +i\|Uh_1+iUh_2\|_\mathscr{K} - i\|Uh_1 - Uih_2\|_\mathscr{K} \bigg) \\&=& \langle Uh_1, \ Uh_2\rangle_\mathscr{K} \end{eqnarray*} [/imath] But what I'm wondering is where surjectivity comes into play? |
1314831 | If [imath]\vec{v}[/imath] is an eigenvector of [imath]A[/imath], then also [imath]B\vec{v}[/imath], when [imath]AB = BA[/imath]
I have the following problem: Let [imath]V[/imath] be a finite dimensional vector space. Let [imath]A[/imath], [imath]B[/imath] be linear maps of [imath]V[/imath] into itself. Assume that [imath]AB = BA[/imath]. Show that if [imath]\vec{v}[/imath] is an eigenvector of [imath]A[/imath], with eigenvalue [imath]\lambda[/imath], then [imath]B\vec{v}[/imath] is an eigenvector of [imath]A[/imath] with eigenvalue [imath]\lambda[/imath] also if [imath]B\vec{v} \neq \vec{0}[/imath]. Now, I have to show that if [imath]\vec{v}[/imath] is an eigenvector of [imath]A[/imath] with eigenvalue [imath]\lambda[/imath], then also [imath]B\vec{v}[/imath] with the same eigenvalue. So I started by assuming [imath]A\vec{v} = \lambda \vec{v}[/imath] Then I multiplied both sides by [imath]B[/imath]: [imath]BA\vec{v} = B \lambda \vec{v}[/imath] Since [imath]AB = BA[/imath], then [imath]A(B\vec{v}) = \lambda (B \vec{v})[/imath] Which should prove what I needed to prove. Is my proof correct? | 1022082 | [imath]AB=BA[/imath] with same eigenvector matrix
I read in G. Strang's Linear Algebra and its Applications that, if [imath]A[/imath] and [imath]B[/imath] are diagonalisable matrices of the form such that [imath]AB=BA[/imath], then their eigenvector matrices [imath]S_1[/imath] and [imath]S_2[/imath] (such that [imath]A=S_1\Lambda_1S_1^{-1}[/imath] and [imath]B=S_2\Lambda_2 S_2^{-1}[/imath]) can be chosen to be equal: [imath]S_1=S_2[/imath]. How can it be proved? I have found a proof here, but it is not clear to me how to see that [imath]C[/imath] is diagonalisable as [imath]DCD^{-1}=Q[/imath]. Matrix [imath]C[/imath] obviously is the matrix with the coordinates of [imath]B\mathbf{x}[/imath] with respect to the basis [imath]\{\mathbf{x}_1,...,\mathbf{x}_k\}[/imath] of the eigenspaceof [imath]V_\lambda (A)[/imath], but I do not see how we can know that it is diagonalisable. Thank you very much for any explanation of the linked proof or other proof!!! EDIT: Corrected statement of the lemma I am interested in. See comments below by the users whom I thank for what they have noticed. |
1315131 | How to prove that [imath](C[a,b], \|\cdot\|_\infty)[/imath] is not a reflexive Banach Space
The tag line basically says it all...this is a question in Luenberger's Optimization book (5.14.4 on p.138). Clearly I don't expect someone to deliver a full proof if it's tedious, but a sketch or theorem or tactic would help...I tried: characterizing the dual of [imath]BV[a,b].[/imath] finding a sequence [imath]\{f_n\}[/imath] of continuous functions without a weakly convergent subsequence. Using the Hahn-Banach theorem to extend a functional on [imath]BV[a,b][/imath]. Any help? Thanks! | 153494 | The space [imath]C^0([0;1],\mathbb{R})[/imath] of all continuous, real-valued functions on [imath][0,1][/imath] is not reflexive.
How does one prove that [imath]C^0([0;1],\mathbb{R})[/imath] equipped with the supremum norm is not reflexive? I don't understand how to show that the [imath]J[/imath] mapping is not surjective. |
1314920 | How does one show that [imath]\int_0^\infty \left|\frac{\sin x} x\right| \, dx=\infty[/imath]?
In many place one finds accounts of how to evaluate [imath] \int_0^\infty \frac{\sin x} x\,dx = \underbrace{\lim_{a\to\infty}\int_0^a}_{\text{Why view it this way?}} \frac{\sin x} x\, dx. [/imath] And it gets asserted that the reason why the integral must first be evaluated over a bounded interval and then afterward the bound is allowed to go to [imath]\infty[/imath] is that [imath] \int_0^\infty \left|\frac{\sin x} x\right| \, dx\ \underbrace{{}\ =\infty\ {}}_{\text{That's why.}} \tag 1 [/imath] and therefore the ways of thinking about integrals introduced by Henri Lebesgue are not applicable on the unbounded interval. If it is asked how we know that [imath](1)[/imath] is true, my first thought it that one ought to compare it somehow with a harmonic series because it ultimately declines to [imath]0[/imath] at the same rate. It seems as if this question is worth having among our stock of questions and answers, but a couple of days ago I found the answer posted as a mere comment. Hence this present posting. I'll notify the person who posted the comment. Doubtless some variety of ways to prove this result exists, so others should post their own versions. | 25586 | How to prove absolute summability of sinc function?
We know that [imath]\int_0^\infty \left(\frac{\sin x}{x}\right)^2 dx=\int_0^\infty \frac{\sin x}{x} dx=\frac{\pi}{2}.[/imath] How do I show that [imath]\int_0^\infty \left\vert\frac{\sin x}{x}\right\vert dx[/imath] converges? |
1315429 | counting total no combinations
I want to count total no of combinations in sets of size n each,which give a fixed sum,say [imath]k[/imath]..for example if [imath]k=1[/imath] and and [imath]n=3[/imath],then the possible combinations are [imath](0,0,1),(0,1,0)[/imath] and [imath](1,0,0)[/imath].So answer will be [imath]3[/imath]. Note:All numbers in each set must be greater than or equal to [imath]0[/imath]. One more example,say if [imath]k=2[/imath] and [imath]n=3[/imath],then there will be [imath]6[/imath] combinations. [imath](0,1,1),(1,0,1),(1,1,0),(0,0,2),(0,2,0),(2,0,0)[/imath] Calculating every possibilities is tedious,so is there any formula which can calculate the required number of combinations in terms of [imath]n[/imath] and [imath]k[/imath]? | 868785 | How many positive integer solutions are there to the equality [imath]x_1+x_2+...+x_r= n[/imath]?
The original problem is there are [imath]r[/imath] identical boxes and [imath]n[/imath] identical balls. Every box is nonempty. Then how many ways of putting balls in boxes? It is equivalent to the problem of finding integer solutions for the equality: [imath]x_1+x_2+...+x_r= n[/imath] and [imath]x_i>0[/imath] for all [imath]1\leq i\leq r[/imath]. Editted by @Sil: Replaced with equality to avoid confusion that this question deals with [imath]x_1+x_2+\dots+x_r\leq n[/imath], which it does not! |
1315484 | Square root of [imath]-1[/imath] over a finite field
It is known that the equation [imath]x^2 \equiv -1 \pmod{p}[/imath], where [imath]p[/imath] is an odd prime number, has a solution iff [imath]p = 4k +1[/imath] for some natural [imath]k[/imath]. Does it exist a similar characterization for a general finite field [imath]F_q[/imath], where [imath]q = p^n[/imath]? Thanks a lot for your help! | 1315001 | Describe all [imath]p^{n}[/imath] (in terms of congruence conditions of [imath]p[/imath] and [imath]n[/imath]) for which [imath]x^{2}+1[/imath] irreducible over [imath]\mathbb{F}_{p^{n}}[/imath].
So I've said [imath]x^{2}+1[/imath] is reducible over [imath]\mathbb{F}_{p^{n}} \iff \mathbb{F}_{p^{n}}[/imath] contains a root [imath]\alpha[/imath]. Hence if [imath]\alpha[/imath] is such a root then [imath]\alpha^2 = -1[/imath] so that [imath]\alpha^4 =1[/imath] and hence [imath]|\alpha|=4[/imath] in [imath]\mathbb{F}^{\times}_{p^{n}}[/imath] and hence [imath]4\mid p^n -1 \implies p^n \equiv 1\bmod 4[/imath]. Hence if [imath]p^{n}[/imath] is congruent to [imath]0,2,3[/imath] mod [imath]4[/imath] then [imath]x^{2}+1[/imath] is irreducible over [imath]\mathbb{F}_{p^{n}}[/imath]. My problem is with [imath]p^{n}[/imath] congruent to [imath]1[/imath] mod [imath]4[/imath] being sufficient to conclude that [imath]x^{2}+1[/imath] is reducible. How can I show this? |
1315935 | Find [imath](f^{-1})'(a) = f(x) = 2x^3 + 3x^2+7x+4, a=4 [/imath]
Find [imath](f^{-1})'(a) = f(x) = 2x^3 + 3x^2+7x+4, a=4 [/imath] How do I go about solving a problem like this? What are the steps? | 726940 | Finding the derivatives of inverse functions at given point of c
Hoping someone can help me the understand the steps to solve a problem like this. I'm guessing it involves the formula: [imath]\frac{d}{dx}f^{-1}(f(x))=1/f'(x)[/imath]. Am I right in this assumption? I would post some work that I've tried, but I'm not sure where to even start. My professor is bad at explaining things so I'm turning to the site for supplemental instruction. "For each of the given functions [imath]f(x)[/imath], find the derivative [imath]\left. \frac{d}{dx}f^{-1}(x) \right|_{x=c}[/imath] at the given point [imath]c[/imath], first finding [imath]a=f^{-1}(c)[/imath]. [imath] f(x)= 3 x + 9 x^{13}; \ \ c = -12, \ \ \ a=? \ \ \ (f^{-1})'(c)=? [/imath] [imath]f(x)= x^2 - 13 x + 58[/imath] on the interval [imath][6.5,\infty);[/imath] [imath]c = 18[/imath], [imath]a=?[/imath] [imath](f^{-1})'(c) = ?[/imath]" I've been able to find the inverse of other functions not involving higher powers like the [imath]9x^{13}[/imath], and then finding the derivative of that inverse, but this particular kind of question stumps me. |
1315820 | If [imath]A[/imath] is an associative algebra Show that [imath]End_{A}(A)=A^{op}[/imath] the algebra with opposite multiplication
Let [imath]A[/imath] be an associative algebra, And let [imath]V[/imath] be a representation of [imath]A[/imath]. By [imath]End_{S}(V)[/imath] one denotes the algebra of homomorphisms of representations [imath]V \to V[/imath] Show that [imath]End_{A}(A)=A^{op}[/imath] the algebra with opposite multiplication. | 270614 | For a ring [imath]R[/imath], does [imath]\operatorname{End}_R(R)\cong R^{\mathrm{op}}[/imath]?
Given a ring [imath]R[/imath], how to prove that [imath]\operatorname{End}_R(R)\cong R^{\mathrm{op}}[/imath], where [imath]R^{\mathrm{op}}[/imath] is the opposite ring of [imath]R[/imath]. I read this proposition somewhere, but I think it is wrong. Because for any given [imath]f\in\operatorname{End}_R(R) [/imath] and any [imath]r[/imath] in [imath]R[/imath], we can get [imath]f(r)[/imath] by [imath]r f(1)[/imath], and [imath]f(1)[/imath]can only be [imath]1[/imath], so [imath]\operatorname{End}_R(R)[/imath] is isomorphic to the trivil group. Is there anything wrong with my logic? Please help me identify where I am wrong in my understanding of [imath]\operatorname{End}_R(R)[/imath]. Thanks! |
1265843 | Vector Subspace for Differential Equation
We are given the following problem: Let [imath]S[/imath] be the set of all functions [imath]y[/imath] that satisfy the following differential equation [imath]2\dfrac{d^2y}{dx^2} - 3\dfrac{dy}{dx} + y = 0.[/imath] Show that [imath]S[/imath] is a subspace of the vector space [imath]A[/imath], where [imath]A[/imath] is the set of all functions [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath]. I do not know how to approach the problem. | 414132 | Is [imath]f'(x)-3f(x) = 0[/imath] subspace of differentiable functions [imath]f\colon (0,1)\to \mathbb{R}[/imath]
[imath]V[/imath] is space of differentiable functions [imath]f(0,1) \to \mathbb{R}[/imath] and [imath]W[/imath] is a subset of [imath]f[/imath] that meets [imath]f'(x) - 3f(x) = 0[/imath] for all [imath]x\in (0,1).[/imath] Is subset [imath]W[/imath] a subspace of [imath]V[/imath]? I know that I have to prove that it's closed under scalar multiplication and under addition. But I forgot all the stuff about differentiable functions, so I don't know how to do this. But my intuition tells me that if there's a zero it's valid subspace :) If anyone could elaborate on this I would be grateful. |
1315125 | Calculus of variations - unilateral constraints
Question about Evans states, chapter 8.4.2! We have [imath]I[w] := \int_U \frac{1}{2}|Dw|^2 - fw\, dx[/imath], among all functions [imath]w[/imath] belonging to the set [imath]\mathcal{A} : = \{w \in H_0^1(U) : w \geq h \, \mbox{ a.e. in } U\}[/imath] with smooth [imath]h[/imath] and [imath]f[/imath]. And we have: Theorem: Assume the admissible set [imath]\mathcal{A}[/imath] is nonempty. Then there exists a unique function [imath]u \in \mathcal{A}[/imath] satisfying [imath]I[u] = min_{w \in \mathcal{A}} I[w][/imath] I don't understand how we can say that [imath]L[/imath] is weakly lower semicontinuous. And I think we need to have this for the proof of the existence. | 1313629 | The functional [imath]I[w] = \int_U \frac{1}{2} |Dw|^2 - fw \, dx[/imath] is weakly lower semicontinuous
I am studying calculus of variation, and I need to prove that [imath]I[w] = \int_U \frac{1}{2} |Dw|^2 - fw \, dx[/imath] with [imath]f \in L^2(U)[/imath] is weakly lower semicontinuous on [imath]H_0^1(U)[/imath]. In classes, I only learned that Assume L is bounded below, and in addition [imath]L(p,z,x)[/imath] is convex in [imath]p[/imath], for each [imath]z \in \mathbb{R}[/imath], [imath]x \in U[/imath]. Then [imath]I[.][/imath] is weakly lower semicontinuous on [imath]W^{1,q}(U)[/imath], where [imath]I[w] = \int_U L(Dw, w, x)[/imath]. And I think my [imath]L[/imath] isn't bounded below. I don't know how I can solve the problem. |
1317263 | intersection of two subspaces is not [imath]\{0\}[/imath]
If V and W are 3-dimensional subspace of [imath]R^5[/imath] then prove that V and W must have a nonzero common vector. | 565838 | Given [imath]3[/imath]-dimensional subspaces [imath]V, W \subset \Bbb R^5[/imath], there is a nonzero vector in [imath]V \cap W[/imath]
Prove that if [imath]V[/imath] and [imath]W[/imath] are three-dimensional subspaces of [imath]\Bbb R^{5}[/imath], then [imath]V[/imath] and [imath]W[/imath] must have a non-zero vector in common. So far I got [imath]V = \{v_{1}, v_{2}, v_{3}\}[/imath] and [imath]W = \{ w_{1}, w_{2}, w_{3}\}[/imath] because they are 3-dimensional subspaces. Then [imath]V \cup W = \{v_{1}, v_{2}, v_{3}, w_{1}, w_{2}, w_{3}\}[/imath] My prof said that now I need to prove that they are linearly independent so this is a contradiction and the statement is wrong because we are in [imath]\Bbb R^{5}[/imath] and we have 6 elements on it. I don't know how to express them as a linearly independent proof, so I hope anyone can give me a good explanation on it. |
1317247 | Why can't a closed interval on R be written as the disjoint union of countably infinite many closed intervals?
As stated in the title: Why can't a closed interval [imath][a,b][/imath] on R be written as the disjoint union of countably infinite many closed intervals [imath][a_i,b_i][/imath]? | 1195179 | The interval [imath][0,1][/imath] is not the disjoint countable union of closed intervals.
The following proof was suggested: suppose [0,1] was the disjoint countable union of closed intervals. Write the intervals as [imath][a_n,b_n][/imath]. Start by showing the set of endpoints [imath]a_n, b_n[/imath] is closed. At first I thought this was obvious since it seems like the complement is just the union of [imath](a_n,b_n)[/imath] which is open but then I thought the complement was the infinite intersection of [imath](0,a_n) \cup (a_n,b_n) \cup (b_n,1)[/imath] which is not necessarily open any more. This comes from Taylor's proof in Is [imath][0,1][/imath] a countable disjoint union of closed sets? |
1316864 | Graph: Prove that [imath]\chi(G)+\chi(G^c)\leq n+1[/imath]
Let [imath]G[/imath] a graph on [imath]n[/imath] vertices. Prove that [imath]\chi(G)+\chi(G^c)\leq n+1,[/imath] where [imath]chi[/imath] denote the minimum vertex coloring such that no two adjacent vertex has that same color. My proof (which is doesn't work): By induction, for [imath]n=0[/imath], there is nothing to prove. Suppose the assumption true. Let [imath]G[/imath] a graph on [imath]n+1[/imath] vertices [imath]\{v_1,...,v_{n+1}\}[/imath] and let [imath]\tilde G=G\backslash \{v_{n+1}\}[/imath]. By hypothesis, [imath]\chi(\tilde G)+\chi\big(\tilde G^c\backslash \{v_{n+1}\}\big)\leq n+1.[/imath] Suppose that [imath]\deg_G v_{n+1}=m\leq n[/imath], then [imath]\chi(G)= \chi(\tilde G)+m+1.[/imath] Moreover, [imath]\deg_{G^c}v_{n+1}=n-m[/imath] and thus [imath]\chi(G^c)=\chi\big(\tilde G^c\backslash \{v_{n+1}\}\big)+n-m.[/imath] We conclude that [imath]\chi(G)+\chi(G^c)\leq \chi(\tilde G)+\chi\big(\tilde G^c\backslash \{v_{n+1}\}\big)+n+n-m\leq n+1+2n-m=3n+1-n[/imath] and thus it doesn't work. Question : How can I prove he claim ? (any proof is welcome, but I need an induction proof). | 670770 | [imath]\chi(G)+\chi(G')\leq n+1[/imath]
How to prove, that the sum of chromatic numbers of graph and it's complement is smaller then the number of vertices incremented by one? [imath]\chi(G)+\chi(G')\leq n+1[/imath] The notes from my classes say to color the first graph using sequential algorithm with vertices ordered by their degrees (the ones with lowest appear first). We would get [imath]k_1[/imath] colors, then we color the graph [imath]G'[/imath] in the backwards order and get [imath]k_2[/imath] colors, and the sum [imath]k_1+k_2[/imath] in my notes is magically smaller than [imath]n+1[/imath], but I just don't see why... |
1316829 | [imath]\mathrm{O}_n(\mathbb{Q})[/imath] is dense in [imath]\mathrm{O}_n(\mathbb{R})[/imath]
Prove that [imath]\mathrm{O}_n(\mathbb{Q})[/imath] is a dense subset of [imath]\mathrm{O}_n(\mathbb{R})[/imath]. Recall that [imath]\mathrm{O}_n(\Bbbk)[/imath] is the set of [imath]n \times n[/imath] matrices with coefficients in the field [imath]\Bbbk[/imath] and such that the columns form an orthonormal system in [imath]\Bbbk^n[/imath] endowed with the usual euclidian structure. I managed to prove that [imath]\mathrm{GL}_n(\mathbb{Q})[/imath] is dense in [imath]\mathrm{GL}_n(\mathbb{R})[/imath], but I couldn't go further. Any ideas ? | 941943 | Density of orthogonal matrices with rational coefficients
Is it true that [imath]{O}_n({\mathbb R}) \cap {\mathbb Q}^n[/imath] is dense in [imath]{\cal O}_n({\mathbb R})[/imath] for any [imath]n\geq 2[/imath] ? It obviously suffices to consider the density of [imath]{SO}_n({\mathbb R}) \cap {\mathbb Q}^n[/imath] in [imath]SO_n({\mathbb R})[/imath]. This is easy for [imath]n=2[/imath] : consider the matrices of the form [imath] \frac{1}{p^2+q^2}\left(\begin{array}{cc} p & -q \\ q & p \end{array}\right) [/imath] where [imath]p,q[/imath] are integers. |
1316832 | How to determine the number removed from the list
One number is removed from a set of integers from 1 to n,the average of the remaining numbers is [imath]\large{\frac{163}{4}}[/imath]. Which number was removed? I tried to find the mean of [imath]\frac{1+2+....+n-1}{n-1}=\frac{163}{4}[/imath] but I can`t find the value from there. | 165013 | Which number was removed from the first [imath]n[/imath] naturals?
A number is removed from the set of integers from [imath]1[/imath] to [imath]n[/imath]. Now, the average of remaining numbers turns out to be [imath]40.75[/imath]. Which integer was removed? By some brute force, I got [imath]61[/imath]. I want to know if there's any analytic approach? |
994752 | For any real matrix [imath]M[/imath] let, [imath]\lambda^{+}(M)[/imath] be the number of positive eigenvalues of [imath]M[/imath] counting multiplicities.
For any real matrix [imath]M[/imath], let [imath]\lambda^{+}(M)[/imath] be the number of positive eigenvalues of [imath]M[/imath] counting multiplicities. Let [imath]A[/imath] be an [imath]n\times n[/imath] real symmetric matrix and [imath]Q[/imath] be an [imath]n\times n[/imath] real invertible matrix . Then which of the followings are correct? (1) [imath]rank(A)=rank(Q^{T}AQ)[/imath] (2) [imath]rank(A)=rank(Q^{-1}AQ)[/imath] (3) [imath]\lambda^{+}(A)=\lambda^{+}(Q^{T}AQ)[/imath] (4) [imath]\lambda^{+}(A)=\lambda^{+}(Q^{-1}AQ)[/imath] | 619372 | For any real square matrix which are correct?
For any real square matrix [imath]M[/imath] let [imath]\lambda^+(M)[/imath] be the number of positive eigenvalues of [imath]M[/imath] counting multiplicity. Let [imath]A[/imath] be an [imath]n\times n[/imath] real symmetric matrix and [imath]Q[/imath] be an [imath]n\times n[/imath] real invertible matrix. Then which are correct? Rank [imath](A)[/imath] = Rank [imath]Q^TAQ[/imath] Rank [imath](A)[/imath] = Rank [imath]Q^{-1}AQ[/imath] [imath]\lambda^+(A)[/imath] = [imath]\lambda^+(Q^TAQ)[/imath] [imath]\lambda^+(A)[/imath] = [imath]\lambda^+(Q^{-1}AQ)[/imath] I can see 2 is correct but what about the rest? |
1313484 | prove that if [imath]a^n+1[/imath] is prime then [imath]a[/imath] is even and [imath]n=2^k[/imath]
i managed to show that [imath]a[/imath] is a even (suppose it is, and then show that [imath]2|a^n+1[/imath]) for the second part, I understand it has something to do with fermat numbers but couldn't solve. please help | 491830 | Fermat primes relation to [imath]2^n+1[/imath]
If [imath]2^n+1[/imath] is prime how to show that [imath]n[/imath] must be a power of [imath]2[/imath] Being at elementary level I am at a loss what to begin with? |
1318436 | using Fibonacci numbers prove that if [imath]d|n[/imath] then [imath]F_d|F_n[/imath]
The first question was to prove that [imath]\gcd(F_{n+1},F_n) = 1[/imath] So i tried to use it but with no success. any help or clue will appreciated thanks | 944233 | Question regarding the Fibonacci sequence
Given the Fibonacci sequence [imath](F_1, F_2,F_3, ...)[/imath] how do I prove that if [imath]m|n[/imath] then [imath]F_m|F_n[/imath]? Can this be proven with mathematical induction? |
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