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2529933 | Proofs involving a [imath]d[/imath]-regular bipartite graph
Suppose that [imath]G[/imath] is a [imath]d[/imath]-regular bipartite graph with bipartition [imath](A, B)[/imath], and that [imath]d ≥ 1[/imath]. a) Prove that [imath]∣A∣ = ∣B∣[/imath]. b) Prove that [imath]G[/imath] contains a perfect matching. So I understand that a [imath]d[/imath]-regular graph will have every vertex of degree [imath]d[/imath]. But I am not sure how that would apply here. Will I have to insert [imath]d[/imath] into my use of Hall's condition and then move from there. And also, how exactly would I prove that [imath]|A| = |B|[/imath]? Would that also involve Hall's condition? Thank you! | 431104 | d-regular bipartite graph
I'm looking to prove that any [imath]d[/imath]-regular bipartite graph satisfies the HALL condition and contains a perfect matching. I'm having trouble seeing why, if [imath]G=(V,W,E)[/imath], for any [imath]H \subseteq V[/imath], [imath]|H|\leqslant |N(H)|[/imath]. Would appreciate an answer! Edit: For degree [imath]\geqslant 1[/imath] obviously. |
2530328 | Let [imath]p[/imath] be a prime and let [imath]F[/imath] be a field with [imath]p^k[/imath] elements and [imath]K[/imath] a subfield. Show that [imath]Gal(F : K)[/imath] is a cyclic group of order [imath][F : K][/imath].
My attempt: Consider the map [imath]σ : F→ F[/imath] such that [imath]∀a ∈ F, σ(a) = a^p[/imath]. Then [imath]σ[/imath] is a field automorphism of [imath]F[/imath]. [imath]ker(σ) = 0[/imath] and since [imath]F[/imath] is a finite field, [imath]σ ∈ Gal(F : K)[/imath], the galois group of order [imath]k[/imath]. Since the order of the galois group is [imath]k[/imath] it is generated by [imath]σ[/imath] so we can deduce that [imath]Gal(F : K)[/imath] is a cyclic group [imath]Z_n[/imath]. By the Fundamental Theorem of Galois Theory, [imath]|G| = |Gal(F : K)| = [F : K][/imath], for the field extension [imath]F : K[/imath] of finite degree. [imath]Gal(F : K)[/imath] is a cyclic group of order [imath][F : K][/imath]. | 1582076 | Compute Galois group of [imath]\mathbb F_q/\mathbb F_p[/imath]
I have to compute Galois group of [imath]\mathbb F_q/\mathbb F_p[/imath] where [imath]q=p^n[/imath]. I know already that [imath]\mathbb F_q/\mathbb F_p[/imath] is galois, so I don't need to prove it. Moreover, I know that \begin{align*}\sigma :\mathbb F_q&\longrightarrow \mathbb F_q\\ x&\longmapsto x^p\end{align*} is an automorphism that fix elements of [imath]\mathbb F_p[/imath] and thus [imath]\sigma \in Gal(\mathbb F_q/\mathbb F_p)[/imath]. Now, I know that if [imath]k<p^n[/imath] then [imath]\sigma ^k(x)=(x^{p})^k=x^{pk}\neq 1[/imath] and thus [imath]\sigma [/imath] is of order [imath]p^n[/imath]. Now if I can prove that [imath]|Gal(\mathbb F_q/\mathbb F_p)|=p^n[/imath], then [imath]Gal(\mathbb F_q/\mathbb F_p)=\left<\sigma \right>[/imath] and it would be finish, but I can't prove it since even [imath]X^q-X[/imath] split over [imath]\mathbb F_q[/imath], theis polynomial is not irreducible, and thus, I can't conclude. So how can I find an irreducible polynomial of degree [imath]q[/imath] that split over [imath]\mathbb F_q[/imath] ? |
2528959 | Show convergence of measurable functions is not induced by convergence of topology
Let X be the set of all measurable functions from [imath][0,1] \rightarrow \mathbb{R}[/imath]. On this set we can define pointwise convergence up to a [imath]\mu[/imath] null set. Now I want to show that there is no topology, where the definition of convergence in said topology is equivalent to the pointwise convergence up to a [imath]\mu[/imath] null set. I want to do this by constructing a sequence of measurable functions, where every subsequence converges to the same function, but the original sequence does not converge to the same function or not at all ( all up to a [imath]\mu[/imath] null set of course ). If I can construct such an example, this would prove that no such topology can exist, because topological convergence has to satisfy this property. But I'm having trouble coming up with an example and am looking for hints on how to construct such a sequence of functions! | 9063 | Characterization of Almost-Everywhere convergence
Given a [imath]\sigma[/imath]-finite measure [imath]\mu[/imath] on a set [imath]X[/imath] is it possible to formulate a topology on the space of functions [imath]f:X \rightarrow \mathbb{R}[/imath] that gives convergence [imath]\mu[/imath]-almost everywhere? I can't seem to find any way to write this and am suspecting that no such topology exists! Is this true? If so, is there some generalisation of a topological space where one can make sense of convergence without having open sets? Any comments, references or tips would be greatly appreciated. |
2529956 | Prove the maximum of convex functions is also convex using this def.
Could someone assist me in proving that the maximum of a convex function is also convex using the below definition? I cannot figure out how this proof would work and would appreciate some help! [imath] f(\lambda x +(1-\lambda)y) \le \lambda f(x)+(1-\lambda)f(y) [/imath] [imath] \lambda \in[1,0] [/imath] Thank you! | 147475 | Proving that the maximum of two convex functions is also convex
Here's a homework question I'm struggling with: Let [imath]f,g[/imath] two convex functions. Prove that [imath]h(x)=\max\{f(x),g(x)\}[/imath] is also convex I don't know where to begin. The only thing I had in mind was was to try proving that if a function is convex on two sets [imath]A[/imath] and [imath]B[/imath], it is also convex on their union. That does not seem right though, for example if I glue together [imath]f(x)=x^2, g(x)=\frac{x^2}{1000}[/imath] where [imath]f[/imath] is defined on [imath][0,1][/imath] and [imath]g[/imath] on [imath](1,2][/imath]. Anyway, that was the only thing I thought about. Any better ideas? thanks! |
2530814 | Is it correct to say the interval of real numbers from 0 to 1 is more dense than that from 0 to n?
If the cardinality of the set of real numbers in the interval [imath][0, 1][/imath] is [imath]c[/imath], and the cardinality of another set of real numbers, say those in the interval [imath][0, n][/imath] where [imath]n>1[/imath] is also [imath]c[/imath], then is it correct to say that the first interval must be more dense than the second? | 1331680 | Are there fewer reals on [imath](0, 1)[/imath] than on [imath](1,\infty)[/imath]?
I know that the cardinality of the sets of real numbers [imath](0, 1)[/imath] and [imath](1, \infty)[/imath] are equal. So what is the fallacy in this argument? For every real on [imath](0, 1)[/imath], we can add any integer [imath]n[/imath] to it and get a number on [imath](1, \infty)[/imath], with the number from the first set as the fractional part of our new number. However, this can be applied to every real on [imath](0, 1)[/imath], with every integer greater than one... seeming to suggest a larger cardinality of the second set than the first. |
2530347 | probability of having all types in a 5 cards hand
I was asked for the probability of having at least one card from each type (hearts/spades/etc.) in a 5 cards hand chosen randomly from a 52 cards normal deck. my solution was [imath]\frac{{{13}\choose{1}}^4{{48}\choose{1}}}{{52}\choose{5}}[/imath] as in, choose one card from each type and then one from the 48 cards left, then divide it all by all the possible hands. my solution is wrong, and I don't understand why, can someone please point out where have I gone wrong? | 223073 | What is the probability one card from each suit will be represented when 5 cards are dealt?
If each suit is represented then we will have two cards from one suit and one card each from the remaining suits. So I am counting the ways this can happen like so - [imath]{52 \choose 1}[/imath] ways of selecting the first card. [imath]{12 \choose 1}[/imath] ways of selecting a card from the same suit as the first. [imath]{13 \choose 1}[/imath] ways of selecting a card for each of the 3 remaining suits. So I have - [imath]{52 \choose 1}{12 \choose 1}{13 \choose 1}{13 \choose 1}{13 \choose 1}[/imath] Then I have the sample space as [imath]52 P 5[/imath] as we are counting with regard to order on the numerator so we need to do the same for the denominator. So I have [imath]\frac{{52 \choose 1}{12 \choose 1}{13 \choose 1}{13 \choose 1}{13 \choose 1}}{52P5}[/imath] Does that look correct? |
2530702 | The real plane, R^2 , is a vector space over R. Describe, in geometrical terms, all subspaces of R^2
The real plane, [imath]R^2[/imath] , is a vector space over [imath]R[/imath]. Describe, in geometrical terms, all subspaces of [imath]R^2[/imath]. Can anyone help me with exercise? How can I do that in geometrical terms? | 289128 | Geometric interpretation of a vector space and subspace?
I understand how to manipulate vector spaces and subspaces and how to prove various statements about them, but I still don't fully understand what they represent geometrically. I just need an intuitive grasp as to what these are. Is a vector space a geometric area that contains all possible vectors of the field [imath]\Bbb F[/imath]? For example, if [imath]V[/imath] is a vector space over [imath]\Bbb R^3[/imath] then does that mean [imath]V[/imath] contains all vectors in three-dimensions that are part of [imath]\Bbb R[/imath]? But then what is a subspace of [imath]V[/imath]? Would that perhaps be a plane? Would it be a vector space in [imath]\Bbb R^2[/imath]? |
2531253 | How can one prove that [imath]\lim_{n\to\infty}(1 +\frac k n)^n = e^k[/imath]?
The equality in question is widely used to solve for limits like [imath]\lim_{n\to\infty} \frac{n+1}{n-2}[/imath], but how do you actually prove that [imath]\lim_{n\to\infty} \left(1+\frac{k}{n}\right)^n = e^k.[/imath] Is it possible to prove it using the other formula: [imath]\lim(1+\frac 1 n)^n = e?[/imath] EDIT: I think I already got it: [imath](1+ \frac{k}{n})^n = (1+ \frac{1}{\frac{n}{k}})^{k \cdot \frac{n}{k}} \rightarrow e^k[/imath] Is this correct? | 568268 | Is it trivial to say [imath]\mathop {\lim }\limits_{n \to \infty } {(1 + {k \over n})^n} = e^{k}[/imath]
Is it trivial to say [imath]\mathop {\lim }\limits_{n \to \infty } {(1 + {k \over n})^n} = e^{k},[/imath] considering the fact that we know [imath]\mathop {\lim }\limits_{n \to \infty } {(1 + {1 \over n})^n} = e?[/imath] |
2530662 | Mean of ratio of minimum and maximum of two standard normal variables
I'm working on the following problem. Let [imath]X,Y[/imath] independent, standard normally distributed random variables and define [imath]U=\max\{|X|,|Y|\}[/imath] and [imath]V=\min\{|X|,|Y|\}[/imath]. Find the mean of [imath]V/U[/imath]. What approach would work here? I don't really see where to start. Using integrals doesn't seem to work. Any hint or (partial) solution is much appreciated! | 1030946 | What is the expected value of [imath]\min\{|X|,|Y|\}/\max\{|X|,|Y|\}[/imath] assuming [imath]X[/imath] and [imath]Y[/imath] are independent?
So I need to compute [imath]E\left[\frac{\min\{|X|,|Y|\}}{\max\{|X|,|Y|\}}\right][/imath] given [imath]X,Y \sim[/imath] Normal[imath](0,1)[/imath] and independent. What I am having trouble seeing is whether [imath]\min\{|X|,|Y|\}[/imath] and [imath]\max\{|X|,|Y|\}[/imath] are independent. If so I can factor the expectation into [imath]E\left[\min\{|X|,|Y|\}\right]\cdot E\left[\frac{1}{\max\{|X|,|Y|\}}\right][/imath] how I should find the expected value of [imath]1/\max\{|X|,|Y|\}[/imath]. |
2531380 | Limit of a trig function. (Without using L'Hopital)
I'm having trouble figuring out what to do here, I'm supposed to find this limit: [imath]\lim_{x\rightarrow0} \frac{x\cos(x)-\sin(x)}{x^3}[/imath] But I don't know where to start, any hint would be appreciated, thanks! | 2189466 | Limit without l'Hopital or Taylor series: [imath]\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}[/imath]
find the limit without l'Hôpital and Taylor rule : [imath]\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}=?[/imath] My Try : [imath]\lim\limits_{x \to 0} \frac{x\cos x- \sin x}{x^3}\\=\lim\limits_{x \to 0}\frac{x\cos x \sin x- \sin x\sin x}{x^3\sin x}=\\\lim\limits_{x \to 0}\frac{x\sin 2x- \sin^2 x}{2x^3\sin x}=[/imath]? what now ? |
2529772 | [imath]l[/imath]-adic expansion of 1/2
I try to understand how to find a [imath]l[/imath]-adic expansion of rational numbers. So I tried with [imath]\frac{1}{2}[/imath]. Unfortunately I failed. What I know\think: We have two cases. What if [imath]l[/imath] is odd? What if [imath]l[/imath] is even? I have to find:[imath]\frac{1}{2} = \sum_{i=\infty}^{\infty} z_i\cdot l^{-i},[/imath] [imath]z_i\in\{ 1,...,l-1\}[/imath] and [imath]l \ge 2[/imath]. | 576838 | P-adic expansion construction
Can anyone teach me about p-adic expansion? especially the case where we have to expand a square root. I need to know how to construct them. for example: the 7-adic expansion of [imath]\sqrt{305}[/imath]. This is a general question and not an assignment or anything, so i cannot post the progress. That is just a random example, please use any other example to explain if it's easier to understand. edit: I'm stuck with the cases where i have to expand the square roots, expanding rational numbers is fine though |
2531016 | Convergence of [imath](1-\frac{1}{kn})^n[/imath].
I want to examine the convergence of [imath](1-\frac{1}{kn})^n[/imath] for [imath]k\in\mathbb{N}[/imath]. My intuition is that for [imath]k=1[/imath], the sequence converges towards [imath]1/e[/imath] and for all other value for [imath]k[/imath], it should converge towards [imath]0[/imath]. I'm just unsure about how to prove this. | 1186210 | How to show [imath]\lim_{k\rightarrow \infty} \left(1 + \frac{z}{k}\right)^k=e^z[/imath]
I need to show the following: [imath]\lim_{k\rightarrow \infty} \left(1 + \frac{z}{k}\right)^k=e^z[/imath] For all complex numbers z. I don't know how to start this. Should I use l'Hopitals rule somehow? |
2532181 | How to prove [imath]\lim\limits_{n\to\infty}\int_{0}^{\frac{\pi}{2}}\sin{(x^n)}dx=0[/imath]?
[imath]\lim\limits_{n\to\infty}\int_{0}^{\frac{\pi}{2}}\sin{(x^n)}dx=0[/imath] How to prove it? It seems that the integral cannot be evaluated. | 656184 | Proving that [imath]\lim_{n \rightarrow\infty} \int_{0}^{\frac{\pi}{2}} \sin(t^n) dt =0[/imath]
It is clear that [imath]\lim_{n \rightarrow\infty} \int_{0}^{1} \sin(t^n) dt =0[/imath]. (which is not what is to be proved here) I don't know how to proceed with the remaining part of the integral ie [imath]\lim_{n \rightarrow\infty} \int_{1}^{\frac{\pi}{2}} \sin(t^n) dt[/imath] (I tried substitution and partial integration). Can anybody give me a hint ? |
2532200 | Help needed for the given question
Given that m,n are natural numbers, when can [imath]3^m + 3^n +1[/imath] be a perfect square? This question was asked in today's exam. I am stuck. Can anyone help? | 419935 | Show [imath]3^m + 3^n +1[/imath] cannot be a perfect square for [imath]m,n[/imath] being positive integers.
So, I decided to work with mod [imath]8[/imath] to help develop some intuition on how to generalize the proof. I noticed that taking [imath]a^2[/imath] to clearly be a perfect square, [imath]a^2[/imath] is always congruent to [imath]0,1,-4,4 \mod 8[/imath], at least for a good finite set of cases. Also, I played around with solutions to [imath]3^m + 3^n + 1 = b[/imath] and seen again for a finite set of cases with [imath]m,n[/imath] varying these were always, [imath]b[/imath] is congruent to [imath]-1,5,-5,3[/imath]. Now, I see that well only perfect squares produce a set [imath]\{0,1,-4,4\} \mod 8[/imath] and none of the cases I tried to [imath]m,n[/imath] gave me any of the numbers in this set. And now I'm stuck. |
2532249 | Can anyone help me in finding this integral? Without using differentiation under integral sign.
Solve the integral [imath]I[/imath] = [imath]\int_0^{\infty} \frac{\sin x}{x} dx[/imath] | 2506224 | Confirm that [imath]\int_{0}^{\infty}t^{-1}\sin t dt=\pi/2[/imath]
Confirm that [imath]\int_{0}^{\infty}t^{-1}\sin t dt=\pi/2[/imath]. The guide book I am using gives the following help: Consider [imath]\int_{\gamma}z^{-1}e^{iz}dz[/imath], where for [imath]0<s<r<\infty[/imath] the contour of integration is described by [imath]\gamma=[s,r]+\gamma_r+[-r,-s]-\gamma_s[/imath], with [imath]\gamma_r(t)=re^{it}[/imath] and [imath]\gamma_s(t)=se^{it}[/imath] on [imath][0,\pi][/imath]. Recall Exercise IV[imath].4.20.[/imath] Exercise IV[imath].4.20.[/imath] For [imath]r[/imath] with [imath]0<r<\infty[/imath] let [imath]I(r)=\int_{\gamma_r}z^{-1}e^{iz}dz[/imath] with [imath]\gamma_r(t)=re^{it}[/imath] on [imath][0,\pi][/imath]. Show that [imath]I(r)\rightarrow 0[/imath] as [imath]r\rightarrow \infty[/imath] and also that [imath]I(r)\rightarrow \pi i[/imath] as [imath]r\rightarrow 0[/imath] Using the hint, I know that [imath]\int_{\gamma}z^{-1}e^{iz}dz=0[/imath] for the Cauchy theorem, with which [imath]\int_{[s,r]}z^{-1}e^{iz}dz+\int_{\gamma_r}z^{-1}e^{iz}dz+\int_{[-r,-s]}z^{-1}e^{iz}dz-\int_{\gamma_s}z^{-1}e^{iz}dz=0[/imath], but I do not know what else to do here, could someone help me please? Thank you very much. |
2531209 | representing f as a sum of symmetrical and antisymmetrical bilinear forms
i've tried to solve this following question without any success. can you help me with it please? let [imath]V=M_{2 \times 2}^\mathbb{R}[/imath] and [imath]f: V \times V \to \mathbb{R}[/imath] so that [imath]f(A,B)=tr(A^tMB)[/imath] for every [imath]A,B \in V[/imath]. when n=2, [imath]V=M_{2 \times 2}^{\mathbb{R}}[/imath] and [imath]M=\begin{pmatrix}1&2\\ 3&5\end{pmatrix}[/imath] find a representation of f as a sum of a symmetrical bilinear form and an anti-symmetrical bilinear form. what i tried: since from the details we can infer that in order for f to be symmetric, M must be equal to [imath]M^t[/imath] ([imath]M=M^t[/imath]). now for a bilinear form to be symmetrical we need that [imath]B(u,v)=B(v,u)[/imath] for every [imath]u,v \in V[/imath], and for a bilinear form to be anti symmetrical it must follow [imath]B(u,v)=-B(v,u)[/imath]. but i don't get to finding a representation of f as a sum of a symmetrical and anti symmetrical bilinear forms using the given values. please help me if you can. thank you very much. (edit: this is not a theoritical question about proving that every biliear map can be written as a sum of biliner symmetric map and a bilinear anti-symmetric map. This question is about implementing the theory using the given details in order to obtain the needed representation. additionally, on that thread the answer is incomplete and not enough for me in order to solve this question. thank you very much). | 2130352 | Prove that every bilinear map can be written as a sum of bilinear symmetric map and a bilinear anti-symmetric map.
Show that every bilinear map [imath]\phi: E \times E \longrightarrow F[/imath] can be written uniquely as a sum of a bilinear symmetric map ([imath]f(u,v) = f(v,u)[/imath], for all [imath]u,v \in E[/imath]) and a bilinear anti-symmetric map ([imath]f(u,v) = -f(v,u)[/imath], for all [imath]u,v \in E[/imath]). My attempt: I defined [imath]L_2(E,F) := \{ \phi: E \times E \longrightarrow F \ ; \ \phi \ is \ bilinear \},[/imath] [imath]W_1 := \{f \in L_2(E,F) \ ; \ f \ is \ a \ bilinear \ symmetric \ map \},[/imath] [imath]W_2 := \{g \in L_2(E,F) \ ; \ g \ is \ a \ bilinear \ anti-symmetric \ map \}[/imath] and I thought to proof that [imath]L_2(E,F) = W_1 \ \oplus \ W_2[/imath]. I got to prove that [imath]W_1 \ \bigcap \ W_2 = \{ 0 \}[/imath], but I dind't get prove that [imath]L_2(E,F) = W_1 + W_2[/imath]. How can I show this? I tried to prove by contradiction unsuccessfully. Thanks in advance! |
2532403 | How to prove that [imath]f[/imath] surjective, if [imath]g[/imath] is injective and [imath]g \circ f[/imath] is surjective?
[imath]X, Y[/imath] and [imath]Z[/imath] are non-empty sets. [imath]f : X \to Y[/imath] and [imath]g: Y \to Z[/imath] are functions. Prove that [imath]f[/imath] is surjective, if [imath]g[/imath] is injective and [imath]g \circ f[/imath] is surjective. I'm aware of the surjective and injective definitions, but how would I prove that [imath]f[/imath] is surjective under the given conditions? | 1725979 | [imath]g\circ f[/imath] is Surjective and [imath]g[/imath] is Injective then Prove that [imath]f[/imath] is surjective
Let [imath]f : A \to B[/imath] and [imath]g : B \to C[/imath] be functions such that [imath]g\circ f: A \to C[/imath] is a surjection and [imath]g[/imath] is an injection , Then prove that [imath]f[/imath] is a surjection. Since [imath]g[/imath] is a function [imath]\forall y \in B[/imath] we have [imath]g(y)=z \tag{1}[/imath] for some [imath]z \in C[/imath] Since [imath]g\circ f[/imath] is a surjection [imath]\forall z \in C \exists x \in A[/imath] such that [imath]g\circ f(x)=z[/imath] that is [imath]g(f(x))=z \tag{2}[/imath] from [imath](1)[/imath] and [imath](2)[/imath] we have [imath]g(y)=g(f(x))[/imath] and since [imath]g[/imath] is an injection we get [imath]f(x)=y[/imath] so [imath]f[/imath] is surjective. Is my proof correct? |
2532287 | is multinomial expansion required to solve this infinite sum?
I cannot solve this infinite series by any means using polynomial ascending power expansions... [imath]1+\sum_{i=1}^\infty(1/2)^i \prod_{k=1}^i\frac{3k-1}{3k}[/imath] Could somebody help me? | 2342250 | Evaluate : [imath]1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot 5\cdot 8}{6\cdot12\cdot 18}+\ldots[/imath]
Evaluate : [imath]1+\frac{2}{6}+\frac{2\cdot 5}{6\cdot 12}+\frac{2\cdot 5\cdot 8}{6\cdot12\cdot 18}+\ldots[/imath] thought: I have tried to manipulate the series so that the sum upto n terms , can be simplified. But I can't figure out any profitable result. |
1459582 | The dual vector space is always complete.
If [imath]N[/imath] is a normed a linear space, then its dual vector space [imath]N^*[/imath] is always complete. Attempt: Let [imath]\{f_n\}[/imath] be a Cauchy sequence in [imath]N^*[/imath]. Then, for some [imath]\varepsilon > 0[/imath], there exists [imath]m,n \in \mathbb{N}[/imath] such that [imath]\|f_n - f_m \| < \varepsilon[/imath]. The way to show that the limit lies in [imath]N^*[/imath] would be to show that [imath]f_n(x)[/imath] converges in [imath]K = \mathbb{C}[/imath] or [imath]\mathbb{R}[/imath]. How do I show that the limit of linear functionals is still a linear functional? | 1026961 | Proof of "Dual normed vector space is complete"
http://en.wikipedia.org/wiki/Dual_norm As in the introduction of dual norm by Wiki, it says dual normed space [imath]X'[/imath] is always complete. How to prove that? or at least explain that? We all know the normed vector space is not always complete; if complete, all Cauchy sequences convergent. However, dual normed vector space is complete? |
2532907 | calculate :[imath]\sum _{n=1}^{oo}\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}[/imath]
How can I calculate [imath]\sum_{n=1}^{+\infty}\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}[/imath] ? Can someone help and explain too? | 454896 | Compute [imath]\sum_{n=1}^\infty \frac{1}{(n+1)\sqrt[]{n}+n\sqrt[]{n+1}}[/imath]
Compute [imath]\sum_{n=1}^\infty \frac{1}{(n+1)\sqrt[]{n}+n\sqrt[]{n+1}}[/imath] |
2532733 | Suppose that [imath]A[/imath] is a normal matrix. Prove that one can write: [imath]A = G^4[/imath] for some self-adjoint matrix [imath]G[/imath]
Let [imath]A \in M_n(\mathbb{C})[/imath] be a matrix of order [imath]n \geq 2[/imath]. Let [imath] p_A(x) = (x - \lambda_1)\cdots(x - \lambda_n) [/imath] be the characteristic polynomial of [imath]A[/imath] such that all [imath]\lambda_i[/imath] are positive real numbers. (a) Suppose that [imath]A[/imath] is a normal matrix. Prove that one can write: (ai) [imath]A = G^4[/imath] for some self-adjoint matrix [imath]G[/imath] Well i can at most prove until the point where there exists an unitary matrix [imath]U[/imath] such that [imath]U^{*}AU = D = G^4[/imath] since [imath]D[/imath]'s diagonal contain all its eigenvalues and since they are all real and positive we can split them into [imath]\lambda^{1/4}[/imath] and hence we let [imath]G[/imath] equals to the self adjoint diagonal matrix with diagonals to be [imath]\lambda_i^{1/4}[/imath]. HOWEVER, i cannot make it until that [imath]A = G^4[/imath] for some self adjoint [imath]G[/imath] which is very annoying. I mean [imath]A[/imath] and [imath]D[/imath] are similar so im sure we can do it, but i just cant seem to find a way! Pls help! | 2232624 | Prove normal matrix is 4th power of some self-adjoint matrix
Let [imath]A\in\mathbb{M}_n(\mathbb{C})[/imath] be a matrix of order [imath]n\geq 2[/imath]. Let [imath]p_A(x) = (x-\lambda_1)\cdots (x-\lambda_n)[/imath] be the characteristic polynomial of [imath]A[/imath] such that all [imath]\lambda_i[/imath] are positive real numbers. Suppose that [imath]A[/imath] is a normal matrix. Prove that one can write [imath]A=G^4[/imath] for some self-adjoint matrix [imath]G[/imath]. My thinking is that since [imath]A[/imath] is normal, by Principal Axis Theorem, we have [imath]A=UDU^\ast[/imath], where [imath]U[/imath] is a unitary matrix and [imath]D[/imath] a diagonal matrix. However, I fail to see how to proceed. Could anyone help me out? |
2533171 | Equivalent definition of normal subgroups
A subgroup [imath]N<G[/imath] is called normal if [imath]gN=Ng ,\forall g \in G[/imath]. My book then says this is equivalent to [imath]gN\subset Ng ,\forall g \in G[/imath]. How do I see this? One way is obvious, but how do I show that if [imath]gN\subset Ng ,\forall g \in G[/imath], then [imath]gN=Ng, \forall g\in G[/imath]? I have tried to say, let [imath]g \in G, n \in N[/imath]. Then [imath]gn= n' g[/imath] for some [imath]n' \in N[/imath] From there I have to somehow show [imath]ng= g n''[/imath] for some [imath]n'' \in N[/imath]. Could someone help? Thanks. | 72792 | equivalent definition of normal subgroup
Let [imath]G[/imath] be a group and [imath]N[/imath] a subgroup of [imath]G[/imath]. I read that the following two definitions are equivalent: 1) [imath]\forall g\in G[/imath], [imath]gNg^{-1}=N[/imath] 2)[imath]\forall g\in G[/imath], [imath]gNg^{-1}\subset N[/imath] does this mean that we have always [imath]N\subset gNg^{-1}[/imath] ? my guess is no because take [imath]n\in N[/imath] then if [imath]n=gn'g^{-1}[/imath] then [imath]n'=g^{-1}ng[/imath] and there is no reason that [imath]g^{-1}ng\in N[/imath] |
1717863 | Complex analysis: Calculating [imath]\int_{-\infty}^{\infty} \frac{\sin x}{x} dx[/imath] by using [imath]f(z) = \frac{e^{iz} - 1}{iz}[/imath]
So I integrate the holomorphic function [imath]\frac{e^{iz} -1}{iz}[/imath] over the half disk in the upper half plane, let me name it [imath]\Gamma[/imath]. By using Cauchy's theorem, it equals 0. [imath]0 = \int_{\Gamma} \frac{e^{iz} -1}{iz} = \int_{0}^{\pi}(e^{iRe^{i\theta}}-1)d\theta + \int_{-R}^{R} \frac{e^{it}-1}{it}dt[/imath] ([imath]Re^{[0, \pi]}[/imath] and [imath]t: [-R, R][/imath] being the respective parametrizations.) By splitting the left integral in two, I get [imath]\int_{0}^{\pi} e^{iRe^{i\theta}}d\theta \rightarrow 0[/imath] by dominated convergence, and [imath]\int_{0}^{\pi} -1d\theta = -\pi[/imath]. By making [imath]R \rightarrow \infty[/imath], I'm left with [imath]\pi = \int_{-\infty}^{\infty} \frac{e^{it} - 1}{it}dt = \int_{-\infty}^{\infty}\frac{\sin t}{t}dt + \int_{-\infty}^{\infty}\frac{\cos t -1}{it}dt.[/imath] How do I get rid of the integral on the right? I welcome all feedback. | 980970 | How to calculate [imath]\int\limits_{-\infty}^\infty\frac{e^{ix}}{x}dx[/imath]
I need to calculate [imath]I_0=\int_{-\infty}^\infty\frac{e^{ix}}{x}dx=\lim_{R\rightarrow\infty}\int_{-R}^R\frac{e^{ix}}{x}dx=\lim_{R\rightarrow 0}(\lim_{\varepsilon\rightarrow 0}(\int_{-R}^\varepsilon\frac{e^{ix}}{x}dx+\int_{-\varepsilon}^R\frac{e^{ix}}{x}dx))=\lim_{R\rightarrow 0}(\lim_{\varepsilon\rightarrow 0}I(R,\varepsilon))[/imath] I have tried to close the contour of the complex integral on the complex plain. Let [imath]C[/imath] be the semicircle with radius R above the real axis and [imath]\gamma[/imath] be the semicircle under the real axis with radius [imath]\varepsilon[/imath], both centered at [imath]O[/imath]. Let [imath]L[/imath] be the closed contour contained [imath]C,[-R,-\varepsilon],\gamma,[\varepsilon,R][/imath]. Denote the [imath]+[/imath] direction by the [imath]+[/imath] direction of [imath]L[/imath]. Then we have: [imath]\int_{L^+}\frac{e^{iz}}{z}dz=I(R,\varepsilon)+\int_{C^+}\frac{e^{iz}}{z}dz+\int_{\gamma^+}\frac{e^{iz}}{z}dz[/imath] By Cauchy Integral Formula we have [imath]\int_{L^+}\frac{e^{iz}}{z}dz=2\pi i[/imath] so [imath]I(R,\varepsilon)=2\pi i-\int_{C^+}\frac{e^{iz}}{z}dz-\int_{\gamma^+}\frac{e^{iz}}{z}dz[/imath] Take [imath]\varepsilon\rightarrow 0[/imath] we have: [imath]I(R,0)=\pi i-\int_{C^+}\frac{e^{iz}}{z}dz=\pi i-\int_0^{\pi}\frac{e^{iRe^{i\phi}}}{Re^{i\phi}}iRe^{i\phi}d\phi[/imath] [imath]=\pi i-i\int_0^\pi\exp(iR(\cos\phi+i\sin\phi))d\phi=\pi i-i\int_0^\pi e^{iR\cos\phi}e^{-R\sin\phi}d\phi[/imath] Now I must take the limit [imath]R\rightarrow\infty[/imath]. I evaluate: [imath]\left|\int_0^\pi e^{iR\cos\phi}e^{-R\sin\phi}d\phi\right|\le\int_0^\pi e^{-R\sin\phi}d\phi[/imath] Is there anything wrong in my argument so far? How can I proceed further? EDIT1: Looking at it for quite long I think I can write [imath]\int_0^\pi e^{-R\sin\phi}d\phi=2\int_0^{\pi/2} e^{-R\sin\phi}d\phi\le 2\int_0^{\pi/2} e^{-R\frac{\phi}{2}}d\phi=\frac{4}{R}(1-e^{-R\frac{\pi}{4}})\le\frac{4}{R}[/imath] So in the end take [imath]R\rightarrow\infty[/imath] we get [imath]I_0=i\pi[/imath]. Is it true? |
2533169 | A complex ordinary differential equation
I’m trying to see what instruments I could use to analyse the following ODE in the complex plane: [imath] \dot{z} = \exp(it)\cdot \bar{z}[/imath]. Where [imath]z[/imath] is a function of real valued time. [imath]{{}}[/imath] | 2530649 | Complex differential equation
How do I solve [imath]x'=e^{it}\overline{x}?[/imath] This is a complex differential equation, but I don't see how to solve it. Edit: the original ODE is given by [imath](x', y')=\begin{pmatrix}\cos t& \sin t\\ \sin t&-\cos t\end{pmatrix}(x,y)[/imath] I want to show that solutions of this ODE do not remain bounded for all [imath]t[/imath], and the idea was to solve the complex equation. |
2520262 | Show that [imath]a_{n+1}= 1 + \frac{1}{a_n}[/imath] converges.
Let [imath]\{a_n\}[/imath] be defined by [imath]a_1 =1 [/imath] and [imath]a_{n+1} = 1 + {\dfrac{1}{a_n}}[/imath] with [imath]n \in N[/imath]. Show that [imath]a_{n+1}= 1 + {\dfrac{1}{a_n}}[/imath] converges. I know the limit but how can I show that is a Cauchy sequence or that this sequence converges. | 518739 | Prove that the sequence [imath](a_n)[/imath] defined by [imath]a_0 = 1[/imath], [imath]a_{n+1} = 1 + \frac 1{a_n}[/imath] is convergent in [imath]\mathbb{R}[/imath]
I will post the exercise below: Prove that the sequence [imath](a_n)[/imath] defined by [imath]a_0 = 1[/imath], [imath]a_{n+1} = 1 + \frac 1{a_n}[/imath] for [imath]n \in \mathbb N[/imath] is convergent in [imath]\mathbb R[/imath] with the Euclidean metric, and determine afterwards is limit. Can you intepret the limit geometrically (hint: Golden ratio)? So I need to prove that the sequence is convergent in [imath]\mathbb{R}[/imath] with the Euclidean metric, and how do I prove that? The limit must be [imath]1[/imath], but how to interpret it geometrically? |
1611596 | Prove that [imath]\int fg=\lim_{n\to\infty}\int f_ng[/imath]
Let [imath](f_n)[/imath] be a sequence in [imath]L_p[0,1][/imath] ([imath]1<p<\infty[/imath]) such that [imath]f_n\to f\in L_p[0,1][/imath] pointwise almost everywhere. If there exists [imath]M>0[/imath] such that [imath]\|f_n\|_p\le M[/imath] for all [imath]n[/imath], and [imath]g\in L^q[0,1][/imath], where [imath]\dfrac{1}{p}+\dfrac{1}{q}=1[/imath], then: [imath]\int fg=\lim_{n\to\infty}\int f_ng.[/imath] I tried this with the Hölder's inequality: [imath]\left|\int fg-\int f_ng\right|\le \|f_n-f\|_p\|g\|_q.[/imath] I'm not sure we can say directly that [imath]\|f_n-f\|_p\to 0[/imath]. Perhaps with uniformly convergence/Egorov's theorem? By Egorov's theorem, if [imath]\epsilon>0[/imath] there exists [imath]A\subseteq[0,1][/imath] such that [imath]\lambda(A^c)<\epsilon[/imath] and [imath]f_n\to f[/imath] uniformly on [imath]A[/imath]. It is clear then [imath]\int _Afg=\lim_{n\to\infty}\int_A f_ng,[/imath] since uniformly convergence implies [imath]L_p[/imath]-convergence. But I don't know how to do it on [imath]A^c[/imath]. Also, why is [imath]\|f_n\|_p\le M[/imath] required? Thank you. Edited wrong title. Sorry. | 1588034 | Proving that:[imath]\int_X f_n g \, d\mu \to \int_X fg \, d\mu[/imath] for all [imath]g[/imath] in [imath]\mathscr{L}^q (X)[/imath]
I found the following exercise and I'd like to know if my answer is correct. Let [imath](X, \mathscr A, \mu)[/imath] a finite measure space. Let [imath]\{f_n\}[/imath] a sequence of measurable functions such that [imath]\|f_n\|_p\le M[/imath] for a real constant [imath]M[/imath] [imath](1<p<\infty)[/imath] and suppose that [imath]f_n \xrightarrow{\text{a.e.}} f[/imath]. Then [imath]\int_X f_n g \,d\mu \to \int_X fg \,d\mu[/imath] for all [imath]g[/imath] in [imath]\mathscr{L}^q (X)[/imath], where [imath]q^{-1}=1- p^{-1}[/imath]. Since [imath]\int |f_n|^p \,d\mu\le M^p [/imath] by Fatou's lemma it follows that [imath]\int |f|^p \,d\mu\le M^p<\infty[/imath], so [imath]f[/imath] belongs to [imath]\mathscr{L}^p (X)[/imath]. Let [imath]g[/imath] in [imath]\mathscr{L}^q[/imath] arbitrary but fixed. Then by Hölder inequality it follows that [imath]\|f_n g\|_1 \le \|f_n\|_p\|g\|_q\le M \|g\|_q<\infty[/imath] and [imath]\|f g\|_1 \le \|f\|_p\|g\|_q \le M \|g\|_q<\infty[/imath] thus [imath]f_n g[/imath] and [imath]fg[/imath] belongs to [imath]\mathscr L ^1(X)[/imath]. Now since [imath]g[/imath] is in [imath]\mathscr L ^q (X)[/imath], then [imath]g[/imath] is finite a.e.,without loss of generality we may assume that [imath]g[/imath] is real-valued so [imath]f_n g \xrightarrow{ a.e} fg[/imath]. Let [imath]\nu (A) = \int_A |g|^q \,d\mu [/imath], so [imath]\nu [/imath] is absolutely continuous with respect to [imath]\mu[/imath], also [imath]\nu[/imath] is a finite measure on [imath](X, \mathscr A )[/imath]. We can use the [imath]\epsilon-\delta[/imath] definition of absolutely continuous, so given [imath]\epsilon>0[/imath] there is a [imath]\delta>0[/imath] such that for [imath]\mu(A)<\delta[/imath] then [imath]\nu (A) <\epsilon^q[/imath] for any [imath]A[/imath] in [imath]\mathscr A[/imath]. Now by Egoroff's thm exists a [imath]B[/imath] in [imath]\mathscr A[/imath] such that [imath](f_ng) (x)\xrightarrow{\text{uniformly}} (fg) (x)[/imath] for [imath]x[/imath] in [imath]B[/imath] and [imath]\mu(X\setminus B) <\delta[/imath]. Let [imath]N[/imath] such that for all [imath]n\ge N[/imath], [imath]|(f_ng)(x)-(fg)(x)|<\epsilon[/imath] for all [imath]x\in B[/imath]. Thus \begin{align*}\left |\int_X (f_n-f) g \, d\mu \right|&\le \int_B |f_ng-fg| \, d\mu +\int_{X-B}|f_n g|\, d\mu +\int_{X-B}|fg|\,d\mu \\[6pt] &\le \int_B |f_ng-fg| d\mu + 2M \left( \int_{X-B}|g|^qd\mu \right)^{1/q}\\[6pt] &\le\epsilon \mu(X)+2M (\nu(X\setminus B))^{1/q}\\[6pt] &\le \epsilon (\mu(X)+2M)\end{align*} Since [imath]\mu(x)<\infty[/imath] and [imath]M<\infty[/imath] the result follows. |
2533756 | For a bounded linear operator [imath]\phi[/imath], show that [imath]||\phi(x)|| \leq ||\phi|| \cdot ||x||[/imath], where [imath]||\phi|| = sup_{||x|| = 1} ||\phi(x)||[/imath]
In the book of Linear Algebra by Wernet Greub, at page 206, it is given that [imath]\phi[/imath] is bounded iff for any [imath]x\in E[/imath], [imath]\exists M[/imath] s.t [imath]||\phi(x)|| \leq M \cdot ||x||[/imath] Now let [imath]\phi : E \to E[/imath] is a bounded linear transformation.Then we denote [imath]||\phi|| = sup_{||x|| = 1} ||\phi(x)||,[/imath] and it follows that [imath]||\phi(x)|| \leq ||\phi|| \cdot ||x||, \quad x\in E.[/imath] However, I couldn't figure out how did he arrived to that conclusion. I mean for [imath]||x|| = 1[/imath], we have [imath]||\phi(x)|| \leq M \cdot ||x|| = M[/imath], hence by defition of supremum, we should have [imath]sup_{||x|| = 1} ||\phi(x)|| = ||\phi|| \leq M,[/imath] hence for an arbitrary [imath]y \in E[/imath] [imath]||\phi|| \cdot ||y|| \leq M \cdot ||y||,[/imath] so I couldn't derive the same conclusion. | 1944895 | Proof that [imath]\|fx\| \leq \|f\|\cdot\|x\|[/imath]
From the wiki article on the dual of a norm: [imath]X[/imath] and [imath]Y[/imath] are normed spaces, and we associate with each [imath]f\in L(X,Y)[/imath] (the space of bounded linear operators from [imath]X[/imath] to [imath]Y[/imath]) the number [imath]\|f\| = \sup\{|f(x)|:x\in X, \|x\| \leq 1\}.[/imath] At some point in the proof that [imath]L(X,Y)[/imath] is bounded, it seems like they use the inequality [imath]\|fx\| \leq \|f\|\cdot\|x\|,[/imath] where [imath]\|x\|\leq 1[/imath]. I can't see why this should holds. [imath]\|f|\|[/imath] is the supremum of what [imath]|f(x)|[/imath] can be, given that [imath]\|x\|\leq 1[/imath]. So, given that [imath]\|x\| \leq 1[/imath], [imath]|fx|[/imath] should be bounded by [imath]\|f\|[/imath]. But why should it be bounded by [imath]\|f\|\cdot\|x\|[/imath]? |
2532715 | Differential application between spheres and fixed point problem
Prove that every differential application [imath]f:\mathbb{S}^{n}\rightarrow\mathbb{S}^{n}[/imath] such that [imath]deg(f)\neq(-1)^{n+1}[/imath] admits, at least, one fixed point. Here it is worth mentioning that [imath]deg(f)[/imath] denotes the degree of the application [imath]f[/imath] and [imath]\mathbb{S}^{n}[/imath] is the unit sphere in [imath]\mathbb{R}^{n+1}[/imath]. Thank you in advance for any help. | 806476 | Fixed point theorem on spheres
In Milnor's book Topology from the Differentiable Viewpoint there's the following problem: Problem [imath]6[/imath] (Brouwer). Show that any map [imath]S^n\to S^n[/imath] with degree different from [imath](-1)^{n+1}[/imath] must have a fixed point. My solution: Assume that the map [imath]f:S^n\to S^n[/imath] has no fixed points. Let [imath]a:S^n\to S^n[/imath] denote the antipodal map [imath]a(x)=-x[/imath]. Then the map [imath]a\circ f[/imath] is homotopic to the identity as follows: Since [imath]x[/imath] is never mapped to [imath]-x[/imath] (by assumption), there exist a unique shortest great circle arc from [imath]a(f(x))[/imath] to [imath]x[/imath], simply take the straight line homotopy flowing along such arches to get a homotopy [imath]a\circ f\simeq\operatorname{id}[/imath]. Now we have: [imath]1=\deg(\operatorname{id})=\deg(a\circ f)=\deg(a)\deg(f)=(-1)^{n+1}\deg(f)[/imath] and thus [imath]\deg(f)=(-1)^{n+1}[/imath]. My questions: Is this solution correct? How to show that the homotopy is continuous? It seems intuitively true to me, but I'm not sure about how to proceed to show it. Maybe saying that it is the flow of some good tangent field? Are there other (elegant/short/interesting) proofs for this fact? |
2533257 | Suppose [imath]f[/imath] is continuous on R such that [imath]\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{h} = 0[/imath] for all [imath]x\in\mathbb R[/imath]. Prove that [imath]f[/imath] is constant.
I'm having a bit of trouble with the title problem out of Davidson and Donsig's Real Analysis. I'll state it again: Suppose [imath]f[/imath] is continuous on [imath]\mathbb{R}[/imath] such that [imath]\lim_{h\to 0} \frac{f(x+h)-f(x-h)}{h} = 0\ \forall x\in\mathbb{R}.[/imath] Prove that [imath]f[/imath] is constant. They provide the following hint which I have been trying to apply. HINT: Fix [imath]\epsilon > 0[/imath]. For each [imath]x[/imath], find a [imath]\delta > 0[/imath] so that [imath]|f(x+h)-f(x-h)| \leq \epsilon h[/imath] for [imath]0\leq h \leq \delta[/imath]. Let [imath]\Delta[/imath] be the supremum of all such [imath]\delta[/imath]. Show that [imath]\Delta = \infty[/imath]. Here's how I've started. Fix [imath]\epsilon > 0,\ x\in\mathbb{R}[/imath]. By the definition of the limit, [imath](\forall x\in\mathbb{R})(\forall\epsilon>0)(\exists\delta>0)(0<|h|\leq\delta\implies\Bigg|\frac{f(x+h)-f(x-h)}{h}\Bigg|<\epsilon).[/imath] Therefore, we immediately get a [imath]\delta>0[/imath] for our [imath]x,\epsilon[/imath] such that [imath]|f(x+h)-f(x-h)|\leq\epsilon h,\ 0\leq h \leq\delta.[/imath] I don't know how to proceed from here. I'm not even sure conceptually how showing that [imath]\Delta=\infty[/imath] would give us that [imath]f[/imath] is constant. Any help would be appreciated! | 720243 | Are constants the only continuous functions with "symmetric derivative" zero?
Put precisely, if [imath]f:\mathbb R\to\mathbb R[/imath] is continuous with [imath] \lim_{s\to 0^+}s^{-1}(f(t+s)-f(t-s))=0 \kern20mm (*) [/imath] for all [imath]t\in\mathbb R[/imath] , is [imath]f[/imath] then constant? I guess it is but I cannot see the proof. Obviously, if instead of ([imath]*[/imath]) we had [imath]\lim_{s\to 0^+}s^{-1}(f(t+s)-f(t))=0[/imath] , i.e. the right-handed derivative zero, then the assertion could be proved indirectly by a simple compactness argument applied to closed intervals [imath][a,b][/imath] with [imath]f(a)\not=f(b)[/imath] . However, I cannot see how this argument could be modified to get the above result. Is there possibly a reference to this, or is the result even false? |
2532787 | Let [imath]T:V\rightarrow W[/imath] a linear transformation, [imath]i_w:W\rightarrow W^{**}[/imath] and [imath]i_v:V\rightarrow V^{**}[/imath] canonical morphism of biduality.
Let [imath]T:V\rightarrow W[/imath] a linear transformation, [imath]i_w:W\rightarrow W^{**}[/imath] and [imath]i_v:V\rightarrow V^{**}[/imath] canonical morphism of biduality. Prove [imath]i_w\circ T=T^{**}\circ i_v[/imath] I' m very very confused with this exercise. Suppose i need to prove this: [imath]i_w\circ T\subset T^{**}\circ i_v[/imath] and [imath]T^{**}\circ i_v\subset i_w\circ T [/imath] but i don't have idea of how to prove this. Can somone give me a hint? | 2021791 | Double dual mappings
[imath]\newcommand{\Hom}{\operatorname{Hom}}[/imath]Let [imath]U,V[/imath] be vector spaces. Denote the dual space of [imath]U[/imath] with [imath]U^\intercal[/imath] and the dual mapping of a linear map [imath]\Phi[/imath] be [imath]\Phi^\intercal[/imath]. Define the double dual embedding on [imath]U[/imath] to be [imath]\iota_U:U \to U^{\intercal\intercal}[/imath]. ([imath]u \in U, f \in U^\intercal[/imath]) [imath] \iota_U(u)(f) = f(u) [/imath] If [imath]\Phi \in \Hom(V,U)[/imath], then [imath]\Phi^{\intercal\intercal} \circ \iota_V = \iota_U \circ \Phi:V \to U^{\intercal\intercal}[/imath] Proof: Let [imath]v \in V, f \in U^\intercal[/imath]. Then [imath] \iota_U \circ \Phi(v)(f) = f(\Phi(v)) [/imath] but how do I show that [imath] \Phi^{\intercal\intercal} \circ \iota_V(v)(f) = f(\Phi(v)) [/imath] ? |
2533938 | Prove the identity: [imath]\sum_{k=0}^{\frac{n}{2}}(-1)^k\binom{n-k}{k}2^{n-2k}=n+1[/imath]
Prove the identity: [imath]\sum_{k=0}^{\frac{n}{2}}(-1)^k\binom{n-k}{k}2^{n-2k}=n+1[/imath] To prove this by induction is useless. I guess the best one solution would be to prove this by generating functions. Does anybody know how can I prove this equation? | 747102 | Proving combinatorial identity [imath]\sum_{k=0}^{n}(-1)^k \binom{2n-k}k 2^{2n-2k} = 2n+1[/imath] "directly"
This is a homework problem. In the first part of the problem, I managed to use a combinatorial problem to prove the following identity: [imath]\sum_{k=0}^{n}(-1)^k {2n-k \choose k} 2^{2n-2k} = 2n+1[/imath] But I actually have problem with the second part of the problem which asked me to prove this identity "directly", probably using some form of generating functions/algebra method. There was a hint given but I got stuck: The hint says to calculate [imath]\Sigma_{n=0}^\infty\Sigma_{k=0}^{n}(-1)^k {2n-k \choose k} 2^{2n-2k} x^{2n}[/imath] and that it would be useful to consider the fact that [imath] \Sigma(2n+1)x^{2n}= \frac{d}{dx}\frac{x}{1-x^2} [/imath] and [imath](1-x)^{-a-1}=\Sigma_{j=0}^\infty {a+j \choose j}x^j[/imath]. I am able to prove both these "possibly useful facts" in the hint, but I don't see how to calculate the suggested double sum or prove the identity. [I would have thought that the combinatorial proof is harder to find!] [You may assume that I am familiar with the formal treatment of power series.] Any help would be greatly appreciated. |
1837906 | Inequality : [imath]\sum_{k=1}^n x_k\cdot \sum_{k=1}^n \frac{1}{x_k} \geq n^2[/imath]
I have to show the inequality of [imath]\left(\sum_{i=1}^n x_i\right)*\left(\sum_{i=1}^n \frac{1}{x_i}\right) \geq n^2.[/imath]For [imath]x_1, ... x_n \in \mathbb{R_{>0}}[/imath] and [imath] n \geq 1[/imath]. I wanted to show this inequality by induction. The basis is clear, but I am not sure how to do the inductive step. Can someone help me with that? | 2873324 | Show that:[imath]\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}} \geq n^2 [/imath]
Show that:[imath]\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}} \geq n^2 [/imath] The following hints are also given: [imath]\left(\frac{x}{y} + \frac{y}{x} \geq 2 \right) \land x,y \gt 0[/imath] Base Case: For n = 2 [imath]\left(1+2\right) \cdot \left(\frac{1}{1}+\frac{1}{2}\right) \geq 2^2[/imath] Inductive hypothesis: [imath]\sum_{i=1}^{n+1} x_{i} \cdot \sum_{i=1}^{n+1} \frac{1}{x_{i}} \geq \left(n+1\right)^2 = n^2+2n+1[/imath] Inductive step: [imath]\sum_{i=1}^{n+1} x_{i} \cdot \sum_{i=1}^{n+1} \frac{1}{x_{i}} = \left(\sum_{i=1}^n x_{i}+(n+1)\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_{i}}+\frac{1}{n+1}\right) = \left(\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) + \left((n+1) \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) + \left(\sum_{i=1}^n x_{i} \cdot \frac{1}{n+1}\right) + (n+1) \cdot \frac{1}{n+1}[/imath] Final words: I came to the conclusion that: [imath]\left(\sum_{i=1}^n x_{i} \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) = n^2[/imath] I inserted for n = 1 so that [imath]\left((n+1) \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) = \frac{2}{1} \land \left(\sum_{i=1}^n x_{i} \cdot \frac{1}{n+1}\right) = \frac{1}{2}[/imath] Since [imath]\left(\frac{x}{y} + \frac{y}{x} \geq 2 \right) \land x,y[/imath] was given as a hint in the beginning I thougt I can say that [imath]\left((n+1) \cdot \sum_{i=1}^n \frac{1}{x_{i}}\right) + \left(\sum_{i=1}^n x_{i} \cdot \frac{1}{n+1}\right) = 2n[/imath] Furthermore it's obvious that: [imath](n+1) \cdot \frac{1}{n+1} = 1[/imath] It's pretty standard proof by induction and I hope you can maybe give me some advices on what I could have done differently and verify the legitimacy of this proof. |
2534360 | For what value of n this inequality is true and how to find it step by step?
The inequality is [imath]1000n^3<2^n[/imath] I'm trying to see for what value of [imath]n[/imath] an algorithm A who takes ([imath]1000n^3[/imath]) steps takes less time than another algorithm B who takes([imath]2^n[/imath]) steps. Any idea how to do that step by step? | 2534316 | exponential inequality with two variables on each side but the variable is places on the exponent on one side only and the other exponent is defined
I'm new to this website. I've checked the other topics and couldn't find the answer to my question. I've also tried to find the result with TI-CAS but the only thing it does is simplify the inequality which really doesn't help at all. the equation is : [imath]1000n^3 < 2^n[/imath] Now I've been trying to use log but it gets me nowhere (I haven't seen this form before), the answer is supposed to be 24 but I'm not sure. So my question is how to solve step by step this inequality for n. Could you guys help me with this one? |
2533809 | A nilpotent matrix is similar to a strictly upper triangular matrix
Let [imath]n>1[/imath] and [imath]A\in M_n(F).[/imath] Suppose [imath]A[/imath] is nilpotent. Prove that [imath]A[/imath] is similar to a strictly upper triangular matrix by induction on [imath]n[/imath]. How should I do this? I don't know how to use the assumption to relate it to the definition of similarity of matrices that's why I'm stuck. Please help. Thanks. | 1003525 | How to prove that a nilpotent operator has a basis representation that is strictly upper triangular?
I wish to prove the following: if [imath]T:X\to X[/imath] is a nilpotent (and the linear space [imath]X[/imath] is finite-dimensional), then there exists a basis of [imath]X[/imath] such that the matrix representation of [imath]T[/imath] is upper triangular with zero diagonal elements. I was trying to prove [imath]\{x_1,...,x_{k+1},v, Tv, ..., T^{n-k-2}v\}[/imath] is a basis for [imath]X[/imath] where [imath]x_i[/imath] are basis for [imath]N(T)[/imath] and [imath]v,..., T^{n-k-2}v[/imath] are basis for [imath]R(T)[/imath] where [imath]T^{n-k}=0[/imath] and [imath]n-k[/imath] is minimal. However the matrix representation w.r.t this basis is diagonal. Is there an easy way to prove this ? I know that for all [imath]T:X\rightarrow X[/imath], there exists a basis that the matrix representation is upper triangular. How is this useful? |
2535265 | Finding a tricky limit using intergral sums
Find the limit, interpreting this as as the limit of the integral sum of a suitably chosen function. [imath]\lim_{n\to\infty} \frac{1}{n} \sqrt[n]{(n+1)(n+2)... (n+n)}.[/imath] A hint that has been given: find the logarithm of equation and then find a limit of logarithm. | 2525939 | Limit of [imath]\frac1n\sqrt[n]{(n+1)(n+2)(n+3)...(n+n)}[/imath]
In my homework I have to find limit (by interpreting it as an appropriately chosen function as the limit of integral sums): [imath]\lim_\limits{n\to \infty} \frac{1}{n}\sqrt[n]{(n+1)(n+2)(n+3)...(n+n)} (1)[/imath]. [imath] [/imath] Solution: [imath]\lim_\limits{n\to \infty} \frac{1}{n}\sqrt[n]{(n+1)(n+2)(n+3)...(n+n)} = \lim_\limits{n\to \infty} ln(\frac{1}{n}\sqrt[n]{(n+1)(n+2)(n+3)...(n+n)}) = \lim_\limits{n\to \infty} ln(\frac{1}{n}\sqrt[n]{n^n(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n})}) = \lim_\limits{n\to \infty} ln(\frac{1}{n} n\sqrt[n]{(1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n})}) = \lim_\limits{n\to \infty} ln((1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n}))^{1/n} = \lim_\limits{n\to \infty} \frac{1}{n} ln((1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})...(1+\frac{n}{n})) = \lim_\limits{n\to \infty} \frac{1}{n} \sum_{k=1}^{n} ln(1+\frac{k}{n}) = \lim_\limits{n\to \infty} \sum_{k=1}^{n} \frac{1}{n} ln(1+\frac{k}{n})[/imath] [imath] [/imath] Note that function [imath]f(x)=ln(1+x)[/imath] is continuous in [imath][-1,1][/imath]. Find integral [imath]\int ln(1+x) dx = (x+1) ln(1+x)-x+C[/imath]. Then [imath]\int_{-1}^{1} ln(1+x)dx = ln(4)-2[/imath]. [imath] [/imath] Assume that that limit [imath](1)[/imath] is A. We show that [imath]\forall \epsilon>0[/imath] [imath]\exists N\in \mathbb{N}[/imath]: [imath]\forall n\in \mathbb{N}[/imath] [imath]n\geq N \Longrightarrow [/imath] [imath]|\sum_{k=1}^{n} \frac{1}{n} ln(1+\frac{k}{n})-A|<\epsilon[/imath]. Suppose that [imath]\mathfrak{T}[/imath] is [imath][-1,1][/imath] set of subdivisions. Fix. [imath]\epsilon > 0[/imath]. That [imath]f[/imath] is integrative [imath][-1,1][/imath], then exist [imath]\delta >0[/imath]: [imath]\forall[/imath] [imath]T = T[x_0,...,x_n]\in \mathfrak{T}[/imath] [imath]\forall \xi=(\xi_1,...,\xi_n)\in[x_0,x_1]\times...\times[x_{n-1},x_n][/imath] [imath]\lambda(T)<\delta \Longrightarrow |\sum_{k=1}^{n} ln(1+\xi_k) (x_k-x_{k-1})-A|<\epsilon[/imath]. Which [imath]N[/imath] to choose? Is this right solution so far? |
2533698 | If [imath]a,b,c[/imath] are three complex numbers Find possible values of [imath]\lvert a+b+c \rvert[/imath]
Given three complex numbers [imath]a,b,c[/imath] such that [imath]\lvert a \rvert=\lvert b \rvert=\lvert c \rvert=1[/imath] and [imath]\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=-1[/imath] Find which of the following are possible values of [imath]\lvert a+b+c \rvert[/imath] A)0 B)2 C)1.5 D)3 My try: I assumed [imath]a=e^{ix}[/imath],[imath]b=e^{iy}[/imath], [imath]c=e^{iz}[/imath] Then we have [imath]\cos (2x-y-z)+\cos (2y-x-z)+\cos (2z-x-y)=-1[/imath] [imath]\sin(2x-y-z)+\sin (2y-x-z)+\sin (2z-x-y)=0[/imath] Squaring and adding we get [imath]3+2(\cos(3x-3y)+\cos(3y-3z)+\cos (3z-3x)=1[/imath] So [imath]\cos(3x-3y)+\cos(3y-3z)+\cos (3z-3x)=-1[/imath] any clue here? | 1026998 | Find the possible values of |A + B + C |
[imath] |A |= |B | = |C | = 1 [/imath] ,where A B and C are complex nos [imath] \frac{A^2}{BC}+ \ \frac{B^2}{ \ {CA}} \ +\ \frac{C^2}{ \ {AB}} + 1 = 0[/imath] Find the possible values of [imath] |A + B + C |[/imath] Tried substituting cos(theta) + i sin(theta) |
2534940 | For [imath]0 \to N' \to N \to M \to 0 [/imath] with [imath]N[/imath] finitely generated and [imath]M[/imath] finitely presented then [imath]N^{'}[/imath] is also finitely generated.
Let [imath]M[/imath] be a finitely presented [imath]R[/imath] module and [imath]0 \to N' \to N \to M \to 0 [/imath] be an exact sequence with [imath]N[/imath] a finitely generated [imath]R[/imath] module. Prove that [imath]N'[/imath] is also a finitely generated [imath]R[/imath] module. By [imath]M[/imath] finitely presented [imath]R[/imath] module we mean there is a ses [imath]0 \to K \to F \to M \to 0[/imath] where [imath]F[/imath] is free of finite rank and [imath]K[/imath] is finitely generated [imath]R[/imath] module. I proved the case when [imath]N[/imath] is free of finite rank but can't show it in this case. Any help will be appreciated. Thanks. | 1201049 | Is the kernel of a map to a finitely presented module finitely generated?
Let [imath]R[/imath] be a ring (unital, not necessarily commutative), [imath]M[/imath] a finitely presented left [imath]R[/imath] module. Suppose [imath]m_1,\ldots,m_n\in M[/imath] generate [imath]M[/imath]. This determines a surjection [imath]f:R^n\to M[/imath]. Must the kernel of [imath]f[/imath] be finitely generated? This is true by definition if the generating set [imath]m_1,\ldots,m_n[/imath] is the one coming from the finite presentation of [imath]M[/imath]. |
2535390 | Determine if the series 1/5 + 1/8 + 1/11 + 1/14 + 1/17 ... is convergent or divergent
I am trying to determine if the series [imath]\frac{1}{5} + \frac{1}{8} + \frac{1}{11} + \frac{1}{14} + \frac{1}{17} [/imath] is convergent or divergent? I can see that the terms in the denominator differ by 3 each time. | 1882204 | Decide if the series converges and prove it using comparison test: [imath]\sum_{k=1}^{\infty}\frac{1}{2k+3}[/imath]
Decide if the series converges and prove it using comparison test: [imath]\sum_{k=1}^{\infty}\frac{1}{2k+3}[/imath] [imath]\frac{1}{2k+3} <\frac{1}{2k}<\frac{1}{k}[/imath] and since [imath]\sum_{k=1}^{\infty}\frac{1}{k}[/imath] diverges (we have defined that in our readings, so I can claim it here), the complete series will diverge, too. Did I do everything correctly? What makes me feel a bit unsure is that there is no "[imath]\leq[/imath]" sign anywhere and when I look up the comparison test on the internet, I don't see any "<" or ">" in the definition. Is it correct anyway? |
2534227 | An example to show that [imath]\left(I \cap J\right)^{e} = I^{e} \cap J^{e} [/imath] is not true in general.
Let [imath]I[/imath] and [imath]J[/imath] be ideals of a commutative ring with identity [imath]R[/imath] and [imath]I^{e}[/imath] be the extension of the ideal to some other ideal of another ring. I have already proven that [imath]\left(I \cap J\right)^{e} \subset I^{e} \cap J^{e}[/imath] and I am trying to look for a counterexample to show that the equality doesn't hold in general. Any help finding this example would be much appreciated! | 1787780 | Extension of intersection of ideals
Let [imath]f:A \rightarrow B[/imath] be a ring homomorphism and [imath]\mathfrak{a}_1,\mathfrak{a_2}[/imath] be ideals of [imath]A[/imath]. Let [imath]\mathfrak{a}^e[/imath] denote the extension of an ideal [imath]\mathfrak{a}[/imath] of [imath]A[/imath] in [imath]B[/imath]. An exercise shows that [imath](\mathfrak{a}_1 \cap \mathfrak{a}_2)^e \subseteq \mathfrak{a}_1^e \cap \mathfrak{a}_2^e.[/imath] It seems that [imath](\mathfrak{a}_1 \cap \mathfrak{a}_2)^e =\mathfrak{a}_1^e \cap \mathfrak{a}_2^e[/imath] is wrong in general, but I can not find a counter-example. |
758297 | Continuous compactly supported real valued functions on a locally compact and [imath]\sigma[/imath] compact space is separable
I already know that if [imath]X[/imath] is a compact metric space then the space of continuous real valued functions [imath]C(X \to \mathbb R)[/imath] are separable. What I'm trying to prove that if [imath]X[/imath] is a locally compact metric space and is [imath]\sigma[/imath]-compact. The space of continuous compactly supported functions [imath]C_c( X \to \mathbb R)[/imath] are separable. Here's my approach: As [imath]X[/imath] is [imath]\sigma[/imath]-compact, we can write [imath]X = \bigcup_{ n=1}^\infty K_n[/imath], where [imath]K_n[/imath] are compact and increasing. Then since each [imath]C(K_n \to \mathbb R)[/imath] is separable, it suffices to show that the set [imath] \{ f \in C_c (X \to \mathbb R) : \operatorname{supp}(f) \subset K_n \}[/imath] is dense in [imath]C(X \to \mathbb R)[/imath] (Not sure). Then get I stuck. Any help is appreciated. | 757053 | Separability of functions with compact support
Let [imath]X[/imath] be a locally compact metric space which is also [imath]\sigma[/imath]-compact. Let [imath]C_{c}(X)[/imath] be the continuous functions on [imath]f[/imath] from [imath]X[/imath] to [imath]\mathbb{R}[/imath] with compact support. Is [imath]C_{c}(X)[/imath] separable? My work so far: If [imath]X[/imath] is a compact metric space, then by Urysohn's Lemma and Stone-Weierstrass, the continuous functions [imath]C(X)[/imath] on [imath]X[/imath] are separable and hence the result follows as [imath]C_{c}(X) = C(X)[/imath]. Suppose [imath]X = \mathbb{R}[/imath]. Write [imath]\mathbb{R} = \bigcup_{N = 1}^{\infty}[-N, N][/imath]. Let [imath]f \in C_{c}(\mathbb{R})[/imath]. Then [imath]f[/imath] is supported on a compact set [imath]K \subset [-N, N][/imath] for some [imath]N[/imath]. Thus [imath]C_{c}(\mathbb{R}) =\bigcup_{N = 1}^{\infty}C([-N, N])[/imath]. Each [imath]C([-N, N])[/imath] has a countable dense subset [imath]\{\psi_{N, n}\}_{n = 1}^{\infty}[/imath] and so [imath]\bigcup_{N, n = 1}^{\infty}\{\psi_{N, n}\}[/imath] is a countable dense subset of [imath]C_{c}(\mathbb{R})[/imath]. In the general case, [imath]X = \bigcup_{i = 1}^{\infty}X_{i}[/imath] where each [imath]X_{i}[/imath] is compact and [imath]X_{1} \subset X_{2} \subset \cdots[/imath]. Let [imath]f \in C_{c}(X)[/imath]. Then [imath]f[/imath] is supported on a compact set [imath]K = \bigcup_{i = 1}^{\infty}K \cap X_{i}[/imath]. Is it still true that [imath]C_{c}(X) = \bigcup_{i = 1}^{\infty}C(X_{i})[/imath]? |
2534872 | How to prove [imath]\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1[/imath] with out Squeezing?
The same question have been asked here. But almost all the answers given there use the idea of squeezing one way or another, even this geometric proof uses the idea of squeezing. So, here is my question; how to prove [imath] \lim_{x\to 0}\frac{\sin x}{x}=1 [/imath] without the idea of squeezing involved. Edited: No offense, but proofs of limits using integrals or derivative(or any other concepts that are defined using limits) to me is like building a house starting from the roof. So I am not looking for those kind of proofs. | 2048922 | Possible road-maps for proving [imath]\lim_{x\to 0}\frac{\sin x}{x}=1[/imath] in a non-circular way
[imath]\lim_{x\to 0}\frac{\sin x}{x}=1\tag{1}[/imath] Poofs for the limit above have been asked many many times in MSE. Here are a few of them: How to prove that [imath]\lim\limits_{x\to0}\frac{\sin x}x=1[/imath]? Finding functions for the squeeze theorem for [imath]\lim_{x \to 0}{\frac {x}{\sin x}}[/imath] Non-circular proof of [imath]\lim_{\theta \to 0}\frac{\sin\theta}{\theta} = 1[/imath] Here is my question: What are possible "road maps" for developing a proof for (1) in a non-circular rigorous way from a few axioms? One possible "road map" is as follows: ZFC --- natural numbers --- rational numbers --- real numbers --- limits and derivatives --- power series --- power series definition of the sine function --- proof of (1) (In such an approach, one can derive the basic properties of the trigonometric functions from the power series definition without any appeal to the geometric notion of angle.) [Added] My question is essential this: are there other "road maps" than the one given above? If one is going to give an geometric argument about (1) like most introductory calculus textbooks do, then I would like to add one more question: What axioms in geometry would you need in order to develop a satisfactory "rigorous" definition of the sine function? |
2535170 | How to evaluate the integral [imath]\int_0^{+\infty} \frac{\sin^4{x}}{x^4}dx[/imath]?
As we know that [imath]\int_0^{+\infty} \frac{\sin{x}}{x}dx=\pi/2[/imath],but how to evaluate the integral [imath]\int_0^{+\infty} \frac{\sin^4{x}}{x^4}dx[/imath]? | 307510 | A sine integral [imath]\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x[/imath]
The following question comes from Some integral with sine post [imath]\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x[/imath] but now I'd be curious to know how to deal with it by methods of complex analysis. Some suggestions, hints? Thanks!!! Sis. |
2536130 | [imath]\mathbb{C}[x,y]/\langle y^2-g(x) \rangle[/imath] is integrally closed in its fraction field.
Let [imath]g(x)=(x-a)(x-b)(x-c)[/imath] with [imath]a\neq b\neq c[/imath]. Then [imath]\mathbb{C}[x,y]/\langle y^2-g(x) \rangle[/imath] is integrally closed in its fraction field. I manage to do with the following theorem: A smooth function [imath]f[/imath] is non-singular if and only if [imath]\mathbb{C}[x,y]/\langle f(x,y) \rangle[/imath] is integrally closed in its fraction field. But how will I do without using this? Like an example has done here. | 2408031 | Why is [imath]\mathbb{C}[x,y]/(y^2 - x^3 + 1)[/imath] normal?
A problem on an algebra qual reads Show that the ring [imath]R = \mathbb{C}[x,y]/(y^2 - x^3 +1)[/imath] is a Dedekind domain. (Hint: compare [imath]R[/imath] with the subring [imath]\mathbb{C}[x][/imath].) [imath]R[/imath] is clearly Noetherian. It is an integral extension of [imath]\mathbb{C}[x][/imath], so inherits its dimension, which is one. How do I know it is normal? |
2536389 | Finding a curious limit
Find the limit: [imath]\lim_{x\to 0} \frac{1}{x} \int_{0}^{x} (1+ \sin(t))^{1/t} dt.[/imath] A hint that has been given is to use L'Hospital's Rule. Moreover, it must be substantiated why we can use L'Hospital' Rule in this situation. It is also important to mention that [imath](1 + \sin(t))[/imath] has a removable point of discontinuity at [imath]0[/imath] and therefore the integral can be interpreted as a Riemann integral. | 2535213 | Finding a tricky limit
I have a homework in real analysis and I'm very confused about it. I would be very thankful, if you could give me any ideas or solutions how to get this task done. The task is as follows: Find the limit: [imath]\lim_{x\to 0} \frac{1}{x} \int_{0}^{x} (1+ \sin(t))^{1/t} dt.[/imath] A hint that has been given is to use L'Hospital's Rule. Moreover, it must be substantiated why we can use L'Hospital' Rule in this situation. It is also important to mention that [imath](1 + \sin(t))[/imath] has a removable point of discontinuity at [imath]0[/imath] and therefore the integral can be interpreted as a Riemann integral. |
316006 | Creating a sequence that does not have an increasing or a decreasing sequence of length 3 from a set with 5 elements
Today my friend asked me following question: Consider the set [imath] A= \{1,2,3,4,5\}[/imath] By using the elements of this set, can you find a permutation that neither has an increasing sequence of length 3, nor has a decreasing sequence of length 3? for example: [imath]3,4,2,1,5[/imath] has an increasing sequence of length [imath]3[/imath], namely [imath]3,4,5[/imath] [imath]5,3,1,2,4[/imath] has a decreasing sequence of length [imath]3[/imath], namely [imath]5,3,2[/imath] I feel like I cannot find such a sequence. I guess problem is that this set has 5 elements and therefore I cannot create such a sequence. If it had 4 elements, I would find lots of sequences with that property. (With the same number of elements, length 4 would be OK too.) But how can I prove it to him? Could you please explain to me? Best Regards | 2578815 | Application of Pigeonhole Principle - non-constructive
Claim: every sequence of [imath]m^2 + 1[/imath] distinct numbers has an increasing or decreasing subsequence of length [imath]m+1[/imath]. Proof: Idea Associate every position [imath] 1 <= l <= m^2 +1 [/imath] with a pair of lengths of longest increasing (decreasing) subsequences [imath](i_{l}, d_{l})[/imath] beginning at position [imath]l[/imath]. ==> Assume claim is false. ==> There are at most [imath] m^2 [/imath] pairs [imath](i_{l}, d_{l})[/imath] that can occur ==> [imath]m^2+1 - m^2 = 1[/imath] ==> [imath]1[/imath] pair two indices [imath]s < t[/imath] with an equal [imath](i,d)[/imath] pair. ==> Since numbers are distinct. 2 cases If [imath]a_s < a_t[/imath]: Then longer subsequence [imath] i_{t} [/imath] can be built with sequence up to [imath]s + a_s + a_t[/imath] sequence contradicting that pairs are equal. ==> Analogous for [imath]a_t < a_s[/imath] ==> Claim is true My problem is seeing how one goes from Claim is false to There are at most [imath]m^2[/imath] pairs that can occur. |
2536778 | Limit of an integral [imath]\lim_{n\to \infty}\int_{1385}^{2017}f(nx)\,dx [/imath]
Suppose [imath]f(x)[/imath] is continuous function [imath]f:[0,+\infty)\mapsto\mathbb{R}[/imath] and [imath]\lim_{x\to \infty}f(x)=1[/imath] how to find [imath]\lim_{n\to \infty}\int_{1385}^{2017}f(nx)\,dx [/imath] Honestly, I have no idea for a start. can someone help me? thanks in advance. | 485267 | Undergraduate math competition problem: find [imath]\lim \limits_{n \to \infty} \int \limits^{2006}_{1385}f(nx)\, \mathrm dx[/imath]
Suppose [imath]f\colon [0, +\infty) \to \mathbb{R}[/imath] is a continuous function and [imath]\displaystyle \lim \limits_{x \to +\infty} f(x) = 1[/imath]. Find the following limit: [imath]\large\displaystyle \lim \limits_{n \to \infty} \int \limits^{2006}_{1385}f(nx)\, \mathrm dx[/imath] |
2536890 | Matrix inequality proof from Munkres' analysis text
Why is this theorem true? If matrix [imath]A[/imath] has size [imath]n[/imath] by [imath]m[/imath] and matrix [imath]B[/imath] has size [imath]m[/imath] by [imath]p[/imath] then [imath]|A\cdot B|\leq m|A||B|[/imath] I found that somebody already provided this proof at the link below so don't bother writing an answer here. Linear Algebra munkres analysis on manifolds question. | 263942 | Linear Algebra munkres analysis on manifolds question.
If [imath]A[/imath] is an [imath]n[/imath] by [imath]m[/imath] matrix and [imath]B[/imath] is an [imath]m[/imath] by [imath]p[/imath] matrix, then [imath] |AB| \leq m|A||B|[/imath] where [imath]|A| = \max\{|a_{ij}| : i = 1,\ldots,n \text{ and} j = 1,\ldots,m\}[/imath] Attempt: [imath] |AB| = \max\{| \sum_{j=1}^{m}a_{ij}b_{jk} |: i = 1,\ldots,n \text{ and } k = 1,\ldots,p\} \leq \max\{ \sum_{j=1}^{m}|a_{ij}b_{jk}| : i = 1,\ldots,n \text{ and } k = 1,\ldots,p\} \leq m\max\{|a_ij|\}\max\{|b_{jk}|\} = m|A||B| [/imath] Is this correct? |
2474895 | If product of finite elements in a ring is a unit then can we say each element as unit
In a ring If product of finite elements is a unit then can we say each element as unit For ex., if [imath]q_1 q_2 ... q_m[/imath] is a unit, then can we say each of [imath]q_i[/imath] [imath]\forall i[/imath] is a unit how to prove this? | 2543273 | Units in a Commutative Ring
Suppose we have a commutative ring [imath]R[/imath] with unity and [imath]a, b \in R[/imath], if [imath]ab[/imath] is a unit, are [imath]a[/imath] and [imath]b[/imath] units as well? My intuition is that they must be, but I'm not sure how to see that. |
2533706 | Triangle Inequality for quotients
Problem states the [imath]p(z)[/imath] and [imath]q(z)[/imath] are complex polynomials with [imath]deg(p)=m[/imath] and [imath]deg(q)=n[/imath]. I need to show that [imath]\mid \frac {p(z)}{q(z)}\mid \leq R^{m-n}C[/imath] when [imath]\mid z \mid[/imath] = [imath]R[/imath] I am tempted to use the triangle inequality to get the individual powers of [imath]R[/imath] but I am not sure if it will work because of the denominator if [imath]\mid a + b\mid \leq \mid a \mid + \mid b \mid[/imath] then the inequality is reversed for the quotient | 2448646 | To prove [imath]\left|\frac{p_n(z)}{q_m(z)}\right|\leq \frac{M}{|z|^{m-n}}[/imath] for some [imath]M>0[/imath]
To prove there exist [imath]M>0[/imath] and [imath]a_0>0[/imath] such that for [imath]|z|>a_0[/imath], [imath]\left|\frac{p_n(z)}{q_m(z)}\right|\leq \frac{M}{|z|^{m-n}}[/imath] where [imath]p_n[/imath] and [imath]q_m[/imath] are the polynomials of degree [imath]n[/imath] and [imath]m[/imath] respectively with [imath]n<m[/imath]. I encountered this question in this post. According to the hint, it is enough to show that for large enough [imath]R[/imath] and for every [imath]|z|>R[/imath], we have [imath]\left|\frac{z-z_1}{z-z_2}\right|<M[/imath] In fact, [imath]\left|\frac{z-z_1}{z-z_2}\right|<\frac{|z-z_2|+|z_2-z_1|}{|z-z_2|}=1+\frac{|z_2-z_1|}{|z-z_2|}<M.[/imath] |
2537301 | how to prove [imath]2\sqrt{5} + \sqrt{11}[/imath] is irrational?
I tried to let [imath]2\sqrt{5} + \sqrt{11} = \frac{a}{b}[/imath] and find contradictions (I set [imath]b \ne 0[/imath], [imath]a[/imath] and [imath]b[/imath] are in their simplest form) but I cannot find any | 2533270 | prove that [imath]2\sqrt5 +\sqrt{11}[/imath] is irrational
how would you prove that [imath]2\sqrt5 +\sqrt{11}[/imath] is irrational? I started with a proof by contradiction that assumes that [imath]2\sqrt5 +\sqrt{11}[/imath] is rational and therefore there exist integers [imath]a[/imath] and [imath]b[/imath] such that [imath]\frac{a}{b}=2\sqrt5 +\sqrt{11}[/imath] and squaring both sides yields [imath]\frac{a^2}{b^2}=31 +4\sqrt5\sqrt{11}[/imath] and from this point im stuck as i dont know how to continue to arrive at a contradiction. |
2537662 | Please Explain Baby Rudin Theorem 1.20 (b) By Using Statements on Baby Rudin
I have a question in Baby Rudin Theorem 1.20 (b). I have checked other Q and A's of this theorem in mathstackexchange (and I can understand this theorem). But, those answers did not explain statements in Baby Rudin (such as Baby Rudin Theorem 1.20 (b) Proof). That theorem states that "If [imath]x\in\mathbb{R}[/imath], [imath]y\in\mathbb{R}[/imath], and[imath]x<y[/imath], then there exists a [imath]p\in\mathbb{Q}[/imath] such that [imath]x<p<y[/imath]". The proof of this theorem is below. Since [imath]x<y[/imath], we have [imath]y-x>0[/imath], and Archimedian property furnishes a positive integer [imath]n[/imath] such that [imath]n(y-x)>1[/imath]. Apply Archimedian property again, to obtain positive integers [imath]m_1[/imath] and [imath]m_2[/imath] such that [imath]m_1>nx[/imath], [imath]m_2>-nx[/imath]. Then [imath]-m_2<nx<m_1[/imath]. Hence there is an integer m(with [imath]-m_2\leq m\leq m_1[/imath]) such that [imath]m-1\leq nx<m[/imath]. If we conbine these inequalities, we obtain [imath]nx<m\leq 1+nx<ny[/imath]. Since n>0, it follows that [imath]x<\frac{m}{n}<y[/imath]. This proves this theorem, with [imath]p=m/n[/imath]. I want to know how Rudin gets a consequence that "Hence there is an integer m(with [imath]-m_2\leq m\leq m_1[/imath]) such that [imath]m-1\leq nx<m[/imath]". | 476689 | Baby Rudin Theorem 1.20 (b) Proof
I have a question about Rudin's proof of Theorem 1.20 (b) in his book Principles of Mathematical Analysis. Theorem 1.20 is stated as follows: (a) If [imath]x\in R, y\in R[/imath], and [imath]x>0[/imath], then there is a positive integer [imath]n[/imath] such that [imath]nx>y.[/imath] (b) If [imath]x\in R, y\in R[/imath], and [imath]x<y[/imath], then there exists a [imath]p\in Q[/imath] such that [imath]x<p<y[/imath]. I understand Rudin's proof of (a). The beginning of Rudin's proof of (b) is given below: Since [imath]x<y[/imath], we have [imath]y-x>0[/imath], and (a) furnishes a positive integer [imath]n[/imath] such that [imath]n(y-x)>1.[/imath] Apply (a) again, to obtain positive integers [imath]m_1[/imath] and [imath]m_2[/imath] such that [imath]m_1>nx[/imath], [imath]m_2>-nx[/imath]. Then [imath]-m_2<nx<m_1.[/imath] Hence there is an integer [imath]m[/imath] (with [imath]-m_2\leq m\leq m_1[/imath]) such that [imath]m-1\leq nx<m.[/imath] I don't understand the justification for this last sentence beginning "Hence...." How is [imath]m[/imath] found, and why are [imath]m_1[/imath] and [imath]m_2[/imath] needed to find [imath]m[/imath]? |
2499681 | [imath] \mathbb{H}[/imath] is division algebra of quaternions
Let [imath]A[/imath] be an algebra. For [imath]a, b \in A [/imath] we define maps [imath]L_{a}, R_{b} : A \longrightarrow A[/imath], called the left multiplication map and right multiplication map, by [imath]L_{a} (x) := ax, R_{b} (x) := xb[/imath]. Note that for all [imath]a, b \in A, \lambda , \mu \in F ( field) [/imath] we have [imath]L_{a}b = L_{a}Lb, R_{ab} = R_{b}R_{a} [/imath], [imath]L_{a}R_{b} = R_{b}L_{a}[/imath], [imath]L_{\lambda a+\mu b} = \lambda L_{a} + \mu L_{b}, \quad R_{\lambda a+\mu b} = \lambda R_{a} + \mu R_{b}[/imath] so , is there an element in [imath]M ( \mathbb{H} ) [/imath] ( [imath] \mathbb{H}[/imath] is division algebra of quaternions ) that cannot be written as [imath] L_{a} R_{b}[/imath], [imath] a, b \in \mathbb{H} [/imath]? | 2537560 | quaternions algebra
Let [imath] \mathbb{H}[/imath] be a division algebra of quaternions. Is there an element in [imath]M ( \mathbb{H} ) [/imath] that cannot be written as the left multiplication map and right multiplication map? Let [imath]D[/imath] be an algebra. For [imath]a, b \in D [/imath], maps [imath]L_{a}, R_{b} : D \longrightarrow D[/imath] are the left multiplication map and right multiplication map, by [imath]L_{a} (x) := ax[/imath] [imath] R_{b} (x) := xb[/imath]. |
1499965 | A problem related to rank and nullity of matrices.
Let [imath]A[/imath] be a 4[imath]\times[/imath] 7 real matrix and B be a 7[imath]\times[/imath]4 real matrix such that [imath]AB=I_4[/imath], where [imath]I_4[/imath] is the [imath]4 \times 4[/imath] identity matrix.Which of the followings are true? [imath]\DeclareMathOperator{\rank}{rank}[/imath] [imath]\rank(A)=4[/imath] [imath]\rank(B)=7[/imath] nullity[imath](B)=0[/imath] [imath]BA=I_7[/imath] I know [imath]B[/imath] is a 7[imath]\times[/imath]4 matrix ,then its row rank and column rank are same. rank(B) [imath]\leq[/imath] 4. So (2) is false, but I have no idea about others.please help. Thanks. | 2280028 | Finding rank-nullity of a given matices
Let [imath]A[/imath] be a [imath]4 \times 7[/imath] real matrix and [imath]B[/imath] be a [imath]7 \times 4[/imath] real matrix such that [imath]AB=I_4[/imath]. Which of the following are true? 1) rank[imath](A)=4[/imath] 2) rank[imath](B)=7[/imath] 3) nullity[imath](B)=0[/imath] 4) [imath]BA=I_7[/imath]. My attempt: [imath]4=\operatorname{rank}(AB) \leq \min\{{\operatorname{rank}(A),\operatorname{rank}(B)}\}[/imath]. So [imath]\operatorname{rank}(A)[/imath] must be [imath]4[/imath]. It shows that 1) is true 2) is false by Dimension theorem How to check 3) and 4) ? |
2538051 | Let [imath]a_1 = 1[/imath] and [imath]a_{n+1} = \frac{a_n}2 + \frac2{a_n}[/imath], [imath]n ≥ 1[/imath]. Find the limit of the sequence [imath]\left \{a_n \right \}[/imath] if it exists.
Should monotonic sequence theorem be used? Thanks in advance. | 1786553 | Proving a sequence defined by a recurrence relation converges
How can I prove that this iterative sequence converges to [imath]2[/imath]? Can I use the convergence definition? [imath]a_{n+1} = \frac{a_n}{2} + \frac{2}{a_n},\qquad a_0 = 4 [/imath] Thanks for the help. |
2538415 | complex number [imath]\sum_{k=1}^{n-1} (n-k) cos\frac{2πk}{n}=\frac{-n}{2}[/imath]
Prove that [imath]\sum_{k=1}^{n-1} (n-k) cos\frac{2πk}{n}=\frac{-n}{2}[/imath] where n [imath]\ge[/imath]3 is an integer . My approach, we need to find real part of the equation. Where k is increasing, i am getting confused. Is there any other method of approaching this question. | 1011773 | Tricky trigonometric sum evaluation
Prove that the sum [imath]\sum_{k=1}^{n-1} (n-k)\cdot\cos\left(\frac{2k\pi}{n}\right) [/imath] Is an integer for any [imath]n\geq 3[/imath]. I found this in my textbook but am unable to evaluate this sum. Any help would be appreciated. |
2538550 | Repetition with restrictions
Your local grocery store just received a large shipment of apples, oranges, pears, and bananas---there are only 5 of each fruit. You are shopping at the store and will purchase your fruit for the week. How many ways can you select [imath]10[/imath] pieces of fruit from your store's supply of apples, oranges, pears, and bananas, if you need at least [imath]2[/imath] oranges and [imath]1[/imath] apple? I understand that there are 17 spots to choose from because 3 have been chosen and there are a total of 20 fruits. Would I use the combination by repetition formula? I can't quite wrap my head around what to do. Any help would be appreciated. | 2537542 | counting with max and min restrictions
[imath]x_1 + x_2 + x_3 + x_4 = 10[/imath] But [imath]x_1 \ge 1[/imath] and [imath]x_2 \ge2[/imath]. Finally, [imath]x_1<6, x_2<6, x_3<6[/imath] and [imath]x_4<6[/imath]. I started off by finding the total amount that you can get for the original equation. That would be [imath]C (13, 3)[/imath]. Then i calculated for the cases of numbers being greater than or equal to 5. I dont know how to incorporate [imath]x_1\ge1[/imath] and [imath]x_2\ge2[/imath]. Any help? [imath]x_1, x_2, x_3, x_4[/imath] are all integers that could be >= 0(But [imath]x_1[/imath] and [imath]x_2[/imath] have their own restriction). |
2303734 | Symmetric billinear form
Let [imath]f[/imath] be non zero symmetric bilinear form on [imath]\Bbb{R^3}[/imath]. Suppose that there exists linear transformation [imath]T_i:\Bbb{R}^3\to\Bbb{R}[/imath] ,[imath]i=1,2[/imath] such that [imath]\forall a,b \in \Bbb{R}^3[/imath] , [imath]f(a,b)=T_1(a)T_2(b)[/imath] Then I) rank [imath]f[/imath]=1 II) dim[imath]\{b\in\Bbb{R}^3 :f(a,b)=0 \forall a\in\Bbb{R}^3\}=2[/imath] III) [imath]f[/imath] is positive semi-definite or negative semi -definite IV) [imath]\{a:f(a,a)=0\}[/imath] is a linear subspace of dimension [imath]2[/imath] | 1545799 | Properties of non-zero symmetric bilinear form
Let [imath]f[/imath] be a non-zero symmetric bilinear form on [imath]\Bbb R^3[/imath]. Suppose that there exist linear transformations [imath]T_i:\Bbb R^3\to\Bbb R, i=1,2[/imath] such that for all [imath]\alpha,\beta\in\Bbb R^3,f(\alpha,\beta)=T_1(\alpha)T_2(\beta)[/imath]. Then rank [imath]f=1[/imath] [imath]\dim\{\beta\in\Bbb R^3:f(\alpha,\beta)=0\text{ for all }\alpha\in\Bbb R^3\}=2[/imath] [imath]f[/imath] is a positive semi-definite or negative semi-definite [imath]\{\alpha:f(\alpha,\alpha)=0\}[/imath] is a linear subspace of dimension [imath]2[/imath] (CSIR December 2014) My attempt: (1), (2) and (4) are wrong if [imath]\ker(T_1)=\{0\},\ker(T_2)=\{0\}[/imath] All information I could find about positive/negative semi-definiteness of bilinear form is this post about positive definiteness. Edit: (question has already been answered in comments; "intermediate link" Travis talks about is here. Can someone please guide me? |
2537931 | Caccioppoli type inequality for harmonic function
I'm reading a paper "The Inhomogeneous Dirichlet Problem in Lipschitz domains" written by David Jerison and Carlos E. Kenig. Let [imath]u[/imath] be a harmonic function in [imath]\Omega[/imath]. In the page 183, the author states an inequality without proof. [imath] \int_Q |\delta \nabla u|^p dx \leq \int_{Q^{*}} |u|^p dx.[/imath] Here [imath]\delta(x)=\mathrm{dist} (x,\partial \Omega)[/imath] and [imath]Q[/imath] is a Whitney cube for [imath]\Omega[/imath] and [imath]Q^*[/imath] denotes the double of [imath]Q[/imath] which also lies in [imath]\Omega[/imath]. So [imath]\mathrm{diam}(Q) \approx \mathrm{dist} (Q,\partial\Omega)[/imath]. When [imath]p=2[/imath], allowing some constants, I can prove this estimate using some standard technique when we deduce Caccioppoli inequality. However, I have no idea when [imath]p\neq 2[/imath]. I tried some iteration argument when we prove the Moser estimate. In this case, the problem occurs due to constant. I tried several approach and searched several books on harmonic function, but I failed it. How can I prove this? | 2060303 | [imath]L^p[/imath]-norm of the gradient of a harmonic function.
Let [imath]1<p<\infty[/imath], [imath]\Omega \subsetneq U \subset \mathbb{R}^n[/imath] be bounded open sets in [imath]\mathbb{R}^n[/imath] with smooth boundary. The claim is that there is [imath]c>0[/imath] such that the following inequality holds [imath] \| \nabla u \|_{L^p(\Omega^\prime)} \leq c \| u \|_{L^p(U)}[/imath] for all [imath]u \in L^p(U)[/imath] with [imath]u[/imath] being a weak solution of [imath]\Delta u = 0[/imath], I believe that it can be proved by the mean value property of harmonic functions that such a [imath]u[/imath] is actually smooth and so [imath]\nabla u[/imath] is defined in the usual sense. My question is how can we prove this inequality? The proof I'm reading just states that applying the mean value property to [imath]\nabla u[/imath] and so the [imath]c[/imath] exists. Also, I'm kind of suspicious of for the case [imath]p \in (1,2)[/imath] because I feel we might need to apply Jensen's inequality to [imath]\frac{p}{2}[/imath] somehow. |
2538202 | Prove the following property of a measure: [imath]\mu ((0,x]) = cx.[/imath]
Let [imath]\mu :\mathcal{B}(\mathbb{R}) \to [0,\infty][/imath] be a measure such that for all [imath]h\in \mathbb{R}[/imath] and [imath]A\in \mathcal{B}(\mathbb{R})[/imath], [imath]\mu(A+h) = \mu(A)[/imath]. Let [imath]c= \mu((0,1])[/imath]. I want to prove the property that [imath]\mu((0,x]) = cx[/imath] for any [imath]x\geq 0[/imath]. I have already proven that [imath]\forall x\geq 0, \forall q\in\mathbb{N}: \mu((0,qx]) = q\mu((0,x])[/imath]. As well as [imath]\forall p,q\in \mathbb{N}: \mu((0,\frac{p}{q}]) = c\frac{p}{q}[/imath]. However now I'm stuck. I know that [imath]\mu[/imath] is a measure and therefore [imath]\sigma[/imath]-additive, however I am not quite sure how to use this because [imath]x[/imath] is now a real number instead of an integer or a fraction. | 147089 | Translation invariant measures on [imath]\mathbb R[/imath].
What are all the translation invariant measures on [imath]\mathbb{R}[/imath]? Except Lebesgue measure on [imath]\mathbb R[/imath] I didn't find any translation invariant measure. So I put this question? I know that if [imath]\mu[/imath] is a measure then [imath]c \times \mu[/imath] is again a measure where [imath]c>0[/imath]. |
2539275 | How to evaluate [imath]\int_{0}^{\infty} \frac{\log(x^{2}+1)}{x^{2}+1}[/imath]
I tried to find [imath]f(a) = \int_{0}^{\infty} \frac{\log(x^{2}+a^{2})}{x^{2}+b^{2}}[/imath]. After differentiating I get : [imath]f(a) = \frac{\pi \log(a+b)}{b} + C[/imath]. But it's not easy to find this constant. I represent constant as [imath]\int_{0}^{\infty} \frac{\log(x^{2}+1)}{x^{2}+1} - \pi \log(2)[/imath]. Any hints ? | 1911996 | Solution of [imath]\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}[/imath]
How do I solve this? [imath]\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}[/imath] I know the answer is [imath]\pi\ln2[/imath]. |
2539484 | If [imath]M\cong M_1\oplus M_2[/imath] is a torsion-free module, does it follow that [imath]M_1[/imath] is a torsion-free module?
Suppose [imath]M[/imath] is a module over some ring [imath]R[/imath] such that [imath]M\cong M_1\oplus M_2[/imath]. If [imath]M,\,M_2[/imath] are both torsion-free modules, does it follow that [imath]M_1[/imath] is also? Intuitively I would expect this to be the case, for if [imath]M_1[/imath] had a torsion element, then surely that means [imath]M[/imath] would also? | 495280 | Prove that direct sum of torsion module is torsion
Suppose that [imath]R[/imath] is a ring and [imath]M[/imath] and [imath]N[/imath] are [imath]R[/imath]-module. Prove that a) If [imath]R[/imath] is integral domain, then [imath]Tor(M \bigoplus N) \cong Tor(M) \bigoplus Tor(N)[/imath] b) If [imath]P[/imath] is a submodule of [imath]M[/imath] and [imath]Q[/imath] is a submodule of [imath]N[/imath], then [imath]M \bigoplus N /P \bigoplus Q \cong M/P \bigoplus N/Q[/imath]. For the first part, I kind of confused of the direct sum. The direct sum here means external direct sum or internal direct sum? For the second part, I define a map [imath]f: M \bigoplus N \rightarrow M/P \bigoplus N/Q[/imath] where [imath](m,n) \rightarrow (m+P,n+Q)[/imath] Then I manage to show the map is bijective module homomorphism. Just want to confirm with the surjectivity, is it trivial in this case? Like I have [imath](m+P,n+Q)[/imath], then its preimage is [imath](m,n)[/imath] right? EDIT: Part a): Let [imath](m,n) \in Tor(M) \bigoplus Tor(N)[/imath]. Then there exists [imath]q,s \in R[/imath], where [imath]q,s \neq 0[/imath] such that [imath](qm,sn)=(0,0)[/imath] Note that since [imath]R[/imath] is integral domain and [imath]q,s \neq 0 \Rightarrow qs \neq 0[/imath]. Hence, [imath]qs(m,n)=(qsm,qsn)=(s(qm),q(sn))=(0,0)[/imath], which implies that [imath](m,n) \in Tor(M \bigoplus N)[/imath] Let [imath](m,n) \in Tor(M \bigoplus N)[/imath]. Then there exists [imath]r \in R,r \neq 0[/imath] such that [imath]r(m,n)=(0,0)=(rm,rn) \Rightarrow rm=0,rn=0 \Rightarrow (m,n) \in Tor(M) \bigoplus Tor(N)[/imath]. Hence, [imath]Tor(M \bigoplus N)=Tor(M) \bigoplus Tor(N)[/imath] |
2523016 | Chain rule misunderstanding : two different use of it give two different answers...
I am studying G.R an I have a mistake by using chain rules (yes probably very basic mistake) that I don't get. Take the following quantity (there is Einstein convention here, the [imath]X[/imath] and [imath]V[/imath] are vectors and [imath]\{x\}[/imath] and [imath]\{y\}[/imath] are two different system of coordinates.): [imath] X^{\beta} V^{\sigma} (\frac{\partial y^{\mu}}{\partial x^{\beta}}\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial^2 y^{\rho}}{\partial x^{\alpha} \partial x^{\sigma}})[/imath] I need to "simplify" it. I can do it in two different ways. First, I can use the summation over [imath]\mu[/imath] : [imath] \frac{\partial y^{\mu}}{\partial x^{\beta}}\frac{\partial x^{\alpha}}{\partial y^{\mu}}=\frac{\partial x^{\alpha}}{\partial x^{\beta}}=\delta^{\alpha}_{\beta} [/imath] I also could use the summation over [imath]\alpha[/imath] (other "method" to simplify it): [imath]\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial^2 y^{\rho}}{\partial x^{\alpha}x^{\sigma}}=\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial}{\partial x^{\alpha}}\frac{\partial y^{\rho}}{\partial x^{\sigma}}=\frac{\partial y^{\rho}}{\partial y^{\mu} \partial x^{\sigma}}=\frac{\partial}{\partial x^{\sigma}}\frac{\partial y^{\rho}}{\partial y^{\mu}}=\frac{\partial}{\partial x^{\sigma}}\delta^{\rho}_{\mu}=0[/imath] Thus, if I use the first thing I will end up with : [imath] X^{\beta} V^{\sigma} (\frac{\partial y^{\mu}}{\partial x^{\beta}}\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial^2 y^{\rho}}{\partial x^{\alpha} \partial x^{\sigma}})=X^{\beta} V^{\sigma} (\frac{\partial^2 y^{\rho}}{\partial x^{\beta} \partial x^{\sigma}}) \neq 0[/imath] And if I use the second method I end up with : [imath] X^{\beta} V^{\sigma} (\frac{\partial y^{\mu}}{\partial x^{\beta}}\frac{\partial x^{\alpha}}{\partial y^{\mu}}\frac{\partial^2 y^{\rho}}{\partial x^{\alpha} \partial x^{\sigma}})=0[/imath] It is probably a very stupid mistake or a more deep incomprehension thing but I really don't see it. Could you help me figure this out ? | 2532711 | Are these mixed partial derivatives equal?
Let's say I have a set of [imath]m[/imath] functions depending on [imath]n[/imath] variables [imath]x^{i} = x^{i}(s^{1}, ... , s^{n})[/imath] I use latin indices to denote [imath](1, ..., m)[/imath] and greek letters to denote [imath](1, ..., n)[/imath]. I know [imath]\frac{\partial x^{i}}{\partial x^{j}} = \delta^{i}_{j}[/imath] [imath] \implies \frac{\partial^{2} x^{i}}{\partial s^{\alpha} \partial x^{j}} = \frac{\partial (\delta^{i}_{j})}{\partial s^{\alpha}} = 0[/imath] What I'd like to know is whether commutativity of partial derivatives allows me to do this: [imath]\frac{\partial^{2} x^{i}}{\partial x^{j} \partial s^{\alpha}} = \frac{\partial^2 x^{i}}{\partial s^{\alpha} \partial x^{j}} = 0[/imath] And if I'm not allowed to do so, why? Thanks a lot, and I'm sorry if it is a silly question. Edit: For example, I get troubles when I'm in the simplest case, [imath]m=1, n=1[/imath]. Let's say [imath]x=s^2[/imath] [imath]\frac{dx}{dx}=1[/imath] [imath]\implies \frac{d}{ds}\left( \frac{dx}{dx} \right) = 0[/imath] But when I do it in the opposite order: [imath]\frac{dx}{ds}=2s[/imath] [imath]\implies \frac{d}{dx}\left( \frac{dx}{ds} \right) = \frac{d}{dx}\left( 2s \right)[/imath] substituting [imath]s = \sqrt{x}[/imath] [imath]\frac{d(2s)}{dx} = \frac{d(2\sqrt{x})}{dx} = \frac{1}{\sqrt{x}} = \frac{1}{s}[/imath] What am I doing wrong? |
2539211 | Test the series [imath]\sum_{n=1}^\infty \frac{1}{n^{1+\frac{1}{n}}}[/imath] for convergence or divergence.
I am trying to test [imath]\sum_{n=1}^\infty \frac{1}{n^{1+\frac{1}{n}}}[/imath] for convergence or divergence. I thought to compare [imath]\sum_{n=1}^\infty \frac{1}{n^{1+\frac{1}{n}}}[/imath] to [imath]\sum_{n=1}^\infty \frac{1}{n^{2}}[/imath] as i reasoned that for the first series the term [imath]\frac{1}{n}[/imath] in the exponent can be at most [imath]1[/imath] and for all [imath]n\gt1[/imath] that term will be a fraction, hence the exponent of [imath]n^{1+\frac{1}{n}}[/imath] will be at most [imath]n^2[/imath]. The second series is a convergent p-series as [imath]p=2 \gt1[/imath]. However, this does not really help me as i see now that [imath]{n^{1+\frac{1}{n}}} \lt n^2[/imath] and a smaller denominator means that [imath]\frac{1}{n^{1+\frac{1}{n}}}[/imath] will actually be bigger than [imath]\frac{1}{n^2}[/imath] Any advice appreciated. | 1673312 | Does [imath]\sum_{n=1}^{\infty} \frac{1}{n^{1 + 1/n}}[/imath] converge?
Does [imath]\sum_{n=1}^{\infty} \frac{1}{n^{1 + 1/n}}[/imath] converge? If yes, to where? I searched this specific series but couldn't find a solution. |
2537008 | Stuck on matrix exponential problem
I want to show that the property [imath]e^{tA}e^{tB}=e^{t(A+B)}[/imath] implies that [imath]AB=BA[/imath]. Here [imath]A,B[/imath] denotes matrices and [imath]t\in \mathbb{R}[/imath]. Im stuck, have tried to expand both sides with their taylor series but not sure if this is the way to go. Could anyone give me a hint? | 1991471 | If [imath]\exp(t(A + B)) = \exp(tA) \exp(tB)[/imath] for all [imath]t \geq 0[/imath] then [imath]A,B[/imath] commute
Let [imath]A,B[/imath] be complex valued square matrices. If [imath]\exp(t(A + B)) = \exp(tA) \exp(tB)[/imath] for all [imath]t \geq 0[/imath] then [imath]A,B[/imath] commute. The converse of this statement can be an easy application of the Cauchy product rule and the binomial theorem. Note that this statement doesn't hold, if we restrict ourselves to [imath]t = 1[/imath]. So far I have been trying to use the fact, that [imath]A[/imath] and [imath]B[/imath] are infinitesimal generators to the semigroups [imath]\{\exp(tA)\}[/imath] and [imath]\{\exp(tB)\}[/imath] but I have had no success. Do you have any other hints? Based on the idea of @Did, I came up with the following: Series expansions give me: [imath] \sum_{n = 0}^\infty \frac{t^n(A + B)^n}{n!} = I + tA + tB + \frac{t^2(AB + BA)}{2} + \sum_{n = 3}^\infty \frac{t^n(A + B)^n}{n!} [/imath] and [imath] \left(\sum_{n = 0}^\infty \frac{t^n(A)^n}{n!} \right) \left(\sum_{n = 0}^\infty \frac{t^n(B)^n}{n!} \right) = I + tA + tB + \frac{t^2A^2}{2} + t^2AB + \frac{t^2B^2}{2} + \sum_{n = 3}^\infty t^n c_n, [/imath] where [imath] c_n := \sum_{k = 0}^n \frac{A^k B^{n - k}}{k! n!}. [/imath] The comparison of both expansions gives [imath] \frac{t^2(AB + BA)}{2} + \sum_{n = 3}^\infty \frac{t^n(A + B)^n}{n!} = t^2AB + \sum_{n = 3}^\infty t^n c_n. [/imath] Division by [imath]t > 0[/imath] yields: [imath] \frac{(AB + BA)}{2} + \sum_{n = 3}^\infty \frac{t^{n-2}(A + B)^n}{n!} = AB + \sum_{n = 3}^\infty t^{n-2} c_n. [/imath] But I can't quite see, why the two sums [imath]\sum_{n = 3}^\infty \dots[/imath] should go to [imath]0[/imath] for [imath]t \to 0[/imath] yielding the desired equality [imath] \frac{(AB + BA)}{2} = AB . [/imath] |
2521788 | if one linear operator of a set of linear operators sends a vector to zero then the product of linear operators must send the vector to zero?
This question is related to linear algebra. Suppose we have a finite set of linear operators called [imath]T[/imath]: [imath]T= \{ T_1, T_2, T_3, ...,T_k \}[/imath] How can I prove the following claim? [imath]\exists T_i \in T \; such \; that \; (T_i) \alpha=0 \implies (T_1T_2...T_k) \alpha=0 [/imath] I mean if there exists some linear operator that sends a vector to zero then the multiplication of all linear operators also sends the vector to zero. ---------------------------------------------MY EDIT------------------------------------------- I ask this question because I saw something like this in a linear algebra book. I state what was exactly in the book: Let [imath]T[/imath] be a diagonizable linear operator and let [imath]c_1,...,c_k[/imath] be the distinct characteristic values of [imath]T[/imath]. Then it is easy to see that the minimal polynomial fot [imath]T[/imath] is the polynomial [imath]p=(x-c_1)...(x-c_k).[/imath] If [imath]\alpha[/imath] is a characteristic vector, then one of the operators [imath]T-c_1I,...,T-c_kI[/imath] sends [imath]\alpha[/imath] into [imath]0[/imath]. Therefore [imath](T-c_1I)...(T-c_kI)\alpha=0[/imath] for every characteristic vector [imath]\alpha[/imath]. Please explain if I asked the wong question and explain why the last claim is true i.e. [imath](T-c_1I)...(T-c_kI)\alpha=0[/imath] for every characteristic vector [imath]\alpha[/imath]. | 973186 | Minimal polynomial for T is a product of distinct linear factors
There proof given in my textbook is as follows: Let T be a diagonalisable linear operator and let [imath]c_1, ..., c_k[/imath] be the distinct eigenvalues of T. Then it is easy to see that the minimal polynomial for T is the polynomial [imath] p = (x-c_1)...(x-c_k)[/imath] If [imath]\alpha[/imath] is an eigenvector, then one of the operators [imath]T-c_1I,...,T-c_kI[/imath] sends [imath]\alpha[/imath] into 0. Therefore [imath] (T-c_1 I)...(T-c_kI)\alpha = 0[/imath] for every eigenvector [imath]\alpha[/imath]. There is a basis for the underlying space which consists of eigenvectors of T and hence [imath] p(T) = (T -c_1I)...(T-c_kI) = 0[/imath] I cannot understand this proof at all. I understand that one of the operators [imath]T-c_1I,...,T-c_kI[/imath] sends [imath]\alpha[/imath] into 0, but I can't see why we we multiply them together, it still works. And I also don't understand why there is a basis for the underlying space. Can someone kindly explain to me please. Thanks |
2539639 | Proof that the cardinality of the Cantor Set is c
I have proven that the Cantor set is uncountable using Cantors diagonalization argument in base [imath]3[/imath]. Now I have to prove that it has cardinality [imath]\mathfrak{c}[/imath] due to the fact that it is uncountable but I am struggling to even get started. Any help would be appreciated. | 996387 | Cantor Sets/nonempty/cardinality
Let [imath]S_0=[0,1][/imath] and define every [imath]S_k[/imath] for [imath]k\geq 1[/imath] \begin{align*} S_1&=\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3}, 1\right],\\ S_2&=\left[0,\frac{1}{9}\right]\cup\left[\frac{2}{9}, \frac{3}{9}\right]\cup\left[\frac{6}{9}, \frac{7}{9}\right]\cup\left[\frac{8}{9},1\right],\\ S_3&=\left[0, \frac{1}{27}\right]\cup\left[ \frac{2}{27}, \frac{3}{27}\right]\cup\left[ \frac{6}{27}, \frac{7}{27}\right]\cup\left[ \frac{8}{27}, \frac{9}{27}\right]\cup\left[ \frac{18}{27}, \frac{19}{27}\right]\cup\left[ \frac{20}{27}, \frac{21}{27}\right]\cup\left[ \frac{24}{25}, \frac{25}{27}\right]\cup\left[ \frac{26}{27}, 1\right]\\ \vdots \end{align*} Now let [imath]C=\bigcap_{k=0}^\infty S_k[/imath]. This is known as a Cantor set. I have to prove that C is non-empty, by showing that 1/4 is in C. I also have find the cardinal of C and prove my answer. I have some ideas how to prove non-emptiness, but I am stuck as to how to specifically prove that 1/4 is in C. Also I think that the cardinal is equal to the cardinal of the reals [imath]2^{\aleph_0}[/imath]. I probably need to find a bijection to prove this right? |
2539734 | Proving existence of a root in second derivative
Let [imath]f:[0, 1] \to \mathbb{R} [/imath] be double differentiable function such that line segment between points [imath](0,f(0))[/imath] and [imath](1,f(1))[/imath] pierces function at point [imath](a,f(a)) [/imath] where [imath]0<a<1[/imath]. So now i have to prove that there exists such [imath]x_0[/imath] that [imath]f''(x_0)=0[/imath] So i know that the function has a second derivative, but what can i say about it, since i only have the function, not second derivative. My first idea to somehow to use that line for describing first derivative. And i know that if i do that i will just have to show that f'(0)=f'(1) and will use Rolle's theorem to prove the statement. Any help or tips are welcome. Thank you in advance. | 2328629 | Help proving existence of an specific [imath]x_0[/imath] for which [imath]f''(x_0)=0[/imath]
I have a function [imath]f: [0,1]\rightarrow \mathbb{R}[/imath] that is twice differentiable. For this function is stated that there is a line segment with endpoints [imath]A=(0,f(0))[/imath] and [imath]B=(1,f(1))[/imath] that cuts function [imath]f[/imath] at [imath](a,f(a))[/imath] where [imath]0<a<1[/imath]. I have to prove that there exists such [imath]x_0[/imath] that [imath]f''(x_0)=0[/imath]. Should I use some theorem for solving this? Like Rolle or Lagrange? Somehow it seems like I should formulate a function between these two points or something, but I don't know whether it's linear or not. Any help would be appreciated. Thank you in advance. |
1221527 | Show that a nonnegative harmonic function is constant
Suppose that [imath]f(z): \mathbb{C} \rightarrow \mathbb{R}[/imath] is harmonic and that [imath]f(z) \ge 0 [/imath] for all complex [imath]z[/imath]. Then [imath]f[/imath] must be constant. I imagine that I will need to make use of Liousville's theorem or otherwise show that [imath]f'[/imath] is zero. However the problem is that the book has not mentioned anything about harmonic functions other then the definition that the second partials of the real and imaginary parts add to zero. It is clear that an analytic function is harmonic from the Cauchy Riemann equations, is the converse true? Thanks in advance | 561818 | Positive harmonic function on [imath]\mathbb{R}^n[/imath] is a constant?
Is it true that a positive harmonic function on [imath]\mathbb{R}^n[/imath] must be a constant? How might we show this? The mean value property seems not to be the way...for that we would need boundedness. |
2535645 | Proof for an equality in a triangle
[imath]O[/imath] is the point inside triangle [imath]ABC[/imath] . The lines joining the three vertices [imath]A,B,C[/imath] to [imath]O[/imath] cut the opposite sides in [imath]K,L[/imath], and [imath]M[/imath] respectively. A line through [imath]M[/imath] parallel to [imath]KL[/imath] cuts the line [imath]BC[/imath] at [imath]V[/imath] and [imath]AK[/imath] at [imath]W[/imath]. Prove that [imath]VM=MW[/imath]. I tried using the ceva's theorem in triangle [imath]ABC[/imath] and equating it with the relation obtained by Menellau's theorem . Then I used basic proportionality theorem and after using them I just got one relation. [imath]\frac {BK}{BV}=\frac {AL}{AP}[/imath] Now I am stuck please help me. | 2384809 | What is a "harmonic quadruple"?
Can anyone tell me what a "harmonic quadruple" is? I had a problem in Australian mathematical Olympiad paper. The solution uses something like harmonic quadruple to prove that two sides are parallel. So can this concept be used to prove the following? [imath]O[/imath] is the point inside triangle [imath]ABC[/imath]. The lines joining the three vertices [imath]A[/imath], [imath]B[/imath], [imath]C[/imath] to [imath]O[/imath] cut the opposite sides in [imath]K[/imath],[imath]L[/imath] and [imath]M[/imath] respectively. A line through [imath]M[/imath] parallel to [imath]KL[/imath] cuts the line [imath]BC[/imath] at [imath]V[/imath] and [imath]AK[/imath] at [imath]W[/imath]. Prove that [imath]VM=MW[/imath]. |
1518133 | [imath]\dim (f^{-1}(K \cap \operatorname{im} f))= \dim (\ker f)+ \dim(K \cap \operatorname{im} f)[/imath]
I've a problem with this: Let [imath]f:V \to W[/imath] be a linear function between two vector spaces, let [imath]K[/imath] be a subspace of [imath]W,[/imath] then [imath]\dim (f^{-1}(K \cap \operatorname{im} f))= \dim (\ker f)+ \dim(K \cap \operatorname{im} f)[/imath] How can I prove that? I can't understand this completely, it is similar to the rank-nullity theorem, is there any link between them? Thanks a lot in advice | 68807 | Question about an application of the Rank-Nullity theorem
Let [imath]V,W[/imath] be two vector spaces over a field [imath]F[/imath] and let [imath]f:V \to W[/imath] be a linear mapping. I have going over some facts about vector spaces (not necessarily finite dimensional) and I have been trying to apply the rank nullity theorem to derive a simple lemma. I have not been able to complete the proof and was wondering if I need to use something other then [imath]\ker f + \mathrm{Im}\; f = \dim V[/imath]. How do we show that for ever vector subspace [imath]K \subset W[/imath], [imath] \dim(f^{-1}(K)) = \dim(K \cap \mathrm{Im}\; f) + \dim( \mathrm{Ker}\; f)[/imath]? |
779784 | Show that an integer of the form [imath]8k + 7[/imath] cannot be written as the sum of three squares.
I have figured out a (long, and tedious) way to do it. But I was wondering if there is some sort of direct correlation or another path that I completely missed. My attempt at the program was as follows: A number of the form, [imath]8k + 7 = 7 (mod 8)[/imath]. That is, we are looking for integers a, b, c such that [imath]a^2 + b^2 + c^2 = 7 (mod 8)[/imath]. LONG and TEDIOUS way: [imath](8k)^2 = 0 (mod 8)[/imath] [imath](8k+1)^2 = 1 (mod 8)[/imath] [imath](8k+2)^2 = 4 (mod 8)[/imath] [imath](8k+3)^2 = 1 (mod 8)[/imath] [imath](8k+4)^2 = 0 (mod 8)[/imath] [imath](8k+5)^2 = 1 (mod 8)[/imath] [imath](8k+6)^2 = 4 (mod 8)[/imath] [imath](8k+7)^2 = 1 (mod 8)[/imath] That is, using three of these modulo there is no way to arrive at [imath]a^2 + b^2 + c^2 = 7 (mod 8)[/imath] | 2559544 | sum of 3 squares
I want to show that any integer of the form [imath]n=4^m(8k+7)[/imath] with [imath]m,k\ge0[/imath] cannot be expressed as a sum of 3 squares. The case for [imath]m=0[/imath] is easy to prove since the sum of 3 squares cannot be [imath]\equiv7\bmod8[/imath]. However, I am very confused with the proof for the general case, because if [imath]m=1[/imath] then [imath]n\equiv4\bmod8\not\equiv7\bmod8[/imath], so can be expressed as a sum of 3 squares if [imath]m>1[/imath], [imath]8\mid n[/imath], so is also a sum of 3 squares I might be overlooking something here, can anyone shed some light? |
2539167 | bounded julia set [imath]f(z)=z^2 + c[/imath]
prove that [imath]|f(z)|>|z|+1[/imath] where [imath]z\in V[/imath] and [imath]V=\left\{z\in\Bbb C\mid|z|> \frac 1 2 + \sqrt{\frac{5}{4} +|c|}\,\right\}[/imath] and [imath]f(z)=z^2+c[/imath] and [imath]c\in \Bbb C[/imath] I can't seem to get anywhere needing to prove [imath]|z^2+c|>|z|+1[/imath] over [imath]V[/imath]. I tried putting in [imath]z= a+bi[/imath] , [imath]c = p + qi[/imath] but it just gets complicated. this is later used to prove that the julia set of [imath]f[/imath] is bounded. any help would be great. thanks! | 2488001 | For [imath]x[/imath] and [imath]\lambda[/imath] complex, if [imath]|x| \gt \frac{1}{2} + \sqrt{\frac{5}{4} + |\lambda|}[/imath] then [imath]|x^2+\lambda| \gt |x| +1[/imath]
Define [imath]f(z) = z \ ^ 2 + \lambda[/imath] and [imath]V = \left\{x \in \Bbb C : |x| \gt \frac{1}{2} + \sqrt{\frac{5}{4} + |\lambda|} \right\}[/imath], for some [imath]\lambda \in \Bbb C[/imath]. Then, if [imath]x \in V[/imath] then [imath]|f(x)| \gt |x| +1[/imath]. I tried the following : Let [imath]x\in V[/imath] then [imath]|f(x)| = |x \ ^ 2 +\lambda | \ge |x | \ ^ 2 - | \lambda| \gt (\frac{1}{2} + \sqrt{\frac{5}{4} + |\lambda|}) \ ^ 2 - |\lambda| = 1.5 + \sqrt{\frac{5}{4} + |\lambda|} [/imath]. But this doesn't get me anywhere. |
2112653 | Proving that [imath]K\otimes_{\mathbb{Q}}\mathbb{C}\simeq \prod_{\tau}\mathbb{C}[/imath]
I'm taking a course in algebraic number theory and we are starting to discuss lattices and Minkowski theory. We were given this introductory exercise: Let [imath]K[/imath] be a field of numbers of order [imath]n[/imath] and embbedings [imath]\tau_1, ..., \tau_n:K\to\mathbb{C}[/imath]. Show that [imath]\phi:K\otimes_{\mathbb{Q}}\mathbb{C}\to \prod_{\tau}\mathbb{C}[/imath] defined by [imath]\phi(a\otimes z)=(z\,\tau_1(a), ..., z\,\tau_n(a))[/imath] is an isomorphism between [imath]\mathbb{C}[/imath]-vector spaces. I've already seen a solution, which uses linear independence of characters (which I'm familiar with) to prove injectivity and then uses a dimension argument to conclude bijectiviy. I know I'm able to follow the outline of that proof, but actually my problem is more basic: I just don't know what the sets [imath]K\otimes_{\mathbb{Q}}\mathbb{C}[/imath] and [imath]\prod_{\tau}\mathbb{C}[/imath] actually mean. I'm familiar with the tensor product [imath]U\otimes V[/imath], when [imath]U[/imath] and [imath]V[/imath] are both [imath]\mathbb{F}[/imath]-vector fields (i.e., both have the same base field), so when I see the notation [imath]U\otimes_{\mathbb{Q}} V[/imath], I assume implicitly that [imath]U[/imath] and [imath]V[/imath] are both [imath]\mathbb{Q}[/imath]-vector spaces, but that is not the case with [imath]K\otimes_{\mathbb{Q}}\mathbb{C}[/imath]. That is really confusing me: take for example [imath]K=\mathbb{Q}(i)[/imath]. What is the canonical basis for [imath]\mathbb{Q}(i)\otimes_{\mathbb{Q}} \mathbb{C}[/imath]? How am I supposed to write [imath]1\otimes \sqrt{2}[/imath] in the canonical basis? Like [imath]\sqrt{2}(1\otimes 1)[/imath]? But am I allowed to "pull out" an irrational number from the second coordinate like that? Then what does [imath]\mathbb{Q}[/imath] in "[imath]\otimes_{\mathbb{Q}}[/imath]" stand for? I can't make sense of it. Second, I thought [imath]\prod_{\tau}\mathbb{C}[/imath] should be the set of [imath]n[/imath]-uples [imath](\tau_1(a), ..., \tau_n(a))[/imath] where [imath]a\in K[/imath]. But then how do I know that [imath]\phi(a\otimes z)=(z\,\tau_1(a), ..., z\,\tau_n(a))[/imath] is of the form [imath](\tau_1(b), ..., \tau_n(b))[/imath] for some [imath]b\in K[/imath]? If there is no such [imath]b[/imath], [imath]\phi[/imath] would not be well defined... Any help would be useful, thanks in advance! | 2070412 | Neukirch’s Number Theory – why is [imath]ℂ \otimes_ℚ K → K_ℂ,~z \otimes a ↦ (j(z)a)_τ[/imath] an isomorphism?
In §5 of chapter 1 of Neukirch’s Algebraic Number Theory on Minkowski theory, it is claimed that [imath]ℂ \otimes_ℚ K → K_ℂ,~z \otimes a ↦ zj(a)[/imath] yields an isomorphism. Here, [imath]K[/imath] is an algebraic number field, [imath]K_ℂ = \prod_τ ℂ[/imath] (where [imath]τ[/imath] runs through all field embeddings [imath]K → ℂ[/imath]) and the map [imath]j[/imath] is given by [imath]j \colon K → K_ℂ,~a ↦ (τa)_τ.[/imath] I’m having a hard time seeing this isomorphism. What’s the inverse map? |
2540792 | Let [imath]R[/imath] be a commutative ring with [imath]1[/imath] such that [imath]R[x][/imath] is a PID. Show that [imath]R[/imath] is a field.
Let [imath]R[/imath] be a commutative ring with [imath]1[/imath] such that [imath]R[x][/imath] is a principal ideal domain. Prove [imath]R[/imath] is a field. So I thought to take [imath]a\in R[/imath] and look at [imath]aR[x][/imath] which is a principal ideal and obviously [imath]a\in R[x][/imath] is a polynomial of degree [imath]0[/imath] so [imath]\langle a\rangle=aR[x][/imath]. I don't have a clue how to continue. I wanted to somehow show that [imath]a^{-1}\in R[/imath] and so every element is invertible (because we took a random one) so [imath]R[/imath] is a field. But I can't seem to find a way to prove that [imath]a^{-1}\in R[/imath]. Any ideas? | 873088 | If [imath]F[/imath] be a field, then [imath]F[x][/imath] is a principal ideal domain. Does [imath]F[/imath] have to be necessarily a field?
If [imath]F[/imath] be a field, then [imath]F[x][/imath] is a principal ideal domain. Does [imath]F[/imath] have to be necessarily a field? My Thoughts: Suppose instead of [imath]F[/imath], we take the set of polynomials [imath]R[x][/imath] over a commutative ring [imath]R[/imath] with unity. Then, Suppose [imath]I[/imath] is an ideal of [imath]R[x][/imath]. Let [imath]g(x) \in I[/imath] such that [imath]g(x)[/imath] is the polynomial of the lowest degree in [imath]I[/imath]. Then: [imath] \langle g(x) \rangle \subseteq I .......... (1)[/imath] Let [imath]f(x) \in I[/imath]. Then [imath]f(x) = p(x)g(x) + r(x)~~|~~p(x),r(x) \in R[x], \deg r(x) < \deg g(x)[/imath] Since, [imath]I[/imath] is an ideal [imath]\implies f(x) - p(x)g(x) = r(x) \in I[/imath] But, [imath]g(x)[/imath] is of the lowest degree in [imath]I \implies r(x) = 0 \implies f(x) \in \langle g(x) \rangle \implies I \subseteq \langle g(x) \rangle ......(2)[/imath] Then from [imath](1),(2) : I = \langle g(x) \rangle[/imath] Does the Presence of zero divisors in [imath]R[x][/imath] really make a difference? The only advantage I see is that if there are no zero divisors in [imath]R[x][/imath] then [imath]I=\{0\} \implies I= \langle 0 \rangle[/imath]. But, every ideal contains the zero element. Why is there the condition of a field specifically given in textbooks for [imath]F[x][/imath] to be a principal ideal domain? Thank you for your help. |
2540488 | [imath]\frac{(2n − 1)!!}{(2n)!!} \leq \frac{1}{\sqrt{2n+1}}[/imath] by induction?
Let [imath](2n)!![/imath] be the product of all positive even integers less than or equal to [imath]2n[/imath]. Let [imath](2n − 1)!![/imath] be the product of all odd positive integers less than or equal to [imath](2n − 1)[/imath]. Prove that [imath]\frac{(2n − 1)!!}{(2n)!!} \leq \frac{1}{\sqrt{2n+1}}.[/imath] I am up to the inductive step, where am I stuck. I can't find a way to substitute in from the assumption because of the square root. | 1131552 | Proving the inequality [imath]\frac12\frac34....\frac{2n-1}{2n}<\frac1{\sqrt{2n+1}}[/imath]
How to show this inequality: [imath]\dfrac{1}{2}\dfrac{3}{4}....\dfrac{2n-1}{2n}<\dfrac{1}{\sqrt{2n+1}}[/imath] Using induction the inequality is verified for [imath]n=1[/imath] now assume that that the inequality holds for [imath]n[/imath],to show it for [imath]n+1[/imath] Then [imath]\dfrac{1}{2}.\dfrac{3}{4}....\dfrac{2n-1}{2n}\dfrac{2n+1}{2(n+1)}<\dfrac{1}{\sqrt{2n+1}}.\dfrac{2n+1}{2(n+1)}=\dfrac{\sqrt{2n+1}}{2(n+1)}<\dfrac{\sqrt{2(n+1)}}{2(n+1)}=\dfrac{1}{\sqrt{2(n+1)}}[/imath] but I have to make it less than [imath]\dfrac{1}{\sqrt{2n+3}}[/imath] which is not coming.Any help |
2541411 | How to show that the composite function of two injective functions is injective
I have 2 injective functions,[imath]X:A \to B[/imath] and [imath]Y:B \to C[/imath]. How do I show that [imath]Y \circ X[/imath] is injective. In the form of a proof. | 2049511 | Is the composition of two injective functions injective?
Consider [imath]f:A \to B[/imath] and [imath]g: B \to C[/imath]. If I know that both [imath]f[/imath] and [imath]g[/imath] are injective, then can I state that [imath]g \circ f:A \to C[/imath] is injective? |
2541035 | Find all real solutions for the system: [imath]x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}[/imath] given that [imath]xyz=1[/imath]
Find all real solutions for the system: [imath]x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}[/imath] given that [imath]xyz=1[/imath]. This is from a math olympiad (OBM 2009). By inspection it is easy to spot that a possible solution is [imath]x=y=z=1[/imath] but are there others? It looks like it is not a difficult question but I'm not finding a nice approach. Hints and solutions are appreciated. Sorry if this is a duplicate. Edit 1: Pointed out by helpful comments below, another answer presents a complete solution for this problem, for all possible cases. From that mentioned answer we can conclude that when [imath]xyz=1[/imath] there are two possible solutions: [imath](x,y,z)=(1,1,1)~~\text{and}~~(x,y,z)=(a,\frac{-1}{a+1},\frac{-(a+1)}{a}),\forall a \in \mathbb R\setminus\{0;-1\}.[/imath] Answers using other techniques or arguments are welcomed. | 2078126 | [imath]x + \frac1y = y + \frac1z = z + \frac1x[/imath], then value of [imath]xyz[/imath] is?
If [imath]x,y,z[/imath] are distinct positive numbers, such that [imath]x + \frac1y = y + \frac1z = z + \frac1x [/imath] then value of [imath]xyz[/imath] is? [imath]A)\ 4\quad B)\ 3\quad C)\ 2\quad D)\ 1[/imath] My attempt: 1.I equaled the equation to '[imath]k[/imath]'. Using the AM-GM inequality, I found that [imath]k>2[/imath] (the equality does not hold because all are distinct). However, this, I couldn't put to much use. 2. For the next attempt, I substituted the values of [imath]x[/imath] and [imath]y[/imath] in terms of [imath]z[/imath] and [imath]k[/imath] in [imath]xyz[/imath]. What I am getting is [imath]xyz=zk^2 - k - z[/imath]. That's the farthest I could do..Please, help. |
2541498 | Find all real solutions for the system: [imath] x+y-z=-1[/imath], [imath]x^2-y^2+z^2=1,[/imath] [imath]-x^3+y^3+z^3=-1.[/imath]
Find all real solutions for the following system: [imath] \left\{ \begin{array}{l} x+y-z=-1 \\ x^2-y^2+z^2=1\\ -x^3+y^3+z^3=-1 \end{array} \right. [/imath] From a math olympiad (IberoAmericana, 1992). This was taken from a book, with no answer provided. My usual tricks using symmetric polynomials appear not to work here (or I'm don't know how to apply them). A possible solution (by inspection) is [imath](x,y,z)=(-1,-1,-1)[/imath], but there might be others. Hints and full answers demonstrating all possible solutions are appreciated. | 2497260 | Equation System with 3 variables
"Find all the possible solutions [imath](x,y,z) \in[/imath] [imath]\Bbb R[/imath] that satisfies the following equation system". My teacher gave me this problem and i got some clues in how to solve it. I evaluated [imath]1[/imath] and [imath]-1[/imath] and i got these solutions [imath](1,-1,1)[/imath] and [imath](-1,-1,-1)[/imath]. I think these are the only ones but i can't prove it. Am i wrong?, are there more solutions?. Thanks [imath] \left\{ \begin{array}{c} x+y-z=-1 \\ x^2-y^2+z^2=1\\ -x^3+y^3+z^3=-1 \end{array} \right. [/imath] |
2541595 | Is the inverse of a continuous monotonic function defined on a compact also continuous?
Let [imath]K \subset \mathbb{R}[/imath] be a compact set, and [imath]f:K \rightarrow \mathbb{R}[/imath] a continuous and strictly monotonic function. Is it true that its inverse function, [imath]f^{-1}: f(K) \rightarrow K[/imath], is also continuous? I know that this result is true if [imath]K[/imath] is an interval, so if [imath]K=[a,b][/imath] this result holds, but does it still hold if [imath]K[/imath] is an arbitrary compact set? | 1523622 | Continuous function with compact domain has continuous inverse
Let [imath]A,B\subset\mathbb R[/imath] and [imath]f:A\to B[/imath] be an invertible function (so 1-1 and onto). Prove that if [imath]A[/imath] is compact and [imath]f[/imath] is continuous, then the inverse [imath]f^{-1}:B\to A[/imath] is continuous. And give a counterexample when [imath]A[/imath] is not compact (no transcendentals). If [imath]f[/imath] is continuous, then for every sequence [imath]\{x_n\}[/imath] in [imath]A[/imath] that converges to [imath]L[/imath], [imath]\lim_{n\to\infty}f(x_n)=f(L)[/imath]. We need to prove that for every sequence [imath]\{y_n\}[/imath] in [imath]B[/imath] that converges to [imath]K[/imath], [imath]\lim_{n\to\infty}f^{-1}(y_n)=f^{-1}(K)[/imath] (Is this sufficient to prove that [imath]f^{-1}[/imath] is continuous?), I'm not sure how to proceed from there, and I'm having trouble coming up with a counterexample as well. |
2541617 | Prove that any element in a ring A can be written as a product of irreducible but A is not a UFD
Let A = [imath]Z[/imath][[imath]\sqrt{10}[/imath]] = {a+b[imath]\sqrt{10}[/imath][imath]\mid[/imath]a,b [imath]\in[/imath] [imath]Z[/imath]}. Prove that any element in A an be written as a product of irreducible, but A is not a UFD. Also a small question, I've proved that [imath]4[/imath]+[imath]\sqrt{10}[/imath] and [imath]4[/imath]-[imath]\sqrt{10}[/imath] are irreducible, but how to show that they cannot be associates? Thank you so much! | 1448942 | Proving [imath]\mathbb{Z}[\sqrt {10}][/imath] is not a UFD
I am wondering how to show that [imath]\mathbb{Z}[\sqrt {10}][/imath] is not a UFD. My only idea is to show that there are two factorizations of [imath]10[/imath], say, [imath]ab, uv[/imath] such that [imath]a[/imath] is not a unit times [imath]u[/imath] or [imath]v[/imath]. In this ring [imath]10=2\cdot5=\sqrt {10}\cdot \sqrt {10}[/imath], so it suffices to show [imath]2[/imath] is not a unit times [imath]\sqrt {10}[/imath]. Suppose [imath]2=\sqrt {10}(a+b\sqrt{10})=a\sqrt{10}+10b[/imath]. Then [imath]a=0[/imath] since [imath]\sqrt{10}[/imath] is not rational. So [imath]10b=2[/imath], which has no integer solutions. So [imath]\mathbb{Z}[\sqrt {10}][/imath] is not a UFD. Is my reasoning correct? What are the flaws? |
2461245 | What is wrong in this proof that any derivative function must be continuous?
[imath]f[/imath] is differentiable on [imath](a, b)[/imath]. (1) Let [imath]\alpha \in (a, b)[/imath]. [imath]f'(\alpha) = \lim_{h \to 0}\frac{f(\alpha + h) - f(\alpha)}h[/imath]. (1) [imath]\implies f[/imath] is differentiable between [imath]\alpha[/imath] and [imath]\alpha + h[/imath], inclusive. By the mean value theorem, [imath]\exists c_n[/imath] between [imath]\alpha[/imath], [imath]\alpha + h[/imath]: [imath]f'(c_n) = \frac{f(\alpha + h) - f(\alpha)}{\alpha + h - \alpha}[/imath]. So, [imath]f'(\alpha) = \lim_{h \to 0}f'(c_n)[/imath]. As [imath]h \to 0[/imath], [imath]c_n \to \alpha[/imath]. So, [imath]f'(\alpha) = \lim_{c_n \to \alpha}f'(c_h) \implies f'[/imath] is continuous at [imath]\alpha \forall\alpha\in(a, b)[/imath]. [imath]\therefore f[/imath] is differentiable on [imath](a, b) \implies f'[/imath] is continuous on [imath](a, b)[/imath]. But clearly the statement must not be true as for some functions like the following, the derivative exists, but is not continuous, at zero: [imath]f(n) = \begin{cases} n^2 \sin(\frac{1}{n^2}) & n \in \mathbb R \\ 0 & n = 0. \end{cases}[/imath] [imath]f'(0) = 0[/imath]. But [imath]\lim_{n \to 0}f'(n)[/imath] does not exist. | 876974 | Where is the error in my proof that all derivatives are continuous?
I know that this can not be true due to counter-examples but I don't know where the error in my reasoning is. Assumption: If [imath]f(x)[/imath] is differentiable in [imath]\mathbb{R}[/imath] then the derivative [imath]f'(x)[/imath] is continuous in [imath]\mathbb{R}[/imath]. Faulty Proof: For every [imath]c \in \mathbb{R}[/imath], using the mean value theorem for [imath]f(x),[/imath] on the interval [imath]x \in [c, c + h] [/imath] where [imath]h[/imath] is positive. [imath] \frac{f(c + h) - f(c)}{h} = f'(\xi(h)) [/imath] Where [imath]\xi(h) \in (c,c+h)[/imath]. Because this equation holds for every [imath]h>0[/imath]. It must hold in the limit as [imath]h \rightarrow 0^+[/imath]. [imath] \lim_{h\to 0^+}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^+}f'(\xi(h)) [/imath] But the left side of the equation is the right one sided derivative. [imath] f'_{+}(c) = \lim_{h\to 0^+}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^+}f'(\xi(h)) [/imath] The same can be done for [imath]h[/imath] being negative, but because of differentiability at every point the left and right derivatives must be equal. [imath] f'(c) = f'_{+}(c) = f'_{-}(c) = \lim_{h\to 0^-}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^-}f'(\xi(h)) [/imath] As [imath]h \rightarrow 0^+[/imath], [imath]\xi(h) \rightarrow c[/imath]. So because the limit [imath]\lim_{h\to 0^+}f'(\xi(h))[/imath] exists and [imath]\xi(h) \neq c[/imath], it is equal to [imath]\lim_{x\to c^+}f'(x)[/imath] It follows that [imath]\lim_{x\to c^+}f'(x) = \lim_{x\to c^-}f'(x) = f'(c)[/imath] so the function [imath]f'(x)[/imath] is continuous. |
2541620 | Proving Existence of Multiplicative Inverse
I am given 2 sets. Set [imath]R = \{a+b\sqrt2: a,b \in \Bbb{Z}\}[/imath] and Set [imath]U = \{\alpha \in R: N(\alpha) = 1\}[/imath] where [imath]N(\alpha) = a^2 - 2b^2[/imath] I am supposed to show that [imath]U[/imath] is a subgroup of [imath]\Bbb{R}^x[/imath] which means the set of units over all real numbers. I was able to show that [imath]1 \in U[/imath] and set [imath]U[/imath] is closed under multiplication, but I still have to show that every element of [imath]U[/imath] has a multiplicative inverse, and I have no clue how to do so. If you could let me know, I would greatly appreciate it. Thanks! | 940690 | 1) Show that [imath](\Bbb Z[\sqrt2]^*, .)[/imath] is infinite.
1) Show that [imath](\Bbb Z[\sqrt2]^*, .)[/imath] is infinite. 2) Classify [imath](\Bbb Z[\sqrt2]^*, .)[/imath], where [imath]\Bbb Z[\sqrt2]^*[/imath] is the group of units of [imath]\Bbb Z[\sqrt2][/imath] What I have done so far that for [imath]a+b\sqrt2[/imath] its inverse will be [imath]\frac {a-b\sqrt2}{a^2-2b^2}[/imath] Now [imath]\frac {a}{a^2-2b^2} \in \Bbb Z[/imath] & [imath]\frac {b}{a^2-2b^2} \in \Bbb Z[/imath] What conclusion can be drawn from it about the cardinality of [imath]\Bbb Z[\sqrt2]^*[/imath]? One can post the answer also for the 1st part only. |
2541519 | How do I think about a natural transformation between categories of groups?
From Mac Lane: I'm trying to prove the question at the bottom true, but I'm not exactly understanding how to jump between the different levels of abstraction. If [imath]B[/imath] and [imath]C[/imath] are categories with one object each (say [imath]B[/imath] and [imath]C[/imath]), then to satisfy the definition of a natural transformation it must be the case that if [imath]g : B \rightarrow B[/imath] is an arrow in [imath]B[/imath] (or an element of [imath]B[/imath]) then [imath]T(g) \circ \tau_B = \tau_B \circ S(g)[/imath], where [imath]\tau_B : S(B) \rightarrow T(B)[/imath] is the natural transformation for each object (only one) in [imath]B[/imath]. It seems obvious that the above equality should imply the conjugacy relationship, but I'm not exactly sure where to bridge the gap between the "category with one element called a group" and the "group" itself. Is the correct way of thinking this?: Because the "group" [imath]B[/imath] with homomorphisms [imath]T,S[/imath] satisfies the axioms for the "category with one element called a group" and because we have the naturality relationship between the functors of the two "categories with one element called a group", we can replace the equation [imath]T(g) \circ \tau_B = \tau_B \circ S(g)[/imath] (that is in terms of arrows and objects) with their respective elements and products? | 1730468 | Group theoretic meaning of natural isomorphisms between certain functors
So imagine you have a group [imath]G[/imath] and we consider the set of group homomorphisms from [imath]\mathbb{Z}[/imath] to [imath]G[/imath] specified by [imath]\forall g[/imath] [imath]\in G[/imath] [imath]\exists[/imath] [imath]\phi(1)=g[/imath]. Each of these homomorphisms is in 1-1 relation with the elements of [imath]G[/imath]. Consider groups for a moment in the context of category theory, viewing the groups as one object categories and thereof group homomorphism as functors. What would a natural isomorphism mean in group theory terms for G? My thoughts are that the natural transformations equate to maps between homomorphisms and since the homomorphisms are in 1-1 relation with the elements of G, then a natural transformation is a map between group elements, i.e another element of the group. I'm not entirely convinced by this. Anyway continuing my reasoning a natural isomorphism is a map between the homomorphisms that has an inverse, the thing is in relation to the elements of G this is every element? Furthermore I'm looking to find an equivalence class defined by this natural isomorphism, is this just a collection of the elements that are each others inverses? Thanks in advance for any assistance. |
2327987 | [imath]x^4 -ax^3 +2x^2 -bx +1[/imath] has real root [imath]\implies[/imath] [imath]a^2+b^2 \ge 8[/imath]
it is requested to show that if the quartic polynomial [imath]f(x) \in \mathbb{R}[x][/imath], defined by: [imath] f(x) = x^4 -ax^3 +2x^2 -bx +1, [/imath] has a real root, then [imath] a^2 +b^2 \ge 8 [/imath] this question was asked by @medo, then deleted a few minutes ago. however having spent a little time on it, i think the problem seems sufficiently instructive to be worth resuscitating. it is not deep or difficult, but to find the right way of rewriting the polynomial to demonstrate the result is an interesting coffee-break challenge. | 1151480 | Prove that [imath]a^2 + b^2 \geq 8[/imath] if [imath] x^4 + ax^3 + 2x^2 + bx + 1 = 0 [/imath] has at least one real root.
If it is known that the equation [imath] x^4 + ax^3 + 2x^2 + bx + 1 = 0 [/imath] has a (real) root, prove the inequality [imath] a^2 + b^2 \geq 8. [/imath] I am stuck on this problem, though, it is a very easy problem for my math teacher. Anyway, I can't figure out. |
2542386 | Why is [imath]a=e[/imath] the smallest number such that [imath]a^x\ge 1+x[/imath] for all [imath]x[/imath]?
Calculus book: Find all numbers [imath]a[/imath] such that [imath]\forall x, a^x \ge 1+x[/imath] I immediately thought of the inequality [imath]e^x\ge 1+x[/imath] and guessed that the answer was any number [imath]a[/imath] in [imath][e,\infty)[/imath]. After playing around with a graphing app I can see this is definitely true, although I can't explain why. Why is [imath]e[/imath] specifically the smallest number such that the inequality holds? To my untrained eye, a quick scan of the equation does not give rise to anything involving the constant [imath]e[/imath]. Maybe because [imath]1+x[/imath] is only a tangent to the equation [imath]a^x[/imath] when [imath]a=e[/imath]? Thanks. | 2553471 | For which positive numbers b is it true that [imath]b^x \geq 1+x[/imath] for all [imath]x[/imath]?
For which positive numbers [imath]b[/imath] is it true that [imath]b^x \geq 1+x,\forall x ?[/imath] My calculus teacher gave this question on a test recently, and I’m not sure how to solve it. Any help is appreciated. |
2542735 | Prove that cardinality of [imath]\lvert D \rvert=2^b[/imath] when [imath]D=\{f \mid f:B \to A\} [/imath]
Let [imath]A,B[/imath] sets and [imath]B[/imath] infinite set , [imath]\lvert A\rvert=a , \lvert B\rvert=b[/imath] and [imath]1<a\leq b[/imath], and let [imath]D=\{f \mid f:B \to A\} [/imath]. Prove that [imath]\lvert D\rvert=2^b[/imath]. I know that [imath]\lvert D\rvert=\lvert A\rvert^{\lvert B\rvert}[/imath] and there is a subset of [imath]B[/imath] that has the same cardinality as [imath]A[/imath], but how to prove [imath]\lvert D\rvert=2^b[/imath]? | 2542195 | Question regarding exponentiation of cardinal numbers
Let [imath]2\leq \kappa\leq \gamma[/imath] where [imath]\gamma[/imath] is an infinite cardinal number and [imath]\kappa[/imath] is any cardinal. Prove that [imath]2^\gamma = \kappa^\gamma[/imath]. One direction is obvious but I'm stuck with the other. i.e. showing [imath]\kappa^\gamma\leq 2^\gamma[/imath]. Any hints or solutions will be appreciated. |
2542828 | A question of range.
If [imath]a,b,c,d,e[/imath] are positive reals, such that [imath]a+b+c+d+e=8[/imath] and [imath]a^2+b^2+c^2+d^2+e^2=16[/imath],then the range of [imath]e[/imath] is? I dont even know how to proceed with problems like these. I have an idea that [imath]A.M-G.M[/imath] might be helpful. Is it true? If yes, then how to go about it? | 1698058 | [imath]a,b,c,d,e[/imath] are positive real numbers such that [imath]a+b+c+d+e=8[/imath] and [imath]a^2+b^2+c^2+d^2+e^2=16[/imath], find the range of [imath]e[/imath].
[imath]a,b,c,d,e[/imath] are positive real numbers such that [imath]a+b+c+d+e=8[/imath] and [imath]a^2+b^2+c^2+d^2+e^2=16[/imath], find the range of [imath]e[/imath]. My book tells me to use tchebycheff's inequality [imath]\left(\frac{a+b+c+d}{4}\right)^2\le \frac{a^2+b^2+c^2+d^2}{4}[/imath] But this not the Chebyshev's inequality given in wikipedia. Can someone state the actual name of the inequality so I can read more about it? (I got [imath]e\in\left[0,\frac{16}{5}\right][/imath] using the inequality) |
2543012 | Proving a given inequality [imath]\frac{2}{9}[/imath]
Edit 1: I did not know that the question has already been asked. Sorry for inconvenience. let [imath]x,y \in \mathbb R[/imath] such that [imath]1<x^2-xy+y^2<2[/imath]. Prove that [imath]\frac{2}{9}<x^4+y^4<8[/imath]. This question is duplicate of question that was originally asked by Hatim Blilet but the question was put on hold. But I am still interested in question. [imath]\mathbf{My}[/imath] [imath]\mathbf{approach}[/imath] : I tried proving the inequality by adding the [imath]xy[/imath] term to all sides and squaring all sides of inequality. After that using the maximum value of the quadratic expression obtained on the rightmost part; I got the term 8. But now I have the problem of finding minimum value of leftmost part. The expression I got was [imath] -(x^2y^2-2xy-1)\lt x^4+y^4 \lt -(x^2y^2-4xy-4)[/imath]. Also to square both sides we need to prove that [imath]1\gt xy[/imath] Second approach : I multiplied all sides by 2 to get [imath] 2\lt {(x -y)}^2+x^2+y^2 \lt 4[/imath].but couldn't proceed further with this method. Can somebody please tell me how to solve this problem | 1145506 | How can I prove that [imath]2/9?[/imath]
[imath]x[/imath] and [imath]y[/imath] are real numbers. Given that [imath]1<x^2-xy+y^2<2[/imath], how can I show that [imath]\frac 29<x^4+y^4<8[/imath] ? Then can I use that to prove that for any natural number [imath]n>3[/imath] [imath]x^{2^n}+y^{2^n}>\frac 2{3^{2^n}} \text{?}[/imath] |
2542996 | Chain complexes as functors and chain morphisms as natural transformations
Definition Let [imath]\mathcal{C}[/imath] be any category. A graded object in [imath]\mathcal{C}[/imath] is a sequence [imath](C_n)_{n\in\mathbb{Z}}[/imath] of objects of [imath]\mathcal{C}[/imath] or, equivalently, a functor [imath]C:\mathbb{Z}\to\mathcal{C}[/imath], where the group of integers is considered as a discrete category. A graded morphism [imath]f:A\to B[/imath] of degree [imath]k[/imath] from a graded object [imath]A[/imath] to a graded object [imath]B[/imath] is a collection [imath](f_n:A_n\to B_{n+k})_{n\in\mathbb{Z}}[/imath] of morphisms of [imath]\mathcal{C}[/imath]. Thai is to say, [imath]f[/imath] is a natural transformation from [imath]A[/imath] to [imath]B\circ s^k[/imath], where [imath]s:\mathbb{Z}\to\mathbb{Z}[/imath] sends [imath]n[/imath] to [imath]n+1[/imath]. Of course, a morphism [imath]f:A\to B[/imath] of degree [imath]k[/imath] composes with a morphism [imath]g:B\to C[/imath] of degree [imath]l[/imath] to a morphism [imath]g\circ f:A\to C[/imath] of degree [imath]k+l[/imath]. Graded objects in [imath]\mathcal{C}[/imath] and morphisms between them form a category we will denote by [imath]\operatorname{Gr}(\mathcal{C})[/imath]. Definition Now let [imath]\mathcal{C}[/imath] be a pointed category and let [imath]0[/imath] be the constant graded object [imath](0)_{n\in\mathbb{Z}}[/imath]. We shall also write [imath]0[/imath] for any morphism (of whichever degree) that factors through [imath]0[/imath]. A chain complex [imath](C,d)[/imath] in [imath]\mathcal{C}[/imath] is a graded object [imath]C[/imath], together with a morphism [imath]d:C\to C[/imath] of degree [imath]-1[/imath] (its differential) such that [imath]d\circ d=0[/imath]. A chain morphism [imath]f:(C,d)\to (C',d')[/imath] is a morphism [imath]f:C\to C'[/imath] of degree [imath]0[/imath] such that [imath]d'\circ f=f\circ d'[/imath]. Chain complexes in [imath]\mathcal{C}[/imath] and their chain morphisms form a category we will denote by [imath]\operatorname{Ch}(\mathcal{C})[/imath]. I want to view chain complexes as functors. The only thing I could say was this. Consider [imath]\mathbb{Z}[/imath] as a poset category, where [imath]m\geq n[/imath] means there exists exactly one morphism from [imath]m[/imath] to [imath]n[/imath]. Then consider the functor category [imath]\operatorname{Funct}(\mathbb{Z},\mathcal{C})[/imath]. Then the category [imath]\operatorname{Ch}(\mathcal{C})[/imath] is the full subcategory of [imath]\operatorname{Funct}(\mathbb{Z},\mathcal{C})[/imath] with the property that [imath]d_n\circ d_{n+1}=0[/imath] for each [imath]n[/imath]. First question: is it correct what I said in the last few lines? Other ideas I had, but I couldn't formalize, were: since differentials [imath]d[/imath] of chain complexes are themselves graded morphisms of degree [imath]-1[/imath], maybe I can define a chain complex as a functor on [imath]\mathbb{Z}[/imath] to the category [imath]\operatorname{Gr}(\mathcal{C})[/imath] of graded objects in [imath]\mathcal{C}[/imath]. Here the main problem for me was that, if to an integer [imath]n[/imath] I associate a pair [imath](C,d)[/imath], then I don't know what to associate to each morphism in [imath]\mathbb{Z}[/imath] coming from the order relation [imath]\geq[/imath]. Probably this approach is wrong. Last approach I am trying to formalize was the one consisting in using the arrow category of [imath]\mathcal{C}[/imath] or of [imath]\operatorname{Gr}(\mathcal{C})[/imath], but again I am stuck and don't know how to proceed. For instance, I consider the category [imath]\operatorname{Arr}(\operatorname{Gr(\mathcal{C})})[/imath]. An object here is a graded morphism. Then I take the full subcategory of graded endomorphisms of degree [imath]-1[/imath] with the property that any two of them compose to the zero morphisms. Do you have any suggestion or just comments? | 2139250 | Can the category of chain complexes be realized as a functor category?
As the title says, is there some sort of category [imath]\mathsf{C}[/imath] which can be thought of as the "walking chain complex", so that the category of chain complexes in some other category [imath]\mathsf{A}[/imath] can be realized as [imath]\mathsf{A}^\mathsf{C}[/imath] (or perhaps some subcategory thereof)? |
2543350 | For what orders are there only abelian groups?
It's about that [imath]A_5[/imath] is the smallest non abelian Group. The task is to determine for what orders < 60 there are only abelian groups. My idea was to go through all those orders and use the main theorem about abelian groups. But.. There must be a more elegant solution right? Thanks in advance :) | 1556811 | Does there exist an [imath]n[/imath] such that all groups of order [imath]n[/imath] are Abelian?
I know that all groups of order [imath]\leq[/imath] 5 are Abelian and all groups of prime order are Abelian. Are there any other examples? If so is there something special about the orders of these groups? |
588281 | Does "[imath]\exists \delta >0[/imath] S.T [imath]||x(0)-x_e||<\delta\Rightarrow \displaystyle \lim_{t\rightarrow \infty}||x(t)-x_e||=0[/imath]" imply stability?
Recall the definition of stable and Asymptotically stable: A fixed point [imath]x_e[/imath] of a vector field is called (Lyapunov) stable if [imath]\forall \varepsilon>0,\exists \delta(\varepsilon)[/imath] such that [imath]\forall ||x(0)-x_e||<\delta \Rightarrow ||x(t)-x_e||<\varepsilon,\forall t\geq 0[/imath] A fixed point [imath]x_e[/imath] of a vector field is called asymptotically stable if it's stable and the condition mentioned in the title "[imath]\exists \delta >0[/imath] S.T [imath]||x(0)-x_e||<\delta\Rightarrow \displaystyle \lim_{t\rightarrow \infty}=||x(t)-x_e||=0[/imath]" is true. So, my question is the that if the second condition in asymptotically stable implies stable? If not, what counter example can we take? I guess the second condition doesn't imply stable. "stable" means I start my flow close to the fixed point [imath]x_e[/imath] and it will be always close but not necessarily convergent to [imath]x_e[/imath]. And "asymptotically stable" means, besides the stability, the flow will not only be just close to [imath]x_e[/imath] but also converges to it in the end. Thus, from this point of view, the second condition cannot imply "stable", because there definitely exists a flow that it will finally converge to [imath]x_e[/imath] but it won't necessarily be close to [imath]x_e[/imath] all the time, which means it might detour for a certain period and come back. However, I have hard time finding such flow. I would really appreciate if you can help me find an example of such flow. | 389136 | stability and asymptotic stability: unstable but asymptotically convergent solution of nonlinear system
Consider nonlinear systems of the form [imath]X(t)'=F(X(t))[/imath], where [imath]F[/imath] is smooth (assume [imath]C^\infty[/imath]). Is it possible to construct such a system (preferably planar system) so that [imath]X_0[/imath] is an unstable equilibrium, but all nearby solution curves tend to [imath]X_0[/imath] as [imath]t \to \infty[/imath]? If so, how? A conceptual construction is enough. What will the phase portrait look like? |
2542528 | How to solve [imath]kx=e^x[/imath]
How do I solve equations of the form [imath]kx=e^x[/imath] for [imath]x[/imath]? E.g. how would I solve [imath]3x=e^x[/imath]? I have tried using logs [imath] \begin{align*} 3x&=e^x \\ \ln3x&=x\\ \ln3+\ln{x}&=x\\ x-\ln{x}&=\ln3 \end{align*} [/imath] but it doesn't look like this is going to get me anywhere. | 1699104 | Solving [imath]4x = e^x[/imath] without graphing and looking for intersection
If I want to solve the equation [imath]4x = e^x[/imath], is there a way to solve for [imath]x[/imath] without graphing and looking for intersection? |
2533423 | When does a polynomial interpolated from [imath]n+1[/imath] points have exactly degree [imath]n[/imath]?
It is proven that for any a set of [imath]n+1[/imath] points, there exists a unique polynomial of degree at most [imath]n[/imath] that satisfies the points. When is it the case that for a set of [imath]n+1[/imath] points, the unique polynomial that satisfies the points has exactly degree n? | 1554596 | Degree of interpolation polynomial
Let [imath]n[/imath] be a positive integer, [imath]x_0, x_1,\ldots,x_n[/imath] be [imath](n+1)[/imath] pairwise distinct real numbers, and [imath]y_0, y_1,\ldots,y_n[/imath] be [imath](n+1)[/imath] real numbers. We have know that there exists a unique polynomial [imath]P(x)[/imath] where its degree is not greater than [imath]n[/imath]. What is the necessary and sufficient condition guaranteeing [imath]P(x)[/imath] has a degree of [imath]n[/imath]. Thank you for all comments and helping. |
2543452 | What are the "strong" and "weak" in mathematics?
If [imath]A→B[/imath] is [imath]A[/imath] strong and [imath]B[/imath] weak? For instance, the "strong inequality" [imath]5<6[/imath] implies the "weak inequality" [imath]5≤6[/imath]. Is there anything more to it than that? I've seen the terms "stronger/weaker condition" or "stronger/weaker tests" used a few times. But I'm not sure if what I understand is entirely correct. | 429568 | Meaning of "strong" and "weak" (formulas?) in propositional logic
I was doing some review of propositional logic from Enderton's book. In one section(pg. 26 of the 2nd edition), he explains the idea that given wffs [imath]\sigma_1, \sigma_2, \cdots, \sigma_k[/imath] and [imath]\tau[/imath], one can use truth tables to check whether or not [imath]\{ \sigma_1, \cdots, \sigma_k \} \models \tau[/imath]. Then he gives some examples of how one might in an educated way avoid checking all combinations of assignments of the propositional variables used in the sigmas and in [imath]\tau[/imath]. This part looks o.k to me. But then Enderton says "The stronger the antecedent(the expression on the left side), the weaker the conditional." He then gives the examples \begin{align} (P \wedge Q) &\models P \\ (P \rightarrow R) &\models ((P \wedge Q) \rightarrow R) \\ (((P \wedge Q) \rightarrow R) \rightarrow S) &\models ((P \rightarrow R) \rightarrow S) \end{align} What does he mean by this? Which antecedent is stronger, the first or the third? Maybe a question to start out with is: what do "strong" and "weak" even mean here? Thanks for any help/clarification! Sincerely, Vien |
2544080 | Proof or disproof that for composite [imath]n[/imath], [imath](2^n)-1[/imath] is also composite
While trying to solve a problem my friend asked me, "is [imath](2^{50})-1[/imath] prime?", I found a pattern where if [imath]n[/imath] is composite, [imath](2^n)-1[/imath] is also composite, and in many cases if [imath]n[/imath] is prime, the term is also prime. However this fails at a few prime numbers, the first of which is [imath]11[/imath]. I was able to check for composite results up to [imath]80[/imath] before having to work on other things. I am wondering if anybody can prove or disprove the assertion above. While playing around with the pattern, I also found that for composite [imath]n[/imath], [imath](a^n)-(a-1)[/imath] is also composite. I am less sure about this one, but it is much more exciting if true. | 74285 | Can [imath]x^{n}-1[/imath] be prime if [imath]x[/imath] is not a power of [imath]2[/imath] and [imath]n[/imath] is odd?
Are there any solutions to [imath]x^{n}-1=p[/imath] with p prime, integers [imath]x,n>1[/imath] and [imath]x[/imath] not a power of [imath]2[/imath]? [imath]x[/imath] must be even. [imath]n[/imath] is odd since if [imath]n=2m[/imath] then [imath]p=x^{n}-1=(x^{m}+1)(x^{m}-1)[/imath] hence [imath]p=x^{m}+1[/imath] and [imath]1=x^{m}-1[/imath], which has solution [imath]p=3[/imath] given by [imath]x=2^{2}[/imath]. This generalises; for any solution to [imath]x^{n}-c^{2}=p[/imath], [imath]n[/imath] must be odd. And for any solution to [imath]x^{n}-c=p[/imath], [imath]x[/imath] is odd iff [imath]c[/imath] is odd unless [imath]p=2[/imath] and, if [imath]a[/imath] divides [imath]c[/imath] and [imath]p[/imath] does not divide [imath]c[/imath] then [imath]x[/imath] divides [imath]c[/imath]. |
2544190 | Group Epimorphism from [imath]\mathbb{R}[/imath] to [imath]\mathbb{Q}[/imath]
Is there a group epimorphism [imath]f[/imath] from [imath](\mathbb{R},+)[/imath] to [imath](\mathbb{Q},+)[/imath]? Surely [imath]f(0)[/imath] goes to [imath]0[/imath], but my problem is where to send elements of [imath]\mathbb{R}\smallsetminus\mathbb{Q}[/imath]. | 1637405 | Does there exist a surjective homomorphism from [imath](\mathbb R,+)[/imath] to [imath](\mathbb Q,+)[/imath] ?
Does there exist a surjective homomorphism from [imath](\mathbb R,+)[/imath] to [imath](\mathbb Q,+)[/imath] ? ( I know that there 'is' a 'surjection' , but I don't know whether any surjective homomrophism from [imath]\mathbb R[/imath] to [imath]\mathbb Q[/imath] exist or not . Please help . Thanks in advance ) |
2544375 | Prove that [imath]\left|\left(0,\infty\right)\right|=\left|\mathbb{R}\right|[/imath]
So this is pretty obvious in what it is asking: To prove that anything between [imath]0[/imath] and infinity are part of the positive reals. I feel kind of stupid for asking but I just don't even know where to start. The topic this question comes under is countable sets and bijections, not sure if that helps or not. | 573794 | Prove that [imath]\mathbb{R}[/imath] and the interval [imath](0, \infty)[/imath] have the same cardinality.
Prove that Real Numbers and the interval [imath](0,\infty)[/imath] have the same cardinality. Attempt: Consider the function [imath]f(x) = e^x[/imath]. The domain of this function is all real numbers. The range of this function is from [imath]0[/imath] to infinity. Let [imath]e^a = e^b[/imath]. Then [imath]\ln(e^a) = \ln(e^b)[/imath] Then [imath]a\ln(e) = b\ln(e)[/imath] This means that [imath]a = b[/imath] Hence, [imath]f[/imath] is injective. Let [imath]c > 0[/imath] Then [imath]e^{ln(c)} = c[/imath] Since [imath]c > 0, \ln(c)[/imath] is defined, so [imath]f(\ln(c)) = c[/imath] Therefore, f is surjective. Then f is bijective. Hence, [imath]\mathbb{R}[/imath] and [imath](0, \infty)[/imath] have the same cardinality. |
2544158 | Localization of Hom of finitely presented module
When I'm reading Matsumura's Commutative Algebra, I find the following statement on the 7th page. When [imath]M[/imath] is of finite presentation, i.e. when there is an exact sequence of the form [imath]A^m\to A^n\to M\to 0[/imath], we have also [imath]S^{-1}(Hom_A(M,N))=Hom_{S^{-1}A}(S^{-1}M,S^{-1}N)[/imath]. I think it for hours, but still don't understand why we need the assumption that [imath]M[/imath] need to be of finite presentation. What will happen if we only assume [imath]M[/imath] is finitely generated? (I think it's enough when [imath]M[/imath] is f.g.) | 75812 | Does localisation commute with Hom for finitely-generated modules?
Question. Let [imath]R[/imath] be a ring, [imath]\mathfrak{p}[/imath] a prime, [imath]M[/imath] a finitely-generated [imath]R[/imath]-module, and [imath]N[/imath] any [imath]R[/imath]-module. Is the natural map [imath]\textrm{Hom}_R(M, N)_\mathfrak{p} \to \textrm{Hom}_{R_\mathfrak{p}}(M_\mathfrak{p}, N_\mathfrak{p})[/imath] an isomorphism (of [imath]R_\mathfrak{p}[/imath]-modules)? I can prove this in the case when [imath]M[/imath] is finitely presented: indeed, let [imath]S[/imath] be any flat [imath]R[/imath]-algebra, and let [imath]R^m \to R^n \to M \to 0[/imath] be a right-exact sequence; tensoring with [imath]S[/imath] gives another right-exact sequence [imath]S^m \to S^n \to M \mathbin{\otimes_R} S \to 0[/imath] and applying hom functors, we get left-exact sequences [imath]0 \to \textrm{Hom}_R(M, N) \to N^n \to N^m[/imath] [imath]0 \to \textrm{Hom}_S(M \mathbin{\otimes_R} S, N \mathbin{\otimes_R} S) \to (N \mathbin{\otimes_R} S)^n \to (N \mathbin{\otimes_R} S)^m[/imath] and [imath]S[/imath] is flat, so tensoring the first sequence yields [imath]0 \to \textrm{Hom}_R(M, N) \mathbin{\otimes_R} S \to (N \mathbin{\otimes_R} S)^n \to (N \mathbin{\otimes_R} S)^m[/imath] but extending the sequences by [imath]0[/imath] to the left, and putting in vertical maps between the last two, we conclude that [imath]\textrm{Hom}_R(M, N) \mathbin{\otimes_R} S \cong \textrm{Hom}_S(M \mathbin{\otimes_R} S, N \mathbin{\otimes_R} S)[/imath] by the five lemma (modulo checking commutativity of diagrams). This is essentially the proof Eisenbud gives [Commutative Algebra, Prop. 2.10]. The trouble with extending it to a proof for finitely-generated modules is that we have to replace [imath]R^m[/imath] with a potentially arbitrary submodule [imath]K[/imath] of [imath]R^n[/imath], and that may not be free or even finitely-generated without some additional assumptions on [imath]R[/imath]. I can't see an abstract nonsense proof of the claim, but I admit I haven't tried a bare-hands proof. However, is the claim even true? |
2166351 | Egoroff's Theorem continues to hold if the convergence is point-wise a.e. and [imath]f[/imath] is finite a.e. on [imath]E[/imath].
To show that Egoroff's Theorem continues to hold if the convergence is point-wise a.e. and [imath]f[/imath] is finite a.e. on [imath]E[/imath]. And Egoroff's Theorem states that if [imath]E[/imath] is a set of finite measure and [imath]\{f_n\}[/imath] is a sequence of measurable functions that converges pointwise on [imath]E[/imath] to the real valued function [imath]f[/imath], then there exists a closed set [imath]F[/imath] that is roughly the same size as [imath]E[/imath] on which [imath]\{f_n\}[/imath] converges uniformly. | 2542176 | Relax Egoroff's Theorem to pointwise convergence a.e. and bounded a.e. pointwise limit
The following question is taken from Royden's Real Analysis [imath]4[/imath]th edition, Chapter [imath]3,[/imath] question [imath]28,[/imath] page [imath]67:[/imath] Question: Show that Egoroff's Theorem continues to hold if the convergence is pointwise a.e. and [imath]f[/imath] is finite a.e., that is. Assume [imath]E[/imath] has finite measure. Let [imath]\{f_n\}[/imath] be a sequence of measurable functions on [imath]E[/imath] that converges pointwise almost everywhere on [imath]E[/imath] to the real-valued function [imath]f[/imath] which is finite almost everywhere. Then for each [imath]\varepsilon>0,[/imath] there is a closed set [imath]F[/imath] contained in [imath]E[/imath] for which [imath]\{f_n\}\to f \text{ uniformly on }F \text{ and }m(E\setminus F)<\varepsilon.[/imath] My attempt: Let [imath]A = \{ x\in E: f_n\not\to f \text{ pointwise} \}[/imath] and [imath]B=\{ x\in E: |f|=\infty \}.[/imath] By assumption, [imath]m(A)=m(B)=0.[/imath] So [imath]A[/imath] and [imath]B[/imath] are measurable. Note that [imath]m[E\setminus (A\cup B)] = m(E) - m(A\cup B) = m(E).[/imath] Fix [imath]\varepsilon>0.[/imath] Since [imath]\{f_n\}[/imath] converges to [imath]f[/imath] pointwise on [imath]E\setminus (A\cup B)[/imath] to the real-valued function [imath]f,[/imath] by Egoroff's Theorem, there exists a closed set [imath]F[/imath] contained in [imath](E\setminus (A\cup B))\subseteq E[/imath] for which [imath]f_n\to f \text{ uniformly on } F \text{ and }m[(E\setminus(A\cup B)) \setminus F] < \varepsilon.[/imath] Since [imath]E[/imath] has finite measure and [imath]F\subseteq E,[/imath] by monotonicity, [imath]F[/imath] has finite measure. As [imath]F[/imath] is closed, it is also measurable. Therefore, by Excision property, we have [imath]m(E\setminus F) = m(E) - m(F) = m[E\setminus(A\cup B)] - m(F) = m[(E\setminus(A\cup B)) \setminus F]<\varepsilon.[/imath] Is my proof correct? I ask for verification because in this post Kenny's comment about countable union of null sets. I do not use this anywhere in my proof. So I wonder whether my proof miss out something. |
2543802 | If [imath]A[/imath] and [imath]B[/imath] are [imath]n×n[/imath] matrices, [imath]AB = -BA[/imath] , and [imath]n[/imath] is odd, show that either [imath]A[/imath] or [imath]B[/imath] has no inverse.
If [imath]A[/imath] and [imath]B[/imath] are [imath]n×n[/imath] matrices, [imath]AB = -BA[/imath] , and [imath]n[/imath] is odd, show that either [imath]A[/imath] or [imath]B[/imath] has no inverse. I have no clue how to do this and any help/guidance would be appreciated! Thanks in advance! [imath]det(AB) = det(-BA)[/imath] [imath]det(AB)= det(-B)det(A)[/imath] [imath]det(AB) = (-1)^ndet(BA)[/imath] since n is odd [imath]det(AB) = -det(BA)[/imath] [imath]det(A)det(B) = -det(B)det(A)[/imath] [imath]2det(A)det(B) = 0[/imath] Therefore [imath]det(A)=0[/imath] or [imath]det(B)=0[/imath] | 1049326 | Proof if [imath]AB+BA=0[/imath] Then atleast one of the matrices are singular.
I have this problem : [imath]A,B \in M_n(\Bbb R)[/imath] matrices while [imath]n[/imath] odd number. Proof if [imath]AB+BA=0[/imath] then atleast one of [imath]A,B[/imath] is singular. I assume that [imath]A,B[/imath] invertible. Its clear that [imath]\det(AB) \neq 0[/imath] and [imath]\det(BA) \neq 0[/imath]. If I could show that, [imath]\det(AB) \neq -\det(BA)[/imath] then I can conclude that [imath]\det(AB)+\det(BA) \neq 0[/imath]. But I don't seem to find a way to show that, I guess it has something to do with [imath]n[/imath] odd number. Any ideas? Thanks! |
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