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263586 | Is noetherianity a local property?
Let [imath]R[/imath] be a ring with finitely many maximal ideals such that [imath]R_{\mathfrak m}[/imath] ([imath]\mathfrak m[/imath] maximal ideal) is noetherian ring for all [imath]\mathfrak m[/imath]. Is [imath]R[/imath] noetherian? I think [imath]R[/imath] has to be noetherian. Let [imath]p_1 \subset p_2 \subset \cdots \subset p_n \subset \cdots[/imath] be an infinite ascending chain of prime ideals in [imath]R[/imath], then I claim that there exist a maximal ideal [imath]m[/imath] which will contain all the prime ideals in this chain (from finiteness of the maximal ideals), this will give a chain of prime ideals in [imath]R[/imath] localised at [imath]m[/imath], but since that has to be finite (the local ring [imath]R_m[/imath] is noetherian) so the chain pulled back will terminate in [imath]R[/imath]. | 641925 | Semilocal Rings and Noetherianity
How to prove that a semilocal ring such that all of its localizations at any maximal ideal are Noetherian is a Noetherian ring? For example, for the easier example, [imath]\operatorname{Max}(R)=\{m_1,m_2\}[/imath], and [imath]R_{m_1}[/imath] and [imath]R_{m_2}[/imath] are noetherian rings. I believe that the ring [imath]R[/imath] is Noetherian, but I don't know how to prove it. Thanks in advance. |
556099 | Prove that every odd natural number divides some number of the form [imath]2^n - 1[/imath]
Suppose that [imath]m[/imath] is an odd natural number. Prove that there is a natural number [imath]n[/imath] such that [imath]m[/imath] divides [imath]2^n -1[/imath]. I have absolutely no idea how to tackle this; any assistance would be welcome. | 555107 | pigeonhole principle divisibility proof
Let [imath]n[/imath] be some positive odd number. Prove, in terms of the pigeonhole principle, that there exists some positive integer [imath]k[/imath] such that [imath]n\mid 2^k-1[/imath]. |
556247 | Inverse of a general nonfactorizable multivector
For a n-dimensional geometric algebra over [imath]\mathbb{R}[/imath] which allows mixed signature, if B is a non-null blade the inverse of B with respect to the geometric product will be: [imath]B^{-1} = \frac{B^\dagger}{B B^\dagger}[/imath] This is valid for any non-null blade, including versors, which are an special type of blades/multivectors. Does anybody know of a general expression for the inverse of a multivector with respect to the geometric product that applies also to the multivectors which can not be factorized as versors? Is it at all possible to obtain such inverse? | 443555 | Calculating the inverse of a multivector
Given a multivector, what is the easiest way to compute its inverse? To take a concrete example, consider a bivector [imath] B = e_1(e_2 + e_3) [/imath]. To compute [imath] B^{-1} [/imath], I can use the dual of [imath] B [/imath]: [imath] B = e_1e_2e_3e_3 + e_1e_2e_2e_3 = I(e_3-e_2) = Ib [/imath] [imath] BB^{-1} = 1 = Ib B^{-1} [/imath] [imath] B^{-1} = -b^{-1}I = -\frac{b}{b^2}I[/imath] But this won't work for a bivector in 4-dimensions for example. Is there a more general/easier way? |
315050 | Using Pigeonhole Principle to prove two numbers in a subset of [imath][2n][/imath] divide each other
Let [imath]n[/imath] be greater or equal to [imath]1[/imath], and let [imath]S[/imath] be an [imath](n+1)[/imath]-subset of [imath][2n][/imath]. Prove that there exist two numbers in [imath]S[/imath] such that one divides the other. Any help is appreciated! | 2210146 | If [imath]A\subseteq \{ 1, 2, ... 2n \}[/imath] has [imath]n+1[/imath] elements, then it contains [imath]a[/imath], [imath]b[/imath] such that [imath]a\mid b[/imath]
Let [imath] A [/imath] be subset of of [imath] \{ 1, 2, ... 2n \} [/imath], such that: [imath] Card(A) = n + 1[/imath]. Prove that there exist at least two distinct elements [imath]a,b[/imath] in [imath] A [/imath] such that: [imath] a \mid b [/imath]. We have: [imath] \{ 1, 2, ... 2n \} = \{ 1, 2, ... n \} \cup \{ n+1, n+2, ... 2n \} [/imath] Intuitively, I know that every subset [imath] A [/imath] of [imath] \{ 1, 2, ... 2n \} [/imath] with [imath] n + 1 [/imath] elements contains at least 2 elements [imath] a [/imath] and [imath] b [/imath] such: [imath] b = 2a [/imath]. I don't know how to prove it mathematically. Thank you for your help. |
556302 | Relationship between [imath]\operatorname{ord}(ab), \operatorname{ord}(a)[/imath], and [imath]\operatorname{ord}(b)[/imath]
Another homework problem from my Group Theory class. Let [imath]a,b[/imath] be elements of a group, [imath]G[/imath]. Let [imath]\operatorname{ord}(a)=m[/imath] and [imath]\operatorname{ord}(b)=n[/imath]. Let [imath]a[/imath] and [imath]b[/imath] commute. Prove: If [imath]m[/imath] and [imath]n[/imath] are relatively prime, then the [imath]\operatorname{ord}(ab)=mn[/imath]. So I am having trouble starting off on this problem. Here is what I believe I know: [imath]a^m=e[/imath] and [imath]b^n=e[/imath] [as defined by the order of the elements] which leads [imath]a^m=b^n[/imath] [imath]\operatorname{ord}(ab)=mn=\operatorname{ord}(ba)=nm[/imath] as defined by commuting. [imath]\gcd(m,n)=1[/imath] as defined by [imath]m,n[/imath] being relatively prime. I don't know how to start this proof. Do I want to to go the route of [imath]a^mb^n=e[/imath]? Then [imath](ab)^{mn}=e[/imath]? Any tips to start this would be greatly appreciated!!! | 370004 | If [imath]a[/imath] and [imath]b[/imath] commute and [imath]\text{gcd}\left(\text{ord}(a),\text{ord}(b)\right)=1[/imath], then [imath]\text{ord}(ab)=\text{ord}(a)\text{ord}(b)[/imath].
Prove if [imath]\operatorname{ord}(a)=m[/imath], [imath]\operatorname{ord}(b)=n[/imath], and [imath]\operatorname{gcd}(m,n)=1[/imath], then [imath]\operatorname{ord}(ab)=mn[/imath]. I was reading this and was thinking how this proof would look like. I tried to do it and am not sure if this is correct. Here is what I did: If [imath]a[/imath] and [imath]b[/imath] commute then [imath](ab)^{mn} = a^{mn} * b^{mn} = (a^m)^n * (b^n)^m = 1 * 1 = 1.[/imath] So [imath]ord(ab) | mn[/imath]. Now, take [imath]k = \operatorname{ord}(ab) = m^\prime * n^\prime[/imath] where m' is relatively prime with [imath]n[/imath] and [imath]n'[/imath] is relatively prime with [imath]m[/imath]. By the result above [imath]m' | m[/imath] and [imath]n' | n[/imath]. Now we have [imath]((ab)^k)^{m/m'} = 1[/imath] since [imath](ab)^k=1[/imath]. But on the other hand, by the commutativity: [imath]((ab)^k)^(m/m') =[/imath] [imath]((ab)^(m' * n'))^(m/m') =[/imath] [imath] a^{n'm} * b^{n'm} = [/imath] [imath](a^m)^{n'} * b^{n'm'}=[/imath] [imath]b^{n'm'} = 1[/imath] This implies that [imath]n' * m'[/imath] is divisible by [imath]n[/imath]. But [imath]m'[/imath] is relatively prime with [imath]n[/imath], so we must have [imath]n' = n.[/imath] By symmetry, [imath]m' = m[/imath]. So [imath]ord(ab) = mn[/imath]. Just to say this again, I want to prove if [imath]a[/imath] and [imath]b[/imath] commute and [imath]m[/imath] and [imath]n[/imath] are relatively prime then [imath]ord(ab)=mn.[/imath] The comments are vague. I guess this must be the answer then and there probably is not another way to do this. |
557236 | Show that [imath]y_n=\frac{1}{n}\sum ^n_{k=1}x_k \overset{n}{\rightarrow} x[/imath] if [imath]x_n \overset{n}{\rightarrow} x[/imath]
This seems obvious, but I can't imagine the formal outline of the proof. I just need a start here. Sorry for being a total noob. | 233838 | [imath]\lim_{n\to \infty}\frac{1}{n} \sum_{k=1}^n x_k =x\;[/imath] given [imath]\;\lim_{n\to \infty} x_n= x\;?[/imath]
I have a question which is giving me a hard time. I want to show that [imath] \lim_{n\to \infty}\frac{1}{n} \sum_{k=1}^n x_k =x[/imath] given that [imath]\lim_{n\to \infty} x_n= x[/imath]. |
557246 | Prove that a set is countable
Let [imath]A\subseteq \Bbb R[/imath], A is an infinite set of positive numbers. Suppose, [imath]\exists k \in \Bbb Z[/imath] for all finite [imath]B\subseteq A[/imath]. [imath]\sum_{b\in B}b<k[/imath] Prove [imath]A[/imath] is countable. I have a hint: [imath]A_n=\lbrace a \in A | a>\frac1n\rbrace[/imath] I understand that [imath]B \in \Bbb Q[/imath] and [imath]k[/imath] can be the last element of [imath]A[/imath] but I'm not really sure if that's right or what to do with it. Any help would be appreciated. | 554469 | Show A is countable infinity
One more question about set theory: [imath]A\subseteq R[/imath] is an infinite set of positive numbers. Assume there is a value [imath]k \in Z[/imath] such that for any [imath]B \subseteq A[/imath]: [imath]\sum_{i=0}^\infty b(i) \le k[/imath] where [imath]b \in B[/imath] Show that A is of countable cardinality Hint: Look at the sets [imath]A(n) = \{a\in A | a>\frac{1}{n}\}[/imath] What I tried doing: I understand that when n goes to infinity, A(n) gets closer and closer to A. And if I could show that for all values of n, A(n) is countable, then I could show that the unification: [imath]A(1) \cup A(2) \cup ... \cup A(n) \cup A(n+1) \cup...=A[/imath] is countable. But why can you say that if there exists such a value k, such that the summary of all values of A(n) is less than k, then A(n) is countable? |
557323 | Prove that if Y is a closed subspace of a normed linear space X and x \nin Y, the span of x and Y is closed.
The question is stated in the title, but I don't see how to prove it. Each element of [imath]span\{x,Y\}[/imath] is of the form [imath]cx + y[/imath] where c [imath]\in \mathbb{R}[/imath] and [imath]y \in Y[/imath]. If I could show that for any convergent sequence in span{x,Y} [imath]\{c_nx + y_n\}[/imath], [imath]c_n[/imath] converges to some [imath]c \in \mathbb{R}[/imath] and [imath]y_n[/imath] converges to some [imath]y \in Y[/imath], I would be done. But I'm prevented from doing that because [imath]c_n[/imath] and [imath]y_n[/imath] are interdepedent. Any ideas? | 524675 | Show that [imath]F+G[/imath] is closed when [imath]G[/imath] a closed subspace of normed space [imath]E[/imath] and [imath]F[/imath] a finite dimensional subspace of [imath]E[/imath].
Question: Let [imath]E[/imath] be a normed space. Let [imath]G[/imath] be a closed subspace of [imath]E[/imath] and let [imath]F[/imath] be a finite dimensional subspace of [imath]E[/imath]. Show that [imath]F+G[/imath] is a subspace of [imath]E[/imath] and is closed. I'm having trouble in showing [imath]F+G[/imath] to be closed. I know that [imath]F[/imath] is itself closed and complete, as it is a finite dimensional subspace of a normed space, and that if [imath]F[/imath] were compact that [imath]G+F[/imath] would be closed. I also know that the closed unit ball of any finite dimensional normed space is compact. I tried two methods. One was to take a convergent sequence [imath](x_n)_{n \in \mathbb N}[/imath] in [imath]G+F[/imath]. Then we can write [imath]x_n = f_n + g_n[/imath] where [imath]f_n[/imath] and [imath]g_n[/imath] are sequences in [imath]F[/imath] and [imath]G[/imath] respectively, and I attempted to find a way to force the individual components [imath]f_n[/imath] and [imath]g_n[/imath] inside the unit ball which would enable me to say that they had convergent subsequences. I couldn't see how to do this, however. The other thought was to try and show that [imath]F[/imath] is compact, but I don't see a way to do this as I can't imagine it to be simply true without some other conditions on [imath]F[/imath]. Are one of these methods the right way to go? Or should I go another direction? I would appreciate any help I can get, although I would prefer not to be presented with a full proof so that I can do some work for myself. Thanks! |
557380 | If [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] is continuous, 1-1, and on-to, then f maps Borel Sets to Borel Sets
Let [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] be continuous, on-to, and 1-1. Prove that if [imath]A[/imath] is a Borel set, then [imath]f(A)[/imath] is a Borel set. | 6268 | Does a continuous and 1-1 function map Borel sets to Borel sets?
Suppose [imath]f: \mathbb{R} \to \mathbb{R}[/imath] is a continuous function which is 1-1, then does [imath]f[/imath] map Borel sets onto Borel sets? |
452078 | Prove that [imath]\sqrt 2 + \sqrt 3[/imath] is irrational
I have proved in earlier exercises of this book that [imath]\sqrt 2[/imath] and [imath]\sqrt 3[/imath] are irrational. Then, the sum of two irrational numbers is an irrational number. Thus, [imath]\sqrt 2 + \sqrt 3[/imath] is irrational. My first question is, is this reasoning correct? Secondly, the book wants me to use the fact that if [imath]n[/imath] is an integer that is not a perfect square, then [imath]\sqrt n[/imath] is irrational. This means that [imath]\sqrt 6[/imath] is irrational. How are we to use this fact? Can we reason as follows: [imath]\sqrt 6[/imath] is irrational [imath]\Rightarrow \sqrt{2 \cdot 3}[/imath] is irrational. [imath]\Rightarrow \sqrt 2 \cdot \sqrt 3[/imath] is irrational [imath]\Rightarrow \sqrt 2[/imath] or [imath]\sqrt 3[/imath] or both are irrational. [imath]\Rightarrow \sqrt 2 + \sqrt 3[/imath] is irrational. Is this way of reasoning correct? | 726421 | Proving that [imath]\sqrt{2}+\sqrt{3}[/imath] is irrational
This is from Spivak. Prove that [imath]\sqrt{2}+\sqrt{3}[/imath] is irrational. So far, I have this: If [imath]\sqrt{2}+\sqrt{3}[/imath] is rational, then it can be written as [imath]\frac{p}{q}[/imath] with integral [imath]p, q[/imath] and in lowest terms. [imath]\sqrt{2}+\sqrt{3}=\frac{p}{q}[/imath] [imath]2\sqrt{6}+5 =\frac{p^2}{q^2}[/imath] [imath](2\sqrt{6}+5)q^2=p^2[/imath] And that's about where I get stuck. In a similar question (prove that [imath]\sqrt{2}+\sqrt{6}[/imath] is irrational,) I was able to show that both [imath]p,q[/imath] had to be even which is impossible. I obviously can't apply this trick here. Any hints? |
557872 | prove that the sequence [imath]x_{n+1} = \frac3{4-x_n}[/imath] is converging and find its limit
prove that this sequence is converging and find its limit [imath]x_1 = \frac32[/imath] [imath]x_{n+1} = \frac3{4-x_n}[/imath] i believe that the solution entails proving the sequence is monotone descending and its infimum is 1, but i don't know to to show that. i also tried messing with triangle inequality, but with no success so far any help into solving this will be very appreciated | 554378 | Prove that a given recursion sequence converges
I'm given: [imath]\begin{align*} x_1&=\frac32\\\\ x_{n+1}&=\frac3{4-x_n} \end{align*}[/imath] How do I go about to formally prove the sequence converges and show it? Thanks in advance. |
278555 | Prove that if for every [imath]x \in \mathbb{R}^N[/imath] [imath]Ax=Bx[/imath] then [imath]A=B[/imath]
How can I quickly prove that if for every [imath]x \in \mathbb{R}^N[/imath] [imath]Ax=Bx[/imath] then [imath]A=B[/imath] ? Where [imath]A,B\in \mathbb{R}^{N\times N}[/imath]. Normally I would multiply both sides by inverse of [imath]x[/imath], however vectors have no inverse, so I am not sure how to prove it. | 41958 | If $Ax = Bx$ for all [imath]x \in C^{n}[/imath], then $A = B$.
Let [imath]A[/imath] and [imath]B[/imath] are [imath]$n\times n$[/imath] matrices and [imath]x \in C^{n}[/imath]. If [imath]Ax = Bx[/imath] for all [imath]x[/imath] then [imath]A = B[/imath]. To prove this I have selected [imath]x[/imath] from Euclidean basis B = {[imath]e_{1},e_{2},...,e_{n}[/imath]}. Then [imath]Ae_{i} = Be_{i}[/imath] implies [imath]i^{th}[/imath] column of A = [imath]i^{th}[/imath] column of B for all [imath]1 \leq i \leq n[/imath]. Hence [imath]A = B.[/imath] Is this proof complete? |
560369 | [imath]\mathcal{l}^\infty[/imath] contains all separable normed spaces
Let [imath]S[/imath] be any set. By [imath]\mathcal{l}^{\infty}(S)[/imath] we denote the space of all bounded functions [imath]S\longrightarrow \mathbb{C}[/imath] endowed with the [imath]\sup[/imath]-norm. It can be shown that each normed space [imath]X[/imath] can be isometrically embedded into [imath]\mathcal{l}^{\infty}(S)[/imath] for some [imath]S[/imath]. Really, if [imath]X[/imath] is some normed space, consider the set [imath]S[/imath] of all bounded linear functionals [imath]\{T: X\longrightarrow \mathbb{C}: \|T\|=1\}[/imath] whose norm is [imath]1[/imath]. Now we build an embedding [imath]F[/imath] as follows: [imath]F(x): T\mapsto Tx\in \mathbb{C}[/imath]. It is easy to see that [imath]F[/imath] is linear, that [imath]T(x)[/imath] is a bounded function [imath]S\longrightarrow\mathbb{C}[/imath], and that [imath]F[/imath] respects the norm (i.e. [imath]\|x\|=\sup\{|Tx|: T\in X^{*}\text{ and }\|T\|=1\}[/imath]). Well, how to prove that each separable space can be isometrically embedded into [imath]\mathcal{l}^\infty =\mathcal{l}^\infty(\mathbb{N})[/imath]? We need something the same, really? | 112619 | Isometric Embedding of a separable Banach Space into [imath]\ell^{\infty}[/imath]
The problem is: Let [imath]X[/imath] be a separable Banach space then there is an isometric embedding from [imath]X[/imath] to [imath]\ell^{\infty}[/imath]. My efforts: I showed that there is an isometry from [imath]X^*[/imath] (topological dual) to [imath]\ell^\infty[/imath] in the following way: Let [imath](e_{i})_{i=1}^{\infty}[/imath] be a dense sequence in [imath]B_{X}[/imath] then define [imath]\Phi:X^*\rightarrow\ell^\infty[/imath] by [imath]\Phi(f)=(f(e_{i}))_{i=1}^\infty[/imath]. It is clear that [imath]\Phi[/imath] is an isometry. An initial idea and a secondary question Is there any canonical isometry from [imath]X[/imath] to [imath]X^*[/imath] since [imath]X[/imath] is separable (or not)? |
553603 | Rolle's Theorem [imath]x^3 - 3x +b[/imath]
Use Rolle's Theorem to prove that the equation [imath]x^3 - 3x + b = 0[/imath] has at most one root in the interval [imath][-1,1][/imath]. Rolle's Theorem: Suppose [imath]f[/imath] is a continuous real-valued function on [imath][a,b][/imath] with [imath]f(a) = f(b)[/imath], and that [imath]f[/imath] is differentiable on [imath](a,b)[/imath]. Then there exists [imath]c[/imath] in [imath](a,b)[/imath] such that [imath]f'(c) = 0[/imath]. Im not exactly sure what to do because [imath]f(1) = -2 + b[/imath] and [imath]f(-1) = 2 + b[/imath], unless we are allowed to have different [imath]b[/imath] values. | 556159 | Rolle's Theorem Contradiction
Rolle's Theorem [imath]x^3 - 3x +b[/imath] Use Rolle's Theorem to prove that the equation [imath]x^3 - 3x + b = 0[/imath] has at most one root in the interval [imath][-1,1][/imath]. Rolle's Theorem : Suppose f is a continuous real-valued function on [imath][a,b][/imath] with [imath]f(a) = f(b)[/imath], and that f is differentiable on [imath](a,b)[/imath]. Then there exists c in [imath](a,b)[/imath] such that [imath]f'(c) = 0[/imath]. Here is my proof: Suppose there are 2 roots in [imath][-1,1][/imath]. [imath]f'(x) = 3(x^2-1)[/imath], which is equal to zero at the endpoints -1 and 1. This results in a contradiction because there are are no roots in the interval [imath](-1,1)[/imath]. Therefore the original hypothesis that there are two roots in [imath][-1,1][/imath] is false, implying that there is at most 1 root. I know that is incorrect because I never used the hypothesis, but I think its the right idea. Help would be more than appreciated. |
561780 | Math Analysis (differentiability)
Please I need help for this exercise : Let [imath]f : \mathbb R → \mathbb R[/imath], and suppose that [imath]|f(x)−f(y)| \leq (x−y)^2[/imath] for all [imath]x, y[/imath] ∈ [imath]\mathbb R[/imath]. Prove that [imath]f[/imath] is constant. First if [imath]a<b[/imath] does [imath]lim a < lim b[/imath]? If that is correct I can prove that lim of[imath] |f(x)−f(y)|[/imath] as [imath]x \rightarrow y[/imath] is [imath]0[/imath] by sandwich theorem and hence [imath] f[/imath] is constant. But this is a chapter about differentiability I think I should proof that [imath]f'(0)=0[/imath] but I don't know how. What do you think? | 537216 | Proving a function is constant, under certain conditions?
The problem: Assume [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] satisfies [imath]|f(t) - f(x)| \leq |t - x|^2[/imath] for all [imath]t, x[/imath]. Prove [imath]f[/imath] is constant. I believe I have some intuition about why this is the case; i.e. if [imath]t[/imath] and [imath]x[/imath] are very close, ([imath]|t - x| = \epsilon[/imath]), then [imath]\sqrt{\epsilon}[/imath] will converge to 0, so [imath]|f(t) - f(x)|[/imath] will also converge to zero if you keep taking [imath]t[/imath] and [imath]x[/imath] closer and closer to each other. However, how do I formalize this argument? Thanks. |
561305 | Question on relative homology
i have that [imath]H_p(X,Y)[/imath] is isomorphic to [imath]Z_p(X,Y)/(B_p(X)+C_p(Y))[/imath], where [imath]Z_p(X,Y)=\lbrace \sigma\in C_p(X), \partial\sigma\in C_{p-1}(Y)\rbrace[/imath] and i want to deduce that [imath]H_0(X,Y)[/imath] is the free module generated by the path connected components of [imath]X[/imath] that do not contain points of [imath]Y[/imath] but i don't know how to prove it ? can someone help me ? Thank you | 551575 | Question on relative homology
I have this question and I'd like an idea to solve it: If [imath]Z_p(X,Y)=\lbrace \sigma\in C_p(X), \partial\sigma\in C_{p-1}(Y)\rbrace[/imath], [imath]1)[/imath]prove that [imath]H_p(X,Y)[/imath] is isomorphic to [imath]Z_p(X,Y)/(B_p(X)+C_p(Y))[/imath], [imath]2)[/imath] deduce that [imath]H_0(X,Y)[/imath] is the free module generated by the path connected components of [imath]X[/imath] that do not contain points of [imath]Y[/imath] [imath]Z_p(X,Y)=\ker(\partial_p: C_p(X,Y)\rightarrow C_{p-1}(X,Y))[/imath] [imath]B_p(X,Y)=Im(\partial_{p+1}:C_{p+1}(X,Y)\rightarrow C_p(X,Y))[/imath] [imath]C_p(X,Y)=C_p(X)/C_p(Y)[/imath] [imath]H_p(X,Y)=Z_p(X,Y)/B_p(X,Y)[/imath] Please help me Thank you. |
561791 | [imath]D := U(z_0,r)\setminus \bigcup_{n=1}^{\infty} \{z_n\}[/imath], [imath]z_n[/imath] poles, prove [imath]f(D)[/imath] is dense in [imath]\mathbb{C}[/imath].
I'm looking at the following complex analysis problem. Suppose that [imath]f[/imath] is analytic in [imath]D := U(z_0,r)\setminus \bigcup_{n=1}^\infty \{z_n\}[/imath], all the points [imath]z_n[/imath] are poles, and [imath]z_n \to z_0[/imath]. Prove [imath]f(D)[/imath] is dense in [imath]\mathbb{C}[/imath]. I feel like I have to define some function and manipulate it somehow, but I haven't been able to come up with a function that works. Can you help? | 135458 | Limit point of poles is essential singularity? Am I speaking nonsense?
The following is exercise 15 in section V.1 of Conway's Functions of One Complex Variable ("Classification of Singularities"). I'm currently studying for a complex analysis qualifying exam and this has appeared in the past. Let [imath]f[/imath] be analytic in [imath]G=\{z:0<|z-a|<r\}[/imath] except that there is a sequence of poles [imath]\{a_n\}[/imath] in [imath]G[/imath] with [imath]a_n\rightarrow a[/imath]. Show that for any [imath]w[/imath] in [imath]\mathbb{C}[/imath] there is a sequence [imath]\{z_n\}[/imath] in [imath]G[/imath] with [imath]a=\lim z_n[/imath] and [imath]w=\lim f(z)[/imath]. The conclusion makes me want to apply the Casorati-Weirstrass theorem. However, the singularity at [imath]a[/imath] is not isolated. As far as I know, an essential singularity is a particular type of isolated singularity. Am I wrong about this? Any help would be greatly appreciated. |
433581 | Let [imath]p\colon X \to Y[/imath] be a quotient map then if each set [imath]p^{-1}({y})[/imath] is connected, and if [imath]Y[/imath] is connected, then [imath]X[/imath] is connected.
Let [imath]p\colon X \to Y[/imath] be a quotient map. Show that if each set [imath]p^{-1}({y})[/imath] is connected, and if [imath]Y[/imath] is connected, then [imath]X[/imath] is connected. I am totally stuck on this problem. Can I get some help how to tackle this problem | 302059 | How to prove this result involving the quotient maps and connectedness?
Given topological spaces [imath]X[/imath] and [imath]Y[/imath], where [imath]Y[/imath] is connected, let [imath]p \colon X \to Y[/imath] be a quotient map. If, for each point [imath]y \in Y[/imath], the set [imath]p^{-1}(\{y\})[/imath] is connected, then how to prove that [imath]X[/imath] is connected also? By the map [imath]p[/imath] being a quotient map is meant the following: The map [imath]p[/imath] is surjective, and, for any subset [imath]V[/imath] of [imath]Y[/imath], the set [imath]V[/imath] is open in [imath]Y[/imath] if and only if [imath]p^{-1}(V)[/imath] is open in [imath]X[/imath]. |
562055 | Let [imath]H[/imath] and [imath]K[/imath] be normal subgroups of [imath]G[/imath] such that [imath]H \cap K = \langle e \rangle[/imath] and [imath]HK = G[/imath]. Prove that [imath]G[/imath] is isomorphic to [imath]H \times K[/imath].
I can't figure out how to do this. Can anyone give me some hints? | 363036 | [imath]G[/imath] a group and [imath]H, K\mathrel{\unlhd}G[/imath]. Assuming that [imath]H \cap K = \{1_G\}[/imath] and [imath]G = \langle H, K \rangle[/imath], prove that [imath]G \cong H \times K[/imath]
I am trying to prove the following statement: Let [imath]G[/imath] be a group and [imath]H, K\mathrel{\unlhd}G[/imath]. Assuming that [imath]H \cap K = \{1_G\}[/imath] and [imath]G = \langle H, K \rangle[/imath], prove that [imath]G \cong H \times K[/imath]. From this, we get that [imath]HK \cong H \times K[/imath]. Naturally, it would be ideal if I could prove that [imath]HK \cong G[/imath]. One issue that I am facing here is the meaning of [imath]G = \langle H, K \rangle[/imath]. How is this defined (I mean, more explicitly)? Is my strategy correct or |
563062 | If [imath]\lim f(x)[/imath] exists and [imath]\lim g(x)[/imath] do not, when [imath]x[/imath] approaches [imath]a[/imath] , why [imath]\lim[f(x)+g(x)][/imath] does not exist?
If [imath]\lim f(x)[/imath] exists and [imath]\lim g(x)[/imath] do not, when [imath]x[/imath] approaches [imath]a[/imath], why [imath]\lim[f(x)+g(x)][/imath] does not exist? need some help over here, to prove the above statement | 1459559 | Limit of the Sum
If [imath] \lim_{x \to x_1}(f(x))[/imath] exists but [imath]\lim_{x \to x_1}(g(x))[/imath] doesn't, then can we say whether or not [imath]\lim_{x \to x_1}(f(x)+g(x))[/imath] exists? |
562960 | Prove that [imath]\int_a^bf(x)dx+ \int_{f(a)}^{f(b)} f^{-1}(y)dy=bf(b)-af(a)[/imath]
Let f' be continuous and positive on [a,b]. Prove that: [imath]\int_a^bf(x)dx+ \int_{f(a)}^{f(b)} f^{-1}(y)dy=bf(b)-af(a)[/imath] I've got the later steps of this covered using integration by parts, and I know that I need to use substitution on the integral containing the inverse. My question is: how do I set up the substitution on [imath]\int_{f(a)}^{f(b)} f^{-1}(y)dy[/imath] to get it in terms of f(x)? | 215650 | Integrating the inverse of a function.
[imath]f(x)[/imath] is a strictly increasing, and continuous, function on [imath][0,+\infty)[/imath]. Consider the integral, [imath]\int_0^a f^{-1}(x) dx.[/imath] Reasoning The interval [imath][0,a][/imath] is "reflected" across the line [imath]y=x[/imath] onto the interval [imath]0 \leq y \leq a.[/imath] [imath]f(x)[/imath] and [imath]f^{-1}(x)[/imath] are symmetric across the line [imath]y=x[/imath]. So, the integral is equivalent to the area bounded by [imath]y = 0[/imath], [imath]y=a[/imath], and [imath]f(x)[/imath]. So, "integrating [imath]f[/imath] along an interval on the [imath]y[/imath]-axis" is the same as integrating [imath]f^{-1}[/imath] along the same interval on the [imath]x[/imath]-axis? Question Is my reasoning correct? |
563748 | Prove that if [imath]3|(a^2+b^2)[/imath], then [imath]3|a[/imath] and [imath]3|b[/imath], where [imath]a, b[/imath] are integers
I would like to know how to prove the above statement by contradition. Somebody said that one should prove it by this method but I have no idea what it is. | 25301 | If [imath]3[/imath] divides [imath]a^2 + b^2[/imath], then [imath]3[/imath] divides [imath]a[/imath] and [imath]3[/imath] divides [imath]b[/imath]
This is a part from a homework. I solved all examples apart from this one. So the task is: We know that [imath]3[/imath] divides [imath]a^2 + b^2[/imath]. Prove that [imath]3[/imath] divides [imath]a[/imath] and [imath]3[/imath] divides [imath]b[/imath]. I cannot think of anything useful. I know that [imath]a^2 + b^2 = (a + b)^2 - 2ab[/imath], but I don't see how it can help me :( Best regards, Petar |
564442 | Finding a matrix where the column space is a subset of the null space
Let [imath]A_{3x3}[/imath] be a matrix [imath] \ne 0[/imath] such that the column space of [imath]A[/imath] is a subset of the null space of [imath]A[/imath]. I need to find [imath]A[/imath]. Here's my process so far: let [imath]v_1, v_2, v_3[/imath] be the column vectors of [imath]A[/imath] [imath]Col(A)=c_1 v_1 + c_2 v_2 + c_3 v_3[/imath] is a subset of [imath]Null(A)[/imath] let [imath]b[/imath] be the column space of A such that [imath]Ax=b[/imath], then assume [imath]b[/imath] is also in the null space of [imath]A[/imath] so [imath]Ax=0=A(Ab)[/imath] How do I go about finding a particular matrix that satisfies this? Am I even heading on the right direction? | 513347 | Can a matrix have a null space that is equal to its column space?
I had a question in a recent assignment that asked if a [imath]3\times 3[/imath] matrix could have a null space equal to its column space... clearly no, by the rank+nullity theorem... but I have a hard time wrapping my head around the concept of such a matrix, no matter what size, even existing. How could this be possible, and does anybody have an example of such a [imath]m\times n[/imath] matrix? |
564003 | If f is differentiable then f is uniformly continuous?
Assume [imath]f:\mathbb{R} \to \mathbb{R}[/imath] is a differentiable function such that [imath]\left|f '(x) \right |< M[/imath] for all [imath]x \in \Bbb R[/imath]. Show that [imath]f[/imath] is uniformly continuous. I have no idea on how to approach this question, can someone help me? | 377531 | Prove that a function whose derivative is bounded is uniformly continuous.
Suppose that [imath]f[/imath] is a real-valued function on [imath]\Bbb R[/imath] whose derivative exists at each point and is bounded. Prove that [imath]f[/imath] is uniformly continuous. |
564380 | Show [imath]\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1[/imath]
I have a homework problem which asks me to show [imath] \lim\limits_{x \to 0}\frac{e^x-1}{x} = 1 [/imath] Any help is appreciated. | 152605 | Proving that [imath]\lim\limits_{x \to 0}\frac{e^x-1}{x} = 1[/imath]
I was messing around with the definition of the derivative, trying to work out the formulas for the common functions using limits. I hit a roadblock, however, while trying to find the derivative of [imath]e^x[/imath]. The process went something like this: [imath]\begin{align} (e^x)' &= \lim_{h \to 0} \frac{e^{x+h}-e^x}{h} \\ &= \lim_{h \to 0} \frac{e^xe^h-e^x}{h} \\ &= \lim_{h \to 0} e^x\frac{e^{h}-1}{h} \\ &= e^x \lim_{h \to 0}\frac{e^h-1}{h} \end{align} [/imath] I can show that [imath]\lim_{h\to 0} \frac{e^h-1}{h} = 1[/imath] using L'Hôpital's, but it kind of defeats the purpose of working out the derivative, so I want to prove it in some other way. I've been trying, but I can't work anything out. Could someone give a hint? |
563944 | Prove that there is a y that [imath]f(y) = [f(x_1)+f(x_2)+ \cdots+ f(x_n) ]/ n[/imath]
let [imath]f : [a,b] \to \Bbb R[/imath] a continuous function and [imath]x_1,x_2,x_3,\cdots,x_n\in[a,b][/imath] I need to prove/show that there is a [imath]y \in [a,b][/imath] that: [imath]f(y) = \frac{f(x_1)+f(x_2)+ .... + f(x_n)}{n}[/imath] Any ideas about how to do it? | 554056 | Prove there is a point [imath]z\in[a,b][/imath] at which [imath]f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}[/imath]
Suppose that the function [imath]f:[a,b]\rightarrow \mathbb{R}[/imath] is continuous. For a natural number [imath]k[/imath], let [imath]x_1,\cdots,x_k[/imath] be points in [imath][a,b][/imath]. Prove there is a point [imath]z\in[a,b][/imath] at which [imath]f(z)=\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}[/imath]. This can be proved by showing [imath]\hspace{130pt}f(a)<\frac{f(x_1)+f(x_2)+\cdots+f(x_k)}{k}<f(b)[/imath] and using the intermediate value theorem to say there exists a point [imath]z\in [a,b][/imath] such that the statement is true. I'm not sure how to show this though. Any suggestions? |
564740 | What does the symbol [imath]\subset\subset[/imath] mean?
In some texts (mainly complex analysis or harmonic analysis) I sometimes see the following double subset symbol [imath]\subset\subset[/imath] for inclusion relation of two regions, e.g., [imath]\Omega[/imath] and [imath]\Omega'[/imath] are two regions in [imath]\mathbb{C}[/imath] such that [imath]\Omega \subset\subset \Omega'[/imath]. I never figured out what it means exactly; I always interpreted it as the closure [imath]\overline{\Omega}[/imath] is contained in [imath]\Omega'[/imath] (so that some nasty boundary effects can be avoided). Is that right? Or does [imath]\subset\subset[/imath] mean some other kind of inclusion? Thanks. | 56872 | The meaning of notation [imath]\subset\subset[/imath] in complex analysis
I have read the book Function Theory of Several Complex Variables of Krantz. But there is a notation the meaning of which I don't know. The notation is [imath]\subset\subset[/imath]. For example, "let [imath]\Omega\subset\subset \mathbb{R}^{n}[/imath] be a connected open set". Can somebody give me a definition? |
564889 | Given positive integers [imath]m[/imath] and [imath]n[/imath], if [imath]m | n[/imath] then [imath]2^m -1 | 2^n -1[/imath]
I've been struggling with this problem for a while now: Prove that, given positive integers [imath]m[/imath] and [imath]n[/imath], if [imath]m | n[/imath] then [imath]2^m-1 | 2^n-1[/imath]. I can't seem to get any traction with this problem. Most of my approaches to the problem depend on knowing something about [imath]\gcd(m,n)[/imath], and all I know from the problem is that [imath]\gcd(m,n) \neq 1[/imath] when [imath]m,n \neq 1[/imath]. Can anyone give me a hint or suggest an approach I'm not seeing? Thanks | 529111 | Why if [imath]n \mid m[/imath], then [imath](a^n-1) \mid (a^m-1)[/imath]?
My Number Theory book says that for [imath]n, m[/imath] be positive integers and [imath]a>1[/imath], then [imath](a^n -1)\mid(a^m -1)[/imath] if and only if [imath]n\mid m[/imath]. I understand the proof for only if part, but in if part the autor says "it is clear". However a tried to prove that but a get stuck. Can you give a hint? |
565163 | [imath]\displaystyle 3^x+4^x=5^x[/imath]
Show that the equation [imath]\displaystyle 3^x+4^x=5^x[/imath] has exactly one root I proved it graphically but I am in need of an analytic solution | 2840050 | How do I solve the exponential equation?
I have this equation: [imath]3^x + 4^x = 5^x[/imath], according to fermat theorem, [imath]x <= 2[/imath], so the answer is [imath]2[/imath]. But how can I get to the result through an elementary algebraic procedure? I have tried many things, but I have not come up with any concrete results, arriving only at: First try, [imath]log(3^x+4^x) = 3log5[/imath] Second try, [imath](\frac{3}{5})^x + (\frac{4}{5})^x = 1[/imath] I have also tried to search several programs, such as symbolab, mathway, etc. But, they have not been able to solve it, the only program that has given me an answer is wolframAlpha, but I can not visualize it step by step, thanks in advance. |
565173 | [imath]G[/imath] satisfying [imath](ab)^n=a^nb^n[/imath]
Let [imath]G[/imath] be a group in which, for some integer [imath]n>1[/imath], [imath](ab)^n=a^nb^n[/imath] for all [imath]a,b\in G[/imath]. Show that [imath]G^{(n)}=\{x^n|x\in G\}[/imath] is a normal subgroup of [imath]G[/imath]. [imath]G^{(n-1)}=\{x^{n-1}|x\in G\}[/imath] is a normal subgroup of [imath]G[/imath]. [imath]a^{n-1}b^n=b^na^{n-1}[/imath] for all [imath]a,b\in G[/imath]. [imath]\left( aba^{-1}b^{-1}\right)^{n(n-1)}=e[/imath] for all [imath]a,b\in G[/imath]. | 107089 | problems in group theory
I. Let [imath]G[/imath] be a group in which, for some integer [imath]n\gt 1[/imath], [imath](ab)^n=a^n b^n[/imath] for all [imath]a,b\in G[/imath]. Show that [imath]G^{(n)}=\{x^{n}|x\in G\}[/imath] is a normal subgroup of [imath]G[/imath]. [imath]G^{(n−1)}=\{x^{n−1}|x\in G\}[/imath] is a normal subgroup of [imath]G[/imath]. II. Let [imath]G[/imath] be as in the problem above. Show that [imath]a^{n−1}b^n=b^n a^{n−1}[/imath] for all [imath]a,b\in G[/imath]. [imath](aba^{−1}b^{−1})^{n(n−1)}=e[/imath] for all [imath]a,b\in G[/imath]. |
564136 | Relations between matrices and linear transformations, diagonalization.
This question bothered me for a while and hopefully someone can shed some light on the issue. A matrix [imath]A[/imath] is said to be diagonalizable if there is an invertible matrix [imath]P[/imath] and a diagonal matrix [imath]D[/imath] such that [imath]A=P^{-1}DP[/imath]. That is the definition of a matrix being diagonalizable. Now, a linear transformation is said to be diagonalizable if there exists a basis [imath]C[/imath] of eigenvectors. Why is that the same thing? Let's say that [imath]A[/imath] is diagonalizable, I define a linear mapping: \begin{gather*} T\colon V \to V \\ T(v)=Av \end{gather*} Why does this say that [imath]T[/imath] is diagonalizable? Why does this mean that there is a basis of eigenvectors? And also the other way around: let's say that the transformation [imath]T[/imath] is diagonalizable. Why does this mean that [imath]A[/imath] is diagonalizable? | 475763 | A basic question on diagonalizability of a matrix
I am following a book where the "diagonalizability" has been introduced as follows: Consider a basis formed by a linearly independent set of eigen vectors [imath]\{v_1,v_2,\dots,v_n\}[/imath]. Then it is claimed that with respect to this basis, the matrix [imath]A[/imath] is diagonal. I am confused at the word "basis" here. In some other books it is said that a matrix [imath]A[/imath] is called diagonalizable if there exists matrix [imath]P[/imath] such that [imath]P^{-1}AP[/imath] is a diagonal matrix. I don't think that diagonalizability has anything to do with basis. It just happened that the set of eigen vectors form a basis of [imath]\Bbb R^n[/imath]. Also I don't think having a set of [imath]n[/imath] linearly independent eigen vectors is a necessary condition for a matrix to be diagonalizable. |
565276 | If [imath]G[/imath] is a finite group then ...
Let [imath]G[/imath] be a finite group. Let [imath]I(G)[/imath] be the set of elements of [imath]G[/imath] that have order 2. Suppose [imath]|I(G)| \ge \frac 3 4 |G|[/imath]. Let [imath]x \in I(G)[/imath]. We note [imath]I_x(G)[/imath] the subset [imath]\{xg \in I(G)\}_{g \in I(G)}[/imath]. Show that [imath]|I_x(G)| \ge \frac 1 2 |G|[/imath]. I have shown that [imath]I_x(G) \subset C_G(x)[/imath] but I have no idea how to continue from there.. Can somebody please give me a hint (I'm not asking for the answer, I'd like to find that myself). | 277127 | Involutions and Abelian Groups
Suppose that [imath] G [/imath] is a finite group where at least three-fourths of the elements are involutions, i.e., [imath] |I(G)| \geq \frac{3}{4} |G|. [/imath] (Here, [imath] I(G) [/imath] denotes the set of all involutions of [imath] G [/imath], where an involution of [imath] G [/imath] is defined as a group element of order [imath] 2 [/imath].) (1) Suppose that [imath] A [/imath] and [imath] B [/imath] are subsets of a finite set [imath] S [/imath]. Show that [imath] |A \cap B| \geq |A| + |B| - |S| [/imath]. (2) Let [imath] x \in I(G) [/imath]. Show that [imath] |I(G) \cap x[I(G)]| \geq \dfrac{1}{2} |G| [/imath]. (3) Let [imath] x \in I(G) [/imath]. Show that [imath] \{ 1_{G} \} \cup \{ g \in I(G) ~|~ xg \in I(G) \} \subseteq {C_{G}}(x) [/imath]. (4) Use your answers to (1)-(3) to conclude that [imath] G [/imath] is an abelian group. |
563840 | Why is [imath]2^n[/imath] the maximum number of subsets of a set of size [imath]n[/imath]?
There is a set with [imath]n[/imath] elements. Why is the maximum number of subsets that can be formed out of it [imath]2^n[/imath]? | 546414 | What is the proof that the total number of subsets of a set is [imath]2^n[/imath]?
What is the proof that given a set of [imath]n[/imath] elements there are [imath]2^n[/imath] possible subsets (including the empty-set and the original set). |
566291 | Finding the value of [imath]\sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}}{n}[/imath]
Does anyone know how to find the exact sum of [imath] \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1}}{n} [/imath] I've only taken second semester calculus and don't see how to go about computing this sum. The only way that I know how to find the sum of an infinite series is if it is a geometric series. Using WolframAlpha, I found that the sum for this series is [imath]\log(2)[/imath]. | 716 | Sum of the alternating harmonic series [imath]\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots [/imath]
I know that the harmonic series [imath]\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}[/imath] diverges, but what about the alternating harmonic series [imath]\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots + \frac{(-1)^{n+1}}{n} + \cdots \text{?} \tag{II}[/imath] Does it converge? If so, what is its sum? |
555742 | Generalizations of the quadratic formula
The quadratic formula can be used to find the roots of any quadratic polynomial of the form [imath]ax^2 + bx + c[/imath]: [imath]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/imath] The derivation is simple enough and uses a technique called completing the square. Is there a formula to solve cubic equations of the form: [imath]ax^3 + bx^2 + cx + d[/imath] Or a general formula to solve an [imath]n[/imath]th degree polynomial? EDIT: According to the answers below, and some mathematicians, such a formula isn't possible. But why is that? I mean, consider an [imath]n[/imath]th degree polynomial [imath]p(x)[/imath] with roots [imath]r_1, r_2, \dots, r_n[/imath]: [imath]p(x) = ax^n - \left(\sum r_i\right)ax^{n-1} + \left(\sum_{i, j} r_ir_j\right)ax^{n-2} + \dots+(-1)^n(r_1r_2\dots r_n)[/imath] We have [imath]n[/imath] equations with [imath]n[/imath] unknowns, surely this can be solved using a general algorithm? | 243063 | Solutions of the equations of the degree 5 and more
I know that there is a precise solution of the equations: \begin{align} x + a & = 0 \\ x^2 + a\cdot x + b & = 0 \\ x^3 + a\cdot x^2 + b\cdot x + c & = 0 \\ x^4 + a\cdot x^3 + b\cdot x^2 + c\cdot x + d & = 0 \end{align} Why there is no solutions for equation of degree [imath]\ge 5[/imath]? |
566478 | Proving [imath]\kappa^{\lambda} = |\{X: X \subseteq \kappa, |X|=\lambda\}|[/imath]
Let [imath]\kappa , \lambda[/imath] be cardinals with [imath]\omega \leq \lambda \leq \kappa.[/imath] Prove [imath]\kappa^{\lambda} = |\{X: X \subseteq \kappa, |X|=\lambda\}|[/imath]. i.e could anyone advise me on how to construct a bijection between [imath]\{f \mid \text{[/imath]f[imath] is a function},~{\rm dom}(f) = \lambda,~ {\rm ran}(f) \subseteq \kappa\}[/imath] and [imath]\{X: X \subseteq \kappa, |X|=\lambda\}[/imath]? Thank you. | 311848 | Proving that for infinite [imath]\kappa[/imath], [imath]|[\kappa]^\lambda|=\kappa^\lambda[/imath]
Initially assume ZFC. Let [imath]\binom{\kappa}{\lambda}[/imath] denotes [imath]\left|[\kappa]^{\lambda}\right|[/imath] where [imath][\kappa]^{\lambda}[/imath] is the collection of all subsets of [imath]\kappa[/imath] with cardinality [imath]\lambda[/imath]. That is, the number of subsets of [imath]\kappa[/imath] of size [imath]\lambda[/imath]. Let [imath]\kappa[/imath] be any infinite cardinal number. Then it is easy to see that [imath]\binom{\kappa}0=1[/imath] [imath]\binom{\kappa}n=\kappa[/imath] for all positive natural number [imath]n[/imath]. [imath]\binom{\aleph_0}{\aleph_0}=\beth_1[/imath] [imath]\binom{\kappa}{\lambda}\le \kappa^\lambda[/imath] when [imath]1\le \lambda \le \kappa[/imath] 3 is due to [imath]\beth_1=\sum_{\lambda=0}^{\aleph_0}\binom{\aleph_0}{\lambda}[/imath] and others besides [imath]\binom{\aleph_0}{\aleph_0}[/imath] sums [imath]\aleph_0[/imath](since countable infinite countable infinites sums to countable infinite) [imath]\left({}^{\kappa}_{\lambda}\right)\le \kappa^\lambda[/imath] is due to an injection can be made from [imath][\kappa]^{\lambda}[/imath] into [imath]\kappa^{\lambda}[/imath](maps [imath]\{a_i|i<\lambda\}[/imath] to [imath]\langle a_i\rangle_{i<\lambda}[/imath] where [imath]a_j<a_k[/imath] iff [imath]j<k[/imath]) I hypothesize that [imath]\binom{\kappa}{\lambda}=\begin{cases} \kappa^\lambda,&\lambda \le \kappa \\ 0,& \lambda>\kappa \end{cases}[/imath] It holds when [imath]\kappa=\aleph_0[/imath], but hard to induct for larger cardinals without assuming GCH. So I need some helps. |
566826 | Limit of product of cosines: [imath]\lim\limits_{n \to \infty} \cos(\frac{x}{2})\cos(\frac{x}{4})\cos(\frac{x}{6})\cdots\cos(\frac{x}{2^n})[/imath]
We've got this problem [imath] y= \lim_{n \to \infty} \cos\left(\frac{x}{2}\right)\cos\left(\frac{x}{4}\right)\cdots\cos\left(\frac{x}{2^n}\right)[/imath] I have got stuck in this problem of evaluating the limits. What I was thinking was to convert [imath]\cos\left(\frac{x}{2^n}\right)[/imath] somehow to [imath]\sin[/imath] and thus term might fully get dissolved for a shorter expression and we can evaluate value of limit. | 455995 | Finding the limit [imath]\lim \limits_{n \to \infty}\ (\cos \frac x 2 \cdot\cos \frac x 4\cdot \cos \frac x 8\cdots \cos \frac x {2^n}) [/imath]
This limit seemed quite unusual to me as there aren't any intermediate forms or series expansions which are generally used in limits. Stuck on this for a while now .Here's how it goes : [imath] \lim \limits_{n \to \infty} \left[\cos\left(x \over 2\right)\cos\left(x \over 4\right) \cos\left(x \over 8\right)\ \cdots\ \cos\left(x \over 2^{n}\right)\right] [/imath] |
564755 | find a limit of [imath]\lim_{n\rightarrow\infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n[/imath]
[imath]\lim_{n\rightarrow\infty}\left(\frac{a^{1/n}+b^{1/n}}{2}\right)^n[/imath] No idea how to do it please help step by step | 130497 | Proving that: [imath]\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}[/imath]
Let [imath]a[/imath] and [imath]b[/imath] be positive reals. Show that [imath]\lim\limits_{n\to\infty} \left(\frac{a^{\frac{1}{n}}+b^{\frac{1}{n}}}{2}\right)^n =\sqrt{ab}[/imath] |
567536 | Determine all monic irreducible polynomials of degree [imath]4[/imath] in [imath]\mathbb{Z_2[x]}[/imath]
Determine all monic irreducible polynomials of degree [imath]4[/imath] in [imath]\mathbb{Z_2[x]}[/imath] Well these polynomials will be of the form - [imath]a_0 + a_1x + a_2x^2 + a_3x^3 + x^4[/imath] So we have four coefficients that can each have values of either [imath]0[/imath] or [imath]1[/imath]. So we have [imath]2^4 = 16[/imath] monic polynomials of degree [imath]4[/imath] in [imath]\mathbb{Z_2[x]}[/imath]. Now to determine the irreducible polynomicals is it necessary to write them all out and manually check if they are irreducible? Or is there some lemma I can apply here? | 32197 | Find all irreducible monic polynomials in [imath]\mathbb{Z}/(2)[x][/imath] with degree equal or less than 5
Find all irreducible monic polynomials in [imath]\mathbb{Z}/(2)[x][/imath] with degree equal or less than [imath]5[/imath]. This is what I tried: It's evident that [imath]x,x+1[/imath] are irreducible. Then, use these to find all reducible polynomials of degree 2. There ones that can't be made are irreducible. Then use these to make polynomials of degree 3, the ones that can't be made are irreducible. Repeat until degree 5. Doing this way takes way too long and I just gave up during when I was about to reach degree 4 polynomials. My question is: is there any easier way to find these polynomials? P.S.: this is not exactly homework, but a question which I came across while studying for an exam. |
567425 | Prove that one of [imath]2^n +1[/imath] and [imath]2^n-1[/imath] must be composite for [imath]n[/imath] is a natural number greater than [imath]2[/imath]
For natural numbers [imath]n>2[/imath] prove that at least one of [imath]2^n-1[/imath] and [imath]2^n+1[/imath] must be composite. How would one conduct this proof? I had the idea of trying to prove by contradiction since for this to be true these would have to be twin primes however this did not seem to go anywhere. Thanks | 402603 | Proving that either [imath]2^n-1 [/imath] or [imath] 2^n+1[/imath] is not prime
Not sure what approach to take with this: Prove that at least [imath]2^n-1 [/imath] or [imath] 2^n+1[/imath] is composite [imath]\forall[/imath] [imath]n>2[/imath] |
567407 | Construction of Borel set with given Lebesgue density.
Let [imath]\alpha \in (0,1)[/imath]. How can I construct a Borel set [imath]A[/imath] such that [imath]\lim_{r \to 0+ } \frac{m(A \cap [-r,r])}{2r}=\alpha[/imath]? Thanks. | 567756 | Find a Borel subset satisfying the condition
Let [imath]\alpha \in (0,1)[/imath]. Find a fixed Borel subset [imath]E[/imath] of [imath][-1,1][/imath] such that [imath] \lim_{r \rightarrow 0^+} \frac{m(E \cap [-r,r])}{2r} = \alpha [/imath] I think it is the trickiest problem for studying Lebesgue measure Differentiability. Firstly, I came up with setting [imath]E = [-\alpha r, \alpha r][/imath], but it does not meet the condition of ''fixed'' subset. It is so tricky that even my TA said this problem is "hard as hell". Anybody can suggest any idea? Thanks in advance. |
568253 | Let [imath]H[/imath] be a subgroup of [imath]G[/imath] and [imath]gH[/imath] is in [imath]Hg[/imath] for some [imath]g[/imath]. Can we get that [imath]gH=Hg[/imath]?
I know it is true if H is finite,but how about H is infinite?Any ideas are welcome .Thanks. | 312131 | Does [imath]gHg^{-1}\subseteq H[/imath] imply [imath]gHg^{-1}= H[/imath]?
Let [imath]G[/imath] be a group, [imath]H<G[/imath] a subgroup and [imath]g[/imath] an element of [imath]G[/imath]. Let [imath]\lambda_g[/imath] denote the inner automorphism which maps [imath]x[/imath] to [imath]gxg^{-1}[/imath]. I wonder if [imath]H[/imath] can be mapped to a proper subgroup of itself, i.e. [imath]\lambda_g(H)\subset H[/imath]. I tried to approach this problem topologically. Since every group is the fundamental group of a connected CW-complex of dimension 2, let [imath](X,x_0)[/imath] be such a space for [imath]G[/imath]. Since [imath]X[/imath] is (locally) path-connected and semi-locally simply-connected, there exists a (locally) path-connected covering space [imath](\widetilde X,\widetilde x_0)[/imath], such that [imath]p_*(\pi_1(\widetilde X,\widetilde x_0))=H[/imath]. The element [imath]g[/imath] corresponds to [imath][\gamma]\in\pi_1(X,x_0))[/imath], and its lift at [imath]\widetilde x_0[/imath] is a path ending at [imath]\widetilde x_1[/imath]. By hypothesis, [imath]H\subseteq g^{-1}Hg[/imath], which leads to the existence of a unique lift [imath]f:\pi_1(\widetilde X,\widetilde x_0)\to\pi_1(\widetilde X,\widetilde x_1)[/imath] such that [imath]p=p\circ f[/imath]. This lift turns out to be a surjective covering map itself, and it is a homeomorphism iff [imath]H=g^{-1}Hg[/imath]. I was unsuccessful in showing the injectivity. If [imath]x_1[/imath] and [imath]x_2[/imath] have the same image under [imath]f[/imath], then [imath]x_1[/imath], [imath]x_2[/imath], and [imath]f(x_1)=f(x_2)[/imath] are all in the same fiber. I took [imath]\lambda[/imath] to be a path from [imath]x_1[/imath] to [imath]x_2[/imath]. I have been playing around with [imath]\lambda[/imath], [imath]p\lambda[/imath], and [imath]f\lambda[/imath], but got nowhere. Of course, there could also be a direct algebraic proof. On the other hand, if the statement is not true then someone maybe knows of a counterexample. |
567879 | For all positive integers n,m,k where [imath]n\ge m\ge k[/imath] , [imath]\binom {n}{m}\binom {m}{k}=\binom {n}{k}\binom {n-k}{n-m}[/imath]
For all positive integers n,m,k where [imath]n\ge m\ge k[/imath] , [imath]\binom {n}{m}\binom {m}{k}=\binom {n}{k}\binom {n-k}{n-m}[/imath] Prove the following statements using combinatorial proofs. I can't come up combinatorial proofs. Also, [imath]\sum_{i=0}^n {n \choose i} = 2^n[/imath] how would you proof [imath]2^n[/imath]? | 534202 | Prove [imath]\binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a}[/imath]
I want to prove this equation, [imath] \binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a} [/imath] I thought of proving this equation by prove that you are using different ways to count the same set of balls and get the same result. But I'm stuck. Help me please. (Presumptive) Source: Theoretical Exercise 1.14(a), P19, A First Course in Pr, 8th Ed, by S Ross. |
548476 | Formula for determinant of block matrix with commuting blocks
On Wikipedia, I saw the following formula [imath]\det\begin{bmatrix}A & B\\ C & D\end{bmatrix} = \det(AD-BC)[/imath] if [imath]C[/imath] and [imath]D[/imath] commute. Is this always true? Or is there a good counter example for each [imath]2 \times 2[/imath] block matrices? | 1460407 | Is this problem statement for proving the determinant of a block-matrix possibly incorrect?
Let [imath]A, B,C,D[/imath] be square matrices, of same size [imath]n\times n[/imath]. Assume [imath]A[/imath] is invertible and [imath]AC = CA.[/imath] Prove that [imath] det \begin{bmatrix} A & B \\ C & D \\ \end{bmatrix} = det[AD - BC] [/imath] I am getting my final answer as [imath]det[AD - CB][/imath] instead, and my work looks fine -- just a simple factorization of the matrix, and using the multiplicativity of the determinant, [imath]AA^{-1}=I[/imath], and [imath]AC=CA[/imath]. Is the problem statement wrong? Or perhaps my answer of [imath]det[AD - CB][/imath] is actually equal to [imath]det[AD - BC][/imath], but I doubt it -- there are no extra assumptions on the relations of [imath]B[/imath] and [imath]C[/imath] other than that they are of the same size. So we can't assume that [imath]B[/imath] and [imath]C[/imath] commute. Thanks, |
568608 | Is a [imath]1\times 1 [/imath] matrix a scalar?
Intuitively I used to think that a [imath]1\times 1[/imath] matrix is simply a scalar number, I also saw this statement in books. However when I think about it now it doesn't make sense to me because of one problem. Let [imath]I[/imath] be the [imath]2\times 2 [/imath] identity matrix, now [imath]\lambda I[/imath] is well defined if [imath]\lambda [/imath] is a scalar or if [imath]\lambda [/imath] is a [imath]1 \times 2[/imath] matrix. However if we want to make the statement that scalars and [imath]1 \times 1[/imath] matrices are the same then the [imath]\lambda I[/imath] would have to be defined for a matrix [imath]\lambda[/imath] which is not [imath]1 \times 2[/imath]. More generally, let [imath]M \in \Bbb R^{n \times m}[/imath], now linear algebra says that if [imath]N[/imath] is another matrix, the product [imath]NM[/imath] only exists if [imath]N \in \Bbb R^{m \times n}[/imath]. However if [imath]n[/imath] and [imath]m[/imath] are not zero then [imath]NM[/imath] is not defined for a [imath]1\times 1[/imath] matrix [imath]N[/imath], while [imath]NM[/imath] is defined if [imath]N[/imath] is simply a scalar... So in conclusion we can simply state that [imath]1\times 1[/imath] matrices and scalars are not the same thing right? Is there any interesting theory behind this? | 219434 | is a one-by-one-matrix just a number (scalar)?
I was wondering. Clearly, we cannot multiply a (1x1)-matrix with a (4x3)-matrix; However, we can multiply a scalar with a matrix. This suggests a difference. On the other hand, I was, for example, in an econometrics lecture today, where we had for a (Tx1)-vector [imath]\underline{û}=\left( \begin{array}{c} û_1\\ \vdots\\ û_T\end{array}\right)[/imath]: [imath]S_{ûû}:= \sum_{i=1}^T û_i^2[/imath] shall be minimized. We see see that [imath]S_{ûû}=\underline{û}^T\underline{û}[/imath]. Well, formally, shouldn't it be [imath](S_{ûû})=\underline{û}^T\underline{û}[/imath] or [imath]S_{ûû}=\det(\underline{û}^T\underline{û})[/imath], to ensure that we stay in the space of matrices and not suddenly go to the space of scalars? So here, the professor (physicist) not only treats [imath]\underline{û}^T\underline{û}[/imath] like a scalar, but also calls it a scalar. Is this formally legit or a wrong simplification (though it does not seem to have any impact, and surely makes life easier)? |
567404 | Does the function $\frac1{1+|z|^2}$ have a power series at $z=1$?
Does [imath]\frac1{1+|z|^2}[/imath] have a power series at [imath]$z=1$[/imath]? If not, why not? How can we generally say that a function has a power series? | 568532 | How can I show that $\frac{1}{1+|z|^2}$ isn't analytic at $z=1$?
How can I show that [imath]\frac{1}{1+|z|^2}[/imath] isn't analytic at [imath]z=1[/imath]? I thought: [imath]\frac{1}{1+|z|^2}=\frac{1}{1+x^2+y^2}[/imath] (substituting [imath]z=x+yi[/imath]). We calculate the Cauchy-Riemann equations and we get that [imath]u_x=-2x[/imath], [imath]u_y=-2y[/imath], [imath]v_x=v_y=0[/imath]. So this function is only analytic at [imath]z=0[/imath]. Is this right? |
568012 | Prove if true and find a counterexample if false (disproofs/algebraic proofs)
For all sets [imath]A, B, C[/imath] [imath]A- (B-C) = (A-B) -C[/imath] Now my books says it is false and begins showing a counter example, but how do you know it is false by just looking at it? Would you go about proving it first and then if a proof can't be found, make up a counter example? The proof I think would start out... [imath]x[/imath] in [imath]A[/imath] and not ([imath]x[/imath] in [imath]B[/imath] and [imath]x[/imath] not in [imath]C[/imath])= | 172279 | Proving [imath]A\setminus(B\setminus C)[/imath] is not equal to [imath](A\setminus B)\setminus C[/imath]
I have several of these types of problems, and it would be great if I can get some help on one so I have a guide on how I can solve these. The question is: Prove [imath]A \setminus (B\setminus C) \neq (A\setminus B) \setminus C[/imath] I know I must prove both sides are not equivalent to each other to complete this proof. Here's my shot: We start with left side. if [imath]x[/imath] is in [imath]A[/imath], then [imath]x[/imath] is not in [imath]B[/imath], not not in [imath]C[/imath] so [imath]x[/imath] is in [imath]A[/imath] and [imath]C[/imath] if [imath]x[/imath] is in [imath]A[/imath], then it's not in [imath]B[/imath] [imath]\rightarrow[/imath] [imath](A\setminus B)[/imath] but if [imath]x[/imath] is in [imath]A\setminus B[/imath], then it must not be in [imath]C[/imath] however, earlier we stated [imath]x[/imath] is in both [imath]A[/imath] and [imath]C[/imath] we see that the two sides are not equal Is this the right idea? Should I then reverse the proof to prove it the other way around, or is that unnecessary? Should it be more formal? Thanks! |
561804 | How to prove that f is one-to-one
Let [imath]g\colon \mathbb{R} → \mathbb{R}[/imath] be differentiable with bounded derivative. i.e. [imath]|g'(x)| < M[/imath] for all [imath]x ∈ \mathbb{R}[/imath]. Let [imath]\epsilon[/imath] be a positive number satisfying [imath]0 < \epsilon < 1/M[/imath]. Let [imath]f(x) = x+\epsilon g(x)[/imath]. How can I prove that [imath]f[/imath] is one-to-one (injective)? I really have no idea how to solve hint. Any hint would be appreciated. Thank you :) | 458939 | For what values will f(x) be necessarily one-one?
Let [imath]g:\mathbb{R}\to\mathbb{R}[/imath] be a differentiable function such that [imath]|g'(x)|\le M[/imath] for all [imath]x\in \mathbb{R}[/imath]. For what values of [imath]\epsilon[/imath] will the function [imath]f(x)=x+\epsilon g(x)[/imath] will be necessarily one-one? Not getting any idea how to proceed, maybe Mean value theorem will work. |
567930 | Prove the following [imath]A \setminus (B\setminus C) = (A\setminus B)\cap (A \cup C)[/imath]
I want to prove the following so I decide to rewrite the RHS [imath]A \setminus (B\setminus C) = (A\setminus B)\cap (A \cup C)[/imath] [imath] A\setminus(B\setminus C) = A\setminus (B \cap C^c) = A\cap (B\cap C^c)^c = A\cap(B^c\cup C) = (A\cap B^c)\cup(A\cap C) [/imath] there is way that [imath](A\cap B^c)\cup(A\cap C)\rightarrow (A\setminus B)\cup(A\cap C) = (A\setminus B)\cap (A \cup C)[/imath]? thanks. EDIT LHS: [imath]x\in A \vee x\in A\cap C \rightarrow x\in A \wedge x\in C[/imath] RHS: [imath]x\in A \wedge x\in A\cup C \rightarrow x\in A \vee x\in C[/imath] how it can help me? | 67861 | Show that [imath]A \setminus ( B \setminus C ) \equiv ( A \setminus B) \cup ( A \cap C )[/imath]
Show that [imath]A \setminus ( B \setminus C ) \equiv ( A \setminus B) \cup ( A \cap C )[/imath] This is an exercise I was trying to do ( not homework ) and I got stuck as follows: Working from [imath]A \setminus ( B \setminus C )[/imath]: [imath]x \in A \wedge x \notin ( B \setminus C )[/imath] [imath]x \in A \wedge ((x \in B \wedge x \in C) \vee (x \notin B \wedge x \in C) \vee (x \notin B \wedge x \notin C) )[/imath] At this point I'm not sure how to proceed. Working from the other side [imath] ( A \setminus B) \cup ( A \cap C ) [/imath]: [imath] x \in ( A \setminus B) \cup ( A \cap C ) [/imath] [imath] x \in ( A \setminus B) \vee x \in ( A \cap C ) [/imath] [imath] ( x \in A \wedge x \notin B ) \vee ( x \in A \wedge x \notin C ) [/imath] [imath] x \in A \wedge (x \notin B \vee x \notin C ) [/imath] [imath] x \in A \wedge x \notin ( B \cap C ) [/imath] [imath] A \setminus ( B \cap C ) [/imath] I could also use some formating tips like how display logical not and how to get the lines to display properly without blank lines inbetween. :-) |
570200 | Noetherianty of ring of continuous functions
Let [imath]X[/imath] be a topological space. Is there any topological property on [imath]X[/imath] that be equivalent to [imath]C(X,\mathbb R)[/imath] be noetherian ring? | 570030 | When is the ring of continuous functions Noetherian?
Let [imath]X[/imath] be a topological space. Is there any topological property on [imath]X[/imath] that be equivalent to [imath]C(X,\mathbb R)[/imath] being a noetherian ring? |
568893 | A question about tensor product of algebras of compact operators.
Let [imath]\cal{H}[/imath] be a separable Hilbert space and [imath]\cal{K(\cal{H})}[/imath] the algebra of compact operators acting on [imath]\cal{H}[/imath]. Then [imath]\cal{K(\cal{H})}\otimes\cal{K}(\cal{H})\cong\cal{K}(\cal{H}\otimes H).[/imath] How to prove it? I can find a mapping [imath]\phi[/imath] from [imath]\cal{K(\cal{H})}\otimes\cal{K}(\cal{H})[/imath] to [imath]\cal{B}(H\otimes\cal{H})[/imath], but there is some difficult to prove the range of [imath]\phi[/imath] is justly [imath]\cal{K}(\cal{H}\otimes H)[/imath]. P.S.: Is [imath]\mathbb{M}_n(\mathbb{C})\otimes\mathbb{M}_n(\mathbb{C})\cong\mathbb{M}_{n^2}(\mathbb{C})[/imath]? Thanks a lot! | 212209 | [imath]\mathcal{K}(L^2(\mathbb{R}^m \times \mathbb{R}^n)) = \mathcal{K}(L^2(\mathbb{R}^m)) \otimes \mathcal{K}(L^2(\mathbb{R}^n))[/imath]?
QUESTION: Is it true that for the algebra of compact operators: [imath]\mathcal{K}(L^2(\mathbb{R}^m \times \mathbb{R}^n))[/imath] is as a [imath]C^{\ast}[/imath]-algebra isomorphic to [imath]\mathcal{K}(L^2(\mathbb{R}^m)) \otimes \mathcal{K}(L^2(\mathbb{R}^n))[/imath]? The latter tensor product is any [imath]C^{\ast}[/imath]-tensor product (because the compact operators are nuclear it doesn't matter). On [imath]L^2[/imath] we use the ([imath]\sigma[/imath]-finite) Lebesgue measure, but of course the algebra [imath]\mathcal{K}(L^2)[/imath] no longer depends on the measure. Clearly, [imath]L^2(\mathbb{R}^{m} \times \mathbb{R}^n)[/imath] can be identified with [imath]L^2(\mathbb{R}^m) \otimes L^2(\mathbb{R}^n)[/imath] by Fubini's theorem. This makes me think that [imath]L^2[/imath] has a better chance than other Hilbert spaces to make [imath]\mathcal{K}(\mathcal{H}_1 \otimes \mathcal{H}_2) = \mathcal{K}(\mathcal{H}_1) \otimes \mathcal{K}(\mathcal{H}_2)[/imath] hold. Thanks for your help. EDIT: Of course [imath]\mathcal{K}(\mathcal{H})[/imath] denotes the compact operators on the Hilbert space [imath]\mathcal{H}[/imath]. |
570444 | Find [imath]\lim _{x\rightarrow \infty }\left[ \sqrt [6] {x^{6}+x^{5}}-\sqrt [6] {x^{6}-x^{5}}\right] [/imath]
How to find [imath]\lim _{x\rightarrow \infty }\left[ \sqrt [6] {x^{6}+x^{5}}-\sqrt [6] {x^{6}-x^{5}}\right]\;?[/imath] | 45970 | Evaluating [imath]\lim\limits_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}[/imath]
How would you evaluate the following limit: [imath]\lim_{x\to \infty}\sqrt[6]{x^{6}+x^{5}}-\sqrt[6]{x^{6}-x^{5}}[/imath] I tried to use this formula: [imath]a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})[/imath], It didn't work. Any hints? |
570340 | Is [imath]\mathbb{R}/\mathbb{Q}[/imath] an interesting group?
Inspired by the construction of the non-mesurable Vitali set I thought about the group [imath]\mathbb{R}/\mathbb{Q}[/imath] (the additive group of the real numbers modulo the rationals). There must be some literature for this, but I haven't found any and it is hard to google. Does anyone know where to find results about this group? | 355852 | What is known about the quotient group [imath]\mathbb{R} / \mathbb{Q}[/imath]?
Let [imath]G = \mathbb{R} / \mathbb{Q}[/imath]. Is this an interesting group to study? Is it isomorphic to any more natural mathematical objects? |
571403 | If [imath]a_n\leq b_n[/imath], then [imath]\lim \sup a_n \leq \lim b_n[/imath]
Hello I want to make sure my work is correct! Suppose that [imath]\{a_n\}[/imath] and [imath]\{b_n\}[/imath] are sequences such that [imath]a_n \leq b_n[/imath] for all [imath]n [/imath] and [imath]b_n \to b[/imath]. Prove that [imath]\lim \sup a_n \leq b.[/imath] Let [imath]\epsilon > 0[/imath] be given. Choose an [imath]N[/imath] s.t. [imath]|b_n-b|<\epsilon[/imath] for [imath]n\geq N[/imath] where [imath]b = \lim \sup b_n[/imath]. [imath]a = \lim \sup a_n[/imath]. Proof by contradiction: If [imath]a \geq b[/imath], then, since [imath]a_n\leq b[/imath], [imath]b\leq a\leq b_n[/imath], meaning [imath]|b_n-a|<|b_n-b|<\epsilon [/imath], implying [imath]b_n \to a[/imath]. This is a contradiction, therefore, [imath]a\leq b[/imath]. | 213719 | Prove: [imath]a_n \leq b_n \implies \limsup a_n \leq \limsup b_n[/imath]
Is my proof correct? Prove: [imath]a_n \leq b_n \implies \limsup a_n \leq \limsup b_n[/imath] Proof: Let [imath]a_n[/imath] and [imath]b_n[/imath] be sequences such that [imath]a_n \leq b_n \forall_n[/imath]. Suppose [imath]\limsup a_n \nleq \limsup b_n[/imath]. That is: [imath]\limsup a_n > \limsup b_n[/imath]. From this we know: [imath]\forall_{\epsilon > 0} \exists_N \forall_{n>N} \implies |b_n - b| <\epsilon[/imath]. Where [imath]b[/imath] is the [imath]\limsup b_n[/imath]. [imath]\forall_{\epsilon_1 > 0} \exists_{N_1} \forall_{n > N_1} \implies |a_n - a| < \epsilon_1[/imath]. Where [imath]a[/imath] is the [imath]\limsup a_n[/imath] So, let [imath]a^* = a + \dfrac{\epsilon_1}{2}[/imath] and let [imath]b^* = b - \dfrac{\epsilon}{2}[/imath]. Hence, [imath]a^* \in |a_n - a| <\epsilon_1[/imath] and [imath]b^* \in |b_n - b| < \epsilon[/imath]. And clearly we see that [imath]b^* < a^*[/imath].Thus, we have found an element of [imath]b_n[/imath] namely [imath]b^* < a^*[/imath] an element of [imath]a[/imath]. This is contradiction since we are given [imath]a_n \leq b_n \forall_n[/imath]. Therefore, the supposition is false, and [imath]\limsup a_n \leq \limsup b_n[/imath]. |
511662 | Proving [imath]f(C) \setminus f(D) \subseteq f(C \setminus D)[/imath] and disproving equality
Let [imath]f: A\longrightarrow B[/imath] be a function. 1)Prove that for any two sets, [imath]C,D\subseteq A[/imath] , we have [imath]f(C) \setminus f(D)\subseteq f(C\setminus D)[/imath]. 2)Give an example of a function [imath]f[/imath], and sets [imath]C[/imath],[imath]D[/imath], for which [imath]f(C) \setminus f(D) \neq f(C\setminus D)[/imath] First time exposed to sets.. how would I go about proving this? What assumptions should I make? | 2148294 | Give an example for [imath]f(A)\setminus f(F) \neq f(A\setminus F)[/imath].
I came across this, [imath]f(A)\setminus f(F) \neq f(A\setminus F)[/imath]. I'm not sure that I completely understand why they cannot be equal. So could someone give an example for which [imath]f(A)\setminus f(F) \neq f(A\setminus F)[/imath]? I think an example of this might help me to better understand. |
571319 | Prove that is [imath]A[/imath] is skew-symmetric, then [imath]X^TAX = 0[/imath] for all [imath]X = [x_1 x_2 \cdots x_n]^T[/imath]
Recall that a matrix [imath]A[/imath] is skew-symmetric if and only if [imath]A^T = -A[/imath]. Prove that if [imath]A[/imath] is skew-symmetric, then [imath]X^TAX = 0[/imath] for all [imath]X = [x_1 x_2 \cdots x_n]^T[/imath] | 358281 | [imath]x^TAx=0[/imath] for all [imath]x[/imath] when [imath]A[/imath] is a skew symmetric matrix
Let [imath]A[/imath] be an [imath]n\times n[/imath] skew symmetric matrix. Show that [imath]x^TAx =0 \ \forall x \in \mathbb R^n[/imath]. How to prove this? |
571696 | Minimal polynomial of Galois extension
Let [imath]K[/imath] be a Galois extension of [imath]F[/imath] and let [imath]a\in K[/imath]. Let [imath]n=[K:F][/imath], [imath]r=[F(a):F][/imath] and [imath]H=Gal(K/F(a))[/imath]. Let [imath]\tau_{1},\dots,\tau_{r}[/imath] be left coset representatives of [imath]H[/imath] in [imath]G[/imath]. Show that [imath]\min(F,a) = \prod_{i=1}^{r}\left(x-\tau_{i}\left(a\right)\right)[/imath]. Conclude that [imath]\prod_{\sigma\in Gal(K/F)}\left(x-\sigma\left(a\right)\right)=\min\left(F,a\right)^{n/r}[/imath]. Help me a hint. Thanks in advanced. | 110122 | Minimal polynomial of intermediate extensions under Galois extensions.
Let [imath]K[/imath] be a Galois extension of [imath]F[/imath], and let [imath]a\in K[/imath]. Let [imath]n=[K:F][/imath], [imath]r=[F(a):F][/imath], and [imath]H=\mathrm{Gal}(K/F(a))[/imath]. Let [imath]z_1, z_2,\ldots,z_r[/imath] be left coset representatives of [imath]H[/imath] in [imath]G[/imath]. Show that [imath]\min(F,a)=\prod_{i=1}^r\left( x - z_i(a)\right).[/imath] In this product [imath]\min(F,a)[/imath] is of degree [imath]r[/imath]. That is true from fundamental theorem. And if one of [imath]z_i[/imath] is the identity, then [imath]a[/imath] satisfies the polynomial. My question is: What is the guarantee that [imath]z_i(a)[/imath] belongs to [imath]F[/imath] for all [imath]i[/imath]? What if I choose a representative which is not the identity of [imath]\mathrm{Gal}(K/F)[/imath]? |
571871 | Limit of [imath]\left[\frac{a_1^x+a_2^x+\cdots+a_n^x}{n}\right]^{1/x}[/imath] when [imath]x\to0[/imath]
I have to calculate the following limit: [imath]\lim_{x \to 0} \left[\frac{a_1^x+a_2^x+\cdots+a_n^x}{n}\right]^{\frac{1}{x}}[/imath] I said that as [imath]x→0[/imath] the [imath]a_1^x+a_2^x+\cdots+a_n^x[/imath] are approaching [imath]n[/imath] since we have [imath]n[/imath] terms, so we will get: [imath]\lim_{x \to 0} \left[\frac{n}{n}\right]^{\frac{1}{x}}[/imath] [imath]\therefore \lim_{x \to 0} \left(1\right)^{\frac{1}{x}}[/imath] as [imath]x→0[/imath], [imath]1/x→\infty[/imath] [imath]\therefore \left(1\right)^{\infty}=1[/imath] But one friend said that it was a false way to calculate that limit, where is my error and what is it's solution? (where [imath]a_i[/imath] are positive real numbers) | 397376 | Find the limit [imath] \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n[/imath]
For [imath]a,b,c>0[/imath], Find [imath] \lim_{n \to \infty}\left(\frac{a^{1/n}+b^{1/n}+c^{1/n}}{3}\right)^n[/imath] how can I find the limit of sequence above? Provide me a hint or full solution. thanks ^^ |
572012 | why the Vitali set has size [imath]2^{\aleph_0}[/imath]?
why the Vitali set has size [imath]2^{\aleph_0}[/imath] and therefore there is a bijection between [imath]\mathbb R[/imath] and the Vitali set? thanks | 231236 | Cardinality of Vitali sets: countably or uncountably infinite?
I am a bit confused about the cardinality of the Vitali sets. Just a quick background on what I gather about their construction so far: We divide the real interval [imath][0,1][/imath] into an uncountable number of disjoint classes in such a way that two numbers [imath]x[/imath] and [imath]y[/imath] are in the same class if their difference x-y is a rational number. Therefore the equivalence relation on the real numbers would be: [imath]x\sim y\iff x-y\in\mathbb Q[/imath]. The cardinality of our classes is equivalent to the cardinality of [imath]\mathbb Q[/imath], so the classes have countably infinite elements. How many of these classes are there? As each only has countably infinite elements, but each element of [imath][0,1][/imath] lies in one class, there are uncountably infinite classes. Now, using the axiom of choice, we take one element out of each class and put it in the set V. This is a Vitali set. Hopefully this is all good at this point. So each Vitali set should have uncountably many elements (because we have uncountably many classes), but there should be countably many Vitali sets (because each class has countably many elements) correct? But then why does Wikipedia state that "There are uncountably many Vitali sets..."? |
572537 | Suppose a, b and n are positive integers. Prove that (a^n) | (b^n) if and only if a | b.
Suppose [imath]a, b[/imath] and [imath]n[/imath] are positive integers. Prove that [imath]a^n\mid b^n[/imath] if and only if [imath]a \mid b[/imath]. I have: [imath]a^n\mid b^n[/imath] [imath]\implies b^n = a^n \cdot k[/imath] [imath]\implies \sqrt[n]{b^n}=\sqrt[n]{a^n}\cdot k[/imath] [imath]\implies a=b\cdot k[/imath] [imath]\implies a\mid b[/imath] Is it really this simple? | 351623 | Does [imath]a^n \mid b^n[/imath] imply [imath]a\mid b[/imath]?
Does [imath]a^n \mid b^n[/imath] imply [imath]a\mid b[/imath]? I think it does but haven't been able to prove it. I don't know much number theory so an elementary answer would be great. |
523716 | Harmonic function vanishes with its normal derivative on a part of boundary; can Green's formula be applied to broken boundary?
Let [imath]\Omega[/imath] be an open domain, and let [imath]\Sigma[/imath] be a smooth and nonempty portion of the boundary. Let [imath]u[/imath] be a harmonic function in [imath]\Omega[/imath] and [imath]u=D_\nu u=0[/imath] on [imath]\Sigma[/imath]. ([imath]D_\nu[/imath] is the derivative along the outward vector). How to prove [imath]u=0[/imath] in [imath]\Omega[/imath]? Can I use Green's formula to this "portion" of boundary? | 482576 | A PDE exercise from Gilbarg Trudinger. problem 2.2
Prove that if [imath]\Delta u=0[/imath] in [imath]\Omega\subset \mathbb{R}^n[/imath] and [imath]u=\partial u/\partial\nu=0[/imath] on an open smooth portion of [imath]\partial \Omega[/imath], then [imath]u[/imath] is identically zero. I found a proof here, but I'm not sure if it's the correct proof. Does anyone have any idea? Any hint would be appreciated. |
572918 | Find [imath]M^n[/imath] of a matrix [imath]M[/imath] whenever diagonalization is possible
Let [imath]M = \begin{bmatrix} -7 & 8 \\ -8 & -7 \end{bmatrix}.[/imath] Find formulas for the entries of [imath]M^n[/imath] where [imath]n[/imath] is a positive integer. (Your formulas should not contain complex numbers.) Your answer should be in the form of a matrix. I diagonalized to the form [imath]M = P D P^{-1}[/imath] and [imath]M^n = P D^n P^{-1}[/imath] where [imath]P[/imath] is my matrix of eigenvectors and [imath]D[/imath] is my matrix of eigenvalues. My final answer after diagonalization was [imath]M^n = \begin{bmatrix} > .5((-7+8i)^n+(-7-8i)^n) & (i/2)(-(-7+8i)^n+(-7-8i)^n) \\ >(.5/i)(-(-7+8i)^n+(-7-8i)^n) & .5((-7+8i)^n+(-7-8i)^n) \end{bmatrix}[/imath] I had a hard time finding an answer not in terms of complex numbers. Can someone show me what I'm missing? I asked this question here a couple of days ago and I finally had time to correct it. Can someone check my solution, Every value I can think of works for it but it was an online homework assignment and my solution isn't being accepted. After using De Moivre's formula as BaronVT suggested I came up with the solution [imath]M^n = \begin{bmatrix} \sqrt{113^n}\cdot\cos\left(n\cdot\left(\tan^{-1}\left(\frac{8}{-7}\right)+\pi\right)\right) & \sqrt{113^n}\cdot\sin\left(n\cdot\left(\tan^{-1}\left(\frac{8}{-7}\right)+\pi\right)\right) \\ -\sqrt{113^n}\cdot\sin\left(n\cdot\left(\tan^{-1}\left(\frac{8}{-7}\right)+\pi\right)\right) & \sqrt{113^n}\cdot\cos\left(n\cdot\left(\tan^{-1}\left(\frac{8}{-7}\right)+\pi\right)\right) \end{bmatrix}[/imath] | 568628 | Where M is a matrix calculate a formula for M^n
Let [imath]M = \begin{bmatrix} -7 & 8 \\ -8 & -7 \end{bmatrix}.[/imath] Find formulas for the entries of [imath]M^n[/imath] where [imath]n[/imath] is a positive integer. (Your formulas should not contain complex numbers.) Your answer should be in the form of a matrix. I diagonalized to the form [imath]M = P D P^{-1}[/imath] and [imath]M^n = P D^n P^{-1}[/imath] where [imath]P[/imath] is my matrix of eigenvectors and [imath]D[/imath] is my matrix of eigenvalues. My final answer after diagonalization was [imath]M^n = \begin{bmatrix} .5((-7+8i)^n+(-7-8i)^n) & (i/2)(-(-7+8i)^n+(-7-8i)^n) \\ (.5/i)(-(-7+8i)^n+(-7-8i)^n) & .5((-7+8i)^n+(-7-8i)^n) \end{bmatrix}[/imath] I can't seem to find an answer not in terms of complex numbers. Can someone show me what I'm missing? |
573080 | Indefinite Integral [imath]\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx[/imath]
How can I evaluate this integral? [imath]\int\frac{3\sin(x)+2\cos(x)}{2\sin(x)+3\cos(x)}dx[/imath] | 571054 | Integrating trigonometric function problem [imath]\int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx[/imath]
\begin{eqnarray*} \int \frac{3\sin x+2\cos x}{2\sin x+3\cos x}dx &=& \int \frac{(3\sin x+2\cos x)/\cos x}{(2\sin x+3\cos x)/\cos x}dx\\ \\ &=& \int \frac{3\tan x +2}{2\tan x +3} dx\\ && u = \tan x \text{ and } du = \sec^2 x \ dx \end{eqnarray*} Am I going in the right direction with this one? It seems like not. |
572903 | Proving an inequality by induction and figuring out intermediate inductive steps?
I'm working on proving the following statement using induction: [imath] \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} [/imath] Fair enough. I'll start with the basis step: Basis Step: (n=1) [imath] \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} [/imath] [imath] \frac{1}{1^2} \le \frac{2}{1+1} [/imath] [imath] 1 \le 1 \checkmark [/imath] Inductive Step: [imath] \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} [/imath] [imath] \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} [/imath] [imath] \sum_{r=1}^n \frac{1}{r^2} + \frac{1}{(n+1)^2} \le \frac{2n}{n+1} + \frac{1}{(n+1)^2} [/imath] [imath] \sum_{r=1}^{k+1} \le \frac{2n}{n+1} + \frac{1}{(n+1)^2} [/imath] My goal is to prove [imath]\forall_{n\ge1} s(n) \implies s(n+1) [/imath] or that this inequality holds true for all [imath]n\ge1[/imath]. I'm not quite sure to go from here on the inductive step. I understand that I need to basically work some clever substitution and manipulation into the problem to end up with: [imath] \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} [/imath] However, I'm not quite sure what needs to done to obtain this after attempting a few times. | 569976 | Proving an inequality by induction, how to figure out intermediate inductive steps?
I'm working on proving the following statement using induction: [imath] \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} [/imath] Fair enough. I'll start with the basis step: Basis Step: (n=1) [imath] \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} [/imath] [imath] \frac{1}{1^2} \le \frac{2}{1+1} [/imath] [imath] 1 \le 1 \checkmark [/imath] Inductive Step: [imath] \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} [/imath] [imath] \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{n+2} [/imath] My goal is to prove [imath]\forall_{n\ge1} s(n) \implies s(n+1) [/imath] or that this inequality holds true for all [imath]n\ge1[/imath]. I'm not quite sure to go from here on the inductive step. I understand that I need to basically work some clever substitution and manipulation into the problem to end up with: [imath] \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2n}{n+1} [/imath] However, I'm not quite sure what needs to done to obtain this after attempting a few times. |
320520 | What is [imath]X[/imath] in this problem?
I captured this picture from the movie A Beautiful Mind. I see this is a problem solving for the dimension of a quotient space, so I guess [imath]W[/imath] is assumed to be a subspace of [imath]V[/imath], and let's also assume [imath]V[/imath] is a vector space over [imath]\mathbb{Q}[/imath]. I cannot figure out what [imath]X[/imath] is here. Also, how do I solve this? | 284788 | What is the solution to Nash's problem presented in "A Beautiful Mind"?
I was watching the said movie the other night, and I started thinking about the equation posed by Nash in the movie. More specifically, the one he said would take some students a lifetime to solve (obviously, an exaggeration). Nonetheless, one can't say it's a simple problem. Anyway, here it is [imath]V = \{F:\mathbb{R^3}/X\rightarrow \mathbb{R^3} \text{ so } \hspace{1mm}\nabla \times F=0\}[/imath] [imath]W = \{F = \nabla g\}[/imath] [imath]\dim(V/W) = \; 8[/imath] I haven't actually attempted a solution myself to be honest, but I thought it would be an interesting question to pose. I have done a quick search on this site and Google, but there were surprisingly few results. In any case, I was curious if anyone knew the answer aside from the trivial. |
573125 | Proving [imath]\lim_{x \to +\infty} f(x + \sqrt{x}) - f(x)=0[/imath] if [imath]|f'(x)|\le \frac1x[/imath] for [imath]x > 1[/imath]
Given a differentiable function [imath]{\rm f}:\left(1,+\infty\right) \to \mathbb{R},\quad[/imath] and [imath]\quad\left\vert\,{\rm f}'\left(x\right)\,\right\vert \leq \frac1x\,,\ \forall\ x>1[/imath]. [imath] \mbox{Prove that}\quad \lim_{x \to +\infty}\left[\vphantom{\Large A}% {\rm f}\left(x + \sqrt{x\,}\,\right) - {\rm f}\left(x\right) \right]=0 [/imath] | 573010 | Limit of [imath]f(x+\sqrt x)-f(x)[/imath] as [imath]x \to\infty[/imath] if [imath]|f'(x)|\le 1/x[/imath] for [imath]x>1[/imath]
Given a differentiable function [imath]f:(1,+\infty) \to \mathbb{R}[/imath], and [imath]|f'(x)| \le 1/x[/imath] for every [imath]x>1[/imath]. Prove that: [imath]\lim_{x\to +\infty} \left(f(x+\sqrt{x})-f(x) \right)=0.[/imath] |
573373 | In characteristic [imath]p[/imath], the field extension [imath]k(X,Y)[/imath] over [imath]k(X^p,Y^p)[/imath] is not simple
Let [imath]k[/imath] be a field with characteristic [imath]p>0[/imath], [imath]L=k(X,Y)[/imath] be the field of rational fractions of two variables over [imath]k[/imath]. Let [imath]K=k(X^p,Y^p)[/imath]. Then [imath][L:K]=p^2[/imath] (for a proof see In characteristic [imath]p[/imath], the field extension [imath]k(X,Y)[/imath] over [imath]k(X^p,Y^p)[/imath] has degree [imath]p^2[/imath]). Show that [imath]L^{p}\subseteq K[/imath] and [imath]L[/imath] is not a simple extension over [imath]K[/imath]. Thank in advance. | 333392 | If [imath]F=K(u,v)[/imath] with [imath]u^p[/imath],[imath]v^p\in K[/imath] and [imath][F:K]=p^2[/imath], [imath]\operatorname{char} K=p>0[/imath], then [imath]F[/imath] is not a simple extension of [imath]K[/imath].
Greetings I'm trying to show this exercise from Hungerford's Algebra Chapter five section 6 exercise 15; but I'm stuck. the exercise says the following: Let [imath]\operatorname{char} K=p>0[/imath] and assume [imath]F=K(u,v)[/imath] where [imath]u^p,v^p\in K[/imath] and [imath][F:K]=p^2[/imath], then [imath]F[/imath] is not a simple extension of [imath]K[/imath]. Exhibit an infinite number of intermediate fields. Thank you. |
573439 | Subring of rationals is a PID
How to show that [imath]R[/imath] is a PID if [imath]\mathbb{Z}\subset R\subset \mathbb Q[/imath]? I tried as follows: Let [imath]I[/imath] be an ideal in [imath]R[/imath]. Then [imath]\mathbb Z\cap I[/imath] is an ideal in [imath]\mathbb Z[/imath]. But I couldn't proceed further. Please help. | 137876 | A subring of the field of fractions of a PID is a PID as well.
Let [imath]A[/imath] be a PID and [imath]R[/imath] a ring such that [imath]A\subset R \subset \operatorname{Frac}(A)[/imath], where [imath]\operatorname{Frac}(A)[/imath] denotes the field of fractions of [imath]A[/imath]. How to show [imath]R[/imath] is also a PID? Any hints? |
573808 | Inverse eigenvalue of a linear transformation
T is a linear transformation, and [imath]\lambda[/imath] is N eigenvalue of T. How do I prove that [imath]\lambda^{-1}[/imath] is an eigenvalue for [imath]T^{-1}[/imath]? I know for a matrix, I can use the fact that [imath]Av=\lambda v[/imath], but how does a linear transformation work? | 572429 | If A is invertible, prove that [imath]\lambda \neq 0[/imath], and [imath]\vec{v}[/imath] is also an eigenvector for [imath]A^{-1}[/imath], what is the corresponding eigenvalue?
If A is invertible, prove that [imath]\lambda \neq 0[/imath], and [imath]\vec{v}[/imath] is also an eigenvector for [imath]A^{-1}[/imath], what is the corresponding eigenvalue? I don't really know where to start with this one. I know that [imath]p(0)=det(0*I_{n}-A)=det(-A)=(-1)^{n}*det(A)[/imath], thus if both [imath]p(0)[/imath] and [imath]det(0) = 0[/imath] then [imath]0[/imath] is an eigenvalue of A and A is not invertible. If neither are [imath]0[/imath], then [imath]0[/imath] is not an eigenvalue of A and thus A is invertible. I'm unsure of how to use this information to prove [imath]\vec{v}[/imath] is also an eigenvector for [imath]A^{-1}[/imath] and how to find a corresponding eigenvalue. |
573839 | How many string of numbers [imath]1,...,n[/imath] of length [imath]2n[/imath] where two of the same numbers are not consecutive.
I need to find the number of strings of length [imath]2n[/imath] that consist of numbers [imath]1,...,n[/imath] each appearing exactly twice and not next to each other. Let [imath]f(n)[/imath] denote the function that gives me back that information: [imath]f(1)=0[/imath], because [imath]11[/imath] is not correct. [imath]f(2)=2[/imath] there are two possibilities- [imath]2121[/imath] i [imath]1212[/imath]. Now let's consider [imath]f(n)[/imath]. Let's take out two of the [imath]n's[/imath]. The string without [imath]n[/imath]'s can be created [imath]f(n-1)[/imath] different ways, and since we have to put two [imath]n[/imath]'s somewhere between the numbers in that string, we can do that [imath]{{2n-1}\choose{2}}=(n-1)(2n-1)[/imath] ways, so the answer is [imath]f(n)=(n-1)(2n-1)*f(n)[/imath]. And now I have to solve this recurrence by annihilating it for example. Correct? | 391734 | In how many ways can n couples (husband and wife) be arranged on a bench so no wife would sit next to her husband?
In how many ways can n couples (husband and wife) be arranged on a bench so no wife would sit next to her husband? I thought about this: (Total amount of ways to sit 2n people in 2n sits)-(Using inclusion and exclusion to find that at least 1 wife sits next to her husband) And i get: Let [imath]A_1[/imath] be the attribute where at least 1 wife sits with her husband, Then we "merge" up the husband and wife into a one person. We have [imath]\binom n1[/imath] ways to choose [imath]1[/imath] couple out of [imath]n[/imath] couples, And we are left with [imath]2n-1[/imath] to place [imath]2n-1[/imath] 'people' so we get [imath](2n-1)![/imath] and so on, And on a general note: [imath](2n)!-(2\binom n1(2n-1)!-2^2\binom n2(2n-2)!+...2^k(-1)^k\binom nk(2n-k)!)[/imath] And a bit simplified: [imath](2n)!-(\sum_{k=1}^n2^k(-1)^k\binom nk(2n-k)![/imath]) Since i don't have answers to this question i wanna know if i did something wrong? Did i even look at the question right? |
573966 | Why has [imath]3^x+4^y=5^z[/imath] has only one solution (2,2,2) in positive integers?
First, do we have to exclude the cases, where [imath](x,y,z)[/imath] are not all even or odd and then show the only possibility ? or is there a geometric solution maybe ? | 571622 | Positive integral solutions of [imath]3^x+4^y=5^z[/imath]
Are there more integral solutions for [imath]3^x+4^y=5^z[/imath], than [imath]x=y=z=2[/imath] ? If not, how do I show that? I could show that for [imath]3^x+4^x=5^x[/imath], but I'm stuck at the general case? Any ideas, maybe graphs, logarithms or infinite descent? |
574235 | Any primitive root modulo [imath]p^m[/imath] is also a primitive root modulo [imath]p[/imath]
This is the last of the stream of number theory problems I have been looking at that I would like to discuss. Let [imath]p[/imath] be an odd prime number and let [imath]m[/imath] be a positive integer. Prove that any primitive root modulo [imath]p^m[/imath] is also a primitive root modulo [imath]p[/imath]. My idea is as follows: Suppose [imath]r[/imath] is a primitive root modulo [imath]p^m[/imath]. Then [imath]r^{p^{m-1}(p-1)} \equiv 1 \mod{p^m}[/imath], and therefore [imath]\left( r^{p-1} \right)^{p^{m-1}} \equiv 1 \mod p.[/imath] Can we conclude from this that [imath]r^{p-1}\equiv \pm 1\mod p[/imath], and since [imath]p^{m-1}[/imath] is odd, that [imath]r^{p-1}\equiv 1 \mod p[/imath]? How do I finish? EDIT: Already answered at Prove that , any primitive root [imath]r[/imath] of [imath]p^n[/imath] is also a primitive root of [imath]p[/imath] . The chosen solution is very simple: If [imath]r[/imath] is a primitive root modulo [imath]p^m[/imath], then there exists an integer [imath]k[/imath] such that [imath]r^k \equiv a \mod{p^m}[/imath] for any [imath]a[/imath] relatively prime to [imath]p^m[/imath]. Therefore [imath]r^k \equiv a \mod p[/imath]. This can only be true if [imath]r[/imath] is a primitive root modulo [imath]p[/imath]. | 170648 | Prove that , any primitive root [imath]r[/imath] of [imath]p^n[/imath] is also a primitive root of [imath]p[/imath]
For an odd prime [imath]p[/imath], prove that any primitive root [imath]r[/imath] of [imath]p^n[/imath] is also a primitive root of [imath]p[/imath] So I have assumed [imath]r[/imath] have order [imath]k[/imath] modulo [imath]p[/imath] , So [imath]k|p-1[/imath].Then if I am able to show that [imath]p-1|k[/imath] then I am done .But I haven't been able to show that.Can anybody help me this method?Any other type of prove is also welcomed. |
574179 | Commutative integral domain with d.c.c. is a field
If [imath]R[/imath] is a commutative integral domain and it also satisfies descending chain condition on its ideals then how will we show that such ring [imath]R[/imath] will be a field? | 463498 | Let [imath]R[/imath] be a Artinian commutative ring with [imath]1 \neq 0[/imath]. If [imath]I[/imath] is prime, then [imath]I[/imath] is maximal.
Prove: Let [imath]R[/imath] a Artinian commutative ring with [imath]1 \neq 0[/imath]. If [imath]I[/imath] is prime, then [imath]I[/imath] is maximal. I got stuck on this. I understand that every ideal is generated by finitely many elements. Here is my approach: It's enough to prove the following implication: [imath] R/I \ \text{is a domain} \quad \Rightarrow \quad R/I \ \text{is a field}[/imath] Then only finding an inverse is left. So let [imath]x \in R[/imath] in arbitrary. We have to find an inverse element of [imath]x+I \in R/I[/imath], that as an element y so that [imath]x \cdot y \in I+1[/imath]. We denote [imath]I = (x_1, x_2, \cdots, x_n)[/imath]. Now I have to find y so that [imath]xy -1 = r_1 x_1+ r_2 x_2 + \cdots + r_n x_n[/imath]. This is where I no more knew what to do. Your advise will be appreciated. |
574792 | Counting graphs on 5 vertices
If we have the set A with [imath]\#A=5[/imath], how much graphs can be made over A? The solution says [imath]2^{5^2}[/imath], that is exactly the same number of binary relations on A. I initially thought that the solution would be: [imath]2^\binom 52[/imath], but i was worng. Any tip about counting graphs would be appreciated. | 100560 | Enumerate non-isomorphic graphs on n vertices
In the following the graphs are assumed to be undirected and simple. 1.Enumerate the number of non-isomorphic graphs on [imath]n[/imath] vetrices where [imath]n[/imath] is fixed. Here are some ideas I had: The number of labeled graphs is [imath] 2^{\frac{n(n-1)}{2}} [/imath]. So it is enough to find the number unlabeled graphs on [imath]n[/imath] vertices.I have no idea for this. 2.Enumerate the number of non-isomorphic graphs on [imath]n[/imath] vertices and [imath]m[/imath] edges where [imath]n,m[/imath] are fixed. Can we find a closed formula for each of this? Any help? Thank you! |
575753 | Convergence of [imath]\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{2}}[/imath]
How I check if the series [imath]\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{2}}[/imath] is convergent or divergent?? I tried few tests, but I didn't success to discover if the series is convergent or is divergent... I need to use one of the tests to show it... [imath]\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{2}}[/imath] Thank you! | 274586 | [imath]\sum_{n=2}^\infty \frac{1}{(\ln\, n)^2}[/imath] c0nvergence
[imath]\sum_{n=2}^\infty \frac{1}{(\ln\, n)^2}[/imath] The series converge? Please verify my solution below |
576224 | Prove that [imath]\alpha[/imath] is algebraic over [imath]K[/imath].
Let [imath]\alpha[/imath] be a transcendental element over a field [imath]E[/imath], and [imath]F=E(\alpha)[/imath]. Prove that for any subfield [imath]K[/imath] of [imath]F[/imath] containing [imath]E[/imath] as a proper subset, [imath]\alpha[/imath] is algebraic over [imath]K[/imath]. Can anyone give me some hint? I have no idea how to start. | 164055 | For field extensions [imath]F\subsetneq K \subset F(x)[/imath], [imath]x[/imath] is algebraic over [imath]K[/imath]
Let [imath]x[/imath] be an element not algebraic over [imath]F[/imath], and [imath]K \subset F(x)[/imath] a subfield that strictly contains [imath]F[/imath]. Why is [imath]x[/imath] algebraic over [imath]K[/imath]? Thanks a lot! |
445477 | Sigma of factorial function?
How would you find this sum mathmatically?[imath]\sum_{i=0}^\infty \left( \dfrac{1}{2} \right) ^{i!}[/imath] What techniques would you use to solve this as well? | 279626 | Compute [imath]\sum_{k=0}^{\infty}\frac{1}{2^{k!}}[/imath]
How could the series below be computed ? [imath]\sum_{k=0}^{\infty}\frac{1}{2^{k!}}[/imath] It's not a series from a book, but a series I thought of many times, and I didn't manage to figure out what I should do here. I'm just curious to know if there are some known ways for approaching such a series. Thanks! |
577015 | prove that [imath](\frac{n}{3})^n[/imath]
prove that [imath](\frac{n}{3})^n<n!<e\cdot(\frac{n}{2})^n[/imath] I tried to prove by the induction that [imath](\frac{n}{3})^n<n![/imath] and [imath]n!<e\cdot(\frac{n}{2})^n[/imath], but I failed my assumption [imath]n^n<n!*3^n[/imath] [imath](n+1)^{n+1}<(n+1)*3^{n+1}[/imath] [imath]\frac{(n+1)^n}{9}<n!*3^{n-1}[/imath] | 144176 | Factorial Inequality problem [imath]\left(\frac n2\right)^n > n! > \left(\frac n3\right)^n[/imath]
I met an inequality, I ask, do not mathematical induction to prove that: Prove \[ \left(\frac n2\right)^n > n! > \left(\frac n3\right)^n \] without using induction |
576989 | How to prove that [imath]\sum_{p \leq x} {\log p \over p} = \log x + O(1)[/imath]?
Problem Prove that [imath] \sum_{p \leq x} {\log p \over p} = \log x + O(1) [/imath] as [imath]x \to \infty[/imath]. Notes: [imath]p[/imath] ranges over primes, [imath]\log[/imath] is natural Progress Using Riemann-Stieltjes integration and integration by parts, I think you can write the sum as [imath] \int_2^x {\log t \over t} d\pi(t) = O(1) - \int_2^x {1 - \log t \over t^2} \pi(t) dt. [/imath] From here, I tried using the prime number theorem, specifically [imath] \pi(x) = \operatorname{li}(x) + O(xe^{-c\sqrt{\log x}}), [/imath] together with [imath] \operatorname{li}(x) = {x \over \log x} + {x \over \log^2 x} + O\Big({x \over \log^3 x}\Big). [/imath] All the terms cancel out perfectly except for the integrals of the error terms from the prime number theorem, which diverge. If they converged, they could go into the [imath]O(1)[/imath] term, which would finish the proof. Please note that this is tagged as homework. Those integrals actually do converge. In fact, [imath] O(xe^{-c\sqrt{\log x}}) = O\Big({x \over \log^3 x}\Big), [/imath] so there is no need to even consider them separately. I will submit this as an answer to the question this is a duplicate of now. | 33980 | How to prove Chebyshev's result: [imath]\sum_{p\leq n} \frac{\log p}{p} \sim\log n [/imath] as [imath]n\to\infty[/imath]?
I saw reference to this result of Chebyshev's: [imath]\sum_{p\leq n} \frac{\log p}{p} \sim \log n \text{ as }n \to \infty,[/imath] and its relation to the Prime Number Theorem. I'm looking into an information-theory proof by Kontoyiannis I was wondering if anyone could give me a sense for how difficult or involved the usual proof is. I don't really need to see the whole thing in detail, more wondering about the general difficulty/complexity of the result. Thanks! |
577347 | Cool simple solution to: [imath]-A^2[/imath] is not the identity matrix
this is not a question per se, just a simple cool solution to a potentially difficult question, that I want to share. I liked it. The question is: Let [imath]A[/imath] be a [imath]3\times3[/imath] matrix with real values. Show that [imath]A^2 \neq -I_3[/imath] There are probably many solutions for this problem, but i really thought this one was simple. Solution is posted down below. | 388643 | Does there exist a matrix [imath]\mathbf{A}\in\mathbb{R}^{3\times3}[/imath] such that [imath]\mathbf{A}^{2}=-\mathbf{I}[/imath]?
Is it possible for a matrix [imath]\mathbf{A}\in\mathbb{R}^{3\times3}[/imath], [imath]\mathbf{A}^2=-\mathbf{I}[/imath] I know that It is possible for [imath]2\times2[/imath] matrix, but is it possible for [imath]3\times3[/imath] matrix ? |
150579 | Question about Riemann integral and total variation
Let [imath]g[/imath] be Riemann integrable on [imath][a,b][/imath], [imath]f(x)=\int_a^x g(t)dt [/imath] for [imath]x \in[a,b][/imath]. Can I show that the total variation of [imath]f[/imath] is equal to [imath]\int_a^b |g(x)| dx [/imath]? | 1270852 | Function of variation equals integral of absolute value
Exercise 11, chapter 6 from second edition of Baby Rudin: Suppose [imath]g[/imath] is Riemann integrable on [imath][a,b][/imath], put [imath]\int_a^x g(t)dt,[/imath] and define [imath]g^+(t)=\max(g(t),0)[/imath], [imath]g^-(t)=-\min(g(t),0)[/imath]. Prove that [imath]f[/imath] is of bounded variation on [imath][a,b][/imath] and that its variation functions are given by [imath]v(x)=\int_a^x \left| g(t)\right|dt[/imath], [imath]p(x)=\int_a^x g^+(t)dt[/imath] and [imath]q(x)=\int_a^x g^-(t)dt[/imath] where [imath]p,q[/imath] are positive and negative variation. I'm aware of this question but I still have doubts, because he says "I also know that if F is BV, then F′ exists a.e.(almost everywhere)" I don't have the concept of almost everywhere yet. Also, I can't follow the proof because I don't know if [imath]f[/imath] is differentiable, that depends on [imath]g[/imath] being continuous! I did this: [imath]\left| f(x_i)-f(x_{i-1}) \right| = \left| \int_{x_{i-1}}^{x_i} g(t)dt \right| [/imath] My intention was to take sum at both sides and then limit as the diameter of the partition goes to zero but...If [imath]g[/imath] is continuous it's super easy using mean value theorem for integrals! Otherwise...I don't know which road to take here... Also, if I could prove the last two equalities it'd be done, since: [imath]|g|=g^-+g^+[/imath], but doing that first was way harder... |
577514 | Show that [imath]X = \{ (x,y) \in\mathbb{R}^2\mid x \in \mathbb{Q}\text{ or }y \in \mathbb{Q}\}[/imath] is path connected.
How do I show that [imath]X = \left\{ (x,y) \in \mathbb{R}^2 \mid x \in \mathbb{Q}\text{ or }y \in \mathbb{Q}\right\}[/imath] is path connected? Note that [imath]X[/imath] is a topological space with subspace topology [imath]\tau = \left\{ U \cap X \mid U\text{ is open in }\mathbb{R}^2\right\}[/imath]. I only know two things (with respect to path connectedness): A topological space [imath]X[/imath] is path connected iff for every [imath]p,q \in X[/imath], there is a continuous function [imath]f: [a,b] \to X[/imath] with [imath]f(a)=p[/imath] and [imath]f(b) = q[/imath]. Continuity [imath]f[/imath] is defined as for every open set in [imath]X[/imath], its preimage is open in [imath][a,b][/imath] (with subspace topology induced by [imath]\mathbb{R}[/imath] (with standard topology) ) Suppose [imath]X[/imath] and [imath]Y[/imath] are two topological spaces and [imath]f: X \to Y[/imath] is a continuous surjection. If [imath]X[/imath] is path connected, then [imath]Y[/imath] is path connected. | 516916 | [imath]\mathbb{R}[/imath] \ [imath]\mathbb{Q}[/imath] and [imath]\mathbb{R}^2\setminus\mathbb{Q}^2[/imath] disconnected?
If I want to prove that [imath]\mathbb{R} \setminus \mathbb{Q}[/imath] is disconnected, does it suffice to say that there are two open disjoint sets that cover [imath]\mathbb{R}\setminus\mathbb{Q}[/imath], namely: [imath](- \infty, 0), (0, \infty)\text{ ?}[/imath] Along the same lines, I want to prove or disprove that [imath]S = \mathbb{R}^2 \setminus \mathbb{Q}^2[/imath] (points [imath](x, y) \in S[/imath] that have at least one irrationais connected. I feel that it is also disconnected; does it suffice to say that there are two open disjoint sets that cover [imath]S = \mathbb{R}^2 \setminus \mathbb{Q}^2[/imath], namely [imath]((- \infty, - \infty), (0, 0))[/imath] and [imath]((0, 0), (\infty, \infty))[/imath]? (Very iffy on my assertion and my notation, sorry.) Thanks! |
577639 | Prove [imath]x + y[/imath] is divisible by [imath]11[/imath]. Is my solution correct?
If [imath]x[/imath] & [imath]y[/imath] are natural numbers, and [imath]56 x = 65 y[/imath], prove that [imath]x + y[/imath] is divisible by [imath]11[/imath]. Solution) [imath]56[/imath] and [imath]65[/imath] are relatively prime So, [imath]65∣x[/imath] and [imath]56∣y[/imath] Let [imath]x = 65m[/imath] and [imath]y = 56n[/imath] Then, [imath]56x = 65y[/imath] [imath]56.65m = 65.56n[/imath], [imath]m = n[/imath] Thus, the solutions are of the form [imath]x = 65k[/imath],[imath]y = 56k[/imath] for integers [imath]k[/imath], and [imath]x+y = (65+56)k = 121k = 11(11k)[/imath]. Thus, [imath]x+y[/imath] is even divisible by [imath]11[/imath] | 571215 | Prove if [imath]56x = 65y[/imath] then [imath]x + y[/imath] is divisible by [imath]11[/imath]
If [imath]x[/imath] and [imath]y[/imath] are natural numbers, and [imath]56x = 65y[/imath], prove that [imath]x + y[/imath] is divisible by [imath]11[/imath]. I tried taking the [imath]\gcd(56x,65y)[/imath] using the Euclidean algorithm, but I got nowhere with it and do not know where to head. |
570373 | Prove there are infinitely many *primitive* solutions to [imath]x^2 + y^2 = z^4[/imath]
For x, y, z [imath]\in \mathbb N [/imath] where [imath]\gcd(x,y) = 1[/imath] These solutions must also be primitive. If we let [imath] w = z^2 [/imath] so that [imath] x^2 + y^2 = w^2[/imath] I have that for r, s [imath]\in \mathbb N[/imath] where [imath]\gcd(r, s) = 1[/imath] (coprime), with opposite parity: Let [imath] x = r^2 - s^2\\ y = 2rs \\ w = r^2 + s^2 [/imath] Now for [imath]t, u \in \mathbb N[/imath] where [imath]\gcd(t, u) = 1[/imath] (coprime), with opposite parity: let [imath] r = t^2 - u^2\\ s = 2tu\\ z = t^2 + u^2 [/imath] Where [imath]r,s,z[/imath] are coprimes with [imath]r[/imath] odd and [imath]s[/imath] even (by definition). So we have: [imath] x = ( t^2 - u^2)^2 -(2tu)^2\\ y = 2( t^2 - u^2)(2tu)\\ w = z^2 = ( t^2 - u^2)^2 + (2tu)^2 [/imath] Hence [imath]x, y[/imath] and [imath]z^2 [/imath] is a primitive solution. Any mistakes here? | 172539 | prove that [imath]x^2 + y^2 = z^4[/imath] has infinitely many solutions with [imath](x,y,z)=1[/imath]
Prove that [imath]x^2 + y^2 = z^4[/imath] has infinitely many solutions with [imath](x,y,z)=1[/imath]. Do I use the terms [imath]x= r^2 - s^2[/imath], [imath]y = 2rs[/imath], and [imath]z = r^2 + s^2[/imath] to prove this problem? Thanks for any help. |
578021 | How to prove that if [imath]\sum_{n=0}^{\infty}a_n[/imath] absolutely convergent [imath]\Rightarrow \sum_{n=0}^{\infty}(a_n)^2[/imath] convergent
I need to prove that if [imath]\sum_{n=0}^{\infty}a_n[/imath] absolutely convergent [imath]\Rightarrow \sum_{n=0}^{\infty}(a_n)^2[/imath] convergent. Do you have any idea haw can I prove it? Thank you! | 493756 | Prove that if [imath]\sum{a_n}[/imath] converges absolutely, then [imath]\sum{a_n^2}[/imath] converges absolutely
I'm trying to re-learn my undergrad math, and I'm using Stephen Abbot's Understanding Analysis. In section 2.7, he has the following exercise: Exercise 2.7.5 (a) Show that if [imath]\sum{a_n}[/imath] converges absolutely, then [imath]\sum{a_n^2}[/imath] also converges absolutely. Does this proposition hold without absolute convergence? I'm posting about this here because my answer to that last question about whether the proposition holds without absolute convergence is "Yes", but my suspicions are raised by him merely asking the question. Usually questions like this are asked to point out that certain conditions are necessary in the statement of propositions, theorems, etc. I just want to see if I'm missing something here. Anyway, here's how I prove the absolute convergence of [imath]\sum{a_n^2}[/imath], and note that I never use the fact that [imath]\sum{a_n}[/imath] is absolutely convergent: Proof: Let [imath]s_n = \sum_{i=1}^n{a_n^2}[/imath]. I want to show that [imath](s_n)[/imath] is a Cauchy sequence. So, let [imath]\epsilon > 0[/imath] and [imath]n > m[/imath], and consider, \begin{equation} \begin{aligned} |s_n - s_m| &= |(a_1^2 + a_2^2 + \cdots a_n^2) - (a_1^2 + a_2^2 + \cdots a_m^2)|\\ &= |(a_{m+1})^2 + (a_{m+2})^2 + \cdots + a_n^2|\\ &= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\ \end{aligned} \end{equation} Now, since [imath]\sum{a_m}[/imath] converges, the sequence [imath](a_m)[/imath] has limit 0. Therefore I can choose an [imath]N[/imath] such that [imath]|a_m| < \sqrt{\epsilon/n}[/imath] for all [imath]m > N[/imath]. So for all [imath]n > m> N[/imath], we have, \begin{equation} \begin{aligned} |s_n - s_m| &= |a_{m+1}|^2 + |a_{m+2}|^2 + \cdots + |a_n|^2\\ &< \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n}}\right)^2\\ &< \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2 + \cdots + \left(\sqrt{\frac{\epsilon}{n-m}}\right)^2\\ &= \epsilon \end{aligned} \end{equation} Therefore [imath](s_n)[/imath] is Cauchy and [imath]\sum{a_n^2}[/imath] converges. And since [imath]\sum{a_n^2} = \sum{|a_n^2|}[/imath], the series [imath]\sum{a_n^2}[/imath] is absolutely convergent. ∎ Am I doing something wrong here? Am I right in thinking that [imath]\sum{a_n}[/imath] need not be absolutely convergent for [imath]\sum{a_n^2}[/imath] to be absolutely convergent? |
577814 | [imath]n^x[/imath] are rationals for all positive integers n, show [imath]x[/imath] must be an integer.
I wanna show that if [imath]n^x[/imath] are rational for all positive integers n, then we may conclude [imath]x[/imath] is an integer. First of all, WLOG, we may assume [imath]x[/imath] is a real number between 0,1. I think it suffices to show that for R={[imath]log_p q[/imath], where p,q are coprime positive integers bigger than 1}, then the elements of R+R contained in R must be two pairs with same p. But I stuck at proving my conjectural statement... | 570218 | If [imath]n^c\in\mathbb N[/imath] for every [imath]n\in\mathbb N[/imath], then [imath]c[/imath] is a non-negative integer?
Supposing that a real number [imath]c[/imath] is given, is the following true? "If [imath]n^c[/imath] is a natural number for every natural number [imath]n[/imath], then [imath]c[/imath] is a non-negative integer." Though this seems true, I can't prove that. Can anyone help? |
578141 | The inverse of [imath]f(x)=\sqrt[3]{1-x^3}[/imath] is itself
As I was studying this function, [imath]f(x)=\sqrt[3]{1-x^3}[/imath], I checked that the function is one-to-one, and so is invertible. Then: [imath]y=\sqrt[3]{1-x^3}[/imath] [imath]y^3=1-x^3[/imath] [imath]y^3-1=-x^3[/imath] [imath]1-y^3=x^3[/imath] [imath]x=\sqrt[3]{1-y^3}[/imath] [imath]f^{-1}(y)=\sqrt[3]{1-y^3}[/imath] What kind of functions are inverse of itselfs? Thanks. | 46635 | Examples of involutions on [imath]\mathbb{R}[/imath]
Just recently I read up something about involutions (functions [imath]f: A \rightarrow A[/imath] such that [imath]f(f(x))=x[/imath], for all [imath]x[/imath] in the domain of [imath]f[/imath]), and was wondering how many (if there is a small set of general functions) such involutions exist for [imath]A = \mathbb{R}[/imath], or maybe [imath]A = \mathbb{R} - S[/imath], where S is some set of points that would make the involution work if they were left out of [imath]A[/imath]. In general I'm interested in real functions that are involutions, continuous or otherwise. There were some examples I found here, but any more would be certainly very interesting! |
578231 | Computing the integration [imath]\int_0^{+\infty}{\frac{\sin(x)}{\sqrt{x}}}dx[/imath].
For any [imath]A>0[/imath],[imath]|\int_0^A{\sin(x)}|\le2[/imath] and [imath]{\frac{1}{\sqrt{x}}}[/imath] is decreasing to [imath]0[/imath] as [imath]x[/imath] tend to [imath]+\infty[/imath].By Dirichlet's test,the integration [imath]\int_0^{+\infty}{\frac{\sin(x)}{\sqrt{x}}}dx[/imath] makes sense. Moreover,I know the method of computing [imath]\int_0^{+\infty}{\frac{\sin(x)}{x}}dx[/imath] which is that considering the integration [imath]I(\beta)=\int_0^{+\infty}f(x,\beta)dx=\int_0^{+\infty}e^{-\beta{x}}{\frac{\sin(x)}{x}}dx[/imath] and [imath]I^{'}(\beta)=\int_0^{+\infty}f_\beta(x,\beta)dx=-\int_0^{+\infty}e^{-\beta{x}}{\sin(x)}dx.[/imath] We can compute [imath]\int_0^{+\infty}e^{-\beta{x}}{\sin(x)}dx[/imath] easily thourgh integral by parts.If we use the same method,we will obtain[imath]I^{'}(\beta)=-\int_0^{+\infty}e^{-\beta{\sqrt{x}}}{\sin(x)}dx.[/imath] Integral by parts is out of use since it will produce many [imath]\sqrt{x}[/imath].Can you tell me how to integrate this integration and more general case [imath]\int_0^{+\infty}{\frac{\sin(x)}{x^a}}dx[/imath],where [imath]0<a<1[/imath].Thank you in advance! | 171970 | Proof of [imath]\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}[/imath]
Numerically it seems to be true that [imath] \int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}. [/imath] Any ideas how to prove this? |
34767 | [imath]\int_{-\infty}^{+\infty} e^{-x^2} dx[/imath] with complex analysis
Inspired by this recently closed question, I'm curious whether there's a way to do the Gaussian integral using techniques in complex analysis such as contour integrals. I am aware of the calculation using polar coordinates and have seen other derivations. But I don't think I've ever seen it done with methods from complex analysis. I am ignorant enough about complex analysis to believe it can somehow be done without knowing how it would be done. | 1628962 | Residue calculus: [imath]\int_{-\infty}^\infty e^{-x^2} \mathrm{d}x[/imath]
I am pretty sure I have read the answer somewhere on this site, but sadly I am unable to find the question. How to evaluate [imath]\int_{-\infty}^\infty e^{-x^2} \mathrm{d}x[/imath] using the residue theorem? |
578536 | How to show that [imath]f(x) = 0[/imath] for all [imath]x[/imath] in [imath][a, b][/imath]
Assume that [imath]a \lt b[/imath] and that the continuous function [imath]{\rm f} : \left[a, b\right] \to {\mathbb R}[/imath] has the following two properties: [imath]\quad\left(\rm a\right)~{\rm f}\left(x\right) \geq 0\,, \forall\ x \in \left[a, b\right]\quad[/imath] and [imath]\quad\left(\rm b\right)~\displaystyle{\int_{a}^{b}{\rm f}\left(x\right)\,{\rm d}x = 0}[/imath]. How do I show that [imath]{\rm f}\left(x\right) = 0\,, \forall\ x \in \left[a, b\right]\ {\large\rm }[/imath]? | 82839 | Prove the integral of [imath]f[/imath] is positive if [imath]f ≥ 0[/imath], [imath]f[/imath] continuous at [imath]x_0[/imath] and [imath]f(x_0)>0[/imath]
Prove that [imath]\int_a^b f(x)\,dx \gt 0[/imath] if [imath]f \geq 0[/imath] for all [imath]x \in [a,b][/imath] and [imath]f[/imath] is continuous at [imath]x_0 \in [a,b][/imath] and [imath]f(x_0) \gt 0[/imath] EDIT. Please ignore below. It is very confusing actually -.- Note: After typing this all out, I think I realized that my proof is complete, but I'll just post it to make sure :) Attempt: Find a partition [imath]P = \{t_0,\ldots,t_n\}[/imath] of [imath][a,b][/imath] s.t. [imath]f(x)\gt f(x_0)/2[/imath] for any [imath]x \in [t_{i-1},t_i][/imath] Thus, the lower sum, [imath]L(f,P)\geq x_0 (t_{i}-t_{i-1})/2 > 0[/imath] [Basic Idea: since f is continuous at [imath]x_0[/imath], in the worst case scenario (i.e. [imath]f=0[/imath] at all points except in a nbhd of [imath]x_0[/imath]) there must be some "bump" at [imath]x_0[/imath], which prevents the integral from actually equaling [imath]0[/imath]. If we can find just one lower sum of a partition to be [imath]> 0[/imath], we will be done (I think...)] |
541179 | If integral is zero and function is continuous and non negative, then what about the function?
If [imath]f[/imath] is continuous on [imath][a,b][/imath], [imath]f(x)≥0[/imath] on [imath][a,b][/imath] and [imath]\int_{a}^{b} f(x) =0[/imath] then prove that [imath]f(x)=0[/imath] for all [imath]x \in [a,b][/imath]. I tried with Riemann's definite integral definition but couldn't proceed | 2771764 | Show that if [imath]\int_{a}^{b}g(t)dt=0[/imath] then [imath]g(t)=0[/imath] for all [imath]t\in[a,b][/imath]
Suppose [imath]g[/imath] is a continuous function on [imath][a,b][/imath] and that [imath]g(t)\ge0[/imath] for all [imath]t\in[a,b][/imath]. Show that if [imath]\int_{a}^{b}g(t)dt=0[/imath] then [imath]g(t)=0[/imath] for all [imath]t\in[a,b][/imath] I know that there exists some function [imath]G(x)= \int_{a}^{b}g(t)dt[/imath] where [imath]G(x)[/imath] is differentialable at any [imath]c[/imath] in the interval where [imath]G'(c)=g(c)[/imath]. Now since [imath]g(t)[/imath] is always positive it should follow that [imath]G'(x)[/imath] is always positive therefore [imath]G(x)[/imath] would be increasing. Now given [imath]G(x)= \int_{a}^{b}g(t)dt =0[/imath] would mean that [imath]G(b)-G(a)=0[/imath] thus [imath]G(b)=G(a)[/imath] now since [imath]G(x)[/imath] is increasing that would mean that [imath]g(t)=0[/imath] for all [imath]t[/imath] in the interval. |
309791 | Sequences in Banach spaces
I am very bad with proofs that ask you to show that "[imath]\exists[/imath] something..." because in most of them you have to explicitly show the something. The following question is one such. Any help will be greatly appreciated! Let [imath]\{u_n\}[/imath] a sequence in a Banach space [imath]X[/imath]. Suppose that [imath]\sum_{n=1}^{\infty} \|u_n\| < \infty[/imath]. Prove that there exists some [imath]x \in X[/imath] such that [imath]\lim_{n \to \infty} \sum_{k=1}^{n} u_k = x[/imath] Thank you very much in advance! | 14980 | Converging series in Banach space
Does someone know if the following is true: Let [imath]\mathbb{X}[/imath] be some arbitrary Banach space. [imath]\{x_k \}_{k=1}^{\infty} \in \mathbb{X}[/imath] is a sequence chosen from [imath]\mathbb{X}[/imath]. Now, if the series [imath]\sum_{k=1}^\infty \|x_k\|_X[/imath] converges, would the "more generic" series [imath]\sum_{k=1}^\infty x_k[/imath] also converge? If yes, could you please give the proof (or just mark the proof steps) ? Thank you. |
578653 | [imath] (\mathbb{R}^3, \|.\|_1) [/imath] and [imath] (\mathbb{R}^3, \|.\|_\infty) [/imath] cannot be isometric
Can anyone prove that the spaces [imath] (\mathbb{R}^3, \|.\|_1) [/imath] and [imath] (\mathbb{R}^3, \|.\|_\infty) [/imath] cannot be isometric? Thanks. | 179887 | How to show that [imath]\mathbb R^n[/imath] with the [imath]1[/imath]-norm is not isometric to [imath]\mathbb R^n[/imath] with the infinity norm for [imath]n>2[/imath]?
Could you please give me a hint to prove that [imath]\mathbb{R}^n[/imath] with the 1-norm [imath]\lvert x\rvert_1=\lvert x_1\rvert+\cdots+\lvert x_n\rvert[/imath] is not isometric to [imath]\mathbb{R}^n[/imath] with the infinity-norm [imath]\lvert x\rvert_\infty = \max_{i=1,\ldots,n}(\lvert x_i\rvert)[/imath] for [imath]n\gt2[/imath]. I do not see how the critical case [imath]n=2[/imath] enters the picture. Thank you! |
578462 | A sufficient condition for irreducibility of a polynomial in an extension field.
Here is another problem of the book Fields and Galois Theory by Patrick Morandi, page 37. Let [imath]f(x)[/imath] be an irreducible polynomial over [imath]F[/imath] of degree [imath]n[/imath], and let [imath]K[/imath] be a field extension of [imath]F[/imath] with [imath][K:F]=m[/imath]. If [imath]\gcd(n,m)=1[/imath], show that [imath]f[/imath] is irreducible over [imath]K[/imath]. I'm preparing for my midterm exam so I'm trying to solve as many as this book problems. Thanks for your helps. | 155319 | An irreducible polynomial of degree coprime to the degree of an extension is irreducible over this extension
I'm having a hard time showing this: If [imath]K[/imath] is an extension of [imath]\mathbb{Q}[/imath] with degree [imath]m[/imath] and [imath]f(x)[/imath] an irreducible polynomial over the rationals with degree [imath]n[/imath], such that [imath]\gcd(m, n)=1[/imath], then [imath]f(x)[/imath] is irreducible over [imath]K[/imath]. I have tried it by writing [imath]f(x)=a(x)b(x)[/imath] and then looking at the coefficients of those polynomials (some of them must belong to [imath]K-\mathbb{Q}[/imath] which could possibly result in a contradiction) to no success. I have no idea where to use the hypothesis of m and n being relatively prime. Any help would be appreciated. |
578465 | Integral mean value theorem
Let [imath]f:[a,b]\to\mathbb{R}[/imath] be continuous and suppose that [imath]\int_a^bf(x)dx=\int_a^bxf(x)dx=\int_a^bx^2f(x)dx=0[/imath] Show that there are distinct points [imath]x_1,x_2,x_3\in(a,b)[/imath] such that [imath]f(x_1)=f(x_2)=f(x_3)=0[/imath] I think this requires mean value theorem for integrals. But this can only show the existence of one such point, how to see there are at least two more? | 443266 | Prove the exsistence of 3 zero points of a function
Assuming [imath]a<b[/imath], function f(x) is continous at [a,b], and we have [imath]\int_a^bf(x)dx=\int_a^bxf(x)dx=\int_a^bx^2f(x)dx=0[/imath] Prove that [imath]\exists \ x_1,x_2,x_3\text {(different from each other)} \in(a,b)[/imath] satisfying [imath]f(x_1)=f(x_2)=f(x_3)=0[/imath]. I have proved the existence of one zero point by differential mean value theorem, but have no idea about going on. Thanks for any solution or hint. |
578711 | Orthogonal projection.
I have found this question in a book, but I don't know how to use that [imath]\left\Vert P\right\Vert =1[/imath]. Question: If [imath]P\in\mathcal{L}(H)[/imath] is a projection and [imath]\left\Vert P\right\Vert =1[/imath], show that [imath]P[/imath] is an orthogonal projection. Where [imath]\mathcal{L}(H)[/imath] is the set of linear bounded operators on a Hilbert space [imath]H[/imath]. Thanks. | 415426 | Prove that [imath]T[/imath] is an orthogonal projection
Let [imath]T[/imath] be a linear operator on a finite-dimensional inner product space [imath]V[/imath]. Suppose that [imath]T[/imath] is a projection such that [imath]\|T(x)\| \le \|x\|[/imath] for [imath]x \in V[/imath]. Prove that [imath]T[/imath] is an orthogonal projection. I can't understand well. The definition of orthogonal operator is [imath]\|T(x)\| = \|x\|[/imath]. But why that [imath]\|T(x)\| \le \|x\|[/imath] for [imath]x \in V[/imath] means orthogonal projection? |
577211 | A question using mean value theorem
Let [imath]f:[a,b]\to \mathbb{R}[/imath] be twice-differentiable (i.e. [imath]f'[/imath] and [imath]f''[/imath] exist) on [imath][a,b][/imath] and [imath]f'(a)=f'(b)=0[/imath]. Prove that there exists [imath]c \in (a,b)[/imath] such that [imath]4\cdot|f(b)-f(a)|/(b-a)^2 \leq |f''(c)|[/imath]. I guess the prove uses the mean value theorem but I cannot figure out how. | 557599 | Prove that [imath]|f''(\xi)|\geqslant\frac{4|f(a)-f(b)|}{(b-a)^2}[/imath]
Let [imath]{\rm f}:\left[a, b\right]\to\mathbb{R}[/imath] be twice differentiable, and suppose [imath]\lim_{x\to a^{+}} \frac{{\rm f}\left(x\right) - {\rm f}\left(a\right)}{x - a} = \lim_{x\to b^{-}}\frac{{\rm f}\left(x\right) - {\rm f}\left(b\right)}{x - b} =0 [/imath] Show that there exists [imath]\xi \in \left(a, b\right)[/imath] such that [imath]\displaystyle{% \left\vert\vphantom{\Large A}\,{\rm f}''\left(\xi\right)\right\vert \geq \frac{4\left\vert\vphantom{\Large A}% \,{\rm f}\left(a\right) - {\rm f}\left(b\right)\right\vert} {\left(b - a\right)^{2}}}[/imath]. I don't know how to start. Any hints ?. |
572166 | How to prove something about the order of element at a group
[imath]G[/imath] is a cyclic group. the order of [imath]G[/imath] is [imath]n[/imath] ([imath]|G|=n[/imath]). [imath]m\mid n[/imath], I have to prove that there is [imath]b\in G[/imath] that [imath]ord(b)=m[/imath]. Know, here is one part of the proof: if [imath]m\mid n[/imath] that means that [imath]n=mk[/imath]. Lets mark: [imath]b=a^k[/imath] then: [imath]b^m=(a^k)^m=a^n=e\Rightarrow ord(b)=m[/imath] Now I need to prove that [imath]m[/imath] is minimal. How I'm prove this? Thank you! | 41731 | A kind of converse of Lagrange's Theorem
Let [imath]G[/imath] a finite group. If [imath]a\in G[/imath], a consequence of the Lagrange's Theorem is that the order of [imath]a[/imath] divides the order of [imath]G[/imath]. Let [imath]p=|G|[/imath]. If [imath]p[/imath] is prime, it is well known that [imath]G[/imath] is cyclic, and therefore, there exists [imath]a\in G[/imath] such that the order of [imath]a[/imath] is [imath]p[/imath] and, of course, exist [imath]b[/imath] in [imath]G[/imath] such that the order of [imath]b[/imath] is [imath]1[/imath]. If [imath]n[/imath] is a composite number, namely [imath]n=p_1^{m_1}p_2^{m_2}\ldots p_n^{m_n},[/imath] where [imath]p_i[/imath] are primes, my question is this: Is there, for each [imath]i[/imath], an element [imath]a_i[/imath] in [imath]G[/imath] such that the order of [imath]a_i[/imath] is [imath]p_i[/imath]? Is there, for each [imath]i[/imath], [imath]b_i\in G[/imath], such that the order of [imath]b_i[/imath] is [imath]p_i^{m_i}[/imath]? |
578917 | Finding determinant using properties of determinant without expanding
show that determinant [imath]\left|\matrix{ x^2+L & xy & xz \\ xy & y^2+L & yz \\ xz & yz & z^2+L \\ }\right| = L^2(x^2+y^2+z^2+L)[/imath] without expanding by using the appropriate properties of determinant. All i can do is LHS [imath]x^2y^2z^2\left|\matrix{ 1+L/x^2 & 1 & 1 \\ 1 & 1+L/y^2 & 1 \\ 1 & 1 & 1+L/z^2 \\ }\right|[/imath] Is it a must to relate to eigenvalue problem? | 578322 | Finding determinant of matrix without expanding
show that determinant [imath]\left|\matrix{ x^2+L & xy & xz \\ xy & y^2+L & yz \\ xz & yz & z^2+L \\ }\right| = L^2(x^2+y^2+z^2+L)[/imath] without expanding by using the appropriate properties of determinant. All i can do is LHS [imath]x^2y^2z^2\left|\matrix{ 1+L/x^2 & 1 & 1 \\ 1 & 1+L/y^2 & 1 \\ 1 & 1 & 1+L/z^2 \\ }\right|[/imath] |
579046 | Let [imath]f[/imath] be Lebesgue integrable on [imath]\mathbb{R}[/imath]. Show that [imath] \sum_{n=1}^{\infty} f(x+n) [/imath] converges almost everywhere.
Let [imath]f[/imath] be Lebesgue integrable on [imath]\mathbb{R}[/imath]. Show that [imath] \sum_{n=1}^{\infty} f(x+n) [/imath] converges almost everywhere. I was thinking that maybe the finite sum could be compared to a simple function and maybe this could lead somewhere, but i've doubt this since I don't know what the level sets look like for the [imath]f(i)'s[/imath]. The result seems a bit obvious but I don't have a good idea. Any helpful hints or help would be great | 334999 | convergence of infinite sum of integer values of integrable function
Assume [imath]f:\mathbb R\rightarrow {\mathbb R} [/imath] is Lebesgue integrable, prove [imath]\sum_{n\in\mathbb N } f(x+n)[/imath] converges for almost everywhere x. This is a question discussed with friends. Thanks |
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