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960736 | Evaluation of [imath]\int_1^\infty \frac{1}{x(x^2+1)} dx[/imath]
Evaluate the integral [imath]\int_1^\infty \frac{1}{x(x^2+1)} dx[/imath] Update: I know that this can be solved with many methods and I do know some of them. I'm just searching for the (agreeably) most efficient way to do this. | 2533478 | Evaluating [imath]\int_1^\infty \frac{1}{x(x^2+1)}\ dx[/imath]
[imath]I=\int_1^\infty \frac{1}{x(x^2+1)}\ dx[/imath] I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect. [imath]\frac{1}{x(x^2+1)} = \frac{1}{2} \left(\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i}\right)[/imath] So [imath]I = \frac{1}{2}\int_1^\infty\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i} \ dx \\ = \frac{1}{2}\left[\log |x| - \log(|x+i|) - \log(|x-i|)\right]_1^\infty [/imath] which evaluates to be [imath]\infty[/imath]. |
3035308 | Questioning the commutative property in complex powers
We have:[imath]e^{2\pi\cdot i(1+\frac{1}{x})}[/imath] Power properties state that: [imath]a^{b\cdot c}=(a^{b})^{c}=(a^{c})^{b}[/imath]. Thus we could re-write the above power as: [imath](e^{2\pi\cdot i})^{1+\frac{1}{x}}=1[/imath] Which should be equal to: [imath](e^{1+\frac{1}{x}})^{2\pi\cdot i}\neq1[/imath] But this is not the case. Also the first result is a Real number while the second is a Complex one. I'm not really questioning anything, I just couldn't find the right title. Can anyone explain me where the mistake is? | 2550145 | For complex numbers [imath]a,b,c[/imath], explain why [imath]a^{b\cdot c}=(a^b)^c[/imath] is not necessarily true.
For complex numbers [imath]a,b,c[/imath], explain why [imath]a^{b\cdot c}=(a^b)^c[/imath] is not necessarily true. I know that complex powers are really sets of complex numbers. But coming from real analysis the above seems confusing. Can anyone give a simple explanation of what is going on. |
3035213 | Is the set of proposition symbols necessarily a countable set?
I am busy latedays with "Logic and Structure" of Van Dalen in order to get some familiarity with logic. Proposition symbols are introduced (on page 7) as members of an infinitely countable set [imath]\{p_0,p_1,p_2,\dots\}[/imath]. My first thought was: why not just a set without any further specification? Uptil now I could not find a satisfying answer. Let me mention that I did find a proof (on page 44) that every (syntactically) consistent set of propositions is contained in a set of propositions that is maximally consistent, and in that proof it is used that the set of proposition symbols is countable. That however can also be proved if there is no countability, but then by application of the lemma of Zorn. So I wondered if the countability is there because this (i.e. Zorn/axiom of choice) must be avoided for some reason. I really don't know... So my question is: Is there a special reason/motivation to go for an (infinitely) countable set of proposition symbols? | 2490770 | Why do first order languages have at most countably many symbols?
Every proof that I read seems to assume that [imath]|L|\leq\aleph_0[/imath]. But then how do you model things like field over [imath]\mathbb{R}[/imath] without running out of variable symbols? More importantly, how can I prove that [imath]|L|\leq\aleph_0[/imath]? |
3035401 | Quotient map [imath]f[/imath] from the [imath]n[/imath]-sphere to the [imath]n[/imath]-disk
I'm trying to construct a quotient map [imath]f[/imath] from the [imath]n[/imath]-sphere [imath]S^n[/imath] to the [imath]n[/imath]-disk [imath]D^n[/imath], in which [imath]f(x_1,x_2,\ldots,x_n,x_{n+1})=f(x_1,x_2,\ldots,x_n,-x_{n+1})[/imath]. The goal is to show that [imath]S^n/\sim_f[/imath] is homeomorphic to [imath]D^n[/imath]. (Is this the right way to go about it?) I've been looking at the stereographic projection for inspiration, but haven't come up with anything. I only just started learning about topology. Any hints? | 3035259 | Show that the quotient space is homeomorphic to the n-disc
Let [imath]\sim[/imath] denote the equivalence relation on the [imath]n[/imath]-sphere [imath]S^n[/imath] defined via [imath] (x_1,\dots,x_n,x_{n+1})\sim(x_1,\dots,x_n,−x_{n+1})\:\text{ for all }\: (x_1,\dots,x_{n+1})\in S^n. [/imath] Show that the quotient space [imath]S^n/\sim[/imath] is homeomorphic to the [imath]n[/imath]-disc [imath]D^n[/imath]. I know that Ii have to show that there exists a bijective function [imath]f: (S^n/\sim) \to D^n[/imath] that is continuous and that the pre-image [imath]f^{-1}[/imath] is continuous as well, but I can't advance from here, any tips? |
3035687 | How to show convex hull of finitely many vectors in [imath]\mathbb{R^n}[/imath] is a compact set?
Suppose [imath]x_1,x_2,\cdots,x_k[/imath] is any finite set of vectors in [imath]\mathbb{R^n}[/imath]. A convex combination of these vectors is [imath]\sum_{i=1}^{k}\lambda_ix_i[/imath] where [imath]\sum_{i=1}^{k}\lambda_i=1[/imath] and [imath]\lambda_1,\lambda_2,\cdots,\lambda_k\geq 0[/imath]. The set [imath]C[/imath] which contains all the convex combinations is called convex hull. Show that [imath]C[/imath] is compact in [imath]\mathbb{R^n}[/imath]. I know that since the underlying space is [imath]\mathbb{R^n}[/imath], using Heine-Borel theorem we need to show the boundedness and closedness. For the boundedness if I assume the 2-norm of [imath]x_i[/imath]'s is bounded, i.e., [imath]\|x_i\| \leq \|M\|[/imath] then [imath] x=\sum_{i=1}^{k}\lambda_ix_i \leq \sum_{i=1}^{k}\lambda_i \|M\|=\|M\| [/imath] But for the closedness we need a sequence [imath](x_k)[/imath] in [imath]C[/imath] and show this converges in [imath]C[/imath]. How to show this part? | 749797 | Is convex hull of a finite set of points in [imath]\mathbb R^2[/imath] closed?
Is the convex hull of a finite set of points in [imath]\mathbb R^2[/imath] closed? Intuitively, yes. But not sure how to show that. Thanks! |
3035582 | Prove this is a cauchy sequence
Let [imath]{a_n}[/imath] be a sequence such that there exists an M > 0 such that for all n ∈ N one has [imath]|a_{n+1} − a_n|[/imath] ≤ M/[imath]2^n[/imath] Prove that [imath]{a_n}[/imath] is a Cauchy sequence. My attempt: I tried to use the triangle inequality to prove that for all m[imath]>[/imath]n, [imath]|a_m - a_n|[/imath]<[imath]\epsilon[/imath] but I am unable to get anything useful out of the inequalities I'm getting. | 699077 | Proving that [imath](x_n)_{n=1}^{\infty}[/imath] is a Cauchy sequence.
Let [imath](x_n)_{n=1}^{\infty}[/imath] be a sequence such that [imath]|x_{n+1} - x_n| < r^n[/imath] for all [imath]n \geq 1[/imath], for some [imath]0 < r < 1[/imath]. Prove that [imath](x_n)_{n=1}^{\infty}[/imath] is a Cauchy sequence. I understand that a Cauchy sequence means that for all [imath]\varepsilon > 0[/imath] [imath]\exists N[/imath] so that for [imath]n,m \ge N[/imath] we have [imath]|a_m - a_n| \le \varepsilon[/imath]. But this one is really giving me a headache. I tried doing something like: let [imath] m > n[/imath]. Therefore [imath]x_n - x_m[/imath] = [imath](x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + ... + (x_{m+1} - x_m) [/imath] and then somehow using the triangle inequality to compute some sum such that [imath]x_n - x_m[/imath] < sum which would be epsilon? any help is appreciated, thank you. |
3035412 | Taylor series of [imath]\ln\frac{1+x}{1-x}[/imath]
Let [imath]f(x)=\ln\frac{1+x}{1-x}[/imath] for [imath]x[/imath] in [imath](-1,1)[/imath]. Calculate the Taylor series of [imath]f[/imath] at [imath]x_0=0[/imath] I determined some derivatives: [imath]f'(x)=\frac{2}{1-x^2}[/imath]; [imath]f''(x)=\frac{4x}{(1-x^2)^2}[/imath]; [imath]f^{(3)}(x)=\frac{4(3x^2+1)}{(1-x^2)^3}[/imath]; [imath]f^{(4)}(x)=\frac{48x(x^2+1)}{(1-x^2)^4}[/imath]; [imath]f^{(5)}(x)=\frac{48(5x^2+10x^2+1)}{(1-x^2)^5}[/imath] and their values at [imath]x_0=0[/imath]: [imath]f(0)=0[/imath]; [imath]f'(0)=2[/imath]; [imath]f''(0)=0[/imath]; [imath]f^{(3)}(0)=4=2^2[/imath] [imath]f^{(4)}(0)=0[/imath]; [imath]f^{(5)}(0)=48=2^4.3[/imath]; [imath]f^{(7)}(0)=1440=2^5.3^2.5[/imath] I can just see that for [imath]n[/imath] even, [imath]f^{(n)}(0)=0[/imath], but how can I generalize the entire series? | 2784182 | Taylor Expansion [imath]\log(\frac{1+z}{1-z})[/imath]
Taylor Expansion of [imath]\log(\frac{1+z}{1-z})[/imath] around [imath]z=0[/imath] But [imath]\log(1+z)=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{z^n}{n}[/imath] How can I continue from here? |
3036217 | Expected Value and Random variables for Uniformly Random Permutation Sets
Question Let [imath]n[/imath] and [imath]k[/imath] be integers such that n is even, [imath]n\ge2[/imath] and [imath]1\le k\le n[/imath]. You are having a party where [imath]n[/imath] students attended. a) [imath]k[/imath] of these [imath]n[/imath] students are politically correct and, thus, refuse to say Merry Christmas. Instead, they say Happy Holidays. b) [imath]n - k[/imath] of these [imath]n[/imath] students do not care about political correctness and, thus, they say Merry Christmas. Consider a uniformly random permutation of these n students. The positions in this permutation are numbered as [imath]1,2,…,n[/imath]. Define the random variable [imath]X[/imath], [imath]X[/imath] = the number of positions with [imath]i[/imath] with 1<=[imath]i[/imath]<=[imath]\frac{n}{2}[/imath] such that both students at positions [imath]i[/imath] and [imath]2i[/imath] are politically correct. What is the expected value [imath]E(X)[/imath] of the random variable [imath]X[/imath]? (Use indicator variables) Options: a) [imath]n[/imath] [imath].[/imath] [imath]\frac{k(k-1)}{n(n-1)}[/imath] b) [imath]n[/imath] [imath].[/imath] [imath]\frac{(k-1)(k-2)}{n(n-1)}[/imath] c) [imath]\frac{n}{2}[/imath] [imath].[/imath] [imath]\frac{k(k-1)}{n(n-1)}[/imath] d) [imath]\frac{n}{2}[/imath] [imath].[/imath] [imath]\frac{(k-1)(k-2)}{n(n-1)}[/imath] answer is c). Attempt: Indicator Variable: [imath]X[/imath] [imath]= 1[/imath] if [imath]i[/imath] with 1<=[imath]i[/imath]<=[imath]\frac{n}{2}[/imath] such that both students at positions [imath]i[/imath] and [imath]2i[/imath] are politically correct. [imath]X=0[/imath] for all other cases We need [imath]E(X)[/imath] = [imath]\sum_{k=0}^{n/2} k . p(k)[/imath] We have [imath]\frac{n}{2}[/imath] positions? but I can’t seem to find [imath]p(k)[/imath] There’s so much information given in this question that Im confused on how to break it down beyond the basic initial expected value steps. | 3019475 | Expected Value and Random variables
Question Let [imath]n[/imath] and [imath]k[/imath] be integers such that n is even, [imath]n\ge2[/imath] and [imath]1\le k\le n[/imath]. You are having a party where [imath]n[/imath] students attended. a) [imath]k[/imath] of these [imath]n[/imath] students are politically correct and, thus, refuse to say Merry Christmas. Instead, they say Happy Holidays. b) [imath]n - k[/imath] of these [imath]n[/imath] students do not care about political correctness and, thus, they say Merry Christmas. Consider a uniformly random permutation of these n students. The positions in this permutation are numbered as [imath]1,2,…,n[/imath]. Define the random variable [imath]X[/imath], [imath]X[/imath] = the number of positions with [imath]i[/imath] with 1<=[imath]i[/imath]<=[imath]\frac{n}{2}[/imath] such that both students at positions [imath]i[/imath] and [imath]2i[/imath] are politically correct. What is the expected value [imath]E(X)[/imath] of the random variable [imath]X[/imath]? (Use indicator variables) Options: a) [imath]n[/imath] [imath].[/imath] [imath]\frac{k(k-1)}{n(n-1)}[/imath] b) [imath]n[/imath] [imath].[/imath] [imath]\frac{(k-1)(k-2)}{n(n-1)}[/imath] c) [imath]\frac{n}{2}[/imath] [imath].[/imath] [imath]\frac{k(k-1)}{n(n-1)}[/imath] d) [imath]\frac{n}{2}[/imath] [imath].[/imath] [imath]\frac{(k-1)(k-2)}{n(n-1)}[/imath] I think the answer is c). Attempt: Indicator Variable: [imath]X[/imath] [imath]= 1[/imath] if [imath]i[/imath] with 1<=[imath]i[/imath]<=[imath]\frac{n}{2}[/imath] such that both students at positions [imath]i[/imath] and [imath]2i[/imath] are politically correct. [imath]X=0[/imath] for all other cases We need [imath]E(X)[/imath] = [imath]\sum_{k=0}^{n/2} k . p(k)[/imath] We have [imath]\frac{n}{2}[/imath] positions? but I can’t seem to find [imath]p(k)[/imath] There’s so much information given in this question that Im confused on how to break it down beyond the basic initial expected value steps. |
3028074 | Prove this A∆B=C <=> B∆C=A
[imath]A∆B=C <=> B∆A=C[/imath] I don't idea. Is this correct task? Maybe the <=> means something else i don't know? | 188103 | Show that [imath]A\Delta B = C[/imath] if and only if [imath]A = B\Delta C[/imath]
For any three sets [imath]A, B[/imath] and [imath]C[/imath], show that [imath]A\Delta B = C \iff A = B\Delta C.[/imath] I am a student and wish some more information on the above. Kindly help. |
3033908 | Prove that there are infinitely many prime numbers [imath]p[/imath] such that [imath]\left(\frac{a}{p}\right)=1[/imath] for fixed [imath]a[/imath].
I already proved this is true for all prime numbers and clearly see how this is true for all perfect squares, I'm just having trouble expanding it to any prime factorization. If we let [imath]a[/imath] have prime factorization [imath]a=p_1^{a_1}p_2^{a_2}...p_n^{a_n}[/imath], then since the Legendre Symbol is multiplicative, we know that: [imath] \left(\frac{a}{p}\right)=\left(\frac{p_1}{p}\right)^{a_1}\left(\frac{p_2}{p}\right)^{a_2}...\left(\frac{p_n}{p}\right)^{a_n} [/imath] I don't, however, understand where to go from here. | 226563 | Infinitely many primes for quadratic residues
Let [imath]a \in \mathbb{N}.[/imath] Prove there are infinitely many primes [imath]p[/imath] satisfying [imath]\left(\frac{a}{p}\right) =1.[/imath] Remark: One may need to use the Dirichlet theorem, which states that if [imath]a,m[/imath] are coprime then there exists infinitely many primes [imath]p[/imath] such that [imath]p \equiv a \pmod{m}.[/imath] |
3036051 | Find all values of [imath]n[/imath] such that [imath]\varphi(n) = n/6[/imath].
Using the product formula (the formula with the prime factors of [imath]n[/imath]), I got [imath]1=6\frac{(P_1-1)}{P_1}\frac{(P_2-1)}{P_2}\cdots\frac{(P_k-1)}{P_k}\,.[/imath] | 1328909 | Is there [imath]\phi(n)=n/6[/imath]
I know how to find for which [imath]n[/imath] [imath]\phi(n)=n/2[/imath] or [imath]\phi(n)=n/3[/imath], my method for finding those was simply to find primes [imath]p[/imath] that satisfy $\Pi_p[imath]_|[/imath]_n[imath]1-1/p$ $ = 1/2$ or $1/3$.[/imath] However, I don't know how to find $\Pi_p_|[imath]_n[/imath]1-1/p = n/6$. Intuitively it seems that if I combine results for both [imath]\phi(n) = n/2[/imath] and [imath]\phi(n) = n/3[/imath] I'll get [imath]\phi(n) = n/6[/imath] but it does not work, cause I get number of the form [imath]2^a3^b[/imath] which gives [imath]\phi(n) = n/3[/imath] again. Is there a way to find [imath]n[/imath] for which [imath]\phi(n)=n/6[/imath]? Or do such numbers exist at all? |
3036522 | [imath]\vert\mathbb{Q}^m\vert = \vert\mathbb{N}\vert[/imath]. Proof
To start with I'm not completely understand what [imath]\mathbb{Q}^m[/imath] is. Is it the number of all elements in [imath]\mathbb{Q}[/imath] that was taken to some power [imath]m[/imath]? To show that two sets have the same cardinality we must introduce a map [imath]f: \mathbb{N} \longrightarrow \mathbb{Q}^m[/imath] that would be a bijection, but I got stuck with it. Can someone please show me a proof or give a hint? | 476383 | Does There exist a continuous bijection [imath]\mathbb{Q}\to \mathbb{Q}\times \mathbb{Q}[/imath]?
Does there exist a continuous bijection [imath]\mathbb{Q}\to \mathbb{Q}\times \mathbb{Q}[/imath]? I am not able to find out how to proceed. |
3036634 | [imath]R[/imath] is commutative ring with unity, prove [imath]I,J\trianglelefteq R \wedge I+J=R \Rightarrow IJ=I\cap J[/imath]
Let [imath]R[/imath] a commutative ring with unity, and let [imath]I,J\trianglelefteq R[/imath] ideals such that [imath]I+J=R[/imath]. Prove that [imath]IJ=I\cap J[/imath]. Recall [imath]IJ=\{\sum_{k=1}^n i_kj_k:i_k\in I,j_k\in J, k\in \mathbb{N} \}[/imath]. I have shown that [imath]IJ\subseteq I\cap J[/imath]. The rest of the solution does not use the commutativity of [imath]R[/imath] so my question is where is the mistake: Let [imath]a\in I\cap J[/imath]. [imath]a=a\cdot 1\in IJ[/imath] thus [imath]I\cap J=IJ[/imath]. | 356684 | For a ring R, and ideals [imath]A[/imath], [imath]B[/imath], then [imath]AB=A \cap B[/imath] if [imath]A + B = R[/imath]
[imath]AB \subseteq A \cap B[/imath] is clear. I have seen reverse inclusion proven thus, Let [imath]x \in A\cap B[/imath]. Since [imath]A+B=R[/imath], there exist [imath]a \in A[/imath], [imath]b \in B[/imath], such that [imath]a+b=1[/imath]. Then [imath]x= axa + axb + bxa + bxb[/imath]. Therefore, [imath]A \cap B \subseteq AB[/imath]. My problem: I cannot see why [imath]bxa \in AB[/imath]. Edit: If this does not hold for noncommutative rings, give a counterexample |
3036857 | Show [imath](B_t )^2[/imath] i.e. square of a Brownian motion is a Markov process.
Problem: Show [imath](B_t )^2[/imath] i.e. square of a Brownian motion is a Markov process. To do this, I want to show [imath]P( B_t ^2 | B_{t_1} ^2 , ... , B_{t_n} ^2 )= P( B_t ^2 | B_{t_n} ^2 ) [/imath] where [imath]0<t_1<...<t_n<t[/imath]. My attempt: [imath] B_t ^2 = (B_t -B_{t_n})^2 +2 (B_t -B_{t_n})B_{t_n} + B_{t_n} ^2[/imath] and [imath](B_t -B_{t_n})[/imath] is independent with [imath] \sigma (B_{t_1} ^2 , ... , B_{t_n} ^2) [/imath] while [imath]B_{t_n} ^2[/imath] is measurable w.r.t. [imath] \sigma (B_{t_1} ^2 , ... , B_{t_n} ^2)[/imath]. However, [imath] B_{t_n}[/imath] is not measurable w.r.t. [imath]\sigma (B_{t_1} ^2 , ... , B_{t_n} ^2)[/imath] so I cannot apply the standard lemma which states that I can integrate the "independent part". I have also tried to calculate the joint p.d.f of [imath]B_{t_1} ^2 , ... , B_{t_n} ^2 , B_t ^2[/imath] by change of variables from the marginal distribution of Brownian motion, but it is not very easy. Any hint is appreciated. | 1004864 | Is [imath](B_t^2)[/imath] Markov where [imath](B_t)[/imath] is Brownian motion?
I am pretty sure [imath](B_{t}^{2})[/imath] not Markov because the squared random walk is not. Showing the square of a Markov process is or isn't Markov I guess I can repeat the method since to be Markov it must satisfy the discrete. Thanks |
921241 | Intuition behind the definition of Measurable Sets
I started studying "Measure Theory and Integration" and went through the first section which talks about Lebesgue Outer Measure of a set. All was well until I started with the second section which starts with the definition of Measurable sets : ** The set [imath]E[/imath] is Lebesgue Measurable if for each set [imath]A[/imath] we have** [imath]m^*(A)=m^*(A\bigcap E)+m^*(A\bigcap E^c)[/imath] I know the [imath]\leq[/imath] inequality comes from subadditivity. So it all boils down to showing the [imath]\geq[/imath] inequality. Though I read the definition and am able to solve questions on the topic. I don't quite understand it intuitively. Any help would be appreciated. Also give me an example where [imath]\geq[/imath] inequality is not satisfied. Thank you. | 485815 | Intuition behind the definition of a measurable set
This week I saw the definition of a measurable set for an outer measure. Let [imath]\mu^*[/imath] be an outer measure on a set [imath]X[/imath]. We call [imath]A \subseteq X[/imath] measurable if [imath]\mu^*(E) = \mu^*(A\cap E) + \mu^*(A^c\cap E)[/imath] for every [imath]E \subseteq X[/imath]. This is not the first time I've seen this definition. Unlike most other things in mathematics, over time I have gained absolutely no intuition as to why this is the definition. The only explanation I've ever seen is that a set is measurable if it 'breaks up' other sets in the way you'd want. I don't really see why this is the motivation though. One reason I am not comfortable with it is that you require a measurable set to break up sets which, according to this definition, are non-measurable; why would you require that? Of course, you can't say what a non-measurable set is without first defining what it means to be measurable so I suppose no matter what your condition is, it will have to apply to all subsets of [imath]X[/imath]. Is there an intuitive way to think about the definition of measurable sets? Is there a good reason why we should use this definition, aside from "it works"? |
3037237 | Suppose that [imath]f_k[/imath] is a sequence of differentiable functions that converges uniformly to a function [imath]f[/imath]. Must [imath]f[/imath] be differentiable?
Suppose that [imath]f_k: (0,1) \rightarrow \mathbb{R}[/imath] is a sequence of differentiable functions that converges uniformly to a function [imath]f: (0,1) \rightarrow \mathbb{R}[/imath]. Must [imath]f[/imath] be differentiable? So I'm pretty sure this isn't true but struggling to find a simple counterexample. | 1461277 | A sequence of differentiable functions that converge uniformly to a non differentiable function - is my example correct?
I was trying to find a series of differentiable functions that converge uniformly to a non differentiable function, to demonstrate the fact that uniform convergence doesn't imply anything about the differentiability of the function the sequence converges to. I came up with this example - [imath]f_n(x)=\left\{ \begin{array}{c} |x| &|x|\gt\frac 1 n \\ \frac x n &|x|\leq\frac1 n \end{array} \right. [/imath] The sequence of functions converges uniformly to [imath]|x|[/imath], and all [imath]f_n[/imath] are differentiable, but [imath]|x|[/imath] isn't differentiable at [imath]x=0[/imath]. Does this seem right? Edit- people here helped me understand that my functions are not continuous at [imath]x=\frac 1n[/imath], which I forgot to check. So a correct example will be: [imath]f_n(x)=\left\{ \begin{array}{c} |x| &|x|\gt\frac 1 n \\ \frac {nx^2}2+ \frac1{2n} &|x|\leq\frac1 n \end{array} \right. [/imath] Which is continuous at [imath]|x|=\frac1n[/imath] and differentiable there. |
1527972 | Give an example of a finite extension of fields that is neither separable nor normal.
The title says it all. This is not homework. Haven't made much progress - I know that with [imath]K=F_2(t)[/imath], [imath]L=K(\sqrt t)[/imath] (i.e. adjoining the square root of [imath]t[/imath] to the function field with coefficients in the finite field on two elements), [imath]L/K[/imath] is not a separable extension. This suggests looking at extensions [imath]M[/imath] of [imath]L[/imath] that are not normal over [imath]K[/imath]. Does such an extension exist? Or is there a better example? | 1242542 | Example of an non-normal inseparable field extension
I was asked to prove or disprove that if a field extension is not normal then it is separable. I can't see why this would be true so I want to disprove it with an example of a non-normal inseparable field extension. I am wondering if there is a simple example of such an extension, or even any examples at all. I know that any inseparable extension [imath]L/K[/imath] has to be such that [imath]K[/imath] is infinite and of positive characteristic, so a good place to start would be [imath]K=\mathbb F_p(T)[/imath], but I'm unsure how to prove that an extension of this is inseparable, and even unsure how to come up with a non-normal example, since my usual approach is to exploit that [imath]K[/imath] is a subfield of [imath]\mathbb R[/imath] (so e.g. [imath]\mathbb Q(\sqrt[3]2)/\mathbb Q[/imath] is not normal as [imath]x^3-2[/imath] has [imath]\sqrt[3]2[/imath] as a root but the other two roots are in [imath]\mathbb C \setminus \mathbb R[/imath] and [imath]\mathbb Q(\sqrt[3]2)\subset \mathbb R[/imath]). However with [imath]K[/imath] as above, [imath]K \nsubseteq \mathbb R. [/imath] |
3038008 | How to prove [imath]\lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0 [/imath]
I need to prove the limit of: [imath]\lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0 [/imath] I have tried bounding it by a pairwise bigger sequence but I cannot find one that is suitable. I also cannot figure out an [imath]\epsilon[/imath]-proof, since I don't know how to adequately handle the factorial to get a substitution for n such that the convergence criterion is fulfilled. | 1319925 | Prove [imath]\lim\limits_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0[/imath]
I used [imath](n!)^{\frac{1}{n}}=e^{\frac{1}{n}\ln(n!)}=e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln(n!)}[/imath] Then using Stirling's approximation and L'Hospital's rule on [imath]\lim\limits_{n\to\infty}\frac{\ln(n!)}{n}[/imath] I get [imath]\lim\limits_{n\to\infty}\frac{\ln(n!)}{n}=\lim\limits_{n\to\infty}(\ln(n)+\frac{n+\frac{1}{2}}{n}-1)=\infty[/imath] Now, [imath]e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln(n!)}=e^{\infty}=\infty[/imath] Thus [imath]\lim\limits_{n\to\infty}\frac{1}{\sqrt[n]n}=\frac{1}{\infty}=0[/imath] Is this correct approach and what other methods could be used? |
3035197 | Show that [imath]−g[/imath] is also a primitive root of [imath]p[/imath] if [imath]p\equiv 1 \pmod{4}[/imath], but that [imath]ord_p(−g) = \frac{p−1}{2}[/imath] if [imath]p \equiv 3 \pmod{4}[/imath].
Let [imath]p[/imath] be an odd prime and let [imath]g[/imath] be a primitive root [imath]\pmod{p}[/imath]. Show that [imath]−g[/imath] is also a primitive root of [imath]p[/imath] if [imath]p \equiv 1 \pmod{4}[/imath], but that [imath]ord_p(−g) = \frac{p−1}{2}[/imath] if [imath]p \equiv 3 \pmod{4}[/imath]. So far, I have shown as [imath]p \equiv1 \pmod{4}[/imath] I can use Fermat's Little Theorem. [imath]g \equiv g^{p} \equiv -(-g)^{p} \pmod{p}[/imath] Since [imath]p \equiv 1 \pmod{4}[/imath], [imath]x^2 \equiv -1 \pmod{p}[/imath]. ([imath]-1[/imath] is a QR of [imath]p[/imath]) There [imath]\exists k \in \mathbb{Z}[/imath] such that [imath]-1 \equiv g^{2k} \equiv (-g)^{2k} \pmod{p}[/imath] Thus, [imath]g \equiv (-g)^{2k}(-g)^{p} \pmod{p}[/imath]. As [imath]g[/imath] is congruent to [imath]-g^{p}[/imath], [imath]-g[/imath] is a primitive root of [imath]p[/imath]. Is this enough to show the first part of the question, also how do I begin to show the 2nd part? | 3010827 | When g and -g are both primitive roots
The question states: Let [imath]g[/imath] by a primitive root of the odd prime [imath]p[/imath]. Show that [imath]-g[/imath] is a primitive root , or not, according as [imath]p \equiv 1 \pmod 4[/imath] or not. For me, I cannot see any connection between the type of primes and the primitive root. Any Hint is highly appreciated. Thanks |
3037392 | Relation between partitions of [imath]n[/imath] into [imath]k[/imath] distinct parts and partitions of [imath]n[/imath] into at most [imath]k[/imath] parts
I'm working on a problem that I'm completely stuck on: Let [imath]Q(n,k)[/imath] be the number of partitions of [imath]n[/imath] into [imath]k[/imath] distinct (unequal) parts. Prove that the number of partitions of [imath]n[/imath] into at most [imath]k[/imath] parts (parts can be equal here) is [imath]Q(n + \binom{k+1}{2} , k)[/imath] Any hints? | 211836 | Combinatorics problem based on Ferrers graph
Need help with this proof using Ferrers' graph or otherwise. Show that the number of partitions of [imath]r+k[/imath] into [imath]k[/imath] parts is equal to The number of partitions of [imath]r + {k+1 \choose 2}[/imath] into [imath] k [/imath] distinct parts The number of partitions of [imath]r[/imath] into parts of size at most [imath]k[/imath] |
3038476 | In a group of order [imath]m p^n[/imath] for [imath]p[/imath] prime, if [imath]k, is there an element of order p^k?[/imath]
Let [imath]G[/imath] a group of order [imath]mp^n[/imath] where [imath]p[/imath] is prime. Let [imath]k\leq n[/imath]. Is there an element of order [imath]p^k[/imath] ? Since [imath]p[/imath] divide [imath]|G|[/imath], by Cauchy theorem, there is [imath]g\in G[/imath] s.t. [imath]g[/imath] has order [imath]p[/imath]. I can't do better. I tried to use the fact that there is a [imath]p-[/imath]Sylow group, but it just confirm the fact that there is an element of order [imath]p[/imath], not of order [imath]p^k[/imath]. | 3037841 | Existence of elements of order [imath]p^k[/imath] in a finite group
By Cauchy's theorem, if a prime number [imath]p[/imath] divides the order of a group [imath]G[/imath], then there is an element [imath]g[/imath] in [imath]G[/imath] whose order is [imath]p[/imath]. In addition, if [imath]|G|[/imath] admits a prime decomposition with a factor [imath]p^k[/imath], then Sylow's theorem tells us that a subgroup of order [imath]p^k[/imath] must exist. My question is the following. If [imath]|G|[/imath] admits a prime decomposition with a factor [imath]p^k[/imath], does this imply that an element of order [imath]p^k[/imath] must exist in [imath]G[/imath]? Update: As explained in Mindlack's answer (among others), the implication does not hold and simple counterexamples exist. In the comments, Mindlack discusses special cases where such an element exists. |
3038487 | Integral [imath]\int_{-1}^{1} \frac{1}{(e^x+1)(x^2+1)}[/imath]
I do not know how to approach this integration problem. [imath] \int_{-1}^{1} \frac{1}{(e^x+1)(x^2+1)} [/imath] I tried some trigonometric change of variable and also tried to break it into two fractions but I failed! | 217762 | Evaluate the integral [imath]\int_{-1}^1\frac{dx}{(e^x+1)(x^2+1)}[/imath].
Evaluate: [imath]\int_{-1}^1\frac{dx}{(e^x+1)(x^2+1)}[/imath] |
3038653 | Is the set of [imath]n \times n[/imath] diagonalizable matrices with real entries dense in [imath]M_n(\mathbb R)[/imath]?
We can prove that diagonalizable matrices with complex values are dense in set of [imath]n \times n[/imath] complex matrices, as it was previously answered. I don't think there is many chances but can we show that the set of diagonalizable matrices over [imath]\mathbb R[/imath] with real entries is dense in [imath]M_n(\mathbb R)[/imath]? | 2184461 | The diagonalizable matrices are not dense in the square real matrices
Suppose that [imath]n \ge 2[/imath]. How to prove that the set [imath]\mathcal D \subset M_n(\mathbb R)[/imath] of the diagonalizable real matrices is not dense in [imath]M_n(\mathbb R)[/imath]? |
3038424 | [imath]K[/imath] finite field, [imath]P'(X) = 0[/imath] then P is reductible
In this exercise I've shown first that if we have [imath]P'(X)=0[/imath] then the [imath]Car K[/imath] different [imath] 0[/imath]. Now they are specifing that if[imath] K[/imath] is a finite field than [imath]P[/imath] is reductible. I've been searching for hours, tried to write the derivative of [imath]P[/imath] but I get nothing intersting. Any Hint ? Thanks for reading, | 127353 | Show [imath]f[/imath] can't be irreducible over a finite field if [imath]f^\prime[/imath] is the zero polynomial.
I'm hoping someone can give me a nudge in the right direction... Let [imath]F[/imath] be a finite field, and let [imath]f(x)[/imath] be a nonconstant polynomial whose derivative is the zero polynomial. Prove that [imath]f[/imath] cannot be irreducible over [imath]F[/imath]. I've got that every root of [imath]f[/imath] is a multiple root and that for [imath]F=\mathbb{F}_{p^r}[/imath], the exponent of every term of [imath]f[/imath] is a multiple of [imath]p[/imath]. |
3038797 | Suppose we have continuous [imath]f_{n} : D \rightarrow \mathbb{R}[/imath] converging to [imath]f : D \rightarrow \mathbb{R}[/imath] uniformly on [imath]D[/imath]. Then, is [imath]f[/imath] continuous?
Suppose we have continuous [imath]f_{n} : D \rightarrow \mathbb{R}[/imath] converging to [imath]f : D \rightarrow \mathbb{R}[/imath] uniformly on [imath]D[/imath]. Then, is [imath]f[/imath] continuous? I know that if we have [imath]f_{n}[/imath] converging to [imath]f[/imath] and [imath]f_{n}[/imath] are all continuous then it is not necessary for [imath]f[/imath] to be continuous. This is shown by [imath]x^{n} \hspace{1cm} 0 < x \leq 1 [/imath] for [imath]n \geq 0[/imath]. But, it is the uniformly part that gets me here. Is this statement true? | 2854796 | [imath]f[/imath] is continuous, if [imath]f_n[/imath] continuous and [imath]f_n\to f[/imath] uniformly
Let [imath](X,d_X)[/imath] and [imath](Y,d_Y)[/imath] be metric spaces. [imath]f_n: X\to Y[/imath] with [imath]n\in\mathbb{N}[/imath] and [imath]f:X\to Y[/imath] functions. [imath]f_n[/imath] is continuous for every [imath]n[/imath] and [imath]f_n\stackrel{n\to\infty}{\longrightarrow} f[/imath] uniformly. Then [imath]f[/imath] is continuous. Here is my proof: Let [imath]x\in X[/imath] and [imath]\epsilon >0[/imath] be arbitrary. Since [imath]f_n\to f[/imath] uniformly it exists [imath]N\in\mathbb{N}[/imath] such that [imath]d_Y(f_N(x), f(x))<\epsilon/3[/imath] for every [imath]x\in X[/imath]. As [imath]f_N[/imath] is continuous we have for [imath]x_0\in X[/imath] and [imath]\delta > 0[/imath], that [imath]d_Y(f_N(x),f_N(x_0))<\epsilon/3[/imath] if [imath]d_X(x,x_0)<\delta[/imath]. This gives us: [imath]\begin{align}d_Y(f(x),f(x_0)&\leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f(x_0))\\ &\leq d_Y(f(x), f_N(x))+d_Y(f_N(x), f_N(x_0))+d_Y(f_N(x_0), f(x_0))\\ &< \epsilon/3+\epsilon/3+\epsilon/3=\epsilon\end{align}[/imath] Thanks in advance for your correction. |
3038505 | Proof of the associated Lie algebra is isomorphic to [imath]T_e(G)[/imath]
I´m looking for a book with the proof of the result that says the associated Lie algebra of a Lie group [imath]G[/imath], let be [imath]G_{Lie}[/imath], is isomorphic to the tangent space [imath]T_e(G)[/imath], where [imath]e \in G[/imath] is the neutral element. Let me said the chapter or theorem, not only the book, please. Thank you all | 2274555 | Lie algebra and tangent space at the identity are isomorphic
Define the Lie algebra [imath]\mathfrak{g}[/imath] of a Lie group G to be the set of all left-invariant vector fields of [imath]G[/imath]. I want to prove that [imath] f: \mathfrak{g} \rightarrow T_eG \\ X \mapsto X(e) [/imath] is a linear isomorphism. The only thing that I have left to prove is surjectivity.Letting [imath]v\in T_eG[/imath], we set [imath]X\in \mathcal{X}(G)[/imath] such that [imath]\forall x\in G,\,X(x)=L_{x_{*,e}}(v)[/imath] (I still have to show that such an [imath]X[/imath] is indeed a vector field on [imath]G[/imath]). We get that [imath] f(X)=X(e)=L_{e_{*,e}}(v)=v [/imath] where the last equality holds because [imath] L_{e_{*,e}}(v)f = v(f \circ L_e) = v(f \circ id_G) = v(f) [/imath] for any [imath]f\in C^\infty(G)[/imath]. (I am thinking of the tangent vectors as derivations). Now all we have left to do is prove that such an [imath]X[/imath] is left-invariant, and it is therefore an element of [imath]\mathfrak{g}[/imath], but I do not know how. I tried to prove that [imath] \forall f\in C^\infty(G),\, L_{{x}_*,y}(X(y))f=X(L_x(y))f [/imath] but with no success. |
3028239 | Proving a logical implication using modus ponens and metatheorems
Using the law of inference the axiomatic system and metatheorems prove that [imath]{(\neg A > B),(A > C),(B > D)} \vdash (\neg C > D)[/imath] Where > is 'implies' and ~ negation. I know how to use the Laws of Logic to prove logical equivalent, but no idea about logical implication. Another question i have is using all logic laws to this impication legit?Whats the steps (a general understanding for solving this)? | 3039194 | Propositional Logic - Deduction
So i have to prove that: [imath]\{\neg A\to B,A\to C,B\to D\}\vdash \neg C\to D[/imath] I can use logical axioms, modus ponens and 'metatheorems'. Logical axioms: φ→(ψ→φ) (φ→(ψ→χ))→((φ→ψ)→(φ→χ)) (¬φ→¬ψ)→(ψ→φ) Also i can use modus ponens(the only rule i can use) and metatheorems Some thoughts:So i started experimenting with all [imath]3[/imath] tools i have, started asking myself is any of the hypotheses can give as something new using the logic axioms but then i stalled, and modus ponens can't do much on it's own knowing these hypotheses atleast.My next thought was that i have to use those 2 metatheorems in order to actualy prove one part of [imath]\neg C\to D[/imath] (based on metatheorem 2) meaning i use as a hypothesis [imath]\neg C[/imath] to prove [imath]D[/imath] but i am stuck and i don't undestand even how to start. |
3039777 | If a sequence of polynomials converges uniformly to a continuous function on the real line, then this function is a polynomial
I'm trying to prove that if [imath]f\colon\mathbb{R}\to\mathbb{R}[/imath] is continuous function and there is a sequence of polynomials [imath]p_n[/imath] that converges uniformly to [imath]f[/imath] on [imath]\mathbb{R}[/imath], then [imath]f[/imath] is a polynomial itself. It smells like somewhat near the Weierstrass approximation with polynomials, but I have no idea how to use uniform convergence on the entire real line, since this statement is wrong on any bounded interval. Any hint or solution will be appreciated. | 961796 | Show that [imath]f[/imath] is a polynomial if it's the uniform limit of polynomais
Let [imath]f:\Bbb R\to \Bbb R[/imath] be a function which is the uniform limit of polynomials. I want to show that [imath]f[/imath] is a polynomial. I mean this seems a bit trivial... If it's the uniform limit of the set of polynomials doesn't that guarantee it's a polynomial? If I define [imath]f[/imath] to be the set of all polynomial do I define a uniform limit [imath]L[/imath] s.t [imath]d(f,L)<\epsilon[/imath]? or approach via contradiction? any help would be great |
3039546 | Proving the product of an odd number and the power of 2 is bijective
How do I prove this function is bijective? [imath]v(s,p)=2^{p-1}(2s-1). [/imath] The domain is the natural numbers and the codomain is also the natural numbers So I have to somehow show that every natural number can be written as a product of the power of 2 and an odd number. | 3039432 | Proving [imath]v(s,p)=2^{p-1}(2s-1)[/imath] is a bijection of natural numbers and [imath]f(s)=2s-1[/imath] is a bijection between natural numbers and odd numbers.
How do I prove this function is bijective? [imath] v(s,p)=2^{p-1}(2s-1). [/imath] The domain is natural numbers and the codomain is also the natural numbers. And this one: [imath] f(s)=2s-1. [/imath] The domain is the natural numbers, and the codomain is the odd numbers in the natural numbers. With this one I would do this to show it's injective: [imath] \begin{align} v(s)&=v(s_1)\\ \implies 2s-1&=2s_1-1\\ \implies (2s_1)/2&=(2s_2)/2\\ \implies s=s_1 \end{align} [/imath] So it's injective since if [imath]v(s)=v(s_1)[/imath] then [imath]s=s_1[/imath]. And to show it's surjective [imath]f(s)=y[/imath]: [imath] \begin{align} y&=2s-1\\ \implies s&=(y+1)/2 \end{align} [/imath] Then the function must be surjective since every [imath]y[/imath] is the same as the codomain for [imath]f[/imath]. Am I correct? |
3024118 | Space of orientation-preserving diffeomorphisms on the circle.
Diff[imath]_+(\mathbb{T})[/imath] is the space of orientation-preserving diffeomorphisms on the circle. Then is it true that it is connected? I have looked through the similar questions on here, but seem to find no clear motivation, so far by reading through the sources of some of the post I have concluded that this statement is true. But I cannot seem to find any good motivation as to why. If somebody could please refer me to a text where this is shown or describe it that would be great. Thanks! | 88386 | Diffeomorphism group of the unit circle
I am given to understand that the group of diffeomorphisms of the unit circle, [imath]\operatorname{Diff}(\mathbb{S}^1)[/imath], has two connected components, [imath]\operatorname{Diff}^+(\mathbb{S}^1)[/imath] and [imath]\operatorname{Diff}^-(\mathbb{S}^1)[/imath], the diffeomorphisms that preserve or reverse the canonical (counterclockwise) orientation, respectively. Question 1: How does one prove that? Question 2: Given [imath]\Phi, \Psi \in \operatorname{Diff}^+(\mathbb{S^1})[/imath], can one construct an explicit path joining them? I'd be satisfied already with a proof that we can join [imath]\Psi, \Phi \in \operatorname{Diff}^+ (\mathbb{S}^1)[/imath], without giving the path explicitly. I tried the obvious path, [imath]t \mapsto \dfrac{t\Phi + (1-t)\Psi}{|t\Phi + (1-t)\Psi|}[/imath], but that doesn't seem to work. Thanks. |
3040146 | let [imath]f:[a,b]\rightarrow\Bbb{R}[/imath] be such that [imath]f(x)=0[/imath] except at finitely many points. Show the integral of [imath]f(x)[/imath] equals zero.
I'm working on the following problem: Let [imath]f:[a,b]\rightarrow\Bbb{R}[/imath] be such that [imath]f(x)=0[/imath] except at finitely many points. Show that [imath]\int_a^b{f(x)}dx=0[/imath]. Am I pointed in the wrong direction for wanting to use contradiction? If I assume [imath]\int_a^b{f(x)}dx\neq0[/imath] then can I assume [imath]f[/imath] must contain a variable x such that [imath]f[/imath] is not a constant? Is it that simple? I'm missing a lot of connections. | 2871367 | Riemann integration of a function with finite number of non zero points.
Proving that, if a function [imath]f:[a,b] \rightarrow \mathbb{R}[/imath] be zero except at finitely many points, it is Riemann-integratable with integral [imath]0[/imath] : Let the function attain non-zero values at [imath]x_i [/imath], where [imath]i={1, 2, 3,...n}[/imath], where [imath]x_i<x_{i+1}[/imath], with respective values being [imath]a_i[/imath]. Let us now split the domain into finite number almost disjoint intervals, [imath]I_1, I_2, ..., I_{\omega}[/imath] , thus forming a finite partition of [imath][a,b][/imath] . Let us now assume a finite subsequence of intervals, [imath]I_{n_1}, I_{n_2},..., I_{n_k}[/imath], containing the respective non zero values. It is to be noted, with proper "refinement" of partitions, we can make [imath]max\{k\}=n[/imath] [as [imath]n[/imath] is finite]. Further, assume the length of the respective subintervals, containing the numbers be [imath] \delta_i[/imath], where [imath]i[/imath] runs from [imath]1[/imath] to [imath]k[/imath]. When [imath]k<n[/imath], we have [imath]sup(f)_{I_{n_i}}=max\{a_{t_i}\}=\alpha_{t_i}[/imath] [ where [imath]a_{t_i} \in I_{n_i}[/imath] ] and [imath]inf(f)_{I_{n_i}}=0[/imath]. For this type of partition modes, we have [imath]U(f;P)= \sum_{i=1}^{k} \delta_i \alpha_{t_i} [/imath] and [imath]L(f;P)=0[/imath]. As the subintervals containing the points containing [imath]x_i[/imath] becomes fine enough, we can have [imath]x_i \notin I_{n_a} \cap I_{n_b}[/imath] for any [imath]i, a[/imath] and [imath]b[/imath]. Now, we have [imath]U(f;P')= \sum_{i=1}^{n} \delta_i x_i \leq U(f;P) [/imath] with [imath]L(f;P')=0[/imath] as usual. Now, as the norm of the partition [imath]P'[/imath] approaches [imath]0[/imath], i.e. [imath] ||P'| | \rightarrow 0[/imath], [imath]\delta_i \rightarrow 0 [/imath]. Hence [imath] \mathbf{inf} \ U(f; P')=0[/imath]. So, [imath] \mathbf{inf} \ U(f; P')=0= \mathbf{sup} \ L(f; P')[/imath]. QED Is the above proof adequate and valid? |
2916616 | relaxation on Dini's Theorem
Dini's Theorem states: Let [imath][a,b][/imath] be a compact intervall. Let [imath]f,f_{n}: [a,b] \xrightarrow{} \mathbb{R}[/imath], [imath] n \in \mathbb{N}[/imath], functions with [imath]f[/imath] and [imath]f_{n}[/imath] are continuos for all [imath]n[/imath] [imath]\lim_\limits{n \to \infty} f_{n}(x) = f(x)[/imath] [imath]\forall x \in [a,b][/imath] [imath]f_{n}(x) \leq f_{n+1}(x)[/imath] [imath]\forall x \in [a,b][/imath] and [imath]n \in \mathbb{N}[/imath]. Then [imath]\{f_{n}\}_{n \in \mathbb{N}}[/imath] converges uniformly to [imath]f[/imath]. Suppose only conditions [imath]1[/imath] and [imath]2[/imath] are satisfied. Are in this case [imath]\{f_{n}\}_{n \in \mathbb{N}}[/imath] also converges uniformly to [imath]f[/imath]? The reason why I'm asking this quesion ist that I found a variant of Dini's theorem that states if condition [imath]1[/imath] and [imath]2[/imath] are satisfied and we have that [imath]f_{n}[/imath] is a so called commute sequence, i.e. f_{n} has a monotonically increasing and a monotonically decreasing part, then Dini's Theorem also holds | 82766 | Dini's Theorem and tests for uniform convergence
Suppose [imath]f_n[/imath] is a sequence of functions defined a set [imath]K[/imath] with pointwise limit function [imath]f[/imath]. I am confused about the following. If the following conditions are satisfied: [imath]f_n[/imath] is continuous on [imath]K[/imath] for all [imath]n[/imath]. The pointwise limit [imath]f[/imath] is continuous on [imath]K[/imath]. [imath]K[/imath] is a compact interval (i.e., a closed and bounded interval in [imath]\mathbb{R}[/imath]). The convergence of [imath]f_n[/imath] to [imath]f[/imath] is increasing or decreasing. Then does this imply that [imath]f_n[/imath] is uniformly convergent to [imath]f[/imath]? Now a different problem: If one of these conditions is not satisfied, does this imply that [imath]f_n[/imath] is not uniformly convergent to [imath]f[/imath]? If [imath]f_n[/imath] is not uniformly convergent to f, does this mean that one these four conditions doesn't hold? In general: what is the logical relationship between uniform convergence and these four conditions? Please, I need answers to all the above questions because this theorem always confuses me when I solve the problems and I don't know how to use it properly. Thanks for your help in advance. |
164178 | Is [imath]\mathbb{R}^\omega[/imath] a completely normal space, in the box topology?
Basically, what the title says. Is [imath]\mathbb{R}^\omega[/imath] a completely normal space in the box topology ? ([imath]\mathbb{R}^\omega[/imath] is the space of sequences to [imath]\mathbb{R}[/imath]) Thanks ! | 3039316 | Is [imath]\mathbb{R}^\omega[/imath] endowed with the box topology completely normal (or hereditarily normal)?
Just out of curiosity, I'd like to know more properties of box topology. I found Is [imath]\mathbb{R}^\omega[/imath] a completely normal space, in the box topology? quite interesting, but unfortunately, it hasn't attracted too much attention. I also searched it in MathOverflow, the comment by Ramiro de la Vega in Is it still an open problem whether [imath]ℝ^[/imath] is normal in the box topology? asserted it's been known the answer is negative. However, I can't obtain any further information on the Internet. Could somebody provide a disproof, or at least offer some useful links? |
3040302 | Solving for a variable in a modular arithmetic equation
[imath]\fbox{$13x + 1 \equiv 0 \pmod {100}$}[/imath] I solved the equation above by trying different multiples to isolate [imath]x[/imath] until I found something that worked. I have two questions: [imath]\fbox{$1.$}\ [/imath] What if there was no solution for [imath]x[/imath]? How would I be able to prove it? [imath]\fbox{$2.$}\ [/imath] Are there a set of steps that I could program a computer to follow and get an answer if other similar modular equations are inputted? My solution is below: [imath]13x +1 \equiv 0 \pmod {100}[/imath] [imath]13x \equiv 99 \pmod {100}[/imath] (added [imath]99[/imath] to both of equation and applied the [imath]\mod 100[/imath] to the left side) [imath]104x \equiv 792 \pmod {100}[/imath] (multiplied both sides by [imath]8[/imath]) [imath]4x \equiv 792 \pmod {100}[/imath] (removed a [imath]100[/imath] from the left side) [imath]x \equiv 198 \pmod {100}[/imath] (divided both side by [imath]4[/imath]) Like I said, I believe I got the right solution but only through trial and error. I was wondering if there is a more systematic way of solving these problems. Thank you for any help. | 318423 | How to solve [imath] 13x \equiv 1 ~ (\text{mod} ~ 17) [/imath]?
How to solve [imath] 13x \equiv 1 ~ (\text{mod} ~ 17) [/imath]? Please give me some ideas. Thank you. |
1308936 | Analytic function defined on unit disc with co-domain except negative real axis.
I am trying to solve the following problem. Let [imath]f[/imath] be an analytic function defined on [imath]\mathbb{D}=\{z:|z| <1 \}[/imath] such that range of [imath]f[/imath] is contained in [imath]\mathbb{C}[/imath] \ [imath](-\infty,0][/imath].Then there exist an analytic function [imath]g[/imath] such that [imath]Re g(z) \ge 0[/imath] and [imath]g[/imath] is a square root of [imath]f[/imath] on [imath]\mathbb{D}[/imath] and there exist an analytic function [imath]g[/imath] such that [imath]Re g(z) \le0[/imath] and [imath]g[/imath] is a square root of [imath]f[/imath] on [imath]\mathbb{D}[/imath]. Clearly fuction [imath]Log(f(z))[/imath] and function [imath]f(z)^{1/n} [/imath]are well defined according to given co-domain of [imath]f[/imath] . But these functions are not required function according to me. Then how to find required function [imath]g[/imath]. Since [imath]f[/imath] is non zero so [imath]f^{'}/f[/imath] is also analytic and there is a function [imath]g[/imath] such that [imath]g=Log(f(z))[/imath]. Now i am stuck please some one help. | 297741 | [imath]f[/imath] be an analytic function defined on [imath]\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}[/imath]
I came across the following problem that says: Let [imath]f[/imath] be an analytic function defined on [imath]\mathbb{D}=\{z\in\mathbb{C}:|z|<1\}[/imath] such that the range of [imath]f[/imath] is contained in the set [imath]\mathbb{C}\setminus (-\infty,0][/imath]. Then [imath]f[/imath] is necessarily a constant function. there exists an analytic function [imath]g[/imath] on [imath]\mathbb{D}[/imath] such that [imath]g(x)[/imath] is a square root of [imath]f(z)[/imath] for each [imath]z\in\mathbb{D}[/imath]. there exists an analytic function [imath]g[/imath] on [imath]\mathbb{D}[/imath] such that [imath]\operatorname{Re}g(z)\geq 0[/imath] and [imath]g(z)[/imath] is a square root of [imath]f(z)[/imath] for each [imath]z\in\mathbb{D}[/imath]. there exists an analytic function [imath]g[/imath] on [imath]\mathbb{D}[/imath] such that [imath]\operatorname{Re}g(z)\leq 0[/imath] and [imath]g(z)[/imath] is a square root of [imath]f(z)[/imath] for each [imath]z\in\mathbb{D}[/imath]. I have to determine which options are correct. Can someone help in the right direction? Thanks in advance for your time. |
3041819 | Understanding Baire Theorem Proof
If [imath](X,d)[/imath] complete and [imath]X= \cup_{n=1}^{\infty} F_n[/imath] where [imath]F_n[/imath] is closed in [imath]X[/imath] for all [imath]n[/imath], then there exist at least one [imath]F_k[/imath] which the interior is non empty. The proof start by setting [imath]U_n = X \backslash F_n[/imath] open for all [imath]n[/imath]. By the DeMorgan Law, [imath]\cap_{n=1}^{\infty}U_n=(\cup_{n=1}^{\infty} F_n)^c=X \backslash \cup_{n=1}^{\infty} F_n = \emptyset[/imath] Then it follow that at least one of the open [imath]U_k[/imath] is not dense in [imath]X[/imath] . I dont understand this part of the proof. Can anyone help me please? thx. | 221423 | Proving Baire's theorem: The intersection of a sequence of dense open subsets of a complete metric space is nonempty
The following is problem 3.22 from Rudin's Princples of Mathematical Analysis: Suppose [imath]X[/imath] is a nonempty complete metric space, and [imath]\{G_n\}[/imath] is a sequence of dense open subsets of [imath]X[/imath]. Prove Baire's theorem, namely, that [imath]\bigcap_{n=1}^\infty G_n[/imath] is not empty. Hint: find a shrinking sequence of neighbourhoods [imath]E_n[/imath] such that [imath]\overline{E}_n\subset G_n[/imath]. Here's what I've tried so far: Let [imath]\{r_n\}[/imath] be a Cauchy sequence of positive real numbers converging to [imath]0[/imath]. Fix [imath]x\in X[/imath] and define [imath]E_i=\{g\in G_i:d(g,x)<r_i\}[/imath], which is nonempty since [imath]G_i[/imath] is dense. I would like to show that for all [imath]i[/imath], [imath]\overline{E}_i\subset G_i[/imath] (I had convinced myself that this would be true but I am now having doubts). Let [imath]e\in \overline{E}_i[/imath]. Then either [imath]e\in E_i[/imath] or [imath]e[/imath] is a limit point of [imath]E_i[/imath]. If [imath]e\in E_i[/imath] then [imath]e\in G_i[/imath]. Otherwise, every neighbourhood of [imath]e[/imath] contains a point in [imath]E_i[/imath]. I thought that I should be able to then choose some point [imath]e'\in E_i[/imath] in a neighbourhood of [imath]e[/imath] and, since [imath]G_i[/imath] is open, it'll have a neighbourhood [imath]N\subset G_i[/imath] which contains [imath]e[/imath], but this is proving to be difficult and I'm worried that it's not true. If I can show that this is true then the rest will follow from results I've already proven. Does my approach make any sense? Incidentally, as a secondary question, what type of a thing would [imath]G_n[/imath] be? A sequence of dense open subsets seems weird to me—at first I was thinking of some sequence of infinite subsets of rational numbers in the real numbers but I realized that those aren't open. Is there anything which would be familiar to my little undergrad brain which would be analogous to this problem? |
3042563 | Why is a polynomial with infinite zeropoints the zeropolymomial?
This was given us as a fact, but why is this true? The zeropolynomial is the polynomial where all the coefficients are equal to [imath]0[/imath] if [imath]R(x)[/imath] is a polynomial over [imath]\mathbb{C}[/imath] and every [imath]x\in \mathbb{N}_0[/imath] is a zeropoint then one can rewrite the polynomial [imath]R(x)=x(x-1)(x-2)…(x-n)…[/imath] but if we put an [imath]x\in \mathbb{C}-\mathbb{N}_0[/imath] in [imath]R(x)[/imath], how does one know that [imath]R(x)=0[/imath]? | 25822 | How to prove that a polynomial of degree [imath]n[/imath] has at most [imath]n[/imath] roots?
How can I prove, that a polynomial function [imath]f(x) = \sum_{0\le k \le n}a_k x^k\qquad n\in\mathbb N,\ a_k\in\mathbb C[/imath] is zero for at most [imath]n[/imath] different values of [imath]x[/imath], unless all [imath]$a_0,a_1,\ldots,a_n$[/imath] are zero? |
3042635 | Let [imath]x \in \mathbb{N}[/imath] and let [imath]p[/imath] be a prime divisor of [imath]x^4+x^3+x^2+x+1[/imath], prove that [imath]p = 5[/imath] or [imath]p \equiv 1 \mod{5}[/imath]
Let [imath]x \in \mathbb{N}[/imath] and let [imath]p[/imath] be a prime divisor of [imath]x^4+x^3+x^2+x+1[/imath], prove that [imath]p = 5[/imath] or [imath]p \equiv 1 \mod{5}[/imath] Note that [imath](x^4+x^3+x^2+x+1)(x-1) = x^5 - 1[/imath]. I tried reducing the equation [imath]\mod{5}[/imath], which gave me some information on [imath]x[/imath], but I was not able to utilize this information. | 3041891 | Show that any prime divisor of [imath]x^4+x^3+x^2+x+1[/imath], with [imath]x\in\mathbb{N}[/imath], is [imath]5[/imath] or [imath]1[/imath] mod [imath]5[/imath]
We can write the "polynomial" as follows: [imath]x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}.[/imath] For even [imath]x=2y[/imath], we have that [imath]x^5-1=(2y)^5-1=32y^5-1\equiv1[/imath] mod [imath]5[/imath]. For odd [imath]x=2y+1[/imath], we have that [imath](2y+1)^5-1\equiv_532y^5\equiv2[/imath] mod [imath]5[/imath]. I find it hard to believe my computations above. I guess this makes me rather clueless. I was hoping there was some way out with Legendre symbols. Any help is appreciated. |
3042719 | Combinatorial proof [imath]\sum_{i=k}^n {i-1 \choose k-1} = {n \choose k}[/imath]
Could someone help me as I am stuck with coming up with a proof for this? Assume n is the total number of people in a town. Assume k is the number of possible ways to select a chief of the town. So the RHS is saying that there are k ways to choose a chief from n people. on the LHS, From [imath]i=k[/imath], and [imath]k=n[/imath], it is referring to from k to n, which is the sum of the remaining people in the town who were not selected [imath](n-k)[/imath], that there is [imath]k-1[/imath] ways to choose from [imath]i-1[/imath] objects. Since [imath]i=k[/imath], i could be the number of ways to possibly select a chief. If one person is chosen from i, who also belongs to [imath]k, k-1[/imath]. But how does this lead to [imath]{n \choose k}[/imath]? | 1451745 | Prove [imath]{n - 1 \choose k - 1} + {n - 2 \choose k - 1} + {n - 3 \choose k - 1} + \dots + {k - 1 \choose k - 1} = {n \choose k}[/imath]
Can someone check my logic here. Question: How many ways are there to choose a an [imath]k[/imath] person committee from a group of [imath]n[/imath] people? Answer 1: there are [imath]{n \choose k}[/imath] ways. Answer 2: condition on eligibility. Assume the creator of the committee is already in the committee. This leaves us with choosing [imath]k - 1[/imath] people from a group of [imath]n - 1[/imath] potentially eligible people. If all remaining people are eligible, there are [imath]{n - 1 \choose k - 1}[/imath] possible committees, if there are [imath]n - 2[/imath] eligible people, there are [imath]{n - 2 \choose k - 1}[/imath] committees, if there are [imath]n - 3[/imath] eligible people, there are [imath]{n - 3 \choose k - 1}[/imath] committees,..., if there are [imath]k - 1[/imath] eligible people there are [imath]{k - 1 \choose k - 1}[/imath] committees. Therefore,[imath]{n - 1 \choose k - 1} + {n - 2 \choose k - 1} + {n - 3 \choose k - 1} + \dots + {k - 1 \choose k - 1} = {n \choose k}[/imath]. |
3042610 | Check my answer: Shortest sequence from [imath]1[/imath] to [imath]2^{2018}[/imath], where each step either doubles or squares the previous value.
My previous question was about the meaning of the problem. But now, I present a solution. Can MSE verify my answer? Given a number, you can perform two operations on it: double it or turn it into its square. Your initial number is one. How many operations at least do you need to turn it into [imath]2^{2018}?[/imath] Is my solution correct? [imath]\begin{align}2^{2018}\longrightarrow 2^{1009}\longrightarrow 2^{1008}\longrightarrow 2^{504}\longrightarrow 2^{252}\longrightarrow 2^{126}\longrightarrow 2^{63}\longrightarrow 2^{62}\longrightarrow 2^{31}\longrightarrow 2^{30}\longrightarrow 2^{15}\longrightarrow 2^{14}\longrightarrow 2^{7}\longrightarrow 2^{6}\longrightarrow 2^{3}\longrightarrow 2^{2}\longrightarrow 2^{1}\longrightarrow 1\end{align}[/imath] I find answer is [imath]17[/imath]. Is it correct? | 3042072 | If the number [imath]"1"[/imath] is written at the beginning, at least how many steps should be taken to reach [imath]2^{2018}?[/imath]
I have a problem understand this math problem: Write a number on the board. This number is either multiplied by [imath]2[/imath] or raised to a square. If the number [imath]"1"[/imath] is written at the beginning, at least how many steps should be taken to reach [imath]2^{2018}?[/imath] A) [imath]15[/imath] B) [imath]16[/imath] C) [imath]17[/imath] D) [imath]18[/imath] E) [imath]12[/imath] I can't solve this problem. Because I don't understand the question. Now, I need to understand the question. Then maybe I can. Is there a problem in the question? The question unclear for me...Can you explain me, what is the meaning of the question? |
3040423 | rearrangement of the alternating harmonic series
The alternating harmonic series [imath]a:=\sum_{n=1}^{\infty}{\frac{(-1)^{n-1}}n}[/imath] converges by the Leibnitz-criteria. Now consider the following rearrangement of this series given by: [imath]\begin{align} S:=1+\frac{1}{3}-\frac{1}{2}-\frac{1}{4}+\frac{1}{5}+\frac{1}{7}-...+\frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{4n-2}-\frac{1}{4n}+... \end{align}[/imath] I want to see that this series converges aswell and has the same value [imath]a[/imath]. I am stuck with this one. Does someone have a hint how to start this? Edit: this not a duplicate to the question in the link below since I am asking on how to get the limit value of this series. | 885329 | Why does the order of summation of the terms of an infinite series influence its value?
I was looking through my lecture notes and got puzzled by the following fact: if we want to find the value of some infinite series we are allowed to rearrange only the finite number of its terms. To visualize this consider the alternating harmonic series: [imath]\sum_{n=1}^\infty(-1)^{k-1}\frac1k=1-\frac12+\frac13-\frac14+\frac15-+\dots=0.693147...[/imath] But if we rearrange the terms as follows the value of the series gets influenced by this action: [imath]1+\frac13-\frac12+\frac15+\frac17-\frac14+\dots=1.03972...[/imath] So commutativity of addition isn't true on infinity? How was it obtained and how can it be proved? |
3043093 | Prove that this function is not integrable: [imath]f(x)[/imath] is [imath]1-x[/imath] for [imath]x[/imath] rational, and [imath]1/x[/imath] for [imath]x[/imath] irrational
How do I prove that the function [imath]f(x)= \begin{cases} 1-x &,\text{when}\; x \;\text{is rational} \\ 1/x &, \text{when}\; x \;\text{is irrational} \end{cases}[/imath] is not integrable on every interval [imath][a,b][/imath]? | 1550808 | Proving a function is not Riemann integrable
Prove that the bounded function [imath]f[/imath] defined by [imath]f(x)=0[/imath] if [imath]x[/imath] is irrational and [imath]f(x)=1[/imath] if [imath]x[/imath] is rational is not Riemann integrable on [imath][0,1][/imath]. I was given the hint to use the inverse definition of Riemann integrable and consider the cases of the partition being all rationals, and all irrationals between [imath][0,1][/imath], but I'm not too sure how to go about it. |
1447255 | When [imath]k[/imath] times composition of a function is a _contraction_
Let [imath](X,d)[/imath] be a complete metric space and [imath]f \ :\ X\rightarrow X[/imath] be a map such that , for some positive integer [imath]k[/imath] , [imath]f\circ f\circ .....\circ f(\ k\ \ fold\ \ composition\ \ with\ \ itself\ )[/imath] is a contraction. Then [imath]f[/imath] has a unique fixed point . How should I approach this problem [imath]?[/imath] Sorry for the lack of efforts here . Please give me some hints as to how to begin the thinking [imath]?[/imath] | 360795 | Generalization of Banach's fixed point theorem
I wanted to show that if [imath]f:X\to X[/imath] is a function from a complete metric space to itself and if [imath]f^k[/imath] is a contraction, then [imath]f[/imath] has a unique fixed point (say [imath]p[/imath]) and for any [imath]x[/imath] in [imath]X[/imath] [imath]f^n(x)\longrightarrow p[/imath]. What I did is use what I know from Banach as follows: Take [imath]x[/imath] arbitrary in [imath]X[/imath]. Then consider the sequence [imath]f^n(x)[/imath], which I denote by [imath]x_n[/imath]. Let [imath]x^{(i)}[/imath] be the sequence [imath]x_i, x_{i+k}, x_{i+2k}...[/imath] for [imath]i[/imath] between 0 and [imath]k - 1[/imath]. As [imath]f^k[/imath] is a contraction, we can apply Banach and so it has a fixed point (say [imath]l[/imath]) and all the [imath]x^{(i)}[/imath] are convergent to [imath]l[/imath]. Then [imath]x_n[/imath] is convergent to [imath]l[/imath] (for any [imath]\epsilon[/imath] exists [imath]N[/imath] such that [imath]d_X(x_n, l) < \epsilon[/imath], this [imath]N[/imath] being the maximum of the corresponding N's obtained from each of the [imath]x^{(i)}[/imath]). Since [imath]x[/imath] was arbitrary and [imath]l[/imath] is unique, this means that [imath]p=l[/imath] and the proof is completed. Am I doing something wrong or missing anything? I just want to ensure I understand this properly, this is not a homework or something similar. Thank you! |
3043644 | How to prove the Inequality [imath]\sqrt[n]{b}-\sqrt[n]{a}<\sqrt[n]{b-a}[/imath]?
Let [imath]0<a<b[/imath] using Lagrange mean value theorem I want to prove the following inequality [imath]\sqrt[n]{b}-\sqrt[n]{a}<\sqrt[n]{b-a}[/imath] Any help is appreciated. | 2992987 | How do you prove the follwing [imath]|\sqrt[n]{x}-\sqrt[n]{y}| \le \sqrt[n]{|x-y|}[/imath]
how do you prove this inequality? [imath]|\sqrt[n]{x}-\sqrt[n]{y}| \le \sqrt[n]{|x-y|}[/imath] At first glance I thought the triangle inequality would be useful but it's not form what I see. If I raise everything to the power of n I get the following: Possibility 1 (n uneven) : [imath]|x + ... -y| \le |x-y|[/imath] I guess I would have to prove that ... [imath]\le[/imath] 0 but I don't know how to do that. Possibility 2 (n even) : [imath]|x + ... +y| \le |x-y|[/imath] Same as above but now I have to prove that ... [imath]\le[/imath] -y. Am I even on the right track here? Any help would be appreciated. Thanks in advance. |
3043664 | Interesting application of mean value theorem
Suppose [imath]f[/imath] is continuous on [imath][a, b][/imath], differentiable on [imath](a, b)[/imath] and satisfies[imath] f^2(a)−f^2(b)= a^2−b^2[/imath]Then show that the equation [imath]f′(x)f(x) = x[/imath] has at least one root in [imath](a, b)[/imath]. | 390464 | If [imath]f^2(b) - f^2(a) = b^2 - a^2[/imath], then the equation [imath]f'(x)f(x) = x[/imath] has at least one root in (a, b).
Suppose [imath]f[/imath] is continuous on [a, b] and differentiable on the open interval (a, b). How to that if [imath]f^2(b) - f^2(a) = b^2 - a^2[/imath], then the equation [imath]f'(x)f(x) = x[/imath] has at least one root in (a, b). |
3043662 | Is there an arithmetic proof to the AM GM Inequality for 3 variables?
As the question states, I am interested in proving the AM GM inequality in 3 variables, namely: [imath]\sqrt[3]{xyz}\le \frac{(x+y+z)}{3}[/imath] Clearly, I expanded everything into the following: [a] [imath] xyz \le \frac{x^3 + y^3 + z^3 + 3(x^2y + x^2z + y^2x + y^2z + z^2x + z^2y) + 6xyz}{27}[/imath] I have been given that we know the proof of the AM GM inequality in 2 variables, so I attempted from the base case of the mean between xy and z: [b] [imath] \sqrt{(xy)z} \le \frac {xy + z}{2} [/imath] Thus: [c] [imath]xyz \le \frac {x^2y^2 +2xyz + z^2}{4}[/imath] If I can prove that the right hand side of c is less than the right hand side of a, I will have completed my proof. I conjecture that my proof will look like: [imath]xyz \le \frac {x^2y^2 +2xyz + z^2}{4} \le something \le \frac{x^3 + y^3 + z^3 + 3(x^2y + x^2z + y^2x + y^2z + z^2x + z^2y) + 6xyz}{27}[/imath] However, this is proving challenging arithmetically. I know that this is possible with calculus, but I am interested in solving the problem without calculus, and from either a geometric, factorial, or arithmetic perspective. Any insight would be appreciated. | 2576966 | Elementary proof for [imath]\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}[/imath]
I am searching for an elementary proof of the AM-GM inequality in three variables: [imath]\sqrt[3]{xyz} \leq \dfrac{x+y+z}{3}[/imath] The inequality of the geometric mean vs the arithmetic mean of two variables can be proven elementarily via [imath]x - 2 \sqrt{xy} + y \geq 0[/imath] whenever [imath]x,y > 0[/imath]. I am searching for a proof that uses a similar technique based on elementary arithmetic. |
3043685 | Compute [imath]\sum\frac1{2-A_k}[/imath] for [imath](A_k)[/imath] the [imath]n[/imath]th roots of unity
If [imath]1,A_1,A_2,A_3....A_{n-1}[/imath] are the [imath]n^{th}[/imath] roots of unity then prove that [imath]\dfrac{1}{2-A_1} + \dfrac{1}{2-A_2}+\cdots+ \dfrac{1}{2-A_{n-1}} = \dfrac{2^{n-1}(n-2) + 1}{2^n-1}[/imath] What I did: I tried to use some of the following formulas: [imath]1+ A_1 +A_2+A_3+\cdots+A_{n-1} = 0[/imath] [imath]\dfrac{2^n - 1}{2-1} = (2 -A_1)(2-A_2)\cdots(2-A_{n-1})[/imath] and the fact that [imath]|A_i| = 1[/imath] for [imath]i =1,2,3,\cdots,n-1[/imath]. | 3038472 | If [imath]x_1,x_2,\ldots,x_n[/imath] are the roots for [imath]1+x+x^2+\ldots+x^n=0[/imath], find the value of [imath]\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}[/imath]
Let [imath]x_1,x_2,\ldots,x_n[/imath] be the roots for [imath]1+x+x^2+\ldots+x^n=0[/imath]. Find the value of [imath]P(1)=\frac{1}{x_1-1}+\frac{1}{x_2-1}+\ldots+\frac{1}{x_n-1}[/imath] Source: IME entrance exam (Military Institute of Engineering, Brazil), date not provided (possibly from the 1950s) My attempt: Developing expression [imath]P(1)[/imath], replacing the 1 by [imath]x[/imath], follows [imath]P(x)=\frac{(x_2-x)\cdots (x_n-x)+\ldots+(x_1-x)\cdots (x_{n-1}-x)}{(x_1-x)(x_2-x)\cdots (x_n-x)}[/imath] As [imath]x_1,x_2,\ldots,x_n[/imath] are the roots, it must be true that [imath]Q(x)=(x-x_1)\cdots(x-x_n)=1+x+x^2+\ldots+x^n[/imath] and [imath]Q(1)=(1-x_1)\cdots(1-x_n)=n+1[/imath] Therefore the denominator of [imath]P(1)[/imath] is [imath](-1)^{n} (n+1).[/imath] But I could not find a way to simplify the numerator. Another fact that is probably useful is that [imath]1+x^{n+1}=(1-x)(x^n+x^{n-1}+\ldots+x+1)[/imath] with roots that are 1 in addition of the given roots [imath]x_1,x_2,\ldots,x_n[/imath] for the original equation, that is [imath]x_k=\text{cis}(\frac{2k\pi}{n+1}),\ \ k=1,\ldots,n.[/imath] This is as far as I could go... Hints and full answers are welcomed. |
3044264 | If [imath]a_n > 0[/imath] prove that [imath]\sum_{n=1}^{\infty} \frac{a_n}{(a_1+1)(a_2+1)\cdots(a_n+1)}[/imath] converges
I have an interesting task: If [imath]a_n > 0[/imath], prove that [imath]\sum_{n=1}^{\infty} \frac{a_n}{(a_1+1)(a_2+1)\cdots(a_n+1)}[/imath] converges. I thought that it will be simple because ratio test gives me: [imath]\frac{u_{n+1}}{u_n}= \frac{a_{n+1}}{a_{n+1}+1}\cdot a_n^{-1} < 1 \cdot a_n^{-1} = \frac{1}{a_n}[/imath] and [imath]a_n[/imath] should be in [imath][0,1][/imath]. But... In my opinion it can be over that... why need I assume that [imath] a_n \rightarrow g \in [0,1] [/imath]? There is similar topic on this forum, but It was not solved there... @edit I saw that: [imath]\sum_{n=1}^{N}\frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-\frac{1}{(1+a_1)(1+a_2)...(1+a_N)} < 1 [/imath] So if series of partial sum is bounded from up, the sum converges, that is right? @edit2 but It is good? Look at that: [imath] \sum_{n=1}^{N}\frac{a_n+1-1}{(1+a_1)(1+a_2)...(1+a_n)} = \sum_{n=1}^{N}\frac{1}{(1+a_1)(1+a_2)...(1+a_{n-1})}-\frac{1}{(1+a_1)(1+a_2)...(1+a_n)} [/imath] why somebody changed first part into [imath]1[/imath]? @edit3 Ok, I think that I have understood, thanks for your time ;) | 721987 | Prove the convergence of a series.
If [imath]a_n >0[/imath] for all [imath]n\geq1[/imath], show that the series [imath]\sum_{n=1}^{\infty} \frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)}[/imath] converges. Can someone check if my solution is correct: [imath]\sum_{n=1}^{M}\frac{a_n}{(1+a_1)(1+a_2)...(1+a_n)} = 1-\frac{1}{(1+a_1)(1+a_2)...(1+a_M)} < 1 [/imath] Since the partial sum of the series is bounded, the series is convergent. |
3044348 | ring theory and ideals of them
Let [imath]R[/imath] be a commutative ring and [imath]I[/imath] and [imath]J[/imath] be two ideals of it. We know that [imath]I+J[/imath] is a subset of [imath]R[/imath] and moreover, [imath]I+J[/imath] is an ideal of [imath]R[/imath]. Question: When [imath]I+J[/imath] can be [imath]R[/imath]? | 1068990 | [imath]I[/imath] and [imath]J[/imath] are coprime ideals iff [imath]x \to (x + I, x + J)[/imath] is surjective.
I'm stuck on this exercise and any help would be well appreciated: Let [imath]R[/imath] be a commutative ring with ideals [imath]I,J[/imath]. Show that [imath]R=I+J[/imath] if and only if [imath]\phi(x)= (x + I, x + J)[/imath] is surjective from [imath]R[/imath] to [imath]R/I \times R/J[/imath]. Assuming surjectivity I got as far as to realize that I need to show [imath]R\subset I+J[/imath] since [imath]I+J\subset R[/imath] always holds (as they are both subrings of R). Now I'm not sure how to proceed. Also, I don't even know where to begin in the direction [imath]R=I+J[/imath] implies [imath]\phi[/imath] is onto. Again, any help, pointers or references are much appreciated. Thanks in advance. |
3023892 | Slight difference in definitions of Glassers Master Theorem
I’m probably misreading but there seems to be a slight difference in the definition of Glasser’s Master Theorem between Wikipedia and Wolfram Mathworld. The difference lies in the substitution [imath]u[/imath] and [imath]\phi(x)[/imath] with [imath]u[/imath] having [imath]x - a - \sum[/imath] over [imath]x - \sum[/imath]. Are these equivalent? Is there an error with one? | 2863440 | Does Wikipedia misstate Glasser's master theorem
Citing from Wikipedia and one of its references, MathWorld, following hold: Glasser's master Theorem For [imath]f[/imath] integrable, [imath]\Phi(x) = |a|x - \sum_{i=1}^N \frac{|\alpha_i|}{x-\beta_i}[/imath] and [imath]a[/imath], [imath]\alpha_i[/imath], [imath]\beta_i[/imath] arbitrary real constants the identity \begin{equation} \mathrm{PV}\int_{-\infty}^\infty f(\Phi(x)) dx = \mathrm{PV} \int_{-\infty}^\infty f(x) dx \label{Glasser} \tag{1} \end{equation} holds. Now consider \begin{align*} \Phi_1(x) &= |a|x - \sum_{i=1}^N \frac{|\alpha_i|}{x-\beta_i} \\ \Phi_2(x) &= x - \sum_{i=1}^N \frac{|a\alpha_i|}{x-|a|\beta_i} \end{align*} Then, by Glasser's theorem \ref{Glasser} [imath]\mathrm{PV}\int_{-\infty}^\infty f(\Phi_1(x)) dx = \mathrm{PV} \int_{-\infty}^\infty f(x) dx = \mathrm{PV}\int_{-\infty}^\infty f(\Phi_2(x)).[/imath] However, under the change of variables [imath]y = |a| x[/imath] \begin{equation} \mathrm{PV}\int_{-\infty}^\infty f(\Phi_1(x)) dx = \frac{1}{|a|}\mathrm{PV} \int_{-\infty}^\infty f(\Phi_2(y)) dy. \label{my Idea} \tag{2} \end{equation} Thus, I assume Glasser's theorem only holds for [imath]|a| = 1[/imath]; a quick numerical check seems to support Eq. \ref{my Idea}. Is Wikipedia and MathWorld wrong about this? |
3044362 | Suggestion for a functional equation [imath]f'\left(\frac{a}{x}\right)=\frac{x}{f(x)}[/imath] where [imath]f:(0,\infty)\to(0,\infty)[/imath] is differentiable
I want to solve the following functional equation: Find all differentiable functions [imath]f : (0,\infty) \rightarrow (0,\infty)[/imath] for which there is a positive real number [imath]a[/imath] such that [imath]f'\left(\frac{a}{x}\right)=\frac{x}{f(x)}[/imath] for all [imath]x > 0[/imath]. I have noted that the function [imath]f[/imath] is increasing but i cannot go any further Any suggestions? | 2994207 | Functional Identity
If we know that [imath] f'(k/x)f(x) = x\tag{ * } [/imath] Then what can we say about [imath]f(k/x)f'(x) ?[/imath] Originally I tried substituting [imath]x=k/x[/imath] into (*), to give [imath]f'(x)f(k/x) = k/x[/imath] But is this valid? I'm guessing not. Ultimately I'm trying to find a solution for [imath]f(x)[/imath], and the above doesn't seem to working. Any thoughts appreciated. |
3043009 | To prove that the limit of a bi-variate function is nonexistent at a point
It has been asked to evaluate [imath]\lim_{(x,y)\to(0,0)} \frac{x^3 + y^3}{x-y}[/imath] if it exists at all and otherwise to disprove it. After much thoughts , I came up with an idea of substituting [imath]y[/imath] with [imath]x-mx^3[/imath] which devolved the limit to [imath](2/m)[/imath] and thus served my purpose of disproving the existence of a limit. But, thinking of such half-weird substitutions take some reasonable amount of time; the luxury of which is seldom available at examinations. So, what are other better methods to disprove the existence of this limit? Any general algorithm (??) on disproving the existence of bi-variate limits (without indulging into trick-substitutions) will be also appreciated. | 1033235 | Non-existence of [imath]\lim \limits_{(x, y) \to (0,0)} \frac{x^3 + y^3}{x - y} [/imath]
How to show that [imath]\lim \limits_{(x, y) \to (0,0)} f(x, y)[/imath] does not exist where, [imath]f(x, y) = \begin{cases} \dfrac{x^3 + y^3}{x - y} \; ; & x \neq y \\ 0 \; \;\;\;\;\;\;\;\;\;\;\; ; & x = y \end{cases} [/imath] I tried bounding the value of the function as [imath](x, y)[/imath] approaches [imath](0,0)[/imath] but was not successful. I graphed the function and saw that it was actually approaching [imath]0[/imath] but Microsoft Math failed to render some points around the [imath]x = y[/imath] plane. So I imagine I need to find points [imath](x, y)[/imath] such that [imath]x[/imath] is extremely close to [imath]y[/imath] but [imath](x, y)[/imath] is not as close to [imath](0,0)[/imath] to show that the function does not approach [imath]0[/imath] around the origin. But was unable to think of any. Hints or suggestions would be awesome. Any help is appreciated. |
3044945 | sum of the infinite series [imath]\cot^{-1}(2.1^2)+\cot^{-1}(2.2^2)+\cot^{-1}(2.3^2)+\cot^{-1}(2.4^2)+...[/imath]
Find the sum of the infinite series [imath]\cos^{-1}(2.1^2)+\cos^{-1}(2.2^2)+\cos^{-1}(2.3^2)+\cos^{-1}(2.4^2)+......[/imath] My Attempt [imath]\begin{align} \cot^{-1}x&=\tan^{-1}\frac{1}{x}\text{, }x>0\\ \cot^{-1}(2.1^2)&+\cot^{-1}(2.2^2)+\cot^{-1}(2.3^2)+\cot^{-1}(2.4^2)+.....=\\ &=\tan^{-1}\frac{1}{2.1^2}+\tan^{-1}\frac{1}{2.2^2}+\tan^{-1}\frac{1}{2.3^2}+\tan^{-1}\frac{1}{2.4^2}+.....\\ &=\tan^{-1}\frac{}{}+\tan^{-1}\frac{}{}+\tan^{-1}\frac{}{}+\tan^{-1}\frac{}{}+..... \end{align}[/imath] The solution given in my reference is [imath]\dfrac{\pi}{4}[/imath], but I am stuck as in my attempt not able to identify any common property among the terms to simplify the series. | 2427602 | Find the value of infinite series [imath]\sum_{n=1}^{\infty} \tan^{-1}(2/n^2)[/imath]
Find the value of infinite series [imath] \sum_{n=1}^{\infty} \tan^{-1}\left(\frac{2}{n^2} \right) [/imath] I tried to find sequence of partial sums and tried to find the limit of that sequence. But I didn't get the answer. |
3045499 | Proving [imath]\mathbb{E}\max\{\xi^2,\eta^2\}\leq 1 + \sqrt{1-\rho^2}[/imath]
Let [imath]\xi[/imath] and [imath]\eta[/imath] be random variables and [imath]\mathbb{E}\xi=\mathbb{E}\eta=0[/imath] also [imath]\mathbb{D}\xi=\mathbb{D}\eta=1[/imath]. Here [imath]\rho=\rho(\xi,\eta)[/imath] is a correlation coefficient. Need to show [imath]\mathbb{E}\max\{\xi^2,\eta^2\}\leq 1 + \sqrt{1-\rho^2}.[/imath] So, I know that [imath]\rho=\rho(\xi,\eta)=\frac{\mathbb{E}(\xi-\mathbb{E}\xi)(\eta-\mathbb{E}\eta)} {\sqrt{\mathbb{D}\xi \mathbb{D}\eta}}[/imath]. But how to start and what qualities to use? | 1394275 | Prove that [imath]\mathrm E\left({\max}\left\{X^2 , Y^2\right\} \right) \leq 1 + \sqrt{1 - \rho^2}[/imath] where [imath]\rho[/imath] is their correlation.
I have 2 random variables [imath]X,Y[/imath] with mean 0 and variance 1, their correlation is [imath]\rho[/imath]. I need to prove this inequality [imath]\mathrm E\left({\max}\left\{X^2 , Y^2\right\} \right) \leq 1 + \sqrt{1 - \rho^2}[/imath] I need some pointers as to how to solve this problem. Thanks! |
3046179 | Isomorphism between two versions of [imath]GF(2^3)[/imath]
I have [imath]GF(2^3)[/imath] generated by [imath]\Pi_1(\alpha)=x^3+x+1[/imath] and [imath]GF(2^3)[/imath] generated by [imath]\Pi_2(\alpha)=x^3+x^2+1[/imath]. [imath]\bullet[/imath] [imath]\Pi_1(\alpha)=x^3+x+1[/imath] [imath]000=0, 100=1,010=\alpha,001=\alpha^2,110=\alpha^3, 011=\alpha^4,111=\alpha^5,101=\alpha^6[/imath] [imath]\bullet[/imath] [imath]\Pi_2(\alpha)=x^3+x^2+1[/imath] [imath]000=0, 100=1,010=\lambda,001=\lambda^2,101=\lambda^3, 111=\lambda^4,110=\lambda^5,011=\lambda^6[/imath] The exampla say that [imath]\alpha[/imath] and [imath]\lambda^3[/imath] both have minimal polynomial [imath]\Pi_1[/imath] and thus [imath]\alpha \iff \lambda^3[/imath] form an isomorphism between the two version. How can I see in a practical way this fact? Someone can explain me this concept? | 2058958 | Finding an isomorphism between these two finite fields
Let [imath]\alpha[/imath] be a root of [imath]x^3 + x + 1 \in \mathbb{Z}_2[x][/imath] and [imath]\beta[/imath] a root of [imath]x^3 + x^2 + 1 \in \mathbb{Z}_2[x][/imath]. Then we know that [imath]\mathbb{Z}_2(\alpha) \simeq \frac{\mathbb{Z}_2[x]}{(x^3 + x + 1)} \simeq \mathbb{F}_8 \simeq \frac{\mathbb{Z}_2[x]}{(x^3 + x^2 + 1)} \simeq \mathbb{Z}_2(\beta). [/imath] I need to find an explicit isomorphism [imath]\mathbb{Z}_2(\alpha) \to \mathbb{Z}_2(\beta). [/imath] I was thinking of finding a basis for [imath]\mathbb{Z}_2(\alpha)[/imath] and [imath]\mathbb{Z}_2(\beta)[/imath] over [imath]\mathbb{Z}_2[/imath]. I let [imath]\left\{1, \alpha, \alpha^2\right\}[/imath] and [imath]\left\{1, \beta, \beta^2\right\}[/imath] be these two bases. Now suppose I have a field morphism [imath] \phi: \mathbb{Z}_2(\alpha) \to \mathbb{Z}_2(\beta) [/imath] which maps [imath]1[/imath] to [imath]1[/imath]. how can I show that the image of [imath]\alpha[/imath], i.e. [imath]\phi(\alpha)[/imath], completely determines this map? |
3045215 | Probability that a random variable X belongs to the set of rational nos.
The question is a Multiple choice question Let [imath]X[/imath] be a random variable with the M.G.F. [imath]M_{X}(t) = \frac{6}{\pi^{2}}\sum_{n\ge1}\frac{e^{\frac{t^2}{2n}}}{n^2}\,,\;t\in R[/imath] Then [imath]P(X \in Q)[/imath], where [imath]\,Q[/imath] is the set of rational nos., equals The options are 0 1/4 1/2 3/4 The correct option is option 1.) i.e. 0 I'm looking for an explanation as to how did they deduce the answer to zero. | 2675877 | To find the probability whether [imath]X[/imath] is rational
I was thinking whether I can replace this sum by an integral , because I can't think of a distribution whose mgf looks like this. Please Help! |
3031336 | Prove: [imath]\alpha \rightarrow \beta, (\alpha \rightarrow \lambda) \rightarrow \beta [/imath] proves [imath]\beta[/imath] in HPC
I want to prove [imath]\alpha \rightarrow \beta, (\alpha \rightarrow \lambda) \rightarrow \beta [/imath] proves [imath]\beta[/imath] in HPC I use MP and the axioms listed here: https://en.wikipedia.org/wiki/Hilbert_system I guess that the end of the proof should look something like this: [imath](\alpha\rightarrow(\beta\rightarrow\lambda)) \rightarrow ((\alpha \rightarrow \beta) \rightarrow (\alpha \rightarrow\lambda))[/imath] [imath](\alpha \rightarrow \beta) \rightarrow (\alpha \rightarrow\lambda)[/imath] [imath]\alpha \rightarrow \lambda[/imath] [imath]\beta[/imath] but I can't find out how I can get to [imath](\alpha\rightarrow(\beta\rightarrow\lambda))[/imath] given my premises would be happy for some help | 337065 | Prove: [imath](A\rightarrow B),(A\rightarrow C)\rightarrow B, \mapsto_{HPC} B [/imath]
I'd really like your help proving: [imath](A\rightarrow B),(A\rightarrow C)\rightarrow B, \mapsto_{HPC} B [/imath] Where [imath]HPC[/imath] is the Hilbert's system proof which contains the following relevant axioms: [imath]A\rightarrow(B \rightarrow A)[/imath] [imath](A\rightarrow(B\rightarrow C)) \to ((A\rightarrow B)\rightarrow(A\rightarrow C))[/imath] [imath](A\rightarrow B)\rightarrow ((A\rightarrow\bar{B})\rightarrow \bar{A})[/imath] [imath]\bar{\bar{A}} \rightarrow A[/imath] In addition tried to use these following lemmas: [imath]\bar{A} \rightarrow (A \rightarrow C) [/imath] and [imath](A\rightarrow B)\rightarrow (\bar{B} \rightarrow \bar{A})[/imath]. Any suggestions? |
3046837 | [imath]n \in \mathbb N[/imath], find all values of [imath]n[/imath] for which [imath]3^{2n+1}-2^{2n+1}-6^n[/imath] is a prime number.
Find all values of [imath]n \in \mathbb N[/imath] for which [imath]3^{2n+1}-2^{2n+1}-6^n[/imath] is a prime number. Let [imath]k=3^{2n+1}-2^{2n+1}-6^n[/imath]. By testing with numbers I got that [imath]k[/imath] cannot be prime for any value of [imath]n[/imath] (tested to [imath]n=20[/imath], so I assumed it's same for every value of [imath]n[/imath]). For [imath]n=1[/imath], [imath]k<0[/imath] [imath]\implies[/imath] [imath]k[/imath] is not prime. (EDIT: oh... it's actually prime if [imath]n=1[/imath]. Stupid me calculated [imath]3^3=9[/imath]) For [imath]n>1[/imath]: I managed to prove that if [imath]n[/imath] is even, [imath]k[/imath] is not a prime: If [imath]n[/imath] is even, then [imath]3^{2n+1}\equiv3\ (mod\ 10),\quad 2^{2n+1}\equiv2\ (mod\ 10),\quad 6^n\equiv6\ (mod\ 10)[/imath] [imath]\implies k\equiv5\ (mod\ 10)[/imath] So [imath]k[/imath] is divisible by 5 and therefore not prime. Using the same method for if [imath]n[/imath] is odd, I get: [imath]k\equiv3\ (mod\ 10)[/imath], which doesn't prove anything. How do I solve for when [imath]n[/imath] is odd? | 2039846 | Show that [imath]3^{2n+1}-4^{n+1}+6^n[/imath] is never prime for natural n except 1.
Show that [imath]3^{2n+1}-4^{n+1}+6^n[/imath] is never prime for natural n except 1. I tried factoring this expression but couldn't get very far. It is simple to show for even n but odd n was more difficult, at least for me. |
3047015 | Cauchy Schwarz inequality for [imath]\langle f,g\rangle = \int_0^1f(x)g(x)\mathrm{d}x[/imath]
For the vector space of continuous functions on [imath][0,1][/imath] Define the inner product as [imath]\langle f,g\rangle = \int_0^1f(x)g(x)\mathrm{d}x[/imath] Please help me to prove the Cauchy Schwarz inequality for this given inner product. Cauchy Schwarz Inequality: [imath]|\langle v,u\rangle|\leq \lVert v\rVert\lVert u\rVert[/imath] for the elements [imath]v,u[/imath] in the inner product space. | 1357968 | Cauchy-Schwarz inequality proof (but not the usual one)
Before you downvote/vote-to-close, I am not asking for a proof of: [imath]\sum^n_{i=1}a_ib_i\le\sqrt{\sum_{i=1}^na_i^2}\sqrt{\sum^n_{i=1}b_i^2 } [/imath] Which is what EVERY link I've found assumes is the inequality (for a proof of that see: http://www.maths.kisogo.com/index.php?title=Cauchy-Schwarz_inequality&oldid=692 ) I am reading a book that claims the Cauchy-Schwarz inequality is actually: [imath]\vert\langle x,y\rangle\vert\le\Vert x\Vert\Vert y\Vert[/imath] where [imath]\Vert x\Vert :=\sqrt{\langle x,x\rangle}[/imath] with the additional claim: equality holds [imath]\iff\ x,y[/imath] are linearly dependent I cannot find a proof of this claim (only proofs for the dot product inner product). My question is: what is the simplest way to prove this (I define simplest to mean "using as few definitions outside of the inner product itself as possible" - so that's not including Gramm-Schmitt orthonormalisation, for example.) I've just found some possible answers in the links on the right (annoyingly) but fortunately I have a second, albeit softer, question: What inequalities that are common for say [imath]\mathbb{R}^n[/imath] are actually based on the inner product? (In future I will check this site first, I was targeting lecture notes and such in my search) Addendum: I want to use this on not-finite dimensional vector spaces. I've found one proof that relies on finite-ness. |
3047896 | Solution to [imath]\int_0^1 \left(\frac{\ln(x)}{1-x}\right)^2dx[/imath] in a closed form.
I'm looking for the solution to the integral [imath]\int_0^1 \left(\frac{\ln(x)}{1-x}\right)^2dx[/imath] I solved and know that the solution to [imath]-\int_0^1 \frac{\ln(x)}{1-x}dx = \frac{\pi^2}{6}[/imath] through a taylor series argument, and am wondering if a similar approach is the best way to go. | 75934 | A proof of [imath]\int_{0}^{1}\left( \frac{\ln t}{1-t}\right)^2\,\mathrm{d}t=\frac{\pi^2}{3}[/imath]
What is the proof of the following: [imath]\int_{0}^{1} \left(\frac{\ln t}{1-t}\right)^2 \,\mathrm{d}t=\frac{\pi^2}{3} \>?[/imath] |
3047881 | Prove that [imath]\lim_{x \to \infty} f(x)/x=\lim_{x \to \infty}[f(x+1)-f(x)][/imath]
Let [imath]f:(a,\infty)\rightarrow \mathbb{R}[/imath] be a function such that [imath]f[/imath] is bounded in any finite interval [imath](a,b][/imath]. Prove that [imath]\lim_{x \to \infty} f(x)/x=\lim_{x \to \infty}[f(x+1)-f(x)][/imath], provided that the right limit exists. Thanks a lot. | 1642662 | Prove that [imath]\lim_{x \to +\infty} \frac{f(x)}{x} = L[/imath] if [imath]\lim_{x \to +\infty} [f(x+1) - f(x)] = L \space[/imath]
Let [imath]f:[0, +\infty) \rightarrow \mathbb{R} [/imath] be a bounded function in each bounded interval. If [imath]\lim_{x \to +\infty} [f(x+1) - f(x)] = L[/imath] then [imath]\lim_{x \to +\infty} \frac{f(x)}{x} = L[/imath] I tried using the definition on the first limit; then, I attempted to use that inequality to apply the triangular inequality to arrive at the definition that gives me the second limit. It is supposed to be simple, but I'm not feeling safe as in how I should write it. |
3047132 | A different way of thinking of numbers
I was more or less day dreaming about whole numbers, and I had an idea which seemed novel to me, but which may in fact be previously known and/or studied. I'm not even sure what to call it, or how to look for previous reports of it. My first question will be: Have I merely rediscovered a way of looking at things that is already known, and if so, where can I find some discussion of it? I imagine an infinite dimension space, with each orthogonal axis corresponding to a prime number. By the fundamental theorem of arithmetic, natural numbers each have a unique representation of the form [imath]\prod p_i^{\alpha_i}[/imath]. Numbers of this form can be located in [imath]n[/imath]-dimensional subspaces as follows: For each prime [imath]p_j[/imath] which has a corresponding exponent [imath]>0[/imath], move [imath]\alpha_1[/imath] units from the origin along the [imath]p_1[/imath] axis, then [imath]\alpha_2[/imath] units parallel to the [imath]p_2[/imath] axis, then [imath]\alpha_3[/imath] units parallel to the [imath]p_3[/imath] axis, etc., for as many primes as are present as factors of that number. The origin would correspond to the number [imath]1[/imath], where all axes intersect and [imath]\alpha_i=0[/imath] for all [imath]i[/imath]. Each axis could be extended through the origin, and fractions could be represented as negative integral distances along the axes so extended, allowing the representation of rational numbers. Interestingly, moving "infinitely" out along these negative axes, one approaches [imath]0[/imath]. Next, by moving non-integral distances along axes, non-integral exponents could be represented and algebraic rational numbers could be represented and located in appropriate subspaces. I'm not certain that transcendental numbers can be represented, but neither are they representable in the form [imath]\prod p_i^{\alpha_i}[/imath]. One concern I have is that if one is allowed to move irrational distances along axes, it might conceivably be possible to represent a particular number in two different ways, and locate it in two different subspaces, i.e. if exponents can be identified such that [imath]p_1^{\alpha_1}=p_2^{\alpha_2}[/imath]. I do not think arithmetic addition/subtraction can be readily performed with numbers so represented, but multiplication/division can be. In this light, this kind of representation bears some kinship to logarithms, but with a unique (prime) base (rather than [imath]e[/imath] or [imath]10[/imath]) as the metric along each axis. However, I can think of no particular utility to representing numbers in this way. So my second question is: Can anyone see any merit or utility to thinking of numbers in this way? | 2879 | Mapping natural numbers into prime-exponents space
Take any natural number [imath]n[/imath], and factor it as [imath]n=2^{e_1} 3^{e_2} 5^{e_3} ... p^{e_i}[/imath], where [imath]i[/imath] is the [imath]i[/imath]-th prime. Now map [imath]n[/imath] to the point [imath]n \mapsto (e_1,e_2,\ldots,e_i,0,\ldots)[/imath], where [imath]i[/imath] is the last prime in the factorization of [imath]n[/imath]. For example, [imath]n=123456789 \mapsto (0,2,\ldots,1,\ldots,1,\ldots)[/imath] because [imath]123456789=2^0 3^2 \cdots 3607^1 \cdots 3803^1[/imath]. So every number in [imath]\mathbb{N}[/imath] is mapped to a point in an arbitrarily high dimensional space. This mapping has the property that addition of the vectors corresponds to multiplication of the numbers. My (extremely vague!) question is: Does this viewpoint helps gain any insights into the structure/properties of natural numbers? Do line/planes/curves in this space mark out numerically interesting regions? Perhaps allowing real or complex numbers? The numbers in [imath]\mathbb{N}[/imath] fill this infinite-dimensional space very sparsely. This is (very!) far from my research expertise, so any comments/references/links would be appreciated. |
3045839 | Prove [imath]x^2=y^3[/imath] where [imath]x,y\in \mathbb{Z}[/imath] implies [imath]x=a^3[/imath] and [imath]y=b^2[/imath] where [imath]a,b\in\mathbb{Z}[/imath]
I feel like I'm meant to use the uniqueness of prime factorization and show that since [imath]\text{lcm}(3,2)=6[/imath] each prime should appear 6 times, but how would I justify this? | 208535 | Are there any integer solutions to [imath]a^3=b^2[/imath]?
I was wondering if there were any two integers [imath]a[/imath] and [imath]b[/imath] where [imath]a^3=b^2[/imath]. |
3045990 | If [imath]xy[/imath] divides [imath]x^2 + y^2[/imath] show that [imath]x=\pm y[/imath]
Problem statement : Let [imath]x,y[/imath] be integers, show that if [imath]xy[/imath] divides [imath]x^2 + y^2[/imath] then [imath]x=\pm y.[/imath] What I have tried: I can reduce this to the case where [imath]\gcd(x,y)=1[/imath], since if [imath]x[/imath] and [imath]y[/imath] have a common factor, [imath]d[/imath] say, then [imath]d^2[/imath] divides through both [imath]xy[/imath] and [imath]x^2 + y^2[/imath] This then allows me to introduce another equation [imath]1=ax+by[/imath] for some [imath]a, b.[/imath] But I then get stuck ... | 622538 | Show that a number is not an integer
Show that [imath]X+\dfrac{1}{X}[/imath] is not an integer number for any rational [imath]X[/imath] and [imath]X \neq 1, X \neq -1[/imath] I think we can substitue [imath]X=\dfrac{P}{Q}[/imath] but I don't know if I can now assume that [imath]\gcd(P,Q)=1[/imath] |
3047827 | Value of integral involving [imath]\tan x[/imath]
[imath] \int_{0} ^ {\pi/4} \frac {\tan^2x}{1+x ^2} dx[/imath] I have tried using [imath]x=\tan \theta[/imath]. But in my opinion i am finding that this might not be closed form. Please help. | 2138866 | MIT 2015 Integration Question
So one of the question on the MIT Integration bee has baffled me all day today [imath]\int_{0}^{\frac{\pi}{4}}\frac{\tan^2 x}{1+x^2}\text{d}x[/imath] I have tried a variety of things to do this, starting with Integration By Parts Part 1 [imath]\frac{\tan x-x}{1+x^2}\bigg\rvert_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\frac{-2x(\tan x -x)}{\left (1+x^2 \right )^2}\text{d}x[/imath] which that second integral is not promising, so then we try Integration By Parts Part 2 [imath]\tan^{-1} x\tan^2 x\bigg\rvert_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}2\tan^{-1} x\tan x\sec^2 x\text{d}x[/imath] which also does not seem promising Trig Substitution [imath]x=\tan\theta[/imath] which results [imath]\int_{0}^{\tan^{-1}\frac{\pi}{4}}\tan^2 \left (\tan\theta\right )\text{d}\theta[/imath] which I think too simple to do anything with (which may or may not be a valid reason for stopping here) I had some ideas following this like power reducing [imath]\tan^2 x=\frac{1-\cos 2x}{1+\cos 2x}[/imath] which didn't spawn any new ideas. Then I thought maybe something could be done with differentiation under the integral but I could not figure out how to incorporate that. I also considered something with symmetry somehow which availed no results. I'm also fairly certain no indefinite integral exists. Now the answer MIT gave was [imath]\frac{1}{3}[/imath] but wolfram alpha gave [imath]\approx[/imath] .156503. Note The integral I gave was a simplified version of the original here is the original in case someone can do something with it [imath]\int_{0}^{\frac{\pi}{4}}\frac{1-x^2+x^4-x^6...}{\cos^2 x+\cos^4 x+\cos^6 x...}\text{d}x[/imath] My simplification is verifiably correct, I'd prefer no complex analysis and this is from this Youtube Video close to the end. |
3047963 | Pigeonhole principle problem 4
If we select [imath]1001[/imath] numbers from the set [imath]\{1,2,3,…,2000\}[/imath] there will be two numbers selected such that one divides the other. We need to prove this fact by noting that every number in the given set can be expressed in the form [imath]2^k⋅m[/imath] where m is an odd number and using the pigeonhole principle. Thank you. | 2933553 | Elementary number theory in sets
I'm back again So there's another problem that I can't get to prove If we take 21 numbers randomly from [imath]1, 2, 3, ..., 40[/imath] then between those [imath]21[/imath] numbers we will be able to find two numbers, of which the smaller one will divide the bigger one I've been reading james hein "discrete structures, logic and computability" but still can't get to think logically myself. I would be very grateful if I could get some directions/tips/hints or anything, thanks in advance |
3048318 | How to solve such a modular equation?
I have an equation as follows: [imath]27217 = 5s [/imath] mod [imath]42547[/imath] Using this website https://www.dcode.fr/modular-equation-solver, the correct result for s is 39481 as shown below however it does not list what steps are being done. Solving modular equation using dCode How would one go about to find the value of s in this modular equation? | 30458 | modular arithmetic, solving [imath]ax + b \equiv c \pmod d[/imath]?
For example [imath]151x - 294\equiv 44\pmod 7 [/imath]. How would I go about solving that? The answer says to simplify it into the [imath]ax\equiv b\pmod c [/imath] form first, but I have no clue on how to get rid of the [imath]294[/imath]... help? |
3048088 | Associativity of smash product for compact spaces
The following is Problem 2.2.14 in Tammo tom Dieck's Algebraic Topology: Let [imath]Y,Z[/imath] be compact or [imath]X,Z[/imath] be locally compact. Then the canonical bijection [imath](X\wedge Y)\wedge Z\to X\wedge(Y\wedge Z)[/imath] is a homeomorphism. I'm not working in the category of compactly generated spaces. I want to use this exercise to show that the iterated suspension is [imath]\Sigma^nX\cong X\wedge S^n[/imath] for any space [imath]X[/imath]. So it remains to prove this exercise. I could prove this for the locally compact case (using the fact that the product of a quotient map with the identity map of a locally compact space is again a quotient map), but for the case [imath]Y,Z[/imath] compact I have no idea how to do this. Any hints will be appreciated! | 2899753 | Associativty of smash product on compact spaces
Let [imath]Y,Z[/imath] be compact or [imath]X,Z[/imath] locally compact. Then the canonical bijection [imath] (X \wedge Y) \wedge Z \rightarrow X \wedge (Y \wedge Z) [/imath] is a homeomoprhism. I can prove the case when [imath]X,Z[/imath] are locally compact using exponential law. But I don't see how one approaches the case when [imath]Y,Z[/imath] compact. Hints? |
3048203 | Invent Binary Operator [imath]*[/imath] on Reals that Can Create [imath]+[/imath], [imath]-[/imath], [imath]\times[/imath], [imath]\div[/imath]
Exact Question: Invent a single binary operator [imath]*[/imath] such that for every real numbers [imath]a[/imath] and [imath]b[/imath], the operations [imath]a + b[/imath], [imath]a - b[/imath], [imath]a \times b[/imath], [imath]a \div b[/imath] can be created by applying [imath]*[/imath] (multiple times), starting with only [imath]a[/imath]'s and [imath]b[/imath]'s From my interpretation, you can apply [imath]*[/imath] recursively some number of times with carefully selected parameters to produce the desired outcome. I thought the operator should be a combination of [imath]a - b[/imath] and [imath]a \cdot b^{-1}[/imath] Since [imath]-[/imath] and [imath]\div[/imath] can produce [imath]+[/imath] and [imath]\times[/imath] respectively Please do not tell me the full answer. Give me a hint to point me towards the right path | 338370 | Is there an numeric arithmetic with a single operator?
I remember reading a while back about arithmetics with only a single operator that did all the work of the familiar [imath]+ - \times /[/imath] operators. If there is such a thing, could someone point me to it? As background: I have a side project where I'm playing with neural nets and genetic search algorithms. Right now it's the good old kind, with matrix multiplication and addition. I thought it would be interesting to implement a sort of neural network that used a combined or "all in one" operator. I've googled like heck and can't pick up the trail of where I thought I saw that before. Is there study in that area, or am I mis-remembering, or worse, just making it up? I'd appreciate any pointers. |
3049000 | adding a constant to a polynomial so all its roots are real.
If I have a polynomial with real cofficients [imath]p_x \in \mathbb{R}[x][/imath] , is it always possible to pick a constant [imath]c \in \mathbb{R}[/imath] so that [imath]p_x + c[/imath] only has real roots✱? Equivalently, does the following hold? [imath]p_x + c \propto \prod_{j=1}^{\deg(p_x)} \left(x-a_j\right) \;\;\;\;\text{where}\;\; a \in \mathbb{R} [/imath] All constant polynomials are proportional to [imath]1_x[/imath] . All linear polynomials are proportional to themselves. A quadratic polynomial [imath]ax^2+bx+c[/imath] has only real roots if and only if [imath]b^2 - 4ac[/imath] is non-negative, which can be ensured by picking [imath]c=0[/imath] . Geometrically, picking a constant [imath]c[/imath] is like picking where the x-axis is. It's definitely possible to pick a [imath]c[/imath] so that the line [imath]y = c[/imath] intersects the graph of [imath]p_x[/imath], but I don't have a good intuition for what the multiplicity is for the root I've created by choosing to intersect the graph of [imath]p_x[/imath] at some specific point. For instance, [imath]x^3[/imath] has a triple root, but [imath]x^3-1[/imath] only has a single root. The graph of [imath]x^3[/imath] flattens out near the origin, but I'm not sure how to turn that into an actual argument. Is it always possible to pick a [imath]c[/imath] so that the resulting polynomial has only real roots? ✱ I'm thinking of an element of [imath]\mathbb{R}[x][/imath] as a formal expression with a degree and as function drawn from [imath]\mathbb{C} \to \mathbb{C}[/imath] that's constrained to send reals to reals, so it makes sense to talk about its complex roots. The graph of [imath]p_x[/imath] is just a figure in [imath]\mathbb{R}^2[/imath], as usual. | 2790821 | Can *any* real polynomial be factored linearly (plus a constant) over [imath]\mathbb{R}[/imath]?
It seems to be true that every real polynomial [imath]p_n[/imath] of degree [imath]n[/imath] can be factored in the following (not unique) way [imath] p_n = \sum_{i=0}^n a_ix^i = s \left\{\prod_{i=1}^n(x-b_i)\right\} +t \tag{$\ast$}[/imath] with [imath]a_i, s, b_i, t [/imath] all in [imath]\mathbb{R}[/imath]. For example: [imath]x-1=(x-1)+0[/imath] [imath]x^2-5x+7=(x-2)(x-3)+1=(x-5)(x-0)+7[/imath] [imath]x^2+1 = (x-0) \cdot (x-0) +1[/imath] I am having a hard time coming up with a rigorous proof. I have tried using the fact that every polynomial [imath]p_n[/imath] can be written as [imath](x-a)g(x)+b[/imath], where [imath]g[/imath] is a polynomial function and [imath]a[/imath] is an arbitrary real number, but with no success. Another way it could be done is by expanding the right-hand side of [imath](\ast)[/imath] and showing that the linear system of equations with the coefficients of matching powers of [imath]x[/imath] has always a real solution. But I feel there must be a more simple proof out there, e.g. by induction, if possible without the Fundamental Theorem of Algebra. Would be glad if someone could enlighten me! |
3049066 | Find [imath]a_1/a_2[/imath] in the Laurent series expansion of [imath]\frac{1}{2z^2-13z+15}[/imath]
Let [imath]\sum a_n z^n [/imath] be the laurent series expansion of [imath]f(z)=\frac{1}{2z^2-13z+15}[/imath] in the annulus [imath]3/2 <|z|<5[/imath]. I am interested in finding [imath]a_1/a_2[/imath]. Efforts: [imath]f(z)=1/(2z-3)(z-10)[/imath] So using partial fraction decomposition [imath]f(z)=\frac{1}{7} (\frac{1}{z-10}-\frac{1}{2z-3})[/imath] How should I proceed from here? Thanks in advance. | 2285139 | Laurent series expansion of a given function
Let [imath]\sum_{-\infty}^{\infty} a_n z^n[/imath] be the Laurent series expansion of [imath]f(z)=\frac{1}{2z^2-13z+15}[/imath] in an annulus [imath]\frac{3}{2}< \vert z \vert < 5[/imath]. What is the value of [imath]\frac{a_1}{a_2}[/imath]? My attempt: First note that [imath]f[/imath] is analytic in the given annulus. [imath]f(z)=\frac{1}{2z^2-13z+15}=\frac{1}{(2z-3)(z-5)}[/imath] , [imath]\quad[/imath] [imath]\quad[/imath][imath]\quad[/imath][imath]\quad[/imath]$\quad$$\quad$$\quad$ [imath]=\frac{\frac{-2}{7}}{2z-3}+\frac{\frac{1}{7}}{z-5}[/imath] , by Partial Fraction. How to goes further to find out the required function in the power series form and find [imath]\frac{a_1}{a_2}[/imath]? I'm Confused with which term can be taken outside of the expression. Any help? |
3049217 | If [imath]f[/imath] has no non trivial fixed points and [imath]f\circ f[/imath] is the identity then [imath]f(x)=x^{-1}[/imath] and [imath]G[/imath] is abelian for [imath]f[/imath] an automorphism of [imath]G[/imath]
Let [imath]f[/imath] be an automorphism of the finite group [imath]G[/imath] such that [imath]f\circ f=id[/imath] and [imath]f(x)=x\implies x=e[/imath] Prove that [imath]f(x)=x^{-1}~\forall x\in G[/imath] If we can prove that [imath]f(x)[/imath] and [imath]x[/imath] commute for any given [imath]x\in G[/imath] then we're done with the proof because it would imply [imath]xf(x)=f(x)x[/imath] [imath]f(f(x))f(x)=f(x)x[/imath] Because [imath]f\circ f=id[/imath]. And since [imath]f[/imath] is a homomorphism: [imath]f(f(x)x)=f(x)x\implies f(x)x=e\implies f(x)=x^{-1}[/imath] I don't know how to proceed The fact that [imath]G[/imath] is abelian can be deduced once we have [imath]f(x)=x^{-1}[/imath] because that function is a homomorphism if and only if [imath]G[/imath] is abelian | 179512 | Finite [imath]G[/imath] with involutory automorphism [imath]\alpha[/imath] with no nontrivial fixed points. Proving properties of [imath]\alpha[/imath].
The question at hand is: Let G be a finite group and [imath]\alpha[/imath] an involutory automorphism of G, which doesn't fixate any element aside from the trivial one. 1) Prove that [imath] g \mapsto g^{-1}g^{\alpha} [/imath] is an injection 2) Prove that [imath]\alpha[/imath] maps every element to its inverse 3) Prove that G is abelian I think I've found 1). I assume [imath] g_1^{-1}g_1^{\alpha} = g_2^{-1}g_2^{\alpha} [/imath] and from this I get [imath] (g_2g_1^{-1})^\alpha = g_2g_1^{-1} [/imath], so [imath]g_2 = g_1[/imath] (is this correct?) For 2) I'm sort of stumped though, not sure how to start proving that, so any help please? |
2788169 | Properties of composition of fields extensions in Galois Theory.
Problem. Let [imath]K[/imath] and [imath]L[/imath] be subfields of a common field, both which contain a field [imath]F[/imath]. Prove the following statements. (a) If [imath]K = F(X)[/imath] for some set [imath]X \subset K[/imath], then [imath]KL = L(X)[/imath]. (b) [imath][KL:F] \leq [K:F][L:F][/imath]. (c) If [imath]K[/imath] and [imath]L[/imath] are algebraic over [imath]F[/imath], then [imath]KL[/imath] is algebric over [imath]F[/imath]. (d) Prove that the previous statement remains true when [imath]``[/imath]algebraic[imath]"[/imath] is replaced by [imath]``[/imath]normal[imath]"[/imath], [imath]``[/imath]separable[imath]"[/imath], [imath]``[/imath]purely inseparable[imath]"[/imath], or [imath]``[/imath]Galois[imath]"[/imath]. For (a). If [imath]K = F(X)[/imath], since [imath]F \subset L[/imath] and [imath]KL = K(L) = L(K)[/imath] we should have [imath]KL = L(X)[/imath]. For (b). I tried two different ways: Take [imath]\lbrace \alpha_{i} \rbrace[/imath] a basis for [imath]K[/imath] and [imath]\lbrace \beta_{j} \rbrace[/imath] a basis for [imath]L[/imath]. How to find a basis for [imath]KL[/imath] from these? The product generates [imath]KL[/imath]? I still get a little confused with the [imath]``[/imath]appearance[imath]"[/imath] of elements of [imath]KL[/imath] (maybe, this is a problem in (c)). Write [imath][KL:F] = [KL:L][L:F][/imath] and try to show that [imath][KL:L] \leq [K:F][/imath]. Intuitively this seems right, but I couldn't prove For (c). Take [imath]\alpha \in KL[/imath], so exists [imath]k_{1},...k_{n}; k'_{1},...,k'_{m};l_{1},...,l_{n};l'_{1},...,l'_{m}[/imath] such that [imath]\alpha = \frac{k_{1}l_{1} + ... + k_{n}l_{n}}{k'_{1}l'_{1} + ... + k'_{m}l'_{m}}.[/imath] Since [imath]k_{i}, k'_{j} \in K[/imath] and [imath]l_{i}, l'_{j} \in K[/imath], they are algebraic, then [imath]\alpha[/imath] is algebraic, because the set of algebraics is a field. This make sense? For (d). I have not tried it yet. Since I have not resolved the previous ones, I do not think it makes sense to try to. I think my difficulty in this problem is due to the fact that I have not understood very well how to explicitly say who the elements of [imath]KL[/imath]. I would like any hint! Thanks for the advance. | 1780882 | About some properties of composites of field extesions
When I'm self-studying Parick Morandi's book Field and Galois Theory,I came across some problems,which I can't work out fully. Let [imath]K[/imath], [imath]L[/imath], be two extension fields of base field [imath]F[/imath]. If [imath]K/F[/imath] and [imath]L/F[/imath] are both Galois extension ,how to proof that their composite [imath]KL[/imath] is also Galois over [imath]F[/imath]. And why we have the same conclusion when "Galois "is replaced by "normal","separable","purely inseparable",or "algebraic"? I worked on these problems and searched this website for two days but could only proof the cases for "separable","purely inseparable",and "algebraic".I have no idea how to proof the "normal" and "Galois" cases. In addition ,why the degree inequality holds for [imath] [KL:F]\leq[K:F][L:F][/imath] If you know how to proof these ,please tell me.I am here waiting for the answers.Thank you! |
3049629 | How to prove the following by Cauchy-Schwarz?
If [imath]u(x) \in C([a, b]), u(a) = 0,\; u(x) = \int_{a}^{x}u^{'}(t)dt[/imath] then [imath]\int_{a}^{b} |u|^{2} dx \le \frac{1}{2}(b - a)^{2}\int_{a}^{b}|u^{'}(t)|^{2}dt[/imath] The book said it can be proved using cauchy-schwarz-inequality, but I cannot make it. | 2228711 | Let [imath]f:[0;1]\to\mathbb R[/imath] be continiously differentiable
Let [imath]f:[0;1]\to\mathbb R[/imath] be continiously differentiable and [imath] f(0)=0[/imath]. Prove that [imath]\int\limits_0^1 |f(x)|^2\, dx\leq \dfrac{1}{2}\int\limits_0^1 |f'(x)|^2\, dx.[/imath] |
3050358 | Functional Equation involving Golden Ratio
Find all [imath]h:\mathbb{N}\rightarrow \mathbb{N}[/imath] such that [imath]h(h(n)) + h(n+1) = n+2[/imath] I tried this, but wasn't able to make any progress after a while. So, in vain, I looked at the solution. The solution basically obtained [imath]h(1)=1[/imath] and [imath]h(2)=2[/imath] after some casework, and then claimed that [imath]h(n) = \lfloor n\alpha \rfloor -n + 1[/imath] (where [imath]\alpha[/imath] denotes the golden ratio) without explaining at all why one would come at such a conclusion. How did they jump to this conclusion? (Source: Functional Equations: A Problem Solving Approach, Venkatachala.) | 1097958 | Functions satisfying [imath]f:\mathbb{N}\rightarrow\ \mathbb{N}[/imath] and [imath]f(f(n))+f(n+1)=n+2[/imath]
Find all functions [imath]f[/imath] such that [imath]f:\mathbb{N}\rightarrow\ \mathbb{N}[/imath] and [imath]f(f(n))+f(n+1)=n+2[/imath] Let us plug in [imath]n=1[/imath] [imath]f(f(1))+f(2)=3[/imath] Since the function is from [imath]\mathbb{N}[/imath] to [imath]\mathbb{N}[/imath], [imath]f(2)[/imath] can only take the values [imath]1,2[/imath]. Now we divide the problem into cases. Case-1: [imath]f(f(1))=2,f(2)=1[/imath] We can assume that [imath]f(1)=c[/imath] for the time being. Then plugging in [imath]n=3[/imath] and using [imath]f(2)=1[/imath] gives [imath]f(3)=4-c[/imath] and again since the range of the function is positive integers,then [imath]4-c[/imath] has to be positive and hence [imath]c[/imath] belongs to {[imath]1,2,3[/imath]}. Now, [imath]f(1)=c[/imath][imath]\implies f(f(1))=f(c)[/imath][imath]\implies 2=f(c)[/imath] by the assumption that [imath]f(f(1))=2[/imath] . Now,since [imath]c[/imath] can only take the values [imath]1,2,3[/imath],we start treating cases. If [imath]c=1[/imath],[imath]f(c)=2[/imath][imath]\implies f(1)=2[/imath] but we know from the deinition of [imath]c[/imath] that [imath]f(1)=c=1[/imath],a contradiction.If [imath]c=2[/imath],then [imath]2=f(c)=f(2)[/imath] but [imath]f(2)=1[/imath] by assumption. Finally,if [imath]c=3[/imath] [imath]2=f(c)=f(3)[/imath] but [imath]f(3)=4-c=4-3=1[/imath] which is once again a contradiction. Therefore there are no such functions in this case. Case-2: [imath]f(f(1))=1,f(2)=2[/imath] Again assuming [imath]f(1)=c[/imath] and using [imath]f(n)\le n[/imath] along with plugging [imath]n=c-1[/imath] will give us that [imath]f(1)=1[/imath] and then it is easy to prove that such a function exists by recursion. I can only give a "sort of recursive" way to define the function. Here it goes [imath]f:\mathbb{N}\rightarrow \mathbb{N}[/imath][imath]f(1)=1[/imath][imath]f(n)=n+1-f(f(n-1))[/imath] But this case is harder to deal with.Some help will be appreciated. |
3050197 | [imath]E(X(t)X(t))=\sigma^2\delta(\tau)[/imath] or [imath]E(X(t)X(t))=\sigma^2[/imath]
Let's say we have a white noise process [imath]x(t)[/imath] such that: [imath]E(X(t)X(t+\tau))=N\delta(\tau)[/imath] [imath]E(X(t))=0[/imath] In particular, with [imath]\tau=0[/imath], [imath]E(X(t)X(t))=E(X^2(t))[/imath] is infinite. Now, I want [imath]X(t)[/imath] at each time [imath]t[/imath] to have a normal distribution of 0 mean and [imath]\sigma^2[/imath] variance. That is: [imath]E(X^2(t))=\sigma^2[/imath] This is not consistent. I guess the expectations mean something different in both cases, but I don't find an explanation. This prevents me from moving ahead in a study of the mean, variance and autocorrelation of a process [imath]y(t)[/imath] defined as [imath]y(t)=1[/imath] if [imath]a<x(t)<b[/imath] and 0 otherwise. | 134193 | What is meant by a continuous-time white noise process?
What is meant by a continuous-time white noise process? In a discussion following a question a few months ago, I stated that as an engineer, I am used to thinking of a continuous-time wide-sense-stationary white noise process [imath]\{X(t) \colon -\infty < t < \infty\}[/imath] as a zero-mean process having autocorrelation function [imath]R_X(\tau) = E[X(t)X(t+\tau)] = \sigma^2\delta(\tau)[/imath] where [imath]\delta(\tau)[/imath] is the Dirac delta or impulse, and power spectral density [imath]S_X(f) = \sigma^2, -\infty < f < \infty[/imath]. At that time, several people with very high reputation on Math.SE assured me that this was an unduly restrictive notion, and that no difficulties arise if one takes the autocorrelation function to be [imath]E[X(t)X(t+\tau)] = \begin{cases}\sigma^2, & \tau = 0,\\ 0, & \tau \neq 0. \end{cases}[/imath] What engineers like to call a white noise process is a hypothetical beast that is never observed directly in any physical system, but which can be used to account for the fact that the output of a linear time-invariant system whose input is thermal noise is well-modeled by a wide-sense-stationary Gaussian process whose power spectral density is proportional to [imath]|H(f)|^2[/imath] where [imath]H(f)[/imath] is the transfer function of the linear system. Standard second-order random process theory says that the input and output power spectral densities [imath]S_X(f)[/imath] nd [imath]S_Y(f)[/imath] are related as [imath]S_Y(f) = S_X(f)|H(f)|^2.[/imath] Thus, pretending that thermal noise is a white Gaussian noise process in the engineering sense and pretending that the second-order theory extends to white noise processes (even though their variance is not finite) allows us to get to the result that the output power spectral density is proportional to [imath]|H(f)|^2[/imath]. My query about the definition of a white noise process is occasioned by a more recent question regarding the variance of a random variable [imath]Y[/imath] defined as [imath]Y = \int_0^T h(t)X(t)\ \mathrm dt[/imath] where [imath]\{X(t)\}[/imath] is a white Gaussian noise process. The answer given by Nate Eldredge leads to [imath]\operatorname{var}(Y) = \sigma^2 \int_0^T |h(t)|^2\ \mathrm dt[/imath] (as I pointed out in a comment on the answer) if the autocorrelation function is taken to be [imath]R_X(\tau) = \sigma^2\delta(\tau)[/imath] (the engineering definition). However, the OP on that question specified [imath]R_X(0) = \sigma^2[/imath], not [imath]\sigma^2\delta(\tau)[/imath], that is, the definition accepted by mathematicians. For this autocorrelation function, the variance is [imath]\int_0^T \int_0^T E[X(t)X(s)]h(t)h(s)\mathrm dt\mathrm ds = 0[/imath] since the integrand is nonzero only on a set of measure [imath]0[/imath]. So, what is the variance of the random variable [imath]Y[/imath]? and what do readers of Math.SE understand by the phrase white noise process? Perhaps this question should be converted to a Community wiki? |
3050844 | Evaluate [imath]\int_{0}^{\frac{\pi}{2}} x\ln(\tan(x))dx [/imath]
How do I evaluate: [imath]\int_{0}^{\frac{\pi}{2}} x\ln(\tan(x))dx [/imath] I know it equals [imath]\frac{7\zeta(3)}{8}[/imath], but I'm not sure how to get there. I've been trying to convert it into a sum somehow, but I'm not getting anywhere. Equivalent integrals are: [imath]\frac{1}{2}\int_{-\infty}^\infty x\operatorname{sech}(x)\arctan(e^x)dx[/imath] [imath]\int_0^\infty \frac{\arctan(x)\ln(x)}{1+x^2} dx[/imath] I've tried differentiating under the integral but to no avail. | 1634265 | How to evaluate [imath]\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2} \, \mathrm{d}\theta[/imath]
I have some trouble in how to evaluate this integral: [imath] \int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right) \,\mathrm{d}\theta [/imath] I think it maybe has another form [imath] \int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right) \,\mathrm{d}\theta = \sum_{n=1}^{\infty}{1 \over n^{2}} \left[\psi\left(n + {1 \over 2}\right) - \psi\left(1 \over 2\right)\right] [/imath] |
1768738 | Induced topologies by Metric Spaces continuous
Let [imath](X,d)[/imath] be a metric space and [imath]T[/imath] the corresponding topology on [imath]X[/imath]. Fix [imath]a[/imath] as an element in X. Prove that the map [imath]f: (X,T) \to \mathbb{R}[/imath] defined by [imath]f(x) = d(a,x)[/imath] is continuous. I know I need to show that the inverse image of an open set in [imath]\mathbb{R}[/imath] is an open set in [imath]X[/imath]. I know that an open set in [imath]X[/imath] is a subset of [imath]X[/imath] whose distance from [imath]a[/imath] is r because of the definition of an open ball in [imath]X[/imath]. The inverse image of an open set in [imath]\mathbb{R}[/imath] (call it [imath]O[/imath]) is the set of all points in [imath]X[/imath] whose distance from a is an [imath]r[/imath] that's an element of the open set [imath]O[/imath] in [imath]r[/imath]. I feel like this thinking is right, but I don't know how to prove this further. Anyone have any ideas? | 1635688 | Distance function is continuous in topology induced by the metric
The question is (from Topology without tears) that: Let [imath](X,d)[/imath] be a metric space and [imath]\tau[/imath] the corresponding topology on [imath]X[/imath]. Fix [imath]a \in X[/imath]. Prove that the map [imath]f:(X,\tau) \rightarrow \mathbb{R}[/imath] defined by [imath]f(x) = d(a,x)[/imath] is continuous. My first attempt was that let an open ball, [imath]B_{\epsilon}(a)\subset \mathbb{R}[/imath] and show that [imath]f^{-1}(B_{\epsilon}(a)) \subset \tau[/imath]. Since [imath]f(x) = d(a,x)[/imath], [imath]f^{-1}(B_{\epsilon}(a))=B_{\epsilon}(a) \in \tau[/imath]. I am pretty sure the last sentence is wrong. I am not sure if every open ball is an open set in topology induced by metric space. Please give me some hint and direction. Thanks |
1531826 | The roots of [imath]x^2-2x+3=0[/imath] are [imath]\alpha[/imath] and [imath]\beta[/imath]. Find the equation whose roots are: [imath]\alpha+2[/imath], [imath]\beta+2[/imath]. Not sure of answer in book.
The roots of [imath]x^2-2x+3=0[/imath] are [imath]\alpha[/imath] and [imath]\beta[/imath]. Find the equation whose roots are: [imath]\alpha+2[/imath], [imath]\beta+2[/imath]. Not sure of answer in book. My working: [imath]\alpha+\beta=2, \alpha\beta=3[/imath] [imath](\alpha+2)+(\beta+2)=\alpha+\beta+4=6[/imath] [imath](\alpha+2)(\beta+2)=\alpha\beta+2(\alpha+\beta)=7[/imath] Therefore, equation whose roots are [imath]\alpha+2, \beta+2[/imath] is [imath]x^2-6x+7=0[/imath] Answer in book is [imath]x^2-6x+11=0[/imath] | 1187355 | If [imath]a,b[/imath] are the roots of the equation [imath]2 x^2 -3 x +1 = 0[/imath], find an equation whose roots are [imath]a/(2b +3)[/imath], [imath]b/(2a +3)[/imath]
If [imath]a,b[/imath] are the roots of the equation [imath]2 x^2 -3 x +1 = 0[/imath], find an equation whose roots are [imath]a/(2b +3)[/imath], [imath]b/(2a +3)[/imath] I was practicing quadratic equation questions online but I am stuck on this question. I found the values of roots [imath]a= 1[/imath], [imath]b= 1/2[/imath] |
3051474 | Equation of a line reflected in a plane
The line [imath]l_1[/imath] has the equation [imath]r=(6i+2−2k)+λ(4i+5j−k)[/imath] and the plane [imath]π_1: 2x−y+4z=4[/imath], the line [imath]l_2[/imath] is the reflection of [imath]l_1[/imath] in the plane π1. Find the exact vector equation of line [imath]l_2[/imath] So the line intersects the plane when [imath]λ=−2[/imath], giving the point [imath](−2,−8,0)[/imath] which will be common on [imath]l_1[/imath] and [imath]l_2[/imath]. I have managed to get a direction vector of [imath](92/21i + 206/21j +26/21k)[/imath] but I don't think this is right. Any help would be appreciated. | 3047370 | Reflection of a line in a plane
The line [imath]l_1[/imath] has the equation [imath]r=(6i+2-2k)+\lambda(4i+5j-k)[/imath] and the plane [imath]\pi_1[/imath]: [imath]2x-y+4z=4[/imath], the line [imath]l_2[/imath] is the reflection of [imath]l_1[/imath] in the plane [imath]\pi_1[/imath]. Find the exact vector equation of line [imath]l_2[/imath] So the line intersects the plane when [imath]\lambda=-2[/imath], giving the point [imath](-2,-8,0)[/imath] which will be common on [imath]l_1[/imath] and [imath]l_2[/imath]. But I am unsure on how to find the direction vector for [imath]l_2[/imath]. Any help would be appreciated. |
3051872 | What is the unit digit of [imath] 1^1+2^2+3^3+...+2018^{2018} [/imath]?
I want to find the unit digit of [imath] 1^1+2^2+3^3+...+2018^{2018} [/imath] . I have find out the unit digit of [imath] 2^2,3^3,4^4,....259^{259},....2018^{2018}[/imath] . But I cant find out the unit digit of sum of all of this term . Is there any way to find out the digit of sum of all of this term ? Please help me . | 325529 | Find [imath]x[/imath] such that [imath]\sum_{k=1}^{2014} k^k \equiv x \pmod {10}[/imath]
Find [imath]x[/imath] such that [imath]\sum_{k=1}^{2014} k^k \equiv x \pmod {10}[/imath] I knew the answer was [imath]3[/imath]. |
3052257 | Why [imath]L_{p}(\mathbb{R})[/imath] is separable?
It's easy to prove that [imath]L_{p}(K)[/imath] is separable , using Stone theorem. But how can we show the same result for real line ? I thought about considering : [imath] \mathbb{R} = \cup (a_i , b_i] [/imath] and [imath]\mathbb{1}((a_i , b_i])[/imath]. But my teacher said that they are not in [imath]L_{p}[/imath]. | 1368174 | Separability of [imath]l^{p}[/imath] spaces
How can I prove that the space [imath]l^{p}[/imath] equipped with the norm (for [imath]x=(x_{n}) \in l^{p})[/imath]: [imath]||x||_{p}=(\displaystyle\sum_{n}|x_{n}|^{p})^{1/p}[/imath] Is a separable space? (i.e. showing that there is a countable dense set in [imath]l_{p}[/imath]). I saw a proof in which they started with first letting [imath]x=(x_{n})[/imath] be an element of [imath]l^{p}[/imath] and for any [imath]\epsilon >0[/imath], then they chose [imath]N[/imath] such that: [imath](\displaystyle\sum_{n>N}|x_{n}|^{p})<(\frac{\epsilon}{2})^{p}[/imath] However, I'm not sure where does this come from. As well, I'm not sure how to proceed from this step. Thank you for your help! |
3052863 | [imath]f[/imath] is continuous on [imath][0,1][/imath], [imath]f(0)=f(1)[/imath],[imath]n\in \mathbb N[/imath] constant [imath]\Rightarrow[/imath] There exists [imath]x\in[0,1-\frac{1}{n}][/imath] : [imath]f(x+\frac{1}{n})=f(x).[/imath]
I've been trying to prove this statement: Let [imath]f[/imath] be a continuous function on [imath][0,1][/imath] such that [imath]f(0)=f(1):=c[/imath], and let [imath]n[/imath] be a natural constant. Prove that there exists [imath]x\in[0,1-\frac{1}{n}], x\in \mathbb R[/imath] such that [imath]f(x+\frac{1}{n})=f(x).[/imath] I tried to use the Intermediate Value Theorem by defining [imath]g(x)=f(x+\frac{1}{n})-f(x)[/imath]. I could see that [imath]g(0)=f(\frac{1}{n})-c[/imath] and [imath]g(1-\frac{1}{n})=c-f(1-\frac{1}{n})[/imath] yet I could not figure out how to prove that [imath]\mathrm s\mathrm i\mathrm g\mathrm n (g(0)\cdot g(1-\frac{1}{n}))=-1.[/imath] Edit: Please notice that this statement is not (necessarily) true for every n. The statement is about some constant n. Thank you and have a good day!. | 1433031 | If [imath]f[/imath] is continuous, then there exists [imath]x+\frac{1}{n}\in[0,1][/imath] and [imath]f\left(x+\frac{1}{n}\right)=f(x)[/imath]
Let [imath]f:[0,1]\to \mathbb{R}[/imath] is continuous, such that [imath]f(0)=f(1)[/imath]. For each [imath]n\in\mathbb{N}[/imath], prove that [imath]\exists x\in [0,1][/imath] such that [imath]x+\dfrac{1}{n}\in[0,1][/imath] and [imath]f\left(x+\dfrac{1}{n}\right)=f(x)[/imath]. My approach: The interval [imath][0,1][/imath] is connected, then if [imath]f[/imath] is continuous, we have [imath]f([0,1])\subset\mathbb{R}[/imath] is a interval. And I suppose that, I can define a closed path such that [imath]f(0)=f(1)[/imath]. Any hint pls, |
3053080 | Relation between GCD and LCM
When I was learning about GCD and LCM I found this relation somewhere. But I didn't get proof for this relation. [imath]\frac{\gcd(a,b,c)^2}{\gcd(a,b)*\gcd(b,c)*\gcd(c,a)}=\frac{\operatorname{lcm}(a,b,c)^2}{\operatorname{lcm}(a,b)*\operatorname{lcm}(b,c)*\operatorname{lcm}(c,a)} [/imath] where GCD represents Greatest Common Divisor and LCM represents lowest common multiple. | 853771 | Identity involving LCM and GCD: $\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$
Let [imath](a,b)[/imath] denote the GCD of [imath]a[/imath] and [imath]b[/imath], and let [imath][a,b][/imath] denote the LCM of [imath]a[/imath] and [imath]b[/imath]. Prove [imath]\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}[/imath]. |
3052156 | Is the natural order on an idempotent semiring total/linear?
We know that on an idempptent semiring [imath]R[/imath], the natural order relation is defined as: for all [imath]x, y\in R[/imath], [imath]x\leq y[/imath] when [imath]x+y=y[/imath], which is clearly a partial order relation. I am unable to point out whether this relation is a total order relation too? i.e., does it satisfy Comparability (trichotomy law)? | 3052091 | Is the natural order relation on an idempotent semiring total/linear?
We know that on an idempotent semiring [imath]R[/imath], the natural order relation is defined as: for all [imath]x, y\in R[/imath], [imath]x\leq y[/imath] when [imath]x+y=y[/imath], which is clearly a partial order relation. I am unable to point out whether this relation is a total order relation too? i.e., does it satisfy Comparability (trichotomy law)? |
3053305 | Why is [imath]a^{-x}[/imath] defined to be equal to [imath]\frac{1}{a^x}[/imath]?
I have searched the reason behind this definition in two textbooks and haven't found any. They just state that this is the definition but don't ever give any motivation for why this is truth. Edit: I figured out why that relation must be true about 10 seconds after posting the question. Instead of looking for explanations I should have thought for a while. Still, thank you everyone. | 1885177 | If [imath]x^a[/imath] is [imath]x[/imath] multiplied [imath]a[/imath] times, then how does [imath]x^{-1}[/imath] make sense?
What is the meaning of [imath]x[/imath] raised to any non-positive value? We know that [imath]x^{-a} = \dfrac{1}{x^a}[/imath] and [imath]x^0 = 1[/imath], but where does that come from? What is the proof? Why is this true? What about [imath]x[/imath] raised to a fraction, like say [imath]\frac{1}{3}[/imath]? How do you multiply [imath]x[/imath] [imath]\frac{1}{3}[/imath] times? |
3053859 | solutions for Diophantine [imath]2x^3+1=y^3[/imath]
Consider Diophantine equation [imath]2x^3+1=y^3[/imath] There are two solutions [imath]x=(0,-1), \, y=(1,-1) [/imath] Are there anymore integer solutions for [imath]x[/imath] and [imath]y[/imath]? I've tried the following argument but am not satisfied with it. Identity: [imath][(B^3)(x^6)+3(B^2)(x^3)+3(B)]x^3+1=(Bx^3+1)^3[/imath] Looks like: [imath][ 2 ]x^3+1=( y )^3[/imath] Let [imath][(B^3)(x^6)+3(B^2)(x^3)+3(B)]=2[/imath] and [imath](Bx^3+1)=y[/imath] Since [imath]2[/imath] divides [imath]B[/imath] we get possible values for [imath]B[/imath] are [imath]B=\{-2,-1,1,2\}[/imath] Testing the [imath]B[/imath] set in equation [imath][(B^3)(x^6)+3(B^2)(x^3)+3(B)]=2[/imath] we can see that the only possible integer solution is [imath]B=2[/imath] and [imath]x=-1[/imath] and so [imath]y=-1[/imath]. The other solution is trivial [imath]x=0,\,y=1[/imath]. So from this argument there are no other integer solutions for [imath]x[/imath] and [imath]y[/imath]. My problem with the above argument is that by me saying: "Let [imath][(B^3)(x^6)+3(B^2)(x^3)+3(B)]=2[/imath] and [imath](Bx^3+1)=y[/imath]" I am indirectly saying from the start that there and no other solutions. What I would have to do is prove that: all possible solutions for [imath]y[/imath] are of the form [imath]Bx^3+1[/imath] and equation [imath][(B^3)(x^6)+3(B^2)(x^3)+3(B)]=2[/imath] is true for all possible solutions for [imath]x[/imath] But I cannot prove it. thank you | 133754 | diophantine equations [imath]x^3-2y^3=1[/imath]
I'm not familiar with diophantine equations. At most my approaches doesn't give results. I need to solve the following equation [imath]x^3-2y^3=1[/imath] Where [imath]x,y,z\in\mathbb{Z}[/imath] I know [imath]x=-1,y=-1.x=1,y=0,[/imath] Aware of any other integer solutions. Prove |
3053901 | Proving the Identity Theorem for polynomials
Bob tells me that the Identity Theorem is the three following statements If a polynomial has infinitely many roots, then it is equal to [imath]0[/imath]. If two polynomials satisfy [imath]P(x)=Q(x)[/imath] for infinitely many [imath]x[/imath], then the two polynomials are equal. If two polynomials of at most degree [imath]n[/imath] satisfy [imath]P(x)=Q(x)[/imath] for [imath]n+1[/imath] values of [imath]x[/imath], then the two polynomials are equal. I do not know how to prove these; I think the Factor, Remainder, and Division Theorems will be useful. I tried using Fundamental Theorem of Algebra, but it did not get me anywhere. | 37517 | Show that if [imath] R [/imath] is an integral domain then a polynomial in [imath] R[X] [/imath] of degree [imath] d [/imath] can have at most [imath] d [/imath] roots
Show that if [imath] R [/imath] is an integral domain then a polynomial in [imath] R[X] [/imath] of degree [imath] d [/imath] can have at most [imath] d [/imath] roots. Thoughts so far: I feel like I might be missing something here. If [imath] R [/imath] is an integral domain, then so is [imath] R[X] [/imath]. The factor theorem gives a correspondence between roots and factors. Clearly (is this clear?) a polynomial of degree [imath] d [/imath] can't have more than [imath] d [/imath] irreducible factors. |
3051997 | Diagonal entries are zero, others are [imath]1[/imath]. Find the determinant.
[imath]\det\begin{vmatrix} 0 & \cdots & 1& 1 & 1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ 1 & \cdots & 1 & 1 & 0 \end{vmatrix}=?[/imath] Attempt: First I tried to use linearity property of the determinants such that [imath]\det\binom{ v+ku }{ w }=\det\binom{v }{ w }+k\det \binom{ u }{ w }[/imath] [imath]v,u,w[/imath] are vectors [imath]k[/imath] is scalar. I have tried to divide it into [imath]n[/imath] parts and tried to compose with sense but didn't acomplish. Second I tried to make use of "Row Reduction" i.e. adding scalar multiple of some row to another does not change the determinant, so I added the all different rows into other i.e. adding [imath]2, 3,4,\dots,n[/imath]th row to first row and similarly doing for all rows we got [imath]\det\begin{vmatrix} 0 & \cdots & 1& 1 & 1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ 1 & \cdots & 1 & 1 & 0 \end{vmatrix}=\det\begin{vmatrix} n-1 & \cdots & n-1& n-1 & n-1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ n-1 & \cdots & n-1 & n-1 & n-1 \end{vmatrix}=0[/imath] The last determinant is zero (I guess) so the given determinant is zero? I don't have the answer this question, so I am not sure. How to calculate this determinant? | 563011 | induction proof of a determinant [imath]n \times n[/imath]
I have to proof the following property: Can somebody help my with a few steps for n=n+1? Thanks in advance. Cheers. |
3054177 | If G solvable and S a subgroup, then S is solvable
Let [imath]G[/imath] be a solvable group, and [imath]S[/imath] a subgroup. Prove that [imath]S[/imath] is solvable. Let [imath]G[/imath] be solvable. Then there exists a sequence of subgroups [imath] \{ e\}\triangleleft H_r\triangleleft H_{r-1}\cdots \triangleleft H_1 \triangleleft H_0=G [/imath] such that [imath]H_{i+1}\triangleleft H_i[/imath] for [imath]i=0, ...r-1,[/imath] and [imath]H_i/H_{i+1}[/imath] is abelian. If [imath]S=H_i[/imath], where [imath]H_i[/imath] is as above for [imath]i\neq r[/imath], then it is trivial. So we need to assume that [imath]S\neq H_i.[/imath] Let [imath]N_S[/imath] be the normalizer of [imath]S[/imath] in [imath]G[/imath]. We know that [imath]S\triangleleft N_S[/imath] and [imath]N_S[/imath] is a subgroup of [imath]G.[/imath] Here I am stuck. Can somebody propose a solution or give me some hint. Thanks. | 144812 | Subgroups of finite solvable groups. Solvable?
I am attempting to prove that, given a non-trivial normal subgroup [imath]N[/imath] of a finite group [imath]G[/imath], we have that [imath]G[/imath] is solvable iff both [imath]N[/imath], [imath]G/N[/imath] are solvable. I was able to show that if [imath]N,G/N[/imath] are solvable, then [imath]G[/imath] is; also, that if [imath]G[/imath] is solvable, then [imath]G/N[/imath] is. I am stuck showing that [imath]N[/imath] must be solvable if [imath]G[/imath] is. It seems intuitive that any subgroup of a finite solvable group is necessarily solvable, as well. Is this true in general? For normal subgroups? How can I go about proving this result? Edit: By solvable, I mean we have a finite sequence [imath]1=G_0\unlhd ... \unlhd G_k=G[/imath] such that [imath]G_{j+1}/G_j[/imath] is abelian for each [imath]1\leq j<k[/imath]. |
3054226 | Is the following statement is True/false?[imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath]
Is the following statement is True/false ? A continuous function [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath] is uniformly continuous if it maps Cauchy sequences into Cauchy sequences. i thinks its true take take [imath]f(x) = x[/imath] | 709626 | Uniform continuity of a function and cauchy sequences
So I'm pretty sure this is almost immediate from the definitions, please tell me if I am incorrect.. Consider two cauchy sequences in D, [imath]\{x_n\}[/imath] and [imath]\{y_m\}[/imath]. Since [imath]f[/imath] is uniform continuous we know that there exists [imath]\delta > 0[/imath] for all [imath] \epsilon > 0[/imath] such that for [imath]x_n , y_m[/imath] ([imath]\forall n,m)[/imath] [imath]|x_n - y_m| < \delta \implies |f(x_n) - f(y_m)| < \epsilon[/imath] [imath]f(x_n), f(y_n)[/imath] are clearly sequences, and since we have this inequality for all [imath]n,m[/imath] we conclude that [imath]f(x_n) , f(y_n)[/imath] are cauchy sequences. |
3053821 | Total subsets in incomplete inner product spaces
I'm reading Induced Representations of locally compact groups by Kaniuth and Taylor and I don't understand how total subsets work (in particular in Lemma 2.24). I know that a total subset of an inner product space is a set such that the closure of its linear span coincide with the entire space. Shortly, in this lemma the authors want to prove that a particular subset [imath]D[/imath] of an Hilbert space [imath]H[/imath] is total. To prove this, they are showing that its linear span is dense in [imath]V[/imath], an inner product space which is dense in [imath]H[/imath] (actually [imath]H[/imath] was constructed as the completion of [imath]V[/imath]). Letting [imath]v \in V[/imath], it happens that [imath]\langle d,v \rangle = 0[/imath] for every [imath]d\in D[/imath] implies [imath]v = 0[/imath]. By this last fact, the authors say that the span of [imath]D[/imath] is a linear dense subspace of [imath]V[/imath]: this is what I can't understand. I looked for some informations on other books and it seems that in Hilbert spaces this can be viewed as an equivalent fact to being a total subset, but here we're considering [imath]D[/imath] a total subset of [imath]V[/imath] and not of [imath]H[/imath]. Is completeness necessary to prove this equivalence? I tried to prove it by myself, but I didn't manage to do it, nor I found books about this topic, since many of them (like Rudin's Real and complex analysis) introduce inner product spaces and switch to Hilbert spaces after a few pages. How does the fact that the orthogonal space of [imath]D[/imath] is trivial imply that [imath]D[/imath] is total? | 1023061 | An inner product space and its proper closed subspace with trivial orthogonal complement
I am looking to do the following: Construct an inner product space [imath]X[/imath] (with inner product [imath]\langle \cdot, \cdot \rangle[/imath] and a proper, closed subspace [imath]Y[/imath] of [imath]X[/imath] such that [imath]Y^\perp = \{0\}[/imath], ie [imath]\langle x, y \rangle = 0 \ \forall y \in Y \iff x = 0[/imath]. I see that we need [imath]X[/imath] to be infinite dimensional, hence isomorphic to [imath]\Bbb R^n[/imath] with the standard dot product, and then clearly not possible, as a proper (closed) subspace has a smaller dimension. If someone could give me a hint as to how to start this construction, then I'd be most grateful, as I'm fairly stuck beyond this! As always, please make sure that the hint is reasonably minor, ie doesn't give away too much - I still want to learn from this question, not just be told the answer! The question is on a course in linear analysis. |
3053941 | Why doesn't L'hopitals Rule work for [imath]\lim\limits_{x \to \infty} \frac{x+ \sin x}{x+ 2 \sin x}[/imath]?
This is how I would evaluate [imath]\lim\limits_{x \to \infty} \dfrac{x+ \sin x}{x+ 2 \sin x}[/imath] [imath]=\lim\limits_{x \to \infty} \dfrac{x \left( 1+ \frac{\sin x}{x} \right)}{x \left(1+ 2 \cdot \frac{ \sin x}{x} \right)}[/imath] [imath]= \dfrac{1+0}{1+2 \cdot 0} = 1[/imath] But now applying L'hopitals Rule, I get [imath]\lim\limits_{x \to \infty} \dfrac{1+ \cos x}{1+ 2 \cos x}[/imath] Since [imath]\cos x [/imath] just oscillates between [imath][-1,1][/imath] I think we can conclude the limit doesn't exist. What is going on here? | 1342202 | Why doesn't L'Hopital's rule work in this case?
I have a very simple question. Suppose I want to evaluate this limit: [imath]\lim_{x\to \infty} \frac{x}{x-\sin x}[/imath] It is easy to evaluate this limit using the Squeeze theorem (the answer is [imath]1[/imath]). But here both the numerator and the denominator are going to infinity as [imath]x\to \infty[/imath] so I tried using L'Hospital's rule: [imath]\lim_{x\to \infty} \frac{x}{x-\sin x}=\lim_{x\to \infty} \frac{1}{1-\cos x}[/imath] However there's no finite [imath]L[/imath] such that [imath]\lim_{x\to \infty} \frac{1}{1-\cos x}=L[/imath] which is a contradiction. I don't understand why in this case L'Hopital's rule doesn't work. Both the numerator and the denominator are differentiable everywhere and both are tending to infinity - which is all we need to use this rule. |
3053595 | Proof Verification: Show that if [imath]f(z)[/imath] is a non-constant entire function , then [imath]g(z)=exp(f(z))[/imath] has essential singularity in [imath]z=\infty[/imath].
Show that if [imath]f(z)[/imath] is a non-constant entire function , then [imath]g(z)=exp(f(z))[/imath] has essential singularity in [imath]z=\infty[/imath]. This question has been answered in the link Thanks! My approach: Let [imath]f[/imath] analytic (then [imath]f[/imath] has a series), then [imath]exp(f(z))[/imath] is analytic. Then [imath]exp(f(\frac{1}{w}))=\sum_{n=0}^{\infty}{\frac{1}{n!}(f(\frac{1}{w}))^{n}}=\sum_{n=0}^{\infty}{\frac{1}{n!}(\sum_{m=0}^{\infty}{\frac{a_{m}}{w^m}})^{n}}[/imath] We can see that [imath]exp(f(\frac{1}{w}))[/imath] has essential singularity in [imath]w=0[/imath], since there exist negative coefficients. So, [imath]exp(f(z))[/imath] has essential singularity in [imath]w=0[/imath]. Is this right? Thanks! | 127168 | Essential singularity
This is an exercise from Gamelin. If [imath]f(z)[/imath] is a complex function with a not removable singularity in [imath] z_{0} \ [/imath], then [imath]e^{f(z)} \ [/imath] has an essential singularity in [imath]z_{0} [/imath]. Any hint? |
3053887 | Limit points of [imath]\left\{ \frac{m}{2^{n}} \right\}[/imath] where [imath]m,n[/imath] are natural numbers.
Let [imath]E = \left\{\frac{m}{2^{n}}\right\}[/imath] where [imath]m,n[/imath] are integers, then [imath]0[/imath] is the only limit point of this set. [imath]([/imath]true[imath]/[/imath]false[imath]) ?[/imath] I think the [imath]0[/imath] is the only limit point of this set. Since, as [imath]n \to \infty[/imath] terms of this sequence come nearer to [imath]0[/imath]. But the statement is false according to answer key. | 268195 | The set of limit points of the sequence [imath]1,\frac12,\frac14,\frac34,\frac18,\frac38,\frac58,\frac78,\frac1{16},\frac3{16},\ldots[/imath]
I came across the following problem that says: The set of limit points of the sequence [imath]1,\dfrac12,\dfrac14,\dfrac34,\dfrac18,\dfrac38,\dfrac58,\dfrac78,\dfrac1{16},\dfrac3{16},\dfrac5{16},\dfrac7{16},\dfrac9{16},\ldots[/imath] is which of the following? (a) [imath][0,1][/imath] (b) [imath](0,1][/imath] (c) the set of all rational numbers in [imath][0,1][/imath] (d) the set of all rational numbers in [imath][0,1][/imath] and of the form [imath]m/2^n[/imath] where [imath]m[/imath] and [imath]n[/imath] are integers. Please help. Thanks in advance for your time. |
3054848 | If [imath]G[/imath] solvable and [imath]f[/imath] a surjective homomorphism between [imath]G,G'[/imath], then [imath]G'[/imath] is solvable
Let [imath]G[/imath] be a solvable group, [imath]f: G \rightarrow G'[/imath] a surjective homomorphism. I need to show that [imath]G'[/imath] is solvable. Let [imath]G[/imath] be solvable. Then there exists a sequence of subgroups [imath]\{e\}=H_r\triangleleft H_{r-1}\triangleleft\cdots \triangleleft H_{1}\triangleleft H_0=G[/imath] such that [imath]H_{i+1}\triangleleft H_i[/imath] for all [imath]i=0,1,...,r-1, [/imath] and such that [imath]H_i/H_{i+1} [/imath] is abelian. We know that [imath]f(H_i)\triangleleft f(G)[/imath], but since [imath]f[/imath] is surjective, [imath]f(H_i)\triangleleft G', \forall i=0,1,...,r-1.[/imath] Also, [imath]f(H_i)\subset f(H_{i-1}),\, [/imath]thus [imath]f(H_i)\triangleleft f(H_{i-1}), \forall i=0,1,...,r-1.[/imath] Thus we get [imath]\{e\}=f(H_r) \triangleleft f(H_{r-1})\triangleleft \cdots \triangleleft f(H_1)\triangleleft f(H_0=G)=G'.[/imath] It remains to be shown that [imath]f(H_i)/f(H_{i+1})[/imath] is abelian. Here I can not go further. I need a solution proposal or some hint. Thanks. | 1890324 | Conditions under which group solvability is invariant under an homomorphism.
Let [imath]G[/imath] be a solvable group and [imath]f: G \rightarrow H[/imath] an isomorphism. My question is if [imath]H[/imath] is a solvable group. Let [imath]1=G_0 \trianglelefteq G_1 \trianglelefteq ... \trianglelefteq G_n = G[/imath] be a normal series of [imath]G[/imath] whose factors are abelian groups. Clearly, [imath]1=f(G_0) \trianglelefteq f(G_1) \trianglelefteq ... \trianglelefteq f(G_n) = H[/imath] is a normal series of [imath]H[/imath] but are the quotients abelian? Is there a weaker condition for invariance? |
3054886 | Indefinite integration of a function with a trigonometric function raised to a high power
I need to integrate indefinitely the expression [imath] \frac{\csc^{2}x - 2017}{\cos^{2017}x} [/imath] I am unable to handle this high power of [imath]\cos(x)[/imath] . A hint would be useful for me. | 884906 | Evaluate [imath]\int\frac {\csc^2{x}-2005}{\cos^{2005}{x}} dx [/imath]
Evaluate the indefinite integral [imath]\int\frac {\csc^2{x}-2005}{\cos^{2005}{x}} dx[/imath] I tried multiplying and dividing by [imath]\sec^2 {x} [/imath] and then setting [imath]\tan{x}=y[/imath] but no good. Then I set [imath]\cos {x}=t [/imath] and tried to create [imath]\sin {x} [/imath] in the numerator. But the integral which came was also a difficult one. Please Help! Thanks! P.S. I am a high school student so kindly use elementary methods only. Thanks again! |
2997625 | Injective ring endomorphism and determinant
I'm trying to prove this: Let [imath]R[/imath] be a ring, and [imath]A \in M_n(R)[/imath]. Write [imath]L_A[/imath] for the linear map [imath]L_A : R^n \to R^n[/imath] determined by left multiplication by [imath]A[/imath]. Shows that if [imath]L_A[/imath] is injective, then det([imath]A[/imath]) is not a zero divisor. | 161523 | Does an injective endomorphism of a finitely-generated free R-module have nonzero determinant?
Alternately, let [imath]M[/imath] be an [imath]n \times n[/imath] matrix with entries in a commutative ring [imath]R[/imath]. If [imath]M[/imath] has trivial kernel, is it true that [imath]\det(M) \neq 0[/imath]? This math.SE question deals with the case that [imath]R[/imath] is a polynomial ring over a field. There it was observed that there is a straightforward proof when [imath]R[/imath] is an integral domain by passing to the fraction field. In the general case I have neither a proof nor a counterexample. Here are three general observations about properties that a counterexample [imath]M[/imath] (trivial kernel but zero determinant) must satisfy. First, recall that the adjugate [imath]\text{adj}(M)[/imath] of a matrix [imath]M[/imath] is a matrix whose entries are integer polynomials in those of [imath]M[/imath] and which satisfies [imath]M \text{adj}(M) = \det(M).[/imath] If [imath]\det(M) = 0[/imath] and [imath]\text{adj}(M) \neq 0[/imath], then some column of [imath]\text{adj}(M)[/imath] lies in the kernel of [imath]M[/imath]. Thus: If [imath]M[/imath] is a counterexample, then [imath]\text{adj}(M) = 0[/imath]. When [imath]n = 2[/imath], we have [imath]\text{adj}(M) = 0 \Rightarrow M = 0[/imath], so this settles the [imath]2 \times 2[/imath] case. Second observation: recall that by Cayley-Hamilton [imath]p(M) = 0[/imath] where [imath]p[/imath] is the characteristic polynomial of [imath]M[/imath]. Write this as [imath]M^k q(M) = 0[/imath] where [imath]q[/imath] has nonzero constant term. If [imath]q(M) \neq 0[/imath], then there exists some [imath]v \in R^n[/imath] such that [imath]w = q(M) v \neq 0[/imath], hence [imath]M^k w = 0[/imath] and one of the vectors [imath]w, Mw, M^2 w,\dots, M^{k-1} w[/imath] necessarily lies in the kernel of [imath]M[/imath]. Thus if [imath]M[/imath] is a counterexample we must have [imath]q(M) = 0[/imath] where [imath]q[/imath] has nonzero constant term. Now for every prime ideal [imath]P[/imath] of [imath]R[/imath], consider the induced action of [imath]M[/imath] on [imath]F^n[/imath], where [imath]F = \overline{ \text{Frac}(R/P) }[/imath]. Then [imath]q(\lambda) = 0[/imath] for every eigenvalue [imath]\lambda[/imath] of [imath]M[/imath]. Since [imath]\det(M) = 0[/imath], one of these eigenvalues over [imath]F[/imath] is [imath]0[/imath], hence it follows that [imath]q(0) \in P[/imath]. Since this is true for all prime ideals, [imath]q(0)[/imath] lies in the intersection of all the prime ideals of [imath]R[/imath], hence If [imath]M[/imath] is a counterexample and [imath]q[/imath] is defined as above, then [imath]q(0)[/imath] is nilpotent. This settles the question for reduced rings. Now, [imath]\text{det}(M) = 0[/imath] implies that the constant term of [imath]p[/imath] is equal to zero, and [imath]\text{adj}(M) = 0[/imath] implies that the linear term of [imath]p[/imath] is equal to zero. It follows that if [imath]M[/imath] is a counterexample, then [imath]M^2 \mid p(M)[/imath]. When [imath]n = 3[/imath], this implies that [imath]q(M) = M - \lambda[/imath] where [imath]\lambda[/imath] is nilpotent, so [imath]M[/imath] is nilpotent and thus must have nontrivial kernel. So this settles the [imath]3 \times 3[/imath] case. Third observation: if [imath]M[/imath] is a counterexample, then it is a counterexample over the subring of [imath]R[/imath] generated by the entries of [imath]M[/imath], so We may assume WLOG that [imath]R[/imath] is finitely-generated over [imath]\mathbb{Z}[/imath]. |
3054937 | Galois correspondence of the extension [imath]\mathbb{Q}(\zeta_{28})/\mathbb{Q}[/imath]
How to determine the Galois correspondence of the extension [imath]\mathbb{Q}(\zeta_{28})/\mathbb{Q}[/imath], where [imath]\zeta_{28}[/imath] is a primitive 28-th root of unity? It seems that in link only [imath]\mathbb{Q}(\zeta_n)/\mathbb{Q}[/imath] for [imath]n=p^k[/imath] is discussed. | 132169 | A complete picture of the lattice of subfields for a cyclotomic extension over [imath]\mathbb{Q}[/imath].
Is there a good general purpose algorithm (batch of theorems) allowing one to determine the intermediate fields between [imath]\mathbb{Q}(\zeta)[/imath] and [imath]\mathbb{Q}[/imath], where [imath]\zeta[/imath] is some primitive root of unity? Let [imath]p[/imath] be a prime. Consider the case where [imath]\zeta=\zeta_{p}[/imath] is a primitive [imath]p[/imath]-th root of unity. Then the Galois extension is cyclic of order [imath]p-1[/imath] and [imath]1,\zeta,\dots,\zeta^{p-1}[/imath] is a [imath]\mathbb{Q}[/imath]-basis for the extension. In this case for any subgroup [imath]H[/imath] of [imath]G=\mathbb{Z}/(p-1)[/imath], by considering the sum [imath]\alpha_H=\sum_{\sigma\in H}\sigma\zeta,[/imath] we can observe that [imath]\alpha_H[/imath] lies in the fixed field for [imath]H[/imath], and any automorphism [imath]\tau[/imath] not in [imath]H[/imath] (note automorphisms are identified with subgroups of [imath]\mathbb{Z}/(p-1)[/imath] in the natural way), [imath]\tau[/imath] does not fix [imath]\alpha_H[/imath]. Therefore we can conclude that [imath]\mathbb{Q}(\alpha_H)[/imath] is the fixed field of [imath]H[/imath]. In this way we can get all of the intermediate fields of [imath]\zeta_p[/imath] for all odd primes [imath]p[/imath]. We also have a theorem that says if we have [imath]n=p^sq^t[/imath], then [imath]\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})\simeq \text{Gal}(\mathbb{Q}(\zeta_{p^s})/\mathbb{Q})\times\text{Gal}(\mathbb{Q}(\zeta_{q^t})/\mathbb{Q}).[/imath] So what I have yet to understand is How can one generally find the intermediate fields between [imath]\mathbb{Q}(\zeta_{p^s})[/imath] and [imath]\mathbb{Q}[/imath] for [imath]s\ge 1[/imath]? I would like to also understand the case where [imath]p=2, s>1[/imath] though this might turn out to be a separate case. EDIT: Even the case [imath]n=pq[/imath] is a little murky to me. Even given the isomorphism given by the Chinese Remainder Theorem I don't see a priori how to get all the "product" subfields. My idea is that you can consider the separate subfields under [imath]\text{Gal}(\mathbb{Q}(\zeta_{p})/\mathbb{Q})[/imath] and [imath]\text{Gal}(\mathbb{Q}(\zeta_{q})/\mathbb{Q})[/imath] separately and then consider the pairwise product of the generators of various subfields to see if you get anything new, but my idea is too inchoate. |
2948951 | What exactly does it mean to say "[imath]dv=4\pi r^2\,dr[/imath] can be thought of as the spherical volume element between [imath]r[/imath] and [imath]r+dr[/imath]"?
In my textbooks and lectures (I'm a second year physics student) I often come across statements such as [imath]``dV=4\pi r^2dr\text{ can be thought of as the spherical volume element between }r\text{ and }r+dr."[/imath] Is this meant only symbolically, or is it a precise statement? For instance, I get [imath]\begin{align}\int_r^{r+dr}4\pi r^2dr&=\frac{4}{3}\pi\left[r^3\right]_r^{r+dr}\\&=\frac{4}{3}\pi(r^3+dr^3+3rdr^2+3r^2dr-r^3)\\&=\frac{4}{3}\pi(dr^3+3rdr^2+3r^2dr)\\&=4\pi r^2dr+\text{increasingly smaller terms.}\end{align}[/imath] Moreover, a Taylor expansion of the spherical volume function at a point [imath]r'[/imath] around a general point [imath]r[/imath] gives [imath]V(r')\approx\frac{4}{3}\pi r^3+4\pi r^2(r'-r)+\cdots.[/imath] Since the volume of a spherical shell between [imath]r[/imath] and [imath]r+dr[/imath] is generally (?) the volume of a sphere with radius [imath]r+dr[/imath] minus the volume of a sphere with radius [imath]r[/imath], we get [imath]\text{volume between }r\text{ and }r+dr\approx\frac{4}{3}\pi r^3+4\pi r^2dr-\frac{4}{3}r^3\approx4\pi r^2dr.[/imath] This leads me to think that what [imath]4\pi r^2dr[/imath] really is, is a good approximation (in fact, a first order Taylor expansion) of the volume between [imath]r[/imath] and [imath]r+dr[/imath], which can furthermore be assumed to be perfectly valid in the limit [imath]dr\to 0[/imath]. Maybe ... I'm just being pedantic? | 1828885 | Volume of spherical shell with [imath]dr[/imath] thickness
Let's consider two spheres in the [imath](x,y,z)[/imath] 3D-space, both centered in the origin: the inner with radius [imath]r[/imath] and the outer with radius [imath]r + dr[/imath]. To compute the volume of the spherical shell between their two surfaces, one should simply proceed as follows: [imath]dV = \frac{4}{3} \pi (r + dr)^3 - \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (r^3 + 3 r^2 dr + 3 r dr^2 + dr^3 - r^3)[/imath] [imath]dV = \frac{4}{3} \pi (3 r^2 dr + 3 r dr^2 + dr^3)[/imath] but usually the last two terms between the brackets are neglected. I know [imath]dr[/imath] is infinitesimal, but can infinitesimals (raised to [imath]n[/imath]-th power, [imath]n>1[/imath]) be neglected and still obtain an exact result? How can this be justified? |
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