qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
2482248 | Show that a metric space [imath]M[/imath] is compact if and only if all infinite open cover of [imath]M[/imath] have a proper subcover.
Let [imath](M,d)[/imath] be a metric space. Show that [imath]M[/imath] is compact if and only if all infinite open cover of [imath]M[/imath] have a proper subcover. We remember: If [imath]\mathcal{U}[/imath] is a cover of [imath]M[/imath], then a proper subcover is a proper subset [imath]\widehat{\mathcal{U}}[/imath] of [imath]\mathcal{U}[/imath] such that it is cover of [imath]M[/imath]. Remark: The diection [imath](\Rightarrow)[/imath] is trivial. The problem is the direction [imath](\Leftarrow)[/imath], I think hypotheses are missing, but I have not been able to find a counterexample or prove it. This exercise I found in a book that for me is not very reliable since I have found other exercises with errors. | 83386 | Some equivalent formulations of compactness of a metric space
Let [imath](X, d)[/imath] be a metric space. I have to prove the following statements are equivalent. [imath](X,d)[/imath] is complete (i.e., every Cauchy sequence is convergent) and totally bounded (i.e., for every [imath]\epsilon>0[/imath], [imath](X,d)[/imath] has a finite [imath]\epsilon[/imath]-net). [imath](X,d)[/imath] is sequentially compact (i.e., every sequence has a convergent subsequence) . [imath](X,d)[/imath] is Bolzano - Weierstrass compact (i.e., every infinite subset of [imath]X[/imath] has a limit point). [imath](X,d)[/imath] is compact (i.e., every open cover of [imath]X[/imath] has a finite subcover). Every infinite open cover of [imath](X,d)[/imath] has a proper subcover. I have proved the following: 1[imath]\Longleftrightarrow[/imath]2, 2[imath]\Longleftrightarrow[/imath]3, 4[imath]\implies[/imath]3, 2[imath]\implies[/imath]4, 4[imath]\implies[/imath]5 and 5[imath]\implies[/imath]3. Although they are enough to prove the equivalence of all five statements, I tried to prove it from the other possible directions. But I'm stuck with the proof of 3[imath]\implies[/imath]4 and 5[imath]\implies[/imath]4. Does anybody know how to prove them without using any of the above statements? Please help. |
2483182 | Does there exist non-trivial linear functional on [imath](C[0,1], ||\cdot||_2)[/imath] which doesn't come from the inner product on [imath]L^2[0,1][/imath]?
Let [imath]X=C[0,1][/imath] i.e. the set of continuous functions on [imath][0,1][/imath] with [imath]L^2[/imath] norm. Then there exists an [imath]F\in X^*[/imath] i.e. dual of [imath]X[/imath] such that there exists no [imath]g \in X[/imath] so that [imath] F(f) = \langle f,g \rangle[/imath] holds for all [imath]f \in X[/imath]. My effort: Since [imath](X,|| \cdot ||_{2})[/imath] is not complete. Hence it is not a Hailbert space. We can't apply the Riesz Representation Theorem. Any help in this regard would be highly appreciated! Thanks in advance! | 678656 | Necessity of completeness of the inner product space in Riesz representation theorem
I wanted to find a counter example to show that the completeness of the inner product space is necessary in Riesz representation theorem. Please give an example of a bounded linear functional [imath]T[/imath] on an incomplete inner product space [imath]X[/imath] which do not have any inner product representation i.e. there does not exist any [imath]z[/imath] in [imath]X[/imath] s.t. [imath]T(x)= \langle ,z\rangle[/imath] for all [imath]x[/imath] in [imath]X[/imath]. |
2475049 | Prime number theorem [imath]n^{\rm th}[/imath] prime bounds
In this Math SE post, the accepted answer begins with: "Because of the Prime Number Theorem, ultimately we have: [imath].9\,n\,\log n \leq p_n \leq 1.1 \,n\,\log n.[/imath] where [imath]p_n[/imath] is the [imath]n[/imath]th prime". Where do the bounds of [imath].9[/imath] and [imath]1.1[/imath] come from? Wikipedia states that: [imath] 0.921<\frac{\pi(n)}{n\log(n)}<1.018, [/imath] for [imath]10<n<10^{25}[/imath], but other than experimental evidence, how can one be sure that these are really the correct bounds? | 434229 | Is the [imath]n[/imath]-th prime smaller than [imath]n(\log n + \log\log n-1+\frac{\log\log n}{\log n})[/imath]?
Let [imath]p_n[/imath] be the [imath]n[/imath]-th prime. Wikipedia gives the following known bounds on [imath]p_n/n[/imath] when [imath]n\geq6[/imath]: [imath] \log n+\log\log n-1 \leq \frac{p_n}{n} \leq \log n+\log\log n. [/imath] If I take the first few terms in the asymptotic expansion for [imath]p_n/n[/imath], like so: [imath] \frac{p_n}{n} = \log n+\log\log n-1+\frac{\log\log n}{\log n} - \frac{2}{\log n} + O\left(\frac{(\log\log n)^2}{\log^2 n}\right), [/imath] it follows that [imath] \frac{p_n}{n} < \log n+\log\log n-1+\frac{\log\log n}{\log n} + \frac{c}{\log n}, [/imath] for [imath]c>-2[/imath] and large enough [imath]n[/imath]. For at least those [imath]n[/imath] that I've checked however ([imath]p_n\leq 10^{11}[/imath]), I find that this inequality holds with [imath]c=0[/imath] and [imath]p_n>347[/imath], and also with [imath]c=-1[/imath] and [imath]p_n > 5393[/imath]. Is this actually correct? Is there a sharper inequality? |
2482708 | If [imath]U \subset V[/imath] are vectors spaces, then [imath]V = U \oplus X[/imath] for some vector space
Is it true that if [imath]U \subset V[/imath] are vectors spaces, one can always find a vector space [imath]X \subset V [/imath] so that [imath]V = U \oplus X[/imath]? I know this it true when [imath]V[/imath] is finite dimensional, but is it true also in the infinite dimension case? Thanks. | 383764 | Can infinite-dimensional vector spaces be decomposed into direct sum of its subspaces?
I'm reading Axler "Linear agebra done right" and in Chapter 1 he discusses subspaces and direct sum. My question is, are there subspaces of the infinite-dimensional vector spaces, e.g. a functional Banach space with sup norm [imath]V[/imath] that directly sum to the entire space [imath]V = U_1\oplus U_2[/imath] (except for trivial [imath]U_1=\{0\}[/imath], [imath]U_2 = V[/imath]). |
2482959 | Simple extension [imath]K( \alpha)[/imath] over [imath]K[/imath] of degree 2, show [imath] \alpha^2 \in K[/imath]
I am trying to show that if [imath]K( \alpha)[/imath] over [imath]K[/imath] is an extension of degree 2, then [imath] \alpha^2 \in K[/imath] if the characteristic of [imath]K[/imath] is not [imath]2[/imath]. Let's suppose the minimal polynomial of [imath] \alpha[/imath] is [imath]f(t) = t^2 +bt + c[/imath]. I am able to use the quadratic formula, and I can show that both roots of [imath]f[/imath] are in this extension (i.e. [imath]K (\alpha) = K( \alpha, \beta)[/imath] where [imath]\alpha, \beta[/imath] are the roots of [imath]f[/imath]), but I don't see how this helps. The degree of the extension only tells me immediately [imath]1, \alpha, \alpha^2[/imath] are not linearly independent, but that only immediately says they satisfy a quadratic, not that [imath]\alpha \in K[/imath]. How can I proceed? | 2474191 | Extension of degree 2 with characteristic not 2
I am told that [imath]K \leq L[/imath] is a field extension of degree 2 and that [imath]K[/imath] has characteristic not 2. I am then asked to show that [imath]\exists \: \alpha \in L[/imath] such that [imath]L = K(\alpha), \; \alpha^2\in K [/imath] So far I have that: [imath]\exists \; \alpha_1, \alpha_2 \in L[/imath] such that [imath]L = K( \alpha_1, \alpha_2)[/imath] [imath]\Rightarrow |L:K| = |L:K(\alpha_1)| |K(\alpha):K| [/imath] so [imath]|K(\alpha):K| = 1[/imath] or [imath]2[/imath] [imath]|K(\alpha_1):K| = 2 \Rightarrow L = K(\alpha_1)[/imath] and [imath]deg(f_{\alpha_1}(t)) = 2 [/imath] where [imath]f_{\alpha_1}(t)[/imath] is the minimal polynomial of [imath]\alpha_1[/imath] over [imath]K[/imath] So [imath]f_{\alpha_1}(t) = t^2 + at + b[/imath] for some [imath]a,b \in K[/imath] Also, [imath]|K(\alpha_1):K|=1 \Rightarrow K(\alpha_1) = K[/imath] and [imath]L = K(\alpha_2)[/imath] which means [imath]degf_{\alpha_2}(t) = 2[/imath] where [imath]f_{\alpha_2}(t)[/imath] is the minimal polynomial of [imath]\alpha_2[/imath] over [imath]K[/imath]. So [imath]f_{\alpha_2}(t) = t^2 + at + b[/imath] for some [imath]a,b \in K[/imath] Given that we consider the same polynomial each time, relabel to consider: [imath]f_{\alpha}(t) = t^2 + at+ b [/imath] for some [imath]a,b \in K[/imath] We have already demonstrated that [imath]L=K(\alpha)[/imath] for some [imath]\alpha \in L[/imath] so now we much show that [imath]\alpha^2 \in K[/imath] [imath]\alpha^2 \in K \Leftrightarrow a = 0_K[/imath] I now have no idea how to show that this coefficient of [imath]f_{\alpha}(t)[/imath] is necessarily [imath]0[/imath]. I know that [imath]f_{\alpha}(t)[/imath] has two roots in [imath]L[/imath] and that because [imath]K[/imath] and thus [imath]L[/imath] has characteristic not 2, they cannot be the same root, but I don't know how to then conclude that they must sum to [imath]0[/imath] Any help would be greatly appreciated. |
2484180 | Does [imath]\mathop{\LARGEΣ}_{k = 1}^{n}k^α = \left(\mathop{\LARGEΣ}_{k = 1}^{n}k\right)^β[/imath] for which [imath]α \neq 3, \ β \neq 2[/imath] and [imath]α \neq β[/imath]?
I figured out that: [imath]\mathop{\LARGEΣ}_{k = 1}^{n}k^3 = \bigg(\mathop{\LARGEΣ}_{k = 1}^{n}k\bigg)^2[/imath] Are there any other solutions for the following equation, for which [imath]α \neq 3, \ β \neq 2[/imath] and [imath]α \neq β[/imath]? [imath]\mathop{\LARGEΣ}_{k = 1}^{n}k^α = \bigg(\mathop{\LARGEΣ}_{k = 1}^{n}k\bigg)^β[/imath] P.S. The Sigmas in the equations look different to how its font usually looks like on the MSE and/or MO, (it's typical look appearing like this [imath]\rightarrow \sum[/imath]) because I am using \mathop{\LARGEΣ}_{k = 1}^{n} just to test how it looks like, and because of its boldness, I don't mind its appearance. Edit: There is another answer that solves my problem here, mentioned by mathlove. | 1869843 | Are there [imath]a,b \in \mathbb{N}[/imath] that [imath]{(\sum_{k=1}^n k)}^a = \sum_{k=1}^n k^b [/imath] beside [imath]2,3[/imath]
We know that: [imath]\left(\sum_{k=1}^n k\right)^2 = \sum_{k=1}^n k^3 [/imath] My question is there other examples that satisfies: [imath]\left(\sum_{k=1}^n k\right)^a = \sum_{k=1}^n k^b [/imath] |
2484186 | Prove that [imath] \ \forall x \exists y \ P(x,y) \ [/imath] and [imath] \ \exists y \forall x \ P(x,y) \ [/imath] are not equivalent
Prove that [imath] \ \forall x \exists y \ P(x,y) \ [/imath] and [imath] \ \exists y \forall x \ P(x,y) \ [/imath] are not logically not equivalent in the domain [imath] \ \{-1,0,1 \} \ [/imath]. Answer: Let, [imath] x=\{-1,0,1 \} \\ y=\{-1,0,1 \} [/imath] Let [imath] P(x,y) \ [/imath] be the property such that [imath] x +y \ [/imath] is even number. Then [imath] \forall x \exists y \ P(x,y) \ [/imath] is true. Because, If [imath] x=-1 \ [/imath] , then take [imath] y=1 \ [/imath] such that [imath] x + y=0 \ [/imath] is even If [imath] x=0 \ [/imath] , then take [imath] y=0 \ [/imath] , such that [imath] x + y=0 \ [/imath] is even If [imath] x=1 \ [/imath] , then take [imath] y=1 \ [/imath] such that [imath] x + y =2 \ [/imath] is even But [imath] \ \exists y \forall x \ P(x,y) \ [/imath] is [imath] False [/imath] Am I right ? I need help. | 2483777 | Show that the logic [imath] \ \forall x \exists y \ P(x,y) \ [/imath] is not logically equivalent to [imath] \ \exists x \forall y \ P(x,y) \ [/imath]
Show that the logic [imath] \ \forall x \exists y \ P(x,y) \ [/imath] is not logically equivalent to [imath] \ \exists x \forall y \ P(x,y) \ [/imath] in the domain [imath] \ \{-1,0,1 \} \ [/imath] by giving a counter example . Answer: Let [imath] x=\{0,1 \} \\ y=\{-1,0 ,1 \} \ [/imath] Let [imath] P(x,y) \ [/imath] be the property such that [imath] \ x \geq y \ [/imath] Then , [imath] \forall x \exists y \ P(x,y) \ [/imath] is true. But, [imath] \ \exists x \forall y \ P(x,y) \ [/imath] is False. Because for [imath] x=0 \ [/imath] , we have [imath] \ 0 \ngeqslant y=1 \ [/imath]. I need confirmation of my work. |
2483841 | Self-intersection of exceptional sphere
Let [imath]M[/imath] be a smooth close [imath]4-[/imath]dimensional manifold and let [imath]p\in M[/imath] be a point. Denote by [imath]\hat{M}[/imath] the blow-up of [imath]M[/imath] at [imath]p[/imath] and let [imath]E \cong \mathbb{C}P^{1}[/imath] denote the exceptional sphere which is embedded. How can one show that the self-intersection number of [imath]E[/imath] is [imath]-1[/imath], i.e. [imath]E \cdot E = -1[/imath]? | 1315703 | Formula for top self intersection of exceptional divisor
Let [imath]X[/imath] be a projective variety over [imath]\mathbf{C}[/imath] of dimension [imath]n[/imath]. Let [imath]\pi: Y \to X[/imath] be the blow-up of a smooth point [imath]x \in X[/imath]. Is there a nice formula for the intersection number [imath]E^n[/imath]? |
2000714 | If [imath]a_0,....a_n[/imath] are the set of residues modulo [imath]n[/imath] and [imath]\gcd(a,n)=1[/imath], prove that [imath]aa_0,aa_1,....,aa_n[/imath] is also a complete set of residue modulo [imath]n[/imath].
Problem: If [imath]a_0,a_1,a_2,a_3,....a_n[/imath] are the set of residues modulo [imath]n[/imath] and [imath]\gcd(a,n)=1[/imath], prove that [imath]aa_0,aa_1,aa_2,aa_3,....,aa_n[/imath] is also a complete set of residue modulo [imath]n[/imath]. I have trouble in identifying what exactly should I show in order to prove this statement. In my opinion, the condition for having a complete set of residues is: [imath]aa_i\neq aa_j[/imath] for any [imath]0\leq i,j\leq n.[/imath] [imath]aa_i[/imath] is not congruent to [imath]aa_j[/imath] for any [imath]0\leq i,j\leq n.[/imath] It is trivial to prove [imath]1[/imath] and [imath]2[/imath], but I would like to know whether both the conditions are sufficient or not. | 1460678 | Number Theory: Complete set of residues modulo [imath]n[/imath]
I have this problem assigned for homework and I'm struggling with the proof of it: If [imath]a_1,a_2,\dotsc,a_n[/imath] is a complete set of residues modulo [imath]n[/imath] and [imath]\gcd(a,n)=1[/imath], prove that [imath]aa_1,aa_2,\dotsc,aa_n[/imath] is also a complete set of residues modulo [imath]n[/imath]. (Hint: It suffices to show that the numbers in question are incongruent modulo [imath]n[/imath].) I'm in elementary number theory so I'm not allowed to use an high-level theorems to prove this, I pretty much have to use the basics of modulo. Thanks! |
1579952 | [imath]f[/imath] holomorphic on [imath]D\setminus \{0\}[/imath] and takes no values in [imath](-\infty,0],[/imath] then [imath]0[/imath] removable
If [imath]f[/imath] is holomorphic on [imath]D\setminus \{0\}[/imath] and takes no values in [imath](-\infty,0][/imath] then [imath]0[/imath] is a removable singularity. I thought to prove this by elimination, but I can't really tell anything about the behavior of [imath]f[/imath] around [imath]0[/imath]. How can one translate the information about the definition of [imath]f[/imath] in semi-open interval [imath](-\infty,0][/imath]. | 1731448 | Show [imath]f[/imath] has a removable singularity or a pole at [imath]0[/imath]
Suppose [imath]f[/imath] is analytic in [imath]\{0 < |z| < R\}[/imath] and that the range of [imath]f[/imath] does not include the negative real axis. Prove that [imath]f[/imath] has either a removable singularity or a pole at [imath]0[/imath]. (Don’t use the Picard theorem). Hint: In this case we can define [imath]\sqrt{f(z)}[/imath] (Why?). Look at the range of [imath]\sqrt{f(z)}[/imath] and notice it omits a disc. Use the Casorati–Weierstrass theorem. I feel like I'm being given most of the problem via this hint, but I'm still not seeing what I should be doing here. Why does omitting the negative real axis make a difference to whether I can define [imath]\sqrt{f(z)}[/imath]? And why would the range of this function omit a disk? I know that once I can see a disk is omitted from the range, this means that [imath]\sqrt{f(z)}[/imath] cannot have an essential singularity at [imath]0[/imath] because this would contradict the Casorati-Weierstrass theorem, I just don't see why this is true to begin with. I don't want somebody to just do the problem for me, I just want help seeing whatever it is that I'm missing here. |
2484893 | Finding the limit of a sequence.
[imath](x_n)[/imath] is a positive sequence and [imath]\lim_{n\to\infty}\frac{x_n}{n}=0\hspace{6mm}\limsup_{n\to\infty} \frac{x_1+x_2+\ldots+x_n}{n}\in\mathbb{R}[/imath] and we need to find [imath] \lim_{n\to\infty} \frac{x^2_1+x^2_2+\ldots+x^2_n}{n^2}[/imath] I have no idea how to approach this task. I was thinking that I could maybe [imath] \frac{x^2_1+x^2_2+\ldots+x^2_n}{n^2}=\frac{\frac{x^2_1}{2}+\frac{x^2}{2}+\ldots+\frac{x^2_n}{2}}{n^2}+\frac{\frac{x^2_1}{2}+\frac{x^2}{2}+\ldots+\frac{x^2_n}{2}}{n^2}=\frac{\frac{x^2_1}{3}+\frac{x^2}{3}+\ldots+\frac{x^2_n}{3}}{n^2}+\ldots[/imath] But that doesn't lead anywhere. How should I approach it? What does the second limit ([imath]\limsup\ldots\in\mathbb{R}[/imath]) mean and how should I use it? | 858134 | Prove that [imath]\lim_{n\rightarrow\infty}\frac{x_1^2+x_2^2+\cdots+x_n^2}{n^2}=0[/imath]
Let's [imath]x_n\ge0[/imath] and [imath]\overline{\lim}_{n\rightarrow\infty}\frac{x_1+x_2+\cdots+x_n}{n}\lt+\infty,~~\lim_{n\rightarrow\infty}\frac{x_n}{n}=0.[/imath]Prove that [imath]\lim_{n\rightarrow\infty}\frac{x_1^2+x_2^2+\cdots+x_n^2}{n^2}=0.[/imath] I have no idea how I can start the proof. Thanks! |
2485214 | Solving only using modular arithmetic [imath]5x+7y=1234[/imath]
Someone asked this on mse: Suppose x, y are positive integers. What is the number of all solutions to [imath] x,y \in \mathbb{N} : \, 5x+7y=1234 [/imath] Here is my attempt, I don't know if it is correct: \begin{align} 1)5x+7y=1234 \pmod {5}\\ 2)5x+7y=1234 \pmod {7}\\ \end{align} Then: \begin{align} 1)y=2 \pmod {5}\\ 2)x=6 \pmod {7}\\ \end{align} Which means that: \begin{align} 1)x=7k+6\\ 2)y=5n+2 \end{align} Plugging that in the equation: [imath]5x+7y=1234[/imath] [imath]k+n=34 \, k,n \in \mathbb{N}[/imath] It means there are 35 couples of (x,y) solutions... Is that correct ? Are there other ways for solving this using modular arithmetics ? Thanks for your help and suggestions. | 2485108 | How many pairs of positive integers are solutions to the equation [imath]5x+7y=1234[/imath]
How many pairs of positive integers are solutions to the equation [imath]5x+7y=1234[/imath] My idea was that since [imath]5[/imath] and [imath]7[/imath] are both odd and [imath]1234[/imath] is even then that forces [imath]x[/imath] and [imath]y[/imath] to both be even or both be odd. Case [imath]1[/imath] if both [imath]x[/imath] and [imath]y[/imath] are even then [imath]5x=1234-7y[/imath] since [imath]x[/imath] is even then [imath]2\leq x \leq 246[/imath] so in case [imath]1[/imath] there are [imath]123[/imath] pairs of solutions. Case [imath]2[/imath] if both [imath]x[/imath] and [imath]y[/imath] are odd then [imath]5x=1234-7y[/imath] since [imath]x[/imath] is odd then [imath]1\leq x \leq 245[/imath] which again yields [imath]123[/imath] pairs of solutions for a total of [imath]246[/imath] solutions. |
2460209 | Proof that a circle can be circumscribed around a quadrilateral
I'm solving a problem you can see above: [imath]PQRS[/imath] is a trapezoid, [imath]HM[/imath] and [imath]KN[/imath] are legs bisectors. I'm supposed to prove that [imath]\alpha = \beta[/imath], which as you look closely can be reduced to proving [imath]\measuredangle SPM = \measuredangle NQR[/imath] (equivalently [imath]\measuredangle PSM = \measuredangle NRQ[/imath]). Hypothesis These angles somehow resemble inscribed angles on arc [imath]NM[/imath]. A circle could be circumsribed around [imath]PQMN[/imath] or [imath]NMRS[/imath]. But I'm struggling with proving that one can actually circumsribe a circle around either of these. I'd appreciate some help from you. Here you can find a brilliant drawing by User Raffaele who's done a great deal of work on the problem. However, a rigorous solution has not been provided by anyone. | 2459118 | Leg bisectors in a trapezoid and angle equality
Let [imath]PQRS[/imath] be a trapezoid. The bisectors of legs [imath]PS[/imath] and [imath]QR[/imath] intersect the opposite legs at [imath]M[/imath] and [imath]N[/imath] so that triangles [imath]PMS[/imath] and [imath]QNR[/imath] are formed. Prove that [imath]\measuredangle PMS= \measuredangle QNR[/imath]. The question may seem easy but I fail to move on, playing with angles didn't help. I'd appreciate some hint from somebody with an eye for geometry and triangles. |
2485785 | definite integral [imath]\int_{0}^{1}\sqrt[3]{x\log\frac 1x} [/imath]
Please solve: [imath]\int_{0}^{1}\sqrt[3]{x\log\frac 1x} [/imath] I have tried substituting but every time it is getting more complicated. Please help. | 2396605 | Computing: [imath]\int_0^1 \sqrt[3]{x\log{\frac{1}x}} \, \mathrm dx[/imath]
How can i integrate following definite integral? I tried integrating by parts which obviously did not work. [imath]I=\int_0^1 \sqrt[3]{x\log{\frac{1}x}} \, \mathrm dx[/imath] |
2482497 | Winding number is an integer proof.
Can anybody give me some link which proves that winding number is an integer in complex analysis. In most of the links I found they approached in assuming that [imath]s \mapsto \int_{0}^{s} \frac {\gamma'(t)} {\gamma (t)-a}\ dt[/imath] is continuous on [imath][0,1][/imath] and differentiable on [imath][0,1][/imath] except possibly at junction points where [imath]\gamma[/imath] is closed contour with parametric interval [imath][0,1][/imath] and [imath]a \notin \{\gamma \}[/imath]. I don't understand this thing clearly. So I want a proof which has some clear discussion. Please help me by giving it if there is any. Thank you in advance. | 1197670 | Winding number (demonstration)
How could I explain mathematically, that the winding number of a closed curve [imath]\gamma[/imath] around [imath]a[/imath] ([imath]a \notin \gamma[/imath]) gives always an integer value. [imath] W(\gamma,a)=\frac{1}{2\pi i} \int_{\gamma} \frac{dw}{w-a} [/imath] where [imath]W(\gamma,a)\in \mathbb{Z}[/imath] |
2485693 | The inverse matrix of [imath]a_{ij}=i^{j}[/imath]
This is my first question, so I’m sorry if I made mistakes. [imath]\left(\begin{array}{ccccc}1&1&1&\cdots&1\\2&2^2&2^3&\cdots&2^n\\3&3^2&3^3&\cdots&3^n\\\vdots&\vdots&\vdots&\ddots&\vdots\\n&n^2&n^3&\cdots&n^n\end{array}\right)[/imath] In order to solve the above inverse matrix, I tried to solve the below liner simultaneous equations in n-th unknowns. [imath]kx_1+k^2x_2+\cdots+k^nx_n=1+2^{n-1}+\cdots+k^{n-1}[/imath] [imath]k=1,2,\cdots,n[/imath] Inductively, I know the solution of the liner simultaneous equations. [imath]x_k=\frac{B_{n-k}}{n-k} {n-1 \choose n-k-1} (1≦k<n-1)[/imath] [imath]x_{n-1}=\frac{1}{2}[/imath] [imath]x_n=\frac{1}{n}[/imath] But even though I solved the solutions of the equations, I have no ideas how to use it. I thought Vandermonde’s déterminant may be effective, but I have no ideas how to use it. | 698254 | Proof of Vandermonde Matrix Inverse Formula
I'm working through Exercise 40 from section 1.2.3 of Knuth's The Art of Computer Programming volume 1, but am finding myself unable to produce a rigorous proof, and the one here is suspect and not quite clear enough (for me, at least) in some of the steps; in particular, how to "[identify] the [imath]k[/imath]th order coefficient in [the] two polynomials." The problem is this: given a Vandermonde matrix [imath][x_j^i]_n[/imath], show that the inverse is given by [imath] [b_{ij}]_n = \left[ \frac{ \sum_{\substack{1 \leq k_1 < \dotsc < k_{n-j} \leq n\\k_1,\dotsc,k_{n-j} \neq i}} (-1)^{j-1} x_{k_1} \dotsc x_{k_{n-j}} }{ x_i \prod_{\substack{1 \leq k \leq n\\k \neq i}} (x_k - x_i) } \right]_n\text{.} [/imath] The author gives a hint by stating that the sum in the above numerator is just the coefficient of [imath]x^{j-1}[/imath] in the polynomial [imath](x_1-x)\dotsc(x_n-x)/(x_i-x)[/imath]; and he gives an intermediate result showing the explicit multiplication of the matrix and its inverse as [imath] \sum_{1 \leq t \leq n}b_{it}x_j^t = \frac{ x_j \prod_{\substack{1 \leq k \leq n\\k \neq i}} (x_k - x_j) }{ x_i \prod_{\substack{1 \leq k \leq n\\k \neq i}} (x_k - x_i) } = \delta_{ij}\text{.} [/imath] The only other hint as to the type of solution he was expecting is a reference to A. de Moivre's The Doctrine of Chances, 2nd edition, pp. 197-199, which deals with polynomial recurrence relations and difference products (available here). At the least, I was just hoping someone could either verify the proof is correct at proofwiki and possibly fill in exactly how one identifies the [imath]k[/imath]th order coefficient in the proof; or perhaps explain a proof strategy as to what steps to take where the intermediate result is obtained at some point before the final result. Thanks so much for any help. |
2484835 | Bilinear form as a product of two linear functionals
I have seen a few answers to this question in this site, one of them deals with a very special case, while the other one only one side of the problem is proved, so I wanted a better answer and asked this question. Let [imath]f[/imath] be a bilinear form on a finite dimensional vector space [imath]V[/imath] over some field [imath]\mathbb F[/imath]. I was trying to prove that [imath]f[/imath] can be expressed as a product of two linear functionals i.e. [imath]f(u,v) = L_{1}(u) L_{2}(v) \text{for} L_{1},L_{2} \in V^*[/imath] if and only if [imath]f[/imath] has rank [imath]1[/imath]. I just know the definitions. If someone suggests an answer it will be very much helpful. | 2308339 | Prove that bilinear form can be presented as product of linear forms if and only if it has rank one
Let [imath]f: V \rightarrow F[/imath] a bilinear form such that [imath]f \ne 0[/imath]. Prove that [imath]f[/imath] can be presented as a product of two linear forms [imath]f(X,Y) = \left(\sum\limits_i^nb_ix_i \right)\ast \left(\sum\limits_j^nc_jy_j\right)[/imath] if and only if [imath]rank(f)=1[/imath] I saw this question and it's answer: Prove that the bilinear form can be presented as a product of two linear forms But the answer seems to ignore the fact that [imath]rank(f)[/imath] must be [imath]1[/imath]. I don't know how to use that fact to solve this. Thanks in advance |
2486073 | Compute the trace and the determinant of (this) operator
Consider the linear operator of left multiplication of an [imath]n\times n[/imath] matrix [imath]A[/imath] on the space [imath]F^{n\times n}[/imath] of all [imath]n\times n[/imath] matrices. Compute the trace and the determinant of this operator. I know that I have to use the formula of characteristic polynomial to find trace and determinant, but I find hard to understand the question. The linear operator, [imath]T[/imath] is such that, [imath]T: F^{n\times n}\to F^{n\times n}[/imath] such that, [imath]T(X)=AX[/imath], where, [imath]X\in F^{n\times n}[/imath] Basis of [imath]F^{n\times n}[/imath] is [imath]B=\{ E_{ij}|1\leq i,j\leq n\}[/imath], [imath]E_{ij}[/imath] contain all [imath]0[/imath] except [imath](i,j)[/imath]-th element is [imath]1[/imath]. Now, [imath]T(E_{ij})=AE_{ij}[/imath] Then I need to find the coefficient of [imath]E_{ij}[/imath] (To make the operator matrix) But I do not understand how I get a general solution. Is their a easy way to solve this? Any help appreciated. | 339263 | Finding trace and determinant of linear operator
I've got the following question Consider the linear operator of left multiplication by an [imath]m \times m[/imath] matrix [imath]A[/imath] on the vector space of all [imath]m \times m[/imath] matrices. Determine the trace and determinant of this operator. I'm a bit stuck as to how to even begin, I know this is going to involve eigenvalues/vectors and that if [imath]\lambda_1, \lambda_2, ... ,\lambda_m[/imath] are the [imath]m[/imath] roots of the characteristic polynomial of an [imath]m \times m[/imath] matrix [imath]A[/imath], then: [imath]\det(A) = \lambda_1 ... \lambda_m[/imath] and [imath]\text{trace}(A) = \lambda_1 + ... + \lambda_m[/imath] But obviously not all [imath]m \times m[/imath] matrices have [imath]m[/imath] eigenvalues so I'm really stuck on this question. Thanks! |
2486263 | prove [imath]G=HN[/imath], if [imath]N \lhd G[/imath] with prime index, and [imath]H[/imath] is not contained in [imath]N[/imath]
Let [imath]G[/imath] be a group with the following properties G has a normal subgroup [imath]N[/imath] with [imath][G;N]=p[/imath] where [imath]p[/imath] is a prime number; G has a subgroup [imath]H[/imath] such that [imath]H[/imath] is not contained in [imath]N[/imath] Prove [imath]G=HN[/imath] attempt1 [imath]\frac{|G|}{|N|} =p[/imath] is a quotient group so [imath]n=p|N|[/imath] Theorem 8.7 hungerford [imath]p[/imath] is prime every group of order [imath]p[/imath] is cylcic and isomorphic to [imath]Z_p[/imath] call that [imath]<g_p> =\frac{G}{N} \cong Z_p[/imath] [imath]\frac{G}{N}=\{g_p, \dots , g_p^p\} [/imath] and [imath] \frac{G}{N} = g_1 N \cup \dots \cup g_p N= \cup ^{p}_{i=1} (g_p)^i N[/imath] lead 2 [imath]K \not \geq H [/imath] so [imath][G:HN]<p[/imath] and it divides [imath]p[/imath] so [imath] \frac{|G|}{|HN|}=1[/imath] and the order of [imath]|G|=|HN| [/imath] that means that [imath]HN=G[/imath] since they have the same number of elements need [imath]HN[/imath] to be normal?? anything wrong/corrections?? | 2468738 | Prove for any [imath]K \leq G[/imath], either [imath]K \leq N[/imath] or [imath]KN=G[/imath] and [imath][K:K \cap N]=p[/imath] where N is normal in G and [imath][G:N]=p[/imath]
The full question: suppose N is normal in G and [imath][G:N]=p[/imath] is a prime. Prove that for any [imath]K \leq G[/imath], either (1) [imath]K \leq N[/imath] or (2)[imath]KN=G[/imath] and [imath][K:K \cap N]=p[/imath]. Since [imath]N,K \leq G[/imath], it is of course possible that [imath]K \leq N[/imath] could happen. but for (2), we can show [imath]KN \subseteq G[/imath] but how can we show [imath]KN=G[/imath]? And what information does "[imath][G:N]=p[/imath]" give us? |
2486079 | Minimum number of cycles in a connected graph.
Please answer Want proof Let [imath]G[/imath] be a connected graph with [imath]m[/imath] edges and [imath]n[/imath] vertices. Prove that the minimum number of cycles in [imath]G[/imath] is [imath]m-n-1[/imath]. (Where one cycle is a path that starts that begins and ends at the same vertex.) | 225570 | Prove that the minimum number of cycles is [imath]m-n+1[/imath]
The question I have is: Prove that the minimum number of cycles is [imath]m-n+1[/imath] in a connected graph. Where one cycle is a path that starts that begins and ends at the same vertex. Where [imath]m[/imath] is edges and [imath]n[/imath] is the vertices. I have no idea where to start. A few hints would be appreciated. Please do not provide me with the answer. |
2486552 | Prove that the series [imath]\sum_{n =2}^{\infty} \frac{1}{(\ln n)^{\ln n}}[/imath] converges.
I am trying to study the convergence of the series: [imath]\sum_{n =2}^{\infty} \frac{1}{(\ln n)^{\ln n}}[/imath] First I have failed to use the D'Alembert test [imath] \lim_{n\to \infty} \frac{(\ln (n+1))^{\ln (n+1)}}{(\ln n)^{\ln n}} =1.[/imath] So I can Conclude anything from here. I have also tried the integral Comparaison test but it was not soo clear. | 1318737 | Convergence of [imath]\frac{1}{(\ln n)^{\ln n}}[/imath]
Could I have a hint for testing the convergence of the following series please? [imath]\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}[/imath] I am very appreciative for your help. |
2486628 | Proving [imath](a_n)[/imath] diverges
For a given sequence [imath](a_n)[/imath], where [imath]a_1>0[/imath] and for each natural number [imath]n\ge 1[/imath] : [imath]a_{n+1}= \dfrac{a_n^2 + 1}{a_n}[/imath] Prove that a sequence [imath](a_n)[/imath] diverges. Proof: (objectionable evidence) Lower barrier 0, prove. [imath]a_n>0[/imath] [imath]\forall n\in N[/imath]. [imath]N[/imath] is natural number and [imath]N>1[/imath]. [imath]n=1[/imath] [imath]a_{1+1}=\frac{a_1\cdot a_1+1}{a_1}=a_1+\frac{1}{a_1} >0[/imath] [imath]a_n>0[/imath] [imath]\Rightarrow a_{n+1}>0[/imath] : [imath]a_{n+1}=a_n+\frac{1}{a_n}>0[/imath], because [imath]a_n > 0[/imath] Monotonics: [imath]a_n\geq a_{n+1}[/imath] [imath]a_n-a_{n+1}=a_n-a_n-\frac{1}{a_n} < 0[/imath], because [imath]\frac{1}{a_n}>0[/imath] Sequence is growing. lim [imath]a_{n+1} = A[/imath] lim [imath]a_{n+1} =[/imath] lim [imath](a_n+\frac{1}{a_n}) =[/imath] lim [imath]a_n +[/imath] lim [imath]\frac{1}{a_n} = A + \frac{1}{A} = \frac{A\cdot A+1}{A}[/imath] .... Is this proof correct? Where are mistakes? The correct end of proof? | 213469 | Let [imath]a_n[/imath] be a sequence s.t [imath]a_1 > 0 \land a_{n+1} = a_n + \dfrac{1}{a_n}[/imath]. Prove that [imath]a_n[/imath] is increasing and tends to infinity
Let [imath]a_n[/imath] be a sequence s.t [imath]a_1 > 0 \land a_{n+1} = a_n + \frac{1}{a_n}[/imath]. Prove that [imath]a_n[/imath] is increasing and tends to infinity. Proof: Consider [imath]a_{n+1} - a_n[/imath]: [imath]a_{n+1} - a_n = a_n + \frac{1}{a_n} - a_n = \frac{1}{a_n}[/imath] This is greater than [imath]0[/imath]. Thus, [imath]a_n[/imath] is increasing. Now this is where I need some help. I would like to say that [imath]a_n[/imath] is unbounded and then conclude that monotone and unbounded implies tending to infinity. Maybe by contradiction? |
2486298 | Show that [imath]S_n[/imath] has elements of order [imath]p^t \Longleftrightarrow n \geq p ^t[/imath], being [imath]n, t[/imath] positive integers and [imath]p[/imath] a prime number.
I am having trouble with the following problem: Show that [imath]S_n[/imath] has elements of order [imath]p^t \Longleftrightarrow n \geq p ^t[/imath], being [imath]n, t[/imath] positive integers and [imath]p[/imath] a prime number. Here is what I thought: Let [imath]\sigma \in S_n[/imath] have order [imath]p^t[/imath]. We can write [imath]\sigma[/imath] as a product of disjoint cycles of lenght [imath]>1[/imath]: [imath] \sigma = \sigma_1 \dots \sigma_r [/imath] Then [imath]p^t = [/imath] lcm [imath](o(\sigma_1), \dots, o(\sigma_r))[/imath], being [imath]o(\sigma_i)[/imath] the order of [imath]\sigma_i[/imath]. Since [imath]p[/imath] is a prime and we are only considering nontrivial cycles, we must have [imath]o(\sigma_i) = p^{n_i}[/imath]. Now, since the maximum length in [imath]S_n[/imath] is [imath]n[/imath], we must have [imath] p^{n_1} + \dots + p^{n_r} \leq n [/imath] And here I am stuck. How can I relate [imath]p^t[/imath] to the above sum? Or is there another way? Thanks in advance. | 2475788 | Order of a element on a Symmetric Group
I'm studying and I can't solve this question. Let be [imath]p, t, n \in \mathbb{Z}^+, p[/imath] prime. Show that [imath]\exists \sigma \in S_n[/imath] of order [imath]p^t \Leftrightarrow p^t \leq n[/imath]. The only thing that I conclude is that if [imath]\sigma[/imath] has order [imath]p^t[/imath] then [imath]p^t[/imath] divides [imath]n![/imath]. |
2487222 | Analysis techinque in proving [imath]e^x>1+x[/imath]
To prove that [imath]e^x>1+x[/imath], for all [imath]x<0[/imath]. If I try to write [imath]e^x[/imath] as [imath]1+x+\frac{1}{2}x^2+\frac{e^{\xi}}{3!}x^3[/imath], for some [imath]\xi\in (x,0)[/imath]. Then I get [imath]e^x-(1+x)=\frac{1}{2}x^2+\frac{e^{\xi}}{3!}x^3[/imath]. How do I show that [imath]\frac{1}{2}x^2+\frac{e^{\xi}}{3!}x^3[/imath] is necessarily positive? I'd tried thinking for a long time, but can't find the way. Can't it be done by using Taylor to exactly order [imath]3[/imath]?? | 2383019 | Proof of [imath]1+x\leq e^x[/imath] for all x?
Does anyone provide proof of [imath]1+x\leq e^x[/imath] for all [imath]x[/imath]? What is the minimum [imath]a(>0)[/imath] such that [imath]1+x\leq a^x[/imath] for all x? |
2487716 | Addition and "subtraction" of polytopes
Suppose [imath]\mathcal{A}, \mathcal{B}\subset\mathbb{R}^2[/imath] are polytopes. Suppose also that I know that there exists a polytope [imath]\mathcal{X}[/imath] such that [imath]\mathcal{A}+\mathcal{X}=\mathcal{B}[/imath]. Is [imath]\mathcal{X}[/imath] defined uniquely? (How about uniquely "up to translation"?) This means that the idea of "subtraction", as the opposite of addition, makes sense. (I know that there is a subtraction defined for polytopes, but I am having some trouble connecting it to the above.) The solution to my question follows from the "duplication" link as if [imath]\mathcal{A}+\mathcal{X}=\mathcal{B}[/imath] and [imath]\mathcal{A}+\mathcal{Y}=\mathcal{B}[/imath] then [imath]\mathcal{A}+\mathcal{X}=\mathcal{A}+\mathcal{Y}[/imath], and hence [imath]\mathcal{X}=\mathcal{Y}[/imath] as required. | 182565 | Cancellation of addition on convex sets
I recently found a question about a property of the Minkowski sums. However the question was not properly answered (it used a projection argument which might not be true in a general Banach space). I was wondering whether the following (weaker) statement holds: Let [imath]X[/imath] be a Banach space and suppose [imath]A,B,C_0\subset X[/imath] are bounded, closed, convex and non-empty subset. Do we then have [imath]A+C_0=B+C_0\implies A=B?[/imath] |
2487693 | What seat will I be in x steps later?
I need help with the following question regarding modular arithmetic. "A seating plan consists of 7 chairs in a circle. If I am currently sat in chair 3, what chair will I be in [imath]3^{453}[/imath] steps later? Assume that for each step, you move one chair forward in the anticlockwise direction and the chairs are numbered from 1 to 7." Obviously, the number of steps is too large to be computed, so I assume there must be a trick to do this? Thanks! | 202382 | How to solve and see resolution of [imath]13^{53} \pmod 7[/imath] using Fermat little Theorem?
How to solve and see resolution of [imath]13^{53} \pmod 7[/imath] using Fermat little Theorem? Using Fermat's Little Theorem, I know it gives me 6 as an answer to this problem..., but why? How is the resolution? Thanks, |
1025961 | How can I find the answer to [imath]\varphi(2^{400} + 2)[/imath]?
How can I find the answer to [imath]\varphi(2^{400} + 2)[/imath] ? So far I have [imath]\varphi(2^{400}) = 2^{400} - 2^{399} = 2^{399}(2 - 1) = 2^{399}[/imath] I'm guessing the answer will be in the form of an exponent like this But I do not know how to use the added [imath]2[/imath] in the original equation. Any help is welcomed, thanks EDIT: As pointed out in the comments, this may be possible to factor. Could it be factored into prime numbers and use the formula [imath]\varphi(p^{k}) = p^{k}-p^{k-1}[/imath] ? | 1022120 | Determine [imath]\phi(2^{399}+1)[/imath]
This is a question I had on an exam so there is no access to calculators or software. I checked the answer using Maple after and as expected, the answer is very large and some of the prime factors are also very large, with over 10 digits. Is there any reasonable way to break this down and get an answer of some sort? |
2388148 | Solutions to [imath]x^n \equiv a (2^e)[/imath] where [imath]e \geq 3[/imath] and [imath]a[/imath] is odd
I've recently starting reading about number theory, and came across this proposition in A Classical Introduction to Modern Number Theory: The hint is to start writing [imath]x[/imath] and [imath]a[/imath] as [imath](-1)^y5^z[/imath] and [imath](-1)^s5^t[/imath] respectively. For the first case where [imath]n[/imath] is odd, I have: [imath]x^n \equiv (-1)^{yn}5^{zn} \equiv (-1)^{y2k}(-1)^y5^{zn} \equiv (-1)^y5^{zn}[/imath] where the third step occurs from writing [imath]n = 2k + 1[/imath] for some [imath]k[/imath]. Thus we are left with the relation: [imath](-1)^y5^{nz} \equiv (-1)^s5^t (2^e)[/imath] but I'm not quite sure where to go from there, any advice would be great, thanks! | 2385495 | Uniqueness of [imath]k[/imath]th root mod [imath]m[/imath] if [imath](k, \phi(m)) = 1[/imath].
I'm trying to prove a fact about [imath]k[/imath]th roots that says that if [imath](k, \phi(m)) = 1[/imath] and [imath](b,m) = 1[/imath], then there is a unique [imath]k[/imath]th root modulo [imath]m[/imath] (composite [imath]m[/imath]). I'm not sure how to go about it but here's an attempt; Suppose towards a contradiction that there are two solutions [imath]x_1 = b^r[/imath] and [imath]x_2 = b^s[/imath] to the congruence [imath]x^k \equiv b \bmod m[/imath]. Then [imath]x_1^k \equiv x_2^k \equiv b \bmod m[/imath] so that [imath]b^{kr} \equiv b^{ks} \equiv b \bmod m[/imath], or equivalently, [imath]b^{kr - 1} \equiv b^{ks - 1} \bmod m[/imath]. We can then conclude that [imath]m \mid b^{kr - 1} - b^{ks - 1}[/imath]. Now we can factor to obtain [imath]b^{kr - 1} - b^{ks - 1} = b^{kr - 1}(b^{k(s-r)} - 1)[/imath] to conclude that [imath]m \mid b^{kr - 1}(b^{k(s-r)} - 1)[/imath]. But since [imath](b, m) = 1[/imath], we must have [imath]m\mid b^{k(s-r)} - 1[/imath], or equivalently, [imath]b^{k(s-r)} \equiv 1 \bmod m[/imath] which would imply [imath]k(s-r) = \phi(m)v[/imath] for some [imath]v \in \mathbb Z[/imath]. I feel like there could be a way to extract a contradiction from this last step but I can't see it, would somebody be able to check if this is a good way to go? |
2487587 | How do I solve [imath]\sin z - \cos z = 3[/imath]
The original one is: [imath]\sin z - \cos z = 3[/imath] by defenition: [imath]\left(\frac{e^{iz}-e^{-iz}}{2i}-\frac{e^{iz}+e^{-iz}}{2}\right) = 3[/imath] [imath]\left(\frac{e^{iz}-e^{-iz}}{i} - e^{iz}+e^{-iz}\right) = 6[/imath] [imath]\left( \frac{e^{iz}-ie^{iz}-ie^{-iz}}{i} \right) = 6[/imath] What am I suposed to do next? I am not sure that it is right direction at all, i wass thinking of represention of [imath]e[/imath] via eulur's formula but it seems it makes the proccess even more complicated... | 2399959 | solving [imath]\cos z + \sin z = i[/imath]
I found a solution but when testing it, it does not work....? [imath]\cos z + \sin z = i \\ \frac{e^{iz} + e^{-iz}}{2} + \frac{e^{iz} - e^{-iz}}{2i} = i \\ e^{2iz}(1+i) + 2e^{iz} + i - 1 = 0 \\ e^{iz} = \frac{-1 \pm \sqrt{2-i}}{1+i} \\ z = -i \mathrm{Log}\left(\frac{-1\pm\sqrt{2-i}}{1+i}\right) + k2\pi[/imath] Upon testing it in W|A with [imath]k=0[/imath], I don't get the required result [imath](=i)[/imath]. I'm not sure where I went wrong and I did this a few times already. |
2488011 | Polynomial of degree 2015 is equal to [imath]1/x[/imath] for [imath]x=1,2,...,2016[/imath]
Degree of [imath]W(x)[/imath] is 2015. [imath]W(x)=\frac{1}{x}[/imath] for [imath]x=1,2,...,2016[/imath]. Calculate [imath]W(2017)[/imath]. I think that [imath]W(2017)=0[/imath] but how to prove this? I thought of [imath]W(x)=\frac{1-(x-1)(x-2)...(x-2016)}{x}[/imath], but I don't know how to factor [imath]x[/imath] out of counter. | 967664 | Difficult Polynomial Question
Let [imath]P(x)[/imath] be a polynomial whose degree is 1996. If [imath]P(n) = \frac{1}{n}[/imath] for [imath]n = 1, 2, 3, . . . , 1997[/imath], compute the value of [imath]P(1998).[/imath] I don't even know where to begin... Any and all help would be appreciated, thanks! |
2487489 | Evaluating the integral [imath]\int_{0}^{\infty} \frac{\cos ax}{1+x^2}dx[/imath]
I encountered this problem while calculating the Fourier Cosine Integral for the function [imath]f(x) = \frac{1}{1+x^2}[/imath]. I know that [imath]\int \frac{1}{1+x^2} = \arctan x + c[/imath]. I just can't figure out what to do with [imath]cosax[/imath] term combined. If I do integration by parts, I get: [imath]\left [ \cos ax \times \tan^{-1}x \right ]_{0}^{\infty} +\int_{0}^{\infty}a \sin ax \times \tan^{-1}xdx[/imath] I don't know how to proceed after that. Perhaps a hint could help. | 393281 | Complex-valued Fourier integral: [imath] \int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx} [/imath]
I'm working on the Fourier transform, but I don't know how to evaluate the integral: [imath]I = \int_{ - \infty }^{ + \infty } {\frac{{\cos (ax)}}{{{x^2} + 1}}{e^{ - ibx}}\,\mathrm dx} [/imath] |
2487402 | How to prove [imath]1+{x\over 2}-{x^2 \over 8}<\sqrt{1+x}[/imath] for all [imath]x>0[/imath]?
How to prove [imath]1+(1/2)x-(1/8)x^2<\sqrt{1+x}[/imath] for [imath]x>0[/imath] by Taylor Expansions up to [imath]n[/imath] order or Mean Value Theorem? I tried to apply MVT on [imath]\sqrt{1+x}[/imath] and get [imath]\displaystyle \sqrt{1+x}=1+\frac{1}{2\sqrt{1+\xi}}x[/imath] for [imath]\xi\in (0,x)[/imath]. How to do next? | 1061325 | Taylor Series type inequality
I need to show that for [imath]x>0[/imath], [imath]1+\frac{x}{2} \ge \sqrt{1+x} \ge 1+\frac{x}{2} - \frac{x^2}{8} [/imath]. I used the geometric/arithmetic mean inequality to show that [imath]1+\frac{x}{2} \ge \sqrt{1+x}[/imath] is indeed true. My issue lies in the second part of this. In trying to show that [imath]\sqrt{1+x} \ge 1+\frac{x}{2} - \frac{x^2}{8} [/imath], I determined that this is a partial taylor series expansion and I attempted to play around with both inequalities to show that this is true but to no avail. I would appreciate any help in solving this. Ideally I would like to do this analytically but any solutions using more complicated theorems are welcome. |
2487976 | Fourier Series Problem w/the Ramp Function
In a book on Fourier Analysis: Question: This assertion follows without proof. Why is it so? I see that if [imath]n = 0[/imath], then [imath]e^{-i 0 \theta} = 1[/imath], so that [imath] {1 \over 2 \pi} \int_{- \pi}^\pi \theta e^{-in\theta} d\theta = {1 \over 2 \pi} 0 = 0 [/imath] But what about when [imath]n \ne 0[/imath]? | 2336083 | Justifying Calculations on [imath]f(n)=\frac{1}{2 \pi}\int_{-\pi}^{\pi}\theta e^{-in \theta} d \theta[/imath]
In Stein's Fourier Analysis i'm having trouble Justifying the following calculations for the Fourier Coefficients in [imath](1.)[/imath] [imath](1.)[/imath] Let [imath]f(\theta) = \theta[/imath] for [imath] -\pi \leq \theta \leq \pi[/imath] in the case when [imath]n \neq 0[/imath]: [imath]f(n)=\frac{1}{2 \pi}\int_{-\pi}^{\pi}\theta e^{-in \theta} d \theta = \frac{1}{2 \pi}[-\frac{\theta}{in}e^{-in \theta}]_{-\pi}^{\pi} + \frac{1}{2 \pi in} \int_{-\pi}^{\pi}e^{-in\theta}d \theta = \frac{(-1)^{n+1}}{in} [/imath] Then the Fourier Series of [imath]f[/imath] is given by: [imath] f(n) = \sum_{n \neq 0} \frac{(-1)^{n+1}}{in}[/imath] |
2488093 | Proof of an inequality by induction
Prove using induction that [imath]\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{n^2} \le 2-\frac{1}{n}[/imath] for all positive whole numbers [imath]n[/imath]. I began by showing that it is true for [imath]n=1[/imath] I then assumed that it is true for [imath]n=p[/imath] [imath]\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} = \sum_{k=1}^p \frac{1}{k^2} \le 2-\frac{1}{p}[/imath] I now want to show that it is true for [imath]n=p+1[/imath] [imath]\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} + \frac{1}{(p+1)^2}= \sum_{k=1}^{p+1} \frac{1}{k^2} [/imath] If I add [imath]\frac{1}{(p+1)^2}[/imath] to [imath]\sum_{k=1}^{p} \frac{1}{k^2}[/imath], I will then get [imath]\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots+ \frac{1}{p^2} + \frac{1}{(p+1)^2} \le 2-\frac{1}{p} + \frac{1}{(p+1)^2}[/imath] If this is true then [imath]2-\frac{1}{(p+1)}=2-\frac{1}{p} + \frac{1}{(p+1)^2}[/imath] or [imath]-\frac{1}{(p+1)}=-\frac{1}{p} + \frac{1}{(p+1)^2}[/imath] [imath]0= \frac{1}{(p+1)} -\frac{1}{p} + \frac{1}{(p+1)^2}[/imath] [imath]0= \frac{p(p+1)}{p(p+1)^2} -\frac{(p+1)^2}{p(p+1)^2} + \frac{p}{p(p+1)^2}[/imath] [imath]0= \frac{p(p+1)}{p(p+1)^2} -\frac{(p+1)^2}{p(p+1)^2} + \frac{p}{p(p+1)^2}[/imath] [imath]0= \frac{p(p+1)-(p+1)^2 + p}{p(p+1)^2}[/imath] [imath]0= \frac{(p^2 + p) - (p^2 + 2p + 1) + p}{p(p+1)^2}[/imath] [imath]0= \frac{-1}{p(p+1)^2}[/imath] This is invalid. I am not sure where I have made a mistake but I think it is [imath]2-\frac{1}{(p+1)}=2-\frac{1}{p} + \frac{1}{(p+1)^2}[/imath] It then must be that [imath]-\frac{1}{(p+1)} < -\frac{1}{p} + \frac{1}{(p+1)^2}[/imath] [imath]0 \le \frac{1}{(p+1)} - \frac{1}{p} + \frac{1}{(p+1)^2}[/imath] [imath] 0< \frac{-1}{p(p+1)^2}[/imath] which is true for all positive whole numbers [imath]p[/imath]. I am pretty sure it is proved now but I would be happy if someone can confirm this. | 351166 | Prove that for every positive integer [imath]n[/imath], [imath]1/1^2+1/2^2+1/3^2+\cdots+1/n^2\le2-1/n[/imath]
Base case: n=1. [imath]1/1\le 2-1/1[/imath]. So the base case holds. Let [imath]n=k\ge1[/imath] and assume [imath]1/1^2+1/2^2+1/3^2+\cdots+1/k^2\le 2-1/k[/imath] We want to prove this for [imath]k+1[/imath], i.e. [imath](1/1^2+1/2^2+1/3^2+\cdots+1/k^2)+1/(k+1)^2\le 2-\frac{1}{k+1}[/imath] This is where I get stuck. Any help appreciated. |
2488277 | Radius of convergence of [imath]\sum_{n=0}^\infty a_nz^{n^2}[/imath]
Suppose that the power series [imath]\sum_{n=0}^\infty a_nz^n[/imath] has a radius of convergence of [imath]R\in (0,\infty)[/imath]. How can we find the radius of convergence of [imath]\sum_{n=0}^\infty a_nz^{n^2}[/imath] | 344483 | Radius of convergence of power series [imath]\sum c_n x^{2n}[/imath] and [imath]\sum c_n x^{n^2}[/imath]
I've got a start on the question I've written below. I'm hoping for some help to finish it off. Suppose that the power series [imath]\sum_{n=0}^{\infty}c_n x^n[/imath] has a radius of convergence [imath]R \in (0, \infty)[/imath]. Find the radii of convergence of the power series [imath]\sum_{n=0}^{\infty}c_n x^{2n}[/imath] and [imath]\sum_{n=0}^{\infty}c_n x^{n^2}[/imath]. From Hadamard's Theorem I know that the radius of convergence for [imath]\sum_{n=0}^{\infty}c_n x^n[/imath] is [imath]R=\frac{1}{\alpha}[/imath], where [imath]\alpha = \limsup_{n \to \infty} |a_n|^{\frac{1}{n}}.[/imath] Now, applying the Root Test to [imath]\sum_{n=0}^{\infty}c_n x^{2n}[/imath] gives [imath]\limsup |a_nx^{2n}|^{\frac{1}{n}}=x^2 \cdot \limsup |a_n|^{\frac{1}{n}}=x^2 \alpha[/imath] which gives a radius of convergence [imath]R_1 = \frac{1}{\sqrt{\alpha}}[/imath]. Now for the second power series. My first thought was to take [imath]\limsup |a_nx^{n^2}|^{\frac{1}{n^2}}=|x| \cdot \limsup |a_n|^{\frac{1}{n^2}}[/imath] but then I'm stuck. I was trying to write the radius of convergence once again in terms of [imath]\alpha[/imath]. Any input appreciated and thanks a bunch. |
2488502 | Prove that [imath]A =\{\omega\in\Omega: X(\omega)\neq Y(\omega)\}[/imath] is a event of [imath]\mathcal{F}[/imath].
Let [imath]X[/imath] and [imath]Y[/imath] be [imath]\mathbb{R}[/imath]-valued random variables dened on the same probability space [imath](\Omega, \mathcal{F}, \mathbb{P})[/imath] and consider the subset of [imath]\Omega[/imath] defined by [imath]A=\{\omega\in\Omega: X(\omega)\neq Y(\omega)\}[/imath]. Prove that [imath]A[/imath] is a event of [imath]\mathcal{F}[/imath]. | 139853 | How to prove [imath]\left\{ \omega|X(\omega)=Y(\omega)\right\} \in\mathcal{F}[/imath] is measurable, if [imath]X[/imath] and [imath]Y[/imath] are measurable?
Given a probability space [imath](\Omega ,\mathcal{F} ,\mu)[/imath]. Let [imath]X[/imath] and [imath]Y[/imath] be [imath]\mathcal{F}[/imath]-measurable real valued random variables. How would one proove that [imath]\left\{ \omega|X(\omega)=Y(\omega)\right\} \in\mathcal{F}[/imath] is measurable. My thoughts: Since [imath]X[/imath] and [imath]Y[/imath] are measurable, it is true, that for each [imath]x\in\mathbb{R}:[/imath] [imath]\left\{ \omega|X(\omega)<x\right\} \in\mathcal{F}[/imath] and [imath]\left\{ \omega|Y(\omega)<x\right\} \in\mathcal{F}[/imath]. It follows that [imath]\left\{ \omega|X(\omega)-Y(\omega)\leq x\right\} \in\mathcal{F}[/imath] Therefore [imath]\left\{ -\frac{1}{n}\leq\omega|X(\omega)-Y(\omega)\leq \frac{1}{n} \right\} \in\mathcal{F}[/imath], for [imath]n\in\mathbb{N}[/imath]. Therefore [imath]0=\bigcap_{n\in\mathbb{N}}\left\{ -\frac{1}{n}\leq\omega|X(\omega)-Y(\omega)\leq \frac{1}{n} \right\} \in\mathcal{F}[/imath]. Am working towards the correct direction? I appreciate any constructive answer! |
2488861 | bracket of intervals
I know [imath][0,1][/imath] denotes the interval between 0 and 1 with the boundary. And [imath](0,1)[/imath] denotes the interval between 0 and 1 without the boundary. Today, I encounter some expressions as [imath]]0,1][/imath], [imath]]-\infty,+\infty][/imath]. What does [imath]]0,1][/imath] mean? Thanks. | 2571747 | weird brackets in unit interval
I have found during looking at the book Linear Algebra and its Applications (K. Nordstrom) some weird (for me) notation for belonging to the unit interval, namely, [imath]\lambda \in \ ] 0,1 [[/imath]. Does it mean as always that [imath]\lambda[/imath] belongs to [imath][0,1][/imath] or something different? |
2488948 | Is there a formula to count the number of solutions for [imath]x_1 x_2 ... x_m=n[/imath]
I know there is a formula to count the number of solutions of [imath]x_1+x_2+...x_m=n[/imath] and [imath]x_i>0[/imath] for [imath]i=1,2...m[/imath], but i was wondering if there is a (nice looking) formula for the number of solutions of [imath]x_1 x_2...x_m=n[/imath] for [imath]x_i>0[/imath], [imath]x_i \in \mathbb{Z}[/imath], [imath]i=1,2...m[/imath] . | 223223 | The number of possible factorizations of a positive integer.
Given a positive integer [imath]n>1[/imath] with prime factorization [imath]n=\prod_{p_i \text{ prime}}p_i^{k_i}, \space i\ge1, \space k_i \in \mathbb N^*[/imath] how can I compute the number of factorizations of [imath]n[/imath], [imath]\text F(n)[/imath] (multiplications by [imath]1[/imath] are excluded) ? [imath]5\times 24[/imath] and [imath]4\times 5\times 6[/imath] are two different factorizations of [imath]120[/imath]. The prime factorization of a number is of course one of its factorizations. [imath]\text F(p) = 0[/imath] for any prime number [imath]p[/imath]. If there is a no formula, an algorithm will be appreciated. |
2011159 | Using Newton's Law of Universal Gravitation to find distance (r), with the question not giving sufficient information to substitute into the equation.
The problem is described as follows: Newton's Law of Universal Gravitation: [imath]F=G \frac{m_1 m_2}{r^2}[/imath] for any point masses [imath]m_1,m_2[/imath] with the distance between them [imath]r[/imath]. The factor [imath]G≈6.67∙10^{-11} N(m/kg)^2[/imath] is a universal constant. Denote the mass of Earth as [imath]m_E≈5.972∙10^{24} kg[/imath], the mass of the Moon as [imath]m_M≈7.348∙10^{22} kg[/imath]. The unknown mass of the object will be [imath]m[/imath] and the distance in question will be [imath]R[/imath]. Then the distance between the object and Earth is [imath]\frac{1}{10} R[/imath] and the distance between the object and the Moon is [imath]\frac{9}{10} R[/imath]. The force from the Moon and Earth acting on the object is equal. Now, I worked on the problem and found that from the law we obtain: [imath]G \frac{mm_E}{\left(\frac{9}{10} R\right)^2} =G \frac{mm_M}{\left(\frac{1}{10} R\right)^2}[/imath] It seems like there is no solution to the question as not enough information is given. However, I was given this question as an assignment in math class and believe that steps should be shown towards a solution whether it exists or not. I fail to see how to solve this problem, at least to an approximate value, which is what is needed. If anybody knows, help will be immensely appreciated. Thank you. | 2012751 | Using Newton's Law of Universal Gravitation to find distance, with the question not giving sufficient information to substitute into the equation.
Problem: [imath]F = \frac{(G m_1 m_2)}{r^2}[/imath] An object of unknown mass is placed directly between the Earth and the Moon such that the forces of gravity acting on the object are equal from both the Earth and the Moon. The distance from the Moon to the object is exactly one-tenth of the distance from the Moon to Earth. Using the above formula, determine the distance from the Moon to the Earth. Given is: The factor [imath]G≈6.67∙10−11\mathbf{N(\frac{m^2}{kg^2})}[/imath] is a universal constant. Mass of Earth [imath]≈5.972∙10^{24}\mathbf{kg}[/imath] Mass of the Moon as [imath]≈7.348∙10^{22}\mathbf{kg}[/imath] Mass of unknown object = [imath]m[/imath] My solution: [imath]F = \frac{(6.67 *10^-11) m (7.348*10^{24})}{R^2}[/imath] [imath]F = \frac{(6.67 *10^-11) (5.92* 10^22) m}{10R^2}[/imath] As gravitational force acting on the object is equal [imath]\frac{(5.92\cdot 10^22)}{100r^2} = \frac{(7.348*10^{24})}{r^2}[/imath] [imath]R = 1.24 \cdot 10^4 KM[/imath] Is this a viable solution to the problem? Please provide feedback. Thank you! |
2488320 | How to write the metric transformation of conformally related [imath]S^3[/imath] and [imath]R^3[/imath]
This should be a simple problem, yet I'm having trouble seeing the answer. Given two Riemannian metrics related by a conformal transformation [imath]e^{2\sigma(x^{i})}[/imath] . The relationship between them is simply: [imath]\bar{g}_{\mu\nu}=e^{-2\sigma(x)}g_{\mu\nu}[/imath] Yet when it comes to the sphere [imath]S^{3}[/imath] and the flat space [imath]R^{3}[/imath], I know they are conformally related, and yet I can't find a way to write them in the above form. I'm supposing it may have to do with the fact that [imath]R^{3}[/imath] has had a point removed (ie. one point compactification). Is there actually a way to write this? As an aside I study physics, but would like to understand the deeper mathematical relationships that are often “swept under” the proverbial rug. | 402102 | What is the metric tensor on the n-sphere (hypersphere)?
I am considering the unit sphere (but an extension to one of radius [imath]r[/imath] would be appreciated) centered at the origin. Any coordinate system will do, though the standard angular one (with 1 radial and [imath]n-1[/imath] angular coordinates) would be preferable. I know that on the 2-sphere we have [imath]ds^2 = d\theta^2+\sin^2(\theta)d\phi^2[/imath] (in spherical coordinates) but I'm not sure how this generalizes to [imath]n[/imath] dimensions. Added note: If anything can be discovered only about the determinant of the tensor (when presented in matrix form), that would also be quite helpful. |
2488794 | showing [imath]\exp : \mathfrak{gl}(n, \mathbb{C}) \rightarrow GL(n,\mathbb{C})[/imath] is surjective.
I want to prove [imath]\exp : \mathfrak{gl}(n, \mathbb{C}) \longrightarrow GL(n,\mathbb{C})[/imath] is surjective. The textbook gives hint as "using Jordan canonical form" So my guess is expressing matrix in terms of similarity transformation, but how does it guarantee that the exponential map is surjective? Can you give me some proof of above statement. | 1242986 | Proving that the matrix exponential map is surjective onto the general linear group
Let [imath]M_n(\mathbb{F})[/imath] be the set of all [imath]n\times n[/imath] with entries in [imath]\mathbb{F}[/imath] and let [imath]\exp:M_n(\mathbb{C})\to M_n(\mathbb{C})[/imath] be defined by [imath] \exp(A)=\sum_{k=0}^{\infty}\frac{A^k}{k!},[/imath] for all [imath]A\in M_n(\mathbb{C}).[/imath] I want to prove that [imath]\exp[/imath] is a surjective map from [imath]M_n(\mathbb{C})[/imath] to [imath]GL(n,\mathbb{C})=\left\{A\in M_n(\mathbb{C})\,\middle| \det(A)\neq0\right\}[/imath], how do I go about that? I mean saying that [imath]\exp:M_n(\mathbb{R})\to GL(n,\mathbb{R})[/imath] is an analogous to saying [imath]\exp:\mathbb{R}\to \mathbb{R}_{>0}[/imath] and this is also pretty intuitive, since, in analogy with the case of numbers, [imath]A^0=I\;\forall A[/imath], so [imath]\exp(0)=I+0+\frac{0^2}{2!}+\dots=I[/imath], so even for [imath]A=0[/imath] we get [imath]\det\left(\exp(A)\right)\neq0[/imath] and so because of the first term we can never get a zero determinant. But I have no idea how to prove the subjectiveness. Thanks in advance. |
2489259 | Where is [imath]f(z) = z^z[/imath] holomorphic?
I've been messing around with this complex function visualizer: davidbau.com/conformal/ The picture it shows for [imath]f(z) = z^z[/imath] is pretty crazy, so I was wondering: is this function holomorphic? In the real case we'd find the derivative of [imath]f(x) = x^x[/imath] by taking the log of both sides and applying the chain rule. Can we still do that here? | 1580019 | Where is this function holomorphic?
I've never really had to think about this problem before, and to be honest complex analysis isn't my strongest suit, so when I suddenly needed to know where [imath]z^z[/imath] is holomorphic, I didn't know where to begin to start proving my hunches. My guess is that it's holomorphic away from [imath]0[/imath]. More generally, given two entire functions [imath]f[/imath] and [imath]g[/imath], where is [imath]f(z)^{g(z)}[/imath] holomorphic? Again, my hunch is that it is away from the zeroes of [imath]f[/imath]. This isn't a time sensitive question, so any help you can provide would be appreciated, from proofs to thoughts to hints! |
2489405 | Prove that if in a certain set with associative operation [imath]*[/imath] defined, [imath]ax = b[/imath] and [imath]ya = b[/imath] have solutions, then that structure is a group.
Prove that if in a certain set with associative operation [imath]*[/imath] defined, [imath]ax = b[/imath] and [imath]ya = b[/imath] have solutions, then that structure is a group. My attempt: [imath]ax = b \Rightarrow (\exists e )(axe = x \land be =b)[/imath] [imath]ya = b \Rightarrow(\exists n)(nya = y \land nb = b)[/imath] Now, I'm stuck - I don't know how to prove that [imath]e[/imath] and [imath]n[/imath] are the same thing and that if [imath]be = nb = b[/imath] then [imath]e=n[/imath]. | 656881 | Proving that [imath]G[/imath] is a group if [imath]a*x=b[/imath] and [imath]y*a=b[/imath] have solutions.
(Reference : Fraleigh, A first course in abstract algebra) Prove that a nonempty set [imath]G[/imath], together with an associative binary operation * on [imath]G[/imath] such that [imath]a*x=b[/imath] and [imath]y*a=b[/imath] have solutions in G, [imath]\forall a,b\in G[/imath], is a group. Because * is an associatie operation, we have to prove that: there exists [imath]e\in G[/imath] such that [imath]e*a=a=a*e[/imath] for any [imath]a\in G[/imath], and that for any [imath]a\in G[/imath] there exists [imath]\alpha \in G[/imath] such that [imath]a*\alpha=e=\alpha *a[/imath]. This is very confusing, first I tried to prove the identity exists. If we consider a=b, then [imath]a*x=a=y*a[/imath], and because of the hypothesis [imath]x,y\in G[/imath] always exists, now we have to show that [imath]x=y[/imath], but I just can't do this with out inverses, so I tried to prove that the inverses exists, however I have no idea how the identity looks like, I'm stuck. And I tried to get unstuck, so I thought particular cases, first if [imath]G[/imath] had only one element, then what would happend if it had just two elements... but I remembered that we don't know if G is finite or not, and I think this makes things a little bit more complicated. What can I do? |
2489815 | Show that [imath]f(x,y)[/imath] is continuous at the origin but not differentiable at the origin
Show that [imath]f(x,y) = \frac{xy^2}{x^2+y^2}[/imath] (where [imath](x,y) \neq0[/imath], and [imath]f(0,0) = 0[/imath]) is continuous but not differentiable at [imath](0,0)[/imath]. I have shown continuity but am stuck on differentiability. I cannot seem to find a way in my text to show that the above is not differentiable. Thanks | 445175 | A function that is continuous at the origin but not differentiable there
Show that [imath]f(x,y)=\dfrac{xy^2}{x^2+y^2}[/imath] (with [imath](x,y)\not=(0,0)[/imath] and [imath]f(0,0)=0[/imath]) is continuous but not differentiable at [imath](0,0)[/imath]. I tried to show continuity with an [imath]\epsilon -\delta[/imath] argument but I don't know how to factorize the expression so that I can have something useful. For differentiability I think I should show that the partials are not continuous at [imath](0,0)[/imath]. But finding the partials is also painful. |
579054 | There exists function sequence [imath]\{f_{n}\}[/imath] converges to [imath]0[/imath] such that [imath]\{a_{n}f_{n}\}[/imath] not converges to [imath]0[/imath]
Let [imath]X[/imath] be the vector space of all complex functions on the unit interval [imath][0,1][/imath], topologized by the family of seminorms [imath]p_{x}(f) = |f(x)|, \quad (0 \le x \le 1).[/imath] Show that there exists a function sequence [imath]\{f_{n}\}[/imath] in [imath]X[/imath] converging to [imath]0[/imath], such that [imath]\{a_{n}\,f_{n}\}[/imath] does not converge to [imath]0[/imath] for all sequences of scalars [imath]\{a_{n}\}[/imath] converging to [imath]\infty[/imath]. Hi everybody. This problem supports the point that the metrizability is necessary for [imath]X[/imath] to make this theorem true: If a sequence of functions [imath]\{f_{n}\}[/imath] in a metrizable topological vector space [imath]X[/imath] converges to [imath]0[/imath], then there is a scalar sequence [imath]\{a_{n}\}[/imath] converging to [imath]\infty[/imath] such that [imath]\{a_{n}\,f_{n}\}[/imath] converges to [imath]0[/imath]. But I got no clue how to find that sequence like above. Also, topologize a vector space by family of seminorms makes me confused so much. So I hope someone can help me. Thanks. | 1972114 | Proving the existence of a sequence of functions converging to 0 with an added property
This is a problem from Rudin's Functional Analysis: Show that there is a sequence of functions [imath]\{ f_n \}[/imath] in [imath]X[/imath] (the vector space of all complex functions) such that [imath]f_n \to 0[/imath] pointwise, but if any sequence [imath]\{ \gamma_n \}[/imath] tends to [imath]\infty[/imath], [imath]\gamma_n f_n \not\to 0[/imath]. The hint says to use the fact that the collections of all complex sequences converging to [imath]0[/imath] has the same cardinality as [imath][0,1][/imath]. Honestly, this confused me more than it helped me. My main issue is through the method of showing existence: I am not sure if I should attempt to construct an explicit sequence [imath]f_n[/imath], or if I need to appeal to other nonconstructive means such as Zorn's Lemma, etc. I have been trying to construct clever choices of sequences for a good bit now, but it always seems like I will be able to find a sequence that diverges slowly enough to allow that [imath]\gamma_n f_n \to 0[/imath]. Thanks! |
2484011 | Schauder basis in the space of convergent sequences
Could someone help me to prove that {[imath]e_k[/imath] : [imath]k\geq{0}[/imath] } (where [imath]e_k[/imath] (if [imath]k>0[/imath]) is the sequence with [imath]1[/imath] in the place [imath]k[/imath] and [imath]0[/imath] in the rest of sequence and [imath]e_0 [/imath] is the constant sequence equal to [imath]1[/imath]) is a Schauder basis of sequences which converge?. Thanks. | 1689436 | Schauder basis of [imath](c,\|\cdot\|_{\infty})[/imath]
Show that [imath]\{e_{n}\}_{n=0}^{\infty}[/imath] is a Schauder basis of [imath](c,\|\cdot\|_{\infty})[/imath] under the . Here, [imath]c[/imath] is the collection of all convergent sequences. Also, let [imath]e_{0} = \{1,1,1,\ldots\}[/imath] and [imath]e_{n} = \{0,0,\ldots,1,0,\ldots\}[/imath] where the [imath]1[/imath] is in the [imath]n[/imath]th position. The fact that [imath]e_{0}[/imath] is defined this way is a source of confusion for me. Starting with the typical arguments for a Schauder basis proof, I can't find a way to reconcile this. Does anyone have any insight into this proof? |
2490028 | Show that the sum of zeros and poles of elliptic function is lattice point
I try to solve the following problem: Let [imath]f[/imath] be an elliptic function with respect to a lattice [imath]\Lambda[/imath]. Let [imath]z_1,\cdot,s z_2[/imath] be the zeroes and poles of [imath]f[/imath] inside a fundamental parallelogram [imath]\Pi[/imath], of degree [imath]d_1,\cdots,d_n[/imath]. Show that [imath]\sum_{k=1}^{n}d_kz_k\in \Lambda[/imath] (Hint: considering the integral [imath]\int\limits_{\partial ~\Pi} z\cdot\frac{f'(z)}{f(z)}~\mathrm{d}z[/imath] I have no clue to prove this with the hint. How to use the hint? Also, the sum seems like the divisor of [imath]f[/imath]. So is this result related to the theory of Riemann surface and even algebraic geometry? Thanks a lot! | 1917451 | Show that [imath]\int_{\partial P}z\frac {f'(z)} {f(z)} dz [/imath] is on the lattice [imath]\Lambda[/imath]
Problem: Let [imath]f(z)[/imath] be a meromorphic function on the complex torus [imath]\mathbb C/\Lambda[/imath] that as a function on [imath]\mathbb C[/imath] has no zeros and no poles on [imath]\partial P[/imath], the boundary of the fundamental parallelogram [imath]P[/imath]. Show that \begin{equation} \frac{1}{2 \pi i}\int_{\partial P}z\frac {f'(z)} {f(z)} dz \in \Lambda \, . \end{equation} Thoughts: By the Residue Theorem \begin{equation} \frac{1}{2 \pi i}\int_{\partial P}z\frac {f'(z)} {f(z)} dz = \sum_{z_0 \in \text{ Int }P} v_{z_0}(f)z_0. \end{equation} I don't know why the latter is on the lattice. Thanks! |
2490014 | Proving Kummer Theorem
Here is the theorem that way it was given to me... Show that the highest power of a prime [imath]p[/imath] dividing the binomial coefficient [imath]{n\choose m}[/imath] equals the number of carries when [imath]m[/imath] is added to [imath]n-m[/imath], [EDIT] *in base [imath]p[/imath]. How do you go about proving this? I looked around and it seems the proof requires Legendre's formula but I'm not familiar with it. I'm not good with binomial coefficients and identities to begin with and I was given an example of this using binary code but I don't have any experience with that so I'm still lost. If this is a duplicate question than I apologize in advance. | 833706 | highest power of prime [imath]p[/imath] dividing [imath]\binom{m+n}{n}[/imath]
How to prove the theorem stated here. Theorem. (Kummer, 1854) Let [imath]p[/imath] be a prime. The highest power of [imath]p[/imath] that divides the binomial coefficient [imath]\binom{m+n}{n}[/imath] is equal to the number of "carries" when adding [imath]m[/imath] and [imath]n[/imath] in base [imath]p[/imath]. So far, I know if [imath]m+n[/imath] can be expanded in base power as [imath]m+n= a_0 + a_1 p + \dots +a_k p^k[/imath] and [imath]m[/imath] have to coefficients [imath]\{ b_0 , b_1 , \dots b_i\}[/imath] and [imath]n[/imath] can be expanded with coefficients [imath]\{c_0, c_1 ,\dots , c_j\}[/imath] in base [imath]p[/imath] then the highest power of prime that divides [imath]\binom{m+n}{n}[/imath] can be expressed as [imath]e = \frac{(b_0 + b_1 + \dots b_i )+ (c_0 + c_1 + \dots c_j )-(a_0 + a_1 + \dots a_k )}{p-1}[/imath] It follows from here page number [imath]4[/imath]. But how does it relate to the number of carries? I am not being able to connect. Perhaps I am not understanding something very fundamental about addition. |
2491439 | Prove that if [imath]f_{xx}[/imath] + [imath]f_{yy}[/imath] + [imath]f_{zz}[/imath] [imath]= 0[/imath], then [imath]g(r) = a/r + b[/imath], where [imath]a[/imath] and [imath]b[/imath] are constants.
Let [imath]f(x,y,z) = g(\sqrt{x^2 + y^2 +z^2})[/imath], where [imath]g(r)[/imath] is twice differentiable. Calculate [imath]f_{xx}[/imath] + [imath]f_{yy}[/imath] + [imath]f_{zz}[/imath] in terms of the derivatives of [imath]g[/imath]. Prove that if [imath]f_{xx}[/imath] + [imath]f_{yy}[/imath] + [imath]f_{zz}[/imath] [imath]= 0[/imath], then [imath]g(r) = a/r + b[/imath], where [imath]a[/imath] and [imath]b[/imath] are constants. I've been able to do the first part of the question and I got that [imath]f_{xx}[/imath] + [imath]f_{yy}[/imath] + [imath]f_{zz}[/imath] [imath]=[/imath] [imath]\frac{g''(r)(x + y + z) + 3g'(r) - 1}{\sqrt{x^2 + y^2 + z^2}}[/imath] But I cannot figure out how to prove that [imath]g(r) = a/r + b[/imath]. I've tried everything and can't seem to be able to show it. Can some one please give me a hand? Any assistance would be really appreciated as I have been working on this question for a few days. | 2487902 | How to find out the following function f which is a function of r?
If [imath]\nabla^2 f(r) = 0[/imath], show that [imath]f(r) = a\log r + b[/imath], where [imath]a,b[/imath] are constants and [imath]r ^2 = x^2+y^2[/imath]. I know [imath]\nabla[/imath] for cartesian coordinates. But i am not getting idea how to solve above? |
2491491 | How to prove [imath]A^n = 0[/imath] if [imath]A[/imath] is an [imath]n\times n[/imath] strictly upper triangular matrix
I would like to prove the statement on the title. I tried to prove it using the definition of multiply of matrices, however it seems like weak to me. Can't decide how to make a correct proof of the statement above. Edit: Those ns are the same ns. So that, the dimension of the Matrix have to match the power to which it is raised. | 123666 | How to prove a matrix is nilpotent?
Given an [imath]n\times n[/imath] upper triangular matrix [imath]A[/imath] with zero on main diagonal, show that [imath]A^n = 0[/imath]. I did some matrix operation and noticed that the diagonal moves up, ultimately all entries will be zero. Is there a nicer way to do it? |
2491560 | Does [imath](s_{n+1}-s_n)\overset{n\to\infty}{\longrightarrow}0[/imath] in [imath]\mathbb R[/imath] imply that [imath](s_n)[/imath] converges?
Given a sequence [imath](s_n)[/imath] in [imath]\mathbb R[/imath] such that [imath]\lim \limits_{n \to \infty}( s_{n+1}-s_n)=0,[/imath] I am asked to prove [imath](s_n)[/imath] converges. I know all Cauchy sequences converge in [imath]\mathbb R^k[/imath]. So I want to prove that [imath](s_n)[/imath] is Cauchy. I am stuck as to how to show the given sequence is a Cauchy. Thank you. | 2426766 | The limit of the difference of two consecutive sequence members is equal to [imath]0[/imath]. Can we conclude that the sequence itself has a limit?
Let [imath]a_n[/imath] be an infinite sequence. The limit of the difference of two consecutive members is equal to [imath]0[/imath]. Can we conclude that the sequence itself has a limit? My attempt: We have [imath] \lim_{n\rightarrow\infty}{a_n} - \lim_{n\rightarrow\infty}{a_{n-1}} = 0 [/imath] since as [imath]n[/imath] approaches infinity [imath]a_{n-1}[/imath] gets arbitrarily close to [imath]a_n[/imath] the sequence cannot diverge or be bounded but have no limit. Is my proof correct and how would I be able to formalize the last sentence? Thanks |
2491932 | Is this true: any matrix after a random perturbation, it is diagonalizable with probability 1?
As someone points out, we could change the matrix as arbitrary small as possible to make it diagnolizable, but for any random perturbation, say [imath]A+E[/imath], E is random matrix, is the summation must be diagozable with probability 1? | 2125534 | Making a non-diagonalizable matrix diagonalizable with an small perturbations
For arbitrary non-diagonalizable square matrix [imath]J[/imath], can we always find a arbitrarily small perturbations matrix [imath]\varepsilon A[/imath] that [imath]J+\varepsilon A[/imath] is diagonalizable? Using Jordan form as following, we can obtain that arbitrarily small pertubations matrix following a certain structure can make a matrix diagonalizable. But can we relax the form of the pertubations matrix? Give any [imath]J[/imath], let [imath]B[/imath] be the Jordan form. That is, [imath]J=U B U^{-1}[/imath]. For a pertubations matrix [imath]\frac{1}{k} U\Delta U^{-1}[/imath] where [imath]\Delta[/imath] is a diagonalizable matrix with different diagonal value, [imath]J+\frac{1}{k} U\Delta U^{-1}[/imath] is diagonalizable. Can we relax the form of the pertubations matrix? |
2491282 | Prove for all all [imath]x\in\mathbb{R}: \exp(x-1) \geq x[/imath]
i just tried to solve this question, which is a small part of a bigger one. Prove for all all [imath]x \in \mathbb{R}: \exp(x-1) \geq x[/imath] My first attempt was to simplify: [imath] e^{x-1} \geq x [/imath] [imath] \Leftrightarrow \ln(e^{x-1}) \geq \ln(x) [/imath] [imath] \Leftrightarrow x-1 \geq \ln(x) [/imath] [imath] \Leftrightarrow x \geq \ln(x)+1 [/imath] [imath] \Leftrightarrow e^x \geq e^{\ln(x)}+e^1 [/imath] [imath] \Leftrightarrow e^x \geq x+e [/imath] My idea is that [imath]e[/imath] power [imath]x[/imath] is of course greater equal [imath]x+e[/imath]. Is this correct and if yes is it enough? | 1263454 | Prove that [imath]ex \leq e^x[/imath] for all [imath]x \in \mathbb{R}[/imath]
This is easy to prove for negative [imath]x[/imath] but what about positive [imath]x[/imath]? Should I use MVT? |
985851 | Compute [imath]\sum_{k=0}^{n}\frac{1}{\binom{n}{k}}[/imath]
I want to calculate [imath]\sum_{k=0}^{n}\frac{1}{\binom{n}{k}}[/imath]. No idea in my mind. Any help? Context I want to calculate the expected value of bits per symbols in adaptive arithmetic coding when the number of symbols goes to infinity. | 2478893 | Calculate the sum of inverse values of [imath]{n\choose 0}, {n\choose 1}, ... {n\choose n}[/imath]
Calculate [imath]A={1\over {n\choose 0}}+ {1\over {n\choose 1}}+ ...+{1\over {n\choose n}}[/imath] and [imath]B={1\over {n\choose 0}}- {1\over {n\choose 1}}+ ...+{(-1)^n\over {n\choose n}}[/imath] My idea for [imath]A[/imath] is some probabilistic reasoning. Color the sets [imath]\{\}, \{1\}, \{1,2\}, \{1,2,3\},...\{1,2,...,n\}[/imath] and ask our self what is the probability that I choose colored set among all sets. Clearly this is exactly [imath]{n+1\over 2^n}[/imath] and on the other side it is [imath]A[/imath]: probability that I take empty set is [imath]{1\over {n\choose 0}}[/imath], probability that I take colored set with 1 element is [imath]{1\over {n\choose 1}}[/imath] probability that I take colored set with 2 elements is [imath]{1\over {n\choose 2}}[/imath] and so on ... So [imath]A ={n+1\over 2^n}[/imath]. But I have no idea how to attack [imath]B[/imath]. |
2492101 | Find the sum of the following series using
The sum of series [imath]\frac{601}{50}(\frac{1}{1+1^2+1^4} + \frac{2}{1+2^2+2^4} + \frac{3}{1+3^2+3^4} +....+\frac{24}{1+24^2+24^4})[/imath] is | 865442 | Find expression for : [imath] S_n =\sum_{i=1}^{n} \frac{i}{i^4+i^2+1} [/imath]
I want to find a formula for the sum of this series using its general term. How to do it? Series [imath] S_n = \underbrace{1/3 + 2/21 + 3/91 + 4/273 + \cdots}_{n \text{ terms}} [/imath] General Term [imath] S_n = \sum_{i=1}^{n} \frac{i}{i^4+i^2+1} [/imath] |
2492240 | Induction for [imath]3^n\geq n^2[/imath]
I need to prove by induction that [imath]3^n\geq n^2[/imath] What I have so far: Basis: [imath]n = 1 \rightarrow 3^1 \geq 1^2 \rightarrow 3 \geq 1[/imath] Condition: [imath]3^n\geq n^2[/imath] Assumption: [imath]3^{n+1}\geq (n+1)^2[/imath] Basis: [imath]3^n\geq n^2[/imath] [imath]3^{n+1}\geq (n+1)^2 = 3(3^n) \geq n^2+2n+1 [/imath] That's where I got stuck. I have no idea how to continue from there. | 1870539 | How to prove this inequality [imath]3^{n}\geq n^{2}[/imath] for [imath]n\geq 1[/imath] with mathematical induction?
Prove this inequality [imath]3^{n}\geq n^{2}[/imath] for [imath]n\geq 1[/imath] with mathematical induction. Base step: When [imath]n=1[/imath] [imath]3^{1}\geq1^{2}[/imath], statement is true. Inductive step: We need to prove that this statement [imath]3^{n+1}\geq (n+1)^{2}[/imath] is true. So, to get the left side of this statement is easy. We can get it by multiplying [imath]3^{n}\geq n^{2}[/imath] with [imath]3[/imath]. After this step we have [imath]3^{n+1}\geq 3n^{2}[/imath]. What we now have to get is the right side and we can transform it like this: [imath]3n^{2}= (n^{2}+2n+1)+(2n^{2}-2n-1)[/imath] which is same as [imath](n+1)^{2}+(2n^{2}-2n-1)[/imath]. So now we have [imath](n+1)^{2}[/imath] and [imath](2n^{2}-2n-1)[/imath] and my question is how should i use this to prove inequality? |
2492544 | If a is nilpotent then a-1 and a+1 are invertible
In a ring with unity prove that if [imath]a[/imath] is nilpotent, then [imath]a+1[/imath] and [imath]a-1[/imath] are both invertible. In a ring with unity we have: [imath]1-a^n=(1-a)(1+a+a^2+...+a^{n-1})[/imath]. I believe this proves [imath]a-1[/imath] is invertible but I'm not sure why. Then [imath]1-(-1)^na^n=(1+a)(1-a+a^2-...+(-1)^{-1}a^{n-1})[/imath]. Again I think this proves it but I don't know how. I apologize as we haven't discussed nilpotent rings and yet I was asked to answer this question. | 637423 | [imath]a+1,a-1[/imath] invertible for nilpotent element
Given a ring [imath]R[/imath] (with unity), prove that if [imath]a \in R[/imath] is nilpotent, then [imath]a+1,a-1[/imath] are both invertible. Suppose [imath]a^n=0[/imath]. Then [imath]1=1-a^n=(1-a)(1+a+\ldots+a^{n-1})[/imath], so [imath]1-a[/imath] is invertible. If [imath]n[/imath] is odd we can use [imath]1+a^n=(1+a)(1-a+\ldots+a^{n-1})[/imath]. But what if [imath]n[/imath] is even? |
2481210 | Prove the inequality [imath]|\mathrm e^{i\theta_1}-\mathrm e^{i\theta_2}|\le|\theta_1-\theta_2|[/imath].
[imath] |\mathrm e^{i\theta_1}-\mathrm e^{i\theta_2}|\le|\theta_1-\theta_2| [/imath] This seems like a simple inequality but I cannot prove it simply. Do I have to consider Taylor expansions? Could anyone help me how to prove this inequality? | 924147 | Prove that [imath]|e^{i\theta_1}-e^{i\theta_2}| \leq |\theta_1 - \theta_2|[/imath]
I'm trying to prove the inequality [imath]|e^{i\theta_1}-e^{i\theta_2}| \leq |\theta_1 - \theta_2|[/imath] I have tried to use Taylor's formula and got this [imath]|e^{i\theta_1}-e^{i\theta_2}| = |(1+i\theta_1 - \frac{\theta_1^2}{2} + \ldots) -(1 +i\theta_2 - \frac{\theta_2^2}{2} + \ldots)| = |i(\theta_1-\theta_2) + \frac{\theta_2^2-\theta_1^2}{2}+\ldots|[/imath] The first term looks right, but how do I proceed? |
2490621 | Cosine function and Riemann integral
[imath]\int_0^{n\pi} f(\cos^2 x )\, dx = n \int_0^\pi f(\cos^2 x )\, dx[/imath] I am studying the chapter on Riemann integral so wondering if this equality requires the concept of Riemann integration to prove or there are other methods? | 2488518 | Prove that: [imath]\int_0^{n\pi}f(\cos^2(x))\,dx=n\int_0^{\pi}f(\cos^2(x))\,dx?[/imath]
If [imath]f[/imath] is continuous on [imath][0,1][/imath], how can I show that [imath]\int_0^{n\pi}f(\cos^2(x))\,dx=n\int_0^{\pi}f(\cos^2(x))\,dx?[/imath] I know how to show that both integrals exist. I believe that [imath]n[/imath] is a multiple of [imath]\frac12[/imath]. I thought that the identity [imath]\cos^2(x)=\frac{\cos(2x)+1}{2}[/imath] would be useful, but maybe not. I think I should use some change of variable. |
2492683 | Why it is called "Linear" Differential Equation?
[imath]y'+P(x) \cdot y=Q(x)[/imath] Why it's called Linear Differential Equation? The function y might be curved. | 948926 | How to explain to a high school student why a linear differential equation is linear?
My mother is teaching a high school course on multivariable calculus, and they were studying linear differential equations of the form [imath]y' + P(x) y = Q(x),[/imath] and the question of why this equation is called "linear" came up. In terms that these students are familiar with, since they haven't been exposed to linear algebra yet, my thought was to say that the equation for a line, [imath]y = mx+b[/imath], is "linear" in [imath]x[/imath] (ignoring the technicality that it's actually an affine equation, not a linear one), because it's in the form "coefficient times [imath]x[/imath]", and then we allow another term which is just a lonely coefficient. And then we extend this notion to saying that the above differential equation is "linear" in [imath]y[/imath] and [imath]y'[/imath], but this time the coefficients are allowed to be functions of [imath]x[/imath]. That's probably a good enough hand-wavy explanation to help students remember the definition, at the very least. I couldn't really think of a good reason why it "should", a priori, be kosher to allow coefficients to be functions of [imath]x[/imath] here. At that point it seems to me like you just have to get into the linear algebraic definition of linearity, which, being completely foreign to the students... it just seems to be a bit too deep of a rabbit hole for this purpose. So my question is: does anyone have a better way of approaching this? And if you think my hand-wavy explanation above is largely acceptable, is there a way you can explain why multiplying by non-constant functions of [imath]x[/imath] "should" still be considered linear in [imath]y[/imath]? |
2490508 | Prove inequalities
Define the sequence of numbers [imath]A_i[/imath] by [imath]A_0 = 2[/imath] [imath]A_{n+1} = \frac{A_n}{2} + \frac{1}{A_n} \quad \text{for} \ n \geq 1[/imath] Prove that [imath] A_n \leq \sqrt{2} + \big(\frac{1}{2}\big)^n \quad \forall n \geq 0. [/imath] I already tried induction but it didn't help in this case. The question is : beside induction what is other approach to solve this kind of problem ? | 1802813 | Prove upper bound for recurrence
I am working on problem set 8 problem 3 from MIT's Fall 2010 OCW class 6.042J. This is covered in chapter 10 which is about recurrences. Here is the problem: [imath]A_0 = 2[/imath] [imath]A_{n+1} = A_n/2 + 1/A_n, \forall n \ge 1[/imath] Prove [imath]A_n \le \sqrt2 + 1/2^n, \forall n \ge 0[/imath] I have graphed the recurrence and the upper bound and they seem to both converge on [imath]\sqrt2[/imath]. Also, if you ignore the boundary condition [imath]A_0 = 2[/imath] then you find that [imath]\sqrt2[/imath] is a solution to the main part of the recurrence. i.e. [imath]\sqrt2 = \sqrt2/2 + 1/\sqrt2[/imath]. The chapter and videos on recurrences have a lot to say about a kind of cookbook solution to divide and conquer recurrences which they call the Akra-Bazzi Theorem. But this recurrence does not seem to be in the right form for that theorem. If it were in the form [imath]A_{n+1} = A_n/2 + g(n)[/imath] then the theorem would give you an asymptotic bound. But [imath]1/A_n[/imath] is not a simple function of [imath]n[/imath] like a polynomial. Instead it is part of the recurrence. Also, the chapter has a variety of things to say about how to guess the right solution and plug it into an inductive proof, but I haven't had much success. I have tried possible solutions of various forms like [imath]a_n = \sqrt2+a/b^n[/imath] and tried solving for the constants [imath]a[/imath] and [imath]b[/imath] but to no avail. So, if someone can point me in the right direction that would be great. I always assume that the problem sets are based on something taught in the videos and in the text of the book but I am having trouble tracking this one down. Bobby |
2492415 | An elementary approach using the divisor function definition:
Proof conclusion: Let m,n,s [imath]\in[/imath] [imath]\mathbb{N} [/imath]\ [imath]\{0\}[/imath]. Prove that [imath]s| \tau(m^{s}) - \tau(n^{s})[/imath], where [imath]\tau[/imath] is the divisor function. Hello so my question is that, I've gotten to the point in the proof where I had used the divisor function definition and have [imath]\tau(m^{s})[/imath] = [imath]\prod[/imath] something +1 , and [imath]\tau(n^{s})[/imath] = [imath]\prod[/imath] something +1. I know, that since the last terms of both polynomials are equal to 1. Somehow this results that [imath]s| \tau(m^{s}) - \tau(n^{s})[/imath]. But I don't understand why the +1 would indicate that: [imath]s| \tau(m^{s}) - \tau(n^{s})[/imath] holds. | 2491966 | Working with elementary type proofs of divisor functions
Proof conclusion to: Let m,n [imath]\in[/imath] [imath]\mathbb{N}[/imath] [imath]\backslash[/imath] [imath]\{0\}[/imath] . Prove that [imath]s |\tau (m^{s}) - \tau(n^{s})[/imath], where [imath]\tau[/imath] is a divisor function. Hello, so I have a tiny question about the summarization of my proof. I got to the point where I defined my [imath]\tau[/imath] divisor function and got them both [imath]\tau(m^s)[/imath]= [imath]\prod[/imath] something +1 and [imath]\tau(n^{s})[/imath] = [imath]\prod[/imath] something +1. I've noticed that the last terms of the polynomial are both one. Why is it with the +1 part, I can conclude with that: s| [imath]\tau(m^s)[/imath] - [imath]\tau(n^s)[/imath]. The line for why that is, isn't quite clear. |
2492876 | Finite group [imath](G, \cdot)[/imath] and H a subgroup of G. Prove that if [imath]x \notin H, y \notin H[/imath] then [imath]xy \in H[/imath].
Given the finite group [imath](G, \cdot)[/imath] with [imath]2n[/imath] elements, [imath]n \in N^*[/imath] and also H a subgroup of G with [imath]n[/imath] elements. Prove that if [imath]x \notin H, y \notin H[/imath] then [imath]xy \in H[/imath]. I'm not really sure how to do this since I've barely started working with subgroups. Could I have some hints on how to approach this type of exercise? Thank you. | 2451208 | Let [imath]G[/imath] be a group and [imath]H[/imath] be a subgroup of [imath]G[/imath] of index [imath]2[/imath] .Then show that if [imath]a,b\notin H[/imath] then [imath]ab\in H[/imath].
Let [imath]G[/imath] be a group and [imath]H[/imath] be a subgroup of [imath]G[/imath] of index [imath]2[/imath] .Then show that if [imath]a,b\notin H[/imath] then [imath]ab\in H[/imath]. [imath]a,b\notin H\implies aH,bH\neq H\implies [/imath] [imath]aH=bH[/imath] since [imath]H[/imath] has only two distinct left cosets in [imath]G[/imath]. [imath]\implies a^{-1}b\in H[/imath] How to show that [imath]ab\in H[/imath] from here? |
1061477 | Prove by induction that an expression is divisible by 11
Prove, by induction that [imath]2^{3n-1}+5\cdot3^n[/imath] is divisible by [imath]11[/imath] for any even number [imath]n\in\Bbb N[/imath]. I am rather confused by this question. This is my attempt so far: For [imath]n = 2[/imath] [imath]2^5 + 5\cdot 9 = 77[/imath] [imath]77/11 = 7[/imath] We assume that there is a value [imath]n = k[/imath] such that [imath]2^{3k-1} + 5\cdot 3^k[/imath] is divisible by [imath]11[/imath]. We show that it is also divisible by [imath]11[/imath] when [imath]n = k + 2[/imath] [imath]2^{3k+5} + 5\cdot 3^{k+2}[/imath] [imath]32\cdot 2^3k + 5\cdot 9 \cdot3^k[/imath] [imath]32\cdot 2^3k + 45\cdot 3^k[/imath] [imath]64\cdot 2^{3k-1} + 45\cdot 3^k[/imath] (Making both polynomials the same as when [imath]n = k[/imath]) [imath](2^{3k-1} + 5\cdot 3^k) + (63\cdot 2^{3k-1} + 40\cdot 3^k)[/imath] The first group of terms [imath](2^{3k-1} + 5\cdot 3^k)[/imath] is divisible by [imath]11[/imath] because we have made an assumption that the term is divisible by [imath]11[/imath] when [imath]n=k[/imath]. However, the second group is not divisible by [imath]11[/imath]. Where did I go wrong? | 2497585 | Prove by induction that if this is divisible by 11, then this is also divisible by 11
How would I do this? I tried using contradiction, but got stuck halfway through. This unit is about induction, but I have no idea how I would use it to do this as it is not a sequence 4.6.15 (a) show that for any [imath]k\in N[/imath], if [imath]2^{3k-1}+ 5\cdot 3^k[/imath] is divisible by 11, then [imath]2^{3(k+2)-1} + 5\cdot 3^{k+2}[/imath] is also divisible by 11 [imath](\forall k\in N)(\exists p\in Z)(\exists w\in Z)(2^{3k-1}+5\cdot 3^k = 11p)\Rightarrow(2^{3(k+2)-1}+5\cdot 3^{k+2})[/imath] we can prove this by contradiction. |
2494159 | Determining if the series $\sum_{n=2}^\infty {1\over {n\ln(n)\ln(\ln(n))}}$ is convergent or divergent and justifying the answer.
By far, these infinite series and sequences have been my biggest problem in calculus 2, and this problem really puts my limited knowledge to the test. I have found that posting these pictures helps show everyone what I have learned so far, and then I'll try to solve it myself. Chapter and section names: Rules that my book gives me to solve these problems. This shows more of what I should and do know: Looking at those rules, it doesn't really even seem to help me with this problem. The only way I could see this being solved with what I know is with the integral test. Also, note that it starts at 2 (I almost missed that myself). If I'm going to try to use the integral test, then here is my book's definition of it: And here is the question again: Determining if the series [imath]\sum_{n=2}^\infty {1\over {nln(n)ln(ln(n))}}[/imath] is convergent or divergent and justifying the answer. I honestly don't know where to start with this one, I would try but there are a lot of changes from previous questions, and it makes it a lot harder. Any help is greatly appreciated. Thank you. | 280257 | Another simple series convergence question: [imath]\sum\limits_{n=3}^\infty \frac1{n (\ln n)\ln(\ln n)}[/imath]
I'm being asked to determine if [imath]\displaystyle\sum\limits_{n=3}^\infty \frac1{n (\ln n)\ln(\ln n)}[/imath] converges. So, using Cauchy's Condensation Test, I reduced the problem to one of determining the convergence of [imath]\displaystyle\sum\limits_{n=3}^\infty\frac 1{n\ln (n\ln 2)}[/imath]. Am I on the right path, and how do I proceed from here? |
2493656 | Why [imath]e^0=I[/imath] for exponential operator?
From definition of [imath]e^T=\sum_{k\geq 0} \frac{T^k}{k!}[/imath], so [imath]e^0=0^0[/imath]... So, why [imath]e^0=I[/imath] for exponential operator? Any help would be appreciated. | 1827556 | Derivative of matrix exponential at [imath]0[/imath]
I have to show that the derivative of 'the matrix exponential' [imath]exp: \mathbb{C}^{n\times n}\mapsto\mathbb{C}^{n\times n}[/imath] at the zero matrix [imath]0[/imath] is [imath]id_{C^{n\times n}}[/imath], i.e. [imath]exp(0)=id[/imath]. The above map isn't given explicitly, so maybe someone can tell me what it does to a matrix [imath]A[/imath]? I think the obvious guess would be [imath]A\mapsto e^A:=\sum_{k=0}^\infty \frac{A^k}{k!}[/imath], but I wouldn't know how to find the derivation of such a map since there is no variable. So far I tried considering [imath]Ax\mapsto e^Ax:=\sum_{k=0}^\infty \frac{(Ax)^k}{k!}[/imath] instead, which is a map [imath]\mathbb{C}\rightarrow\mathbb{C}[/imath] I suppose. As derivative I get [imath]Ax\mapsto Ae^{Ax}[/imath], which gives the identity for [imath]x=0[/imath], but [imath]0[/imath] for [imath]A=0[/imath]. Can somebody tell me what I got wrong, and how it is true that [imath]exp(0)=id[/imath]? |
216360 | Counting the number of surjections.
How many functions from set [imath]\{1,2,3,\ldots,n\}[/imath] to [imath]\{A,B,C\}[/imath] are surjections? [imath]n \geq 3[/imath] Attempt I was hoping to count the number of surjections by treating [imath]A,B,C[/imath] like bins, and counting the number of ways to fill them with [imath]1,2,3,\ldots,n[/imath] such that no bins are empty. This is equivalent to putting two "separator" bars in between the [imath]n[/imath] numbers or [imath]n-1 \choose 2[/imath]. However, I think I am missing a step here: prior to putting in the separator bars, I should permutate the [imath]n[/imath] elements. So, multiply by [imath]n![/imath]. The order of elements in the same bins do not matter. So I should be divided by the number of permutations of the elements in each bins. This is where I am stuck. Additional Info I made the question up myself. Feel free to modify the question into a more solvable form if it is too difficult/complicated. | 2611130 | Number of ways to select [imath]m[/imath] items out of [imath]k[/imath] categories, where each category is selected at least once and selection order matters?
Suppose I ask the following question: If we have [imath]k[/imath] categories and [imath]m[/imath] selections of an item, where each item can belong to one of the categories, how many possible enumerations of each category can we have? This is a common combinatorics question which is answered by the stars-&-bars analogy. If you make a sequence with [imath]m[/imath] stars and [imath]k-1[/imath] bars, every possible orientation of stars and bars represents a completely unique enumeration of categories, and the set of possible orientations you can make is exhaustive. Therefore, the number of possible orientations you can make with those stars and bars, which is [imath]m+k-1 \choose m[/imath], is the answer to the question. Similarly, if you specify that each category must have been chosen at least once, this is equivalent to having [imath]k-1[/imath] bars and [imath]m-k[/imath] stars. Thus the answer to that question is [imath]m-1 \choose k-1[/imath]. However, suppose we asked the following question: How many ways are there to select [imath]m[/imath] items out of [imath]k[/imath] categories where order of item selection matters? For example, if we have [imath]3[/imath] categories, and [imath]5[/imath] selections, one possible [imath]k[/imath]-tuple is {2,0,3} (that is, 2 items of category 1 and 3 items of category 3). However this can be done in multiple orders of choosing categories ([imath]5 \choose 2[/imath] to be exact). For example, you can make the above choice with order 3-1-3-1-3 or 1-1-3-3-3. So what if we do a weighted sum of selecting each possible tuple, where the weight is given by the number of possible ways of selecting that tuple? Well if each category is allowed to be selected any number of times, including [imath]0,[/imath] then the answer is actually simple. This is just all possible ways of selecting [imath]m[/imath] items, where each item can be one of [imath]k[/imath] categories. So the answer is [imath]k^m.[/imath] Now here's where I'm stuck. What is the answer to the above problem, with the constraint that every category is chosen at least once? As far as I understand, to do this we would have to sum over every possible tuple, with the summand being the number of possible ways to select that tuple. So there are [imath]m-1 \choose k-1[/imath] tuples. But for each tuple, the number of ways of selecting that tuple is a multinomial coefficient, where the top term is [imath]m[/imath] and the bottom terms are the individual values in the tuple. A very simple example is the following: Suppose there are [imath]3[/imath] categories and [imath]4[/imath] selections. Each category must be chosen at least once. Then there are [imath]{3 \choose 2} = 3[/imath] possible enumerations of categories, and we would sum over them using: [imath] N = {4 \choose 2,1,1} + {4 \choose 1,2,1} + {4 \choose 1,1,2} = 36. [/imath] However, I don't know how to do this sum in general. For [imath]m[/imath] selections and [imath]k[/imath] categories I think we would have to set up the following sum, [imath] N = \sum_{n_1=1}^{m-k+1} \sum_{n_2=1}^{m-k +2-n_1} \sum_{n_3=1}^{m - k + 3 - n_1 - n_2} \dots \sum_{n_k = 1}^{m - n_1 - n_2 - \dots - n_{k-1} } {m \choose n_1, n_2, \dots, n_k}. [/imath] (I believe I set up the upper bounds properly, but I could be wrong.) I don't know how to do such a sum, or if it even has a closed form solution. |
2426376 | Polynomial interpolation of an analytic function
Let [imath]f\in A(\Omega)[/imath] be an analytic function on a bounded domain [imath]\Omega \subset \mathbb C[/imath] and let [imath]a_0,a_1\cdots a_m\in \omega\Subset \Omega[/imath] where [imath]\omega[/imath] is a smooth open set compactly contained in [imath]\Omega[/imath]. Show that [imath]p(z) = f(z)- \frac{1}{2i\pi}\int_{\partial \omega}\prod_{j=0}^{m}\left(\frac{z-a_j}{\zeta-a_j}\right)\frac{f(\zeta)d\zeta}{\zeta-z}[/imath] is the unique polynomial of degree at most m that interpolates [imath]f[/imath] at [imath]a_0,a_1\cdots a_m\in \omega [/imath] i.e such that [imath]f(a_j)=P(a_j)[/imath], [imath]j=0,1,\dots ,m .[/imath] My problem is that I am not able to prove that the above defined [imath]P(z)[/imath] is a polynomial. But the fact [imath]f(a_j)=P(a_j)[/imath] follows immediately from Cauchy formula. | 2229110 | Show [imath]P(z) [/imath] is a polynomial of degree [imath]n-1[/imath] interpolating an Analytic function.
Let [imath]C[/imath] be a regular curve enclosing the distinct points [imath]ω_1,ω_2,...ω_n[/imath] and let [imath]p(ω) = (ω −ω_1)(ω −ω_2) \cdots (ω −ω_n)[/imath]. Suppose that [imath]f (ω)[/imath] is analytic in a region that includes [imath]C[/imath]. Show that [imath]P(z) = \frac{1}{2 \pi i} \int_C \frac{f(ω)}{p(ω)} . \frac{p(ω) -p(z)}{ω -z} dω[/imath] is a polynomial of degree [imath]n-1[/imath] and [imath]P(ω_i) = f(ω_i)[/imath]. |
2424839 | Interval of convergence of [imath]\sum_{n=1}^\infty x^{\ln(n)}[/imath].
What is the interval of convergence of this series or for what values of [imath]x[/imath] does it converge? [imath]\sum_{n=1}^\infty x^{\ln(n)}[/imath] I tried the ratio and root test but they were inconclusive, any help would be appreciated. | 2392088 | Find the set of [imath]x>0[/imath] such that the series [imath]\sum\limits_n x^{\ln{n}}[/imath] converges
If [imath]x>0[/imath], find the set of all values of [imath]x[/imath] such that series is convergent[imath]\sum_{n=1}^{\infty} x^{\ln{n}}[/imath] My attempt:- I used Ratio test for finding the set of all values of [imath]x[/imath] such that series is convergent. [imath]\lim_{x\to\infty}\frac{x^{\ln{n+1}}}{x^{\ln{n}}}[/imath] [imath]=\lim_{x\to\infty}x^{\ln\frac{n+1}{n}}[/imath] This quantity must be less than one for getting a convergent series, I am not able to judge. Can you please help me to find the interval of convergence? |
2493948 | Proving pairwise independence is equivalent to uncorrelated
The problem states: "Let [imath]X = (X_1, X_2, \dots, X_n)^T[/imath] be a multivariate normal random vector. Show that [imath]X_i[/imath] and [imath]X_j[/imath], [imath]i \neq j[/imath], are independent if and only if [imath]X_i[/imath] and [imath]X_j[/imath] are uncorrelated." Going from pairwise independence to uncorrelated is easy: Just use the fact that [imath]E(X_i X_j) = E(X_i)E(X_j)[/imath] to show that [imath]\mathrm{Corr}(X_i X_j)=0[/imath]. But how to prove the converse? Uncorrelated doesn't imply independence, does it? | 609732 | Jointly Gaussian uncorrelated random variables are independent
Let [imath]X,Y[/imath] be jointly normally distributed and uncorrelated. Why are they independent? |
2495382 | Problem on Riemann Integration: [imath]f(x)=\frac{1}{n}[/imath] for [imath]\frac1n>x\geq \frac{1}{n+1}[/imath].
Problem on Riemann Integration: [imath]f[/imath] is defined on [imath][0,1][/imath] by [imath]\,f(x)=1/n[/imath] for [imath]1/n>x\geq 1/(n+1)[/imath] and [imath]f(x)=0[/imath] for [imath]x=0[/imath], where [imath]n=1,2,3,....[/imath] Find [imath]\int_{0}^1f(x)dx[/imath]. First I tried to find out [imath]\int_{1/(n+1)}^1f(x)dx[/imath]. But I got a series sum:[imath]\sum_{k=1}^{n}\cfrac{1}{k^2(k+1)}[/imath] How to proceed? | 2396444 | Obtain the value of [imath]\int_0^1f(x) \ dx[/imath], where [imath]f(x) = \begin{cases} \frac {1}{n}, & \frac{1}{n+1}[/imath]
Is the function Riemann integrable? If yes, obtain the value of [imath]\int_0^1f(x) \ dx[/imath] [imath]f(x) = \begin{cases} \frac {1}{n}, & \frac{1}{n+1}<x\le\frac{1}{n}\\ 0, & x=0 \end{cases}[/imath] My attempt [imath]f[/imath] is bounded and monotonically increasing on [imath][0,1][/imath]. Also, [imath]f[/imath] has infinite discontinuities but only one limit point. Therefore [imath]f[/imath] is Riemann integrable. Now, to calculate the integration [imath]\int_0^1f(x) \ dx=\int_{1/2}^{1}1 \ dx + \int_{1/3}^{1/2}\frac{1}{2} \ dx + \int_{1/4}^{1/3}\frac{1}{3} \ dx+...[/imath] [imath]=\sum_{n=1}^\infty \frac{1}{n^2}-\frac{1}{n}+\frac{1}{n+1}[/imath] How do I proceed from here? How do I calculate these summations? I know [imath]\sum \frac{1}{n}[/imath] is [imath]\log 2[/imath], but not the other two summations. |
2495600 | Inequality involing square-root
I want to show for [imath]x>0[/imath] the inequality [imath] \sqrt{x+1}-\sqrt{x} \leq \frac{1}{\sqrt{x}} , [/imath] which gets suprisingly sharp for large values of [imath]x[/imath]. I guess some kind of binomial theorem for non-integer exponents would be helpful. But maybe there is something simpler? Any hints? thanks. | 2243976 | Prove [imath]\sqrt{n+1}-\sqrt{n}\lt \frac{1}{2\sqrt{n}}[/imath]
I'm having some trouble figuring out where to start for this proof. Prove for all positive integers [imath]$n$[/imath], [imath]\sqrt{n+1}-\sqrt{n}\lt \frac{1}{2\sqrt{n}}[/imath] Any help will be greatly appreciated. |
2495525 | For every element [imath]x[/imath] in a given set [imath]A[/imath], [imath]x[/imath] is an element of the corresponding equivalence class [imath][x]_R[/imath] with R an equivalence relation. Why?
On page 217 of Velleman's "How to Prove It", Velleman states a lemma. It goes like this (note that [imath][x]_R = \{y \in A : y \, R \, x \})[/imath]: Lemma 4.6.5. Suppose R is an equivalence relation on A. Then: For every [imath]x \in A[/imath], [imath]x \in [x]_R[/imath] For every [imath]x \in[/imath] and [imath]y \in A[/imath], [imath]y \in [x]_R[/imath] iff [imath][y]_R=[x]_R.[/imath] I don't understand why (1) should be true. After all, an equivalence class is a set [imath][a]_R = \{ x \in A : x \ R\ a \}[/imath], where [imath]R[/imath] is an equivalence relation - i.e., a relation that is reflexive, symmetric and transitive. As I didn't understand it, I turned to the commentary on the page, and found the author had written: According to the definition of equivalence classes, [imath]x \in [x]_R[/imath] means [imath]x R \ x[/imath]. This is what leads us to apply the fact that R is reflexive. ...so? Yes, [imath]x \in [x]_R[/imath] means that [imath]x R \ x[/imath], but that isn't the only hard requirement for it to be in [imath][x]_R[/imath] - we just established that an eq. class is one defined using an equivalence relation, that is to say, a relation that is not only reflexive but also symmetric and transitive! How can we say that [imath]x \in [x]_R[/imath] by just looking at one of 3 requirements? So I turned to another book. Cunningham's "A Logical Introduction to Proof", pg. 216. I found an almost identical list on that page, and the first pointer was yet again For every [imath]x \in A[/imath], [imath]x \in [x]_R[/imath] So I turned to the proof, and in the proof the author goes To prove (1), let [imath]x \in A[/imath]. Clearly, [imath]x \in [x]_R[/imath], as [imath]R[/imath] is reflexive. What? How can you conclude that [imath]x \in [x]_R[/imath] by just considering one of three requirements of the eq. relation R? The same thing in Lay's "Analysis with an Introduction to Proof": Given an equivalence relation [imath]R[/imath] on a set [imath]S[/imath] [...] we define the equivalence class (with respect to [imath]R[/imath]) of [imath]x \in S[/imath] to be the set: [imath]E_x = \{y \in S : y \, R \,x \}[/imath] Since [imath]R[/imath] is reflexive, each element of [imath]S[/imath] is in some equivalence class. Why! Why would [imath]R[/imath] being reflexive have anything to do with each element of [imath]S[/imath] being in some equivalence class! To be in an eq. class, they're supposed to be reflexive, symmetric and transitive - why does everybody keep repeating they only need to be reflexive! | 2495403 | Why is it always the case that, given an equivalence relation R on a set X and a [imath]x \in X[/imath], that x is equivalent to itself?
Why is it always the case that, given an equivalence relation R on a set X and a [imath]x \in X[/imath], that x is equivalent to itself? According to my book, this is due to the reflexivity of [imath]R[/imath] and consequently [imath]x \in [x]_R[/imath], but isn't an equivalence relation supposed to be reflexive, symmetric and transitive - not just reflexive? I've cross-checked this with some other books, and they all say the same thing. I don't get it, how can [imath]x \in [x][/imath] if x only satisfies reflexivity? |
2496148 | Equation of the parabola which touches [imath]x[/imath] axis at [imath](a,0)[/imath] and [imath]y[/imath] axis at [imath](0,b)[/imath]
Prove that the equation of parabola which touches [imath]x[/imath] axis at [imath](a,0)[/imath] and [imath]y[/imath] axis at [imath](0,b)[/imath] is [imath]\sqrt{\frac{x}{a}}+\sqrt{\frac{y}{b}}=1.[/imath] | 2225828 | Is [imath]\sqrt{x/a}+\sqrt{y/b}=1[/imath] the equation of a parabola tangent to the coordinate axes?
Is the below equation represents a parabola that touches the axes of coordinates? [imath]\sqrt{x/a}+\sqrt{y/b}=1[/imath] I know it is very stupid to ask this type of easy question here in the forum, but I'm very curious to know. I have searched many places and found nothing. My professor is not here, so I can't ask him. Suspense would have killed me. Note from @Blue. Months later, I have edited the original problem to move the "[imath]a[/imath]" and "[imath]b[/imath]" under the radical signs. (This is because a duplicate problem recently appeared and I wanted to minimize confusion.) Most answers assumed this was the intention and proceeded accordingly. Those answers that use "[imath]\sqrt{x}/a[/imath]" and "[imath]\sqrt{y}/b[/imath]" should not be penalized for this after-the-fact notational change. |
1714017 | In how many ways can you choose [imath]k[/imath] numbers out of [imath]\{1,2,3,\dots,n\}[/imath] so none of them is consecutive?
I am a newbie in combinatorics and still don't have enough tools to handle this kind of problems. Assuming I have a set of integers: [imath] \{1,2,3,4,\dots,n\} [/imath] In how many ways, can I choose [imath]k[/imath] numbers out of those [imath]n[/imath], such that none of them are consecutive? For instance, for the following set [imath] \{1,2,3,4,5\} [/imath] For [imath] k=3 [/imath] I have only 1 option: [imath]\{1,3,5\}[/imath] My purpose is first of all to understand the way you would think about this problem, not necessarily the solution itself (I have the final answer for it). Thanks. | 2851604 | In how many ways can [imath]4[/imath] books be removed from a bookshelf with [imath]15[/imath] books such that no two adjacent books are chosen?
A bookshelf has [imath]15[/imath] books. in how many ways can [imath]4[/imath] books be removed such that no two adjacent books are chosen? I started to solve the question by saying that the first book can be selected in [imath]15[/imath] ways, the second one can be selected in [imath]13[/imath] ways,the third one in 11 ways and the fourth one in [imath]9[/imath] ways. Total number of ways:[imath]15\times13\times11\times9=19305[/imath]. However the correct answer must be [imath]495[/imath]. Can you explain why my counting technique is false and provide me with hints about the correct one? Thanks you for your help |
2496076 | Show that a metric space [imath](X,d)[/imath] is totally bounded(ball compact) if and only if every sequence in X has a Cauchy subsequence.
Show that a metric space [imath](X,d)[/imath] is totally bounded(ball compact) if and only if every sequence in X has a Cauchy subsequence. I already proved that if [imath](X,d)[/imath] is ball compact, then every sequence in X has a Cauchy subsequence. But I have no idea how to the other part. | 556150 | Metric space is totally bounded iff every sequence has Cauchy subsequence
Prove that a metric space is totally bounded if and only if every sequence has a Cauchy subsequence. I think I proved the Cauchy subsequence part: Let [imath]a_{0},a_{1}, a_{2}, a_{3}, a_{4},...\in X[/imath] be a sequence. For each [imath]k[/imath], let [imath]F \subseteq X[/imath] be a finite [imath]\frac1k[/imath]-net. Given [imath]I \subseteq \Bbb N_{\ge0}[/imath] and [imath]k>1[/imath] you find an infinite [imath]J \subseteq I[/imath] such that: [imath]\exists p\in F:\forall n\in J: d(x_n,p)<\frac1k[/imath] |
2495754 | prove a general inequality between radicals
I've been trying to work on a proof for this for a few days now, but can't seem to see the answer. It might be trivial but I would be grateful if anyone could let me know their strategy (if not a proof) for this: [imath]\sqrt[n]{n!} \lt \sqrt[n+1]{(n+1)!}[/imath] Thanks in advance! | 580432 | Prove that [imath]\sqrt[n]{n!}[/imath] is increasing and diverges
I have to prove that [imath]a_n[/imath] is (strictly) increasing and diverges [imath]a_n = \sqrt[n]{n!}[/imath] ; n [imath]\in[/imath] [imath]\ \mathbb{N}[/imath] From sequence I see that [imath]a_n[/imath] increasing to infinitive. [imath]\sqrt[1]{1!}=1 ,\ \sqrt[2]{2!} \approx 1.41, \ \sqrt[3]{3!} \approx 1.81 ,..., \sqrt[n]{n!} [/imath] |
2496921 | A matrix [imath]R[/imath] where [imath]R\neq I[/imath] but [imath]R^7=I[/imath]
I'm having trouble trying to find a [imath]2 \times 2[/imath] matrix [imath]R[/imath] in which [imath]R[/imath] is not equal to [imath]I[/imath], but [imath]R^7=I[/imath]. I thought to start with the matrix [imath] \begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x\end{bmatrix}[/imath] was a good place to start but what I got when trying a sample run on my calculator was that it comes out to the same matrix as [imath]R[/imath]. Any suggestions of how I should approach this? Or should I use a very specific value for [imath]x[/imath]? The value I used on my calc was [imath]\pi/3[/imath] just to avoid it being equal to the identity matrix. | 2491555 | Need a matrix to seventh power which is the identity but the original is not or the negative identity
The question on the homework wants a [imath]2\times2[/imath] matrix [imath]R[/imath], but [imath]R[/imath] not equal to positive or negative identity, that when you raise to seventh power you get the identity. [imath]R^7=\operatorname{Id}[/imath]. I thought it would have to be some ones and negative ones, it no luck, then tried some ones and some [imath].5[/imath]. Just using an online matrix calculator to do brute force. We just started talking about eigenvalues with visual example, so assuming wouldn't need that since we have not talked about how to find them yet. Any hints or ideas to get me back on track? |
2495424 | Let [imath]A[/imath] be nonempty set, check whether [imath]\subseteq[/imath] is a partial order on [imath]P(A)[/imath].
How can I show that this is a partial order? did not really help because I got confused I found it to be *Transitive since if [imath]A\subseteq A[/imath] and [imath]A\subseteq B[/imath] then [imath]A\subseteq B[/imath] *Antisymmetric since [imath]A\subseteq A[/imath] and [imath]A\subseteq A[/imath] so [imath]A=A[/imath] *Reflective since [imath]A\subseteq A[/imath] and [imath]A\subseteq A[/imath] But I am not sure if my way of solving this or the answer is whether true or not, can anyone confirm? | 371536 | How can I show that this is a partial order?
Let [imath]S[/imath] be an arbitrary amount and define the relation [imath]R \subseteq \mathcal{P}(S) \times \mathcal{P}(S)[/imath] so that [imath](A,B) \in R[/imath] if and only if [imath]A \supseteq B[/imath]. Here [imath]\mathcal{P}(S)[/imath] is a spelling for the power set to [imath]S[/imath]. Show that [imath](\mathcal{P}(S),R)[/imath] is a partially ordered quantity. |
2496674 | Is the integral of the product of separable functions equal to the product of their integrals
For two continuous functions [imath]f(x)[/imath] and [imath]g(y)[/imath] let their integrals be given by: [imath]I_1 = \int f(x)\,\mathrm{d}x[/imath] [imath]I_2 = \int g(y)\,\mathrm{d}y[/imath] Then under which conditions is it true that [imath]\int \int f(x)\cdot g(y)\,\mathrm{d}y\,\mathrm{d}x = I_1 \cdot I_2[/imath] That is to say that the integral of their product is equal to the product of their integrals? | 2250993 | When the integral of products is the product of integrals.
I'm self-studying and was doing the following integral: [imath]I = \int \frac{e^{\frac{1}{x}+\tan^{-1}x}}{x^2+x^4} dx [/imath] I solved it fine by letting [imath] u = \frac{1}{x} + \tan^{-1}x[/imath]. My question is about an alternative method I saw in which it seems the product rule was not applied: [imath] I = \int \left(\frac { e^{\frac{1}{x}}} {x^2}\right) \left( \frac{e^{\tan^{-1}x}}{x^2+1}\right) dx [/imath] [imath] = \int \frac {e^{\frac{1}{x}}}{x^2} dx \cdot \int \frac{e^{\tan^{-1}x}}{x^2+1}dx[/imath] Completing the work following this step leads to the same solution as I originally found. It is this step that has confused me. I have checked using Wolfram and the two statements are equivalent but I do not understand why. Why are we able to write the integral of products as the product of integrals here, and not apply the product rule? Thanks in advance. |
2496666 | which of the following, inequalities is correct?
let n ≥ 3 be an integer . then the statement [imath] ({n!}) ^{\frac{1} {n} } [/imath] ≤ [imath]\frac{n+1}{2}[/imath] is A) true for every n ≥ 3 B)true if and only if n ≥ 5 C) not true for n ≥ 10 D) true for even integers n≥ 6,not true for odd n ≥ 5 my attempts ; i was using the stirling formula [imath] n! = e^{-n} {n^{n}} [/imath],, now im getting [imath] ({n!}) ^{\frac{1} {n} } [/imath] [imath] = (e^{-n} {n^{n}})^\frac{1}{n} [/imath] = [imath] \frac{n}{e} [/imath] ≤ [imath]\frac{n+1}{2}[/imath]..which is true for every n ≥ 3....so correct option is option A). Is my answer is correct or not .pliz verified and tell me the solution i would be more thankful.. | 1552666 | Prove, [imath]\sqrt{n} \le (n!)^\frac{1}{n} \le \frac{n+1}{2} [/imath]
Prove the following inequality, [imath]\sqrt{n} \le (n!)^\frac{1}{n} \le \frac{n+1}{2} \ \ \ \ \forall \ n\in \mathbb{N} [/imath] |
2250189 | [imath]\operatorname{rank}(g\circ f)\le \min\{\operatorname{rank}(f), \operatorname{rank}(g)\}[/imath] proof?
I have these two linear transformations [imath]f:V \to W[/imath] and [imath]g: W \to Z[/imath], where [imath]V, W, Z[/imath] are vector spaces and [imath]V[/imath] and [imath]W[/imath] are subspaces of [imath]Z[/imath]. Now I need to prove that [imath]\operatorname{rank}(g \circ f) \le \min\{\operatorname{rank}(f), \text{rank}(g)\}[/imath] I kind of see it intuitively, but have no idea how to prove it. Any ideas? Thanks. | 1564579 | Prove that rank [imath](g \circ f)\le \min[\operatorname{rank}(f),\operatorname{rank}(g)][/imath]; [imath]g[/imath] and [imath]f[/imath] are linear mappings
Prove that rank [imath](g \circ f)\le \min[\operatorname{rank}(f),\operatorname{rank}(g)][/imath], where [imath]g:V \to W[/imath] and [imath]f:U \to V[/imath] are linear mappings and [imath]U,V,W[/imath] are vector spaces. I do not understand how it can be less than? Can someone explain the proof to me. |
2487159 | Find the mean and variance of a random variable that is the absolute value of a normal distribution
Let [imath]X \sim \text{Normal}(\mu, \sigma^2)[/imath]. Let [imath]Y = |X|[/imath]. Find the pdf of [imath]Y[/imath], as well as [imath]\mathsf E(Y)[/imath] and [imath]\mathsf{Var}(Y)[/imath]. I'm assuming if I can just find the pdf, it'll be simple to find the expected value and variance. I've done method of transformations before, but the absolute value is just throwing me off. EDIT: I found the pdf to be [imath]\frac 1{\sigma\sqrt{2\pi}}e^\frac {-(y-\mu)^2}{\sigma^2} - \frac 1{\sigma\sqrt{2\pi}}e^\frac {-(y+\mu)^2}{\sigma^2}[/imath], which is the folded normal. However, I still cannot figure out how to get [imath]E(Y)[/imath] and [imath]Var(Y)[/imath]. | 1248334 | P.d.f of the absolute value of a normally distributed variable
I came across this question as an exercise, had a brief idea, but didn't know how to proceed. Let [imath]X \sim N(0, 1)[/imath]. What is the p.d.f of [imath]|X|[/imath] ? I know the final p.d.f looks just like the right half of the original pdf, but extended vertically for a factor of 2. Could someone please show mathematically what the p.d.f of variable |X| is going to be, and explain briefly? Thank you. |
2489494 | [imath]AA^H[/imath] and [imath]A^HA[/imath] has same eigen values and any else they don't share is zero
If [imath]A[/imath] is any [imath]m\times n[/imath] matrix then [imath]AA^H[/imath] and [imath]A^HA[/imath] are respectively [imath]m\times m[/imath] and [imath]n \times n[/imath] matrix. Then I observed that both the matrix share the same eigenvalue and if anyone has any other eigenvalue, that is [imath]0[/imath]. I could not figure out the reason. Thanks in advance for any help. | 1087064 | Non-zero eigenvalues of [imath]AA^T[/imath] and [imath]A^TA[/imath]
As a part of an exercise I have to prove the following: Let [imath]A[/imath] be an [imath](n \times m)[/imath] matrix. Let [imath]A^T[/imath] be the transposed matrix of [imath]A[/imath]. Then [imath]AA^T[/imath] is an [imath](n \times n)[/imath] matrix and [imath]A^TA[/imath] is an [imath](m \times m)[/imath] matrix. [imath]AA^T[/imath] then has a total of [imath]n[/imath] eigenvalues and [imath]A^TA[/imath] has a total of [imath]m[/imath] eigenvalues. What I need to prove is the following: [imath]AA^T[/imath] has an eigenvalue [imath]\mu \not = 0[/imath] [imath]\Longleftrightarrow[/imath] [imath]A^TA[/imath] has an eigenvalue [imath]\mu \not = 0[/imath] In other words, they have the same non-zero eigenvalues, and if one has more eigenvalues than the other, then these are all equal to [imath]0[/imath]. How can I prove this? Thanks and regards. |
2496146 | Let [imath]f(x) =\dfrac{\sin x}{x}[/imath] prove that : [imath]|f^{(n)}(x)|<\frac{1}{n+1}[/imath]
let [imath]x>0[/imath] And [imath]f(x) =\dfrac{\sin x}{x}[/imath] prove that for every [imath]n[/imath] : [imath]|f^{(n)}(x)|<\frac{1}{n+1}[/imath] for [imath]f^{(1)}[/imath] we have : [imath]\frac{x\cos x-\sin x}{x^2}<\frac{1}{2}[/imath] for [imath]f^{(2)}[/imath] we have : [imath]\frac{x^2(-x\sin x)-2x(x\cos x-\sin x)}{x^2}<\frac{1}{3}[/imath] Now what ? | 2496101 | Proving that: [imath]\forall n \in \mathbb{N} :|f^{(n)}(x)|\leq \frac{1}{n+1}[/imath]
Suppose [imath]x>0[/imath] and we have function [imath]f(x)=\frac{\sin x}{x}[/imath] how can we show [imath]\forall n \in \mathbb{N} :|f^{(n)}(x)|\leq \frac{1}{n+1}[/imath] I need a hint to show this property .Thanks in advance . I tried for [imath]n=1 ,2[/imath] by finding maximum of [imath]|f'| ,|f''|[/imath] but I get stuck to show for [imath]n[/imath] |
2496891 | Vertices of [imath]G[/imath] covered entirely by disjoint [imath]k[/imath]-stars
Let [imath]k \in \mathbb{N}[/imath] with [imath]k \geq 2[/imath], and consider a bipartite graph with vertex classes [imath]A,B[/imath] such that [imath]\mid B \mid = k\mid A \mid[/imath]. Using Hall's theorem on a suitable graph [imath]G'[/imath], show that if [imath] \mid N(S)\mid \geq k\mid S \mid[/imath] for all [imath]S \subseteq A[/imath] then the vertices of [imath]G[/imath] can be covered entirely by disjoint [imath]k[/imath]-stars (a [imath]k[/imath]-star is a tree with [imath]k+1[/imath] vertices and [imath]k[/imath] leaves) | 2494948 | Use Hall's Theorem to show if [imath]|N(S)|\geq k|S|[/imath] for all [imath]\emptyset \neq S \subseteq A[/imath], vertices of [imath]G[/imath] can be covered entirely by disjoint [imath]k[/imath]-stars.
Let [imath]k \in \mathbb N[/imath] with [imath]k\ge2[/imath]. Consider a bipartite graph [imath]G[/imath] with vertex classes [imath]A,B[/imath], where [imath]|B| = k|A|[/imath]. Use Hall's Theorem applied to a suitable graph [imath]G'[/imath] to show that if [imath]|N(S)|\geq k|S|[/imath] for all [imath]\emptyset \neq S \subseteq A[/imath], then the vertices of [imath]G[/imath] can be covered entirely by disjoint [imath]k[/imath]-stars. (A [imath]k[/imath]-star is a tree with [imath]k+1[/imath] vertices and [imath]k[/imath] leaves.) |
2493768 | Conditional expectation inequality (short)
I am stumped with starting this problem. Could I please get some hints/ advice on tackling it: Show [imath] \mathbb{E}[X\mid X \leqslant x] \leqslant \mathbb{E}[X] [/imath] Thanks in advance! EDIT (Solution): Following Marcus's hint: [imath]\mathbb{E}[X]=\mathbb{P}[X≤x]\mathbb{E}[X|X≤x]+\mathbb{P}[X>x]\mathbb{E}[X|X>x][/imath] and noting [imath]\mathbb{E}[X|X≤x] \leqslant x < \mathbb{E}[X|X>x][/imath] it follows that [imath]\mathbb{E}[X] \geqslant \mathbb{P}[X≤x]\mathbb{E}[X|X≤x]+\mathbb{P}[X>x]\mathbb{E}[X|X≤x] = \mathbb{E}[X|X≤x]. [/imath] Note: It has been brought to my attention that there is an existing question on here almost identical to mine. It does not however have such a complete answer and so I leave it to the operators to remove this post. | 585214 | Showing that [imath]E[X|X is smaller or equal than E[X] for all x[/imath]
I would like to show that: [imath]\hspace{2mm} E[X|X<x] \hspace{2mm} \leq \hspace{2mm} E[X] \hspace{2mm} [/imath] for any [imath]x[/imath] X is a continuous R.V. and admits a pdf. I'm guessing this isn't too hard but I can't come up with a rigorous proof. Thanks so much. |
2497338 | proof check, ([imath] E(X^p)=\int_0^\infty px^{p-1}\mathbb{P}\{ X>x \}dx [/imath])
Problem: If [imath]X ≥ 0,[/imath] then for every [imath]p \in R_+,[/imath] [imath] E(X^p)=\int_0^\infty px^{p-1}\mathbb{P}\{ X>x \}dx [/imath] Show this, using Fubini's theorem with the product measure [imath] \mathbb{P}\times Leb, [/imath] after noting that [imath] X^{p}(\omega)=\int_0^\infty px^{p-1}dx=\int_0^\infty px^{p-1} \chi_{\{ X>x \}}(\omega) dx. [/imath] Actually, I have problems with the notations. I have passed measure theory but I have problems with this probability theory. This a problem from Erhan Çinar's book. Here is what I have done Let [imath] \mu [-\infty , x]= P(X\leq x) [/imath] so [imath] \mu [x , +\infty]= P(X\geq x) [/imath] then [imath] \int_0^\infty d\mu(y)=\int_x^\infty \mathbb{P}\{ X>y \}dy, [/imath] Now I want to write [imath]\int_0^\infty px^{p-1}\mathbb{P}\{ X>x \}dx=\int_0^\infty px^{p-1}(\int_x^\infty d\mu(y))dx=\int_0^\infty\int_0^y px^{p-1}\mathbb{P}\{ X>y \}dxd\mu(y)= \int_0^\infty y^{p}\mathbb{P}\{ X>y \}d\mu(y)=E(X^p)[/imath] Please let me know where my problem is? | 985099 | Explain why [imath]E(|X|^p)=p\int_{0}^{\infty}y^{p-1}P(|X|>y) dy[/imath]
I'm trying to understand a passage from the book: A Basic Course in Probability Theory, Rabi Bhattacharya Edward C. Waymire, in the page 21. The calculation is the following: If [imath]X[/imath] is a random variable on [imath](\Omega,\mathcal{ F}, \mathbb{P} )[/imath], then for any [imath]p > 0, x ≥ 0[/imath], writing [imath]x^p=p\int_0^x y^{p−1} dy[/imath] in the formula [imath]\mathbb{E}|X|^p = \int_Ω |X(\omega)|^p \mathbb{P} (d\omega)[/imath] we obtains: [imath] \mathbb{E}|X|^p=\int_{\Omega}\left(p\int_0^{|X(\omega)|}y^{p-1}dy \right)d \mathbb{P} ~~ {\color{red}=^{??}}~ p\int_{0}^{\infty}y^{p-1}\mathbb{P}(|X|>y) dy [/imath] I do not understand how the left side of the above equation became the right hand side. I know that the Fubini's theorem should be used but do not know the complete details. |
2497174 | Quotient Ring has No Zero Divisors
Let [imath]R[/imath] be a ring without identity and with no zero divisors. Let [imath]S[/imath] be the ring whose additive group is [imath]R \times \Bbb{Z}[/imath]. Let [imath]A = \{(r,n)\in S \mid rx + nx = 0 \text{ for every } x \in R\}[/imath]. (a) [imath]A[/imath] is an ideal in [imath]S[/imath]. (b) [imath]S/A[/imath] has an identity and contains a subring isomomorphic to [imath]R[/imath]. (c) [imath]S/A[/imath] has no zero divisors. Multiplication on [imath]R \times \Bbb{Z}[/imath] is defined like so: [imath](r,n) \cdot (s,k) = (rs + kr + ns,nk)[/imath]. I was able to prove parts (a) and (b), but I am having trouble with part (c). I could use a hint. On a somewhat related note (perhaps I might stretch the definition of "related"), to prove that the quotient of a principal ideal ring is a principal ideal ring, wouldn't I just use the fact that [imath]f : R \to R/I[/imath] defined by [imath]f(r) = r + I[/imath] is a surjective ring homomorphism, and so if [imath]R[/imath] is a principal ideal ring, so is [imath]f(R) = R/I[/imath]? I asked this in chat but didn't receive any responses; and I didn't think this question merited its own post. | 244752 | How to show that the ring [imath]S/A[/imath] has no zero divisors? (Hungerford, Algebra, Problem 12, Chapter III, Section 2)
I am having trouble with another homework problem (Hungerford, Algebra, Problem 12, Chapter III, Section 2). Let [imath]R[/imath] be a ring without identity and with no zero divisors. Let [imath]S[/imath] be the ring whose additive group is [imath]R \times \mathbb{Z}[/imath] with multiplication given by [imath](r_1,n_1)(r_2,n_2)=(r_1r_2+n_2r_1+n_1r_2,n_1n_2).[/imath] Let [imath]A=\{(r,n) \in S \mid rx+nx=0 \text{ for all } x \in R\}[/imath]. (a) [imath]A[/imath] is an ideal in [imath]S[/imath]. (b) [imath]S/A[/imath] has an identity and contains a subring isomorphic to [imath]R[/imath]. (c) [imath]S/A[/imath] has no zero divisors. I've done (a) and (b), but haven't been able to crack (c). Here's what I have so far: (c) Suppose [imath]((r, n) + A)((s, m) + A) = (0, 0) + A[/imath]. Then [imath](r, n)(s, m) \in A[/imath], so [imath](rs + mr + ns)x + (nm)x = 0[/imath] for all [imath]x \in R[/imath]. But [imath](rs + mr + ns)x + (nm)x = rsx + mrx + nsx + nmx = r(sx + mx) + n(sx + mx)[/imath]. Now here we have an expression involving [imath]r(something) + n(something)[/imath] and [imath]s(something) + m(something)[/imath], and I want to say that if it's zero for all [imath]x[/imath] then [imath]rx + nx = 0[/imath] for all [imath]x[/imath] or [imath]sx + mx = 0[/imath] for all [imath]x[/imath], but all I seem to be able to say is that [imath]rx + nx = 0[/imath] for all [imath]x = sx' + mx'[/imath], which is not (or not obviously) what I want. I appreciate any help! |
2497679 | Prove nonexistence of a countable basis.
Let us consider space [imath]R = \mathbb{R}^{\mathbb{N}}[/imath] of infinite sequences of real numbers. I have to prove two claims: Let [imath]a_i=(0,0,\ldots,1,0,0,\ldots)[/imath]; [imath]1[/imath] at i-th place. Then [imath]B=\cup_{i=1}^{\infty}a_i[/imath] is not a basis of [imath]R[/imath]. It can proved with the Zorn lemma that there is basis in [imath]R[/imath]. Prove that there is no countable basis in [imath]R[/imath]. My attempt: Let us consider sequence [imath]A=(1, 1, \ldots)[/imath] (all [imath]1[/imath]). Since linear combination is by definition finite then it is clear that we can not express [imath]A[/imath] as finite linear combination of sequences from [imath]B[/imath], because any linear combination of sequences from [imath]B[/imath] would have only finite number of nonzero elements. Well, honestly I do not know to solve this. I think we should do something like this - assume that there is countable basis [imath]B'[/imath]. Then we should construct a sequence that does not belong to [imath]\operatorname{span}(B')[/imath] explicitly or use some set-theoretic theorems about cardinalities that [imath]\operatorname{span}(B')[/imath] has smaller size compared to [imath]R[/imath]. However while I am pretty sure about the approach we should take, I do not see how to execute it. So my questions are: Is [imath]1[/imath] is correct/make sense? How to approach [imath]2[/imath]? Thanks a lot for your time! | 214984 | Uncountability of basis of [imath]\mathbb R^{\mathbb N}[/imath]
Given vector space [imath]V[/imath] over [imath]\mathbb R[/imath] such that the elements of [imath]V[/imath] are infinite-tuples. How to show that any basis of it is uncountable? |
1205086 | [imath]\mathrm{Aut}(D_4)[/imath] is isomorphic to [imath]D_4[/imath]
Problem statement: I need to find out if [imath]\mathrm{Aut}(D_4)[/imath] is isomorphic to [imath]D_4[/imath] and explain my answer. I already know that it is isomorphic, so now all I need to do is to prove it. I assume that first we need to look where it sends [imath]s[/imath] and [imath]t[/imath] and it must send them to elements that satisfy the same relations() And then need to show that those two are conjugation by some element in [imath]D_4[/imath](?) Any help is appreciated. I was hoping for a duplicate post, but couldn't find it. Thank you! | 3036084 | Automorphism of [imath]D_8[/imath]
I am trying to prove that [imath]Aut(D_8) \equiv D_8[/imath]. It is not hard to see that [imath]\lvert Aut(D_8)\rvert = 8[/imath]. Indeed, it is at most [imath]8[/imath] as [imath]r[/imath] (canonical rotation) has order [imath]4[/imath] and [imath]s[/imath] (canonical reflection) has order [imath]2[/imath]. On the other hand, [imath]D_8[/imath] is normal in [imath]D_{16}[/imath] which acts by conjugation on [imath]D_8[/imath]. So, the order is exactly [imath]8[/imath]. However, I am having troubles to show that [imath]Aut(D_8) \equiv D_8[/imath] (I could check all possible solutions and stop once I have 8 different automorphisms but I wonder if there is a simpler solution). |
2497753 | Unary union proof
given some [imath]n \in \mathbb{N}[/imath], prove that [imath]A \cap ( B_1 \cup \cdots \cup B_n) = (A \cap B_1) \cup \cdots\cup (A \cap B_n)[/imath] Using unary union and intersection and basic set theory. My approach : Let [imath]B = \{ B_i \mid 1\leq i \leq n \}[/imath] And let [imath]C = \{ A \cap B_i \mid 1\leq i \leq n \}[/imath], so now i need to prove that [imath]A \cap \bigcup B = \bigcup C[/imath] but i don't know how to proceed, i know i have to let [imath]x[/imath] be anything and assume that [imath]x \in \bigcup C[/imath] and prove it must be in [imath]x \in A \cap \bigcup B[/imath], but this formal step i don't know how to do it, please help. Update : one should use unary union | 2494972 | Proof in logic for arbitrary number of sets
given [imath]n \in \mathbb{N}[/imath] ,prove that [imath]A \cap (B_1 \cup B_2 \cup \cdots \cup B_n) = (A \cap B_1) \cup (A \cap B_2) \cup \cdots \cup (A \cap B_n) [/imath] ? (As HomeWork) Now i know how to prove using induction or using "without loss of generality", But unfortunately we did not learn them in class so "We don't know them", is there another way to prove the above equality ! |
2498545 | Finding [imath]U(\mathbb{Z}[i])[/imath].
I will put my way here, the way that I'm not sure about, and wait for good suggestions :) Let [imath]a+bi[/imath] be a nonzero element of [imath]\mathbb{Z}[i][/imath]. For [imath]a+bi[/imath] to be a unit, there should be another nonzero element [imath]c+di[/imath] in [imath]\mathbb{Z}[i][/imath] where [imath](a+bi)(c+di)= 1[/imath]. so this system is then obtained: [imath]ac-bd=1[/imath] [imath]ad+bc=0[/imath] Since we have that [imath]a+bi\neq 0[/imath], so [imath]a\neq 0[/imath] or [imath]b\neq0[/imath]. We have 3 cases here: First one: if [imath]a=0[/imath], then [imath]bd=1[/imath] and so [imath]b=\pm1[/imath] and [imath]d=\pm 1[/imath]. Second one: if [imath]b=0[/imath], then [imath]ac=1[/imath] and so [imath]a= \pm1[/imath] and [imath]c= \pm1[/imath]. Third one: if [imath]a\neq 0[/imath] and [imath]b\neq0[/imath]: We have from the second equation of the system above that [imath]d=\frac{-bc}{a}[/imath], but [imath]d[/imath] is an integer, so we have 2 cases here, first is that [imath]a=b[/imath] then [imath]d=-c[/imath], and this will give, from the first equation of the system above that [imath]a=\frac{1}{2c}[/imath] which is not an integer, so this case is excluded. Second one is that [imath]a=c[/imath], then [imath]d=-b[/imath] and so, from the first equation of the system [imath]c^2 + b^2 =1[/imath] which gives either [imath]c=0[/imath] and [imath]b=\pm1[/imath] or [imath]b=0[/imath] and [imath]c=\pm1[/imath]. Then finding [imath]a[/imath] and [imath]d[/imath] from the system above, we conclude that [imath]U(\mathbb{Z}[i])=[/imath] {[imath]\pm1, \pm i[/imath]}. In the third case I put, I feel like it's not correct to decide that if [imath]d=\frac{-bc}{a}[/imath] and [imath]d[/imath] is an integer, then either [imath]b=a[/imath] or [imath]c=a[/imath] without knowing that this fraction can't be simplified more. Any suggestions for improvements are welcomed. | 527711 | Prove that [imath]ℤ[i]^*= \{1,-1,i,-i\}[/imath]
Prove that [imath]ℤ[i]^*= \{1,-1,i,-i\}[/imath]. [imath]\{1,-1,i,-i\} ⊂ ℤ[i]^*[/imath] is trivial. But I'm not sure about the other inclusion. Let [imath](a+bi) \in ℤ[i]^*[/imath]. Then there exist [imath]c,d \in ℤ[/imath] such that [imath](a+bi)(c+di)=1[/imath]. Then [imath]ac+bci+adi-bd=1[/imath]. Then [imath]bc-ad=0[/imath] and [imath]ac=1+bd[/imath]. What can I conclude from this ? |
2498501 | Classifying non-abelian group(s) of order [imath]18[/imath]
Problem: Classify the non-abelian group(s) of order [imath]18[/imath]. I am currently working on the problem above. I think the answer can be [imath]C_{18}[/imath], [imath]D_{18}[/imath], [imath]C_{3} \times C_{3} \times C_{2}[/imath], [imath]S_{3} \times C_{3}[/imath]. Can someone show me what the correct answer is? Thank you. | 1301063 | Classifying groups of order 18
I am trying to classify groups of order 18. So far, I have shown that a group [imath]G[/imath] of order 18 is given by [imath]G\cong C_9 \rtimes_{\varphi} C_2[/imath] or [imath]G\cong (C_3 \times C_3)\rtimes_{\varphi} C_2[/imath]. If [imath]G\cong C_9 \rtimes_{\varphi} C_2[/imath], then [imath]\mid \varphi(1) \mid[/imath] divides [imath]2[/imath], so that [imath]\varphi(1)[/imath] is trivial or inverts a generator of [imath]C_9[/imath]. I concluded that [imath]G \cong C_{18}[/imath] in the former case and [imath]G \cong D_{18}[/imath] in the latter case. I am now considering the case [imath]G\cong (C_3 \times C_3)\rtimes_{\varphi} C_2[/imath], but I am stuck. I have found this article, but they do not do it the same way as me (finding the image of [imath]\varphi(1)[/imath] in [imath]Aut(C_3 \times C_3)[/imath]). Is there a way to do this using my method? My goal is to be able to do questions like this on my algebra qual without having to fiddle around. I was messing around with the fact that [imath]Aut(C_3 \times C_3)\cong GL_2(C_3)[/imath]. |
2498675 | Distribution function of a random variable defined by a cumulative distribution function
I have been given a question: Suppose that [imath]F(x)=P(X\leq x)[/imath]. Show that if [imath]F[/imath] is continuous and [imath]Y:=F(X)[/imath], then [imath]P(Y\leq y)=y[/imath] for all [imath]y\in [0,1][/imath]. I have been having difficult even starting to prove anything, since the definiton of [imath]Y[/imath] seems to me somewhat self refrencing. I will be thankful for any clues. | 2374343 | Proof of [imath]Y=F_X(X)[/imath] being uniformly distributed on [imath][0,1][/imath] for arbitrary continuous [imath]F_X[/imath]
This question is related to Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X, with the difference being that [imath]F_X[/imath] (the probability distribution function of random variable [imath]X[/imath]) is an arbitrary continuous distribution function, not necessarily strictly increasing. I think the proof is similar, but we have to take care of the possibility that [imath]F_X[/imath] may not be [imath]1[/imath]-to-[imath]1[/imath]. I list my attempt below, and would appreciate it if someone can confirm if it's correct, and, in particular, if it can be improved. Thanks a lot! The goal is to show [imath]F_Y(y)=y[/imath] for any [imath]y \in [0,1][/imath]. To do so, note that [imath]F_Y(y)\triangleq\mathbb P(\{Y\le y\})[/imath], and [imath]\{Y\le y\}=\{F_X(X)\le y\}=\{X\in F_X^{-1}([0, y])\}.[/imath] Since [imath]F_X[/imath] is continuous, [imath]F_X^{-1}([0,y])[/imath] must be closed. So it follows that [imath]\sup F_X^{-1}([0, y])=\max F_X^{-1}([0, y])=\max F_X^{-1}(\{y\}),[/imath] which let's denote by [imath]a[/imath]. Therefore, [imath]F_Y(y)=\mathbb P(\{X\le a\})=F_X(a)=y.\quad[/imath] (Q.E.D.) |
2498095 | Decomposition of measure in probability measures
Let [imath]m[/imath] be a [imath]\sigma[/imath]-finite measure on sigma-algebra [imath]A[/imath] of subsets of [imath]X[/imath]. How do I prove that there exist probability measures [imath](m_n)_{n\in \mathbb{N}}[/imath] on [imath]A[/imath] and coefficients [imath]c_n>0[/imath] such that [imath]m(U)=\sum_n c_nm_n(U)?[/imath] I know that a [imath]\sigma[/imath]-finite measure means that there exist subsets [imath](X_n)_{n\in \mathbb{N}}[/imath] such that [imath]X=\cup_n X_n[/imath] and [imath]m(X_n)<\infty[/imath]. And a probability measure is such that [imath]m_n(X)=1[/imath]. But how can I use these to prove the statement? | 2497906 | probability measure on [imath]\sigma[/imath]-algebra
Let [imath]\mu[/imath] be a [imath]\sigma[/imath]-finite measure on a [imath]\sigma[/imath]-algebra [imath]\mathcal{A}[/imath] of subsets of [imath]\Omega[/imath]. Show that there exists probability measures [imath]\mu_n[/imath] on [imath]\mathcal{A}~(n \in \mathbb{N})[/imath] and [imath]a_n>0[/imath] such that for every set [imath]A \in \mathcal{A}:[/imath] [imath]\mu (A)= \sum_{n=1}^{\infty} a_n \mu_n (A).[/imath] Is there anyone can give me hints or show how to start proving this question? |
563219 | Show that there exists a constant [imath]c[/imath] such that [imath]\left|\int_0^b\frac{\sin ax}{x}dx\right|\le c[/imath]
Show that there exists a constant [imath]c[/imath] such that [imath]\left|\int_0^b\frac{\sin ax}{x}dx\right|\le c[/imath] In fact, show that the smallest such number is [imath]c=\int_0^\pi\frac{\sin x}xdx[/imath]. Well, I'm thinking of a change of variable [imath]y=ax[/imath], and we want to show [imath]\left|\int_0^B\frac{\sin x}xdx\right|[/imath] is bounded. Then consider the intervals [imath][n\pi,(n+1)\pi)[/imath] for natural number n. Thank you. | 563483 | Real Analysis, Existence of a Limit Boundedly:[imath]|\int_0^b\frac{sin(ax)}{x}dx | \le c[/imath]
I need to show that there exists a [imath]c[/imath] such that [imath]|\int_0^b\frac{sin(ax)}{x}dx | \le c[/imath], where [imath]a[/imath] is any real number, and [imath]b[/imath] is any real number greater than 0. Also I need to show that the smallest [imath]c[/imath] possible is [imath]c=\int_0^\pi\frac{sinx}{x}dx[/imath]. I know that [imath]\int_0^\infty \frac{sin(ax)}{x}dx = sig(a)\frac{\pi}{2}[/imath] where [imath]sig(a)[/imath] literally just means positive one or negative one, depending on the sign of [imath]a[/imath]. I think the fourier tranform of the characteristic function on the interval [imath](0,a)[/imath] is [imath]\frac{sin(ax)}{x}[/imath]. After that, I am unsure about how to proceed. Thanks for your help! |
2498951 | How to show that [imath]\sum_{n=1}^\infty r^n\cos n\theta=\frac{r\cos\theta-r^2}{1-2r\cos\theta+r^2}[/imath]?
I'm supposed to prove the following series expansion: [imath]s_n=\sum_{n=1}^\infty r^n\cos n\theta=\frac{r\cos\theta-r^2}{1-2r\cos\theta+r^2}[/imath] I first note the following: [imath]s_n=\sum_{n=1}^\infty \Re\left[re^{i\theta}\right]^n=\Re\left[\sum_{n=1}^\infty z^n\right][/imath] Using geometric series definition: [imath]s_n=\Re\left[\frac{1}{1-z}\right][/imath] [imath]s_n=\Re\left[\frac{1-z}{1-2z+z^2}\right][/imath] Substituting polar coordinates of [imath]z[/imath]: [imath]s_n=\Re\left[\frac{1-r\exp(i\theta)}{1-2r\exp(i\theta)+r^2\exp(2i\theta)}\right][/imath] Taking real component: [imath]s_n=\frac{1-r\cos\theta}{1-2r\cos\theta+r^2\cos2\theta}[/imath] I can't really seems to do anything beyond here, could someone point out where I've gone wrong? | 2229735 | Show that [imath]\sum_{n=1}^\infty r^n\cos(n\theta)=\dfrac{r\cos\theta -r^2}{1-2r\cos\theta+r^2}[/imath] whenever [imath]0.[/imath]
Note I have seen that this question has already been posted but I believe my concerns with the question have yet to be answered. Question: Write [imath]z=re^{i\theta}[/imath], where [imath]0<r<1[/imath], in the summation formula [imath]\sum_{n=0}^\infty z^n=\dfrac{1}{1-z}[/imath] whenever [imath]|z|<1[/imath]. Then, with the aid of the following theorem, Suppose that [imath]z_n=x_n+iy_n[/imath] ([imath]n=1,2,\dots[/imath]) and [imath]S=X+iY[/imath]. Then [imath]\sum_{n=1}^\infty z_n=S \text{ if and only if }\sum_{n=1}^\infty x_n=X \text{ and } \sum_{n=1}^\infty y_n=Y[/imath] show that [imath]\sum_{n=1}^\infty r^n\cos(n\theta)=\dfrac{r\cos\theta -r^2}{1-2r\cos\theta+r^2} \text{ and } \sum_{n=1}^\infty r^n\sin(n\theta)=\dfrac{r\sin\theta}{1-2r\cos\theta+r^2}[/imath] whenever [imath]0<r<1[/imath]. Proof: Let [imath]z=re^{i\theta}[/imath], where [imath]0<r<1[/imath]. Recall [imath]\sum_{n=0}^\infty z^n=\dfrac{1}{1-z}[/imath] whenever [imath]|z|<1[/imath]. Replace [imath]z[/imath] by [imath]re^{i\theta}[/imath] in the summation. [imath]\sum_{n=0}^\infty \left(re^{i\theta}\right)^n=\sum_{n=0}^\infty r^ne^{i\theta n}=\dfrac{1}{1-re^{i\theta}}[/imath] whenever [imath]\left|re^{i\theta}\right|<1[/imath]. Replace [imath]e^{i\theta}[/imath] by [imath]\cos\theta +i\sin\theta[/imath]. \begin{equation*} \sum_{n=0}^\infty r^ne^{i\theta n} =\dfrac{1}{1-r\cos\theta-ir\sin\theta} =\dfrac{1-r\cos\theta+ir\sin\theta}{((1-r\cos\theta)-ir\sin\theta)((1-r\cos\theta)+ir\sin\theta)}=\dfrac{1-r\cos\theta+ir\sin\theta}{(1-r\cos\theta)^2+(r\sin\theta)^2}=\dfrac{1-r\cos\theta+ir\sin\theta}{1-2r\cos\theta+r^2\cos^2\theta+r^2\sin^2\theta}=\dfrac{1-r\cos\theta+ir\sin\theta}{1-2r\cos\theta+r^2} \end{equation*} whenever [imath]\left|re^{i\theta} \right|<1[/imath]. Replace [imath]e^{i\theta n}[/imath] by [imath]\cos(n\theta)+i\sin(n\theta)[/imath]. [imath]\sum_{n=0}^\infty r^n(cos(\theta n)+i\sin(\theta n))=\dfrac{1-r\cos\theta}{1-2r\cos\theta+r^2}+i \cdot \dfrac{r\sin\theta}{1-2r\cos\theta+r^2}[/imath] whenever [imath]\left|re^{i\theta}\right|<1[/imath]. By the theorem, we have the next two sums: [imath]\sum_{n=1}^\infty r^n\cos(n\theta)=\dfrac{1-r\cos\theta}{1-2r\cos\theta+r^2}[/imath] and [imath]\sum_{n=1}^\infty r^n\sin(n\theta)=\dfrac{r\sin\theta}{1-2r\cos\theta+r^2}[/imath] whenever [imath]\left|re^{i\theta}\right|<1[/imath]. So my questions are: How can I get [imath]1-r\cos\theta[/imath] to become [imath]r\cos\theta -r^2[/imath]? What do I do with the [imath]n=0[/imath] term of both sums? |
2094601 | Proving that [imath]\lim\limits_{R\to\infty}\int_0^\pi e^{-R\sin\theta}d\theta=0[/imath]
I want to show that the limit below is zero. [imath]\lim_{R\to\infty}\int_0^\pi e^{-R\sin\theta}d\theta[/imath] Wolframalpha and my intuition say that the limit is truly zero but I cannot approach. Any hints will be appreciated. Thanks. | 442950 | Show that: [imath]\lim\limits_{r\to\infty}\int\limits_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0[/imath]
I would like to show [imath]\lim\limits_{r\to\infty}\int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta=0[/imath]. Now, of course, the integrand does not converge uniformly to [imath]0[/imath] on [imath]\theta\in [0, \pi/2][/imath], since it has value [imath]1[/imath] at [imath]\theta =0[/imath] for all [imath]r\in \mathbb{R}[/imath]. If [imath]F(r) = \int_{0}^{\pi/2}e^{-r\sin \theta}\text d\theta[/imath], we can find the [imath]j[/imath]th derivative [imath]F^{(j)}(r) = (-1)^j\int_{0}^{\pi/2}\sin^{j}(\theta)e^{-r\sin\theta}\text d\theta[/imath], but I don't see how this is helping. The function is strictly decreasing on [imath][0,\pi/2][/imath], since [imath]\partial_{\theta}(e^{-r\sin\theta})=-r\cos\theta e^{-r\sin \theta}[/imath], which is strictly negative on [imath](0,\pi/2)[/imath]. Any ideas? |
2499104 | Show that [imath]\|T\|=\|T^*\|[/imath]
Let [imath]T:H \to H[/imath] be a bounded linear transformation between Hilbert spaces. Let [imath]T^*[/imath] be the adjoint of T. Show that [imath]\|T\|=\|T^*\|[/imath]. I know that [imath]\langle Tx,y\rangle=\langle x,T^*y\rangle[/imath], but I don't know how to use it to prove [imath]\|T\|=\|T^*\|[/imath]. | 73670 | Norm of adjoint operator in Hilbert space
Suppose [imath]H[/imath] is a Hilbert space and let [imath]T \in B(H,H)[/imath] where in our notation [imath]B(H,H)[/imath] denotes the set of all linear continuous operators [imath]H \rightarrow H[/imath]. We defined the adjoint of [imath]T[/imath] as the unique [imath]T^* \in B(H,H)[/imath] such that [imath]\langle Tx,y \rangle = \langle x, T^*y\rangle[/imath] for all [imath]x,y[/imath] in [imath]H[/imath]. I proved its existence as follows: Fix [imath]y \in H[/imath]. Put [imath]\Phi_y: H \rightarrow \mathbb{C}, x \mapsto \langle Tx,y \rangle[/imath]. This functional is continuous since [imath]|\langle Tx, y\rangle | \leq ||Tx||\; ||y|| \leq ||T||\; ||x||\; ||y||[/imath]. Therefore we can apply the Riesz-Fréchet theorem which gives us the existence of a vector [imath]T^*y \in H[/imath] such that for all [imath]x \in H[/imath] we have [imath]\langle Tx, y\rangle = \langle x, T^* y\rangle[/imath]. I now have to prove that [imath]||T^*|| = ||T||[/imath]. I can show [imath]||T^*|| \leq ||T||[/imath]: Since the Riesz theorem gives us an isometry we have [imath]||T^*y|| = ||\Phi_y||_{H*} = \sup_{||x||\leq 1} |\langle Tx, y\rangle| \leq ||T||\;||y||[/imath] and thus [imath]||T^*|| \leq ||T||[/imath]. However, I do not see how to prove the other inequality without using consequences of Hahn-Banach or similar results. It seems like I am missing some quite simple point. Any help is very much appreciated! Regards, Carlo |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.