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812191 | Limit with logarithm.
[imath]\lim_{x \rightarrow 0}\left( \frac{1}{\ln(1+x)}-\frac{1}{\ln(x+\sqrt{1+x^2})}\right)[/imath] Can any give the process how to do this problem? | 552661 | Proving [imath]\lim\limits_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right) =-\frac12[/imath]
How can I prove that [imath]\lim_{x\to0}\left(\frac{1}{\log(x+\sqrt{1+x^2})}-\frac{1}{\log(1+x)}\right)=-\frac{1}{2}[/imath] |
464816 | Isomorphism between [imath]B_2[/imath] and [imath]C_2[/imath]
For small values of [imath]l[/imath], isomorphisms occur among certain of the classical algebras. Show that [imath]B_2[/imath] is isomorphic to [imath]C_2[/imath]. Well, both [imath]B_2[/imath] and [imath]C_2[/imath] have dimension [imath]10[/imath]. [imath]B_2[/imath] consists of [imath]5\times 5[/imath] matrices, while [imath]C_2[/imath] consists of [imath]4\times 4[/imath] matrices. It seems to take too much work to exhibit an explicit isomorphism, and then computing to show that [imath]\phi([x,y])=[\phi(x),\phi(y)][/imath], where [imath][x,y]=xy-yx[/imath]. How can we show that the two algebras are isomorphic? | 29043 | Why Lie algebras of type [imath]B_2[/imath] and [imath]C_2[/imath] are isomorphic?
both of Lie algebras of type [imath]B_2[/imath] and [imath]C_2[/imath] have dimension 10 and we can find two basis of them on page 3 in the book: Introduction to Lie algebras and representation theory . How could we show that these two Lie algebras are isomorphic by constructing an explicit correspondence between two basis of them. Thank you very much. |
818981 | Antipodal map commutes with antipodal map?
Suppose we have a closed form [imath]d\omega[/imath] on [imath]S^{n}[/imath], and antipodal map [imath]i: S^{n} \to S[/imath] n i.e [imath]i:x \to −x[/imath]. How to see that the external differential commutes with antipodal map? | 818096 | Antipodal mapping of the sphere
Suppose we have a closed form [imath]d\omega=0[/imath] on [imath]S^{n}[/imath]. If [imath]i: S^{n} \to S^{n}[/imath] is the antipodal map, it induces a decomposition [imath]\Omega^{n}(S^{n})=\Omega^{n}_{+}(S^{n})\oplus \Omega^{n}_{-}(S^{n})[/imath], where [imath]\Omega^{n}_{\pm}(S^{n})[/imath] are invariant or anti-invariont from antipodal map, i.e. [imath]i^{*}\omega=\pm\omega [/imath]. I don't understand why form is contained either in [imath]\Omega^{n}_{+}(S^{n})[/imath] or [imath]\Omega^{n}_{-}(S^{n})[/imath], why there is no such forms on the sphere which does not fall within these classes. And if I'm right, the fact that [imath]i^{*}(d\omega)=d\omega[/imath] or [imath]i^{*}(d\omega)=-d\omega[/imath] induces a decomposition between cohomology [imath]H^{n}(S^{n})=H^{n}_{+}(S^{n})\oplus H^{n}_{-}(S^{n})[/imath]. |
819232 | Algebra - How do I answer this?
Let [imath]T[/imath] be a linear map from [imath]\mathbb{R}^n \to \mathbb{R}^m[/imath] , and let [imath]L[/imath] be a line in [imath]\mathbb{R}^n[/imath] . Show that [imath]T(L)[/imath] is also a line. (Is it not enough to just consider lines in the plane?) | 811778 | Linear transformations map lines to lines
I need help with this question: Let [imath]T[/imath] be a linear map from [imath]\mathbb{R}^n \to \mathbb{R}^m[/imath], and let [imath]L[/imath] be a line in [imath]\mathbb{R}^n[/imath]. Show that [imath]T(L)[/imath] is also a line (is it not enough to just consider lines in the plane). |
516062 | Absolute and Conditional Convergence of the integral [imath]\frac{\sin(x)}{x^p}[/imath] for real values of [imath]p[/imath]
I need to determine the values of p for which this integral converges conditionally and absolutely. [imath]\int_{0}^{\infty} \dfrac{\sin(x)}{x^p} dx [/imath] I think the interval for conditional convergence is [imath]0 < x < 2[/imath] and for absolute convergence the interval is probably [imath]0 < x < 1[/imath]. I'm guessing that I need to somehow compare it with the [imath]\dfrac{1}{x^p}[/imath] integral, but I am not sure exactly where to begin and how to logically proceed. I first thought about dividing the integral into two separate improper integrals, with one of them integrating from [imath]0[/imath] to [imath]1[/imath] and the other from [imath]1[/imath] to infinity, but I do not know how to continue from that point on. | 793595 | Convergence [imath]I=\int_0^\infty \frac{\sin x}{x^s}dx[/imath]
Hi I am trying to find out for what values of the real parameter does the integral [imath] I=\int_0^\infty \frac{\sin x}{x^s}dx [/imath] (a) convergent and (b) absolutely convergent. I know that the integral is convergent if [imath]s=1[/imath] since [imath] \int_0^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}. [/imath] For [imath]s=0[/imath] it is easy to see divergent integral since [imath]\int_0^\infty \sin x\, dx[/imath] is divergent. However I am stuck on figuring out when it is convergent AND or absolutely convergent. I know to check for absolute convergence I can determine for an arbitrary series [imath]\sum_{n=0}^\infty a_n[/imath] by considering [imath] \sum_{n=0}^\infty |a_n|. [/imath] If it helps also [imath]\sin x=\sum_{n=0}^\infty \frac{(-1)^{2n+1}}{(2n+1)!} {x^{2n+1}}[/imath]. Thank you all |
818986 | Infinite abelian group counterexample
A finite group [imath]G[/imath] is abelian iff all its irreducible representation [imath]\rho[/imath] have dimension 1. I'm looking for a counter-example when [imath]G[/imath] is an infinite group. Are there any? EDIT We're dealing with finite representations over [imath]\mathbb C[/imath]. | 358104 | Nonabelian group with all irreducible representations one-dimensional
All irreducible representations of an abelian group are one-dimensional. For a finite group, the coverse is also true - if all irreducible representations are one-dimensional then the group is abelian. Is there a nonabelian group [imath]G[/imath] such that all of its finite-dimensional complex irreducible representations are one-dimensional (such group [imath]G[/imath] is necessarily infinite of course)? |
819892 | Matrix with real entries question
Is there a [imath]2\times2[/imath] matrix [imath]X[/imath] with real entires such that [imath]X^2 +2X = -I_2[/imath] other than [imath]X=I_2[/imath]? If the answer is yes, can someone explain it to me? | 818628 | Real [imath]2\times 2[/imath] matrix [imath]X[/imath] such that [imath]X^2 + 2X= -5I[/imath]
Find a real [imath]2\times 2[/imath] matrix [imath]X = \left(\begin{matrix} a& b\\ c & d\end{matrix}\right)[/imath] such that [imath]X^2 + 2X = -5I.[/imath] With this question I'm kinda lost with the [imath]2X[/imath] part but a full explanation or a little bit of a lead would be greatly appreciated. |
820613 | Rolling [imath]n[/imath] times with an [imath]m[/imath]-sided dice. Closed, finite formula for the distribution of the sum?
My current idea is the following: practically we want to get the distribution of the sum of [imath]n[/imath]-times of a discrete uniform distribution between [imath]1,...,m[/imath] . It is practically the discrete convolution of these, which is a multiplication in the space of the Laplace/Fourier/[imath]z[/imath]-transformations. Going in this direction I found complicated formulas whose de-transformation was already hopeless. (I could write them here on need.) Was it a good direction? Maybe somebody knows the formula, or can show me a better way? | 107329 | Ways of getting a number with [imath]n[/imath] dice, each with [imath]k[/imath] sides
Assume the dice are numbered from [imath]1[/imath] to [imath]k[/imath]. My hunch is that this will form a normal distribution with a median at [imath]n\cdot\frac{k}{2}[/imath]. However, I have no idea as to turn this fact into an answer (I have a minimal knowledge of stats, but I know that I am missing the standard distribution) and this is probably the wrong approach How can I approach and solve this problem? (Aside, this is not for a class, stats or other, so any and all approaches welcome). *Edit: * I want to find the number of ways that the sum of the numbers that are rolled has a particular value, if [imath]n[/imath] dice are rolled, and each has [imath]k[/imath] sides, numbered [imath]1[/imath] to [imath]k[/imath]. |
756620 | Integral [imath]I=\int_0^1 \frac{\arctan\big(\sqrt{x^2 + 2}\big)}{\sqrt{x^2 + 2}(x^2 + 1)}dx[/imath]
Hi I'm trying to show that [imath] I=\int_0^1 \frac{\arctan\big(\sqrt{x^2 + 2}\big)}{\sqrt{x^2 + 2}(x^2 + 1)}dx=\frac{5\pi^2}{96}. [/imath] We can try the substitution [imath]u=(x^2+2)^{1/2}, du=x(2+x^2)^{-1/2}dx[/imath] but that didn't help me much because of the (x^2+1) piece. Any ideas? Thanks. | 580521 | Generalizing [imath]\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{\operatorname dx}{x^{2}+1} = \frac{5\pi^{2}}{96}[/imath]
The following integral \begin{align*} \int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{dx}{x^{2}+1} = \frac{5\pi^{2}}{96} \tag{1} \end{align*} is called the Ahmed's integral and became famous since its first discovery in 2002. Fascinated by this unbelievable closed form, I have been trying to generalize this result for many years, though not successful so far. But suddenly it came to me that some degree of generalization may be possible. My conjecture is as follows: Define the (generalized) Ahmed integral of parameter [imath]p[/imath], [imath]q[/imath] and [imath]r[/imath] by \begin{align*} A(p, q, r) := \int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1}}{q \sqrt{p^{2}x^{2} + 1} } \, \frac{pqr \, dx}{(r^{2} + 1)p^{2} x^{2} + 1}. \end{align*} Now suppose that [imath]p q r = 1[/imath], and define its complementary parameters as \begin{align*} \tilde{p} = r \sqrt{\smash{q}^{2} + 1}, \quad \tilde{q} = p \sqrt{\smash{r}^{2} + 1}, \quad \text{and} \quad \tilde{r} = q \sqrt{\smash{p}^{2} + 1}, \tag{2} \end{align*} Then my guess is that \begin{align*} A(p, q, r) = \frac{\pi^{2}}{8} + \frac{1}{2} \left\{ \arctan^{2} (1 / \tilde{p} ) - \arctan^{2} ( \tilde{q} ) - \arctan^{2} ( \tilde{r} ) \right\}. \end{align*} Plugging the values [imath](p, q, r) = (1/\sqrt{2}, \sqrt{2}, 1)[/imath], the corresponding complementary parameters become [imath](\tilde{p}, \tilde{q}, \tilde{r}) = (\sqrt{3}, 1, \sqrt{3})[/imath]. Then for these choices, the original Ahmed's integral [imath]\text{(1)}[/imath] is retrieved: \begin{align*} \int_{0}^{1} \frac{\arctan \sqrt{x^{2} + 2} }{\sqrt{x^{2} + 2} } \, \frac{dx}{x^{2} + 1} &= \frac{\pi^{2}}{8} + \frac{1}{2} \left\{ \arctan^{2} \frac{1}{\sqrt{3}} - \arctan^{2} 1 - \arctan^{2} \sqrt{3} \right\} \\ &= \frac{5\pi^{2}}{96}. \end{align*} In fact, I have a more generalized conjecture involving dilogarithms depending on complementary parameters. But since this specialized version is sufficiently daunting, I won't deal with it here. Unfortunately, proving this relation is not successful so far. I just heuristically calculated and made some ansatz to reach this form. Can you help me improve the situation by proving this or providing references to some known results? EDIT. I finally succeeded in proving a general formula: let [imath]k = pqr[/imath] and complementary parameters as in [imath]\text{(2)}[/imath]. Then whenever [imath]k \leq 1[/imath], we have \begin{align*} A(p, q, r) &= 2\chi_{2}(k) - k \arctan (\tilde{p}) \arctan \left( \frac{k}{\tilde{p}} \right) \\ &\quad + \frac{k}{2} \int_{0}^{1} \frac{1}{1-k^{2}x^{2}} \log\left( \frac{1+\tilde{p}^{2}x^{2}}{1+\tilde{p}^{2}} \times \frac{1+\tilde{q}^{2}x^{2}}{1+\tilde{q}^{2}} \times \frac{1+\tilde{r}^{2}x^{2}}{1+\tilde{r}^{2}} \right) \, dx. \end{align*} Then the proposed conjecture follows as a corollary. I'm planning to gather materials related to the Ahmed's integrals and put into a combined one. You can find an ongoing proof of this formula here. |
821187 | History of the Coefficients of Elliptic Curves -- Why [imath]a_6[/imath]?
I would like to know what is the motivation behind the naming convention of the Weierstrass form of elliptic curves given as [imath]E:y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6.[/imath] I can see that [imath]a_1,a_2,a_3,a_4[/imath] are named by sorting monomials by lex order with [imath]x<y[/imath]. But why go from [imath]a_5[/imath] to [imath]a_6[/imath]? | 743473 | Reason behind standard names of coefficients in long Weierstrass equation
A long Weierstrass equation is an equation of the form [imath]y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6[/imath] Why are the coefficients named [imath]a_1, a_2, a_3, a_4[/imath] and [imath]a_6[/imath] in this manner, corresponding to [imath]xy, x^2, y, x[/imath] and [imath]1[/imath] respectively? Why is [imath]a_5[/imath] absent? |
815750 | Linear Transformation - linear algebra question
[imath]T:\mathbb{R}_2[x] \mapsto \mathbb{R}_2[x][/imath] s.t.: [imath] \begin{array}{l} T(1) = 3+2x+4x^2, \\ T(x) = 2+2x^2, \\ T(x^2) = 4+2x+3x^2. \end{array} [/imath] Is there base [imath]B[/imath] of [imath]\mathbb{R}_2[x][/imath] that [imath][T]_B = \begin{pmatrix} 2 & 0 & 0 \\-1 & 1 & 0 \\ 5 & 0 & 4 \end{pmatrix}[/imath]? | 819486 | algebra - matrices and polynoms
[imath]T:\mathbb P_2[x]↦\mathbb P_2[x][/imath] s.t.: [imath]T(1)=3+2x+4x^2[/imath], [imath]T(x)=2+2x^2[/imath], [imath]T(x^2)=4+2x+3x^2[/imath]. Is there base [imath]B[/imath] of [imath]\mathbb P_2[x][/imath] that [imath][T]_B[/imath]= \begin{bmatrix} 2 & 0 & 0\\ -1 & 1 & 0\\ 5 & 0 & 4\\ \end{bmatrix} Please help. Thanks. |
821338 | [imath]\sum_{k=1}^{\infty} \frac{a_1 a_2 \cdots a_{k-1}}{(x+a_1) \cdots (x+a_k)}[/imath]
Hey guys I was reading Alfred van der Poortens paper regarding Apery's constant and I came across this pretty equality. For all [imath]a_1, a_2, \ldots[/imath] \begin{align} \large\sum_{k=1}^{\infty}\frac{a_1a_2\cdots a_{k-1}}{(x+a_1)\cdots (x+a_k)} = \frac 1x \end{align} I tried fiddling around with the the summands by trying to use a partial fraction decomposition but it just got messy. link to paper: http://maths.mq.edu.au/~alf/Humid%20Summer/45.pdf | 131004 | Proof that [imath]\sum\limits_{k=1}^\infty\frac{a_1a_2\cdots a_{k-1}}{(x+a_1)\cdots(x+a_k)}=\frac{1}{x}[/imath] regarding [imath]\zeta(3)[/imath] and Apéry's proof
I recently printed a paper that asks to prove the "amazing" claim that for all [imath]a_1,a_2,\dots[/imath] [imath]\sum_{k=1}^\infty\frac{a_1a_2\cdots a_{k-1}}{(x+a_1)\cdots(x+a_k)}=\frac{1}{x}[/imath] and thus (probably) that [imath]\zeta(3)=\frac{5}{2}\sum_{n=1}^\infty {2n\choose n}^{-1}\frac{(-1)^{n-1}}{n^3}[/imath] Since the paper gives no information on [imath]a_n[/imath], should it be possible to prove that the relation holds for any "context-reasonable" [imath]a_1[/imath]? For example, letting [imath]a_n=1[/imath] gives [imath]\sum_{k=1}^\infty\frac{1}{(x+1)^k}=\frac{1}{x}[/imath] which is true. The article is "A Proof that Euler Missed..." An Informal Report - Alfred van der Poorten. |
821788 | How prove this Rāmā ujan Aiya kār identity [imath]\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}=3[/imath]
show that [imath]\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}=3[/imath] I know [imath]3=\sqrt{1+8}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+15}}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot6}}}=\cdots[/imath] | 7204 | Evaluating the nested radical [imath] \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots}}} [/imath].
How does one prove the following limit? [imath] \lim_{n \to \infty} \sqrt{1 + 2 \sqrt{1 + 3 \sqrt{1 + \cdots \sqrt{1 + (n - 1) \sqrt{1 + n}}}}} = 3. [/imath] |
537167 | Please help me prove: [imath]v(a+b)\leq v(a)+v(b)[/imath], and [imath]v(ab)\leq v(a)v(b)[/imath] where [imath]v(x)=\inf{\{\vert x^n \vert}^{1/n}: n\in\mathbb{N}\}[/imath]
I'm reading a book functional analysis, and reading and have seen an example of somebody please help me if you can. The example that I've seen is the following: If [imath]A[/imath] is a normed algebra and [imath]a,b\in A[/imath] so that [imath]ab=ba[/imath], then [imath]v(a+b)\leq v(a)+v(b),[/imath] and [imath]v(ab)\leq v(a)v(b).[/imath] Remarks: Let [imath]A[/imath] is a normed algebra whith unit. For [imath]x\in A[/imath] number [imath]v(x)=\inf{\{\vert x^n \vert}^{1/n}: n\in\mathbb{N}\}[/imath] is spectral radius of element [imath]x[/imath]. Please help me. Thanks for your attention. | 1160581 | an inequality in Banach algebra
Let [imath](V, \| \ \|)[/imath] be a Banach algebra. Given two elements [imath]x,y\in V[/imath] satisfying [imath]xy=yx[/imath], prove that [imath] \lim_{n\to\infty}\|(x+y)^n\|^{1/n}\le\lim_{n\to\infty}\|x^n\|^{1/n}+\lim_{n\to\infty}\|y^n\|^{1/n} [/imath] I just do not know where to start. I want to use this inequality to prove a relationship between the spectrum of [imath]x+y, x, y[/imath]. |
822001 | How to prove continuity of the retraction map in proof of Brouwer's fixed point theorem?
I was reading Allen Hatcher's proof of the Brouwer's fixed point theorem using the no retraction theorem. That is, we want to prove that every continuous [imath]f: \mathbb{D}^n \to \mathbb{D}^n[/imath] has a fixed point [imath]f(x) = x[/imath], by contradiction. The no retraction theorem says there is no continuous [imath]r: \mathbb{S}^{n-1} \to \mathbb{D}^n[/imath] such that [imath]r \mid_{\mathbb{S}^{n-1}}(y) = y[/imath]. i.e. [imath]r[/imath] restricted to [imath]\mathbb{S}^{n-1}[/imath] is the identity functon. Suppose a continuous [imath]f[/imath] exists with no fixed point. Then, define [imath]r(x)[/imath] by the point on the line [imath]tx + (1-t)f(x)[/imath] intersecting [imath]\mathbb{S}^{n-1}[/imath]. The book, and lecture notes that I looked up, say this is clearly continuous. Question: How do we proof that [imath]r(x)[/imath] is continuous? | 803499 | Continuous function on closed unit ball
Take a continuous mapping [imath]f: \bar{B^{n}} \rightarrow \bar{B^{n}}[/imath], where [imath]\bar{B^{n}}[/imath] is a closed unit ball in [imath]\mathbb{R}^{n}[/imath]. Assume that [imath]f(x) \neq x[/imath] for every [imath]x \in \bar{B^{n}}[/imath]. Define another function [imath]r[/imath] by following the directed line segment from [imath]f(x)[/imath] through [imath]x[/imath] to its intersection with [imath]\partial B^{n}[/imath], and let the intersection point be [imath]r(x)[/imath]. Is it immediately evident that [imath]r[/imath] is a continuous function? Is so how does it follow? Thanks. |
639068 | Sum of odd numbers always gives a perfect square.
[imath]1 + 3 = 4[/imath] (or [imath]2[/imath] squared) [imath]1+3+5 = 9[/imath] (or [imath]3[/imath] squared) [imath]1+3+5+7 = 16[/imath] (or [imath]4[/imath] squared) [imath]1+3+5+7+9 = 25[/imath] (or [imath]5[/imath] squared) [imath]1+3+5+7+9+11 = 36[/imath] (or [imath]6[/imath] squared) you can go on like this as far as you want, and as long as you continue to add odd numbers in order like that, your answer is always going to be a perfect square. But how to prove it? | 606483 | How to derive the formula for the sum of the first [imath]n[/imath] odd numbers: [imath]n^2=\sum_{k=1}^n(2k-1).[/imath]
How to derive this formula? [imath]n^2=\sum_{k=1}^n(2k-1).[/imath] |
822542 | Integral with trig functions and substitution
How to integrate: [imath]\int_0^T\cos(T-s)\sin(s)ds[/imath]? I was trying to use [imath]\cos(a+b)=\cos a\cos b-\sin a\sin b[/imath] and substitute [imath]\cos(s)=u \Rightarrow \sin(s)ds=du[/imath] but it does't help. | 813987 | Convolution integral [imath]\int_0^t \cos(t-s)\sin(s)\ ds[/imath]
How can I calculate the following integral? [imath]\int_0^t \cos(t-s)\sin(s)\ ds[/imath] I can't get the integral by any substitutions, maybe it is easy but I can't get it. |
561845 | Coker of powers of an endomorphism
Let [imath]F\in\operatorname{End}_R(M)[/imath], where [imath]M[/imath] is a Noetherian [imath]R[/imath]-module. If [imath]\operatorname{Coker}F[/imath] is of finite length, is Coker and Ker of all powers of [imath]F[/imath] of finite length? Is the condition of being noetherian necessary? | 1074298 | [imath]M[/imath] noetherian, [imath]f[/imath] endomorphism of [imath]M[/imath], [imath]\operatorname{coker}f[/imath] has finite length, then [imath]\operatorname{coker}f^n[/imath] and [imath]\ker f^n[/imath] have finite length.
Let [imath]M[/imath] be noetherian and let [imath]f[/imath] be an endomorphism of [imath]M[/imath]. Suppose that [imath]\operatorname{coker}f[/imath] has finite length. Prove that both [imath]\operatorname{coker}f^n[/imath] and [imath]\ker f^n[/imath] have finite length ([imath]n=1,2,...[/imath]). If [imath]M[/imath] is noetherian, then the intersection of [imath]\operatorname{im}f^n[/imath] and [imath]\ker f^n[/imath] equals [imath]0[/imath] for some [imath]n[/imath]. On the other hand, by Jordan-Holder theorem, [imath]\operatorname{coker}f[/imath] is both noetherian and artinian, but I have no idea to continue. I needs some hints to prove it. Thank you in advance. |
631278 | An equivalence for [imath]\operatorname{grade}(I,M)\ge 2[/imath].
Let [imath]R[/imath] be a Noetherian ring, [imath]M[/imath] a finite [imath]R[/imath]-module and [imath]I[/imath] an ideal of [imath]R[/imath]. Show that [imath]\operatorname{grade}(I,M)\ge 2[/imath] iff the canonical homomorphism [imath]M \mapsto\operatorname{Hom}_R(I,M)[/imath] is an isomorphism. This question is Exercise 1.2.24 in the book: Winfried Bruns and Jürgen Herzog, Cohen-Macaulay Rings, Cambridge University Press, 1998. | 19098 | On the grade of an ideal
I need to prove the following statment (actually a special case of it). Let [imath]R[/imath] be a Noetherian ring, [imath]M[/imath] a finite [imath]R[/imath]-module and [imath]I[/imath] an ideal of [imath]R[/imath]. Then [imath]\operatorname{grade}(I,M)\geq 2[/imath] if and only if the homomorphism [imath]M\rightarrow[/imath]Hom[imath]_R(I,M)[/imath] given by [imath]m\mapsto(i\mapsto im)[/imath] is an isomorphism. This is Exercise 1.2.24 in Bruns and Herzog, Cohen-Macaulay Rings. |
822820 | How to find matrix [imath]A[/imath] given [imath]Ax=b[/imath]. Also [imath]det(A)[/imath] & [imath]sum(A)[/imath] are known.
Here is an example: [imath]A = \begin{bmatrix} 2 & 3 \\ 5 & 1 \end{bmatrix}[/imath] [imath]x = \begin{bmatrix} 3 \\ 8 \end{bmatrix}[/imath] [imath]b = Ax[/imath] so [imath]b = \begin{bmatrix} 30 \\ 23 \end{bmatrix}[/imath] Now i want to find [imath]A[/imath] using [imath]x[/imath] and [imath]b[/imath] matrices. How can i do that ? Some says there is a way to invert a non-square matrix (something like right and left) so i can find [imath]A[/imath], but i dunno how to do that. Additional: What if i had [imath]det(A)[/imath] and sum of entries in [imath]A[/imath] (which is [imath]-13[/imath] & [imath]11[/imath] in this case). | 778421 | Solving for [imath]A[/imath] in [imath]Ax = b[/imath]
How does one solve for the matrix [imath]A[/imath] in the system [imath] Ax = b[/imath] when say, [imath]A[/imath] is a [imath]3\times 3[/imath] matrix and [imath]x[/imath] and [imath]b[/imath] are both [imath]3 \times 1[/imath]. [imath]x[/imath] and [imath]b[/imath] are obviously known. The obvious approach would be to Multiply [imath]A[/imath] and [imath]x[/imath] and compare coefficients. How does one solve this in a general setting. |
823132 | Let [imath]p[/imath] be a prime number such that [imath]p \equiv 3 \pmod 4[/imath]. Show that [imath]x^2 \equiv -1 \pmod p[/imath] has no solutions.
Let [imath]p[/imath] be a prime number such that [imath]p \equiv 3 \pmod 4[/imath]. Show that [imath]x^2 \equiv -1 \pmod p[/imath] has no solutions. I noticed that this is equivalent to proving [imath]x^2\equiv 2(2k+1) \pmod p[/imath]. I also know that [imath]x^2 \neq 2(2k+1)[/imath]. But I still can't prove it. Any help would be greatly appreciated. | 142007 | Prove that [imath]x^{2} \equiv -1 \pmod p[/imath] has no solutions if prime [imath]p \equiv 3 \pmod 4[/imath].
Assume: [imath]p[/imath] is a prime that satisfies [imath]p \equiv 3 \pmod 4[/imath] Show: [imath]x^{2} \equiv -1 \pmod p[/imath] has no solutions [imath]\forall x \in \mathbb{Z}[/imath]. I know this problem has something to do with Fermat's Little Theorem, that [imath]a^{p-1} \equiv 1\pmod p[/imath]. I tried to do a proof by contradiction, assuming the conclusion and showing some contradiction but just ran into a wall. Any help would be greatly appreciated. |
823241 | If [imath]H[/imath] is a normal subgroup of group [imath]G[/imath] of odd order and [imath]|H|=5[/imath] . show that [imath]H \subseteq Z(G)[/imath]
If [imath]H[/imath] is a normal subgroup of group [imath]G[/imath] of odd order and [imath]|H|=5[/imath] . show that [imath]H \subseteq Z(G)[/imath] Attempt: If [imath]|H|=5[/imath] and if [imath]H[/imath] is normal, then [imath]H[/imath] must be a normal cyclic subgroup. [imath]\implies H = \{e,a,a^2,a^3,a^4\}[/imath] for some [imath]a \in G[/imath] Since, [imath]H[/imath] is normal, [imath]\implies gH = Hg \implies g a^i = a^j g[/imath] And now, we need to prove that [imath]i=j[/imath]. [imath]G[/imath] is of odd order and [imath]|H|=5 \implies[/imath] index of [imath]H[/imath] in [imath]G = [G:H] = |G|/|H|=2m-1[/imath] is odd [imath]\implies g^{2m-1} \in H~~\forall~ g \in H[/imath] How do I proceed further? Help will be appreciated. Thank you Attempt 2: By the [imath]N/C[/imath] Theorem : Since [imath]H[/imath] is a normal subgroup of [imath]G \implies N(H) =G.[/imath] Now, [imath]H[/imath] is cyclic of order [imath]5 \implies H \subseteq C(H)[/imath] and [imath]|Aut(H)|=1,2[/imath] or [imath]4[/imath] CASE [imath]1[/imath] : [imath]|Aut(H)|=1 \implies |G|=|C(H)| \implies C(H)=G[/imath] CASE [imath]2[/imath]: [imath]|Aut(H)|=2 \implies C(H)[/imath] is normal in [imath]G ~~ i.e.~~ C(H) \vartriangleleft G[/imath]. CASE [imath]3[/imath]: [imath]|Aut(H)|=4 \implies G/C(H) \approx Z_2 \bigoplus Z_2 [/imath] or [imath]~ G/C(H) \approx Z_4[/imath] We need to show that [imath]\forall h \in H,~~ hg~=~gh ~~\forall g \in G[/imath]. I am kind of stuck again. Help will be appreciated. Thank you | 267980 | [imath] N [/imath] normal in a finite group [imath] G [/imath], [imath] |N| = 5 [/imath] and [imath] |G| [/imath] odd. Why is [imath] N \subseteq Z(G) [/imath]?
Suppose that [imath] N [/imath] is a normal subgroup of a finite group [imath] G [/imath]. If [imath] |N| = 5 [/imath] and [imath] |G| [/imath] is odd, why is [imath] N [/imath] contained in [imath] Z(G) [/imath], the center of [imath] G [/imath]? I know how to do this when [imath] |N| = 2 [/imath] and [imath] |G| [/imath] is even, but am not sure what to do with this one. Sylow's theorems maybe? Thank you guys! |
823704 | Combinatorics - binary - very interesting question
I tried to solve this problem without any success. I really hope that you will know the answer. [imath]f(n,m)[/imath] presents the number of binary strings (empty string included) that include at most n times '1', and at most m times '0'. Prove: [imath]f(n,m) = \binom{n+m+2}{n+1}-1[/imath] | 823333 | Combinatorica - binary order
i tried to solve this question without success. do anyone know the answer? f(n,m) presents the number of binary strings (empty string included) that include at most n-1 times '1', and at most m times '0'. prove: f(n,m) = [imath]\binom{n+m+2}{n+1}-1[/imath] |
823897 | Find a closed form for [imath]\sum_{k=0}^{n} k^3[/imath]
Find a closed form for [imath]\sum_{k=0}^{n} k^3[/imath]. I would appreciate ideas for approaching questions like this in general as well. Thanks. | 320985 | How to determine equation for [imath]\sum_{k=1}^n k^3[/imath]
How do you find an algebraic formula for [imath]\sum_{k=1}^n k^3[/imath]? I am able to find one for [imath]\sum_{k=1}^n k^2[/imath], but not [imath]k^3[/imath]. Any hints would be appreciated. |
823884 | Wedge product of Lie algebra valued differential forms
Let [imath]\mathfrak{g}[/imath] be the Lie algebra of a matrix Lie group. Furthermore, let us consider the following [imath]\mathfrak{g}[/imath]-valued [imath]p[/imath]-form and [imath]\mathfrak{g}[/imath]-valued [imath]q[/imath]-form: \begin{equation} \begin{array}{cc} \zeta = \zeta^\alpha \otimes T_\alpha \; ,& \eta = \eta^\alpha \otimes T_\alpha \end{array} \end{equation} where [imath]\zeta^\alpha \in \Omega^p(P)[/imath], [imath]\eta^\alpha \in \Omega^q(P)[/imath] and [imath]\{ T_\alpha \}[/imath] is a basis of [imath]\mathfrak{g}[/imath]. Then I'm trying to show that: \begin{equation} \zeta \wedge \eta = T_\alpha T_\beta \zeta^\alpha \wedge \eta^\beta \tag{1} \end{equation} Can someone help me or give me a nudge in the right direction? The only possibility I see is using the definition: \begin{equation} (\zeta \wedge \eta)(X_1,\ldots,X_{p+q}) = \frac{1}{p!q!} \sum\limits_{P \in S_r} \mathrm{sgn}(P) \zeta(X_{P(1)},\ldots,X_{P(p)}) \eta (X_{P(p+1)},\ldots,X_{P(p+q)}) \end{equation} but is don't really understand how to manipulate this in order to get equation [imath](1)[/imath]. I'm not sure if the notation used above is conventional notation for mathematicians, so please let me know if I need to clarify anything. | 315235 | Reference for Lie-algebra valued differential forms
I am learning about vector-valued differential forms, including forms taking values in a Lie algebra. On Wikipedia there is some explanation about these Lie algebra-valued forms, including the definition of the operation [imath][-\wedge -][/imath] and the claim that "with this operation the set of all Lie algebra-valued forms on a manifold M becomes a graded Lie superalgebra". The explanation on Wikipedia is a little short, so I'm looking for more information about Lie algebra-valued forms. Unfortunately, the Wikipedia page does not cite any sources, and a Google search does not give very helpful results. Where can I learn about Lie algebra valued differential forms? In particular, I'm looking for a proof that [imath][-\wedge -][/imath] turns the set of Lie algebra-valued forms into a graded Lie superalgebra. I would also appreciate some information about how the exterior derivative [imath]d[/imath] and the operation [imath][-\wedge -][/imath] interact. |
823733 | Ito's process and martingale
Let [imath]{W_t}[/imath] be 1 dim Brownian motion and [imath]X_t:=\exp(t/2)\cos W_t[/imath] [imath]t\in[0,T][/imath]. Show that [imath]X_t[/imath] is martingale. My try is below. I understood [imath]df(t,W_t)=-\exp(t/2)\sin xdW_t[/imath] , but I don't know why it become [imath]X_t=1-\int_0^t \sin X_sdW_s[/imath]. [imath]f(0,X_0)=e^0 \sin X_0=1[/imath]? [imath]X_0=?[/imath] Please tell me. | 820068 | martingale and stochastic Integral
Let [imath]{W_t}[/imath] be 1 dimension Brownian motion and [imath]X_t:=\exp(t/2)\cos W_t[/imath] [imath]t\in[0,T][/imath]. Show that [imath]X_t[/imath] is martingale. I understood [imath]df(t,W_t)=-\exp(t/2)\sin xdW_t[/imath] , but I don't know why it become [imath]X_t=1-\int_0^t \sin X_sdW_s[/imath]. [imath]f(0,X_0)=e^0 \sin X_0=1[/imath]? [imath]X_0=?[/imath] Please tell me. |
824279 | How does one calculate [imath]\int_0^1 \frac {\arcsin(x)}{x}dx[/imath]?
How can I evaluate the following? [imath]\int_0^1 \frac {\arcsin(x)}{x}dx.[/imath] Could not find a primitive, so I went for some other methods like arranging it as a double integral or introducing a parameter, but it was in vain... | 823923 | Compute [imath]\int_0^1 \frac{\arcsin(x)}{x}dx[/imath]
[imath]\int_0^1 \frac{\arcsin(x)}{x}dx[/imath] This is a proposed for a Calculus II exam, and I have absolutely no idea how to solve it. Tried using Frullani or Lobachevsky integrals, or beta and gamma functions, but I can't even find a way to start it. Wolfram Alpha gives a kilometric solution, but I know that cannot be the only answer. Any help appreciated! |
825498 | If [imath]\lim_{x\to \infty} f(x) + f'(x) = L[/imath]...
How can I prove [imath]\lim_{x\to \infty} f(x) = L[/imath] ? I tried to prove by L'Hospital's Rule, but I only proved when the limit exists... How can I prove when I am not provided that the limit exists? | 648652 | If [imath]\lim_{x\to\infty}(f(x)+f'(x))=L[/imath] show that [imath]\lim_{x\to\infty} f(x) = L[/imath] and [imath]\lim_{x\to\infty} f'(x) = 0[/imath]
Let [imath]f:(0,\infty) \to R[/imath] be differentiable. Suppose that [imath]\lim_{x\to\infty}(f(x)+f'(x))=L[/imath]. Show that [imath]\lim_{x\to\infty} f(x) = L[/imath] and [imath]\lim_{x\to\infty} f'(x) = 0[/imath]. (Hint: Write [imath]f(x) = e^xf(x)/e^x[/imath] and use l’Hopital’s Rule.) My working for [imath]\lim_{x\to\infty}f'(x)=0[/imath]: For [imath]\lim_{x\to\infty}f'(x) = 0[/imath], I let [imath]f(x) = e^xf(x)/e^x[/imath] and applied quotient rule which then cancels off [imath]e^{2x}[/imath] and I'm left with [imath]\lim_{x\to\infty}(f(x)+f'(x)-f(x))[/imath]. Can I then equate this with [imath]\lim_{x\to\infty} \left(f(x)+f'(x)\right) - \lim_{x\to\infty}f(x)[/imath] which then gives [imath]L-L=0[/imath]? Is this step correct? |
825519 | Convergence of real sequence [imath]{\frac{x_n}{n}}[/imath] as [imath]n[/imath] tends to [imath]\infty[/imath]
Let [imath]{x_n}[/imath] be a real sequence such that [imath]\lim_{n\to\infty}(x_{n+1}-x_n)=c[/imath]. Then, talk about convergence of the sequence [imath]{\frac{x_n}{n}}[/imath] My try: I did not understand how to proceed. I thought of integers and even integers. The former's seq [imath]\frac{x_n}{n}[/imath] converges to 1 and latter to 2. So i think, maybe, the sequence should converge to [imath]c[/imath]. But i have no clue how to prove it. Any hwlp would be appreciated. | 800411 | How to prove these two limits [imath]\lim_{n\to\infty}\frac{a_{n}}{n}[/imath] and [imath]\lim_{n\to\infty}\frac{a_{n+1}-a_{n}}{n}[/imath] exist?
Assume that [imath]\lim_{n\to\infty}(a_{n+2}-a_{n})=A[/imath] show that [imath]\displaystyle \lim_{n\to\infty}\dfrac{a_{n}}{n}[/imath]and [imath]\displaystyle\lim_{n\to\infty}\dfrac{a_{n+1}-a_{n}}{n}[/imath] exist and find these limits. maybe this problem have some methods, Thank you. since [imath]\lim_{n\to\infty}(a_{n+2}-a_{n})=A[/imath] so there exists [imath]N[/imath], and for [imath]\forall \varepsilon>0[/imath], such [imath]|a_{n+2}-a_{n}-A|\le\varepsilon[/imath] |
825437 | Combinatorial identity
Find a formula for: [imath] \binom n0 + 2\binom n1 + 4\binom n2 + 8\binom n3 + \cdots + 2^{n-1}\binom n{n-1} + 2^n\binom nn [/imath] using a counting argument. I just used the binomial t. to show it's [imath]3^n[/imath] but I can't find a combinatorial way. Any help? | 388532 | Weird [imath]3^n[/imath] in an identity to be combinatorially proved
Give a combinatorial proof of the following identity: [imath]3^n=\sum_{i=0}^{n}\binom{n}{i}2^{n-i}[/imath] I can't see any counting argument that would yield [imath]3^n[/imath], and the right hand side is also pretty opaque. For some reason I really really suck at doing these proofs - I just started my first combinatorics course. |
825942 | [imath]\sqrt{X}[/imath] where [imath]X[/imath] is a positive definite matrix is smooth [imath]C^{\infty}[/imath]
I'm trying to prove the following statement. Let [imath]P_n \subset Mat_{nxn}(\mathbb R)[/imath] be the set of all symmetric positive definite matrices with real entries of size [imath]n[/imath]x[imath]n[/imath]. Let [imath]\sqrt{}:P_n \to P_n[/imath] such that [imath]\sqrt{A}=B[/imath] where [imath]B^2=A[/imath] (From Linear Algebra 2, we know that there is such a matrix and there is only one, so this function is well defined). Show that [imath]\sqrt{}[/imath] is smooth [imath]C^{\infty}[/imath], meaning, it is differentiable [imath]\infty[/imath] times and each differential (which can be seen as a linear transformation) is continuous. Would appreciate any push in the right direction. This is not homework, I am preparing for an exam. | 117011 | The square root of positive definite matrix
Let [imath]M[/imath] be the manifold of real positive definite [imath]n \times n[/imath] matrices, define a mapping [imath]i:A \to \sqrt A[/imath] (where [imath]A\in M[/imath] and [imath]\sqrt A[/imath] means the unique positive definite square root of [imath]A[/imath]). Please show that [imath]i[/imath] is smooth. |
827293 | Show that, given an element [imath]\bar{x}\in\mathbb{Q}/\mathbb{Z}[/imath], there is an integer [imath]n \ge 1[/imath] such that [imath]n\bar{x} = 0[/imath].
Consider [imath]\mathbb{Z}[/imath] as a subgroup of the additive group [imath]\mathbb{Q}[/imath] of rational numbers. >Show that, given an element [imath]\bar{x}\in\mathbb{Q}/\mathbb{Z}[/imath], there is an integer [imath]n \ge 1[/imath] such that [imath]n\bar{x} = 0[/imath]. I know that [imath]\mathbb{Q}/\mathbb{Z} = \left \{ \frac{a}{b} + \mathbb{Z} \mid a,b \in Z, b \neq 0 \right \}[/imath]. From that, [imath] \bar{x} = \frac{a}{b} + c[/imath] where [imath]a,b, c \in Z[/imath] and [imath] b \neq 0 [/imath], so [imath]n\left ( \frac{a}{b} + c \right ) = 0[/imath]. But when I try to solve [imath]n\left ( \frac{a}{b} + c \right ) = 0[/imath] the result is always [imath]n=0[/imath]. Any help in solving this is welcome. | 423094 | An integer [imath]n[/imath], such that [imath]nx = 0[/imath], where [imath]x[/imath] belongs to the quotient group [imath]\Bbb Q/\Bbb Z[/imath]
Well, first, let [imath]x[/imath] be an element of the factor group [imath]\mathbb Q/\mathbb Z[/imath] ([imath]\mathbb Q[/imath] and [imath]\mathbb Z[/imath] are the additive groups, by the way). Now, how do we find an integer n, such that, [imath]nx = 0[/imath], [imath]n \ge 0[/imath]? I managed to figure out that each x can be uniquely represented as [imath]x= \dfrac{a}{b} + \mathbb Z[/imath] for [imath]0 \le a< b < 1 , \; \text{gcd}(a,b)=1[/imath]. Now, I've considered that perhaps, I'm not understanding the question. When [imath]nx = 0[/imath], I assume it means, essentially the same thing as [imath]nx =\mathbb Z[/imath], which is the identity element of Q/Z, right? The obvious choice for [imath]n[/imath] would be a multiple of [imath]b[/imath]. Trying [imath]n=b[/imath], we have, [imath]nx = a + b \mathbb Z [/imath], but this is just a single coset of [imath]b \mathbb Z[/imath]. A particular equivalence class [imath](a \mod b)[/imath]. Which brings me to another question: Consider [imath]x+x = \dfrac{a}{b}+ \mathbb Z + \dfrac{a}{b} + \mathbb Z = 2 \dfrac{a}{b}+ \mathbb Z [/imath], But, [imath]2x = \frac{2a}{b} + 2 \mathbb Z [/imath]. I know, I must have slipped with the permissible algebra, but somehow, I just seem to be unable to reconcile the [imath]2[/imath], or choose between them. Which of these is correct? (Perhaps the first one, given that addition is actually the operation of [imath]\mathbb Q/\mathbb Z[/imath]?) Also, how would I go about finding the integer [imath]n[/imath]? I would really appreciate it if, instead of actually finding [imath]n[/imath], someone could hand me a pointer in the right direction, and let me wrangle it myself. Thank You! |
827540 | Proving trigonometric equation [imath]\cos(36^\circ) - \cos(72^\circ) = 1/2[/imath]
Please help me to prove this trigonometric equation. [imath]\cos \left( 36^\circ \right)-\cos \left( 72^\circ \right) = \frac{1}{2}[/imath] Thank you. | 130817 | How do we prove [imath]\cos(\pi/5) - \cos(2\pi/5) = 0.5[/imath] ?.
How do we prove [imath]\cos(\pi/5) - \cos(2\pi/5) = 0.5[/imath] without using a calculator. Related question: how do we prove that [imath]\cos(\pi/5)\cos(2\pi/5) = 0.25[/imath], also without using a calculator |
828024 | how to prove that [imath]n(n+1)[/imath] can't be square?
Prove: for [imath]n[/imath] is positive integer, it's impossible that: [imath]\exists k \in \mathbb Z, n(n+1)=k^{2}[/imath] I know that [imath](n,n+1)=1[/imath], but the process seems odd using it. | 787520 | Show that the product of two consecutive natural numbers is never a square.
I'd like to have my proof verified and if possible, to see other solutions that are interesting. Proof: Suppose [imath]n(n+1)[/imath] is a square. Then we write [imath]n(n+1) = \prod_{p} p^{c(p)}[/imath] where [imath]c(p) = a(p) + b(p)[/imath] are such that \begin{align*} n &= \prod_p p^{a(p)} \\ n+1 &= \prod_p p^{b(p)} \end{align*} Now, by our hypothesis, [imath]c(p)[/imath] is even for all primes [imath]p[/imath]. As [imath](n,n+1)=1[/imath] for all [imath]n[/imath], it must be that [imath]a(p)[/imath] and [imath]b(p)[/imath] are even for all primes [imath]p[/imath] and moreover, [imath]a(p) = 0[/imath] whenever [imath]b(p)>0[/imath] and reversely. This indicates that both [imath]n[/imath] and [imath]n+1[/imath] are squares. This is impossible as there are no consecutive squares in the natural numbers. |
828280 | Product of more than two subgroups
Let [imath]A_1,A_2,...,A_n[/imath] be subgroups of [imath]G[/imath] and [imath]H=A_1A_2...A_n[/imath]. Is there any sufficient and necessary condition for [imath]H[/imath] to be subgroup ? When [imath]n=2[/imath], [imath]H[/imath] is a subgroup if and only if [imath]A_1A_2=A_2A_1[/imath]. I do not know if there is any generalization of this principle ? Note: if necessary you may assume [imath]G[/imath] is finite. Edit: This question seems to be already asked here with not accepted answer. | 191690 | When is the product of [imath]n[/imath] subgroups a subgroup?
Let [imath]G[/imath] be any group. It's a well-known result that if [imath]H, K[/imath] are subgroups of [imath]G[/imath], then [imath]HK[/imath] is a subgroup itself if and only if [imath]HK = KH[/imath]. Now, I've always wondered about a generalization of this result, something along the lines of: Theorem: If [imath]H_1, \ldots, H_n[/imath] are subgroups of [imath]G[/imath], then [imath]H_1H_2\dots{H_n}[/imath] is a subgroup if and only if ([imath]\star[/imath]) holds, where [imath](\star)[/imath] is some condition on [imath]H_1, \ldots, H_n[/imath], preferably related to how the smaller products [imath]H_{m_1}\ldots{}H_{m_k}[/imath], for [imath]k < n[/imath], behave. Question 1: Is there such a theorem? I do know, and its easy to prove, that if [imath]H_iH_j = H_jH_i[/imath] for every [imath]i, j[/imath], then the big product is a group, but this is not satisfying since it's far from necessary (just take one of the groups to be [imath]G[/imath], and you need no commutativity at all). Also, I've been told that there is no really satisfactory answer; if that is indeed the case, then my question would be why? In particular: Question 2: Are there really problematic counterexamples where you can see that the behavior of the smaller products has nothing to do with the big product, so that no such a theorem can ever exist? I would appreciate even an answer for the particular case [imath]n = 3[/imath]. Thanks. |
819048 | Evaluating the quotient of Dedekind Eta function
This expression I found in some research paper, which connects quotient of Dedekind eta function and ray class field of conductor N, which in turn gives the value of j-invariants. For [imath]K=\mathbb Q(i)[/imath] and N=3[imath] (conductor), the discriminant of the order is -36.[/imath] Now, j_{1,3}(\tau)=\dfrac{\eta(\tau)^{12}}{\eta(3\tau)^{12}}[imath], where [/imath]\tau=i$. I evaluated this expression in Pari/Gp, and I got the answer in decimals which I found wrong, if some one knows their reply will be of great help | 819074 | How to evaluate the quotient of Dedekind eta function in Pari/Gp
This expression I found in some research paper, which connects quotient of Dedekind eta function and ray class field of conductor N, which in turn gives the value of j-invariants. For [imath]K=\mathbb Q(i)[/imath] and [imath]N=3[/imath] (conductor), the discriminant of the order is -36. Now,[imath]j_{1,3}(\tau)=\frac{\eta(\tau)^{12}}{\eta(3\tau)^{12}}[/imath] where [imath]\tau=i[/imath]. I evaluated this expression in Pari/Gp, and I got the answer in decimals which I found wrong, if some one knows how to evaluate the above quotient, their reply will be of great help |
829082 | Be [imath]G[/imath] a group of order [imath]|G|=2p[/imath] where [imath]p[/imath] is a prime number odd, prove that either [imath] G [/imath] is cyclic or dihedral [imath]G\simeq D_p[/imath] the group of order 2[imath]p[/imath].
Be [imath]G[/imath] a group of order [imath]|G|=2p[/imath] where [imath]p[/imath] is a prime number odd, prove that either [imath] G [/imath] is cyclic or dihedral [imath]G\simeq D_p[/imath] the group of order 2[imath]p[/imath]. I thought the question a little difficult and advanced for my current level, but I have to present it at the University on Monday next, and I'm trying to understand it ... So please help me with a little more details ... | 739237 | Let [imath]G[/imath] be a group of order [imath]2p[/imath] , where [imath]p[/imath] is a prime greater than [imath]2[/imath]. Then, G is isomorphic to [imath]\mathbb{Z}_{2p}[/imath] or [imath]D_p[/imath]
Let [imath]G[/imath] be a group of order [imath]2p[/imath] , where [imath]p[/imath] is a prime greater than [imath]2[/imath]. Then, G is isomorphic to [imath]Z_{2p}[/imath] or [imath]D_p[/imath] . Gallian gives a proof as follows : They prove that G = [imath]\langle a \rangle \bigcup b \langle a \rangle[/imath] where [imath]a[/imath] is an element of order [imath]p[/imath] and [imath]b[/imath] is an element of order [imath]2[/imath].... [imath](1)[/imath] Now consider [imath]ab[/imath]. Since [imath]ab \notin \langle a \rangle[/imath], it is proved that [imath]ab[/imath] has order [imath]2[/imath]. Hence , [imath]ab = (ab)−1 = b^{−1}a^{−1} = ba^{−1} [/imath](as [imath]b[/imath] has order [imath]2[/imath]). Furthermore, [imath]a^ib = ba^{−i} ~\forall ~i \geq 1[/imath] , so this relation completely determines the multiplication table for [imath]G[/imath]. Since the multiplication tables of all non cyclic groups of order [imath]2p[/imath] are uniquely determined , they must all be isomorphic to each other. Attempt : I have trouble understanding the above statement in bold. I do have an intuition that [imath]a^ib = ba^{−i} ~\forall ~i \geq 1[/imath] , might be able to completely determine the multiplication table for [imath]G[/imath] because : Order of [imath]ab = 2[/imath] { as proved in Gallian} Any element in G is of the form [imath]a^l b a^m [/imath] But to prove that the mutliplication tables for all non cyclic groups of order [imath]2p[/imath] are uniquely determined, if we consider the multiplication of two elements [imath]x,y[/imath] [imath]\in G[/imath] , we [imath]x y = a^{l_1} b a^{m_1} a^{l_2} b a^{m_2}[/imath] How is this uniquely determined? And if they are uniquely determined, why should all non cyclic groups of order [imath]2p[/imath] be isomorphic to each other? Attempt : If they are uniquely determined, then i can think of a one to one correspondence between the cayley tables of two groups [imath]G_1[/imath] and [imath]G_2[/imath] . They will surely be onto as well as both [imath]G_1[/imath] and [imath]G_2[/imath] are of order [imath]2p[/imath] . But to prove isomorphism, how do i prove the group function operation preservation? |
829105 | A frightening sum
Let [imath]x_1,\ldots,x_r,y_1,\ldots,y_p,z_0,\ldots,z_r,t_0,\ldots,t_p[/imath] be complex numbers. Let [imath]A[/imath] be the ring generated by these numbers. Prove the following holds in [imath]\mathbb C(A)[/imath]. [imath]\begin{multline*}\sum_{i=1}^r \left(\prod_{j=0}^r(z_j-x_i)\prod_{1\leq j\leq r; j\neq i}\left({x_j-x_i}\right)^{-1}\prod_{j=1}^r(y_j-x_i)\prod_{j=0}^p\left({t_j-x_i} \right)^{-1}\right)\\-\sum_{i=0}^p \left(\prod_{j=0}^r(t_i-z_j)\prod_{j=1}^r {(t_i-x_j)}^{-1}\prod_{j=1}^p(t_i-y_j)\prod_{0\leq j \leq p; j\neq i}\left({t_i-t_j}\right)^{-1}\right) \\=-\left(\sum_{i=1}^rx_i-\sum_{i=0}^rz_i \right)+\left(\sum_{i=1}^py_i-\sum_{i=0}^pt_i \right)\end{multline*}[/imath] This was asked at an oral examination. I post it out of curiosity (no idea how it should be approached). EDIT It is a duplicate of An awful identity | 735770 | An awful identity
We take place on [imath]\mathbb C(x_1,...,x_r,x'_1,...,x'_p,u_0,...,u_r,u'_0,...,u'_p)[/imath] with [imath]r,p\in \mathbb N[/imath] Show that : [imath]\displaystyle{\sum_{i=1}^r \left( \frac{\prod_{j=0}^r (u_j-x_i) \prod_{j=1}^p (x'_j-x_i) }{ \prod_{1\leq j\leq r, j\neq i }(x_j-x_i) \prod_{j=0}^p (u'_j-x_i)} \right)}-\displaystyle{\sum_{i=0}^p \left( \frac{\prod_{j=0}^r (u'_i-u_j) \prod_{j=1}^p (u'_i-x'_j) }{ \prod_{0\leq j\leq p, j\neq i }(u'_i-u'_j) \prod_{j=1}^r (u'_i-x_j)} \right)}=\displaystyle{\sum_{i=1}^p x'_i +\sum_{i=0}^r u_i -\sum_{i=1}^r x_i-\sum_{i=0}^p u'_i}[/imath] I have seen this exercise in an old book at the library. Unfortunately there was no indication how can I solve the problem. Anyway if someone can give me some ideas it will be greatful. NB: Sorry for the title, I have not had other ideas |
829654 | Use induction to prove that [imath]2^n[/imath] divides [imath](2n)![/imath] for any [imath]n\in\mathbb{Z}_{\geq0}.[/imath]
Use induction to prove that [imath]2^n[/imath] divides [imath](2n)![/imath] for any [imath]n\in\mathbb{Z}_{\geq0}.[/imath] How would you do the inductive step for this proof? I have the base case done. | 164611 | Proof that [imath]\frac{(2n)!}{2^n}[/imath] is integer
I am trying to prove that [imath]\dfrac{(2n)!}{2^n}[/imath] is integer. So I have tried it by induction, I have took [imath]n=1[/imath], for which we would have [imath]2/2=1[/imath] is integer. So for [imath]n=k[/imath] it is true, so now comes time to proof it for [imath]k+1[/imath], [imath](2(n+1))!=(2n+2)![/imath], which is equal to [imath]1 \times 2 \times 3 \times \cdots \times (2n) \times (2n+1) \times (2n+2),[/imath] second term would be [imath]2^{n+1}=2 \times 2^n[/imath] Finally if we divide [imath](1 \times 2 \times 3 \times \cdots \times (2n) \times (2n+1) \times (2n+2)[/imath] by [imath]2^{n+1}=2 \times 2^n[/imath],and consider that,[imath](2n)!/(2^n)[/imath] is integer, we get [imath](2n+1) \times (2n+2)/2=(2n+1) \times 2 \times (n+1)/2[/imath], we can cancel out [imath]2[/imath], we get [imath](2n+1)(n+1)[/imath] which is definitely integer. I am curious this so simple? Does it means that I have proved correctly? |
829683 | Finding limit [imath]\lim_{n\to\infty}\frac{x_1+x_2+\dots+x_n}{n}[/imath]
I don't know how to solve the following: Let [imath](x_n)_{n=1}^\infty[/imath] be a sequence with property [imath]\sum_{n=1}^\infty |\frac{x_n}{n}|<\infty[/imath]. Find a limit [imath]\lim_{n\to\infty}\frac{x_1+x_2+\dots+x_n}{n}[/imath]. Any hint is welcome. Thanks in advance. | 793540 | Dirichlet series
Suppose that the series [imath]\sum \limits_{n=1}^{\infty}\dfrac{a_n}{n^{\sigma}} \quad(\sigma>0)[/imath] converges. Prove that [imath]\lim \limits_{n\to \infty}\dfrac{a_1+a_2+\dots+a_n}{n^{\sigma}}=0[/imath]. I applied the Abel transformation, but unsuccessfully. |
829776 | Calculate the number [imath]o(\mathbb{R})[/imath] of open subsets of the real line.
Calculate the number [imath]o(\mathbb{R})[/imath] of open subsets of the real line. I know that the answer is [imath]\mathfrak{c}[/imath] but I don't know how my lecturer got this. I am doing an introductory topology course, so I know a few topological concepts. Such as that the set of all open intervals forms a base for the real numbers. Is this somehow relevant in showing that [imath]o(\mathbb{R}) = \mathfrak{c}[/imath]? | 786678 | Prove that the family of open sets in [imath]\mathbb{R}[/imath] has cardinality equal to [imath]2^{\aleph_0}[/imath]
Let [imath]\mathcal{T}[/imath] be the family of all open sets in [imath]\mathbb{R}[/imath]. Show that [imath]| \mathcal{T}|=2^{\aleph_0}[/imath] [imath]\textbf{My Attempt:}[/imath] I know that [imath]\forall A \in \mathcal{T}[/imath]. [imath]A[/imath] is the countable union of open intervals with rational end points. I want to use the Cantor-Bernstein Theorem. That is I need to find injective functions [imath]f[/imath] and [imath]g[/imath] such that [imath]f: 2^{\aleph_0} \to \mathcal{T}[/imath] and [imath]g: \mathcal{T} \to 2^{\aleph_0}[/imath]. I know each [imath]A \in \mathcal{T}[/imath] is of the form [imath]A = \bigcup_{x \in A} (r_x,s_x)[/imath] where [imath]r_x,s_x \in \mathbb{Q}[/imath]. How can I use this fact to find the injective function [imath]f[/imath] and [imath]g[/imath]? |
828398 | Natural Deduction [imath](∀x∃y (P(x) → Q(y))) → (∃y∀x(P(x) → Q(y)))[/imath]
I am having trouble with this Natural Deduction question [imath](∀x∃y (P(x) → Q(y))) → (∃y∀x(P(x) → Q(y)))[/imath] | 89584 | Prove or disprove validity: [imath](\forall x \exists y (P(x) \supset Q(y))) \supset(\exists y \forall x (P (x) \supset Q(y)))[/imath]
I have working on this formula [imath](\forall x \exists y (P(x) \supset Q(y))) \supset (\exists y \forall x (P (x) \supset Q(y)))[/imath] to either prove or disprove it. First, I tempted to disprove it, but I changed my mind. I wrote down "for all x that there exists some y satisfies corresponding condition", and "there exists some y that for all x satisfies corresponding condition." I think these statements refer to the same idea. Any suggestions? |
829718 | Question about factorial function
Show that [imath]n!=1+\left(1−{1 \over 1!}\right)n+\left(1−{1 \over 1!}+ {1 \over 2!}\right)n(n−1)+\cdots[/imath] I can't figure out how this can be solved . I tried to use binomial theorem but I couldn't prove it . I think that it can be solved using the gamma function but it's not immediately obvious | 829340 | Factorial identity [imath]n!=1+(1-1/1!)n+(1-1/1!+1/2!)n(n-1)+\cdots[/imath]
Show that [imath]\displaystyle{n!=1+\left(1-\frac1{1!}\right)n+\left(1-\frac1{1!}+\frac1{2!}\right)n(n-1)+\cdots}[/imath]. I can't figure out how this can be solved. I tried to use the binomial theorem but I couldn't prove it. Any help will be greatly appreciated. |
829853 | If [imath]a,b[/imath] is an [imath]R[/imath]-sequence, then [imath](ax-b)[/imath] is prime
If [imath]R[/imath] is an integral domain, [imath]a, b\in R, a\neq 0[/imath] and [imath]\bar b[/imath] is not a zero divisor in [imath]R/(a)[/imath]. I'm trying to prove [imath](ax-b)\in R[x][/imath] is prime. This question seems easy but I couldn't prove it, maybe there is some trick which I could solve this problem easily? Thanks | 798439 | If [imath]a,b[/imath] is an [imath]R[/imath]-sequence, then [imath]ax-b[/imath] is prime (Eisenbud, Exercise 10.4)
This is the exercise mentioned above: Let [imath]a,b[/imath] be regular sequence over a domain [imath]R[/imath]. Prove that [imath]ax-b[/imath] is a prime of [imath]R[x][/imath]. Thank you for your answer! |
830508 | Show [imath]\det(A + B) \geq 0[/imath] implies [imath]\det(A^k + B^k) \geq 0[/imath] if [imath]A, B[/imath] commute
Let [imath]A, B[/imath] be two real [imath]n\times n[/imath] matrices that commute. Assume [imath]\det(A + B) \geq 0[/imath]. Show then that [imath]\det(A^k + B^k) \geq 0[/imath] for all [imath]k \geq 1[/imath]. I believe what must happen is find some factorization of [imath]A^k + B^k[/imath] using commutativity of [imath]A, B[/imath]. then we may find some squared terms, which would imply that the determinant is greater than zero. Not sure what factorizations would help. Or perhaps use the fact the [imath]A, B[/imath] can be simultaneously put in upper triangular form by commutativity. | 830376 | Prove or disprove : [imath]\det(A^k + B^k) \geq 0[/imath]
This question came from here. As the OP hasn't edited his question and I really want the answer, I'm adding my thoughts. Let [imath]A, B[/imath] be two real [imath]n\times n[/imath] matrices that commute and [imath]\det(A + B)\ge 0[/imath]. Prove or disprove : [imath]\forall k\in \mathbb{N}^*[/imath] [imath]\det(A^k + B^k) \geq 0[/imath] What I think about it I think we can consider two cases : k even and k odd. For k even, the result is pretty easy considering [imath]\det(A^{2k}+B^{2k})=\det(A^{k}+iB^{k}){\det(A^{k}-iB^{k})}=\det(A^{k}+iB^{k})\overline {\det(A^{k}+iB^{k})}\ge 0[/imath] For k odd, I thought about this equality but I don't know if it can help [imath]A^{2k+1}+B^{2k+1}=(A+B)\left(\sum_{i=0}^{2k} (-1)^i A^{2k-i}B^{i}\right)[/imath] so [imath]\det(A^{2k+1}+B^{2k+1})\ge 0 \iff \det\left(\sum_{i=0}^{2k} (-1)^i A^{2k-i}B^{i}\right)\ge 0[/imath] |
830577 | Show that it doesn't exist any of natural number [imath] n = 4m + 3[/imath] that [imath] n= x^2+y^2 [/imath] for any natural x and y
Show that it doesn't exist any of natural number [imath] n = 4m + 3[/imath] that [imath] n= x^2+y^2 [/imath] for any natural x and y Show that every prime number in form [imath] p=4m+1 [/imath] could be showed as [imath] p = x^2+y^2[/imath] (x and y are natural) I checked it and it | 163519 | [imath]p=4n+3[/imath] never has a Decomposition into [imath]2[/imath] Squares, right?
Primes of the form [imath]p=4k+1\;[/imath] have a unique decomposition as sum of squares [imath]p=a^2+b^2[/imath] with [imath]0<a<b\;[/imath], due to Thue's Lemma. Is it correct to say that, primes of the form [imath]p=4n+3[/imath], never have a decomposition into [imath]2[/imath] squares, because sum of the quadratic residues [imath]a^2+b^2[/imath] with [imath]a,b\in \Bbb{N}[/imath] [imath] a^2 \bmod 4 +b^2 \bmod 4 \le 2? [/imath] If so, are there alternate ways to prove it? |
496255 | Can an integer of the form [imath]4n+3[/imath] written as a sum of two squares?
Let [imath]u[/imath] be an integer of the form [imath]4n+3[/imath], where [imath]n[/imath] is a positive integer. Can we find integers [imath]a[/imath] and [imath]b[/imath] such that [imath]u = a^2 + b^2[/imath]? If not, how to establish this for a fact? | 1076198 | primes of the form [imath]4k+3[/imath] and sums of squares
It is well-known that if [imath]p[/imath] is a prime of the form [imath]4k+3[/imath] and [imath]p|x^2+y^2[/imath] then [imath]p|x[/imath] and [imath]p|y[/imath]. I forget what is the name of this result, and where can I find a proof (please provide a link). |
830706 | Is this set A equipollent to [imath]\mathbb{R}[/imath] or to [imath]\mathbb{R}^\mathbb{R}[/imath]?
I have to find out whether the set A consisting of all bijective functions from [imath]\mathbb{R}[/imath] to [imath]\mathbb{R}[/imath] is equipollent to [imath]\mathbb{R}[/imath] or to [imath]\mathbb{R}^\mathbb{R}[/imath] and prove it. I think it is equipollent to [imath]\mathbb{R}^\mathbb{R}[/imath], but I don't how to prove it. I'm not allowed to use the axiom of choice. Is there someone who knows how to? Thank you. | 388785 | Factorial of Infinite Cardinal
I have been thinking about the following problem: Let [imath]A[/imath] be a set of cardinality [imath]k[/imath] and denote [imath]\sum_A[/imath] the set of all bijection from [imath]A[/imath] to [imath]A[/imath]. Also denote [imath]k! = \mathrm{card}\left(\sum_A \right)[/imath]. Prove that [imath]k!=2^k[/imath]. My proof consists of finding a bijection [imath]F:\sum_A\to P(A)[/imath] which associates each bijection from the left to the set of its fixed points. Then the result would follow. ([imath]P(A)[/imath]=the power set of [imath]A[/imath]). Since this proof seems quite easy I am afraid it is wrong. Can someone enlighten me? Thank you very much! |
831178 | Intuitively, why does [imath]\int_{-\infty}^{\infty}\sin(x)dx[/imath] diverge?
According to Wolfram Alpha, [imath]\int_{-\infty}^{\infty}\sin(x)dx[/imath] does not converge. This makes no sense to me, intuitively, which I'll justify with a plot: As we see, the positive and negative areas 'cancel out', so, for any [imath]\alpha \in \mathbb{R}[/imath], [imath]\int_{-\alpha}^{\alpha}\sin(x)dx=0[/imath] (I'm just thinking geometrically- in no way is this supposed to be a rigourous justification). So, why is is that [imath]\int_{-\infty}^{\infty}\sin(x)dx[/imath] diverges? A natural and intuitive reason, more than a rigourous one, would be best. Thanks | 725025 | Is [imath]\int_{-\infty}^{\infty} \sin x \, \mathrm{dx}[/imath] divergent or convergent?
I was determining whether [imath]\int_{-\infty}^{\infty} \sin x \, \mathrm{dx}[/imath] was divergent or convergent. So, I did the following steps: [imath]\begin{align} \int_{-\infty}^{\infty} \sin x \, \mathrm{dx} &= \int_{0}^{\infty}\sin x \, \mathrm{dx}+\int_{-\infty}^0\sin x \, \mathrm{dx} \\ &=\lim_{t\rightarrow\infty} \left(-\cos x |^{t}_{0}\right) + \lim_{a\rightarrow-\infty} \left(-\cos x |^{0}_{a}\right)\\ &=\lim_{t\rightarrow\infty} -\cos (t) + \cos 0 + \lim_{a\rightarrow-\infty} -\cos 0 + \cos a\\ &=\lim_{t\rightarrow\infty}1 - \cos t + \lim_{a\rightarrow-\infty} -1+\cos a \end{align}[/imath] Now, at this point, it would be reasonable to say that both the limits are undefined and therefore, the integral is divergent but then if I try something like the following \begin{align} \quad\quad&=\lim_{t\rightarrow\infty}1 -\cos t + \lim_{a\rightarrow\infty} -1+\cos a \\ \quad\quad&=\lim_{t\rightarrow\infty}-1 -\cos t + \lim_{a\rightarrow\infty} \cos a+1 \\ \quad\quad&=\lim_{b\rightarrow\infty}-1 +\cos b - \cos b+1 \\ \quad\quad&= 0 \end{align} So, as you can see, it was shown before that the integral is divergent but with some manipulation, we came at an answer of [imath]0[/imath] but is that valid? I assume, a similar technique can be applied to [imath]\int_{-\infty}^{\infty} \frac{1}{x} \, \mathrm{dx}[/imath]. |
134152 | Prove that every set with more than one element has a permutation without fixed points
I cannot prove this statement so need help. This problem is one of exercises right after the chapter about Hausdorff's maximal principle and Zorn's Lemma. Thus, you cannot use the concept of cardinal number. I found the exact same question on this website, but the proof is done by using the concept of cardinal number. Please make sure that this question is at right after introducing the axiom of choice and Zorn's lemma. Let [imath]A[/imath] be any set with more than one element. Prove that there exists a bijective function [imath]f:A \to A[/imath] such that [imath]f(x)\neq x[/imath] for every element [imath]x[/imath] in [imath]A[/imath]. | 1008228 | Does every set have a derangement?
A derangement of a set [imath]A[/imath] is a bijection from [imath]A[/imath] to itself with no fixed points. Is it the case that every infinite set has a derangement? |
832023 | Let S be a subring of R, I and J are ideals of R. Prove that if [imath]S \subseteq I \cup J[/imath] then [imath]S \subseteq I[/imath] or [imath]S \subseteq J[/imath]
The question is this: Let [imath]S[/imath] be a subring of [imath]R[/imath] ([imath]R[/imath] is a commutative ring). [imath]I[/imath] and [imath]J[/imath] are ideals of [imath]R[/imath]. Prove that if [imath]S \subseteq I \cup J[/imath], then [imath]S \subseteq I[/imath] or [imath]S \subseteq J[/imath] My thought is let [imath]a,b \in S \subseteq I \cup J [/imath], let assume [imath]a \in I[/imath] then I want to show that [imath]b \in I[/imath] After that I don't know how to start. | 498463 | if an ideal is contained in union of two ideals then it is wholly contained in one of them.
If an ideal is contained in union of two ideals then it is wholly contained in one of them. Let [imath]A,B,C[/imath] are ideals of a ring [imath]R[/imath] such that [imath]C[/imath] is contained in [imath]A\cup B[/imath]. Then required to prove that either [imath]C\subset A[/imath] or [imath]C\subset B[/imath]. I can't understand how to do it. If [imath]C\subset A[/imath], then nothing to prove so assume [imath]C \not\subset A[/imath] then there exists some [imath]c \in C[/imath] such that [imath]c \notin A[/imath]. Now we have to prove that all [imath]x \in C[/imath] are in [imath]B[/imath]. How to proceed next? Please someone give me hint. |
832553 | how to prove the solution to brainteaser "Urban planning"
"Urban planning" is a brainteaser asking: There are four towns positioned on the corners of a square. The towns are to be joined by a system of roads such that the total road length is minimized. What is the shape of the road? well, the solution and question is as attached: A,B,C,D are the 4 towns, dotted lines show the sequare, solid lines are the solution, [imath]\angle AEB=\angle BFD = \frac{2}{3} \pi[/imath]. It sounds convincing, but, how to prove mathematically the existence of the solution and, the valid of the solution? | 705844 | Connecting square vertexes with minimal road
I have four cities in [imath]A=(0,0),B=(1,0),C=(1,1),D=(0,1)[/imath]. I am asked to build the shortest motorway to connect the cities. How can I do that? I was thinking that first I need some compactness argument to prove that there really is the shortest way to connect cities. Or if one can prove that one can travel via finitely many line segments, the existence of minimal way migth be possible to prove via some olympiad-level inequalities. Then I think I need to find points [imath]E,F[/imath] such that [imath]\angle AEB=\angle CFD=120^\circ[/imath]. But I think it is difficult to prove that such a configuration is really a minimal. Could anyone give hints/solution how to fill the details? Bonus: Has anyone generalized this problem to [imath]n[/imath] cities in a regular [imath]n[/imath]-gon? |
832661 | Finding if a linear transformation exist
I have to following question which I find hard to solve. Is there a linear transformation from [imath]\mathbb{R}^5[/imath] to [imath]\mathbb{R}^4[/imath] such that: [imath]T:\mathbb{R}^5→\mathbb{R}^4[/imath] Such that: [imath]\text{ker}T=\{(x,y,z,w)\in \mathbb{R}^5 \mid x=2y [/imath] and [imath] z=2t=3w\}[/imath] Define it on a basis of [imath]\mathbb{R}^5[/imath] Now I understand that I need to complete the basis to [imath]\mathbb{R}^5[/imath] with the standard basis, but I don't know how. From there, after I found that basis, I can prove that a linear transformation exists (and only one, such that [imath]T(Xi)=Yi[/imath]. Your help is appreciated, thank you. | 813823 | Linear transformations, [imath]R^5,R^4[/imath]
Is there a Linear Transformation [imath]T : R^5 \rightarrow R^4 [/imath] that its [imath]KerT[/imath] is [imath]KerT = \{( x,y,z,t,w) \in R^5 | x = 2y ,and, z = 2t = 3w\}[/imath] Well, I tried to prove that by first saying : [imath]KerT = \{2y,y,2t,t,w\}[/imath] and [imath]x=2,y=1,z=6,t=3,w=2[/imath] Thus [imath]KerT = Sp\{2,1,6,3,2\} [/imath]Thus [imath]dimKerT = 1 [/imath] Well, to show that it just a linear transformation from [imath]R^5[/imath] to [imath]R^4[/imath] with that given [imath]KerT[/imath] I need to first find a base in [imath]R^5[/imath], so I did: [imath]B = \{(2,1,6,3,2),(0,1,0,0,0),(0,0,1,0,0),(0,0,0,1,0),(0,0,0,0,1)\}[/imath] So B is a base in R^5 now. Well, now it's still not enough so I tried to define [imath]ImT[/imath] and use [imath]ImT = \{T(v1),...,T(v_n)\}[/imath] to prove it's linear transformation. So I took ImT like that [imath]T(2,1,6,3,2) = (0,0,0,0,0)[/imath] It's obviously true because [imath]Ker(2,1,6,3,2)= (0,0,0,0,0)[/imath] And then, defining: [imath]T(0,1,0,0,0) = (1,1,1,1,1)[/imath] [imath]T(0,0,1,0,0) = (2,4,5,6,7)[/imath] [imath]T(0,0,0,1,0) = (3,5,6,7,8)[/imath] [imath]T(0,0,0,0,1) = (4,6,7,8,9)[/imath] Does it prove that [imath]T : R^5 \rightarrow R^4 [/imath] is a Linear Transformation when [imath]KerT = \{( x,y,z,t,w) \in R^5 | x = 2y ,and, z = 2t = 3w\}[/imath]? If not, is it even possible to prove it? |
833129 | Calculate [imath]\int_0^\infty\frac{\sin x}xdx[/imath] by integration of a suitable function along given paths
How can I calculate [imath]\int_0^\infty\frac{\sin x}xdx[/imath] by integration of a suitable function along the following paths: where [imath]R[/imath] and [imath]\varepsilon[/imath] are the radius of the shown outer and inner semicircle, respectively. This seems to be quite interesting. How would one do that? | 594641 | Computing [imath]\int_{-\infty}^\infty \frac{\sin x}{x} \mathrm{d}x[/imath] with residue calculus
This refers back to the integral of [imath]\frac{\sin(x)}x = \frac\pi2[/imath] already posted. How do I arrive at [imath]\frac\pi2[/imath] using the residue theorem? I'm at the following point: [imath]\int \frac{e^{iz}}{z} - \int \frac{e^{iz}}{z},[/imath] and I would appreciate any help. |
320441 | Standard deviation of the weighted mean
How do you find the standard deviation of the weighted mean? The weighted mean is defined: [imath]\bar{x}_w = \frac{\sum{wx}}{\sum{w}}[/imath] The weighted standard deviation (since it is not specified, I take it as of the distribution) is defined: [imath]s_w = \sqrt{\frac{N'\sum_{i=1}^N {w_i(x_i-\bar{x}_w)^2}}{(N'-1)\sum_{i=1}^N{w_i}}},[/imath] where [imath]N'[/imath] is the number of nonzero weights, and [imath]\bar x_w[/imath] is the weighted mean of the sample (source) For an unweighted sample, calculating the standard deviation of the mean from the standard deviation of the distribution is described on Wikipedia. How do I calculate it for the weighted mean, and how is the expression derived? | 823125 | Sampling error with weighted mean
I am studying statistics and I am wondering when it comes to standard error or a sampling if the calculation changes when there are weights added. I have a weighted mean: [imath]\mu_{w} = \dfrac{\sum_{i=1}^Nw_ix_i}{\sum_{i=1}^Nw_i}[/imath] and a weighted variance calculated by: [imath]s^2_{w}=\dfrac{\sum_{i=1}^Nw_i}{(\sum_{i=1}^Nw_i)^2-\sum_{i=1}^Nw_i^2}\cdot \sum_{i=1}^N(x_i-\mu)^2[/imath] is the sampling error still calculated as [imath]\text{SE}=\sqrt{\dfrac{s^2_{w}}{n}}[/imath] |
833323 | Why this is true?[imath]\exists x \in U, [P(x) \land Q(x)] \Leftrightarrow [(\exists x \in U, P(x)) \land (\exists x \in U, Q(x))][/imath]
[imath] [\exists x \in U, P(x) \land Q(x)] \Leftrightarrow [(\exists x \in U, P(x)) \land (\exists x \in U, Q(x))] [/imath] | 652415 | Mixing and Distributing Qualifiers ([imath]\forall x[/imath], [imath]\exists x[/imath])
Context I'm having trouble understanding the limited situations in which qualifiers can be distributed. I am given that the rules are: [imath]\forall x\left[P(x)\land Q(x)\right]\equiv\forall xP(x) \land\forall xQ(x) \tag{1}[/imath] [imath]\exists x\left[P(x)\lor Q(x)\right]\equiv\exists xP(x) \lor\exists xQ(x) \tag{2}[/imath] [imath]\forall x\left[P(x)\lor Q(x)\right]\not\equiv\forall xP(x) \lor\forall xQ(x) \tag{3}[/imath] [imath]\exists x\left[P(x)\land Q(x)\right]\not\equiv\exists xP(x) \land\exists xQ(x) \tag{4}[/imath] Questions What are examples of [imath]P(x)[/imath] and [imath]Q(x)[/imath] that illustrate when these statements are true? How can I show these statements always hold using proofs? Why can't "[imath]\forall x[/imath]" be distributed in example 3 or "[imath]\exists x[/imath]" in example 4? |
827431 | What is [imath]f(x)[/imath]?
I have the following question. When [imath]x > 0[/imath], [imath]f(x)[/imath] is differentiable and satisfies that [imath]f(x)=1+\frac{1}{x}{\int_{1}^{x}}f(t)dt.[/imath] Then What is [imath]f(x)[/imath]? Thanks for your help. | 416944 | Find a continuous function [imath]f[/imath] that satisfies...
Find a continuous function [imath]f[/imath] that satisfies [imath] f(x) = 1 + \frac{1}{x}\int_1^x f(t) \ dt [/imath] Note: I tried differentiating with respect to [imath]x[/imath] to get an ODE but you get one that contains integrals - likely difficult to solve. |
833625 | Suppose [imath]G[/imath] is a group generated by elements [imath]x[/imath] and [imath]y[/imath] where [imath]xy^2 = y^3x[/imath] and [imath]yx^3 = x^2y[/imath] What can you prove about [imath]G[/imath]?
Suppose [imath]G[/imath] is a group generated by elements [imath]x[/imath] and [imath]y[/imath] where [imath]xy^2 = y^3x[/imath] and [imath]yx^3 = x^2y[/imath] What can you prove about [imath]G[/imath]? I've just been playing around with the relations but I can't seem to get anywhere. It seems as if I'm going in circles. I don't know what to look for in terms of how the equations should look in order to give me an insight into the structure of the group. So given a group presentation, are there any general strategies of what to look for in the relations (as in how to rearrange them) in order to obtain insight about the group? | 512143 | Elementary manipulation with elements of group
Let [imath](G,*)[/imath] be a group with identity [imath]e[/imath] , let [imath]a,b∈G[/imath] such that [imath]a*b^3*a^{-1}=b^2[/imath] and [imath]b^{-1}*a^2*b=a^3[/imath] , then how do we prove that [imath]a=b=e[/imath] ? |
198575 | Explicit norm on [imath]\mathcal{C}^0(\mathbb{R},\mathbb{R})[/imath]
Do you know an explicit norm on [imath]\mathcal{C}^0(\mathbb{R},\mathbb{R})[/imath]? Using the axiom of choice, every vector space admits a norm but have you an explicit formula on [imath]\mathcal{C}^0(\mathbb{R},\mathbb{R})[/imath]? A related question is: Can we proved that [imath]\mathcal{C}^0(\mathbb{R},\mathbb{R})[/imath] has a norm without the axiom of choice? | 2917697 | Norm in [imath]\mathcal{C}[\Bbb{R},\Bbb{R}][/imath]
I’m interested in expliciting a norm in the space of continuous functions from [imath]\Bbb{R}[/imath] to itself. It does not need to induce a complete metric. |
811393 | Find all entire functions such that [imath]|f(z)|\leq |\sin(z)|[/imath] for all [imath]z\in\mathbb C[/imath].
Find all entire functions such that [imath]|f(z)|\leq |\sin(z)|[/imath] for all [imath]z\in\mathbb C[/imath]. Half an hour ago I asked the same question for [imath]|f(z)|\leq |z^2-1|[/imath] here but this time I cannot say something like [imath]|f(z)|\leq M|z|^m[/imath] so I don't know how to start. The problem is that [imath] \sin(z) = \frac{e^{iz}-e^{-iz}}{2i} [/imath] gets really large if [imath]z[/imath] gets large. | 856565 | Exercise involving Rouché's theorem
Let [imath]f[/imath] be an entire function such that [imath]f(0)=\dfrac{1}{2}[/imath] and [imath]|f(z)|\leq |e^z-\dfrac{1}{2}|[/imath] for all [imath]z \in \mathbb C[/imath]. Prove that [imath]f(z)=e^z-\dfrac{1}{2}[/imath] for all [imath]z \in \mathbb C[/imath]. Suppose [imath]f \neq e^z-\dfrac{1}{2}[/imath], note that since [imath]f(0)=\dfrac{1}{2}[/imath], then [imath]f(z) \neq \dfrac{1}{2}-e^z[/imath]. It follows [imath]|f(z)|<|e^z-\dfrac{1}{2}|[/imath]. Since we are under the hypothesis of Rouché's theorem, we can affirm that [imath]e^z-\dfrac{1}{2}[/imath] and [imath]e^z-\dfrac{1}{2}+f(z)[/imath] have the same number of zeros in all [imath]\mathbb C[/imath]. I am trying to arrive to an absurd but I can't, I would appreciate if someone could help me to arrive to the conclusion [imath]f(z)=e^z-\dfrac{1}{2}[/imath]. |
833187 | Affine [imath]K[/imath]-algebra is Hilbert ring!?
We know that when [imath]F[/imath] is a field, the ring [imath]F[x_1,...,x_n][/imath] is a Hilbert ring, because the field [imath]F[/imath] is a Hilbert ring. My questions: Is any (non-trivial) affine algebra over an algebraically closed field a Hilbert ring? If this is so, is the algebraically closedness necessary?" Grateful! | 416231 | Every finitely generated algebra over a field is a Jacobson ring
Knowing that the polynomial ring in [imath]n[/imath] variables over a field [imath]k[/imath] is a Jacobson ring, how can we prove from it that every finitely generated [imath]k[/imath]-algebra is a Jacobson ring? EDIT: We define a ring to be Jacobson if every prime ideal is an intersection of maximal ideals. |
834984 | 74th derivative
I have this HW where I have to calculate the [imath]74[/imath]th derivative of [imath]f(x)=\ln(1+x)\arctan(x)[/imath] at [imath]x=0[/imath]. I have no Idea where to begin with. Someone in previous post suggested using Taylor, but I can't because we haven't learned it yet | 834949 | second derivative at point where there is no first derivative
I have this HW where I have to calculate the [imath]74[/imath]th derivative of [imath]f(x)=\ln(1+x)\arctan(x)[/imath] at [imath]x=0[/imath]. And it made me think, maybe I can say (about [imath]\arctan(x)[/imath] at [imath]x=0[/imath]) that there is no limit for the second derivative, therefore, there are no derivatives of degree grater then [imath]2[/imath]. Am I right? |
832658 | Homeomorphisms of product spaces: an example
In the first of these lectures (http://www.mpim-bonn.mpg.de/node/4436) given by M. Freedman he says that there exists (compact metric) spaces [imath]X[/imath] and [imath]Y[/imath] such that [imath]X\times S^{1}[/imath] is homeomorphic to [imath]Y\times S^{1}[/imath] but [imath]X\times \mathbb{R}[/imath] is not homeomorphic to [imath]Y\times\mathbb{R}[/imath]. Does anyone know an example of such spaces? | 396608 | Does [imath]X\times S^1\cong Y\times S^1[/imath] imply that [imath]X\times\mathbb R\cong Y\times\mathbb R[/imath]?
This question came up in a recent video series of lectures by Mike Freedman available through Max Planck Institut's website. He proves the "difficult" converse direction, that [imath]X\times \mathbb R\cong Y\times \mathbb R[/imath] implies [imath]X\times S^1\cong Y\times S^1[/imath] using a subtle "push-pull" argument, when [imath]X[/imath] and [imath]Y[/imath] are compact. He goes on to make the remark that the converse holds in the simply connected case by taking universal covers, but it is not obvious in general and there may even be a counterexample. So my question is whether anyone has any ideas about this question. Maybe someone can spot a counterexample in the non-simply connected case? Edit: Here's a link to the video series. If I recall correctly, the push-pull argument is in the first video. |
835254 | Does [imath]\sum_{j = 1}^{\infty} \sqrt{\frac{j!}{j^j}}[/imath] converge?
I need to solve [imath]\sum_{j = 1}^{\infty} \sqrt{\frac{j!}{j^j}}[/imath] Does this converge or diverge and why? | 835311 | Does this Sum converge?
I need to solve [imath]\displaystyle \sum_{j=0}^\infty \sqrt{\frac{j!}{j^j}}[/imath] Does this converge or diverge? |
368448 | Zero-dimensional ideals and finite-dimensional algebras
I encountered in the literature the term zero-dimensional ideal, however I can not find a relevant definition anywhere in Atiyah-MacDonald or Matsumura. In fact, I encounted the statement: [imath]I[/imath] is a zero-dimensional ideal of [imath]k[x_1,\dots,x_n][/imath] [imath]\Leftrightarrow[/imath] [imath]k[x_1,\dots,x_n]/I[/imath] is a finite-dimensional [imath]k[/imath]-vector space, where we assume that [imath]k[/imath] is algebraically closed. Question 1: What is the definition of the dimension of an ideal? Question 2: How do we prove the above statement? Question 3: Is [imath]k[/imath] being algebraically closed a necessary condition for the statement to be true? | 340527 | Artinian affine [imath]K[/imath]-algebra
Let [imath]K[/imath] be a field and [imath]A[/imath] an affine [imath]K[/imath]-algebra. Show that [imath]A[/imath] has (Krull) dimension zero (is artinian) if and only if it is finite dimensional over [imath]K[/imath]. |
835939 | If [imath]\sum a_{n}[/imath] is convergent, prove [imath]\sum \frac{a_{n}}n[/imath] is also convergent
If [imath]\sum [/imath][imath]a_{n}[/imath] is convergent, prove [imath]\sum \frac{a_n}{n}[/imath] is also convergent. I would say that [imath]a_{n}\geq \frac{a_{n}}{n}[/imath] so being [imath]a_{n}[/imath] convergent, [imath]\frac{a_n}{n}[/imath] also has to be convergent. However, I think it is not enough... | 800490 | If [imath]\sum{a_n}[/imath] converges does that imply that [imath]\sum{\frac{a_n}{n}}[/imath] converges?
I know if [imath]\sum{a_n}[/imath] converges absolutely then [imath]\sum{\frac{a_n}{n}}[/imath] converges since [imath]0\le \frac{|a_n|}{n} \le |a_n| [/imath] for all [imath]n[/imath] so it converges absolutely by the basic comparison test and therefore converges. However, I cannot prove the convergence of [imath]\sum \frac{a_n}{n}[/imath] if [imath]\sum{a_n}[/imath] converges but not absolutely even though I suspect it to be true. Can you give me a proof or a counterexample for this? |
835865 | Prove superpolynomial growth rate
Let [imath]p(n)[/imath] be the number of partitions of [imath]n[/imath]. Prove that growth rate of [imath]p(n)[/imath] is superpolynomial, meaning that for every given [imath]k[/imath] there is [imath]p(n)= \omega (n^k)[/imath]. | 564907 | Help understanding solution to growth of partition function
I'm currently a Combinatorics student trying to parse through this solution. I do not understand the proof currently. Any help understanding it is greatly appreciated. Question Let the number of partitions be given by the function [imath]p(n)[/imath], i.e. the number of ways of writing the integer [imath]n[/imath] as a sum of positive integers. Prove that that if [imath]g[/imath] is any polynomial function in [imath]n[/imath], then there exists an integer [imath]N[/imath] such that [imath]g(n)<p(n)[/imath] for all [imath]n>N[/imath]. Do not use the asymptotic Hardy-Ramanujan formula. Provided Proof. Enough to show [imath]n^a<p(n)[/imath] for fixed positive integer [imath]a[/imath]. (This is just saying that we want to show [imath]p(n)[/imath] grows faster than any degree polynomial of [imath]n[/imath], right?) We have that [imath](i_1,i_2,\ldots,i_{a+1},1^{n-i_1-\cdots -i_{a+1}})[/imath] is a partition of [imath]n[/imath] as long as [imath]i_{a+1}\leq i_a\leq \cdots i_1 \leq \lfloor\frac{n}{a+1}\rfloor[/imath]. (I really don't understand this step. What is this partition? Of the coefficients or terms of polynomial [imath]p(n)[/imath]? And how do we know this is [imath]\leq \lfloor\frac{n}{a+1}\rfloor[/imath]?) The number of partitions [imath]n[/imath] of this form is multiset binomial coefficient [imath]\left(\dbinom{\lfloor \frac{n}{a+1}\rfloor}{a+1}\right)=\frac{\lfloor\frac{n}{a+1}\rfloor(\lfloor\frac{n}{a+1}\rfloor+1)\cdots(\lfloor\frac{n}{a+1}\rfloor+a-1)}{(a+1)!}[/imath] (I'm assuming that if I can understand (2.) then I can understand this line) This means [imath]p(n)>\left(\binom{\lfloor \frac{n}{a+1}\rfloor}{a+1}\right)[/imath] for all [imath]n>0[/imath]. Since [imath]a[/imath] is constant, the given fraction grows with order [imath]n^{a+1}[/imath], which means [imath]p(n)>n^a[/imath] for [imath]n[/imath] sufficiently large. (Why does it grow with order [imath]n^{a+1}[/imath]?) Thanks all again in advance! |
836066 | How to show that if [imath]A, B[/imath], and [imath]A + B[/imath] are invertible matrices with the same size, then [imath]A(A^{ −1} + B^{ −1} )B(A + B)^ {−1} = I[/imath]
Show that if [imath]A[/imath], [imath]B[/imath], and [imath]A + B[/imath] are invertible matrices with the same size, then [imath]A(A^{-1} + B^{-1})B(A + B)^{-1} = I[/imath] What does the result in part [imath]1[/imath] tell you about the matrix [imath]A^{-1} + B^{-1}[/imath]? Ok I never came across any identities that would allow me to cancel these values out in the book so far... So I have no idea how I'm supposed to "show" that they are equal... So how do I solve/show this? Note: for those who say this is a duplicate, read both questions again, this question asks how to solve the first part, the other question asks and only answers how to solve the second part. the first part isn,t explained in the other question, only the second part. | 835586 | Properties of invertible matrices
Show that if [imath]A[/imath], [imath]B[/imath], and [imath]A + B[/imath] are invertible matrices with the same size, then [imath]A(A^{-1}+B^{-1})B(A+B)^{-1}=I[/imath] What does the result in the first part tell you about the matrix [imath](A^{-1}+B^{-1})[/imath]? I get the first part. Help me with the second part. |
836640 | How do i prove this is the second derivative
Let [imath]f[/imath] be a real function defined on [imath][a,b][/imath]. Assume [imath]f[/imath] is twice differentiable. How do i prove that [imath]\lim_{h\to 0} \frac{f(a+h)-2f(a)+f(a-h)}{h^2}=f''(a)[/imath]? I know that it can be divided into [imath]f(a+h)-f(a)-(f(a)-f(a-h))[/imath], but next i don't have any idea.. | 491248 | How to prove that [imath]\displaystyle f''(a)=\lim_{h \to 0}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}[/imath]
Let [imath]f:U\longrightarrow \mathbb{R}[/imath] where [imath]U\subset\mathbb{R}[/imath] open set and [imath]f[/imath] twice-differentiable in [imath]a\in U[/imath]. How to prove that [imath]f''(a)=\lim_{h \to 0}\frac{f(a+2h)-2f(a+h)+f(a)}{h^2}[/imath] without using Taylor or L'Hopital's rule. Any hints would be appreciated. |
837346 | General expanded form of [imath](x+y+z)^k[/imath]
(hope it doesn't seem so weird), I'm looking for a general expanded form of [imath](x+y+z)^k, k\in\mathbb{N}[/imath]. [imath]k=1: x+y+z[/imath] [imath]k=2: x^2+y^2+z^2+2xy+2xz+2yz[/imath] [imath]k=3: x^3+y^3+z^3+3xy^2+3xz^2+3yz^2+3x^2y+3x^2z+3y^2z+6xyz[/imath] [imath]k=4: x^4+y^4+z^4+4xy^3+4x^3y+4xz^3+4x^3z+4yz^3 +4y^3z+6x^2y^2+6y^2z^2+6x^2z^2+12x^2yz+12xy^2z+12xyz^2[/imath] The elements are obviously determined by combinations of their powers, whose sum is always [imath]k[/imath]. I just cannot find the algorithm for element's constants | 592907 | find the formula of trinomial expansion
I wonder as if there exist a equivalent forumla to newton binomial [imath](x+y)^n=\sum_{n=1}^{k} n {n\choose k} x^{n-k}y^k[/imath] for three coefficients [imath](a+b+c)^n[/imath] ? |
837372 | Hyberbolic and Circular (Trig) Functions: Why no parabolic?
There are circular (trig) functions which determine all the points on a unit circle: and which relate to the area swept out by an angle subtended on the circle. -- These functions can of course be extended to relations to ellipses as well. There are also hyperbolic functions which determine all the points on a hyperbola: My question is why there are no analogs of these functions for parabolas (the other type of conic section): Here I have defined [imath]\mathrm {sinp}(\theta)[/imath] and [imath]\mathrm {cosp}(\theta)[/imath] to be the x- and y-coordinates of points on a "unit parabola". Is there any good reason why we should have these extremely useful transcendental functions (sin, cos, sinh, cosh, etc), but we can't (or don't) define analogous functions for parabolas? NOTE: I recommended that this post get deleted because Henning's response in Do "Parabolic Trigonometric Functions" exist? explained theoretically why a "parabolic trigonometric function" is different than the circular and hyperbolic trigonometric functions. However johannesvalks' answer to this was very interesting, as well, and probably shouldn't be deleted. | 134906 | Do "Parabolic Trigonometric Functions" exist?
The parametric equation [imath]\begin{align*} x(t) &= \cos t\\ y(t) &= \sin t \end{align*}[/imath] traces the unit circle centered at the origin ([imath]x^2+y^2=1[/imath]). Similarly, [imath]\begin{align*} x(t) &= \cosh t\\ y(t) &= \sinh t \end{align*}[/imath] draws the right part of a regular hyperbola ([imath]x^2-y^2=1[/imath]). The hyperbolic trigonometric functions are very similar to the standard trigonometric function. Do similar functions exist that trace parabolas (because it is another conic section) when set up as parametric equations like the above functions? If so, are they also similar to the standard and hyperbolic trigonometric functions? |
838267 | Let [imath]F[/imath] be a field. If [imath]F^*[/imath] is a cyclic group, must [imath]F[/imath] be a finite field?
Let [imath]F[/imath] be a field. If [imath]F^*[/imath] is a cyclic group, must [imath]F[/imath] be a finite field? (It's well-known that if [imath]F[/imath] is a finite field, [imath]F^*[/imath] is a cyclic group). Thank you in advanced. | 753437 | Why must a field with a cyclic group of units be finite?
Let [imath]F[/imath] be a field and [imath]F^\times[/imath] be its group of units. If [imath]F^\times[/imath] is cyclic show that [imath]F[/imath] is finite. I'm a bit stuck. I know that I can represent [imath]F^\times = \langle u \rangle[/imath] for some [imath]u \in F^\times[/imath] and that we must have that [imath]|F^\times| = o(u)[/imath], I tried assuming [imath]o(u) = \infty[/imath], but I'm not sure exactly where to go from there. I was wondering if I could get a hint. |
838350 | Proof of the derivative of [imath]a^x[/imath]
I've tried for a while myself from first principles and applying various rules, but always end up going in circles. I've gotten as far as [imath] y = a^x [/imath] [imath] \frac{dy}{dx} = a^x \left( \lim_{x \rightarrow 0} \frac{a^h-1}{h} \right) [/imath] but I have no idea how I should go about cancelling the [imath]h[/imath] in the denominator. Any help is appreciated. | 678309 | Derivative of [imath]a^x[/imath] from first principles
I've been trying to think for the past few days how one could differentiate [imath]a^x[/imath] based on the definition that [imath]a^n[/imath] is repeated multiplication, [imath]a^{n/m}=(\sqrt[m]a)^n[/imath], and [imath]a^x[/imath] is the completion of the above function by continuity. With a bit of algebra, the problem quickly reduces to finding the derivative at [imath]0[/imath]: [imath]\lim_{h\to0}\frac{a^h-1}{h}\tag{1}[/imath] And that limit's really got me stumped. Since you'll obviously have to use the definition in some way, I thought I'd replace [imath]h[/imath] with [imath]\frac{1}{n}[/imath] and use the [imath]n[/imath]-th root definition: [imath]\lim_{n\to\infty}n(\sqrt[n]a-1)\tag{2}[/imath] Of course, just because [imath](2)[/imath] exists doesn't automatically imply [imath](1)[/imath] exists, but it might be a first step. Even [imath](2)[/imath] has me stumped, though. |
838120 | area of triangle in terms of sides ratio
In [imath]\triangle ABC[/imath], [imath]X[/imath] and [imath]Y[/imath] are points on the sides [imath]AC[/imath] and [imath]BC[/imath] respectively. If [imath]Z[/imath] is on the segment [imath]XY[/imath] such that [imath]\frac{AX}{XC}=\frac{CY}{YB}=\frac{XZ}{ZY}[/imath], prove that the area of [imath]\triangle ABC[/imath] is given by: [imath] \text{area of }\triangle ABC=\left((\text{area of }\triangle AXZ)^{1/3} +(\text{area of }\triangle BYZ)^{1/3}\right)^3 [/imath] | 836875 | Area of a triangle in terms of areas of certain subtriangles
In triangle [imath]ABC[/imath] , [imath]X[/imath] and [imath]Y[/imath] are points on sides [imath]AC[/imath] and [imath]BC[/imath] respectively . If [imath]Z[/imath] is on the segment [imath]XY[/imath] such that [imath]\frac{AX}{XC} = \frac{CY}{YB} = \frac{XZ}{ZY}[/imath] , then how to prove that the area of triangle [imath]ABC[/imath] is given by [imath][ABC]=([AXZ] ^{1/3} + [BYZ]^{1/3})^3.[/imath] |
837742 | Why is [imath] \mathrm{Frac} ( A / \mathfrak{p} ) = A_{\mathfrak{p}} / \mathfrak{p} A_{\mathfrak{p}} [/imath]?
[imath] A [/imath] is a commutative ring, [imath] \mathfrak{p} \in \mathrm{Spec} A [/imath], [imath] A_{\mathfrak{p}} = ( A \backslash \mathfrak{p} )^{-1} A [/imath], [imath] \mathrm{Frac} ( A / \mathfrak{p} )[/imath] is the field of fractions of [imath] A / \mathfrak{p} [/imath]. I would like to know why is [imath] \mathrm{Frac} ( A / \mathfrak{p} ) = A_{\mathfrak{p}} / \mathfrak{p} A_{\mathfrak{p}} [/imath] ? Thanks a lot. | 239649 | Quotient ring of a localization of a ring
Let [imath]A[/imath] be a commutative ring. Let [imath]P[/imath] be a prime ideal of [imath]A[/imath]. Let [imath]I[/imath] be an ideal of [imath]A[/imath] such that [imath]I \subset P[/imath]. Let [imath]\bar A = A/I[/imath]. Let [imath]\bar P = P/I[/imath]. Is [imath]\bar A_{\bar P}[/imath] isomorphic to [imath]A_P/IA_P[/imath]? |
838536 | show that [imath]f'(x)=0[/imath]
Let, [imath]f[/imath] : $\mathbb{R}[imath]\rightarrow[/imath]\mathbb{R}$ be a twice differentiable function such that [imath]f[/imath]([imath]\frac{1}{2^n}[/imath])= [imath]0[/imath] [imath]\forall[/imath] [imath]n[/imath] [imath]\in[/imath][imath]\mathbb{N}[/imath]. Show that, [imath]f'(0)=0[/imath]=[imath]f''(0)[/imath]. Help me how to proceed. from definition of [imath]f'[/imath] I am unable to do anything! | 802926 | [imath]f:\mathbb{R} \to \mathbb{R}[/imath] st [imath]f(\frac{1}{2^n})=0 \forall n \in \mathbb{N}[/imath] , show [imath]f'(0)=f''(0)=0[/imath]
Let[imath]f:\mathbb{R} \to \mathbb{R}[/imath] be twice differentiable function such that [imath]f(\frac{1}{2^n})=0 \forall n \in \mathbb{N}[/imath]. Show that [imath]f'(0)=f''(0)=0[/imath] Applying mean value theorem on [imath](\frac{1}{2^n},\frac{1}{2^{n-1}})[/imath] we see that [imath]\exists c_n[/imath] st [imath]f'(c_n)=\frac{f(\frac{1}{2^{n-1}})-f(\frac{1}{2^{n}})}{\frac{1}{2^{n-1}}-\frac{1}{2^{n}}}[/imath] , thus [imath]f'(c_n) =0[/imath] [imath]\implies[/imath] [imath]\lim_{n\to \infty}f'(c_n)=0 \implies f'(0)=0[/imath] But I can't use similar argument for [imath]f''[/imath] since it is not known whether it is continuous or not! |
838704 | Definite Integral: [imath]\int_0^1\frac{x^{2} -1}{\log x}\,dx[/imath]
How to figure out the following integral? I have not been able to solve it from some time. [imath]\int_0^1\frac{x^{2} -1}{\log x}\,dx[/imath] | 778278 | Hints on calculating the integral [imath]\int_0^1\frac{x^{19}-1}{\ln x}\,dx[/imath]
I would be happy to get some hints on the following integral: [imath] \int_0^1\frac{x^{19}-1}{\ln x}\,dx [/imath] |
839347 | Collection of sets with a given cardinality [imath]\kappa[/imath] is not set
Show that collection of all sets with cardinality [imath]\kappa\neq0[/imath], is not set. I'll state my approach and I need to see whether this idea is precise/precisable or not : First let [imath]K[/imath] be the set with [imath]card(K)=\kappa.[/imath] Then let [imath]C[/imath] be the collection of all sets with cardinality [imath]\kappa[/imath]. By contrary, suppose it's a set. Now it can be shown that there's a bijection [imath]\varphi:C\rightarrow F[/imath], which [imath]F[/imath] is the collection of all one-to-one function like [imath]f[/imath], with [imath]dom(f)=K[/imath]. I don't know whether [imath]F[/imath] is now a set or not ! But if [imath]F[/imath] is a set, then I think [imath]\displaystyle\bigcup ran(f)[/imath] would be a good set to be considered and to construct set of all sets. Every guidance (even a completely new approach) or correction to my idea is very appreciated. | 21973 | Class of sets of a given infinite cardinality
This question was inspired by the question on examples of classes that are not sets. From the discussion in the comments there, it seems there does not exist a set of all sets of a given cardinality. To me, it seems easy to see that this is true for any nonzero finite cardinality [imath]n[/imath]. Given any set [imath]x[/imath], you can always make a set of size [imath]n[/imath] by adding [imath]n-1[/imath] other sets to [imath]x[/imath] to make a set of size [imath]n[/imath], which exists by repeated use of the pairing axiom. But how would you do this for infinite cardinalities? For suppose [imath]\kappa[/imath] is some given infinite cardinality. To show that the set of all sets of cardinality [imath]\kappa[/imath] does not exist, it seems you would have to show that for any set [imath]x[/imath], there exists a set of that size with [imath]x[/imath] as an element, and then you could take the union of that set to find the set of all sets, and thus a contradiction. But it doesn't seem reasonable to simply say, for a given set of size [imath]\kappa[/imath] if [imath]x[/imath] is in the set, we have no problem. If not, just take an element out of the set and put [imath]x[/imath] in. Something about that seems like it would not be allowed. So how would you do this for infinite cardinalities? |
839213 | Prove that [imath]\cos(z)[/imath] and [imath]\sin(z)[/imath] are surjective over the complex numbers.
I have an exercise that says: (a) Prove that [imath]\cos(z)[/imath] and [imath]\sin(z)[/imath] are surjective functions from [imath]\mathbb C \to \mathbb C[/imath]. (b) Find the solutions of the equation [imath]\cos(z)=\dfrac{5}{4}[/imath]. As far as part (a) goes, I have no idea how to show surjectivity. I know by definition that [imath]\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}[/imath] and [imath]\sin(z)=\dfrac{e^{iz}-e^{-iz}}{2i}[/imath]. I want to show that given [imath]w \in \mathbb C[/imath], there are [imath]z_1, z_2 \in \mathbb C[/imath] such that [imath]\cos(z_1)=w[/imath] and [imath]\sin(z_2)=w[/imath]. I've tried to play with the expressions from above but I couldn't get to anything. I need help on this. For part (b), once I've proved (a), then I know (b) makes sense, i.e., there exist solutions for that equation. [imath]\cos(z)=\dfrac{5}{4}[/imath] iff [imath]\dfrac{e^{iz}+e^{-iz}}{2}=\dfrac{5}{4}[/imath] Again, I have no idea what can I get out from this equation. | 623775 | Proving surjectivity of [imath]\cos(z)[/imath] and [imath]\sin(z)[/imath] and find all [imath]z : \cos(z) \in \mathbb R[/imath] and all [imath]z: \sin(z) \in \mathbb R[/imath]
I am trying to solve the following two problems: 1) Prove that the functions [imath]\cos(z)[/imath], [imath]\sin(z)[/imath] are surjective over the complex numbers. 2) Find all [imath]z \in \mathbb C[/imath]: [imath]cos(z) \in \mathbb R[/imath] and find all [imath]z \in \mathbb C[/imath]: [imath]\sin(z) \in \mathbb R[/imath]. For 1),I've tried to prove it for the function [imath]\cos(z)[/imath] (I suppose the other one is analogue) so I've used the fact that [imath]\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}[/imath]. Let [imath]w \in \mathbb C[/imath], I want to show there exists [imath]z \in \mathbb C : f(z)=w[/imath], i.e., [imath]\dfrac{e^{iz}+e^{-iz}}{2}=w[/imath], multiplying by [imath]2[/imath] and then by [imath]e^{iz}[/imath] yields [imath]e^{2iz}+1=2we^{iz}[/imath] iff [imath]e^{2iz}-2we^{iz}+1=0[/imath]. I don't know if this approach is the correct one but here I've replaced [imath]e^{iz}[/imath] by [imath]x[/imath], so the solutions of the equation would be the roots of the polynomial [imath]p(x)=x^2-2wx+1[/imath], by the quadratic formula, I get that [imath]x \in \{\dfrac{2w+w_0}{2},\dfrac{2w-w_0}{2}\}[/imath], where [imath]w_0^2=4w^2-4[/imath]. Then, [imath]e^{iz} \in \{\dfrac{2w+w_0}{2},\dfrac{2w-w_0}{2}\}[/imath]. At this point I got lost, I would like to explicitly show that [imath]z[/imath] exists and I can't see existence directly from the fact that [imath]e^{iz}=\dfrac{2w+w_0}{2}[/imath] or [imath]e^{iz}=\dfrac{2w-w_0}{2}[/imath]. First I thought of taking logarithm of both sides of the equation ir order to solve for [imath]z[/imath], but this is not a legitimate operation unless [imath]e^{iz} \in \mathbb R[/imath]. I couldn't go any farther. For point 2) I have no idea what to do, should I use the identity [imath]\cos(z)=\dfrac{e^{iz}+e^{-iz}}{2}[/imath]? I would appreciate some help with the two points (specially with point 2), at least in 1) I could do something). Btw, Happy new year! |
837399 | [imath]M_{\mathfrak{p}} \otimes_{R_{\mathfrak{p}}} N_{\mathfrak{p}} = 0[/imath] implies [imath]M_{\mathfrak{p}} = 0[/imath] or [imath]N_{\mathfrak{p}} = 0[/imath]
Studying commutative algebra I've found this statement: If [imath]M[/imath] and [imath]N[/imath] are finitely generated [imath]R[/imath]-modules, with [imath]R[/imath] a commutative ring, and [imath]\mathfrak{p} \subset R [/imath] is a prime ideal, then [imath]M_{\mathfrak{p}} \otimes_{R_{\mathfrak{p}}} N_{\mathfrak{p}} = 0[/imath] implies [imath]M_{\mathfrak{p}} = 0[/imath] or [imath]N_{\mathfrak{p}} = 0[/imath]. Is it true ? Why ? | 321374 | Tensor product of two finitely-generated modules over a local ring is zero
If [imath]R[/imath] is a local ring and [imath]M[/imath] and [imath]N[/imath] are finitely generated [imath]R[/imath]-modules such that [imath]M\otimes N=0[/imath] then how does it follow from Nakayama's lemma that either [imath]M=0[/imath] or [imath]N=0[/imath]? This is an exercise in Atiyah and Macdonald. The part I could not show in the hints is [imath]({M{\otimes}_R N)}_{k}=0[/imath] implies [imath]M_{k}{\otimes }_{k} N_{k}=0[/imath], where [imath]k=R/\mathfrak m[/imath] and [imath]\mathfrak m[/imath] is the maximal ideal of [imath]R[/imath]. |
839555 | [imath]S^{-1}A \cong A[x]/(1-ax)[/imath]
If [imath]A[/imath] is a commutative ring with unit, [imath]a \in A [/imath] and [imath]S = \lbrace a^n \mid n \geq 0 \rbrace [/imath] then there is an isomorphism [imath]S^{-1}A \cong A[x]/(1-ax).[/imath] In fact we can consider the homomorphism [imath]\phi : A[x] \to S^{-1}A[/imath] [imath]p(x) \mapsto p\left(\frac{1}{a} \right)[/imath] Then obviously [imath](1-ax) \subseteq \ker(\phi) [/imath], but why [imath]\ker(\phi) \subseteq (1-ax) [/imath] ? | 767765 | Localisation isomorphic to a quotient of polynomial ring
Let [imath]R[/imath] be a commutative ring and [imath]A=\{1,a,a^2,\dots\}[/imath] for some [imath]a\in R[/imath]. Prove that [imath]A^{-1}R[/imath] is isomorphic to [imath]R[T]/(aT-1)[/imath]. I guess I'm meant to find a surjective homomorphism between [imath]A^{-1}R[/imath] and [imath]R[T][/imath] and then use first isomorphism theorem. What homomorphism should I use? |
839889 | How to solve this equation in general [imath]ax^3+bx^2+cx+d=0 ?[/imath]
For example if you have [imath]ax^2+bx+c=0[/imath] then solution given by formula [imath]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.[/imath] Is there formula like this for [imath]ax^3+bx^2+cx+d=0 ?[/imath] | 61725 | Is there a systematic way of solving cubic equations?
In my text book, to solve cubic equations, I need to find by trial & error what [imath]f(a)[/imath] will make the equation 0. The factor will be [imath](x-a)[/imath] then the other factor will be [imath]Ax^2+Bx+C[/imath] then I can solve using compare coefficient But for te first part, is there any other way apart from trail & error? Or is there some technique to guess the factor of an equation? The question I am currently doing happens to be [imath]2x^3+3x+4=9[/imath] UPDATE: [imath]2x^3+3x+4=9[/imath] should be [imath]2x^3+3x+4=9x^2[/imath] UPDATE 2 So I am going to try substituting [imath]x=\pm{1}, \pm\frac{1}{2}, \pm{2}, \pm{4}[/imath] When [imath]x=-\frac{1}{2}[/imath], one of the provided answers, [imath]2(−0.5)^3−9(−0.5)^2+3(−0.5)+4=-18.75\neq 0[/imath] did I make a mistake? |
839967 | Complex roots of a complex number
I know how to find the roots to the equation [imath]z^n=w[/imath], for [imath]n \in \mathbb{R\setminus\{ 0\}}[/imath] (by writing [imath]w[/imath] as [imath]re^{i(\theta+2k\pi)}[/imath]), and taking the nth root of both sides, which I'm perfectly happy with, since [imath]r^{1/n}[/imath] is easy to find (with a calculator), as is [imath]\frac{\theta+2k\pi}{n}[/imath]. But what happens when we want to solve the equation [imath]z^t=w \tag{*}[/imath] for [imath]t \in \mathbb{C}[/imath]? e.g. how would one go about finding all roots of [imath]z^{2+3i}=1+i?[/imath] Obviously, it's not simply a case of writing [imath]1+i[/imath] in polar form and then taking the [imath](2+3i)[/imath]th root of [imath]1+i[/imath], since we don't know what [imath]\left| 1+i\right|^{1/({2+3i})}=\sqrt{2}^{1/(2+3i)}[/imath] is (or, even, that it exists). Any ideas? Short version: how do I find [imath]t[/imath]th roots of a complex number for [imath]t \in \mathbb{C}[/imath]? Also, is there any way that we can determine how many roots this [imath](*)[/imath] will have? | 476968 | Complex power of a complex number
Can someone explain to me, step by step, how to calculate all infinite values of, say, [imath](1+i)^{3+4i}[/imath]? I know how to calculate the principal value, but not how to get all infinite values...and I'm not sure how to insert the portion that gives me the other infinity values. |
840059 | Is [imath]O_L[/imath] a free [imath]O_K[/imath] module?
This must have been known. Let [imath]L/K[/imath] be a finite number field extension. Then is [imath]O_L[/imath] a free [imath]O_K[/imath] module? How to prove it if so? So far, I know how to prove the following: Let [imath]\alpha_1,\dotsc,\alpha_n\in O_L[/imath] be a [imath]K[/imath]-basis of [imath]L[/imath] and let [imath]d=d_{L/K}(\alpha_1,\dotsc,\alpha_n)[/imath]. Then [imath]dO_L\subset O_K\alpha_1\oplus\dotsm\oplus O_K\alpha_n[/imath]. | 245607 | Integral basis of an extension of number fields
Let [imath]K\subseteq F[/imath] be number fields with ring of integers [imath]\mathcal{O}_K\le \mathcal{O}_F[/imath]. Question: Is [imath]\mathcal{O}_F[/imath] a free [imath]\mathcal{O}_K[/imath]-module ? By the integral basis theorem this is true when [imath]K=\mathbb{Q}[/imath] but I don't know about the general case. |
840122 | Finding limit of cube root
I'm trying to evaluate this limit, but I don't think it's coming out correctly. Could someone please offer me some assistance? Evaluate limit analytically [imath]\lim_{h\to 0}\frac{\sqrt[3]{x + h} - \sqrt[3]{x}}{h}.[/imath] What I did was multiply [imath](x+h)^{2/3} + x^{2/3}[/imath] top and bottom to get [imath]\lim_{h\to 0}\frac{(x+h)-x}{h((x+h)^{2/3} + x^{2/3})}.[/imath] I end up getting [imath]\dfrac{1}{2x^{2/3}}[/imath]. The reason why I don't think I did this write is because isn't the limit above the definition of a derivative? And if so, then isn't the derivative of [imath]\sqrt[3]{x}[/imath] equal to [imath]\dfrac{1}{3x^{2/3}}[/imath]? I would really appreciate any kind of help. Thanks. | 112865 | Finding derivative of [imath]\sqrt[3]{x}[/imath] using only limits
I need to finding derivative of [imath]\sqrt[3]{x}[/imath] using only limits So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator [imath]\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}[/imath] [imath]= \lim_{h \to 0} \frac{x+h-x}{h(\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2})}[/imath] [imath]= \lim_{h \to 0} \frac{1}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}[/imath] [imath]= \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x^2}}[/imath] [imath]= \frac{1}{2 \sqrt[3]{x^2}}[/imath] But I think it should be [imath]\frac{1}{3 \sqrt[3]{x^2}}[/imath] (3 instead of 2 in denominator?) UPDATE I found that I am using the wrong conjugate in step 1. But this (wrong) conjugate gives the same result when I multiply the numerator by it. So whats wrong with it? (I know its wrong, but why?) |
839989 | Criteria for a sequence in [imath]C[0,1][/imath] to converge weakly
Is there a criterion for a sequence in [imath]C[0,1][/imath] to converge weakly? Let [imath]\{f_n\}[/imath] be a sequence in [imath]C[0,1][/imath], [imath]f\in C[0,1][/imath]. Suppose [imath]f_n\to f[/imath] (weakly). Then for each [imath]x\in [0,1][/imath] [imath]ev_x(f_n)\to ev_x(f)[/imath], i.e. [imath]f_n(x)\to f(x)[/imath]. Thus, pointwise convergence is necessary. I guess it is not enough. Could you give any reference? | 184083 | Characterization of weak convergence in [imath]\ell_\infty[/imath]
Is there some simple characterization of weak convergence of sequences in the space [imath]\ell_\infty[/imath]? If yes, is there some similar claim for nets? I was only able to come up with a characterization of sequential weak convergence using limit along ultrafilters, which I will describe below. I wonder whether there is some insight into this characterization. (E.g. whether there is some simple reformulations which does not use ultralimits.) Moreover, I do not know whether at least this characterization works for nets, too. We have the following result describing weak convergence in [imath]C(K)[/imath], see e.g. Corollary 3.138, p.140 in Banach Space Theory by Fabian, Habala et al. (It is a consequence of Rainwater theorem and characterization of extreme points of unit ball in [imath]C(K)^*[/imath].) Let [imath]K[/imath] be a compact topological space. Let [imath]\{f_n\}[/imath] be a bounded sequence in [imath]C(K)[/imath] and [imath]f\in C(K)[/imath]. Then, if [imath]f_n\to f[/imath] pointwise, we have [imath]f_n\overset{w}\to f[/imath]. Moreover we have isometric isomorphism between [imath]\ell_\infty[/imath] and [imath]C(\beta\mathbb N)[/imath], which is described e.g. in the Wikipedia article on Stone–Čech compactification or in Chapter 15 of Carothers' book A short course on Banach space theory. This isomorphism assigns to each bounded sequence [imath](x_n)[/imath] the continuous function [imath]\overline x[/imath] on [imath]\beta\mathbb N[/imath] defined by [imath]\overline x(\mathscr U) = \operatorname{\mathscr U-lim} x_n,[/imath] where [imath]\operatorname{\mathscr U-lim} x_n[/imath] denotes the ultralimit of [imath]x_n[/imath] w.r.t the ultrafilter [imath]\mathscr U[/imath]. Combining the above results we get the following characterization: Let [imath]f^{(n)},f\in\ell_\infty[/imath]. The sequence [imath]f^{(n)}[/imath] converges to [imath]f[/imath] weakly if and only if for every ultrafilter [imath]\mathscr U[/imath] [imath]\lim_{n\to\infty} \operatorname{\mathscr U-lim} f^{(n)}= \operatorname{\mathscr U-lim} f.[/imath] (The above claim for principal ultrafilters is just a pointwise convergence. But in the above claim the equality is required for free ultrafilters, too.) |
840666 | How can I show that [imath]\phi(m) \mid \phi(n)[/imath]?
I want to prove that: [imath]\text{ if } m,n \geq 1 \text{ and } m \mid n,\text{ then } \phi(m) \mid \phi(n).[/imath] How can I show this? I thought the following: [imath]m \mid n \Rightarrow \exists k \in \mathbb{Z} \text{ such that } n=km[/imath] [imath]\text{We write } m \text{ like that : }m=p_1^{a_1} \cdots p_k^{a_k}, \text{ where } p_i \text{ are primes and } a_i>0[/imath] Then,we have [imath]n=k p_1^{a_1} \cdots p_k^{a_k}[/imath] We could show that [imath]\phi(m) \mid \phi(n)[/imath],using the fact that as [imath]\phi[/imath] is multiplicative and if [imath](m,n)=1[/imath] ,then [imath]\phi(mn)=\phi(m) \phi(n)[/imath] But..it is possible that [imath]k \mid p_1^{a_1 \cdots p_k^{a_k}}[/imath]. So,do I have to show maybe that [imath]\gcd(kp_1^{a_1} \cdots p_{k-1}^{a_{k-1}},p_k^{a_k})=1[/imath] ? | 719575 | Prove that if [imath]d \mid n \in \mathbb{N}[/imath], then [imath]\varphi(d) \mid \varphi(n)[/imath].
I want to prove that if [imath]d \mid n \in \mathbb{N}[/imath], then [imath]\varphi(d) \mid \varphi(n)[/imath]. It's given that [imath]d \mid n[/imath], so we know that [imath]n = dm[/imath], for some [imath]m \in \mathbb{Z}[/imath]. Now, I want to show that [imath]\varphi(d) \mid \varphi(n)[/imath], i.e. that [imath]\varphi(n) = k\varphi(d)[/imath], for some [imath]k \in \mathbb{Z}[/imath]. Drawing a blank of where to go from here. |
840308 | How can I solve this problem without having to do it by hand?
I'm dealing with the following problem in computational programming. I'm trying to find a way to build an algorithm that can quickly resolve the following problem statement without forcing me to do it by hand. Is there any way to group the following relations below by some pattern or identity in order to make finding a possible values of c much more efficiently? Problem: Given any integer x, where x > 2, is there any integer c, where c > 1 that satisfies one of the following relations for some integer k, where k >= 0. [imath]\frac{x}{11 + 10k} = c \Rightarrow x = 11c + 10ck[/imath] [imath]\frac{x}{3 + 10k} = c \Rightarrow x = 3c + 10ck[/imath] [imath]\frac{x}{7 + 10k} = c \Rightarrow x = 7c + 10ck[/imath] [imath]\frac{x}{9 + 10k} = c \Rightarrow x = 9c + 10ck[/imath] Example [imath]\#1[/imath]: [imath]x = 3[/imath] (Does not Satisfy) From [imath]\#1[/imath]: [imath]c = \frac{3}{11 + 10k}[/imath], so for any [imath]k \ge 0[/imath], the denominator will always be larger than [imath]3[/imath]. No integer value of [imath]c \gt 1[/imath] will ever satisfy this equation. From [imath]\#2[/imath]: [imath]c = \frac{3}{3 + 10k}[/imath], so for [imath]k = 0[/imath], c would be 1, but the restriction on c is that it must be greater than 1. For [imath]k \ge 1[/imath], the denominator will always be larger than [imath]3[/imath]. No integer value of [imath]c \gt 1[/imath] will ever satisfy this equation. From [imath]\#3[/imath]: [imath]c =\frac{3}{7 + 10k}[/imath], so for any [imath]k \ge 0[/imath], the denominator will always be larger than [imath]3[/imath]. No integer value of [imath]c \gt 1[/imath] will ever satisfy this equation. From [imath]\#4[/imath]: [imath]c = \frac{3}{9 + 10k}[/imath], so for any [imath]k \ge 0[/imath], the denominator will always be larger than [imath]3[/imath]. No integer value of [imath]c \gt 1[/imath] will ever satisfy this equation. Example [imath]\#2[/imath]: [imath]x = 4[/imath] (Does not Satisfy) From [imath]\#1[/imath]: [imath]c = \frac{4}{11 + 10k}[/imath], so for any [imath]k \ge 0[/imath], the denominator will always be larger than [imath]4[/imath]. No integer value of [imath]c \gt 1[/imath] will ever satisfy this equation. From [imath]\#2[/imath]: [imath]c = \frac{4}{3 + 10k}[/imath], so for [imath]k = 0[/imath], the denominator does not divide evenly into [imath]4[/imath]. For [imath]k \gt 1[/imath], the denominator will always be larger than [imath]4[/imath]. No integer value of [imath]c \gt 1[/imath] will ever satisfy this equation. From [imath]\#3[/imath]: [imath]c =\frac{4}{7 + 10k}[/imath], so for any [imath]k \ge 0[/imath], the denominator will always be larger than [imath]4[/imath]. No integer value of [imath]c \gt 1[/imath] will ever satisfy this equation. From [imath]\#4[/imath]: [imath]c = \frac4{9 + 10k}[/imath], so for any [imath]k \ge 0[/imath], the denominator will always be larger than [imath]4[/imath]. No integer value of [imath]c \gt 1[/imath] will ever satisfy this equation. Example [imath]\#3[/imath]: [imath]x = 37[/imath] (Does not Satisfy) No integer value of [imath]c \gt 1[/imath] will ever satisfy this equation. Example [imath]\#4[/imath]: [imath]x = 91[/imath] (Satisfies a Relation Above) From [imath]\#2[/imath]: [imath]c =\frac{91}{3 + 10k}[/imath], so for [imath]k = 1[/imath], [imath]c = 7[/imath] will satisfy this equation. Example [imath]\#5[/imath]: [imath]x = 147[/imath] (Satisfies a Relation Above) From [imath]\#3[/imath]: [imath]c =\frac{147}{7 + 10k}[/imath], so for [imath]k = 0[/imath], [imath]c = 21[/imath] will satisfy this equation. | 839958 | How can I solve this problem without doing it by hand?
I'm dealing with the following problem in computational programming. I'm trying to find a way to build an algorithm that can quickly resolve the following problem statement without forcing me to do it by hand. Is there any way to group the following relations below by some pattern or identity in order to make finding a possible values of c much more efficiently? Problem: Given any integer x, where x > 2, is there any integer c, where c >= 1 that satisfies the following relations for some integer k, where k >= 0. [imath]\frac{x}{11 + 10k} = c \Rightarrow x = 11c + 10ck[/imath] [imath]\frac{x}{3 + 10k} = c \Rightarrow x = 3c + 10ck[/imath] [imath]\frac{x}{7 + 10k} = c \Rightarrow x = 7c + 10ck[/imath] [imath]\frac{x}{9 + 10k} = c \Rightarrow x = 9c + 10ck[/imath] Example [imath]\#1[/imath]: [imath]x = 3[/imath] From [imath]\#2[/imath]: [imath]3 = 3c + 10ck[/imath], so [imath]k = 0, c = 1[/imath] satisfy this relation. Example [imath]\#2[/imath]: [imath]x = 4[/imath] From [imath]\#1[/imath]: [imath]c = \frac{4}{11 + 10k}[/imath], so for any [imath]k \ge 0[/imath], the denominator will always be larger than [imath]4[/imath]. No integer value of [imath]c \ge 1[/imath] will ever satisfy this equation. From [imath]\#2[/imath]: [imath]c = \frac{4}{3 + 10k}[/imath], so for [imath]k = 0[/imath], the denominator does not divide evenly into [imath]4[/imath]. For [imath]k \ge 1[/imath], the denominator will always be larger than [imath]4[/imath]. No integer value of [imath]c \ge 1[/imath] will ever satisfy this equation. From [imath]\#3[/imath]: [imath]c =\frac{4}{7 + 10k}[/imath], so for any [imath]k \ge 0[/imath], the denominator will always be larger than [imath]4[/imath]. No integer value of [imath]c \ge 1[/imath] will ever satisfy this equation. From [imath]\#4[/imath]: [imath]c = \frac4{9 + 10k}[/imath], so for any [imath]k \ge 0[/imath], the denominator will always be larger than [imath]4[/imath]. No integer value of [imath]c \ge 1[/imath] will ever satisfy this equation. |
840108 | Prove that for every [imath]f[/imath], a bilinear form, there exists a basis [imath]{v_1,...,v_n}[/imath] so that [imath]f(v_i,v_j) = -f(v_j,v_i)[/imath]
I didn't want to bloat the title, i'll also add that the basis is in [imath]{\mathbb{R}^n}[/imath]. This question seems kind of easy, but its from a test so I assume there is a catch here somewhere. I tried to prove it by contradiction. Assume by contradiction that there exists no basis that satisfies the condition. But as known, every bilinear form can by identified by a matrix so that [imath]A_{ij} = f(v_i,v_j)[/imath]. If there is no basis so [imath]f(v_i,v_j) = -f(v_j,v_i)[/imath] it means there is no skew-symmetric matrix which can identify a bilinear form over [imath]\mathbb{R}[/imath] which is obviously not true( I guess an example can be given here, I think ). This proof just seems to simple to be true, but I can't seem to figure what could be wrong with it. I'll greatly appreciate any tips, thanks! | 841067 | Question about bilinear form
Prove that every bilinear form [imath]f:\mathbb R^n \times \mathbb R^n\rightarrow \mathbb R[/imath] has a basis [imath]\{v_1,\ldots,v_n\} \subset \mathbb R^n[/imath] such that [imath]f(v_i,v_j)=-f(v_j,v_i)[/imath] for every [imath]i\neq j[/imath]. I have no idea how to start thinking about this one. Should I actually "build" the required basis or what? Any ideas ? thanks |
842662 | An unsolvable Number Theory Question
This is a question I was sent by a friend, I really have zero idea on how to solve the question. I'm not even sure where to begin solving it. Though I did make an attempt to solve it, I always reached a dead end. So here's the question: Let [imath]n[/imath] be a positive integer greater than [imath]1[/imath] and let [imath]p_1, p_2, ... p_t[/imath] be the primes not exceeding [imath]n[/imath]. Show that [imath]p_1p_2...p_t < 4^n[/imath] | 15902 | Show that product of primes, [imath]\prod_{k=1}^{\pi(n)} p_k < 4^n[/imath]
This an interesting problem my friend has been working on for a while now (I just saw it an hour ago, but could not come up with anything substantial besides some PMI attempts). Here's the full problem: Let [imath]x_{1}, x_{2}, x_{3}, \cdots x_{y}[/imath] be all the primes that are less than a given [imath]n>1,n \in \mathbb{N}[/imath]. Prove that [imath]x_{1}x_{2}x_{3}\cdots x_{y} < 4^n[/imath] Any ideas very much appreciated! |
842667 | Random 0-1 matrices
For [imath]n,r∈N[/imath], [imath]1<r<n[/imath], let [imath]z(r,n)[/imath] be the largest possible number of 0 entries in an [imath]n×n[/imath] matrix which has no [imath]r×r[/imath] submatrix whose entries are all [imath]0[/imath]. (Here a submatrix is obtained by selecting any [imath]r[/imath] rows and any [imath]r[/imath] columns; the rows/columns need not be consecutive.) Consider a random matrix in which each entry is [imath]0[/imath] with probability [imath]p[/imath] and [imath]1[/imath] with probability [imath]1−p[/imath], independently. Deduce that [imath]z(r,n)>pn^{2}−p^{r^{2}}n^{2r}[/imath]. My solution: Let [imath][n]_{2}[/imath] denote the set of all [imath]n×n[/imath] matrices, with only [imath]0−1[/imath] entries. For each [imath]σ∈[n]_{2}[/imath] let the random variable [imath]X(σ)[/imath] denote the number of [imath]0[/imath] entries in [imath]σ[/imath] and the random variable [imath]Y(σ)[/imath] denote the number of [imath]r×r[/imath] submatrices filled with entirely with [imath]0[/imath]'s. It follows that [imath]\mathbb{E}(X)=pn^{2}[/imath] and [imath]\mathbb{E}(Y)=\binom{n}{r}^{2}p^{r^{2}}<n^{2r}p^{r^{2}}[/imath] . Further more consider the random variable [imath]X−Y[/imath] for which [imath]\mathbb{E}(X-Y)>pn^{2}-n^{2r}p^{r^{2}}[/imath]. It follows that there exists a [imath]σ∈[n]_{2}[/imath] such that [imath]X(σ)−Y(σ)>pn^{2}-n^{2r}p^{r^{2}}[/imath]. Given such a [imath]σ[/imath] construct [imath]σ′∈[n]_{2}[/imath] as follows. For each of the all [imath]0[/imath], [imath]r×r[/imath] sub matrices of [imath]σ[/imath] remove at random a single [imath]0[/imath] from each to ensure [imath]σ′[/imath] has no [imath]r×r[/imath] such sub-matrices. Then we have removed at most [imath]\binom{n}{r}^{2}<n^{2r}[/imath] such zero's and so we can conclude that [imath]σ′[/imath] has at least [imath]pn^{2}-n^{2r}p^{r^{2}}[/imath] zeros thus we can conclude that [imath]z(n,r)>pn^{2}-n^{2r}p^{r^{2}}[/imath]. Id appreciate any comments on the validity of the proof. | 842468 | Random [imath]0-1[/imath] matrices
I'm working my way through the Oxford notes in Probabilistic Combinatorics and came across this question in one of the question sheets; I'd like to stress that this is not my homework: I'm simply working through the notes for my own pleasure. For [imath]n, r ∈ \mathbb{N}[/imath], [imath]1 < r < n[/imath], let [imath]z(r, n)[/imath] be the largest possible number of [imath]0[/imath] entries in an [imath]n × n[/imath] matrix which has no [imath]r × r[/imath] submatrix whose entries are all [imath]0[/imath]. (Here a submatrix is obtained by selecting any [imath]r[/imath] rows and any [imath]r[/imath] columns; the rows/columns need not be consecutive.) Consider a random matrix in which each entry is [imath]0[/imath] with probability [imath]p[/imath] and [imath]1[/imath] with probability [imath]1−p[/imath], independently. Deduce that [imath]z(r, n) > pn^{2}-p^{r^{2}}n^{2r}.[/imath] My solution: Let [imath][n]_{2}[/imath] denote the set of all [imath]n \times n[/imath] matrices, with only [imath]0-1[/imath] entries. For each [imath]\sigma \in [n]_{2}[/imath] let the random variable [imath]X(\sigma)[/imath] denote the number of [imath]0[/imath] entries in [imath]\sigma[/imath] and the random variable [imath]Y(\sigma)[/imath] denote the number of [imath]r \times r[/imath] submatrices filled with entirely with [imath]0[/imath]'s. It follows that [imath]\mathbb{E}(X)=pn^{{2}}[/imath] and [imath]\mathbb{E}(Y)=\binom{n}{r}^{2}p^{r^{2}}<n^{2r}p^{r^{2}}[/imath]. Further more consider the random variable [imath]X-Y[/imath] for which [imath]\mathbb{E}(X-Y)>pn^{2}-n^{2r}p^{r^{2}}[/imath]. It follows that there exists a [imath]\sigma \in [n]_{2}[/imath] such that [imath]X(\sigma)-Y(\sigma)>pn^{2}-n^{2r}p^{r^{2}}[/imath]. Given such a [imath]\sigma[/imath] construct [imath]\sigma' \in [n]_{2}[/imath] as follows. For each of the all [imath]0[/imath], [imath]r \times r[/imath] sub matrices of [imath]\sigma[/imath] remove at random a single [imath]0[/imath] from each to ensure [imath]\sigma'[/imath] has no [imath]r \times r[/imath] such sub-matrices. We have removed at most [imath]\binom{n}{r}^{2}<n^{2r}[/imath] such zero's and so we can conclude that [imath]\sigma'[/imath] has at least [imath]pn^{2}-p^{r^{2}}n^{2r}[/imath] zeros thus we can conclude that [imath]z(n,r)>pn^{2}-p^{r^{2}}n^{2r}[/imath]. I'd appreciate any comments on the validity of the proof. |
842852 | Extending a power series of a holomorphic function when the function extends continuously...?
If [imath]f : \mathbb{C} \to \mathbb{C}[/imath] is represented by a power series in [imath]D[/imath] (the unit disc), and [imath]f[/imath] extends continuously into [imath]\bar D[/imath], does the same power series represent [imath]f[/imath] in [imath]\bar D[/imath]? I suspect that the answer is no, since this would imply that [imath]f[/imath] then extends holomorphically into [imath]\bar D[/imath]. However, this is not that great of a reason, and I can't think that of a specific counter example. (This question arose while trying to solve a different analysis exercise.) I should also add that, after looking at the power series expansion theorem again, I was reminded that one only has a power series expansion for f in an open disc whose closure is contained in the region where f is holomorphic (otherwise the Cauchy integral formula doesn't apply). However, I'm not sure if this means that one cannot find a single power series expansion for [imath]f[/imath] in the whole disc, even if [imath]f[/imath] is not holomorphic on the boundary. This is possibly a separate question. Duplicate here: Continuity of analytic function implies convergence of power series? | 286119 | Continuity of analytic function implies convergence of power series?
Suppose that [imath]f(z)[/imath] is analytic in the unit disk [imath]\Delta:\,|z|<1[/imath]. Then [imath]f(z)[/imath] has a Taylor series [imath]\sum=\sum a_nz^n[/imath] in the unit disk. One may assume that [imath]\sum[/imath] has [imath]R=1[/imath] as its radius of convergence for convenience. Question 1 (local version): If [imath]f(z)[/imath] is continous at some point, say [imath]z=1[/imath], on the unit circle [imath]S^1[/imath], then its Taylor series [imath]\sum[/imath] is covergent at [imath]z=1[/imath]? Question 2 (global version): Suppose that [imath]f(z)[/imath] is continous on the closed unit disk [imath]\overline\Delta[/imath]. Is the Taylor series [imath]\sum[/imath] covergent at every poin of [imath]S^1[/imath]? Of course, if the answer to the local version is "yes", then so is the answer to the global version. EDIT. It seems that Question 1 obviously has a negative answer. So I modify it into a new question. Question 3 (New local version): Suppose that [imath]f(z)[/imath] is continous at some point, say [imath]z=1[/imath], on the unit circle [imath]S^1[/imath] and the coefficients of [imath]\sum[/imath] satisfies [imath]a_n\to 0[/imath]. Is the Taylor series [imath]\sum[/imath] covergent at [imath]z=1[/imath]? Note. If [imath]f[/imath] is analytic at [imath]z=1[/imath], then the answer to Question 3 is yes. This is Fatou's Theorem. |
842958 | A set is compact if and only if every continuous function is bounded on the set?
I was asked to prove the following statement: Let [imath]K \subseteq R^n[/imath]. show that [imath]K[/imath] is compact (meaning closed and bounded) if and only if every continuous function is bounded on [imath]K[/imath]. What I did: Suppose [imath]K[/imath] is not bounded, and so, it is not compact. Then the function [imath]\sum |x_i|[/imath] is a continuous unbounded function on [imath]K[/imath]. Via contrapositive, this shows that if every function is bounded, then [imath]K[/imath] is also bounded. What I need help with: Assume [imath]K[/imath] is not closed. I need to find a continuous and unbounded function on [imath]K[/imath]. that will prove that if every continuous function is bounded on [imath]K[/imath], then [imath]K[/imath] is compact. after that, i still need to show that if [imath]K[/imath] is compact then every continuous function [imath]f: K \to \mathbb R[/imath] is bounded. Would someone point me in the right direction? Clarification: it's not homework. I am preparing for an exam. | 99637 | [imath]K\subseteq \mathbb{R}^n[/imath] is a compact space iff every continuous function in [imath]K[/imath] is bounded.
I need to prove that [imath]K\subseteq \mathbb{R}^n[/imath] is a compact space iff every continuous function in [imath]K[/imath] is bounded. One direction is obvious because of Weierstrass theorem. How can i prove the other direction? I tried to assume the opposite but it didn't work for me. Thanks a lot. |
260876 | What exactly is infinity?
On Wolfram|Alpha, I was bored and asked for [imath]\frac{\infty}{\infty}[/imath] and the result was (indeterminate). Another two that give the same result are [imath]\infty ^ 0[/imath] and [imath]\infty - \infty[/imath]. From what I know, given [imath]x[/imath] being any number, excluding [imath]0[/imath], [imath]\frac{x}{x} = 1[/imath] is true. So just what, exactly, is [imath]\infty[/imath]? | 288905 | Is ∞ considered defined?
[imath]\infty[/imath] (Infinity) is not a number, but infinity is considered to be defined, right? There are expressions in mathematics such as: [imath]\frac x0,0^0, \frac\infty\infty,[/imath] which are not defined because they do not have a certain place on any domain, and sometimes because there is no single value that satisfies all functions yielding such expression in a limit. How can infinity be defined? Maybe the argument would hinge on how 'defined' is defined? Perhaps this requires a circular(infinite) argument? Or is my assumption that infinity is considered to be defined; false? |
843230 | Counter example of monotone union
I saw this exercise in "Elements of Abstract and Linear Algebra" by E. H. Connell: Suppose [imath]G[/imath] is a group. Suppose [imath]T[/imath] is an index set and for each [imath]t \in T[/imath], [imath]H_t[/imath] is a subgroup of [imath]G[/imath]. Furthermore, if [imath]\{H_t\}[/imath] is a monotonic collection, show that [imath]\bigcup_{t \in T} H_t[/imath] is a subgroup of [imath]G[/imath]. Suppose "monotonic" is defined as follows: For arbitrary [imath]k[/imath], [imath]l \in T[/imath], either [imath]H_k \subseteq H_l[/imath] or [imath]H_l \subseteq H_k[/imath] (or both). Then here is my proof: Suppose [imath]a[/imath], [imath]b \in \bigcup_{t \in T}H_t[/imath], then [imath]a \in H_k[/imath] for some [imath]k[/imath] and [imath]b \in H_l[/imath] for some [imath]l[/imath]. So either [imath]a[/imath], [imath]b \in H_k[/imath] or [imath]a[/imath], [imath]b \in H_l[/imath] (or both). Then either [imath]a \cdot b \in H_k[/imath] or [imath]a \cdot b \in H_l[/imath] (or both). Hence [imath]a \cdot b \in \bigcup_{t \in T}H_t[/imath]. Therefore [imath]\bigcup_{t \in T}H_t[/imath] is a subgroup. My question is: Is the above proof correct? I think it looks too trivial. In my opinion, if you have a monotonic collection of sets, all satisfying certain criterion, then "obviously" the union also satisfies that criterion. Can someone give a counter-example? | 842628 | Monotone Union of subgroups being subgroup
I saw this exercise in a book: Suppose [imath]G[/imath] is a group. Suppose [imath]T[/imath] is an index set and for each [imath]t \in T[/imath], [imath]H_t[/imath] is a subgroup of [imath]G[/imath]. Furthermore, if [imath]\{H_t\}[/imath] is a monotonic collection, show that [imath]\bigcup_{t \in T} H_t[/imath] is a subgroup of [imath]G[/imath]. My question is if [imath]T[/imath] is not [imath]\mathbb N[/imath], how to define "monotonic"? If it is defined by what Jyrki Lahtonen suggested: For arbitrary [imath]k[/imath], [imath]l \in T[/imath], either [imath]H_k \subseteq H_l[/imath] or [imath]H_l \subseteq H_k[/imath] (or both). Then here is my proof: Suppose [imath]a[/imath], [imath]b \in \bigcup_{t \in T}H_t[/imath], then [imath]a \in H_k[/imath] for some [imath]k[/imath] and [imath]b \in H_l[/imath] for some [imath]l[/imath]. So either [imath]a[/imath], [imath]b \in H_k[/imath] or [imath]a[/imath], [imath]b \in H_l[/imath] (or both). Then either [imath]a \cdot b \in H_k[/imath] or [imath]a \cdot b \in H_l[/imath] (or both). Hence [imath]a \cdot b \in \bigcup_{t \in T}H_t[/imath]. Therefore [imath]\bigcup_{t \in T}H_t[/imath] is a subgroup. My question is: Is the above proof correct? I think it looks too trivial. In my opinion, if you have a monotonic collection of sets, all satisfying certain criterion, then "obviously" the union also satisfies that criterion. Can someone give a counter-example? |
643322 | Prove not an integral domain
Let [imath]R[/imath] be a commutative ring with identity 1, such that [imath]R[/imath] contains exactly three ideals:[imath]\{0\},J,R[/imath]. Prove that [imath]R[/imath] is not an integral domain. I was thinking start with supposing that R is an integral domain, then for [imath]ab=0, a=0,b \ne0[/imath] for [imath]b \in R-J[/imath], it can found a contradiction of proving [imath]Rb[/imath] is also an ideal. But I cannot go on anymore, may be I was thinking wrong and hope some body can help me with it. Thanks | 639870 | Questions about a commutative ring with exactly three ideals
Let [imath]R[/imath] be a commutative ring with identity. Assume that [imath]R[/imath] has exactly three distinct ideals: [imath]\{0\},I, R.[/imath] 1) Show that if [imath]a \in R-I[/imath], then [imath]a[/imath] is a unit in [imath]R[/imath]. 2) Let [imath]a,b\ne0[/imath] in [imath]I[/imath]. Show that [imath]ab=0[/imath]. |
844485 | how to prove the limit of the sequences exist?
suppose [imath]{x_n}[/imath] is a numerical sequence such that [imath]0<x_{n+m}\leq x_n+x_m[/imath] for all [imath]n[/imath] and [imath]m\in \mathbb{N}[/imath] and [imath]x_1>0[/imath]. Prove that [imath]lim_{n\rightarrow \infty }\frac{x_n}{n}[/imath] exists. | 760484 | Sequence Limit Problem: If [imath]0 \leq x_{m+n} \leq x_n + x_m[/imath] then limit of [imath]x_n/n[/imath] exists
If the sequence [imath]\{x_n\}[/imath] satisfies the property that [imath]0 \leq x_{m+n} \leq x_n + x_m[/imath] for all [imath]n[/imath], [imath]m \in \mathbb{N}[/imath] , show that the limit of the sequence [imath]\left\{\frac{x_n}{n}\right\}_n[/imath] exists. Provide an example of such a sequence. I can show that it is bounded, but I don't know how to show that it is monotone increasing (I think it is monotone increasing, not sure). |
844720 | Hochschild (co)-homology of a formal quantization of an associative algebra
Let [imath]A[/imath] be a commutative associative [imath]k[/imath]-algebra and let [imath]A[[\hbar]][/imath] be the formal deformation of [imath]A[/imath]. I would like to know if there is a relation between the Hochschild co-homologies [imath]HH^{\bullet}(A,A)[/imath] and [imath]HH^{\bullet}(A[[\hbar]],A[[\hbar]])[/imath], respectively. In particular, is it true that [imath]HH^{\bullet}(A[[\hbar]],A[[\hbar]])=HH^{\bullet}(A,A)[[\hbar]][/imath]? | 844855 | Hochschild cohomology of a formal quantization of an associative algebra
Let [imath]A[/imath] be a commutative associative [imath]k[/imath]-algebra and let [imath]A[[\hbar]][/imath] be the formal deformation of [imath]A[/imath]. I would like to know if there is a relation between the Hochschild co-homologies [imath]HH^{\bullet}(A,A)[/imath] and [imath]HH^{\bullet}(A[[\hbar]],A[[\hbar]])[/imath], respectively. In particular, is it true that [imath]HH^{\bullet}(A[[\hbar]],A[[\hbar]])=HH^{\bullet}(A,A)[[\hbar]][/imath]? |
845151 | Integral defining expectation of [imath]\chi^2[/imath] distribution
If U has a [imath]\chi^2[/imath] distribution with v df, find E(U) and V(U). By definition, [imath]E(U) =\int^{\infty}_{0} u\frac{1}{\gamma(\frac{v}{2})2^\frac{v}{2}}u^{\frac{v}{2}-1} e^\frac{-u}{2}\,du =\int^{\infty}_{0} \frac{1}{\gamma(\frac{v}{2})2^\frac{v}{2}}u^\frac{v}{2} e^\frac{-u}{2}\,du[/imath]. How do I integrate this? NOTE: I already know that this question has been asked originally however there were not steps on how to integrate it. Just a short hand solution. I would like someone to show me not only the steps but also what the do with the Gamma function as well. Thanks! | 651018 | Help with an integral for: If U has a [imath]\chi^2[/imath] distribution with v df, find E(U) and V(U)
If U has a [imath]\chi^2[/imath] distribution with v df, find E(U) and V(U). By definition, [imath]E(U) =\int^{\infty}_{0} u\frac{1}{\gamma(\frac{v}{2})2^\frac{v}{2}}u^{\frac{v}{2}-1} e^\frac{-u}{2}\,du =\int^{\infty}_{0} \frac{1}{\gamma(\frac{v}{2})2^\frac{v}{2}}u^\frac{v}{2} e^\frac{-u}{2}\,du[/imath]. How do I integrate this? Note: This isn't a homework problem. |
845613 | An inequality involving Möbius function
For any positive integer [imath]n[/imath] show the inequality holds : [imath]\left|\sum_{i=1}^{n}\frac{\mu(i)}{i}\right|\le 1[/imath] I tried induction. when [imath]\mu(n+1)=0[/imath] it is trivial. But what if [imath]\mu(n+1)\ne 0[/imath]? I am stuck there. Can someone help me? Thanks. | 557998 | Simple Divisor Summation Inequality (with Moebius function)
Show that [imath]\left| \sum_{k=1}^{n} \frac {\mu(k)}{k} \right| \le 1 [/imath] where [imath]\mu[/imath] is Moebius function and n is a positive integer. The hard thing here is that the sum is not directly divisor sum; it's just a normal summation. What I know is that when [imath]F(n)=\sum_{d|n}f(d)[/imath], [imath]\sum_{k=1}^N F(k)=\sum_{k=1}^N f(k)\left[\frac{N}{k}\right][/imath] But it was hardly useful since the given inequality doesn't contain the Guass function, so it became really complicated(and I'm not very confident in using the doulbe-summation). One thing I also know is that [imath]\frac{\phi(n)}{n}=\sum_{d|n}\frac{\mu(d)}{d}[/imath] I also tried to use this using Moebius Inversion Formula, but it also became quite complicated. For example, we can derive that [imath]\frac{\mu(k)}{k}=\sum_{d|k}\frac{\phi(d)}{d} \mu \left(\frac{k}{d}\right)[/imath] I couldn't make it more compact. So by substitution, we get [imath]\sum_{k=1}^{n} \frac {\mu(k)}{k}=\sum_{k=1}^{n}\sum_{d\mid k}\frac{\phi(d)}{d} \mu \left(\frac{k}{d}\right)[/imath] If it were not for the [imath]k[/imath] in the right term of Moebius function, we could use the first lemma I'd written down, but we can't. So I'm just stuck. Could anybody help me solving this problem? It would be really grateful if someone could also give me general methods to tackle divisor-sum problems since I'm having a difficult time in changing the order of doulbe-summations. Thanks! |
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