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1356585 | Show that the letters X and I (thought of as topological spaces) are not homeomorphic
I am reading Hajime Sato's: Algebraic Topology, an Intuitive Approach. His Sample Problem 1.3 is: Show that the topological spaces X and I are not homeomorphic. (Note that this requires a font where I appears as a simple straight vertical line.) Sato argues by contradiction. Brief sketch of his argument: Suppose there existed a homeomorphism [imath]f: X \to I[/imath]. Remove the point [imath]x_0[/imath] which lies at the crossing point of X. Then the restriction of [imath]f[/imath] to the new domain is still homeomorphic. But [imath]X-x_0[/imath] consists of four disjoint line segments (each half open), and [imath]I-f(x_0)[/imath] consists of two disjoint line segments (each half open). Therefore, he concludes, the spaces aren't homeomorphic. I don't understand how the conclusion (in the final sentence) follows. I know that two spaces are homeomorphic if there exists a bijective continuous map between them with a continuous inverse. And I know the characterization of continuity in terms of open sets. But somehow I am not seeing it. I feel like I'm missing something really obvious. Any ideas? | 1072206 | What does "removing a point" have to do with homeomorphisms?
I am self-studying topology from Munkres. One exercise asks, in part, to show that the spaces [imath](0,1)[/imath] and [imath](0,1][/imath] are not homeomorphic. An apparent solution is as follows: If you remove a point, [imath]x[/imath], from [imath](0,1)[/imath], you get a disconnected space; however, you can remove the point [imath]\{1\}[/imath] from the space [imath](0,1][/imath] and the space will still be connected. This apparently means the spaces aren't homeomorphic. I don't quite see the connection to the definition of homeomorphic spaces (that two spaces are homeomorphic if there is a bicontinuous function between them). So, what is the connection between the "bicontinuous" definition of homeomorphisms and the "removing a point" procedure? Sorry for missing something obvious. |
1349953 | construction of linear independent local section in vector bundle
Let [imath]X_1,\ldots,X_k[/imath] be lineary independent local sections of a vector bunlde [imath]V[/imath], [imath]rank(V)>k[/imath], defined on some open set [imath]U[/imath]. Can I construct a local section [imath]X_{k+1}[/imath] on a (maybe smaller) open subset of [imath]U[/imath], such that [imath](X_1,\ldots,X_{k+1})[/imath] are lineary independent on the smaller subset? edit: I know that answer, however I would be interested in a more "elementary" argument. | 594118 | "Basis extension theorem" for local smooth vector fields
Let [imath]\pi: E \to M[/imath] be a smooth vector bundle of rank [imath]n[/imath], and suppose [imath]s_1, \ldots, s_m[/imath] are independent smooth local sections over an open subset [imath]U \subset M[/imath]. Can I prove the "basis extension theorem" for smooth sections? That is, for each [imath]p \in U[/imath], there are smooth sections [imath]s_{m+1},\ldots,s_n[/imath] defined on some neighborhood of [imath]V[/imath] of [imath]p[/imath] such that [imath](s_1, \ldots, s_n)[/imath] is a smooth local frame for [imath]E[/imath] over [imath]U \cap V[/imath]. |
1357687 | Finding nth roots of complex number
If I have a complex number of the form [imath]z = a+bi[/imath], how would I find the complex roots? I know that each root will be equidistant from each other and will form a circle, but I'm not sure how to solve for the complex roots. I also know that for a complex number with [imath]n[/imath] roots, each root will be rotated [imath]2\pi/n[/imath] radians. It's not an issue of graphing the complex roots, but it is for solving for them. | 101518 | The [imath]n[/imath] complex [imath]n[/imath]th roots of a complex number [imath]z[/imath]
Suppose [imath]z[/imath] is a nonzero complex number, so [imath]z=re^{i\theta}[/imath]. Show that there are only [imath]n[/imath] distinct complex [imath]n[/imath]-th roots, given by [imath]r^{1/n}e^{i(2\pi k+\theta)/n}[/imath] for [imath]0\leq k\leq n-1[/imath]. My proof: Let [imath]z=re^{i\theta}\in\mathbb{C}[/imath] and [imath]a\in\mathbb{C}[/imath] s.t. [imath]a^{n}=z[/imath]. Let [imath]a=\rho e^{i\varphi}[/imath]. Then, [imath]a^{n}=(\rho e^{i\varphi})^{n}[/imath]. Setting [imath]\rho^{n}e^{i\varphi n}=re^{i\theta}[/imath] and rewriting in polar form, we get [imath]\rho^{n}(\cos(n\varphi)+i\sin(n\varphi))=r(\cos(ΞΈ)+i\sin(\theta)).[/imath] Thus [imath]\rho^{n}=r[/imath] [imath]\rightarrow[/imath] [imath]\rho=r^{1/n}[/imath] and [imath]n\varphi=\theta\rightarrow\varphi=\theta/n[/imath]. Thus the roots of [imath]a\in\mathbb{C}[/imath] have the form [imath]r^{1/n}(\cos(\tfrac{\theta}{n})+i\sin(\tfrac{\theta}{n}))=r^{1/n}(\cos(\tfrac{\theta}{n}+2\pi j)+i\sin(\tfrac{\theta}{n}+2\pi j))[/imath] for some [imath]j\in\mathbb{Z}[/imath] due to the periodicity of trig functions. Then, [imath]r^{1/n}(\cos(\tfrac{\theta}{n}+2\pi j)+i\sin(\tfrac{\theta}{n}+2\pi j))=r^{1/n}e^{i\left(2\pi j+\tfrac{\theta}{n}\right)}=r^{1/n}e^{i(\theta+2\pi k)/n}[/imath] for [imath]0\leq k\leq n-1 ,j=nk\in\mathbb{Z} [/imath]. I feel like there could be a more concise way to put this. |
1357168 | Continuity of increasing function
If [imath]f[/imath] is an increasing function over the reals, given a number [imath]M[/imath], is it always possible to find some [imath]x \ge M[/imath] such that [imath]f[/imath] is continuous at [imath]x[/imath]? This seems like it should be intuitively true but I can't find a proof. | 56831 | Is there an everywhere discontinuous increasing function?
Does there exist a function [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath] that is strictly increasing and discontinuous everywhere? My line of thought (possibly incorrect): I know there are increasing functions such as [imath]f(x) = x[/imath], and there are everywhere-discontinuous functions such as the Dirichlet function. I also know that when there is a discontinuity at a point [imath]c[/imath], there is a finite gap [imath]\epsilon[/imath] such that there are points [imath]d[/imath] arbitrarily close to [imath]c[/imath] such that [imath]|f(d) - f(c)| > \epsilon[/imath]. This is where my thinking gets unclear - does it make sense to have a "gap" at every real number? |
1358089 | A problem in Algebra
If [imath]a+b+c+d=0=a^7+b^7+c^7+d^7[/imath], Then prove [imath]a(a+b)(a+c)(a+d)=0[/imath] and all are real numbers. I am confused and don't even know which to tag for help. | 1328236 | How can I apply Newton's sums to solve this problem?
Given [imath]x_1,x_2,x_3,x_4[/imath] real numbers such that [imath]x_1+x_2+x_3+x_4 = 0[/imath] and [imath]x_1^7+x_2^7+x_3^7+x_4^7 = 0,[/imath] how can I use symmetric functions and Newton's sums to prove that [imath]x_1(x_1+x_2)(x_1+x_3)(x_1+x_4)=0[/imath]? This is what I have so far: Note that [imath]x_1 = -(x_2+x_3+x_4)[/imath] and the relation between Newton's sums, elementary symmetric polynomials, and power sums gives us that [imath]P_7-S_1 P_6+S_2 P_5-S_3 P_4+S_4P_3=0,[/imath] where [imath]P_k[/imath] denotes the [imath]kth[/imath] power sum and [imath]S_k[/imath] denotes the [imath]kth[/imath] elementary symmetric polynomial of [imath]x_1,x_2,x_3,x_4.[/imath] We want to show that [imath](x_2+x_3+x_4)(x_2+x_3)(x_2+x_4)(x_3+x_4) = 0.[/imath] But how? |
1357274 | [imath]G[/imath], group. [imath]a,b\in G[/imath]. Show that [imath]|a|=|a^{-1}|; |ab|=|ba|[/imath], and [imath]|a|=|cac^{-1}|, \forall c\in G[/imath].
Isn't it obvious for [imath]|a|=|a^{-1}|[/imath], since [imath]\langle a\rangle = \langle a^{-1} \rangle[/imath]? For [imath]|ab|=|ba|[/imath], I think we should go like this: [imath]e=(ab)^n\Rightarrow e=(ab)(ab)^{n-1}\Rightarrow (ab)^{-1}=(ab)^{n-1}\Rightarrow b^{-1}a^{-1}=(ab)^{n-1}\Rightarrow a^{-1}=b(ab)^{n-1}[/imath] But I get stuck here, since I don't know whether to get from here to [imath]e=(ba)^n[/imath]. For [imath]|a|=|cac^{-1}|[/imath], do I have to show that [imath]\langle a\rangle = \langle cac^{-1}\rangle[/imath]? | 633757 | Order of conjugate of an element given the order of its conjugate
Let [imath]G[/imath] is a group and [imath]a, b \in G[/imath]. If [imath]a[/imath] has order [imath]6[/imath], then the order of [imath]bab^{-1}[/imath] is... How to find this answer? Sorry for my bad question, but I need this for my study. |
1358290 | Prove for [imath] \forall n \in \mathbb{N}, \exists x,y,z[/imath] ( [imath]0 \leq x < y < z[/imath] ) such that [imath] n = \binom{x}{1} + \binom{y}{2} + \binom{z}{3}[/imath]
I'm trying to solve a problem from the combinatorics book. Prove or disprove for [imath] \forall n \in \mathbb{N}, \exists x,y,z \in \mathbb{N} [/imath] ([imath]0 \leq x < y < z[/imath]) such that [imath] n = \binom{x}{1} + \binom{y}{2} + \binom{z}{3}[/imath] I was trying to substitute binomial coefficients with: [imath] n = \binom{x}{1} + \binom{y}{2} + \binom{z}{3} = x + \frac{y*(y-1)}{2} + \frac{z*(z-1)*(z-2)}{3}[/imath] However, I've stucked after that. I know that any binomial coefficient is integer-valued. So, it's likely equation hold. | 1310541 | integer as sum of three binomials
Prove that for any nonnegative integer [imath]n[/imath] [imath]\exists x,y,z \in \mathbb{N}[/imath] and [imath]0\leq x<y<z[/imath] so [imath]n=\binom{x}{1}+\binom{y}{2}+\binom{z}{3}[/imath] Please give me a hint, I don't have any idea. |
1358194 | question on identity of sums with exponent.
I want to show that for [imath]x>0[/imath]: [imath]\sum_{n=-\infty}^\infty e^{-n^2\pi x}= \frac{1}{\sqrt{x}}\sum_{n=-\infty}^\infty e^{-n^2\pi / x}[/imath] It doesn't seem that a simple change of variables will do, like [imath]n\mapsto n^2 x[/imath], so how to show this identity? | 1332831 | Proving the functional equation [imath]\theta (x) = x^{-\frac{1}{2}} \theta (x^{-1})[/imath] from the Poisson summation formula
We have the relationship [imath]\theta (x) = x^{-\frac{1}{2}} \theta (x^{-1})[/imath] Now I know one uses the Poisson summation formula to prove this. The Poisson summation formula comes from Fourier Transform and series and satisfies, \begin{equation} \sum_{n=-\infty}^{+\infty} f(n) = \sum_{n=-\infty}^{+\infty} \hat f (n) \end{equation} So do we say [imath]f(n)= \theta(x)[/imath]? (where [imath]\theta(x) = \sum_{-\infty}^\infty e^{-\pi n^2 x}[/imath]) So filling in we have, \begin{equation} \sum_{n= - \infty}^{+\infty} e^{-\pi n^2 x}= \sum_{m= - \infty}^{+\infty} \int_{- \infty}^{\infty} e^{-\pi y^{2}x}e^{-2 \pi i y m} dy \end{equation} Now what is the best way to keep going? Do I rearrange with the goal of getting a Gaussian integral (and hence the relationship)? Or is there a more elegant way of achieving the desired result? What I've done so far is, after completing the square in the power of the integrand, we have \begin{equation} =\sum_{m=-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-\pi x((y+i\frac{m}{x})^{2}-i^{2} \frac{m^{2}}{x^{2}})}dy \end{equation} Then separating the integrand, breaking it up into a product of exponentials again we have, \begin{equation} =\sum_{m=-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-\pi m^{2} \frac{1}{x}} e^{-\pi x (y+i \frac{m}{x})^{2}}dy \end{equation} The left factor of the integrand is a constant as it does not contain [imath]y[/imath], so it can be brought outside of the integral. [imath] \sum_{n = -\infty}^{+\infty} e^{-\pi n^{2} x} = \sum_{m=-\infty}^{+\infty}e^{- \pi m^{2} \frac{1}{x}} \int_{- \infty}^{+ \infty} e^{ -\pi x (y+i \frac{m}{x})^{2}} dy [/imath] What would be the correct approach from here? Should I make a substitution? Any help or suggestions will be appreciated! |
1358419 | Non-martingale with respected to the natural filtration, and satisfies [imath]E[M_{n+1} | M_n]=M_n[/imath]
I am thinking about the exercise: Exercise 5. Give an example of a random sequence ([imath]M_n[/imath]) such that [imath]E[ M_{n+1} | M_n ] = M_n[/imath] for all [imath]n\ge0[/imath], but which is not a martingale w.r.t. the filtration [imath]F_n = \sigma(M_0, \dots , M_n)[/imath]. From Exercise 5 of Link : http://www.math.bme.hu/~gabor/oktatas/SztoM/StModHW_2014.pdf | 1186859 | Example: satisfying [imath]E(X_{n+1}\mid X_n)=X_n[/imath] but not a martingale
I am wondering if there is such a sequence of random variables [imath](X_n)_{n=0}^\infty[/imath] such that [imath]\mathbb{E}(X_{n+1}\mid X_n)=X_n[/imath] for all [imath]n\geq0[/imath] but which is not a martingale with respect to the filtration [imath]\mathcal{F_n}=\sigma(X_0, \dots, X_n)[/imath]. I would really appreciate if you could show me such an example. |
1358429 | When do we say independence is the probability of intersection being equal to the product of probabilities?
This is something I never really got in either Elementary Probability Theory or Advanced Probability Theory because my professors mainly discussed independence between 2 objects. Please tell me if my understanding is right: Events [imath]A_1, A_2,\dots,A_n[/imath] are independent: For any distinct indices [imath]i_1, i_2, \dots, i_n[/imath] [imath]P(A_{i_1}, A_{i_2}, \dots, A_{i_n}) = \prod_{i = i_1}^{i_n} P(A_i).\tag{A}[/imath] This is not the same as [imath]P(A_1, \dots, A_n) = \prod_{i = 1}^{n} P(A_i)[/imath]. Sigma-algebras or Pi-systems [imath]\mathscr{A}_1, \mathscr{A}_2, \dots, \mathscr{A}_n[/imath] are independent: For any distinct indices [imath]i_1, i_2, \dots, i_n[/imath] [imath]P(A_{i_1}, A_{i_2}, \dots, A_{i_n}) = \prod_{i = i_1}^{i_n} P(A_i)\mbox{ where } A_{i_1} \in \mathscr{A}_{i_1}, A_{i_2} \in \mathscr{A}_{i_2}, \dots, A_{i_n} \in \mathscr{A}_{i_n}.\tag{B}[/imath] I'm guessing this is not the same as [imath]P(A_1, \dots, A_n) = \prod_{i = 1}^{n} P(A_i)[/imath] for similar reasons (where [imath]A_{1} \in \mathscr{A}_{1}, A_{2} \in \mathscr{A}_{2}, \dots, A_{n} \in \mathscr{A}_{n}[/imath]). However, I saw that in Stochastic Calculus when random variables [imath]Y_1, Y_2, ... Y_n[/imath] are independent, we CAN say that for all Borel sets [imath]B_i[/imath], [imath]P\left(\bigcap_{i=1}^{n} (Y_i \in B_i)\right) = \prod_{i=1}^{n} P(Y_i \in B_i).\tag{C}[/imath] Apparently, that is equivalent to saying for any distinct indices [imath]i_1, i_2, \dots, i_n[/imath] and for all Borel sets [imath]B_i[/imath], [imath]P\left(\bigcap_{i=i_1}^{i_n} (Y_i \in B_i)\right) = \prod_{i=i_1}^{i_n} P(Y_i \in B_i)\tag{$C_1$}.[/imath] I was surprised because I thought [imath]P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)[/imath] does not establish [imath]k[/imath]-wise independence, but apparently it does. Is there an analogue of [imath]C_1[/imath] for A or B? Note: I acknowledge the title may not be very good. Please suggest a better title if needed. | 956869 | Mutual Independence Definition Clarification
Let [imath]Y_1, Y_2, ..., Y_n[/imath] be iid random variables and [imath]B_1, B_2, ..., B_n[/imath] be Borel sets. It follows that [imath]P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)[/imath]...I think? If so, does the converse hold true? My Stochastic Calculus professor says it does (or maybe misinterpreted him somehow?), but I was under the impression that independence of the n random variables was equivalent to saying for any indices [imath]i_1, i_2, ..., i_k[/imath] [imath]P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)[/imath]. So, if the RVs are independent, then we can choose [imath]i_j=j[/imath] and k=n to get [imath]P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)[/imath], but given [imath]P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)[/imath], I don't know how to conclude that for any indices [imath]i_1, i_2, ..., i_n[/imath] [imath]P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)[/imath], if that's even the right definition. p.17 here seems to suggest otherwise. idk Help please? Also this: or So, this answer is to use the Omega part to establish pairwise independence and ultimately conclude independence. Without that assumption, we cannot conclude independence. Is that right? Why does that not contradict the definition of independence: [imath]P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)[/imath] ? |
1358651 | How to integrate [imath]\int_0^{\infty} \frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1+ e^{bx})}dx[/imath] where [imath]a,b > 0[/imath].
This [imath]\ \int_0^{\infty} \frac{e^{ax} - e^{bx}}{(1 + e^{ax})(1+ e^{bx})}dx \text{ where } a,b > 0. [/imath] is a problem that showed up on a GRE practice test. I believe you're supposed to use complex contour integration, but I'm not sure which contour to use. I think it's the keyhole contour, but I wasn't able to get anything useful. | 1885616 | How do you evaluate this integral: [imath]\int_0^\infty \frac {e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx[/imath]?
This is from GRE Math subject test, Question #55, from https://www.ets.org/s/gre/pdf/practice_book_math.pdf If [imath]a,b > 0[/imath], then what is the value of [imath] \int_0^\infty \frac {e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx [/imath] I'm not sure if it's Calc II integration or Residue Theorem from complex analysis, but I have no idea where to start ... |
1358207 | Any epi with codomain [imath]P[/imath] is split implies [imath]P[/imath] is projective
I'm struggling to prove that if any epi with codomain [imath]P[/imath] splits, then [imath]P[/imath] is a projective object. The converse direction I proved by factoring the identity to give a right inverse of the pi. How can I prove this claim? | 135777 | Different definitions of projective objects
There are various characterizations for an [imath]R[/imath]-module to be projective. Two of them can be generalized to any category: i) [imath]P[/imath] is an object such that given morphisms [imath]\alpha: A \rightarrow B[/imath] and [imath]\varphi: P \rightarrow B[/imath] such that [imath]\alpha[/imath] is an epimorphism, there exists a morphism [imath]\widetilde{\varphi}: P \rightarrow A[/imath] such that [imath]\alpha \circ \widetilde{\varphi} = \varphi[/imath]. (Lifting property) In other words, the functor [imath]\text{Hom}_{\mathsf{C}}(P,-): \mathsf{C} \rightarrow \mathsf{Set}[/imath] preserves epimorphisms. ii) [imath]P[/imath] is an object such that every epimorphism with codomain [imath]P[/imath] splits. It is easy to see that (i) [imath]\Rightarrow[/imath] (ii) holds in general, by taking [imath]\varphi = \text{id}_P[/imath]. Also if the category has the property that for every object [imath]X[/imath] there is an epimorphism [imath]\pi:P \rightarrow X[/imath] where [imath]P[/imath] satisfies (i) (an abelian category with this property is said to have enough projectives), we can show the reverse implication (ii) [imath]\Rightarrow[/imath] (i). My first question: Does (ii) [imath]\Rightarrow[/imath] (i) hold in general? I suspect the answer is no. If so, under which conditions does it hold? |
1359175 | Finding the acute angle between the planes [imath]x-3y+2z=14[/imath] and [imath]-x+y+z=10[/imath]
The two planes in this image form an acute angle. What is its value? Well, that's a confusing drawing. I don't even know why are there two [imath]\theta[/imath]s there (or what do [imath]S_1[/imath] and [imath]S_2[/imath] even mean). Anyway, here are the equations: [imath]x-3y+2z=14\\ -x+y+z=10[/imath] The angle is [imath]\cos^{-1}\left(\frac{(1,-3,2)\cdot(-1,1,1)}{||(1,-3,2)||\cdot||(-1,1,1)||}\right)[/imath] That's like [imath]107.97[/imath] degrees. That's clearly not acute. Did I do something wrong? | 1359161 | Having a plane equation equaled to zero for calculating its angle with another plane
I got to calculate the angle between two planes. [imath]x-3y+2z=14\\ -x+y+z=10[/imath] As far as I am concerned, that's simply the angle between their normal vectors. And such vectors are [imath](1,-3,2)[/imath] and [imath](-1,1,1)[/imath]. Therefore [imath]\cos^{-1}\left(\frac{(1,-3,2)\cdot(-1,1,1)}{||(1,-3,2)||\cdot||(-1,1,1)||}\right)[/imath] Which is like [imath]1.88[/imath] radians. Now then, to verify this I looked for online calculators, and I couldn't help but notice that all the calculators I found required to input the planes like this: [imath]x+y+z=\color{red}{0}[/imath] See, they ask for plane equations to be equaled to zero. My questions are, then: Do I need my plane equation to be equaled to zero to calculate such angle? Is there an advantage/disadvantage to having an equation equaled to zero? Is there a way to simplify my equation to equal it to zero? |
1359115 | invertibility of [imath]A^{-1} + B^{-1}[/imath]
Let [imath]A[/imath] and [imath]B[/imath] be two invertible [imath]n\times n[/imath] real matrices. Assume that [imath]A+B[/imath] is invertible. Show that [imath]A^{-1} + B^{-1}[/imath] is also invertible. | 782526 | How to prove that [imath]A^{-1} + B^{-1}[/imath] is invertible given the conditions
If [imath]A[/imath] and [imath]B[/imath] be two invertible [imath]n \times n[/imath] real matrices and [imath]A + B[/imath] is invertible, how to prove that [imath]A^{-1} + B^{-1}[/imath] is also invertible? |
1359631 | How many different functions are there that are equal to their own inverse?
I know that functions can be their own inverse such as [imath]f(x)=x[/imath] however I thought there were only two [imath]f(x)=x[/imath] and [imath]f(x)=-x[/imath]. Is there more? | 1356095 | Functions that are their own inversion.
What are the functions that are their own inverse? (thus functions where [imath] f(f(x)) = x [/imath] for a large domain) I always thought there were only 4: [imath]f(x) = x , f(x) = -x , f(x) = \frac {1}{x} [/imath] and [imath] f(x) = \frac {-1}{x} [/imath] Later I heard about a fifth one [imath]f(x) = \ln\left(\frac {e^x+1}{e^x-1}\right) [/imath] (for [imath]x > 0[/imath]) This made me wonder are there more? What are conditions that apply to all these functions to get more, etc. |
1359964 | Gamma function is too difficult
During my study to gamma function I find this form but I try to prove but I can't [imath]\Gamma(x) = \lim_{n\to \infty} \frac {n! n^{x-1}}{x(x-1)(x-2)........(x+n-1)} [/imath] So I want to know how I can prove this limit ? | 1345797 | Gamma function still hard for me
During my study I find a form for gamma function it was [imath]\Gamma (x) = \lim_{n\to\infty} \frac{n! n^{x-1}}{x(x-1)(x-2)........(x+n-1)}[/imath] And then by simplify this limit I get [imath]\lim_{n\to\infty} \frac{n^{x-1}}{{x+n-1\choose n}}[/imath] then I use Stirling's approximation and evaluate the limit to [imath]\ (x-1)! * e^{x-1} * (\frac{n}{n+x-1})^{x+n-1{\over2}}[/imath]$ What next I should do to get Gamma function from this limit ? |
728874 | What is the cardinality of the set of infinite cardinalities?
I am currently aware of only two infinite cardinalities: [imath]\aleph_0 = |\Bbb N|[/imath] [imath]\aleph_1 = |\Bbb R|[/imath] Questions: Is there an infinite number of infinite cardinalities? If yes, is this set of cardinalities countable or uncountable? I know that [imath]\aleph_0<\aleph_1[/imath] and I would tend to guess that the set of infinite cardinalities is countable at most (or possibly even finite), since there is no known element [imath]K[/imath] such that [imath]\aleph_0<K<\aleph_1[/imath]. BTW, are there any other acceptable operations (such as [imath]+,-,\times,/[/imath]) between elements in this set? Thanks | 2677165 | Understanding proper classes vs sets: why the set of all cardinal numbers cannot exist?
I'm trying to understand the difference between sets and proper classes. I found this PDF. I'm having trouble in understand the following at page 115: It basically says that if a set of all cardinal numbers exist, then it has a cardinal number [imath]k[/imath]. Then [imath]k[/imath] must be bigger than all cardinal numbers in the set of which it is the cardinal number, that is, the set of all cardinal numbers. Why [imath]k[/imath] must be bigger? As I know, cardinal numbers are numbers related to sizes of sets. There's nothing about size in this definition. |
1360001 | Elementary inequality: [imath]x^x+y^y \geq x^y +y^x[/imath]?
Let [imath]x,y>0[/imath], it seems (with numerical simulations) that [imath]x^x+y^y \geq x^y +y^x[/imath]. If this is true, it has to be well known by some people. Does this inequality have a name? several proofs? | 1302691 | [imath]\forall x,y>0, x^x+y^y \geq x^y + y^x[/imath]
Prove that [imath]\forall x,y>0, x^x+y^y \geq x^y + y^x[/imath] A friend of mine told me none of the teachers in my school have succeeded in proving this seemingly simple inequality (it was asked at an oral exam last year). I tried it myself, but I made no significant progress. |
1360283 | Cardinality of a set of non-continuous functions
I have to find the cardinality of the set of the non-continuous functions [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath]. I think we should look for function that at least have a point of discontinuity, but i don't have a clue, really. Any hint? | 1119166 | Cardinality of the set of all (real) discontinuous functions
This is a question from the book Introduction to Set Theory (Hrbacek and Jech), chapter 5, question 2.6. (Show that) The cardinality of the set of all discontinuous functions is [imath]2^{2^{\aleph_0}}[/imath]. [Hint: Using exercise 2.5, show that [imath]|\mathbb{R}^\mathbb{R}-C|=2^{2^{\aleph_0}}[/imath] whenever [imath]|C|\leq 2^{\aleph_0}[/imath].] Exercise 2.5, as referred to in the hint, provides the following result: For [imath]n>0[/imath], [imath]n \cdot 2^{2^{\aleph_0}} = \aleph_0 \cdot 2^{2^{\aleph_0}} = 2^{\aleph_0} \cdot 2^{2^{\aleph_0}} = 2^{2^{\aleph_0}} \cdot 2^{2^{\aleph_0}} = (2^{2^{\aleph_0}})^n = (2^{2^{\aleph_0}})^{\aleph_0} = (2^{2^{\aleph_0}})^{2^{\aleph_0}}=2^{2^{\aleph_0}}[/imath]. The only way I could answer this question was by making use of two theorems stated later in the same textbook. First, it is easy to prove that [imath]|\mathbb{R}^\mathbb{R}-C|[/imath] is an infinite set, since if it was finite we would have \begin{equation}2^{2^{\aleph_0}}=|\mathbb{R}^\mathbb{R}|=|\mathbb{R}^\mathbb{R}-C|+|C| \leq n+2^{\aleph_0}=2^{\aleph_0}< 2^{2^{\aleph_0}},\end{equation}which is a contradiction. Now I make use of a theorem requiring the axiom of choice: For every infinite set [imath]S[/imath] there exists a unique aleph [imath]\aleph_\alpha[/imath] such that [imath]|S|=\aleph_\alpha[/imath]. So I let [imath]|\mathbb{R}^\mathbb{R}-C|=\aleph_\alpha[/imath] for some ordinal [imath]\alpha[/imath]. Now assume [imath]\aleph_\alpha < 2^{2^{\aleph_0}}[/imath]. Here I make use of another theorem, which i am not sure I am allowed to do: For every [imath]\alpha[/imath] and [imath]\beta[/imath] such that [imath]\alpha \leq \beta[/imath], we have [imath]\aleph_\alpha+\aleph_\beta=\aleph_\beta[/imath]. So I basically then say either [imath]|\mathbb{R}^\mathbb{R}-C|+|C|=\aleph_\alpha[/imath] or [imath]|\mathbb{R}^\mathbb{R}-C|+|C|=2^{\aleph_0}[/imath], depending on whether [imath]\aleph_\alpha \leq 2^{\aleph_0}[/imath] or vice-versa. The assumption I am making is that [imath]2^{\aleph_0}=\aleph_1[/imath], which as I understand it is basically equivalent to the continuum hypothesis, as we are saying that the least uncountable number is [imath]\aleph_1[/imath] (but by the previous theorem requiring the axiom of choice this seems reasonable?). So anyway assuming I can do that, we again have a contradiction since then we are saying [imath]|\mathbb{R}^\mathbb{R}|=2^{\aleph_0}<2^{2^{\aleph_0}}[/imath] or [imath]|\mathbb{R}^\mathbb{R}|=\aleph_\alpha < 2^{2^{\aleph_0}}[/imath]. My main problem with my proof is that I am not making use of the result given in exercise 2.5 as suggested in the hint. Can anyone help with a proof making use of this result (stated in the second block quote above). |
1360411 | How to prove [imath]A=(A\setminus B)\cup (A\cap B)[/imath]
How to prove [imath]A=(A\setminus B)\cup (A\cap B)[/imath]. I have seen this problem and the solution is clear to me. Initially I was satisfied by my prove but now I think it is wrong. How I have proved [imath](A\cup B)=(A\setminus B)\cup(A\cap B)\cup(B\setminus A)\\ (A\cup B)\cap A=[(A\setminus B)\cup(A\cap B)\cup(B\setminus A)]\cap A=[(A\setminus B)\cap A]\cup[(A\cap B)\cap A]\cup[(B\setminus A)\cap A]\\ A=(A\setminus B)\cup(A\cap B)\cup\phi=(A\setminus B)\cup(A\cap B)[/imath] Have I proved it correctly. If yes then how? I mean that I have used [imath]A=(A\setminus B)\cup (A\cap B)[/imath] and [imath]B=(B\setminus A)\cup (A\cap B)[/imath] to get the [imath](A\cup B)=(A\setminus B)\cup(A\cap B)\cup(B\setminus A)[/imath] and then I'm using it to prove [imath]A=(A\setminus B)\cup (A\cap B)[/imath]. This means that I'm using a statement to prove itself! Kindly help me. | 843081 | Prove [imath]A = (A \setminus B) \cup (A \cap B)[/imath]
Prove [imath]A = (A \setminus B) \cup (A \cap B)[/imath] Logically, this is clearly true. I can explain why: start with [imath]A[/imath], remove all elements in [imath]B[/imath] and then add in any elements in both [imath]A[/imath] and [imath]B[/imath], which restores you back to [imath]A[/imath]. That's an explanation, but AFAIK, it's not a proof in the formal proof sense. I submitted a proof by a truth table which considers the four possible scenarios where an element is in/not in [imath]A[/imath]/[imath]B[/imath]. My professor asked me to redo the proof without a truth table. My question is what mechanisms and strategies can I use to prove this in a acceptable formal proof sense? |
1360648 | Computing some fundamental groups
I'm studying algebraic topology and got stuck. a. [imath]X_n\in \mathbb{R}^3[/imath] is the union of [imath]n[/imath] distinct lines through the origin. Find [imath]\pi_1(\mathbb{R}^3-X_n)[/imath] for each [imath]n[/imath]. b. Let [imath]X[/imath] be the sum of two tori [imath]S_1\times S_1[/imath] by identifying a circle [imath]S_1\times x_0[/imath] in a torus with [imath]S_1\times x_0[/imath] of the other torus. Find [imath]\pi_1(X)[/imath]. For (a), [imath]\pi_1(\mathbb{R}^3-X_1)=\mathbb{Z}.[/imath] But I cannot compute [imath]n\ge 2[/imath] cases.. for (b), I guess the group is [imath]<a,b,c|ab=ba,bc=cb>[/imath] from Van Kampen. Is it correct? | 929969 | Fundamental group of two tori with a circle ([imath]S^1β[/imath]{[imath]x_0[/imath]}) identified
Compute the fundamental group of the space obtained from two tori [imath]S^1βS^1[/imath] by identifying a circle [imath]S^1β[/imath]{[imath]x_0[/imath]} in one torus with the corresponding circle [imath]S^1β[/imath]{[imath]x_0[/imath]} in the other. Using van Kampen's theorem, I can fairly quickly show that if [imath]T_1[/imath] is one torus and [imath]T_2[/imath] is the other, then the group has to be isomorphic to [imath]((Ο_1(T_1) β Ο_1(T_2)))/N β
(\mathbb{Z} β \mathbb{Z})β (\mathbb{Z} β \mathbb{Z})/N[/imath], where N is generated by elements of the form [imath]i_{12}(w)i_{21}(w)^{-1}[/imath], w is in [imath]\pi_1(S^1β[/imath]{[imath]x_0[/imath]}[imath]) β
\mathbb{Z}[/imath] and [imath]i_{12}[/imath], [imath]i_{21}[/imath] are the maps from [imath]\pi_1(S^1β[/imath]{[imath]x_0[/imath]}[imath])[/imath] to [imath]T_1[/imath], [imath]T_2[/imath] respectively determined by inclusion. My problem is, as it seems to always be with van Kampen-related problems, figuring out what the elements of N look like. My attempt to do this in a pure-algebra way, as far as I can tell, failed me: I tried examining the case where w represents a loop that goes n (some integer) times around the circle, but then it seems like [imath]i_{12}(w)[/imath] "[imath]=[/imath]" [imath](n, 0)[/imath] in [imath]T_1[/imath] and [imath]i_{21}(w)[/imath] "[imath]=[/imath]" [imath](n, 0)[/imath] in [imath]T_2[/imath], which says to me that [imath]i_{12}(w)i_{21}(w)^{-1} [/imath]"[imath]=[/imath]"[imath] (n, 0) + (-n, 0) = 0[/imath]. The notation here isn't right, since [imath](n, 0)[/imath] in [imath]T_1[/imath] is not the same as [imath](n, 0)[/imath] in [imath]T_2[/imath] and their addition (composition?) shouldn't be written quite that way, hence my quotes around the equals signs, but certainly no loop in [imath]S^1β[/imath]{[imath]x_0[/imath]} is going to magically include anything but the 0 loop in either torus's second component, right? But this doesn't seem right to me, since if a, b are the generators of the circles in [imath]T_1[/imath] and d, e are the generators of the circles in [imath]T_2[/imath], and c is the constant path then the identification of the circle should make, say, [imath]a = d[/imath], so for example [imath](a, b)*(d, e)[/imath] [imath] = (a^2,b)*(c, e)[/imath] [imath] = (c, b)*(d^2,e)[/imath]. This doesn't match with what I just worked out N might be, but I also can't figure out what N should look like to make this equivalence make sense. My attempt to understand this geometrically has gone even worse. What I know for sure is that the resulting figure isn't the "2-torus", 2 tori glued together at a disc. What I don't know for sure is whether this identification should actually create the hypertorus [imath]S^1βS^1βS^1[/imath] or something completely different. |
1361173 | In [imath](X,d)[/imath] metric space, an intersection of finite families of open sets is open.
Can anyone help prove this? I am allowed to used the fact that [imath]X[/imath] and [imath]\emptyset[/imath] are open and that a union of an arbitrary family of open sets is open. I tried to understand what it says in my textbook, but it makes no sense to me. It goes like this: If the sets [imath]U_1, U_2,...,U_k[/imath] are open and [imath]x[/imath] is a point which lays in their intersection, then there exist a finite number of balls [imath]B_{r_1}(x),..,B_{r_k}(x)[/imath] such that [imath]B(x,r_i)\subset U_i[/imath]. The ball [imath]B(x,r)[/imath] with the radius [imath]r=\min\{r_1,..r_k\}[/imath] lays in [imath]\bigcap_{i=1}^{k}U_i[/imath] | 1209533 | Unions and Intersections of Open Sets are Open
Let [imath](X,d)[/imath] be a metric space. Prove: the union of any open sets in [imath]X[/imath] is open in [imath]X[/imath] the intersection of a finite number of open sets in [imath]X[/imath] is open in [imath]X[/imath] I could prove the first one but how do we prove the second one? |
1360674 | trigonometry solution
[imath]\frac{\sin A - \cos A}{\sin A+\cos A}=\frac{X}{Y}[/imath] Then prove that [imath]X^2 + Y^2 = 2[/imath]. Answer : [imath]\frac{\sin A-\cos A}{\sin A+\cos A}=\frac{x}{y}[/imath] [imath]\implies x=k(\sin A-\cos A)[/imath] and [imath]y=k(\sin A+\cos A)[/imath]. Squaring on both sides and adding [imath]x^2+y^2=k^2(\sin A-\cos A)^2+k^2(\sin A+\cos A)^2[/imath] [imath]\implies x^2+y^2=k^2[\sin^2A+\cos^2A-2\sin A\cos A+\sin^2A+\cos^2A+2\sin A\cos A][/imath] [imath]\implies x^2+y^2=2 k^2[/imath] But how can eliminate k.means by what logic we can say k=+1 or -1. Please suggest. Thanks. | 1360652 | trigonometry query
[imath]\frac{\sin A - \cos A}{\sin A+\cos A}=\frac{X}{Y}[/imath] Then prove that [imath]X^2 + Y^2 = 2[/imath]. |
1341154 | A geometric interpretation of a geometric series
The following code is compiled by TikZ to draw a right triangle so that the enclosed area is partitioned by infinitely many triangles similar to itself. I saw on this site an image of this from an old textbook. The post indicated that the display is a geometric interpretation of the geometric series. I would like to either find that post or know how this display gives a geometric interpretation of the geometric series. \documentclass{amsart} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{tikz} \usetikzlibrary{calc,angles,positioning,intersections,quotes,decorations.markings} \usepackage{pgfplots} \pgfplotsset{compat=1.11} \begin{document} \noindent \hspace*{\fill} \begin{tikzpicture} %Rays k and [imath]\ell[/imath] have a common endpoint at the origin, which is labeled A. With respect to %the horizontal line through A, k is inclined at an angle of 50 degrees, and [imath]\ell[/imath] is %inclined at an angle of 15 degrees. To place the label for A, the rays are extended 15pt by an %invisible path command "before" A. The endpoints of these extensions are called "label_A_below" %and "label_A_above" and the midpoint of the line segment between label_A_below and label_A_above %is called "label_A". A node command typesets "A" at the point 7.5pt from A on the invisible %line segment between A and label_A . \coordinate (A) at (0,0); \draw[fill] (A) circle (1.5pt); %B' is a point on ray k and C' is a point on [imath]\ell[/imath]. These points are used to draw two %invisible paths from A. On these paths, two sides of a triangle will be drawn. \coordinate (B') at (50:9.25); \path[name path=ray_k] (A) -- (B'); \coordinate (C') at (15:7.5); \path[name path=ray_ell] (A) -- (C'); \path (A) -- (C'); %These commands label a point "B_3" 3.75cm from A on ray [imath]k[/imath]. (Two more points B_2 and B_1 on k %that are further from A than B_3 are to be defined, and three more points B_4, B_5, and B_6 %on k that are closer to A than B_3 are to be defined.) \coordinate (B_3) at (50:{15/4}); %These commands label the projection of AB_3 onto \ell "C_3." A line segment is drawn between %B_3 and C_3, and a right-angle mark is placed at C_3. (Commands for the label for C_3 are %issued after the point [imath]B_{4}[/imath] on k is defined.) \coordinate (C_3) at ([imath](A)!(B_3)!(C')[/imath]); \draw (B_3) -- (C_3); %The following command make the right-angle mark at C_3. \coordinate (U_3) at ([imath](C_3)!3mm!45:(C')[/imath]); \draw (U_3) -- ([imath](C_3)!(U_3)!(C')[/imath]); \draw (U_3) -- ([imath](C_3)!(U_3)!(B_3)[/imath]); %These commands draw a line segment perpendicular to ray k from B_3 to ray [imath]\ell[/imath]. %The intersection is labeled C_2. (The label for C_2 is placed after the point %B_2 is defined.) \path[name path=perpendicular_line_segment_from_B_3] (B_3) -- ([imath](B_3)!3.5cm!90:(A)[/imath]); \coordinate[name intersections={of=perpendicular_line_segment_from_B_3 and ray_ell,by={C_2}}]; \draw (B_3) -- (C_2); %The following commands make the right-angle mark at B_3. \coordinate (V_3) at ([imath](B_3)!3mm!-45:(B')[/imath]); \draw (V_3) -- ([imath](B_3)!(V_3)!(B')[/imath]); \draw (V_3) -- ([imath](B_3)!(V_3)!(C_2)[/imath]); %The following commands define B_2 as the intersection of k and a line segment %perpendicular to [imath]\ell[/imath] through C_2. \path[name path=perpendicular_line_segment_from_C_2] (C_2) -- ([imath](C_2)!4cm!90:(C')[/imath]); \coordinate[name intersections={of=perpendicular_line_segment_from_C_2 and ray_k,by={B_2}}]; \draw (C_2) -- (B_2); %The following commands make the right-angle mark at C_2. \coordinate (U_2) at ([imath](C_2)!3mm!45:(C')[/imath]); \draw (U_2) -- ([imath](C_2)!(U_2)!(C')[/imath]); \draw (U_2) -- ([imath](C_2)!(U_2)!(B_2)[/imath]); %These commands draw a line segment perpendicular to ray k from B_2 to ray [imath]\ell[/imath]. %The intersection is labeled C_1. (The label for C_1 is placed after the point %B_1 is defined.) \path[name path=perpendicular_line_segment_from_B_2] (B_2) -- ([imath](B_2)!5cm!90:(A)[/imath]); \coordinate[name intersections={of=perpendicular_line_segment_from_B_2 and ray_ell,by={C_1}}]; \draw (B_2) -- (C_1); %The following commands make the right-angle mark at B_1. \coordinate (V_2) at ([imath](B_2)!3mm!-45:(B')[/imath]); \draw (V_2) -- ([imath](B_2)!(V_2)!(B')[/imath]); \draw (V_2) -- ([imath](B_2)!(V_2)!(C_1)[/imath]); %The following commands define B_1 as the intersection of k and a line segment %perpendicular to [imath]\ell[/imath] through C_1. \path[name path=line_segment_from_C_1_to_ray_k] (C_1) -- ([imath](C_1)!5.25cm!90:(C')[/imath]); \coordinate[name intersections={of=line_segment_from_C_1_to_ray_k and ray_k,by={B_1}}]; \draw (C_1) -- (B_1); %These commands label the projection of AC_3 onto k "B_4." \coordinate (B_4) at ([imath](A)!(C_3)!(B')[/imath]); \draw (B_4) -- (C_3); %These commands label the projection of AB_4 onto [imath]\ell[/imath] "C_4." \coordinate (C_4) at ([imath](A)!(B_4)!(C')[/imath]); \draw (B_4) -- (C_4); %The following commands make the right-angle mark at B_4. \coordinate (V_4) at ([imath](B_4)!3mm!-45:(B')[/imath]); \draw (V_4) -- ([imath](B_4)!(V_4)!(B')[/imath]); \draw (V_4) -- ([imath](B_4)!(V_4)!(C_3)[/imath]); %These commands label the projection of AC_4 onto k "B_5." \coordinate (B_5) at ([imath](A)!(C_4)!(B')[/imath]); \draw (B_5) -- (C_4); %The following commands make the right-angle mark at C_4. \coordinate (U_4) at ([imath](C_4)!3mm!45:(C')[/imath]); \draw (U_4) -- ([imath](C_4)!(U_4)!(C')[/imath]); \draw (U_4) -- ([imath](C_4)!(U_4)!(B_4)[/imath]); %The following commands make the right-angle mark at B_5. \coordinate (V_5) at ([imath](B_5)!3mm!-45:(B')[/imath]); \draw (V_5) -- ([imath](B_5)!(V_5)!(B')[/imath]); \draw (V_5) -- ([imath](B_5)!(V_5)!(C_4)[/imath]); %These commands label the projection of AB_5 onto [imath]\ell[/imath] "C_5." \coordinate (C_5) at ([imath](A)!(B_5)!(C')[/imath]); \draw (B_5) -- (C_5); %The following commands make the right-angle mark at C_5. \coordinate (U_5) at ([imath](C_5)!3mm!45:(C')[/imath]); \draw (U_5) -- ([imath](C_5)!(U_5)!(C')[/imath]); \draw (U_5) -- ([imath](C_5)!(U_5)!(B_5)[/imath]); %The following commands draw five more smaller, similar triangles, with dashed lines, %to suggest that the process continues. \coordinate (B_6) at ([imath](A)!(C_5)!(B')[/imath]); \draw[dashed] (C_5) -- (B_6); \coordinate (C_6) at ([imath](A)!(B_6)!(C')[/imath]); \draw[dashed] (B_6) -- (C_6); \coordinate (B_7) at ([imath](A)!(C_6)!(B')[/imath]); \draw[dash dot] (C_6) -- (B_7); \coordinate (C_7) at ([imath](A)!(B_7)!(C')[/imath]); \draw[dash dot] (B_7) -- (C_7); \coordinate (B_8) at ([imath](A)!(C_7)!(B')[/imath]); \draw[dotted] (C_7) -- (B_8); \coordinate (C_8) at ([imath](A)!(B_8)!(C')[/imath]); \draw[dotted] (B_8) -- (C_8); %These commands draw the two sides of the triangle with the common endpoint A. \draw (A) -- (C_1); \draw (A) -- (B_1); \end{tikzpicture} \end{document} | 483575 | Question on geometrical proof of Geometric Series
The following image is from Geometric Series Proofs: An Annotated Bibliography. Please explain why it is said that: "[imath]ON[/imath] is the limit of the sum [imath]1+x+\dots[/imath]." Thank you. Edit: I guess what through me off was the word "limit"! |
1356973 | Generate all multisets of length k for n symbols
I am trying to generate a list of all multisets of length [imath]k[/imath] in a set with [imath]n[/imath] symbols. For example, if I had the set [imath]S = {A, B, C}[/imath] I would expect the following output for [imath]k = 2[/imath] and [imath]n = 3[/imath]: [imath]O = {(A,A),(A,B),(A,C),(B,B),(B,C),(C,C)}[/imath] What is the proper way to go about generating this list? | 17083 | How to find unique multisets of n naturals of a given domain and their numbers?
Let's say I have numbers each taken in a set [imath]A[/imath] of [imath]n[/imath] consecutive naturals, I ask myself : how can I found what are all the unique multisets, which could be created with [imath]k[/imath] elements of this set [imath]A[/imath]? For example I've got [imath]A=[1,2,\dots,499][/imath]. If I wanted to create unique multiset of 3 elements, I would search all the multisets [imath]\{a,b,c\}[/imath] such as [imath]a\leq b\leq c[/imath] and then have all the unique multisets. As such for each elements [imath]a[/imath] there is [imath]499-(a-1)=500-a[/imath] possibilities for [imath]b[/imath] and [imath]500-b[/imath] possibilities for [imath]c[/imath]. Unfortunately I'm stumped here and can't find the number of possible combination, as I didn't do any maths for years. I know that I should have some kind of product, but I don't know how to find the product anymore. So first, I would like to know if I am right in my original assumption, or if I am looking in the wrong direction. Second I would like to approach that as if I were doing an homework, as I would like to understand the logic behind it: What would be the formula to give the numbers of multisets of [imath]k[/imath] elements from a set [imath]A[/imath] of [imath]n[/imath] consecutive naturals? P.S. I'm totally new to math.SE, as such, I'm not sure I tagged the post appropriately. |
1361256 | Can I have a logical explanation for why this number is so ridiculously close to a whole number?
[imath]e^ {\pi\sqrt{163}}=262537412640768743.9999999999992\cdots[/imath] Why does this number run so incredibly close to a whole number? Can I have a logical explanation for why this finding? I know how to calculate it, but I want an answer that would explain it to me by intuition only. Thank you very much for any help. It is not the same as the other similar problem because I am looking for a logical reason and the other is wondering whether or not it is concidence. | 4544 | Why is [imath]e^{\pi \sqrt{163}}[/imath] almost an integer?
The fact that Ramanujan's Constant [imath]e^{\pi \sqrt{163}}[/imath] is almost an integer ([imath]262 537 412 640 768 743.99999999999925...[/imath]) doesn't seem to be a coincidence, but has to do with the [imath]163[/imath] appearing in it. Can you explain why it's almost-but-not-quite an integer in layman's terms (I'm not a mathematician)? |
1362098 | how to solve [imath]\int_{-\infty}^\infty e^{-x^2-x{\tau}} \cdot x\ dx[/imath]?
I took [imath]{-x^2-x{\tau}} = k [/imath] on differentiating it [imath]dk=(-2x-\tau)dx[/imath] but this substitution doesn't work how can i proceed further . | 1272402 | How do I solve this improper integral: [imath]\int_{-\infty}^\infty e^{-x^2-x}dx[/imath]?
I'm trying to solve this integral: [imath]\int_{-\infty}^\infty e^{-x^2-x}dx[/imath] WolframAlpha shows this to be approximately [imath]2.27588[/imath]. I tried to solve this by integration by parts, but I just couldn't get there. I'd be glad if someone could show me how to do it. I've included my attempt at the problem below: Note [imath]\int_{-\infty}^\infty e^{-x^2-x} = \int_{-\infty}^\infty e^{-x^2}e^{-x}dx[/imath]. Then we can integrate by parts. Let [imath]u(x) = e^{-x^2}[/imath]. Then [imath]u'(x) = -2xe^{-x}[/imath]. Let [imath]v'(x) = e^{-x}[/imath]. Then [imath]v(x) = -e^{-x}[/imath]. Then [imath]u(x)v'(x) = u(x)v(x) - \int v(x)u'(x)dx[/imath], i.e. [imath]\int_{-\infty}^\infty e^{-x^2}e^{-x}dx = -e^{-x^2}e^{-x} - \int e^{-x}2xe^{-x}dx = -e^{-x^2-x} - 2\int e^{-2x}xdx[/imath] Then we integrate [imath]\int e^{-2x}xdx[/imath] by parts. We pick [imath]u(x) = x[/imath], [imath]u'(x) = dx[/imath], [imath]v'(x) = e^{-2x}[/imath], and [imath]v(x) = \int e^{-2x} dx = \frac {-1}{2} e^{-2x}[/imath] by u-substitution. Skipping some steps, it follows that [imath]\int e^{-2x}xdx = -\frac{1}{4}(e^{-2x})(2x+1)[/imath] Then [imath] -e^{-x^2-x} - 2\int e^{-2x}xdx = -e^{-x^2-x} + \frac{1}{2}(e^{-2x})(2x+1)[/imath] Which, when evaluated numerically, doesn't yield the desired result. So there's a problem here. Maybe someone else knows how to do this. |
1361825 | Easy proof of Cayley Hamilton theorem
What is wrong with this proof of Cayley hamilton? If [imath]A[/imath] is [imath]n \times n[/imath] matrix and [imath]P[/imath] is its characteristic polynomial then [imath]P(A) = 0.[/imath] Proof: [imath]P(x) = \det(A - xI) \implies P(A) = \det(A - A) = \det(0) = 0.[/imath] | 1289711 | A bad CayleyβHamilton theorem proof
Given [imath]A\in M_{n \times n}(\mathbb{F})[/imath] and [imath]p_{A}(x)=\det(xI-A)[/imath] why saying that [imath]\det(AI-A)=0[/imath] is not valid? |
1362106 | Why do we require differential manifolds to be Hausdorff?
Among the requirements for a differential manifold [imath]M[/imath] is that it be connected and Hausdorff. What fails if a manifold is not Hausdorff? | 472998 | Why do we need Hausdorff-ness in definition of topological manifold?
Suppose [imath]M^n[/imath] is a topological manifold, then [imath]M^n[/imath] locally looks like [imath]\mathbb{R}^n[/imath]. [imath]M^n[/imath] is locally Hausdorff, since [imath]\mathbb{R}^n[/imath] is Hausdorff and Hausdorff-ness is a topological invariant. All I want to understand is: 1) Does locally Hausdorff-ness imply Hausdorff-ness? (I can not imagine a locally Hausdorff topological space that is not globally Hausdorff) 2) Why do we need Hausdorff-ness in definition of the topological manifold? Is locally Hausdorff-ness not sufficient? If not, why? Can anyone say anything that might be helpful? |
1081241 | Build a bijection [imath]\mathbb{R} \to \mathbb{R}\setminus \mathbb{N}[/imath]
Build a bijection [imath]f: \mathbb{R} \to \mathbb{R}\setminus \mathbb{N}[/imath]. What about [imath]f(x)=\pi \cdot x?[/imath] | 1068949 | Find a bijection from [imath]\mathbb R[/imath] to [imath]\mathbb R-\mathbb N[/imath]
Find a bijection from [imath]\mathbb R \to \mathbb R-\mathbb N[/imath]. I want to set my function up such that all natural numbers get mapped to [imath]n+.1[/imath] and reals of the form [imath]n+.1[/imath] to [imath]2n+.1[/imath] is this correct? |
1361996 | A question about polynomial in two variables
Let [imath]f:\mathbb R^2 \to \mathbb R[/imath] be a function such that for any [imath]b \in \mathbb R[/imath] , the function [imath]f_b : \mathbb R \to \mathbb R[/imath] defined as [imath]f_b(x):=f(x,b) , \forall x \in \mathbb R[/imath] , is a polynomial in [imath]x[/imath] and for any [imath]a \in \mathbb R[/imath] , the function [imath]f_a : \mathbb R \to \mathbb R[/imath] defined as [imath]f_a(y):=f(a,y) , \forall y \in \mathbb R[/imath] , is a polynomial in [imath]y[/imath] . Then is the function [imath]f(x,y)[/imath] a polynomial in [imath]x,y[/imath] i.e. is [imath]f \in \mathbb R [x,y][/imath] ? | 606523 | Real polynomial in two variables
I have problems proving the following result: Each [imath]f: \mathbb{R}^2 \rightarrow \mathbb{R}[/imath] such that [imath]\forall a,b \in \mathbb{R} \ : \ f_a(y) := f(a,y), \ f_b(x) := f(x,b) [/imath] are polynomials is a polynomial with two variables. If I consider [imath]f[/imath] as a function of [imath]x[/imath], then its derivative [imath]f'(x) = \frac{\partial f}{\partial b}(x)[/imath]. Similarly if we treat [imath]f[/imath] as a function of [imath]y[/imath]. I assume that [imath]f_a(y) = \frac{\partial f}{\partial a} (y) [/imath] and [imath]f_b(x)=\frac{\partial f}{\partial b}(x)[/imath]. But I am not sure if we can assume that, because the degree of the derivative should be smaller than the degree of the original function (and it isn't). Actually, I'm not even sure if what I'm trying to prove is true, because in the original formulation of the problems there is written [imath]f_a(y) := (a,y), \ f_b(x):=(x,b)[/imath]. But that didn't make sense. Could you help me here? Thank you. |
1362500 | euler function property [imath]\varphi(n)\geq\sqrt{n}[/imath]
How to prove that [imath]\varphi(n)\geq \sqrt{n}[/imath] for every positive integer [imath]n[/imath] distinct of [imath]2[/imath] and [imath]6[/imath]. | 308686 | How to justify [imath]\phi(n) \ge \sqrt{n}[/imath]
If [imath]\phi(n)[/imath] is the Euler-totient function, how can I show that [imath]\phi(n) \ge \sqrt{n}[/imath]? |
1362523 | Why is the Gamma function off by 1 from the factorial?
Why didn't they define it as [imath] \tilde \Gamma(x) = \int_0^\infty t^x e^{-t} \, dt ?[/imath] Then the definition would have two less characters than the standard definition of [imath]\Gamma(x)[/imath], and we would have [imath]\tilde \Gamma(n) = n![/imath] for [imath]n[/imath] a non-negative integer. And this would save a lot of confusion. | 18356 | Why isn't the gamma function defined so that [imath]\Gamma(n) = n! [/imath]?
As a physics student, I have occasionally run across the gamma function [imath]\Gamma(n) \equiv \int_0^{\infty}t^{n-1}e^{-t} \textrm{d}t = (n-1)![/imath] when we want to generalize the concept of a factorial. Why not define the gamma function so that [imath]\Gamma(n) = n![/imath] instead? I realize either definition is equally good, but if someone were going to ask me to choose one, I would choose the second option. Are there some areas of mathematics where the accepted definition looks more natural? Are there some formulas that work out more cleanly with the accepted definition? |
1362588 | Let [imath]a_n \to +\infty[/imath], and let [imath]b_n \to +\infty[/imath]. Show that [imath]a_n + b_n \to +\infty[/imath]
I know how to show [imath]a_n[/imath] diverges, or [imath]b_n[/imath]. I know the sum of two convergent limits is the sum of the limits. But how do we show that two divergent limits are divergent? | 554147 | if [imath]A_n \longrightarrow \infty [/imath] and [imath]B_n \longrightarrow \infty [/imath] then [imath](A_n+B_n) \longrightarrow \infty[/imath]
if [imath]A_n \longrightarrow \infty [/imath] and [imath]B_n \longrightarrow \infty [/imath] then [imath](A_n+B_n) \longrightarrow \infty[/imath]. How do you prove it? |
1362728 | How do I prove that the standard definition of prime ideal is equivalent to that of Krull's?
Definition Let [imath]R[/imath] be a commutative ring and [imath]I[/imath] be a proper ideal of [imath]R[/imath]. Then [imath]I[/imath] is prime if and only if [imath]\forall a,b\in I, a\in I[/imath] or [imath]b\in I[/imath]. Let [imath]R[/imath] be a commutative ring and [imath]P[/imath] be a prime ideal of [imath]R[/imath]. Let [imath]I,J[/imath] be ideals of [imath]R[/imath] such that [imath]IJ\subset P[/imath]. How do I prove that [imath]I\subset P[/imath] or [imath]J\subset P[/imath]? (which is the Krull's definition)? | 73213 | Equivalence of definitions of prime ideal in commutative ring
Could someone prove that in a commutative ring [imath]R[/imath] the following two definitions of prime ideal are equivalent: 1) An ideal [imath]P[/imath] is prime if [imath]P\neq R[/imath] and if for all ideals [imath]A,B[/imath] of [imath]R[/imath] with [imath]AB\subseteq P[/imath] we have that [imath]A \subseteq P[/imath] or [imath]B \subseteq P[/imath]. 2) An ideal [imath]P[/imath] is prime if [imath]P\neq R[/imath] and if [imath]ab \in P \implies a \in P[/imath] or [imath]b\in P[/imath]. I have read (in Artin as well as online) that these are equivalent, but I cannot find a suitable proof. Artin's proof of [imath]1) \implies 2)[/imath] is one line: Take [imath]A=(a)[/imath] and [imath]B=(b)[/imath]. I do not see how this is a proof. If someone could explain the implication as if I were a complete novice to ring theory (which I am), that would be most helpful. Note: I am not assuming that [imath]R[/imath] is unital, only that it is commutative. |
1363067 | [imath]\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{k^2 + 1}[/imath]
Please, explain, how to calculate this limit? [imath]\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{k}{k^2 + 1}[/imath] | 1278311 | [imath]\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k}{k^{2}+1}[/imath]
Could please give me a hint. [imath]\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{k}{k^{2}+1}[/imath] It looks like an integral, but I failed to figure out the function. |
1363141 | IMO 2015 problem 2
Determine all triples [imath](a,b,c)[/imath] of positive integers such that each of the numbers [imath]ab-c, \quad bc-a, \quad ca-b[/imath] is a power of [imath]2[/imath]. (A power of [imath]2[/imath] is an integer of the form [imath]2^n[/imath], where [imath]n[/imath] is a non-negative integer.) -- So far I only figured out that if either two of [imath]a,b,c[/imath] is the same value, there are only 2 triples: [imath](2,2,2)[/imath] and [imath](2,2,3)[/imath]. But for [imath]a\ne b\ne c\ne a[/imath], I'm a bit lost as how to proceed. By trial and error there's a triple [imath](3,5,7)[/imath] but how to get the complete triples? Could someone pls gimme some hint? | 1356164 | Finding [imath](a, b, c)[/imath] with [imath]ab-c[/imath], [imath]bc-a[/imath], and [imath]ca-b[/imath] being powers of [imath]2[/imath]
This is a 2015 IMO problem. It seems difficult to solve. Find all triples of positive integers [imath](a, b, c)[/imath] such that each of the numbers [imath]ab-c[/imath], [imath]bc-a[/imath], and [imath]ca-b[/imath] is a power of [imath]2[/imath]. Four such triples are [imath](a,b,c)=(2,2,2),(2,3,2),(3,5,7),(2,6,11)[/imath]. |
1361032 | Why does dZ(t)dt=(dt)^2=0
I am having trouble with this question: [imath]\mbox{Let Y(t)}=\begin{cases} 0 & -1\le t \le 0\\ Z(t)-tZ(t) & 0 < t < 1 \\ 1-t & 1 \le t \le 2\end{cases}[/imath] Find the quadratic variations of [imath]Y(t)[/imath] over [-1,2]. Z(t) is standard brownian motion. The solution to this question says: [imath]dY(t)=\begin{cases} 0 & -1\le t \le 0\\ dZ(t)-Z(2) dt & 0 < t < 1 \\ -dt & 1 \le t \le 2\end{cases}[/imath] [imath][dY(t)]^2=\begin{cases} 0 & -1\le t \le 0\\ [dZ(t)]^2 & 0 < t < 1 \\ 0 & 1 \le t \le 2\end{cases}[/imath] ...by ItΓ΄'s lemma because [imath]dZ(t)dt=(dt)^2=0[/imath]. So, the quadratic variation is [imath]\int_{-1}^2 [dY(t)]^2 = \int_0^1 dt = 1[/imath][imath][/imath] I do not understand why dZ(t)dt=(dt)^2=0$. I understand how to apply ItΓ΄'s lemma to stochastic differential equations, but I don't know much about it beyond that. Hopefully someone can clear this up in a relatively simple way. | 81865 | Wiener Process [imath]dB^2=dt[/imath]
Why is [imath]dB^2=dt[/imath]? Every online source I've come across lists this as an exercise or just states it, but why isn't this ever explicitly proved? I know that [imath]dB=\sqrt{dt}Z[/imath], but I don't know what squaring a Gaussian random variable means. |
908894 | Infinite Series [imath]\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots[/imath]
How do I find the sum of the following infinite series: [imath]\left(\frac12+\frac14-\frac23\right)+\left(\frac15+\frac17-\frac26\right)+\left(\frac18+\frac{1}{10}-\frac29\right)+\cdots[/imath] The series definitely seems to be convergent. | 657241 | Infinite Series [imath]1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots[/imath]
Was given the following infinite sum in class as a question, while we were talking about Taylor series expansions of [imath]\ln(1+x)[/imath] and [imath]\arctan(x)[/imath]: [imath]1+\frac12-\frac23+\frac14+\frac15-\frac26+\cdots[/imath] The question isn't homework or anything, just a thought tease. I tried for a long while but couldn't find anything remotely close. Thanks in advance for the help. |
774814 | Prove that [imath]n[/imath] divides [imath]\phi(a^n-1)[/imath], where [imath]\phi[/imath] is Euler's [imath]\phi[/imath]-function.
Let a, n be positive integers. Prove that n divides [imath]\phi(a^n-1)[/imath], where [imath]\phi[/imath] is Euler's [imath]\phi[/imath]-function. I know this problem can be done using number theory approaches, however I am rusty on those concepts can someone help me? | 2267481 | Number Theory: Prove that [imath]m | \phi(a^m - 1)[/imath]
[imath]a[/imath] and [imath]m[/imath] are natural numbers and [imath]a>1[/imath]. Prove that [imath]m | \phi(a^m - 1)[/imath]. Any hints how can I prove this statement? |
1363287 | Behaviour of [imath]L^2[/imath] functions at infinity
Is it possible to prove that if [imath]f\in L^2(\mathbb {R}) [/imath] then [imath]\exists\lim_{x\to\pm\infty}\lvert f\rvert^2[/imath] and [imath]\lim_{x\to\pm\infty}\lvert f\rvert^2=0[/imath]? If not, is it easy to find a counterexample? | 85975 | Integrable function [imath]f[/imath] on [imath]\mathbb R[/imath] does not imply that limit [imath]f(x)[/imath] is zero
1) Construct a continuous function [imath]f[/imath] on [imath]\mathbb{R}[/imath] that is integrable on [imath]\mathbb{R}[/imath] but [imath]\displaystyle\limsup_{x \to \infty} f(x) = \infty[/imath]. I took the function that is equal to [imath]n[/imath] on [imath][n, n + 1/n^{3})[/imath] and made it continuous by saying that [imath]f[/imath] is the line segment joining [imath]n[/imath] and [imath]n+1[/imath] on [imath][n + 1/n^{3}, n+1)[/imath]. But I am failing to prove this integrable. For this, [imath]\lim f(x) = \infty[/imath] but how do you prove in general, if limit does not exist that [imath]\limsup[/imath] is infinity? 2) Prove that if [imath]f[/imath] is uniformly continuous and integrable on [imath]\mathbb{R}[/imath] we have [imath]\displaystyle\lim_{|x| \to \infty} f(x) = 0[/imath]. Any help is appreciated. Thanks |
1364133 | Show that [imath]\mathbb{Z}_4\rightarrow \mathbb{Z}_8\oplus \mathbb{Z}_2\rightarrow \mathbb{Z}_4[/imath] is exact
I want to know whether [imath]0\rightarrow \mathbb{Z}_4\stackrel{f}\rightarrow \mathbb{Z}_8\oplus \mathbb{Z}_2\rightarrow \mathbb{Z}_4\rightarrow 0[/imath] is exact wrt group homomorphism under addition. Since [imath]\mathbb{Z}_2^2\neq \mathbb{Z}_4[/imath], so [imath] f(1)=(x,1) [/imath] Since [imath]\mathbb{Z}_4[/imath] is cyclic of order [imath]4[/imath], so [imath] 4x=0,\ kx\neq 0\ (1\leq k\leq 3)[/imath] Hence [imath]x=2[/imath] or [imath] 6[/imath] so that [imath] {\rm image}\ f = \{ (2,1),\ (4,0),\ (6,1),\ (0,0)\}[/imath] (That is, image of [imath]f[/imath] is unique) Hence by direct computation, [imath] (\mathbb{Z}_8\oplus \mathbb{Z}_2)/f(\mathbb{Z}_4) = ((1,1)) = \{ (0,0),\ (1,1),\ (2,0),\ (3,1) \} [/imath] Here element is a representative. That is [imath](1,1)[/imath] is corresponded to [imath]\{ (1,1),\ (3,0),\ (5,1),\ (7,0) \}[/imath], whose order is [imath]4[/imath]. Hence it is exact. Here we have a question : Question : To prove this, we must compute directly ? Does there exists more systematic way ? Thank you in anticipation. | 61817 | Determining whether there is a short exact sequence
I'm doing some exercises in Hatcher: Determine whether there is a short exact sequence [imath] 0 \rightarrow \mathbb{Z}_4 \xrightarrow{f} \mathbb{Z}_8 \oplus \mathbb{Z}_2 \xrightarrow{g} \mathbb{Z}_4 \rightarrow 0[/imath]. More generally, which abelian groups [imath]A[/imath] fit into a short exact sequence [imath]0 \rightarrow \mathbb{Z}_{p^m} \xrightarrow{f} A \xrightarrow{g} \mathbb{Z}_{p^n} \rightarrow 0 [/imath] where [imath]p[/imath] prime. What about [imath]0 \rightarrow \mathbb{Z}Β \xrightarrow{f} A \xrightarrow{g} \mathbb{Z}_n \rightarrow 0[/imath]. Can you tell me if this is right: In the first case, because [imath]f[/imath] injective, it has to map the generator, [imath]1[/imath], to an element of order [imath]4[/imath]. There are [imath]2[/imath] essentially different elements of order [imath]4[/imath]: [imath](2,1)[/imath] and [imath](2,0)[/imath]. In both cases I get a contradiction if they are in [imath]ker g[/imath], so there cannot be such an exact sequence. In the next case, [imath]\mathbb{Z}_{p^m}[/imath] is cyclic, let's say it's generated by [imath]c[/imath]. Then [imath]f(c)[/imath] has order [imath]p^m[/imath]. So [imath]ker g[/imath] has order [imath]p^m[/imath]. Therefore [imath]A[/imath] has to have at least order [imath]p^{n+m}[/imath]. Now I'm not sure how to proceed. Can I deduce anything else about [imath]A[/imath]? Many thanks for your help! |
703874 | Please can you check my proof of this about twice differentiable function
Let [imath]g: [0,1] \to \mathbb R[/imath] be twice differentiable with [imath]g''(x) > 0[/imath] for all [imath]x \in [0,1][/imath]. If [imath]g(0) > 0[/imath] and [imath]g(1) = 1[/imath] show that [imath]g(d) = d[/imath] for some [imath]d \in (0,1)[/imath] if and only if [imath]g'(1) > 1[/imath]. Please can you check my proof: [imath]\implies[/imath]: Assume there exists [imath]c \in (0,1)[/imath] such that [imath]g(c) = c[/imath]. If [imath]g'(1) \le 1[/imath] and [imath]g(1) = 1[/imath] then since [imath]g'' > 0[/imath] it follows that [imath]g(x) > x[/imath] for all [imath]x \in [0,1)[/imath] which would contradict [imath]g(c) = c[/imath] where [imath]c \in (0,1)[/imath]. [imath]\Longleftarrow[/imath]: Assume [imath]g'(1) > 1[/imath] and [imath]g(1) = 1[/imath]. Then there exists [imath]\varepsilon > 0[/imath] such that [imath]x \in (1-\varepsilon,1) [/imath] implies [imath]g(x) < x[/imath]. On the other hand, [imath]g(0) > 0[/imath]. Hence by the intermediate value theorem there exists [imath]c \in (0,1)[/imath] such that [imath]g(c) = c[/imath]. | 9728 | Twice differentiable function, show there is a fixed point
Let [imath]g:[0,1] \rightarrow \mathbb{R}[/imath] be twice differentiable with [imath]g''(x)\gt 0[/imath] for all [imath]x\in[0,1][/imath]. If [imath]g(0)>0[/imath] and [imath]g(1)=1[/imath], show that [imath]g(d)=d[/imath] for some point [imath]d\in(0,1)[/imath] if and only if [imath]g'(1)\gt 1[/imath]. Proof. Suppose there exits [imath]d\in(0,1)[/imath] such that [imath]g(d)=d[/imath]. Then by the MVT applied on [imath][d,1][/imath], [imath]f'(c)(d-1)=g(d)-g(1) [/imath]for some [imath]c\in(d,1)[/imath]. But [imath]g(d)=d[/imath] and [imath]g(1)=1[/imath] so [imath]f'(c)=1[/imath]. Now, since [imath]g''(x)\gt 0[/imath] for [imath]x\in[0,1][/imath], [imath]g'[/imath] is increasing on this interval, therefore [imath]1\gt d[/imath] implies [imath]g'(1)-g'(d)\gt 0[/imath]. Assume [imath]g'(1)\gt 1[/imath] and let [imath]f(x)=g(x)-x[/imath]. Then [imath]f(0)=g(0)-0\gt 0[/imath] by hypothesis. I want to show [imath]f(x)\lt 0[/imath] for some [imath]x[/imath] in a neighborhood of [imath]1[/imath]. [imath]g'(1)= \lim\limits_{x\to1}\frac{g(x)-g(1)}{x-1}[/imath] so there exists [imath]\delta\gt 0[/imath] such that given [imath]\epsilon\gt 0[/imath] with [imath]x\in(1-\delta,1)[/imath] implies [imath]|g'(1)-\frac{g(x)-1}{x-1}|\lt \epsilon[/imath]. If [imath]\epsilon=g'(1)-1\gt 0[/imath], then [imath]g'(1)-\epsilon\lt \frac{g(x)-1}{x-1}[/imath] which implies [imath]g(x)\lt x[/imath] for [imath]x\in(1-\delta,1)[/imath]. The result follows by the IVT. Any comments, corrections or different solutions are welcome. |
1364262 | For all integers [imath]x[/imath] and [imath]y[/imath], if [imath] x^3 + x = y^3 + y[/imath] then [imath]x = y[/imath].
For all integers [imath]x[/imath] and [imath]y[/imath], if [imath]x^3 + x = y^3 + y[/imath] then [imath]x = y[/imath]. This is what I have done so far: Proof: Suppose [imath]x[/imath] and [imath]y[/imath] are arbitrary integers. We know that [imath]x^3 + x = y^3 + y[/imath], we want to prove that [imath]x = y[/imath]. So, this is logically making sense, so it is true. any hints please? | 172261 | Proof: For all integers [imath]x[/imath] and [imath]y[/imath], if [imath]x^3+x = y^3+y[/imath] then [imath]x = y[/imath]
I need help proving the following statement: For all integers [imath]x[/imath] and [imath]y[/imath], if [imath]x^3+x = y^3+y[/imath] then [imath]x = y[/imath] The statement is true, I just need to know the thought process, or a lead in the right direction. I think I might have to use a contradiction, but I don't know where to begin. Any help would be much appreciated. |
112935 | Notation for: all subsets of size 2
How would one denote the set of all subsets of [imath]A[/imath] which have size [imath]2[/imath]? Would [imath]\binom{A}{2}[/imath] be a good idea? | 884348 | Notation: the set of two-element subsets of [imath]\Bbb N[/imath]
Let [imath]\{a,b\}\subseteq \Bbb N[/imath]. Is there a special name or notation for sets of this type, for example [imath]\Bbb N^{2\ge}[/imath]? Any subset size may be used, but the specific size and denoting that order does not matter is part of my interest in this notation, i.e., subsets of size [imath]n[/imath] of set [imath]X[/imath]. Particularly inspired by this question: Produce unique number given two integers. |
1364618 | Repeated roots of biquadratic equation
What is condition for repeated roots of the fourth order polynomial [imath] x^4 + a x^3 + b x^2 + c x + d = 0 ?[/imath] | 140797 | How to tell if a quartic equation has a multiple root.
Is there any way to tell whether a quartic equation has double or triple root(s)? [imath]x^4 + a x^3 + b x^2 + c x + d = 0[/imath] |
1365000 | Question on primitive element for extensions of finite fields
I was given this question in field theory class which I am stumped on as after some effort I still cannot tackle. I am given a general natural number [imath]n>1[/imath] and am asked to prove or disprove: There exists a prime [imath]p[/imath] such that the finite field extension [imath] F_{p^n}/F_p [/imath] has a primitive element [imath] \alpha \in F_{p^n} [/imath] satisfying [imath]\alpha^n \in F_p [/imath]. I have tried to solve it using cyclotomic extensions but cannot really proceed further that way. I also know that the condition [imath] \alpha^n \in F_{p} [/imath] and [imath]\alpha[/imath] being primitive is equivalent to the minimal polynomial of [imath]\alpha[/imath] of degree n being of the form [imath]x^n-c[/imath] for [imath] c \in F_p [/imath]. I have no idea how to proceed or for that matter if the claim is ture or false (they are asking me to prove or disprove). Any help appreciated Thank you very much to all Edit: I figured I have the following theorem from Dummit and Foote which might be helpful but I cannot use it straightaway. Any cyclic extension E/F of degree n over a field F containing all nth roots of unity, of characteristic not dividing n has an element [imath] a \in F [/imath] such that [imath] E = F(\sqrt[n]{a})[/imath] I still cannot incorporate the theorem Thanks | 1363091 | primitive element [imath]a[/imath] of [imath]\mathbb F_{p^n}/\mathbb F_p[/imath] such that [imath]a^n\in\mathbb F_p[/imath]
Is it true that for every [imath]n\in \mathbb N[/imath] there exists a prime [imath]p[/imath] such that the extension [imath]\mathbb F_{p^n}/\mathbb F_p[/imath] has a primitive element [imath]a\in \mathbb F_{p^n}[/imath] and [imath]a^n\in\mathbb F_p[/imath]? I tried finding a counter example but didn't mange to think of one. on the other hand I tried to prove it. From Artin's primitive element theorem, for every prime [imath]p[/imath] there exists an element [imath]a\in \mathbb F_{p^n}[/imath] such that \begin{equation*}\mathbb F_{p^n}=\mathbb F_p (a)\end{equation*} and\begin{equation*}[\mathbb F_{p^n}:\mathbb F_p]=n\end{equation*}implies that \begin{equation*}\textrm{deg}(m_{a,\mathbb F_p })=n\end{equation*}Now I dont know how to choose [imath]p[/imath] so \begin{equation*}m_{a,\mathbb F_p }=x^n+a^n\end{equation*}i.e every coefficient of [imath]x^k[/imath] for [imath]k=1,...,n-1[/imath] is [imath]0[/imath] |
1364745 | For which integers [imath]r[/imath] is [imath]\sigma ^r[/imath] also a [imath]k[/imath]-cycle?
Let [imath]\sigma[/imath] be a [imath]k[/imath]-cycle in [imath]S_n[/imath]. For which integers [imath]r[/imath], is [imath]\sigma ^r[/imath] also a [imath]k[/imath]-cycle? I think I managed to prove that this is true iff [imath](k,r)=1[/imath], but my proof was too long and not elegant so I'm looking for a shorter proof. | 878625 | When is a power of [imath]m[/imath]-cycle is also an [imath]m[/imath]-cycle?
I have a question taken from Abstract Algebra by Dummit and Foote ([imath]pg.33[/imath], [imath]q.11[/imath]): Let [imath]\sigma\in S_{n}[/imath] be an [imath]m[/imath]-cycle. Show that [imath]\sigma^{k}[/imath] is also an [imath]m[/imath]-cycle iff [imath]\gcd(k,m)=1[/imath]. My efforts: By considering a few examples I believe that [imath]\sigma^{k}[/imath] decomposition is a multiplication of cycles where each cycle is of length [imath]\frac{m}{\gcd(k,m)}[/imath] I have tried proving this by showing that [imath] \frac{m}{\gcd(k,m)}+k\equiv_{m}1 [/imath] which shows that [imath]\sigma^{k}[/imath] maps [imath]\frac{m}{\gcd(k,m)}[/imath] back to [imath]1[/imath] hence is the cycle length (at least for the first cycle in the decomposition, but I imagine that proving similar claim for the other circles can be done analogy) . I have tried showing this by writing the equivalence as [imath] m\mid\frac{m}{\gcd(k,m)}+k-1 [/imath] and trying to manipulate the above expression to see the divisibility, but I haven't managed to do so. Can someone please help me continue with this exercise ? |
1364751 | How can I prove the following equality
I have the following equality : [imath]I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}[/imath] [imath]I_2=-\frac{ab}{2\pi}\int_0^\pi \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=0[/imath] where [imath]0 <b \leq a[/imath]. I used the residues but I could not prove this equality | 1306776 | integral [imath]\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt[/imath]
I want to compute this integral [imath]\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt[/imath] where [imath]0<b \leq a[/imath]. I have this results [imath]I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2sin^2(t)+b^2cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}[/imath] But I don't know how to prove this equality. Which can help me, Thanks for all. |
1364887 | Number of abelian groups of order 108
What is the number of abelian groups of order 108 upto isomorphism ? To answer this I wrote explicitly the possible abelian groups of order 108 as follows : [imath]\Bbb Z_{108}[/imath] [imath]\Bbb Z_{4}\times\Bbb Z_{3}\times\Bbb Z_{9}[/imath] [imath]\Bbb Z_{2}\times\Bbb Z_{2}\times\Bbb Z_{27}[/imath] [imath]\Bbb Z_{4}\times\Bbb Z_{3}\times\Bbb Z_{3}\times\Bbb Z_{3}[/imath] [imath]\Bbb Z_{2}\times\Bbb Z_{2}\times\Bbb Z_{3}\times\Bbb Z_{9}[/imath] [imath]\Bbb Z_{2}\times\Bbb Z_{2}\times\Bbb Z_{3}\times\Bbb Z_{3}\times\Bbb Z_{3}[/imath] And I found the answer to be 6. But my problem is that what if I was given a much bigger number? Is this the only way to find abelian groups of a certain order? If there are better ways to find the exact answer to such question please let me know. | 201690 | Computing the number of nonisomorphic finite abelian groups of order [imath]n[/imath]
Let's say I'm given a number [imath]n = p^{3}q^{4}r^{2}s[/imath] and I want to find the number of (non-isomorphic) abelian groups of order [imath]n[/imath]. How I'm computing the number is basically partitioning, that is, I'm just looking at it and saying that I have [imath]\mathbb{Z}/p^{3}q^{4}r^{2}s\mathbb{Z}[/imath], [imath]\mathbb{Z}/p^{2}q^{4}r^{2}s\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}[/imath], and so on. Then I count the number of groups I've written down. However, I've found that I'm quite prone to error when I do this. Is there a faster way of doing this? If not, is there a systematic trick/way I could use to make this easier? |
1365268 | Part A: Prove that [imath](k, n+k) = 1[/imath] if and only if [imath](k, n)= 1[/imath]
Part A is in the title, Part B is here: Is it true that [imath](k, n+k)= d[/imath] if and only if [imath](k, n)=d[/imath]? I am still working on the Part A. What I have so far: if [imath](k, n)= 1[/imath] then [imath]1|k[/imath], [imath]1|n[/imath] and [imath]1|(n-k)[/imath] if [imath](k, n+k)=1[/imath] then [imath]1|k[/imath], [imath]1|n+k[/imath] and [imath]1|((n+k)- k) \to 1|n[/imath] I was under the impression that if [imath]d|a[/imath] and [imath]d|b[/imath], that [imath]d|(b-a)[/imath]. Is this false? What more should I be doing to tackle part A? | 110233 | Prove [imath](k,n+k)=1[/imath] iff [imath](k,n)=1[/imath]
(a) Prove [imath](k,n+k)=1[/imath] iff [imath](k,n)=1[/imath]. (b) Is it true that [imath](k,n+k)=d[/imath] iff [imath](k,n)=d[/imath]? (all variables are integers, (a,b) means gcd(a,b)) For (a), I can only get so far. [imath](k,n+k)=1\ \Rightarrow\ kq_1+(n+k)q_2=k(q_1+q_2)+nq_2=1[/imath]. Note that we don't yet have that theorem (from a different class) that says something like if [imath]d=(a,b)[/imath], then [imath]d[/imath] is the smallest (or is it biggest) number which can be made by [imath]ak_1+bk_2[/imath], (which I'm not quite sure how it would help, anyway). So I'm stuck. |
1365751 | How do you determine if a number is a even Fibonacci number or not?
Rather than computing out the whole Fibonacci sequence and check if [imath]n[/imath] is even and in there, is there a more straightforward way to compute if [imath]n[/imath] is a even Fibonacci number? | 9999 | Checking if a number is a Fibonacci or not?
The standard way (other than generating up to [imath]N[/imath]) is to check if [imath](5N^2 + 4)[/imath] or [imath](5N^2 - 4)[/imath] is a perfect square. What is the mathematical logic behind this? Also, is there any other way for checking the same? |
1365778 | Limit of definite integral
I found this problem on an old math contest: [imath]\lim_{b\to1^+}\int_1^b\frac{dx}{\sqrt{x(x-1)(b-x)}}.[/imath] I've tried substituting [imath]u=\sqrt{x}[/imath], but that hasn't helped much. I've also looked for ways to interpret this geometrically, since numerical estimation suggests the answer is [imath]\pi[/imath]. I've also had the idea to try expressing the integrand as a series to get a better handle on how fast terms are growing, but I can't make that work either. Mainly, the difficulty seems to come from the fact that the integrand is undefined at both boundaries, and that the width of the interval is also going to 0. I'm not sure how to work around either of these, but any help is much appreciated! | 1363246 | [imath]\text{Evaluate:} \lim_{b \to 1^+} \int_1^b \frac{dx}{\sqrt{x(x-1)(b-x)}}[/imath]
I'm trying to solve the following problem [imath]\text{Evaluate:} \lim_{b \to 1^+} \int_1^b \frac{dx}{\sqrt{x(x-1)(b-x)}}[/imath] I'm tempted to think it's 0 because the bounds would be about equal, but that's not the correct answer. But I don't know what integration techniques to use, so any hints are greatly appreciated. I don't know how to solve for the indefinite integral itself. |
1333848 | Variance of number of tails minus heads
Given [imath]X_n[/imath] is a random variable that equals the number of tails minus the number of heads when n fair coins are flipped, what is the Variance of [imath]X_n[/imath] ? I tried to solve it | 887894 | Variance of number of tails in a coin-toss experiment
Let X be the random variable that equals the number of tails minus the number of heads when n fair coins are flipped. What is the variance of X? I've run a simulation and the answer seems to be n, but I can't get at it myself. [imath]V(X) = E(X^2) - E(X)^2[/imath] and E(X) is zero, so [imath]V(X) = E(X^2)[/imath]. Now [imath]X = (2X_t - n)[/imath] where [imath]X_t[/imath] is number of tails. [imath]X^2 = 4X_t^2 + n^2 - 2X_tn[/imath]. Taking E() on both sides: [imath]E(X^2) = 4E(X_t^2) + n^2 - 2nE(X_t)[/imath] we know that [imath]E(X_t)=n/2[/imath] so, [imath]E(X^2) = 4E(X_t^2) - n^2[/imath] I tried to get [imath]E(X_t^2)[/imath] from a simulation and got the right answer again. So the question boils down to: How to find the expectation of [imath]E(X_t^2)[/imath] where [imath]X_t[/imath] is the number of tails in n coin tosses. |
1366570 | What was the original motivation for matrix multiplication?
When I took linear algebra class in my freshman year, the multiplication operation for matrices was defined without any apparent motivation. Given an [imath]m[/imath]-times-[imath]n[/imath] matrix [imath]A[/imath] and an [imath]n[/imath]-times-[imath]p[/imath] matrix [imath]B[/imath], define [imath](AB)_{ij} := \sum_{k=1}^n A_{ik}B_{kj}[/imath]. I recently read a book which treats linear algebra more abstractly, introducing the notion of a matrix after it introduces linear maps. (The book is Linear Algebra Done Right by Sheldon Axler.) Let [imath]V[/imath] and [imath]U[/imath] be vector spaces with bases [imath]\{v_1,\ldots,v_n\}[/imath] and [imath]\{u_1,\ldots,u_m\}[/imath]. Notice that once we know how a linear map [imath]T:U\to V[/imath] acts on the basis elements of [imath]U[/imath], there is a unique way to extend it to the map on the whole space. The author then goes on to argue that we can thus associate with each linear map a matrix of that map. They are a very compact way of specifying linear maps relative to a given pair of bases. We can define operations of addition and scalar multiplication on matrices so that they coincide with how these two operations act on the set of linear maps: [imath]M(S+T)=M(S)+M(T)[/imath] and [imath]M(cT)=cM(T)[/imath] for a scalar [imath]c[/imath] and linear maps [imath]S,T:U\to V[/imath]. What I like the most about the book is the next step. Let [imath]M(T)[/imath] denote the matrix of the linear map [imath]T:U\to V[/imath] (relative, say, to standard basis). We know how to compose two linear maps [imath]S:U\to V[/imath] and [imath]T:V\to W[/imath] to get a linear map [imath]ST:U\to W[/imath]. What is more natural than to define matrix multiplication so that the equation [imath]M(ST)=M(S)M(T)[/imath] is satisfied? And we get the non-obvious rule for matrix multiplication. Despite not being simple it feels natural. I like it. I know where it came from. It makes sense. Was this the original motivation for matrix multiplication? Who introduced it and when? (If not, when did people realize that is satisfies this rule?) Edit (July 19th 2015). There is a substantial overlap with the question Why, historically, do we multiply matrices as we do? but that answer does not have a satisfactory answer. In particular, the URL in the accepted answer is dead. | 271927 | Why, historically, do we multiply matrices as we do?
Multiplication of matrices β taking the dot product of the [imath]i[/imath]th row of the first matrix and the [imath]j[/imath]th column of the second to yield the [imath]ij[/imath]th entry of the product β is not a very intuitive operation: if you were to ask someone how to mutliply two matrices, he probably would not think of that method. Of course, it turns out to be very useful: matrix multiplication is precisely the operation that represents composition of transformations. But it's not intuitive. So my question is where it came from. Who thought of multiplying matrices in that way, and why? (Was it perhaps multiplication of a matrix and a vector first? If so, who thought of multiplying them in that way, and why?) My question is intact no matter whether matrix multiplication was done this way only after it was used as representation of composition of transformations, or whether, on the contrary, matrix multiplication came first. (Again, I'm not asking about the utility of multiplying matrices as we do: this is clear to me. I'm asking a question about history.) |
950707 | Calculate Brieskorn Manifold?
I need show that Brieskorn Manifold is submanifold with dimension [imath]2n-1[/imath] and calculate specifically for [imath]d=2[/imath] and [imath]n=1[/imath] [imath]W(d)=\lbrace (z_{0},z_{1},...,z_{n})\in \mathbb{C}^{n+1}\vert[/imath] [imath] z_{0}^{d}+z_{1}^{2}+...+z_{n}^{2}=0[/imath] and [imath]\vert z_{0}\vert^{2}+\vert z_{1}\vert^{2}+...+\vert z_{n}\vert^{2}=2\rbrace[/imath] The second cuestion I don't know what it means. | 1101325 | Proof a [imath](2n-1)[/imath]-compact manifold
I have no idea how prove that [imath]\{(z_0,\ldots,z_n)\in\mathbb{C}^{n+1} \quad| \quad z_0^d+z_1^2\ldots+z_n^2=0, \quad |z_0|^2+|z_1|^2\ldots+|z_n|^2=2\}[/imath] is a [imath](2n-1)[/imath]-compact manifold. How give the charts. For [imath]n=1[/imath] and [imath]d=2[/imath] not distinguish which 1- manifdold is? |
1366449 | Finding the error in this induction proof
Claim: If [imath]n[/imath] belongs to [imath]\mathbb{N}[/imath], and [imath]p[/imath] and [imath]q[/imath] are natural numbers with maximum [imath]n[/imath], then [imath]p=q[/imath]. Let [imath]S[/imath] be the subset of the natural numbers for which the claim is true. [imath]1[/imath] belongs to [imath]S[/imath], since if [imath]p[/imath] and [imath]q[/imath] belong to [imath]N[/imath] and their maximum is [imath]1[/imath], then [imath]p=q=1[/imath]. Now assume [imath]k[/imath] belongs to [imath]S[/imath], and that the maximum of [imath]p[/imath] and [imath]q[/imath] is [imath]k+1[/imath]. Then the maximum of [imath]p-1[/imath], [imath]q-1[/imath] is [imath]k[/imath]. But [imath]k[/imath] is in [imath]S[/imath], so [imath]p-1=q-1[/imath], thus [imath]p=q[/imath] and [imath]k+1[/imath] is in [imath]S[/imath], so the assertion is true for all [imath]n[/imath] in [imath]N[/imath]. | 342321 | All natural numbers are equal.
I saw the following "theorem" and its "proof". I can't explain well why the argument is wrong. Could you give me clear explanation so that kids can understand. Theorem: All natural numbers are equal. Let [imath]a, b \in \mathbb{N}[/imath], then a=b. Proof by induction. Let [imath]m=\max\{a, b\}[/imath]. We will prove that the theorem holds for all [imath]m\in \mathbb{N}[/imath]. If [imath]m=1[/imath], then [imath]\max\{a,b\}=1[/imath], so [imath]a=b=1[/imath]. Now assume that it holds for [imath]m=k[/imath] for some number [imath]k[/imath]. Now let [imath]\max\{a, b\}=k+1[/imath]. Then [imath]\max\{a-1, b-1\}=k[/imath] and thus by assumption [imath]a-1=b-1[/imath], so [imath]a=b[/imath]. Therefore, the proof is complete. |
1366649 | If [imath]x_1, x_2,...,x_{10}[/imath] are such that [imath]\sum_{i=1}^{10} \sin^2(x_i) = 1[/imath], prove that [imath]3 \sum_{i=1}^{10} \sin(x_i) \leq \sum_{i=1}^{10} \cos(x_i)[/imath]
Take [imath]x_1, x_2,...,x_{10}[/imath] such that [imath]\sum_{i=1}^{10} \sin^2(x_i) = 1[/imath] with [imath]x_1, x_2,...,x_{10}[/imath] on [imath]\left[0,\frac{\pi}{2}\right][/imath], prove that [imath]3 \sum_{i=1}^{10} \sin(x_i) \leq \sum_{i=1}^{10} \cos(x_i)[/imath]. I tried to do this: [imath]\sum_{i=1}^{10} \sin^2(x_i) = \frac{ n-\sum_{i=1}^{10} \cos(2 x_i)}{2} = 1 \rightarrow \sum_{i=1}^{10} \cos(2x_i) = 8[/imath] So i said [imath]\sum_{i=1}^{10} \cos(2x_i) = A[/imath] and [imath]\sum_{i=1}^{10} \sin(2x_i) = B[/imath]. Therefore: [imath]A + Bi = \sum_{i=1}^{10} \operatorname{cis}(2x_i) = \sum_{i=1}^{10} \operatorname{cis}^2(x_i)[/imath] But I make no idea about how I will get the inequality. Anybody can help me? | 180885 | Prove : [imath]\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3[/imath]
If we assume that: [imath]0\le x_1,x_2,\ldots,x_{10}\le\frac{\pi}{2} [/imath] such that: [imath]\sin^2 (x_1) +\sin^2 (x_2)+\cdots+\sin^2(x_{10})=1[/imath] How to prove that: [imath]\frac{\cos(x_1) +\cos(x_2) +\cdots+\cos(x_{10})}{\sin(x_1) +\sin(x_2) +\cdots+\sin(x_{10})} \ge 3[/imath] |
1367639 | Two topologies are coincide
Let [imath](X,\tau)[/imath] and [imath](X,\tau')[/imath] are both metrizable topological vector space, and let [imath](x_{n})\subset X[/imath] and [imath]x\in X[/imath] and [imath]x_{n}\rightarrow x[/imath] in [imath]\tau[/imath] topology if and only if [imath]x_{n}\rightarrow x[/imath] in [imath]\tau'[/imath] topology . Prove then two topologies coincide on [imath]X[/imath]. | 1367445 | Do the topologies with the same convergent sequences coincide?
Do the same sequential convergence in two topologies the result are the same topologies? Let [imath]X[/imath] be topological space with topologies [imath]\tau[/imath], [imath]\tau'[/imath]. Let [imath](x_{n})\in X[/imath] and [imath]x\in X[/imath] and [imath](x_{n})\rightarrow x[/imath] with [imath]\tau[/imath] if and only if [imath](x_{n})\rightarrow x[/imath] with [imath]\tau'[/imath]. Do the two topologies on [imath]X[/imath] coincide? |
1366576 | [imath]A[/imath] and [imath]B[/imath] be two orthogonal matrices with [imath]|A|+|B|=0[/imath]. Then prove that [imath](A+B)[/imath] is singular
Let , [imath]A[/imath] and [imath]B[/imath] be two orthogonal matrices with [imath]|A|+|B|=0[/imath]. Then prove that [imath](A+B)[/imath] is singular. From the relation [imath]|A|+|B|=0[/imath] it is clear that [imath]|AB|=-1[/imath]. Also , [imath]|A|=|B|=\pm 1[/imath]. But how I find out [imath]|A+B|[/imath] ? | 491344 | Is sum of two orthogonal matrices singular?
I am trying to solve following problem. Let [imath]$A,B\in \mathbb{R}^{n\times n}$[/imath] be an orthogonal matrices and [imath]$\det(A) = -\det(B)$[/imath]. How can it be proven that [imath]A+B[/imath] is singular? I could start with implication: [imath]$\det(A)=-\det(B)\RightarrowB$[/imath] is created from [imath]A[/imath] by swapping two lines or columns. But I am not sure if this implication is correct. |
1367856 | Improper integral of [imath]\sqrt{x^4 + 1} - x^2[/imath]
I'm having a little trouble with this integral: [imath]\int^\infty_0 (\sqrt{x^4+1} - x^2)\,dx[/imath]. Using the likes of Maple, I can easily find that it takes the form [imath]-\frac{2}{3}\sqrt{2}(1+i)K(i) - \frac{1}{3}\sqrt{2}(1+i)K(\sqrt{2})[/imath], where [imath]K(m) = \int_0^{\pi/2} \frac{d\theta}{\sqrt{1 - m\sin^2\theta}}[/imath] is the complete elliptic integral of the first kind. However, I am looking at a textbook which tells me the solution is simply [imath]\frac{\Gamma^2(1/4)}{6\sqrt{\pi}}[/imath], with [imath]\Gamma[/imath] denoting the usual gamma function [imath]\Gamma(t) = \int^\infty_0 e^{-u}u^{t - 1}du[/imath]. I've thrown a fair amount at this: trying to reduce it via various substitutions to, for instance, a combination of beta functions which can be rewritten as gamma functions via [imath]B(m,n)= \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}[/imath], or just a combination of elliptic [imath]K[/imath] functions. Even after a fair amount of time consulting Gradshteyn and Ryzhik, I can't seem to get anywhere... I'd like to eventually get to the gamma function solution, and I'd be most grateful to anyone who can help me get there, even if it's just via a nudge in the right direction. | 867676 | Prove [imath]\int_0^{\infty} \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}[/imath]
I have in trouble for evaluating following integral [imath]\int_0^{\infty} \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}[/imath] It seems really easy, but I don't know how to handle it at all. (The results are well known, here I tried to evaluate it but I failed) I tried to use the relation [imath]\sqrt{1+x^{4}}-x^{2}=\frac{1}{\sqrt{1+x^{4}}+x^{2}}[/imath] but I couldn't find the desired results. |
1368040 | Discrete subspace of a Hausdorff space
Every infinite Hausdorff space has a countably infinite discrete subspace. Now I know [imath]R[/imath] under usual topology has [imath]Z[/imath] and under "lower limit topology" also has [imath]Z[/imath] as such. So goes on for [imath]R^{n}[/imath] as long as [imath]n[/imath] is finite it will be [imath]Z^{n}[/imath] .But what about general Hausdorff Spaces? Please some hints. I need to know what such discrete subspaces might look like for [imath]R^{inf}[/imath] since [imath]Z^{\inf}[/imath] or [imath]N^{\inf}[/imath] are not countable. | 533051 | Every infinite Hausdorff space has an infinite discrete subspace
I want to show that any infinite Hausdorff space contains an infinite discrete subspace. I am motivated by the role of [imath]\mathbb N[/imath] in [imath]\mathbb R[/imath]. We know that if a Hausdorff space is finite, then it is a discrete space, but an infinite subspace of a Hausdorff space is obviously not necessarily discrete. |
1367867 | Finding Surface area of Paraboloid
I am having trouble finding the surface area of the part of the paraboloid that lies in the first octant of [imath]z=5-x^2-y^2[/imath]. So, I realized that the first octant refers to when, [imath]x,y,z \ge 0[/imath]. How do I proceed with this? | 1367939 | Solving a double integration in parametric form
I need help to find the surface area of the part of the paraboloid that lies in the first octant, meaning [imath]x,y,z\ge 0[/imath], of [imath]z=5-x^2-y^2[/imath]. So I found that I need to parametrize the paraboloid using cylindrical coordinates: [imath]{\bf x}(u,v) = (u \cos v, u \sin v, 5 - u^2),[/imath]with [imath]0 \leq u \leq \sqrt{5}[/imath] and [imath]0 \leq v \leq \pi/2[/imath]. Then compute: [imath]\int_0^{\pi/2} \int_0^\sqrt{5}\|{\bf x}_u \times {\bf x}_v\|\,{\rm d}u\,{\rm d}v.[/imath] But I need help doing this. Can someone please help me continue on with this? |
1367973 | Write integral in different order
How can I rewrite, the following triple integrals into different orders?: [imath]\int^1_0\int^{z^2}_0\int^y_0{f} \, \mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z[/imath] In the orders [imath]\mathrm{d}z \, \mathrm{d}x \, \mathrm{d}y[/imath] and [imath]\mathrm{d}y\,\mathrm{d}z\,\mathrm{d}x[/imath]? Currently the limits are: [imath]\begin{split} &0 \le x \le y \\ &0 \le y \le z^2 \\ &0 \le z \le 1 \end{split}[/imath] | 1367910 | Rewriting Triple Integral in different order
How can I rewrite, say: [imath]\int^1_0\int^{z^2}_0\int^y_0{f}dxdydz[/imath] In the orders [imath]dzdxdy[/imath] and [imath]dydzdx[/imath]? Would I have to solve this triple integral all the way or is there a simple way to do this? |
1368345 | Function Domain of cubic root
Does the domain of the function [imath]y=\sqrt[3]{x^3+1}[/imath] include [imath]x<-1[/imath]? If yes, why is Mathematica and Wolfram Alpha not plotting that part of the function? | 25528 | cubic root of negative numbers
Excuse my lack of knowledge and expertise in math,but to me it would came naturally that the cubic root of [imath]-8[/imath] would be [imath]-2[/imath] since [imath](-2)^3 = -8[/imath]. But when I checked Wolfram Alpha for [imath]\sqrt[3]{-8}[/imath], real it tells me it doesn't exist. I came to trust Wolfram Alpha so I thought I'd ask you guys, to explain the sense of that to me. |
1368415 | Is a function [imath]f:\mathbb R\to\mathbb R[/imath] such that [imath]f(x+y)=f(x)+f(y)[/imath] always continuous?
Is there a function [imath]f:\mathbb R\to\mathbb R[/imath] such that [imath]f(x+y)=f(x)+f(y)[/imath] which is not continuous? I have proved that if it's continuous in one point [imath]a\in\mathbb R[/imath] then it's continuous on all [imath]\mathbb R[/imath], but I didn't find such a function which is not continuous everywhere. Therefore I tried to prove that all function of this form is continuous at [imath]x=0[/imath] but with no success. I think that if such a function exist it would be of the form [imath]f(x)=...[/imath] if [imath]x\in\mathbb Q[/imath] and [imath]f(x)=...[/imath] if [imath]x\in\mathbb R\backslash\mathbb Q[/imath] but I didn't find it. | 269434 | non-continuous function satisfies [imath]f(x+y)=f(x)+f(y)[/imath]
As mentioned in this link, it shows that For any [imath]f[/imath] on the real line [imath]\mathbb{R}^1[/imath], [imath]f(x+y)=f(x)+f(y)[/imath] implies [imath]f[/imath] continuous [imath]\Leftrightarrow[/imath] [imath]f[/imath] measurable. But how to show there exists such an non-measurable function satisfying [imath]f(x+y)=f(x)+f(y)[/imath]? I guess we may use the uniform bounded principal and the fact that [imath]f[/imath] is continuous iff it is continuous at zero under the above assumption. Thanks in advance! |
1368337 | If [imath]f[/imath] and [imath]g[/imath] are measurable and well-defined a.e., show [imath]f+g[/imath] is measurable.
The question is If [imath]f[/imath] and [imath]g[/imath] are measurable and well-defined a.e., show [imath]f+g[/imath] is measurable. I can show that the case when [imath]f+g[/imath] are well-defined everywhere. Let [imath]\Omega_{f+g>a}=\{x\in \Omega:f(x)+g(x)>a\}[/imath], then [imath]\Omega_{f+g>a}=\Omega_{f>a-g}[/imath]. There is a theorem in my text book saying that if [imath]f[/imath] is measurable then [imath]\Omega_{f+a}[/imath] is measurable for any real number [imath]a[/imath]. Another theorem states that if both [imath]f,g[/imath] are measurable, then [imath]\Omega_{f>g}[/imath] is measurable. As a result [imath]\Omega_{f>a-g}[/imath] is measurable, hence [imath]\Omega_{f+g>a}[/imath] is measurable, then [imath]f+g[/imath] is measurable. Can anyone help with the case when [imath]f,g[/imath] are well-defined a.e. respectively? Thank you! | 541118 | Proving that sum of two measurable functions is measurable.
Suppose [imath]f[/imath] and [imath]g[/imath] are Lebesgue measurable, we want to show [imath]f+g[/imath] is measurable. So, the hint is to consider the continuous functions [imath]F : \mathbb{R}^2 \to \mathbb{R} [/imath] given by [imath]h(x) = F(f ,g ) [/imath]. If we can show [imath]F[/imath] Is measurable, then Taking [imath]F = f +g [/imath] would solve our problem. In other words, I want to show that the set [imath]R = \{ (f,g) : F(f,g) > a [/imath] } is lebesgue measurable.. But this set is just a rectangle in the plane. And since [imath]F[/imath] is continuous, then [imath]R[/imath] must be open, and hence a union of open rectangles which are measurable and hence [imath]R[/imath] must be measurable. Is this a correct approach to the problem? Can someone help me to make this formal? thanks |
1369017 | Are there sets [imath]S\subseteq\Bbb N[/imath] which are provably non-empty, but we don't know what is [imath]\min S[/imath]?
I was wondering if there is a property that is known to be satisfied for certain "things" but for which we do not know any explicit example. More explicitly (also more restrictive but possibly easier): Is there a set [imath]S \subseteq \mathbb{N}[/imath] for which one can prove that [imath]S \neq \varnothing[/imath] and yet there is no known (to us, humans, at this point in time) element of [imath]S[/imath]? | 1315615 | What is the smallest unknown natural number?
There are several unknown numbers in mathematics, such as optimal constants in some inequalities. Often it is enough to some estimates for these numbers from above and below, but finding the exact values is also interesting. There are situations where such unknown numbers are necessarily natural numbers, for example in Ramsey theory. For example, we know that there is a smallest integer [imath]n[/imath] such that any graph with [imath]n[/imath] vertices contains a complete or an independent subgraph of 10 vertices, but we don't know the exact value of [imath]n[/imath]. What kinds of unknown small (less than 100, say) integers are there? What are the smallest unknown constants which are known to be integers? Or, more rigorously, what is the smallest upper bound for an unknown but definable number that is known to be an integer? I know that asking for the smallest unknown integer is ill-defined since we do not know the exact values. The more rigorous version of the question is well-posed, but I do not want to keep anyone from offering interesting examples even if they are clearly not going to win the race for the lowest upper bound. An answer should contain a definition of an integer quantity (or a family of them) and known lower and upper bounds (both of which should be integers, not infinite). Conjectures about the actual value are also welcome. I have given one example below to give an idea of what I'm looking for. |
1369038 | Techniques of Integration
Evaluate the following definite integral by letting [imath]u= \pi/2 - x[/imath]. The integral is [imath]\int_0^{\pi/2} \frac{\sin(x)}{\cos(x)+\sin(x)}dx.[/imath] | 364923 | Evaluate [imath] \int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x}\,dx [/imath] using substitution [imath]t=\frac{\pi}{2}-x[/imath]
This problem is given on a sample test for my calculus two class. [imath] \int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x}\,dx [/imath] I can find the value of this integral using other substitutions which lead to partial fractions but the prof added a hint to use the substitution [imath]t=\frac{\pi}{2}-x [/imath] so I've been trying to figure out how to do it his way but am pretty lost. Any ideas? |
1368642 | Does group inverse commute with multiplication for all groups?
Is the following a property of [imath]\textbf{all}[/imath] groups? [imath]a^{-1} \circ b^{-1} = (a \circ b)^{-1}[/imath] As far as I can tell it is true for addition and multiplication, but in the notes that I have come across, I have not seen it stated as a property. | 326589 | f(a) = inverse of a is an isomorphism iff a group G is Abelian
[imath]G[/imath] is a group and [imath]f:G \rightarrow G[/imath] is a function defined as [imath]f(a)=a^{-1}[/imath] where [imath]a^{-1}[/imath] is the inverse of [imath]a[/imath] under the group operation. Prove that [imath]f[/imath] is an isomorphism if and only if [imath]G[/imath] is abelian. I understand that I have to prove [imath]f(ab)=(ab)^{-1}=b^{-1}a^{-1}[/imath]. How might I do that? Reference: Fraleigh p. 49 Question 4.40 in A First Course in Abstract Algebra |
1310076 | You have to draw 10 cards. What is the probability that you will draw at least one repeated card?
Assume you have 52 cards in a deck, and you have to draw 10 cards. Assuming you have to put back the card after each draw, what is the probability that you will draw at least one repeated card? Here's what I was thinking... Draw 1: [imath]\frac{52}{52}[/imath] since first card is allowed to be any card. Draw 2: [imath]\frac{1}{52}[/imath] since you want to try to draw the first card. ... Draw 9: [imath]\frac{8}{52}[/imath] since you want to try to draw one of the last 8 cards. Draw 10: [imath]\frac{9}{52}[/imath] since you want to try to draw one of the last 9 cards. So [imath]\frac{52}{52} \times \frac{1}{52} \times ... \times \frac{8}{52} \times \frac{9}{52} = 1.31 \times 10^{-10}[/imath] or [imath]0\%[/imath] chance of getting at least one repeat. Is this the correct answer? | 1308718 | You have to draw 10 cards. What is the probability you will not drawn any repeated cards?
Assume you have 52 cards in a deck, and you have to draw 10 cards. Assuming you have to put back the card after each draw, what is the probability that you will not draw any repeated cards? Here's what I was thinking... Draw 1: [imath]\frac{52}{52}[/imath] since first card is allowed to be any card. Draw 2: [imath]\frac{51}{52}[/imath] since you don't want to draw the first card, you're now allowed to draw 51 of the remaining cards. ... Draw 9: [imath]\frac{44}{52}[/imath] since you don't want to draw the last 8 cards, you're now allowed to draw 44 of the remaining cards. Draw 10: [imath]\frac{43}{52}[/imath] since you don't want to draw the last 9 cards, you're now allowed to draw 43 of the remaining cards. So [imath]\frac{52}{52} \times \frac{51}{52} \times ... \times \frac{44}{52} \times \frac{43}{52} = 0.397[/imath] or [imath]39.7\%[/imath] chance of not getting a repeat. Just wanted to check if this is the correct approach to the problem or not. |
1370039 | Prove that $N$ normal and $TN=NT$ implies [imath]T^*N=NT^*[/imath]
Let [imath]T,N:V\to V[/imath] such that [imath]N[/imath] is a normal operator and [imath]TN=NT[/imath]. Prove that [imath]T^*N=NT^*[/imath] and [imath]N^*T=TN^*[/imath]. I'd be glad for help because I don't even have an idea how to approach this | 1311571 | If [imath]A[/imath] is normal and [imath]A[/imath] and [imath]B[/imath] commute, then [imath]A^*[/imath] and [imath]B[/imath] commute
Let [imath]A[/imath] is a normal matrix: [imath]A^*\! A = A A^*\!\![/imath],[imath]\,[/imath] and [imath]AB = BA[/imath]. Prove that [imath]A^*\!B=BA^*\!\![/imath]. I can prove that if [imath]\det A\ne 0[/imath] by multiplication [imath]AB=BA[/imath] by [imath]A^*[/imath] left and right and using some manipulation. But I have no idea what to do if [imath]\det A = 0[/imath]. |
82030 | Size of a union of two sets
We were ask to prove that [imath]|A \cup B| = |A| + |B| - |A \cap B|β«βͺ[/imath]. It was easy to prove it using a Venn diagram, but I think we might be expected to do if more formally. Is there a formal way? | 1849071 | Prove [imath]|A\cup B| = |A| + |B| - |A \cap B|[/imath]
Problem: Let [imath]A[/imath] and [imath]B[/imath] be finite sets. Prove [imath]|A\cup B| = |A| + |B| - |A \cap B|[/imath] where [imath]|A|[/imath] denotes the cardinality of set [imath]A[/imath]. I've tried breaking up [imath]A \cup B[/imath] in various ways, and then using the theorem that if [imath]A[/imath] and [imath]B[/imath] are finite and disjoint then [imath]|A \cup B| = |A| + |B|[/imath], but it didn't seem to work for me. Can someone help me with this? |
1370730 | Find A such that [imath]A^2 \neq I[/imath] but [imath]A^4 = I[/imath]
Find a [imath]3 \times 3[/imath] matrix A such that [imath]A^2 \neq I[/imath] but [imath]A^4 = I[/imath], where [imath]I[/imath] is the [imath]3 \times 3[/imath] identity matrix. Is there a simpler way to solve this problem rather than bashing it out by substituting a variable matrix value for A? | 1312131 | Find 3x3 real matrix A such that [imath]A^2 \neq I[/imath] and [imath]A^4 = I[/imath], where I is identity matrix.
Find [imath]3 \times 3[/imath] real matrix A such that [imath]A^2 \neq I[/imath] and [imath]A^4 = I[/imath], where [imath]I[/imath] is identity matrix. I first thought that there is no such matrix and tried to show that using determinants, but all I get is that A has to have [imath]detA=1[/imath] or [imath]-1[/imath]. Next I actually found such Matrix A= [imath]\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{matrix} \right)[/imath], but I did it through random manipulations and do not have any idea how to coherently write the answer. Any help would be appreciated. Thanks! |
1358496 | Perfect set without rational numbers
Sorry if this problem is repeated. Is there a nonempty perfect set in [imath]\mathbb{R}^1[/imath] which contains no rational number? Proof sketch: This set must be uncountable because any nonempty perfect set in [imath]\mathbb{R}^k[/imath] is uncountable. We know that Cantor set (C) is perfect set on a line. If we take [imath]E=C+\sqrt{2}=\{x+\sqrt{2}: x\in C\}[/imath]. Will this set be an answer to above problem? | 1730292 | Prob. 18, Chap. 2 in Baby Rudin: Any non-empty perfect set of real numbers which contains no rationals?
Here's Prob. 18 in the Exercises after Chapter 2 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition. Is there a non-empty perfect set in [imath]\mathbb{R}^1[/imath] which contains no rational number? Here's Definition 2.18 (h) in Baby Rudin. Let [imath]X[/imath] be a metric space. A subset [imath]E[/imath] of [imath]X[/imath] is said to be perfect if [imath]E[/imath] is closed and if every point of [imath]E[/imath] is a limit point of [imath]E[/imath]. And, here's Definition 2.18 (d) in Rudin. Let [imath]X[/imath] be a metric space. A subset [imath]E[/imath] of [imath]X[/imath] is said to be closed if every limit point of [imath]E[/imath] is a point of [imath]E[/imath]. Moreover, I know that the Cantor set --- discussed in 2.44 in Rudin --- is an example of a perfect set in [imath]\mathbb{R}^1[/imath] which contains no segment (i.e. no open interval). Of course, any non-empty perfect set in a metric space must necessarily be infinite. How to go about finding an answer to Rudin's question? |
464031 | Find the sum : [imath]\frac{1}{\cos0^\circ\cos1^\circ}+\frac{1}{\cos1^\circ \cos2^\circ} +\frac{1}{\cos2^\circ \cos3^\circ}+......+[/imath]
Find the sum of the following : (i) [imath]\frac{1}{\cos0^\circ \cos1^\circ}+\frac{1}{\cos1^\circ\cos2^\circ} +\frac{1}{\cos2^\circ \cos3^\circ}+......+\frac{1}{\cos88^\circ \cos89^\circ}[/imath] I tried : [imath]\frac{1}{\cos1^\circ}\left[\frac{\cos(1^\circ-0^\circ)}{\cos0^\circ\cos1^\circ} + \frac{\cos(2^\circ-1^\circ)}{\cos1^\circ\cos2^\circ}+...\right][/imath] = [imath]\frac{1}{\cos1^\circ}\left[\frac{\cos1^\circ\cos0^\circ}{\cos0^\circ\cos1^\circ} - \frac{\sin1^\circ \sin0^\circ}{\sin0^\circ\cos1^\circ} + \frac{\cos2^\circ \cos1^\circ}{\cos1^\circ\cos2^\circ} -\frac{\sin2^\circ \sin1^\circ}{\cos1^\circ\cos2^\circ}...\right][/imath] For this, as well, I am not getting any pattern to solve further. Please suggest, thanks. | 1506051 | prove that [imath]\sum_{k=1}^{n=90}\frac{1}{\sin(k-1)\sin(k)} =\frac{\cos1^{\circ}}{\sin^21^{\circ}}[/imath]
how do I prove that [imath]\sum_{k=1}^{n=90}\frac{1}{\sin(k-1)\sin(k)} =\frac{\cos1^{\circ}}{\sin^21^{\circ}}[/imath] or [imath]{1\over \sin1^{\circ}\sin2^{\circ}}+{1\over \sin2^{\circ}\sin3^{\circ}}+{1\over \sin3^{\circ}\sin4^{\circ}}+...+{1\over \sin89^{\circ}\sin90^{\circ}} = \frac{\cos1^{\circ}}{\sin^21^{\circ}}[/imath] Any ideas about how to go about doing this ? |
446093 | Generate Correlated Normal Random Variables
I know that for the [imath]2[/imath]-dimensional case: given a correlation [imath]\rho[/imath] you can generate the first and second values, [imath] X_1 [/imath] and [imath]X_2[/imath], from the standard normal distribution. Then from there make [imath]X_3[/imath] a linear combination of the two [imath]X_3 = \rho X_1 + \sqrt{1-\rho^2}\,X_2[/imath] then take [imath] Y_1 = \mu_1 + \sigma_1 X_1, \quad Y_2 = \mu_2 + \sigma_2 X_3[/imath] So that now [imath]Y_1[/imath] and [imath]Y_2[/imath] have correlation [imath]\rho[/imath]. How would this be scaled to [imath]n[/imath] variables? With the condition that the end variables satisfy a given correlation matrix? I'm guessing at least n variables will need to be generated then a reassignment through a linear combination of them all will be required... but I'm not sure how to approach it. | 2287778 | Generating c identically correlated standard normals
I want to generate [imath]c[/imath] correlated standard normals where the correlation matrix M is something like \begin{matrix} 1 & \rho & \rho \\ \rho & 1 & \rho \\ \rho & \rho & 1 \\ .... \end{matrix} so ones along the diagonal but otherwise all variables have the same correlation [imath]\rho[/imath] where [imath]\rho \geq 0[/imath]. What's the pattern in generating this using linear combinations of the independent standard normal variables? |
1370765 | Projection of a discrete subgroup of [imath]R^n[/imath]
Let [imath]A[/imath] be a discrete subgroup of [imath]\Bbb R^n[/imath] and let [imath]V[/imath] be a [imath]m<n[/imath] dimensional [imath]\Bbb R[/imath]-subspace of [imath]\Bbb R^n[/imath]. Is the projection of [imath]A[/imath] onto [imath]V[/imath] a discrete subgroup? I am most interested in the case of [imath]m=1[/imath]. Edit: I want the subspace to contain a non trivial element of [imath]A[/imath], you can assume this element has minimal norm if necessary(any of the norms that generate the usual topology will do). Note that the result is not true for arbitrary discrete subsets of [imath]\Bbb R^n[/imath]. [imath]S = \{(2^{1/n},n)\subset \Bbb R^2\}[/imath] projected onto [imath]V = \{(x,0)|x \in \Bbb R\}[/imath] is a counter example in this case. This has been answered at : Subgroups of [imath]\Bbb{R}^n[/imath] that are closed and discrete | 186998 | Subgroups of [imath]\Bbb{R}^n[/imath] that are closed and discrete
I am trying to prove that every closed discrete subgroup of [imath]\Bbb{R}^n[/imath] under addition is a free abelian group of finite rank. I have tried to do this by induction on the dimension [imath]n[/imath]. Base case: We claim that a subgroup [imath]H \subset \Bbb{R}[/imath] has a least positive element [imath]\alpha[/imath]. The completeness axiom tells us that there is a least positive real number [imath]\alpha[/imath] among all positive reals in [imath]H[/imath], and being closed implies that [imath]\alpha \in H[/imath]. Now assume the inductive hypothesis and consider a closed, discrete subgroup [imath]H[/imath] of [imath]\Bbb{R}^n[/imath]. Choose some element [imath]x \in H[/imath] of least positive distance to the origin. This can be done because otherwise we get a contradiction like the above. Now let [imath]L[/imath] be the subspace spanned by [imath]x[/imath] and write [imath]\Bbb{R}^n = L \oplus W[/imath] for some complement [imath]W[/imath]. Consider the projection (as a linear map) [imath]p : \Bbb{R}^n \to W[/imath]. Now this is also a group homomorphism and so we have [imath]\textrm{Im} (H) \subset W[/imath] being a subgroup and so by the induction hypothesis is isomorphic to [imath]\Bbb{Z}^l[/imath] for some [imath]1 \leq l \leq n-1[/imath]. However from here I am having some trouble concluding that [imath]H[/imath] itself must be isomorphic to [imath]\Bbb{Z}^l \oplus \Bbb{Z}x[/imath]. If I know the existence of a map [imath]l : \textrm{Im}(H) \to H[/imath] such that [imath]p \circ l = \textrm{id}_{\textrm{Im}(H)}[/imath] then by the splitting lemma I can conclude my problem. However, I don't have such a map so how do I conclude the problem? Please do not post complete solutions. Thanks. Edit: Perhaps I should add some context. I am trying to conclude that the kernel of the exponential map [imath]\exp : \mathfrak{g} \to G[/imath] where [imath]G[/imath] is a connected abelian matrix Lie group is a free abelian group of finite rank. We know that the assumptions on our matrix Lie group mean that [imath]\exp[/imath] is a group homomorphism, so that [imath]\ker \exp[/imath] is a subgroup of the Lie algebra [imath]\mathfrak{g}[/imath]. Edit: The problem is reduced to showing that the image of [imath]H[/imath] under the projection is indeed discrete, because otherwise we cannot apply the inductive hypothesis. |
1310429 | How to solve non-linear differential equation
How to solve non-linear differential equation [imath]y'(x) = y(y(x)), \quad y\colon\mathbb{R}\to\mathbb{R}?[/imath] Of course, [imath]y(x)\not\equiv 0[/imath]. If we substitute [imath]y(x) = Ax^n[/imath], we get complex [imath]n[/imath] and [imath]A[/imath]. Any numerical solution doesn't work because we can't calculate [imath]y(y(x))[/imath]. Series expansion for [imath]y(x)[/imath] is also failed. Is there a theory of such equations? Numerical solution is not what I want. | 916967 | How do you solve [imath]f'(x) = f(f(x))[/imath]?
A friend told me to solve the following differential equation: [imath]f'(x)=f(f(x))[/imath] I have no idea how to solve this! This doesn't seem to be an ordinary differential equation and I can't even solve this numerically! I think my friend is trolling me. |
1372055 | Finding probability density of [imath]Y=X^2[/imath], assuming [imath]X\sim U(-1,1)[/imath]
Suppose [imath]X[/imath] has a uniform distribution on [imath](-1,1)[/imath], and let [imath]Y = X^2[/imath]. How do we find [imath]f_Y[/imath]? | 305997 | Does the square of uniform distribution have density function?
[imath]X\sim U[0,1][/imath] and [imath]Y\sim U[-1,1][/imath] are two uniform-distributed R.V.'s. Are [imath]X^2[/imath] and [imath]Y^2[/imath] still uniform? Do they have explicit probability density funtion? |
1372323 | A sequence of polynomials
I posted this question a while back, and I think I may not have made myself clear or shown what I got so far. So let me post this question again with more information and clarification. I have a sequence of polynomials [imath]Q_k(x, y)[/imath], [imath]k\geq 1[/imath] defined recursively as follows: [imath]Q_1=x[/imath]. There is a sequence of polynomials [imath]p_j(y)[/imath] of degree [imath]j[/imath] such that [imath]Q_{2m}[/imath] is of the form \begin{eqnarray}\frac{p_{0}(y)}{(2m)!}x^{2m}+\frac{p_{1}(y)}{(2m-2)!}x^{2m-2}+\cdots+\frac{p_{m-1}(y)}{2!}x^2+p_{m}(y)\end{eqnarray} and [imath]Q_{2m+1}[/imath] \begin{eqnarray} \frac{p_{0}(y)}{(2m+1)!}x^{2m+1}+\frac{p_{1}(y)}{(2m-1)!}x^{2m-1}+\cdots+\frac{p_{m-1}(y)}{6}x^3+p_{m}(y)x \end{eqnarray} [imath]Q_k(k+1-2i, k+1)[/imath] as a polynomial in [imath]i[/imath] has roots [imath]1, 2, \cdots, k[/imath]. Find a general formula of [imath]Q_k(x, y)[/imath]. The following are the first 4 polynomials in the sequence: \begin{align*} Q_1&=x\\ Q_2&=\frac{x^2}{2}-\frac{y}{6}\\ Q_3&=\frac{x^3-xy}{6}\\ Q_4&=\frac{x^4-2x^2y}{24}+\frac{y(5y+2)}{360} \end{align*} In fact, it suffices to find the sequence of polynomials [imath]p_j(y)[/imath]. Here's what I've got so far. By condition (2) and (3), we have \begin{eqnarray} \sum_{j=0}^m \frac{p_j(2m+1)}{(2m-2j)!}(2i-2m-1)^{2m-2j}=\frac{2^{2m}}{(2m)!}(i-1)(i-2)\cdots(i-2m) \end{eqnarray} If we let [imath]m=j+k[/imath], differentiate both sides with respect to [imath]i[/imath] [imath]2k[/imath] times and put [imath]\displaystyle i=\frac{2j+2k+1}{2}[/imath], we have \begin{eqnarray} p_j(2j+2k+1)=\left.\frac{d^{2k}}{di^{2k}}\right|_{i=\frac{2j+2k+1}{2}}\frac{2^{2j}}{(2(j+k))!}(i-1)(i-2)\cdots(i-2(j+k)) \end{eqnarray} Letting [imath]k=0, 1, \cdots, j[/imath], we get the [imath]j+1[/imath] values taken by [imath]p_j[/imath] at [imath]2j+1, 2j+3, \cdots, 4j+1[/imath]. [imath]p_j[/imath] then can be computed using Lagrangian interpolation. It seems to me that, though the above algorithm can be implemented on a computer to get a few polynomials in the sequence, it does not yield directly a general formula I want. Is there another better way to go about getting a general formula? Edit: The first 5 members in the polynomial sequence [imath]p_j(y)[/imath] are the following: \begin{align*} p_0(y)&=1\\ p_1(y)&=-\frac{y}{6}\\ p_2(y)&=\frac{y(5y+2)}{360}\\ p_3(y)&=-\frac{y(35y^2+42y+16)}{45360}\\ p_4(y)&=\frac{y(5y+4)(35y^2+56y+36)}{5443200} \end{align*} | 1360476 | A polynomial sequence
I have a sequence of polynomials [imath]Q_k(x, y)[/imath], [imath]k\geq 1[/imath] defined recursively as follows: [imath]Q_1=x[/imath]. There is a sequence of polynomials [imath]p_j(y)[/imath] of degree [imath]j[/imath] such that [imath]Q_{2m}[/imath] is of the form \begin{eqnarray}\frac{p_{0}(y)}{(2m)!}x^{2m}+\frac{p_{1}(y)}{(2m-2)!}x^{2m-2}+\cdots+\frac{p_{m-1}(y)}{2!}x^2+p_{m}(y)\end{eqnarray} and [imath]Q_{2m+1}[/imath] \begin{eqnarray} \frac{p_{0}(y)}{(2m+1)!}x^{2m+1}+\frac{p_{1}(y)}{(2m-1)!}x^{2m-1}+\cdots+\frac{p_{m-1}(y)}{6}x^3+p_{m}(y)x \end{eqnarray} [imath]Q_k(k+1-2i, k+1)[/imath] as a polynomial in [imath]i[/imath] has roots [imath]1, 2, \cdots, k[/imath]. Find a general formula of [imath]Q_k(x, y)[/imath]. The following are the first 4 polynomials in the sequence: \begin{align*} Q_1&=x\\ Q_2&=\frac{x^2}{2}-\frac{y}{6}\\ Q_3&=\frac{x^3-xy}{6}\\ Q_4&=\frac{x^4-2x^2y}{24}+\frac{y(5y+2)}{360} \end{align*} In fact, it suffices to find the sequence of polynomials [imath]p_j(y)[/imath]. Here's what I've got so far. By condition (2) and (3), we have \begin{eqnarray} \sum_{j=0}^m \frac{p_j(2m+1)}{(2m-2j)!}(2i-2m-1)^{2m-2j}=\frac{2^{2m}}{(2m)!}(i-1)(i-2)\cdots(i-2m) \end{eqnarray} If we let [imath]m=j+k[/imath], differentiate both sides with respect to [imath]i[/imath] [imath]2k[/imath] times and put [imath]\displaystyle i=\frac{2j+2k+1}{2}[/imath], we have \begin{eqnarray} p_j(2j+2k+1)=\left.\frac{d^{2k}}{di^{2k}}\right|_{i=\frac{2j+2k+1}{2}}\frac{2^{2j}}{(2(j+k))!}(i-1)(i-2)\cdots(i-2(j+k)) \end{eqnarray} Letting [imath]k=0, 1, \cdots, j[/imath], we get the [imath]j+1[/imath] values taken by [imath]p_j[/imath] at [imath]2j+1, 2j+3, \cdots, 4j+1[/imath]. [imath]p_j[/imath] then can be computed using Lagrangian interpolation. It seems to me that, though the above algorithm can be implemented on a computer to get a few polynomials in the sequence, it does not yield directly a general formula I want. Is there another better way to go about getting a general formula? Edit: The first 6 members in the polynomial sequence [imath]p_j(y)[/imath] are the following: \begin{align*} p_0(y)&=1\\ p_1(y)&=-\frac{y}{6}\\ p_2(y)&=\frac{y(5y+2)}{360}\\ p_3(y)&=-\frac{y(35y^2+42y+16)}{45360}\\ p_4(y)&=\frac{y(5y+4)(35y^2+56y+36)}{5443200}\\ p_5(y)&=-\frac{y(385y^4+1540y^3+2684y^2+2288y+768)}{359251200} \end{align*} It is noteworthy that the denominators happen to be the first 6 numbers in this sequence, as pointed out by Solomonoff's Secret. |
1372459 | One divided by infinity is not zero?
I know that [imath]\frac{1}{\infty}[/imath] is undefined. But my question is - can we say that [imath]\frac{1}{\infty}\neq0[/imath] ? I've got some idea how to explain that: Let's say we have a random-number generator that generates numbers in interval [imath](0, \infty)[/imath]. What is the probability that it generates 5? [imath]\frac{1}{\infty}[/imath]. So it cannot equals zero because if it was zero, it wouldn't generate any number (every number has the same probability - [imath]\frac{1}{\infty}[/imath]). Am I right? Or is that wrong idea? Thanks | 156473 | Is [imath]\frac{1}{\infty}[/imath] equal zero?
After reading this paragraph: A simpler version of this distinction might be more palatable: flip a coin infinitely many times. The probability that you flip heads every time is zero, but it isn't impossible (at least, it isn't more impossible than flipping a coin infinitely many times to begin with!). From here: Is it generally accepted that if you throw a dart at a number line you will NEVER hit a rational number? I thought to myself, why is the probability of flipping heads every time zero? If we have a perfect coin and we flip it twice we get the probability to equal this: [imath]\frac 12 \times \frac 12 = \frac 14[/imath] or for four flips we get this: [imath]\frac 12 \times \frac 12 \times \frac 12 \times \frac 12 = \frac 1{16}[/imath] So as we approach an infinite amount of coin flips, the probability gets smaller and smaller, but it shouldn't ever reach zero, so the probability shouldn't be zero, it should be [imath]1/+\infty[/imath]. Does this mean that [imath]1/+\infty[/imath] equals zero or have I misunderstood the question? |
1372546 | Prove eigenvalue for [imath]A^2 + I[/imath]
This is a proof I've been trying to figure out since the problem was presented to me. We are given that [imath]\lambda[/imath] is an eigenvalue for a matrix [imath]A[/imath] and the vector [imath]u[/imath] is the eigenvector corresponding to [imath]\lambda[/imath]. The problem asks to us to prove that [imath]{\lambda}^2 + 1[/imath] is an eigenvalue for [imath]A^2 + I[/imath], where [imath]I[/imath] is the identity matrix. I am not quite sure where to start. My first instinct was to somehow manipulate [imath]Au = {\lambda}u[/imath], but I'm not quite sure what to do with it. Any ideas for this one? It's certainly an interesting problem, in my opinion. | 1372522 | Eigenvalue Problem -- prove eigenvalue for [imath]A^2 + I[/imath]
This is a proof I've been trying to figure out since the problem was presented to me. We are given that [imath]\lambda[/imath] is an eigenvalue for a matrix [imath]A[/imath] and the vector [imath]u[/imath] is the eigenvector corresponding to [imath]\lambda[/imath]. The problem asks to us to prove that [imath]\lambda^2 + 1[/imath] is an eigenvalue for [imath]A^2 + I[/imath], where [imath]I[/imath] is the identity matrix. I am not quite sure where to start. My first instinct was to somehow manipulate [imath]Au = \lambda u[/imath], but I'm not quite sure what to do with it. Any ideas for this one? It's certainly an interesting problem, in my opinion. |
1372487 | [imath]A v = \lambda v \implies A^* v = \bar{\lambda} v[/imath] if [imath]A[/imath] is normal
I want to show that if [imath]A[/imath] is normal then [imath] A v = \lambda v \implies A^* v = \bar{\lambda} v [/imath] I can show that [imath]A^*v[/imath] is also an eigenvector of [imath]A[/imath], using the fact that [imath]A[/imath] and [imath]A^*[/imath] commute, but I know that this doesn't imply [imath]A^* v \propto v[/imath]. So I'm not sure where to go from here. Thanks. EDIT: The top answer solves the problem, but I would like to know how to prove this result using the polarisation identity as suggested below. This is not the approach taken in the other questions on this site, linked to in the comments below. | 436318 | Adjoint matrix eigenvalues and eigenvectors
I just wanted to make sure that the following statement is true: Let [imath]A[/imath] be a normal matrix with eigenvalues [imath]\lambda_1,...,\lambda_n[/imath] and eigenvectors [imath]v_1,...,v_n[/imath]. Then [imath]A^*[/imath] has the same eigenvectors with eigenvalues [imath]\bar{\lambda_1},..,\bar{\lambda_n}[/imath], correct? |
1372166 | A = UL factorization
How do I calculate [imath]A=UL[/imath] factorization where [imath]U[/imath] is upper triangular matrix with 1's along the diagonal and [imath]L[/imath] is lower triangular matrix? How is this similar to the [imath]LU[/imath] factorization? | 248579 | Using permutation matrix to get LU-Factorization with [imath]A=UL[/imath]
Let [imath]Q[/imath] be the [imath]n[/imath]x[imath]n[/imath] permutation matrix [imath]Q= \begin{bmatrix} 0&0&...&0&1\\ 0&0&...&1&0\\ .& \\ .&\\ .&\\ 0&0&...&0&0\\ 1&0&...&0&0\\ \end{bmatrix}[/imath] If [imath]L \in \mathbb{R}[/imath] is lower triangular, what is the structure of [imath]QLQ[/imath]? Use this to show that one can factorize [imath]A=UL[/imath] where [imath]U[/imath] is unit upper triangular and [imath]L[/imath] is lower triangular. I can see that [imath]QLQ = L^T[/imath] And [imath]Q^2=I[/imath] So here's what I am doing [imath]A=LU[/imath] [imath]QAQ=QLUQ = QLQ^2UQ =QLQQUQ = UL[/imath] But now I am left with [imath]QAQ = UL[/imath] rather than [imath]A=UL[/imath] But does that matter? It seems like it does as the factorization I get would be for solving [imath]A^Tx=b[/imath] rather than [imath]Ax=b[/imath] So have I missed something, is there a way to get the factorization [imath]A=UL[/imath] or have I actually got it but it's the case that I am misinterpreting my answer? |
1372381 | How should I read formulas using multiple variables with the same name?
I am reading computer science texts, which tend to include a lot of summations and set logic. In set logic, formulas commonly use a capitalized variable to reference a whole set, i.e. [imath]\Omega[/imath], and a lowercase variable to represent a given member, i.e. [imath]\omega[/imath]. Given that, how does one read something like: [imath] 0\leq Pr(\omega)\leq 1\mathrm{\quad and \quad}\sum_{\omega\in\Omega}{Pr(\omega)} = 1 [/imath] Perhaps it's due to my [poor] command of math, but I tend to read this out loud to myself, and it's odd to say "The sum of the probability of omega, where omega is a member of omega..." | 284201 | How to read letters such as [imath]\mathbb A[/imath], [imath]\mathbb B[/imath], etc., or [imath]\mathfrak A[/imath], [imath]\mathfrak B[/imath], etc.?
This is a soft question. We can read the letters [imath]\bf A[/imath], [imath]\bf B[/imath], etc. as bold A, bold B, etc. We can read the letters [imath]\textit{A}[/imath], [imath]\textit{B}[/imath], etc. as italic A, italic B, etc. We can read the letters [imath]\mathcal A[/imath], [imath]\mathcal B[/imath], etc. as calligraphic A, calligraphic B, etc. But how do we read letters such as [imath]\mathbb A[/imath], [imath]\mathbb B[/imath], etc., or [imath]\mathfrak A[/imath], [imath]\mathfrak B[/imath], etc.? |
1372943 | Do analytic functions on open subsets of [imath]\mathbb{C}[/imath] with an analytic square root form a sheaf?
I'm trying to learn algebraic geometry and am trying to think about what kinds of things are presheafs but not sheafs. One exercise I had was to show that bounded holomorphic functions on open subsets of [imath]\mathbb{C}[/imath] do not form a sheaf. This is pretty obvious: [imath]D_1(0), D_2(0), ...[/imath] form an open cover of [imath]\mathbb{C}[/imath], and [imath]f_n: D_n(0) \rightarrow \mathbb{C}[/imath] given by [imath]f_n(z) = z[/imath] is holomorphic on [imath]D_n(0)[/imath] and bounded, and the restriction of any [imath]f_n, f_m[/imath] to [imath]D_n(0) \cap D_m(0)[/imath] gives you the same thing. If we had a sheaf structure, then there would be a bounded holomorphic function [imath]g[/imath] on [imath]\mathbb{C}[/imath] (necessarily [imath]g(z) = z[/imath]) which restricts to [imath]f_n[/imath] on [imath]D_n(0)[/imath] for each [imath]n[/imath], which is absurd. What about the presheaf of holomorphic functions on open subsets of [imath]\mathbb{C}[/imath] which have holomorphic square roots? This shouldn't be a sheaf, but I'm having trouble seeing why, mostly because I'm not really comfortable with saying with certainty that a given function has no holomorphic square root. | 38423 | Presheaf which is not a sheaf -- holomorphic functions which admit a holomorphic square root
I'm thinking about a problem in Ravi Vakil's algebraic geometry notes http://math.stanford.edu/~vakil/216blog/ (Exercise 3.2B) and I'm having trouble with the second part of the exercise as my understanding of complex analysis is quite rudimentary. If to every open set [imath]U\subset \mathbb{C}[/imath] we associate the ring of holomorphic functions which admit a holomorphic square root, this defines a presheaf on [imath]\mathbb{C}[/imath] (I can see this). Apparently however this is not a sheaf -- supposedly it fails to satisfy the gluing property. To prove this, we need to take an open cover [imath]\{U_i\}[/imath] of some open [imath]U\subseteq \mathbb{C}[/imath], a holomorphic function [imath]f_i:U_i\to \mathbb{C}[/imath] for each [imath]i[/imath] which has a holomorphic square root (i.e. there exists a holomorphic function [imath]g_i[/imath] for each [imath]i[/imath] such that [imath]f_i(z)=g_i(z)^2[/imath] for all [imath]z\in U_i[/imath]), which all agree nicely on the intersections of the various open sets, but such that it is impossible to find a holomorphic function [imath]f:U\to \mathbb{C}[/imath] with a holomorphic function which restricts to [imath]f_i[/imath] on each [imath]U_i[/imath]. Any help would be great -- my problem is in showing that such a function cannot exist. |
1372956 | Decompose finitely generated modules and use Krull-Schmidt theorem
I'm trying to show that if [imath]R[/imath] is an Artinian ring, then for finitely generated modules [imath]M,N,N'[/imath], we have that [imath]M\oplus N\cong M\oplus N'[/imath] implies that [imath]N\cong N'[/imath]. I'm supposed to do this by decomposing [imath]M,N,N'[/imath] and then using the Krull-Schmidt Theorem. For the first part, this means writing [imath]M,N,N'[/imath] as direct sums of non-zero submodules. As [imath]R[/imath] is Artinian, finitely generated modules are of finite length (from Hopkins' Theorem) and hence these decompositions will be of finite length. Therefore, we can use the Krull-Schmidt Theorem, which says that these indecomposable components are uniquely determined up to isomorphism. However, this only gives us information about the components of the decompositions of these modules - not about the relations between them (specifically, between [imath]N[/imath] and [imath]N'[/imath]). How can I use the Krull-Schmidt Theorem to get the desired result? | 158316 | In which case [imath]M_1 \times N \cong M_2 \times N \Rightarrow M_1 \cong M_2[/imath] is true?
Usually for modules [imath]M_1,M_2,N[/imath] [imath]M_1 \times N \cong M_2 \times N \Rightarrow M_1 \cong M_2[/imath] is wrong. I'm just curious, but are there any cases or additional conditions where it gets true? James B. |
1373103 | [imath]|f(x)g(x)| = |(f(x)||g(x)|[/imath]
I was wondering if [imath]|f(x)g(x)| = |f(x)| |(g(x)|[/imath] is true all the time as in the case of real numbers. I was not convinced enough that that was true. But I can't think of any counterexample. Thank you. | 633642 | Prove that the absolute value of a product is the product of the absolute values of factors.
Theorem. [imath]|a||b|=|ab|[/imath] Proof. Applying the definition of absolute value, the left hand side of the equation could be either [imath]a\times(-b)[/imath] or [imath](-a)\times(b)[/imath] or [imath]a\times b[/imath] or [imath](-a)\times(-b)[/imath]. For this reason we could have either [imath]-ab[/imath] or [imath]ab[/imath]. If we apply the definition of absolute value to the right hand side of our equation, we discover that we could have either [imath]-ab[/imath] or [imath]ab[/imath]. So the theorem is proved. Is this proof valid? Is it elegant from a mathematical viewpoint? Thank you. Edit: The definition of absolute value I am using above is the following: [imath]|x|=\begin{cases}x, & \text{if $x\ge0$}\\-x,&\text{if $x<0$}\end{cases}[/imath] |
1373306 | Show that in a finite abelian group [imath]G=\{a_1,\dots, a_n\}[/imath], and [imath]x=a_1\dots a_n[/imath], then [imath]x^2=e[/imath]
Let [imath]G=\{a_1, a_2, \dots, a_n\}[/imath] be a finite abelian group and [imath]x=a_1a_2\dots a_n[/imath]. Then show that [imath]x^2=e[/imath]. Let [imath]a_i\in G[/imath], then [imath]a^{-1}_i\in G[/imath]. Suppose [imath]a_i^{-1}=a_j[/imath]. How can show the required result? | 768246 | If G is a finite abelian group and [imath]a_1,...,a_n[/imath] are all its elements, show that [imath]x=a_1a_2a_3...a_n[/imath]must satisfy [imath]x^2=e[/imath].
I have already tried with [imath]S_3[/imath], and indeed, the product is [imath](13)[/imath], and [imath](13)^2=e[/imath] But what about this: I define + in this way:[imath]45=2[/imath],[imath]26=3[/imath], [imath]1[/imath] is the identity. therefore, [imath]123456=2326=233=3[/imath], and as you can see from the picture, the inverse of [imath]3[/imath] is not itself, thus contradicts the statement. My question: What is wrong with my argument? And How do you prove the statement? it seems that my set of numbers perfectly forms a group under the operation I defined via the table above. first it's closed, second there is an identity, third every elements have an inverse. By the way it's commutitive. I tried (45)2=22=6=42=4(52), (45)6=26=3=43=4(56), so associativity seems to be satisfied. |
1373241 | Proof of [imath]P(A'\mid B)=1-P(A\mid B)[/imath]
Could someone help me understand how to prove [imath]P(A'\mid B) = 1-P(A \mid B)[/imath]? I tried to make it so: [imath]P(A'\mid B)= \cfrac{P(A'\cap B)}{P(B)}[/imath] but I'm not sure how to continue. (I see that there is a question asking to prove [imath]P(A|B) = 1-P(A'|B) [/imath] How is it possible that P(Not A |B) would yield the same proof as P(A|B)? | 825698 | Is this always true: [imath]P(A|B) = 1-P(A^c|B)[/imath]?
Does this identity hold for all events? [imath] P(A|B) = 1-P(A'|B) [/imath] Logically speaking, if the probability of [imath]A[/imath] given [imath]B[/imath] occurred is [imath]X[/imath], shouldn't the probability that [imath]A[/imath] does not occur, [imath]A'[/imath], given [imath]B[/imath], be similarly [imath]1-X[/imath]? There is a related question here. This is the closest that I could get to proving (or disproving) it: [imath]P(A\cap B)=P(A)-P(A' \cap B)[/imath] [imath]P(A|B)P(B)=P(A)-P(A'|B)P(B)[/imath] [imath]\therefore P(A'|B)= \frac {P(A)} {P(B)}-P(A|B)[/imath] Are there are certain formulae which can be used to prove this? Or does the identity only hold under certain situations, and if so, what kind of situations? Thanks. |
1373427 | Prob. 3, Sec. 3.2 in Kreyszig's Functional Analysis Book: Is the space of all polynomials of a fixed degree complete?
Let [imath]n[/imath] be a given natural number, and let [imath]X[/imath] denote the vector space consisting of the zero polynomial and of all polynomials of degree at most [imath]n[/imath], with real or complex numbers as co-efficients, and defined on a given closed interval [imath][a,b][/imath] of the real line, with the inner product defined by [imath]\langle x, y \rangle \ \colon= \ \int_a^b \ x(t) \ \overline{y(t) } \ \mathrm{d} t \ \ \ \mbox{ for all } \ x, y \in X.[/imath] Then is [imath]X[/imath] complete with respect to the norm induced by the above inner product? I know that the space of all continuous functions is not complete in the above norm. Let [imath]x_m \colon= \sum_{j=0}^n \alpha_{jm} t^j[/imath], where [imath]t \in [a,b][/imath] and [imath]m \in \mathbb{N}[/imath], be a Cauchy sequence in [imath]X[/imath]. Then, given [imath]\epsilon > 0[/imath], there is a natural number [imath]N = N(\epsilon)[/imath] such that [imath]\Vert x_m - x_k \Vert < \epsilon \ \ \ \mbox{ for all } \ m, k \in \mathbb{N} \ \mbox{ such that } \ m > N \ \mbox{ and } \ k > N.[/imath] Or, [imath]\sqrt{\int_a^b \ \vert x_m(t) - x_k(t) \vert^2 \ \mathrm{d} t } < \epsilon \ \ \ \mbox{ for all } \ m, k \in \mathbb{N} \ \mbox{ such that } \ m > N \ \mbox{ and } \ k > N.[/imath] That is, [imath]\sqrt{\int_a^b \ \left\vert \sum_{j=0}^n \left( \alpha_{jm} - \alpha_{jk} \right) \ t^j \ \right\vert^2 \ \mathrm{d} t } < \epsilon \ \ \ \mbox{ for all } \ m, k \in \mathbb{N} \ \mbox{ such that } \ m > N \ \mbox{ and } \ k > N.[/imath] What next? Our aim should be to achieve the "Cauchy-ness" of the sequence [imath]\alpha_{jm}[/imath] of real or complex numbers, for each [imath]j = 0, 1, \ldots, n[/imath]. Am I right? If so, how to achieve this goal? | 42663 | Why is it true that every finite-dimensional inner product space is a Hilbert space?
A Hilbert space is, by definition, a complete inner product space. If [imath](V,|.|)[/imath] is finite dimensional inner product space of dimension [imath]n[/imath] then it is (topologically) isomorphic to [imath]\mathbb{R}^n[/imath] which is of course complete. My instinct here is to say "and therefore, [imath]V[/imath] is also a Hilbert space". I'm not sure about this step however since completeness depends on the norm and the norm depends on the selected inner product (assuming the induced norm [imath]||x|| = (x | x)^{1/2}[/imath] is used). So, must we place some condition on the inner product or is the topological isomorphism between [imath]\mathbb{R}^n[/imath] enough to guarantee that V is Hilbert? |
1364072 | If [imath]n^x\in\Bbb Z,[/imath] for every [imath]n\in\Bbb Z^+,[/imath] then [imath]x\in\Bbb Z[/imath]
Let [imath]x[/imath] is a real number such that [imath]n^x\in\Bbb Z,[/imath] for every positive integer [imath]n.[/imath] Prove that [imath]x[/imath] is an integer. I got that problem here and it looks difficult, I tried writing [imath]x[/imath] as [imath]\lfloor x\rfloor+\{x\}[/imath] (where [imath]\lfloor x\rfloor[/imath] and [imath]\{x\}[/imath] are the floor and the fractional part of [imath]x,[/imath] respectively), then [imath]n^x=n^{\lfloor x\rfloor}n^{\{x\}}.[/imath] Since [imath]n^{\lfloor x\rfloor}>0,[/imath] then [imath]n^{x-\lfloor x\rfloor}=n^{\{x\}},[/imath] so that [imath]n^{\{x\}}[/imath] must be a rational number for every positive integer [imath]n,[/imath] but I don't know how to prove that [imath]\{x\}[/imath] must necessarily be [imath]0[/imath] ([imath]0\leq\{x\}<1[/imath]). Any help is really appreciated! | 858893 | If any integer to the power of [imath]x[/imath] is integer, must [imath]x[/imath] be integer?
My apologies if this has been asked already, I've searched but couldn't find it... Let [imath]x[/imath] such that for every [imath]y \in N[/imath], [imath]y^x[/imath] is an integer. Does that necessarily mean that [imath]x[/imath] is an integer? |
1374159 | [imath]|g(x)| \leq K \int_a^x|g| \ \ \forall x \in I[/imath]
Let [imath]I:=[a,b][/imath] and let [imath]g: I \to \Bbb R[/imath] be continuous on [imath]I[/imath]. Suppose that there exists [imath]K > 0[/imath] such that [imath]|g(x)| \leq K \int_a^x|g| \ \ \forall x \in I.[/imath] Then [imath]g(x) = 0\ \ \forall x \in I [/imath]. I am stuck with the problem please help! | 1355925 | Inequality of continuous functions
Let [imath]u, v[/imath] be continuous functions on [imath][a,b][/imath] and let [imath]c>0[/imath]. Suppose that for all [imath]x \in [a,b][/imath] we have the following inequality: [imath]|u(x)-v(x)| \leq c \int_a^x |u(t)-v(t)| dt[/imath] Show that [imath]u(x)=v(x)[/imath] for all [imath]x \in [a,b][/imath] My first thought was to consider [imath]h(x)=|u(x)-v(x)|[/imath] and try to show that [imath]h=0[/imath], but I got stuck. Also, I proved the inequality considering the case [imath]c(b-a) \leq 1[/imath], but I'm not sure how to continue. |
1374336 | Continuity of translation property
Let [imath]u \in L^{p}(U)[/imath] where [imath]1 \leq p \lt \infty[/imath] & [imath]U \subseteq \mathbb R^{n}[/imath] . Define : [imath]F : \mathbb R^{n} \to L^{p}(U) [/imath] by [imath] F(y) := u(x+y)[/imath] . Prove that: as a function of [imath]y[/imath] ; [imath]F(y) [/imath] is CONTINUOUS . What I am thinking: I am to show that as [imath]\epsilon \to 0 [/imath] ; we will have : [imath]|| F(y+\epsilon) - F(y)||_{p}^{p} \to 0[/imath] While computing; I obtain : [imath]|| F(y+\epsilon) - F(y)||_{p}^{p} = || u(x+y+\epsilon) - u(x+y)||_{p}^{p} = \int_{\mathbb R^{n}} |u(x+y+\epsilon) - u(x+y)|^{p}[/imath] Here , I am thinking to use the inequality: [imath](a+b)^{p} \leq 2^{p}(a^{p} + b^{p}) [/imath] i.e. to obtain: [imath]\int_{\mathbb R^{n}} |u(x+y+\epsilon) - u(x+y)|^{p} \leq [/imath]C[imath][ |\int_{\mathbb R^{n}} |u(x+y+\epsilon)|^{p} + \int_{\mathbb R^{n}}|u(x+y)|^{p}][/imath] Now, using the fact that: [imath]u \in L^{p}[/imath] can we say that: [imath]F[/imath] is continuous?? If my thinking is wrong, please let me know the correct solution!! Thanking you.. | 842937 | Show that [imath]\lim _{r \to 0} \|T_rfβf\|_{L_p} =0.[/imath]
I am having a hard time with the following real analysis qual problem. Any help would be awesome. Thanks. Suppose that [imath]f \in L^p(\mathbb{R}),1\leq p< + \infty.[/imath] Let [imath]T_r(f)(t)=f(tβr).[/imath] Show that [imath]\lim_{r \to 0} \|T_rfβf\|_{L_p} =0.[/imath] |
1374859 | E or natural log problem, solve the equation
If someone could explain to me the first step or two so I could solve this that would be great. All the e's are confusing me Solve the equation. (Round your answer to four decimal places.) [imath]e^x β 6e^β{^x} β 1 = 0[/imath] | 1366723 | Solve the equation. e and natural logs
[imath]e^x β 6e^{-x} β 1 = 0[/imath] No idea how to solve this. If someone could show me the first one or two steps to push me in the right direction that would be great. |
1374813 | Existence of formulae for sines/cosines of products of angles in terms of sines/cosines of original angles?
There was something that I was getting a little curious about. We know that there are the so-called compound-angle formulae for calculating sines and cosines of sums of angles in terms of those of the original angles: [imath]\sin (\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta[/imath] [imath]\cos (\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.[/imath] Now, do there exist any sort of formulae for products of angles, that is, for [imath]\sin(\alpha\beta)[/imath] and [imath]\cos(\alpha\beta)[/imath]? If so, how can one derive them? (I had a play and tried to get something, and it ended up being rather nasty; only holding for small-ish angles [imath]\beta[/imath] and resulting in some infinite series that may or may not converge to anything well-known.) | 1280906 | How is [imath] \cos (\alpha / \beta) [/imath] expressed in terms of [imath]\cos \alpha [/imath] and [imath] \cos \beta [/imath]?
If it is possible to express [imath] \cos n \alpha [/imath] in terms of [imath] \cos \alpha [/imath] as a power series for integer [imath]n[/imath] ... I like to see an expression for the quotient angle that obviously tallies when [imath] (\alpha , \beta) [/imath] are swapped. EDIT1: Something like: [imath] \cos (\alpha - \beta)= \cos \alpha \cos \beta + \sin \alpha \sin \beta [/imath] EDIT2: Like to know why [imath] \cos ( \alpha / \beta )[/imath] cannot be expressed in terms of [imath] \cos \alpha, \cos \beta [/imath], but [imath] \cos ( \alpha + \beta) [/imath] can be expressed in terms of [imath]\cos \alpha [/imath] and [imath] \cos \beta. [/imath] |
706837 | Cardinality with property [imath]x_1^2+x_2^2+...+x_n^2<1[/imath]
{[imath]x_1,x_2,...,x_n[/imath]} of T (with no two of [imath]x_1,x_2,...,x_n[/imath] equal) has the property that [imath]x_1^2+x_2^2+...+x_n^2<1[/imath], then prove that T is a countable set. I do it in this way, in interval [imath][\frac{1}{k+1},\frac{1}{k})[/imath], for [imath]k=1[/imath], i.e. [imath][\frac{1}{2},1)[/imath], let [imath]x_1=\frac{1}{2}[/imath] and [imath]x_1^2+x_2^2+x_3^2+x_4^2<4x_1^2=1[/imath], so there are 4 elements at most. for [imath]k=2[/imath], there are at most 9 elements. so for every big [imath]n\in \mathbb{N}[/imath], there are at most [imath](n+1)^2[/imath] elements on the interval [imath][\frac{1}{n+1},\frac{1}{n})[/imath], so the set T is countable, am I right? there is a hint said we have to use Archimedean principle somewhere, but I don't know where to use it? | 704144 | Prove that T is a countable set
Let [imath]T[/imath] be a nonempty subset of the interval [imath](\,0, 1)[/imath]. If every finite subset [imath]\{x_1,x_2,β¦,x_n\}[/imath] of [imath]T[/imath] (with no two of equal) has the property that [imath]x_1^2+x_2^2+β―+x_n^2<1[/imath], then prove that [imath]T[/imath] is a countable set. |
1375115 | Given a real [imath]x[/imath] and an integer [imath]N \gt 1[/imath], prove that there exist integers [imath]h[/imath] and [imath]k[/imath] with [imath]0 \lt k \le N[/imath] such that [imath]|kx-h|\lt 1/N[/imath].
Given a real [imath]x[/imath] and an integer [imath]N \gt 1[/imath], prove that there exist integers [imath]h[/imath] and [imath]k[/imath] with [imath]0 \lt k \le N[/imath] such that [imath]|kx-h|\lt 1/N[/imath]. Hint. Consider the [imath]N+1[/imath] numbers [imath]tx-[tx][/imath] for [imath]t=0,1,2,\dots, N[/imath] and show that some pair differs by at most [imath]1/N[/imath]. I'm trying to prove this hint. Let [imath]a_t=tx-[tx][/imath] for [imath]t=0,1,2,\dots, N[/imath], and assume that for every different pair [imath]|a_i-a_j| \ge 1/N[/imath]. I can't lead to a contradiction from this. I would greatly appreciate any help. | 922629 | Given [imath]x \in \mathbb{R}[/imath] and [imath]N \in \mathbb{N}, N>1,[/imath] find integers [imath]h,k[/imath] with [imath]0 < k \leq N [/imath] that satisfy [imath]\lvert{kx - h\rvert} < 1/N.[/imath]
This is a problem from Apostol's [imath]\textit{Mathematical Analysis}[/imath]. He provides a hint: Consider the [imath]N+1[/imath] numbers [imath]tx-[tx][/imath] for [imath]t=0,1,\ldots,N[/imath] and show that some pair differs by at most [imath]1/N[/imath], where [imath][x][/imath] is defined as the greatest integer [imath]\leq x[/imath]. I have defined [imath]a_t:= tx-[tx][/imath]. I know [imath]a_t \in [0,1)[/imath] for every [imath]t=0,1,2,\ldots,N[/imath]. How can I show that there exists [imath]i,j \in \lbrace{0,1,\ldots,N\rbrace}, i \neq j[/imath] such that [imath]N\lvert{a_i-a_j\rvert}[/imath] < 1? I've been stuck on this problem for a quite a while. A hint would be much appreciated. |
358356 | Solving the improper integral [imath]\int_0^{\infty} {e^{-ax}\cos{(bx)}} dx[/imath]
[imath]\int_0^{\infty} {e^{-ax}\cos{(bx)}} dx[/imath] I know I need to use integration by part method. But I'm not sure how does the improper integral take place? | 663131 | Integral [imath]\int_{0}^{\infty}e^{-ax}\cos (bx)\operatorname d\!x[/imath]
I want to evaluate the following integral via complex analysis [imath]\int\limits_{x=0}^{x=\infty}e^{-ax}\cos (bx)\operatorname d\!x \ \ ,\ \ a >0[/imath] Which function/ contour should I consider ? |
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