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407212 | Proving that [imath]\mathbb{Z}[/imath] is not a field.
This question is related to this answer. I always thought that we can simply say since [imath]2 \in \mathbb{Z}[/imath], and [imath]2^{-1} = 1/2 \notin \mathbb{Z}[/imath], [imath]\mathbb{Z}[/imath] is not a field. Can someone explain why this proof is wrong as the answer claims? | 407163 | Clearly, [imath]3 \in \mathbb{Z}[/imath] is not a unit, because [imath]1/3 \notin \mathbb{Z}[/imath]. What theorem does this kind of reasoning appeal to?
Intuitively, we may conclude that [imath]3[/imath] is not a unit in [imath]\mathbb{Z}[/imath] simply by observing that [imath]1/3 \notin \mathbb{Z}[/imath]. However, what does this reasoning actually appeal to??? Is it true that: Conjecture. For all rings [imath]R[/imath] and all [imath]r \in R[/imath], if there exists a ring [imath]S[/imath] and a homomorphism [imath]\varphi : R \rightarrow S[/imath] such that [imath]\varphi(r)[/imath] has a multiplicative inverse [imath]s' \in S[/imath] such that [imath]s' \notin \mathrm{im}\, \varphi[/imath], then [imath]r[/imath] is not a unit in [imath]R[/imath]. If not, what is the correct statement of the theorem? |
407288 | Topology of [imath](S^1 \times S^1)/C_n[/imath]
Prepare two [imath]S^1[/imath]. Let a cyclic group [imath]C_n[/imath] of order [imath]n[/imath] acts on each [imath]S^1[/imath] by rotation. The actions may be distinct and may not be faithful. What is the topology of the quotient space [imath](S^1 \times S^1)/C_n[/imath]? I think this is again [imath]S^1 \times S^1[/imath] as topological space. Am I right? | 204273 | What is [imath](S^1\times S^1)/C_{n}[/imath] topologically?
Assume that the cyclic group [imath]C_{n}[/imath] of order [imath]n[/imath] acts on [imath]T^2=S^1\times S^1[/imath] by rotating each factor, i.e. a generator of [imath]C_{n}[/imath] acts as [imath] (x,y)\mapsto (e^{\frac{2\pi i}{n}}x,e^{\frac{2\pi i}{n}}y). [/imath] What is the quotient [imath]T^2/C_{n}[/imath] topologically? I initially thought it would be again [imath]T^2[/imath], but things seem not that easy. |
407350 | character of a group is constant on each conjugacy class in the group
I'm trying to solve this problem and I'm not quite sure what I need to prove here. Can you guys please help?. Here is the question. Definition: Character of a group [imath]G[/imath] is defined to be a group homomorphism [imath]\chi: G \rightarrow \mathbb{C^*}[/imath]. Prove that the character of a group [imath]G[/imath] is constant on each conjugacy class in [imath]G[/imath]. My answer: Let [imath]x\in G[/imath] be fixed. Then [imath]\chi(gxg^{-1})=\chi(g) \chi(x) \chi(g) ^{-1}[/imath] since [imath]\chi[/imath] is a homomorphism. But [imath]\mathbb{C^*}[/imath] is abelian so [imath]\chi(gxg^{-1})=\chi(x)[/imath]. Have I proved what I needed to prove? :-). I guess my question is what exactly do they mean by "constant on each conjugacy class in [imath]G[/imath]". Thanks for all your help. | 332005 | Character on conjugacy classes
Let [imath]V_j[/imath], [imath]j = 1,2[/imath] be finite dimensional representations of a group [imath]G[/imath]. Show: [imath]\chi_{V_j}[/imath] is a constant on each conjugacy class of [imath]G[/imath], where [imath]\chi_{V_j}[/imath] is the character of the representation. I've just started with group theory and have a really hard time so I'd like someone to confirm what I did so far was correct: Per definition: [imath]\chi_{V_j} = Tr(\rho(g))[/imath] where [imath]\rho[/imath] is the grouphomomorphism [imath]G \rightarrow GL(V)[/imath] which represents [imath]G[/imath]. The conjugacy class of [imath]G[/imath] is defined as [imath]\{ ghg^{-1} | g \in G \}[/imath] so I'm just plugging in: [imath]\chi_{V_j}(ghg^{-1}) = Tr(\rho(ghg^{-1}))[/imath] which is the same as (because it's a group hom.) [imath]Tr(\rho(g) \rho(h) \rho(g^{-1}))[/imath] and since [imath]Tr(AB) = Tr(BA)[/imath] we get: [imath]Tr(\rho(g) \rho(h) \rho(g^{-1})) = Tr(\rho(h) \rho(g^{-1}) \rho(g)) = Tr(\rho(h))[/imath] Now since [imath]h[/imath] is a constant element of [imath]G[/imath] this is a constant function and I'm done. Is this correct? Thanks a lot in advance! Cheers |
407550 | Is the union of a connected set [imath]Y[/imath] and a clopen in its complement's subspace [imath]X - Y[/imath] connected?
Let [imath]X[/imath] be connected topological space, [imath]Y\subset X[/imath] be connected subspace. Suppose [imath]A\subset X-Y[/imath] is clopen in subspace [imath]X-Y[/imath]. Is [imath]Y\cup A[/imath] connected? | 302054 | How to prove this result about connectedness?
Let [imath]X[/imath] and [imath]Y[/imath] be connected, and let [imath]Y \subseteq X[/imath]. If [imath]A[/imath] and [imath]B[/imath] are two non-empty, disjoint open sets (open in the subspace [imath]X-Y[/imath]) whose union is [imath]X-Y[/imath], or in other words if [imath]A[/imath] and [imath]B[/imath] form a separation of [imath]X-Y[/imath], then how to prove that [imath]Y \cup A[/imath] and [imath]Y \cup B[/imath] are connected? |
407938 | The relation between minimal ideals and zero divisors
How we can prove this Theorem. Let [imath]R[/imath] be a reduced ring. Then [imath]a\in ZD(R)[/imath] (the zero divisors of [imath]R[/imath]) if and only if [imath]a\in P[/imath] for some minimal prime ideal [imath]P[/imath]. | 518 | Why does a minimal prime ideal consist of zerodivisors?
Let [imath]A[/imath] be a commutative ring. Suppose [imath]P \subset A[/imath] is a minimal prime ideal. Then it is a theorem that [imath]P[/imath] consists of zero-divisors. This can be proved using localization, when [imath]A[/imath] is noetherian: [imath]A_P[/imath] is local artinian, so every element of [imath]PA_P[/imath] is nilpotent. Hence every element of [imath]P[/imath] is a zero-divisor. (As Matt E has observed, when [imath]A[/imath] is nonnoetherian, one can still use a similar argument: [imath]PA_P[/imath] is the only prime in [imath]A_P[/imath], hence is the radical of [imath]A_P[/imath] by elementary commutative algebra.) Can this be proved without using localization? |
408104 | Is the square root of [imath]4[/imath] only [imath]+2[/imath]?
Why is [imath]4^{1/2}=+2[/imath]? It should also be [imath]-2[/imath] since both squared just give two only. Also why do we always represent root of [imath]x[/imath] on the right side of the number line? | 158566 | Is it wrong to say [imath] \sqrt{x} \times \sqrt{x} =\pm x,\forall x \in \mathbb{R}[/imath]?
Is it wrong to say [imath] \sqrt{x} \times \sqrt{x} =\sqrt{x^2}= \pm x[/imath] I am quite sure that [imath]\sqrt{(x)^2} = \pm(x)[/imath] But, does [imath]\sqrt{x } \times \sqrt{x} =- (x)[/imath] doesn't holds in [imath]\mathbb{R}[/imath] but if we assume [imath]\mathbb{C}[/imath] it holds right? |
93618 | Irreducibility issue
This is a homework question. Given [imath]f(x)=x^{p-1}+x^{p-2}+\cdots+x+1[/imath], where [imath]p[/imath] is any prime. Prove that [imath]f(x)[/imath] is irreducible over [imath]\mathbb{Z}[x][/imath]? Any idea, hint, etc? Hint given by my book was to use Eisenstein's Irreducibility Criterion. But I see that the coefficients of each term is 1 which is not divisible by any prime number, so how can the criterion be satisfied? | 215042 | Irreducibility of [imath]X^{p-1} + \cdots + X+1[/imath]
Can someone give me a hint how to the irreducibility of [imath]X^{p-1} + \cdots + X+1[/imath], where [imath]p[/imath] is a prime, in [imath]\mathbb{Z}[X][/imath] ? Our professor gave us already one, namely to substitute [imath]X[/imath] with [imath]X+1[/imath], but I couldn't make much of that. |
408282 | Entire function and odd/even function
f is an entire function.If [imath]f(R)\subset R,f(iR)\subset iR,[/imath]then f(z) is an odd function;Similarly,if[imath]f(R)\subset R,f(iR)\subset R,[/imath]then f is an even function. | 103074 | Entire function invariant on the coordinate axes (as sets).
From old qualifying exam: Let [imath]E[/imath] be the union of the two coordinate axes, i.e. [imath]E = \{z=x+iy : xy=0\}[/imath]. Describe all entire functions satisfying [imath]f(E) \subset E[/imath]. I feel like the best approach is to consider the power series of [imath]f[/imath]. My first approach was to write down constraints by considering the function applied to the real or imaginary axis. When I didn't get anywhere with this, I began thinking of the function geometrically: on [imath]E[/imath] it's allowed to scale by a real constant, and rotate by [imath]k\pi/2[/imath]. But again, I couldn't see how to usefully translate this to produce information about the power series. Thanks! As an example, [imath]z^2[/imath] has this property. In fact, so does [imath]az^2+bz^4[/imath] (with [imath]a,b \in \mathbb{R}[/imath]) since each term maps the imaginary axis to the real axis, which ends up back on the real axis when added. A similar argument shows real odd polynomials work, too. |
199415 | Infinite number of rationals between any two reals.
Let [imath]a[/imath] and [imath]b[/imath] be reals with [imath]a<b[/imath]. Show that there are infinitely many rationals [imath]x[/imath] such that [imath]a<x<b[/imath]. My plan of action was to assume that [imath]x[/imath] is the smallest such rational and find another rational in the interval [imath](a, x)[/imath], but I am struggling to make it work. A hint will be much preferred to a full solution. | 1077192 | If [imath]a,b\in\mathbb R[/imath] with [imath]a, then there is some rational r with a.[/imath]
How do you prove this question? I was thinking proving contrapositive. But I was stuck..Thanks guys. |
408435 | If [imath]a \mid c, b \mid c, \gcd (a,b)=1[/imath] then [imath]ab \mid c.[/imath]
If [imath]a \mid c, b \mid c, \gcd (a,b)=1[/imath] then [imath]ab \mid c.[/imath] I understand that given problem is true. however im struggling with writing to prove. I let A=2 , B= 3 , C= 6 2 l 6= 3 3 I 6=2 3*2 l 6=1 I have shown my work to prove that the theorem is true however I can't write in words. how can you write it in words to prove it?? I think proving the theorem or proof is the hardest part of the course | 407540 | If [imath]\gcd(a,b)=1[/imath] and [imath]a[/imath] and [imath]b[/imath] divide [imath]c[/imath], then so does [imath]ab[/imath]
Using divisibility theorems, prove that if [imath]\gcd(a,b)=1[/imath] and [imath]a|c[/imath] and [imath]b|c[/imath], then [imath]ab|c[/imath]. This is pretty clear by UPF, but I'm having some trouble proving it using divisibility theorems. I was looking at this Math.SE problem, and if this could be proven, then the solution would follow immediately. |
408503 | How to calculate the integral of [imath]x^x[/imath] between [imath]0[/imath] and [imath]1[/imath] using series?
How to calculate [imath]\int_0^1 x^x\,dx[/imath] using series? I read from a book that [imath]\int_0^1 x^x\,dx = 1-\frac{1}{2^2}+\frac{1}{3^3}+\dots+(-1)^n\frac{1}{(n+1)^{n+1}}+\cdots[/imath] but I can't prove it. Thanks in advance. P.S: I found some useful materials here and here. | 410609 | How to prove that [imath]\int^1_0 \frac{1}{x^x} dx = \sum^{\infty}_{n=1} \frac{1}{n^n} [/imath]?
The task is: Prove, that [imath]\int^1_0 \frac{1}{x^x} dx = \sum^{\infty}_{n=1} \frac{1}{n^n}[/imath] I completly don't have an idea, how to prove it. It seems very interesting, I will be glad if someone share a proof. My initial thoughts are to use generating function to calculate the series, but I can't find a suitable function. Thanks in advance for help! |
399844 | ¿Can you help me with axiom of regularity?
I understand that if we define a set such that [imath]A=\{A\}[/imath] we have then a contradiction because of regularity axiom, thus, a set can not be a member of itself. My question is; how can it be so, if I define a set [imath]A=\{A,b\}[/imath] ([imath]b[/imath] disjoint from [imath]A[/imath])? Thanks. | 387467 | A basic doubt on axiom of foundation of Zermelo-Fraenkel set theory
I have read in a book that the "axiom of foundation prevents anomalies such as a set being an element of itself". Now, axiom of foundation says that there exist an element in every set which is disjoint with the set. Now if [imath]S[/imath] be a set then according to the above axiom [imath]S=\{S\}[/imath] can't be a set. But, [imath]S=\{\{1\},S\}[/imath] satisfies axiom of foundation and contains [imath]S[/imath]. So, is the statement mentioned in the book correct ? |
409332 | Calculus - Indefinite integration Find [imath]\int \sqrt{\cot x} +\sqrt{\tan x}\,dx[/imath]
Problem : Find [imath]\int \sqrt{\cot x} +\sqrt{ \tan x}\,dx[/imath] My Working : Let [imath]I_1 = \int \sqrt{\cot x}\,dx[/imath] and [imath]I_2 = \int \sqrt{\tan x}\,dx[/imath] By using integration by parts: Therefore , [imath]I_1 = \sqrt{\cot x}.\int1 dx - \int\{(d\sqrt{\cot x}\int 1.dx\}[/imath] [imath]\Rightarrow I_1= x\sqrt{\cot x} + 2x\sqrt{\cot x}-2\int \sqrt{\cot x}\,dx[/imath] [imath]\Rightarrow I_1= x\sqrt{\cot x} + 2x\sqrt{\cot x}-2 I_1[/imath] [imath]\Rightarrow 3I_1= x\sqrt{\cot x} + 2x\sqrt{\cot x} = 3x\sqrt{\cot x} [/imath] [imath]\Rightarrow I_1 = x\sqrt{\cot x}[/imath] Similarly we can find [imath]I_2 = \int\sqrt{\tan x}\,dx[/imath] [imath]I_2 = x\sqrt{\tan x}[/imath] Please suggest whether this is wrong or correct... | 338591 | Calculate [imath]\int\left( \sqrt{\tan x}+\sqrt{\cot x}\right)dx[/imath]
How to calculate following integration? [imath]\int\left( \sqrt{\tan x}+\sqrt{\cot x}\right)dx[/imath] |
410019 | Determine if [imath]\sum\limits_{n=0}^{\infty}\frac{1}{2^{\sqrt{n}}}[/imath] converges
Apparently it can be proven with a comparison but I've tried to compare it to [imath]\frac{1}{n^{p}}[/imath] with not results. I've also tried comparing [imath]\sqrt{n}[/imath] with [imath]\ln{n}[/imath] but [imath]\frac{1}{2^{\ln n}}[/imath] diverges so that doesn't give me anything useful. | 386878 | Does the series [imath]\sum_{n=1}^{\infty}|x|^\sqrt n[/imath] converge pointwise? If it then what would be the sum?
Does the series [imath]\sum_{n=1}^{\infty}|x|^\sqrt n[/imath] converge pointwise? |
408716 | How to find [imath]p[/imath] when [imath] ({\frac{1}{2}})^p + ({\frac{1}{4}})^p + ({\frac{1}{8}})^p - 1 = 0. [/imath]
Kindly mention solution-techniques along with solution | 409925 | Not able to solve [imath]({\frac{1}{2}})^p + ({\frac{1}{3}})^p + ({\frac{1}{7}})^p - 1 = 0.[/imath]
I'm not able to solve [imath]({\frac{1}{2}})^p + ({\frac{1}{3}})^p + ({\frac{1}{7}})^p - 1 = 0.[/imath] If you put values of [imath]p[/imath] (like [imath]\frac{1}{2}[/imath] or 2) back in the equation it doesn't satisfy! So please check your values also. What I found is: If [imath]p = 1[/imath] then the equation becomes [imath] - \frac{1}{42}[/imath]. That means if [imath]p >1[/imath] then values of all the fractions will decrease more and the equation will become more negative. So surely [imath]p < 1[/imath]. Similarly putting [imath]p=0[/imath], the equation becomes [imath]-1[/imath] i. e. < [imath] - \frac{1}{42}[/imath]. Hence [imath] 0 < p < 1. [/imath] Now I'm wondering if there is certain generic techniques to evaluate [imath]p[/imath] ? [ ie. without guessing technique like " Lets assume [imath]p =\frac14(1\pm\sqrt{13}[/imath]) " ] |
407228 | Measure of a Union of different translates of a set
I'm self-studying for prelims and a certain question came up that I have been unable to answer: Given a sequence [imath]\{a_n\}[/imath] dense in [imath]\mathbb{R}[/imath], and [imath]F[/imath] a positive (Lebesgue) measure set, show that the set [imath]\cup_N F + a_n [/imath] is all of [imath]\mathbb{R}[/imath] up to a set of measure zero. So it's clearly true if [imath]F[/imath] is an open interval, but I'm not quite sure how to extend. Any hints/suggestions? Thanks! | 410434 | Sum of a countable dense set and a set of positive measure
Assume [imath]A[/imath] is a countable dense set in [imath]\mathbb{R}[/imath], and set [imath]B[/imath] has positive (Lebesgue) measure. Prove that [imath]A+B=\{a+b:a\in A, b\in B\}=\mathbb{R}\backslash N[/imath], where [imath]N[/imath] is a set of measure zero. I haven't come up with a good idea. Thanks in advance! |
132039 | Laplacian in polar coordinates
I am stuck with an exercise that requires me to find the Laplacian [imath]\Delta u=(D_x^2u+D_y^2u)[/imath] of a 2d-function [imath]u[/imath] in polar coordinates (in the standard Euclidean plane). I found the following article on the net, and tried to follow its logic, but I could not understand two steps: http://www.sci.brooklyn.cuny.edu/~mate/misc/laplacian_polarcoord_higherdim.pdf at first, the representation of [imath]D_x[/imath] and [imath]D_y[/imath] in terms of [imath]r, \theta[/imath], at the bottom of page 2: [imath]D_y=\sin\theta D_r+\frac{\cos\theta}{r}D_\theta[/imath] and [imath]D_x=\cos\theta D_r-\frac{\sin\theta}{r}D_\theta[/imath] . When I draw a sketch of the plane with a circle and all the coordinates, I get that it should be [imath]D_y=\sin\theta D_r+r\ \cos\theta D_\theta[/imath], because the larger the radius is, the greater will be the impact of a change in [imath]\theta[/imath] on a change in [imath]y[/imath]. What am I making wrong here? And then, secondly, what looks like an easy multiplication, namely taking the square of the above terms (on the top of page 3 in the link): [imath]D_y^2=(\sin\theta D_r+\frac{\cos\theta}{r}D_\theta)(\sin\theta D_r+\frac{\cos\theta}{r}D_\theta)[/imath] [imath]=\sin^2\theta D_r^2+\frac{2\sin\theta \cos\theta}{r}D_\theta D_r+\frac{\sin^2\theta}{r^2}D_\theta^2 \\ -\frac{\cos\theta}{r^2}D_\theta + \frac{\cos^2\theta}{r}D_r - \frac{\cos\theta \sin\theta}{r^2}D_\theta[/imath] and similarly for [imath]D_x[/imath]. I don't see from where the last three summands come, what I see is: [imath](a+b)(a+b)=a^2+2ab+b^2+c+d+e[/imath] and I cannot see the context of [imath]c,d,e.[/imath] It would be great if you could explain those issues to me! | 1054768 | Use the chain rule to convert the Laplace equation in (x,y) coordinates into an equivilent differental equation in (r,theta) coordinates.
use the equations [imath]r=\sqrt{x^2 +y^2}[/imath] and [imath]\theta=\arctan(\frac{y}{x})[/imath]. I was able to get the partial derivative of of [imath]r[/imath] with respect to [imath]x[/imath] and [imath]y[/imath] and the partial derivative of [imath]\theta[/imath] with respect to [imath]x[/imath] and [imath]y[/imath]. I have spent hours trying to figure this problem out by looking at the polar form of the Laplace equation as a reference but everything I try does not get me to that point. If anyone is up to the challenge please help me out! I apologize for not putting in Latex format, I have not quite gotten the hang of it yet. |
410707 | Definition of functions
A function is a type of relation. That is, a function [imath]f[/imath] from [imath]X[/imath] to [imath]Y[/imath] is a subset of [imath]X \times Y[/imath] where for each [imath]x \in X[/imath], there is exactly one [imath]y \in Y[/imath] such that [imath](x, y) \in f[/imath]. Suppose [imath]Y \subsetneq Y'[/imath] and [imath]g: X \to Y'[/imath] is defined by [imath]g(x) = f(x)[/imath]. Since functions are sets, and sets are completely determined by there elements [imath]f = g[/imath]. However, [imath]f[/imath] can be surjective while [imath]g[/imath] is not, and in that case, [imath]f \ne g[/imath]. How does one resolve this? | 274967 | It's in my hands to have a surjective function
Let [imath]f[/imath] be any function [imath]A \to B[/imath]. By definition [imath]f[/imath] is a surjective function if [imath]\space \forall y \in B \space \exists \space x \in A \space( \space f(x)=y \space)[/imath]. So, for any function I only have to ensure that there doesn't "remain" any element "alone" in the set [imath]B[/imath]. In other words, the range set of the function has to be equal to the codomain set. The range depends on the function, but the codomain can be choose by me. So if I chose a codomain equal to the range I get a surjective function, regardless the function that is given. M'I right? |
160056 | What is a good book to study linear algebra?
I'm looking for a book to learn Algebra. The programme is the following. The units marked with a [imath]\star[/imath] are the ones I'm most interested in (in the sense I know nothing about) and those with a [imath]\circ[/imath] are those which I'm mildly comfortable with. The ones that aren't marked shouldn't be of importance. Any important topic inside a unite will be boldfaced. U1: Vector Algebra. Points in the [imath]n[/imath]-dimensional space. Vectors. Scalar product. Norm. Lines and planes. Vectorial product. [imath]\circ[/imath] U2: Vector Spaces. Definition. Subspaces. Linear independence. Linear combination. Generating systems. Basis. Dimesion. Sum and intersection of subspaces. Direct sum. Spaces with inner products. [imath]\circ[/imath] U3: Matrices and determinants. Matrix Spaces. Sum and product of matrices. Linear ecuations. Gauss-Jordan elimination. Range. Roché Frobenius Theorem. Determinants. Properties. Determinant of a product. Determinants and inverses. [imath]\star[/imath] U4: Linear transformations. Definition. Nucleus and image. Monomorphisms, epimorphisms and isomorphisms. Composition of linear transformations. Inverse linear tranforms. U5: Complex numbers and polynomials. Complex numbers. Operations. Binomial and trigonometric form. De Möivre's Theorem. Solving equations. Polynomials. Degree. Operations. Roots. Remainder theorem. Factorial decomposition. FTA. Lagrange interpolation. [imath]\star[/imath] U6: Linear transformations and matrices. Matrix of a linear transformation. Matrix of the composition. Matrix of the inverse. Base changes. [imath]\star[/imath] U7: Eigen values and eigen vectors Eigen values and eigen vectors. Characteristc polynomial. Aplications. Invariant subspaces. Diagonalization. To let you know, I own a copy of Apostol's Calculus [imath]\mathrm I [/imath] which has some of those topics, precisely: Linear Spaces Linear Transformations and Matrices. I also have a copy of Apostol's second book of Calc [imath]\mathrm II[/imath]which continues with Determinants Eigenvalues and eigenvectors Eigenvalues of operators in Euclidean spaces. I was reccommended Linear Algebra by Armando Rojo and have Linear Algebra by Carlos Ivorra, which seems quite a good text. What do you reccomend? | 1626093 | Recommended Algebra Books to read?
Can somebody recommend me any books to read that cover the following topics? Chapter 1. Vector Spaces 1.1. Solutions of Simultaneous Linear Equations 1.2. Fields and Vector Spaces 1.3. Product of Sets and of Vector Spaces 1.4. Vector Subspaces 1.5. Linear Independence and Bases 1.6. Dimension of a Vector Space 1.7. Linear Mappings 1.8. Rank-Nullity Theorem Chapter 2. Linear Mappings and Matrices Linear Mappings [imath]F^m → F^n[/imath] and Matrices Basic Properties of Matrices 27 Abstract Linear Mappings and Matrices 30 Change of a Matrix by Change of Basis 33 Chapter 3. Rings and Modules 37 Chapter 4. Determinants and Eigenvalues Redux 63 Chapter 6. Jordan Normal Form Statement of the Jordan Normal Form and Strategy of Proof The proof of Jordan Normal Form Example of Jordan Normal Form A Brief Explanation of the Final Step in PageRank as an Application of the Jordan |
171259 | Equivalent Definitions of the Operator Norm
How do you prove that these four definitions of the operator norm are equivalent? [imath]\begin{align*} \lVert A\rVert_{\mathrm{op}} &= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\ &=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\ &=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\ &=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}. \end{align*}[/imath] | 1951829 | Norms on vector spaces
I would appreciate it if you could give me a few hints as to how I should go about solving this problem. Suppose [imath]T:\mathfrak{X}\rightarrow \mathfrak{Y}[/imath] is bounded. Show that [imath]\left\|T\right\|=\inf\{K\in\mathbb{R}:\left\|Tx\right\|_{\mathfrak{Y}}\leq K\left\|x\right\|_{\mathfrak{X}} \}[/imath]. I know that [imath]T[/imath] is bounded if [imath]\left\|T\right\|:=\sup\{\left\|Tx\right\|_{\mathfrak{Y}}:\left\|x\right\|_{\mathfrak{X}}\leq 1 \}[/imath]. How can I go from [imath]\sup[/imath] to [imath]\inf[/imath]? Thanks in advance |
410995 | A question of finite Group
Let [imath]G = A_5[/imath], the alternating group of degree five,. Let [imath]\pi = \{2,3\}.[/imath] Prove that [imath]M[/imath] is a maximal [imath]\pi[/imath]-subgroup of [imath]G[/imath] if, and only if, [imath]M\cong A_4[/imath] or [imath]M\cong S_3[/imath], where [imath]A_4[/imath] is the alternating group of degree four and [imath]S_3[/imath] is the symmetric group [imath]3[/imath]. | 387777 | An question of [imath]\pi[/imath]-groups for the alternating group [imath]A_5[/imath]
Let [imath]G=A_5[/imath], the alternating group of degree 5. Please help me show that for [imath]\pi = \{2,3\}[/imath] we have that [imath]M \leq G[/imath] is a maximal [imath]\pi[/imath]-group if and only if [imath]M \cong A_4[/imath] or [imath]M \cong S_3[/imath] (the symmetric group). Here a [imath]\pi[/imath]-subgroup is a subgroup all of whose elements are [imath]\pi[/imath]-elements, a [imath]\pi[/imath]-element is an element whose order is a [imath]\pi[/imath]-number, and a [imath]\pi[/imath]-number is a number all of whose prime factors are in [imath]\pi[/imath]. |
411350 | Does there exist a function whose derivative is equal to it's self composition
Not homework, but I was just curious whether or not there exists a function which satisfies: [imath]\ f'(x)=f(f(x))[/imath] I tried differentiating both sides and working from there, but just arrived at nonsense. Thanks. | 408004 | Does a non-trivial solution exist for [imath]f'(x)=f(f(x))[/imath]?
Does [imath]f'(x)=f(f(x))[/imath] have any solutions other than [imath]f(x)=0[/imath]? I have become convinced that it does (see below), but I don't know of any way to prove this. Is there a nice method for solving this kind of equation? If this equation doesn't have any non-trivial solutions, do you know of any similar equations (i.e. involving both nested functions and derivatives) that do have interesting solutions? If we assume [imath]f[/imath] is analytic (which I will be doing from this point onward), then it must also be injective (see alex.jordan's attempted proof), and therefore has at most one root (call it [imath]x_0[/imath].) We know [imath]f'(x_0)=f(f(x_0))=f(0)[/imath]. Claim: [imath]f[/imath] cannot have a positive root. Suppose [imath]x_0[/imath] is positive. If [imath]f(0)[/imath] is negative, then for some sufficiently small [imath]\delta>0[/imath], [imath]f(x_0-\delta)>0[/imath]. This implies there must be another root between [imath]x_0[/imath] and [imath]0[/imath], but [imath]f[/imath] has at most one root. The same reasoning applies if [imath]f(0)[/imath] is positive. If [imath]f(0)=0[/imath], then both [imath]x_0[/imath] and [imath]0[/imath] are roots. Thus, we conclude that [imath]x_0[/imath] cannot be positive. Claim: [imath]f[/imath] cannot have zero as a root. Suppose [imath]x_0=0[/imath]. Since [imath]f[/imath] has at most one root, we know [imath]f[/imath] will be of constant sign on each of the positive and negative halves of the [imath]x[/imath] axis. Let [imath]a<0[/imath]. If [imath]f(a)<0[/imath], this implies [imath]f'(a)=f(f(a))<0[/imath], so on the negative half of the real line, [imath]f[/imath] is negative and strictly decreasing. This contradicts the assumption that [imath]f(0)=0[/imath]. Therefore, [imath]a<0\implies f(a)>0[/imath], which then implies [imath]f'(a)<0[/imath]. But since [imath]f'(a)=f(f(a))[/imath], and [imath]f(a)>0[/imath], this implies [imath]f(b)<0[/imath] when [imath]b>0[/imath]. Moreover, we know [imath]f'(b)=f(f(b))[/imath], and [imath]f(b)<0 \implies f(f(b))>0[/imath], so we know [imath]f[/imath] is negative and strictly increasing on the positive half of the real line. This contradicts the assumption that [imath]f(0)=0[/imath]. Claim: [imath]f[/imath] is bounded below by [imath]x_0[/imath] (which we've proved must be negative if it exists) We know [imath]f(x_0)=0[/imath] is the only root, so [imath]f'(x)=0[/imath] iff [imath]f(x)=x_0[/imath]. And since [imath]f[/imath] is injective, it follows that [imath]f[/imath] is either bounded above or bounded below by [imath]x_0[/imath] (if [imath]f[/imath] crossed [imath]y=x_0[/imath], that would correspond to a local minimum or maximum.) Since [imath]f(x_0)=0[/imath] and [imath]x_0<0[/imath], we know [imath]x_0[/imath] must be a lower bound. Claim: [imath]f[/imath] is strictly decreasing. Question B5 from the 2010 Putnam math competition rules out strictly increasing functions, so we know [imath]f[/imath] must be strictly decreasing. Claim: [imath]f[/imath] has linear asymptotes at [imath]\pm \infty[/imath] Since [imath]f[/imath] is strictly decreasing and bounded below by [imath]x_0[/imath], we know [imath]\lim_{x\rightarrow\infty}f(x)[/imath] is well defined, and [imath]\lim_{x\rightarrow\infty}f'(x)=0[/imath]. Since [imath]f'(x)=0[/imath] iff [imath]f(x)=x_0[/imath], it follows that [imath]\lim_{x\rightarrow\infty}f(x)=x_0[/imath]. [imath]f''(x)=\frac{d}{dx}f'(x)=\frac{d}{dx}f(f(x))=f'(f(x))f'(x)[/imath]. Since [imath]f'(x)<0[/imath], we know [imath]f[/imath] is concave up. Thus, [imath]\lim_{x\rightarrow -\infty}f(x)\rightarrow\infty[/imath]. This in turn implies [imath]\lim_{x\rightarrow -\infty}f'(x)=\lim_{x\rightarrow -\infty}f(f(x))=\lim_{x\rightarrow \infty}f(x)=x_0[/imath]. So [imath]f[/imath] goes to [imath]x_0[/imath] when [imath]x\rightarrow\infty[/imath], and approaches the asymptote [imath]y=x_0\cdot x[/imath] when [imath]x\rightarrow-\infty[/imath]. Claim: [imath]x_0<1[/imath] Consider the tangent line at [imath]f(x_0)[/imath]. We know [imath]f'(x_0)=f(0)[/imath], so this line is given by [imath]y=f(0)x-f(0)x_0[/imath]. Since [imath]f[/imath] is concave up, we know [imath]f(x) > f(0)x-f(0)x_0[/imath] for [imath]x\neq x_0[/imath], so [imath]f(0) > -f(0)x_0[/imath]. And we can conclude [imath]x_0<-1[/imath]. Claim: [imath]f[/imath] must have a fixed point, [imath]x_p[/imath] (i.e. [imath]f(x_p)=x_p[/imath]) We know [imath]f(x_0)=0[/imath], [imath]x_0<0[/imath], and [imath]f(0)<0[/imath]. Therefore, [imath]f(x)-x[/imath] has a root in the interval [imath](x_0,0)[/imath]. This is all I have been able to prove. However, the existence of a fixed point turns out to be very useful in constructing approximate solutions. Consider the following: [imath]f(x_p)=x_p[/imath] [imath]f'(x_p)=f(f(x_p))=x_p[/imath] [imath]f''(x_p)=f'(f(x_p))f'(x_p)=f(f(f(x_p)))f(f(x_p))=x_p^2[/imath] [imath]f'''(x_p)=\cdots=x_p^4+x_p^3[/imath] If we are willing to put in the work, we can evaluate any derivative at the fixed point. I wrote a python program that computes these terms (unfortunately it runs in exponential time, but it's still fast enough to compute the first 20 or so terms in a reasonable amount of time). It leverages the following bit of information. Suppose [imath]f^{[n]}[/imath] represents the nth iterate of [imath]f[/imath]. e.g. [imath]f^{[3]}=f(f(f(x)))[/imath]. Then we can derive the following recursive formula. [imath]\frac{d}{dx}f^{[n]}=f'(f^{[n-1]})\frac{d}{dx}f^{[n-1]}=f^{[n+1]}\frac{d}{dx}f^{[n-1]}[/imath] And since we know the base case [imath]\frac{d}{dx}f^{[1]}=f^{[2]}[/imath], this lets us determine [imath](f^{[n]})'=f^{[n+1]}f^{[n]}\cdots f^{[3]}f^{[2]}[/imath]. So, if we choose a fixed point, we can calculate the expected Taylor series around that point. Here's the graph for the Taylor series with 14 terms, calculated with fixed point [imath]-0.6[/imath] You can clearly see the points where the series starts to fail (the radius of convergence doesn't seem to be infinite), but elsewhere the approximation behaves just as we would expect. I computed [imath](P'(x)-P(P(x)))^2[/imath], where [imath]P[/imath] is the Taylor polynomial, over the range where the series seems to converge, and the total error is on the order of [imath]10^{-10}[/imath]. Moreover, this error seems to get smaller the more accurately you compute the derivative (I used [imath]P'(x)\approx\frac{P(x+0.001)-P(x-0.001)}{0.002}[/imath]). |
411175 | Differentiability of [imath]\operatorname{dist}(x,\partial \Omega)[/imath] function
Let [imath]\Omega[/imath] be a bounded open set in [imath]\mathbb{R}^n[/imath] with smooth boundary and set [imath]\phi(x)=\operatorname{dist} (x,\partial \Omega) =\inf_{y\in \partial \Omega} |x-y|[/imath] for [imath]x\in \overline{\Omega}[/imath]. Is there a subset of [imath]\Omega[/imath] where [imath]\phi[/imath] is differentiable? I'm thinking a subset like [imath]N_s=\{x \in \Omega \mid \operatorname{dist}(x,\partial \Omega) < s\}[/imath]. (for example, in the unitary ball, [imath]\phi[/imath] is not differentiable in the center but it is in [imath]N_s[/imath] for [imath]s<1[/imath]) | 119136 | The differentiability of distance function
Let [imath]M[/imath] be a submanifold of [imath]\mathbb R^n[/imath], then is there an open set [imath]\Omega[/imath] in [imath]\mathbb R^n[/imath] such that function [imath]d(x,M)[/imath] (distance function) is smooth on [imath]\Omega[/imath]? |
407633 | Expectation and Uniqueness
Let Y and Z be two nonnegative random variables (RVs). Let [imath]f(x) \in [0,1][/imath] for all [imath]x\geq0[/imath], and it is monotonically decreasing from 1 to 0 as its argument goes from 0 to [imath]\infty[/imath]. If [imath]E_Y\{f(aX)\}=E_Z\{f(aZ)\}[/imath] for all [imath]a\geq 0[/imath], does this imply Y and Z have the same PDF? More precisely, if the PDFs are [imath]\phi_Y(x)[/imath] and [imath]\phi_Z(x)[/imath], and if [imath]\int_0^{\infty}f(ax)[\phi_Y(x)-\phi_Z(x)]dx=0[/imath] for all [imath]a\geq 0[/imath], does this imply [imath]\phi_Y=\phi_Z[/imath]? note: [imath]E_Y\{\cdot\}[/imath] denotes expectation over the RV Y. | 407377 | Uniqueness for Integral Transform
What can be said about the uniqueness of the following integral transformation: [imath] (Tf)(u) = \int_0^{\infty} f(t)G(tu)dt[/imath] defined for all [imath]u\geq 0[/imath], where the kernel [imath]G(z) \in [0,1][/imath] for all [imath]z\geq0[/imath], and it is monotonically decreasing from 1 to 0 as its argument goes from 0 to [imath]\infty[/imath]. Let [imath]f(t)[/imath] and [imath]h(t)[/imath] are two probability distributions. Suppose they have the same transform, i.e., [imath] (Tf)(u) = (Th)(u)[/imath] for all [imath]u\geq 0[/imath]? Does this imply uniqueness, i.e. [imath]f(t)=g(t)[/imath] for all [imath]t\geq0[/imath]? |
412169 | Proving that the [imath]n[/imath]th derivative satisfies [imath](x^n\ln x)^{(n)} = n!(\ln x+1+\frac12+\cdots+\frac1n)[/imath]
Question: Prove that [imath](x^n\ln x)^{(n)} = n!(\ln x+1+\frac 12 + ... + \frac 1n)[/imath] What I tried: Using Leibnitz's theorem, with [imath]f=x^n[/imath] and [imath]g=\ln x[/imath]. So [imath]f^{(j)}=n\cdots(n-j+1)x^{n-j} , g^{(j)}=(-1)^{j+1} \dfrac 1{x^{n-j}}[/imath] But somehow I get stuck on the way... | 94018 | Finding the [imath]n[/imath]-th derivatives of [imath]x^n \ln x[/imath] and [imath]\frac{\ln x}{x}[/imath].
How can I prove the following identities: [imath] \left( x^n \ln x \right)^{(n)}= n! \left(\ln x + 1 + \frac{1}{2} +\cdots +\frac{1}{n} \right), \quad x>0, \quad n\ge 1, \tag{a}[/imath] [imath] \left( \frac{\ln x}{x} \right)^{(n)}= (-1)^n \ n! \ x^{-n-1} \left( \ln x - 1 - \frac{1}{2} - \cdots -\frac{1}{n} \right), \quad x>0,\quad n\ge 1, \tag{b}[/imath] Here [imath](\cdot )^{(n)}[/imath] indicates the [imath]n[/imath]-th derivative with respect to [imath]x[/imath]. I don't know where to start! |
85153 | How do I choose an element from a non-empty set?
Suppose I have a non-empty set [imath]A[/imath]. How do I choose an element [imath]x\in A[/imath]? More precisely, I believe I would like to find a formula [imath]P(x,y)[/imath] of ZF such that for every non-empty set [imath]y[/imath] there is exactly one [imath]x\in y[/imath] such that [imath]P(x,y)[/imath] is true. I have thought about this for a while without success and I am beginning to doubt such a formula is even possible. But in mathematical proofs it is quite common to "choose a fixed element" from a non-empty set. So how exactly do we achieve such a thing? Am I missing something? Thank you in advance. | 2011721 | Does this choice require Axiom of Choice?
This theorem is from Munkres Topology 2nd ed.: In the last para, it is said that 'choose an element' [imath]n[/imath] of [imath]D[/imath]. Why we do not need axiom of choice here? (Here [imath]D[/imath] may not be finite.) |
412406 | Need to show that [imath]\lim_{x\to\infty}\left(\sum_{n\le x}^{}\frac{1}{n}-\ln x \right)[/imath] exist and is less than [imath]1[/imath]
Need some help here. I need prove that the following limit exist and is less than [imath]1[/imath] [imath]\lim_{x\to\infty}\left(\sum_{n\le x}^{}\frac{1}{n}-\ln x\right)[/imath] I feel a little lost here, this is my first exercise of this type, and I don't know what to use. Any help is welcome! | 344314 | Showing that [imath]\lim_{n\to\infty}\sum^n_{k=1}\frac{1}{k}-\ln(n)=0.5772\ldots[/imath]
How to show that [imath]\lim_{n\to\infty}\left[\sum^n_{k=1}\frac{1}{k}-\ln(n)\right]=0.5772\ldots[/imath] No clue at all. Need help! Appreciated! |
413397 | show AB has n real eigenvalues
Let [imath]n[/imath] be a postive integer. Assume that [imath]A,B\in{\mathcal{R}^{n\times n}}[/imath] are two symmetric real matrices, and one of them is symmetric postive definite. Show that the matrix AB has [imath]n[/imath] real eigenvalues. This is a problem from qualified exam. | 134884 | Eigenvalues of product of two hermitian matrices
I know that if [imath]A[/imath] and [imath]B[/imath] are hermitian matrices, then it doesn't follow that the eigenvalues of [imath]AB[/imath] are real, because of the following counter-example: [imath]A=\begin{bmatrix} 0 &1 \\ 1& 0 \end{bmatrix}[/imath] and [imath]B=\begin{bmatrix} 1 & 0\\ 0& -1 \end{bmatrix}[/imath] On the other hand, I came across the following problem, which says that if [imath]A[/imath] is hermitian and positive definite, and [imath]B[/imath] is hermitian, then [imath]AB[/imath] has real eigenvalues. Why if we add the property "positive definite" to [imath]A[/imath], the eigenvalues of [imath]AB[/imath] become real? The proof I read in the book says: Let [imath]\lambda [/imath] be an eigenvalue of the hermitian matrix [imath]AB[/imath] with non zero eigenvector [imath]x[/imath]. Then: [imath]\left \langle BABx,x \right \rangle=\left \langle ABx,Bx \right \rangle=\left \langle \lambda x,Bx \right \rangle=\lambda \left \langle x,Bx \right \rangle[/imath] Since [imath] \left \langle BABx,x \right \rangle[/imath] and [imath]\left \langle x,Bx \right \rangle,[/imath] then [imath]\lambda[/imath] is real. However, I can't see where in the proof the fact that [imath]A[/imath] is positive definite is used. Can anyone explain, please? |
413443 | Angle preserving linear maps
In Spivak's Calculus On Manifolds, in part (c) of question 1-8, he asks the following question: What are all angle preserving [imath]T:\mathbf{R}^n \to \mathbf{R}^n[/imath]? I already showed that if [imath]T[/imath] is diagonalizable with a basis [imath]\{x_1,\ldots,x_n\}[/imath] where [imath]Tx_i = \lambda_i[/imath], then [imath]T[/imath] is angle preserving [imath]\iff[/imath] [imath]|\lambda_i| = |\lambda_j|[/imath] for all [imath]i,j[/imath] (this was part (b)). Perhaps this can be used? Thanks. | 177005 | Question about Angle-Preserving Operators
This an exercise out of Spivak's "Calculus on Manifolds". Edit: There was a typo in the exercise as is noted below in the answers. The statement has been edited to reflect this. Given [imath]x,y\in\mathbb{R}^{n}[/imath], the angle between [imath]x[/imath] and [imath]y[/imath] is defined by [imath]\angle(x,y) = \arccos\left(\frac{\langle x,y \rangle}{|x|\cdot |y|}\right),[/imath] where [imath]\langle x,y \rangle[/imath] denotes the standard Euclidean inner product. A linear operator [imath]T:\mathbb{R}^{n}\to\mathbb{R}^{n}[/imath] is said to be angle-preserving if [imath]\angle(T(x),T(y)) = \angle(x,y)[/imath] for every [imath]x,y\in\mathbb{R}^{n}[/imath]. The exercise as stated: Let [imath]\{x_{1},\dots, x_{n}\}[/imath] be a basis for [imath]\mathbb{R}^{n}[/imath]. Then suppose that [imath]\lambda_{1}, \dots, \lambda_{n}\in \mathbb{R}[/imath] are such that [imath]Tx_{j} = \lambda_{j}x_{j}[/imath] for each [imath]j = 1,\dots, n[/imath]. Then [imath]T[/imath] is angle-preserving only if (not if and only if!)[imath]|\lambda_{i}| = |\lambda_{j}|[/imath] for every [imath]1\leq i\leq j\leq n[/imath]. I'm having problems with the [imath](\Rightarrow)[/imath] direction. My best attempt (which seems to lead nowhere) is to suppose that [imath]|\lambda_{j}|\neq |\lambda_{k}|[/imath]. Then by assumption, \begin{align*} \angle(Tx_{j},Tx_{k}) & = \arccos\left(\frac{\langle Tx_{j},Tx_{k} \rangle}{|Tx_{j}|\cdot |Tx_{k}|}\right)\\ & = \arccos\left(\frac{\langle \lambda_{j}{x_{j}},\lambda_{k}{x_{k}} \rangle}{|\lambda_{j}{x_{j}}|\cdot |\lambda_{k}{x_{k}}|}\right)\\ & = \arccos\left(\frac{\lambda_{j}\lambda_{k}\langle {x_{j}},{x_{k}} \rangle}{|\lambda_{j}|\cdot|\lambda_{k}|\cdot|{x_{j}}|\cdot |{x_{k}}|}\right)\\ & = \arccos\left(\text{sign}(\lambda_{j})\text{sign}(\lambda_{k})\frac{\langle {x_{j}},{x_{k}} \rangle }{|{x_{j}}|\cdot |{x_{k}}|}\right)\\ \end{align*} may also be calculated as \begin{align*} \angle(Tx_{j},Tx_{k}) & = \angle(x_{j},x_{k})\\ & = \arccos\left(\frac{\langle x_{j},x_{k} \rangle}{|x_{j}|\cdot |x_{k}|}\right). \end{align*} Then since [imath]\arccos[/imath] is injective, I believe I can make the jump that [imath]\text{sign}(\lambda_{j})\text{sign}(\lambda_{k}) = 1[/imath], which does not resemble the conclusion that I should arrive at. Note: I wasn't sure what tag to put this under, so anyone who knows better please feel free to adjust. Thanks for any help you can give. |
413773 | Representing [imath]A \rightarrow B[/imath] as [imath]A \supseteq B[/imath]
I know that many people like to think of elementary logic in terms of Venn diagrams, i.e., elementary set theory. I have never found this representation useful, because I can never remember whether implication is supposed to be represented by the relation "contains" or to the relation "is contained in". IOW, do we represent [imath]A \rightarrow B[/imath] as [imath]A \supseteq B[/imath] or as [imath]A\subseteq B\;[/imath]? For all I know, both representations could result in useful interpretations, depending on the situation. Whenever I come across an exposition that resorts to such a representation of logic through Venn diagrams, for some reason that is (to me at least) very obscure, my initial gut-reaction is that [imath]A \rightarrow B[/imath] corresponds to [imath]A \supseteq B[/imath]. This is annoying, because I eventually come to realize that my instinct is wrong: the intended representation is the one where [imath]A \rightarrow B[/imath] corresponds to [imath]A \subseteq B[/imath]. (I want to stress that I have no problem at all understanding the correspondence between [imath]A \rightarrow B[/imath] and [imath]A \subseteq B[/imath]. My problem is only that this correspondence is not at all intuitive: I always need to think it through, or "compute" it, so to speak, and this makes this representation more of a hindrance than an aid to my thinking.) I rack my brains trying to figure out why my instinct here is so backwards (and apparently incurably so). The only possible explanation I can come up with (and I'm definitely "grasping at straws" here) is that maybe there is some situation in which the representation "[imath]A \rightarrow B[/imath] is [imath]A\supseteq B\,[/imath]" is actually useful and used, and maybe I learned it first somehow? My question is: does anyone know of a reasonably common application of representing the implication [imath]A \rightarrow B[/imath] as [imath]A \supseteq B\;[/imath]? Conversely, does anyone know of a good reason for why this representation would rarely, if ever, be useful? | 391217 | Using [imath]p\supset q[/imath] instead of [imath]p\implies q[/imath]
I saw that a use for the notation [imath]p\supset q[/imath] instead of [imath]p\implies q[/imath] that got me a bit confused. One occurrences is in this Wikipedia link. It seems to me opposite than what it should be, let me explain what I mean: If [imath]A,B[/imath] are sets s.t [imath]A\subset B[/imath], [imath]p[/imath] is for [imath]x\in A[/imath], and [imath]q[/imath] is for [imath]x\in B[/imath] then we can identify (in some way) [imath]A\text{ with }p[/imath] [imath]B\text{ with }q[/imath] We have it that [imath]p\implies q[/imath], since [imath]A\subset B[/imath], but in the above notation we have [imath]q\subset p[/imath] which to me looks like [imath]B\subset A[/imath] which is the opposite of what we wanted to express. Can someone please explain to me the logic behind this notation ? |
392654 | A sequence of embedded closed balls that have empty intersection
I'm reading soviet textbook "Elements of theory of functions and functional analysis" by Kolmogorov and Fomin. There is an exercise is in it: show example of complete metric space and a sequence of embedded closed balls, that have empty intersection". I understand, that sequence of their diameters must not converge, but I could not find the example. I was thinking about space of infinite sequences [imath](x_1,x_2, x_3, \dots): |x_i|<=1[/imath]with distance between x and y [imath]\rho(x,y)=max(|x_i-y_i|)_i[/imath], and sequence of balls with radius 1 and centers [imath]x_0=(0,0,\dots), x_1=(1,0,0,\dots), x_2=(1,1,0,\dots),etc[/imath]. But is it correct? | 388930 | Can the intersection of open or closed balls be empty, if their radii are bounded from below?
I am wondering about the following question: Given a (countable) sequence of nested open balls: [imath] B_1 \supseteq B_2 \supseteq \cdots [/imath] Not necessarily having the same same center. All having radius bounded from below, say by [imath]r > 0[/imath]. Then can we say that [imath]\bigcap_{i=1}^{\infty} B_i \neq \varnothing[/imath] it is certainly true in the reals, as one can simply go to the point where the radii are close to [imath]r[/imath] then take the center of that ball. However, I'm having trouble seeing whether or not it is true in the general case. So thanks in advance for proof or counterexample. |
396967 | Prime numbers problem - discrete math
Show that natural numbers of the form [imath]n^2+1[/imath] are not divisible by primes of the form [imath]p=4k-1[/imath]. I can't really find a place to start. Thank you very much in advance, Yaron. | 122048 | [imath]-1[/imath] is a quadratic residue modulo [imath]p[/imath] if and only if [imath]p\equiv 1\pmod{4}[/imath]
I came across this problem and I believe Lagrange's theorem is the key to its solution. The question is: Let [imath]p[/imath] be an odd prime. Prove that there is some integer [imath]x[/imath] such that [imath]x^2 \equiv −1 \pmod p[/imath] if and only if [imath]p \equiv 1 \pmod 4[/imath]. I appreciate any help. Thanks. |
414345 | a function [imath]f[/imath] being not monotone---statement
As we know, a function [imath]f[/imath] is monotone if for each pair [imath]x<y[/imath], we have [imath]f(x)\leq f(y)[/imath]; or for each pair [imath]x<y[/imath], we have [imath]f(x)\geq f(y)[/imath]. Then [imath]f[/imath] is not monotone iff we can find a pair [imath]x<y,s<t[/imath] such that [imath]f(x)>f(y)[/imath], [imath]f(s)<f(t)[/imath]. But in some circumstances, I see that "if [imath]f[/imath] is not monotone, then we can find [imath]x<y<z[/imath] such that [imath]f(x)<f(y)>f(z)[/imath], or [imath]f(x)>f(y)<f(z).[/imath]" How to prove this? Thanks. | 203061 | Existence of [imath]\vee[/imath] or [imath]\wedge[/imath] for non-monotonic functions
This question is inspired by a discussion in chat with wj32. We allow for equality in the definition of increasing and decreasing and call a function monotonic if it is increasing or decreasing. If [imath]f:\mathbb R\to \mathbb R[/imath] is not monotonic, are there three points [imath]x<y<z[/imath] such that [imath]f(y)<f(x),f(z)[/imath] or [imath]f(y)>f(x),f(z)[/imath]? For convenience, call these the [imath]\vee[/imath] and [imath]\wedge[/imath] formations respectively. |
265634 | When [imath]K[/imath] is compact, if [imath]S\subset C_b(K)[/imath] is closed,bounded and equicontinuous, then [imath]S[/imath] is compact? (ZF)
Let [imath]K[/imath] be a compact metric space and [imath]S\subset C(K,\mathbb{C})[/imath]. Let [imath]S[/imath] be closed,bounded and equicontinuous. The usual proof for this is, using Arzela-Ascoli Theorem and Axiom of countable choice, showing that [imath]S[/imath] is limit point compact. (The problem is, under ZF, additional hypothesis "[imath]K[/imath] is separable" should be added to make Arzela-Ascoli Theorem true) I think, in order to prove [imath]S[/imath] is compact without AC, one should start directly with an infinite subcover. Assuming above hypotheses, I have proved [imath]S[/imath] is totally bounded and complete, and it really does seem provable that [imath]S[/imath] is compact under ZF. Is it unprovable? Or if there is an argument proving this without choice please let me know help! Thank you in advance | 265802 | Totally bounded, sequentially compact, complete, bounded, closed, equicontinuous [imath]\Rightarrow[/imath] compact?
Related; When [imath]K[/imath] is compact, if [imath]S\subset C_b(K)[/imath] is closed,bounded and equicontinuous, then [imath]S[/imath] is compact? (ZF) I just edited my whole question since i think it was a bit messy. Here is my question. Let [imath]K[/imath] be a separable compact metric space and [imath]S\subset C(K,\mathbb{C})[/imath]. Let [imath]S[/imath] be closed,bounded,uniformly equicontinuous on [imath]K[/imath], sequentially compact, totally bounded and complete. Then is [imath]S[/imath] compact? (in ZF) Thank you in advance! |
414535 | Why [imath]\cap \emptyset \ne \emptyset[/imath]?
If [imath]X[/imath] is a collection of sets, define [imath]\bigcap X[/imath] = [imath]\bigcap_{U \in X} U[/imath]. According to this definition, [imath]\bigcap \emptyset \ne \emptyset[/imath] Too see this, note that if [imath]x \notin \bigcap \emptyset[/imath], there exists [imath]U \in \emptyset[/imath] such that [imath]x \notin U[/imath]. However, by definition, that's impossible. But what set would [imath]\bigcap \emptyset[/imath] be? Everything? | 348668 | intersection of the empty set and vacuous truth
Let [imath]\mathbb S = \varnothing[/imath]. Then from the definition: [imath] \bigcap \mathbb S = \left\{{x: \forall X \in \mathbb S: x \in X}\right\}[/imath] Consider any [imath]x \in \mathbb U[/imath]. Then as [imath]\mathbb S = \varnothing[/imath], it follows that: [imath]\forall X \in \mathbb S: x \in X[/imath] from the definition of vacuous truth. It follows directly that: [imath]\bigcap \mathbb S = \left\{{x: x \in \mathbb U}\right\}[/imath] That is: [imath]\bigcap \mathbb S = \mathbb U[/imath]. (http://www.proofwiki.org/wiki/Intersection_of_Empty_Set) Proofwiki uses the above "proof" to "prove" that intersection of the empty set is the whole universe. My question is, is the use of vacuous truth really allowed in axiomatic set theory, like ZFC? I don't see how the use of vacuous truth is justified. The next problem I can think of is that we cannot really "define" the elements of empty set (to my knowledge, there is no element in empty set) so how can we then prove as the above proof did? This seems to contradict the use of vacuous truth. And of course, there is issue of using the whole universe as a set, and I don't think this is allowed.... (Maybe proof above is using a different axiomatic set theory, as I am using ZF-minded thoughts...) |
370188 | Empty intersection and empty union
If [imath]A_\alpha[/imath] are subsets of a set [imath]S[/imath] then [imath]\bigcup_{\alpha \in I}A_\alpha[/imath] = "all [imath]x \in S[/imath] so that [imath]x[/imath] is in at least one [imath]A_\alpha[/imath]" [imath]\bigcap_{\alpha \in I} A_\alpha[/imath] = "all [imath]x \in S[/imath] so that [imath]x[/imath] is in all [imath]A_\alpha[/imath]" It is the convention that [imath]\bigcup_{\alpha \in \emptyset}A_\alpha = \emptyset[/imath] and [imath]\bigcap_{\alpha \in \emptyset} A_\alpha = S[/imath]. But if [imath]x[/imath] is in [imath]\bigcap_{\alpha \in \emptyset} A_\alpha = S[/imath] then [imath]x[/imath] is in all [imath]A_\alpha[/imath] with [imath]\alpha \in \emptyset[/imath] and therefore [imath]x[/imath] is certainly in at least one [imath]A_\alpha[/imath] with [imath]\alpha \in \emptyset[/imath]. But then [imath]x \in \bigcup_{\alpha \in I}A_\alpha[/imath]. Can someone help me and tell me what is wrong with this? Thank you. | 800261 | How to show that [imath]\bigcup_{i\in \emptyset} A_i=\emptyset[/imath] and [imath]\bigcap_{i\in \emptyset}A_i=X[/imath]?
It was probably asked multiple times before which is about basic logic. But I couldnt find any. let X a set and I is index set. for [imath]i\in I[/imath], [imath]A_i[/imath] is a subset of X. if [imath]I=\emptyset[/imath], how can we show [imath]\bigcup_{i\in \emptyset} A_i=\emptyset\quad\text{and}\quad\bigcap_{i\in \emptyset}A_i=X[/imath] is it because from the definition if [imath]x\in\bigcap_{i\in \emptyset}A_i\iff\forall i(x\in \emptyset\Rightarrow x\in A_i) [/imath] which equals to [imath]\forall i(x\notin \emptyset \lor x\in A_i)[/imath] which is true for all [imath]x\in X[/imath] because [imath]x\notin \emptyset[/imath] is always true. similarly if [imath]x\in\bigcup_{i\in \emptyset} A_i\iff\exists i(i\in \emptyset \land x\in A_i) [/imath] right statement is wrong because [imath]i\in \emptyset[/imath] is wrong always. that is, for all [imath]x \in X, x\notin\bigcup_{i\in \emptyset} A_i[/imath] so [imath]\bigcup_{i\in \emptyset} A_i=\emptyset[/imath] is this sufficient? |
399657 | What is the largest number such that the number formed by the first [imath]n[/imath] digits is divisible by [imath]n[/imath]?
What is the largest number such that the number formed by the first [imath]n[/imath] digits is divisible by [imath]n[/imath]? For example, if we have a number [imath]abcdefghijklm,[/imath] and all of these leters stand for digits, then [imath]a[/imath] is divisible by [imath]1[/imath], [imath]ab[/imath] is divisible by [imath]2[/imath], [imath]abc[/imath] is divisible by [imath]3[/imath], and so on. Also, what is allowed is (besides [imath]a[/imath]) the digits can be [imath]0[/imath] and digits can repeat. | 411897 | Find the largest number having this property.
The [imath]13[/imath]-digit number [imath]1200549600848[/imath] has the property that for any [imath]1 \le n \le 13[/imath], the number formed by the first [imath]n[/imath] digits of [imath]1200549600848[/imath] is divisible by [imath]n[/imath] (e.g. 1|2, 2|12, 3|120, 4|1200, 5|12005, ..., 13|1200549600848 using divisor notation). Question 1: Find the largest computed number having this property. Question 2: Is there a theoretical upper bound on the largest possible number with this property? Edit: Added Question 2 as I believe it is more insightful as compared to brute force computer calculations. |
103124 | An example of a non-abelian group [imath]G[/imath] where, for all [imath]a,b\in G[/imath], the equality [imath](ab)^{n}=a^{n}b^{n}[/imath] holds for two consecutive integers [imath]n[/imath]
I am looking for an example of a group [imath]G[/imath] where the equality [imath](ab)^{n}=a^{n}b^{n}[/imath] holds for two consecutive integers [imath]n[/imath], but [imath]G[/imath] is not an abelian group. I've started do some calculations in the group [imath]D_{4}[/imath] (I gave up!) Do you know where I can find that example? Is it possible find such example without doing a lot of calculations? Thanks for your help! | 2387803 | If [imath](ab)^k = a^k b^k[/imath] for two consecutives integers [imath]k[/imath], is [imath]G[/imath] abelian?
Suppose [imath]G[/imath] is a group such that there exists [imath]k\in\mathbb{Z}[/imath] satisfying [imath](ab)^k = a^k b^k[/imath] and [imath](ab)^{k+1} = a^{k+1} b^{k+1}[/imath]. Is [imath]G[/imath] abelian? I've tried to use the same argument in this post, but it doesn't work. Is there a non-abelian group [imath]G[/imath] with this property? |
414257 | Special case of the hodge decomposition theorem
I am trying to prove the following special case of the hodge decomposition theorem in differential geometry for a n component vector field [imath]V_i[/imath] in [imath]\mathbb{R}^n[/imath]. any vector can be written as the following combination [imath]V_i = −\partial_i \phi + \epsilon^{ii_2 i_3···i_n}\partial_{i_2}F_{i_3i_4···i_n}[/imath] where [imath]F[/imath] is a rank [imath]n-2[/imath] anti-symmetric tensor. I have read the proof of the Helmholtz theorem, but I don't know how I can generalize it. Do I have to assume some Ansatz? What would that be? It would be great if somebody could show a stepwise method showing tensor calculations, and stating the necessary theorems? | 410111 | Special case of the Hodge decomposition theorem
I am trying to prove the following special case of the Hodge decomposition theorem in differential geometry for an [imath]n[/imath] component vector field [imath]V_i[/imath] in [imath]\mathbb{R}^n[/imath]. I have very little knowledge of differential geometry. any vector can be written as the following combination [imath]V_i = −\partial_i \phi + \epsilon^{ii_2 i_3···i_n}\partial_{i_2}F_{i_3i_4···i_n}[/imath] where [imath]F[/imath] is a rank [imath]n-2[/imath] anti-symmetric tensor. I have read the proof of the Helmholtz theorem, but I don't know how I can generalize it. More specifically the Helmholtz theorem uses the identity [imath]\bar{\nabla} \times (\bar{\nabla} \times \bar{V})=\bar{\nabla}(\bar{\nabla} \cdot \bar{V} )-\nabla^2 \bar{V}[/imath] What is the generalization of this identity to [imath]n[/imath] dimensions? Do I have to assume some Ansatz? What would that be? It would be great if somebody could show a stepwise method showing tensor calculations, and stating the necessary theorems? Lastly, how do I prove the uniqueness that given the curl and divergence ([imath]\partial_i V_i=s[/imath] and [imath]\partial_iV_j-\partial_jV_i=c_{ij}[/imath]), and the normal component at the boundary, the vector is uniquely specified. |
415611 | Proving when Gram's determinant is equal to zero
Prove that Gram's determinant [imath]G(x_1,\dots, x_n)=0[/imath] if and only if [imath]x_1, \dots, x_k[/imath] are linearly dependent. So I know that [imath]G(x_1,\dots, x_n)=\det \begin{vmatrix} \xi( x_1,x_1) & \xi( x_1,x_2) &\dots & \xi( x_1,x_n)\\ \xi( x_2,x_1) & \xi( x_2,x_2) &\dots & \xi( x_2,x_n)\\ \vdots&\vdots&\ddots&\vdots\\ \xi( x_n,x_1) & \xi( x_n,x_2) &\dots & \xi( x_n,x_n)\end{vmatrix}[/imath] (where [imath]\xi[/imath] denotes an inner product) which means that if [imath]G(x_1,\dots, x_n)=0[/imath], then the determinant has to be equal to 0 as well. Why does it happen only with [imath]x_1,\dots, x_k[/imath] being linearly dependent, though? | 408307 | Proof: [imath]\det\pmatrix{\langle v_i , v_j \rangle}\neq0[/imath] [imath]\iff \{v_1,\dots,v_n\}~\text{l.i.}[/imath]
Let [imath]V[/imath] be a real inner product space and [imath]S=\{v_1,v_2, \dots, v_n\}\subset V[/imath]. How am I to prove that [imath]S[/imath] is linearly independent if and only if the determinant of the matrix [imath] a_{ij}=\pmatrix{\langle v_i , v_j \rangle}[/imath] is nonzero? Just to be clear, the matrix we're talking about is this one: [imath]\pmatrix{\langle v_1,v_1\rangle & \langle v_1,v_2\rangle &\langle v_1,v_3\rangle & \cdots & \langle v_1,v_{n-1}\rangle & \langle v_1, v_n\rangle \\\langle v_2,v_1\rangle & \langle v_2,v_2\rangle &\langle v_2,v_3\rangle & \cdots & \langle v_2,v_{n-1} \rangle & \langle v_2,v_n\rangle \\\langle v_3,v_1\rangle & \langle v_3,v_2\rangle &\langle v_3,v_3\rangle & \cdots & \langle v_3,v_{n-1}\rangle & \langle v_3,v_n \rangle \\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\\langle v_{n-1}, v_1\rangle & \langle v_{n-1},v_2\rangle &\langle v_{n-1},v_3\rangle & \cdots & \langle v_{n-1},v_{n-1}\rangle & \langle v_{n-1},v_n\rangle \\\langle v_n,v_1\rangle & \langle v_n,v_2\rangle &\langle v_n,v_3\rangle & \cdots & \langle v_n,v_{n-1}\rangle & \langle v_n,v_n\rangle \\ }[/imath] I highly doubt that anybody here has Roman's Advanced Linear Algebra, or maybe you do, but I think on page [imath]261[/imath] there is a small note on something which looks similar. Should anybody need the code (C++) in their research, here is a gadget which streams LaTeX code to a file named "matrix.txt" for an [imath]n \times n[/imath] matrix such as this with some value for [imath]n[/imath]: ofstream fout; fout.open("matrix.txt"); int n; cout << "Enter your desired n: "; cin >> n; fout << endl; fout << "[imath]\\begin{pmatrix}" << endl; for( int j = 1 ; j <= n ; j++ ) { for( int i = 1 ; i<= n ; i++ ) { if( j == i && i == n ) { fout << "\\langle v_" << j << "," << "v_" << i << " \\rangle" << endl; } else { if( i == n ) { fout << "\\langle v_" << j << "," << "v_" << i << " \\rangle\\\\" << endl; } else { fout << "\\langle v_" << j << "," << "v_" << i << " \\rangle&" << endl; } } } //fout << endl; } fout << "\\end{pmatrix}[/imath]" << endl << endl; For n equal to 5 you get this result: [imath]\begin{pmatrix} \langle v_1,v_1 \rangle& \langle v_1,v_2 \rangle& \langle v_1,v_3 \rangle& \langle v_1,v_4 \rangle& \langle v_1,v_5 \rangle\\ \langle v_2,v_1 \rangle& \langle v_2,v_2 \rangle& \langle v_2,v_3 \rangle& \langle v_2,v_4 \rangle& \langle v_2,v_5 \rangle\\ \langle v_3,v_1 \rangle& \langle v_3,v_2 \rangle& \langle v_3,v_3 \rangle& \langle v_3,v_4 \rangle& \langle v_3,v_5 \rangle\\ \langle v_4,v_1 \rangle& \langle v_4,v_2 \rangle& \langle v_4,v_3 \rangle& \langle v_4,v_4 \rangle& \langle v_4,v_5 \rangle\\ \langle v_5,v_1 \rangle& \langle v_5,v_2 \rangle& \langle v_5,v_3 \rangle& \langle v_5,v_4 \rangle& \langle v_5,v_5 \rangle \end{pmatrix}[/imath] As is lengthily explained here. |
414052 | Existence of arbitrarily large ordinal subgroups in a group structure on a regular cardinal
Suppose [imath]\kappa[/imath] is an uncountable regular cardinal, and [imath](\kappa, \cdot, ^{-1}, e[/imath]) is a group. Prove that that [imath]C = \{\alpha < \kappa: \alpha\, \textrm{is a subgroup of}\, \kappa)[/imath] is unbounded in [imath]\kappa[/imath]. It's easy to see that [imath]C[/imath] is actually closed: if [imath]\alpha_\beta[/imath] are subgroups of [imath]\kappa[/imath] for [imath]\beta < \theta[/imath], then [imath]\bigcup_{\beta < \theta} \alpha_\beta[/imath] is also a subgroup, as a union of increasing sequence of groups is again a group. I fail to see how to exhibit arbitrarily large ordinal subgroups in [imath]\kappa[/imath], though. My problem is that if we take some [imath]\alpha_0 < \kappa[/imath] and close [imath]\alpha[/imath] under group operations, we'll get some subgroup [imath]U_0[/imath] of [imath]\kappa[/imath] with cardinality [imath]< \kappa[/imath]. The thing is that [imath]U_0[/imath] is not necessarily ordinal, so we consider [imath]\alpha_1 = \sup U_0[/imath], close [imath]\alpha_1[/imath] under group operations to obtain some subgroup [imath]U_1[/imath] with [imath]|U_1| < \kappa[/imath], and then transfinitely continue this process until some [imath]U_\beta[/imath] will turn out to be ordinal. The problem is that we may get [imath]\beta \geq \kappa[/imath], and so [imath]U_\beta = \kappa[/imath] which is not a proper subgroup. | 229183 | Existence of elementary substructures of a uncountable structure over a countable language
Let [imath]\kappa[/imath] be an regular uncountable cardinal carrying a [imath]\tau[/imath]-structure for some countable language [imath]\tau[/imath]. What can be said regarding the existence of ordinals [imath]\alpha <\kappa[/imath] carrying elementary substructures of [imath]\kappa[/imath] ? This is an (altered) question from an exercise which I am having difficulties solving. The Löwenheim–Skolem theorem doesn't seem to provide any direct insights here. |
416034 | A principal ideal domain that is not a Euclidean domain.
Somebody can to help me in the following problem: Let R be the following subring of the complex numbers: [imath]R = \left\{\frac{z_1}{2}+\frac{z_2\sqrt{-19}}{2} : z_1,z_2\in\mathbb{Z}, \;\text{with the same parity}\right\}.[/imath] Then R is a principal ideal domain that is not a Euclidean domain. Note: A ring for me is commutative and with identity. | 23844 | A ring that is not a Euclidean domain
Let [imath]\alpha = \frac{1+\sqrt{-19}}{2}[/imath]. Let [imath]A = \mathbb Z[\alpha][/imath]. Let's assume that we know that its invertibles are [imath]\{1,-1\}[/imath]. During an exercise we proved that: Lemma: If [imath](D,g)[/imath] is a Euclidean domain such that its invertibles are [imath]\{1,-1\}[/imath], and [imath]x[/imath] is an element of minimal degree among the elements that are not invertible, then [imath]D/(x)[/imath] is isomorphic to [imath]\mathbb Z/2\mathbb Z[/imath] or [imath]\mathbb Z/3\mathbb Z[/imath]. Now the exercise asks: Prove that [imath]A[/imath] is not a Euclidean Domain. Everything hints to an argument by contradiction: let [imath](A, d)[/imath] be a ED and [imath]x[/imath] an element of minimal degree among the non invertibles we'd like to show that [imath]A/(x)[/imath] is not isomorphic to [imath]\mathbb Z/2\mathbb Z[/imath] or [imath]\mathbb Z/3\mathbb Z[/imath]. How do we do that? My problem is that, since I don't know what this degree function looks like, I don't know how to choose this [imath]x[/imath]! I know that the elements of [imath]A/(x)[/imath] are of the form [imath]a+(x)[/imath], with [imath]a[/imath] of degree less than [imath]x[/imath] or zero. By minimality of [imath]x[/imath] this means that [imath]a\in \{0, 1, -1\}[/imath]. Now I'm lost: how do we derive a contradiction from this? |
221816 | Showing that projections [imath]\mathbb{R}^2 \to \mathbb{R}[/imath] are not closed
Consider [imath]\mathbb{R}^2[/imath] as [imath]\mathbb{R} \times \mathbb{R}[/imath] with the product topology. I am simply trying to show that the two projections [imath]p_1[/imath] and [imath]p_2[/imath] onto the first and second factor space respectively are not closed mappings. It seems like this should be easy, but I have not been able to come up with a closed set in [imath]\mathbb{R}^2[/imath] whose projection onto one of the axes is not closed. I don't really have any work to show...I've really just tried the obvious things like closed rectangles and unions of such, the complement of an open rectangle or union of open rectangles, horizontal and vertical lines, unions of singletons, etc., and haven't come up with anything non-obvious, which I hope is where the answer lies. It's bothering me that I can't come up with an answer, and I'd appreciate some help. Thanks. | 436164 | Projection of closed set
Set [imath]A \subset R^2[/imath], set B is projection of A on x-axis. Do you know a counterexample to the statement: if A is closed, then B is closed. |
411717 | Group of inner automorphisms of a group [imath]G[/imath]
Let [imath]G[/imath] be a group. By an automorphism of [imath]G[/imath] we mean an isomorphism [imath]f: G\to G[/imath] By an inner automorphism of [imath]G[/imath] we mean any function [imath]\Phi_a[/imath] of the following form: For every [imath]x\in G[/imath], [imath]\Phi_a(x)=a x a^{-1}[/imath]. Prove that every inner automorphism of [imath]G[/imath] is an automorphism of [imath]G[/imath] which means I should prove [imath]\Phi_a[/imath] is isomorphism? any suggestion? thanks | 364864 | Inner Automorphisms with Groups
Let [imath]G[/imath] be a group, and let [imath]g \in G[/imath] . Prove that the function [imath]\gamma_g: G \to G[/imath] defined by [imath](\forall a \epsilon g): \gamma_g(a)=g a^{-1} g [/imath] is an automorphism of G. The automorphisms [imath]\gamma_g[/imath] are called 'inner' automorphisms. |
416123 | Why do we write second derivatives like [imath]\frac{d^2x}{dt^2}[/imath]
Why do we write the second derivative of [imath]x[/imath] with respect to [imath]t[/imath] as [imath]\frac{d^2x}{dt^2}[/imath]? It's never been explained to me, and I've never found a particularly good explanation. What's up with this weird derivative notation? | 407747 | What is the use of, and intuition behind, writing [imath]\frac{d^2}{dx^2}[/imath] for the second derivative?
Is it possible to take a second derivative without taking the first derivative before? Why do we multiply the [imath]d[/imath] and [imath]dx[/imath] operators? Like, does [imath]\dfrac{d^2}{dx^2}[/imath] really mean [imath]\dfrac{d}{dx} \cdot \dfrac{d}{dx}[/imath]? What's the intuitive understanding about this? Can it be represented in a graph? Like... 'Little change squared in [imath]y[/imath] over little change squared in [imath]x[/imath]'? |
289971 | Given two subspaces [imath]U,W[/imath] of vector space [imath]V[/imath], how to show that [imath]\dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W)[/imath]
Let [imath]U,W[/imath] be subspaces of a vector space [imath]V[/imath]. Show that [imath]\dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W)[/imath] Hint: Show that the map given by [imath]L:U×W\to V[/imath] given by [imath]L(u,w)=u-w[/imath] is linear. I can show that [imath]L:U×W\to V[/imath] given by [imath]L(u,w)=u-w[/imath] is a linear map. I also know that the dimension of [imath]U×W[/imath] is [imath]\dim(U)+\dim(W)[/imath]. What do I do next? Any hints? | 545152 | Show that [imath]\dim(U + V) = \dim(U) + \dim(V) - \dim(U \cap V)[/imath]
Let [imath]W[/imath] be a vector space and let [imath]U[/imath] and [imath]V[/imath] be finite dimensional subspaces. Not sure how to go about solving this. |
416312 | Get several X from a matrix-equation
I need to find [imath]\det X[/imath] where [imath]8GX=XX^T,\quad G=\left(\begin{matrix}5 & 4\\3 & 2\\\end{matrix}\right).[/imath] My answer is that the determinant of [imath]X[/imath] is [imath]-128[/imath] and that is correct but there is one more value of [imath]\det X[/imath] that can solve the equation and I belive that it is 0 but how do I write it a matrix that is equal to 0? | 377162 | Find [imath]\det X[/imath] if [imath]8GX=XX^T[/imath]
I need to find [imath]\det X[/imath] where [imath]8GX=XX^T,\quad G=\left(\begin{matrix}5 & 4\\3 & 2\\\end{matrix}\right).[/imath] My answer is that the determinant of [imath]X[/imath] is [imath]-128[/imath] and that is correct but there is one more value of [imath]\det X[/imath] that can solve the equation. |
416466 | How to show [imath]d(x,A)=0[/imath] iff [imath]x[/imath] is in the closure of [imath]A[/imath]?
This is a problem form Topology by Munkres: Let [imath]X[/imath] be a metric space with metric [imath]d[/imath] and [imath]A[/imath] is a nonempty subset of [imath]X[/imath]. Show that [imath]d(x,A)=0[/imath] if and only if [imath]x[/imath] is in the closure of [imath]A[/imath]. I think this problem is quite easy to understand emotionally but I don't know how to express the proof in standard math language. Thanks in advance! | 385127 | In a metric [imath](X,d)[/imath], prove that for each subset [imath]A[/imath], [imath]x\in\bar{A}[/imath] if and only if [imath]d(x,A)=0.[/imath]
In a metric space [imath](X,d)[/imath], prove that for each subset [imath]A[/imath], [imath]x\in\bar A[/imath] if and only if [imath]d(x,A)=0[/imath] I feel like this isn't necessarily true. For example, let [imath]X[/imath] equal the reals and [imath]A[/imath] be some open subset of [imath]X[/imath], let's say [imath](0,1)[/imath]. Then [imath]0.5[/imath] is an element of [imath]\bar{A}[/imath], but [imath]d(x,A) = d(0.5,0) \neq 0[/imath] This question seems simple enough; I don't know why it's giving me such trouble. |
417126 | A question on a symmetric derivative
Suppose [imath]f:J→\mathbb R[/imath] is a function on an open interval [imath]J[/imath] satisfying following inequality: [imath]∣f(x)+f(y)−2f(\frac{x+y}{2})∣≤C(f)|x−y|^{1+\delta}[/imath] for all [imath]x,y\in J,[/imath] and for some [imath]\delta \in(0,1][/imath]. Does it follow from this fact that [imath]f∈C^1[/imath] or [imath]C^{1+δ}[/imath]? | 416067 | A question on second symmetric derivatives
Suppose function [imath]f:J→R[/imath] on an open interval [imath]J[/imath] satisfies following inequality: [imath]\left|f(x)+f(y)−2f\Big(\frac{x+y}{2}\Big)\right|\leq C(f)\, |x−y|^{1+\delta}[/imath] for all [imath]x,y∈J[/imath], and for some [imath]\delta \in (0,1].[/imath] Is follow from this fact that [imath]f\in C^{1}[/imath] or [imath]C^{1+\delta}[/imath]? |
417120 | Maclaurin series for [imath]f(x)=\frac {1}{(1+x+x^2)} [/imath]
Find the Maclaurin series of [imath]f(x)=\dfrac {1}{(1+x+x^2)} [/imath]and the radius of convergence of the series. [imath]f(x) = \dfrac {(1-x)}{(1-x^3)} = (1-x) \sum _{0} ^\infty x^{(3n)}[/imath] and so? How this can be the shape of [imath]\sum_0 ^\infty \dfrac {f^n}{n!}x^n[/imath] ? And I don't know how to get a radius. | 416736 | Find the Maclaurin series of f(x)
Find the Maclaurin series of [imath]f(x)=\dfrac{1}{1+x+x^2}[/imath] and the radius of convergence of the series. I can't solve this problem. |
391784 | How to show: if [imath]b \mid a[/imath] and [imath]c \mid a[/imath] and [imath]\mathrm{gcd}(b,c) = 1[/imath], then [imath]bc \mid a[/imath]?
A little stumped on this problem, any help would be greatly appreciated. Show that for all [imath]a,b,c \in \mathbb{Z}[/imath], if [imath]b \mid a[/imath] and [imath]c \mid a[/imath] and [imath]\mathrm{gcd}(b,c) = 1[/imath], then [imath]bc \mid a[/imath]. | 173471 | show that if [imath]a | c[/imath] and [imath]b | c[/imath], then [imath]ab | c[/imath] when [imath]a[/imath] is coprime to [imath]b[/imath].
Given two numbers [imath]a[/imath] and [imath]b[/imath], where [imath]a[/imath] is co-prime to [imath]b[/imath], Show that for any number [imath]c[/imath], if [imath]a|c[/imath] and [imath]b | c[/imath] then [imath]ab| c[/imath]. Is the reverse also true? In other words, if [imath]ab |c[/imath] and [imath]a[/imath] is co-prime to [imath]b[/imath], then do we have [imath]a | c[/imath] as well as [imath]b|c[/imath]? |
414541 | Prove that If [imath]f[/imath] is polynomial function of even degree [imath]n[/imath] with always [imath]f\geq0[/imath] then [imath]f+f'+f''+\cdots+f^{(n)}\geq 0[/imath].
I can't solve this problem: Suppose [imath]f[/imath] is polynomial function of even degree [imath]n[/imath] with always [imath]f\geq0[/imath]. Prove that [imath]f+f'+f''+\cdots+f^{(n)}\geq 0[/imath]. | 88260 | Sum of derivatives of a polynomial
Let [imath]p(x)[/imath] be a polynomial of degree [imath]n[/imath] satisfying [imath]p(x)\geq 0[/imath] for all [imath]x[/imath]. That is, for all [imath]x[/imath], [imath]p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \geq 0[/imath], [imath]a_n\neq 0[/imath]. Show that [imath]p(x)+p'(x)+p''(x)+\cdots+p^{(n)}(x)\geq 0[/imath] for all [imath]x[/imath] where [imath]p^{(i)}(x)[/imath] is the [imath]i^\text{th}[/imath] derivative. My interest: I know that, we can rewrite the LDE as follows [imath] p(x)+p'(x)+p''(x)+\cdots+p^{(n)}(x) = Lp(x) [/imath] where [imath]L := I + D + D^2 + \cdots + D^{(n)}[/imath]. Can we say anything about a linear operator of this kind so that it does not change the sign of the input it takes? I can try to solve the question by writing for all the derivatives and factoring them using [imath]p(x)[/imath], but I think there should be a clever way of showing this by the properties of [imath]L[/imath]. Can I figure out a solution by just looking at [imath]L[/imath] and the sign of [imath]p(x)[/imath] as in the question? Where should I look for that? |
417936 | Least Upper Bounds based problem
I am having difficulty with this problem from chapter 8 in Spivak's Calculus, any help is appreciated. (a) Suppose that [imath]y - x > 1[/imath]. Prove that there is an integer [imath]k[/imath] such that [imath]x < k < y[/imath]. Hint: Let [imath]l[/imath] be the largest integer satisfying [imath]l \le x[/imath], and consider [imath]l + 1[/imath]. (b) Suppose [imath]x < y[/imath]. Prove that there is a rational number [imath]r[/imath] such that [imath]x < r < y[/imath]. Hint: If [imath]1/n < y-x[/imath], then [imath]ny - nx > 1[/imath] (c) Suppose that [imath]r < s[/imath] are rational numbers. Prove that there is an irrational number between [imath]r[/imath] and [imath]s[/imath]. Hint: As a start, you know that there is an irrational number between [imath]0[/imath] and [imath]1[/imath]. (d) Suppose that [imath]x < y[/imath]. Prove that there is an irrational number between [imath]x[/imath] and [imath]y[/imath]. Hint: It is unnecessary to do any more work; this follows from (b) and (c). | 149350 | Density of a set. Exercise from Spivak.
I'm trying to do a series of exercises from Spivak's Calculus, in chapter 8, Least Upper Bounds. I'm trying to tackle these two exercises, [imath]5.[/imath] and [imath]^*.6[/imath] From [imath]5.[/imath] I have proven the first claim [imath](a)[/imath] Let [imath]x-y>1[/imath]. Prove there is an integer [imath]k[/imath] such that [imath]x<k<y[/imath]. P Let [imath]\ell[/imath] be the greatest integer such that [imath]\ell \leq x[/imath]. Then [imath]y-x >1[/imath] [imath]y-\ell >1[/imath] [imath]y>1+\ell [/imath] Thus the integer [imath]\ell +1[/imath] is between [imath]y[/imath] and [imath]x \text{ }\blacktriangle[/imath]. [imath](b)[/imath] Let [imath]x<y[/imath]. Show there is a rational [imath]r[/imath] such that [imath]x<r<y[/imath]. P If [imath]x<y[/imath] then there is [imath]\epsilon>0[/imath] such that [imath]x+\epsilon=y[/imath]. Thus [imath]\epsilon=y-x[/imath]. But from T3, we have that there is an [imath]n[/imath] such that [imath]1/n < \epsilon[/imath], thus [imath]\frac{1}{n} < y-x[/imath] [imath]1<ny-nx \text{ }\blacktriangle[/imath] and from the last theorem we have that there is a integer [imath]k[/imath] such that [imath]nx<k<ny[/imath] [imath]x< \frac{k}{n}<y \text{ }\blacktriangle[/imath] [imath](c)[/imath] Let [imath]r<s[/imath] be rational numbers. Prove there is an irrational number between [imath]r[/imath] and [imath]s[/imath]. Hint: it is known there is an irrational number between [imath]0[/imath] and [imath]1[/imath]. Ok, this is a proof based on your answers. P Since [imath]\sqrt 3 [/imath] is irrational and [imath]\sqrt{3}<3[/imath], then [imath]\ell = \sqrt{3}/3<1[/imath] and thus it is in [imath][0,1][/imath]. Now, [imath]r<s \Rightarrow 0<s-r[/imath]. Then [imath]0<\ell < 1[/imath] [imath]0<\ell(s-r) < s-r[/imath] [imath]r<r+\ell(s-r) < s[/imath] And since [imath]\ell[/imath] is irrational [imath]r+\ell(s-r)[/imath] is irrational. [imath]\blacktriangle [/imath] [imath](d)[/imath] Show that if [imath]x<y[/imath] then there is an irrational number between [imath]x[/imath] and [imath]y[/imath]: There is no need to work here, this is consequence of [imath](b)[/imath] and [imath](c)[/imath] This one was quite straightforward, but thanks anyways. P [imath]x<y[/imath] [imath](b)\Rightarrow x<r<y,r<y \Rightarrow r<q<y \Rightarrow x<r<q<y[/imath] Then by [imath](c)[/imath], there is an irrational [imath]\ell[/imath] such that [imath] x<r<\ell<q<y \text{ } \blacktriangle [/imath] This will let me conclude [imath]\mathbb Q[/imath] is dense on any [imath][a,b]\subset \Bbb R[/imath] [imath]\mathbb I[/imath] is dense on any [imath][a,b] \subset \Bbb R [/imath] and will let me move on into [imath]^*6.[/imath] which is [imath](a)[/imath] Show that [imath]f[/imath] is continuous and [imath]f(x)=0[/imath] for all [imath]x[/imath] in a dense set [imath]A[/imath], then [imath]f[/imath] is [imath]f(x)=0[/imath] for all [imath]x[/imath]. [imath](b)[/imath] Show that [imath]f[/imath] and [imath]g[/imath] are continuous and [imath]f(x)=g(x)[/imath] for all [imath]x[/imath] in a dense set [imath]A[/imath], then [imath]f(x)=g(x)[/imath] for all [imath]x[/imath]. [imath](c)[/imath] If we suppose [imath]f(x)\geq g(x)[/imath] for all [imath]x[/imath] in [imath]A[/imath], then [imath]f(x)\geq g(x)[/imath] for all [imath]x[/imath]. ¿Can [imath]\geq[/imath] be substituted with [imath]>[/imath] everywhere? I'm not asking for solutions for this last problems (which will be asked separately), but for [imath](c)[/imath] and [imath](d)[/imath] in [imath]5.[/imath] The chapter has several important proofs, which might or might not be relevant here, but I think it is important you know what tools we have at hand: THEOREM 7-1 If [imath]f[/imath] is continuous on [imath][a,b][/imath] and [imath]f(a)<0<f(b)[/imath], then there is [imath]x \in [a,b]:f(x)=0[/imath] THEOREM 1 If [imath]f[/imath] is continuous in [imath]a[/imath], then there exists a [imath]\delta>0[/imath] such that [imath]f[/imath] is bounded above in [imath](a-\delta,a+\delta)[/imath]. THEOREM 7-2 If [imath]f[/imath] is continuous on [imath][a,b][/imath] then [imath]f[/imath] is bounded on [imath][a,b][/imath]. THEOREM 7-3 If [imath]f[/imath] is continuous on [imath][a,b][/imath] then there is an [imath]y[/imath] in [imath][a,b][/imath] such that [imath]f(y)\geq f(x)[/imath] for all [imath]x[/imath] in [imath][a,b][/imath]. THEOREM 2 [imath]\Bbb N[/imath] is not bounded above. THEOREM 3 If [imath]\epsilon >0[/imath], there is an [imath]n \in \Bbb N[/imath] such that [imath]1/n < \epsilon[/imath]. |
417657 | Is there any "superlogarithm" or something to solve [imath]x^x[/imath]?
Is there any "superlogarithm" or something to solve an equation like this: [imath]x^x = 10?[/imath] | 786629 | how to solve equation [imath]x^x=5[/imath]
How can I calculate the equation [imath]x^x=5[/imath] Is it an exponential function? Thank you. |
418024 | Expressing Complex Number in terms of its conjugate
Given a complex number [imath]z[/imath] , is it possible to express its conjugate [imath]\bar z[/imath] in terms of [imath]z[/imath] using only operations of addition , subtraction , multiplication , division and exponentiation on [imath]z[/imath] as a whole . In other words , expressing [imath]z[/imath] as [imath]x + iy[/imath] and then manipulating [imath]x[/imath] and [imath]y[/imath] is not allowed . If it is not possible , can we prove that in some elementary way ? | 28594 | Complex conjugate of [imath]z[/imath] without knowing [imath]z=x+i y[/imath]
Is it possible to determine (and if so, how) the complex conjugate [imath]\bar{z}[/imath] of [imath]z[/imath], if you don't already know that [imath]z = x + i y[/imath]? I think you can use [imath]\log(z)[/imath] to get the angle, and therefore the ratio of [imath]y[/imath] and [imath]x[/imath]. But how do you get [imath]|z|[/imath], the radius? How then to get [imath]r[/imath] (so that [imath]x = {\rm Re}(z) = r \cos(\log(z))[/imath] and [imath]y = {\rm Im}(z) = r \sin(\log(z))[/imath])? (this is related to How to express in closed form?, namely you can compute Re and Im using the conjugate, but then how do you reduce the conjugate itself fully to elementary functions (if at all)) |
413473 | Showing that [imath]a^n - 1 \mid a^m - 1 \iff n \mid m[/imath]
Let [imath]a\ge 2[/imath] be an integer. Show that for positive integers [imath]m,n[/imath], we have [imath]a^n - 1[/imath] divides [imath]a^m - 1[/imath] if and only if [imath]n[/imath] divides [imath]m[/imath]. I am having trouble showing this. I've seen a similar problem on here but it only shows one direction for the proof. I would like to see both directions. This is what I have so far: Let [imath]m = nq + r[/imath], then [imath]a^m - 1 = a^{nq + r}- 1 = a^{nq}a^r - 1 = a^{nq}a^r - a^r + a^r - 1 = a^r(a^{nq} - 1) + a^r - 1[/imath]. Since [imath]0\le r\lt n[/imath], [imath]r = 0[/imath] then [imath]a^m - 1[/imath] = [imath]a^{nq} - 1[/imath] so [imath]a^m[/imath] = [imath]a^{nq}[/imath] . Thus [imath]m = nq[/imath] hence [imath]n\mid m[/imath]. | 704004 | How can I prove that [imath]a^m-1|a^n-1 \Leftrightarrow m|n[/imath]?
Suppose that [imath](a,m,n)\in \mathbb{N}^{3}[/imath], as [imath]a>1[/imath] How can I prove that [imath]a^m-1|a^n-1 \Leftrightarrow m|n[/imath]? I proved that [imath]m|n \implies a^m-1|a^n-1[/imath], but I couldn't prove the reciproque. |
418345 | If [imath]f'(x)[/imath] tends to [imath]\infty[/imath] then [imath]f(x)[/imath] does the same
I noticed that when [imath]f'(x)[/imath] tends to [imath]+\infty[/imath] as [imath]x[/imath] tends to [imath]+\infty[/imath], then [imath]f(x)[/imath] must tend to [imath]+\infty[/imath] as [imath]x[/imath] tends to [imath]+\infty[/imath] as well. I'm stuck at the proof though. If you implement the mean value theorem for [imath]f(x)[/imath] in [imath](x,x+1)[/imath] you'll get: [imath]f(x+1) - f(x)= f'(b)[/imath] where [imath]b>x[/imath] now if u take limits for [imath]x\rightarrow+\infty[/imath], b must tend to [imath]+\infty[/imath] as well. So, [imath]\lim_{x\to\infty}[/imath] [imath]f(x+1) - f(x)[/imath] = [imath]+\infty[/imath], as [imath]x[/imath] tends to [imath]+\infty[/imath] Also, [imath]f(x)[/imath] must be increasing as [imath]x[/imath] tends to [imath]+\infty[/imath]. Can somebody continue this or show me another way of proving this? | 51596 | If [imath]f'[/imath] tends to a positive limit as [imath]x[/imath] approaches infinity, then [imath]f[/imath] approaches infinity
Some time ago, I asked this here. A restricted form of the second question could be this: If [imath]f[/imath] is a function with continuous first derivative in [imath]\mathbb{R}[/imath] and such that [imath]\lim_{x\to \infty} f'(x) =a,[/imath] with [imath]a\gt 0[/imath], then [imath]\lim_{x\to\infty}f(x)=\infty.[/imath] To prove it, I tried this: There exist [imath]x_0\in \mathbb{R}[/imath] such that for [imath]x\geq x_0[/imath], [imath]f'(x)\gt \frac{a}{2}.[/imath] There exist [imath]\delta_0\gt 0[/imath] such that for [imath]x_0\lt x\leq x_0+ \delta_0[/imath] [imath]\begin{align*}\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)&\gt -\frac{a}{4}\\ \frac{f(x)-f(x_0)}{x-x_0}&\gt f'(x_0)-\frac{a}{4}\\ &\gt \frac{a}{2}-\frac{a}{4}=\frac{a}{4}\\ f(x)-f(x_0)&\gt \frac{a}{4}(x-x_0)\end{align*}.[/imath] We can assume that [imath]\delta_0\geq 1[/imath]. If [imath]\delta_0 \lt 1[/imath], then [imath]x_0+2-\delta_0\gt x_0[/imath] and then [imath]f'(x_0+2-\delta_0)\gt \frac{a}{2}.[/imath] Now, there exist [imath]\delta\gt 0[/imath] such that for [imath]x_0+2-\delta_0\lt x\leq x_0+2-\delta_0+\delta[/imath] [imath]f(x)-f(x_0+2-\delta_0)\gt \frac{a}{4}(x-(x_0+2-\delta_0))= \frac{a}{4}(x-x_0-(2-\delta_0))\gt \frac{a}{4}(x-x_0).[/imath] It is clear that [imath]x\in (x_0,x_0+2-\delta_0+\delta][/imath] and [imath]2-\delta_0+\delta\geq 1[/imath]. Therefore, we can take [imath]x_1=x_0+1[/imath]. Then [imath]f'(x_1)\gt a/2[/imath] and then there exist [imath]\delta_1\geq 1[/imath] such that for [imath]x_1\lt x\leq x_1+\delta_1[/imath] [imath]f(x)-f(x_1)\gt \frac{a}{4}(x-x_1).[/imath] Take [imath]x_2=x_1+1[/imath] and so on. If [imath]f[/imath] is bounded, [imath](f(x_n))_{n\in \mathbb{N}}[/imath] is a increasing bounded sequence and therefore it has a convergent subsequence. Thus, this implies that the sequence [imath](x_n)[/imath]: [imath]x_{n+1}=x_n+1,[/imath] have a Cauchy's subsequence and that is a contradiction. Therefore [imath]\lim_{x\to \infty} f(x)=\infty[/imath]. I want to know if this is correct, and if there is a simpler way to prove this. Thanks. |
8996 | Hom of finitely generated modules over a noetherian ring
This is an exercise from Rotman, An Introduction to Homological Algebra, which I've been thinking now and then for a few days and I haven't solved it yet. I've decided to ask here because it is bugging me and I don't have any friends also studying this subject (in case it is of any use, I'm an undergraduate). Let [imath]R[/imath] be a commutative noetherian ring. If [imath]A[/imath], [imath]B[/imath] are finitely generated [imath]R[/imath]-modules, then [imath]\operatorname{Hom}_R(A,B)[/imath] is a finitely generated [imath]R[/imath]-module. Here's what I've thought as of yet (maybe the problem is that I haven't been trying to apply the useful theorems...): First of all, since [imath]R[/imath] is noetherian, it suffices to inject [imath]\operatorname{Hom}_R(A,B)[/imath] into some finitely generated module (in noetherian rings, submodules of f.g. are f.g.). Now, since [imath]R[/imath] is commutative (or since [imath]R[/imath] is noetherian), a finitely generated [imath]R[/imath]-module is a quotient of [imath]R^n[/imath] for some [imath]n[/imath]. Then let us write [imath]A\simeq \frac{R^n}{I}[/imath] and [imath]B\simeq \frac{R^m}{J}[/imath]. Now we've got: [imath]\operatorname{Hom}_R(A,B)\simeq \operatorname{Hom}_R\left(\frac{R^n}{I},\frac{R^m}{J}\right)[/imath]. If I had [imath]\operatorname{Hom}_R\left(\frac{R^n}{I},\frac{R^m}{J}\right) \hookrightarrow \operatorname{Hom}_R(R^n, R^m)[/imath] (this is what I tend to think is the wrong direction, but a interesting question-rising path nevertheless) then it would be over since [imath]\operatorname{Hom}_R(R^n, R^m)\simeq R^{nm}[/imath] which is finitely generated. So I ask myself: can I see [imath]\frac{R^n}{I}[/imath] as a submodule of [imath]R^n[/imath]? Because if I could, then doing the same thing for [imath]R^m[/imath] and passing it to the hom, then it would be over. Now, this would be true if the sequence [imath]0\to I\hookrightarrow R^n\to \frac{R^n}{I} \to 0[/imath] split. But this (of course) doesn't always happen. But if I somehow had the third module to be projective, or the first one to be injective, then it would happen. But why would, for example, be [imath]\frac{R^n}{I}[/imath] be projective? And this is where I got stuck. I'm sorry if this is overly detailed, but I remember reading that it is always good, when asking about a textbook question, to add what you've thought up to that moment. Also, even if my thoughts don't lead to the solution, I would like to know if they are correct! | 814655 | [imath]\hom_R(A,B)[/imath] is finitely generated if [imath]R[/imath] is noetherian
This is part of an exercise I'm doing, from Rotman Introduction to homological algebra. Let [imath]R[/imath] be a commutative ring, and let [imath]A[/imath] and [imath]B[/imath] be finitely generated [imath]R[/imath]-modules. Then if [imath]R[/imath] is noetherian, prove that [imath]\hom_R(A, B)[/imath] is a finitely generated [imath]R[/imath]-module. I'm stuck on it, any hint ? |
418459 | Sturm boundary value problem
I am having difficulty with the following question: [imath]y''+ky=0[/imath] [imath]y(-\pi)=y(\pi)[/imath] [imath]y'(-\pi)=y'(\pi)[/imath] For each eigenvalue [imath]k[/imath], there correspond(s): Only one eigenfunction Two eigenfunction Two linearly independent eigenfunctions. Two orthogonal eigenfunctions. If someone could help me, thanks in advance for your time. | 262124 | A problem For the boundary value problem, [imath]y''+\lambda y=0[/imath], [imath]y(-π)=y(π)[/imath] , [imath]y’(-π)=y’(π)[/imath]
For the boundary value problem, [imath]y''+\lambda y=0[/imath] [imath]y(-π)=y(π)[/imath] , [imath]y’(-π)=y’(π)[/imath] to each eigenvalue [imath]\lambda[/imath], there corresponds Only one eigenfunction Two eigenfunctions Two linearly independent eigenfunctions Two orthogonal eigenfunctions I have tried to solve the problem but could not get my calculations right. Can somebody help? |
418907 | Why is it that, [imath]\forall x \in \mathbb{Z},\ x^5 \equiv x \pmod{10}[/imath]?
Just playing around, I realized that [imath]x^{5}\equiv x \pmod{10}[/imath] for all integral [imath]x[/imath]. Euler's theorem explains this for [imath]x[/imath] coprime with [imath]10[/imath], as for such [imath]x[/imath], [imath]x^{4}\equiv 1[/imath], but I don't know why this happens for other [imath]x[/imath]. Also, how would this be generalized to other bases and powers? | 404157 | [imath]n^5-n[/imath] is divisible by [imath]10[/imath]?
I was trying to prove this, and I realized that this is essentially a statement that [imath]n^5[/imath] has the same last digit as [imath]n[/imath], and to prove this it is sufficient to calculate [imath]n^5[/imath] for [imath]0-9[/imath] and see that the respective last digits match. Another approach I tried is this: I factored [imath]n^5-n[/imath] to [imath]n(n^2+1)(n+1)(n-1)[/imath]. If [imath]n[/imath] is even, a factor of [imath]2[/imath] is guaranteed by the factor [imath]n[/imath]. If [imath]n[/imath] is odd, the factor of [imath]2[/imath] is guaranteed by [imath](n^2+1)[/imath]. The factor of [imath]5[/imath] is guaranteed if the last digit of [imath]n[/imath] is [imath]1, 4, 5, 6,[/imath] [imath]or[/imath] [imath]9[/imath] by the factors [imath]n(n+1)(n-1)[/imath], so I only have to check for [imath]n[/imath] ending in digits [imath]0, 2, 3, 7,[/imath] [imath]and[/imath] [imath]8[/imath]. However, I'm sure that there has to be a much better proof (and without modular arithmetic). Do you guys know one? Thanks! |
406955 | A loss and gain problem
This is a very simple but confusing puzzle. A customer buys goods worth [imath]200[/imath] rupees from a shop. The shopkeeper selling these goods makes zero profit from this purchase. The lady gives him a [imath]1000[/imath] rupee note. The shopkeeper has no change, so he goes next door to another shopkeeper to get change for the [imath]1000[/imath] rupee note. He keeps [imath]200[/imath] for himself and returns [imath]800[/imath] to the customer. Later, the second shopkeeper from next door comes back with the [imath]1000[/imath] rupee note with a stamp on it saying "counterfeit" and takes his money back. How much loss does the first shopkeeper face? | 694764 | How much the shopkeeper loses?
I struck with this tricky math question A girl went to a shop and bought a Rs.[imath]200[/imath] show piece. She gave a Rs.[imath]1000[/imath] note to shopkeeper. Shopkeeper didn't have any change so he went outside and took loan of Rs.[imath]800[/imath] from other shop and gave it to the girl. After sometime he realized that the Rs.[imath]1000[/imath] note was fake. So how much the shopkeeper loses? Rs is Indian currency. I think that the shopkeeper loses the show piece price [imath]200[/imath] + he borrowed [imath]800[/imath] + fake note of [imath]1000[/imath] = Rs.[imath]2000[/imath] Am I right? |
418880 | If [imath]M \otimes M \simeq M[/imath] is there anything we can say about [imath]M[/imath]?
Over a commutative (and unital) ring, if [imath]M \otimes M \simeq M[/imath] can we say anything about [imath]M[/imath]? If we base change to a point, ie tensor with a map from the ring into a field, then [imath]M[/imath] becomes a vector space and we know that it must be one-dimensional. Is there anything else we can say? I'd like to work as generally as possible, and I'm willing to assume that [imath]M[/imath] is of finite presentation. | 413316 | Modules which are isomorphic to their tensor product.
Suppose that we have a commutative ring [imath]R[/imath]. I am interested in finding the (finitely generated and projective, if you want) [imath]R[/imath]-modules [imath]M,[/imath] such that [imath]M\cong M\otimes_R M[/imath] as [imath]R[/imath]-modules. I know that it is true for [imath]M=R[/imath] or when [imath]M[/imath] is generated by a finite set of orthogonal idempotents. I want to know if whether or not those are all the modules which that property. Thanks. |
414824 | [imath]C[0,1][/imath] is incomplete with the norm [imath]\|f\|_{2}=\sqrt{\int_{0}^{1}|f(x)|^2}[/imath]?
Could any one give me an explicit example of a sequence which shows that [imath]C[0,1][/imath] is incomplete with the norm [imath]\|f\|_{2}=\sqrt{\int_{0}^{1}|f(x)|^2}[/imath]? Thank you for the help | 253034 | The vector space formed on C[0,1] and the norm [imath](\int_{0}^{1}|f(t)|^2 dt)^{1/2} [/imath]
How does one show that [imath](C[0,1], ||.||_{2})[/imath] where [imath]||.||_2=(\int_{0}^{1}|f(t)|^2dt)^{1/2}[/imath] and [imath]C[0,1][/imath] is the space of all continous function which are mapped from [imath][0,1]\rightarrow \mathbb{R}[/imath], is not complete? First to show that [imath](C[0,1], ||.||_{2})[/imath] is a normed vector space: [imath]||.||_2[/imath] fulfills [imath]||f(t)||=0 \Rightarrow f(t)=0[/imath] [imath]||.||_2[/imath] fulfills the scalar multiplication property: [imath]||\lambda f(t)||_{2} = (\int_{0}^{1}|\lambda f(t)|^2dt)^{1/2}=|\lambda|(\int_{0}^{1}| f(t)|^2dt)^{1/2}=|\lambda|||f(t)||_2[/imath] Thirdly, it also fulfills the triangle inequality:[imath]||x(t)+y(t)||_2^2 = \int_{0}^{1}|x(t)+y(t)|^2dt \le \int_{0}^{1}|x(t)|^2dt + 2\int_{0}^{1}|x(t)y(t)|dt +\int_{0}^{1}|y(t)|^2dt \le ||x(t)||_2^2 + 2||x(t)||_2||y(t)||_2+||y(t)||_2^2 = (||x(t)||_2+||(y(t)||_2)^2[/imath] this also shows the closure under addition in [imath]C[0,1][/imath] so , [imath](C[0,1],||.||_2)[/imath] is a normed vector space , but how can we show that it is not complete? |
419334 | Prove this equality [imath]\frac{x}{y^2+5}+\frac{y}{z^2+5}+\frac{z}{x^2+5}\le\frac{1}{2}[/imath]
let [imath]x^3+y^3+z^3=3,x,y,z>0[/imath] show that [imath]\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2}[/imath] I have show that let [imath]x,y,z[/imath] be positive numbers,such that [imath]x+y+z=3[/imath],prove that [imath]\dfrac{x}{1+y^3}+\dfrac{y}{1+z^3}+\dfrac{z}{1+x^3}\ge\dfrac{3}{2}[/imath] pf: use [imath]AM-GM[/imath] we have [imath]\dfrac{x}{1+y^3}=x-\dfrac{xy^3}{1+y^3}\ge x-\dfrac{xy^3}{2y^{3/2}}=x-\dfrac{xy^{3/2}}{2}[/imath] and,similarly [imath]\dfrac{y}{1+z^3}\ge y-\dfrac{yz^{3/2}}{2},\dfrac{z}{1+x^3}\ge z-\dfrac{zx^{3/2}}{2}[/imath] Thus,it suffices to show that [imath]xy^{3/2}+yz^{3/2}+zx^{3/2}\le 3[/imath] and it is known that [imath](a^3b^2+b^3c^2+c^3a^2)^2\le\dfrac{1}{3}(a^2+b^2+c^2)^3[/imath] seting [imath]x=a^2,y=b^2,z=c^2[/imath],by done! But for this problem : [imath]y^2+5=y^2+1+1+1+1+1\ge 6y^{1/3}[/imath] and similarly [imath]z^2+5\ge 6z^{1/3}, x^2+5\ge 6x^{1/3}[/imath] it suffices prove that [imath]xy^{-1/3}+yz^{-1/3}+zx^{-1/3}\le 3[/imath] with [imath]x^3+y^3+z^3=3[/imath],I use maple find this is ([imath]xy^{-1/3}+yz^{-1/3}+zx^{-1/3}\le 3[/imath],with [imath]x^3+y^3+z^3=3[/imath]) not true!,But after I use maple find this [imath]\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2},x^3+y^3+z^3=3[/imath] is true! and my other idea [imath]\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2}[/imath] [imath]\Longleftrightarrow 2\sum x^2y^3+10\sum x^2y+50\sum x-5\sum x^2y^2-25\sum y^2-x^2y^2z^2\le 95[/imath] so I think my methods can't prove this problem, can someone use other methods show it? Thank you everyone. | 306024 | Prove that [imath]\dfrac{a}{b^2+5}+ \dfrac{b}{c^2+5} + \dfrac{c}{a^2+5} \le \dfrac 12[/imath]
Let [imath]a,b,c>0[/imath] and [imath]a^3+b^3+c^3=3[/imath]. Prove that [imath]\dfrac{a}{b^2+5}+ \dfrac{b}{c^2+5} + \dfrac{c}{a^2+5} \le \dfrac 12[/imath] I have an ugly solution for this solution. |
419703 | Let [imath]1=b_1 be integers relatively prime with n. Prove that B_n =b_1\cdots b_{\phi(n)} is congruent to \pm1 modulo n.[/imath]
Let [imath]1=b_1<b_2<\cdots<b_{\phi(n)}<n[/imath] be integers relatively prime with [imath]n[/imath]. How do I show that [imath]B_n =b_1\cdots b_{\phi(n)}\equiv\pm1\bmod n.[/imath] | 266351 | If [imath]\{a_1,\dots,a_{\phi(n)}\}[/imath] is a reduced residue system, what is [imath]a_1\cdots a_{\phi(n)}[/imath] congruent to?
I'm trying work on problem 11 of Chapter 3 in Ireland and Rosen's Number Theory. Suppose [imath]\{a_1,\dots,a_{\phi(n)}\}[/imath] is a reduced residue system modulo [imath]n[/imath]. Let [imath]N[/imath] be the number of solutions to [imath]x^2\equiv 1\pmod{n}[/imath], then [imath] a_1\cdots a_{\phi(n)}\equiv (-1)^{N/2}\pmod{n}. [/imath] How can this be shown? I know that if [imath]n=p_1^{\alpha_1}\cdots p_m^{\alpha_m}[/imath], then [imath]N=\prod N(p_i^{\alpha_i})[/imath] where [imath]N(p_i^{\alpha_i})[/imath] denotes the number of solutions to [imath]x^2\equiv 1\pmod{p_i^{\alpha_i}}[/imath]. Also, if [imath]x_0[/imath] is a solution, then it is also a solution to [imath]x^2\equiv 1\pmod{p_i}[/imath] for any prime in the factorization of [imath]n[/imath]. This implies that [imath]x_0\equiv \pm 1\pmod{p_i}[/imath], so that [imath]x_0[/imath] is coprime to all prime factors of [imath]n[/imath], and thus coprime to [imath]n[/imath]. But I'm not seeing a connection between the left and right hand side products of the congruence in question. Thanks. |
419732 | What does [imath]\lim \limits_{n\rightarrow \infty }\sum \limits_{k=0}^{n} {n \choose k}^{-1}[/imath] converge to (if it converges)?
How we can show if the sum of [imath]\lim_{n\rightarrow \infty }\sum_{k=0}^{n} \frac{1}{{n \choose k}}[/imath] converges and then find the result of the sum if it converges? Thanks for any help. | 4783 | Finding [imath]\lim\limits_{n \to \infty} \sum\limits_{k=0}^n { n \choose k}^{-1}[/imath]
We know that [imath] 2^n= (1+1)^n = \sum_{k=0}^n {n \choose k}[/imath] I was asked to solve this limit, [imath]\lim_{n \to \infty} \ \sum_{k=0}^n {n \choose k}^{-1}=? \quad \text{for} \ n \geq 1[/imath] |
419992 | Class of finite groups a Fraïssé Class?
Is the class of finite groups a Fraïssé class? Calling this class [imath]K[/imath], does [imath]K[/imath] satisfy the following: Joint embedding property Amalgamation property Hereditary property: if [imath]G \in K[/imath] and [imath]H \le G[/imath], then [imath]H \in K[/imath]) (1) holds because if [imath]G, H \in K[/imath], then [imath]G \times H \in K[/imath] as well. Obviously [imath]G[/imath] and [imath]H[/imath] are both subgroups of [imath]G \times H[/imath]. (2) seems to hold. Suppose [imath]G_1, G_2 \in K[/imath], with [imath]H = G_1 \cap G_2[/imath]. Then the amalgamated free product [imath]G_1 *_H G_2[/imath] contains both [imath]G_1[/imath] and [imath]G_2[/imath] as subgroups. However, is it true that [imath]G_1 *_H G_2[/imath] is finite? (3) holds because substructures of finite groups are again finite groups. By definition, substructures must contain [imath]0[/imath] and must be closed under the group operation. Furthermore, every element has an inverse, because they all have finite order (you will eventually reach the inverse by multiplying an element by itself successively). Associativity is universal, so also holds. So every substructure is also a (sub)group. Is my reasoning correct? I would appreciate some kind of proof that amalgamated free products of finite groups are also finite, if this is true, or some counterexample if not, because I don't know much about amalgamated free products. | 88169 | Fraïssé limits and groups
I was recently reading up on Fraïssé limits in Hodges' "A shorter model theory." I was trying to think of some examples and wanted to see if I could take the Fraïssé limit on the category of finite groups. It is clear that this category has the hereditary and joint embedding property. However, I am having trouble showing that this category has the amalgamation property, by which I mean that if there are finite groups [imath]G, H, K[/imath] such that there are embeddings [imath]\varphi: G\hookrightarrow H[/imath] and [imath]\psi: G\hookrightarrow K[/imath], then there exists a finite group [imath]J[/imath] and embeddings [imath]\vartheta: H\hookrightarrow J[/imath] and [imath]\eta: K\hookrightarrow J[/imath] such that [imath]\vartheta\circ \varphi = \eta\circ \psi[/imath]. One simplification is that we can, by taking isomorphic copies if necessary, assume that [imath]H\cap K = G[/imath], and that [imath]\varphi [/imath] and [imath]\psi[/imath] are just inclusion maps. I am not even sure what group [imath]J[/imath] should be. Any help would be appreciated. |
420280 | eigenvector and operator
Let [imath]V[/imath] be a finite-dimensional vector space over [imath]\mathbb{C}[/imath], let [imath]T:V\rightarrow V[/imath] be a linear operator, and let [imath]M[/imath] be a nontrivial subspace of [imath]V[/imath] that is invariant under [imath]T[/imath]. Prove that [imath]M[/imath] contains an eigenvector of [imath]T[/imath]. I just have no idea about it. | 415008 | invariant space and eigenvector
Let [imath]V[/imath] be a finite-dimensional vector space over [imath]C[/imath], let [imath]T:V\rightarrow V[/imath] be a linear operator, and let [imath]M[/imath] be a nontrivial subspace of [imath]V[/imath] that is invariant under [imath]T[/imath]. Prove that [imath]M[/imath] contains an eigenvector of T. I am trying to find the eigenvector, but I can not. |
419854 | Question about the number of elements of order 2 in [imath]D_n[/imath]
[imath]\text{Given}\;\; D_n = \{ a^ib^j \mid \text{ order}(a)=n, \text{ order}(b)=2, a^ib = ba^{-i} \}[/imath] [imath]\text{ how many elements does $D_n$ contain that have order $2$ ?}[/imath] My answer would be: We can write any element as [imath]a^ib^j[/imath]. If [imath]j[/imath] is odd, then we can write it as [imath]a^ij[/imath], whereas if [imath]j[/imath] is even, then [imath]b[/imath] just cancels out. So to count: [imath]b[/imath] is one element of order [imath]2[/imath]. If [imath]j[/imath] is odd we have [imath]a^ib[/imath]. [imath](a^ib)^2= a^i(ba^i)b = a^i a^{-i} bb = e[/imath]. So all elements of the form [imath]a^ib[/imath] have order [imath]2[/imath]. These are [imath]n[/imath] elements. If [imath]j[/imath] is even we can write the remaining elements as follows: [imath]a^i[/imath] . This element has order [imath]2[/imath] only if [imath]i=n/2[/imath]. This can only be true if [imath]n[/imath] is even. So my final answer would be [imath]n+1[/imath] if [imath]n[/imath] is odd, [imath]n+2[/imath] if [imath]n[/imath] is even... Do you think this is correct? Can I improve this somewhere? Thanks ! | 126639 | Dihedral group - elements of order [imath]2[/imath]
If [imath]D_{n}[/imath] is the Dihedral group of degree [imath]n[/imath] (order [imath]2n[/imath]) where n is an odd number, can I then conclude that the numbers of elements of order 2 in [imath]D_{n}[/imath] is equal [imath]n?[/imath] I suppose there are the following types of elements in [imath]D_{n}[/imath]: b, [imath]a^{i}[/imath], [imath]a^{i}b[/imath] for some [imath]0\leq i < n[/imath] . |
420389 | Is a totally ordered set well-ordered, provided that its countable subsets are?
Let [imath](X,⪯[/imath]) be totally ordered set, prove that if every non empty countable subset of [imath]X[/imath] is well ordered then X is well ordered. It does seem obvious that any subset should have a minimum but I am not sure of how to prove it. Any help would be appreciated, and sorry I put up the same problem earlier but forgot to include the 'countable' subset. | 433650 | Well ordered sets
[imath](X,\le)[/imath] is totally ordered. How do you prove that if every non empty countable subset of [imath]X[/imath] is well ordered then [imath](X,\le)[/imath] is well ordered? |
419765 | Limit. [imath]\lim_{n \to \infty}\frac{1^p+2^p+\ldots+n^{p}}{n^{p+1}}[/imath].
Have you any idea how to find the limit of the following sum: [imath]\lim_{n \to \infty}\frac{1^p+2^p+\ldots+n^{p}}{n^{p+1}}.[/imath] Stolz-Cesaro? any more ideas? | 149142 | Evaluating [imath]\lim\limits_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)[/imath]
Evaluate [imath]\lim_{n\to\infty} \left(\frac{1^p+2^p+3^p + \cdots + n^p}{n^p} - \frac{n}{p+1}\right)[/imath] |
420400 | Sets question, without Zorn's lemma
Is there any proof to [imath]|P(A)|=|P(B)| \Longrightarrow |A|=|B|[/imath] that doesn't rely on Zorn's lemma (which means, without using the fact that [imath]|A|\neq|B| \Longrightarrow |A|<|B|[/imath] or [imath]|A|>|B|[/imath] ) ? Thank you! | 74477 | Does [imath]2^X \cong 2^Y[/imath] imply [imath]X \cong Y[/imath] without assuming the axiom of choice?
A friend of mine told me that [imath]X \cong Y \Rightarrow 2^X \cong 2^Y[/imath] ([imath]X[/imath] and [imath]Y[/imath] being sets), which is very easy to prove, but he was wondering about the converse in ZF, i.e., can one take logarithms? Since the (apparently) simpler question of whether it is possible to divide by a natural number is not particularly trivial without assuming the axiom of choice (see Doyle, Conway: Division by three), I would imagine that this problem doesn't have an easy answer either. |
420618 | Explanation of why [imath]\frac{d}{dx} e^x=e^x[/imath]
I'm taught that all the way back when I'm in high school that [imath]\dfrac{de^x}{dx}=e^x[/imath] and [imath]\int e^x dx=e^x[/imath]. Can someone explain why this is the case? | 296666 | How do I prove that the exponential function [imath]e^x[/imath] has gradient [imath]e^x[/imath] from first principles?
Furthermore why is it that [imath]e^x[/imath] is used in exponential modelling? Why aren't other exponential functions used, i.e. [imath]2^x[/imath], etc.? Thank you! |
171904 | Limit of a sequence involving root of a factorial: [imath]\lim_{n \to \infty} \frac{n}{ \sqrt [n]{n!}}[/imath]
I need to check if [imath]\lim_{n \to \infty} \frac{n}{ \sqrt [n]{n!}}[/imath] converges or not. Additionally, I wanted to show that the sequence is monotonically increasing in n and so limit exists. Any help is appreciated. I had tried taking log and manipulating the sequence but I could not prove monotonicity this way. | 448324 | How to solve [imath] \lim_{n\to\infty} \frac {(n!)^\frac{1}{n}}{n} [/imath]?
I need to find the limit for: [imath] \lim_{n\to\infty} \frac {(n!)^\frac{1}{n}}{n} [/imath] I know the answer is [imath]\frac {1}{e}[/imath] but I have no idea how to get that answer. I'd appreciate some help. |
421040 | Calculate [imath]irr(a,\mathbb{Q})[/imath] where [imath]a=\sqrt{2}+\sqrt{3}\in\mathbb{R}[/imath]
Consider the element [imath]a=\sqrt{2}+\sqrt{3}\in\mathbb{R}[/imath]. Calculate [imath]irr(a,\mathbb{Q})[/imath]. What I did: Calculate powers of [imath]a[/imath]. [imath]a^2=5+2\sqrt{6},a^3=11\sqrt{2}+9\sqrt{3},a^4=49+20\sqrt{6}[/imath]. I wish to find some relationships between powers of [imath]a[/imath], but I have no idea where to begin. I believe there should be a better way than just trial-and-error. It is shown in the hint that we need to consider vectors [imath]v_0=(1,0,0,0), v_1=(0,1,1,0), v_2=(5,0,0,2),v_3=(0,11,9,0), v_4=(49,0,0,20)[/imath], but I have no clue how to use the given hints and how can we even use vectors in our problem. Thanks in advance for the helps! Edit: I have made some attempt and I observed that [imath]v_4=-v_0+10v_2[/imath] and by following the pattern (a very naive way) to replace [imath]v_4[/imath] by [imath]a^4[/imath], [imath]v_2[/imath] by [imath]a^2[/imath] and [imath]v_0[/imath] by 1, I got [imath]a^4-10a^2+1=0[/imath]. Is this progress useful in solving the problem? Is the "pattern" just a coincidence or is there any explanation to it? How can then I proceed in finding [imath]irr(a,\mathbb{Q})[/imath]? | 359054 | Constructing a degree 4 rational polynomial satisfying [imath]f(\sqrt{2}+\sqrt{3}) = 0[/imath]
Goal: Find [imath]f \in \mathbb{Q}[x][/imath] such that [imath]f(\sqrt{2}+\sqrt{3}) = 0[/imath]. A direct approach is to look at the following [imath] \begin{align} (\sqrt{2}+\sqrt{3})^2 &= 5+2\sqrt{6} \\ (\sqrt{2}+\sqrt{3})^4 &= (5+2\sqrt{6})^2 = 49+20\sqrt{6} \\ \end{align} [/imath] Putting those together gives [imath] -1 + 10(\sqrt{2}+\sqrt{3})^2 - (\sqrt{2}+\sqrt{3})^4 = 0, [/imath] so [imath]f(x) = -1 + 10x^2 - x^4[/imath] satisfies [imath]f(\sqrt{2}+\sqrt{3}) = 0[/imath]. Is there a more mechanical approach? Perhaps not entirely mechanical, but something more abstract. |
139842 | In how many ways can a number be expressed as a sum of consecutive numbers?
All the positive numbers can be expressed as a sum of one, two or more consecutive positive integers. For example [imath]9[/imath] can be expressed in three such ways, [imath]2+3+4[/imath], [imath]4+5[/imath] or simply [imath]9[/imath]. In how many ways can a number be expressed as a sum of consecutive numbers? In how many ways can this work for [imath]65[/imath]? Here, for [imath]9[/imath] answer is [imath]3[/imath], for [imath]10[/imath] answer is [imath]3[/imath], for [imath]11[/imath] answer is [imath]2[/imath]. | 1207077 | Number of ways to write n as a sum of consecutive integers
Allow [imath]d(n)[/imath] to be the number of divisors of [imath]n[/imath]. Show that there are [imath]d(m)[/imath] ways to write [imath]n[/imath] as the sum of consecutive integers where [imath]m[/imath] is the largest odd divisor on [imath]n[/imath]. I have absolutely no idea where to go. The hint given to me is that [imath]\ x + (x+1) +... + (x+r-1) + (x+r) = (r+1)x + \frac{1}{2}r(r+1)[/imath] I can set the above expression equal to [imath]n[/imath], but this still confuses me. |
421237 | problem on Continuity of a function
Let [imath] \{r_n\}[/imath] be an enumeration of the set of rational numbers such that [imath] r_n\neq r_m [/imath] if [imath]n\neq m [/imath] define [imath] f :\Bbb R \to \Bbb R [/imath] by [imath] f(x) =\sum_{r_n \leq x} [/imath] [imath]1/2^{n} x \in \Bbb R [/imath] Prove that [imath]f[/imath] is continuous at each point of [imath]\Bbb Q^{c} [/imath] and discontinuous at each point of [imath] \Bbb Q [/imath] I couldn't first understand the function and had little idea on proving either continuity or discontinuity | 421220 | Proving a function is continuous on all irrational numbers
Let [imath]\langle r_n\rangle[/imath] be an enumeration of the set [imath]\mathbb Q[/imath] of rational numbers such that [imath]r_n \neq r_m\,[/imath] if [imath]\,n\neq m.[/imath] [imath]\text{Define}\; f: \mathbb R \to \mathbb R\;\text{by}\;\displaystyle f(x) = \sum_{r_n \leq x} 1/2^n,\;x\in \mathbb R.[/imath] Prove that [imath]f[/imath] is continuous at each point of [imath]\mathbb Q^c[/imath] and discontinuous at each point of [imath]\mathbb Q[/imath]. I find this question very challenging and have no idea even how to start off with the proof. Please suggest a proof or any hint. |
421442 | Function to map a range of [imath][-1,1][/imath] to a range of [imath][0,1][/imath]
I'm not capable of 'rigorously' defining the problem I have but this is the best I can do. If I have a set of points that range from [imath]-1[/imath] to [imath]1[/imath] inclusive and I have to transform the data so that it fits onto a range of [imath]0[/imath] to [imath]1[/imath]. What's the simple function to do this? | 135461 | Change of coordinate codomain from [imath][-1,1][/imath] to [imath][0,1][/imath]
How does one translate coordinates from [imath][-1,1][/imath] to [imath][0,1][/imath]? That is, suppose we have an ordered pair [imath](x,y)[/imath] which lies between [imath][-1,1][/imath] and want to push into the range delimited by [imath][0,1][/imath]. A lot of places does the transformation something like this: [imath]f(u) = ((u_1 + 1)/2)i + ((u_2 +1)/2)j[/imath] I'm not even sure if I may express it this way (?), here's the usual expression: [imath]x = (c_x + 1) / 2[/imath] [imath]y = (c_y + 1) / 2[/imath] I can see that that the addition of [imath]1[/imath] will force all the possible inputs into the [imath][0,2][/imath] range and that the division by the maximum of the range will force it to be expressed as a part of the whole, the whole being the dimensionless [imath]1[/imath], effectively expressing all values in the range of [imath][0,1][/imath]. Is this all there is to this? Just a heuristic, trial and error technique until you get what you want? Or is there a more obvious, better, formalized way? |
422057 | [imath]G_1/H\cong G_2\implies G_1\cong H\times G_2[/imath]?
Lagrange's lets us write the deceptively tidy relation: [imath]\left|\frac{G}{H}\right|=\frac{|G|}{|H|}[/imath] and from this we can do neat things like, in the proof of the Orbit-stabiliser theorem, [imath]\frac{G}{\text{stab}(s)}\cong \text{orb}(s)\implies\left|\frac{G}{\text{stab}(s)}\right|=|\text{orb}(s)|\implies\frac{|G|}{|\text{stab}(s)|}=|\text{orb}(s)|\\\implies|G|=|\text{orb}(s)|\times|\text{stab}(s)|[/imath] Which leads to my question: is it always/sometimes/ever possible to use naive reasoning and say that [imath]\frac{G_1}{H}\cong G_2\implies G_1\cong H\times G_2[/imath] | 321176 | If [imath]H[/imath] is a normal subgroup of [imath]G[/imath], is [imath]G/H \times H \cong G[/imath]?
If [imath] H [/imath] is a normal subgroup of [imath] G [/imath], is [imath] G/H \times H \cong G [/imath]? For example, I think [imath] \mathbb{Z}/2 \mathbb{Z} \times 2 \mathbb{Z} \cong \mathbb{Z} [/imath]. I would construct a map as follows: \begin{align} \phi: \mathbb{Z}/2 \mathbb{Z} \times 2 \mathbb{Z} &\longrightarrow \mathbb{Z}; \\ (a,b) &\longmapsto a + b. \end{align} If it is not true in general, then are there a few criteria to show when it is true? Thanks! |
421408 | [imath]a^{\log_b(c)} = c^{\log_b(a)}[/imath]
I'm not sure how to start. My questions is how do you prove: [imath]a^{\log_b(c)} = c^{\log_b(a)}[/imath] where [imath]a,b,c > 0[/imath]. | 10866 | How [imath]a^{\log_b x} = x^{\log_b a}[/imath]?
This actually triggered me in my mind from here. After some playing around I notice that the relation [imath]a^{\log_b x} = x^{\log_b a}[/imath] is true for any valid value of [imath]a,b[/imath] and [imath]x[/imath]. I am very inquisitive to see how this holds ? |
422224 | counting more problem continue
i have asked but no one was able to help so i am re-posting, hoping someone can help me. i did the computation and i could be wrong but i have provided my answer. Given problem: How many ways can 5 rings be placed 4 fingers if the rings are all the same? What if the rings are all distinct? Answer: I used the formula [imath]\tbinom{n+k-1} k[/imath] because I assumed that the question is saying that order doesn't matter and repetition is allowed; therefore I',m using that formula. By using that formula I have come up with this: [imath]\tbinom{5+4-1}4 = \tbinom 8 4=\tfrac {8!}{4!(8-4)!}[/imath] Since the rings are distinct (meaning different), i used [imath]n^k[/imath] to compute it: [imath]5^4[/imath] | 422136 | How many ways can [imath]5[/imath] rings be placed on [imath]4[/imath] fingers?
I've been trying to solve this problem and I am kind of struggling with it and with other combinatorics problems. Could you check and see if i did it right? Given problem: How many ways can 5 rings be placed 4 fingers if the rings are all the same? What if the rings are all distinct? Answer: I used the formula [imath]\tbinom{n+k-1} k[/imath] because I assumed that the question is saying that order doesn't matter and repetition is allowed; therefore I',m using that formula. By using that formula I have come up with this: [imath]\tbinom{5+4-1}4 = \tbinom 8 4=\tfrac {8!}{4!(8-4)!}[/imath] Since the rings are distinct (meaning different), i used [imath]n^k[/imath] to compute it: [imath]5^4[/imath] Could anyone check if I did it right or wrong? If wrong, could you correct it and tell me how to solve it correctly? Because I really want to understand what I did wrong and know how to solve it correctly. Thank you so much. |
422196 | [imath]\left(-\frac{1}{2}\right)! = \sqrt{\pi}?[/imath]
I recently learned that [imath]\left(-\frac{1}{2}\right)! = \sqrt{\pi}[/imath] but I don't understand how that makes sense. Can someone please explain how this is possible? Thanks! | 215352 | Why is [imath]\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}[/imath] ?
It seems as if no one has asked this here before, unless I don't know how to search. The Gamma function is [imath] \Gamma(\alpha)=\int_0^\infty x^{\alpha-1} e^{-x}\,dx. [/imath] Why is [imath] \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}\ ? [/imath] (I'll post my own answer, but I know there are many ways to show this, so post your own!) |
422428 | Is there a bijection [imath]f:\mathbb R^+\to\mathbb R^+[/imath] s.t. [imath]f'(x)=f^{-1}(x)[/imath]?
Is there a bijection [imath]f:\mathbb R^+\to\mathbb R^+[/imath] s.t. [imath]f'(x)=f^{-1}(x)[/imath]? And if there is, can I prove uniqueness? This problem troubles me. | 279517 | Does there exist [imath]f:(0,\infty)\to(0,\infty)[/imath] such that [imath]f'=f^{-1}[/imath]?
Recently the following question was posed: does there exist a differentiable bijection [imath]f:\mathbb R\to\mathbb R[/imath] such that [imath]f'=f^{-1}[/imath]? (Here, [imath]f^{-1}[/imath] is the inverse of [imath]f[/imath] with respect to composition of functions.) The answer, as it turns out, is negative, because such a bijection is monotonic, which implies that the derivative cannot be bijective. There seem to be no such problems if we instead search for a bijection [imath]f:(0,\infty)\to(0,\infty)[/imath] such that [imath]f'=f^{-1}[/imath], so the following variation of the question seems interesting: Does there exist a differentiable bijection [imath]f:(0,\infty)\to(0,\infty)[/imath] such that [imath]f'=f^{-1}[/imath]? (Edit: as mentioned at the end of this question, the question of existence has been resolved. Is the solution unique?) Here's the first idea: if there is a function [imath]f:(0,\infty)\to(0,\infty)[/imath] such that [imath]f'(x)=f^{-1}(x)[/imath] holds for all [imath]x\in(0,\infty)[/imath], then [imath]f'(f(x))=f^{-1}(f(x))=x[/imath] also must hold for all [imath]x[/imath], since [imath]f[/imath] is bijective. This immediately reminds us of the chain rule, so we multiply by [imath]f'(x)[/imath] to yield [imath]f'(f(x))f'(x)=xf'(x)[/imath] which is equivalent to [imath]f'(f(x))f'(x)=xf'(x)+f(x)-f(x).[/imath] This implies (integrate from [imath]1[/imath] to [imath]x[/imath], for instance) that there is a primitive function [imath]F[/imath] of [imath]f[/imath] such that [imath]f(f(x)) = x f(x) - F(x)[/imath] holds for all [imath]x\in(0,\infty)[/imath], so we may try solving this equation instead. I don't know if this is any easier than the original problem, though. Maybe some kind of fixed point principle might work to show existence? It would be very nice if it turns out that there is a nice characterization of such functions. Edit: At the link pointed out by Christian Blatter in the comments there is an explicit solution ([imath]f(x)=\frac{x^\phi}{\phi^{\phi-1}}[/imath], where [imath]\phi=\frac{1+\sqrt5}{2}[/imath] is the golden ratio), so maybe it would also be interesting to know: Is this solution unique? |
422549 | Weak Topology determined by identity maps
Let [imath](X,τ_1)[/imath] and [imath](X,τ_2)[/imath] be two tychonoff spaces. Let [imath]τ[/imath] be the smallest topology on X such that identity maps [imath]id_1:(X,τ)→(X,τ_1)[/imath] and [imath]id_2:(X,τ)→(X,τ_2)[/imath] are continuous. If both [imath](X,τ_1)[/imath] and [imath](X,τ_2)[/imath] are separable topological spaces. Then (X,τ) will be separable? | 420027 | Weak topology generated by identity maps
Let [imath](X, \tau_1)[/imath] and [imath](X,\tau_2)[/imath] be two topological spaces having the same underlying set X. Let [imath]\tau[/imath] be the smallest topology on [imath]X[/imath] such that identity maps [imath]I_1 : (X,\tau) \rightarrow (X,\tau_1)[/imath] and [imath]I_2 : (X,\tau) \rightarrow (X,\tau_2)[/imath] are continuous. Then if both [imath](X,\tau_1)[/imath] and [imath](X,\tau_2)[/imath] are separable topological space. Does it imply that [imath](X,\tau)[/imath] will be separable? |
423008 | Generating function of partition with restriction
Let [imath]c(m,n)[/imath] denotes number of partitions [imath]n[/imath] into parts not greater than [imath]m[/imath], where order of elements does matter (so they are not classic partitions). Prove that: [imath]\sum_{n\ge 0}c(m,n)x^n = \frac{1-x}{1-2x+x^{m+1}}[/imath] I don't know how to approach. Generating function of such partitions is (if I'm not crazy): [imath](x+x^2+x^3+...+x^m)(x+x^2+x^3+...+x^m)(x+x^2+x^3+...+x^m)...[/imath] but I don't see how it can help here. | 392787 | Compositions of [imath]n[/imath] with largest part at most [imath]m[/imath]
This is a problem from Stanley's Enumerative Combinatorics that I'm failing at a bit (lot): Let [imath]\bar{c}(m,n)[/imath] denote the number of compositions of [imath]n[/imath] with largest part at most [imath]m[/imath]. Show that [imath]\sum_{n\geq 0}\bar{c}(m,n)x^n={{1-x}\over{1-2x+x^{m+1}}}[/imath] Some definitions: A composition of [imath]n[/imath] is an [imath]ordered[/imath] list of positive integers that equals [imath]n[/imath]. If [imath]\{a_1,...,a_k\}[/imath] is one such composition, we say that the composition has [imath]k[/imath] [imath]parts[/imath]. I know it's pretty traditional to list a "what you've done so far" but I'm really about as blindly stuck as can be. |
423002 | On a question of closed property of morphism of schemes
Let [imath]f: X \rightarrow Y[/imath] be a morphism of schemes of finite type over an algebraically closed field [imath]k[/imath]. This means [imath]X[/imath], [imath]Y[/imath] are schemes of finite type over [imath]k[/imath] and [imath]f[/imath] is a [imath]k[/imath] -morphism. Let [imath]M, N[/imath] be the set of closed points in [imath]X, Y[/imath] respectively. Then it is clear that [imath]f(M) \subset N[/imath] and for any closed subset [imath]C[/imath] of [imath]X[/imath] we have [imath]f(C) \cap N = f(C\cap M)[/imath]. So if [imath]f[/imath] is a closed map then the restriction [imath]f|_M : M \rightarrow N[/imath] is so. Is the converse true? We know that [imath]f(C)[/imath] is constructible subset of [imath]Y[/imath] for any closed set [imath]C[/imath] in [imath]X[/imath]. Can I claim that [imath]f(C)[/imath] is closed in [imath]Y[/imath] if [imath]f(C) \cap N[/imath] is closed in [imath]N[/imath]? I can see this if [imath]f(C)[/imath] is locally closed but not in general i.e. when [imath]f(C)[/imath] is union of finite number of locally closed sets. | 245317 | Is a morphism between schemes of finite type over a field closed if it induces a closed map between varieties?
This is the converse of this question. Let [imath]X[/imath](resp. [imath]Y[/imath]) be a scheme of finite type over a field [imath]k[/imath]. Let [imath]f\colon X \rightarrow Y[/imath] be a morphism. Let [imath]X_0[/imath](resp. [imath]Y_0[/imath]) be the set of closed points of [imath]X[/imath](resp. [imath]Y[/imath]). Then [imath]f[/imath] induces a map [imath]f_0\colon X_0 \rightarrow Y_0[/imath]. We consider [imath]X_0[/imath](resp. [imath]Y_0[/imath]) as a subspace of [imath]X[/imath](resp. [imath]Y[/imath]). Suppose [imath]f_0[/imath] is a closed map. Is [imath]f[/imath] closed(*)? (*) A morphism is closed if the image of any closed subset is closed(Hartshorne, p.100). |
417815 | Four integer numbers to express all integers from 1 to 40
Let [imath]a[/imath], [imath]b[/imath], [imath]c[/imath], and [imath]d[/imath] four integers such that [imath]0 <a <b <c <d[/imath]. We can get all integers from [imath]1[/imath] to [imath]40[/imath] by expressions containing or not only the numbers [imath]a, b, c[/imath] and [imath]d[/imath]. In these expressions [imath]a, b, c[/imath] and [imath]d[/imath] do not appear more than once and in those expressions that use more than one of these numbers only addition and / or subtraction are allowed. Determine [imath]a, b, c[/imath] and [imath]d[/imath]. I have found out these numbers but not in an elegant way. I wonder if anyone knows any more appropriate method to tackle that problem. Edit: It's not sufficiently clear to me why {1,3, 9, 27} is the only set that solves the problem. If it was stated in the problem "get all integers from 1 to 38", then there would be at least three solutions for this problem: {1, 3, 9, 25}, {1, 3, 9, 26} and {1, 3 , 9, 27}. | 119606 | Represent every Natural number as a summation/subtraction of distinct power of 3
I have seen this in a riddle where you have to chose 4 weights to calculate any weight from 1 to 40kgs. Some examples, [imath]8 = {3}^{2} - {3}^{0}[/imath] [imath]12 = {3}^{2} + {3}^{1}[/imath] [imath]13 = {3}^{2} + {3}^{1}+ {3}^{0}[/imath] Later I found its also possible to use only 5 weights to calculate any weight between 1-121. [imath]100 = {3}^{4} + {3}^{3} - {3}^{2} + {3}^{0}[/imath] [imath]121 = {3}^{4} + {3}^{3} + {3}^{2} + {3}^{1} + {3}^{0}[/imath] Note: It allows negative numbers too. how I represent 8 and 100. I want to know if any natural number can be represented as a summation of power of 3. I know this is true for 2. But is it really true for 3? What about the other numbers? Say [imath]4, 5, 6, ... [/imath] |
423328 | If [imath]G[/imath] is a group, [imath]H[/imath] is a subgroup of [imath]G[/imath] and [imath]g\in G[/imath], is it possible that [imath]gHg^{-1} \subset H[/imath]?
If [imath]G[/imath] is a group, [imath]H[/imath] is a subgroup of [imath]G[/imath] and [imath]g\in G[/imath], is it possible that [imath]gHg^{-1} \subset H[/imath] ? This means, [imath]gHg^{-1}[/imath] is a proper subgroup of [imath]H[/imath]. We know that [imath]H \cong gHg^{-1}[/imath], so if [imath]H[/imath] is finite then we have a contradiction since the isomorphism between the two subgroups implies that they have the same order so [imath]gHg^{-1}[/imath] can't be proper subgroup of [imath]H[/imath]. So, what if [imath]H[/imath] is infinite is there an example for such [imath]G , H , g[/imath] ? Edit: I suppose that [imath]H[/imath] has a subgroup [imath]N[/imath] such that [imath]N[/imath] is a normal subgroup of [imath]G[/imath]. | 107862 | Conjugate subgroup strictly contained in the initial subgroup?
Let [imath]G[/imath] be a group, [imath]H\subseteq G[/imath] a subgroup and [imath]a\in G[/imath] an element of the group. Is it possible that [imath]aHa^{-1} \subset H[/imath], but [imath]aHa^{-1} \neq H[/imath]? If [imath]H[/imath] has finite index or finite order, this is not possible. |
423578 | Joining finite sequences
How do I describe the joining of two finite sequences in mathematical notation? For example, suppose the following: [imath] A=(a_i)_{i=1,2}=(4,2)\\ B=(b_i)_{i=1,2}=(9,5)\\ C=(c_i)_{i=1,...,4}=(4,2,9,5) [/imath] Sequence [imath]C[/imath] can be considered sequence [imath]A[/imath] with sequence [imath]B[/imath] attached to the end. How do I describe sequence [imath]C[/imath] in terms of sequence [imath]A[/imath] and [imath]B[/imath]? | 298648 | Is there a common symbol for concatenating two (finite) sequences?
Say we have two finite sequences [imath]X = (x_0,...,x_n)[/imath] and [imath]Y = (y_0,...,y_n)[/imath]. Is there a more or less common notation for the concatenation of these sequences, like [imath]\sum (X,Y) = (x_0,...,x_n,y_0,...,y_n)[/imath]? |
423765 | To show a set [imath](X\times Y)-(A\times B)[/imath] is connected .
Let [imath]A[/imath] be a proper subset of of [imath]X[/imath], [imath]B[/imath] be a proper subset of [imath]Y[/imath]. If [imath]X[/imath] and [imath]Y[/imath] are connected, show that [imath](X\times Y)-(A\times B)[/imath] is connected. | 44807 | If [imath]X[/imath] and [imath]Y[/imath] are connected, then [imath](X\times Y)\setminus(A\times B)[/imath] is connected for any proper subsets [imath]A,B[/imath]
I meet these two exercises: Q1: let [imath]A[/imath] be a proper subset of [imath]X[/imath], and [imath]B[/imath] be a proper subset of [imath]Y[/imath]. If [imath]X[/imath] and [imath]Y[/imath] are connected, show that [imath](X\times Y)\setminus(A\times B)[/imath] is connected. Q2: Let [imath]Y\subset X[/imath]. Assume that [imath]X[/imath] and [imath]Y[/imath] be connected. Show that if [imath]A[/imath] and [imath]B[/imath] form a separation of [imath]X\setminus Y[/imath], then [imath]Y\cup A[/imath] and [imath]Y\cup B[/imath] are connected. My attempt for Q2 I think to prove it by contradiction, assume [imath]Y\cup A[/imath] and [imath]Y\cup B[/imath] are not connected then for [imath]P[/imath] and [imath]Q[/imath] disjoint [imath]Y\cup A=P\coprod Q[/imath] and for [imath]M[/imath] and [imath]N[/imath] disjoint [imath]Y\cup B=M\coprod N[/imath] [imath](Y\cup A)\cup (Y\cup B)=(P\coprod Q)\cup (M\coprod N)[/imath] The left side will give [imath]X[/imath],and the right side can be written as a disjoint union , this contradicts the fact that [imath]X[/imath] is connected, so [imath]Y\cup A[/imath] and [imath]Y\cup B[/imath] must be connected. I need help for Q1. |
423874 | Do negative dimensions make sense?
Some time ago I read in a popular physics book that in M-theory, there are some "things" which can be said to have dimension [imath]-1[/imath]. Probably, the author was vastly exaggerating, but this left me wondering: Are there mathematical theories which contain a notion that can be regarded as some sort of generalization of the classical notion of dimensionality and which allows negative values? For example the Hausdorff dimension can assume, as far as I know, only nonnegative real values. I think that stable homotopy groups can be defined for arbitrary integer dimensions, but this doesn't really count, since one is not dealing with negative dimensional objects per se. | 100883 | Has the notion of having a complex amount of dimensions ever been described? And what about negative dimensionality?
The notion of having a number [imath]a \in \mathbb{R}_{\geq 0} [/imath] associated to any metric space is described by the definition of a "Hausdorff Dimension". I was wondering if work has been done on spaces which (seem to) have a complex amount dimensions associated with it? Does this concept exist? If so, when is it useful, if at all? Inspired by the comments, I am also interested as to whether the concept of negative dimensionality has been explored already. Thanks in advance. |
423940 | Distance function on metric space is continuous
In showing that the diameter of a compact set [imath]A[/imath] is attainable, one approach is to consider a function [imath]f:A\times A\rightarrow\mathbb{R}[/imath] such that [imath]f(x,y)=d(x,y)[/imath]. The key is to show that the function [imath]f[/imath] is continuous on [imath]A[/imath]. I proved the continuity by using the definition: that for [imath]\delta=\epsilon[/imath], if [imath]d(a_1,b_1)+d(a_2,b_2)<\delta[/imath], then [imath]|d(a_1,a_2)-d(b_1,b_2)|<\epsilon[/imath]. This follows from repeated applications of the triangle inequality: [imath]d(a_1,a_2)-d(b_1,b_2)<(d(a_1,b_1)+d(b_1,a_2))-(d(b_1,a_2)-d(a_2,b_2)) = d(a_1,b_1)+d(a_2,b_2)<\epsilon[/imath]. Similarly, [imath]d(b_1,b_2)-d(a_1,a_2)<\epsilon[/imath], and hence [imath]|d(a_1,a_2)-d(b_1,b_2)|<\epsilon[/imath]. Are there other ways to see that [imath]f[/imath] is continuous? (I was hoping there's perhaps a simpler way...) | 315306 | show that a distance function is continuous
Let [imath]X[/imath] be a metric space with metric [imath]d[/imath]. Define [imath]d: X \times X \to \mathbb{R}[/imath], show that [imath]d[/imath] is continuous. I would like to show that the function is continuous the topology way (since it is a course on topology). Let [imath](a,b)[/imath] be a basic open set in [imath]\mathbb{R}[/imath], then [imath]d^{-1}(a,b) = \{(x,y): a<d(x,y)<b\}[/imath]. Define [imath]A= \bigcup_{x \in X} B_{b} (x)[/imath], where [imath]B_b (x) = \{y \in X| d(x,y)<b[/imath]}. Clearly [imath]A[/imath] is open. Now define [imath]C = \{(x,y):d(x,y)>a \}.[/imath] Essentially I hope that [imath]A \cap C[/imath] is open. But I am stuck in showing that [imath]C[/imath] is open. Any help please? :) |
424031 | Is this fact about matrices and linear systems true?
Let [imath]A[/imath] be a [imath]m[/imath]-by-[imath]n[/imath] matrix and [imath]B=A^TA[/imath]. If the columns of [imath]A[/imath] are linearly independent, then [imath]Bx=0[/imath] has a unique solution. If is true, can you help me prove it? If is false, could you give a counterexample? Thanks. | 424092 | Can we conclude that this matrix is definite positive?
Let [imath]A[/imath] be a [imath]n\text{-by-}m[/imath] matrix. Suppose that columns of [imath]A[/imath] are linearly independent. Can we conclude that [imath]A^TA[/imath] is definite positive? Could you help me with proof? Thanks. |
296035 | Martingale inequality related to Kolmogorov's maximal inequality
The problem This is a homework question from Durrett's Probability: Theory and Examples. Hints would be appreciated. Let [imath]\xi_i[/imath] be independent with expected value zero and variances [imath]\sigma^2_i < \infty[/imath] (not necessarily all identical). Let [imath]S_m = \xi_1 + \dots + \xi_m[/imath]. From Doob's inequality we can obtain Kolmogorov's maximal inequality, [imath]P\left(\max_{1 \leq m \leq n} |S_m| \geq x\right) \leq x^{-2} \operatorname{var} S_n.[/imath] I feel like I understand this derivation, which is given in the book. Now the homework problem. Everything is as defined above, except we additionally impose the constraint [imath]|\xi_i| \leq K\; a.s.[/imath] I am asked to prove the following inequality: [imath]P\left(\max_{1 \leq m \leq n} |S_m| \leq x\right) \leq \frac{(x+K)^2}{ \operatorname{var} S_n}.[/imath] This differs from Doob's inequality in three ways: The inequality inside the probability is switched [imath]x[/imath] is replaced by [imath]x+K[/imath], which is the highest we can go if we stop as soon as we exceed [imath]x[/imath]. The fraction on the right hand side is inverted. I am told to use the fact that for a submartingale [imath]X_m[/imath] and a stopping time [imath]N[/imath] which is bounded by [imath]n \; a.s.[/imath], [imath]E X_0 \leq E X_N \leq E X_n.[/imath] There is also a hint to consider the martingale given by [imath]X_m = S_m^2 - \operatorname{var} S_m[/imath]. What I tried Following the derivation of Doob's inequality from the submartingale inequality given above, I defined the set [imath]A = \{\max_{i \leq m \leq n} X_m \leq x^2 - \operatorname{var}S_n\} \supset \{\max |S_m| \leq x\}[/imath] and the stopping time [imath]N = \inf_m \{X_m \geq x^2 - \operatorname{var}{S_m}\text{ or }m = n\},[/imath] as well as several variations on the theme. In all my attempts, though, I'm unable to get an inverted fraction (realtive to Kolmogorov's maximal inequality) on the right-hand-side of the inequality. Has anyone seen this? Any guidance? | 313415 | Martingale and bounded stopping time
A theorem of submartingale and bounded stopping time says: Theorem 5.4.1. If [imath]X_n[/imath] is a submartingale and [imath]N[/imath] is a stopping time with [imath]\mathbb P (N \le k) = 1[/imath] then [imath]\mathbb EX_0 ≤ \mathbb EX_N ≤ \mathbb EX_k[/imath]. An exercise for this theorem is Example 5.4.1. Random walks. If we let [imath]S_n = \xi_1 + · · · + \xi_n[/imath] where the [imath]ξ_m[/imath] are independent and have [imath]\mathbb E \xi_m = 0[/imath], [imath]\sigma_m^2 = \mathbb E \xi_m^2 < \infty[/imath]. Suppose we have that [imath]|\xi_m | \le K[/imath] and let [imath]s^2_n = \sum_{m \le n} \sigma^2_m[/imath]. Note that [imath]S_n^2 − s^2_n[/imath] is a martingale. Use this fact and Theorem 5.4.1 to conclude [imath]\mathbb P \left(\max_{1 \le m \le n} |S_m| ≤ x \right) ≤ (x + K)^2/ \mathbb E(S_n^2)[/imath] Let [imath]A = \{\max_{1 \le m \le n} |S_m| ≤ x\}[/imath]. Let [imath]X_n = S^2_n - s^2_n[/imath]. Let [imath]N = \inf\{m:|S_m| \ge x~\text{or}~n+1\}[/imath]. So [imath]N[/imath] is a bounded stopping time. Thus by the previous theorem we have [imath] 0 = \mathbb E{X_1} = \mathbb E{X_N} = \mathbb E{X_{n+1}}. [/imath] Since [imath]X_{n+1} = X_N[/imath] on [imath]A^c[/imath], we have [imath]\mathbb E (X_{n+1} 1_A) = \mathbb E (X_N 1_A)[/imath]. Therefore, as long as we have [imath]\mathbb E (X_N 1_A) \ge 0[/imath], (which I don't know how to prove), we have [imath]\mathbb E (X_{n+1} 1_A) \ge 0[/imath]. It follows that [imath] \mathbb E(s_n^2 1_A) \le \mathbb E(s_{n+1}^2 1_A) \le \mathbb E(S_{n+1}^2 1_A) \le (x + K)^2. [/imath] But I have problem to show that [imath]\mathbb E(X_N 1_A) \ge 0[/imath]. Am I on the right direction? |
424532 | Maximum value of the vandermonde polynomial
Let [imath]n[/imath] be an integer greater than 1. How must [imath]n[/imath] numbers [imath]a_i[/imath] in the interval [imath][0,1][/imath] be chosen that the vandermonde polynomial [imath]\prod_{i\ne j} (a_i-a_j) [/imath] has the maximum possible value ? My conjecture, that the numbers must be equidistant, seems to be false due to numerical analysis. | 15366 | Spreading points in the unit interval to maximize the product of pairwise distances
This is prompted by question 15312, but moved to the reals. It must be solved already. Pick n points [imath]x_i \in [0,1][/imath] to maximize [imath]\prod_{i < j} (x_i - x_j)[/imath]. A little playing shows you don't want them evenly distributed-they need to push out to the ends. With four points, Alpha says to use [imath]\{0,\frac{1}{2}\pm\frac{1}{2\sqrt{5}},1\}[/imath] and with five, [imath]\{0,\frac{1}{2}-\frac{\sqrt{\frac{3}{7}}}{2},\frac{1}{2},\frac{1}{2}+\frac{\sqrt{\frac{3}{7}}}{2},1\}[/imath] |
425323 | Why do we use 'this' Gamma Function.
The Gamma function is a generalization of the factorial defined by Euler as: [imath]\Gamma(z)=\int\limits_{0}^{\infty}t^{z-1}e^{-t}\,dt[/imath] for [imath]z\in\mathbb{C}[/imath] with positive real part. It satisfies, [imath]\Gamma(n)=(n-1)![/imath] for all [imath]n\in\mathbb{N}[/imath]. My question is why we choose this particular function from all the functions that satisfy the previous property. And, how is the Gamma function deduced? | 1537 | Why is Euler's Gamma function the "best" extension of the factorial function to the reals?
There are lots (an infinitude) of smooth functions that coincide with f(n)=n! on the integers. Is there a simple reason why Euler's Gamma function [imath]\Gamma (z) = \int_0^\infty t^{z-1} e^{-t} dt[/imath] is the "best"? In particular, I'm looking for reasons that I can explain to first-year calculus students. |
56523 | rank of the free group: [imath]\mathrm{rank} F_X=|X|[/imath]
This question is an extension of this one. Let [imath]F_X[/imath] denote the free group on the set [imath]X[/imath]. For any group [imath]G[/imath] and subset [imath]S\!\subseteq\!G[/imath], [imath]\langle S\rangle[/imath] denotes the subgroup generated by [imath]S[/imath] and [imath]\mathrm{rank}(G) :=\min\{|S|;\:S\!\subseteq\!G, \langle S\rangle\!=\!G\}[/imath]. PROPOSITION: a) [imath]F_X\cong F_Y\:\Leftrightarrow\:|X|=|Y|[/imath] b) [imath]\mathrm{rank}(F_X)=|X|[/imath] Thus for every cardinal number [imath]c[/imath], there is (up to isomorphism) exactly one free group of rank [imath]c[/imath]. Proof: a) [imath](\Leftarrow)[/imath]: If [imath]f\!:X\rightarrow Y[/imath] is the bijection, then [imath]\varphi(x_1\ldots x_k):=f(x_1)\ldots f(x_k)[/imath] is the isomorphism. [imath](\Rightarrow)[/imath]: [imath]F_X\!\cong\!F_Y[/imath] [imath]\Rightarrow[/imath] [imath]\mathrm{Ab} F_X\!\cong\!\mathrm{Ab} F_Y[/imath] [imath]\Rightarrow[/imath] [imath]\oplus_{x\in X}\mathbb{Z}\!\cong\!\oplus_{y\in Y}\mathbb{Z}[/imath], so [imath]|X|\!=\!|Y|[/imath], since rank is known to be an invariant of free modules. Alternatively, [imath]\big(\oplus_{x\in X}\mathbb{Z}\big)\otimes_\mathbb{Z}\mathbb{Q}[/imath] [imath]\cong[/imath] [imath]\oplus_{x\in X}\big(\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q}\big)[/imath] [imath]\cong[/imath] [imath]\oplus_{x\in X}\mathbb{Q}[/imath], so [imath]\oplus_{x\in X}\mathbb{Q}[/imath] [imath]\cong[/imath] [imath]\oplus_{y\in Y}\mathbb{Q}[/imath], even as [imath]\mathbb{Q}[/imath]-modules, but isomorphic vector spaces are known to have equipollent bases. b) Since [imath]\langle X\rangle\!=\!F_X[/imath], [imath]\mathrm{rank}(F_X)\leq|X|[/imath]. Suppose we have [imath]Y\!\subseteq\!F_X[/imath], [imath]\langle Y\rangle\!=\!F_X[/imath], [imath]|Y|\!<\!|X|[/imath]. QUESTION: how can I finish the proof of b), i.e. prove that [imath]F_X[/imath] can't be generated by a subset with smaller cardinality than [imath]|X|[/imath]? | 936801 | Generating sets of the free group [imath]F_k[/imath] on [imath]k[/imath] generators
Is it true that the free group [imath]F_k[/imath] on [imath]k<\infty[/imath] generators requires at least [imath]k[/imath] elements to generate. I.e. does every set which generates [imath]F_k[/imath] have cardinality at least [imath]k[/imath]? |
131475 | Why is the radical of a Clifford algebra generated by the kernel of the associated symmetric form?
I was recently reading through Jacobson's Basic Algebra. I got to the section on Clifford algebras, and have the following question. Let [imath]Cl_\omega[/imath] be the Clifford algebra with bilinear symmetric form [imath]\omega[/imath] on a vector space [imath]V[/imath]. I hear that the Jacobson radical [imath]\text{rad}(Cl_\omega)[/imath] is generated by [imath]\ker\omega[/imath], but it's not immediately clear to my why it is. How can one see this? Thanks. | 131953 | Quotient of a Clifford algebra by its radical is a Clifford algebra?
I'm fumbling a bit in my reading on Clifford algebras. I'm hoping someone can shed some light on the following isomorphism. Suppose you have a symmetric bilinear form [imath]G[/imath] over a vector space [imath]V[/imath], and let [imath]\mathrm{Cl}_G(V)[/imath] be the corresponding Clifford algebra. I'll denote it by [imath]C_G[/imath] for short when the vector space is clear. Now [imath]G[/imath] induces a form [imath]\hat{G}[/imath] on [imath]V/\ker(G)[/imath], and apparently [imath]C_G/\mathrm{rad}(C_G)\cong C_{\hat{G}}[/imath], where [imath]\mathrm{rad}(C_G)[/imath] is the radical of [imath]C_G[/imath]. I thought this will fall out easily from some application of the isomorphism theorems, but some epimorphism [imath]C_G\to C_{\hat{G}}[/imath] whose kernel conveniently happens to be [imath]\mathrm{rad}(C_G)[/imath]. However, I can't quite find such a map. Does anyone see what the trick is here? Thanks. Later. I believe I now understand that there is a surjective map [imath]p:C(\beta)\to C(\bar{\beta})[/imath] induced by the quotient map [imath]V\to V/\ker\beta[/imath] described below. If [imath]I[/imath] is the ideal generated by [imath]\ker\beta[/imath], then since [imath]p(\ker\beta)=0[/imath] in [imath]C(\bar{\beta})[/imath], it follows that [imath]I\subset\ker p[/imath]. However, I don't understand why [imath]I[/imath] is nilpotent. What is the explanation for this last bit? |
425699 | Prove that [imath] \bigcap _{i=1}X_i[/imath] is connected.
Let [imath]X[/imath] be a compact Hausdorff space and let [imath]X_1 \subset X_2 \subset X_3 \subset \cdots[/imath] be a sequence of closed, connected subspaces. Prove that [imath]\bigcap_{i=1}^\infty X_i[/imath] is connected. Give an example showing that the compactness of [imath]X[/imath] is necessary | 420401 | Does the limit of a descending sequence of connected sets still connected?
Given a descending sequence of sets [imath] F_1\supset F_2\supset\cdots F_n\supset\cdots [/imath] in which each [imath]F_i[/imath] is connected. I wonder if the limit set [imath] F=\bigcap_{i=1}^\infty F_i [/imath] is still connected? I believe it is, but cannot make a proof. Anyone can help? Updated: Samuel has showed a counter example. Thus now I wonder can I add some constraints such that the conclusion holds? I ask this problem because when I look up the The Princeton Companion to Mathematics，chapter IV.14. Dynamics, section 2.8 The Mandelbrot Set M, there is the following words: It follows from the above that as [imath]t[/imath] approaches zero, the equipotential of potential [imath]t[/imath], together with its interior, gets closer and closer to M: that is, M is the intersection of all such sets. Hence, M is a connected, closed, bounded subset of the plane. I wonder why such argument shows [imath]M[/imath], the Mandelbrot set, is connected. |
425645 | Let [imath]f:[a,b]\to\mathbb R[/imath]. Evaluate [imath]\lim_{n\to\infty}\int_a^bf(x)\sin(nx)\,dx[/imath]
Let [imath]f:[a,b]\to\mathbb R[/imath]. Evaluate [imath]\lim_{n\to\infty}\int_a^bf(x)\sin(nx)\,dx[/imath]. [imath]f[/imath] is continuously differentiable. I'm told this can be done using basic calculus. It's difficult for me to see where I should begin. I'd like some hints. | 387060 | How to prove that [imath]\lim\limits_{n\to\infty}\int\limits _{a}^{b}\sin\left(nt\right)f\left(t\right)dt=0\text { ? }[/imath]
Let [imath]f:\left[a,b\right]\to\mathbb{R}[/imath] be a function that is derivative so that [imath]f'[/imath] is continuous then [imath] \lim_{n\to\infty}\int\limits _{a}^{b}\sin\left(nt\right)f\left(t\right)dt=0 [/imath] My attempt: I want to show that [imath]\forall\epsilon>0 \ \exists \ n_0 \in \mathbb{N}[/imath] so that [imath]\forall n>n_0 \ \ \ \left|\int\limits _{a}^{b}\sin\left(nt\right)f\left(t\right)dt\right|<\epsilon[/imath]. Let there be [imath]\epsilon>0[/imath] because [imath]f[/imath] and [imath]\sin[/imath] are derivative and [imath]f'[/imath] is continuous we can use integration by parts so that [imath] \left|\int\limits _{a}^{b}f'\left(t\right)\sin\left(nt\right)dt\right|=\left|\left[f\left(t\right)\sin\left(nt\right)\right]_{a}^{b}-n\int\limits _{a}^{b}f\left(t\right)\cos\left(nt\right)\right|\leq\left|\left[f\left(t\right)\sin\left(nt\right)\right]_{a}^{b}\right|+\left|n\int\limits _{a}^{b}f\left(t\right)\cos\left(nt\right)\right| [/imath] That is my main idea. any suggestions? |