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113705 | The only 1-manifolds are [imath]\mathbb R[/imath] and [imath]S^1[/imath]
I recall having heard somewhere that the only 1-manifolds (second countable, Hausdorff, connected spaces locally homeomorphic to [imath]\mathbb R[/imath]) are [imath]\mathbb R[/imath] and [imath]S^1[/imath]. Is this true? If so, is there a reasonably elementary proof? | 1304107 | Characterization of 1-dimensional manifolds.
My intuition tells me that the only connected 1-dimensional topological manifolds are the real line [imath]\mathbb{R}[/imath] and the circle [imath]S^1[/imath]. Is this true? If yes, is it possible to prove it from first principles, or is it something that needs some highly technical theorems? If no, do we have an example of connected 1-dimensional manifold not homeomorphic to [imath]\mathbb{R}[/imath] or [imath]S^1[/imath]? |
548 | Explanation of method for showing that [imath]\frac{0}{0}[/imath] is undefined
(This was asked due to the comments and downvotes on this Stackoverflow answer. I am not that good at maths, so was wondering if I had made any basic mistakes) Ignoring limits, I would like to know if this is a valid explanation for why [imath]\frac00[/imath] is undefined: [imath]x = \frac00[/imath] [imath]x \cdot 0 = 0[/imath] Hence There are an infinite number of values for [imath]x[/imath] as anything multiplied by [imath]0[/imath] is [imath]0[/imath]. However, it seems to have got comments, with two general themes. Once is that you lose the values of [imath]x[/imath] by multiplying by [imath]0[/imath]. The other is that the last line is: [imath]x \cdot 0 = \frac00 \cdot 0[/imath] as it involves a division by [imath]0[/imath]. Is there any merit to either argument? More to the point, are there any major flaws in my explanation and is there a better way of showing why [imath]\frac00[/imath] is undefined? | 792996 | Why does [imath]0/0[/imath] have to be undefined?
Why can't it no be [imath]\pm[/imath] Infinity? If [imath]x/1[/imath] is [imath]x[/imath] then [imath]x/0[/imath] should be [imath]\pm[/imath] Infinity. |
6942 | Finding an angle within an 80-80-20 isosceles triangle
The following is a geometry puzzle from a math school book. Even though it has been a long time since I finished school, I remember this puzzle quite well, and I don't have a nice solution to it. So here is the puzzle: The triangle [imath]ABC[/imath] is known to be isosceles, that is, [imath]AC=BC[/imath]. The labelled angles are known to be [imath]\alpha=\gamma=20°[/imath], [imath]\beta=30°[/imath]. The task is to find the angle labelled "?". The only solution that I know of is to use the sine formula and cosine formula several times. From this one can obtain a numerical solution. Moreover this number can be algebraically shown to be correct (all sines and cosines are contained in the real subfield of the 36th cyclotomic field). So in this sense I solved the problem, but the solution is kind of a brute force attack (for example, some of the polynomials that show up in the computation have coefficients > 1000000). Since the puzzle originates from a book that deals only with elemetary geometry (and not even trigonometry if I remember correctly) there has to be a more elegant solution. | 548292 | Determinating the angle in a triangle
Determine the value of the angle [imath]\alpha[/imath] of the figure below. |
58097 | Proof that [imath]C\exp(x)[/imath] is the only set of functions for which [imath]f(x) = f'(x)[/imath]
I was wondering the following. And I probably know the answer already: NO. Is there another number with similar properties as [imath]e[/imath]? So that the derivative of [imath]\exp(x)[/imath] is the same as the function itself. I can guess that it's probably not, because otherwise [imath]e[/imath] wouldn't be that special, but is there a proof of it? | 376330 | How to prove that if [imath] f(x) = f'(x) \Rightarrow f(x) = c e^x[/imath]
In calculus, If I know that a derivative of a function is the function itself, how do I prove that the function equals [imath]e^x[/imath]: [imath] f(x) = f'(x) \Rightarrow f(x) = ce^x [/imath] Why is this always true? Thank you |
299076 | If a holomorphic function [imath]f[/imath] on the punctured unit disc satisfies that for some fixed [imath]N[/imath] [imath]f(z)=w[/imath] has at most [imath]N[/imath] solutions for any [imath]w[/imath]
Show that [imath]0[/imath] is not an essential singularity. This is a homework question. I know that there is a theorem saying that if [imath]f[/imath] has an essential singularity [imath]a[/imath], then for all but two [imath]w[/imath] in [imath]\mathbb{C}[/imath], in any neighborhood of [imath]a[/imath], [imath]f[/imath] attains the value [imath]w[/imath]. Using this theorem the question is quite simple, but I am not sure if I am meant to quote this theorem, because it was only mentioned in my lectures without a proof. I wonder if this can be shown without using the theorem I quoted. | 287683 | Proof that a certain entire function is a polynomial
Let [imath]n\in\mathbf{N}[/imath] be fixed, and [imath]f[/imath] entire and [imath]|f^{-1}(\left\lbrace w\right\rbrace)|\leq n[/imath] for every [imath]w\in\mathbf{C}[/imath]. Then [imath]f[/imath] is a polynomial of degree at most [imath]n[/imath]. I try to prove this statement, and I think one can prove it as follows: consider [imath]f(1/z)[/imath]. [imath]0[/imath] cannot be an essential singularity of [imath]f(1/z)[/imath], for the big Picard theorem would imply that on any neighborhood of [imath]0[/imath] [imath]f(1/z)[/imath] takes on all possible complex values (with at most one exception) infinitely often, but this is contrary to [imath]|f^{-1}(\left\lbrace w\right\rbrace)|\leq n[/imath]. Then [imath]f(1/z)[/imath] has a pole of order [imath]k[/imath] say, and since [imath]f[/imath] is holomorphic it is a polynomial of degree [imath]k[/imath] (for its principal part vanishes). By the fundamental theorem of algebra, [imath]k\leq n[/imath]. Is it possible here to avoid using Picard's theorem? |
10616 | Order of elements is lcm-closed in abelian groups
How can I prove that if [imath]G[/imath] is an Abelian group with elements [imath]a[/imath] and [imath]b[/imath] with orders [imath]m[/imath] and [imath]n[/imath], respectively, then [imath]G[/imath] contains an element whose order is the least common multiple of [imath]m[/imath] and [imath]n[/imath]? It's an exercise from Hungerford's book, but it's not homework. I could not solve it when I take the course on groups and I think it should be easy. There is a hint that says to first prove when [imath]m[/imath] and [imath]n[/imath] are coprimes. I did this part. But I have no idea how to solve the general case. Thanks. | 255894 | Abelian Group Element Orders
I want to show that if a finite abelian group has elements of order [imath]m[/imath] and [imath]n[/imath] then it will have an element of order [imath]\text{lcm}(m,n)[/imath]. First I proved the lemma if [imath]a[/imath] has order [imath]m[/imath] and [imath]b[/imath] has order [imath]n[/imath] with [imath]m,n[/imath] coprime, then [imath]ab[/imath] has order [imath]mn[/imath]. This is because the subgroups [imath]\langle a \rangle[/imath] and [imath]\langle b \rangle[/imath] have trivial intersection (because the order of their intersection must divide their orders [imath]m[/imath] and [imath]n[/imath], which is a consequence of it being a subgroup of both), which implies [imath]a^{i} \in \langle b \rangle[/imath] iff [imath]i \equiv 0 \pmod n[/imath], and similarly [imath]b^{i} \in \langle a \rangle[/imath] iff [imath]i \equiv 0 \pmod m[/imath]. Using that we deduce that [imath](ab)^i = a^i b^i = 1[/imath] iff [imath]i \equiv 0 \pmod {mn}[/imath]. So if [imath]a[/imath] was an element of order [imath]m[/imath] and [imath]b[/imath] and element of [imath]n[/imath] with [imath]g = \gcd(m,n) \not = 1[/imath] I thought that [imath]\text{lcm}(m,n) = \frac{m}{g}n[/imath] so and [imath]\frac{m}{g},n[/imath] are coprime so I should construct an element (from [imath]a[/imath]) with order [imath]\frac{m}{g}[/imath] then conclude the theorem by the lemma. For the construction I think it's just [imath]a^g[/imath]. I just have a nagging doubt about the correctness of the second proof, did I miss some important detail? Also if there are any neater ways to prove this (which don't depend on the structure theorem) I would like to learn them too. |
151700 | if [imath]f[/imath] is entire and [imath]|f(z)| \leq 1+|z|^{1/2}[/imath], why must [imath]f[/imath] be constant?
How can we prove that if [imath]f:\mathbb{C}\rightarrow\mathbb{C}[/imath] is holomorphic (analytic) and [imath]|f(z)| \leq 1+|z|^{1/2} \forall z[/imath], then [imath]f[/imath] is constant? Liouville's theorem springs to mind, but I can't see how to use it since [imath]1+|z|^{1/2}[/imath] is not holomorphic. The maximum modulus principle doesn't seem easily usable either. And the principle of isolated zeroes can't really be applied since all we know is an inequality, not an equation. Many thanks for any help with this! | 367954 | Proving an entire function is bounded
I'm having trouble with proving the following: Let [imath]f(z)[/imath] be an entire function such that for all [imath]z[/imath], [imath]|f(z)| \leq \sqrt{|z|}[/imath]. Show that [imath]f(z) = 0[/imath] for all [imath]z[/imath]. I know you need Cauchy's formula and take the first derivative for [imath]f[/imath] and then prove if we take [imath]R[/imath] sufficiently large then we have a function that is constant or equal to [imath]0[/imath]. But I'm sure how to get to that stage with the above entire function. |
69532 | Partial sum of rows of Pascal's triangle
I'm interested in finding [imath]\sum_{k=0}^m \binom{n}{k}, \quad m<n[/imath] which form rows of Pascal's triangle. Surely [imath]\sum\limits_{k=0}^n \binom{k}{m}[/imath] using addition formula, but the one above involves hypergeometric functions and I don't know how to approach it. EDIT: if possible, please don't solve it, just a few hints will do. | 596397 | Find the sums of the series (combinatorial series)
I would appreciate if somebody could help me with the following problem Q: Find ? [imath](n,k\in\mathbb{N},k\leq n)[/imath] [imath]\binom{n}{0} +\binom{n}{1}+\binom{n}{2}+\ldots+\binom{n}{k}=\,?[/imath] |
11849 | [imath](W_1+W_2+\cdots+W_n)^a \leq W_1^a +\cdots + W_n^a[/imath] for [imath]n[/imath] integer, [imath]n\geq 2[/imath], [imath]W\gt 0[/imath] and [imath]a[/imath] constant, real, [imath]0\lt a\lt 1[/imath]
I am looking for a proof that this inequality: [imath](W_1+W_2+\cdots+W_n)^a \leq W_1^a +\cdots + W_n^a[/imath] is valid. I have a power function [imath]f(W)=W^a[/imath] where [imath]a[/imath] is a real number, constant but usually [imath]0\lt a\lt 1[/imath]. The [imath]W_i[/imath] are real positive numbers. Then I want to check if [imath]\left(\sum W_i\right)^a \leq \sum(W_i^a)[/imath] For example for [imath]n=2[/imath] : [imath](x+y)^a \leq x^a + y^a[/imath]. Thanks. | 1183435 | How to prove inequality similar to that of Minkowski
I am struggling with this problem. How to prove the below inequality [imath]|x+y|^r[/imath] [imath]\leq[/imath] [imath]|x|^r+|y|^r[/imath] for [imath]0<r\leq1[/imath]. A little guidance please. |
298575 | In direct products of [imath]n[/imath] groups, do we also prove conditions for a group to be abelian or not?
If we let [imath]G_1,...,G_n[/imath] be groups, When proving that the direct product [imath]G_1 \times .... \times G_n[/imath] is abelian if and only if each of [imath]G_1,...,G_n[/imath] is abelian, can someone please help me Im concerned about whether it should also prove that it holds for the conditions of a group to be abelian(inverse, unit element ...) or just prove straightforward that left hand side is true iff right hand side is? Can someone please help clarify this. Thanks | 292004 | Proving that Direct product [imath]G_1 \times ... \times G_n[/imath] is abelian if and only if each of [imath]G_1, ...,G_n[/imath] is abelian
If we let [imath]G_1,...,G_n[/imath] be groups, how can we prove that the direct product [imath]G_1 \times .... \times G_n[/imath] is abelian if and only if each of [imath]G_1,...,G_n[/imath] is abelian. Please if someone can help and guide with this question... |
302511 | Funcional Equations:I'm confused
I need help with this: "Find functions [imath]f[/imath], [imath]g : \mathbb{Z} \rightarrow \mathbb{Z}[/imath], knowing that [imath]g[/imath] is injective and such that: [imath]f(g(x)+y) = g(f(x)+x), \mbox{ for all } x, y \in \mathbb{Z}.[/imath] Or : [imath]f(g(x)+y) = g(f(y)+x), \mbox{ for all } x, y \in \mathbb{Z}.[/imath] | 302354 | What's the solution of the functional equation
I need help with this: "Find all functions [imath]f[/imath], [imath]g : \mathbb{Z} \rightarrow \mathbb{Z}[/imath], with [imath]g[/imath] injective and such that: [imath]f(g(x)+y) = g(f(x)+y), \mbox{ for all } x, y \in \mathbb{Z}.[/imath] |
440 | Why isn't reflexivity redundant in the definition of equivalence relation?
An equivalence relation is defined by three properties: reflexivity, symmetry and transitivity. Doesn't symmetry and transitivity implies reflexivity? Consider the following argument. For any [imath]a[/imath] and [imath]b[/imath], [imath]a R b[/imath] implies [imath]b R a[/imath] by symmetry. Using transitivity, we have [imath]a R a[/imath]. Source: Exercise 8.46, P195 of Mathematical Proofs, 2nd (not 3rd) ed. by Chartrand et al | 1129605 | There is relation that is symmetric and transitive but not reflexive?
Let [imath]L=\left\{R\right\}[/imath] be a language with only one relation symbol. Consider these formulas: [imath]\Psi _1\:=\:\forall x\left(R\left(x,x\right)\right)[/imath] [imath]\Psi _2\:=\:\forall x\forall y\left(R\left(x,y\right)\rightarrow R\left(y,x\right)\right)\:[/imath] [imath]\Psi _3\:=\:\forall x\forall y\forall z\left(R\left(x,y\right)\wedge R\left(y,z\right)\rightarrow R\left(x,z\right)\right)\:[/imath] I need to show [imath]Mod\left\{\Psi _1,\Psi _2,\Psi _3\right\}\ne Mod\left\{\Psi \:_2,\Psi \:_3\right\}[/imath]. So far, i understand the meaning of each formula, but i wonder which relation on any domain, can show that [imath]R[/imath] is symmetric and transitive but not reflexive. Any ideas? EDITED: Well, i looked at If a relation is symmetric and transitive, will it be reflexive? and it really helpfull! but i want to find other examples just from curiosity .. |
148679 | Why aren't the graphs of [imath]\sin(\arcsin x)[/imath] and [imath]\arcsin(\sin x)[/imath] the same?
(source for above graph) (source for above graph) Both functions simplify to x, but why aren't the graphs the same? | 2151364 | Difference between [imath]\sin(\arcsin(x))[/imath] and [imath] \arcsin(\sin x)[/imath]
What's the difference between [imath]\sin(\arcsin(x))[/imath] and [imath] \arcsin(\sin x)[/imath] ? I think these are equal each other and equals to [imath]x[/imath] . Also , what's the plot of these functions ? I'm really confused about inverse function and its concept that is appearing in many situations and problems like this. |
676 | Probability that a stick randomly broken in two places can form a triangle
Randomly break a stick (or a piece of dry spaghetti, etc.) in two places, forming three pieces. The probability that these three pieces can form a triangle is [imath]\frac14[/imath] (coordinatize the stick form [imath]0[/imath] to [imath]1[/imath], call the breaking points [imath]x[/imath] and [imath]y[/imath], consider the unit square of the coordinate plane, shade the areas that satisfy the triangle inequality edit: see comments on the question, below, for a better explanation of this). The other day in class*, my professor was demonstrating how to do a Monte Carlo simulation of this problem on a calculator and wrote a program that, for each trial did the following: Pick a random number [imath]x[/imath] between [imath]0[/imath] and [imath]1[/imath]. This is the first side length. Pick a random number [imath]y[/imath] between [imath]0[/imath] and [imath]1 - x[/imath] (the remaning part of the stick). This is the second side length. The third side length is [imath]1 - x - y[/imath]. Test if the three side lengths satisfy the triangle inequality (in all three permutations). He ran around [imath]1000[/imath] trials and was getting [imath]0.19[/imath], which he said was probably just random-chance error off [imath]0.25[/imath], but every time the program was run, no matter who's calculator we used, the result was around [imath]0.19[/imath]. What's wrong with the simulation method? What is the theoretical answer to the problem actually being simulated? (* the other day was more than [imath]10[/imath] years ago) | 865429 | Probability of their's constituting a triangle?
We have a line segments with length [imath]l[/imath] then you choose two random points and cut it from these points so that we have three piece of line segments. What is the probability that these piece constitute a triangle ? We need to use triangle inequality but I could not manage it. Thanks. |
94233 | An integrable and periodic function [imath]f(x)[/imath] satisfies [imath]\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx[/imath].
I want to prove: For an integrable function [imath]f(x)[/imath] and periodic with period [imath]T[/imath], for every [imath]a \in \mathbb{R}[/imath], [imath]\int_{0}^{T}f(x)\;dx=\int_{a}^{a+T}f(x)\;dx.[/imath] I tried to change the values and define [imath]y=a+x[/imath] so that [imath]dy=dx[/imath] and the limits of the integrals are as we want, but I'm not sure how to use the fact that [imath]f(x)[/imath] is periodic. Thanks a lot! | 383616 | integration of a periodic function by substitution
Given a periodic function [imath]f[/imath] with period [imath]L[/imath] I want to prove the following: [imath]\int_a^{a+L}f(x) \,dx=\int_0^Lf(x)\,dx[/imath] I've found some nice proofs here but I have tried by my own doing the following: [imath]\begin{align*}\int_a^{a+L}f(x)dx&=\int_a^Lf(x)dx+\int_L^{a+L}f(x)dx\\ &=\int_a^Lf(x)dx+\int_0^af(y)dy\\ &=\int_0^Lf(x)dx \end{align*}[/imath] by using the substitution [imath]y=x-L[/imath]. So I was wondering if this is enough? Because all the proofs there say that this only holds for [imath]a\in[0,L)[/imath] but not for [imath]a[/imath] in general. So my question is why is this not enough and only holds for [imath]a\in[0,L)[/imath] but not [imath]a[/imath] in general? |
180744 | Compute [imath]\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx[/imath]
I'm having trouble computing the integral: [imath]\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.[/imath] I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as [imath]u=\sin(x)[/imath] and [imath]u=\cos(x)[/imath], but it was not very effective. Any suggestions are welcome. Thanks. | 298461 | Evaluate [imath]\int\frac{\cos x}{\sin x + \cos x}\,\text{d}x[/imath].
I'm trying to figure out how to take this indefinite integral: [imath] \int\frac{\cos x}{\sin x + \cos x}\,\text{d}x.[/imath] I tried simplifying and rearranging it, and this is the best I got: [imath]\int\frac{1}{\tan x + 1 }\,\text{d}x.[/imath] But I still can't figure out how to integrate from there. I know that it's integrable, as Wolfram Alpha indicates that the integral is [imath] \frac{1}{2}\big(x+\ln{(\sin x + \cos x)}\big)+C[/imath], but I can't figure out the steps to deriving it. Does anyone know how to evaluate this integral? |
55616 | Is there a free subgroup of rank 3 in [imath]SO_3[/imath]?
There are known free subgroups of rank 2 in the set of rotations about the origin in [imath]\mathbb{R}^3[/imath], [imath]SO_3[/imath]. For instance, the rotations by angle [imath]\arccos \frac {1}{3}[/imath] about the [imath]z[/imath]- and [imath]x[/imath]-axis generate such a free subgroup. Are there free subgroups of rank 3 (or higher) in [imath]SO_3[/imath]? | 1666265 | Free group on 3 generators is a subgroup of free group on 2 generators
I think this true but can it be proven in an explicit way? A group, G, is called Free if there is a subset S of G such that any element of G can be written in one and only one way as a product of finitely many elements of S and their inverses. Clearly, the free group on 3 generators has a larger order than the free group on 2 generators [imath]F_3[/imath] is generated by [imath]<a, b, c>[/imath] Nielsen–Schreier theorem states that every subgroup of a free group is itself free, so this holds but does not show what the subgroups are. |
29357 | Prove [imath](a_1+b_1)^{1/n}\cdots(a_n+b_n)^{1/n}\ge \left(a_1\cdots a_n\right)^{1/n}+\left(b_1\cdots b_n\right)^{1/n}[/imath]
consider positive numbers [imath]a_1,a_2,a_3,\ldots,a_n[/imath] and [imath]b_1,b_2,\ldots,b_n[/imath]. does the following in-equality holds and if it does then how to prove it [imath]\left[(a_1+b_1)(a_2+b_2)\cdots(a_n+b_n)\right]^{1/n}\ge \left(a_1a_2\cdots a_n\right)^{1/n}+\left(b_1b_2\cdots b_n\right)^{1/n}[/imath] | 2842364 | Show that [imath](\prod_{i=1}^{n} (a_i+b_i))^{\frac{1}{n}} \geq (\prod_{i=1}^{n} {a_i})^{\frac{1}{n}}+(\prod_{i=1}^{n} {b_i})^{\frac{1}{n}}[/imath]
Assume [imath]a_i, b_i[/imath] are positive reals for [imath]i = 1,2,...,n.[/imath] Show that [imath](\prod_{i=1}^{n} (a_i+b_i))^{\frac{1}{n}} \geq (\prod_{i=1}^{n} {a_i})^{\frac{1}{n}}+(\prod_{i=1}^{n} {b_i})^{\frac{1}{n}}.[/imath] I tried applying AM-GM to this product in some way, seeing that there is a product of [imath]n[/imath] terms and an exponent of [imath]\frac{1}{n}.[/imath] However, applying it directly only gives [imath]\frac{1}{n}\sum_{i=1}^{n} (a_i+b_i) \geq (\prod_{i=1}^{n} (a_i+b_i))^{\frac{1}{n}}[/imath] and [imath]\frac{1}{n}\sum_{i=1}^{n} a_i + \frac{1}{n}\sum_{i=1}^{n} b_i\geq (\prod_{i=1}^{n} {a_i})^{\frac{1}{n}}+(\prod_{i=1}^{n} {b_i})^{\frac{1}{n}},[/imath] which isn't helpful whatsoever. I can't think of other helpful inequalities that deal with sequence products like this, so I'm currently stuck trying to manipulate the expression and apply AM-GM somehow, but can't find a good way. |
4603 | Series converges implies [imath]\lim{n a_n} = 0[/imath]
I'm studying for qualifying exams and ran into this problem. Show that if [imath]\{a_n\}[/imath] is a nonincreasing sequence of positive real numbers such that [imath]\sum_n a_n[/imath] converges, then [imath]\lim_{n \rightarrow \infty} n a_n = 0[/imath]. Using the definition of the limit, this is equivalent to showing \begin{equation} \forall \varepsilon > 0 \; \exists n_0 \text{ such that } |n a_n| < \varepsilon \; \forall n > n_0 \end{equation} or \begin{equation} \forall \varepsilon > 0 \; \exists n_0 \text{ such that } a_n < \frac{\varepsilon}{n} \; \forall n > n_0 \end{equation} Basically, the terms must be bounded by the harmonic series. Thanks, I'm really stuck on this seemingly simple problem! | 741342 | A convergent series property
This came up in a friend's exam and it must be one of those [imath]{\epsilon},N(\epsilon)[/imath] arguments I could do in a snapshot in my twenties but now I can't figure out how the proof should go: For a positive non-increasing sequence [imath]\{a_k, {k \in \mathbb{N}}\}[/imath] if [imath]\sum_{k=1}^{\infty}a_k[/imath] converges then [imath]b_n=na_n{\to}0[/imath]. Thanks in advance! |
111314 | Choose a random number between [imath]0[/imath] and [imath]1[/imath] and record its value. Keep doing it until the sum of the numbers exceeds [imath]1[/imath]. How many tries do we need?
Choose a random number between [imath]0[/imath] and [imath]1[/imath] and record its value. Do this again and add the second number to the first number. Keep doing this until the sum of the numbers exceeds [imath]1[/imath]. What's the expected value of the number of random numbers needed to accomplish this? | 843528 | Expected value of series of uniformly converges random variables
Let [imath]X_1,X_2,X_3,...[/imath] a series of i.i.d. variables with [imath]X_i \sim \mathcal{U}(0,1)[/imath]. Let [imath]N=\inf\{n\mid \sum_{i=1}^{n}X_i\geq1\}[/imath] Prove that [imath]E(N)=e[/imath]. I don't really have a clue how to even start proving that. Can someone please help? Thanks. |
60698 | If a field [imath]F[/imath] is such that [imath]\left|F\right|>n-1[/imath] why is [imath]V[/imath] a vector space over [imath]F[/imath] not equal to the union of [imath]n[/imath] proper subspaces of [imath]V[/imath]
If [imath]U_1[/imath], [imath]U_2,\ldots,U_n[/imath] are proper subspaces of a vector space [imath]V[/imath] over a field [imath]F[/imath], and [imath]|F|\gt n-1[/imath], why is [imath]V[/imath] not equal to the union of the subspaces [imath]U_1[/imath], [imath]U_2,\ldots,U_n[/imath]? | 1113384 | Prove that [imath]\mathbb{R}^n[/imath] cannot be a finite union of its hyperplanes .
Prove that [imath]\mathbb{R}^n[/imath] cannot be a finite union of its hyperplanes . I want a prove using linear algebra only and not functional analysis i tried by contradiction we know R^n is a vector space over R. let R^n = U Wi (i from 1-k) Wi's are hyperplanes so are proper subspaces, let x belongs to W1. and take y belongs to R^n-W1. so there are infinitely many x+ay for a belongs to R. x+ay doesnt belong to W1 as R^n=U Wi so x+ay belongs to some Wj, j not equal to 1. so Wj contains x and y. so W1 is a subset of U Wi (i from 2 to k) now applying induction we get R^n=Wk which is a contradiction as Wk is a proper subspace. but my prof says there are gaps in the proof which i am not able to find. |
155952 | Arcwise connected part of [imath]\mathbb R^2[/imath]
Here's a question that I share: Show that if [imath]D[/imath] is a countable subset of [imath]\mathbb R^2[/imath] (provided with its usual topology) then [imath]X=\mathbb R^2 \backslash D [/imath] is arcwise connected. | 428944 | The complement of every countable set in the plane is path connected
I'm trying to show that if [imath]A[/imath] is a countable subset of [imath]\mathbb{R}^{2}[/imath] then [imath]\mathbb{R}^{2}\backslash A[/imath] is path connected. I already have the idea of the proof. It is clear to me that for every two points in the plane there is an uncountable collection of paths between them that intersect only at the edges. Thus given two points not in [imath]A[/imath] since [imath]A[/imath] is countable it is impossible that all these paths intersect [imath]A[/imath]. Thus for every two points in [imath]\mathbb{R}^{2}\backslash A[/imath] there is path between them that doesn't intersect [imath]A[/imath] What I want is an explicit construction of such a collection of paths given two points [imath]\left(x_{1},y_{1}\right),\left(x_{2},y_{2}\right)\in\mathbb{R}^{2}[/imath]. Help would be appreciated. |
67198 | Does [imath] \int_0^{\infty}\frac{\sin x}{x}dx [/imath] have an improper Riemann integral or a Lebesgue integral?
In this wikipedia article for improper integrals, [imath] \int_0^{\infty}\frac{\sin x}{x}dx [/imath] is given as an example for the integrals that have an improper Riemann integral but do not have a (proper) Lebesgue integral. Here are my questions: Why does this one have an improper Riemann integral? (I don't see why [imath]\int_0^a\frac{\sin x}{x}dx[/imath] and [imath]\int_a^{\infty}\frac{\sin x}{x}dx[/imath] converge.) Why doesn't this integral have a Lebesgue integral? Is it because that [imath]\frac{\sin x}{x}[/imath] is unbounded on [imath](0,\infty)[/imath] and Lebesgue integral doesn't deal with unbounded functions? | 416284 | Absolute integrability
Bartle has a statement: Although the absolute value of a proper Riemann integrable function is Riemann integrable, this may no longer be the case for a function which has an improper Riemann integral (for example, consider [imath]f(x)=x^{-1} \sin(x)[/imath] on the interval [imath]1\le x\lt +\infty[/imath]). Could I get a little help with this statement? I am assuming I need to show that the improper integral is integrable, but the absolute value is not. Thanks. |
118518 | Is the set of polynomial with coefficients on [imath]\mathbb{Q}[/imath] enumerable?
Using the definition of enumerability of sets: A non-empty set B is enumerable iff there is a bijection [imath]f:\mathbb{N}\supset L \rightarrow B[/imath]. So, I have to prove that the set of polynomial of one variable with coefficients on [imath]\mathbb{Q}[/imath] is enumerable. What I thought is that I can write every polynomial [imath]p_0=A_0^0+A_1^0X+\dots +A_n^0X^n[/imath] as [imath]p_0=(A_0^0,A_1^0,\dots ,A_n^0,\dots ),[/imath] where the [imath]A_i's[/imath] are 0 except for a finite number of them. To say that this set is enumerable is the same that to say (denoting this set by B): [imath]B=\{p_1,p_2,\dots\}[/imath] Now define the polynomial [imath] p=(x_0 \neq A_0^0, x_1 \neq A_1^1, \dots), [/imath] where every [imath]x_i \neq A_i^i[/imath]. This polynomial is obviously not in the list B. My question is: my approach is right to conclude that the set B is non-enumerable? | 2213925 | Countability of sets with countable coefficients
Let me know about this two examples. 1) set of all polynomials with rational coefficients. 2) set of all polynomials whose coefficients belongs to set [imath]{\{1, 2\}}[/imath] I know the first one is countable set. Because we can write first set as union of countable sets [imath]P_n[/imath] which is set of polynomial with rational coefficients having degree at most n" But, I am stuck at second one, as I see in the first one coefficient belongs to countable set [imath]Q[/imath] so the given set is countable and in second one also, coefficient belongs to set [imath]\{1, 2\}[/imath] which is also countable set and hence set given in 2) is also countable? Is am I right? |
16831 | Nonzero [imath]f \in C([0, 1])[/imath] for which [imath]\int_0^1 f(x)x^n dx = 0[/imath] for all [imath]n[/imath]
As the title says, I'm wondering if there is a continuous function such that [imath]f[/imath] is nonzero on [imath][0, 1][/imath], and for which [imath]\int_0^1 f(x)x^n dx = 0[/imath] for all [imath]n \geq 1[/imath]. I am trying to solve a problem proving that if (on [imath]C([0, 1])[/imath]) [imath]\int_0^1 f(x)x^n dx = 0[/imath] for all [imath]n \geq 0[/imath], then [imath]f[/imath] must be identically zero. I presume then we do require the [imath]n=0[/imath] case to hold too, otherwise it wouldn't be part of the statement. Is there ay function which is not identically zero which satisfies [imath]\int_0^1 f(x)x^n dx = 0[/imath] for all [imath]n \geq 1[/imath]? The statement I am attempting to prove is homework, but this is just idle curiosity (though I will tag it as homework anyway since it is related). Thank you! | 1048287 | Proof that a continuous function is zero
Let [imath]f:[0, 1] \to \mathbb{R}[/imath] be a continuous function such that [imath]\int^1_0 f(x)x^ndx=0[/imath] [imath]\forall n \in \mathbb{N}[/imath]. Show that [imath]f \equiv 0[/imath] in [0, 1]. I need help. |
12320 | Proof that if group [imath]G/Z(G)[/imath] is cyclic, then [imath]G[/imath] is commutative
I am looking for a correct proof of this statement: If [imath]G[/imath] is a group such that [imath]G/Z(G)[/imath] is cyclic, then [imath]G[/imath] is commutative. Proof: [imath]G/Z(G)[/imath] is isomorphic to [imath]\operatorname{Inn}(G)[/imath] and is cyclic, and then for every [imath]a[/imath] and [imath]b[/imath] in [imath]G[/imath] the inner isomorphisms [imath]\gamma_a[/imath] and [imath]\gamma_b[/imath] satisfy [imath]\gamma_a \gamma_b = \gamma_{ab} = \gamma_{ba} = \gamma_b \gamma_a[/imath], and therefore for every [imath]a,b \in G[/imath], [imath]ab = ba[/imath]. Is that proof complete, or am I missing something? Thanks a lot for the help. | 364519 | Groups - Prove that if [imath]G/Z(G)[/imath] is cyclic then [imath]G[/imath] is abelian
Prove that if [imath]G/Z(G)[/imath] is cyclic then [imath]G[/imath] is abelian. Using this fact and [imath]G[/imath] is a nontrivial group of prime power order, deduce that a group of order [imath]p^2[/imath] , [imath]p[/imath] prime, is abelian. |
63087 | If [imath]G/Z(G)[/imath] is cyclic, then [imath]G[/imath] is abelian
Continuing my work through Dummit & Foote's "Abstract Algebra", 3.1.36 asks the following (which is exactly the same as exercise 5 in this related MSE answer): Prove that if [imath]G/Z(G)[/imath] is cyclic, then [imath]G[/imath] is abelian. [If [imath]G/Z(G)[/imath] is cyclic with generator [imath]xZ(G)[/imath], show that every element of [imath]G[/imath] can be written in the form [imath]x^az[/imath] for some [imath]a \in \mathbb{Z}[/imath] and some element [imath]z \in Z(G)[/imath]] The hint is actually the hardest part for me, as the quotient groups are somewhat abstract. But once I have the hint, I can write: [imath]g, h \in G[/imath] implies that [imath]g = x^{a_1}z_1[/imath] and [imath]h = x^{a_2}z_2[/imath], so \begin{align*}gh &= (x^{a_1}z_1)(x^{a_2}z_2)\\\ &= x^{a_1}x^{a_2}z_1z_2\\\ & = x^{a_1 + a_2}z_2z_1\\\ &= \ldots = (x^{a_2}z_2)(x^{a_1}z_1) = hg. \end{align*} Therefore, [imath]G[/imath] is abelian. 1) Is this right so far? 2) How can I prove the "hint"? | 1390076 | Quotient group G/Z is cyclic
Let [imath]Z[/imath] be the center of a group [imath]G[/imath]. Prove that if [imath]G/Z[/imath] is a cyclic group, then [imath]G[/imath] is abelian. This is from Michael Artin's algebra chapter 7 section 3. I'm quite unsure as to how I should start. My first thought was to prove that [imath]G/Z[/imath] is isomorphic to some subgroup of [imath]G[/imath]. Then we would have that [imath]G \cong G/Z \times Z[/imath], which is Abelian, assuming that [imath]G/Z \cap Z = \{0\}[/imath]. However, I don't see a way to prove that such a subgroup of [imath]G[/imath] exists. A second idea was to note that [imath]G/Z = \{Z, xZ, (xZ)^2, \cdots, (xZ)^{n-1} \}[/imath] for some [imath]x\not \in Z[/imath]. Since [imath]Z[/imath] is the center, which is a normal group, we can say that [imath](xZ)^k = xZxZxZ\cdots xZ = x^kZ[/imath] for all [imath]k[/imath]. Sadly, this only guarantees an [imath]x[/imath] such that [imath]x^n[/imath] is in [imath]Z[/imath], rather than guaranteeing a subgroup, which would allow the first idea to follow. Some help would be appreciated. |
114841 | Proof of a formula involving Euler's totient function: [imath]\varphi (mn) = \varphi (m) \varphi (n) \cdot \frac{d}{\varphi (d)}[/imath]
The third formula on the wikipedia page for the Totient function states that [imath]\varphi (mn) = \varphi (m) \varphi (n) \cdot \dfrac{d}{\varphi (d)} [/imath] where [imath]d = \gcd(m,n)[/imath]. How is this claim justified? Would we have to use the Chinese Remainder Theorem, as they suggest for proving that [imath]\varphi[/imath] is multiplicative? | 754341 | Prove that if d = gcd(m,n) then [imath]\phi(mn)=\phi(m)*\phi(n)/d[/imath]
So if m and n are relatively prime, then the [imath]\phi(mn)=\phi(m)*\phi(n)[/imath] but what happens when [imath]d > 1[/imath]? |
61366 | Surface of genus [imath]g[/imath] does not retract to circle (Hatcher exercise)
I'm trying exercise 9 on page 53 in Hatcher but I need some help with it. The exercise is: In the surface [imath]M_g[/imath] of genus [imath]g[/imath], let [imath]C[/imath] be a circle that separates [imath]M_g[/imath] into two compact subsurfaces [imath]M_h^\prime [/imath] and [imath]M_k^\prime[/imath] obtained from the closed surfaces [imath]M_h[/imath] and [imath]M_k[/imath] by deleting an open disk from each. Show that [imath]M_h^\prime[/imath] does not retract onto its boundary circle [imath]C[/imath], and hence [imath]M_g[/imath] does not retract onto [imath]C[/imath]. [Hint: abelianize [imath]\pi_1[/imath].] But show that [imath]M_g[/imath] does retract onto the nonseparating circle [imath]C^\prime[/imath] in the figure. My first question is: assume there wasn't the hint, how would I think of abelianising? What does it mean exactly? I thought I could do this by contradiction: if it retracts the induced map [imath]i_\ast : \pi_1(C) \rightarrow \pi_1(M_h^\prime) [/imath] is injective. I know [imath]\pi_1(C) \cong \mathbb{Z}[/imath], then I computed [imath]\pi_1(M_h^\prime) \cong \mathbb{Z}[/imath] and then I think I'm stuck. Right? Do you agree with [imath]\pi_1(M_h^\prime) \cong \mathbb{Z}[/imath] and being stuck after that? What do I need to know to make progress? Many thanks for your help! | 625513 | Prove [imath]S_g[/imath] retracts to [imath]C[/imath]
Let [imath]S_g[/imath] be the closed oriented surface of genus g. Let [imath]C[/imath] be a simple closed curve in [imath]S_g[/imath]. Prove that [imath]S_g[/imath] retracts to [imath]C[/imath] if and only if [imath]C[/imath] does not separate [imath]S_g[/imath]. If [imath]S_g[/imath] retracts to [imath]C[/imath], by Alexander-Lefschetz duality, I can get [imath]H_0(S_g-C)=\mathbb{Z}[/imath], therefore [imath]C[/imath] does not separate [imath]S_g[/imath]. But I have no idea how to prove another direction. |
4323 | Are all algebraic integers with absolute value 1 roots of unity?
If we have an algebraic number [imath]\alpha[/imath] with (complex) absolute value [imath]1[/imath], it does not follow that [imath]\alpha[/imath] is a root of unity (i.e., that [imath]\alpha^n = 1[/imath] for some [imath]n[/imath]). For example, [imath](3/5 + 4/5 i)[/imath] is not a root of unity. But if we assume that [imath]\alpha[/imath] is an algebraic integer with absolute value [imath]1[/imath], does it follow that [imath]\alpha[/imath] is a root of unity? I know that if all conjugates of [imath]\alpha[/imath] have absolute value [imath]1[/imath], then [imath]\alpha[/imath] is a root of unity by the argument below: The minimal polynomial of [imath]\alpha[/imath] over [imath]\mathbb{Z}[/imath] is [imath]\prod_{i=1}^d (x-\alpha_i)[/imath], where the [imath]\alpha_i[/imath] are just the conjugates of [imath]\alpha[/imath]. Then [imath]\prod_{i=1}^d (x-\alpha_i^n)[/imath] is a polynomial over [imath]\mathbb{Z}[/imath] with [imath]\alpha^n[/imath] as a root. It also has degree [imath]d[/imath], and all roots have absolute value [imath]1[/imath]. But there can only be finitely many such polynomials (since the coefficients are integers with bounded size), so we get that [imath]\alpha^n=\sigma(\alpha)[/imath] for some Galois conjugation [imath]\sigma[/imath]. If [imath]\sigma^m(\alpha) = \alpha[/imath], then [imath]\alpha^{n^m} = \alpha[/imath]. Thus [imath]\alpha^{n^m - 1} = 1[/imath]. | 1438184 | If a complex number with unit modulus is a root of a monic polynomial with integer coefficients, is it a root of unity?
The title says pretty much everything: We have [imath]z\in\mathbb{C}[/imath] such that [imath]\|z\|=1[/imath] and a monic polynomial [imath]p\in\mathbb{Z}[x][/imath] such that [imath]p(z)=0[/imath]. Is it true that for some [imath]n\in\mathbb{N}[/imath], [imath]z^n=1[/imath]? My thoughts about it: assume that [imath]z=e^{\pi i\alpha}[/imath] and [imath]z[/imath] is not a root of unity, i.e. [imath]\alpha\not\in\mathbb{Q}[/imath]. Then [imath]\alpha[/imath] has to be a trascendental number: otherwise, by the Gelfond-Schneider theorem, [imath]z[/imath] is a trascendental number. But obviously this is far from closing the question: for instance, [imath]\frac{3+4i}{5}[/imath] is a complex number with unit modulus, and [imath]\frac{1}{\pi}\arctan\left(\frac{4}{3}\right)[/imath] is a trascendental number, but there is no monic polynomial with integer coefficients having [imath]\frac{3+4i}{5}[/imath] as a root. So, what is the good way to go for proving the claim above, since I believe it is true? Counter-examples are welcome as well, obviously. |
5 | How can you prove that the square root of two is irrational?
I have read a few proofs that [imath]\sqrt{2}[/imath] is irrational. I have never, however, been able to really grasp what they were talking about. Is there a simplified proof that [imath]\sqrt{2}[/imath] is irrational? | 71237 | How to prove [imath]\sqrt3[/imath] is irrational?
How to prove [imath]\sqrt3[/imath] is irrational using Fermat's infinite descent method? Like says in Carl Benjamim Boyer's book. Isnt the same prove to [imath]\sqrt2[/imath], in Boyer's book says something like this. [imath]\sqrt3=a1/b1[/imath] [imath]1/(\sqrt3-1)=(\sqrt3+1)/2[/imath] [imath]\sqrt3=(3b1-a1)/(a1-b1)[/imath] |
298910 | Binary Connective Proofs
I’m working to understand proofs that involve showing the completeness (or incompleteness) of a set of binary connectives and I have run into some confusion. Alright, so I believe I understand how to show a set of binary connectives is complete; you just need to show that this set is equivalent to a set you know is complete. For example, in order to show that {|} (i.e. Sheffer’s Stroke ) is complete we note the following: [imath] ¬ \alpha \Leftrightarrow \alpha | \alpha[/imath] [imath] \alpha \vee \beta \Leftrightarrow (¬ \alpha) | (¬ \beta) [/imath] And since we know that {¬, [imath]\vee[/imath]} is complete and this sets behavior can be simulated with only {|}, we know that {|} is complete. But how would we show that a particular set of binary connectives is not complete? For instance how would we show that {[imath]\rightarrow[/imath]}, {[imath]\vee[/imath]}, or any other single binary connective except {[imath]\downarrow[/imath] } is not complete? Any help would be appreciated | 278759 | How to prove that a set of logical connectives is functionally complete(incomplete)?
How to prove that a set of logical connectives is functionally complete(incomplete)? For example, we are given this set: [imath] \left\{\begin{matrix} f = (01101001) \\ g = (1010) \\ h = (01110110) \\ \end{matrix}\right. [/imath] I would appreciate any examples, helpful information or links\books where this subject is accessibly explained. |
17966 | How can we sum up [imath]\sin[/imath] and [imath]\cos[/imath] series when the angles are in arithmetic progression?
How can we sum up [imath]\sin[/imath] and [imath]\cos[/imath] series when the angles are in arithmetic progression? For example here is the sum of [imath]\cos[/imath] series: [imath]\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)[/imath] There is a slight difference in case of [imath]\sin[/imath], which is: [imath]\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2 a + (n-1)\cdot d}{2}\biggr)[/imath] How do we prove the above two identities? | 831159 | Evaluating [imath]\sum_{n=1}^{99}\sin(n)[/imath]
I'm looking for a trick, or a quick way to evaluate the sum [imath]\displaystyle{\sum_{n=1}^{99}\sin(n)}[/imath]. I was thinking of applying a sum to product formula, but that doesn't seem to help the situation. Any help would be appreciated. |
183361 | Examples of bijective map from [imath]\mathbb{R}^3\rightarrow \mathbb{R}[/imath]
Could any one give an example of a bijective map from [imath]\mathbb{R}^3\rightarrow \mathbb{R}[/imath]? Thank you. | 309282 | Bijection between [imath]\mathbb{R}[/imath] and [imath]\mathbb{R}^2[/imath]
I have been thinking for a while whether its possible to have bijection between [imath]\mathbb{R}[/imath] and [imath]\mathbb{R}^2[/imath], but I cant think of a solution. So my question is: is there a bijection between [imath]\mathbb{R}[/imath] and [imath]\mathbb{R}^2[/imath] (with proof)? |
29599 | Is the set of all finite sequences of letters of Latin alphabet countable/uncountable? How to prove either?
Today in Coding/Cryptography class, we were talking about basic definitions, and the professor mentioned that for a set [imath]A=\left \{ \left. a, b, \dots, z \right \} \right.[/imath] (the alphabet) we can define a set [imath]A^{*}=\left \{ \left. a, ksdjf, blocks, coffee, maskdj, \dots, asdlkajsdksjfs \right \} \right.[/imath] (words) as a set that consists of all finite sequences of the elements/letters from our [imath]A[/imath]/alphabet. My question is, is this set [imath]A^{*}[/imath] countably or uncountably infinite? Does it matter how many letters there are in your alphabet? If it was, say, [imath]A=\left \{ \left. a \right \} \right.[/imath], then the words in [imath]A^{*}[/imath] would be of form [imath]a, aa, aaa, \dots[/imath] which, I think, would allow a bijection [imath]\mathbb{N} \to A^{*}[/imath] where an integer would signify the number of a's in a word. Can something analogous be done with an alphabet that consists of 26 letters (Latin alphabet), or can countability/uncountability be proved otherwise? And as mentioned before, I am wondering if the number of elements in the alphabet matters, or if all it does is change the formula for a bijection. P.S. Now that I think of it, maybe we could biject from [imath]\underset{n}{\underbrace{\mathbb{N}\times\mathbb{N}\times\mathbb{N}\times\dots\times\mathbb{N}}}[/imath] to some set of words [imath]A^{*}[/imath] whose alphabet [imath]A[/imath] has [imath]n[/imath] elements? Thanks! | 810889 | Number of all finite sequences from a set?
Given a set [imath]\Sigma[/imath] of letters, apply the Kleene star operation to it, and we get [imath]\Sigma^*[/imath], the set of all finite-length sequences from [imath]\Sigma[/imath], called strings (allowing a letter appearing more than once in a string). If [imath]\Sigma[/imath] is empty, then [imath]\Sigma^*[/imath] consists only one string, the empty string. If [imath]\Sigma[/imath] is not empty, then [imath]\Sigma^*[/imath] has an infinite cardinality. I wonder if [imath]\Sigma^*[/imath] can be countably infinite? When will that be true? When [imath]\Sigma[/imath] is finite? When will its cardinality is uncountably infinite? Thanks! |
158982 | Circular Permutation
Can someone please explain how to solve circular permutation sums. I just cannot seem to understand them. eg. [imath]\text{(4)}[/imath] The number of ways in which [imath]6[/imath] men and [imath]5[/imath] women can dine at a round table if no two women are to sit together is given by [imath]\text{(a)}[/imath] [imath]6!*5![/imath] [imath]\text{(b)}[/imath] [imath]50[/imath] [imath]\text{(c)}[/imath] [imath]5!*4![/imath] [imath]\text{(d)}[/imath] [imath]7!*5![/imath] | 1461017 | Cycle permutation problem, no women sit next to each other
I have a problem : There are 6 men and 5 women who will sit in a circular table. How many ways they sit where no two women sit next to each other? I have tried to work for it and get the result [imath]5!*5![/imath], where the first [imath]5![/imath] represents the number of ways men sit and the second [imath]5![/imath] is the number of ways to permute the women, but I am not so sure. Help please |
9141 | About irrational logarithms
Could someone provide, please, a proof of the theorem below? "Being [imath]x[/imath] and [imath]b[/imath] integers greater than [imath]1[/imath], which can not be represented as powers of the same basis (positive integer) and integer exponent, then the logarithm of [imath]x[/imath], in base [imath]b[/imath], is an irrational number." Well... Assuming that the logarithm is a rational number [imath]p/q[/imath] ([imath]p[/imath] and [imath]q[/imath] are relatively prime integers), I know I can write [imath]x^q = b^p[/imath], but I can not conclude from this fact that [imath]b[/imath] and [imath]x[/imath] are powers of a same integer (and with integer exponent). | 430538 | How to show that [imath]\log_{10} n[/imath] is not a rational number if [imath]n[/imath] is any integer not a power of [imath]10.[/imath]
How to show that [imath]\log_{10} n[/imath] is not a rational number if [imath]n[/imath] is any integer not a power of [imath]10.[/imath] If not, let [imath]\log_{10}n=\dfrac{p}{q}[/imath] for some [imath]p,q(\ne0)\in\mathbb Z[/imath] where [imath](p,q)=1.[/imath] Then [imath]q\log_{10}n=p\implies\log_{10}n^q=p\implies\log_{10}n^q=\log_{10}10^p\implies n^q=10^p[/imath]. I don't know what to do next. Added: I can see [imath]5,2[/imath] are the only prime factors of [imath]n.[/imath] But I cant get why the same number of 5 and 2 would occur in [imath]n?[/imath] |
297682 | What is the value of [imath]i^i[/imath]?
I understand that when you raise any number [imath]x[/imath] to a power, you multiply [imath]x[/imath] by itself the number of times indicated in the power. However, what happens when [imath]i^i[/imath] is performed? How can a number be multiplied an imaginary amount of times? Wolfram Alpha says that it is equal to [imath]e^{{-\pi}/{2}}[/imath], but how would you arrive at that answer? Any response will be appreciated, thanks! | 191572 | Prove that [imath]i^i[/imath] is a real number
According to WolframAlpha, [imath]i^i=e^{-\pi/2}[/imath] but I don't know how I can prove it. |
83035 | How to prove [imath](1+1/x)^x[/imath] is increasing when [imath]x>0[/imath]?
Let [imath]F(x)=(1+\frac{1}{x})^x[/imath]. How do we prove [imath]F(x)[/imath] is increasing when [imath]x>0[/imath]? | 1614390 | Proove that the sequence [imath]\left (1+\frac{1}{x} \right )^{x}[/imath] increases.
I want to show that [imath]\left (1+\frac{1}{x} \right )^{x}[/imath] increases. I have to show that [imath]\left (1+\frac{1}{x+1} \right )^{x+1} > \left (1+\frac{1}{x} \right )^{x}[/imath] [imath]\left (1+\frac{1}{x}-\frac{1}{x(x+1)} \right )^{x+1} > \left (1+\frac{1}{x} \right )^{x}[/imath] I triend to define [imath] \left (1+\frac{1}{x} \right ) = k[/imath] [imath]\left (k-\frac{1}{x(x+1)} \right )^{x+1} > k^x[/imath] Now im pretty stuck. |
233999 | How to find the roots of [imath]x^4 +1[/imath]
I'm trying to find the roots of [imath]x^4+1[/imath]. I've already found in this site solutions for polynomials like this [imath]x^n+a[/imath], where [imath]a[/imath] is a negative term. I don't remember how to solve an equation when [imath]a[/imath] is a positive term as the equation above. Thanks | 247312 | What are the solutions to [imath]z^4+1=0[/imath]?
I can't seem to find the solutions to [imath]z^4+1=0 [/imath]. [imath]z[/imath] is in the complex plane. The solutions show four roots; however, how do I find them once [imath]z^4 = -1[/imath]? |
62548 | Why is a finite integral domain always field?
This is how I'm approaching it: let [imath]R[/imath] be a finite integral domain and I'm trying to show every element in [imath]R[/imath] has an inverse: let [imath]R-\{0\}=\{x_1,x_2,\ldots,x_k\}[/imath], then as [imath]R[/imath] is closed under multiplication [imath]\prod_{n=1}^k\ x_i=x_j[/imath], therefore by canceling [imath]x_j[/imath] we get [imath]x_1x_2\cdots x_{j-1}x_{j+1}\cdots x_k=1 [/imath], by commuting any of these elements to the front we find an inverse for first term, e.g. for [imath]x_m[/imath] we have [imath]x_m(x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k)=1[/imath], where [imath](x_m)^{-1}=x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k[/imath]. As far as I can see this is correct, so we have found inverses for all [imath]x_i\in R[/imath] apart from [imath]x_j[/imath] if I am right so far. How would we find [imath](x_{j})^{-1}[/imath]? | 358184 | Problem on a finite commutative ring with no zero divisors
This is a problem from Dummit & Foote. Prove that a non-zero finite commutative ring that has no divisor is a field. (Do not assume the ring has a 1) Evidently, one has to use the theorem proved earlier, that a finite integral domain is a field. So, I have to prove that any non-zero finite commutative ring has a multiplicative identity. I am stumped. I had proved that the only ideals this ring has are the trivial ideals, [imath]0[/imath] and [imath]R[/imath]. This is seen easily by defining the homomorphism [imath]\phi(r): R \to R = ar [/imath] where [imath]a[/imath] is a non-zero element of a non-zero ideal [imath]I[/imath]. The surjectivity of this map (pigeon hole principle), shows that the ideals are trivial. Is this going to help me prove it has a [imath]1[/imath]. How do I approach the problem? |
47395 | The Duals of [imath]l^\infty[/imath] and [imath]L^{\infty}[/imath]
Can we identify the dual space of [imath]l^\infty[/imath] with another "natural space"? If the answer is yes, what can we say about [imath]L^\infty[/imath]? By the dual space I mean the space of all continuous linear functionals. | 2765338 | Dual of l_inf space of bounded sequences in R
So, I was searching for ways of finding the dual of [imath]l_\infty(\mathbb{R})[/imath], (the space of bounded sequences of real numbers, under the infinity norm, i.e., [imath]||x|| = sup_{n\in\mathbb{N}} |x_n|.[/imath] I tried to approach the same way I did for [imath]l_1, l_p, c_0[/imath] and other similar spaces, however, I was not able to find any Schauder Basis for [imath]l_\infty[/imath]. So I started searching, but I didn't find anything concrete about it. Does anybody knows an explicit "calculation" of [imath](l_\infty)^*[/imath], or a book or paper that shows a way to obtain it? |
296265 | How to find the number of elements of the general linear group.
Yesterday i encountered the following question: let [imath]G[/imath] be the group [imath]G \ =\ \{\left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \mid a,b,c,d\in \mathbb{Z}_{3},\ ad-bc\neq 0\}.[/imath] The question was to evaluate the order of [imath]G[/imath]. By brute-force I figured out the order to be [imath]48[/imath]. But it took so much time. My question is that is there any formula or rule to find the orer of such groups in general case like when all the entries of the matrix are from [imath]\mathbb{Z}_{p}[/imath] where [imath]p[/imath] is any prime. | 296047 | Order of general linear group of [imath]2 \times 2[/imath] matrices over [imath]\mathbb{Z}_3[/imath]
From problem 2.3.25 in Topics in Algebra, 2[imath]\varepsilon[/imath] by I. N. Herstein: Let [imath]G[/imath] be the group of all [imath]2 \times 2[/imath] matrices [imath]\left(\begin{array}{c c}a & b \\ c & d\end{array}\right)[/imath] where [imath]ad-bc \ne 0[/imath] and [imath]a,b,c,d[/imath] are integers modulo 3, relative to matrix multiplication. Show that [imath]o(G) = 48[/imath]. I know that [imath]o(G) \le 3^4 = 81[/imath], since [imath]a,b,c,d[/imath] can each take one of 3 values (mod 3). I attempted to tighten this bound by finding the number of matrices such that [imath]ad=bc[/imath] (mod 3): Suppose [imath]ad=bc=0[/imath] (mod 3). Then ([imath]a = 0[/imath] or [imath]d = 0[/imath]) and ([imath]b = 0[/imath] or [imath]c = 0[/imath]), leading to 36 possible values for [imath](a,b,c,d)[/imath]. Suppose [imath]ad=bc=1[/imath] (mod 3). Then ([imath]a=d=1[/imath] or [imath]a=d=2[/imath]) and ([imath]b=c=1[/imath] or [imath]b=c=2[/imath]), leading to 4 possible values for [imath](a,b,c,d)[/imath]. Suppose [imath]ad=bc=2[/imath] (mod 3). Then ([imath](a,d)=(1,2)[/imath] or [imath](a,d)=(2,1)[/imath]) and ([imath](b,c)=(1,2)[/imath] or [imath](b,c)=(2,1)[/imath]), leading to 4 possible values for [imath](a,b,c,d)[/imath]. So, there are in total [imath]36+4+4 = 44[/imath] such [imath]\left(\begin{array}{c c}a & b \\ c & d\end{array}\right)[/imath] where [imath]ad-bc=0[/imath] (mod 3). That means there are at most [imath]81-44 = 37[/imath] such [imath]\left(\begin{array}{c c}a & b \\ c & d\end{array}\right)[/imath] where [imath]ad-bc\ne 0[/imath], i.e., [imath]o(G) \le 37[/imath]. However, this contradicts the problem. Where did I go wrong? Can someone set me on the right path? |
16611 | Bijecting a countably infinite set [imath]S[/imath] and its cartesian product [imath]S \times S[/imath]
From Herstein's Topics in Algebra (exercise 14, section 1.2): If [imath]S[/imath] is infinite and can be brought into one-to-one correspondence with the set of integers, prove that there is one-to-one correspondence between [imath]S[/imath] and [imath]S \times S[/imath]. So far I know there exists some bijection [imath]\sigma : S \rightarrow \mathbb{Z}[/imath]. If I can define a bijection [imath]\tau : \mathbb{Z} \rightarrow \mathbb{Z} \times \mathbb{Z}[/imath], then a one-to-one correspondence between [imath]S[/imath] and [imath]S \times S[/imath] is given by [imath]\mu[/imath] such that [imath]s_\mu = (a_{\sigma^{-1}},b_{\sigma^{-1}})[/imath], with [imath]a,b\in \mathbb{Z}[/imath] given by [imath]s_{\sigma \circ \tau} = (a,b)[/imath]. I'm having trouble defining [imath]\tau[/imath]. I think a possible way to do so would be to create some kind of spiral (like the Ulam spiral), and assign each point to a different integer. I suppose this would be a one-to-one correspondence but I'm at a loss on how to prove it. Thanks a lot! | 467034 | A bijective map from [imath]\bf N \times N[/imath] to [imath]\bf N[/imath]
I want a bijective map from [imath]\bf N \times N[/imath] to [imath]\bf N[/imath]. I can think of injective map easily. But, bijective seems to be hard to think. |
54158 | The cartesian product [imath]\mathbb{N} \times \mathbb{N}[/imath] is countable
I'm examining a proof I have read that claims to show that the Cartesian product [imath]\mathbb{N} \times \mathbb{N}[/imath] is countable, and as part of this proof, I am looking to show that the given map is surjective (indeed bijective), but I'm afraid that I can't see why this is the case. I wonder whether you might be able to point me in the right direction? Indeed, the proof begins like this: "For each [imath]n \in \mathbb{N}[/imath], let [imath]k_n, l_n[/imath] be such that [imath]n = 2^{k_n - 1} \left(2l_n - 1 \right)[/imath]; that is, [imath]k_n - 1[/imath] is the power of [imath]2[/imath] in the prime factorisation of [imath]n[/imath], and [imath]2 l_n - 1[/imath] is the (necessarily odd) number [imath]\frac{n}{2^{k_n - 1}}[/imath]." It then states that [imath]n \mapsto \left(k_n , l_n \right)[/imath] is a surjection from [imath]\mathbb{N}[/imath] to [imath]\mathbb{N} \times \mathbb{N}[/imath], and so ends the proof. I can intuitively see why this should be a bijection, I think, but I'm not sure how to make these feelings rigorous? I suppose I'd say that the map is surjective since given any [imath]\left(k_n , l_n \right) \in \mathbb{N} \times \mathbb{N}[/imath] we can simply take [imath]n[/imath] indeed to be equal to [imath]2^{k_n - 1} \left(2l_n - 1 \right)[/imath] and note that [imath]k_n - 1 \geq 0[/imath] and thus [imath]2^{k_n - 1}[/imath] is both greater or equal to one so is a natural number (making the obvious inductive argument, noting that multiplication on [imath]\mathbb{N}[/imath] is closed), and similarly that [imath]2 l_n - 1 \geq 2\cdot 1 - 1 = 1[/imath] and is also a natural number, and thus the product of these two, [imath]n[/imath] must also be a natural number. Is it just as simple as this? I suppose my gut feeling in the proving that the map is injective would be to assume that [imath]2^{k_n - 1} \left(2 l_n - 1 \right) = 2^{k_m - 1} \left(2 l_m - 1 \right)[/imath] and then use the Fundamental Theorem of Arithmetic to conclude that [imath]n = m[/imath]. Is this going along the right lines? The 'implicit' definition of the mapping has me a little confused about the approach. On a related, but separate note, I am indeed aware that if [imath]K[/imath] and [imath]L[/imath] are any countable sets, then so is [imath]K \times L[/imath], so trivially, taking the identity mapping we see trivially that this map is bijective and therefore that [imath]\mathbb{N}[/imath] is certainly countable (!), and thus so is [imath]\mathbb{N} \times \mathbb{N}[/imath]. Hence, it's not really the statement that I'm interested in, but rather the exciting excursion into number theory that the above alternative proof provides. | 490324 | Defining bijective function [imath]f:\mathbb{N}\times\mathbb N\to\mathbb N[/imath]
I want to prove that [imath]\mathbb N\times \mathbb N[/imath] is countable set using cantor first diagonal method: where every-time we count the elemnts on the digonal with the direction of the arrow ([imath](1,1)\mapsto1,(2,1)\mapsto 2,(1,2)\mapsto3,(3,1)\mapsto4[/imath] etc). I know that if I had same problem with [imath]\mathbb N \cup{0}\times \mathbb N \cup{0}[/imath] I could define cantor's function by [imath]f(x,y)=\frac {(x+y+1)(x+y)}{2}+y[/imath] which I'm not sure its' bijective. How can I define a function according to my scheme? EDIT: The question is NOT about proving [imath]\mathbb N\times \mathbb N[/imath] is countable but on writing appropriate bijective function for the described diagonal method. |
43964 | If [imath]f(xy)=f(x)f(y)[/imath] then show that [imath]f(x) = x^t[/imath] for some t
Let [imath]f(xy) =f(x)f(y)[/imath] for all [imath]x,y\geq 0[/imath]. Show that [imath]f(x) = x^p[/imath] for some [imath]p[/imath]. I am not very experienced with proof. If we let [imath]g(x)=\log (f(x))[/imath] then this is the same as [imath]g(xy) = g(x) + g(y)[/imath] I looked up the hint and it says let [imath]g(x) = \log f(a^x) [/imath] The wikipedia page for functional equations only states the form of the solutions without proof. Attempt Using the hint (which was like pulling a rabbit out of the hat) Restricting the codomain [imath]f:(0,+\infty)\rightarrow (0,+\infty)[/imath] so that we can define the real function [imath]g(x) = \log f(a^x)[/imath] and we have [imath]g(x+y) = g(x)+ g(y)[/imath] i.e [imath]g(x) = xg(1)[/imath] as [imath]g(x)[/imath] is continuous (assuming [imath]f[/imath] is). Letting [imath]\log_a f(a) = p[/imath] we get [imath]f(a^x) =a^p [/imath]. I do not have a rigorous argument but I think I can conclude that [imath]f(x) = x^p[/imath] (please fill any holes or unspecified assumptions) Different solutions are invited | 609603 | Solutions of [imath]f(x)\cdot f(y)=f(x\cdot y)[/imath]
Can anyone give me a classification of the real functions of one variable such that [imath]f(x)f(y)=f(xy)[/imath]? I have searched the web, but I haven't found any text that discusses my question. Answers and/or references to answers would be appreciated. |
255067 | Using Fermat's Little Theorem Prove if [imath]p[/imath] is prime, prove [imath]1^p + 2^p + 3^p +...+(p-1)^p \equiv 0 \bmod{p}[/imath]
Using Fermat's Theorem prove if [imath]p[/imath] is prime, prove [imath]1^p + 2^p + 3^p +...+(p-1)^p \equiv 0 \bmod{p}[/imath] The two definitions of Fermat's Little Theorem is [imath]a^p \equiv a \bmod{p}[/imath] and [imath]a^{p-1} \equiv 1 \bmod{p}[/imath] but I don't know how to use this solve the problem | 255843 | Using Fermat's Little Theorem, prove [imath]1^{p-1} + 2^{p-1} + 3^{p-1} +\ldots+(p-1)^{p-1} \equiv -1 \pmod p[/imath]
Using Fermat's Little Theorem, prove [imath]1^{p-1} + 2^{p-1} + 3^{p-1} +\ldots+(p-1)^{p-1} \equiv -1 \pmod p[/imath] where [imath]p[/imath] is a prime. So we would use Fermat's theorem: [imath]a^{p-1}\equiv 1\pmod p[/imath] Would the proof go something like this....? [imath]1^{p-1} + 2^{p-1} + 3^{p-1} +\ldots+(p-1)^{p-1} \equiv 1+\ldots+1[/imath] There would be [imath]p-1[/imath] many [imath]1[/imath]'s so (i.e. [imath]p-1=5-1=4[/imath] so [imath]1+1+1+1=4[/imath]) [imath]p-1 \equiv -1\pmod p[/imath] [imath]p \equiv 0 \pmod p[/imath] Would that be right? |
302179 | Proof Regarding Property of Odd Integers
The question I am working on is: "Use a direct proof to show that every odd integer is the difference of two squares." Proof: Let n be an odd integer: [imath]n = 2k + 1[/imath], where [imath]k \in Z[/imath] Let the difference of two different squares be, [imath]a^2-b^2[/imath], where [imath]a,b \in Z[/imath]. Hence, [imath]n=2k+1=a^2-b^2[/imath]... As you can see, this a dead-end. Appealing to the answer key, I found that they let the difference of two different squares be, [imath](k+1)^2-k^2[/imath]. I understand their use of [imath]k[/imath]--[imath]k[/imath] is one number, and [imath]k+1[/imath] is a different number--;however, why did they choose to add [imath]1[/imath]? Why couldn't we have added [imath]2[/imath]? | 263101 | Prove every odd integer is the difference of two squares
I know that I should use the definition of an odd integer ([imath]2k+1[/imath]), but that's about it. Thanks in advance! |
305923 | Question about interpretation of the divergence of a vector field
I've been explained that a vector field, when seen as "arrows" in the plane, has 0 divergence when its magnitude doesn't change, i.e. when the "arrows" keep same length. But the following examples puzzle me: [imath]F(x)=x/|x|[/imath] has always norm 1 but its divergence is not 0 [imath]F(x)=x/|x|^2[/imath] has not constant norm but its divergence is 0 Is there some contradiction or do I have a wrong/incomplete picture? | 305738 | About the divergence of vector fields
I've been explained that a vector field, when seen as "arrows" in the plane, has 0 divergence when its magnitude doesn't change, i.e. when the "arrows" keep same length. But the following examples puzzle me: [imath]F(x)=x/|x|[/imath] has always norm 1 but its divergence is not 0 [imath]F(x)=x/|x|^2[/imath] has not constant norm but its divergence is 0 Is there some contradiction or do I have a wrong/incomplete picture? |
9933 | Why is negative times negative = positive?
Someone recently asked me why a negative [imath]\times[/imath] a negative is positive, and why a negative [imath]\times[/imath] a positive is negative, etc. I went ahead and gave them a proof by contradiction like so: Assume [imath](-x) \cdot (-y) = -xy[/imath] Then divide both sides by [imath](-x)[/imath] and you get [imath](-y) = y[/imath] Since we have a contradiction, then our first assumption must be incorrect. I'm guessing I did something wrong here. Since the conclusion of [imath](-x) \cdot (-y) = (xy)[/imath] is hard to derive from what I wrote. Is there a better way to explain this? Is my proof incorrect? Also, what would be an intuitive way to explain the negation concept, if there is one? | 495109 | Product of '-' and '-' is '+'!
In School, students were taught that "if [imath]m[/imath] and [imath]n[/imath] are two non-negative integer, then their product [imath]m \times n[/imath] is equal to the sum of [imath]n[/imath], [imath]m[/imath]-times or sum of [imath]m[/imath], [imath]n[/imath]-times". If one of them is negative, then the principle still work. After this they were taught that "[imath] - \times - = +[/imath]", so the product of two negative integers is well defined. Does this follow from the above principle? Some people told me that it follows from the above principle, but i don't know how it follow. If someone ask you that how "[imath] - \times - = +[/imath]", what will be your answer? Thanks in advance |
65011 | How to prove that the Binet formula gives the terms of the Fibonacci Sequence?
This formula provides the [imath]n[/imath]th term in the Fibonacci Sequence, and is defined using the recurrence formula: [imath]u_n = u_{n − 1} + u_{n − 2}[/imath], for [imath]n > 1[/imath], where [imath]u_0 = 0[/imath] and [imath]u_1 = 1[/imath]. Show that [imath]u_n = \frac{(1 + \sqrt{5})^n - (1 - \sqrt{5})^n}{2^n \sqrt{5}}.[/imath] Please help me with its proof. Thank you. | 405189 | How do I prove Binet's Formula?
My initial prompt is as follows: For [imath]F_{0}=1[/imath], [imath]F_{1}=1[/imath], and for [imath]n\geq 1[/imath], [imath]F_{n+1}=F_{n}+F_{n-1}[/imath]. Prove for all [imath]n\in \mathbb{N}[/imath]: [imath]F_{n-1}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n\right)[/imath] Which, to my understanding, is Binet's Formula. I came up with a proof strategy similar to that described here. But as you can see, it is considering the case of [imath]F_n[/imath], not [imath]F_{n-1}[/imath]. So, my question is: does the same strategy hold for [imath]F_{n-1}[/imath]? If not, how is it different? |
25983 | When is the vector space of continuous functions on a compact Hausdorff space finite dimensional?
I know that the vector space of all real valued continuous functions on a compact Hausdorff space can be infinite dimensional. When will it be finite dimensional? And how will I identify that vector space with [imath]\mathbb{R}^n[/imath] for some [imath]n[/imath]? | 144090 | space of continuous functions
How to prove this fact: If [imath]X \subset \mathbb{R}[/imath] is finite, then the space of continuous functions [imath]C(X)[/imath] is finite dimensional (isomorphic to [imath]\mathbb{C}^n[/imath])? Thanks for help! |
131678 | Positive series problem: [imath]\sum\limits_{n\geq1}a_n=+\infty[/imath] implies [imath]\sum_{n\geq1}\frac{a_n}{1+a_n}=+\infty[/imath]
Let [imath]\sum\limits_{n\geq1}a_n[/imath] be a positive series, and [imath]\sum\limits_{n\geq1}a_n=+\infty[/imath], prove that: [imath]\sum_{n\geq1}\frac{a_n}{1+a_n}=+\infty.[/imath] | 395292 | If [imath]\sum a_n [/imath] is a positive series that diverges, does [imath]\sum \frac{a_n}{1+a_n}[/imath] diverge?
Let [imath] \{a_n\}_{n=1}^\infty [/imath] be a sequence such that [imath]\displaystyle \sum a_n [/imath] that is divergent to [imath]+\infty[/imath]. What can be said about the convergence of [imath]\displaystyle \sum \frac{a_n}{1 + a_n} [/imath]? Any hints, thoughts or leads would be greatly appreciated. Thanks! |
299099 | What is the value of [imath]i+i^2+i^3+\cdots+i^{23}[/imath]?
Can anyone help me with this question and show me a step by step solution please? The imaginary number is [imath]i[/imath] is defined such that [imath]i^2=-1[/imath]. What is [imath]i+i^2+i^3+\cdots+i^{23}[/imath]? | 4062 | Adding powers of [imath]i[/imath]
I've been struggling with figuring out how to add powers of [imath]i[/imath]. For example, the result of [imath]i^3 + i^4 + i^5[/imath] is [imath]1[/imath]. But how do I get the result of [imath]i^3 + i^4 + ... + i^{50}[/imath]? Writing it all down would be pretty mundane... It has to do something with division by 4, since the "power cycle" of [imath]i[/imath] repeats every fourth power. Thank you for any clues. |
20913 | Are continuous self-bijections of connected spaces homeomorphisms?
I hope this doesn't turn out to be a silly question. There are lots of nice examples of continuous bijections [imath]X\to Y[/imath] between topological spaces that are not homeomorphisms. But in the examples I know, either [imath]X[/imath] and [imath]Y[/imath] are not homeomorphic to one another, or they are (homeomorphic) disconnected spaces. My Question: Is there a connected topological space [imath]X[/imath] and a continuous bijection [imath]X\to X[/imath] that is not a homeomorphism? For the record, my example of a continuous bijection [imath]X\to X[/imath] that is not a homeomorphism is the following. Roughly, the idea is to find an ordered family of topologies [imath]\tau_i[/imath] ( [imath]i\in \mathbb Z[/imath]) on a set [imath]S[/imath] and use the shift map to create a continuous bijection from [imath]\coprod_{i\in \mathbb Z} (S, \tau_i)[/imath] to itself. Let [imath]S = \mathbb{Z} \coprod \mathbb Z[/imath]. The topology [imath]\tau_i[/imath] is as follows: if [imath]i<0[/imath], then the left-hand copy of [imath]\mathbb Z[/imath] is topologized as the disjoint union of the discrete topology on [imath][-n, n][/imath] and the indiscrete topology on its complement, while the right-hand copy of [imath]\mathbb Z[/imath] is indiscrete. The space [imath](S, \tau_0)[/imath] is then indiscrete. For [imath]i>0[/imath], the left-hand copy of [imath]\mathbb Z[/imath] is indiscrete, while the right-hand copy is the disjoint union of the indiscrete topology on [imath][-n, n][/imath] with the discrete topology on its complement. Now the map [imath]\coprod_{i\in \mathbb Z} (S, \tau_i)\to \coprod_{i\in \mathbb Z} (S, \tau_i)[/imath] sending [imath](S, \tau_i) \to (S, \tau_{i+1})[/imath] by the identity map of [imath]S[/imath] is a continuous bijection, but not a homeomorphism. | 1524976 | Example of a metrizable space and a continuous self-bijection that is not a homeomorphism
I'm looking for an example of a metrizable space [imath](X, \tau)[/imath] and a continuous bijection [imath]f : (X, \tau) \to (X, \tau)[/imath] that is not a homeomorphism. One hint I have is to let [imath]Y[/imath] be a fixed set and to let [imath]\{ \tau_n : n \in \mathbb{Z} \} [/imath] be a family of topologies on [imath]Y[/imath] such that [imath]\tau_n \subseteq \tau_{n+1}[/imath] for all [imath]n \in \mathbb{Z}[/imath]. Then let [imath]Y_n[/imath] denote [imath]Y[/imath] equipped with the topology [imath]\tau_n[/imath], and let [imath](X, \tau)[/imath] be [imath]\Pi_{n \in \mathbb{Z}} Y_n[/imath] with the Tychonoff topology. Any hints or answers would be appreciated. |
3315 | What is [imath]\sqrt{i}[/imath]?
If [imath]i=\sqrt{-1}[/imath], is [imath]\large\sqrt{i}[/imath] imaginary? Is it used or considered often in mathematics? How is it notated? | 411174 | Square roots of complex numbers
I know that the square root of a number x, expressed as [imath]\displaystyle\sqrt{x}[/imath], is the number y such that [imath]y^2[/imath] equals x. But is there any simple way to calculate this with complex numbers? How? |
97229 | How many rationals of the form [imath]\large \frac{2^n+1}{n^2}[/imath] are integers?
This was Problem 3 (first day) of the 1990 IMO. A full solution can be found here. How many rationals of the form [imath]\large \frac{2^n+1}{n^2},[/imath] [imath](n \in \mathbb{N} )[/imath] are integers? The possible values of [imath]n[/imath] that i am able to find is [imath]n=1[/imath] and [imath]n=3[/imath], so there are two solutions and this seems to be the answer to this problem. But now we have to prove that no more of such [imath]n[/imath] exists, and thus the proof reduces to: Proving that [imath]n^2[/imath] does not divides [imath]2^n+1[/imath] for any [imath]n \gt 3[/imath]. Does anybody know how to prove this? | 362721 | When is [imath]\frac{2^n+1}{n^2}[/imath] an integer?
Can anyone see how to solve this number puzzle? Find all integers [imath]n>1[/imath] such that [imath]\frac{2^n+1}{n^2}[/imath] is an integer. |
108962 | Steps to solve this system of equations: [imath]\sqrt{x}+y=7[/imath], [imath]\sqrt{y}+x=11[/imath]
I want to solve this system of equations, I have been out of Maths for a long!! [imath]\sqrt{x}+y=7[/imath] [imath]\sqrt{y}+x=11[/imath] Just wondering easiest step to find values for [imath]x[/imath] and [imath]y[/imath] from the above equations? | 594083 | Finding the values of [imath]a[/imath] and [imath]b[/imath]
We are given two equations: [imath] \left\{ \begin{array}{c} \sqrt{a}+b=11 \\ a+\sqrt{b}=7 \\ \end{array} \right. [/imath] How to find the value of [imath]a[/imath] and [imath]b[/imath] by solving the equation? All I could do was use the hit and trial method. Help me with this! |
164074 | How to evaluate [imath] \lim \limits_{n\to \infty} \sum \limits_ {k=1}^n \frac{k^n}{n^n}[/imath]?
I can show that the following limit exists but I am having difficulties to find it. It is [imath]\lim_{n\to \infty} \sum_{k=1}^n \frac{k^n}{n^n}[/imath] Can someone please help me? | 493537 | Finding [imath]\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k^n}{n^n}[/imath] if it exists
I've known the following: [imath]\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k^k}{n^n}=1.[/imath] Then, I got interested in the following similar limitation: [imath]\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k^n}{n^n}[/imath] By using computer, it seems that [imath]\lim_{n\to\infty}\sum_{k=1}^{n}\frac{k^n}{n^n}=\frac{e}{e-1}.[/imath] However, I haven't been able to prove this. I need your help. |
83358 | How do I prove [imath]n[/imath] is a Carmichael number?
I am having some trouble with this proof and I need some help heading down the right direction: Suppose [imath]n = p_1p_2 \cdots p_k[/imath] where [imath]p_i[/imath] are distinct primes and that [imath]p_i - 1 \mid n - 1[/imath]. Show that [imath]n[/imath] is a Carmichael number, that is, that [imath]a^{n - 1} \equiv 1 \pmod n[/imath] for all [imath]a[/imath] with [imath]\gcd(a, n) = 1[/imath]. Thank you for your help. | 1694813 | Suppose that [imath]n[/imath] is a composite, squarefree integer such that for every prime divisor [imath]p[/imath] of [imath]n[/imath]...
Suppose that [imath]n[/imath] is a composite, squarefree integer such that for every prime divisor [imath]p[/imath] of [imath]n[/imath], we have [imath](p - 1) | (n - 1)[/imath]. Prove that [imath]n[/imath] is a Carmichael number. Having a lot of trouble with this problem, looking through my notes couldn't really find anything useful apart from the definition of a Carmichael number. After going back to my professor his two hints were to use Fermat's Little Theorem and to consider what the prime factorization of n looks like. This wasn't much help, and any further assistance is greatly appreciated. |
155941 | Evaluate the integral: [imath]\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx[/imath]
Compute [imath]\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx[/imath] | 653204 | Integral of [imath]8\int_{0}^{1}\frac{\log (1+x)}{1+x^2}dx[/imath]
So we got this problem [imath]8\int_{0}^{1}\frac{\log (1+x)}{1+x^2}dx[/imath] I have been stuck on this problem for days basically I tried everything I could think of to solve this integral i tried substituting [imath]x=\tan p[/imath] and integral became [imath]8\int_{0}^{1}\frac{\log (1+\tan p)}{1+\tan^2 p}(\sec^2 p)\, dp[/imath] then I was stuck at [imath]8\int_{0}^{1}{\log (1+\tan p)}\,dp [/imath] now nowhere to go from here can you suggest me another way to approach this problem ? |
7793 | Bounding the integral [imath]\int_{2}^{x} \frac{\mathrm dt}{\log^{n}{t}}[/imath]
If [imath]x \geq 2[/imath], then how do we prove that [imath]\int_{2}^{x} \frac{\mathrm dt}{\log^{n}{t}} = O\Bigl(\frac{x}{\log^{n}{x}}\Bigr)?[/imath] | 944637 | How to prove that [imath] \int_{2}^{x} \frac{dt}{(\log(t))^{k}} = O \Big{(} \frac{x}{(\log(x))^{k}} \Big{)} [/imath] as [imath]x \to \infty[/imath]?
For a homework exercise, we are asked to prove that [imath] \int_{2}^{x} \frac{dt}{(\log(t))^{k}} = O \Big{(} \frac{x}{(\log(x))^{k}} \Big{)} \quad \text{, as } x \to \infty . [/imath] The following hint is given: "Split the integral into [imath]\int_{2}^{f(x)} + \int_{f(x)}^{x} [/imath] for a well-chosen function [imath]f(x)[/imath] with [imath] 2\leq f(x) \leq x[/imath] and estimate both parts from above. I tried to use the functions [imath]f(x) = \sqrt{x}[/imath] and [imath]f(x) = \log(x)[/imath], but these functions don't seem to work. Do you know which function [imath]f[/imath] I ought to choose in order to be able to prove what was asked? |
57317 | Construction of a Borel set with positive but not full measure in each interval
I was wondering how one can construct a Borel set that doesn't have full measure on any interval of the real line but does have positive measure everywhere. To be precise, if [imath]\mu[/imath] denotes Lebesgue measure, how would one construct a Borel set [imath]A \subset \mathbb{R}[/imath] such that [imath]0 < \mu(A \cap I) < \mu(I)[/imath] for every interval [imath]I[/imath] in [imath]\mathbb{R}[/imath]? Moreover, would such a set necessarily have to contain infinite measure? | 551908 | Two exercises regarding measures and outer measures
I'm doing some self educating about Measure Theory and I've come across two exercises I haven't managed to solve: Given [imath]\mathcal{L}[/imath] the Lebesgue measure on [imath]\mathbb{R}[/imath] construct a Borel-Set [imath]\mathbb{E}\subseteq\mathbb{R}[/imath] such that [imath]0<\mathcal{L}\left(E\cap\left[a,b\right]\right)<\mathcal{L}\left(\left[a,b\right]\right)[/imath] for all [imath]-\infty<a<b<\infty[/imath] Show that if [imath]A\subseteq\mathbb{R}[/imath] is a measurable set for which there is an [imath]\alpha\in\left[0,1\right)[/imath] such that [imath]\mathcal{L}\left(A\cap\left[a,b\right]\right)\leq\alpha\mathcal{L}\left(\left[a,b\right]\right)[/imath] for all [imath]\left[a,b\right]\subseteq \mathbb{R}[/imath] then [imath]\mathcal{L}\left(A\right)=0[/imath] Help would be very appreciated as I am quite stumped. |
28751 | Proof of upper-tail inequality for standard normal distribution
[imath]X \sim \mathcal{N}(0,1)[/imath], then to show that for [imath]x > 0[/imath], [imath] \mathbb{P}(X>x) \leq \frac{\exp(-x^2/2)}{x \sqrt{2 \pi}} \>. [/imath] | 805453 | limiting behavior of standard normal survivor function
How do you show that [imath]\lim_{x\to \infty} 1-\Phi(x) \sim \phi(x)/x[/imath]? In the previous, I'm using [imath]\Phi[/imath] to refer to the standard normal CDF and [imath]\phi[/imath] to refer to the standard normal pdf. Thanks!! |
120 | Why is 1 not a prime number?
Why is [imath]1[/imath] not considered a prime number? Or, why is the definition of prime numbers given for integers greater than [imath]1[/imath]? | 1073378 | Why [imath]1[/imath] isn't a prime?
I was wondering the reason behind defining the Prime Numbers in a manner of which [imath]1[/imath] isn't an example. I read in Rotman's A First Course in Abstract Algebra that one reason that [imath]1[/imath] is not called a prime is that many theorems involving primes would otherwise be more complicated to state. So, here are my questions, Can anyone give examples of many theorems involving primes would be complicated to state had [imath]1[/imath] been considered a prime? What are other reasons for not considering [imath]1[/imath] a prime apart from what Rotman said? |
4787 | Motivation behind standard deviation?
Let's take the numbers 0-10. Their mean is 5, and the individual deviations from 5 are -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 And so the average (magnitude of) deviation from the mean is [imath]30/11 \approx 2.72[/imath]. However, this is not the standard deviation. The standard deviation is [imath]\sqrt{10} \approx 3.16[/imath]. The first mean-deviation is a simpler and by far more intuitive definition of the "standard-deviation", so I'm sure it's the first definition statisticians worked with. However, for some reason they decided to adopt the second definition instead. What is the reasoning behind that decision? | 460940 | Why the definition of Variance is such.
Why we define the variance of a random variable [imath]X[/imath] as [imath]\text{var}[X]=\text{E}[(X-\mu)^2][/imath] instead of [imath]\text{var}[X]=\text{E}[\left|X-\mu\right|][/imath]. Normally we understand the standard deviation [imath]\sigma=\sqrt{\text{var}[X]}[/imath] as a measure of the average distance of a sample from the mean. If this is the case, isn't it more reasonable to use the absolute value instead of squaring as a measure of distance? Then this way we don't have to square afterwards to obtain [imath]\sigma[/imath]. |
171073 | How to evaluate these integrals by hand
How can I evaluate[imath]\int_{-\infty}^\infty \frac{\cos x}{\cosh x}\,\mathrm dx\text{ and }\int_0^\infty\frac{\sin x}{e^x-1}\,\mathrm dx.[/imath] Thanks in advance. | 977036 | Evaluate [imath] \int_0^\infty \cos(x)/\cosh(x) dx [/imath]
I need to evaluate [imath]\int_0^\infty \frac{\cos(x)}{\cosh(x)} dx [/imath] through Residue Theory, but it has an infinite number of poles in x=(2n+1)pi/2, I couldn't find the way to the solution that is [imath]\frac{\pi\ }{2 \cosh \ (\pi\ /2)}.[/imath] I tried also the Method of the Fourier Integral, but couldn't find the answer too. I'm in this question for over 1.5h and have a test tomorrow, can someone help me? |
797 | What's so "natural" about the base of natural logarithms?
There are so many available bases. Why is the strange number [imath]e[/imath] preferred over all else? Of course one could integrate [imath]\frac{1}x[/imath] and see this. But is there more to the story? | 1755386 | Why is Euler's number used as a base for logarithms?
Is there some special property of '[imath]e[/imath]' which makes it suitable to be used as a base for logarithms? Moreover, does the natural logarithm possess some advantage over the common logarithm? I don't understand why there is a need to choose an irrational number, '[imath]e[/imath]', for a base. Isn't it much simpler to use 10 as a base? |
231596 | Induction: [imath]\sum_{k=0}^n \binom nk k^2 = n(1+n)2^{n-2}[/imath]
I found crazy (for me at least) induction example, in fact it just would be nice to prove. (Even have problems with starting) Any hints are highly valued: [imath]0^2\binom{n}{0}+1^2\binom{n}{1}+2^2\binom{n}{2}+\cdots+n^2\binom{n}{n}=n(1+n)2^{n-2} [/imath] | 305317 | Prove that [imath]\sum_{k=0}^n \binom{n}{k} k^2 = n(n+1) 2^{n-2}[/imath]
[imath]\sum_{k=0}^n \binom{n}{k} k^2 = n(n+1) 2^{n-2}[/imath] |
58560 | Elementary central binomial coefficient estimates
How to prove that [imath]\quad\displaystyle\frac{4^{n}}{\sqrt{4n}}<\binom{2n}{n}<\frac{4^{n}}{\sqrt{3n+1}}\quad[/imath] for all [imath]n[/imath] > 1 ? Does anyone know any better elementary estimates? | 329446 | Theta asymptotic for [imath]\binom{2m}{m}[/imath]
Show that [imath]\binom{2m}{m} = \Theta\left(\frac{2^{2m}}{\sqrt{m}}\right)[/imath] without using Stirling's approximation. |
270064 | Does the series [imath] \sum\limits_{n=1}^{\infty} \frac{1}{n^{1 + |\sin(n)|}} [/imath] converge or diverge?
Does the following series converge or diverge? I would like to see a demonstration. [imath] \sum_{n=1}^{\infty} \frac{1}{n^{1 + |\sin(n)|}}. [/imath] I can see that: [imath] \sum_{n=1}^{\infty} \frac{1}{n^{1 + |\sin(n)|}} \leqslant \sum_{n=1}^{\infty} \frac{1}{n^{1 + {\sin^{2}}(n)}} \leqslant \sum_{n=1}^{\infty} \frac{1}{n^{1 + {\sin^{2n}}(n)}}. [/imath] | 406625 | Is the series [imath]\sum_{n=1}^\infty n^{-1-|\!\sin n|}[/imath]convergent?
As in the title, is the series [imath] \sum_{n=1}^{+\infty}\frac{1}{n^{1+|\!\sin(n)|}} [/imath] convergent or not? This is a variant of a simpler exercise in a first year curse in analysis, so I don't know if there is a simple answer (in any case wasn't able to find one). Thank you! |
164874 | [imath]|2^x-3^y|=1[/imath] has only three natural pairs as solutions
Consider the equation [imath]|2^x-3^y|=1[/imath] in the unknowns [imath]x \in \mathbb{N}[/imath] and [imath]y \in \mathbb{N}[/imath]. Is it possible to prove that the only solutions are [imath](1,1)[/imath], [imath](2,1)[/imath] and [imath](3,2)[/imath]? | 2245417 | Finding integer solutions of [imath] 2^{m} + 1 = 3^{n} [/imath].
How do you prove that the only integer solutions of [imath]2^m+1=3^n[/imath] are at [imath]m=1, n=1[/imath] and [imath]m=3, n=2[/imath]? I have observed the following: [imath]9\equiv1[/imath] (mod [imath]4[/imath]), so any [imath]n>1[/imath] that is part of an integer solution must be even. [imath]3^n-1=2^m[/imath], so [imath]2^{m-1}=\sum_{k=0}^{n-1}3^k[/imath]. [imath]8\equiv-1[/imath] (mod [imath]9[/imath]), so any [imath]m>1[/imath] that is part of an integer solution must be odd divisible by 3. However, none of these observations have gotten me anywhere. I have no idea how to continue and would appreciate any help you can give. It would also be interesting to know if there are ways to solve more generalized versions of these kinds of problems (such as finding integer solutions to [imath]2^m+1=k^n[/imath], [imath]2^m+k=3^n[/imath], etc.). |
69125 | Inequality between [imath]\ell^p[/imath]-norms
Suppose that a sequence [imath]x=(x_n)[/imath] belongs both to [imath]\ell^p[/imath] and [imath]\ell^q[/imath] ([imath]p,q>1[/imath], [imath]p\neq q[/imath]). Is there any inequality between [imath]\|x\|_p[/imath] and [imath]\|x\|_q[/imath]. Can one [imath]\ell^p[/imath] be continuously embedded into another [imath]\ell^q[/imath]? | 405838 | Convergent sequences
If [imath]l^p=\{\langle x_k \rangle \in R^n | \sum_{k=1}^{\infty} |x_k|^p < \infty \}[/imath] and [imath]1\leq p<q[/imath], then [imath]l^p\subset l^q[/imath]. Prove. I know that I have to show that if [imath]x=\langle x_k \rangle\in l^p\rightarrow x\in l^q[/imath] and I can use a fact that [imath]l^p\subset c_0[/imath], where [imath]c_0=\{x=\langle x_k \rangle \in R^n | \displaystyle \lim_{k \to +\infty} x_k=0\}[/imath]. |
298063 | Show if the elements of [imath]B[/imath] are integral and [imath]\Delta_{B}(K)[/imath] is square-free, then [imath]\mathcal{O}_K = \mathbb{Z}b_1 + \cdots + \mathbb{Z}b_d[/imath].
Let [imath]K[/imath] be a number field of degree [imath]d[/imath], and let [imath]B = \{b_1,\ldots,b_d\}[/imath] be a subset of [imath]K[/imath] of cardinality [imath]d[/imath] such that the matrix [imath](\mathrm{Tr}(b_ib_j))^d_{i,j=1}[/imath] has non-zero determinant [imath]\Delta_{B}(K)[/imath]. i) Show that [imath]B[/imath] is a basis for [imath]K[/imath] as a vector space over [imath]\mathbb{Q}[/imath]. ii) Show if the elements of [imath]B[/imath] are integral and [imath]\Delta_{B}(K)[/imath] is square-free, then [imath]\mathcal{O}_K = \mathbb{Z}b_1 + \cdots + \mathbb{Z}b_d[/imath]. | 297184 | Show that [imath]B[/imath] is a basis for [imath]K[/imath] as a vector space over [imath]\mathbb{Q}[/imath]
Let [imath]K[/imath] be a number field of degree [imath]d[/imath], and let [imath]B = \{b_1,\ldots,b_d\}[/imath] be a subset of [imath]K[/imath] of cardinality [imath]d[/imath] such that the matrix [imath](\mathrm{Tr}(b_ib_j))^d_{i,j=1}[/imath] has non-zero determinant [imath]\Delta_{B}(K)[/imath]. i) Show that [imath]B[/imath] is a basis for [imath]K[/imath] as a vector space over [imath]\mathbb{Q}[/imath]. ii) Show if the elements of [imath]B[/imath] are integral and [imath]\Delta_{B}(K)[/imath] is square-free, then [imath]\mathcal{O}_K = \mathbb{Z}b_1 + \cdots + \mathbb{Z}b_d[/imath]. |
83383 | Midpoint-Convex and Continuous Implies Convex
Given that [imath]f\left(\frac{x+y}{2}\right)\leqslant \frac{f(x)+f(y)}{2}~,[/imath] how can I show that [imath]f[/imath] is convex. Thanks. Edit: I'm sorry for all the confusion. [imath]f[/imath] is assumed to be continuous on an interval [imath](a,b)[/imath]. | 721524 | Convex continuous functions
\begin{equation} w \left (\dfrac{x+y}{2} \right ) \le \dfrac{1}{2}(w(x) +w(y)) \quad \mbox{for all} \quad x,y \in \Omega, \end{equation} is a sufficient condition to a continuous function [imath]w \in C^0(\Omega)[/imath] be convex? |
8385 | Prove that [imath]\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}[/imath]
Using [imath]\text{n}^{\text{th}}[/imath] root of unity [imath]\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1[/imath] Prove that [imath]\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}[/imath] | 1001278 | how to prove [imath]\prod_{k=1}^{p-1} \sin(\frac{\pi k}{p}) = \frac{p}{2^{p-1}}[/imath]?
i found this relation whilst trying to evaluate the norm (over [imath]\mathbb{Q}[/imath]) of [imath]1-\zeta[/imath] for [imath]\zeta[/imath] a primitive [imath]p[/imath]-th root of unity ([imath]p[/imath] supposed prime) [imath] \prod_{k=1}^{p-1} \sin(\frac{\pi k}{p}) = \frac{p}{2^{p-1}} [/imath] as yet i have no means of proving it. any suggestions? |
65454 | The last two digits of [imath]9^{9^9}[/imath]
I tried to calculate the last two digits of [imath]9^{9^9}[/imath] using Euler's Totient theorem, what I got is that it is same as the last two digits of [imath]9^9[/imath]. How do I proceed further? | 657131 | Find the last two digits of [imath]9^{9^{9}}[/imath]
I want to find the last two digits of [imath]9^{9^9}[/imath] or [imath]9^{81}[/imath]. I tried using Euler's theorem but I can't make anything of it. Any hint or a guide? Thanks! |
199289 | Combinatorially prove that [imath]\sum_{i=0}^n {n \choose i} 2^i = 3^n [/imath]
So I'm not sure at all how to prove things using a combinatorial proof. Where to do I start? What do i need to think about etc. For example how would i prove [imath]\sum_{i=0}^n {n \choose i} 2^i = 3^n [/imath] | 1549998 | Prove [imath]\sum_{i=0}^n {{n \choose i} \times 2^i} = 3^n [/imath]
I have to prove that [imath]\sum_{i=0}^n {{n \choose i} \times 2^i} = 3^n [/imath] Such that [imath]{n \choose i} = \frac{n!}{i!(n-i)!} [/imath] and [imath]n [/imath] is some arbitrary int I proved we can expand 2^I in a way such that [imath]2^i = {i \choose 0} +{i \choose 1} + ... + {i \choose i} [/imath] But this doesn't seem to approach the proof is want so I tried to expand out [imath]{n \choose i}[/imath] and again this got ugly and led no where. I can't quite seem to find any approach that would lead to [imath]3[/imath] of any power. I know this proof is true for the base case[imath]\sum_{i=0}^0 {{0 \choose i} \times 2^i} = 1 => {0 \choose 0} \times 2^0 = 1 => 1 = 1 [/imath] Could anyone give me a hint to lead me in the right direction? EDIT: Wow, I always feel so stupid when I miss tiny identities such as that! Thank you everyone! I got my answer, I wish I could give all of you checks for the advice. |
15521 | Making Change for a Dollar (and other number partitioning problems)
I was trying to solve a problem similar to the "how many ways are there to make change for a dollar" problem. I ran across a site that said I could use a generating function similar to the one quoted below: The answer to our problem (293) is the coefficient of [imath]x^{100}[/imath] in the reciprocal of the following: [imath](1-x)(1-x^5)(1-x^{10})(1-x^{25})(1-x^{50})(1-x^{100})[/imath] But I must be missing something, as I can't figure out how they get from that to [imath]293[/imath]. Any help on this would be appreciated. | 203864 | How many ways you can make change for an amount
I am looking for a formula or at least something to use when trying to compute how many ways I can make change for an amount. Example: there are [imath]3[/imath] ways to give change for [imath]4[/imath] if you have coins with denomination [imath]1[/imath] and [imath]2[/imath]: [imath]1+1+1+1, 1+1+2, 2+2[/imath]. What formula should I use? |
622 | Importance of Representation Theory
Representation theory is a subject I want to like (it can be fun finding the representations of a group), but it's hard for me to see it as a subject that arises naturally or why it is important. I can think of two mathematical reasons for studying it: The character table of a group is packs a lot of information about the group and is concise. It is practically/computationally nice to have explicit matrices that model a group. But there must certainly be deeper things that I am missing. I can understand why one would want to study group actions (the axioms for a group beg you to think of elements as operators), but why look at group actions on vector spaces? Is it because linear algebra is so easy/well-known (when compared to just modules, say)? I am also told that representation theory is important in quantum mechanics. For example, physics should be [imath]$\mathrm{SO}(3)$[/imath] invariant and when we represent this on a Hilbert space of wave-functions, we are led to information about angular momentum. But this seems to only trivially invoke representation theory since we already start with a subgroup of [imath]$\mathrm{GL}(n)$[/imath] and then extend it to act on wave functions by [imath]\psi(x,t) \mapsto \psi(Ax,t)[/imath] for [imath]$A$[/imath] in [imath]$\mathrm{SO}(n)$[/imath]. This Wikipedia article on particle physics and representation theory claims that if our physical system has [imath]$G$[/imath] as a symmetry group, then there is a correspondence between particles and representations of [imath]$G$[/imath]. I'm not sure if I understand this correspondence since it seems to be saying that if we act an element of G on a state that corresponds to some particle, then this new state also corresponds to the same particle. So a particle is an orbit of the [imath]$G$[/imath] action? Anyone know of good sources that talk about this? | 2122571 | Representation theory's usability after one course
What are the things one can expect to understand after studying the representation of a group? I have seen many answers on pages regarding the importance and philosophy of representation theory but I am unable to understand what good it will do as a short term goal (meaning I'm not interesting in things it will enable me to study later). Suppose I have studied the representation of [imath]SO(n)[/imath] or [imath]SU(n)[/imath]. What are the questions I will be able to answer about it that I can't (or not easily) without the knowledge of representation theory? Edit: I don't understand why even after explaining that what I'm looking for is not available on that page, this question is marked as duplicate. To quote, it says now that "This question was marked as an exact duplicate of an existing question." which according to me is rather absurd. I'd argue that this is much more specific and will benefit a lot of people getting introduced to the subject. |
208740 | Cayley tables for two non-isomorphic groups of order 4.
Not sure how to make tables, but: For a binary operation [imath]*[/imath], and set [imath]\{e,a,b,c\}[/imath], in the Cayley table, [imath]a*a[/imath] can be filled with either the identity or an element different from both [imath]e[/imath] and [imath]a[/imath]. If in the table the place for [imath]a*a[/imath] is filled with [imath]e[/imath], then the rest of the table can be filled out in two different ways. What are the two different ways? Thanks | 2195277 | Let [imath]G_1[/imath] and [imath]G_2[/imath] be groups, each with two/three/four elements. Is it true that [imath]G_1\cong G_2[/imath]?
a) Let [imath]G_1[/imath] and [imath]G_2[/imath] be groups, each with exactly one element. Show [imath]G_1\cong G_2[/imath]. b) Let [imath]G_1[/imath] and [imath]G_2[/imath] be groups, each with two elements. Show [imath]G_1\cong G_2[/imath]. Does this statement also hold for three elements? And for four? a) Let [imath]g_1[/imath] be the element in [imath]G_1[/imath], and [imath]g_2[/imath] the element in [imath]G_2[/imath]. Let [imath]f\colon G_1\to G_2[/imath] be defined as f(g_1)=g_2. We know that [imath]g_1[/imath] and [imath]g_2[/imath] are the neutral elements in [imath]G_1[/imath] and [imath]G_2[/imath] respectively. So [imath]f(g_1g_1)=f(g_1)=g_2=g_2g_2=f(g_1)f(g_1)[/imath]. b) The same kind of reasoning can be applied for the case of 2 elements (where we have a neutral element, and an element [imath]a\in g_1[/imath] which is its own inverse). For three elements, we could consider [imath]e,a,b\in g_1[/imath], with either [imath]a^{-1}=b[/imath], or [imath]a^2=b^2=e[/imath]. This still seems doable for me to check. However, how am I supposed to consider the case of 4 elements? Should I still list all possibilities and check if there exists an isomorphism? Or can this be done more quickly? EDIT Based on the comments below and some help from the chat: There is only one group with three elements, because we can't have that [imath]aa=e[/imath], otherwise we would have that [imath]aa=ae=e[/imath], which would mean that [imath]a=b[/imath]. I've considered the groups [imath]\mathbb Z/4\mathbb Z[/imath] and [imath](\mathbb Z/2\mathbb Z)\times(\mathbb Z/2\mathbb Z)[/imath]. I won't write out their Cayley tables, but it is clear that these are not isomorphic groups. I think the most telling property is that for each [imath]\overline x\in(\mathbb Z/2\mathbb Z)\times(\mathbb Z/2\mathbb Z)[/imath], we have that [imath]\overline x+\overline x=e[/imath], however, for [imath]\mathbb Z/4\mathbb Z[/imath] this is not the case (consider: [imath]\overline 1+\overline 1=\overline 2\neq\overline 0[/imath]). Back to the Cayley table: these two groups are never the same on the diagonal, no matter how you order their elements. |
98065 | How many ways can [imath]b[/imath] balls be distributed in [imath]c[/imath] containers with no more than [imath]n[/imath] balls in any given container?
I think there are [imath]\binom{b+ c - 1}{c-1}[/imath] ways to distribute [imath]b[/imath] balls in [imath]c[/imath] containers. (Please correct me if that's a mistake.) How does this change if I am not allowed to place more than [imath]n[/imath] balls in any given container? If we call this number [imath]N(b,c,n)[/imath] then I can come up with [imath]N(b,c,n) = \sum_{k=0}^n N(b-k, c-1, n)[/imath] with the base cases [imath]N(b,c,n) = \binom{b+ c - 1}{c-1}[/imath] for [imath]n\geq b[/imath], and [imath]N(b,c,n) = 0[/imath] for [imath]n<b/c[/imath]. Is there some better way of writing this than that recursion relation? | 1546739 | How to solve Frog-stair problem with combinatorics instead of recursion?
The problem about a frog climbing a stair of size N, in steps of size 1 or 2, can be formulated as Fibonacci; and if it can jump in steps of size [imath]1,2,...k[/imath] there must be [imath]k[/imath] recursive calls. I want to know if its possible to solve this problem using combinations with repetition instead of using recursion? Im trying to use combinations with repetition formula to give number of positive integer solutions to the restricted equation representing the problem (for example k=3 and N=4, the equation [imath]a+b+c+d = 4[/imath] with [imath]1 \le a,b,c,d \le k[/imath] ) and after that subtracting the impossible solutions (i.e. solutions wich contain value of zero for one or more variables), but I'm having trouble restricting the equation and counting the impossible solutions. Wonder if Im walking in the right direction here? |
72690 | liminf and limsup with characteristic (indicator) function
So first let me state my homework problem: Let [imath]X[/imath] be a set, let [imath]\{A_k\}[/imath] be a sequence of subsets of [imath]X[/imath], let [imath]B = \bigcup_{n=1}^{+\infty} \bigcap_{k=n}^{+\infty} A_k[/imath], and let [imath]C = \bigcap_{n=1}^{+\infty} \bigcup_{k=n}^{+\infty} A_k[/imath]. Show that (a) [imath]\liminf_k\; {\xi_A}_{_k} = \xi_B[/imath], and [imath](b)[/imath] [imath]\limsup_k \;{\chi_A}_{_k} = \chi_C.[/imath] I know that, in the context I am familiar with, that [imath]\liminf_{k\rightarrow +\infty}\; X_k = \bigcup_{k=1}^{+\infty} \bigcap_{n=k}^{+\infty} X_n[/imath] and [imath]\limsup_{k\rightarrow +\infty}\;X_k = \bigcap_{k=1}^{+\infty} \bigcup_{n=k}^{+\infty} X_n.[/imath] I also know that the characteristics (indicator) function is defined as [imath]\chi_A(x) = \begin{cases} 1, & x \in A \\ 0, & x \notin A .\end{cases} [/imath] So I wrote out [imath]B[/imath] in some of its `glory': [imath]B= (A_1 \cap A_2 \cap A_3 \cap \cdots) \cup (A_2 \cap A_3 \cap \cdots) \cup (A_3 \cap A_4 \cap \cdots) \cup \cdots[/imath], and as the first argument is the smallest, with increasing size to the right, the last term in the expression for [imath]B[/imath] would be [imath]B[/imath], which would be the largest. So if I replace the [imath]X_k[/imath]'s above with [imath]\chi[/imath]'s I still don't see how I can get the correct answer - though it looks pretty clear from the definition of [imath]B[/imath] and that of [imath]\liminf[/imath] being basically the same, except in this case for the [imath]\chi[/imath]. Any direction would be greatly appreciated. By the way, I have checked out limsup and liminf of a sequence of subsets of a set but I was somewhat confused by the topology, the meets/joins, etc. Thanks much, Nate | 1084308 | trying to prove the characteristic function commutes with limsup
Let [imath](A_n)[/imath] be a seqeunce of sets, I am trying to show that [imath] \limsup_{n \to \infty} \chi_{A_n} = \chi_{\limsup_{n \to \infty}A_n}[/imath] [imath] \liminf_{n \to \infty} \chi_{A_n} = \chi_{\liminf_{n \to \infty}A_n}[/imath] I am stuck trying to show these identities. I was thinking if let [imath]A = \limsup A_n[/imath], then a good strategy to prove this would be to bound [imath]| \chi_{A_n} - \chi_A | [/imath]?? Any ideas would be greatly appreacited. thanks. |
47868 | Vandermonde's Identity: How to find a closed formula for the given summation
I've stumbled upon the following challenging question. Find a closed formula for the following summation [imath] S(a,b,n) = \sum_{k=0}^n {a \choose k} {b \choose n-k}[/imath] for all possible parameters [imath]a,b[/imath]. Anyone happens to see how to solve this one? | 337923 | How to prove Vandermonde's Identity: [imath]\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}[/imath]?
How can we prove that [imath]\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}?[/imath] (Presumptive) Source: Theoretical Exercise 8, Ch 1, A First Course in Probability, 8th ed by Sheldon Ross. |
64982 | Is every quotient of a finite abelian group [imath]G[/imath] isomorphic to some subgroup of [imath]G[/imath]?
I'm having difficulty with exercise 1.43 of Lang's Algebra. The question states Let [imath]H[/imath] be a subgroup of a finite abelian group [imath]G[/imath]. Show that [imath]G[/imath] has a subgroup that is isomorphic to [imath]G/H[/imath]. Thinking about this for a bit, the only reasonable approach I could think of was to construct some surjective homomorphism [imath]\phi\colon G\to K[/imath] for [imath]K\leq G[/imath], and [imath]\ker\phi=H[/imath], and then just use the isomorphism theorems to get the result. After a while of trying, I've failed to come up with a good map, since [imath]H[/imath] seems so arbitrary. I'm curious, how can one construct the desired homomorphism? This is just the approach I thought of, if there's a better one, I wouldn't mind seeing that either/instead. Thank you. | 3012053 | Subgroup of finite abelian group [imath]G[/imath] isomorphic to [imath]G/H[/imath]
Let [imath]H[/imath] be a subgroup of a finite abelian group [imath]G[/imath]. Show that [imath]G[/imath] has a subgroup that is isomorphic to [imath]G/H[/imath]. My approach: Let's try to prove by induction on [imath]o(G)[/imath]. If [imath]o(G)=1[/imath] then the result is obvious. Suppose it is true for all groups with order less than [imath]n[/imath]. Suppose that [imath]o(G)=n[/imath] and [imath]H\neq \{e\},G[/imath] because otherwise the result is trivial. Let [imath]x\in H[/imath] s.t. [imath]x\neq e[/imath] and consider subgroup generated by [imath]x[/imath], i.e. [imath]\langle x\rangle[/imath] then [imath]H/\langle x\rangle\leq G/ \langle x\rangle[/imath]. Note that [imath](G/\langle x\rangle)/(H/\langle x\rangle)\cong G/H[/imath]. Since [imath]o(G/\langle x\rangle)<o(G)=n[/imath] then by induction hypothesis there exists subgroup [imath]\bar{K}\leq G/\langle x\rangle[/imath] such that [imath]\bar{K}\cong (G/\langle x\rangle)/(H/\langle x\rangle)[/imath] [imath]\Rightarrow[/imath] [imath]\bar{K}\cong G/H[/imath]. Then I was trying to find subgroup of [imath]G[/imath] which is isomorphic to [imath]G/H[/imath] but no results. I've considered the surjective mapping [imath]\pi: G\to G/\langle x\rangle[/imath] which does give any useful results. I would be very thankful if anyone can show how to complete my approach? |
115822 | How to show that [imath]\lim_{n \to +\infty} n^{\frac{1}{n}} = 1[/imath]?
I've spent the better part of this day trying to show from first principles that this sequence tends to 1. Could anyone give me an idea of how I can approach this problem? [imath] \lim_{n \to +\infty} n^{\frac{1}{n}} [/imath] | 419492 | Limit of the root [imath]\sqrt[k]{k}[/imath]
Haow can I calculate this limit? [imath]L=\lim_{k\to\infty}\sqrt[k]{k}[/imath] I suppose its value is one, but how can I prove it? Thanks |
261326 | How many [imath]N[/imath] digits binary numbers can be formed where [imath]0[/imath] is not repeated
How many [imath]N[/imath] digits binary numbers can be formed where [imath]0[/imath] is not repeated. Note - first digit can be [imath]0[/imath]. I am more interested on the thought process to solve such problems, and not just the answer. If anyone can cite some resources for learning how to solve such problems would be great. | 1470414 | find recurrence relation such that you have n digit sequence of [imath]1[/imath]'s and [imath]2[/imath]'s such that you have at least one instance of consecutive 2's
I let [imath]a_n[/imath] be the different sequences with [imath]n[/imath] digits such that there is at least one instance of consecutive [imath]2[/imath]'s. This is what I did, if I place a a [imath]1[/imath] first, I have [imath]n-1[/imath] digits left and by definition, I have [imath]a_{n-1}[/imath] different sequences with [imath]n-1[/imath] digits...one instance of consecutive [imath]2[/imath]'s. Next case: Now instead of placing one [imath]2[/imath] I place two [imath]2[/imath]'s such that I have my met condition. What follows in the next [imath]n-2[/imath] digits may or may not have an instance of consecutive [imath]2[/imath]'s but I don't care anymore. With that being said, there are [imath]2[/imath] to the [imath](n-2)[/imath] different sequences left. So my recurrence relation is [imath]a_n = a_{n-1} + 2^{n-2}[/imath]. What's wrong with my recurrence relation? Thank you in advance for your time. |
13094 | Significance of [imath]\sqrt[n]{a^n} [/imath]?!
There is a formula given in my module: [imath] \sqrt[n]{a^n} = \begin{cases} \, a &\text{ if $n$ is odd } \\ |a| &\text{ if $n$ is even } \end{cases} [/imath] I don't really understand the differences between them, kindly explain with an example. | 1170206 | Question regarding the square root of a squared number.
I've learnt that the square root of a number squared is equal to the absolute value of that number, but I haven't really understood why. I have looked through other questions on MSE but didn't really find a good answer. As an example: for me there are two ways to arrive at a solution for [imath]\sqrt{(-5)^2}[/imath] First: [imath]\sqrt{(-5)^2}= [/imath] [imath]\sqrt{25}=5[/imath] Second: [imath]\sqrt{(-5)^2}= [/imath] [imath]\ (-5)^{\frac{2}{2}}[/imath][imath]\ =(-5)^1=-5[/imath] but according to the [imath]\sqrt{x^2} = | x | [/imath] rule 5 is the only solution. What is the flaw in my logic in getting -5 as a solution. I would really appreciate a comprehensive explanation that clears this up. Thanks in advance. |
114403 | 3rd iterate of a continuous function equals identity function
If [imath] f: \mathbb{R} \to \mathbb{R} [/imath] is continuous, and [imath]\forall x \in \mathbb{R} :\;(f \circ f \circ f)(x) = x [/imath], show that [imath] f(x) = x [/imath]. The condition that [imath]f[/imath] is continuous on [imath]\mathbb{R}[/imath] is crucial for the proof. I can find functions such as [imath]\displaystyle\frac{x-3}{x+1}[/imath] that satisfies [imath] (f \circ f \circ f)(x) = x [/imath]. I have tried to negate the conclusion to see if there's a contradiction, but got stuck. | 731275 | If [imath]f^3=\rm id[/imath] then it is identity function
Let [imath]f:\mathbb{R}\to \mathbb{R}[/imath] be a continuous function such that [imath]f^3(x)=f\circ f\circ f(x)=x[/imath] for all [imath]x[/imath]. How can I prove [imath]f[/imath] is the identity function? |
118655 | The simplest way of proving that [imath]|\mathcal{P}(\mathbb{N})| = |\mathbb{R}| = c[/imath]
What is the simplest way of proving (to a non-mathematician) that the power set of the set of natural numbers has the same cardinality as the set of the real numbers, i.e. how to construct a bijection from [imath]\mathcal{P}(\mathbb{N})[/imath] to [imath]\mathbb{R}[/imath]? | 152786 | Cardinality of the real numbers
Why is cardinality the of real numbers [imath]2^{\aleph_0}[/imath]? While I know that real numbers can be constructed using sets of natural numbers, that solely does not mean that the real numbers has cardinality [imath]2^{\aleph_0}[/imath]. So, what makes the real numbers cardinality [imath]2^{\aleph_0}[/imath]? I can't find why it is like this in my textbook... |
145310 | Proving that surjective endomorphisms of Noetherian modules are isomorphisms and a semi-simple and noetherian module is artinian.
I am revising for my Rings and Modules exam and am stuck on the following two questions: [imath]1.[/imath] Let [imath]M[/imath] be a noetherian module and [imath] \ f : M \rightarrow M \ [/imath] a surjective homomorphism. Show that [imath]f : M \rightarrow M [/imath] is an isomorphism. [imath]2[/imath]. Show that if a semi-simple module is noetherian then it is artinian. Both these questions seem like they should be fairly straightforward to prove but I cannot seem to solve them. | 722468 | Ascending chain of ideals
Let [imath]R[/imath] be a commutative ring with identity such that every ascending chain of ideals terminate. Let [imath]f:R \to R[/imath] be a surjective homomorphism. Prove that it is an isomorphism. |
20567 | Irrationality proofs not by contradiction
Per now, I have basically come upon proofs of the irrationality of [imath]\sqrt{2}[/imath] (and so on) and the proof of the irrationality of [imath]e[/imath]. However, both proofs were by contradiction. When thinking about it, it seems like the definition of irrationality itself demands proofs by contradiction. An irrational number is a number that is not a rational number. It seems then that if we were to find direct irrationality proofs, this would rely on some equivalent definition of irrational numbers, not involving rational numbers themselves. Are there any irrationality proofs not using contradiction? | 395500 | Direct proof for the irrationality of [imath]\sqrt 2[/imath].
Prove that [imath]\sqrt 2[/imath] is irrational using direct proof. I have seen TONS indirect proofs (e.g. proof by contradiction) for it, and people say that it's difficult to proof this directly. So is this impossible? Thank you. |
39269 | How does Cantor's diagonal argument work?
I'm having trouble understanding Cantor's diagonal argument. Specifically, I do not understand how it proves that something is "uncountable". My understanding of the argument is that it takes the following form (modified slightly from the wikipedia article, assuming base 2, where the numbers must be from the set [imath] \lbrace 0,1 \rbrace [/imath]): [imath]\begin{align} s_1 &= (\mathbf{0},1,0,\dots)\\ s_2 &= (1,\mathbf{1},0,\dots)\\ s_3 &= (0,0,\mathbf{1},\dots)\\ \vdots &= (s_n \text{ continues}) \end{align}[/imath] In this case, the diagonal number is the bold diagonal numbers [imath](0, 1, 1)[/imath], which when "flipped" is [imath](1,0,0)[/imath], neither of which is [imath]s_1[/imath], [imath]s_2[/imath], or [imath]s_3[/imath]. My question, or misunderstanding, is: When there exists the possibility that more [imath]s_n[/imath] exist, as is the case in the example above, how does this "prove" anything? For example: [imath]\begin{align} s_0 &= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\ s_1 &= (\mathbf{0},1,0,\dots)\\ s_2 &= (1,\mathbf{1},0,\dots)\\ s_3 &= (0,0,\mathbf{1},\dots)\\ s_4 &= (0,1,1,\mathbf{1},\dots)\\ s_4 &= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\ s_5 &= (1,0,0,0,\dots)\\ s_6 &= (1,0,0,1,\dots)\\ \vdots &= (s_n \text{ continues}) \end{align}[/imath] In other words, as long as there is a [imath]$\dots \text{ continues}$[/imath] at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it: For each [imath]m[/imath] and [imath]n[/imath] let [imath]s_{n,m}[/imath] be the [imath]m^{th}[/imath] element of the [imath]n^{th}[/imath] sequence on the list; so for each [imath]n[/imath], [imath]$$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$[/imath] ...snip... Otherwise, it would be possible by the above process to construct a sequence [imath]s_0[/imath] which would both be in [imath]T[/imath] (because it is a sequence of 0s and 1s which is by the definition of [imath]T[/imath] in [imath]T[/imath]) and at the same time not in [imath]T[/imath] (because we can deliberately construct it not to be in the list). [imath]T[/imath], containing all such sequences, must contain [imath]s_0[/imath], which is just such a sequence. But since [imath]s_0[/imath] does not appear anywhere on the list, [imath]T[/imath] cannot contain [imath]s_0[/imath]. Therefore [imath]T[/imath] cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable. I'm not sure this definition is correct, because if we assume that [imath]m = (1, \dots)[/imath], then this definition says that "[imath]s_n[/imath] is equal to itself"&mdadshthere is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible [imath]T[/imath] list above. An attempt to correct the "diagonalization" and "flipping" problem: [imath]$$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where m is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 & \mathrm{if\ } s_{m,m} = 1\\1 & \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$[/imath] This definition doesn't quite work either, as we immediately run in to problems with just [imath]s_1 = (0),[/imath] which is impossible because by definition [imath]s_1[/imath] must be [imath] = (1)[/imath] if [imath]s_1 = (0)[/imath], which would also be impossible because... it's turtles all the way down!? Or more generally, with the revised definition there is a contradiction whenever [imath]n = m[/imath], which would seem to invalidate the revised formulation of the argument / proof. Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable". What am I missing? | 517151 | Can somebody explain to me Cantor's diagonalization argument?
Like..can somebody explain this to me as if I was a 5 year old or something? Every explanation I read repeats the same exact thing that I simply do not understand. This is what my book says: "The real numbers between 0 and 1 can be listed in some order, say, [imath]r_1, r_2, r_3, ...[/imath]Let the decimal representation of these real numbers be $r_1 = 0.d_{11}[imath]d_{12}[/imath]d_{13}[imath]d_{14}$... $r_2 = 0.d_{21}[/imath]d_{22}[imath]d_{23}[/imath]d_{24}$... $r_3 = 0.d_{31}[imath]d_{32}[/imath]d_{33}[imath]d_{34}$... $r_4 = 0.d_{41}[/imath]d_{42}[imath]d_{43}[/imath]d_{44}$... Where [imath]d_{ij}[/imath] is an element of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Then for a new real number with decimal expansion r = $d_{1}[imath]d_{2}[/imath]d_{3}$[imath]d_{4}[/imath]... where the decimal digits are determined by the following rule: [imath]d_{i}[/imath] = {4 if [imath]d_{ii}[/imath] does not equal 4, 5 if [imath]d_{ii}[/imath] = 4}. And I'm sorry but..what? What in the world is any of this trying to get at? What is the whole r1, r2, r3 thing even mean? Why do we have to create a "new real number"? What is the point? Why? Why are we doing any of this? I don't understand any of the process behind it and I don't understand how it all leads to the conclusion that the real numbers are uncountable. I have absolutely no idea what is going on here. |
22367 | Sum and product of Martingale processes
Given two Martingale processes [imath](X_t)[/imath] and [imath](Y_t)[/imath], are their sum [imath](X_t+Y_t)[/imath] and their product [imath](X_t \times Y_t)[/imath] also Martingale? If not, will the two [imath](X_t)[/imath] and [imath](Y_t)[/imath] being independent grant their sum and product Martingale? Thanks! | 2801644 | Given two martingals sum also martingal?
Let [imath]X_{n}, Y_{n}[/imath] be independent martingals. Is it also the case that [imath]X_{n} + Y_{n}[/imath] is a martingale? And if we drop the independence condition does ist also define a martingal? |
165118 | How does partial fraction decomposition avoid division by zero?
This may be an incredibly stupid question, but why does partial fraction decomposition avoid division by zero? Let me give an example: [imath]\frac{3x+2}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}[/imath] Multiplying both sides by [imath]x(x+1)[/imath] we have: [imath]3x+2=A(x+1)+Bx[/imath] when [imath]x \neq -1[/imath] and [imath]x \neq 0[/imath]. What is traditionally done here is [imath]x[/imath] is set to [imath]-1[/imath] and [imath]0[/imath] to reveal: [imath]-3+2=-B \implies 1=B[/imath] and [imath]2=A[/imath] so we find that [imath]\frac{3x+2}{x(x+1)}=\frac{2}{x}+\frac{1}{x+1}[/imath] Why can [imath]x[/imath] be set equal to the roots of the denominator (in this case, [imath]0[/imath] and [imath]-1[/imath]) without creating a division by zero problem? | 2333589 | Why can I decompose fractions this way?
In order to be able to find an antiderivative, I have been taught to decompose fractions by first observing an identity, like [imath]\frac{2x+1}{(3x-2)(x+1)} \equiv \frac{A}{3x-2}+\frac{B}{x+1}[/imath] wherever the function is defined. Then I would expand and substitute, like so: [imath]A(x+1)+B(3x-2) \equiv 2x+1[/imath] letting [imath]x=-1[/imath]: [imath]-5B=-1 \implies B=\frac{1}{5}[/imath] and then letting [imath]x=\frac{2}{3}[/imath]: [imath]\frac{5}{3}A=\frac{7}{3} \implies A= \frac{7}{5}[/imath] Why are these substitutions justified? In both cases the original function is undefined for these [imath]x[/imath] values as the denominator would be zero - does this mean my claim "wherever the function is defined" is unnecessary? I can't see why this gives the right answer, but it does (as I can verify by matching coefficients instead of substituting)! |
118592 | Limit of measures is again a measure
Given a sequence [imath](\mu_n)_{n\in \mathbb N}[/imath] of finite measures on the measurable space [imath](\Omega, \mathcal A)[/imath] such that for every [imath]A \in \mathcal A[/imath] the limit [imath]\mu(A) = \lim_{n\to \infty} \mu_n(A)[/imath] exists. I want to show that [imath]\mu[/imath] is a measure on [imath]\mathcal A[/imath]. What I managed to figure out: [imath]\mu[/imath] is monotone, additive and - if [imath]\lim_n \mu_n(\Omega)[/imath] is supposed to be taken in [imath]\mathbb R[/imath] - then [imath]\mu[/imath] is also finite (I'll assume this). Also [imath]\mu(\varnothing) = 0[/imath]. So we can wlog assume that [imath]\mu(\Omega) = 1[/imath] and that [imath]\mu_n(\Omega) \le 2[/imath] for all [imath]n[/imath]. Now what remains to be shown is that [imath]\mu[/imath] is [imath]\sigma[/imath]-additive, or equivalently that for [imath]A_n\downarrow \varnothing[/imath], we have [imath]\mu(A_n) \to 0[/imath] (since [imath]\mu[/imath] is finite). All attempts at proving this have been futile so far. I don't seem to see the right approach. If possible, I would like to only receive a hint rather than a full answer. But of course, I'd be quite happy with a full answer, too; if a good hint is hard to find! Thanks a lot in advance for your help! =) | 344711 | "pointwise convergence"of measures
Let [imath](\Omega , \mathcal{A})[/imath] be any measurable space, and let [imath](\mu_n)_n[/imath] be a sequence of probability measures on this space, such that the limit [imath]\mu(A) = \lim \mu_n(A)[/imath] exists for any [imath]A \in \mathcal{A}[/imath]. Then [imath]\mu[/imath] is a finitely additive measure of total mass [imath]1[/imath] on [imath](\Omega , \mathcal{A})[/imath]. Is it always true that [imath]\mu[/imath] is [imath]\sigma[/imath]-additive? |
298046 | [imath]K[/imath] is the fraction field of [imath]R[/imath], but [imath]R[/imath] is not an order in [imath]K[/imath].
Give an example of non-trivial number field [imath]K[/imath] (so [imath]K \not= \mathbb{Q}[/imath]) and a proper subring [imath]R[/imath] of [imath]K[/imath] such that [imath]K[/imath] is the fraction field of [imath]R[/imath], but [imath]R[/imath] is not an order in [imath]K[/imath]. Given that there exist solutions where [imath]K[/imath], [imath]L_1[/imath], [imath]L_2[/imath] are quadratic, give an example of two number fields [imath]L_1[/imath] and [imath]L_2[/imath] contained in a bigger number field [imath]M[/imath] such that [imath]\mathcal{O}_{L_1L_2}\neq \mathcal{O}_{L_1}\mathcal{O}_{L_2}[/imath]. | 296571 | Example of non-trivial number field
(i) Give an example of non-trivial number field [imath]K[/imath] (so [imath][K:\mathbb Q]<\infty[/imath]) and a proper subring [imath]R[/imath] of [imath]K[/imath] such that [imath]K[/imath] is the fraction field of [imath]R[/imath], but [imath]R[/imath] is not an order in [imath]K[/imath]. (ii) Given that there exist solutions where [imath]K[/imath], [imath]L_1[/imath], [imath]L_2[/imath] are quadratic, give an example of two number fields [imath]L_1[/imath] and [imath]L_2[/imath] contained in a bigger number field [imath]M[/imath] such that [imath]O_{L_1L_2}\neq O_{L_1}O_{L_2}[/imath]. |
83830 | How to show that [imath]C=C[0,1][/imath] is a Banach space
Let [imath]C=C[0,1][/imath] be the space of all continuous functions on [imath][0,1][/imath]. Define [imath]\|f \|=\max \ |f(x)|[/imath]. I want to show that [imath]C[/imath] is a Banach space. Below is my attempt and I was wondering if it's ok. I know I have to show that [imath]C[/imath] is a complete normed space. Clearly, [imath]\|f\| \geqslant 0[/imath] and [imath]\|f\|=0 \Leftrightarrow f=0[/imath]. [imath]\|cf \|=\max~|cf(x)|=|c|\max |f(x)|=|c| \cdot \|f\|[/imath]. [imath]\|f+g\|=\max~|f(x)+g(x)|\leq \max~|f(x)|+\max~|g(x)|=\|f\|+ \|g\|[/imath]. So [imath]C[/imath] is a normed space. Next, I show that every Cauchy sequence in [imath]C[/imath] is convergent. Let [imath]\{f_n\}[/imath] be a Cauchy sequence in [imath]C[/imath]. Let [imath]\varepsilon \gt 0.[/imath] Then [imath]\exists[/imath] an [imath]N_1[/imath] such that [imath] \max~|f_n(x)-f_m(x)| \lt \frac{\varepsilon}{2}[/imath] for [imath]n, m \gt N_1[/imath] and [imath]x\in[0,1][/imath]. But there is a subsequence [imath]f_{k_n} [/imath], which converges to [imath]f[/imath]. So [imath]\exists[/imath] an [imath]N_2[/imath] such that [imath] \max~\left|f_{k_n} - f\right|\lt \frac{\varepsilon}{2}[/imath] for each [imath]n\gt N_2[/imath]. Now Let [imath]N = \max\{N_1, N_2\}[/imath], if [imath]n \gt N[/imath] then [imath]k_n \geqslant n\gt N[/imath]. So we have [imath] \max~\left|f_n(x) - f(x)\right| \leqslant \max~\left|f_n - f_{k_n}\right| + \max~\left| f_{k_n} - f\right| \lt\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.[/imath] Thus, [imath]\|f_n-f\| \to 0[/imath] as [imath]n\to \infty[/imath]. [imath]\quad \square[/imath] Thanks. | 803247 | Stating whether a space is complete
I'm asked to state whether or not [imath](X, d_u)[/imath] and [imath](X, d_L)[/imath] are complete where [imath]d_u[/imath] is the uniform metric and [imath]d_L[/imath] is the [imath]L^1[/imath] metric. All I need is to give the name of a supporting theorem or counterexample. [imath]X[/imath] is taken to be the space of continuous real-valued functions over the interval [imath][0, 1][/imath]. My thoughts are if I can show compactness then this holds (using perhaps heine-borel) arzela-ascoli directly completeness directly But since all I'm required is a theorem name - I'm supposing that it should be obvious. So how would one go about showing completeness or lack of for these metric spaces? |
304973 | combinatorial proof of Fibonacci identities
Give a combinatorial proof to each of the Fibonacci identities: [imath]nF_0+(n-1)F_1+\dots\dots+2F_{n-2}+F_{n-1}=F_{n+3}-(n+2)[/imath] and [imath] F_2+F_5+\dots\dots+F_{3n+1}=\frac{F_{3n+1}-1}{2} [/imath] Assume that [imath]F_0=0, F_1=F_2=1, F_n=F_{n-1}+F_{n-2},\ \ \ for \ \ n\geqslant3[/imath]. | 15469 | Combinatorial proof of a Fibonacci identity: [imath]n F_1 + (n-1)F_2 + \cdots + F_n = F_{n+4} - n - 3.[/imath]
Does anyone know a combinatorial proof of the following identity, where [imath]F_n[/imath] is the [imath]n[/imath]th Fibonacci number? [imath]n F_1 + (n-1)F_2 + \cdots + F_n = F_{n+4} - n - 3[/imath] It's not in the place I thought it most likely to appear: Benjamin and Quinn's Proofs That Really Count. In fact, this may be a hard problem, as they say the similar identity [imath] F_1 + 2F_2 + \cdots + nF_n = (n+1)F_{n+2} - F_{n+4} +2[/imath] is "in need of a combinatorial proof." For reference, here (from Benjamin and Quinn's text) are several combinatorial interpretations of the Fibonacci numbers. |
215118 | How do you integrate [imath]e^{x^2}[/imath]?
I know that [imath]\int{\frac{1}{x}}dx[/imath] is simply [imath]\ln{(x)}+c[/imath] (-which is clearly unrelated to the problem but I just thought I would share anyway) but I am not sure how to approach [imath]e^{x{^2}}[/imath]. Perhaps a substitution? | 912766 | Antiderivative of [imath]\exp(x^2)[/imath]
Can you please provide a step by step solution for next integral. I don't have any idea of how this can be solved [imath]\displaystyle\int e^{x^2}\,dx[/imath]. |