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1174186 | How to solve [imath]\int \tan^2(x) \sec (x) \ dx[/imath]?
How should one solve the following integral? [imath]\int \tan^2(x) \sec (x) \ dx[/imath] I can't think of any substitutions to be made involving [imath]\tan^2(x)=\sec^2 (x)-1[/imath] or [imath]\sec^2(x)=\tan^2(x)+1[/imath], which is how I've been solving most of the similar problems in my book until now. What should I do? | 1173256 | Integrating [imath]\sec\theta\tan^2\theta d\,\theta[/imath]
I'm stuck in this problem. I can not integrate [imath]\int \sec\theta\tan^2\theta \,d\theta[/imath]. Can anyone help me. |
688599 | Limit of p-norms of functions
[imath]f[/imath] is a continuous function on the closed interval [imath][a,b][/imath] such that [imath]f \geq 0[/imath] on [imath][a,b][/imath]. Let [imath] M_n = \left( \int_a^b f (x)^n dx \right)^{1/n}. [/imath] Show [imath]\lim_n M_n = \sup \{ f(x) \mid x \in [a,b] \}[/imath]. | 181592 | If [imath]f[/imath] is nonnegative and continuous on [imath][a,b][/imath], then [imath]\left(\int_a^b f(x)^n \ dx\right)^{1/n}\to\max\limits_{[a,b]} f[/imath]
I've been working on the following problem: Show that if [imath]f\in C[a, b][/imath] , [imath]f\ge 0[/imath] on [imath][a, b][/imath], then [imath]\left(\int_a^b f(x)^n \,dx\right)^{1/n}[/imath] converges when [imath]n\to\infty[/imath] and the limit is [imath]\max_If[/imath] with [imath]I=[a, b][/imath]. This is my solution: For Weierstrass [imath]f[/imath] has maximum, [imath]\exists \ \xi : f(\xi)=M[/imath]; and as [imath]f[/imath] is defined on [imath][a,b][/imath], [imath]f[/imath] is U.C., then: [imath]\forall \epsilon >0 \ \exists \delta >0: \forall x,y \in [a,b]: |x-y|< \delta \Rightarrow |f(x)-f(y)|< \epsilon[/imath] Let [imath][a,b]=\bigcup_{k=1}^m I_k[/imath], with [imath]mis(I_k)<\delta[/imath], and [imath]M_k=max_{\bar(I_k)} f(x)[/imath] [imath](\int_a^b f(x)^n \ dx)^{1/n}=(\sum_{k=1}^m M_h^n mis(I_k))^{1/n}=(M_1^n mis(I_1)+...+M^n mis(I_j)+...M_m^n mis(I_m))^{1/n}=[/imath] =[imath]M ((M_1/M)^n mis(I_1)+...+mis(I_j)+...+(M_m/M)^n mis(I_m))^{1/n}[/imath] Then: [imath](\int_a^b f(x)^n \ dx)^{1/n} \ge M[/imath] [imath](\int_a^b f(x)^n \ dx)^{1/n} \le M(b-a)^{1/n}[/imath] [imath]\Rightarrow \exists \ \lim_n \ (\int_a^b f(x)^n \ dx)^{1/n}=M=\max_I \ f[/imath] |
68505 | Is there a rule of integration that corresponds to the quotient rule?
When teaching the integration method of u-substitution, I like to emphasize its connection with the chain rule of integration. Likewise, the intimate connection between the product rule of derivatives and the method of integration by parts comes up in discussion. Is there an analogous rule of integration for the quotient rule? Of course, if you spot an integral of the form [imath]\int \left (\frac{f(x)}{g(x)} \right )' = \int \frac{g(x) \cdot f(x)' - f(x) \cdot g(x)'}{\left [ g(x)\right ]^2 }[/imath], then the antiderivative is obvious. But is there another form/manipulation/"trick"? | 1589938 | New Integration Rule?
As is well known pretty much all integration rules come from differentiation rule, so the substitution rule comes from the chain rule, and the integration by parts comes from the product rule, indeed [imath](uv)'=uv'+vu'\implies\int uv'dx=uv-\int vu'dx.[/imath] But I never saw a rule that used the Over Rule [imath](u/v)'=\frac{u'v-v'u}{v^2}=\frac{u'v}{v^2}-\frac{v'u}{v^2}=\frac{u'}{v}-\frac{uv'}{v^2}\implies \int \frac{u'}{v}dx=\frac uv+\int\frac{uv'}{v^2}dx[/imath] Is this like a new rule or does it already exist? |
412107 | Is any subset of a [imath]A[/imath] of a separable norm linear space [imath]X[/imath] is separable?
I just read the theorem "If X is norm linear space and if [imath]X^*[/imath] is separable [imath]X[/imath] is also separable" from Bachman and Narichi, page 201. To prove it they consider the set [imath]S=\{f \in X^* :\lVert f\rVert=1\}[/imath]. And the writer said that [imath]S[/imath] is separable as [imath]S[/imath] is subset of [imath]X^*[/imath] which is separable. How it is possible? Can anyone explain it? I know it is true for open subsets. | 407339 | If [imath]X^\ast [/imath] is separable [imath]\Longrightarrow[/imath] [imath]S_{X^\ast}[/imath] is also separable
Let [imath]X[/imath] be a Banach space such that [imath]X[/imath]* (Dual space of [imath]X[/imath]) is separable How can we prove that [imath]S_{X^\ast}[/imath] (Unit sphere of [imath]X[/imath]*) is also separable Any hints would be appreciated. |
1174856 | Binomial Theorem and Summation
All positive integers [imath]m_1, m_2, n[/imath] satisfy: [imath] \sum_{k=0}^n {m_1 \choose k}{m_2 \choose n-k} = {m_1 + m_2 \choose n} [/imath] Prove using the binomial theorem and the fact that [imath](1+x)^{m_1}(1+x)^{m_2} = (1+x)^{m_1+m_2}[/imath] To my understanding, [imath] (1+x)^{m_1} = \sum_{k=0}^n {m_1 \choose k}x^k [/imath]. Continuing with this, wouldn't the left side be a summation of a summation? Where does [imath] x^k[/imath] go? Using the given, I imagine that [imath](1+x)^{m_1+m_2}[/imath] could be used to retrieve the right side but then I don't know where sigma goes or how it becomes "choose n" Any hints would be appreciated. Edit: I feel really dumb. Thanks everyone! | 490741 | Algebraic proof of [imath]\sum_{i=0}^k{{n \choose i}{m \choose {k-i}}}= {{m+n}\choose k}[/imath]
I can't figure out an algebraic proof for the following identity: [imath]\sum_{i=0}^k{{n \choose i}{m \choose {k-i}}}= {{m+n}\choose k}[/imath] Combinatorical solution: We can see that as choosing some from [imath]n[/imath] and the rest of [imath]k[/imath] from [imath]m[/imath], thus [imath]k[/imath] in total. Or we could just choose [imath]k[/imath] from the union. |
1174965 | Generators of symmetric group [imath]S_n[/imath]
How can you prove that [imath]S_n[/imath] is generated by [imath](1\space 2)[/imath] and [imath](1\space 2\space 3\space ... \space n))[/imath] for [imath]n\geq 2[/imath]? | 772294 | How do i prove how [imath]S_5[/imath] is generated by a two cycle and a five cycle?
How do I prove that [imath]S_5[/imath] (the permutation group on five letters) can be generated by a two-cycle [imath](12)[/imath] and a five cycle [imath](12345)[/imath]? |
569274 | Prove inequality [imath]|\sqrt{x}- \sqrt{y}| \le \sqrt{|x-y|}[/imath]
How to prove that [imath]|\sqrt{x}- \sqrt{y}| \le \sqrt{|x-y|}[/imath] ? Of course, I know that [imath]|\sqrt{x}- \sqrt{y}| \le \sqrt{|x-y|} \Leftrightarrow ... \Leftrightarrow x \le y \quad \text{or} \quad x \ge y[/imath], but for me it isn't a good proof. | 946804 | Proof that [imath]|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0[/imath]
Any hints on how I can prove the inequality: [imath]|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|},\quad x,y \geq 0[/imath] Thank you. |
1175428 | [imath]1+ab[/imath] is a unit if and only if [imath]1+ba[/imath] is a unit.
Let [imath]R[/imath] be a ring with identity and [imath]a,b \in R[/imath] then prove that [imath]1+ab[/imath] is a unit if and only if [imath]1+ba[/imath] is a unit and find the inverse. Then there exist an element say [imath]s \in R[/imath] such that [imath](1+ab)s =1[/imath] then from here how will I approach to find the inverse of [imath]1+ba[/imath]?? | 675034 | Finding inverse in non-commutative ring
Let [imath]a,b,c,x[/imath] be elements of a unital non-commutative ring. Assume [imath]c[/imath] is an inverse of [imath]1-ab[/imath]: [imath] c(1-ab) = 1[/imath] How can I find an inverse for [imath]1-ba[/imath]? What I tried: Denote the unknown by [imath]x[/imath]. Then [imath]x (1-ba) = 1 = x - xba[/imath]. I tried to replace [imath]1[/imath] with [imath]c(1-ab)[/imath] but it didn't help because I cannot solve for [imath]x[/imath]. I also tried to subtract [imath]x (1-ba) [/imath] from [imath] c(1-ab)[/imath] but couldn't solve for [imath]x[/imath] either. Any suggestions? |
1175109 | Show that [imath]F(x) = f(\|x\|)[/imath] is differentiable on [imath]\mathbb{R}^n[/imath].
Let an even function [imath]f:\mathbb{R}\to\mathbb{R}[/imath] which is even and differentiable. We define [imath]F:\mathbb{R}^n\to\mathbb{R}[/imath] as [imath]F(x) = f(\|x\|)[/imath]. Show that [imath]F(x)[/imath] is differentiable on [imath]\mathbb{R}^n[/imath]. My Work: Let [imath]x_0\in\mathbb{R}^n[/imath]. If [imath]x_0\ne 0[/imath] then the partial derivatives of the norm are well-defined and: [imath] \frac{\partial \|.\|}{\partial x_i} = \frac{x_i}{\|x\|}[/imath] [imath]\frac{x_i}{\|x\|}[/imath] continuous and since [imath]f[/imath] is differentiable it's partial derivatives are continuous. So we can conclude that [imath]\frac{\partial F}{\partial x_i}[/imath] is continuous and therefore [imath]F[/imath] is differentiable. If [imath]x_0=0[/imath] then the [imath]\|x_0\| = 0[/imath] and since [imath]f[/imath] is even [imath]f(0)=0[/imath] and it must be an extremum point. Therefore, [imath]F'(0)=0[/imath]. I'd like to get a critique of my work. Am I being right/rigorous? Thanks. | 658660 | Multivariate differentiability verification
I have tried an attempt on the following: Let [imath]F:U\in \Bbb{R}^n\to \Bbb{R}[/imath] and [imath]f:\Bbb{R}\to \Bbb{R}[/imath] where [imath]f[/imath] is an even function. Now [imath]F(\mathbf{x})=f(|\mathbf{x}|)[/imath], where [imath]|\ . |[/imath] is the Euclidean norm. Given [imath]f[/imath] is [imath]C^r[/imath], I have to show that [imath]F[/imath] is [imath]C^r[/imath]. My attempt was to argument by induction. I showed that [imath]F[/imath] is [imath]C^1[/imath], basically using univariate chain rule on [imath]\frac{\partial f\left(\sqrt{x_1^2+\dots+x_n^2}\right)}{\partial x_1}[/imath] and arguing that it would be continuous using the fact that [imath]f[/imath] is [imath]C^r[/imath]. Since the argument works for all [imath]x_i[/imath], it establishes first-differntiability for [imath]F[/imath] and continuity of the operator [imath]DF(\mathbf{x})[/imath]. Then I moved on to show [imath]F[/imath] is [imath]C^r[/imath] assuming [imath]F[/imath] is [imath]C^{r-1}[/imath] and [imath]f[/imath] is [imath]C^r[/imath]. Now what I did was partially differentiate: [imath] {\partial \over \partial x_2} \left[\frac{\partial ^{k-1}f}{\partial x_1^{k-1}}\right] = \underbrace{{\partial \over \partial |\mathbf{x}|} \left[\frac{\partial ^{k-1}f}{\partial x_1^{k-1}}\right]}_{(1)} . \underbrace{{\partial |\mathbf{x}| \over \partial x_2}}_{(2)} [/imath] My argument runs as follows: this partial derivative is continuous since [imath]f[/imath] is [imath]C^r[/imath] and hence [imath](1)[/imath] is continuous. Continuity of [imath](2)[/imath] is establishd in a similar manner, since [imath]\sqrt{\mathbf{x}}[/imath] is [imath]C^r[/imath] if [imath]\mathbf{x} \neq \mathbf{0}[/imath]. Then the product would be continuous. I am uneasy on multivariate arguments and dont know if my approach is completely off track. I didnt use the even function property, so I am sure I might me missing something. Any hints and references would be welcome. Thanks! |
1170791 | Show a group to be a subgroup
Let [imath]H[/imath] be a subgroup of [imath]G[/imath]. Let [imath]K = \{x \in G: xax^{-1} \text { iff } a \in H\}.[/imath] Prove [imath]H[/imath] is a subgroup of [imath]K[/imath]. Suppose [imath]a \in H.[/imath] Then [imath]aaa^{-1} = a \in H.[/imath] So, [imath]a \in K.[/imath] Thus, [imath]H \subseteq K.[/imath] Since [imath]H[/imath] is a group, [imath]H[/imath] s a subgroup of [imath]K[/imath]. Would that work? | 1168396 | Let [imath]H[/imath] be a subgroup of [imath]G[/imath]. Let [imath]K = \{x \in G: xax^{-1} \in H \iff a \in H\}[/imath]. Prove that [imath]K[/imath] is a subgroup of [imath]G[/imath].
Closure under multiplication: [imath]\to[/imath] Let [imath]a, x, y \in H.[/imath] Since [imath]H[/imath] is closed under inverses and multiplication, [imath]xax^{-1} \in H[/imath] and [imath]yay^{-1} \in H.[/imath] Since [imath]H[/imath] is closed under multiplication, [imath]xyax^{-1}y^{-1} \in H.[/imath] [imath]\leftarrow[/imath] Suppose [imath]xyax^{-1}y^{-1} \in H[/imath], then [imath]a \in H[/imath] since [imath]H[/imath] is closed under multiplication. So, [imath]xy \in K.[/imath] Closure under inverses: Since [imath]H[/imath] is closed under multiplication and inverses, if [imath]a \in H,[/imath] then [imath]x^{-1}ax \in H.[/imath] If [imath]x^{-1}ax \in H,[/imath] then [imath]a \in H[/imath] because [imath]H[/imath] is a group. So, [imath]x^{-1} \in K.[/imath] Identity: Suppose [imath]a \in H.[/imath] Then [imath]a = eae^{-1} \in H.[/imath] So, [imath]e \in H.[/imath] I'd like to see how I can fix/improve this proof. |
1175634 | Set of two cycles generate [imath]S_n[/imath]
How do I show that [imath]S_n[/imath] is generated by the set [imath]\{(1\space i), i=2,3,...,n\}[/imath]? Is it correct to say that since each product of elements of [imath]\{(1\space i), i=2,3,...,n\}[/imath] is an element of [imath]S_n[/imath], we know that [imath]\{(1\space i), i=2,3,...,n\}\subset S_n[/imath], and how can I prove that [imath]S_n\subset\{(1\space i), i=2,3,...,n\}[/imath]? | 520615 | $(12)$ and $(123\dots n)$ are generators of [imath]S_n[/imath]
Show that [imath]S_n[/imath] is generated by the set [imath] \{ (12),(123\dots n) \} [/imath]. I think I can see why this is true. My general plan is (1) to show that by applying various combinations of these two cycles you can get each transposition, and then (2) to show that each cycle is a product of transpositions. I'm just having trouble on the first step. Any ideas? |
1175767 | [imath]n(n+1)(n+2)[/imath] is not a perfect power
We have [imath]n,n+1, n+2 \in \mathbb Z^+[/imath] Their product can't be a whole exponentiation. Why? I noticed that [imath]gcd(n,n+1)=1[/imath] and [imath]gcd(n+1,n+2)=1[/imath] This could be a good starting point in the proof. But how do I proceed? Thanks | 266042 | Product of three consecutive positive integers is never a perfect power
I am trying to prove that the product of three consecutive positive integers is never a perfect power. Can anyone point to gaps in my proof and/or post an alternate solution? Let the three positive consecutive integers be [imath]n - 1[/imath], [imath]n[/imath] and [imath]n + 1[/imath] and let [imath](n - 1)n(n + 1) = h^k, k \ge 2[/imath]. Note that [imath]\gcd(n - 1, n) = 1[/imath] and [imath](n, n + 1) = 1[/imath] implies that [imath]n[/imath] itself must be a perfect power and of the form [imath]z^k[/imath] (which is apparent once we look at the canonical representation of [imath]h^k[/imath]). That means [imath]n^2 - 1[/imath] must be a perfect power itself and of the form [imath]a^k[/imath]. If [imath]n[/imath] is odd, [imath]n = 2m + 1[/imath] for some [imath]m \in \mathbb{N}[/imath] i. e. [imath](n - 1)(n + 1) = 2m(2m + 2) = 2^2m(m + 1) = a^k[/imath]. Using the fact that [imath](m, m + 1) = 1[/imath], [imath]m[/imath] and [imath]m + 1[/imath] must be perfect powers themselves and so [imath]k\leq 2[/imath]. Coupled with fact that [imath]k > 1[/imath], we infer [imath]k = 2[/imath]. So, [imath]n^2 - 1 = a^2[/imath] which implies [imath]n[/imath] is not a natural number. We are left with the case when [imath]n[/imath] is even. Let [imath]n = 2t + 1[/imath]. [imath](n + 1, n - 1) = 1[/imath] as both are of (by the Euclidean algorithm). That means [imath]n - 1[/imath] and [imath]n + 1[/imath] are perfect powers. So, let [imath](n - 1) = b^k[/imath] and [imath](n + 1) = l^k[/imath]. So, [imath]l^k - b^k = 2[/imath] for natural [imath]l[/imath] and [imath]p[/imath] and [imath]k>1[/imath]. I will prove that the diophantine equation has no solutions. Consider the function [imath]f(k) = l^k - b^k - 2[/imath]. [imath]f'(k) = \frac{1}{l}e^{k\log l} - \frac{1}{b}e^{k\log b}>0[/imath] if and only if [imath]e^{k(log\frac{l}{b})}>\frac{l}{b}[/imath]. Taking logarithm again, we get an equivalent condition [imath](k - 1)\log\frac{l}{b}>0[/imath] which is true as [imath]k>1[/imath] and [imath]l>b[/imath] as log is a monotonically increasing function. Is my proof correct? Please feel free to chip in with your own solutions! |
1153853 | Real analysis: Show a continuous and nonnegative function on [imath][a, b][/imath] = 0
Suppose [imath]f[/imath] is continuous and nonnegative on [imath][a, b][/imath], and [imath]\int_{a}^{b}f dx=0[/imath]. Show that [imath]f=0[/imath] on [a,b]. My idea: The conditions, continuous and nonnegative, remind me to use the theorem to show they are Rienman integrable. My question: It seems that we do not need to show they are Rienman integrable, since we already know their integration. It seems it does not help if I have already showed their Rienman integrable. I can come up two approaches, but I am not sure whether they are right. Suppose there is a point larger than 0, then this point must fall into one of the partitions, then its integration cannot be 0. We take [imath][a,b][/imath] as the partition, then the difference between [imath]U(p,f)[/imath] and [imath]L(p,f)[/imath] must be less than arbitrary small positive number. Hence [imath]\sup f[/imath] and [imath]\inf f[/imath] must be 0. | 349641 | If [imath]f:[a,b]\to \mathbb{R}[/imath] is continuous and nonnegative and [imath]\int_a^b{f}=0[/imath], then [imath]f(x)=0[/imath] for all [imath]x\in [a,b][/imath]
Suppose that a function [imath]f:[a,b]\to\mathbb{R}[/imath] is continuous and nonnegative. Prove that if [imath]\int_a^b{f}=0[/imath], then [imath]f(x)=0[/imath] for all [imath]x\in [a,b][/imath]. I've been trying to prove it using the extreme value theorem, continuity and the upper and lower sums but can't come up with something tight enough. More specifically, I've been trying to relate [imath]|f(x)-f(t)|<\epsilon[/imath] to the sum of [imath]M_k-m_k(x_k-x_{k-1})[/imath] |
425663 | equivalent norms in Banach spaces of infinite dimension
Suppose [imath] X [/imath] is a Banach space with respect to two different norms, [imath] \|\cdot\|_1 \mathrm{ e } \|\cdot\|_2 [/imath]. Suppose there is a constant [imath] K > 0 [/imath] such that [imath] \forall x \in X, \|x\|_1 \leq K\|x\|_2 .[/imath] show then that these two norms are equivalent | 1652197 | If [imath]\|\cdot\|_{1}\le\|\cdot\|_{2}[/imath] then [imath]\|\cdot\|_{2}\le M\|\cdot\|_{1}[/imath]
Let [imath](X,\|\cdot\|_{1})[/imath] and [imath](X,\|\cdot\|_{2})[/imath] be complete normed vector spaces and [imath]\|x\|_{1}\le\|x\|_{2}[/imath] [imath]\forall x\in X[/imath]. I want to prove that [imath]\exists M>0[/imath] such that [imath]\|x\|_{2}\le M\|x\|_{1}[/imath]. We know that any Cauchy sequence [imath](x_{n})_{n\in\mathbb{N}}\subset X[/imath] converges. And of course, [imath]\frac{1}{M}\|x\|_{1}\le\|x\|_{1}\le M\|x\|_{1}[/imath]. Does anyone have any hints on how I can proceed? I assume that I have to use this completeness property somehow. |
1174439 | Partial differential equation problem1
I'm supposed to solve [imath]u_{xx}-3u_{xt}-4u_{tt}=0[/imath] with initial conditions [imath]u(x,0)=x^2[/imath] and [imath]u_t(x,0)=e^x[/imath]. So I factored the problem into [imath](u_x-4u_t)(u_x + u_t)[/imath] and set each equal to 0 and found the 2 solutions to be [imath](x,t)=f(x+t/4)[/imath] and [imath]u(x,t)=g(x-t)[/imath] Then I get [imath]u(x,t)=1/2[\phi(x+t/4) + \phi(x-4)][/imath] + 1/2c [imath]\int_{x-t}^{x+t/4} e^{s} ds[/imath] =1/2[[imath](x+t/4)^2 + (x-t)^2[/imath]] + 1/2c[[imath]e^{x+t/4}-e^{x-t}[/imath]] I multiply out the first part then and get a solution of: 1/2[[imath]5/2xt-15/16t^2[/imath]]+1/2c[[imath]e^{x+t/4}-e^{x-t}[/imath]] But the book has a solution of [imath]x^2 + t^2/4 + 4/5[e^{x+t/4}-e^{x-t}][/imath] I don't understand how they get 4/5 in front of the exponentials and how [imath]1/2[\phi(x+t/4) + \phi(x-4)][/imath] turns into [imath]x^2 + t^2/4[/imath] . Can anyone see where I went wrong? | 306250 | Solve [imath]u_{xx}-3u_{xt}-4u_{tt}=0[/imath] where [imath]u(x,0)=x^2[/imath] and [imath]u_t(x,0)=e^x[/imath]
Solve [imath]u_{xx}-3u_{xt}-4u_{tt}=0[/imath] where [imath]u(x,0)=x^2[/imath] and [imath]u_t(x,0)=e^x[/imath]. My workings so far: I have factored the differential equation in the following way: [imath](\delta_x-4\delta_t)(\delta_x+\delta_t)=0[/imath] where [imath]\delta_x=\frac{\delta}{\delta x}[/imath] etc. Now if we let [imath]v[/imath] be the solution to [imath](\delta_x+\delta_t)u[/imath] then we have the following two equations: \begin{eqnarray*} (\delta_x+\delta_t)u=u_x+u_t=v\\ (\delta_x-4\delta_t)v=v_x-4v_t=0 \end{eqnarray*} Now for [imath]v[/imath] we simply find [imath]v=h(t+4x)[/imath] where [imath]h[/imath] is an arbitrary function of one variable. Now what remains is find [imath]u[/imath] such that [imath]u_x+u_t=h(t+4x)[/imath] I am stuck here, I thought about making a change of variables [imath]\zeta=x+t[/imath] and [imath]\eta=x-t[/imath] and thus using the product rule to show that [imath]u_x=u_{\zeta}+u_{\eta}[/imath] and [imath]u_t=u_{\zeta}-u_{\eta}[/imath] and thus [imath]u_x+u_t=2u_{\zeta}[/imath] and we need to solve [imath]u_{\zeta}=h(t+4x)[/imath] (I left out the factor 2 because [imath]h[/imath] is an arbitrary function). Do I simply integrate now and conclude [imath]u=f(\zeta)h(t+4x)+g(\eta)=f(x+t)h(t+4x)+g(x-t)[/imath] This seems wrong to me... Some help would be greatly appreciated! |
1176794 | Number of ways to write a prime [imath]p \equiv 1 \bmod 4[/imath] as a sum of two squares
My intuition, from checking many cases, is that there is only ONE way to write a prime [imath]p \equiv 1 \bmod 4[/imath] as a sum of two squares. But how can I formally show this? | 112313 | If [imath]p\equiv1\mod{4}[/imath], then [imath]p=a^{2}+b^{2}[/imath] where a and b are uniquely determined
From Ireland,Rosen's Number Theory book, ex.8.12. Let [imath]p\equiv1\mod{4}[/imath], then we can write [imath]p=a^{2}+b^{2}[/imath], where [imath]a,b\in\mathbb{Z}[/imath]. Now, let [imath]a[/imath] be odd and [imath]b[/imath] be even and [imath]a,b>0[/imath]. How can we show that [imath]a[/imath] and [imath]b[/imath] are uniquely determined? The hint says to use the fact that [imath]\mathbb{Z[i]}[/imath] is a UFD and [imath]p=a^{2}+b^{2}\implies a+bi[/imath] is prime in [imath]\mathbb{Z[i]}[/imath] |
727999 | [imath]G[/imath] is Topological [imath]\implies[/imath] [imath]\pi_1(G,e)[/imath] is Abelian
Hypothesis: Let [imath]G[/imath] be a topological group with identity element [imath]e[/imath]. Let [imath]\mu[/imath] denote the multiplication mapping in [imath]G[/imath]. Goal: Show that [imath]\pi_1(G,e) = \pi(G)[/imath] is an abelian group via the hint below. Hint: There are two products on [imath]\pi(G)[/imath]. The usual product [imath]\circ[/imath] defined for the fundamental group and the product [imath]\ast[/imath] induced by [imath] \ast: \pi(G) \times \pi(G) \cong \pi(G \times G) \overset{\pi(\mu)}{\rightarrow} \pi(G). [/imath] Show that there is a common two sided unit, [imath]u \in \pi(G)[/imath] for both products and there is a distributive law [imath] (f \ast g) \circ (a \ast b) = (f \circ a) \ast (g \circ b) [/imath] Attempt: Suppose the distributive law in the hint holds. Suppose further that [imath]1[/imath] -- the identity element in [imath]\pi(G)[/imath] with respect to [imath]\circ[/imath] -- serves also as the identity element in [imath]\pi(G)[/imath] with respect to [imath]\ast[/imath]. Then for [imath]a,b \in \pi(G)[/imath], we would have the following relationship: [imath] (1 \ast a) \circ (b \ast 1) = (1 \circ b) \ast (a \circ 1) [/imath] so that [imath] a \circ b = b \ast a [/imath] Then if we can show that for all [imath]f,g \in \pi(G)[/imath] that [imath]f \ast g \iff f \circ g[/imath], we could complete the above relation to [imath] a \circ b = b \ast a = b \circ a [/imath] so that [imath]\pi(G)[/imath] is abelian as desired. Question: Am I on the right track? | 1820880 | Fundamental group of a topological group
Given [imath](G,\cdot)[/imath] a topological group with identity [imath]e[/imath], is it always true that [imath]\pi_1(G,e)[/imath] is abelian? For what it's worth: I have already shown that if [imath]f,g\in\Omega(X,e)[/imath], then [imath][f \times g]=[f\cdot g][/imath], where "[imath]\times[/imath]" is the ordinary path multiplication (left to right), "[imath]\cdot[/imath]" is the group multiplication (that's to say, [imath]f\cdot g(t)=f(t)g(t)[/imath]) and [imath][\cdot][/imath] is the homotopy class. So in view of this property, if [imath]G[/imath] is abelian itself then [imath]\pi_1(X,e)[/imath] is trivially abelian too. However, I have no clue how to proceed when [imath]G[/imath] is nonabelian. |
1177077 | relating to [imath]L_{p}[/imath]-norm
Let [imath](X, \Sigma, \mu)[/imath] be a measure space and [imath]f[/imath] be a measurable function. For each [imath]1\le p<+\infty[/imath], setting [imath]\|f\|_{p}:=\left(\int\limits_{X}|f|^{p}\, \mathrm{d}\mu\right)^{\frac{1}{p}} \text{(not necessary finite)}.[/imath] The question is that [imath]\|f\|_{p}\to \|f\|_{p_{0}}\, \text{ as }\, p\to p_{0}?[/imath] | 135243 | Is the [imath] L^{p}[/imath][imath][0,1][/imath] norm continuous in p?
I ran into the following problem when I was doing my homework, and I have no thoughts on where I should start with: (1) If [imath]f\in L^{2}[/imath], show that [imath]\displaystyle \lim_{p \rightarrow 1^{+}}\int_{[0,1]}|f|^{p}=\int_{[0,1]}|f|[/imath] (2) If [imath]0<p[/imath], show that [imath]\displaystyle \lim_{q\rightarrow p^{-}}||f||_{q}=||f||_{p}[/imath] My first thought was Generalized LDCT, but it didn't seem to work. I also made some other attempts but none of them were successful... Can anybody give me some hints on how I should look at this question? Also, I know if [imath]p\rightarrow\infty[/imath] then [imath]||f||_{p}\rightarrow||f||_{\infty}[/imath] on [imath][0,1][/imath], but does similar continuity in p holds for other [imath]L^{p}[0,1][/imath] norms in general? Thank you! Edit: Sorry if I did not make it clear enough in the question. All [imath]L^{p}[/imath] refers to [imath]L^p[0,1][/imath]. The first question is found here (thanks to t.b.), but the second question remains, mainly because [imath]f[/imath] is not guaranteed to be in any [imath]L^{p}[/imath]. |
1008727 | Conditional expectation equals random variable almost sure
Let [imath]X[/imath] be in [imath]\mathfrak{L}^1(\Omega,\mathfrak{F},P)[/imath] and [imath]\mathfrak{G}\subset \mathfrak{F}[/imath]. Prove that if [imath]X[/imath] and [imath]E(X|\mathfrak{G})[/imath] have same distribution, then they are equal almost surely. I know what I have to show, that [imath]X[/imath] is [imath]\mathfrak{G}[/imath] measurable, but I don't know how... | 1233876 | Problem on "Distribution" of conditional expectation w.r.t a sigma field
Let [imath]Y[/imath] be a random variable s.t. [imath]E|Y|<\infty[/imath]. Let [imath]E[Y|\mathcal{G}][/imath] and [imath]Y[/imath] have the same distribution. I need to prove that [imath]E[Y|\mathcal{G}]=Y[/imath] a.s. How does one use the fact that a random variable and its conditional expectation has the same distribution ? |
1177415 | How to find a probability density function.
Suppose [imath]X[/imath] and [imath]Y[/imath] are random variables that are independent also [imath]X[/imath] and [imath]Y[/imath] are uniformly distributed on the interval [imath](0, 1)[/imath]. If [imath]Z=\max \{ X, Y \}[/imath]. Then find the probability that [imath]Z \leq z[/imath] and then determine the probability density function of [imath]Z[/imath]. So far I have said that both [imath]X[/imath] and [imath]Y[/imath] are [imath]1[/imath] in the interval [imath](0,1)[/imath] and [imath]0[/imath] elsewhere. I then said that [imath]P(Z \leq z)=P(x \leq z~\cap~y \leq z)[/imath] so we can split these up and then work through to get this to be equal to. [imath] \int_{-\infty}^zdx \times \int_{-\infty}^zdy=z^2[/imath] but then I got really confused at to what [imath]z[/imath] even wasand how to proceed with the question. Also the [imath]z^2[/imath] can't be right as this gives values greater than [imath]1[/imath]. Any help? | 1177269 | How to get started on this statistics problem?
Suppose [imath]X[/imath] and [imath]Y[/imath] are random variables that are independent also [imath]X[/imath] and [imath]Y[/imath] are uniformly distributed on the interval [imath](0, 1)[/imath]. If [imath]Z=\max \{ X, Y \}[/imath]. Then find the probability that [imath]Z \leq z[/imath] and then determine the probability density function of [imath]Z[/imath]. Okay so I must admit I an trumped by this question but I think it's mainly to do with not actually understanding what I'm meant to do. Here are my issues: 1) Okay so [imath]X[/imath] and [imath]Y[/imath] are uniformly distributed on [imath](0,1)[/imath] does that mean that it is [imath]0[/imath] elsewhere? 2) What exactly is [imath]Z[/imath] is it just the maximum value that either [imath]X[/imath] or [imath]Y[/imath] can take and how exactly do I interpret this? At the moment I'm just thinking of two graphs and [imath]Z[/imath] being the highest point that is taken on both graphs in [imath]X[/imath] or [imath]Y[/imath] since this is a uniform distribution then it will be the same highest point for all values in the interval on either [imath]X[/imath] or [imath]Y[/imath] whatever is highest. This is just intuition again I'm not sure. 3) What is [imath]P(Z \leq z)[/imath] actually meaning? I'm guessing this is because I'm confused on what [imath]Z[/imath] is and so I can't get this either. If anyone could help me on this it would be greatly appreciated thanks! |
1177559 | Some hint on this question, prime number theorem
for all prime number [imath]x,y,z[/imath], [imath]x^2 + y^2 \ne z^2[/imath]. I assume [imath]x^2+y^2 = z^2[/imath], then [imath]x^2 = z^2 - y^2[/imath], so we have [imath]x^2 = (z+y)(z-y)[/imath] but I don't know how to continue because I don't know any number theory theorem. | 991947 | Why can't prime numbers satisfy the Pythagoras Theorem? That is, why can't a set of 3 prime numbers be a Pythagorean triplet?
Suppose [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath] are three prime numbers. How to prove that [imath]a^2 + b^2 \neq c^2[/imath]? |
1176861 | Why is not the ring [imath]\mathbb{Z}[2\sqrt{2}][/imath] a unique factorization domain?
Why is not the polynomial ring [imath]R[x][/imath] a unique factorization domain, where [imath]R[/imath] is the quadratic integer ring [imath]\mathbb{Z}[2\sqrt{2}][/imath]? I'm trying to find a irreducible nonprime element or something but I don't know where to start. | 1294158 | Is [imath]\mathbb{Z}[2\sqrt{2}][/imath] a PID?
I am practicing for my algebra qual and I would like to know if [imath]\mathbb{Z}[2\sqrt{2}][/imath] is a PID. I had no intuition at first except the fact that [imath]\mathbb{Z}[i\sqrt{2}][/imath] is a ED with norm [imath]N(a+i\sqrt{2}b)=a^2+2b^2[/imath]. I tried proving that [imath]\mathbb{Z}[2\sqrt{2}][/imath] is a ED with norm [imath]N(a+b2\sqrt{2})=a^2-8b^2[/imath] using the standard proof for [imath]\mathbb{Z}[i\sqrt{2}][/imath] and [imath]\mathbb{Z}[i][/imath]. This led nowhere. This may not be a PID, but I'm not sure how to prove it. Any tips? Thanks in advance. |
1178199 | prove that [imath]\binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+\cdots+\binom{n}{n}^2=\binom{2n}{n}[/imath]
How can i prove that : [imath]\binom{n}{0}^2+\binom{n}{1}^2+\binom{n}{2}^2+\cdots+\binom{n}{n}^2=\binom{2n}{n}[/imath] i tried to prove it : by [imath]\binom{2n}{n} = \frac{(2n)!}{n!n!} = \frac{2^n (1\cdot3\cdot5\cdots(2n-1))}{n!}[/imath] but i can’t take advantage of it. | 957804 | Proving [imath] \binom n 0 ^2 + \binom n 1 ^2 + \dots + \binom n n ^2 = \binom { 2n} n [/imath] without induction
I have to prove that: [imath] \binom n 0 ^2 + \binom n 1 ^2 + \dots + \binom n n ^2 = \binom { 2n} n [/imath] I don't want a complete solution, but only a hint. |
1178459 | Primes of the form [imath]a^2 + 5b^2[/imath]
How do I check if a number [imath]m \bmod 20[/imath] can be written in the form [imath]a^2 + 5b^2[/imath]? I am trying to prove that if the prime [imath]p = a^2 + 5b^2[/imath] then [imath]p \equiv 1\text{ or }9 \pmod{20}[/imath] and the only way I know how is to check each case from [imath]p\equiv 1 \pmod{20}[/imath] to [imath]19 \pmod{20}[/imath] and show that only [imath]1[/imath] and [imath]9[/imath] works. If there's a simpler way please share. Thanks! | 1175930 | Number Theory: which primes are a sum of two squares?
Problem: show that if the prime [imath]p = a^2 + 5b^2[/imath] then [imath]p \equiv 1 \textrm{ or } 9 \pmod{20}[/imath]. Is there any other way to prove this other than to check each case [imath]p = 1, \ldots, 19 \pmod{20}[/imath] and show that only 1 and 9 work? |
1178266 | Justify [imath]E(X|Y)=E(E(X|Z,Y)|Y)[/imath]
Why [imath]E(X|Y)=E(E(X|Z,Y)|Y)[/imath]? I know that [imath]E(U)=E(E(U|V))[/imath]. So, [imath]E(U|W)[/imath] should be [imath]E(E(U|V,W))[/imath]. But the latter expression is free from [imath]W[/imath] so it is not possible. I can get a intuitive idea that right hand side of the above expression should be a function of [imath]Y[/imath], so we are conditioning two times. But how to prove it rigorously? | 153677 | Tower property of conditional expectation
I'm trying to prove the "tower property" of conditional expectations, [imath] E[V\mid W] = E[\ E[V\mid U,W]\ \mid W\ ], [/imath] where [imath]U[/imath], [imath]V[/imath] and [imath]W[/imath] are any random variables. [imath]E[X \mid Y][/imath] is itself a random variable [imath]f(Y)[/imath] where [imath]f(y) = E[X \mid Y = y) = \sum_x x\cdot Pr[X=x\mid Y=y].[/imath] Keeping this observation in mind, I still don't see why [imath]U[/imath] is "averaged out" when moving from the right hand side to the left side. |
1175803 | the [imath]\sigma[/imath] algeba generated by the class of open intervals with rational end points coincide with the borel [imath]\sigma[/imath] algebra on the real line.
Show that the [imath]\sigma[/imath] algeba generated by the class of open intervals with rational end points coincide with the borel [imath]\sigma[/imath] algebra on the real line. I tried to solve the question but I cannot do proper solution, thus i cannot write here. please show me this question. thank you. | 166082 | What is the minimum [imath] \sigma[/imath]-algebra that contains open intervals with rational endpoints
What are the minimum [imath]\sigma[/imath]-ring and [imath]\sigma[/imath]-algebra on [imath]\mathbb R[/imath] which contain the open intervals with rational endpoints? Is there a relation between this [imath]\sigma[/imath]-algebra and Borels? |
1178692 | Additional properties of closure
Definitions: [imath]A'[/imath] is the set of all accumulation or limit points. [imath]\bar{A} = A \cup A'[/imath] - this is known as the closure of [imath]A[/imath]. Let [imath]A[/imath] be a subset of [imath]\mathbb{R}[/imath]. A point [imath]p\in\mathbb{R}[/imath] is an accumulation point of [imath]A[/imath] if and only if every open set [imath]G[/imath] containing [imath]p[/imath] contains a point of [imath]A[/imath] different from [imath]p[/imath]. Prove or disprove: [imath](\overline{A\bigcup B}) = \overline{A}\bigcup \overline{B}[/imath] proof: let [imath]p\in (\overline{A\bigcup B})[/imath] be an accumulation of the union, then there exists some open set [imath]S_{p}[/imath] containing [imath]p[/imath] which also contains a point of [imath](\overline{A\bigcup B})[/imath] different from [imath]p[/imath]. Let's call this point [imath]q[/imath], where [imath]p\neq q[/imath] such that [imath]q\in S_{p} \subset (\overline{A\bigcup B})[/imath]. This is where I am lost, if anyone can help that would be greatly appreciated. | 657644 | Elementary properties of closure
Hi everyone I'd like to know if the following is correct. I really appreciate any suggestion. (Honestly the only one that matters me is the second property the others are easy, I think) Thanks. Definition: Let [imath]X\subset \mathbb{R}[/imath] and let [imath]x'\in \mathbb{R}[/imath], we say that [imath]x'[/imath] is an adherent point of [imath]X[/imath] iff [imath]\,\forall\varepsilon>0\,\exists x\in X \text{ s.t.} \; d(x',x)\le \varepsilon[/imath]. We say that [imath]\overline{X}[/imath] is the closure of [imath]X[/imath] iff contain all the adherent points of [imath]X[/imath]. Lemma (Elementary properties of Closures): Let [imath]X[/imath] and [imath]Y[/imath] be arbitrary subset of [imath]\mathbb{R}[/imath]. Then [imath]X\subset \overline{X}, \; \overline{X\cup Y}=\overline{X}\cup\overline{Y},\; \overline{X\cap Y}\subset\overline{X}\cap\overline{Y}[/imath]. If [imath]X\subset Y[/imath] then [imath]\overline{X}\subset \overline{Y}[/imath]. Proof: Clearly if [imath]x\in X[/imath], [imath]x[/imath] is [imath]\varepsilon[/imath]-adherent to [imath]x[/imath] (indeed is [imath]0[/imath]-close to itself) for every [imath]\varepsilon[/imath], hence [imath]x\in \overline X[/imath]. Let [imath]z'[/imath] be an adherent point of [imath]X\cup Y[/imath], then there exists some [imath]z\in X\cup Y[/imath] such that [imath]d(z',z)[/imath] for a given [imath]\varepsilon>0[/imath]. Suppose [imath]z\in X[/imath], hence [imath]z'[/imath] is an adherent point of [imath]X[/imath], i.e., [imath]z'\in \overline{X}\subset \overline{X}\cup \overline{Y}[/imath]. Similarly when [imath]z\in Y[/imath], [imath]z'\in \overline{Y}\subset \overline{X}\cup \overline{Y}[/imath]. Thus [imath] \overline{X\cup Y}\subset\overline{X}\cup \overline{Y}[/imath]. We know that [imath]X\cup Y \subset\overline{X\cup Y}\subset\overline{X}\cup \overline{Y}[/imath]. Then [imath]\overline{\overline{X\cup Y}}=\overline{X}\cup \overline{Y}[/imath]. Since [imath]\overline{\overline{X\cup Y}} = \overline{X\cup Y}[/imath] we're done. Let [imath]z'\in \overline{X\cap Y}[/imath], i.e., [imath]z'[/imath] is an adherent point of [imath]X\cap Y[/imath]. So, there is some [imath]z\in X\cap Y \subset X[/imath] such that [imath]d(z',z)\le \varepsilon[/imath] for any given [imath]\varepsilon>0[/imath]. It follows that [imath]z'[/imath] is an adherent point of [imath]X[/imath], i.e., [imath]z'\in \overline{X}[/imath]. Similarly since [imath]z\in X\cap Y \subset Y[/imath] we have [imath]z'\in \overline{Y}[/imath]. Hence [imath]z'\in \overline{X}\cap\overline{Y}[/imath]. Let [imath]x'\in \overline{X}[/imath] we'd like to show that [imath]x'\in \overline{Y}[/imath]. Let [imath]\varepsilon>0[/imath] be arbitrary then [imath]\exists x\in X\subset Y[/imath] such that [imath]d(x',x)\le \varepsilon[/imath], so [imath]x'[/imath] is an adherent point of [imath]Y[/imath], i.e., [imath]x'\in \overline{Y}[/imath] as desired. [imath]\Box[/imath] Claim 1: If [imath]X\subset Y \subset \overline{X}[/imath]. Then [imath]\overline{Y}=\overline{X}[/imath]. Proof claim: Let [imath]y'[/imath] be an adherent point of [imath]Y[/imath], i.e., [imath]y'\in \overline{Y}[/imath] and let [imath]\varepsilon>0[/imath] be given. Thus, [imath]\exists y \in Y\subset \overline{X}[/imath] s.t. [imath]d(y,y')\le \varepsilon[/imath]. So, [imath]y[/imath] is an adherent point of [imath]X[/imath]. Thus [imath]\exists x\in X[/imath] s.t. [imath]d(x,y)\le \varepsilon[/imath] and therefore we must have [imath]d(x,y')\le 2\varepsilon[/imath] which shows that [imath]y'[/imath] is an adherent point of [imath]X[/imath], i.e., [imath]\overline{Y}\subset \overline{X}[/imath]. Conversely if [imath]x'[/imath] is an adherent point of [imath]X[/imath]. Then [imath]\exists x\in X\subset Y[/imath] s.t. [imath]d(x,x')\le \varepsilon[/imath], and so is adherent to [imath]Y[/imath]. [imath]\Box[/imath] Claim 2: [imath]\overline{\overline{X}} = \overline{X}[/imath] Proof: It will suffice to show that if [imath]x''\in \overline{\overline{X}}[/imath], then [imath]x''[/imath] is an adherent point of [imath]X[/imath]. Let [imath]\varepsilon >0[/imath] be arbitrary and let [imath]x''[/imath] be an adherent point of [imath]\overline{X}[/imath]. Then there is some [imath]x' \in \overline{X}[/imath] such that [imath]d(x'',x')\le \varepsilon[/imath]. Since [imath]x' \in \overline{X}[/imath] by definition is an adherent point of [imath]X[/imath], so there is some [imath]x\in X[/imath] for which [imath]d(x,x')\le \varepsilon[/imath]. Thus [imath]d(x,x'')\le d(x,x')+d(x',x'')\le 2\varepsilon[/imath] it follows that [imath]x''[/imath] is an adherent point of [imath]X[/imath] as desired. [imath]\Box[/imath] |
1134410 | Show that the group is trivial.
Show that the following group is identity: [imath]G=\langle x,y,z \mid xyx^{-1}=y^{2}\, , \, yzy^{-1}=z^{2}\, , \, zxz^{-1}=x^{2} \rangle.[/imath] This group is its own derived group. So all I get is group is perfect. Am I correct? What else I missed? Should I use tietze transformation, I am not getting anywhere with them. | 1023341 | Presentation of group equal to trivial group
Problem: Show that the group given by the presentation [imath]\langle x,y,z \mid xyx^{-1}y^{-2}\, , \, yzy^{-1}z^{-2}\, , \, zxz^{-1}x^{-2} \rangle [/imath] is equivalent to the trivial group. I have tried all sorts of manners to try to show that the relations given by the presentation above imply that [imath]x=y=z=e[/imath]. However, I am stuck and would appreciate any hints as to how I should move forward. |
546823 | If [imath]G[/imath] is isomorphic to [imath]H[/imath], then [imath]{\rm Aut}(G)[/imath] is isomorphic to [imath]{\rm Aut}(H)[/imath]
If a group [imath]G[/imath] is isomorphic to [imath]H[/imath], prove that [imath]{\rm Aut}(G)[/imath] is isomorphic to [imath]{\rm Aut}(H)[/imath]. Can someone provide a step by step solution? Explaining along the way our strategy in proving this. | 790997 | Automorphism groups of isomorphic groups are isomorphic
Say [imath]G \cong H[/imath] are isomorphic groups. Show [imath]Aut(G) \cong Aut(H)[/imath] I just made this up so I'm not sure if actually [imath]Aut(G) \cong Aut(H)[/imath] is true but I'm [imath]99.9\%[/imath] sure this should be true I'm having trouble with the proof :'( Let [imath]\theta:G \rightarrow H[/imath] be the isomorphism and [imath]\phi\in Aut(G)[/imath] Let [imath]g_1,g_2\in G[/imath] and let [imath]h_1,h_2 \in H \ \ with \ \ h_1 = \theta(g_1) \ \ and \ \ h_2 = \theta(g_2)[/imath] If [imath]\phi(g_1) = g_2[/imath] I want to show that [imath]\exists \alpha\in Aut(H)[/imath] such that [imath]\alpha(h_1) = h_2[/imath] I tried composing [imath]\theta[/imath] and [imath]\phi[/imath] to get into [imath]H[/imath] but that doesn't get me anywhere. Is there something wrong with this approach? Thanks ! :D Wait a minute.... [imath]\theta(\phi(g_1)) = \theta(g_2) = h_2[/imath] [imath]\theta(\phi(g_2)) = \theta(g_1) = h_1[/imath] I think I see something here, [imath]h_1[/imath] and [imath]h_2[/imath] were permuted in the same way [imath]g_1[/imath] and [imath]g_2[/imath] were but [imath]\theta \phi : G\rightarrow H[/imath] so it cant be in [imath]Aut(H)[/imath] |
1175055 | Show that [imath]\frac{1}{n} \sum_{k=1}^n f_k[/imath] converges uniformly to [imath]f[/imath].
Let [imath]f_n:[0,1]\to [0,1][/imath] continuous functions and let [imath]f:[0,1]\to [0,1][/imath] such that [imath]f_n[/imath] converges uniformly to [imath]f[/imath]. Show that [imath]\frac{1}{n} \sum_{k=1}^n f_k[/imath] also converges uniformly to [imath]f[/imath]. Now, I've seen a proof which starts with: [imath]\frac{1}{n} \sum_{k=1}^n (f_k - f) = \frac{1}{n} \left[ f-f_1 + f-f_2 + \ldots + f_n -f \right] \le \ldots \le \varepsilon[/imath] BUT, why is it showing uniform converges? I mean, shouldn't it start with: [imath] \left( \frac{1}{n} \sum_{k=1}^n f_k \right) -f [/imath] | 848904 | [imath]f_n:[0,1]\to [0,1][/imath] continuous, [imath]f_n\to f[/imath] uniformly, prove: [imath]{\frac1n}\sum_{k=1}^{n}{f_k} \to f[/imath] uniformly
I was able to prove, hopefully correctly, that the sum converges uniformly. But, I'm not sure how to show it converges uniformly specifically to [imath]f[/imath]. The way I proved it converges uniformly was by using the Dirichlet convergence test, by taking [imath]a_k(x) = \frac1n[/imath] and [imath]b_k(x) = f_k(x)[/imath]. [imath]\sum_{k=1}^n{b_k(x)}[/imath] is uniformly bounded on [imath][0,1][/imath] by [imath]n+2[/imath] for instance, and [imath]a_n[/imath] is monotone and uniformly convergent to 0 because it does not depend on the value of [imath]x[/imath]. I got rather stuck after this. I guess I have to use the continuity of the functions and the uniform convergence of both, but I can't figure out how to show that they converge to the same thing. Any help will be greatly appreciated. |
1179491 | Explicit bijection between [imath]\mathbb N[/imath] and [imath]\mathbb N \times \mathbb N[/imath]
We can consider the quadratic scheme above for a possible explicit bijection between [imath]\mathbb N[/imath] and [imath]\mathbb N \times \mathbb N[/imath]. The part [imath]\mathbb N \times \mathbb N \to \mathbb N[/imath] is easy via [imath](m,n) \mapsto m + \frac{(n + m)\cdot(n + m + 1)}{2}[/imath] for [imath]m[/imath] going down (as row index) and [imath]n[/imath] going right (as column index). Consider [imath](0,0) \mapsto 0[/imath] and [imath](2,1) \mapsto 8[/imath] and [imath](3,2) \mapsto 18[/imath] as examples. What ist the other map [imath]\mathbb N \to \mathbb N \times \mathbb N[/imath]? | 222709 | Inverting the Cantor pairing function
Towards the end of this, how do I get [imath] \frac{\sqrt{8z + 1} - 1}{2} < w + 1 [/imath] Here [imath]w = x + y \geq 0, t = (w^2 + w)/2 , z = t + w, x,y \in \mathbb N_{\geq 0}[/imath]. Thanks. |
1179497 | To prove that [imath](F,+)[/imath] and [imath](F-\{0\},\cdot)[/imath] are not isomorphic as groups.
Let [imath](F,+,\cdot)[/imath] be a field. Then to prove that [imath](F,+)[/imath] and [imath](F-\{0\},\cdot)[/imath] are not isomorphic as groups. I am facing difficulty in finding the map to bring a contradiction!! | 1080499 | Existence of a group isomorphism between [imath](\mathbb K,+)[/imath] and [imath](\mathbb K^\times,\cdot)[/imath]
Let [imath](\mathbb K,+,\cdot)[/imath] be a field. Is there a group isomorphism between [imath](\mathbb K,+)[/imath] and [imath](\mathbb K^\times,\cdot) [/imath] ? The answer should clearly be negative. I tried to proceed via contradiction, but this has not led me very far. Since it's some kind of "trick problem", I'm merely looking for hints. |
778233 | Is there any perfect squares that are also binomial coefficients?
Examining the Tartaglia's triangle, I have observed that all the squares were the trivial cases, that is, [imath]\binom{n^2}1[/imath] or [imath]\binom{n^2}{n^2-1}[/imath]. More formally: Conjecture: If [imath]\binom nm=k^2[/imath] then [imath]n=k^2[/imath]. Is it known to be true? I have tried to use the formula [imath]\nu_p\left(m!\right)=\sum_{k=1}^\infty\left\lfloor \frac m{p^\alpha}\right\rfloor[/imath] to prove that the exponents of the factorization of the binomial coefficients are odd, but I realized that this cannot be proved, because the binomial coefficients needn't be square-free: [imath]\binom 63=20[/imath], for example. Any ideas? | 1345739 | To find positive integers [imath]n[/imath] such that [imath]\dfrac {n(n+1)(n+2)}6[/imath] is a perfect square
How many positive integers [imath]n[/imath] are there such that [imath]\dfrac {n(n+1)(n+2)}6[/imath] is a perfect square ? I know [imath]n=1 , 2[/imath] works ; are there any more ? Are there only finitely many such [imath]n[/imath] ? |
1078781 | Proof If [imath]AB-I[/imath] Invertible then [imath]BA-I[/imath] invertible.
I have these problems : Proof If [imath]AB-I[/imath] invertible then [imath]BA-I[/imath] invertible. Proof If [imath]I-AB[/imath] invertible then [imath]I-BA[/imath] invertible. I think I solve it correctly, But I'm not so sure, I'll be glad to receive feedback. If [imath]AB-I[/imath] invertible then : [imath]\det|AB-I| \neq 0 \implies \\ \det|A-I||B| \neq 0 \implies \\ \det|B||A-I| \neq 0 \implies\\ \det|BA-I| \neq 0[/imath] Therefore [imath]BA-I[/imath] invertible. If [imath]I-AB[/imath] invertible then : [imath]\det|I-AB| \neq 0 \implies \\ \det|I-B||A| \neq 0 \implies \\ \det|I-BA| \neq 0[/imath] Therefore [imath]I-BA[/imath] invertible. | 1247091 | Let [imath]R[/imath] be a ring with identity. Prove that if [imath]1-ab[/imath] is invertible for some [imath]a,b \in R[/imath], then [imath]1-ba[/imath] is also invertible.
Let [imath]R[/imath] be a ring with identity. Prove that if [imath]1-ab[/imath] is invertible for some [imath]a,b \in R[/imath], then [imath]1-ba[/imath] is also invertible. Ok, si if [imath]R[/imath] is a ring with unity, then we have [imath]R[/imath] with [imath]1 \ne 0[/imath] We have to show that [imath]\forall a,b \in R[/imath], [imath](1-ab)[/imath] is invertible, so then [imath](1-ba)[/imath] is invertible as well. My thought process if to maybe show that if [imath](1-ab)x=1, x(1-ab)=1[/imath] for some [imath]x \in R[/imath]. Then we would have to show that [imath](1-ba)y=1,[/imath] so then [imath]y(1-ba)=1[/imath] for some [imath]y \in R[/imath]???? |
1176195 | Would you ever stop rolling the die?
You have a six-sided die. You keep a cumulative total of your dice rolls. (E.g. if you roll a 3, then a 5, then a 2, your cumulative total is 10.) If your cumulative total is ever equal to a perfect square, then you lose, and you go home with nothing. Otherwise, you can choose to go home with a payout of your cumulative total, or to roll the die again. My question is about the optimal strategy for this game. In particular, this means that I am looking for an answer to this question: if my cumulative total is [imath]n[/imath], do I choose to roll or not to roll in order to maximize my cumulative total? Is there some integer [imath]N[/imath] after which the answer to this question is always to roll? I think that there is such an integer, and I conjecture that this integer is [imath]4[/imath]. My reasoning is that the square numbers become sufficiently sparse for the expected value to always be in increased by rolling the die again. As an example, suppose your cumulative total is [imath]35[/imath]. Rolling a [imath]1[/imath] and hitting 36 means we go home with nothing, so the expected value of rolling once is: [imath]E(Roll|35) = \frac 0 6 + \frac {37} 6 + \frac {38} 6 + \frac{39} 6 + \frac {40} {6} + \frac{41}{6} = 32.5[/imath] i.e. [imath]E(Roll|35) = \frac 1 6 \cdot (37 + 38 + 39 + 40 + 41) = 32.5[/imath] But the next square after [imath]35[/imath] is [imath]49[/imath]. So in the event that we don't roll a [imath]36[/imath], we get to keep rolling the die at no risk as long as the cumulative total is less than [imath]42[/imath]. For the sake of simplification, let's say that if we roll and don't hit [imath]36[/imath], then we will roll once more. That die-roll has an expected value of [imath]3.5[/imath]. This means the expected value of rolling on [imath]35[/imath] is: [imath]E(Roll|35) = \frac 1 6 \cdot (40.5 + 41.5 + 42.5 + 43.5 + 44.5) = 35.42[/imath] And since [imath]35.42 > 35[/imath], the profit-maximizing choice is to roll again. And this strategy can be applied for every total. I don't see when this would cease to be the reasonable move, though I haven't attempted to verify it computationally. I intuitively think about this in terms of diverging sequences. I recently had this question in a job interview, and thought it was quite interesting. (And counter-intuitive, since this profit-maximizing strategy invariably results in going home with nothing.) | 977679 | Toss a fair die until the cumulative sum is a perfect square-Expected Value
Suppose we keep tossing a fair dice until we want to stop, at which point the game ends and our score is the cumulative sum, or until the cumulative sum is a perfect square, in which case we lose and end up with a score of [imath]0[/imath]. We can set up a Markov-style recurrence. Define [imath]f(k)[/imath] to be the expected value of the game if the current cumulative sum is [imath]k[/imath]. Clearly, [imath]f(k)[/imath] is [imath]0[/imath] if [imath]k[/imath] is a perfect square. In addition, [imath]f(k)=\text{max}\{k,\sum_{i=1}^6 f(k+i)\}[/imath]. How can we solve such a recurrence to find a good estimate/bounds of the expected score ([imath]f(0)[/imath]) and the optimal strategy? Can these methods be extended to a game where we lose if the cumulative sum is an even number, or a perfect cube? |
1179786 | Find the limit. (If an answer does not exist, enter DNE.) [imath]lim_{x \to \infty} {\sqrt{9x^2 + x}− 3x}[/imath]
Is my process is correct? Also, why is one to allowed to divide the variable in the radical by its highest power? [imath] \lim_ {x-> \infty} \sqrt { 9 x^2 +x} - 3 x [/imath] | 1169437 | Find the limit. (If an answer does not exist, enter DNE.) [imath]\lim _{x→∞} (\sqrt{9x^2 + x} − 3x)[/imath]
I was following this explanation until the 5th step. The most important misunderstanding is if [imath]1x/1x[/imath] is [imath]1[/imath], and one can multiply a quotient by [imath]1[/imath] and not change its value. However from what I can see the [imath]9x^2[/imath] is being divided by [imath]x^2[/imath]. How is that going to work? And why is that allowed? |
1179918 | How could I prove whether the infinite series [imath]n!/n^n[/imath] converges or diverges?
How would you go about proving the infinite series [imath]\frac{n!}{n^n}[/imath] converges or diverges? | 81113 | Is [imath]\sum\limits_{n=1}^{\infty}\frac{n!}{n^n}[/imath] convergent?
I can clearly see that [imath]\dfrac{n!}{n^n}\to 0[/imath] when [imath]n\to\infty[/imath]. But how do I know if the sum [imath]\sum_{n=1}^{\infty}\frac{n!}{n^n}[/imath] is convergent or not? I know this might be basic, but thank you if anyone can help me. |
1180319 | Open and Closed Sets in the Euclidean Space R^n
Prove that if [imath]F_1, F_2, ..., F_k[/imath] are closed sets in [imath]R^{n},[/imath] then [imath]\cup_{i = 1}^{k}F_i[/imath] is also a closed set in [imath]\mathbb{R}^{n}.[/imath] | 107692 | Prove that a finite union of closed sets is also closed
Let [imath]X[/imath] be a metric space. If [imath]F_i \subset X[/imath] is closed for [imath]1 \leq i \leq n[/imath], prove that [imath]\bigcup_{i=1}^n F_i[/imath] is also closed. I'm looking for a direct proof of this theorem. (I already know a proof which first shows that a finite intersection of open sets is also open, and then applies De Morgan's law and the theorem "the complement of an open set is closed.") Note that the theorem is not necessarily true for an infinite collection of closed [imath]\{F_\alpha\}[/imath]. Here are the definitions I'm using: Let [imath]X[/imath] be a metric space with distance function [imath]d(p, q)[/imath]. For any [imath]p \in X[/imath], the neighborhood [imath]N_r(p)[/imath] is the set [imath]\{x \in X \,|\, d(p, x) < r\}[/imath]. Any [imath]p \in X[/imath] is a limit point of [imath]E[/imath] if [imath]\forall r > 0[/imath], [imath]N_r(p) \cap E \neq \{p\}[/imath] and [imath]\neq \emptyset[/imath]. Any subset [imath]E[/imath] of [imath]X[/imath] is closed if it contains all of its limit points. |
1180705 | Explicit well-ordering of [imath]\mathbb Q[/imath]
Starting from the axiom of choice and passing the lemma of Zorn we arrive at the well-ordering theorem which states that for every set [imath]X[/imath] there exists a well-ordering with domain [imath]X[/imath]. I am aware that possibly nobody will ever be able to construct an explicit well-ordering of [imath]\mathbb R[/imath]. There are in fact enough questions to be found here on math.stackexchange.com But what about the "next best thing": an explicit (=constructed) well-ordering of [imath]\mathbb Q[/imath]? | 424654 | List of explicit enumerations of rational numbers
A well-known mathematical fact is that the rational numbers are countable, i.e. there is a bijective function [imath]f:\mathbb{N}\rightarrow \mathbb{Q}[/imath] I am interesting in making a list of all explicit such bijections since each one that I know have a different philosophy behind it. This is one of the more counterintuitive facts about infinite, at least when one enters inside set theory. When I explain this in the first time to anyone, he/she is surprised. Thus I think it will be useful for showing the fact in a more clear manner or as possible exercises to show alternatives ways with respect to the standard one. |
215452 | Show that there exists infinitely many numbers that are coprime pairwise, in the set defined as following
The set [imath]A[/imath] = {[imath]X_n\mid n\in \Bbb N[/imath]} where [imath]X_n = a^{n+1} + a^{n} - 1[/imath], with [imath]a \gt 1, a \in \Bbb Z[/imath]. Show that there are infinitely many numbers that are pairwise coprime. | 1178134 | Infinite subset with pairwise comprime elements
I'm looking for solution for this problem Let [imath]a[/imath] be an integer greater then 1. Then, the set [imath]A\mathrel{:=}\{a^n(a+1)-1\mid n\in\mathbb{Z}_{+}\}[/imath] a has an infinite subset b such that it's elements are pairwise coprime. I found similar question here but proposed solutions are wrong: Author of solution claims that " Then [imath]d[/imath] is also a multiple of [imath]m[/imath] and using the well-known fact that [imath]x−y[/imath] divides [imath]x^k−y^k[/imath], we see that [imath]a^m−1[/imath] is a multiple of [imath]a^d−1[/imath] " Here we can find a small but critical mistake: from [imath]m\mid d[/imath] follows that [imath]a^m-1 \mid a^d-1[/imath] but no [imath]a^d-1\mid a^m-1[/imath]. Also author states that if [imath]m\mid n[/imath] that [imath]\gcd(X_m,X_n)=1[/imath]. It's also false because we can take [imath]X_1=a(a+1)-1[/imath] and it's prime factor [imath]p[/imath]. But then [imath]X_p=a^p(a+1)-1[/imath] is also divided by [imath]p[/imath] because of Fermat's little theorem. |
1181087 | Continuity of Product Topology
Let [imath]X_1, X_2, Y[/imath] be topological spaces and let [imath]X_1 \times X_2[/imath] be the topological space obtained by furnishing the Cartesian product set with the product topology. Let [imath]f: X_1 \times X_2 \to Y[/imath] be a given map. Then f is continuous iff for each [imath]U \in Open(Y)[/imath] and for each [imath](x_1, x_2) \in X_1 \times X_2[/imath] such that [imath]f(x_1, x_2) \in U[/imath], there exists [imath](U_1, U_2) \in [Open(X_1) \times Open(X_2)][/imath] such that [imath](x_1, x_2) \in U_1 \times U_2[/imath] and [imath]f(U_1 \times U_2) \subseteq U[/imath]. Now, for fixed [imath]x_1[/imath], define a map [imath]f_2^{x_1}:X_2\to Y[/imath] by [imath]x_2 \mapsto f(x_1,x_2)[/imath]. For fixed [imath]x_2[/imath], define a map [imath]f_1^{x_2}:X_1\to Y[/imath] by [imath]x_1 \mapsto f(x_1, x_2)[/imath]. Is there a way to formulate the continuity of [imath]f[/imath] in terms of continuity of [imath]f_1^{x_2}[/imath] for all [imath]x_2[/imath] and continuity of [imath]f_2^{x_1}[/imath] for all [imath]x_1[/imath]? If not, what additional assumptions are required? I could prove that if these functions are uniformly continuous (that is, in [imath]x_1[/imath] and [imath]x_2[/imath] respectively) then [imath]f[/imath] is continuous. | 256906 | Does factor-wise continuity imply continuity?
Let [imath]f[/imath] denote a map from a product space [imath]X \times Y[/imath] to [imath]Z[/imath]. If for every [imath]x\in X[/imath], the map [imath]f(x,-)[/imath] is continuous, and the same holds for every [imath]y \in Y[/imath], then is [imath]f[/imath] continuous in general? If not, is there any condition to be imposed to make [imath]f[/imath] continuous? |
903117 | Integral of Sinc Function Squared Over The Real Line
I am trying to evaluate [imath]\int_{-\infty}^{\infty} \frac{\sin(x)^2}{x^2} dx [/imath] Would a contour work? I have tried using a contour but had no success. Thanks. Edit: About 5 minutes after posting this question I suddenly realised how to solve it. Therefore, sorry about that. But thanks for all the answers anyways. | 1557394 | how to evaluate this definite integral [imath]\int_0^\infty\frac{\sin^2(x)}{x^2}dx[/imath]?
For [imath]\int_{0}^{\infty}\frac{\sin^2(x)}{x^2}dx[/imath]. I considered using residue theorem. But since the function inside is holomorphic except for a removable singularity at the origin. So whatever contour I take, the integral would always be zero. Is there a way that we can get away with the situation and apply residue theorem to do the work? |
1181631 | How do you prove the set [imath]G = \{(x, f(x)) \mid x \in \mathbb R\}[/imath] is closed?
Let [imath]f : \mathbb R \to \mathbb R[/imath] continuous. Prove that graph [imath]G = \{(x, f(x)) \mid x \in \mathbb R\}[/imath] is closed. I'm a little confused on how to prove [imath]G[/imath] is closed. I get the general strategy is to show that every arbitrary convergent sequence in [imath]G[/imath] converges to a point in [imath]G[/imath]. Here is what I tried so far: Let [imath]x_k[/imath] be a sequence which converges to [imath]x[/imath]. Since [imath]f[/imath] is continuous, this implies that [imath]f(x_k)[/imath] converges to [imath]f(x)[/imath]. At this point, can you say every [imath](x_k, f(x_k))[/imath] converges to a [imath](x, f(x))[/imath], so [imath]G[/imath] is closed? | 722545 | Graph of a continuous function is closed
Let [imath]f: \mathbb{R} \to \mathbb{R}[/imath] be continuous. Then [imath]G = \{ (x, f(x) ) : x \in \mathbb{R} \} [/imath] is a closed set. My try: Suppose [imath](z_n) = (x_n, f(x_n) ) [/imath] is sequence in [imath]G[/imath] with limit [imath](x,y)[/imath]. We must show [imath](x,y) \in G[/imath]. Since [imath]x_n \to x[/imath] and since [imath]f[/imath] is continuous, then we must have that [imath]f(x_n) \to f(x) [/imath]. Since limits are unique. Then [imath]y = f(x) [/imath]. Is this enough to conlude that [imath](x,y) \in G [/imath] ? Hence showing [imath]G[/imath] is closed ? thanks |
627667 | Every compact metric space is complete
I need to prove that every compact metric space is complete. I think I need to use the following two facts: A set [imath]K[/imath] is compact if and only if every collection [imath]\mathcal{F}[/imath] of closed subsets with finite intersection property has [imath]\bigcap\{F:F\in\mathcal{F}\}\neq\emptyset[/imath]. A metric space [imath](X,d)[/imath] is complete if and only if for any sequence [imath]\{F_n\}[/imath] of non-empty closed sets with [imath]F_1\supset F_2\supset\cdots[/imath] and [imath]\text{diam}~F_n\rightarrow0[/imath], [imath]\bigcap_{n=1}^{\infty}F_n[/imath] contains a single point. I do not know how to arrive at my result that every compact metric space is complete. Any help? Thanks in advance. | 2308241 | Proof Review: Every compact set is complete
[imath]X[/imath] is a compact metric space. Let [imath]A \subseteq X[/imath] be a compact set. Let [imath]\{C_n\} \in A[/imath] be a Cauchy sequence. Because it is Cauchy, [imath]\{C_n\}[/imath] must converge somewhere (although it doesn't need to converge in [imath]A[/imath]). Let the limit point for [imath]\{C_n\}[/imath] be denoted by [imath]C[/imath]. We also have that [imath]A[/imath] is compact [imath]\implies A[/imath] is closed and bounded [imath]\implies[/imath] [imath]C \in A[/imath] (because closed sets include the limit points for all its convergent sequences) [imath]\implies[/imath] A is complete. |
1181142 | [imath]\mathbb{Z}[x_{1},\dots,x_{n}]/I[/imath] is a field therefore it's finite
I'd spent much time for this but didn't get any results.. Could u give me only the idea but not a full proof | 148745 | Fields finitely generated as [imath]\mathbb Z[/imath]-algebras are finite?
Suppose [imath]k[/imath] is a field that is finitely generated as a [imath]{\mathbb Z}[/imath]-algebra. (That is, [imath]k[/imath] is a quotient of [imath]{\mathbb Z}[X_1,\dots,X_n][/imath] for some [imath]n[/imath]). Does it follow that [imath]k[/imath] is finite? |
1182577 | Why is "All horses have the same color" considered a false proof by induction?
Upon reading of All horses have the same color "paradox", I began to wonder a couple of things. First of all, to me the inductive step seems flawed. Just because I have [imath]n[/imath] white horses, does not mean that the [imath]n+1[/imath] will be white, too. The only explanation to make the inductive step work seems to me that the actual statement we are assuming as base case it's not "there are at least [imath]n[/imath] horses of the same color", but rather "[imath]P(n)[/imath] = any grouping of [imath]n[/imath] horses is such all the horses in that grouping have the same color". Now this makes the inductive step works, but it's completely useless, as [imath]P(2)[/imath] already implies that all the horses have the same color. And of course the base case [imath]P(2)[/imath] is not true because it's equivalent to our problem. Setting to prove [imath]P(2)[/imath] is the same as proving that all horses have the same color. So I can't understand why this has been considered a "false" proof by induction; either the inductive step does not work or it's completely useless. The wikipedia page linked before explains the problem as The problem in the argument is the assumption that because each of these two sets contains only one color of horses, the original set also contained only one color of horses. Because there are no common elements (horses) in the two sets, it is unknown whether the two horses share the same color. Which does not make any sense to me.. It's like looking at the finger instead that at the moon. Is there something wrong with my understanding of induction? Edit Admittedly, the base case is [imath]P(1)[/imath] and not [imath]P(2)[/imath]. But the point holds;the only implication we need is [imath]P(1) \implies P(2)[/imath]. All the subsequent implications are useless; a proof of this type is seldom called a proof by "induction". Otherwise all proofs are by induction! :) What is even the point of establishing what [imath]P(n)[/imath] is? Just prove [imath]P(2)[/imath] and use as known fact [imath]P(1)[/imath] as you would do with any other known fact. Edit 2 Since there has been some confusion about what I asked, I'll explain my thought process: 1) Begin reading the proof. Imagine that the statement is "P(n) = at least n horses have the same color", and if by induction on [imath]n[/imath] this hold for every [imath]n[/imath], then QED. 2) Realize the inductive step does not work 3) Understand that the proposition to be proved has been (silently) changed and now it's "P(n) = any grouping of [imath]n[/imath] horses is such all the horses in that grouping have the same color"." 4) Realize that [imath]P(2)[/imath] is what we want to prove. 5) Wonder why on earth we are trying to show [imath]P(n)[/imath] for [imath]n > 2[/imath]. Induction proof usually works because the final statement to be proved come from the fact that it works for all [imath]n[/imath]. Here [imath]n=2[/imath] suffice, so it is really odd. 6) Starting to think I've missed something here. But hey, it's a false proof, let's see what was wrong. 7) Turns out that the outline of the proof is fine but here's the catch: of all the implications [imath]P(1) \implies P(2)[/imath] is not valid hence the whole [imath]P(n)[/imath] cannot be valid. 8) Thinking (again) that this seems weird. Wonder if I've missed something . By extended discussion it turns out that my understanding of induction was right. This lead me to believe that is is a bad example because it "seems" like a proof by induction but really some essential elements are not present. It's not that is wrong (apart from the failed implication), is that it gives a very bad idea of what a proof by induction is | 1351821 | The pencils in a box of crayons always have the same color
I retrieved an old math book and I'm delighted to share following exercise. The pencils in a box of crayons always have the same color. Proof by induction on the number [imath]n[/imath] of pencils in the box: This is true when [imath]n=1[/imath]. All the pencils, only one, havs the same color. Suppose the result true for a box having [imath]n+1[/imath] crayons. Remove one. Remains [imath]n[/imath] crayons. By induction hypothesis they all have the same color. Put back the crayon you removed and remove another one. The [imath]n[/imath] remaining crayons in the box have again the same color by induction hypothesis. Hence the two crayons that were removed have the same color. Therefore all the [imath]n+1[/imath] crayons have the same color. Can you find what's wrong here? |
1183225 | How does one find the reduced Singular Value Decomposition of a row or column vector?
If we treat a column vector [imath]a[/imath] as an [imath]n \times 1[/imath] matrix, or a row vector [imath]a^T[/imath] as a [imath]1 \times n[/imath] matrix, how would one write out the reduced singular value decomposition of [imath]a[/imath]? | 1181800 | Singular value decomposition of column & row vectors
(a) Given the column vector [imath]a[/imath] as an [imath]n\times 1[/imath] matrix. Write out its singular value decomposition, showing the matrices [imath]U, \sum[/imath], and [imath]V[/imath] explicitly. (b) Given the row vector [imath]a^T[/imath] as an [imath]1\times n[/imath] matrix. Write out its singular value decomposition, showing the matrices [imath]U, \sum[/imath], and [imath]V[/imath] explicitly. I currently get stucked on this question, since I couldn't find the eigenvalues of the [imath]n\times n[/imath] matrix [imath]aa^T[/imath] in part (a), and [imath]a^Ta[/imath] in part (b) to find the columns of [imath]U[/imath] and [imath]V[/imath], respectively. Can someone please help me on this problem? |
1183587 | P-value hypthesis test(my bad - found same question in math stack)
textbook exercise How can I calculate P-value at a test of [imath]H_0: u=3.2[/imath] against [imath]H_1: u \neq 3.2[/imath] if we have the observations [imath]2.0[/imath], [imath]3.2[/imath], [imath]3.8[/imath], [imath]2.5[/imath], [imath]3.3[/imath], [imath]2.8[/imath], [imath]3.0[/imath] and [imath]3.4[/imath] which is a sample from [imath]N(u,o^2[/imath]), there [imath]u[/imath] and [imath]o^2[/imath] are espected value respectively variance. I have calculated the confidence interval to [imath]I_u=(2.58,3.42)[/imath] rejection region to [imath]C=(|T|\geq1.96)[/imath] where [imath]T=\frac{u^*-u_0}{\frac{o}{n^{1/2}}}[/imath] [imath]u^*[/imath]=[imath]m[/imath]=mean value of the observations. EDIT: Finding P value | 695677 | Finding P value
I have these observations [imath](2,3.2,3.8,2.5,3.3,2.8,3.0,3.4)[/imath] from [imath]X \sim N(\mu,\sigma^2)[/imath] and i want to calculate the [imath]P[/imath]-value testing [imath]H_0: \mu =3.2[/imath] against [imath]H_1 \neq 3.2[/imath] with [imath]\sigma = 0.6[/imath] should i calculate [imath]P r(X > 3.3) [/imath] and [imath]Pr( X < 3)[/imath] and add these togheter ? I tried it using that [imath]X = Y\sigma + \mu \sim N(0,1)[/imath] but the answer don't seem to be right according to the solution which is [imath]P = 0.347[/imath] Anyone can tell me what Iam doing wrong? |
1183920 | What is wrong with this argument that the [imath]\left(0,1\right)[/imath] interval is countable?
Take each number [imath]n \in \mathbb{N}[/imath] and reverse the digits (as if it is a string). This is a bijection. For example, [imath]123000[/imath] becomes [imath]000321[/imath], [imath]456[/imath] becomes [imath]654[/imath] and so on. Now prepend the string "0." in all these string-numbers and one has all the numbers in the [imath]\left(0,1\right)[/imath] interval, this is a bijection from [imath]\mathbb{N}[/imath] to [imath]\left(0,1\right)[/imath], thus the [imath]\left(0,1\right)[/imath] interval is countable. Of course similar questions have been posted on this site and let me give an answer that addresses this argument, here (A set [imath]\,\{0,1\}^*[/imath] is countable , but its subset [imath]\,\{0,1\}^{\Bbb N}\,[/imath] is uncountable?). The answer is that while the set [imath]\{0,1\}^*[/imath] is in fact countable, it represents sequencses (or string-numbers) of finite length, while the construction here would correspond to the set [imath]\{0,1\}^\mathbb{N}[/imath] which represents sequences of infinite length and is uncountable. Then the question could be phrased as follows: Each number [imath]n \in \mathbb{N}[/imath] can have an arbitrary length [imath]\|n\|[/imath] (if seen as a sequence of digits/symbols), more precisely, an arbitrarily large length (effectively infinite) why the counter-arguments does not apply here? What is wrong with the set [imath]\{0,1\}^*[/imath] of finite-length sequences [imath]n[/imath], since [imath]n[/imath] can vary and take arbitrary large values? This question is not duplicate, as the proposed duplicates do not address the later part of this question (but only hint to them as arguments to the contrary). Editing after [ON HOLD] so lets take this to the point: Does the sequence [imath]11111...[/imath] (repeated indefinately) [imath]\in \{0,1\}^*[/imath] or not? | 329282 | A set [imath]\,\{0,1\}^*[/imath] is countable , but its subset [imath]\,\{0,1\}^{\Bbb N}\,[/imath] is uncountable?
I think [imath]\{0,1\}^*[/imath] represents all [imath]0,1[/imath]-sequences, and [imath]\{0,1\}^{\mathbb{N}}[/imath] is the [imath]0,1[/imath]-sequences with infinite length. So [imath]\{0,1\}^{\mathbb{N}}[/imath] is a subset of [imath]\{0,1\}^*[/imath]. [imath]\{0,1\}^*[/imath] is countable, while [imath]\{0,1\}^\mathbb{N}[/imath] is uncountable. It's really strange, because I can count the whole set and cannot when it comes to its subset! I don't know how to understand it. Who can save me? Thanks in advance. |
1183924 | Why is [imath]\lim(n!)^{1/n}=\infty[/imath]
Why is [imath]\lim\limits_{n\to\infty}(n!)^{1/n}=\infty[/imath] It is more or less clear that the sequence is increasing by ratio test [imath]\frac{((n+1)!)^{1/(n+1)}}{(n!)^{1/n}}=\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n+1]{(n)!}}\cdot\frac{\sqrt[n+1]{(n)!}}{\sqrt[n]{(n)!}}=\sqrt[n+1]{n+1}\cdot(n!)^{\frac{1}{n(n+1)}}[/imath] but left multiplicand tends to [imath]1[/imath], what about the right one ? Is there convergence/divergence tests for sequences, I only found some for series ? | 706461 | Calculating the limit [imath]\lim((n!)^{1/n})[/imath]
Find [imath]\lim_{n\to\infty} ((n!)^{1/n})[/imath]. The question seemed rather simple at first, and then I realized I was not sure how to properly deal with this at all. My attempt: take the logarithm, [imath]\lim_{n\to\infty} \ln((n!)^{1/n}) = \lim_{n\to\infty} (1/n)\ln(n!) = \lim_{n\to\infty} (\ln(n!)/n)[/imath] Applying L'hopital's rule: [imath]\lim_{n\to\infty} [n! (-\gamma + \sum(1/k))]/n! = \lim_{n\to\infty} (-\gamma + \sum(1/k))= \lim_{n\to\infty} (-(\lim(\sum(1/k) - \ln(n)) + \sum(1/k)) = \lim_{n\to\infty} (\ln(n) + \sum(1/k)-\sum(1/k) = \lim_{n\to\infty} (\ln(n))[/imath] I proceeded to expand the [imath]\ln(n)[/imath] out into Maclaurin form [imath]\lim_{n\to\infty} (n + (n^2/2)+...) = \infty[/imath] Since I [imath]\ln[/imath]'ed in the beginning, I proceeded to e the infinity [imath]= e^\infty = \infty[/imath] So am I write in how I approached this or am I just not on the right track? I know it diverges, I was just wanted to try my best to explicitly show it. |
1184689 | Positive continuous [imath]f_n:\mathbb{R}\to\mathbb{R}[/imath] such that [imath]\{f_n(x)\}[/imath] is unbounded [imath]\iff[/imath] [imath]x\in\mathbb{Q}[/imath].
Does there exist a sequence of continuous positive functions [imath]f_n[/imath] on [imath]\mathbb{R}[/imath] such that the sequence [imath]\{f_n(x)\}_{n=1}^\infty[/imath] is unbounded if and only if [imath]x\in\mathbb{Q}[/imath]? I was thinking of letting [imath]\mathbb{Q}=\{r_n\}_{n=1}^\infty[/imath] and for each [imath]n[/imath], letting [imath]f_n\in C(\mathbb{R})[/imath] be such that [imath]f_n(x)=n[/imath] for [imath]x=r_1,\ldots,r_n[/imath] and [imath]0[/imath] outside small disjoint intervals around [imath]r_1,\ldots,r_n[/imath]. We would have [imath]f_n(x)\to\infty[/imath] for all [imath]x\in\mathbb{Q}[/imath]. But maybe [imath]f_n(x)[/imath] is not necessarily bounded for [imath]x\notin\mathbb{Q}[/imath]. | 1127141 | Existence of a sequence of positive continuous functions on [imath]\mathbb R[/imath] which is unbounded precisely on [imath]\mathbb Q[/imath]
Is it possible to find a sequence of positive continuous functions [imath]g_n[/imath] on the real numbers such that [imath]( g_n(x) )[/imath] is unbounded if and only if [imath]x \in \mathbb{Q}[/imath] ? |
1184059 | The series [imath]n^{-1-1/n}[/imath] diverges, but how do I show this?
I have problems with the following sum [imath]n^{-1-1/n}[/imath]. I need to show that the sum diverges, I know that I have to use the comparison test after spending too much time with the wrong tests. My problem is now that I can't see what sequence [imath]a_n[/imath] is less than the sequence [imath] b_n=n^{-1-1/n}[/imath]. Is there any hints on how I can do this? I have looked through my notes and previous exercises, but I can't find anything that helps. Best regards | 1171184 | Why doesn't [imath]\sum_{n=1}^\infty \frac{1}{n^{1+\frac{1}{n}}}[/imath] converge?
[imath]\sum_{n=1}^\infty \frac{1}{n^{1+\frac{1}{n}}} = \infty[/imath]. Is there a comparison that works well to prove this? |
1184452 | Is the Russell paradox the only possible contradiction to the axiom schema of comprehension due to Frege (1893)? [imath]\{x:P(x)\}[/imath]
Is the Russell paradox the only possible contradiction to the axiom schema of comprehension due to Frege (1893)? The axiom that says that if [imath]\varphi[/imath] is a property, then there exists a set [imath]Y = \{X: \varphi(X)\}[/imath] of all elements having property [imath]\varphi[/imath]. If not then what are other paradoxes that result from that axiom? | 139936 | Paradox of General Comprehension in Set Theory, other than Russell's Paradox
As is well known, the General Set Comprehension Principle (any class is a set) leads to the Russell Paradox (the class [imath]x \notin x[/imath] cannot be a set). As a result, set theories must restrict the Comprehension Principle to avoid self-reference. For example, in the case of ZFC, this is done by enumerating a small list of "safe" comprehension schema such as Separation. Does General Comprehension lead to other types of paradox than Russell's? Or is this basically the only thing that can go wrong? |
1184961 | Proof using strong induction
I need to prove/show that [imath]n^3 \leq 3^n[/imath] for all natural numbers [imath]n[/imath] by strong induction. I have no clue where to begin!!!! :( I know how to do the beginning steps of showing that it's true for [imath]k = 0[/imath] and [imath]k = 1[/imath], etc but get suck on how to start the strong inductive step. | 1180201 | Prove by mathematical induction that: [imath]\forall n \in \mathbb{N}: 3^{n} > n^{3}[/imath]
Prove by mathematical induction that: [imath]\forall n \in \mathbb{N}: 3^{n} > n^{3}[/imath] Step 1: Show that the statement is true for [imath]n = 1[/imath]: [imath]3^{1} > 1^{3} \Rightarrow 3 > 1[/imath] Step 2: Show that if the statement is true for [imath]n = p[/imath], it is true for [imath]n = p + 1[/imath] The general idea I had was to start with [imath](p+1)^{3}[/imath] and during the process substitute in [imath]3^{p}[/imath] for [imath]p^{3}[/imath] as an inequality. [imath](p+1)^{3} = p^{3} + 3p^{2} + 3p + 1 < 3^{p} + 3p^{2} + 3p + 1[/imath] Now, if it can be shown that: [imath]\forall p \in \mathbb{N}: 3p^{2} + 3p + 1 \leq 2 \cdot 3^{p}[/imath] ...the proof is complete. This is because [imath]3^{p+1} = 3 \cdot 3^{p}[/imath] and one of those three have already been used. We do this by mathematical induction. First, the base case of [imath]n = 1[/imath]: [imath]3\cdot 1^{2} + 3 \cdot 1 + 1 \not \leq 2 \cdot 3^{1}[/imath] ..which turns out to be false. What are some more productive approaches to this step? |
1185054 | issue in figuring out how to calculate probability
A fair, 6-sided die is rolled 6 times independently. Assume that the results of the different rolls are independent. Let [imath](a_1,\ldots,a_6)[/imath] denote a typical outcome, where each [imath]a_i[/imath] belongs to [imath]\{1,\ldots,6\}[/imath]. For any outcome [imath]\omega=(a_1,\ldots,a_6)[/imath], let [imath]R(\omega)[/imath] be the set [imath]\{a_1,\ldots,a_6\}[/imath]; this is the set of numbers that showed up at least once in the different rolls. For example, if [imath]\omega=(2,2,5,2,3,5)[/imath], then [imath]R(\omega)=\{2,3,5\}[/imath]. Find the probability that [imath]R(\omega)[/imath] has exactly two elements. (Answer with at least 3 decimal digits.) Find the probability that [imath]R(\omega)[/imath] has exactly three elements. can anyone clarify or make hints for answer | 1177721 | A fair 6-sided die is rolled 6 times, what's the probability the outcome has exactly 2 or 3 elements?
A fair [imath]6[/imath]-sided die is rolled [imath]6[/imath] times independently. For any outcome, this is the set of numbers that showed up at least once in the different rolls. For example, the outcome is [imath](2,3,3,3,5,5)[/imath], the element set is [imath]\{2,3,5\}[/imath]. What is the probability the element set has exactly [imath]2[/imath] elements? how about [imath]3[/imath] elements? I know the sample space is [imath]6^6[/imath]. The counting for [imath]2[/imath] elements is [imath]6C2 \cdot 6![/imath]? I would really appreciate the help! :) |
1185180 | Finding [imath]y[/imath] In Calculus(Area) Problem?
Find the number b such that the line [imath]y=b[/imath] divides the region bounded by the curves [imath]y = x^2[/imath] and [imath]y = 4[/imath] into two regions with equal area. | 482557 | Find the number [imath]b[/imath] such that the line [imath]y = b[/imath] divides the region bounded by the curves [imath]y = x^2[/imath] and [imath]y = 4[/imath] into two regions with equal area.
Find the number [imath]b[/imath] such that the line [imath]y = b[/imath] divides the region bounded by the curves [imath]y = x^2[/imath] and [imath]y = 4[/imath] into two regions with equal area. I could find the area of this region but i have no clue how to split it parallel to the [imath]x[/imath] axis perfectly... this is too abstract for me... I found the whole area and got [imath]\dfrac{32}{3}[/imath]... I tried plugging that back into the definite integral but that definitely wasn't right... |
1185284 | How would I go about evaluating [imath]\int_{0}^{\infty}e^{{-u}^{2}}\cos(ux)du[/imath]?
So I had this question as extra credit on my exam and no one was able to solve it. I tried to do integration by parts but that just keep on expanding with no solution in sight. I have also tried to take the derivative of the integral but that didn't seem to help either. How would I integrate [imath]\int_{0}^{\infty}e^{{-u}^{2}}\cos(ux)du[/imath]? | 317249 | Gaussian-like integral : [imath]\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x[/imath]
Prove that : [imath] \frac{\sqrt{\pi}}{2} e^{-\frac{a^2}{4} } =\int_0^{\infty} e^{-x^2} \cos( a x) \ \mathrm{d}x[/imath] the only thing I can think of is differentiating the RHS and trying to get : [imath] -2 f'(a) =a f(a) [/imath] But I couldn't do it. Can anyone show me how to do this ? |
488416 | Change of variable problems in probability
If [imath]X[/imath] is a continuous random variable with positive values,find it's Cumulative distribution function (CDF) and it's probability density function (PDF) for [imath]Y = \sqrt{X}[/imath] This was a detable question on another topic so I made it a seperate one. Can someone explain how do we solve this problem to it's end? (putting some theory/ explanations in between would be appreciated. My goal is to understand the concepts that are behind this and be prepared for problems of equal difficulty). Thank you. Added by A.F. This was originally the "third pattern" of this question. Possible solutions may exist on the linked question. | 488049 | Given a continuous random variable [imath]X[/imath] and the pdf find....
There is a pattern excercise: First pattern: We have [imath]X[/imath] and the probability density function is usually: [imath] f(x) = \begin{cases} {ax+bx^2} & {0\lt x \lt 1} \\ 0 & \text{else} \end{cases} [/imath] Then we are given [imath]E|X| = 0.6[/imath] And usually asks to find: a) [imath]P\{X>0.5\}[/imath] b) [imath]Var(X)[/imath] Second pattern: [imath] f(x) = \begin{cases} {cx^4} & {0\lt x \lt 2} \\ 0 & \text{else} \end{cases} [/imath] Where c is a constant. Find: a) Mean value of [imath]X[/imath] b) Variance of [imath]X[/imath] Third pattern (it's more like a reverse problem): We are given [imath]Y= \sqrt{X}[/imath], find it's cumulative distribution function (CDF) and it's probability density function (pdf). Update: The third one is transferred to a new seperate question. This is not homework. I'm self-studying and in need of some help. Thank you. |
13641 | An orthonormal set cannot be a basis in an infinite dimension vector space?
I'm reading the Algebra book by Knapp and he mentions in passing that an orthonormal set in an infinite dimension vector space is "never large enough" to be a vector-space basis (i.e. that every vector can be written as a finite sum of vectors from the basis; such bases do exist for infinite dimension vector spaces by virtue of the axiom of choice, but usually one works with orthonormal sets for which infinite sums yield all the elements of the vector space). So my question is - how can we prove this claim? (That an orthonormal set in an infinite dimension vector space is not a vector space basis). Edit (by Jonas Meyer): Knapp says in a footnote on page 92 of Basic algebra: In the infinite-dimensional theory the term "orthonormal basis" is used for an orthonormal set that spans [imath]V[/imath] when limits of finite sums are allowed, in addition to finite sums themselves; when [imath]V[/imath] is infinite-dimensional, an orthonormal basis is never large enough to be a vector-space basis. Thus, without explicitly using the word, Knapp is referring only to complete infinite-dimensional inner product spaces. | 2426714 | Does every infinite-dimensional inner product space have an orthonormal basis?
For a finite-dimensional inner product space [imath]V[/imath], since it has a finite basis, then we can do the Gram-Schimidt process to produce an orthonormal basis. However, the Gram-Schmidt process does not work with infinitely many vectors. So it is natural to ask, does every infinite-dimensional inner product space have an orthonormal basis? If the answer is yes, how to prove it? PS: For "basis", I mean the Hamel basis. |
1186001 | Show (0,1) is not compact
Let [imath]I_n=\left(\frac{1}{n},1\right)[/imath]. Show that [imath](0,1)[/imath] is not compact: show that any finite collection of [imath]\{I_n\}[/imath] will not cover [imath](0,1)[/imath]. Give me a hint. | 252389 | Proof: in [imath]\mathbb{R}[/imath], [imath]((0,1),|\cdot|)[/imath] is not compact.
Let [imath](M,d)[/imath] be a metric space, and [imath]A\subset M[/imath]. By definition, [imath]A[/imath] is said to be compact if every open cover of [imath]A[/imath] contains a finite subcover. What is wrong with saying that, in [imath]\mathbb{R}[/imath], if [imath]I=(0,1)[/imath], we can choose [imath]G=\{(0,\frac{3}{4}), (\frac{1}{4}, 1)\}[/imath], which satisfies [imath]I \subset \bigcup_{U\in G} U[/imath], but we can't extract a finite subcover, so [imath]I[/imath] is not compact. Is [imath]G[/imath] a finite subcover of [imath]G[/imath], so it is not a valid cover for proving this? I would take [imath]\cup_{n\in\mathbb{N}} (\frac{1}{n},1)[/imath] in order to prove this, can we conclude that every open cover is necessarily a infinite union of open sets [imath]\neq \emptyset[/imath]? |
1175559 | Automorphism group of a simple graph
I am trying to understand an exercise in Bondy and Murty's book "Graph Theory with Applications". The exercise is as follows: "Consider the permutation group [imath]\Lambda[/imath] with elements [imath](1)(2)(3)[/imath], [imath](1,2,3)[/imath] and [imath](1,3,2)[/imath]. Show that there is no simple graph [imath]G[/imath] with vertex set [imath]{1,2,3}[/imath] such that [imath]\Gamma(G)=\Lambda[/imath]" Here [imath]\Gamma(G)[/imath] means the automorphism group of [imath]G[/imath]. I don't see how to start this proof. Edit: It looks like this question is a duplicate of a question posted last year: https://math.stackexchange.com/a/1000292/221021 | 635406 | Example of a simple graph isomorphic to a permutation group.
I'm taking a first course in graph theory this semester and I'm working trough Graph Theory with Applications by J.A. Bonday and U.S.R. Murty. I can't find an answer to question 1.2.12(f): (a) Show, using execise 1.2.5 that an automorphism of a simple graph [imath]G[/imath] can be regarded as a permutation on [imath]V[/imath] which preserves adjacency,and that the set of such permutations form a group [imath]\Gamma(G)[/imath] (the automorphism group of [imath]G[/imath]) under the usual operation of composition. (e) Consider the permutation group [imath]\Lambda [/imath] with elements (1)(2)(3), (1, 2, 3) and (1, 3, 2). Show that there is no simple graph [imath]G[/imath] with vertex set {1, 2, 3} such that [imath]\Gamma(G)=\Lambda[/imath]. (f) Find a simple graph G such that [imath]\Gamma(G)\cong\Lambda[/imath]. |
240515 | Odd Corollary in Baby Rudin Chapter 2, Question 28
I've finished all the questions in Chapter 2 of Principles of Mathematical Analysis by Walter Rudin (self study), but I have a question about Q.28, which reads: Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable. (Corollary: Every countable closed set in [imath]\mathbb{R}^k[/imath] has isolated points.) It's easy to answer the question, given what's proved in Q.27, but the corollary is a bit weird. It looks to me as though it is just an immediate consequence of the fact that non-empty perfect sets in [imath]\mathbb{R}^k[/imath] are uncountable, which is proved in the main text. I don't see what it has to do with what is proved in this question. Given how meticulous the book is, I suspect the apparent non-sequiteur means I'm missing something. | 469200 | Is every set in a separable metric space the union of a perfect set and a set that is at most countable?
I'm reading Baby Rudin and exercise 28 of chapter 2 reads "Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable." I don't understand why the set must be closed in order for this to be true. My proof is as follows, and I don't make any reference to the closedness of the set: Let [imath]E[/imath] be a set in a separable metric space. Let [imath]B[/imath] be a countable base of that metric space. Let [imath]P[/imath] be the set of condensation points of [imath]E[/imath]. I claim that [imath]P[/imath] is perfect and [imath]E-P[/imath] is at most countable. If [imath]p[/imath] is a limit point of [imath]P[/imath], then every neighborhood of [imath]p[/imath] includes at least one point [imath]q[/imath] of P. For a neighborhood [imath]N_{p}[/imath] of [imath]p[/imath], choose a neighborhood [imath]N_{q} \subset N_{p}[/imath] of [imath]q[/imath]. Since [imath]q[/imath] is a condensation point of [imath]E[/imath], uncountably many points of [imath]E[/imath] are in [imath]N_{q}[/imath], and thus in [imath]N_{p}[/imath]. So [imath]p\in P[/imath] and [imath]P[/imath] is closed. If [imath]p \in P[/imath], then [imath]p[/imath] is a condensation point of [imath]E[/imath], so every neighborhood [imath]N_{p}[/imath] of [imath]p[/imath] must include uncountably many points of [imath]E[/imath]. If [imath]p[/imath] is not a limit point of [imath]P[/imath], then there exists a neighborhood [imath]N_{p}[/imath] of [imath]p[/imath] such that [imath]N_p[/imath] includes no condensation points of [imath]E[/imath] except for [imath]p[/imath]. This means that every point [imath]q\in N_{p}-\{p\}[/imath] must have a neighborhood [imath]N_{q}[/imath] that does not have uncountably many points of [imath]E[/imath]. For each [imath]b_{k}\in B \cap \mathscr{P}(N_{p})[/imath], choose such a neighborhood [imath]N_{q_{k}} \supset b_{k}[/imath] of some point [imath]q\in N_{p}-\{p\}[/imath]. Now [imath]\{N_{q_{k}}\}[/imath] covers [imath]N_{p}[/imath], and [imath]\bigcup_{k=1}^{\infty}N_{q_{k}}\supset N_{p}[/imath] has at most countably many points of [imath]E[/imath], for it is a union of countably many sets that have at most countably many points of [imath]E[/imath]. This contradicts the assumption that [imath]p\in P[/imath]. So [imath]P[/imath] is perfect. Let [imath]S=E-P[/imath]. There are no condensation points in [imath]S[/imath]. So there exists an open cover [imath]\{N_s\}[/imath], where each [imath]N_s[/imath] is a neighborhood of [imath]s \in S[/imath] with at most countably many elements of [imath]S[/imath]. For each [imath]b_{k}\in B \cap \mathscr{P}(S)[/imath], choose such a neighborhood [imath]N_{s_{k}} \supset b_{k}[/imath] of some point [imath]s\in S[/imath]. Then we have a countable subcover [imath]\{N_{p_{k}}\}[/imath] of [imath]S[/imath]. As [imath]N_{p_{k}} \cap S[/imath] is at most countable for all [imath]k[/imath], [imath]S[/imath] must be at most countable. So what am I doing wrong? Or is Rudin's assumption of closedness extraneous? |
1187102 | Common divisor of [imath]a+b[/imath] and [imath]ab[/imath].
If [imath]\gcd(a,b) =1[/imath]. Why does [imath]\gcd(a+b,ab)=1[/imath] ? I know that if [imath]\gcd(a,b)=1[/imath] then there exists [imath]u[/imath] and [imath]v[/imath] where [imath]au+bv=1[/imath]. But I can't seem to relate it to [imath]a+b[/imath] and [imath]ab[/imath]. | 696432 | Prove that if the [imath]\gcd(a,b)=1[/imath], then [imath]\gcd(a+b,ab)=1[/imath]
I need to prove that: If [imath]\gcd(a,b)=1[/imath], then [imath]\gcd(a+b,ab)=1.[/imath] So far I used what's given so I have: [imath]ax+by=1[/imath] (I wrote the gcd of a and b as a linear combination) and [imath](a+b)u+ab(v)=1[/imath] (I also wrote this as a linear combination) where do I go from here? |
274594 | [imath]\epsilon[/imath]-[imath]\delta[/imath] proof that [imath]f(x) = x \sin(1/x)[/imath], [imath]x \ne 0[/imath], is continuous
I'm doing an exercise that asks me to prove that [imath]f[/imath] is continuous using a [imath]\epsilon[/imath]-[imath]\delta[/imath] proof. I have that [imath] f(x) = \begin{cases} x\cdot \sin \frac1x,&x\neq 0 \\ 0,&x = 0 \end{cases} [/imath] I've already managed to show this property for [imath]x=0[/imath]. How can I show it for [imath]x \ne 0[/imath], also using a [imath]\epsilon[/imath]-[imath]\delta[/imath] proof? Thank you very much. | 2758396 | Proof Verification: [imath]x\sin(1/x)[/imath] is continuous on [imath](0,1)[/imath]
Prove that [imath]x\sin(1/x)[/imath] is continuous on [imath](0,1)[/imath] Proof: Let [imath]\epsilon>0[/imath] be given. Then [imath]\exists \delta>0[/imath] such that if [imath]x,c \in (0,1)[/imath] and if [imath]|x-c|<\delta \implies |f(x)-f(c)|<\epsilon[/imath] Then, [imath]|x\sin(\frac{1}{x}) - c\sin(\frac{1}{c})|\le |x|\frac{1}{x}| -c|\frac{1}{c}||=|1-1|=0<\epsilon[/imath] Then, [imath]x\sin(1/x)[/imath] is continuous on [imath](0,1)[/imath] Note: Inequality used is [imath]\sin(x) \le |x|[/imath] Can anyone please verify this proof for me and let me know if there are any loopholes in my argument? Thank you. |
1177873 | To show that the only nonempty subset of [imath]\Bbb R[/imath] which is both open and closed in [imath]\Bbb R[/imath] is [imath]\Bbb R[/imath].
To show that the only nonempty subset of [imath]\Bbb R[/imath] which is both open and closed in [imath]\Bbb R[/imath] is [imath]\Bbb R[/imath]. Let [imath]A[/imath] be a non empty subset of [imath]\Bbb R[/imath] which is both open and closed. Let [imath]x \in A[/imath]. Then [imath](x- \epsilon ,x+ \epsilon) \subset A[/imath] for some [imath]\epsilon >0[/imath]. Then how to bring the contradiction by putting in the fact that [imath]A[/imath] cannot be simultaneously open and closed!! | 70415 | Showing that [imath]\mathbb{R}[/imath] is connected
So I know that [imath]\mathbb{R}[/imath] is both open and closed. But given a set, [imath]X\subset \mathbb{R}[/imath], [imath]X\ne \emptyset [/imath] that is both open and closed, how does one show that [imath]X=\mathbb{R}[/imath]? Here is my attempt: Using the definitions I was given, [imath]X[/imath] open implies that for every [imath]x\in X[/imath] there is a [imath]\delta \gt 0[/imath], [imath](x-\delta,~ x+\delta)\subset X[/imath]. Also, since [imath]X[/imath] is closed , [imath]B=\mathbb{R}[/imath]\ [imath]X[/imath] is open. However, I don't know how to proceed. NB: This is not a homework problem. |
1173225 | Linking surface integral of a gradient field to a contour integral
I have a vector field [imath]F[/imath] deriving from a scalar potential [imath]f[/imath], i.e. [imath]F=\text{grad}(f)[/imath]. I want to compute the integral of [imath]F[/imath] over a surface (To evaluate the flux of [imath]F[/imath]). I think there exists a theorem linking this integral over a surface to the integral of [imath]f[/imath] over the edge of the surface but I can't find it. What I find is : -Green theorem linking integration of [imath]\text{div}(F)[/imath] over a surface to [imath]F[/imath] over a line -Stokes theorem linking integration of [imath]\text{curl}(F)[/imath] to integration of [imath]F[/imath]. But no theorem linking integration of [imath]F[/imath] over a surface to [imath]f[/imath] over a contour. Can someone help me ? Thanks | 130081 | Does the Divergence Theorem Work on a Surface?
The divergence theorem in [imath]\mathbb{R}^3[/imath] says that the integral of the divergence of a vector field over a solid [imath]\Omega[/imath] in [imath]\mathbb{R}^3[/imath] equals the flux through the surface of [imath]\Omega[/imath] denoted by [imath]\partial \Omega[/imath]. My question is that can you use the divergence theorem in [imath]\mathbb{R}^3[/imath] where you're integrating the divergence of a vector field over a surface [imath]S[/imath] in [imath]\mathbb{R}^3[/imath] to get the flux through the edges ([imath]\partial S[/imath]) of the surface? So if the surface [imath]S[/imath] is closed in [imath]\mathbb{R}^3[/imath] such as a sphere, then clearly [imath]\partial S = 0[/imath] so the integral of the divergence over this particular [imath]S[/imath] is [imath]0[/imath]? |
1189498 | Let [imath]f: X \to X[/imath] be such that [imath]d(f(x), f(y)) = d(x, y)[/imath] for all [imath]x, y \in X[/imath]. To show that [imath]f[/imath] is onto.
Let [imath](X, d)[/imath] be a compact metric space. Let [imath]f: X \to X[/imath] be such that [imath]d(f(x), f(y)) = d(x, y)[/imath] for all [imath]x, y \in X[/imath]. To show that [imath]f[/imath] is onto. Since the function [imath]f[/imath] satisfies [imath]d(f(x), f(y)) = d(x, y)[/imath] for all [imath]x, y \in X[/imath] we can say that the function is uniformly continuous. Now let us assume that [imath]f[/imath] is not onto then [imath]f(X) \subset X[/imath], a proper subset. Then how can I bring a contraction to show that [imath]f[/imath] will be onto?? | 1146685 | Let [imath](X,d)[/imath] be a compact metric space. Let [imath]f: X \to X[/imath] be such that [imath]d(f(x),f(y)) = d(x,y)[/imath] for all [imath]x,y \in X[/imath]. Show that [imath]f [/imath] is onto (surjective).
Let [imath](X,d)[/imath] be a compact metric space. Let [imath]f: X \to X[/imath] be such that [imath]d(f(x),f(y)) = d(x,y)[/imath] for all [imath]x,y \in X[/imath]. Show that [imath]f [/imath] is onto (surjective). If [imath]f[/imath] is not onto then there exist a [imath]p \in X[/imath] such that there does not exist any [imath]y \in X[/imath] such that [imath]f(y) =p[/imath]. Then there exist [imath]x \in X[/imath] such that [imath]d(p,f(x)) = d(p,x)[/imath]. I am finding difficulty to do the proof please help!! Is the result true if [imath]X[/imath] is not compact?? |
14113 | Series that converge to [imath]\pi[/imath] quickly
I know the series, [imath]4-{4\over3}+{4\over5}-{4\over7}...[/imath] converges to [imath]\pi[/imath] but I have heard many people say that while this is a classic example, there are series that converge much faster. Does anyone know of any? | 2410944 | What is the most accurate method of the approximation of [imath]\pi[/imath]
Question What is the most effective way to approximate [imath]\pi[/imath]. There are many a way that we can approximate [imath]\pi[/imath]. The way I am looking for the most effective approximation of [imath]\pi[/imath] is via the equations in forms of products or sums. This means that there will be an exclusion of the complex realm as you would have to approximate [imath]e[/imath] first as well as the [imath]cos(x)[/imath] and [imath]sin(x)[/imath] functions due to the involvement of [imath]e[/imath]. So far I am currently aware of 2 methods of the approximation of [imath]\pi[/imath] being. [imath]{{\pi}\over{2}} = \prod_{n=1}^{\infty}{{4n^2}\over{4n^2-1}}[/imath] [imath]{\pi^2\over 6}={\zeta (2)}=\sum_{n=1}^{\infty}{1\over n^2}[/imath] I have been able to calculate that the first product method is more accurate as when you apply a variable to the [imath]\infty[/imath] of both you are able to see that the product method approaches [imath]\pi[/imath] at a faster rate. I am more so interested in what other methods there are of the approximation of forumlae of [imath]\pi[/imath] and I do not require anyone to make the calculations of which is more accurate. If anyone has any ideas, please leave them down below. Thank you. |
972333 | Closed or open subsets of [imath]C[a,b][/imath]?
[imath]C[a,b][/imath] denotes the space of continuous real-valued functions on [imath][a,b][/imath]. The metric associated with [imath]C[a,b][/imath] here is [imath]d(f,g)=sup[|f(x)-g(x)|][/imath] where the supremum is taken over [imath][a,b][/imath]. [imath]C^1[a,b][/imath] denotes the one of continuously differentiable functions on [imath][a,b][/imath]. I think the second is a closed subset of the first but I can't get my head around a proof! Can someone give me a plan, a hint or anything, so I can try this out again? It would be very helpful and clarify what I (probably) didn't understand. Thank you very much! | 531150 | [imath]C^1[a,b][/imath] is closed in [imath]C[a,b][/imath]
So, I have the following question: Let [imath]C[a,b][/imath] denote the space of continuous real-valued functions on [imath][a,b][/imath] with the sup-metric. Let [imath]C^1[a,b][/imath] denote the space of continuously differentiable functions on [imath][a,b][/imath]. Is [imath]C^1[a,b][/imath] a closed subset of [imath]C[a,b][/imath]? So, what I've done so far. The sup metric is [imath]d_{\infty}(f,g) = \sup \left\{|f(x) - g(x)| : a \leq x \leq b \right\}[/imath] for [imath]f,g\in C[a,b][/imath] Clearly [imath]C^1[a,b] \subset C[a,b][/imath] [imath]C^1[a,b][/imath] is a closed subset of [imath]C[a,b][/imath] if [imath]C[a,b]\backslash C^1[a,b][/imath] is open in [imath]C[a,b][/imath] - i.e. [imath]\forall h \in C[a,b]\backslash C^1[a,b][/imath], [imath]\exists \epsilon > 0[/imath] such that [imath]\left\{ f \in C[a,b] : d_{\infty}(f,h) < \epsilon \right\} \subseteq C[a,b]\backslash C^1[a,b][/imath] Which seems to be saying that, essentially, for any non-differentiable continuous function [imath]h[/imath] we can find an [imath]\epsilon > 0[/imath] such that when the supremum of the distance between [imath]h[/imath] and all other continuous functions on [imath][a,b][/imath] is less than [imath]\epsilon[/imath], then they will also not be differentiable. This is where I get stuck, not least because I'm not convinced whether that is true or not. Clearly, it's easily possible to create further non-differentiable functions by, for example, moving one point of [imath]h[/imath] less than [imath]\epsilon[/imath]. My instinct is that you're not going to be able to turn a non-differentiable function into a differentiable one with an arbitrarily small [imath]\epsilon[/imath], so I think such an [imath]\epsilon[/imath] exists - but this obviously isn't formal (or even likely to be correct) Thanks. |
1189697 | Prove that [imath]\forall n\ge 1[/imath], there are [imath]n[/imath] successive, non-prime,natural numbers.
Prove that [imath]\forall n\ge 1[/imath], there are [imath]n[/imath] successive, non-prime,natural numbers. How can I turn to such a question? I have never dealt with such before. I started to course this week. I would really appreciate it if you led me or assisted me. | 982332 | Is it possible to find [imath]n-1[/imath] consecutive composite integers
Given an integer [imath]n\geq 2[/imath] ,can we always find an integer [imath]m[/imath] such that each of the [imath]n-1[/imath] consecutive integers [imath]m+2,m+3,.....,m+n[/imath] are composite? |
454443 | Continuous mapping problem
Question: Let f and g be continuous mappings of a metric space X into a metric space Y and let E be a dense subset of X. Prove that f(E) is dense in f(X). If [imath]f(p) = g(p)[/imath] for all p [imath]\in[/imath] E prove that [imath]f(p) = g(p)[/imath] for all X. Answer: Let's show that every point of f(X) is either a point of f(E) or a limit point of f(E) or both. Suppose that y [imath]\in[/imath] f(X). Then, there exists a point p [imath]\in[/imath] X such that f(p) = y. Since E is dense in X, p [imath]\in[/imath] E U E[imath]'[/imath]. If p [imath]\in[/imath] E then y [imath]\in[/imath] f(E) and we are done. If p [imath]\in[/imath] E[imath]'[/imath] and p is not in E, then there exists a sequence p[imath]_n[/imath] [imath]\in[/imath] E such that p[imath]_n[/imath] -> p. We then have f(p[imath]_n[/imath]) -> f(p). That is , f(p) [imath]\in[/imath] f(E). Now, suppose f(p) = g(p) for all p [imath]\in[/imath] E. Let x [imath]\in[/imath] X\E. Since E is dense in X we have a sequence q[imath]_n[/imath] [imath]\in[/imath] E such that q[imath]_n[/imath] -> x. So, f(x) = f(lim q[imath]_n[/imath]) = lim f(q[imath]_n[/imath]) = lim g(q[imath]_n[/imath]) = g(lim q[imath]_n[/imath]) = g(x). Thus, f(x) = g(x) for all x [imath]\in[/imath] X Could you tell me if the bold line is legitimate and explain me why it is legitimate if it is? Thank you so much! | 610123 | A continuous mapping is determined by its values on a dense set
Let f and g be continuous mappings of a metric space [imath]X[/imath] into a metric space [imath]Y[/imath] and let [imath]E[/imath] be a dense subset of [imath]X[/imath]. Prove that [imath]f(E)[/imath] is dense in [imath]f(X)[/imath]. If [imath]g(p)=f(p)[/imath] for all [imath]p \in E[/imath], prove that [imath]g(p)=f(p)[/imath] for all [imath]p \in X[/imath]. |
1190496 | Relation Reflexive?
Suppose [imath]R[/imath] is a relation on [imath]N_4=\{1,2,3,4\}[/imath] such that [imath]R\circ R=R[/imath]. How would I prove that [imath]R[/imath] is reflexive? I am geting this statement as false, Please Let me know , How to prove this ? | 1190121 | Is this relation reflexive?
Suppose [imath]R[/imath] is a relation on [imath]N_4=\{1,2,3,4\}[/imath] such that [imath]R\circ R = R[/imath]. How would I prove that [imath]R[/imath] is reflexive? Could you give me some tips about how to start off? |
1189153 | Convex Convolution
Let [imath]f: R^n\to R\cup\{\infty\}[/imath] to be a convex function. Let [imath]f_{\epsilon}(x)=\frac{|x|^2}{2\epsilon}[/imath]. Show that: [imath]\lim_{\epsilon\to 0}\inf_{x=y+z}f_{\epsilon}(y)+f(z) =f(x)[/imath] The [imath]\leq[/imath] part is trivial. But how to prove the other part? | 162502 | Some properties of Yosida-Moreau transform
Let [imath]f(x)[/imath] be a continuous function on [imath]\mathbb{R}^n[/imath], [imath]f(x) \geqslant 0[/imath] for any [imath]x[/imath]. Define [imath] f_{\alpha}(x) = \inf\limits_{y}\left( f(y) +\frac{|x-y|^2}{2\alpha} \right) [/imath] where [imath]\alpha > 0[/imath]. How to show that [imath]f_{\alpha}(x)[/imath] is locally Lipschitz continuous and [imath]f_{\alpha} \to f[/imath] when [imath]\alpha \to 0+0[/imath] uniformly on any compact [imath]K[/imath] in [imath]\mathbb{R}^n[/imath]? |
1190581 | Limit of [imath]y[/imath] if limit of [imath]y+y'[/imath] goes to [imath]0[/imath]?
Let [imath]y[/imath] be a differentiable function on [imath]\mathbb{R}[/imath]. If [imath]\lim_{x\rightarrow \infty}(y(x)+y'(x))=0[/imath] Then how does one show that [imath]\lim_{x\rightarrow \infty}y(x)=0[/imath] I'd appreciate some help on this problem. Thanks a lot. | 1861202 | [imath]\lim_{x\to\infty}(f(x)+f'(x))=0 \rightarrow \lim_{x\to\infty}f(x)=0[/imath]?
Suppose that [imath]f:[0,\infty)\to\Bbb R[/imath] is differentiable, and [imath]\lim_{x\to\infty}(f(x)+f'(x))=0[/imath]. Prove that [imath]\lim_{x\to\infty}f(x)=0[/imath]. I tried to show that [imath]\lim_{x\to\infty}f(x)\neq0\rightarrow \lim_{x\to\infty}(f(x)+f'(x))\neq0[/imath] If [imath]\lim_{x\to\infty}f(x)\neq0, \exists e>0, \forall N\in\Bbb N, \exists x_1, x_2, ...>N,|f(x_i)|\ge e[/imath]. So that the possibility for [imath]\lim_{x\to\infty}(f(x)+f'(x))=0[/imath] is only when [imath]f(x_i)+f'(x_i)=0[/imath] is true for any [imath]i[/imath]. So if [imath]f(x_1)\gt0,[/imath] it must be decreasing, to less than [imath]e[/imath] But there exists [imath]x_2\gt x_1 s.t. |f(x_2)|\ge e[/imath] In conclusion, [imath]f(x)[/imath] has to decrease when [imath]x[/imath] is large enough and [imath]f(x)\ge e[/imath] but there must exists infinitely many points whose function value is no less than [imath]e[/imath]. But it cannot happen. Is my idea of proof valid? I don't know how to formally write my idea...please teach me.. |
533772 | A formula for the power sums: [imath]1^n+2^n+\dotsc +k^n=\,[/imath]?
Is there explicit formula for the expression [imath]1^n + 2^n + \dotsc + k^n\,[/imath]? I know that for [imath]n=1[/imath] the explicit formula becomes [imath]S=k(k+1)/2[/imath] and for [imath]n=3[/imath] the formula becomes [imath]S^2[/imath]. But what about general [imath]n[/imath]? I know there is a way using the Taylor expansion of [imath]f(x)=1/(1-x)=1+x+x^2+\dotsc\;[/imath], by differentiating it and then multiplying by [imath]x[/imath] and then differentiating again. Repeating this [imath]n[/imath] times, we get [imath]\frac{d}{dx}(x\frac{d}{dx}(\dots x\frac{d}{dx}f(x))\dots )=1+2^nx^n+3^nx^n\dots.[/imath] Now do the same process but with the function [imath]g(x)=x^{k+1}f(x)[/imath]. Then subtract them and we get [imath]1+2^nx^n+\dots k^nx^n[/imath]. Because we have the explicit formulas [imath]f(x)[/imath] and [imath]g(x)[/imath] we can find the explicit formula by this process for arbitrary [imath]n[/imath]. A big problem is that as [imath]n[/imath] grows, it is going take a lot of time finding the explicit formula. My question is therefore: are there other ways? | 1190741 | Need to evaluate the series [imath]\sum_{i = 0}^n i^5[/imath]
I need help in evaluating the following sum: [imath]\sum_{i = 0}^n i^5 [/imath] I can evaluate series when they are arithmetic or geometric but I don't know how to solve this one. |
875509 | Prove [imath]\sum_{k = 1}^n \mu(k)\left[ \frac nk \right] = 1[/imath]
I need to prove the identity [imath]\sum_{k = 1}^n \mu(k)\left[ \frac{n}{k} \right] = 1[/imath] where [imath]n[/imath] is a natural number, and [imath][n][/imath] denotes the floor function. The proof also should not use the Möbius Inversion Formula; I'm looking for something elementary. I've tried to give a counting argument, and to give a proof using the identity [imath]\sum_{d|n}^n\mu(d)= 0[/imath] for [imath]n > 1,[/imath] but I haven't succeeded. What's a good way to solve this problem? | 332588 | Prove [imath]\sum_{d \leq x} \mu(d)\left\lfloor \frac xd \right\rfloor = 1 [/imath]
I am trying to show [imath]\sum_{d \leq x} \mu(d)\left\lfloor \frac{x}{d} \right\rfloor = 1 \;\;\;\; \forall \; x \in \mathbb{R}, \; x \geq 1 [/imath] I know that the sum over the divisors [imath]d[/imath] of [imath]n[/imath] is zero if [imath]n \neq 1[/imath]. So we can rule out integers that are divisors of [imath]x[/imath] (if [imath]x > 1[/imath]). I am not sure if I am on the right track to prove this. Any hints would be helpful. This is homework. |
570428 | Prison problem: locking or unlocking every [imath]n[/imath]th door for [imath] n=1,2,3,...[/imath]
I have a problem called "The Prison Problem" that I need to explain to my 9-year-old cousin. I would think that he has just started learning about divisors and perfect squares, and as such, I have a proposed solution. Any input from you guys would be welcome, as to what is the best way to go about this. Problem: There was a jail with 100 cells in it, all in a long row. The warden was feeling very jolly one night and told his assistant that he wanted to give all the prisoners a wonderful surprise. While they were sleeping, he wanted the assistant to unlock all the cells. This should be done, he told the assistant, by putting the key in each lock, turning it once. The assistant did this, then came back to report that the job was done. Meanwhile, however, the warden has second thoughts. "Maybe I shouldn't let all the prisoners free," he said. "Go back and leave the first cell open, but lock the second one (by putting the key in and turning it once). Then leave the third open, but lock the fourth, and continue in this way for the entire row." The assistant wasn't very surprised at this request as the warden often changed his mind. After finishing this task, the assistant returned, and again the warden had other thoughts. "Here's what I really want you to do," he said. "Go back down the row. Leave the first two cells as they are, and put your key in the third cell and turn it once. Then leave the fourth and fifth cells and turn the key in the sixth. Continue down the row this way." The assistant again did as instructed. Fortunately, the prisoners were still asleep. As a matter of fact, the assistant was getting pretty sleepy, but there was no chance for rest yet. The warden changed his mind again, and the assistant had to go back again and turn the lock in the fourth cell and in every fourth cell down the row. This continued all through the night, next turning the lock in every fifth cell, and then in every sixth, and on and on, until on the last trip, the assistant just had to turn the key in the hundredth cell. When the prisoners finally woke up, which ones could walk out of their cells? Proposed solution: Ten cells are left open after this process. Every cell that is a perfect square will remain open (1, 4, 9, 16, 25, 36, 49, 64, 81, and 100). If a number is not a perfect square, then it has an even number of divisors, therefore it will be "toggled" an even number of times and end up where it started (closed). Perfect squares have an odd number of divisors, so they will end up the opposite of where they started (open). | 1394951 | Ghosts closing and opening doors
There are [imath]1000[/imath] doors [imath]D_1,D_2,D_3,\dots,D_{1000}[/imath] and [imath]1000[/imath] persons [imath]P_1,P_2,\dots,P_{1000}[/imath]. Initially all the doors were closed. Person [imath]P_1[/imath] goes and opens all the doors. Then person [imath]P_2[/imath] closes doors [imath]D_2,D_4,\dots,D_{1000}[/imath] and leaves the odd-numbered doors open. Next, [imath]P_3[/imath] changes the state of every third door, that is [imath]D_3,D_6,\dots,D_{999}[/imath].(For instance, [imath]P_3[/imath] closes the open door [imath]D_3[/imath] and opens the closed door [imath]D_6[/imath], and so on.) This goes on. Finally, person [imath]P_{1000}[/imath] opens [imath]D_{1000}[/imath] if it were closed or closes if it were opened. At the end, how many doors remain open[imath]?[/imath] |
695086 | Show that [imath]T_\infty=\{U : X-U \text{ is infinite or empty or all of [/imath]X[imath]} \}[/imath] is a topology on [imath]X[/imath]
Let [imath]X[/imath] be a set. Show that the collection [imath]T_\infty=\{U : X-U \text{ is infinite or empty or all of $X$} \}[/imath] is a topology on [imath]X[/imath]. Well [imath]X-X=\emptyset[/imath], so [imath]X \in T_\infty[/imath]. And [imath]X-\emptyset= X[/imath] and therefore in [imath]T_\infty[/imath]. Let assume that [imath]\{U_\alpha\}[/imath] is indexed family of sets of [imath]T_\alpha[/imath] that are not empty or [imath]X[/imath]. Than [imath]X-\cup U_\alpha =\cap X- U_\alpha[/imath] it seems strange to me that this intersection must be infinite. I don't know [imath](-\infty,0][/imath] and [imath][0,\infty)[/imath] have a finite intersection. Maybe I oversee something. | 877773 | Is the collection [imath]\tau_\infty = \{U:X-U[/imath] is infinite or empty or all of [imath]X\}[/imath] a topology on [imath]X[/imath]?
Can someone please verify my proof? Is the collection [imath]\tau_\infty = \{U:X-U[/imath] is infinite or empty or all of [imath]X\}[/imath] a topology on [imath]X[/imath]? No. Let [imath]X = \mathbb{R}[/imath]. Clearly, [imath]\{x\} \in \tau_\infty[/imath] for all [imath]x \in X[/imath]. Set [imath]\displaystyle{B = \cup_{x \neq 0} \{x\}}[/imath]. Clearly, [imath]X - B = \{0\}[/imath], but [imath]\{0\} \notin \tau_\infty[/imath]. Since [imath]\tau_\infty[/imath] is not closed under arbitrary unions, it does not define a topology on [imath]X[/imath]. |
1191987 | How to guarantee existence of a finite field
Existence of a finite field: Solution: I can understand that if I have a finite field [imath]F[/imath] of characteristic [imath]p[/imath] where [imath]p[/imath] is prime then I can consider [imath]\mathbb Z_p[/imath] as its prime field and hence [imath]F[/imath] will have [imath]p^n[/imath] elements But how to find such a field [imath]F[/imath] ? | 565135 | Finite fields, existence of field of order [imath]p^n[/imath],proof help
I have to prove the following theorem : Let [imath]p[/imath] be a prime number and let [imath]n \ge 1[/imath],be any integer, then there exists a field of order [imath]p^n[/imath]. My attempt I started off by considering the polynomial [imath]f(x)[/imath]=[imath]x^{p^n}-x \in \Bbb Z_p[x][/imath]. I took [imath]F[/imath] to be the splitting field of [imath]f(x)[/imath] over [imath]\Bbb Z_p[/imath]. Since [imath]F[/imath] is the splitting field of [imath]f(x)[/imath] over [imath]\Bbb Z_p[/imath] therefore [imath]f(x)[/imath] has exactly [imath]p^n[/imath] zeros in [imath]F[/imath] counting multiplicity. Also since [imath]f'(x)[/imath] = [imath]p^nx^{p^n-1}-1[/imath] , therefore [imath]f(x)[/imath] will not have any multiple zeros and hence it will have [imath]p^n[/imath] distinct zeros in [imath]F[/imath]. Then I took [imath]K =\{k \in F | k^{p^n} = k\}[/imath]. I was trying to prove that [imath]K[/imath] is a sub-field of [imath]F[/imath]. [imath]1 \in K[/imath] thus [imath]K[/imath] is non-empty. For [imath]a,b \in K[/imath] [imath]a^{p^n}=a[/imath] and [imath]b^{p^n}=b[/imath] So [imath](a+b)^{p^n}[/imath] (should be) = [imath]a^{p^n}[/imath]+[imath]b^{p^n}[/imath], this is where I am currently stuck! I know that this is true when [imath]p[/imath] is the characteristic of a field, but why would this expansion be true here? |
1186540 | Prove that the sum of two balls is a ball
If [imath]B(x,r)[/imath] be a ball with center [imath]x[/imath] and radius [imath]r[/imath], for [imath]t \in [0,1][/imath] prove that : [imath]tB(x,r)+(1-t)B(y,r)=B(tx+(1-t)y,r)[/imath] please help me. i can proof [imath] tB(x,r)+(1-t)B(y,r)[/imath] is in [imath] B(tx+(1-t)y,r)[/imath]. but for equal i dont show that [imath] B(tx+(1-t)y,r)[/imath] has in [imath] tB(x,r)+(1-t)B(y,r)[/imath]. | 444845 | Minkowski sum of two disks
An open disk with radius [imath]r[/imath] centered at [imath]\mathbf{p}[/imath] is [imath]D(\mathbf{p}, r)=\{\mathbf{q} \mid d(\mathbf p, \mathbf q) < r\}[/imath], and the Minkowski sum of two sets [imath]A[/imath] and [imath]B[/imath] is [imath]A \oplus B=\{\mathbf p + \mathbf q \mid \mathbf p \in A, \mathbf q \in B \}[/imath]. How can you show that [imath]D(\mathbf{a}, r_a) \oplus D(\mathbf{b}, r_b) = D(\mathbf{a} + \mathbf{b}, r_a + r_b)[/imath]? Attempt: \begin{align} D(\mathbf{a}, r_a) \oplus D(\mathbf{b}, r_b) &= \{\mathbf p + \mathbf q \mid \mathbf p \in D(\mathbf{a}, r_a), \mathbf q \in D(\mathbf{b}, r_b) \} \\&= \{\mathbf p + \mathbf q \mid \mathbf p \in \{\mathbf{x} \mid d(\mathbf a, \mathbf x) < r_a\}, \mathbf q \in \{\mathbf{y} \mid d(\mathbf b, \mathbf y) < r_b\} \\&= \{\mathbf p + \mathbf q \mid d(\mathbf a, \mathbf p) < r_a, d(\mathbf b, \mathbf q) < r_b \} \end{align} And here I got stuck. As best as I can tell, now I would need to prove that [imath]d(\mathbf a, \mathbf p) < r_a, d(\mathbf b, \mathbf q) < r_b \iff d(\mathbf a + \mathbf b, \mathbf p + \mathbf q) < r_a + r_b[/imath] but this seems false to me. I tried adding the two inequalities together, but that doesn't seem to give me that condition unless [imath]\mathbf a - \mathbf p[/imath] and [imath]\mathbf b - \mathbf q[/imath] are parallel. |
1185318 | Finding the basis of a vector space [imath]\Bbb{W}[/imath]
Given that [imath]\Bbb{W}=\{(a,b)|a,b\in\Bbb{R}\}[/imath] with addition defined by [imath](a,b)\oplus(c,d)=(a+c+1, b+d)[/imath] and scalar multiplication defined by [imath]k\odot(a,b)=(ka-k+1, kb)[/imath] is a vector space, find a basis for [imath]\Bbb{W}[/imath] and hence determine the dimension of [imath]\Bbb{W}[/imath] | 1183995 | Find a basis given a vector space
The following question is what I want to solve: Given that V = [imath]{(a,b)}[/imath] is a vector space, and addition is defined as [imath](a,b) + (c,d)[/imath] = [imath](a + c - 1, b + d)[/imath], and multiplication is defined as [imath]r • (a,b)[/imath] = [imath](ra - r + 1, rb)[/imath]. Find a basis for V. So far, I know that I have to build up a basis. I must start with a point, then add to it, until it spans V, and is linearly independent, but I don't know what to do from there. |
1191949 | Finite Field with no Order
Question: Let F be a field, that is, a set with operations [imath]+[/imath] and [imath]\cdot[/imath] which satisfy the axioms of the definition of an "ordered field". Prove that if [imath]F[/imath] is finite (i.e. has only finitely many elements), then there does not exist an ordering [imath]<[/imath] which makes [imath]F[/imath] into an ordered field. How can I go about proving a finite field has no order? | 616828 | A finite field cannot be an ordered field.
I am reading baby Rudin and it says all ordered fields with supremum property are isomorphic to [imath]\mathbb R[/imath]. Since all ordered finite fields would have supremum property that must mean none exist. Could someone please show me a proof of this? Thank you very much, Regards. |
1192271 | Is there a polynomial which vanishes at only one point
Assume [imath](a,b) \in \mathbb C^2[/imath]. Is there a polynomial [imath]f(x,y)[/imath] which vanishes only at [imath](a,b)[/imath] ? Thanks. | 25556 | Why are there no discrete zero sets of a polynomial in two complex variables?
Why is the zero set in [imath]\mathbb{C}\times\mathbb{C}[/imath] of a polynomial [imath]f(x,y)[/imath] in two complex variables always non-discrete (no zero of [imath]f[/imath] is isolated)? |
1192462 | Mathematical Analysis Question: Cauchy sequences proof
Let [imath](s_n)_n[/imath] and [imath](a_n)_n[/imath] be Cauchy sequences. Demonstrate that [imath](s_na_n)_n[/imath] is a Cauchy sequence. | 376324 | Proving that product of two Cauchy sequences is Cauchy
Given that [imath]x_n[/imath] and [imath]y_n[/imath] are Cauchy sequences in [imath]\mathbb{R} [/imath], prove that [imath]x_n y_n[/imath] is Cauchy without the use of the Cauchy theorem stating that Cauchy [imath]\Rightarrow[/imath] convergence. Attempt: Without that condition on not been able to use the theorem, the question becomes trivial. Instead: For all [imath]\epsilon > 0[/imath] there exists an [imath]N \in \mathbb{N}[/imath] such that for [imath]n,m \geq N, -\frac{\epsilon}{2} \leq x_n - x_m \leq \frac{\epsilon}{2}[/imath] and similar statment for [imath]y_n[/imath]. Multiply the above by [imath]y_n[/imath] and the equivalent statement for [imath]y_n[/imath] by [imath]x_m[/imath]. Then add these together. The result is: [imath]|x_ny_n - x_my_m| < \frac{\epsilon}{2}(x_m + y_n)[/imath] I have proved in a previous question that [imath]x_n + y_n[/imath] is Cauchy so could I apply that here and say for [imath]n,m \geq N[/imath], [imath]x_n + y_n[/imath] is Cauchy and hence convergent so tends to a finite limit for [imath]n,m \geq N[/imath]. This would mean my upper bound is a multiple of [imath]\epsilon[/imath] and since [imath]\epsilon[/imath] is arbritarily small, so is this upper bound. Hence Cauchy. I don't think this would warrant a full proof in any case since by multiplying by [imath]x_m[/imath] and [imath]y_n[/imath], I am assuming they are positive so as to not reverse the inequality signs. Nonetheless, I would appreciate some feedback on what I have done. Many thanks |
104530 | How do we prove that [imath]\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor[/imath]?
Are the floor functions of [imath]0.999\cdots[/imath] and 1 equal? It is true that [imath]0.999\cdots=1[/imath] but how does one justifies the integer part of [imath]0.999\cdots[/imath] being 1 , where it is not, or alternatively without using [imath]0.999\cdots=1[/imath] how can we show that [imath]\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor[/imath] ? | 1363785 | What is [imath]\left\lfloor0.\overline{9}\right\rfloor[/imath]?
We know that [imath]0.\overline{9} = 1[/imath] but then what is [imath]\left\lfloor0.\overline{9}\right\rfloor[/imath]? My thought process went: [imath]0.\overline{9} = 1[/imath] so therefore [imath]\left\lfloor0.\overline{9}\right\rfloor = \left\lfloor1\right\rfloor[/imath] But also [imath]\left\lfloor0.9\right\rfloor = 0[/imath] and it shouldn't change no matter how many numbers you add on to the back of it. So what is the right answer, if there is one, and why? |
493176 | Smallest topology containing all topologies
Let [imath]\{T_\alpha\}[/imath] be a family of topologies on [imath]X[/imath]. Show that there is a unique smallest topology on [imath]X[/imath] containing all the collections [imath]T_\alpha[/imath], and a unique largest topology contained in all [imath]T_\alpha[/imath]. We can check that [imath]\bigcap T_\alpha[/imath] is a topology on [imath]X[/imath], so it is the unique largest topology contained in all [imath]T_\alpha[/imath]. Now, a topology containing all [imath]T_\alpha[/imath] must contain [imath]\bigcup T_\alpha[/imath]. It must also contain arbitrary unions and finite intersections of sets in [imath]\bigcup T_\alpha[/imath]. Since the union of the sets in [imath]T_\alpha[/imath] is [imath]X[/imath], this is the topology generated by the subbasis [imath]\bigcup T_\alpha[/imath]. How can we prove that it is the unique smallest one containing all the collections [imath]T_\alpha[/imath]? | 695132 | What is meant with unique smallest/largest topology?
I'm doing this exercise: Let [imath]\{T_\alpha\}[/imath] be a family of topologies on [imath]X[/imath]. Show that there is a unique smallest topology on [imath]X[/imath] containing all the collections [imath]T_\alpha[/imath], and a unique largest topology contained in all [imath]T_\alpha[/imath]. I have proved everything except the unique part.. I just can't get my head around what is meant with unique here. Which may sounds silly. I have proved that the intersection is a topology. And if you are a topology that is also contained in every [imath]T_\alpha[/imath], than you surely are contained in the intersection, so you are not larger. But I don't see from what it follows that this intersection is the unique largest topology contained in all [imath]T_\alpha[/imath]. One part of my head say it is trivial, the other part gets confused. Like it is redundant to talk about unique in this context. The same for proving the uniqueness of the smallest topology. Edit Should I read topology [imath]A[/imath] larger than topology [imath]B[/imath] as, [imath]A[/imath] has more elements than [imath]B[/imath] ? I thought that, because the author uses the word finer for [imath]A \supset B[/imath]. |
995493 | Modular arithmetic Proofs
For all [imath]a[/imath], [imath]m \in\mathbb{Z}[/imath], prove that For all [imath]x \in [a][/imath], where [imath][a][/imath] is the congruence class of [imath]a \pmod m[/imath], [imath]\quad\gcd(x,m)=\gcd(a,m)[/imath]. I have no idea where to start for this. | 995384 | Congruence class [imath][a][/imath] modulo [imath]m[/imath], [imath]\gcd(x, m) = \gcd(a, m)[/imath]
I'm currently stumped on this question: Let [imath]a[/imath] and [imath]m[/imath] be integers such that [imath]m\ge1[/imath]. Consider the congruence class of [imath]a[/imath], i.e., [imath][a][/imath] modulo [imath]m[/imath]. Prove that: For all [imath]x\in[a][/imath], [imath]\gcd(x,m)=\gcd(a,m)[/imath]. From this, I know that the congruence class of [imath][a][/imath] modulo [imath]m[/imath] is the set of integers [imath][a]={x\in Z|x\equiv a\pmod m}[/imath]}. And by the Linear Congruence Theorem, [imath]x\equiv a\pmod m[/imath] if and only if [imath]\gcd(x,m)|a[/imath]. I'm not sure where I can go from knowing this. Any ideas? |
1193163 | How do I prove the unique hermitian decomposition [imath]A=A_1+i A_2[/imath]?
If [imath]A \in M_n(\mathbb{C})[/imath], then there exist unique hermitian matrices [imath]A_1,A_2[/imath] such that [imath]A=A_1+iA_2[/imath] | 380788 | Show that exist a unique expression for [imath] A [/imath] of the form [imath] A = (A_1 + A_2) + i (B_1 + B_2) [/imath]
Let [imath] A [/imath] be a [imath] n × n [/imath] matrix with real or complex elements. Show that exist a unique expression for [imath] A [/imath] of the form [imath] A = (A_1 + A_2) + i (B_1 + B_2) [/imath] where [imath] A_1,B_1[/imath] are real symmetric matrices and [imath] A_2,B_2 [/imath] are real antisymmetric matrices |
1187741 | How to solve a quintic congruence equation?
My textbook has this quadratic equation that I have to solve, any ideas how I could show that? [imath]15 | (21n^5+10n^3+14n),\;\forall n\in\mathbb{Z}[/imath] | 1193224 | Show that [imath]15\mid 21n^5 + 10n^3 + 14n[/imath] for all integers [imath]n[/imath].
I'm not sure if it's correct, but what I have so far is; [imath]21n^5 + 10n^3 + 14n ≡ (1 + 0 - 1) ≡ 0 \mod 5[/imath] but I'm having trouble solving it in [imath]\bmod 3[/imath]. I have: [imath]21n^5 + 10n^3 + 14n ≡ (0 + (?) + 2),[/imath] I'm not sure how to solve the [imath]10n^3[/imath] part of the congruence in [imath]\mod 3[/imath]. Also, if I have any mistakes please let me know. |
1191880 | Proof regarding harmonic functions and boundaries
I've been given the following proof as an exercise; Let [imath]\Omega \subset \Re^3 \setminus \{0\}[/imath], [imath]u : \Omega \rightarrow \Re[/imath] be harmonic. Show that; [imath]v(x_1, x_2, x_3) = \frac{1}{|x|}u\left(\frac{x_1}{|x|^2}, \frac{x_2}{|x|^2}, \frac{x_3}{|x|^2}\right)[/imath] is harmonic in the region; [imath]\Omega' = \left\{x \in \Re^3 \mid \left(\frac{x_1}{|x|^2}, \frac{x_2}{|x|^2}, \frac{x_3}{|x|^2}\right) \in \Omega\right\}[/imath] I don't really know where to start with this. I was going to attempt to use the mean value property, but I'm not sure where I'd be aiming for if I did. Any insight in getting started would be great. | 601384 | Show that the Kelvin-transform is harmonic
First, I have to give you our definitions: Consider [imath]\Omega:=\mathbb{R}^n\setminus\overline{B}_R(0)[/imath] with [imath]R>0[/imath] and [imath]n>1[/imath]. The function [imath]\phi\colon\Omega\to G:=B_R(0)\setminus\left\{0\right\}[/imath] with [imath] y=\phi(x):=\frac{R^2}{\| x\|^2}x\text{ for }\| x\|>R [/imath] is called Kelvin-transformation (relating to the sphere [imath]S_R(0)[/imath]). Let [imath]u\colon\Omega\to\mathbb{R}[/imath]. The function defined by [imath] v(y):=\frac{\| x\|^{n-2}}{R^{n-2}}u(x)\text{ for }\lVert x\rVert >R~~~(*) [/imath] is called Kelvin-transform of u. Alternatively, the Kelvin-transform of [imath]u[/imath] can be definied by [imath] v(z):=\frac{R^{n-2}}{\lVert z\rVert^{n-2}}u\left(\frac{R^2}{\lVert z\rVert^2}z\right)\text{ for }\lVert z\rVert <R. [/imath] Let [imath]u[/imath] be harmonic in [imath]\Omega[/imath]. Then its Kelvin-transform is harmonic in [imath]G[/imath]. Show that for the case [imath]n=2[/imath]. To be honest, I do not know, what is to show here, because if I take the alternative definition of the Kelvin-transform, for [imath]n=2[/imath] it is [imath] v(z)=u\left(\frac{R^2}{\lVert z\rVert^2}z\right), z\in G [/imath] and then [imath] \Delta v(z)=\Delta u\left(\frac{R^2}{\lVert z\rVert^2}z\right)=0, [/imath] because [imath]x:=\frac{R^2}{\lVert z\rVert^2}z[/imath] is in [imath]\Omega[/imath] and [imath]u[/imath] is harmonic in [imath]\Omega[/imath]. So I only had to use that the inverse of the Kelvin-transformation is given by [imath] \phi^{-1}\colon G\to\Omega, y\longmapsto\frac{R^2}{\lVert y\rVert^2}y. [/imath] Is that the proof? It seems TOO easy... |
1193531 | [imath]\sum_{n=1}^{\infty} \frac{n^2}{ n!}[/imath] equals
[imath] \sum_{n=1}^{\infty} \frac{n^2}{ n!} [/imath] equals I'm not able to convert in any standard series? Any hints? | 1098845 | Calculate sum of series [imath]\sum \frac{n^2}{n!}[/imath]
I have to calculate sum of series [imath]\sum \frac{n^2}{n!}[/imath]. I know that [imath]\sum \frac{1}{n!}=e[/imath] but I dont know how can I use that fact here.. |
1194115 | When does function composition commute?
I've read that function composition "generally does not commute." Not counting compositions involving the identity function, and compositions of a function and its inverse, are there examples of functions on the reals (for example) [imath]f, g[/imath] where [imath]fg = gf[/imath] outside of these cases? | 11431 | When functions commute under composition
Today I was thinking about composition of functions. It has nice properties, its always associative, there is an identity, and if we restrict to bijective functions then we have an inverse. But then I thought about commutativity. My first intuition was that bijective self maps of a space should commute but then I saw some counter-examples. The symmetric group is only abelian if [imath]n \le 2[/imath] so clearly there need to be more restrictions on functions than bijectivity for them to commute. The only examples I could think of were boring things like multiplying by a constant or maximal tori of groups like [imath]O(n)[/imath] (maybe less boring). My question: In a euclidean space, what are (edit) some nice characterizations of sets of functions that commute? What about in a more general space? Bonus: Is this notion of commutativity important anywhere in analysis? |
86434 | What lies beyond the Sedenions
In the construction of types of numbers, we have the following sequence: [imath]\mathbb{R} \subset \mathbb{C} \subset \mathbb{H} \subset \mathbb{O} \subset \mathbb{S}[/imath] or: [imath]2^0 \mathrm{-ions} \subset 2^1 \mathrm{-ions} \subset 2^2 \mathrm{-ions} \subset 2^3 \mathrm{-ions} \subset 2^4 \mathrm{-ions} [/imath] or: "Reals" [imath]\subset[/imath] "Complex" [imath]\subset[/imath] "Quaternions" [imath]\subset[/imath] "Octonions" [imath]\subset[/imath] "Sedenions" With the following "properties": From [imath]\mathbb{R}[/imath] to [imath]\mathbb{C}[/imath] you gain "algebraic-closure"-ness (but you throw away ordering). From [imath]\mathbb{C}[/imath] to [imath]\mathbb{H}[/imath] we throw away commutativity. From [imath]\mathbb{H}[/imath] to [imath]\mathbb{O}[/imath] we throw away associativity. From [imath]\mathbb{O}[/imath] to [imath]\mathbb{S}[/imath] we throw away multiplicative normedness. The question is, what lies on the right side of [imath]\mathbb{S}[/imath], and what do you lose when you go from [imath]\mathbb{S}[/imath] to one of these objects ? | 1300956 | Complex Number, Quaternions and Octonions
There are complex [imath]\mathbb C[/imath], quaternions [imath]\mathbb H[/imath] and octonions [imath]\mathbb O[/imath]. Is there any higher dimensional generalization of them, such in the [imath]\mathbb R^{16}[/imath]? Or why do we just study three kinds of numbers in Mathematics? Any advice is helpful. Thank you. |
1192344 | Show that [imath]\neg[/imath] and [imath]\wedge[/imath] form a functionally complete collection of logical operators
Show that [imath]\neg[/imath] and [imath]\wedge[/imath] form a functionally complete collection of logical operators Can someone give a hint? | 1192755 | Show that [imath]\neg[/imath] and [imath]\wedge[/imath] form a functionally complete collection of logical operators.
Earlier this day I ask about the assignmet: Show that [imath]\neg[/imath] and [imath]\wedge[/imath] form a functionally complete collection of logical operators. I was given the hint that I could use De Morgan law to show that [imath]p \vee q[/imath] is logically equivalent to [imath]\neg (\neg p \wedge \neg q)[/imath]. [imath]\neg (\neg p \wedge \neg q) \equiv \neg (\neg p) \vee \neg (\neg q)[/imath] [imath]\neg (\neg p) \vee \neg (\neg q) \equiv p \vee q[/imath] But I cannot figure out? am I done? |
1195966 | About Cantor set: Cantor set + Cantor set
Let us have the well-known Cantor set. We know that the Cantor set has continuum amount of numbers in the interval [imath][0,1][/imath]. Here is the thing that I need to prove: Prove, that every [imath]x \in [0,2][/imath] can be written in the form [imath]a+b[/imath], where [imath]a[/imath] and [imath]b[/imath] are part of the Cantor set. Edit: If I prove that Cantor set + Cantor set [imath]= [0,2][/imath], is that a good solution as well? My idea: We can write all the elements of Cantor set instead of decimal, in 3-based number system. In this way, every number of the Cantor set has the form [imath]0,abc...[/imath] but it doesn't contain the digit 1, since I deleted that part of the numbers. Now, with this form, I am able to make the Cantor set [imath] \Rightarrow\ [0,1][/imath] bijection, which means I am able to write down every [imath]x \in [0,1][/imath] number with the elements of Cantor set. From this part, I can pick any [imath]a,b\in[0,1][/imath] and with them I can cover the interval [imath][0,2][/imath] as well. Is my proof correct, or what you think about this task? Anything can help! :) | 309080 | Cantor set + Cantor set =[imath][0,2][/imath]
I am trying to prove that [imath]C+C =[0,2][/imath] ,where [imath]C[/imath] is the Cantor set. My attempt: If [imath]x\in C,[/imath] then [imath]x= \sum_{n=1}^{\infty}\frac{a_n}{3^n}[/imath] where [imath]a_n=0,2[/imath] so any element of [imath]C+C [/imath] is of the form [imath]\sum_{n=1}^{\infty}\frac{a_n}{3^n} +\sum_{n=1}^{\infty}\frac{b_n}{3^n}= \sum_{n=1}^{\infty}\frac{a_n+b_n}{3^n}=2\sum_{n=1}^{\infty}\frac{(a_n+b_n)/2}{3^n}=2\sum_{n=1}^{\infty}\frac{x_n}{3^n}[/imath] where [imath]x_n=0,1,2, \ \forall n\geq 1[/imath]. Is this correct? |
1196242 | Finding ring isomorphism
Find isomorphism from [imath]F_5[x]/(x^2+x+2) \rightarrow F_5[x]/(x^2+4x+2) [/imath] I realise both polynomials are irreducible therefore form fields, not sure how to form a isomorphism from one to the other, help please | 1193074 | Ring isomorphism (polynomials in one variable)
Find an explicit isomorphism of the map [imath]h: \Bbb F_5[x]/(x^2+x+2) \to \Bbb F_5[x]/(x^2+4x+2)[/imath] I take the question to mean showing the 2 rings are isomorphic by finding an explicit mapping of each element. I think [imath]x^2+x+2[/imath] and [imath]x^2+4x+2[/imath] are irreducible over [imath]\Bbb F_5[/imath], so these are both fields. Also [imath]x^2=4x+3[/imath] in the first case and [imath]x^2=x+3[/imath] in the second. But beyond these basic facts I'm not sure how I should proceed. |
1196549 | [imath]\sup A =-\inf(-A)[/imath] Proof
Let [imath]A\neq\emptyset[/imath] and bounded above. Let [imath]-A=\{-x\ | \ x\in A\}.[/imath] Prove that [imath]\sup A=-\inf -A.[/imath] We need to show that [imath]-\inf -A[/imath] is the least upper bound of [imath]A.[/imath] Suppose [imath]x \in A.[/imath] Then [imath]-x \in -A,[/imath] and, hence [imath]-x\ge \inf -A. [/imath] This implies that [imath]x \le -\inf -A.[/imath] Thus [imath]-\inf -A[/imath] is an upper bound of [imath]A.[/imath] Let [imath]\beta[/imath] be any upper bound of [imath]A.[/imath] Thus [imath]\beta \ge x[/imath] for every [imath] x \in A.[/imath] This implies that [imath]-\beta \le -x[/imath] for every [imath]x \in A[/imath] which tells us that [imath]y \ge -\beta[/imath] for all [imath]y \in -A.[/imath] Hence [imath]-\beta[/imath] is a lower bound of [imath]-A.[/imath] Hence [imath]-\beta \le \inf -A,[/imath] [imath]\beta \ge -\inf -A,[/imath] and so [imath]-\inf -A[/imath] is the least upper bound of [imath]A.[/imath] | 522259 | Prove that [imath]\sup \{-x \mid x \in A\} = -\inf\{x\mid x \in A\}[/imath]
I need to prove that [imath]\sup \{-x \mid x \in A\} = -\inf\{x \mid x \in A\}[/imath] and am having trouble moving the [imath]-x[/imath] out of the [imath]\sup[/imath] to [imath]\inf[/imath]. Another thing is that I don't quite know how to prove [imath]b = \inf[/imath] (see below). Any help would be appreciated! Here my attempt: Let [imath]a = \sup\{-x \mid x \in A\}[/imath] Then by definition [imath]a \geq -x[/imath] for [imath]x \in A \implies -a\leq x[/imath] for [imath]x \in A[/imath] Thus [imath]-a[/imath] is a lower bound of [imath]A[/imath]. Assume [imath]b[/imath] is a lower bound of [imath]A[/imath] Then [imath]b \leq -a \implies a \geq -b[/imath] and [imath]a = \sup \{−x \mid x\in A\} = -\inf(A)[/imath] |
1192802 | Number of solutions of arithmetic function's equation
Say, an equation is given below \begin{equation} 2\pi(x) - \pi(2x)=\omega(x) \end{equation} where [imath]x[/imath] is a positive integer, [imath]\pi(x)[/imath] is the prime-counting function, and [imath]\omega(x)[/imath] is the number of distinct prime factors of [imath]x[/imath]. I would like to know if this equation holds for finitely many values of [imath]x[/imath] or infinitely many. If one uses asymptotic formula (from number theory) it can be written that \begin{equation} \omega(x) \sim \ln x / \ln \ln(x) \end{equation} (from http://www.math.uiuc.edu/~hildebr/ant/main3.pdf) and \begin{equation} \pi(x) \sim x/\ln(x), \end{equation} so the above equation can be rewritten (after simplification) as \begin{equation} 2x\ln(2)=\ln(x)\ln\ln(x)\ln(2x) +f(x) \end{equation} where [imath]f(x)[/imath] is an error term. If I plot the left side (which is linear, supposed to be a straight line) and the right side (a curve), assuming [imath] f(x) [/imath] is 100% accurate, then the equation implies(?!) that the right will intersect left side. Since line can intersect a curve ([imath]ln [/imath] function) finite times, so there are finite intersection points. So, can I say that there are only finitely many values of [imath]x[/imath]? | 1186835 | Number of solutions of an arithmetic function's equation
Say, an equation is given below \begin{equation} 2\pi(x) - \pi(2x)=\omega(x) \end{equation} where [imath]x[/imath] is a positive integer, [imath]\pi(x)[/imath] is the prime-counting function, and [imath]\omega(x)[/imath] is the number of distinct prime factors of [imath]x[/imath]. I would like to know if this equation holds for finitely many values of [imath]x[/imath] or infinitely many. If one uses asymptotic formula (from number theory) it can be written that \begin{equation} \omega(x) \sim \ln x / \ln \ln(x) \end{equation} (from http://www.math.uiuc.edu/~hildebr/ant/main3.pdf) and \begin{equation} \pi(x) \sim x/\ln(x), \end{equation} so the above equation can be rewritten (after simplification) as \begin{equation} 2x\ln(2)=\ln(x)\ln\ln(x)\ln(2x) +f(x) \end{equation} where [imath]f(x)[/imath] is an error term. If I plot left side (which is linear, supposed to be a straight line) and right side(a curve),assuming [imath] f(x) [/imath] is 100% accurate, then the equation implies(?!) that the right will intersect left side. Since line can intersect a curve ([imath]ln [/imath] function) finite times, so there are finite intersection point. So, can I say that there are only finitely many values of [imath]x[/imath]? graph example- here f(x) is assumed as x + sinx without the f(x)- |
566139 | Should [imath]0[/imath] be considered a prime?
Typically, a prime is defined as follows: [imath]p[/imath] is prime iff [imath](p \mid xy \implies p \mid x[/imath] or [imath]p \mid y)[/imath] and [imath]p[/imath] is not a unit or zero. But for ideals, we say the zero ideal is prime. There is a strong correspondence between statements about primes and statements about prime ideals: "A non-unit is prime if [imath]p \mid ab \implies p \mid a[/imath] or [imath]p \mid b[/imath]" vs. "A proper ideal is prime if [imath]P \ni ab \implies P \ni a[/imath] or [imath]P \ni b[/imath]" "All non-zero primes are irreducible" vs. "all non-zero prime ideals are maximal" "Primes are only divisible by themselves and units" vs. "prime ideals are only contained by themselves and the whole ring (generated by a unit)" "Zero is divisible in every ring element vs. "the zero ideal is contained in every ideal" "In a UFD, non-zero elements can be uniquely factored into primes" vs. "in a Dedekind ring, non-zero ideals can be uniquely factored into prime ideals" So, I feel that zero should be prime iff the zero ideal is a prime ideal. Why is there this discrepancy? I lean towards "zero is not a prime", but what are the consequences of rejecting the zero ideal as prime as well? | 3698 | Why doesn't [imath]0[/imath] being a prime ideal in [imath]\mathbb Z[/imath] imply that [imath]0[/imath] is a prime number?
I know that [imath]1[/imath] is not a prime number because [imath]1\cdot\mathbb Z=\mathbb Z[/imath] is, by convention, not a prime ideal in the ring [imath]\mathbb Z[/imath]. However, since [imath]\mathbb Z[/imath] is a domain, [imath]0\cdot\mathbb Z=0[/imath] is a prime ideal in [imath]\mathbb Z[/imath]. Isn't [imath](p)[/imath] being a prime ideal the very definition of [imath]p[/imath] being a prime element? (I know that this would violate the Fundamental Theorem of Arithmetic.) Edit: Apparently the answer is that a prime element in a ring is, by convention a non-zero non-unit (see wikipedia). This is strange because a prime ideal of a ring is, by convention, a proper ideal but not necessarily non-zero (see wikipedia). So, my question is now: Why do we make this awkward convention? |
1197070 | Which symbol to use for composition of a sequence of functions
I know how to write the composition of two functions: [imath]f\circ g[/imath] but I don't know whether there's a standard symbol for a sequence (similar to [imath]\sum_i{f_i}[/imath], [imath]\prod_i{f_i}[/imath] or [imath]\bigotimes_i{f_i}[/imath], etc.). How would you write it? edit: Would you understand [imath]\bigcirc_i{f_i}[/imath] without explanation? | 926247 | Notation for repeated composition of functions
I have a repeated composition of functions [imath]{T_n}(z) = {\tau _0} \circ {\tau _1} \circ {\tau _2} \circ \cdots \circ {\tau _n}(z)[/imath] By analogy with [imath]\sum\limits_{i = 1}^n {} ,\prod\limits_{i = 1}^n {} ,\bigcup\limits_{i = 1}^n {} ,\bigcap\limits_{i = 1}^n {} ,[/imath] I want to write [imath]{T_n}(z) = \left( {\mathop \circ \limits_{i = 0}^n {\tau _i}} \right)(z)[/imath] or even [imath]{T_n}(z) = {\mathop \circ \limits_{i = 0}^n {\tau _i}} (z)[/imath]. Can I do this? |
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