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481138

A difficult integral: [imath]\int_0^{\pi/2} (\log \sin x)^2 \ dx [/imath] I tried integration by parts but I Ended up with intetgral log sinx*log cosx from 0 to pi/2 I tried to use exponential form for integration but I couldn't integrate further

58654

Integrate square of the log-sine integral: [imath]\int_0^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx[/imath] [imath]\displaystyle \int_{0}^{\frac{\pi}{2}} \ln(\sin(x))dx=-\frac{\pi}{2}\ln(2)[/imath] is an integral that is common. But, how can we show [imath]\displaystyle\int_{0}^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx=\frac{{\pi}^{3}}{24}+\frac{\pi}{2}\ln^{2}(2)[/imath]?. Does anyone have any ideas on how to approach [imath]\displaystyle\int_{0}^{\frac{\pi}{2}}\ln^{2}(\sin(x))dx[/imath]?. Thank you very much.

481133

Equilateral triangle in complex plane Prove that the points [imath]a_1,a_2,a_3[/imath] are vertices of an equilateral triangle if and only if [imath]a_1^2+a_2^2+a_3^2=a_1a_2+a_2a_3+a_3a_1[/imath]. I rewrite the equation as [imath]2a_1^2+2a_2^2+2a_3^2-2a_1a_2-2a_2a_3-2a_3a_1=0[/imath], which is [imath](a_1-a_2)^2+(a_2-a_3)^2+(a_3-a_1)^2=0[/imath]. This looks quite nice, but I'm not sure how to relate it to the equilateral triangle.

102458

How does this equality on vertices in the complex plane imply they are vertices of an equilateral triangle? I've read that if the complex numbers [imath]a_1[/imath], [imath]a_2[/imath] and [imath]a_3[/imath] are the vertices of a triangle in the complex plane such that [imath] a_1^2+a_2^2+a_3^2=a_1a_2+a_2a_3+a_1a_3 [/imath] then the vertices are actually those of an equilateral triangle. I tried to see why this by first shifting the vertex [imath]a_1[/imath] to the origin, and considering the triangle with vertices [imath]0[/imath], [imath]a_2-a_1[/imath] and [imath]a_3-a_1[/imath]. The given equation holds if and only if [imath] (a_2-a_1)^2+(a_3-a_1)^2=(a_2-a_1)(a_3-a_1) [/imath] which is equivalent to saying [imath] \frac{a_2-a_1}{a_3-a_1}+\frac{a_3-a_1}{a_2-a_1}=1. [/imath] Letting [imath]\frac{a_2-a_1}{a_3-a_1}=x[/imath], this gives the quadratic [imath]x+\frac{1}{x}=1[/imath], and so I solve for [imath]x[/imath] to be [imath] \frac{a_2-a_1}{a_3-a_1}=x=\frac{1\pm\sqrt{-3}}{2}=e^{\pm i\pi/3}.[/imath] I'm not sure where I'm going now. How can I conclude that such a relation implies the vertices form an equilateral triangle? Thank you.

481128

Does this limit exist? if so what is it? Given the set [imath]A_n =\{(i,j)∈ Z^2:(i,j)=1, 0≤i,j≤n\}[/imath] I am having problems proving that the limit [imath]\lim_{N\to \infty}[/imath] [imath]|A_n|\over N[/imath] exist and to calculate it, any tips?

461108

The limit of [imath]\frac{|A_n|}{n^2}[/imath] Let [imath]A_n=\{(i,j)\in\mathbb{Z}^2:\gcd(i,j)=1, \ \ 0 \leq i,j\leq n \}[/imath]. How to prove the existence of [imath]\lim_{n\to\infty}\frac{|A_n|}{n^2}[/imath], and how calculate this limit? Thank you!

142572

If cancellation laws hold, then a finite semi-group is a group Show that if both cancellation laws i.e [imath]w.a = w.b \implies a = b[/imath] and [imath]a.w = b.w \implies a = b[/imath] holds then a finite semi-group (a finite set with associative binary operation) is a group. I have seen some proofs which uses the alternative definition of group to prove it i.e. [imath]a.x = b[/imath] and [imath]y.a =b[/imath] have unique solutions for [imath]x[/imath] and [imath]y[/imath]. I am not interested in such proofs. How to prove this statement starting with cancellation laws and then showing that all axioms of group can be derived from them? EDIT : As pointed out in one of the answer. This is only true when underlying set is finite. Edited accordingly.

203023

A finite, cancellative semigroup is a group Let [imath]G[/imath] be a finite, nonempty set with an operation [imath]*[/imath] such that [imath]G[/imath] is closed under [imath]*[/imath] and [imath]*[/imath] is associative Given [imath]a,b,c \in G[/imath] with [imath]a*b=a*c[/imath], then [imath]b=c[/imath]. Given [imath]a,b,c \in G[/imath] with [imath]b*a=c*a[/imath], then [imath]b=c[/imath]. I want to prove that [imath]G[/imath] is a group, but I don't know how to show that there exists an identity [imath]e\in G[/imath] such that [imath]e*x=x[/imath] and [imath]x*e=x[/imath] [imath]\forall x \in G[/imath]. I also don't know how to show that [imath]\forall[/imath] x [imath]\in G[/imath] there exists a [imath]y \in G[/imath] such that [imath]y*x=e[/imath] and [imath]x*y=e[/imath]. How do I do this?

463754

Solving for the closed term solution of a third order recurrence relation with real constant coefficients How would you solve for the closed term form of [imath]a(n)[/imath] given the general form of the third order linear homogenous recurrence relation with real constant coefficients. [imath]a(n)-P\,a(n-1)-Q\,a(n-2)-R\,a(n-3)=0[/imath] with the initial terms of a1, a2, and a3 and given that the roots of the characteristic equations have two repeated roots and a real root three repeated roots (can you give answers for both cases please) For second order recurrence relations I know that you can use generating functions to deduce a closed form because it is then expressed as a arithmetic series which can be converted into a closed form. However in the case of the general term of the third order recurrence relations if I follow the same steps what I did with the second order recurrence relation, instead of getting a simple arithmetic series I seemed to get a second order recurrence relation inside the series. What am I doing wrong? or is there a different method of approach in this case? When I search the web I get these results S(n) = nAx1^n + Bx1^n + Cx2^n,for the case when there are two repeated roots and S(n) = n^2Ax^n + nBx^n + Cx^n, for the case when there are three repeated roots I just don't know how to get to these results Please help

478927

Finding the closed form solution of a third order recurrence relation with constant coefficients How would you solve for the closed form solution of a(n) given the general form of the third order linear homogenous recurrence relation with real constant coefficients. [imath]a_n=Pa_{n-1}+Qa_{n−2}+Ra_{n−3}[/imath] with the initial terms of a_1, a_2, and a_3 given that the roots of the characteristic equations have 1)two repeated roots and a real root 2)three repeated roots (can you give answers for both cases please) When I search the web I get these results 1)[imath]a_n = nAx_1^n + Bx_1^n + Cx_2^n[/imath],for the case when there are two repeated roots and 2)[imath]a_n = n^2Ax^n + nBx^n + Cx^n[/imath], for the case when there are three repeated roots Can anyone help derive the closed form of each case in order to get such results? Please help I'm new to the system so i didn't quite know how to get the symbols right (sorry) if you're uncertain about anything please ask

481240

Idea of the Proof : Existence of a & b so that (Any integer greater than 8) = 3a + 5b Claim: Prove that for every integer [imath]n \geq 8[/imath], there exist nonnegative integers [imath]a[/imath] and [imath]b[/imath] such that [imath]n = 3a + 5b.[/imath] Proclaimed solution : Let [imath]n ∈ \mathbb{Z}[/imath] with [imath]n ≥ 8.[/imath] [imath]\text{ Then } n = 3q, \text{ where } q ≥ 3, \quad \text{ or } n = 3q+1, \text{ where } q ≥ 3, \quad \text{ or } n = 3q+2, \text{ where } q ≥ 2.[/imath] We consider these three cases. (Rest of proof omitted) I do not apprehend the above parts of the proof BEFORE (Rest of proof omitted). However, I understand and completed the rest of the proof that I omitted. I recognise that it uses the result that division of any integer by [imath]n[/imath] must yield a remainder of [imath]0, 1, 2, ..., n - 2, \text{ or } n - 1[/imath]. [imath]1.[/imath] However, what is the idea or strategy of the proof? [imath]2.[/imath] What motivates or propounds the claim? [imath]3.[/imath] Is there a more natural or easier proof? I referenced 1. Source: P102, Problem 4.9 on P101 (related to P90, Result 4.8) of Mathematical Proofs, 2nd ed by Chartrand et al.

69961

Determine the Set of a Sum of Numbers I want to determine the set of natural numbers that can be expressed as the sum of some non-negative number of 3s and 5s. [imath]S=\{3k+5j∣k,j∈\mathbb{N}∪\{0\}\}[/imath] I want to check whether that would be: 0,3, 5, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and so on. Meaning that it would include 0, 3, 5, 8. Then from 9 and on, every Natural Number. But how would I explain it as a set? or prove that these are the numbers in the set?

461810

Problem on abelian group Let [imath]G[/imath] be an abelian group, and [imath]\Phi:G\to \mathbb{R}[/imath] is a function with the following property: [imath]\forall a,b\in G,~~ |\Phi(a+b)-\Phi(a)-\Phi(b)|<c[/imath] The problem asks to prove the existence of homomorphism, [imath]\Phi^*[/imath], where [imath]\Phi^*:G\to\mathbb{R}[/imath] and [imath]\forall a\in G, |\Phi(a)-\Phi^*(a)|<c[/imath]

823092

Additive like function representation Let's [imath]f:R\rightarrow R,~r\gt0.[/imath] And we have that for every [imath]x,y\in R\Rightarrow |f(x+y)-f(x)-f(y)|\le r.[/imath] Prove that there are [imath]h:R \rightarrow R[/imath] additive and [imath]g:R \rightarrow [-r,r][/imath] functions such that [imath]f(x)=g(x)+h(x).[/imath] I have no idea how I can approach this problem.

481869

For [imath]N\unlhd G[/imath] , with [imath]C_G(N)\subset N[/imath] we have [imath]G/N[/imath] is abelian Question is that : let [imath]N\unlhd G[/imath] such that every subgroup of [imath]N[/imath] is Normal in [imath]G[/imath] and [imath]C_G(N)\subset N[/imath]. Prove that [imath]G/N[/imath] is abelian. what could be the possible first thought (though for me it took some time :)) is to use that [imath]C_G(N)[/imath] is Normal subgroup (As in general centralizer is a subgroup). one reason to see this is that [imath]C_G(N)[/imath] is not Normal in General and [imath]C_G(N)[/imath] is not subset of [imath]N[/imath] in general. As [imath]C_G(N)\subset N[/imath], we have [imath]G/N\leq G/C_G(N)[/imath] I some how want to say that [imath]G/C_G(N)[/imath] is abelian and by that conclude that [imath]G/N[/imath] is abelian. I would like someone to see If my way of approach is correct/simple?? I have not yet proved that [imath]G/C_G(N)[/imath] is abelian. I would be thankful if someone can give an idea. Thank You.

189694

Abelian quotient group I'm stuck on the following practice problem. Any hints would be appreciated. Suppose [imath]N[/imath] is a normal subgroup of [imath]G[/imath] such that every subgroup of [imath]N[/imath] is normal in [imath]G[/imath] and [imath]C_{G}(N) \subset N[/imath]. Prove that [imath]G/N[/imath] is abelian. I'm not sure how to use the fact that [imath]C_{G}(N) \subset N[/imath]. Thanks

482294

[imath]n! =[/imath] the product of consecutive integers. Can [imath]n![/imath] be the product of [imath]k[/imath] consecutive integers for [imath]k > 1[/imath]? (Not including the degenerate cases such as when [imath]k = 2[/imath], then [imath]1\cdot2 = 2![/imath] and [imath]2\cdot 3 = 3![/imath], and so on.) I am asking not for [imath]n![/imath] to be divisible by [imath]n[/imath] consecutive integers, I am asking for [imath]n![/imath] to equal the product of [imath]k[/imath] consecutive integers, implying that [imath]n[/imath] is not necessarily equal to [imath]k[/imath] (And when [imath]n = k[/imath], then clearly there exists an answer, namely [imath]n! = (1)(2)(3)...(k)[/imath])

112670

On the factorial equations [imath]A! B! =C![/imath] and [imath]A!B!C! = D![/imath] I was playing around with hypergeometric probabilities when I wound myself calculating the binomial coefficient [imath]\binom{10}{3}[/imath]. I used the definition, and calculating in my head, I simplified to this expression before actually calculating anything [imath] \frac {8\cdot9\cdot10}{2\cdot3} = 120 [/imath] And then it hit me that [imath]8\cdot9\cdot10 = 6![/imath] and I started thinking about something I feel like calling generalized factorials, which is just the product of a number of successive naturals, like this [imath] a!b = \prod_{n=b}^an = \frac{a!}{(b-1)!},\quad a, b \in \mathbb{Z}^+, \quad a\ge b [/imath] so that [imath]a! = a!1[/imath] (the notation was invented just now, and inspired by the [imath]nCr[/imath]-notation for binomial coefficients). Now, apart from the trivial examples [imath](n!)!(n!) = n![/imath] and [imath]a!1 = a!2 = a![/imath], when is the generalized factorial a factorial number? When is it the product of two (non-trivial) factorial numbers? As seen above, [imath]10!8[/imath] is both.

470807

Prove that [imath]\gcd(M, N)\times \mbox{lcm}(M, N) = M \times N[/imath]. I'm not sure how to go about this proof. I just need help getting started. Is there a way to prove it algebraically?

2141336

Let [imath]a[/imath] and [imath]b[/imath] be positive integers. Prove that [imath](a,b)|[a,b][/imath]. Hi everyone here is my question; So, let [imath]a[/imath] and [imath]b[/imath] be positive integers. Prove that [imath](a,b)|[a,b][/imath]. Here what I have so far; [imath](a,b)=d[/imath] [imath][a,b]=m[/imath] so I need an equation of the form [imath]m=d()[/imath]. I just need some hints on how to proceed. Thank you so much in advance.

482743

Bijection between [imath]P(\mathbb{N})[/imath] and set [imath]\{f | f:\mathbb{N} \to \{0,1\}\}[/imath] I need to find a bijection between the power set [imath]P(\mathbb{N})[/imath] and the set of functions [imath]J=\{f|f:\mathbb{N}\to \{0,1\}\}[/imath] I considered the map [imath]g:P\to J[/imath] as [imath]g(A)=f [/imath] where [imath]f[/imath] is a function who is mapping [imath]A[/imath] to {0} and [imath]\mathbb{N}\setminus A[/imath] to [imath]{1}[/imath], then i think this will be a required bijection!

368143

Binary sequences and [imath]{2}^{\mathbb{N}}[/imath] have the same cardinality I recently got the book "selected problems in real analysis", and I'm stuck solving the very first problem [imath](u_n)[/imath] is a binary sequence iff it only contains [imath]0[/imath] and [imath]1[/imath] in the sequence Let [imath]A[/imath] be the set of all binary sequences I have to prove that [imath]A[/imath] and [imath]{2}^{\mathbb{N}}[/imath] have the same cardinality, that is to say there exists a 1-1 function from one set to another I've thought about maybe considering integers as base-2 numbers Thanks for your help

483034

A relation on 2 countable sets Let [imath]R[/imath] be a relation on two countable sets [imath]A[/imath] and [imath]B[/imath], where [imath]R\subset A\times B[/imath], with the following properties: [imath]\forall a\in A[/imath] the set [imath]\{b\in B: (a,b)\in R\}[/imath] is finite. For any finite set [imath]A_0\subset A[/imath]: [imath]|\{b\in B : \exists a_0\in A_0, \text{such that} \ \ (a_0,b)\in R \}|\leq n|A_0|[/imath] where [imath]n\in\mathbb{N}[/imath]. I need to show then, that there exist [imath]n[/imath] disjoint sets, [imath]B_1,B_2,\dots, B_n[/imath], where [imath]B_i\subset B\ \ \ \ \ \ \forall \ \ 1 \leq i\leq n[/imath], and there exists [imath]n[/imath] one to one and onto functions [imath]f_1,f_2,\dots, f_n[/imath] such that [imath]f_i:B_i\to A \ \ \ \ \ \ \ \forall \ \ 1 \leq i\leq n [/imath]? Thank you for your help.

478917

Relation on countable sets Let [imath]R[/imath] be a relation on two countable sets [imath]A[/imath] and [imath]B[/imath], where [imath]R\subset A\times B[/imath], with the following properties: [imath]\forall a\in A[/imath] the set [imath]\{b\in B: (a,b)\in R\}[/imath] is finite. For any finite set [imath]A_0[/imath]: [imath]|\{b\in B : \exists a_0\in A_0, \text{such that} \ \ (a_0,b)\in R \}|\leq n|A_0|[/imath] where [imath]n\in\mathbb{N}[/imath]. How can I show then, that there exist [imath]n[/imath] disjoint sets, [imath]B_1,B_2,\dots, B_n[/imath], where [imath]B_i\subset B\ \ \ \ \ \ \forall \ \ 1 \leq i\leq n[/imath], and there exists [imath]n[/imath] one to one and onto functions [imath]f_1,f_2,\dots, f_n[/imath] such that [imath]f_i:B_i\to A \ \ \ \ \ \ \ \forall \ \ 1 \leq i\leq n [/imath]? Edited: [imath]B_i\in B[/imath] replaced with [imath]B_i\subset B[/imath]

483249

[imath]S[/imath] is a compact subset of [imath]\mathbb{R}[/imath] and [imath]T[/imath] is a closed subset of [imath]S[/imath] [imath]\implies[/imath] [imath]T[/imath] compact If [imath]S[/imath] is a compact subset of [imath]\mathbb{R}[/imath] and [imath]T[/imath] is a closed subset of [imath]S[/imath], then [imath]T[/imath] is compact. How can I show this using the definition of compactness, and separately showing this by the Heine-Borel theorem.

229868

Is a closed subset of a compact set (which is a subset of a metric space [imath]M[/imath]) compact? Is there a way to prove this using sequential compactness instead of open cover definitions? My first gut reaction was that the fact was obvious since we can show that the closed subset [imath][a,b][/imath] is compact in [imath]\mathbb R[/imath].

483016

Bounded Infinite Set: Infinitely Many Points How is it that if [imath]S\subseteq\mathbb{R}[/imath] is a bounded infinite set, where [imath]x=\sup(S)[/imath], then every neighborhood of [imath]x[/imath] contains infinitely many points of [imath]S[/imath]?

126105

Question about limit points of a Subset of [imath]\mathbb{R}[/imath] The question : Let [imath]D[/imath] be a nonempty subset of the reals that is bounded above. Is the supremum of [imath]D[/imath] a limit point of [imath]D[/imath]? My Reasoning: I think this is false for these two cases. Case 1:If I look at [imath]D = \{n \in \mathbb{Z} | n \le 0\}[/imath] the supremum is [imath]0[/imath]. And since I need a convergent sequence [imath]\{x_n\} \subset D/\{0\}[/imath] the converges to [imath]0[/imath] for it to be a limit point I can say in this case if I look at [imath]\epsilon = \frac{1}{2}[/imath] for the converges of the sequence it will fail to converge and so [imath]0[/imath] isn't a limit point. Case 2: Also if I look at [imath]D = {0}[/imath] then the supremum is [imath]0[/imath]. And [imath]D[/imath] is a subset of the reals. So if I look for a sequence [imath]\{x_n\} \subset D/\{0\}[/imath] I can't make one because [imath]D/\{0\}[/imath] is the empty set. My question is this. Since the problem asked about an arbitrary subset of the reals [imath]D[/imath], can I define [imath]D[/imath] to give a counterexample like above or have I misunderstood the question? --Thanks in advance.

483373

Infinite series with prime number I know the [imath]\sum_{n=1}^\infty \frac{1}{\text{Prime[[/imath]n[imath]]}}[/imath] does not converge, but what about the following series? [imath]\sum_{n=1}^\infty\frac{1}{\text{Prime[Prime[$n$]]}}[/imath] (Where [imath]\text{Prime[[/imath]n[imath]]}[/imath] is the [imath]n[/imath]th prime number.) I ran some calculation on Mathematica, but I am not sure in the answer gets very close to 1. Can anyone help with that? NB: I an not a professional mathematician. I am just an amateur.

430369

What is the value of [imath]\sum\limits_{i=1}^\infty\frac{1}{p_{p_i}}[/imath] where [imath]p_{i}[/imath] is the [imath]i[/imath]th prime? What is the value of [imath]\sum\limits_{i=1}^\infty\dfrac{1}{p_{p_i}}[/imath] where [imath]p_n[/imath] is the nth prime (and so [imath]p_{p_n}[/imath] is the [imath]k[/imath]th prime, where [imath]k[/imath] is the [imath]n[/imath]th prime) ? Thus [imath]\frac{1}{3}+\frac{1}{5}+\frac{1}{11}+...=[/imath] ?? How do I efficiently decide if [imath]\sum_{i=1}^\infty\dfrac{1}{p_{p_i}}\gt1[/imath] is true without using a computer ? I assume this infinite sum is an irrational number. Has this already been proven ?

248245

Exactly half of the elements of [imath]\mathcal{P}(A)[/imath] are odd-sized Let [imath]A[/imath] be a non-empty set and [imath]n[/imath] be the number of elements in [imath]A[/imath], i.e. [imath]n:=|A|[/imath]. I know that the number of elements of the power set of [imath]A[/imath] is [imath]2^n[/imath], i.e. [imath]|\mathcal{P}(A)|=2^n[/imath]. I came across the fact that exactly half of the elements of [imath]\mathcal{P}(A)[/imath] contain an odd number of elements, and half of them an even number of elements. Can someone prove this? Or hint at a proof?

489069

Bits and counting problem I'm stuck in figuring out a problem, and I can't quite figure out a way out right now. any suggestion is highly appreciated. we have a 10bit number(made up of 1s and 0s), and we do not know the number. what is the probability of having even number of 1s? So, what i did: -0 ones -2 ones -4 ones -6 ones -8 ones -10 ones I started to look at each case. The first and the last can only be one number, I also found that the second has [imath]90[/imath] numbers possibilities of construction. When we have [imath]4[/imath] ones, i calculated we get 10*9*8*7 numbers etc. I'm getting really weird numbers though at the final step of adding it all in. Is there an easy way to go about this counting problem? I feel my solution is very error-prone and tedious. :/

480068

How to reverse digits of an integer mathematically? Is there a mathematical way to reverse digits of an integer mathematically? for example: [imath]18539 \rightarrow 93581[/imath]

472981

Formula for reversing digits of positive integer [imath]n[/imath] I was able to work out the cases for [imath]n[/imath] having up to [imath]4[/imath] digits and was wondering if someone could verify my generalization to [imath]m[/imath] digits. Here I am assuming that when a reversal results in there being a leading [imath]0[/imath] it get ignored (e.g. [imath]76130[/imath] gets reversed to [imath]3167[/imath], [imath]998700[/imath] to [imath]7899[/imath] etc.) I believe the function can be expressed as [imath]r(n): \mathbb{N} \rightarrow \mathbb{N}[/imath] [imath]r(n) = \left \lfloor \frac{n}{10^m}\right \rfloor +\sum_{k=0}^{m-1}10^{m-k} \left( \left \lfloor\frac{n}{10^k} \right \rfloor-10 \left \lfloor \frac{\left \lfloor \frac{n}{10^k}\right \rfloor}{10}\right \rfloor \right)[/imath] where again, [imath]n[/imath] has [imath]m[/imath] digits. It seems like this combination of the floor function and powers of 10 is the only way to achieve this but is this true?

483862

How do I prove [imath]\gcd(a,b) = \gcd(a+b, b)[/imath] How do I prove [imath]\gcd(a, b) = \gcd(a+b, b)[/imath]. I know that by the euclidean algorithm, I can obtain the following equations [imath]ax_1 + by_1 = \gcd(a, b)\tag{1}[/imath] [imath](a+b)(x_2) + (b)(y_2) = \gcd(a+b, b)\tag{2}[/imath] I tried some algebraic manipulation but I can't seem too prove that [imath]\gcd(a, b) = \gcd(a+b, b)[/imath].

23316

Prove: [imath]\gcd(a,b) = \gcd(a, b + at)[/imath]. I know that [imath]\gcd(a,b)[/imath] divides [imath]a[/imath] and [imath]b[/imath], and must also then divide [imath](a)(t)[/imath] ([imath]t[/imath] being some integer). This makes sense to me, but how do I prove it? It seems that the addition of [imath](a)(t)[/imath] is a continuation of the linear combination of [imath]\gcd(a,b) = av + bu[/imath] for some [imath]v[/imath], [imath]u[/imath] being elements of [imath]\mathbb{Z}[/imath]. Any help?

484131

What is [imath]\sup(\sin(n))[/imath]? My classmate asked a question during lecture about our discussion of bounded sequences, particularly the sequence [imath]\sin(n)[/imath]. His question was, What is [imath]\sup(\sin(n))[/imath]?

63526

Showing [imath]\sup \{ \sin n \mid n\in \mathbb N \} =1[/imath] how to prove [imath]\sup \{ \sin n \mid n\in \mathbb N \} =1[/imath]

268873

A finite group of even order has an odd number of elements of order 2 Need some help with this, i'm a bit stuck: Show that if [imath]G[/imath] is a finite group of even order, then [imath]G[/imath] has an odd number of elements of order [imath]2[/imath]. Note that [imath]e[/imath] is the only element of order [imath]1[/imath].

42034

Group of even order contains an element of order 2 I am working on the following problem from group theory: If [imath]G[/imath] is a group of order [imath]2n[/imath], show that the number of elements of [imath]G[/imath] of order [imath]2[/imath] is odd. That is, for some integer [imath]k[/imath], there are [imath]2k+1[/imath] elements [imath]a[/imath] such that [imath]a \in G,\;\; a*a = e[/imath], where [imath]e[/imath] is the identity element of [imath]G[/imath].

484691

Evaluating [imath]\lim_{x\to0}{x^x}[/imath] Find the limit: [imath]\lim_{x\to0^+}{x^x}[/imath] *This question appeared in a calculus exam, and I would like to see different approaches and solutions to the problem.

473535

Why does this limit exist [imath]x^{x}[/imath] I was wondering what is the value of [imath]\lim\limits_{x\to 0+} x^{x}[/imath]. Assuming that the limit exists, I could show using the usual logarithm techniques that the limit is [imath]1[/imath]. However, I am not able to show that the limit exists. Could some one help on that? EDIT: It would be so nice if we could do it without l'Hopital...

484837

Finding every [imath]n[/imath] such that there exists a [imath]n[/imath]-th degree polynomial which satisfies [imath]f(x^2+1)={f(x)}^2+1[/imath] I'm interested in functional equation. I've been thinking about the following functional equation: [imath]f(x^2+1)={f(x)}^2+1\ \ \ \cdots(\star).[/imath] I found several functions such as [imath]f(x)=x, x^2+1, (x^2+1)^2+1,\cdots[/imath]. Then, I got interested the following question: Find every natural number [imath]n[/imath] such that there exists a [imath]n[/imath]-th polynomial which satisfies [imath](\star).[/imath] Then, I got the following fact: Fact: There exists a [imath]2^k[/imath]-th polynomial which satisfies [imath](\star)[/imath] for any non-negative integer [imath]k[/imath]. Proof: Let us define [imath]\{f_k(x)\}[/imath] as the following: [imath]f_0(x)=x, f_{k+1}(x)={f_k(x)}^2+1\ \ \ \cdots(\star\star).[/imath] Supposing that both [imath]f_0(x)[/imath] and [imath]f_k(x)[/imath] satisfy [imath](\star)[/imath], then [imath]f_{k+1}(x)[/imath] also satisfies [imath](\star)[/imath] because [imath]f_{k+1}(x^2+1)=\{f_k(x^2+1)\}^2+1=\{{f_k(x)}^2+1\}^2+1={f_{k+1}(x)}^2+1.[/imath] Hence, [imath]\{f_k(x)\}[/imath] satisfy [imath](\star)[/imath]. Since we know that the degree of [imath]f_k(x)[/imath] is [imath]2^k[/imath] inductively from [imath](\star\star)[/imath], now the proof is completed. However, I can't prove that there is no other [imath]n[/imath] such that there exists a [imath]n[/imath]-th polynomial which satisfies [imath](\star)[/imath]. Then, here is my question. Question: Could you show me how to solve the above question?

271337

Find all polynomials [imath]P[/imath] such that [imath]P(x^2+1)=P(x)^2+1[/imath] Find all polynomials [imath]P[/imath] such that [imath]P(x^2+1)=P(x)^2+1[/imath]

485485

Evaluate [imath] \int \cfrac { \sec^2x}{ ( \sec x + \tan x)^{9/2}}dx [/imath] Problem : Evaluate [imath] \int \cfrac { \sec^2x}{ ( \sec x + \tan x)^{9/2}}dx [/imath] Solution : [imath] \int \cfrac { \sec^2x}{ ( \sec x + \tan x)^{9/2}}dx [/imath] [imath] \int \cfrac { \sec^2x}{ \left(1+ \sin x \over \cos x \right)^ {9/2}}dx [/imath] [imath] \int \cfrac { \sec^2x}{ (\sin {x \over 2} + \cos {x \over 2})^9 \over (\cos^2 {x \over 2} - \sin^2 {x \over 2})^ {9/2}}dx [/imath] [imath] \int \cfrac { \sec^2x}{ \left(\cos {x \over 2} + \sin {x \over 2} \over\cos {x \over 2} - \sin {x \over 2}\right)^ {9/2}}dx [/imath] [imath] \int \cfrac { \sec^2x}{ \left(1 + \tan {x \over 2} \over 1 - \tan {x \over 2}\right)^ {9/2}}dx [/imath] [imath] \int \cfrac { \sec^2x}{[tan({ \pi \over 4}+ { x \over 2})]^ {9/2}}dx [/imath] Am I doing right ?

130987

Evaluate [imath]\int \frac{\sec^{2}{x}}{(\sec{x} + \tan{x})^{5/2}} \ \mathrm dx[/imath] I am unable to solve the following integral: [imath]\int \frac{\sec^{2}{x}}{(\sec{x} + \tan{x})^{5/2}} \ \mathrm dx.[/imath] Tried putting [imath]t=\tan{x}[/imath] so that numerator has [imath]\sec^{2}{x}[/imath] but that doesn't help.

485471

How to prove that [imath]\varphi\left(mn\right)=\varphi\left(m\right)\varphi\left(n\right)[/imath] Of course we are familiar with the notion that if [imath]n=p_{1}^{k_{1}}\cdots p_{r}^{k_{r}}[/imath] ([imath]p_{i}[/imath] distinct primes; [imath]k_{i}>0[/imath]), then [imath]\varphi\left(n\right)=n\left(1-1/p_{1}\right)\left(1-1/p_{2}\right)\cdots\left(1-1/p_{r}\right),[/imath] where [imath]\varphi[/imath] is the Euler totient function. The goal is to prove this using the fact that [imath]\varphi[/imath] is multiplicative, that is, [imath]\varphi\left(mn\right)=\varphi\left(m\right)\varphi\left(n\right),[/imath] given that [imath]m[/imath] and [imath]n[/imath] are coprime. Proving the former notion from the multiplicative property. But how do we prove the multiplicative property of [imath]\varphi[/imath] without using that notion?

192452

What's the proof that the Euler totient function is multiplicative? That is, why is [imath]\varphi(AB)=\varphi(A)\varphi(B)[/imath], if [imath]A[/imath] and [imath]B[/imath] are two coprime positive integers? It's not just a technical trouble—I can't see why this should be, intuitively: I bellyfeel that its multiplicativity should be an approximation at most. And why the minimum value for [imath]\varphi (n)[/imath] should be [imath] \frac{4n}{15} [/imath] completely passes me by.

288068

How variance is defined? The variance of a random variable [imath]X[/imath] is defined as [imath]E[(x-\mu )^2][/imath]. Why can't it be defined as [imath]E[|x-\mu |][/imath]. i.e., What is the basic idea behind this definition. Thank you.

1556457

Was it really necessary to have the variance be calculated with [imath]\Sigma (X-\mu)^2 [/imath]? From what I understand, in calculating [imath]\sigma^2 = \frac{\sum (X - \mu)^2}{N}[/imath] the only reason why the squaring is done is to make distance from mean always positive. So my question, why couldn't we do something like this [imath]\sigma = \frac{\sum \left|X-\mu \right|}{N}[/imath] instead?

486424

[imath]A_n[/imath] contains an isomorphic copy of [imath]S_{n-2}[/imath] Question is to prove that : [imath]A_n[/imath] contains an isomorphic copy of [imath]S_{n-2}[/imath] I have no idea from where to start. I do not even believe this can be true (expect for cardinality conditions). But as this is an exercise.it has to be true. please provide some hints/suggestions to attack this problem. Thank you :)

83160

Can the symmetric group [imath]S_n[/imath] be imbedded as a subgroup in [imath]A_{n+1}[/imath]? I'm working through Rotman (1994). It's the 2nd chapter where my curiosity lead me to a question that is not from the book. Here's the problem that inspired the question: 2.8. Imbed [imath]S_n[/imath] as a subgroup of [imath]A_{n+2}[/imath], but show, for [imath]n\ge2[/imath], that [imath]S_n[/imath] cannot be imbedded in [imath]A_{n+1}.[/imath] Since the alternate subgroup has the property of even parity, I thought that a little investigation of parity for the second and third symmetric groups was in order. So I constructed a table for each of them, grouped by even and odd number of disjoint products. I'll forego writing the actual table and simply provide the answers. In symmetric groups with an even idegree, two even permutations come together under composition to produce an even permutation. Two odds also make an even. Two different parities produce odd elements. When the symmetric group has an odd degree, e.g. [imath]S_3[/imath], two odds make an odd, two evens make an odd, and two different parities make an even element. Exactly the opposite results from the symmetric groups with even index. Two questions: is this enough to prove the [imath]A_{n+1}[/imath] case? And can this be generalized modulo 2 e.g. [imath]A_{2n+1}[/imath]?

486695

Relation of Supremum and Infimum Let [imath]\mathcal{S}[/imath] be a non-empty bounded set of real numbers, and define for a real number [imath]a[/imath] the set [imath]aS=\left\{ax|\ \ x \in\mathcal{S}\right\}[/imath]. Prove that [imath]\sup(aS)=a \inf S[/imath] and [imath]\inf(aS)=a \sup S[/imath] if [imath]a<0[/imath].

476684

Infinum & Supremum: An Analysis on Relatedness [imath]\require{color}[/imath] Question: I need some help in proving that [imath]\color{green}{\text{if [/imath]k\geq 0[imath], then [/imath]\sup (kS) = k\sup(S)[imath] and [/imath]\inf (kS) = k\inf (S)[imath]}}[/imath], and also that [imath]\color{red}{\text{if [/imath]k<0[imath], then [/imath]\sup(kS)=k\inf(S)[imath] and [/imath]\inf(kS)=k\sup(S)[imath]}}[/imath]. Thoughts: As far as I can tell, [imath]S[/imath] is a non-empty, bounded subset of [imath]\mathbb{R}[/imath], and [imath]kS=\{ks:s\in S\}[/imath], where [imath]k\in \mathbb{R}[/imath], so [imath]kS\subseteq\mathbb{R}[/imath]. I suspect that we are working with the ordered set [imath](\mathbb{R},\leq)[/imath], where, again, [imath]kS\subseteq \mathbb{R}[/imath], so if an element [imath]\alpha\in \mathbb{R}[/imath] is the supremum of [imath]kS[/imath] in [imath]\mathbb{R}[/imath], then [imath]\alpha[/imath] is an upper bound of [imath]kS[/imath] in [imath]\mathbb{R}[/imath], and [imath]\alpha\leq \beta[/imath] for all upper bounds [imath]\beta[/imath] of [imath]kS[/imath] in [imath]\mathbb{R}[/imath], and if and element [imath]\alpha'\in\mathbb{R}[/imath] is the infimum of [imath]kS[/imath] in [imath]\mathbb{R}[/imath], then [imath]\alpha'[/imath] is a lower bound of [imath]kS[/imath] in [imath]\mathbb{R}[/imath], and [imath]\beta'\leq\alpha'[/imath] for all lower bounds [imath]\beta'[/imath] of [imath]kS[/imath] in [imath]\mathbb{R}[/imath]. [imath]\star\star\star\star\star\star\star\star\star\star[/imath] Here, an upper bound is an element [imath]\gamma\in\mathbb{R}[/imath] such that for all [imath]\zeta\in kS[/imath] we have [imath]\zeta\leq\gamma[/imath]. Further, a lower bound in this case is an element [imath]\gamma'[/imath] such that for all [imath]\zeta'\in kS[/imath] we have that [imath]\gamma'\leq \zeta'[/imath]. Comment: Please let me know if I have the foundational ideas here down, so I can begin working on this. I want to get it solid. Edit: So it turns out that this problem was pushed to the next homework set, so I get to keep working on it. So far this is what I've got: Consider the case where [imath]k=0[/imath]. This means [imath]kS=0S=\{0\}[/imath], and so we have that [imath]\sup(kS)=\sup\{0\}=0=0\sup(S)=k\sup(S).[/imath] Now consider the case where [imath]k>0[/imath]. Note that [imath]\sup(kS)[/imath] is the element in [imath]\mathbb{R}[/imath] such that [imath]\hspace{0.25cm}[/imath](a) [imath]\sup(kS)[/imath] is an upper bound for [imath]kS[/imath], and [imath]\hspace{0.25cm}[/imath](b) if [imath]\alpha[/imath] is any upper bound for [imath]kS[/imath], then [imath]\sup(kS)\leq \alpha[/imath], so if is sufficient to show that [imath]k\sup(S)[/imath] is the element in [imath]\mathbb{R}[/imath] such that [imath]\hspace{0.25cm}[/imath](a') [imath]k\sup(S)[/imath] is an upper bound for [imath]kS[/imath], and [imath]\hspace{0.25cm}[/imath](b') if [imath]\alpha[/imath] is any upper bound for [imath]kS[/imath], then [imath]k\sup(S)\leq \alpha[/imath], namely that properties (a) and (b) hold for [imath]\sup(kS)=k\sup(S)[/imath]. Thus, if [imath]k\sup(S)[/imath] is an upper bound for [imath]kS[/imath], then there is an [imath]s\in S[/imath] such that [imath]\beta=ks[/imath], where [imath]\beta\in kS[/imath]. This would mean [imath]s\leq \sup (S)[/imath] because [imath]\sup(S)[/imath] is an upper bound for [imath]S[/imath], which means [imath]\beta=ks\leq k\sup(S)[/imath], and so [imath]k\sup(S)[/imath] is an upper bound for [imath]kS[/imath], as [imath]\beta[/imath] was arbitrary. Now in order to show that if [imath]\alpha[/imath] is any upper bound for [imath]kS[/imath], then [imath]k\sup(S)\leq \alpha[/imath], first suppose that [imath]\alpha[/imath] is an upper bound for [imath]kS[/imath]. If this were the case, then [imath]ks\leq \alpha[/imath] for all [imath]s\in S[/imath], and so [imath]s\leq\frac{\alpha}{k}[/imath] for all [imath]s\in S[/imath], which therefore means that [imath]\frac{\alpha}{k}[/imath] is an upper bound for [imath]S[/imath]; hence [imath]\sup(S)\leq\frac{\alpha}{k}[/imath]. Multiplication by [imath]k[/imath] on both side of this inequality gives us that [imath]k\sup(S)\leq \alpha[/imath]; quod erat demonstrandum. Similarly, to show [imath]\inf(kS)=k\inf(S)[/imath] there are two case to consider, namely [imath]k=0[/imath], and [imath]k>0[/imath]. If [imath]k=0[/imath], then [imath]kS=0S=\{0\}[/imath], and so we have that [imath]\inf(kS)=\inf\{0\}=0=0\inf(S)=k\inf(S).[/imath] Suppose now that [imath]k>0[/imath]. In this case, just as above, is is sufficient to show that [imath]\hspace{0.25cm}[/imath](a) [imath]k\inf(S)[/imath] is a lower bound for [imath]kS[/imath], and [imath]\hspace{0.25cm}[/imath](b) if [imath]\alpha[/imath] is any lower bound for [imath]kS[/imath], then [imath]\alpha\leq k\inf(S)[/imath]. Thus if [imath]k\inf(S)[/imath] is a lower bound for [imath]kS[/imath], then there is an [imath]s\in S[/imath] such that [imath]\beta=ks[/imath], where [imath]\beta\in kS[/imath]. Therefore [imath]s\geq\inf(S)[/imath] because [imath]\inf(S)[/imath] is a lower bound for [imath]S[/imath], and so [imath]\beta=ks\leq k\inf(S)[/imath], which means [imath]k\inf(S)[/imath] is a lower bound for [imath]kS[/imath] since [imath]\beta[/imath] was arbitrary. Now to show (b), suppose that [imath]\alpha[/imath] is a lower bound for [imath]kS[/imath]. In this case we have that [imath]ks\leq\alpha[/imath] for all [imath]s\in S[/imath], and so [imath]s\leq\frac{\alpha}{k}[/imath] for all [imath]s\in S[/imath]; hence [imath]\frac{\alpha}{k} [/imath] is a lower bound for [imath]S[/imath], and so we have that [imath]\inf(S)\leq\frac{\alpha}{k}[/imath]. If we multiply by [imath]k[/imath] on both side of this inequality we get that [imath]k\inf(S)\leq \alpha[/imath]; quod erat demonstrandum. Comment: [imath]\color{green}{\text{Above}}[/imath] is the essence of the kind of proof I'd like to exhibit for the [imath]\color{red}{\text{second part}}[/imath]. I still haven't got the second part, so any help would be appreciated. Note: If you are from UCLA, or anywhere, please don't just copy and paste this as your answer.

487164

For what integer [imath]e[/imath] is a group [imath]G[/imath] necessarily Abelian? If [imath]e[/imath] is the smallest positive integer such that [imath]x^e=1[/imath] for every [imath]x[/imath] in group [imath]G[/imath], and if [imath]e=2[/imath], then it is easy to proof that [imath]G[/imath] is Abelian. [imath]\forall x,y\in G[/imath], [imath]x^2=y^2=(xy)^2=1 \rightarrow x^{-1}=x[/imath], [imath]y^{-1}=y[/imath] and [imath](xy)^{-1}=xy[/imath], so [imath]xy=(xy)^{-1}=y^{-1}x^{-1}=yx[/imath]. My question is, for what other integer [imath]e[/imath] is a group [imath]G[/imath] necessarily Abelian? If the integer [imath]p\ge2[/imath] such that [imath]x^p=1[/imath] for every [imath]x[/imath] in group [imath]G[/imath], by some similar steps, I can proof [imath](xy)^{p-1}=y^{p-1}x^{p-1}[/imath]. But I can not know for what integer p the G is also Abelian.

459250

Exponent and abelian groups What exponent [imath]e[/imath] can guarantee the group to be abelian? Are there any known results except the case [imath]e = 2[/imath]? Thanks for help.

487288

Cardinality/Set Theory For [imath]a,b \in \mathbb{R}[/imath] with [imath]a<b[/imath], find an explicit bijection of [imath]A=\{x: a < x < b\}[/imath] onto [imath]B=\{y: 0<y<1 \}[/imath] Can someone please help? I don't understand what is meant by 'explicit'. Would sinx, cosx or tanx classify?

367296

Let [imath](a,b)[/imath] and [imath](c,d)[/imath] be intervals in [imath]\Bbb R[/imath], and find an injective and surjective function from [imath](a,b)[/imath] to [imath](c,d)[/imath] So here is this question I got stuck on: Let [imath](a,b)[/imath], [imath](c,d)[/imath] be intervals (not sure if that's the correct term) on [imath]\Bbb R[/imath], so that [imath]a<b[/imath], [imath]c<d[/imath]. Find an injective and surjective function [imath]f:(a,b)\rightarrow(c,d)[/imath]. Thanks in advance! P.S. I get these question a lot. Is that some way of thinking, or some general method I should follow to make these question easier for me to solve?

485959

Primary decomposition of an ideal (exercise 7.8 in Reid, Undergraduate Commutative Algebra) I would like to understand how to use geometry to solve the problem 7.8 from Reid's book Undergraduate Commutative Algebra. The problem is the following Let [imath]k[/imath] be a field and consider the ideal [imath]I = (xy, x - yz) \subset k[x,y,z][/imath]. Find a primary decomposition of [imath]I[/imath]. So the vanishing locus [imath]V(I)[/imath] is the two axis [imath]Y[/imath] and [imath]Z[/imath], and if [imath]I = q_1 \cap \cdots \cap q_n[/imath] with all [imath]q_i[/imath] being [imath]p_i[/imath]-primary, we could guess that we can take [imath]p_1 = (x,z)[/imath] and [imath]p_2 = (x,y)[/imath]. Moreover, the vanishing of [imath]xy[/imath] is the [imath]Z[/imath]-axis, but with double multiplicity (is this true ?), so we could guess to take [imath]q_2 = (x,y)^2[/imath], but we need [imath]x - yz[/imath] to be in there, so I think that [imath]q_2 = (xy, y^2, x - yz)[/imath] and [imath]q_1 = (x,z)[/imath] would give a decomposition into primary ideals ([imath]x^2[/imath] is automatically in [imath]q_2[/imath], don't need it). Question 1 : How to make sure that these primes [imath]p_1[/imath] and [imath]p_2[/imath] are the only ones ? We can compute them by [imath]\operatorname{Ass}(k[x,y,z]/(xy,x - yz))[/imath], but wasn't able to do it. Question 2 : Is [imath](x,z) \cap (xy, y^2, x - yz)[/imath] a primary decomposition of [imath]I[/imath] ? Question 3 : What is a better way to get to this answer with geometric intuition and not algebraic brute-force ? Thanks

385001

What is a primary decomposition of the ideal [imath]I = \langle xy, x - yz \rangle[/imath]? Given the ring [imath]k[x,y,z][/imath], where [imath]k[/imath] is a field, and an ideal [imath]I=(xy,x-yz)[/imath], find a primary decomposition of [imath]I[/imath]. I tried to draw the graph of the variety of [imath]I[/imath] and get a decomposition of [imath](x,y)\cap(x,z)[/imath] (the two prime ideals corresponds to the irreducible components of the variety), but apparently these two ideals are not the primary decomposition of [imath]I[/imath], and I don't know how to fix this. I would really appreciate it if someone can help me. thx

487702

Set of commuting Matrices [imath]\Rightarrow [/imath] Common Eigenvector I am trying to prove that if we have an arbitrary set of commuting matrices in [imath]M_n(\mathbb C)[/imath] then they have a common eigenvector. Well, if we have only 2 matrices, the answer is easy and it has been already answered here: Matrices commute if and only if they share a common basis of eigenvectors? (acutally in this case we find a common eigenbasis, in case they are diagonalizable) For the general case of an arbitrary set I am trying to find a subspace of [imath]M_n(\mathbb C) [/imath] which contains that arbitrary set of commuting matrices. Then Prove the statement for the basis ( is going to have [imath]\leq n^2[/imath] elements) by induction. So if [imath]Y [/imath] is that arbitary set, take [imath]span(Y)[/imath] which has dimension [imath]\leq n^2[/imath] so find a basis [imath]Y_1 ,Y_2,\dots Y_k,k\leq n^2[/imath] Then prove by induction that we can find a common eigenvector [imath]v\neq 0 [/imath] for the vectors of the basis. So, if [imath]A\in Y\Rightarrow [/imath] [imath]A=\sum_{m=1}^k a_mY_m\Rightarrow Av=\sum_{m=1}^k a_mY_mv\Rightarrow[/imath] [imath]Av=\sum_{m=1}^k \left( a_m\lambda_m\right) v[/imath] i.e. [imath]v[/imath] is an eigenvector for [imath]A[/imath]. What do you think about that solution? P.S. I asked if you could check my solution. I have already seen the post which is supposed to answer my question but I am not satisfied ( see comment).

197331

How do I prove that in every commuting family there is a common eigenvector? The proof given by my textbook is highly non-satisfying. The author adopted some magic-like "reductio ad absurdum" and the proof (although is correct) didn't reveal the nature of this problem. I made my own effort into it and tried a different approach. Yet I can't finish it. Let [imath]\mathscr{F}[/imath] be a commuting family in [imath]M_n(\mathbb{C}^n)[/imath], and [imath]A\in\mathscr{F}[/imath], then [imath]A[/imath] has [imath]n[/imath] eigenvalues. We pick one, say [imath]\lambda[/imath]. Let [imath]x[/imath] be one of its eigenvector. We can easily prove that, if [imath]A[/imath] has no other eigenvector with eigenvalue [imath]\lambda[/imath] that linearly independent with [imath]x[/imath], which means that [imath]\{cx|c\in\mathbb{C}\}[/imath] are the only vectors satisfying [imath]Ax=\lambda x[/imath], then [imath]x[/imath] is a common eigenvector. Because [imath]\forall B\in\mathscr{F}[/imath] and [imath]\forall y\in \{cx|c\in\mathbb{C}\}[/imath], [imath]A(Bx)=ABx=BAx=B(Ax)=B(\lambda x)=\lambda (Bx)[/imath], so that [imath]Bx[/imath] has to be in [imath]\{cx|c\in\mathbb{C}\}[/imath], that is, [imath]Bx=c_0x[/imath] for some [imath]c_0\in\mathbb{C}[/imath], which means [imath]x[/imath] is a eigenvector of [imath]B[/imath] too. But what if there are vectors satisfying [imath]Ax=\lambda x[/imath] that's not in [imath]\{cx|c\in\mathbb{C}\}[/imath]? Well, then we should have a set of linearly independent eigenvectors [imath]\{x_1,x_2,...,x_k\}[/imath], that [imath]\{c_1x_1+c_2x_2+...+c_kx_k|c_i\in\mathbb{C}\}[/imath] are the only vectors satisfying [imath]Ax=\lambda x[/imath]. Now, I have a reasonable hypothesis that there exists some [imath]x=c_1x_1+c_2x_2+...+c_kx_k[/imath], that can be proven to be a common eigenvector of [imath]\mathscr{F}[/imath]. I've tried some approaches to prove it but all failed. Do you guys believe it's true? And if it is true then how do I prove it?

487309

Some property of an ideal in commutative ring Let [imath]R[/imath] be a Dedekind domain, and let [imath]\mathfrak{m}[/imath] be a maximal ideal in [imath]R[x][/imath] is of the form [imath]\mathfrak{m} = (\mathfrak{p},f(x))[/imath] where [imath]\mathfrak{p}[/imath] is a maximal ideal in [imath]R[/imath], and [imath]f[/imath] is a polynomial in [imath]R[x][/imath] which is irreducible in [imath](R/\mathfrak{p})[x][/imath]. Let [imath]g(x) \in \mathfrak{m}[/imath]. I am know that exist a ideal [imath]\mathfrak{a}[/imath] in [imath]R[/imath] such that [imath](\mathfrak{p},\mathfrak{a}) = 1 [/imath], i.e [imath]\mathfrak{p} + \mathfrak{a} = R[/imath] and [imath]\mathfrak{p}\mathfrak{a} = \mathfrak{p}\cap \mathfrak{a} = (p)[/imath]. (chapter 16, corollary 19, page 768, Abstract Algebra, Dummit Foote...) Easily, since [imath]g(x)\in \mathfrak{m}[/imath], we have [imath] g(x) = l(x) + f(x).k(x)[/imath] where [imath]l(x) \in \mathfrak{p}[x][/imath]. The question is: Can we make a "strong" form as show that there exists an ideal [imath]\mathfrak{a}[/imath] in [imath]R[/imath] such that [imath]g[/imath] can be show as the form [imath]ag(x) = p h(x) + a f(x) k(x)[/imath] where [imath]p\in \mathfrak{p}[/imath] such that [imath](p) = \mathfrak{p} \mathfrak{a}[/imath], [imath](\mathfrak{p},\mathfrak{a}) = 1[/imath] and [imath]a\in \mathfrak{a} - \mathfrak{a}\mathfrak{p}[/imath].

480724

Maximal ideal in commutative ring Let [imath]R[/imath] be a Dedekind domain, [imath]\mathfrak{m}[/imath] be a maximal ideal in [imath]R[x][/imath] is of the form [imath]\mathfrak{m} = (\mathfrak{p},f(x))[/imath] where [imath]\mathfrak{p}[/imath] is a maximal ideal in [imath]R[/imath], and [imath]f[/imath] is a polynomial in [imath]R[x][/imath] which is irreducible in [imath](R/\mathfrak{p})[x][/imath]. Let [imath]g(x) \in \mathfrak{m}[/imath]. I am know that exist a ideal [imath]\mathfrak{a}[/imath] in [imath]R[/imath] such that [imath](\mathfrak{p},\mathfrak{a}) = 1 [/imath], i.e [imath]\mathfrak{p} + \mathfrak{a} = R[/imath] and [imath]\mathfrak{p}\mathfrak{a} = \mathfrak{p}\cap \mathfrak{a} = (p)[/imath]. (chapter 16, corollary 19, page 768, Abstract Algebra, Dummit Foote...) Easily, since [imath]g(x)\in \mathfrak{m}[/imath], we have [imath] g(x) = l(x) + f(x).k(x)[/imath] where [imath]l(x) \in \mathfrak{p}[x][/imath]. The questions is: Can we make a "strong" form as to show that there exists an ideal [imath]\mathfrak{a}[/imath] in [imath]R[/imath] such that [imath]g[/imath] can be show as the form [imath]ag(x) = p h(x) + a f(x) k(x)[/imath] where [imath]p\in \mathfrak{p}[/imath] such that [imath](p) = \mathfrak{p} \mathfrak{a}[/imath], [imath](\mathfrak{p},\mathfrak{a}) = 1[/imath] and [imath]a\in \mathfrak{a} - \mathfrak{a}\mathfrak{p}[/imath].

488199

Does the formula [imath](x^n)' = nx^{n-1}[/imath] fail for some value of [imath]n \in \mathbb{Z}[/imath]? It is well known that the formula [imath]\int x^n dx = \frac{x^{n+1}}{n+1}+C[/imath] breaks down at [imath]n=-1[/imath] (we get a logarithm). My question is, does there exist a value of [imath]n \in \mathbb{Z}[/imath] for which the formula [imath]\frac{d}{dx}x^n = nx^{n-1}[/imath] breaks down?

318555

>Prove that [imath]\frac d {dx} x^n=nx^{n-1}[/imath] for all [imath]n \in \mathbb R[/imath]. Prove that [imath]\frac d {dx} x^n=nx^{n-1}[/imath] for all [imath]n \in \mathbb R[/imath]. I saw some proof of [imath]\frac d {dx} x^n=nx^{n-1}[/imath] using binomial theorem, which is only available for [imath]n \in\mathbb N[/imath]. Do anyone have the proof of [imath]\frac d {dx} x^n=nx^{n-1}[/imath] for all real [imath]n[/imath]? Thank you.

488186

Math inequality proof If [imath]a, b[/imath] are positive real numbers and [imath]a + b = 1[/imath], prove that [imath] \left(a +\frac{1}{a}\right)^2 + \left(b +\frac{1}{b}\right)^2 \geq \frac{25}{2} [/imath] Thank you.

487486

Proving inequality [imath](a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}[/imath] for [imath]a+b=1[/imath] If [imath]a, b[/imath] are positive real numbers and [imath]a+b = 1[/imath], prove that : [imath]\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}[/imath] I can see that the value [imath]\frac{25}2[/imath] is attained for [imath]a=b=\frac12[/imath]. But I do not know how to show that this is the minimal possible value. Thank you.

488783

uniqueness of solutions of [imath]ax=b[/imath] and [imath]ya=b[/imath] in a semigroup . Suppose [imath]G[/imath] is a semigroup in which every equation of the form [imath]ax=b[/imath] or [imath]ya=b[/imath] has a solution. Does this solution have to be unique?

346595

Prove that [imath](G, \circ)[/imath] is a group if [imath]a\circ x = b[/imath] and [imath]x\circ a = b[/imath] have unique solutions I have some difficulties with a task in algebra. I guess it's trivial and really easy but I can't figure out how to solve it. I have a set [imath]G[/imath] and a binary operation on it, let it be [imath]\circ[/imath]. I have that the operation is associative and that the equations [imath]a\circ x = b[/imath] and [imath]x\circ a = b[/imath] have unique solutions. I have to prove that [imath](G, \circ)[/imath] is a group. I already have that the operation is binary and associative, so I have to prove that there is unique identity element and unique inverse element and it will come from the equations, but how exactly?

488982

[imath](0,1)[/imath] is an open subset of [imath]\Bbb R[/imath] but not of [imath]\Bbb R^2[/imath] when we think of [imath]\Bbb R[/imath] as the [imath]x[/imath]-axis in [imath]\Bbb R^2[/imath]. [imath](0,1)[/imath] is an open subset of [imath]R[/imath] but not of [imath]R^2[/imath] when we think of [imath]R[/imath] as the x-axis in [imath]R^2[/imath]. How do I prove the above statement?

481667

[imath](0,1)[/imath] is an open subset of [imath]\mathbb{R}[/imath] but not of [imath]\mathbb{R}^2[/imath], when we think of [imath]\mathbb{R}[/imath] as the x-axis in [imath]\mathbb{R}^2[/imath]. Prove this. It's problem #1 of Chapter 2 in Pugh's book. I think I got the first part by taking [imath]r=1[/imath] and pointing out that [imath]d(p,q) < 1 \implies q \in (0,1)[/imath] for all [imath]p \in (0,1)[/imath] but I have no idea how to show it isn't a subset of [imath]\mathbb{R}^2[/imath]. I tried proof by contradiction and got no where. I've been working at this problem for hours and not getting anywhere. How should I do it?

488983

[imath](\Bbb Z[x]/(x^{n+1}))^\times\cong\Bbb Z/2\Bbb Z \times \prod _{i=1}^n \Bbb Z[/imath] I'm trying to prove the group isomorphism [imath](\Bbb Z[x]/(x^{n+1}))^\times\cong\Bbb Z/2\Bbb Z \times \prod _{i=1}^n \Bbb Z[/imath]. Obviously I tried to establish a ring isomorphism from [imath]\Bbb Z[x]/(x^{n+1})[/imath] to some ring [imath]R[/imath], a direct product of easier rings, and prove that [imath]R^\times[/imath] equals the RHS of the original isomorphism. In the solutions of similar problems I've seen, one defines a surjective ring homomorphism [imath]\phi : R^\prime[x]\rightarrow R[/imath] of substituting [imath]x[/imath] with an element of [imath]R[/imath] such that the kernel of [imath]\phi[/imath] equals the dividing ideal, and uses the fundamental homomorphism theorem and CRT to show the ring isomorphism. However this doesn't work for this particular problem; since [imath]x^{n+1}[/imath] has a (unique) multiple root, the kernel of homomorphism of plugging in does not equal to the dividing ideal. Moreover, since [imath]\Bbb Z[x][/imath] is not an Euclidean domain, I have no idea what the units of [imath]\Bbb Z[x][/imath] modulo some ideal are like. I would appreciate your help.

483642

Show that [imath](\mathbb{Z}[x]/(x^{n+1}))^{\times}\cong \mathbb{Z}/2\mathbb{Z}\times\Pi_{i=1}^n\mathbb{Z}[/imath] Show that [imath](\mathbb{Z}[x]/(x^{n+1}))^{\times}\cong \mathbb{Z}/2\mathbb{Z}\times\Pi_{i=1}^n\mathbb{Z}[/imath]. Anyway how and what method is used to prove this. I still have no idea now. really thanks for your help

490067

Proving a graph must be connected Let [imath]G[/imath] be a graph with [imath]n[/imath] vertices and [imath]e[/imath] edges such that [imath]e>\binom{n-1}{2}[/imath]. Then [imath]G[/imath] must be connected. As usual, hints would be greatly appreciated. If it where[imath]\binom{n}{2}[/imath] then wouldn't everything be adjacent to itself? I'll try to answer the question myself at the end if I'm able to.

34212

Show that if [imath]G[/imath] is simple a graph with [imath]n[/imath] vertices and the number of edges [imath]m>\binom{n-1}{2}[/imath], then [imath]G[/imath] is connected. I'm trying to pick up a little graph theory out of Bondy and Murty's Graph Theory as suggested here. Problem 1.1.12 has given me a little hitch. Let [imath]G[/imath] be a simple graph of order [imath]n[/imath] and size [imath]m[/imath]. (So there are [imath]n[/imath] vertices and [imath]m[/imath] edges). If [imath]m>\binom{n-1}{2}[/imath], then [imath]G[/imath] is connected. I'm following a hint in an appendix which says if [imath]G[/imath] is not connected, we can partition the vertices into parts [imath](X,Y)[/imath] such that no edge joins a vertex in [imath]X[/imath] to a vertex in [imath]Y[/imath]. What is the largest number of edges in [imath]G[/imath] if [imath]|X|=r[/imath] and [imath]|Y|=n-r[/imath]? I suppose the graphs on [imath]X[/imath] and [imath]Y[/imath] are then complete graphs, for a total of [imath]\binom{r}{2}+\binom{n-r}{2}[/imath] edges. Simplifying, this is [imath]\frac{2r^2+n^2-2rn-n}{2}[/imath], so I'm trying to show [imath] \frac{2r^2+n^2-2rn-n}{2}\leq\frac{(n-1)(n-2)}{2} [/imath] to get the contrapositive. The above is equivalent, if my algebra is correct, to showing [imath]r\geq\frac{n-1}{n-r}[/imath] for any [imath]0\lt r\lt n[/imath]. This seems reasonable, but I can't quite show it. How would I do so? Thanks. I'd also be happy to see a solution that doesn't necessarily make use of the hint.

490127

A group that needs to be proven by definition? Let X be a set and let [imath]P(X)[/imath] denote the set of all subsets of X, that is [imath]P(X)= \{A|A\subseteq X\}[/imath]. given [imath]A,B\in P(X),[/imath] define [imath]A \triangle B:=(A\setminus B)\cup(B\setminus A)[/imath]. Prove that [imath](P(X), \triangle)[/imath] is a group. Using the definition of a group I assume I have to show that there exist an associativity, identity, and inverse.

267708

Does the Symmetric difference operator define a group on the powerset of a set? [imath]G[/imath] is the set of all subsets of a set [imath]A[/imath], under the operation of [imath]\triangle\;[/imath]: Symmetric Difference of sets. [imath]A[/imath] has at least two different elements. I need to check if this is a group, and if it does to show if the group is abelian and/or finite. Associativity - easy from Symmetric Difference. Identity element - empty group. Inverse element - each element is inverse to itself. Am I right? abelian? finite?

490146

How do I solve [imath]\frac{-1}{3}|6−5x|+2<1[/imath]? How do i solve this problem? [imath]\frac{-1}{3}|6−5x|+2<1[/imath] so far I took the 2 to the other side, multiplied both sides by -3 to cancel out the -1/3 leaving me with [imath]|6-5x|>3[/imath] is this correct? by flipping the signs since i multiplied by a negative?but then what?

490058

How do I solve these absolute value equations? How do I solve these? [imath]|-2w-4|<12[/imath] and [imath](-1/3)|6-5x|+2<1[/imath] for the first problem i got [imath]-8<w<4[/imath] is that right?

490340

Determining all subgroups of [imath]S_3[/imath] Related to : How would I prove what elements [imath]S_3[/imath] contains, and what its subgroups are? I don't know Lagrange's theorem, and I don't want to prove that there are no subgroups of size [imath]4[/imath] or [imath]5[/imath], is there a better way to argue that there aren't any, with basic abstract algebra knowledge?

490294

How would I prove what elements [imath]S_3[/imath] contains, and what its subgroups are? Given that [imath]S_3[/imath] is a symmetric group of size three, how would I find all elements of it, and all subgroups?

355049

Let [imath]A[/imath] be any uncountable set, and let [imath]B[/imath] be a countable subset of [imath]A[/imath]. Prove that the cardinality of [imath]A = A - B [/imath] I am going over my professors answer to the following problem and to be honest I am quite confused :/ Help would be greatly appreciated! Let [imath]A[/imath] be any uncountable set, and let [imath]B[/imath] be a countable subset of [imath]A[/imath]. Prove that [imath]|A| = |A - B|[/imath] The answer key that I am reading right now follows this idea: It says that [imath]A-B[/imath] is infinite and proceeds to define a new denumerable subset [imath]A-B[/imath] as [imath]C[/imath]. Of course since [imath]C[/imath] is countably infinite then we can write [imath]C[/imath] as [imath]{c1,c2,c3...}[/imath] Once we have a set [imath]C[/imath], we know that the union of [imath]C[/imath] and [imath]B[/imath] must be denumerable (from another proof) since [imath]B[/imath] is countable and [imath]C[/imath] is denumerable. This is where I start to have trouble. The rest of the solution goes like this... Since the union of [imath]C[/imath] and [imath]B[/imath] is denumerable, there is a bijective function [imath]f[/imath] that maps the union of [imath]C[/imath] and [imath]B[/imath] to [imath]C[/imath] again. The solution then proceeds to define another function [imath]h[/imath] that maps [imath]A[/imath] to [imath]A-B[/imath]. I am just so lost. The thing is I don't even understand the point of constructing a new subset [imath]C[/imath] or defining functions like [imath]f[/imath] or [imath]h[/imath]. So I suppose my question is in general, how would one approach this problem? I am not mathematically inclined unfortunately, and a lot of the steps in almost all of these problems seems arbitrary and random. Help would be really appreciated on this problem and some general ideas on how to solve problems like these!!! Thank you so very much!

948021

If [imath]B[/imath] is an uncountable set and [imath]A[/imath] is a countable set, then prove that [imath]B[/imath] is similar to [imath]B-A[/imath]. If [imath]B[/imath] is an uncountable set and [imath]A[/imath] is a countable set, then prove that [imath]B[/imath] is similar to [imath]B-A[/imath]. Attempt: Two sets [imath]A[/imath] and [imath]B[/imath] are called similar [imath]\iff[/imath] thee exists a one to one function [imath]F[/imath] whose domain is the set [imath]A[/imath] and whose range is the set [imath]B[/imath]. [imath]B[/imath] is an uncountable set and [imath]A[/imath] is a countable set, then [imath]B-A[/imath] must also be uncountable and hence, an infinite set. To prove this, let us suppose [imath]B-A[/imath] is countable. Since, [imath]A[/imath] is countable, hence, [imath](B-A) \cup A[/imath] should also be countable. But [imath]B \subseteq (B-A) \cup A[/imath] should be countable as well which is a contradiction. Hence, [imath]B-A[/imath] is an uncountable set as well. Now, we need to show that [imath](B-A) \sim B[/imath]. How do I move ahead? I think we need to define a one-one onto function from [imath](B-A) \rightarrow B[/imath] but I am not able to think of such a function. Thank you for your help.

491412

Probability and perfect squares An integer [imath]n[/imath] is randomly chosen from [imath]1[/imath] to [imath]k^2[/imath], where [imath]k[/imath] is an integer. What is the probability that [imath]n[/imath] is a perfect square? I know I have to first figure out the probability of getting a perfect square and then take the complement of this. How would I go about? Thank you!!

490205

Rational, integers and probability Consider [imath]\sqrt{n}[/imath], where [imath]n[/imath] is an integer and [imath]1\le n \le 10[/imath]. How many of these ten numbers are irrational? An integer [imath]n[/imath] is randomly chosen from [imath]1[/imath] to [imath]k^2[/imath], where [imath]k[/imath] is an integer. What is the probability that [imath]n[/imath] is a perfect square? For the first one I had figured out it was [imath]7[/imath] irrationals numbers but did not know how to prove the second question

492181

How to show this equality of [imath]\mu[/imath] since we know that X and Y are independent, then the pair [imath](XY^c)[/imath], [imath](X^cY)[/imath], [imath](X^cY^c)[/imath] also independent.

491926

How to show this equality of [imath]\rho(X,Y)[/imath] Let I have [imath](S,\Sigma,\mu)[/imath] be a probability space, then [imath]X,Y \in \Sigma[/imath]. Define [imath]\rho (X,Y)[/imath] by [imath]\rho (X,Y)[/imath] = correlation between random variable [imath]I_X[/imath] and [imath]I_Y[/imath], where [imath]I_X[/imath] and [imath]I_Y[/imath] are the indicator function of [imath]X[/imath] and [imath]Y[/imath]. Express [imath]\rho (X,Y)[/imath] in term of [imath]\mu (X)[/imath], [imath]\mu (Y)[/imath], [imath]\mu(XY)[/imath]. Conclude that [imath]\rho(X,Y)=0[/imath], if and only if [imath]X[/imath] and [imath]Y[/imath] are independent. How to show that: [imath]\rho(X,Y)= \frac{\mu(XY)\,\mu(X^cY^c)-\mu(XY^c)\,\mu(X^cY)}{(\mu(X) \,\mu(X^c)-\mu(Y) \,\mu(Y^c))^{1/2}};0<\mu(X)<1;\,0<\mu(Y)<1[/imath]

492257

What are some good subtly incorrect proofs of obviously incorrect results? I'm interested in compiling a list of proofs that look logically correct at a glance, but "prove" something obviously incorrect. Here are some examples. [imath]e^{i \pi} = -1[/imath] [imath]e^{2i\pi} = 1[/imath] [imath]2i\pi = \ln 1[/imath] [imath]2i\pi = 0[/imath] [imath]-4\pi^2 = 0[/imath] Let [imath]a = b[/imath]. Then: [imath]a = b[/imath] [imath]a^2 = ab[/imath] [imath]a^2 - b^2 = ab - b^2[/imath] [imath](a+b)(a-b) = b(a-b)[/imath] [imath]a + b = b[/imath] [imath]a = 0[/imath] The Two Envelopes Paradox: You are on a game show, in which the host fills two indistinguishable envelopes with random amounts of money, such that one envelope contains [imath]x[/imath] dollars and the other contains [imath]10x[/imath] dollars. You pick an envelope at random, but then you are offered a chance to switch envelopes (intuitively, it shouldn't matter whether or not you choose to switch). You reason: there is a 50/50 chance that I currently hold the higher-valued or lower-valued envelope in my hands. If I keep this envelope, my expected return is [imath]x[/imath]. If I switch, then my expected return is [imath]\frac{1}{2}(\frac{1}{10}x) + \frac{1}{2}(10x) = 5.05x[/imath]. Therefore, I should switch. Any other good ones?

113

What are some classic fallacious proofs? If you know it, also try to include the precise reason why the proof is fallacious. To start this off, let me post the one that most people know already: Let [imath]a = b[/imath]. Then [imath]a^2 = ab[/imath] [imath]a^2 - b^2 = ab - b^2[/imath] Factor to [imath](a-b)(a+b) = b(a-b)[/imath] Then divide out [imath](a-b)[/imath] to get [imath]a+b = b[/imath] Since [imath]a = b[/imath], then [imath]b+b = b[/imath] Therefore [imath]2b = b[/imath] Reduce to [imath]2 = 1[/imath] As @jan-gorzny pointed out, in this case, line 5 is wrong since [imath]a = b[/imath] implies [imath]a-b = 0[/imath], and so you can't divide out [imath](a-b)[/imath].

448098

If [imath]ax^2-bx+c=0[/imath] has two distinct real roots lying in the interval [imath](0,1)[/imath] [imath]a,b,c[/imath] belongs to natural prove that [imath]\log_5 {abc}\geq2[/imath] If [imath]ax^2-bx+c=0[/imath] has two distinct real roots lying in the interval [imath](0,1)[/imath] with [imath]a, b, c\in \mathbb N[/imath], prove that [imath]\log_5 {abc}\geq2[/imath]. The equations I could form are: 1) [imath]f(0)>0[/imath] and [imath]f(1)>0[/imath] 2) [imath]\frac{b}{2a}[/imath] lies between [imath]0[/imath] and [imath]1[/imath], because: [imath]\frac{b}{2a}-\frac{\sqrt{\Delta}}{2a}<\frac{b}{2a}<\frac{b}{2a}+\frac{\sqrt{\Delta}}{2a} [/imath]. 3) [imath]\Delta=b^2-4ac > 0[/imath]

2619761

If [imath]ax^2+bx+c=0[/imath] has [imath]2[/imath] different solutions in [imath](0,1)[/imath] then prove [imath]a\geq 5[/imath]. Say [imath]a,b,c[/imath] are integers, [imath]a>0[/imath]. Suppose [imath]ax^2+bx+c=0[/imath] has [imath]2[/imath] different solutions in [imath](0,1)[/imath] then prove [imath]a\geq 5[/imath]. Find an example for [imath]a=5[/imath]. I am struggling with this for some time with no success. I try Vieta's formula [imath]0<x_1+x_2 = -{b\over a}<2[/imath] and [imath]x_1x_2 ={c\over a}<1[/imath] so [imath]c<a[/imath] and [imath]0<-b<2a[/imath].

492788

Finding two rationals that satisfy an equation So the homework question goes as follows: Find rational numbers such that [imath] \sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}[/imath] The homework only asks for one pair of [imath]\alpha[/imath] and [imath]\beta[/imath] and it is almost trivial to find the solution [imath]\alpha=\beta = 1[/imath] Solution: [imath]\sqrt[3]{7+5\sqrt{2}}=\alpha+\beta\sqrt{2}[/imath] [imath]7+5\sqrt{2}=(\alpha+\beta\sqrt{2})^3[/imath] [imath]7+5\sqrt{2}=\alpha^3+6\beta^2\alpha +\sqrt{2}(3 \alpha^2 \beta +2\beta^3)[/imath] By identification of coefficients: [imath]7=\alpha^3+6\beta^2\alpha \space and \space 5=3 \alpha^2 \beta +2\beta^3[/imath] So I just "saw" that [imath]\alpha=\beta = 1[/imath] is a solution My question is, how do we find other solutions? Thank you!

491464

Finding Rational numbers Please help with the following question: Find rational numbers a and b such that: [imath]\left(7 + 5\sqrt2\right)^{\frac13} = a + b \sqrt2[/imath] Thank you

492984

Existence of certain non-commutative finite monoid Is there a finite monoid with two elements [imath]a[/imath] and [imath]b[/imath] such that [imath]ab = e[/imath] but [imath]ba \neq e[/imath]?

216187

In a monoid, does [imath]x \cdot y=e[/imath] imply [imath]y \cdot x=e[/imath]? A monoid is a set [imath]S[/imath] together with a binary operation [imath]\cdot:S \times S \rightarrow S[/imath] such that: The binary operation [imath]\cdot[/imath] is associative, that is, [imath](a\cdot b) \cdot c=a\cdot (b \cdot c)[/imath] for all [imath]a,b,c \in S[/imath]. There is an identity element [imath]e \in S[/imath], that is, there exists [imath]e \in S[/imath] such that [imath]e \cdot a=a[/imath] and [imath]a \cdot e=a[/imath] for all [imath]a \in S[/imath]. Question: Suppose, [imath]x,y \in S[/imath] such that [imath]x \cdot y=e[/imath]. Does [imath]y \cdot x=e[/imath]? This question was motivated by the question here, where the author attempts to prove a special case of the above in the context of matrix multiplication. It was subsequently proved, but the proofs require the properties of the matrix. I attempted to use Prover9 to prove the statement. Here's the input: formulas(assumptions). % associativity (x * y) * z = x * (y * z). % identity element a x * a = x. a * x = x. end_of_list. formulas(goals). x * y = a -> y * x = a. end_of_list. and it returned sos_empty, which, I guess, implies that no proof of the above statement is possible from the axioms of monoids alone. I ran Mace4 on the same input, and found no counter-examples for monoids of sizes [imath]1,2,\ldots,82[/imath]. A comment by Martin Brandenburg here regarding K-algebras might also apply here. For example, the property might be true for finite monoids, but not all infinite monoids. A counter-example would (obviously) need to be non-commutative.

487195

Is there a bijective continuous function [imath]f: \Bbb R \to \Bbb R[/imath] whose inverse [imath]f^{-1}[/imath] is not continuous? Looking at the definition of an homeomorphism, this question came to my mind.

960299

Example of a continuous bijection on [imath]\mathbb R^n[/imath] whose inverse is not continuous For [imath]n \ge 2[/imath] give example of a bijective continuous map [imath]f: \mathbb R^n \to \mathbb R^n[/imath] whose inverse is not continuous ; example of such a function is also an example of Does there exist a bijection of [imath]\mathbb{R}^n[/imath] with itself such that the forward map is connected but the inverse is not?

493435

Finding [imath]\lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}[/imath] if it exists Does there exist the following limitation? If the answer is yes, could you show me how to find that? [imath]\lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}[/imath] In the following, I'm going to write what I've done. By Taylor expansion, [imath]e^x=\sum_{k=0}^n \frac{x^k}{k!}+\frac{1}{n!}\int_{0}^x (x-t)^ne^tdt.[/imath] Letting [imath]x=n[/imath], [imath]e^n=\sum_{k=0}^n \frac{n^k}{k!}+\frac{1}{n!}\int_{0}^n (n-t)^ne^tdt.[/imath] Dividing the both sides by [imath]e^n[/imath], [imath]1=e^{-n}\sum_{k=0}^n \frac{n^k}{k!}+\frac{e^{-n}}{n!}\int_{0}^n (n-t)^ne^tdt.[/imath] Hence, I know that what I need to do is to find the following limitation: [imath]\lim_{n\to\infty}\frac{e^{-n}}{n!}\int_{0}^n (n-t)^ne^tdt[/imath] However, I'm facing difficulty. I need your help.

121099

Limit using Poisson distribution Show using the Poisson distribution that [imath]\lim_{n \to +\infty} e^{-n} \sum_{k=1}^{n}\frac{n^k}{k!} = \frac {1}{2}[/imath]

493601

where is the mistake with my fake proof? I tried to show that [imath]\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac {(2n)!\pi}{2^{2n}(n!)^2}[/imath] using contour integration so [imath]\int_{C} \frac{dz}{(z^2+1)^{n+1}}=\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}} +\int_{\gamma}..=2\pi i\frac{1}{n!}Res_{z=i}\frac{1}{(z^2+1)^{n+1}}[/imath] [imath]\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}} +0=2\pi i\frac{1}{n!}Res_{z=i}\frac{1}{(z^2+1)^{n+1}}[/imath] [imath]\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac{2\pi i}{n!}(\frac {d^n}{dz^n}(z+i)^{-n-1})_{z=i}[/imath] [imath]\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac{2\pi i}{n!}(-1)^np(n+1,n)(2i)^{-2n-1}[/imath] [imath]\int_{-\infty}^{+\infty} \frac{dx}{(x^2+1)^{n+1}}=\frac{\pi }{2^n}[/imath] where is [imath]p(n,r)=(n)(n-1)......(n-r+1)[/imath] so where is the mistake !

432849

Contour integration of [imath]\int \frac{dx} {(1+x^2)^{n+1}}[/imath] I want to compute [imath] \int_{-\infty}^\infty \frac 1{ (1+x^2)^{n+1}} dx [/imath] for [imath]n \in \mathbb N_{\geq 1}[/imath]. If I let [imath] f(z) := \frac 1 {(z+i)^{n+1}(z-i)^{n+1}} [/imath] then I see that [imath]f[/imath] has poles of order [imath]n+1[/imath] in [imath]\pm i[/imath]. Initially I thought that [imath] [-R,R] \cup \{R \exp(i \theta) \mid \theta \in [0,\pi] \} [/imath] would be a good conture but for the residue in [imath]i[/imath] I get [imath](-1)^n (n+1) 2 i[/imath]. I know that the result must be [imath] \frac {1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdots 2 n} \pi [/imath] which looks quite different than my residue. What goes wrong here ?

493711

Show that [imath]n>(\log n)^k[/imath] for sufficiently large values of [imath]n[/imath] How do you show that [imath]n>(\log n)^k[/imath], [imath]k \in Z_+[/imath] for sufficiently large values of [imath]n[/imath]? This is a part of a larger problem that I want to solve, so I would be thankful if someone could show me the method.

164243

How to find [imath]\lim\limits_{n\rightarrow \infty}\frac{(\log n)^p}{n}[/imath] How to solve [imath]\lim_{n\rightarrow \infty}\frac{(\log n)^p}{n}[/imath]

258776

Set of limit points of a subset of a Hausdorff space is closed. Let [imath]X[/imath] be a Hausdorff space and [imath]A\subset X[/imath]. Define [imath]A'=\{x\in X\mid x\text{ is a limit point of }A\}[/imath]. Prove that [imath]A'[/imath] is closed in [imath]X[/imath]. Relevant information: (1.) Every neighborhood of a point [imath]x\in A'[/imath] contains a point [imath]y\in A'[/imath] distinct from [imath]x[/imath] (in fact, in a Hausdorff space, every neighborhood of [imath]x[/imath] contains [imath]\infty[/imath]-many points of [imath]A[/imath] distinct from [imath]x[/imath]) (2.) In a Hausdorff space, every sequence has a unique limit. On first glance, it should be an easy proof, but I've made little progress. I was planning on showing [imath]\overline{A'}=A'[/imath]. Firstly, [imath]A'\subset \overline{A'}[/imath] trivially. To show [imath]\overline{A'}\subset A'[/imath], proceed by contradiction. Assume there exists [imath]x\in\overline{A'}[/imath] such that [imath]x\not\in A'[/imath]. This should yield an easy contradiction but I don't see it. In particular, I'm unsure if the fact that [imath]x\in\overline{A'}[/imath] implies that there actually exists a sequence in [imath]A'[/imath] converging to [imath]x[/imath]. By (2) we know all sequences have unique limits, but do we know that elements in the closure are limits of sequences? If this is true, it should yield an easy contradiction. Any help?

1287933

The set of limit points in a [imath]T_1[/imath] space is closed Prove the set of limit points in a [imath]T_1[/imath] space is closed. Definitions: Limit Point: [imath]x[/imath] is a limit point of a set if every open set that contains [imath]x[/imath] contains at least one other point of the set that is not [imath]x[/imath]. [imath]T_1[/imath]-space: All pairs of disjoint points have neighborhoods not containing each others' points. I wrote that all points in the set of all limit points are disjoint from the complement of the set of all limit points, so all the points in the complement of the set of all limit points have neighborhoods that are disjoint from all the neighborhoods of all the points in the set of all limit points, but this was marked incorrect. How should I go about proving this correctly?

493493

Finite group in [imath]GL(n,\mathbb{Q})[/imath] is conjugate to finite group in [imath]GL(n,\mathbb{Z})[/imath]? I can't solve this problem: Finite group in [imath]GL(n,\mathbb{Q})[/imath] is conjugate to finite group in [imath]GL(n,\mathbb{Z})[/imath]. Could any one help me? Thanks a lot!

459554

How can one convert rational matrices into integer ones? Let [imath]G[/imath] be a finite subgroup of [imath]GL_n(\Bbb{Q})[/imath]. I want to show the existence of a matrix [imath]A\in GL_n(\Bbb{Q})[/imath] with the property that [imath]AGA^{-1} \subseteq M_n(\Bbb{Z})[/imath].

474754

[imath]f(z)[/imath] and [imath]\overline{f(\overline{z})}[/imath] simultaneously holomorphic Prove that the functions [imath]f(z)[/imath] and [imath]\overline{f(\overline{z})}[/imath] are simultaneously holomorphic. I take this to mean that [imath]f(z)[/imath] is holomorphic if and only if [imath]\overline{f(\overline{z})}[/imath] is holomorphic. Let [imath]g(z)=\overline{f(\overline{z})}[/imath]. Note that [imath]\overline{g(\overline{z})}=f(z)[/imath]. So it suffices to prove that if [imath]f(z)[/imath] is holomorphic, then [imath]g(z)[/imath] is holomorphic. Write [imath]f(z)=u(z)+iv(z)[/imath]. Since [imath]f(z)[/imath] is holomorphic, the real and imaginary parts satisfy the Cauchy-Riemann equations: [imath]\frac{\partial{u(z)}}{\partial{x}} = \frac{\partial{v(z)}}{\partial{y}}, \frac{\partial{u(z)}}{\partial{y}} = -\frac{\partial{v(z)}}{\partial{x}}.[/imath] We have [imath]g(z) = u(\overline{z})+i(-v(\overline{z}))[/imath]. To prove that [imath]g(z)[/imath] is holomorphic, we must prove that its real and imaginary parts satisfy the Cauchy-Riemann equations: [imath]\frac{\partial{u(\overline{z})}}{\partial{x}} = \frac{\partial{(-v(\overline{z}))}}{\partial{y}}, \frac{\partial{u(\overline{z})}}{\partial{y}} = -\frac{\partial{(-v(\overline{z}))}}{\partial{x}}.[/imath] How can we obtain this from the above relations?

1689734

Show that [imath]\overline{f(\overline{z})}[/imath] is holomorphic on the domain [imath]D^*:=\{\overline z: z\in D\}[/imath] using Cauchy Riemann equation. Please do not vote to close it as I want to find errors in my proof, which cannot be rectified on previously answered question. I want a different proof using Cauchy Riemann equation. Let [imath]D\subset \mathbb C[/imath] be a domain and suppose f is holomorphic on [imath]D[/imath]. Show that [imath]\overline{f(\overline{z})}[/imath] is holomorphic on the domain [imath]D^*:=\{\overline z: z\in D\}[/imath]. Attempt: let [imath]z= x+i y[/imath] and [imath]f(z)=u(x,y)+iv(x,y)[/imath] [imath]f[/imath] is holomorphic on [imath]D \Rightarrow u_x=v_y[/imath] and [imath]u_y=-v_x[/imath] To show: [imath]\overline{f(\overline{z})}[/imath] is holomorphic on the domain [imath]D^*[/imath] Let [imath]w\in D^* \Rightarrow w=\overline z[/imath] for some [imath]z \in D[/imath] To show: [imath]\overline{f(\overline{w})}[/imath] satisfy Cauchy Riemann equation. i.e. To Show: [imath]\overline{f({z})}[/imath] satisfy Cauchy Riemann equation. [imath]\overline{f({z})}= u(x,y)-iv(x,y)[/imath] Let [imath]v_1=-v[/imath] [imath]\overline{f({z})}= u(x,y)+iv_1(x,y)[/imath] i.e. To show: [imath]u_x={v_1}_y[/imath] and [imath]u_y=-{v_1}_x[/imath] But [imath]-v_y={v_1}_y[/imath] and [imath]-v_x=-{v_1}_x[/imath] [imath]\Rightarrow u_x=-v_y[/imath] and [imath]u_y=v_x[/imath] which is not what I want. Where I go wrong ?

494683

Interior of closure of open set in topological space If [imath]U[/imath] is an open set in a topological space, is it true that [imath]U[/imath] is the interior of the closure of itself? If [imath]U[/imath] is open, it must be the interior of itself. But is it the interior of the closure of itself? In the closure, we include all points [imath]x[/imath] such that any open set containing [imath]x[/imath] also contains a point in [imath]U[/imath]. In the interior of the closure, we take the union of all open sets contained in the closure.

348569

Interior of closure of an open set The question is is the interior of closure of an open set equal the interior of the set? That is, is this true: [imath](\overline{E})^\circ=E^\circ[/imath] ([imath]E[/imath] open) Thanks.

494671

Equalities in Groups, How prove this? Let [imath] H,K≤ G[/imath], for all [imath]g ∈ G [/imath] , [imath] \frac {|H||K|}{|H∩(gKg^{-1})|}= \frac {|H||K|}{|(g^{-1}Hg)∩ K|} [/imath] I try to show this but I do Know how to attack this exercise.

312318

How to prove [imath]|H|[K:K \cap (g^{-1}Hg)]=|K|[H:H \cap (gKg^{-1})][/imath]? Let [imath]G[/imath] be a finite group and [imath]H \leq G[/imath], [imath]K \leq G[/imath]. For an element [imath]g \in G[/imath], prove [imath]|H|[K:K \cap (g^{-1}Hg)]=|K|[H:H \cap (gKg^{-1})].[/imath] I am not sure where to start proving this. A hint would be helpful please.

495180

Let [imath]G[/imath] be a group with order [imath]p[/imath] a prime number. Show [imath]G[/imath] is cyclic. Let [imath]G[/imath] be a group with order [imath]p[/imath] a prime number. Show [imath]G[/imath] is cyclic. I know that if [imath]G[/imath] is a cyclic group and if [imath]|G|=n[/imath], then [imath]G\cong\mathbb{Z}_n[/imath]. But I don't where to go from there. Any help/hints would be welcome. ^_^

106163

Show that every group of prime order is cyclic Show that every group of prime order is cyclic. I was given this problem for homework and I am not sure where to start. I know a solution using Lagrange's theorem, but we have not proven Lagrange's theorem yet, actually our teacher hasn't even mentioned it, so I am guessing there must be another solution. The only thing I could think of was showing that a group of prime order [imath]p[/imath] is isomorphic to [imath]\mathbb{Z}/p\mathbb{Z}[/imath]. Would this work? Any guidance would be appreciated.

483797

Count of numbers with the given prime factors in a range Given two primes: [imath]p[/imath] and [imath]q[/imath], [imath]p \neq q[/imath] and [imath]n \in N[/imath] find count of numbers [imath]u[/imath], so that [imath]u \leq n[/imath] and [imath]u = p^k q^l[/imath]; [imath]k, l \in N[/imath]. If we'd given with just one prime [imath]p[/imath] this count would be obvious - [imath][log_p{n}][/imath] but it seems to be harder to find similar simple closed form for two primes. UPDATE 1: The second question related to the problem I'm particularly curious in: whether it is possible to find count of these numbers [imath]u[/imath] in [imath]O(1)[/imath] space and [imath]O(log (n))[/imath] time for two (or better arbitrary [imath]m[/imath]) different primes. UPDATE 2: Very good estimation can be found here: Ramanujan's First Letter to Hardy and the Number of [imath]3[/imath]-Smooth Integers

15966

Ramanujan's First Letter to Hardy and the Number of [imath]3[/imath]-Smooth Integers A positive integer is [imath]B[/imath]-smooth if and only if all of its prime divisors are less than or equal to a positive real [imath]B[/imath]. For example, the [imath]3[/imath]-smooth integers are of the form [imath]2^{a} 3^{b}[/imath] with non-negative exponents [imath]a[/imath] and [imath]b[/imath], and those integers less than or equal to [imath]20[/imath] are [imath]\{1,2,3,4,6,8,9,12,16,18\}[/imath]. In Ramanujan's first letter to G. H. Hardy, Ramanujan emphatically quotes (without proof) his result on the number of [imath]3[/imath]-smooth integers less than or equal to [imath]N > 1[/imath], \begin{eqnarray} \frac{\log 2 N \ \log 3N}{2 \log 2 \ \log 3}. \end{eqnarray} This is an amazingly accurate approximation, as it differs from the exact value by less than 3 for the first [imath]2^{1000} \approx 1.07 \times 10^{301}[/imath] integers, as shown by Pillai. Question: Knowing full well that Ramanujan only gave proofs of his own claims while working in England, I wonder if a proof of this particular estimate appears somewhere in the literature. Is this problem still open? If not, what is a reference discussing its proof? Thanks!

495863

How to show that a sequence does not converge if it is not bounded above I am trying to show that this sequence [imath]\{a_n\}[/imath] = (2n+1)/(3n+5) does not converge to [imath]42[/imath] if it is not bounded above. I have already showed that it converges to [imath]2/3[/imath]. For this I want to use a proof by contradiction,i.e, I assume that the sequence does converge to [imath]42[/imath], which will lead to a contradiction to the initial assumption that is the sequence will be bounded above. Any ideas or theorems that can help me solve this question?

495784

Proof to sequences in real analysis I need some verification for my proof in part a) and help to get me started on part b) a) Prove that the sequence [imath]a_n = (2n+1)/(3n+5)[/imath] converges to [imath]2/3[/imath] directly from the definition of convergence of a sequence. Solution: Some rough work first: Given [imath]\epsilon>0[/imath], we want [imath]|a_n-(2/3)|< \epsilon[/imath]. Need [imath]|(2n+1)/(3n+5) - (2/3)|< \epsilon[/imath] which after simplifying gives [imath]7/(9n+15)< \epsilon[/imath]. Solving for [imath]n[/imath] gives [imath]n> (7-15\epsilon)/9\epsilon[/imath]. Then we write our formal proof: Given [imath]\epsilon>0[/imath], set [imath]N= (7-15\epsilon)/9\epsilon[/imath]. Let [imath]n>N[/imath]. Then [imath]|a_n-(2/3)| = 7/(9n+15) < 7/(9((7-15\epsilon)/9\epsilon)+15) =\epsilon[/imath]. Thus by definition the sequence converges to 2/3. b) Suppose that the sequence [imath]\{a_n\}[/imath] is not bounded above (that is, for any real number [imath]x[/imath], there is some [imath]n[/imath] so that [imath]a_n>x[/imath]). Prove that [imath]\{a_n\}[/imath] does not converge to [imath]42[/imath]. To solve this I am assuming that we have to suppose that [imath]\{a_n\}[/imath] does converge to [imath]42[/imath], and then we will get a contradiction, that is the sequence will be bounded above. So can anyone help me get started on this?

496068

Why is Cauchy-Riemann equation not sufficient for differentiablity In my undergraduate complex analysis textbook, it claims that Cauchy Riemann equations is not a sufficient condition for the existence of derivative. Intuitively, I do not understand why this is true, as if you satisfied [imath]u_x=v_y[/imath] [imath]u_y=-v_x[/imath] it implies that both [imath]f_x \;\; and\;\; f_y[/imath] exist and therefore [imath]f[/imath] should be differentiable Any hint would be much appreciated

420212

Is following contradictory? Can you give an example? Is following contradictory? "Then [imath]f = u + iv[/imath] is complex-differentiable at that point if and only if the partial derivatives of [imath]u[/imath] and [imath]v[/imath] satisfy the Cauchy–Riemann equations (1a) and (1b) at that point. The sole existence of partial derivatives satisfying the Cauchy–Riemann equations is not enough to ensure complex differentiability at that point." - 1[1, second paragraph] They say that [imath]f[/imath] is complex-differentiable iff partial derivatives of [imath]u[/imath] and [imath]v[/imath] satisfy C-R equations, but still it is not enought to ensure complex differentiability at that point. So do you need extra conditions as wikipedia says or not for [imath]f[/imath] to be complex differentiability? Can you give me example, where function satisfy C-R equations, but is not Complex differentiable at certain point?

496116

Is there a partial sum formula for the Harmonic Series? There is a partial sum formula for [imath]\sum_{x=1}^n x^1 = \frac{n(n+1)}{2}[/imath] and even one when the exponent of [imath]x[/imath] is [imath]0[/imath]: [imath]\sum_{x=1}^n x^0 = n[/imath] but I cannot find one for exponent [imath]-1[/imath]: [imath]\sum_{x=1}^n x^{-1} = ?[/imath] I tried [imath]\frac2{n(n+1)},[/imath] but that failed miserably.

451558

How to find the sum of this series : [imath]1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}[/imath] Problem : How to find the sum of this series : [imath]1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}[/imath] This is a Harmonic progression : So is this formula correct to sum the series : [imath]\frac{(number ~of ~terms)^2}{sum~ of~ all ~the~ denominators}[/imath] [imath]\Rightarrow [/imath] if [imath]\frac{1}{A} + \frac{1}{B} +\frac{1}{C}[/imath] are in H.P. Therefore the sum of the series can be written as : [imath]\Rightarrow \frac{(3)^3}{(A+B+C)}[/imath] Is this correct please suggest.

496709

[imath]\mathbb{Z}/\mathbb{nZ}[/imath] admits idempotents I need to show that if [imath]n[/imath] is not a prime power,then [imath]\mathbb{Z}/\mathbb{nZ}[/imath] admits idempotents [imath]\neq 0,1[/imath] I noticed this thing for [imath]\mathbb{Z_6}[/imath] and [imath]\mathbb{Z_{12}}[/imath] and few more but how do we show that there will always exists such [imath]a[/imath]?

175963

Idempotents in [imath]\mathbb Z_n[/imath] An element [imath]a[/imath] of the ring [imath](P,+,\cdot)[/imath] is called idempotent if [imath]a^2=a[/imath]. An idempotent [imath]a[/imath] is called nontrivial if [imath]a \neq 0[/imath] and [imath]a \neq 1[/imath]. My question concerns idempotents in rings [imath]\mathbb Z_n[/imath], with addition and multiplication modulo [imath]n[/imath], where [imath]n[/imath] is natural number. Obviously when [imath]n[/imath] is a prime number then there is no nontrivial idempotent. If [imath]n[/imath] is nonprime it may happen, for example [imath]n=4, n=9[/imath], that also there is no. Is it known, in general, for what [imath]n[/imath] there are nontrivial idempotents and what is a form of such idempotents?

496892

Induction proof [imath]n^2 < 2^n[/imath] for [imath]n > 4[/imath] I need to prove that [imath]n^2 < 2^n[/imath] for all natural numbers [imath]n[/imath] greater than [imath]4[/imath]. I understand that you start by proving the base case of [imath]n = 5[/imath] and then prove the inequality substituting the inductive hypothesis for [imath]n + 1[/imath], but I am unsure about how to do this.

263825

Proof of [imath]n^2 \leq 2^n[/imath]. I am trying to prove that [imath]n^2 \leq 2^n[/imath] for all natural [imath]n[/imath] with [imath]n \ne 3[/imath]. My steps are: induction base case: [imath]n=0:[/imath] [imath]0² \leq 2⁰[/imath] which is okay. inductive step: [imath]n \rightarrow n+1:[/imath] [imath](n+1)²\leq2^{n+1}[/imath] [imath](n+1)^2 = n^2 + 2n + 1 = ...help...\leq 2^{n+1}[/imath] I know the bernoulli inequality but don't know where to use it, if I even need to. I have problems when it comes to proving things which are based on orders..

294213

Prove that [imath]1^3 + 2^3 + ... + n^3 = (1+ 2 + ... + n)^2[/imath] This is what I've been able to do: Base case: [imath]n = 1[/imath] [imath]L.H.S: 1^3 = 1[/imath] [imath]R.H.S: (1)^2 = 1[/imath] Therefore it's true for [imath]n = 1[/imath]. I.H.: Assume that, for some [imath]k \in \Bbb N[/imath], [imath]1^3 + 2^3 + ... + k^3 = (1 + 2 +...+ k)^2[/imath]. Want to show that [imath]1^3 + 2^3 + ... + (k+1)^3 = (1 + 2 +...+ (k+1))^2[/imath] [imath]1^3 + 2^3 + ... + (k+1)^3[/imath] [imath] = 1^3 + 2^3 + ... + k^3 + (k+1)^3[/imath] [imath] = (1+2+...+k)^2 + (k+1)^3[/imath] by I.H. Annnnd I'm stuck. Not sure how to proceed from here on.

973456

Prove by Induction : [imath]\sum n^3=(\sum n)^2[/imath] I am trying to prove that for any integer where [imath]n \ge 1[/imath], this is true: [imath] (1 + 2 + 3 + \cdots + (n-1) + n)^2 = 1^3 + 2^3 + 3^3 + \cdots + (n-1)^3 + n^3[/imath] I've done the base case and I am having problems in the step where I assume that the above is true and try to prove for [imath]k = n + 1[/imath]. I managed to get, [imath](1 + 2 + 3 + \cdots + (k-1) + k + (k+1))^2 = (1 + 2 + 3 + \cdots + (k-1) + k)^2 + (k + 1)^3[/imath] but I'm not quite sure what to do next as I haven't dealt with cases where both sides could sum up to an unknown integer.

497092

Proof by induction: [imath]2^n > n^2[/imath] for all integer [imath]n[/imath] greater than [imath]4[/imath] I am a CS undergrad and I'm studying for the finals in college and I saw this question in an exercise list: Prove, using mathematical induction, that [imath]2^n > n^2[/imath] for all integer n greater than [imath]4[/imath] So I started: Base case: [imath]n = 5[/imath] (the problem states "[imath]n[/imath] greater than [imath]4[/imath]", so let's pick the first integer that matches) [imath]2^5 > 5^2 \implies 32 > 25[/imath] - ok! Now, Inductive Step: [imath]2^{n+1} > (n+1)^2[/imath] now expanding [imath]2 * 2^n > n^2 + 2n + 1[/imath] Well, I can't go on, I don't know how to finish this proof. Could anyone help me with this one? Also, I know that the statement (inductive hypothesis) is true. Simply, I'll have a number to an exponent [imath]n[/imath] twice, next to [imath]n[/imath] squared and [imath]2n[/imath] (which could be seen as quadratic and linear functions) which "grows" much lower than the exponential one. It could be easily "proved by example" with any [imath]n[/imath] greater than [imath]4[/imath]. I want to know how to prove it algebraically , so my professor does not freak out. thanks in advance.

319913

Proof that [imath]n^2 < 2^n[/imath] How do I prove the following statement by induction? [imath]n^2 \lt 2^n[/imath] [imath]P(n)[/imath] is the statement [imath]n^2 \lt 2^n[/imath] Claim: For all [imath]n \gt k[/imath], where [imath]k[/imath] is any integer, [imath]P(n)[/imath] (since [imath]k[/imath] is any integer, I assume I have to prove this for positive and negative integers) So let base case be [imath]P(1)[/imath], and I have to prove [imath]P(n)[/imath] for [imath]n \ge 1[/imath] and [imath]n \lt 1[/imath] [imath]P(1)[/imath] is [imath]1^2 \lt 2^1[/imath] which is clearly true. Induction hypothesis: [imath]k^2 \lt 2^k[/imath] Inductive step [imath](k+1)^2 \lt 2^{(k+1)}[/imath] [imath](k+1)^2 = k^2 + 2k + 1[/imath] [imath]k^2 + 2k + 1 \lt 2^k + 2k + 1[/imath] by inductive hypothesis Not sure how to proceed. Is my previous intuition that I have to prove this for [imath]n \ge 1[/imath] and [imath]n \lt 1[/imath] correct?

497380

[imath]P_n[0,1][/imath] be the set of all polynomial of degree atmost [imath]n[/imath] with supnorm is it a closed in [imath]C[0,1][/imath]? [imath]P_n[0,1][/imath] be the set of all polynomial of degree atmost [imath]n[/imath] with supnorm is it a closed in [imath]C[0,1][/imath]? and [imath]P[0,1][/imath] is set of all polynomials in [imath]C[0,1][/imath] I know which is dense in [imath]C[0,1][/imath] as we know by Weirstrass Polynomial approximation theorem any continous function can be uniformly approximated by sequence of polynomials. Thank you for help.

236003

Is the set of polynomials of degree less than or equal to [imath]n[/imath] closed? This question is in relation to the space [imath]C(I)[/imath], [imath]I = [a, b][/imath]. Define [imath]P_n =\{ a_0+\dots+a_nx^n \mid a_i \in \mathbb{R}\}[/imath] (any or all [imath]a_i[/imath] could be zero); clearly [imath]P_n \subset C(I)[/imath]. The norm I'm using is [imath]\lVert f\rVert_I = \sup_I |f(x)|[/imath]. Is [imath]P_n[/imath] closed under [imath]\lVert\cdot\rVert_I[/imath]? I am almost sure the answer is "yes", but I can't seem to prove it. My first instinct was to biject [imath]P_n[/imath] to [imath]\mathbb{R}^{n+1}[/imath], using coefficients as coordinates, and prove that sequences of degree-[imath]n[/imath] polynomials converge to degree-[imath]n[/imath] polynomials, but I can't prove that the metric [imath]\lVert\cdot\rVert_I[/imath] is equivalent to the standard metric on [imath]\mathbb{R}^{n+1}[/imath]. Next, intuitively given a function [imath]f \in C(I), \notin P_n[/imath] I should be able to find some [imath]\epsilon > 0[/imath] such that there are no low-degree polynomials "nearby", then that function was not a limit point of [imath]P_n[/imath] so [imath]P_n[/imath] is closed. Again I have no idea how to prove this. What should I do?

498538

Why is this a field? I have a trouble. I hope someone can do it. How to prove that [imath]\left\{a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6} \mid (a,b,c,d)\in \mathbb{Q}^4\right\}[/imath] is a field? I can't prove the existence of multiplicative inverse. Thanks.

496334

Showing that [imath]\mathbb{Q}[\sqrt{2}, \sqrt{3}][/imath] contains multiplicative inverses Why must [imath]\mathbb{Q}[\sqrt{2}, \sqrt{3}][/imath] -- the set of all polynomials in [imath]\sqrt{2}[/imath] and [imath]\sqrt{3}[/imath] with rational coefficients -- contain multiplicative inverses? I have gathered that every element of [imath]\mathbb{Q}[\sqrt{2}, \sqrt{3}][/imath] takes form [imath]a + b\sqrt{2} + c\sqrt{3} + d\sqrt{2}\sqrt{3}[/imath] for some [imath]a,b,c,d \in \mathbb{Q}[/imath], and I have shown that [imath]\mathbb{Q}[\sqrt{2}][/imath] and [imath]\mathbb{Q}[\sqrt{3}][/imath] are both fields, but it's not clear to me why this allows for general multiplicative inverses to exist in [imath]\mathbb{Q}[\sqrt{2}, \sqrt{3}][/imath].

497532

Sum of infinite series How to find sum of the following series [imath]\frac{1}{6}+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\cdots={1\over 6} + \sum_{n=1}^\infty{\Pi_{i=1}^n{3i+2}\over (n+1)!6^{n+1}}[/imath] Please give me some hints. Thanks in advance.

1091479

Sum [imath]\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots[/imath] A series is given as follows [imath]\frac{1}{6} + \frac{5}{6\cdot 12} + \frac{5\cdot 8}{6\cdot 12\cdot 18} + \frac{5\cdot 8\cdot 11}{6\cdot 12\cdot 18\cdot 24}+\ldots[/imath] Can you give me hints to get started finding its value? Thanks.

499771

Rephrase the proof "For all odd n, there exists a Group G ..." I am trying to construct a proof and would like to know if I have started it correctly. The proof is as follows. "Prove that for every odd integer n, there is a group with exactly n elements of order 2." Written more compactly I have it as: "[imath]\forall n \in \mathbb{Z^{+odd}}, \exists[/imath] a group G with exactly n elements of order 2." To prove this universal statement, I figure I should disprove the negation. Thus I have: Assume: "[imath]\exists n \in \mathbb{Z^{+odd}} , \nexists[/imath] a group G with exactly n elements of order 2." I am new to proofs and would like to know if the beginnings of this proof are sound. Thanks in advance.

498833

Prove that for every odd integer n, there is a group... Prove that for every odd integer n, there is a group with exactly n elements of order 2. I am new to abstract algebra and proofs in general. Proving existence is difficult for me, but I will show what I have. [imath]n = 2k + 1[/imath] [imath]k \in \mathbb{Z^+}[/imath] [imath]\exists G \supset H = (a_1 , a_2 , ..., a_n, e)[/imath] [imath]s.t. |a_1| = |a_2| = ... = |a_n| = 2[/imath] My instinct is to use Lagrange's Theorem, although I would like to try another method to prove this as we have not gone over cosets in class yet. If someone could just give me a hint of how to start, it would be very beneficial. Thank you.

501095

What are complex numbers, actually? What are complex numbers, actually? You can prove [imath]1=-1[/imath] and a complex cosine function can have value greater than [imath]1[/imath] and so on, there are many unexpected results when we use complex numbers. So, what are they actually? Do, they have any physical meaning or are they just a method in mathematics to manipulate numbers?

199676

What are imaginary numbers? At school, I really struggled to understand the concept of imaginary numbers. My teacher told us that an imaginary number is a number that has something to do with the square root of [imath]-1[/imath]. When I tried to calculate the square root of [imath]-1[/imath] on my calculator, it gave me an error. To this day I still do not understand imaginary numbers. It makes no sense to me at all. Is there someone here who totally gets it and can explain it? Why is the concept even useful?

190040

how to solve [imath] {\partial u \over \partial t} - k {\partial ^2 u \over \partial x^2} =0[/imath] How do I solve the following PDE for it's general solution? [imath] {\partial u \over \partial t} - k {\partial ^2 u \over \partial x^2} =0[/imath] How do I determine the general the solution of this equation will be [imath]u(x, t) = X(x)T(t) [/imath]? I tried Monge's method but couldn't get it. My textbook only deals with particular solution using boundary condition.

5689

Solution to 2nd order PDE What is the general solution to the differential equation: [imath]\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial x}[/imath] I'm a little stuck because all the techniques I know are unable to solve it.

501130

Prove that [imath]\det A = 1[/imath] with [imath]A^T M A = M[/imath] and [imath]M = \begin{bmatrix} 0 & I \\ -I &0 \end{bmatrix}[/imath]. Prove that [imath]\det A = 1[/imath] with [imath]A^T M A = M[/imath] and [imath]M = \begin{bmatrix} 0 & I \\ -I &0 \end{bmatrix}[/imath] ([imath]I[/imath] is the identity matrix of order n).

242091

Why is the determinant of a symplectic matrix 1? suppose [imath]A \in M_{2n}(\mathbb{R})[/imath]. and[imath]J=\begin{pmatrix} 0 & E_n\\ -E_n&0 \end{pmatrix}[/imath] where [imath]E_n[/imath] represents identity matrix. if [imath]A[/imath] satisfies [imath]AJA^T=J[/imath] How to figure out [imath]\det(A)=1[/imath] My approach: I have tried to separate [imath]A[/imath] into four submartix:[imath]A=\begin{pmatrix}A_1&A_2 \\A_3&A_4 \end{pmatrix}[/imath] and I must add a assumption that [imath]A_1[/imath] is invertible. by elementary transfromation:[imath]\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}\rightarrow \begin{pmatrix}A_1&A_2 \\ 0&A_4-A_3A_1^{-1}A_2\end{pmatrix}[/imath] we have: [imath]\det(A)=\det(A_1)\det(A_4-A_3A_1^{-1}A_2)[/imath] from[imath]\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}\begin{pmatrix}0&E_n \\ -E_n&0\end{pmatrix}\begin{pmatrix}A_1&A_2 \\ A_3&A_4\end{pmatrix}^T=\begin{pmatrix}0&E_n \\ -E_n&0\end{pmatrix}[/imath] we get two equalities:[imath]A_1A_2^T=A_2A_1^T[/imath] and [imath]A_1A_4^T-A_2A_3^T=E_n[/imath] then [imath]\det(A)=\det(A_1(A_4-A_3A_1^{-1}A_2)^T)=\det(A_1A_4^T-A_1A_2^T(A_1^T)^{-1}A_3^T)=\det(A_1A_4^T-A_2A_1^T(A_1^T)^{-1}A_3^T)=\det(E_n)=1[/imath] but I have no idea to deal with this problem when [imath]A_1[/imath] is not invertible... Thanks

501683

Isomorphism Between Gaussian Integers Modulo Some Element and the Ring of Integers Modulo it's Norm. Show that for [imath]\theta\in\mathbb{Z}[i][/imath], if [imath]\theta=a+bi[/imath] for [imath]\mathrm{gcd}(a,b)=1[/imath], we have [imath]\mathbb{Z}[i] / (\theta)\cong\mathbb{Z}/{(N(\theta))}.[/imath]

373073

Quotient rings of Gaussian integers I used this isomorphism today but now I'm having trouble justifying it. The norm function isn't additive so I can't come up with a ring isomorphism to prove the following: For any [imath]\,a+bi\in\Bbb Z[i],\,\gcd(a,b)=1[/imath], we have a ring isomorphism [imath]\Bbb Z[i]/\langle a+bi\rangle\,\cong\Bbb Z/(a^2+b^2)\Bbb Z=:\Bbb Z_{a^2+b^2}.[/imath] Could someone show me an isomorphism between these rings to prove this?

501634

Submodules of a Noetherian module are finite intersections of irreducible submodules If [imath]M[/imath] is a Noetherian [imath]R[/imath]-module then every submodule of [imath]M[/imath] is a finite intersection of irreducible submodules. Please show me the way how to get the proof of this statement.

27018

Proper submodules of Noetherian modules are intersections of indecomposable modules If [imath]M[/imath] is Noetherian, any submodule [imath]N \neq M[/imath] can be written as a finite intersection of intersection-indecomposable ones. I just don't know how to begin with the proof of this proposition. Very kind of you for reading or answering.

501758

Let [imath]x_0[/imath] [imath]\in \space G[/imath], G is open, and put [imath]F=[x_0,\infty) \cap G^c.[/imath] (a) Prove that [imath]F[/imath] is a non-empty set. Let G be a non-empty open subset of [imath] \mathbb{R}[/imath]. Let [imath]x_0[/imath] [imath]\in \space G[/imath] and put [imath]F=[x_0,\infty) \cap G^c.[/imath] Here [imath][x_0,\infty)=\lbrace x \in \mathbb{R}:x \geq x_0 \rbrace [/imath]. (a) Prove that [imath]F[/imath] is a non-empty set. My attempt: I found a counter example. If [imath]G= \mathbb{R}-\lbrace 0 \rbrace [/imath] then [imath]G[/imath] is clearly open. However, if [imath]x_0 =5[/imath] for example, then we have F=[imath]\emptyset[/imath]. Yet, the exercise claims that this is never true. Please help! Thank you very much

501148

How can I prove these statements? I would be very glad if you helped me solving this problem. It's about topology. Okay let's begin : Let [imath]A[/imath] be a non-empty open subset of [imath]\mathbb R[/imath]. Let [imath]x_0 \in A[/imath] and [imath]B = [x_0,\infty) ∩ A^c[/imath] (complement of [imath]A[/imath]) First I should prove that [imath]B[/imath] is a non-empty, closed set. Then I should prove that : [imath]B[/imath] is bounded below. [imath]\inf B > x_0[/imath] [imath]\inf B[/imath] doesn't belong to [imath]A[/imath]. [imath][x_0, \inf B) \subseteq A[/imath]. Then I should explain how to define a number [imath]\beta[/imath] so that [imath](\beta,\inf B) \subseteq A[/imath], where [imath]\beta[/imath] doesn't belong to [imath]A[/imath]. I really can't get up with proofs for these question. Can you please help me solving this or at least give me some hint or sketch? I would be very grateful. Thank you :)

501851

prove if ST=TS, T is a scalar multiple of the identity Suppose that [imath]V[/imath] is finite dimensional and [imath]T \in L(V)[/imath]. Prove that [imath]T[/imath] is a scalar multiple of the identity if and only if [imath]ST = TS[/imath] for every [imath]S \in L(V)[/imath]. Suppose [imath]T=\lambda I[/imath] for some [imath]\lambda \in \mathbb{F}[/imath], showing [imath]ST = TS[/imath] is straightforward, but I don't know how to prove the converse.

181761

linear transformation [imath]T[/imath] such that [imath]TS = ST[/imath] Let [imath]V[/imath] be a finite-dimensional vector space over [imath]F[/imath]. Let [imath]T:V \rightarrow V[/imath] be a linear transformation such that [imath]ST=TS[/imath] for all linear transformations [imath]S:V \rightarrow V[/imath]. Show that [imath]T = cI_v[/imath] for some [imath]c \in F[/imath].

501936

Show that the sequence [imath]1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}[/imath] is Cauchy Show that the sequence [imath]1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}[/imath] is Cauchy. I'm not sure where to start with this problem, I know that if I can show that the sequence is convergent I can manipulate the inequality definition of the limit to have [imath]|X_m-X_n|<\epsilon[/imath] but I don't get how to prove the limit exists.

455997

How to prove that [imath]x_n=1+\frac{1}{2!}+\cdots+\frac{1}{n!}[/imath] is a Cauchy sequence? Let [imath](x_n)_{n\in\mathbb{N}}[/imath] be a real sequence given by [imath]x_n=1+\frac{1}{2!}+\cdots+\frac{1}{n!}[/imath] I would like to prove that, for each [imath]\varepsilon>0[/imath], there exists [imath]n_0\in\mathbb{N}[/imath] such that if [imath]n>m>n_0[/imath] then [imath]|x_n-x_m|=\frac{1}{(m+1)!}+\cdots+\frac{1}{n!}<\varepsilon[/imath] In other words, I want to prove that [imath](x_n)[/imath] is a Cauchy sequence. I know this sequence is a Cauchy sequence because it's convergent, but I want to prove it by definition of Cauchy sequence. Thanks.

502243

Describe the curve in the complex plane determined by [imath]z = e^{(-1-i)t}[/imath] Describe the curve in the complex plane determined by [imath]z = e^{(-1-i)t}[/imath] where [imath]0\leq t \leq 2\pi[/imath].

312488

Sketch complex curve [imath]z(t) = e^{-1t+it}[/imath], [imath]0 \le t \le b[/imath] for some [imath]b>0[/imath] Sketch complex curve [imath]z(t) = e^{-1t+it}[/imath], [imath]0 \le t \le b[/imath] for some [imath]b>0[/imath] I tried plotting this using mathematica, but I get two curves. Also, how do I find its length, is it just the integral? This equation doesn't converge right? Edit: I forgot the [imath]t[/imath] in front of the [imath]1[/imath] so it's not a circle of radius [imath]e[/imath]

502553

Homomorphism and isomorphism. Please help me to find an example of a group epimorphism [imath]f:G\to G[/imath] which is not an isomorphism. I understand that [imath]G[/imath] should an infinite group.

79852

Does [imath]G\cong G/H[/imath] imply that [imath]H[/imath] is trivial? Let [imath]G[/imath] be any group such that [imath]G\cong G/H[/imath] where [imath]H[/imath] is a normal subgroup of [imath]G[/imath]. If [imath]G[/imath] is finite, then [imath]H[/imath] is the trivial subgroup [imath]\{e\}[/imath]. Does the result still hold when [imath]G[/imath] is infinite ? In what kind of group could I search for a counterexample ?

502686

Proving an [imath]m\times n[/imath] matrix has left and right inverse Let [imath]A[/imath] be and [imath]m\times n[/imath] matrix and [imath]B[/imath] be an [imath]n\times m[/imath] so that [imath]AB=I_m[/imath] and [imath]BA=I_n[/imath]. Need to show that [imath]m=n[/imath]. Now I thought I could just show that [imath]AB=BA[/imath]. The way I approached this problem is the following: [imath]AABB=ABAB \Longrightarrow (BA)ABB=(BA)BAB[/imath] [imath]I_nABB=I_mBAB \Longrightarrow I_nAB(BA)=I_nBA(BA)[/imath] [imath]I_nABI_n=I_nBAI_n\Longrightarrow I_nABI_n=BA.[/imath] I have noticed several problems with this though specifically [imath]AABB[/imath] because [imath]AABB=A(AB)B=AI_mB[/imath] since [imath]A[/imath] is [imath]m\times n[/imath] and [imath]I_m[/imath] [imath]m\times m[/imath] they can not be multiplied similarly for [imath]B[/imath] [imath]n\times m[/imath]. Also for [imath]I_n[/imath] times [imath]A[/imath] and [imath]B[/imath] times [imath]I_n[/imath]. I have also tried [imath]BBAA=BABA[/imath] and run into a similar problem. Any help would be appreciated.

468685

If [imath]AB=I_n[/imath] and [imath]BA=I_m[/imath] then [imath]n=m[/imath]. Let [imath]A, B[/imath] matrices [imath]n\times m[/imath] and [imath]m\times n[/imath] respectively so that [imath]AB=I_n[/imath] and [imath]BA=I_m[/imath]. Show that [imath]n=m[/imath].

502865

Summablity and countability Let [imath]A[/imath] be an infinite set. For each [imath]a \in A[/imath], let [imath]x_a[/imath] be a non-negative number. Let the value of the series [imath]\sum_{a \in A} x_a[/imath] be the value of the supremum of its finite partial sums, i.e.: [imath]\sum_{a \in A} x_a = \sup_{n \geq 0} \sup_{(a_1, \ldots , a_n) \subset A} \left(x_{a_1} + x_{a_1} + \ldots + x_{a_n}\right)[/imath] (Supremum is well defined even when its value is infinite) If [imath]\sum_{a \in A} {x_a}[/imath] is finite, prove that the set [imath]A' = \left\{{a \in A | a > 0}\right\}[/imath] is countable. My guess is that we have to write [imath]A[/imath] as a union of finitely many sets.

372125

How many positive numbers need to be added together to ensure that the sum is infinite? The question in the title is naively stated, so let be make it more precise: Let [imath]\sum_{n\in\alpha}a_n[/imath] be an ordinal-indexed sequence of real numbers such that [imath]a_n>0[/imath] for each [imath]n\in\alpha[/imath], where [imath]\alpha[/imath] is an ordinal number. What is the smallest [imath]\alpha[/imath] which guarantees that [imath]\sum_{n\in\alpha}a_n[/imath] diverges? Since bijections correspond to rearrangements of the sum, and since the [imath]a_n[/imath] are positive, the sum is either absolutely convergent or diverges to [imath]+\infty[/imath] regardless of the order, so it follows that [imath]\alpha[/imath] is a cardinal number. My intuition tells me that [imath]\alpha=\omega_1[/imath], but I can only prove that [imath]\alpha\le\frak{c}^+[/imath], as follows: Let [imath]\sum_{n\in\frak{c}^+}a_n[/imath] be a sum of positive reals with [imath]\frak{c}^+[/imath] terms, and let [imath]s_\beta=\sum_{n\in \beta}a_n[/imath] be the sequence of partial sums, so that [imath]s_{\beta+1}=s_\beta+a_\beta[/imath]. Then if [imath]\beta<\gamma[/imath], [imath]s_\beta<s_\beta+a_\beta=s_{\beta+1}\le s_\gamma[/imath], so in particular, the [imath]\{s_\beta\}[/imath] are all distinct, and [imath]|\{s_\beta\}|=\frak{c}^+[/imath]. If [imath]\sum_{n\in\frak{c}^+}a_n=A[/imath] is finite, then every partial sum is less than [imath]A[/imath], so [imath]\{s_\beta\}\subseteq [0,A]\subseteq\mathbb{R}[/imath], so [imath]\frak{c}^+=|\{s_\beta\}|\le|\mathbb{R}|=\frak{c}[/imath], a contradiction. Thus [imath]\sum_{n\in\frak{c}^+}a_n[/imath] is not finite. As indicated above, [imath]\alpha\ge\omega_1[/imath] is obvious because [imath]\alpha[/imath] is a cardinal, and [imath]\sum_{n\in\omega}2^{-n}=2[/imath] is finite, so [imath]\alpha>\omega[/imath]. Can anyone prove that [imath]\alpha\le\frak{c}[/imath] or that [imath]\alpha=\frak{c}^+[/imath]? I can't imagine any set of positive numbers indexed by reals whose sum could possibly be finite, so I lean strongly toward [imath]\alpha\le\frak{c}[/imath], but I don't know how to prove it.

502923

Modular Arithmetic Question mod I am having problems solving two-variable mod equations. How would one solve this? [imath] \left\{\begin{aligned} x + 3y &\equiv 1 \pmod{11} \\ 2x + y &\equiv 7 \pmod{11} \end{aligned}\right. [/imath]

502925

Solving modular arithmetic questions I am having trouble finding mod arithmetic questions. Can you show how to solve these? [imath]x + 30 \equiv 1 \pmod {12}[/imath] [imath]30x \equiv 1 \pmod {12}[/imath] [imath]x + 3y \equiv 1 \pmod {12}[/imath] and [imath]2x + y \equiv 7 \pmod {12}[/imath]

503125

Does monotonicity imply measurability? Let [imath]f \colon \mathbb{R} \to \mathbb{R}[/imath] be an increasing function. Prove or disprove: [imath]f[/imath] is Lebesgue measurable. Thank you all for the great comments I'll receive! :D

252421

Are Monotone functions Borel Measurable? Could you guide me how to prove that any monotone function from [imath]R\rightarrow R[/imath] is Borel measurable? Should we separate the functions into continuous and non-continuous? How to prove for not continuous points? Thanks for your help

503184

Suppose that [imath]f:[a,b] \to [a,b][/imath] is continuous. Prove that there is at least one fixed point in [imath][a,b][/imath] - that is, [imath]x[/imath] such that [imath]f(x) = x[/imath]. Suppose that [imath]f:[a,b] \to [a,b][/imath] is continuous. Prove that there is at least one fixed point in [imath][a,b][/imath] - that is, [imath]x[/imath] such that [imath]f(x) = x[/imath]. I haven't a clue where to even start on this one.

416126

Prove that [imath]f[/imath] has a fixed point . For [imath]f:[a,b]\rightarrow [a,b][/imath] is a continuous . Prove that [imath]f[/imath] has a fixed point . Is that true if we change [imath][a,b][/imath] by [imath][a,b)[/imath] or [imath](a,b)[/imath].

501703

Logical Problems (Tautologies) When addressing the questions, which are featured below, I use the following definition and two lemmas. Definition: [imath]\phi[/imath] is a tautology if [imath][[\phi]]_{v}=1[/imath] for all valuations [imath]v[/imath]. Moreover, [imath]\models \phi [/imath] stands for [imath]\phi[/imath] is a tautology. Let [imath]\Gamma[/imath] be a set of propositions, then [imath]\Gamma \models \phi[/imath] if and only if for all [imath]v[/imath]: ([imath][[\psi]]_{v}=1[/imath] for all [imath]\psi \in \Gamma[/imath]) [imath]\implies [[\phi]]_{v}=1[/imath]. Lemma 1. [imath]\phi_{1}, \dots, \phi_{n} \models \phi \iff [[\phi_{1}]] \land \dots \land [[\phi_{n}]] \leq [[\phi]] [/imath] Proof sketch: This result follows immediately from the definition above. Lemma 2. [imath]\phi_{1}, \dots, \phi_{n}, \phi \models \psi \iff \phi_{1}, \dots, \phi_{n} \models \phi \to \psi[/imath] Proof sketch: Use lemma 1 to transform the problem into [imath][[\phi_{1}]] \land \dots \land [[\phi_{n}]] \ [/imath] Then, note that this form satisfy a Galois connection. Now, here are the problems and my answer. (a) For every set of formulas [imath]\Gamma[/imath], every formula [imath]\phi[/imath] and every formula [imath]\psi[/imath] we have that if [imath]\Gamma \models \phi \land \psi[/imath], then [imath]\Gamma \models \phi[/imath] and [imath]\Gamma \models \psi[/imath]. True. We may interprete [imath][[\phi \land \psi]]=min([[\phi]],[[\psi]])[/imath] and so [imath]\Gamma \leq \phi[/imath] and [imath]\Gamma \leq \psi[/imath], which shows the proposition. (b) For every set of formulas [imath]\Gamma[/imath], every formula [imath]\phi[/imath] and every formula [imath]\psi[/imath] we have that if [imath]\Gamma \models \phi \lor \psi[/imath], then [imath]\Gamma \models \phi[/imath] or [imath]\Gamma \models \psi[/imath]. False (?). We may interprete [imath][[\phi \lor \psi]]=max([[\phi]],[[\psi]])[/imath] and so [imath]\Gamma \models \phi \lor \psi[/imath] really means [imath]\Gamma \leq \text{max}([[\phi]],[[\psi]])[/imath]. Here I am stuck... (maybe a sign that my reasoning is incorrect?) (c) [imath]P_{1} \land P_{2}, \neg P_{2} \models \neg P_{1}[/imath] True (?). Combining lemma 2 and lemma 1 we see that the proposition really means [imath]P_{1} \land P_{2} \leq P_{2} \lor \neg P_{1}[/imath], which holds. (d) [imath]\bot \models \phi[/imath] for any formula [imath]\phi[/imath] True. This is a consequence of lemma 1. (e) [imath]\phi \models \top[/imath] for any formula [imath]\phi[/imath]. True. Again this is also a consequence of lemma 1. Am I on the right track, or is there something wicked in the reasoning? Any feedback is very much appreciated.

502348

Simple proof theory - Propositional Logic When addressing the questions, which are featured below, I use the following definition and two lemmas. Definition: [imath]\phi[/imath] is a tautology if [imath][[\phi]]_{v}=1[/imath] for all valuations [imath]v[/imath]. Moreover, [imath]\models \phi [/imath] stands for [imath]\phi[/imath] is a tautology. Let [imath]\Gamma[/imath] be a set of propositions, then [imath]\Gamma \models \phi[/imath] if and only if for all [imath]v[/imath]: ([imath][[\psi]]_{v}=1[/imath] for all [imath]\psi \in \Gamma[/imath]) [imath]\implies [[\phi]]_{v}=1[/imath]. Lemma 1. [imath]\phi_{1}, \dots, \phi_{n} \models \phi \iff [[\phi_{1}]] \land \dots \land [[\phi_{n}]] \leq [[\phi]] [/imath] Proof sketch: This result follows immediately from the definition above. Lemma 2. [imath]\phi_{1}, \dots, \phi_{n}, \phi \models \psi \iff \phi_{1}, \dots, \phi_{n} \models \phi \to \psi[/imath] Proof sketch: Use lemma 1 to transform the problem into [imath][[\phi_{1}]] \land \dots \land [[\phi_{n}]] \ [/imath] Then, note that this form satisfy a Galois connection. Now, here are the problems and my answer. (a) For every set of formulas [imath]\Gamma[/imath], every formula [imath]\phi[/imath] and every formula [imath]\psi[/imath] we have that if [imath]\Gamma \models \phi \land \psi[/imath], then [imath]\Gamma \models \phi[/imath] and [imath]\Gamma \models \psi[/imath]. True. We may interprete [imath][[\phi \land \psi]]=\min([[\phi]],[[\psi]])[/imath] and so [imath]\Gamma \leq \phi[/imath] and [imath]\Gamma \leq \psi[/imath], which shows the proposition. (b) For every set of formulas [imath]\Gamma[/imath], every formula [imath]\phi[/imath] and every formula [imath]\psi[/imath] we have that if [imath]\Gamma \models \phi \lor \psi[/imath], then [imath]\Gamma \models \phi[/imath] or [imath]\Gamma \models \psi[/imath]. True (?). We may interprete [imath][[\phi \lor \psi]]=\max([[\phi]],[[\psi]])[/imath] and so [imath]\Gamma \models \phi \lor \psi[/imath] really means [imath]\Gamma \leq \text{max}([[\phi]],[[\psi]])[/imath]. Now, it is clear that for every instance, either [imath][[\Gamma]] \leq [[\phi]][/imath] or [imath][[\Gamma]] \leq [[\psi]][/imath], which, together with lemma 1, shows the claim. (c) [imath]P_{1} \land P_{2}, \neg P_{2} \models \neg P_{1}[/imath] True (?). Combining lemma 2 and lemma 1 we see that the proposition really means [imath]P_{1} \land P_{2} \leq P_{2} \lor \neg P_{1}[/imath], which holds. (d) [imath]\bot \models \phi[/imath] for any formula [imath]\phi[/imath] True. This is a consequence of lemma 1. (e) [imath]\phi \models \top[/imath] for any formula [imath]\phi[/imath]. True. Again this is also a consequence of lemma 1. Am I on the right track, or is there something wicked in the reasoning? Any feedback is very much appreciated.

503782

Show that any complex roots must occur in conjugate pairs. Let [imath]p(\lambda)=c_0 + c_1\lambda + \dots + c_n\lambda^n[/imath] be a polynomial with real coefficients. Show that any complex roots of [imath]p(\lambda)=0[/imath] must occur in conjugate pairs, i.e. if [imath]p(\alpha+i\beta)=0[/imath] then [imath]p(\alpha-i\beta)=0[/imath]. So I am kind of having trouble doing this for a general expression. Where should I start? I'd be able to figure it out for some small fixed [imath]n[/imath]... Any hints would be helpful

121631

Roots of polynomial with real coefficients appear in conjugate pairs. How to prove most simply that if a polyonmial [imath]f[/imath], has only real coefficients and [imath]f(c)=0[/imath], and [imath]k[/imath] is the complex conjugate of [imath]c[/imath], then [imath]f(k)=0[/imath]?

503790

i cant find this integral [imath]\int_0^1 x^{-x} \mathrm{dx}[/imath] [imath]\int_0^1 x^{-x} \mathrm{dx} [/imath]

238199

Showing that [imath]\int_0^\infty x^{-x} \mathrm{d}x \leq 2[/imath]. This integral is very closely related to the sophmores dream that states [imath] \int_0^1 x^{-x}\mathrm{d}x = \sum_{n=1}^\infty n^{-n} = 1.27\ldots [/imath] For example here http://en.wikipedia.org/wiki/Sophomore%27s_dream Now I want to bound the integral, and showing that is less that 2. For the interval [imath][0,1][/imath] a good bound is rewriting it to [imath]\exp(x\log x)[/imath] and using the expansion [imath] 1 - x \log(x) + \frac12 (-x \log(x))^2[/imath] but how does one handle [imath][1,\infty)[/imath] ? In this answer here How to evaluate [imath] \int_0^\infty {1 \over x^x}dx[/imath] in terms of summation of series? gives bounds to the integral, but they are not tight enough.. So to taste my question again, how does one prove that [imath] \int_0^\infty \frac{\mathrm{d}x}{x^x} \leq 2 [/imath]

504063

Show [imath]\sum\limits_{d|n}\phi(d) = n[/imath]. Show [imath]\sum\limits_{d|n}\phi(d) = n[/imath]. Example : [imath]\sum\limits_{d|4}\phi(d) = \phi(1) + \phi(2) + \phi(4) = 1 + 1 + 2 = 4[/imath] I was told this has a simple proof. Problem is, I can not think of a way to show this in a very simple and straightforward way.

194705

Is there a direct, elementary proof of [imath]n = \sum_{k|n} \phi(k)[/imath]? If [imath]k[/imath] is a positive natural number then [imath]\phi(k)[/imath] denotes the number of natural numbers less than [imath]k[/imath] which are prime to [imath]k[/imath]. I have seen proofs that [imath]n = \sum_{k|n} \phi(k)[/imath] which basically partitions [imath]\mathbb{Z}/n\mathbb{Z}[/imath] into subsets of elements of order [imath]k[/imath] (of which there are [imath]\phi(k)[/imath]-many) as [imath]k[/imath] ranges over divisors of [imath]n[/imath]. But everything we know about [imath]\mathbb{Z}/n\mathbb{Z}[/imath] comes from elementary number theory (division with remainder, bezout relations, divisibility), so the above relation should be provable without invoking the structure of the group [imath]\mathbb{Z}/n\mathbb{Z}[/imath]. Does anyone have a nice, clear, proof which avoids [imath]\mathbb{Z}/n\mathbb{Z}[/imath]?

428151

Questions on "All Horse are the Same Color" Proof by Complete Induction I'm bugged by the following that's summarized on p. 109 of this PDF. False theorem: All horses are the same color. Proof by induction: [imath]\fbox{[/imath]P(n)[imath] is the statement: In every set of horses of size [/imath]n[imath], all [/imath]n[imath] horses are the same color.}[/imath] [imath]\fbox{Base Case or [/imath]P(1)[imath]:}[/imath] One horse is the same color as itself. This is true by inspection. [imath]\fbox{Induction Step:}[/imath] Assume [imath]P(k)[/imath] for some [imath]k \geq 1[/imath]. [imath]\fbox{Proof of [/imath]P(k + 1) :[imath]}[/imath] Since [imath]\{H_1, H_2, ... , H_n\}[/imath] is a set of [imath]n[/imath] horses, the induction hypothesis applies to this set. Thus, all the horses in this set are the same color. Since [imath]\{H_2, H_3, ... , H_{n+1}\}[/imath] is also a set of [imath]n[/imath] horses, the induction step likewise holds for this set. Thus, all the horses in this set are the same color too. Therefore, all [imath]n +1[/imath] horses in [imath]\{H_1, H_2, H_3, ... , H_n , H_{n+1}\}[/imath] are the same color. QED. The issue the instructor was trying to point out is clearly valid. For the case [imath]n = 1[/imath], there is only [imath]{H_1}[/imath]. So this case says nothing about possible overlapping elements of each set of [imath](n + 1)[/imath], for instance [imath]H_2[/imath] in the above proof. But it was proposed in the class discussion that this was the only problem. Had you been able to prove [imath]P(2)[/imath] true, then a proof of the above format would have been fine. My interpretation is that yes, you could prove all horses are the same color, if you can prove that any set of two horses will be the same color. But this format would not work. Why not? The problem I see is that the above proof is for the existence of at least one particular pair of sets of horses of sizes [imath]n[/imath] and [imath]n + 1[/imath], such that in each set, all horses are the same color. Particularly when the set of size [imath]n[/imath] is a subset of the set of size [imath]n + 1[/imath]. In order to prove the induction step, don't you need to prove that sets of sizes [imath]n[/imath] and [imath]n + 1[/imath], do not necessarily contain the same, overlapping elements? You could prove that any horse can be added to a set of 2 horses. Take the last two, and they must be the same color, and so on. Wwhatever color the first two happen to be, all other horses must thus be the same color. Am I misinterpreting the example, or am I making a logical error? Thanks in advance.

652429

Fallacious Induction Proof that All Integers are Equal We're currently learning about induction in my real analysis course. I came up with the following proof that is obviously false but cannot quite figure out why it fails. Here it is... "Any set [imath]S[/imath] of [imath]n \in \mathbb{N}[/imath] integers are equal." Proof: For [imath]n = 1[/imath], we only have a single integer in [imath]S[/imath], and thus it is equal to itself. Assume our statement is true for [imath]n[/imath] integers. Now consider any set of [imath]n + 1[/imath] integers. Without loss of generality they are integers [imath]\left\{ 1, 2, \dots, n, n + 1\right\}[/imath]. Consider the sets [imath]\left\{ 1, 2, \dots, n \right\}[/imath] and [imath]\left\{ 2, 3, \dots, n + 1\right\}[/imath]. Each of these two sets have only [imath]n[/imath] integers, so they are all equal. But this means [imath]1 = 2 = 3 = \cdots = n = n + 1[/imath], so all integers are equal.

324724

How can I prove: if [imath]p [/imath] is prime and [imath]n>1[/imath], then [imath] p^{\frac1n} [/imath] is irrational? Please see this question's title.

520355

If [imath]p[/imath] is prime and any integer [imath]k>1[/imath], then [imath]p^{\frac 1k} [/imath] is irrational. Prove this by assuming [imath]p^{\frac 1k}[/imath] rational I've tried setting [imath]p^{\dfrac 1k}= \dfrac a b[/imath], and then raising [imath]p^{\dfrac 1k}[/imath] to the [imath]k^\text{th}[/imath] power, but I'm stuck.

502125

Proving a boolean algebra question Let [imath]\sqsubseteq[/imath] be a boolean ordering of the boolean algebra [imath]X[/imath], which means that for each [imath]x[/imath] and [imath]y[/imath] the following applies: [imath]x \sqsubseteq y[/imath] if [imath]x \sqcap y = x[/imath]. Let [imath]v, w, a, b \in X[/imath] with [imath]v \sqsubseteq a[/imath] and [imath]w \sqsubseteq b[/imath]. Show/prove the following: that [imath]v \sqcup w \sqsubseteq a \sqcup b[/imath] and that [imath]v \sqcap w \sqsubseteq a \sqcap b[/imath] Now, I can answer this question with the help of set theory: if [imath]V \subset A[/imath], then it follows that [imath]V \subset A\cup B[/imath]. Also, from [imath]W \subset B[/imath] it follows that [imath]W \subset A\cup B[/imath]. It then follows that if both [imath]V[/imath] and [imath]W[/imath] are subsets of [imath]A\cup B[/imath], it also follows that [imath]V\cup W \subset A \cup B[/imath]. The same story goes for [imath]V \subset A[/imath] and [imath]V \subset A \cap B[/imath]. But how can I show/express this in boolean algebra notation? How can I use [imath]x \sqsubseteq y[/imath] if [imath]x \sqcap y = x[/imath] to prove [imath]v \sqcup w \sqsubseteq a \sqcup b[/imath] and [imath]v \sqcap w \sqsubseteq a \sqcap b[/imath]?

493583

How to prove boolean ordering question Let [imath]\sqsubseteq[/imath] be the boolean ordering of [imath]X[/imath], so for every [imath]x[/imath] and [imath]y[/imath] applies [imath]x \sqsubseteq y[/imath] if [imath]x \sqcap y = x[/imath]. Let [imath]v, w, a, b \in X[/imath] with [imath]v \sqsubseteq a[/imath] and [imath]w \sqsubseteq b[/imath]. Show that [imath]v \sqcup w \sqsubseteq a\sqcup b[/imath] and [imath]v \sqcap w \sqsubseteq a\sqcap b[/imath]. Should this be solved algebraically, or in a different way? And if so, where would be my starting point?

504761

How Is This a Bijection from [imath]P(X)[/imath] to [imath]2^X[/imath]? Let [imath]X[/imath] be a set, [imath]P(X)[/imath] the power set of [imath]X[/imath], and [imath]2^X[/imath] the set of all function from [imath]X[/imath] to [imath]\{0, 1\}[/imath]. Now define [imath]J: P(X) \rightarrow 2^X[/imath] by [imath]J(A) = \chi_A[/imath], where [imath]\chi_A[/imath] is the characteristic function. Show that [imath]J[/imath] is a bijection from [imath]P(X)[/imath] to [imath]2^X[/imath]. I can demonstrate that [imath]J[/imath] is a bijection from [imath]P(X)[/imath] to [imath]\{\chi_A | A \subset X\}[/imath]. But how do I know for sure that [imath]\{\chi_A | A \subset X\} = 2^X[/imath]? Or, is there a function in [imath]2^X[/imath] that does not take the form of [imath]\chi_A[/imath]? I am not seeing how the answer is no, but I can't think of any counter examples either.

275904

Bijection [imath]f:\mathcal{P}(A)\to(A\to \{0,1\})[/imath] How can I formalize that the function [imath]f:\mathcal{P}(A)\to(A\to \{0,1\})[/imath] is bijective? It seems clear to me but I don't know how to write a proof. Thanks in advance.

504845

Open subgroups of [imath]\mathbb{R}[/imath] Let [imath]G[/imath] be a nonempty open subset of [imath]\mathbb{R}[/imath] (with usual topology on [imath]\mathbb{R}[/imath]) such that [imath]x,y\in G[/imath] implies that [imath]x-y\in G[/imath]. Show that [imath]G=\mathbb{R}[/imath]. Clearly [imath]0\in G[/imath]. Now how to show that all real numbers are there in [imath]G[/imath]? Please help.

327073

what are all the open subgroups of [imath](\mathbb{R},+)[/imath] I am not able to find out what are all the open subgroups of [imath](\mathbb{R},+)[/imath], open as a set in usual topology and also subgroup.

262828

Using proof by contradiction vs proof of the contrapositive What is the difference between a "proof by contradiction" and "proving the contrapositive"? Intuitive, it feels like doing the exact same thing. And when I compare an exercise, one person proves by contradiction, and the other proves the contrapositive, the proofs look almost exactly the same. For example, say I want to prove: [imath]P \implies Q[/imath] When I want to prove by contradiction, I would say assume this is not true. Assume [imath]Q[/imath] is not true, and [imath]P[/imath] is true. Blabla, but this implies [imath]P[/imath] is not true, which is a contradiction. When I want to prove the contrapositive, I say. Assume [imath]Q[/imath] is not true. Blabla, this implies [imath]P[/imath] is not true. The only difference in the proof is that I assume [imath]P[/imath] is true in the beginning, when I want to prove by contradiction. But this feels almost redundant, as in the end I always get that this is not true. The only other way that I could get a contradiction is by proving that [imath]Q[/imath] is true. But this would be the exact same things as a direct proof. Can somebody enlighten me a little bit here ? For example: Are there proofs that can be proven by contradiction but not proven by proving the contrapositve?

2051775

Contradiction vs. contrapositive I am having trouble in understanding the difference between a contrapositive and a contradiction. For example, on one of my practice exam, there was a question that asked to prove that [imath]n[/imath] is even if [imath]n[/imath] is an integer and [imath]n^2 + 5[/imath] is odd using both a contrapositive and a contradiction. When I proved it, I assumed [imath]n[/imath] is odd and assigned it to [imath]2k + 1[/imath]. What method am I using right now? How would the other method look like.

505714

How to evaluate [imath]\int_{0}^{\pi} \log(2+\cos x)dx[/imath]? I tried integrating it by part but it didn't work. I konw it can be solved by the Gauss Mean Value theorem .Is there some elementary method for evaluate it or just some others? It seems there are at least four differential ways(page 4 (17)) but I cannot think out any(the link only says there are four but not give any way).

114401

A problem about parametric integral How to solve the following integral. [imath]I(\theta) = \int_0^{\pi}\ln(1+\theta \cos x)dx[/imath] where [imath]|\theta|<1[/imath]

505901

Compute [imath]\phi^{-1}(k)[/imath], [imath]\phi[/imath] Euler's totient function? Given a positive integer [imath]k[/imath], I'd like to be able to compute the set of positive integers [imath]m[/imath] such that [imath]m[/imath] is prime to precisely [imath]k[/imath] positive integers less than [imath]m[/imath]. In other words, I'd like to compute the set [imath]\phi^{-1}(k)[/imath], where [imath]\phi[/imath] is Euler's totient function. We could compute this by brute force using one of the lower bounds for [imath]\phi[/imath] to find an [imath]N[/imath] so that [imath]n \geq N[/imath] implies [imath]\phi(n) > k[/imath] and then we could test all [imath]m < N[/imath] to see if [imath]\phi(m)=k[/imath], but is there are smarter way to compute [imath]\phi^{-1}(k)[/imath]?

23947

How to solve the equation [imath]\phi(n) = k[/imath]? Let [imath]\phi(n) [/imath] is the numbers of number that are relatively prime to n. Then, how could we solve the equation [imath]\phi(n) = k, k > 0?[/imath] For example: [imath]\phi(n) = 8 [/imath] I can use computer program to check all numbers that are relatively prime to [imath]n[/imath], but I think there must be an easier way to approach this problem. Base on this formula: [imath]\prod_{i=0}^{k} p_{i}a^{a_i} [/imath] The only thing I can see is n must not have a prime factor > 9, otherwise [imath]\phi(n) > 8 [/imath]. I really don't know where to start :( ? A hint would be greatly appreciated.

477364

Prove that [imath]\tan A + \tan B + \tan C = \tan A\tan B\tan C,[/imath] [imath]A+B+C = 180^\circ[/imath] I want to prove \begin{equation*} \tan A + \tan B + \tan C = \tan A\tan B\tan C \quad\text{when } A+B+C = 180^\circ \end{equation*} We know that \begin{equation*} \tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}~\text{and that}~A+B = 180^\circ-C. \end{equation*} Therefore [imath]\tan(A+B) = -\tan C.[/imath] From here, I get stuck. Please help.

506002

Prove that : [imath]\tan 40 + \tan 60 + \tan 80 = \tan 40 \cdot \tan 60 \cdot \tan 80[/imath] I started from Left hand side as 3^1/2 + tan 2(20) +tan 4(20). But that brought me a lot of terms to solve which ends (9 tan 20 - 48 tan^3 20 -50 tan^5 20 - 16 tan^7 20 + tan^9 20)/(1- 7 tan^2 20 + 7 tan^4 20 - tan^6 20), which is very huge to solve. If someone can help me in teaching me how to begin, that would be very greatful.

505967

The mapping of a Set onto the characteristic function is a bijection Let [imath]X[/imath] be a Set. For all Subsets [imath] A \subset X[/imath] the characteristic function of A is defined as: \begin{align} \chi_A(x)= \begin{cases} 1 \iff x \in A \\ 0 \iff x \notin A \end{cases} \end{align} Let [imath]\lbrace0,1\rbrace^X[/imath] be the Set of functions [imath] X \longrightarrow \lbrace 0,1\rbrace [/imath]. Further let [imath]P(X)[/imath] be the power set of [imath]X[/imath]. Show that the following function is a bijection: \begin{align} P(X) & \longrightarrow \lbrace 0,1\rbrace ^X \\ A & \longmapsto \chi_A \end{align} I have struggled with this problem for a few days now and I believe one of my biggest issues is to create the desired set [imath]\lbrace0,1\rbrace^X[/imath].

496429

Showing [imath]\mathcal{P}(X)[/imath] is isomorphic to [imath]2^X[/imath] I have a homework question that plays off my las question about the characteristic function [imath]\chi_A(x)[/imath]. It is: Prove that there is a function [imath]f[/imath] that gives a one-to-one correspondence between [imath]\mathcal{P}(X)[/imath] and [imath]2^X[/imath]. Here the power set [imath]\mathcal{P}(X)=\{A|A\subseteq{X}\}[/imath] and [imath]2^X[/imath] is the set of all functions that map elements into the set [imath]\{0,1\}[/imath] Now i know that [imath]|\mathcal{P}(X)|=2^{|X|}[/imath], but the hint in the example mentions this: "Now let us define a function [imath]f[/imath] on [imath]\mathcal{P}(X)[/imath] into [imath]2^X[/imath] by taking as the image a subset [imath]A[/imath] of [imath]X[/imath] the characteristic function of [imath]A[/imath]," so then this mapping looks like this [imath]f:\mathcal{P}(X)\rightarrow{2^X}[/imath] [imath]f(A)=\chi_A{x}[/imath] From here I must show injectivity and surjectivity. So [imath]f(A)=f(B)\Rightarrow{A=B}[/imath] So by the definition of the characteristic function we can show that injectivity holds since if [imath]x\in{A}, \chi_A(x)=1=\chi_B(x) \text{ if } x\in{B}[/imath] and [imath]x\in{X-A}, \chi_A(x)=0=\chi_B(x) \text{ if } x\in{X-B}[/imath] But how do we create a surjectivity argument?

505382

Complex differentiable function Let [imath]U[/imath] be an open set in [imath]\mathbb{C}[/imath] and [imath]f[/imath] differentiable in the real sense. Prove that if [imath]c[/imath] is such that the following limit exists: [imath]\lim_{h→0}=\frac{|f(c+h)−f(c)|}{|h|}[/imath] Then [imath]f[/imath] is differentiable in the complex sense or [imath]\bar{f}[/imath] is differentiable in the complex sense. [imath]h[/imath] is a complex number.

29615

if a complex function [imath]f[/imath] is real-differentiable, then [imath]f[/imath] or [imath]\overline{f}[/imath] are complex-differentiable This is an exercise from Remmert's Theory of Complex functions. Let [imath]D\subset \mathbb{C}[/imath] be a domain and [imath]f:D\rightarrow \mathbb{C}[/imath] a real-differentiable function. Assume that the following limit exists: [imath] \mathrm{lim}_{h\rightarrow 0} \left| \frac{f(c+h) - f(c)}{h} \right|.[/imath] Show that either [imath]f[/imath] or [imath]\overline{f}[/imath] is complex-differentiable. I've tried showing that [imath]\frac{\partial f}{\partial \overline{z}} = 0[/imath] or [imath]\frac{\partial \overline{f}}{\partial z} = 0[/imath] by using the fact that there exist continuous functions [imath]g[/imath] and [imath]h[/imath] such that in [imath]D[/imath] one can write [imath]f(z) = f(c) + (z-c)g(z) + (\overline{z} - \overline{c})h(z)[/imath] and that [imath]g(c)= f_{z}(c)[/imath] and [imath]h(c) = f_{\overline{z}}[/imath] and then plugging this into the limit above. Does this approach works and I just can´t see how to do it? Can someone give a hint or a guideline solution to this?