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460996 | [imath]\mathbb{Q}(\sqrt[3]{2})[/imath] is not in any cyclotomic extension of [imath]\mathbb{Q}[/imath]
Question is to prove that [imath]\mathbb{Q}(\sqrt[3]{2})[/imath] is not in any cyclotomic extension of [imath]\mathbb{Q}[/imath]. As i was not so sure how to proceed i did the following thing(which may possibly be not so relevant to the Question but hope it would give me some hint to proceed). As [imath]\mathbb{Q}(\sqrt[3]{2})[/imath] is not Galois over [imath]\mathbb{Q}[/imath], I tried Computing Splitting Field and corresponding galois Group and ended up in concluding that Galois Group is isomorphic to [imath]S_3[/imath]. As [imath]S_3[/imath] is not abelian Group there could possibly no Sub field [imath]K[/imath] of Cyclotomic field with [imath]Gal(\frac{K}{\mathbb{Q}})\cong S_3[/imath]. I am helpless after this. Please let me know am i going in a correct path?? Any suggestion/hint would be appreciated :) Thank You | 460397 | Is [imath]\sqrt[3]{2}[/imath] contained in [imath]\mathbb{Q}(\zeta_n)[/imath]?
Is [imath]\sqrt[3]{2}[/imath] contained in [imath]\mathbb{Q}(\zeta_n)[/imath] for some [imath]n[/imath], where [imath]\zeta_n=e^{2\pi i/n}[/imath]? I think the answer is no, but I can't give a full proof. Assume the contrary, we then have [imath]\mathbb{Q}(\sqrt[3]{2}) \subset \mathbb{Q}(\zeta_n)[/imath], can I say [imath]\mathbb{Q}(\zeta_n)/ \mathbb{Q}[/imath] is a cyclic extension but [imath]\mathbb{Q}(\sqrt[3]{2}) / \mathbb{Q}[/imath] is not, so this is a contradiction? |
463190 | Show that [imath]\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2[/imath]
How to show that [imath]\gcd(a + b, a^2 + b^2) = 1\mbox{ or } 2[/imath] for coprime [imath]a[/imath] and [imath]b[/imath]? I know the fact that [imath]\gcd(a,b)=1[/imath] implies [imath]\gcd(a,b^2)=1[/imath] and [imath]\gcd(a^2,b)=1[/imath], but how do I apply this to that? | 307545 | How can I find the possible values that [imath]\gcd(a+b,a^2+b^2)[/imath] can take, if [imath]\gcd(a,b)=1[/imath]
If [imath]\gcd(a,b)=1[/imath], how can I find the values that [imath]\gcd(a+b,a^2+b^2)[/imath] can possibly take? I can't find a way to use any of the elemental divisibility and gcd theorems to find them. |
463247 | construction of a sequence of smooth function
Consider [imath]r>0[/imath] , [imath]1<p< \infty [/imath]. Consider [imath]K [/imath] a compact set in [imath]R^n[/imath] with [imath]K \subset B(x_0 , r)[/imath]. Define [imath] B = \{ u \in C^{\infty}_{0}(B(x_0 , r)) ; u=1 \ on \ K , 0 \leq u \leq 1\}[/imath] [imath] A= \{ u \in C^{\infty}_{0}(B(x_0 , r)) ; u=1 \ \textit{in a open neighborhood of K } \}[/imath] A natural question is : given [imath]u \in B[/imath] , exists a sequence [imath]v_n \subset A[/imath] with [imath]v_n \rightarrow u[/imath] in [imath]H^{1,p}( B(x_0 , r))[/imath] ? I tried to prove this fact , but nothing... I draw a picture and I belive the result is true. Someone can give me a help ?If i prove the affirmation , then I understand a theorem in the book that i am studying. thanks in advance =) | 463085 | aproximation in Sobolev Spaces
consider [imath]r>0 , p>1[/imath] and [imath]K \subset B(x_0 , 2r) \subset R^n[/imath] . [imath]K[/imath] compact. Define the sets : [imath]A = \{ u \in C^{\infty}_{0} (B(x_0 , 2r)); \textit{ such that } \ u=1 \textit{ in a open neighborhood of K} \}[/imath] [imath] B = \{ u \in H^{1,p}_{0} (B(x_0 , 2r)) \cap C (B(x_0 , 2r)) : u=1 \ on \ K , \ 0 \leq u \leq 1\}[/imath] I want to show that [imath] \displaystyle\inf_{u \in A} \displaystyle\int_{B(x_0 , 2r)} |\nabla u|^p \ dx= \displaystyle\inf_{u \in B} \displaystyle\int_{B(x_0 , 2r)} |\nabla u|^p \ dx[/imath] For this i am trying to aproximate (in the sense of the norm of [imath]H^{1,p}[/imath]) the function of B , by function of [imath]A[/imath]. Somenone can give me a hint ? thanks in advance |
463470 | simplifying an equation
I gave details here of my last question...I hope this helps I am having doubt over an equation. That is my calculation. Can anybody check and find the error, if any. Specially in the last line. I am confused. Thanks a lot. NOTE : please check only last two equations. Here I am asking whether we can write [imath]min\{d(g,g'),d(h,h')\}[/imath] as [imath]min\{ecc(g),ecc(h)\}[/imath] where [imath]d(g,g')[/imath] runs over every vertex [imath]g \in V(G)[/imath]. We know that [imath]ecc(g,h)[/imath]= max{d(g,g'),d(h,h')} can be written as [imath]max\{ecc(g),ecc(h)\}[/imath]. Am I right in writing [imath]min\{d(g,g'),d(h,h')\}[/imath] as [imath]min\{ecc(g),ecc(h)\}[/imath]. If I am wrong please help me there. If any other details needed, please mention that. I will do that too. We took vertex as [imath](g,h)[/imath] because its a vertex of product graph and for such graphs vertex set is the cartesian product of [imath]V(G)[/imath] and [imath](VH)[/imath]. http://en.wikipedia.org/wiki/Glossary_of_graph_theory | 463417 | doubt over an equation
I am having doubt over an equation. That is my calculation. Can anybody check and find the error, if any. Specially in the last line. I am confused. Thanks a lot. NOTE : please check only last two equations. Here I am asking whether we can write [imath]min\{d(g,g'),d(h,h')\}[/imath] as [imath]min\{ecc(g),ecc(h)\}[/imath] where [imath]d(g,g')[/imath] runs over every vertex [imath]g \in V(G)[/imath]. We know that max{d(g,g'),d(h,h')} can be written as [imath]max\{ecc(g),ecc(h)\}[/imath]. Am I right in writing [imath]min\{d(g,g'),d(h,h')\}[/imath] as [imath]min\{ecc(g),ecc(h)\}[/imath]. If I am wrong please help me there. If any other details needed, please mention that. I will do that too. We took vertex as [imath](g,h)[/imath] because its a vertex of product graph and for such graphs vertex set is the cartesian product of [imath]V(G)[/imath] and [imath](VH)[/imath]. |
463607 | Even dimensionality for skew-symmetric form
From Humphreys' Introduction to Lie Algebras and Representation Theory, written in parentheses: (It can be shown that even dimensionality is a necessary condition for existence of a non-degenerate bilinear form satisfying [imath]f(v,w)=-f(w,v)[/imath].) Why is this? Suppose we have odd dimensionality. So [imath]n[/imath] is odd. We have an [imath]n\times n[/imath] matrix [imath]S[/imath] with independent rows such that [imath]v^TSw=-w^TSv[/imath] for all vectors [imath]v,w[/imath]. How does this provide a contradiction? | 165116 | Odd-dimensional complex skew-symmetric matrix has eigenvalue [imath]0[/imath]
There is the standard proof using [imath]\det(A)=\det( A^{T} ) = \det(-A)=(-1)^n \det(A)[/imath] I would like a proof that avoids this. Specifically, there is the proof that for [imath]A[/imath] a [imath]\bf{real} [/imath] matrix, the transpose is the same as the adjoint, which gives (using the complex inner product) [imath]\lambda \|x\|^2 =\langle Ax, x \rangle= \langle x, -Ax \rangle=-\overline{\lambda } \|x\|^{2}[/imath], so any eigenvalue is purely imaginary. Then we conclude that, since any odd-dimensional real matrix has a real eigenvalue, that eigenvalue must be zero. This argument doesn't work for a general complex skew-symmetric matrix. Is there something I'm missing, is there a way to modify this argument to get that zero is an eigenvalue for the complex case? Also, can somebody please give a geometric reason why odd-dimensional skew-symmetric matrices have zero determinant (equiv., a zero eigenvalue)? Thanks! |
463194 | How do I show that [imath]\gcd(a^2, b^2) = 1[/imath] when [imath]\gcd(a,b)=1[/imath]?
How do I show that [imath]\gcd(a^2, b^2) = 1[/imath] when [imath]\gcd(a,b)=1[/imath]? I can show that [imath]\gcd(a,b)=1[/imath] implies [imath]\gcd(a^2,b)=1[/imath] and [imath]\gcd(a,b^2)=1[/imath]. But what do I do here? | 460213 | prove that if [imath]\gcd(a,b)=1[/imath] then [imath]\gcd(a^2, b^2)=1[/imath]
Using Bezout's identity I have shown [imath](a,b^2)=(b,a^2)=1[/imath] but what should be the next step? Thanks. |
463808 | Strange property of polynomial functions. Is this a typo in the book?
There's one Apostol's Exercise that asks to show that if [imath]f : \mathbb{C} \to \mathbb{C}[/imath] is a polynomial function with real coefficients, then [imath]f(z)=f(i)[/imath] for every complex number [imath]z \in \mathbb{C}[/imath]. But I'm certain that this is false, because it would be saying that every polyonomial is constant. And indeed, finding a counterexample is pretty straightforward. Let [imath]f(z)=1+z+z^2[/imath] and compute [imath]f(1+i) =2+3i[/imath] and on the other hand [imath]f(i)=i[/imath] which are clearly different. Since the seccond item of the exercise asks to show that using this fact we can prove that the complex zeroes of [imath]f[/imath] occur at pairs of complex conjugates, I've managed to replace the first question by proving that [imath]f(z)=f(\bar{z})[/imath] for every [imath]z[/imath]. Is my conclusion correct? Is this really a typo? Thanks very much in advance. | 431136 | How to show that [imath]f(z) = f(i)[/imath]
Exercise problem. I do not need a full solution because I am trying to solve myself. Just a hint would be great. Let [imath]f[/imath] be a polynomial with real coefficients. How to show that [imath]f(z) = f(i)[/imath] for every complex [imath]z[/imath]? |
463574 | Operations research - summation notation
Outline: Hermione has been thinking about the imminent return of the Dark Lord, so she has been busy packing her bag with all the items required for her survival. Because she has so many different items, it is impossible to list them all here; however she knows that she can formulate the problem even without knowing those (trivial) details. She has [imath]N[/imath] items indexed from [imath]1[/imath] to [imath]N[/imath]; each item [imath]x_i[/imath] is associated with a value [imath]c_i[/imath], weight [imath]w_i[/imath] and volume [imath]v_i[/imath]. She cannot carry more than [imath]W[/imath] in weight, and the bag can only hold up to [imath]V[/imath] in volume. Items must either be in the backpack or not; i.e. we cannot put half a book in the bag! She needs to maximize the value of the items that she is carrying, because she knows she will not be able to replenish these for a very long time. How would I formulate this problem using summation notation? | 460143 | Summation notation problem
Any help is greatly appreciated! Outline: Hermione has been thinking about the imminent return of the Dark Lord, so she has been busy packing her bag with all the items required for her survival. Because she has so many different items, it is impossible to list them all here; however she knows that she can formulate the problem even without knowing those (trivial) details. She has [imath]N[/imath] items indexed from [imath]1[/imath] to [imath]N[/imath]; each item [imath]x_i[/imath] is associated with a value [imath]c_i[/imath], weight [imath]w_i[/imath] and volume [imath]v_i[/imath]. She cannot carry more than [imath]W[/imath] in weight, and the bag can only hold up to [imath]V[/imath] in volume. Items must either be in the backpack or not; i.e. we cannot put half a book in the bag! She needs to maximize the value of the items that she is carrying, because she knows she will not be able to replenish these for a very long time. How would I formulate this problem using summation notation? |
464313 | Why can't a [imath]C^1[/imath]-class mapping with nonzero derivative fill a square?
Let [imath]f: [0,1] \rightarrow \mathbb R^2[/imath] be of class [imath]C^1[/imath] with [imath]f'(t)\neq (0,0)[/imath]. Why can't [imath]f[/imath] be a Peano-type curve, i.e. [imath]f(I) \neq I\times I[/imath]? | 385165 | Why isn't there a continuously differentiable surjection from [imath]I \to I \times I[/imath]?
I was asked this question recently in an interview. Why can't there be a a continuously differentiable map [imath]f \colon I \to I \times I[/imath], which is also surjective? In contrast to just continuous, where we have examples of space filling curves. I have one proof which is direct via Sard's theorem. But the statement seems simple enough to have a proof using at most implicit (and inverse) function theorem. |
464499 | The irreducibility of determinant
Consider the determinant of the matrix [imath]A = [x_{ij}]_{n\times n}[/imath] as a polynomial in the ring [imath]\Bbb{R}[x_{11},\dots,x_{nn}][/imath]. How can I show that it is irreducible ? | 147073 | Slick proof the determinant is an irreducible polynomial
A polynomial [imath]p[/imath] over a field [imath]k[/imath] is called irreducible if [imath]p=fg[/imath] for polynomials [imath]f,g[/imath] implies [imath]f[/imath] or [imath]g[/imath] are constant. One can consider the determinant of an [imath]n\times n[/imath] matrix to be a polynomial in [imath]n^2[/imath] variables. Does anyone know of a slick way to prove this polynomial is irreducible? It feels like this should follow quite easily from basic properties of the determinant or an induction argument, but I cannot think of a nice proof. One consequence of this fact is that [imath]GL_n[/imath] is the complement of a hypersurface in [imath]M_{n}[/imath]. Thanks. |
464767 | Show that [imath]19\mid 12^{2013}+7^{2013}[/imath]
Show that [imath]19\mid 12^{2013}+7^{2013}[/imath][imath]$[/imath]$No use of modular arithmetic, have not come this content. | 448828 | Number Theory divisibilty
How can I check if [imath]12^{2013} + 7^{2013}[/imath] is divisible by [imath]19[/imath]? Also, how can I format my questions to allow for squares instead of doing the ^ symbol. |
464784 | Idempotent elements in [imath](\mathbb{Z}_n,+,\cdot)[/imath]
Can we find the idempotents in [imath](\mathbb{Z}_n,+,\cdot)[/imath] for any [imath]n[/imath]? Is there a general rule? Note: Trying to consider the prime factors! | 45747 | How many idempotent elements does the ring [imath]{\bf Z}_n[/imath] contain?
Let [imath]R[/imath] be a ring. An element [imath]x[/imath] in [imath]R[/imath] is said to be idempotent if [imath]x^2=x[/imath]. For a specific [imath]n\in{\bf Z}_+[/imath] which is not very large, say, [imath]n=20[/imath], one can calculate one by one to find that there are four idempotent elements: [imath]x=0,1,5,16[/imath]. So here is my question: Is there a general result which tells the number of the idempotent elements of [imath]{\bf Z}_n[/imath]? |
464824 | Integrating [imath]\int_{0}^{\infty} \frac{(\log x)^2}{x^2+x+1}[/imath] using residue theorem
Just out of curiosity, how does one integrate something like this using residue theory? [imath]\int_{0}^{\infty}\frac{(\log x)^2}{x^2+x+1} dx[/imath] According to Wolfram Alpha, the answer is [imath]\dfrac{16\pi^3}{81\sqrt{3}}[/imath]. I have seen similar integrals before, like here and here, and they all require seem some sort of ingenuity. I am sure local Ramanujans will come to rescue soon. :) | 294541 | Definite integral, quotient of logarithm and polynomial: [imath]I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x[/imath]
I was thinking this integral : [imath]I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x[/imath] What I do is use a Reciprocal subsitution, easy to show that: [imath]I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{\lambda ^2x^2+\lambda x+1}\text{d}x =\frac{1}{\lambda}\int_0^{\infty}\frac{(\ln x-\ln \lambda)^2}{x^2+x+1}\text{d}x \\ =\frac{\ln ^2\lambda}{\lambda}\int_0^{\infty}\frac{1}{x^2+x+1}\text{d}x+\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2x}{x^2+x+1}\text{d}x \\ =\frac{\ln ^2\lambda}{\lambda}\frac{2\pi}{3\sqrt{3}}+\frac{2}{\lambda}\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x[/imath] But I hav no idea on process the remaining integral:( Anyone knows how to solve it? THX guys! I came with another method might work for this: Recall the handy GF [imath]\frac{1}{x^2+x+1}=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k[/imath] Then we have : [imath]\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\frac{2}{\left(k+1\right)^3}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k[/imath] With [imath]\sum_{k=1}^{\infty}\frac{\sin (kx)}{k^3}=\frac{\pi ^2}{6}x-\frac{\pi}{4}x^2+\frac{x^3}{12}[/imath] What can we arrived? |
217516 | Let [imath]X[/imath] be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.
Let [imath]X[/imath] be an infinite dimensional Banach space. Prove that every basis of [imath]X[/imath] is uncountable. Can anyone help how can I solve the above problem? | 659646 | Question about Hamel's basis.
I'd like to prove that if I consider [imath]X[/imath] Banach space with [imath]\dim X =\infty[/imath], then [imath]X[/imath] can't have a countable Hamel's basis. Someone can give me a hint? Even a counterexample is enough. I think this can be linked with Baire's theorem about Banach's space, but I'm not really sure... Thanks! |
465013 | Solve covariance matrix of multivariate gaussian
This is a practical, and basic question. I have a multivariate Gaussian in [imath]M[/imath] dimensions with center [imath]\mu[/imath] (known, lets assume [imath]0[/imath]) and some points [imath]p[/imath] where I have the value of [imath] \ln(L)= -\frac{1}{2}\mathbf{x}^{\rm T}\boldsymbol\Sigma^{-1}\mathbf{x} [/imath] My questions is how do I find [imath]\Sigma[/imath] exactly, given [imath]\ln(L)[/imath] and [imath]\mathbf{x}[/imath] values? I.e. which quantities go into a linear matrix equation solver [imath]Ax=b[/imath]. In particular, I am confused how to make a system of linear equations built from this one matrix multiplication. | 465306 | system of matrix equations
I have the equation [imath] \mathbf{x}^T A \mathbf{x} = b [/imath], where [imath]b[/imath] is a scalar, [imath]\mathbf{x}[/imath] a vector of size [imath]M[/imath], and A a matrix of size [imath]M\times M[/imath]. [imath]b[/imath] and [imath]\mathbf{x}[/imath] are given. How many such equations do I need to find the (common) [imath]A[/imath], and how do I do it? |
465547 | How Prove this [imath] \sum\limits_{k=1}^{2n} { 2n \choose k } { 2n \choose 2n-k } = 2^{4n} - 1[/imath]
[imath] \sum_{k=1}^{2n} { 2n \choose k } { 2n \choose 2n-k } = \binom{4n}{2n} - 1[/imath] This equation have nice methods? Thank you everyone, This problem is from this Proving [imath]\binom{2n}{n}\ge\frac{2^{2n-1}}{\sqrt{n}}[/imath] | 148583 | Combinatorial proof of summation of [imath]\sum\limits_{k = 0}^n {n \choose k}^2= {2n \choose n}[/imath]
I was hoping to find a more "mathematical" proof, instead of proving logically [imath]\displaystyle \sum_{k = 0}^n {n \choose k}^2= {2n \choose n}[/imath]. I already know the logical Proof: [imath]{n \choose k}^2 = {n \choose k}{ n \choose n-k}[/imath] Hence summation can be expressed as: [imath]\binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + \cdots + \binom{n}{n}\binom{n}{0}[/imath] One can think of it as choosing [imath]n[/imath] people from a group of [imath]2n[/imath] (imagine dividing a group of [imath]2n[/imath] into [imath]2[/imath] groups of [imath]n[/imath] people each. I can get [imath]k[/imath] people from group [imath]1[/imath] and another [imath]n-k[/imath] people from group [imath]2[/imath]. We do this from [imath]k = 0[/imath] to [imath]n[/imath]. |
463331 | well ordered and totally ordered set
Can anyone please help me in this? Prove that every well ordered set is a totally ordered set. To start this proof will this be sufficient: Let [imath]x_1, x_2 \in X.[/imath] Let [imath]A=\{ x_1, x_2 \}[/imath]. If [imath]x_1[/imath] is the least element, then [imath]x_1\le x2[/imath]. If [imath]x_2[/imath] is the least element, then [imath]x_1\ge x_2[/imath]. How do I end the proof? | 463514 | Well-ordered sets, properties of an element that is not largest in the set
Can anyone please help me in this question: [imath](X,\leq )[/imath] is a well-ordered set. [imath]\forall x\in X[/imath], either [imath]x[/imath] is the largest element of [imath]X[/imath] or there exists [imath]y\in X[/imath] such that (i) [imath]x<y[/imath] and (ii) if [imath]x_0\in X[/imath] such that [imath]x\leq x_0\leq y[/imath], then either [imath]x = x_0[/imath] or [imath]x_0 = y[/imath]. Will this be sufficient to start the proof: If [imath]x[/imath] is the largest element then there is nothing to prove. Suppose [imath]x[/imath] is not the largest element. [imath]A=\{ z\in X \mid x<z\}\neq \emptyset[/imath] How do I proceed? |
461243 | Prove that [imath]F_{n}^{2}+F_{n+1}^{2}=F_{2n+1}[/imath]
Prove that [imath]F_{n}^{2}+F_{n+1}^{2}=F_{2n+1}[/imath] This identity holds for [imath]n>=1[/imath] Instead of using induction, how do I prove it in a geometry approach? | 136645 | Showing that [imath]f_{2n+1}=f_{n+1}^2+f_n^2[/imath].
I am trying to solve the following exercise: Let [imath]f_1=1[/imath], [imath]f_2=1[/imath], [imath]f_{n+1}=f_n+f_{n-1}[/imath], where [imath]n\in\mathbb{N}[/imath]. Show that [imath]f_{2n+1}=f_{n+1}^2+f_n^2[/imath]. I have not had much progress, but this is what I have managed to observe: [imath]\sum_{k=1}^nf_{2k}=f_{2n+1}-1.[/imath] However, this is not of much help when trying to show the above. Also, [imath]f_{n+1}^2=(f_n+f_{n-1})^2=f_n^2+2f_{n+1}+f_{n-1}^2[/imath], and [imath]f_n^2=(f_{n-1}+f_{n-2})^2=f_{n-1}^2+2f_n+f_{n-2}^2[/imath]. But when I add these, I do not get anything that resembles the LHS above. Do you guys have any ideas? |
466511 | Give an example to show that a factor ring of an integral domain may be a field
A) Give an example to show that a factor ring of an integral domain may be a field B) Give an example to show that a factor ring of an integral domain have divisors of 0. C) Give an example to show that a factor ring of a ring with divisors of 0 may be integral domain. A) [imath]\mathbb{Q}[X]/\langle x+1\rangle[/imath] and [imath]\mathbb{Z}/4\mathbb{Z}[/imath] B) [imath]\mathbb{Z}/4\mathbb{Z}[/imath], C) [imath]\mathbb{Z}_2/\mathbb{Z}_6[/imath] Are my answers for A), B), C) correct? And can you give a few more examples for A), B) and C)? | 459924 | Give an example to show that a factor ring of a ring with divisors of 0 may be an integral domain
Fraleigh Section 23 12.Give an example to show that a factor ring of an integral domain may be a field 13.Give an example to show that a factor ring of an integral domain may have divisors of [imath]0[/imath] 14.Give an example to show that a factor ring of a ring with divisors of 0 may be an integral domain. For 12, [imath]Z/2Z[/imath] works For 13, [imath]Z/4Z[/imath] works For 14, [imath]4Z/8Z[/imath] ( isomorphic to Z[imath]_2[/imath]?) works. -> I just figured out 4Z is not a ring with divisors of 0... I'm unsure about my answer for 14. Is it correct? And any other examples for 12,13,14? |
466709 | [imath]\Sigma[/imath]-product and Tychonoff product
I want to say that the below example is countable compact but not closed, [imath]X[/imath] a Tychonoff product [imath][0,1]^{ω_1}[/imath] and as [imath]Y[/imath] the [imath]\Sigma[/imath]-product [imath] \{ x ∈[0,1]^{ω_1}:|\{α<ω_{1} :x(α)≠0 \}|≤ω \}\;.[/imath] The space [imath]X[/imath] is compact by Tychonoff Theorem. If we want to show that Y is countably compact, will we prove that there ia an infinite subspace s.t. has a accumulation point? and for non-closed , do we prove that, this accumulation point do not belong to [imath]Y[/imath]? But I fail to complet it, please guide me. | 465599 | About [imath] \{ x \in[0,1]^{\omega_1}:|\{\alpha<\omega_{1} :x(\alpha)\ne 0 \}|\le\omega \}[/imath]
Take [imath]X[/imath] a Tychonoff product [imath][0,1]^{\omega_1}[/imath] and as [imath]Y[/imath] the [imath]\Sigma[/imath]-product [imath] \{ x ∈[0,1]^{\omega_1}:|\{\alpha<\omega_{1} :x(\alpha)\ne 0 \}|\le\omega \}\;.[/imath] The space [imath]X[/imath] is compact by Tychonoff Theorem and [imath]Y[/imath] is a countably compact and not closed subspace of [imath]X[/imath]. why [imath]Y[/imath] is a countably compact and not closed subspace of [imath]X[/imath]? |
466743 | Prove that [imath]\bigcap_{k = 1}^\infty C_k[/imath] is also compact and connected.
Let [imath]X[/imath] be a Hausdorff space and let [imath]C_0 \supset C_1 \supset ...[/imath] be a decreasing sequence of compact connected subsets of [imath]X[/imath]. Prove that [imath]\bigcap_{k = 1}^\infty C_k[/imath] is also compact and connected. I have the compact part. Just don't know how to show it is connected. | 310455 | Intersection of Connected Sets
This is an old exam question that I don't have a solution to: Let [imath]X[/imath], a compact Hausdorff (T2) space, and let [imath]\phi[/imath] a family of closed, non-empty, and connected subsets of [imath]X[/imath], such that for every [imath]A, B \in \phi[/imath], [imath]A \subset B[/imath] or [imath]B \subset A[/imath]. Prove that [imath]Y:= \cap \{A: A \in \phi\}[/imath] is connected. I tried to solve this question with a friend, and this is what we came up with: Obviously, [imath]X[/imath] is normal space (T4). We tried to see what happens if [imath]Y = U_1 \cup U_2[/imath], disjoint and open sets (and?) We tried to use nets and maybe see if we can create a net that converges to [imath]x \neq y[/imath] (and contradict X being Hausdorff space). We tried to work with continuous functions, but this idea didn't lead us anywhere either. I feel like there is a simple observation that we are missing. Any ideas? Thanks! |
29231 | Arbitrary intersection of closed, connected subsets of a compact space connected?
Let [imath](B_i)_{i\in I}[/imath] be an indexed family of closed, connected sets in a compact space X. Suppose [imath]I[/imath] is ordered, sucht that [imath]i < j \implies B_i \supset B_j[/imath]. Is [imath]B = \bigcap_i B_i[/imath] necessarily connected? I can prove it, if I assume [imath]X[/imath] to be Hausdorff as well: If [imath]B[/imath] is not connected, then there are two disjoint, closed, nonempty sets [imath]C[/imath], [imath]D[/imath] in [imath]B[/imath], such that [imath]C \cup D = B[/imath]. Now these sets are also closed in [imath]X[/imath], hence by normality there exist open disjoint neighborhoods [imath]U[/imath], [imath]V[/imath] of [imath]C[/imath] and [imath]D[/imath], respectively. Then for all [imath]i[/imath]: [imath]B_i \cap U^c \cap V^c \ne \emptyset[/imath], since [imath]B[/imath] is contained in [imath]B_i[/imath] and [imath]B_i[/imath] is connected. Thus we must also have [imath] B \cap U^c \cap V^c = \bigcap_i B_i \cap U^c \cap V^c \ne \emptyset [/imath] by compactness and the fact that the [imath]B_i[/imath] satisfy the finite intersection property. This is a contradiction to the choice of [imath]U[/imath] and [imath]V[/imath]. I can neither see a counterexample for the general case, nor a proof. Any hints would be greatly appreciated! Thanks, S. L. | 1284713 | Let [imath]K_1 \supset K_2 \supset... [/imath] be a sequence of connected compact subsets of [imath] \Bbb R^2 [/imath]. Is [imath] K = \cap_{i=1}^\infty K_i [/imath] is connected?
I have managed to write down two proofs showing the connectedness of [imath]K[/imath]. But still shaky about both of them. Here are the proofs: 1)Suppose [imath]K[/imath] is disconnected. Then we write it's separation as [imath] K = U \cup V [/imath]. Hence [imath] K [/imath] is open since U and V are open. Now, since each [imath]K_i[/imath] is compact in [imath]\Bbb R^2[/imath], hence by Heine-Borel theorem it is closed and bounded. Hence [imath]K[/imath] is also closed and bounded and non-empty. We have got that [imath]K[/imath] is open as well as closed. But the only open and closed sets in [imath]\Bbb R^2[/imath] are [imath]{\emptyset}[/imath] and the whole [imath]\Bbb R^2[/imath]. But because [imath]K[/imath] is non-empty and bounded, hence it is not one of them. Hence our assumption was wrong. Hence K must be connected. another proof I thought was like this: 2)Let [imath]f : K_1 \to \{0,1\} [/imath] be a continuous function. Then it's constant by connectedness of [imath]K_1[/imath] . Say [imath]f(K_1)=0[/imath]. Then restricting the function to K, we get that [imath]f(K)=0[/imath]. Can any other proof be given? are these proofs valid? |
466670 | Question on the existence of finite open subsets in [imath]\mathbb{R}^{k}[/imath]
Dear reader of this post, can any subset [imath]S \subset \mathbb{R}^{k}[/imath], with the euclidean distance for [imath]x,y \in \mathbb{R}^{k}[/imath], be open and contain only finitely many elements? I am looking forward for your reply! | 326154 | Is every open subset of [imath] \mathbb{R} [/imath] uncountable?
Is every open subset of [imath]\mathbb{R}[/imath] uncountable? I was crafting a proof for the theorem that states every open subset of [imath]\mathbb{R}[/imath] can be written as the union of a countable number of disjoint intervals when this question came up. I feel like the answer is yes, but I'm not sure how to go about proving it or whether there is a crazy construction (like the Cantor Set) that creates a countable, open subset of [imath]\mathbb{R}[/imath]. Any ideas? |
294383 | Evaluate [imath]\int_{0}^{+\infty }{\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\frac{1}{{{x}^{2}}}\text{d}x}[/imath]
Evaluate : [imath]\int_{0}^{+\infty }{\left( \frac{x}{{{\text{e}}^{x}}-{{\text{e}}^{-x}}}-\frac{1}{2} \right)\frac{1}{{{x}^{2}}}\text{d}x}[/imath] | 729140 | How to calculate the improper integral [imath]\int_{0}^{\infty}\Big(\frac{x}{\mathrm{e}^x-\mathrm{e}^{-x}}-\frac{1}{2}\Big)\frac{\mathrm{d}x}{x^2}[/imath]
How to calculate the improper integral [imath]\int_{0}^{\infty}\Big(\frac{x}{\mathrm{e}^x-\mathrm{e}^{-x}}-\frac{1}{2}\Big)\frac{\mathrm{d}x}{x^2}[/imath] Do you have same idea? |
446631 | Number of draws required for ensuring 90% of different colors in the urn with large populations
My problem is: An urn contains [imath]10000000[/imath] ([imath]10^7[/imath]) different colored balls, namely [imath]K_1, K_2,\dots,K_n (n=10^7)[/imath] with [imath]K_1=1000, K_2=1000,\dots,K_n=1000[/imath]. My question is: How many balls do I need to extract to ensure to obtain [imath]90\%[/imath] ([imath]1000000[/imath] or [imath]10^6[/imath]) of all the colors? Many thanks in advance. | 446622 | drawing at least 90% of colors from urn with large populations
My problem is: suppose I have an urn containing balls of [imath]n = 10000000[/imath] (i.e., [imath]10^7[/imath]) different colors, with [imath]1000[/imath] balls of each color (so the total number of balls is [imath]1000n = 10^{10}[/imath]). Suppose I draw [imath]100000000[/imath] (i.e. [imath]10^8 = 10n[/imath]) balls. My question is: how do I calculate the probability that I have drawn at least 90% (in this case [imath]1000000[/imath] or [imath]10^6[/imath]) of the different colors? Many thanks in advance |
467525 | Regarding uses of [imath]i[/imath] (square root of [imath]-1[/imath])
Are there any uses of 'square root of [imath]-1[/imath]' in practical life ; like in Physics ? | 138325 | Simple applications of complex numbers
I've been helping a high school student with his complex number homework (algebra, de Moivre's formula, etc.), and we came across the question of the "usefulness" of "imaginary" numbers - If there not real, what are they good for? Now, the answer is quite obvious to any math/physics/engineering major, but I'm looking for a simple application that doesn't involve to much. The only example I've found so far is the formula for cubic roots applied to [imath]x^3-x=0[/imath], which leads to the real solutions by using [imath]i[/imath]. Ideally I'd like an even simpler example I can use as motivation. Any ideas? |
467805 | Solution to over damped harmonic spring
I'm trying to programmatically model a damped harmonic spring for use in mobile UI animations (physics isn't my background, please pardon any misconceptions). Having derived the parameters for the general case equation, I can iteratively calculate values until I reach a suitable threshold, though because this is bound to "simple" C math trig and exp functions on the CPU, the 4000-some-odd steps can cause about 0.25 seconds on slow devices lag while it calculates. I'd like to speed this up using my platform's super-optimized vector and BLAS/LAPACK functions. The requirement for doing this is precalculating the number of steps necessary to reach my threshold value. In the underdamped case, where the roots of the characteristic function of the differential equation are non-real, I can use algebraic tricks to get my values: [imath]x(t) = c_{1}e^{r_{1}}\cos(i_{1}t) + c_{2}e^{r_{2}}\sin(i_{2}t)[/imath] (Given [imath]r_{1}[/imath], [imath]i_{1}[/imath], [imath]r_{2}[/imath], and [imath]i_{2}[/imath] are the real and irrational components of my two roots, respectively.) Knowing that [imath]r_{1} = r_{2}[/imath] and [imath]i_{1} = -i_{2}[/imath], I can simplify to: [imath]x(t) = c_{1}e^{r_{1}}\cos(i_{1}t)[/imath] And get my desired value of [imath]t[/imath] for my threshold [imath]a[/imath]: [imath]t = \arccos(a / c_{1} / e^{r_{1}}) / i_{1}[/imath] When the roots are real, the equation is a lot simpler: [imath]x(t) = c_{1}e^{r_{1}} + c_{2}e^{r_{2}}[/imath] But I don't have my trig functions floating around to help me solve it (even if I did, the irrational components being 0 would cause problems, of course). Take the concrete example on pages 3-4 of this document (my bible during this process), since they at least solve cleanly: [imath]x(t) = 1.5e^{-t} - 0.5e^{-3t}[/imath] I know how I would solve for [imath]t[/imath] to get my for when [imath]x(t) = a[/imath] on paper, by setting [imath]x=e^{t}[/imath], solving, and back substituting, but I don't have that luxury here. How would I go about solving for my threshold value in this (supposedly) simplified case? | 467804 | Solution to over-damped harmonic spring
(A kind soul at physics.stackexchange suggested I post here as well, sorry if out of bounds.) I'm trying to programmatically model a damped harmonic spring for use in mobile UI animations (physics mathematics isn't my background, please pardon any misconceptions). Having derived the parameters for the general case equation, I can iteratively calculate values until I reach a suitable threshold, though because this is bound to "simple" trigonometric and [imath]e^{x}[/imath] functions on the CPU, the 4000-some-odd steps can cause about 0.25 seconds lag on slow devices while it calculates. I'd like to speed this up using my platform's super-optimized vector and BLAS/LAPACK variants. The requirement for doing this is precalculating the number of steps necessary to reach my threshold value. In the underdamped case, where the roots of the characteristic function of the differential equation are non-real, I can use algebraic tricks to get my values: [imath]x(t) = c_{1}e^{r_{1}}\cos(i_{1}t) + c_{2}e^{r_{2}}\sin(i_{2}t)[/imath] (Given [imath]r_{1}[/imath], [imath]i_{1}[/imath], [imath]r_{2}[/imath], and [imath]i_{2}[/imath] are the real and irrational components of my two roots, respectively.) Knowing that [imath]r_{1} = r_{2}[/imath] and [imath]i_{1} = -i_{2}[/imath], I can simplify to: [imath]x(t) = c_{1}e^{r_{1}}\cos(i_{1}t)[/imath] And get my desired value of [imath]t[/imath] for my threshold [imath]a[/imath]: [imath]t = \arccos(a / c_{1} / e^{r_{1}}) / i_{1}[/imath] When the roots are real, the equation looks a lot simpler: [imath]x(t) = c_{1}e^{r_{1}} + c_{2}e^{r_{2}}[/imath] However, I don't have my trig functions floating around to help me solve it (even if I did, the irrational components being 0 would cause problems, of course). Take the concrete example on pages 3-4 of this document (my bible during this process), since they at least solve cleanly: [imath]x(t) = 1.5e^{-t} - 0.5e^{-3t}[/imath] I know how I would solve for [imath]t[/imath] to get my for when [imath]x(t) = a[/imath] on paper, by setting [imath]x=e^{t}[/imath], solving, and back substituting, but I don't have that luxury here. I can make a few assumptions: the roots and constants are all real. I'm always going to be looking for the smallest, first, positive value of [imath]t[/imath]. Obviously, the iterative solution is the simplest for this case, but in the end that would involve more steps and therefore be slower no matter what my other optimizations would be. How, then, would I go about solving for my threshold value algorithmically in this (supposedly) simplified case? Addendum The underdamped solution presents an extra requirement. The motion curve will oscillate back and forth a few times across the endpoint. Therefore, "first and lowest" [imath]t[/imath] requirement is not necessarily true. In my current, iterative code, the threshold value is both checked against the distance from the current [imath]x(t)[/imath] to the endpoint, as well as to the distance from the previous [imath]x(t)[/imath] as well to allow for a number of oscillations. This might make a more efficient solution nearly impossible. |
467824 | Proving a subset of the natural numbers has a minimal element
Given [imath]A \subseteq \mathbb{N}, A \neq \emptyset[/imath], prove that A has a minimum element. Can anyone help me with this problem? | 379775 | Natural Numbers and Well ordering
I have to show that in any non empty subset of [imath]N[/imath] there is least element. Note: This is not a homework question. So this is how my incomplete proof looks like. And I tried this by induction: let [imath]S[/imath] be nonempty subset of [imath]N[/imath]. I defined [imath]M = \{m \in N \mid m\le s \text{ for all } s \in S\}.[/imath] Then I showed [imath]1[/imath] is in [imath]M[/imath]. I did this using proof by contradiction. Now if [imath]s[/imath] is in [imath]S[/imath] then [imath]s < s^+[/imath]. So [imath]s^+[/imath] is not in [imath]M[/imath]. Thus [imath]M[/imath] is not equal to [imath]N[/imath]. Thus there exists [imath]z[/imath] in [imath]M[/imath] such that such that [imath]z^+[/imath] is not in [imath]M[/imath]. Now I have to show [imath]z[/imath] is the least element of [imath]S[/imath]. Since [imath]z[/imath] is in [imath]M[/imath] by the way [imath]M[/imath] is defined [imath]z[/imath] must be least element of [imath]S[/imath]. But what if [imath]z[/imath] is not in [imath]S[/imath]? What should I do? Help please. |
467941 | [imath]10^n+1[/imath] is never prime?
I tested a few numbers of the form [imath]10^n+1[/imath] (i.e. [imath]100000001[/imath]) besides [imath]11[/imath] and [imath]101[/imath] and they were all composite. Clearly some of them if we have an even number of [imath]0[/imath]'s in between are multiples of [imath]11[/imath], but is it just a coincidence that the ones with odd number of [imath]0[/imath]'s are also composite? | 34877 | Are there infinitely many primes and non primes of the form [imath]10^n+1[/imath]?
Prove that there are infinitely many primes and non-primes in the numbers [imath]10^n+1[/imath], where [imath]n[/imath] is a natural number. So numbers are 101, 1001, 10001 etc. |
468598 | Intersection of a closed set and compact set is compact
I've been stuck on the following problem for several days: Let [imath](M,d)[/imath] be an arbitrary metric space and [imath]S, T[/imath] be subsets of [imath]M[/imath]. If [imath]S[/imath] is closed and [imath]T[/imath] is compact, then [imath]S \cap T[/imath] is compact. I know that if [imath]T[/imath] is compact, [imath]T[/imath] is closed and bounded. That would imply that [imath]S \cap T[/imath] is also closed and bounded since [imath](S \cap T) \subseteq T[/imath]. Also since [imath]S[/imath] is closed, [imath]S[/imath] contains all its accumulation points. Other than writing down the definitions, I really don't know how to proceed. Could someone give me a hint? | 35038 | Is the intersection of a closed set and a compact set always compact?
I am going through Rudin's Principles of Mathematical Analysis in preparation for the masters exam, and I am seeking clarification on a corollary. Theorem 2.34 states that compact sets in metric spaces are closed. Theorem 2.35 states that closed subsets of compact spaces are compact. As a corollary, Rudin then states that if [imath]L[/imath] is closed and [imath]K[/imath] is compact, then their intersection [imath]L \cap K[/imath] is compact, citing 2.34 and 2.24(b) (intersections of closed sets are closed) to argue that [imath]L \cap K[/imath] is closed, and then using 2.35 to show that [imath]L \cap K[/imath] is compact as a closed subset of a compact set. Am I correct in believing that this corollary holds for metric spaces, and not in general topological spaces? |
468718 | [imath]f[/imath] extends to a smooth map [imath]\tilde{f} : S^2 \to S^2[/imath]. - Is the proof legit?
Let [imath]f : \{\mathbb{R}^2 - (1, 0) - (-1, 0)\} \to \mathbb{R}^2[/imath] be the function [imath]f(z) = \frac{1}{z-1} - \frac{1}{\bar{z}+1}[/imath]. Show that [imath]f[/imath] extends to a smooth map [imath]\tilde{f} : S^2 \to S^2[/imath]. We define [imath]\hat{f}(\infty) = 0, \hat{f}(\pm 1,0) = \infty[/imath]. And we show \begin{align*} \lim_{z \to \infty} f(z) & = \frac{1}{z-1} - \frac{1}{\bar{z}+1} = 0.\\ \lim_{z \to \pm 1 + 0i} f(z) & = \frac{1}{z-1} - \frac{1}{\bar{z}+1} = \infty. \end{align*} Hence, [imath]f[/imath] extends smoothly into [imath]\hat{f}[/imath]. Thanks~very much! | 466300 | Show that [imath]f[/imath] extends to a smooth map.
Identify [imath]\mathbb{R}^2[/imath], with coordinates [imath]x, y[/imath], with [imath]\mathbb{C}[/imath], with coordinate [imath]z = x + iy[/imath]. Likewise, identify a copy of [imath]\mathbb{R}^2[/imath] with coordinates [imath]u, v[/imath] with [imath]\mathbb{C}[/imath] with coordinate [imath]w = u+iv[/imath]. Let [imath]f : \{\mathbb{R}^2 - (1, 0) - (-1, 0)\} \to \mathbb{R}^2[/imath] be the function [imath]f(z) = \frac{1}{z-1} - \frac{1}{\bar{z}+1}[/imath]. Show that [imath]f[/imath] extends to a smooth map [imath]\tilde{f} : S^2 \to S^2[/imath], where [imath]S^2[/imath] (or [imath]\mathbb{CP}^1[/imath] is the one-point compactification of [imath]\mathbb{R}^2[/imath] (or [imath]\mathbb{C})[/imath]. Following FBD's comment below, I want to show: 1, [imath]f[/imath] is continuous, which is obvious; 2, [imath]f[/imath] is surjective; 3, [imath]f^{-1}[/imath] maps compact set to compact set Correct? Thank you very much. |
388988 | Formal proof that a complete graph [imath]K_n[/imath] is Class 1 if n is even and it is Class 2 if n is odd.
Recall that from Vizing's theorem, the chromatic index [imath]\chi'(G)[/imath] can only be either [imath]\Delta(G)[/imath] (we call the graph Class 1) or [imath]\Delta(G) +1[/imath] (Class 2). My question is regard to determining the class of a given complete graph [imath]K_n[/imath]. I don't know how complicated this question is, so it is very appreciable if there is a suggestion for me to tackle this. | 226664 | Coloring Graph Problem
If G is a graph containing no loops or multiple edges, then the edge-chromatic number [imath]X_e(C)[/imath] of G is defined to be the least number of colours needed to colour the edges of G in such a way that no two adjacent edges have the same colour. How to find edge-chromatic number : 1)Complete Graph([imath]K_n[/imath]) [I know [imath]k_n[/imath] is (n-1)regular and if n is even then edge-chromatic number of [imath]k_n[/imath] is n-1 and if n is odd then edge-chromatic number of [imath]k_n[/imath] is n,but how to prove it??] 2)Wheel([imath]W_n[/imath]) |
322481 | Principal [imath]n[/imath]th root of a complex number
This is really two questions. Is there a definition of the principal [imath]n[/imath]th root of a complex number? I can't find it anywhere. Presumably, the usual definition is [imath][r\exp(i\theta)]^{1/n} = r^{1/n}\exp(i\theta/n)[/imath] for [imath]\theta \in [0,2\pi)[/imath], but I have yet to see this anywhere. Is this because it has bad properties? For instance, according to this definition is it true that for all complex [imath]z[/imath] it holds that [imath](z^{1/a})^b = (z^b)^{1/a}[/imath]? | 2866185 | principal root confusion
Im quite confused over the use of the radical. I can understand its definition in [imath]\mathbb{R}[/imath], with [imath]\sqrt[n]{x}[/imath] being defined for [imath]x>0[/imath] if [imath]n[/imath] an even integer greater than one and [imath]x[/imath] any real number if [imath]n[/imath] an odd integer greater than one. I believe its referred to as the principal [imath]n[/imath]th root for even [imath]n[/imath] as it outputs the positive root rather than the negative root. In [imath]\mathbb{R}[/imath] there are no square roots of negative numbers, so on this domain it's undefined. I also know it defines the same function of [imath]x[/imath] as [imath]x^{1/n}[/imath] for [imath]n[/imath] integer greater than [imath]1[/imath]. In [imath]\mathbb{C}[/imath], a principal [imath]n[/imath]th root function can be defined using the same radical notation which for me is confusing. The fact there are so many ways to define the principal [imath]n[/imath]th root function in [imath]\mathbb{C}[/imath] makes it worse, and is [imath]x^{1/2}= \sqrt{x}[/imath] in [imath]\mathbb{C}[/imath] or do you have to define it that way after you have defined exactly what you mean by [imath]\sqrt{x}[/imath]. Should this principal [imath]n[/imath]th root function in [imath]\mathbb{C}[/imath] be viewed as a different function altogether from the one in [imath]\mathbb{R}[/imath], or just an extension of the domain? Thanks |
470007 | Prove that invertible metrices set is an open set in a given space, and the determinant is continuous
Given a matrix [imath]M_{n\times m}[/imath], we can think about it as a vector in [imath]\mathbb{R}^{n\times m}[/imath] (How come?). How can I prove that the set of all the invertible metrices of size [imath]n\times n[/imath] is an open set in [imath]\mathbb{R}^{n^2}[/imath], and that the function [imath]f:M\mapsto M^{-1}[/imath] is continuous? I understand that [imath]f[/imath] is the determinant of [imath]M[/imath], and [imath]f = \det(M)\ne 0[/imath], and becuase [imath]f[/imath] is a linear function, it's a continuous function. But I'm not 100% sure why [imath]f[/imath] is a linear. The invertible metrices set is [imath]\mathbb{R}^{n^2}\setminus\{0\}[/imath] (am I wrong?), and that's why the set is an open set? Thank you! | 1422770 | [imath]GL_n(R)[/imath] is open set in [imath]M_n(R)[/imath]
Show that set of all invertible [imath]n\times n[/imath] matrices with real entries (denoted by [imath]GL_n(R)[/imath]) is open set in [imath]M_n(R)[/imath]. My attempt: by open set I think it means neighborhood of every point in set is contained in set, but this definition doesn't seem to help solve this question. Any other approach? |
470021 | How can one go from "no roots" to "is irreducible" in this case?
This problem is paraphrased from an old version of an exam that I will be taking, and I have no idea how one would do solve it. Let [imath]p[/imath] be a prime number, let [imath]F[/imath] be a field of characteristic [imath]p[/imath], and let [imath]c[/imath] be a field element such that [imath]\:x^{\hspace{.025 in}p}\hspace{-0.04 in}+(-c)\:[/imath] has no roots in [imath]F[/imath]. [imath]\;\;\;[/imath] Show that [imath]\:x^{\hspace{.025 in}p}\hspace{-0.04 in}+(-c)\:[/imath] is irreducible over [imath]F[/imath]. How would one solve that problem? | 88440 | Inseparable polynomial over a non-perfect field
Assume that [imath]F[/imath] is a field and [imath]\operatorname{char}(F)=p[/imath]. Let [imath]a[/imath] be an element in [imath]F[/imath] without [imath]p[/imath]th root, then the polynomial [imath]x^{p^n}-a[/imath] is irreducible and inseparable over [imath]F[/imath] for all [imath]n[/imath]. I have proved the inseparable part by considering the derivative of the polynomial, but I'm having trouble with the irreducible part. Any help? |
470159 | Matrices: left inverse is also right inverse?
If [imath]A[/imath] and [imath]B[/imath] are square matrices, and [imath]AB=I[/imath], then I think it is also true that [imath]BA=I[/imath]. In fact, this Wikipedia page says that this "follows from the theory of matrices". I assume there's a nice simple one-line proof, but can't seem to find it. Nothing exotic, here -- assume that the matrices have finite size and their elements are real numbers. This isn't homework (if that matters to you). My last homework assignment was about 45 years ago. | 1969006 | Matrices: Example where [imath]AB = I \ne BA[/imath]
For the inverse of a matrix, we learned that to prove two matrices are inverses of each other, you must show that [imath]AB = I = BA[/imath]. However, today in class someone claimed you only have to show either [imath]AB = I[/imath] or [imath]BA = I[/imath] to prove they are inverse. I know the commutative property doesn't apply to matrices, so can someone provide an example where [imath]AB = I \ne BA[/imath]? Edit: The question type goes as follows: Given matrix [imath]A[/imath] and matrix [imath]B[/imath], show whether or not the two are inverses. -For this question, do we need to show both [imath]AB = I[/imath] and [imath]BA = I[/imath] or will one or the other suffice? |
470278 | Probability of last ball is Blue.
There are [imath]B[/imath] blue balls and [imath]R[/imath] red balls in a bag. Now, randomly two balls are removed . If both of them are of different color , then blue ball is added to the bag . If both of them are of same color , then red ball is added to the bag . Also the two drawn balls are discarded. This is done until one ball is left. What is the probability that the last ball is blue? Now, I could not even start a solution on this , so could anyone provide help in solving this ? | 445919 | black and white balls in the box
A box contains [imath]731[/imath] black balls and [imath]2000[/imath] white balls. The following process is to be repeated as long as possible. (1) arbitrarily select two balls from the box. If they are of the same color, throw them out and put a black ball into the box. (We have sufficient black balls for this). (2) if they are of different colors, place the white ball back into the box and throw the black ball away. What will happen at last? Will the process stop with a single black ball in the box or a single white ball in the box or with an empty box? I am unable to decide how to start and in which direction? should we apply probability or what? |
470626 | [imath](A\times B)\cap (C\times D)=(A\cap C)\times (B\cap D)[/imath]: Proof Strategy for Cross Products of Intersected Sets
How can I prove that [imath](A\times B)\cap (C\times D)=(A\cap C)\times (B\cap D)?[/imath] | 326240 | Proof of Cartesian product intersection
How to prove [imath]( A \times B ) \cap ( C \times D ) = ( A \cap C ) \times ( B \cap D )[/imath] ? |
470649 | If [imath]G \times G \cong H \times H[/imath] must [imath]G \cong H[/imath]?
If [imath]G[/imath] and [imath]H[/imath] are both groups and [imath]G \times G \cong H \times H[/imath], is it required that [imath]G \cong H[/imath]? | 27744 | Does [imath]G\oplus G \cong H\oplus H[/imath] imply [imath]G\cong H[/imath] in general?
In this question, The Chaz asks whether [imath]G\times G\cong H\times H[/imath] implies that [imath]G\cong H[/imath], where [imath]G[/imath] and [imath]H[/imath] are finite abelian groups. The answer is to his question is yes, by the structure theorem for finite abelian groups, as noted in the answer by Anjan Gupta. Even though I don't know the first thing about categories -- except for the things that I do know -- I'm wondering if and how such property could be expressed and proven in terms of universal properties, without actually manoevring inside the objects. For instance one may attempt to create a morphism [imath]G\to H[/imath] somehow appealing to the universal property of [imath]\oplus[/imath], and subsequently show this morphism is an isomorphism by chasing diagrams. But it seems likely that the existence of a structure theorem of some sort will be required. This question may be considered trivial or weird for someone who's fluent with categories, I don't know. This topic contains quite a few references. I haven't really worked through any of them (yet) but I couldn't find anything helpful at first sight. -- edit, So a more precise title would have been "Under what conditions that can be expressed in a universal way does [imath]G\oplus G\cong H\oplus H[/imath] imply [imath]G\cong H[/imath] ?" but I don't like long titles. -- edit2, By the comment by Alexei Averchenko, maybe It's more natural to ask this question with the product instead of the coproduct. An answer to my question with the 'real' product would be appreciated too, obviously. |
470854 | Classification of radical ideals of [imath]\mathbb{Z}[X][/imath]
The prime ideals of [imath]\mathbb{Z}[X][/imath] are [imath]0[/imath] [imath](p)[/imath] for prime integer [imath]p[/imath] [imath](f)[/imath] for irreducible polynomial [imath]f[/imath] [imath](p, f)[/imath] for prime integer [imath]p[/imath] and irreducible polynomial [imath]f[/imath] that remains irreducible mod [imath]p[/imath]. The radical ideals are intersections of prime ideals. It is easy to compute the intersection when finitely many prime integers are involved using Chinese remainder theorem: they are of the form [imath](\prod^{n}_{i=1} p_i, f)[/imath] where [imath]f[/imath] is zero or square-free mod [imath]p_i[/imath] for distinct prime integers [imath]p_i[/imath]. However, I could not compute the intersection when infinitely many prime integers are involved. Related question: if there is no simple form for all the radical ideals, is there a simple description of the Zariski-closed sets of the prime spectrum (which correspond to radical ideals)? | 300170 | Ideals of [imath]\mathbb{Z}[X][/imath]
Is it possible to classify all ideals of [imath]\mathbb{Z}[X][/imath]? By this I mean a preferably short enumerable list which contains every ideal exactly once, preferably specified by generators. The prime ideals are well-known, but I'm interested in all the ideals. I couldn't find any literature about this. I don't even know the number of generators we need (Edit: Hurkyl has pointed out that any number appears!). An ideal of [imath]\mathbb{Z}[X][/imath] restricts to some ideal of [imath]\mathbb{Z}[/imath], say [imath]n \mathbb{Z}[/imath]. For [imath]n>0[/imath] the ideal corresponds to some ideal in [imath]\mathbb{Z}/n [X][/imath], and by CRT we may assume that [imath]n[/imath] is a prime power, say [imath]n=p^k[/imath]. For [imath]k=1[/imath] we have the PID [imath]\mathbb{F}_p[X][/imath], whose ideal structure is well-understood. What happens for [imath]k=2[/imath]? The case [imath]n=0[/imath] seems to be even more complicated. Since [imath]\mathbb{Z}[X][/imath] is noetherian, every ideal is an intersection of primary ideals, and we should classify them first? What about other special classes of ideals, for example radical ones? If the question is too naive in this generality, what about special cases, or slight weakenings? For example, is it possible to classify all (reduced, finite, coprimary, ...) commutative rings which are generated by a single element? This comes down to the classification of the ideals of [imath]\mathbb{Z}[X][/imath] modulo isomorphic quotient rings. |
471509 | Matrix properties and the smallest eigenvalue
Let [imath]\lambda_{\min}[/imath] be the smallest eigenvalue of the positive definite matrix [imath]\mathbf{S}[/imath], and [imath]\|\mathbf{a}\|=r[/imath]. Then [imath] \mathbf{a}^T \mathbf{S}\mathbf{a} > \frac{1}{2}\lambda_{\min} r^2 [/imath] What properties of matrices were used to obtain the result? Thanks! | 245866 | Minimum and Maximum eigenvalue inequality from a positive definite matrix.
I got a positive definite matrix [imath]B,[/imath] that is, [imath]V(x) = x^TBx > 0[/imath] for any vector [imath] x \neq 0.[/imath] I want to show that [imath] \lambda_\min \|x\|_2^2 \leq V(x) \leq \lambda_\max \|x\|_2^2[/imath] for any [imath]x \neq 0,[/imath] where [imath]\lambda_\min[/imath] and [imath] \lambda_\max[/imath] are defined by [imath] \lambda_\min = \min \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B \rbrace[/imath] and [imath] \lambda_\max = \max \lbrace | \lambda|: \lambda \text{ is an eigenvalue of } B \rbrace[/imath] Any hint please? |
471592 | Interior of a convex set is convex
A set [imath]S[/imath] in [imath]\mathbb{R}^n[/imath] is convex if for every pair of points [imath]x,y[/imath] in [imath]S[/imath] and every real [imath]\theta[/imath] where [imath]0 < \theta < 1[/imath], we have [imath]\theta x + (1- \theta) y \in S[/imath]. I'm trying to show that the interior of a convex set is convex. If [imath]x, y \in[/imath] int [imath]S[/imath], then I know there exists open balls such that [imath]B(x) \subseteq S[/imath] and [imath]B(y) \subseteq S[/imath]. I need to show that there exists a ball [imath]B(\theta x + (1- \theta) y) \subseteq S[/imath]. Other then writing down the definitions, I don't really see how to proceed. Could someone give me a hint? | 427721 | interior points and convexity
Question: Let If [imath]P\subseteq \mathbb{R}^n[/imath] be a convex set. show that [imath]int(P)[/imath] is a convex set. I know that a point [imath]x[/imath] is said to be the interior point of the set [imath]P[/imath] if there is an open ball centred at [imath]x[/imath] that is contained entirely in [imath]P[/imath]. The set of all interior points of [imath]P[/imath] is denoted by [imath]int(P)[/imath]. Also a set is convex if [imath]x,y \in P[/imath] [imath]\implies[/imath] [imath](tx+(1-t)y) \in P[/imath] for all [imath]t \in (0,1)[/imath]. How to go about the above proof? |
471887 | Cantor set homeomorphism
I am trying to prove that [imath]\{0,1\}^\mathbb{N}[/imath] is homeomorphic to the cantor set. Consider the mapping [imath]f:\{0,1\}^\mathbb{N}\to[0,1][/imath] defined as[imath]f(x)=2\sum_{n=0}^\infty 3^{-n-1}x(n)[/imath] I think that [imath]f[/imath] being continuous and injective would be enough. how does one show it is injective? | 69905 | The Cantor set is homeomorphic to infinite product of [imath]\{0,1\}[/imath] with itself - cylinder basis - and it topology
I know the Cantor set probably comes up in homework questions all the time but I didn't find many clues - that I understood at least. I am, for a homework problem, supposed to show that the Cantor set is homeomorphic to the infinite product (I am assuming countably infinite?) of [imath]\{0,1\}[/imath] with itself. So members of this two-point space(?) are things like [imath](0,0,0,1)[/imath] and [imath](0,1,1,1,1,1,1)[/imath], etc. Firstly, I think that a homeomorphism (the 'topological isomorhism') is a mapping between two topologies (for the Cantor sets which topology is this? discrete?) that have continuous, bijective functions. So I am pretty lost and don't even know what more to say! :( I have seen something like this in reading some texts, something about [imath]f: \sum_{i=1}^{+\infty}\,\frac{a_i}{3^i} \mapsto \sum_{i=1}^{+\infty}\,\frac{a_i}{2^{i+1}} ,[/imath] for [imath]a_i = 0,2[/imath]. But in some ways this seems to be a 'complement' of what I need.... Apparently I am to use ternary numbers represented using only [imath]0[/imath]'s and [imath]1[/imath]'s in; for example, [imath]0.a_1\,a_2\,\ldots = 0.01011101[/imath]? Thanks much for any help starting out! Here is the verbatim homework question: The standard measure on the Cantor set is given by the Cantor [imath]\phi[/imath] function which is constant on missing thirds and dyadic on ternary rationals. Show the Cantor set is homeomorphic to the infinite product of [imath]\{0,1\}[/imath] with itself. How should we topologize this product? (Hint: this product is the same as the set of all infinite binary sequences) Fix a binary [imath]n[/imath]-tuple [imath](a_1,\ldots, a_n)[/imath] (for e.g., [imath](0,1,1,0,0,0)[/imath] if [imath]n = 6[/imath]). Show that the Cantor measure of points ([imath]b_k[/imath]) with [imath]b_k=a_k[/imath] for [imath]k \leq n[/imath] and [imath]b_k \in \{0,1\}[/imath] arbitrary for [imath]k>n[/imath], is exactly [imath]1/2^n[/imath]. These are called cylinders. (They are the open sets, but also closed!) |
471961 | Proving [imath]\sum_{k = 0}^n k\binom nk^2 = n \binom{2n-1}{n-1}[/imath] with combinatorial proof
Prove that [imath]\sum_{k = 0}^n k\binom nk^2 = n \binom{2n-1}{n-1}[/imath] I have to prove it in a combinatorically. No induction or algebra. | 469559 | Proving [imath]\sum_{k=0}^{n}k{n\choose k}^2 = n{2n-1 \choose n-1} [/imath]
[Corrected question] I'm struggling at proving the following combinatorical identity: [imath]\sum_{k=0}^{n}k{n\choose k}^2 = n{2n-1 \choose n-1} [/imath] I would like to see a combinatorical (logical) solution, or an algebraic solution. |
472347 | Prove, that [imath] K [/imath] needn't be normal subgroup of [imath] G [/imath]
Suppose, that [imath] K [/imath] is normal subgroup of [imath] N [/imath], and [imath] N [/imath] is also normal subgroup of [imath] G [/imath]. Prove, that [imath] K [/imath] needn't be normal subgroup of [imath] G [/imath]. I can give only counter-example? | 381035 | Normal subgroup of a normal subgroup
Let [imath]F,G,H[/imath] be groups such that [imath]F\trianglelefteq G \trianglelefteq H[/imath]. I am asked whether we necessarily have [imath]F\trianglelefteq H[/imath]. I think the answer is no but I cannot find any counterexample with usual groups. Is there a simple case where this property is not true? |
472532 | Why is the supremum of the empty set [imath]-\infty[/imath] and the infimum [imath]\infty[/imath]?
I read in a paper on set theory that the supremum and the infimum of the empty set are defined as [imath]\sup(\{\})=-\infty[/imath] and [imath]\inf(\{\})=\infty[/imath]. But intuitively I can't figure out why that is the case. Is there a reason why the supremum and infimum of the empty set are defined this way? | 432295 | Infimum and supremum of the empty set
Let [imath]E[/imath] be an empty set. Then, [imath]\sup(E) = -\infty[/imath] and [imath]\inf(E)=+\infty[/imath]. I thought it is only meaningful to talk about [imath]\inf(E)[/imath] and [imath]\sup(E)[/imath] if [imath]E[/imath] is non-empty and bounded? Thank you. |
472579 | evaluating sum of limits
I was given this problem in definite integral and limits chapter [imath]\lim_{n \to \infty}\left( \frac{n}{n^2 + 1^2}+ \frac{n}{n^2 + 2^2} + \frac{n}{n^2 + 3^2 }+........+\frac{n}{n^2 + n^2}\right)[/imath] which method is used to evaluate these kind of problems and how do i evaluate this sum of limits ? I have heard some Riemann's sum is used but i do not know it and how to solve by that method if you can explain me how its done | 469885 | The limit of a sum [imath]\sum_{k=1}^n \frac{n}{n^2+k^2}[/imath]
How to compute this limit: [imath]\lim_{n\to\infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\cdots+\frac{n}{n^2+n^2}\right)[/imath] Please give me some hint. |
473366 | [imath]I(X\times Y)=I(X)+I(Y)[/imath] for affine varieties [imath]X[/imath] and [imath]Y[/imath]?
Let [imath]X\subset\mathbb{A}^{n},Y\subset\mathbb{A}^{m}[/imath] be affine varieties over [imath]k[/imath] algebraically closed. Then, [imath]X\times Y\subset \mathbb{A}^{m+n}[/imath] may be checked to be an affine variety as well; furthermore, it's just the product in the category of varieties. Then, because the functor [imath]A(-)[/imath] is an arrow-reversing equivalence of categories, it follows that [imath]A(X\times Y)=A(X)\otimes_{k} A(Y)[/imath], as tensor product is the coproduct in the category of finitely generated [imath]k[/imath]-algebras which are domains. Unless I have made a mistake, one can check directly from the universal property of tensor product that we also have \begin{equation*} k[x_{1},\ldots,x_{n}]/I(X)\otimes_{k}k[y_{1},\ldots,y_{m}]/I(Y)=k[x_{1},\ldots,x_{m},y_{1},\ldots,y_{m}]/(I(X)+I(Y)), \end{equation*} and of course [imath]I(X)+I(Y)\subset I(X\times Y)[/imath], so it should be true that we in fact have equality [imath]I(X)+I(Y)=I(X\times Y)[/imath]. Is there a direct way to see this? In particular, it seems like it should be true for arbitrary Zariski-closed [imath]X,Y[/imath]. (Source: Hartshorne, Exercise I.3.15) | 297014 | Is it true that for algebraic sets [imath]V,W[/imath] we have [imath]I(V \times W ) =I(V) + I(W)[/imath]?
This is a follow up question to my previous question here. Let [imath]k[/imath] be a field and [imath]V \subseteq \Bbb{A}^n[/imath] and [imath]W \subseteq \Bbb{A}^m[/imath] be algebraic sets. Then it should be true that [imath]I(V \times W ) = I(V) + I(W)[/imath] where by [imath]I(V)[/imath] here we mean the extension of [imath]I(V)k[x_1,\ldots,x_{m+n}][/imath]. Now I believe I have proven this (see the proof at the bottom of my question) but when I look at Martin's answer here, it is instead claimed that we actually have [imath]I(V \times W )= \sqrt{I(V) + I(W)},[/imath] and for [imath]I(V) + I(W)[/imath] to be a radical ideal we need [imath]k[/imath] to be algebraically closed. My question is: What's going on here? I believe my claim is true even without the assumption that [imath]k[/imath] is algebraically closed. Here's a proof user Sanchez told me of, which I have simplified: First it is clear that we always have [imath]I(V) + I(W) \subseteq I(V \times W)[/imath]. For the reverse inclusion consider a polynomial [imath]f \in I(V \times W)[/imath]. Then we can always write [imath]f = \sum_{i=1}^n f_ig_i[/imath] where [imath]f_i \in k[x_1,\ldots,x_n][/imath] and [imath]g_i \in k[x_{n+1},\ldots,x_{m+n}][/imath]. Now take any [imath]b' \in W[/imath]. If for all [imath]i[/imath] we have [imath]g_i(b') = 0[/imath] then since [imath]b'[/imath] is arbitrary, [imath]g_i \in I(W)[/imath] for all [imath]i[/imath]. Then [imath]f \in I(W) \subseteq I(V ) + I(w)[/imath] and we are done. Otherwise suppose there exists [imath]b \in W[/imath] and [imath]i[/imath] such that [imath]g_i(b) \neq 0[/imath]. Then wlog we may suppose that [imath]g_1(b) \neq 0[/imath]. Next, [imath]\sum f_ig_i(b) = 0 [/imath] on all of [imath]V[/imath] by assumption of [imath]f \in I(V \times W)[/imath]. So [imath]\sum f_i g_i(b) = p[/imath] for some [imath]p \in I(V)[/imath]. Now write [imath]f_1 = \frac{ p - g_2(b) f_2 + \ldots g_n(b)f_n}{g_1(b)}.[/imath] Substituting this for [imath]f_1[/imath] in [imath]\sum f_ig_i[/imath], followed by taking things mod [imath]I(V)[/imath] we get an expression with only [imath]n-1[/imath] terms [imath]\mod{I(V)}[/imath]. Continuing this process we will finally get an expression with [imath]0[/imath] terms [imath]\mod{I(V)}[/imath] so that [imath]f \in I(V)[/imath]. This shows [imath]I(V \times W) \subseteq I(V) + I(W)[/imath] which completes the proof. |
473431 | Why does the volume of a hypersphere decrease in higher dimensions?
First let us define an [imath]n[/imath]-ball as the euclidean sphere in [imath]\mathbb{R}^n[/imath] including its interior and its surface where [imath]n[/imath] refers to the number of coordinates needed to describe the object (the geometer's notation), NOT the topologist's notation which refers to the dimension of the manifold. All [imath]n[/imath]-balls considered here are of radius 1 centered at the origin. So an [imath]n[/imath]-ball is the set of points [imath]\{x=(x_1,x_2,...,x_n)\in \mathbb{R}^n : \sum_{i=1}^n x_i^2 \leq 1\}.[/imath] It is well known that the volume of the unit [imath]n[/imath]-ball is given by [imath]V(n)=\frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}[/imath] and before anyone points it out, we will consider all volumes to be unit-less so that they can be compared with each other. So for example [imath]\pi=V(2)<V(3)=\frac{4}{3}\pi.[/imath] Now, my question is why does [imath]V(n)\rightarrow 0[/imath] as [imath]n\rightarrow\infty[/imath]? This behavior is independent of the radius. Depending on the radius, [imath]V(n)[/imath] may (or may not) increase at first, hit a peak at some [imath]n[/imath], and then monotonically decrease and converge to zero. Considering [imath]n\in\mathbb{N}[/imath] to take on only discrete values, for the unit [imath]n[/imath]-ball, the max volume is achieved at [imath]n=5[/imath]. I have read all I could find (here/wikipedia and elsewhere) and I do see that the [imath]n[/imath]-ball occupies a smaller and smaller portion of the circumscribing cube [imath][-1,1]^n[/imath] and I have seen the arguments that the diameter should be used as the fundamental quantity instead of the radius. This way, the volume monotonically decreases to zero for all [imath]n[/imath] at least for the unit ball. I also see the analytic reason why the function converges to zero. The gamma function in the denominator grows much faster than the numerator and eventually the whole fraction converges to zero even if the radius is a googol. But my question is, intuitively speaking (as if intuition is a good guide in higher dimensions), [imath]V(n)[/imath] should be monotonically increasing or at the very least non-decreasing. The way I see it, a 2-ball is contained in a 3-ball and a 2-ball can be rotated in [imath]\mathbb{R}^3[/imath] to create a 3-ball. Similarly, you can rotate any [imath](n-1)[/imath]-ball in [imath]\mathbb{R}^n[/imath] around the appropriate axis to create an [imath]n[/imath]-ball. I know that any [imath](n-1)[/imath]-ball in [imath]\mathbb{R}^n[/imath] has Lebesgue measure zero but inside any [imath]n[/imath]-ball, an [imath](n-1)[/imath]-ball can be rotated around any of the [imath]n[/imath] axes so [imath]V(n)[/imath] should be larger than [imath]V(n-1)[/imath]. This specific "argument" hasn't been addressed on any of the previous questions that I could find. Any geometric interpretation of what's happening? Bonus points for something intuitive and/or 1-2-3 dimensional examples to show me the fallacy of my argument. Thanks! Addendum: I have seen this thread (and even this one and many others on stack exchange and math overflow) and like I said, they don't address my argument presented in this question. They do mention other arguments such as comparison with the circumscribing cube or analytically looking at fraction to see why it goes to zero. But those I already know...namely by reading these very threads. | 67039 | Why does the volume of the unit sphere go to zero?
The volume of a [imath]d[/imath] dimensional hypersphere of radius [imath]r[/imath] is given by: [imath]V(r,d)=\frac{(\pi r^2)^{d/2}}{\Gamma\left(\frac{d}{2}+1\right)}[/imath] What intrigues me about this, is that [imath]V\to 0[/imath] as [imath]d\to\infty[/imath] for any fixed [imath]r[/imath]. How can this be? For fixed [imath]r[/imath], I would have thought adding a dimension would make the volume bigger, but apparently it does not. Anyone got a good explanation? |
473487 | Bijection between [imath]\mathbb{R}^n[/imath] and [imath]\mathbb{R}^m[/imath]
I've heard that it is possible to present a bijection [imath]\phi : \mathbb{R}^n \to \mathbb{R}^m[/imath] due to Cantor that although not continuous show that those sets have the same cardinality independent of the dimension as a vector space. The problem is that I couldn't find about it anywhere. Can someone tell me where I can find about this bijection and how to show it's existance? Thanks very much in advance! | 829855 | Bijective Functions between Multiple Dimensions
Do bijective functions exist that map from a function of one dimension to a function of another dimension? For example, does there exist a function [imath]f : \mathbb{R^2} \rightarrow \mathbb{R^3}[/imath] that is bijective? If so, would you please provide an example? If not, how is it possible to prove why this is not possible? |
473514 | On the existence of certain one-to-one analytic function
I am wondering if there exists any one-to-one analytic function mapping annulus to punctuated disk? i.e. if we let [imath]D_1=\{1/2<|z|<1\}, D_2=\{0<|z|<1\}[/imath], is there a one-to-one analytic function [imath]f[/imath] maps [imath]D_1[/imath] to [imath]D_2[/imath]? | 248985 | Non-existence of a bijective analytic function between annulus and punctured disk
Suppose [imath]A=\{z\in \mathbb{C}: 0<|z|<1\}[/imath] and [imath]B=\{z\in \mathbb{C}: 2<|z|<3\}[/imath]. Show that there is no one -to-one analytic function from A to B. Any hints? Thanks! |
454715 | If [imath]x^{2}-x\in Z(R)[/imath] for all [imath]x\in R[/imath], then [imath]R[/imath] is commutative.
If [imath]x^{2}-x\in Z(R)[/imath] for all [imath]x\in R[/imath], then [imath]R[/imath] is commutative. I need to proof this theorem and I have something like this below. However, I do not know how to continue this proof. [imath](x+y)^{2}-(x+y)=x^{2}+xy+yx+y^{2}-x-y=(x^{2}-x)+(y^{2}-y)+xy+yx\in Z(R)[/imath] so [imath]xy+yx\in Z(R)[/imath] | 906866 | If [imath]\forall x \in R, x^2-x \in Z(G)[/imath], than [imath]R[/imath] is commutative
Let [imath]R[/imath] be a ring such that for every [imath]x\in R[/imath] we have [imath]x^2-x \in Z(G)[/imath]. Show that [imath]R[/imath] is a commutative ring. My thoughts What should I do? I could show that every [imath]y \in R[/imath] could be written in the shape [imath]x^2-x[/imath] but I don't know if that is true. We know that [imath]y \cdot(x^2-x) \ = \ (x^2-x)\cdot y[/imath] for any [imath]x,y \in R[/imath], but that doesn't mean that [imath]xy = yx[/imath]. Please give me a hint |
473611 | Metric Space (Elementary Analysis)
Let [imath]X \subseteq \mathbb{R}^{n}[/imath] be given by [imath] X = \left\{ (b_{1}, \ldots, b_{n}) \in \mathbb{R}^{n} \mid \sum\limits_{i=1}^{n} \frac{b_{i}}{i} = 0 \right\}[/imath] Then prove that [imath]X[/imath] is closed in [imath]\mathbb{R}^{n}[/imath]. | 473567 | Proving a particular subset of [imath]R^n[/imath] is closed
Let [imath]S,X[/imath] be subsets of [imath]R^n[/imath] given by [imath]S=\{(a_1,a_2,\dotsc,a_n)\in R^n|\sum a_i^2=1\}[/imath] [imath]X=\{(b_1,b_2,\dotsc,b_n)\in R^n|\sum\frac{b_i}{i}=0\}[/imath] Then prove that [imath]S+X[/imath] is a closed set in [imath]R^n[/imath]. |
473762 | Proving [imath]\frac{1}{n^2}[/imath] infinite series converges without integral test
Just out of curiosity, I was wondering if anybody knows any methods (other than the integral test) of proving the infinite series where the nth term is given by [imath]\frac{1}{n^2}[/imath] converges. | 2620961 | Another solution.
I need to show that, for every [imath]n\in\mathbb{N}[/imath], we obtain: [imath] 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}\leq2. [/imath] My proof is the following: We know that [imath]\sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{\pi^2}{6}[/imath], so [imath] \sum_{i=1}^n\frac{1}{i^2}\leq\sum_{i=1}^{\infty}\frac{1}{i^2}=\frac{\pi^2}{6}<2. [/imath] But I want to know if exists another proof for this problem. I try induction, but I can't do it. Thanks for advance. |
42308 | Continuous bijection from [imath](0,1)[/imath] to [imath][0,1][/imath]
Does there exist a continuous bijection from [imath](0,1)[/imath] to [imath][0,1][/imath]? Of course the map should not be a proper map. | 945403 | Whether there is a continuous bijection from [imath](0,1)[/imath] to closed interval [imath][0,1][/imath].
Is there a continuous bijection from open interval [imath](0,1)[/imath] to [imath][0,1][/imath]. The answer is not. How to prove? I think it may proceed by contradiction and apply open mapping theorem. However, [imath](0,1)[/imath] is not complete. I get stuck. |
473987 | Polynomial [imath]p(z)[/imath] in [imath]\mathbb{C}[/imath] has no zero whose modulus does not exceed [imath]1[/imath]
For [imath]n>1[/imath] consider real numbers [imath]c_0>c_1>.....>c_n>0[/imath]. Prove that the polynomial [imath]p(z) = c_0+c_1z+.....+c_nz^n[/imath] in [imath]\mathbb{C}[/imath] has no zero whose modulus does not exceed [imath]1[/imath]. | 328850 | All the zeroes of [imath]p(z)[/imath] lie inside the unit disk
Let [imath]p(z) = c_0 + c_1z + c_2z^2 + \dots + c_nz^n[/imath] where [imath]0 \le c_0 \le c_1 \le \dots \le c_n[/imath]. I would like to show that all zeroes of this polynomial lie inside the unit disk by applying Rouche's theorem to the polynomial [imath](1-z)p(z)[/imath]. I'm not completely sure how to do this. Using the given, information I can deduce that [imath]|(1 - z)p(z)| \le |c_nz^{n+1}|[/imath] on the unit circle, but this doesn't really match the assumptions of Rouche's theorem. Help would be appreciated. |
473956 | Lagrange finite elements
We consider in [imath]\mathbb{R}^2[/imath] the set of points [imath]\{M_1(-1,1),M_2(0,1), M_3(2,1),M_4(-1,0),M_5(1,0),M_6(2,0)\}[/imath] Let [imath]\Omega[/imath] a rectangular structure consisting of the heads [imath]\{M_4(-1,0),M_6(2,0), M_3(2,1),M_1(-1,1)\}[/imath] Let the two polygones [imath]\sum_1 = \{M_4,M_5,M_2,M_1\}[/imath] et [imath]\sum_2 = \{M_5,M_6,M_3,M_2\}[/imath] let the mesh [imath]\{(e_1,\sum_1,Q_1),(e_2,\sum_2,Q_2)\}[/imath] such [imath]Q_1=\{f : \mathbb{R}^2 \rightarrow \mathbb{R} , f(x,y) = a + bx + cy + d xy , a , b, c , d \in \mathbb{R}\}[/imath] 1- Proove that [imath](e_1,\sum_1,Q_1)[/imath] and [imath](e_2,\sum_2,Q_1)[/imath] are two finite elements of Lagrange. 2- Let [imath]V[/imath] the finite-dimensional space that corresponds to the mesh. Proove that [imath]V \nsubseteqq H^1(\Omega)[/imath] Where i can found the definition of V? (book, or paper...?) please and my second question is: dimension of [imath]\partial \sum_1 \cap \partial \sum_2[/imath] is 1 or 2? | 388892 | finite elements-exercice
We consider in [imath]\mathbb{R}^2[/imath] the set of points [imath]\{M_1(-1,1),M_2(0,1), M_3(2,1),M_4(-1,0),M_5(1,0),M_6(2,0)\}[/imath] Let [imath]\Omega[/imath] a rectangular structure consisting of the heads [imath]\{M_4(-1,0),M_6(2,0), M_3(2,1),M_1(-1,1)\}[/imath] Let the two polygons [imath]\sum_1 = \{M_4,M_5,M_2,M_1\}[/imath] and [imath]\sum_2 = \{M_5,M_6,M_3,M_2\}[/imath] let the mesh [imath]\{(e_1,\sum_1,Q_1),(e_2,\sum_2,Q_2)\}[/imath] such [imath]Q_1=\{f : \mathbb{R}^2 \rightarrow \mathbb{R} , f(x,y) = a + bx + cy + d xy , a , b, c , d \in \mathbb{R}\}[/imath] 1- Prove that [imath](e_1,\sum_1,Q_1)[/imath] and [imath](e_2,\sum_2,Q_1)[/imath] are two finite elements of Lagrange. 2- Let [imath]V[/imath] the finite-dimensional space that corresponds to the mesh. Prove that [imath]V \nsubseteqq H^1(\Omega)[/imath] My problem is to prove 2. |
474610 | Is it possible to evaluate [imath]\sum_{r=1}^{20}\frac{1}{r(r+1)}[/imath] using [imath]\sum_{r=1}^{n}\frac12n(n+1)[/imath]
Evaluate [imath]\sum_{r=1}^{20}\frac{1}{r(r+1)}[/imath] It splits into [imath]\sum_{r=1}^{20}\frac{1}{r}-\sum_{r=1}^{20}\frac{1}{r+1}[/imath] I'm stuck on how to apply the standard result [imath]\sum_{r=1}^{n}\frac12n(n+1)[/imath] to it. It may be simpler to use the method of differences but is that the only straightforward way? | 447990 | Find a formula for this sequence (and prove it).
This is a 2 part problem. Part I I need help finding a formula for this sequence of numbers: [imath]\frac{1} {1\times 2} + \frac {1} {2\times3} + \cdots + \frac {1} {n(n+1)}[/imath] Part II I need to prove the formula conjectured in Part I. |
474825 | P-groups and homomorphisms...
If [imath]f: G \rightarrow H[/imath] is a surjective homomorphism, [imath]G[/imath] finite, prove the following: 1) If [imath]P[/imath] is a p-subgroup of [imath]G[/imath], then [imath]f(P)[/imath] is a p-subgroup of [imath]H[/imath]. 2) If [imath]S[/imath] is a Sylow-p-subgroup of [imath]G[/imath], then then [imath]f(S)[/imath] is a Sylow-p-subgroup of [imath]H[/imath]. EDIT: I have been doing this problem for several days now on and off and I have no idea how to start...I'd rather skip the hints honestly, as I've been given a few and still am missing a key piece. | 474804 | Normality of subgroups and Sylow groups under homomorphisms...
Prove or disprove: if [imath]f:G \rightarrow H[/imath] is a surjective group homomorphism, then 1) [imath]f(A)[/imath] normal implies [imath]A[/imath] normal. 2) [imath]S[/imath] is a p-Sylow subgroup of [imath]G[/imath] implies [imath]f(S)[/imath] is a Sylow-p subgroup of [imath]H[/imath]. I believe 1 is false since I believe we need injectivity, but I cannot think of a counterexample... I believe 2 is true but do not know how to show it... |
474993 | How to solve the integral [imath]\int \frac{1}{e^x}\,dx[/imath] step by step?
I'm primitive in integrals and derivatives and I'm trying to solve the integral [imath]\int \frac{1}{e^x}\,dx[/imath], but especially this integral was hard to me to solve it. So I tried: [imath]\begin{align} \int\frac{1}{e^x}\,dx&=\int \frac{1}{\color{Red}{e^x}}\color{Blue}{e^x\,dx}\\&=\int \frac{1}{\color{Red}{u}}\,\color{Blue}{du}\\&=\ln\left(|u|\right)\\&=\ln \left(|e^x|\right)+C \end{align}[/imath] But my solution is wrong while I used the integration by substitution method ?! Correct answer: [imath]\begin{align} \int\frac{1}{e^x}\,dx&=\left(-e^{-x}\right)+C \end{align}[/imath] | 474683 | How to solve the definite integral [imath]\int_{-4}^{-2}e^{-x}\,dx[/imath]?
I'm trying to find the value of the integral [imath]\int_{-4}^{-2}e^{-x}\,dx[/imath] but I just couldn't solve it. Actually I found in a List of integrals that [imath]\int e^x\,dx=e^x+C[/imath] so I concluded: [imath] \int e^{-x}\,dx=\int\frac{1}{e^x}\,dx=\ln|e^x|[/imath] [imath]\int\limits_{-4}^{-2}e^{-x}\,dx=\left(\ln|e^{-2}|\right)-\left(\ln|e^{-4}|\right)=-2+4=2[/imath] I know the solution is wrong, but how can I solve this integral or any another integral like this? |
475059 | Number of graphs with vertices of only even degree.
This is an exercise I don't know how to solve, as I am preparing for an exam it would be great if you could help me with it. Show that for [imath]n > 0[/imath] a number of graphs with vertices from set [imath]\left\{1, 2, ..., n\right\}[/imath] in which every vertice has even degree equals [imath]2^{\binom{n-1}{2}}[/imath] Thanks in advance! | 361650 | The number of labelled graphs with all vertices of even degree
I have a graph exam tomorrow, and there is a problem that said number of graph with labeled vertices and all of them of even degree is [imath] 2^{n-1 \choose 2} [/imath]. According to that topic, it means the number of cut spaces is [imath] 2^{n-1} [/imath]. Well, I don't understand what is cut space yet, and why its count is [imath] 2^{n-1} [/imath] And don't know why we should divide the total number of graphs by this! |
44374 | Intermediate ring between a field and an algebraic extension.
This is an exercise in some textbooks. Let [imath]E[/imath] be an algebraic extension of [imath]F[/imath]. Suppose [imath]R[/imath] is ring that contains [imath]F[/imath] and is contained in [imath]E[/imath]. Prove that [imath]R[/imath] is a field. The trouble is really with the inverse of [imath]r[/imath], where [imath]r\in R[/imath]. How to prove that [imath]r^{-1}\in R[/imath], in apparent lack of a characterization of [imath]R[/imath]. It occurred to me to use the smallest field containing [imath]R[/imath] ([imath]R[/imath] is easily shown to be an integral domain), that's the field of quotients, and proving that it's [imath]R[/imath] itself, but I don't really know how to proceed. A not-too-weak, not-too-strong hint will be much appreciated. Beware [imath] [/imath] Readers seeking only hints should beware that there is now a complete answer. | 397733 | Ring Inside an Algebraic Field Extension
Let [imath]E|F[/imath] be an algebraic field extension and a ring [imath]K[/imath] such that [imath]F\subseteq K\subseteq E[/imath]. It is true that [imath]K[/imath] is a field? |
475391 | Find Fourier Series of the function [imath]f(x)= \sin x \cos(2x) [/imath]
Find Fourier Series of the function [imath]f(x)= \sin x \cos(2x) [/imath] in the range [imath] -\pi \leq x \leq \pi [/imath] any help much appreciated I need find out [imath]a_0[/imath] and [imath]a_1[/imath] and [imath]b_1[/imath] I can find [imath]a_0[/imath] which is simply integrating something with respect to the limits I can get as far as [imath]\frac{1}{2} \int_{-\pi}^\pi \ \frac12 (\sin (3x)-\sin(x)) dx[/imath] How would I integrate the above expression ? secondly how would I calculate [imath]a_1[/imath] and [imath]b_1[/imath] but despite knowing the general formula to find the fourier series Im having trouble applying them to this question | 382656 | What is the odd Fourier extension of [imath]\sin x \cos(2x)[/imath]?
odd half range extension of [imath]f(x) = \sin x \cos(2x)\text{ with limits $0$ to $\pi$}[/imath] |
475405 | About the cardinality of the Borelians in the real line.
Is true that the set of the Borelians in [imath]\mathbb{R}[/imath] has the same cardinality of [imath]\mathbb{R}[/imath]? I need of the Continuum Hypothesis for to prove this? | 70880 | Cardinality of Borel sigma algebra
It seems it's well known that if a sigma algebra is generated by countably many sets, then the cardinality of it is either finite or [imath]c[/imath] (the cardinality of continuum). But it seems hard to prove it, and actually hard to find a proof of it. Can anyone help me out? |
475081 | Upper bound for [imath]T(n) = T(n - 1) + T(n/2) + n[/imath] with recursion-tree
I'm reading through Introduction to Algorithms, 3rd ed. and I got stuck on the following recurrence (exercise 4.4-5): [imath]T(n) = T(n - 1) + T(n/2) + n[/imath] The exercise asks you to find the upper bound using the recurrence-tree method. I tried drawing one. It would have [imath]2^n[/imath] leafs if it was a complete binary tree (the height is [imath]n[/imath]), but it is not. Furthermore, I cannot find a formula that expresses the complexity of each level like they do in the book. After writing some code to calculate [imath]T(n)[/imath] and comparing it with different functions, I end up thinking that the complexity is exponential. Using [imath]c2^n[/imath] with the substitution method works out, but I'm not certain that this is a tight bound. I'm not certain that the tight bound is exponential either. Can somebody help me out through the reasoning? P.S.: If it matters, I'm not in school/university and this is not homework. I'm a (let's say) Ruby programmer who is self-studying to fill in his gaps in Computer Science. | 518682 | solve [imath]T(n)=T(n-1)+T(\frac{n}{2})+n[/imath]
Using the recursion tree i tried solving this: [imath]T(n)=T(n-1)+T(\frac{n}{2})+n[/imath]; the tree has two parts (branches) one that of [imath]T(n-1)[/imath] and other branch is of [imath]T(\frac{n}{2})[/imath]. But as the term T(n-1) reach higher depth, I solved that part of the tree. Finally it resulted in O[imath](2^n)[/imath] is it correct or is it [imath]O(n^2)[/imath]? How can we solve it by substitution method? |
475995 | Can anyone derive the formula for the expansion [imath](x + \Delta x)^{n}[/imath] that uses Big O notation?
There is a formula that describes this expansion using big O notation, I'm very curious on how this is derived. I also understand that the order term may very depending on what [imath]\Delta x[/imath] approaches to in limit ? [imath](x + \Delta x)^{n} = x^{n} + nx^{n-1}\Delta x + O((\Delta x)^{2})[/imath] | 475251 | Explanation of the binomial theorem and the associated Big O notation
I'm currently following the MIT Single Variable lectures online and the professor states that the binomial theorem for the expansion [imath](x + \Delta x)^{n} = x^{n} + nx^{n-1}\Delta x + O((\Delta x)^{2})[/imath] How is this derived, and what does the big O term of the expansion represent in terms of the binomial theorem. Just to put this in context, this expression was used when computing the derivative of [imath]x^{n}[/imath] using the limit definition. |
476011 | How to show that [imath]\{\sqrt m - \sqrt n : m,n \in \mathbb N\}[/imath] is dense in [imath]\mathbb R[/imath]?
Show that [imath]S=\{\sqrt m - \sqrt n : m,n \in \mathbb N\}[/imath] is dense in [imath]\mathbb R[/imath]. I know the definition of dense as: A set S is dense in [imath]\mathbb R[/imath] if there exists [imath]a,b \in \mathbb R[/imath] such that [imath]S \cap(a,b) \neq \emptyset.[/imath] I don't understand how to proceed. Please help. | 275047 | Prove that the following set is dense
It is hard for me to show that the set [imath]\{\sqrt{m}-\sqrt{n}; m,n\in \Bbb N\}[/imath] is dense in [imath]\Bbb R[/imath]. Please help me. |
475823 | Find matrices [imath]X[/imath] such that for any matrix [imath]Y[/imath] we have [imath]\det(X^2 + Y^2) \geq 0[/imath]
What is the characterization of real matrices [imath]X \in \mathbb{R}^{n\times n}[/imath] such that for any real matrix [imath]Y \in \mathbb{R}^{n\times n}[/imath]: [imath]\det(X^2 + Y^2) \geq 0?[/imath] | 438744 | Is it always true that [imath]\det(A^2+B^2)\geq0[/imath]?
Let [imath]A[/imath] and [imath]B[/imath] be real square matrices of the same size. Is it true that [imath]\det(A^2+B^2)\geq0\,?[/imath] If [imath]AB=BA[/imath] then the answer is positive: [imath]\det(A^2+B^2)=\det(A+iB)\det(A-iB)=\det(A+iB)\overline{\det(A+iB)}\geq0.[/imath] |
476329 | Identify the ring [imath]\mathbb{Z}[x]/(x^2-3, 2x+4)[/imath]
Consider the quotient ring [imath]\mathbb{Z}[x]/(x^2-3, 2x+4)[/imath]. Sometimes we can get identifications of rings like these with other more familiar rings if we first kill [imath]x^2-3[/imath] in the ring [imath]\mathbb{Z}[x][/imath] and then kill [imath]2x+4[/imath] or vise versa. The substitution homomorphism [imath]\mathbb{Z}[x] \rightarrow \mathbb{Z}[-2][/imath] doesn't seem to be relevant since its ideal is [imath]x + 2[/imath] and not [imath]2x+4[/imath]. Working with [imath]x^2 -3[/imath] though I got that [imath]\mathbb{Z}[x]/(x^2-3)\cong \mathbb{Z}[\sqrt{3}][/imath] and since the substitution homomorphism is surjective, we have that [imath]\mathbb Z[x](x^2-3, 2x+4) \cong \mathbb{Z}[\sqrt{3}]/(2\sqrt{3} + 4)[/imath], making use of the correspondence theorem. But my use isn't good enough since I thought the identification would be a more familiar ring. | 468631 | Cardinality of the quotient ring [imath]\mathbb{Z}[x]/(x^2-3,2x+4)[/imath]
This problem is from a practice exam I was working on. What is the cardinality of the quotient [imath]\mathbb{Z}[x]/(x^2-3,2x+4)[/imath] ? Thoughts. If I find a ring that is easier to handle then this then I can go from there. So I think this is isomorphic to [imath]\mathbb{Z}_2[x]/(x^2-3,x+2)[/imath]. And then I am stuck. Am I correct so far? What should I be looking to do from here? Thanks for your time and your answers. |
476531 | Special subgroup of a group of order [imath]n[/imath]
I apologize if this seems elementary but I don't know how to deal with. Let [imath]G[/imath] be a group of order [imath]n[/imath] and let [imath] \emptyset \ne S \subseteq G[/imath]. Is it true that [imath]S^n :=\lbrace s_1\cdots s_n \; | \; s_i \in S\rbrace[/imath] is a subgroup of [imath]G[/imath] !? For example if [imath]S[/imath] contains a single element then by Lagrange's theorem [imath]S^n = \lbrace e \rbrace[/imath]. | 475657 | If [imath]S[/imath] is a nonempty subset of group [imath]G[/imath], then [imath]S^{|G|}[/imath] is a subgroup of [imath]G[/imath].
Let [imath]G[/imath] be a group with [imath]|G| = n[/imath] and let [imath] \emptyset \ne S \subseteq G[/imath]. I want to show that [imath]S^n[/imath] is a subgroup of [imath]G[/imath] where by [imath]S^n[/imath] I mean the set [imath]\lbrace s_1\cdots s_n \; | \; s_i \in S\rbrace[/imath]. |
476728 | If a is less than 1 in absolute value, than n times a to the n'th power converges to zero
Prove that if [imath]|a|<1[/imath] then [imath]\lim_{n\to\infty}{n\cdot a^n}=0[/imath] Lopital's rule won't help here, and this problem appears in a book before lopital has been taught. I noticed that: [imath]\frac{a_{n+1}}{a_n}=a\frac{n+1}{n}\to a[/imath] So, can I conclude that it behaves like a geometric series for large n, so it converges to zero? *Other solution methods are welcome. | 173061 | Sequence Limit: [imath]\lim\limits_{n \rightarrow \infty}{n\,x^n}[/imath]
If [imath]-1<x<1[/imath] show that [imath]\lim\limits_{n \to \infty}{n\,x^n} = 0[/imath]. I don't have idea. I only prove that [imath]n\,x^n[/imath] is decreasing. |
474003 | Bijectivity of T(z) = λ z + μ z*
Let [imath]T = \lambda z + \mu \bar z[/imath] where [imath]\lambda, \mu \in \mathbb{C}[/imath]. Show that [imath]T[/imath] is bijective exactly when [imath]\lambda \bar\lambda \ne \mu \bar\mu[/imath]. | 476070 | How can I find such a [imath]z[/imath]? (Injectivity of [imath]T(z) = \lambda z + \mu\bar{z}[/imath])
We consider [imath]T:\mathbb{C}\to\mathbb{C}[/imath] defined by [imath]T(z)=\lambda z+\mu\overline{z}[/imath], where [imath]\lambda ,\mu\in\mathbb{C}[/imath]. I want to prove that if [imath]T[/imath] is injective, then [imath]\lambda\cdot\overline{\lambda}\neq\mu\cdot\overline{\mu}[/imath]. Clearly, if [imath]\mu =0[/imath], then [imath]\lambda\neq 0[/imath] because [imath]T[/imath] is injective. Now, if [imath]\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}[/imath] and [imath]\mu\neq 0[/imath], then I wanted to find a [imath]z\in\mathbb{C}\setminus\{0\}[/imath] such that [imath]z=-\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}[/imath], because we would have that: [imath]T(z)=-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}+\mu\overline{z}=\overline{z}(-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}+\mu)=0[/imath] ('Cause this would prove that [imath]\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}[/imath] implies [imath]T[/imath] is not injective.) So, the question is, can we always find such a [imath]z\in\mathbb{C}\setminus\{0\}[/imath]? |
261769 | Power Series Expansion
I was flicking through a book on perturbation methods and saw a simple question asking the reader to expand the following expression for [imath]f[/imath] in a power series (up to the first 2 terms): [imath]f = (1 + \epsilon \,x)^{1/\epsilon}[/imath], where [imath]\epsilon[/imath] is a small parameter. I'm sure this is very simple, but I wasn't certain about the best way to approach this. A quick look at mathematica tells me the solution is [imath]e^x - \frac{1}{2} (e^x x^2) \,\epsilon + ...[/imath]. How would I go about getting this answer - and more importantly, how would I systematically find series expansions for problems similar to this one? | 261741 | Simple Power Series Expansion for Problems similar to [imath]f = (1 + \epsilon \,x)^{1/\epsilon}[/imath]
I was flicking through a book on perturbation methods and saw a simple question asking the reader to expand the following expression for [imath]f[/imath] in a power series (up to the first 2 terms): [imath]f = (1 + \epsilon \,x)^{1/\epsilon}[/imath], where [imath]\epsilon[/imath] is a small parameter. I'm sure this is very simple, but I wasn't certain about the best way to approach this. A quick look at mathematica tells me the solution is [imath]e^x - \frac{1}{2} (e^x x^2) \,\epsilon + ...[/imath]. How would I go about getting this answer - and more importantly, how would I systematically find series expansions for problems similar to this one? |
243142 | What is the general formula for calculating dot and cross products in spherical coordinates?
I was writing a C++ class for working with 3D vectors. I have written operations in the Cartesian coordinates easily, but I'm stuck and very confused at spherical coordinates. I googled my question but couldn't find a direct formula for vector product in the search results. Assume that I have [imath] \overrightarrow{V_1} [/imath] and [imath] \overrightarrow{V_2} [/imath] vectors in shperical coordinates: [imath] \overrightarrow{V_1} = r_1\hat{u_r} + \theta_1\hat{u_\theta} + \phi_1\hat{u_\phi} \\ \overrightarrow{V_2} = r_2\hat{u_r} + \theta_2\hat{u_\theta} + \phi_2\hat{u_\phi} \\ \hat{u_r}: \mbox{the unit vector in the direction of radius} \\ \hat{u_\theta}: \mbox{the unit vector in the direction of azimuthal angle} \\ \hat{u_\phi}: \mbox{the unit vector in the direction of polar angle} [/imath] [imath] \theta [/imath] and [imath] \phi [/imath] angles are as represented in the image below: What is the general formula for taking dot and cross products of these vectors? [imath] \overrightarrow{V_1} \bullet \overrightarrow{V_2} = ? \\ \overrightarrow{V_1} \times \overrightarrow{V_2} = ? [/imath] If you need an example, please work on this one: [imath] \overrightarrow{V_1} = 2\hat{u_r} + \frac{\pi}{3}\hat{u_\theta} + \frac{\pi}{4}\hat{u_\phi} \\ \overrightarrow{V_2} = 3\hat{u_r} + \frac{\pi}{6}\hat{u_\theta} + \frac{\pi}{2}\hat{u_\phi} [/imath] | 1685531 | Vectors and dot product in spherical coordinate system
Let [imath]\vec{v_1}[/imath] and [imath]\vec{v_2}[/imath] given: [imath]\overrightarrow{V_1} = r_1\hat{u_r} + \theta_1\hat{u_\theta} + \phi_1\hat{u_\phi} \\ \overrightarrow{V_2} = r_2\hat{u_r} + \theta_2\hat{u_\theta} + \phi_2\hat{u_\phi} \\[/imath] where [imath]r_n\theta_n\phi_n[/imath] are known constants. My first question is whether [imath]\vec{v_1}[/imath] and [imath]\vec{v_2}[/imath] are functions? I think they are functions since for each point in coordinate system, the result of [imath]\vec{v_1}[/imath] and [imath]\vec{v_2}[/imath] is different since [imath]\hat{u_r} \hat{u_\theta} \hat{u_\phi}[/imath] changes direction. This is little weird when compared to cartesian coordinate system. Second, Can we dot product in a similar manner done in cartesian coordinate system, i.e does dot product equal to [imath]\vec{v_1} \cdot \vec{v_2}=r_1 \cdot r_2 + \theta_1 \cdot \theta_2 + \phi_1 \cdot \phi_2[/imath]? By similar arguments at question 1 It seems to me that it is only possible when they are defined at same point. Otherwise, i think we need to convert vectors to cartesian coordinate system. |
376500 | Prime ideal and nilpotent elements
If [imath]\mathfrak p \subset R[/imath] is a prime ideal, prove that for every nilpotent [imath]r \in R[/imath] it follows that [imath]r \in \mathfrak p[/imath]. The only hint that my tutor gave me was to use induction. Can someone explain what he means by this? Thanks for the help! | 2288225 | How to proof [imath]W\subseteq P\left( R\right)[/imath]
R is a ring. [imath]W=\left\{ x\in R|\text{x is a nilpotent}\right\}[/imath] and [imath]P\left( R\right) =\cap \{ P|P\unlhd A, \text{P is prime}\}[/imath] .How to proof [imath]W\subseteq P\left( R\right)[/imath] |
477353 | Forget about the sin and cos functions, show that [imath](x-x^3/3!+x^5/5!-x^7/7!+...)^2+ (1-x^2/2!+x^4/4!-x^6/6!+...)^2=1[/imath].
Forget about the [imath]\sin[/imath] and [imath]\cos[/imath] functions, are there possibly some brilliant way to show that [imath]\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^2+ \left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)^2=1[/imath] ? I've thought for a long time, without making much progress. Can someone help me? Thanks. | 374420 | Could we show [imath]1-(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots)^2=(1-\frac{x^2}{2!}+\frac{x^4}{4!}- \dots)^2[/imath] if we didn't know about Taylor Expansion?
Suppose that humanity haven't discovered Taylor Series Expansion of trigonometric functions or of any function that would help us on this. Which means we are not allowed to replace the given infinite series sums with the corresponding [imath]\sin x[/imath] and [imath]\cos x[/imath] functions. Could we still show that the following identity is true for all real [imath]x[/imath]? [imath] 1 - \left( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \dots \right)^2 = \left( 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \dots \right)^2 [/imath] |
477486 | A problem out of reach for forever
Given any really large number [imath]x[/imath], such as the busy beaver number [imath]x=BB(BB(99))[/imath], can we construct a proposition such that we know it has a proof or a disproof, but we also know the shortest such proof or disproof is longer then [imath]x[/imath] symbols? | 151831 | Lower bound on proof length
Given any positive integer [imath]n[/imath], is there a way to quickly construct a statement [imath]S[/imath], such that the shortest proof of [imath]S[/imath], if it exists, must have length at least [imath]n[/imath]? And such that one out of [imath]S[/imath] or not [imath]S[/imath], is provable. And question 2: If S, then a proof of S must have length atleast n. If not S, then a proof of not S must have length atleast n. Edit: I had in mind that the number of symbols in S be of the same order as the number of symbols required to specify n. |
477499 | If [imath]x^2+ax+b=0[/imath] has a rational root, show that the root is an integer.
So far I have Assume [imath]a,b,x,y[/imath] are integers; [imath]\frac xy[/imath] is rational; and that [imath]\frac xy[/imath] is in simplest form. WTS: [imath]\frac xy[/imath] is integer So [imath](\frac xy)^2 + a(\frac xy) + b = 0[/imath]; [imath]{x^2\over y^2} = -b - a(\frac xy)[/imath]; [imath]x^2 = -by^2- axy[/imath]; [imath]x^2 = y(-by - ax)[/imath]; so [imath]y[/imath] divides [imath]x^2.[/imath] I am having trouble on how to show that [imath]\text{gcd}(x,y)=1.[/imath] | 432023 | If [imath]x^2+ax+b=0[/imath] has a rational root, show that it is in fact an integer
I have tried as follows. Please help to double check the proof! Thank you! Since [imath]x=p/q[/imath] ([imath]p[/imath], [imath]q[/imath] are integers), [imath](p/q)^2+(p/q)a+b=0[/imath] So, [imath](p/q)^2=-b-a(p/q)[/imath] then, [imath]p^2=-bq^2-a(p/q)q^2[/imath] and, [imath]p = \dfrac {q(-bq-a(p/q)q)}{p}[/imath] now, it is clear that [imath]q \mid p[/imath], thus [imath]p/q[/imath] must be an integer. |
477587 | How to show that [imath]x_n[/imath] is bounded by [imath]2[/imath] and converges?
Let [imath]x_n=\sqrt{1+\sqrt{2+\sqrt{3+\dots+\sqrt{n}}}}[/imath] Show that [imath]x_n<x_{n+1}[/imath] and [imath]x_{n+1}^2\le 1+\sqrt{2}x_n[/imath]. Hence show that [imath]x_n[/imath] is bounded by [imath]2[/imath] and converges. Trials:I can show that [imath]x_n<x_{n+1}[/imath] and [imath]x_{n+1}^2= 1+\sqrt{2+\sqrt{3+\dots+\sqrt{n}}}\\=1+\sqrt 2[1+\sqrt{3/4+\dots+\sqrt{n/2^{n-1}}}]\ \\\le1+\sqrt 2 x_n[/imath] Then I can't show [imath]x_n[/imath] is bounded by [imath]2[/imath] and converges. Please help. | 437209 | How can I show that [imath]\sqrt{1+\sqrt{2+\sqrt{3+\sqrt\ldots}}}[/imath] exists?
I would like to investigate the convergence of [imath]\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt\ldots}}}}[/imath] Or more precisely, let [imath]\begin{align} a_1 & = \sqrt 1\\ a_2 & = \sqrt{1+\sqrt2}\\ a_3 & = \sqrt{1+\sqrt{2+\sqrt 3}}\\ a_4 & = \sqrt{1+\sqrt{2+\sqrt{3+\sqrt 4}}}\\ &\vdots \end{align}[/imath] Easy computer calculations suggest that this sequence converges rapidly to the value 1.75793275661800453265, so I handed this number to the all-seeing Google, which produced: OEIS A072449 "Nested Radical Constant" from MathWorld Henceforth let us write [imath]\sqrt{r_1 + \sqrt{r_2 + \sqrt{\cdots + \sqrt{r_n}}}}[/imath] as [imath][r_1, r_2, \ldots r_n][/imath] for short, in the manner of continued fractions. Obviously we have [imath]a_n= [1,2,\ldots n] \le \underbrace{[n, n,\ldots, n]}_n[/imath] but as the right-hand side grows without bound (It's [imath]O(\sqrt n)[/imath]) this is unhelpful. I thought maybe to do something like: [imath]a_{n^2}\le [1, \underbrace{4, 4, 4}_3, \underbrace{9, 9, 9, 9, 9}_5, \ldots, \underbrace{n^2,n^2,\ldots,n^2}_{2n-1}] [/imath] but I haven't been able to make it work. I would like a proof that the limit [imath]\lim_{n\to\infty} a_n[/imath] exists. The methods I know are not getting me anywhere. I originally planned to ask "and what the limit is", but OEIS says "No closed-form expression is known for this constant". The references it cites are unavailable to me at present. |
477795 | A Variation of "rational is dense in [imath]\Bbb R[/imath]"
I know that the theorem that the set [imath]\Bbb Q[/imath] of rationals is dense in [imath]\Bbb R[/imath] says: For every [imath]x\in \Bbb R[/imath] and every [imath]\epsilon>0[/imath], there exist [imath]a[/imath], [imath]b\in \Bbb Z[/imath] with [imath]b\ne0[/imath] such that [imath]|x-{a\over b}|<\epsilon.[/imath] But what about if I change the condition "[imath]a,b\in \Bbb Z[/imath] with [imath]b\ne0[/imath]" by "[imath]a[/imath] and [imath]b[/imath] are both primes". Does the theorem still hold? In other words, I wonder whether the following is true: For every [imath]x\in \Bbb R^+[/imath] and every [imath]\epsilon>0[/imath], there exist [imath]a[/imath], [imath]b\in \Bbb P[/imath], where [imath]\Bbb P[/imath] is the prime numbers set, such that [imath]|x-{a\over b}|<\epsilon.[/imath] | 4066 | Rationals of the form [imath]\frac{p}{q}[/imath] where [imath]p,q[/imath] are primes in [imath][a,b][/imath]
Consider the closed interval [imath][0,1][/imath], there is [imath]\frac{2}{3} \in [0,1][/imath] where [imath]p=2[/imath] and [imath]q=3[/imath]. Similarly consider [imath][2,3][/imath], one can have [imath]\frac{5}{2} \in [2,3][/imath] where [imath]p=5[/imath] and [imath]q=2[/imath]. Does every interval of the form [imath][a,b][/imath], where [imath]a,b \in \mathbb{R}[/imath] contain a rational of this kind. If yes, how can we prove it? |
477841 | Second derivative of class [imath]C^2[/imath] expressed as limit
Let [imath]g:\mathbb{R}\rightarrow\mathbb{R}[/imath] be a function of class [imath]C^2[/imath]. Show that [imath]\lim_{h\rightarrow 0}\dfrac{g(a+h)-2g(a)+g(a-h)}{h^2}=g''(a).[/imath] It seems like an application of mean-value theorem might help: Since [imath]g[/imath] is differentiable, for any [imath]h>0[/imath] there exists [imath]h_1\in(0,h)[/imath] such that [imath]g(a+h)-g(a)=h\cdot g'(a+h_1).[/imath] And then since [imath]g'[/imath] is differentiable, there exists [imath]h_2\in(0,h_1)[/imath] such that [imath]g'(a+h_1)-g'(a)=h_1\cdot g''(a+h_2).[/imath] Combining the two we have [imath]g(a+h)-g(a)=hg'(a)+hh_1g''(a+h_2).[/imath] which doesn't quite get to the desired expression. | 210264 | Second derivative "formula derivation"
I've been trying to understand how the second order derivative "formula" works: [imath]\lim_{h\to0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2}[/imath] So, the rate of change of the rate of change for an arbitrary continuous function. It basically feels right, since it samples "the after [imath]x+h[/imath] and the before [imath]x-h[/imath]" and the [imath]h^2[/imath] is there (due to the expected /h/h -> /h*h), but I'm having trouble finding the equation on my own. It's is basically a derivative of a derivative, right? Newtonian notation declares as [imath]f''[/imath] and Leibniz's as [imath]\frac{\partial^2{y}}{\partial{x}^2}[/imath] which dissolves into: [imath](f')'[/imath] and [imath]\frac{\partial{}}{\partial{x}}\frac{\partial{f}}{\partial{x}}[/imath] So, first derivation shows the rate of change of a function's value relative to input. The second derivative shows the rate of change of the actual rate of change, suggesting information relating to how frequenly it changes. The original one is rather straightforward: [imath]\frac{\Delta y}{\Delta x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{x + h - x} = \lim_{h\to0} \frac{f(x+h) - f(x)}{h}[/imath] And can easily be shown that [imath]f'(x) = nx^{n-1} + \dots[/imath] is correct for the more forthcoming of polynomial functions. So, my logic suggests that to get the derivative of a derivative, one only needs to send the derivative function as input to finding the new derivative. I'll drop the [imath]\lim_{h\to0}[/imath] for simplicity: [imath]f'(x) = \frac{f(x+h) - f(x)}{h}[/imath] So, the derivative of the derivative should be: [imath]f''(x) = \lim_{h\to0} \frac{f'(x+h) - f'(x)}{h}[/imath] [imath]f''(x) = \lim_{h\to0} \frac{ \frac{ f(x+2h) - f(x+h)}{h} - \frac{ f(x+h) - f(x)}{h} }{h}[/imath] [imath]f''(x) = \lim_{h\to0} \frac{ \frac{ f(x+2h) - f(x+h) - f(x+h) + f(x)}{h} }{h}[/imath] [imath]f''(x) = \lim_{h\to0} \frac{ f(x+2h) - f(x+h) - f(x+h) + f(x) }{h^2}[/imath] [imath]f''(x) = \lim_{h\to0} \frac{ f(x+2h) - 2f(x+h) + f(x) }{h^2}[/imath] What am I doing wrong? Perhaps it is the mess of it all, but I just can't see it. Please help. |
477889 | [imath]h(x)={f(x)\over x}[/imath] is decreasing or increasing or both over [imath][0,\infty)[/imath]
[imath]f[/imath] is real valued function on [imath][0,\infty)[/imath] such that [imath]f''(x)>0[/imath] for all [imath]x[/imath] and [imath]f(0)=0[/imath] Then [imath]h(x)={f(x)\over x}[/imath] is decreasing or increasing or both over [imath][0,\infty)[/imath] [imath]h'(x)={xf'(x)-f(x)\over x^2}[/imath] What I can conclude from here? | 475267 | A secant inequality for convex functions
Suppose [imath]f(0) =0 [/imath] and [imath]0<f''(x)<\infty (\forall[/imath] [imath]x>0)[/imath], then [imath]\frac{f(x)}{x}[/imath] strictly increases as [imath]x[/imath] increases. I have shown that [imath]f'(x)-\frac{f(x)}{x} = \frac{1}{2}xf''(c)[/imath], for some [imath]c\in (0,x)[/imath]. How do I proceed from here? |
477382 | I need to calculate [imath]x^{50}[/imath]
[imath]x=\begin{pmatrix}1&0&0\\1&0&1\\0&1&0\end{pmatrix}[/imath], I need to calculate [imath]x^{50}[/imath] Could anyone tell me how to proceed? Thank you. | 267492 | for a [imath]3 \times 3[/imath] matrix A ,value of [imath] A^{50} [/imath] is
I f [imath]A= \begin{pmatrix}1& 0 & 0 \\ 1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}[/imath] then [imath] A^{50} [/imath] is [imath] \begin{pmatrix}1& 0 & 0 \\ 50 & 1 & 0\\ 50 & 0 & 1 \end{pmatrix}[/imath] [imath]\begin{pmatrix}1& 0 & 0 \\ 48 & 1 & 0\\ 48 & 0 & 1 \end{pmatrix}[/imath] [imath]\begin{pmatrix}1& 0 & 0 \\ 25 & 1 & 0\\ 25 & 0 & 1 \end{pmatrix}[/imath] [imath]\begin{pmatrix}1& 0 & 0 \\ 24 & 1 & 0\\ 24 & 0 & 1\end{pmatrix}[/imath] I am stuck on this problem. Can anyone help me please............... |
103786 | Direct proof that [imath]\pi[/imath] is not constructible
Is there a direct proof that [imath]\pi[/imath] is not constructible, that is, that squaring the circle cannot be done by rule and compass? Of course, [imath]\pi[/imath] is not constructible because it is transcendental and so is not a root of any polynomial with rational coefficients. But is there a simple direct proof that [imath]\pi[/imath] is not a root of polynomial of degree [imath]2^n[/imath] with rational coefficients? The kind of proof I seek is one by induction on the height of a tower of quadratic extensions, one that ultimately relies on a proof that [imath]\pi[/imath] is not rational. Does any one know of a proof along these lines or any other direct proof? I just want a direct proof that [imath]\pi[/imath] is not constructible without appealing to transcendence. | 1008993 | Is there a direct proof that pi is not the root of an algebraic equation whose degree is a power of 2
All known proofs that the circle cannot be squared are based on Lindemann's theorem that [imath]\pi[/imath] is not analgebraic number. But this seems to be a case of using an atomic bomb to kill a fly. What really has to be proved is that [imath]\pi[/imath] is not the root of an algebraic equation with rational coefficients whose degree is a power of [imath]2[/imath]. Lindemann proves it for all degrees, which is much much more than what is needed for the classical construction problem. It is not too hard to prove that [imath]\pi[/imath] is not the root of a quadratic equation, and perhaps some clever induction argument could carry the day. The point is that a direct proof for degree [imath]2^n[/imath] has not only never been published, it does not even seem to have been noticed that this much less powerful result is all that is needed. |
478274 | Connectedness of set with [imath]\sin(1/x)[/imath] function
Let [imath]A[/imath] be the set of points [imath](x,y)\in\mathbb{R}^2[/imath] with [imath]x=0,|y|\leq 1[/imath], and let [imath]B[/imath] be the set with [imath]x>0,y=\sin 1/x[/imath]. Is [imath]A\cup B[/imath] connected? From the picture, I think it should be connected, because the points in [imath]B[/imath] oscillates between [imath]y=-1[/imath] and [imath]y=1[/imath] as [imath]x[/imath] approaches [imath]0[/imath], and the points in [imath]A[/imath] cover all of the range [imath][-1,1][/imath] for [imath]x=0[/imath]. However, to prove it rigorously, I must show that there doesn't exist disjoint, nonempty, open sets [imath]C, D[/imath] such that [imath]A\cup B=C\cup D[/imath]. How can I show that? | 317125 | Topologist's sine curve is connected
I just came across the example of the topologist's sine curve that is connected but not path-connected. The rigorous proof of the non-path-connectedness can be found here. But how can I prove that the curve is connected? To be honest, even intuitively I am not being able to see that the curve is connected. I am thinking if it is proved that the limit point of [imath]\sin(1/x)[/imath] as [imath]x \to 0=0[/imath], then it would be proved. But, why is this true? IMO, this limit doesn't exist. Intuitively also, it seems that the graph would behave crazily and not approach a particular value as a tends to [imath]0.[/imath] EDIT (Brett Frankel): There are a few different working definitions of the topologist's since curve. For the sake of clarity/consistency, I have copied below the definition used in the linked post: [imath] y(x) = \begin{cases} \sin\left(\frac{1}{x}\right) & \mbox{if $0\lt x \lt 1$,}\\\ 0 & \mbox{if $x=0$,}\end{cases}[/imath] |
428068 | Right and left inverse
If [imath]AR=I[/imath] ([imath]R[/imath] is the right inverse) and [imath]LA=I[/imath] (L is the left inverse), how can we show that [imath]L=R[/imath]? I am a bit skeptic about the statement but stuck at the moment in terms of showing that [imath]L=R[/imath]. Any help would be great! | 2493639 | left inverse and right inverse of matrices
If [imath]A \in R^{m×n}[/imath] has a left inverse, [imath]A^{T}[/imath] has a right inverse. How I can prove this property? I hope someone can help me. |
478429 | The difference between [imath]\mathbb{C}[/imath] and [imath]\mathbb{R}^2[/imath]
I am learning complex analysis now, but I still can't tell the difference between [imath]\mathbb{C}[/imath] and [imath]\mathbb{R}^2[/imath]. If we define a map:[imath]F:\mathbb{C}\rightarrow\mathbb{R^2}[/imath], [imath]x+iy\rightarrow (x,y),x,y\in R.[/imath] Further, we define a product in [imath]\mathbb{R}^2[/imath]:[imath](a,b)\cdot(c,d)=(ac-bd,ad+bc)[/imath].So why we still need complex number？ | 364044 | Difference between [imath]\mathbb C[/imath] and [imath]\mathbb R^2[/imath]
What are the basic differences between [imath]\mathbb C[/imath] and [imath]\mathbb R^2[/imath]? The points in these two sets are written as ordered pairs, I mean the structure looks similar to me. So what is the reason to denote these two sets differently? |
444475 | What's the difference between [imath]\mathbb{R}^2[/imath] and the complex plane?
I haven't taken any complex analysis course yet, but now I have this question that relates to it. Let's have a look at a very simple example. Suppose [imath]x,y[/imath] and [imath]z[/imath] are the Cartesian coordinates and we have a function [imath]z=f(x,y)=\cos(x)+\sin(y)[/imath]. However, now I change the [imath]\mathbb{R}^2[/imath] plane [imath]x,y[/imath] to complex plane and make a new function, [imath]z=\cos(t)+i\sin(t)[/imath]. So, can anyone tell me some famous and fundamental differences between complex plane and [imath]\mathbb{R}^2[/imath] by this example, like some features [imath]\mathbb{R}^2 [/imath] has but complex plane doesn't or the other way around. (Actually I am trying to understand why electrical engineers always want to put signal into the complex numbers rather than [imath]\mathbb{R}^2[/imath], if a signal is affected by 2 components) Thanks for help me out! | 701826 | Complex plane and [imath]\mathbb{R}^2[/imath].
What differences -if any- are there between the complex numbers [imath]\mathbb{C}[/imath] and [imath]\mathbb{R}^2[/imath]? I am taking multi variable analysis now and I was wondering what possible changes there might be from the Analysis of two dimensions vs Complex Analysis. A short summary of the differences would suffice. |
478328 | If [imath](ab)^3=a^3 b^3[/imath], prove that the group [imath]G[/imath] is abelian.
If in a group [imath]G[/imath], [imath](ab)^3=a^3 b^3[/imath] for all [imath]a,b\in G[/imath], amd the [imath]3[/imath] does not divide [imath]o(G)[/imath], prove that [imath]G[/imath] is abelian. I interpreted the fact that [imath]3[/imath] does not divide [imath]o(G)[/imath] as saying [imath](ab)^3\neq e[/imath], where [imath]e[/imath] is the identity of the group. As for proving [imath]ab=ba[/imath], I did not get anywhere useful. I got the relation [imath](ba)^2=a^2 b^2[/imath], but could not proceed beyond that. I got a lot of other relations too which I could not exploit- like [imath]a^2 b^3=b^3 a^2[/imath] A helpful hint instead of a solution would be great! Thanks in advance! | 282597 | Problem from Herstein on group theory
The problem is: If [imath]G[/imath] is a finite group with order not divisible by 3, and [imath](ab)^3=a^3b^3[/imath] for all [imath]a,b\in G[/imath], then show that [imath]G[/imath] is abelian. I have been trying this for a long time but not been able to make any progress. The only thing that I can think of is: [imath]ab\cdot ab\cdot ab=aaa\cdot bbb\implies(ba)^2=a^2b^2=aabb=(\text{TPT})abba.[/imath] Now, how can I prove the last equality? If I write [imath]aabb=abb^{-1}abb[/imath], then in order for the hypothesis to be correct, [imath]b^{-1}abb=ba\implies ab^2=b^2a[/imath]. Where am I going wrong? What should I do? |
461850 | mean of two consecutive number helps proving both number equals..
Well I was wondering how is this possible. let say: 4=4 -> R.H.S [imath]=4-(9/2)+(9/2)[/imath] [imath]=\sqrt{ (4 - ( 9/2 )) ^ 2}+ (9/2)[/imath] [imath]=\sqrt{ 16 - 36 +( 9/2 )^2}+ (9/2)[/imath] [imath]=\sqrt{ - 20 +( 9/2 )^2}+ (9/2)[/imath] [imath]=\sqrt{ -45 + 25 +( 9/2 )^2}+ (9/2)[/imath] [imath]=\sqrt{ 5^2 - 2x(9/2)x5 +( 9/2 )^2 }+ (9/2)[/imath] [imath]=\sqrt{ (5 - (9/2))^2 }+ (9/2)[/imath] [imath]=5- (9/2) + (9/2)[/imath] =5 going through this I found that every number can be proven equal to any number?????? still scratching my head..... | 170676 | Where's the problem in this equation? Resulting in [imath]4 = 5[/imath]
I just saw this equation and I can't find out where's the problem: [imath]25-45 = 16-36[/imath] [imath]25- 2 \cdot 5 \cdot \frac{9}{2} = 16- 2\cdot4\cdot\frac{9}{2}[/imath] [imath]25 - 2\cdot 5\cdot \frac{9}{2} + \frac{81}{4} = 16 - 2\cdot 4 \cdot \frac{9}{2} + \frac{81}{4}[/imath] [imath]\left( 5-\frac{9}{2} \right) ^2 = \left (4-\frac{9}{2} \right) ^2[/imath] [imath]5-\frac{9}{2} = 4 - \frac{9}{2}[/imath] [imath]5=4[/imath] |
478394 | How to show no analytic function with absolute value greater than negative power of absolute value z?
I am taking an exam tomorrow on Complex Analysis, and this is a problem I have come across multiple times on old exams and I've been stumped. "Show that there is no function f analytic on the punctured plane [imath]\mathbb{C}-\{0\}[/imath] that satisfies [imath]|f(z)| \ge \frac{1}{\sqrt{|z|}} \,\,\, \forall z \neq 0."[/imath] I probably tried a number of things back in June when I first saw it, but can't find my work on that right now. When I was going through again today, my initial approach was to assume by contradiction that there is such an f. I set [imath]g(z)=f(1/z)[/imath]. Then g would have a pole at infinity. Bringing that information back to f, there's a pole at 0. Which I probably should have seen before doing that. Then there's an entire function h and some integer n such that [imath]f(z)=\frac{h(z)}{z^n}[/imath]. Right? So I was playing with h. From the inequality, [imath]|h(z)| \ge |z|^{n-1/2}[/imath]. That doesn't look like a contradiction to me, but maybe I'm missing something. I asked somebody who has passed this exam before, and they suggested setting [imath]g(z)=\frac{1}{f(z)}[/imath]. That gives the inequality [imath]|g(z)| \le \sqrt{|z|}[/imath]. Again, I look at singularities. Here, the only singularity is at infinity, the same as my other g. I feel like it's just giving the same information as my first attempt, and I am still seeing no contradiction. I attempted to search for this on here, but couldn't find a way to formulate it that got me any relevant results through the search bar. If this is a repeat, just point me to the old one and trash mine. | 463729 | Does there exist an holomorphic function such that [imath]|f(z)|\geq \frac{1}{\sqrt|z|}[/imath]?
I have some trouble solving this problem: Does there exist an holomorphic function [imath]f[/imath] on [imath]\mathbb C\setminus \{0\}[/imath] such that [imath]|f(z)|\geq \frac{1}{\sqrt|z|}[/imath] for all [imath]z\in\mathbb C \setminus \{0\}[/imath]? I don't know where to start. My intuition is that you would get a problem with the singularity near [imath]0[/imath], but I am not sure how to prove it. Any help would be appreciate! Thanks! |
479778 | Find [imath]\frac{\sum\limits^{2499}_{k=1}{\sqrt{10+\sqrt{50+\sqrt{k}}}}}{\sum\limits^{2499}_{k=1}{\sqrt{10-\sqrt{50+\sqrt{k}}}}}[/imath]
Find the ratio,[imath]\frac{\sum\limits^{2499}_{k=1}{\sqrt{10+\sqrt{50+\sqrt{k}}}}}{\sum\limits^{2499}_{k=1}{\sqrt{10-\sqrt{50+\sqrt{k}}}}}[/imath] | 460331 | Simplify [imath]\left({\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt{k}}}}}\right)\left({\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt{k}}}}}\right)^{-1}[/imath]
Simplify [imath]\frac{\displaystyle\sum_{k=1}^{2499}\sqrt{10+{\sqrt{50+\sqrt{k}}}}}{\displaystyle\sum_{k=1}^{2499}\sqrt{10-{\sqrt{50+\sqrt{k}}}}}[/imath] I don't have any good idea. I need your help. |
479671 | Let [imath]p[/imath] and [imath]q[/imath] be two distinct primes. Prove that [imath]p^{q-1} + q^{p-1} =1 \mod pq[/imath]
Let [imath]p[/imath] and [imath]q[/imath] be two distinct primes. Prove that [imath]p^{q-1} + q^{p-1} =1 \mod pq[/imath] I try to used Fermat little theorem and I obtain the congruence [imath]p^q +q^p=0 \mod pq[/imath]. From this I don know how to relate to the congruence on the question. | 380510 | Proving the congruence [imath]p^{q-1}+q^{p-1} \equiv 1 \pmod{pq}[/imath]
If [imath]p[/imath] and [imath]q[/imath] are distinct primes , prove that [imath]p^{q-1}+q^{p-1} \equiv 1 \pmod {pq}[/imath] Using Fermat's Theorem we can get [imath]p^{q-1} \equiv 1 \pmod q, \qquad q^{p-1} \equiv 1 \pmod p.[/imath] After that I have no idea at all. Can anyone help me please |
474959 | Solving ODE with negative expansion power series
I am solving a series of ODE, such that each DE is equal to some degree of term that I'm expanding to. For instance, one DE is this: [imath]\xi^r\partial_r g_{rr}+2g_{tt}\partial_t\xi^t=\mathcal{O}(r)[/imath] [imath]g_{ii}[/imath] are given, from metric, but that's not important. I need to assume that the solution (since I'm looking for components of [imath]\xi^\mu[/imath], which is a vector with components [imath]\xi^t,\xi^r,\xi^\phi[/imath]) is given with power series of the form: [imath]\xi^\mu=\sum\limits_{n}\xi^\mu_n(t,\phi)r^n[/imath], and this is to be seen as expansion around [imath]1/r[/imath] (expansion around [imath]r=\infty[/imath]). Now when I plug this in the ODE I get this [imath]\frac{2}{l^2}\sum_n\xi^r_nr^{n+1}+2\sum_n\xi^t_{n,t} r^n+\frac{2}{l^2}\sum_n\xi^t_{n,t}r^{n+2}=\mathcal{O}(r)[/imath], where [imath]\xi^\mu_{n,i}[/imath] is the derivative with the respect to i-th component. What troubles me is, how to expand this? Do I set n=0,-1,-2,... until my [imath]\mathcal{O}(r)[/imath] terms cancel each other out? Or? I'm kinda stuck :\ | 478684 | Solving ODE with negative expansion power series
Since on Mathematics stackexchange I didn't get an answer, I'll try it here, since people here are more familiar with this topic (general relativity related). I am reading a dissertation of Porfyriadis "Boundary Conditions, Effective Action, and Virasoro Algebra for [imath]AdS_3[/imath]", and I am trying to solve a system of DE to get the appropriate diffeomorfism (page 31 onward). I am trying to solve a system of ODE, such that each DE is equal to some degree of term that I'm expanding to. For instance, one DE is this: [imath]\xi^r\partial_r g_{rr}+2g_{tt}\partial_t\xi^t=\mathcal{O}(r)[/imath] which you get by taking a Lie derivative of background metric and setting it equal to certain [imath]\mathcal{O}(r^n)[/imath] terms. [imath]g_{ij}[/imath] are given, from metric ofc. I need to assume that the solution (since I'm looking for components of [imath]\xi^\mu[/imath]( which is a vector with components [imath]\xi^t, \xi^r, \xi^\phi[/imath]) is given with power series of the form: [imath]\xi^\mu=\sum\limits_{n}\xi^\mu_n(t,\phi)r^n[/imath], and this is to be seen as expansion around 1/r (expansion around [imath]r=\infty[/imath]). Now when I plug this in the ODE I get this [imath]\frac{2}{l^2}\sum_n\xi^r_nr^{n+1}+2\sum_n\xi^t_{n,t} r^n+\frac{2}{l^2}\sum_n\xi^t_{n,t}r^{n+2}=\mathcal{O}(r)[/imath], where [imath]\xi^\mu_{n,i}[/imath] is the derivative with the respect to i-th component. What troubles me is, how to expand this? Do I set n=0,-1,-2,... until my O(r) terms cancel each other out? Or? I'm kinda stuck, at this seemingly easy point. In the thesis he gets 6 equations with coefficients, first one should be: [imath]\xi^r_{n-1}+l^2\xi^t_{n,t}+\xi^t_{n-2,t}=0,\ n\ge 2[/imath], But I am not getting this. What am I doing wrong? EDIT: For further clarity: The metric is that of [imath]AdS_3[/imath] given with line element: [imath]ds^2=-\left(1+\frac{r^2}{l^2}\right)dt^2+\left(1+\frac{r^2}{l^2}\right)^{-1}dr^2+r^2 d\phi^2[/imath], and the differential equations in question are given by solving [imath]\mathcal{L}_\xi g_{\mu\nu}=\mathcal{O}(r^n)[/imath], where [imath]\mathcal{O}(r^n)[/imath] are the fall off conditions. In the dissertation, he took the deviation of nonzero components of the metric to be subleading, that is: [imath]\mathcal{L}_\xi g_{tt}=\mathcal{O}(r)[/imath] [imath]\mathcal{L}_\xi g_{rr}=\mathcal{O}(r^{-3})[/imath] [imath]\mathcal{L}_\xi g_{\phi\phi}=\mathcal{O}(r)[/imath], while others are [imath]\mathcal{O}(1)[/imath]. Solving Lie derivative gives me 6 equations, which I should solve by plugging in the above ansatz ([imath]\xi^\mu=\sum\limits_{n}\xi^\mu_n(t,\phi)r^n[/imath]), but this is the part I get stuck. ADDENDUM: I was looking at other components, and have noticed that I have [imath]g_{rr}[/imath] factor with some of them. That term in metric is: [imath]g_{rr}=\left(1+\frac{r^2}{l^2}\right)^{-1}[/imath] Now, is it legitimate thing to expand this around [imath]r=\infty[/imath] so that I can put [imath]r[/imath] terms inside the sums (assumed solution)? |
326172 | The [imath] l^{\infty} [/imath]-norm is equal to the limit of the [imath] l^{p} [/imath]-norms.
If we are in a sequence space, then the [imath] l^{p} [/imath]-norm of the sequence [imath] \mathbf{x} = (x_{i})_{i \in \mathbb{N}} [/imath] is [imath] \displaystyle \left( \sum_{i=1}^{\infty} |x_{i}|^{p} \right)^{1/p} [/imath]. The [imath] l^{\infty} [/imath]-norm of [imath] \mathbf{x} [/imath] is [imath] \displaystyle \sup_{i \in \mathbb{N}} |x_{i}| [/imath]. Prove that the limit of the [imath] l^{p} [/imath]-norms is the [imath] l^{\infty} [/imath]-norm. I saw an answer for [imath] L^{p} [/imath]-spaces, but I need one for [imath] l^{p} [/imath]-spaces. Besides, I didn’t really understand the [imath] L^{p} [/imath]-answer either. Thanks for your help! | 669051 | Limit of [imath]\|x\|_p[/imath] as [imath]p\rightarrow\infty[/imath]
I am not sure how to start this hw problem. Here it is: Let [imath]x[/imath] be a given vector in [imath]\ell^p[/imath] with [imath]1\le p\lt\infty[/imath]. Show that [imath]x \in \ell^\infty[/imath] and [imath]\lim_{p \to \infty} \|x\|_p = \|x\|_\infty.[/imath] This seems like it would be obvious, but trying to prove it is different. I am not sure how to even start this one. Thanks. EDIT: Would the triangle inequality be useful here? |
480652 | conditioning by independent sigma-algebras
Let [imath]X[/imath] a random variable, [imath]\mathcal{A}[/imath], [imath]\mathcal{B}[/imath], two [imath]\sigma[/imath]-algebras such that [imath]\sigma(X,\mathcal{A})[/imath] is independent of [imath]\mathcal{B}[/imath]. Is it always true that [imath]\mathbb{E}[X|\sigma(\mathcal{A}, \mathcal{B})]=\mathbb{E}[X|\mathcal{A}][/imath]? I can show it for [imath]\mathcal{A}[/imath], [imath]\mathcal{B}[/imath] finite algebras, but I do not know how to extend it to general sigma-algebras. | 365310 | Conditional expectation on more than one sigma-algebra
I'm facing the following issue. Let [imath]X[/imath] be an integrable random variable on the probability space [imath](\Omega,\mathcal{F},\mathbb{P})[/imath] and [imath]\mathcal{G},\mathcal{H} \subseteq \mathcal{F}[/imath] be two sigma-algebras. We assume that [imath]X[/imath] is independent of [imath]\mathcal{G}[/imath], i.e. [imath]\sigma(X)[/imath] is independent of [imath]\mathcal{G}[/imath]. Can I say (and can I prove) that [imath] E(X \mid \sigma(\mathcal{G} \cup \mathcal{H})) = E(X\mid \mathcal{H}) ?[/imath] Thank you very much for your help! |
480734 | Proving that doesn't exist subgroup H with order 6 in [imath]A_4[/imath]
Let [imath]G=A_4[/imath]. Prove that does not exist subgroup [imath]H\le G[/imath] s.t [imath]|H|=6[/imath]. I don't know from where to start (maybe I need to prove that if so [imath]H\triangleleft A_4 [/imath]?) Any hint will be appreciated. | 94137 | Existence of subgroup of order six in [imath]A_4[/imath]
Show that the alternating group [imath]A_4[/imath] of all even permutations of [imath]S_4[/imath] does not contain a subgroup of order [imath]6[/imath]. For me am thinking to write all elements of [imath]A_4[/imath] and trying to find every cyclic subgroup generated by each element of [imath]A_4[/imath], then I have to check whether there exist such a subgroup or not! This is a long procedure for me, I ask if there is a short way to do this. |
480632 | Prove that [imath]2^n>2n[/imath] for all integral values of n greater than 2
Prove [imath]2^n >2n[/imath] for all integral values of n greater than 2. Let [imath]p_n[/imath] be the statement: [imath]2^n>2n\ \forall\ n\gt2[/imath] If the inequality is valid for [imath]n=k[/imath] where [imath]k>2[/imath]: [imath]p_k: 2^k>2k[/imath] Then for [imath]n=k+1[/imath]: [imath]p_{k+1} = 2^{k+1}>2(k+1)[/imath] I don't know how to do the inductive step itself, I have only done series/recurrence relations inductions. Have I used the correct layout/notation? Is there more cool notation I could add to improve the mathematical-ness of the proof? Thanks | 169859 | Prove by mathematical induction that [imath]2n ≤ 2^n[/imath], for all integer [imath]n≥1[/imath]?
I need to prove [imath]2n \leq 2^n[/imath], for all integer [imath]n≥1[/imath] by mathematical induction? This is how I prove this: Prove:[imath]2n ≤ 2^n[/imath], for all integer [imath]n≥1[/imath] Proof: [imath]2+4+6+...+2n=2^n[/imath] [imath]i.)[/imath] Let [imath]P(n)=1 P(1): 2(1)=2^1\implies 2=2[/imath]. Hence, [imath]P(1)[/imath] is true. [imath]ii.)[/imath] Assume that [imath]P(n)[/imath] is true for [imath]n=k[/imath], i.e, [imath]2+4+6+...+2k=2^k[/imath], and prove that [imath]P(n)[/imath] is also true for [imath]n=k+1[/imath], i.e, [imath]2+4+6...+2(k+1)=2^{(k+1)}[/imath] from the assumption add [imath]2(k+1)[/imath] on both sides so we have [imath]2+4+6...2k+2(k+1)=2^k+2(k+1)[/imath] I'm confused with [imath]2^k+2(k+1)[/imath], I don't know how to make [imath]2^k[/imath] be equivalent to [imath]2^{k+1}[/imath]. I feel i'm doing something wrong. Any help would be appreciated! |
480807 | Question about sum of l'p spaces
Is this true or false: [imath]\bigcup_{1 \leq p < q} \ell^p = \ell^q?[/imath] I've tried to give an counter example, but i did not get any one. | 111743 | About the [imath]p[/imath] summable sequences
How to prove that [imath]\ell^p \setminus \cup_{1\leq q <p}\ell^q[/imath] is not empty? Here [imath]\ell^p=L^p(\mathbb{N})[/imath]. |
481038 | Prove [imath]\max\{x,y\}=\frac {(x-y)}{2} + \frac {(x+y)}{2}[/imath]
Problem: prove [imath]\max\{x,y\}=\frac {(x-y)}{2} + \frac {(x+y)}{2}[/imath] [imath]x[/imath] and [imath]y[/imath] are max elements in two sets. Here is what i have thought of so far as a concrete problem: [imath]\max \{4,5\}=d(4,5)= 1[/imath] I know that the [imath]\begin{align} \max\{4,5\} &= \frac{4-5}{2}+ \frac{4+5}{2} \\ &= \frac{-1}{2}+\frac{1}{2} \\ &= 0\end{align}[/imath] I am thinking of breaking it down to trichotomy: [imath]x=y, x>y[/imath], and [imath]x<y[/imath]. for [imath]x=y[/imath] what i got is.. [imath] \begin{align} \max\{x,y\} &= \frac{(x-y)}{2}+\frac{(x+y)}{2} \\ &= 0+x \\ &= x \end{align} [/imath] (which is the max) but why isn't this 0? and am i even thinking in the right direction or am i mixing up two different concepts? | 429622 | Show that the [imath]\max{ \{ x,y \} }= \frac{x+y+|x-y|}{2}[/imath].
Show that the [imath]\max{ \{ x,y \} }= \dfrac{x+y+|x-y|}{2}[/imath]. I do not understand how to go about completing this problem or even where to start. |