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1864656
|
A necessary and sufficient condition for [imath]B \subset \ell^1[/imath] to be compact
I want to show that: For any [imath]B\subset \ell^1[/imath], [imath]B[/imath] is compact if and only if [imath]B[/imath] is bounded, closed and satisfies [imath] \forall \epsilon>0, \; \exists N\in \mathbb{N}, \; \forall \{x_n\}_{n=1}^\infty \in B, \; \sum_{n=N}^\infty |x_n| < \epsilon. [/imath]
|
1206280
|
A set [imath]A \subset l_1[/imath] is compact if and only if closed, bounded, and one other condition
A set [imath]A \subset \ell_1[/imath] is compact if and only if [imath]A[/imath] is closed and bounded and given any [imath]\epsilon >0[/imath], there exists [imath]n_0[/imath] such that [imath]\sum_{k=n}^{\infty} |x_k| < \epsilon[/imath] for all [imath]n> n_0[/imath] and [imath]x \in A[/imath]. It's easy to prove the part assuming that [imath]A \subset l_1[/imath] is compact, but I am finding difficulty to prove the other part.
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1864555
|
Last digit on [imath]3^{100}[/imath]
How to find the last digit on [imath]3^{100}[/imath]? Is there any proper method to solve such questions without calculator of course?
|
194482
|
Last two digits of [imath]3^{400}[/imath]
I am trying to calculate the last two digits of [imath]3^{400}[/imath], but working [imath]\bmod {100}[/imath] is proving intractable. I did the last digit case pretty handily, and now I'm stuck at last two.
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1864768
|
Smooth path lifting on [imath]S^1[/imath]
Given a smooth map [imath]f:S^1\to S^1[/imath], there is a smooth map [imath]g:\Bbb R\to\Bbb R[/imath] such that [imath]f(\cos t,\sin t)=(\cos g(t),\sin g(t))[/imath] and [imath]g(2\pi)=g(0)+2\pi q[/imath] for some [imath]q\in\Bbb Z[/imath]. I'm fairly sure one can prove this without using algebraic topology/path lifting. The idea is that [imath]g[/imath] measures the "signed distance along [imath]S^1[/imath]" that the image of [imath]f[/imath] travels (at least on [imath][0,2\pi][/imath]). One then extends [imath]g[/imath] to [imath]\Bbb R[/imath] by setting [imath]g(t+2\pi)=g(t)+2\pi q[/imath]. I don't know how to measure that in a smooth manner (I can't even get it to be continuous). Is there an elementary proof of this?
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981339
|
Prove that there exists a smooth map [imath]g\colon R\to R[/imath] such that [imath]f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))[/imath] and satisfying [imath]g(2π)=g(0)+2\pi q[/imath].
Let [imath]f\colon S^1\to S^1[/imath] be any smooth map. Prove that there exists a smooth map [imath]g\colon\mathbb{R}\to\mathbb{R}[/imath] such that [imath]f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))[/imath] and satisfying [imath]g(2π)=g(0)+2\pi q[/imath]. The book told me to show that [imath]g(2\pi)=g(0)+2\pi q[/imath], then extend [imath]g(t+2\pi) =g(t)+2\pi q[/imath]. But I don't understand what they mean. How can I get function [imath]g[/imath] from [imath]f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))[/imath], and what is [imath]g[/imath]?
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556482
|
What do we mean by [imath]CP^2[/imath] with reverse orientation?
[imath]CP^2[/imath] and [imath]\bar{CP^2}[/imath] are not diffeomorphic since they have non-isomorphic intersection forms. so why do we call the latter [imath]CP^2[/imath] with reverse orientation? it seems like we are not just reversing a given orentation on [imath]CP^2[/imath] to get [imath]\bar {CP^2}[/imath].
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131130
|
Orientation on [imath]\mathbb{CP}^2[/imath]
I am confused by the orientation of a topological manifold. My understanding is: An orientation of a topological manifold is a choice of generator of the [imath]H^n(M,\mathbb Z)[/imath]. So given a manifold, we could have 2 orientation defined on the manifold. For [imath]\mathbb{CP}^2[/imath] on orientation is determined by the complex structure, the other orientation is denoted by [imath]\overline{\mathbb{CP}}^2[/imath]. And it is well known that there is no orientation reversing map from [imath]\mathbb{CP}^2[/imath] to itself. My question is: are they homeomorphic? I guess they are not homeomorphic. But I am confuesed, doesn't the orientations defined on the same manifold, how come after reversing the orientation they become not homeomorphic?
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1865041
|
Showing that [imath]\sigma[/imath]-algebra is uncountable
Suppose [imath]\mathcal{A}[/imath] is a [imath]\sigma[/imath]-algebra with the property that whenever [imath]A \in \mathcal{A}[/imath] is nonempty, there exist [imath]B[/imath], [imath]C \in \mathcal{A}[/imath] with [imath]B \cap C = \emptyset[/imath], [imath]B \cup C = A[/imath], and neither [imath]B[/imath] nor [imath]C[/imath] is empty. How do I see that [imath]\mathcal{A}[/imath] is uncountable?
|
928344
|
Uncountable [imath]\sigma[/imath]-algebra
I'm stuck on the following problem (Source: Real Analysis for Graduate Students; Exercise 2.6; Bass): Suppose [imath]\mathcal A[/imath] is a [imath]\sigma[/imath]-algebra with the property that whenever [imath]A \in \mathcal A[/imath], there exists [imath]B, C \in \mathcal A[/imath] with [imath]B \cap C = \emptyset[/imath], [imath]B \cup C = A[/imath], and neither [imath]B[/imath] nor [imath]C[/imath] is empty. Prove that [imath]\mathcal A[/imath] is uncountable. I think there is the added assumption that this is only true for [imath]A \in \mathcal A[/imath] having at least two elements, so it doesn't hold for singletons and the emptyset? I tried to show this by way of contradiction and say that [imath]\mathcal A = \{A_k\}[/imath] is countable, but I didn't see how this could get me to a contradiction. The other approach I tried was to look at [imath]X = B_1 \cup C_1[/imath] where [imath]B_1 \cap C_1 = \emptyset[/imath], then look at [imath]B_1 = B_2 \cup C_2[/imath] where [imath]B_2 \cap C_2 = \emptyset[/imath] and so on. If I look at the [imath]C_k[/imath]'s they are pairwise disjoint and I either have a finite number of them or an infinite number of them in which case I've created a countable sequence of pairwise disjoint nonempty elements of [imath]\mathcal A[/imath]. This path looked promising, but I couldn't see what to do next. Any ideas? As a secondary question: I think this result is supposed to be used to show that if [imath]\mathcal A[/imath] is a [imath]\sigma[/imath]-algebra with infinitely many elements, then it's uncountable, but I wasn't able to show that the property mentioned above (the one I'm trying to show) was satisfied in this case.
|
1392247
|
How do ideal quotients behave with respect to localization?
Suppose [imath]R[/imath] is commutative ring with unity. For ideals [imath]I[/imath], [imath]J \subseteq R[/imath], the ideal quotient [imath](J:I)[/imath] is [imath](J:I) := \{x\in R \, : \, xI \subseteq J\}[/imath] Let [imath]S\subset R[/imath] be a multiplicative set. When does localization at [imath]S[/imath] commute with taking quotients, i.e. when does the equality [imath](S^{-1} J : S^{-1} I) = S^{-1} (J:I)[/imath] hold? More generally, when is it true that [imath]S^{-1} \text{Ann}_{R} (M) = \text{Ann}_{S^{-1} R} (S^{-1} M)[/imath] for an [imath]R[/imath]-module [imath]M[/imath]?
|
1942533
|
Localization and annihilator of modules
I have read that for a given finitely generated module [imath]M[/imath] over a ring [imath]R[/imath] and a multiplicative set [imath]S[/imath] in [imath]R[/imath], [imath]\mathrm{Ann}(S^{-1}M)=S^{-1}\mathrm{Ann}(M)[/imath] as ideals in [imath]S^{-1}R. [/imath] Is it also true for an infinitely generated module? If not, what is a counterexample? Here, [imath]\mathrm{Ann}(M)=\{r\in R:rm=0 \ \forall \ m\in M\}[/imath] and similarly [imath]\mathrm{Ann}(S^{-1}M)[/imath] is defined.
|
1756545
|
Linearity of Lebesgue measure: [imath]\mu(AV)=|\det A|\mu(V)[/imath]
Suppose [imath]\mu[/imath] is the Lebesgue measure defined on [imath]\Bbb R^k[/imath], I want to show that [imath]\mu[/imath] has some kind of linearity, which seems intuitively correct: Suppose [imath]A[/imath] is a linear transformation on [imath]\Bbb R^k[/imath] (namely, a square matrix), then given any Lebesgue measurable subset [imath]V[/imath], [imath]\mu(AV)=|\det A|\mu(V),[/imath] where [imath]AV:=\{Av\mid v\in V\}[/imath]. I'm almost sure of it. But I couldn't find a proof about this property in any literature at hand. I don't think the proof should be hard, though, if the elementary case has been proved. But that's exactly where I get stuck: how am I supposed to, for example, prove the [imath]V=[0,1]^k[/imath] case? Now if this can be proved, then I may just use elementary sets (unions of [imath]k[/imath]-cells each two of which share an at most measure zero set) to approach [imath]V[/imath] and the result follows. Could anyone help me on this part or give me other insights about the proof? Thanks in advance.
|
52161
|
Measure of Image of Linear Map
I am trying to work my way through the proof of the change of variables theorem for Lebesgue integrals. A key lemma in this context is as follows: If [imath]T:\mathbb{R}^n \rightarrow \mathbb{R}^n[/imath] is a linear map and [imath]A \subset \mathbb{R}^n[/imath] is Lebesgue measurable then [imath]\lambda(T(A)) = |\det T \ | \cdot \lambda (A)[/imath], where [imath]\lambda(X)[/imath] denotes the Lebesgue measure of [imath]X[/imath]. Can anyone provide a reference for a proof of this lemma that clearly references the facts from linear algebra that are necessary to effect the proof? The source I have for this lemma refers to German texts that I am incapable of reading and I have been unsuccessful at finding an alternate proof. Added For the Benefit of Future Readers: In addition to the excellent references I received in response to this question, I have managed to find an additional reference that also provides a good proof of this fact: Aliprantis and Burkinshaw's Principles of Real Analysis, Third Edition, Lemma 40.4 pp 389-390.
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1865744
|
Difference between | and /
"If we find a prime [imath]p[/imath] such that [imath]p\mid n[/imath] , then [imath]n/p[/imath] is a positive integer that's smaller than [imath]n[/imath]." I understand [imath]n/p[/imath] is [imath]n[/imath] divided by [imath]p[/imath] but what is [imath]n\mid p[/imath]?
|
1207404
|
Mid '|' in math?
What does this equation mean? What does the [imath]|[/imath] mean? [imath]446617991732222310 | mn(m^k - n^k)[/imath] Here is the complete question for reference - What is the smallest positive integer [imath]k[/imath], such that for every ordered pair of integers [imath](m, n)[/imath], we have [imath]446617991732222310 | mn(m^k - n^k)[/imath]? Please help
|
1866421
|
Show that if a function is nonnegative and continuous on [a,b] then f(x)=0 for all x in [a,b]
Show that if [imath]f : [a, b] \to \mathbb{R}[/imath] is non-negative and continuous on [imath][a, b][/imath] and [imath]\int_{a}^{b} f(x)dx = 0[/imath] then [imath]f(x)=0[/imath], for all [imath]x ∈ [a, b][/imath]. I'm having difficulties in proving things to do with integrals, how do I start this?
|
1224799
|
If [imath]f[/imath] is continuous, nonnegative on [imath][a, b][/imath], show that [imath]\int_{a}^{b} f(x) d(x) = 0[/imath] iff [imath]f(x) = 0[/imath]
If [imath]f[/imath] is continuous, nonnegative on [imath][a, b][/imath], show that [imath]\int_{a}^{b} f(x) dx = 0[/imath] iff [imath]f(x) = 0[/imath] "[imath]\Rightarrow[/imath]" Assume by contradiction that [imath]f(x) \neq 0[/imath] for some [imath]x_0 \in [a, b][/imath]. Without loss of generality, assume that [imath]f(x_0) > 0[/imath]. Since [imath]f[/imath] is continuous, [imath]\exists \delta > 0[/imath] such that [imath]f(x) > 0, \forall x \in (x_0 - \delta, x_0 + \delta)[/imath]. Since [imath]f[/imath] is continuous on [imath][a, b][/imath], then it is Riemann integrable. Therefore [imath]\int_{a}^{b} f(x) dx = \sup L(f, p)[/imath] for some partition [imath]p[/imath]. Now what I want to do is to show that the lower sum, [imath]L(f, p) > 0[/imath], in that case its supremum is [imath]>0[/imath], and that contradicts the condition that [imath]\int_{a}^{b} f(x) dx = 0[/imath]. However, I don't know how to articulate it. "[imath]\Leftarrow[/imath]" If [imath]f(x) = 0, \forall x \in [a, b][/imath], then [imath]\int_{a}^{b} f(x) dx = \int_{a}^{b} 0 dx = 0[/imath]
|
1866614
|
Matrix identity with symmetry conditions
Let [imath]A,B,C,D[/imath] be four complex [imath]n\times n[/imath] matrices such that [imath]AB^T[/imath] and [imath]CD^T[/imath] are symmetric and [imath]AD^T-BC^T=I[/imath]. Show that [imath]A^TD-C^TB=I[/imath]. The point is that I can't see how any theory can be used here. It appears to be a mere exercise of calculus which, either way, I'm not able to solve. I think that some multiplications must be done, in order to use the fact that the identity is involved (and not just show that [imath]A^TD-C^TB=AD^T-BC^T)[/imath]. I tried to get help from particular cases, but found nothing.
|
463791
|
Prove that [imath]A^TD-C^TB=I[/imath]
Let A,B,C,D be complex matrices [imath]n \times n[/imath] such that [imath]AB^T,CD^T[/imath] are symmetric and [imath]AD^T-BC^T=I[/imath]. Prove that [imath]A^TD-C^TB=I[/imath]. Can anyone give me any idea? Thank you.
|
1866733
|
A question on inner product space
I don't have much idea about inner product space. So, plz help me to understand this question; Let [imath]A[/imath] be a [imath]n\times n[/imath] matrix with real entries. Define [imath]\langle x,y\rangle _A:=\langle Ax,Ay\rangle, \quad x,y\in \mathbb{R}^n[/imath]. Then [imath]\langle x,y\rangle_A[/imath] defines an inner-product if and only if (a) Ker A={0} (b) Rank A=n (c) All eigenvalues of are positive. (d) All eigenvalues of are non-negative More than one option may be true.
|
1833492
|
Show that [imath]\langle x,y\rangle_A = \langle Ax,Ay\rangle[/imath] is an inner product on [imath]\mathbb R^n[/imath]
Let [imath]A[/imath] be an [imath]n \times n[/imath] matrix with real enteries. Define [imath]\langle x,y\rangle_A = \langle Ax,Ay\rangle, \quad x,y \in \mathbb R^n[/imath] , where [imath]\langle,\rangle[/imath] is a standard inner product on [imath]\mathbb R^n[/imath]. Then [imath]\langle x,y\rangle_A[/imath] is an inner product iff 1) [imath]\ker A = \{0\}[/imath]. 2) [imath]\operatorname{rank} A= n[/imath]. 3) all eigenvalues of [imath]A[/imath] are positive. 4) all eigen value of [imath]A[/imath] are non negative. Clearly if [imath]\ker A \neq \{0\}[/imath], then there exist [imath]x \neq 0[/imath] such that [imath]Ax =0[/imath], then [imath]\langle Ax, Ax\rangle_A = 0[/imath], then [imath]\langle ,\rangle_A[/imath] is not inner product, so [imath]\ker A = \{0\}[/imath], thus rank of [imath]A[/imath] is [imath]n[/imath]. I am unable to check the 3) and 4) option. Please help to check the other option. Thank you
|
1866369
|
How many subsspaces and how calculate them in a Subspace
Let [imath]n>0[/imath]. Let [imath]V\subset \mathbb{F}_2^n[/imath] a vectorial space. Let [imath]U[/imath] a subspace of [imath]V[/imath]. How many subspaces [imath]W_i \subset V[/imath], at most, I can describe (for example using the basis) such that [imath]U \subset W_i[/imath]?
|
1865610
|
Number of Subspaces that contains other Space
In [imath]GF(2)[/imath], How Can I calculate the number of subspaces of dimension [imath]k<w[/imath] that contains a fixed subspace of dimension [imath]k'<w[/imath]:
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1867250
|
[imath]V[/imath] be a vector space , [imath]T:V \to V[/imath] be a linear operator , then is [imath](\ker(T) \cap R(T) ) \times R(T^2) \cong R(T)[/imath]?
Let [imath]V[/imath] be a vector space , [imath]T:V \to V[/imath] be a linear operator , then is it true that [imath](\ker(T) \cap R(T) ) \times R(T^2) \cong R(T)[/imath] ? (note that the direct product is well-defined as both the spaces are subspaces of [imath]V[/imath] so have same ground field ) See [imath]V[/imath] be a vector space , [imath]T:V \to V[/imath] be a linear operator , then is [imath]\ker (T) \cap R(T) \cong R(T)/R(T^2) [/imath]? , note that the isomorphism holds when [imath]V[/imath] is finite dimensional as it is known that if [imath]U,W[/imath] are finite dimensional vector spaces over a same field , then [imath]\dim (U \times W)=\dim U + \dim W[/imath] . Also note that the example given in the linked question doesn't violate the stated isomorphism here . Please help . Thanks in advance . EDIT : One general thing that can be noted is that [imath]R(T)/R(T)\cap \ker (T) \cong R(T^2)[/imath] holds , but I don't know whether it can be used to prove or disprove the claim in question NOTE : If [imath]V,W[/imath] are vector spaces over a field [imath]F[/imath] then [imath]V \times W[/imath] is a vector space with addition and multiplication defined as [imath](v,w)+(v',w'):=(v+v',w+w') [/imath] ; [imath] k.(v,w):=(kv,kw)[/imath] . It can be seen that if [imath]B_V , B_W[/imath] are bases for [imath]V , W[/imath] repectively , then [imath](B_V \times \{O_W\})\cup (\{O_V\}\times B_W)[/imath] is a basis for [imath]V \times W[/imath] , so that if [imath]V,W[/imath] are finite dimensional then so is [imath]V \times W[/imath] and [imath]\dim (V \times W)=|(B_V \times \{O_W\}) \cup (\{O_V\}\times B_W)|=|B_V \times \{O_W\}|+|\{O_V\}\times B_W|=|B_V|+|B_W|=\dim V + \dim W[/imath] So this "direct product" is substantially different from "direct sum"
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1867207
|
[imath]V[/imath] be a vector space , [imath]T:V \to V[/imath] be a linear operator , then is [imath]\ker (T) \cap R(T) \cong R(T)/R(T^2) [/imath]?
Let [imath]V[/imath] be a vector space , [imath]T:V \to V[/imath] be a linear operator , then is it true that [imath]\ker (T) \cap R(T) \cong R(T)/R(T^2) [/imath] ( where [imath]R(T)[/imath] denotes the range of [imath]T[/imath] ) ? I know that the statement is true when [imath]V[/imath] is finite dimensional as I can show that if [imath]V[/imath] is finite dimensional , then [imath]\dim (\ker (T) \cap R(T))=rank (T) - rank (T^2)=\dim R(T)/R(T^2)[/imath] , but I don't know what happens in general . Please help . Thanks in advance
|
1867953
|
If [imath]AB = BA[/imath] for [imath]A,B \in \mathcal{L}(V,V)[/imath], then [imath]A[/imath] and [imath]B[/imath] have these properties
There is a base such [imath]A[/imath] and [imath]B[/imath] are both upper triangular on these base, and if [imath]A[/imath] and [imath]B[/imath] are diagonalizable, then [imath]A[/imath] and [imath]B[/imath] are diagonalizable simultaneously. For the first I have no idea. To the second one I am trying to show something like: Iv [imath]v[/imath] is an eigenvector for [imath]A[/imath] then [imath]Bv[/imath] is an eigenvector for [imath]A[/imath]. To the second I proceeded as: [imath]v \in \ker (A - \lambda * Id)[/imath] for same [imath]\lambda \neq 0[/imath], then [imath]Bv = B\frac{1}{\lambda}Av = \frac{1}{\lambda}BAv = \frac{1}{\lambda}ABv[/imath] Then [imath]A(Bv) = \lambda Bv[/imath] and then [imath]Bv[/imath] is an eigenvector for [imath]A[/imath] also with eigenvalue [imath]\lambda[/imath]. But I am stuck here. I mean, I don't know what to do with this. I do appreciate any hint or answer.
|
1054886
|
Prove two commutative linear transformations on a vector space over an algebraically closed field can be simultaneously triangularized
Prove two commutative linear transformations on a finite-dimensional vector space [imath]V[/imath] over an algebraically closed field can be simultaneously triangularized. It is equivalent to show if [imath]AB=BA[/imath], then there exists a basis [imath]X[/imath] such that both [imath][A:X][/imath] and [imath][B:X][/imath] are triangular, where [imath]A[/imath] and [imath]B[/imath] are both linear transformations. Hint: Find a subspace [imath]W[/imath] of [imath]V[/imath] that is invariant under both by [imath]A[/imath] and [imath]B[/imath], with this in mind, consider any proper value [imath]λ[/imath] of [imath]A[/imath] and examine the set of all solutions of [imath]Ax=λx[/imath] for the role of [imath]W[/imath]. Notation:Let [imath]A[/imath] be a linear transformation on an [imath]n[/imath]-dimensional vector space, and let [imath]X={x_1,x_2,...,X-n}[/imath] be a basis in that space, then [imath][A]=[A:X]=(a_{ij})[/imath] be the matrix of [imath]A[/imath] in the coordinate system [imath]X[/imath], so that [imath]Ax_{j}=\sum_{i}a_{ij}x_{i}[/imath]. Source: Halmos, Finite dimensional vector spaces, section 56 Triangular form exercise #2. Could anyone please help, I have not idea how to do this. Any help is appreciated. Thanks a lot.
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1868306
|
An Extreme Point of a closed ball of [imath]\ell^\infty[/imath]
I am trying to prove that all "closed unit ball" of [imath] c_0 = \{ \{x_n\}_{n=1}^\infty \in \ell^\infty : \lim_{n\to\infty} x_n = 0\} [/imath] do not have any extreme point. (Extreme Point) Let [imath]X[/imath] be a vector space and [imath]A \subset X[/imath] be convex. We say [imath]x\in A[/imath] is "Extreme Point" if for [imath]x = (1-t)y + tz,\; y,z,\in A, \;t\in(0,1)[/imath] then [imath]y = z = x[/imath]. What I tried is as follows: Let [imath]B[/imath] be a closed unit ball of [imath]c_0[/imath], that is, [imath]B = \{\{x_n\}_{n=1}^\infty \in \ell^\infty : \lim_{n\to \infty} x_n = 0 \text{ and } \|x\|_{\ell^\infty}\le 1\}.[/imath] If there is a extreme point [imath]b = \{b_n\}_{n=1}^\infty\in B[/imath], then we have for [imath] b = (1-t)y + tz, \quad y,z\in B,\quad t\in (0,1) [/imath] implies [imath] y = z = b. [/imath] But I cannot do anymore here. Would you please help me?
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1866546
|
Extreme points of the unit ball of the space [imath]c_0 = \{ \{x_n\}_{n=1}^\infty \in \ell^\infty : \lim_{n\to\infty} x_n = 0\}[/imath]
I want to prove that all "closed unit ball" of [imath] c_0 = \{ \{x_n\}_{n=1}^\infty \in \ell^\infty : \lim_{n\to\infty} x_n = 0\} [/imath] do not have any extreme point. Would you please help me? (Extreme Point) Let [imath]X[/imath] be a vector space and [imath]A \subset X[/imath] be convex. We say [imath]x\in A[/imath] is an extreme point if for [imath]x = (1-t)y + tz,\; y,z,\in A, \;t\in(0,1)[/imath] then [imath]y = z = x[/imath]. What I tried is as follows: Let [imath]B[/imath] be a closed unit ball of [imath]c_0[/imath], that is, [imath]B = \{\{x_n\}_{n=1}^\infty \in \ell^\infty : \lim_{n\to \infty} x_n = 0 \text{ and } \|x\|_{\ell^\infty}\le 1\}.[/imath] If there is a extreme point [imath]b = \{b_n\}_{n=1}^\infty\in B[/imath], then we have for [imath] b = (1-t)y + tz, \quad y,z\in B,\quad t\in (0,1) [/imath] implies [imath] y = z = b. [/imath] But I cannot do anymore here. Would you please help me?
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1868576
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[imath]\Bbb Z[\sqrt{-5}][/imath] is not a PID
I want to show: In a PID [imath]R[/imath] two elements [imath]a,b\in R[/imath] always have a greatest common divisor. Therefore [imath]\Bbb Z[\sqrt{-5}][/imath] is not a PID. For the first part: [imath]I=\{ax+by:x,y\in R\}[/imath] is an ideal, so since [imath]R[/imath] is a PID we have [imath]I=(g)[/imath] for some [imath]g\in R[/imath]. In particular, [imath]a=gr[/imath] and [imath]b=gs[/imath] for some [imath]r,s\in R[/imath], which says that [imath]g[/imath] divides both [imath]a[/imath] and [imath]b[/imath]. If [imath]d\in R[/imath] is any other divisor of both [imath]a[/imath] and [imath]b[/imath], then [imath]d[/imath] divides any linear combination of [imath]a[/imath] and [imath]b[/imath]. In particular, since [imath]I=(g)[/imath], [imath]g[/imath] is such a linear combination. This proves [imath]d\mid g[/imath], hence [imath]g[/imath] is a [imath]\gcd[/imath]. For the second part: I have a hunch that [imath]6[/imath] and [imath]2(1+\sqrt{-5})[/imath] have no [imath]\gcd[/imath] since [imath]6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})[/imath]. How can I finish the argument?
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103593
|
A question about proving that there is no greatest common divisor
I have to answer this question: Prove that in the ring [imath]\mathbb Z[\sqrt{-5}][/imath] there's no gcd to [imath]6[/imath] and [imath]2\cdot (1+\sqrt{-5})[/imath]. I have no clue how to do this but however I've tried to prove that their sum [imath]6+2\cdot (1+\sqrt{-5}) = 8+2\cdot\sqrt{-5}[/imath] is not in [imath]\mathbb Z[\sqrt{-5}][/imath] but I was not able to do this. How should I proceed?
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1868664
|
A problem concerning compact operators in [imath]\ell^p[/imath]
Let [imath]\alpha_n \in \mathbb C[/imath] and [imath]\lim_{n\to\infty}\alpha_n = 0[/imath]. Let [imath]T[/imath] be a linear continuous operator from [imath]\ell^p \to \ell^p (1\le p\le \infty)[/imath] defined by [imath] T((x_1, x_2, \ldots)) = (\alpha_1 x_1, \alpha_2 x_2 , \ldots). [/imath] Then I want to show that [imath]T[/imath] is compact operator on [imath]\ell^p[/imath], which means any bounded sequence in [imath]\ell^p[/imath], there exists a subsequence [imath]\{x_{n_k}\}[/imath] such that [imath]\{Tx_{n_k}\}[/imath] converges in the sense of norm.
|
1764477
|
Problem about a compact operator [imath]T:l^p\rightarrow l^p[/imath]
I have to solve this problem. Let [imath]\{\lambda_n\}[/imath] be a sequence of real number such that [imath]\lim_{n\rightarrow\infty}\lambda_n=0[/imath] and consider the operator [imath]T:l^p\rightarrow l^p[/imath], [imath]1\leq p\leq \infty[/imath], defined by [imath]T(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_n x_n,\ldots\} [/imath] Prove that [imath]T[/imath] is compact. My attempt is the following. I have considered the operator [imath] T_N:l^p\rightarrow l^p,\quad T_N(\{x_1,\ldots x_n,\ldots\})=\{\lambda_1 x_1,\ldots,\lambda_N x_N,0,0,\ldots\}. [/imath] [imath]T_N[/imath] is clearly linear and it is also compact. In fact given any bounded sequence [imath]\{x^{(n)}\}_{n\geq 1}[/imath] in [imath]l^p[/imath], the sequence [imath]\{T_Nx^{(n)}\}_{n\geq 1}[/imath] is bounded in [imath]\mathbb{R}^N\subset l^p[/imath]. Then, since every bounded sequence in [imath]\mathbb{R}^N[/imath] has a convergent subsequence, it follows that [imath]T_N[/imath] is compact. Now, I have to prove that [imath]T[/imath] is compact. For this I proved that [imath]||T_N-T||\rightarrow 0[/imath] as [imath]N\rightarrow\infty[/imath]. In fact we have: [imath] ||T_N-T||=\sup_{||x||=1}||(T_N-T)x||=\sup_{||x||=1}||T_Nx-Tx||=\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}} [/imath] and, given [imath]\epsilon>0[/imath], there exists [imath]\bar n[/imath] such that [imath]|\lambda_n|<\epsilon[/imath], for every [imath]n>\bar n[/imath]. So if we take [imath]N>\bar n[/imath], we get [imath] \sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|\lambda_i x_i|^p\right)^{\frac{1}{p}}\leq\epsilon\sup_{||x||=1}\left(\sum_{i=N+1}^{\infty}|x_i|^p\right)^{\frac{1}{p}}\leq\epsilon [/imath] Therefore [imath]||T_N-T||\rightarrow 0[/imath] as [imath]N\rightarrow\infty[/imath] implies that [imath]T[/imath] is compact. My questions are: (1) is the proof of the compactness of the operator [imath]T_N[/imath] correct? (2) is the conclusion correct? Thanks
|
1868836
|
Every alternating permutation is a product of 3-cycles
Show that every element in [imath]A_n[/imath] (alternating group of degree [imath]n[/imath]) for [imath]n \ge 3[/imath] can be expressed as a [imath]3[/imath]-cycle or a product of three cycles. I understand that if [imath]n[/imath] is odd, then any element can be written as (let the numbers represent arbitrary elements) [imath](1,2,3)(3,4,5)(5,6,7)...(n-2,n-1,n)[/imath] as a product of [imath]3-[/imath]cycles, but when [imath]n[/imath] is even I can't figure out how to write it as a product of three cycles.
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914338
|
The alternating group is generated by three-cycles
Prove that, for [imath]n \geq 3[/imath], the three-cycles generate the alternation group [imath]A_n[/imath] Proof: We multiply on the left by 3-cycles to "reduce" an even permutation [imath]p[/imath] to the identity, using induction on the number of indices fixed by a permutation. How the indices are numbered is irrelevant. If [imath]p[/imath] contains a [imath]k[/imath]-cycle with [imath]k \geq 3[/imath], we may assume that it has the form [imath]p=(123\dots k)\dots[/imath] Multiplying on the left by [imath](321)[/imath] gives [imath]p'= (321)(123 \dots k)\dots=(1)(2)(3\dots k)\dots[/imath] More fixed indices. What do you think ?
|
1868789
|
Even harmonic sums?
How do we calculate this? [imath] \displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{H_{2n}}{2n} [/imath] I am stuck that the integrals isn't converging for harmonic (even) numbers . somebody please help .
|
1000140
|
Evaluating [imath]\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}[/imath]
I would appreciate to understand the main steps giving the evaluation of this series: [imath] S=\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}[/imath] where [imath]H_n[/imath] is the harmonic number. I've tried with no success to obtain this sum with the help of Wolfram Alpha.
|
1869210
|
[imath]L^p([0,1])[/imath] stricly convex
Exercise: For which [imath]p\in [1,\infty][/imath] is [imath]L^p([0,1])[/imath] strictly convex? Solution: For strict convexity we have two equivalent definition: If [imath]x\neq 0\neq y[/imath] and [imath]\Vert x+y\Vert=\Vert x\Vert+\Vert y\Vert[/imath] then [imath]x=\lambda y[/imath] for some [imath]\lambda[/imath]. If [imath]x\neq y[/imath] with [imath]\Vert y\Vert=\Vert x\Vert=1[/imath] then [imath]\Vert\frac{x+y}{2}\Vert<1[/imath] So I found that [imath]L^p([0,1])[/imath] is strictly convex for [imath]p=2[/imath]: \begin{equation} \int_0^1(\frac{f(t)+g(t)}{2})^2dt=0.5\int_0^1(f(t)^2+g(t)^2+2f(t)g(t))dt\leq \frac{\Vert f\Vert^2_2+\Vert g\Vert^2_2}{2}<1 \end{equation} Question: Are there other [imath]p[/imath] for which [imath]L^p([0,1])[/imath] is strictly convex?
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80139
|
Why is the [imath]L_p[/imath] norm strictly convex for [imath]1[/imath]
Let [imath]x,y \in L_p[/imath] such that [imath]\|x\|_p=\|y\|_p=1[/imath] , [imath]1< p<\infty[/imath] and [imath]x\neq y.[/imath] Why is [imath]\|x+y\|_p<2[/imath] ? I'm not sure how to start the proof.. I don't know how to handle integral of [imath](x+y)^p[/imath] and it seems that using the binomial theorem won't be a great success.
|
823766
|
[imath]X,Y\sim U(0,1)[/imath] are independent, [imath]W=\max (X,Y)[/imath]
[imath]X,Y\sim U(0,1)[/imath] are independent, [imath]W=\max (X,Y)[/imath]. How do I find the PDF of [imath]W[/imath]? How do I find the expectation of [imath]W[/imath] at two ways: 1. with the PDF of [imath]W[/imath] and without the PDF of [imath]W[/imath]. I'd like to get any idea how to solve it... Thank you!!!
|
832570
|
[imath]X,Y\sim U(0,1)[/imath] are independent, [imath]W=\max (X,Y)[/imath].
[imath]X,Y\sim U(0,1)[/imath] are independent, [imath]W=\max (X,Y)[/imath]. How do I find the PDF of [imath]W[/imath]? How do I find the expectation of [imath]W[/imath] at two ways: 1. with the PDF of [imath]W[/imath] and without the PDF of [imath]W[/imath]. I ask this Q before few days, but now I want show you me solution and ask if it's right (and ask to tell me where I wrong): We assume that [imath]0<t<1[/imath] becuase otherwise the function [imath](f)[/imath] returns 0. 1. Lets say: [imath]W=\max(X,Y)\Rightarrow W(t)=P(t>X)\cdot P(t>Y)=(1-P(t<X))(1-P(t<Y))[/imath] because they are independent. We know that [imath]P(t<X)=F_X(t)=t[/imath] (Because [imath]X\sim U(0,1)[/imath]), and this is why: [imath]F_W(t)=(1-t)^2[/imath] (if [imath]0<t<1[/imath] otherwise is [imath]0[/imath]). So [imath]f_W(t)=F'_W(t)=2t-2[/imath]. 2: a. [imath]E(W(t))=\int _{0}^1 t\cdot f_W dt=\int_0^12t^2-2tdt=\left[\frac{2t^3}{3}-\frac{2t^2}2\right]^1 _0[/imath] I dot here something negative and I don't know why... Please tell me where my mistake here.... b. Here I'll show you the idea, and I'd like to know if I'm right... [imath]\iint_{D_1}\max(X,Y)dxdy+\iint_{D_2}\max(X,Y)dxdy \\= \iint_{D_1}xdxdy+\iint_{D_2}ydxdy[/imath] I'm right? If I'm wrong please tell me where are my mistakes (at all the question). Thank you!
|
1869233
|
What is this space called in general topology?
I am self-studying the general topology these day and find that the third axiom of the topological space [imath](X,\tau)[/imath] defined by open set is: For any finite collection of [imath]U_i \in \tau[/imath], the intersection of [imath]U_i[/imath] is still a member of [imath]\tau[/imath]. Now I am wondering is there any space[imath](X,\tau)[/imath] such that satisfying the following conditions: [imath]1)\emptyset , X \in \tau[/imath] [imath]2)[/imath]For any finite or infinite collection of [imath]U_i \in \tau[/imath], the union of [imath]U_i[/imath] is still a member of [imath]\tau[/imath]. [imath]3)[/imath]For any finite or infinite collection of [imath]U_i \in \tau[/imath], the intersection of [imath]U_i[/imath] is still a member of [imath]\tau[/imath]. I come up with space [imath](X,2^X)[/imath] satisfying this condition but I cannot find non-trivial one satisfying such condition. Does such a space exists? What name is it?
|
1867861
|
In which topologies do open sets maintain open under countable or arbitrary intersection?
We know that in the usual topology, countable or arbitrary intersection of open sets can zoom into a singleton, hence is not in the topology. I am curious if there is well known classes of topologies, whose open sets maintains their openness after countable or arbitrary intersection. I know two examples: 1) In the discrete topology, [imath]\mathcal{T} = \mathcal{P}(X)[/imath]. The power set is a [imath]\sigma[/imath]-algebra, hence "closed" under countable intersections, so a countable intersection of open sets in the discrete topology is open. 2) In the particular point topology, given [imath]x_o \in \mathbb{R}[/imath], [imath]\mathcal{T}_{x_o} = \{U \subseteq \mathbb{R}| x_o \in U\}\cup\{\varnothing\}[/imath], an arbitrary intersection of open sets will always contain [imath]x_o[/imath], in particular [imath]x_o[/imath] is open. Hence a countable intersection of open sets in the particular point topology topology is open. Are there a lot more well known classes of topology whose condition on intersections of open set can be extended?
|
1868366
|
simplify factorials: [imath]\frac{(k-1)!}{(k+2)!}[/imath]
Question: simplify [imath]\frac{(k-1)!}{(k+2)!}[/imath] What I did was: [imath]\frac{(k - 1)!k!}{(k + 2)! \cdot (k + 1)!}[/imath] This I did following the rule [imath]n! = n \times (n - 1)![/imath]. can this be simplified further? Thanks.
|
1868348
|
Simplifying factorials: [imath]\frac{(n-1)!}{(n-2)!}[/imath]
Question: simplify [imath]\frac{(n-1)!}{(n-2)!}[/imath] What I did was: [imath]\frac{(n - 1)!}{(n - 2)! \times (n - 3)!}[/imath] This I did following the rule [imath]n! = n \times (n - 1)![/imath]. But my answer just doesn't look correct and I don't have a solution guide that tells me the correct answer. This is why I don't know whether my answer is correct or not.
|
1868951
|
Convolution of a function with itself n times convergence to bell curve
If we have a piecewise function defined as [imath]f(x) = \begin{cases} 1, & \text{0 [/imath]\le[imath] [/imath]x[imath] [/imath]\le[imath] 1} \\ 0, & \text{otherwise} \end{cases}[/imath] Explain how the convolution of [imath]f[/imath] with itself for [imath]n[/imath] times i.e. [imath](f*f*f.......*f)[/imath] [imath]n[/imath] times might begin to look like bell curve of the central limit theorem. I have no idea how to proceed. Any help would be appreicated.
|
1173103
|
Convolution of a function with itself
Function [imath]\phi (x)[/imath] is defined as: [imath]\phi(x) = \begin{cases} 1 & \text{ if } 0 \leq x \leq 1\\0 & \text{otherwise} \end{cases} [/imath] How do I find the convolution of [imath]\phi(x)[/imath] with itself? I tried to take the Fourier transform of [imath]\phi(x)[/imath] and square it, then take the inverse Fourier transform. However, in the latter step I couldn't figure out what the limits on the integral are (I think it would be integral in all space, but I'm not sure). More importantly, I don't know how to do that integral for the inverse Fourier step. Thank you!
|
1869372
|
Type of irrational number in [imath]R[/imath]
How can we prove that all irrational numbers in real line can not be represented as union of countable closed sets. The result intuitively makes sense and we all know the irrational numbers are uncountable and dense on [imath]R[/imath]. And I don't know how to use these facts to prove the result in strict math language.
|
1321874
|
The set of irrational numbers is not a [imath]F_{\sigma}[/imath] set.
I want to proove that the set of irrational numbers is not a [imath]F_{\sigma }[/imath] set and also the set of rational numbers is not a [imath]G_{\delta}[/imath] set using Baire theorem. I started with saying that [imath]\mathbb{R}[/imath]\ [imath]\mathbb{Q}[/imath] is a [imath]G_{\delta}[/imath] set because [imath]\mathbb{R}[/imath]\ [imath]\mathbb{Q}[/imath] = [imath]\bigcap_{ q\in\mathbb{Q}}^{}[/imath][imath]\mathbb{R}[/imath]\ {q}. So, if [imath]\mathbb{R}[/imath]\ [imath]\mathbb{Q}[/imath] were also [imath]F_{\sigma }[/imath] then [imath]\mathbb{Q}[/imath], as a complementary, would be a [imath]G_{\delta }[/imath] set. Both [imath]\mathbb{R}[/imath]\ [imath]\mathbb{Q}[/imath] and [imath]\mathbb{Q}[/imath] are dense in [imath]\mathbb{R}[/imath] and ([imath]\mathbb{R}[/imath]\ [imath]\mathbb{Q}[/imath]) [imath]\cup [/imath] [imath]\mathbb{Q}[/imath] = [imath]\mathbb{R}[/imath], so the whole space can be written as a sum of two dense and [imath]G_{\delta }[/imath] sets. I've read that this is a contradiction with Baire theorem, but I can't see this.
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1869134
|
A prime ideal [imath]\mathfrak{p} \subset \mathcal{O}_K[/imath] lies above/over [imath]p[/imath] if [imath]\mathfrak{p}\cap \mathbb{Z} = p\mathbb{Z}[/imath]
In the concrete case that [imath]\mathcal{O}_K = \mathbb{Z}[i][/imath], and [imath]\mathfrak{p} = (1+i)[/imath], how to make sense of [imath]\mathfrak{p}\cap \mathbb{Z}[/imath]? I want to know if (1+i) lies above/over 2.
|
1600449
|
How do we prove the "lying over" property for integral extensions?
Let [imath]R \subset S[/imath] be an integral extension of commutative rings. Then if [imath]P \subset R[/imath] is prime, there exists a prime ideal [imath]Q \subset S[/imath] such that [imath]Q \cap R = P[/imath]. My D&F book says look at Corollary 50, but I cannot find it in the book! Is the proof easy or does it require much build-up lemmas? Hints, or location of Corollary 50, please.
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914309
|
Solve [imath]2tx'(t)-x(t)=\ln x'(t)[/imath]
Solve [imath]2tx'(t)-x(t)=\ln \left[x'(t)\right][/imath] That would be an easy Clairaut's equation if [imath]tx'(t)[/imath] wasn't multiplied by [imath]2[/imath]. But unfortunately it is, and I have no idea what to do here.
|
851289
|
Second-order non-linear ODE
[imath]2tx'-x=lnx'[/imath] I differentiated both sides with respect to x: [imath]x'+2tx''=\frac {x''}{x'}[/imath] Substituting [imath]p=x'[/imath], [imath]p+2tp'=\frac{p'}{p}[/imath] But I have no clue what can I do from here on. EDIT: [imath]t[/imath] is the non-dependent variable.
|
1869590
|
How to prove that [imath]A-(B\cap C)=(A-B)\cup (A-C)[/imath]
Prove that [imath]A-(B\cap C)=(A-B)\cup (A-C)[/imath]. I have tried to prove it but I can't get precious proof. I am grateful if anyone give correct proof.
|
1261951
|
Does set difference distribute over set intersection?
I am asked to prove that, if [imath]A, B[/imath] and [imath]C[/imath] are sets, then [imath]A-(B\cap C)=(A-B)\cap(A-C).[/imath] However, I think that either I have made an error in my working, or the wording of the problem contains a typographical error. My working so far is as follows: [imath]\begin{align}x\in(A-(B\cap C)) &\iff (x\in A)\wedge(x\not\in(B\cap C))\\ &\iff(x\in A)\wedge((x\not\in B)\vee(x\not\in C))\\ &\iff((x\in A)\wedge(x\not\in B))\vee((x\in A)\wedge(x\not\in C))\\ &\iff(x\in(A-B))\vee(x\in(A-C))\\ &\iff x\in((A-B)\cup(A-C)) \end{align}[/imath] First line to second line: by De Morgan's Law. Second line to third line: by distributivity of set intersection over set union. Third line to fourth line: by definitions of intersection and set difference. If my working is correct, then I have shown that [imath]A-(B\cap C)=(A-B)\cup(A-C).[/imath] Am I correct?
|
1869634
|
[imath]\mathbb{C}/\mathbb{Z}[/imath] is isomorphic to multiplicative group [imath]\mathbb{C}\setminus\{0\}[/imath]
I have to show that [imath]\mathbb{C}/\mathbb{Z}[/imath] is isomorphic to the multiplicative group [imath]\mathbb{C} \setminus \{0\}[/imath]. Proof. Let [imath]f:\mathbb{C} \setminus \{0\} \rightarrow \mathbb{C}/\mathbb{Z}[/imath] be the map [imath] f(\alpha) = \alpha \mathbb{Z}.[/imath] This map has inverse [imath]f^{-1}(\alpha \mathbb{Z}) = \alpha[/imath] so it is bijective. Furthermore, [imath]f(\alpha \beta) = (\alpha \beta)\mathbb{Z} = \alpha \mathbb{Z} \beta \mathbb{Z} =f(\alpha) f(\beta)[/imath] and [imath]f(1) = \mathbb{Z}[/imath], so the map [imath]f[/imath] is also a homomorphism. Question. Is this correct? It feels a bit like I am cheating.
|
1294898
|
Complex numbers modulo integers
Is there a "nice" way to think about the quotient group [imath]\mathbb{C} / \mathbb{Z}[/imath]? Bonus points for [imath]\mathbb{C}/2\mathbb{Z}[/imath] (or even [imath]\mathbb{C}/n\mathbb{Z}[/imath] for [imath]n[/imath] an integer) and how it relates to [imath]\mathbb{C} / \mathbb{Z}[/imath]. By "nice" I mean something like: [imath]\mathbb{R}/\mathbb{Z}[/imath] is isomorphic to the circle group via the exponential map [imath]\theta \mapsto e^{i\theta}[/imath], and [imath]\mathbb{C}/\Lambda[/imath] is a complex torus for [imath]\Lambda[/imath] an integer lattice (an integer lattice is a discrete subgroup of the form [imath]\alpha\mathbb{Z} + \beta\mathbb{Z}[/imath] where [imath]\alpha,\beta[/imath] are linearly independent over [imath]\mathbb{R}[/imath].) Intuitively, it seems like it should be something like a circle or elliptic curve.
|
1870197
|
Prove a composition of two functions is meaurable
Let [imath]f,g:[0,1]\rightarrow [0,1][/imath] be measurable functions.Is [imath]g\circ f[/imath] measurable or not? The composition is definitely measurable from the axiom definition of measurable function. But if we want to prove it from the classic definition of measurable function,we have to prove that preimage of an open set of this composition is measurable. In order to do that,we have to prove that the preimage of a mesurable set by a measurable function is a measurable set.Now I don't know what to do with the mesurable sets in [imath][0,1][/imath] except for the Borel sets.
|
283443
|
Is composition of measurable functions measurable?
We know that if [imath] f: E \to \mathbb{R} [/imath] is a Lebesgue-measurable function and [imath] g: \mathbb{R} \to \mathbb{R} [/imath] is a continuous function, then [imath] g \circ f [/imath] is Lebesgue-measurable. Can one replace the continuous function [imath] g [/imath] by a Lebesgue-measurable function without affecting the validity of the previous result?
|
867533
|
For what powers [imath]k[/imath] is the polynomial [imath]n^k-1[/imath] divisible by [imath](n-1)^2[/imath]?
How do you prove this? [imath]\left(n-1\right)^2\mid\left(n^k-1\right)\Longleftrightarrow\left(n-1\right)\mid k[/imath]
|
2272145
|
Prove that [imath](n - 1)^2 \mid n^k -1[/imath] if and only if [imath](n - 1) \mid k[/imath]
I need help! I need to prove that for any [imath]2 \le n,k[/imath] positive integers [imath](n - 1)^2 \mid n^k -1[/imath] if and only if [imath](n - 1) \mid k[/imath] Thanks!
|
1870918
|
Equivalence Relation Proof Question
Let [imath]R[/imath] be the relation on [imath]N\times (N\setminus\{0\})[/imath] defined by [imath]((a, b),(c, d)) \in R[/imath] if [imath]ad = bc[/imath]. Prove that [imath]R[/imath] is an equivalence relation. I'm pretty confused with this problem, mainly because I don't understand the significance of [imath](N\setminus\{0\})[/imath]. Any ideas on how to solve this?
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1765538
|
Prove the relation [imath]R[/imath] in [imath]N \times N[/imath] defined by [imath](a,b) \simeq (c,d)[/imath] iff [imath]ad=bc[/imath] is an equivalence relation.
If [imath]N[/imath] is the set of all natural numbers, [imath]R[/imath] is a relation on [imath]N \times N[/imath], defined by [imath](a,b) \simeq (c,d)[/imath] iff [imath]ad=bc[/imath], how can I prove that [imath]R[/imath] is an equivalence relation ?
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1127308
|
Limit of sequence [imath]n!\left(\frac{e}{n}\right)^n[/imath]
Find the limit of [imath] \lim_{n\to +\infty} n!\left(\frac{e}{n}\right)^n. [/imath] I have shown that [imath]u_{n+1}>u_n[/imath], but I am not sure where to go from here.
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494776
|
Elementary proof for [imath]\lim\limits_{n \to\infty}\frac{n!e^n}{n^n} = +\infty[/imath]
As seen in this question, the series [imath]\sum\limits_{n=1}^{\infty} \dfrac{n!e^n}{n^n}[/imath] diverges. (One way to see this is by noting that the terms of the sum are greater than [imath]1[/imath] and, therefore, don't converge to zero.) However, more is true: not only are the terms greater than [imath]1[/imath], they blow up to [imath]+\infty[/imath]! To see this, one can apply Stirling's approximation and get [imath]\dfrac{n!e^n}{n^n} \sim \sqrt{2\pi n}[/imath]. My question is: is there an elementary proof for [imath]\lim_{n \to\infty}\dfrac{n!e^n}{n^n} = +\infty[/imath]? The reason I'm asking this is that I'm taking a Calculus course and I was assigned the exercise of proving the divergence of the series above and I would feel more satisfied if I could prove that its terms go to [imath]+\infty[/imath] (as opposed to merely proving the terms are greater than [imath]1[/imath].) Obviously, things like Stirling's approximation are outside the scope of the course, so I can't really use them. (To be clear, you can use anything one learns in [imath]4[/imath] semesters of standard Calculus courses. I hope this restriction is not too obscure.)
|
1871228
|
want to check answer given in my book is correct or not
The problem is to find coefficient of [imath]x^n[/imath] using binomial theorem for rational index in the expansion of [imath]\frac{1}{1-x+x^2-x^3}.[/imath] In my book the answer is given as [imath]\frac14+\frac{n+1}{2}+\frac{(-1)^n}{4}.[/imath] I think the answer in book is wrong. I will be grateful if anyone check and tell this answer is correct or not with explanation. thanks
|
1868757
|
Find the coefficient of [imath]x^n[/imath] simply by using binomial theorem
Find the coefficient of [imath]x^n [/imath] in the expansion of [imath]\frac{1}{1-x+x^2-x^3}[/imath].
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1871220
|
The notation [imath] \otimes [/imath] (Tensor product)
Related to the question Riemannian manifolds isometry, could anyone be able to explain to me what the notation [imath] \otimes [/imath]? Some examples were given in the question : [imath]dx\otimes dx+dy\otimes dy = \frac12(dz\otimes d\bar z + d\bar z\otimes dz) = |dz|^2[/imath] or [imath]dw \otimes d\bar{w}[/imath]. Clarification : I think my question is different of What does this symbol [imath]\otimes[/imath] mean?, because I would like some details on a specific example.
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1271853
|
What does this symbol [imath]\otimes[/imath] mean?
I came across the symbol [imath]\otimes[/imath] as below and I would like to know what this symbol [imath]\otimes[/imath] means: [imath]\text{.... the projection operator P is given by: }[/imath] [imath]P = I_nd - \nabla G^T(\nabla G \nabla G^T)^{-1} \nabla G= I_{nd} - I_d \otimes uu^T,[/imath] where [imath]I_x[/imath] denotes the identity matrix of size [imath]x\times x[/imath] and [imath]\mathbf{u}[/imath] is the unit vector , [imath]\mathbf{u} = (1,1,1,\dots,1)/\sqrt{n})[/imath] in [imath]\mathbb{R}^n[/imath]
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1870782
|
Prove that [imath]{2^n-1\choose k}[/imath] and [imath]{2^n-k\choose k}[/imath] ar always odd.
How can I prove that [imath]{2^n-1\choose k}[/imath] and [imath]{2^n-k\choose k}[/imath] always returns odd numbers? It is possible to prove this by congruence? by the way : [imath]0 \leq k \leq (2^n-1)[/imath]
|
148017
|
Odd Binomial Coefficients?
By Newton's Formula: [imath](a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k [/imath] Proof that every [imath]\dbinom{n}{k}[/imath] is odd if and only if [imath]n=2^r-1[/imath]. I have already shown that if [imath]n[/imath] is of the form [imath]2^r-1[/imath], having used the property [imath]\binom{n-1}{k} = \binom{n}{k}-\binom{n}{k-1}+ \binom{n}{k-2} - \cdots \pm \binom{n}{0}.[/imath] But I have not been able to demonstrate "[imath]\Rightarrow[/imath]". Please, help me! Thanks.
|
951522
|
Trig sum: $\tan ^21^\circ+\tan ^22^\circ+\cdots+\tan^2 89^\circ = \text{?}$
As the title suggests, I'm trying to find the sum [imath]$$\tan^21^\circ+\tan^2 2^\circ+\cdots+\tan^2 89^\circ$$[/imath] I'm looking for a solution that doesn't involve complex numbers, or any other advanced branch in maths. The solution can involve techniques such as induction, telecoping, etc, but preferably only ideas from precalculus, e.g. trig identities, polynomials, etc. EDIT: I know that the sum is a rational number.
|
2149604
|
how to find this trigonometric sum?
[imath]\sum_{r=1}^{45} \tan^2(2r-1)°[/imath] I tried simplifying it to Cosecant terms and use complex numbers and binomial theorem... I know the answer but not the solution so I can verify your answer once you post it... I am using the Euler form for Sin but then I am stuck..
|
1871953
|
Simplify the expression [imath]\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +\binom{n+k}{k}[/imath]
Simplify the expression [imath]\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +\binom{n+k}{k}[/imath] My attempt: Using the formula [imath]\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}[/imath] [imath]\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots \binom{n+k-1}{k-1}+\binom{n+k}{k}[/imath] =[imath]\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +\binom{n+k-1}{k-1}+(\binom{n+k-1}{k}+\binom{n+k-1}{k-1})[/imath] =[imath]\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +2\binom{n+k-1}{k-1}+\binom{n+k-1}{k}[/imath] I can again use the same formula for the term [imath]2\binom{n+k-1}{k-1}[/imath], and in the next to step to the term [imath]3\binom{n+k-2}{k-2}[/imath]. But I don't think this way the expression will get simplified. Any help is appreciated.
|
2034004
|
Prove that [imath]C(n,0) + C(n+1,1) +\dots+ C(n+r,r) = C(n+r+1,r).[/imath]
I am not even sure where to start with this question. Any help would be much appreciated! Thanks! We are not allowed to use Mathematical Induction!
|
1871771
|
Integral that makes square root of [imath]\frac{\pi}{2}[/imath]
My question is regarding a integral that´s giving me a huge headache. I want to show [imath]\int_{0}^{\infty}y^2e^{-\frac{y^2}{2}}dy=\sqrt{\frac{\pi}{2}}[/imath] I'm studying for an exam. I'm suppose to find the integral of [imath](x^2/c^2)\, e^{-x^2/(2c^2)}[/imath] between [imath]0[/imath] and [imath]\infty[/imath]. So first i substituted [imath]x/c = y[/imath] which makes the integral simpler. Then I'm stuck, I'm guessing that the equation that appears when substituting like this, is well know(a standard integral used a lot) which is equal to integral of pi/2.
|
738940
|
Gaussian integral evaluation
Asked a question to evaluate the Gaussian Integral, [imath]\dfrac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty x^2 \exp(-x^2/2) dx [/imath] using the the following approximation, [imath]J=\Bbb E[X^2] \sim J_N = 1/N \sum_1^NX_n^2 [/imath] where [imath]x_n \sim N(0, 1)[/imath] Examin [imath]J_N[/imath] for different values of [imath]N[/imath] and plot an error graph (difference between exact and numerical values) I can simulate and generate the approximations but I am stuck at getting the exact value to compare to. As far as I am aware there are no analytical solutions for gaussian integrals. The other way I can generate comparable numbers is to go use numerical integration techniques but those are approximate values too. Any help will be greatly appreciated!
|
1871992
|
Solve [imath]28^x = 19^y+87^z[/imath]
Solve the equation [imath]28^x = 19^y+87^z[/imath], where [imath]x,y,z[/imath] are integers. This is related to Beal's conjecture (and it turns out there are no integer solutions to the equation), but I am wondering how to first prove there are no integer solutions when [imath]x,y,z[/imath] are positive integers. For positive integers we can use the following argument: Taking the equation modulo [imath]19[/imath] gives [imath]28^x-(19^y+87^z) \equiv 9^x-11^z \equiv 0 \pmod{19}[/imath]. Thus, since the complete set of residues of [imath]9^x[/imath] modulo [imath]19[/imath] is [imath]\{9,5,7,6,16,11,4,17,1\}[/imath] and the complete set of residues of [imath]11^z[/imath] modulo [imath]19[/imath] is [imath]\{11,7,1\}[/imath]. Therefore, [imath]x = 3+9k_1[/imath] and [imath]y = 2+3k_2[/imath], or [imath]x = 6+9k_1[/imath] and [imath]y = 1+3k_2[/imath], or [imath]x = 9+9k_1[/imath] and [imath]y = 3+3k_2[/imath] where [imath]k_1, k_2 \in \mathbb{N}_0[/imath]. How can we complete this argument and also extend it to all integers?
|
1871591
|
Solve the equation [imath]28^x = 19^y+87^z[/imath] in integers
Solve the equation [imath]28^x = 19^y+87^z[/imath], where [imath]x,y,z[/imath] are integers. I am confused how to find solutions for all integers and not just positive ones. How should we do that?
|
1871826
|
Two questions on the Gaussian integers
I have two questions on the Gaussian integers. Is any element in [imath]\mathbb{Z}[i][/imath] the root of a monic polynomial with coefficients in [imath]\mathbb{Z}[/imath]? Conversely, does any element in [imath]\mathbb{Q}(i)[/imath] that is the root of a monic polynomial with coefficients in [imath]\mathbb{Z}[/imath] lie in [imath]\mathbb{Z}[i][/imath]?
|
655222
|
The ring of integers of [imath]\mathbf{Q}[i][/imath]
Is there a relatively "simple" (in the sense that it does not require knowledge of algebraic number theory) proof that the ring of integers of the algebraic number field [imath]\mathbf{Q}[i][/imath] is [imath]\mathbf{Z}[i][/imath]? One can assume a one year course in algebra, covering the usual topics on ring and field theory, e.g. Gauss's lemma etc
|
1861583
|
Let [imath]a[/imath] and [imath]b[/imath] belong to a group. If [imath]|a|[/imath] and [imath]|b|[/imath] are relatively prime, show that [imath]\langle a \rangle \cap \langle b \rangle = \{e \}[/imath]
Let [imath]a[/imath] and [imath]b[/imath] belong to a group. If [imath]|a|[/imath] and [imath]|b|[/imath] are relatively prime, show that [imath]\langle a \rangle \cap \langle b \rangle = \{e \}[/imath]. The most I can figure to do is to let [imath]a^k = b^j[/imath] for some [imath]a^k[/imath] and [imath]b^j[/imath] in [imath]\langle a \rangle[/imath] and [imath]\langle b \rangle[/imath], respectively. I'd assume that I'd need to show that the only possible element either could be would be [imath]e[/imath] (the identity element) but I'm not seeing a way. Would it make sense to say that [imath]aa^k \neq bb^j[/imath] because then otherwise [imath]\langle a \rangle = \langle b \rangle[/imath], therefore the only option is to have [imath]a^k = e[/imath] and [imath]b^j = e?[/imath]
|
1227800
|
Let [imath]H[/imath] have order [imath]m[/imath] and [imath]K[/imath] have order [imath]n[/imath], where [imath]m[/imath] and [imath]n[/imath] are relatively prime. Then [imath]H \cap K=\{e\}[/imath]
Let [imath]H[/imath] and [imath]K[/imath] be subgroups of [imath]G[/imath]. Let [imath]H[/imath] have order [imath]m[/imath] and [imath]K[/imath] have order [imath]n[/imath], where [imath]m[/imath] and [imath]n[/imath] are relatively prime. Then [imath]H \cap K=\{e\}[/imath] My proof: Let [imath]H[/imath] and [imath]K[/imath] be subgroups where the [imath]ord(H)=m[/imath] and [imath]ord(K)=n[/imath] and [imath]m,n[/imath] are relatively prime. We know that [imath]H \cap K[/imath] is a subgroup of [imath]H[/imath] and [imath]K[/imath] since it contains the elements of both in [imath]H[/imath] and [imath]K[/imath]. We shall let the ord([imath]H \cap K[/imath])=d. Then, by Lagrange's theorem, the order of [imath]H[/imath] is a multiple of the order of [imath]H \cap K[/imath]. In other words, [imath]m[/imath] is a multiple of [imath]d[/imath] or [imath]d|m[/imath]. Similarly, by Lagrange's theorem, the order of [imath]K[/imath] is a multiple of the order of [imath]H \cap K[/imath]. In other words, [imath]n[/imath] is a multiple of [imath]d[/imath] or [imath]d|n[/imath]. Since [imath]d[/imath] divides both [imath]m[/imath] and [imath]n[/imath] and we know [imath]m[/imath] and [imath]n[/imath] are relatively prime then, the order of [imath]H \cap K[/imath] must be [imath]1[/imath]. Since [imath]H \cap K[/imath] is a subgroup, then by properties of subgroup, it contains an inverse which means it must also contain an identity and since [imath]H \cap K[/imath] contains one element, it must contain the identity. As a result, [imath]H \cap K=\{e\}[/imath] Hopefully, someone can confirm or correct any mistakes I made. Thanks!
|
1872380
|
Derivative of a analytic function at its fixed point
Let [imath]D[/imath] be a bounded domain, and let [imath]f(z)[/imath] be an analytic function from [imath]D[/imath] to [imath]D[/imath].Show that if [imath]z_{0}[/imath] is fixed point for [imath]f(z)[/imath],then [imath]|f'(z_{0})|\leq 1[/imath] All the conditions above make me think about Schwartz Lemma to solve this problem.But I don't know how to construct a proper function satisfying all the conditions in Schwartz Lemma.
|
955691
|
Estimate on the derivative at fixed point
Let [imath]D[/imath] be a bounded domain and let [imath]f[/imath] be analytic function from [imath]D[/imath] into [imath]D[/imath]. Show that if [imath]z_{0}[/imath] [imath]\in D[/imath] is a fixed point for [imath]f[/imath] , then [imath]|f' (z_{0} )| \leq 1[/imath]. WHAT I WAS THINKING: To have a conformal mapping from [imath]D[/imath] to the unit disc & after composing it with [imath]f[/imath] ; I was thinking to use Pick's Lemma. So, can anyone tell me any conformal map from any bounded domain to the unit disc?? OR Is there any other way out to solve this problem??
|
1872842
|
Show that no integer of the form [imath]a^3 +1[/imath] is a prime for [imath]a>1[/imath]
Can someone please solve this and explain the steps taken to reach this solution ?
|
28572
|
factorization of a^n+1?
I'm trying to prove that [imath]a^n+1[/imath] can only be prime if [imath]n[/imath] is a power of [imath]2[/imath]. Is there a general factorization of [imath]a^n+1[/imath]?
|
1234859
|
First uncountable ordinal
I am a beginner of ordinals and I don't know any powerful techniques in it. I come across with a problem about the first uncountable ordinal like this. Let [imath]X[/imath] be a set of uncountable cardinality. Using the Principle of Well Order we have a well ordering [imath]\le[/imath] on X(and [imath]<[/imath] means [imath]\le[/imath] but not euqual). By adding an element, which we denote by [imath]∞[/imath], and introducing the convention that [imath]x < ∞[/imath] for all [imath]x ∈ X[/imath], we will assume that X has a maximum with respect to [imath]\le[/imath]. We define [imath]ω_1 = \text{min} \{ x ∈ X : \{y ∈ X : y < x \} \ \text{is uncountable}\} [/imath] Clearly such a [imath]ω_1[/imath] exists. we define [imath]0 = \text{min} \ X[/imath] and intervels [imath][0,x], [x,y), (a,b][/imath] etc. in the usual sense. Here are my questions: (1)For any countable [imath]A ⊂ [0, ω_1)[/imath] there is an [imath]x < ω_1[/imath] so that [imath]A ⊂ [0, x][/imath].(I don't know how to make use of the countability and uncountablity here) (2)Equipped with the topology generated by open intervals, [imath][0, ω_1][/imath] is compact. (3)A famous application of the first uncountable ordinal is to find an example of a Borel measure (with respect to the topology in (2)) that is not Radon. So how to construct a finite Borel measure [imath]µ[/imath] on [imath][0, ω_1][/imath] which is not a Radon measure? Any solutions or elementary references will be appreciated!
|
3009947
|
Showing that [imath][0, \omega_1][/imath] is compact.
I want to show that [imath][0, \omega_1][/imath] is compact, where [imath]\omega_1[/imath] is the least uncountable ordinal, and I have just been introduced to the concepts of ordinals. The tips I have seen to showing this is that [imath][0, \omega_1][/imath]: contains a maximal element there are no infinite strictly decreasing sequences of ordinals. Take an open cover of the desired set. My thoughts are that some open set must contain [imath]\omega_1[/imath], which in turn contains all ordinals smaller than it by the definition of an ordinal, and so this open set is a cover of the entire space. Any further insights appreciated.
|
1873246
|
Infinite product [imath]\prod_{n=1}^\infty\frac{2}{1+\pi^{2^{-n}}}[/imath] calculation problem
To calculate the infinite product [imath]\prod_{n=1}^{\infty}\frac{2}{1+\pi^{\frac{1}{2^n}}},[/imath] I found that [imath]\sum_{n=1}^{\infty}\ln\left(\frac{2}{1+\pi^{\frac{1}{2^n}}}\right)[/imath] converges, so as [imath]\prod_{n=1}^\infty\frac{2}{1+\pi^{\frac{1}{2^n}}}.[/imath] The task though is asking to calculate the product and not just verify the convergence. Is there a mathematical way to calculate this product, or calculate this limit : [imath]\lim_{n\rightarrow\infty}\frac{2^n}{\left(1+\pi^{\frac{1}{2}}\right)\left(1+\pi^{\frac{1}{2^2}}\right)\cdots\left(1+\pi^{\frac{1}{2^n}}\right)}[/imath] Thank you for your time
|
1873233
|
Evaluate an infinite product in a closed form $\prod\limits_{n=1}^{\infty} \frac{1}{1+\pi^{1/{2^n}}}$
I run into this infinite product today. I don't have that much experience evaluating products therefore I don't have any idea of how to tackle this. Here is the question. Evaluate (if possible) in a closed form the product: [imath]\Pi = \prod_{n=1}^{\infty} \frac{1}{1+\pi^{{\large \frac{1}{2^n}}}}[/imath] The numerical value seems to be [imath]\Pi= 0.534523[/imath] which is very close to [imath]\frac{\pi}{6}[/imath] taking into account that [imath]\frac{\pi}{6} \approx 0.523598[/imath]. In the mean time W|A evaluates it to [imath]0[/imath]. I'm lost. Can the community help? Edit: Based on the answer we have two products. The product I asked tends to zero and the bonus product provided by @you're in my eye (thanks for that) is [imath]\frac{\ln \pi}{\pi-1}[/imath]. Thanks for the quick response.
|
1874540
|
calculate area of a region defined by holomorphic function
Let [imath]f[/imath] be a holomorphic function on unit disk [imath]D[/imath]. (a). express the Jacobian of the map [imath]f[/imath] in terms of [imath]f[/imath] or [imath]f'[/imath] (b).Give a formula for the area of [imath]f(D)[/imath] in terms of the Taylor Coefficients of [imath]f[/imath]. (c).If [imath]f(x)=z+\dfrac{z^{2}}{2}[/imath],find the area of [imath]f(D)[/imath]. My solotion: (a). According to Cauchy-Riemann equation,I compute the Jacobian of the map [imath]f[/imath] as [imath]|f'(x)|^{2}[/imath]. （b).From the elementary calculus,the area of [imath]f(D)[/imath] can be computed according to coordinate transform.So [imath]\iint_{f\left(D\right)}dw=\iint_{D}\left|f'\left(D\right)\right|^{2}dz[/imath].But I don't know how to express it in terms of Taylor coefficients. (c).Followed from the result in (b), Area[imath]f(D)[/imath]=[imath]\iint_{D}\left|1+re^{i\theta}\right|^{2}drd\theta=\dfrac{4}{3}[/imath] Are my results valid and what about answer for the second question?
|
1764827
|
Let [imath]f[/imath] be an analytic isomorphism on the unit disc [imath]D[/imath], find the area of [imath]f(D)[/imath]
Let [imath]f[/imath] have power series [imath]f(z) = \sum_{n=1}^\infty a_n z^n[/imath] in [imath]D[/imath], then prove that [imath]\mathrm{area}\, f(D) = \sum_{n=1}^\infty n \,|a_n|^2[/imath]. Note: We define [imath]\mathrm{area}\, S = \iint_S \mathrm{d}x\,\mathrm{d}y[/imath]. I presume the way to do this is to take the integral [imath]\iint_{f(D)} \mathrm{d}x\,\mathrm{d}y = \iint_D \mathbf{J}_f (x+iy) \,\mathrm{d}x\,\mathrm{d}y = \int_0^1 \int_0^{2\pi} r \mathbf{J}_f (r e^{i \theta}) \mathrm{d}\theta\,\mathrm{d}r,[/imath] and letting [imath]\gamma_r : [0,2\pi]\to\Bbb{C}, \,\gamma_r (\theta) = r e^{i\theta},\, \gamma_r '(\theta) = i\gamma_r (\theta),[/imath] then we get [imath]\mathrm{area}\,f(D) = \int_0^1 \int_{\gamma_r}\frac{\lvert f'(z)\rvert^2 \bar{z}}{ir}\mathrm{d}z\,\mathrm{d}r.[/imath] Unfortunately, this approach seems to be a dead end. I think I'm meant to use Cauchy's Formula somewhere, but I can't see where that might be useful in this kind of question.
|
1875023
|
showing [imath]k[x,y]\ncong k[u,v,w]/(uw-v^2)[/imath]
Let [imath]k[/imath] be a field. I want to show that [imath]k[x,y]\ncong k[u,v,w]/(uw-v^2)[/imath] as [imath]k[/imath]-algebras, but can't find a way to do it. The dimension of the [imath]k[/imath]-vector space generated by the degree 1 monomials are different on both sides, but then it's possible that an isomorphism doesn't preserve the graded parts, right? Any help will be appreciated.
|
350489
|
Show that [imath]k[x,y,z]/(xz-y^2)[/imath] is not a UFD.
I am trying to show that [imath]k[x,y,z]/(xz-y^2)\not\cong k[x,y][/imath]. The latter is a UFD, so I am trying to show the former is not. Clearly [imath]x[/imath] is not prime, since [imath]x\mid xz[/imath] which implies [imath]x\mid y^2[/imath], but [imath]x\mid y[/imath]. So if I can show that [imath]x[/imath] is irreducible then [imath]k[x,y,z]/(xz-y^2)[/imath] is a not a UFD, because irreducible implies prime in UFDs. Attempt: Assume [imath]x[/imath] is reduced into [imath]ab[/imath] in [imath]k[x,y,z]/(xz-y^2)[/imath], then we know there exists [imath]c\in k[x,y,z][/imath] such that [imath]x=ab+c(xz-y^2)[/imath] in [imath]k[x,y,z][/imath], or more usefully as [imath]x-ab=(xz-y^2)c[/imath]. Someone suggested that I write [imath]a,b[/imath] such they are degree at most 1 in [imath]y[/imath], that is, replace every [imath]y^n[/imath] using [imath]xz[/imath]. This means the LHS is degree at most 2 in [imath]y[/imath] and the RHS is degree at least two in [imath]y[/imath] in [imath]x-ab=(xz-y^2)c[/imath]. I am not sure where to go from here just by knowing that [imath]c[/imath] has no [imath]y[/imath] terms though. Ultimately I want to show that this forces [imath]x=ab[/imath] in [imath]k[x,y,z][/imath] which, since [imath]x[/imath] is irreducible there, shows that either [imath]a,b[/imath] is a unit, which means [imath]x[/imath] is irreducible in [imath]k[x,y,z]/(xz-y^2)[/imath].
|
1875209
|
How to solve the question related to continuity.
The question is : Let [imath]f : \mathbb {R} \longrightarrow \mathbb {R}[/imath] be a continuous function such that [imath]f(x) = f(x^2)[/imath].Then show that [imath]f(x) = f(0)[/imath] , [imath]\forall x \in \mathbb {R}[/imath]. How can I solve it?Please help me.Thank you in advance.
|
338802
|
Suppose that [imath]f(x)[/imath] is continuous on [imath](0, \infty)[/imath] such that for all [imath]x > 0[/imath],[imath]f(x^2) = f(x)[/imath]. Prove that [imath]f[/imath] is a constant function.
Suppose that [imath]f(x)[/imath] is continuous on [imath](0, +\infty)[/imath] such that for all [imath]x > 0[/imath],[imath]f(x^2) = f(x)[/imath]. Prove that [imath]f[/imath] is a constant function. My attempt is to show that for any point [imath]a \neq b[/imath] , we have [imath]f(a)=f(b)[/imath]. But I have no idea on how to get this. Anyone can help?
|
1875374
|
Inequalities based on arithmetic progression
Let [imath]m_1<m_2<m_3<\cdots <m_{k-1}<m_k[/imath] are positive integers where their reciprocals are in arithmetic progression. Show that [imath]k<m_1+2[/imath]
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1495367
|
Sequence of positive integers such that their reciprocals are in arithmetic progression
Let [imath]m_1 < m_2 < \ldots < m_k[/imath] be [imath]k[/imath] distinct positive integers such that their reciprocals [imath]\dfrac{1}{m_i}[/imath] are in arithmetic progression. Show that [imath]k < m_1 + 2[/imath]. Give an example of such a sequence of length [imath]k[/imath] for any positive integer [imath]k[/imath]. Any kind of help would be appreciated.
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1871825
|
How solve for x in an Infinite exponent
How would one solve for x in the following equation: [imath]x^{x^{x^{x^{\cdots}}}} = 4[/imath] The exponent continues forever... So what is the value of x? Thank you for helping
|
1855000
|
Why do [imath]x^{x^{x^{\dots}}}=2[/imath] and [imath]x^{x^{x^{\dots}}}=4[/imath] have the same positive root [imath]\sqrt 2[/imath]?
What is [imath]x[/imath] when is satisfies [imath]x^{x^{x^{\dots}}}=2[/imath] ? I am really confused with this; the root is [imath]\sqrt{2}[/imath], but why does the equation [imath]x^{x^{x^{\dots}}}=4[/imath] have the same root?
|
1876095
|
What is the Area in shaded Region?(Tricky one)involves Circles,Rectangle,Triangle
I have solved so far this Image is in the form of link Rectangle area[imath]= 10*20=200cm^2[/imath] Circle diameter=[imath]10cm[/imath] Radius=[imath]5cm[/imath] 1 circle area=[imath]3.1456*5^=78.64cm^2[/imath] 2 circle area=[imath]2*78.64=157.28[/imath] Area outside circle boundry=[imath]200-157.2=42.72cm^2[/imath] Area in the lower triangle=[imath]42.72/2=21.36cm^2[/imath] Area of Square if we draw line in between=[imath]100cm^2[/imath] Area outside of circle in square=[imath]100-78.64=21.36cm^2[/imath] Area of each of size outside of circle halved=[imath]21.36/8=2.67cm^2[/imath] But the triangle line deosn't properly half the area on circle sides in squares what to do? Estimated could be this=[imath]18.69cm^2[/imath] But this is not true because the line cutting doesn't properly half it. Answer Plz!
|
1874736
|
Any smart ideas on finding the area of this shaded region?
Don't let the simplicity of this diagram fool you. I have been wondering about this for quite some time, but I can't think of an easy/smart way of finding it. Any ideas? For reference, the Area is: [imath]\bbox[10pt, border:2pt solid grey]{90−18.75\pi−25\cdot \arctan\left(\frac 12\right)}[/imath]
|
1685236
|
Calculating the stiefel whitney class of Tangent Bundle of projective space
I was reading the proof of the fact that whitney sum of the tangent bundle of [imath]RP^n[/imath] with the trivial bundle is isomorphic to whitney sum of [imath]n+1[/imath] tautological line bundles on [imath]RP^n[/imath] from Hatcher's Vector Bundles and K theory. However I am confused at this particular point in which he says In the normal bundle [imath]NS^n [/imath] the identification [imath](x, v) ∼ (−x, −v)[/imath] can be written as [imath](x, tx) ∼ (−x, t(−x)) [/imath]. This identification yields the product bundle [imath]RP^n × R[/imath] since the section [imath]x \to (−x, −x)[/imath] is well-defined in the quotient. Now let us consider the identification [imath](x, v) ∼ (−x, −v)[/imath] in [imath]S^n × R^{n+1}[/imath] . This identification respects the co- ordinate factors of [imath]R^{n+1}[/imath] , so the quotient is the direct sum of [imath]n + 1[/imath] copies of the line bundle [imath]E[/imath] over [imath]RP^n[/imath] obtained by making the identifications [imath](x, t) ∼ (−x, −t)[/imath] in [imath]S^n × R[/imath] . The claim is that E is just the canonical line bundle over [imath]RP^n [/imath]. To see this, let us identify [imath]S^n × R [/imath] with [imath]NS^n[/imath]by the isomorphism [imath](x, t)\to(x, tx)[/imath] ,hence [imath](−x, −t)\to ((−x, (−t)(−x)) = (−x, tx)[/imath] . Thus we have the identification[imath] (x, tx) ∼ (−x, tx) [/imath]in [imath]NS^n[/imath] . The quotient is the canonical line bundle over [imath]RP^n[/imath] since the identifications [imath]x ∼ −x[/imath] in the first coordinate give lines through the origin in [imath]R^{n+1}[/imath] , and in the second coordinate there are no identifications so we have well-defined vectors [imath]tx[/imath] in these lines. So it seems like he is proving that [imath](x,v) \sim (-x,-v)[/imath] is trivial in the first three lines and later he seems to be proving that it is the tautological line bundle which is non trivial. Also I could not find this same proof anywhere else(all had the proof from the book characteristic classes by Milnor-Stasheff). If anyone can point out where I am going wrong or give some other reference it would be great. Thanks
|
1850825
|
Relating the normal bundle and trivial bundles of [imath]S^n[/imath] to the tautological and trivial line bundles of [imath]\mathbb{R}P^n[/imath]
On page [imath]10[/imath] of Hatcher's Vector Bundles and K Theory, he gives a proof that the Whitney sum of the trivial line bundle over [imath]\mathbb{R}P^n[/imath] and the tangent bundle is equal to the Whitney sum of copies of the tautological line bundle. A summary of the proof is 1) The image of the tangent bundle of the sphere under the quotient is the tangent bundle of real projective space, and the image of the normal bundle is the trivial line bundle over real projective space. 2) He says the sum of the tangent bundle and normal bundle of the sphere is trivial. 3) He then proves that the trivial line bundle over the sphere is isomorphic to the normal bundle and proves that the normal bundle under the quotient is mapped to the tautological line bundle over real projective space. But this doesn't make any sense because he claimed earlier that the image of the normal bundle is trivial and the tautological bundle is NOT trivial. Could someone please clarify what has happened? Is the image of the normal bundle of the sphere under the quotient the tautological bundle or the trivial bundle?
|
1876806
|
Show that sum obtained in this chessboard is already 260
Consider the square of an [imath]8 \times 8 [/imath] chessboard filled with the number [imath]1[/imath] to [imath]64[/imath] as in given figure .If we choose [imath]8[/imath] squares with the property that is exactly one from each row and exactly one from each column , and add up the numbers in the chosen squares , show that sum obtained is always [imath]260[/imath] . My work I can just observer that it is happening but I don't know how to prove it . [imath]1+10+29+28+37+46+55+64[/imath] [imath]2+11+20+29+38+47+56+57[/imath] I know many pair like are existing but how I will show this ?
|
787856
|
Sum of numbers on chessboard.
Consider the squares of an [imath]8 \times 8[/imath] chessboard filled with the numbers [imath]1,2,3,4 \ldots ,64[/imath] in sequential order. If we choose [imath]8[/imath] squares with the property that there is exactly [imath]1[/imath] from each row and exactly [imath]1[/imath] from each column, and add up the numbers in the chosen squares, show that the sum obtained is always [imath]260[/imath]. Please provide full solution and explanation.
|
1877081
|
Computing [imath]\int\limits_{-\pi/2}^{\pi/2} \cos(\tan x)\,dx[/imath]
How to compute [imath]\int\limits_{-\pi/2}^{\pi/2} \cos(\tan x)\,dx[/imath]? I tried the usual Riemann integral method but could not figure out and thought of contour integral and seemed working. But whenever I tried to find the residue at the essential singularity of [imath]f(z)=\cos (\tan z)[/imath] at [imath]\frac{\pm \pi}{2}[/imath], I got stuck there.
|
1840598
|
How do I prove [imath]\int_{-\infty}^{\infty}{\cos(x+a)\over (x+b)^2+1}dx={\pi\over e}{\cos(a-b)}[/imath]?
How do I prove these? [imath]\int_{-\infty}^{\infty}{\sin(x+a)\over (x+b)^2+1}dx={\pi\over e}\color{blue}{\sin(a-b)}\tag1[/imath] [imath]\int_{-\infty}^{\infty}{\cos(x+a)\over (x+b)^2+1}dx={\pi\over e}\color{blue}{\cos(a-b)}\tag2[/imath] I am trying to apply the residue theorem to [imath](2)[/imath] [imath]f(x)={\cos(x+a)\over (x+b)^2+1}[/imath] [imath](x+b)^2+1[/imath]=[imath](x+b-i)(x+b+i)[/imath] [imath]2\pi{i}Res(f(x),-b-i)=2\pi{i}\lim_{x\rightarrow -b-i}{\cos(a-b-i)\over -2i}=-\pi\cos(a-b-i)[/imath] [imath]2\pi{i}Res(f(x),-b+i)=2\pi{i}\lim_{x\rightarrow -b+i}{\cos(a-b+i)\over 2i}=\pi\cos(a-b+i)[/imath] How do I suppose to evaluate [imath]\cos(a-b-i)[/imath] and [imath]\cos(a-b+i)[/imath]?
|
1877476
|
Possible rank of closed orientable manifolds
List all [imath]i[/imath] for which there is a closed orientable [imath]6[/imath]-manifold [imath] M[/imath] with [imath]H_i(M) = \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}[/imath]. So, obviously this has to do with Poincare Duality. We can't have [imath]i=0[/imath] or [imath]i=6[/imath] by definition of a manifold. So we are left with the cases [imath]i=1[/imath] through [imath]5[/imath], which reduce to [imath]i=1, 2,[/imath] and [imath]3[/imath], since the rank of [imath]H_1[/imath] is equal to the rank of [imath]H_5[/imath] by Poincare Duality (and thus for [imath]H_2[/imath] and [imath]H_4[/imath], too). I am thinking that setting up the exact sequences here gives me some kind of equation which will lead to a contradiction in some of those cases, but I'm not able to see why yet. Can I get a push in the right direction, and an evaluation of my thoughts so far?
|
1750441
|
Which homology groups of a closed orientable 6-manifold can be isomorphic to [imath]\mathbb{Z}^3[/imath]?
List all [imath]i[/imath] for which there is a closed orientable [imath]6[/imath]-manifold [imath]M[/imath] with [imath]H_i(M) =\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}[/imath]. I am working on an old exam problem and this one stumped me. Progress so far: I can take [imath]Y=\mathbb{T}^3 \times \mathbb{S}^3[/imath] and get the desired group for [imath]i =1,5,2,4[/imath] I think. But this construction I believe gives [imath]H_3(Y)=\mathbb{Z}^2[/imath]. What to do in this case?
|
1877277
|
[imath]f_n,f_n^{\prime}\rightarrow 0[/imath] pointwise. But [imath](f_n^{\prime}) [/imath] is not uniformly convergent. Find an example of such a sequence of functions.
I was studying uniform convergence of sequence and series. I encounter this problem. Trying to find an example. Any help will be appreciated. The domain of the function can be any closed interval. For example, [0,1] will suffice. I am looking for an example to show that the conditions of the following theorems are necessary but not sufficient: Let [imath]f_n\colon [a,b]\rightarrow \mathbb{R}[/imath] be a sequence of differentiable functions on [imath][a,b][/imath], and let [imath]f_n\rightarrow f[/imath] pointwise on [imath][a,b][/imath]. If [imath]f_n^{\prime}\rightarrow g[/imath] uniformly on [imath][a,b][/imath], then [imath]f[/imath] is differentiable on [imath][a,b][/imath] and [imath]f^{\prime} = g[/imath].
|
1039207
|
Find sequence of differentiable functions [imath]f_n[/imath] on [imath]\mathbb{R}[/imath] that converge uniformly, but [imath]f'_n[/imath] converges only pointwise
Question: Find a sequence of differentiable functions [imath]f_n[/imath] on [imath]\mathbb{R}[/imath] that converge uniformly to a differentiable function [imath]f[/imath], such that [imath]f'_n[/imath] converges pointwise but not uniformly to [imath]f'[/imath]. Attempt: I have tried a number of possibilities, such as [imath]f_n=x^n[/imath] or [imath]f_n=\frac{x^n}{n}[/imath] but I don't know what the right approach is to construct the function. I am initially thinking that it's easiest to construct such a sequence of functions on the interval [imath][0,1][/imath] so that in the limit of [imath]n[/imath], part of the function goes to [imath]0[/imath] and the other part goes to [imath]1[/imath]. However, this would make the resulting [imath]f[/imath] non-differentiable.
|
1877434
|
If [imath]K[/imath] is closed and bounded, then it is compact.
Let [imath]K \subseteq \mathbb{R}[/imath]. We want to show that if [imath]K[/imath] is closed and bounded, then it is a compact set. B-W THM: Every bounded sequence contains a convergent subsequence. Theorem: Subsequences of a convergent sequence converge to the same limit as the original sequence. Theorem: [imath]x[/imath] is a limit point of a set [imath]A \iff[/imath] [imath]x = \lim a_n[/imath] for some [imath](a_n) \subset A[/imath] s.t: [imath]a_n \neq x, \, \forall n \in \mathbb{N}[/imath] Proof: Let [imath](x_n) \subset K[/imath] and [imath]\lim x_n = x[/imath], so that [imath]x[/imath] is a limit point of [imath]K[/imath]. If [imath]K[/imath] is bounded, then by the Bolzano-Weierstrass Theorem, [imath](x_n)[/imath] is bounded so that it contains a convergent subsequence [imath](x_{n_k}) \subset K[/imath]. Since subsequences of a convergent sequence converge to the same limit, we have that [imath]x_{n_k} \longrightarrow x[/imath]. But, [imath]K[/imath] is closed, so it contains all its limit points. Therefore, [imath]x \in K[/imath]. Hence, we have shown [imath]K[/imath] is compact since every sequence in [imath]K[/imath] has a subsequence that converges to a limit that is also in [imath]K[/imath]. QED. How did I do?
|
1877413
|
[imath]K \subset \mathbb{R}^n[/imath] is compact iff it is closed and bounded
I want to prove [imath]\Leftarrow[/imath]. I just need one more step, now. If I find a compact set that contains K, then it will finish the proof. Rudin take a k-cell to do that, but before it he proves that every k-cell is compact, and I think I can take a more simple compact to do this step. I has thinked in the closed ball, but don't know how to prove that it is compact. Can anyone help me to find this set, if it really exists? Thanks.
|
1390531
|
Estimate for [imath]\frac{1}{\Delta}\sqrt{abc(a+b+c)}[/imath]
If the triangle ABC has sides [imath]a,b,c[/imath] opposite to the vertices A,B,C respectively and [imath]\Delta[/imath] is the area.The expression [imath]\frac{1}{\Delta}\sqrt{abc(a+b+c)}[/imath] is [imath](A)\leq16\hspace{1 cm}(B)\geq16\hspace{1 cm}(C)\leq4\hspace{1 cm}(D)\geq4[/imath] Using Herons formula, [imath]\Delta=\sqrt{s(s-a)(s-b)(s-c)}[/imath] and [imath]s=\frac{a+b+c}{2}[/imath], given expression reduces to [imath]4\sqrt{\frac{abc}{(a+b-c)(b+c-a)(a+c-b)}}[/imath] and I could not solve further to find the answer. Please help....
|
621182
|
proving the inequality [imath]\triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}[/imath]
If [imath]\triangle[/imath] be the area of [imath]\triangle ABC[/imath] with side lengths [imath]a,b,c[/imath]. Then show that [imath]\displaystyle \triangle\leq \frac{1}{4}\sqrt{(a+b+c)\cdot abc}[/imath] and also show that equality hold if [imath]a=b=c[/imath]. [imath]\bf{My\; Try}::[/imath] Here we have to prove [imath]4\triangle\leq \sqrt{(a+b+c)\cdot abc}[/imath] Using the formula [imath]\triangle = \sqrt{s(s-a)(s-b)(s-c)},[/imath] where [imath]2s=(a+b+c)[/imath] So [imath]4\triangle = \sqrt{2s(2s-2a)(2s-2b)(2s-2c)}=\sqrt{(a+b+c)\cdot(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}[/imath] Now using [imath]\bf{A.M\geq G.M}[/imath] for [imath](b+c-a)\;,(c+a-b)\;,(a+b-c)>0[/imath] [imath]\displaystyle \frac{(b+c-a)+(c+a-b)+(a+b-c)}{3}\geq \sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}[/imath] So we get [imath]\displaystyle (a+b+c)\geq 3\sqrt[3]{(b+c-a)\cdot(c+a-b)\cdot(a+b-c)}[/imath] But I did not understand how can I prove above inequality help Required Thanks
|
1874244
|
What integrability of Hamiltonian system says about non compact orbits?
Suppose that we have an [imath]n[/imath]-dimensional manifold [imath]M[/imath] and consider a Hamiltonian system on cotangent bundle which is integrable in the following way: there exist [imath]n[/imath] functions [imath]f_1,...,f_n[/imath], which are in an involution (their Poisson brackets vanish) and [imath]f_1=H[/imath]. For compact phase spaces properties of such system (at least away from points where gradients of [imath]f_i[/imath] become collinear) are neatly characterized by the classical Liouville theorem. I am curious if something interesting can be said in non-compact setting. I am aware that sometimes this question reduces to compact case if in some region of [imath]T^* M[/imath] isosurfaces of [imath]f_i[/imath] are compact, but I am interested in properties of noncompact orbits. Particular system which I am investigating now actually has no compact orbits at all.
|
1855956
|
On the Liouville-Arnold theorem
A system is completely integrable (in the Liouville sense) if there exist [imath]n[/imath] Poisson commuting first integrals. The Liouville-Arnold theorem, anyway, requires additional topological conditions to find a transformation which leads to action-angle coordinates and, in these set of variables, the Hamilton-Jacobi equation associated to the system is completely separable so that it is solvable by quadratures. What I would like to understand is if the additional requirement of the Liouville-Arnold theorem (the existence of a compact level set of the first integrals in which the first integrals are mutually independent) means, in practice, that a problem with an unbounded orbit is not treatable with this technique (for example the Kepler problem with parabolic trajectory). If so, what is there a general approach to systems that have [imath]n[/imath] first integrals but do not fulfill the other requirements of Arnold-Liouville theorem? Are they still integrable in some way?
|
1878027
|
Are real numbers a subset of the complex numbers?
I am having an argument with a friend. I think that in a sense, the answer is no. My reasoning is that in linear algebra, a vector [imath](a, b)[/imath] is not the same as a vector [imath](a, b, 0)[/imath] because the first one is in [imath]\mathbb{R}^2[/imath], while the second is in [imath]\mathbb{R}^3[/imath]. However I am not sure if a similar argument can be made for real vs complex numbers.
|
389897
|
The set of real numbers is a subset of the set of complex numbers?
So, I was taught that [imath]\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}[/imath]. But since the set of complex numbers is by definition [imath]\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\},[/imath] doesn't this mean [imath]\mathbb{R}\subseteq\mathbb{C}[/imath], since for each [imath]x \in \mathbb{R}[/imath] taking [imath]z = x + 0i[/imath] we have a complex number which equals [imath]x[/imath]?
|
1878202
|
Convergence of [imath]\sum\frac{a_n}{b_n} [/imath] and [imath] \sum (\frac{a_n}{b_n})^2 [/imath] implies convergence of [imath] \sum\frac{a_n}{a_n+b_n}[/imath]
Let [imath]a_n,b_n\in \mathbb R[/imath] such that for each [imath]n[/imath] we have [imath](a_n+b_n)b_n\neq 0[/imath] and both the series [imath]\sum_1^\infty \frac{a_n}{b_n}[/imath] and [imath]\sum_1^\infty (\frac{a_n}{b_n})^2[/imath] convergent then prove [imath]\sum_1^\infty\frac{a_n}{a_n+b_n}[/imath] is convergent. I tried writing [imath]\frac{a_n}{a_n+b_n}[/imath] like [imath]\frac{\frac{a_n}{b_n}}{\frac{a_n}{b_n}+1}[/imath] but I don't think if this is enough and I don't know how [imath](\frac{a_n}{b_n})^2[/imath] appears.
|
296442
|
how prove [imath]\sum_{n=1}^\infty\frac{a_n}{b_n+a_n} [/imath]is convergent?
Let[imath]a_n,b_n\in\mathbb R[/imath] and [imath](a_n+b_n)b_n\neq 0\quad \forall n\in \mathbb{N}[/imath]. The series [imath]\sum_{n=1}^\infty\frac{a_n}{b_n} [/imath] and [imath]\sum_{n=1}^\infty(\frac{a_n}{b_n})^2 [/imath] are convergent. How to prove that[imath]\sum_{n=1}^\infty\frac{a_n}{b_n+a_n} [/imath] is convergent. Thanks in advance
|
1878282
|
Which of the following are quadratic integers in [imath]Q[√−5][/imath]?
Which of the following are quadratic integers in [imath]Q[√−5][/imath]? EDIT: THANKS TO THE COMMUNITY FOR DIRECTING ME TO THE ORIGINAL QUESTION! I DID NOT KNOW THIS QUESTION ALREADY EXISTED ON MATH.SE (a) [imath]3/5[/imath] (b) [imath]2 + 3√−5[/imath] (c) [imath](3 + 8√−5)/2[/imath] (d) [imath](3 + 8√−5)/5[/imath] (e) [imath]i√−5[/imath] Prove your result in each case. Here's what I've got so far (I'm not even sure if this is half right): [imath]\{1,i\sqrt{5}\}[/imath] forms a basis for the field [imath]\mathbb{Q}(\sqrt{-5})[/imath] viewed as a vector space over [imath]\mathbb{Q}[/imath]. We note that, [imath]\mathbb{Q}\leq \mathbb{Q}(\sqrt{-5})[/imath] Meaning, that the rationals are contained in this simple extension. a)Yes, as just explained all rationals are contained in this field. d)Yes, it is [imath]\frac{3}{5}+\frac{8}{5}\sqrt{-5}[/imath] so if we select [imath]a=3/5[/imath], [imath]b=8/5[/imath] in [imath]\mathbb{Q}[/imath] as a linear combination for [imath]a+b\sqrt{-5}[/imath] then it works! e)No, unless we are asking for [imath]i\sqrt{5}[/imath]. I am having trouble on how to explain (b) and (c).
|
1219077
|
Quadratic Integers in [imath]\mathbb Q[\sqrt{-5}][/imath]
Can someone tell me if [imath]\frac{3}{5}[/imath], [imath]2+3\sqrt{-5}[/imath], [imath]\frac{3+8\sqrt{-5}}{2}[/imath], [imath]\frac{3+8\sqrt{-5}}{5}[/imath], [imath]i\sqrt{-5}[/imath] are all quadratic integers in [imath]\mathbb Q[\sqrt{-5}][/imath]. And if so why are they in [imath]\mathbb Q[\sqrt{-5}][/imath].
|
1878697
|
Find [imath]\int \sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx[/imath]
How to find [imath]\int \sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx[/imath] ? I could get rid of of [imath](1+\sin x)[/imath] by multiplying numerator and denominator by [imath](1-\sin x)[/imath].What about the [imath](2+\sin x)[/imath] term? Any other method possible?
|
1396045
|
Evaluate the integration : [imath]\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx[/imath]
[imath]\int{\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx}[/imath] [imath]\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx=\int \frac{(1-\sin x)(2-\sin x)}{\sqrt{(1-\sin x)(2-\sin x)(1+\sin x)(2+\sin x)}}dx[/imath] I am stuck. Please help me....
|
1878732
|
Sequence of Holomorphic functions.
Let [imath]\Omega[/imath] be open and connected. Suppos that a sequence [imath]\{f_n : \Omega \to \mathbb{C} : f_n \mbox{ is holomorphic and injective in $\Omega$ }\}[/imath] Converges uniformly to a function [imath]f[/imath] on every compact subset of [imath]\Omega[/imath]. Prove that if [imath]f[/imath] is not a constant in [imath]\Omega[/imath], then [imath]f[/imath] is injective in [imath]\Omega[/imath]. First I proved that [imath]f[/imath] is holomorphic. Suppose that [imath]f[/imath] is not injective. There are distinct point [imath]z_1,z_2[/imath] in [imath]\Omega[/imath]. Then there are path from [imath]z1[/imath] to [imath]z2[/imath] lying in [imath]\Omega[/imath]. Then... I can't tell you How to solve. Could you help me?
|
1737017
|
Show that [imath]f[/imath] is either injective or a constant function.
Let [imath]\Omega[/imath] be a domain in [imath]\mathbb{C}[/imath] and let [imath]\{f_n\}_{n \in \mathbb{N}}[/imath] be a sequence of injective functions that converge in [imath]O(\Omega)[/imath] to [imath]f[/imath] . Show that [imath]f[/imath] is either injective or a constant function. How does the conclusion change if, instead of a domain, we allow [imath]\Omega[/imath] to be an arbitrary open set ? I know that [imath]f[/imath] is holomorphic as an almost uniform limit. But I dont know how to proceed.
|
551137
|
[imath] \frac1{bc-a^2} + \frac1{ca-b^2}+\frac1{ab-c^2}=0[/imath] implies that [imath] \frac a{(bc-a^2)^2} + \frac b{(ca-b^2)^2}+\frac c{(ab-c^2)^2}=0[/imath]
[imath]a ,b , c[/imath] are real numbers such that [imath] \dfrac1{bc-a^2} + \dfrac1{ca-b^2}+\dfrac1{ab-c^2}=0[/imath] , then how do we prove (without routine laborious manipulation) that [imath]\dfrac a{(bc-a^2)^2} + \dfrac b{(ca-b^2)^2}+\dfrac c{(ab-c^2)^2}=0[/imath]
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581884
|
How prove this [imath]\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0[/imath]
let [imath]a,b,c\in R[/imath],if such [imath]\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0[/imath] show that [imath]\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0[/imath] this problem have nice methos? My idea:let [imath](ca-b^2)(ab-c^2)+(bc-a^2)(ab-c^2)+(bc-a^2)(ca-b^2)=0[/imath] then I can't.Thanks
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1878746
|
My proof of "Is [imath]2^{2n} = O(2^n)[/imath]?"
I was wondering if my proof for this problem was correct. Let [imath]x = 2^n[/imath]. Then the problem reduces to is [imath]x^2 = O(x)[/imath], which is clearly false.
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131420
|
Is [imath]2^{2n} = O(2^n)[/imath]?
Is [imath]2^{2n} = O(2^n)[/imath]? My solution is: [imath]2^n 2^n \leq C_{1}2^n[/imath] [imath]2^n \leq C_{1}[/imath], TRUE. Is this correct?
|
1838535
|
Representation of Frey's curve.
I read that Frey's curve is a semi-stable elliptic curve. What doe this mean ? I can find two dimensional representations of [imath]y^2 = x^3 + ax + b[/imath] in Wikipedia. What does [imath]y^2 = x(x-a)(x+b)[/imath] look like if [imath]a[/imath] and [imath]b[/imath] are solutions to [imath]FLT[/imath] ?
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433973
|
transform into weierstrass-form
How can I transform the elliptic curve [imath]E/\mathbb{C}[/imath] of the form [imath]y^2=4(x-e_1)(x-e_2)(x-e_3)[/imath] with [imath]e_1>e_2>e_3\in\mathbb{R}[/imath] roots of [imath]E[/imath] into a Weierstrass-Form [imath]y^2=x^3+ax+b[/imath]?
|
1879125
|
Diagonal Quadric Form
I have the equation: [imath]Q(x,y) = 5x^2 - 6xy + 5y^2[/imath]. The first question is to write this in [imath] Q(z) = z^{T}Az[/imath], in Matrixnotation. The matrix I calculated is: [imath]\begin{bmatrix} 5 & -3 \\ -3 & 5 \end{bmatrix} [/imath] Then I have to diagonalize it, such that [imath]A = UDU^{T}[/imath] in order to use another Basis (from [imath]z \to \overline{z}[/imath]). [EDIT] I solved this part using the diagonalization, and obtaining [imath]\begin{bmatrix} 2 & 0 \\ 0 & 8 \end{bmatrix} [/imath] as the matrix for the new quadric. Now how can I solve the next question? At the end I have to draw the new curve given by [imath]z^{T}Az = 8[/imath]. I tried to read some documentation but I don't understand how to procede, can someone help? Thanks!
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438164
|
diagonalize quadratic form
I have this quadratic form [imath]Q= x^2 + 4y^2 + 9z^2 + 4xy + 6xz+ 12yz[/imath] And they ask me: for which values of [imath]x,y[/imath] and [imath]z[/imath] is [imath]Q=0[/imath]? and I have to diagonalize also the quadratic form. I calculated the eigenvalues: [imath]k_{1}=0=k_{2}, k_{3}=14[/imath], and the eigenvector [imath]v_{1}=(-2,1,0), v_{2}=(1,2,3), v_{3}=(3,6,-5)[/imath] I don't know if this is usefull in order to diagonalize or to see when is [imath]Q=0[/imath]
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1838562
|
Elliptic curve, different forms of.
[imath]y^2 = x^3 + mx + c[/imath] An elliptic curve in the form defined in Wikipedia [imath]y^2 = x(x-A)(x+B) = x^3 +(B-A)x^2 + ABx[/imath] Frey's curve has no term in [imath]x^2[/imath], but [imath]2[/imath]. does because from Fermat, [imath]A=a^n[/imath] not equal to [imath]B=b^n[/imath] Question: Is there a simple way to transform Frey's curve with given [imath]A,B[/imath] into form [imath]1[/imath]. (Has it got anything to do with Weierstrass ?)
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1838101
|
Derivation of Frey equation from FLT
I understand, on a layman's level, Fray's motivation to write an elliptic equation corresponding to an assumed solution to FLT. My question is, how technically is Frey's equation derived? [imath]1.[/imath] FLT : Contradiction. There exist integers [imath]A,B,C,n>2[/imath] such that [imath]A^n + B^n = C^n[/imath] [imath]2.[/imath] Frey: [imath]y^2 = x(x-A^n)(x+B^n)[/imath] [imath]3.[/imath] Elliptic curve [imath]y^2= x^3 + ax + b[/imath] [imath]4.[/imath] The curve is not modular. => not elliptic => FLT proved. I need is the algebra to go from 1. to 3. above. Assume Numbers A,B,C such that A^n + B^n = C^n contradicting FLT. [imath]y^2 = x^3 + ax + b[/imath] an elliptic curve in Weirstrass form. I am guessing A and B go unchanged into x(x-A^n)(x+B^n) Is that correct ? Expanding 2. I get a term in x^2 , there is none in 3. This is my problem. What does a [imath]2[/imath] dimensional graph of the Frey curve look like ?
|
1880065
|
Prove inequality [imath]\frac{1}{1+2b^2c}+\frac{1}{1+2c^2a}+\frac{1}{1+2a^2b}\ge1[/imath]
Let [imath]a,b,c>0[/imath] and [imath]a+b+c=3[/imath]. Prove that [imath]\frac{1}{1+2b^2c}+\frac{1}{1+2c^2a}+\frac{1}{1+2a^2b}\ge1.[/imath] My attempts: 1) I use Titu's Lemma: [imath]\frac{1}{1+2b^2c}+\frac{1}{1+2c^2a}+\frac{1}{1+2a^2b}\ge\frac{(1+1+1)^2}{3+2(a^2b+b^2c+c^2a)}[/imath] 2) I use AM-GM: [imath]\frac{1}{1+2b^2c}=1-\frac{2b^2c}{1+2b^2c}\ge1-\frac{bc}{\sqrt{2c}}[/imath]
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168704
|
Proving :[imath]\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1[/imath]
Let [imath]a,b,c>0[/imath] be real numbers such that [imath]a+b+c=3[/imath],how to prove that? : [imath]\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1[/imath]
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1880442
|
If [imath]U[/imath] has uniform distribution over the interval [imath](0, 1)[/imath] what is the density function of [imath]X = -Kln(U)[/imath] for some constant [imath]K > 0[/imath]?
I am aware of he uniform distribution density function however I am not entirely sure how I would find out the distribution of X. Thanks very much, any help or hints in the right direction would be gladly appreciated.
|
199614
|
Distribution of [imath]-\log X[/imath] if [imath]X[/imath] is uniform.
For [imath]X[/imath] and [imath]Y[/imath] random variables; [imath]X[/imath] follows the uniform distribution. (1): if [imath]Y=-\log X[/imath] (2): then it can be shown that [imath]-\log X[/imath] is distributed as [imath]\exp(1)[/imath] {i.e. exponential with mean 1}. Why is this so? Intuitively statement (2) make sense to me. But i'd like a mathematical proof. -Probably wrong working: (1) seems to imply [imath]\exp(-Y) = X[/imath] (which is like saying [imath]X[/imath] is exponentially distributed, which is a contradiction, since its actually uniform!); or is it incorrect for me to do this since [imath]X[/imath] and [imath]Y[/imath] are random variables? Ultimately how do I prove (2)? Thanks
|
1880473
|
Finding the value of K
[imath]π1:( 1+k , -3 , 6 )[/imath] [imath]π2: ( 1 , 5+k , 3k)[/imath] Does anyone know the values of k that would make these parallel and perpendicular, ive been trying for hours but nothing seems to work The original equation is [imath]π1:(1+k)x−3y+6z−4=0[/imath] [imath]π2:x+(5+k)y+3kz+1=0[/imath]
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1880300
|
Finding the value of [imath]k[/imath] for parallel/orthogonal planes
Given two planes [imath]\pi_1: (1+k)x-3y+6z-4=0[/imath] [imath]\pi_2: x+(5+k)y+3kz+1=0[/imath] Find the value of [imath]k[/imath] such that [imath]\pi_1[/imath] is parallel to [imath]\pi_2[/imath] and when [imath]\pi_1[/imath] is perpendicular to [imath]\pi_2[/imath]
|
1876920
|
Evaluate given limit
Let the function [imath]f(x)[/imath] be differentiable and [imath]f'(x)[/imath] be continuous in [imath]\left(-\infty,\infty \right)[/imath] with [imath]f'(2)=14[/imath] then evaluate the limit [imath]\lim_{x\to 0}\frac{f(2+\sin x)-f(2+x\cos x)}{x-\sin x}[/imath] My attempt: [imath]\lim_{x\to 0}\frac{f(2+\sin x)-f(2+x\cos x)}{x-\sin x}[/imath] [imath]\lim_{x\to 0}\left(\frac{f(2+\sin x)-f(2+x\cos x)}{\sin x-x\cos x}\right)[/imath] [imath]\frac{\sin x-x\cos x}{x-\sin x}[/imath][imath]=\left(f'(2)\right)(2)=28[/imath] Is the method used here correct
|
1870604
|
Doubt regarding a limit which is related to MVT
Let the function [imath]f(x)[/imath] be differentiable and [imath]f'(x)[/imath] be continuous in [imath]\left(-\infty,\infty \right)[/imath] with [imath]f'(2)=14[/imath] then evaluate the limit [imath]\lim_{x\to 0}\frac{f(2+\sin x)-f(2+x\cos x)}{x-\sin x}[/imath] My attempt: [imath]\lim_{x\to 0}\frac{f(2+\sin x)-f(2+x\cos x)}{x-\sin x}[/imath] [imath]\lim_{x\to 0}\left(\frac{f(2+\sin x)-f(2+x\cos x)}{\sin x-x\cos x}\right)[/imath] [imath]\frac{\sin x-x\cos x}{x-\sin x}[/imath][imath]=\left(f'(2)\right)(2)=28[/imath] Is the method used here correct
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1882929
|
Diophantine equation of [imath]a^2+b^2=c^2+d^2[/imath]
Is there a known solution to: [imath]a^2+b^2=c^2+d^2[/imath] Hopefully the question is clear, if not I do apologize.
|
153603
|
Diophantine equation [imath]a^2+b^2=c^2+d^2[/imath]
I was reasonably certain I've seen this before, but I was wondering how to solve the Diophantine equation [imath]a^2+b^2=c^2+d^2[/imath] I tried a web search and found nothing on this one. I'm trying to avoid another library trip to a less than local library (maybe I should have taken better notes on that chapter...). I'm not quite sure how to handle this one. The only thing I can figure out with this equation, if I remember correctly, is that the sum on either side may only contain prime factors of 2 or odd primes congruent to 1 mod 4. And if I don't want a and b equal to c and d, the sum can't be prime as I believe a prime congruent to 1 mod 4 can be represented as the sum of 2 squares in exactly one way. But that doesn't give me any insight into actually solving this problem.
|
1880641
|
Radon measure on locally compact Hausdorff spaces
Suppose [imath]\mu[/imath] is a Radon measure on a locally compact Hausdorff space [imath]X[/imath] such that [imath]\mu(\{x\})=0[/imath] for all [imath]x\in X[/imath] and [imath]A \in B_X[/imath] satisfies [imath]0<\mu(A)<\infty[/imath]. Then for any [imath]\alpha[/imath] such that [imath]0<\alpha<\mu(A)[/imath], there is a Borel set [imath]B\subset A[/imath] such that [imath]\mu(B)=\alpha[/imath]. I am reviewing measure theory for my exam and got stuck on this question. Can any one help me ? Thank you very much.
|
1735558
|
Folland's Real Analysis 7.11
Suppose [imath]\mu[/imath] is a Radon measure on [imath]X[/imath] such that [imath]\mu(\left\lbrace x \right\rbrace)=0[/imath] for all [imath]x \in X[/imath], and [imath]A \in \mathbb{B}_X[/imath] satisfies [imath]0 < \mu(A) < \infty[/imath]. Then for any [imath]\alpha[/imath] such that [imath]0 < \alpha < \mu(A)[/imath] there is a Borel set [imath]B \subset A[/imath] such that [imath]\mu(B) = \alpha[/imath] I'm not even sure how to start this problem. It seems like a nice result, but any hints on how to start?
|
1165539
|
Factoring A Matrix Polynomial
I'm working my way through this paper: Down with Determinants! In Section 2 (pretty much right off the bat) he gives his determinant-less proof that every finite-dimensional complex linear operator has an eigenvalue. First he says that a vector and its images under the transformation, repeated n times, can't be linearly independent, and defines [imath]a_i[/imath] to be the coefficients of the linear combination. [imath] a_0v + a_1Tv + \cdots + a_nT^nv=0 [/imath] Then he makes a (non-matrix) polynomial of those coefficients and factors it. [imath] a_0 + a_1z + \cdots + a_nz^n = c(z-r_1)\cdots(z-r_m) [/imath] The above holds for any complex number [imath]z[/imath]. I understand everything so far, but then he has this step, which seems to use the first equation as a polynomial of matrices: [imath] 0=(a_oI + a_1T + \cdots + a_nT^n)v = c(T-r_1I)\cdots(T-r_mI)v [/imath] This seems totally non-obvious to me. I can sort of see this as an analogy with the factorization of the regular polynomial, and I verified it by hand with a generic 2-dimensional T, but I'm not sure why it works in the general case. My main questions: What justifies the last step? Does it necessarily use the 2nd equation, or is that just sort of by analogy? In the last part, we've basically factored a polynomial of matrices. The "roots" are all constants times the identity matrix. Is this always true, or could the roots be any matrix?
|
1882069
|
Why does polynomial factorization generalize to matrices
I'm reading about linear algebra and I came across with the following theorem where I have a problem convincing myself: Theorem 2.1 [imath]\,[/imath] Every linear operator on a finite-dimensional complex vector space has an eigenvalue. Proof: To show that [imath]T[/imath] (our linear operator on [imath]V[/imath]) has an eigenvalue, fix any nonzero vector [imath]v \in V[/imath]. The vectors [imath]v, Tv, T^2v,..., T^nv[/imath] cannot be linearly independent, because [imath]V[/imath] has dimension [imath]n[/imath] and we have [imath]n + 1[/imath] vectors. Thus there exist complex numbers [imath]a_0,...,a_n[/imath], not all [imath]0[/imath], such that [imath]a_0v + a_1Tv + ··· + a_nT^nv = 0.[/imath] Make the [imath]a[/imath]’s the coefficients of a polynomial, which can be written in factored form as [imath]a_0 + a_1z + ··· + a_nz^n = c(z − r_1)\cdots(z − r_m),[/imath] where [imath]c[/imath] is a non-zero complex number, each [imath]r_j[/imath] is complex, and the equation holds for all complex [imath]z[/imath]. We then have [imath]{\color{red}{ 0=(a_0I + a_1T + ··· + a_nT^n)v= c(T − r_1I)\cdots(T − r_mI)v}},[/imath] which means that [imath]T − r_j[/imath] I is not injective for at least one [imath]j[/imath]. In other words, [imath]T[/imath] has an eigenvalue.[imath]\;\blacksquare[/imath] I'm having trouble with the factorized form of the matrix polynomial (in red). I understood that the factorization holds for a polynomial by the fundamental theorem of algebra but why does this also hold for matrices? In other words, why is the the part I highlighted true? Does such an factorization always exist? Could I have some help to see this? Thank you =) P.S. here is my reference (page 3). UPDATE: Someone else also has asked the same question before it seems.
|
1883918
|
Proving Pascal's Identity algebraically, stuck with simplification
I'm trying to prove Pascal's Identity algebraically but I'm getting stuck... I'm ashamed to say I spent an hour trying to do this with no luck. The various solutions I've seen seem to do steps that I don't understand. Would appreciate if someone could do a detailed walkthrough of this. Here's what I have so far: [imath]\frac{(n−1)!}{k!(n−1−k)!} + \frac{(n−1)!}{(k−1)!(n−1−(k−1))!}[/imath] [imath]= \frac{(n−1)!}{k!(n−1−k)!} + \frac{(n−1)!}{(k−1)!(n−k)!}[/imath] [imath]= \frac{((n−1)!(k−1)!(n−k)! + (n−1)!k!(n−1−k)!}{k!(n−1−k)!(k−1)!(n−k)!}[/imath] (I don't even know if this part is right... sigh) Edit: Yes, there are other solutions, but I do not understand them. I'm looking for a more detailed step-by-step solution. I'm clearly having trouble with even basic algebra so sticking with that would be appreciated.
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461194
|
Prove Pascal's Rule Algebraically
I am trying to prove Pascal's Rule algebraically but I'm stuck on simplifying the numerator. This is the last step that I have, but I'm not sure where to go from here [imath]=\frac{\left[(k-1)(n-k)!+k(n-1-k)!\right]\times(n-1)!}{k!(n-k)!}[/imath]
|
1877764
|
Orthonormal basis of [imath]\{(x_1, \dots, x_n) \mid x_1+x_2+\cdots+x_n=0\}[/imath]
How can we find the orthonormal basis of [imath]\{(x_1,\dots,x_n) \in \Bbb R^n \mid x_1+x_2+\cdots+x_n=0\}[/imath]? It is easy to find a basis, but using Gram-Schmidt procedure seems difficult to obtain an orthonormal one.
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602550
|
Find an orthonormal basis for the subspace of [imath]\mathbb R^4[/imath]
Find an orthonormal basis for the subspace of [imath]\mathbb{R}^4[/imath] that consists of vectors perpendicular to [imath]u = (1, -1, -1, 1)[/imath]. I know the components of the vector [imath]u[/imath] is [imath]u_1 = 1, u_2 = -1, u_3 = -1, u_4 = 1[/imath]. I managed to find a vector [imath]v[/imath] that is perpendicular to the [imath]u[/imath]. This is done by [imath]u_1v_1+ u_2v_2+ u_3v_3 + u_4v_4 = 0[/imath]. This means [imath]v_1 = -1, v_2 = 1, v_3= 1, v_4 = -1[/imath] because [imath](1)(-1) + (-1)(1) + (-1)(1) + (1)(-1) = 0[/imath]. So I found a vector [imath]v[/imath] perpendicular to the given vector [imath]u[/imath] which is [imath](-1, 1, 1, -1)[/imath]. So my question would I be able to call [imath](-1, 1, 1, -1)[/imath] my basis and perform Gram Schmidt process on it?
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1588776
|
Subadditivity of square root function
Any hint to prove [imath] \sqrt{x+y} \le \sqrt{x} + \sqrt{y}, \qquad \forall x,y \ge 0 [/imath]
|
408177
|
Proving [imath]\sqrt{x + y} \le \sqrt{x} + \sqrt{y}[/imath]
How do I prove [imath]\sqrt{x + y} \le \sqrt{x} + \sqrt{y}[/imath]? for [imath]x, y[/imath] positive? This should be easy, but I'm not seeing how. A hint would be appreciated.
|
1884552
|
Is a reciprocal partial derivative equal to the reciprocal of the partial derivative?
In particular, I have [imath]\frac{\partial r}{\partial x} = \frac{\partial (x^2 + y^2 + z^2)^{1/2}}{\partial x} = \frac{x}{(x^2 + y^2 + z^2)^{1/2}}= \frac{x}{r}[/imath] does this mean [imath]\frac{\partial x}{\partial r} = \frac{r}{x}[/imath] ? If it is, can we safely assume that this is so for all partial derivatives?
|
1090061
|
Manipulating Partial Derivatives of Inverse Function
In lectures we're told: [imath]\dfrac {\partial y} {\partial x} = \dfrac 1 {\dfrac {\partial x} {\partial y}}[/imath] as long as the same variables are being held constant in each partial derivative. The course is 'applied maths', i.e non-rigorous, so don't confuse me. But anyway, if we have: [imath]\xi = x - y \qquad \eta = x[/imath] Then [imath]\dfrac {\partial x} {\partial \xi} = 0[/imath] and [imath]\dfrac {\partial \xi} {\partial x} = 1[/imath]. The rule presumably fails because one of the partial derivatives is [imath]0[/imath]. But then isn't this rule useless? Since I cannot use it without checking that the partial derivative isn't in fact [imath]0[/imath], but then I've worked out the partial derivative manually anyway.
|
1884758
|
Estimated value of [imath]\sum \limits_{i=1}^{50} \frac{1}{2i-1}[/imath]?
Anyone has any idea the estimated result(no need to be accurate) of [imath]\sum \limits_{i=1}^{50} \frac{1}{2i-1}[/imath]? I think this problem is not hard but I don't have a generalized method to solve! And I also remember this problem related to a very famous probability problem, any one has any idea? Thank you!
|
141224
|
Finite sum of reciprocal odd integers
Mathematica tells me that [imath]\sum\limits_{i=1}^n \frac1{2i-1}[/imath] is equal to [imath]\frac12 H_{n-1/2}+\log\,2[/imath], where [imath]H_n[/imath] is a harmonic number. Why is this true? Is there a general strategy for evaluating sums of the form [imath]\sum\limits_{i=1}^n \frac1{ai+b}[/imath]?
|
1885106
|
[imath]\lim (3^n+1)^{\frac{1}{n}}[/imath] anyone?
[imath]\lim \limits _{n \to \infty} (3^n+1)^{\frac{1}{n}} = ?[/imath] The help tip says I should use squeeze theorem but somehow still can't figure it out :/
|
1875291
|
Squeeze/Sandwich Theorem Involving [imath]n^{th}[/imath] root: [imath]\lim _{ n\rightarrow \infty }{\left(3^n+1\right)}^{\frac 1n}[/imath]
Find [imath]\lim _{ n\rightarrow \infty }{ { \left( { 3 }^{ n }+1 \right) }^{ \frac { 1 }{ n } } } [/imath] using the squeeze theorem I have come across ways to do this but none mention the squeeze (or sandwich) theorem. I know I need to find [imath]2[/imath] functions which squeeze the given function but can only think of using [imath](3^n)^{1/n}[/imath] i.e. [imath]3[/imath] as the [imath]\le [/imath] function
|
1882982
|
Why are we allowed to multiply a 1x1 matrix by any matrix?
So, in order to multiply 2 matrices, there must be the same number of columns in the left matrix as there are rows in the right matrix. So if [imath]A[/imath] is an [imath]m \times n[/imath] matrix and [imath]B[/imath] is a [imath]p \times q[/imath] matrix, must not [imath]n = p[/imath] in order for [imath]AB[/imath] to exist and mustn't [imath]q = m[/imath] in order for [imath]BA[/imath] to exist. By this logic, we should only be allowed to multiply a [imath]1 \times 1[/imath] matrix by either a [imath]1 \times n[/imath] matrix on the right or a [imath]n \times 1[/imath] matrix on the left. However, if [imath]C[/imath] is a [imath]1 \times 1[/imath] matrix and [imath]D[/imath] is a [imath]m \times n[/imath] matrix, where neither [imath]m[/imath] nor [imath]n = 1[/imath], we're allowed to multiply the 2 matrices simply by multiplying each entry in [imath]D[/imath] by the entry in [imath]C[/imath]. Why? Shouldn't the rules for a [imath]1 \times 1[/imath] matrix be the same as for all the other matrices?
|
143116
|
Multiplying by a [imath]1\times 1[/imath] matrix?
For matrix multiplication to work, you have to multiply an [imath]m \times n[/imath] matrix by an [imath]n \times p[/imath] matrix, so we have [imath]\bigg(m \times n\bigg)\bigg( n\times p \bigg).[/imath] But what about a [imath]1 \times 1[/imath] matrix? Is this just a scalar? But every matrix can be mulitplied by a scalar; so do [imath]1 \times 1[/imath] matrices break the rule?
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1885239
|
Multivariable limit [imath]\lim_{(x, y) \to (0, 0)} \frac{x \sin(x^2 y^2)}{x^2 + y^2}[/imath]
I'm learning multivariable calculus, specifically multivariable limits and continuity, and need help to understand the solution to the following problem: Let [imath]f(x,y) = \begin{cases} \frac{x \sin(x^2 y^2)}{x^2 + y^2}, & (x,y) \neq (0, 0) \\ 0, & (x,y) = (0, 0). \end{cases}[/imath] Show that [imath]f[/imath] is continuous at [imath](0, 0)[/imath]. So we need to show that [imath]\lim_{(x, y) \to (0, 0)} f(x, y) = f(0, 0)[/imath] that is [imath]\lim_{(x, y) \to (0, 0)} \frac{x \sin(x^2 y^2)}{x^2 + y^2} = 0.[/imath] Solution: (from textbook) We have that [imath]\left| \frac{x \sin(x^2 y^2)}{x^2 + y^2} \right| = \frac{\left|x\right| \left|\sin(x^2 y^2)\right|}{x^2 + y^2} \leq \frac{\left|x\right|x^2 y^2}{x^2 + y^2} \leq \frac{\left|x\right|^3 y^2}{x^2 + y^2} \leq \left|x\right|^3.[/imath] Question: I don't understand how the author found the upper bound for the sine function. Why does [imath]\left|\sin(x^2y^2)\right| \leq x^2y^2[/imath]? When I first tried to solve this problem I used [imath]\left|\sin(x^2y^2)\right| \leq 1[/imath] without success. Can someone explein my error?
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858230
|
Prove that [imath]x\sqrt{1-x^2} \leq \sin x \leq x[/imath]
Use the mean value theorem to prove that if [imath]0 \leq x \leq 1[/imath], then [imath]x\sqrt{1-x^2} \leq \sin x \leq x[/imath] The theorem guarantees the existence of a point, but not an inequality, so I don't know how to begin.
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1885881
|
How to prove [imath]R[/imath] is an equivalence relation where: [imath]\forall x, y \in A[/imath], [imath]xRy \iff 4|(x − y)[/imath]
[imath]A = \{ 2, 4, 6, 8, 10 \}[/imath] I'm unsure how to 'show' that [imath]R[/imath] is an equivalence relation. My conclusion is it is not as [imath]x-x[/imath] reflexive as given any number in the set will equal [imath]0[/imath] therefore [imath]4[/imath] divides [imath]0[/imath] Am I Wrong?
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366614
|
Proving that [imath]4 \mid m - n[/imath] is an equivalence relation on [imath]\mathbb{Z}[/imath]
I have been able to figure out the the distinct equivalence classes. Now I am having difficulties proving the relation IS an equivalence relation. [imath]F[/imath] is the relation defined on [imath]\Bbb Z[/imath] as follows: For all [imath](m, n) \in \Bbb Z^2,\ m F n \iff 4 | (m-n)[/imath] equivalence classes: [imath]\{-8,-4,0,4,8\}, \{-7,-3,1,5,9\} \{-6,-2,2,6,10\}[/imath] and [imath]\{-5, -1, 3, 7, 11\}[/imath]
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1886288
|
[imath]x^2 + x + 41[/imath] is composite
As you know, [imath]x^2+x+41[/imath] is composite when [imath]x=41[/imath] but, how can I claim that [imath]41[/imath] is the smallest positive integer satisfying condition ? Please explain why
|
142133
|
Does this polynomial evaluate to prime number whenever [imath]x[/imath] is a natural number?
I am trying to prove or disprove following statment: [imath]x^2-31x+257[/imath] evaluates to a prime number whenever [imath]x[/imath] is a natural number. First of all, I realized that we can't factorize this polynomial using its square root like [imath]ax^2+bx+c=a(x-x_1)(x-x_2)[/imath] because discriminant is negative, also I used excel for numbers from 1 to 1530 (just for checking), and yes it gives me prime numbers, unfortunately I dont know what the formula is, which evaluates every natural number for given input, maybe we may write [imath]x_{k+1}=k+1[/imath] for [imath]k=0,\ldots\infty[/imath], but how can I use this recursive statment? I have tried instead of [imath]x[/imath], put [imath]k+1[/imath] in this polynomial, and so I got [imath](k+1)^2-31k+257=k^2+2k+1-31k+257=k^2-29k+258[/imath] but now this last polynomial for [imath]k=1[/imath] evaluates to [imath]259-29=230[/imath], which is not prime, but are my steps correct? Please help me.
|
1886683
|
Let [imath]T: V \rightarrow W [/imath] be a linear mapping such that [imath]T^2=T[/imath], show that [imath]V = M \oplus N[/imath].
Let [imath]T: V \rightarrow V [/imath] be a linear mapping such that [imath]T^2=T[/imath]. If [imath]M = Image(T)[/imath] and N = [imath]Ker(T)[/imath], I want to show that [imath]V = M \oplus N[/imath]. Not sure how to proceed.
|
261704
|
Show that the direct sum of a kernel of a projection and its image create the originating vector space.
I got the following question as my homework. Given [imath]V[/imath] is a vector space with [imath]P \in \operatorname{End} V[/imath]. [imath]P \circ P = P[/imath] ("P is idempotent"). Show that [imath]V = \operatorname{Ker} P \oplus \operatorname{Im} P[/imath]. One [imath]P[/imath] I can imagine is a projection from 3d-space to plane, that just sets some coordinates to zero. For example [imath]\begin{pmatrix} x \\ y \\ z\end{pmatrix} \mapsto \begin{pmatrix}x \\ y \\ 0\end{pmatrix}[/imath]. Then [imath]\operatorname{Ker} P[/imath] would give the line [imath]\begin{pmatrix} 0 \\ 0 \\ z\end{pmatrix}[/imath] and [imath]\operatorname{Im} P[/imath] would contain all [imath]\begin{pmatrix}x \\ y \\ 0\end{pmatrix}[/imath]. So the result of [imath]\operatorname{Ker} P \oplus \operatorname{Im} P[/imath] is of course [imath]V[/imath]. But how do I prove that in a mathematical way?
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1885700
|
Cardinality of power set of [imath]\mathbb N[/imath] is equal to cardinality of [imath]\mathbb R[/imath]
How to prove that; The cardinality of power set of [imath]\mathbb N[/imath] is equal to cardinality of [imath]\mathbb R[/imath] I need a reference or proof outline. I looked up but didn't find results thanks.
|
2978624
|
Find a bijection [imath]^\mathbb{N}\{0,1\} \to \mathbb{R}[/imath]
Let [imath]^\mathbb{N}\{0,1\}[/imath] be defined as [imath]\{f~|~ f:\mathbb{N}\to\{0,1\}\}[/imath], i.e., all the functions that map [imath]\mathbb{N}[/imath] to the set [imath]\{0,1\}[/imath]. We want to show that this set is in bijection with [imath]\mathbb{R}[/imath]. I've been having trouble figuring this out. If we can show there is an injection from [imath]^\mathbb{N}\{0,1\}\to\mathbb{R}[/imath] and an injection [imath]\mathbb{R}\to ~^\mathbb{N}\{0,1\}[/imath], then we can invoke the Schroder-Bernstein theorem to complete the argument that the two sets are in bijection with one another. We can represent each [imath]f\in~^\mathbb{N}\{0,1\}[/imath] as the dyadic expansion of a real number [imath]r_j\in[0,2][/imath], i.e., [imath] r_j = \sum_{k=0}^\infty \frac{f(k)}{2^k}[/imath] But [imath]r_j[/imath] may be mapped to by more than one function, so I am unsure as to how to proceed. Any help is appreciated.
|
172471
|
Solving [imath]E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}[/imath]
[imath]E=\frac{1}{\sin10^\circ}-\frac{\sqrt3}{\cos10^\circ}[/imath] I got no idea how to find the solution to this. Can someone put me on the right track? Thank you very much.
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2139054
|
To prove [imath]\frac{1}{\sin 10^\circ}-\frac{\sqrt 3}{\cos 10^\circ}=4[/imath]
To prove: [imath]\frac{1}{\sin 10^\circ}-\frac{\sqrt 3}{\cos 10^\circ}=4[/imath] I tried taking lcm but could not get to anything.
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1887310
|
[imath]\lim_{x\to \infty}f(x)[/imath] exists. [imath]\lim_{x\to \infty}f'(x)[/imath] does not exist.
Let [imath]f : \mathbb{R} \to \mathbb{R}[/imath] be differentiable function. If [imath]\lim_{x\to \infty}f(x) = 0[/imath] , then [imath]\lim_{x \to \infty} f^{'}(x) [/imath] exists? I think the answer is No. Because I think there may exists function [imath]f[/imath] alternanting its sign as [imath]x \to \infty[/imath]. Could you give me a such function or Prove above statement?
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851327
|
Show that it is possible that the limit [imath]\displaystyle{\lim_{x \rightarrow +\infty} f'(x)} [/imath] does not exist.
Let [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] a differentiable function with continuous derivative and the limit [imath]\displaystyle{\lim_{x \rightarrow +\infty} f(x) }[/imath] exists. Show with an example that it is possible that the limit [imath]\displaystyle{\lim_{x \rightarrow +\infty} f'(x)} [/imath] does not exist. My attempt: [imath]f(x)=\int_{-\infty}^{x} e^{t^2}dt[/imath] [imath]\lim_{x \to +\infty} f(x)=\lim_{x \to +\infty} \int_{-\infty}^{+\infty} e^{t^2}dt=\frac{\sqrt{\pi}}{2}[/imath] [imath]\lim_{x \to +\infty} f'(x) =\lim_{x \to +\infty} e^{x^2}= +\infty \notin \mathbb{R}[/imath] Is my attempt right?
|
1887077
|
[imath]G[/imath] is abelian group of order [imath]pq[/imath], where [imath]p[/imath] and [imath]q[/imath] prime. Prove that [imath]G[/imath] cyclic
Let [imath]G[/imath] be an abelian group of order [imath]pq[/imath], where [imath]p[/imath] and [imath]q[/imath] are two distinct prime numbers. Prove that [imath]G[/imath] is cyclic. (Hint: use a theorem of Sylow: if [imath]G[/imath] is a group of order [imath]n[/imath] and [imath]p[/imath] is prime, and if [imath]p[/imath] divides [imath]n[/imath], then [imath]G[/imath] has en element of order [imath]p[/imath].) Attempt at proof: Since [imath]p[/imath] divides [imath]pq[/imath], Sylow says that [imath]G[/imath] has an element [imath]g_1[/imath] with order [imath]p[/imath]. Since [imath]q[/imath] divides [imath]pq[/imath], [imath]G[/imath] has also an element [imath]g_2[/imath] with order [imath]q[/imath]. Now consider the element [imath]g = g_1 g_2[/imath]. Then this element has order [imath]pq[/imath], since [imath]G[/imath] is abelian we have [imath]g^{pq} = (g_1 \ast g_2)^{pq} = (g_1^{pq} \ast g_2^{pq}) = (e^q \ast e^p) = (e \ast e) = e. [/imath] So I have shown that [imath]G[/imath] has an element with the same order as the order of the group. Can I conclude from this that [imath]G[/imath] is cyclic? Or do I still need to show explicitly that [imath]g[/imath] is a generator for [imath]G[/imath]?
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1502186
|
Structure of groups of order [imath]pq[/imath], where [imath]p,q[/imath] are distinct primes.
I don't know about the Sylow Theorems. But I have been wondering about a proof of the fact that a group or order [imath]pq[/imath] where [imath]p[/imath] and [imath]q[/imath] are distinct primes must be cyclic. I can't quite work out the details, but here is the general idea. I would like help with filling in details. I assume that it is already known that [imath]G[/imath] has subgroup(s) of order [imath]p[/imath] and subgroup(s) of order [imath]q[/imath]. If [imath]G[/imath] is a group of order [imath]pq[/imath] ([imath]p\neq q[/imath]), then I know that [imath]G[/imath] has a subgroup [imath]H[/imath] of order [imath]p[/imath] and a subgroup [imath]K[/imath] of order [imath]q[/imath]. Then [imath]H\simeq \mathbb{Z}_p[/imath] and [imath]K\simeq \mathbb{Z}_q[/imath]. But then [imath]H\oplus K \simeq \mathbb{Z}_{pq}[/imath], so I would think that [imath]H\oplus K \simeq G[/imath]. I guess one could do an internal direct product instead of an external direct product, but I don't know that [imath]H[/imath] and [imath]K[/imath] are normal subgroups. I am asking for help completing this argument. Edit: I see from the comments below that I might need to assume that the smaller prime does not divide the larger prime minus [imath]1[/imath]. Or maybe it is enough to assume that the primes are greater than or equal to [imath]3[/imath] (Still distinct).
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1887265
|
Prove that if [imath]A=BQ[/imath] then [imath]A=B[/imath]
I'm having some trouble with a question in my linear algebra textbook: Let [imath]A[/imath] and [imath]B[/imath] be positive definite matrices, and let [imath]Q[/imath] be a unitary matrix. Prove that if [imath]A=BQ[/imath], then [imath]A=B[/imath]. I tried to play around with the given equality, but I'm not getting any results. Tips would be appreciated:)
|
555347
|
Given [imath]A[/imath] and [imath]B[/imath] positive-definite matrices and [imath]Q[/imath] unitary matrix, prove that if [imath]A = BQ[/imath], then [imath]A=B[/imath].
Given [imath]A[/imath] and [imath]B[/imath] positive-definite matrices and [imath]Q[/imath] unitary matrix, prove that if [imath]A = BQ[/imath], then [imath]A=B[/imath]. [imath]Q[/imath] is unitary, so [imath]QQ^*=I[/imath] If [imath]A[/imath] and [imath]B[/imath] are positive-definite, than [imath]A=A^*[/imath] and [imath]B=B^*[/imath]. [imath]A^*=(BQ)^*=Q^*B^*[/imath] [imath]A^2=AA=AA^*=(BQ)(Q^*B^*)=B(QQ^*)B^*=BIB^*=B^2[/imath] [imath]A^2=B^2[/imath]. I don't know how to use the fact that A and B are positive-definite to finish the proof.
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329975
|
What are the Complex (Non-Real) Eigenvectors of [imath]3\times 3[/imath] Rotation Matrices?
A [imath]3\times 3[/imath] rotation matrix [imath]R[/imath] that rotates [imath]\mathbb{R}^3[/imath] around the unit vector [imath]v\in\mathbb{R}^3[/imath] by angle [imath]\theta[/imath] (as defined by Rodrigues' rotation formula) satisfies the eigendecomposition [imath] R=W\Sigma W^\mathrm{*} \enspace, [/imath] where [imath] W=\left(\begin{matrix}v \;|\; x \;|\; y \end{matrix}\right) [/imath] is a unitary matrix of eigenvectors, and [imath] \Sigma=\mathrm{diag}\left(1,e^{i\theta},e^{-i\theta}\right) [/imath] is the matrix of the corresponding eigenvalues. What expressions define the non-real eigenvectors [imath]x[/imath] and [imath]y[/imath]?
|
2571316
|
How to show eigenvalues of any rotation matrix of [imath]3 \times 3[/imath] are [imath]\{ 1, e^{i\theta}, e^{-i\theta}\}[/imath]
Eigenvalues of any rotation matrix of [imath]3 \times 3[/imath] are [imath]\{ 1, e^{i\theta}, e^{-i\theta}\}[/imath] where [imath]\theta[/imath] is the rotation angle, and corresponding eigenvectors are [imath]{a,I,J}[/imath] which are rotation axis and circular points for the plane orthogonal to [imath]a[/imath] respectively. In a 2-dimensional projective plane [imath]\Bbb P^2[/imath], [imath]I,J[/imath] can be written as [imath]I=[1,i,0]^T,J=[1,-i,0]^T[/imath] To my understanding, [imath]Ra=a[/imath] accounts for eigenvalue 1 and corresponding eigenvector [imath]a[/imath], but I cannot figure out other cases. Can anyone give a brief proof? Original problem comes from R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision at page 628.
|
1887460
|
Angle Bisectors
A circle is drawn that intersects all three sides of [imath]\triangle PQR[/imath] as shown below. Prove that if [imath]AB = CD = EF[/imath], then the center of the circle is the incenter of [imath]\triangle PQR[/imath].
|
905555
|
Proof of Incircle
A circle is drawn that intersects all three sides of [imath]\triangle PQR[/imath] as shown below. Prove that if AB = CD = EF, then the center of the circle is the incenter of [imath]\triangle PQR[/imath]. Designate the center of the circle [imath]G[/imath]. Thinking about it a bit, we realize if we could prove the incircle of the triangle and the circle given are concentric, we would know that the center of the circle is the incenter, for they have the same center. If we shrink the circle given down to the incircle, the incircle should intersect the triangle at the midpoints of [imath]AB[/imath], [imath]CD[/imath], and [imath]EF[/imath]. We therefore denote the midpoint of [imath]AB M[/imath], the midpoint of [imath]CD L[/imath], andd the midpoint of [imath]EF N[/imath]. We know the incircle is concentric to the given circle if and only if [imath]NG[/imath] and [imath]LG[/imath] are congruent, because a circle is a circle if and only if the radii are congruent. Thus, we must prove [imath]NG\cong LG[/imath]. When proving lengths equal, congruent triangles are always a good idea. We see that [imath]NGE\cong LGD[/imath] seems likely, and it would also prove our answer if we could prove it. We know that [imath]GD\cong GE[/imath] because they are both radii of the larger circle, and we also know [imath]DL\cong EN[/imath] because they are both half of two lines that are congruent. We know [imath]\angle GLD[/imath] and [imath]\angle GNE[/imath] are both right angles, because that is where the smaller circle is tangent to the sides of the triangle. Thus, [imath]NGE\cong LGD[/imath] by [imath]HL[/imath] congruency. Because congruent parts of congruent triangles are congruent, we know [imath]NG\cong LG[/imath], which proves [imath]G[/imath] is the center of the incircle. Is this proof sound?
|
1887675
|
Any Implications of Fermat's Last Theorem?
In our discourse FLT is Fermat's Last Theorem. I am unaware of any theorems or conjectures that begin assuming FLT is true, or otherwise use FLT as a starting point or tool. The small amount of literature review I've done on this question reveals nothing. My question is: Where can I find a work requiring FLT, or some useful implication of FLT? Even an implication of a polynomial inequality, that may not be FLT, may be a good answer to this question, as I'll likely try to use it to find something regarding FLT. The following is not acceptable as an answer to this question: [imath] (a^{x_{1}}_{1} + b^{x_{1}}_{1} - c^{x_{1}}_{1}) \ldots (a^{x_{n}}_{n} + b^{x_{n}}_{n} - c^{x_{n}}_{n}) \not= 0 : x_{i} > 2 [/imath] and it's expansions imply (something trivial) Another acceptable answer to this question would be a proof requiring FLT to be false. Thanks and please let me know if I can ask this question in a way more fitting math.se (new user).
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47761
|
Fermat's Last Theorem: implications (there is no new proof)
I am not experienced in Number Theory but what I know is that some results of this filed are applicable in other areas, e.g. algebra. For sure FLT made (and makes) people be interested in Number Theory leading to the development of new methods which can be themselves applied not only for the proof of FLT (like financial problems motivated somehow development of stochastic analysis). I am interested - if there are applications or implications of FLT itself? More precisely: if the fact "for each [imath]n\geq3[/imath] there are no integer solutions of [imath]a^n+b^n=c^n[/imath]" leads to solutions of problems which are not in the field of Number Theory? I would specify that I wonder about some problems which are already formulated: since FLT is known for more then 300 years I am pretty sure that there were formulated hypothesis which follow from FLT directly (if there are such hypothesis not in the field of Number Theory).
|
1887395
|
Show that [imath]h[/imath] is entire and [imath]f[/imath] is a constant function.
Let [imath]D=\{Z\in\mathbb{C}:|z|<1\}[/imath]. Suppose [imath]f:\overline D\to\mathbb{C}[/imath] is continuous, holomorphic in [imath]D[/imath], and [imath]f(\partial D)\subset \mathbb{R}[/imath]. Define [imath]h:\mathbb{C}\to\mathbb{C}[/imath] by [imath] h(z)=\begin{cases} f(z) &\text{if}\ z\in \overline D\\ \overline{f(1/\overline z)} &\text{if}\ z\in\mathbb{C}\setminus \overline{D} \end{cases}[/imath] Show that [imath]h[/imath] is entire and [imath]f[/imath] is a constant function. My work: I can show that [imath]h[/imath] is holomorphic in [imath]\mathbb{C}\setminus \overline{D}[/imath]. To show [imath]h[/imath] is entire, by using Morera's theorem, considering arbitrary triangular path, the integral of [imath]h[/imath] is zero if it does not cross the boundary of the unit disc, however, what if the triangular path cross the unit disc? How to show in this case the integral is also zero. Could anyone kindly help? Thanks!
|
363288
|
Function [imath] f [/imath] defined on unit disk [imath] D [/imath] has real values on [imath] \partial D [/imath], why is [imath] f [/imath] real valued?
I have a function [imath] f [/imath] that is analytic on the unit disc and real valued on the unit circle. I want to prove that [imath] f [/imath] is real valued on the unit disk. I know that I need to use the identity theorem, but I am unsure how. The only thing I can think to do is: Let [imath] g [/imath] be some real valued function s.t. [imath] g(e^{i\theta}) = f(e^{i\theta}) [/imath] then [imath] f = g [/imath], but this argument obviously isn't sound. Any help would be greatly appreciated.
|
1888053
|
What does the symbol [imath]\sqsubset[/imath] mean? Is it the same as [imath]\subset[/imath]?
I have tried looking for this symbol but I found the definition for latter only. Hence, I have this doubt if these two are same. If not, can anyone please let me know the difference? Thanks a lot.
|
1569400
|
Does [imath]\sqsubset[/imath] have any special meaning?
What is the meaning of [imath]\sqsubset[/imath] and [imath]\sqsubseteq[/imath]? Does it have any special meaning, or is it just an alternative to writing [imath]\subset[/imath] and [imath]\subseteq[/imath] respectively (for proper subsets and subsets)? I have been looking for an explanation everywhere, but so far I could not find it. This may have to do with the fact that I am not even sure what this symbol is called (makes it difficult to search for it), but I have tried several things (like searching for [imath]\sqsubset[/imath] on this site), and nothing came up, other than lists of mathematical symbols for LaTeX without any explanation. I have seen it used in papers (e.g., http://www.cril.univ-artois.fr/~marquis/everaere-konieczny-marquis-ecai10.pdf on page 4, footnote 5), but never explained. I am starting to think that [imath]\sqsubset[/imath] and [imath]\sqsubseteq[/imath] are equivalent to [imath]\subset[/imath] and [imath]\subseteq[/imath]. However, sometimes there are subtle differences, so I want to be certain about this. I want to be sure that I understand the intended meaning when reading future papers, to avoid any misunderstandings. Thanks in advance.
|
1888002
|
About the topology of the Heisenberg group
By definition, if [imath]A[/imath] is simply-connected then [imath]A[/imath] is connected. I would like to know why in the Wikipedia we find this phrase: "The Heisenberg group is a connected, simply-connected Lie group ...", because it is sufficient to say "The Heisenberg group is a simply-connected Lie group ..." only. My problem is, why connected and simply-connected (both at the same time), it is sufficient to write just simply-connected. Thank you in advance
|
936707
|
Is "connected, simply connected" Redundant?
Here are my definitions of "connected" and "simply connected." A topological space [imath]X[/imath] is connected if and only if it is not the union of two nonempty disjoint open sets. A topological space [imath]X[/imath] is simply connected if and only if it is path-connected and has trivial fundamental group (i.e. [imath]\pi_1(X)\simeq\{\mathrm{e}\}[/imath] and [imath]|\pi_0(X)|=1[/imath]). It is a classic and elementary exercise in topology to show that, if a space is path-connected, then it is connected. Thus, if a space is simply connected, then it is connected. Yet, despite this implication, I've read several cases where the words "connected, simply connected" appear together. For example, in Kobayashi and Nomizu's Foundations of Differential Geometry, Volume 1, page 252, the following is written: "Let [imath]M[/imath] be a connected, simply connected analytic manifold with an analytic linear connection." In this paper on [imath]T^3[/imath] actions and this paper on rotationally symmetric manifolds (Links to Journal Storage, also known as JSTOR), they use "connected, simply connected" on their first page. The first of these takes it a step further, referring to "closed, compact, connected, simply connected [imath]4[/imath]-manifolds," where a closed manifold (unfortunate terminology) is defined as a compact manifold without boundary. My question is simple: Why use the words "connected, simply connected" when being simply connected implies connectedness? Thank you.
|
1884277
|
Improper Integral from Gradshteyn and Ryzhik
This is the integral one can find in the Introduction of 'Special Integrals of Gradshteyn and Ryzhik the Proofs - Volume I' by Victor H. Moll: [imath]\int_0^{+\infty} \frac{dx}{(1+x^2)^{3/2} \left[ \phi(x) + \sqrt{\phi(x)} \right]^{1/2}}, \quad \phi(x) = 1 + \frac{4x^2}{3(1+x^2)^2} \;.[/imath] The author doesn't know the final answer. It is claimed that it is [imath]\pi / 2 \sqrt{6}[/imath], though numerical integration contradicts this. Any ideas how to solve it or where to find clues?
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879618
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An incorrect answer for an integral
As the authors pointed out in this paper (p. 2), the following evaluation which was in Gradshteyn and Ryzhik (sixth edition) is incorrect (and has been removed). [imath] ''\int_{0}^{\infty}\frac{1}{\left(1+x^{2}\right)^{3/2}} \frac{1}{\sqrt{1+ \frac{4 x^{2}}{3\left(1+x^{2}\right)^{2}}+\sqrt{1+ \frac{4 x^{2}}{3\left(1+x^{2}\right)^{2}}}}} \mathrm{d}x = \frac{\pi}{2\sqrt{6}}'' \qquad (*) [/imath] A numerical evaluation gives [imath]0.6663771 \cdots[/imath] on the left hand side and [imath]0.64127491 \cdots [/imath] on the right hand side. I have not succeeded to correct [imath](*)[/imath]. Do you have any idea on how to evaluate the above integral?
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