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1907198
|
Isomorphism between general linear groups
Let [imath]GL_{n}(\mathbb{F})[/imath] be the set of all [imath]n \times n[/imath] invertible matrices over a field [imath]\mathbb{F}[/imath] of characteristic [imath]0[/imath]. Whether [imath]GL_{r}(\mathbb{F})[/imath] is isomorphic to [imath]GL_{s}(\mathbb{F})[/imath] for [imath]r \neq s[/imath]? I think that [imath]GL_{r}(\mathbb{F}) \not\cong GL_{s}(\mathbb{F})[/imath] for [imath]r \neq s[/imath], but I don't know how to prove it. I need to a strict proof.
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1686695
|
Is there a simple way of proving that [imath]\text{GL}_n(R) \not\cong \text{GL}_m(R)[/imath]?
Letting [imath]\mathbb{F}_{1}[/imath] and [imath]\mathbb{F}_{2}[/imath] be fields, and letting [imath]n \geq 3[/imath] and [imath]m[/imath] be natural numbers, it is known that [imath]\text{GL}_{m}(\mathbb{F}_{1})[/imath] and [imath]\text{GL}_{n}(\mathbb{F}_{2})[/imath] are elementarily equivalent if and only if [imath]m=n[/imath] and [imath]\mathbb{F}_{1} \equiv \mathbb{F}_{2}[/imath] (as proven in "Elementary Properties of Linear Groups" in the collection "The Metamathematics of Algebraic Systems — Collected Papers: 1936–1967"). So, given a field [imath]\mathbb{F}[/imath], if [imath]n \neq m[/imath], then [imath]\text{GL}_{m}(\mathbb{F}) \not\equiv \text{GL}_{n}(\mathbb{F})[/imath], and thus [imath]\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})[/imath]. Letting [imath]R[/imath] be a commutative ring (with unity), and letting [imath]n, m \in \mathbb{N}[/imath] be such that [imath]n \neq m[/imath], is there a simple "algebraic" way of proving that [imath]\text{GL}_{m}(R)[/imath] and [imath]\text{GL}_{n}(R)[/imath] are not isomorphic (as groups)? Is there a simple group-theoretic way of showing that [imath]\text{GL}_{m}(\mathbb{F}) \not\cong \text{GL}_{n}(\mathbb{F})[/imath] for a field [imath]\mathbb{F}[/imath]? Certain special cases of this problem trivially hold, for example in the case whereby [imath]\mathbb{F}[/imath] is finite, in which case [imath]|\text{GL}_{m}(\mathbb{F})| \neq |\text{GL}_{n}(\mathbb{F})|[/imath].
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1907341
|
singular cohomology groups of the torus
Let [imath]X:=S^1 \times S^1[/imath] be the torus. Is the zero and second singular cohomology group [imath]\mathbb{Z}[/imath], the first [imath]\mathbb{Z}^2[/imath] and otherwise vanishing? To show this: One first computes singular homology which gives that the zero and second homology group equals [imath]\mathbb{Z}[/imath] and first homology group equals [imath]\mathbb{Z}^2[/imath]. Using universal coefficient theorem and that homology groups are free over [imath]\mathbb{Z}[/imath], we get the desired. I was just wondering, because I wanted to compute cohomology groups of the sheaf of locally constant function with values in [imath]\mathbb{Z}[/imath], which should coincide with the singular cohomology group, but in this case I get that the first cohomology group equals [imath]\mathbb{Z}^4[/imath].
|
41284
|
Homology groups of torus
I computed the homology groups of the torus, can someone tell me if this is correct? The calculation, not the result that is. Thanks! The cells of [imath]T^2[/imath] are [imath]e^0, e^1_a, e^1_b, e^2[/imath] The chain groups are [imath] C_0(T^2) = \{ k e^0 | k \in \mathbb{Z} \} = \mathbb{Z}[/imath] [imath] C_1(T^2) = \{ k_1 e^1_a + k_2 e^1_b | k_1 , k_2 \in \mathbb{Z} \} = \mathbb{Z} \oplus \mathbb{Z}[/imath] [imath] C_2(T^2) = \mathbb{Z}[/imath] [imath] C_k(T^2) = \{ 0 \} , k > 2[/imath] Now the homology groups: [imath] H_0(T^2) = \ker \partial_0 / im \partial_1 = \mathbb{Z} / 0 = \mathbb{Z}[/imath] where [imath]im \partial_1 = 0[/imath] because there is no chain in [imath]C_1(T^2)[/imath] whose boundary is a zero-chain in [imath]C_0(T^2)[/imath]. (Is this reasoning correct?) [imath] H_1(T^2) = \ker \partial_1 / im \partial_2 = \mathbb{Z} \oplus \mathbb{Z}[/imath] where [imath]\ker \partial_1 = \mathbb{Z} \oplus \mathbb{Z} [/imath], i.e. again everything maps to zero because there is no element in [imath]C_1[/imath] whose boundary maps to an element in [imath]C_0[/imath]. [imath]im \partial_2 = 0[/imath] again because there is no element in [imath]C_2[/imath] whose boundary is an element of [imath]C_1[/imath]. I don't want to use Hurewicz to do [imath]H_1[/imath]. [imath] H_2(T^2) = \mathbb{Z}[/imath] using similar arguments as above. Thanks for your help. Edit I posted the computations as an answer below. I got two up votes but I don't know by who so I'm not yet sure I can trust them...
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1907251
|
If [imath]A=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)[/imath], find the rank of [imath]A+A^2[/imath]
If [imath]A=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)[/imath], find the rank of [imath]A+A^2[/imath] Note: Ans is given as Rank=2. [imath]A(e_1) = e_3, A(e_2) = e_1, A(e_3) = e_2[/imath]. Thus [imath](A+A^2)(e_1) = e_3+e_2[/imath], [imath](A+A^2)(e_2) = e_1+e_3[/imath] and [imath](A+A^2)(e_3) = e_2+e_1[/imath]. Thus [imath]A+A^2[/imath] has rank .....??????.
|
1907112
|
If [imath]A=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)[/imath], find the rank of [imath]A+A^2+A^3[/imath]
If [imath]A=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right)[/imath], find the rank of [imath]A+A^2+A^3[/imath] We have [imath]\rho(A+A^2+A^3)\leq\rho(A)+\rho(A^2)+\rho(A^3) [/imath] where [imath]\rho(A)[/imath] means rank of A. Here [imath]\rho(A)=3.[/imath] Also [imath]\rho(A^2)=3[/imath] and [imath]\rho(A^3)=3[/imath] as [imath]\det A\neq 0[/imath] and [imath]\det(A^2)=\det(A)\det(A)\neq 0[/imath] and so on. But how to get the desired result.
|
1907934
|
Prove that [imath]\pi[/imath] is irrational
I haven't found a way, except using advanced methods, to prove the irrationality of [imath]\pi[/imath]. It is possible to prove that [imath]\pi[/imath] is irrational, using only elementary algebra?
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1364970
|
Help with proving that [imath]\pi[/imath] is irrational
I was trying to prove that [imath]\pi[/imath] is irrational, just to see if I could do it. So far, I've tried to do this by using the fact that the sum [imath]S=\sum\limits_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}[/imath] and arguing that [imath]S[/imath] is irrational instead, and thus implying that [imath]\pi[/imath] is also irrational. To do this, I thought I could use partial sums [imath]S_n[/imath] of [imath]S[/imath]: [imath]S_n=\sum\limits_{k=1}^n \frac{1}{k^2}=\frac{A_n}{B_n}[/imath] where the fraction [imath]A_n/B_n[/imath] is written in lowest terms. We can note that the sequence [imath]\lbrace S_n \rbrace[/imath] is a strictly increasing sequence, with [imath]\pi^2/6[/imath] as its lowest upper bound, so I thought that maybe the sequences [imath]\lbrace A_n \rbrace[/imath] and [imath]\lbrace B_n \rbrace[/imath] both have no upper bound, and the sequences would tend to infinity, and if I were able to prove this, I thought I could use it to argue that [imath]S[/imath] is an irrational number, using the fact that every rational number can be written as the quotient between two finite integers. We note that [imath]A_n\geq B_n[/imath] for all natural [imath]n[/imath], as [imath]S_n\geq 1[/imath], so all we would really need to prove in that case is that [imath]\lbrace B_n \rbrace \rightarrow \infty[/imath](if this assertion is even correct). At first I thought that both [imath]\lbrace A_n \rbrace[/imath] and [imath]\lbrace B_n \rbrace[/imath] would be increasing sequences, but after checking with Maple I noticed that they weren't, sadly enough([imath]S_9=\frac{9778141}{6350400}[/imath] and [imath]S_{10}=\frac{1968329}{1270080}[/imath]). However, they do indeed seem to get very large very quickly, so I'm thinking that my hypothesis about [imath]\lbrace A_n \rbrace[/imath] and [imath]\lbrace B_n \rbrace[/imath] is correct. But I have trouble proving my hypothesis, and I'm kind of stuck, no knowing what to do. Is there a way to prove that [imath]\lbrace B_n \rbrace \rightarrow \infty[/imath]? And of course, is this approach to prove that [imath]\pi[/imath] is irrational logically sound, or is it fundamentally flawed in some important aspect? In the latter case, what idea should I try next?
|
1907905
|
Calculate [imath]\int_0^1 f(x) dx[/imath]
Function [imath]f(x)[/imath] is defined as: [imath]f(x) =0[/imath] when [imath]x=0[/imath], and [imath]f(x) = n[/imath] when [imath]x \in (\frac{1}{n+1}, \frac{1}{n}][/imath]. Calculate [imath]\int_0^1 f(x) dx[/imath]. Assuming that we can safely ignore when [imath]x=0[/imath], then when [imath]x\in (\frac{1}{n+1},\frac{1}{n}][/imath], we essentially have [imath]1\times\left(1 - \frac{1}{2}\right)+2\times\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots=\sum_{n \ge 1} n\left(\frac{1}{n}-\frac{1}{n+1}\right)=\sum_{n \ge 1}\frac{1}{n+1}[/imath] But then the series diverges by integral test. So I am confused.
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984839
|
Calculate [imath]\int_0^1f(x)dx[/imath]
Calculate [imath]\int_0^1f(x)dx[/imath],where [imath]\ f(x) = \left\{ \begin{array}{l l} 0 & \quad \text{if $x=0$ }\\ n & \quad \text{if $x\in(\frac{1}{n+1},\frac{1}{n}]$} \end{array} \right.[/imath] How we can calculate this integral? Is this simply [imath]\int_0^1f(x)dx=\int_{\frac{1}{n+1}}^{\frac{1}{n}}ndx=n\left(\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1}{n+1}[/imath]?
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1907949
|
Question from Herstein's Topics in Algebra on Sylow subgroups
Let [imath]G[/imath] be a finite group and suppose that [imath]f[/imath] is an automorphism of [imath]G[/imath] with order 3 that fixes only the identity element of [imath]G[/imath]. Show that [imath]G[/imath] has a single (i. e. normal) [imath]p[/imath]-Sylow subgroup for each prime [imath]p[/imath] dividing [imath]|G|[/imath]. Been working for hours and haven't made much progress. I did note that since [imath]f[/imath] considered as a permutation factors as the product of disjoint 3-cycles, 3 divides [imath]|G|[/imath]. So my next step is showing that the 3-Sylow subgroup is normal. Edit: sorry, since [imath]f[/imath] fixes the identity it actually factors into a bunch of 3-cycles and a 1-cycle. So 3 does not divide [imath]|G|[/imath].
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1182699
|
To show every sylow p-subgroup is normal in G
Let [imath]G[/imath] be a finite group and suppose that [imath]\phi[/imath] is an automorphism of [imath]G[/imath] such that [imath]\phi^3[/imath] is the identity automorphism. Suppose further that [imath]\phi(x)=x[/imath] implies that [imath]x=e[/imath]. Prove that for every prime [imath]p[/imath] which divides [imath]o(G)[/imath], the [imath]p[/imath]-Sylow subgroup is normal in G. Its an problem from Herstein, prob 19 in 2.12,2nd edition.I need some hint to solve this.Thank you.
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1908163
|
Suppose G is a finite abelian group with a nontrivial subgroup H contained in every subgroup of G. Show that G is cyclic.
The prompt is from Herstein's Topics in Algebra. The proof is to be done without using the structure theorem for finite abelian groups. I showed that [imath]H[/imath] was cyclic of order [imath]p[/imath] a prime and G was a [imath]p[/imath]-group. I then set [imath]|G| = p^n[/imath] and proceeded by induction on n. The case [imath]n=1[/imath] is clear. Then, by induction every proper subgroup of [imath]G[/imath] is cyclic. By the Sylow theorems [imath]G[/imath] therefore contains an element [imath]a[/imath] of order [imath]p^{n-1}[/imath]. We have [imath]G = \langle a\rangle \langle b\rangle [/imath], where [imath]b[/imath] is any element of [imath]G[/imath] not in [imath]\langle a\rangle [/imath] (here [imath]\langle a\rangle [/imath] denotes the subgroup generated by [imath]a[/imath]). Quotient by [imath]K[/imath], the intersection of [imath]\langle a\rangle [/imath] and [imath]\langle b\rangle [/imath], to see that [imath]G/K[/imath] is isomorphic to [imath]\Bbb{Z}_{p}^{2}[/imath], since [imath]G/K[/imath] is the direct product of two nontrivial cyclic p-groups ([imath]\langle a\rangle /K[/imath] and [imath]\langle b\rangle /K[/imath]) and by homomorphism every one of [imath]G/K[/imath]'s subgroups is cyclic. Therefore taking the preimage of each of the cyclic subgroups of [imath]G/K[/imath], each of which is a distinct maximal proper subgroup of [imath]G[/imath], [imath]G[/imath] must have [imath]p+1[/imath] distinct subgroups of order [imath]p^{n-1}[/imath]. Also, note that since [imath]b[/imath] was an arbitrary element not in [imath]\langle a\rangle [/imath] and [imath]\langle a\rangle /K[/imath] and [imath]\langle b\rangle /K[/imath] have the same number of elements, [imath]\langle a\rangle [/imath] contains every element of order less than [imath]p^{n-1}[/imath]. Since [imath]\langle a\rangle [/imath] has a unique subgroup of order [imath]p^m[/imath] where [imath]m<n-1[/imath], [imath]G[/imath] has a unique cyclic subgroup of order [imath]p^m[/imath]. This is as far as I've got so far. Not really sure where to go from here. Any ideas? Let me know if you need more details.
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948722
|
[imath]p[/imath] prime, Group of order [imath]p^n[/imath] is cyclic iff it is an abelian group having a unique subgroup of order [imath]p[/imath]
I've just read from An introduction to the theory of groups from Rotman's the following theorem: Theorem 2.19. Let [imath]p[/imath] be a prime. A group [imath]G[/imath] of order [imath]p^n[/imath] is cyclic if and only if it is an abelian group having a unique subgroup of order [imath]p[/imath]. Proof Necessity follows at once from Lemma 2.15. For the converse, let [imath]a \in G[/imath] have largest order, say [imath]p^k[/imath] (it follows that [imath]g^{p^k} = 1[/imath] for all [imath]g \in G[/imath]). Of course, the unique subgroup [imath]H[/imath] of order [imath]p[/imath] is a subgroup of [imath]\langle a\rangle[/imath]. If [imath]\langle a\rangle[/imath] is a proper subgroup of [imath]G[/imath], then there is [imath]x \in G[/imath] with [imath]x \not \in \langle a\rangle[/imath] but with [imath]x^p \in \langle a\rangle[/imath]; let [imath]x^p = a^l[/imath]. If [imath]k = 1[/imath], then [imath]x^p = 1[/imath] and [imath]x \in H \subset \langle a\rangle[/imath], a contradiction; we may, therefore, assume that [imath]k>1[/imath]. Now [imath]1 = x^{p^k}=(x^p)^{p^{k-1}}=a^{lp^{k-1}},[/imath] so that [imath]l = pm[/imath] for some integer [imath]m[/imath], by Exercise 2.13. Hence, [imath]x^p = a^{mp}[/imath], and so [imath]1 = x^{-p}a^{mp}[/imath]. Since [imath]G[/imath] is abelian, [imath]x^{-p}a^{mp} = (x^{-1}a^m)^p[/imath], and so [imath]x^{-1}a^m \in H \subset \langle a\rangle[/imath]. This gives [imath]x \in \langle a\rangle[/imath], a contradiction. Therefore, [imath]G = \langle a\rangle[/imath] and hence is cyclic. As usual, there is some trivial part of the proof which I don't understand: I don't see why "If [imath]\langle a\rangle[/imath] is a proper subgroup of [imath]G[/imath], then there is [imath]x \in G[/imath] with [imath]x \not \in \langle a\rangle[/imath] but with [imath]x^p \in \langle a\rangle[/imath]", I understand the rest of the proof. I would appreciate if someone could explain to me that statement.
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1908524
|
Congruence with a power of 2 as exponent
Anyone knows any elementary way of solving this? What's the least integer [imath]a>1[/imath] for some [imath]n[/imath] positive integer such that [imath]a^{2^n}-1[/imath] is a multiple of 2015?
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1908499
|
Solving a congruence without Fermat's little theorem
Given [imath]n\in\Bbb N[/imath], what is the least [imath]a>1[/imath] with [imath]a^{2^n}\equiv1\bmod2015[/imath]? Is there a solution not using Fermat's little theorem or the Chinese remainder theorem, any ideia?
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1908607
|
If [imath]f^2[/imath] is analytic in [imath]\Omega[/imath] then so is [imath]f[/imath]-Problem in proof Understanding
Problem in understanding the following proof in my book: If [imath]f^2[/imath] is analytic in [imath]\Omega[/imath] then so is [imath]f[/imath] where [imath]f[/imath] is continuous in domain [imath]\Omega[/imath] Zeros of [imath]f[/imath] are isolated . Let [imath]z_0[/imath] be a zero of [imath]f[/imath] then [imath]f(z_0)=0\implies f^2(z_0)=0[/imath].Now zeros of an analytic function are isolated hence zeros of [imath]f^2[/imath] are also isolated. And so zeros of [imath]f[/imath] are also isolated.Done [imath]f[/imath] has a derivative at each point at each point where it does not vanish. Unable to understand this fact Using Riemann's singularity theorem we have that [imath]f[/imath] is analytic in [imath]\Omega[/imath]. I am unable to understand that how the fact that If [imath]f[/imath] has a removable singularity at [imath]z_0[/imath],then [imath]f[/imath] is bounded in a ngbd of [imath]z_0[/imath] proves that [imath]f[/imath] is analytic in [imath]\Omega[/imath].
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307084
|
If [imath]f^2[/imath] is an analytic function then so is [imath]f[/imath]
I want to show the following: If [imath]f(z) [/imath] is a continuous function on a connected open subset of the complex plane and [imath]f(z)^2[/imath] is an analytic function, then [imath]f(z)[/imath] is analytic. Clearly if [imath]f(z) \neq 0[/imath] then [imath]\frac{f(z+h)-f(z)}{h}=\frac{f(z+h)^2-f(z)^2}{h}.\frac{1}{f(z+h)+f(z)}[/imath] which shows [imath]f^\prime(z) [/imath] exists if [imath]f(z)\neq0[/imath]. What do I do when [imath]f(z) =0[/imath]?
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1908639
|
A bogus proof of countable power set
I know that [imath]2^\mathbb{N}[/imath] is not countable but what is wrong with the following "proof"? Consider the following map [imath]f: \mathbb{N} \to 2^\mathbb{N}[/imath], which maps a natural number [imath]n[/imath] to a set of integers which correspond to the index of non-zero bits in [imath]n[/imath]'s binary representation: [imath]f(n)=\{i ~|~ n = \sum_{j=0}^\infty k_j2^j \text{ and } k_i=1\}[/imath] [imath]f(n)[/imath] is clearly "onto" since for any [imath]S\in 2^\mathbb{N}[/imath], there exists a natural number [imath]n = \sum_{i\in S} 2^i[/imath] s.t. [imath]f(n)=S[/imath]. However, if [imath]f(n)[/imath] is onto, then this is effectively proving that [imath]2^\mathbb{N}[/imath] is countable, which is, of course, false. Where was the mistake?
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1685222
|
What is wrong with this proof that the power set of the natural numbers is countable?
Here I offer a "proof" of why the power set of the natural numbers is countable by creating a supposed bijection. I know that my proof is wrong, but where? If anyone could point it out, that'd be awesome. Consider the power set of the natural numbers. We can form a bijection [imath]f:\mathbb{N}\to\mathcal{P}(\mathbb{N})[/imath] as follows. For any [imath]f(x)[/imath] convert [imath]x[/imath] to its binary representation, which is a string of [imath]0[/imath]s and [imath]1[/imath]s. If the [imath]i[/imath]-th digit is [imath]1[/imath], then include [imath]i[/imath] in the output set [imath]S=f(x)[/imath]. Otherwise do not include [imath]i[/imath]. Note that [imath]i[/imath] is [imath]0[/imath]-indexed and starts from the right-most digit. So in this case [imath]f(5)=\{1,3\}[/imath]. Supposedly we can construct any set in [imath]\mathcal{P}(\mathbb{N})[/imath] with an appropriate choice of [imath]x[/imath]. Where does this proof blow up?
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1909173
|
Let [imath]z[/imath] be a point in a metric space [imath](X, p)[/imath]. Why is [imath]f: X\to\mathbb{R}[/imath], [imath]f(x)=p(x,z)[/imath] uniformly continuous?
Let [imath]z[/imath] be a point in a metric space [imath](X, p)[/imath]. Why is [imath]f: X\to\mathbb{R}[/imath], [imath]f(x)=p(x,z)[/imath] uniformly continuous? I could draw a picture that illustrates the idea, please help me with the proof.
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8066
|
Is the distance function in a metric space (uniformly) continuous?
Let [imath](X, d)[/imath] be a metric space. Is the function [imath]x\mapsto d(x, z)[/imath] continuous? Is it uniformly continuous?
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1908341
|
A bijective function [imath]f:\Bbb R\to [0, \infty)[/imath] has infinitely many discontinuities.
A bijective function [imath]f:\Bbb R\to [0, \infty)[/imath] has infinitely many discontinuities. Suppose that [imath]f[/imath] has finitely many discontinuities say [imath]\{x_1,x_2,\ldots ,x_n\}[/imath].Then [imath]f[/imath] is continuous on [imath](-\infty,x_1)\cup (x_1,x_2)\cup \ldots \cup (x_{n-1},x_n)\cup (x_n,\infty)[/imath]. So [imath]f[/imath] is monotone on each of the sub-intervals. Obviously [imath]f(x_i,x_j)\cap f(x_j,x_k)=\emptyset[/imath] since [imath]f[/imath] is bijective. Now [imath]f(x_i,x_j)[/imath] will be an interval since [imath]f[/imath] is injective.Also it will be an open interval since [imath]f[/imath] is continuous. Now [imath][0,\infty)\setminus\cup \{(f(x_i,x_j))\}\cup f(-\infty ,x_1)\cup f(x_n,\infty)[/imath] must have [imath]n+1[/imath] points since [imath]f(x)\neq 0\forall x[/imath] and [imath]f(x)\neq f(x_1),f(x_2)\ldots f(x_n)\forall x[/imath] But [imath]\Bbb R\setminus \{x_i\}[/imath] has [imath] n[/imath] points. So we have a contradiction. Is the proof correct?Please suggest required edits
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8149
|
Points of discontinuity of a bijective function [imath]f:\mathbb{R} \to [0,\infty)[/imath]
We know that the points of discontinuity of a monotone function on an interval [imath][a,b][/imath] are countable. Using this can we prov that: Any bijection [imath]f: \mathbb{R} \to [0,\infty)[/imath] has infinitely many points of discontinuity. If yes, how or otherwise how to prove the above result?
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1909942
|
Redefining multiplication so that [imath]\mathbb{Z}[/imath] is a field?
My book (Finite Dimensional Vector Spaces by Halmos) says to redefine (if possible) addition and/or multiplication in order to make [imath]\mathbb{Z}[/imath] a field. The only field proprety that [imath]\mathbb{Z}[/imath] lacks is the existence of a multiplicative inverse. I defined multiplication as normal, with the exception that [imath]\forall ( a \not = 0) a \cdot 2 = 1)[/imath]. Since [imath]a \cdot 2 = a \cdot (1+1) = a+a[/imath], I also redefined addition as normal, except in the case of [imath]a+a[/imath], which is always equal to [imath]1[/imath]. But I noticed that this gives me a contradiction because we cannot have [imath]0+0=1[/imath]. What is a better way to solve this problem, and more importantly, how can I know that it won't give me a contradiction somewhere?
|
399703
|
Is it possible to make integers a field?
Is it possible to define addition and/or multiplication on the set of a) natural numbers (including [imath]0[/imath]: [imath]0,1,2,3,...[/imath]) b) integers [imath](..., -2, -1, 0, 1, 2, ...)[/imath] in such way that they will become fields? Thanks in advance.
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1910296
|
What does the symbol [imath]\ll[/imath] mean?
I came across, If [imath]|y| \ll 1[/imath], then [imath]x/y[/imath] may have large relative and absolute errors. I'm not sure what the symbol [imath]\ll[/imath] means. I looked it up on Google and nothing came up.
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1516976
|
Much less than, what does that mean?
What exactly does [imath]\ll[/imath] mean? I am familiar that this symbol means much less than. ...but what exactly does "much less than" mean? (Or the corollary, [imath]\gg[/imath]) On Wikipedia, the example they use is that [imath]1\ll 9999999999[/imath] But my thought on that is that [imath]10^{10^{10^{11}}}\ll 10^{10^{10^{11}}}+9999999999[/imath], based on the same logic. But I am confused because the numbers are comparitively. Logically, I am really kinda spent dealing with this issue of mine. Thanks for any input.
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1830274
|
How can I proceed to find a maximal principal ideal in [imath]\mathbb Z_{(2)}[x][/imath]?
How can I proceed to find a maximal principal ideal in [imath]\mathbb Z_{(2)}[x][/imath]? I know the answer in the sense that i know that [imath](2x+1)[/imath] is a maximal principal ideal of that polynomial ring. But if i didn't, how could i reach that result, i.e. how could i reason to answer?
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752342
|
Checking the maximality of a principal ideal in [imath]R[x][/imath]
Let [imath]R = \mathbb{Z}_{(2)}[/imath] be the localization of [imath]\mathbb{Z}[/imath] at the prime ideal generated by [imath]2[/imath] in [imath]\mathbb{Z}[/imath]. Then prove that the ideal generated by [imath](2x-1)[/imath] is maximal in [imath]R[x][/imath].
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1909669
|
How to prove that [imath]|HK| = \dfrac{|H| \; |K|}{|H \cap K|}[/imath]?
I am wondering how to prove the formula [imath]|HK| = \dfrac{|H| \; |K|}{|H \cap K|}[/imath] where [imath]H[/imath] and [imath]K[/imath] are finite groups. Let [imath]|H|[/imath] denote the order of [imath]H[/imath] and so on. I have seen proofs from these websites: (1) and (2). But I do not really understand the idea of the proof. Can someone explain the proof to me, please? Thanks so much.
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1359815
|
Why is it true that [imath]|AB:A|=|B:A\cap B|[/imath] even if [imath]A[/imath] is not normal in [imath]AB[/imath]? (Second Isomorphism Theorem)
I just read about the First and Second Isomorphism Theorems in the book Abstract Algebra by Dummit and Foote. After proving the Second Isomorphism Theorem, they said: Proposition 13 isn't really important for my question (I guess) but anyway it is the one that states that if [imath]H[/imath] and [imath]K[/imath] are two finite subgroups of [imath]G[/imath] then [imath]|HK|=\dfrac{|H||K|}{|H\cap K|}[/imath]. I didn't understand why [imath]|AB:A|=|B:A\cap B|[/imath]. It is obvious if [imath]AB[/imath] is finite because using the Second Isomorphism theorem we have [imath]|AB:B|=|A:A\cap B|[/imath] because [imath]AB/ B \cong A/ A\cap B[/imath]. Then [imath] |AB:B|=\dfrac{|AB|}{|B|}\quad\textsf{and}\quad|A:A\cap B|=\dfrac{|A|}{|A\cap B|}\\[0.3in] \implies\dfrac{|AB|}{|B|}=\dfrac{|A|}{|A\cap B|}\\[0.3in] \implies\dfrac{|AB|}{|A|}=\dfrac{|B|}{|A\cap B|} [/imath] which proves that [imath]|AB:A|=|B:A\cap B|[/imath], but this doesn't prove it if [imath]A[/imath] or [imath]B[/imath] is infinite and I believe there's a simple general proof that under the asssumptions in Theorem 18 we get [imath]|AB:A|=|B:A\cap B|[/imath]. Could anyone please help me? If the proof is easy (and I guess so because they stated the relation in the book like if it was obvious) then could you give me hints? Thank you in advance!
|
68099
|
Question about Real Representations
I am stuck on a problem in Fulton's Representation Theory: A First Course. Exercise 3.39 states: Let [imath]V_0[/imath] be a real vector space on which [imath]G[/imath] acts irreducibly, [imath]V=V_0 \otimes \Bbb C[/imath] the corresponding real representation of [imath]G[/imath]. Show that if [imath]V[/imath] is not irreducible, then it has exactly two irreducible factors, and they are conjugate complex representations of [imath]G[/imath]. I had originally misread the problem and taken [imath]V_0[/imath] to be not only a real [imath]G[/imath]-invariant vector space, but a representation of [imath]G[/imath] itself. Proceeding from there, I showed that [imath]V= V_0 \oplus iV_0[/imath], but this is clearly wrong, as these [imath]V_0[/imath] is not complex and these two representations are not conjugate. However, given that [imath]V_0[/imath] is only a real [imath]G[/imath]-invariant vector space, I'm not sure how to proceed.
|
1906087
|
Exercise 3.39 of Fulton & Harris
I would like some help with exercise 3.39 from Fulton & Harris' 'Representation Theory A First Course': Let [imath]V_0[/imath] be a real vector space on which [imath]G[/imath] acts irreducibly, [imath]V=V_0 \otimes \mathbb{C}[/imath] is a complex representation of [imath]G[/imath]. Show that if [imath]V[/imath] is not irreducible, then it has two irreducible factors, and they are conjugate complex representations of [imath]G[/imath]. First of all I would like to remark that conjugate complex representations were not covered before in the book, but after a bit of research I believe the statement in the exercise is equivalent to: [imath]V=W \oplus W^*[/imath] for some (complex) irreducible sub-representation [imath]W \subset V[/imath]. If I am mistaken please do correct me! This question was asked here: Question about Real Representations, but the post got bogged down with notations and since it was so old I decided to ask it again here. It was suggested in the comments of that post that we consider a irreducible (proper) sub-representation [imath]W \subset V[/imath], let [imath]V=W \oplus U[/imath], and show that [imath]U=W^*[/imath], but I don't really see how we can go on... The only way I have learnt to compute the amount of irreducible factors inside a representation [imath]V[/imath] is by computing [imath](\chi_V,\chi_V)=\frac{1}{|G|} \sum_{g \in G} |\chi_V(g)|^2[/imath], but since we don't have a corresponding formula (not one that I am aware of) in the real case, I'm not sure how we can prove there are 2 irreducible factors in [imath]V[/imath] here. I am also failing to see how the second statement about the relation between the two factors can follow from the first part. Any help is appreciated!
|
1909893
|
What is the real part of [imath](i+1)^{20}[/imath]?
Do I have to calculate it all out? That would be a pretty long calculation and boring... Is there an easier ways to find the real part? Thanks in advance
|
762775
|
Converting complex number raised to a power to polar form
How would one convert [imath](1+i)^n[/imath] to polar form? I've heard about de Morgan's law but I don't know how to apply it here.
|
1910447
|
Self-adjoint compact operator on a Hilbert space cannot be surjective
I am trying to prove that a self-adjoint compact operator on a Hilbert space ([imath]T \colon \mathcal H \to \mathcal H[/imath]) cannot be surjective. The hint says that I need to use open mapping theorem and spectral theorem. Could someone help?
|
74063
|
Compact operators on an infinite dimensional Banach space cannot be surjective
I am reading a book about functional analysis and have a question: Let [imath]X[/imath] be a infinite-dimensional Banach-space and [imath]A:X \rightarrow X[/imath] a compact operator. How can one show that [imath]A[/imath] can not be surjective?
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1910732
|
Sum of connected subsets is connected
Let [imath]A,B[/imath] be connected subsets of the Real Numbers. Show that [imath]A+B[/imath] is a connected subset of the Real Numbers.
|
258517
|
sum of two connected subset of [imath]\mathbb{R}[/imath]
[imath]A[/imath] and [imath]B[/imath] are two connected subset of [imath]\mathbb{R}[/imath] define [imath]A+B=\{x+y:x\in A,y\in B\}[/imath], then is [imath]A+B[/imath] also connected? naturally I was thinking about two disjoint connected subsets of [imath]\mathbb{R}[/imath], say [imath]A=[0,1][/imath] and [imath]B=[4,5][/imath] then [imath]A+B=[4,6][/imath]? so is it ingeneral true?
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1909744
|
Does there exist a positive irrational number [imath]\alpha[/imath], the number [imath]\lfloor n\alpha \rfloor[/imath] is never a prime?
Does there exist a positive irrational number [imath]\alpha,[/imath] such that for any positive integer [imath]n[/imath] the number [imath]\lfloor n\alpha \rfloor[/imath] is not a prime?
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1812785
|
Does there exist a positive irrational number [imath]\alpha [/imath], such that for any positive integer [imath]n[/imath] the number [imath]\lfloor n\alpha \rfloor[/imath] is not a prime?
Does there exist a positive irrational number [imath]\alpha [/imath], such that for any positive integer [imath]n[/imath] the number [imath]\lfloor n\alpha \rfloor[/imath] is not a prime? My try if [imath]\alpha=\sqrt{17}[/imath] then [imath]\lfloor n\alpha \rfloor=4n[/imath]
|
960407
|
[imath]f[/imath] increasing , [imath]c \in I[/imath] ; [imath]a_n and \lim \Big(f(a_n)-f(b_n) \Big)=0 , then f is continuous at c ?[/imath]
Let [imath]f:I \to \mathbb R[/imath] be an increasing function where [imath]I[/imath] is an open real interval ; how to show that if for [imath]c \in I[/imath] there exist sequences [imath](a_n)[/imath] and [imath](b_n)[/imath] such that [imath]a_n<c<b_n , \forall n\in \mathbb N[/imath] and [imath]\lim \Big(f(a_n)-f(b_n) \Big)=0 [/imath] , then [imath]f[/imath] is continuous at [imath]c[/imath] ? ( I can prove the converse but having trouble with this one ; Please Help )
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1903463
|
Monotone functions and continuity
I want to show that if [imath]f:(a,b)\to \mathbb R[/imath] is a strictly monotone function and if [imath]x_0\in (a,b)[/imath] such that there exist two sequences [imath](a_n)[/imath] and [imath](b_n)[/imath] with [imath]a_n<x_0<b_n[/imath] and [imath]\lim\limits_{n\to \infty}(f(b_n)-f(a_n))=0[/imath], then [imath]f[/imath] is continuous at [imath]x_0[/imath]. By monotonicity we can say that [imath]\lim\limits_{n\to \infty}f(a_n)=f(x_0)=\lim\limits_{n\to \infty}f(b_n)[/imath]. Let [imath](x_n)[/imath] be a sequence in [imath](a,b)[/imath] such that [imath]x_n\to x[/imath]. Here I got stuck. How to show that [imath]f(x_n)\to f(x_0)[/imath]? Please help!
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1912455
|
Handwriting for Ideals
I am currently studying ring theory, and with that ideals. A typical occurrence in my textbook might be something like Let [imath]R[/imath] be a ring, and [imath]\mathfrak{a}\subset R[/imath] an ideal in [imath]R[/imath], with [imath]a\in\mathfrak{a}[/imath], etc.. My question is, how do I write this by hand (on paper)? How do you guys manage to write the "ideal" a ([imath]\mathfrak{a}[/imath]) different from the element [imath]a[/imath]? Mine always seems to end up looking the same..
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1899330
|
How deal with Gothic letters, like [imath]\mathfrak{ A,B,C,D,a,b,c,d}\dots[/imath], when writing by hand?
In mathematics, I sometimes encounter Gothic letters, I mean the letters [imath]\mathfrak A, \mathfrak B, \mathfrak C, \mathfrak D, \dots, \mathfrak a, \mathfrak b, \mathfrak c, \mathfrak d, \dots[/imath]. To get them in [imath]\LaTeX[/imath] one would use [imath]\mathfrak{A}[/imath] etc. For example, in the book Model theory by Chang and Keisler, structures are denoted [imath]\mathfrak A = (A, \dots)[/imath], [imath]\mathfrak B = (B, \dots)[/imath] and so on. I would like to know how to write this by hand.
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1910697
|
Example of non complete space
I have to prove that the space of continuous functions on a (I suppose closed) interval [imath]I[/imath] (let's say [imath][0,1][/imath]) with distance [imath]d(f,g)[/imath] [imath]=[/imath] [imath]\int _I[/imath] [imath]|f-g|[/imath] is not complete. The Cauchy function to be used is [imath]f_n[/imath] = [imath]n[/imath] in [[imath]0[/imath], [imath]e[/imath] [imath]^-[/imath] [imath]^n[/imath]] and [imath]\log(1/x)[/imath] in [[imath]e[/imath] [imath]^-[/imath] [imath]^n[/imath], [imath]1[/imath]] Why is it not complete though? Thanks!
|
560622
|
A normed space of continuous functions with norm [imath]\int_{0}^{1}|f(t)|dt[/imath] is not complete
Suppose [imath]E[/imath] is a normed space of all continuous functions on [imath][0,1][/imath] with norm [imath]\int_{0}^{1}|f(t)|dt[/imath]. Prove that [imath]E[/imath] is not complete I know that we must do is to find a Cauchy sequence of continous functions that doesn't converge in [imath]E[/imath], but I can't find that sequence. Can any one help me? Thanks.
|
1912321
|
Prove Without Induction: [imath]\sum\limits_{k=2}^{n} \frac{1}{k(k-1)} = 1 - \frac{1}{n}[/imath]
everybody. I'm suppose to prove this without induction: Prove Without Induction: [imath]\sum\limits_{k=2}^{n} \frac{1}{k(k-1)} = 1 - \frac{1}{n}[/imath] I'm not sure how to do it. I tried a bit of algebraic manipulation, but I'm not sure how to do it. It's suppose to be basic. I did get a hint of factorizing [imath]\frac{1}{k(k-1)}[/imath] but that didn't get me anywhere. A hint or any directions would be much appreciated!
|
1742825
|
Showing [imath]\sum_{i = 1}^n\frac1{i(i+1)} = 1-\frac1{n+1}[/imath] without induction?
I oversaw a high-school mathematics test the other day, and one of the problems was the following Show, using induction or other means, that [imath]\sum_{i = 1}^n\frac1{i(i+1)} = 1-\frac1{n+1}[/imath] The induction proof is very standard, where the induction step relies on the fact that [imath]\frac{1}{n+1} + \frac{1}{n(n+1)} = \frac{1}{n}[/imath], and I'm sure it's been answered on this site before. However, I got intrigued by the "or other means" part of the question. I don't know whether the teacher who wrote the test even considered any alternative solutions (he may just have written it so that if anyone has a crazy idea that works out, then they should get a full score for it), but I tried to find one. For instance, we may do the following telescope-ish argument: [imath] \frac{1}{1(1+1)} + \frac{1}{2(2+1)} + \cdots + \frac{1}{(n-1)n} + \frac{1}{n(n+1)} + \frac{1}{n+1}\\ = \frac{1}{1(1+1)} + \frac{1}{2(2+1)} + \cdots + \frac{1}{(n-1)n} + \frac1n\\ \vdots\\ = \frac{1}{1(1+1)} + \frac12\\ = 1 [/imath] However, I feel that this is just an induction proof in disguise (hidden in the vertical dots). If one uses the mechanics of the induction proof to check whether the formula is true for a specific [imath]n[/imath], then one certainly does the exact same calculations as I have done here. Is there a proof of this fact that clearly does not use induction (or at least hides it better)? The more elementary the better, and the ultimate goal would be to do it within the syllabus of the students taking the test (or at least not far from it). For reference, they should be familiar with the summation formula of arithmetic and geometric series and their derivations (so techniques resembling those would be well within specifications). If there is a solution using calculus, then the students should be able to integrate elementary trigonometric functions, as well as exponential functions, logarithms and rational functions. They are familiar with integration by parts, substitution and partial fractions. I welcome more advanced solutions as well, of course.
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1912443
|
Is there a relationship between [imath]a_1\cdot a_2\cdot\dots\cdot a_n[/imath] and [imath](a_1+1)\cdot(a_2+1)\cdot\dots\cdot(a_n+1)[/imath]?
I want to find an inequality between the to terms... Assuming all the [imath]a_i [/imath] to be positive, the one where [imath]1[/imath] is added to all the [imath]a_i[/imath] is bigger, but is there a real relationship? It doesn't have to be an inequality because I need something stronger than that...
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1911849
|
Prove that [imath](1+a_1) \cdot (1+a_2) \cdot \dots \cdot (1+a_n) \geq 2^n[/imath]
Let [imath]a_1,a_2,\dots,a_n[/imath] [imath]\in[/imath] [imath]\mathbb{R}^+[/imath] and [imath]a_1\cdot a_2\cdots a_n=1[/imath], , prove that [imath](1+a_1) \cdot (1+a_2) \cdot \dots \cdot (1+a_n) \geq 2^n[/imath] I have tried factorising but it just lead me to extremily complicated equation that were extremily difficult to understand... Could someone help me?
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1912894
|
[imath]V[/imath] is a simple [imath]\text{End}_kV[/imath]-module?
Let [imath]V[/imath] be a finite dimensional vector space over [imath]k[/imath] and [imath]A = \text{End}_k V[/imath]. How do I see that [imath]V[/imath] is a simple [imath]A[/imath]-module?
|
1415627
|
Is [imath]V[/imath] a simple [imath]\text{End}_kV[/imath]-module?
Let [imath]V[/imath] be a finite-dimensional vector space over [imath]k[/imath] and [imath]A = \text{End}_k V[/imath]. Is [imath]V[/imath] a simple [imath]A[/imath]-module?
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1913021
|
Strong Counterexample to MVT on Q
I initially asked this question on MSE, but then thought it would appropiate to ask it here as well ( correct me if I'm wrong): Strong Counterexample to MVT on Q A well-known application of the MVT is to prove that any [imath]f: \mathbb{R} \to \mathbb{R}[/imath] with [imath]f'= 0[/imath] is constant. But of course, the MVT relies fundamentally on the properties of [imath]\mathbb{R}[/imath], and in particular does not hold in [imath]\mathbb{Q}[/imath]. A standard counter-example is the function [imath]f :\mathbb{Q} \to \mathbb{Q}[/imath] defined by [imath] f(x)=0 [/imath] if [imath] x <\sqrt{2}[/imath], and [imath]f(x)=1[/imath] if [imath] x > \sqrt{2}[/imath], for this function is locally constant. There are several ways of thinking about the difference between [imath]\mathbb{R}[/imath] and [imath]\mathbb{Q}[/imath] : [imath]\mathbb{R}[/imath] is connected while [imath]\mathbb{Q}[/imath] is totally disconnected, closed bounded intervals are compact in [imath]\mathbb{R}[/imath], but not in [imath]\mathbb{Q}[/imath], [imath]\mathbb{R}[/imath] satisfies the least upper bound axiom, [imath]\mathbb{Q}[/imath] does not. Now, I wanted to see how far one could push this counterexample - is there a function [imath]f :\mathbb{Q} \to \mathbb{Q}[/imath] such that [imath]f'=0[/imath] and [imath]f[/imath] is strictly increasing? Clearly, any slight modification of the usual counterexample where [imath]f[/imath] is locally constant will not do. Also, one can show that if [imath]f[/imath] is uniformly differentiable with derivative zero, it must be constant, even in [imath]\mathbb{Q}[/imath], so I knew I was looking for non-uniformly differentiable but still differentiable [imath]f[/imath]. Eventually, I came up with a construction, albeit not a very explicit one: the idea is that, since [imath]\mathbb{Q}[/imath] is countable, one can enumerate it [imath]q_{1}, q_{2}, q_{3}, \cdots[/imath], and then define [imath]f: \mathbb{Q} \to \mathbb{Q}[/imath] inductively and recursively. If one could define [imath]f[/imath] in such a way that [imath] \forall \ q_{n} \in \mathbb{Q} \ \exists \ c_{n} \in \mathbb{Q} : \forall \ x \in \mathbb{Q} \ |f(x)-f(q_{n})| \leq c_{n}|x-q_{n}|^2[/imath], and [imath]f[/imath] is strictly increasing, we would be done. The recursive definition essentially defines [imath]f(q_{n+1})[/imath], after having defined [imath]f(q_{1}),f(q_{2}),f(q_{3}), \cdots , f(q_{n})[/imath], in such a way that these inequalities hold and [imath]f[/imath] is strictly increasing, and at each step the new [imath]c_{n}[/imath] is chosen large enough so that these inequalities do not become incompatible at subsequent steps ( I am willing to provide more details of this construction if asked). Note that, as [imath]n \to \infty[/imath], we must have [imath]c_{n} \to \infty[/imath], so that [imath]f[/imath] is not uniformly differentiable. Interestingly, the only key fact about [imath]\mathbb{Q}[/imath] used in the construction is that it is countable, so in a sense the construction provides a proof that [imath]\mathbb{R}[/imath] is uncountable, given the MVT. While the construction works, it is not very simple nor explicit, which is perhaps to be expected, and the function must be evaluated point by point in the order given by the enumeration of [imath]\mathbb{Q}[/imath]. So I was wondering if anyone else knows or can come up with a simpler construction of a function [imath] f : \mathbb{Q} \to \mathbb{Q}[/imath] such that [imath]f'=0[/imath] and [imath]f[/imath] is strictly increasing. Any thoughts and ideas are welcome.
|
1911440
|
Strong Counterexample to MVT on Q
A well-known application of the MVT is to prove that any [imath]f: \mathbb{R} \to \mathbb{R}[/imath] with [imath]f'= 0[/imath] is constant. But of course, the MVT relies fundamentally on the properties of [imath]\mathbb{R}[/imath], and in particular does not hold in [imath]\mathbb{Q}[/imath]. A standard counter-example is the function [imath]f :\mathbb{Q} \to \mathbb{Q}[/imath] defined by [imath] f(x)=0 [/imath] if [imath] x <\sqrt{2}[/imath], and [imath]f(x)=1[/imath] if [imath] x > \sqrt{2}[/imath], for this function is locally constant. There are several ways of thinking about the difference between [imath]\mathbb{R}[/imath] and [imath]\mathbb{Q}[/imath] : [imath]\mathbb{R}[/imath] is connected while [imath]\mathbb{Q}[/imath] is totally disconnected, closed bounded intervals are compact in [imath]\mathbb{R}[/imath], but not in [imath]\mathbb{Q}[/imath], [imath]\mathbb{R}[/imath] satisfies the least upper bound axiom, [imath]\mathbb{Q}[/imath] does not. Now, I wanted to see how far one could push this counterexample - is there a function [imath]f :\mathbb{Q} \to \mathbb{Q}[/imath] such that [imath]f'=0[/imath] and [imath]f[/imath] is strictly increasing? Clearly, any slight modification of the usual counterexample where [imath]f[/imath] is locally constant will not do. Also, one can show that if [imath]f[/imath] is uniformly differentiable with derivative zero, it must be constant, even in [imath]\mathbb{Q}[/imath], so I knew I was looking for non-uniformly differentiable but still differentiable [imath]f[/imath]. Eventually, I came up with a construction, albeit not a very explicit one: the idea is that, since [imath]\mathbb{Q}[/imath] is countable, one can enumerate it [imath]q_{1}, q_{2}, q_{3}, \cdots[/imath], and then define [imath]f: \mathbb{Q} \to \mathbb{Q}[/imath] inductively and recursively. If one could define [imath]f[/imath] in such a way that [imath] \forall \ q_{n} \in \mathbb{Q} \ \exists \ c_{n} \in \mathbb{Q} : \forall \ x \in \mathbb{Q} \ |f(x)-f(q_{n})| \leq c_{n}|x-q_{n}|^2[/imath], and [imath]f[/imath] is strictly increasing, we would be done. The recursive definition essentially defines [imath]f(q_{n+1})[/imath], after having defined [imath]f(q_{1}),f(q_{2}),f(q_{3}), \cdots , f(q_{n})[/imath], in such a way that these inequalities hold and [imath]f[/imath] is strictly increasing, and at each step the new [imath]c_{n}[/imath] is chosen large enough so that these inequalities do not become incompatible at subsequent steps ( I am willing to provide more details of this construction if asked). Note that, as [imath]n \to \infty[/imath], we must have [imath]c_{n} \to \infty[/imath], so that [imath]f[/imath] is not uniformly differentiable. Interestingly, the only key fact about [imath]\mathbb{Q}[/imath] used in the construction is that it is countable, so in a sense the construction provides a proof that [imath]\mathbb{R}[/imath] is uncountable, given the MVT. While the construction works, it is not very simple nor explicit, which is perhaps to be expected, and the function must be evaluated point by point in the order given by the enumeration of [imath]\mathbb{Q}[/imath]. So I was wondering if anyone else knows or can come up with a simpler construction of a function [imath] f : \mathbb{Q} \to \mathbb{Q}[/imath] such that [imath]f'=0[/imath] and [imath]f[/imath] is strictly increasing. Any thoughts and ideas are welcome. Edit: I've asked the same question on MO: https://mathoverflow.net/questions/248913/strong-counterexample-to-mvt-on-q
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1913125
|
Prove that [imath]A \triangle B=B\triangle C\implies A=C[/imath]
Prove that [imath]A \triangle B=B\triangle C\implies A=C[/imath] We have [imath]A \triangle B=B\triangle C\implies A=C\\ \implies (A\cap B')\cup (B\cap A') = (B\cap C')\cup (C\cap B') [/imath] How can I proceed for the result?
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382773
|
Prove that if [imath]A\triangle B = C\triangle B[/imath], then [imath]A = C[/imath]
I am working with proofs in discrete math. Help to prove: For the sets [imath]A[/imath] and [imath]B[/imath], we define the symmetric difference of [imath]A[/imath] and [imath]B[/imath] to be [imath]A \triangle B = (A-B)\cup(B-A).[/imath] Prove that if [imath]A \triangle B = C \triangle B,[/imath] then [imath]A = C.[/imath]
|
1913151
|
[imath]P(x)=\frac{x}{x+1}[/imath] for [imath]x=1,2,...,n[/imath], find the value of [imath]P(n+1)[/imath].
I've recently come across this question and have no idea how to do it. Let [imath]P(x)[/imath] be a 11-degree polynomial such that [imath]P(x)=\frac{1}{x+1}[/imath] for [imath]x=0,1,2,...11[/imath] Find [imath]P(12)[/imath]. I've tried using the general form of a polynomial but it's turning to long. I think I'm missing some simple trick. After this comes a general form of the problem. If [imath]P(x)[/imath] is a polynomial of degree [imath]n[/imath] such that [imath]P(x)=\frac{x}{x+1}[/imath] for [imath]x=1,2,...,n[/imath], find the value of [imath]P(n+1)[/imath].
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769938
|
Find [imath]P(12)[/imath] when [imath]P(x)={1\over(x+1)}[/imath], for [imath]x=0,1,2,......11[/imath]
let [imath]P(x)[/imath] be a polynomial of degree [imath]11[/imath] such that [imath]P(x)={1\over(x+1)}[/imath], for [imath]x=0,1,2,......11[/imath] then find value of [imath]P(12)[/imath]
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1913354
|
Does [imath]c\cdot f(x) = f(cx) \implies f[/imath] is linear?
A linear functions [imath]f[/imath] is a function that has the following two properties: [imath]c\cdot f(x) = f(cx)[/imath] [imath]f(x+y) = f(x) + f(y)[/imath] But does [imath]c\cdot f(x) = f(cx) \implies f[/imath] is linear? Can we assume property [imath]2[/imath] above, if we have a function [imath]f[/imath] satisfying property [imath]1[/imath]?
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192396
|
Are all multiplicative functions additive?
Suppose [imath]cf(x)=f(cx)[/imath] and [imath]f:\mathbb{R}\to\mathbb{R}[/imath]. I believe it follows that [imath]f(x+y)=f(x)+f(y)[/imath]. Proof: There is some [imath]c[/imath] such that [imath]y=cx[/imath]. Then [imath]f(x+y)=f\left((1+c)x\right)=(1+c)f(x)=f(x)+cf(x)=f(x)+f(cx)=f(x)+f(y)[/imath] QED. I wonder if the same thing holds for when [imath]f:\mathbb{R}^n\to\mathbb{R}^n[/imath]? I can't use the same trick, because all vectors are not scalar multiples. I tried thinking of it in terms of basis units, but didn't get anywhere.
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1913093
|
Taylor's theorem without L'Hospital
Let [imath]f:D \to \mathbb{R}[/imath] be a real function with [imath]U \subset D[/imath] an open interval. Suppose that [imath]f[/imath] is [imath]k[/imath] times differentiable at [imath]a \in U[/imath]. Then [imath]R_k(x) \in o(|x-a|^k) \ \text{as} \ x \to a[/imath] where [imath]R_k(x)=f(x)-P_k(x)[/imath] and [imath]P_k[/imath] is the [imath]k[/imath]th order Taylor polynomial of [imath]f[/imath] evaluated at [imath]a[/imath]. Like many, I am somewhat averse to the usage of L'Hospital's rule. Is there a proof of the above without it? Note the minimal conditions on [imath]f[/imath].
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1809293
|
Taylor's Theorem with Peano's Form of Remainder
The following form of Taylor's Theorem with minimal hypotheses is not widely popular and goes by the name of Taylor's Theorem with Peano's Form of Remainder: Taylor's Theorem with Peano's Form of Remainder: If [imath]f[/imath] is a function such that its [imath]n^{\text{th}}[/imath] derivative at [imath]a[/imath] (i.e. [imath]f^{(n)}(a)[/imath]) exists then [imath]f(a + h) = f(a) + hf'(a) + \frac{h^{2}}{2!}f''(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a) + o(h^{n})[/imath] where [imath]o(h^{n})[/imath] represents a function [imath]g(h)[/imath] with [imath]g(h)/h^{n} \to 0[/imath] as [imath]h \to 0[/imath]. One of the proofs (search "Proof of Taylor's Theorem" in this blog post) of this theorem uses repeated application of L'Hospital's Rule. And it appears that proofs of the above theorem apart from the one via L'Hospital's Rule are not well known. I have asked this question to get other proofs of this theorem which do not rely on L'Hospital's Rule and instead use simpler ideas. BTW I am also posting one proof of my own as a community wiki.
|
1913482
|
How do I prove that [imath](A + B)^2 = A^2 + 2AB + B^2[/imath]
Assume that [imath]A[/imath] and [imath]B[/imath] are both [imath]n\times n[/imath] matrices, how would I prove that [imath](A + B)^2 = A^2 + 2AB + B^2[/imath] for any [imath]n\times n[/imath] matrices?
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1889593
|
Why in these matrices are [imath]AB=BA[/imath] not equal? What is the logic behind them?
We know that in matrices AB=BA.Why in this Matrices [imath]A=\begin{bmatrix} -1 & 3\\ 2 & 0\end{bmatrix}[/imath], [imath]B=\begin{bmatrix} 1 & 2\\ -3 & -5\end{bmatrix}[/imath] are not equal to [imath]AB=BA[/imath]. WHY? This is matrix of order [imath]2\times 2[/imath] for both [imath]A[/imath] and [imath]B[/imath].
|
699505
|
For all square matrices [imath]A[/imath] and [imath]B[/imath] of the same size, it is true that [imath](A+B)^2 = A^2 + 2AB + B^2[/imath]?
The below statement is a true/false exercise. Statement: For all square matrices A and B of the same size, it is true that [imath](A + B)2 = A^2 + 2AB + B^2[/imath]. My thought process: Since it is not a proof, I figure I can show by example and come to a valid conclusion based on such example. My work: Come up with a square matrix A and B let both be a 2 by 2 matrix(rows and cols must be same). Matrix [imath]A[/imath]: [imath]A = \begin{array}{ccc} 3 & 5 \\ 4 & 6 \\ \end{array} [/imath] Matrix [imath]B[/imath]: [imath]B = \begin{array}{ccc} 5 & 8 \\ 9 & 4 \\ \end{array} [/imath] [imath]A + B = \begin{array}{ccc} 8 & 13 \\ 13 & 10 \\ \end{array}[/imath] [imath](A + B)^2 = \begin{array}{ccc} 233 & 234 \\ 234 & 264 \\ \end{array}[/imath] [imath]A^2 = \begin{array}{ccc} 29 & 45 \\ 36 & 56 \\ \end{array}[/imath] [imath](AB) = \begin{array}{ccc} 60 & 44 \\ 74 & 56 \\ \end{array}[/imath] [imath]2(AB) = \begin{array}{ccc} 120 & 88 \\ 234 & 112 \\ \end{array}[/imath] [imath]B^2 = \begin{array}{ccc} 97 & 72 \\ 81 & 88 \\ \end{array}[/imath] [imath]A^2 + 2AB + B^2 = \begin{array}{ccc} 246 & 205 \\ 265 & 256 \\ \end{array}[/imath] Based my above work, the answer is false. Is there another way to approach the problem? It seems like a lot of work needed to be done for a true/false question which raised my suspicion about whether there is a better way to look at the problem.
|
2124519
|
Give an example in which [imath](A + B)^2 \neq A^2 + 2AB + B^2[/imath]
Other provided information: 'Let A and B be square matrices of the same size.' The last question for my Linear Algebra assignment involves providing matrices [imath]A[/imath] and [imath]B[/imath] that satisfy the above requirement, as well as 'Give a valid identity for [imath](A + B)^2[/imath]' and I do not see any theorem or example from my text that helps me break this problem down. So far I have been guessing random matrices to no avail and I know there must be a methodical solution to this. What do I need to know in order to approach this problem?
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1913592
|
Generating a subgroup with two permutations
I have what appears to be a simple question but I do not know if my method is correct. I have to find what the subgroup of [imath]S_5[/imath] generated by [imath](123)[/imath] and [imath](12345)[/imath] is. So I have worked out that the product of the above permutations is [imath](52134)[/imath] so now do I compute the 5th power of the product I calculated? Then somehow find out what subgroup that is?
|
955760
|
Find the order of a subgroup of [imath]S_5[/imath] generated by two elements
The Question I am trying to solve reads : Find the order of a subgroup of [imath]S_5[/imath] generated by the elements [imath]s=(123)[/imath] and [imath]r=(12345)[/imath]. Is this group sovable ? Is it nilpotent ? I tried to compute a couple of terms but then I realized this subgroup is way large, and computing the elements one by one might not be a good idea (Is there another way to compute the order ?) My calculation shows : [imath]r, r^2, r^3, r^4, rs, r^3s, sr [/imath] are all [imath]5-[/imath]cycles but there are elements like [imath]r^2s=(14)(25)[/imath] and also [imath]r^4s=(354)[/imath]...It seems that all sorts of things can happen in this subgroup...I do not even know how to approach the solvability and nilpotency before I know what the subgroup is... I already appreciate your hints/ideas/answers
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1914166
|
Direct sum of groups vs product
I've got a problem with the finish of the answer of the following problem: Homology of connected sum of real projective spaces. How is that [imath](\mathbb{Z}^{n-1}\oplus\mathbb{Z})/(2,\dots,2)\mathbb{Z}[/imath] with [imath]n[/imath] twos is equal to [imath]\mathbb{Z}^{n-1}\oplus\mathbb{Z}_2[/imath]? [imath]\mathbb{Z}^{n-1}\oplus\mathbb{Z}=\mathbb{Z}^n[/imath],since [imath]n[/imath] is finite and we have finite sequences, right? Or am I wrong? Thanks in advance!
|
1047076
|
[imath]\mathbb Z^n/\langle (a,...,a) \rangle \cong \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle[/imath]
I am trying to show the isomorphism [imath]\mathbb Z^n/\langle (a,...,a) \rangle \cong \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle.[/imath] I've tried to define [imath]\psi:\mathbb Z^n \to \mathbb Z^{n-1} \oplus \mathbb Z/\langle a \rangle[/imath] an epimorphism with [imath]\ker(\psi)=\langle (a,...,a) \rangle[/imath] so to apply the first isomorphism theorem, but I couldn't come up with an appropiate morphism. Any hints or suggestions would be greatly appreciated.
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1913984
|
Prove "integral convergence" for a periodic function
Let [imath]A[/imath] be the compact annulus in the real plane with radii [imath]2[/imath] and [imath]3[/imath], and [imath](r,\phi)[/imath] its polar coordinates. Define [imath]F(r,\phi)=(r,r\phi)\ (\mod\ 2\pi)[/imath]. Is it true that, for a continuous function [imath]g:A\to \mathbb R[/imath], the integral [imath]\int_A g(F^n(x))dx[/imath] converges as [imath]n\to+\infty[/imath]? ([imath]F^n[/imath] is the n-th iteration of [imath]F[/imath].) My attempt was to prove that [imath]F^n(x)[/imath] is Cauchy for almost every [imath]x[/imath]. This would allow us to apply Lebesgue's dominated convergence theorem, since [imath]g[/imath] must be bounded on the compact [imath]A[/imath]. But I just can't obtain the desired result, so I'm beginning to suspect that this can be false, and the convergence (if it is true) comes from weaker properties. Thank you in advance.
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1906754
|
Prove or the disprove the existence of a limit of integrals
Let [imath]A \subset \mathbb{R}^2[/imath] be the annulus defined, in polar coordinates, as [imath] A = \{ (r, \theta) \in [0, \infty) \times [0, 2\pi) : 2 \leq r \leq 3 \} \, .[/imath] Let [imath]F: A \to A[/imath], [imath]F:(r, \theta) \mapsto (r, r \theta \text{ (mod } 2\pi))[/imath]. Let [imath]g : A \to \mathbb{R}[/imath] be a continuous function. Discuss the existence of the limit [imath] \lim_{n \to \infty} \int_A g \circ F^{(n)}(x) \, \text{d}x \, ,[/imath] where [imath]F^{(n)} = F \circ F \circ \dots \circ F[/imath] denotes the composition of [imath]F[/imath] with itself [imath]n[/imath] times. I cannot find a way to attack this problem. Here's what I tried: if [imath]g \circ F^{(n)}[/imath] converged, I think we could take the limit under the integral sign. So I tried to show the convergence of [imath]F^{(n)}[/imath]. If it converged, then the limit function must send every point to a fixed point of [imath]F[/imath]. So I determined the fixed points of [imath]F[/imath]. But I don't think [imath]F^{(n)}[/imath] converges.
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1914565
|
How to prove that : [imath]\sum \limits_{k=0}^{n-1} \lfloor x+\frac{k}{n}\rfloor = \lfloor nx\rfloor[/imath]?
for all [imath]x\in \mathbb{R}[/imath] and [imath]n\in\mathbb{N}^*[/imath] without using euclidian division. First I re-write the sum in this way : [imath]\sum \limits_{k=0}^{n-1} \lfloor x+\frac{k}{n}\rfloor = \lfloor x\rfloor+\lfloor x+\frac{1}{n}\rfloor+... +\lfloor x+\frac{i-1}{n}\rfloor+ \lfloor x+1-\frac{i}{n}\rfloor+...+\lfloor x+1-\frac{1}{n}\rfloor[/imath] Now we can see that there is two possibilities for [imath]x[/imath] between two natural numbers [imath][q,q+1][/imath]. Indeed we have the part [imath]\sum \limits_{k=0}^{i-1}\lfloor x+\frac{k}{n}\rfloor=ji[/imath] with [imath]0\le j\le n[/imath].And the other part [imath]\sum \limits_{k=i}^{n-1}\lfloor x+1-\frac{k}{n}\rfloor=(j+1)(n-i)[/imath]. Finally the floor will be [imath]nj+(n-i)[/imath]. Now for [imath]\lfloor nx \rfloor[/imath] it will be between [imath][nq,nq+q][/imath] which lets [imath]n[/imath] possibilities for its values but it would be perfect if it was [imath]nj+(n-i)[/imath] . Thanks in advance !
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1815289
|
Prove [imath]\Sigma_{k=0}^{n-1}\lfloor x+\frac{k}{n}\rfloor=\lfloor nx\rfloor[/imath] , n is a Natural Number
Prove the following identity: [imath]\lfloor x\rfloor +\lfloor x+\frac{1}{n}\rfloor +\lfloor x+\frac{2}{n}\rfloor +\lfloor x+\frac{3}{n}\rfloor+...+\lfloor x+\frac{n-1}{n}\rfloor =\lfloor nx\rfloor[/imath] where n is a Natural Number. [imath]$$At first I thought of splitting it into 2 cases: when x is an integer, and when x isnn't an integer. The case of $x$ being an integer is quite simple: $\lfloor x+\frac{k}{n}\rfloor=x$ for $0\le k\le n-1$. Thus the LHS becomes $n\lfloor x\rfloor$ which is equal to the RHS. [/imath][imath]$However I do not know how to go about the case of [/imath]x[imath] not being an integer.[/imath] Lastly, I would actually prefer a proof where it is not necessary to make cases based on the values of x[imath], but to have one general proof which satisfies all [/imath]x[imath].[/imath] Could somebody please show me how to complete the proof in both ways (ie first, the case where x$ isn't an integer, and secondly the general proof which doesn't involve breaking into cases)? Many thanks in advance!
|
1914576
|
Prove that if [imath]n[/imath] is an odd integer, then [imath]n^2-1[/imath] is divisible by [imath]8[/imath].
I have it proved for when k is odd, but when k is even, I get 8[q(2q+1)]. Is it correct to state that 8(Integer) proves that n^2-1 is divisible by 8?
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242543
|
Prove that if [imath]x[/imath] is odd then [imath]x^2 -1[/imath] is divisible by [imath]8[/imath].
If [imath]x[/imath] is odd then prove that [imath]x^2-1[/imath] is divisible by [imath]8[/imath]. I start by writing: [imath]x = 2k+1 [/imath] where [imath]k\in\mathbb{N}[/imath]. Then it follows that: [imath](2k+1)^2 -1 = 4k^2 +4k + 1 -1 [/imath] Therefore: [imath]\frac{4k^2 +4k}{8} = \frac{k(k+1)}{2}[/imath] At the end part I can see that for what [imath]k[/imath] is, the number on top is divisible by 2. I was expecting the end result to be a number, not a fraction. Or is it "divisible" from the definition [imath]\text{even} = 2k[/imath] that completes the proof?
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1915088
|
What is a [imath]k[/imath]-algebra, what is a [imath]k[/imath]-algebra morphism?
Another question on algebras. My question is, can anyone explain what a [imath]k[/imath]-algebra is and what a [imath]k[/imath]-algebra morphism is to me as if I'm a dummy? What is the intuition for working with them?
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1367667
|
Intuition behind [imath]k[/imath]-algebra, [imath]k[/imath]-algebra morphisms?
I will state the definition of a [imath]k[/imath]-algebra and [imath]k[/imath]-algebra morphisms. A ring [imath]A[/imath] equipped with a ring homomorphism [imath]k \to Z(A)[/imath] is called a [imath]k[/imath]-algebra. More explicitly, this means that [imath]A[/imath] has a structure of vector space over [imath]k[/imath] and also a ring structure such that: (1) The operations "[imath]+[/imath]" coming from the vector space structure and the ring structure, respectively, are the same. (2) The ring multiplication [imath]\cdot[/imath]: [imath]A \times A \to A[/imath] is a [imath]k[/imath]-bilinear map. One defines [imath]k[/imath]-algebra morphisms as [imath]k[/imath]-linear ring morphisms. I understand these definitions formally, but I have quite a poor grasp on the intuition behind the definition of [imath]k[/imath]-algebra and [imath]k[/imath]-algebra morphisms. Could someone provide me with their intuitions on such?
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1915026
|
Find the largest positive integer [imath]n[/imath] such that [imath]n^3 + 100[/imath] is divisible by [imath]n+10[/imath]
I'm still kinda new to number theory so could someone tell me if my solution is right? Reducing [imath]\mod n +10[/imath] [imath]n \equiv -10 \pmod{n+10} \implies n^3 + 100 \equiv -900 \pmod{n+10}[/imath] So we want [imath](n+10) \mid 900[/imath] and the largest number would be [imath]n = 890[/imath]
|
124927
|
What is the largest positive [imath]n[/imath] for which [imath]n^3+100[/imath] is divisible by [imath]n+10[/imath]
What is the largest positive [imath]n[/imath] for which [imath]n^3+100[/imath] is divisible by [imath]n+10[/imath] I tried to factorize [imath]n^3+100[/imath], but [imath]100[/imath] is not a perfect cube. I wish it were [imath]1000[/imath].
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1915694
|
Is it true that [imath]k[\alpha,\beta]=k(\alpha,\beta)[/imath], [imath]\alpha[/imath] and [imath]\beta[/imath] are algebraic over [imath]k[/imath].
Is there a generalization to this question? Namely, if [imath]k \subset K[/imath] are fields and [imath]\alpha, \beta \in K[/imath] are algebraic over [imath]k[/imath], is it true that [imath]k[\alpha,\beta]=k(\alpha,\beta)[/imath]?
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1715450
|
Is [imath]K[a,b]=K(a,b)[/imath] for algebraic [imath]a, b[/imath]?
Consider a field extension [imath]L/K[/imath] and [imath]a,b \in L[/imath] algebraic over [imath]K[/imath]. Is it true that in this setting [imath]K[a,b][/imath] is already a field? I know that [imath]K[a]=K(a)[/imath] if and only if [imath]a[/imath] is algebraic over [imath]K[/imath].
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1915975
|
If [imath]A^n[/imath] = [imath]I[/imath] for some [imath]n>0[/imath], then [imath]A[/imath] can be diagonalized?
Let [imath]A[/imath] be a [imath]n\times n[/imath] matrix over [imath]\mathbb{C}[/imath]. If [imath]A^n[/imath] = [imath]I[/imath] for some [imath]n>0[/imath], then how to show that [imath]A[/imath] can be diagonalized? The hint I have is to show that the characteristic polynomial has no repeated root.
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1013429
|
If [imath]A[/imath] is a complex matrix of size [imath]n[/imath] of finite order then is [imath]A[/imath] diagonalizable ?
Let [imath]A[/imath] be a complex matrix of size [imath]n[/imath] if for some positive integer [imath]k[/imath] , [imath]A^k=I_n[/imath] , then is [imath]A[/imath] diagonalizable ?
|
1840801
|
Why is [imath]A^TA[/imath] invertible if [imath]A[/imath] has independent columns?
How can I understand that [imath]A^TA[/imath] is invertible if [imath]A[/imath] has independent columns? I found a similar question, phrased the other way around, so I tried to use the theorem [imath] rank(A^TA) \le min(rank(A^T),rank(A)) [/imath] Given [imath]rank(A) = rank(A^T) = n[/imath] and [imath]A^TA[/imath] produces an [imath]n\times n[/imath] matrix, I can't seem to prove that [imath]rank(A^TA)[/imath] is actually [imath]n[/imath]. I also tried to look at the question another way with the matrices [imath] A^TA = \begin{bmatrix}a_1^T \\ a_2^T \\ \ldots \\ a_n^T \end{bmatrix} \begin{bmatrix}a_1 a_2 \ldots a_n \end{bmatrix} = \begin{bmatrix}A^Ta_1 A^Ta^2 \ldots A^Ta_n\end{bmatrix} [/imath] But I still can't seem to show that [imath]A^TA[/imath] is invertible. So, how should I get a better understanding of why [imath]A^TA[/imath] is invertible if [imath]A[/imath] has independent columns?
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2044381
|
Linearly Independent columns and invertibility of transpose
If a matrix [imath]A[/imath] has linearly independent columns, why is [imath]A^TA[/imath] invertible? I think it has to do with [imath]A^TA[/imath] being square and that [imath]A^TAx[/imath]=0 only has the trivial solution, but I don't really know how to explain that properly.
|
1915570
|
Equilateral triangle using complex analysis
Let [imath]\Delta ABC[/imath] be a triangle. The vertices [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] are denoted by the complex numbers [imath]\alpha[/imath], [imath]\beta[/imath] and [imath]\gamma[/imath]. Let [imath]\omega=e^{\frac{2\pi}{3}i}[/imath]. Prove that [imath]\alpha+\omega\beta+\omega^2 \gamma =0[/imath] or [imath]\alpha+\omega\gamma+\omega^2\beta=0[/imath] iff [imath]\Delta ABC[/imath] is equilateral. In previous steps, we need to prove that [imath]1-e^{\frac{\pi}{3}i}+e^{\frac{2\pi}{3}i}=0[/imath] and [imath]1+\omega+\omega^2=0[/imath]. I know why this is true, but I don't see the connection to the problem..
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947813
|
Condition for 3 complex numbers to represent an equilateral triangle
[imath]z_1[/imath], [imath]z_2[/imath], and [imath]z_3[/imath] are 3 complex numbers. Prove that if they represent the vertices of an equilateral triangle then [imath]z_1 + \omega z_2 + \omega^2 z_3 = 0[/imath] where [imath]\omega[/imath] is a 3rd root of unity. Any help would be thoroughly appreciated.
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1916306
|
Fields closed up to degree [imath]n[/imath]
Are there fields that are not algebraically closed fields but such that any quadratic polynomial has a root? Suppose now [imath]n[/imath] is a fixed natural number, [imath]n\geq 2[/imath]. Is there a field [imath]K_n[/imath] such that every polinomial of degree [imath]\leq n[/imath] with coefficients in [imath]K_n[/imath] has a root in [imath]K_n[/imath], but [imath]K_n[/imath] is not algebraically closed? I was thinking that perhaps a tower of algebraic extensions of [imath]\mathbb{Q}[/imath] might work, but then I realized that in order to be "quadratically closed" I might have to close under all extensions of degree [imath]2^k[/imath] for some [imath]k[/imath]. Is [imath]K_2=\bigcup \left\{\mathbb{Q}[\alpha]: [\mathbb{Q}(\alpha):\mathbb{Q}]=2^k\text{ for some [/imath]k[imath]}\right\}[/imath] going to work as a field that is quadratically closed but not algebraically closed?
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93744
|
Non-algebraically closed field in which every polynomial of degree [imath] has a root[/imath]
My problem is to build, for every prime [imath]p[/imath], a field of characteristic [imath]p[/imath] in which every polynomial of degree [imath]\leq n[/imath] ([imath]n[/imath] a fixed natural number) has a root, but such that the field is not algebraically closed. If I'm not wrong (please correct me if I am) such a field cannot be finite, by counting arguments. But on the other hand, the union of all finite fields (or of any ascending chain of finite fields) of characteristic [imath]p[/imath], which is what I get if I start with [imath]F_p[/imath] and add a root to each polynomial of degree [imath]\leq n[/imath] in each step, is the algebraic closure of [imath]F_p[/imath], hence algebraically closed. I don't see how I can control this process so that in the end I get a field that is not algebraically closed. Any hint will be welcome. Thanks in advance.
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1916527
|
A faster way than tedious multiplying
Show that [imath](a+b+c)(a+b\epsilon +c\epsilon^2)(a+b\epsilon^2 + c\epsilon) = a^3 + b^3 + c^3 - 3abc[/imath] If [imath]\epsilon^2 + \epsilon + 1 =0[/imath] The solution in the back of the book is given as Proved by a direct check, taking into consideration that [imath]\epsilon^2 = -\epsilon -1 \: \:[/imath] and [imath]\epsilon^3 = 1[/imath] However the solutions still feels like tedious multiplication, doesn't it? Is there a faster, more elegant way to do this?
|
475354
|
How to show that [imath]A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)[/imath] indirectly?
I found this amazingly beautiful identity here. How to prove that [imath]A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)[/imath] without directly multiplying the factors? (I've already verified it that way). Moreover, how could someone possibly find such a factorization using complex numbers? Is it possible to find such a factorization because [imath]A^3+B^3+C^3 - 3ABC[/imath] is a symmetric polynomial in [imath]A,B,C[/imath]?
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1917029
|
Prove that in an ordered field, [imath]0\cdot x=0[/imath] for every [imath]x[/imath]
Prove that in an ordered field [imath]0\cdot x=0[/imath] for every [imath]x[/imath]. I'm not sure how to go about proving something is true in an ordered field. What are the steps involved?
|
911075
|
Fields - Proof that every multiple of zero equals zero
This is one of my first proofs about fields. Please feed back and criticise in every way (including style and details). Let [imath](F, +, \cdot)[/imath] be a field. Non-trivially, [imath]\textit{associativity}[/imath] implies that any parentheses are meaningless. Therefore, we will not use parentheses. Therefore, we will not use [imath]\textit{associativity}[/imath] explicitly. By [imath]\textit{identity element}[/imath], [imath]F \ne \emptyset[/imath]. Now, let [imath]a \in F[/imath]. It remains to prove that [imath]0a = 0[/imath]. \begin{equation*} \begin{split} 0a &= 0a + 0 && \quad \text{by }\textit{identity element }(+ ) \\ &= 0a + a + -a && \quad \text{by }\textit{inverse element }(+ ) \\ &= 0a + 1a + -a && \quad \text{by }\textit{identity element }(\cdot) \\ &= (0 + 1)a + -a && \quad \text{by }\textit{distributivity } \\ &= (1 + 0)a + -a && \quad \text{by }\textit{commutativity }(+ ) \\ &= 1 a + -a && \quad \text{by }\textit{identity element }(+ ) \\ &= a + -a && \quad \text{by }\textit{identity element }(\cdot) \\ &= 0 && \quad \text{by }\textit{inverse element }(+ ) \end{split} \end{equation*} QED PS: Is "Let [imath](F, +, \cdot)[/imath] be a field." ok? Besides, I would not want to call [imath]F[/imath] a field, because [imath]F[/imath] is just a set. Also, what do you think about using adverbs like "Now"? How would you have said the associativity-thing?
|
1917605
|
[imath]mn[/imath] People standing in a rectangular array
Suppose that [imath]mn[/imath] people of a marching band are standing in a rectangular formation of [imath]m[/imath] rows and [imath]n[/imath] columns in such a way that in each row each person is taller than the one to his or her left. Suppose that the leader rearranges the people in each column in increasing order of height from front to back. Show that the rows are still arranged in increasing order of height from left to right. EDIT: Thanks. I didn't know that this question was already here, only differently phrased.
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181615
|
[imath]m \times n[/imath] persons stand in [imath]m[/imath] rows and [imath]n[/imath] columns
[imath]m \times n[/imath] persons stand as a rectangle of [imath]m[/imath] rows and [imath]n[/imath] columns. Each one is taller than the one next to his left. Now, if commander orders each column to sort by height, prove that after such sorting, each one is still taller than the one next to his left. I'm reading a Combinatorics book on my own, but some exercises I can't solve... many thanks!
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1917662
|
If [imath]A/I \cong A/J[/imath] as rings and [imath]I\subseteq J,[/imath] then [imath]I=J.[/imath]
Let [imath]A[/imath] be a commutative Noetherian ring and let [imath]I[/imath] and [imath]J[/imath] be ideals of [imath]A.[/imath] Suppose that [imath]I\subseteq J[/imath] and that [imath]A/I \cong A/J[/imath] as rings. I want to prove that [imath]I=J.[/imath] Observations so far: 1) If we drop the hypothesis [imath]I\subseteq J,[/imath] then the result is false: [imath]\mathbb{Q}[X,Y]/(X)\cong \mathbb{Q}[X,Y]/(Y) \text{ but } (X)\neq (Y).[/imath] 2) If [imath]A/I\cong A/J[/imath] as [imath]A[/imath]-modules, then [imath]I=J[/imath]: [imath]I=\mathrm{Ann}_A(A/I)=\mathrm{Ann}_A(A/J)=J.[/imath] I'm now a bit stuck as to how to proceed. Any hints would be most helpful! Many thanks :)
|
1905186
|
Let [imath]R[/imath] be a commutative Noetherian ring (with unity), and let [imath]I[/imath] be an ideal of [imath]R[/imath] such that [imath]R/I \cong R[/imath]. Then is [imath]I=(0)[/imath]?
Let [imath]R[/imath] be a commutative Noetherian ring (with unity), and let [imath]I[/imath] be an ideal of [imath]R[/imath] such that [imath]R/I \cong R[/imath]. Then is it true that [imath]I=(0)[/imath] ? I know that surjective ring endomorphism on Noetherian ring is injective also, and since there is a natural surjection of [imath]R[/imath] onto [imath]R/I[/imath] so we get a surjection of [imath]R[/imath] onto [imath]R[/imath] but the problem is I can not determine the map explicitly and I am not sure about the positiveness of the statement. Please help. Thanks in advance.
|
1525666
|
Definition of the norm of a function in [imath]L^2(\mathbb{R^n})[/imath]?
Have I got the following correct - Let [imath]f \in L^2(\mathbb{R^n})[/imath], with [imath]n = 3[/imath] for the sake of example. Is the the norm of [imath]f[/imath] in this space as follows? [imath]||f||_{L^2(\mathbb{R^3})} = (\int_{\mathbb{R^n}} |f(x)|^2 dx)^{\frac{1}{2}}[/imath] [imath]= (\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} |f(x,y,z)|^2 dxdydz)^{\frac{1}{2}}[/imath]
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1918356
|
Norm [imath]L^2[/imath] of a multivariate function
I know that the [imath]2[/imath]-norm of a function [imath]f(x)[/imath] of one real variable is: [imath]||f(x)||_2 = \bigg (\int_x f(x)^2 dx \bigg )^{\frac{1}{2}}[/imath] How does this extend to a multivariate function [imath]f : \mathbb{R}^n \rightarrow \mathbb{R}[/imath]?
|
1918016
|
Prove: [imath]|x-y|\leq |x|+|y|[/imath]
Can you help me to prove that [imath]|x-y|\leq |x|+|y|[/imath] I get a proff of this equality, but it's very short and I don't know if it's correct.
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3044768
|
How to prove [imath]|x − y| ≤ |x| + |y|[/imath], proof and reasoning
Prove that, for all [imath]x, y ∈ \mathbb{R}[/imath], we have [imath]|x − y| ≤ |x| + |y|[/imath]. Can I say . [imath]|x − y|^2 ≤ (x − y)^2[/imath], and work from there? Thank you
|
1917646
|
[imath]\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2[/imath], for all integers [imath]n\ge 1[/imath].
Let [imath]x_1,\ldots, x_n[/imath] be positive integers. Use mathematical induction to prove that [imath]\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2[/imath] for all integers [imath]n \ge 1[/imath]. (Given Hint: For all positive integers [imath]a[/imath] and [imath]b[/imath], [imath]\frac{a}{b}+\frac{b}{a} \ge 2[/imath].) Can anyone help? Thank you.
|
504240
|
Proof that [imath]\left(\sum^n_{k=1}x_k\right)\left(\sum^n_{k=1}y_k\right)\geq n^2[/imath]
If [imath]x_1,...,x_n[/imath] are positive real numbers and if [imath]y_k=1/x_k[/imath], prove that [imath]\left(\sum^n_{k=1}x_k\right)\left(\sum^n_{k=1}y_k\right)\geq n^2.[/imath] I've been learning induction, and I've come across this problem that I really can't even break down and begin to think about. I've been told it has something to do with Cauchy-Schwarz, but I cannot figure out how to apply it. I would appreciate help figuring out how to go about and formulate this proof. Thanks!
|
1918055
|
Meaning of [imath]GL(n)[/imath] in this expression [imath]f: GL(n) \subset Mat_n (R) \rightarrow R^n, A \mapsto f(A) = A^{-1}b[/imath]
I was reading about condition number of a matrix (from Numerical Analysis in Modern Scientific Computing: An Introduction, 2nd edition, by Andreas Hohmann and Peter Deuflhard), and it's stated: Next we take perturbations in [imath]A[/imath] into account, too. For the purpose of consider matrix [imath]A[/imath] as input quantity [imath]f: GL(n) \subset Mat_n (R) \rightarrow R^n, A \mapsto f(A) = A^{-1}b[/imath] What does [imath]GL(n)[/imath] actually mean? What does [imath]G[/imath] and [imath]L[/imath] (or [imath]GL[/imath]) stand for?
|
534602
|
General linear group of a vector space
I am reading a text on Lie groups. There is a whole chapter devoted to the group of invertible real or complex matrices of degree [imath]n[/imath], which are called the general linear groups (complex and real). Further in the text, there is a reference to a general linear group of a vector space, in general. What is meant exactly by that? And why do we use the same terminology?
|
1918801
|
Show that there is a [imath]\xi\in(a,b)[/imath] s.t. [imath]\int_a^bf(x)g(x)\,dx=f(\xi)\int_a^b g(x)\,dx[/imath]
If [imath]f[/imath] and [imath]g[/imath] are continuous on an interval [imath][a,b][/imath] and [imath]g[/imath] is nonnegative, show that there is a number [imath]\xi \in (a,b)[/imath] such that [imath]\int_a^b f(x)g(x) \; dx = f(\xi) \int_a^b g(x) \; dx.[/imath] I understand the proof for the case when [imath]\xi \in[a,b],[/imath] however, I'm not sure how to proceed when [imath]\xi \in(a,b).[/imath]
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1916300
|
Question about condtion of MVT for integral.
For [imath]f \in C[a,b], g\ge0[/imath] (or [imath]g\le0[/imath]) on [imath][a,b][/imath], then there exists [imath]c\in(a,b)[/imath] [imath]\int _a^b f(x)g(x)dx = f(c)\int_a^bg(x)dx[/imath]. It is MVT for integral. Why [imath]c \in (a,b)[/imath] ? How about [imath]c\in[a,b][/imath]? Is it wrong? I am wondering this because, in proving progress, by IVT, there exist [imath]m,M[/imath] s.t [imath]m\le f(x)\le M[/imath] so [imath]m\int_a^bg(x)dx[/imath] [imath]\le[/imath] [imath]\int_a^b f(x)g(x)dx[/imath] [imath]\le[/imath] [imath]M\int_a^bg(x)dx[/imath]. when [imath]\int_a^bg(x)dx>0[/imath] then [imath]m\le[/imath] [imath]\frac{\int_a^b f(x)g(x)dx} {\int_a^bg(x)dx}[/imath] [imath]\le[/imath] [imath]M[/imath] and [imath]\frac{\int_a^b f(x)g(x)dx} {\int_a^bg(x)dx}[/imath] becomes [imath]f(c)[/imath], so it seems [imath]c\in [a,b][/imath] might work. So, I want to know why [imath]c\in(a,b)[/imath], or [imath]c\in[a,b] [/imath] would be okay.
|
1918374
|
Why is : [imath] \mathrm{dim}_{ \mathbb{C} } V(d,n) = \binom{d+n-1}{n-1} [/imath]?
Let [imath] V(d,n) [/imath] be the vector space over [imath] \mathbb{C} [/imath], of forms of degree : [imath]d[/imath] , in [imath] \mathbb{C} [X_1 , \dots , X_n ] [/imath]. How to show that : [imath] \mathrm{dim}_{ \mathbb{C} } V(d,n) = \binom{d+n-1}{n-1} [/imath] ? I think [imath] \mathrm{dim}_{ \mathbb{C} } V(d,n) = \mathrm{Card} \big( \{ (i_1 , \dots , i_n ) \in \{ 1 , \dots , n \}^n \ | \ i_1 + \dots + i_n = d \} \big) [/imath] but, why is : [imath] \mathrm{Card} \big( \ \{ (i_1 , \dots , i_n ) \in \{ 1 , \dots , n \}^n \ | \ i_1 + \dots + i_n = d \ \} \big) = \binom{d+n-1}{n-1} [/imath] ? This is in reality a combinatoric question, so we need to take in account repetition, without order. but, i still not understand why it's equal to : [imath] \binom{d+n-1}{n-1} [/imath]. Thank you in advance.
|
516850
|
Show that [imath]\dim H^0(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(m)) = {n + m \choose n}[/imath] if [imath]m \geq 0[/imath], and [imath]0[/imath] otherwise.
Show that [imath]\dim H^0(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(m)) = {n + m \choose n}[/imath] if [imath]m \geq 0[/imath], and [imath]0[/imath] if [imath]m < 0[/imath]. This statement came up in an algebraic geometry text with no explanation provided, and I'm trying to understand why it's true. Thoughts so far: Since [imath]H^0(\mathbb{P}^n, \mathcal{O}_{\mathbb{P}^n}(m))[/imath] is the vector space of regular sections of [imath]\mathcal{O}_{\mathbb{P}^n}(m)[/imath] over [imath]\mathbb{P}^n[/imath], we can try to find a basis. Since the regular sections are the degree [imath]m[/imath] polynomials in [imath]x_0, \dots, x_n[/imath], it seems like the collection of degree [imath]m[/imath] monomials in [imath]x_0, \dots, x_n[/imath] should work. But I don't see how there are [imath]{n + m \choose n}[/imath] of these. Is it just a basic counting argument that I'm struggling with? On a side note, I think I understand what global sections of [imath]\mathcal{O}_{\mathbb{P}^n}(m)[/imath] look like, but how can we describe the regular sections over a distinguished open set [imath]D(f)[/imath]? Is it something like degree [imath]m[/imath] polynomials with denominator [imath]f[/imath]?
|
1919219
|
Prove that there is an integer [imath]k[/imath] such that [imath]2^k[/imath] is starting with [imath]999[/imath]
As the title says, I want to prove that there is a natural number [imath]k[/imath] such that [imath]2^k[/imath] is starting with [imath]999[/imath]. Can you help me please ?
|
789043
|
decimal representation of [imath]2^m[/imath] starts with a particular finite sequence of decimal digits
Given a finite sequence of decimal digits [imath]a_1,a_2,...,a_n[/imath] prove that there exists a natural number [imath]m[/imath] such that decimal representation of [imath]2^m[/imath] starts with that sequence of digits. Thanks for your help :)
|
1919229
|
Prove: [imath]|x|-|y|\leq |x-y|[/imath]
Can you help me proving this equality? [imath]|x|-|y|\leq |x-y|[/imath] I think that it's possible to give a too short proof if the things are written correctly.
|
877096
|
[imath]|x| - |y| \leq |x-y|?[/imath]
Is there a clever way to show that [imath]|x| - |y| \leq |x-y|[/imath] I believe I can think of a way to solve this using cases, but the book I'm working out of said that "A very short proof is possible if you write things in the right way" which has motivated me to figure out this quick way, but I've failed thus far and any help would be appreciated
|
1919550
|
If [imath]\frac{a^n-1}{b^n-1} [/imath] is a natural number for every [imath]n[/imath] then [imath]a = b^m[/imath]
Consider two natural numbers [imath]a,b[/imath]. Prove that [imath]a = b^m[/imath] if [imath] \frac{a^n-1}{b^n-1} [/imath] is natural for all [imath]n \in N[/imath]. I tried to assume that there is exist some [imath]k [/imath], such : [imath]a = b^m + k[/imath], so i get [imath]\displaystyle \frac{\sum_{0}^{n}b^{n-i+m}k^i -1}{b^n-1}[/imath] but I guess I just waste my time. Need some good idea.
|
417340
|
Do there exist two primes [imath]p such that p^n-1\mid q^n-1 for infinitely many n?[/imath]
We can prove that there is no integer [imath]n>1[/imath] such that [imath]2^n-1\mid 3^n-1[/imath]. This leads to the following question: Is it true that for every pair of primes [imath]p<q[/imath] there are only finitely many integers [imath]n[/imath] such that [imath]p^n-1\mid q^n-1[/imath]? Are there two primes [imath]p<q[/imath] and an integer [imath]n>p+q[/imath] such that [imath]p^n-1\mid q^n-1[/imath]?Is it true that if [imath]n>6[/imath] then [imath]2^n-1\nmid 5^n-1[/imath]? Edit: Here are some examples: \begin{array}{ll} 2^{36}-1\mid 41^{36}-1, &3^{12}-1\mid 97^{12}-1,\\ 5^{6}-1\mid 37^{6}-1, &7^{4}-1\mid 151^{4}-1. \end{array} Now I prove there is no integer [imath]n>1[/imath] such that [imath]2^n-1\mid 3^n-1.[/imath] Proof: Denote [imath]A=2^n-1[/imath] and [imath]B=3^n-1[/imath]. If [imath]n[/imath] is even then [imath]3[/imath] divides [imath]A[/imath] but not [imath]B[/imath], a contradiction. If [imath]n[/imath] is odd then [imath]A\equiv -5\pmod {12}[/imath]. Since every prime greater than [imath]3[/imath] is [imath]\equiv \pm1,\pm5 \pmod {12}[/imath], some prime factor [imath]p[/imath] of [imath]A[/imath] must be congruent to [imath]\pm5 \pmod {12}[/imath]. As [imath]p \mid B[/imath], we have [imath]3^{n+1}\equiv 3 \pmod p[/imath]. Since [imath]n+1[/imath] is even, we get [imath](\frac{3}{p})=1[/imath], a contradiction again.
|
1918065
|
Prove that [imath]f: \mathbb{R} \ni x \mapsto \left(x-x_{0}\right) \cdot g(x) \in \mathbb{R}[/imath] is continuous
Let [imath]g \cdot \mathbb{R} \rightarrow \mathbb{R}[/imath] with [imath]|g(x)| \leq 1 [/imath] for all [imath]x \in \mathbb{R}[/imath]. Prove that [imath]f: \mathbb{R} \ni x \mapsto \left(x-x_{0}\right) \cdot g(x) \in \mathbb{R}[/imath] is continuous. Ok I have trouble because not any specific [imath]x_{0}[/imath] is given. Also I think the inequality [imath]|g(x)| \leq 1[/imath] will cause trouble / no easy birth. We will have to prove it for [imath]|g(x)| = 1[/imath] [imath]|g(x)| < 1[/imath] is that right so far? [imath]f(x) = (x-x_{0}) \cdot 1 = (x-x_{0})[/imath] So in this case, the limit of function [imath]f[/imath] would be [imath]x-x_{0}[/imath]. Using squeeze theorem on this we get: [imath](x-x_{0}) \leq (x-x_{0}) \cdot 1 \leq (x-x_{0})[/imath] I think doesn't make much sense I wrote here but the limit should be [imath](x-x_{0})[/imath] if [imath]|g(x)| = 1[/imath]. 2.If [imath]|g(x)| < 1[/imath]: [imath]0 \leq (x-x_{0}) \cdot |g(x)| \leq (x-x_{0})[/imath] Hmm.. I don't know what to do don't know what [imath]x_{0}[/imath] is ... :( I have task from exam 2005 published by someone who wrote it already, old student. Maybe he write wrong? I can't imagine how show continuity here, we don't know [imath]x_{0}[/imath] and not much info about [imath]g(x)[/imath] too?
|
1916546
|
Prove that the function is continuous and differentiable (not as easy as it sounds imo)
From an old exam: The not-necessary-continuous function [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] is given. We know that [imath]|f(x)| < 1[/imath] for all [imath]x \in \mathbb{R}[/imath]. Let [imath]a(x) = (x-3)f(x)[/imath] and [imath]b(x) = (x-3)^{2}f(x)[/imath]. Prove that [imath]a[/imath] is continuous at [imath]x_{0}=3[/imath] Prove that [imath]b[/imath] is differentiable at [imath]x_{0}=3[/imath] 1.We need to show that left and right side limit are the same. If that's the case, the function is continuous: [imath]\lim_{x\rightarrow 3^{-}}\left((x-3)f(x)) \right )= (3-3)\cdot f(3)= 0 \cdot f(3)= 0[/imath] [imath]\lim_{x\rightarrow 3^{+}}\left((x-3)f(x)) \right )= (3-3)\cdot f(3)= 0 \cdot f(3)= 0[/imath] Thus the function [imath]a[/imath] is continuous at [imath]x_{0}=3[/imath]. 2.I will use difference-quotient for proving differentiability: [imath]\lim_{x\rightarrow 3^{-}}\left ( \frac{b(x)-b(x_{0})}{x-x_{0}} \right )= \lim_{x\rightarrow 3^{-}} \left ( \frac{(x-3)^{2}f(x)-(3-3)^{2}f(3)}{x-3} \right ) = \lim_{x\rightarrow 3^{-}} \frac{(x-3)^{2}f(x)-0 \cdot f(3)}{x-3}= \lim_{x\rightarrow 3^{-}}\frac{(x-3)^{2}f(x)}{x-3}= \lim_{x\rightarrow 3^{-}}(x-3)f(x) = 0[/imath] Apply same for the right side and get [imath]0[/imath] as well. Thus [imath]b[/imath] is differentiable at [imath]x_{0} = 3[/imath]. Did I do this task correctly or it's completely wrong..? :o
|
1920034
|
Prove the set [imath]\Bbb Z^+ \times\Bbb Z^+ \times \Bbb Z^+[/imath] is countable.
Prove the set [imath]\Bbb Z^+ \times \Bbb Z^+ \times \Bbb Z^+[/imath] is countable. I know you can use diagonal processing to prove that [imath]\Bbb Z^+ \times \Bbb Z^+[/imath] is countable, but how would you go from there?
|
127695
|
Proving that a set is countable by finding a bijection
[imath]Z[/imath] is the set of non-negative integers including [imath]0[/imath]. Show that [imath]Z \times Z \times Z[/imath] is countable by constructing the actual bijection [imath]f: Z\times Z\times Z \to \mathbb{N}[/imath] ([imath]\mathbb{N}[/imath] is the set of all natural numbers). There is no need to prove that it is a bijection. After searching for clues on how to solve this, I found [imath](x+y-1)(x+y-z)/z+y[/imath] but that is only two dimensional and does not include [imath]0[/imath]. Any help on how to solve this?
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1920354
|
Example of a function [imath]f : A \times B \to X[/imath] that isn't continuous but every [imath]f_a : B \to X, f_b : A \to X[/imath] is
Let's say that we have a function [imath]f : A \times B \to X[/imath]. For every [imath]a \in A[/imath] we get a function [imath] f_a : B \to X, \ f_a(b) = f(a, b),[/imath] and for every [imath]b \in B[/imath] we get [imath]f_b : A \to X[/imath] in similar fashion. It isn't hard to show that continuity of [imath]f: A \times B \to X[/imath] (in product topology) implies continuity of every [imath]f_a, f_b[/imath]. But I suspect that converse isn't true, just because I have never seen such theorem. I'll be grateful for some easy examples of such behavior (or a theorem disproving my hunch).
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393111
|
Continuity of a function of product spaces
Let [imath]f: X \times Y \rightarrow Z [/imath] such that. [imath] \forall x_0 \in X , f_{x_0} : Y \rightarrow Z, y\mapsto f(x_0,y)[/imath] is continuous. [imath] \forall y_0 \in Y , f_{y_0} : X \rightarrow Z, x \mapsto f(x,y_0)[/imath] is continuous. How i can prove that [imath]f[/imath] is continuous? Thanks in advance.
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1750061
|
If [imath]V[/imath] is finite-dimensional with [imath]J : V \to V[/imath] such that [imath]J^2 = -id[/imath], then [imath]V[/imath] has even dimension
Let [imath]V[/imath] be a [imath]\Bbb R[/imath]-vector space, with [imath]J[/imath] being an endomorphism [imath]J: V \to V[/imath] with [imath]J^2=-id[/imath] (identity). I already had to show that [imath]V[/imath] became a [imath]\Bbb C[/imath]-vector space with the scalar multiplication: [imath](a+bi)\cdot v=av+bJ(v).[/imath] Now [imath]V[/imath] is set to be finite-dimensional. The problem is the following: Show that [imath]\operatorname{dim}_{\Bbb R} V[/imath] is even. Any ideas on how to show this? (I've seen the post If [imath]V[/imath] is a vector space, then, proving that... which has the same problem but I don't really get the solution that is offered there, they write [imath]J[/imath] as a matrix, so far so good, but from there I don't get it.)
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249924
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Let [imath]T:V→V[/imath] be a linear transformation satisfying [imath]T^2(v)=-v[/imath] for all [imath]v\in V[/imath]. How can we show that [imath]n[/imath] is even?
Let [imath]V[/imath] be a real n dimensional vector space & let [imath]T:V→V[/imath] be a linear transformation satisfying [imath]T^2(v)=-v[/imath] for all [imath]v\in V[/imath]. Then how can we show that [imath]n[/imath] is even? I am completely stuck on it. Can anybody help me please?
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1920807
|
Conjecture for a closed form of a parametric integral
Let us assume that [imath]n \in \mathbb{N}[/imath]. It is quite easy to prove that: [imath]\int_{0}^{1} x^{n-1} \log (1-x) \, {\rm d}x = - \frac{\mathcal{H}_n}{n}[/imath] since if we use the series representation of [imath]\mathcal{H}_m[/imath] we have that: \begin{equation} \mathcal{H}_m = \sum_{k=1}^{\infty} \left [ \frac{1}{k} - \frac{1}{m+k} \right ] \tag{1}\end{equation} and it follows that our initial integral , call that [imath]\mathcal{J}[/imath] , \begin{align*} \mathcal{J} &=\int_{0}^{1} x^m \log(1-x) \, {\rm d}x \\ &= - \int_{0}^{1}x^m \sum_{n=1}^{\infty} \frac{x^n}{n} \, {\rm d}x \\ &= - \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1}x^{m+n} \, {\rm d}x\\ &= - \sum_{n=1}^{\infty} \frac{1}{n \left ( n+m+1 \right )}\\ &= - \sum_{n=1}^{\infty}\left[\frac{1}{(m+1)n} - \frac{1}{(m+1) \left ( m+n+1 \right )} \right]\\ &= - \frac{1}{m+1} \sum_{n=1}^{\infty} \left [ \frac{1}{n} - \frac{1}{n+m+1} \right ]\\ &\overset{(1)}{=} - \frac{\mathcal{H}_{m+1}}{m+1} \end{align*} Now let us raise the exponent of [imath]\log (1-x)[/imath] by one. That is we now consider the integral [imath]\int_{0}^{1} x^{n-1} \log^2 (1-x) \, {\rm d}x = \frac{1}{n} \left [ \mathcal{H}_n^2 + \mathcal{H}_n^{(2)} \right ][/imath] The derivation is pretty much easy. What can we say for the integral [imath]\mathcal{J}(n, m) =\int_{0}^{1} x^{n-1} \log^m (1-x) \, {\rm d}x [/imath] I have a conjecture that states that it will involve a sum of [imath]\mathcal{H}_n[/imath] raised to the corresponding power of the log and a sum of [imath]\mathcal{H}_n^{(m)}[/imath]. However, I am unable to make any progress for the general case. Diffing the Beta won't help us here. This is what I've done for the [imath]m=2[/imath] case. I just diffed the Beta twice. Any clever approach? Also what we can say for the integral: [imath]\mathcal{J}^*(n, m) =\int_{0}^{1} x^{n-1} \log^m (1+x) \, {\rm d}x [/imath] Well, one last thing one should note is that both integrals actually connect to Stirling Numbers of first kind. Maybe we could use the Taylor series expansion of [imath]\log^m (1 \pm x)[/imath] and reduce the problem to Stirling numbers. I am not that familiar though.
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1839100
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Is there a closed form for the integral [imath]\int_0^1 x^n \log^m (1-x) \, {\rm d}x[/imath]?
Let [imath]n \in \mathbb{N}[/imath]. We know that: [imath]\int_0^1 x^n \log(1-x) \, {\rm d}x = - \frac{\mathcal{H}_{n+1}}{n+1}[/imath] Now, let [imath]m , n \in \mathbb{N}[/imath]. What can we say about the integral [imath]\int_0^1 x^n \log^m (1-x) \, {\rm d}x[/imath] For starters we know that [imath]\displaystyle \log^m (1-x)=m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{k!} x^k[/imath] where [imath]s(k, m)[/imath] are the Stirling numbers of first kind. Thus \begin{align*} \int_{0}^{1} x^n \log^m (1-x) \, {\rm d}x &=m! \int_{0}^{1}x^n \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{k!} x^k \\ &= m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{m!} \int_{0}^{1}x^{n+m} \, {\rm d}x\\ &= m! \sum_{k=m}^{\infty} (-1)^k \frac{s(k, m)}{m!} \frac{1}{m+n+1} \end{align*} Can we simplify? I know that Striling numbers are related to the Harmonic number but I don't remember all identities.
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1920288
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Proof using induction: [imath]r![/imath] divides the product of [imath]r[/imath] consecutive integers
Prove by induction that the product [imath]n(n+1)...(n+r-1)[/imath] of any [imath]r[/imath] consecutive numbers is divisible by [imath]r![/imath]. In the inductive case i've[imath]n(n+1)...(n+k-1)(n+k)[/imath] I was not able to find a way to prove this is divisible by (k+1)!. I try to use distributive property in (n+k).
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663548
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Proving [imath]r![/imath] divides the product of r succesive positive integers
I have to prove the following theorem: Prove that the product of [imath]r[/imath] consecutive positive integers in divisible by [imath]r![/imath] I am having a hard time getting a generalization down for the full set of real numbers, if I start from 1 and work up to r, I have the following: [imath]r!k=\prod_{i=1}^{r}n_i[/imath] Can easily prove the base case of this, (n=1), and then go in to prove: [imath](r+1)!k=\prod_{i=1}^{r+1}n_i[/imath] Expand that out and get: [imath](r+1)r!k=n(n+1)(n+2).....(n+r)(n+r+1)[/imath] Can say that the product of the first [imath]r[/imath] elements in equal to [imath]r!k[/imath] by our base case. Leaving using with: [imath](r+1)k=(n+r+1)[/imath] Not sure where I can go from here, n is the integer that we start at, so how can I get it to work out to be equal to our induction hypothesis?
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1920504
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Convergence test for: [imath]\sum\limits_{n = 1}^{\infty} \frac{n}{e^n}[/imath]
I know, for example, that the series: [imath]\sum\limits_{n= 1}^{\infty} \frac{1}{e^n}[/imath] is a geometric series of the form [imath]\sum\limits_{n = 1}^{\infty} k x^n[/imath], where [imath]k = 1[/imath] and [imath]x = \frac{1}{e}[/imath] and it is convergent1. But when I have, as in my case: [imath]\sum\limits_{n = 1}^{\infty} \frac{n}{e^n}[/imath] a ratio of functions I'm struggling to find a method to find whether the series is convergent/ divergent and to find a value in the former case. Thus the following question arises: What series convergence method to use in case of ratio of functions? 1. And its value is: [imath]\frac{k}{1 - x} - 0^{th}term[/imath]
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1736830
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Determining convergence of [imath]\frac{n}{e^n}[/imath]
I am trying to determine if [imath]\sum_{i=1}^\infty \frac{n}{e^n}[/imath] is converging This is what I have so far [imath]\sum_{i=1}^\infty \frac{n}{e^n}[/imath] converges by the geometric series test since [imath]\frac{1}{e}[/imath] < 1 This seems too simple of answer and was wondering if this was even a geometric sequence.
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1921249
|
What are the conditions for one-one and onto of a given liner transformation?
Let [imath]T:\Bbb R^4\longrightarrow\Bbb R^4[/imath] be a linear transformation satisfying [imath]T^3+3T^2=4I[/imath], where [imath]I[/imath] is the identity transformation. Then the linear transformations [imath]S=T^4+3T^3-4I[/imath] is a). one-one but not onto. b). onto but not one-one. c). Invertible. d). Non-invertible. on solving i get [imath]S=4T-4I[/imath], but how does this suggest invertibilty? Also, Logically thinking option [imath]a,b[/imath], and [imath]d[/imath] are interlinked. So if only one choice is correct then i could simply say that yes option [imath]c[/imath] is correct. But can someone give the general aproach?
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1411730
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Let [imath]T,S[/imath] be linear transformations, [imath]T:\mathbb R^4 \rightarrow \mathbb R^4[/imath], such that [imath]T^3+3T^2=4I, S=T^4+3T^3-4I[/imath]. Comment on S.
Let [imath]T,S[/imath] be linear transformations, [imath]T:\mathbb R^4 \rightarrow \mathbb R^4[/imath], such that [imath]T^3+3T^2=4I, S=T^4+3T^3-4I[/imath]. Then S is: one-one but not onto onto but not one one invertible non-invertible (One or more correct options) My attempt: [imath]S=T^4+3T^3-4I=T(T^3+3T^2)-4I=T(4I)-4I=4T-4I[/imath] How do I go about proving or disproving my options? My thoughts on options # 1. and 2.: I'm guessing [imath]S[/imath] is one-one if [imath]Ker(S)=\theta[/imath] is one way to go, but how do I obtain the kernel when I don't know what the transformation is? [imath]Ker(S)=\theta[/imath] if [imath]T=I[/imath], which does satisfy [imath]T^3+3T^2=4I[/imath]. But so can other [imath]T[/imath]s. My thoughts on options # 3. and 4.: [imath]S[/imath] is invertible if [imath]det(S)\neq0[/imath], which is possible [imath]det(T-I)\neq0[/imath], i.e. [imath]T\neq I[/imath]. How do I show that? Please help!
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1141219
|
Algebraic field extensions: Why [imath]k(\alpha)=k[\alpha][/imath].
If [imath]K[/imath] and [imath]k[/imath] are fields, [imath]K\supset k[/imath] is a field extension and [imath]\alpha \in K[/imath] is algebraic over [imath]k[/imath], then we denote by [imath]k[\alpha][/imath] the set of elements of [imath]K[/imath] which can be obtained as polynomial expressions of [imath]\alpha[/imath]. [imath]k[\alpha] = \left\{P(\alpha): \quad P\in k[X] \right\} [/imath] Also, we denote by [imath]k(\alpha)[/imath] the smallest subfield of [imath]K[/imath] containing both [imath]k[/imath] and [imath]\{\alpha\}[/imath]. This is easily seen to be equal to the set of fractions of polynomial expressions in [imath]\alpha[/imath] (just thinking about the way the "smallest" subfield must be generated): [imath]k(\alpha) = \left\{\dfrac{P(\alpha)}{Q(\alpha)}: \quad P,Q\in k[X] \text{ and } Q(\alpha)\neq 0 \right\} [/imath] If [imath]M[/imath] is the minimal polynomial of [imath]\alpha[/imath], it is easily shown that [imath]k[\alpha]\simeq k[X]/\langle M\rangle[/imath] is a field and therefore a subfield of [imath]k(\alpha)[/imath] containing both [imath]k[/imath] and [imath]\{\alpha\}[/imath]. Therefore [imath]k(\alpha)=k[\alpha][/imath], because [imath]k(\alpha)[/imath] is the smallest. It follows that all quotients [imath]\dfrac{P(\alpha)}{Q(\alpha)}[/imath] are equal to [imath]H(\alpha)[/imath] for some polynomial [imath]H\in k[X][/imath]. What I want to see is a more direct proof of this fact. Given [imath]P[/imath] and [imath]Q[/imath], how do you produce this [imath]H[/imath]?
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2372433
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Field extension and its inverse elements
Let [imath]\mathbb{Q}[\alpha]=\{g(\alpha); g\in \mathbb{Q}[X]\}[/imath] be a field extension of the rational numbers. The field extension [imath]\mathbb{Q}[\alpha][/imath] is the smallest possible subring of [imath]\mathbb{C}[/imath] containing the element [imath]\alpha[/imath] and the rational numbers. It can be seen by using the undamental theorem on homomorphisms that [imath]\mathbb{Q}[\alpha][/imath] is even a field itself. While I understand the proof using the homomorphism theorem, and see that [imath]\mathbb{Q}[\alpha][/imath] is a ring, I can't see why it satisfies the inverse element (for multiplication) condition needed to be a field. In general [imath]g(\alpha)[/imath] can be something like [imath]g(\alpha)=\sum_{i=0}^\infty c_i\alpha^i, c_i\in \mathbb{Q}[/imath]. So how would I construct the inverse element, for example to [imath]\alpha[/imath]? I can only do it for the addition but not for multiplication.
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1921435
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[imath]\lim_{x \rightarrow 0}\left(\frac{1}{\sin^2(x)} -\frac{1}{x^2}\right)[/imath] and the problem of [imath]\infty-\infty[/imath]
I'm trying to calculate [imath]\lim_{x \rightarrow 0}\left(\frac{1}{\sin^2(x)} -\frac{1}{x^2}\right)[/imath] but I'm having the problem of getting [imath]\infty - \infty[/imath] which I read to be indeterminate. I cannot figure out how could I manipulate the above to get rid of the indeterminacy.
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1338411
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Why doesn't using the approximation [imath]\sin x\approx x[/imath] near [imath]0[/imath] work for computing this limit?
The limit is [imath]\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)[/imath] which I'm aware can be rearranged to obtain the indeterminate [imath]\dfrac{0}{0}[/imath], but in an attempt to avoid L'Hopital's rule (just for fun) I tried using the fact that [imath]\sin x\approx x[/imath] near [imath]x=0[/imath]. However, the actual limit is [imath]\dfrac{1}{3}[/imath], not [imath]0[/imath]. In this similar limit, the approximation reasoning works out.
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1921473
|
Number of vertices after taking out an edge from a tree
I have a tree (v is number of vertex and d is the maximum degree)whose maximum degree is [imath]\geq 2[/imath]. I want to show that, if I take out an specific edge from this tree, I'll get [imath]2[/imath] trees whose number of the vertex in each tree is no more than [imath]\frac{v(d-1)}d[/imath]
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1918980
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Graph theory: Tree holds two sub-tree with an amount of vertices
I was asked the following question: Let [imath]T=\left(V,E\right)[/imath] be a tree, and let [imath]d[/imath] be the maximum degree of a vertex in that tree. Assume that [imath]d \geq 2[/imath]. Prove there an edge in the tree such that, when we remove it from [imath]T[/imath], we will get two trees each of which has at most [imath]\left\lceil \dfrac{d-1}{d}\left|V\right|\right\rceil[/imath] vertices. I tried going by induction on [imath]d[/imath] the maximum degree while I set the number of vertices and it didn't worked. also tried to do induction on the number of vertices and it didn't worked too. Thanks for the help, Yoav
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1701616
|
If [imath]E[/imath] is a finitely presented left [imath]A[/imath]-module, then tensor product by E commutes with direct product.
Please can anyone help me to prove this statement: If [imath]E[/imath] is a finitely presented left [imath]A[/imath]-module, then for every family [imath](F_i)[/imath] of right [imath]A[/imath]-modules the canonical homomorphism [imath]\phi: E\bigotimes_A \prod F_i \rightarrow \prod (E\bigotimes _A F_i)[/imath] is an isomorphism. I have proved that the map is surjective, and I want anyone to help me to prove that the map is injective. Thank you!
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1916457
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direct product commutes with tensor product?
Let [imath](A_i)_{i\in I}[/imath] be a family of right [imath]R[/imath]-modules and [imath]M[/imath] be a left [imath]R[/imath]-module, where [imath]I[/imath] is an index set. The natural homomorphism [imath]\varphi:(\prod_{i\in I}A_i)\otimes_RM\to \prod_{i\in I}(A_i\otimes_RM)[/imath] given by [imath](a_i)\otimes m\mapsto(a_i\otimes m)[/imath] is not always a bijection. It is easy to obtain [imath]\varphi[/imath] is a surjection provided [imath]M[/imath] is finitely generated. If we assume that [imath]M[/imath] is finitely presented, can we prove [imath]\varphi[/imath] is a bijection?
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1921833
|
[imath]{\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c}+ \sqrt{c}\over b+a} \ge {9-3\sqrt{3}\over2\sqrt{a+b+c}}[/imath]
[imath]{\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge {9+3\sqrt{3}\over2\sqrt{a+b+c}}[/imath] I tried AM-GM, which only gives large terms without any answer. [imath]{\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge 3\sqrt[3]{\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right) \left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)}[/imath] Now this will never simplify to anything. using a different approach i got, [imath]{\sqrt{a+b+c} + \sqrt{a}\over b+c} + {\sqrt{a+b+c} + \sqrt{b}\over a+c} +{\sqrt{a+b+c} + \sqrt{c}\over b+a} \ge \sqrt{\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right) \left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)}+\sqrt{\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)\left({\sqrt{a+b+c} + \sqrt{b}\over a+c}\right)}+\sqrt{\left({\sqrt{a+b+c} + \sqrt{c}\over b+a}\right)\left({\sqrt{a+b+c} + \sqrt{a}\over b+c}\right)}[/imath] which also does not simplify further. breaking the individual terms on the LHS also does not help nor does multiplying each term on LHS by its conjugate. :< This was a introductory problem, so i guess there must exist a easy solution which i can't find. Any hints will be helpful.
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1078777
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A problem from KVS 2014
I was doing the the following problem- Prove that [imath]\frac { \sqrt { a+b+c } +\sqrt { a } }{ b+c } +\frac { \sqrt { a+b+c } +\sqrt { b } }{ c+a } +\frac { \sqrt { a+b+c } +\sqrt { c } }{ a+b } \ge \frac { 9+3\sqrt { 3 } }{ 2\sqrt { a+b+c } }. [/imath] I normalized this to [imath]a+b+c=1[/imath] and simplified to get- [imath]\frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9+3\sqrt { 3 } }{ 2 }.[/imath] By Titu's lemma, [imath]\Rightarrow \quad \frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9 }{ 3-(\sqrt { a } +\sqrt { b } +\sqrt { c } ) } .[/imath] However the RHS is maximized when [imath]\sqrt { a } +\sqrt { b } +\sqrt { c } [/imath] is maximized which is at [imath]\sqrt { 3 } [/imath], [imath]\Rightarrow \quad \frac { 1 }{ 1-\sqrt { a } } +\frac { 1 }{ 1-\sqrt { b } } +\frac { 1 }{ 1-\sqrt { c } } \ge \frac { 9 }{ 3-(\sqrt { a } +\sqrt { b } +\sqrt { c } ) } \le \frac { 9 }{ 3-\sqrt { 3 } } =\frac { 9+3\sqrt { 3 } }{ 2 } .[/imath] What went wrong? You can view the rest of the problems here.
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1921810
|
[imath]\int_{0}^{\pi/2} \frac{\arctan^2 \left ( \sin^2 x \right )}{\sin^2 x} \, {\rm d}x[/imath]
Does anyone have any ideas on how to attack this: [imath]\int_{0}^{\pi/2} \frac{\arctan^2 \left ( \sin^2 x \right )}{\sin^2 x} \, {\rm d}x[/imath] I cannot think of something that could work. Wolfram Alpha does not evaluate it either.
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1741089
|
Closed form of [imath]\int_0^{\pi/2} \frac{\arctan^2 (\sin^2 \theta)}{\sin^2 \theta}\,d\theta[/imath]
I'm trying to evaluate the closed form of: [imath]I =\int_0^{\pi/2} \frac{\arctan^2 (\sin^2 \theta)}{\sin^2 \theta}\,d\theta[/imath] So far I've tried introducing a parameters, [imath]\displaystyle I(a,b) = \int_0^{\pi/2} \frac{\arctan (a\sin^2 \theta)\arctan (b\sin^2 \theta)}{\sin^2 \theta}\,d\theta[/imath] but that doesn't lead to an integral I can manage. Expanding the series for [imath]\arctan^2 x[/imath] leads to the sum: [imath]I = \frac{\pi}{2}\sum\limits_{n=0}^{\infty}\frac{(-1)^n}{(n+1)4^{2n+1}}\binom{4n+2}{2n+1}\left(\sum\limits_{k=0}^{n}\frac{1}{2k+1}\right)[/imath] and using [imath]\displaystyle \int_0^1 x^{n-\frac{1}{2}}\log (1-x)\,dx = \frac{-2\log 2 + 2\sum\limits_{k=0}^{n}\dfrac{1}{2k+1}}{n+\frac{1}{2}}[/imath] leads to an even uglier integral: [imath]\displaystyle \Im \int_{0}^{1} \frac{1}{1+\sqrt{1-i\sqrt{x}}}\frac{\log(1-x)}{x}\,dx[/imath] among others. I got the non-square version, which seems to have a nice closed form [imath]\displaystyle \int_0^{\pi/2} \frac{\arctan (\sin^2 \theta)}{\sin^2 \theta}\,d\theta = \frac{\pi}{\sqrt{2}\sqrt{1+\sqrt{2}}}[/imath] but the squared version seems difficult. Any help is appreciated. (P.S. - I'm not familiar with Hypergeometric identities, so it would be very helpful if a proof or a reference to a proof was provided, should the need arise to use them.)
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1922478
|
[imath]0=1[/imath] through integrals?
Let's consider an indefinite integral [imath]\int \frac{dx}{x\ln x}[/imath] It can be easily calculated to be [imath]\ln(\ln x)+C[/imath], e.g. via substitution [imath]\ln x=t[/imath] or directly from [imath]\int\frac{f'(x)}{f(x)}dx=\ln|f(x)|+C[/imath]. So far so good. But when integration by parts is employed: [imath]u'=\frac{1}{x}[/imath], [imath]v=\frac{1}{\ln x}[/imath], one gets [imath]\int \frac{dx}{x\ln x}=1+\int \frac{dx}{x\ln x}[/imath] from which [imath]0=1[/imath]. Even if we plug an arbitrary constant of integration in the r.h.s. of the last equality, we'll just get that [imath]C[/imath] should be [imath]-1[/imath] for the equality to be an identity (but in general the constant of integration can be put in at the very last step of integration; like here and here), and we'll still know nothing about the integral. So, my question is: why does integration by parts fail in this case? Are there some assumptions not fulfilled that I have overseen here? An explanation that "it doesn't work so one has to use different methods" is no explanation.
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1371551
|
What is the mistake in doing integration by this method?
Integration Of a given function can be found out in many ways, For a specific function ∫1/xlogx, if we do integration by parts (∫f(x) g(x)= f(x) ∫ g(x)- ∫ [d/dx (f(x)) ∫g(x)] dx ) we get this way Let ∫1/xlogxdx = A If we do integration by parts, we have [imath]\int \frac{1}{x\log x} \, dx= \frac{1}{\log x}\cdot\int \frac{1}{x} \, dx - \int \frac{d}{dx}\left(\frac{1}{\log x}\right)\,dx\cdot\int \frac{1}{x} \, dx+c[/imath] [imath]=\frac{1}{\log x}\cdot \log x+\int \frac{1}{x\log x}\,dx+c[/imath] i.e, A = 1 + A + c Which tells us that the value of "[imath]c[/imath]" in this specific integral is [imath]-1[/imath]. So does this mean that the Integral of function [imath]\dfrac{1}{x\log x}[/imath] has only one value.! So this means only two things, either I am wrong somewhere or i am missing a point somewhere. Edit: I do know we can simple get it by writing logx as u and ¹/x as du, we directly get it as log(logx) +c. But I wanted to know the reason why we can't apply By parts when we can integrate it by substitution.
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1922151
|
Reference request: definition of [imath]H^{1/2}(\partial\Omega)[/imath].
In this note on PDE, which is based on Evans's Partial Differentail Equations, it is said that the range of the trace map on [imath]H^1(\Omega)[/imath] for a smooth domain [imath]\Omega[/imath] is [imath]H^{1/2}(\partial \Omega)[/imath]. Could anybody come up with a reference regarding a precise definition of [imath]H^{1/2}(\partial \Omega)[/imath] and the result why it is the range of the trace map?
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1095246
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The Sobolev Space [imath]H^{1/2}[/imath]
This is a very stupid question. In my course on linear PDEs, the professor used [imath]H^{1/2}[/imath] without defining it, and I have been looking on google trying to find a definition, but the only related thing I found was [imath]H^{-1/2}[/imath] as being the dual space to [imath]H^{1/2}[/imath] which does not really help. Plugging in the one half in the defintion of the standard Sobolev spaces [imath]H^m[/imath] does not make any sense. Could someone quickly help me out there? Thank you.
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1922601
|
Find the sum of digits of [imath]999...999^3[/imath] (12 nines)
What will be the sum of digits of the following expression? [imath]999999999999^3[/imath] Note that the base has [imath]12[/imath] nines. Without doing it using brute force, is there an easy and simple method to find the sum of digits. Also, in finding sum of digits of expressions like those, is there a trick that is used? Thanks in advance!
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1915130
|
Prove that the sum of digits of [imath](999...9)^{3}[/imath] (cube of integer with [imath]n[/imath] digits [imath]9[/imath]) is [imath]18n[/imath]
Someone had posted a question on this site as to what would be sum of digits of [imath]$999999999999^3$[/imath] (twelve [imath]$9s$[/imath] ) equal to? I did some computation and found the pattern that sum of digits of [imath]9^3 = 18[/imath], [imath]99^3 = 36[/imath], [imath]999^3 = 54[/imath] and so on. So, I had replied that sum of digits of [imath]999999999999^3 = 12 \cdot 18 = 216[/imath]. Can anybody help me prove this, that sum of digits of [imath](\underbrace{999\dots9}_{n\text{ times}})^3 = 18n.[/imath]
|
1921852
|
Prove the following Combinatorial identity
For any m [imath]\ge[/imath] 1. Prove that [imath]\sum_{j=0}^m\frac{1}{j+1}{m \choose j} = \frac{1}{m+1}(2^{m+1} -1 )[/imath] Attempt: I tried using subsets after rearranging the equation. Then even though the R.H.S gives the no of substes of n+1 excluding empty. I am unable to get a interpretation for L.H.S
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1847437
|
Express [imath]1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}[/imath] in a simplifed form
I need to express [imath]1 + \frac {1}{2} \binom{n}{1} + \frac {1}{3} \binom{n}{2} + \dotsb + \frac{1}{n + 1}\binom{n}{n}[/imath] in a simplified form. So I used the identity [imath](1+x)^n=1 + \binom{n}{1}x + \binom{n}{2}x^2 + \dotsb + \binom{n}{n}x^n[/imath] Now on integrating both sides and putting [imath]x=1[/imath]. I am getting [imath]\frac{2^{n+1}}{n+1}[/imath] is equal to the given expression.But the answer in my book is [imath]\frac{2^{n+1}-1}{n+1}.[/imath] Where does that -1 term in the numerator come from?
|
1922857
|
Proof that monomorphism is injective
In category theory, a monomorphism is a situation where [imath]g: A \rightarrow B \;\;[/imath] , [imath]\;\;h: A \rightarrow B\;\;[/imath] and [imath]\;\;f:B \rightarrow C[/imath] [imath]f \circ g = f \circ h \;\text{ imples in } \; g = h[/imath] I heard recently something about monomorphism to be an injective homomorphism , but it didn't make sense to me and I still couldn't prove it. How can I prove that monomorphism is an injective homomorphism (at least for monoids)?
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510169
|
Monos in [imath]\mathsf{Mon}[/imath] are injective homomorphisms.
Let [imath]f: M \to N[/imath] be a mono in [imath]\mathsf{Mon}[/imath]. I want to prove that the underlying function [imath]U(f): U(M) \to U(N)[/imath] is injective. How can this be done ? Thanks in advance.
|
1923038
|
Combinatorial proof by demonstrating bijection
I need to prove that [imath]n^2 = {n \choose 2} + {n+1 \choose2}[/imath]. I have already proved this using algebra, but I am required to use both algebra and a formal combinatorial proof which demonstrates a bijection between the right and left hand sides. If someone could show me how to get started with this proof and the general steps to take and why, I would greatly appreciate it.
|
762325
|
Combinatorial explanation for why [imath]n^2 = {n \choose 2} + {n+1 \choose 2}[/imath]
An exercise in the first chapter of Discrete Mathematics, Elementary and Beyond asks for a proof of the following identity: [imath] {n \choose 2} + {n+1 \choose 2} = n^2 [/imath] The algebraic solution is obvious to me, but less so the combinatorial logic. I think the right-hand side relates to choosing one of n elements twice in a row to produce a set of two elements, including the possibility of duplicate elements (for instance, one could choose the nth element twice in a row). However, given that the sets discussed thus far do contain duplicate elements, it's odd to then interpret n^2 as a representation of such a set. I'm not too sure what the two parts of the left side mean when taken together, either. How can I think about this intuitively based on the combinatorial meanings of the individual terms?
|
1922636
|
Closure of an open ball is equal to the closed ball (Euclidean space)
Consider the Metric space [imath](\mathbb{R}^n,\|\cdot\|)[/imath]. Show that [imath]Cl(B(x,r))=C(x,r)[/imath] with [imath]C(x,r)[/imath] the closed ball around [imath]x\in X[/imath] with radius [imath]r[/imath]. [imath]\operatorname{Cl}(B(x,r))\subseteq C(x,r)[/imath] is trivially true (the closure is the smallest closed set containing [imath]B(x,r)[/imath]). I, however, have difficulties proving it the other way around. I am able to prove it when I introduce the concept of a boundary point and show that [imath]\operatorname{Cl}(B(x,r))=B(x,r) \cup \partial(B(x,r))[/imath], but the concept of a boundary hasn't been introduced yet in the lectures. Can anyone give me a hint on how to prove this without using the concept of a boundary? Definition: Let [imath]x\in X[/imath]. We say [imath]x[/imath] is a closure point of [imath]A\subseteq X[/imath] when [imath]\forall r>0[/imath] [imath]B(x,r)\cap A\neq\emptyset[/imath]. We denote the set of all closure points as [imath]\operatorname{Cl}(A)[/imath]
|
1749818
|
What is the closure of an open ball [imath]B_X(\mathbf{a},r)[/imath] in [imath]X=\mathbb{R}^n[/imath]?
Suppose we have the open ball [imath]B_{X}(\mathbf{a},r)[/imath] and the closed ball [imath]\bar{B}_{X}(\mathbf{a},r)[/imath] of radius [imath]r[/imath] about [imath]\mathbf{a}\in\mathbb{R}^n=X[/imath] with the Euclidean metric [imath]d_2[/imath]. What is the closure of an open ball [imath]B_{X}(\mathbf{a},r)[/imath] in [imath](\mathbb{R}^n,d_2)[/imath]? First, I have defined the sets like this: [imath]B_X(\mathbf{a},r)=\{\mathbf{y}\in\mathbb{R}^n:d_2(\mathbf{a},\mathbf{y})<r\}[/imath] [imath]\bar{B}_X(\mathbf{a},r)=\{\mathbf{y}\in\mathbb{R}^n:d_2(\mathbf{a},\mathbf{y})\le r\}[/imath] Then, let [imath]U=B_X(\mathbf{a},r)[/imath] and claim the closure is [imath]\bar{U}=\{\mathbf{x}\in\mathbb{R}^n:\exists (\mathbf{x}_n)\in U\text{s.t}~\mathbf{x}_n\to \mathbf{x}\}=\bar{B}_X(\mathbf{a},r).[/imath] Since closed balls are closed then we have [imath]\bar{U}\subset\bar{B}_X(\mathbf{a},r).[/imath] But, how can I prove the other direction [imath]\bar{B}_X(\mathbf{a},r)\subset\bar{U}[/imath]? Since it's a containment argument I know that I have to take any element from [imath]\bar{B}_X(\mathbf{a},r)[/imath] and show it is in [imath]\bar{U}=\{\mathbf{x}\in\mathbb{R}^n:\exists (\mathbf{x}_n)\in U\text{s.t}~\mathbf{x}_n\to \mathbf{x}\}[/imath]. But I do not know how to get an expression for [imath]\mathbf{x}_n[/imath]. I would appreciate any help.
|
1923710
|
What is the meaning of symbols: [imath]\Omega[/imath] is [imath]C^2[/imath] and [imath]u(t,x) \in L^p(0,T;L^q(\mathbb{R}^n))[/imath]
What does it mean when we say the domain [imath]\Omega[/imath] is [imath]C^2[/imath]? (For example, in enter link description here) And what does it mean when we say the function [imath]u(t,x) \in L^p(0,T;L^q(\mathbb{R}^n))[/imath]?
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236903
|
Regularity of a domain - definition
What does it mean when we say that a domain is [imath]C^k[/imath], [imath]C^{k,\alpha}[/imath], Lipschitz, or smooth? Is there an intuitive understanding?
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1923318
|
Proof that [imath]n=250[/imath] is the largest value for [imath]n[/imath] for which [imath]1005![/imath] is divisible by [imath]10^n[/imath] where [imath]n\in\Bbb{N}[/imath]
Proof that [imath]n=250[/imath] is the largest value for [imath]n[/imath] for which [imath]1005![/imath] is divisible by [imath]10^n[/imath] where [imath]n\in\Bbb{N}[/imath]. My attempt was to find a way for counting the number of last zeroes for any permutation [imath]k![/imath] where [imath]k\in\Bbb{N}[/imath], however I haven't arrived at any useful result. I don't know how to proceed with this one.
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1384419
|
Number of zeros at the end of [imath]k![/imath]
For how many positive integer [imath]k[/imath] does the ordinary decimal representation of the integer [imath]k\text { ! }[/imath] end in exactly [imath]99[/imath] zeros ? By inspection I found that [imath]400\text{ !}[/imath] end in exactly [imath]99[/imath] zeros , but [imath]399\text{ !}[/imath] does NOT end in [imath]99[/imath] zeros ; using the formula [imath]\text{ number of zeros at the end }=\sum_{n=1}^{\infty}\left[\frac{k}{5^n}\right][/imath]where , [imath][x][/imath] denotes the greatest integer part not exceeding [imath]x[/imath]. I also found that, for [imath]k=401,402,403,404[/imath] the number of zeros is same, but for [imath]k=405[/imath] the number of zeros increase by [imath]1[/imath] ; as [imath]405[/imath] is divisible by [imath]5[/imath] again , after [imath]400[/imath]. Thus I got that there are only [imath]5[/imath] integers satisfying the condition which are [imath]400,401,402,403,404[/imath]. The question is possible duplicate of this or this but my question is different from these two questions. Does there exist any other rule or easy formula from where I can get how many integers are there
|
1920037
|
Prove that if [imath] \gcd(a,m) = 1 [/imath] and [imath] a b \equiv a c ~ (\operatorname{mod} ~ m) [/imath], then [imath] b \equiv c ~ (\operatorname{mod} ~ m) [/imath].
Problem. Prove that if [imath] \gcd(a,m) = 1 [/imath] and [imath] a b \equiv a c \pmod m [/imath], then [imath] b \equiv c \pmod m [/imath]. I have [imath] m | a b - a c [/imath], or [imath] m | a (b - c) [/imath], and I know that if [imath] \gcd(a,m) = 1 [/imath], then [imath] n | b - c [/imath]. I just don’t know what lemma I can use here.
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1877649
|
To prove that if [imath]ab\equiv ac\bmod n[/imath] and [imath](a,n)=1[/imath] then [imath]b\equiv c\bmod n[/imath]
To prove that if [imath]ab \equiv ac\bmod n[/imath] and [imath](a,n)=1[/imath] then [imath]b\equiv c \bmod n[/imath] I write as [imath]1=ar+ns[/imath] [imath]ac=ab+nq[/imath] I have to prove [imath]c=b+nr[/imath] How do I manipulate equations to reach conclusion Thanks
|
1922819
|
Stars and Bars with bounds
This question is related to Error solving “stars and bars” type problem I have what I thought is a fairly simple problem: Count non-negative integer solutions to the equation [imath]x_1 + x_2 + x_3 + x_4 + x_5 = 23[/imath] such that [imath]0 \leq x_1 \leq 9[/imath]. The difference here is on the constraint. It bounds all the [imath]x_i[/imath] under 10 : [imath]\forall i\le5[/imath] , [imath]0 \leq x_i \leq 9[/imath]. [imath]8+8+0+0+7=23[/imath] is accepted but not [imath]18+3+0+0+2=23[/imath] or [imath]11+0+0+0+12=23[/imath] . In other words all the [imath]x_i[/imath] must be usual digits. Note that bad solutions may include one or two bad [imath]x_i[/imath]. It is the main difficulty. What is the count of combinations ? Edit : this question includes a double bounds and a supplemental difficulty to find the right solution.
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1649039
|
How many solutions for an equation with simple restrictions
I'm working on an assignment in which I have to count the number of solutions for this particular equation: [imath]x_1+x_2+x_3+x_4=20[/imath]for non negative integers with [imath]x_1<8 [/imath] and [imath]x_2<6[/imath] I'm aware that this kind of a task isn't that complicated, but I don't get combinatorics in general that well. So I've tried two following approaches to get this done. Firstly I tried to substitute the variable x: [imath]x_1+x_2+x_3+x_4=20 \Leftrightarrow y_1+y_2+y_3+y_4=34[/imath] in which [imath]y_1=x_1+8[/imath] and [imath]y_2=x_2+6[/imath] (casue [imath]x_1=y_1-8[/imath] and [imath]x_2=y_1-6[/imath]) Following this approach the total number of possible solutions would be [imath]{34+3 \choose 3} [/imath] But I'm not sure if its the right solution. The second approach is to sum all of the possible values that [imath]x_1[/imath] and [imath]x_2[/imath] could possibly take, also [imath]x_1=0,1,2,3,4,5,6,7[/imath] and [imath]x_2=0,1,2,3,4,5,6[/imath] And then count all the possibilities for each of the variables [imath]{20 -x_1-x_2+1\choose 1}[/imath] and sum them together like this: [imath]{21\choose 1}+{20\choose 1}+{19\choose 1}+{18\choose 1}+... [/imath] and so on... I'm sure I'll get the correct number with this one, but I'm not feeling like summing all of this possibilities. There's got to be a better, more elegant way to deal with this. My professor gave me a hint that I should do it using the complement.
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1922928
|
A doubt regarding polynomials
Suppose [imath]F(x)[/imath] is a polynomial having coefficients in the field of reals ( [imath]\mathbb{R}[X][/imath] ) . [imath]F(x)[/imath] has the property that [imath]y \geq 0 \implies F(y)[/imath] is nonzero.( [imath]y[/imath] is a real number ) Can we find a polynomial [imath]T(x) \in \mathbb{R}[X][/imath] so that [imath]T(x) \cdot F(x)[/imath] has all its coefficients non-negative ?
|
1547305
|
Property of a polynomial with no positive real roots
The following is an exercise (Exercise #3 (a), Chapter 3, page 28) from Richard Stanley's Algebraic Combinatorics. Let [imath]P(x)[/imath] be a nonzero polynomial with real coefficients. Show that the following two conditions are equivalent: There exists a nonzero polynomial [imath]Q(x)[/imath] with real coefficients such that all coefficients of [imath]P(x) Q(x)[/imath] are nonnegative. There does not exist a real number [imath]a > 0[/imath] such that [imath]P(a) = 0[/imath]. That the first item implies the second is straightforward (since if [imath]a[/imath] is a positive real root of [imath]P(x)[/imath] then it is a positive real root of [imath]P(x)Q(x)[/imath], but since [imath]P(x)Q(x)[/imath] is nonzero with all coefficients nonnegative, this is impossible). However, I can't seem to find a way to prove that the second item implies the first. I would very much like to see a proof if someone can provide it.
|
1925239
|
How to show [imath]\langle x,y \mid x^3=y^3=(xy)^3=1\rangle[/imath] is presentation of an infinite group?
How to show [imath]\langle x,y \mid x^3=y^3=(xy)^3=1 \rangle[/imath] is presentation of an infinite group? This is a result in textbook, but I do not understand the rationale.
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1897367
|
Trying to prove that [imath]H=\langle a,b:a^{3}=b^{3}=(ab)^{3}=1\rangle[/imath] is a group of infinite order.
I'm trying to prove that the following group has infinite order: [imath]H=\langle a,b\mid a^{3}=b^{3}=(ab)^{3}=1\rangle.[/imath] Currently I'm checking on some cases using the relations, but my problem is the reducibility for large products. Naively I started to check that [imath]ab[/imath] is different than [imath]1,a,b[/imath] and then [imath]ba[/imath] than [imath]1,a,b,ab[/imath], just to understand [imath]H[/imath] in some extent. Now I'm wondering about some more effective method to prove that [imath]|H|=\infty[/imath], I'm tempted to look for an injection from some group of infinite order into [imath]H[/imath], but I'm still stuck. More than asking for a solution I'd rather appreciate some hints or thoughts about it. Thanks a lot.
|
1925430
|
How many subsets does a set [imath]S[/imath] with [imath]n[/imath] elements have?
How many subsets does a set [imath]S[/imath] with [imath]n[/imath] elements have? I think the product principle can be used to answer this question. But I am not sure how it applies. I tried to do this by finding all of the subsets for [imath]n=1, 2, 3[/imath] but I couldn't find a pattern because I was probably doing it wrong. For example when [imath]n=2[/imath] we could have a subset with just the first element, just the second element, both elements, or the empty set, this gives us [imath]4[/imath] subsets when [imath]n=2[/imath], is this correct?
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936036
|
How many elements does [imath]\mathcal{P}(A)[/imath] have?
Let [imath]A[/imath] be a set of size fifteen. Let [imath]\mathcal{P}(A)[/imath] denote the power set of [imath]A[/imath], that is the set of all the subsets of [imath]A[/imath]. How many elements does [imath]\mathcal{P}(A)[/imath] contain? This is the same as being [imath]0[/imath] or [imath]1[/imath], binary sets of size [imath]1\to 14[/imath]? Does this mean the answer is [imath]{15\choose 0} + {15\choose 1} + {15\choose 2} +\dots +{15\choose 13}+{15\choose 14}+{15\choose 15}[/imath]? Or is there something I am not considering?
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1924081
|
Derivative of Adjugate Matrix, [imath]\frac{d^i}{dx^i}\operatorname {Adj}(x \mathbf I - \mathbf A)[/imath]
Let [imath]\mathbf A[/imath] denote an [imath]n \times n[/imath] matrix with [imath]r=\operatorname{rank}\mathbf A[/imath] and define [imath] \mathbf B_i(x) \equiv \frac{d^i}{dx^i}\operatorname {Adj}(x \mathbf I - \mathbf A). [/imath] Conjecture: If [imath]n - r \ge 2[/imath], then [imath] \mathbf B_i(0)= \mathbf 0, \quad \text{for all } i \in \{0,...,n-r-2\}. [/imath]
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1925406
|
[imath]\operatorname{Adj} (\mathbf I_n x-\mathbf A)[/imath] when [imath]\operatorname{rank}(\mathbf A)\le n-2[/imath]
Let [imath]\mathbf B[/imath] denote an [imath]n \times n[/imath] matrix with [imath]r\equiv\operatorname{rank}(\mathbf B)[/imath]. I need to prove the following conjecture: If [imath]r \le n - 2[/imath], then there exists a polynomial matrix [imath]\mathbf P(x)[/imath] such that [imath] \operatorname{Adj} (\mathbf I_n x-\mathbf B)= x^{n-r-1}\mathbf P(x). [/imath]
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1925773
|
Counterexample to viewing derivative as a quotient of differentials
The derivative of a function [imath]y = f(x)[/imath], [imath]\frac{dy}{dx}[/imath] seems to behave like a quotient in many cases: [imath] dy = \frac{dy}{dx} dx,[/imath] or [imath] u = h(x) [/imath] [imath]\int f(x) dx = \int h(f(x)) \frac{du}{dx} du[/imath] Yet we're often told that it's not correct to view it that way. For example, the wikipedia article on Leibniz' notation says The expression dy/dx should not be read as the division of two quantities dx and dy. Is there an example of a situation in which viewing [imath]\frac{dy}{dx}[/imath] as a quotient will lead to an incorrect result?
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1784671
|
When can we not treat differentials as fractions? And when is it perfectly OK?
Background I am a first year calculus student so I would prefer if answers remained in Layman's terms. It is common knowledge and seems to me a mantra that I keep hearing over and over again to "not treat differentials/derivatives as fractions". I am of course, in particular, referring to Leibniz notation. However, aside from a quick response such as "oh, it's because its not a fraction but rather a type of operator", I never really got a full answer as to why we can't treat it as such. It just kind of sits at the edge of taboo in my mind where it sometimes gets used and sometimes doesn't. Confusion is further compounded when a lot of things seem to just work out if we treat them just as fractions (e.g. u-substitution / related-rates) Example Air is being pumped into a balloon at a rate of [imath]100cm^3/s[/imath]. We want the rate of change of radius when the radius is at [imath]25cm[/imath]. [imath]\text{we are given}\ \frac{dv}{dt}=100cm^3/s[/imath] [imath]\text{we want}\ \frac{dr}{dt}\ \text{when}\ r=25cm[/imath] Thus we will solve this by using the relation [imath]v=\frac{4}{3}\pi r^3[/imath] [imath]\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}[/imath] [imath]\frac{dv}{dt}\frac{dr}{dv}=\frac{dr}{dt}[/imath] [imath]100\frac{1}{4\pi r^2}=\frac{1}{25\pi}[/imath] So the answer is [imath]\frac{dr}{dt}=\frac{1}{25\pi}[/imath] when [imath]r=25cm[/imath] *Note the manipulation of derivatives just as if they were common fractions using algebra. Question When exactly can I treat differentials/derivatives as fractions and when can I not? Please keep in mind that at the end of the day, I am a first year college student. An answer that is easy to understand is preferred over one that is more mathematically rigorous but less friendly to a beginner such as me.
|
1925786
|
Given [imath]p[/imath] prime and [imath]0, show that p divides {p \choose k}[/imath]
Suppose that [imath]p[/imath] is prime and [imath]0<k<p[/imath]. Prove that [imath]p|[/imath][imath]{p}\choose{k}[/imath]. I am asked to prove a Lemma beforehand, which follows from Euclid's Lemma. It states: Let [imath]m,n \in\mathbb{Z}[/imath], [imath]p[/imath] be prime, [imath]m|n[/imath], [imath]p|n[/imath], and [imath]p\nmid m[/imath]. Then [imath]p|\frac{n}{m}[/imath]. After I have done this, I return to the normal problem and try to fulfill the Lemma's assumptions with the current variables. First, [imath]p|p!=p*(p-1)*...*2*1[/imath] since [imath]p|p[/imath], always. Then, all we have left to show is [imath]1[/imath]) [imath]p\nmid k!(p-k)![/imath] and [imath]2[/imath]) [imath]k!(p-k)!|p![/imath] and apply the Lemma. I attempt to show [imath]1[/imath]) by noting that [imath]p\nmid k![/imath] and [imath]p\nmid (p-k)![/imath] since [imath]p[/imath], as a prime, will only divide its multiples which are greater than or equal to [imath]p[/imath]. Since both [imath](p-k)[/imath] and [imath]k[/imath] are strictly less than [imath]p[/imath], they will not contain any multiples of [imath]p[/imath]. Then, [imath]p\nmid k!(p-k)![/imath] by the contrapositive of Euclid's Lemma. My concerns are these: Is the argument above for [imath]p\nmid k!(p-k)![/imath] rigorous enough? And, how can I show that [imath]k!(p-k)! | p![/imath] ?
|
51469
|
Prime dividing the binomial coefficients
It is quite easy to show that for every prime [imath]p[/imath] and [imath]0<i<p[/imath] we have that [imath]p[/imath] divides the binomial coefficient [imath]\large p\choose i[/imath]; one simply notes that in [imath]\large \frac{p!}{i!(p-i)!}[/imath] the numerator is divisible by [imath]p[/imath] whereas the denominator is not (since it is a product of numbers smaller than [imath]p[/imath] and [imath]p[/imath] is prime). My problem is with generalizing this argument for [imath]q=p^n[/imath]. I'm looking for the most elegant and simple way to prove that [imath]p[/imath] still divides [imath]\large q\choose i[/imath].
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1925889
|
Simplify the subtraction of two summations.
[imath]\sum_{i=1}^n (3i^2 +4) - \sum_{j=2}^{n+1} (3j^2 +1) [/imath] I thought I would use the distributive rule but the question stipulates that my answer should not include a summation sign...
|
1924800
|
How do i simplify the following: [imath]\sum_{i=1}^n (3i^2+4) - \sum_{j=2}^{n+1} (3j^2+1)[/imath]
I want to simplify [imath]\sum_{i=1}^n (3i^2+4) - \sum_{j=2}^{n+1} (3j^2+1)[/imath] without a summation sign in the answer. How can I do this? I can't work out how to simplify this expression. I appreciate any help.
|
1925816
|
Why [imath]\langle 1,0\rangle[/imath] is a limit point of the subset [imath]x\times 0[/imath] fo the ordered square?
This is a question following this one. I pictured all the points of: [imath]C = \{x\times 0 \mid 0<x<1\}[/imath] as a subset of the ordered square. (the question asks me to find the closure) I calculated its limit points, which are [imath][0,1)\times 1[/imath], andthe points of [imath]C[/imath], which are [imath](0,1)\times 0[/imath]. In order to calculate the closure (which is what the exercise asks, I need to unite the limit points with the set). Brian told me that [imath]\langle 1,0\rangle[/imath] is also a limit point, but I can't see why: Look at the image below: If I take an open set around [imath]\langle 1,0\rangle[/imath], there's no intersection with [imath]C[/imath], which are the points that lie on the ground, that is, the points on [imath](0,1)\times 0[/imath]
|
1923358
|
Finding the closure of some subsets of the ordered square
I need to find the closure of these sets on the ordered square: [imath]A = \{(1/n)\times 0 | n\in \mathbb{Z}_+\}[/imath] [imath]B = \{(1-1/n)\times \frac{1}{2}| n\in \mathbb{Z}_+\}[/imath] [imath]C = \{x\times 0 | 0<x<1\}[/imath] [imath]D = \{x\times \frac{1}{2}| 0<x<1\}[/imath] [imath]E = \{\frac{1}{2}\times y | 0<y<1\}[/imath] I know that the closure is just the set itself united with its limit points. For [imath]A[/imath], the set of limit points, I imagine, are [imath]\{<1,0>\}[/imath], that is, the [imath]x[/imath] coordinate is [imath]1[/imath] and the [imath]y[/imath] is [imath]0[/imath]. But here is says the reverse. What am I getting wrong? Following the same reasoning, [imath]B' = \{<1,\frac{1}{2}>\}[/imath] [imath]C' = \{<0,0>,<0,1>\}[/imath] What am I doing wrong?
|
1925910
|
Let G be simple, |G| is not 2, and ϕ a homomorphism from G to H. If H contains a normal subgroup A of index 2, then ϕ(G) ≤ A .
Let [imath]G[/imath] be simple group, [imath]|G|[/imath] is not 2, and [imath] ϕ [/imath] a homomorphism from [imath]G[/imath] to [imath]H[/imath]. If [imath]H[/imath] contains a normal subgroup A of index 2, then [imath] ϕ(G) ≤ A [/imath] . I have been trying the following: [imath]A[/imath] is normal and maximal in [imath]H[/imath] since its index is 2(basic fact + correspondence theorem). [imath]G[/imath] is simple so it is isomorphic to its image so its image is a simple subgroup of [imath]H[/imath]. Now I have to show somehow that its image must be a subgroup of [imath]A[/imath] and I can't find any reason it has to. I would very appreciate if someone could help me with this.
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1428631
|
Problem from "The Theory of Finite Groups" by Kurzweil and Stellmacher
I am currently working my way through The Theory of Finite Groups by Kurzweil and Stellmacher and came across the following question in the first chapter: Let [imath]G[/imath] be simple, [imath]|G| \ne 2[/imath], and [imath]f : G \to H[/imath] be a homomorphism. If [imath]A \unlhd H[/imath] is a normal subgroup of index [imath]2[/imath], then [imath]\text{range}(f) \le A[/imath]. Please could someone help me to prove this statement? It doesn't seem like it should be too difficult but having spent some time on it I'm still getting nowhere. Thanks!
|
1926031
|
Let [imath]s_i=a_1+a_2+\dots+a_n[/imath] diverge, does [imath]\sum_{i=1}^\infty\frac{a_i}{s_i^2}[/imath] converge?
Let [imath]a_1,a_2\dots[/imath] be a sequence of positive reals such that [imath]s_n=a_1+a_2+\dots+a_n[/imath] diverges. Does it follow that [imath]\sum\limits_{i=1}^\infty \frac{a_i}{s_i^2}[/imath] converges? I am really stumped at this problem. I tried finding counterexamples like [imath]a_n=\frac{1}{n\log n}[/imath] and got some series which I was unsure whether they diverged or not. But at then every example I have tried converges. Thanks in advance.
|
55670
|
Convergence of [imath]\sum \frac{a_n}{S_n ^{1 + \epsilon}}[/imath] where [imath]S_n = \sum_{i = 1} ^ n a_n[/imath]
Let [imath]a_n[/imath] be a sequence of positive reals, such that the partial sums [imath]S_n = \sum_{i = 1} ^ n a_i[/imath] diverge to [imath]\infty[/imath]. For given [imath]\epsilon > 0[/imath] do we have [imath]\sum_{n = 1} ^ \infty \frac{a_n}{S_n^{1 + \epsilon}} < \infty?[/imath] For [imath]\epsilon \ge 1[/imath] we can resolve this quickly by noting [imath]\frac{a_n}{S_n ^ 2} \le \frac 1 {S_{n - 1}} - \frac 1 {S_n}[/imath] so for sufficiently large [imath]n[/imath] we can bound [imath]\frac{a_n}{S_n^{1 + \epsilon}}[/imath] by [imath]\frac 1 {S_{n - 1}} - \frac 1 {S_n}[/imath] as well. I'm wondering if this is true for arbitrary [imath]\epsilon > 0[/imath]. I know that the series in question diverges for [imath] \epsilon = 0[/imath], so all that is missing is what happens in [imath](0, 1)[/imath].
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1923833
|
Lie algbra so(2n)?
I am currently reading the book ``Symmetries, Lie Algebras and Representations: A graduate course for physicists'', by Jurgen Fuchs & Christoph Schweigert. On page 75 (unfortunately, the google book does not cover this page), I read: Another simple subalgebra of [imath]gl(2n)[/imath] consists of those matrices of [imath]M[/imath] which fulfill the relation [imath]M^tK+KM=0[/imath], where [imath]K[/imath] is the [imath]2n\times 2n[/imath] -matrix [imath]K:=\begin{pmatrix} 0_n & 1_n \\ 1_n & 0_n \end{pmatrix}.[/imath] My question is how can such a definition be consistent with the normal definition of the lie algebra [imath]so(2n)[/imath] which consist of all [imath]2n\times 2n[/imath] antisymmetric matrices? The authors also note after some sentences that Note that, here the symbols [imath]sp(n)[/imath] and [imath]so(n)[/imath] refer to Lie algebras over complex numbers. I am not sure is this remark relevant to my question. Let's take [imath]n=1[/imath] for example. Let [imath]M=\begin{pmatrix} a & b \\ c & d \end{pmatrix}[/imath], then [imath]M^tK+KM=0[/imath] implies that [imath]\begin{pmatrix} a & c \\ b & d \end{pmatrix}\equiv M^t=-KMK^{-1}=-KMK=-\begin{pmatrix} d & c \\ b & a \end{pmatrix}.[/imath] That is, [imath]a=-d[/imath], [imath]c=b=0[/imath], meaning that [imath]M=\begin{pmatrix} a & 0 \\ 0 & -a \end{pmatrix}[/imath] which is inconsistent with the normal one [imath]M=\begin{pmatrix} 0 & b \\ -b & 0 \end{pmatrix}[/imath].
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323801
|
Two Definitions of the Special Orthogonal Lie Algebra
I am encountering two definitions of the special orthogonal lie algebra, and I would like to know if they are equivalent, and if there are advantages to working with one over the other. If we begin with an [imath]n[/imath]-dimensional vector space [imath]V[/imath] over a field [imath]k[/imath] and a chosen basis, we can define a bilinear form on [imath]V[/imath] by a matrix [imath]S\in M_n(k)[/imath], ie, let [imath]\langle v,w\rangle=v^tSw[/imath] for all [imath]v,w\in V[/imath]. Now [imath]g\in GL_n(k)[/imath] preserves the form ([imath]\langle g(v),g(w)\rangle=\langle v,w\rangle[/imath]) if and only if [imath]g^tSg=S[/imath], so all such [imath]g[/imath] form a linear algebraic group [imath]G[/imath]. The tangent space at the identity of [imath]G[/imath] will be contained in that of [imath]GL_n(k)[/imath], so [imath]T_eG\subset T_eGL_n(k)=M_n(k)[/imath], and in fact, [imath]T_eG=\{B\in M_n(k)\mid B^tS+SB=0\}[/imath]. [imath]T_eG[/imath] becomes a lie algebra, [imath]Lie(G)[/imath], if we define the bracket to be the commutator of two matrices. Now, if [imath]S=I_n[/imath], it follows that [imath]G=O_n(k)[/imath] is the orthogonal group of matrices satisfying [imath]g^tg=I_n[/imath], and [imath]Lie(G)=\mathfrak{so}_n[/imath] is the lie algebra of antisymmetric matrices. In Humphrey's Introduction to Lie Algebras and Representation Theory, he defines [imath]\mathfrak{so}_n[/imath] to be all matrices [imath]B[/imath] satisyfing [imath]B^tS+SB=0[/imath], where [imath] S=\begin{pmatrix} 1&0&0\\ 0&O&I_l\\ 0&I_l&O \end{pmatrix} \hspace{.5in}\text{or}\hspace{.5in} S=\begin{pmatrix} O&I_l\\ I_l&O \end{pmatrix} [/imath] depending on the parity of [imath]n[/imath]. The matrices obtained in this way are not antisymmetric, nor is the group [imath]G[/imath] preserving the form defined by [imath]S[/imath] the orthogonal group [imath]O_n(k)[/imath]. Are the two groups obtained from considering different [imath]S[/imath] isomorphic? Are the two lie algebras isomorphic? If so, why would one prefer one form to the other?
|
1926382
|
Comparison of Linear dependency between [imath]2[/imath] families of vectors in [imath]\Bbb R^n[/imath].
Let [imath]T:\Bbb R^n\longrightarrow\Bbb R^n[/imath] be a linear transformation, where [imath]n\geq 2[/imath]. For [imath]k\leq n[/imath], let [imath]E=\{v_1,v_2,\dots,v_k\}[/imath] contained in, equal to [imath]R^n[/imath] and [imath]F=\{Tv_1,Tv_2,\dots,Tv_k\}[/imath]. Then a). If [imath]E[/imath] is linearly independent, then [imath]F[/imath] is linearly independent. b). If [imath]F[/imath] is linearly independent, then [imath]E[/imath] is linearly independent. c). If [imath]E[/imath] is linearly independent, then [imath]F[/imath] is linearly dependent. d). If [imath]F[/imath] is linearly independent, then [imath]E[/imath] is linearly dependent. a) part can be correct if [imath]T[/imath] is one-one. And since our transformation is finite dimensional [imath]n[/imath]-to-[imath]n[/imath]. Therefore, it is one-one, which implies that [imath]F[/imath] is linearly independent if [imath]E[/imath] is linearly independent. I thought of part c) as follows: supposingly I take [imath]E=\{(1,0,0),(0,1,0)\}[/imath] contained in [imath]\Bbb R^3[/imath] and [imath]F=\{T(1,0,0),T(0,1,0)\}[/imath]. Now [imath]E[/imath] is linearly independent, if I define [imath]T(x_1,x_2,x_3)=((x_1+x_2),0,x_3)[/imath] then [imath]T(1,0,0)=(1,0,0)[/imath] and [imath]T(0,1,0)=(1,0,0)[/imath] , this shows that [imath]T(v_1)[/imath] and [imath]T(v_2)[/imath] are linearly dependent. This implies that both part a) and c) are correct? can someone please give the explaination for part b) and d)?
|
1630139
|
Let [imath]T:\mathbb{R}^n\to \mathbb{R}^n[/imath] be a linear transformation...
Let [imath]T:\mathbb{R}^n\to \mathbb{R}^n[/imath] be a linear transformation, where [imath]n\ge2[/imath]. For [imath]k\le n[/imath], let [imath]E=\{v_1,v_2,\dots,v_k\}\subset \mathbb{R}^n[/imath] and let [imath]F=\{Tv_1,Tv_2,\dots,Tv_k\}[/imath], then which of the following is true. [imath]A.[/imath] If [imath]E[/imath] is linearly independent, then [imath]F[/imath] is linearly independent [imath]B.[/imath] If [imath]F[/imath] is linearly independent, then [imath]E[/imath] is linearly independent. [imath]C.[/imath] If [imath]E[/imath] is linearly independent, then [imath]F[/imath] is linearly dependent [imath]D.[/imath] If [imath]F[/imath] is linearly independent, then [imath]E[/imath] is linearly dependent. Please help me out with this problem. And if some theorem is responsible for it's solution please mention that.
|
1926743
|
X people line up to get into Y different clubs.
I have a combinatorics problem, and I don't know my answer is correct or not. The problem is the following : Assume that X people line up to get into Y different clubs. How many ways are there to do it? (The people are distinguishable and the order people are in line matters.) My attempt : we have Y different kinds of clubs. So, we have [imath]{X+Y-1 \choose X}[/imath]. But, [imath]X[/imath] people are distinguishable, we have to permute them, [imath]X![/imath]. Thus, the final answer is [imath]X!{X+Y-1 \choose X}[/imath]. Is that right answer?
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1924133
|
Combinatorics problem: [imath]n[/imath] people line up to [imath]m[/imath] clubs
Assume that there are [imath]n[/imath] people line up to get into [imath]m[/imath] different clubs. How many ways to do it if each people is different and order of people in the line matters. Sol. Clearly, each people has [imath]m[/imath] different choices for clubs. So there are [imath]m^n[/imath] ways of choosing clubs. Now we count number of different lines. Let [imath]x_i[/imath] be the number of people lining to get to the club [imath]i[/imath], [imath]1 \leq i \leq m[/imath], [imath]x_i \geq 0[/imath]. Then [imath]x_1 + x_2 + ... + x_m = n[/imath] with [imath]x_i[/imath] non-negative integers. So there are [imath]{m+n-1}\choose{n-1}[/imath] ways, ignore order in each line. So there are [imath]{m+n-1}\choose{n-1}[/imath][imath] x_1!x_2!...x_n![/imath] possible lines. The answer is weird since it is in [imath]x_1, ...,x_n[/imath] terms, and I think it is wrong. However, I am not sure how to fix it. Any help please ?
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1927411
|
Number of Divisors Given a Prime Factorization
This could be a classic problem or my wording might be totally off, but I can't seem to find an answer to this problem: Given [imath]n\in\mathbb{Z}[/imath] with prime factorization [imath]n=p_1^{q_1}\dots p_m^{q_m}[/imath], how many integers divide [imath]n[/imath]? For the sake of this problem, let's ignore [imath]1[/imath] and [imath]n[/imath] until the end. First off, we have that there are at least [imath]q_1\dots q_n[/imath] divisors, since there are [imath]q_1[/imath] divisors made up of [imath]p_1[/imath] (namely [imath]p_1,p_1^2,\dots,p_1^{q_1}[/imath]), thus the same holds for [imath]p_2,\dots,p_n[/imath] I've never done much with combinatorics, so I'm not quite sure how to begin counting unique divisors by mixing and matching distinct primes. Any thoughts?
|
433848
|
Prime factors + number of Divisors
I know that one way to find the number of divisors is to find the prime factors of that number and add one to all of the powers and then multiply them together so for example [imath]555 = 3^1 \cdot 5^1 \cdot 37^1[/imath] therefore the number of divisors = [imath](1+1)(1+1)(1+1) = 2 \cdot 2 \cdot 2 = 8[/imath] What I do not know (and can't seem to find when I look) is WHY this relationship exists or a formula which shows its proof. Has anyone seen such a formula?
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1927506
|
Prove that if [imath]p[/imath] and [imath]q[/imath] are odd numbers then [imath]x[/imath] is not a rational number in [imath]x^2 + 2px +2q=0[/imath]
How do I prove that if [imath]p[/imath] and [imath]q[/imath] are odd numbers then [imath]x[/imath] is not a rational number in [imath]x^2 + 2px +2q=0[/imath]? So I don't know a whole bunch about quadratic functions behaviour but I know that we can get the solution for [imath]x[/imath] with the quadratic formula. I tried comparing the general formula with a fraction in which either [imath]m[/imath] or [imath]n[/imath] is an irrational number. \begin{equation*} \qquad \frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{m}{n} \end{equation*} We know that n is not the irrational number since a equals [imath]1[/imath] . Therefore the upper part of the quadratic formula should equal an irrational number. I'm stuck on that part though, which function property could I use to prove this? Note: I'm new to this site and I am learning to correctly use the features of stack exchange. If there is something I can improve in I'll be more than happy to take some constructive criticism
|
1923356
|
Prove: For odd integers [imath]a[/imath] and [imath]b[/imath], the equation [imath]x^2 + 2 a x + 2 b = 0[/imath] has no integer or rational roots.
If [imath]a[/imath] and [imath]b[/imath] are odd integers, prove that the equation [imath]x^2 + 2ax + 2b = 0[/imath] has no integer or rational roots.
|
1409562
|
Derivatives of ordinal order
This question actually arises from this answer to another question, which contains the sentence A function is smooth is it has derivatives of infinite order. While the author surely didn't actually mean [imath]\frac{\mathrm d^\infty f}{\mathrm dx^\infty}[/imath], a comment on that post got me to think about that. I've come to the following definition: Be [imath]\alpha[/imath] an arbitrary ordinal. Then the [imath]\alpha[/imath]th derivative of [imath]f(x)[/imath] is defined as follows, provided the used derivatives/limits exist: [imath]\frac{\mathrm d^\alpha f}{\mathrm dx^\alpha} = \begin{cases} f & \text{if }\alpha = 0\\ \frac{d}{dx}\,\frac{\mathrm d^\beta f}{\mathrm dx^\beta} & \text{if } \alpha=\beta+1\\ \lim_{\beta\to\alpha}\frac{\mathrm d^\beta f}{\mathrm dx^\beta} & \text{if } \alpha \text{ is a limit ordinal} \end{cases}[/imath] Here the limit is meant pointwise. Here are a few cases: If [imath]f(x)[/imath] is an arbitrary polynomial, [imath]\frac{\mathrm d^\omega f}{\mathrm dx^\omega}=0[/imath]. If [imath]f(x) = \exp(a x)[/imath], then [imath]\frac{\mathrm d^\omega f}{\mathrm dx^\omega}= \begin{cases} 0 & \text{if } -1 \le a < 1\\ \exp(x) & \text{if } a = 1\\ \text{does not exist} & \text{otherwise} \end{cases}[/imath] [imath]\sin(x)[/imath] and [imath]\cos(x)[/imath] don't have a [imath]\omega[/imath]th derivative. Also the [imath]C^n[/imath] classes can be generalized in a straightforward way: A function is in [imath]\tilde C^\alpha[/imath] if for every [imath]\beta\in\alpha[/imath], the [imath]\beta[/imath]th derivative of [imath]f[/imath] exists and is continuous. For finite [imath]n[/imath], we have [imath]C^n=\tilde C^{n+1}[/imath] ([imath]\tilde C^0[/imath] would be all functions). Moreover, [imath]C^\infty = \tilde C^\omega[/imath]. If the [imath]\omega[/imath]th derivative exists and is continuous, the function is in [imath]\tilde C^{\omega+1}[/imath]. Now my questions: Has anyone already considered such derivatives? Are they even useful somewhere? Is there an easy way to characterize the functions in [imath]\tilde C^{\omega+1}[/imath]? Are there functions in [imath]\tilde C^{\omega+1}[/imath] whose [imath]\omega[/imath]th derivative is neither [imath]0[/imath] nor [imath]\exp(x)[/imath]? (Otherwise the ordinal order derivatives would be rather boring).
|
905712
|
Is it possible to iterate a function transfinite times?
Let [imath]f:A\rightarrow A[/imath] be a function. We can simply define [imath]f\circ f[/imath], [imath]f\circ f\circ f[/imath], etc., for each given natural number inductively. [imath]f^{(0)}=id_A[/imath] [imath]\forall n\in\omega \qquad f^{(n+1)}=f\circ f^{(n)}[/imath] How to define [imath]f^{(\omega)}[/imath] such that it be a function from [imath]A[/imath] to [imath]A[/imath]? Is it possible to define [imath]f^{(\alpha)}[/imath] for arbitrary large ordinal number [imath]\alpha[/imath]?
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