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2977206 | Is the canonical morphism of sheaves [imath]f^{-1}f_*\mathcal F\to \mathcal F[/imath] an isomorphism?
Let [imath]f:X\to Y[/imath] be a closed immersion of topological spaces. Let [imath]\mathcal F[/imath] be a sheaf of rings on [imath]X[/imath]. Is the canonical morphism of sheaves [imath]\varphi: f^{-1}f_*\mathcal F\to \mathcal F[/imath] an isomorphism? | 228735 | Sheafs and closed immersion
Let [imath]f:X \rightarrow Y[/imath] be a continuous map of topological spaces, such that it is closed immersion. Let [imath]\mathfrak{F}[/imath] and [imath]\mathfrak{G}[/imath] be sheafs on [imath]X[/imath] and [imath]Y[/imath] respectively. How to show, that canonical morphisms [imath] \mathfrak{G} \rightarrow f_* f^{-1} \mathfrak{G}, \; f^{-1} f_* \mathfrak{F} \rightarrow \mathfrak{F} [/imath] are isomorphisms? Is it true, that taking stalk doesn't commute with direct image, but commutes with inverse image? |
2977375 | Theorem of compound probability - intuition
What is the intuition behind [imath]P(A\cap B) = P(A)P(B)[/imath], given A and B are independent events? I saw the derivation but wasn't able to to think of it intuitively. Is there any combinatorial reasoning behind it? I have even seen it for cases like tossing a coin and it works! But I am not convinced that it should be true always. I am looking for a strong foundation of it. | 206980 | Intuition behind independence and conditional probability
I have a good intuition that [imath]A[/imath] is independent of [imath]B[/imath] if [imath]P(A \vert B) = P(A)[/imath], and I see how you can easily derive from this that it must hold that [imath]P(A,B) = P(A)P(B)[/imath]. But the first statement is not normally taken as a definition; instead the second is. What is the intuition, or even derivation behind defining [imath]A[/imath] and [imath]B[/imath] as independent iff [imath]P(A, B) = P(A)(B)[/imath]? The kind of explanation I am looking for would be one similar to that given by Jaynes for the definition of conditional probability in the first chapter of Probability: The Logic of Science, or even a Kolmogorov axiomatic explanation would help. |
2977274 | Differentiating a vector function
For real number functions (if they are differentiable) [imath]f[/imath] and [imath]d[/imath] it holds: [imath]\frac{\mathbf{d}}{\mathbf{d}t}d(f(t)) = d'(f(t))f'(t).[/imath] Now let [imath]t[/imath] be a real number but [imath]v[/imath], [imath]f(t)[/imath], and [imath]d(v)[/imath] be vectors. The above formula may not make sense because [imath]d[/imath] is now a function from a vector. Please help to "save" the formula, that is define how it can be demonstrated a similar formula for vectors. Maybe it may be described with something like gradient? | 1781183 | How does the chain rule work for functions from vectors to vectors?
Suppose I have a function: [imath] \vec{s} = \vec{f}\left(\vec{\theta}\right)[/imath] and a derivative: [imath] \vec{v} = \frac{\mathrm{d} \vec{s}}{\mathrm{d} t}[/imath] How do I apply the chain rule? For simplicity lets call [imath]\omega = \frac{\mathrm{d} \vec{\theta}}{\mathrm{d} t}[/imath] I think the chain rule should be something along the line of: [imath] \vec{v} = \vec{\omega} \times \nabla_{\theta} \vec{f}\left(\theta\right) [/imath] but I don't know the exact rule. I think I may have to use matrices and more complicated derivatives like the Jacobian. |
2977209 | If [imath]F[/imath] and [imath]G[/imath] are adjoint functors, why are [imath]\operatorname{Nat}(G,G)\simeq\operatorname{Nat}(F,F)[/imath] as algebras?
Suppose [imath]F\colon\mathcal{A}\to\mathcal{B}[/imath] and [imath]G\colon\mathcal{B}\to\mathcal{A}[/imath] are adjoint functors between some [imath]R[/imath]-linear abelian categories ([imath]R[/imath] a ring), via a fixed counit-unit adjunction [imath]\epsilon\colon GF\Rightarrow 1[/imath] and [imath]\eta\colon 1\Rightarrow FG[/imath]. Why is the following an isomorphism of [imath]R[/imath]-algebras? [imath] \operatorname{Nat}(G,G)\to\operatorname{Nat}(F,F):\alpha\mapsto (F\ast\epsilon)\circ (F\ast\alpha\ast F)\circ (\eta\ast F) [/imath] My primary issue is I don't see how to invert it, since the counit-unit equations [imath]F\epsilon\circ \eta F=1[/imath] and [imath]\epsilon G\circ G\eta=1[/imath] don't seem to allow me to "undo" the [imath]F\ast\epsilon[/imath] and [imath]\eta\ast F[/imath] from the left and right. | 88334 | Pair of adjoint functors and showing that two certain maps are inverses
I'm currently self-studying "Categories and Sheaves"by Schapira and Kashiwara, and I've been stuck on problem 1.19 all day today, so I was hoping that someone could help me out. Let [imath]\mathcal{C}[/imath], [imath]\mathcal{C'}[/imath] be categories and [imath]L_v:\mathcal{C} \rightarrow \mathcal{C'}[/imath], [imath]R_v:\mathcal{C'} \rightarrow \mathcal{C}[/imath] be functors such that [imath](L_v,R_v)[/imath] is a pair of adjoint functors (v=1,2). Let [imath]\epsilon_v: id_\mathcal{C} \rightarrow R_v L_v[/imath] and [imath]\eta_v: L_v \circ R_v \rightarrow id_{\mathcal{C'}}[/imath] be the adjunction morphisms. Prove that the two maps [imath]\lambda[/imath],[imath]\mu[/imath] : [imath]Hom_{Fct(\mathcal{C},\mathcal{C'})}(L_1,L_2) \rightleftarrows^{\lambda}_{\mu} Hom_{Fct(\mathcal{C'},\mathcal{C})}(R_2,R_1)[/imath] given by : [imath]\lambda(\varphi):R_2 \rightarrow^{\epsilon_1 \circ R_2} R_1 \circ L_1 \circ R_2 \rightarrow^{R_1 \circ \varphi \circ R_2} R_1 \circ L_2 \circ R_2 \rightarrow^{R_1 \circ \eta_2} R_1[/imath] for [imath]\varphi \in Hom_{Fct(\mathcal{C},\mathcal{C'})}(L_1,L_2)[/imath] [imath]\mu(\psi) : L_1 \rightarrow^{L_1 \circ \epsilon_2} L_1 \circ R_2 \circ L_2 \rightarrow^{L_1 \circ \psi \circ L_2} L_1 \circ R_1 \circ L_2 \rightarrow^{\eta_1 \circ L_2} L_2[/imath] for [imath]\psi \in Hom_{Fct(\mathcal{C}',\mathcal{C})}(R_2,R_1)[/imath] are inverse to eachother. So far, I've been trying to play around with the zig-zag identity, but it doesn't seem to lead me anywhere. Any help will be greatly appreciated! |
2974420 | Showing that [imath]0 < \cos (\theta)<\frac {\sin (\theta)}{\theta}<\frac {1}{\cos(\theta)} [/imath]
[imath]0 < \cos (\theta)<\frac {\sin (\theta)}{\theta}<\frac {1}{\cos(\theta)} [/imath] for [imath]\theta\in(0,\pi/2)[/imath]. | 817683 | If [imath]0 \lt x \lt \frac{\pi}{2}[/imath], is [imath]\cos(x) \le \frac{\sin(x)}{x} \le \frac{1}{\cos(x)}[/imath]?
How can I find the following product using elementary trigonometry? Suppose [imath]0 \lt x \lt \frac{\pi}{2}[/imath] is an angle measured in radians. Use the trigonometric circle and show that [imath]\cos(x) \le \frac{\sin(x)}{x} \le \frac{1}{\cos(x)}[/imath]. I have been trying to solve this question. I can't figure out whether or not the solution requires a trigonometric circle or if it can be done using another method. |
2978596 | How do I prove that, if [imath]a[/imath] divides [imath]b[/imath], then [imath]a^n[/imath] divides [imath]b^n[/imath]?
How would I prove the following? If [imath]a[/imath] divides [imath]b[/imath], then [imath]a^n[/imath] divides [imath]b^n[/imath]. | 346034 | Proof by induction; [imath]a^n[/imath] divides [imath]b^n[/imath] implies [imath]a[/imath] divides [imath]b[/imath]
I want to prove by induction that [imath]a^n \mid b^n[/imath] implies that [imath]a \mid b[/imath] holds for all integers [imath]n\geq 1[/imath]. Clearly for [imath]n=1[/imath] this is true, since if [imath]a \mid b[/imath], then [imath]a \mid b[/imath]. Suppose this is true for some [imath]n = k[/imath]. Then [imath]a^k \mid b^k[/imath], so [imath]a|b[/imath]. [imath]a^k \mid b^k[/imath] means there exists some integer [imath]m[/imath] such that [imath]b^k = ma^k[/imath], and [imath]a \mid b[/imath] means there exists an integer [imath]r[/imath] such that [imath]b=ra[/imath]. Then we can multiply both sides of the above equality by any constant, namely [imath]b[/imath]. Then [imath]b \cdot b^k = bma^k[/imath] so [imath]b^{k+1} = (ra)ma^k = rma^{k+1}[/imath] which means that [imath]a^{k+1} \mid b^{k+1}[/imath] completing the induction step, and thus proving that for all integers [imath]n \geq1[/imath] the original statement holds. Are there any logical holes in the proof? |
2978515 | How many elements π in [imath]S_n[/imath] such that [imath]π^2 = e[/imath] , (the order of π is 1 or 2)?
For example in [imath]S_3[/imath], there are 4, namely: e , (01)(2), (02)(1), and (12)(0). In [imath]S_4[/imath] , i think there are 1 + 6 + 6 = 13. The identity e: gives 1, Permutation in the form of (ab)(c)(d) : 6, In the form of (ab)(cd) : 6 But i had a feeling it must be 14 or something even... (Because 4! - 13 = 11 which is cannot be partitioned in pairs) What am i missing ? Also, Is there a way to dertermine how many such element in [imath]S_n[/imath] ? Tnk u in advance | 288597 | Cycles in permutations of a given size
How would I find the number of permutations of [imath][n][/imath] in which all cycles have length [imath]1[/imath] or [imath]2[/imath]? I know how to do this if I have, say, [imath]8[/imath] numbers from [imath]{{1,2,3,...,8}}[/imath] and I want them to be broken into Count[imath](4, 3, 1)[/imath]. But not when I don't know how many cycles of [imath]2[/imath] and [imath]1[/imath] I want. Note: I'm looking for a way to derive a formula or something. I need to use it to prove by induction, I'm assuming, [imath]r(n + 1) = r(n) + n · r(n − 1)[/imath]. |
2978540 | Combinatorics Question Involving Arranging Letters in INTELLIGENT with Restriction
How many ways are there to arrange the letters in INTELLIGENT with at least two consecutive pairs of identical letters? I know that we would use the inclusion exclusion principle here and that there are [imath]\dfrac{11!}{2!2!2!2!2!1!}[/imath] ways to arrange the word with no restrictions. My first guess would be to assign two pairs of identical letters as a single block which we could choose two pairs from five possible pairs to get the pairs themselves. Then these pairs could be arranged in such a block in two ways. Then we would have to figure the number of ways to arrange this block in the entire word. I'm just confused where to go from here... | 1285459 | Inclusion-Exclusion: INTELLIGENT permutations
How many ways are there to arrange the letters in INTELLIGENT with at least two consecutive pairs of identical letters? I got an answer of [imath]C(5,2)\cdot\frac{9!}{2!^3}-C(5,3)\cdot\frac{8!}{2!^2}+C(5,4)\cdot\frac{7!}{2!}-6!\;,[/imath] but the solution manual says [imath]C(5,2)\cdot\frac{9!}{2!^3}-2\cdot C(5,3)\cdot\frac{8!}{2!^2}+3\cdot C(5,4)\cdot\frac{7!}{2!}-4\cdot 6!\;.[/imath] Where am I going wrong? can someone explain where the [imath]2\cdot,3\cdot[/imath], and [imath]4\cdot[/imath] are coming from in the solution manual? |
2978962 | Does [imath]S[/imath] converge or diverge?
Does [imath]S = \sum_{n=1}^{\infty}(-1)^n (e- (1+ \frac{1}{n})^n)[/imath] converge or diverge? My attempt : I know that [imath]e = \lim_{n\rightarrow \infty}( 1+ \frac{1}{n})^n[/imath]. Now put the value e in given series [imath]S[/imath] , I got [imath]\sum_{n=1}^{\infty}(-1)^n (e- (1+ \frac{1}{n})^n)= \sum_{n=1}^{\infty}(-1)^n (e- e)=0[/imath] so the given series is converges is it correct???? | 71285 | Series - Apostol Calculus Vol I, Section 10.20 #24
I am having a lot of trouble with these series questions. Up until this point, I had relatively little trouble with all the questions in the book. These seem to require knowledge about approximations of functions and other external experience-based knowledge, which I just don't have yet. Determine convergence or divergence of the given series. In the case of convergence, determine whether the series converges absolutely or conditionally. [imath]\sum_{n=1}^\infty (-1)^n\left[e-\left(1+\frac 1 n \right)^n\right][/imath] It's easy to see that [imath]\lim_{n\to\infty}\left[e-\left(1+\frac 1 n\right)^n\right]=0[/imath] however, in order to apply Leibniz's Rule and show conditional convergence I need to show that the sequence is monotonically decreasing. This doesn't seem doable with straight inequalities, so I tried taking the derivative, which just resulted in an uninterpretable mess. This doesn't even begin to address the question of absolute convergence/divergence. There are 54 of these questions... I must be missing something really fundamental if they all take this long. |
2979633 | Prove that [imath]8p^2+2p+1[/imath] is prime number
If [imath]p[/imath] and [imath]8p^2+1[/imath] is prime number than [imath]8p^2+2p+1[/imath] is prime number proof. I know that prime number can write as a multiply 1 and [imath]8p^2+2p+1[/imath], so if I show that this number is a multiply [imath]a*b[/imath] where [imath]a\not=1[/imath] and [imath]b\not=1[/imath] than that is not a prime number, can someone help me? | 1454539 | Show that [imath]p^3+4[/imath] is prime
If [imath]p[/imath] and [imath]p^2 +8[/imath] are both prime number, prove that [imath]p^3 +4[/imath] is also prime. I was thinking using two cases for even and odd. So [imath]p[/imath] is even thus there is no prime, the statement does not hold. So, [imath]p[/imath] is odd, [imath]p[/imath] has to be of the form of [imath]2k+1[/imath]. let [imath]p=2k+1[/imath] which is prime [imath](2k+1)^2 + 8 = 4k^2+4k+1+8 = 2(2k^2+2k+4)+1[/imath] which is prime so [imath](2k+1)^3 +4 = 8k^3+12k^2+6k+1 +4= 2(4k^3+6k^2+3k+2)+1[/imath] which is also prime. is that all that i need to show |
2979666 | The dihedral group [imath]D_8[/imath] isn't Hamiltonian
Let [imath]D_8=\{a^ib^j:i\in\{0,1\},j\in\{0,...,3\}\} [/imath] be a dihedral group, where [imath]a=\begin{pmatrix} -1 &0 \\ 0 & 1 \end{pmatrix}\qquad\text{ and }\qquad b=\begin{pmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{pmatrix},[/imath] with [imath]\theta=\frac{2\pi}{4}[/imath]. Prove that [imath]D_8[/imath] is not a Hamiltonian group. | 1889610 | Prove whether or not a subgroup is normal in the dihedral group [imath]D_4[/imath]
Let [imath]H[/imath] be the subgroup of the dihedral group [imath]D_4[/imath] generated by the elements [imath]\left<I\ (\text{identity}),p^2\epsilon\ (\text{flip across the horizontal diagonal})\right>[/imath]. I have to tell whether or not this is a normal subgroup of the dihedral group [imath]D_4[/imath] (the regular polygon with 4 sides). I know that [imath]H = \{I,p^2\epsilon\}[/imath]. After that I made the multyplicative tale of D_4 and I saw that that [imath]xH = Hx, \forall x \in D_4[/imath], therefor H is normal in [imath]D_4[/imath]. But I'm not sure if this is the way it should be done. I think that there is another way which does not imply the multyplicative table. |
2979707 | If [imath]\phi:G\to H[/imath] is an isomorphism, prove that [imath]o(x)=o(\phi(x))[/imath] for all [imath]x\in G[/imath]
If [imath]\phi:G\to H[/imath] is an isomorphism, prove that [imath]o(x)=o(\phi(x))[/imath] for all [imath]x\in G[/imath] My current thought: [imath]\phi(g^n)=\phi(g)^n[/imath]. Thus,[imath]\phi(g^n)=\phi(e)=\phi(g)^n[/imath]. However, I don't know how to proceed from here. Also, does my previous steps work when the order of [imath]g[/imath] is infinite? | 2353831 | Isomorphism [imath]f[/imath] preserves the order of an element?
Let's say [imath]f[/imath] is an isomorphism [imath]f:G \rightarrow G'[/imath], where [imath]G[/imath] and [imath]G'[/imath] are multiplicative groups. Then if [imath]f(x \in G) = x' \in G'[/imath]. Do we have always have [imath]o(x) = o(x')[/imath]? Why? |
2978872 | Product of non-invertible elements
Let [imath]R[/imath] be a unital ring (not necessarily commutative) and [imath]x,y\in R[/imath] be non-invertible elements. Then is [imath]xy[/imath] non-invertible? And how about the case [imath]xy=yx[/imath]? (postsclipt) Can the product of two non invertible elements in a ring be invertible? : This page gives a counterexample of the case [imath]xy\neq yx[/imath]. | 627562 | Can the product of two non invertible elements in a ring be invertible?
Let [imath]A[/imath] be a unitary ring. The question is simply: can the product of two non invertible elements in [imath]A[/imath] be invertible? I proved that the answer is negative if [imath]A[/imath] does not have zero divisors, because if you have [imath]a,b[/imath] non invertible elements in [imath]A[/imath] and [imath]abx=xab=1[/imath], then [imath]a(bx)=1[/imath] so [imath]a[/imath] is right invertible and if we put [imath]d=bxa \implies db=b(xab)=b1=b \implies d=1[/imath] (because obviously [imath]b[/imath] is not [imath]0[/imath]), so [imath]a[/imath] is left invertible, as desired. If [imath]A[/imath] does have zero divisors, I don't see how to adapt the proof... Thank you! |
713597 | Schur's Lemma and the Center of a Group
If I have a group [imath]G[/imath] and a complex irreducible representation [imath]g:G\rightarrow GL_n(\mathbb{C})[/imath]. I am trying to use schur's lemma to show that for [imath]x\in Z(G)[/imath] we have that [imath]g(z)=\lambda_z I_n[/imath]. Now if we define a map [imath]f_z:\mathbb{C}\rightarrow \mathbb{C}[/imath] to be a homomorphism of represenations from [imath](g,\mathbb{C})[/imath] to [imath](g,\mathbb{C})[/imath] given by [imath]f(c)=g(z)c[/imath] Then from Schur's Lemma we have that this is an isomorphism. Am I on the right track? Where do I go from here? | 142317 | A problem regarding to Schur's lemma
Let [imath]\rho: G \to GL_n(\mathbb{C})[/imath] be an irreducible representation, and [imath]g\in Z(G)[/imath]. Show [imath]\rho(g)[/imath] is a scalar multiple of the identity matrix [imath]I[/imath]. I think I have it, here is my solution: Since [imath]\rho(g) \in Hom_G(\mathbb{C}, \mathbb{C})[/imath] and [imath]\rho[/imath] is irreducible, consider a nonzero eigenvalue of [imath]\rho(g)[/imath], say [imath]\lambda[/imath], we have [imath]\rho(g) -\lambda [/imath] is a zero map by Schur's lemma, as the map contains a non-trivial kernel (i.e. the eigenvectors associated to [imath]\lambda[/imath] are in the kernel). But I didn't use the condition that [imath]g\in Z(G)[/imath], so are there something wrong with my solution? |
2980496 | Are the epsilon numbers the ones greater than [imath]\omega[/imath] unreachable from below with exponentiation?
Let [imath]\varepsilon[/imath] be an epsilon number, i.e., an ordinal such that [imath]\omega^{\varepsilon} = \varepsilon[/imath]. Is it true that then, for each pair [imath]\alpha,\beta[/imath] of ordinals, with [imath]\alpha,\beta < \varepsilon[/imath], we have [imath]\alpha^\beta < \varepsilon[/imath]? Does the converse also hold? If [imath]\varepsilon[/imath] is an infinite ordinal unreachable from below with exponentiation, is it an epsilon number? I suspect this is the case (I haven't seen it written anywhere) but I'm having trouble to prove it. Ordinal exponentiation doesn't behave so nicely. If there exist [imath]\alpha, \beta < \varepsilon[/imath] with [imath]\alpha^\beta = \varepsilon[/imath], we can show that the Cantor Normal Form of [imath]\varepsilon[/imath] has at least two terms, but I don't know what to do if we have [imath]\alpha^\beta \geq \varepsilon[/imath]... | 2820508 | How symmetric is the hierarchy of indecomposable ordinals?
How symmetric is the hierarchy of indecomposable ordinals? The process of ordinal arithmetic follows a repetitive, inductive pattern from addition to multiplication to exponentiation etc., in which the morphism from one operation to the next looks pretty much the same - in the respect that every step up is an induction beyond the transinfinite limit of the previous operation. I'm calling this a symmetry. We have the concepts of additively indecomposable ordinals, and beyond that, multiplicatively indecomposable ordinals etc. Does this hierarchy enjoy the same inductive symmetry infinitely far in the same way - i.e. to exponentiation, tetration and beyond? [imath]1,\omega^\beta[/imath] are additively indecomposable (for all ordinals [imath]\beta[/imath]). [imath]2,\omega^{\omega^\beta}[/imath] are multiplicatively indecomposable. How does this list continue and what does it look like? It looks like [imath]0^0=1[/imath] and [imath]1^1=1[/imath] but [imath](n-1)^{n-1}>n\forall n>2\implies [/imath] so the first two exponentially indecomposable ordinals would be: [imath]2,\omega[/imath] Is it [imath]2,\omega,\omega^{\omega^{\omega^\beta}}[/imath] for all ordinals [imath]\beta[/imath] and so on? Or does the list of special cases at the start get ever more complicated, and do the power towers just get taller or do they change? Tetration is a complicated matter (at least over regular numbers) as the orders of exponents matter and the rule to always apply powers from the right isn't necessarily well-justified, so I imagine there's the potential to get into considering Dyck words etc. How messy does it get? |
2980675 | The set of continuous functions on [imath][0,1][/imath] is a vector space.
I'm a beginner in the course of Linear Algebra; please bear with me if the question seems too trivial. The set of all continuous functions on interval [imath][0,1][/imath] is a vector space. I have trouble in understanding this. What does a continuous function on [imath][0,1][/imath] mean? That the range lies within [imath][0,1][/imath]? For it to be a vector space, it needs to satisfy vector additivity. Say, we take a vector with continuous function [imath]f(t)=0.9[/imath] which belongs to [imath]V[/imath]. Another vector belonging to [imath]V[/imath] has continuous function [imath]g(t)=0.8[/imath]. For vector additivity, we add the elements of vector (in this case I'm considering only 1 element in the vector). Here the new vector would give us [imath]0.8+0.9[/imath] which us not in [imath][0,1][/imath]. And yet this is a valid vector space. I'm sure I'm missing something. I'm probably not able to understand what the question demands. | 1732877 | Show that C[0,1] is a vector space
The set [imath]C[0,1][/imath] is the set of all continuous functions [imath]f:[0,1]\to \mathbb{R}[/imath]. Show that [imath]C[0,1][/imath] is a vector space. Note: for [imath]f,g[/imath] (as elements of) [imath]C[0,1][/imath], we define [imath](f+g)(x) = f(x)+g(x)[/imath] and for a scalar [imath]c,[/imath] [imath](cf)(x) = cf(x)[/imath]. The first step in showing that something is a vector space is to show that vector [imath]u[/imath] + vector [imath]v[/imath] is in [imath]V[/imath]. I don't know what I'm supposed to assign as [imath]u[/imath] and [imath]v[/imath], though. Please help. Thanks. |
669487 | Prove that the product of some numbers between perfect squares is [imath]2k^2[/imath]
Here's a question I've recently come up with: Prove that for every natural [imath]x[/imath], we can find arbitrary number of integers in the interval [imath][x^2,(x+1)^2][/imath] so that their product is in the form of [imath]2k^2[/imath]. I've tried several methods on proving this, but non of them actually worked. I know, for example, that the prime numbers shouldn't be in the product. I was also looking for numbers [imath]x[/imath] so that between [imath]x^2[/imath] and [imath](x+1)^2[/imath] there is actually the number [imath]2k^2[/imath] for some natural [imath]k[/imath]. If we find all of these numbers, then we should prove the case only for the numbers which are not in this form. These [imath]x[/imath]s have this property: [imath]x^2<2k^2<(x+1)^2[/imath] leading to [imath]x<k\sqrt 2<x+1[/imath] and [imath]x\frac{\sqrt 2}{2}<k<(x+1)\frac{\sqrt 2}{2}[/imath]. This means there should be a natural number between [imath]x\frac{\sqrt 2}{2}[/imath] and [imath](x+1)\frac{\sqrt 2}{2}[/imath]. I've checked some of the numbers that aren't like that with computer, and they were: [imath]3,6,10,13,17,...[/imath]. the thing i noticed was that the difference between the two consecutive numbers of that form, is either [imath]3[/imath] or [imath]4[/imath]. I think this has something to do with the binary representation of [imath]\frac{\sqrt 2}{2}[/imath] but I don't know how to connect it with that. I would appreciate any help :) | 624251 | product is twice a square
For every positive integer [imath]n[/imath], there exists a set [imath]S\subset \{n^2+1,n^2+2,\dotsc,(n+1)^2-1\}[/imath], such that [imath]\prod_{k\in S}k=2m^2[/imath] for some positive integer [imath]m[/imath] I have no clue about it. Could anyone help me? Thanks a lot. p.s. Whether or not the problem was a conclusion of an paper is unknown. |
2980283 | Geometric intuition of the dimension of Grassmannians and flag manfolds
I wish to understand geometrically (not just algebraically) why the dimension of the Grassmanian [imath]G(k,n)[/imath] is [imath]k(n-k)[/imath] and the dimension of a flag manifold [imath]F(k_{1},k_{2},...,k_{n},N)[/imath] is [imath]\sum_{i=1}^{n}k_{i}(k_{i-1}-k_{i})+Nk_{n}[/imath] (in fact with understanding the Grassmanian case it would be enough because the flag are just "nested" Grassmanians). I am thinking in a spatial way in the well known [imath]G(2,5)[/imath] but I am unable to see geometrically how the space of all [imath]2[/imath]-planes in [imath]\mathbb{P}^{5}[/imath] can be [imath]6[/imath]-dimensional... | 2980359 | Geometric intuition of the dimension of Grassmannians and flag manfolds
I wish to understand geometrically (not just algebraically) why the dimension of the Grassmanian [imath]G(k,n)[/imath] is [imath]k(n-k)[/imath] and the dimension of a flag manifold [imath]F(k_{1},k_{2},...,k_{n},N)[/imath] is [imath]\sum_{i=1}^{n}k_{i}(k_{i-1}-k_{i})+Nk_{n}[/imath] (in fact with understanding the Grassmanian case it would be enough because the flag are just "nested" Grassmanians). I am thinking in a spatial way in the well known [imath]G(2,5)[/imath] but I am unable to see geometrically how the space of all [imath]2[/imath]-planes in [imath]\mathbb{P}^{5}[/imath] can be [imath]6[/imath]-dimensional. |
2980835 | Prime elements in the ring of integers.
Let [imath]\mathbb{Z}[\frac{{1+\sqrt{-3}}}{2}][/imath] be the ring of integers where [imath]\alpha=\frac{{1+\sqrt{-3}}}{2}[/imath]. I wish to show that [imath]1-\alpha[/imath] is a prime and check whether 3 is a prime or not. To show that [imath]1-\alpha[/imath] is prime, suppose [imath]1-\alpha[/imath] divides [imath](a+b\alpha)(c+d\alpha)[/imath]. Then I wish to show that it divides one of them. I wrote down the division algorithm for [imath]1-\alpha[/imath] divides [imath](a+b\alpha)(c+d\alpha)[/imath] but it's a big mess that I can't seem to make sense of. Any help is appreciated. | 1134338 | Factor into irreducibles in [imath]\mathbb{Z}[\omega][/imath] (Eisenstein Integers).
[imath]\mathbb{Z}[\omega] = \{a + b\omega : a, b\text{ are integers}\}[/imath] [imath]\omega = e^{2πi/3} = -\frac12 + \frac{\sqrt3}2i[/imath] My task is to factor the numbers [imath]2, 3, 4, 5, 6, 7, 8, 9, 10, 11[/imath] into irreducibles in [imath]\mathbb{Z}[\omega][/imath]. I don't know if I'm overthinking this, but I do not even know where to start. I'm used to doing this in [imath]\mathbb{Z}[i][/imath] and [imath]\mathbb{Z}[\sqrt2][/imath], but I have no idea how to approach [imath]\mathbb{Z}[\omega][/imath]. EDIT: I am not supposed to use the norm in this exercise. My biggest struggle right now is just figuring out how to factor these numbers. Using guess and check I was able to find: [imath]3 = (2 + \omega)(1 - \omega)[/imath] [imath]7 = (3 + \omega)(2 - \omega)[/imath] I'm plugging and chugging into Wolfram Alpha right now. Is there any better way to do this? |
2981127 | [imath]\frac{d^n}{dx^n}x^x=[/imath]?
I am looking for a general formula for [imath]D_n=\frac{d^n}{dx^n}x^x[/imath] Here's what I've gotten so far: [imath]n=0[/imath]: [imath]D_0=x^x[/imath] [imath]n=1[/imath]: [imath]D_1=x^x(\log x+1)[/imath] [imath]n=2[/imath]: [imath]D_2=x^{x-1}+x^x(1+2\log x+\log^2 x)[/imath] And then I gave up. But I would be surprised if there wasn't a general formula. In any case, I am trying to find [imath]D_n[/imath] so that I can create a Taylor series for [imath]x^x[/imath]. If this isn't the way to go, please point me in the right direction or give me a series representation for [imath]x^x[/imath] | 1607520 | How should I calculate the [imath]n[/imath]th derivative of [imath]f(x)=x^x[/imath]?
What would be the [imath]n[/imath]th derivative of [imath] f (x) = x ^ x[/imath] I have reached the fifth derivative, very long indeed but I see no pattern that will help me find a general expression. \begin{align*} \frac{df}{dx} &= x^x(1+\ln(x))\\ \frac{d^2f}{dx^2} &= x^x\left(\frac{1}{x}+1+2\ln(x)+\ln(x)^2\right)\\ \frac{d^3f}{dx^3} &= x^x\left( \frac{-1}{x^2}+\frac{3}{x} + \ln(x)^3 + 3 \ln(x)^2 + \frac{3\ln(x)}{x} + 3\ln(x) + 1 \right)\\ \frac{d^4f}{dx^4} &= x^x\left( \frac{2}{x^3}-\frac{1}{x^2}-\frac{4\ln(x)}{x^2}+\frac{6}{x}+\ln(x)^4+4\ln(x)^3\right.\\ &\qquad\qquad\left.+\frac{6\ln(x)^2}{x}+6\ln(x)^2+\frac{12\ln(x)}{x}+4\ln(x)+1 \right)\\ \frac{d^5f}{dx^5} &= x^x\left( \frac{-6}{x^4}+\frac{10\ln(x)}{x^3}+\frac{5}{x^2}-\frac{10\ln(x)^2}{x^2}+\frac{10}{x}+\ln(x)^5+5\ln(x)^4\right.\\ &\qquad\qquad\left.+\frac{10\ln(x)^3}{x}+10\ln(x)^3+\frac{30\ln(x)^2}{x}+10\ln(x)^2+\frac{30\ln(x)}{x}+5\ln(x)+1 \right) \end{align*} |
2980906 | Showing an orthogonalisation process
Can anyone show that: [imath]\mathbf{a_{\perp}}=\mathbf{a}-\frac{\mathbf{x}\mathbf{x}^T}{\mathbf{x}^T\mathbf{x}}\mathbf{a}[/imath], [imath]\mathbf{a}\in\mathbb{R}^N[/imath], [imath]\mathbf{x}=(1,1,\dots,1)^T\in\mathbb{R}^N[/imath] results in [imath]\mathbf{a}[/imath] becoming orthogonal to [imath]\mathbf{x}[/imath]? Thank you for your help. | 1785187 | Proving the two given vectors are orthogonal
I am given the vectors [imath]\mathbf{w}, \mathbf{v}, \mathbf{u}[/imath] in [imath]\mathbb{R}^n[/imath] such that [imath]\mathbf{u} \neq 0[/imath] and [imath]\mathbf{w} = \mathbf{v} - \frac{\mathbf{u}\,\bullet\,\mathbf{v}}{\|\mathbf{u}\|^2}\bullet\mathbf{u}[/imath]. I am asked to show that the vector [imath]\mathbf{w}[/imath] is orthogonal to [imath]\mathbf{u}[/imath]. So far, I have written out the definition of orthogonal: two vectors are orthogonal if and only if their dot product is zero. So what we need to prove is [imath]\mathbf{w}\bullet\mathbf{u} = 0[/imath] where [imath]\mathbf{w}\bullet\mathbf{u}[/imath] is defined as [imath]\mathbf{w}^T\bullet\mathbf{u}[/imath]. However, I have been stuck on this problem for about an hour and haven't made any significant progress from here. How do we go about proving that the vectors are orthogonal? |
2981463 | [imath]|\int \limits_a^b f(x) dx|\leq\int \limits_a^b |f(x)|dx[/imath] for f continuous
How to prove [imath]|\int \limits_a^b f(x) dx|\leq\int \limits_a^b |f(x)|dx[/imath] for f continuous? This is a step in the solution of a problem from Mendelson's introduction to topology. This book assumes the reader has only a background in first-year calculus, not measure theory or advanced calculus. | 284322 | The triangle inequality for integrals
Let [imath]f[/imath] be a continuous and integrable function over [imath][a;b][/imath]. Prove or disprove that [imath]\int_a^b |f(x)|\ \mathrm{d}x\geq \left | \int_a^b f(x)\ \mathrm{d}x\right| [/imath] |
2981791 | Eigenvalue and singular value of a matrix
Let [imath]A[/imath] be a matrix. Let [imath]\sigma_{\text{min}}(A)[/imath] be the minimal singular value of [imath]A[/imath], and [imath]\lambda_{\text{min}}(A)[/imath] be the minimal eigenvalue of [imath]A[/imath]. Show we have this inequality : [imath]\sigma_{\text{min}}(A) < |\lambda_{\text{min}}(A)|[/imath] | 380802 | Minimum eigenvalue and singular value of a square matrix
How to show that the relationship [imath]\left | \lambda_{min} \right | \geq \sigma_{min}[/imath] holds between the minimum eigenvalue and singular value of a square matrix [imath]A \in \mathbb{C}^{n \times n}[/imath]? |
2981900 | Proving closed interval [0,1] has same cardinality as Real Numbers
I want to prove that the closed interval [0, 1] has same cardinality as Real Numbers. I was able to figure out (0, 1), but need help with proving the closed part. What I have so far is: the function [imath]y = tan x[/imath] gives us a bijection between [imath](−π/2, π/2)[/imath] and [imath]R[/imath], and that the function [imath]y = (2x − 1)π/2[/imath] gives a bijection between [imath](0, 1)[/imath] and [imath](−π/2, π/2)[/imath]. Therefore [imath]y = tan (2x − 1)π/2[/imath] gives a bijection between (0, 1) and [imath]R[/imath]. How do I finish this off for the closed aspect? | 2121250 | Cardinality of [imath][0,1][/imath]
Assuming we know that the cardinality of [imath](0,1)[/imath] is [imath]c[/imath]. Can we from this determine the cardinality of [imath][0,1][/imath]? Is it valid to say that [imath][0,1][/imath] also has cardinality c because adding a finite number of elements to [imath](0,1)[/imath] (i.e. including the boundaries) does not affect the cardinality? |
2980560 | For any [imath]a \in \Bbb{Z}[/imath], can we always find two prime numbers [imath]p, q[/imath], such that [imath]p - q \in (a)[/imath]?
This is a major weakening of many prime sum / difference existence questions. Let [imath]a \in \Bbb{Z}[/imath] and [imath](a)[/imath] the ideal generated by [imath]a[/imath]. Then do there exist two primes [imath]p, q[/imath] such that [imath]p - q \in (a)[/imath] at least? Thanks. | 1364540 | primes whose difference is a multiple of [imath]n[/imath]
Hello everyone this is my first question here so if there are any suggested edits feel free to participate! Is it true that for every [imath]n[/imath] there two prime numbers [imath]p[/imath] and [imath]q[/imath] with [imath]n\mid p-q[/imath]? I have verified this empirically but am not sure if there is a proof for this |
2981394 | Expectation of sample mean, given maximum and minimum order statistics
Let [imath]X_1, · · · , X_n[/imath] be i.i.d. [imath]\mathrm{Uniform}[\alpha, \beta][/imath], where [imath]\alpha[/imath] and [imath]\beta[/imath] are unknown. Show that [imath] E(X_\ast|X(1), X(n)) = \frac{X(1) + X(n)}{2} [/imath] where [imath]X_\ast[/imath] is the sample mean, and [imath]X(n)[/imath] is the [imath]n[/imath]th order statistic. I know that [imath]X(1),X(n)[/imath] are the sufficient statistics for this family (minimal in fact) and intuitively it makes sense that the expected value of the sample mean will lie in the middle of the max and min. But how do I show it rigorously? | 2001780 | Show that [imath]\mathbb{E}\left(\bar{X}_{n}\mid X_{(1)},X_{(n)}\right) = \frac{X_{(1)}+X_{(n)}}{2}[/imath]
Let [imath]X_{1},\ldots,X_{n}[/imath] be i.i.d. [imath]U[\alpha,\beta][/imath] r.v.s., and let [imath]X_{(1)}[/imath] denote the [imath]\min[/imath], and [imath]X_{(n)}[/imath] the [imath]\max[/imath]. Show that [imath] \mathbb{E}\left(\overline{X}_{n}\mid X_{(1)},X_{(n)}\right) = \frac{X_{(1)}+X_{(n)}}{2}. [/imath] I know that [imath]\displaystyle\mathbb{E}\left(\overline{X}_{n}\mid X_{(1)},X_{(n)}\right)=\mathbb{E}\left({X}_{1}\mid X_{(1)},X_{(n)}\right)[/imath] but not much more. |
2971608 | How do Hrbacek and Jech derive the class function [imath]F[/imath] in Theorem 4.12 from Theorem 4.11?
My question is about a proof of Theorem 4.12 given in the text Introduction to Set Theory by Hrbacek and Jech. Let [imath]V[/imath] be the class of all sets, [imath]\operatorname{Ord}[/imath] be the class of all ordinals, and [imath]G,G_1,G_2,G_3[/imath] be class functions from [imath]V[/imath] to [imath]V[/imath]. Theorem 4.11: There exists a class function [imath]F:V\times\operatorname{Ord}\to V[/imath] such that, for all [imath]z\in V[/imath] [imath]F(z,0)=G_1(z,\emptyset)[/imath] [imath]F(z,\alpha+1)=G_2(z,F_z(\alpha))[/imath] for all [imath]\alpha\in\operatorname{Ord}[/imath], with [imath]F_z(\alpha):=F(z,\alpha)[/imath] [imath]F(z,\alpha)=G_3(z,F_z\restriction\alpha)[/imath] for all [imath]\alpha\neq 0[/imath] limit, with [imath]F_z\restriction \alpha:=\{\langle\beta,F(z,\beta)\rangle\mid\beta<\alpha\}[/imath] Theorem 4.12: For any set [imath]a[/imath], there is a unique infinite sequence [imath](f_n\mid n\in \Bbb N)[/imath] such that (1) [imath]f_0=a[/imath] (2) [imath]f_{n+1}=G(f_n,n)[/imath] for all [imath]n\in \Bbb N[/imath] The proof for Theorem 4.12 is given in the text as follows: Let [imath]G:V\to V[/imath] be a class function. We want to find, for every set [imath]a[/imath], a sequence [imath](f_n\mid n\in \Bbb N)[/imath] such that [imath]f_0=a[/imath] and [imath]f_{n+1}=G(f_n,n)[/imath] for all [imath]n\in \Bbb N[/imath]. By Theorem 4.11, there is class function [imath]F[/imath] such that [imath]F(0)=a[/imath] and [imath]F(n+1)=G(F(n),n)[/imath] for all [imath]n\in \Bbb N[/imath]. Now we apply the Axiom of Replacement: There exists a sequence [imath](f_n\mid n\in \Bbb N)[/imath] that is equal to [imath]F\restriction \omega[/imath] and the Theorem follows. I don't understand how Theorem 4.11 is used to derive the class function [imath]F[/imath] in the proof of Theorem 4.12. Can someone elaborate on how to obtain the desired [imath]F[/imath]? Thank you for your help! | 2964007 | How do the authors deduce this weaker version of Transfinite Recursion?
My question is about a proof of Theorem 4.12 given in the text Introduction to Set Theory by Hrbacek and Jech. The authors first prove Theorems 4.9, 4.10, and 4.11. For reference, I quote these theorems below. Let [imath]V[/imath] be the class of all sets, [imath]\operatorname{Ord}[/imath] be the class of all ordinals, and [imath]G:V\to V[/imath] be a class function. Theorem 4.9: Transfinite Recursion Theorem There exists a class function [imath]F:\operatorname{Ord}\to V[/imath] such that [imath]F(\alpha)=G(F\restriction \alpha)[/imath] for all [imath]\alpha\in\operatorname{Ord}[/imath]. Theorem 4.10: Transfinite Recursion Theorem, Parametric Version There exists a class function [imath]F:\operatorname{Ord}\to V[/imath] such that [imath]F(z,\alpha)=G(z,F_z\restriction \alpha)[/imath] for all [imath]\alpha\in\operatorname{Ord}[/imath] and [imath]z\in V[/imath] where [imath]F_z\restriction \alpha:=\{\langle\beta,F(z,\beta)\rangle\mid\beta<\alpha\}[/imath]. Theorem 4.11: Let [imath]G_1,G_2,G_3[/imath] be class functions from [imath]V[/imath] to [imath]V[/imath]. There exists a class function [imath]F:\operatorname{Ord}\to V[/imath] such that (1) [imath]F(0)=G_1(\emptyset)[/imath] (2) [imath]F(\alpha+1)=G_2(F(\alpha))[/imath] for all [imath]\alpha\in\operatorname{Ord}[/imath] (3) [imath]F(\alpha)=G_3(F\restriction\alpha)[/imath] for all limit [imath]\alpha\neq 0[/imath] Then they say A parametric version of Theorem 4.11 is straightforward and we leave it to the reader. From my attempt, the parametric version of Theorem 4.11 is as follows (Honestly, I'm not sure if my attempt is correct or not) Let [imath]G_1,G_2,G_3[/imath] be class functions from [imath]V[/imath] to [imath]V[/imath]. There exists a class function [imath]F[/imath] such that, for all [imath]z\in V[/imath] [imath]F(z,0)=G_1(z,\emptyset)[/imath] [imath]F(z,\alpha+1)=G_2(z,F_z(\alpha))[/imath] for all [imath]\alpha\in\operatorname{Ord}[/imath] [imath]F(z,\alpha)=G_3(z,F_z\restriction\alpha)[/imath] for all limit [imath]\alpha\neq 0[/imath] Next they make use of this parametric version of Theorem 4.11 in the proof of Theorem 4.12. Theorem 4.12: For any set [imath]a[/imath], there is a unique infinite sequence [imath](f_n\mid n\in \Bbb N)[/imath] such that (1) [imath]f_0=a[/imath] (2) [imath]f_{n+1}=G(f_n,n)[/imath] for all [imath]n\in \Bbb N[/imath] The proof for Theorem 4.12 given in the text is as follows: Let [imath]G:V\to V[/imath] be a class function. We want to find, for every set [imath]a[/imath], a sequence [imath](f_n\mid n\in \Bbb N)[/imath] such that [imath]f_0=a[/imath] and [imath]f_{n+1}=G(f_n,n)[/imath] for all [imath]n\in \Bbb N[/imath]. By the parametric version of the Transfinite Recursion Theorem 4.11, there is class function [imath]F[/imath] such that [imath]F(0)=a[/imath] and [imath]F(n+1)=G(F(n),n)[/imath] for all [imath]n\in \Bbb N[/imath]. Now we apply the Axiom of Replacement: There exists a sequence [imath](f_n\mid n\in \Bbb N)[/imath] that is equal to [imath]F\restriction \omega[/imath] and the Theorem follows. I don't understand how the parametric version of Theorem 4.11 is used to derive the class function [imath]F[/imath] in the proof of Theorem 4.12. Can someone please help me fill in the blanks? |
1195601 | Posed in regional mathematics Olympiad 2005
Let a, b, c be three positive real numbers such that [imath]a+b+c = 1[/imath]. Let [imath]\Delta= \min( a^{3} + a^{2}bc, b^{3}+ab^{2}c, c^{3}+abc^{2} )[/imath]. Prove that the roots of the equation [imath]x^{2} + x + 4 \Delta = 0[/imath] are real. The last line is equivalent to [imath]\Delta \leq \frac{1}{16} [/imath]. So I tried to prove by contradiction. I assumed all of them are [imath] > \frac{1}{16} [/imath]and tried to draw a contradiction with the fact [imath]a+ b+ c=1[/imath] but failed | 767465 | proving roots of equation [imath]x^2+x+4f=0[/imath] are real.
Let [imath]a,b,c[/imath] be three positive real numbers such that [imath]a+b+c=1[/imath]. Let [imath]f=\min(a^3+a^2bc, b^3+ab^2c,c^3+abc^2)[/imath]. Prove that the roots of the equation [imath]x^2+x+4f=0[/imath] are real. |
2982515 | lframConvergence of [imath]\sum_{n=2}^\infty \frac1{\log(n!)}[/imath]
How do I show that this sum diverges/converges? [imath]\sum_{n=2}^\infty \frac1{\log(n!)}[/imath] I want to use the comparison test, but I do not know how to approach. Also, Wolfram says this diverges by the comparison test, but Mathematica gives me a numerical answer using the NSum function. | 2312845 | Sum of [imath]\sum^\infty_{n=2} \frac{1}{\ln{(n!)}}[/imath]
I am sorry that my question is not whether [imath]\sum^\infty_{n=2} \frac{1}{\ln{(n!)}}[/imath] convergent or not. My question is how to compute the sum of series: [imath]\sum^\infty_{n=2} \frac{1}{\ln{(n!)}},[/imath] It is a question in a textbook, and I have worked on it for days, but fail. |
2982524 | What is wrong with the followig proof that [imath]\mathbb{Z}[/imath] is isomorphic to [imath]\mathbb{Z} \times \mathbb{Z}[/imath]
Since [imath] G =( \prod_{1}^{\infty} \mathbb{Z} )\times \mathbb{Z} \times \mathbb{Z} \cong (\prod_{1}^{\infty} \mathbb{Z}) \times \mathbb{Z} [/imath], by taking quotients we get [imath]\mathbb{Z \times Z} \cong G/ \prod_{1}^{\infty} \mathbb{Z} \cong \mathbb{Z}[/imath]. Therefore [imath]\mathbb{Z} \cong \mathbb{Z \times Z}[/imath]. But [imath]\mathbb{Z}[/imath] is indecomposable! What's wrong with above proof? | 2519951 | Cancellation Law for Direct Sums - What is wrong with this argument?
Let [imath]G[/imath], [imath]H[/imath], and [imath]K[/imath] be abelian groups. Assume [imath]G \oplus H \cong G \oplus K[/imath]. If each are finitely generated, then the structure theorem easily gives us that [imath]H \cong K[/imath]. However, there are counterexamples in the general case, namely [imath]H=\mathbb{Z}[/imath], [imath]K=\mathbb{Z} \oplus \mathbb{Z}[/imath], and [imath]G=\bigoplus_{i=1}^\infty\mathbb{Z}[/imath]. I am wondering what goes wrong with the following argument: Form the split short exact sequence [imath]0 \to G \xrightarrow{i} G \oplus H \xrightarrow{\pi} H \to 0[/imath] where [imath]i[/imath] and [imath]\pi[/imath] are the inclusion and projection, respectively. Then don't we immediately get from the first isomorphism theorem that [imath]H \cong (G \oplus H)/G[/imath], which would contradict the counterexample above? If we run this for the counterexample given, we get the sequence [imath]0 \to \bigoplus_{i=1}^\infty\mathbb{Z} \to \bigoplus_{i=1}^\infty\mathbb{Z} \to \mathbb{Z} \to 0[/imath] which seems to still be short exact, even though here [imath]G \cong G \oplus H[/imath]. But of course here we can see directly that [imath](G \oplus H)/G \cong G/G[/imath], which is trivial. So what went wrong? It seems to me to be something along these lines: In this case, the kernel of the projection is not the same as the original group, it is just, say, [imath]0 \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \dots[/imath], i.e. nontrivial from the second coordinate on, which just happens to be isomorphic to the original group. So if we form the other short exact sequence with [imath]K[/imath] in place of [imath]H[/imath], then the difference seems to be that we have two different projection maps with different (but isomorphic) kernels. But I'm not really sure how to explain this in the general case, if my reasoning is correct to begin with. Thanks in advance for any help! |
2982596 | Characteristic and Minimal Polynomial
Duplicate Let [imath]A[/imath] be the complex [imath]3 \times 3[/imath] matrix [imath]A = \begin{bmatrix} 2 & 0 & 0 \\ a & 2 & 0 \\ b & c & -1 \end{bmatrix}[/imath] Find all triples [imath](a,b,c)[/imath] for which the characteristic and minimal polynomials of [imath]A[/imath] are different. | 2762855 | find all the value of x,y z?
Let [imath]A[/imath] be the complex [imath]3 × 3[/imath] matrix [imath]A = \begin{bmatrix}2&0&0 \\ x&2 & 0 \\y&z&-1\end{bmatrix}[/imath] Find all triples [imath](x, y, z)[/imath] for which the characteristic and minimal polynomials of [imath]A [/imath] are different. My attempts : the Characteristic polynomial of [imath]A = (\lambda +1)(\lambda -2)^2[/imath] Case [imath]1[/imath]: if [imath]x=y= z \neq 0[/imath] ,then minimial polynomial of [imath]A = (\lambda +1)(\lambda -2)^2[/imath] Case [imath]2[/imath] : if [imath]x=y= z =0[/imath],then minimial polynomial of [imath]A = (\lambda +1)(\lambda -2)[/imath] Here im confused that how can i find all triples [imath](x, y, z)[/imath] for which the characteristic and minimal polynomials of [imath]A [/imath] are different. |
2061981 | Prove that [imath](\frac{bc+ac+ab}{a+b+c})^{a+b+c} \geq \sqrt{(bc)^a(ac)^b(ab)^c}[/imath]
Prove that [imath]\left( \frac{bc+ac+ab}{a+b+c} \right)^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}[/imath] where [imath]a,b,c>0[/imath] My attempt: I couldn't proceed after that. Please help me in this regard, thanks! | 621253 | Prove that [imath](\frac{bc+ac+ab}{a+b+c})^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}[/imath]
Prove that [imath](\frac{bc+ac+ab}{a+b+c})^{a+b+c} \ge \sqrt{(bc)^a(ac)^b(ab)^c}[/imath] I tried it to do using [imath]AM \ge GM[/imath] but don't know how to proceed. Please help. |
2982656 | Prove a series [imath]\sum_{n=1} ^\infty \frac{(-1)^n}{n}[/imath] is convergent
I was asked to prove that an infinite series [imath]\sum_{n=1} ^\infty \frac{(-1)^n}{n}[/imath] is a convergent series. I tried using ratio test but the limit results in 1 which is inconclusive. I am stuck at this point. Can you give me some hint on how to approach this question? Thank you edit : My professor only taught ratio test, root test and comparison test where if [imath]|b_i| \leq a_i[/imath] for all i = 1, 2, ... and [imath]\sum a_i[/imath] converges then the sum of [imath]b_i[/imath] converges absolutely. Is there any way other than alternating series test to prove this problem? | 368293 | [imath]\sum (-1)^n/n[/imath] fails the p-series test, but passes the alternating series test?
P-Series Reference Alternating Series Test Reference [imath] \sum_{i=0}^\infty \frac{(-1)^n}{n} [/imath] This alternating series fails the p-series test because the exponent of n = 1. Yet it seems to pass the alternating series test. 1 - [imath]a_n[/imath] must be positive. True. 2 - Terms must be decreasing. [imath]\frac{d}{dn} 1/n = -n^{-2}[/imath], which is < 1. True. 3 - [imath] \lim_{n\rightarrow\infty} 1/n = 0 [/imath] True. [imath](-1)^n/n[/imath] is clearly a divergent series, so why does it pass the AST? |
2983500 | Use Eisenstein´s criterion to show that [imath]P(x)=x^n+5x^{n-1}+3[/imath] is irreducible.
Since Eisenstein´s criterion ist not directly applicable we look at the polynomial P (mod 5), which then reduces to [imath]P(x) = x^n + 3[/imath] (mod 5) ... is this correct? Now I can apply the Eisenstein Criterion with prime number 3: The leading coefficient of [imath]x^n[/imath] is 1 and not divisible by 3. The other coefficients, 0 and 3 are divisible by the prime 3. Further, the constant, 3, ist not divisible by the square of the chosen prime, [imath]3^2 = 9[/imath]. Therefore, [imath]P[/imath] is an irreducible polynomial. Is this correct? Have I understood the E.C. correctly? | 164043 | [imath]f(x)=x^n+5x^{n-1}+3[/imath] can't be expressed as the product of two polynomials
Let [imath]f(x)=x^n+5x^{n-1}+3[/imath] where [imath]n\geq1[/imath] is an integer.Prove that [imath]f(x)[/imath] can't be expressed as the product of two polynomials each of which has all its coefficients integers and degree[imath]\geq1[/imath]. If the condition that each polynomial must have all its coefficients integers was not there, then I needed to show only that [imath]f(x)[/imath] is irreducible over real numbers.But since this condition is given,therefore,we can't exclude the case when [imath]f(x)[/imath] is reducible over real numbers but not with polynomials of integer coefficients.Anyone with idea how to proceed?Thanks in advance!! |
2983777 | If [imath]|z+w|=|z-w|[/imath], prove that [imath]\arg z-\arg w =\pm\pi/2[/imath]
If [imath]|z+w|=|z-w|[/imath], how to prove that [imath]\arg z-\arg w =\pm\frac{\pi}{2}[/imath] I squared the modulus so that I can use properties of conjugates to simplify it yielding [imath]zw*=-wz*[/imath]. | 1446618 | If [imath]z[/imath] and [imath]w[/imath] are complex numbers such that [imath]|(z+w)| = |(z-w)|[/imath], prove that [imath]\arg(z)-\arg(w) = \pm (\pi/2)[/imath]
If z and w are complex numbers such that [imath]|z+w|[/imath] = [imath]|z-w|[/imath], prove that [imath]\arg(z)-\arg(w)= \pm\pi/2[/imath]. Can this be solved algebraically or would a graphic interpretation be better. Both methods would be grateful. |
2983042 | Mathematical Proof By Induction of [imath]1 + 2 +\dots + n = (n + 1)n/2[/imath]
Trying to prove that [imath]1 + 2 +\dots + n = (n + 1)n/2[/imath]. I have let [imath]n = 1[/imath] in the basis step, which lead to [imath]1=1[/imath], so it is true for [imath]n = 1[/imath]. For the induction step, I have assumed that [imath]k\geq 1[/imath] and let [imath]n = k[/imath] leading to: [imath]1 + 2 + . . . + k = (k + 1)k/2[/imath] For which we then have to use to prove that the statement is valid for [imath]n = k + 1[/imath], which is where I get confused because this the statement is supposed to become: [imath]1 + 2 + . . . + k + (k + 1) = (k + 2)(k + 1)/2[/imath] While I thought it would have just been [imath]1 + 2 + . . . + (k + 1) = (k + 2)(k + 1)/2 [/imath] Via the substitution of [imath]k = k + 1[/imath]. Why is there still a singular [imath]k[/imath] in the [imath]k + 1[/imath] statement? | 1206645 | Showing [imath]1+2+\cdots+n=\frac{n(n+1)}{2}[/imath] by induction (stuck on inductive step)
This is from this website: Use mathematical induction to prove that [imath]1 + 2 + 3 +\cdots+ n = \frac{n (n + 1)}{2}[/imath] for all positive integers [imath]n[/imath]. Solution to Problem 1: Let the statement [imath]P(n)[/imath] be [imath]1 + 2 + 3 + \cdots + n = \frac{n (n + 1)}{2}.[/imath] STEP 1: We first show that [imath]P(1)[/imath] is true. Left Side [imath]= 1[/imath] Right Side [imath]= \frac{1 (1 + 1)}{2} = 1[/imath] Both sides of the statement are equal hence [imath]P(1)[/imath] is true. STEP 2: We now assume that [imath]P(k)[/imath] is true [imath]1 + 2 + 3 + \cdots + k = \frac{k (k + 1)}{2}[/imath] and show that [imath]P(k + 1)[/imath] is true by adding [imath]k + 1[/imath] to both sides of the above statement [imath] \begin{align} 1 + 2 + 3 + \cdots + k + (k + 1) &= \frac{k (k + 1)}{2} + (k + 1) \\ &= (k + 1)\left(\frac{k}{2} + 1\right) \\ &= \frac{(k + 1)(k + 2)}{2} \end{align} [/imath] The last statement may be written as [imath]1 + 2 + 3 + \cdots + k + (k + 1) = \frac{(k + 1)(k + 2)}{2}[/imath] Which is the statement [imath]P(k + 1)[/imath]. My question is how in the very last line is the statement [imath]P(k + 1)[/imath] equal to [imath]\frac{(k + 1)(k + 2)}{2}[/imath]. I don't get the last step. |
2984535 | Take [imath]f[/imath] and entire function such that [imath]f(z) = f(z+1)=f(z+\sqrt{2})[/imath], then it is constant
I am working through some practice problems and I have this one which is stumping me: Take [imath]f[/imath] and entire function such that [imath]f(z) = f(z+1)=f(z+\sqrt{2})[/imath], then it is constant. I was thinking, given the function is entire, then it has a Taylor series expansion about the origin, with an infinite radius of convergence: [imath]f(z) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}z^n=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(z+1)^n=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(z+\sqrt{2})^n[/imath] Which we can also write: [imath]\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}z^n = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\sum_{k=0}^{n}{n\choose k}z^k = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\sum_{k=0}^{n}{n\choose k}z^k\sqrt{2}^{n-k}[/imath] But I am not quite sure this is the correct place to go. Any ideas as to how I can move forward on this problem? Thank you! | 2538873 | Doubly periodic entire function with periods that are linearly independent over [imath]\mathbb{Q}[/imath] is constant.
Question Let [imath]f[/imath] be entire and suppose there exists [imath]\lambda_1,\lambda_2 \in \mathbb{C}[/imath] linearly independent over [imath]\mathbb{Q}[/imath] such that [imath]f(z+\lambda_1)=f(z)=f(z+\lambda_2)[/imath]. Then [imath]f[/imath] is constant. Attempt A similar question was asked here but I don't think the answer is satisfactory. I want to (hopefully) show that for any [imath]\epsilon>0[/imath] I can find integers [imath]m_1,m_2[/imath] such that [imath]|m_1\lambda_1+m_2\lambda_2|<\epsilon[/imath]. Then I will be able prove the result as follows: Let [imath]z_0\in \mathbb{C}[/imath]. Then [imath]g(z)=f(z)-z_0[/imath] has a zero at [imath]z_0[/imath] and this zero must be isolated. Consequently we can find an [imath]\epsilon>0[/imath] such that the epsilon ball about [imath]z_0[/imath] contains only one zero ([imath]z_0[/imath]) of [imath]g(z)[/imath]. However, by the above, we can find [imath]m_1,m_2[/imath] so that [imath]|m_1\lambda_1+m_2\lambda_2|<\epsilon[/imath] and by assumption [imath]g(z_0+m_1\lambda_1+m_2\lambda_2)=g(z_0)=0[/imath] which is only possible if [imath]f[/imath] had been constant. |
1691724 | Show that a borelien outer measure on a metric space is a metric outer mesure.
Let [imath](X,d)[/imath] a metric space and [imath]\mu[/imath] an outer measure on [imath](X,d)[/imath] s.t. every borel set are [imath]\mu-[/imath]Caratheodory measurable. Show that [imath]\mu[/imath] is a meatric outer measure on [imath](X,d)[/imath]. It mean that I have to prove that [imath]\mu^*(A\cup B)=\mu^*(A)+\mu^*(B)[/imath] if [imath]d(A,B)=\inf\{d(x,y)\mid x\in A,y\in B\}>0[/imath], but I don't have any idea on how prove it. The fact that [imath]\mu^*(A\cup B)\leq \mu^*(A)+\mu^*(B)[/imath] is obvious. But since I don't have a precise definition of [imath]\mu^*[/imath] (like [imath]\mu^*(E)=\inf\{...\}[/imath]) I don't know how to proceed... Any help would be appreciated. | 2370244 | If every open set of a metric space [imath](X, \rho)[/imath] is [imath]\mu^*[/imath]-measurable, then [imath]\mu^*[/imath] is a metric outer measure.
Let [imath](X, \rho)[/imath] be a metric space and [imath]\mu^*[/imath] and outer measure on [imath]X[/imath]. Then, if every open subset of [imath]X[/imath] is [imath]\mu^*[/imath]-measurable, [imath]\mu^*[/imath] is a metric outer measure. I'm stuck with this problem. I'm supposed to prove that for every [imath]A[/imath], [imath]B[/imath] subsets of [imath]X[/imath] with [imath]d(A, B) = \inf\{d(a,b) \mid a \in A, b \in B \} > 0[/imath], [imath]\mu^*(A \cup B) = \mu^*(A) + \mu^*(B)[/imath] and the hypothesis says that if [imath]U \subset X[/imath] is open, then [imath]\mu^*(E) = \mu^*(E\cap U) + \mu^*(E\cap U^c)[/imath] for every set [imath]E[/imath] but I don't know how to do it. I thought of letting [imath]E = A \cup B[/imath] and then try to find some choice of [imath]U[/imath]. With that choice of [imath]E[/imath], we have that [imath] \begin{align*} \mu^*(A\cup B) &= \mu^*((A\cup B) \cap U)+\mu^*((A\cup B)\cap A^c)\\ & =\mu^*((A\cap U) \cup (B \cap U))+ \mu^*((a \cap U^c)\cap (B\cap U^c)) \end{align*}[/imath] So at first sight, it looks like the open set [imath]U[/imath] we seek for has to be of one of the following two forms, [imath]U \subset A[/imath] and [imath]B \subset U^c[/imath] or the inverse, [imath]U \subset B[/imath] and [imath]A \subset U^c[/imath]. The problem is that, even though the existence of an open set [imath]U \subset A[/imath] is always guaranteed, I don't know why it should also satisfy the property that [imath]B \subset U^c[/imath] as well. My guess is that there is some property of metric spaces that I'm not aware off (or maybe I am really off the track here). Any help would be appreciated. Thank you! |
2984699 | Setting up the triple integrals for a solid given by [imath]y + x = 4[/imath] and [imath]16 - 4z^2 = y^2[/imath]?
I'm trying to set up all six triple integrals to find the volume of the solid that lies in the first octant bounded by the coordinate planes, the plane [imath]y + x = 4[/imath], and the cylinder [imath]16 - 4z^2 = y^2[/imath]. [imath]3[/imath]D graph I've been able to set up the triple integrals for every other combination except for [imath]dydxdz[/imath] and [imath]dydzdx[/imath], which I am struggling to find the bounds for. I know the volume is [imath]8\pi - 32/3[/imath], but I keep getting an incorrect answer when I set up and calculate dydxdz and dydzdx. Here's my equation for the projection onto the [imath]xz[/imath]-plane, but I'm not sure it is correct/complete: [imath]xz[/imath]-plane projection? I'd appreciate any help regarding how to find the bounds of dydxdz and dydzdx. | 2984584 | Setting up the triple integrals for a solid given by [imath]y+z=2[/imath] and [imath]x=4-y^2[/imath]?
I'm trying to set up all six triple integrals to find the volume of the solid that lies in the first octant bounded by the coordinate planes, the plane [imath]y+z=2[/imath], and the cylinder [imath]x=4-y^2[/imath]. [imath]3[/imath]D-Graph: I've been able to set up the triple integrals for every other combination except for [imath]dydxdz[/imath] and [imath]dydzdx[/imath], which I am struggling to find the bounds for. I know the volume of the solid is [imath]20/3[/imath], but no matter what I try I haven't been able to come up with [imath]dydxdz[/imath] and [imath]dydzdx[/imath] to produce that result. Looking at the [imath]3[/imath]D graph, the projection onto the [imath]xz[/imath]-plane appears to be the equation [imath]z=2-(4-x)^{1/2}[/imath] and [imath]x=4[/imath] (see image below), but I don't think that is correct since it isn't producing the correct volume. [imath]xz[/imath]-plane projection? If that isn't correct, what does the projection look like? I'd appreciate any help. Thank you! |
2984835 | If X is an exponentially distributed variable with lambda = 1, how can we find the distribution of Y~ Log(X)?
My first thought was to take the CDF of X, which we know is [imath]1-e^{-x}[/imath] for values of [imath]x > 0[/imath] and [imath]0[/imath] for values of [imath]x[/imath] less than[imath] 0[/imath], and then solve the [imath]\log(x)[/imath] for those values, however I am not sure that this is correct. Any thoughts? I also attempted to graph [imath]\log(1-e^{-x})[/imath] and of course this gave me a straight line. Not sure where to go from here. Any help would be appreciated. Thanks! | 1579261 | Let [imath]X[/imath] have the exponential distribution with [imath]\lambda=2[/imath] Find the density function of the random variable [imath]Y=\ln(X)[/imath]
I have an answer but I'm not sure if I'm doing it right: [imath]f(x)[/imath] is the pdf [imath]f(x)=2e^{-2x}[/imath], then, if the random variable [imath]Y=\ln(X)[/imath], then we simply substitute: [imath]P(Y \leq y)[/imath]= [imath]P(2e^{-2X} \leq y)[/imath] [imath]P(-4X \leq y)[/imath] since [imath]y=ln(x)[/imath] [imath]P(X \leq - \dfrac {1} {4} y)[/imath] This is the distribution function, right? Thanks! |
2985209 | Show that there exists a group of order [imath]p^{3}[/imath], with p prime.
I am a bachelor student in mathematics and studying group theory at home I came across this exercise where there is need to show the existence of this group, because given any prime, is it possible to ensure that there is a group of order [imath]p^{3}[/imath] ? | 516505 | Prove that for each prime [imath]p[/imath] there exists a nonabelian group of order [imath]p^3[/imath]
Can anyone helo me out with this question: "Prove that for each prime there exists a nonabelian group of order [imath]p^3[/imath]". This is my attempt: I know how to show that [imath]|Z(G)|= p[/imath]. So, since [imath]C(x)[/imath] contains [imath]Z(G)[/imath] for every [imath]x[/imath] not in [imath]G[/imath] then [imath]|C(x)|> |Z(G)|[/imath] i.e. [imath]|C(x)|> p[/imath]. But [imath]C(x)[/imath] is always a subgroup of [imath]G[/imath] so it’s order must divide the order of [imath]G[/imath] and so [imath]|C(x)= p^2[/imath]. (If [imath]|C(x)|= p^3[/imath] then [imath]|G:C(x)|= |G|/|C(x)| = p^3/p^3 = 1[/imath], so the number of elements in every conjugacy class would be 1, which would imply that every element is in the center of [imath]G[/imath], which in turn would mean that [imath]G[/imath] is abelian, which is not what we're after.) If [imath]|C(x)|= p^2[/imath] then [imath]|G:C(x)|= |G|/|C(x)| = p^3/p^2= p[/imath]. Here’s where I don’t know how to go forward. I know that if there would be [imath]p^2 – 1[/imath] conjugacy classes then [imath]|G |= |Z(G)|+ (p^2-1) \cdot|G:C(x)|= p + (p^2-1)p = p + p^3 – p = p^3[/imath]. But how do I know that there actually exists a group with exactly [imath]p^2 – 1[/imath] conjugacy classes? Greatful for input on my reasoning and also other ways of proving this, including how to construct such a group. |
2985437 | How to show [imath]\ln{(\frac{x+1}{x})}\geq \frac{1}{x+1}[/imath]?
I used mathematica to check the values of [imath]\ln{(\frac{x+1}{x})}- \frac{1}{x+1}[/imath] are positive for any [imath]x\geq 0[/imath]. How do I analytically show the statement is true for all [imath]x\geq 0[/imath]? Any idea? | 1592906 | Prove that [imath]\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0[/imath]
Prove that [imath]\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0[/imath] First inequality using MVT: [imath]\frac{1}{a+1}<\ln \frac{a+1}{a}:[/imath] [imath]f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}[/imath] [imath]f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}>f(1)\Rightarrow f(a)>f(1)[/imath] [imath]\frac{1}{a+1}-\ln \frac{a+1}{a}-\frac{1-2\ln 2}{2}>0[/imath] This is not the starting inequality. Is there something wrong in this method? |
2985673 | Find the inverse of [imath]I_n-uv^T[/imath]
There are two [imath]n[/imath]-components column vectors [imath]u[/imath] and [imath]v[/imath]. [imath]I_n[/imath] is the [imath]n\times n[/imath] identity matrix. Find the inverse of [imath]I_n – uv^T[/imath]. | 2300320 | In a proof of the Sherman-Morrison formula why does [imath](I+wv^T)^{-1}=I-\frac{wv^T}{1+v^Tw}[/imath]
I'm trying to understand the alternate proof of the formula [imath](A+uv^T)^{-1} = A^{-1} - {A^{-1}uv^T A^{-1} \over 1 + v^T A^{-1}u}[/imath] found on wikipedia here. The closest I can find on stack exchange is the use of the identity but not its proof. Does someone know how [imath](I+wv^T)^{-1}=I-\frac{wv^T}{1+v^Tw} \tag{1}[/imath] is easily proven? I don't see it. You can mark the question as duplicate if you want but all the proofs are proving the Sherman-Morrison formula, not the intermediate identity given in (1). The two are similar as I see now that someone proved it, but it still is worth showing explicitely how to prove the identity. |
2985771 | How can I derive the function?
My function is [imath]f(x)=\int_{0.5}^{\sin(x)} e^{\arcsin(t)} dt[/imath] for [imath]x∈(−π/2,π/2)[/imath]. I'm stuck whether the derivative is [imath]f'(x) = e^{\arcsin(x)}[/imath] or [imath]f'(x) = e^{\arcsin(\sin(x))}[/imath]? | 2981581 | How can I find the domain and derivative of the function?
How can I find the domain of the following function [imath]D(f)[/imath] [imath]\int_{1/2}^{\sin(x)}e^{\arcsin t} dt[/imath] As well as the derivative for [imath] x \in (\frac{-\pi}{2}, \frac{\pi}{2})[/imath] |
2985961 | Show a group cannot have order [imath]2n[/imath] for some odd [imath]n\gt 1[/imath] and be simple (without Cauchy's Theorem)
We were asked to prove the following today: Let [imath]G[/imath] be a group. Prove that if [imath]G[/imath] has order [imath]2n[/imath] for some odd integer [imath]n[/imath] greater than [imath]1[/imath], then [imath]G[/imath] contains a proper non-trivial normal subgroup and cannot be a simple group. We were told that Cayley's Theorem would be very beneficial to the proof but all I managed to get was to say that with a group [imath]G[/imath], we have a map [imath]\varphi[/imath] such that [imath]\varphi (g) = T_g [/imath] where [imath]T_g[/imath] is simply a function [imath]G \to G[/imath] where [imath]T_g(x) = gx \ \ \ \forall x \in G[/imath] Next I am told we need to show that [imath]G[/imath] definitely has an element of some order. Naturally I wanted to used LaGrange for this but I was told this isn't the correct approach. Searching for help after this I see that Cauchy's Theorem is used but we haven't covered that in class so we can't use that. Is there a way that I can continue from here using only Cayley's Theorem? As for a bit of reference, the exact proof of Cayley's Theorem we went over is "Every group is isomorphic to a group of permutations" After proving we were supposed to use it to show that a group of order [imath]216[/imath] cannot be simple. Assuming the theorem is true, I still don't know how this can be, seeing that [imath]216 = 2*108[/imath]. [imath]108[/imath] is an even number so does the theorem we are supposed to prove even play a role in this case? This has been bugging me all morning and I'm not getting any further along looking at it by myself. So any help would greatly be appreciated. | 1270724 | Group of [imath]2n[/imath] elements, [imath]n[/imath] odd, is not simple
Problem Let [imath]n \geq 3[/imath] be an odd number and let [imath]G=\{1,...,2n\}[/imath] be a group of order [imath]2n[/imath]. Let [imath]\phi:G \to S_{2n}[/imath] be the morphism defined by [imath]\phi(g_i)(g_j)=g_ig_j[/imath] and let [imath]H=\phi^{-1}(A_{2n})[/imath]. Show that [imath]H[/imath] is a proper normal subgroup of [imath]G[/imath]. So basically the result of this problem is that every group of order [imath]2n[/imath], with [imath]n[/imath] odd, is not simple. The morphism [imath]\phi[/imath] can be seen as the action of [imath]G[/imath] on itself by left multiplication. Since [imath]A_{2n} \lhd S_{2n}[/imath], then [imath]H=\phi^{-1}(A_{2n}) \lhd G[/imath]. I am stuck at showing that [imath]H \neq \{1\},G[/imath]. All I could think of is that [imath]G[/imath] has an element [imath]g[/imath] of order [imath]2[/imath], then [imath]\phi(g)[/imath] is of order [imath]1[/imath] or [imath]2[/imath], but it is easy to see that [imath]\phi(g)=1[/imath] if and only if [imath]g=1[/imath], so [imath]ord(\phi(g))=2[/imath]. I don't know which are the elements of order two in [imath]S_{2n}[/imath]. I would appreciate some help to solve the problem, thanks in advance. |
2986230 | Prove [imath]{a+b+c\over abc}\leq {1\over a^2}+{1\over b^2}+{1\over c^2}[/imath]
I need to prove that [imath]\frac{a+b+c}{abc} \le \frac1{a^2}+\frac1{b^2}+\frac1{c^2}[/imath] I started with this [imath]\frac 1b\frac1c+\frac1a\frac1c+\frac1a\frac1b\le\frac1a\frac1a+\frac1b\frac1b+\frac1c\frac1c[/imath] then I stick here what can I do? Thank you | 1840148 | Prove [imath]\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}[/imath].
So I have to prove [imath] \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.[/imath] I rearranged it [imath] a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .[/imath] My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas? Thanks |
2985700 | Finding Global Inverse of [imath] f(x,y,z)= \left( \frac{x}{a+x+y+z}, \frac{y}{a+x+y+z}, \frac{z}{a+x+y+z} \right).[/imath]
Let [imath]a\in \mathbb{R}[/imath], [imath]a\neq 0[/imath], and [imath]E=\{(x,y,z)\in \mathbb{R}^3:a+x+y+z\neq 0\}[/imath] and [imath]f:E\rightarrow \mathbb{R}^3[/imath] defined by [imath] f(x,y,z)= \left( \frac{x}{a+x+y+z}, \frac{y}{a+x+y+z}, \frac{z}{a+x+y+z} \right).[/imath] Compute the inverse of [imath]f[/imath]. I know I can use the Inverse Function Theorem to show local invertibility, however, that does not necessarily imply global invertibility. Thus, even though I can invert the Jacobian of the function, it does not follow that I can recover the inverse function from its Jacobian. Therefore, I have no idea what to try for this problem. Could you please provided a small hint on something to try for this problem? Thank you in advance. | 2985671 | Proving [imath]f(x,y,z)=\big(\frac{x}{a+x+y+z}, \frac{y}{a+x+y+z}, \frac{z}{a+x+y+z} \big)[/imath] is injective
Let [imath]a\in \mathbb{R}[/imath], [imath]a\neq 0[/imath], and [imath]E=\{(x,y,z)\in \mathbb{R}^3:a+x+y+z\neq 0\}[/imath] and [imath]f:E\rightarrow \mathbb{R}^3[/imath] defined by [imath] f(x,y,z)=\big(\frac{x}{a+x+y+z}, \frac{y}{a+x+y+z}, \frac{z}{a+x+y+z} \big) [/imath] Show that [imath]f[/imath] is injective. I've been trying to prove it directly by showing that if [imath]f(x_1,y_1,z_1)=f(x_2,y_2,z_2)\Rightarrow[/imath] [imath]x_1=x_2, y_1=y_2,z_1=z_2[/imath]. But I get a system of equations which I'm find difficult to deal with. Any help would be greatly appreciated. |
2986333 | Are there graphs for which [imath]G \cong Cay(Aut(G), S) [/imath]?
I'm looking for graph [imath]G[/imath], [imath]V(G) >2[/imath], such that[imath]G \cong Cay(Aut(G), S) [/imath], where [imath]Cay[/imath] is the Cayley graph (considered as undirected graph) and [imath]S[/imath] is minimal generator of [imath]G[/imath]. I've written a GAP code that looped through [imath]\Gamma[/imath] groups, and formed [imath]G'=Cay(\Gamma)[/imath] and [imath]G''=Cay(Aut(G')) [/imath]. After it checked if [imath]G' \cong G''[/imath] is true. It happened to be false for the first 1000 groups, except when [imath]\Gamma[/imath] is the trivial group and when it is isomorphic to [imath]C_2[/imath]. Now, I am not even sure what generator set was used by GAP. Is it possible that for other generator set it would work? Maybe if it isn't minimal? Thanks in advance! | 1098115 | When is the automorphism group of the Cayley graph of [imath]G[/imath] just [imath]G[/imath]?
Let [imath]G[/imath] be a finite group and [imath]S[/imath] a generating set of [imath]G[/imath]. We can draw the Cayley graph [imath]C(G,S)[/imath] by putting each element of [imath]G[/imath] as a vertex, and drawing an edge between two elements [imath]g[/imath], [imath]h\in G[/imath] iff [imath]g^{-1} h \in S[/imath]. Choose [imath]x\in G[/imath]. Note that the map [imath]g \mapsto xg[/imath] is a graph automorphism of [imath]C(G,S)[/imath]. This allows us to embed [imath]G \le \text{Aut}(C(G,S))[/imath]. My question is: when is it in fact the case that [imath]G = \text{Aut}(C(G,S))[/imath]? Is there a nice criterion to determine this? |
2986097 | For any positive integer [imath]n[/imath], prove the following inequality.
Prove that for any positive integer n, the following inequality is true. [imath]\left(1+ \frac{1}{n}\right)^n < \left(1+\frac{1}{n+1}\right)^{n+1}[/imath] Attempt Not a good attempt but this is my thinking It will not be a problem when checking whether this is true for smaller integers. Only for larger integers, it will be difficult to prove. So, I took the equation LHS-RHS < [imath]0[/imath]. I took limit for n tending to infinity and then applied L'Hospital Rule( though the equation was difficult to handle) & therefore could not do anything further. Edit: I, now, want to know about @Rebellos' last equation and his answer (though at first my intention was to know about the question). Please see the comments of @Rebellos ' answer. Adding graph for @Rebellos 's last equation | 508967 | show that if [imath]n\geq1[/imath], [imath](1+{1\over n})^n<(1+{1\over n+1})^{n+1}[/imath]
I have derived the inequality if [imath]k>1[/imath], [imath]{n(n-1)⋯(n-k+1)\over k!} ({1\over n})^k<{(n+1)n⋯(n-k+2)\over k!} ({1\over n+1})^k[/imath] But, my problem is how to use this inequality to prove that if [imath]n\geq1[/imath], [imath](1+{1\over n})^n<(1+{1\over n+1})^{n+1}[/imath] |
2986690 | Minimize the following term under the given constraint
If [imath]f(x)[/imath] is a real valued continuous and differentiable function in [imath][0, 1][/imath] and [imath]\int_{1/3}^{2/3}f(x)dx = 0[/imath] minimize [imath]\frac{\int_{0}^{1}f'(x)^2 dx}{(\int_{0}^{1}f(x)dx)^2}[/imath] | 2985972 | Integral inequality with a strange condition
Let [imath]f[/imath] be a continuously differentiable real valued function on [imath][0,1][/imath]. It is given that [imath]\displaystyle \int_{\frac{1}{3}}^{\frac{2}{3}}f(x) dx=0[/imath] Find the minimum value of [imath]\dfrac{\int_{0}^{1} (f'(x))^2 dx}{\left( \int_{0}^{1} f(x) dx \right)^2}[/imath] I tried to use Cauchy-Schwarz to show that [imath]\frac{\int_{0}^{1} (f'(x))^2 dx}{\left( \int_0^1 f(x) dx \right)^2} \ge \frac{\left( \int_0^1 \bigl| f(x)f'(x) \bigr| dx \right)^2}{ \left( \int_0^1 f(x) dx \right)^2} \ge \frac{ f(1)^2 - f(0)^2}{2 \left( \int_0^1 f^2 (x) dx \right)^2}[/imath] But I can't proceed from here. Also, I don't know how to use the condition [imath]\int_{1/3}^{2/3}f(x) dx=0[/imath] |
2655363 | Prime number theorem for finite fields
This comes from Lang's algebra, chapter [imath]5[/imath] exercise [imath]23[/imath] [imath](b)[/imath]. Part [imath](a)[/imath] asks to find a rational function to represent [imath]Z(t) := (1-t)^{-1} \prod_p (1-t^{\deg p} )^{-1}[/imath] Where [imath]p[/imath] ranges through all monic irreducible polynomials of a finite field of cardinality [imath]q[/imath] (which I will denote [imath]F_q[/imath]). I can solve this - you find [imath]Z(t) = \frac{1}{1-qt}[/imath] [imath]\textbf{Main question:}[/imath] Part [imath](b)[/imath] asks the following: Let [imath]\pi_q (m)[/imath] denote the number of monic irreducibles with degree [imath]\leq m[/imath]. Prove [imath]\pi_q (m) \sim \frac{q}{q-1} \cdot \frac{q^m}{m}[/imath] I assume that it is worthwhile to use the result of part [imath](a)[/imath]; use the notation [imath]\psi (d)[/imath] to denote the number of monic irreducibles of degree [imath]d[/imath] so that [imath]\pi_q (m) = \sum_{d=1}^m \psi(d)[/imath]. If I take the natural log of both sides I see [imath]\log (1-qt) = \sum_{d=1}^\infty \psi(d) \log(1-t^d)[/imath] I did this because I get pretty close to having the sum of my [imath]\psi (d)[/imath] terms occur, but if I continue doing standard power series manipulations I essentially do the solution for part [imath](a)[/imath] backwards and get nowhere. Another attempt is to "brute force" it. I can look at [imath]\sum_{d=1}^m \psi (d)[/imath], and using the standard formula for [imath]d \psi (d) = \sum_{d \ | \ n} \mu (d) q^{n/d}[/imath], I can manipulate my sum to get it into the form [imath]\pi_q (m) = \sum_{d=1}^m \frac{\mu (d)}{d} \sum_{k=1}^{[m/d]} \frac{q^k}{k}[/imath] where [imath][\cdot][/imath] denotes floor function. I confess that I am not terribly familiar with asymptotic properties of the Mobius function, so that this form is not particularly enlightening to me, either. Is there any other way to see this relation more clearly? Any help is appreciated. | 2927385 | The prime number theorem over a finite field - Lang's *Algebra*, Chapter V, Exercise 23(b)
This is Exercise 23(b) of Chapter V (Algebraic Extensions) from Lang's Algebra. Let [imath]k[/imath] be finite field with [imath]q[/imath] elements, and let [imath]\pi_q(n)[/imath] be the number of monic irreducible polynomials [imath]p \in k[X][/imath] of degree [imath]\leq n[/imath]. Prove that [imath] \pi_q(m) \sim \frac{q}{q-1} \frac{q^m}{m} \quad \text{for} \quad m \to \infty. [/imath] I have tried a few things but I'm not making any progress at all. The hint given in class was to take the logarithmic derivative of the zeta function that was defined in part (a) of the same problem. We defined the zeta function to be [imath] Z(t) = (1-t)^{-1} \prod_p (1-t^{\deg p})^{-1}. [/imath] I computed this to be equal to the rational function [imath] (1-t)^{-1}(1-qt)^{-1} [/imath] on the region [imath]|t| < q^{-1}[/imath]. Taking the logarithmic derivative of [imath]Z(t)[/imath], I get [imath] \frac{Z'(t)}{Z(t)} = \frac{1}{1-t} + \frac{q}{1-qt} = \frac{1+q-2qt}{(1-t)(1-qt)} = (1+q-2qt)Z(t). [/imath] I am not getting any further ideas at this point in how to use this to describe [imath]\pi_q(n)[/imath]. From Exercise 22 I know that if [imath]\psi(d)[/imath] denotes the number of monic irreducible polynomials of degree [imath]d[/imath], then the total number of polynomials of degree [imath]n[/imath], which is [imath]q^n[/imath], can be expressed as [imath] q^n = \sum_{d \mid n} d \psi(d). [/imath] Using the Möbius inversion formula, I can deduce that [imath] n\psi(n) = \sum_{d \mid n} \mu(d) q^{n/d}, [/imath] where [imath]\mu[/imath] is the Möbius function. Since [imath]\pi_q(m) = \sum_{k=1}^m \psi(k)[/imath], I can use the above equation to write [imath] \pi_q(m) = \sum_{k=1}^m \frac{1}{k} \sum_{d \mid k} \mu(d) q^{k/d}. [/imath] My intuition is that the highest power of [imath]q[/imath] will dominate the sum, so the RHS is approximately [imath] \frac{q^m}{m}, [/imath] so I get roughly what I'm asked to show in the problem. I'm having trouble refining my ideas any further. I have also looked at this question, which asks the same problem, but no answers or comments have been posted there. Lang remarks after stating the problem, "This is the analogue of the prime number theorem in number theory, but it is essentially trivial in the present case, because the Riemann hypothesis is trivially verified." I have tried looking at the proof of the prime number theorem on the internet, but I haven't really understood it; and I certainly can't see how to deduce it from the Riemann hypothesis even in this case. Any help in solving this problem would be appreciated. |
2986647 | Find angle of line along a circle's perimeter?
If I know coordinates of point [imath]A[/imath], coordinates of circle center [imath]B[/imath] and [imath]r[/imath] is the radius of the circle, is it possible to calculate the angle of the lines that are passing through point A that are also tangent to the circle? [imath]A[/imath] is the green point, [imath]B[/imath] is the center of the red circle and I am trying to find out the angle of the blue lines. | 2985042 | Find a line along a circle's perimeter?
If I know coordinates of point [imath]A[/imath], coordinates of circle center [imath]B[/imath] and r as the radius of the circle, is it possible to calculate lines that are passing through point [imath]A[/imath] that are also tangent to the circle? A is the green point, B is the center of the red circle and I am trying to find out the blue lines. |
2987010 | If [imath]G/Z(G)[/imath] is nilpotent then [imath]G[/imath] is nilpotent
Notation : [imath]Z_{i+1}(G)=[/imath]{[imath]g\in G : [g,x]\in Z_i(G) \forall x\in G[/imath]} where [imath]Z_0(G)=1[/imath]. Definition : [imath]G[/imath] is said to be nilpotent if [imath]Z_n(G)=G[/imath] for some [imath]n\in \Bbb N[/imath]. Let [imath]Z_n(G/Z(G))=G/Z(G)[/imath]. I have that [imath]Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G))[/imath]. Then [imath]\frac{Z_{i+1}(G)/Z(G)} {Z_i(G)/Z(G)}≈Z(\frac{G/Z(G)}{Z_i(G)/Z(G)})[/imath]. Now the L.H.S is a subgroup of [imath]\frac{G/Z(G)}{Z_i(G)/Z(G)}[/imath] and the R.H.S is a characteristic subgroup of [imath]\frac{G/Z(G)}{Z_i(G)/Z(G)}[/imath]. Since L.H.S is isomorphic to R.H.S, I conclude that L.H.S.=R.H.S. At this stage I am stuck. What can I conclude from this? | 626423 | If [imath]G / Z(G)[/imath] nilpotent then G is nilpotent.
Let [imath]$G$[/imath] be a group. If [imath]G / Z(G)[/imath] nilpotent, then prove that [imath]$G$[/imath] is nilpotent. The way I have worked out this question is fairly tedious, and I would be interested if there is a more straightforward approach here. I solved this problem by showing that if [imath]$G / Z(G)$[/imath] is of nilpotence class [imath]k[/imath], then [imath]G / Z(G) = Z_{k}(G / Z(G)) = Z_{k+1}(G) / Z(G)[/imath], the latter equality shown using a tedious induction proof. Then the conclusion readily follows. Is there a different approach to this problem, or is there a short way to prove that [imath]Z_{k}(G / Z(G)) = Z_{k+1}(G) / Z(G)[/imath]? |
2987132 | Cauchy implies monotone + bounded
In my whole problem, I was stuck in one direction, which is to show (i) implies (ii). (i) Every Cauchy sequence in [imath]\mathbb{R}[/imath] converges to a limit in [imath]\mathbb{R}[/imath]. (ii) Every monotone and bounded sequence in [imath]\mathbb{R}[/imath] converges to a limit in [imath]\mathbb{R}[/imath]. Explaining in english, my idea is: Cauhcy is a bounded sequence. Given that (i) is true, i.e: this bounded cauchy sequence is convergent. Then, since not every bounded sequence is convergent, it requires some conditions so that it is convergent. One of a condition is "monotone". So, I try to use this to show (i) implies (ii). But my TA think "this is saying nothing". If I would like to use some maths to explain, like [imath]\epsilon[/imath], a sequence [imath]\{x_n\}[/imath] etc.... How could I use it?? | 566635 | Show that every monotonic increasing and bounded sequence is Cauchy.
The title is kind of misleading because the task actually to show Every monotonic increasing and bounded sequence [imath](x_n)_{n\in\mathbb{N}}[/imath] is Cauchy without knowing that: Every bounded non-empty set of real numbers has a least upper bound. (Supremum/Completeness Axiom) A sequence converges if and only if it is Cauchy. (Cauchy Criterion) Every monotonic increasing/decreasing, bounded and real sequence converges to the supremum/infimum of the codomain (not sure if this is the right word). However, what is allowed to use listed as well: A sequence is called covergent, if for [imath]\forall\varepsilon>0\,\,\exists N\in\mathbb{N}[/imath] so that [imath]|\,a_n - a\,| < \varepsilon[/imath] for [imath]\forall n>N[/imath]. (Definition of Convergence) A sequence [imath](a'_k)_{k≥1}[/imath] is called a subsequence of a sequence [imath](a_n)_{n≥1}[/imath], if there is a monotonic increasing sequence [imath](n_k)_{k≥1}\in\mathbb{N}[/imath] so that [imath]a'_{k} = a_{n_{k}}[/imath] for [imath]\forall k≥1[/imath]. (Definition of a Subsequence) A sequence [imath](a_n)_{n≥1}[/imath] is Cauchy, if for [imath]\forall\varepsilon>0\,\,\exists N=N(\varepsilon)\in\mathbb{N}[/imath] so that [imath]|\,a_m - a_n\,| < \varepsilon[/imath] for [imath]\forall m,n>N[/imath]. (Definition of a Cauchy Sequence) (Hint) The sequence [imath](\varepsilon\cdot\ell)_{\ell\in\mathbb{N}}[/imath] is unbounded for [imath]\varepsilon>0[/imath]. (Archimedes Principle) Would appreciate any help. |
2986692 | Foci Product of Distances Ellipse
Show that the product of the distances from the foci to the tangent at any point of an ellipse is [imath]b^2[/imath] where [imath]b[/imath] is the semi-minor axis. My process of proof so far: Let [imath]F(c, 0)[/imath] and [imath]F'(-c, 0)[/imath] be 2 points that lie on the x-axis of the plane and [imath]P(x_0, y_0)[/imath] be the point that lies on the tangent to the ellipse. I simply started by simplifying the equation [imath]\sqrt{(x+c)^2+y^2}\sqrt{(x-c)^2+y^2}[/imath] However, this got me to a dead end. Plus, I feel as if there's some property about the tangent of an ellipse that I'm not using. If anyone could guide (not answer) me towards the answer, that would be extremely beneficial to my sanity (and knowledge, of course). Thanks again. | 1073870 | For an ellipse with minor radius [imath]b[/imath], show that the product of distances from the foci to any tangent line is [imath]b^2[/imath]
Consider the ellipse with equation: [imath]\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1[/imath] How do we prove the product of distances from the foci to any tangent line is [imath]b^2[/imath]? [imath]F_1A \cdot F_2B = b^2[/imath] |
2985798 | How would I show that there is no holomorphic function [imath]f(z)[/imath] on [imath]C\setminus\{0\}[/imath] such that [imath]\exp f(z) = z[/imath]?
I'm not sure how to show it's not holomorphic. I can find a general equation for [imath]f(z)[/imath] but can only seem to show it's not one to one, and am not sure how to show it is not holomorphic. | 2098036 | No holomorphic logarithmn
There doesnt exist a holomorphic [imath]f: \mathbb C\setminus0 \rightarrow \mathbb C\setminus 0 [/imath], so that [imath]e^{f(z)} = z [/imath] for all [imath]z \in \mathbb C \setminus 0 [/imath] I tried to work with the derivative of f(z) but i didnt manage to get anything out of it. |
2986865 | Finding example of a convergent series [imath]\sum a_n [/imath] such that [imath]\sum a_n^2 [/imath] does not converge
I'm feeling that examples for this would arise from some alternating series, in particular of the form [imath]\sum (-1)^nx_n[/imath]. But I'm having trouble finding a concrete example. Thanks! | 2984967 | [imath]\sum a_{n}[/imath] converges but [imath]\sum a_{n}^2 [/imath] diverges?
I have to give an example of a convergent series [imath]\sum a_{n}[/imath] for which [imath]\sum a_{n}^2 [/imath] diverges. I think that such a series cannot exist because if [imath]\sum a_{n}[/imath] converges absolutely then [imath]\sum a_{n}^2 [/imath] will always converge right? |
2988197 | Inverse Trigonometric Functions and Summation of Series
I am aware that these types of questions are dealt with by the properties of [imath]\tan^{-1} x_r-\tan^{-1} x_{r-1}[/imath] but here I am not able to proceed by that approach...PLEASE HELP! [imath] \sum_{r=1}^\infty\tan^{-1}\left(\frac{2r}{1-r^2+r^4}\right) [/imath] | 339486 | finding [imath]\lim_{n\rightarrow\infty}\sum_{r=1}^{n}\arctan\left(\frac{2r}{1-r^2+r^4}\right)[/imath]
I want to find [imath]L=\lim_{n\rightarrow\infty}\sum_{r=1}^{n}\arctan\left(\frac{2r}{1-r^2+r^4}\right)[/imath] I already know that I need to split the expression [imath]\frac{2r}{1-r^2+r^4}[/imath] of the form [imath]\frac{a_r+a_{r-1}}{1-a_ra_{r-1}}[/imath] or [imath]\frac{a_r-a_{r-1}}{1+a_ra_{r-1}}[/imath] so that the main expression gives [imath]\lim_{n\rightarrow\infty}\sum_{r=1}^{n}\arctan\left(\frac{a_r-a_{r-1}}{1+a_ra_{r-1}}\right)=\lim_{n\rightarrow\infty}\sum_{r=1}^{n}(\arctan(a_r)-\arctan(a_{r-1}))=\arctan(a_\infty)-\arctan(a_0)[/imath] But I am not able to split up that thing. Any hints on how to start? Thanks in advance. |
2985452 | If [imath]p \equiv 3 \ (\text{mod} \ 4)[/imath] is a prime, show [imath](\frac{p-1}{2})! \equiv (-1)^{t} \ (\text{mod} \ p)[/imath]
If [imath]p \equiv 3 \ (\text{mod} \ 4)[/imath] is a prime, show [imath](\frac{p-1}{2})! \equiv (-1)^{t} \ (\text{mod} \ p),[/imath] where [imath]t[/imath] is number of positive integers less than [imath]\frac{p}{2}[/imath] that are nonquadratic residues of [imath]p.[/imath] My attempt: Since [imath]p-1 \equiv -1, \ p-2 \equiv -2, ..., p-(\frac{p-1}{2}) \equiv \frac{p-1}{2} \ (\text{mod} \ p)[/imath] and [imath]1 \equiv 1, \ 2\equiv 2,..., \frac{p-1}{2} \equiv \frac{p-1}{2} \ (\text{mod} \ p),[/imath] it follows that [imath]((\frac{p-1}{2})!)^2 \equiv-(p-1)! \equiv 1,[/imath] by Wilson's theorem. I am not sure how to continue from here. How do I relate it to [imath]t?[/imath] Appreciate if anyone could advise. Thank you. | 1339292 | [imath]r! \equiv (−1)^k \pmod p[/imath]
Suppose that p ≡ 3 (mod 4) and [imath]r = \frac {p-1}2[/imath] Show that [imath]r! \equiv (−1)^k \pmod p[/imath] where k is the number of non-quadratic residues modulo p which are smaller than [imath]\frac p2[/imath] I know from Wilson's Theorem for a prime p: [imath](p−1)! \equiv −1 \pmod p[/imath] I don't know where to start from .. I managed to show that [imath]r! \equiv \pm 1 \pmod p[/imath] any help or hint will be appreciated |
2985322 | Prove [imath]\mid[/imath]x-1[imath]\mid[/imath]+[imath]\mid[/imath]x+5[imath]\mid[/imath] [imath]\ge[/imath] 6 for all real numbers
I need to prove it with proof by cases. I graphed it out and it seems that I need to put into 2 cases which is when x[imath]>[/imath]1 and x[imath]>[/imath]-5. But this where my problem is, how do plug that back into the statement? Would it just be calculating (x-1)+(x+5)? | 2981975 | Proving Inequalities with Absolute values
Say the following expression is to be proven for all real [imath]x[/imath]: [imath]|x-1| + |x+1| \geq 2[/imath] Is it sufficient to just look at the cases where [imath]|x-1|[/imath] and [imath]|x+1|[/imath] are minimum? Or is it necessary to show that the expression it true using other algrebraic manipulations? |
1835360 | the uniform convergence of the sequence of functions
Let [imath]f_1:[a,b]\rightarrow \mathbb{R}[/imath] be a Riemann integrable function. Define the sequence of functions [imath]f_n:[a,b] \rightarrow \mathbb{R}[/imath] by [imath]f_{n+1}(x)=\int_a^x f_n(t)dt,[/imath] for each [imath]n\ge 1[/imath] and each [imath]x\in [a,b][/imath]. Prove that the sequence of functions [imath]g_n(x)=\sum\limits_{k=1}^n f_k(x)[/imath] converges uniformly on [a,b]. I try to use Ascoli Theorem to prove this problem by considering [imath]|g_n(x)-gn(y)|\le\sum\limits_{k=1}^n |f_k(x)-f_k(y)|[/imath]. However, [imath]|f_k(x)-f_k(y)|\le M |b-a|^{k-2} |x-y|, [/imath] where [imath]M>0[/imath] is a constant achieving by the condition that [imath]f_1[/imath] is an integrable function. However, this estimation depends on [imath]k[/imath], so I cannot apply Ascoli-Azela Theorem. Thank you in advance for your help. | 411184 | [imath]f_{n+1}(x):= \int_a ^x f_n(t)dt[/imath], [imath]\sum_{m=1} ^{\infty} f_m(x)[/imath] is uniformly convergent
Let [imath]f_1 : [a,b] \rightarrow \mathbb{R}[/imath] be an integrable function. Let's define a sequence [imath](f_n)[/imath], [imath] \ \ f_n : [a,b] \rightarrow \mathbb{R}[/imath] as [imath]f_{n+1}(x):= \int_a ^x f_n(t)dt[/imath].' Prove that [imath]\sum_{m=1} ^{\infty} f_m(x)[/imath] is uniformly convergent on [imath][a,b][/imath]. Could you help me with that? Thank you. |
241752 | Prove that a set [imath]S \subseteq \mathbb{R}[/imath] is closed and bounded if and only if every sequence in [imath]S[/imath] has a sub sequence converging to a point of [imath]S[/imath].
[imath](\to)[/imath] Assume that [imath]S \subseteq R[/imath] is bounded. If [imath](x_n)[/imath] is a sequence in [imath]S[/imath] , then [imath](x_n)[/imath] is bounded and thus it has a convergent subsequence. [imath](\leftarrow)[/imath] Assume that [imath]S[/imath] is not bounded. Then for any integer [imath]n[/imath] there is an [imath]x_n[/imath] in [imath]S[/imath] such that [imath]|x_n| > n[/imath]. Since any convergent sequence is bounded, the sequence [imath](x_n)[/imath] cannot have a convergent subsequence. Can someone check to see whether my proof is missing something. thank you. | 2988572 | Let [imath]E[/imath] be a subset of [imath]\mathbb{R}^k[/imath]. Show that [imath]E[/imath] is compact if and only if every sequence in [imath]E[/imath] has a converging subsequence to a point in [imath]E[/imath].
Pretty much just what the title says. So if a set is compact, then it is closed and bounded, correct? Since the set E is bounded, the sequence [imath]\{x_n\}[/imath] is also bounded. Meaning that it has a limit point (I can not remember the theorem stating that). I think that is how I should begin? But I am not sure how to proceed. I know there was a previous post on this, but there was no complete answer. It was essentially just saying what you need to do to prove an if and only if. |
2986824 | Sequence of fractions [imath]c_n = a_n / b_n[/imath] converges to irrational [imath]x[/imath], prove [imath]a_n[/imath] and [imath]b_n[/imath] diverges to infinity.
[imath]a_n[/imath] and [imath]b_n[/imath] are sequences of natural numbers, [imath]\lim_{n->\infty}\frac{a_n}{b_n} = x[/imath], where [imath]x[/imath] is irrational. Prove that [imath]a_n[/imath] and [imath]b_n[/imath] diverge to infinity. I've proved that if a sequence of natural numbers converges to some number, then this sequence has to be constant for big [imath]n[/imath]. | 1465812 | Sequence of rationals with an irrational limit have denominators going to infinity
Let [imath]\alpha[/imath] be an irrational real number and let [imath]a_j[/imath] be a sequence of rational numbers converging to [imath]\alpha[/imath]. Suppose that each [imath]a_j[/imath] is a fraction expressed in lowest terms: [imath]a_j = \alpha_j / \beta_j[/imath]. Prove that the [imath]\beta_j[/imath] tend to [imath]\infty[/imath] Attempt: AFSOC that [imath]\beta_j[/imath] does not tend to infinity, then it is bounded by some [imath]M[/imath]. We can also find an interval [imath]\alpha \in (k, k+1)[/imath]. Let [imath]\epsilon > 0[/imath], [imath]\exists N_0[/imath] such that for all [imath]N > N_0[/imath] we have [imath]|a_N - \alpha| < \epsilon[/imath]. So [imath]|\frac{p_N}{q_N} - \alpha| < \epsilon[/imath]. We try to use the fact that [imath]q[/imath] is bounded but fail to derive a contradiction. I do not know how to use the fact that [imath]\alpha[/imath] is between two integers, although it might not be relevant at all. Ideas? |
2985707 | Second order ODE with initial condition
How can I solve the following second order differential equation [imath] \frac{d^2\theta \:}{dt^2\:}\:=-\frac{g\theta \:\:}{L} [/imath] with initial conditions [imath]θ(0) = θ_0 , v(0) = v_0 [/imath] I know that I can rewrite the equation as [imath] \frac{d^2\theta \:}{dt^2\:}\:+\frac{g\theta \:\:}{L} = 0 [/imath] , find characteristic polynomial and solve the equation. I am just confused on how to solve this problem because of all the constants. Thanks for help in advance | 1089575 | Differential euqation for a pendulum: [imath] {d^2\alpha \over dt^2} + {g \over L} \cdot \alpha = 0 [/imath]
The differential equation [imath] {d^2\alpha \over dt^2} + {g \over L} \cdot \alpha = 0 [/imath] describes a 1-dimensional mathematical pendulum, where [imath]\alpha [/imath] is the angle, [imath] g = 9.82 [/imath], and [imath] L = 0.2 [/imath] is the lentgh of the string. What is the position of the pendulum after 1 second if the velocity at time equals zero is [imath] 0 [/imath] m/s and the angle at time equals zero is [imath] {\pi \over 60} [/imath] radians. To my understanding this is a homogenous equation, ergo [imath] y'' + Cy = 0[/imath] And with the help of the characteristic equation I get the complex roots [imath] Ci \, [/imath] and [imath] -Ci [/imath]. This is a farily simple differential equation, but I have a feeling my calculations so far are wrong (mainly because I didn't expect to see complex roots for this equation). |
2309200 | Semisimple Lie subalgebra closed under JC decomposion
I have a question about the tagged argument in Prop. 8.5.1: Let [imath]g[/imath] be a semisimple Lie subalgebra of [imath]gl(V )[/imath]. For any [imath]x \in g[/imath] we have J-C decomposion [imath] x= x_s +x_n[/imath] but that's not clear to me why [imath] x_s ,x_n \in n_{gl(V)}(g) [/imath] where [imath] n_{gl(V)}(g) = \{y \in gl(V) | [y,x] \in g [/imath] for every [imath] x \in g \} [/imath]. By the way: it would be enough to prove [imath]ad(x_s)(g) ⊆ g, ad(x_n)(g) ⊆ g[/imath], but I don't now how... | 2308807 | Jordan-Chevalley decomposition
Good evening everybody, I have a question about the red tagged argument in Prop. 8.5.1: Let [imath]g[/imath] be a semisimple Lie subalgebra of [imath]gl(V )[/imath]. For [imath]x \in g[/imath] we have J-C decomposion [imath] x= x_s +x_n[/imath] but that's not clear to me why [imath] x_s ,x_n \in n_{gl(V)}(g) [/imath] where [imath] n_{gl(V)}(g) = \{y \in gl(V) | [y,x] \in g [/imath] for every [imath] x \in g \} [/imath]. My idea: We know that [imath] x_s = P(x) ,x_n = S(x) [/imath] for appropriate polynomials without constant terms. Thats clear that [imath]x \in n_{gl(V)} [/imath] so because [imath] g[/imath] is a subspace it would be enough to show that [imath][z,x^n] \in g [/imath] for every [imath]z \in g[/imath]. But how? |
2989241 | Need to check one to one and onto functions
a) Give an example of a function from [imath]\Bbb{Z}^+[/imath] to [imath]\Bbb{Z}^+[/imath] that is neither one-to-one nor onto. b) Let [imath]g:A\to B[/imath] and [imath]f : B \to C[/imath] be functions. Let [imath]f\circ g[/imath] be onto. Are both [imath]f[/imath] and [imath]g[/imath] necessarily onto? I was thinking for a) [imath]f(n) = 1[/imath] I clearly see that this is not one to one, since there is no unique codomain for each input. but im not sure about onto.. Wouldn't all values point to 1 which would make it onto? since the codomain is 1? or would it? im confused if there is no x value in the function. | 2987784 | Let [imath]g : A \to B[/imath] and [imath]f : B \to C[/imath] be functions. Let [imath]f \circ g[/imath] be onto. Are both [imath]f[/imath] and [imath]g[/imath] necessarily onto?
What would the steps be for someone new trying to learn how to solve this? I feel like I understand the concepts and what they mean but not how to prove it. Thanks |
2989189 | What's an example of a finite, non-abelian, non-simple group that is *not* semidirectly reducible?
Say I want to classify all groups of a given order. The abelian case is completely understood by the structure theorem for finitely generated abelian groups. Assume our group is non-abelian, and we somehow managed to find a normal subgroup [imath]N[/imath] (e.g., by considering the core of some Sylow subgroup). It would be pretty nice if we could go on to construct all semidirect products of [imath]N[/imath] with groups of the remaining order – but [imath]G[/imath] need not be semidirectly reducible*. The only counterexample that comes to mind is [imath]0 \to C_2 \to C_4 \to C_2 \to 0[/imath], which does not admit an appropriate section, i.e. a subgroup complementing our normal subgroup with trivial intersection. What are counterexamples in the non-abelian case? | 1504422 | Can every non-simple group [imath]G[/imath] be written as a semidirect product?
When people want to classify a group with certain (small) order, they seem to find a normal subgroup [imath]H[/imath] and a subgroup [imath]K[/imath], and then consider how many distinct [imath]H \rtimes_\varphi K[/imath] are there. My question is: For any group [imath]G[/imath], does it always exist normal subgroup [imath]H[/imath], subgroup [imath]K[/imath] and [imath]\phi : K \rightarrow Aut(H)[/imath] s.t. [imath]G\cong H \rtimes_\varphi K [/imath]? Sorry if this question is too basic... Thanks the comment and the answers. Then if the [imath]G[/imath] is not simple, can we have can [imath]G\cong H\rtimes_\varphi K[/imath] for proper subgroup [imath]H[/imath], [imath]K[/imath], with [imath]H[/imath] normal? EDIT: If we have a group [imath]G[/imath], that is not simple (then [imath]G[/imath] has at least one normal subgroup), are there always a non-trivial normal subgroup [imath]H[/imath] and a proper subgroup [imath]K[/imath] s.t. [imath]G \cong H \rtimes K[/imath]. (Note that if [imath]G[/imath] is simple this is clearly not possible). But the quaternion group and [imath]Z_4[/imath] given in answers have shown that we cannot always have it. |
2986848 | Let [imath]\alpha,\gamma[/imath] be ordinals and [imath]A[/imath] be a set of ordinals. Then [imath]\gamma<\alpha+\sup_{\beta\in A}\beta\implies\exists\beta\in A:\gamma<\alpha+\beta[/imath]
Let [imath]\alpha,\gamma[/imath] be ordinals and [imath]A[/imath] be a set of ordinals. Then [imath]\gamma<\alpha+\sup\limits_{\beta\in A}\beta\implies\exists\beta\in A:\gamma<\alpha+\beta[/imath] This problem arises when I prove the theorem: Let [imath]\alpha,\beta[/imath] be ordinals and [imath]A[/imath] be a set of ordinal. Then [imath]\sup\limits_{\beta\in A}(\alpha+\beta)=\alpha+\sup\limits_{\beta\in A}(\beta)[/imath] I'm almost done except for one minor point: [imath]\gamma<\alpha+\sup_\limits{\beta\in A}\beta\implies\exists\beta\in A:\gamma<\alpha+\beta[/imath]. This result seems quite obvious, but I don't know how to prove in spite of several attempts. Could you please shed me some lights on how to prove [imath]\gamma<\alpha+\sup\limits_{\beta\in A}\beta\implies\exists\beta\in A:\gamma<\alpha+\beta[/imath]? Thank you very much! | 2985740 | Let [imath]α[/imath] be an ordinal and [imath]A[/imath] be a set of ordinals. Then [imath]\sup\limits_{β∈A} (α+β) = α+\sup\limits_{β∈A}(β)[/imath]
My idea is to prove that [imath]\alpha+ \sup\limits_{β∈A}(β)[/imath] is the supremum of [imath]\{α+β\mid β∈A\}[/imath]. While I'm able to prove that [imath]\alpha+ \sup\limits_{β∈A}(β)[/imath] is a upper bound of [imath]\{α+β\mid β∈A\}[/imath], I failed to show that [imath]\alpha+ \sup\limits_{β∈A}(β)\le\gamma[/imath] where [imath]\gamma[/imath] is an upper bound of [imath]\{α+β\mid β∈A\}[/imath]. The addition of ordinals is defined as follows [imath]\alpha+0=\alpha[/imath], [imath]\alpha+(\beta+1)=(\alpha+\beta)+1[/imath], and [imath]\alpha+\beta=\sup\limits_{\gamma<\beta}(\alpha+\gamma)[/imath] if [imath]\beta[/imath] is limit. My attempt: For all [imath]\beta\in A,\beta\le\sup\limits_{β∈A}(β)[/imath], then [imath]\alpha+\beta\le\alpha+ \sup\limits_{β∈A}(β)[/imath] and thus [imath]\alpha+ \sup\limits_{β∈A}(β)[/imath] is a upper bound of [imath]\{α+β\mid β∈A\}[/imath]. Assume that [imath]\gamma[/imath] is an upper bound of [imath]\{α+β\mid β∈A\}[/imath], then [imath]α+β\le\gamma[/imath] for all [imath]β∈A[/imath]. My questions: Please leave me some hints to complete my proof! I meet another proof of this theorem on the Internet: Could you please elaborate on the sentence Since set inclusion is the ordering, the two are equal. I'm unable to understand what this sentence means. Thank you so much for your help! For an ordinal [imath]α[/imath] and a set [imath]A[/imath] of ordinals: [imath]\sup\limits_{β∈A} (α+β) = α+\sup\limits_{β∈A}(β)[/imath] [imath]\sup\limits_{β∈A} (α\cdot β)=α\cdot \sup\limits_{β∈A}(β)[/imath] |
2990136 | Is [imath]\Bbb Q[/imath] intersect [imath][0,1][/imath] in [imath]\Bbb R[/imath] compact?
A subset is compact iff it is closed and bounded. I know [imath][0,1][/imath] is closed in [imath]\Bbb R[/imath], and the intersection of two closed sets is closed. I don’t know if [imath]\Bbb Q[/imath] is open or closed in [imath]\Bbb R[/imath] and have seen proofs for both. | 1437023 | Why is [imath][0, 1] \cap \mathbb{Q}[/imath] not compact in [imath]\mathbb{Q}[/imath]?
Statement: [imath][a, b] \cap \mathbb{Q}[/imath] in [imath]\mathbb{Q}[/imath] is not compact. Thus the interior of all compact subsets of [imath]\mathbb{Q}[/imath] is [imath]\emptyset[/imath]. I am trying to understand the first sentence. I read that a closed subspace of a compact space is compact, so for example, consider the unit interval [imath][0, 1][/imath] which is a compact space. Take a closed subspace [imath][0, 1] \cap \mathbb{Q}[/imath] of [imath][0, 1][/imath]. This set is closed since it just consists of all the rational numbers in between [imath]0[/imath] and [imath]1[/imath], including [imath]0[/imath] and [imath]1[/imath]. So it is a closed subspace of a compact space. But why isn't this compact? |
2989508 | Lindelöf and separable metric space
Let [imath](X,d)[/imath] be a metric space. How to prove that every lindelöf metric space is separable? | 2667647 | Let [imath](X, \mathcal{T})[/imath] be a metrizable space. Prove that [imath]X[/imath] Lindelöf implies that [imath]X[/imath] is separable.
Let [imath](X, \mathcal{T})[/imath] be a metrizable space. Prove that [imath]X[/imath] Lindelöf implies that [imath]X[/imath] is separable. My attempt: Let [imath]d[/imath] be the metric that induces the topology [imath]\mathcal{T}[/imath]. For every [imath]n \geq 1[/imath], consider the open cover: [imath]\{B_X(x,1/n)\mid x \in X\}[/imath] Because the space is Lindelöf, this implies that for every [imath]n \geq 1[/imath], there is a countable (or finite) subcover given by [imath]\mathcal{B}_n:= \{B_X(x,1/n) \mid x \in I_n, |I_n| \leq |\mathbb{N}|\}[/imath] Now, define [imath]D:= \bigcup_{n\geq 1} I_n[/imath]. Let [imath]y \in X[/imath]. Let [imath]\epsilon > 0[/imath]. Choose [imath]m[/imath] a positive integer so large that [imath]1/m < \epsilon[/imath]. Because [imath]y\in X =\bigcup\mathcal{B}_m[/imath], it follows that there exists [imath]x \in D[/imath] such that [imath]y \in B_X(x,1/m)[/imath], meaning that [imath]d(x,y) < 1/m < \epsilon[/imath], such that [imath]x \in B_X(y, \epsilon) \cap D[/imath]. Hence, [imath]y \in \overline{D}[/imath] and it follows that [imath]D[/imath] is a dense countable set in [imath]X[/imath], such that [imath]X[/imath] is separable. Is this correct? |
2990353 | Find angle [imath] \angle AED [/imath] in the following triangle.
Find angle [imath] \angle AED [/imath] in the following triangle. In the above triangle we have : [imath]CA=CB ,CE=DB=BA ,\angle ACB =20^° , \angle CAB=\angle CBA=80^°[/imath] now find [imath] \angle AED [/imath]. I think if we draw line [imath]AD[/imath] we have [imath]AB=BD[/imath] then [imath]\angle BAD= \angle BDA =50^°[/imath] then in the triangle [imath] AED[/imath] we have [imath]\angle EAD=30^°[/imath] but I can't find [imath] \angle AED [/imath] | 2192439 | Problem about angle in isosceles triangle
[imath]ABC[/imath] is an isosceles triangle, [imath]AB=AC[/imath], and [imath]\measuredangle A=20^{\circ}[/imath]. For point [imath]M[/imath] on [imath]AB[/imath] and [imath]N[/imath] on [imath]AC[/imath], we have [imath]AM=NC=BC[/imath]. Compute the [imath]\measuredangle BMN[/imath]. In a similar problem (with only one point on one side) the key step was to construct an equilateral triangle on the other side. I tried using that in this problem too, but I didn't manage to solve it. |
2990757 | Prove that if f is continous on R and the limit as x approaches infinity and negative infinity is infinity, then f obtains its minimum value in R?
Problem: Suppose [imath]f[/imath] is a continous function defined on [imath]\mathbb{R}[/imath] s.t. [imath]\lim\limits_{x\to -\infty}= \lim\limits_{x\to\infty} = \infty[/imath]. Then [imath]f[/imath] obtains its minimum value for some [imath]x\in\mathbb{R}[/imath]. My attempt: Suppose [imath]f[/imath] is continous on [imath]\mathbb{R}[/imath] and [imath]\lim\limits_{x\to -\infty}= \lim\limits_{x\to\infty} = \infty[/imath]. Let [imath][a,b]\in\mathbb{R}[/imath] Since [imath]f[/imath] is continous on [imath]\mathbb{R}[/imath] and [imath][a,b]\in\mathbb{R}[/imath], then [imath]f[/imath] is continous on [imath][a,b][/imath]. Since [imath]f[/imath] is continous on [imath][a,b][/imath] and [imath][a,b][/imath] is closed and bounded, then [imath]f[/imath] is bounded on [imath][a,b][/imath]. Thus, [imath]f[/imath] obtains its maximum and minimum value at some [imath]x\in [a,b][/imath] Here is where I am stuck because I know that the minimum may be to the left or right of this closed interval, but I do not know how to connect that idea and the idea of the limits that are given. Thanks for your help. | 1189439 | Continuous function on R which tends to infinity on open interval must have a minimum
Suppose we have [imath]f : \mathbb{R} \to \mathbb{R}[/imath] continuous and [imath]\lim\limits_{x \to \pm \infty} f(x) = \infty.[/imath] Then must [imath]f[/imath] have a minimum on [imath]\mathbb{R}[/imath]? Intuitively it seems so, and it's easy to prove for some arbitrary closed interval in [imath]\mathbb{R}[/imath] with the Extreme Value Theorem, but how can we apply that to an open interval like [imath](-\infty, \infty)[/imath]? |
2989746 | Find all accumulation points of the set whose elements in form [imath]2^{-n}+5^{-m}, m,n\in\mathbb N^+[/imath]
Attempt: I thought that by fixing one of the index and doing it conversely I can find accumulation points as follows: For every fix [imath]n_0\in \mathbb N[/imath] [imath]2^{-n_0}+5^{-m}[/imath] converges to [imath]2^{-n_0}[/imath] and For every fix [imath]m_0\in \mathbb N[/imath] [imath]2^{-n}+5^{-m_0}[/imath] converges to [imath]5^{-m_0}[/imath] and letting free both indices we have a accumulation point [imath]0[/imath] So all accumulation points are [imath] {2^{-n},n\in\mathbb N^+ } \cup {5^{-n},m\in\mathbb N^+} \cup 0 [/imath]$ is it my reasoning right? | 1353520 | Accumulation points of [imath]\{ 2^{-n} + 5^{-m} : n,m \geq 1 \}[/imath]
Determine all of the accumulation points of the following sets in [imath]\mathbb{R}^1[/imath] and decide whether the sets are open or closed or neither. The set [imath]X[/imath] of all numbers of the form [imath]2^{-n} + 5^{-m}[/imath] where [imath]m,n = 1,2,\ldots[/imath]. Claim: The set of accumulation points is the following: [imath]\left\{\frac{1}{2^n}: n \in \mathbb{N} \right\} \cup \left\{\frac{1}{5^n}: n \in \mathbb{N}\right\} \cup \{0\}.[/imath] Let [imath]n \in \Bbb{N}[/imath] and [imath]r>0[/imath], and consider the open neighbourhood [imath]B\left(\frac{1}{2^n},r\right) = \left(\frac{1}{2^n} - r,\frac{1}{2^n} + r\right)[/imath]. For the given [imath]r[/imath] we can find [imath]m[/imath] large enough such that [imath]5^{-m} < r \implies \frac{1}{2^n} + \frac{1}{5^m} < \frac{1}{2^n} + r[/imath], and [imath]\frac{1}{2^n} - \frac{1}{5^m} > \frac{1}{2^n} - r[/imath]; therefore [imath]5^{-m} + 2^{-n} \in B\left(\frac{1}{2^n},r\right)[/imath]. A similar argument shows that each element of [imath]\{5^{-m}: m \in \mathbb{N}\} \cup \{0\}[/imath] is an accumulation point of [imath]X[/imath]. Now consider [imath]y \in X[/imath]. [imath]B(y,r)[/imath] isn't a subset of [imath]X[/imath] since for all [imath]r > 0[/imath], [imath]B(y,r) \subset (\mathbb{R} - \mathbb{Q}) \not\subset \Bbb{Q}[/imath], hence [imath]B(y,r) \not\subset X[/imath]. |
2990791 | Show the only homomorphism [imath]F:\Bbb Q\rightarrow \Bbb Q\setminus\{0\}[/imath] is the trivial map
I've been working on this problem for a long time and am stuck. I was thinking I could use Lagrange's theorem and the fact that the image of any homomorphism must have an order that divides both the order of the domain group and the codomain group. Since they differ by 1, their orders must be coprime meaning that the order of the image must be 1, namely the trivial map. I'm not exactly sure if this is right, if I could get some feedback that would be great. Thanks!!! | 2989531 | Every group homomorphism from [imath](\mathbb{Q}, +)[/imath] to [imath](\mathbb{Q}, \times)[/imath] is the trivial map.
How do you show that every group homomorphism from [imath](\mathbb{Q}, +)[/imath] to [imath](\mathbb{Q}, \times)[/imath] is the trivial map? I am trying to use proof by contradiction by assuming there is an element [imath]\frac{a}{b}[/imath] such that some homomorphism [imath]\phi[/imath] has [imath]\phi(\frac{a}{b}) = \frac{c}{d} \ne 1[/imath]. But I cannot seem to deduce any contradictions from here. Maybe using direct proof is a better approach? Any help is appreciated. |
2990836 | A directed complete graph with equal number of incoming and outgoing edges
In a complete graph with [imath]n[/imath] vertices, every vertex has [imath]n-1[/imath] edges. Assuming [imath]n[/imath] is odd, the number of edges from each vertex is even. If we now give every edge an orientation (making the graph a tournament), is it always possible to arrange the orientations such that every vertex has the same number of incoming and outgoing edges (apparently called a balanced directed graph)? It is easy to give edges an orientation where the above is not true. E.g. orient all edges to a given vertex as incoming edges: But it is also relatively easy to orient the edges so the above is true, at least for [imath]n=3, 5, 7, 9[/imath], which is what I've tried so far. An example for [imath]n=5[/imath] is given below: But is it always possible for any odd [imath]n[/imath]? | 938947 | Prove that there exist regular tournament of every odd order but there is no regular tournament of even order.
Prove that there exist regular tournament of every odd order but there is no regular tournament of even order. Here is what I got so far. Let [imath]T[/imath] be our regular tournament of order [imath]n[/imath]. Since [imath]T[/imath] is regular tournament [imath]id(u)=od(u)[/imath] for every [imath]u \in V(T)[/imath]. Since [imath]t[/imath] is a tournament [imath]deg(u)=n-1[/imath], thus [imath]id(u)=id(u)=\frac{n-1}{2}[/imath]. If [imath]n[/imath] is even then [imath]\frac{n-1}{2}[/imath] isn't a whole number, which can't be degree of [imath]u[/imath], so there is no egular tournament of even order. For [imath]n[/imath] is odd, [imath]od(u)=id(u)= k[/imath] for [imath]n=2k+1[/imath]. But i'm not sure how to prove this for every odd number. By induction ? |
2991873 | Solution of, an ODE related to the modified bessel equation.
I would like to know the general solution of [imath]x^2 y^{''}+x y^{'} -k^2 x^2 y=0\tag{1}[/imath] for [imath]y(x)[/imath]. (1) is related to the equation [imath]\quad x^2 y^{''}+x y^{'} - x^2 y=0[/imath] , which is the [imath]n=0[/imath] form of the modified Bessel equation [imath]x^2 y^{''}+x y^{'} -( x^2+n^2) y=0[/imath] Other Information. The material of this question turns up in fluid mechanics, in the theory of the instability of jets. There is plenty of material on Wikipedia about the Bessel and related equations but I have not found there, the general solution I am looking for. | 1575650 | Modified Bessel differential equation
The modified Bessel differential equation is always presented as [imath]r^2 \frac{\partial^2 f(r)}{\partial r^2} + r\frac{\partial f(r)}{\partial r} - (r^2 + n^2)f(r) = 0[/imath] with solutions [imath]f(r) = AI_n(r) + BK_n(r)[/imath] But if I had [imath]r^2 \frac{\partial^2 f(r)}{\partial r^2} + r\frac{\partial f(r)}{\partial r} - (\alpha^2 r^2 + n^2)f(r) = 0[/imath] What form should have the solutions and how to prove it? It seems not to be a simple substitution [imath]r' = \alpha r[/imath], because [imath]\alpha[/imath] appears just one time in the second equation and not in all the [imath]r[/imath] terms (as I would expect instead). |
2992229 | Evaluate: [imath]\lim_{n\to \infty}\sqrt{n}\int_{0}^1 \frac{1}{(1+x^2)^{n}}dx[/imath]
Evaluate: [imath]\lim_{n\to \infty}\sqrt{n}\int_{0}^1 \frac{1}{(1+x^2)^{n}}dx[/imath] | 2984468 | Finding [imath]\lim_{n\to \infty}\sqrt n \int_0^1 \frac{\,dx}{(1+x^2)^n}[/imath]
[imath] \lim_{n\to\infty} n^{1/2} \int_{0}^{1} \frac{1}{(1+x^2)^n}\mathrm{d}x=0 [/imath] Is my answer correct? But I am not sure of method by which I have done. |
2992345 | True or false: question about generated ideal and his powers.
Let [imath]R[/imath] be a commutative ring with identity and [imath]a\in R[/imath]. Let [imath]n\in\mathbb{N}[/imath]. The principal ideal generated by [imath]a[/imath] is [imath]\langle{}a\rangle{}=aR=\{ar\;|\;r\in R\}.[/imath] Question. Is true or false that [imath]\langle{}a\rangle{}^n=\langle{}a^n\rangle{}[/imath]? My attempt It's true for me: in fact by definition [imath]\langle{}a^n\rangle{}=\{a^nr\;|\;r\in R\},[/imath] and [imath]\langle{}a\rangle{}^n=\bigg\{\sum_{\text{finite}}(a_{1}\cdots a_{n})\;|\;a_i\in\langle{}a\rangle{}\bigg\}=\bigg\{\sum_{\text{finite}}a^n(r_1\cdots\ r_n)\;|\;r_i\in R\bigg\}.[/imath] We observe that [imath](a^n)\subseteq(a)^n[/imath], as if [imath]\tilde{a}\in (a^n)[/imath] exists [imath]r\in R[/imath] such that [imath]\tilde{a}=a^nr[/imath], accordingly [imath]\tilde{a}=a^n(r\cdot \underbrace{1_R\cdot 1_R\cdots 1_R}_{n-1\;\text{times}})\in\langle{}a\rangle{}^n[/imath]. Vice versa we prove that [imath](a^n)\supseteq(a)^n[/imath]. If [imath]\overline{a}\in\langle{a}\rangle^n[/imath], then [imath]\overline{a}=\sum_{\text{finite}}\overline{a}^n(t_1\cdots\ t_n)[/imath], where [imath]t_i\in R[/imath]. Therefore [imath]\overline{a}=a^n[\underbrace{(s_1\cdots \ s_n)\underbrace{+\cdots+}_{\text{finite}}(v_1\cdots v_n)]}_{:=r}[/imath], then [imath]\overline{a}=a^nr[/imath], where [imath]r\in R[/imath]. Therefore [imath]\overline{a}\in\langle{}a\rangle{}^n[/imath]. Thanks! | 1267777 | Product of principal ideals: [imath](a)\cdot (b) = (a b)[/imath]
In which kinds of rings [imath]R[/imath] does the following hold: [imath](a)\cdot (b) = (ab) \; ?[/imath] With [imath]a, b\in R[/imath], [imath](a)[/imath] denoting the (two-sided) ideal generated by [imath]a[/imath] and the multiplication of ideals [imath]I, J\subset R[/imath] defined as [imath] I\cdot J = \biggl\{\sum_{i=1}^n x_i y_i : n\in\mathbb{N}, x_i \in I, y_i \in J \biggr\}\, .[/imath] It seems to me that it only holds for commutative rings with [imath]1[/imath]. Is that right? Ok, I'm trying a proof: Let [imath]R[/imath] be commutative with [imath]1\in R[/imath]. Then [imath](a)\cdot (b) = (a b)[/imath] for any [imath]a, b\in R[/imath]. First let [imath]I_a := \{ra : r\in R\}[/imath], we're going to show that [imath](a) = I_a\, .[/imath] [imath]I_a[/imath] is obviously an ideal. Since [imath]1 \in R[/imath] we have [imath]1 \cdot a \in I_a[/imath], so [imath](a) \subset I_a[/imath]. On the other hand, any [imath]x \in I_a[/imath] can be written as [imath]x = ra[/imath] and so must be an element of [imath](a)[/imath]. This proves that [imath](a) = I_a[/imath]. Then [imath](a)\cdot (b) = I_a \cdot I_b = \biggl\{\sum_{i=1}^n x_i y_i : n\in\mathbb{N}, x_i \in I_a, y_i \in I_b \biggr\} = \biggl\{\sum_{i=1}^n (r_i a) (s_i b) : n\in\mathbb{N}, r_i, s_i \in R \biggr\} = \biggl\{a b \sum_{i=1}^n r_i s_i : n\in\mathbb{N}, r_i, s_i \in R \biggr\} = \biggl\{a b r : r \in R \biggr\} = I_{ab} = (ab)\, .[/imath] We obviously had to use that [imath]R[/imath] is commutative. In the last step we also used that every [imath]r \in R[/imath] can be written as [imath]r=\sum_{i=1}^n r_i s_i[/imath]. This is because [imath]1 \in R[/imath], so with [imath]n=1[/imath] we have [imath]r = 1\cdot r[/imath]. |
2992228 | Prove that, [imath]\int^{\infty}_0\frac{x\sin(rx)}{a^2+x^2}dx=\frac{\pi}{2}e^{-ar}[/imath]
The question is: prove that [imath]\int^{\infty}_0\frac{x\sin(rx)}{a^2+x^2}dx=\frac{\pi}{2}e^{-ar}[/imath] This is what I've got so far: Let [imath]I(r)=\int^{\infty}_0\frac{x\sin(rx)}{a^2+x^2}dx[/imath] [imath]I'(r)=\int^{\infty}_0\frac{x^2\cos(rx)}{a^2+x^2}dx[/imath] = [imath]\int^{\infty}_0\cos(rx)dx-\int^{\infty}_0\frac{a^2\cos(rx)}{a^2+x^2}dx[/imath] [imath]I''(r)=\int^{\infty}_0-x\sin(rx)dx+\int^{\infty}_0\frac{xa^2\sin(rx)}{a^2+x^2}dx[/imath] [imath]I''(r)=\int^{\infty}_0-x\sin(rx)dx+a^2I(r)[/imath] I think I'm supposed to form a differential equation and solve it by letting [imath]I(r)=c_1e^r+c_2e^{-r}[/imath], but the problematic part is this: [imath]\int^{\infty}_0-x\sin(rx)dx[/imath]. I don't believe it simplifies to any constant and the differential equation doesn't solve nicely. Hints/suggestions appreciated! | 2699274 | show [imath]\int_0^{\infty} \frac{x\sin ax}{x^2+t^2}dx = \frac{\pi}{2}e^{-at}[/imath]
I'm asked to show [imath]\int_0^{\infty} \frac{x\sin ax}{x^2+t^2}dx = \frac{\pi}{2}e^{-at}[/imath] when [imath]t,a > 0[/imath]. I tried using integration by parts integrating [imath]\frac{x}{x^2+t^2}[/imath]. But it seems that [imath]\frac{x\sin ax}{x^2+t^2}[/imath] has no antiderivative. What should I try? |
1594728 | [imath]\lim_{n\to \infty} \sqrt n(A_{n+1}-A_n) [/imath] when [imath] A_n = \frac{1}{n}(a_1+a_2+\cdots+a_n)[/imath] and [imath]|a_n| \leq 1[/imath]
Let [imath]a_n[/imath], [imath]n \geq 1[/imath] be a sequence of real numbers satisfying [imath]|a_n| \leq 1[/imath] for all [imath]n[/imath]. Define [imath] A_n = \frac{1}{n}(a_1+a_2+\cdots+a_n)[/imath] for [imath]n\geq 1[/imath]. Then [imath]\lim_{n\to \infty} \sqrt n(A_{n+1}-A_n) [/imath] is A. [imath]0[/imath] B. [imath]-1[/imath] C. [imath]1[/imath] D. None of them ATTEMPT I tok sequence [imath]a_n=1[/imath] and answer came out to be 0. But how do i choose between A and D. Thanks | 172305 | Compute: [imath]\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)[/imath] where [imath]A_n = \frac{1}{n}(a_1 + a_2 + \cdots + a_n)[/imath]
Let [imath]\{a_n\}, {n\geq 1}[/imath], be a sequence of real numbers satisfying [imath]|a_n|\leq 1[/imath] for all [imath]n[/imath]. Define [imath]A_n = \frac{1}{n}(a_1 + a_2 + \cdots + a_n),[/imath] for [imath]n\geq 1[/imath]. Then find [imath]$\displaystyle\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)$[/imath] . I proceed in this way [imath]\lim_{n \rightarrow \infty}\sqrt{n}(A_{n+1} − A_n)=\lim_{n \rightarrow \infty}\sqrt{n}\left[\frac{1}{n+1}(a_1 + a_2 + \cdots + a_n+a_{n+1})-\frac{1}{n}(a_1 + a_2 + \cdots + a_{n})\right]=\lim_{n \rightarrow \infty}\left[{(na_{n+1}-a_1 - a_2 - \cdots - a_n})\frac{1}{\sqrt{n}(n+1)}\right][/imath] Please help me to complete from here |
2988284 | Regarding the value of [imath]X+Y+Z[/imath] if [imath]\frac1X + \frac1Y + \frac1Z = 1[/imath]
Given [imath]X, Y[/imath] and [imath]Z[/imath] are three positive and non-equal natural numbers. If [imath]\frac1X + \frac1Y + \frac1Z = 1[/imath], then what's the value of [imath]X + Y + Z[/imath]? This is a question from my nephew, I cannot think of any such triples. Any idea guys? | 616639 | Find all solutions of [imath]{\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1[/imath], where [imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] are positive integers
Find all solutions of [imath]{\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1[/imath] , where [imath]x,y,z[/imath] are positive integers. Found ten solutions [imath](x,y,z)[/imath] as [imath]{(3,3,3),(2,4,4),(4,2,4),(4,4,2),(2,3,6),(2,6,3),(3,6,2),(3,2,6),(6,2,3),(6,3,2)}[/imath]. Are these the only 10 solutions? First, none of [imath]x[/imath], [imath]y[/imath] or [imath]z[/imath] can be [imath]1[/imath] ([imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] are positive integers) If I let [imath]x=2[/imath], then finding all solutions to [imath]1/y+1/z = 1/2[/imath] leads to [imath](4,4), (3,6)[/imath] and [imath](6,3)[/imath] which gives me [imath](x,y,z)[/imath] as [imath](2,4,4), (2,3,6), (2,6,3)[/imath] but this also means [imath](4,4,2), (4,2,4), (3,2,6), (3,6,2), (6,2,3), (6,3,2)[/imath] are all valid triples for this equation. If I let [imath]x=3[/imath], the only different values of [imath]y[/imath] and [imath]z[/imath] are [imath](3,3)[/imath] How do I prove these are the only ten solutions? (without using any programming) Known result: If we denote [imath]d(n^2)[/imath] as the number of divisors of [imath]n^2[/imath], then the number of solutions of [imath]{\frac {1} {x} }+{\frac {1} {y} } = {\frac {1} {n} }[/imath] = [imath]d(n^2)[/imath] (For positive [imath]x[/imath], [imath]y[/imath]) For [imath]{\frac {1} {x} } + {\frac {1} {y} } +{\frac {1} {z}}=1[/imath] [imath]z = \frac{xy}{y(x-1)-x}[/imath] where [imath]xy \neq 0[/imath] What happens after that? Question is: how do we sho there are the only ten solutions? I'm not asking for a solution. Assuming [imath]x \le y \le z[/imath] [imath]1 \le y \le \frac{xy}{y(x-1)-x}[/imath] [imath]\Longrightarrow 1 \le x \le y \le \frac{2x}{x-1} [/imath] Got the answer. I'll probably call @mathlove's answer. (Any additional answers I'll view later) Liked @user44197 answer. |
1480776 | Embedding of two extensions of a field
It is well known that if we have two groups/rings/vector spaces (over same field), then we can embed them in a common corresponding object. It is natural to consider for fields. However, if we take two fields of different characteristics, then we can not embed both in a common field. With this remark, I will shorten my problem in a simple case. Question: If [imath]F[/imath] is a field, and [imath]E_1,E_2[/imath] are any two extensions of [imath]F[/imath], then does there always exists a field [imath]E[/imath] which is extension of both [imath]E_1[/imath] and [imath]E_2[/imath]? | 2902479 | Do extension fields always belong to a bigger field?
Let [imath]F[/imath] be a field, [imath]E_1[/imath] and [imath]E_2[/imath] are two distinct extension fields of [imath]F[/imath]. Is it the case that we can always somehow find a field [imath]G[/imath] that contains both [imath]E_1[/imath] and [imath]E_2[/imath]? In other words, could extensions of fields have different 'direction's such that they are incompatible? Edit: I began to think about this problem while reading a proof. [imath]F[/imath] is a field. [imath]a[/imath] and [imath]b[/imath] are algebraic over [imath]F[/imath]. [imath]p(x)[/imath] and [imath]q(x)[/imath] are two polynomials in [imath]F[x][/imath] of minimum degree that respectively make [imath]a[/imath] and [imath]b[/imath] a zero. The proof claims that there is an extension [imath]K[/imath] of [imath]F[/imath] such that all distinct zeros of [imath]p(x)[/imath] and [imath]q(x)[/imath] lie in [imath]K[/imath]. For a single polynomial, I know this kind of field exists because of the existence of splitting field, why it is true for two polynomials? |
2992651 | Löwenheim-Skolem and proper class models of ZFC.
Let [imath]N[/imath] be a proper class model of ZFC and [imath]x \subset N[/imath] a set. Show that there is a set [imath]y \in N[/imath] such that [imath]x \subset y[/imath]. If [imath]x \subset N[/imath], I think that by the downward's part of Löwenheim-Skolem, we can find an elementary submodel [imath]y[/imath] of [imath]N[/imath] such that [imath]x \subset y[/imath]. But how would I conclude that [imath]y \in N[/imath]? It seems like I'm missing something trivial but I can't figure out what exactly. Note: A hint to point me to the right direction would be more appreciated than a complete answer. | 2964048 | Almost universal class
I so stuck with a problem of set theory. But first a recursive definition: Define [imath]R_0=\emptyset[/imath] If [imath]R_\alpha[/imath] is defined, then [imath]R_{\alpha+1}=\mathcal{P}(R_\alpha)[/imath] (the power set). For a limit ordinal [imath]\gamma[/imath], if [imath]R_\alpha[/imath] is defined for all [imath]\alpha<\gamma[/imath], then define [imath]R_\gamma=\displaystyle\bigcup_{\alpha<\gamma} R_\alpha[/imath] We define [imath]\text{BF}=\displaystyle\bigcup_{\alpha\in\text{OR}} R_\alpha[/imath] (the class of well founded sets). Next, my problem Take [imath]A\subseteq \text{BF}[/imath] a proper transitive class such that [imath](A,\in)\models\text{ZF}[/imath]. Prove that [imath]A[/imath] is almost universal. I think that the exercise is false because if it was true, then we could conclude that the strongly inaccessible cardinals doesn't exists. This because if [imath]\kappa[/imath] is strongly inaccessible then [imath]R_\kappa[/imath] satisfies the hypothesis but [imath]R_\kappa[/imath] isn't almost universal. I really appreciate any hint or/and suggestion. Edit: my counterexample is wrong. But, then, how can I solve the exercise? |
2992915 | Proof of [imath]{n \choose 0},{n \choose 1},{n \choose 2},...,{n \choose \lfloor n/2\rfloor}[/imath] is monotonically increasing
Prove that for [imath]n>1[/imath], sequence [imath]{n \choose 0},{n \choose 1},{n \choose 2},...,{n \choose \lfloor n/2\rfloor}[/imath] is strictly monotonically increasing. If I will show, that [imath]{n \choose \lfloor n/2\rfloor}-{n \choose \lfloor n/2-1\rfloor} >0[/imath] will that be the end of proof? | 2984998 | Proving that for [imath]n \equiv 0 \pmod{2}[/imath], we get [imath]{n \choose 0} < {n \choose 1} <\ldots< {n \choose n/2-1}<{n\choose n/2}[/imath] etc.
How can one prove that for [imath]n \equiv 0[/imath] mod [imath]2[/imath] we have [imath]{n \choose 0} < {n \choose 1} <\ldots< {n \choose n/2-1}<{n \choose n/2}>{n\choose n/2+1}>\ldots>{n\choose n-1}>{n\choose n}\,?[/imath] Can I say that for fixed [imath]n[/imath], the binomial coefficients [imath]n \choose k[/imath] increase with [imath]k[/imath] for [imath]k < n/2[/imath]? If n is even (like in our case), then the central binomial coefficient [imath]n \choose n/2[/imath] is the largest one. So [imath]n \choose k+1[/imath] is greater than, equal to, or less than [imath]n \choose k[/imath] according as [imath]n-k[/imath] is greater than, equal to, or less than [imath]k+1[/imath], that is according as [imath]k[/imath] is less than, equal to, or greater than [imath](n−1)/2[/imath]. Is that correct? |
2993225 | True/false question Regarding closed subsets
Is the following statement is True/false For any closed subset [imath]A ⊂ \mathbb{R}[/imath], there exists a continuous function [imath]f[/imath] on [imath]\mathbb{R}[/imath] which vanishes exactly on [imath]A[/imath]. | 2564798 | Existence of continuous function [imath]f[/imath] on [imath]\Bbb R[/imath] which vanishes exactly on [imath]A\subset \Bbb R[/imath]
Question For any closed subset [imath]A \subset \Bbb R[/imath], does there exist a continuous function [imath]f[/imath] on [imath]\Bbb R[/imath] which vanishes exactly on [imath]A[/imath]? If we take [imath]A=[a,\infty)[/imath] or [imath](-\infty,a][/imath] or [imath]\{a_1,a_2,\cdots,a_n\}[/imath] or [imath]\bigcup_{i=1}^{k} [a_i,b_i][/imath], then indeed we can have such a continuous function. But when I thought about cantor set or the set [imath]\{\frac 1n : n \in \Bbb N\} \cup \{0\}[/imath], which are closed in [imath]\Bbb R[/imath], I couldn't think of a continuous map vanishing exactly at these two sets. How should I go about this problem? |
2993800 | Do compacta in [imath]\mathbb R^n[/imath] have finite Hausdorff measure?
For [imath]A\subset \mathbb R^n[/imath] let [imath]\mathcal H^s(A)[/imath] be the [imath]s[/imath]-dimensional Hausdorff measure with respect to the Euclidean metric. The Hausdorff dimension of [imath]A[/imath] is given by [imath]\dim_H(A) = \inf\{s>0\mid\mathcal H^s(A)=0\}.[/imath] One of the go-to curiosities in measure theory courses is the fact the middle-third Cantor [imath]C[/imath] set has Hausdorff dimension [imath]\log2/\log3[/imath]. This value is easily determined using a self-similarity argument assuming that [imath]0<\mathcal H^s(C)<\infty[/imath] for some [imath]s[/imath]. However, verifying this assumption for [imath]s=\log2/\log3[/imath] is not quite as easy. Thinking about this and other self-similar objects got me wondering about a more general question: For non-empty, compact [imath]K\subset\mathbb R^n[/imath] of Hausdorff dimension [imath]d[/imath], is it automatic that [imath]0<\mathcal H^d(K)<\infty[/imath]? In other words, do non-empty compacta in [imath]\mathbb R^n[/imath] have finite, non-zero Hausdorff measure in their Hausdorff dimension? At first, I thought that the answer must be yes for the extreme cases [imath]d=0[/imath] and [imath]d=n[/imath] where the relevant Hausdorff measures are the counting measure and the Lebesgue measure (up to scaling). However, at some point I realized that I had implicitly assumed that Hausdorff dimension 0 or n would imply discreteness or non-empty interior (which may or may not be true, probably not). I'm now at the point where I've realized that I have already spent much more time thinking about this than I should have. Hence, I resort to the infinite wisdom of the internet. I believe that the answer to the above question is no in the stated generality, but I'm basically at a loss. Maybe someone on here has something up their sleeve. | 1583133 | Sets with Hausdorff-Measure 0
The [imath]\alpha[/imath]-Dimensional Hausdorff-Measure of a Set A is defined as [imath]H^\alpha (A)=\inf_{A\text{ is countable covering}}\sum_{A'\in A} diam(A')^\alpha[/imath]. It is easy to show, that for every set [imath]E\subseteq\mathbb{R}^d[/imath] there exists a unique [imath]\beta\in \mathbb{R}[/imath], so that [imath]H^\alpha(E)=0[/imath] for [imath]\alpha>\beta[/imath] and [imath]H^\alpha(E)=\infty[/imath] for [imath]\alpha < \beta[/imath]. This [imath]\beta[/imath] is called the Hausdorff-Dimension [imath]dim(E)[/imath] of E. It is easy to show, that [imath]dim(\mathbb{R}^d)=d[/imath] and [imath]H^d(\mathbb{R}^d)=\infty[/imath]. I wonder if there is any Set A with [imath]dim(A)=\alpha[/imath] and [imath]H^\alpha(A)=0[/imath]. Does anybody have an idea how to approach this? |
2994116 | Integration Inequalities
The question we need to prove is [imath]1.222\le 1+3^{-2}+5^{-2}+\cdots\le 1.252.[/imath] I know how to prove the first part [imath]1+3^{-2}+...+43^{-2}\ge 1.222.[/imath] But I do not know how to prove the second part. If I upper bound it using [imath]1.222 + 45^{-2} + \text{integration from $45$ to $\infty$}[/imath], I found that it is about [imath]1.277777[/imath] which is too large. How should I do this question? Thank you very much. | 2993925 | Show that [imath]1.222 \le 1 + 3^{-2} + 5^{-2} + ... \le 1.252[/imath]
Show that [imath]1.222 \le 1 + 3^{-2} + 5^{-2} + ... \le 1.252[/imath] without calculate the value of [imath]1 + 3^{-2} + 5^{-2} + ...[/imath] Can I do in this way? [imath]\lim\limits_{n \to \infty} \int_2^n (2x+3)^{-2} \,dx \le 7^{-2} + 9^{-2} + 11^{-2} + ... \le \lim\limits_{n \to \infty} \int_2^n (2x+1)^{-2} \,dx[/imath] [imath]\lim\limits_{n \to \infty} \int_2^n (2x+3)^{-2} \,dx = \lim\limits_{n \to \infty}\frac{-1}2(2n+3)^{-1}+\frac12(7)^{-1} = \frac1{14}[/imath] [imath]\lim\limits_{n \to \infty} \int_2^n (2x+1)^{-2} \,dx = \lim\limits_{n \to \infty}\frac{-1}2(2n+1)^{-1}+\frac12(5)^{-1} = \frac1{10}[/imath] [imath]\therefore \frac1{14} \le 7^{-2} + 9^{-2} + 11^{-2} + ... \le \frac1{10}[/imath] [imath]\therefore 1.22253... \le 1 + 3^{-2} + 5^{-2} + ... \le 1.25111... [/imath] [imath]\therefore 1.222 \le 1 + 3^{-2} + 5^{-2} + ... \le 1.252 [/imath] |
2993719 | Limit as [imath]z[/imath] approaches [imath]0[/imath] for [imath]e^{1/z^4}[/imath]
I want to work out if the limit as [imath]z[/imath] approaches [imath]0[/imath] for [imath]e^{1/z^4}[/imath] exists and if not then why. I worked out that the left and right sided limits both equal to +∞ so I thought that was enough to conclude that the limit therefore existed. But when I checked the solutions it said that the limit did not exist, because you get different limits when you approach ([imath]0,0[/imath]) along different rays, ([imath]x,y[/imath]) = ([imath]at,bt[/imath]) as [imath]t[/imath] [imath]\rightarrow[/imath] [imath]0[/imath]. I don't understand this though can anybody help explain? What does it mean by the rays? Like gradients? Edit [imath]z[/imath] is a complex number | 1477231 | Does the limit of [imath]e^{-1/z^4}[/imath] as [imath]z\to 0[/imath] exist?
The question is as follows (note that [imath]z = x + yi[/imath]): Does [imath]\lim_\limits{z \to 0} e^{-1/z^4}[/imath] exist? I've shown that [imath]\lim_\limits{x \to 0} e^{-1/x^4} = 0[/imath] is equal to [imath]\lim_\limits{y \to 0} e^{-1/(iy)^4} = 0[/imath] To prove both parts, but I'm not sure this is how it is done. I've been away from class for surgery, so all I have to work off of is what I've read online and in the textbook. The problem is, it seems that with problems such as this one: [imath]\lim_\limits{z \to 2} \frac{z^2 + 3}{iz} = \frac{-7i}{2}[/imath] All you have to do is substitute in 2 directly. I can't seem to figure out how to approach these problems, so hopefully all I have explained makes sense. Thanks for any explanations! |
2993689 | Prove that if [imath]a+b+c=1[/imath] then [imath]\frac{ab}{c+1}+\frac{ac}{b+1}+\frac{bc}{a+1} \le \frac{1}4[/imath] [imath]a,b,c[/imath] are positive real numbers
I simplified the inequality , but I didn't get anything that may seem useful. I got [imath]^\sum_{cyc} 4a^2b+8ab+4ab^2\le abc+ab+ac+ab+a+b+c+1[/imath] | 779330 | Show [imath]a^2+b^2+c^2=1[/imath] [imath]\implies[/imath] [imath] \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 [/imath]
If [imath]a[/imath], [imath]b[/imath] and [imath]c[/imath] are real numbers such that [imath]a^2+b^2+c^2=1[/imath] , then how to prove that [imath] \dfrac {b^2c^2}{1+a^2} +\dfrac {c^2a^2}{1+b^2}+\dfrac {a^2b^2}{1+c^2} \le \dfrac 14 [/imath] ( don't apply Schur's inequality )? Thanks. |
2994305 | Specific parabola parametrization.
The equation of a parabola with vertex at the origin is given by [imath]Ax^2+Bxy+Cy^2+Dx+Ey+F=0[/imath] with the contraints, [imath]B^2-4AC=0[/imath] and [imath]F=0[/imath]. There exists a parametrization as follows: [imath]r(t)=\left(\begin{array}{cc}at^2+bt\\ct^2+dt\end{array}\right)[/imath] I believe this is true because Geogebra will show me the numeric parametrization going either direction. That is, I can put in any [imath]A,B,C,D,E[/imath] or I can put in [imath]a,b,c,d[/imath] and the program will show me the other set. I just have to play by the constraint rules above. How do I find variables [imath]a,\;b,\;c[/imath] and [imath]d[/imath]? What I have tried? Brute force. Use a CAS and substitute [imath]x=at^2+bt[/imath] into the general conic form and solve for [imath]y[/imath]. From the mess of terms, I collected the ones where [imath]B^2-4AC[/imath] could be factored away thus reducing the complexity. What remains is [imath]r(t)=\left(\begin{array}{cc}at^2+bt\\\frac{-B(at^2+bt)-E\pm\sqrt{(2BE-4CD)(at^2+bt)+E^2}}{2C} \end{array}\right)[/imath] This is a parametrization and it will half-graph but it isn't in the correct form. The [imath]y[/imath] term didn't become [imath]ct^2+dt[/imath]. I did notice in passing that the root term is somehow related to the eigenvectors. It was elusive. Discovery using numerical examples showed me something. In the desired parametrization let the parabola's focal point be [imath](f_x,f_y),[/imath], then [imath]a=f_x[/imath] and [imath]b=2f_y[/imath] and [imath]c=f_y[/imath] and [imath]d=2f_x[/imath]. I can get the focal point by rotating the graph to the vertical with [imath]\lambda x^2+E_1y=0[/imath] (eigenval/vec rotation method) and using [imath]x^2=4py[/imath] to get p and then rotating back again. I know that "discovery" is sort of an answer to my question, but I really want to know how to derive [imath]a,b,c,d[/imath] rather than my didactic method for finding them. | 2044922 | Parametric Form for a General Parabola
It is well known that a parametric form of the parabola [imath]y^2=4ax[/imath] is [imath](at^2, 2at)[/imath]. What are possible parametric forms of the general parabola [imath](Ax+Cy)^2+Dx+Ey+F=0[/imath] ? |
2995319 | Is [imath](-1)^{1/3} = -1[/imath] , but [imath](-1)^{2/6} = 1[/imath]. Why aren't these the same?
So if you try to solve [imath](-1)^{1/3}[/imath] you can do [imath](-1)^{1/3} = \sqrt[3]{-1} = -1[/imath] (cubic root of [imath]-1[/imath]) But what if I write [imath]1/3[/imath] as [imath]2/6[/imath]? [imath](-1)^{1/3} = (-1)^{2/6}[/imath] So [imath](-1)^{2/6} = \sqrt[6]{(-1)^2}[/imath] (sixth root of [imath](-1)^2[/imath]) So we have [imath]\sqrt[6]{(-1)^2}= \sqrt[6]{1} = + 1[/imath] Then we have: [imath]+1 = -1[/imath] How is that possible? | 1887177 | Dealing with non-integral powers on negative numbers
If we are given an expression [imath](-8)^\frac26[/imath], how do we solve it? If it is [imath](-8)^\frac13[/imath], we can find the cube root of -8 which is -2. However, if we square it first and find the sixth root, we get +2 (or maybe [imath]\pm2[/imath], which still isn't the same as just -2. Also realised that when solving logarithmic and exponential questions, I had often made the assumption that if [imath]b^a=b^c[/imath], then [imath]a=c[/imath], either explicitly, or via some formula. I don't know if this is really true now. So is there some convention by which we solve non-integral powers or is it just undefined to do so or something? P.s. Talking of non-integral powers, will a quantity like [imath](-8)^{√3}[/imath], will it be positive or negative. (Basically that's an irrational power now.) |
2995079 | Can we get a graph not containing [imath]K_n[/imath] with [imath]\chi(G)\geq n[/imath]
It is known that the chromatic number of a complete graph is equal to the number of its vertices, that is [imath]\chi(K_n)=n[/imath]. My question is can we get a graph [imath]G[/imath] not containing [imath]K_n[/imath] but with [imath]\chi(G)\geq n[/imath], for [imath]n\geq 3[/imath]? | 579892 | Construction of a triangle-free graph of chromatic number [imath]1526[/imath]
I found this exercise in Bollobas: Modern Graph Theory "Construct a triangle-free graph of chromatic number 1526" It is added not to use results from the chapter about Ramsey Theory. Now my qestions: (1) Is there a nice way to construct such a graph? (2) If yes then what is so special about this specific graph? My ideas: Grötzsch's theorem states that every triangle-free planar graph may be 3-colored, so we know that we can exclude planar grpahs and focus on non-planar graphs. I read something about the Mycielskian on http://en.wikipedia.org/wiki/Mycielskian I think this is the key to the solution but I do not see how to follow a concrete construction. |
2994436 | Prove that the sequence [imath]a_n = \sum_{k=0}^n \frac{1}{k!}[/imath] is convergent.
Prove that the sequence [imath]a_n = \sum_{k=0}^n \frac{1}{k!}[/imath] is convergent. I am using the theorem: All bounded monotone sequences converge. So i need to prove it is bounded and monotone. [imath]a_1=1,a_2=1.5, a_3=1.667, a_4=1.708, a_5=1.717, a_6=1.7181, a_7=1.7182[/imath] I can see that it is bounded below by 1 and increasing, but I'm not sure how to go about proving so a little tip in the right direction there would be great. As far as finding where it is bounded above I think I need to take the limit using a geometric series? but I'm not sure how to do that with the "!" in the problem. Is the geometric series thing the right direction to go in? I apologize if I'm asking too much, please do not solve the problem for me. I only want tips so I know how to go about it. | 166310 | Proving [imath]\mathrm e <3[/imath]
Well I am just asking myself if there's a more elegant way of proving [imath]2<\exp(1)=\mathrm e<3[/imath] than doing it by induction and using the fact of [imath]$\lim\limits_{n\rightarrow\infty}\left(1+\frac1n\right)^n=\mathrm e$[/imath], is there one (or some) alternative way(s)? |
2995802 | [imath]\sum_{n=1}^{\infty}a_n^2[/imath] converges, so does [imath]\sum_{n=1}^{\infty}b_n^2[/imath]. Prove that [imath]\sum_{n=1}^{\infty}|a_nb_n|[/imath] converges.
[imath]\sum^{\infty}a_n^2[/imath] converges, so does [imath]\sum^{\infty}b_n^2[/imath]. Prove that [imath]\sum^{\infty}|a_n\cdot b_n|[/imath] converges. I don't know how to use the information I am given in the hypothesis to make any conclussion. What I know: (1) [imath]\sum^{\infty}a_n^2[/imath] converges [imath]\Longrightarrow \lim_{n\to\infty}a_n^2 = 0 \Longrightarrow \lim_{n\to\infty}a_n = 0[/imath] Same for [imath]b_n[/imath]: [imath]\sum^{\infty}b_n^2[/imath] converges [imath]\Longrightarrow \lim_{n\to\infty}b_n^2 = 0 \Longrightarrow \lim_{n\to\infty}b_n = 0[/imath] Then: [imath]\lim_{n\to\infty}|a_n\cdot b_n|=\lim_{n\to\infty}|a_n|\cdot |b_n|=\lim_{n\to\infty}|a_n|\cdot \lim_{n\to\infty}|b_n|=0[/imath] Which doesn't help because (1) is a one-way implication. Any hints? | 939884 | Prove [imath]\sum_{n=1}^{\infty}|a_{n}b_{n}|[/imath] converges if [imath]\sum_{n=1}^{\infty}a_{n}^{2}[/imath] and [imath]\sum_{n=1}^{\infty}b_{n}^{2}[/imath] converge
This is a homework problem for an undergrad topology course. Let [imath]l^{2}[/imath] be the set of all real-valued sequences [imath](c_{n})[/imath] where [imath]\sum_{n=1}^{\infty}c_{n}^{2}[/imath] converges. Let [imath](a_{n}),(b_{n})\in l^{2}[/imath]. Claim: [imath]\sum_{n=1}^{\infty}|a_{n}b_{n}|[/imath] converges. I've done a couple of examples where [imath]a_{n}[/imath] and [imath]b_{n}[/imath] are general harmonic series. I've also written out the definition of convergence for [imath]a_{n}[/imath] and [imath]b_{n} [/imath]. My problem is that thus far I do not understand on any level why my claim is true. |
2995955 | Let [imath]R[/imath] be a finite commutative ring with unity. Show that every prime ideal of [imath]R[/imath] is maximal.
Attempt: Suppose [imath]P[/imath] a prime ideal, then [imath]R/P[/imath] is an integral domain, and so [imath]R/P[/imath] is a field, which implies that [imath]R[/imath] is maximal. I am not sure if this is a sufficient proof. Furthermore, I do not understand how to use the condition "finite". | 350047 | Let [imath]R[/imath] be a finite commutative ring. Show that an ideal is maximal if and only if it is prime.
Let [imath]R[/imath] be a finite commutative ring. Show that an ideal is maximal if and only if it is prime. My attempt: Let [imath]I[/imath] be an ideal of [imath]R[/imath]. Then we have [imath]I[/imath] is maximal [imath]\Leftrightarrow[/imath] [imath]R/I[/imath] is a finite field [imath]\Leftrightarrow[/imath] [imath]R/I[/imath] is a finite integral domain [imath]\Leftrightarrow[/imath] [imath]I[/imath] is a prime ideal. Is my proof valid ? |
2993178 | Intuition to understanding why homotopic attaching maps of n-cell yield homotopy equivalent space
I know that there's already another post that asked the same question. But I am trying to verify my understanding of the correct approach to this problem. The problem: let [imath]Y[/imath] and [imath]Y'[/imath] be spaces obtained from attaching an n-cell to [imath]X[/imath] by homotopic attaching maps. Prove that [imath]Y[/imath] and [imath]Y'[/imath] are homotopy equivalent. The approach I have in mind, described pictorially: Let [imath]f[/imath] and [imath]g[/imath] be the corresponding attaching maps. The author gives the hint of considering the adjunction space [imath]Z =:X \cup_{H} (\bar B^n \times I)[/imath], where [imath]H[/imath] is the homotopy from [imath]f[/imath] to [imath]g[/imath]. This space is basically the disjoint union of different spaces obtained from attaching an n-cell, with the attaching map being a [imath]H(u,t)[/imath] with a fixed [imath]t[/imath]. We try to find a retraction from [imath]Z[/imath] to [imath]X \cup_{H} (\bar B^n \times {0})[/imath] which can be identified with [imath]Y[/imath] (Assuming [imath]H(u,0) = f[/imath]). And to define such a map, my idea was, every points that are in [imath]\bar B^n \times I[/imath] that are not on the boundary of [imath]\bar{B^n}[/imath], are not identified with any points in [imath]X[/imath], So any points that are either on the boundary, or on [imath]X[/imath], are identified with some points in [imath]X[/imath]. so for every fixed [imath]t[/imath], can we not just map [imath](x, (u,t)) \in Z[/imath] to the points on [imath]x \in X[/imath] they are identified with in [imath]X \cup_{H} (\bar B^n \times {0})[/imath]? so in some sense, can we not define a retraction from [imath]X \cup_{H} (\bar B^n \times I)[/imath] to [imath]X \cup_{H} (\bar B^n \times {0})[/imath] by defining an "inclusion map" for each [imath]X \cup_{H} (\bar B^n \times {t})[/imath] into [imath]X \cup_{H} (\bar B^n \times {0})[/imath]? Sorry for the vague explanation of my approach, this is because I don't even know how to begin to construct such a map, or even know if this approach makes any sense. | 608908 | Prove homotopic attaching maps give homotopy equivalent spaces by attaching a cell
Prove: If [imath]f,g:S^{n-1} \to X[/imath] are homotopic maps, then [imath]X\sqcup_fD^n[/imath] and [imath]X\sqcup_gD^n[/imath] are homotopy equivalent. I think it can be proved by showing they are both deformation retracts of [imath]X\sqcup_H(D^n\times I)[/imath] where [imath]H[/imath] is the homotopy between [imath]f[/imath] and [imath]g[/imath]. However, I have difficult in proving that the deformation retracts are continuous map. In fact, I have difficulty in representing a map in quotient spaces like [imath]X\sqcup_fD^n[/imath]. I think a map from [imath]X\sqcup_fD^n[/imath] to [imath]W[/imath] can be represented by two maps: [imath]m_1: X\to W[/imath], [imath]m_2: D^n\to W[/imath], where for [imath]x\in S^{n-1}[/imath], [imath]m_1\circ f(x)=m_2\circ i(x)[/imath]. Then I construct the deformation retract this way: [imath]m_1: X\to X[/imath]. For [imath]x\in H(S^{n-1},t)[/imath], [imath]m_1(x)=H(S^{n-1},0)[/imath], otherwise [imath]m_1(x)=x[/imath]. [imath]m_2: D^n\times I\to D^n\times {0}[/imath]: [imath]m_2((D^n,t))=(D^n,0)[/imath]. It is easy to verify that [imath]m_1[/imath] and [imath]m_2[/imath] define a map from [imath]X\sqcup_H(D^n\times I)[/imath] to [imath]X\sqcup_fD^n[/imath]. As long as this is a continous map, obvioulsy then we find a deformation retract. But it seems such a map is not continous? |
2995384 | How do I prove the statement below by the method of mathematical induction?
If [imath]n ∈ N[/imath] and [imath]n \ge 4[/imath] , how do I prove by the method of induction that [imath]n! > n^2[/imath]? | 1403015 | mathematical induction proof of a square vs factorial
So lets say I have [imath] n^{2} \le n! [/imath] For what positive integers is this not true? [imath]n=2[/imath] and [imath] 3[/imath] Base case? [imath]n=4 \implies 16 \le 24 [/imath] What is the inductive hypothesis and how do I show the inductive proof? Thanks |
2996382 | Solving a homogeneous recurrence relation
How to solve [imath]a_n=4a_{n-1}-4a_{n-2}[/imath] [imath]a_0=6[/imath] and [imath]a_1=8[/imath] I found first few terms as [imath]6,8,8,0,... [/imath] But I don't know how to proceed. | 218762 | showing a genereral function from a generating function and recursive function
I have an assignment problem that i have been fighting with for a while now.. I have this recursive function: [imath]a_n=\begin{cases} 3,&\text{if }n=0\\ 5,&\text{if }n=1\\ 4a_{n-1}-4a_{n-2},&\text{if }n\ge 2\;. \end{cases}[/imath] We define the generating function [imath]P(n)=\sum_{n=0}^\infty (a_n+a_{n+1})x^n[/imath] Now I need to use the definition of [imath]a_n[/imath] in the recursive function to show that [imath]P(n)=\frac{8-19x}{(1-2x)^2}[/imath] I cant really get to this result, and I have been trying all sorts of things by now, nothing really leading me anywhere.. I hope some of you can help me! Thanks |
2995771 | Let [imath]G = (V, E)[/imath] be a simple undirected graph with [imath]n = |V | \ge 1[/imath] vertices.
Let [imath]G = (V, E)[/imath] be a simple undirected graph with [imath]n = |V | \ge 1[/imath] vertices. A subset [imath]U \subseteq V[/imath] of the vertices is called a vc-set if for every edge [imath]\{i, j\} \in E[/imath] either [imath]i \in U[/imath] or [imath]j \in U[/imath] (or both). Let [imath]U^∗[/imath] be a vc-set for [imath]G[/imath] that has the smallest size possible (meaning there does not exist any other vc-set [imath]U'[/imath] such that ([imath]|U'| < |U^∗|[/imath]). A subset [imath]W \subseteq V[/imath] of the vertices of [imath]G[/imath] is called an in-set if for all [imath]i, j \in W, \{i, j\} \not\in E[/imath]. Let [imath]W^∗[/imath] be a largest possible in-set in [imath]G[/imath] (meaning there is no other in-set, [imath]W′[/imath], such that [imath]|W^∗| < |W^{'}|[/imath]). Prove that [imath]|W^∗| = n − |U^∗|[/imath]. I tried to use the basics of graph theory but was of no use as I got stuck and back to the starting point. If you could help me I would be really grateful thanks a ton! | 2987919 | Let G = (V, E) be a simple undirected graph with n = |V | ≥ 1 vertices...
Let G = (V, E) be a simple undirected graph with n = |V | ≥ 1 vertices. A subset U ⊆ V of the vertices is called a vc-set if for every edge {i, j} ∈ E either i ∈ U or j ∈ U (or both). Let [imath]U^∗[/imath] be a vc-set for G that has the smallest size possible (meaning there does not exist any other vc-set [imath]U^{'}[/imath] such that (|[imath]U^{'}[/imath]| < |[imath]U^∗[/imath]|). A subset W ⊆ V of the vertices of G is called an in-set if for all i, j ∈ W, {i, j} [imath]\notin[/imath] E. Let [imath]W^∗[/imath] be a largest possible in-set in G (meaning there is no other in-set, [imath]W^{'}[/imath], such that |[imath]W^∗[/imath]| < |[imath]W^{'}[/imath]|). Prove that |[imath]W^*[/imath]| = n − |[imath]U^∗[/imath] |. I tried to use the basics of graph theory but was of no use as I got stuck and back to the starting point. If you could help me I would be really grateful thanks a ton! |
2996626 | Sum of fractions Inequality
Prove that: [imath] {a\over b} + {b \over c} + {c \over a} \ge 3 [/imath] Assuming [imath]a, b, c > 0[/imath]. I was able to prove that this is true: [imath] {a\over b} + {b \over a} \ge 2 [/imath] by just rearranging it to get: [imath] a^2 + b^2 > 2ab [/imath] I figured if I would then repeat this with a,c and then b,c; I could sum the three inequalities to get closer to the third; so I get to: [imath] {a\over b} + {b \over c} + {c \over a} + {b\over a} + {c \over b} + {a \over c} \ge 6 [/imath] Now because of the symmetry; it certainly seems like the top inequality should be true; but I'm having trouble arguing past this point. | 770341 | Show [imath]\frac{b}{c} + \frac{c}{a} + \frac{a}{b} \ge 3[/imath] for [imath]a,b,c > 0[/imath].
Disclaimer: The statement may be false, but for now I'm operating under the assumption it's true and trying to prove it. My workings: I got a common denominator and expressed it as: [imath]\frac{ab^2 + bc^2 + ca^2}{abc} \ge 3[/imath] [imath]ab^2 + bc^2 + ca^2 \ge 3abc[/imath] This is where I'm stuck. I feel as if I keep encountering this same result in various different problems and always am stumped on how to show it is true. I tried to look at similar problems and it seems AM-GM inequality is the usual line of attack, so using that I reasoned, [imath]a^2 + b^2 + c^2 \ge 3abc[/imath], and now I need to establish [imath]ca^2 + ab^2 + bc^2 \ge a^2 + b^2 + c^2[/imath] However, this doesn't seem like it'd be true for general positive [imath]a,b,c[/imath]. |
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