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685812 | Why is a quotient mapping not necessarily open?
I read that a quotient mapping is not necessarily open. I wonder why that is. Say we have a quotient mapping [imath]f[/imath] between [imath](X,\mathfrak{A})[/imath] and [imath](Y,\mathfrak{B})[/imath]. Let us take an open set [imath]O\subset X[/imath]. If [imath]f(O)\neq\emptyset_Y[/imath], then [imath]f(O)[/imath] is open in [imath]Y[/imath]. If [imath]f(O)=\emptyset_Y[/imath], then, as [imath]\emptyset_Y[/imath] is open in [imath]Y[/imath], we still have an open mapping. Thanks in advance! | 655797 | Example of quotient mapping that is not open
I have the following definition: Let ([imath]X[/imath],[imath]\mathcal{T}[/imath]) and ([imath]X'[/imath], [imath]\mathcal{T'}[/imath]) be topological spaces. A surjection [imath]q: X \longrightarrow X'[/imath] is a quotient mapping if [imath]U'\in \mathcal{T'} \Longleftrightarrow q^{-1}\left( U'\right) \in \mathcal{T} \quad \text{i.e. if } \mathcal{T'}=\{ U' \subset X' : q^{-1}\left( U' \right) \in \mathcal{T} \}[/imath] and the properties: [imath]q[/imath] is a bijective quotient mapping [imath]\Leftrightarrow[/imath] [imath]q[/imath] is a homeomorphism In general, [imath]q[/imath] quotient [imath]\not \Rightarrow q[/imath] open. If [imath]U \in \mathcal{T}, q(U)\subset X'[/imath] is open if [imath]q^{-1}\left( q\left( U \right) \right) \in \mathcal{T}[/imath] but not in general. I could not find an example of quotient mapping for which [imath]q^{-1}\left( q\left( U \right) \right)[/imath] is not open. I would understand the idea better if you could show me one. |
685891 | Periodic sequence
[imath](x_n)_n[/imath] is a sequence defined by the relation: [imath]x_{n+2}=|x_{n+1}-x_{n-1}|[/imath] for [imath]n\geq1[/imath] and [imath]x_0,x_1,x_2[/imath] are non-negative integers, not all three equal 0. I think this sequence is periodic, so here is my question: is it true and if yes which is the period? Thank you! | 653531 | Sequence is periodic [imath]x_{n+2}=|x_{n+1}-x_{n-1}|[/imath]
How to show that the sequence [imath]x_n, n \geq 0, x_{n+2}=|x_{n+1}-x_{n-1}|, n \geq 1[/imath] with [imath]x_0, x_1, x_2[/imath] positive integers, not all null, is periodic? I tried to pick up the square but obtained equalities not helped. Thanks so much for any suggestion. |
686420 | How to use Parseval' identity( Plancherel)?
(May be this is very basic question for MO) Let [imath]f\in L^{2} (\mathbb R)[/imath] with [imath]\lim_{t\to \pm \infty} f(t)=0.[/imath] Put [imath]F_{n} (x)= \frac{1}{2\pi} \int_{-n}^{n}e^{itx} f(t) dt \ (n=1,2,...)[/imath] Fix [imath]\alpha \in (0, \infty)[/imath] and we define [imath]H_{n}(x)[/imath] as follows: [imath]\frac{1}{2\pi}\int_{-n}^{n} e^{itx} (f(t+\alpha) -f(t-\alpha))dt = (e^{-i\alpha x}- e^{i\alpha x})F_{n}(x) + H_{n}(x)[/imath] My Question: Can we expect to prove: [imath]H_{n}(x) \to 0[/imath] as [imath]n\to \infty[/imath] in [imath]L^{2}(\mathbb R)[/imath] ? I guess some where we need to use Parseval' identity( Plancherel); but I am bit confused, how to use it. Thanks | 685680 | How to use Parseval' s( Plancherel' s) identity?
Let [imath]f\in L^{2} (\mathbb R)[/imath] with [imath]\lim_{t\to \pm \infty} f(t)=0.[/imath] Put, [imath]F_{n} (x)= \frac{1}{2\pi} \int_{-n}^{n}e^{itx} f(t) dt, \ (n=1,2,...).[/imath] Fix [imath]\alpha \in (0, \infty)[/imath] and we define [imath]H_{n}(x)[/imath] as follows: [imath]\frac{1}{2\pi}\int_{-n}^{n} e^{itx} (f(t+\alpha) -f(t-\alpha))dt = (e^{-i\alpha x}- e^{i\alpha x})F_{n}(x) + H_{n}(x).[/imath] My Question: How to show [imath]H_{n}(x) \to 0[/imath] as [imath]n\to \infty[/imath] in [imath]L^{2}(\mathbb R)[/imath] ? (I guess some where we need to use Parseval' identity( Plancherel); but I am bit confused, how to use it.) Edit: @LutzL, pointed out below, and in MO, for the same, I have tried little in that direction: [imath]H_{n}(x)= \frac{1}{2\pi}\int_{-n}^{n} e^{itx} (f(t+\alpha) -f(t-\alpha))dt -(e^{-i\alpha x}- e^{i\alpha x}) \frac{1}{2\pi} \int_{-n}^{n}e^{itx} f(t) dt[/imath] we may write, [imath]H_{n}(x)= \frac{1}{2\pi} e^{-i\alpha x} \int_{-n+\alpha}^{n+\alpha}e^{itx} f(t) dt- \frac{1}{2\pi} e^{i\alpha x} \int_{-n-\alpha}^{n-\alpha} e^{itx} f(t) dt-(e^{-i\alpha x}- e^{i\alpha x}) \frac{1}{2\pi} \int_{-n}^{n}e^{itx} f(t) dt;[/imath] Or, we may write, [imath]H_{n}(x)= \frac{1}{2\pi} e^{-i\alpha x} \int_{-n+\alpha}^{n+\alpha}e^{itx} f(t) dt- \frac{1}{2\pi} e^{i\alpha x} \int_{-n-\alpha}^{n-\alpha} e^{itx} f(t) dt- \frac{1}{2\pi}\int_{-n-\alpha}^{n+\alpha} e^{itx} f(t+\alpha) dt + \int_{-n+\alpha}^{n+\alpha} e^{itx} f(t-\alpha) dt= I_{1}-I_{2}-I_{3}+I_{4};[/imath] But then,(using one of the above form), we need to show: [imath]\lim_{n\to \infty}\int_{\mathbb R} |H_{n}(x)|^{2} dx=0.[/imath] Now If I interchange the limit and integral (this may not be possible, here, I am doing without rigours), and then I can take limit inside the square, and then after taking limit, each integrals, [imath]I_{i}, (i=1,2,3,4)[/imath], will be becomes over [imath]\mathbb R[/imath]; then I think, it will cancels each other; Is this o.k or am I missing something, still, I don't see, where we need to use asymptotic behaviours of [imath]f[/imath] (and what does it mean ?, here is [imath]f[/imath] is arbitrary member of [imath]L^{2}[/imath]) ? I am wondering, is it possible to get solution without, Plancherel or parsevel somewhere ? |
687253 | How do the first three axioms of an abelian category imply that hom-sets are enriched over the monoidal category of abelian groups?
According to this article in Wikipedia the following first three axioms in the definition of of an abelian category imply that hom-sets are enriched over the monoidal category of Abelian groups: existence of a zero object existence of binary products and binary coproducts existence of all kernels and cokernels (see this image for a saved copy of the relevant section) My intuition is that it should follow from a diagram like this: [imath]A\to A\otimes A \to B \oplus B \to B[/imath] where the first map is the diagonal map, the second is the product of the two maps one wishes to sum, i.e. [imath]\phi\times \psi[/imath] and the third map is the sum. But how could I use the first and third axioms to prove the existence of the diagonal and sum maps? | 625477 | I've read that abelian categories can naturally be enriched in [imath]\mathbf{Ab}.[/imath] How does this actually work?
Wikipedia defines the notion of an abelian category as follows (link). A category is abelian iff it has a zero object, it has all binary products and binary coproducts, and it has all kernels and cokernels. all monomorphisms and epimorphisms are normal. It later explains that an abelian category can naturally be enriched in [imath]\mathbf{Ab},[/imath] as a result of the first three axioms above. How does this actually work? |
687413 | How does [imath]\frac{\sigma(n)}{n}[/imath] behave as [imath]n[/imath] tends to infinity?
Is [imath]\frac{\sigma(n)}{n}[/imath] unbounded as n tends to infinity? ([imath]\sigma(n)[/imath] is sum of proper divisors of [imath]n[/imath]) | 571751 | Is something similar to Robin's theorem known for possible exceptions to Lagarias' inequality?
Robin's theorem says that if [imath]\sigma(n)<e^\gamma n\log\log n[/imath] holds for all [imath]n>5040[/imath], where [imath]\sigma(n)[/imath] is the sum of divisors of [imath]n[/imath], then the Riemann hypothesis is true, but if there are any counterexamples, then they are colossally abundant numbers, and there are infinitely many counterexamples. Lagarias' theorem says [imath]\sigma(n)\leq H_n+e^{H_n}\log(H_n)[/imath] holding for all natural numbers [imath]n[/imath] is equivalent to the Riemann hypothesis, where [imath]H_n[/imath] is the [imath]n[/imath]th harmonic number, the sum of the reciprocals of the first [imath]n[/imath] positive integers. It looks like Lagarias' inequality is sharper than Robin's, so any counterexamples to Robin's inequality must also be counterexamples to Lagarias'. Is known to be impossible for integers that aren't colossally abundant numbers or of some similar more general type to be exceptions to Lagarias' inequality? |
687728 | Proof of de Morgans' law: For any sets [imath]X[/imath] and [imath]Y[/imath], [imath]\overline{X\cup Y}= \overline{X}\cap\overline{Y}[/imath]
For any sets [imath]X[/imath] and [imath]Y[/imath], [imath]\overline{X\cup Y}= \overline{X}\cap\overline{Y}[/imath] [Sorry, the hat should be a horizontal bar ] Same as the last two questions - this set theory stuff just isn't my walk in the park. Last question in my proofs homework, similar to both other questions I posted, not exactly the same How would I approach it? | 687695 | Prove/Disprove: For any sets [imath]X[/imath] and [imath]Y[/imath], [imath]\overline{X\cap Y} = \bar{X}\cup\bar{Y}[/imath]
Prove/Disprove: For any sets [imath]X[/imath] and [imath]Y[/imath], [imath]\overline{X\cap Y} = \bar{X}\cup\bar{Y}[/imath] Extra question in my proofs homework, similar to another question I posted, but not exactly the same I don't think. How would I start with this? |
689325 | Solve for x in [imath]\tan x=2x[/imath]
Is there a way to solve for [imath]x[/imath] for [imath]\tan x=2x[/imath]? My cousin asked me about this and I wondering if there was some sort of trig identity I had to know. | 302362 | Solve [imath]x = \frac{1}{2}\tan(x)[/imath]
I did this using trial and error, but I was just wandering if there is an algebraic way of solving this? I thought about double angle formula but that doesn't work properly does it? I then tried writing it in the form [imath]x \cdot 2 \cot(x) = 1[/imath] but even then, I can't solve it properly by re-writing [imath]\cot(x)[/imath] because of the [imath]x[/imath] outside the brackets. Anybody know how I would do this algebraically? |
689625 | Is [imath]\left((-1)^2\right)^\frac12 = (-1)^\left(2\cdot\frac12\right)[/imath]?
I'm feeling confused. If I square 1 and -1, the answers should be equal: [imath]1^2 = (-1)^2[/imath] Then I take both sides to the power of [imath]\frac12[/imath]: [imath]\left(1^2\right)^\frac12 = \left((-1)^2\right)^\frac12[/imath] This next step seems to make sense according to the simple arithmetic rule about multiplying exponents: [imath]1^\left(2\cdot\frac12\right) = (-1)^\left(2\cdot\frac12\right)[/imath] And then comes the weirdness: [imath]1^\left(\frac22\right) = (-1)^\left(\frac22\right)[/imath] [imath]1^1 = (-1)^1[/imath] [imath]1 = -1[/imath] Obviously I did something wrong... Every step seems perfectly reasonable except going from step 2 to step 3. That seems reasonable too, that's what I was taught about exponents, but that's the only step which I could conceive has special constraints I violated. Is [imath]\left((-1)^2\right)^\frac12 = (-1)^\left(2\cdot\frac12\right)[/imath]? | 664383 | What is the condition for example [imath](number^c)^b=number^{cb}[/imath] to be true?
What is the condition for [imath](\text{number}^c)^b=\text{number}^{cb}[/imath] to be true? I can't find it on google.ze so I asked it here on math.stackexchange.com. |
687715 | If [imath]|A| > \frac{|G|}{2} [/imath] then [imath]AA = G [/imath]
I'v found this proposition. If [imath]G[/imath] is a finite group , [imath] A \subset G [/imath] a subset and [imath]|A| > \frac{|G|}{2} [/imath] then [imath]AA = G [/imath]. Why this is true ? | 1105137 | Show that [imath]S^2=G[/imath].
I try to solve a question from Grove's Algebra book. The question is that: Suppose [imath]S[/imath] is a subset of a finite group [imath]G[/imath], with [imath]|S| > \dfrac{|G|}{2}[/imath]. If [imath]S^2[/imath] is defined to be [imath]\{xy:x, y \in S\}[/imath] show that [imath]S^2 = G[/imath] Solution should be easy but i couldn't solve it. Thanks for your help |
690294 | Why can't the base of a logarithm be negative?
I understand why the base of a logarithm can't be 0 or 1, but why negative? What I found out is that when the base is negative we get imaginary results when the powers are rational numbers with odd denominators, for example: [imath]-2^{\frac{1}{2}}[/imath] is undefined, thus the logarithm is not defined either? Is this true? and if it's, is that the only reason? Thanks. | 690024 | Why must the base of a logarithm be a positive real number not equal to 1?
Why must the base of a logarithm be a positive real number not equal to 1? and why must [imath]x[/imath] be positive? Thanks. |
690679 | if [imath]x^2+ax+b=0[/imath] has an integer root, show that it divides b
I don't know where to start. can anyone help me please ? if [imath]x^2+ax+b=0[/imath] has an integer root, show that it divides b | 205196 | finding rational roots of polynomials
Could someone please explain me how to apply the Descartes's Criterion? For example , how do I find the rational roots of [imath] x^3 -x +1[/imath]. I've been looking at some examples, but I get confused. |
690791 | Classification of Rings
I am trying to classify rings of order 10. I believe the only possible ring is [imath]\mathbb{Z_{10}}[/imath]. Thus I am trying to find a map from my ring [imath]R[/imath] to [imath]\mathbb{Z_{10}}[/imath]. The most obvious map [imath]f: R \rightarrow \mathbb{Z_{10}}[/imath] such that [imath]f(1)=1[/imath] and [imath]f(n)=[n]_{10}[/imath]. I was able to prove that it was a homomorphism and surjective. However, I am stuck at proving injectivity. Can anyone please give me some hints | 110271 | Ring with 10 elements is isomorphic to [imath]\mathbb{Z}/10 \mathbb{Z} [/imath]
How do I prove that a finite ring [imath]R[/imath] of order 10 is isomorphic to the ring [imath]\mathbb{Z}/10 \mathbb{Z}[/imath]? I know that as a group under addition, [imath](R,+)[/imath] is isomorphic to the group [imath](\mathbb{Z}/10 \mathbb{Z}, + )[/imath], but the multiplication is rather mysterious to me. |
691042 | [imath]\displaystyle\lim_{x\to\infty}(f(x)-f(-x))=0[/imath] show that [imath]f'(x_0)=0[/imath]
Let [imath]f:\mathbb R\to\mathbb R[/imath] [imath]f[/imath] is derivable in all of [imath]\mathbb R[/imath]. Let [imath]\displaystyle\lim_{x\to\infty}(f(x)-f(-x))=0[/imath] Show that there exists [imath]x_0\in\mathbb R[/imath] such that [imath]f'(x_0)=0[/imath]. From intuiton I see that [imath]f[/imath] is an even function because [imath]f(x)=f(-x)[/imath] so it can either be like for example [imath]f(x)=x^2[/imath] which has an extramum point where [imath]f'(x_0)=0[/imath] or [imath]f(x)=c, \ c\in\mathbb R[/imath] which is also an even function and the derivative is always 0. I'm not really sure how to formally prove it. Note: we can't use integrals. | 688635 | formal proof from calulus
Given [imath]f:R \to R[/imath], [imath]f[/imath] is differentiable on [imath]R[/imath] and [imath]\lim_{x \to \infty}(f(x)-f(-x))=0[/imath]. I need to show that there is [imath]x_0 \in R[/imath] such that [imath]f'(x_0)=0[/imath] I am trying to prove it by contradiction .... so i assume there is no [imath]x_0 \in R[/imath] such that [imath]f'(x_0)=0[/imath] this means this function is strictly monotonic because the derivative "respects" the mean value theorem so it cannot be negative and positive without passing a zero value .... (from what i understand the derivative function of a continuous/differentiable functions isn't necessarily continuous but it still respect the mean value theorem i think it's called darboux's theorem ) now from [imath]\lim_{x \to \infty}(f(x)-f(-x))=0[/imath] ,we get [imath]\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)[/imath] now the last step is the contradiction its very intuitive and even maybe obvious that we can't get [imath]\lim_{x \to \infty}f(x)=\lim_{x \to \infty}f(-x)[/imath] for strictly monotonic function but i cant think of a formal way to prove it . |
690064 | Converge of Random Variables Using Characteristic Function
Assume [imath]X_n[/imath] Converges in distribution to [imath]X[/imath] and [imath]a_n \rightarrow a[/imath] ,[imath]b_n \rightarrow b[/imath] where [imath]a_n[/imath] and [imath]b_n[/imath] are deterministic sequence, prove [imath]a_nX_n+b_n \rightarrow aX+b[/imath] using characteristic function? I have started this question as the following but I am not sure [imath] \phi_{a_nX_n+b_n}(t) = E\{e^{it(a_nX_n+b_n)}\} = e^{itb_n} E\{e^{ita_nX_n}\} \rightarrow e^{itb} E\{e^{itaX}\} = \phi_{aX+b}(t) [/imath] how can I show that [imath]a_nX_n[/imath] converges in distribution to [imath]aX[/imath]? | 317706 | If [imath]X_n \stackrel{d}{\to} X[/imath] and [imath]c_n \to c[/imath], then [imath]c_n \cdot X_n \stackrel{d}{\to} c \cdot X[/imath]
Let [imath]X_n[/imath], [imath]X[/imath] random variables on a probability space [imath](\Omega,\mathcal{A},\mathbb{P})[/imath] and [imath](c_n)_n \subseteq \mathbb{R}[/imath], [imath]c \in \mathbb{R}[/imath] such that [imath]c_n \to c[/imath] and [imath]X_n \stackrel{d}{\to} X[/imath]. Then [imath]c_n \cdot X_n \stackrel{d}{\to} c \cdot X[/imath]. Here is my proof: By Lévy's continuity theorem it suffices to show that [imath]\forall \xi: \quad \hat{\mu}_n(c_n \cdot \xi) \to \hat{\mu}(c \cdot \xi) \qquad (n \to \infty)[/imath] where [imath]\hat{\mu}_n[/imath] (resp. [imath]\hat{\mu}[/imath]) denotes the characteristic function of [imath]X_n[/imath] (resp. [imath]X[/imath]). Using the tightness of the distributions one can show that [imath]\{\hat{\mu}_n; n \in \mathbb{N}\}[/imath] is uniformly equicontinuous. We have [imath]|\hat{\mu}_n(c_n \cdot \xi) - \hat{\mu}(c \cdot \xi)| \leq |\hat{\mu}_n(c_n \cdot \xi)- \hat{\mu}_n(c \cdot \xi)| + |\hat{\mu}_n(c \cdot \xi)- \hat{\mu}(c \cdot \xi)|[/imath] The first addend converges to zero as [imath]n \to \infty[/imath] by the uniform equicontinuity and the second one converges to zero since [imath]X_n \stackrel{d}{\to} X[/imath]. I was wondering whether there is an easier proof - my own proof seems rather like overkill to me. Thanks! |
691451 | Suppose p(x) is a polynomial with integer coefficients. Show that if p(a) =1 for some integer a then p(x) has at most two integer roots.
Problem : Suppose [imath]p(x)[/imath] is a polynomial with integer coefficients. Show that if [imath]p(a) =1[/imath] for some integer [imath]a[/imath] then [imath]p(x)[/imath] has at most two integer roots. Let [imath]p(x)[/imath] be a cubic polynomial such that [imath]p(x) =ax^3+bx^2+cx +d =0 [/imath] Let [imath]p(\alpha) =a\alpha^3 +b \alpha^2 + c\alpha +d =1 [/imath] ( as per the question) . where [imath]a,b,c[/imath] are integers. Now how to prove the given problem please suggest thanks..... | 680102 | Polynomial [imath]p(a) = 1[/imath], why does it have at most 2 integer roots?
The question that I am trying to answer is : Suppose is [imath]p(x)[/imath] is a polynomial with integer coefficients. Show that if [imath]p(a) = 1[/imath] for some integer a then [imath]p(x)[/imath] has at most two integer roots. I have no idea how to get started. Any help would be awesome! |
691542 | Infinetly many primes of form [imath]4k+3[/imath]
Prove that there are infinitely many primes of the form [imath]4k + 3[/imath] (where [imath]k[/imath] is an integer). Note that it is a special case of "Theorem 6 (Dirichlet). Let a and b be positive coprime integers. Then the sequence [imath]b[/imath], [imath]b + a[/imath], [imath]b + 2a[/imath], [imath]b + 3a[/imath], [imath]b + 4a[/imath], [imath]b + 5a[/imath], ....," contains infinitely many prime numbers So far I got that suppose there are a finite number of primes [imath]p......p[/imath] and if [imath]4(p.....p)+3[/imath] is prime it's a contradiction so the initial statement is proven? | 21333 | Help understand the proof of infinitely many primes of the form [imath]4n+3[/imath]
This is the proof from the book: Theorem. There are infinitely many primes of the form [imath]4n+3[/imath]. Lemma. If [imath]a[/imath] and [imath]b[/imath] are integers, both of the form [imath]4n + 1[/imath], then the product [imath]ab[/imath] is also in this form. Proof of Theorem: Let assume that there are only a finite number of primes of the form [imath]4n + 3[/imath], say [imath]p_0, p_1, p_2, \ldots, p_r.[/imath] Let [imath]Q = 4p_1p_2p_3\cdots p_r + 3.[/imath] Then there is at least one prime in the factorization of [imath]Q[/imath] of the form [imath]4n + 3[/imath]. Otherwise, all of these primes would be of the form [imath]4n + 1[/imath], and by the Lemma above, this would imply that [imath]Q[/imath] would also be of this form, which is a contradiction. However, none of the prime [imath]p_0, p_1,\ldots, p_n[/imath] divides [imath]Q[/imath]. The prime [imath]3[/imath] does not divide [imath]Q[/imath], for if [imath]3|Q[/imath] then [imath]3|(Q-3) = 4p_1p_2p_3\cdots p_r,[/imath] which is a contradiction. Likewise, none of the primes [imath]p_j[/imath] can divides [imath]Q[/imath], because [imath]p_j | Q[/imath] implies [imath]p_j | ( Q - 4p_1p_2\cdots p_r ) = 3[/imath], which is absurd. Hence, there are infinitely many primes of the form [imath]4n +3[/imath]. END From "however, none of the prime ...." to the end, I totally lost! My questions: Is the author assuming [imath]Q[/imath] is prime or is not? Why none of the primes [imath]p_0, p_1,\ldots, p_r[/imath] divide [imath]Q[/imath]? Based on what argument? Can anyone share me a better proof? Thanks. |
414309 | For subspaces, if [imath]N\subseteq M_1\cup\cdots\cup M_k[/imath], then [imath]N\subseteq M_i[/imath] for some [imath]i[/imath]?
I have a vector space [imath]V[/imath] over a field of characteristic [imath]0[/imath]. If [imath]M_1,\dots,M_k[/imath] are proper subspaces of [imath]V[/imath], and [imath]N[/imath] is a subspace of [imath]V[/imath] such that [imath]N\subseteq M_1\cup\cdots\cup M_k[/imath], how can you tell [imath]N\subseteq M_i[/imath] for some [imath]i[/imath]? I was first trying to show it just in the case [imath]N\subseteq M_1\cup M_2[/imath], and hoping to extend it to finitely many [imath]M_i[/imath]. If either of the [imath]M_i[/imath] contains the other, the claim follows, so I suppose neither [imath]M_i[/imath] contains the other. In hopes of a contradiction, I suppose [imath]N\not\subseteq M_1[/imath] and [imath]N\not\subseteq M_2[/imath], Picking [imath]x_1\in N\setminus M_1[/imath] and [imath]x_2\in N\setminus M_2[/imath], I'd have [imath]x_1+x_2\in N\subseteq M_1\cup M_2[/imath]. The only thing I could conclude was that actually [imath]x_1,x_2\notin M_1[/imath] and [imath]x_1,x_2\notin M_2[/imath], which seems like a dead end. What's the right way to do this? | 1838023 | If a vector subspace is in a union of other subspaces, then it's contained in one of them
Problem: Let [imath]V[/imath] be a finite dimensional vector space and [imath]V_1,\ldots,V_n\subset V[/imath] vector subspaces. Show that if [imath]W\subset V[/imath] is a vector subspace and [imath]W\subset V_1\cup\cdots\cup V_n,[/imath] then [imath]W\subset V_k[/imath] for some [imath]1\le k\le n[/imath]. I get the intuitive idea of why this should hold by drawing pictures, but I can't prove it rigorously. I know that if [imath]V_1,V_2[/imath] are two subspaces and [imath]V_1\cup V_2[/imath] is also a subspace, then [imath]V_1\subset V_2[/imath] or [imath]V_2\subset V_1[/imath]. Does that help? |
692332 | Mary needs to go to work
This was on my combinatorics exam. I definitely didn't answer it correctly, but I'll explain what I did. Mary has to go to work or some bs. She needs to travel 6 blocks east and 7 blocks north. How many different routes can she take? Honestly, Mary, pick one route and stick to it. You don't need to know all the different ways you can walk to the same place. Then they throw a wrench in the gears and tell you she has to go 2 blocks east and 4 blocks north first before she takes her daughter to preschool, and then go to work. How many ways can she do this? First I tried 13! which seemed too large. Then I started thinking like a decision tree. I tried to imagine the scale of the tree in considering my answer. I was somehow led to thinking in terms of binary decisions. So I ended up writing [imath]2^{13}[/imath] because I figured she had 2 decisions to make at each block, but clearly, that's just not true. By my incorrect logic, she could go north 13 times. | 534735 | Possible ways to walk to school
I am not sure how to approach this problem. Every day, a dance student walks from her home to dance-school, which is located [imath]12[/imath] blocks east and [imath]16[/imath] blocks north from home. She always takes the shortest walk of [imath]28[/imath] blocks. [imath]a[/imath]) How many different walks are possible? [imath]b[/imath]) Suppose that [imath]5[/imath] blocks east and [imath]6[/imath] blocks north of her home lives her dance friend, whom she meets each day on her way to dance school. Now how many walks are possible? For [imath]a[/imath] I was thinking [imath]P(28,12)\cdot P(28,16)[/imath], but not sure if that would work because you can't walk the furthest block north before the closest block north. And for [imath]b[/imath] I'm not really sure where to start. |
286855 | triangle free graph [imath] \sum_{v \in V(G)} deg ^{2} v \leq mn [/imath]
I am studying for an exam in a graph theory class. Here is the exercise I need some help with. Let [imath]G[/imath] a triangle free graph with [imath]n[/imath] vertices and [imath]m[/imath] edges. (a) Prove that there exists vertices [imath]x,y[/imath] in [imath]G[/imath] such that [imath] \displaystyle{ \text{deg(x)} + \text{ deg(y)} \leq n}[/imath]. (b) Prove that: [imath]\displaystyle{ \sum_{v \in V(G)} \text{ deg } ^{2} v \leq mn }[/imath] I have done (a) but I have some trouble with (b). | 218292 | Proof about Graph with no triangle
For graph G without a triangle(with no triangle) with n vertices and m edges Prove that: 1)[imath] \forall xy \in E , d(x)+d(y) \le n[/imath] 2)[imath](\sum_{v\in_V}^{ }d(v)^2)\le mn[/imath] |
602988 | Finding max values
I have been using Manipulate[Plot[{(LogIntegral[x])^(1/2), (((x*E^s)/Log[x*E^s]) ((((Log[Log[x*E^s]])^(w - 1))/((w - 1)!))))/ RiemannR[x]}, {x, 2, 5000}, PlotStyle -> {Blue,Red}, ImageSize -> 700], {w, 33.34, 40, 0.01}, {s, 43.2, 50, 0.01}] in Mathematica to play with the plot I am trying to find the max values of w and s where the red curve is at no point greater in value that the blue curve. Is there a better way of doing this? What would the mathematical approach be? I am guessing I will have to find the derivative of [imath] {{\rm Li}^{1/2}\left(x\right)\,y \left[\log\left(\log\left(y\right)\right)\right]^{z} \over {\rm R}\left(x\right)\log\left(y\right)\left(z!\right)}[/imath] where [imath]{\rm R}[/imath] is the Riemann counting function, but am at a bit of a loss as to how to proceed. (I have started w and s at a reasonable estimate - but when values become much higher than 1000, manipulation is not really feasible.) NB My best guess so far at the relationship between s and w is s=[N[Log[((w/5) + 1)!]], but this is clearly way off. | 602816 | Finding max value
Which approach would be the best to take in order to calculate the max value of [imath]a[/imath], where [imath]a\ \log(x+1)[/imath] (blue) at no point exceeds [imath]\sqrt{x}[/imath] (red)? |
269159 | Linear algebra, modules annihilator
[imath]N[/imath] and [imath]K[/imath] are submodules of [imath]M[/imath] with [imath]I=Ann(N)[/imath] and [imath]J=Ann(K)[/imath], then show that annihilator of intersection of [imath]N[/imath] and [imath]K[/imath] contains [imath]I+J[/imath]. Give example to show that the inclusion may be strict. | 268075 | Let [imath]N[/imath], [imath]K[/imath] be sub-modules of [imath]M[/imath] with [imath]I=\mathrm{Ann}(N)[/imath], [imath]J=\mathrm{Ann}(K)[/imath]. Show [imath]I+J[/imath] is a proper subset of [imath]\mathrm{Ann}(N \cap K)[/imath].
Let [imath]N[/imath] and [imath]K[/imath] be sub-modules of [imath]M[/imath] with [imath]I=\operatorname{Ann}(N)[/imath] and [imath]J=\operatorname{Ann}(K)[/imath]. Show that [imath]I+J[/imath] is a proper subset of [imath]\operatorname{Ann}(N \cap K)[/imath]. |
693348 | Proof question using Rolle's Theorem
Let [imath]f:\mathbb R \rightarrow \mathbb R[/imath] have derivatives of all orders. Suppose that [imath]a<b[/imath] and that [imath]f(a) = f(b) = f'(a) = f'(b) = 0[/imath]. Prove that [imath]f'''(c) = 0[/imath] for some [imath]c \in (a,b)[/imath] [imath]f(a) = f(b)[/imath] and since f is continuous and differentiable, we can use Rolle's Theorem. [imath]\exists c \in (a,b)[/imath] such that [imath]f'(c) = 0[/imath] So surely, differentiating twice more, [imath]f'''(c) = 0[/imath] for some [imath]c \in (a,b)[/imath]? Or am I misunderstanding something? Why would the question ask for [imath]f'''[/imath] ? Thanks in advance. | 692828 | Rolle's Theorem and the Mean Value Theorem
Let [imath]f\colon \mathbb{R} \to \mathbb{R}[/imath] have derivatives of all orders. Suppose that [imath]a < b[/imath] and that [imath]f(a)=f(b)=f'(a)=f'(b) =0[/imath]. Prove that [imath]f'''(c) = 0[/imath] for some [imath]c[/imath] in [imath](a,b)[/imath]. |
694273 | Prove that [imath]-1 \cdot x=-x[/imath]
While working on a proof for class, I came to a point where I couldn't go any further without knowing that [imath]-1 \cdot x=-x[/imath]. Is there a way to prove this using the axioms of a field? | 118033 | Proving that [imath]-a=(-1)\cdot a[/imath]
As the title reveals, I want to prove (based on the axioms of field) that [imath]-a=(-1)\cdot a[/imath] I've been trying for a while now, but couldn't think of a way to do it and it got me thinking that maybe its does not require a proof, but it doesn't feel right. Any ideas? |
693474 | If [imath]x=1\mod3[/imath] and [imath]x=0\mod2[/imath], what is it [imath]\mod6[/imath]?
If you know what a number mod two different primes is (3 and 2) in this case, how can you tell what the mod is of the two products? | 693441 | If [imath]x \equiv 1 \pmod 3[/imath] and [imath]x \equiv 0 \pmod 2[/imath], what is [imath]x \pmod 6[/imath]?
If you know what a number mod two different primes is (3 and 2) in this case, how can you tell what the mod is of the two products? |
694684 | How to intuivitely see that [imath]e^{i \pi}+1=0[/imath] is true?
Gauss is reported to have said that if Euler's formula [imath]e^{i \pi} + 1 = 0\,\![/imath] "was not immediately apparent to a student upon being told it, that student would never be a first-class mathematician." How is someone supposed to see that this is true? | 72113 | Has anyone talked themselves into understanding Euler's identity a bit?
What does the ratio of the circumference of a circle to its diameter have to do with the base of the natural logarithm and [imath]\sqrt{-1}[/imath]? |
694593 | Filling up space with irrational fractional parts
While trying to generalise a mechanics exercise with a friend, we came up with this question, in an attempt to understand wether sine curves with irrational period defined inside an annulus will end up 'filling up' the space in the annulus. Mathematically, let [imath]x \in \mathbb{R} \setminus \mathbb{Q}[/imath]: Is [imath]S_x=\{\mathrm{frac}(n x): n \in \mathbb{N}\}[/imath] dense in [imath][0,1] \subset\mathbb{R}[/imath]? This is equivalent to asking wether sequences of fractional parts of multiplies of one fixed irrational number can approximate every number in [imath][0,1][/imath]. My intuition says this may be true, but I see no reason why there shouldn't be some 'holes' in [imath]S[/imath]. If that's the case, which kind of properties of [imath]x[/imath] determine the how [imath]S_x[/imath] is filled up? | 272545 | Multiples of an irrational number forming a dense subset
Say you picked your favorite irrational number [imath]q[/imath] and looking at [imath]S = \{nq: n\in \mathbb{Z} \}[/imath] in [imath]\mathbb{R}[/imath], you chopped off everything but the decimal of [imath]nq[/imath], leaving you with a number in [imath][0,1][/imath]. Is this new set dense in [imath][0,1][/imath]? If so, why? (Basically looking at the [imath]\mathbb{Z}[/imath]-orbit of a fixed irrational number in [imath]\mathbb{R}/\mathbb{Z}[/imath] where we mean the quotient by the group action of [imath]\mathbb{Z}[/imath].) Thanks! |
694798 | Continuous function, mapping of a set to itself
Let [imath]f: [0,1] \rightarrow [0,1][/imath] be continuous. Any idea on how we can prove that it is not possible for [imath]f[/imath] to map [imath][0,1][/imath] onto [imath][0,1][/imath] exactly two-to-one. That is, there is no continuous [imath]f[/imath] as above such that for each [imath]y \in [0,1][/imath], there are exactly two values [imath]x_1[/imath] and [imath]x_2[/imath] such that [imath]y = f(x_1) = f(x_2)[/imath] | 677085 | For each [imath]y \in \mathbb{R}[/imath] either no [imath]x[/imath] with [imath]f(x) = y[/imath] or two such values of [imath]x[/imath]. Show that [imath]f[/imath] is discontinuous.
This is from Apostol's Mathematical Analysis. Let [imath]f[/imath] be a function defined in [imath][0, 1][/imath] such that for each real number [imath]y[/imath] either there is no value of [imath]x \in [0, 1][/imath] such that [imath]f(x) = y[/imath] or there are exactly two such values of [imath]x[/imath]. Prove that a) [imath]f[/imath] must be discontinuous on [imath][0, 1][/imath]. b) [imath]f[/imath] has an infinite number of discontinuities in [imath][0, 1][/imath]. Long back I had tried to solve the problem. I reasoned for a) as follows: Suppose that [imath]f[/imath] is continuous and let [imath]m, M[/imath] be its minimum and maximum values on [imath][0, 1][/imath]. Clearly if [imath]m = M[/imath] then [imath]f[/imath] is constant and clearly for every [imath]x \in [0, 1][/imath] we have [imath]f(x) = m = M[/imath] which contradicts given property of [imath]f[/imath]. So we must have [imath]m < M[/imath]. Clearly there are now four distinct points [imath]a, b, c, d[/imath] in [imath][0, 1][/imath] such that [imath]f(a) = f(b) = m, f(c) = f(d) = M[/imath]. By checking possible linear orderings of [imath]a, b, c, d[/imath] I could find a find a value of [imath]y \in (m, M)[/imath] for which there exist more than 2 values of [imath]x[/imath] such that [imath]f(x) = y[/imath]. However this approach looks very clumsy and I suspect much simpler and easier proof is possible. Part b) seemed bit complex and my only guess was that probably [imath]f[/imath] was discontinuous on each subinterval of [imath][0, 1][/imath] but I could not prove this claim (may be the claim is wrong!). Perhaps there is a simpler solution to this part of the problem. Apostol's also asks to provide an example of such a function. Any hints on the overall problem will be appreciated. Update: After looking at Marc's answer, I think I have found a way to show that there are infinitely many discontinuities. Earlier the solution was a part of the question itself but it made the question look quite lengthy hence I moved it to a separate answer. |
628909 | How prove this [imath]f'(c)=2c(f(c)-f(0))[/imath]
let [imath]f(x)[/imath] is continuous on [imath][0,1/2][/imath], and derivative on [imath](0,1/2)[/imath],such [imath]f'(1/2)=0[/imath] show that there exsit [imath]c\in(0,1/2)[/imath], such [imath]f'(c)=2c(f(c)-f(0))[/imath] My try: let [imath]F(x)=e^{-x^2}[f(x)-f(0)]\Longrightarrow F'(x)=e^{-x^2}[f'(x)-2x(f(x)-f(0))][/imath] then I can't | 575127 | Prove the existence of [imath]c[/imath] such that [imath]f'(c) = 2c(f(c) - f(0))[/imath]
Let [imath]f:\mathbb{R}\to\mathbb{R}[/imath] be a differentiable function s.t. [imath]f'[/imath] is continuous. Suppose [imath]f'\left(\frac{1}{2}\right)=0[/imath], prove that there is [imath]c\in\left(0,\frac{1}{2}\right)[/imath] s.t. [imath]f'(c)=2c(f(c)-f(0))[/imath] |
696136 | Composition of Functors which have adjoints has also an adjoint
The exercise is the following: Suppose [imath]F[/imath] has right adjoint [imath]G[/imath] and [imath]H[/imath] has right adjoint [imath]J[/imath] ([imath]F:\mathbb{A}\rightarrow\mathbb{B}[/imath] and [imath]H:\mathbb{B}\rightarrow\mathbb{C}[/imath] with [imath]\mathbb{A,B,C}[/imath] categories). Prove that [imath]GJ[/imath] is the right adjoint of [imath]HF[/imath]. What is the best way to start with? I want to do this by the two-way-rule, thus by bijective correspondence, but i don't see how to start. Someone who can help me? Thank you very much. | 682136 | Composition of adjoint functors
Does the composition of adjoint functors again form an adjunction? Say [imath]\langle F_1,G^1\rangle[/imath] is an adjunct pair between two categories A and B and [imath]\langle F_2,G^2\rangle[/imath] is also an adjoint pair between the categories B and C, then would [imath]\langle F_2\circ F_1,G^1\circ G^2\rangle[/imath] be an adjoint pair between the categories A and C? If not is there some way of obtain an adjunction given the assumptions made on the 4 functors, or by adding extra properties to the categories? Particularly, here I am assuming A is the category [imath]_RMod[/imath] for some (apriori) arbitrary ring. |
696317 | Finding a permutation [imath] \alpha [/imath] given [imath] \alpha^4 [/imath]
I have the following question: Find a permutation [imath]\alpha ∈ S_7 [/imath] such that [imath]\alpha^4 = (2 1 4 3 5 6 7)[/imath]. Is [imath]\alpha[/imath] unique? How should I go about this? I've tried a few different trial and error type solutions but to no avail. Thanks | 329593 | Finding a permutation, and number of, from powers of the permutation
Sorry for the vagueness of the title, I couldn't think of a better way to put it. I just wanted to run a couple of simple questions past SE to check my reasoning is correct etc. Find a permutation [imath]\alpha \in S_7[/imath] such that [imath]\alpha^4 = (2143567)[/imath] Since any permutation can be expressed as a product of disjoint cycles, [imath]\alpha[/imath] cannot be comprised of any disjoint transpositions as then [imath]\alpha^4[/imath] maps the elements of the transposition back to itself. [imath]\alpha[/imath] cannot have any disjoint [imath]3[/imath]-cycles, [imath]p_i[/imath] as then [imath]p_i^4 = p_i[/imath]. We can argue like this to see that [imath]\alpha[/imath] must be a [imath]7[/imath]-cycle. So since [imath]\alpha^4 = (2143567)[/imath] in [imath]\alpha[/imath] there are three other elements between the consecutive elements in [imath]\alpha^4[/imath]. So between elements [imath]2[/imath] and [imath]1[/imath] in [imath]\alpha[/imath] there are [imath]3[/imath] other elements, between [imath]1 [/imath] and [imath]4[/imath] in [imath]\alpha[/imath] there are [imath]3[/imath] other elements, there is only one such permutation and [imath]\alpha = (1362457)[/imath] Clearly we could write [imath]\alpha = (3624571)[/imath], which satisfies the condition above, but this is the identical permutation. Therefore, [imath]\alpha = (1362457)[/imath] and is unique. Find all permutation [imath]\alpha \in S_7[/imath] such that [imath]\alpha^3 = (1234)[/imath] Clearly, there is more than one such permutation as [imath]\alpha[/imath] could be a [imath]4[/imath]-cycle with elements [imath]1,2,3,4[/imath] or [imath]\alpha[/imath] could be a [imath]4[/imath]-cycle of elements [imath]1,2,3,4[/imath] and a [imath]3[/imath]-cycle of elements [imath]5,6,7[/imath]. [imath]\alpha[/imath] cannot be a product of any disjoint transpositions as [imath](ij)^3 = (ij)[/imath]. Also [imath]\alpha[/imath] cannot have any disjoint [imath]k[/imath]-cycles where [imath]k\geq4[/imath] as there would be the same length cycle in [imath]\alpha^3[/imath]. So one possible permutation is [imath]\alpha_1 = (1432)[/imath] There are no other possible permutations comprised of just one [imath]4[/imath]-cycle for reasons discussed in the previous question. Now, there are [imath]2[/imath] possible ways to arrange the elements [imath]5,6,7[/imath] in a [imath]3[/imath]-cycle, [imath](567)[/imath] and [imath](576)[/imath] so two further permutations are [imath]\alpha_2 = (1432)(567)[/imath] and [imath]\alpha_3 = (1432)(576)[/imath] In total there are [imath]3[/imath] permutations that satisfy [imath]\alpha^3 = (1234)[/imath] If there are any errors with this, or if there is a better method to exhaust possibilities, I'd appreciate your input. Thanks. |
693391 | Linear transformations and change of basis
Let [imath]T:\in \Bbb R^3 \rightarrow \Bbb R^3[/imath] be the linear transformation given by reflecting across the plane [imath]-x_1+x_2+x_3=0[/imath] (...) If [imath]S=\{x:-x_1+x_2+x_3=0\} \implies S=\gen[(1,1,0),(1,0,1)][/imath] But how can I get the reflection matrix across [imath](1,1,0)[/imath] and [imath](1,0,1)[/imath]? Thanks! | 693414 | Reflection across the plane
Let [imath]T: \Bbb R^3 \rightarrow \Bbb R^3[/imath] be the linear transformation given by reflecting across the plane [imath]S=\{x:-x_1+x_2+x_3=0\}[/imath] (...) Then, [imath]$S={\rm gen}[(1,1,0),(1,0,1)].$[/imath] But how can I get the matrix [imath]R_v[/imath] such that reflects across [imath]S[/imath]? Thanks! |
696671 | Prove that if [imath]d[/imath] is a common divisor of [imath]a[/imath] & [imath]b[/imath], then [imath]d=\gcd(a,b)[/imath] if and only if [imath]\gcd(a/d,b/d)=1[/imath]
Prove that if [imath]d[/imath] is a common divisor of [imath]a[/imath] & [imath]b[/imath], then [imath]d=\gcd(a,b)[/imath] if and only if [imath]\gcd(\frac{a}{d},\frac{b}{d})=1[/imath] I know I already posted this question, but I want to know if my proof is valid: So for my preliminary definition work I have: [imath]\frac{a}{d}=k, a=\frac{dk b}{d}=l,b=ld [/imath] so then I wrote a linear combination of the [imath]\gcd(a,b)[/imath], [imath]ax+by=d[/imath] and substituted: [imath]dk(x)+dl(y)=d d(kx+ly)=d kx+ly=1 a/d(x)+b/d(y)=1[/imath] Is this proof correct? If not, where did I go wrong? Thanks! | 696544 | Prove that if d is a common divisor of a and b, then [imath]d=\gcd(a,b)[/imath] if and only if [imath]\gcd(a/d,b/d)=1[/imath]
Prove that if [imath]$d$[/imath] is a common divisor of two integers [imath]$a$[/imath] and [imath]$b$[/imath], then [imath]d=\gcd(a,b)[/imath] if and only if [imath]\gcd(a/d,b/d)=1[/imath]. So far I used what was given so I have [imath]a=dk[/imath], [imath]b=ld[/imath] and [imath]\gcd(a,b)=d[/imath] can be written as a linear combination of [imath]ax+by=d[/imath] but I am unsure how to use the information. Where do I go from here? Can someone show me how to solve this using Bezout's Identity if possible? |
696527 | Cech cohomology
There are 2 complexes computing Cech cohomology. The difference between them is that in the second one we require skew symmetry when you change the order of indices. How to show that they are quasi-isomorphic? Let [imath]C(U,F)[/imath] be the first and [imath]C'(U,F)[/imath] the second. Open sets are indexed by the set [imath]I[/imath] with some perfect order. There is inclusion [imath]i: C'(U,F) \to C(U,F)[/imath]. There is projection [imath]h: C(U,F) \to C'(U,F)[/imath], defined by [imath] (hc)_{i_0...i_n}=sgn(\varepsilon) c_{i_{\varepsilon_0}i_{\varepsilon_1}...} [/imath] I failed to construct a homotopy [imath]k:C(U,F) \to C(U,F)[1][/imath] such that [imath] 1-ih=dk+kd. [/imath] Could you help me? | 157864 | About the definition of Cech Cohomology
Let [imath]X[/imath] be a topological space with and open cover [imath]\{U_i\}[/imath] and let [imath]\mathcal F[/imath] be a sheaf of abelian groups on [imath]X[/imath]. A [imath]n[/imath]-cochain is a section [imath]f_{i_0,\ldots,i_n}\in U_{i_0,\ldots,i_n}:= U_{i_0}\cap\ldots\cap U_{i_n}[/imath]; we can costruct the following abelian group (written in additive form): [imath] \check C^n(\mathcal U,\mathcal F):=\!\!\prod_{(i_o,\ldots,i_n)}\!\!\mathcal F(U_{i_0,\ldots,i_n}) [/imath] Now my question is the following: we consider oredered sequences [imath](i_o,\ldots,i_n)[/imath] ? Because in this case in the direct product we have each group repetead [imath](n+1)![/imath] times, that is the number of permutations of the set [imath]\{i_o,\ldots,i_n\}[/imath]. |
695890 | Nilpotent Subring
The question is : Show that the nilpotent elements of a commutative ring form a subring. Here is my unsuccessful take on it: Let [imath]R[/imath] be a commutative ring and let [imath]S = \{a \in R | a^n = 0 \}[/imath] be the set of nilpotent elements from [imath]R[/imath]. Now, [imath]0 \in S[/imath], since [imath]0^1 =0[/imath]. Hence [imath]S[/imath] is nonempty. Let [imath]x,y \in S[/imath], then for some positive integers [imath]n[/imath] and [imath]m[/imath] we have [imath]x^n=0[/imath] and [imath]y^m=0[/imath]. So far so good. Now we try to show that [imath](x + y)^k = 0[/imath], for some positive integer k. This is where I am stuck. | 690656 | Nilpotent elements in a commutative ring
Let [imath]R[/imath] be a commutative ring. Show that for any [imath]a,b \in R[/imath] nilpotent that [imath]a+b[/imath] is also nilpotent in [imath]R[/imath]. We know [imath]a^n = 0[/imath] for some n and [imath]b^m = 0 [/imath] for some m, so consider [imath](a+b)^{m+n} = (a+b)(a+b)...(a+b)[/imath] [imath]n+m[/imath] times. I don't see how [imath]R[/imath] being commutative helps me here. |
696872 | Proving density of [imath]\mathbb{Q}^n[/imath] in [imath]\mathbb{R}^n[/imath]
I'm looking for a relatively simple way to prove that [imath]\Bbb {Q}^n[/imath] is dense in [imath]\Bbb{R}^n[/imath] given [imath]\Bbb Q[/imath] is dense in [imath]\Bbb R[/imath] Thanks | 366986 | Density of [imath]\mathbb{Q}^n[/imath] in [imath]\mathbb{R}^n[/imath]
[imath]\mathbb{Q}[/imath] is dense in [imath]\mathbb{R}[/imath] (with the standard topology). I'm pretty sure that [imath]\mathbb{Q}^n[/imath] is dense in [imath]\mathbb{R}^n[/imath] too. Is there an easy argument to prove that without reproducing the same argument to show "[imath]\mathbb{Q}[/imath] dense in [imath]\mathbb{R}[/imath]"? |
696978 | Let [imath]f(x)[/imath] belong to [imath]\mathbb{Z}_p[x][/imath]. Prove that if [imath]f(b)=0[/imath], then [imath]f(b^p)=0[/imath].
Let [imath]f(x)[/imath] belong to [imath]\mathbb{Z}_p[x][/imath]. Prove that if [imath]f(b)=0[/imath], then [imath]f(b^p)=0[/imath]. Not sure how to proceed with this problem. I usually use Chegg, but Chegg doesn't have the solution for this problem. This is what I have so far, "Notice that [imath]\mathbb{Z}_p[/imath] is a field (Corollary of Thm 13.2), and let [imath]f(b)=0[/imath] then [imath]b[/imath] is a zero of [imath]f(x)[/imath] so [imath](x-b)[/imath] is a factor of [imath]f(x)[/imath]. Then [imath]f(x)=(x-b)q(x)[/imath] ([imath]r(x)=0[/imath] because [imath]f(b)=0[/imath]." | 674580 | Roots of polynomials in field of prime characteristic p
Let [imath]F[/imath] be a field of characteristic [imath]p[/imath], and let [imath]\alpha \in F[/imath]. Let [imath]f \in \mathbb{Z}_p[x][/imath] be such that [imath]f(\alpha) = 0[/imath]. Apparently, we have [imath]f(\alpha^p) = 0[/imath]. This was mentioned but not proved in my lectures. Can someone explain why please? Thanks. |
697318 | Formal proof for an easy calculus result
I have the following situation: Suppose [imath]x,y[/imath] are real numbers with [imath]y > 0[/imath]. Then, I would like to show that for small values of [imath]t[/imath], we have [imath] (tx)^2 + (ty - 1)^2 \leq 1 [/imath] How can I show this in a formal manner?? For instance, I find myself struggling because I don't know that is the definition of small values of [imath]t[/imath]. Can someone help me? thanks | 696063 | Inequality for small values of [imath]t[/imath]
Suppose [imath]x,(y > 0)[/imath] are real numbers. I want to know if it is true that for small [imath]t[/imath], we have [imath] (tx)^2 + (ty)^2 \leq 2ty [/imath] |
697083 | Question about Finite Abelian Groups
Let [imath](G, .)[/imath] be a finite abelian group, [imath]G=\{x_1, ..., x_n\}[/imath] and let [imath]x=x_1. \cdots. x_n[/imath]. Show that [imath]x^2=e[/imath] Suppose [imath]G[/imath] has no element of order [imath]2[/imath] or that [imath]G[/imath] has more than one element of order [imath]2[/imath]. Show that [imath]x=e[/imath]. Suppose [imath]G[/imath] has a single element [imath]y[/imath] of order [imath]2[/imath]. Show that [imath]x=y[/imath]. Progress: I have done part [imath]1, 3[/imath], and I shown that if [imath]G[/imath] has no element of order [imath]2[/imath] then [imath]x=e[/imath]. For this part, It's easy since as [imath]x^2=e[/imath] and [imath]o(x)\neq 2[/imath] then [imath]x=e[/imath], but I'm stuck in the second part of part 2. Can someone help me? | 474214 | Product of elements of a finite abelian group
Suppose [imath]G=\{a_1,...,a_n\}[/imath] is a finite abelian group, and let [imath]x=a_1a_2\dotsm a_n[/imath]. Prove that if there is more than one element of order [imath]2[/imath] then [imath]x=e[/imath]. What I've done so far: (#1 is just for illustration, There should be at least [imath]3[/imath] such elements) [imath]G[/imath] has an even number of elements with order [imath]>2[/imath] (They are paired so: [imath](a,a^{-1})[/imath] ). And then there are [imath]e[/imath] and [imath]\{b|o(b)=2\}[/imath]. Also, if [imath]G[/imath] has at least one element of order 2 then [imath]|G|[/imath] is even (Lagrange). Thus [imath]|\{b|o(b)=2\}|[/imath] must be odd. If there is exactly one element [imath]b[/imath] of order [imath]2[/imath] then [imath]x=b[/imath]. That's because [imath]G[/imath] is abelian, and we can write: [imath]x=e\cdot b\cdot(a_1 a_{1}^{-1}\dotsm a_n a_{n}^{-1})=b\cdot(e\dotsm e)=b[/imath]. If there are exactly [imath]3[/imath] such elements [imath]a,b,c\in G[/imath], then, as shown above, [imath]x=abc[/imath] and thus [imath]x=e[/imath] (That's because [imath](abc)^2=a^2b^2c^2=e[/imath] which means [imath]abc\in \{a,b,c,e\}[/imath]. If [imath]abc=a[/imath] then [imath]b=c^{-1}=c[/imath] which is false, same goes for [imath]abc=b,abc=c[/imath].) I haven't managed to find a reason why the claim must hold for [imath]5, 7, ...[/imath] (i.e. All other odd integers). Does anyone have an idea? Thanks in advance. Please excuse my English. edit: Not a canonical answer but a simple one (if such even exists). |
697851 | Algebra & Re-arrangement
I need to make this: [imath]K(K+1)(2K+7) + 6(k+1)[(k+1)+2]/6[/imath] Equal to this: [imath](K+1)[(k+1)+1][(2(k+1)+7]/6[/imath] By using algebra and re-arrangement. From the initial equation there should be just one more step before the two can equal each-other. | 697817 | Mathematical Induction
I've gotten to the final step and believe my problem lies within my algebra. Prove the following: [imath]1 \times 3 + 2 \times 4 + 3 \times 5 + ... + N(N+2) = \frac{N(N+1)(2N+7)}6[/imath] Here is my show that it is true answer: [imath]\displaystyle\frac{(K+1)[(k+1)+1][(2(k+1)+7]}6[/imath] Here is where I am stuck: K(K+1)(2K+7) + 6 (k+1) [(k+1)+2]/6 So all that needs to be done is algebra and moving around the above to show that it is equal to the show that it is true answer above. I just cannot figure it out. I am pretty sure I have the right equation. |
698331 | Taylors theorem application
I posted this question yesterday, and, despite getting answers, I am still confused how to solve it: Use Taylor's theorem to prove that [imath]\displaystyle\lim_{n \to \infty} n \ln\left(1+\frac{1}{n}\right)=1[/imath] I know the Taylor expansion of ln(1+x) is [imath]x-\frac{1}{2}x^2+ \frac{1}{3}x^3 ...[/imath], but I don't see what this gives me. Also, one respondent posted: ln(1+t)=t+o(t), and I don't understand how this follows by the Theorem... | 697203 | Proof using Taylor's theorem
Use Taylor's theorem to prove that [imath]\displaystyle\lim_{n \to \infty} n \ln\left(1+\frac{1}{n}\right)=1[/imath] I don't understand how to apply Taylor's theorem to a limit, especially one with a product of two functions... |
698455 | Proving [imath]1 + 1 = 2[/imath]
How do you break down the theory of [imath]1 + 1 = 2[/imath]? How do you provide a proof, please be precise. This is for one of my discrete math courses and I don't know how this is relevant to the course. And remember, please be detailed and precise. | 348889 | Is 1+1 =2 a theorem?
A theorem is defined to be a mathematical statement that is proven to be true. The statement [imath]1+1=2[/imath] has definitely been proven in the history of mankind (Russel and Whitehead had once proven it in the book Principia Mathematica). So can it be considered as a theorem? What determines something to be a theorem (besides it being proven to be true)? |
698713 | Limit of Ratio of Adjacent Fibonacci numbers [imath]\to \phi[/imath]
We define the [imath]n^{th}[/imath] Fibonacci number as [imath]a_1 = a_2 = 1[/imath] and [imath]a_n = a_{n-1} + a_{n-2}[/imath] for [imath]n \geq 3[/imath]. Consider [imath] \lim_{n \to \infty} \frac{a_{n+1}}{a_n}. [/imath] I wrote a script and found that this limit converges to the golden ratio [imath]\phi \approx 1.61803[/imath]. However, I'm having trouble giving a rigorous proof for this. Here's what I have so far: [imath] \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{a_n + a_{n-1}}{a_n} = \lim_{n \to \infty} \left( 1 + \frac{a_{n-1}}{a_n} \right). [/imath] I'm unsure of how to proceed after this. I know that expressing the limit as a certain continued fraction would imply that it converges to [imath]\phi[/imath] but I'm unsure of how to get there. Any ideas? | 1953942 | Convergence of [imath]r_n[/imath] = [imath]a_n[/imath] / [imath]a_{n+1}[/imath] where [imath]a_n[/imath] is Fibonacci sequence
Let [imath]a_n[/imath] be the Fibonacci sequence defined by [imath]a_{n+1} = a_n + a_{n-1} \ \ \text{with} \ \ a_0 = a_1 = 1 [/imath] Let [imath]r_n = \frac{a_n}{a_{n+1}}[/imath] I have already found two subsequences which are [imath]\{r_{2k-1}\}[/imath] bounded above by [imath]2[/imath] and increasing (thus is convergent), and [imath]\{r_{2k}\}[/imath], bounded below by [imath]1[/imath] and decreasing (thus is convergent,too) Besides, [imath] \lim_{k \to \infty} r_{2k-1} = \ell_1 [/imath] I found that [imath]l_1^2 - l_1 - 1 = 0[/imath] from above equation. However, i cannot find [imath]l_2[/imath] where [imath] \lim\limits_{k \to \infty} r_{2k} = \ell_2.[/imath] |
556186 | How many irreducible monic quadratic polynomials are there in [imath]\mathbb{F}_p[X][/imath]?
Can some of you help me with my homework? I had to count the irreducible, quadratic, monic polynomials in [imath]\mathbb{F}_p[X][/imath] for arbitrary [imath]p[/imath]. I will show you what I tried myself. Research effort First of all the set [imath]\{X^ 2 + bX +c : b, c \in \mathbb{F}_p \}[/imath] contains [imath]p^2[/imath] elemenst. We need to substract the amount of elements that is reducible. These elements are exactly the set [imath]A = \{(X+c)(X+d) : c, d \in \mathbb{F}_p \}[/imath]. There are [imath]p(p-1)[/imath] pairs of [imath](a,b) \in \mathbb{F}_p \times \mathbb{F}_p[/imath] such that [imath]a \neq b[/imath], and there are [imath]p[/imath] pairs such that [imath]a=b[/imath]. When [imath]a[/imath] and [imath]b[/imath] are unequal, whe know that [imath]a[/imath] and [imath]b[/imath] can be swapped without changing anything to the polynomial [imath](X+a)(X+b)[/imath], so I obtained that [imath]|A| \leq p +(p-1)- \frac{1}{2}(p-1)=p-\frac{1}{2}(p-1)[/imath]. In other to find out these are the elements, or that there are still less, I thought I had to assume that [imath](X+c)(X+d) = (X+\gamma)(X+\delta)[/imath], to deduce some information about the realation between [imath](c,d)[/imath] and [imath](\gamma, \delta)[/imath], but this led to taking square roots in [imath]\mathbb{F}_p[/imath], and I am not familiar with that. Could you please provide me some more information? | 1665598 | Finding the number of irreducible quadratics in [imath]\Bbb Z_p[x][/imath], where [imath]p[/imath] is a prime
The problem is to find the number of irreducible quadratics in [imath]\Bbb Z_p[x][/imath], where [imath]p[/imath] is a prime number. To solve this, I wish to find first the number of reducible quadratics of the form [imath]x^2+ax+b[/imath], then the number of reducible quadratics, and subtract this from the total number of quadtratics. I know that each reducible quadratic that is of the form [imath]x^2 +ax+b[/imath] is a product [imath](x+c)(x+d)[/imath] for [imath]c,d\in \Bbb Z_p[/imath]. So, I don't know what to do next. Please help me solve this problem. Thanks for the help! |
699638 | Why a false statement can imply anything?
According to the truth table, If [imath]P[/imath] is false,then [imath]P->Q[/imath] is true. if pigs fly, then [imath]1+1=3[/imath]. Why is this implication true? How do you prove it? | 137890 | Why is it sensical for a proposition with a false antecedent to validate to true?
In propositional logic, the statement "If pigs can fly, then elephants can lay eggs." validates to true because the antecedent validates to false. In other words, given [imath]a \rightarrow b[/imath], if a is false, the entire statement is true. Why? Just because the antecendent is false doesn't mean that another fact depends on it, right? |
699685 | Proof that a median minimizes 1-norm.
I was wondering whether there is an easy way to show the following: We have a data set [imath]x_1,...,x_n[/imath] and [imath]m[/imath] is a median if for at least half of the n data points we have that [imath]x_i \le m[/imath] and for the other half we have [imath]x_i \ge m[/imath]. Now I want to show that [imath]m[/imath] is a median iff [imath]m[/imath] minimizes the following 1 norm, such that: [imath]||(x_1-m,...,x_n-m)||_1 = \text{inf }_x ||(x_1-x,...,x_n-x)||_1[/imath]. Unfortunately it seems so, as if calculus would not work here, so is there an easy way to show this? | 113270 | The Median Minimizes the Sum of Absolute Deviations (The [imath] {L}_{1} [/imath] Norm)
Suppose we have a set [imath]S[/imath] of real numbers. Show that [imath]\sum_{s\in S}|s-x| [/imath] is minimal if [imath]x[/imath] is equal to the median. This is a sample exam question of one of the exams that I need to take and I don't know how to proceed. |
699788 | Prove that x_n tends to 0 (real-analysis)
Suppose [imath](x_n)[/imath] is a sequence of positive terms and that the sequence [imath]y_n[/imath]=[imath](x_{n+1})/(x_n)[/imath] converges to a limit L<1. Prove that the sequence x_n converges to 0 I'm not sure exactly how to start it but I know that if [imath]x_n[/imath] tends to 0 then [imath]1/(x_n)[/imath] tends to [imath]infinity[/imath]. I know it can be solved doing two parts. The first part being induction and the second part is the squeeze theorem. For example say [imath]x_2 < Lx_1[/imath] and then [imath]x_3<Lx_2[/imath]. If say L=1/2 then L is monotone decreasing. I just don't know how to start the proof. | 697511 | Help proving exercise on sequences in Bartle's Elements
Self learning Analysis and found the following exercise in Bartle's Elements of Real Analysis: Let [imath]X = (x_n)[/imath] be a sequence of strictly positive real numbers such that [imath]\lim \left({\frac {x_{n + 1}} {x_n}} \right) \lt 1.[/imath]Show that for some [imath]r[/imath] with [imath]0\lt r \lt 1[/imath] and some [imath]C \gt 0 [/imath] we have [imath] \ 0 \lt x_n \lt Cr^n[/imath] for all sufficiently large [imath]n \in \Bbb N[/imath]. Use this to show that [imath]\lim(x_n) = 0.[/imath] Now I haven't proven any significant results concerning sequences and hence am required to solve this problem with basically nothing but the definition of a limit. I am not getting too far with this because I cannot see a way to get there with a direct implication and I cannot for the life of me negate "for all sufficiently large [imath]n[/imath]" in the context of the sentence above. Would be extremely grateful if someone could help me out. What I've done so far: We know that [imath]l = \lim \left({\frac {x_{n + 1}} {x_n}} \right) \lt 1[/imath]. If [imath]l \lt 0[/imath] we can find a neighbourhood of [imath]l[/imath] entirely contained in [imath](-\infty, 0)[/imath] which would contain no elements of the sequence [imath] \left({\frac {x_{n + 1}} {x_n}} \right) [/imath] all of whose elements are positive leading to a contradiction. This implies [imath]l \ge 0.[/imath] We can pick [imath] 0 \lt \epsilon \lt 1 - l[/imath]. There is [imath]m \in \Bbb N[/imath] such that [imath]n \ge m \implies \left|{\frac {x_{n + 1}} {x_n} - l}\right| \lt \epsilon \implies \frac {x_{n + 1}} {x_n} \lt \epsilon + l \lt 1 \implies x_{n + 1} \lt x_n[/imath] Also can someone tell me if [imath]\lim (x_n) = 0[/imath] necessarily? Doesn't [imath]\left({2 + \frac 1 n}\right) [/imath] satisfy the given conditions?? |
700269 | How find this sum [imath]\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)(n+3)\cdots(n+2014)}[/imath]
Find the sum [imath]\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(n+2)(n+3)\cdots(n+2014)}[/imath] My idea: since [imath]\dfrac{1}{n(n+1)(n+2)\cdots(n+2014)}=\dfrac{1}{2014}\left(\dfrac{1}{n(n+1)(n+2)\cdots(n+2013)}-\dfrac{1}{(n+1)(n+2)\cdots(n+2014)}\right)?[/imath] | 477174 | Infinite Series [imath]\sum\limits_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}[/imath]
How to prove the following equality? [imath]\sum_{n=1}^{\infty}\frac{1}{\prod\limits_{k=1}^{m}(n+k)}=\frac{1}{(m-1)m!}.[/imath] |
445321 | [imath]20[/imath] hats problem
I've seen this tricky problem, where [imath]20[/imath] prisoners are told that the next day they will be lined up, and a red or black hat will be place on each persons head. The prisoners will have to guess the hat color they are wearing, if they get it right the go free. The person in the back can see every hat in front of him, and guesses first, followed by the person in front of him, etc. etc. The prisoners have the night to think of the optimal method for escape. This method ends up allowing [imath]19[/imath] prisoners to always escape. The person in the back counts the number of red hats, and if its even, says red, if its odd, says black. This allows the people in front to notice if its changed, and can determine their hat color, allowing the person in front to keep track as well. What I'm wondering is what is the equivalent solution for [imath]3[/imath] or more people, and how many people will go free. If possible, a general solution would be nice. | 338088 | line of mathematicians guess their own hat color out of c colors
There is a common problem in which a long line of N mathematicians are each given a hat that is either red or blue. They cannot see their own hat but can see all in front of time and can hear any response. They must state their own hat color from the back of the line forwards and no information may be passed besides their guess. They may have a planning session in advance. What is the maximum number of correct guesses can they guarantee? The answer is (look away if you don't know) [imath]N-1[/imath]. If, instead of 2 colored hats, we use [imath]c[/imath] different colored hats, how many can we guarantee? Assuming [imath]N>>c[/imath] I believe the answer to be [imath]N-c+1[/imath] and have confirmed this for [imath]c=2, 3[/imath], and [imath]4[/imath]. I believe this should be simple to prove by using the concept of "[imath]x[/imath] formulas allow you to solve for [imath]x[/imath] variables" but I am not well versed enough in pure mathematics to answer this. |
700967 | If G is a group of order [imath]pq[/imath], where [imath]p[/imath] and [imath]q[/imath] are primes. How do I prove that any nontrivial subgroup of [imath]G[/imath] must be cyclic?
This makes sense since [imath]p[/imath] and [imath]q[/imath] are primes but I'm not sure how I can prove that the subgroup will be cyclic? Will the generator of the subgroup be [imath]pq[/imath]? | 697823 | G is group of order pq, pq are primes
Problem. Let [imath]G[/imath] be a group of order [imath]pq[/imath] such that [imath]p[/imath] and [imath] q[/imath] are prime integers. I am to show that every proper subgroup of [imath]G[/imath] is cyclic. My attempt. What I know: Any element [imath]a[/imath] divides [imath]pq[/imath] and [imath]a^{pq} = e[/imath]. The order of subgroups [imath]H[/imath] divide [imath]pq[/imath] by Lagrange. If I could show that [imath]G[/imath] is cyclic, then all subgroups must be cyclic. If I can show that [imath]G[/imath] is a group of prime order, then I can show that it is cyclic. I'm not sure what more I can do here...I've tried looking at Fermat's Little Theorem but I can't seem to properly understand it and how it could affect my problem.. |
701564 | Prove the statement for definite integral
We have positive continuous function [imath]f(x)[/imath] defined on [imath]\mathbb{R}[/imath], such as [imath]\int_{-\infty}^{+\infty} f(x) dx = 1[/imath] let [imath]\alpha \in (0,1)[/imath] and [imath][a,b][/imath] is an interval of minimal length amongst intervals for those holds: [imath]\int_{a}^{b} f(x) dx = \alpha[/imath]. Task is to prove that [imath]f(a) = f(b)[/imath]. I managed to sum up following statements: [imath]F(\infty) - F(-\infty) = 1[/imath] [imath]F(b) - F(a) = \alpha[/imath] And one to prove is: [imath]f(b) - f(a) = 0[/imath] I am looking for hints to build a proof here. | 680162 | Integration and theorems on continuous functions
[imath]f(x)[/imath] is positive and continuous function on [imath]\mathbb{R}[/imath] and, moreover, [imath]\int_{-\infty}^{+\infty}f(x)dx=1[/imath]. [imath]\alpha\in(0;1)[/imath] and [imath][a;b][/imath] is the interval having a minimum length such that the [imath]\int_{a}^{b}f(x)dx=\alpha[/imath]. Prove [imath]f(a)=f(b)[/imath]. With mean value theorem easy to show, that for every [imath]\alpha\in(0;1)[/imath], exist such [imath][a;b][/imath], so [imath]\int_{a}^{b}f(x)dx=\alpha[/imath]. The statement looks like Rolle's theorem, but I have no idea. I will be grateful for help. |
701578 | Proving inverse of two matrices
Given the identity matrix [imath]I_n[/imath] and a matrix [imath]J_n[/imath] with only 1's as every element. I must prove that if [imath]n>0[/imath] then [imath](I_n - J_n)^{-1} = I_n - \frac{1}{n-1} J_n[/imath] I'm trying to understand if there is some sort of theorem that states that [imath](I_n - J_n)^{-1} = I_n^{-1} - J_n^{-1}[/imath] or something like that? | 441505 | Proving the inverse of a matrix equals [imath]I_n-\frac{1}{n-1}A[/imath]
Question: Let [imath]A[/imath] be a matrix whose elements are all [imath]1[/imath]. Prove that [imath](I_n-A)^{-1}=I_n - \frac 1{n-1}A.[/imath] Thought: I tried using this identity, but couldn't get any further (computing the adjoint looks pretty nasty): [imath](I_n-A)^{-1}=\frac 1{\det(I_n-A)}\operatorname{Adj}(I_n-A)[/imath] |
701615 | Prove that every element of a group G can be represented as [imath]g = x^{-1}(xT)[/imath] for some x [imath]\in[/imath] G?
Let G be a finite group and T be an automorphism on G with the property that T(x) = x for [imath] \ \ [/imath] x [imath]\in[/imath] G iff x = e. Prove that every element of G can be represented as [imath]g = x^{-1}(xT)[/imath]. Suppose further that [imath]T^2 = I[/imath] . Then G must be abelian . I have tried : Since G is finie . Then G = [imath]\{e,a_1,a_2,..........,a_n \}[/imath] and T be an automorphism any [imath]a_j \in[/imath] G , then [imath]\exists \ \ a_i[/imath] such that T([imath]a_i) = a_j[/imath] [imath]\Rightarrow a_i^{-1}T(a_i) = a_i^{-1}a_j[/imath]. Further how to solve | 305284 | [imath]xT=x[/imath] iff [imath]x=e[/imath] implies that every [imath]g[/imath] may be written as [imath]x^{-1}(xT)[/imath] for some [imath]x\in G[/imath].
Let [imath]G[/imath] be a finite group, [imath]T[/imath] an automorphism of [imath]G[/imath] with the property that [imath]xT=x[/imath] for [imath]x[/imath] in [imath]G[/imath] if and only if [imath]x=e[/imath]. Prove that every [imath]g[/imath] in [imath]G[/imath] can be represented as [imath]g=x^{-1}(xT)[/imath] for some [imath]x[/imath] in [imath]G[/imath]. I don't understand how I should start to pursue a solution. Can you please give me a hint instead of a whole solution? Which properties of automorphisms should I use? Thanks. |
696388 | Random operators
Let [imath](\Omega, \mathcal F,P)[/imath] be a probability spaces and [imath]H[/imath] be a Hilbert space. By a random operator [imath]A[/imath] from [imath]H[/imath] to [imath]H[/imath] we mean a linear continuous mapping from [imath]H[/imath] into the Frechet space [imath]L_0^H (\Omega, \mathcal F,P)[/imath] of all [imath]H[/imath]-valued random variables. My question is: How Can we define [imath]|A|[/imath] and the adjoint [imath]A^*[/imath]? | 700554 | Random operators
Let [imath](\Omega, \mathcal F,P)[/imath] be a probability spaces and [imath]H[/imath] be a Hilbert space. By a random operator [imath]A[/imath] from [imath]H[/imath] to [imath]H[/imath] we mean a linear continuous mapping from [imath]H[/imath] into the Frechet space [imath]L_0^H (\Omega, \mathcal F,P)[/imath] of all [imath]H[/imath]-valued random variables. My question is: How Can we define [imath]|A|[/imath] and the adjoint [imath]A^*[/imath]? |
701717 | A inequality of calculus
Let [imath]f \in C^2[a,b][/imath] and [imath]f(a) = f(b) = 0[/imath], [imath]f'(a) = 1[/imath],[imath]f'(b) = 0[/imath], prove that [imath]\int_a^b|f''(x)|^2\,dx \geq \frac{4}{b-a}[/imath] Remark: This question is in the book functional analysis of Peking University; We have[imath]u(x) = \int_a^xu'(t)\,dt[/imath]so [imath]|u(x)|^2 \leq (b-a)\int_a^bu'(x)\,dx[/imath] by applying the Cauthy-Schwartz inequality. but I cannot get the number 4 I have construct a function of which satisfies the condition using quadratic function，and the infimum is attained, and [imath]4[/imath] is got from differentiating and squaring. | 499416 | How prove this [imath]\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}[/imath]
let [imath]f[/imath] on [imath][a,b][/imath] two continuously differentiable functions,such [imath]f(a)=f(b)=0, f'(a)=1,f'(b)=0,b>a>0[/imath] show that [imath]\int_{a}^{b}[f''(x)]^2dx\ge\dfrac{4}{b-a}[/imath] My idea: use Cauchy-Schwarz inequality [imath]\int_{a}^{b}[f''(x)]^2dx\cdot\int_{a}^{b}g^2(x)dx\ge\left(\int_{a}^{b}f''(x)g(x)dx\right)^2[/imath] How find the [imath]g(x)[/imath]? then I think \begin{align} \int_{a}^{b}f''(x)g(x)dx &=\int_{a}^{b}g(x)df'(x)=f'(x)g(x)|_{a}^{b}-\int_{a}^{b}f'(x)g(x)dx\\ &=f'(b)g(b)-f'(a)g(a)-\int_{a}^{b}g(x)df(x)\\ &=-g(a)+\int_{a}^{b}g'(x)dF(x)\\ &=-g(a)+F(x)g'(x)|_{a}^{b}-\int_{a}^{b}F(x)g''(x)\\ &=-g(a)+F(b)g'(b)-F(a)g'(a)-\int_{a}^{b}F(x)g''(x)dx \end{align} where [imath]F(x)=\int_{a}^{x}f(t)dt[/imath] Now following I can't find the [imath]g(x)[/imath]? can you help me or use other methods solve it? Thank you and some hours ago: I ask this integral inequality:How prove this [imath]\int_{a}^{b}f^2(x)dx\le (b-a)^2\int_{a}^{b}[f'(x)]^2dx[/imath] |
159167 | Traces of all positive powers of a matrix are zero implies it is nilpotent
Let [imath]A[/imath] be an [imath]n\times n[/imath] complex nilpotent matrix. Then we know that because all eigenvalues of [imath]A[/imath] must be [imath]0[/imath], it follows that [imath]\text{tr}(A^n)=0[/imath] for all positive integers [imath]n[/imath]. What I would like to show is the converse, that is, if [imath]\text{tr}(A^n)=0[/imath] for all positive integers [imath]n[/imath], then [imath]A[/imath] is nilpotent. I tried to show that [imath]0[/imath] must be an eigenvalue of [imath]A[/imath], then try to show that all other eigenvalues must be equal to 0. However, I am stuck at the point where I need to show that [imath]\det(A)=0[/imath]. May I know of the approach to show that [imath]A[/imath] is nilpotent? | 1232774 | If [imath]A [/imath] is a square matrix of size [imath]n[/imath] with complex entries such that [imath]Tr(A^k)=0 , \forall k \ge 1[/imath] , then is it true that [imath]A[/imath] is nilpotent ?
If [imath]A[/imath] is a square matrix of size [imath]n[/imath] with complex entries and is nilpotent , then I can show that all the eigenvalues of [imath]A^k[/imath] , for any [imath]k[/imath] , is [imath]0[/imath] , so [imath]Tr(A^k)=0 , \forall k \ge 1[/imath] . Now conversely if [imath]A [/imath] is a square matrix of size [imath]n[/imath] with complex entries such that [imath]Tr(A^k)=0 , \forall k \ge 1[/imath] , then is it true that [imath]A[/imath] is nilpotent ? |
235296 | An application of Poincare inequality [solved]
I am woking on Evans PDE problem 5.10. #15: Fix [imath]\alpha>0[/imath] and let [imath]U=B^0(0,1)\subset \mathbb{R}^n[/imath]. Show there exists a constant [imath]C[/imath] depending only on [imath]n[/imath] and [imath]\alpha[/imath] such that [imath] \int_U u^2 dx \le C\int_U|Du|^2 dx, [/imath] provided that [imath]u\in W^{1,1}(U)[/imath] satisfies [imath] |\{x\in U\ |\ u(x)=0\}|>\alpha. [/imath] I would appreciate your helping me with this problem. | 247682 | Evans PDE Problem 5.15: Poincaré inequality for functions with large zero set
I have trouble proving the following problem (Evans PDE textbook 5.10. #15). Could anyone kindly help me solving the problem? I know that I should somehow use Poincaré inequality but I still cannot solve it. Fix [imath]\alpha>0[/imath] and let [imath]U=B^0(0,1)\subset \mathbb{R}^n[/imath]. Show there exists a constant [imath]C[/imath] depending only on [imath]n[/imath] and [imath]\alpha[/imath] such that [imath] \int_U u^2 dx \le C\int_U|Du|^2 dx, [/imath] provided that [imath]u\in W^{1,2}(U)[/imath] satisfies [imath]|\{x\in U\ |\ u(x)=0\}|>\alpha[/imath]. |
28243 | Is there a proof that [imath]\pi \times e[/imath] is irrational?
A little reading suggests: It is known that either [imath]\pi + e[/imath] or [imath]\pi \times e[/imath] is transcendental (or possibly both), but no proof is known that one of those two numbers in particular is transcendental. If we just want irrationality rather than transcendence, is a proof known? Can we prove [imath]\pi+e[/imath] is irrational? Can we prove [imath]\pi \times e[/imath] is irrational? | 1860711 | Is [imath]\pi e[/imath] irrational?
During our ongoing research, we need to prove that [imath]\pi e<\lceil \pi e\rceil[/imath]. Is [imath]\pi e[/imath] irrational? How to prove it? Thanks- mike |
702156 | Clarification regarding hint given for showing [imath]p(x)[/imath] is irreducible over [imath]\mathbb{Q}(\sqrt{-2})[/imath].
I want to prove that the polynomial [imath]p(x) = x^{4} - 4x^{2} + 8x + 2[/imath] is irreducible over [imath]\mathbb{Q}(\sqrt{-2})[/imath]. Following a series of hints given by my textbook, I start by using the fact that [imath]\mathbb{Z}(\sqrt{-2})[/imath] is a unique factorization domain. So then by a corollary to Gauss' Lemma, I observe that [imath]p(x)[/imath] is irreducible over [imath]\mathbb{Q}(\sqrt{-2})[/imath] iff it is irreducible over [imath]\mathbb{Z}[\sqrt{-2}][/imath]. Suppose [imath]p(x)[/imath] has a linear factor [imath]x - \alpha[/imath] in [imath]\mathbb{Z}[\sqrt{-2}][/imath]. My hint says that I should be able to infer that [imath]\alpha[/imath] is a divisor of [imath]2[/imath] in [imath]\mathbb{Z}[\sqrt{-2}][/imath] (this is obvious) and then conclude that [imath]\alpha \in \{\pm 1, \pm 2, \pm \sqrt{-2}\}[/imath]. What I don't get is how it even makes sense to consider that [imath]\alpha = \pm \sqrt{-2}[/imath] in the integer field? | 661803 | Irreducibility of [imath]p(x)=x^4-4x^2+8x+2[/imath] over [imath]\mathbb{Q}(\sqrt{-2})[/imath]- Dummit Foote Abstract algebra [imath]9.4.10[/imath]
Question is : Prove that the polynomial [imath]p(x)=x^4-4x^2+8x+2[/imath] is irreducible over the quadratic field [imath]F=\mathbb{Q}(\sqrt{-2})[/imath]. [Hint : first use the method of proposition [imath]11[/imath] for the U.F.D [imath]\mathbb{Z}[\sqrt{-2}][/imath](cf.Exercise [imath]8[/imath], Section [imath]8.1[/imath]) to show that if [imath]\alpha \in \mathbb{Z}[\sqrt{-2}][/imath] is a root of [imath]p(x)[/imath] then [imath]\alpha[/imath] is a divisor of [imath]2[/imath] in [imath]\mathbb{Z}[\sqrt{-2}][/imath] . Conclude that [imath]\alpha[/imath] must be [imath]\pm 1,\pm \sqrt{-2}[/imath] or [imath]\pm 2[/imath] and hence show that [imath]p(x)[/imath] has no linear factor over [imath]F[/imath]. Show similarly that [imath]p(x)[/imath] is not the product of quadratics with coefficients in [imath]F[/imath].] What I have done so far is : Suppose [imath]\alpha \in\mathbb{Z}[\sqrt{-2}][/imath] is a root of [imath]p(x)=x^4-4x^2+8x+2[/imath] we would then have : [imath]p(\alpha)=\alpha^4-4\alpha^2+8\alpha+2=0\Rightarrow 2=\alpha(-\alpha^3+4\alpha-8)[/imath] i.e., [imath]\alpha[/imath] is a divisor of [imath]2[/imath] in [imath]\mathbb{Z}[\sqrt{-2}][/imath]. so, I have used the method of proposition [imath]11[/imath] for the U.F.D [imath]\mathbb{Z}[\sqrt{-2}][/imath](cf.Exercise [imath]8[/imath], Section [imath]8.1[/imath]) to show that if [imath]\alpha \in \mathbb{Z}[\sqrt{-2}][/imath] is a root of [imath]p(x)[/imath] then [imath]\alpha[/imath] is a divisor of [imath]2[/imath] in [imath]\mathbb{Z}[\sqrt{-2}][/imath] I do not understand why does he mentioned that [imath]\mathbb{Z}[\sqrt{-2}][/imath] is U.F.D and all... I do not use that at all... It is unnecessarily confusing me or i am unnecessarily getting confused.. hint is actually misleading me :( Now, I have to prove that [imath]\alpha[/imath] must be [imath]\pm 1,\pm \sqrt{-2}[/imath] or [imath]\pm 2[/imath] i.e., suppose I have [imath]2=ab[/imath] in [imath]\mathbb{Z}[\sqrt{-2}][/imath] then, [imath]N(2)=N(ab)\Rightarrow 4=N(a)N(b)[/imath] i.e.,[imath]N(a)=1\text{ or }2\text{ or } 4[/imath] i.e., [imath]p^2+2q^2=1\text{ or }2\text{ or } 4[/imath] for [imath]a=p+\sqrt{-2}q[/imath] [imath]p^2+2q^2=1\Rightarrow p=\pm 1 \Rightarrow a=\pm 1[/imath] [imath]p^2+2q^2=2\Rightarrow q=\pm 1\Rightarrow a=\pm\sqrt{-2}[/imath] [imath]p^2+2q^2=4\Rightarrow p=\pm 2\Rightarrow a=\pm 2[/imath] Once I prove that those are the only divisors then I would consider : [imath]p(1)=(1)^4-4(1)^2+8(1)+2\neq 0[/imath] [imath]p(-1)=(-1)^4-4(-1)^2+8(-1)+2\neq 0[/imath] [imath]p(\sqrt{-2})=(\sqrt{-2})^4-4(\sqrt{-2})^2+8(\sqrt{-2})+2\neq 0[/imath] [imath]p(-\sqrt{-2})=(-\sqrt{-2})^4-4(-\sqrt{-2})^2+8(-\sqrt{-2})+2\neq 0[/imath] [imath]p(2)=(2)^4-4(2)^2+8(2)+2\neq 0[/imath] [imath]p(-2)=(-2)^4-4(-2)^2+8(-2)+2\neq 0[/imath] So, no divisor of [imath]2[/imath] is a root.. Thus [imath]p(x)[/imath] do not have a root in [imath]\mathbb{Z}[\sqrt{-2}][/imath] suppose I have something like : [imath]x^4-4x^2+8x+2=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd[/imath] But then, [imath]bd=2\Rightarrow \text { b,d are a divisors of 2 in [/imath]\mathbb{Z}\sqrt{-2}[imath]}[/imath] But then we have seen that only divisors of [imath]2[/imath] in [imath]\mathbb{Z}\sqrt{-2}[/imath] are [imath]\pm 1,\pm \sqrt{-2}[/imath] or [imath]\pm 2[/imath] So, only possibilities are [imath]x^4-4x^2+8x+2=(x^2+ax\pm 1)(x^2+cx\mp 2)[/imath] [imath]x^4-4x^2+8x+2=(x^2+ax\pm \sqrt{-2})(x^2+cx\mp \sqrt{-2})[/imath] But these are not possible... This only tells that [imath]p(x)[/imath] is irreducible in [imath]\mathbb{Z}[\sqrt{-2}][/imath] but then how do i show this is irreducible in [imath]\mathbb{Q}(\sqrt{-2})[/imath] I was expecting gauss lemma to help but it only works for integers and rationals... So, please help me to clear this.. Thank you... P.S : Proposition [imath]11[/imath] is Rational root theorem and Exercise [imath]8[/imath] is that ring of integers of [imath]\mathbb{Q}(\sqrt{-2})[/imath] is an Euclidean domain. |
702065 | Real analysis based on rings and ideals
Let [imath]R[/imath] be the ring of all the real valued continuous functions on the closed unit interval. Show that [imath] M=\{ f\in R:f(1/3)=0 \} [/imath] is a maximal ideal | 630198 | Ideals in ring of continuous functions [imath]\mathcal{C}[0,1][/imath] ... NBHM- Algebra
I would like to compile all questions I have encountered with Ideals in the ring [imath]\mathcal{C}[0,1][/imath] of all continuous real valued functions and ask if there are any gaps. Question is to see if : the ideal [imath]\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=0\}[/imath] is maximal ideal in the ring [imath]\mathcal{C}[0,1][/imath] of all continuous real valued functions. What I have done so far is : consider [imath]\eta :\mathcal{C}[0,1]\rightarrow \mathbb{R}[/imath] where [imath]f\rightarrow f(0)[/imath] [imath]Ker (\eta) = \{ f\in \mathcal{C}[0,1] : f(0)=0\}[/imath] I could see that for each [imath]r\in \mathbb{R}[/imath], I would set [imath]f[/imath] such that [imath]f(x)=x+r[/imath] So, [imath]\eta[/imath] is surjective. So, I would get [imath]\mathcal{C}[0,1]/\mathcal{I}\cong \mathbb{R}[/imath] As [imath]\mathbb{R}[/imath] is a field so ia the quotient [imath]\mathcal{C}[0,1]/\mathcal{I}[/imath] which means that [imath]\mathcal{I}[/imath] is an ideal in [imath]\mathcal{C}[0,1][/imath]. Now Another Question on same idea : Question is to see if : the ideal [imath]\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=f(1)=0\}[/imath] is maximal ideal in the ring [imath]\mathcal{C}[0,1][/imath] of all continuous real valued functions. I thought of using same ideas as above but i am unable to choose perfect [imath]\eta[/imath] I believe that this is not even a prime ideal.. I set [imath]f(x)=x[/imath] and [imath]g(x)=x-1[/imath] and consider [imath]f(x)g(x)=(x)(x-1)[/imath] then, [imath](fg)(0)=f(0)g(0)=0.(-1)=0[/imath] [imath](fg)(1)=f(1)g(1)=1.0=0[/imath] So, [imath]fg\in \mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=f(1)=0\}[/imath] but the neither [imath]f[/imath] nor [imath]g[/imath] is in [imath]\mathcal{I}[/imath]. I have two questions: I see that [imath]0[/imath] does not play any role in the question [imath]\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=0\}[/imath] I mean I would have same case if i replace [imath]0[/imath] by any real number. So, I would like to say that [imath]\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(r)=0\}[/imath] is a maximal ideal in the ring [imath]\mathcal{C}[0,1][/imath] of all continuous real valued functions. And one more thing I would claim is that If the collection [imath]\mathcal{I}[/imath] is such that all elements of [imath]\mathcal{I}[/imath] has more than [imath]1[/imath] common zero then [imath]\mathcal{I}[/imath] is not a maximal ideal and not even a prime ideal.. I would like to learn more about this kind of ideals in [imath]\mathcal{C}[0,1][/imath]. I would be thankful if some one can assure that what have done is sufficient to conlcude what i have concluded and I would be thankful if some one can suggest some material to read regarding this. Thank you |
703214 | Lebesgue outer measure satisfies [imath]\lambda^{*}([a,b]) \leq b-a[/imath]
Aaagain, I fail to understand the trivial: Using compactness argument it is straightforward to show: [imath]\lambda^{*}([a,b]) \geq b-a[/imath] And everything is OK. But, regarding [imath]\lambda^{*}([a,b]) \leq b-a[/imath], Cohn says: For any closed bounded interval [imath][a,b] \subset \mathbb{R}[/imath] it is easy to see that [imath]\lambda^{*}([a,b]) \leq b-a[/imath] (cover [imath][a,b][/imath] with sequences of open intervals in which the first interval is barely larger than [imath][a,b][/imath], and the sum of the lengths of the other intervals is very small). From this hint I should be able to arrive at [imath]\lambda^{*}([a,b]) < b-a + \epsilon[/imath], for arbitrary [imath]\epsilon >0[/imath], but I don't understand how.. Help? | 174517 | On Lebesgue Outer Measure of an interval
We define the Lebesgue Outer Measure of an interval [imath][a,b][/imath] by [imath]\lambda([a,b])= \inf \left\{\sum_{j=1}^\infty |I_j|: [a,b]\subset \bigcup_{j=1}^{\infty} I_j\right\}[/imath] where [imath]\{I_j\}[/imath] is a sequence of open intervals that cover [imath][a,b][/imath], and we define the length of an open interval [imath]|(a,b)|= b-a[/imath]. I want to show [imath]\lambda([a,b])= b-a[/imath] The proof of the book that I am using starts with the following: Let [imath]\epsilon >0[/imath]. Because [imath][a,b] \subset \left(a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4}\right) \cup \bigcup_{n=2}^\infty \left(- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n}\right)[/imath], we obtain [imath]\lambda([a,b]) < b-a +\epsilon[/imath] which implies [imath]\lambda([a,b])\le b-a[/imath]. I can understand [imath][a,b] \subset (a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4})[/imath], but I do not get the part [imath]\cup \bigcup_{n=2}^{\infty} (- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n})[/imath] Since [imath](a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4})[/imath] already covers [imath][a,b][/imath], and I do not think that [imath]\bigcup_{n=2}^{\infty} (- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n})[/imath] will necessarily cover an arbitrary closed interval [imath][a,b][/imath], so I do not really know why you need that union of arbitrary small intervals. (My guessing is that you add the countable union so that it coincides with the definition of Lebesgue outer measure) In addition, I do not know how can we arrive at the inequality [imath]\lambda([a,b]) < b-a +\epsilon[/imath]. I do know that [imath]b-a + \epsilon[/imath] is essentially the length of [imath](a- \frac{\epsilon}{4}, b+ \frac{\epsilon}{4}) \cup \bigcup_{n=2}^{\infty} (- \frac{\epsilon}{2 \cdot 2^n},\frac{\epsilon}{2 \cdot 2^n})[/imath] since I already computed it out. Taking the infimum of the length, we get [imath]b-a[/imath] and this is the outer measure (I think this statement should be wrong though) Sidenote: I have not officially encountered Lebesgue Measure yet. I encountered this outer measure thing when the book is trying to show function continuous a.e is Riemann-integrable. I really hope someone can shed some light since I have been puzzled by this the entire day. |
704086 | Find integers [imath]t,z[/imath] satisfying [imath](a+b)z+abt=1[/imath]
Let be [imath]a,b,u,v[/imath] integers and [imath]au+bv=1[/imath]. I have to find integers [imath]t,z[/imath] satisfying [imath](a+b)z+abt=1[/imath]. I think I can solve it by congruence. What do you think? I realised [imath]bz\equiv 1 \quad (\text{mod}\, a)\Longrightarrow z\equiv v \quad (\text{mod}\, a)[/imath] [imath]az\equiv 1 \quad (\text{mod}\, b)\Longrightarrow z\equiv u \quad (\text{mod}\, b)[/imath] Any ideas? Thank you. | 257434 | If [imath](a,b)=1[/imath] then prove [imath](a+b, ab)=1[/imath].
Let [imath]a[/imath] and [imath]b[/imath] be two integers such that [imath]\left(a,b\right) = 1[/imath]. Prove that [imath]\left(a+b, ab\right) = 1[/imath]. [imath](a,b)=1[/imath] means [imath]a[/imath] and [imath]b[/imath] have no prime factors in common [imath]ab[/imath] is simply the product of factors of [imath]a[/imath] and factors of [imath]b[/imath]. Let's say [imath]k\mid a+b[/imath] where [imath]k[/imath] is some factor of [imath]a[/imath]. Then [imath]ka=a+b[/imath] and [imath]ka-a=b[/imath] and [imath]a(k-l)=b[/imath]. So [imath]a(k-l)=b, \ a\mid a(k-1)[/imath] [[imath]a[/imath] divides the left hand side] therefore [imath]a\mid b[/imath] [the right hand side]. But [imath](a,b)=1[/imath] so [imath]a[/imath] cannot divide [imath]b[/imath]. We have a similar argument for [imath]b[/imath]. So [imath]a+b[/imath] is not divisible by any factors of [imath]ab[/imath]. Therefore, [imath](a+b, ab)=1[/imath]. Would this be correct? Am I missing anything? |
672855 | i^i^i^i^... Is there a pattern?
I was messing around with [imath]i[/imath] and I (haha) noticed that certain progressions arise when I keep on raising [imath]i[/imath] to [imath]i[/imath] to [imath]i[/imath] and so forth. Though, I am not really quite sure what is going on (and I don't have time to explore further). In other words, is there an interesting pattern in the sequence: [imath]i[/imath] , [imath]i^i[/imath], [imath]i^{\left(i^i\right)}[/imath], [imath]i^{\left(i^{\left(i^i\right)}\right)}[/imath], etc. | 431145 | Fixed Point of [imath]x_{n+1}=i^{x_n}[/imath]
For [imath]x \in \Bbb C[/imath], let [imath]f(x)=i^x = \exp(i\pi x)[/imath], where [imath]i^2=-1[/imath]. Then find the fixed points for [imath]f[/imath]. EDIT: Let for all [imath]n\geq 1[/imath] [imath]\large a_n=\underbrace{i^{i^{\cdots i}}}_{\text{$n$ times}}[/imath] My question is, does the sequence of tetrations [imath]\{a_n\}_{n\geq1}[/imath] converge to some complex number? If yes, then what is it? |
704245 | Proofs of [imath]\sum_{n=1}^\infty \frac{1}{n^n}=\int_0^1 \frac{1}{x^x}dx[/imath]
Prove that [imath]\sum_{n=1}^\infty \frac{1}{n^n}=\int_0^1 \frac{1}{x^x}dx[/imath]. This comes from this question Evaluate [imath]\sum_0^\infty \frac{1}{n^n}[/imath]. | 237513 | Series as an integral (sophomore's dream)
I need help with this exercise. I need to prove [imath]$$\int_{0}^{1}x^{-x}\ dx=\sum_{n=1}^{\infty}n^{-n}$$[/imath] I think I should use some convergence theorem, but I'm stuck. Thanks a lot! |
704466 | solving a sequence problem
[imath] a_n = (1+ \frac{1} {n}) ^n [/imath] where n is a natural number then which one is greater [imath] a_{2013} [/imath] or [imath] a_{2014} ?[/imath] I tried using binomial expansion but could not solve it. I also tried using monotonicity but expression is complicated. | 64860 | Proving : [imath] \bigl(1+\frac{1}{n+1}\bigr)^{n+1} \gt (1+\frac{1}{n})^{n} [/imath]
How could we prove that this inequality holds [imath] \left(1+\frac{1}{n+1}\right)^{n+1} \gt \left(1+\frac{1}{n} \right)^{n} [/imath] where [imath]n \in \mathbb{N}[/imath], I think we could use the AM-GM inequality for this but not getting how? |
167843 | Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing
Show that [imath]$U_n:=\left(1+\dfrac{1}{n}\right)^n$[/imath], [imath]$n\in\Bbb N$[/imath], defines a monotonically increasing sequence. I must show that [imath]U_{n+1}-U_n\geq0[/imath], i.e. [imath]$$\left(1+\dfrac{1}{n+1}\right)^{n+1}-\left(1+\dfrac{1}{n}\right)^n\geq0.$$[/imath] I am trying to go ahead of this step. | 2324662 | Showing the exponential sequence is strictly increasing
My goal is to show that the sequence [imath]\left(\frac{1}{n} +1\right)^n = e^1[/imath] is strictly increasing. My attempt was to show this by induction. After checking for [imath]n=1[/imath] and [imath]n=2[/imath] I assumed it to be true for [imath]n-[/imath] and [imath]n[/imath]: [imath]\left(\frac{1}{n-1} +1\right)^{n-1} < \left(\frac{1}{n} +1\right)^n = [/imath] And then to show the inequality for [imath]n[/imath] and [imath]n+1[/imath]. This led me nowhere. Another attempt was to secretly assume that I can view [imath]n[/imath] as a real parameter and then differentiate with respect to [imath]n[/imath]. My thought was that if the resulting derivative was strictly positive for positive [imath]n[/imath] or rather [imath]n\geq 1[/imath] then the statement would hold. But I strongly suspect that this is not so easy as it seems. Can anyone offer help? |
704241 | Existence of countable ordinal
Prove that there exist countable ordinal [imath]\xi[/imath] such that [imath]\xi=\omega^\xi[/imath]. The [imath]\xi= \sup \{b^i \mid b_1=w, b_{i+1}=w^{b^i}\}[/imath] should work. But how to prove that [imath]\xi[/imath] is countable? | 181424 | [imath]\varepsilon[/imath]-number countability without choice
Let [imath]\alpha\mapsto\varepsilon_\alpha[/imath] be the enumeration of the [imath]\varepsilon[/imath]-numbers--that is, those [imath]\alpha[/imath] such that [imath]\omega^\alpha=\alpha[/imath]--by the ordinals. If we know that countable unions of countable sets are countable, (slightly weaker than countable choice), then we can show fairly easily that [imath]\varepsilon_\alpha[/imath] is countable iff [imath]\alpha[/imath] is countable. I think I saw a result at some point that [imath]\varepsilon_0[/imath] is countable without relying on any choice principle (though I can't recall where I saw that). Is this true? More generally, for which [imath]\alpha[/imath] can we conclude that [imath]\varepsilon_\alpha[/imath] is countable in ZF? Update: I have been able to prove (in ZF) that if [imath]\varepsilon_\alpha[/imath] is countable, then [imath]|\alpha|<\aleph_1[/imath], so that's one direction. I'm still a bit uncertain about the answers I've got so far. |
705076 | [imath]a,b\in\mathbb{N}[/imath] and [imath]ab>2[/imath].Suppose,lcm[imath](a,b)=L,\gcd(a,b)=G[/imath] and [imath]a+b\mid L+G[/imath].Prove that [imath]\dfrac{(a+b)}{4}(a+b)\ge (L+G)[/imath] and [imath]a,b[/imath] are two odd...
[imath]a,b\in\mathbb{N}[/imath] and [imath]ab>2[/imath].Suppose,[imath]\text{lcm}(a,b)=L,\gcd(a,b)=G[/imath] and [imath]a+b\mid L+G[/imath].Prove that [imath]\dfrac{(a+b)}{4}\cdot(a+b)\ge (L+G)[/imath]. Also prove that equality occurs when [imath]a,b[/imath] are consecutive odd integers. Hint: There is an exception when [imath]a,b[/imath] are both equal to 2. I could not approach this problem at all. Please help. | 675211 | L.C.M. and H.C.F. problem
Let [imath]a[/imath] and [imath]b[/imath] be natural numbers with [imath]ab>2[/imath]. Suppose that the sum of their highest common factor and least common multiple is divisible by [imath]a+b[/imath]. Prove that the quotient is at most [imath]\frac{a+b}4[/imath]. When is this quotient exactly equal to [imath]\frac{a+b}4[/imath]. |
418126 | A function continuous on all irrational points
Let [imath]$h:[0,1]\to\mathbb R$[/imath] [imath]$h(x)=\begin{cases}0&\text{if }x=1\\\frac{1}{n}& \text{otherwise if }x\in\mathbb Q,x=\frac{m}{n},\;m,n\in\mathbb N,\gcd(m,n)=1\\0&\text{otherwise if }x\in\mathbb R\setminus\mathbb Q\end{cases}$[/imath] How do you prove that [imath]h[/imath] is continuous on all irrational points within [imath][0,1][/imath]? | 1105591 | Continous function iff [imath]x \notin \mathbb{Q}[/imath] or [imath]x = 0[/imath]
Let [imath]f : \mathbb{R} \rightarrow \mathbb{R}[/imath] be given as [imath]f(x) = \begin{cases} 0 & \textrm{x = 0} \\ 0 & x \notin \mathbb{Q} \\ \frac{1}{q} & x = \frac{p}{q} \in \mathbb{Q} \\ \end{cases} [/imath] where [imath]\frac{p}{q}[/imath] is irreducible fraction. I have to prove that [imath]f[/imath] is continous in [imath]x[/imath] iff [imath]x \notin \mathbb{Q}[/imath] or [imath]x = 0[/imath]. If [imath]x = 0[/imath] then [imath]f(x)[/imath] and for every [imath]a_n \to 0[/imath] we have [imath]f(a_n) \to 0[/imath] because we know that [imath]\forall_{\epsilon > 0}{\exists_{N \in \mathbb{N}}{\forall_{n > N}{|a_n|<\epsilon}}}[/imath] so we take [imath]a_n[/imath] for [imath]n > N[/imath] and if [imath]a_n = 0[/imath] or [imath]a_n \notin\mathbb{Q}[/imath] then [imath]f(a_n) =0[/imath] by definition, otherwise [imath]|f(a_n)| < |a_n|[/imath] so [imath]|f(n)| < \epsilon[/imath]. How to make proof for other cases? |
705078 | Taylor series expansion for [imath]e^{\sin{x}}[/imath]
Given the function [imath]f(x)= e^{\sin{x}}[/imath] I have to write it without using the exponential or sine function. I came to this point [imath]f(x) = \sum_{k=0}^{\infty} \frac{\sin^k{x}}{k!}[/imath] How can I get rid of that [imath]\sin^k(x)[/imath]? Thanks in advance! | 704250 | Taylor series expansion and the radius of convergence
Hello I have some problems concerning Taylor series. Given the function [imath]f(x)=e^{\sin{x}} [/imath] I concluded that the Taylor series expansion would be [imath]f(x) = \sum^\infty_{n=0}\frac{1}{n!}f^{(n)}(x)(x-x_0)^n[/imath] But Wolfram wrote a much simpler form [imath]f(x)=\sum^\infty_{n=0}\frac{\sin^k{x}}{k!}[/imath] My question is: how come? Secondly, could somebody explain to me how to get the second sum's radius of convergence? Thank you very much for your time! |
701873 | Homology of A Compact Manifold
I am working on a compact manifold [imath]M[/imath] of dimension [imath]m[/imath]. Moreover, I suppose that [imath]M[/imath] is oriented of fundamental class [imath] [M] \in H_m (M)[/imath]. I want to show that I have an isomorphism [imath]H_\ast (M, \mathbb{Q}) \cong H_\ast (M) \otimes \mathbb{Q}[/imath]. I figured I'd use the poincaré duality which tells me that [imath] H^i (M) \cong H_{m-i} (M)[/imath] via the map [imath] - \cap [M][/imath] ? Any help would be greatly appreciated! | 622158 | Singular homology [imath]S^{n}[/imath] with rational coefficients
What does [imath]H_{i} (S^{n} , \mathbb{Q} )[/imath] equal? Where this denotes singular homology with rational coefficients. |
706011 | Why is cross product only defined in 3 and 7 dimensions?
Why [imath]3[/imath] and [imath]7[/imath]? I know from some reading that Hurwitz's Theorem explains this, but can someone help me build some intuition behind this or perhaps provide a simpler explanation? It still seems mysterious to me. | 185991 | Is the vector cross product only defined for 3D?
Wikipedia introduces the vector product for two vectors [imath]\vec a[/imath] and [imath]\vec b[/imath] as [imath] \vec a \times\vec b=(\| \vec a\| \|\vec b\|\sin\Theta)\vec n [/imath] It then mentions that [imath]\vec n[/imath] is the vector normal to the plane made by [imath]\vec a[/imath] and [imath]\vec b[/imath], implying that [imath]\vec a[/imath] and [imath]\vec b[/imath] are 3D vectors. Wikipedia mentions something about a 7D cross product, but I'm not going to pretend I understand that. My idea, which remains unconfirmed with any source, is that a cross product can be thought of a vector which is orthogonal to all vectors which you are crossing. If, and that's a big IF, this is right over all dimensions, we know that for a set of [imath]n-1[/imath] [imath]n[/imath]-dimensional vectors, there exists a vector which is orthogonal to all of them. The magnitude would have something to do with the area/volume/hypervolume/etc. made by the vectors we are crossing. Am I right to guess that this multidimensional aspect of cross vectors exists or is that last part utter rubbish? |
306744 | Can the definition of continuity be said both of these ways?
So if the definition of continuity is: [imath]\forall[/imath] [imath]\epsilon \gt 0[/imath] [imath]\exists[/imath] [imath]\delta \gt 0:|x-t|\lt \delta \implies |f(x)-f(t)|\lt \epsilon[/imath]. However, I get confused when I think of it this way because it's first talking about the [imath]\epsilon[/imath] and then it talks of the [imath]\delta[/imath] condition. Would it be equivalent to say: [imath]\forall[/imath] [imath]\delta \gt 0[/imath] [imath]\exists[/imath] [imath]\epsilon \gt0[/imath] [imath]:|x-t|\lt \delta \implies|f(x)-f(t)|\lt \epsilon[/imath]. I guess what I'm asking is whether there is a certain order proofs or more formal statements need to follow. I know I only changed the place where I said there is a [imath]\delta[/imath] but is that permissable in a "formal" way of writing? | 192883 | What's wrong with this "backwards" definition of limit?
Can anyone please give an example of why the following definition of [imath]\displaystyle{\lim_{x \to a} f(x) =L}[/imath] is NOT correct?: [imath]\forall[/imath] [imath]\delta >0[/imath] [imath]\exists[/imath] [imath]\epsilon>0[/imath] such that if [imath]0<|x-a|<\delta[/imath] then [imath]|f(x)-L|<\epsilon[/imath] I've been trying to solve this for a while, and I think it would give me a greater understanding of why the limit definition is what it is, because this alternative definition seems quite logical and similar to the real one, yet it supposedly shouldn't work. |
706846 | If [imath]0 < r < 1[/imath] show that [imath]r^n[/imath] goes to [imath]0[/imath] as [imath]n \to ∞[/imath].
If [imath]0 < r < 1[/imath] show that [imath]r^n[/imath] goes to [imath]0[/imath] as [imath]n \to ∞[/imath]. If [imath]|r| < 1[/imath] then [imath]r^2 < r[/imath] similarly [imath]r^4 < r^3 < r^2 < r[/imath] so [imath]r^n[/imath] as [imath]n \to +\infty[/imath] will be equal to [imath]∞[/imath] Let [imath]r=\frac1m , m>1 \to[/imath] you need to show [imath]lim_{n\to +\infty} r^n = ∞[/imath] Since [imath]0<r<1[/imath] [imath]lim_{n\to +\infty} r^n = lim_{n\to +\infty} \left(\frac1m\right)^n=lim_{n\to +\infty}\frac{1}{m^n} = \frac{1}{+\infty} = ∞[/imath] | 706800 | If [imath]0 show that r^n goes to 0 as n goes to 1.[/imath]
If [imath]0 < r < 1[/imath] show that [imath]r^n[/imath] goes to [imath]0[/imath] as [imath]n \to 1[/imath]. If [imath]|r| < 1[/imath] then [imath]r^2 < r[/imath] similarly [imath]r^4 < r^3 < r^2 < r[/imath] so [imath]r^n[/imath] as [imath]n \to +\infty[/imath] will be equal to [imath]1[/imath] how can i finish this problem? |
350844 | cardinality of the set of [imath] \varphi: \mathbb N \to \mathbb N[/imath] such that [imath]\varphi[/imath] is an increasing sequence
I know that the set of functions [imath] f:\mathbb N \to \mathbb N[/imath] is uncountable, but what if we consider only [imath]f[/imath] such that [imath]f[/imath] is increasing? I want to know if this set is countable D: and also the case of bijective and increasing (clearly if the firstone holds then this too). I think that it could be possible that it's countable | 2407091 | Prove that set [imath]\{f \in \mathbb{N^N} \: | \:f \: [/imath]is strictly increasing [imath]\}[/imath] has the same cardinality as [imath]\mathbb R[/imath]
Prove, that set [imath]\{f \in \mathbb{N^N} \: | \:f \: [/imath]is strictly increasing [imath]\}[/imath] has the same cardinality as [imath]\mathbb R[/imath]. My attempts: The beginning of this task was quite easy, but then I got stuck on constructing an injection between a set of function (let's call it [imath]X[/imath]) and [imath]\mathbb R[/imath]. I started with proving that [imath]|\mathbb R| \geq |X|[/imath]: [imath]|\mathbb R| \geq |X|[/imath] because if [imath]\forall _f , f\in \mathbb{N^N}[/imath], and [imath]|\mathbb{N^N}|=|\mathbb R[/imath]|, then [imath]X \subset\mathbb R[/imath]. Then I tried to prove that [imath]|\mathbb R| \leq |X|[/imath], but I don't know how to do it. I tried to define a function [imath]g(x)=x^3[/imath], but the result is a number, not a function. Or maybe it is a correct solution? If not, can you explain to me how can I construct an injective function from [imath]\{f \in \mathbb{N^N} \: | \:f \: [/imath]is strictly increasing [imath]\}[/imath] to [imath]\mathbb R[/imath]? Is it even possible? |
699859 | Trust region sub-problem Explicit Formula
Consider the [imath]2 \times 2[/imath] trust region sub-problem. Given [imath]Q[/imath] symmetric [imath]2 \times 2[/imath], vector [imath]\mathbf b[/imath] and [imath]\Delta > 0[/imath], find [imath]\mathbf x[/imath] that minimizes [imath]f(x)=\frac {1}{2} \mathbf x^T Q \mathbf x - \mathbf b^T \mathbf x[/imath] over [imath]\mathbf x[/imath] satisfying [imath]||\mathbf x||_{2}^{2} \leq \Delta[/imath]. We can use the necessary conditions: there is a [imath]\lambda \geq 0[/imath] where [imath]Q + \lambda I[/imath] is positive semi-definite [imath](Q + \lambda I)\mathbf x=\mathbf b[/imath] either [imath]\lambda =0[/imath] or [imath]||\mathbf x||_{2} =\Delta[/imath] Give an explicit formula for [imath]\mathbf x(\lambda) := (Q + \lambda I)^{-1}\mathbf b[/imath] I think that I want to take out det[imath]( Q + \lambda I)^2[/imath] as a common factor but I'm not sure where to go from there. | 699702 | Trust region sub-problem with Jacobi Condition
Consider the [imath]2 \times 2[/imath] trust region sub-problem. Given [imath]Q[/imath] symmetric [imath]2 \times 2[/imath], vector [imath]\mathbf b[/imath] and [imath]\Delta > 0[/imath], find [imath]\mathbf x[/imath] that minimizes [imath]f(x)=\frac {1}{2} \mathbf x^T Q \mathbf x - \mathbf b^T \mathbf x[/imath] over [imath]\mathbf x[/imath] satisfying [imath]||\mathbf x||_{2}^{2} \leq \Delta[/imath]. We can use the necessary conditions: there is a [imath]\lambda \geq 0[/imath] where [imath]Q + \lambda I[/imath] is positive semi-definite [imath](Q + \lambda I)\mathbf x=\mathbf b[/imath] either [imath]\lambda =0[/imath] or [imath]||\mathbf x||_{2} =\Delta[/imath] Use the Jacobi conditions to find the [imath]\lambda[/imath] for which [imath]Q + \lambda I[/imath] is positive definite. (These conditions are that [imath]q_{11} + \lambda > 0[/imath] and det[imath](Q + \lambda I >0[/imath]) Then find the minimum value of [imath]\lambda \geq 0[/imath] for which [imath]Q + \lambda I[/imath] is positive semi-definite. I'm not exactly sure how to approach this problem and any advice would be greatly appreciated! I do know that a matrix B is positive semi-definite if and only if [imath]B+\epsilon I[/imath] is positive definite for all [imath]\epsilon>)[/imath] |
437835 | Calculate sum of squares of first $n$ odd numbers
Is there an analytical expression for the summation [imath]$$1^2+3^2+5^2+\cdots+(2n-1)^2,$$[/imath] and how do you derive it? | 1020031 | Find the exact closed from expression of [imath]1^2 + 3^2 + 5^2 + · · · + (2n + 1)^ 2[/imath]
I know the above expression equals to [imath]\frac{n(2n−1)(2n+1)}{3}[/imath], but how exactly can i come up with something from scratch? |
706974 | Question about identity of Dirac delta function
I am trying to understand an identity of the [imath]\delta[/imath]-function written on this Wikipedia page: \begin{equation} \int \mathrm{d} x \; f(x) \delta[g(x)] = \sum\limits_i \frac{f(x_i)}{\left| \frac{dg(x_i)}{dx}\right|} \tag{1} \end{equation} where [imath]x_i[/imath] are the zeros of [imath]g(x)[/imath] (i.e. [imath]g(x_i)=0[/imath]). Now, my question is about the denominator on the right-hand side of equation [imath](1)[/imath]. Is that supposed to be a Jacobian determinant: \begin{equation} \left|\frac{dg(x_i)}{dx}\right| \overset{?}{=} \mathrm{det}\left(\frac{dg(x_i)}{dx}\right) \end{equation} or is it supposed to mean the modulus? The reason I think it might be a Jacobian is because Wiki mentions that we can use the following identity to change variables of integration: \begin{equation} \int_{\mathbf{R}} \delta\bigl(g(x)\bigr) f\bigl(g(x)\bigr) |g'(x)|\,dx = \int_{g(\mathbf{R})} \delta(u)f(u)\, du \end{equation} | 662226 | How to write [imath]\delta (f(x))[/imath] in terms of [imath]\delta (x)[/imath]?
I've seen this identity in my electrodynamics book: [imath]\delta (f(x))=\sum_i{ \frac{1}{|{df\over dx}(x_i)|}\delta (x-x_i)}[/imath] Where [imath]x_i[/imath] shows the [imath]i[/imath]th zero of [imath]f(x)[/imath]. How can I prove it? I've tried the integral definition of delta function, but doesn't work. |
707559 | Problem of the Week!
This week in Algebra II we are studying the Hanoi tower's. Our assignment was to find what type of formula would give the number of moves it would take to solve the puzzle. After using a T-chart (where [imath]x=[/imath]number of disks and [imath]y=[/imath] shortest number of moves that gives the solution)I found that (0 , 0) (1 , 1) (2 , 3) (3 , 7) (4 , 15) and (5 , 31). Using the information in the T-chart one should be able to find that [imath]y=2^x-1[/imath], however I don't know how to come to this conclusion on my own. My question today is, how do I find this formula with the given T-chart? That is my only question.....(P.S) everything I need to complete the assignment is done so this is more or less of an add on to improve what I know, NOT to get answers. | 666989 | Is it possible to play the Tower of Hanoi with fewer than [imath]2^n-1[/imath] moves?
The Tower of Hanoi game consists of three identical upright pegs and n rings all of different diameters that can be stacked over any of the pegs. Initially, all of the rings are stacked around one of the pegs in order of decreasing diameter with the largest ring on the bottom. The object of the game is to transfer all the rings, one at a time, until they are stacked in the same order around another peg, but at no time may any ring be placed above a ring of smaller diameter. Prove that, for any number of rings, the transfer can be made in exactly [imath]2^n-1[/imath] moves. We need to use induction... Basis Step IF we let [imath]n=1[/imath], then the left hand side must be 1 and the right hand side must be one as well. [imath]1 = 2^1-1[/imath] [imath]1 = 1[/imath] Induction Step. If [imath]P(k)[/imath] is true, then we can move [imath]k[/imath] disks in [imath]2^k-1[/imath] moves. If [imath]P(k+1)[/imath] is true, then we can move [imath]k+1[/imath] disks [imath]2^{k+1}-1[/imath] Let [imath]2^k - 1 [/imath] be the number of moves for [imath]k[/imath] disks. Then, [imath]2(2^k-1)+1[/imath] Using our induction hypothesis, we have [imath]2(2^{k+1}-1)[/imath]. For any [imath] k \geq 1[/imath], [imath]P(k)[/imath] and [imath]P(k+1)[/imath] is true. Edit: This isn't right at all... it's just fluff...I need to start over. Can the Tower of Hanoi be played in less than [imath]2^n-1[/imath] steps? This is where I'm stuck... My guess is that unless you break the rules of the game, then you can't play in fewer steps. Do I use induction to prove that it's impossible or possible? |
707660 | Determination of a connected groupoid by its objects and by a set of automorphisms.
One may readily show that a connected groupoid [imath]G[/imath] is determined up to isomorphism by a group (one of the groups [imath]\hom_G(x,x)[/imath]) and by a set (the set of all objects). This is the nature of the problem I'm trying to understand. Well, I see the [imath]\hom_G(x,x)[/imath] determines all automorphisms because they all are isomorphic each other. I note also that given two arrows in [imath]\hom_G(x,x)[/imath], [imath]f[/imath] and [imath]g[/imath], the composites with the arrow (which must exist because of connectivity) [imath]h \in[/imath] [imath]\hom_G(x,x')[/imath], [imath]hf[/imath] and [imath]hg[/imath], are different morphisms. But there's one thing I don't understand: Because of the determinability of [imath]G[/imath] by the given information (the group of automorphisms and the objects), I thought I had to prove that, given any [imath]h'[/imath], it must exist a morphism [imath]f[/imath] in [imath]\hom_G(x,x)[/imath] such that [imath]h'= hf[/imath]. This fact, however, I cannot see. Thanks in advance for any suggestion. | 419939 | A comparison between the fundamental groupoid and the fundamental group
Are there two path connected topological spaces [imath]X,Y[/imath] such that the fundamental groupoid of [imath]X[/imath] is not isomorphic to the fundamental groupoid of [imath]Y[/imath] but the fundamental group of [imath]X[/imath] is isomorphic to the fundamental group of [imath]Y[/imath] ? I guess that there exists such a pair of topological spaces. I don't know an example though. I am very interested to see such a pair. Edit: The first version of the question was already solved by Zev Chonoles. Here is the second version of the question Are there two path-connected topological spaces [imath]X,Y[/imath] such that: 1) [imath]|X|=|Y|[/imath] 2) The fundamental group of [imath]X[/imath] is isomorphic to the fundamental group of [imath]Y[/imath] 3) The fundamental groupoid of [imath]X[/imath] is not isomorphic to the fundamental groupoid of [imath]Y[/imath] In other words, this is a comparison between the fundamental groupoid functor and the combined use of the fundamental group and the forgetful functor from Top to Set Thank you |
708009 | Can the derivative of a [imath]C^1[/imath] function vanish at the infinity imply it has a limit at the infinity?
Now I'm interesting in the question as follow, Let [imath]f\in C^1(\mathbb{R})[/imath] and [imath]f[/imath] is bounded, [imath]\lim_{x\rightarrow+\infty}f'(x)=0[/imath], then [imath]\lim_{x\rightarrow+\infty}f(x)[/imath] exists. Is this statement true? If not, please give a counterexample! My try: if we remove the boundness condition of the function [imath]f[/imath], I can give a counterexample on the half line, namely, [imath]f(x):=\ln x[/imath]. Any answer will be appreciated! | 348417 | Limit of derivative is zero. Does it imply a limit for f(x)?
I have come across the following question: Say [imath]f(x): [0,\infty) \rightarrow \mathbb{R}[/imath] differentiable. Given [imath]\lim_{x\rightarrow\infty}f'(x) = 0[/imath], does this imply an existence of a limit (finite or not) of [imath]f[/imath] at infinity? From what I tried to do, looking at the definition of the derivative: [imath]\lim_{\substack{h\rightarrow0 \\ x\rightarrow\infty}}\frac{f(x+h)-f(x)}{h} = 0[/imath] you could say that [imath]\lim_{\substack{h\rightarrow0 \\ x\rightarrow\infty}} f(x+h) = \lim_{x\rightarrow\infty} f(x)[/imath]. Is it enough to prove the existence of a limit? if not, is there another way to prove or disprove that? |
708693 | How to prove [imath]n^3 < 4^n[/imath] using induction?
It's true for all Natural numbers. What I've got so far: Prove [imath]P(0) \to [/imath] base case: Let [imath]n = 0[/imath] [imath](0)^3 < 4^0 = 0 < 1[/imath] Then [imath]P(0)[/imath] is true. Part Two: Prove [imath]P(n) \Rightarrow P(n + 1) [/imath] Assume [imath]P(n)[/imath] [imath]= n^3 < 4^n [/imath] [imath]= 4(n + 1)^3 < 4^{(n + 1)}[/imath] im not sure if the last step is right. Where can I go from here? | 706159 | How to prove that for all natural numbers, [imath]4^n > n^3[/imath]?
This is a problem set I have, it's not a homework but it's very important practice... Send me some hints please, I don't want an answer I need to get it by myself but I'm failing miserably... The problem is: Prove that for all natural numbers [imath]n[/imath], [imath]4^n > n^3[/imath] |
708927 | Showing how this infinite sum diverges
[imath]\displaystyle{\sum_{n=1}^{\infty} \left[\left( 1 + {1 \over n}\right)^{n} - {\rm e}\right]}[/imath] I tried both root and ratio tests (for the root test, the expression became way too complex to handle) but both didn't really work out. Would there be an easy way of showing that this diverges? | 703207 | Find the exact value of the infinite sum [imath]\sum_{n=1}^\infty \big\{\mathrm{e}-\big(1+\frac1n\big)^{n}\big\}[/imath]
How can we find the exact value of the infinite sum [imath] \displaystyle\sum_{n=1}^\infty \left\{\mathrm{e}-\Big(1+\frac1n\Big)^n\right\}? [/imath] This problem appears in: T. Andreescu, T. Radulescu & V. Radulescu, Problems in Real Analysis: Advanced Calculus on the real line, p.114. |
708977 | Intro Math Matrices problem
Find all [imath]2 \times 2[/imath] matrices [imath]A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}[/imath] such that [imath]AB= BA [/imath] for all [imath]2\times2 [/imath] matrices B. I am stumped as to how to approach this question overall. I am not sure if its asking for me to use some theorem or if some easier solution is present. | 199210 | How can I prove the IFF condition for this group of matrices?
I'm tasked with showing that matrix [imath]A[/imath] commutes with every [imath]2\times2[/imath] matrix if and only if [imath]A = \begin{bmatrix}a & 0\\0 & a\end{bmatrix}[/imath] for some a. I was able to prove in the first direction, assuming that [imath]A = \begin{bmatrix}a & 0\\0 & a\end{bmatrix}[/imath], by just multiplying the matrix by [imath]\begin{bmatrix}a & b\\c & d\end{bmatrix}[/imath] in AB = BA format, and showing how AB does equal BA. [imath] (=>)[/imath] [imath]\begin{bmatrix}a & 0\\0 & a\end{bmatrix}\begin{bmatrix}a & b\\c & d\end{bmatrix} = \begin{bmatrix}a & b\\c & d\end{bmatrix}\begin{bmatrix}a & 0\\0 & a\end{bmatrix}[/imath] [imath]\begin{bmatrix}a^2 & ab\\ac & ad\end{bmatrix} = \begin{bmatrix}a^2 & ab\\ac & ad\end{bmatrix}[/imath] But now I have to prove in the other direction, assuming that A already does commute with every [imath]2\times 2[/imath] matrix, and I'm not particularly sure how I can do that. Somehow, I have to show that if A commutes with [imath]\begin{bmatrix}a & b\\c & d\end{bmatrix}[/imath], then A will be [imath]\begin{bmatrix}a & 0\\0 & a\end{bmatrix}[/imath], and I don't know how I can carry that out. Where can I start? Ideas are appreciated. |
708120 | The property of positive fourier series.
This is the problem in the book 'Classical and multilinear harmonic analysis, volume 1' Let [imath]f(x)=\sum_{n=0}^{N}[a_{n}\cos{2\pi nx}+b_{n}\sin{2\pi nx}][/imath] be a nonnegative function defiend on [imath][0,1][/imath]. Show that there exists an complex sequence [imath]\{c_{n} \}_{n=0}^{N}[/imath] such that [imath]f(x)=|\sum_{n=0}^{N}c_{n}e^{2\pi inx}|^2[/imath] I have no idea. How can we use the property that [imath]f[/imath] is a positive function? | 316436 | Finding a trigonometric polynomial
I'm trying to solve exercise 5 in chapter 14 of Rudin's Real & Complex Analysis: Suppose [imath]f[/imath] is a trigonometric polynomial, [imath]f(\theta) = \sum_{k=-n}^n a_k e^{ik\theta}[/imath] and [imath]f(\theta) > 0[/imath] for all real [imath]\theta[/imath]. Prove that there is a polynomial [imath]P(z) = c_0 + c_1z + ... + c_nz^n[/imath] such that [imath]f(\theta)= |P(e^{i\theta})|^2[/imath] ([imath]\theta[/imath] real). I've made good progress but I can not finish it. My thoughts: Define [imath]F(z) = \sum_{k=-n}^n a_k z^k[/imath]. This is a rational function that is positive on the unit circle so it must be of the form: [imath]F(z) = c \prod_{j=1}^n\frac{(z-\beta_j)(1-\overline\beta_j z)}{(z-\gamma_j)(1-\overline\gamma_j z)}[/imath] where [imath]c > 0[/imath]. [imath]\beta_j[/imath] are zeros of [imath]F[/imath]. [imath]\gamma_j[/imath] are poles. On the unit circle [imath]f(\theta) = F(e^{i\theta}) = c \prod_{j=1}^n\frac{(e^{i\theta}-\beta_j)(1-\overline\beta_j e^{i\theta})}{(e^{i\theta}-\gamma_j)(1-\overline\gamma_j e^{i\theta})}[/imath] The terms in the product simplify to [imath]\left| \frac{e^{i\theta} - \beta_j}{e^{i\theta} - \gamma_j} \right|^2[/imath] If I can show that this is a polynomial, I'll solve the exercise, but it doesn't look like a polynomial to me. Is there a better way? |
709206 | A basic question on distribution function and stieljes integral
Is it true that [imath]\int_{\Bbb R} G(x)dF(x) + \int_{\Bbb R} F(y)dG(y) = 1[/imath] where [imath]F[/imath] and [imath]G[/imath] are distribution functions of [imath]X[/imath] and [imath]Y[/imath] such that both have no common jump points i.e. [imath]P(X=Y) = 0[/imath] ? How to prove it ? I am unfamiliar with Stieljes integral | 708468 | A basic question on expectation of distribution composed random variables
Suppose that [imath]X[/imath] and [imath]Y[/imath] are random variables with distribution functions [imath]F[/imath] and [imath]G[/imath]. If [imath]F[/imath] and [imath]G[/imath] have no common jumps then I need to show that [imath]E[F(Y)] + E[G(X)] = 1[/imath]. How to proceed here ? Shall I try to show that the LHS is some probability of almost sure events or product of two almost sure events. Clearly, [imath]F(Y)[/imath] and [imath]G(X)[/imath] are random variables with range in [imath][0,1][/imath]. |
709459 | Integral of composition
Prove that if [imath]f,g:[0,1]\rightarrow[0,1][/imath] - continuous functions and f is strictly increasing then [imath]\int\limits_0^1f(g(x))dx\leq\int\limits_0^1f(x)dx+\int\limits_0^1g(x)dx.[/imath] I tried to prove that [imath]f(g(x))\leq f(x)+g(x)[/imath] for all [imath]x\in[0,1][/imath] but realized that it is wrong. For example, [imath]f(x)=\begin{cases}0,0\leq x< \frac{1}{4}\\2x-\frac{1}{2},\frac{1}{4}\leq x < \frac{3}{4} \\1,\frac{3}{4}\leq x\leq 1\end{cases}[/imath], [imath]g(x)=\frac{3}{4}[/imath] but [imath]f(g(0))=1>0+\frac{3}{4}=f(0)+g(0)[/imath]. | 276722 | Prove the following integral inequality: [imath]\int_{0}^{1}f(g(x))dx\le\int_{0}^{1}f(x)dx+\int_{0}^{1}g(x)dx[/imath]
Suppose [imath]f(x)[/imath] and [imath]g(x)[/imath] are continuous function from [imath][0,1]\rightarrow [0,1][/imath], and [imath]f[/imath] is monotone increasing, then how to prove the following inequality: [imath]\int_{0}^{1}f(g(x))dx\le\int_{0}^{1}f(x)dx+\int_{0}^{1}g(x)dx[/imath] |
709517 | [imath]n^x[/imath], [imath]x[/imath] is irrational number what is the result?
if for a real number [imath]n[/imath] , [imath]n^x[/imath] where [imath]x[/imath] is an irrational number. What is nature of number [imath]n^x[/imath] ? Possible values or how to determine nature of number [imath]n^x[/imath] ? value of square root of 2 rays to square root of 2. | 55068 | Can you raise a number to an irrational exponent?
The way that I was taught it in 8th grade algebra, a number raised to a fractional exponent, i.e. [imath]a^\frac x y[/imath] is equivalent to the denominatorth root of the number raised to the numerator, i.e. [imath]\sqrt[y]{a^x}[/imath]. So what happens when you raise a number to an irrational number? Obviously it is not so simple to break it down like above. Does an irrational exponent still have a well formed meaning? The only example that comes to mind is Euler's identity, but that seems likes a pretty exceptional case. What about in general? |
709753 | Different definitions of topological group
Recently I discovered the definition of topological group. So, topological group is an abstract group [imath]G[/imath] endowed with topological structure such that the maps [imath]mult: G\times G\longrightarrow G[/imath] and [imath]inv:G\longrightarrow G[/imath] (multiplication and inversion) are continuous. I'd like to understand if it is enough just to suppose [imath]mult[/imath] to be continuous. Could you give me some slight hint? | 172945 | Can continuity of inverse be omitted from the definition of topological group?
According to Wikipedia, a topological group [imath]G[/imath] is a group and a topological space such that [imath] (x,y) \mapsto xy[/imath] and [imath] x \mapsto x^{-1}[/imath] are continuous. The second requirement follows from the first one, no? (by taking [imath]y=0, x=-x[/imath] in the first requirement) So we can drop it in the definition, right? |
706458 | Calculate limiting distribution of [imath]\frac{\sum_{i=1}^n X_i}{\sum_{i=1}^n Y_i}[/imath]
Let [imath]X_1,X_2,\ldots,X_n[/imath] be a random sample of Bernoulli distribution with parameter [imath]\frac{\theta_1}{\theta_1+\theta_2}[/imath] and [imath]Y_1,Y_2,\ldots,Y_n[/imath] be a random sample of geometric distribution with parameter [imath]\theta_1+\theta_2[/imath]. If the two samples are independent of each other, how can I calculate the limiting distribution of [imath]\dfrac{\sum_{i=1}^n X_i}{\sum_{i=1}^n Y_i}[/imath] ? | 707047 | Limiting distribution [imath]\frac{\sum_{i=1}^n X_i}{\sum_{i=1}^n Y_i}[/imath]
let [imath]X_1,X_2,\ldots,X_n[/imath] be random sample of bernoulli distribution with parameter of [imath]\displaystyle\frac{\theta_1}{\theta_1+\theta_2}[/imath] and let [imath]Y_1,Y_2,\ldots,Y_n[/imath] be random sample of geometric distribution with parameter [imath]\theta_1+\theta_2[/imath]. If the two samples are independent of each other, how can I calculate the limiting distribution [imath]\displaystyle\frac{\sum_{i=1}^n X_i}{\sum_{i=1}^n Y_i}[/imath]. |
709867 | Suppose that matric [imath]AB+A+B=0[/imath], how to prove that [imath]AB=BA[/imath]?
Assume that [imath]A[/imath] and [imath]B[/imath] and [imath]n\times n[/imath] matrices, and [imath] AB+A+B=0. [/imath] How can we prove that [imath]AB=BA[/imath]? Thank you in advance. Any help is much appreciated. | 702256 | Proof involving matrix equation
[imath]A[/imath] and [imath]B[/imath] are [imath](n\times n)[/imath] matrices and [imath]AB + B + A = 0[/imath]. Prove that then [imath]AB=BA[/imath]. How should I approach this problem? |
710146 | If the domain is a compact set, pointwise equicontinuity implies uniform equicontinuity.
Conversely, is it true that if every sequence of pointwise equicontinuous functions from [imath]M[/imath] to [imath]\mathbb{R}[/imath] is uniformly equicontinuous, them [imath]M[/imath] is compact? | 402160 | Does pointwise equicontinuous and uniformly equicontinuous implies compactness?
If every sequence of pointwise equicontinuous functions [imath]M \rightarrow \mathbb{R}[/imath] is uniformly equicontinuous, does this imply that [imath]M[/imath] is compact? |
702292 | Mutlivariable integral, How to compute it?
Can anybody please tell me, how to evaluate a multivariate integral with a gaussian weight function. [imath]\mathcal{Z_{n}}=\int_{-\infty}^{\infty} dx_{1}dx_{2}dx_{3}dx_{4}...dx_{n}\exp(-\frac{a}{2}\sum_{j}x_{j}^2)\times f(x_{1},x_{2},x_{3},x_{4}....x_{n})[/imath] where [imath]f(x_{1},x_{2},x_{3},x_{4}....x_{n})=\Pi_{j} \frac{1}{\sqrt{(1+i\,b(x_{j}^2-x_{j+1}^2)^2)}}[/imath], and [imath]i=\sqrt{-1}[/imath]. Also we have the condition [imath]x_{n+1}=x_{1}[/imath]. I need a hint to solve this integral. This is how I proceeded, \begin{eqnarray} \mathcal{Z_{n}}=\int_{-\infty}^{\infty} \Pi_{j=1}^{n} dx_{j} \exp{\Bigg(-\frac{a}{2}\sum_{j=1}^{n}x_{j}^2-\frac{1}{2}\log{\Big(1+i b(x_{j}^2-x_{j+1}^2)^{2}\Big)\Bigg)}} \end{eqnarray} Now the integral is of the form of the canonical partition function integrated over the configuration space. Hence the integral can be identified as an [imath]n-[/imath]particle partition of the canonical ensemble, which is given by \begin{eqnarray} \mathcal{Z}_{n}= \int_{-\infty}^{\infty} \Pi_{j=1}^{n} dx_{j} e^{-\beta \mathcal{H}}, \end{eqnarray} where [imath]\mathcal{H}=\Bigg(\frac{a}{2}\sum_{j=1}^{n}x_{j}^2+\frac{1}{2}\log{\Big(1+i b(x_{j}^2-x_{j+1}^2)^{2}\Big)\Bigg)}.[/imath] and [imath]\beta=1[/imath]. Then I got stuck ! | 697647 | Multivariable Integral, How to compute it?
Can anybody please tell me, how to evaluate a multivariate integral with a gaussian weight function. [imath] \mathcal{Z_{n}} \equiv\int_{-\infty}^{\infty} \exp\left(-a\sum_{j = 1}^{n}x_{j}^2\right)\, {\rm f}\left(x_{1},x_{2},\ldots,x_{n}\right)\, {\rm d}x_{1}\,{\rm d}x_{2}\ldots{\rm d}x_{n} [/imath] where [imath]\displaystyle{% {\rm f}(x_{1},x_{2},\ldots,x_{n}) =\prod_{j} {1 \over \sqrt{1 + {\rm i}\,b\left(x_{j}^2-x_{j+1}^2\right)^2}}}[/imath]. I need a hint to solve this integral. This is how I proceeded, [imath] \mathcal{Z_{n}}= \int_{-\infty}^{\infty}\prod_{j=1}^{n}{\rm d}x_{j}\, \exp\left(% -\,{a \over 2}\sum_{j = 1}^{n}x_{j}^{2} - {1 \over 2} \log\left(1 + {\rm i}\,b\left[x_{j}^{2} - x_{j + 1}^{2}\right]^{2}\right)\right) [/imath] Now the integral is of the form of the canonical partition function integrated over the configuration space. Hence the integral can be identified as an [imath]n-[/imath]particle partition of the canonical ensemble, which is given by [imath] \mathcal{Z}_{n}= \int_{-\infty}^{\infty}\prod_{j = 1}^{n}{\rm d}x_{j}\, {\rm e}^{-\beta{\cal H}}, [/imath] where [imath]\mathcal{H}=\Bigg(-\frac{a}{2}\sum_{j=1}^{n}x_{j}^2-\frac{1}{2}\log{\Big(1+i b(x_{j}^2-x_{j+1}^2)^{2}\Big)\Bigg)}.[/imath] Then I got stuck. How to proceed? Thanks. |
710361 | Show that [imath]n^5/5 + n^3/3 + 7n/15[/imath] is an integer for every n
So far I have combined the fractions to get [imath] \ (3n^5 + 5n^3 +7n)/15 \ [/imath] and that is equal to [imath][n(3n^4 + 5n^2 + 7)]/15[/imath]. So I need to show that [imath]15\mid n(3n^4 + 5n^2 + 7)[/imath]. Case 1: 15 divides [imath]n[/imath]. Then we are done. Case 2: 15 does not divide n. Then I need to show that [imath]15\mid(3n^4 + 5n^2 + 7)[/imath]. This is where I am stuck. Can someone please tell me if I am going in the right direction or if I need to do it a different way? | 21548 | Using congruences, show [imath]\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n[/imath] is integer for every [imath]n[/imath]
Using congruences, show that the following is always an integer for every integer value of [imath]n[/imath]: [imath]\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n.[/imath] |
710570 | Binomial theorem [imath](a+b)^n=\sum \limits_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}[/imath]
I'm trying to understand the proof by induction of: [imath] (a+b)^n = \sum \limits_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k} [/imath] I'm at the point of deriving the inductive step and am getting next: [imath] (a+b)^{n+1} = (a+b)(a+b)^n=\dotsb = a^{n+1} + b^{n+1} + \sum_{k=1}^{n}\binom{n}{k}a^{n+1-k}b^{k} + \sum_{k=1}^{n}\binom{n}{k-1}a^{n-1-k}b^{k} [/imath] Now, I also have a solution to this problem, and in the solution the two summations are turned into exactly the middle elements of the series (since [imath]a^{n+1}[/imath] and [imath]b^{n+1}[/imath] are the first and the last elements of the series correspondingly) and we get a perfect proof. However, I don't understand how next was achieved: [imath] \sum_{k=1}^{n}\binom{n}{k}a^{n+1-k}b^{k} + \sum_{k=1}^{n}\binom{n}{k-1}a^{n-1-k}b^{k} = \sum_{k=1}^{n}\binom{n+1}{k}a^{n+1-k}b^{k} [/imath] Any advises and hints are welcome. | 502360 | Proof of [imath](a+b)^{n+1}[/imath]
I have to do proof of [imath](a+b)^n[/imath] and [imath](a+b)^{n+1}[/imath] with mathematical induction. I finished the first one [imath](a+b)^n = a^n + na^{n-1}b+\dots+b^n[/imath]. I however have trouble with the second one, I don't know what to start exactly. Any help/hints? |
108010 | When is the closure of an open ball equal to the closed ball?
It is not necessarily true that the closure of an open ball [imath]B_{r}(x)[/imath] is equal to the closed ball of the same radius [imath]r[/imath] centered at the same point [imath]x[/imath]. For a quick example, take [imath]X[/imath] to be any set and define a metric [imath] d(x,y)= \begin{cases} 0\qquad&\text{if and only if $x=y$}\\ 1&\text{otherwise} \end{cases} [/imath] The open unit ball of radius [imath]1[/imath] around any point [imath]x[/imath] is the singleton set [imath]\{x\}[/imath]. Its closure is also the singleton set. However, the closed unit ball of radius [imath]1[/imath] is everything. I like this example (even though it is quite artificial) because it can show that this often-assumed falsehood can fail in catastrophic ways. My question is: are there necessary and sufficient conditions that can be placed on the metric space [imath](X,d)[/imath] which would force the balls to be equal? | 1693875 | [imath]X[/imath] is a normed space and for any [imath]x \in X[/imath] and [imath]r \gt 0[/imath], let [imath]T=\{y\in X: \|y-x\|\le r\}[/imath] and [imath]S=\{y\in X: \|y-x\|\lt r\}[/imath], then [imath]\bar{S}=T[/imath].
Let [imath]X[/imath] be a normed linear space and, for any [imath]x \in X[/imath] and [imath]r \gt 0[/imath], let [imath]T=\{y\in X: \|y-x\|\le r\}[/imath] and [imath]S=\{y\in X: \|y-x\|\lt r\}[/imath]. If [imath]z\in T[/imath] and [imath]z_n=(1-n^{-1})z[/imath], for [imath]n\in \mathbb{N}[/imath], show that [imath]\lim z_n=z[/imath] and hence show that [imath]\bar{S}=T[/imath]. Below is the solution to the problem. But I think it's wrong. It shows that [imath]\|z_n\|\lt r[/imath], but we need [imath]\|z_n-x\|\lt r[/imath], where [imath]x[/imath] is a fixed point in [imath]X[/imath], to show that [imath]z_n \in S[/imath]. How can I show this? Perhaps the problem is wrong? I would greatly appreciate any help. |
675977 | Proof for absolute value inequality of three variables: [imath]|x-z| \leq |x-y|+|y-z|[/imath]
[imath]|x-z| \leq |x-y|+|y-z|[/imath] We know that both LHS and RHS are non negatives. So, I thought of proving this by comparing the squares of both sides but can't advance beyond that step. Any help would be appreciated. | 708431 | Proving the inequality [imath]|a-b| \leq |a-c| + |c-b|[/imath] for real [imath]a,b,c[/imath]
Let [imath]a,b,c[/imath] real numbers. Prove the inequality [imath]|a-b| \leq |a-c| + |c-b|[/imath]. Prove that equality holds if and only if [imath]a \leq c \leq b[/imath] or [imath]b \leq c \leq a[/imath]. I've tried starting with just [imath]a \leq c[/imath] and using field properties to reconstruct the inequality, however I haven't been able to make it work. I also tried making the negatives positive and stripping the inequalities and making something happen but again I don't know if that's a proper rule and it didn't seem to get me anywhere. |
711905 | How to find the exact value of this series?
I found this on wolfram alpha : How does wolfram evaluate this series? How to proof that [imath]\sum_{n=1}^\infty \left(\frac{1}{n}+\frac{1}{2n}+...+\frac{1}{n^2}\right)^2= \frac{17\pi^4}{360} [/imath] ? Thanks in advance. | 554003 | Infinite Series [imath]\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2[/imath]
How can I find a closed form for the following sum? [imath]\sum_{n=1}^{\infty}\left(\frac{H_n}{n}\right)^2[/imath] ([imath]H_n=\sum_{k=1}^n\frac{1}{k}[/imath]). |
710775 | Is [imath]a \le b[/imath] a true statement if [imath]a < b[/imath]?
My question is: Is [imath]a \le b[/imath] true if [imath]a < b[/imath]? For instance: Is [imath]3 \le 4[/imath] a true statement? I think yes, because [imath]a \le b[/imath] is defined as [imath]a < b\vee a = b[/imath] and this should be true, even if [imath]a = b[/imath] is always false, right? Thank you very much for your help! FunkyPeanut | 679451 | Less than or equal sign
If I know for two numbers a and b that [imath]{a < b }[/imath] Then is it correct to say that [imath] a \leq b [/imath] I know that the second statement is true as long as the first one is. It seems OK as it is true but from the other side it seems kinda weird (to me), to say that, if you know, that a is strictly less that b.It's like you lose some extra information. |
712367 | Interesting determinant problem
how to go about computing following determinant? I tried using Gaussian elimination on some special cases and figured there might be some pattern, maybe a recurrence relation involved, but I just can't see it. [imath]\begin{vmatrix} 1 & 2 & 3 & \cdots & n-1 & n \\ 2 & 3 & 4 & \cdots & n & 1 \\ 3 & 4 & 5 & \cdots & 1 & 2 \\ \vdots & \vdots & \ddots & \vdots \\ n & 1 & 2 & \cdots & n-2 & n-1 \end{vmatrix}[/imath] | 386526 | Circular determinant problem
I'm stuck in this question: How calculate this determinant ? [imath]\Delta=\left|\begin{array}{cccccc} 1&2&3&\cdots&\cdots&n\\ n&1&2&\cdots&\cdots& n-1\\ n-1&n&1&\cdots&\cdots&n-2\\ \vdots &\ddots & \ddots&\ddots&&\vdots\\ \vdots &\ddots & \ddots&\ddots&\ddots&\vdots\\ 2&3&4&\cdots&\cdots&1 \end{array}\right|[/imath] Thanks a lot. |
680624 | variational problem: obtain the lagrangian from the PDE equations of motion
given the 2 PDE [imath] \Delta u-au_{tt}+u_{t}=0[/imath] and [imath] \Delta u + Du*Df=0 [/imath] here [imath] \Delta u [/imath] is the Laplacian [imath] Du= grau[/imath] and * means scalar product [imath]u_{t} = \frac{\partial u}{\partial t}[/imath] my doubt is what term should i include to get the linear part of the equations [imath] u_{t} [/imath] and [imath]Du*Df [/imath] from the Euler Lagrange equations in [imath] R^{n} [/imath] thanks | 685608 | A variational problem with a lagrangian , what is the lagrangian?
given the 2 PDE [imath] \Delta u-au_{tt}+u_{t}=0[/imath] and [imath] \Delta u + Du*Df=0 [/imath] here [imath] \Delta u [/imath] is the Laplacian [imath] Du= grad(u) [/imath] is the gradient and * means scalar product [imath]u_{t} = \frac{\partial u}{\partial t}[/imath] my doubt is what term should i include to get the linear part of the equations [imath] u_{t} [/imath] and [imath]Du*Df [/imath] from the Euler Lagrange equations in [imath] R^{n} [/imath] thanks |
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