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2034951
|
General solution of [imath]x[/imath] if [imath]\sin x+\cos x=1[/imath]
General solution to the equation [imath]\sin x+\cos x[/imath]=[imath]1[/imath] is found to be [imath]x=2n\pi[/imath] and [imath]x=2n\pi+\pi/2[/imath], Pls refer Solving cosx+sinx−1=0 My Approach: [imath] \sin x+\cos x=1\implies \sin x\frac{1}{\sqrt{2}}+\cos x\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\\\implies\sin x\cos(\pi/4)+\cos x\sin(\pi/4)=\sin(\pi/4)\implies \sin(x+\pi/4)=\sin(\pi/4)\implies x+\pi/4=n\pi+(-1)^n [\pi/4]\implies x=n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{4} [/imath] How do I check both the results are the same, without inputting the values for [imath]n[/imath] ?
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1307085
|
Solving [imath]\cos x+\sin x-1=0[/imath]
How does one solve this equation? [imath]\cos {x}+\sin {x}-1=0[/imath] I have no idea how to start it. Can anyone give me some hints? Is there an identity for [imath]\cos{x}+\sin{x}[/imath]? Thanks in advance!
|
2035706
|
A non-normal variety that is regular in codimension one
So, for a projective variety [imath]X[/imath] over, say, an algebraically closed field [imath]\Bbbk[/imath], being normal implies being regular in codimension one. However, the converse is not true, in addition you require an extension property. I am looking for an example of an [imath]X[/imath] which is regular in codimension one, but not normal. By common lore, this can not be a curve or a surface in [imath]\Bbb P^3[/imath]. I was hoping that maybe there is an example of a surface in [imath]\Bbb P^4[/imath]. Beating around the bush less: I would like a small and explicit example if possible.
|
143954
|
isolated non-normal surface singularity
I am looking for an isolated non-normal singularity on an algebraic surface. One obvious example occurs to me: the union of two [imath]2[/imath]-dimensional affine subspaces of [imath]\mathbb{A}^4[/imath] which meet in a point. But this seems like "cheating." Can someone provide an irreducible example?
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2035817
|
Prime ideals of [imath]M_{n,n}(R)[/imath]
Let [imath]R[/imath] be a ring. Denote [imath]M_{n,n}(R)[/imath] as the matrix ring over [imath]R[/imath]. I've already proven that [imath]M_{n,n}(I)[/imath] is an ideal of [imath]M_{n,n}(R)[/imath] [imath]\Leftrightarrow[/imath] [imath]I[/imath] is an ideal of [imath]R[/imath]. Is there a short way to prove that [imath]M_{n,n}(I)[/imath] is a prime ideal of [imath]M_{n,n}(R)[/imath] [imath]\Leftrightarrow[/imath] [imath]I[/imath] is a prime ideal of [imath]R[/imath]? note: every ideal is considered to be two-sided.
|
2015769
|
Prime ideals of the matrix ring
I have to do the following question: Given a unital ring [imath]R[/imath] and a natural number [imath]n[/imath] [imath]\geq 1[/imath], identify (with proof) the prime ideals of the matrix ring [imath]\mathbb{M}_{n}(R)[/imath]. I understand the theory, but putting it all together is proving difficult. I understand that the prime ideals of [imath]\mathbb{M}_{n}(R)[/imath] should be [imath]\mathbb{M}_{n}(\mathbb{P})[/imath]. Can anyone suggest any steps as to how to prove this?
|
2035556
|
How to calculate this improper integral [imath]\int_{0}^{+\infty}\frac{x}{\mathrm{e}^x-1}\mathrm{d}x[/imath]?
How to calculate this improper integral [imath]\int_{0}^{+\infty}\frac{x}{\mathrm{e}^x-1}\mathrm{d}x[/imath]
|
472687
|
Prove [imath]\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}[/imath]
I know that [imath]\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}[/imath] For substituting [imath]u=2[/imath] into [imath]\zeta(u)\Gamma(u)=\int^\infty_0\frac{x^{u-1}}{e^x-1}dx[/imath] However, I suspect that there is an easier proof, maybe by the use of complex analysis. I haven't learnt the zeta function yet. All I know is the above formula and [imath]\zeta(2)=\frac{\pi^2}{6}[/imath]. Can we can use the above integral to find out some of the [imath]\zeta[/imath]'s value?
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2035298
|
Hodge star operator and the adjoint of exterior multiplication
Given the adjoint of exterior multiplication by [imath]\xi[/imath], that is [imath]\gamma: \Lambda_{p+1}(V) \rightarrow \Lambda_p(V) [/imath], [imath]\langle \gamma(v), w\rangle = \langle v, \xi\wedge w\rangle[/imath] for [imath]v\in \Lambda_{p+1}(V)[/imath] and [imath]w\in \lambda_p(V)[/imath]. I would like to show that [imath]\gamma(v) = (-1)^{np} \star(\xi\wedge(\star v))[/imath] where [imath]\star[/imath] is the Hodge star operator. I know the basic definition of [imath]\star: \Lambda_p \rightarrow \Lambda_{n-p}[/imath], and given an orthonormal basis [imath]\{e_1,\cdots,e_n\}[/imath] on [imath]V[/imath], we have the computation [imath]\star (e_{i_1} \wedge e_{i_2}\wedge \cdots \wedge e_{i_k})= e_{i_{k+1}} \wedge e_{i_{k+2}} \wedge \cdots \wedge e_{i_n}.[/imath] Also for [imath]u,w \in \Lambda_p(V)[/imath], we have [imath]\star(w\wedge \star v) = \star(v\wedge \star w) = \langle u,w \rangle = \det(u_i, w_j).[/imath]
|
1285693
|
The adjoint of left exterior multiplication by [imath]\xi[/imath] for hodge star operator
As we know, for [imath]V[/imath] vectoral space and a orientation [imath]\mathcal{O}[/imath] on [imath]V[/imath] and [imath]e_{1},...,e_{n}[/imath], the hodge star operator [imath]\ast:\wedge V^*\rightarrow\wedge V^*[/imath] is defined for [imath]\ast(e_{1}\wedge...\wedge e_{n})=\pm 1[/imath](acoording to the orientation preserve/invert for the basis [imath]\{e_{i}\}[/imath]) [imath]\ast(1)=\pm e_{1}\wedge...\wedge e_{n}[/imath](again [imath]+[/imath] if [imath][\{e_{i}\}]=\mathcal{O}[/imath] and [imath]-[/imath] if [imath][\{e_{i}\}]=-\mathcal{O}[/imath]) [imath]e_{1}\wedge...\wedge e_{k}=\pm e_{k+1}\wedge...\wedge e_{n}[/imath] Already prove that [imath]\ast\ast=(-1)^{n(n-k)}\cdot[/imath] and [imath]<T,S>=\ast(T\wedge\ast S)=\ast(S\wedge \ast T)[/imath] (where [imath]T=\eta_{1}\wedge...\wedge \eta_{1}[/imath], [imath]S=\omega_{1}\wedge...\wedge \omega_{1}[/imath] and [imath]<T,S>=det(<\omega_i,\eta_{j}>)[/imath]) For any [imath]\xi\in V[/imath] denote [imath]\gamma:\ast:\wedge^{k+1} V^*\rightarrow \wedge^{k} V^*[/imath] the adjoint of left exterior multiplication by [imath]\xi[/imath] [imath]<\gamma(T),S>=<T,\xi\wedge S>[/imath] for all [imath]T\in\wedge^{k+1} V^*[/imath] and [imath]S\in\wedge^{k} V^*[/imath]. I need to prove that [imath]\gamma(T)=(-1)^{n-k}\ast(\xi\wedge\ast T)[/imath] but I'm a little confused, how prove that [imath]\gamma[/imath] has this expression from this property?
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2036084
|
A doubt about ramification in cyclotomic fields
I'm studying the Algebraic Number Theory Notes of Robert B. Ash. I really like his notes, but I don't understand a suggestion he gives at page 7 of chapter 8 (Factoring of prime ideals in Galois Extensions). He's using all the theory of that chapter to discover new properties of cyclotomic fields. So pick [imath]\zeta[/imath] a primitive [imath]m^{th}[/imath] root of unity and let [imath]L=\mathbb Q(\zeta)[/imath], [imath] A=\mathbb Z[/imath] and [imath]K=\mathbb Q[/imath]. Consider [imath]p[/imath] rational prime that does not divide [imath]m[/imath]. Say [imath]B[/imath] the integral closure of [imath]A[/imath] in [imath]L[/imath] and that [imath](p)[/imath] factors in [imath]B[/imath] as [imath]Q_1....Q_g[/imath]. We know that the relative degree [imath]f[/imath] is the same for all [imath]Q_i[/imath]. He wants to find the Frobenius automorphism [imath]\sigma[/imath] explicitly. We know that [imath]\sigma[/imath] has the property that [imath]\sigma(x)\equiv x^p\pmod {Q_i}[/imath] for all [imath]i[/imath] and for all [imath]x\in B[/imath]. From this, why do we deduce that [imath]\sigma(\zeta)=\zeta^p[/imath]?
|
399802
|
Frobenius element in cyclotomic extension
Let [imath]K=\mathbb{Q}(\zeta_m)[/imath]. Then if [imath]p\nmid m[/imath] is any odd prime, how i can show that Frobenius map is [imath](p,K/\mathbb{Q})(\zeta_m)=\zeta_m^p[/imath]. We know, if [imath]P[/imath] is a prime above [imath]p[/imath] [imath](p,K/\mathbb{Q})(x) \equiv x^p (\mbox{mod } P) \quad\forall x \in \mathbb{Z}_K[/imath] Thanks!
|
2035268
|
Primitive roots and indices
Given the congruence [imath]x^3 \equiv a[/imath] (mod p), where [imath]p \geq 5[/imath] is a prime and gcd([imath]a,p[/imath])=1, prove the following a. If [imath]p \equiv 1[/imath] (mod 6), then the congruence has either no solutions or three incongruent solutions modulo p. b.If [imath]p \equiv 5[/imath] (mod 6), then the congruence has a unique solution modulo p. How to solve it...by using the theory of indices.
|
2013414
|
number theory about index
Given the congruence [imath]x^3 \equiv a \pmod p[/imath], where [imath]p \geq 5[/imath] is a prime and [imath]\gcd(a,p)=1[/imath], prove the following: If [imath]p \equiv1 \pmod 6[/imath], then the congruence has either no solutions or three incongruent solutions modulo [imath]p[/imath] If [imath]p \equiv5 \pmod 6[/imath], then the congruence has a unique solution modulo [imath]p[/imath] Help me... I don't know what I learn because Prof teaches so quickly. So I need your help to organize my thoughts.
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2036685
|
Proof by contradiction that [imath]\sqrt{2}[/imath] is irrational, why assume the fraction is simplified?
As a student with no history of proof-based math courses, I am having some difficulty understanding the proof by contradiction by assuming [imath]\sqrt{2} = a/b[/imath] and that a and b are in least common denominators. Could you explain how this disproves that it is irrational? Doesn't it just merely disprove that our initial assumption that a and b are even, is wrong? I'm speaking of the popular proof that is given here:https://www.math.utah.edu/~pa/math/q1.html
|
718956
|
Incorrect logic in popular proof of the irrationality of [imath]\sqrt2[/imath]?
A popular proof of the irrationality of [imath]\sqrt2[/imath] is to first assume that the number is rational. This means that [imath]\sqrt2=a/b[/imath] where [imath]a[/imath] and [imath]b[/imath] are integers. Another assumption is that [imath]a[/imath] and [imath]b[/imath] are coprime. It turns out that this leads to a contradiction. And the conclusion is that [imath]\sqrt2[/imath] is not rational (because the assumption of rationality led to a contradiction). But what about the second assumption? Logically, the conclusion could also be that [imath]a[/imath] and [imath]b[/imath] are not coprime.. How can this be resolved? Edit: Thank you all for your comments. It cleared up my mind. Three aspects of this kind of proof are: mathematical, logical, and didactical. Mathematically, it is the case that any rational number can be written as a fraction with coprime numerator and denominator. But this point is often described in a way that is either literally a logical assumption, or in a way that is natural to interpret as a hidden logical assumption. In this case, the proof will have a flawed logic, because any of the two assumptions could be the cause of the contradiction. Didactically, this can cause confusion in the mind of the student, who might not be so accepting of the proofs conclusion. I used to be that student. So, how to resolve it? Many of you provided explanations on the mathematical aspect, which are all correct. But my main point is not of not understanding the proof, but of the use of flawed logic which leads to not understanding the proof. I am sure that a lot of teachers think or say something along the lines of WLOG, but for many students, this makes the proof weaker, because it is not so convincing. When we present this proof to our students, we need to have this in mind. Thank you.
|
2036676
|
Infinite series of converging sequence: [imath]\sum_{n=1}^{\infty}\frac{1}{n2^n}=\ln 2[/imath]
How can I prove the following equality; [imath]\sum_{n=1}^{\infty}\frac{1}{n2^n}=\ln 2\,.[/imath]
|
1153499
|
Show [imath]\ln2 = \sum_\limits{n=1}^\infty\frac1{n2^n}[/imath]
Problem: Show that [imath]\ln2 = \sum_\limits{n=1}^\infty\frac1{n2^n}.[/imath] My progress: The problem before this one had me find the Taylor series for [imath]\ln(1-x)[/imath] which was [imath]-\sum\limits_{n=1}^\infty \frac{x^n}n[/imath] so I figured I'd use [imath]x=-1[/imath] and plug that into the Taylor series. However, there was a side note stating that the Taylor series I found is only valid for [imath]x\in(-1, 1)[/imath]. And in any case, my calculation isn't going anywhere, since I end up with the series [imath]1-\frac12+\frac13-\frac14\cdots[/imath] Question 1: How can I use this to solve the problem stated initially, and in the title? Question 2: I can see why [imath]x[/imath] is restricted to be less than 1, to prevent taking the log of zero or a negative. Why is it not valid for x greater than 1?
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2036657
|
Math Induction Proof
I'm having a little bit of trouble with this proof. Use mathematical induction to show that [imath]f_{n−1} \cdot f_{n+1} − f_{2 n} = (−1)^n[/imath] for [imath]n[/imath] in the set of positive integers. I know that in recursive functions, at least the first term is provided and then you could play around with it to come to a conclusion. I was maybe thinking of using the Fibonacci numbers definition as a guide but I'm not really going forward.
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1928285
|
Show [imath]F_{n+1} \cdot F_{n-1} = F_n^2 + (-1)^n[/imath] for all [imath]n \in \mathbb{N}[/imath]
By calculating for [imath]n\in \{1,2,3,4,5,6,7\}[/imath], I've formulated the rule \begin{equation} F_{n+1} \cdot F_{n-1} = F_n^2 + (-1)^n, \end{equation} where [imath]F_n[/imath] is the [imath]n[/imath]th fibonacci number. I want to show that this is true for all [imath]n \in \mathbb{N}[/imath]. I tried using induction, with [imath]n=1[/imath] as the basis step, but didn't get very far: For the induction step, we assume the formula holds for a [imath]n = k[/imath], and checks for [imath]n=k+1[/imath]: \begin{align*} F_{k} \cdot F_{k+2} &= F_{k} \cdot (F_{k+1} + F_{k}) \\ &= F_k \cdot F_{k+1} + F_k^2 \\ \end{align*} If somehow [imath]F_k \cdot F_{k+1} = (-1)^k[/imath], then I would be done. But I don't see how that's possible. Is there a better way of proving this, maybe without using induction? Or am I just going about it the wrong way?
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2036372
|
Distribution of X+Y of a bivariate normally distributed (X,Y)
I need to give the distribution of [imath]X+Y[/imath] and [imath]X-Y[/imath] knowing that [imath](X,Y)[/imath] is bivariate normally distributed with marginal means 1, marginal variances 1 and correlation [imath]p=0.2[/imath].Is it right that the marginal mean are simply [imath]E(X)=1[/imath] and [imath]E(Y)=1[/imath] ? I don't see where I should start knowing that.
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2037252
|
(X, Y) Bivariate normally distribution
If [imath](X, Y)[/imath] is bivariate normally distributed with marginals means [imath]1[/imath], marginal variances [imath]1[/imath] and correlation coefficient [imath]\rho=0.2[/imath], what is the distribution of [imath]X + Y[/imath] ? How can I start the problem ?
|
2037372
|
If [imath]p_n[/imath] is prime does [imath]\frac{p_{n+1}}{p_n} [/imath] or [imath]\frac{p_n}{p_{n+1}}[/imath] tend towards some number for large [imath]n[/imath]?
If [imath]p_n[/imath] is the nth prime number does [imath]\frac{p_{n+1}}{p_n}[/imath] approach any particular number when [imath]n[/imath] is large? What about [imath]\frac{p_n}{p_{n+1}}[/imath]? Is it possible to prove that the limits converge or diverge with elementary methods?
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144539
|
Limit inferior of the quotient of two consecutive primes
I have recently read an article about the prime number theorem, in which Mathematicians Erdos and Selberg had claimed that proving [imath]\lim \frac{p_n}{p_{n+1}}=1[/imath], where [imath]p_k[/imath] is the [imath]k[/imath]th prime, is a very helpful step towards proving the prime number theorem, although I don't know how, primarily because I have not gone through the proof of the theorem even once. Anyway, I was trying to prove the result [imath]\lim \frac{p_n}{p_{n+1}}=1[/imath] by really elementary methods, as is always my habit, and quite recently I learned of a theorem that the sum of the reciprocals of the primes diverges (I am only a beginner). The idea that suddenly struck me is that this theorem implies [imath]\limsup\frac{p_n}{p_{n+1}}=1[/imath], by a simple application of the ratio test and the fact that [imath]\frac{p_n}{p_{n+1}} < 1[/imath] for all [imath]n[/imath]. So, is there any simple way to show that [imath]\liminf\frac{p_n}{p_{n+1}}=1[/imath] or at least [imath]\ge1[/imath], so that the result is proved? Another beautiful result that came to me, follows from the theorem on divergence of [imath]\sum\frac{1}{p}[/imath], that given any [imath]h > 1[/imath] there are infinitely many [imath]n[/imath] such that [imath]p_n < h^n[/imath]. Are there any other results with really simple proofs (easy to understand even for beginners like me) having deep consequences in the theory of the distribution of primes? I'd really like to hear them!
|
2037354
|
Prove that [imath]\|x\|_A[/imath] is a vector norm.
Prove: A is a positive definite symmetric matrix and [imath]\forall x \in \mathbb{R^n}[/imath] define [imath]||x||_A[/imath] by [imath]||x||_A=\sqrt{x^TAx}[/imath] Show [imath]||x||_A[/imath] is a vector norm. My solution: For [imath]||x||_A[/imath] to be a vector norm, [imath]||x||_A=\sqrt{x_1^2+x_2^2+...+x_n^2}=\sqrt{x^Tx}[/imath] must be satisfied. since [imath]A[/imath] is symmetric, [imath]A=A^T[/imath], thus [imath]||x||_A=\sqrt{x^TAx}=\sqrt{x^TA^Tx}=\sqrt{(Ax)^Tx}[/imath]. Since [imath]A[/imath] is positive definite it must have full range. Thus the set of all vectors [imath]b[/imath] for which [imath]Ax=b[/imath] must have a solution. Thus, we can choose [imath]Ax=x[/imath]. Thus, [imath]\sqrt{(Ax)^Tx}=\sqrt{x^Tx}[/imath]. Thus, [imath]||x||_A=\sqrt{x^Tx}[/imath] and [imath]||x||_A[/imath] is a vector norm. I'm not sure if we can choose [imath]Ax=x[/imath]. Any comments are welcome.
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1011563
|
Norm with symmetric positive definite matrix
If B is [imath]n \times n[/imath] real symmetric positive definite matrix, then [imath](x,y) = x^T By[/imath] defines an inner product on [imath]R^n[/imath]. How do you prove that [imath]\|x\|=(x^T B x)^{1/2}[/imath] is a norm on [imath]R^n[/imath]?
|
1953523
|
Does a Zero Variance imply a constant random variable?
I managed to show that if a random variable is constant : [imath]P(X = \mu ) = 1[/imath] then the Variance is zero: [imath]E[X^2] = \sum_{i = 1}^{k} p_iX_i^2 =>[/imath] (given X_i is constant)[imath]=> X^2\sum_{i=1}^kp_i = X^2 [/imath] [imath]E[X]^2 = (\sum_{i=1}^{k}p_iX_i)^2 = (X\sum_{i=1}^{k}p_i)^2 = X^2[/imath] This is relatively straight forward with the for on the variance of [imath]E[X^2]-E[X]^2[/imath] However is the converse true? Does a zero variance necessarily imply a constant random variable?
|
990009
|
If [imath]Var(X)=0[/imath] then is [imath]X[/imath] a constant?
We know that the variance of a constant is [imath]0[/imath]. Is the converse also true? Can we say that if the variance of some random variable is [imath]0[/imath] it is a constant?
|
2037079
|
Fractal dimension of a set of points in [imath]\mathbb{R}^2[/imath]
I have a set of points in [imath]\mathbb{R}^2[/imath], of the form: [imath]\left(\frac{a}{\ell^2},\frac{b}{\ell^3}\right)[/imath] where [imath]\ell>0[/imath] is an integer and [imath]a[/imath] and [imath]b[/imath] are some real positive numbers. I am interested to know the fractal dimension of this set of points as [imath]\ell[/imath] becomes infinite. Is there a simple way of computing this?
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2036524
|
Fractal dimension of set in [imath]\mathbb{R}^2[/imath]
I have a set of points in [imath]\mathbb{R}^2[/imath], of the form: [imath]\Bigg\{\left(\frac{a}{\ell^2},\frac{b}{\ell^3}\right): \ell \in \mathbb{N}^+\Bigg\}[/imath] where [imath]a[/imath] and [imath]b[/imath] are some real positive numbers. I am interested to know the box dimension of this set. Is there a simple way to determine this analytically?
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2038275
|
Using the mean value theorem to solve an inequality
If f ' is increasing, show that [imath]\frac{f(x)-f(x-h)}{h} \le f'(x) \le \frac{f(x+h)-f(x)}{h}[/imath] for all h>0, specifically using the mean value theorem. [imath]f'(x) \le f'(c)[/imath] is obviously true due to the fact that [imath]\frac{f(x+h)-f(x)}{h} = f'(c)[/imath] for some c, but I don't know how the left hand inequality relates at all.
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2034749
|
Showing that [imath]\frac{f(x)-f(x-h)}{h} \le f'(x) \le \frac{f(x+h)-f(x)}{h}[/imath] given that f' is increasing
We are considering that [imath]h > 0[/imath]. I just don't really know how to approach this question; it seems that by taking the limit of the inequality as h goes to 0, each part of the inequality is going to be equal to [imath]f'[/imath], but I don't really understand why the fact that [imath]f'[/imath] is increasing is relevant to finding a solution to this question (and how to answer the question based on that constraint).
|
2038265
|
Prove that if [imath]x,y \in R[/imath], and [imath]1 \leq p < \infty[/imath] then [imath]|x+y|^p \leq 2^p(|x|^p+|y|^p )[/imath]
Prove that if [imath]x,y \in R[/imath], and [imath]1 \leq p < \infty[/imath] then [imath]|x+y|^p \leq 2^p(|x|^p+|y|^p )[/imath] what I'm thinking is [imath]|x+y|\leq |x|^p+|y|\le 2\cdot\max\{|x|,|y|\}[/imath] Without loss of generality , let [imath]\max\{|x|, |y|\}=|x|[/imath] hence and [imath]|x+y|\leq 2|x| \Rightarrow |x+y|^p \le (2|x|)^p = 2^p|x|^p [/imath] [imath]|x+y|^p \le 2^p|x|^p [/imath] and since [imath]|y| \ge 0 [/imath], we have [imath] |x+y|^p \le 2^p(|x|^p+|y|^p )[/imath] Is there any idea I could have used here?
|
1184781
|
Prove that [imath]|a+b|^p \leq 2^p \{ |a|^p +|b|^p \}[/imath]
For any real numbers [imath]a[/imath] and [imath]b[/imath] and [imath]1 \leq p < \infty[/imath], prove that [imath]|a+b|^p \leq 2^p \{ |a|^p +|b|^p \}[/imath] This inequality is given in the the book Real Analysis by Royden, Chapter [imath]7[/imath], page [imath]136[/imath]. I don't understand how the author comes to this inequality. Can anyone provide some hints?
|
2037987
|
Determine all homomorphisms from [imath]S_4[/imath] to [imath]\mathbb Z_2[/imath].
Suppose that [imath]\phi[/imath] is a homomorphism from [imath]S_4[/imath] onto [imath]\mathbb Z_2[/imath]. Determine [imath]\ker\phi[/imath]. Determine all homomorphisms from [imath]S_4[/imath] to [imath]\mathbb Z_2[/imath]. Please help me with this problem. I am stuck and I feel like I am not making any progress... Also, unlike isomorphism, is it okay for one group to be non-cyclic and the other to be cyclic under homomorphisms? Because I think [imath]S_4[/imath] is not cyclic but [imath]Z_2[/imath] is cyclic.
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219333
|
Homomorphisms from [imath]S_4[/imath] to [imath]\mathbb Z_2[/imath]
Suppose [imath]\phi : S_4 \rightarrow \mathbb Z_2[/imath] is a surjective homomorphism. Find [imath]\ker\phi[/imath]. Determine all homomorphisms from [imath]S_4[/imath] to [imath]\mathbb Z_2[/imath]. My solution: since [imath]\phi[/imath] is surjective, then by the first isomorphism theorem, [imath]S_4/\ker\phi \cong \mathbb Z_2[/imath]. Therefore, by computation, [imath]\ker\phi = A_4[/imath]. Now, for the second part, I have some trouble since they ask for all the possible homomorphisms. I know that the map that takes even permutations to [imath]0[/imath] and odd permutations to [imath]1[/imath] is a homomorphism. But, my question is: How to make sure this is the only one? thanks,
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2037963
|
Proving if f has a continuous extension to [a,b] then f is uniformly continuous on (a,b).
I want to show that if a function [imath]f:(a,b) \rightarrow \mathbb{R}[/imath] has a continuous extension to [a,b], then f is uniformly continuous. Also, assume that [imath](X, d_x)[/imath] and [imath](Y, d_y)[/imath] are nonempty metric spaces and all subsets of [imath]\mathbb{R}^k[/imath] are given the Euclidean metric. I know the proof is trivial, but I'm worried about my notation as I don't think I quite understand continuous extension correctly. I understand that for [imath]E \subset F \subset X[/imath], the function [imath]g: F \rightarrow Y[/imath] is an extension to f if [imath]g(x)=f(x) \forall x \in E[/imath]. So, we have [imath](a,b) \subset [a,b] \subset \mathbb{R}[/imath] and [imath]f: (a,b) \rightarrow \mathbb{R}[/imath]. So if f has a continuous extension to a,b, then (f(x)) = f(y)) [imath]\forall x \in (a,b), y \in [a,b][/imath]. And so this then implies that d(f(x), f(y)) = 0, and so [imath]\forall \epsilon > 0, \exists \delta > 0[/imath] s.t. [imath]d(x, y) < \delta[/imath] implies [imath]d(f(x), f(y)) < \epsilon[/imath]. Help and hints would be much appreciated!
|
586378
|
Show that function [imath]f[/imath] has a continuous extension to [imath][a,b][/imath] iff [imath]f[/imath] is uniformly continuous on [imath](a,b)[/imath]
Let [imath]E \subset F \subset X[/imath] and [imath]f:E\rightarrow Y[/imath]. We say that the function [imath]g:F\rightarrow Y[/imath] is an extension of [imath]f[/imath] if [imath]g(x) = f(x)[/imath] for all [imath]x \in E[/imath]. Let [imath]f: (a, b) \rightarrow \mathbb{R}[/imath]. Show that [imath]f[/imath] has a continuous extension to [imath][a, b][/imath] if and only if [imath]f[/imath] is uniformly continuous on [imath](a, b)[/imath]. [imath](X,d)[/imath] is a metric space. Hint: To prove [imath]\Leftarrow[/imath] start by showing that [imath]f[/imath] maps Cauchy sequences to Cauchy sequences. Then how should you define [imath]g(a)[/imath] and [imath]g(b)[/imath]? The [imath]\Rightarrow[/imath] part of the proof was quite simple, but even with the hint, the other way seems a bit difficult. Doesn't the first part of the "hint" follow from the way we defined [imath]f[/imath]? If so, how do I prove this? Help/Hints appreciated!
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2039211
|
[imath]G[/imath] is a finite abelian group. [imath]|G| = p^n[/imath] where [imath]p[/imath] is a prime. Show that [imath]G[/imath] has a subgroup of order [imath]p^r, r \leq n[/imath] without Sylow Theorems.
I know that for [imath]G[/imath] to have a subgroup of order [imath]p^r[/imath] it must have an element of order [imath]p^r[/imath]. My approach to this problem was to use the Fundamental Theorem of finite abelian groups. i.e there are, up to isomorphism, only the following abelian groups of order [imath]p^n[/imath]: [imath]\mathbb Z_{p^{a_1}} \times \mathbb Z_{p^{a_2}} \times \mathbb Z_{p^{a_3}} \times ... \times \mathbb Z_{p^{a_k}}[/imath], where [imath]a_i \in \mathbb N[/imath] and [imath]a_1 + a_2 + a_3 + ... + a_k = n[/imath] Then I was going to show that each group of this form has an element of order [imath]p^r[/imath]. I started off by using Cauchy's theorem to show that each group of the form [imath]\mathbb Z_{p^{m}}[/imath] where [imath]m \in \mathbb N, m \ne 0[/imath] has an element of order [imath]p[/imath]. However, I am stuck now an do not know how to show that each group of the form [imath]\mathbb Z_{p^{a_1}} \times \mathbb Z_{p^{a_2}} \times \mathbb Z_{p^{a_3}} \times ... \times \mathbb Z_{p^{a_k}}[/imath] has an element of order [imath]p^r[/imath], where r is every integer less than [imath]n[/imath]. Does anyone know how to prove this without using Sylow theorems?
|
477500
|
Showing that a finite abelian group has a subgroup of order m for each divisor m of n
I have made an attempt to prove that a finite abelian group of order [imath]n[/imath] has a subgroup of order [imath]m[/imath] for every divisor [imath]m[/imath] of [imath]n[/imath]. Specifically, I am asked to use a quotient group-induction argument to show this. I'd appreciate comments on the validity or lack thereof of my attempted proof below. Let [imath]G[/imath] be a finite abelian group of order [imath]n[/imath] and let [imath]m[/imath] be a divisor of [imath]n[/imath]. The proposition is true for [imath]n=1[/imath], so we'll proceed by induction and assume [imath]n \ge 2[/imath]. Let [imath]p[/imath] be a prime dividing [imath]m[/imath] and let [imath]x[/imath] be an element of order [imath]p[/imath] in [imath]G[/imath] (which exists by Cauchy's Theorem for Abelian Groups). By the induction hypothesis, [imath]G/\langle x \rangle[/imath] has a subgroup of order [imath]\dfrac{m}{p}[/imath]. This subgroup is of the form [imath]H/\langle x \rangle[/imath] for some [imath]H \le G[/imath]. Since [imath]|H/\langle x \rangle| = \dfrac{m}{p}[/imath], it follows that [imath]H \le G[/imath] has order [imath]m[/imath]. I chose to use a prime divisor of [imath]m[/imath], but I don't see why it wouldn't work to use any proper divisor of [imath]m[/imath]. Am I correct on this point? Thanks, I appreciate the help.
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56397
|
Do finite algebraically closed fields exist?
Let [imath]K[/imath] be an algebraically closed field ([imath]\operatorname{char}K=p[/imath]). Denote [imath]{\mathbb F}_{p^n}=\{x\in K\mid x^{p^n}-x=0\}.[/imath] It's easy to prove that [imath]{\mathbb F}_{p^n}[/imath] consists of exactly [imath]p^n[/imath] elements. But if [imath]|K|<p^n[/imath], we have collision with previous statement (because [imath]{\mathbb F}_{p^n}[/imath] is subfield of [imath]K[/imath]). So, are there any finite algebraically closed fields? And if they exist, where have I made a mistake? Thanks.
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2419076
|
Which of the following statesment are true?
Which of the following statesment are true? there exists a finite field in which additive group is not cyclic [imath]F[/imath] is a finite field then there exist a polynomial [imath]p[/imath] over [imath]F[/imath] such that [imath]p(x) \ne0[/imath] for all [imath]x\in F[/imath], where [imath]0[/imath] denotes zeros of [imath]F[/imath] Every finite field is isomorphic to a subfield of the field of complex numbers
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1467918
|
Finding Sylow 2-subgroups of the dihedral group [imath]D_n[/imath]
I am trying to describe the Sylow [imath]2[/imath]-subgroups of an arbitrary dihedral group [imath]D_n[/imath] of order [imath]2n[/imath]. In the case that [imath]n[/imath] is odd, [imath]2[/imath] is the highest power dividing [imath]2n[/imath], so that all Sylow [imath]2[/imath]-subgroups have order [imath]2[/imath], and it is fairly easy to describe them. However, if [imath]n[/imath] isn't odd, we may factor a power of [imath]2[/imath] out and write [imath]|D_n|=2^{k}m[/imath] for some odd integer [imath]m[/imath]. There is a proof that there exist precisely [imath]m[/imath] Sylow [imath]2[/imath]-subgroups, but it does not provide an explicit description of such subgroups in the case [imath]n[/imath] is odd. Additionally, someone has asked a similar question in the past, and they claim to give a description of the Sylow [imath]2[/imath]-subroups in the case [imath]n[/imath] is odd, but I can not find a source or an explanation. The question is here. Can anyone provide a description of how one may determine precisely the Sylow [imath]2[/imath]-subgroups for the case [imath]n[/imath] is odd? Thank you.
|
2040738
|
Finding all Sylow 2-subgroups
I want to find all Sylow 2-subgroups of [imath]D_6 = \{ e, r, r^2, r^3, r^4, r^5, s, sr, sr^2, sr^3, sr^4, sr^5\}[/imath]. [imath]r[/imath] is the rotation element, and [imath]s[/imath] is the reflection element. [imath]|D_6| = 12[/imath], and the order of a Sylow 2-subgroup of [imath]D_6[/imath] must be the highest order of [imath]2[/imath] that divides [imath]12[/imath] which is [imath]4[/imath]. Also, the order of each element of the Sylow 2-subgroup must be a power of [imath]2[/imath]. Using this information I was able to come up with only this Sylow 2-subgroup [imath]\{ e, s, r^3, sr^3 \}[/imath]. However, I do not know if there are more Sylow 2-subgroups of [imath]D_6[/imath]. Lastly, I figured out, by Sylow's Third Theorem, that the number of Sylow 2-subgroups of [imath]D_6[/imath] must divide the order of [imath]D_6[/imath], and also be congruent to 1 mod 2. The first part of this statement suggests that there can only be either [imath]1,2,3,4,6[/imath], or [imath]12[/imath] Sylow 2-subgorups of [imath]D_6[/imath], but I am unsure of what congruent to 1 mod 2 means. I know that 1 mod 2 [imath]= 1[/imath], but what does congruent mean?
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2039063
|
Is a function whose derivative vanishes at rationals constant?
I'm trying to make a problem for my advanced calculus students. I was thinking, if we have a differentiable function [imath]f:\mathbb{R}\to\mathbb{R}[/imath] such that [imath]f'(q)=0[/imath] for all [imath]q\in\mathbb{Q}[/imath], can we say that [imath]f[/imath] is constant?
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161733
|
Let [imath]f:\mathbb{R}\longrightarrow \mathbb{R}[/imath] a differentiable function such that [imath]f'(x)=0[/imath] for all [imath]x\in\mathbb{Q}[/imath]
Let [imath]f:\mathbb{R}\longrightarrow \mathbb{R}[/imath] a differentiable function such that [imath]f'(x)=0[/imath] for all [imath]x\in\mathbb{Q}.[/imath] [imath]f[/imath] is a constant function?
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2040160
|
On the proof that, if [imath](xy)^2=(yx)^2[/imath] for all [imath]x[/imath] and [imath]y[/imath] in a group, then [imath]xy^2=y^2x[/imath] for all [imath]x[/imath] and [imath]y[/imath]
Let G be a group and [imath](xy)^2=(yx)^2[/imath] for all [imath]x, y ∈ G[/imath]. Show that [imath]xy^2=y^2x[/imath] for all [imath]x, y ∈ G[/imath]. I understand the solution for this question but I have seen some people solve it in this way: Choose [imath]y = x^{-1}y[/imath] and substitute in for [imath]y[/imath] [imath]xx^{−1}yxx^{−1}y=x^{−1}yxx^{−1}yx⟹y^2=x^{−1}y^2x⟹xy^2=y^2x[/imath] My issue with this solution is that we're assuming under multiplication closure for any arbitrary element [imath]y ∈ G[/imath], we have [imath]y = x^{-1}y[/imath] for all [imath]x ∈ G[/imath]. This seems incorrect. Can anyone explain why I may be wrong about this?
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2037217
|
Let G be a group such that [imath](xy)^2 = (yx)^2[/imath] for all x, y ∈ G. Show that [imath]xy^2 = y^2x[/imath] for all x, y ∈ G.
I'm not quite sure if I did this right but I had another solution that made no sense to me: This is the solution I do not understand. I did not solve it this way. [imath]=>(xy)^{-1}(xy)^2(yx)^{-1} = (xy)^{-1}(yx)^2(yx)^{-1}[/imath] [imath]=>(xy)(yx)^{-1} = (xy)^{-1}(yx)[/imath] then somehow we have [imath]=> xy^2 = x((x^{-1}y)x)^2[/imath] [imath]=> xy^2 = x(x(yx^{-1}))^2[/imath] [imath]=> xy^2 = (xyx)^{-1}(xyx^{-1})x[/imath] [imath]=> xy^2 = y^2x[/imath] I'm positive that the solution above is wrong I just need confirmation. This is my solution: Since G is a group then for all x,y ∈ G there exists [imath]x^{-1}[/imath] , [imath]y^{-1}[/imath] ∈ G. [imath](xy)^2 = (yx)^2[/imath] [imath]y^{-1}x^{-1}xyxy = y^{-1}x^{-1}yxyx[/imath] [imath]xy = y^{-1}x^{-1}yxyx[/imath] [imath]y = x^{-1}y^{-1}x^{-1}yxyx[/imath] Take the left hand side of [imath]xy^2 = y^2x[/imath] and substitute [imath]y = x^{-1}y^{-1}x^{-1}yxyx[/imath] [imath]xy^2 = x(x^{-1}y^{-1}x^{-1}yxyx)^2[/imath] [imath]xy^2 = x(x^{-1}y^{-1}x^{-1}yxyx)(x^{-1}y^{-1}x^{-1}yxyx)[/imath] [imath]xy^2 = xx^{-1}y^{-1}x^{-1}yxyxx^{-1}y^{-1}x^{-1}yxyx[/imath] [imath]xy^2 = y^{-1}x^{-1}yyxyx[/imath] [imath]xy^2 = y^{-1}x^{-1}y^2xyx[/imath] [imath]xyxy^2 = y^2xyx[/imath] [imath]y^{-1}x^{-1}xyxy^2 = y^2xyxx^{-1}y^{-1}[/imath] [imath]xy^2 = y^2x[/imath].
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2039711
|
Idempotent elements on the quotient ring
I am trying to prove the following result: If [imath]R[/imath] is a commutative ring with identity, [imath]I[/imath] is an ideal of [imath]R[/imath] such that [imath]I^2=\{0\}[/imath] and [imath]a+I[/imath] is an idempotent element in [imath]R/I[/imath] then there are some idempotent element of [imath]R[/imath] in the coset [imath]a+I[/imath]. What I tried to do? Well, suppose that [imath]a+I[/imath] is idempotent in [imath]R/I[/imath], then [imath](a+I)(a+I)=a+I[/imath], so, [imath]a^2+I=a+I[/imath] and then [imath]a^2-a \in I[/imath]. We know also that [imath]1-a+I[/imath] is idempotent since [imath](1-a+I)(1-a+I)=1-2a+a^2+I = 1-2a+a+I=1-a+I[/imath]. And then, from this we know that [imath]a(1-a)+I[/imath] is idempotent. We need to show that there exist some [imath]m \in I[/imath] such that [imath](a+m)^2=a+m[/imath], but [imath](a+m)^2=a^2+2am+m^2=a^2+2am[/imath] since [imath]m^2=0[/imath], so [imath]m[/imath] needs to satysfy [imath]a^2+(2m-1)a-m=0.[/imath] I dont know if this reasoning is right, neither if this is a way to go over it. Can someone give me a hint to finish it?
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2023115
|
Lifting idempotents modulo a nilpotent ideal
The problem is this: Suppose [imath]I \subseteq R[/imath] is a nilpotent ideal and there is [imath]r \in R[/imath] with [imath]r \equiv r^2 \pmod I[/imath]. Show [imath]r \equiv e \pmod I[/imath] for some [imath]e \in R[/imath] idempotent. I have spent a few hours rolling around in abstracta with no destination. I believe that if I could write down a concrete example of this, the example could guide me through the abstract definitions and show me where to look for an idempotent in [imath]R[/imath]. I have looked at 2x2 matrices and could not find any such examples. Might anybody have one?
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2033091
|
Finding the limit as h goes to zero
It seems to be a simple question, but I couldn't figure it out. I want to find [imath]\lim_{h\to 0} \frac{\frac{1}{\sqrt {x+h}}-\frac{1}{\sqrt {x}}}{h}[/imath] I don't know how to control the font size.
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202347
|
Find the derivative of the function
If [imath]f(x) = 1/\sqrt{x}[/imath], find [imath]f^{\prime}(x)[/imath]. Please show the answer by using [imath]\lim_{h\to{0}}\frac{f(x+h)-f(x)}{h}.[/imath] I know the answer by using the shortcut, but my teacher wants me to get the answer using that equation. I am stuck and I got to: [imath]-\sqrt{x}/2x^2[/imath] by using the [imath]\frac{f(x+h)-f(x)}{h}[/imath].
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2040265
|
Find [imath](2^{a}+1,2^{b}+1),[/imath] where [imath]a[/imath] and [imath]b[/imath] are prime numbers greater than [imath]3[/imath] and [imath](x,y)[/imath] represents the [imath]\gcd(x,y).[/imath]
Problem: Find [imath](2^{a}+1,2^{b}+1),[/imath] where [imath]a[/imath] and [imath]b[/imath] are prime numbers greater than [imath]3[/imath] and [imath](x,y)[/imath] represents the [imath]\gcd(x,y).[/imath] My Attempt: Observe that [imath](2^a+1,2^b+1)=(2^{a-b}-1,2^b+1)[/imath] if we assume that [imath]a>b.[/imath] Thus it is obvious that [imath](2^a+1,2^b+1)=2^{(a,b)}\pm1.[/imath] Since [imath](a,b)=1[/imath], we have [imath](2^a+1,2^b+1)=1,3.[/imath] I can't figure out the condition under which [imath](2^a+1,2^b+1)=3.[/imath] Please help.
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78239
|
Prove that if [imath]a[/imath] and [imath]b[/imath] are odd, coprime numbers, then [imath]\gcd(2^a +1, 2^b +1) = 3[/imath]
Prove that if [imath]a[/imath] and [imath]b[/imath] are odd, coprime numbers, then [imath]\gcd(2^a +1, 2^b +1) = 3[/imath]. I was thinking among the lines of: Since [imath]a[/imath] and [imath]b[/imath] are coprime numbers, [imath]\gcd(a,b)=1[/imath]. Then there exist integers [imath]x[/imath] and [imath]y[/imath] such that, [imath]ax+by=1[/imath]. Then, [imath]a=(1-by)/x[/imath], [imath]b=(1-ax)/y[/imath] So if I write [imath]2^a+1[/imath] as: [imath]2^{(1-by)/x}+1[/imath] Then can I say that the above expression is equivalent to 0 (mod 3)?
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2040796
|
Why [imath]\exp(A+B) \ne \exp(A)\exp(B)[/imath]
What is the deep reason why [imath] \exp(A+B) \ne \exp(A)\exp(B)[/imath] where [imath]A,B \in \Re^{N\times N} [/imath] where [imath]\exp[/imath] is function defined via Taylor Series [imath]\exp(M)=\sum_i \frac{M^i}{i!}[/imath] Looks like it is due to non-commutativity of matrix multiplication?
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755360
|
On the proof: [imath]\exp(A)\exp(B)=\exp(A+B)[/imath] , where uses the hypothesis [imath]AB=BA[/imath]?
I was seeing the proof that [imath]\exp(A)\exp(B)=\exp(A+B)[/imath] on link Show that [imath] e^{A+B}=e^A e^B[/imath] where uses the hypothesis [imath]AB=BA[/imath]? Thanks!
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2034986
|
Volume form on unit sphere
The volume form on the unit sphere [imath]S^{n}[/imath] in [imath]\mathbb{R}^{n+1}[/imath] is given by [imath]i_{\bf r}\ dx^1 \wedge \cdots \wedge dx^{n+1}=\sum (-1)^{i-1} x^i \, dx^1 \wedge\cdots \wedge \widehat{dx^i} \wedge \cdots \wedge dx^{n+1}[/imath] Why must the volume form [imath]dx^1 \wedge \dots \wedge dx^{n+1}[/imath] act on the vector [imath]{\bf r}[/imath] to give the volume form on the unit sphere? Also, how do I get the form of the volume form on the right-hand side of the equation?
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1284234
|
Volume form on [imath](n-1)[/imath]-sphere [imath]S^{n-1}[/imath]
Let [imath]\omega[/imath] the (n-1) form on [imath]\mathbb{R}^n[/imath] [imath]\omega=\sum_{j=1}^{n}(-1)^{j-1}x_{j}dx_{1}\wedge\cdots\wedge \hat{dx_{j}}\wedge\cdots dx_{n}[/imath] Show that the restriction of [imath]\omega[/imath] to [imath]S^{n-1}[/imath] in precisely the volume for this sphere. What I did was: [imath]\omega[/imath] never vanish on the sphere, because, defining [imath]\sigma\in \Omega^{n-1}(S)[/imath] for [imath]\sigma_{p}(v_{1},...,v_{n-1})=det(p,v_{1},...,v_{n-1})[/imath] and [imath]i:S^{n-1}\rightarrow \mathbb{R}^{n}[/imath] the inclusion function, then [imath]\omega=i^{\ast}(\sigma)[/imath] then [imath]\omega\not=0[/imath] and therfore is a volume form. How proof that [imath]\omega[/imath] is the volume form? The first thing that comes to mind is show that [imath]\int_{S^{n-1}}\omega=Vol(S^{n-1})[/imath] but I have serious problems with the definition, I think that is to much. How see that [imath]\omega[/imath] is invariant on [imath]\mathbb{R}^{n}[/imath] under action of [imath]O(n)[/imath]
|
2040434
|
The Projection Matrix is Equal to its Transpose
By inspecting its formula one can see easily that the matrix for projection onto a subspace is equal to its transpose. But what is the underlying "geometric" reason for this equality? I have hard time figuring out why it has to be so? EDIT: After reviewing the suggested links, it became more clear, but the reasoning was still escaping my intuition. However, I think that I can (based on the formal arguments in the links) put forward a rough argument. Basically, for any operator [imath]A[/imath] and two vectors [imath]x[/imath], [imath]y[/imath] [imath]<Ax,y>=<x,A^Ty>[/imath] For a vector [imath]y[/imath] in the orthogonal complement to the subspace S we're projecting onto using the [imath]P^S[/imath] projection operator, [imath]<P^Sx,y>=0=<x,(P^S)^Ty>[/imath] Because we took any vector [imath]x[/imath], it means that [imath](P^S)^Ty=0[/imath]. As [imath]y[/imath] is in the orthogonal complement, [imath]P^Sy=0[/imath] this means that [imath](P^S)^Ty=0=P^Sy[/imath] (in plain English, for any vector in the orthogonal complement to [imath]S[/imath] the action of [imath]P^S[/imath] makes it 0, which is a symmetric action). For [imath]y[/imath] in [imath]S[/imath] and any [imath]x[/imath], [imath]<P^Sx,y>=<x,y>[/imath] and [imath]P^Sy=y[/imath] (by the properties of the projection operation), but also [imath]<P^Sx,y>=<x,(P^S)^Ty>[/imath] (true for any operator). Putting these together that yields that:[imath](P^S)^Ty=y=P^Sy[/imath]. (in plain English, for any vector in [imath]S[/imath] the action of [imath]P^S[/imath] is the identity, which is a symmetric action). To sum it up, this shows that the operator [imath]P^S[/imath] acts symmetrically on both elements of S and on elements in its orthogonal complement. Being that it is a linear operator, then it acts symmetrically on all vectors [imath]x[/imath]. A big thing why this works (I think) is that the concept of projection implies (finite) dimension inner product space and that is much more structured than a simple (finite) dimensional vector space. LAST EDIT: Even more clear after digesting the answer of Trial and Error. Given a subspace M to project onto, any vector [imath]z_1[/imath] can be written as the sum of two orthogonal components: [imath]z_1= (z_1-P_{\mathcal{M}}z_1) + P_{\mathcal{M}}z_1,[/imath] But the key geometric thing is that these two components belong to orthogonal subspaces, M and [imath]\perp M[/imath] (any vector from one subspace is perpendicular on any vector from the other subspace). The projection operator will leave the component in M unchanged and annihilate the component in [imath]\perp M[/imath]. Because of the fact that the subspaces themselves are orthogonal the product of ANY two arbitrary vectors [imath]z_1, z_2[/imath] [imath]<z_1,z_2>=<(z_1-P_{\mathcal{M}}z_1) + P_{\mathcal{M}}z_1,(z_2-P_{\mathcal{M}}z_2) + P_{\mathcal{M}}z_2>[/imath] becomes (by the orthogonality of M and [imath]\perp M[/imath]: [imath]<z_1,z_2>=<(z_1-P_{\mathcal{M}}z_1),(z_2-P_{\mathcal{M}}z_2)>+<P_{\mathcal{M}}z_1,P_{\mathcal{M}}z_2>[/imath] But applying [imath]P_M[/imath] annihilates/zeroes the first bracket and leaves unchanged the second bracket. This holds whether we apply it to [imath]z_1[/imath] or to [imath]z_2[/imath]. Ergo the effect of [imath]P_M[/imath] is the same whether applied to the first of second term of the scalar product. So its matrix should be equal to its transpose. The bolded text indicates key issues that I was not fully appreciating before. Thank you for your patience!
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456354
|
Why is a projection matrix symmetric?
I am looking for an intuitive reason for a projection matrix of an orthogonal projection to be symmetric. The algebraic proof is straightforward yet somewhat unsatisfactory. Take for example another property: [imath]P=P^2[/imath]. It's clear that applying the projection one more time shouldn't change anything and hence the equality. So what's the reason behind [imath]P^T=P[/imath]?
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2032920
|
There is given a countable set [imath]A[/imath]. Prove that [imath](x+A)\cap A=\emptyset[/imath] for some [imath]x[/imath]
There is given a countable set [imath]A\subset \Bbb{R}.[/imath] Prove that there exists an [imath]x\in R[/imath] that [imath](x+A)\cap A=\emptyset[/imath], where [imath]x+A=\{x+a: a\in A\}[/imath].
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250882
|
Proof that[imath] (a+A)\cap A=\varnothing[/imath] if [imath]A[/imath] is countable
I finished my test and there is a question I completely failed but that my teacher did not go over, so I was hoping someone could post a correction of it, so that I understand what I was supposed to do for next time. Suppose that A is a countable set of real numbers. Show that there exists a real number a such that [imath](a+A)\cap A=\varnothing[/imath]. (Note: By definition, [imath]a+A= \{a+r\mid r\in A\}[/imath].)
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2041218
|
Undecidable language problems
Let [imath]L_1[/imath] and [imath]L_2[/imath] be two undecidable languages. 1) Is it possible that [imath]L_1 - L_2[/imath] is regular and [imath]L_1 - L_2 \neq \emptyset[/imath] ? Well I know that if I let [imath]L_1 = L_2[/imath] then [imath]L_1 - L_2 = \emptyset[/imath] which is obviously regular. In the case of this question however, I cannot set [imath]L_1 = L_2[/imath] if the outcome is an empty set, which makes me wonder what would [imath]L_1 - L_2[/imath] actually look like if they were not equal to each other. 2) Is it possible that [imath]L_1 \cup L_2[/imath] is decidable and [imath]L_1 \neq \neg L_2[/imath] ? This one is also tricky, since the obvious solution here would be to let [imath]L_1 = \neg L_2 [/imath]. Does anyone here have an idea as to how to finish my solutions here? I need to prove my solutions in both cases.
|
2040900
|
Undecidable language problems.
Let [imath]L_1[/imath] and [imath]L_2[/imath] be two undecidable languages. 1) Is it possible that [imath]L_1 - L_2[/imath] is regular and [imath]L_1 - L_2 \neq \emptyset[/imath] ? Well I know that if I let [imath]L_1 = L_2[/imath] then [imath]L_1 - L_2 = \emptyset[/imath] which is obviously regular. In the case of this question however, I cannot set [imath]L_1 = L_2[/imath] if the outcome is an empty set, which makes me wonder what would [imath]L_1 - L_2[/imath] actually look like if they were not equal to each other. 2) Is it possible that [imath]L_1 \cup L_2[/imath] is decidable and [imath]L_1 \neq \neg L_2[/imath] ? This one is also tricky, since the obvious solution here would be to let [imath]L_1 = \neg L_2 [/imath] Does anyone here have an idea as to how to finish my solutions here? I need to prove my solutions in both cases.
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2041441
|
Divisibility of [imath]\binom{74}{37} -2[/imath] where [imath]\binom{74}{37}[/imath] is a central binomial coefficient
[imath]\binom{74}{37}-2[/imath] is divisible by : a) [imath]1369[/imath] b) [imath]38[/imath] c) [imath]36[/imath] d) [imath]none[/imath] [imath]of[/imath] [imath] these[/imath] I have no idea how to solve this...I tried writing [imath]\binom{74}{37}[/imath] in some useful form but its not helping...any clues?? Thanks in advance!!
|
325559
|
How to show [imath]\binom{2p}{p} \equiv 2\pmod p[/imath]?
how to prove [imath]\forall p[/imath] prime : [imath]\binom{2p}{p} \equiv 2 \pmod p[/imath] we have: [imath]\binom{2p}{p} = \frac{2p (2p-1)(2p-3)...1}{p!p!}[/imath] but how to continue?
|
2041550
|
Construct a continuous onto map from [imath](0,1][/imath] to [imath][0,1][/imath]
What are some example of continuous surjective function in a given interval from [imath](0,1][/imath] to [imath][0,1][/imath]? Any help is appreciated.
|
772075
|
A continuous surjective function from [imath](0,1][/imath] onto [imath][0,1][/imath]
I'm trying to construct a continuous surjection from [imath](0,1][/imath] onto [imath][0,1][/imath], but I'm not getting anywhere. I don't immediately see a contradiction which falsifies the existence of such a function, so my intuition tells me one exists. I feel like an absolute value function would work, but I'm not sure how to arrive at it in the proper way. Thanks for any help.
|
2035463
|
Describe in clear English a Turing machine that semidecides the following language.
The language being L, which is described as follows: [imath]L[/imath] = { [imath]<M>[/imath] | [imath]M[/imath] accepts the binary encodings of at least 4 odd numbers } . I understand that this involves using one turing machine to simulate another. What sort of methodology could I use for these kinds of questions?
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2037631
|
Describe in English a Turing machine that semidecides the following language - Is this solution correct?
The language being L, which is described as follows: [imath]L[/imath] = { [imath]<M>[/imath] | [imath]M[/imath] accepts the binary encodings of at least 4 odd numbers } . My solution: M generates the binary encodings in [imath]\Sigma_M*[/imath] in ascending order and uses dovetailing to interleave the computation of M on those binary encodings. As soon as four computations accept, M halts and accepts. Is this correct? What corrections would have to be made if not? I am not sure if I am providing enough details in my solution which may cause it to be incorrect.
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2041678
|
Finitely generated monoids- axiomatization- compactness theorem.
Let [imath]\mathbb{M}=\langle M,\circ, e\rangle[/imath] be a monoid. We say that [imath]\mathbb{M}[/imath] is finitely generated when there exist finitely many elements in [imath]M[/imath] such that each element of [imath]M[/imath] may be generated using only these elements. For example, [imath]\langle A^*,\circ, \epsilon \rangle [/imath] is finitely generated if only and only alphabet [imath]A[/imath] is finite. Show that the class of finitely generated monoids is not axiomatized. My solution Let assume that there exists a such set of sentences [imath]\Delta[/imath] that [imath]\mathbb{M} \models \Delta \iff \mathbb{M} \in \mathbb{M_K}[/imath] where [imath] \mathbb{M_K}[/imath] is a class of finitely generated monoids. So, let's consider a language [imath]L = \langle A^*, \circ, e \rangle[/imath] where [imath]A = \{a \}[/imath] So [imath]A[/imath] is a finite set, therefore [imath]L[/imath] is a finitely generated monoid. So, [imath]L \models \Delta[/imath]. Let [imath]\Delta' = \Delta \cup \Gamma, \Gamma = \{\exists w w = b_0, \exists w w = b_0 \circ b_1, \exists w w = b_0 \circ b_1 \circ .. \circ b_n | b_0, .., b_n \text { are pairwise different and } n \in \mathbb{N}[/imath]. Now, for every finite [imath]\Delta_0 \subseteq \Delta'[/imath] it is easy to point the such finitely generated [imath]L_2[/imath] that [imath]L_2 \models \Delta_0[/imath]. From the compactness theorem there exists the model ( a language [imath]L'[/imath]) for [imath]\Delta', L' \models \Delta'[/imath]. But, [imath]L' \models \Delta \cup \Gamma [/imath] and it is not possible because [imath]L'[/imath] is not finitely generated while [imath]\Delta[/imath] force that. Contradiction. Ok? I would like to solve it using Ehrenfeucht-Fraisse game, but I don't know how to start. Please help. Especially, what does it mean that two words are in relationship?
|
2039109
|
Finitely generated monoids - how to show that there is not first order set axiomatizing them.
Let [imath]\mathbb{M}=\langle M,\circ, e\rangle[/imath] be a monoid. We say that [imath]\mathbb{M}[/imath] is finitely generated when there exist finitely many elements in [imath]M[/imath] such that each element of [imath]M[/imath] may be generated using only these elements. For example, [imath]\langle A^*, \cdot, \epsilon\rangle[/imath] is finitely generated if only and only alphabet [imath]A[/imath] is finite. Show that the class of finitely generated monoids is not axiomatized. I would like to use Ehrenfeucht-Fraissé games to prove it. I solved some tasks about showing not-axiomatization with these games, however mentioned exercises were about graphs. So my solutions are based on relations (edges, degrees..). Here, I have no idea which side of this exercise should be attacked by me. Help me please. Edit after hints of @Noah Lets suppose that such [imath]\Delta[/imath] exists. Then, lets extend signature by constans: [imath]c_1,c_2,...[/imath] and let [imath]\Delta'=\Delta\cup \Gamma[/imath], where [imath]\Gamma=\{\forall_{x\in A}\forall_{y\in A}x\circ y \neq c_i:i\in\mathbb{N} \}[/imath]. Of course we have infinitely many elements not generated by rest of elements. Using compactness theorem we conclude contradiction. Let [imath]\Delta_0\subseteq\Delta'[/imath] will be finite set. Then it is easy to see that it is satisfable - We can add properly many elements to [imath]A[/imath] generated by nothing and assign to them finitely number of constanst [imath]c_i[/imath]. What do you think about this solution ? If I am wrong, tell me where.
|
2041717
|
Evaluating [imath]\int_{-\infty}^{\infty} \frac{x\sin(kx)}{x^2+a^2} \,\mathrm dx[/imath] using complex analysis
I'm trying to solve the following integral: [imath]\int_{-\infty}^{\infty} \frac{x\sin(kx)}{x^2+a^2} \,\mathrm{d}x[/imath] I don't really have any idea where to start. The previous parts of the question involved complex numbers and Cauchy's Integral Formula, however I can't think how I'd applying that here. Any help would be greatly appreciated :)
|
769438
|
Evaluating the integral [imath]\int_0^\infty \frac{x \sin rx }{a^2+x^2} dx[/imath] using only real analysis
Calculate the integral[imath] \int_0^\infty \frac{x \sin rx }{a^2+x^2} dx=\frac{1}{2}\int_{-\infty}^\infty \frac{x \sin rx }{a^2+x^2} dx,\quad a,r \in \mathbb{R}. [/imath] Edit: I was able to solve the integral using complex analysis, and now I want to try and solve it using only real analysis techniques.
|
2041742
|
How would you prove that this limit exists at the origin
Consider [imath]\lim_{(x,y)\to(0,0)} \frac{xy}{|x|+|y|}[/imath] I tried the following technique but couldn't work it out define [imath]p=(x,y)[/imath] we end up with the following inequalities : [imath]|p|^2=x^2+y^2\\ |x|≤|p| \\ |y|≤ |p|[/imath] But couldn't use them at all couldn't find a boundary. What should I do? What would you recommend?
|
307845
|
Wrong Wolfram|Alpha limit? [imath] f(x,y) = \frac {xy}{|x|+|y|} [/imath] for [imath](x,y)\to(0,0)[/imath]
I have this function: [imath] f(x,y) = \frac {xy}{|x|+|y|} [/imath] And I want to evaluate it's limit when [imath] (x,y) \to (0,0)[/imath] My guess is that it tends to zero. So, by definition, if: [imath] \forall \varepsilon \gt 0, \exists \delta \gt 0 \diagup \\ 0\lt||(x,y)||\lt \delta , \left|\frac{xy}{|x|+|y|}\right| \lt \varepsilon [/imath] Then [imath] \lim_{(x,y)\to(0,0)}\frac {xy}{|x|+|y|} = 0 [/imath] So: [imath] \left|\frac{xy}{|x|+|y|}\right| = \frac{|xy|}{|x|+|y|} = \frac{|x||y|}{|x|+|y|} \le 1 |y| \lt \delta [/imath] So for any [imath]\delta \lt \varepsilon[/imath] the inequality is true. Hence, the limit exists and is equal to zero. Wolfram|Alpha says that the limit does not exist. Am I wrong or is Wolfram|Alpha wrong?
|
1856983
|
Find an irrational number [imath]n[/imath] such that [imath]n^n[/imath] is a rational number.
Find an irrational [imath]n[/imath] such that [imath]n^n[/imath] is a rational number. I have some tries to find this... I have tried so much numbers but no success. How can I find them.
|
1641878
|
Is there an irrational number [imath]a[/imath] such that [imath]a^a[/imath] is rational?
It can be proved that there are two irrational numbers [imath]a[/imath] and [imath]b[/imath] such that [imath]a^b[/imath] is rational (see Can an irrational number raised to an irrational power be rational?) and that for each irrational number [imath]c[/imath] there exists another irrational number [imath]d[/imath] such that [imath]c^d[/imath] is rational (see For each irrational number b, does there exist an irrational number a such that a^b is rational?). My question is: Is there an irrational number [imath]a[/imath] such that [imath]a^a[/imath] is rational (and how could you prove that)?
|
2043306
|
How to calculate [imath]\lim_{n \to \infty} \frac{1}{n}\sqrt[n] {(n+1)(n+2)...(n+n)}[/imath]?
How can I solve this limit: [imath]\lim_{n \to \infty} \frac{1}{n}\sqrt[n] {(n+1)(n+2)...(n+n)}[/imath]? I've tried to do it by Sandwich, but I only obtained this: [imath]\frac{1}{n}\sqrt[n] {(n+1)^n} \leq \frac{1}{n}\sqrt[n] {(n+1)(n+2)...(n+n)}\leq \frac{1}{n}\sqrt[n] {(n+n)^n}[/imath] But in this way I only know that limit value is between 1 and 2.
|
1773275
|
What is the [imath]\lim_{n\to \infty} {\sqrt[n]{(1+1/n)(1+2/n)...(1+n/n)}}[/imath]
Because [imath](1+\frac{k}{n})\leq (1+\frac{1}{n})^k[/imath] [imath] {\sqrt[n]{(1+1/n)(1+2/n)...(1+n/n)}}\leq\sqrt[n]{(1+\frac{1}{n})^{n(n+1)/2}}=(1+\frac{1}{n})^{(n+1)/2}[/imath] and so [imath]\lim_{n\to \infty}{\sqrt[n]{(1+1/n)(1+2/n)...(1+n/n)}}<\sqrt{e}[/imath] But what is the exact limit? Added: I Gave this to Wolfram and the answer was [imath]\infty[/imath] ! but why?
|
2042727
|
How do I evaluate this trigonometric limit: [imath]\lim_{x\to\frac\pi4}\frac{\sqrt2\cos(x)-1}{\cot(x)-1}[/imath]?
[imath]\lim\limits_{x\to\frac\pi4}\frac{\sqrt2\cos(x)-1}{\cot(x)-1}=?[/imath] I tried using many formulae but am not able to cancel out the zero factor. Need help...thanks!!
|
632538
|
How to evaluate this trigonometric limit?
I have been struggling to solve this limit. What is the limit as [imath]x[/imath] approaches [imath]45^0[/imath] of [imath]\frac{\sqrt{2}\cos x-1}{\cot x-1}?[/imath] I know how to use L'Hospital's rule to calculate this limit and got the answer as [imath]0.5[/imath]. But, how do I calculate the limit by manipulating the function? Please provide only some hints to proceed.
|
2044107
|
Is a linear transformation continuous?
Let f :[imath] ℝ^n→ℝ^n[/imath] be a linear transformation. Is f always continuous? Can anyone help how to show this?
|
401605
|
Proof that a linear transformation is continuous
I got started recently on proofs about continuity and so on. So to start working with this on [imath]n[/imath]-spaces I've selected to prove that every linear function [imath]f: \mathbb{R}^n \to \mathbb{R}^m[/imath] is continuous at every [imath]a \in \mathbb{R}^n[/imath]. Since I'm just getting started with this kind of proof I just want to know if my proof is okay or if there's any inconsistency. My proof is as follows: Since [imath]f[/imath] is linear, we know that there's some [imath]k\in \mathbb{R}[/imath] such that [imath]|f(x)|\leq k|x|[/imath] for every [imath]x\in \mathbb{R}^n[/imath], in that case let [imath]a\in \mathbb{R}^n[/imath] and let [imath]\varepsilon >0[/imath]. Consider [imath]\delta = \varepsilon /k[/imath] and suppose [imath]|x-a|<\delta[/imath], in that case we have: [imath]|f(x)-f(a)|=|f(x-a)|\leq k |x-a|<k \frac{\varepsilon}{k}=\varepsilon[/imath] And since [imath]|x-a|<\delta[/imath] implies [imath]|f(x)-f(a)|<\varepsilon[/imath] we have that [imath]f[/imath] is continuous at [imath]a \in \mathbb{R}^n[/imath]. Since [imath]a[/imath] was arbitrary, [imath]f[/imath] is continous in [imath]\mathbb{R}^n[/imath]. Is this proof fine? Or there was something I've missed on the way?
|
2043019
|
Let be [imath]f[/imath] a continuous function. Determine the limit [imath]\lim\limits_{h \to 0} \frac{1}{h} \int_{a-h}^{a+h} f(x)\,dx[/imath]
[imath]\lim\limits_{h \to 0} \frac{1}{h} \int_{a-h}^{a+h} f(x)\,dx[/imath] I think that this kind of limit should I probably calculate with some kind of epsilon-delta definition. And using the limits: [imath]\lim\limits_{h \to 0^+} \frac{1}{h}=\infty [/imath] [imath]\lim\limits_{h \to 0^-} \frac{1}{h}=-\infty [/imath] I appreciate any helps.
|
1529506
|
Let [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] be continuous.Let [imath]t\in\mathbb{R}[/imath]. Evaluate:[imath]\lim_{h\rightarrow0}\frac{1}{h}\int_{t-h}^{t+h}f(s)ds[/imath]
Let [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] be continuous.Let [imath]t\in\mathbb{R}[/imath]. Evaluate:[imath]\lim_{h\rightarrow0}\frac{1}{h}\int_{t-h}^{t+h}f(s)ds[/imath]. [imath]\lim_{h\rightarrow0}\frac{1}{h}\int_{t-h}^{t+h}f(s)ds=\lim_{h\rightarrow0}\frac{1}{h}\int_{t-h}^{t}f(s)ds+\lim_{h\rightarrow0}\frac{1}{h}\int_{t}^{t+h}f(s)ds[/imath]. Intuitively, I know that these both limit tends to [imath]f(t)[/imath]. Therefore answer will be [imath]2f(t)[/imath]. But I don't know the formal way to prove that this limit tends to [imath]2f(t)[/imath]. Help me if there is any formal proof of this (like using any theorem etc.) Thank you
|
2041988
|
How to get inverse of formula for sum of integers from 1 to n?
I know very well that the sum of integers from [imath]1[/imath] to [imath]n[/imath] is [imath]\dfrac{n\times(n+1)}2[/imath]. What I'm interested in today, and cannot find a solution for, is performing the opposite operation. Let [imath]m = \dfrac{n^2 + n} 2[/imath]. Knowing the value of [imath]m[/imath], how do I figure out the value of [imath]n[/imath]? I could easily program a solution but I'd much prefer an algebraic one.
|
1778490
|
Reversing the [imath]T(n) = \frac{n(n+1)}{2}[/imath] formula
Can someone reverse this formula? Sorry if it's too basic, I'm an old guy and my math is rusty :-D [imath]s =\frac{n(n+1)}{2}[/imath] If it ain't clear, by reverse I mean obtaining the value of n by providing s Thanks
|
2042938
|
Let [imath]A[/imath] be a subspace of [imath]M_{n\times n}(F)[/imath], prove that if [imath]\dim(A)> n^2-n[/imath] , [imath]\exists[/imath] invertible matrix [imath]M \in A[/imath].
Let [imath]A[/imath] be a subspace of [imath]M_{n\times n}(F)[/imath], prove that if [imath]\dim(A)> n^2-n[/imath] , [imath]\exists[/imath] invertible matrix M [imath]\in[/imath] [imath]A[/imath].
|
66877
|
Max dimension of a subspace of singular [imath]n\times n[/imath] matrices
I am sure the answer to this is (kind of) well known. I've searched the web and the site for a proof and found nothing, and if this is a duplicate, I'm sorry. The following question was given in a contest I took part. I had an approach but it didn't solve the problem. Consider [imath]V[/imath] a linear subspace of the real vector space [imath]\mathcal{M}_n(\Bbb{R})[/imath] ([imath]n\times n[/imath] real entries matrices) such that [imath]V[/imath] contains only singular matrices (i.e matrices with determinant equal to [imath]0[/imath]). What is the maximal dimension of [imath]V[/imath]? A quick guess would be [imath]n^2-n[/imath] since if we consider [imath]W[/imath] the set of [imath]n\times n[/imath] real matrices with last line equal to [imath]0[/imath] then this space has dimension [imath]n^2-n[/imath] and it is a linear space of singular matrices. Now the only thing there is to prove is that if [imath]V[/imath] is a subspace of [imath]\mathcal{M}_n(\Bbb{R})[/imath] of dimension [imath]k > n^2-n[/imath] then [imath]V[/imath] contains a non-singular matrix. The official proof was unsatisfactory for me, because it was a combinatorial one, and seemed to have few things in common with linear algebra. I was hoping for a pure linear algebra proof. My approach was to search for a permutation matrix in [imath]V[/imath], but I used some 'false theorem' in between, which I am ashamed to post here.
|
2044584
|
Rings of order [imath]p^2[/imath] and some related information
Although group of order [imath]p^2[/imath] are well known, the rings of order [imath]p^2[/imath] may not be so well known; I was feeling that there could be more than two rings of order [imath]p^2[/imath]. I have two questions related to this, and I don't know whether the questions I am posing are trivial. [imath]\mathbb{Z}_{p^2}[/imath] and [imath]\mathbb{Z}_p\oplus\mathbb{Z}_p[/imath] are obvious examples of rings (with unity) of order [imath]p^2[/imath]. Question 1. Are there more than two rings (with unity) of order [imath]p^2[/imath]? Question 2. If there are more than two rings of order [imath]p^2[/imath] (with unity), is the group of units of these rings known?
|
109506
|
Classifying Unital Commutative Rings of Order [imath]p^2[/imath]
I'm trying to classify unital commutative rings of order [imath]p^2[/imath], where [imath]p[/imath] is a prime. At first, I happened to neglect the 'unital' and 'commutative' requirements, and after an arduous route I managed to show that there are [imath]11[/imath] such rings up to isomorphism. Some are non-commutative and non-unital, and the journey to that result is a bit ugly for a class on commutative algebra. This should be easier if we only consider unital rings of order [imath]p^2[/imath]. (It's easy to show that if it's unital, then it's commutative in this case). But so far, I haven't found a solution that doesn't use the fact that I found representations for the [imath]11[/imath] rings. I am certain that a much less technical and messy method to find unital commutative rings of order [imath]p^2[/imath] is possible. For reference on the context, this question comes amidst a review of the Chinese Remainder Theorem, the Structure Theorem on Modules over a PID, tensor products, and algebras. I heavily suspect that if I were more fluent in applying the CRT, I would be able to get there. Do you have any ideas? By the way, I believe there are [imath]4[/imath]. They should look like [imath]\mathbb{F}_{p^2}, \mathbb{Z}_{p^2}, \mathbb{Z}_p [x]/(x^2),[/imath] and (I don't know a convenient name for the fourth, but e.g., the Klein 4-group with standard ring structure on top, which I'm inclined to designate [imath]\mathbb{Z}_{p \times p}[/imath]). In a more convenient designation, 1 is built on the additive group [imath]C_{p^2}[/imath] and 3 are built on [imath]C_p \times C_p[/imath]. I would be content if I could enumerate how many rings are on each additive group, up to homomorphism, instead of actually classifying them.
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2036507
|
Are [imath]I\otimes_{R}J[/imath] and [imath]IJ[/imath] isomorphic as [imath]R[/imath]-modules?
Let [imath]I[/imath] and [imath]J[/imath] be ideals of a commutative ring [imath]R,[/imath] considered as [imath]R[/imath]-modules. The product ideal [imath]IJ=\left\{\sum_{i=1}^{n}a_{i}b_{j}:n\in\mathbb{N},a_{i}\in I,b_{j}\in J\right\}[/imath] is also an [imath]R[/imath]-module. Question: Is it true that [imath]I\otimes_{R}J\cong IJ?[/imath] This seems intuitive to me, since I like to think of the tensor product [imath]M\otimes_{R}N[/imath] as "sums of things of the form [imath]m\otimes n[/imath] where [imath]m\in M[/imath] and [imath]n\in N,[/imath]" and this sounds a lot like a description of [imath]IJ.[/imath] I came up with the following (now seen to be incorrect) argument, which originally I had posted as an answer: Let [imath]f\colon I\times J\to IJ[/imath] be given by [imath](a,b)\mapsto ab.[/imath] This is an [imath]R[/imath]-bilinear map, so there exists a (unique) corresponding [imath]R[/imath]-linear map [imath]h\colon I\otimes_{R} J\to IJ[/imath] with [imath]h(a\otimes b)=ab.[/imath] We would like to prove that this map [imath]h[/imath] is bijective. To prove that [imath]h[/imath] is surjective: the element [imath]\sum_{i=1}^{n}a_{i}b_{j}\in IJ[/imath] is mapped to by [imath]\sum_{i=1}^{n}a_{i}\otimes b_{j}.[/imath] To prove that [imath]h[/imath] is injective, we consider its kernel. Suppose [imath]x=\sum_{i=1}^{n}a_{i}\otimes b_{j}\in I\otimes_{R}J[/imath] is such that [imath]h(x)=0.[/imath] Letting [imath]y=\sum_{i=1}^{n}a_{i}b_{j}\in IJ,[/imath] we see that [imath]x = \sum_{i=1}^{n}(a_{i}b_{j}\otimes1) = \left(\sum_{i=1}^{n}a_{i}b_{j}\right)\otimes1 = y\otimes 1.[/imath] Therefore [imath]h(y\otimes1) = 0.[/imath] But by the defining property of [imath]h[/imath] we have [imath]h(y\otimes 1)=y\cdot1=y,[/imath] whence [imath]y = h(x) =0.[/imath] Therefore we must have had [imath]x=0\otimes 1=0[/imath] all along, so the kernel of [imath]h[/imath] is trivial and this completes the proof. However, the above argument does not work, as pointed out to my by @xyzzyz, since we don't know that [imath]1\in J[/imath] (and indeed this is true if and only if [imath]J=R,[/imath] which changes the nature of the question).
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410396
|
My proof of [imath]I \otimes N \cong IN[/imath] is clearly wrong, but where have I gone wrong?
Ok, I'm reading some thesis of some former students, and come up with this proof, but it doesn't really look good to me. So I guess it should be wrong somewhere. So, here it goes: Let [imath]R[/imath] be a unitary commutative ring, and [imath]I[/imath] be an ideal, and [imath]N[/imath] is an [imath]R[/imath]-module. I'll now show that [imath]I \otimes_R N \cong IN\,.[/imath] First, since [imath]1 \in R[/imath], and [imath]I[/imath] is an ideal of [imath]R[/imath], we must have that [imath]I = RI[/imath]. So then [imath]I \otimes_R N = RI \otimes_R N = R \otimes_R IN \cong IN\,.[/imath] But this must be so wrong, since if this proof is true, I could prove that any [imath]R[/imath]-module is flat. So how come this proof is wrong?
|
2042787
|
Fourier coefficients of [imath]\chi_E[/imath] assuming [imath]m(E)=1.[/imath]
Let [imath]E[/imath] be a Lebesgue measurable set in [imath]\mathbb{R}[/imath] having measure [imath]1[/imath]. Prove that there is an integer [imath]n\neq0[/imath] such that [imath]\int_Ee^{inx}dx\neq0[/imath]. This is an exercise problem. A hint is given: periodize with period [imath]2\pi[/imath]. However, I still do not have an idea how to prove it. Any help will be appreciated.
|
2044513
|
How do I show that the integral of [imath]e^{inx}[/imath] over a set of measure [imath]1[/imath] is nonzero for some nonzero [imath]n[/imath]?
I'm trying to solve a problem from an old qualifying exam that asks to show that [imath]\displaystyle\int_{E}e^{inx}dx[/imath] is nonzero for some integer [imath]n\neq 0[/imath]. Here [imath]E[/imath] is assumed to be a measurable set of measure 1. How do I approach this problem? Do I need to use some result from Fourier analysis or is this a problem from Lebesgue integration theory. So far, the results I know of don't seem to work at all.
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2044629
|
How to prove that the limit of a function is [imath]0[/imath] at every point
I have the following function defined on [imath]\mathbb{R}[/imath]: [imath]f(x) = \begin{cases} 0 & \text{if $x$ irrational} \\ 1/n & \text{if $x = m/n$ where $m, n$ coprime} \end{cases}[/imath] I want to show that [imath]f[/imath] is continuous at every irrational point, and has a simple discontinuity at every rational point. I was able to show the first and partially the second (I showed that [imath]f[/imath] has a discontinuity at every rational point, but got stuck on showing that the discontinuity is simple). However, I've realized that I can simply show that [imath]\lim_{t \rightarrow x}f(t) = 0[/imath] for every [imath]x[/imath], and both of the things I want to show follow from this. How can I show this?
|
1547668
|
Prove continuity/discontinuity of the Popcorn Function (Thomae's Function).
I have to prove that a function [imath]f:]0,1] \rightarrow \Bbb R[/imath] : [imath] f(x) = \begin{cases} \frac1q, & \text{if $x \in \Bbb Q$ with $ x=\frac{p}q$ for $p,q \in \Bbb N$ coprime} \\ 0, & \text{if $x \notin \Bbb Q $} \end{cases} [/imath] is discontinuous in every point [imath]x \in \ ]0,1] \cap\Bbb Q[/imath]. And then to consider [imath]x \in \ ]0,1] \backslash \Bbb Q[/imath] and prove that it is continuous. For now I learned different ways to prove continuity (epsilon-delta, sequences), but I'm never sure what would be better to use in each different case. I wanted to prove the discontinuity by using sequences: [imath]\forall x_n \quad x_n\rightarrow a \quad \Rightarrow \quad f(x_n) \rightarrow f(a)[/imath] I tried creating a sequence [imath] x_n=\frac1n + a[/imath], we know it converges to [imath]a[/imath] but [imath]f(x_n)\ [/imath] doesn't converges to [imath]\ f(a)[/imath] because there would still be some points not in our set (irrational numbers that creates gaps). But I don't think it works, so I'm asking you if you could help me solving the two questions.
|
2045048
|
PDF of X is [imath]f_X(u)=c(1−u^2)[/imath] for some suitable constant c. Find c & PDF of Y where [imath]Y=X^2[/imath].
Let [imath]X[/imath] be a continuous random variable defined on the interval [imath][0,1][/imath] with density function [imath]f_X(u)=c(1−u^2)[/imath] for some suitable constant [imath]c[/imath]. Find: (a) the value of [imath]c[/imath], (b) (i) the expectation and (ii) the variance of [imath]X[/imath]. Also find (c) the density function of the random variable [imath]Y = X^{2}[/imath]. Found [imath]c=\frac{3}{2}[/imath], [imath]E(X)=\frac{1}{4}[/imath] & [imath]E(X^2)=\frac{1}{8}[/imath], so: [imath]Var(X)=\frac{1}{8}-(\frac{1}{4})^2=\frac{1}{16}[/imath] Have I done parts (a) & (b) correctly? If so how do I go about finding the PDF for Y and if not same question but also where did I go wrong? Thank you.
|
2044830
|
Moments of random variable [imath]X[/imath] with PDF [imath]f_X(x)=c\cdot(1 - x^2)[/imath] on [imath](0,1)[/imath], and PDF of [imath]X^2[/imath]
Let [imath]X[/imath] be a continuous random variable defined on the interval [imath][0,1][/imath] with density function [imath]f_X(u)=c(1−u^2)[/imath] for some suitable constant [imath]c[/imath]. Find: (a) the value of [imath]c[/imath], (b) (i) the expectation and (ii) the variance of [imath]X[/imath]. Also find (c) the density function of the random variable [imath]Y = X^{2}[/imath]. Now I've already attempted (a) by integrating [imath](1-u^{2})[/imath] between [imath]0[/imath] and [imath]1[/imath] (like [imath]\pi[/imath] in the Cauchy density function) and got [imath]c=\frac{2}{3}[/imath] and also for (b) (i) I integrated [imath]uc(1-u^{2})[/imath] between [imath]0[/imath] and [imath]1[/imath] which came out to be [imath]c/4 = 1/6[/imath]. Have I done these first [imath]2[/imath] parts correctly? If so how do I go about doing the rest and if not same question but also where did I go wrong?
|
2045417
|
Group with an interesting property
Let [imath] (G,\cdot ) [/imath] be a group and let [imath] a [/imath] be the only element of the group which has order 2. Prove that [imath] a\cdot x=x\cdot a,\forall x\in G [/imath] .
|
821309
|
Prove that if a group contains exactly one element of order 2, then that element is in the center of the group.
I'm stuck at this question. Can someone please help me? Prove that if a group contains exactly one element of order 2, then that element is in the center of the group. Let [imath]x[/imath] be the element of [imath]G[/imath] which has order 2. Let [imath]y[/imath] be an arbitrary element of [imath]G[/imath]. We have to prove that [imath]x \cdot y = y \cdot x[/imath]. Since [imath]x[/imath] has order [imath]2[/imath], \begin{equation} x^2 = e \end{equation} That is, \begin{equation} x^{-1}=x \end{equation} I don't really know how to proceed. I've tried a number of things, but none of them seem to work.
|
2045573
|
Prove there is no "slowest" converging series
I want to show that for any convergent series [imath]\sum_{n=0}^{\infty}a_n[/imath] with [imath]a_n>0[/imath] there exists a sequence [imath](b_n)[/imath], [imath]\lim\limits_{n \to \infty} b_n = \infty[/imath], such that [imath]\sum_{n=0}^{\infty}a_nb_n[/imath] converges. Any hints?
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1060590
|
Can a sequence which decays more slowly still yield a converging series?
In Bergman's companion notes to Rudin, he says that "If a sequence of positive terms has convergent sum, so does every sequence of positive terms which decays more rapidly." So given a sequence [imath]\{a_n\}[/imath] of positive terms such that [imath]\sum_n a_n[/imath] converges, if [imath]\{b_n\}[/imath] is such that [imath] \lim_{n\to\infty} \frac{a_n}{b_n} = +\infty, [/imath] then [imath]\sum_nb_n[/imath] converges. I can prove this given [imath]\{a_n\}[/imath] and [imath]\{b_n\}[/imath]. However, can we find [imath]b_n[/imath] which decays more slowly, i.e. [imath] \lim_{n\to\infty} \frac{b_n}{a_n} = +\infty [/imath] such that [imath]\sum_n b_n[/imath] converges? Similarly, we have the claim "if a sequence of positive terms has divergent sum, then so does every sequence of positive terms which decays more slowly."
|
2045448
|
How to determine the convergence of [imath]\sum_{n\geq 2}{} \frac{(-1)^{n}}{(-1)^n+n}[/imath]?
This is most likely very easy to show, but with the load of midterms I had, my brain just declines to work properly. How do I determine the convergence of [imath]\sum_{n\geq 2}{} \frac{(-1)^{n}}{(-1)^n+n}[/imath]?
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843093
|
Does this series with alternating elements converge: [imath]\sum_{k=10}^{+\infty}\frac{(-1)^k}{k+(-1)^k}[/imath]?
I have to investigate convergence of series [imath]\sum_{k=10}^{+\infty}\frac{(-1)^k}{k+(-1)^k}[/imath] It certainly does not converge absolutely, because it is basically a harmonic series with every two elements flipped and if harmonic series converged, or greater than series [imath]\sum_{k=10}^{+\infty}\frac{1}{k-1}[/imath] However, I do not know, how to prove convergence of this series overall. I thought about the fact, that it is basically a harmonic series(well, alternating harmonic or whatever I should call it) with every two elements flipped and if normal "alternating harmonic series" converges, then this rearangement could technically converge too. However, I am not sure if this is sufficient enough and if I can do this and I am afraid that it is not. I was considering those theorems that say "every divergent series with alternating elements can be rearranged to make arbitrary finite sum" and I am not sure if it holds with convergent series, respectively, if it holds against convergent series and infinity - that you can rearrange convergent series to make divergent... Neverthless, there should be an easy way to prove that this sum is convergent/divergent, since it is in our textbook. Also I thought, since it is certainly lesser or equal to [imath]1/(k-1)[/imath] and greater or equal to [imath]1/(k+1)[/imath] assigning numbers [imath]S_1[/imath] and [imath]S_2[/imath] to sums of these "alternating harmonics" and just saying that it is lower bounded by [imath]S_2[/imath] and upper bounded by [imath]S_1[/imath], since those series are convergent and have to be upper and lower-bounded too. But also, I am not sure if this is enough.
|
2044510
|
How do I compute the limit of this integral: [imath]\lim_{n\to\infty}\int_{0}^{\infty}\frac{\sin(\frac{x}{n})}{(1+\frac{x}{n})^{n}}dx[/imath]?
I'm trying to compute the limit of the integral [imath]\displaystyle\int_{0}^{\infty}\frac{\sin(\frac{x}{n})}{(1+\frac{x}{n})^{n}}dx[/imath] as [imath]n[/imath] goes to infinity. I tried and realized that this is probably not an elementary limit computation. I'm completely lost as to what result to use. Convergence theorems of Lebesgue integrals didn't get me far. Is this a problem of Fourier Analysis? I'd like to know of a way to do this.
|
148529
|
Computing [imath]\lim\limits_{n\to\infty} \int_{0}^{\infty}\frac{\sin(x/n)}{(1+x/n)^{n}}\, dx[/imath]
[imath]\lim_{n\to\infty} \int_{0}^{\infty}\frac{\sin(x/n)}{(1+x/n)^{n}}\, dx[/imath] I've been able to show that the integral is bounded above by 1 (several ways). One of the simplest is just letting [imath]u=x/n[/imath] and doing some bounding above. It must converge to something (if it is monotone increasing.. don't think so), but all my efforts to use Lebesgue Dominated Convergence have failed. I can't seem to dominate this function... However, if I could dominate it the integrand converges pointwise to 0 and so then the integral would be 0, which I think is incorrect. I have found it sufficient to deal with [imath]\lim_{n\to\infty}\int_{0}^{n}\frac{\sin(x/n)}{(1+x/n)^{n}}\, dx[/imath] I tried to turn it into a series and use the converse of the integral test, but that hasn't went well. I've used Egoroff's theorem to try to use uniform convergence but that failed. Some of the inequalities and ideas I've tried to use in a solution include: [imath]sin(x/n)\leq x/n[/imath] after a [imath]u[/imath] sub I've tried to turn it into a riemann sum... failed [imath](1+x/n)^{n}\geq 1+x[/imath] and the binomial theorem [imath](1+x/n)^{n}[/imath] converges to [imath]e^x[/imath] and it is increasing in [imath]n[/imath] so [imath]e^{-x}\leq \frac{1}{(1+x/n)^{n}}[/imath] I've tried to use that the integrand is measurable and subtract out different measurable sets from the integral to make it more convenient.. I've been playing with this on and off for a month or so and no luck. I appreciate any help here. Update: From mixedmath's suggestion splitting it into [imath][0,10][/imath] and [imath][10,\infty][/imath] it is clear that the integrand is dominated on [imath][0,10][/imath] (say by 1) and so lebesgue sends that portion to 0. On the larger interval I am still running into a problem making it bounded nicely. If I try a [imath]u[/imath] sub the bounds get messy. I would like to choose an [imath]n[/imath] so that [imath]2e^{-x} \leq (1+x/n)^n[/imath] for all [imath]x\in [10,\infty][/imath] .. maybe: there exists an [imath]N[/imath] such that for all [imath]n\geq N[/imath] [imath]e^{10}/2 < (1+10/n)^{n}[/imath] but for [imath]x>10[/imath].. [imath]e^{10}/2<(1+10/n)^{n}<(1+x/n)^{n}[/imath] but that's not quite right..
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2045061
|
How to solve this infinite radical
[imath]\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4+\sqrt{5+\sqrt{ \dots }}}}}}[/imath] I don't understand how to solve that. I mean I don't know where to begin. Tell me if this infinite radical has a solution or converge to a number. Thanks.
|
201193
|
Sum and Product of Infinite Radicals
How to compute [imath]\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}[/imath] & [imath]\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}[/imath]? I understand that [imath]\sqrt{(1\sqrt{(2\sqrt{(3\dots)})})}=(1^{1/2})(2^{1/4})(3^{1/8})\cdots[/imath] and [imath]\sqrt{(1+\sqrt{(2+\sqrt{(3+\cdots)})})}=(1+(2+(3+(\cdots))^{1/8})^{1/4})^{1/2} [/imath] How to Proceed further?
|
2045991
|
Prove that a set of nonzero real numbers is a group under the operation [imath]a*b[/imath]
prove that the set of nonzero numbers is a group under the operation [imath]*[/imath] defined by [imath]a*b = \{ab \text{ if } a > 0 \text{ or } \frac{a}{b} \text{ if } a < 0\}[/imath]. Show that it proves closure, identity, inverses and associative properties of a group.
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658168
|
Prove that [imath]a * b[/imath] is a group (Hungerford, 2nd ed., sec [imath]7.2[/imath], Exercise [imath]22[/imath])
Prove that the set of nonzero real numbers is a group under the operation [imath]*[/imath] defined by \begin{align} a*b = \begin{cases} ab &\mbox{if } a > 0 \\ \frac{a}{b} &\mbox{if } a < 0 \end{cases} \end{align} I have trouble proving the associativity propery of a group here. Here is my work so far: If [imath]a > 0, b > 0[/imath], then [imath](a*b)*c = (ab)*c = (ab)c[/imath] and [imath]a*(b*c) = a*(bc) = a(bc)[/imath]. If [imath]a > 0, b < 0[/imath], then [imath](a*b)*c = (a/b)*c = (a/b)/c = a/(bc)[/imath] and [imath]a*(b*c) = a*(b/c) = a/(b/c) = ac/b[/imath]. If [imath]a < 0, b > 0[/imath], then [imath](a*b)*c = (a/b)*c = (a/b)c = ac/b[/imath] and [imath]a*(b*c) = a*(bc) = a/(bc)[/imath]. If [imath]a < 0, b < 0[/imath], then [imath](a*b)*c = (a/b)*c = (a/b)/c = a/(bc)[/imath] and [imath]a*(b*c) =a*(b/c) = a/(b/c) = ac/b[/imath]. I had been unable to prove that [imath](a*b)*c = a*(b*c)[/imath] for all cases but the first one. I cannot understand why...
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2046631
|
[imath]\nabla U=0 \implies U=\mathrm{constant}[/imath] with [imath]A[/imath] connected and [imath]U[/imath] continuous?
I'm looking for suggestions on how to prove that Given a function [imath]U: A \subset \mathbb{R}^n \to \mathbb{R}[/imath] with [imath]U \in C^1 [/imath] (therefore continuous) and [imath]A[/imath] a connected set, [imath]\nabla U=\bar{0} \,\,\,\,\,\,\, \forall \bar{x} \in A\implies U=\mathrm{constant} \,\,\,\,\,\,\, \forall \bar{x} \in A[/imath] In particular, how is the condition of [imath]A[/imath] as a connected necessary?
|
2045698
|
[imath]\nabla U=0 \implies U=\mathrm{constant}[/imath] only if [imath]U[/imath] is defined on a connected set?
How can I prove the following statement? Given a function [imath]U: A \subset \mathbb{R}^n \to \mathbb{R}[/imath] with [imath]A[/imath] connected set, [imath]\nabla U=\bar{0} \,\,\,\,\,\,\, \forall \bar{x} \in A\implies U=\mathrm{constant} \,\,\,\,\,\,\, \forall \bar{x} \in A[/imath] In particular I do not understand how the condition of [imath]A[/imath] as a connected set is necessary to prove the statement.
|
2046338
|
[imath]{2^k-1\choose a}[/imath] is odd?
How can I see that [imath]{2^k-1\choose a}\equiv 1\mod 2[/imath] for [imath]0<a<2^k[/imath]? I tried playing around with [imath]{2^k-1\choose a}=\frac{(2^k-1)!}{a!(2^k-1-a)!}[/imath] but it didn't get me very far.
|
317163
|
Prove: If [imath]n=2^k-1[/imath], then [imath]\binom{n}{i}[/imath] is odd for [imath]0\leq i\leq n[/imath]
Kinda stuck on this one. Help is appreciated. I'm going for either a direct or contrapositive proof. Prove: If [imath]n=2^k-1[/imath], for [imath]k\in\mathbb{N}[/imath], then every entry in Row [imath]n[/imath] of Pascal's Triangle is odd. I've been considering entry [imath]i[/imath] in row [imath]n[/imath] of Pascal's Triangle, so for [imath]0\leq i\leq n[/imath], we have: [imath] \binom{n}{i} =\frac{n!}{i!(n-i)!} =\frac{(2^k-1)!}{i!(2^k-1-i)!} [/imath] I've tried manipulating this in a bunch of ways, including using the fact that [imath]\binom{n}{i}=\binom{n-1}{i-1}+\binom{n-1}{i}[/imath], but nothing's panned out.
|
2046106
|
Prove this Equation is Divisible By...
Prove that among any three distinct integers we can find two, say [imath]a[/imath] and [imath]b[/imath], such that the number [imath]a^3b − ab^3[/imath] is divisible by [imath]10[/imath]. Can anyone help me solve this?
|
1972373
|
Prove [imath]30|(a^3b-ab^3) [/imath]
Prove that if three distinct integers are chosen at random then there will exists two among them, say [imath]a[/imath] and [imath]b[/imath] such that [imath]30|(a^3 b-ab^3)[/imath]
|
258704
|
sequence of differentiable functions
[imath]f_n[/imath] be a sequence of differentiable functions on [imath][a,b][/imath] such that [imath]f_n(x)\rightarrow f(x)[/imath] which is riemann integrable consider the statements [imath]f_n[/imath] converges uniformly. [imath]f_n^{'}[/imath] converges uniformly. [imath]\int_{a}^{b}f_n\rightarrow \int_{a}^{b}f[/imath] [imath]f[/imath] is differentiable. we need to find out which of the statement is not true.
|
153973
|
Sequence of differentiable functions
Let [imath]f_n[/imath] be a sequence of differentaible functions on [imath][0,1][/imath] to [imath]\mathbb{R}[/imath] converging uniformly to a function [imath]f[/imath] on [imath][0,1][/imath], Then [imath]f[/imath] is differentiable and Riemann integrable there [imath]f[/imath] is uniformly continuous and R-integrable [imath]f[/imath] is continuous, need not be differentiable on [imath](0,1)[/imath] and need not be R-integrable on [imath] [0,1][/imath], [imath]f[/imath] need not be continuous. Well I think 2 is only correct statement as every [imath]f_n[/imath] are differentiable on [imath][0,1][/imath] hence uniformly contnous, and as they converge uniformly to [imath]f[/imath] on [imath][0,1][/imath] so [imath]f[/imath] is uniformly continuous and every continuous function is rieman integrable..Am I correct?
|
1049375
|
Arithmetics of cardinalities: if [imath]|A|=|C|[/imath] and [imath]|B|=|D|[/imath] then [imath]|A\times B|=|D\times C|[/imath]
Suppose that [imath]A, B, C[/imath], and [imath]D[/imath] are sets with the cardinalities related as [imath]|A|=|C|[/imath] and [imath]|B|=|D|[/imath]. Prove that the cardinality of [imath]A\times B[/imath] is equal to the cardinality of [imath]D\times C[/imath]. I know that I must prove the bijection of the function [imath]f: A \times B \to C \times D[/imath], but I am not sure how to say this? Can anyone help me?
|
847443
|
Is it true , if [imath]|A|=|B|[/imath] and [imath]|C|=|D|[/imath], then [imath]|A \times C| = |B \times D|[/imath]?
Check my proof, please. Divide into subsets [imath]A \times C[/imath] and [imath]B \times D[/imath] so that , all pairs with the same element belong to the same subset. Each such subset [imath]|A \times C|[/imath] bijective [imath]C[/imath], [imath]|C|=|D|[/imath] and [imath]D[/imath] equipotent to each subset [imath]B \times D[/imath]. This implies , that [imath]|A \times C| = |B \times D|[/imath].
|
2046618
|
Consider the rationals [imath]\mathbb{Q}[/imath] viewed as group under addition and let [imath]G=\mathbb{Q}/\mathbb{Z}[/imath]
Consider the rationals [imath]\mathbb{Q}[/imath] viewed as group under addition and let [imath]G=\mathbb{Q}/\mathbb{Z}[/imath] where [imath] \mathbb{Z} \;\triangleleft \; \mathbb{Q}[/imath] A) What is order of [imath]\frac{3}{5}[/imath] B) Show that G has a cyclic subgroup of order n for each [imath]n>1[/imath] Please help to solve this problem i really stuck
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189335
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consider the group [imath]G=\mathbb Q/\mathbb Z[/imath]. For [imath]n>0[/imath], is there a cyclic subgroup of order n
consider the group [imath]G=\mathbb Q/\mathbb Z[/imath]. Let [imath]n[/imath] be a positive integer. Then is there a cyclic subgroup of order [imath]n[/imath]? not necessarily. yes, a unique one. yes, but not necessarily a unique one. never
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2047956
|
Expected Value Inequality - bounded by sums of probabilitys
I have to show the following inequality: [imath] \sum_{k=0}^{\infty} P[Y>k] \leq E[Y] \leq \sum_{k=1}^{\infty} P[Y>k][/imath] under the condition that Y is a non negative continuous random Variable. My first idea was to rewrite the inequality using integrals, but then i end up having to integrate over a sum, which is nothing im particulary excited about. Any Ideas on how i can do this?
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1795529
|
Why is [imath]\mathbb{E}[X] = 1 + \sum^\infty_{k=1}\mathbb{P}(X > k)[/imath] true?
I'm working through a problem regarding expected values in Markov chains, and at some point it says: Recall from probability that if [imath]X[/imath] is a positive integer valued random variable, then [imath]\mathbb{E}[X] = 1 + \sum^\infty_{k=1}\mathbb{P}(X > k)[/imath]. I know that by definition [imath]\mathbb{E}[X] = \sum_a a\mathbb{P}(X = a)[/imath], but I can't see how the above equality follows from this, nor am I sure if this is how to approach the problem.
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2048089
|
[imath]f:\Bbb R\to \Bbb R[/imath] be a continuous function such that [imath]f(i)=0\forall i\in \Bbb Z[/imath]
Let [imath]f:\Bbb R\to \Bbb R[/imath] be a continuous function such that [imath]f(i)=0\forall i\in \Bbb Z[/imath]. Which of the following is always true: [imath]Image(f)[/imath] is closed in [imath]\Bbb R[/imath]. [imath]Image(f)[/imath] is open in [imath]\Bbb R[/imath]. [imath]f[/imath] is uniformly continuous. My try: 1.I am unable to find a example here. False take [imath]f(x)=0[/imath]. 3.I am unable to find a example here. Please give some hints here.
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2039289
|
What can you say about a continuous function that is zero at all integer values?
Written with StackEdit. Let [imath]f: \Bbb R \to \Bbb R [/imath] be a continuous function such that [imath]f(i) = 0 \ for \ all \ i \in \Bbb Z[/imath]. Then which of the following is true : - A. Image(f) is closed in [imath]\Bbb R[/imath] B. Image(f) is open in [imath]\Bbb R[/imath] C. f is uniformly continuous D. None of the above For the correct option, look at the end of the question. I tried to contradict the options by bringing examples(It was obviously the first step considering 'None of the above was' an option) I wasn't able to think of a single function that contradicts either of the options. However, graphically I would say that these functions satisfy the hypothesis and contradict the options (Image uploaded) Here's what the functions are 1) Just like [imath]sin x,[/imath] the function touches 1 and -1. The major difference would be that this function touches the x-axis at all integral values and no where else. 2) The second graph touches the integers and zero and in each interval [imath](m,m+1)[/imath], the graph's maximum and minimum value tend to increase slightly and they tend to +1 and -1 as [imath]x \to \infty[/imath] Thus, the function gets arbitrarily close to 1 and -1 without touching them. 3) The third graph is just like [imath]x Sin(x)[/imath]. The difference is that this one touches the x-axis at integer values. With each interval [imath](m,m+1)[/imath], our graph has its 'magnitude of difference' increasing. Are my graphs correct? If they, can they explicitly be expressed as familiar functions? If not, which explicitly express-able examples can be used to contradict the first three options? The correct answer is D Source - Tata Institute of Fundamental Research Graduate Studies Exam 2016
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2048096
|
Proving this inequality that involves the Trace of matrices
Let [imath]A, B \in M_n(\mathbb{C})[/imath] be Hermitian. Show that [imath]\text{tr} (AB)^2 \leq \text{tr} (A^2B^2[/imath]). What we can say is that [imath]AB - BA[/imath] is skew-hermitian since [imath]BA = (AB)^*[/imath]. Since it is Skew-hermitian, its eigenvalues are all imaginary. And its square is similar to a diagonal matrix whose diagonal entries are all negative. And so [imath]\text{tr}(AB-BA)^2 \leq 0[/imath]. How do you proceed from here?
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1303910
|
[imath]A,B[/imath] be Hermitian.Is this true that [imath]tr[(AB)^2]\le tr(A^2B^2)[/imath]?
Suppose [imath]A,B \in {M_n}[/imath] be Hermitian.Is this true that [imath]tr[(AB)^2]\le tr(A^2B^2)[/imath]?
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2047919
|
In epsilon delta method, is this possible?
When we use the epsilon delta method, we use the Strictly inequality as follows: function [imath]f[/imath] is continous at [imath]x=a[/imath] if and only if For arbitrary [imath]\epsilon[/imath], there exist a [imath]\delta[/imath] such that If [imath] | x-a | < \delta[/imath] then [imath] | f(x) - f(a) | < \epsilon[/imath] Is it possible to change the strictly inequality([imath]<[/imath] to inequality([imath]\leq[/imath]) in the above expression? If possible is this equivalent?
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643680
|
Can [imath]\le[/imath] be used insted of < in the definition of continuity?
A common definition of a continuous map [imath]T:M_1\to M_2[/imath] is that for every [imath]x\in M_1[/imath] and every [imath]\epsilon>0[/imath] there exists a [imath]\delta >0[/imath] such that for all [imath]y[/imath] in [imath]M_1[/imath] [imath]d_1(x,y)<\delta \implies d_2(T(x),T(y))<\epsilon,[/imath] i.e. we use a strict inequality. Now in a proof it reads that [imath]T[/imath] is continuous if for every [imath]x\in M_1[/imath] and every [imath]\epsilon>0[/imath] there is a [imath]\delta>0[/imath] such that [imath]\|Tx-Ty\|\le\epsilon\ \mbox{for all $y$ in $M_1$ satisfying } \|x-y\|\le \delta.[/imath] Here the norm is given by the metric. Is it correct to use [imath]\le[/imath] instead of the strict inequality <, and can one somehow prove the equality of these definitions? Or is this obviously the same condition? Thanks in advance!
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1277915
|
Example of Markov process not strong Markov
This is again a question about this example (also see here). It seems we can write this process as [imath]X_t := \bigl(t - \text{Exp}(1) \mathbf{1}_{\{X_0 = 0\}}\bigr)^+ + \mathbf{1}_{\{X_0 \neq 0 \}}X_0\, .[/imath] Where [imath]\text{Exp}(\lambda)[/imath] is the exponential distribution (CDF: [imath]1-e^{-\lambda x}[/imath]) with parameter [imath]\lambda[/imath]. Define the stopping time [imath]\tau := \inf \{ t\geq 0 : X_t > 0\}[/imath]. I've read the following explanation why this is not strong Markov: [imath]\mathbf{E}_0 \bigl[f((X_{\tau + t}))|\mathcal{F}_\tau\bigr] = f(t) \neq \int f(y) \kappa_t(0, dy) = e^{-t} f(0) + \int_0^t e^{-s} f(t-s)\, ds\, ,[/imath] but that makes no sense to me... how can it be [imath]f(t)[/imath]? Because [imath]t \in I[/imath] doesn't fit into a function [imath]f\colon E^{\otimes I}\rightarrow \mathbb{R}[/imath] (see here for the definition of strong Markov property. For this example [imath]E = [0, \infty)[/imath])...
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246892
|
Why is the following example of a Markov process not strong Markov
[imath]X(t) := 0 \;\; (t \leq \tau),\;\; t - \tau\;\;(t \geq \tau)[/imath] with [imath]\tau[/imath] exponentially distributed. Then X has the Markov property but not Strong Markov Property. But why ???? Can someone kindly explain in words and maths why the strong markov property does not hold?
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1724246
|
Find last two digits of [imath]33^{100} [/imath]
Find last two digits of [imath]33^{100}[/imath]. My try: So I have to compute [imath]33^{100}\mod 100[/imath] Now by Euler's Function [imath]a^{\phi(n)}\equiv 1\pmod{n}[/imath] So we have [imath]33^{40}\equiv 1 \pmod{100}[/imath] Again by Carmichael Function : [imath]33^{20}\equiv 1 \pmod{100}[/imath] Since [imath]100=2\cdot40+20[/imath] so we have [imath]33^{100}=1\pmod{100}[/imath] So last two digits are [imath]01[/imath] Is it right?
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1342127
|
Find the last two digits of [imath]33^{100}[/imath]
Find the last two digits of [imath]33^{100}[/imath] By Euler's theorem, since [imath]\gcd(33, 100)=1[/imath], then [imath]33^{\phi(100)}\equiv 1 \pmod{100}[/imath]. But [imath]\phi(100)=\phi(5^2\times2^2)=40.[/imath] So [imath]33^{40}\equiv 1 \pmod{100}[/imath] Then how to proceed? With the suggestion of @Lucian: [imath]33^2\equiv-11 \pmod{100}[/imath] then [imath]33^{100}\equiv(-11)^{50}\pmod{100}\equiv (10+1)^{50}\pmod{100}[/imath] By using the binomial expansion, we have: [imath]33^{100}\equiv (10^{50}+50\cdot 10^{49}+ \cdots + 50\cdot 10+1)\pmod{100}[/imath] [imath]\implies 33^{100}\equiv (50\cdot 10+1)\pmod{100}\equiv 01 \pmod{100}[/imath]
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2047901
|
Prime ideals in [imath]F[x][/imath] when [imath]F[/imath] is a field
Let [imath]F[/imath] be a field. Prove that the polynomial ring [imath]F[x][/imath] in one variable over [imath]F[/imath] has infinitely many prime ideals. All I can think of if that if we can show there are infinitely many irreducible polynomials in [imath]F[x][/imath] but I see no reason why that should hold, I have no information about the field [imath]F[/imath].
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80389
|
Ring of polynomials over a field has infinitely many primes
Let [imath]F[/imath] be a field. Why does [imath]F[x][/imath] have infinitely many irreducible elements? For the case F has characteristic 0 Then x-a is irreducible for all a [imath]\in F[/imath] since x satisfies no non-trivial relations in F. Obviously this argument fails for a finite field since there are only finitely many a to choose from. So how may I construct irreducible polynomials in a finite field? I figure it must involve higher powers of x, maybe [imath]x^n-a[/imath] ?
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2048837
|
Counter example to show that [imath] s(X) \le d(X) [/imath]
I am trying to show that there exists a topology in which [imath] s(X) \le d(X) [/imath]. I am working with the co-countable topology with an infinite topological space X. I understand the [imath]d(X)[/imath] is uncountable and the [imath] s(X) [/imath] is countable. For the density, I understand that I have to show that no countable set is dense. For the spread, I know a little less. I know that I have to take an uncountable subset of X and show the spread is countable in the co-countable topology. I understand these concepts however I am having trouble writing them on paper. Any hints and direction would be appreciated!
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2043797
|
Is [imath] d(X) \le s(X)? [/imath]
Is the density of X always less than or equal to the spread of X? If not, can someone help me find a counter example? [imath]s(X) = sup ( \vert A \vert : A \textrm { is a discrete subset of } X ) [/imath] [imath]d(X) = \min ({ \vert D\vert : \overline{D} = X})[/imath]
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2044643
|
Does this have a closed form [imath]\lim_{n\to \infty}\left(\prod_{r=0}^{n/2} \left(\frac{mn+1-mr}{mn+3-mr}\right)\right)[/imath]?
[imath]\lim_{n \to \infty}\left(\prod_{r=0}^{n/2}{\left(\frac{mn+1-mr}{mn+3-mr}\right)}\right)[/imath] Of course if we check out cases where it should telescope like [imath] m = 1, 2, \frac{2}{3}, \cdots[/imath], we get closed forms - [imath]\frac{1}{4}, \frac{1}{2}, \frac{1}{8}, \cdots[/imath]. So, I was thinking if it has a closed form. One can reduce it to [imath]\displaystyle \lim_{n \to \infty} \dfrac{\prod_{r=0}^n{\frac{r + p/m}{r+q/m}}}{\prod_{r=0}^{n/2}{\frac{r + p/m}{r+q/m}}}[/imath], where [imath]p,q[/imath] are [imath]1,3[/imath] in the above problem [imath]\displaystyle = \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{2} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{2} + 1\right)}}[/imath] Now, how to find this limit? Wolfram Alpha now can find the limit quite fast for example for [imath](p,q) = (1,3)[/imath], and [imath]m=3[/imath] it gives the limit [imath]\frac{\sqrt[3]{2}}{2}[/imath].
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2047875
|
Interesting limit involving gamma function
[imath]\displaystyle \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{2} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{2} + 1\right)}}[/imath] This is from here. If we add one more variable - [imath]\displaystyle L(p,q,m,t) = \lim_{n \to \infty}{\dfrac{\Gamma\left(\frac{p}{m}+n+1\right) \Gamma\left(\frac{q}{m} + \frac{n}{t} + 1\right)}{\Gamma\left(\frac{q}{m}+n+1\right) \Gamma\left(\frac{p}{m} + \frac{n}{t} + 1\right)}}[/imath] Then, the results I got from W|A are really interesting. They seem to follow the pattern which says [imath]\displaystyle L(p,q,m,t) = \dfrac{1}{t^{|p-q|/m}}[/imath] If it is indeed true, how to prove it? Factorial approximations did not lead me anywhere.
|
2048726
|
How can I use the shift property of the Fourier transform to calculate the Fourier transform of an impulse train?
As we all know: [imath]\mathcal F\{\delta(t)\} = 1[/imath] and: [imath]\mathcal F\{x(t-k)\}=X(f)e^{-i2\pi fk}[/imath] However when I try to use these properties to calculate the Fourier transform (FT) of an impulse train, [imath]x(t) = \sum_{n=-\infty}^{\infty}\delta(t-nT)[/imath], I get: [imath]\mathcal F\{x(t)\}=\mathcal F\{\sum_{n=-\infty}^{\infty}\delta(t-nT)\} = \sum_{n=-\infty}^{\infty}e^{-i2\pi nTf}[/imath] As far as I know, the FT of an impulse train should be another impulse train, but the RHS in the expression above doesn't look like it. What am I doing wrong? Thanks for your time!
|
1898432
|
The Fourier transform of a "comb function" is a comb function?
Let [imath]f(x) = \sum_{n=-\infty}^{\infty} \delta(x - n)[/imath], where [imath]\delta[/imath] is the Dirac delta function. This function [imath]f[/imath] (a "comb function") is important in signal processing because evenly sampling a function [imath]g[/imath] can be viewed as multiplying [imath]g[/imath] pointwise with [imath]f[/imath]. This idea is the first step towards understanding how to approximate the Fourier transform of [imath]g[/imath], given evenly spaced samples of [imath]g[/imath]. The next step is to note that [imath]\hat{gf} = \hat{g}*\hat{f}[/imath], where [imath]\hat{f}[/imath] is the Fourier transform of [imath]f[/imath] and [imath]*[/imath] denotes convolution. Is it true that [imath]\hat f[/imath] is also a comb function? If that statement isn't quite right, what is the correct statement? I would like to see: 1) an informal, intuitive, nonrigorous but easy derivation of the Fourier transform of [imath]f[/imath] . 2) A rigorous version of the same calculation. (How would you even define [imath]f[/imath] rigorously, is it a distribution?) (I'd also be interested in recommendations of math textbooks that cover this topic, including the Nyquist sampling theorem, even if it's only an exercise or series of exercises in an analysis textbook.)
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2049345
|
About a divergent series
If [imath]\sum_{i=1}^\infty a_i[/imath] diverges, then prove that [imath]\sum_{i=1}^\infty ia_i[/imath] diverges. My English is too bad to express the question clearly. Thank you for your answer.
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603297
|
Is there a sequence such that [imath]\sum{a_n}[/imath] diverges but [imath]\sum{na_n}[/imath]converges?
Is there a (real) sequence such that [imath]\sum{a_n}[/imath] diverges but [imath]\sum{na_n}[/imath]converges?
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2049675
|
[imath]f:[0,1]\rightarrow R[/imath] is continuous, [imath]f[/imath] is differentiable on [imath](0,1)[/imath] , [imath]f(0)=f(1)=0[/imath]. Then [imath]f(x)=f^\prime(x)[/imath] has soln in [imath](0,1)[/imath]
I know we can use Rolle's Theorem to get a point in [imath](0,1)[/imath] such that [imath]f^\prime(x)[/imath] is zero at that point but don't know how to take it from there.
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2021877
|
Rolle's theorem [imath]\beta \cdot f(x)+f'(x)=0[/imath]
Prove that if [imath]f[/imath] is differentiable on [imath][a,b][/imath] and if [imath]f(a)=f(b)=0[/imath] then for any real [imath]\beta[/imath] there is an [imath]x \in (a,b)[/imath] such that [imath]\beta \cdot f(x)+f'(x)=0[/imath]. (Using rolle's theorem) My attempt: Using Rolle's theorem we can say that there exists some [imath]c \in (a,b)[/imath], where [imath]f'(c)=0[/imath] . Therefore one factor in the expression [imath]\beta \cdot f(x)+f'(x)=0[/imath] is [imath]0[/imath] at [imath]c[/imath] but I am unable to prove that the other factor will simultaneously be [imath]0[/imath] at [imath]c[/imath].
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2048413
|
Conditional distribution of multivariate normal
[imath]X \sim N(\mu, \Sigma)[/imath]. How can I find conditional distribution of [imath]X[/imath] given [imath]AX[/imath], where [imath]A[/imath] is a non-random matrix? I know that [imath]AX \sim N(A\mu, A\Sigma A^T)[/imath] but don't know what to do next. P.S - I am preparing for an exam and solving problems from textbook, so should i add hashtag self-study?
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637694
|
Distribution of multivariate Gaussian conditional on value of linear function
Given a Gaussian random vector [imath]X \in \mathbf{R}^p \sim \mathcal{N}(\boldsymbol\mu, \boldsymbol\Sigma)[/imath], a matrix [imath]\mathbf{K} \in \mathbf{R}^{q \times p}[/imath], and a vector [imath]y \in \mathbf{R}^q[/imath], I'd like to find the distribution of [imath]\tilde X = (X \mid \boldsymbol{K} X = y)[/imath] while assuming that [imath]\operatorname{rank}(\boldsymbol\Sigma) = p[/imath] and [imath]\operatorname{rank}(\boldsymbol{K}) \le q < p[/imath]. Ideally without having to (psuedo-)invert any singular matrices.
|
2049566
|
Prove that [imath]x^7+x^5+x^3+1=0[/imath] has only one solution in [imath]\mathbb R[/imath]
Question : The equation [imath]x^7+x^5+x^3+1=0[/imath] is given. Prove that this equation has only one solution on [imath]\mathbb R[/imath]. Note 1 : You should use Rolle's theorem in your proof. That's what the question wants ! its not optional !!! Note 2 : The question is taken from "Real Analysis : A first course " : Russel Gordon Note 2 : I don't have any idea... I've never proved something like that ! Any help would be great !
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575504
|
Prove using Rolle's Theorem that an equation has exactly one real solution.
Prove that the equation [imath]x^7+x^5+x^3+1=0[/imath] has exactly one real solution. You should use Rolle’s Theorem at some point in the proof. Since [imath]f(x) = x^7+x^5+x^3+1[/imath] is a polynomial then it is continuous over all the real numbers, [imath](-\infty,\infty)[/imath]. Since [imath]f(x)[/imath] is continuous on [imath](-\infty,\infty)[/imath] then it is certainly continuous on [imath][-1,0][/imath] and the Intermediate Value Theorem (IVT) applies. The IVT states that since [imath]f(x)[/imath] is continuous on [imath][-1,0][/imath] we can let [imath]C[/imath] be any number between [imath]f(-1)=-2[/imath] and [imath]f(0)=1[/imath], namely [imath]C=0[/imath], then there exists a number [imath]c[/imath] with [imath]-1 < c < 0[/imath] such that [imath]f(c)=C=0[/imath]. Since [imath]f(x)[/imath] is continuous on [imath][-1,0][/imath] and differentiable on [imath](-1,0)[/imath] then Rolle’s Theorem applies. Rolle’s Theorem states that if a function [imath]f\colon [a,b] \to \mathbf{R}[/imath] is continuous on [imath][a,b][/imath] and differentiable on [imath](a,b)[/imath] then if [imath]f(a)=f(b)[/imath], there exists a point [imath]c \in (a,b)[/imath] such that [imath]f'(c)=0[/imath]. We assume that there is more than one real solution for this equation, namely [imath]f(a)=0=f(b)[/imath]. If there exists more than one real solution for [imath]f(x)=0[/imath] then [imath]f(a)=0=f(b)\implies a=b[/imath], and thus there is only one real solution to the equation, as desired. But I feel I am missing something, and I'm not sure how to show that the equation is differentiable on the interval and that Rolle's Theorem applies.
|
2050042
|
Does this limit... exist?
I found this interesting function defined by the following infinite series: [imath]g(x)=\sum_{n=0}^\infty(-1)^nx^{2^n}[/imath] And I noticed that it followed the functional equation [imath]g(x)=1-g(x^2)[/imath]. I was interested, particularly, in the following limit: [imath]L=\lim_{x\to1^-}g(x)[/imath] And assuming it exists, the functional equation yields [imath]L=1-L\implies L=\frac12[/imath] Which, at first look, the limit appears to exist. However, upon closer inspection, convergence gets a little sketchy... Can we determine if the limit converges?
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1080746
|
The limit of the alternating series [imath]x - x^2 + x^4 - x^8 + {x^{16}}-\dotsb[/imath] as [imath]x \to 1[/imath]
I want to calculate the limit of this sum : [imath]\lim\limits_{x \to 1} {\left(x - x^2 + x^4 - x^8 + {x^{16}}-\dotsb\right)}[/imath] My efforts to solve the problem are described in the self-answer below.
|
2050093
|
Proving a sequence is a cauchy
Let's suppose that [imath]a_n[/imath] satisfy [imath]|a_{n+1}-a_n|<2^{-n}[/imath] for all [imath]n[/imath] . How can I prove that [imath]a_n[/imath] is a cauchy sequence?
|
182830
|
Proving that a sequence such that [imath]|a_{n+1} - a_n| \le 2^{-n}[/imath] is Cauchy
Suppose the terms of the sequence of real numbers [imath]\{a_n\}[/imath] satisfy [imath]|a_{n+1} - a_n| \le 2^{-n}[/imath] for all [imath]n[/imath]. Prove that [imath]\{a_n\}[/imath] is Cauchy. My Work So by the definition of a Cauchy sequence, for all [imath]\varepsilon > 0[/imath] [imath]\exists N[/imath] so that for [imath]n,m \ge N[/imath] we have [imath]|a_m - a_n| \le \varepsilon[/imath]. However, questions like this one make me understand that the [imath]2^{-n}[/imath] condition is necessary for this to be a true statement. So I am wondering how to appeal to the Cauchy definition for this proof. Do I prove that every convergent sequence is therefore Cauchy, and then try to prove convergence?
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2050385
|
Help to solve [imath]15x \equiv 20 \pmod{88}[/imath]
Solve [imath]15x[/imath] "congruent to" [imath]20\mod 88[/imath] So far I think I know [imath]15\mod 88[/imath] is [imath]-41[/imath] or if positive [imath]47[/imath]`
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2043364
|
[imath]15x\equiv 20 \pmod{88}[/imath] Euclid's algorithm
Use Euclid's algorithm to find a multiplicative inverse of 15 mod 88, hence solve the linear congruence [imath]15x\equiv 20 \pmod{88}[/imath] So far I have: [imath]88=5\cdot15+13[/imath] [imath]15=1\cdot13+2[/imath] Backwards substitution gives [imath]15v+18w=1[/imath] [imath]1=15-1\cdot13[/imath] [imath]=15-1(88-5\cdot15)[/imath] [imath]=15\cdot15-1\cdot88[/imath] now very stuck what to do from here:(
|
1749231
|
Trying to prove that [imath]\lim_{N \rightarrow \infty} \frac{1}{N} \Sigma_{n=1}^N f(n\alpha) = \int_0^1 f(x) dx[/imath]
Let [imath]\alpha[/imath] be an irrational number. Let [imath]f: \mathbb{R} \rightarrow \mathbb{C}[/imath] be a continuous periodic function with period 1. Show that [imath]\lim_{N \rightarrow \infty} \frac{1}{N} \Sigma_{n=1}^N f(n\alpha) = \int_0^1 f(x)\,dx[/imath] The beginning (but probably not the end) of my confusion with this problem has to do with the irrational inputs. Why would that be necessary? Any help is appreciated!
|
384765
|
Prove that [imath]\lim_{N\rightarrow\infty}(1/N)\sum_{n=1}^N f(nx)=\int_{0}^1f(t)dt[/imath]
Suppose [imath]f[/imath] is continuous and periodic on the reals with period 1. Prove that if [imath]x\in[0,1][/imath] is an irrational number, then [imath]\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^N f(nx)=\int_{0}^1f(t)dt[/imath] Suggestion: First consider [imath]f(t) = e^{2\pi(ikt)}[/imath] where k is an integer. I can see that this is a limit of a weighted average, but the suggestion throws me off. I've seen the suggestion in fourier transforms but it's not clicking at the moment. Any help would be welcome.
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1817018
|
Generalized Poincaré Inequality on H1 proof
let's see if someone can help me with this proof. Let [imath]\Omega\subset\mathbb{R}^n[/imath] be a bounded domain. And let [imath]L^2\left(\Omega\right)[/imath] be the space of equivalence classes of square integrable functions in [imath]\Omega[/imath] given by the equivalence relation [imath]u\sim v \iff u(x)=v(x)\, \text{a.e.}[/imath] being a.e. almost everywhere, in other words, two functions belong to the same equivalence classes if they only are different in a zero meassure set. This space is a Hilbert space and its norm is: [imath] \Vert v \Vert ^2_{L^2 \left( \Omega \right) }= \int_\Omega v(x)^2\,\mathrm{d}x [/imath] Let [imath]H^1\left(\Omega\right)=\left\{ v\in L^2\left( \Omega \right);\,\vert\mathbf{grad}(v)\vert \in L^2\left( \Omega \right ) \right\}[/imath] be also a Hilbert space and its norm : [imath] \Vert v \Vert ^2_{H^1 \left( \Omega \right) }= \Vert v \Vert ^2_{L^2 \left( \Omega \right) } + \Vert \vert\mathbf{grad}(v)\vert \Vert ^2_{L^2 \left( \Omega \right) } [/imath] I've been asked to proove that: [imath] \Vert v \Vert ^2_{H^1 \left( \Omega \right) } \le C\left( \Omega\right) \left( \Vert \vert\mathbf{grad}(v)\vert \Vert ^2_{L^2 \left( \Omega \right) } + \int_\Gamma \left ( \gamma v(x) \right) ^2\,\mathrm{d}\sigma \right) \quad \forall v\in H^1 \left( \Omega \right) [/imath] where [imath]\Gamma = \partial \Omega [/imath], [imath]\Omega [/imath] is smooth enough, [imath]\mathrm{d}\sigma[/imath] is a meassure on the boundary of [imath]\Omega[/imath], and the operator [imath]\gamma: L^2\left( \Omega\right )\mapsto L^2\left( \Gamma\right )[/imath] is the trace operator. ( if it is needed I can define it also ) Sincerely, I just don't know how to begin, and maybe someone could give me some clue about it.(by the way the [imath]C[/imath] is a positive constant depending only on the domain [imath]\Omega[/imath]). Thank you very much!
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1795925
|
Generalized Poincaré Inequality on H1 proof.
Let [imath]\Omega\subset\mathbb{R}^n[/imath] be a bounded domain. And let [imath]L^2\left(\Omega\right)[/imath] be the space of equivalence classes of square integrable functions in [imath]\Omega[/imath] given by the equivalence relation [imath]u\sim v \iff u(x)=v(x)\, \text{a.e.}[/imath] being a.e. almost everywhere, in other words, two functions belong to the same equivalence classes if they only are different in a zero meassure set. This space is a Hilbert space with the norm is: [imath] \Vert v \Vert ^2_{L^2 \left( \Omega \right) }= \int_\Omega v(x)^2\,\mathrm{d}x [/imath] Let [imath]H^1\left(\Omega\right)=\left\{ v\in L^2\left( \Omega \right);\,\vert\mathbf{grad}(v)\vert \in L^2\left( \Omega \right ) \right\}[/imath] be also a Hilbert space and its norm : [imath] \Vert v \Vert ^2_{H^1 \left( \Omega \right) }= \Vert v \Vert ^2_{L^2 \left( \Omega \right) } + \Vert \vert\mathbf{grad}(v)\vert \Vert ^2_{L^2 \left( \Omega \right) } [/imath] I've been asked to prove that: [imath] \Vert v \Vert ^2_{H^1 \left( \Omega \right) } \le C\left( \Omega\right) \left( \Vert \vert\mathbf{grad}(v)\vert \Vert ^2_{L^2 \left( \Omega \right) } + \int_\Gamma \left ( \gamma v(x) \right) ^2\,\mathrm{d}\sigma \right) \quad \forall v\in H^1 \left( \Omega \right) [/imath] where [imath]\Gamma = \partial \Omega [/imath], [imath]\Omega [/imath] is smooth enough, [imath]\mathrm{d}\sigma[/imath] is a meassure on the boundary of [imath]\Omega[/imath], and the operator [imath]\gamma: L^2\left( \Omega\right )\mapsto L^2\left( \Gamma\right )[/imath] is the trace operator. ( if it is needed I can define it also ) Sincerely, I just don't know how to begin, and maybe someone could give me some clue about it.(by the way the [imath]C[/imath] is a positive constant depending only on the domain [imath]\Omega[/imath]). Thank you very much!
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2050557
|
Product of imaginary numbers.
A book stated that [imath]\sqrt{-a}•\sqrt{-b} = -\sqrt{ab}[/imath] where, a and b are positive real numbers. I know the proof of the above equation but why isn't this [imath]\sqrt{-a}•\sqrt{-b} = \sqrt{ab}[/imath]. It also stated that this equality [imath]\sqrt{a}•\sqrt{b} = \sqrt{ab}[/imath] is true only when either of a and b are positive or zero but is false when a and b are negative, but why? Thank you for your help.
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1362654
|
Why is [imath]\sqrt{-x}*\sqrt{-x}=-x?[/imath]
Q1 - Why is [imath]\sqrt{-x}*\sqrt{-x}=-x?[/imath] Q2 - I was thinking it would be: [imath]\sqrt{-x}*\sqrt{-x}=\sqrt{-x*-x}=\sqrt{x^2}[/imath] but apparently not (why not?) Q3 - What are the formal algebra rules to use? Can I calculate this without using i such as in: [imath]\sqrt{-x}*\sqrt{-x}=i\sqrt{x}*i\sqrt{x}=-\sqrt{x^2}=-x[/imath].
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2049315
|
Inequality of multiple integral
How to prove following inequality : [imath] \iint_V \frac{dxdy}{x^{-1} + \mid \ln y \mid -1} \le 1, V = [0, 1]×[0, 1][/imath] Could you say me how to start?
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2043275
|
Prove the inequality involving multiple integrals
Please, help me to prove that: [imath] \int_Q \int \frac{dxdy}{x^{-1} + |\ln y| - 1} \leq 1,[/imath] where [imath] Q = [0; 1] \times [0; 1] [/imath] Any ideas how to start. Thank you.
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2051640
|
What's a differential equation?
What exactly is a differential equation? Pray, is [imath]\frac{dx}{dt}= x(x(t))[/imath] not a DE? Because Arnold thinks not. In a note on page 11 of his Ordinary Differential Equations, he says, 'Differential equations are sometimes said to be equations containing unknown functions and their derivatives. This is false...' And this is -- I remember -- exactly what we were told during our undergraduate ODE classes; I also have found this in many other places -- books, online, etc. So Arnold's comment is strange and seems like nonsense. Who has an idea what he means, or why he's saying this?
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33153
|
Definition of a Differential Equation?
Here is one definition of a differential equation: "An equation containing the derivatives of one or more dependent variables, with respect to one of more independent variables, is said to be a differential equation (DE)" (Zill - A First Course in Differential Equations) Here is another: "A differential equation is a relationship between a function of time & it's derivatives" (Braun - Differential equations and their applications) Here is another: "Equations in which the unknown function or the vector function appears under the sign of the derivative or the differential are called differential equations" (L. Elsgolts - Differential Equations & the Calculus of Variations) Here is another: "Let [imath]f(x)[/imath] define a function of [imath]x[/imath] on an interval [imath]I: a < x < b[/imath]. By an ordinary differential equation we mean an equation involving [imath]x[/imath], the function [imath]f(x)[/imath] and one of more of it's derivatives" (Tenenbaum/Pollard - Ordinary Differential Equations) Here is another: "A differential equation is an equation that relates in a nontrivial way an unknown function & one or more of the derivatives or differentials of an unknown function with respect to one or more independent variables." (Ross - Differential Equations) Here is another: "A differential equation is an equation relating some function [imath]f[/imath] to one or more of it's derivatives." (Krantz - Differential equations demystified) Now, you can see that while there is just some tiny variation between them, calling [imath]f(x)[/imath] the function instead of [imath]f[/imath] or calling it a function instead of an equation but generally they all hint at the same thing. However: "Let [imath]U[/imath] be an open domain of n-dimensional euclidean space, & let [imath]v[/imath] be a vector field in [imath]U[/imath]. Then by the differential equation determined by the vector field [imath]v[/imath] is meant the equation [imath]x' = v(x), x \in U[/imath]. Differential equations are sometimes said to be equations containing unknown functions and their derivatives. This is false. For example, the equations [imath]\frac{dx}{dt} = x(x(t))[/imath] is not a differential equation." (Arnold - Ordinary Differential Equations) This is quite different & the last comment basically says that all of the above definitions, in all of the standard textbooks, are in fact incorrect. Would anyone care to expand upon this point if it is of interest as some of you might know about Arnold's book & perhaps be able to give some clearer examples than [imath]\frac{dx}{dt} = x(x(t))[/imath], I honestly can't even see how to make sense of [imath]\frac{dx}{dt} = x(x(t))[/imath]. The more explicit (and with more detail) the better! A second question I would really appreciate an answer to would be - is there any other book that takes the view of differential equations that Arnold does? I can't find any elementary book that starts by defining differential equations in the way Arnol'd does & then goes on to work in phase spaces etc... . Multiple references welcomed.
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2050263
|
If [imath]A[/imath] is denumerable and [imath]x\in A[/imath], then [imath]A\setminus\{x\}[/imath] is denumerable. (With my proof)
My Question reads: Prove: if [imath]A[/imath] is denumerable, then [imath]A\setminus\{x\}[/imath] is denumerable. I have proposed this: Let [imath]f:\mathbb{N}\to A[/imath] be a bijection. Let [imath]g:\mathbb{N}\to A\setminus\{x\}[/imath] be a function defined by: [imath] g(y)= \begin{cases} y, &y=1\\ y+1, &y\geq 2 \end{cases} [/imath] I just wanted to see if the answer I have provided is a reasonable choice for a bijection. This is an answer I have come up with myself. I want a verification of my answer and/or clarification.
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551630
|
Subset of a countable set is itself countable
How is it proved that if [imath]A \subset B\ \text{with}\ B\ \text{countable} [/imath] then [imath]A[/imath] is either countable, finite, or empty? I think the proof involves a [imath]1-1[/imath] correspondence between [imath]\mathbb{N}[/imath] and [imath]A[/imath] but other than that I do not know how to proceed. EDIT: I have checked the solution and it advises me to proceed as follows. " As a start to a definition of [imath]g: \mathbb{N} \rightarrow A[/imath] set [imath]g(1)=f(n_1) [/imath] Then show how to inductively continue this process to produce a [imath]1-1[/imath] function [imath]g[/imath] from [imath]\mathbb{N}[/imath] onto [imath]A[/imath]." So the proof according to my book involves induction.
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2051720
|
If [imath]E_1\subset E_2[/imath], then [imath]\overline E_1\subset\overline E_2[/imath]
Definitions: Let [imath]E[/imath] be any set of real numbers and let [imath]E'[/imath] denote the set of all accumulation points of [imath]E[/imath]. Then the set [imath]\overline E=E\cup E'[/imath] is called the closure of the set [imath]E[/imath]. Show that if [imath]E_1\subset E_2[/imath], then [imath]\overline E_1\subset\overline E_2[/imath]. My attempt: We need to prove that if [imath]x\in\overline E_1[/imath], then [imath]x\in\overline E_2[/imath]. Let [imath]x\in\overline E_1[/imath] [imath]\implies x\in E_1\cup E_1'[/imath] [imath]\implies x\in E_1[/imath] or [imath]x\in E_1'[/imath] If [imath]x\in E_1[/imath], then [imath]x\in E_2\implies x\in E_2\cup E_2'\implies x\in\overline E_2[/imath] and we are done. But what if [imath]x\in E_1'[/imath]? What property of accumulation points do I use to prove that [imath]x\in\overline E_2[/imath]?
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121236
|
If $A$ is a subset of $B$, then the closure of $A$ is contained in the closure of $B$.
I'm trying to prove something here which isn't necessarily hard, but I believe it to be somewhat tricky. I've looked online for the proofs, but some of them don't seem 'strong' enough for me or that convincing. For example, they use the argument that since [imath]$A\subset \overline{B} $[/imath], then [imath]$ \overline{A} \subset \overline{B} $[/imath]. That, or they use slightly altered definitions. These are the definitions that I'm using: Definition #1: The closure of [imath]$A$[/imath] is defined as the intersection of all closed sets containing A. Definition #2: We say that a point x is a limit point of [imath]$A$[/imath] if every neighborhood of [imath]$x$[/imath] intersects [imath]$A$[/imath] in some point other than [imath]$x$[/imath] itself. Theorem 1: [imath]$ \overline{A} = A \cup A' $[/imath], where [imath]$A'$[/imath] = the set of all limit points of [imath]$A$[/imath]. Theorem 2: A point [imath]$x \in \overline{A} $[/imath] iff every neighborhood of [imath]$x$[/imath] intersects [imath]$A$[/imath]. Prove: If [imath] A \subset B,[/imath] then [imath]$ \overline{A} \subset \overline{B} $[/imath] Proof: Let [imath]$ \overline{B} = \bigcap F $[/imath] where each [imath]$F$[/imath] is a closed set containing [imath]$B$[/imath]. By hypothesis, [imath] A \subset B [/imath]; hence, it follows that for each [imath]$F \in \overline{B} $[/imath], [imath]$ A \subset F \subset \overline{B} $[/imath]. Now that we have proven that [imath]$ A \subset \overline{B} $[/imath], we show [imath]$A'$[/imath] is also contained in [imath]$\overline{B} $[/imath]. Let [imath] x \in A' [/imath]. By definition, every neighborhood of x intersects A at some point other than [imath]$x$[/imath] itself. Since [imath] A \subset B [/imath], every neighborhood of [imath]$x$[/imath] also intersects [imath]$B$[/imath] at some other point other than [imath]$x$[/imath] itself. Then, [imath]$ x \in B \subset \overline{B} $[/imath]. Hence, [imath]$ A \cup A' \subset \overline{B}$[/imath]. But, [imath]$ A \cup A' = \overline{A}$[/imath]. Hence, [imath]$ \overline{A} \subset \overline{B}.$[/imath] Is this proof correct? Be brutally honest, please. Critique as much as possible.
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1184989
|
Finding an independent set given partition of a cycle - Lovasz Local Lemma (Alon, Spencer)
Let [imath]G = (V,E)[/imath] be a cycle of length [imath]4n[/imath] and let [imath]V = V_1 \cup V_2 \cup ... \cup V_n[/imath] be a partition of its [imath]4n[/imath] vertices into [imath]n[/imath] pairwise disjoint subsets, each of cardinality 4. Is it true that there must be an independent set of [imath]G[/imath] containing precisely one vertex from each [imath]V_i[/imath]? (Prove or supply a counter example.) -- This problem appears in Alon, Spencer. After working for a long time, I am starting to believe that the statement is false. However, I have not been able to construct a counter example. If we replace 4 with 11, the problem can be proven by symmetric Lovasz Local Lemma: for each [imath]V_i[/imath], randomly pick a vertex to be [imath]v_i[/imath] with probability [imath]1/11[/imath] and place it in some set [imath]H[/imath]. For [imath]xy \in E(G)[/imath], define [imath]A_{xy}[/imath] to be the event that [imath]x[/imath] and [imath]y[/imath] are chosen to be in [imath]H[/imath]. [imath]Pr(A_{xy}) = 1/11^2[/imath] if [imath]x[/imath] and [imath]y[/imath] are in different [imath]V_i[/imath]'s. Otherwise, [imath]Pr(A_{xy}) = 0[/imath] because they cannot both be picked. Let [imath]x \in V_i[/imath], [imath]y \in V_j[/imath] where [imath]i\neq j[/imath]. Events [imath]A_{xy}[/imath] and [imath]A_{uv}[/imath] are dependent if [imath]u[/imath] or [imath]v[/imath] are in [imath]V_i \cup V_j[/imath]. Consider [imath]V_i[/imath]. We have 11-1 choices for vertices that are not [imath]x[/imath]. Each of these vertices has 2 neighbors (because [imath]G[/imath] is a cycle). Also, if [imath]u[/imath] = [imath]x[/imath] then there is one choice for [imath]v \neq y[/imath]. Therefore, by symmetry, the number of dependent events is at most [imath]d = 2(2(10) + 1)=42[/imath]. Then [imath]e p (d+1) = e 43/121 = 0.966... < 1[/imath]. And so there is a choice of [imath]v_i[/imath]'s such that no edges have both endpoints in [imath]H[/imath], i.e., [imath]H[/imath] is an independent set. -- I might've calculated some things incorrectly, but I think in general this idea works. Any ideas/counterexamples for [imath]4n[/imath] in the original question?
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788705
|
prove or supply counter example about graph
Let [imath]G = (V, E)[/imath] be a cycle of length [imath]4n[/imath] and let [imath]V = V_1 \cup V_2 \cup ... \cup V_n[/imath] be a partition of its [imath]4n[/imath] vertices into n pairwise disjoint subsets, each of cardinality 4. Is it true that there must be an independent set of [imath]G[/imath] containing precisely one vertex from each [imath]V_i[/imath]? (Prove or supply a counter example.) I think this is not right because just consider square with four vertices and four edges,but I have doubt that it is proper counter example.please help with your knowledge,thank you very much. actually this is exercise 4 of chapter 5 from alon spencer probabilistic method.so I also tag this question as mentioned.
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1985505
|
:Duplicate: How to prove that [imath]\cos (\pi * z) = \prod_{n \in \mathbb{Z}_{\text{odd}}} (1-\frac{2 z}{n})e^{2z/n}[/imath]
Duplicate The question was also answered here: How can I deduce [imath]\cos\pi z=\prod_{n=0}^{\infty}(1-4z^2/(2n+1)^2)[/imath]? I found this formula on Wikipedia under Weierstrass factorization theorem, [imath]\cos (\pi z) = \underset{n \in \mathbb{Z}_{\text{odd}}}{\prod} (1-\frac{2 z}{n})e^{2z/n}[/imath] However, I am not able to prove it from [imath]\sin (\pi z) = \pi z \underset{n \neq 0}{\prod} (1-\frac{z}{n})e^{z/n}[/imath] The derivation of [imath]\sin (\pi z)[/imath] on page 175 of "Functions of One Complex Variable I" by Conway may be helpful. Would someone please help me prove it? Thanks! Edit: I would like to derive [imath]\cos (\pi z)[/imath] from [imath]\sin (\pi z)[/imath]. Sorry for the confusion
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366844
|
How can I deduce [imath]\cos\pi z=\prod_{n=0}^{\infty}(1-4z^2/(2n+1)^2)[/imath]?
Using the infinite product of [imath]\sin(\pi z)[/imath], one can find the Hadamard product for [imath]e^z-1[/imath]: [imath]e^z-1 =2ie^{z/2}\sin(-iz/2)= 2i e^{z/2} (-iz/2) \prod_n \left(1+\frac{z^2}{4\pi n^2}\right)\\= e^{z/2} z \prod_n \left(1+\frac{z^2}{4\pi n^2}\right).[/imath] I don't see a way to find the product for [imath]\cos\pi z[/imath]. A naive attempt is letting [imath]\{a_n\}\subset{\Bbb C}[/imath] be all the zeros of [imath]\cos(\pi z)[/imath] and showing the possible convergence of [imath] \prod_{n=1}^\infty\left(1-\frac{z}{a_n}\right) [/imath] Is there an alternative way to find the Hadamard product in the title for [imath]\cos\pi z[/imath]?
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415105
|
Computing limit of [imath](1+1/n^2)^n[/imath]
I want to compute the limit [imath]\left(1+\frac{1}{n^2}\right)^n[/imath]. One way to do this is to take logs. So [imath]x_n=\left(1+\frac{1}{n^2}\right)^n[/imath] Then [imath]\log x_n = n\log\left(1+\frac{1}{n^2}\right) = n\left(\frac{1}{n^2}-\frac{1}{2n^4}+O\left(\frac{1}{n^6}\right)\right)[/imath]So [imath]\log x_n[/imath] converges to [imath]0[/imath], and [imath]x_n[/imath] converges to [imath]1[/imath]. Is there a simpler way to do this, maybe without Taylor log expansion?
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1562459
|
Prove that [imath]( 1 + n^{-2}) ^n \to 1[/imath].
I need to prove that [imath]\left(1 + \frac 1 {n^2} \right)^n \to 1[/imath]. I tried to use Bernoulli's inequality, but that is not very useful since in the original sequence there is a plus sign. I then tried to use the Sandwich Theorem by finding two sequences which would make bounds for the original one. The lower bound is obvious, the upper bound not so much. I tried using the sequence [imath]\left( \frac 1 {n+1} \right) ^{\frac 1 n}[/imath], but I could not show that this sequence is bigger than the original one for all [imath]n[/imath]. Could anyone help me with this?
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2052845
|
A question about limits and derivatives
[imath]f:(0,\infty)\to \mathbb{R}[/imath] is differentiable such that [imath]f'(x)\to l[/imath] as [imath]x\to \infty[/imath]. Prove that [imath] \frac{f(x)}{x}\to l [/imath] as [imath]x\to \infty[/imath].
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62916
|
How to show that [imath]\lim\limits_{x \to \infty} f'(x) = 0[/imath] implies [imath]\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0[/imath]?
I was trying to work out a problem I found online. Here is the problem statement: Let [imath]f(x)[/imath] be continuously differentiable on [imath](0, \infty)[/imath] and suppose [imath]\lim\limits_{x \to \infty} f'(x) = 0[/imath]. Prove that [imath]\lim\limits_{x \to \infty} \frac{f(x)}{x} = 0[/imath]. (source: http://www.math.vt.edu/people/plinnell/Vtregional/E79/index.html) The first idea that came to my mind was to show that for all [imath]\epsilon > 0[/imath], we have [imath]|f(x)| < \epsilon|x|[/imath] for sufficiently large [imath]x[/imath]. (And I believe I could do this using the fact that [imath]f'(x) \to 0[/imath] as [imath]x \to \infty[/imath].) However, I was wondering if there was a different (and nicer or cleverer) way. Here's an idea I had in mind: If [imath]f[/imath] is bounded, then [imath]\frac{f(x)}{x}[/imath] clearly goes to zero. If [imath]\lim\limits_{x \to \infty} f(x)[/imath] is either [imath]+\infty[/imath] or [imath]-\infty[/imath], then we can apply l'Hôpital's rule (to get [imath]\lim\limits_{x \to \infty} \frac{f(x)}{x} = \lim\limits_{x \to \infty} \frac{f'(x)}{1} = 0[/imath]). However, I'm not sure what I could do in the remaining case (when [imath]f[/imath] is unbounded but oscillates like crazy). Is there a way to finish the proof from here? Also, are there other ways of proving the given statement?
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2052831
|
Find the last 3 digits of [imath]2003^{2002^{2001}}[/imath]
help.. Find the last 3 digits of [imath]2003^{2002^{2001}}[/imath] I try.. I need find [imath]2003^{2002^{2001}}\equiv x(mod\ 1000)[/imath] ?? How?
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108868
|
Find The Last 3 digits of the number [imath]2003^{2002^{2001}}[/imath]
Find The Last 3 digits of the number [imath]2003^{2002^{2001}}[/imath] BY number theory or otherwise, Also i would like to ask is there a property observed in the numbers of the form [imath]k^n[/imath], where for some [imath]k, n[/imath] is varied then the digits of [imath]k^n[/imath] are periodic, for example, [imath]2^n[/imath], its last digit is periodic with period 4, its second last digit is periodic [imath]4\cdot 5 = 20[/imath] its third last digit is periodic with periodic with period [imath]20\cdot 5 =100[/imath] I have observed this property with other numbers as well, though period might vary,for different values of [imath]k[/imath].
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1810729
|
Group generated by [imath]x,y[/imath]. With relations [imath]x^3=y^2=(xy)^2=1[/imath].
Let [imath]G[/imath] be a group generated by [imath]x,y[/imath] with the relations [imath]x^3=y^2=(xy)^2=1[/imath]. Then show that the order of [imath]G[/imath] is 6. My attempt: So writing down the elements of [imath]G[/imath] we have [imath]\{1,x,x^2,y,\}[/imath]. Other elements include [imath]\{xy, xy^2, x^2y\}[/imath] it seems I am counting more than [imath]6[/imath]. Are some of these equal? how do I prove that?
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993443
|
How to find the order of a group generated by two elements?
What is the order of a group [imath]G [/imath] generated by two elements [imath]x[/imath] and [imath]y[/imath] subject only to the relations [imath]x^3 = y^2 = (xy)^2 = 1[/imath]? List the subgroups of [imath]G[/imath]. Since the above relation is the 'only' relation, I presume that the order of [imath]x[/imath] is [imath]3[/imath] and the order of [imath]y[/imath] is [imath] 2[/imath]. Also y is the inverse of itself, and xy is the inverse of itself. The inverse of [imath]x[/imath] is [imath]x^2[/imath]. I have calculated the elements of the group manually using the given relation. [imath]G = \{1, x, x^2,y, xy, x^2y \}[/imath] I want to know if there is a formula to calculate the order of [imath]G[/imath]? I came across another question where the above relation is [imath]x^3 = y^2 = (xy)^3 = 1[/imath]. In this case the group $G has order 12. It is very time consuming to calculate the elements manually. Is there a better way?
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2052233
|
How does one get that [imath]1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}[/imath]?
While watching interesting mathematics videos, I found one of the papers of Srinivasa Ramanujan to G.H.Hardy in which he had written [imath]1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}[/imath]. The problem is that every term on the left is more than [imath]\frac{1}{120}[/imath] yet the sum is [imath]\frac{1}{120}[/imath]. How is that ??? I know that there are many and much more interesting things presented by Ramanujan (like [imath]1-1+1-1+1...=\frac{1}{2}[/imath] and [imath]1+2+3+4.....=\frac{-1}{12}[/imath]) but for now I am interested in the summation in the title. Any idea/hint is heartily welcome. Thanks. Here is the video I'm talking about.
|
137530
|
Ramanujan Summation
It seems that under the light of Ramanujan Summation the following is plausible: [imath]1 + {2^{2n - 1}} + {3^{2n - 1}} + \cdots = - \frac{{{B_{2n}}}}{{2n}}(\Re)[/imath] Alas, I can't really find any concrete definition of Ramanujan Summation. Could someone provide a small explanation?
|
2052548
|
Non-negative solutions of linear system
I have a linear system [imath]A x = b [/imath] with [imath]A[/imath] being an [imath]m \times n[/imath] matrix ([imath]m < n [/imath]) and [imath]b [/imath] being an [imath]n [/imath] vector, both over [imath]\mathbb R_{\geq 0} [/imath] (non-negative real numbers). The system is guaranteed to have at least one non-negative solution (not known to us). I want to find the set of all non-negative solutions of the system.
|
715807
|
Finding nonnegative solutions to an underdetermined linear system
Here's the environment of my problem: I have a linear system of 4 equations in 8 unknowns (i.e. [imath]Ax = b[/imath], where [imath]A[/imath] is [imath]4 \times 8[/imath], [imath]x[/imath] is [imath]8 \times 1[/imath], and [imath]b[/imath] is [imath]4 \times 1[/imath], with [imath]A[/imath] given and [imath]b[/imath] varying with some exogenous parameters, i.e. essentially given). So, after reducing the system, the solution space can be described by a linear function [imath]f \, : \, \mathbb{R}^4 \rightarrow \mathbb{R}^4[/imath], which I can find explicitly. Because of the nature of the problem, I am interested only in nonnegative values for each of the 8 unknowns. So, I can restrict the function [imath]f[/imath] to the domain [imath]\mathbb{R}^4_{\geq 0}[/imath] (i.e. examine only nonnegative values for each of the 4 free variables). What I'm interested in knowing (and classifying for different parameter values, which will change [imath]b[/imath]) is whether solutions exist for which all 8 unknowns are nonnegative--or equivalently, whether the function [imath]f[/imath] takes nonnegative values in the nonnegative domain [imath]\mathbb{R}^4_{\geq 0}[/imath]. If there's a way to figure this out analytically, that would be great; alternatively, if there's a good way to do this numerically (I use MATLAB), that would be only slightly less great. Thanks in advance! Thank you all so much for your answers--this has been a huge help. Working out the solution by hand using the LL Dines paper may prove to be too cumbersome, as it requires knowing the sign of the coefficients, and we examine thousands of different sets of coefficients (as we vary several parameters); however, if nothing else, it's really nice to know that there is an algorithm for (analytically) determining the existence of nonnegative solutions. Moreover, all of the software advice given has been extremely helpful. I've looked into all of your suggestions, and for now I think we'll try using lsqnonneg. Thanks again for your help!
|
2053880
|
If [imath]A[/imath] and [imath]B[/imath] are proper ideals of a ring with no zero divisors, show that [imath]A\cap B \neq \{0\}[/imath]
If [imath]A[/imath] and [imath]B[/imath] are proper ideals of a ring with no zero divisors, show that [imath]A\cap B \neq \{0\}[/imath] Ideals: A subset [imath]I[/imath] of a ring [imath]R[/imath] is called an ideal if [imath]I[/imath] is a subring of [imath]R[/imath] For all [imath]a\in I,r\in R[/imath] [imath]ar\in I[/imath] and [imath]ra\in I[/imath]. [imath]A[/imath] and [imath]B[/imath] both satisfy the above. How to show that [imath]A\cap B \neq 0[/imath]?
|
2051145
|
[imath]A[/imath] and [imath]B[/imath] are ideals of a ring [imath]{R}[/imath] such that [imath]A\cap B=\{0\}[/imath]. Prove that [imath]st=0[/imath] for every [imath]s\in A, t\in B[/imath].
[imath]A[/imath] and [imath]B[/imath] are ideals of a ring [imath]{R}[/imath] such that [imath]A\cap B=\{0\}[/imath]. Prove that [imath]st=0[/imath] for every [imath]s\in A, t\in B[/imath]. I have no idea how to solve it. please help.
|
2053598
|
A sequence [imath]\{a_n\}[/imath] that diverges but [imath]\displaystyle\lim_{n\to 0} |a_n-a_{n+1}| =0[/imath]
What is a sequence [imath]\{a_n\}[/imath] that diverges and [imath]\displaystyle\lim_{n\to 0} |a_n-a_{n+1}| =0[/imath]
|
75840
|
A sequence of real numbers such that [imath]\lim_{n\to+\infty}|x_n-x_{n+1}|=0[/imath] but it is not Cauchy
Give an example of a sequence [imath](x_n)[/imath] of real numbers, where [imath]\displaystyle\lim_{n\to+\infty}|x_n-x_{n+1}|=0[/imath], but [imath](x_n)[/imath] is not a Cauchy sequence
|
1862326
|
[imath]\lim_{n \to \infty} \mid a_n + 3(\frac{n-2}{n})^n \mid^{\frac1n} = \frac35[/imath]. Then find [imath]\lim_{n \to \infty} a_n[/imath].
Let [imath]\{a_n\}[/imath] be a sequence of real numbers such that [imath]\lim_{n \to \infty} \mid a_n + 3(\frac{n-2}{n})^n \mid^{\frac1n} = \frac35[/imath] Then find [imath]\lim_{n \to \infty} a_n[/imath]. Tried very hard yet not able to crack it. Help Needed. Anyone able to do the sum??
|
2033032
|
What is the value of lim[imath]_{n\to \infty} a_n[/imath] if [imath]\lim_{n \to \infty}\left|a_n+ 3\left(\frac{n-2}{n}\right)^n \right|^{\frac{1}{n}}=\frac{3}{5}[/imath]?
Let [imath]\{a_n\}[/imath] be a sequence of real numbers such that [imath]\lim_{n \to \infty}\left|a_n+ 3\left(\frac{n-2}{n}\right)^n \right|^{\frac{1}{n}}=\frac{3}{5}.[/imath] What is the value of [imath]\lim_{n\to \infty} a_n[/imath]?
|
2054116
|
Finding a nontrivial normal subgroup of a group of rank 36
Let [imath]G[/imath] be a group such that [imath]|G| = 2^2 \cdot 3^2[/imath]. I want to show that there exists a nontrivial normal subgroup of [imath]G[/imath]. My tries: By the Sylow theorem I find out that te number of [imath]2[/imath]-Sylow subgroups is equal to 1,3 or 9. On the other side the number of [imath]3[/imath]-Sylow subgroups is equal to [imath]1[/imath] or [imath]4[/imath]. Now I don't have any ideas what to do.
|
225367
|
Existence of normal subgroups for a group of order [imath]36[/imath]
Prove that a group of order 36 must have a normal subgroup of order 3 or 9. Let n2 be the number of 2-Sylow subgroups of G (with |G|=36). Then n must be 1 or 3. Let n3 be the number of 3-Sylow subgroups of G. then n3=1 or n3=4 if n3=1 we have 1 3-sylow group of order 9. and it is also a normal group(from sylow theorem ) if n2=1 there is normal group of order 4 but I cant show normal group of order 3.
|
2054082
|
For which value of [imath]n[/imath] the ring [imath]\mathbb{Z}[\sqrt{n}][/imath] is a UFD?
For which value of [imath]n[/imath] the ring [imath]\mathbb{Z}[\sqrt{n}][/imath] is a UFD ? I am literally very confused about it. I know that when [imath]n=2,3,-1,-2[/imath] (there are many other value also) then the ring is UFD but I want to know some generalized principles that help in decision making. Your help will be appreciated. Thanks.
|
726826
|
When is [imath]\mathbb{Z}[\sqrt{d}][/imath] not an UFD (for [imath]d>1[/imath])?
I was wondering if there is a classification for this: For which [imath]d[/imath] is [imath]D=\mathbb{Z}[\sqrt{d}][/imath] a UFD, with [imath]d > 1[/imath]? For [imath]d \equiv 1 [/imath] (mod [imath] 4[/imath]), [imath]D[/imath] is not a UFD (proof here).
|
2053912
|
Proving diverges to infinity
If [imath]a_1[/imath] is positive integer and [imath]a_{n+1} = a_n + \frac {1}{a_n}[/imath] , how can I prove [imath]{a_n}[/imath] diverges to [imath]\infty[/imath] ?
|
843014
|
Showing a recursive sequence isn't bounded [imath]a_{n+1}=a_n+\frac 1 {a_n}[/imath]
Show the sequence isn't bounded: [imath]a_1=1[/imath], [imath]a_{n+1}=a_n+\frac 1 {a_n}[/imath]. Proof by contradiction: Let [imath]M>0[/imath] such that [imath]\forall n: |a_n|< M[/imath]. Let [imath]\epsilon >0 [/imath] and for some [imath]n=N, \epsilon: a_N=M-\epsilon<M [/imath] pluging that in the recursion: [imath]a_{N+1}=M-\epsilon+\frac 1 {M-\epsilon}>M>M-\epsilon[/imath]. Contradiction. I wondered if I could suppose about the boundary that [imath]\forall n: |a_n|\le M[/imath] ? The proof would basically be the same only I could drop the epsilon.
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