qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
619536 | A question concerning cyclic property of a group
For which [imath]n[/imath] is the group [imath]U_n[/imath] (group of all positive integers less than [imath]n[/imath] that are coprime to [imath]n[/imath]) a cyclic group? | 314846 | For what [imath]n[/imath] is [imath]U_n[/imath] cyclic?
When can we say a multiplicative group of integers modulo [imath]n[/imath], i.e., [imath]U_n[/imath] is cyclic? [imath]U_n=\{a \in\mathbb Z_n \mid \gcd(a,n)=1 \}[/imath] I searched the internet but did not get a clear idea. |
619902 | How can I resolve [imath]\sum_{x=0}^{\infty} xe^{-x/\theta}[/imath]?
I stumble on this summation during an exercise. How can I resolve [imath]\sum_{x=0}^{\infty} xe^{-x/\theta}[/imath]? | 516934 | How to solve these series?
Can anyone help me understand how to solve these two series? More than the solution I'm interested in understanding which process I should follow. Series 1: [imath] \sum_{i = 3}^{\infty} i * a^{i-1}, 0 < a < 1. [/imath] Series 2: [imath] \sum_{i = 3}^{\infty} i\sum_{k = 2}^{i-1} a^{i-k} * b^{k-2} , 0 < a < 1, 0 < b < 1. [/imath] These two series come as part of a long mathematical proof which I omitted for brevity, if you think it is relevant I will post it. |
620048 | Is there an identity that links [imath]\pi[/imath] and [imath]\phi[/imath] (the golden ratio)?
Is there some identity that shows a connection between [imath]\pi[/imath] and the golden ratio, [imath]\phi[/imath]? | 454333 | Can the golden ratio accurately be expressed in terms of [imath]e[/imath] and [imath]\pi[/imath]
I was playing around with numbers when I noticed that [imath]\sqrt e[/imath] was very somewhat close to [imath]\phi[/imath] And so, I took it upon myself to try to find a way to express the golden ratio in terms of the infamous values, [imath]\large\pi[/imath] and [imath]\large e[/imath] The closest that I've come so far is: [imath] \varphi \approx \sqrt e - \frac{\pi}{(e+\pi)^e - \sqrt e} [/imath] My question is, Is there a better (more precise and accurate) way of expressing [imath]\phi[/imath] in terms of [imath]e[/imath] and [imath]\pi[/imath] ? |
605691 | Equivalent of a recurrence sequence
Let [imath]x_{0} = 2[/imath] and [imath]x_{n+1} = x_{n} + \ln(x_{n})[/imath], how can I find an asymptotic equivalent of this sequence say, to the third term? (This is not homework, it was a problem in the Oral Examination 2010 of Ecole Centrale). | 390613 | What is a good asymptotic for [imath]f_n = f_{n-1}+\ln(f_{n-1})[/imath]?
Let [imath]f_0=2[/imath] and [imath]f_n=f_{n-1}+\ln(f_{n-1})[/imath]. What is a good asymptotic to the sequence [imath]f_n[/imath]? With good I mean much better than [imath]f_n \sim \dfrac{3n \ln(2)\ln(n)}{2}[/imath]. |
620215 | Why [imath]\prod_{i\in I} \Bbb Z[/imath] is not a projective [imath]\Bbb Z[/imath]-module?
Let [imath]\Bbb Z[/imath] be the ring of integers and [imath]I[/imath] a non-countable index set. Why [imath]\prod_{i\in I} \Bbb Z[/imath] is not a projective [imath]\Bbb Z[/imath]-module? | 320444 | Why isn't an infinite direct product of copies of [imath]\Bbb Z[/imath] a free module?
Why isn't an infinite direct product of copies of [imath]\Bbb Z[/imath] a free module? Actually I was asked to show that it's not projective, but as [imath]\Bbb{Z}[/imath] is a PID, so it suffices to show it's not free. I am stuck here. I saw some questions in SE, but there is no satisfactory answer at all. |
619505 | Is [imath]\log (1 + {x^2})[/imath] uniformly continuous on [imath][0,\infty)[/imath]?
Is [imath]\log (1 + {x^2})[/imath] uniformly continuous? Here is my attempt: Let [imath]\forall\left| {x - y} \right| < \delta[/imath]: [imath]\left| {\log (1 + {x^2}) - \log (1 + {y^2})} \right| = \left| {\log (\frac{{1 + {x^2}}}{{1 + {y^2}}})} \right| < \varepsilon [/imath] because [imath]\log (1) = 0[/imath] it's sufficent to prove: [imath]\left| {\frac{{1 + {x^2}}}{{1 + {y^2}}} - 1} \right| < \varepsilon [/imath] This is where I got stuck. What do you suggest? EDIT: I'd like to prove it without involving derivative. Namely, by the definition of uniform continuity. | 271785 | prove that [imath]f(x)=\log(1+x^2)[/imath] is Uniform continuous with [imath]\epsilon ,\delta[/imath] ...
I have to prove that [imath]f(x)=\log(1+x^2)[/imath] is Uniform continuous in [imath][0,\infty)[/imath] (with [imath]\epsilon ,\delta[/imath] formulas...) I wrote the definition: (what I have to prove): [imath]\forall \epsilon>0 \quad \exists \delta>0 \quad s.t. \quad \forall x,y\in [0,\infty) : \quad |x-y|<\delta \quad \Rightarrow \quad |f(x)-f(y)|<\epsilon [/imath] So I tried developing [imath]|f(x)-f(y)| = |\log(1+x^2)-\log(1+y^2)| = |\log(\frac{1+x^2}{1+y^2})| [/imath]... now this is where I try to make it bigger and simplify the expression so i can choose the right [imath]\delta[/imath] which depends on the [imath]\epsilon[/imath], and then say that if the simplified bigger expression is still smaller than [imath]\epsilon[/imath] then of course the original [imath]|f(x)-f(y)|[/imath] is smaller than [imath]\epsilon[/imath] but what can I do with this expression? any algebraic tricks? |
615493 | Continuous mapping [imath]f: [0,1]\rightarrow (0,1)[/imath] CSIR December [imath]2013[/imath]
Question is : Suppose [imath]f: [0,1]\rightarrow (0,1)[/imath] is Continuous then which of the following is NOT true.. [imath]F\subseteq[0,1][/imath] is closed set implies [imath]f(F)[/imath] is closed in [imath]\mathbb{R}[/imath] If [imath]f(0)<f(1)[/imath] then [imath]f([0,1])[/imath] must be equal to [imath][f(0),f(1)][/imath] There must exist [imath]x\in(0,1)[/imath] such that [imath]f(x)=x[/imath] [imath]f([0,1])\neq (0,1)[/imath] Continuous map need not map closed sets to closed sets.. So, first option is not true... Continuous maps takes connected sets to connected sets ... So [imath]f([0,1])[/imath] must be connected and it is equal to [imath][f(0),f(1)][/imath].. So, Second option is true.. Continuous maps takes compact sets to compact sets... So, [imath]f([0,1])\neq (0,1)[/imath] and so fourth option is true... I guess third option is also false though I can not think of any example.. Continuous map from compact set to itself has a fixed point. But how do i conclude that this would imply third option is true/false. Please help me to clear this... Thank you. | 2321964 | If [imath]f:[0,1]\to (0,1)[/imath] is a continuous mapping then which are NOT correct?
If [imath]f:[0,1]\to (0,1)[/imath] is a continuous mapping then which are NOT correct? [imath]F\subset [0,1][/imath] is a closed set implies that [imath]f(F)[/imath] is a closed set in [imath]\Bbb R[/imath]. If [imath]f(0)<f(1)[/imath] then [imath]f([0,1])[/imath] must be equal to [imath][f(0),f(1)][/imath]. [imath]\exists x[/imath] such that [imath]f(x)=x[/imath]. [imath]f([0,1])\neq (0,1)[/imath]. is false as continuous functions don't take closed sets to closed sets. is false as [imath]f[/imath] is not monotonically increasing. 3.Let [imath]h(x)=f(x)-x\implies h(0)=f(0)>0;h(1)=f(1)-1<0\implies h(k)=0[/imath] for some [imath]k[/imath] .hence true 4.true since continuous image of a compact set must be compact. hence the correct options are 1,2. But the answer is given to be 2 only .Please help. |
457392 | Is a square-integrable continuous local martingale a true martingale?
I am wondering, if the following is true without any other assumptions and if so, how to prove it: Let [imath](M_t)_{t \geq 0}[/imath] be a continuous local martingale on a filtered probability space [imath](\Omega, \mathcal{F}, (\mathcal{F}_t)_{t\geq 0}, P)[/imath] that satisfies the usual conditions. If [imath](M_t)_{t\geq 0}[/imath]is square integrable (that is, for all [imath]t\geq 0[/imath] we have [imath]E[M_t^2] < \infty[/imath]) then [imath](M_t)_{t\geq 0}[/imath] is a true martingale. Thanks for any advice. | 272509 | Is continuous L2 bounded local martingale a true martingale?
I can prove it briefly, but I found a "counter" example. (There must be a mistake in the following words...) I can prove: X is a continuous local martingale, with [imath]X_0=0[/imath] a.s, then X is [imath]L_2[/imath] bounded iff [imath]E[X]_\infty<\infty[/imath] I can prove: X is a continuous local martingale, if [imath]E[X]_\infty<\infty[/imath], then X is a true martingale. It seems that we can get: any [imath]L_2[/imath] bounded continuous local martingale is a true martingale. But there is a "counter" example: If X is a standard Brownian motion in [imath]R^3[/imath], started at [imath]0[/imath], I can prove (i) [imath]Y_t =1/|X_{1+t}|[/imath] is a local martingale; (ii) Y is bounded in [imath]L^2[/imath]; (iii) Y is not a martingale. Can I say: since Y isn't continuous, therefore Y is bounded in [imath]L^2[/imath], but not a martingale??? But I guess that we can prove Y is continuous almost surely. Is continuous L2 bounded local martingale a true martingale??? |
619980 | Combinatorics with n case
Show that if n is a positive integer, then [imath]{2n\choose0} + {2n\choose2}+ \dots + {2n\choose2n} = 2^{2n-1}[/imath] I assume the proof is done by induction but how to I do the k+1 case? | 42797 | Evaluate [imath] \binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots+\binom{n}{2k}+\cdots[/imath]
I need to evaluate, for a certain worded question: If n is even [imath]\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots\binom{n}{n}[/imath] If n is odd [imath]\binom{n}{0}+\binom{n}{2}+\binom{n}{4}+\cdots\binom{n}{n-1}[/imath] I havent been able to reduce this using the binomial expansion. |
619649 | Integral of [imath]\frac{1}{x^4+1}[/imath]
Just doing this for revision, seems much harder than it should be, should I use [imath]x=\tan u[/imath] ? Any help appreciated. | 645872 | Have a question related to indefinte integrals
What is [imath]\int \frac1{1+x^4}\text dx[/imath]? I have tried all standard substitutions and couldn't make them work. PLease help |
615582 | [imath]A\subset \mathbb{R}[/imath] with more than one element and [imath]A/ \{a\}[/imath] is compact for a fixed [imath]a\in A[/imath]
Question is : Suppose [imath]A\subset \mathbb{R}[/imath] with more than one element and [imath]A/ \{a\}[/imath] is compact for a fixed [imath]a\in A[/imath] then [imath]A[/imath] is compact Every subset of [imath]A[/imath] must be compact [imath]A[/imath] must be finite [imath]A[/imath] is disconnected Only compact subsets of [imath]\mathbb{R}[/imath] I can think of are finite union of closed intervals and finite sets. Take a finite union of closed intervals, If i remove one element for that I would end up with something which contains [imath][a,b)[/imath] and this would be not compact so should be the whole set. So, First option is wrong i.e., [imath]A[/imath] is not compact. I can not say anything about second and fourth option but third option is possible I guess. Please help me to see this in detail.. | 2028426 | Let A be a subset of R with more than one element, let a in A. If A-{a} is compact.
Let [imath]A[/imath] be a subset of [imath]\mathbb R[/imath] with more than one element, let [imath]a[/imath] in [imath]A[/imath]. If [imath]A-\{a\}[/imath] is compact, then [imath]A[/imath] is compact Every subset of [imath]A[/imath] must be compact [imath]A[/imath] must be finite set [imath]A[/imath] is disconnected for any open cover of [imath]A[/imath] is also open cover for [imath]A-\{a\}[/imath], since [imath]A-\{a\}[/imath] is compact implies that it has finite sub cover to cover [imath]A-\{a\}[/imath]. with this sub cover union an open set containing [imath]a[/imath] will be the finite sub cover for [imath]A[/imath]. Hence [imath]A[/imath] is compact Given that [imath]A-\{a\}[/imath] is compact, so it is closed, this implies [imath]\{a\}[/imath] is open. But every singleton in [imath]\mathbb{R}[/imath] is closed, so [imath]\{a\}[/imath] is both closed and open. Hence A is disconnected so option (1) and (4) is right. what can we say about option (2) and (3)? |
620498 | If a linear map, [imath]φ: U→V[/imath] is bijective, then [imath]φ(x)=0[/imath] has only the solution [imath]x=0[/imath].
If a linear map, [imath]φ: U→V[/imath] is bijective, then [imath]φ(x)=0[/imath] has only the solution [imath]x=0[/imath]. "an attempt": [imath]φ(0)=φ(0\cdot x)=φ(0)+φ( x)[/imath] | 620439 | Prove: If a linear map, [imath]φ: U→V[/imath] is bijective, then [imath]φ(x)=0[/imath] has only the solution [imath]x=0[/imath].
Prove: If a linear map, [imath]φ: U→V[/imath] is bijective, then [imath]φ(x)=0[/imath] has only the solution [imath]x=0[/imath]. Either a proof or explanation will be appreciated. |
620474 | Perfect Squares of a Specific Form
Are there any perfect squares of the form [imath]1+5+5^2+...+5^n[/imath], [imath]n[/imath] being a natural number? I'm conjecturing not, but how can one prove it? | 372367 | [imath]1+x+\ldots+x^n[/imath] perfect square
Let [imath]p[/imath] be the polynomial [imath]p(x)=1+x+\ldots+x^n[/imath]. For which couples [imath](a, n)\in\mathbb{N}^2[/imath], [imath]p(a)[/imath] is a perfect square? I'm particularly interested in [imath]p(3)[/imath]. |
620387 | how can i evaluate this indefinite-integral?
[imath]\int \frac{x^2}{(x\cos{x}-\sin{x})^2} dx[/imath] I tried to turn it to [imath]\tan[/imath] and [imath]\sec[/imath] but it didn't work out very well. How can I evaluate this indefinite integral using regular methods? | 620143 | Indefinite integral [imath]\int(\sin x)^2/(x\cdot\cos x-\sin x)^2\,dx[/imath]
How to find the antiderivative of [imath]\dfrac{(\sin x)^2}{(x\cdot\cos x-\sin x)^2}[/imath]? I know that series theory can solve the problem. But I wish to solve the problem by the antiderivative way. The numerator and denominator divide the [imath]x^2[/imath] at the same time, don't they? |
620787 | How do I prove [imath]F_{n+1}^2 - F_nF_{n+2} = (-1)^n[/imath] using induction?
[imath]F_n[/imath] refers to the [imath]n[/imath] term of the Fibonacci Sequence. I think I'm supposed to prove this by induction. I already have the base case. I am at: [imath]\text{F}_\text{k+1}^2 - F_k\text{F}_\text{k+2} + \text{F}_\text{k+2}^2 - \text{F}_\text{k+1}\text{F}_\text{k+3} = (-1)^k +\text{F}_\text{k+2}^2 - \text{F}_\text{k+1}\text{F}_\text{k+3}[/imath] So it's probably easier to work backwards. Multiply [imath]-1[/imath] in order to get [imath](-1)^\text{k + 1}[/imath] and then get: [imath]-\text{F}_\text{k+1}^2 + F_k\text{F}_\text{k+2} - \text{F}_\text{k+2}^2 + \text{F}_\text{k+1}\text{F}_\text{k+3} = (-1)^\text{k+1} -\text{F}_\text{k+2}^2 + \text{F}_\text{k+1}\text{F}_\text{k+3}[/imath] So then bring everything to the left side and mess around with the left side. [imath]-\text{F}_\text{k+1}^2 + F_k\text{F}_\text{k+2} - \text{F}_\text{k+2}^2 + \text{F}_\text{k+1}\text{F}_\text{k+3} +\text{F}_\text{k+2}^2 - \text{F}_\text{k+1}\text{F}_\text{k+3} = (-1)^\text{k + 1} [/imath] And so somehow get this to look more like this: [imath]\text{F}_\text{k+1}^2 - F_k\text{F}_\text{k+2} + \text{F}_\text{k+2}^2 - \text{F}_\text{k+1}\text{F}_\text{k+3}[/imath] ...except I don't know how to do that. Thanks | 186040 | Fibonacci numbers and proof by induction
Consider the Fibonacci numbers [imath]F(0) = 0; F(1)=1; F(n) = F(n-1) + F(n-2)[/imath]. Prove by induction that for all [imath]n>0[/imath], [imath]F(n-1)\cdot F(n+1)- F(n)^2 = (-1)^n[/imath] I assume [imath]P(n)[/imath] is true and try to show [imath]P(n+1)[/imath] is true, but I got stuck with the algebra. How do we reach [imath]P(n+1)[/imath] from [imath]P(n)[/imath]? Also, strong induction may be used instead. |
620860 | Product of 5 consecutive integers cannot be a perfect square
Problem: Show that the product of any five consecutive positive integers cannot be a perfect square. Proof: Let [imath]N=n \times (n+1) \times (n+2) \times (n+3) \times (n+4)[/imath], where [imath]n \in \mathbb{N}[/imath]. We know the that N must have at least: one factor is divisible by 5, two or three factors are even, one factor of the even factors, is divisible by [imath]4[/imath], and one factor is divisible by [imath]3[/imath]. To proof the statement above we need to assume that the product of any five consecutive positive integers is a perfect square to lead us to a contradiction. Hence, I know there will be cases for us to test for. However, I am unsure where to start after mentioning everything above. Can anyone help me from here? | 304417 | Product of [imath]5[/imath] consecutive integers cannot be perfect square
How can we prove that the product of [imath]5[/imath] consecutive integers cannot be a perfect square? |
621037 | Let z1=x1+iy1 and z2=x2+iy2 be two complex numbers. The dot product of z1 and z2 is defined by =x1x2+y1y2
Let z1=x1+iy1 and z2=x2+iy2 be two complex numbers. The dot product of z1 and z2 is defined by z1,z2=x1x2+y1y2 For non zero z1 and z2 prove the following [imath]<z1,z2> =|z1||z2|\cosθ ~~\hbox{where $0≤θ≤π$ is the angle between z1 and z2.}[/imath] [imath]⟨z1,z2⟩ = Re(z1¯¯¯z2)[/imath] [imath]⟨z1,z2⟩ = 12(z1¯¯¯z2+z1z2¯¯¯)[/imath] [imath]z1⊥z2⟺θ=π2[/imath] | 613811 | Let [imath]z_1 = x_1 +iy_1 [/imath] and [imath]z_2 = x_2 +iy_2[/imath] be two complex numbers. The dot product of [imath]z_1[/imath] and [imath]z_2[/imath] is defined by [imath]< z_1 , z_2> = x_1x_2+y_1y_2[/imath]
Let [imath]z_1 = x_1 +iy_1[/imath] and [imath]z_2 = x_2 +iy_2[/imath] be two complex numbers. The dot product of [imath]z_1[/imath] and [imath]z_2[/imath] is defined by [imath]\langle z_1 , z_2 \rangle = x_1x_2+y_1y_2[/imath] For non zero [imath]z_1[/imath] and [imath]z_2[/imath] prove the following [imath] \langle z_1, z_2 \rangle = |z_1||z_2|\cos\theta[/imath] (where [imath]0 \le \theta \le \pi[/imath] is the angel between [imath]z_1[/imath] and [imath]z_2[/imath].) [imath]\langle z_1, z_2 \rangle[/imath] = [imath]Re(\overline{z_1}z_2)[/imath] [imath]\langle z_1, z_2 \rangle[/imath] = [imath]\frac{1}{2}(\overline{z_1}z_2 + z_1\overline{z_2})[/imath] [imath]z_1 ⊥ z_2 \iff \theta = \frac{\pi}{2}[/imath] |
620757 | Find a basis for the vector space of symmetric matrices with an order of [imath]n \times n[/imath]
Find a basis for the vector space of symmetric matrices with an order of [imath]n \times n[/imath] This is my thought: by definition of symmetry, [imath]a_{i,j}=a_{j,i}[/imath]. Therefore, the basis should consist [imath]{n^2-n} \over 2[/imath] matrices to determine each symmetric pair. In addition, it should also consist [imath]n[/imath] matrices to determine each term in the diagonal. Therefore, the dimension of the vector space is [imath]{n^2+n} \over 2[/imath]. It's not hard to write down the above mathematically (in case it's true). Two questions: Am I right? Is that the desired basis? Is there a more efficent alternative to reprsent the basis? Thanks! | 2776792 | Determine whether S is a subspace of V, if yes find a basis and its dimension.
So this here questions got me stuck for a bit: Determine whether S is a subspace of V, if yes find a basis and its dimension. Let [imath]S[/imath] be the set of all 3x3 symmetric matrices in [imath]V = M_{3x3}[/imath] And here's why: I checked the appropriate axioms and it turns out S is in fact a subspace of V, but my main problem is that before this exercise, to find a basis, I'd find the augmented matrix and solve it (RREF) and then determine the solution, but here I'm not sure if I can. So I have an approach but not sure which if it works: If I say for e.g. Assume the matrix [imath]A=\begin{pmatrix} a & b & c\\ b & d & e\\ c & e & f\\ \end{pmatrix}[/imath] if det(A) [imath]\neq 0[/imath] => linearly independent => the basis exists. [imath]k_1\begin{pmatrix} a\\ b\\ c\\ \end{pmatrix}[/imath] + [imath]k_2\begin{pmatrix} b\\ d\\ e\\ \end{pmatrix}[/imath] + [imath]k_3\begin{pmatrix} c\\ e\\ f\\ \end{pmatrix} = 0; dim(S) = 3[/imath] Please let me know how to go about this. Thank you. |
621118 | Isomorphism problem involving the Symmetric Group
Let [imath]G[/imath] be a group of order [imath]24[/imath] with no elements of order [imath]6[/imath]. Prove that [imath]G[/imath] isomorphic to the symmetric group [imath]S_4[/imath]. This is what I did and I'm a bit unease about it since I didn't use that G has no element of order 6. Let A = {1,2,3,4}. Suppose that G is acting on A by left multiplication. Then [imath]\phi[/imath] : [imath]G \longrightarrow S_A[/imath] is a Homorphism. By the First Isomorphism Theorem: [imath] | \ker \phi| | |G| =S_A =S_4 [/imath] But this says that [imath] \ker \phi = 1[/imath] and thus [imath]\phi[/imath] is injective. This establishes that G is isomorphic to [imath] S_4[/imath] | 534686 | Group of order 24 with no element of order 6 is isomorphic to [imath]S_4[/imath]
Proposition: Given a group [imath]G[/imath] with [imath]|G|=24[/imath] such that [imath]\nexists g\in G[/imath] with [imath]|g|=6[/imath], then [imath]G\cong S_4[/imath]. I understand methods you can employ to deduce the number of Sylow [imath]p[/imath]-groups in [imath]G[/imath] by counting elements or reasoning about permutation representations. But how can we construct an isomorphism to [imath]S_4[/imath] given that [imath]n_{2-\text{Sylow}}=3[/imath] and [imath]n_{3-\text{Sylow}}=4[/imath]? Or otherwise, rule out all other possible cases (perhaps reasoning with Cayley's Theorem)? I am also interested in finding a proof that is perhaps less direct, but more elegant, in particular one that may involve the irreducible representations of [imath]S_4[/imath]. Any insight appreciated. |
621464 | [imath]x[/imath] and [imath]y[/imath] be distinct elements of order [imath]2[/imath] in [imath]G[/imath] that generate [imath]G[/imath]. Prove that [imath]G \cong D_{2n}[/imath]
Problem Let [imath]G[/imath] be a finite group and let [imath]x[/imath] and [imath]y[/imath] be distinct elements of order [imath]2[/imath] in [imath]G[/imath] that generate [imath]G[/imath]. Prove that [imath]G \cong D_{2n}[/imath], where [imath]\vert xy \vert = n[/imath]. Solution We have [imath]x^2 = 1[/imath] and [imath]y^2=1[/imath]. If we let [imath]t=xy[/imath], we then have [imath]tx = xt^{-1}[/imath]. This is because, [imath]tx = xyx[/imath] and [imath]xt^{-1} = x(xy)^{-1} = xyx[/imath], since [imath]x^{-1} = x[/imath] and [imath]y^{-1} = y[/imath]. And note that [imath]y = xt[/imath]. Hence, instead of [imath]x[/imath] and [imath]y[/imath] generating the group, we can have [imath]x[/imath] and [imath]t[/imath] to generate the group. I proved that setting [imath]t=xy[/imath], we have [imath]tx=xt^{−1}[/imath]. And since [imath]x[/imath] and [imath]y[/imath] generate G, since [imath]y=xt[/imath], [imath]t[/imath] and [imath]x[/imath] also generates [imath]G[/imath]. Is this (along with the fact that [imath]G[/imath] is finite) sufficient to conclude that [imath]G \cong D_{2n}[/imath]? Also, I have another related question, can there be an infinite Dihedral group, i.e., [imath]D_{\infty} = \langle x,y \vert x^2=1, yx=xy^{-1}, y \text{ has infinite order}\rangle[/imath] Or is this isomorphic to some other group? | 487128 | Prove that a group generated by two elements of order [imath]2[/imath], [imath]x[/imath] and [imath]y[/imath], is isomorphic to [imath]D_{2n}[/imath], where [imath]n = |xy|.[/imath]
I am completely stuck at the question Let [imath]G[/imath] be a finite group and let [imath]x[/imath] and [imath]y[/imath] be distinct elements of order 2 in [imath]G[/imath] that generate [imath]G[/imath]. Prove that [imath]G \cong D_{2n}[/imath], where [imath]n = |xy|.[/imath] I have proved that Let [imath]x[/imath] and [imath]y[/imath] be elements of order 2 in any group [imath]G[/imath]. If [imath]t = xy[/imath] then [imath]tx = xt^{-1}[/imath]. Can I get some hints? |
533837 | Non-Relatively Prime Polynomials in a Field Extension are Not Relatively Prime in Original Field Either
Hypothesis: [imath]E[/imath] is a field extension of [imath]F[/imath]. [imath]p, q \in F[x][/imath] s.t. [imath]p[/imath] and [imath]q[/imath] are not relatively prime over [imath]E[/imath]. Goal: Show that [imath]p[/imath] and [imath]q[/imath] are not relatively prime over [imath]F[/imath]. Attempt: Since [imath]p[/imath] and [imath]q[/imath] are not relatively prime over [imath]E[/imath], we have that [imath]\gcd(p,q) \ne 1[/imath] in [imath]E[x][/imath]. Then let [imath]d(x)[/imath] be the greatest common divisor of [imath]p,q[/imath] in [imath]E[x][/imath] which from (1) satisfies [imath]\deg(d(x)) \ge 1[/imath]. And from here I'm not sure how to proceed. Since [imath]E[/imath] is just a field extension over [imath]F[/imath], we don't even know if there exist roots of [imath]d(x)[/imath] in [imath]E[/imath], much less [imath]F[/imath]. If necessary, we can extend [imath]E[/imath] to some field [imath]K[/imath] s.t. [imath]K[/imath] possesses all of the roots of [imath]d(x), p(x)[/imath] and [imath]q(x)[/imath] -- but I'm not sure how this helps us. | 83247 | Given fields [imath]K\subseteq L[/imath], why does [imath]f,g[/imath] relatively prime in [imath]K[x][/imath] imply relatively prime in [imath]L[x][/imath]?
[imath]K,L[/imath] are fields, [imath]K\subseteq L[/imath]. [imath]f,g \in K[x][/imath]. Suppose that [imath]f,g[/imath] are relatively prime as elements of [imath]K[x][/imath]. Prove they remain relatively prime in [imath]L[x][/imath]. I've tried everything I can think of. I feel like working with the contrapositive may be helpful but that's just a feeling. |
438089 | Real valued analytic function defined on a connected set is constant
Let [imath]G[/imath] be a connected set and [imath]f : G \rightarrow \mathbb{C}[/imath] a real valued analytic function. Prove that [imath]f[/imath] is constant. My idea to prove the result is to prove a subset [imath]A \neq \varnothing[/imath] of the connected set [imath]G[/imath] is both open and closed. So [imath]G=A[/imath] Take [imath]f(w) = a[/imath] [imath]A = \{z\colon z \in G, f(z) = a\}[/imath] Now I want to show that [imath]A[/imath] is infinite. How to do it? After that it is easy to prove [imath]A=G[/imath]. | 1173185 | is there an non-constant analytic complex function that receives only real values ?
I want to find something like [imath]f(z) = |z|[/imath] (complex function that get's only real values) the problem here is that [imath]f(z)[/imath] is not analytic. |
622278 | Relation between the convergence of [imath]\sum a_{n}[/imath] and [imath]\prod (1+a_{n})[/imath]
What is the relation between the convergence of [imath]\sum a_{n}[/imath] and [imath]\prod (1+a_{n})[/imath] where [imath]a_{n} \in \mathbb{C} \ \forall n[/imath] ? Where can I find some references about this topic ? | 380094 | sufficiency and necessity of convergence of [imath]\sum a_n[/imath] wrt convergence of [imath]\prod (1 + a_n)[/imath]
Does there exist a sequence [imath]a_n[/imath] of complex numbers such that [imath]\sum _{i = 0}^\infty a_n[/imath] converges and the product [imath]\prod _{i = 0}^\infty (1+a_n)[/imath] does not converge to any complex number(not even 0). And conversely a sequence [imath]a_n[/imath] s.t. [imath]\prod _{i = 0}^\infty (1+a_n)[/imath] converges to a non zero complex number. But [imath]\sum _{i = 0}^\infty a_n[/imath] does not converge. We all know that if [imath]\sum _{i = 0}^\infty a_n [/imath] converges absolutely iff the corresp product converges absolutely. So i tried alternating serieses like [imath] a_n = (-1)^n /n [/imath] and [imath](-1)^n / \sqrt n [/imath] but that did not work. i tried like [imath] \sum _{i} \sum _ {k} \frac{ e^{2(pi)i/k}}{k} [/imath]but i couldn't go furthur. Can anybody help? |
596454 | Any prime ideal of [imath]R[x][/imath] properly containing [imath]M[x][/imath] is a maximal ideal of [imath]R[x][/imath]
Let [imath]M[/imath] be a maximal ideal in a ring [imath]R[/imath]. Prove that any prime ideal of [imath]R[x][/imath] properly containing [imath]M[x][/imath] is a maximal ideal of [imath]R[x][/imath]. Help me some hints | 182297 | Maximal ideals in [imath]R[x][/imath]
I want to prove the following result: Let [imath]R[/imath] be a ring and [imath]M[/imath] a maximal ideal in [imath]R[/imath]. If [imath]P[/imath] is a prime ideal in [imath]R[x][/imath] that (strictly) contains [imath]M[x][/imath], then [imath]P[/imath] is a maximal ideal in [imath]R[x][/imath]. I have an idea how to prove that, but I'm not quite sure if the argument is totally valid; maybe someone has a cleaner proof. My argument is like that: [imath]R[x]/M[x][/imath] is isomorphic to [imath](R/M)[x][/imath], that is a PID (since [imath]R/M[/imath] is a field). So, any prime ideal in this ring is maximal; If [imath]P[/imath] contains [imath]M[x][/imath], then [imath]P[/imath] corresponds to a (prime??) ideal in [imath]R[x]/M[x][/imath] (this needs more clarification); The prime ideal (that corresponds to [imath]P[/imath]) in [imath]R[x]/M[x][/imath] is then maximal, and this guarantees that [imath]P[/imath] is maximal in [imath]R[x][/imath]. I'll appreciate any comment. |
622373 | maximum order of an element in symmetric group
While doing my homework i find out that the maximum order of an element in [imath]S_3[/imath] is 3 (the element [imath](123)[/imath]) and the maximum order of an element in [imath]S_4[/imath] is 4 (the element [imath](1234)[/imath]) Can i generalize that and say that the maximum order of an element in [imath]S_n[/imath] is n? | 251333 | Orders of a symmetric group
Consider the symmetric group [imath]S_5[/imath]. I would like to find how many elements of [imath]S_5[/imath] are of order 5, and how many are of order 6. I would also like to determine what the maximum order of an element in this group would be. Here is what I have so far: elements of order 5 are the 5-cycles. Elements of order 6 (since a 6-cycle is impossible for 5 elements) must have at least an even cycle and a cycle of length divisible by 3, and 2 + 3 = 5, so elements of order 5 are a combination of 2-cycles and 3-cycles. How can I find the total count of such elements of order 5 and order 6, and the maximum order? I'm not entirely sure where to go from here. |
622982 | Determine whether the series [imath]\sum_{n=1}^{\infty }\left ( \frac\pi2-\arctan n \right )[/imath] converges or not.
I want to know whether the series [imath]\displaystyle{% \sum_{n=1}^{\infty }\left[{\pi \over 2} - \arctan\left(n\right)\right ]}[/imath] converges or not. Some series such as [imath]\sum_{n=1}^{\infty}\sin \frac1n[/imath], [imath]\sum_{n=1}^{\infty}\tan \frac1n[/imath] are solved by the comparison test with [imath]\sum_{n=1}^{\infty}\frac1n[/imath]. But the given series is not compared with [imath]\sum_{n=1}^{\infty}\frac1n[/imath]. Is there another way to determine whether the series converges or not? | 308377 | Does the series [imath]\;\sum\limits_{n=0}^{\infty}\left(\frac{\pi}{2} - \arctan(n)\right)[/imath] converge or diverge?
I tried everything that I know but I couldn't solve this series: [imath]\sum_{n=0}^{\infty} \left(\frac{\pi}{2} - \arctan(n)\right)[/imath] Does it diverge or converge? |
623250 | When is an orthogonal projection a unitary operator?
Question: When is an orthogonal projection [imath]P: \Bbb C_n \to \Bbb C_n[/imath] a unitary operator? Thoughts: I thought about using the fact that the only eigenvalues for projections are 1 and 0. Don't really know how to use it though.. | 622583 | Why is this true: The only orthogonal projection that is also unitary from [imath]\Bbb C^n[/imath] to [imath]\Bbb C^n[/imath] is the identity
Can anyone explain me please how to see this statement: the only orthogonal projection that is also unitary from [imath]\Bbb C^n[/imath] to [imath]\Bbb C^n[/imath] is the Identity. how can I prove formally that? or how can I see that only the Identity satisfy it? |
623315 | Smallest subfield containing [imath]F[/imath] and [imath]\alpha[/imath]
Let [imath]F[/imath] be a field and let [imath]K[/imath] be an extension of [imath]F[/imath]. Show that if [imath]\alpha\in K[/imath] is algebraic over [imath]F[/imath], [imath]F[\alpha]=\{p(\alpha)\mid p(x)\in F[x]\}[/imath] is the smallest subfield of [imath]K[/imath] containing [imath]F[/imath] and [imath]\alpha[/imath]. | 622546 | algebraic element over a field
Let [imath]F[/imath] be a subfield of a field [imath]K[/imath], [imath]a[/imath] an element of [imath]K[/imath]. Prove that [imath]a[/imath] is algebraic over [imath]F[/imath] iff [imath]F[a]=F(a)[/imath]. For the first direction I've tried this: [imath]a[/imath] is algebraic over [imath]F[/imath] [imath]\Rightarrow[/imath] there is a polynomail [imath]f[/imath] over [imath]F[/imath] such that [imath]f(a)=0[/imath] [imath]\Rightarrow[/imath] [imath]1,a,a^2,...,a^m[/imath] are linearly dependent [imath]\Rightarrow[/imath] [imath]\deg(F[a])<m[/imath]. but it contains [imath]f(a)[/imath] so they are equal.. is that correct? about the other direction.. i simply have no clue |
623573 | Prove that [imath] n^3 + 5n[/imath] is divisible by 6 for all [imath]n\in \textbf{N}[/imath]
Prove that [imath] n^3 + 5n [/imath] is divisible by 6 for all [imath] n \in \textbf{N} [/imath]. I provide my proof below. | 233094 | Prove 24 divides [imath]u^3-u[/imath] for all odd natural numbers [imath]u[/imath]
At our college, a professor told us to prove by a semi-formal demonstration (without complete induction): For every odd natural: [imath]24\mid(u^3-u)[/imath] He said that that example was taken from a high school book - maybe I din't get something, but I really have no idea to prove that without using complete induction. Any idea of smart demonstration? Thanks for your help. (Excuse my bad English.) |
623651 | Composition of three functions
If [imath]f:W\rightarrow X[/imath], [imath]g:X\rightarrow Y[/imath], and [imath]h:Y\rightarrow Z[/imath], does [imath]h \circ (g \circ f) = (h \circ g) \circ f[/imath]? How can I justify this? | 523906 | Show that Function Compositions Are Associative
My intent is to show that a composition of bijections is also a bijection by showing the existence of an inverse. But my approach requires the associativity of function composition. Let [imath]f: X \rightarrow Y, g: Y \rightarrow Z, h: Z \rightarrow W[/imath] be functions. [imath]((f \circ g) \circ h)(x) = h((f \circ g)(x)) = h(g(f(x)))[/imath], and [imath](f \circ (g \circ h))(x) = (g \circ h)(f(x)) = h(g(f(x)))[/imath]. However, I am having problems in justifying that the two compositions, [imath](f \circ g) \circ h[/imath] and [imath]f \circ (g \circ h)[/imath], have the same domain and range. When I consulted ProofWiki, whose link is at the bottom, I got even more confused. Specifically, for [imath](f \circ g) \circ h = f \circ (g \circ h)[/imath] to be defined, ProofWiki requires that dom[imath]g =[/imath] codom[imath]f[/imath] and dom[imath]h =[/imath] codom[imath]g[/imath]. First of all, I think that it should be dom[imath]g =[/imath] range[imath]f[/imath] .... Moreover, as you can see in the example below, you actually have to adjust domains and ranges of [imath]f, g, h[/imath] for the requirement to hold true. Let [imath]f: \mathbb R \rightarrow \mathbb R[/imath] be [imath]f(x) = 2x[/imath], [imath]g: \mathbb R^+ \rightarrow \mathbb R[/imath] be [imath]g(y) = ln(y)[/imath], [imath]h: \mathbb R \rightarrow \mathbb R[/imath] be [imath]h(z) = z - 10[/imath]. Then [imath]((f \circ g) \circ h)(x) = ln(2x) - 10 = (f \circ (g \circ h))(x)[/imath], with dom[imath]((f \circ g) \circ h) = \mathbb R^+[/imath] = dom[imath](f \circ (g \circ h))[/imath]. As a result, we need to set dom[imath]f = \mathbb R^+[/imath], range[imath]f = \mathbb R^+[/imath]; dom[imath]g[/imath], range[imath]g[/imath], dom[imath]h[/imath], and range[imath]h[/imath] remain the same. Am I allowed to do that? This adjustment implies that when we say dom[imath]f = X[/imath], [imath]f[/imath] must be defined for all elements in [imath]X[/imath], but [imath]X[/imath] may not be the entire set of elements for which [imath]f[/imath] is defined. http://www.proofwiki.org/wiki/Composition_of_Mappings_is_Associative |
624168 | sum of unit roots is also unit root
A complex number [imath]\epsilon[/imath] is a unit root if [imath]\epsilon^n=1[/imath] for some positive integer [imath]n[/imath]. How would one prove that if [imath]\epsilon_1, \epsilon_2,\dots, \epsilon_k[/imath] are all unit roots, and [imath]|\epsilon_1+\epsilon_2+\dots+\epsilon_k|=1[/imath], then [imath]\epsilon_1+\epsilon_2+\dots+\epsilon_k[/imath] is also a unit root ? I have no clue about it. Could anyone help me? Thanks a lot. p.s. I only know the problem was from a math student who got perfect score on the IMO | 39856 | Sums of roots of unity
If the integral linear combination of some [imath]n[/imath]th roots of unity has magnitude 1, does this necessarily imply that this linear combination is some root of unity as well? More precisely, Let [imath]\zeta_1, \ldots \zeta_k[/imath] be [imath]n[/imath]th roots of unity. If [imath]|\sum_{i=1}^k n_i \zeta_i| = 1,[/imath] where [imath]n_i \in \mathbb{Z}[/imath], does this imply that [imath]\sum_{i=1}^k n_i \zeta_i[/imath] is an [imath]n[/imath]th root of unity? What about if the [imath]n_i[/imath] are Gaussian integers? |
623872 | What is the sum of the prime numbers up to a prime number [imath]n[/imath]?
How to find the sum of prime numbers up to a prime number [imath]n[/imath], that is for example: the sum of prime numbers up to 7 is: 2+3+5+7=17. So what is the formula for finding: [imath]\sum_{k=0}^n p_k=????,[/imath] with [imath]p_k[/imath] being the [imath]k[/imath]th prime. Also if we have the sum of an even number of primes then would it be a new prime? Example: 2+3+5+7=17 and 17 is a prime. 2+3+5+7+11+13=41 and 41 is prime. Thank you. | 190666 | how to prove this extended prime number theorem?
A Generalized Prime Number Theorem? Conjecture Let [imath]n[/imath] and [imath]k[/imath] be positive integers with [imath]n - 50 > k^2 > 0[/imath] and [imath]n[/imath] sufficiently large. Then for the odd primes we have, when [imath]p[/imath] is the biggest odd prime [imath]\le n[/imath], [imath] 3^k + 5^k + 7^k + 11^k + ... + p^k \sim \frac{n^{k+1}}{(k+1) ( \log(n) - \log(k) ) } [/imath] I wonder if you guys have seen it before ? How to prove it ? Any useful references for [imath]k > 1[/imath] ? |
624377 | What's wrong with this demonstration? (1 = -1)
What's wrong with this demonstration?: [imath]A \iff 1 = 1^1[/imath] [imath]A \implies 1 = 1^\frac{2}{2}[/imath] [imath]A \implies 1 = (1^2)^\frac{1}{2}[/imath] [imath]A \implies 1 = ((-1)^2)^\frac{1}{2}[/imath] [imath]A \implies 1 = (-1)^\frac{2}{2}[/imath] [imath]A \implies 1 = (-1)^1 = -1[/imath] So finaly we get [imath]1 = -1[/imath] | 2721408 | problem with solving one complex number problem
Is [imath]1= -1[/imath] ? because, [imath]1=\sqrt{1} =\sqrt{-1×-1} =\sqrt{-1}×\sqrt{-1 } = i×i = i^2 = -1 [/imath] this can be done with all natural numbers. how this can be? please help me |
624372 | Relative rotations using quaternions
I have a sensor at some arbitrary orientation (non-zero roll, yaw, pitch) given by quaternion [imath]q_{0}[/imath]. I then pitch the sensor to orientation [imath]q_{1}[/imath]. When I compute the relative rotation between the two as [imath]r = q_{1} q_{0}^{-1}[/imath] Then I notice that the relative rotation has roll, pitch and yaw components. Why is this and how do I get a pure pitch from [imath]r[/imath]? | 615891 | Relative rotation between quaternions
Say I have a quaternion q which describes how to get from frame 0 to frame 1, and a quaternion r which describes how to get from frame 0 to frame 2. To get the "quaternion difference" between q and r, I do [imath] q_{d} = q^{-1} r [/imath] This is however in frame 1. How do I get the quaternion difference in frame 0? |
616656 | A question on a sequence in a Banach algebra
If [imath]\{u_{k}\}_{k=1}^{\infty}[/imath] is a sequence in an Banach algebra (and more specifically, in the set of all the bounded linear operators of a Banach space [imath]X[/imath]). If [imath]\sum_{k=1}^{\infty}\lambda^{k}u_{k}=0,\quad \text{for every $\lambda\in \mathbb{D}(0, 1/3)$},[/imath] where the [imath]\mathbb{D}(0, 1/3)~[/imath] denotes an open disc centered at [imath]0[/imath] and radius [imath]1/3[/imath]. Then can we conclude that [imath]\boldsymbol{ u_{n}=0}[/imath], for every [imath]\boldsymbol{n}[/imath]? | 616551 | An easy question on complex
Let [imath]\{u_{k}\}_{k=1}^{\infty}[/imath] be a complex number sequence. If [imath]\sum_{k=1}^{\infty}\lambda^{k}u_{k}=0[/imath], for each [imath]\lambda\in \mathbb{D}(0, 1/3)[/imath](where the [imath]\mathbb{D}(0, 1/3)~[/imath]denotes an open disc centered at [imath]0[/imath] and radius [imath]1/3[/imath]). Then can we conclude [imath]u_{n}=0[/imath], for [imath]n=1, 2, 3...[/imath] ? |
623594 | derivative calculation involving floor function [imath]\left\lfloor {{x^2}} \right\rfloor {\sin ^2}(\pi x)[/imath]
I was asked to find when the function is differentiable and what is the derivative of: [imath]\left\lfloor {{x^2}} \right\rfloor {\sin ^2}(\pi x)[/imath] Now, I am not sure how to treat the floor function. I'll be glad for help. | 619036 | Limits involving trigonometric functions [imath]f(x)=\lfloor{x^2 \rfloor} \sin^2(\pi x)[/imath]
I was asked to find for what values of x the function is differentiable and write down the derivative. [imath]f(x)=\lfloor{x^2 \rfloor} \sin^2(\pi x)[/imath] for [imath]x \geq 0[/imath]. There are two steps: When [imath]x \in (\sqrt{n},\sqrt{n+1})[/imath] we get [imath]f'(x)=\pi n \sin(2 \pi x)[/imath]. If [imath]x=\sqrt{n}[/imath] then: I need to find [imath]\lim_{x \rightarrow \sqrt{n}^+} \frac{f(x)-f(\sqrt{n})}{x-\sqrt{n}}=\lim_{x \rightarrow \sqrt{n}^+} \frac{n \sin^2(\pi x)-n \sin^2(\pi \sqrt{n})}{x-\sqrt{n}}[/imath] I need to find [imath]\lim_{x \rightarrow \sqrt{n}^-} \frac{f(x)-f(\sqrt{n})}{x-\sqrt{n}}=\lim_{x \rightarrow \sqrt{n}^-} \frac{(n-1) \sin^2(\pi x)-n \sin^2(\pi \sqrt{n})}{x-\sqrt{n}}[/imath] I just don't know how to find the limits :\ We haven't covered L'hopital yet, so I need to use other methods. Of course, after finding both limits, we need to write [imath]f'_{-}(\sqrt{n})=f'_{+}(\sqrt{n})=f'(\sqrt{n})=\pi n \sin(2\pi \sqrt{n})[/imath]. This is true if and only if [imath]\sqrt{n} \in \mathbb{N}[/imath] (limits are all 0), otherwise the function is not differentiable in [imath]x=\sqrt{n}[/imath]. Please show how to continue, thanks! |
623095 | finding a limit of a general function.
[imath]{\rm f}[/imath] is differentiable at [imath]{\rm f}\left(1\right),\,[/imath] [imath]{\rm f}\left(1\right) > 0[/imath]. Calculate the following limit, and here's what I did: [imath] \lim_{n \to \infty }\left[% {\rm f}\left(1 + 1/n\right) \over {\rm f}\left(1\right)\right]^{1/n} = \left[{\rm f}\left(1\right) \over {\rm f}\left(1\right)\right]^{0} = 1 [/imath] Now, On the one hand, I can't see why the above isn't true. On the other hand, I got this hunch something is wrong here. Can you direct me please? | 618169 | Limits and derivatives - two questions
I was asked to find two limits. Let [imath]f[/imath] be differentiable function at [imath]x=1[/imath] and [imath]f(1)>0[/imath]. [imath]\lim_{n \rightarrow \infty}\left(\frac{f\left(1+\frac{1}{n}\right )}{f(1)} \right)^{\frac{1}{n}}[/imath] Let [imath]\frac{1}{n}=h[/imath] so the limit becomes [imath]\lim_{h \rightarrow 0}\left(\frac{f\left(1+h \right)}{f(1)} \right)^h=\lim_{h \rightarrow 0} \left(\frac{f(1+h)-f(1)}{f(1)}+1 \right)^h[/imath] How may I continue? Same for [imath]\lim_{x \rightarrow 1} \left(\frac{f(x)}{f(1)} \right)^{\frac{1}{\log(x)}}[/imath]. I defined [imath]h=\log(x)[/imath] and tried to continue with no luck. Please help, thank you! |
625238 | Does the following have an analitical solution?
I have the following equations: \begin{equation} {1\over f(x,y)} {\partial f(x,y) \over \partial x} \alpha(x,y) + {1\over f(x,y)} {\partial f(x,y) \over \partial y} \beta(x,y) = \gamma(x,y) \end{equation} Where [imath]\alpha(x,y)[/imath], [imath]\beta(x,y)[/imath] and [imath]\gamma(x,y)[/imath] are exactly known functions. What I need to find is [imath]f(x,y)[/imath]. Can anyone help me find a solution? Or are there any assumptions (other than [imath]f(x,y) = g(x)h(y)[/imath] that will allow me to solve this analytically? Is it a standard equation that I haven't come across? If no analytical solution exists can it be solved numerically - if so how would I go about doing that? | 625203 | Does the following have a solution for f(x,y)?
I have the following equations: \begin{equation} {1\over f(x,y)} {\partial f(x,y) \over \partial x} \alpha(x,y) + {1\over f(x,y)} {\partial f(x,y) \over \partial y} \beta(x,y) = \gamma(x,y) \end{equation} Where [imath]\alpha(x,y)[/imath], [imath]\beta(x,y)[/imath] and [imath]\gamma(x,y)[/imath] are exactly known functions. What I need to find is [imath]f(x,y)[/imath]. Can anyone help me find a solution? Or are there any assumptions (other than [imath]f(x,y) = g(x)h(y)[/imath] that will allow me to solve this analytically? Is it a standard equation that I haven't come across? Many thanks guys! |
598874 | Famous Problem of Euler
I believe this is a well known problem of Euler, but I was unable to find its solution anywhere. If [imath]2^{2^m}+1[/imath] is divisible by some prime [imath]p[/imath], then [imath]p\equiv 1\pmod{2^{m+2}}[/imath]. Thanks in advance. | 91413 | How to prove that if a prime divides a Fermat Number then [imath]p=k\cdot 2^{n+2}+1[/imath]?
If a prime [imath]p[/imath] divides a Fermat Number then [imath]p=k\cdot 2^{n+2}+1[/imath]? Does anyone know a simple/elementary proof? |
247385 | Prove that [imath] f(x) [/imath] has at least two real roots in [imath] (0,\pi) [/imath]
Let [imath] f [/imath] be a continuous function defined on [imath] [0,\pi] [/imath]. Suppose that [imath] \int_{0}^{\pi}f(x)\sin {x} dx=0, \int_{0}^{\pi}f(x)\cos {x} dx=0 [/imath] Prove that [imath] f(x) [/imath] has at least two real roots in [imath] (0,\pi) [/imath] | 2072150 | Proof that [imath]f(x)[/imath] has two roots over [imath](0, \pi)[/imath] at least! And other developing question.
Question 1: Let [imath]f(x)[/imath] be a continuous function over [imath][0, \pi][/imath]. Then the definite integration of [imath]f(x)\sin{x}[/imath] and [imath]f(x)\cos{x}[/imath] over [imath][0, \pi][/imath] are all 0. Prove that [imath]f(x)[/imath] has at least two zeros over [imath](0,\pi)[/imath]. It is easy to see that [imath]f(x)[/imath] one root over [imath](0, \pi)[/imath] But it is difficult to show [imath]f(x)[/imath] has at least two roots over [imath](0, \pi)[/imath]. Question 2: Let [imath]f(x)[/imath] be a continuous function over [imath][a, b][/imath]. Then the definite integration of [imath]f(x)[/imath] and [imath]f(x)g(x)[/imath] over [imath][a, b][/imath] are all 0. And g'(x)~=0(over [imath](a, b)[/imath]). Prove that [imath]f(x)[/imath] has at least two zeros over [imath](a,b)[/imath]. |
625748 | How does [imath]E(|X|)=\int_0^{\infty}P[|X|\ge x]dx[/imath]?
As the title states, how does [imath]E(|X|)=\int_0^{\infty}P[|X|\ge x]dx[/imath] ? The only assumption being that [imath]E(|X|)\le \infty[/imath] Mybe I can use the identity function in some way, since [imath]E[1_{X\ge x}]=P[X\ge x][/imath]? Thanks in advance! | 172841 | Explain why [imath]E(X) = \int_0^\infty (1-F_X (t)) \, dt[/imath] for every nonnegative random variable [imath]X[/imath]
Let [imath]X[/imath] be a non-negative random variable and [imath]F_{X}[/imath] the corresponding CDF. Show, [imath]E(X) = \int_0^\infty (1-F_X (t)) \, dt[/imath] when [imath]X[/imath] has : a) a discrete distribution, b) a continuous distribution. I assumed that for the case of a continuous distribution, since [imath]F_X (t) = \mathbb{P}(X\leq t)[/imath], then [imath]1-F_X (t) = 1- \mathbb{P}(X\leq t) = \mathbb{P}(X> t)[/imath]. Although how useful integrating that is, I really have no idea. |
625821 | Finding the improper integral [imath]\int^\infty_0\frac{1}{x^3+1}\,dx[/imath]
[imath]\int^\infty_0\frac{1}{x^3+1}\,dx[/imath] The answer is [imath]\frac{2\pi}{3\sqrt{3}}[/imath]. How can I evaluate this integral? | 556115 | Compute [imath]\int_0^\infty \frac{dx}{1+x^3}[/imath]
Problem Compute [imath]\displaystyle \int_0^\infty \frac{dx}{1+x^3}.[/imath] Solution I do partial fractions [imath]\frac{1}{x^3+1}= \frac{2-x}{3 \left( x^{2}-x+1 \right)}+\frac{1}{3 \left( x+1 \right)}.[/imath] But we could simplify the left one [imath]\frac{2-x}{3\left( x^{2}-x+1 \right)} = \frac{2}{3}\cdot \frac{1}{x^{2}-x+1}-\frac{x}{x^{2}-x+1}[/imath] From here, I do this see images. But I find the wrong primitive functions. Why am I wrong, and how do I find the correct one? |
625713 | Prove that there is a subgroup of index 2
Let [imath]G[/imath] be a group, and [imath]N \lhd G[/imath] so that [imath][G:N]=2014[/imath]. I need to prove that there is a subgroup of index 2. Indeed, since [imath]N \lhd G[/imath], [imath]o(G/N)=2014[/imath], but I couldn't find how to go on. | 225987 | Show group of order [imath]4n + 2[/imath] has a subgroup of index 2.
Let [imath]n[/imath] be a positive integer. Show that any group of order [imath]4n + 2[/imath] has a subgroup of index 2. (Hint: Use left regular representation and Cauchy's Theorem to get an odd permutation.) I can easily observe that [imath]\vert G \vert = 2(2n + 1)[/imath] so [imath]2 \mid \vert G \vert[/imath] and 2 is prime. We have satisfied the hypothesis of Cauchy's Theorem so we can say [imath]G[/imath] contains an element of order 2. This is where I am stuck I am confused about how the left regular representation relates to the group. So my understanding at this point is that every group is isomorphic to a subgroup of some symmetric group. My question: is the left regular representation [imath]\varphi : G \to S_G[/imath] an isomorphism? where [imath]G \cong S_G[/imath] or is [imath]S_G[/imath] the same thing as [imath]S_{\vert G \vert}[/imath] and [imath]\varphi[/imath] is only an injection? I'm using Dummit and Foote for definitions. I saw an argument online that said that since we have an element of order 2, there is a [imath]\sigma \in S_G[/imath] of order 2, but it is a product of [imath]2n + 1[/imath] disjoint 2-cycles. I don't understand how they could claim this and tried working it out on my own but didn't get there. -- They then went on to use the parity mapping [imath]\varepsilon : S_G \to \{\pm1\}[/imath] and since we have an odd permutation [imath]\sigma[/imath], we have [imath][S_G:\text{ker }\varepsilon] = 2[/imath]. I understood their computation but not how that directly shows that [imath]G[/imath] has a subgroup of order 2? unless [imath]G \cong S_G[/imath] because of how left regular representation is defined. (but again, I'm not understanding that concept very well yet.) So, to be clear about my questions: What is meant by left regular representation, is it an isomorphism or just an injection? and how would it be used here. If it is an isomorphism, the argument online starts to make more sense, but how can they say that since [imath]\sigma[/imath] is even, it is made up of [imath]2n + 1[/imath] disjoint transpositions? If you have a full, proof, I'd appreciate it, but good hints are just as good! Thanks you! |
175286 | Velleman exercise 1.5.7a
I've been trying to solve the exercise 7(a) of Velleman's "How To Prove It" and haven't succeeded. It asks the verification of the following equivalence: [imath] (P \to Q) \land (Q \to R) = (P \to R) \land ((P \leftrightarrow Q) \lor (R \leftrightarrow Q)) [/imath] While checking the website for help, I found a question posed by the user "yamad", who, despite his concern with a step futher on the resolution, came to this possible reduced form: [imath](\lnot P \lor Q) \land (\lnot Q \lor R) \land (\lnot P \lor R)[/imath] The problem is I couldn't even get to this step or any other simplfied form. I would appreciate if someone could provide me a hint. | 135623 | Finding minimal form -- Velleman exercise 1.5.7a
I am self-studying out of Velleman's "How to Prove It", and am trying to solve exercise 7, part a from section 1.5. The problem is to show that [imath] (P \to Q) \land (Q \to R) = (P \to R) \land ((P \leftrightarrow Q) \lor (R \leftrightarrow Q)) [/imath] By converting the right-hand side to the equivalent AND/OR statements I was able to simplify the expression to [imath](\lnot P \lor Q) \land (\lnot Q \lor R) \land (\lnot P \lor R)[/imath] which is equivalent to the left-hand side (confirmed through truth tables, Venn diagrams, and Wolfram Alpha), but I am having trouble showing it analytically. Clearly, the first part of this statement is identical to the left-hand side. But I cannot figure out the analytical steps to show that [imath]\lnot P \lor R[/imath] cancels out of the statement. I've tried distributing the statements into each other completely, but I keeping ending up back at the above expression. By converting back to conditional statements, the fact that [imath]\lnot P \lor R[/imath] is dispensable becomes intuitive to me. That is, it makes sense that [imath]P \to Q[/imath] and [imath]Q \to R[/imath] already implies that [imath]P \to R[/imath]. However, I want to be able to show it analytically based on the Boolean laws and I've hit a wall there. I'm sure I'm missing something simple, but it's driving me crazy. Any hints? |
626325 | Calculate the determinant of the [imath]2n \times 2n[/imath] matrix with entries equal to zero on the main diagonal, [imath]1[/imath] below and [imath]-1[/imath] above
Calculate the determinant of the [imath]2n \times 2n[/imath] matrix with entries equal to zero on the main diagonal, equal to [imath]1[/imath] below and equal to [imath]-1[/imath] above. I'll denote this matrix [imath]A_{2n}[/imath]. So for example you have [imath]A_{2} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ \end{bmatrix} [/imath] and [imath]A_{4}= \begin{bmatrix} 0 & -1 & -1 &-1 \\ 1 & 0 & -1 & -1\\ 1 & 1 & 0 & -1 \\ 1 & 1 & 1 & 0\\ \end{bmatrix} [/imath] From trying it out by hand I think you have [imath]\det(A_{2n})=1[/imath] for all odd [imath]n[/imath] , or [imath]=-1[/imath] for all even [imath]n[/imath]. Can anyone come up with some way to prove this? From doing these two examples my algorithm seems to be something like this: Switch the first and second rows (multiplies det by [imath]-1[/imath]) If the matrix is now upper triangular, calculate the determinant by product of main diagonal. If not, make a new pivot and clear that column below. Swap the third and fourth rows (multiplies det by [imath]-1[/imath]). Repeat from 2. | 58935 | Determinant of a special skew-symmetric matrix
Simple calculation show that: [imath] \begin{align} \det(A_2)=\begin{vmatrix} 0& 1 \\ -1& 0 \end{vmatrix}&=1\\ \det(A_4)=\begin{vmatrix} 0& 1 &1 &1 \\ -1& 0 &1&1\\ -1& -1& 0&1\\ -1& -1& -1&0 \end{vmatrix}&=1 \end{align} [/imath] Here is my question: Is it true that [imath]\det(A_{2n})=1[/imath] for all [imath]n\in{\mathbb Z_+}[/imath]? With MAPLE, I tried some large [imath]n[/imath]. And I guess it is true. But temporarily I have no idea how to show it. |
626101 | Can [imath][0,1][/imath] be partitioned into an uncountable union of uncountable sets?
I was thinking about this: [imath][0,1][/imath] can be partitioned into a countable union of uncountable sets. Write [imath][0,1]=(0,1]\cup \{0\}:[/imath] [imath](0,1]=\bigcup_{n=1}^{\infty}\Big(\frac{1}{n+1},\frac{1}{n}\Big][/imath] [imath][0,1][/imath] can be partitioned into an uncountable union of countable sets. Write [imath][0,1]=(0,1]\cup \{0\}:[/imath] [imath](0,1]=\bigcup_{x\,\in \,(1,2]}\Big\{\frac{x}{2^n}:\;n\in\mathbb{N}\Big\}[/imath] I cannot find a partitioning of [imath][0,1][/imath] and into an uncountable union of uncountable sets though. Does one exist? I was maybe thinking of taking a union of Cantor-like sets, but maybe I've missed an easy one. | 463601 | A partition of the unit interval into uncountably many dense uncountable subsets
The title says it all: Is there a partition of [imath][0,1][/imath] into uncountably many dense uncountable subsets ? |
626664 | Why, conceptually, is it that [imath]\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}[/imath]?
Why, conceptually, is it that [imath]\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}?[/imath] I know how to prove that this is true, but I don't understand conceptually why it makes sense. | 86093 | why is [imath]{n+1\choose k} = {n\choose k} + {n\choose k-1}[/imath]?
Can someone explain to me the proof of [imath]{n+1\choose k} = {n\choose k} + {n\choose k-1}[/imath]? |
626925 | If [imath]f(1) = i[/imath] what is [imath]f(i)[/imath] for a function with [imath]|f(z)| \leq K|z|[/imath] ? GATE 2011
Let [imath]f(z)[/imath] be an entire function s.t. [imath]|f(z)| < K|z|[/imath], [imath]\forall[/imath] [imath]z \in \mathbb{C}[/imath], for some [imath]K > 0[/imath]. If [imath]f(1) = i[/imath] what is the value of [imath]f(i)[/imath] and why? The value of [imath]f(i)[/imath] will be one of these four followings. 1 -1 i -i My intuition is [imath]i[/imath] i.e. 3. I am imagining some boundedness property like Liouville's theorem, but I do not know accurate result(s) and proof(s). I am more interested in why part. Please add sufficient reference. Thank you for your help. | 275140 | Let [imath]f (z) [/imath] be an entire function such that [imath]|f (z)|≤K|z|[/imath], [imath]∀z∈\mathbb{C}[/imath], for some [imath]K>0[/imath]. If [imath]f (1) =i[/imath], then[imath]f (i) [/imath] is
Let [imath]f (z) [/imath] be an entire function such that [imath]|f (z)|≤K|z|[/imath], [imath]∀z∈\mathbb{C}[/imath], for some [imath]K>0[/imath]. If [imath]f (1) =i[/imath], the value of [imath]f (i) [/imath] is (A) [imath] 1 [/imath] (B)[imath]-1[/imath] (C) [imath]i[/imath] (D) [imath]-i[/imath] how can I able to solve this problem?totally stuck. |
627177 | Proving that there are [imath]n-1[/imath] roots in [imath]a_1x^{b_1}+a_2x^{b_2}+...+a_nx^{b_n}=0 [/imath] on [imath](0,\infty)[/imath]
We know that: [imath]a_1,...,a_n\in \mathbb R , \forall a_i\neq0 \\ b_1,...,b_n\in \mathbb R : b_j\neq b_k : \forall j\neq k[/imath] Prove that there are [imath]n-1[/imath] roots in [imath](0,\infty)[/imath]: [imath]a_1x^{b_1}+a_2x^{b_2}+...+a_nx^{b_n}=0 [/imath] Using induction, for n=1 it's obvious that there are 0 roots. Suppose the statement is true for n-2 and prove for n-1: From here on I'm not sure how to continue, I noticed that [imath]x^{b_1}(a_1 + \ldots + a_n x^{b_n - b_1})=0[/imath] but at what point can it come into the induction process ? Also, how does the IVT is applied here ? | 623757 | Number of roots of differentiable function
I have two related problems that are causing me big trouble: Let [imath]\;a_1,...,a_n\in\Bbb R\;,\;\;a_i\neq 0\;\;\forall\;i\;[/imath] , and let [imath]\;b_1,...,b_n\in\Bbb R\;,\;\;b_i\neq b_j\;\;\forall\,i\neq j\;[/imath] . (1) Prove that the equation [imath]a_1x^{b_1}+\ldots+a_nx^{b_n}=0[/imath] has at most [imath]\;n-1\;[/imath] different zeros in [imath]\;(0,\infty)\;[/imath] (Hint: Use induction) (2) Prove that the equation [imath]a_1e^{b_1x}+\ldots+a_ne^{b_nx}=0[/imath] has at most [imath]\;n-1\;[/imath] different zeros in [imath]\;\Bbb R\;[/imath] . Now, the following is what I've done so far: for a differentiable function [imath]\;f\;[/imath] , I can prove that if [imath]\;f'(x)=0\;[/imath] has [imath]\;n-1\;[/imath] different zeros, then [imath]\;f(x)\;[/imath] has at most [imath]\;n\;[/imath] different zeros, using the mean value theorem, say. But I can't see how to use induction in this case: if I differentiate in the first problem, I get [imath]0=f'(x)=a_1b_1x^{b_1-1}+\ldots +a_nb_nx^{b_n-1}=\frac1x\sum_{i=1}^na_ib_ix^{b_i}[/imath] ...and now?! How does induction kick in here? I though perhap something like: if we suppose WLOG [imath]\;b_1<b_2<...<b_n\;[/imath] , then perhaps inducting on [imath]\;\lfloor b_n\rfloor\;[/imath], but it gets messy and blurry if [imath]\;b_n<0\;[/imath]... Any help, hints are very much appreciated. |
581112 | Quadratic Reciprocity
For any integer [imath]a[/imath] and any two primes [imath]p,q[/imath] with [imath](p,q)=1[/imath]. Prove that if [imath]p \equiv q[/imath] mod [imath]4a[/imath], then [imath](\frac{a}{p})=(\frac{a}{q})[/imath] I know I need to write [imath]a=(−1)^{e_0} 2^{e_2} p_1 p_2 \cdots p_r[/imath] with [imath]p_i[/imath] odd and possibly equal; and then compute the Legendre symbol and apply the reciprocity laws, I am just unfamiliar with this process and would appreciate any help. | 999544 | Congruence and Legendre symbol problem
Let [imath]p,q[/imath] different odd prime numbers and [imath]a[/imath] a positive odd number, Prove that : [imath]p \equiv - q\ \ \ \ (mod \ \ 4a )\Longrightarrow \Bigg(\frac{a}{p}\Bigg) =\Bigg(\frac{a}{q}\Bigg) [/imath] Where [imath]\Bigg(\frac{a}{p}\Bigg)[/imath] and [imath]\Bigg(\frac{a}{q}\Bigg)[/imath] represent the Legendre Symbol. |
627344 | Show that: [imath]\lim \limits_{n \rightarrow+\infty} \int_{0}^{1}{f(x^n)dx}=f(0)[/imath]
Supose that [imath]f\colon [0,1] \rightarrow \mathbb R[/imath] is a continuous function. Show that: [imath]\lim \limits_{n \rightarrow+\infty} \int_{0}^{1}{f(x^n)dx}=f(0)[/imath] I'm studying the Riemann Integral theory and I have not idea to how do this. I need some hint to start because I do not see how to apply the Riemann sum or other theory concept to be able to help me. | 571845 | proving [imath]\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)[/imath] when f is continous on [0,1]
[imath]\lim\limits_{n\to\infty} \int_{0}^{1} f(x^n)dx = f(0)[/imath] when f is continuous on [imath][0,1][/imath] I know it can be proved using bounded convergence theorem but, I wanna know proof using only basic properties of riemann integral and fundamental theorem of calculus and MVT for integrals ... Thank you. |
627624 | Ideal Generated by [imath]H\cup K[/imath]
If [imath]H[/imath] and [imath]K[/imath] are ideals of a ring [imath]R[/imath], what are the elements of [imath]\langle H\cup K\rangle[/imath]? What I am trying to ask is that how does its element look? | 626649 | Ideal Generated by the Union of Two Ideals
Let [imath]I[/imath] and [imath]J[/imath] be ideals of a ring [imath]R[/imath]. Prove that [imath]I+J[/imath] is an ideal of [imath]R[/imath] and that [imath]I+J=\langle I\cup J\rangle[/imath], the ideal of [imath]R[/imath] generated by [imath]I\cup J[/imath]. |
627885 | Periodic solution of different equation
I am trying the following exercise : Let [imath]f\in C^\infty(\mathbb{R}^2, \mathbb{R}^2)[/imath] and [imath]\forall x\in\mathbb{R}^2[/imath] [imath]f(kx)=k^2f(x)[/imath] for [imath]k\in\mathbb{R}[/imath] Show that any periodic solution of [imath]y'=f(y)[/imath] is constant. Thank you in advance | 627998 | Periodic solution of differential equation y′=f(y)
Let [imath]f∈C^∞(ℝ^2,ℝ^2)[/imath] and [imath]∀x∈ℝ^2[/imath] [imath]f(kx)=k^2f(x)[/imath] for [imath]k∈ℝ[/imath] Show that any periodic solution of [imath]y′=f(y)[/imath] is constant. My attempt : Let [imath]\lambda \in \mathbb{R}[/imath]. Let [imath]g[/imath] a periodic solution defined by [imath]g_{\lambda}(t) = \lambda g(\lambda t)[/imath]. With Cauchy-Lipshitz theorem [imath]g_{\lambda}[/imath] and [imath]g[/imath] do not intersect for [imath]\lambda \neq 1[/imath] and for [imath]\lambda >1[/imath], if [imath]g(0)>0[/imath] we have [imath]g_{\lambda} > g.[/imath] but I do not see how to continue. |
627969 | Induction without a base case
I am looking for an example where you have [imath]P(n)[/imath] implying [imath]P(n+1)[/imath]. However there is no base case. For which there is therefore no solution at all for the induction problem even though the inductive step itself works. | 587560 | What is the purpose of the first test in an inductive proof?
Learning about proof by induction, I take it that the first step is always something like "test if the proposition holds for [imath]n = \textrm{[the minimum value]}[/imath]" Like this: Prove that [imath]1+2+3+...+n = \frac{n(n + 1)}{2}[/imath] for all [imath]n \ge 1[/imath]. Test it for [imath]n = 1[/imath]: [imath]n = \frac{n(n + 1)}{2} \implies 1 = \frac{2}{2}\textrm{, which holds.}[/imath] * The rest of the proof goes here * So, I do it all the time (like a standard). But I never really thought about why. Why do I do such test? I can see that if the test does not hold, it cannot be proven by induction, I guess. But is there another reason we do this? |
622239 | How find this integral [imath]\;\int\frac{\sqrt{\ln{(x+\sqrt{x^2+1})}}}{1+x^2}dx[/imath]
Find the integral [imath]I=\int\dfrac{\sqrt{\ln{(x+\sqrt{x^2+1})}}}{1+x^2}dx[/imath] My try: let [imath]x=\tan{t}[/imath] [imath]I=\int\sqrt{\ln{(\tan{t}+\sec{t})}}dt[/imath] note [imath]\tan{t}+\sec{t}=\dfrac{\sin{t}}{\cos{t}}+\dfrac{1}{\cos{t}}=\dfrac{\sin{\dfrac{t}{2}}+\cos{\dfrac{t}{2}}}{\cos{\dfrac{t}{2}}-\sin{\dfrac{t}{2}}}=\tan{\left(\dfrac{t}{2}+\dfrac{\pi}{4}\right)}[/imath]then I can't,Thank you | 613674 | A question on integration
I want to compute the following integral: [imath]\raise 1ex{\Large\int} \frac{\sqrt{\ln(x+\sqrt{1+x^2}})}{1+x^2}\,dx[/imath] |
628636 | Direct summands in additive categories
Let [imath]A[/imath] be an additive category, with objects [imath]X[/imath] and [imath]Y[/imath]. If there exists direct sum decompositions [imath]X = Y\oplus Y'[/imath] and [imath]Y = X \oplus X'[/imath] in [imath]A[/imath], must [imath]X[/imath] and [imath]Y[/imath] be isomorphic? If one only assumes that there exist monomorphisms [imath]X\to Y[/imath] and [imath]Y\to X[/imath], the answers is No in the category of abelian groups, as explained in answers linked to in the comments. However, the question about direct summands is not resolved there. | 257650 | Analogue of the Cantor-Bernstein-Schroeder theorem for general algebraic structures
The Cantor-Bernstein-Schroeder theorem states that if there are two sets [imath]A[/imath] and [imath]B[/imath] such that there exist injective (alternatively, surjective, assuming choice I think) maps [imath]A \to B[/imath] and [imath]B \to A[/imath], then [imath]A[/imath] and [imath]B[/imath] are in bijection. It then seems natural to try to strengthen the result to other structures. If [imath]A[/imath] and [imath]B[/imath] are, say, groups (or rings, modules, etc) such that there are injective homomorphisms from each to the other, then [imath]A[/imath] and [imath]B[/imath] are isomorphic. I'm wondering if there are any results on this, or if there are known counterexamples. The theorem for sets alone is nontrivial, so I feel like any results about how it holds for other structures would be quite interesting. |
623809 | Express [imath] G_y[/imath] in terms of [imath]G_x[/imath].
A finite group [imath]G[/imath] acts on a finite set [imath]X[/imath], the action of [imath]g \in G[/imath] on [imath]x \in X[/imath] being denoted by [imath]gx[/imath]. For each [imath]x \in X[/imath] the stabilizer of [imath]x[/imath] is the subgroup [imath]G_x = \{g \in G : gx = x\}[/imath]. If [imath]x, y ∈ G[/imath] and if [imath]y = gx[/imath], then express [imath]G_y[/imath] in terms of [imath]G_x[/imath]. How can I able to solve this problem? I am totally stuck on it. | 88285 | Are orbits equal if and only if stabilizers are conjugate?
Let [imath]X[/imath] be a topological space and [imath]G[/imath] a group acting on [imath]X[/imath]. Do we have this property: [imath]\operatorname{orb}(x)=\operatorname{orb}(y)\iff\operatorname{stab}(x)\sim \operatorname{stab}(y)\qquad ?[/imath] where [imath]\operatorname{orb}(x)[/imath] is the orbit of [imath]x[/imath], [imath]\operatorname{stab}(x)=\{g\in G\mid gx=x\}[/imath], and the symbol [imath]\sim[/imath] means conjugate to. One way is obvious: if [imath]\operatorname{orb}(x)=\operatorname{orb}(y)[/imath], then [imath]x=gy[/imath] for some [imath]g\in G[/imath], so [imath]\operatorname{stab}(x)=\operatorname{stab}(gy)=g\operatorname{stab}(y)g^{-1}.[/imath] But the other way is not obvious to me: if [imath]\operatorname{stab}(x)\sim \operatorname{stab}(y)[/imath], then [imath]\exists k\in G[/imath] such that for all [imath]g\in \operatorname{stab}(x)[/imath], [imath] \exists h \in \operatorname{stab}(y)[/imath] such that [imath]g=khk^{-1}[/imath]. Now since [imath]gx=x[/imath] and [imath]hy=y[/imath], then [imath]khk^{-1}x=x[/imath] so [imath]hk^{-1}x=k^{-1}x[/imath] hence [imath]h\in \operatorname{stab}(k^{-1}x)[/imath], but I can't go any further... Thanks for your help. |
628583 | Primary decomposition of a monomial ideal
Can anyone give me an idea about the primary decomposition of the ideal [imath]I=(x^3y,xy^4)[/imath] of the ring [imath]R=k[x,y][/imath]? I am trying to connect the primary decomposition with the set Ass(R/I) which i have find before.. | 1793841 | Easy explanation on primary decomposition of ideals.
The primary decomposition of an ideal [imath](x^2, xy)[/imath] is [imath](x^2, xy) = (x) \cap (x, y)^2[/imath] which can be found on these notes. Could someone explain to me how this can be done? Edited: My question is not restricted to the example I gave [imath](x^2, xy)[/imath], one can use any other example to show me the process. |
609811 | Step-by-step proof of principal ideals.
Could someone go through a step-by-step proof of: Let [imath]\theta:R\to S[/imath] be a surjective ring homomorphism. Show that if every ideal of [imath]R[/imath] is principal, than every ideal of [imath]S[/imath] is also principal. I really just have the assumptions. Take an ideal [imath]I[/imath] in [imath]S[/imath] and show that it is principal. Could someone help please? | 608091 | If every ideal is principal show the same is true for the image of a homomorphism.
I was working on a worksheet for class and came across this problem: Let [imath]\varphi:R\to S[/imath] be an onto ring homomorphism with the property that every ideal of [imath]R[/imath] is principal. Show that the same must be true for [imath]S[/imath]. I honestly do not know how to do this one! |
628295 | How to evaluate [imath]\int_0^{\frac{\pi}{2}}\frac{dx}{1+\tan^{2013}x}[/imath]
How to evaluate the integral [imath]$$\int_0^{\frac{\pi}{2}}\frac{\mathrm dx}{1+\tan^{2013}x}$$[/imath] It is almost impossible to calculate the antiderivative of this function by hand. Are there any tricks? | 605673 | Integrate [imath]\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x[/imath]
Evaluate the integral [imath]\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x[/imath] |
628933 | chain A s.t. [imath]|X|<|A|\leq |P(X)|[/imath]
Can we prove that there exists at least one chain [imath]A[/imath] in P(X), where X is a non-empty set (finite or infinite), s.t. [imath] |X|<|A|\leq |P(X)|[/imath]? If you can't solve it, ideas/possible directions are welcome :) The two links mentioned above do not contain answers. | 90946 | Uncountable chains
[imath]P(\mathbb N)[/imath] = power set of [imath]\mathbb N[/imath]. [imath]A \subset P(\mathbb N)[/imath] is a chain if [imath]a,b \in A \implies[/imath] either [imath]a \subseteq b[/imath] or [imath] b \subseteq a[/imath] That is we have something like this: [imath]\ldots a \subseteq b \subseteq c \subseteq\ldots[/imath] where [imath]a,b,c \in A[/imath] are distinct. We can show easy enough that there is an uncountable chain - this is done by noting [imath]\mathbb N\sim\mathbb Q[/imath] then using Dedekind cuts in [imath]\mathbb Q[/imath] to define [imath]\mathbb R[/imath] we see that a family of (nearly arbitrary) cuts satisfy the condition. For instance the family [imath]L_r=\{q \in \mathbb Q : q < r \}[/imath] for [imath]r > 0[/imath] gives us the sets we need and obviously we can pick others. I tried doing this for [imath] \mathbb R[/imath] and don't seem to be getting anywhere. To be more specific, does there exist a chain in [imath]P(\mathbb R)[/imath] with cardinality [imath]2^ \mathbb R[/imath] ? Given a set of cardinality [imath]X[/imath] (necessarily non-finite), is there a chain in [imath]P(X)[/imath] of the same cardinalty of [imath]P(X)[/imath]? |
76469 | Characteristic polynomial of companion matrix
I have a matrix in companion form, [imath]A=\begin{pmatrix} 0 & \cdots & 0& -a_{0} \\ 1 & \cdots & 0 & -a_{1}\\ \vdots &\ddots & \vdots &\vdots \\ 0 &\cdots & 1 & -a_{n-1} \end{pmatrix}[/imath], [imath]A\in M_{n}[/imath]. I want to prove by induction that the characteristic polynomial [imath]p_{A}=t^{n}+a_{n-1}t^{n-1}+\cdots +a_{0}[/imath]. The part that is confusing me is if we assume this hold for [imath]A_{1}\in M_{n-1}[/imath] how do we transform, or "add" to [imath]A_{1}[/imath] to get the n by n matrix [imath]A\in M_{n}[/imath]? I hope this makes sense. | 10216 | The characteristic and minimal polynomial of a companion matrix
The companion matrix of a monic polynomial [imath]f \in F\left[x\right][/imath] in 1 variable [imath]x[/imath] over a field [imath]F[/imath] plays an important role in understanding the structure of finite dimensional [imath]F[x][/imath]-modules. It is an important fact that the characteristic polynomial and the minimal polynomial of [imath]C(f)[/imath] are both equal to [imath]f[/imath]. This can be seen quite easily by induction on the degree of [imath]f[/imath]. Does anyone know a different proof of this fact? I would love to see a graph theoretic proof or a non inductive algebraic proof, but I would be happy with anything that makes it seem like more than a coincidence! |
629097 | Adding members of the Cantor set to get [imath][0,2][/imath]
Let [imath]C[/imath] be the Cantor tertiary set. Show [imath]C+C \equiv \{x+y : x,y\in C\}=[0,2][/imath] My guess is that I should utilize the standard base-3 representation for [imath]x,y \in C[/imath]. It is immediate that [imath]C+C \subset [0,2][/imath]. I cannot show the other inclusion. Here's where I'm stuck: let [imath]\alpha \in [0,2][/imath] (we assume [imath]\alpha[/imath] has a base-3 representation). We must find an [imath]x,y \in C[/imath] such that [imath]x+y=\alpha[/imath]. My thought is to break up [imath]\alpha[/imath] as such: [imath]\alpha = \sum_{n=0}^{\infty}\frac{x_n}{3^n}+ \sum_{n=1}^{\infty}\frac{y_n}{3^n} \equiv \alpha_1 + \alpha_2[/imath] with [imath]x_n \in \{0,1\}[/imath] and [imath]y_n \in \{0,2\}[/imath]. Certainly, [imath]\alpha_2 \in C[/imath]. I also note that [imath]2*\alpha_1 \in C[/imath]. But I can't get any further insight from this. My other attempt was to approach the problem from a base-2 perspective, but I was not able to advance at all here. Any hint or bump in the right direction would be greatly appreciated! | 2448 | Measure of the Cantor set plus the Cantor set
The Sum of 2 sets of Measure zero might well be very large for example, the sum of [imath]x[/imath]-axis and [imath]y[/imath]-axis, is nothing but the whole plane. Similarly one can ask this question about Cantor sets: If [imath]C[/imath] is the cantor set, then what is the measure of [imath]C+C[/imath]? |
629203 | Criteria for local freeness of a module
Let [imath]M[/imath] be a finite type module over a local ring [imath]R[/imath]. If the minimum number of generators equals the maximum number of independent elements, is [imath]M[/imath] free? If not, do you have a counterexample? | 574653 | Infinite linear independent family in a finitely generated [imath]A[/imath]-module
So I'm stuck with this problem. Let [imath]A[/imath] be a nonzero commutative ring (with unit). I have several questions that are really close to each other. 1) Let [imath]M[/imath] be a finitely generated module over [imath]A[/imath]. Can we have [imath]\{x_n, n \in \mathbb{N}\} \subset M[/imath] a linear independent family in [imath]M[/imath] ? 2) Let [imath]M[/imath] be a module over [imath]A[/imath] generated by [imath]f_1, \ldots, f_n \in M[/imath]. Can we have [imath]x_1,\ldots, x_{n+1} \in M[/imath] a linear independent family ? (Obviously a negative answer to the latter implies a negative answer to the former.) 3) Let [imath]N[/imath] be a free module over [imath]A[/imath], and [imath]M[/imath] a module such that there exists an exact sequence [imath]0 \to N \to M[/imath] and another exact sequence [imath]N \to M \to 0[/imath]. Is [imath]M[/imath] a free module as well (is [imath]M \simeq N[/imath])? If not, can we add [imath]N[/imath] of finite rank to get a positive answer ? 4) Let [imath]M[/imath] be finitely generated by [imath]f_1, \ldots, f_n[/imath], and [imath]x_1, \ldots, x_n[/imath] be a linear independent family in [imath]M[/imath]. Is [imath]M[/imath] free? Thank you for your help! |
372045 | Why is Matrix Multiplication Not Defined Like This?
I'm sure everyone already thought about this at least one time. Why matrix multiplication is not defined the way showed below? [imath]\left( \begin{array}{ccc} a_{11} & a_{12} & \ldots \\ a_{21} & a_{22} & \ldots \\ \vdots & \vdots & \ddots \end{array} \right) \cdot \left( \begin{array}{ccc} b_{11} & b_{12} & \ldots \\ b_{21} & b_{22} & \ldots \\ \vdots & \vdots & \ddots \end{array} \right) = \left( \begin{array}{ccc} a_{11}\cdot b_{11} & a_{12}\cdot b_{12} & \ldots \\ a_{21}\cdot b_{21} & a_{22}\cdot b_{22} & \ldots \\ \vdots & \vdots & \ddots \end{array} \right) [/imath] I know this definition has it's limitations. The product only works with same order matrix and any matrix with some zero entry won't be invertible. But this definition is associative, commutative, has identity element, the distributive works, is simpler and more intuitive. I have no problem about the classic definition, but this definition also has good properties, why it is never used? | 185888 | why don't we define vector multiplication component-wise?
I was just wondering why we don't ever define multiplication of vectors as individual component multiplication. That is, why doesn't anybody ever define [imath]\langle a_1,b_1 \rangle \cdot \langle a_2,b_2 \rangle[/imath] to be [imath]\langle a_1a_2, b_1b_2 \rangle[/imath]? Is the resulting vector just not geometrically interesting? |
629495 | Intuition behind the residue at infinity
The residue at infinity is given by: [imath]\underset{z_0=\infty}{\operatorname{Res}}f(z)=\frac{1}{2\pi i}\int_{C_0} f(z)dz[/imath] Where [imath]f[/imath] is an analytic function except at finite number of singular points and [imath]C_0[/imath] is a closed countour so all singular points lie inside it. It can be proven that the residue at infinity can be computed calculating the residue at zero. [imath]\underset{z_0=\infty}{\operatorname{Res}}f(z)=\underset{z_0=0}{\operatorname{Res}}\frac{-1}{z^2}f\left(\frac{1}{z}\right)[/imath] The proof is just to expand [imath]-\frac{1}{z^2}f\left(\frac{1}{z}\right)[/imath] as a Laurent series and to see that the [imath]1/z[/imath] is the integral mentioned. I can see that we change [imath]f(z)[/imath] to [imath]f(1/z)[/imath] so the variable tends to infinity. But, is there any intutive reason of why we introduce the [imath]-1/z^2[/imath] factor? | 571510 | The residue at [imath]\infty[/imath]
I am stuck on the following problem : [imath]\,\,\,\,[/imath]*Problem*[imath]\quad[/imath]The residue of an entire function at [imath]\infty[/imath] is [imath]0[/imath]. Solution: True. This follows from the definition of the residue at [imath]\infty[/imath] together with the Cauchy-Goursat Theorem. Another way to see this is to take the Taylor expansion for [imath]f[/imath] at [imath]0[/imath] [imath]\displaystyle f(z)=\sum_{n=0}^\infty a_nz^n[/imath] and replace [imath]z[/imath] with [imath]1/z[/imath] to get the Laurent expansion for [imath]f(1/z)[/imath] at [imath]0[/imath]: [imath]\displaystyle f(z)=\sum_{n=-\infty}^0a_{-n}z^n.[/imath] Then we have that the residue at [imath]\infty[/imath] is given by the negative coefficient of the [imath]1/z[/imath] term of [imath] \dfrac1{z^2}f\left(\dfrac1z\right)=\sum_{n=-\infty}^{-2}a_{-n-2}z^n, [/imath] which is clearly [imath]0[/imath]. I am having trouble to understand the last few lines (in italic) in the given solution. Can someone give lucid explanation? Thanks and regards to all. |
124888 | Are the eigenvalues of [imath]AB[/imath] equal to the eigenvalues of [imath]BA[/imath]? (Citation needed!)
First of all, am I being crazy in thinking that if [imath]\lambda[/imath] is an eigenvalue of [imath]AB[/imath], where [imath]A[/imath] and [imath]B[/imath] are both [imath]N \times N[/imath] matrices (not necessarily invertible), then [imath]\lambda[/imath] is also an eigenvalue of [imath]BA[/imath]? If it's not true, then under what conditions is it true or not true? If it is true, can anyone point me to a citation? I couldn't find it in a quick perusal of Horn & Johnson. I have seen a couple proofs that the characteristic polynomial of [imath]AB[/imath] is equal to the characteristic polynomial of [imath]BA[/imath], but none with any citations. A trivial proof would be OK, but a citation is better. | 1173614 | Eigenvalues of [imath]AB[/imath] and [imath]BA[/imath] [imath]{}\qquad{}[/imath]
Are the eigenvalues of the matrices [imath]AB[/imath] and [imath]BA[/imath] identical? If yes, why? From the examples that I have tried I think they are identical but I just can't come up with a formal proof for this. |
629427 | [imath]C[/imath] Hermitian, show [imath]\text{Tr}(C)=0 \iff \exists P,Q \text{ Hermitian s.t.} \,\, PQ-QP=iC[/imath]
Let [imath]C[/imath] be an [imath]n\times n[/imath] Hermitian matrix. Show that [imath]\text{Tr}(C)=0 \iff \exists P,Q \text{ Hermitian s.t.} \,\, PQ-QP=iC.[/imath] Ideas: The right-to-left direction I have no problem with. For the left-to-right direction, I'm going to start with the fact that [imath]\text{Tr}(C)=0 \implies \exists A,B \in M_n \text{ s.t. } C = AB-BA.[/imath] I'm having trouble showing we can choose [imath]A,B[/imath] Hermitian. One method would be to try to show [imath]A,B[/imath] must be Hermitian, but that seems too strong. Along this road... [imath]\text{Tr}(iC)=0[/imath], so we can write [imath]iC=AB-BA[/imath]. [imath]iC[/imath] is diagonalizable with all imaginary eigenvalues, so it is skew-Hermitian. Thus [imath]BA-AB=-(AB-BA)=(AB-BA)^*\\=(AB)^* - (BA)^* = B^*A^*-A^*B^*.[/imath] But this hasn't gotten me anywhere. | 251678 | Is every skew-adjoint matrix a commutator of two self-adjoint matrices
I'm looking to solve some matrix equations. One of the equations involves a commutator, so my question is as follows: let [imath]A[/imath] be a skew-self-adjoint, traceless matrix, does the equation [imath][X,Y] = A[/imath] always have a self-adjoint solution? For every size of matrices. I hope that this is a well-known fact. Perhaps it is related to the fact that the traceless skew-adjoint matrices are [imath]\mathfrak{s}\mathfrak{u}_n[/imath]. |
629706 | 7) Prove that [imath]2n-3 \leq 2^{n-2}[/imath] for all [imath]n \geq 5[/imath] by mathematical induction
Prove that [imath]2n-3\leq 2^{n-2}[/imath] , for all [imath]n \geq 5[/imath] by mathematical induction I have to prove by mathematical induction that: [imath]2n-3\leq 2^{n-2}[/imath] , for all [imath]n \geq 5[/imath] | 629686 | Prove that [imath]2n-3\leq 2^{n-2}[/imath] , for all [imath]n \geq 5[/imath] by mathematical induction
Prove that [imath]2n-3\leq 2^{n-2}[/imath] , for all [imath]n \geq 5[/imath] by mathematical induction I have to prove by mathematical induction that: [imath]2n-3\leq 2^{n-2}[/imath] , for all [imath]n \geq 5[/imath] Thank you for the Review. |
492821 | On a Putnam's 2009 problem
Find all even natural numbers [imath]n[/imath] such that the following is true: There is a non-constant function [imath]f : \Bbb{R}^2 \longrightarrow \Bbb{Z}_2[/imath] such that for any regular [imath]n[/imath]-gon [imath]A_1...A_n[/imath], [imath]f(A_1) + \cdots + f(A_n) =0[/imath]. NOTE: This is the unsolved part of THIS problem. I chose to state it in a separate question first to get more attention and second since some people may want to put a bounty on this part. | 489318 | On the problem [imath]1[/imath] of Putnam [imath]2009[/imath]
(This is adapted from problem [imath]1[/imath] of Putnam [imath]2009[/imath]) Find all values of [imath]n[/imath] such that the following is true: There is a non-constant function [imath]f : \Bbb{R}^2 \longrightarrow \Bbb{Z}_2[/imath] such that for any regular [imath]n[/imath]-gon [imath]A_1...A_n[/imath], [imath]f(A_1) + \cdots + f(A_n) =0[/imath]. |
629760 | How prove this [imath]4A^5+2A^3+A=7I[/imath]
Let [imath]A_{n\times n}[/imath] is Hermitian matrix,and [imath]A_{n\times n}\neq I_{n\times n}[/imath],where [imath]I_{n\times n}[/imath] is Identity matrix, prove or disprove [imath]4A^5+2A^3+A=7I[/imath] my try: if such this condition: then matrix [imath]A[/imath] Characteristic polynomial [imath]4\lambda^5+2\lambda^3+\lambda=7[/imath] [imath](\lambda-1)(4\lambda^4+4\lambda^3+4\lambda^2+4\lambda+7)=0[/imath] then I can't since Hermitian matrix :http://en.wikipedia.org/wiki/Hermitian_matrix | 628336 | Hermitian matrix such that [imath]4M^5+2M^3+M=7I_n[/imath]
[imath]n[/imath] is a positive integer. Besides the identity matrix [imath]I_n[/imath], does there exist other [imath]n\times n[/imath] Hermitian matrix [imath]M[/imath], such that the following equality [imath]4M^5+2M^3+M=7I_n [/imath] hold? I try this: Since [imath]4M^5+2M^3+M=7I_n[/imath], then [imath](M-I_n)(4M^4+4M^3+6M^2+6M+7I_n)=0[/imath] but, What should I do next? Thanks! |
629887 | How [imath]x^2[/imath] increases by [imath]x+1/x[/imath]?
I was going through one of the topic "Introduction to Formal proof". In one example while explaining "Hypothesis" and "conclusion" got confused. The example is as follows: If [imath]x\geq4[/imath], then [imath]2^x \geq x^2[/imath]. While deriving conclusion article said "As [imath]x[/imath] grows larger than [imath]4[/imath], LHS [imath]2^x[/imath] doubles as [imath]x[/imath] increases by [imath]1[/imath] and RHS grows by ratio [imath]x+1/x[/imath]". Thanks in advance. | 629867 | How [imath]x^2[/imath] increases by [imath]x+\frac{1}{x}[/imath]?
I was going through one of the topic "Introduction to Formal proof".In one example while explaining "Hypothesis" and "conclusion" got confused. The example is as follows: If [imath]x\geq 4[/imath] then [imath]2^x \geq x^2[/imath]. While deriving conclusion article said "As [imath]x[/imath] grows larger than [imath]4[/imath], LHS [imath]2^x[/imath] doubles as [imath]x[/imath] increases by [imath]1[/imath] and RHS grows by ratio [imath]x+\frac{1}{x}[/imath]". Thanks in advance |
439412 | A fixed point theorem
Let [imath]\emptyset \not = X\subseteq \Bbb{R}^n[/imath] be convex and compact and let [imath]\cal{A}[/imath] be a family of affine maps from [imath]\Bbb{R}^n[/imath] into [imath]\Bbb{R}^n[/imath] such that [imath]X[/imath] is invariant under each element of [imath]\cal A[/imath] and [imath]f\circ g = g\circ f[/imath] for any [imath]f, g \in \cal A[/imath]. I am trying to show that elements of [imath]\cal A[/imath] have a common fixed point which is in [imath]X[/imath]. Any idea or references would be helpful. | 458162 | A weak version of Markov-Kakutani fixed point theorem
Let [imath]\emptyset \not = X\subseteq \Bbb{R}^n[/imath] be convex and compact and let [imath]\cal{A}[/imath] be a commuting family of affine maps from [imath]\Bbb{R}^n[/imath] into [imath]\Bbb{R}^n[/imath] such that [imath]X[/imath] is invariant under each element of [imath]\cal A[/imath]. How can I show that [imath]\cal{A}[/imath] has a common fixed point. Remark I know this is a consequence of some stronger theorems such as Markov-Kakutani fixed point theorem. But I have no access to its proof. Besides I am looking for a much elementary proof (which I think must exists for this special case). |
629836 | Ratio of coefficients for Laurent series expansions
Let [imath]f[/imath] be analytic in the disk [imath]D(0,2)[/imath] except for a pole of order [imath]1[/imath] at [imath]z=1[/imath], and let [imath]f(z)=\sum_{k=0}^\infty a_k z^k[/imath] be the series expansion for [imath]f[/imath] in the disk [imath]D(0,1)[/imath]. Prove that [imath]\lim_{k \to \infty}\frac{a_{k+1}}{a_k}=1.[/imath] Intuitively, I want to say that the information given yields a Laurent series representation on [imath]D(0,2)-\{1\}[/imath]:[imath]f(z)=\sum_{k=-1}^\infty b_k (z-1)^k[/imath] Expanding instead around [imath]z=0[/imath], we have an issue of a non-removable singularity at [imath]z=1[/imath] with analyticity everywhere else. Thus, the distance from [imath]z=0[/imath] to [imath]z=1[/imath] (the nearest singularity from the point of analyticity) is 1, so the result must follow by the Ratio Test as applied to the radius of convergence. My solution has a feel of lack of rigor, however, and I don't know how to improve on its statement (or whether my approach is even valid). | 243040 | function with isolated singularity on the unit circle and coefficients of its taylor expansion
Let [imath]f[/imath] be a function holomorphic in an open set containing the closed unit disc [imath]D(0,1)[/imath], except at the point [imath]z_0[/imath] with [imath]|z_0|=1[/imath], where [imath]f[/imath] has an isolated singularity. If [imath]a_n[/imath] are the coefficients of the Taylor expansion of [imath]f[/imath] centered at the origin, show that [imath]\displaystyle\lim_{n \to\infty}\frac{a_n}{a_{n+1}}=z_0[/imath] I considered first the case: [imath]z_0[/imath]=simple pole. Then i can write the Laurent series for [imath]f[/imath] at [imath]z_0[/imath] [imath]f(z)=\frac{c}{z-z_0}+c_0+c_1(z-z_0)+\ldots[/imath] I put [imath]g(z)=f(z)-\frac{c}{z-z_0}[/imath] Then [imath]g[/imath] is holomorphic in an open set containing the closed unit disc [imath]\overline{D(0,1)}[/imath] and we have [imath]g(z)=\sum a_n z^n+\frac{c}{z_0}\sum(\frac{z}{z_0})^n=\sum(a_n+\frac{c}{z_0^{n+1}})z^n[/imath] The power series of [imath]g[/imath] has radius of convergence [imath]>1[/imath], hence i can substitute 1 in RHS of last equation, and i get a numeric convergent series. The necessary condition for convergence says that its general term has to be infinitesimal [imath]\displaystyle\lim_{n\to\infty}(a_n+\frac{c}{z_0^{n+1}})=0[/imath] from which follows [imath]\frac{a_n}{a_{n+1}}\rightarrow z_0[/imath] the question is:how can i adapt this argument to deal with cases where z_0 is not a simple pole? |
630042 | Are [imath]\mathbb{Z} \times \mathbb{Z}[/imath] and [imath]\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}[/imath] isomorphic as groups?
To be clear, by [imath]\mathbb Z[/imath] I mean the group [imath](\mathbb Z, +)[/imath]. My intuitive answer would be no, but I haven't been able to find a proof for it. The basic invariants I know of (e.g. order of elements) are not useful here and my knowledge of infinite abelian groups is very limited. In general, is it true that [imath]\mathbb Z^n \cong \mathbb Z^m \iff n=m[/imath]? | 624002 | Determine whether [imath]\mathbb{Z}\times \mathbb{Z}[/imath] and [imath]\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}[/imath] are isomorphic groups or not.
Determine whether [imath]\mathbb{Z}\times \mathbb{Z}[/imath] and [imath]\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}[/imath] are isomorphic groups or not. pf) Suppose that these are isomorphic. Note that [imath]\mathbb{Z}\times \mathbb{Z}[/imath] is a subgroup of [imath]\mathbb{Z}\times \mathbb{Z}[/imath] and [imath]\mathbb{Z}\times \mathbb{Z}\times\left \{ 0 \right \}[/imath] is a subgroup of [imath]\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}[/imath]. Since [imath]\mathbb{Z}\times \mathbb{Z}[/imath] and [imath]\mathbb{Z}\times \mathbb{Z}\times\left \{ 0 \right \}[/imath] are isomorphic, [imath]\mathbb{Z}\times \mathbb{Z}/\mathbb{Z}\times \mathbb{Z}[/imath] and [imath]\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}/\mathbb{Z}\times \mathbb{Z}\times \left \{ 0 \right \}[/imath] are isomorphic. But the first one is isomorphic to the trivial group and the second one is isomorphic to [imath]\mathbb{Z}[/imath]. It is a contradiction. Is my proof right? If not, is there another proof? |
629950 | Why [imath]I=\left\{p(x)\in \mathbb{Z}\left[X\right]:2\mid p(0)\right\}[/imath] is not a principal ideal?
I saw this question but I still do not understand: What is the difference between ideal and principal ideal? At my homework I had to prove to things about [imath]I=\left\{p(x)\in \mathbb{Z}\left[X\right]:2\mid p(0)\right\}[/imath]: 1. That [imath]I[/imath] is an ideal. I prove it... 2. That [imath]I[/imath] is not a principal ideal. Now how can I prove 2.? Why [imath]I[/imath] isn't a principal ideal? Thank you! | 36169 | Show that [imath]\langle 2,x \rangle[/imath] is not a principal ideal in [imath]\mathbb Z [x][/imath]
Hi I don't know how to show that [imath]\langle 2,x \rangle[/imath] is not principal and the definition of a principal ideal is unclear to me. I need help on this, please. The ring that I am talking about is [imath]\mathbb{Z}[x][/imath] so [imath]\langle 2,x \rangle[/imath] refers to [imath]2g(x) + xf(x)[/imath] where [imath]g(x)[/imath], [imath]f(x)[/imath] belongs to [imath]\mathbb{Z}[x][/imath]. |
630490 | How does the sum of the series “[imath]1 + 2 + 3 + 4 + 5 + 6\ldots[/imath]” to infinity = “[imath]-1/12[/imath]”?
(I was requested to edit the question to explain why it is different that a proposed duplicate question. This seems counterproductive to do here, inside the question it self, but that is what I have been asked by the site and moderators. There is no way for me to vote against their votes. So, here I go: Please stop voting this as a duplicate so quickly, which will eventually lead to this question being closed off. Yes, the other question linked to asks the same math, but any newcomer to the problem who was exposed to it via physics, as I was, will prefer this question instead of the one that is purely mathematically. I beg the moderators to not be pedantic on this one. This question spills into physics, which is why I did the cross post to the physics forum as well.) How does the sum of the series “1 + 2 + 3 + 4 + 5 + 6…” to infinity = “-1/12”, in the context of physics? I heard Lawrence Krauss say this once during a debate with Hamza Tzortzis (http://youtu.be/uSwJuOPG4FI). I found a transcript of another debate between Krauss and William Lane Craig which has the same sum. Here is the paragraph in full: Let’s go to some of the things Dr. Craig talked about. In fact, the existence of infinity, which he talked about which is self-contradictory, is not self-contradictory at all. Mathematicians know precisely how to deal with infinity; so do physicists. We rely on infinities. In fact, there’s a field of mathematics called “Complex Variables” which is the basis of much of modern physics, from electro-magnetism to quantum mechanics and beyond, where in fact we learn to deal with infinity; without the infinities we couldn’t do the physics. We know how to sum infinite series because we can do complex analysis. Mathematicians have taught us how. It’s strange and very unappetizing, and in fact you can sum things that look ridiculous. For example, if you sum the series, “1 + 2 + 3 + 4 + 5 + 6…” to infinity, what’s the answer? “-1/12.” You don’t like it? Too bad! The mathematics is consistent if we assign that. The world is the way it is whether we like it or not. -- Lawrence Krauss, debating William Lane Craig, March 30, 2011 Source: http://www.reasonablefaith.org/the-craig-krauss-debate-at-north-carolina-state-university CROSS POST: I'm not sure if I should post this in mathematics or physics, so I posted it in both. Cross post: https://physics.stackexchange.com/questions/92739/how-does-the-sum-of-the-series-1-2-3-4-5-6-to-infinity-1-12 EDIT: I did not mean to begin a debate on why Krauss said this. I only wished to understand this interesting math. He was likely trying to showcase Craig's lack of understanding of mathematics or logic or physics or something. Whatever his purpose can be determined from the context of the full script that I linked to above. Anyone who is interested, please do. Please do not judge him out of context. Since I have watched one of these debates, I understand the context and do not hold the lack of a full breakdown as being ignorant. Keep in mind the debate I heard this in was different from the debate above. | 632750 | How to prove that the numeric series [imath]S := \sum_{n=0}^{\infty} x^n=\frac{1}{1-x}\text{ for any } x<1[/imath]
So... [imath]S := \sum_{n=0}^{\infty} x^n=\frac{1}{1-x}\text{ for any } x<1[/imath] I just found out, and it tested it myself, that if this is true, then: [imath]S_1 := \sum_{n=1}^{\infty} n = -\frac{1}{12}[/imath] Why? Because if the [imath]S[/imath] equality is true, then we derive [imath]S[/imath], and use [imath]x=-1[/imath], then [imath]S_2 := 1-2+3-4+5-... = \frac{1}{4}[/imath], and then [imath]S_1-S_2 = 4\cdot S_1[/imath], and [imath]S_1 = -\frac{1}{12}[/imath] which doesn't make any sense to me. I'm trying to see that everything is correct in this reasoning. Though I could test [imath]S[/imath], because it is so for [imath]x=\frac{1}{2}[/imath], for [imath]x=0[/imath], for [imath]x=-\frac{1}{2}[/imath] (apparently), I can't test it for [imath]-1[/imath] or smaller numbers. So how do I know that [imath]S = \sum_{n=1}^{\infty}x^n=\frac{1}{1-x}[/imath] for any [imath]x<1[/imath]? For my whole adult life I've known [imath]S_1[/imath] to be a divergent sequence. Now it converges? How? |
630221 | second derivative question
Let [imath]f:I\to\mathbb{R}[/imath], [imath]f''(x)=a[/imath] for all [imath]x \in I[/imath]. Show the existance of [imath]b,c \in \mathbb{R}[/imath] such that [imath]f(x)= {a \over 2}x^2 + bx + c [/imath]. So, it's actually just taking the derivative twice. Isn't it? [imath]f'(x) = ax + b[/imath] [imath]f''(x) = a[/imath] Show the statement isn't necessarily true if [imath]f:A \to \mathbb{R}, f''(x)=a[/imath] for all [imath]x \in A[/imath] when [imath]A[/imath] isn't an interval. Can you direct me? I suspect I should give an example such that the derivative doesn't exist. Right? | 626987 | Proving and disproving [imath]\exists b,c\in \mathbb R[/imath] such that [imath]f(x)=\frac a2x^2+bx+c [/imath]
Let [imath]f:I\to\mathbb R[/imath] where [imath]I[/imath] is an interval, [imath]f''(x)=a \ \ \forall x\in I[/imath]. Prove that there exsits such numbers [imath]b,c\in \mathbb R[/imath] such that: [imath]f(x)=\frac a2x^2+bx+c ,\ \forall x\in I[/imath]. Show that the above statement is not necessarily true if [imath]f:A\to\mathbb R[/imath] and [imath]f''(x)=a \ \ \forall x\in A[/imath] where [imath]A[/imath] is not an interval. I'm not sure I know how to prove 1, it's probably isn't enough to just derive [imath]f(x)[/imath] twice and we can't integrate... About 2. it means that the function isn't continuous but we see that it's still derivable so how can I find the relation between [imath]b,c[/imath] and [imath]f[/imath] ? |
630598 | Why is Euler's totient function equal to (p-1)(q-1) when N=pq and p and q are prime?
Why is Euler's totient function equal to [imath](p-1)(q-1)[/imath] when [imath]N=pq[/imath] and [imath]p[/imath] and [imath]q[/imath] are prime? I had my own proof for it but I really don't like it (it feels not intuitive at all because it requires factoring!) and I feel there must be a more direct way to do it by a different counting argument. If anyone has one it would be greatly appreciated! My "proof": Recall: [imath]\phi(N) = [/imath] the number of elements that are relatively prime to [imath]n[/imath] (i.e. have a mod inverse) [imath]N = pq[/imath] where [imath]p[/imath] and [imath]q[/imath] are prime. Let's count the number of elements between [imath]1[/imath] and [imath]N-1[/imath] that are NOT relatively prime to [imath]p[/imath] and [imath]1[/imath]. Those elements must have at least [imath]p[/imath] or [imath]q[/imath] as one of its factors. So let include all number that have [imath]p[/imath] as a factor and [imath]q[/imath] as a factor and are in the given range. Thus: [imath]0p, 1p, 2p, ..., kp, ..., p(q-1)[/imath] and [imath]0q,1q,2q,...,jp,...,q(p-1)[/imath] That gives a count of: [imath]N - p - q + 1 = pq - p - q + 1 = (p -1)(q-1)[/imath] QED as required. The reason I am not happy with this argument is that it seems to me there should be a clear intuitive way of doing it without having to factor or substitute the definition of N. This seems clear because multiplying [imath](p-1)[/imath] by [imath](q-1)[/imath] seems to be a very cute formula and it I would be surprise that its a coincidence because [imath](p-1)[/imath] and [imath](q-1)[/imath] also have clear interpretations. | 629949 | Why is Euler's totient function equal to [imath](p-1)(q-1)[/imath] when [imath]N=pq[/imath] and [imath]p[/imath] and [imath]q[/imath] are prime in a clean intuitive way?
I had my own proof for it but I really don't like it (it feels not intuitive at all because it requires factoring!) and I feel there must be a more direct way to do it by a different counting argument. If anyone has one it would be greatly appreciated! My "proof": Recall: [imath]\phi(N) = [/imath] the number of elements that are relatively prime to [imath]N[/imath] (i.e. have a mod inverse) [imath]N = pq[/imath] where [imath]p[/imath] and [imath]q[/imath] are prime. Let's count the number of elements between [imath]1[/imath] to [imath]N-1[/imath] that are NOT relatively prime to [imath]p[/imath] and [imath]q[/imath]. Those elements must have at least [imath]p[/imath] or [imath]q[/imath] as one of its factors. So let include all number that have [imath]p[/imath] as a factor and [imath]q[/imath] as a factor and are in the given range. thus: [imath]0p, 1p, 2p, ..., kp, ..., p(q-1)[/imath] and [imath]0q,1q,2q,...,jq,...,q(p-1)[/imath] That gives a count of: [imath]N - p - q + 1 = pq - p - q + 1 = (p -1)(q-1)[/imath] QED as required. The reason I am not happy with this argument is that it seems to me there should be a clear intuitive way of doing it without having to factor or substitute the definition of [imath]N[/imath]. This seems clear because multiplying [imath](p-1)[/imath] by [imath](q-1)[/imath] seems to be a very cute formula and it I would be surprise that its a coincidence because [imath](p-1)[/imath] and [imath](q-1)[/imath] also have clear interpretations. I am looking for a more intuitive clear reasoning, because it seems evident from the form of the formula that it should exist. |
631054 | Matrix diagonalizable or not
Let [imath]A[/imath] is in [imath]M_3(\mathbb R^3)[/imath] which is not a diagonal matrix. Pick out the cases when [imath]A [/imath] is diagonalizable over [imath]\mathbb R[/imath]: a. when [imath]A^2=A[/imath]; b. when [imath](A-3I)^2=0[/imath]; c. when [imath]A^2+I=0[/imath]. My attempt is if [imath]A[/imath] is diagonalizable then there is some invertible [imath]P[/imath] s.t. [imath]PAP^{-1}[/imath]=[imath]D[/imath]. Then case a. gives [imath]D^2[/imath]=[imath]D[/imath]. then [imath]D[/imath] is either [imath]0[/imath] or [imath]I[/imath]. Which gives [imath]A[/imath]=[imath]0[/imath] or [imath]I[/imath]..a contradiction to the fact [imath]A[/imath] is not diagonal. But I am not sure about my approach. Similarly I arrive contradiction for other cases. Please help. | 615666 | Diagonalization and eigenvalues
Let [imath]A[/imath] [imath]\in M_3(\mathbb R)[/imath] which is not a diagonal matrix. Pick out the cases when [imath]A[/imath] is diagonalizable over [imath]\mathbb R[/imath]: a. when [imath]A^2 = A[/imath]; b. when [imath](A - 3I)^2 = 0[/imath]; c. when [imath]A^2 + I = 0[/imath]. I could eliminate c. by using the equation [imath]\lambda^2+1[/imath] and showing if a matrix has to satisfy this then it has to be diagonal. I am also in doubt whether all the eigen values of [imath]A[/imath] satisfy this or not. |
631164 | Find a function such [imath]f(f(x))=-x[/imath]
I was looking for an answer to this question for some days now: Find a function f such that [imath]f(f(x))=-x[/imath] Please help, thank you! | 103290 | Is there a real-valued function [imath]f[/imath] such that [imath]f(f(x)) = -x[/imath]?
Is there a function [imath]f\colon \mathbb{R} \to\mathbb{R} [/imath] such that [imath] f(f(x)) = -x[/imath] ? |
631205 | Is it possible to prove that [imath]\lim_{x\to 0} {ln(x)x} = 0[/imath] without L'Hospital's rule?
Could somebody give me the answer: Is it possible to prove that [imath]\lim_{x\to 0} {\ln(x)x} = 0[/imath] without L'Hospital's rule? | 522973 | [imath]\lim_{x\to0^{+}} x \ln x[/imath] without l'Hopital's rule
I have a midterm coming up and on the past exams the hard question(s) usually involve some form of [imath]\lim_{x\to0^{+}} x \ln x[/imath]. However, we're not allowed to use l'Hopital's rule, on this year's exam anyways. So how can I evaluate said limit without l'Hopital's rule? I got somewhere with another approach, don't know if it's useful: [imath]\lim_{x\to0^{+}} x \ln x = \lim_{x\to0^{+}} x^2 \ln (x^2) = L[/imath] [imath]= (\lim_{x\to0^{+}} 2x)(\lim_{x\to0^{+}} x \ln x)[/imath] [imath]= 0 * L[/imath] Then I just need to prove that L is finite/exists (which means it must be 0) |
631204 | Evaluating a series with some given formula
I have a formula for the power series corresponding to the function [imath]\frac{z^{3k}}{(3k)!}[/imath] and I need to evaluate a new series with it but I can't see how to manipulate it even though I've had some ideas... Here it is : the series [imath]\frac{27^k}{(3k+1)!}.[/imath] I've tried to change it into [imath]\frac{3^{3k}}{(3k+1)!}[/imath] and then make the exponent of 3 3k+1 rather than 3k but it gives an expression that I can't use. I've had the idea of writing equalities with sums for multiples of 3, multiples of 3 substracting 1 and adding 1,... but it doesn't seem efficient... Any hint would be of great help, thank you! | 610526 | Infinite Series [imath]\sum\limits_{n=1}^\infty\frac{x^{3n}}{(3n-1)!}[/imath]
How can we prove that? [imath]\sum_{n=1}^\infty\frac{x^{3n}}{(3n-1)!}=\frac{1}{3}e^{\frac{-x}{2}}x\left(e^{\frac{3x}{2}}-2\sin\left(\frac{\pi+3\sqrt{3}x}{6}\right)\right).[/imath] I think if we write the taylor expansion of [imath]\sin(u)[/imath] and [imath]e^u[/imath], we can arrive from RHS to LHS, but I am looking for a way to prove it from LHS. |
511453 | Showing the image of a subgroup is a subgroup.
Let [imath]G[/imath] and [imath]G'[/imath] be groups. Let [imath]\phi: G \to G'[/imath] be a homomorphism. Let [imath]H[/imath] be a group of [imath]G[/imath] and let [imath]H'=\{\phi(h): h\in H\}[/imath]. Want to show that [imath]H'[/imath] is a subgroup. Proof: (i) [imath]H'[/imath] has identity since [imath]\phi(e_G)\in H'[/imath]. This is because [imath]\phi(e_G)=e_{G'}[/imath]. (ii) [imath]H'[/imath] is closed because of the following: Let [imath]h_{1}^{'}, h_{2}^{'} \in H'[/imath]. Then there exists an [imath]h_1,h_2[/imath] in [imath]H[/imath] such that [imath]h_1^{'}=\phi(h_1)[/imath] and [imath]h_2=\phi(h_{2}^{'})[/imath]. Then [imath]h_1^{'}h_2^{'}=\phi(h_1)\phi(h_2)=\phi(h_1h_2)[/imath]. (iii) For each [imath]h^{'} \in H'[/imath] there exists an [imath]h \in H[/imath] such that [imath]\phi(h)=h'[/imath]. But, [imath]h'^{-1}=\phi(h)^{-1}=\phi(h^{-1})[/imath]. Hence it has inverses. Did I prove this correctly? | 459585 | Image of subgroup and Kernel of homomorphism form subgroups
Is my proof ok? Let [imath]f:G\to G^{\prime}[/imath] be a group homomorphism and let [imath]H\lt G[/imath]. [imath]Im(H) = \{f(x):x\in H\}[/imath]. To show that [imath]Im(H)[/imath] is a group, it suffices to show that [imath]f(x)f(y)^{-1}\in Im(H)[/imath]. [imath]f(x)f(y)^{-1}=f(xy^{-1})\in Im(H)\text{ because } xy^{-1}\in H[/imath] [imath]Ker(f)=\{x:f(x)=e^{\prime}\}[/imath] ([imath]e^{\prime}[/imath] is the identity in [imath]G^{\prime}[/imath]). [imath]f(xy^{-1})=f(x)f(y)^{-1}=e^{\prime}(e^{\prime})^{-1}=e^{\prime}\in Ker(f)[/imath]. |
632999 | Integrate [imath]\int \frac{(1+x^2)dx}{(1−x^2)\sqrt{1+x^4}}[/imath]
Integrate [imath]\displaystyle \int \frac{(1+x^2)dx}{(1−x^2)\sqrt{1+x^4}}[/imath] I don't know how to do this one. I need some suggestions. Thank you! | 15719 | How do I integrate the following? [imath]\int{\frac{(1+x^{2})\mathrm dx}{(1-x^{2})\sqrt{1+x^{4}}}}[/imath]
[imath]\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx[/imath] This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go about it? |
633285 | Is the sum of all natural numbers [imath]-\frac{1}{12}[/imath]?
My friend showed me this youtube video in which the speakers present a line of reasoning as to why [imath] \sum_{n=1}^\infty n = -\frac{1}{12} [/imath] My reasoning, however, tells me that the previous statement is incorrect: [imath] \sum_{n=1}^\infty n = \lim_{k \to \infty} \sum_{n=1}^k n = \lim_{k \to \infty}\frac{k(k+1)}{2} = \infty [/imath] Furthermore, how can it be that the sum of any set of integers is not an integer. Even more, how can the sum of any set of positive numbers be negative? These two ideas lead me to think of inductive proofs as to why the first statement is incorrect. Which of the two lines of reasoning is correct and why? Are there any proven applications (i.e. non theoretical) which use the first statement? | 633117 | Is there any mathematical or physical situations that [imath]1+2+3+\ldots\infty=-\frac{1}{12}[/imath] shows itself?
I just saw the proof that [imath]1+2+3+\cdots=-\frac{1}{12}[/imath] and my brain still hurts. I completely understood the proof and my question is NOT actually about the proof itself. At the end of the proof, they said that this answer shows itself in quantum physics (e.g. it has a role in why there are 26 dimensions in string theory). Sadly though, I am no expert in quantum physics and I'm wondering are there any other physical and/or mathematical situations that this counter intuitive result shows itself? P.S. Needless to say, it would be great if your answer doesn't need a very deep background in physics or math. |
633343 | sum of infinite roots?
What is the sum of this infinite series of roots: [imath]\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4 + \cdots+\sqrt{\infty}}}}}[/imath] This is an interesting expression because the increase created by the addition of the next term is offset by the square root introduced with the next term. the key is determining what the relationship between these two is exactly I know it converges on an irrational number around 1.75 as I have tried computing it for a few small iterations. But I was wondering how to algebraically simplify this expression. I have tried many things and ways to manipulate the expression but to no avail. | 532647 | Infinite square-rooting
[imath] \lim_{n\to\infty} {\sqrt{1+{\sqrt{2+{\sqrt{\cdots +\sqrt{n}\ }\ }\ }\ }\ \ }\ } = ? [/imath] Either closed answer or an upper bound would help. |
633448 | Fields and proper subfields.
Specific question: Let [imath]F[/imath] be a field and assume that [imath]\mathbb{Q}[/imath] is a proper subfield of [imath]F[/imath]. Can [imath]F[/imath] be isomorphic to [imath]\mathbb{Q}[/imath]? Studying the foundaments of field theory I have to ask: Can a field be isomorphic to one of its proper subfields? | 137906 | Algebraic closure of [imath]\mathbb{C}(x)[/imath] is isomorphic to [imath]\mathbb{C}[/imath]
Let [imath]x[/imath] be transcendental over [imath]\mathbb{C}[/imath]. Let [imath]K[/imath] be the algebraic closure of [imath]\mathbb{C}(x)[/imath]. How to show that [imath]K[/imath] is isomorphic to [imath]\mathbb{C}[/imath]? |
495848 | The family of analytic functions with positive real part is normal
I'm having difficulty with the following exercise in Ahlfors' text, on page 227. Prove that in any region [imath]\Omega[/imath] the family of analytic functions with positive real part is normal. Under what added condition is it locally bounded? Hint: Consider the functions [imath]e^{-f}[/imath]. Here is what I've tried: I will start with a remark: Apparently, Ahlfors wants us to show that the family is "normal in the classical sense". That is, every sequence of functions in the family has a subsequence which converges uniformly on compact subsets or tends uniformly to [imath]\infty[/imath] on compact subsets. In order to see why this is the right definition, consider the sequence [imath]f_n(z)=n[/imath]. It is contained in the family but has no appropriate subsequence (in the sense of definition 2 in the text with [imath]S=\mathbb C[/imath]). Now, to the attempt itself: Let [imath]\Omega \subset \mathbb C[/imath] be a fixed region, and consider the family [imath]\mathfrak F=\{f: \Omega \to \mathbb C | f \text{ is analytic and } \Re(f) >0 \}. [/imath] We would like to show that [imath]\mathfrak F[/imath] is normal in the classical sense. Following the hint, we examine the family [imath] \mathfrak G=\{e^{-f}:f \in \mathfrak F \}.[/imath] [imath]\mathfrak G[/imath] is locally bounded (since [imath]|e^{-f}|=e^{- \Re (f)}<1[/imath] for every [imath]f \in \mathfrak F[/imath]), thus it is normal with respect to [imath]\mathbb C[/imath] (theorem 15), and obviously, it is normal in the classical sense as well. Let [imath]\{ f_n \}[/imath] be a sequence in [imath]\mathfrak F[/imath], and consider the sequence [imath]\{ g_n \}=\{e^{-f_n} \}[/imath] in [imath]\mathfrak G[/imath]. According to normality it has a convergent subsequence [imath]\{ g_{n_k} \}=\{e^{-f_{n_k}} \}[/imath] which converges uniformly on compact subsets of [imath]\Omega[/imath] to some function [imath]g[/imath] (which is analytic by Weierstrass' theorem). Since each [imath]\{ g_{n_k} \}[/imath] is nonvanishing, the limit function [imath]g[/imath] is either identically zero, or non vanishing as well (Hurwitz's theorem). In the former case it is easy to show that the subsequence [imath]\{ f_{n_k} \}[/imath], obtained by the same indices, tends to [imath]\infty[/imath] uniformly on compact sets. Hence, we will assume from now that [imath]g(z) \neq 0[/imath] for all [imath]z \in \Omega[/imath]. Up until now, I was trying to show that the subsequence [imath]\{ f_{n_k} \}[/imath] works in all cases, but sadly, this is not the case. Consider the sequence [imath]f_n(z) \equiv 1+2 \pi i (-1)^n \in \mathfrak F[/imath]. In that case [imath]g_n(z)=e^{-1}[/imath], and an admissible subsequence is [imath]g_{n_k}=g_k=e^{-1}[/imath]. However, [imath]f_{n_k}=1+2 \pi i (-1)^k[/imath] diverges everywhere. Can anyone please help me finish this proof? Or maybe give me some hints? Thanks! | 317504 | A question about normal families
This is Ahlfors q. 1, p. 227. Prove that in any region the family of analytic functions with positive real part is normal. Under what added condition is it locally bounded? Hint: Consider [imath]e^{-f}[/imath]. Well I proved that [imath]e^{-f}[/imath] is normal, but I don't know what to do next, moreover If I am right normality implies locally boundeness in the analytic case,I'm trying to figure out why asked that question... |
633850 | eigen values and eigen vector of a matrix
Let [imath]A[/imath] be an n *n matrix all of whose entries are 1. Find all the eigenvalues and eigenvectors of [imath]A[/imath]. I have checked for 2*2 ,3*3 matrices and guessing the answer but in general how to show. | 217521 | What are the eigenvalues of matrix that have all elements equal 1?
As in subject: given a matrix [imath]A[/imath] of size [imath]n[/imath] with all elements equal exactly 1. What are the eigenvalues of that matrix ? |
633862 | Rules for [imath](x^a)^b[/imath] - complex number.
How potentiate [imath](x^a)^b[/imath] for complex numbers? | 2914 | How to combine complex powers?
Regarding this thread, it is not possible to combine complex powers in the usual way: [imath] (x^y)^z = x^{yz} [/imath] There was mention of multi-valued functions, is there some theory that makes this all work out correctly? |
633940 | Explanation of [imath]d^{-1}[/imath] in modular arithmetic
I wasnt quite sure what to name this question, so that's what it is. I'me working on an encryption system, and I need modulus. I already asked a question on this, here, and I cannot figure out the answer to this, and I was hoping someone could explain it a little better for me? I said that it made sense, because it seemed to, but then I realized I was doing it wrong and it was too late -_- How does d = [imath]d^{-1} = 103[/imath] BECAUSE [imath]103 \cdot 7 = 721[/imath] and [imath]721 ≡ 1 \pmod{ 120}[/imath]? I understand that [imath]721 ≡ 1 \pmod {120}[/imath] is [imath]6[/imath] remainder [imath]0[/imath], but how does that justify that [imath]d^{-1} = 103[/imath]? and in the second part, [imath]103 \times 7[/imath], why is it [imath]\times 7[/imath]? Any help would be appreciated! Please explain this like talking to a ten year old :D | 627365 | Define ≡ in this situation?
"Determine [imath]d[/imath] as [imath]d^{-1} \equiv e \bmod \phi(n)[/imath], i.e., [imath]d[/imath] is the multiplicative inverse of [imath]e \bmod \phi(n)[/imath]." (number [imath]5[/imath]). I'm looking at this, and i'm not sure what the [imath]\equiv[/imath] means in this instance? I know the rest of the values, but what does [imath]\equiv[/imath] mean right now? Thanks in advance! |
633951 | Proof of convergence test - (Apostol, Calculus, 10.16 #15)
I'm self-studying from Apostol, Calculus, Volume I and am having trouble with the following exercise (Section 10.16, #15, p. 402): Let [imath]\{a_n\}[/imath] and [imath]\{b_n\}[/imath] be two sequences with [imath]a_n > 0[/imath] and [imath]b_n > 0[/imath] for all [imath]n \geq N[/imath], and let [imath]c_n = b_n - \frac{b_{n+1} a_{n+1}}{a_n}[/imath]. Prove that: (a) If there is a positive constant [imath]r[/imath] such that [imath]c_n \geq r > 0[/imath] for all [imath]n \geq N[/imath], then [imath]\sum a_n[/imath] converges. [Hint: Show that [imath]\sum_{k=N}^n a_k \leq \frac{a_N b_N}{r}[/imath].] (b)If [imath]c_n \leq 0[/imath] for [imath]n \geq N[/imath] and if [imath]\sum 1/b_n[/imath] diverges, then [imath]\sum a_n[/imath] diverges. [Hint: show that [imath]\sum a_n[/imath] dominates [imath]\sum 1/b_n[/imath].] For part (a) I cannot seem to figure out how to arrive at the hint, nor how to use it. I looked at a simple test case of [imath]b_n = 1[/imath], so we arrive at [imath] c_n = 1 - \frac{a_{n+1}}{a_n} \geq r > 0 \qquad \forall \; n \geq N[/imath] Hence, we have [imath]\lim_{n \to \infty} \frac{a_{n+1}}{a_n} < 1 - r[/imath], and hence [imath] < 1[/imath]. Therefore, [imath]\sum a_n[/imath] converges. So, from the simple case we see a bit of what's going on. My inclination is then to do a proof by contradiction… Suppose the limit is [imath]\geq 1[/imath] and get a contradiction (which should result since [imath]b_n[/imath] will need to go to [imath]0[/imath], contradicting the positivity of [imath]b_n - b_{n+1} \frac{a_{n+1}}{a_n}[/imath]. Obviously, this doesn't make any use of the hint, though, so maybe is the wrong idea. I do think this contradiction will work, but it seems Apostol envisioned a more direct approach that I do not see. | 115146 | Kummer's test - Calculus, Apostol, 10.16 #15.
I want to prove the following: Let [imath]\{ a_n \}[/imath] and [imath]\{ b_n \}[/imath] be two sequences with [imath]a_n>0[/imath] and [imath]b_n>0[/imath] for all [imath]n \geq N[/imath], and let [imath]c_n = b_n - \frac{a_{n+1}b_{n+1}}{a_n}[/imath] Then If there exists [imath]r>0[/imath] such that [imath]0<r\leq c_n \text{ ; } \forall n\geq N[/imath] then [imath]\displaystyle\sum a_n[/imath] converges. If [imath]c_n\leq0[/imath] for [imath]n\geq N[/imath] and if [imath]\displaystyle\sum \dfrac{1}{b_n}[/imath] diverges, then so does [imath]\displaystyle\sum a_n[/imath]. So far I know how to use and prove Cauchy's, D'Alambert's and the integral criterion, so I'd like a hint on either using those or a new idea. The book suggests that for [imath]1[/imath], I show that [imath]\sum\limits_{k = N}^n {{a_k} \leq \frac{{{a_N}{b_N}}}{r}} [/imath] and for [imath]2[/imath], to prove that [imath]\displaystyle\sum a_n[/imath] dominates [imath]\displaystyle\sum \dfrac{1}{b_n}[/imath] I'd like to prove this with previous theory on series convergence (Cauchy's and/or D'Alambert's preferrably, comparison test) since it is what precedes the problem, and would appreaciate great HINTS rather than answers. I can't seem to understand the inequalities. I mean, Cauchy's and D'Alambert's criteria reveal that the proof relies on the fact that a geometric series of with ratio [imath]|r| < 1[/imath] converges, but I can't understand the motivation in this proof. Summing up the work. For 1: [imath]\eqalign{ & {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{b_{N + 1}} \cr & {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{b_{N + 2}} \cr & {b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{c_{N + 2}} + \frac{{{a_{N + 3}}}}{{{a_N}}}{b_{N + 3}} \cr} [/imath] Induction over [imath]n[/imath] we get [imath]{b_N} = {c_N} + \frac{{{a_{N + 1}}}}{{{a_N}}}{c_{N + 1}} + \frac{{{a_{N + 2}}}}{{{a_N}}}{c_{N + 2}} + \cdots + \frac{{{a_{N + n}}}}{{{a_N}}}{c_{N + n}} + \frac{{{a_{N + n + 1}}}}{{{a_N}}}{b_{N + n + 1}}[/imath] So [imath]\eqalign{ & {b_N} \geqslant r\left( {1 + \frac{{{a_{N + 1}}}}{{{a_N}}} + \frac{{{a_{N + 2}}}}{{{a_N}}} + \cdots + \frac{{{a_{N + n}}}}{{{a_N}}}} \right) + \frac{{{a_{N + n + 1}}}}{{{a_N}}}{b_{N + n + 1}} \cr & {b_N} > r\left( {1 + \frac{{{a_{N + 1}}}}{{{a_N}}} + \frac{{{a_{N + 2}}}}{{{a_N}}} + \cdots + \frac{{{a_{N + n}}}}{{{a_N}}}} \right) \cr & \frac{{{a_N}{b_N}}}{r} > {a_N} + {a_{N + 1}} + {a_{N + 2}} + \cdots + {a_{N + n}} = \sum\limits_{k = N}^n {{a_k}} \cr} [/imath] As desired. |
118889 | Mean distance from origin after [imath]N[/imath] equal steps of Random-Walk in a [imath]d[/imath]-dimensional space.
I am looking for a formula that evaluates the mean distance from origin after [imath]N[/imath] equal steps of Random-Walk in a [imath]d[/imath]-dimensional space. Such a formula was given by "Henry" to a question by "Diego" (q/103170) [imath]\sqrt{\dfrac{2N}{d}} \dfrac{\Gamma(\frac{d+1}{2})}{\Gamma(\frac{d}{2})}[/imath] I will be very gratefull if you can give me reference to an article that show how this formula was derived. Thanks! | 413692 | Reference Request, Random Walk
The expected distance from origin after a random walk of [imath]N[/imath] steps in a [imath]d[/imath] dimensional space, is close to [imath]\sqrt{\dfrac{2N}{d}}\dfrac{\Gamma\left(\dfrac{d+1}{2}\right)}{\Gamma\left(\dfrac{d}{2}\right)}[/imath] for very large [imath]N[/imath]. This was mentioned here. I would be very obliged if someone can mention to me a reference (bibliographic) to an article, book or any publication, where this expression was derived or mentioned. Please restrict to references, and not explanation of the formula itself. |
628704 | [imath]M[/imath] and [imath]N[/imath] are matrices that satisfy [imath]MNMN=0[/imath],what can we say about [imath]NMNM[/imath].
From my personal point of view, we cannot deduce that [imath]NMNM=0[/imath],but I can't find a counterexample. | 628154 | [imath]ABAB=0_{n\times n}[/imath], then must [imath]BABA[/imath] be [imath]0_{n\times n}[/imath]?
Let [imath]$A$[/imath] and [imath]$B$[/imath] be [imath]n\times n[/imath] matrices such that [imath]ABAB=0_{n\times n}[/imath]. Can we conclude that [imath]BABA[/imath] must be [imath]0_{n\times n}[/imath]? I fail to give a counterexample when [imath]n=2[/imath]， so I guess the answer is "yes" |
634470 | How to prove this formula
I want to know how to prove this: [imath] \sum_{k=1}^\infty \arctan\left( \frac{x}{k^2+a} \right) = \pi \left\lfloor \frac{b}{\sqrt{2}} \right\rfloor + \arctan\left( \frac{x}{b^2} \right) - \arctan\left[ \tanh\left( \frac{\pi}{\sqrt{2}} \frac{x}{b} \right) \cot\left( \frac{\pi}{\sqrt{2}} b \right) \right] [/imath] for [imath]0 \le a, b = \sqrt{\sqrt{x^2 + a^2} - a}[/imath]. At least in the special case [imath]x=1[/imath],[imath]a=0[/imath] Thank you .. | 613670 | Prove [imath]\sum_{n=1}^{\infty}\arctan{\left(\frac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\frac{\sqrt{2}-1}{2}}\right)\cdots\right)}[/imath]
show that: [imath]\sum_{n=1}^{\infty}\arctan{\left(\dfrac{1}{n^2+1}\right)}=\arctan{\left(\tan\left(\pi\sqrt{\dfrac{\sqrt{2}-1}{2}}\right)\cdot\dfrac{e^{\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}+e^{-\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}}{e^{\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}-e^{-\pi\sqrt{\dfrac{\sqrt{2}+1}{2}}}}\right)}-\dfrac{\pi}{8}[/imath] This relust is nice.(maybe have some wrong) becasue I know this famous problem [imath]\sum_{n=1}^{\infty}\arctan{\dfrac{2}{n^2}}=\dfrac{3\pi}{4}[/imath] and follow AMM( E3375) problem [imath]\sum_{n=1}^{\infty}\arctan{\dfrac{1}{n^2}}=\arctan{\left(\dfrac{\tan{\frac{\pi}{\sqrt{2}}}-th{\dfrac{\pi}{\sqrt{2}}}}{\tan{\dfrac{\pi}{\sqrt{2}}}+th{\dfrac{\pi}{\sqrt{2}}}}\right)}[/imath] Follow is AMM solution: My try: my problem I want use this methods,But at last failure it. Thank you for you help. This problem is similar this:we konw this [imath]\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}[/imath] and then little hard problem: [imath]\sum_{n=1}^{\infty}\dfrac{1}{n^2+1}=\dfrac{1}{2}(\pi\coth{\pi}-1)[/imath] |
633821 | Why is [imath]x^\alpha-\alpha x\le 1-\alpha[/imath]?
I was reading the Proof of Hölder's Inequality in Measure Theory, and i saw, that our professor used an inequality and didn't explain, why it is true, and the inequality is; For [imath]\alpha\in(0,1)[/imath] and [imath]x\in\mathbb R_+[/imath] [imath]x^\alpha-\alpha x\le 1-\alpha[/imath] [imath]\textbf{I tried ;}[/imath] [imath]x^\alpha-\alpha x\le 1-\alpha[/imath] [imath]\Leftrightarrow x^\alpha\le 1-\alpha+\alpha x=1+\alpha(x-1)[/imath] and [imath]x^\alpha\le 1+\alpha(x-1)[/imath] looks similar to bernoulli inequality, if you take [imath]x'=\alpha(x-1)[/imath] and [imath]r=\frac{1}{\alpha}[/imath] [imath](1+x')^r\ge x[/imath] Is my justification correct, and does Bernoulli hold also for rational exponents ? | 624850 | Proving [imath]x^\alpha-\alpha x \le 1- \alpha [/imath]
Prove: [imath]x^\alpha-\alpha x \le 1- \alpha \\ \forall x\ge 0 \ , \ 0<\alpha <1[/imath] It does resamble Lagrange's MVT, in order to get to the RHS, say [imath]\alpha\in[0,1]\Rightarrow \frac{1-\alpha-0^\alpha-\alpha 0}{1-0}=1-\alpha[/imath]. But I'm not sure what to do about the LHS. I always get a negative value in the LHS and we know that the RHS will always be postive but that doesn't really help with proving it... Edit: maybe define a function: [imath]g(x) = x^\alpha-\alpha x - 1+ \alpha[/imath] Derive it: [imath]g'(x) = \alpha x^{\alpha-1}-\alpha [/imath] but then I get to [imath]x^{\alpha-1}=0[/imath] Also, I had two questions when trying to figure this out: Can we use deriviation of both sides of an inequality to prove it ? Can we use limit of both sides of an inequaltiy to prove it ? Thanks. |
634605 | If [imath]X[/imath] and [imath]Y[/imath] are independent. How about [imath]X^2[/imath] and [imath]Y[/imath]? And how about [imath]f(X)[/imath] and [imath]g(Y)[/imath]?
If [imath]X[/imath] and [imath]Y[/imath] are independent. How about [imath]X^2[/imath] and [imath]Y[/imath]? And how about [imath]f(X)[/imath] and [imath]g(Y)[/imath]? I always have confusion about it. I feel ... yeah of course [imath]f(X)[/imath] and [imath]g(Y)[/imath] are independent, because [imath]X[/imath] and [imath]Y[/imath] are. But .. is it right?? Then how can I prove it? | 220377 | Are squares of independent random variables independent?
If X and Y are independent random variables both with the same mean (0) and variance, how about [imath]X^2[/imath] and [imath]Y^2[/imath]? I tried calculating E([imath]X^2Y^2[/imath])-E([imath]X^2[/imath])E([imath]Y^2[/imath]) but haven't been able to get anywhere. |
634683 | Prove convergence of this integration
[imath]\int_0^\infty\frac1{(x^2+1)(\ln^2(x)+\pi^2)}[/imath] Prove the convergence of this integration. | 633259 | calculate an integration by using residue theorem
[imath]\int_0^\infty\frac1{(x^2+1)(\ln^2(x)+\pi^2)}[/imath] Calculate the following integration using the residue theorem: |
633476 | Compute [imath]\lim_{n \rightarrow \infty} \left(\left(\frac{9}{4} \right)^n+\left(1+\frac{1}{n} \right)^{n^2} \right)^{1/n}[/imath]
may someone show how to compute [imath]\lim_{n \rightarrow \infty} \left(\left(\frac{9}{4} \right)^n+\left(1+\frac{1}{n} \right)^{n^2} \right)^{\frac{1}{n}}[/imath]? According to W|A it's e, but I don't know even how to start... Please help, thank you. | 586141 | Limit calculation
[imath]\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}[/imath] Here's what I did: [imath]\lim_{n\to\infty}((\frac94)^n+(1+\frac1n)^{n^2})^{\frac1n}\\ =(\lim(\frac94)^n+\lim((1+\frac1n)^{n})^n)^{\frac1n}\\ =(\lim(\frac94)^n+\lim e^n)^{\frac1n}\\[/imath] Any hints on how to continue? PS: no logs/integration/derivation because we haven't covered it. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.