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101681 | Equation of straight line
I know, [imath]Ax + By = C[/imath] is the equation of straight line but a different resource says that: [imath]y = mx + b[/imath] is also an equation of straight line? Are they both same? | 26833 | How is [imath]Ax + By = C[/imath] the equation of a straight line?
I know the equation, [imath]y = mx + b[/imath] where [imath]m[/imath] is slope and [imath]b[/imath] is [imath]y[/imath]-intercept, is a straight line. But I know also that [imath]Ax + By = C[/imath] is a straight line equation, but how does it represent a straight line? |
283017 | The limit in law of a sequence of normal distributions is normal
Let [imath] \{ \xi_n \}_{n=1}^{\infty}[/imath] be a sequence of normal random variables, where [imath] \xi_n\sim\mathcal{N}(\alpha_n, \sigma_n^2)[/imath] and [imath]\xi_n \overset{d}{\rightarrow} \xi[/imath]. I need to prove, that [imath]\xi[/imath] is also a normal random variable, can anyone help? | 232540 | The limit of a convergent Gaussian random variable sequence is still a Gaussian random variable
I'm trying to prove this conclusion but have some problems with one of the steps. Assume [imath]X_1,\ldots,X_n,\ldots[/imath] is a sequence of Gaussian random variables, converging almost surely to [imath]X[/imath], prove that [imath]X[/imath] is Gaussian. We use characteristics function here. Since [imath]|\phi_{X_n}(t)|\leq 1[/imath], by dominated convergent theorem, we have for any [imath]t[/imath] [imath] \lim_{n\rightarrow\infty}e^{it\mu_n-t^2\sigma_n^2/2}=\lim_{n\rightarrow \infty}\phi_{X_n}(t) = \lim_{n\rightarrow \infty}\mathbb{E}\left[e^{itX_n}\right] = \mathbb{E}\left[e^{itX}\right] = \phi_X(t) [/imath] this is the step that I cannot figure out: [imath]e^{it\mu_n-t^2\sigma_n^2/2}[/imath] converges for any [imath]t[/imath] if and only if [imath]\mu_n[/imath] and [imath]\sigma_n[/imath] converges. Let [imath]\mu=\lim_n \mu_n[/imath], and [imath]\sigma=\lim_n\sigma_n[/imath], then [imath]\phi_X(t)=e^{it\mu-t^2\sigma^2/2}[/imath], which proves that [imath]X[/imath] is a Gaussian random variable. Why can we get that [imath]\mu_n[/imath] and [imath]\sigma_n[/imath] converge? This looks intuitive for me, but I cannot make a rigorous prove. |
354416 | Proving two summations equivalent
Let [imath]h_n[/imath] be an infinite sequence. I need to show that: \begin{align}\dfrac{1}{1+x}H\left(\dfrac{x}{1+x}\right) = \sum\limits_{k=0}^\infty \sum\limits_{i=0}^k(-1)^{k-j}{k\choose i}x^kh_i \end{align} where [imath]H(x) = \sum\limits_{k=0}^\infty h_kx^k[/imath] Transforming the LHS we get: [imath]\dfrac{1}{1+x}\sum\limits_{k=0}^\infty h_kx^k(1+x)^{-k}[/imath] \begin{align} =& \sum\limits_{k=0}^\infty h_kx^k(1+x)^{-(k+1)} \\ =& \sum\limits_{k=0}^\infty h_kx^k\sum\limits_{i=0}^\infty (-1)^i{k+i\choose i}x^i \end{align} This last step is by Newtons generalized binomial theorem. Now moving the [imath]h_k[/imath] and [imath]x^k[/imath] we get: \begin{equation} \sum\limits_{k=0}^\infty \sum\limits_{i=0}^\infty (-1)^i{k+i\choose i}x^{i+k}h_k \end{equation} Where can I go from here? I typed both of these double sums into a CAS (without [imath]h_n[/imath] and found that they both converge to [imath]1[/imath] for any value of [imath]x[/imath]). Intuition tells me that it is a matter of interpreting the [imath]i+k[/imath] and the subscript on [imath]h[/imath] to find that the LHS = RHS All help is greatly appreciated | 354076 | Show that [imath]\frac{1}{1+x}H(\frac{x}{1+x})=\sum^\infty_{k=0}[\Delta^kh_0]x^k[/imath]
For a sequence [imath]\{h_n\}_{\geq 0}[/imath], let [imath]H(x)=\sum_{n\geq0}h_nx^n[/imath]. Show that: [imath]\frac{1}{1+x}H(\frac{x}{1+x})=\sum^\infty_{k=0}[\Delta^kh_0]x^k[/imath] What I did was that by proving the [imath]\Delta^k h_o=\sum^k_{j=0}(-1)^{k-j}{k \choose j}h_j[/imath] But no clue how to continue. Help appreciated. |
267724 | Is every subalgebra of [imath]F[x][/imath] one-dimensional?
Let [imath]F[/imath] be a field. Let [imath]A[/imath] be a subalgebra of the polynomial ring [imath]F[x][/imath]. Does [imath]A[/imath] necessarily have Krull dimension [imath]\leq 1[/imath]? | 154166 | Polynomial algebra
Let [imath]k[/imath] be a field and [imath]k \subset A \subseteq k[X][/imath] be a [imath]k[/imath]-subalgebra of [imath]k[X][/imath]. Prove that [imath]\dim(A)=1[/imath] (Krull dim) and that [imath]A[/imath] is a finitely-generated [imath]k[/imath]-algebra. My initial thought: Consider the short exact sequence [imath]0 \rightarrow A \hookrightarrow k[X] \rightarrow k[X]/A \rightarrow 0[/imath] and see how the Krull dimension behaves under this sequence. It's just a spark, I don't know how to continue exactly. |
49079 | What does the notation [imath]\binom{n}{i}[/imath] mean?
What do the parentheses next to the summation involving the binomial coefficients mean? Like this: [imath]\sum _{i=0}^{n} \binom{n}{i}a^{(n-i)}b^i=\left(a+b\right)^n [/imath] | 106018 | What does a particular part of this equation mean?
I've seen this a couple times. What does the part where y is over k with a space between them? What does this imply? (Below is the Binomial Theorem equation) [imath](x + a)^y = \sum\limits_{k=1}^{\infty}\binom{y}{k}x^ka^{y-k}[/imath] |
354681 | Proving that a formula is true for [imath]k[/imath]-th order differences of a sequence
How do I use induction to prove that: [imath]\Delta^k h_0=\sum^k_{j=0}(-1)^{k-j}{k\choose j}h_j[/imath] holds for the [imath]k[/imath]-order differences of a sequence [imath]\{h_n\}_{n\leq 0}[/imath] Thanks for the help! | 340385 | Exponential Generating Function of the numbers [imath]r(n)[/imath]
We define [imath]R(x)=\sum_{n\geq 0}r(n)\frac{x^n}{n!}[/imath], the exponential generating function of the numbers [imath]r(n)[/imath]. LEt us multiply both sides of equation by [imath]\frac{x^n}{n!}[/imath]. Then, sum over all positive integers [imath]n[/imath], to get: [imath]\sum_{n\geq 0}r(n+2)\frac{x^n}{n!}=\sum_{n\geq 0}r(n+1)\frac{x^n}{n!}+\sum_{n\geq 0}(n+1)r(n)\frac{x^n}{n!}[/imath] Now note that left hand-side is [imath]R''(x)[/imath], and the first member of the righ-hand side is [imath]R'(x)[/imath]. Te second member of the right-hand side is somewhat harder ro recognize, but ith a little practice, one can see that it is in fact [imath](x\ R(x))'[/imath]. [imath]\therefore R''(x)=R'(x)+(xR(x))'=R'(x)+xR'(x)+R(x)[/imath] By solving this, we will get [imath]R(x)=e^\frac{x+x^2}{2}[/imath] Hey guys, I don't really get this, how do we solve "this"? |
355177 | Multivariate polynomials have infinitely many zeroes
Let [imath]F \in K[x_1, \dotsc, x_n][/imath], [imath]n\ge 2[/imath], with [imath]K[/imath] an algebraically closed field and [imath]\deg F \ge 1[/imath]. Prove that [imath]F[/imath] has infinitely many zeroes. I would consider [imath]F[/imath] as a polynomial in the single variable [imath]x_1[/imath], so that from the algebraic closure of [imath]K, \exists c \in K[/imath] s.t. [imath]F(c) = 0[/imath] considered as a polynomial of the only variable [imath]x_1[/imath]. But then I can vary the remaining variables as I like in [imath]K[/imath], thus obtaining infinite zeroes. However this argument doesn't convince me: can I vary the variables [imath]x_2, \dotsc, x_n[/imath] and keep fixed [imath]x_1[/imath] still obtaining a zero of [imath]F[/imath]? | 286352 | Polynomial with infinitely many zeros.
Can a polynomial in [imath] \mathbb{C}[x,y] [/imath] have infinitely many zeros? This is clearly not true in the one-variable case, but what happens in two or more variables? |
355329 | show that [imath]A \cap B[/imath] is a normal subgroup of [imath]B[/imath]
Suppose [imath]A[/imath] is a normal subgroup of [imath]G[/imath] and [imath]B[/imath] is a subgroup of [imath]G[/imath]. Please help me to show that [imath]A \cap B[/imath] is a normal subgroup of [imath]B[/imath]. I know that since [imath]B[/imath] is subgroup of [imath]G[/imath], it has identity element, [imath]e[/imath] which makes it not to be an empty set. The same applies to [imath]A[/imath], which is a normal subgroup of [imath]G[/imath]. Therefore, [imath]e[/imath] is an element of [imath]A \cap B[/imath]. | 354961 | If [imath]A[/imath] is a normal subgroup of [imath]G[/imath] and [imath]B[/imath] is a subgroup of [imath]G[/imath] , then [imath]A\cap B[/imath] is a normal subgroup of [imath]B[/imath]
Suppose [imath]A[/imath] is a normal subgroup of [imath]G[/imath] and [imath]B[/imath] is a subgroup of [imath]G[/imath]. Show that [imath]A\cap B[/imath] is a normal subgroup of [imath]B[/imath]. |
355323 | [imath]\tau:=\{Y\in P(X) | A\subseteq Y\}\cup\{\emptyset\}[/imath] topological space
Let [imath]X[/imath] be a set and [imath]A\subseteq X[/imath] I would like to show that [imath]\tau:=\{Y\in P(X) | A\subseteq Y\}\cup\{\emptyset\}[/imath]is a topological space on [imath]X[/imath] and afterwards I would like to describe the closure [imath]\bar{E}[/imath] and the inner [imath]E°[/imath] concerning [imath]\tau[/imath] for arbitrary subsets [imath]E\subseteq X[/imath] Well first of all [imath]\emptyset \in\tau, X\in\tau[/imath] is clear.Any union of open sets is open: This is intuitviely clear but I do not know how to show that, the same with The intersection of any finite number of open sets is open. May you could help me with that. | 306761 | Closures and Interiors in a topology
Let [imath]X[/imath] be a set. Let [imath]A[/imath] be a proper non-empty subset of [imath]X[/imath]. Let [imath]\tau = \{\emptyset\} \cup \{U \in P(X): A \subseteq U\}[/imath] Question: Find the interior and closure of [imath]A[/imath] and prove that they are indeed the interior and closure. (i) Interior of [imath]A[/imath]: The interior of [imath]A[/imath] is [imath]A[/imath] since [imath]A[/imath] is an element of [imath]\tau[/imath] and hence it is an open set. (ii) I'm stuck trying to figure out the closure of [imath]A[/imath]. In this topology, since any [imath]U[/imath] containing [imath]A[/imath] is an open set...does this mean that any closed set would be a set that does not contain [imath]A[/imath]. So the closure of [imath]A[/imath] is the empty set. Question: Suppose [imath]B[/imath] is a nonempty proper subset of [imath]X[/imath] such that [imath]A[/imath] is a nonempty proper subset of [imath]B[/imath]. Find the interior and closure of [imath]B[/imath] and prove your answers. Again I think the interior of [imath]B[/imath] is [imath]B[/imath]. Similarly, the closure is empty. |
350415 | Equiv Relation of Orbits - Group Action
Let [imath]G[/imath] be a group that acts on [imath]X[/imath]. I want to show that the orbits of [imath]G[/imath] partition [imath]X[/imath]. I am given the relation [imath]x\sim y \iff x\in Orb(y)[/imath]. Now: [imath]x\sim y\iff x\in Orb(y) \iff x=gy[/imath] for some [imath]g\in G \iff g^{-1}x=y[/imath] for some [imath]g\in G[/imath] [imath]\iff y\in Orb(x) \iff Orb(x)=Orb(y)[/imath]. So I use this to show the three conditions of an equivalence relation: [imath]x\sim x \iff Orb(x)=Orb(x)[/imath] [imath]x\sim y \iff Orb(x)=Orb(y) \iff Orb(y)=Orb(x) \iff y\sim x[/imath] [imath]x\sim y, y\sim z \Rightarrow Orb(x)=Orb(y)=Orb(z) \Rightarrow Orb(x)=Orb(z)\Rightarrow x\sim z[/imath] Does this look okay? | 290793 | Properties of set [imath]\mathrm {orb} (x)[/imath]
Properties of set [imath]\mathrm {orb} (x)[/imath]: [imath]{\displaystyle \bigcup_{x\in X}\mathrm{orb}(x)=X}[/imath]; [imath]\mathrm{orb}(x)\cap\mathrm{orb}(y)=\emptyset[/imath] for all [imath]x,y\in X, x\neq y[/imath] How to prove it? Please help. Appedix: Let [imath]\phi: G \times X \longrightarrow X[/imath] - action of the group G on the non-empty set [imath]X[/imath]. The set [imath]\mathrm {orb} (x) = \{ \phi (g,x) \in X: g \in G \}[/imath] called orbit of [imath]x \in X[/imath] |
355455 | How to prove when Möbius transformation determinant ad - bc < 0, the upper half plane does not map to itself?
I proved the part where [imath]\operatorname{Im} f(z) = \frac {(ad - bc) \operatorname{Im}z} {|cz + d|^2} [/imath] But now I need to show that when [imath] ad - bc < 0 [/imath], then the upper half plane does not map to itself | 334900 | When does a Möbius transformation map [imath]\Im(z)>0[/imath] to itself?
Show Möbius transformation which maps [imath]\Im(z)>0[/imath] to itself iff [imath] f(z)= \frac{az+b}{cz+d}\,,\,\,ad-bc>0[/imath] and [imath]a,b,c,d[/imath] are real. |
355536 | Subspaces: intersection and union
Assume that [imath]A[/imath] and [imath]B[/imath] are subspaces of the vector space [imath]\Bbb{R}^n[/imath]. Show that [imath]H=A \cap B[/imath] is also a subspace of the vector space [imath]\Bbb{R}^n[/imath]. Show by construction of the example that [imath]R=A \cup B[/imath] is not always a subspace of [imath]\Bbb{R}^n[/imath]. | 298955 | Intersection and Union of subspaces
Let [imath]F[/imath] and [imath]G[/imath] be subspaces of a vector space [imath]E[/imath]. When is [imath]F \cup G[/imath] a subspace ? When is [imath]F \cap G[/imath] a subspace? Thank you in advance. |
355458 | Showing that the open ball is homeomorphic to [imath]\mathbb{R}^n[/imath]
I'm trying to prove that an open ball is homeomorphic to [imath]\mathbb{R}^n[/imath] but since this is the first proof of this kind that I try to give I'm having a little doubt on how to begin. I've had some thoughts about working with the inclusion map, however I'm not sure if I'm on the right way. Can someone give just some hint on how to start this proof ? Thanks for your help in advance, and sorry if the question is too basic. | 108268 | How can I find a homeomorphism from [imath]\mathbb{R}^n[/imath] to the open unit ball centered at 0?
I'm trying to prove that the open ball of radius 1 centered at the origin in [imath]\mathbb{R}^n[/imath] is homeomorphic to [imath]\mathbb{R}^n[/imath]. I believe the "shrinking map" from [imath]\mathbb{R}^n[/imath] to the ball given by [imath]x \mapsto \dfrac{x}{1 + |x|}[/imath] does the job, but I'm having trouble showing it's a homeomorphism, particularly the "continuous inverse" part. What's a good way to do this? |
355734 | Compare one real number to one complex number.
I understand that complex numbers can be neither ordered nor compared by 'size', but if mapped one for one by a transformation, then they can be. Latter point aside, can I say that [imath]2-xi < 2 < 2+xi[/imath]? Thanks | 113378 | Comparing real and complex numbers
If I'm correct, a complex number can be interpreted as a set in the following manner: [imath] \forall x, y \in \mathbb{R}, x + yi = \{(x,\ y)\}.\ \mathbf{(1)} [/imath] My question is, is it technically correct to say: [imath] \forall a \in \mathbb{R},\ ( b = 0 \implies a + bi = a)?\ \mathbf{(2)} [/imath] Since a real number and complex numbers are different objects, would it not be "irelevant" to ask something as [imath]\mathbf{(2)}?[/imath] In asking this, I'm taking a cue from the reasoning presented in Whitehead's and Russell's Principia Mathematica. Thank you all in advance. |
356145 | Proving that language is not regular by pumping lemma
Suppose [imath]L = \{0^k \mid \text{[/imath]k[imath] is composite}\}[/imath]. Prove that this language is not regular. What bugs me in this lemma is that when I choose a string in [imath]L[/imath] and try to consider all cases of dividing it into three parts so that in each case it violates lemma, I always find one case that does not violate it. A bit of help will be appreciated. Thanks in advance. My attempt: Suppose that [imath]L[/imath] is regular. Choosing string [imath]x = 0^{2k}[/imath] where [imath]k[/imath] is prime ([imath]2k[/imath] pumping constant) We can divide [imath]x[/imath] into three parts [imath]u, v, w[/imath] such that: [imath]|uv| \le 2k \qquad |v| > 0\qquad uv^iw \in L \text{ for $i \ge 0$}[/imath] If [imath]u[/imath] and [imath]w[/imath] are empty, all conditions are met. It is the same when I change [imath]2[/imath] for any other number. Maybe I'm choosing wrong. | 355900 | Proving that $L = \{0^k \mid \text{$k$ is composite}\}$ is not regular by pumping lemma
Suppose [imath]L = \{0^k \mid \text{[/imath]k[imath] is composite}\}[/imath]. Prove that this language is not regular. What bugs me in this lemma is that when I choose a string in [imath]L[/imath] and try to consider all cases of dividing it into three parts so that in each case it violates lemma, I always find one case that does not violate it. A bit of help will be appreciated. Thanks in advance. My attempt: Suppose that [imath]L[/imath] is regular. Choosing string [imath]x = 0^{2k}[/imath] where [imath]k[/imath] is prime ([imath]2k[/imath] pumping constant) We can divide [imath]x[/imath] into three parts [imath]u, v, w[/imath] such that: [imath]|uv| \le 2k \qquad |v| > 0\qquad uv^iw \in L \text{ for $i \ge 0$}[/imath] If [imath]u[/imath] and [imath]w[/imath] are empty, all conditions are met. It is the same when I change [imath]2[/imath] for any other number. Maybe I'm choosing wrong. |
356174 | Better method for calculating this integral
How is integral calculated? [imath]\int_ 0^\infty e^{-x^2}~dx [/imath] | 138664 | How to integrate $\int e^{-t^{2}} \space \, \mathrm dt $ using introductory calculus methods
Earlier today I stumbled across this when I was doing some practice questions for a physics course: [imath]$$\int e^{-t^2} \space \, \mathrm dt $$[/imath] To expand, the limits of integration were something like [imath]1[/imath] and [imath]4[/imath] (it was just a velocity function that needed to be integrated to find distance - it was not a known integral like [imath]$\int_0^\infty e^{-t^2} \space\, \mathrm dt$[/imath].) Based on Wolfram|Alpha, it appears it cannot be expressed in elementary terms (i.e. it involves the error function.) Note that the questions involved the use of a calculator, so I was able to integrate the function using a CAS with ease, but I am wondering how to do it by hand. Thus, I was wondering if there was possibly a way to evaluate the integral using elementary methods from a calculus one or two course (read: no complex analysis). I thought there may perhaps be an elementary solution (I don't know what kind of algorithm Wolfram uses to evaluate integrals - I have seen them evaluate easy integrals in a lot of steps before.) |
356407 | Modulus question
Hey i am studying for my exam and I was wondering how to solve [imath]2023^{2297}\equiv x \pmod{3953}[/imath]. The example just says it is [imath]20 \bmod{3953}[/imath] but I am unable to arrive at this answer. Thanks so much for your help! The examples solves the answer by using 2297 = 2^11 + 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^0 then 2023^2297=(2023^2)^11 * (2023^2)^7 etc and then arrives at 2023^2297 = 20 mod 3953. | 355615 | Basic Modulo Question
I've been having trouble with this example while studying for my exams. Why is [imath]2023^{2297}\equiv 20 \pmod{3953}\;?[/imath] Thanks so much for any help I can get! The examples solves the answer by using [imath]2297 = 2^{11} + 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^0[/imath] |
356451 | 6 points in a rectangle puzzle
Six points are in a rectangle with dimensions [imath]3[/imath] by [imath]4[/imath]. Prove that the distance between at least two of the six points is smaller than the square root of [imath]5[/imath]. Note that points can be on the edges and corners. General observations: The diagonal of the rectangle has length [imath]5[/imath]. [imath]\sqrt{5}[/imath] is about [imath]2.2[/imath]. Putting a point on each corner, and one in the center of the rectangle will work. The last point can not be put anywhere else. I can see how to do it. I am not really sure how to write genuine proof on the problem though. | 280362 | Distribution of points on a rectangle
Let [imath]R[/imath] be a rectangular region with sides [imath]3[/imath] and [imath]4[/imath]. It is easy to show that for any [imath]7[/imath] points on [imath]R[/imath], there exists at least [imath]2[/imath] of them, namely [imath]\{A,B\}[/imath], with [imath]d(A,B)\leq \sqrt{5}[/imath]. Just divide [imath]R[/imath] into six small rectangles with sides [imath]2[/imath] and [imath]1[/imath] and so at least one such rectangle must contain [imath]2[/imath] points from the seven ones. Thus the result follows. Here is the question: What about six points? I believe that the same is true. How do I prove my belief? ps: I don't want to find such [imath]6[/imath] points. I'd like to show it for any set with [imath]6[/imath] points. |
356154 | Show that [imath]K[x,xy,xy^2,\dots][/imath] is not Noetherian
Here is the problem I am stuck on: Fix a field [imath]K[/imath] and consider the subring [imath]A \leq K[x,y][/imath] generated by [imath]K \cup \{x,xy,\dots,\}[/imath]. Show that [imath]A[/imath] is not Noetherian. I figure that taking ideals [imath]I_n = (x,xy,\dots,xy^n)[/imath] should give an infinite strictly ascending chain, which would establish that [imath]A[/imath] is not Noetherian, but I cannot figure out how to show that each [imath]I_n[/imath] is a proper subset of [imath]I_{n+1}[/imath]. Any help here? | 274590 | How to prove that $k[x, xy, xy^2, \dotsc]$ is not noetherian?
Consider the subring [imath]R=k[x,xy,xy^2,\ldots][/imath] of [imath]k[x,y][/imath]. I want to prove that [imath]R[/imath] is not noetherian. An ascending chain of ideals is the following: [imath](x)\subset(x,xy)\subset(x,xy,xy^2)\subset\cdots[/imath] It is intuitively clearly to me that this is an ascending chain of ideals. But how do I prove it rigorously that [imath]xy^n \notin (x,xy,xy^2,\ldots,xy^{n-1})[/imath] or that this chain of ideals can never stabillize? |
350002 | Total variation norm in [imath]\mathbb{R}^n[/imath]
Let's consider total variation norm ρ( , ) on [imath](\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n)),[/imath] where [imath]\mathcal{B}(\mathbb{R}^n)[/imath] is a Borel [imath]\sigma[/imath]-algebra. Is it true that for probabiblity measures [imath]P[/imath] and [imath]Q[/imath] [imath]ρ(P,Q):=\sup_{A∈\mathcal{B}(\mathbb{R}^n)}|P(A)−Q(A)|=\sup_{I=I_1\times\dots\times I_n}|P(I)−Q(I)|?[/imath] Here [imath]I_k=(a_k, b_k].[/imath] If it is, why? Thanks in advance! | 125984 | Difference between two measures
Suppose that [imath](\mathscr X, \mathscr B)[/imath] is a measurable space and denote [imath]\Omega = \mathscr X^n[/imath] and let [imath]\mathscr F[/imath] be its product [imath]\sigma[/imath]-algebra. Consider two measures [imath]\nu[/imath] and [imath]\tilde \nu[/imath] on [imath]\Omega[/imath] with a condition that [imath] |\tilde \nu(B_0\times\dots\times B_n) - \nu(B_0\times\dots\times B_n)|\leq \epsilon [/imath] for any collection [imath]B_0,\dots,B_n\in \mathscr B[/imath]. I wonder how to prove (if it is indeed true) that [imath] |\tilde\nu(F) - \nu(F)|\leq \epsilon [/imath] where [imath]F\in \mathscr F[/imath] is arbitrary. I appreciate any hint. |
356815 | Solving for x in a mod relation?
How do I approach the problem: [imath]7^{95} \equiv x^3\text{(mod 10)}[/imath] when solving for [imath]x[/imath]? | 356637 | Congruence Relation with exponents and variables
I am currently trying to solve a congruence relation with a constant and a variable, both of which have attached exponents. The relation is as follows: [imath]7^{95}\equiv x^{3} (mod 10)[/imath] How does one go about solving for x when it has an attached exponent? |
357460 | Classifying Algebraic Structures as Fields
I just have a quick question! We got a question asking us which of the following algebraic structures were fields, I got every one of them except the last one. I don't even know how to begin how to attack the question. Is real numbers modulo 2[imath]\pi[/imath] a field? | 355001 | Is it in any way possible to work with reals modulo integer values?
I'm working with a strange and perhaps (mathematically) nonsensical realm. I'd like to know if we can work with [imath]\mathbb{R}/ \mathbb{Z}[/imath]. For example, if we take [imath](\pi \bmod 3)[/imath] we get [imath](\pi - 3)[/imath]. I'm wondering if I always end up with the same equivalence class modulo 3 if I truncate the reals in this way. In general, I'd like to know if I can work this way modulo any integer. I want to know if this will work for addition, subtraction and multiplication. |
357854 | Functions defined on General sets
I am learning how to determined whether a function is well defined. I am doing so by relying on two disticnt reasons that show a not well defined function: (1) There is no y that satisfies the given equation or (2) there are two different values of y that satisfy the quation. I am having difficulties determining if the following function is well defined: On certain computers the integer data type goes from [imath]-2, 147, 483, 648[/imath] through [imath]2, 147, 483, 647[/imath]. Let S be the set of all integers from [imath]-2, 147, 483, 648[/imath] through [imath]2, 147, 483, 647[/imath]. Try to define a function [imath]f:S -> S[/imath] by the rule [imath]f(n) = n^2[/imath] for each n in S. Is f well defined? Why? | 357875 | Determine if function is well defined
I am having difficulties determining if the following function is well defined: On certain computers the integer data type goes from [imath]-2, 147, 483, 648[/imath] through [imath]2, 147, 483, 647[/imath]. Let S be the set of all integers from [imath]-2, 147, 483, 648[/imath] through [imath]2, 147, 483, 647[/imath]. Try to define a function [imath]f:S -> S[/imath] by the rule [imath]f(n) = n^2[/imath] for each n in S. Is f well defined? Why? Side notes: I am learning how to determined whether a function is well defined. I am doing so by relying on two disticnt reasons that show a not well defined function: (1) There is no y that satisfies the given equation or (2) there are two different values of y that satisfy the equation. |
358049 | Let [imath] f:(X, d) \mapsto (Y,d) [/imath] be an mapping such that [imath] Graph (f) [/imath] is connected.
Where [imath] X [/imath] is connected. Does it imply [imath] f [/imath] to be continuous? | 183737 | Is a continuous function simply a connected function?
Intuitively, a function [imath]\mathbb{R}\rightarrow\mathbb{R}[/imath] is continuous if you can draw its graph without taking the pen off the page. This suggests the following theorem: A map [imath]f:X \rightarrow Y[/imath] is continuous if and only if [imath]f[/imath] is connected in the product topology [imath]X \times Y[/imath]. Is this true? And if not, can anyone think of an additional premise or two that would make it true? |
314936 | How to show that a finite commutative ring without zero divisors is a field?
[imath]R[/imath] is finite commutative ring without zero divisors which has at least two elements. How to show that [imath]R[/imath] is field? I'm just starting with abstract algebra and I'd really appreciate if someone could explain it to me. | 1081858 | Prove, that if the commutative ring has no zero divisors, then it is a field
Let [imath]R[/imath] be a commutative finite ring in which [imath]ab = 0[/imath] implies either [imath]a = 0[/imath] or [imath]b = 0[/imath] for any [imath]a,b \in R[/imath]. Then, [imath]R[/imath] is a field. I do not understand how I should act. I tried different ways, but I was not able to prove this assertion. |
358241 | How to show that for [imath]g\in G,u\in U,~gug^{-1}\in U?[/imath]
Let [imath]G[/imath] be a group and [imath]U=\{xyx^{-1}y^{-1}:x,y\in G\}.[/imath] How to show that for [imath]g\in G,u\in U,~gug^{-1}\in U?[/imath] Added: Actually I was trying to show that the commutator subgroup of [imath]G[/imath] is normal in [imath]G[/imath] and I've in hand the result which says for [imath]U\subset G,gug^{-1}\in U~\forall~g\in G,u\in U\implies(U)\lhd G.[/imath] | 141888 | How to show that the commutator subgroup is a normal subgroup
It is suggested as an exercise in Serge Lang's book "Algebra" to show that the commutator subgroup [imath]G^c[/imath] of a group [imath]G[/imath] is a normal subgroup. I'd like to do that but I am afraid I need help, I think the first thing I need to figure out is how a general element in the commutator subgroup looks like, so that I can check that the defining condition for normality is satisfied. That is, supposing for a moment that a general element in [imath]G^c[/imath] is denoted by [imath]g[/imath], I need to show that [imath]aga^{-1} \in G^c,[/imath] for all [imath]a \in G[/imath]. But here I get stuck, first because I am unsure how to write a general element in [imath]G^c[/imath] - a simple product in [imath]G^c[/imath] is of the form [imath]xyx^{-1}y^{-1}aba^{-1}b^{-1}[/imath] where [imath]a,b \in G[/imath]. I cannot see a way to simplify this - I am sure there is one, but somehow I am blind today. The second thing then is, even if one tries out the conjugation of a simple element like [imath]xyx^{-1}y^{-1}[/imath] in [imath]G^c[/imath], again not simplification offers itself easily I think .. what am I missing ? An alternative would be to find a homomorphism of [imath]G[/imath] whose kernel is precisely [imath]G^c[/imath] - here I tried to think of this as a map [imath]G \times G \to G[/imath] but whatever I cook up is not a homomorphism. Thanks for your hints !! |
358324 | number of Ring homomorphism
The number of non-trivial ring homomorphism from [imath]\mathbb Z _{12}[/imath] to [imath]\mathbb Z _{28}[/imath]. Is there any general formula for ring homomorphism between [imath]\mathbb Z _{m}[/imath] to [imath]\mathbb Z _{n}[/imath], like we have for group homomorphism from [imath]\mathbb Z _{m}[/imath] to [imath]\mathbb Z _{n}[/imath] which is [imath]gcd(m,n)[/imath]? | 303 | What are all the homomorphisms between the rings [imath]\mathbb{Z}_{18}[/imath] and [imath]\mathbb{Z}_{15}[/imath]?
Any homomorphism [imath]φ[/imath] between the rings [imath]\mathbb{Z}_{18}[/imath] and [imath]\mathbb{Z}_{15}[/imath] is completely defined by [imath]φ(1)[/imath]. So from [imath]0 = φ(0) = φ(18) = φ(18 \cdot 1) = 18 \cdot φ(1) = 15 \cdot φ(1) + 3 \cdot φ(1) = 3 \cdot φ(1)[/imath] we get that [imath]φ(1)[/imath] is either [imath]5[/imath] or [imath]10[/imath]. But how can I prove or disprove that these two are valid homomorphisms? |
286634 | Non surjectivity of [imath]C^1[/imath] function.
I am trying to figure out how to do the following question, but i seems to not have any success. If [imath]f:\mathbb{R}^1\to \mathbb{R}^2[/imath] is of class [imath]C^1[/imath] mapping, show that [imath]f[/imath] does not carry [imath]\mathbb{R}^1[/imath] onto [imath]\mathbb{R}^2[/imath]. I know that if suppose that [imath]f[/imath] is onto, then let [imath]g[/imath] be a right inverse of [imath]f[/imath] where [imath]f(g(x,y))=x[/imath]. If I differentiate [imath]f(g(x,y))=x[/imath], using the chain rule, i get some expression, but there is a problem because i am not sure if [imath]g(x,y)[/imath] is itself differentiable to begin with. Alternatively, if i let the mapping of [imath]f[/imath] be written as [imath]f(t)=(\phi_1(t), \phi_2(t))[/imath], then the Jacobian matrix [imath]Df(t)[/imath] is a [imath]2[/imath] by [imath]1[/imath] matrix. So the column rank of [imath]Df(t)[/imath] is [imath]1[/imath]. But how can i use this information to conclude that [imath]f[/imath] is not onto, since if [imath]f[/imath] were onto in the first place, won't the column rank of its Jacobian be equal to the dimension of the function's codomain. Thanks in advance | 179290 | There is no [imath]C^1[/imath] function [imath]f[/imath] mapping an open interval in [imath]\mathbb{R}[/imath] onto open ball in [imath]\mathbb{R}^2[/imath]
We know that there are no [imath]C^1[/imath] functions which map open [imath]E\subset \mathbb{R}^2[/imath] INTO [imath]\mathbb{R}[/imath]. (Actually we can drop the [imath]C^1[/imath] requirement and just use continuity). A nice related question is that There is no [imath]C^1[/imath] function [imath]f[/imath] mapping an open interval in [imath]\mathbb{R}[/imath] ONTO open ball in [imath]\mathbb{R}^2[/imath]. But I can't prove it, any suggestion? |
358868 | Show that f is not surjective.
Let [imath]f : S^1 \to \mathbb{R}[/imath] be continuous, where [imath]S^1[/imath] is the unit circle in [imath]\mathbb{R}^2[/imath]. (a) Show that there is a point [imath]z \in S^1[/imath] such that [imath]f(z) = f(−z)[/imath] where [imath]z = (x, y), −z = (−x, −y)[/imath] (b) Show that [imath]f[/imath] is not surjective. | 357570 | general topology exercise 777
Let f : [imath]S^1[/imath] → R be continuous, where [imath]S^1[/imath] is the unit circle in [imath]R^2[/imath]. (a) Show that there is a point z ∈ [imath]S^1[/imath] such that f(z) = f(−z). [z = (x; y), −z = (−x;−y)]\ (b) Show that f is not surjective. |
359122 | Binary Relations and Total Orders
This question has two parts: 1) i) Given a relation [imath]<[/imath], define a relation [imath]\leq[/imath] by setting x [imath]\leq[/imath]y if and only if x[imath]<[/imath]y or x = y. Prove that if < satisfies transitivity and 'trichotomy' then [imath]\leq[/imath] is a total ordering. ii) Given a relation [imath]\leq[/imath], define a relation < by setting x[imath]<[/imath]y if and only if x[imath]\leq[/imath]y and x[imath]\neq[/imath]y. Prove that if [imath]\leq[/imath] is a total ordering then < satisfies transitivity and 'trichotomy'. What I've gathered: So I know from class that the trichotomy is that either x[imath]<[/imath]y, y[imath]<[/imath]x or x[imath]=[/imath]y For a relation to be total ordered, it must be antisymmetric, transitive and total (a[imath]\leq[/imath]b or b[imath]\leq[/imath]a) My thoughts on the question: i) So we are given the relation <, now this is the part where I get stuck, do I proceed with the question assuming that < is defined for transitivity and the trichotomy? So then since < is defined, then say for a, b and c, then since < satisfies trichotomy and transitivity, then that implies that if [imath]a<b[/imath] and [imath]b<c[/imath], then [imath]a<c[/imath] (transitive) and [imath]a<b[/imath] or [imath]a>b[/imath] or [imath]a=b[/imath] (trichotomy), but I don't know how to apply that knowledge to show that '[imath]\leq[/imath]' is a total order. | 358932 | Help on total ordering and partial ordering
Hi guys my final exam is coming up, and I am given a few practice problems and I don't get how to do these questions on total ordering and partial ordering. Here are the question and here is how I did it. 1) On the set [imath]\mathbb{N} \times \mathbb{N}[/imath] define: [imath](x_1 , y_1) \leq (x_2,y_2)[/imath] if either [imath]x_1 < x_2[/imath] or [imath]x_1=x_2[/imath] and [imath]y_1\leq y_2[/imath]. Prove that this relation is a partial ordering. 2) Given a relation [imath]<[/imath] define a relation [imath]\leq[/imath] by setting [imath]x \leq y[/imath] if and only if [imath]x<y[/imath] or [imath]x=y[/imath]. Prove that if [imath]<[/imath] satisfies transitivity and 'trichotomy' then [imath]\leq[/imath] is a total ordering 3) Given a relation [imath]\leq[/imath], define a relation [imath]<[/imath] by setting [imath]x<y[/imath] if and only if [imath]x \leq y[/imath] and [imath]x \neq y[/imath]. Prove that if [imath]\leq[/imath] is a total ordering then [imath]<[/imath] satisfies transitivity and 'trichotomy'. For (1) I know partial ordering means I have to show if it is reflexive, antisymmetric and transitivity. I think I know how to do if it is reflexive and transitive but I am having trouble in how to do if it is anti symmetric. For (2) and (3) I am having trouble understanding in what it means |
359220 | Prove disprove, and find the limit
Can anyone solve the following two for me: 1-Let [imath]a[/imath], [imath]b[/imath], [imath]c[/imath] and [imath]d[/imath] be real numbers satisfying [imath]a>b[/imath] and [imath]c>d[/imath]. Does this imply that [imath]ac>bd[/imath]? Prove or disprove. 2-From analysis and the concepts of limits find [imath] \lim_{n\to\infty}\frac{3^n+n^3}{(2n)^2+2^{2n}}\,. [/imath] I really appreciate your efforts. Thank you. | 359213 | Prove disprove that for [imath]a[/imath], [imath]b[/imath], [imath]c[/imath], [imath]d[/imath] are real numbers
Can any one solve the following two for me: 1- Let [imath]a[/imath], [imath]b[/imath], [imath]c[/imath] and [imath]d[/imath] be real numbers satisfying [imath]a>b[/imath] and [imath]c>d[/imath]. Does this imply that [imath]ac>bd[/imath]? Prove or disprove. 2- From the analysis concepts find [imath] \lim_{n\to\infty}\frac{3^n+n^3}{(2n)^2+2^{2n}}\,. [/imath] I really appreciate your effort. |
11538 | Intuition for uniform continuity of a function on [imath]\mathbb{R}[/imath]
I understand the formal definition of uniform continuity of a function, and how it is different from standard continuity. My question is: Is there an intuitive way to classify a function on [imath]\mathbb{R}[/imath] as being uniformly continuous, just like there is for regular continuity? I mean, that for a "nice" function [imath]f:\mathbb{R} \to \mathbb{R}[/imath], it is usually easy to tell if it is continuous on an interval by looking at or thinking of the graph of the function on the interval, or on all of [imath]\mathbb{R}[/imath]. Can I do the same for uniform continuity? Can I tell that a function is uniformly continuous just by what it looks like? Ideally, this intuition would fit with the Heine-Cantor theorem for compact sets on [imath]\mathbb{R}[/imath]. | 2167785 | What does it mean that a function is uniformly continuous? (Geometrically)
What does it mean that a function is uniformly continuous? (Geometrically) I have this definition but i dont understand the definition geometrically. Can someone help me to understand this? Let [imath]D\subseteq\mathbb{R}[/imath] and [imath]f:D\rightarrow\mathbb{R}[/imath], [imath]f[/imath] is uniformly continuous if [imath]\forall\epsilon>0[/imath] exists [imath]\delta>0[/imath] such that [imath]\forall x,y\in D[/imath] and [imath]\text{|x-y|<}\delta[/imath] then [imath]|f(x)-f(y)|<\epsilon[/imath] |
359042 | Prove that [imath]Tx=x^{-1}~\forall~x\in G.[/imath]
Let [imath]G[/imath] be a finite group and suppose the automorphism [imath]T[/imath] sends more than [imath]\dfrac{3}{4}th[/imath] of the elements of [imath]G[/imath] onto their inverses. Prove that [imath]Tx=x^{-1}~\forall~x\in G.[/imath] | 330138 | If [imath]|\lbrace g \in G: \pi (g)=g^{-1} \rbrace|>\frac{3|G|}{4}[/imath], then [imath]G[/imath] is an abelian group.
Assume that [imath]\pi[/imath] is an automorphism of a finite group [imath]G[/imath]. Let [imath]S[/imath] denote the set [imath]\lbrace g \in G: \pi (g)=g^{-1} \rbrace[/imath]. Show that if [imath]|S|>\frac{3|G|}{4}[/imath], then [imath]G[/imath] is an abelian group. Anyone has any idea on how to solve this ? I have no idea how to start. |
165696 | Your favourite application of the Baire Category Theorem
I think I remember reading somewhere that the Baire Category Theorem is supposedly quite powerful. Whether that is true or not, it's my favourite theorem (so far) and I'd love to see some applications that confirm its neatness and/or power. Here's the theorem (with proof) and two applications: (Baire) A non-empty complete metric space [imath]X[/imath] is not a countable union of nowhere dense sets. Proof: Let [imath]X = \bigcup U_i[/imath] where [imath]\mathring{\overline{U_i}} = \varnothing[/imath]. We construct a Cauchy sequence as follows: Let [imath]x_1[/imath] be any point in [imath](\overline{U_1})^c[/imath]. We can find such a point because [imath](\overline{U_1})^c \subset X[/imath] and [imath]X[/imath] contains at least one non-empty open set (if nothing else, itself) but [imath]\mathring{\overline{U_1}} = \varnothing[/imath] which is the same as saying that [imath]\overline{U_1}[/imath] does not contain any open sets hence the open set contained in [imath]X[/imath] is contained in [imath]\overline{U_1}^c[/imath]. Hence we can pick [imath]x_1[/imath] and [imath]\varepsilon_1 > 0[/imath] such that [imath]B(x_1, \varepsilon_1) \subset (\overline{U_1})^c \subset U_1^c[/imath]. Next we make a similar observation about [imath]U_2[/imath] so that we can find [imath]x_2[/imath] and [imath]\varepsilon_2 > 0[/imath] such that [imath]B(x_2, \varepsilon_2) \subset \overline{U_2}^c \cap B(x_1, \frac{\varepsilon_1}{2})[/imath]. We repeat this process to get a sequence of balls such that [imath]B_{k+1} \subset B_k[/imath] and a sequence [imath](x_k)[/imath] that is Cauchy. By completeness of [imath]X[/imath], [imath]\lim x_k =: x[/imath] is in [imath]X[/imath]. But [imath]x[/imath] is in [imath]B_k[/imath] for every [imath]k[/imath] hence not in any of the [imath]U_i[/imath] and hence not in [imath]\bigcup U_i = X[/imath]. Contradiction. [imath]\Box[/imath] Here is one application (taken from here): Claim: [imath][0,1][/imath] contains uncountably many elements. Proof: Assume that it contains countably many. Then [imath][0,1] = \bigcup_{x \in (0,1)} \{x\}[/imath] and since [imath]\{x\}[/imath] are nowhere dense sets, [imath]X[/imath] is a countable union of nowhere dense sets. But [imath][0,1][/imath] is complete, so we have a contradiction. Hence [imath]X[/imath] has to be uncountable. And here is another one (taken from here): Claim: The linear space of all polynomials in one variable is not a Banach space in any norm. Proof: "The subspace of polynomials of degree [imath]\leq n[/imath] is closed in any norm because it is finite-dimensional. Hence the space of all polynomials can be written as countable union of closed nowhere dense sets. If there were a complete norm this would contradict the Baire Category Theorem." | 987599 | Applications of Baire's Threom
In a lecture on Baire's Theorem (for complete metric spaces), I gave, for a rather advanced undergraduate class in Real Analysis (covering the theory of metric spaces and elements of general topology), I realised that all the applications I provided were either straight-forward or could be proved using more elementary tools. For example: a. [imath]\mathbb Q[/imath] is not complete. b. [imath]\mathbb R[/imath] is uncountable. (Otherwise, if [imath]\mathbb R=\{x_n\}_{n\in\mathbb N}[/imath], then [imath]\bigcap_{n\in\mathbb N} (\mathbb R\setminus\{x_n\})[/imath] would be dense and empty.) c. A complete metric, with the property that every point is an accumulation point, is uncountable. Could you suggest more interesting applications? They do not have to be too easy to prove. |
320035 | If [imath]S[/imath] is an infinite [imath]\sigma[/imath] algebra on [imath]X[/imath] then [imath]S[/imath] is not countable
I am going over a tutorial in my real analysis course. There is an proof in which I don't understand some parts of it. The proof relates to the following proposition: ([imath]S[/imath] - infinite [imath]\sigma[/imath]-algebra on [imath]X[/imath]) [imath]\implies [/imath] [imath]S[/imath] is uncountable. Proof: Assume: [imath]S=\{A_{i}\}_{i=1}^{+\infty}[/imath]. [imath]\forall x\in X: B_{x}:=\cap_{x\in A_{i}}A_{i}[/imath]. [Note: [imath]B_{x}\in S[/imath] [imath]\impliedby[/imath] ([imath]B_{x}[/imath] - countable intersection]. Lemma: [imath]B_{x}\cap B_{y}\neq\emptyset \implies B_{x}=B_{y}[/imath]. Proof(of lemma): [imath]z\in B_{x}\cap B_{y} \implies B_{z}\subseteq B_{x}\cap B_{y}[/imath]. 1.[imath]x\not\in B_{z} \implies B_{x}\setminus B_{z} \subset S \wedge x\in B_{x}\setminus B_{z} \wedge B_{x}\setminus B_{z} \subset S[/imath] (contradiction:[imath]\space[/imath] definition of [imath]B_{x}[/imath]) [imath]\implies[/imath] [imath]B_{z}=B_{x}[/imath] 2.[imath]y\not\in B_{z} \implies y\in B_{y} \setminus B_{z} \space \wedge \space B_{y}\setminus B_{z} \subset S \space\wedge\space B_{y} \setminus B_{z}\subset B_{y} [/imath](contradiction: definition of [imath]B_{y}[/imath]) [imath]\implies[/imath] [imath]B_{z}=B_{y}[/imath] [imath]\implies B_{x}=B_{y} \space \square[/imath] Consider: [imath]\{B_{x}\}_{x\in X}[/imath]. If: there are finite sets of the form [imath]B_{x}[/imath] then: [imath]S[/imath] is a union of a finite number of disjoint sets [imath]\implies[/imath] [imath]S[/imath] is finite [imath]\implies[/imath] there is an infinite number of sets of the form [imath]B_{x}[/imath]. [imath]\implies[/imath] [imath]|\bigcup\limits_{i\in A \subseteq\mathbb{N}}B_{x_{i}}| \geq \aleph_{0}[/imath].(contradiction) [imath]\square[/imath] There are couple of things I don't understand in this proof: Why the fact that we found a set ([imath]B_{x}\setminus B_{z}[/imath]) in [imath]S[/imath] containing [imath]x[/imath] and is strictly contained in [imath]B_{x}[/imath] a contradiction ? Why if there are only a finite number of different sets of the form [imath]B_{x}[/imath] then [imath]S[/imath] is a union of a finite number of disjoint sets and is finite ? | 926736 | [imath]\sigma[/imath]-algebra that is countable, but not finite
I just had a chat with my real analysis teacher today, we talked about how there is no [imath]\sigma[/imath]-algebra such that it has countably many elements, but finitely many. In the end, he said that I can visualize it by drawing out the [imath]\sigma[/imath]-algebra on (0,1) using intervals. But I'm not too sure how to do that, should I draw it out on the real line? Or can I use any type of diagram? |
353166 | Show that this language cannot be accepted by a deterministic push-down automaton
How do you show that there exists no DPDA that accepts [imath] L = \{0^n1^n \} \cup \{ 0^n1^{2n}\}[/imath] ? | 204572 | Context free languages closure property [imath]\{a^n b^n : n\geq 0\} \cup \{a^n b^{2n}: n\geq 0\}[/imath]
I have been working on the following two problems: 1) Given any context free language L, form a new language by taking symbols at the odd positions, i.e. [imath]w=a_1a_2\dots a_n \mapsto w'=a_1 a_3 a_5 \dots a_k[/imath] if n is even then k=n-1 otherwise n. Is the resulting language context free for any given context free language [imath]L[/imath]? 2) How to show [imath]\{a^n b^n : n\geq 0\} \cup \{a^n b^{2n}: n\geq 0\}[/imath] is not deterministic context free? For the second problem, my initial thought was trying to prove its complement (maybe intersecting with some regular languages) is not context free. But it did not work out so well. So I think it might be reasonable to prove that it or its complement is inherently ambiguous then it is not deterministic context free. However, the ideas are quite hard to implement. Any thoughts? |
359872 | Stably free module
Let [imath]M[/imath] be any [imath]A[/imath]-module. Is it true that [imath]M \oplus A \simeq A^{2} [/imath] implies [imath] M \simeq A [/imath] ? | 280180 | [imath]M\oplus A \cong A\oplus A[/imath] implies [imath]M\cong A[/imath]?
Let [imath]A[/imath] be a commutative unital ring and [imath]M[/imath] an [imath]A[/imath]-module. Suppose that [imath]M\oplus A \cong A\oplus A[/imath]. Then is [imath]M\cong A[/imath]? We have that both [imath]M\oplus A[/imath] and [imath]A\oplus A[/imath] are biproduct for [imath](A, A)[/imath] and [imath](M, A)[/imath], so actually short exact sequences. Further [imath]A\oplus A[/imath] is the free [imath]A[/imath]-module of rank two. However I can't conclude. Maybe I have to use some extension stuff from homological theory or there are some apparent counter-examples that I'm not able to figure out. |
360077 | Is this necessarily a basis?
Let [imath]T[/imath] be a linear endomorphism of [imath]\mathbb{R}^{3}[/imath], suppose there exists non-zero vectors [imath]u,v,w[/imath] with [imath]T(u)=u, T(v)=2v, T(w)=3w[/imath]. Is [imath]\{u,v,w\}[/imath] necessarily a basis of [imath]\mathbb{R}^{3}[/imath]? | 359125 | Proving linear independence
Let [imath]A[/imath] be an [imath]n \times n[/imath] matrix and suppose [imath]v_1, v_2, v_3 \in \mathbb{R}^n[/imath] are nonzero vectors that satisfy: [imath] Av_1 = v_1 \\ Av_2 = 2v_2 \\ Av_3 = 3v_3 [/imath] Prove that [imath]\{v_1, v_2, v_3\}[/imath] is linearly independent. (Hint: Start by showing that [imath]\{v_1, v_2\}[/imath] must be linearly independent.) I know if [imath]A[/imath] is non-singular,it is easy. But if [imath]A[/imath] is singular, I have no idea to get that. |
360069 | Meaure theory problem no. 12 page 92, book by stein and shakarchi
Show that there are [imath]f \in L^1(\Bbb{R}^d,m)[/imath] and a sequence [imath]\{f_n\}[/imath] with [imath]f_n \in L^1(\Bbb{R}^d,m)[/imath] such that [imath]\|f_n - f\|_{L^1} \to 0[/imath], but [imath]f_n(x) \to f(x)[/imath] for no [imath]x[/imath]. | 138043 | Does convergence in [imath]L^{p}[/imath] implies convergence almost everywhere?
If I know [imath]\|f_{n}(x) - f(x)\|_{L^{p}(\mathbb{R})} \rightarrow 0[/imath] as [imath]n \rightarrow \infty[/imath], do I know [imath]\lim_{n \rightarrow \infty}f_{n}(x) = f(x)[/imath] for almost every [imath]x[/imath]? |
360339 | Number of irreducible polynomials over [imath]\mathbb Z_p[/imath]
How many irreducible polynomials over [imath]\mathbb Z_p[/imath] of the form [imath]x^2+ax+b[/imath] are there? No idea. | 40811 | Number of monic irreducible polynomials of prime degree [imath]p[/imath] over finite fields
Suppose [imath]F[/imath] is a field s.t [imath]\left|F\right|=q[/imath]. Take [imath]p[/imath] to be some prime. How many monic irreducible polynomials of degree [imath]p[/imath] do exist over [imath]F[/imath]? Thanks! |
360355 | Continous map question
Let [imath]f:[0,1]\rightarrow [0,1][/imath] be continuous s.t. [imath]f(0)=f(1)[/imath]. Let [imath] A = \{ (t,s)\in [0,1] \times [0,1]|: t \neq s\,\,\, \rm and \,\,\, \ f(t)=f(s)\},[/imath] then find the number of elements in [imath]A[/imath]. | 348220 | A problem on continuous functions
[imath]f : S^1 \rightarrow \mathbb{R}[/imath] is a continuous map. Define [imath]A = \{(x, y) \in S^1 \times S^1: x \neq y, f(x) = f(y)\}.[/imath] We want to prove that [imath]A[/imath] has uncountably many points. It seems very evident, but I want a rigorous argument. Thanks in advance. |
348627 | If [imath]f^2[/imath] and [imath]f^3[/imath] are analytic prove that [imath]f[/imath] is analytic at every point of [imath]\mathbb{C}[/imath].
Let [imath]f : \mathbb{C} \to \mathbb{C}[/imath] be continuous. If [imath]f^2[/imath] and [imath]f^3[/imath] are analytic prove that [imath]f[/imath] is analytic at every point of [imath]\mathbb{C}[/imath]. if [imath]f^2[/imath] has no zero then [imath]f=f^3/f^2[/imath] and then it is analytic.but if [imath]f^2[/imath] has zero then how can I able to proceed.help me please. | 298951 | Proving that if [imath]f: \mathbb{C} \to \mathbb{C} [/imath] is a continuous function with [imath]f^2, f^3[/imath] analytic, then [imath]f[/imath] is also analytic
Let [imath]f: \mathbb{C} \to \mathbb{C}[/imath] be a continuous function such that [imath]f^2[/imath] and [imath]f^3[/imath] are both analytic. Prove that [imath]f[/imath] is also analytic. Some ideas: At [imath]z_0[/imath] where [imath]f^2[/imath] is not [imath]0[/imath] , then [imath]f^3[/imath] and [imath]f^2[/imath] are analytic so [imath]f = \frac{f^3}{f^2}[/imath] is analytic at [imath]z_0[/imath] but at [imath]z_0[/imath] where [imath]f^2[/imath] is [imath]0[/imath], I'm not able to show that [imath]f[/imath] is analytic. |
360684 | [imath]\{\sigma\in S_n;\sigma(n)=n\}[/imath] is isomorphic to [imath]S_{n-1}[/imath]
I'm trying to prove that the set [imath]K=\{\sigma\in S_n;\sigma(n)=n\}[/imath] is a subgroup of [imath]S_n[/imath] which is isomorphic to [imath]S_{n-1}[/imath]. In order to prove that this set is a subgroup of [imath]S_n[/imath], I did this: [imath]\alpha[/imath], [imath]\beta\in K[/imath], then [imath]\alpha\beta^{-1}(n)=\sigma(n)=n\implies \alpha\beta^{-1}\in K[/imath], then we can conclude easily that K is a subgroup of [imath]S_n[/imath]. Is that approach correct? And to prove the second part of this question I know intuitively it's true but I couldn't find an isomorphic function between [imath]K[/imath] and [imath]S_{n-1}[/imath]. Thanks a lot | 138384 | Show that a subgroup is isomorphic to [imath]S_{n-1}[/imath]
Q: If [imath] H = \{ \sigma \in S_n : \sigma(n) = n \} [/imath] is a subgroup of [imath]S_n[/imath], then show that [imath]H \simeq S_{n-1}[/imath]. I know any group is isomorphic to a subgroup of the symmetric group. But I don't know how to proceed. |
4744 | Formal power series coefficient multiplication
Given that I have two formal power series: [imath] A(x) = \sum_{k \ge 0} a_k x^k [/imath] [imath] B(x) = \sum_{k \ge 0} b_k x^k [/imath] The Cauchy Product gives a series [imath] C(x) = \sum_{k \ge 0} c_k x^k [/imath] [imath] c_k = \sum_{n=0}^k a_n b_{k-n} [/imath] Which comes from taking the product of the two series [imath]C(x)=A(x)B(x)[/imath]. What then, in terms of [imath]A(x)[/imath] and [imath]B(x)[/imath], is this series? [imath]Y(x) = \sum_{k \ge 0} a_k b_k x^k [/imath] | 688273 | Finding generating function for product of two sequences
If I know generating funcions for sequences [imath]A: a_0, a_1, a_2, a_3, a_4, \dots[/imath] and [imath]B: b_0, b_1, b_2, b_3, b_4, \dots[/imath] and I want to find a new generating function for [imath]C: a_0b_0, a_1b_1, a_2b_2, a_3b_3, a_4b_4,\dots[/imath] How would I go about doing this? I would like an explanation, thanks! |
360945 | Show that if [imath]D[/imath] is a domain but not a field then [imath]D[x][/imath] is not a principal ideal domain.
Show that if [imath]D[/imath] is a domain but not a field then [imath]D[x][/imath] is not a principal ideal domain. Sorry for my english.... | 356924 | A commutative ring [imath]A[/imath] is a field iff [imath]A[x][/imath] is a PID
Wikipedia says that a commutative ring [imath]A[/imath] is a field iff [imath]A[x][/imath] is a PID. The "only if" part is easy: we just apply the Euclidean algorithm. I've stumbled trying to prove the "if" part, though. [imath]\newcommand{\aa}{\mathfrak{a}}[/imath] My best attempt so far: suppose [imath]\aa[/imath] is an ideal of [imath]A[/imath], and [imath]\aa'[/imath] is the ideal of [imath]A[x][/imath] spanned by [imath]i(\aa)[/imath], where [imath]i: A \hookrightarrow A[x][/imath] is the natural embedding. [imath]i^{-1}(\aa') = \aa[/imath] for grading reasons ([imath]i(A) = A[x]^0[/imath], and multiplication by a non-zero degree polynomial takes us out of [imath]i(A)[/imath], because [imath]A \cong i(A)[/imath] is integral). [imath]A[x][/imath] is a PID so [imath]\aa' = (a')[/imath], where [imath]a' = i(a)[/imath] for some [imath]a \in A[/imath]. Therefore, [imath]\aa = (a)[/imath], and thus [imath]A[/imath] is also a PID. That's the best way to use the fact that [imath]A[x][/imath] is a PID that I've found so far. Now I feel like there must be a trick to show that if [imath]a \neq 0[/imath] then [imath]a = 1[/imath], but I don't know how to do this. Any hints? |
361114 | Binomial expansion for solving american put option identity
For american put option I have to prove that: 1) As [imath]D[/imath] tends to [imath]\infty[/imath], [imath]a_n[/imath] tends to [imath]-r/D[/imath] so that [imath]S^*[/imath] tends to [imath]0[/imath]. 2) As [imath]D[/imath] tends to [imath]-\infty[/imath], [imath]a_n[/imath] tends to [imath]2D/ \sigma^2[/imath] so that [imath]S^*[/imath] tends to [imath]E[/imath]. I have simplified the question posted here american put option to the point where [imath]a_n =\frac1{\sigma^2}\lim_{D\to\infty}[-\sqrt{D^2 + 2 r \sigma^2} + D][/imath]. I think to prove 1) and 2) above I have to do a binomial expansion on the square root keeping only the first two terms in the expansion I am not sure how do I go about this. Any help would be greatly appreciated. | 361096 | american put option
For a perpetual american put option [imath]v(s)[/imath], satisfies the following problem: [imath]\frac12\sigma^2S^2\frac{\mathrm d^2V}{\mathrm dS^2}+(r-D)S\frac{\mathrm dV}{\mathrm dS} - rV = 0\quad\text{for }S^*<S<\infty,[/imath] [imath]V(S) = E- S\quad\text{for }0\le S<S^*,[/imath] [imath]V(S^*) = E- S^*, \quad \frac{\mathrm dV}{\mathrm dS}(S^*) = -1,\quad \lim_{S\to\infty} V(S) = 0.[/imath] If [imath]V(S) = S^a[/imath] is the solution of the differential equation, I can derive that [imath]\tag1 \frac12\sigma^2a^2 + \left(r-D-\frac12\sigma^2\right)a - r = 0.[/imath] One of the roots of the equation, [imath]a_n[/imath] is negative and for [imath]S > S^*[/imath], [imath] V(S) =-\frac{S^*}{a_n} \left(\frac S{S^*}\right)^{a_n},\quad S^* = \frac{a_n}{a_n-1}E.[/imath] I am given the following condition holds [imath]\tag2 \sigma^2a_n+ \left(r - D - \frac12 \sigma^2\right)= - \sqrt{\left(r - D - \frac12 \sigma^2\right)^2 + 2 r \sigma^2} < 0.[/imath] Suppose now the parameters [imath]\sigma>0[/imath], [imath]r>0[/imath], and [imath]E>0[/imath] are fixed, but [imath]D[/imath] can vary, so you may regard [imath]a_neg[/imath] as a function of [imath]D[/imath], how can I show the following: As [imath]D[/imath] tends to [imath]\infty[/imath], [imath]a_n[/imath] tends to [imath]-\frac rD[/imath] so that [imath]S^*[/imath] tends to [imath]0[/imath]. As [imath]D[/imath] tends to [imath]-\infty[/imath], [imath]a_n[/imath] tends to [imath]\frac{2D}{\sigma^2}[/imath] so that [imath]S^*[/imath] tends to [imath]E[/imath]. By differentiating (1) above with respect to [imath]D[/imath] and using (2) deduce that [imath]\frac{\mathrm \partial a_n}{\mathrm \partial D} > 0[/imath] and hence [imath]\frac{\mathrm \partial S^*}{\mathrm \partial D} < 0[/imath]. Please can you help me in proving this property? Many thanks |
361049 | Real analysis - converging sequence
My answer Solution 1). [imath]Let\; \epsilon = L/2 > 0 \mbox{thus by definition of}\; x_m→L, \mbox{there exists}\; a \;n_o∈ N \;\mbox{such that }∀m>n_o\; \\ |x_m - L|< ε\\ -ε <|x_m - L| < ε\\ x_m – L > -ε = L - L/2 = ½ L >0[/imath] [imath]2).False\\\mbox {Counter Example Let} x_1=-1,x_2=2, x_3=2……………………..x_n=2[/imath] | 357812 | Positive limit of sequence vs. positive terms
Let [imath]\{x_m\}[/imath] be a sequence in [imath]E_1[/imath] that converges to [imath]L \in E_1[/imath]. a. prove that if [imath]L>0[/imath] and there exists [imath]n \in N[/imath] such that for all [imath]m >n[/imath] holds that [imath]x_m > 0[/imath] b. True or false? If for all [imath]m \in N[/imath] holds that [imath]x_m > 0[/imath], then [imath]L > 0[/imath]. |
361258 | In a principal ideal ring, is every nonzero prime ideal maximal?
Inspired by this question, I was wondering whether from just the hypothesis that [imath]A[X][/imath] is a nontrivial (commutative) principal ideal ring (so without supposing it is a domain) one can deduce that [imath]A[/imath] is a field. One possibility to prove that would be to use the fact, if it is one, that in principal ideal rings nonzero prime ideals are maximal. Namely, if [imath]\def\p{\mathfrak p}\p\subset A[/imath] is a prime ideal, then [imath](A/\p)[X][/imath], which is a quotient of [imath]A[X][/imath] that is an integral domain but not a field, would have to be [imath]A[X][/imath] itself, so [imath]\p=(0)[/imath], and [imath]A[/imath] having no nonzero prime ideals would have to be a field. I think I can prove this, but I don't really like the argument I found, and since commutative algebra with zero divisors is not my speciality, I might have tripped up. So my question is are the proposition and its proof below correct, and if so is there a more elementary proof? Notably I tried, under the hypothesis that [imath]R[/imath] is a principal ideal ring and [imath]\p\subset R[/imath] a prime ideal, and supposing [imath]R\supset I=xR\supset\p=pR[/imath], to use the existence of [imath]y[/imath] with [imath]xy=p[/imath] to arrive at a contradiction; but from [imath]\p[/imath] prime I only get [imath]yR=pR[/imath], and since over rings with zero divisors principal ideals may concide without their generators being associated, I don't see how to conclude. (Proposed but false) Proposition. Any nonzero prime ideal [imath]\p[/imath] of a principal ideal ring [imath]R[/imath] is maximal in [imath]R[/imath]. Proof. According to the Zariski-Samuel theorem [imath]R[/imath] is isomorphic to a finite direct product of rings that are either a principal ideal domain or a special principal ring. Every ideal [imath]I[/imath] of a finite product [imath]R_1\times\cdots\times R_n[/imath] is equal to a product [imath]I_1\times\cdots\times I_n[/imath] of ideals of the respective rings, where [imath]I_k[/imath] is the projection of [imath]I[/imath] to [imath]R_k[/imath] (clearly [imath]I\subseteq I_1\times\cdots\times I_n[/imath], but also [imath](0,\ldots,0,1,0,\ldots,0)I=\{0\}\times\cdots\{0\}\times I_k\times\{0\}\times\cdots\times\{0\}\subseteq I[/imath]) and since a product of more than one nontrivial ring is never an integral domain, [imath]\p[/imath] has to be of the form [imath]R_1\times\cdots\times R_{k-1}\times\p_0\times R_{k+1}\times\cdots\times R_n[/imath] where [imath]\p_0\subset R_k[/imath] is a prime ideal. So we have reduced to proving the proposition in the cases that [imath]R[/imath] is a principal ideal domain or a special principal ring. The former case is well known, and in a nontrivial special principal ring every ideal is a power of the unique maximal ideal [imath]\def\m{\mathfrak m}\m[/imath]; putting [imath]\p=\m^k[/imath] a generator of [imath]\m[/imath] gives a nilpotent element of the integral domain [imath]R/\p[/imath], which must be zero, so [imath]\p=\m[/imath] as desired. QED | 91587 | Ring of Polynomials is a Principal Ideal Ring implies Coefficient Ring is a Field?
I read this proof that if [imath]D[/imath] is an integral domain and [imath]D[X][/imath] is a principal ideal domain, then [imath]D[/imath] is a field. My question is if the requirements can be relaxed a bit, namely: Is it true that if [imath]D[/imath] is a commutative unitary ring and [imath]D[x][/imath] is a principal ideal ring (this allows zero-divisors), then [imath]D[/imath] is a field? I would be very pleased if anyone could give me a counter-example or could sketch a proof, certainly the linked proof would completely break down in this case as one could not use the properties of degree. |
361969 | What does elementary function mean?
I am looking at my double integration example problems and I see a note that says integral of [imath]e^{-x^2}[/imath] is not an elementary function. What does that mean? And why isn't that an elementary function? It actually says observe integral of [imath]e^{-x^2} dx[/imath] is not an elementary function. But why is that? Same goes for integral of (sin(x)/x) dx, why isn't that an elementary function also? | 118113 | What makes elementary functions elementary?
Is there a mathematical reason (or possibly a historical one) that the "elementary" functions are what they are? As I'm learning calculus, I seem to focus most of my attention on trigonometric, logarithmic, exponential, and [imath]n[/imath]th roots, and solving problems that have solutions which are elementary functions. I've been curious why these functions are called elementary, as opposed to some other functions that turn up rather naturally in mathematics. What is the reason that these functions take up most of our attention, and is there a reason that some additional functions are not included amongst the elementary functions? In other words, what property or properties do these functions possess that separates them from non-elementary functions (if there is one)? |
361746 | Number of matrices giving even row and column sum
Count the number [imath]n\times n[/imath] matrices such that every row and column sum to an even number. Each element is either [imath]0[/imath] or [imath]1[/imath]. I know for a there are [imath]2^n[/imath] subsets of [imath]\{1,2,\dots,n\}[/imath]. This gives [imath]2^{n-1}[/imath] subsets that have an even sum. But I don't know how to continue so that the columns are even as well. Any ideas? | 155057 | Counting matrices over [imath]\mathbb{Z}/2\mathbb{Z}[/imath] with conditions on rows and columns
I want to solve the following seemingly combinatorial problem, but I don't know where to start. How many matrices in [imath]\mathrm{Mat}_{M,N}(\mathbb{Z}_2)[/imath] are there such that the sum of entries in each row and the sum of entries in each column is zero? More precisely find cardinality of the set [imath] \left\{A\in\mathrm{Mat}_{M,N}(\mathbb{Z}/2\mathbb{Z}): \forall j\in\{1,\ldots,N\}\quad \sum\limits_{k=1}^M A_{kj}=0,\quad \forall i\in\{1,\ldots,M\}\quad \sum\limits_{l=1}^N A_{il}=0 \right\} [/imath]. Thanks for your help. |
362349 | A question about Cayley–Hamilton theorem
In wikipedia "In linear algebra, the Cayley–Hamilton theorem (named after the mathematicians Arthur Cayley and William Rowan Hamilton) states that every square matrix over a commutative ring (such as the real or complex field) satisfies its own characteristic equation." [imath]\iff[/imath] If [imath]A[/imath] is an [imath]n\times n[/imath] martix and [imath]I_n[/imath] is the [imath]n\times n[/imath] identity martix and [imath]\text{let}[/imath] [imath]f(\lambda)=\text{det}(\lambda I_n-A)[/imath] then we get [imath]f(A)=0[/imath]. In my book and wiki both use a order polynomial expression to prove it. The question is: Why can't we just let [imath]\lambda=A [/imath] [imath]\Rightarrow[/imath] [imath]\lambda I_n=AI_n=A[/imath] [imath]\Rightarrow[/imath] [imath]f(A)=\text{det}(A-A)=0[/imath] Can anyone help me? Thanks | 70991 | Error in argument regarding the Cayley Hamilton theorem
I cannot spot the mistake in the following argument regarding the Cayley Hamilton theorem: Let [imath]A\in M_n[/imath], then, [imath]\begin{align*} P_A(t)&=\det(tI-A)\\ &\implies P_A(A)=\det(AI-A)\\ &\implies P_A(A)=\det(0)\\ &\implies P_A(A)=0 \end{align*}[/imath] |
362509 | Does [imath]a_{n+1} = \frac{4}{a_{n}} + \frac{a_{n}}{2}[/imath] converge?
Suppose [imath]a_{n}[/imath] is a sequence of real numbers then does [imath]a_{n+1} = \frac{4}{a_{n}} + \frac{a_{n}}{2}[/imath] converge? I can find to where it converges, by setting [imath]\lim a_{n+1} = a_{n} =x[/imath] and then solving for [imath]x[/imath], but I don't know how to prove it's convergence. A.M - G.M gives [imath]\frac{4}{a_{n}}+ \frac{a_{n}}{2} \geq 2 \cdot \sqrt{2}[/imath] but that doesn't seem to help :( | 359290 | Prove the converges of the followin sequence and find the limit
I wonder if anyone can solve the following problem for me: The sequence of real numbers [imath]x_n[/imath] is defined inductively by [imath] x_1=4 \quad\text{and}\quad x_{n+1}=\frac4{x_n}+\frac{x_n}2. [/imath] Show that [imath]x_n[/imath] converges and find its limit Thanks to every one contribute in the solution. |
362494 | Tensor product of an irreducible [imath]G[/imath]-representation and a one-dimensional representation
If [imath]G[/imath] is a finite group, [imath]V[/imath] is an irreducible [imath]G[/imath]-representation and [imath]W[/imath] is any 1-dimensional [imath]G[/imath]-representation (both over an algebraically closed field of characteristic zero), show that [imath]V \otimes W[/imath] is an irreducible [imath]G[/imath]-representation. This is the second time I've come across this problem this month (2 different classes), and I'm unsure of the solution I tried in the first class. Essentially my solution was to fix bases [imath]v_1, \ldots, v_k[/imath] of [imath]V[/imath] and [imath]w[/imath] of [imath]W[/imath], show that [imath]v_i \otimes w[/imath] is a basis of [imath]V \otimes W[/imath] and then through brute force show that a submodule of [imath]V \otimes W[/imath] would imply a submodule of [imath]V[/imath], which gives the contradiction. I feel like there is an easier way - any hints would be great! | 355027 | Show [imath]U \otimes V[/imath] is an irreducible G-module
Let [imath]G[/imath] is some group and [imath]U[/imath] is an irriducible [imath]G[/imath]-module over the complex numbers. Now if [imath]V[/imath] is a [imath]G[/imath]-module of dimension 1, I would like to prove [imath]U \otimes V[/imath] is an irriducible [imath]G[/imath]-module. My knowledge of tensor products is lacking.. |
360390 | Safes and keys probability puzzle
I have [imath]100[/imath] keys and [imath]100[/imath] safes. Each key opens only one safe, and each safe is opened only by one key. Every safe contains a random key. 98 of these safes are locked. What's the probability that I can open all the safes? This question is confusing for me. Can you walk me through it step-by-step? | 320485 | Probability of opening all piggy banks
Interesting problem I found, and can't solve it since the morning: We have [imath]n[/imath] keys and [imath]n[/imath] piggy banks. Each key fits only one piggy bank. We randomly put exactly one key in each piggy bank. Then we randomly shatter [imath]1\le k\le n[/imath] piggy banks. What is the probability that we are able to open all other piggy banks? Really interested me. I thought about approach with cycles in permutation. When we number all piggy banks and their keys with numbers [imath]1...n[/imath] and then denote [imath]f(i)[/imath] - the number of key that is in [imath]i[/imath]-th piggy bank then we have permutation [imath]f[/imath]. So to open all piggy banks we need at least one key from each cycle in permutation [imath]f[/imath]. But I completely don't know how to count it. Can anybody help? |
360038 | How to maximize this function?
The function is [imath]lik(\theta)=\theta^n\prod_{i=1}^n x_i^{-2}[/imath] where [imath]\theta<x<\infty[/imath] I am getting [imath]0[/imath] if I just take the derivative of the log of this function. I believe this has to do with the domain of x being dependent on [imath]\theta[/imath]? | 361499 | What is the sufficient statistic for this function?
[imath]f(x)=\theta/x^2[/imath] where [imath]\theta<x<\infty[/imath] Given that the likelihood is [imath]lik(\theta)=\theta^n \prod x_i^{-2}[/imath] It seems to me that [imath]T(X)=\prod x_i^{2}[/imath] is the sufficient statistic, but intuitively, I think it should be [imath]T(X)=\min(x_1...x_n)[/imath] |
362749 | Prove that [imath]13|2^{70}+3^{70}[/imath]
Prove that [imath]13|2^{70}+3^{70}[/imath] Any HINT how I can start solving this task? What can be a first step? | 25701 | Show that 13 divides [imath]2^{70}+3^{70}[/imath]
Show that [imath]13[/imath] divides [imath]2^{70} + 3^{70}[/imath]. My main problem here is trying to figure out how to separate the two elements in the sum, and then use Fermat's little theorem. So how can I separate the two? Thanks! |
363332 | Show that if [imath]G[/imath] is a group of order 6, then [imath]G \cong \Bbb Z/6\Bbb Z[/imath] or [imath]G\cong S_3[/imath]
Show that if [imath]G[/imath] is a group of order 6, then [imath]G \cong \Bbb Z/6\Bbb Z[/imath] or [imath]G\cong S_3[/imath] This is what I tried: If there is an element [imath]c[/imath] of order 6, then [imath]\langle c \rangle=G[/imath]. And we get that [imath]G \cong \Bbb Z/6 \Bbb Z[/imath]. Assume there don't exist element of order 6. From Cauchy I know that there exist an element [imath]a[/imath] with [imath]|a|=2[/imath], and [imath]b[/imath] with [imath]b=3[/imath]. As the inverse of [imath]b[/imath] doesn't equal itself, I have now 4 distinct elements: [imath]e,a,b,b^{-1}[/imath]. As there are no elements of order [imath]6[/imath], we have two options left. Option 1: [imath]c[/imath] with [imath]|c|=3[/imath], and [imath]c^{-1}[/imath] with [imath]|c|=3[/imath]. Option 2: two different elements of order 2, [imath]c,d[/imath]. My intuition says that for option 1, [imath]H= \{e,b,b^{-1},c,c^{-1} \}[/imath] would be a subgroup of order [imath]5[/imath], which would give a contradiction, but I don't know if this is true/how I could prove this. I also don't know how I could prove that option 2 must be [imath]S_3[/imath]. I think that [imath]G= \{e,a,b,b^{-1},c,d \}[/imath]. Looks very similar to [imath]D_3[/imath]. But I don't know how I could prove this rigoursly. Can anybody help me finish my proof ? If there are other ways to see this, I'd be glad to hear them as well! | 347447 | Let [imath]G[/imath] a group of order [imath]6[/imath]. Prove that [imath]G \cong \Bbb Z /6 \Bbb Z[/imath] or [imath]G \cong S_3[/imath].
Let [imath]G[/imath] a group of order [imath]6[/imath]. Prove that: i) [imath]G[/imath] contains 1 or 3 elements of order 2. ii) [imath]G \cong \Bbb Z /6 \Bbb Z[/imath] or [imath]G \cong S_3[/imath]. I haven´t covered Sylow groups and normal groups. This is an exercise from the chapter about group actions. I have covered Lagrange and cosets. |
362727 | Interchange product [imath]\sigma[/imath]-algebra and countable intersection
If [imath](\mathcal F_n)_{n\in\mathbb N}[/imath] is a decreasing sequence of [imath]\sigma[/imath]-algebras and [imath]\mathcal G[/imath] another [imath]\sigma[/imath]-algebra, is it possible to interchange the intersection with the product [imath]\sigma[/imath]-algebra? [imath]\bigcap\limits_{n\in \mathbb N}(\mathcal F_n\otimes \mathcal G)=(\bigcap\limits_{n\in \mathbb N}\mathcal F_n\otimes \mathcal G)[/imath] If not, is there a condition for this equality? | 92546 | Decreasing sequence of product [imath]\sigma[/imath]-algebras
My question is as follows. If [imath](X,\mathcal A)[/imath] and [imath](Y,\mathcal B)[/imath] are two measure spaces and [imath]\mathcal A_n\subseteq \mathcal A[/imath] is a decreasing sequence of [imath]\sigma[/imath]-algebras with [imath]\mathcal A_n\downarrow \mathcal A_\infty[/imath], then we have [imath]\mathcal A_n \times \mathcal B\downarrow \mathcal A_\infty \times \mathcal B[/imath] One direction is easy, but I am stuck at the other direction. Thank you for helping me with this. I only did one direction as follows: take [imath]A\times B\in \mathcal A_\infty \times \mathcal B[/imath], we have that [imath]A\times B\in \mathcal A_n\times \mathcal B[/imath] for all [imath]n[/imath]. So, [imath]A\times B\in \bigcap_n(\mathcal A_n \times \mathcal B)[/imath]. The latter is a [imath]\sigma[/imath]-algebra, so it contains [imath]\mathcal A_\infty \times \mathcal B[/imath], which is generated by the class of sets of the form [imath]A\times B[/imath]. This implies [imath]\mathcal A_\infty \times \mathcal B \subseteq \bigcap_n(\mathcal A_n \times \mathcal B)[/imath]. We need to prove that [imath]\bigcap_n(\mathcal A_n \times \mathcal B)\subseteq \mathcal A_\infty \times \mathcal B[/imath]. Update: I should add another condition, that both [imath](X,\mathcal A,\mu)[/imath] and [imath](Y,\mathcal B,\nu)[/imath] are probability spaces, and all the [imath]\sigma-[/imath] algebras are considered modulo the null sets. But I still want to know what happens if we don't have such a condition. Are there any counterexamples? |
363696 | what is the image set under this linear map?
[imath]f:\mathbb{R}^n\to\mathbb{R}[/imath] be a linear map with [imath]f(0)=0[/imath] then the set [imath]\{f(x_1,\dots,x_n):\sum_{j=1}^{n}x_j^2\le 1\}[/imath] equals [imath]1.[/imath] [imath][-a,a][/imath] for some [imath]a\in\mathbb{R},a\ge 0[/imath] [imath]2.[/imath] [imath][0,a][/imath] for some [imath]a\ge 0[/imath] [imath]3.[/imath] [imath][0,1][/imath] [imath]4.[/imath] [imath][a,b][/imath] Linear map [imath]\Rightarrow[/imath] Continuos so image of a compact set will be compact but what more I can say? Thank you for help. | 296837 | [imath]f\colon\mathbb R^n \rightarrow \mathbb R[/imath] be a linear map with [imath]f(0,0,0,\ldots,0)=0.[/imath]
I was thinking about the following problem : Let [imath]f\colon\mathbb R^n \rightarrow \mathbb R[/imath] be a linear map with [imath]f(0,0,0,\ldots,0)=0.[/imath] Then the set [imath]\{f(x_1,x_2,\ldots,x_n):\sum_{j=1}^{n}x_j^2 \leq 1\}[/imath] equals to which of the following: [imath]1.[-a,a][/imath] for some [imath]a \in \mathbb R,a \geq 0.[/imath] [imath]2.[0,1][/imath]. [imath]3.[0,a][/imath] for some [imath]a \in \mathbb R,a \geq 0[/imath] [imath]4.[a,b][/imath] for some [imath]a,b \in \mathbb R, 0<a \leq b .[/imath] My Attempt: Since [imath]f\colon\mathbb R^n \rightarrow \mathbb R[/imath] is a linear map, [imath](a) f(u+v)=f(u)+f(v), [/imath] for [imath]u,v \in \mathbb R^n[/imath] and [imath](b)[/imath] for [imath]\lambda \in \mathbb R, f(\lambda v)=\lambda f(v)[/imath] .Now Using @David Mitra's suggestion, we see that if [imath]c=(c_1,c_2,\ldots,c_n)[/imath] is in the set then so is [imath]-c.[/imath] Now, I see [imath]f\{c+(-c)\}=f\{(c_1,c_2,\ldots,c_n)+(-c_1,-c_2,\ldots,-c_n)\}=f\{(c_1-c_1,c_2-c_2,\ldots,c_n-c_n)\}=f(0,0,\ldots,0)=0[/imath]. Also,since [imath]f[/imath] is linear,[imath]f\{c+(-c)\}=f(c)+f(-c)[/imath] and hence [imath]f(c)+f(-c)=0[/imath] and so [imath]f(-c)=-f(c).[/imath] So, if for some [imath]c \in \mathbb R^n, f(c)=a[/imath],then [imath]f(-c)=-a.[/imath] So option [imath](1)[/imath] seems to be right . Am I going in the right direction?Please comment.Thanks in advance for your time. |
363806 | Dense linear orderings isomorphism
First of all the definition: A dense linear order is a [imath]\{<\}[/imath]-structure in which the following formulas are valid: Question: How can I prove that any two countable DLO's (dense linear order) are isomorph? Hint (from my logic book): Let [imath]M[/imath] and [imath]N[/imath] be two countable DLOs. Now we should somehow be able to proof that every partial finite isomorphism [imath]f:M\rightarrow N[/imath] (what is the definition for a partial iso.?) can be continued to a finite partial isomorphism [imath]f':M\rightarrow N[/imath] which contains a given element from [imath]M[/imath] (or [imath]N[/imath]) in the domain (range). Using this we can make enumerations from [imath]M[/imath] and [imath]N[/imath] and construct an isomorphism. | 156145 | dense linear orders DLO
I am asked to prove that if I have two models of dense linear orders DLOs WITH the minimum and maximum. must be izomorpic to each other by fining direct izomorphy. I seem to always get stuck because the desity i just cannot understand how to count them im trying something like this: assuming [imath]M_1=\{a_1 ... a_2 ...\}[/imath] assuming [imath]M_1=\{b_1 ... b_2 ...\}[/imath] I choose [imath]a_i[/imath] but if I choose [imath]H(a_1)=b_i[/imath] how can I know its in the same order? sorry for my bad english - translating from hebrew |
363927 | The union of two subspaces is not alwys a subspace
[imath]F[/imath] and [imath]G[/imath] are two subspaces of [imath]E[/imath] over [imath]K[/imath], how can I show that, [imath]F \cup G[/imath] is a subspace of E over K [imath]\Leftrightarrow[/imath] [imath]F \subseteq G[/imath] or [imath]G \subseteq F[/imath] ? [imath]\Leftarrow[/imath] is obvious, but what about [imath]\Rightarrow[/imath] ? | 982582 | Is union of two subspace a subspace too?
Assume that [imath]W[/imath] and [imath]V[/imath] are two subspace of [imath]X[/imath]. Is their union a subspace of [imath]X[/imath] too? I think it is not true unless under certain conditions but I do not know what conditions... |
364078 | Old Qualifying Exam Problem topology
I'm studying for qualifying exam and was struggling with this problem. Thanks for any help! Is it true that if the 1-point compactifications of two locally compact Hausdorff spaces [imath]X[/imath], [imath]Y[/imath] are homeomorphic, then X and Y are necessarily homeomorphic? Give a proof or counterexample, as appropriate. | 357794 | Are locally compact Hausdorff spaces with the homeomorphic one-point compactification necessarily homeomorphic themselves?
When practicing old qualifying exam problems, I had trouble with this one. Thanks for any help! Is it true that if the [imath]1[/imath]-point compactifications of two locally compact Hausdorff spaces [imath]X[/imath], [imath]Y[/imath] are homeomorphic, then [imath]X[/imath] and [imath]Y[/imath] are necessarily homeomorphic? Give a proof or counterexample, as appropriate. |
364547 | How do we derive the number of ways to divide [imath]n[/imath] non-identical objects into [imath]r[/imath] groups such that each group gets [imath]0[/imath] or more number of objects
How do we get [imath]r^n[/imath] as the number of divisions possible. Please give a full description. | 47345 | Number of ways of distributing $n$ identical objects among $r$ groups
I came across this formula in a list: The number of ways of distributing [imath]n[/imath] identical objects among [imath]r[/imath] groups such that each group can have [imath]0[/imath] or more [imath](\le n)[/imath] objects I know that standard way of doing this is to solve the problem of distributing n identical objects and [imath](r-1)[/imath] partitions among themselves which can be done in [imath]C(n+r-1,r-1)[/imath] ways. But I am unable to prove to myself why it is not [imath](r+1)^n[/imath]. Because each of the n objects has r+1 choices, either group1, group2,... group r or none at all. |
364642 | Show that [imath]S_4[/imath] has a unique subgroup [imath]H[/imath] of index 2.
Show that [imath]S_4[/imath] has a unique subgroup [imath]H[/imath] of index 2. What I tried: Of course, [imath]A_4[/imath] has index 2. And since any subgroup of index 2 must be normal, I need to show that [imath]A_4[/imath] is the only normal subgroup of [imath]S_4[/imath]. Let [imath]H≠A_4[/imath] from index 2. Then 6 of the elements must be even and 6 of the elements must be odd. If this would be a subgroup, then it would also be normal, so I think I need to show that this can't be a subgroup. I think I need to show that [imath]\langle H \rangle[/imath] must be [imath]S_4[/imath] or something like that. Any hints ? | 22559 | [imath]A_{4}[/imath] unique subgroup of [imath]S_4[/imath] of order [imath]12[/imath]
I was reading the proof that [imath]A_{4}[/imath] is the unique subgroup of order [imath]12[/imath]. So the author counts the number of conjugates: [imath]1[/imath] cycle of type [imath]()[/imath] , [imath]6[/imath] cycles of type [imath](1 \ 2)[/imath], 8 cycles of type [imath](1 \ 2 \ 3)[/imath], [imath]6[/imath] cycles of type [imath](1 \ 2 \ 3 \ 4)[/imath] and [imath]3[/imath] cycles of type [imath](1 \ 2)(3 \ 4)[/imath]. Now it says, the only possible way to get [imath]12[/imath] elements is [imath]1 + 3 + 8[/imath]. Why is this? can't we have [imath]1[/imath] cycle of type [imath]()[/imath], [imath]2[/imath] cycles of type [imath](1 \ 2)[/imath], [imath]3[/imath] cycles of type [imath](1 \ 2 \ 3)[/imath], [imath]2[/imath] cycles of type [imath](1 \ 2)(3 \ 4)[/imath] and [imath]4[/imath] cycles of type [imath](1 \ 2 \ 3 \ 4)[/imath]. This also gives you [imath]12[/imath]. Why is this impossible though? I don't really understand why the only possibility is [imath]1 + 3 + 8[/imath]. |
364845 | Give an example of a group [imath]G[/imath] with [imath]N,M \triangleleft G[/imath] such that [imath]N \cong M[/imath], but [imath]G/M \not \cong G/N[/imath].
I'm trying to find an example of a group [imath]G[/imath] with [imath]N[/imath] and [imath]M[/imath] normal subgroups such that [imath]N \cong M[/imath], but [imath]G/M \not \cong G/N[/imath]. Can anybody help me with this ? | 151911 | [imath]G[/imath] is a group [imath],H \cong K[/imath], then is [imath]G/H \cong G/ K[/imath]?
[imath]G[/imath] is a group with subgroups [imath]H[/imath] and [imath]K[/imath] such that [imath],H \cong K[/imath], then is [imath]G/H \cong G/ K[/imath]? |
365130 | Direct product [imath]\mathbb Z_p \times A[/imath] and [imath]\mathbb Z_p \times B[/imath]
Suppose that [imath]\mathbb Z_p \times A \simeq \mathbb Z_p \times B[/imath], where [imath]p[/imath] is prime. Is it true, that [imath]A \simeq B[/imath]? | 349855 | Does [imath]G\times K\cong H\times K[/imath] imply [imath]G\cong H[/imath]?
Let [imath]G\times K[/imath] be a finite group. Suppoose [imath]G\times K\cong H\times K[/imath]. Is this sufficient to imply [imath]G\cong H[/imath]? |
365248 | How to find the last two digit of [imath]7^{81}[/imath]?
Could any one tell me how to find the last two digit of [imath]7^{81}[/imath]? I have succeeded in finding the last digit only which is [imath]7[/imath]. Any group theoretic approach or any other approach is welcome. | 362012 | Find the last two digits of [imath] 7^{81} ?[/imath]
I came across the following problem and do not know how to tackle it. Find the last two digits of [imath] 7^{81} ?[/imath] Can someone point me in the right direction? Thanks in advance for your time. |
365254 | Rational and irrational fractions over finite fileds
I've been told that over the field [imath]\mathbb{F}_7[/imath] the square root of [imath]2[/imath] is actually [imath]3[/imath]. How come? Why does it happen? | 364896 | Roots of polynomials over finite fields
I've been trying to find the decomposition of [imath]x^2-2[/imath] to irreducible polynomials over [imath]\mathbf{F}_5[/imath] and [imath]\mathbf{F}_7[/imath]. I know that for some [imath]a[/imath] in [imath]\mathbf{F}_5[/imath] (for example), [imath]x-a[/imath] divides [imath]x^2-2[/imath] iff [imath]f(a) = 0[/imath], i.e [imath]a[/imath] is a root of [imath]x^2-2[/imath]. Over the field [imath]\mathbf{F}_7[/imath], I've found (by trail and error) that one irreducible polynomial is [imath]x-3[/imath]. I've now got two questions - How can I find the other irreducible polynomial? Is there any more efficient method to find roots than trial and error? Thanks in advance |
14282 | Why do we define quotient groups for normal subgroups only?
Let [imath]G \in \mathbf{Grp}[/imath], [imath]H \leq G[/imath], [imath]G/H := \lbrace gH: g \in G \rbrace[/imath]. We can then introduce group operation on [imath]G/H[/imath] as [imath](xH)*(yH) := (xy)H[/imath], so that [imath]G/H[/imath] becomes a quotient group when [imath]H[/imath] is a normal subgroup. But why do we only work with quotient groups by normal subgroups? If we introduce the notion of left quotient group in the above manner, how much good properties of a quotient group do we lose? UPD: fixed confusing notation: from [imath](xH)(yH)[/imath] to [imath](xH)*(yH)[/imath]. | 1878284 | Normal subgroup implies quotient group?
I was reading through a proof of the first isomorphism theorem on proofwiki and I didn't understand a line that said "by kernel is a normal subgroup of domain, [imath]G_1/K[/imath] exists", where [imath]K = ker(\phi)[/imath] and [imath]\phi: G_1 \rightarrow G_2[/imath] is a group homomorphism. I understand the proof of the kernel being a normal subgroup, but does that imply the quotient group [imath]G_1/K[/imath] exist? In general, if you have a normal subgroup, can you always construct a quotient group? I'm guessing that comes from the definition of a normal subgroup but I can't seem to make the connection. |
365569 | what's the difference between Syntactic consequence ⊢ and Semantic consequence ⊨
Can you help me to differentiate the Syntactic consequence [imath]\vdash[/imath] and Semantic consequence [imath]\vDash[/imath] ? I think [imath]A \vdash B[/imath] means, "[imath]A[/imath] proves [imath]B[/imath]" and [imath]A \vDash B[/imath] means , if [imath]A[/imath] is true, then [imath]B[/imath] is true. If so, what is difference [imath]A \to B[/imath] and [imath]A\vDash B[/imath] ? | 286077 | Implies vs. Entails vs. Provable
Consider A [imath]\Rightarrow[/imath] B, A [imath]\models[/imath] B, and A [imath]\vdash[/imath] B. What are some examples contrasting their proper use? For example, give A and B such that A [imath]\models[/imath] B is true but A [imath]\Rightarrow[/imath] B is false. I'd appreciate pointers to any tutorial-level discussion that contrasts these operators. Edit: What I took away from this discussion and the others linked is that logicians make a distinction between [imath]\vdash[/imath] and [imath]\models[/imath], but non-logicians tend to use [imath]\Rightarrow[/imath] for both relations plus a few others. Points go to Trevor for being the first to explain the relevance of completeness and soundness. |
365772 | Constructing an example s.t. [imath]\operatorname{Hom}_R(M,N)[/imath] is not finitely generated
Let [imath]R[/imath] be a commutative ring and [imath]M[/imath] and [imath]N[/imath] be two finitely generated [imath]R[/imath]-modules. I wanna construct an example s.t. [imath]\operatorname{Hom}_R(M,N)[/imath] is not finitely generated. It's well-known that if [imath]R[/imath] be a Noetherian ring and [imath]M[/imath] and [imath]N[/imath] be two finitely generated [imath]R[/imath]-modules then [imath]\operatorname{Hom}_R(M,N)[/imath] is finitely generated. So in this example [imath]R[/imath] must be a Non-Noetherian ring. | 234485 | When [imath]\operatorname{Hom}_{R}(M,N)[/imath] is finitely generated as [imath]\mathbb Z[/imath]-module or [imath]R[/imath]-module?
Assume that [imath]M[/imath] and [imath]N[/imath] are two finitely generated [imath]R[/imath]-modules. Then [imath]\operatorname{Hom}_{R}(M,N)[/imath] is a finitely generated [imath]\mathbb Z[/imath]-module and/or [imath]R[/imath]-module (in this case, assume that [imath]R[/imath] is commutative)? Note that [imath]R[/imath] is not Noetherian ring. Is there any counterexample (for any cases)? |
365827 | how to prove following matrix is invertible?
how to prove A is invertible or [imath]\ detA\neq 0[/imath] [imath]A=\begin{pmatrix} \frac11 & \frac12 & \frac13 & \cdots & \frac1n \\ \frac12 & \frac13 & \frac14 & \cdots & \frac{1}{n+1} \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ \frac1n & \frac{1}{n+1} & \frac{1}{n+2} & \cdots & \frac{1}{2n-1} \end{pmatrix}[/imath] Thanks in advance | 46862 | Prove that a matrix is invertible
Show that the matrix [imath]A = \begin{bmatrix} 1 & \frac{1}{2} & \ldots & \frac{1}{n}\\ \frac{1}{2} & \frac{1}{3} & \ldots & \frac{1}{n+1}\\ \vdots & \vdots & & \vdots \\ \frac{1}{n} &\frac{1}{n+1} &\ldots &\frac{1}{2n-1} \end{bmatrix}[/imath] is invertible and [imath]A^{-1}[/imath] has integer entries. This problem appeared in Chapter one of my linear algebra textbook so I assume nothing more is required to prove it than elementary row operations. I've been staring at it for a while and considered small values of [imath]n[/imath] but there doesn't seem to be a clever way of easily reducing it to the identity matrix. Could someone give me a hint or a poke in the right direction rather than a full solution as I'm still hoping to work it out myself. |
365837 | [imath]I-AB[/imath] be invertible [imath]\Leftrightarrow[/imath] [imath]I-BA[/imath] is invertible
assume [imath]A,B\in M_n(F)[/imath] if [imath]I-AB[/imath] be invertible then how to prove [imath]I-BA[/imath] is invertible and how find inverse of [imath]I-BA[/imath] Thanks in advance | 876565 | Invertibility of [imath]I-AB[/imath]
I got a question in linear algebra: 1) Let A and B be [imath]n\times n[/imath] matrices. If [imath]I - AB[/imath] is an invertible matrix, then prove that [imath]I - BA[/imath] is invertible. Can someone tell me how to solve this question? I've no idea how to start. Thank you! |
365679 | On the commutator subgroup of a group
Prove that there is no group [imath]G[/imath] such that the commutator subgroup of [imath]G[/imath] to be [imath]S_{3}[/imath]. | 361582 | Which groups are derived subgroups?
Let [imath]G[/imath] be a group. When is there a group [imath]H[/imath] such that [imath]G[/imath] is isomorphic to its derived subgroup [imath]H'[/imath]? I only know that there is not always such a [imath]H[/imath]; for instance, no group has its derived subgroup isomorphic to [imath]\mathfrak{S}_n[/imath] for [imath]n \geq 4[/imath] and [imath]n \neq 6[/imath]. |
314645 | Is there a [imath](6,9,3)_2[/imath] code?
Does there exist a [imath]1[/imath]-error-correcting binary code with block length [imath]6[/imath] and [imath]9[/imath] codewords? The Hamming bound says that for any code [imath]C[/imath] with those parameters, [imath]|C| \le \frac{2^6}{1+6} \approx 9.14[/imath]. So, we can't rule out the existence of such a code using the Hamming bound. The Singleton bound says that for any code [imath]C[/imath] with those parameters, [imath]|C| \le 2^{6-3+1} = 16[/imath], so we can't rule out the existence of such a code using the Singleton bound. Also, this can't be a linear code, since [imath]\log_2 9[/imath] isn't an integer. That rules out the possibility of just enumerating the possible generator matrices. I feel a bit silly asking this, but the direct approach (assume [imath]000\ 000[/imath] is in the code, then all others must have Hamming weight [imath]\ge 3[/imath], ...) becomes unmanageable quickly. How else can I go about this? | 1204679 | Show a binary code does not exist.
Show that there is no [imath](6,9,3)[/imath] binary code. I'm pretty sure the way to tackle this problem is to deal with it's generator matrix and then get a condradiction. However I seem to be getting no contradiction and I'm unsure if this is even the right approach. Any help would be appreciated. |
366341 | Complex analysis problem, prove that [imath]f[/imath] is a polynomial of degree [imath]\leq 1[/imath]
Suppose that [imath]f[/imath] is entire and that for each [imath]z[/imath], either [imath]|f(z)|≤1[/imath] or [imath]|f′(z)|≤1[/imath]. Prove that f is a polynomial of degree [imath]\leq 1[/imath]. Hint: Use a line integral to show that [imath]|f(z)|≤A+|z|[/imath] where[imath] A = \max\{1,|f(0)|\}[/imath]. This is an exercise of complex analysis (author : newman, bak) Actually, I found similar question but I don't understand answer. Suppose that [imath] f [/imath] is entire and that for each [imath] z [/imath], either [imath] |f(z)| \leq 1 [/imath] or [imath] |f^\prime (z) |\leq 1 [/imath]. Prove that [imath] f [/imath] is a linear polynomial. Is there someone explaining easily? | 341724 | Suppose that [imath] f [/imath] is entire and that for each [imath] z [/imath], either [imath] |f(z)| \leq 1 [/imath] or [imath] |f^\prime (z) |\leq 1 [/imath]. Prove that [imath] f [/imath] is a linear polynomial.
My question is in the title. I'm a little lost in how to solve this problem. There is a hint associated with the problem that states the following: Use a line integral to show that [imath] |f(z)| \leq A + |z| [/imath] where A = [imath]\max\{1, |f(0)|\}[/imath] Thanks for the help. |
365680 | how to prove that [imath](1 + \frac{1}{n})^{n+1}[/imath] is decreasing?
Please, help me to prove that [imath]x_n=\left(1+\frac{1}{n}\right)^{n+1}[/imath]decreases. I know I must to prove that that [imath]\frac{x_n}{x_{n+1}}> 1[/imath] What to do next? | 306178 | Given [imath]y_n=(1+\frac{1}{n})^{n+1}[/imath] show that [imath]\lbrace y_n \rbrace[/imath] is a decreasing sequence
Given [imath] y_n=\left(1+\frac{1}{n}\right)^{n+1}\hspace{-6mm},\qquad n \in \mathbb{N}, \quad n \geq 1. [/imath] Show that [imath]\lbrace y_n \rbrace[/imath] is a decreasing sequence. Anyone can help ? I consider the ratio [imath]\frac{y_{n+1}}{y_n}[/imath] but I got stuck. |
367373 | Surjective function
Let [imath]G[/imath] be an arbitrary group with identity element e and let [imath]K[/imath] and [imath]N[/imath] be normal subgroups of [imath]G[/imath] with [imath]K∩N=e[/imath]. I know that [imath]nk=nk[/imath] And the representation of an element in [imath]K×N[/imath] in form [imath]kn[/imath] is unique for all [imath]k∈K[/imath] and [imath]n∈N[/imath]. How can I show the function [imath]φ:(K×N)→(K⊕N)[/imath] be defined by [imath]φ(kn)=(k,n)[/imath]. is surjective? | 367335 | well defined function question
Let [imath]G[/imath] be an arbitrary group with identity element [imath]e[/imath] and let [imath]K[/imath] and [imath]N[/imath] be normal subgroups of G with [imath]K∩N={e}[/imath]. I know that [imath]nk=nk[/imath] And the representation of an element in [imath]K×N[/imath] in form [imath]kn[/imath] is unique for all [imath]k∈K[/imath] and [imath]n∈N[/imath]. How can I show the function [imath]φ:(K×N)→(K⊕N)[/imath] be defined by [imath]φ(kn)=(k,n)[/imath]. is well defined? I know that I need to show for all [imath]kn,xy∈K×N[/imath] such that [imath]kn=xy[/imath], [imath]φ(kn)=φ(xy)[/imath] But I don't know how to show that |
367558 | Hermitian matrices
Suppose we have a hermitian matrix [imath]H[/imath], and a matrix [imath]A[/imath] composed of eigenvectors of [imath]H[/imath], such that [imath]\langle A_i \mid A_i \rangle =1[/imath], where [imath]A_i[/imath] is the [imath]i[/imath]-th column of matrix H. How to prove that [imath]HA=AB[/imath], where [imath]B[/imath] is a diagonal matrix whose diagonal elements are eigenvalues of [imath]H[/imath]? Thanks | 363421 | Hermitian matrices are diagonalizable
I am trying to prove that Hermitian Matrices are diagonalizable. I have already proven that Hermitian Matrices have real roots and any two eigenvectors associated with two distinct eigen values are orthogonal. If [imath]A=A^H;\;\;\lambda_1,\lambda_2[/imath] be two distinct eigenvalues and [imath]v_1,v_2[/imath] be two eigenvectors associated with them. [imath]Av_1=\lambda_1v_1\Rightarrow v_1^HAv_1=\lambda_1v_1^Hv_1\Rightarrow \lambda_1=\lambda_1^*.[/imath] Similarly, [imath]Av_1=\lambda_1v_1\Rightarrow v_2^HAv_1=\lambda_1v_2^Hv_1\Rightarrow v_1^H\lambda_2v_2=\lambda_1v_1^Hv_2\Rightarrow v_1^Hv_2=0[/imath] However, I am not aware of Spectral Theorem. Given this circumstances, how can I prove that Hermitian Matrices are diagonalizable? It should follow from above but the only sufficient condition of a matrix being diagonalizable is to have [imath]dim(A)[/imath] distinct eigenvalues, or existance of [imath]P[/imath] such that [imath]A=P^{-1}DP[/imath]. I am not sure how this follows from above conditions. |
367676 | Why are there [imath]2^{\aleph_0}[/imath] injections from [imath]\omega[/imath] to [imath]\omega_1?[/imath]
I have to prove that there are [imath]2^{\aleph_0}[/imath] injections from [imath]\omega[/imath] to [imath]\omega_1.[/imath] I can see that there is a bijection between this set and the set of pairs: (permutation of [imath]\omega[/imath], infinitely countable subset of [imath]\omega_1[/imath]), which means that the standard "finite" formula works, and there are [imath]\aleph_0!\cdot{\aleph_1 \choose\aleph_0}[/imath] of such injections. I know that [imath]\aleph_0!=2^{\aleph_0},[/imath] but I can't see how the other factor is [imath]2^{\aleph_0}[/imath] too. | 312431 | What is the value of [imath]\aleph_1^{\aleph_0}[/imath]?
Is there any neat way to calculate the value of [imath]\aleph_1^{\aleph_0}[/imath]? |
477 | Cardinality of set of real continuous functions
The set of all [imath]\mathbb{R\to R}[/imath] continuous functions is [imath]\mathfrak c[/imath]. How to show that? Is there any bijection between [imath]\mathbb R^n[/imath] and the set of continuous functions? | 403984 | Prove that [imath]C^{0}[/imath] and [imath]\mathbb{R}[/imath] have equal cardinality
How to Prove that [imath]C^{0}[/imath] and [imath]\mathbb{R}[/imath] have equal cardinality ? [imath]C^{0}[/imath] denote the set of all Continuous function [imath]\mathbb{R} \rightarrow \mathbb{R}[/imath] |
272908 | Spivak Chapter 2, problems 27 (and 28)
To be honest, I have no idea how to even start this problem. I'm sorry I don't have any work to show, but I'm just at a blank. Help? Chapter 2: Problem 27: University B, once boasted [imath]17[/imath] tenured professors of mathematics. Tradition prescribed that at their weekly luncheon meeting, faithfully attended by all [imath]17[/imath], any members who had discovered an error in their published work should make an announcement of this fact, and promptly resign. Such an announcement had never actually been made, because no professor was aware of any errors in her or his work. This is not to say that no errors existed, however. In fact, over the years, in the work of every member of the department at least one error had been found, by some other member of the department. This error had been mentioned to all other members of the department, but the actual author of the error had been kept ignorant of the fact, to forestall any resignations. One fateful year, the department was augmented by a visitor from another university, one Prof. X, who had come with hopes of being offered a permanent position at the end of the academic year. Naturally, he was apprised, by various members of the department, of the published errors which had been discovered. When the hoped-for appointment failed to materialize, Prof. X obtained his revenge at the last luncheon of the year. "I have enjoyed my visit here very much", he said, "but I feel that there is one thing that I have to tell you. At least one of you has published an incorrect result, which has been discovered by others in the department." What happened the next year?" Chapter 2: Problem 28: After figuring out, or looking up, the answer to Problem 27, consider the following: Each member of the department already knew that Prof.X asserted, so how could his saying it change anything? | 2418573 | k of n aides to n ministers leak information; ministers only know whether other ministers' aides leaked (not their own)
I know the relationships between the ministers in the below problem can be represented by a [imath]K_n[/imath] graph since each one of the n ministers has connections to the n-1 other ministers/ministries they represent. However, I can't seem to figure out where to go next. Here is the problem: In a certain government there are n cabinet ministers. Each of these has a senior aide. Exactly k of these n aides leak important information to the press. None of the ministers know whether their own aide leaks information or not. However,because of interdepartmental rivalries, every minister has many sources in the other ministries, and thus is aware of all the leaks outside of her ministry. There is an unspoken law in this government that as soon as any minister has irrefutable proof that their aide leaked information, they must re the aide at their first press conference. Every ministry has one press conference per day at exactly 8:00 A.M. For a long time no one is fired, because no minister is aware of the leaks, if any, made by their own aide. One afternoon a newspaper runs a story, and announces that news are leaked by some of the top aides. Exactly k days later, all the guilty aides are fired. What happened? |
367899 | [imath]q[/imath]-norm [imath]\leq[/imath] [imath]p[/imath]-norm
[Ciarlet 1.4-8] If [imath]0 < p < q[/imath], show that [imath]\left(\sum_{i=1}^n|v_i|^q\right)^{1/q}\ \leq\ \left(\sum_{i=1}^n|v_i|^p\right)^{1/p}[/imath] Somebody knows how prove that? Thanks in adavance for the help. | 4094 | How do you show that [imath]l_p \subset l_q[/imath] for [imath]p \leq q[/imath]?
I can't seem to work out the inequality [imath](\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}[/imath] for [imath]p \leq q[/imath] (which I'm assuming is the way to go about it). |
368196 | How to prove that [imath]\displaystyle \lim_{n \to \infty} \sqrt[n]{n}=1[/imath]?
I have seen this fact thrown around a lot and never really stopped to prove it; plugging in a few values convinced me of its truth. But I would like to see the result proved. To be clear, this is not homework, just human curiosity; I'm looking for any nice proof. Thanks. | 28348 | Proof that [imath]\lim_{n\rightarrow \infty} \sqrt[n]{n}=1[/imath]
Thomson et al. provide a proof that [imath]\lim_{n\rightarrow \infty} \sqrt[n]{n}=1[/imath] in this book. It has to do with using an inequality that relies on the binomial theorem. I tried to do an alternate proof now that I know (from elsewhere) the following: \begin{align} \lim_{n\rightarrow \infty} \frac{ \log n}{n} = 0 \end{align} Then using this, I can instead prove: \begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} &= \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \newline & = \exp{0} \newline & = 1 \end{align} On the one hand, it seems like a valid proof to me. On the other hand, I know I should be careful with infinite sequences. The step I'm most unsure of is: \begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} = \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \end{align} I know such an identity would hold for bounded [imath]n[/imath] but I'm not sure I can use this identity when [imath]n\rightarrow \infty[/imath]. If I am correct, then would there be any cases where I would be wrong? Specifically, given any sequence [imath]x_n[/imath], can I always assume: \begin{align} \lim_{n\rightarrow \infty} x_n = \lim_{n\rightarrow \infty} \exp(\log x_n) \end{align} Or are there sequences that invalidate that identity? (Edited to expand the last question) given any sequence [imath]x_n[/imath], can I always assume: \begin{align} \lim_{n\rightarrow \infty} x_n &= \exp(\log \lim_{n\rightarrow \infty} x_n) \newline &= \exp(\lim_{n\rightarrow \infty} \log x_n) \newline &= \lim_{n\rightarrow \infty} \exp( \log x_n) \end{align} Or are there sequences that invalidate any of the above identities? (Edited to repurpose this question). Please also feel free to add different proofs of [imath]\lim_{n\rightarrow \infty} \sqrt[n]{n}=1[/imath]. |
368194 | Is every group of order [imath]21[/imath] cyclic?
solution :- [imath]21= 3 \times 7[/imath] there is only one Sylow [imath]3[/imath] and Sylow [imath]7[/imath] subgroup so, Sylow [imath]3[/imath] and Sylow [imath]7[/imath] subgroup are normal in group [imath]G[/imath] so [imath]G[/imath] is cyclic group of order [imath]21[/imath]. Am I right ? somebody told me that group of order [imath]21[/imath] is not cyclic. he gave me this link. If group of order 21 is not cyclic, then can we understand it by Sylow method ? | 154709 | Showing that a group of order [imath]21[/imath] (with certain conditions) is cyclic
How can i show that if [imath]o(G)=21[/imath] and if [imath]G[/imath] has only one subgroup of order [imath]3[/imath] and only one subgroup of order [imath]7[/imath], then show that [imath]G[/imath] is cyclic. |
76912 | Why does the ideal [imath](a+bi)[/imath] have index [imath]a^2+b^2[/imath] in [imath]\mathbb{Z}[i][/imath]?
In the comments on the question Why does this module structure have [imath]352512[/imath] elements?, it is mentioned that the index of the ideal generated by [imath]a+bi[/imath] in [imath]\mathbb{Z}[i][/imath] has order [imath]a^2+b^2[/imath]. Is there a nice rigorous explanation of why this is so? | 23358 | Quotient ring of Gaussian integers
A very basic ring theory question, which I am not able to solve. How does one show that [imath]\mathbb{Z}[i]/(3-i) \cong \mathbb{Z}/10\mathbb{Z}[/imath]. Extending the result: [imath]\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}[/imath], if [imath]a,b[/imath] are relatively prime. My attempt was to define a map, [imath]\varphi:\mathbb{Z}[i] \to \mathbb{Z}/10\mathbb{Z}[/imath] and show that the kernel is the ideal generated by [imath]\langle{3-i\rangle}[/imath]. But I couldn't think of such a map. Anyhow, any ideas would be helpful. |
52537 | Order of some quotient ring of Gaussian integers
I'm trying to get through a proof of Gauss' that certain primes can be written as the sum of two squares. An assumption is that the order of [imath]\mathbb{Z}[i]/(a+bi)[/imath] is [imath]a^2+b^2[/imath]. I get that [imath](a+bi)(a-bi)=a^2+b^2[/imath], so this places a bound on the order of integers with no imaginary part. But since [imath]b[/imath] isn't a unit, it doesn't seem like this finishes the proof. Any hints? | 1237543 | Is it true that the order of any quotient ring [imath]\mathbb Z[i]/\langle a+ib \rangle [/imath] is [imath]a^2+b^2[/imath] ? (where not both [imath]a,b[/imath] are zero)
Is it true that the order of any quotient ring [imath]\mathbb Z[i]/\langle a+ib \rangle [/imath] is [imath]a^2+b^2[/imath] ? ( I know it is atmost finite ) Please help . Thanks in advance . |
264145 | Uses of Taylor series expansion
In calculus we use taylor series expansion at large number of places.I recently one of the application in number theory(To find solution of polynomial in finite field of order [imath]p^{n}[/imath], where p is prime and n is some natural number). I didnt understand why the procedure should work here except that it works here(i solved some problems based on this procedure). So are there any other places where we can use the taylor approximation and also its limitations(if any). | 73733 | The Power of Taylor Series
I am teaching a Calculus class and we are finishing up power/Taylor series this week. The last section of the chapter is on applications, but the only ones listed there are approximating non-rational numbers like [imath]\sqrt{1.02}[/imath] and computing limits like [imath]\lim_{x\to 0}\frac{\sin x}{x}[/imath]. I would like to find better examples that may or may not have a quick physical application (I cannot assume they know any physics beyond what I can explain). So, my question is, do any of you know some applications of Taylor series that I could spend maybe about half an hour to forty five minutes doing? They needn't be physical applications, just interesting. I have already done Euler's formula. Also, we do not deal with any remainder theorems in this class. Thanks. |
369124 | Number of solutions of an equation in [imath]\mathbb F_p[/imath]
I have no idea on how can we calculate [imath]|\{(x_1,\dots,x_n)\in\mathbb F_p^n: x_1^2+\cdots+x_n^2\equiv 0 \pmod p\}|[/imath] where [imath]n[/imath] is an odd integer and [imath]p[/imath] is prime. | 183097 | Number of solutions of [imath]x^2_1+\dots+x^2_n=0,[/imath] [imath]x_i\in \Bbb{F}_q.[/imath]
Let [imath]\mathbb F[/imath] be field with [imath]q[/imath] element and [imath]\operatorname{char}(\mathbb F)\neq2[/imath]. I want to know about the number of solutions of this equation: [imath]x^2_1+x^2_2+\dots+x^2_n=0 \text{ where } x_i\in \mathbb F.[/imath] Any suggestion? |
366325 | [imath]\chi(G) \cdot \chi(\bar{G})\geq n[/imath]
Prove that [imath]\chi(G) \cdot \chi(\bar{G})\geq n[/imath] [imath]\chi(G)[/imath]: number of colors required for a graph [imath]G[/imath]. Here [imath]\bar{G}[/imath] is a graph that consists of all the edges that are not in [imath]G[/imath]. | 153772 | the Nordhaus-Gaddum problems for chromatic number of graph and its complement
Is there any relation between the chromatic number of a graph [imath]G[/imath] and its complement [imath]G'[/imath] that are always true? I saw these ones: [imath]\chi(G)\chi(G')\geq n[/imath] and [imath]\chi(G)+\chi(G')\geq 2n[/imath], but I'm not pretty sure about them. |
369213 | Problem about convergence in Probability (2)
Let [imath]X_1,X_2,\dots[/imath] be a sequence of random variables with [imath] \lim_{n\rightarrow+\infty}E\left[\left|X_n\right|\right]=0 [/imath] Is it true or false that the sequence [imath]X_n[/imath] must converge to [imath]0[/imath] in probability? If true, prove it. If false, provide a counter example. Thank very much | 368241 | Must the sequence [imath]X_n[/imath] converge to [imath]0[/imath] in probability?
Let [imath]X_1, X_2,\dots[/imath] be a sequence of random variables with [imath]\lim_{n\to +\infty} E[|X_n|] = 0[/imath]. Is it correct or wrong that the sequence [imath]X_n[/imath] must converge to [imath]0[/imath] in probability? |
368990 | Why [imath]f^{-1}(f(A)) \not= A[/imath]
Let [imath]A[/imath] be a subset of the domain of a function [imath]f[/imath]. Why [imath]f^{-1}(f(A)) \not= A[/imath]. I was not able to find a function [imath]f[/imath] which satisfies the above equation. Can you give an example or hint. I was asking for an example function which is not addressed here | 78110 | Is [imath]f^{-1}(f(A))=A[/imath] always true?
If we have a function [imath]f:X\rightarrow Y[/imath] where [imath]A\subset X[/imath], is it true to say that [imath]f^{-1}(f(A))=A[/imath]? |
349688 | Combinatorial Proof of Sum of Binomial Coefficients [imath]\sum_{m=k}^{n-k}\binom{m}k\binom{n-m}k=\binom{n+1}{2k+1}[/imath]
I was curious if anyone might be able to give me a hint as to how one might show the following identity combinatorially: [imath]\sum_{m=k}^{n-k}\binom{m}k\binom{n-m}k=\binom{n+1}{2k+1}[/imath] For the left hand side I reason that we are choosing, for example, one object out of [imath]1[/imath] type, then choose from [imath]4[/imath] other types for another object. Continuing on we take [imath]1[/imath] object from [imath]2[/imath] types and then select another object from the remaining [imath]3[/imath] types. The right hand side seems to be telling us something about picking an odd number of objects, which makes me think my initial attempt might have been off the mark and there is instead I should focus on parity. Perhaps I'm missing something simple? | 369237 | Combinatorial Proof of a Binomial Identity [imath]\sum_k {m\choose k} {n \choose k} = {m+n \choose n}[/imath]
[imath]\sum_k {m\choose k} {n \choose k} = {m+n \choose n}[/imath] In this identity we seem to be choosing subsets that do [imath]\it not[/imath] contain k of type m and type n for all possible k. In the style of Vandermonde's identity we are then choosing n elements of both types to not be in the set. Where might I go from here to form a full proof? |
369285 | Linear Algebra: Linear transformation and eigenvalues
Hi could some one please help. I am having problems proving this. Let [imath]A[/imath] be an [imath]n \times n[/imath] matrix with complex entries and let [imath]f (t) =\det(A - tI)[/imath] be its characteristic polynomial. Prove that [imath]f(A) = 0[/imath]. Thank you!! | 96370 | Cayley-Hamilton theorem on square matrices
Can anyone help me by giving the proof of the Cayley-Hamilton theorem? It states that every square matrix [imath]A[/imath] satisfies its own characteristic equation: [imath]p_{A}(A)=0[/imath]. I could prove it when [imath]A[/imath] has distinct eigenvalues, because then it will have [imath]n[/imath] linearly independent eigenvectors, but I couldn't prove the general case. |
367135 | Why does the harmonic series diverge but the p-harmonic series converge
I am struggling understanding intuitively why the harmonic series diverges but the p-harmonic series converges. I know there are methods and applications to prove convergence, but I am only having trouble understanding intuitively why it is. I know I must never trust my intuition, but this is hard for me to grasp. In both cases, the terms of the series are getting smaller, hence are approaching zero, but they both result in different answers. [imath]\sum_{n=1}^{\infty}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots = \text{diverges}[/imath] [imath]\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots =\text{converges}[/imath] | 1209313 | [imath]1\over n[/imath] is not element of [imath]\ell^1[/imath]?
My teacher said [imath]1\over n[/imath] is not element of [imath]\ell^1[/imath] .I know if [imath]\sum_{i=1}^\infty|{1\over n}| < \infty[/imath] , it is element of [imath]\ell^1[/imath]. Let's consider the series [imath](x_i)=(1,{1\over 2}, {1\over 3} , . . .)[/imath] .It seems to me if i sum all the component of the series up , it will be smaller than [imath]2[/imath] so it is finite. Is it right ? Thank you for your help. |
319044 | Necessary condition for pairwise sufficient statistic
I'm struggling to prove the following. If [imath]T:\left(X,\mathbf{A}\right)\rightarrow\left(Y,\mathbf{B}\right)[/imath] is a pairwise sufficient statistic for a set [imath]\left\{\mu_0,\mu_1,\mu_2\right\}[/imath] of three measures on [imath]\left(X,\mathbf{A}\right)[/imath], then [imath]\frac{d\mu_0}{d\left(\mu_0+\mu_1+\mu_2\right)}[/imath] (the Radon-Nikodym derivative) is [imath]T^{-1}\left(\mathbf{B}\right)[/imath]-measurable modulo [imath]\mu_0+\mu_1+\mu_2[/imath]. It is supposedly proved in the otherwise accessible and irreproachable article "Application of the Radon-Nikodym Theorem to the Theory of Sufficient Statistics" by Halmos and Savage (Lemma 9, page 238), but i'm dissatisfied with the proof, since in my opinion it justifies the claim modulo [imath]\mu_0[/imath] only. I'd appreciate help in either understanding Halmos & Savage's proof or proving it from scratch. | 368491 | Pairwise measurable derivatives imply measurability of combined derivative
I've found the following simple claim in an article. Unfortunately, i don't understand the proof given there nor can i come up with an alternative proof of my own. Maybe math.stackexchange can give me a hand? Claim ([HAL], a combination of Lemma 9 and Corollary 3) Let [imath]S=\left(\Omega,\mathcal{A}\right)[/imath] be a measurable space and let [imath]\mathfrak{M}=\left\{\mu_0,\mu_1,\mu_2\right\}[/imath] be a set of three finite measures on [imath]S[/imath]. Suppose [imath]\mathcal{T}[/imath] is a sub-[imath]\sigma[/imath]-algebra of [imath]\mathcal{A}[/imath] such that for any two measures [imath]\mu,\nu\in\mathfrak{M}[/imath], the Radon-Nikodym derivative [imath]\frac{d\mu}{d\left(\mu+\nu\right)}[/imath] is [imath]\mathcal{T}/\mathfrak{B}[/imath]-measurable. Then [imath]\frac{d\mu_0}{d\left(\mu_0+\mu_1+\mu_2\right)}[/imath] is [imath]\mathcal{T}/\mathfrak{B}[/imath]-measurable. As an alternative to help in proving the claim, i'd appreciate help in understanding the proof given in the article that i reproduce below almost verbatim. The points that baffle me are: Why is [imath](*)[/imath] true? In the last paragraph, why would [imath]f[/imath] need to be redefined? In the last paragraph, why is [imath]f[/imath] redefined on a set of [imath]\mu_0[/imath]-measure [imath]0[/imath]? Shouldn't it be redefined on a set of [imath]\left(\mu_0+\mu_1+\mu_3\right)[/imath]-measure [imath]0[/imath], since [imath]f[/imath] is unique up to a set of [imath]\left(\mu_0+\mu_1+\mu_3\right)[/imath]-measure [imath]0[/imath]? Why is the redefined [imath]f[/imath] [imath]\mathcal{T}/\mathfrak{B}[/imath]-measurable? Proof Define [imath] \begin{array}{lcl} f_1 & := & \frac{d\mu_0}{d\left(\mu_0+\mu_1\right)} \\ f_2 & := & \frac{d\mu_0}{d\left(\mu_0+\mu_2\right)} \end{array} [/imath] Since [imath]d\mu_0=f_1d\left(\mu_0+\mu_1\right)=f_2d\left(\mu_0+\mu_2\right)[/imath], we have [imath]f_1d\mu_0=f_1f_2d\left(\mu_0+\mu_2\right)[/imath] and [imath]f_2d\mu_0=f_1f_2d\left(\mu_0+\mu_1\right)[/imath], so that [imath]\left(f_1+f_2-f_1f_2\right)d\mu_0=f_1f_2d\left(\mu_0+\mu_1+\mu_2\right) \tag{*}[/imath] If we define [imath]f:=\frac{d\mu_0}{d\left(\mu_0+\mu_1+\mu_2\right)}[/imath] then it follows that [imath]\left(f_1+f_2+f_1f_2\right)f=f_1f_2\space\left[\mu_0+\mu_1+\mu_2\right]-\mathrm{a.e.}[/imath] Since [imath]0\leq f_1\leq1[/imath] and [imath]0\leq f_2\leq1[/imath], the equation [imath]f_1+f_2+f_1f_2=0[/imath] is equivalent to [imath]f_1=f_2=0[/imath]. Since [imath]\mu_0\left(\left\{x:f_1(x)=f_2(x)=0\right\}\right)=0[/imath], it follows that [imath]f[/imath] may be redefined, if necessary, to be [imath]0[/imath] on the set [imath]\left\{x:f_1(x)=f_2(x)=0\right\}[/imath] without affecting the relation [imath]d\mu_0=fd\left(\mu_0+\mu_1+\mu_2\right)[/imath]; since outside this set [imath]f=f_1f_2/\left(f_1+f_2-f_1f_2\right)[/imath], the proof is complete. References [HAL] Halmos, Paul R., Savage, L. J. "Application of the Radon-Nikodym Theorem to the Theory of Sufficient Statistics". Annals of Mathematical Statistics, 20, 225-241 (1949) |
369890 | Please help tell me what i am doing wrong for multivariable calculus problem
Suppose [imath]F =(2x−4y)i +(x+3y)j[/imath]. Use Stokes' Theorem to make the following circulation calculations: (a) Find the circulation of [imath]F[/imath] around the circle [imath]C[/imath] of radius [imath]10[/imath] centered at the origin in the [imath]xy[/imath]-plane, oriented clockwise as viewed from the positive [imath]z[/imath]-axis. Circulation = [imath]\int_CF\cdot dr[/imath] Here is my work: Please tell me the correct answer and what I am doing wrong: [imath] \begin{align*} \int_C F\cdot dr&= \int\int_S \text{curl} F · dS & \text{by Stokes' Theorem} \\ &= \int\int\langle 0, 0, 5\rangle \cdot \langle 0, 0, 1\rangle~dA & \text{since the circle lies on }z = 0 \\ &= 5 * (\text{Area of }C) \\ &= 5 * 100π \\ &= 500π. \end{align*} [/imath] | 369885 | Please tell me what I am doing wrong for this multivariable Calculus Problem
Suppose [imath]F =(2x−4y)i +(x+3y)j[/imath]. Use Stokes' Theorem to make the following circulation calculations: (a) Find the circulation of [imath]F[/imath] around the circle [imath]C[/imath] of radius [imath]10[/imath] centered at the origin in the [imath]xy[/imath]-plane, oriented clockwise as viewed from the positive [imath]z[/imath]-axis. Circulation = [imath]\int_CF\cdot dr[/imath] Here is my work: Please tell me the correct answer and what I am doing wrong: [imath] \begin{align*} \int_C F\cdot dr&= \int\int_S \text{curl} F · dS & \text{by Stokes' Theorem} \\ &= \int\int\langle 0, 0, 5\rangle \cdot \langle 0, 0, 1\rangle~dA & \text{since the circle lies on }z = 0 \\ &= 5 * (\text{Area of }C) \\ &= 5 * 100π \\ &= 500π. \end{align*} [/imath] |
369957 | What is the probability that we get more than [imath]\frac{n}{2} + 2\sqrt{nln(n)}[/imath] heads?
Toss [imath]n[/imath] coins. What is the probability that we get more than [imath]\frac{n}{2} + 2\sqrt{n[\ln(n)]}[/imath] heads? How do I apply Chernoff Bounds to this? I really need help understanding Chernoff Bounds. | 361290 | Chernoff Bounds. Solve the probability
Toss [imath]n[/imath] coins. What is the probability that we get more than [imath] n/2 + 2\sqrt{n\cdot \log(n)}. [/imath] I have to use Chernoff Bounds here. If I let [imath]X_i[/imath] indicate whether coin [imath]i[/imath] comes up heads, [imath]X=[/imath] the summation of [imath]X_i[/imath], I found the [imath]E(X)=n/2[/imath]. I can't seem to apply the Chernoff bounds now. The answer to the question is [imath]1/n^{8/3}[/imath]. |
181720 | Proof of Non-Ordering of Complex Field
Let [imath]\mathcal F[/imath] be a field. Suppose that there is a set [imath]P \subset \mathcal F[/imath] which satisfies the following properties: For each [imath]x \in \mathcal F[/imath], exactly one of the following statements holds: [imath]x \in P[/imath], [imath]-x \in P[/imath], [imath]x =0[/imath]. For [imath]x,y \in P[/imath], [imath]xy \in P[/imath] and [imath]x+y \in P[/imath]. If such a [imath]P[/imath] exists, then [imath]\mathcal F[/imath] is an ordered field. Define [imath]x \le y \Leftrightarrow y -x \in P \vee x = y[/imath]. Exercise: Prove that the field of complex numbers [imath]\mathbb C[/imath] cannot be given the structure of an ordered field. My Work So Far: (Edit 1 note: This section and the Question is at the beginning, simply leaving this up for reference as to where I started) Let [imath]i[/imath] be such that [imath]i \in P, i \ne 0 \Rightarrow i > 0[/imath]. But [imath]i^2 = -1 \notin P[/imath]. My Question: I am not sure how much I need to redefine, and how I go about rigorously making this patchwork argument airtight. I am aware that I have not addressed how I assumed that [imath]-1 \notin P[/imath], but I'm not sure how to distinguish between [imath]1[/imath] and [imath]i[/imath] in this proof. Edit #1 1st Step: Showing that [imath]-1 \notin P[/imath], observe that [imath](-1)(-1) = 1[/imath] therefore if [imath]-1 \in P[/imath], both [imath]x, -x \in P[/imath], a contradiction. 2nd Step: To show [imath]i \notin P[/imath], we have that if [imath]i \in P \Rightarrow i^2 \in P[/imath], but [imath]i^2 = -1 \notin P[/imath], so [imath]i \notin P[/imath]. 3rd Step: To show [imath]-i \notin P[/imath], we have [imath](-i)(-i) = i^2 \notin P[/imath], so [imath]-i[/imath] cannot be in [imath]P[/imath]. Conclusion: Since [imath]i \ne 0[/imath], and [imath]i, -i \notin P[/imath], there is no set [imath]P \subset \mathbb C[/imath] that satisfies the above properties, thus [imath]\mathbb C[/imath] is not ordered. Thank you André Nicolas and Eric Stucky for your help! | 312204 | Prove that field of complex numbers cannot be equipped with an order relation
Please guide me in this problem. I am confused about whether its asking that having the relation [imath]z>0[/imath] does not satisfy the order axioms. Any help would be really appreciated. Thanks! |
369970 | integral on [imath][a,x][/imath] is zero for all x implies [imath]f=0[/imath] a.e.
Let [imath]f:[a,b] \to \mathbb{\mathbb{R}\cup\{\pm\infty\}}[/imath] be integrable such that [imath]{\int_{[a,x]}fdm=0}[/imath] for any [imath]x\in [a,b][/imath]. How can I show that [imath]f=0[/imath] almost everywhere? I don't know where to start. | 78029 | Proof of [imath]g[/imath] is integrable on [imath][a,b][/imath] and [imath]\int_a^c g=0~~\implies g(x) = 0[/imath] a.e
I have a proof of a Lemma I don't quite understand and I was wondering if someone would be kind enough to explain it to me. Lemma. If [imath]g[/imath] is integrable on [imath][a,b][/imath] and [imath]\int_{a}^{c} g=0~~\forall~c\in [a,b][/imath], then [imath]g(x) = 0[/imath] almost everywhere on [imath][a,b][/imath]. Proof: Let [imath]\mathcal{C}[/imath] be the collection of all sets over which [imath]\int g =0[/imath]. Then [imath]\mathcal{C}[/imath] contains all the intervals in [imath][a,b][/imath]. Thus, it contains all closed sets in [imath][a,b] [/imath]. Now, suppose to the contrary the [imath]g\neq 0[/imath]. Then either [imath]\{g>0\}[/imath] or [imath]\{g<0\}[/imath] contains a closed set, say, [imath]F[/imath] of positive measure. Then [imath]\int_{F}~ g \neq 0[/imath], which is a contradiction. Hence [imath]\int_{a}^{c} g=0[/imath]. |
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