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2679832 | Divisors are cycles for (singular) homology
I'm sure I am missing something quite simple, but why are divisors on an algebraic variety actually homological cycles? I can see them as chains, but in order to take homology (as one always does), or to integrate along them, I would like to see them as singular cycles ([imath]\partial D=0[/imath]). (Not just algebraic cycles, of course, which they are by definition.) Perhaps I am just a bit confused about definitions. Thank you in advance for any clarifying suggestion. | 2393326 | Top cohomology and irreducible components
Consider a connected (possibily reducible) proper algebraic variety [imath]X[/imath] over [imath]\mathbb{C}[/imath] of pure dimension [imath]n[/imath]. Is it true that [imath]H^{2n}(X,\mathbb{Q})\cong \mathbb{Q}^r[/imath] with [imath]r=[/imath]#irreducible components of [imath]X[/imath], where the left-hand side is the singular cohomology with respect to the analytic topology? |
2451861 | Prove that the equivalence classes of an equivalence relation are disjoint.
I want to prove that an equivalence relation [imath]\sim[/imath] on a set [imath]A[/imath] has disjoint equivalence classes—that is, for any nonequivalent elements [imath]a,b\in A[/imath], we have [imath][a]\cap [b]=\emptyset[/imath]. Here is my attempt at the proof: If [imath]\sim[/imath] is an equivalence relation on [imath]A[/imath], then by transitivity, [imath]a\sim b[/imath] and [imath]b\sim c[/imath] implies [imath]a\sim c[/imath]. Therefore, [imath]\forall a\in A[/imath], [imath]a[/imath] can only be equivalent to an element of the equivalence class [imath][a][/imath]. Therefore, for two distinct, non-equivalent elements [imath]a,b\in A[/imath], [imath][a]\cap [b]=\emptyset[/imath]. I would like to know if [imath]1)[/imath] Are there any corrections to be made, and [imath]2)[/imath] Any better wording for the proof. | 1789051 | How to prove two equivalence classes are disjoint?
I know how to prove when the two equivalence classes are not disjoint, i.e. [imath][a]=[b][/imath]. I see that the proof works for proving a equivalence class is disjoint, but I don't get it. Can someone explain it to me? |
2682248 | [imath]AB = C[/imath], prove if [imath]C[/imath] is invertible, then [imath]A[/imath] and [imath]B[/imath]
Let [imath]A, B, C[/imath] be [imath]n \times n[/imath] matrices and assume [imath]AB = C[/imath]. Prove if [imath]C[/imath] is invertible, then [imath]A[/imath] and [imath]B[/imath] are invertible and specify inverses. | 1504170 | Let [imath]A,B[/imath] be [imath]m\times m[/imath] matrices such that [imath]AB[/imath] is invertible. Show [imath]A,B[/imath] invertible.
Let [imath]A,B[/imath] be [imath]m\times m[/imath] matrices such that [imath]AB[/imath] is invertible. Show [imath]A,B[/imath] invertible. Here is what I have attempted: We have [imath]AB[/imath] invertible so there exists an [imath]m\times m[/imath] matrix [imath]C[/imath] such that [imath]C(AB)=(AB)C=I_{m}[/imath]. Since matrix multiplication is associative we immediately have a right and left inverse respectively: [imath](CA)B=A(BC)=I_{m}[/imath]. I know the fact that I have square matrices needs to play a roll and gaurentees me that the product will be, in general, defined. I have tried to multiply [imath]CA[/imath] and [imath]BC[/imath] on other sides but that didn't get me anywhere. I've asked myself what would happen if they weren't square and its obvious, by definition, that they would not be invertible but I am not seeing a way to use the fact that they are to gaurentee ourselves a general inverse. |
2614354 | Rate of convergence of Newton's method near a double root.
Suppose that [imath]r[/imath] is a double root of [imath]f(x)=0[/imath]; that is, [imath]f(r)=f′(r)=0[/imath] but [imath]f''(r)\ne 0[/imath], and suppose that f and all derivatives up to and including the second are continuous in some neighborhood of [imath]r[/imath]. Show that [imath]e_{n+1} ≈ 1/2 e_n[/imath] for Newton’s method and thereby conclude that the rate of convergence is linear near a double root. (If the root has multiplicity [imath]m[/imath], then [imath]e_{n+1} ≈ [(m − 1)/m]e_n[/imath].) I fully understand Newton's method and its calculation. However, this question is a bit confusing and I do not really understand what I am supposed to do. Thanks for the help. | 2685846 | Newton's method and fixed point
Can anyone explain this problem? I am wondering what it means for evaluate [imath]g'(x^∗)[/imath], isn't that zero? Newton's method can be viewed as a way of transforming a root-finding problem [imath]f(x)=0[/imath] into a fixed-point problem [imath]x=g(x)[/imath], where [imath]g(x)=x−f(x)/f′(x).[/imath] Recall that for simple roots, Newton's method has a quadratic convergence rate since [imath]g'(x^∗)=0[/imath], where [imath]g′(x)=f(x)f''(x)/(f'(x))^2.[/imath] When there is a root with multiplicity, [imath]f'(x^∗)= 0[/imath] which leads to a division by [imath]0[/imath] in [imath]g'(x^∗)[/imath]. We will study how this affects the convergence rate of Newton's Method. Consider the function [imath]f(x)=(x−x^∗)^m * h(x),[/imath] where [imath]m[/imath] is an integer greater than [imath]1[/imath] and [imath]h[/imath] is arbitrary function such that [imath]h(x^∗)≠0[/imath] Please answer the following: What is the fixed point problem obtained by applying Newton's Method? Please give the formula for [imath]g(x)[/imath]. Evaluate [imath]g'(x^∗)[/imath]. Based on that value, what can you conclude about the convergence rate of Newton's Method for a root of multiplicity [imath]m[/imath]? Based on the previous result, can you propose a modified Newton's method that recovers quadratic convergence rate? What is the formula for [imath]g(x)[/imath] in this modified version? Prove that your new method achieves quadratic convergence. |
2682763 | Determining Non-Trivial Ideal 0f L(V)
Prove that If [imath]dim_F[/imath] [imath](V ) > 1[/imath] be finite, prove that [imath]L(V )[/imath] has two sided ideals other than [imath](0)[/imath] and [imath]L(V )[/imath]. Further Prove that the conclusion above is false if [imath]V[/imath] is not finite dimensional over [imath]F[/imath]. If [imath]dim_F[/imath] [imath](V ) > 1[/imath].Then let [imath]v_1,v_2[/imath] be two basis elements of [imath]V[/imath].The I tried making an ideal of [imath]L(V )[/imath] using these two basis elements but I was unsucessful. Please help me with the problem.I tried the last part too but wasn't sucessful Note that the above is false as written. See the dupliacte for the correct assertion. | 2682853 | Determining the existence or non-existence of Non Trivial Ideals of [imath]L(V)[/imath]
Prove that If [imath]\text{dim}\ F (V)>1[/imath] be finite, prove that [imath]L(V)[/imath] has no two sided ideals other than [imath](0)[/imath] and [imath]L(V)[/imath]. Further Prove that the conclusion above is false if [imath]V[/imath] is not finite dimensional over [imath]F[/imath]. I assumed that [imath]I[/imath] is an ideal properly contained in [imath]L(V)[/imath] and I attempted that any element in [imath]I[/imath] is [imath]0[/imath]. Please help me with the problem.I tried too but wasn't sucessful. I am unable to understand the approach they did using simple groups etc..We hadnt been taught simple groups in our course..I will be thankful if someone can give an elemetary proof of this result.. Morever that question was different(shown as possible duplicate).Because that was wrong |
2681806 | Finding the expected value of a nonnegative integer valued random variable
If [imath]X[/imath] is a nonnegative integer valued random variable, show that [imath]\mathbb{E}(X) = \sum_{i=1}^\infty\mathbb{P}(X\geq i) = \sum_{i=0}^\infty \mathbb{P}(X\geq i). [/imath] I'm not sure how to do this. I only know the definition that [imath]\mathbb{E}(X):= \sum_{i=-\infty}^\infty i \mathbb{P}(X=i)\mathbf{1}_{\{X\geq 0\}} = \sum_{i=1}^\infty i \mathbb{P}(X=i). [/imath] | 660185 | Show that [imath]\mathbb{E}(T) = \sum\limits^\infty_{k=1}\mathbb{P}(T \geq k)[/imath] for [imath]T[/imath] nonnegative integer valued and [imath]E[T] < \infty[/imath]
Let T be a non-negative integer-valued random variable with [imath]\mathbb{E}(T) < \infty [/imath]. Prove that [imath]\mathbb{E}(T) = \sum^\infty_{k=1}\mathbb{P}(T \geq k)[/imath]. Had a few attempts, haven't really got anywhere. I'm wondering as I'm typing this if proof by induction is a good way to go. Edit: One major thing I forgot to add, am I correct in thinking that also, [imath]\mathbb{E}(T) = \sum^\infty_{k=1}k\mathbb{P}(T = k)[/imath]? |
2682573 | Examine whether [imath]\sum_{n=1}^\infty{\sin^2\left(\frac{1}{n}\right)}[/imath] converges or not
I have the series [imath]\sum_{n=1}^\infty{\sin^2\left(\frac{1}{n}\right)}[/imath] and I'm trying to examine whether it converges or not. My Attempts: I first tried finding whether it diverges by checking if [imath]\lim_{n\to\infty}{\sin^2\left(\frac{1}{n}\right)} \ne 0[/imath]. [imath] \lim_{n\to\infty}{\sin^2\left(\frac{1}{n}\right)}= \lim_{n\to\infty}{\sin\left(\frac{1}{n}\right)}\cdot\lim_{n\to\infty}{\sin\left(\frac{1}{n}\right)}=0 [/imath] Since I didn't get a confirmation from the first try, I then tried the d'Alembert's Criterion which didn't get me very far. [imath] \frac{a_{n+1}}{a_n}= \frac{\sin^2\left(\frac{1}{n+1}\right)}{\sin^2\left(\frac{1}{n}\right)}= \frac{ -\dfrac{2\cos\left(\frac{1}{n}\right)\sin\left(\frac{1}{n}\right)}{n^2} }{ -\dfrac{2\cos\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n+1}\right)}{\left(n+1\right)^2} }= \frac{ \cos\left(\frac{1}{n}\right)\sin\left(\frac{1}{n}\right)\left(n+1\right)^2 }{ \cos\left(\frac{1}{n+1}\right)\sin\left(\frac{1}{n+1}\right)n^2 }=\ ... [/imath] Finally, I tried Cauchy's Criterion, but I didn't get any conclusive result either. [imath] \sqrt[n]{a_n}= \sqrt[n]{\sin^2\left(\frac{1}{n}\right)}= \sin^{\frac{2}{n}}\left(\frac{1}{n}\right)=\ ... [/imath] Question: I've been thinking for while of using the Comparison Test, but I'm not sure which series to compare mine to. How can I examine whether the series converges or not? | 1358037 | convergence of [imath] \sum_{k=1}^\infty \sin^2(\frac 1 k)[/imath]
convergence of [imath] \sum_{k=1}^\infty \sin^2(\frac 1 k)[/imath] How can I do this? Should I use the Ratio Test (I tried this but it started getting complicated so I stopped)? Or the Comparison test(what should I compare it to?)? |
2683072 | Proving a recursive relationship
this is the actual question Often poetic meters of some fixed beat length (per line) have some rhythmic pattern composed of light and heavy syllables from the source language (for example, English or Sanskrit). It is common to treat the light syllable to measure as being 1 beat in length, while a heavy syllable to measure as 2 beats. In such a framework, given a fixed beat length of n, write a program that can compute the total number of patterns involving light and/or heavy syllables? As an example, a line with a fixed beat length of 5 can be composed of a pattern that is light-light-heavy-light (1+1+2+1=5), or heavy-light-heavy (2+1+2=5), or any one of the other six possible patterns. Now I know this is a programming question but I realise that it is actually a fibonacci recursive relation where [imath]f(N) = f(N-1) + f(N-2)[/imath] My only problem is that I can't seem to prove it | 1543085 | Fibonacci sequence emerging from integer partitioning
Let [imath]a_n[/imath] count the number of ways a sequence of [imath]1[/imath]s and [imath]2[/imath]s will sum to n. For example [imath]a_3 = 3[/imath] since [imath]1 + 1 + 1 = 3 = 1 + 2 = 3 = 2 + 1 = 3[/imath] (The ordering matters so 1 + 2 is different from 2 + 1). Find [imath]a_4[/imath] and [imath]a_5[/imath] I think that [imath]a_4 = 5[/imath] since [imath]1 + 1 + 1 + 1 = 2 + 2 = 1 + 1 + 2 = 1 + 2 + 1 = 2 + 1 + 1[/imath] and [imath]a_5 =8[/imath] since [imath]1 + 1 + 1 + 1 + 1 = 2 + 2 + 1 = 2 + 1 + 2 = 1 + 2 + 2 = 1 + 1 + 1 + 2 = 1 + 1 + 2 + 1 = 1 + 2 + 1+ 1= 2 + 1 + 1 + 1[/imath] Now the pattern is emerging, [imath]a_3 + a_4 = a_5[/imath] because [imath]3 + 5 = 8[/imath] Now I want to give a recurrence relation for [imath]a_n[/imath] with initial conditions. So I would say that [imath]a_n = a_{n-1} + a_{n-2}[/imath], It is basically the fibonacci sequence ? and [imath]a_2 =2[/imath] since [imath]1 + 1 = 2[/imath] and [imath]a_1 = 1[/imath] So the initial conditions is what then ? [imath]a_1 =1[/imath] and [imath]a_2 = 2[/imath] ?? |
2683078 | Let [imath]X \sim exp(\lambda)[/imath] be independent of [imath]Y \sim exp(\lambda)[/imath] compute [imath]E[X|X+Y][/imath]
Let [imath]X \sim exp(\lambda)[/imath] be independent of [imath]Y \sim exp(\lambda)[/imath] compute [imath]E[X|X+Y][/imath] I have tried to do this in a couple different ways and all of them have either led to an indefinite integral or a bad answer (illogically [imath]-\lambda[/imath]). Let [imath]Z := X+Y[/imath] The first way I tried to do it was, [imath] \begin{align} E[X|Z] &= \int_0^{\infty} x p_{x|z}dx\\ \end{align}[/imath] Then I found [imath]p_{x|z}[/imath] to be [imath] \begin{align} p_{x|z} &= \frac{P(X=x, X+Y=z)}{P(Z=z)}\\ &= \frac{P(X=x)P(Y=z-x)}{P(Z=z)}\\ &= \frac{\lambda^2 e^{-\lambda x} e^{-\lambda (z-x)}}{\lambda e^{-\lambda z}}\\ &= \lambda \end{align}[/imath] So, [imath] \begin{align} E[X|Z] &= \int_0^{\infty} x p_{x|z}dx\\ &= \int_0^{\infty} x \lambda dx \end{align}[/imath] Which doesn't coverage, can anyone help? | 843013 | Compute [imath]E(X\mid X+Y)[/imath] for independent random variables [imath]X[/imath] and [imath]Y[/imath] with standard exponential distributions
Being [imath]X[/imath] and [imath]Y[/imath] independent random variables distributed as [imath]\mathrm{Exponential}(1)[/imath] and let [imath]T=X+Y[/imath] Calculate [imath]E(X|T)[/imath]. This is my attempt at solving this: Being [imath]Z=\sum_{i=1}^{n}P_{i}[/imath] where [imath]P_{i}[/imath]~[imath]\mathrm{Exp}(\lambda)[/imath] then using it's MFG you have that [imath]M_{P_i}(t)=\frac{\lambda}{\lambda-t}[/imath] then [imath]M_{Z}(t)=M_{\sum{P_{i}}}(t)=(M_{P_{1}})...(M_{P_{n}})=(\frac{\lambda}{\lambda-t})^n[/imath] with this being the MFG of a [imath]\mathrm{Gamma}[/imath] distribution with parameters [imath](n,\lambda)[/imath] With this result is easy to see that [imath]T[/imath]~[imath]\mathrm{Gamma}(2,1)[/imath] Now [imath]E[X|T]=\int xf_{X|T}(x|t)dx[/imath] Then using Bayes' theorem we can make [imath]f_{X|T}(x|t)=\frac{f_{T|X}(t|x)f_{X}(x)}{f_{T}(t)}[/imath] then [imath]f_{T|X}(t|x)=Pr(T=t|X=x)[/imath] [imath]=Pr(X+Y=t|X=x)[/imath] [imath]=Pr(x+Y=t|X=x)[/imath] [imath]=Pr(Y=t-x)=f_{Y}(t-x)[/imath] Using this result we can now rearrenge [imath]f_{X|T}(x|t)=\frac{f_{Y}(t-x)f_{X}(x)}{f_{T}(t)}[/imath] [imath]=\frac{e^{-(t-x)}.e^{-x}}{\frac{te^{-t}}{\Gamma(2)}}[/imath] [imath]=\frac{1}{t}[/imath] then [imath]E[X|T]=\int (x)(\frac{1}{t})dx[/imath] And this where I got stuck because [imath]1_{x\epsilon(0,\infty})[/imath], I made this explanation step-by-step hoping that someone can se easily check where I went wrong. I would aprecciate a step-by-step answer, thanks. |
2680903 | Is this convexification of non-convex constraint correct?
Original problem setup How to handle equality constraints in this problem? \begin{equation}\tag{1} \begin{array}{c} \min_{\mathbf{b}} \hspace{4mm} \mathbf{b}^{T}_{}\mathbf{A}^{}_{}\mathbf{b}^{}_{} \\ s.t \hspace{5mm} \mathbf{b} \in \mathbb{R}^{4} \\ \hspace{9mm}b_0=1\\ \hspace{15mm}b_3=b_1b_2 \end{array} \end{equation} where [imath]\mathbf{A}[/imath] is a [imath]4\times 4[/imath] positive semi-definite matrix. I Reformulated this to \begin{equation}\tag{2} \begin{array}{c} \min_{\mathbf{b}} \hspace{4mm} \mathbf{b}^{T}_{}\mathbf{A}^{}_{}\mathbf{b}^{}_{} + \lambda(\mathbf{b}^{T}_{}\mathbf{Z}^{}_{}\mathbf{b}^{}_{}-1) \end{array} \end{equation} where \begin{equation} \mathbf{Z} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix} \end{equation} My Question: Are (1) and (2) exactly same? Does (2) result in convexification of non-convex constraint [imath]b_3=b_1b_2[/imath] If (2) is correct, then do we have a closed-form solution for [imath]\mathbf{b}[/imath] (Can someone please edit to place [imath]\mathbf{b}[/imath] exactly below min) | 2680030 | How to handle equality constraints in this problem?
Here is the problem setup \begin{equation} \begin{array}{c} \text{min} \hspace{4mm} \mathbf{b}^{T}_{}\mathbf{A}^{}_{}\mathbf{b}^{}_{} \\ s.t \hspace{5mm} \mathbf{b} \in \mathbb{R}^{N} \\ \hspace{9mm}b_0=1\\ \hspace{62mm}b_i=b_{i-2}b_{i-1} \hspace{5mm} \forall \hspace{5mm} 2<i\leq N-1 \end{array} \end{equation} where [imath]\mathbf{A}[/imath] is a [imath]N\times N[/imath] positive semi-definite matrix To handle [imath]b_0=1[/imath] constrain, I know I can introduce Lagrange multiplier in objective function \begin{equation} \begin{array}{c} \text{min} \hspace{4mm} \mathbf{b}^{T}_{}\mathbf{A}^{}_{}\mathbf{b}^{}_{} + \lambda(\mathbf{b}^{T}_{}\mathbf{u}^{}_{}-1) \end{array} \end{equation} where [imath]\mathbf{u}_{}^{}[/imath] is a [imath]N\times 1[/imath] column vector and [imath]u_0=1[/imath] and [imath]u_i=0 \forall 1<i\leq N-1[/imath]. But I dont know how to handle second set of constraints? EDIT: Example Let [imath]N=4[/imath]. Therefore I am looking for [imath]\mathbf{b}=[b_0 \quad b_1 \quad b_2 \quad b_3]^T = [1 \quad b_1 \quad b_2 \quad b_1b_2]^T[/imath]. How do I include [imath]b_3=b_1b_2[/imath] constraint in objective function |
60015 | Real norms on vector spaces over finite fields
I am interested in functions of the form [imath]\psi: F^n \to \mathbb{R}^+[/imath], where [imath]F[/imath] is a finite field, that have norm-like properties, e.g., [imath]\psi(x+y) \le \psi(x) + \psi(y)[/imath]. Does anybody know if there is any literature on this area? | 2151779 | Normed vector spaces over finite fields
Normed vector spaces are typically defined over the reals or complex numbers. Is there any "standard," well-behaved construction that generalizes this to a vector space over a finite field, such as [imath]\Bbb F_2[/imath]? I'm looking for something kind of like the class of [imath]\ell_p[/imath] norms, except designed with finite fields in mind. Ideally, something that has deep fundamental properties making it well-behaved in the same way that the Euclidean norm is. |
2682933 | Transforming Differential Equation to Bessel ODE
Here is the question I refer to I need help solving a differential equation I am not the author of the question, it is just we face same problem. I was given a differential equation [imath]y'= x^2 + y^2[/imath]. Letting [imath]y(x)=\frac{-f(x)}{f(x)}[/imath] gives us \begin{equation}f''+x^2f=0\end{equation} This is where I got stuck. The answer in the question stated that the ODE can be transformed to ODE of Bessel kind. However, I am not sure on how to do this since the ODE does not actually has form [imath]x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx}+(x^2-v^2)y = 0[/imath] which is a Bessel Equation. Any help appreciated. | 269743 | Convert [imath]\frac{d^2y}{dx^2}+x^2y=0[/imath] to Bessel equivalent and show that its solution is [imath]\sqrt x(AJ_{1/4}+BJ_{-1/4})[/imath]
I have been following the thread " Convert Airy's Equation [imath]y''-xy=0[/imath] to Bessel equation [imath]t^2u''+tu'+(t^2-c^2)u[/imath] " but I can't join the dots to a solve similar equation [imath]y''+x^2y=0[/imath] so as to obtain a solution of the form [imath]\sqrt x\left(AJ_\frac{1}{4}+BJ_{-\frac{1}{4}}\right)[/imath] I actually get an equation that looks this way [imath]t^2\frac{du}{dt}+t\frac{du}{dt}+(t^2+\frac{5}{64})u[/imath] The above equation can not yield the desired solution. Please help me to clearly see this. Thank you. |
2683070 | How do you solve [imath]\dot{X} = UX[/imath]?
Let [imath]X(t) = \left[\begin{matrix}a(t) & b(t) \\c(t) & d(t)\end{matrix}\right][/imath] and let [imath]U[/imath] be a nonsingular matrix. How do you solve [imath]\frac{d}{dt} X(t)=UX(t)[/imath] I presume there is some general method to solve these kinds of ODEs but I cannot find anything about it, online. BTW: ----------------- I know that you can get 4 ODE equations for the four unknown functions of [imath]t[/imath]. The problem is that each ODE equation includes other functions as so [imath]\frac{d a}{d t}=u_{11}a+u_{12}c[/imath] | 694214 | Systems of linear differential equations - eigenvectors
Solve the following system of equations [imath] \begin{cases} x_1^{'}(t)=x_1(t)+3x_2(t) \\ x_2^{'}(t)=3x_1(t)-2x_2(t)-x_3(t) \\ x_3^{'}=-x_2(t)+x_3(t)\end{cases} [/imath]. First, I create the column vectors [imath]X[/imath] and [imath]X^{'}[/imath]. Then the matrix [imath]A= \begin{bmatrix} 1 & 3 & 0 \\ 3 & -2 & -1 \\ 0 & -1 & 1 \\ \end{bmatrix} [/imath] Now, I find the eigenvalues, [imath]-4,3,1[/imath] and their corresponding eigenvectors [imath](-3,5,1)^T (-3,-2,1)^T (1,0,3)^T[/imath]. I'm just not sure how to take it from here and solve the system of differential equations. I want a diagonal matrix [imath]D[/imath] so that I can read the solutions easy, but I'm not sure how to do it. EDIT Building on @Francisco 's answer, I'd have that: [imath]X=c_1 (-3,5,1)^T e^{-4t} + c_2 (-3,-2,1)^T e^{3t} + c_3 (1,0,3)^T e^{t} [/imath]. But I believe this could be written in a simpler form. |
2683136 | Counting irreducible polynomials over finite fields
How many irreducible polynomials in [imath]Z_2[x][/imath] of degree [imath]3[/imath]? I have discussed this with my friend before and we found that [imath]x^3 + x^2 + 1[/imath] and [imath]x^3+x+1[/imath] are the two said polynomials which irreducible. We got this the approachment that polynomials [imath]p(x) = ax^3+bx^2+cx+d[/imath] must satisfy [imath]a\ne 0[/imath] and later [imath]d \neq 0[/imath] and here we only need to take some cases of the value of [imath]b[/imath] and [imath]c[/imath]. Do you have idea to solve this problem clearly without guessing? I am afraid that there will be a new problem asking for polynomials whose degree is higher than this one and here I will be stuck. | 1419677 | Irreducible polynomials of degree 3 and 5
Two part question: 1) There are three irreducible cubic polynomials [imath]\mathbb F_2[/imath] Aproach: Bruteforce, there are not as much cubic polynomials, just 8 of them: [imath]x^3[/imath] [imath]x^3+1[/imath] [imath]x^3+x[/imath] [imath]x^3+x+1[/imath] [imath]x^3+x^2[/imath] [imath]x^3+x^2+1[/imath] [imath]x^3+x^2+x[/imath] [imath]x^3+x^2+x+1[/imath] Only three of them are irreducible(??): [imath]x^3+1[/imath],[imath]x^3+x+1[/imath] and [imath]x^3+x^2+1[/imath] Reality check - do i miss point of irreducible polynomial here, and should consider adding [imath]x^3+x^2+x+1[/imath] to irreducible list? It is reducible, bit different way then other polynomials: [imath]x^3+x^2+x+1=(x+1)^3[/imath] 2) Only irreducible polynomial of degree 2 over [imath]\mathbb F_2[/imath] is [imath]x^2+x+1[/imath]. Does it help, with conjuction of part 1, to find all irreducible polynomials of degree 5? |
1035837 | Solving this trigonometric task: [imath]\sqrt{3}\cos{\theta}-\sin{\theta}=R\cos(\theta+\alpha)[/imath]
Find the values of [imath]R[/imath] and [imath]\alpha[/imath] in the identities below, given that [imath]R>0[/imath] and [imath]\alpha[/imath] is an acute angle. [imath]\sqrt{3}\cos{\theta}-\sin{\theta}=R\cos(\theta+\alpha)[/imath] I'm a bit confused by this task. How should I start? I have [imath] \cos(a+b) = \cos(a)\cos(b)-\sin(a)\sin(b). [/imath] If I square anything, I can use the trig identity [imath] \sin^2(x) + \cos^2(x) =1. [/imath] | 1192729 | Express [imath]\sqrt{3}\sin\theta - \cos\theta[/imath] as: [imath]a\cos (\theta + \alpha) [/imath]
Express [imath]\sqrt{3}\sin\theta - \cos\theta[/imath] as: [imath]a\cos (\theta + \alpha) [/imath] Can someone please explain to me how to go about doing this? |
2679319 | Uniform convergence of power series with decreasing coefficients
Let [imath]\{a_n\}[/imath] be a monotone decreasing sequence of real numbers, with limit [imath]0[/imath]. Prove that, for each [imath]0<\delta<2[/imath], the power series [imath]\sum_{n=0}^{\infty} a_n z^n[/imath] converges uniformly on each [imath]\Omega_\delta = \{z : |z|\leq 1 , |z-1|\geq \delta\}[/imath] I have no idea on how to deal with the points in the boundary. Thanks in advance! | 505165 | Let [imath]b_n[/imath] decrease monotonically to zero, prove [imath]\sum b_nz^n[/imath] converges for [imath]|z|\leq 1[/imath] and [imath]z\neq 1[/imath]
Let [imath]b_n[/imath] be a sequence which decreases monotonically to zero, prove [imath]\sum b_nz^n[/imath] converges for [imath]|z|\leq 1[/imath] and [imath]z\neq 1[/imath]. So I was able to prove it converges on the open unit disk in the complex plane, but except for a few special points on the boundary, such as [imath]-1[/imath] and [imath]i[/imath], where I was able to use the alternating series test, I can't seem to prove it converges in general on the boundary minus [imath]z=1[/imath]. I feel I have to somehow use the fact that all the numbers on the unit circle except [imath]z=1[/imath] rotate around and around as you raise them to higher and higher powers, thus you get the real and imaginary parts of your series to alternate between sets of positive and negative terms, but in a very irregular way for most initial angles. I tried to simplify the problem by breaking the series into its real and imaginary parts, and then trying to prove they each converge separately, but I still can't prove either converges: [imath]\sum_{n=0}^{\infty}b_ne^{n\theta i} = \sum_{n=0}^{\infty}b_n\cos(n\theta) +i\sum_{n=0}^{\infty}b_n\sin(n\theta).[/imath] |
1324248 | Proving multivariable limit exists where [imath]\frac{a}{c} + \frac{b}{d} \gt 1[/imath]
Given that [imath]\frac{a}{c} + \frac{b}{d} > 1[/imath], I am attempting to show that [imath]\lim_{(x,y)\rightarrow(0,0)} \frac{|x|^{a}|y|^{b}}{|x|^{c} + |y|^{d}} = 0[/imath] My attempt at a solution: Let's assume the limit is in fact zero, and prove it with the Squeeze Theorem. We have that \begin{align*} \Bigg|\frac{|x|^{a}|y|^{b}}{|x|^{c} + |y|^{d}}\Bigg| &= \frac{|x|^{a}|y|^{b}}{|x|^{c} + |y|^{d}} \\ &\leq \frac{1}{2}|x|^{a-\frac{c}{2}}|y|^{b-\frac{d}{2}} && \text{using the fact that } a^{2} + b^{2} > 2ab \end{align*} But I'm not quite sure how to proceed from here to apply the inequality given and find a limiting function, if I am even on the right track with the inequality I applied. This isn't homework ( a practice question I found ), any help would be greatly appreciated. | 66226 | Multivariable limit proof: [imath]\lim_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0[/imath]
I have some trouble with this: Show that if [imath]a, b \ge 0[/imath], [imath]c, d > 0[/imath] and [imath]\frac{a}{c} + \frac{b}{d} > 1[/imath] then [imath]\lim_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0[/imath] we were not really shown how to evaluate multivariable limits other than trying different paths, which wouldn't work when the limit exists, or by applying the squeeze theorem which is my initial plan. Unfortunately I cannot find a proper bound for the denominator and so I am quite lost right now. Any help would be appreciated. |
2684012 | Show that if [imath]n^2[/imath] is even then [imath]n[/imath] is even, for [imath]n[/imath] an integer.
I know that this is a common question, but I want to see your take on this. The way I approached this was by contraposition (i'm still new with the proofs but any help would be appreciated, even if it's by contradiction.) Assume [imath]n[/imath] is odd, thus [imath]n^2[/imath] is odd as well. Since [imath]n[/imath] is odd then [imath]n=2k+1[/imath] for some integer [imath]k[/imath]. Then [imath]n^2= (2k+1)^2 = 4k^2+4x+1 \rightarrow 2(2k^2+2k)+1.[/imath] Which clearly shows that [imath]n^2[/imath] is odd. | 1927343 | Proving [imath]a^2[/imath] is even [imath]\implies[/imath] [imath]a[/imath] is even by contradiction
Prove [imath]a^2[/imath] is even [imath]\implies[/imath] [imath]a[/imath] is even. It was proven via contradiction by my friend. Here is the proof in question, which uses proof by contradiction Assume towards a contradiction that [imath]a[/imath] is odd. Let [imath]a = 2k+1[/imath], then [imath]a^2 = 4k^2 + 4k + 1 = 2(2k^2+2k)+1[/imath] Therefore [imath]a^2[/imath] is odd. Contradiction. To me, this proof in question seems incorrect because it shows that if [imath]a[/imath] is odd then [imath]a^2[/imath] is odd. We have shown that the converse of what we want to contradict is a contradiction. I feel like this proof by contradiction is not mathematically correct or vigorous enough, and for it to be correct we would have to start from [imath]a^2[/imath] being even, working our way down to [imath]a[/imath] being odd, and then finding a contradiction. Am I correct? Or am I just confused about something? |
2683895 | Show that [imath]f'(x) = f^2(x)[/imath] and [imath]f(0) = 0[/imath] implies [imath]f[/imath] is the zero function
Let [imath]f : \mathbb{R} \to \mathbb{R}[/imath] differentiable be such that [imath]f'(x)=f^2(x)[/imath] for all [imath]x[/imath] and [imath]f(0)=0[/imath]. Show that [imath]f(x)=0[/imath] for all [imath]x[/imath]. My attempt: Notice that [imath]f'(x)=f^2(x)\geq 0[/imath], thus [imath]f[/imath] is increasing. Since [imath]f(0)=0, f(x)\geq0\; \forall x \geq 0[/imath]. Since [imath]f[/imath] is differentiable, [imath]f[/imath] is continous, therefore given [imath]1> \epsilon >0[/imath], there exists [imath]\delta<1[/imath] such that for [imath]x \in (-\delta,\delta)\implies f(x)<\epsilon<1.[/imath] Take [imath]x=\delta/2[/imath]. Then [imath]f(x) = f'(c)\delta/2=f^2(c)\delta/2[/imath]. Then [imath]f^2(c) \geq f(x)[/imath]. But since [imath]c\in (-\delta,\delta)[/imath], [imath]0\leq f (c)<1 \implies f^2(c) \leq f(c)\implies f(c)\geq f(x)[/imath]. But [imath]c<x[/imath], therefore [imath]f(c)\geq f(x)[/imath] which means [imath]f(c)=f(x)[/imath]. Therefore for all [imath]y \in (c,x)[/imath], [imath]f(y)=f(x) \implies f'(y)=f^2(y)=0 \implies f(y)=0[/imath]. Where do I go from here? | 2533837 | [imath]f'(x) = [f(x)]^{2}[/imath]. Prove [imath]f(x) = 0 [/imath]
Let [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] be a differentiable function such that [imath]f(0) = 0[/imath] and [imath]f'(x) = [f(x)]^{2}[/imath], [imath]\forall x \in \mathbb{R}[/imath]. Show that [imath]f(x) = 0[/imath], [imath]\forall x \in \mathbb{R} [/imath]. I first (unsuccessfully) tried using the Mean Value Theorem, but this is in the Integrals chapter so the solution probably involves them. Can't really see where integrals come in here though. What I've got so far: (i) Since [imath]f[/imath] is differentiable, thus it is continuous and, hence, integrable. Therefore [imath]f^2[/imath] is also integrable and as [imath]f'=f^2[/imath], [imath]f'[/imath] is too. (ii) [imath]f' \geq 0[/imath], [imath]\forall x \in \mathbb{R}[/imath] |
2684219 | General term for fibonacci sequence
Can we define the general term for the Fibonacci Sequence? I mean we can make a generating function for fibonacci sequence which equals [imath]\dfrac1{1-x-x^2}[/imath], but is there any way to find the general term of fibonacci sequence? | 2710881 | Help me understand how to Find f(n)
I have this question: [imath]f(0) = 0[/imath] [imath]f(1) = 1[/imath] [imath]f(n) = f(n-1) + f(n-2)[/imath] help me understand and find [imath]f(n)[/imath]. |
2684546 | Equivalent form of CH in ZF
I know that in [imath]ZFC[/imath] the continuum hypothesis (CH) is equivalent to the statement: "a subset [imath]A\subset\mathbb R[/imath] is either finite, infinite countable or [imath]A[/imath] has the same cardinality of [imath]\mathbb R[/imath], i.e. [imath]|A|=|\mathbb R|[/imath]". We can call this statement [imath]CH(\mathbb R)[/imath], but what's going on if we work in [imath]ZF[/imath]? Clearly [imath]CH\Rightarrow CH(\mathbb R)[/imath], but does the other implication hold? What I want to know is if there is a model of [imath]ZF[/imath] in which [imath]CH(\mathbb R)[/imath] is true but [imath]CH[/imath] not. | 404807 | How to formulate continuum hypothesis without the axiom of choice?
Please correct me if I'm wrong but here is what I understand from the theory of cardinal numbers : 1) The definition of [imath]\aleph_1[/imath] makes sense even without choice as [imath]\aleph_1[/imath] is an ordinal number (whose construction doesnt depend on the axiom of choice) with some minimal property. With or without choice, there is no cardinal number [imath]\mathfrak{a}[/imath] such that [imath]\aleph_0 < \mathfrak{a} < \aleph_1[/imath]. 2) In ZFC, all cardinal numbers are [imath]\aleph[/imath] and are comparable (by the trichotomic property of the ordinals). The continuum hypothesis states that [imath]\aleph_1 = 2^{\aleph_0}[/imath]. 3) In ZF, [imath]2^{\aleph_0}[/imath] need not be an [imath]\aleph[/imath]. However, I don't know if talking about [imath]2^{\aleph_0}[/imath] in ZFC makes sense or not. Does it make sense in ZF to define CH to be the statement that if a set is larger than the natural numbers, then it must contain a copy of the reals (up to a relabeling of the elements) no matter what the cardinality of the reals is ? |
2685060 | Recurrence relation question (generating direct formula)
Solve this recurrence relation by generating its direct formula: [imath]a_n = 3a_{n-1} + 2n, a_0 = 1[/imath] Using the direct formula find the [imath]10th[/imath] term of this recurrence relation. My answer: [imath]a_{10} = 147601[/imath] Is this correct? | 2684316 | Recurrence relation: [imath]a_n = 3a_{n-1} + 2n, a_0 = 1[/imath]
Solve the following recurrence relation by generating its direct formula: [imath]a_n = 3a_{n-1} + 2n, a_0 = 1[/imath] Use the direct formula to find the [imath]10th[/imath] term of the recurrence relation. My attempt: [imath]3(10-1) + 2(10)[/imath] [imath]3(9) + 20[/imath] [imath]27 + 20[/imath] [imath]10th[/imath] term = [imath]47[/imath] Is this correct? |
2684903 | If f:X→R is a continuous non constant function and X is connected then X is uncountable.
If [imath]f:X \to \mathbb{R}[/imath] is a continuous non constant function and [imath](X,d)[/imath] is a connected metric space then [imath]X[/imath] is uncountable. How to proceed this proof I am clueless , I know that if [imath]X[/imath] is connected so [imath]f(X)[/imath] is also connected and [imath]f[/imath] is non constant to [imath]f(X)[/imath] is an interval which is not singleton, also each interval is uncountable, so, [imath]f(X)[/imath] is uncountable does it means that [imath]X[/imath] is also uncountable? Any other methods other than this to understand this question? | 1207777 | [imath]f(X)[/imath] is uncountable and hence [imath]X[/imath] is uncountable.
My question: let [imath]f : X \to \Bbb R[/imath] be a non-constant continuous function on a connected metric space and assume that [imath]f(X)[/imath] is uncountable; then [imath]X[/imath] is uncountable. We know continuous image of a connected metric space is connected. Since connected space in [imath]\Bbb R[/imath] are intervals and the function is non-constant continuous function so [imath]f(X)[/imath] is an interval in [imath]\Bbb R[/imath] and hence uncountable. Is the proof correct?? How does it follows that [imath]X[/imath] is uncountable? |
2685357 | Integrating characterstic function
I want to show that the sequence [imath]a_n(t)=\frac{1}{n}\textbf{1}_{[0,n)}(t)[/imath] does not converge in [imath]L^1(\lambda)[/imath]. [imath]u_n[/imath] converges in [imath]L^p[/imath] implies [imath]u_n[/imath] is Cauchy so I want to use the contrapositive. [imath]\begin{align}||a_{2n}-a_n||_1&=\int|a_{2n}-a_n|~d(\lambda)\\&=\int\left|\frac{1}{2n}\textbf{1}_{[0,2n)}(t)-\frac{1}{n}\textbf{1}_{[0,n)}(t)\right|~d(\lambda)\\&=\int\left|\frac{1}{n}\left(\frac{1}{2}\textbf{1}_{[0,2n)}(t)-\textbf{1}_{[0,n)}(t)\right)\right|~d(\lambda) \end{align} [/imath] not sure how to continue I know the sequence does not converge. I want to prove it doesn't; this is why the question isn't a duplicate. | 2685277 | Does this sequence converge pointwise and/or in [imath]L^1[/imath]?
Let [imath]a_n(t)=\frac{1}{n} \text{1}_{ [0,n)}(t)[/imath]. Does this sequence converge in [imath]L^1/[/imath]pointwise? My ideas: We have for all [imath]t[/imath]: [imath]\mathbf{1}(t)\in\{0,1\}[/imath] and [imath]0 \leq \frac{1}{n}\textbf{1}(t) \leq 1 \Rightarrow 0 \leq \left| \frac{1}{n}\textbf{1}(t)\right|^p \leq 1 \quad \forall p\ge1[/imath] So it's a null sequence and we have dominated convergence with dominated function [imath]g(t)=1[/imath]. [imath]g(t)=1[/imath] is integrable so all [imath]a_n[/imath] is integrable for all [imath]n[/imath]. |
2685399 | Equality of two analytic functions based on a condition
Suppose that [imath]f[/imath] and [imath]g[/imath] are two analytic functions on the set [imath]\phi[/imath] of all complex numbers with [imath]f(\frac{1}{n})=g(\frac{1}{n})[/imath] for [imath]n=1,2,3,\ldots[/imath] Then show that [imath]f(z)=g(z)[/imath] for each [imath]z[/imath] in [imath]\phi[/imath]. I am not able to even start proving it. Can some one please help me? | 1421861 | Complex Analysis analytic function problem
Suppose that [imath]f[/imath] ang [imath]g[/imath] are two analytic functions on the set [imath]\Bbb C[/imath] of all complex numbers with [imath]f(1/n) = g(1/n)[/imath] for [imath]n= 1,2,3,\ldots[/imath], then show that [imath]f(z) = g(z)[/imath] for all [imath]z\in\Bbb C[/imath]. |
2686087 | What is [imath]\int_0^\infty \frac{\sin(x^2)}{x^2}\ \,dx[/imath]?
I was asked this [imath]\int_0^\infty \frac{\sin(x^2)}{x^2}\ \,dx[/imath] I have a solution by letting [imath]I(b) = \int_0^\infty \frac{\sin(x^2)}{x^2}e^{-bx} \,dx[/imath] However, I was told that the question can also be solved via contour integration. Can anyone demonstrate that? You can take a picture of the steps if you are too lazy to deal with mathjax symbols. Thanks! | 306011 | Integral [imath]\int_0^{\infty} \sin(x^2)/x^2\,dx[/imath]
Does anyone have a proof for [imath]\int_0^{\infty}\frac{\sin(x^2)}{x^2}\,dx=\sqrt{\frac{\pi}{2}}.[/imath] I tried to get it from contour integrating [imath]\frac{e^{iz^2}-1}{z^2},[/imath] but failed. Thanks. |
2686094 | Prove [imath] \sin^4(x)[/imath] identity using [imath]z=cis(x)?[/imath]
I cannot prove the [imath]\sin^4(x)[/imath] identity using [imath]z= cis(x)[/imath]. I know that you have to use de Moivre's theorem and compare the real values of [imath]z^4[/imath] but I am stuck at this step: [imath]\sin^4(x)= -\cos^4(x) + 6\sin^2(x)\cos^2(x) + \cos(4x)[/imath] The identity is [imath]\sin^4(x) = 1/8(\cos(4x) - 4\cos(2x) + 3)[/imath] | 696593 | Prove the trigonometric identity [imath]\sin^4{x} = \frac{3-4\cos{2x}+\cos{4x}}{8}[/imath]
I need to show the steps to prove this identity: [imath]\sin^4{x} = \frac{3-4\cos{2x}+\cos{4x}}{8}[/imath] I know that [imath]\cos{2x}=\cos^2{x}-\sin^2{x}[/imath]. From there I do not know what to do. The solution should look like: [imath]\sin^4{x}=sin^4{x}[/imath] I need to prove the right side equals the left side. |
2400859 | Rational value of sine
Given [imath]x[/imath], [imath]\sin(x) \in \mathbb{Q}[/imath], where [imath]x[/imath] is in degrees, we want to find all [imath]x[/imath] in the range [imath](0,90)[/imath]. One trivial solution is [imath]x=30[/imath]. | 2339177 | Rational values of trigonometric functions
I am using extensively trigonometric functions when an angle is given in degrees. Some of these functions like sine or cosine have rational values, for example, the well known example is that [imath]\cos(\theta) =0.6 [/imath] and [imath]\sin(\theta) =0.8 [/imath]. However besides the case of multiplicity of [imath]90^\circ[/imath] it seems there are no rational numbers [imath]\theta[/imath] with simultaneously rational values of sine and cosine. Is it possible somehow to prove that for rational values of an angle given in degrees there are no values simultaneously rational of sine and cosine functions, beside obvious case of multiplicities of [imath]90^\circ[/imath]? |
2686922 | Understanding sets of polynomials as subspaces
I have a homework assignment, that I can't quite wrap my head around, let me state the question, and show my progress. Let [imath]n[/imath] be a natural number, and let [imath]\alpha \in \mathbb{R}[/imath] be a real scalar. Consider the subset \begin{equation*} V=\{p\in P_n(\mathbb{R}):p(\alpha)=0\}, \end{equation*} of the real vector space [imath]P_n(\mathbb{R})[/imath]. First I need to show, that [imath]V[/imath] is a subspace of [imath]P_n(\mathbb{R})[/imath], and as far as I figure, [imath]V[/imath] is the set containing all polynomials of degree [imath]\leq n[/imath] where no matter the variable [imath]\alpha[/imath] the polynomial equals [imath]0[/imath]. I've then showed that since all elements are [imath]0[/imath] it's closed under addition, scalar multiplication and it ofcourse contains [imath]0[/imath] itself. I assume what I have done so far is correct, but am unsure. Next up I need to decide the dimension of [imath]V[/imath] and I have been hinted that I need to use the formular [imath]\text{dim}(v)=\text{dim}(\text{ker}(L))+\text{dim}(L(V))[/imath] on the imaging [imath]L:P_n(\mathbb{R})\rightarrow\mathbb{R}[/imath] defined as [imath]L(p)=p(\alpha)[/imath]. I assume [imath]\ker(L)=0[/imath] since all coefficients in the polynomial has to be [imath]0[/imath], and therefore [imath]\dim(\ker(L))=0[/imath] as well. Same goes for [imath]L(V)=0[/imath] and [imath]\dim(L(V))=0[/imath], so is the answer to this question just [imath]0[/imath]? The reason why I'm unsure of myself is because I feel it's almost too easy with all these zeroes. Sorry if some of the terminology is wrong, english is not my first language. | 2680986 | Show a subset is subspace of a vectorspace and find the dimension
Let [imath]n>1[/imath] be a natural number and let [imath]\alpha\in\mathbb{R}[/imath] be a real scalar. Let [imath]V[/imath] be a subset of the vector space [imath]P_n(\mathbb{R})[/imath]. Define [imath]V[/imath] as [imath]V=\{p\in P_n(\mathbb{R}):p(\alpha)=0\}[/imath] Task: show that V is a sub space of [imath]P_n(\mathbb{R})[/imath] and find the dimension of V. I believe I have the correct definitions but I'm not sure if I've done this correctly. I know V is a subspace if the following holds 1) Vector 0 is in the subset. 2) V is closed under addition. 3) V is closed under scalar multiplication. For (1): I have a hard time showing but from the definition of V it only contains the zero vector. Meaning it satisfies (1). For (2): Any real scalar multiplied with the zero vector gives the zero vector, which is in [imath]V[/imath]. Therefore it satisfies (2). For (3): If you add two zero vectors you will always get a the zero vector. Therefore it satisfies (3). I'm uncertain how to find the dimension but I know how to define it using the dimension formula for V. [imath]L:P_n(\mathbb{R}) \mapsto (\mathbb{R})[/imath] defined by [imath]L(p)=p(\alpha)[/imath]. We can use the formula: [imath]Dim(V)=Dim(Ker(L))+Dim(L(v))[/imath] |
2687678 | Equivalent metrics - proof
Why are [imath]d_1[/imath] and [imath]d_2[/imath] equivalent? [imath]d_1(x,y)=\sqrt{\sum_{i=1}^{n} (x_i-y_i)^2}[/imath] [imath]d_2(x,y)=\max_i|x_i-y_i|[/imath] I'm stuck, I started here: [imath]d_1(x,y)=\sqrt{\sum_{i=1}^{n} (x_i-y_i)^2}=\sum_{i=1}^{n} |x_i-y_i|[/imath] | 1185344 | Show the Euclidean metric and maximum metric are strongly equivalent.
I need to show that the Euclidean metric and maximum metric (or square metric??) are strongly equivalent. I have no idea how to start this proof. Any help? [imath]d_1, d_2[/imath] are called strongly equivalent if there exist positive constants [imath]K, M[/imath] such that for all [imath]x, y\in X[/imath]: [imath]Md_1(x,y)\leq d_2(x,y)\leq Kd_1(x,y)[/imath] |
2688172 | Find all holomorphic functions [imath]f: \mathbb{C} \rightarrow \mathbb{C}[/imath] such that, for all [imath]x,y\ \epsilon \ \mathbb{R} [/imath] , [imath]Re(f(x+iy)) := 2xy[/imath]
Find all holomorphic functions [imath]f: \mathbb{C} \rightarrow \mathbb{C}[/imath] such that, for all [imath]x,y\ \epsilon \ \mathbb{R} [/imath] , [imath]Re(f(x+iy)) := 2xy[/imath] So I figured out thus far that [imath]\frac{\partial u}{\partial x}:= 2y[/imath] and [imath]\frac{\partial u}{\partial y}:= 2x[/imath] with [imath]u(x,y):= 2xy[/imath]. Now by the Cauchy-Rieman equations I can see that [imath]\frac{\partial v}{\partial y}:= 2y\ [/imath] and [imath]\frac{\partial v}{\partial x}:= -2x\ [/imath] for some [imath]v(x,y)[/imath]. However I seem to be stuck finding [imath]v(x,y)[/imath]. Can anybody help me out here, would be very happy. | 1193828 | Find all holomorphic functions with [imath]\Re(f(x+iy))=2xy[/imath]
Find all holomorphic functions [imath]f:\mathbb C\to\mathbb C[/imath] such that for all [imath]x,y\in\mathbb R[/imath] [imath]\Re(f(x+iy))=2xy[/imath] So [imath]f[/imath] must be complex differentiable, which is equivalent to that Cauchy-Riemann equations must be satisfied Let [imath]f(x+iy)=P(x,y)+iQ(x,y)[/imath] (Here [imath]P(x,y)=2xy[/imath]) then [imath]\frac{\partial P}{\partial x}=2y=\frac{\partial Q}{\partial y}[/imath] [imath]\implies Q(x,y)=y^2+Q(x)[/imath] the second equation is: [imath]\frac{\partial P}{\partial y}=2x=-\frac{\partial Q}{\partial x}[/imath] [imath]\implies Q(x)=-y^2[/imath] Hence [imath]Q(x,y)=x^2-y^2\implies f(x+iy)=2xy+i(x^2-y^2)[/imath] Can you verify my steps, and holomorphic means differentiability in the complex sense right ? |
782166 | [imath]p[/imath] is an odd prime number where [imath]p=3k+1\Longleftrightarrow\exists a,b\in\Bbb Z^+[/imath] such that [imath]p=a^2+ab+b^2[/imath]
Let [imath]p[/imath] be odd prime number,show that: [imath]p=3k+1\Longleftrightarrow \exists a,b\in\Bbb Z^+ \textrm{ such that } p=a^2+ab+b^2[/imath] I guess this is true because I find when: [imath]p=7,k=2[/imath],and [imath]7=2^2+2\cdot 1+1^2[/imath] (2) when [imath]p=13,k=4[/imath],and [imath]13=1^2+1\cdot 3+3^2[/imath] and so on. How do I prove this ? | 280551 | Primes congruent to 1 mod 6
I came across a claim that I found interesting, but can't seem to prove for some reason. I have the feeling it should be easy a prime [imath]p[/imath] can be written in the form [imath]p = a^2 -ab +b^2[/imath] for some [imath]a,b\in\mathbb{Z}[/imath] if and only if [imath]p\equiv 1\bmod{6}[/imath] |
2688179 | How do I show this fourier series relationship?
I want to show that [imath]\int^a_0\sin(mx)\sin(nx)\ dx = 0[/imath] for [imath]m, n[/imath] positive integers and [imath]m\neq n[/imath], and [imath]a = k\pi[/imath], i.e. the integer multiple of [imath]\pi[/imath]. I have tried expressing the equation in a fourier series: [imath]f(x) = \frac{a_0}{2} + \sum^\infty_1 a_n \cos(n\pi x)[/imath] since the equation is an even function. However, I'm unable to show that [imath]f(x)[/imath] is equivalent to zero in the interval mentioned. I can reason that for half wavelengths, the summation should end up in zero because of how the cosine wave is like. But how do I remove [imath]a_0 /2[/imath]? | 638308 | Orthogonality of sine and cosine integrals.
How to prove that [imath] \int_{t_0}^{t_0+T} \sin(m\omega t)\sin(n\omega t)\,\mathrm{d}t[/imath] will equal to [imath]0[/imath] when [imath]m\ne n[/imath] and [imath]\frac{T}{2}[/imath] when [imath]m=n\ne 0[/imath]? Besides [imath] \int_{t_0}^{t_0+T} \cos(m\omega t)\cos(n\omega t)\,\mathrm{d}t[/imath] will equal to [imath]0[/imath] when [imath]m\ne n[/imath] and [imath]\frac{T}{2}[/imath] when [imath]m=n\ne 0[/imath] and [imath]T[/imath] when [imath]m=n=0[/imath]? |
2687844 | A detail about Lie algebra?
I am reading GTM 9, and I am trying to compute a structure constants of an abstract Lie algebra using Jacobi identity: [imath][x[yz] + [y[zx]] + [z[xy]] = 0[/imath]. Apply it on [imath][x_l [x_i x_j] + [x_i [x_j x_l]] + [x_j [x_l x_i]] = 0[/imath] and use [imath][x_i x_j] = \sum_k a_{ij}^k x_k[/imath]. In the book the equation is My question is why need not we sum by index m? Just k? | 2317890 | What does the Jacobi identity impose on structure constants?
I am currently self-studying Lie algebras and want to check my answer to the following question from Erdmann and Wildon: Let [imath]L[/imath] be a Lie algebra with basis [imath](x_1, \dots, x_n)[/imath]. What condition does the Jacobi identity impose on the structure constants [imath]a_{ij}^k[/imath]? The structure constants are defined as the scalars [imath]a_{ij}^k[/imath] in the underlying field [imath]F[/imath] such that [imath][x_i, x_j] = \sum_k a_{ij}^k x_k.[/imath] I have already established that [imath]a_{ji}^k = - a_{ij}^k[/imath] and that [imath]a_{ii}^k =0[/imath]. Here's my stab at answering the above question: Let [imath]x_i, x_j, x_k[/imath] be basis elements. Applying the Jacobi identity: \begin{align} [x_i, [x_j, x_k]] + [x_j, [x_k, x_i]] + [x_k, [x_i, x_j]]&=0 \\ \left[x_i, \sum_l a_{jk}^l x_l \right] + \left[ x_j, \sum_l a_{ki}^l x_l \right] + \left[ x_k, \sum_l a_{ij}^l x_l \right] &= 0 \\ \sum_l a_{jk}^l [ x_i, x_l] + \sum_l a_{ki}^l [ x_j, x_l] + \sum_l a_{ij}^l [ x_k, x_l] &= 0 \\ \sum_l a_{jk}^l \sum_m a_{il}^m x_m + \sum_l a_{ki}^l \sum_m a_{jl}^m x_m + \sum_l a_{ij}^l \sum_m a_{kl}^m x_m &= 0 \\ \sum_l \sum_m \left( a_{jk}^l a_{il}^m x_m + a_{ki}^l a_{jl}^m x_m + a_{ij}^l a_{kl}^m x_m \right) &=0 \\ \sum_m x_m \underbrace{\sum_l \left( a_{jk}^l a_{il}^m + a_{ki}^l a_{jl}^m + a_{ij}^l a_{kl}^m \right)} &= 0. \end{align} Since [imath](x_1, \dots, x_n)[/imath] forms a basis, they are linearly independent. Therefore, the indicated term is zero for all [imath]i, j, k, m[/imath]. Questions Is the above correct? Is there a less computationally intensive approach? Note that I have been particularly explicit to try to catch any possible errors; my scratch work compressed this to three lines. Is this the strongest result possible? The conditions that [imath]a_{ji}^k = - a_{ij}^k[/imath] and that [imath]a_{ii}^k =0[/imath] are easy to visualize by analogy with skew-symmetric matrices. Is there a similar interpretation of this result? |
2688250 | How to calculate square matrix to power n with induction
How could i possibly calculate square matrix to power n with induction for lets say this square matrix [imath]A =\begin{bmatrix} 1 & 2\\ 0& 1 \end{bmatrix}[/imath]I cant use Cayley-Hamilton theorem and i need to use induction.Any ideas on what i could do? | 1005446 | Help to compute all powers [imath]A^{n}[/imath] and find is the matrix of [imath]A[/imath] diagonal?
I have a matrix: [imath] A=\begin{bmatrix}1 & a \\ 0 & 1\end{bmatrix} [/imath] which satisfies: [imath]A\left( e_{1}\right)=e_{1},A\left( e_{2}\right)=ae_{1}+e_{2}, a\neq 0[/imath] I need compute all powers [imath]A^{n}, n\in \mathbb{Z}[/imath] and find their matrices. [imath] A^n= \begin{bmatrix}1 & na \\ 0 & 1\end{bmatrix} [/imath] Is it true? Is there a basis of [imath]V[/imath] (if [imath]V[/imath] is a 2-dimensional real vector space with basis [imath]\left\{e_{1},e_{2}\right\}[/imath]) so that the matrix of [imath]A[/imath] is diagonal? |
2689113 | Why is [imath]\frac{\log 8}{\log 3} = \frac{\ln8}{\ln3}[/imath]
Need help clarifying why: [imath]\frac{\log 8}{\log 3} = \frac{\ln8}{\ln3}[/imath] To my understanding [imath]\ln[/imath] implies [imath]\log[/imath] base e and [imath]\log 8[/imath] implies [imath]\log[/imath] base [imath]10[/imath] of [imath]8[/imath]. Kind of confused how that is equivalent. Thanks | 673660 | Why [imath]\frac{\log(x)}{\log(y)}[/imath] gives the same value as [imath]\frac{\ln(x)}{\ln(y)}[/imath]
If [imath]x=16384[/imath] and [imath]y=2[/imath] [imath]\ln(x)=9.704[/imath] [imath]\ln(y)=0.6931[/imath] [imath]\log(x)=4.2144[/imath] [imath]\log(y)=0.3010[/imath] If we divide [imath]\frac{\ln(x)}{\ln(y)}[/imath] we get [imath]14[/imath] and same answer for [imath]\frac{\log(x)}{\log(y)}[/imath]. So can anyone tell me the concept behind this? Why does dividing [imath]\frac{\ln(x)}{\ln(y)}[/imath] give the same result as [imath]\frac{\log(x)}{\log(y)}[/imath]? |
2688616 | Location of the extrema of the sinc function
Wikipedia provides the following expression for the location of the extrema of the sinc function: [imath]x_n=q-q^{-1}-\dfrac{2q^{-3}}{3}-\dfrac{13q^{-5}}{15}-\dfrac{146q^{-7}}{105}...[/imath] with [imath]q =\left(n + \dfrac 12\right)\pi[/imath] However, I have not been able to find this expression in the literature. Can anyone please provide some references where this expansion is derived, or at least mentioned? | 2242007 | Approximating the locations of the extrema of the sinc function
The wikipedia article on the sinc function gives an approximation for the (positive [imath]x[/imath]) locations of the extrema of the sinc function as [imath]x_n \simeq (n + \frac{1}{2})\pi - \frac{1}{(n + \frac{1}{2})\pi}[/imath] but provides no reference or proof. How would I go about deriving such an approximation? |
1437092 | Number Theory Prove [imath]\gcd(a, b, c)=\gcd(\gcd(a,b),c)=\gcd(\gcd(a,c),b)=\gcd(\gcd(b,c),a)[/imath].
Let [imath]a, b, c[/imath] be integers, no two of which are zero, and [imath]d=\gcd(a, b, c)[/imath]. Show that [imath]d=\gcd(\gcd(a,b),c)=\gcd(a,\gcd(b,c))=\gcd(\gcd(a,c),b)[/imath]. Here is what I have tried, but I'm unsure if the part that I prove [imath]d=\gcd(a,b)[/imath] is correct: Proof: [imath]d=\gcd(a,b,c)\implies d\mid a[/imath], [imath]d\mid b[/imath], and [imath]d\mid c[/imath]. Let [imath]d=\gcd(a,b)[/imath]. So, [imath]d=ax+by[/imath], some [imath]x,y\in \mathbb{Z}[/imath], by Theorem 2.3. Let [imath]n\mid a[/imath] and [imath]n\mid b[/imath], [imath]n\in \mathbb{N}[/imath]. Then, [imath]n\mid (ax+by)[/imath] by Theorem 2.2(g) [imath]\implies n\mid d[/imath]. So, [imath]d=\gcd(a,b)[/imath] by Theorem 2.6. Similarly, we can show [imath]d=\gcd(a,c)=\gcd(b,c)[/imath]. Thus, we have [imath]\gcd(\gcd(a,b),c)=\gcd(d,c)[/imath]. Since [imath]d\mid c[/imath], we have [imath]dk=c[/imath], some [imath]k\in \mathbb{Z}[/imath]. So, we have [imath]\gcd(d,c)=d\cdot \gcd(1,k)=d\cdot 1=d[/imath] by Theorem 2.7. Similarly, we can prove that [imath]\gcd(a,\gcd(b,c))=\gcd(\gcd(a,c),b)=d.\blacksquare[/imath] Theorems 2.2(g): If [imath]a\mid b[/imath] and [imath]a\mid c[/imath], then [imath]a\mid (bx+cy)[/imath] for arbitrary integers [imath]x[/imath] and [imath]y[/imath]. 2.3: Given integers [imath]a[/imath] and [imath]b[/imath], not both of which are zero, there exist integers [imath]x[/imath] and [imath]y[/imath] such that [imath]\gcd(a,b)=ax+by[/imath]. 2.6: Let [imath]a,b[/imath] be integers, not both zero. For a positive integer [imath]d[/imath], [imath]d=\gcd(a,b)[/imath] if and only if (a) [imath]d\mid a[/imath] and [imath]d\mid b[/imath]. (b) Whenever [imath]c\mid a[/imath] and [imath]c\mid b[/imath], then [imath]c\mid d[/imath]. 2.7: If [imath]k>0[/imath], then [imath]\gcd(ka,kb)=k\cdot \gcd(a,b)[/imath]. | 1189424 | $\gcd(a,b,c)=\gcd(\gcd(a,b),c)\,$ [GCD Associative Law]
Prove [imath]\gcd(a,b,c)=\gcd(\gcd(a,b),c)[/imath] for [imath]0\ne a,b,c\in \Bbb{Z}[/imath]. I tried solving it with sets but I sense there are some details I am missing. I would truly appreciate your reference. |
2688905 | Prove [imath]\bigcap_{n\in\mathbb N}[n,\infty)\ne\varnothing[/imath]
[imath]\bigcap_{n\in\mathbb N}[n,\infty)\ne\varnothing[/imath] I do not understand how the intersection is not empty if for every [imath]x[/imath] in the intersection there is [imath]n>x[/imath]. | 2397081 | Let [imath]\{K_n \}_{n \in \mathbb{N}}[/imath] be a collection of sets, where [imath]K_n = [n, \infty)[/imath]. Why is [imath]\bigcap_{n \in \mathbb{N}} K_{n}[/imath] empty?
This question is related to Exercise 2.15 in Baby Rudin, in which the reader is instructed to "[s]how that Theorem 2.36 and its Corollary become false (in [imath]R^1[/imath], for example) if the word 'compact' is replaced by 'closed' or by 'bounded.'" The relevant theorem and corollary are as follows: Theorem 2.36: If [imath]\{K_{α}\}[/imath] is a collection of compact subsets of a metric space [imath]X[/imath] such that the intersection of every finite subcollection of [imath]\{K_{α}\}[/imath] is nonempty, then [imath]\{K_{α}\}[/imath] is nonempty. Corollary: If [imath]\{K_{n}\}[/imath] is a sequence of nonempty compact sets such that [imath]K_{n}[/imath] contains [imath]K_{n+1}, (n = 1, 2, 3, ...)[/imath], then [imath]K_n[/imath] is not empty. My Question: I realize this question and similar ones have been asked elsewhere on MSE, but I've yet to see a straightforward (to me, at least) answer to my question specifically. Usually, for the "closed" part of the exercise, the following collection of closed sets, [imath]\{K_n \}_{n \in \mathbb{N}}[/imath] is used: [imath]K_n = [n, \infty), n \in \mathbb{N}.[/imath] For, the intersection of every finite subcollection of [imath]{K_n}[/imath] is clearly nonempty. Indeed, [imath] \bigcap_{i=1}^{m} K_{n_{i}} = [N, \infty) [/imath] where [imath]N = \max\{n_1, n_2, ..., n_m\}[/imath]. Then, the assertion is made that [imath] \bigcap_{n=1}^{\infty} K_n = \emptyset. [/imath] However, I don't see how the same argument can't simply be used, i.e., that, for example, [imath]N \in \bigcap_{n=1}^{\infty} K_n[/imath] (or maybe, say, [imath]n+1[/imath]). Intuitively, this certainly makes sense, I'd say. For, given the archimedean property, it seems that as [imath]n[/imath] approaches infinity, we can always find an [imath]l[/imath] in each increasingly smaller set such that [imath]n < l < \infty[/imath]; and so [imath]l \in \bigcap_{n=1}^{\infty} K_n[/imath]. What is it exactly that I'm misunderstanding or overlooking here? |
1619950 | Distribution of a transformation of normally distributed independent variables.
If [imath]W = \frac{X + YZ}{\sqrt{1 + Z^2}}[/imath] where all variables involved are standard-normally distributed and independent, what is the distribution of [imath]W[/imath]? The solution I am reading begins with [imath]P(W \le w) = \int \int \int 1_{(-\infty, w)} ( (x+yz)/(1+z^2)) \frac{1}{2\pi} \exp^{- (x^2 + y^2)/2} \ dx dy \frac{1}{\sqrt{2\pi}} e^{-z/2} dz [/imath] but I don't find that particularly helpful. What happened? | 1102336 | Find the distribution of sum and product of standard normal random variables
Let [imath]X,Y[/imath] and [imath]Z[/imath] be three independent real valued random variables. All with finite second moment and all with mean [imath]0[/imath] and variance [imath]1[/imath]. Define [imath] W= \frac{X+YZ}{\sqrt{1+Z^2}} [/imath] Find the distribution of [imath]W[/imath] under the additional assumption that [imath]X,Y[/imath] and [imath]Z[/imath] all have a marginal [imath]\mathcal{N}(0,1)[/imath]-distribution Hint: Compute [imath]P(W \le w )[/imath] by Tonellie, integrating out of the joint distribution of [imath](X,Y)[/imath] first. I believe that [imath]P(W \le w )[/imath] can be written as \begin{align} P(W \le w )&=P\left(\frac{X+YZ}{\sqrt{1+Z^2}}\le w\right)\\ & =P(X+YZ \le w\sqrt{1+Z^2})\\ & =P(X \le w\sqrt{1+Z^2})+P(Y\le w\sqrt{1+Z^2})P(Z \le w\sqrt{1+Z^2}) \end{align} But I am not totally sure. Further I do not know how to proceed from here. |
2689567 | Let [imath]p(x) = x^3 + x + 1 \in \mathbb Z_2[x][/imath] and [imath]E = \mathbb Z_2[x]/p(x)[/imath]. Factor [imath]p(x)[/imath] into linear factors in [imath]E[x][/imath].
Let [imath]p(x) = x^3 + x + 1 \in \mathbb{Z}_2[x][/imath] and [imath]E = \mathbb{Z}_2[x]/p(x)[/imath]. Factor [imath]p(x)[/imath] into linear factors in [imath]E[x][/imath]. Note that [imath]p(t) = 0[/imath], where [imath]t = x + ⟨p(x)⟩[/imath]. You might also wish to show that [imath]p(t^2) = 0[/imath]. So far, I have that [imath]E = \{a+bt+ct^2 : a,b,c \in \mathbb{Z}_2, t^3 = 1+t\}[/imath] and I have shown that [imath]p(t^2) = 0[/imath]. But I really do not know how to proceed from here. I have in my notes that the solution is [imath]p = [x + t][x + t^2][x + (t + t^2)] \in E[x][/imath] but I do not have any details on how this solution was obtained. Any help would be appreciated. | 1537229 | Prove that [imath]x^3+x+1[/imath] splits in [imath]Z_2[x]/(x^3+x+1)[/imath]
Let [imath]f(x)=x^3+x+1[/imath] and set [imath]E=z_2[x]/(x^3+x+1)[/imath] prove that [imath]f(x)[/imath] splits in E that is find 3 distinct roots of [imath]f(x)[/imath] in [imath]E[/imath] Questions found [imath][x^2+1][/imath] [imath][x^2+x+1][/imath] on my sloppy work. There are more than 3 roots if my notes are correct. If that is the case do we just use the irreducibles?? Another question I have is that what is [x]?? Is it a variable, some sort of constant. I know the book uses idertimenatn and has special properties. |
2689539 | Let [imath]K=\mathbb{Q}(\sqrt[3]{2})[/imath] then [imath]U(K)=\{\pm (\sqrt[3]{2} -1)^n : n \in \mathbb{Z}\}[/imath]
Let [imath]K=\mathbb{Q}(\sqrt[3]{2})[/imath] I want to show that [imath]U(K)=\{\pm (\sqrt[3]{2} -1)^n : n \in \mathbb{Z}\}[/imath] I know that an integral basis for [imath]K[/imath] is [imath]1, \sqrt[3]{2}, \sqrt[3]{2}^2[/imath] So far I have got: Since the minimal polynomial of [imath]\sqrt[3]{2}[/imath] has one real root and 2 complex roots then K has one real embedding and one complex embedding, so the signature of K is (1,1). The rank of K=1+1-1=1. Then by using Dirichlet's Unit Theorem there is some unit [imath]u[/imath] such that every other unit can be written in the form [imath]\pm u^n[/imath]. Therefore [imath]U(K)=\{\pm u^n\}[/imath] but I don't know how to proceed from here. | 1789643 | What are the units in [imath]\mathbb{Z}[\root 3 \of 2][/imath]?
I asked Wolfram Alpha to tell me the fundamental unit of [imath]\mathbb{Z}[\root 3 \of 2][/imath], it replied [imath]1 - \root 3 \of 2[/imath]. Then I tried asking it for [imath](1 - \root 3 \of 2)^n[/imath] for [imath]-5 \leq n \leq 5[/imath]. If I run [imath]n[/imath] over all the integers, does that give me all the units of this domain? Or does this give me some of the units and some non-units? |
2690659 | Prove [imath]\left\|x\right\|_2\leq\sqrt{n}\left\|x\right\|_\infty[/imath]
I can prove other inequalities like [imath]\|x\|_{1}\leq \sqrt{n}\|x\|_{2}[/imath] and [imath]\|x\|_{1}\leq {n}\|x\|_{\infty}[/imath] using cauchy-schwarz and holder's inequality, but I don't know how to prove this one [imath]\left\|x\right\|_2\leq\sqrt{n}\left\|x\right\|_\infty[/imath]. Can anybody tell me how to prove this one? Also, how do I prove inequalities like these ones: [imath]\|x\|_2\leq\|x\|_1[/imath],[imath]\|x\|_\infty\leq\|x\|_2[/imath], [imath]\|x\|_\infty\leq\|x\|_1[/imath]? | 2687126 | Proof: [imath]\|X\|_2\leq \sqrt{n}\|X\|_\infty[/imath]
(Euclidean Norm) Proof that [imath]\|X\|_2\leq \sqrt{n}\|X\|_\infty[/imath], if [imath]x \in \mathbb{R^n}[/imath]. With that, proof [imath]\|X\|_\infty \to 0 \implies \|X\|_2\to 0[/imath] I'm stuck: [imath]\|X\|_2= \sqrt{x^2_1+x^2_2+...+x^2_n}\leq\sqrt{n}\sqrt{x^2_1+x^2_2+...+x^2_n}[/imath] I don't know how to use the fact: [imath]\|X\|_\infty=\max(|x_1|,|x_2|,...,|x_n|)[/imath] |
2690888 | Let p be prime. If a group has more than [imath]p-1[/imath] elements of order [imath]p[/imath], why can't the group be cyclic?
Let [imath]p[/imath] be prime. If a group has more than [imath]p-1[/imath] elements of order [imath]p[/imath], why can't the group be cyclic? Can I assume that the group must be finite group since there is at least one element with finite order? And I am wondering now if all cyclic groups are finite, or all finite groups are cyclic? | 368376 | Let [imath]p[/imath] be a prime. If a group has more than [imath]p − 1[/imath] elements of order [imath]p[/imath], why can’t the group be cyclic?
I need help to prove the following result Let [imath]p[/imath] be a prime. If a group has more than [imath]p − 1[/imath] elements of order [imath]p[/imath], why can’t the group be cyclic? Thanks for the help. |
2690889 | Is [imath]V[/imath] infinite-dimensional?
[imath]V[/imath] is a vector space and for any positive integer [imath]n[/imath], there exists a linearly independent subset [imath]S_n \subseteq V[/imath] of size [imath]n[/imath]. Is [imath]V[/imath] infinite-dimensional and how do I go about proving this? | 2678507 | Proving a vector space is infinite-dimensional
Let [imath]S[/imath] be a vector space. Suppose that for any positive integer [imath]n[/imath], there exists a linearly independent subset [imath]S_n ⊆ V[/imath] of size [imath]n[/imath]. How do you show that [imath]S[/imath] is not finite dimensional but infinite-dimensional. I understand that to be finite dimensional then it must have a finite then there exists a finite set that such at [imath]\operatorname{Span}(T)=S[/imath]. So would it be right to prove by contradiction that there is not a finite spanning set that satisfies [imath]\operatorname{Span}(T)=S[/imath], and hence I could argue that the vector space [imath]S[/imath] is infinite dimensional. How would I show that [imath]S[/imath] does not have a spanning set [imath]T[/imath], such that [imath]\operatorname{Span}(T)=S[/imath]? |
2689853 | Taylor series coefficients of general rational function
Given two (possibly infinite) sequences of complex numbers [imath]\{a_1, a_2, \ldots\}[/imath] and [imath]\{b_1, b_2, \ldots\}[/imath], I'm looking to find a closed form expression for each element of the sequence of complex numbers [imath]\{c_1, c_2, \ldots\}[/imath] where the sequences are related by [imath]\frac{1+\sum_{n=1}^\infty a_n x^n}{1+\sum_{m=1}^\infty b_m x^m} = 1+\sum_{r=0}^\infty c_r x^r\,.[/imath] Essentially, I am looking for a rapid computational algorithm to find the Taylor series expansion coefficients of a general rational function. Edit: The problem can be simplified to finding a series expansion of [imath](1+\sum_{m=1}^\infty b_m x^m)^{-1}[/imath] since then the [imath]c[/imath] coefficients are given by the series product. | 2050216 | Dividing an infinite power series by another infinite power series
Let's say I have two power series [imath]\,\mathrm{F}\left(x\right) = \sum_{n = 0}^{\infty}\,a_{n}\,x^{n}[/imath] and [imath]\,\mathrm{G}\left(x\right) = \sum_{n = 0}^{\infty}\,b_{n}\,x^{n}[/imath]. If I define the function [imath]\displaystyle{\,\mathrm{H}\left(x\right) = \frac{\mathrm{F}\left(x\right)}{\mathrm{G}\left(x\right)} = \frac{\sum_{n = 0}^{\infty}\, a_{n}\,x^{n}}{\sum_{n = 0}^{\infty}\, b_{n}\, x^{n}}}[/imath], is there a general way to expand [imath]\,\mathrm{H}[/imath] such that [imath]\,\mathrm{H}\left(x\right) = \sum_{n=0}^{\infty}\,c_{n}\,x^{n}[/imath] ?. I guess, what i'm asking is if there is a way to get the first few [imath]c_{n}[/imath] coefficients ?. I'm dealing with a physics problem in which I have two such functions [imath]\,\mathrm{F}[/imath], [imath]\,\mathrm{G}[/imath] and I'd like to get the first few terms in the power series [imath]\,\mathrm{H}[/imath]. |
2691594 | Proving either [imath]T(v)=0[/imath] or [imath]T(x)=v[/imath] has a solution for every [imath]v[/imath] in [imath]V[/imath]?
If [imath]T:V\longrightarrow V[/imath] is a linear transformation where [imath]V[/imath] is finite dimensional, then either: a) [imath]T(v)=0[/imath], for some [imath]v≠0[/imath] in [imath]V[/imath]. b) [imath]T(x)=v[/imath] has a solution [imath]x\in V[/imath] for every [imath]v\in V[/imath]. -- I don't think I properly understand the question. Please help me | 2689940 | Let [imath]T:V → V[/imath] be a linear transformation where [imath]V[/imath] is finite dimensional.
Let [imath]T:V → V[/imath] be a linear transformation where [imath]V[/imath] is finite dimensional. Show that exactly one of (i) and (ii) holds: (i) [imath]T(v)=0[/imath] for some [imath]v ≠ 0[/imath] in [imath]V[/imath] ; (ii) [imath]T(x)=v[/imath] has a solution [imath]x[/imath] in [imath]V[/imath] for every [imath]v[/imath] in [imath]V[/imath] Not sure how to answer this, I'll get any help I can get. Thanks in advance! |
917654 | Good data structure for hyperbolic tiling
Say you're doing something computational where each data point is a tile in a (not necessarily Euclidean) 2-dimensional tiling, for instance, a Life-like cellular automata. You might want a data structure where tiles that are near in distance are also near each other in memory. You might choose your tiling based on what's easy to represent this way. For a tiling of the Euclidean plane, you probably have square tiles backed by a 2d array with array point a[i][j] = tile [imath]a_{ij}[/imath]; you might have hex tiles instead, which can also be backed by a 2d array as explained in http://www.redblobgames.com/grids/hexagons/. For a tiling of the sphere, it's pretty common to start with a sub-triangulation of the icosahedron and subdivide it into 5 2d arrays plus 2 points for the north pole, as explained in http://kiwi.atmos.colostate.edu/BUGS/geodesic/text.html. It looks like a geodesic dome. Is there a 2d hyperbolic tiling that admits a representation in a nice data structure? The only software I've ever seen that uses hyperbolic tiling in this way is the game HyperRogue (a roguelike on the hyperbolic plane). It just has an object for each tile, and each tile keeps a record of its neighbors. It works but it's not pretty. | 85793 | Symbolic coordinates for a hyperbolic grid?
Rephrasing (one year later) (original question is below) Apparently the original question wasn't clear, or nobody knows an answer (or both). So I will try to rephrase it. Look at your favorite hyperbolic grid. This question asks for a labeling system for the points. Of course we could draw the grid on the Poincare disk, and use real numbers (the (x,y) coordinates of each point), but in practice that would raise all sorts of precision issues that could surely be avoided if one had a proper symbolic coordinate system. You shouldn't need real numbers if you're staying on grid points! Just like (2,3)+(0,1) is trivial to do by hand (or in your head) for the square grid, we'd like some form of symbolic coordinates (i.e. consisting of finite sequences of symbols) for a hyperbolic grid, in which simple operations on small grid vectors are similarly easy. If you take some sequence of left and right turns in a hyperbolic grid city, what should you keep track of in your head so you can take the optimal route back to your starting point? In a Euclidean square grid, it is {(int)X, (int)Y, (enum)direction you are facing}, and we all know how to use this. Is there any hyperbolic grid where this question has a nice answer? Original Question If we consider the Euclidean plane, then a square grid works well with using two integers (the coordinates) as a representation for vectors that lie on the grid. Does anybody know of a similarly nice representation for working in the hyperbolic plane? Desirable features of any such system: [imath]-[/imath] It should be easy to add two vectors. [imath]-[/imath] It should be easy to see if two vectors are the same. [imath]-[/imath] It should be easy to see if two vectors are close. [imath]-[/imath] It should be easy to rotate a vector by an angle that is a symmetry of the grid. [imath]-[/imath] Every point in the plane should be near some vector. (The grid shouldn't have large empty regions.) Note that "addition" using continuous coordinates would most naturally be defined in terms of parallel transport, and it is neither commutative nor associative. On a grid, however, the issue of orientation needs to be addressed differently, since parallel transport along arbitrary grid vectors will introduce rotations that are not a symmetry of the grid, so to remain on the grid, addition will need to be defined using something other than parallel transport. Since hyperbolic grids have an exponential number of points (as a function of distance from the origin), our symbolic coordinates will need to have a length proportional to the length of the vector. Don Hatch has a web page which gives some possible grids. ("Easy" means an efficient algorithm that is doable by hand, like how the algorithms for addition and subtraction using the standard digit-sequence representation of integers can be used for the Euclidean square grid. Using hyperbolic functions of real numbers, always keeping track of enough digits to identify the nearest grid point, does not count as easy!) Addendum: Many readers seem to be disturbed by a little voice saying "But vectors don't make sense in the hyperbolic plane!": What that voice is really saying is that the hyperbolic plane doesn't give you all the nice properties you are used to in Euclidean space or in vector spaces. So if you define vectors as objects having specific properties you are used to (such as having a direction that is unaffected by parallel transport), then that definition may not correspond to anything in the hyperbolic world. You could use the term isometry, but that is typically a fixed mapping from the space to itself, and in the hyperbolic plane the isometry corresponding to a vector would depend on the starting point. (However, isometries will be central to any solution, I assume.) And a geodesic is a fixed path within the space, no different from a hyperbolic line or segment. The paper "What is a vector in hyperbolic geometry?" offers one solution. Gyrovectors offer another. I deliberately leave them undefined in this question, as I am looking for any workable system. |
2691666 | How can I compute sum of [imath]i 4^i[/imath]?
How can I compute [imath]\sum_{i=0}^n i 4^i[/imath] this equation? What is the way? | 2207532 | Summation by parts (finite calculus) of [imath]\sum\limits_{i=0}^{n}i\cdot a^i[/imath]
I'm studying the calculus of finite differences and have read about summation by parts: [imath]\sum f(x)\Delta g(x) = f(x)g(x) - \sum g(x+h)\Delta f(x)[/imath]. The tutorials I'm using go a bit woolly at the point of introducing this technique: they use examples where they set [imath]\Delta g(x)[/imath] to [imath]2^x[/imath], and don't really show how to use it for anything a bit harder. I'm at a loss for basic examples. I wondered if it is possible to determine [imath]S.n = \sum\limits_{i=0}^{n}i\cdot a^i[/imath] using summation by parts? If so, how is it done? Thanks in advance. |
2691879 | Proof that for infinite sets, bijections and injections are equivalent
Online I found the following statement that for [imath]X[/imath] an infinite set, if there is a bijection from [imath]X[/imath] to [imath]\mathbb{N}[/imath] this is equivalent to there is an injection from [imath]X[/imath] to [imath]\mathbb{N}[/imath]. How would I prove this? Attempt: The fact that if there is a bijection it is also an injection is true by definition of bijection. Beyond that, I don't know how to demonstrate that an injection is equivalent to a bijection for an infinite set [imath]X[/imath] to [imath]\mathbb{N}[/imath]. | 1703801 | Prove that an infinite subset of positive integers is countable.
[imath]Let\ C\ be\ an\ infinite\ subset\ of\ \mathbb{Z}_+ [/imath] Then definite [imath]h: \mathbb{Z}_+ \rightarrow C [/imath] by [imath]h(n)=smallest\ element\ of\ [C-h(\{1,2,3,\dots ,n-1\})][/imath] It is not hard to see why this is injective. For surjection, in my book, it says that there exists smallest element m such that for any c in C [imath]h(m)\geq c[/imath] [imath]Then\ for\ all\ i<m, we\ must\ have\ h(i)<c[/imath] here I do not understand why Since [imath]h(m)[/imath] is the smallest element of [imath]C-\{h(1),h(2),h(3)...h(m-1)\}[/imath], if [imath]h(m)>c[/imath], shouldn't it follow that [imath]c \in \{h(1),h(2),h(3)...h(m-1)\}[/imath]? |
2692069 | Any hints or answers to figure out the unknown value of C?
Evaluate [imath] \int_{1}^{\infty} \left(\frac{1}{\,\sqrt{\, x^{2} + 4\,}\,} - \frac{C}{x + 2}\right)\,\mathrm{d}x [/imath] I am stuck at this question where I should find the value of C for which the integral converges and evaluate the integral for this value of C. Any hints or answers ??. | 2705916 | Why the integral converges only when C is 1 and why other values of C would give infinity?
Find the value of the constant [imath]C[/imath] for which the following integral converges and evaluate the integral for this value of [imath]C[/imath] [imath]\int_{1}^{\infty}\left[\frac{C}{x+2} - \frac{1}{\sqrt{x^2 +4}} \right]dx = \lim_{t\rightarrow\infty}\int_{1}^{t}\left[\frac{C}{x+2} - \frac{1}{\sqrt{x^2 +4}} \right]dx = \lim_{t\rightarrow\infty} \left(\sinh^{-1}(\frac{t}{2}) - C\ln(t+2)- \text{constant}\right)[/imath] By taking [imath]\ln(t+2)[/imath] common factor [imath] \lim_{t\rightarrow\infty} \ln(t+2)\left(\frac{\sinh^{-1}(\frac{t}{2})}{\ln(t+2)} - C\right)- \text{constant}[/imath] [imath] \text{Since} \lim_{t\rightarrow\infty} \frac{\sinh^{-1}(\frac{t}{2})}{\ln(t+2)} = 1 \text{ then C has to be equal 1} [/imath] and this is where I am confused [imath] \lim_{t\rightarrow\infty} \ln(t+2) = \infty [/imath] and when we multiply this by [imath]0[/imath], we get undefined, so why should the [imath]C[/imath] be [imath]1[/imath]. It is stated that if [imath]C[/imath] was anything but [imath]1[/imath] the answer would be [imath]\infty[/imath] |
2691643 | Analytic expression for [imath]\sum_{k=0}^\infty \frac{\log(k+1)}{k!}[/imath]
I want to find an simple analytic expression for [imath]\sum_{k=0}^\infty \frac{\log(k+1)}{k!}.[/imath] Wolfram alpha just tells me this is approximately 1.55 (for natural log). Equivalently, this is equal to [imath]e \cdot \underset{K \leftarrow \mathsf{Poisson}(1)}{\mathbb{E}} \left[K \log K\right][/imath] (taking the convention that [imath]0 \log 0 = 0[/imath] for continuity). Surely this sort of expectation has been evaluated? This has come up in trying to analyze the entropy of a random distribution. Any help would be much appreciated. Another answer suggests using the fact that [imath]\mathbb{E}[K \log K] = \frac{\mathrm{d}}{\mathrm{d}x} \mathbb{E}[K^x][/imath] evaluated at [imath]x=1[/imath]. However, I don't know if this brings us closer to a simple expression -- I don't have a good formula for [imath]\mathbb{E}[K^x][/imath] unless [imath]x[/imath] is an integer. | 237519 | Information on the sum [imath]\sum_{n=1}^\infty \frac{\log n}{n!}[/imath]
In my personal study of interesting sums, I came up with the following sum that I could not evaluate: [imath]\sum_{n=1}^\infty \frac{\log n}{n!} = 0.60378\dots[/imath] I would be very interested to see what can be done to this sum. Does a closed form of this fascinating sum exist? |
2693678 | \sum_{k=0, k even}^n {n \choose k}*2^k = \frac{(3^n)+((-1)^n)}{2}
Prove the following combinatoric identity: [imath]\sum_{k=0, k even}^n {n \choose k}*2^k = \frac{(3^n)+((-1)^n)}{2}[/imath] | 688969 | Prove the identity [imath]\sum_{{{\underset{k-even}{k=0}}}}^{n}{n \choose k}2^{k}=\frac{3^{n}+(-1)^{n}}{2}[/imath]
I need to prove the following identity: [imath]\sum_{{{\underset{k-even}{k=0}}}}^{n}{n \choose k}2^{k}=\frac{3^{n}+(-1)^{n}}{2}[/imath] I know that - [imath]\sum_{k=0}^{n}{n \choose k}2^{k}=3^{n}[/imath] but don't know how to continue. |
1943503 | What are the formal rules of manipulating calculus symbols?
I'm not sure if this is the right place for this question, but here goes. I am often inclined to treat, without knowing if this is correct, the operators of calculus, namely [imath]d, \frac{d}{dx},\int,\int_a^b[/imath], as if they were "constants" that you can apply the rules of multiplication to. So for example, I would be inclined to "multiply both sides by [imath]dx\cdot dy[/imath]" as follows: [imath]\frac{y^2}{dx}=\frac{x}{dy}\implies x^2dx=ydy [/imath] or to "apply [imath]\int[/imath]" to both sides as follows: [imath]x^2dx=ydy \implies \int x^2dx=\int ydy[/imath] even though I am not sure whether [imath]\int[/imath] is an operator by itself, or whether [imath]\int ...dx[/imath] together form a single operator. So my question is, not about whether the above particular derivations are correct, but what is the topic that one should study to learn the rules of manipulation of these operators? I'd like to really understand them, not just memorize them. (note that I've taken an introduction to real analysis, but this focused very much on general concepts such as continuity, differentiability, integrability, without touching on the implications for the rules of manipulation of these operators) | 1167100 | Valid justification for algebraic manipulation of [imath]\mathrm{d}y/\mathrm{d}x[/imath]?
I've read many question/answer threads here on SE re: justification for the algebraic manipulation of [imath]\mathrm{d}y/\mathrm{d}x[/imath] in the standard formulation of calculus. I worked up my own shot at a justification using the definition of the derivative and would like to get the input of others whether it holds. It doesn't involve pushforward maps, non-standard analysis (two things I don't yet know), or any other tool than simple limit algebra/laws, the definition of the derivative, and (implicitly), the chain rule that makes the substitution possible. If it does hold as a justification as far as it goes, i.e., in one-dimension, it would seem to me, at least, to be a simple and clear justification in the form of an example that could be useful to other beginners. If goes wrong somewhere please let me know for my own understanding. Integrate [imath]\int \frac{x}{1+x^2}\mathrm{d}x[/imath]. We'll do this by substitution. We want to restate the original problem in terms of [imath]u[/imath]. That means that we want to state [imath]x[/imath], [imath]1+x^2[/imath], and [imath]\mathrm{d}x[/imath] in terms of [imath]u[/imath]. Let [imath]u=1+x^2 \rightarrow \frac{\mathrm{d}u}{\mathrm{d}x}=2x \rightarrow \mathrm{d}x=\frac{\mathrm{d}u}{2x}[/imath]. Let's justify that last step as that's where the justification (that I speak of) is required: By definition, [imath]\frac{\mathrm{d}u}{\mathrm{d}x}=2x[/imath] is defined as [imath] \lim_{\Delta x \to 0}\left( \frac{u(x+\Delta x)-u(x)}{\Delta x} \right)=2x [/imath] Recall that the limit of a quotient is equal to the quotient of the limits as long as the denominator is not equal to zero, thus the foregoing becomes [imath] \frac{\lim_{\Delta x \to 0}\left( u(x+\Delta x)-u(x) \right)}{\lim_{\Delta x \to 0} \Delta x } =2x [/imath] Multiplying both sides by [imath]\lim_{\Delta x \to 0}\Delta x[/imath] and then dividing both sides by [imath]2x[/imath] yields [imath] \frac{\lim_{\Delta x \to 0}\left( u(x+\Delta x)-u(x) \right)}{2x} =\lim_{\Delta x \to 0} \Delta x [/imath] which, in the limit, is equal to [imath] \frac{\mathrm{d}u}{2x}=\mathrm{d}x [/imath] Rearranging we have [imath] dx=\frac{\mathrm{d}u}{2x} [/imath] as desired. |
2693636 | [imath]\sum_{i=1}^n P(n,i)[/imath]
We can know clearly from [imath](1+X)^n=\sum\limits_{i=0}^{n}C(n,i)X^n[/imath] that [imath] \sum\limits_{i=0}^{n}C(n,i)=2^n.[/imath] Whereas, I want to know if there are any researched results about permutations in the similar case, i.e., what can we know about [imath] \sum\limits_{i=0}^{n}P(n,i).[/imath] I’m really curious about that, but have found no answers elsewhere. Any help will be sincerely appreciated! | 2019675 | Sum of Permutations from 0 to n
In a question I asked on algorithms with time complexity of [imath]f(x) = n^n[/imath] I was told that enumerating the number of strings that can be formed from a string of length [imath]n[/imath] qualifies. I.e the sum of all permutations of [imath]n[/imath] from [imath]n[/imath] to [imath]0[/imath] is [imath]n^n[/imath] [imath]\sum ^n_{i=0} nPi[/imath] [imath]= n^n[/imath] Can I please see an easy to understand derivation of that formula. EDIT The above identity is wrong. I just tested it. Can I get a derivation of the formula for the sum of permutations. |
2693749 | studying the series [imath]\sum_\limits{n=1}^\infty \frac{1}{n^ {\alpha}(\log n)^ {\beta}}[/imath].
I have to study the character of this series [imath]\sum_{n=1}^\infty \frac{1}{n^{\alpha}(\log n)^{\beta}}.[/imath] [imath]\alpha[/imath] and [imath]\beta[/imath] are two parameters. I'm considering the case [imath]\alpha >1 [/imath] and all real values for [imath]\beta [/imath] . If [imath]\beta \ge 0[/imath], [imath]\exists n_0 \in N [/imath] such that [imath]\forall n\ge n_0, \dfrac{1}{n^ {\alpha}(\log n)^ {\beta}}<\dfrac{1}{n^ {\alpha}}[/imath]. For the comparison test [imath]\dfrac{1}{n^{\alpha} (\log n)^ {\beta}}[/imath] converges. I tried to analyze the case [imath]-1<\beta<0,\beta=-1, \beta<-1[/imath] but I'm not sure. Anyway, for [imath]\beta<-1[/imath] the series is convergent for [imath]\alpha > 1- \beta[/imath], divergent for [imath]\alpha < 1- \beta[/imath]. For [imath]-1<\beta<0[/imath] the series is convergent for [imath]\alpha > 2[/imath], divergent for [imath]\alpha <2[/imath]. For [imath]\beta=-1[/imath] the series is convergent for [imath]\alpha > 2[/imath], divergent for [imath]1<\alpha < 2[/imath]. I've used the fact that [imath]\log n<n[/imath] | 2563446 | On convergence of Bertrand series [imath]\sum\limits_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}[/imath] where [imath]\alpha, \beta \in \mathbb{R}[/imath]
Study the convergence of [imath]\sum_{n=2}^{\infty} \frac{1}{n^{\alpha}\ln^{\beta}(n)}[/imath] where [imath]\alpha, \beta \in \mathbb{R}[/imath] I have proved that: This series diverges when [imath]\alpha \leq 0[/imath]. This series converges when [imath]\alpha > 1, \beta > 0[/imath] This series diverges when [imath]0 < \alpha < 1, \beta > 0[/imath] This series converges when [imath]\alpha = 1, \beta > 1[/imath] Question: What happens when [imath]\alpha > 0[/imath] and [imath] \beta < 0[/imath]? There are other questions on MSE which ask about this series, but this question is distinct because I would like an argument which does not rely on the integral test for series convergence, and this question considers all real [imath]$\alpha$[/imath] and [imath]$\beta$[/imath], while other questions ask only about [imath]\alpha, \beta > 0[/imath], where we can apply Cauchy condensation criterion |
2693231 | How to prove that [imath]\{\land,\lnot\}[/imath] is complete?
I am curious one one would prove that a set of operators is "complete," which to my understanding means it can be used to represent all possible truth tables. So normally if we had [imath]\{\land,\lor, \lnot\}[/imath] and a truth table what I do is look at where the outputs are true and then chain the expressions together with ors. For example in the truth table for XOR the output is true when either [imath]p[/imath] or [imath]q[/imath] is true but not both, so [imath](p \land \lnot q) \lor (\lnot p \land q)[/imath]. Using this approach I think you can do it for any truth table so [imath]\{\land,\lor, \lnot\}[/imath] is complete. Now let's say you want to show that some other set of operators are complete. Does it suffice to show that you can synthesize the [imath]\{\land,\lor, \lnot\}[/imath] operators? For example let's say you asked me to prove: Is [imath]\{\land, \lnot\}[/imath] complete? We can synthesize the missing "or" operation in terms of these two with with [imath]p \lor q = \lnot(\lnot p \land \lnot q)[/imath] Is [imath]\def\nand{\barwedge} \{\nand\}[/imath] (NAND) complete? Assuming [imath]p \nand q[/imath] is the same as [imath]\lnot(p \land q)[/imath] I can say [imath]\lnot p = \lnot p \lor \lnot p = \lnot(p \land p) = p \nand p[/imath], and then [imath]p \land q = \lnot(\lnot(p \land q)) = \lnot(p \nand q) = (p \nand q) \nand (p \nand q)[/imath]. And if we absolutely needed [imath]\lor[/imath] we could just copy our example from earlier in terms of NAND: [imath]p \lor q = \lnot(\lnot p \land \lnot q) = \lnot p \nand \lnot q = (p \nand p) \nand (q \nand q)[/imath]. Am I way off base or is this a valid way to prove completeness? First by showing [imath]\{\land,\lor, \lnot\}[/imath] is complete via the process in my second paragraph, and then in the future, showing that any arbitrary set of operators can be used to synthesize [imath]\{\land,\lor, \lnot\}[/imath] (or if we wish to be minimal, [imath]\{\land, \lnot\}[/imath])? | 2692786 | How do I prove if a set of operations is complete?
For example [imath]\{\land, \lnot\}[/imath] apparently forms a functionally complete set as it can form any logical expression. But I don't really know what this means or how you show it. Do we just show that some set is complete and then in the future if we have a new set, show that we can replicate the known operations with the new ones? Do you have to find these strategically or is there a systematic way? |
2694055 | Solving a square sequence limit
I've started learning sequences and I'm having a hard time calculating the following: [imath]\lim_{n\to ∞}{\sqrt[n]{3^n + 7^n}}[/imath] Using Heine’s Lemma I'm trying to solve it analogous to the corresponding limit definitions for functions, but I get stuck every direction I go. Any help is appreciated. | 2692558 | Calculate the limit: [imath]\lim_{x\to\infty} \sqrt[x]{3^x+7^x}[/imath]
Calculate the limit: [imath]\lim_{x\to\infty} \sqrt[x]{3^x+7^x}[/imath] I'm pretty much clueless on how to approach this. I've tried using the identity of [imath]c^x = e^{x \cdot \ln(c)}[/imath] but that led me to nothing. Also I've tried replacing [imath]x[/imath] with [imath]t=\frac{1}{x}[/imath] such that I would end up with [imath]\lim_{t\to 0} (3^{1/t} + 7^{1/t})^{1/t}[/imath] however I've reached yet again a dead end. Any suggestions or even hints on what should I do next? |
2693689 | Pigeon Hole Principle Cases
Given seven real numbers. Show that we can always choose two of them, say [imath]a, b[/imath] such that [imath]0 < \dfrac{a-b} {ab+1}< \sqrt{3}[/imath] I think this is one of pigeon hole principle problem. I have tried to think further about the cases for some days, but didn't found any logical sense answer. Instead of plugging random real numbers, do you have any argument to prove it?? Please help | 1573849 | Show that given seven real numbers, it is always possible take two of them, such that [imath]\left\vert\frac{a-b}{1+ab}\right\vert<\frac{1}{\sqrt{3}}[/imath]
Show that given seven real numbers, it is always possible take two of them, such that [imath]\left\vert\frac{a-b}{1+ab}\right\vert<\frac{1}{\sqrt{3}}[/imath] The "Pigeonhole principle" states that if [imath]n[/imath] items are put into [imath]m[/imath] containers, with [imath]n > m[/imath], then at least one container must contain more than one item, but how I an define my "container" such that meets the condition of the problem? |
1236684 | Explicit construction of Haar measure on quotient group
Let [imath]G[/imath] be a locally compact Hausdorff group, [imath]H[/imath] a closed normal subgroup. To simplify matters we assume that the underlying topological space of [imath]G[/imath] has a countable base. Suppose a left Harr measure [imath]\mu[/imath] is explicitly given on [imath]G[/imath]. Can we explicitly construct a left Haar measure on [imath]G/H[/imath] using [imath]\mu[/imath]? The motivation is as follows. When [imath]H[/imath] is compact or discrete, the answer seems to be yes. Suppose [imath]H[/imath] is compact. Let [imath]K[/imath] be a compact subset of [imath]G/H[/imath]. Then [imath]\psi^{-1}(K)[/imath] is compact where [imath]\psi[/imath] is the canonical homomorphism [imath]G \rightarrow G/H[/imath]. So we can define [imath]\nu(K) = \mu(\psi^{-1}(K))[/imath]. Then I think [imath]\nu[/imath] defines a left Haar measure on [imath]G/H[/imath]. When [imath]H[/imath] is discrete, [imath]\psi[/imath] is a local homeomorphism. So we can define locally a Borel measure on [imath]G/H[/imath]. Then I think we can get it globally. I have no idea about the answer when [imath]H[/imath] is not compact or discrete. | 1896663 | Disintegration of Haar measures
Suppose I have a locally compact group [imath]G[/imath] and a quotient map [imath]f:G\to G/N[/imath]. Is it true that for every Borel-measurable function [imath]f : G → [0, +∞][/imath], [imath]\int_{G} f(g) \, \mathrm{d} \mu_G (g) = \int_{G/N} \int_{N} f(gn) \, \mathrm{d} \mu_{N} (n) \mathrm{d} \mu_{G/N} (gN),[/imath] where [imath]\mathrm{d} \mu_{G}[/imath], [imath]\mathrm{d} \mu_{N}[/imath] and [imath]\mathrm{d} \mu_{G/N}[/imath] are the Haar measures on [imath]G[/imath], [imath]N[/imath] and [imath]G/N[/imath] respectively? In particular, is it true that if a subset [imath]A\subseteq G[/imath] is such that its intersection with [imath]\mathrm{d}\mu_{G/N}[/imath]-almost every coset [imath]gN[/imath] is of full measure in [imath]gN[/imath] (w.r.t. to the measure on [imath]gN[/imath] obtained by shifting the measure of [imath]N[/imath], which is independent of the choice of [imath]g[/imath]), then [imath]A[/imath] is of full measure in [imath]G[/imath]? (This is implied from the equation above by taking [imath]f[/imath] to be the characteristic function of [imath]A[/imath]). I wanted to use the disintegration theorem in Wikipedia, but I'm not sure if it applies here. I'm not sure I understand the definition of a Radon space and I don't know which locally compact groups satisfy it. I know of a more specific disintegration result, which appears in many/most introductions to Haar measures (e.g. Raghunthan's book), but it is only stated for continuous [imath]f[/imath] with compact support. I suppose it's not hard to get rid of the continuous part by using some Luzin argument (although I am not sure how to do it myself), but the compact support bothers me. In any case, if this is not true for general locally compact groups, for which groups it is true? I have a reference for second-countable compact groups (Halmos's book Measure Theory; his definitions are a bit out-dated, but coincide with the modern ones for second-countable compact groups). I don't mind assuming separability. Compactness is not awful either, but I prefer not to assume metrisability. Thanks. |
2694102 | the sequence [imath]n!+2,...,n!+n[/imath] is made up of only composite numbers
I have found the claim that given that [imath]n\geq2[/imath], we have that the sequence of [imath]n-1[/imath] numbers [imath]n!+2,n!+3,...,n![/imath] is made up of only composite numbers. Is there a proof of this? I found this pretty fascinating but I am not sure how to go around it. It seems to hold for the first few examples [imath]n=2[/imath]: [imath]S=\{4\}[/imath] [imath]n=3[/imath]: [imath]S=\{8,9\}[/imath] [imath]n=4[/imath]: [imath]S=\{26,27,28\}[/imath] | 852598 | If [imath]n = 51! +1[/imath], then find number of primes among [imath]n+1,n+2,\ldots, n+50[/imath]
If [imath]n = 51! +1[/imath], Then find no of primes among [imath]n+1,n+2,\ldots, n+50[/imath] Really speaking, I don't have any clue ... |
1597077 | Haar measure on [imath]G \rtimes_\phi H[/imath]
Let [imath]G, H[/imath] be locally compact, [imath]\sigma[/imath]-compact metric groups equipped with left Haar measures [imath]m_G, m_H[/imath] respectively. Let [imath]\Phi: G \times H \to G[/imath] be continuous such that [imath]\phi: H \to Aut(G), \, h \mapsto \Phi( \cdot, h)[/imath] is a group homomorphism. Denote [imath]\Phi_h = \phi(h)[/imath]. Then, the semidirect product [imath]G \rtimes_\phi H[/imath] is defined as the topological space [imath]G \times H[/imath] equipped with [imath](g_1, h_1) \cdot (g_2, h_2) = (g_1 \Phi_{h_1}(g_2), h_1 h_2)[/imath] for [imath]g_1, g_2 \in G, \, h_1, h_2 \in H[/imath]. This is a topological group. I would like to show that [imath]dm_{G \rtimes_\phi H}(g,h) = mod(\Phi_h) \,dm_H(h)\, dm_G(g) [/imath] defines a left Haar measure on [imath]G \rtimes_\phi H[/imath], where [imath]mod[/imath] is of course the modular function [imath]mod: Aut(G) \to \mathbb{R}_{>0}[/imath] defined by [imath]m_G(\varphi^{-1} B) = mod(\varphi) m_G(B)[/imath] for Borel sets [imath]B \subseteq G[/imath]. How can I do this without any explicit form for [imath]\Phi_h[/imath]? The definition for the semidirect product on wikipedia uses conjugation instead of a general automorphism. Is it obvious from my definition that [imath]\Phi_h[/imath] must be some conjugation? | 2240696 | Haar Measure on Semidirect Product of Unimodular Groups
Let [imath]M, N[/imath] be closed subgroups of a given locally compact topological group such that the product [imath]P = MN[/imath] is semidirect, with [imath]M[/imath] normalizing [imath]N[/imath]. Assume second countable, sigma compact, whatever so that Radon products coincide with product measures. To make life easy, assume [imath]M, N[/imath] are unimodular. How do we construct a Haar measure on [imath]P[/imath]? |
2696036 | Prove that the determinant of a swap matrix is [imath]-1[/imath]
How can I prove that a swap matrix (a matrix which multiplied with another, swaps a pair of its rows/columns) has determinant [imath]-1[/imath]? | 473240 | Determinant of elementary permutation matrix
Why is the determinant of an elementary permutation matrix equal to [imath]-1[/imath]? I am brand new to determinants and I've tried expanding it and using cofactor expansion, but it's messy and complicated. I would prefer if someone could show me using expansion, but alternative methods are welcome. Thanks. |
2696018 | Mittag-Leffler Theorem Exercise - Find Sum of Series
The exercise asks to find the closed form of [imath]\sum_{n=-\infty}^{\infty}\frac{1}{(z+n)^2+a^2}[/imath]where [imath]a[/imath] is a complex number. I know that [imath]\frac{\pi^2}{\sin^2(\pi z)}=\sum_{n=-\infty}^{\infty}\frac{1}{(z+n)^2}[/imath], but don't how to develop this fact. Is this some kind of a shift of this series? Thank you! | 608059 | Closed form for [imath]\sum_{n=-\infty}^\infty \frac{1}{(z+n)^2+a^2}[/imath]
I want to express [imath]\sum_{n=-\infty}^\infty \dfrac{1}{(z+n)^2+a^2}[/imath] in closed form. What comes to mind is the formula [imath]\pi\cot\pi z = \dfrac{1}{z}+\sum_{n\ne 0}\left(\dfrac{1}{z-n}+\dfrac1n\right)=\dfrac{1}{z}+\sum_{n=1}^\infty\dfrac{2z}{z^2-n^2}[/imath] and also [imath]\dfrac{\pi^2}{\sin^2\pi z}=\sum_{n=-\infty}^\infty \dfrac{1}{(z-n)^2}.[/imath] But neither of these gives the term [imath](z+n)^2+a^2[/imath] that we want. Perhaps we can adjust somehow? |
2693963 | Converse of L'Hospital
L'Hospital's Rule: If [imath]{f(x)}\over{g(x)}[/imath] is either [imath]0\over0[/imath] or [imath]\infty\over\infty[/imath], then the $\lim_{x\to \infty}[imath]{f(x)}\over{g(x)}$ $=$ $\lim_{x\to \infty}[/imath]{f'(x)}\over {g'(x)}$. What I want to know is if the converse is true. In other words, if $\lim_{x\to \infty}[imath]{f(x)}\over{g(x)}$ $=$ $\lim_{x\to \infty}[/imath]{f'(x)}\over {g'(x)}$, then is [imath]{f(x)}\over{g(x)}[/imath]equal to one of [imath]0\over0[/imath] or [imath]\infty\over\infty[/imath]? By this I mean that there does not exist a case in which L'Hopital does not apply yet gives the correct answer. It is not a repeat of another question, as many have suggested. I would ask those people to take a look at the question that they have said I have repeated. These are two different questions, and it surprises me that they could have been confused. | 1710786 | Why does L'Hopital's rule fail in calculating $\lim_{x \to \infty} \frac{x}{x+\sin(x)}$?
[imath]\lim_{x \to \infty} \frac{x}{x+\sin(x)}[/imath] This is of the indeterminate form of type [imath]\frac{\infty}{\infty}[/imath], so we can apply l'Hopital's rule: [imath]\lim_{x\to\infty}\frac{x}{x+\sin(x)}=\lim_{x\to\infty}\frac{(x)'}{(x+\sin(x))'}=\lim_{x\to\infty}\frac{1}{1+\cos(x)}[/imath] This limit doesn't exist, but the initial limit clearly approaches [imath]1[/imath]. Where am I wrong? |
2696932 | [imath]R[/imath] is integral domain
Let[imath] S[/imath] be a subring of[imath] R[/imath]. If [imath]R[/imath] is an integral domain, then [imath]1_R[/imath] = [imath]1_S[/imath]? I know that if [imath]S[/imath] is subring of [imath]R[/imath] then [imath]S[/imath] is integral domain. How to prove this problem ? Thanks in advance. | 327999 | Prove that a nontrivial subring of a ring that is a domain has the same identity
Let [imath]R[/imath] be a ring with an identity element [imath]1_R[/imath] which is a domain. Let [imath]S[/imath] be a nontrivial subring of [imath]R[/imath] with identity element [imath]1_S[/imath]. Prove that [imath]1_R = 1_S[/imath]. |
783080 | [imath]a_{n+1}=a_{n-1}-{4\over a_n}[/imath] for [imath]n>1[/imath]
Let [imath]a_1=10, a_2=20[/imath] and [imath]a_{n+1}=a_{n-1}-{4\over a_n}[/imath] for [imath]n>1[/imath] Find the smallest value of [imath]k[/imath] for which [imath]a_k=0[/imath] | 781047 | smallest value of [imath]k[/imath] for which [imath]a_k[/imath] is [imath]0[/imath]
Let [imath]a_1=10[/imath], [imath]a_2=20[/imath] and define [imath]a_{n+1}=a_{n-1}-\frac{4}{a_n}[/imath] for [imath]n \gt 1[/imath]. What is the smallest value of [imath]k[/imath] for which [imath]a_k=0?[/imath] |
2694110 | Computing expectation under a change of measure
Let [imath]X[/imath] be a random variable on a probability space [imath](\Omega,\mathscr F, P)[/imath]. Define a new probability measure [imath]\tilde P(A) = E[1_A X][/imath] for all [imath]A\in\mathscr F[/imath]. Let [imath]\tilde E[/imath] be expectation taken with respect to the new measure [imath]\tilde{P}[/imath]. Suppose now that [imath]Y[/imath] is also a random variable [imath](\Omega,\mathscr F)[/imath]. Then intuitively the expectation should be computed as [imath] \tilde E [1_A Y] = E[1_A YX], [/imath] but I'm not sure how to prove this rigorously using the definition. | 2690208 | Prove [imath]E^{\mathbb Q}[Y]=E^{\mathbb P}[XY][/imath] if [imath]E^{\mathbb P}[X]=\mathbb P(X>0)=1[/imath] and [imath] \mathbb Q(A)=E^{\mathbb Q}[X1_A] [/imath]
How do I prove the following? I don't know where to start. If [imath]X[/imath] is a random variable with [imath]E^{\mathbb P}[X] = \mathbb P(X>0)=1[/imath] and [imath] \mathbb Q[/imath] is the probability measure defined by [imath] \mathbb Q(A)=E^{\mathbb Q}[X1_A] [/imath] then [imath]E^{\mathbb Q}[Y]=E^{\mathbb P}[XY][/imath] |
2697267 | A series that converges to π/3
While surfing on YouTube, I stumbled into this video which gave me a new insight about the well-known series [imath] \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7}+ \ldots [/imath] The idea shown there consists of counting (in a very clever way) how many points of the 2d integer lattice lie on a generic circumference of radius [imath]\sqrt{r}, r \in \mathbb{N}[/imath], centered at the origin. Then, we name [imath]N(R)[/imath] the number of points of this kind that lie inside the circumference of radius [imath]\sqrt{R}[/imath]: as [imath]R[/imath] grows, we can think of [imath]N(R)[/imath] as a fairly good approximation of the area [imath]\pi R[/imath] of the circle, since each of the [imath]N(R)[/imath] points can be thought as the center of a square of area [imath]1[/imath]. From the equality [imath]\pi R = N(R)[/imath] we get the series above. I was fairly amazed by the way this result was obtained, and I started wondering: what if we used the hexagonal lattice (I mean [imath]\mathbb{Z} \times \zeta_3\mathbb{Z}[/imath], where [imath]\zeta_3[/imath] is a non-trivial third root of unity) instead of the integer lattice? Will I get another series to approximate [imath]\pi[/imath]? After some work, following the same kind of argument, I came to this formula: [imath] \frac{\pi}{3} = \sqrt{3} \left (1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{5} + \frac{1}{7} - \frac{1}{8} + \ldots \right ) [/imath] or equivalently [imath] \frac{\pi}{3} = \sum_{k=0}^\infty \frac{\sqrt{3}}{9k^2+9k+2} [/imath] I have some questions: Can anyone give me a proof of this result which does not follow the argument I sketched above? I checked by doing some simple calculations (I also asked Wolfram) and the result seems to hold, but I'd like to be 100% sure... The [imath]\frac{\pi}{4}[/imath] formula relates to the expansion of [imath]\arctan(x)[/imath], but I couldn't find any straightforward connection between the [imath]\frac{\pi}{3}[/imath] formula and [imath]\arctan(x)[/imath]. I actually couldn't find any connection to the "pi facts" I know or I was able to find. Do anybody know anything that can explain "easily" what's happening here? Thanks in advance! | 1632311 | Find the value of [imath]\sum_{n=0}^\infty\frac{1}{9n^2+9n+2}[/imath]
I was doing some problems in algebraic number theory and this series came up [imath]\sum_{n=0}^\infty\frac{1}{9n^2+9n+2}.[/imath] So, I would like to know the value of this series. However, I don't want a full answer, only hints. I have tried to use Fourier's series, but with no success, perhaps an answer can be gotten usign this. Thanks |
2697659 | How can I prove [imath]AB = BA[/imath]
Let [imath]A[/imath] and [imath]B[/imath] be [imath]n \times n[/imath] matrix, each with [imath]n[/imath] distinct Eigenvalues. Prove that [imath]A[/imath] and [imath]B[/imath] have same [imath]n[/imath] eigenvalues iff [imath]AB = BA[/imath]. | 1763307 | prove that [imath]TS = ST[/imath]
Let [imath]V[/imath] be a finite-dimensional vector space over [imath]F[/imath] with [imath]\dim(V) = n[/imath]. Let [imath] T:V \rightarrow V[/imath] be a linear transformation. Assuming that [imath]T[/imath] has [imath]n[/imath] different eigenvalues. prove that :[imath] TS = ST \iff \textrm{each eigenvector of} \ T \ \textrm{is an eigenvector for }S\textrm{ (maybe with a different eigenvalue)}[/imath] I managed to prove the right to left direction. Thanks for helping. |
2365159 | Christoffel Coefficients equality [imath]\rightarrow[/imath] Same Metric?
Let [imath]M_1,M_2[/imath] be two surfaces with the corresponding metrics [imath](g_{ij}),(h_{ij})[/imath] if they both have the same Christoffel coefficients does [imath](g_{ij})=(h_{ij})[/imath]? Intuitively the Christoffel coefficients are expression of the metric of the surface so we may find two different metrics with the same Christoffel coefficients? | 339518 | Does the Levi-Civita connection determine the metric?
Can I reconstruct a Riemannian metric out of its Levi-Civita connection? In other words: Given two Riemannian metrics [imath]g[/imath] and [imath]h[/imath] on a manifold [imath]M[/imath] with the same Levi-Civita connection, can I conclude that [imath]g=h[/imath] up to scalars? If not, what can I say about the relationship between [imath]g[/imath] and [imath]h[/imath]? How rigid is the Levi-Civita-Connection? |
2692790 | In how manys can a programming team and a hacking team be selected if Jack does not serve on both teams?
A club has [imath]12[/imath] members. They need to choose a programming team of [imath]3[/imath] members and a hacking team of [imath]4[/imath] members. Members can be on both teams, but Jack only wants to be on at most one team. How many ways can both teams be chosen? I did two different ways. I solved if Jack chooses programming, [imath]C(12,3)=220[/imath] [imath]C(11,4)=330[/imath]. In total there are [imath]550[/imath] different ways both teams can be picked. But, if Jack chooses hacking, I solved [imath]C(11,3)=165[/imath] and [imath]C(12,4)=495[/imath]. In total there are [imath]650[/imath] ways both teams can be chosen. I was wondering if this is the correct method. | 2697502 | Can permutations be solved by cases?
"A Math Club has 12 members. They have to choose euclid contest team of 3 members and a National Math Competition team with 4 members. Students can be on both teams, but James agrees to only be on at most one team. In how many ways can both teams be chosen?" Here's what I thought should be the answer but doesn't make much sense: Case 1: James is in Euclid team: [imath]\therefore [/imath] [imath]11\choose2[/imath] for Euclid Team This would be because James is in team Euclid assumed so we only have to get 2 other players and we have 11 students left. Also : [imath]11\choose4[/imath] for Math team so total ways would be [imath]11\choose2[/imath] . [imath]11\choose4[/imath] (Product Rule) Case 2: James is in Math team: [imath]\therefore [/imath] [imath]11\choose3[/imath] for Math team This would be because james is in team Math assumed so we only have to get 3 other players and we have 11 students left. Also : [imath]11\choose3[/imath] for Euclid team , as the 11 students can form a team of 3 for Euclid competition here. so total ways would be [imath]11\choose3[/imath] . [imath]11\choose3[/imath] (Product Rule) Case 3: James in no team [imath]\therefore [/imath] [imath]11\choose3[/imath] . [imath]11\choose4[/imath] (Product Rule) as James is in no team so we are left with 11 people for both teams Now the question is did I do it correctly? Because I have never seen cases being used in a permutation questions? Therfore, The total number of ways would be to add all the cases meaning Case 1 + Case 2 + Case 3 : ([imath]11\choose2[/imath] . [imath]11\choose4[/imath]) + ([imath]11\choose3[/imath] . [imath]11\choose3[/imath]) +([imath]11\choose3[/imath] . [imath]11\choose4[/imath]) Is that correct? |
954073 | Derivative of a function with respect to another function.
I want to calculate the derivative of a function with respect to, not a variable, but respect to another function. For example: [imath]g(x)=2f(x)+x+\log[f(x)][/imath] I want to compute [imath]\frac{\mathrm dg(x)}{\mathrm df(x)}[/imath] Can I treat [imath]f(x)[/imath] as a variable and derive "blindly"? If so, I would get [imath]\frac{\mathrm dg(x)}{\mathrm df(x)}=2+\frac{1}{f(x)}[/imath] and treat the simple [imath]x[/imath] as a parameter which derivative is zero. Or I should consider other derivation rules? | 291376 | differentiate with respect to a function
Let's say I have this function [imath]f(x)=x[/imath]. I want to differentiate with respect to [imath]x^2[/imath]. So I want to calculate [imath]\large\frac{df(x)}{dx^2}[/imath]. In general, how can I calculate the derivative of a function [imath]f(x)[/imath] with respect to a function [imath]g(x)[/imath], so [imath]\large\frac{df(x)}{dg(x)}[/imath]? (I dont know whether this is a good notation)? |
2698322 | Proof of equality of independent variables
With [imath]X_1,X_2,..[/imath] as a sequence of i.i.d variables with F as a distribution function and [imath]M_n=max\{X_m:m<=n\}[/imath] for [imath]n=1,2,..[/imath] To prove [imath]P(M_n<=x)=F^n(x)[/imath], I did the following: a) [imath]P(M_n<=x)=P(max(X_1,X_2,...X_n)<=x)=\prod_{i=1}^{n}P(X_i<=x)=P(X_1<=x)\cdot P(X_2<=x)\cdot\cdot\cdot P(X_n<=x)=F\cdot F \cdot F.. = [F(x)]^n= F^n(x)[/imath] b) Given [imath] F(x)=1-x^{-α}[/imath] for [imath]x>=1, α > 0[/imath]. Need to prove that as n approaches infinity [imath]P(\frac{M_n}{n^{1/α}} <= y) -> exp(-y^{-α})[/imath] I proved [imath]P(\frac{M_n}{n^{1/α}} <= y)=P(M_n<=y \cdot n^{1/α})=(F(y \cdot n^{1/α}))^n=(1-\frac{1}{n \cdot y^α})^n -> e^{-y^{-α}}[/imath] as n approaches [imath]\infty[/imath] Are proofs in a) and b) are done correctly? Please let me know if something needs to be added to make it solid. Thank you very much! | 2029040 | Find the cumulative function and the density of [imath]m_n[/imath] and [imath]M_n[/imath], where [imath]m_n=\min(X_1,X_2,...,X_n)[/imath] and [imath]M_n=\max(X_1,X_2,...,X_n)[/imath].
Let be [imath]X_1,X_2,...,X_n[/imath] be uniformly distributed random variables i.i.d. a) Find the cumulative function and the density of [imath]m_n \text{ and } M_n[/imath] , where [imath]m_n=min(X_1,X_2,...,X_n)[/imath] and [imath]M_n=max(X_1,X_2,...,X_n)[/imath]. b) Let [imath]Z_n=n(1-M_n)[/imath]. Show that [imath]Z_n \xrightarrow{d} Z [/imath], where [imath]Z[/imath] is a random variable with cumulative function [imath]F_Z(z)=1-e^{-z}[/imath] |
2682889 | Proving that the counting measure is a measure on [imath]\mathcal A[/imath]
Here is the problem: Let [imath]\Omega[/imath] be a nonempty set and [imath]\mathcal A[/imath] [imath]=[/imath] [imath]\mathcal P[/imath]([imath]\Omega[/imath]). Define [imath]\mu[/imath] on [imath]\mathcal A[/imath] by [imath]\mu[/imath]([imath]E[/imath]) equal to [imath]N[/imath]([imath]E[/imath]) if [imath]E[/imath] is finite and [imath]\infty[/imath] if [imath]E[/imath] is infinite, where [imath]N[/imath]([imath]E[/imath]) denotes the number of elements of E. Prove that [imath]\mu[/imath] is measurable on [imath]\mathcal A[/imath]. I have already proved the first two conditions for [imath]\mu[/imath] to be measurable on [imath]\mathcal A[/imath]. However, I am stuck on proving the third condition [i.e [imath]\mu([/imath][imath]\bigcup_{n=1}^{\infty}E_n[/imath]) [imath]=[/imath] [imath]\sum_{n=1}^{\infty}\mu[/imath]([imath]E_n[/imath])] This is what I'm thinking: Let [imath]\{E_n\}_{n=1}^{\infty}[/imath] be a sequence of pairwise disjoint sets, and consider the following three cases: 1.) Suppose for some [imath]k[/imath] [imath]\in[/imath] [imath]\Bbb N[/imath], we have [imath]N[/imath]([imath]E_k[/imath]) = [imath]\infty[/imath]. 2.) Suppose that for all [imath]n[/imath] [imath]\in[/imath] [imath]\Bbb N[/imath], we have [imath]N[/imath]([imath]E_n[/imath]) [imath]<[/imath] [imath]\infty[/imath], and that there are infinitely many [imath]n[/imath] such that [imath]N[/imath]([imath]E_n[/imath]) [imath]>[/imath] [imath]0[/imath]. 3.) Suppose that for all [imath]n[/imath] [imath]\in[/imath] [imath]\Bbb N[/imath], we have [imath]N[/imath]([imath]E_n[/imath]) [imath]<[/imath] [imath]\infty[/imath] and that there are finitely many [imath]n[/imath] such that [imath]N[/imath]([imath]E_n[/imath]) [imath]>[/imath] [imath]0[/imath]. | 848167 | Counting measure proof
Can somebody tell me why the counting measure (so, if [imath]S=P(X)[/imath], then [imath]\mu(A)[/imath]=infinity if [imath]A[/imath] isn't finite and [imath]\mu(A)=[/imath]#[imath]A[/imath] if [imath]A[/imath] is finite) is a measure? (The second property of a measure isn't clear for me). |
2698997 | Prove that [imath]X>Y>0 \Rightarrow Y^{-1}>X^{-1}>0[/imath]
Let [imath]X,Y[/imath] be some invertible positive definite matrices such that [imath]X>Y>0[/imath], how do we prove that [imath]Y^{-1}>X^{-1}>0[/imath]? Any hint is greatly appreciated! I just need some initial directions. Thanks a lot!! | 688100 | Partial Ordering of Positive Definite Matrices
Assume that [imath]\mathbf{A}[/imath], [imath]\mathbf{B}[/imath], and [imath]\mathbf{A-B}[/imath] are all positive definite Hermitian symmetric matrices of same dimensions. Prove that [imath]\mathbf{B}^{-1}-\mathbf{A}^{-1}[/imath] is positive definite. |
2699129 | Calculating [imath]\int_0^{2\pi} \sin(nt)\cos(mt)dt[/imath]
Is there a way to calculate the 2nd integral quickly, if I already know what the first one is? [imath]\int_0^{2\pi} \exp(int)\exp(-imt)dt[/imath] [imath]\int_0^{2\pi} \sin(nt)\cos(mt)dt[/imath] | 1259698 | Orthogonality lemma sine and cosine
I want to know how much is the integral [imath]\int_{0}^{L}\sin(nx)\cos(mx)dx[/imath] when [imath]m=n[/imath] and in the case when [imath]m\neq n[/imath]. I know the orthogonality lemma for the other cases, but not for this one. |
2698627 | Calculate integral with help the Euler's integrals
It is my first question. In advance please sorry for my bad English! I need to calculate this integral with help the Euler's integrals: [imath] \int_0^{+\infty} \frac{1}{1+x^5} [/imath] I have tried decompose integrand in Taylor Series but I did not get anything. Also I tried use partial fractions and I got crazy expression. I'm here to get an elegant solution of this question. Thank you for help in advance. | 1217830 | [imath]\int_0^\infty \frac{1}{1+x^ 9} \, dx[/imath]
[imath]\int_0^\infty \frac{1}{1+x^9} \, dx[/imath] I tried taking the integral of [imath]\Gamma_R = [0,R] \cup \gamma_R \cup I_R[/imath], where we see that \gamma_R is the circle parametrized by [imath]z = Re^{it}[/imath] with [imath]t\in[0,\frac{\pi}{2}][/imath]. And [imath]I_R[/imath] is the line form [imath]iR[/imath] to [imath]0[/imath] one can note the following three things: 1) [imath]\int_0^\infty \frac{1}{1+x^ 9}\,dz[/imath] = [imath]\int_0^\infty \frac{1}{1+z^ 9}\,dz[/imath]. (Note that [imath]z[/imath] is on the [imath]x[/imath]-axis) 2) [imath]\int_{\gamma_R}\frac{1}{1+z^ 9} = 0[/imath]. (By the ML inequality) the main problem is when i parametrize [imath]I^{-}_{R}[/imath] with [imath]z(x) = xi[/imath] and [imath]x \in [0,R][/imath]. The main problem is that i get that: 3) [imath]\int_{I_R}\frac{1}{1+z^9} = -\int_{I_R^-}\frac{1}{1+z^9} = -\int_{0}^{R} \frac{i}{ix^9 + 1}[/imath]. note that i want to work towards taking [imath]R[/imath] to infinity eventually and then equal the sum of the integrals to the sum of the residuals of [imath]\Gamma_R[/imath] times [imath]2\pi i[/imath]. The main problem here is that my term in 3), does not look like the form [imath]c \cdot \int_{0}^{\infty}\frac{1}{1+x^ 9}[/imath] with c a constant value. If i can get this done i think i'm finished. PS: The singular points i got where [imath]e^{\frac{\pi i}{9}},e^{\frac{3 \pi i}{9}}[/imath]. |
2699076 | Using the Pigeonhole principle
By placing 27 squares with an area of 1 meter inside of a circle with the radius of 2 meters, prove that exist such a point that belongs to 3 different squares. (each one of the squares [imath]\subseteq[/imath] in the circle completely) now I know that the combined area of the squares is 27 and 2 combined area of the circles is only [imath]8\pi<27[/imath] so the point that I need to prove that exist is indeed exist but I can't figure out how to do it formally with the Pigeonhole principle. thanks in advance and sorry about the english edited: I need a solution that involves the Pigeonhole principle | 2692315 | 27 squares of side [imath]1[/imath] inside a circle of radius [imath]2[/imath]: Show that three squares share an interior point.
By placing 27 squares with side [imath]1[/imath] in a circle with radius [imath]2[/imath], such that each square is entirely inside the circle, prove that there exists a point that 3 squares share (inside of their area). I thought about dividing the circle into 9 equal areas and using the pigeonhole principle, but got stuck there. Any hint will help. |
2700635 | Convergence of series implies convergence of product
Suppose [imath]a_n[/imath] is a positive sequence, then the product [imath]\prod_{n=1}^{\infty}(1+a_n)[/imath] converges iff the series [imath]\sum_{n=1}^{\infty}a_n[/imath] converges as well. Any hints please ? | 2027618 | Equivalence of convergence of a series and convergence of an infinite product
Let [imath](a_n)_n[/imath] be a sequence of non-negative real numbers. Prove that [imath]\sum\limits_{n=1}^{\infty}a_n[/imath] converges if and only if the infinite product [imath]\prod\limits_{n=1}^{\infty}(1+a_n)[/imath] converges. I don't necessarily need a full solution, just a clue on how to proceed would be appreciated. Could any convergence test like the ratio or the root test help? |
2700734 | Lower bounds for [imath]{{2n} \choose {n}}[/imath]
What are common lower bounds for [imath]{{2n} \choose {n}}[/imath]? Edit: I made a mistake in my original question. It doesn't change my question but there is no reason for me to include the mistake. | 2714383 | Bounds on [imath]\frac{2n!}{(n!)^2}[/imath]
Here is my problem: Use induction to show that for [imath]n\ge1[/imath], [imath]\frac{4^n}{\sqrt{4n}} \le \frac{(2n)!}{(n!)^2}\le \frac{4^n}{\sqrt{3n+1}}.[/imath] What I have so far: For [imath]n=1[/imath] it is [imath]2≤2≤2[/imath], and I tried to originally do this problem by assuming that the case holds for [imath]n\ge 1[/imath], and if it hold for n+1, than it should hold for all [imath]n\ge 1[/imath]. I started with [imath]n+1[/imath] case: [imath]\frac{4^{n+1}}{\sqrt{4(n+1)}}\le\frac{(2(n+1))!}{((n+1)!)^2}\le\frac{4^{n+1}}{\sqrt{3(n+1)+1}},[/imath] but this did not really lead me anywhere. |
2694690 | A better way to prove this inequality
Exercise 1.1.6. (b) For positive real numbers [imath]a_1, a_2, ... , a_n[/imath] prove that [imath](a_1+a_2+ \ldots +a_n)\Big(\frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_n}\Big) \geq n^2.[/imath] From [imath]AM \geq GM[/imath]: [imath]a_1+a_2+\ldots+a_n \geq n\sqrt[n]{a_1a_2 \ldots a_n}[/imath] [imath]a_2a_3 \ldots a_n + a_1a_3 \ldots a_n + \ldots + a_1a_2 \ldots a_{n-1} \geq n\sqrt[n]{(a_1a_2 \ldots a_n)^{n-1}}[/imath] Multiply, [imath](a_1+a_2+\ldots+a_n)(a_2a_3 \ldots a_n + a_1a_3 \ldots a_n + \ldots + a_1a_2 \ldots a_{n-1}) \geq n^2 \cdot a_1a_2 \ldots a_n[/imath] [imath](a_1+a_2+ \ldots +a_n)\Big(\frac{1}{a_1}+\frac{1}{a_2}+ \ldots +\frac{1}{a_n}\Big) \geq n^2.[/imath] I'm not satisfied with this. I'm looking for a better way to prove it...or at least a better way to write it (notation). | 2134362 | How do I prove this using question using Cauchy-Schwarz inequality?
[imath](a_1 + a_2 + · · · + a_n)( 1/a_1 +1/a_2+ · · · +1/a_n) ≥ n^2[/imath] My TA mentioned he solved this using Cauchy Schwartz inequality, but I don't see how. All I've managed to do is make a summation of [imath]1[/imath], [imath]n[/imath] times, which ends up being [imath]n ≥ n^2[/imath]. |
2701961 | How to show the set {u,v,w} is linearly independent with 2 variables
So the vectors are r 1 1 1 s 2 1 2s 2 For which values of [imath]r,s[/imath] in subset [imath]\mathbb R[/imath] is the set [imath]\{u,v,w\}[/imath] linearly independent? I got if [imath]s(rs-1)(2r-1)[/imath], does not equal [imath]0[/imath], thus [imath]r[/imath] cannot equal [imath]1/2[/imath], [imath]rs [/imath] cannot equal [imath]1[/imath] and [imath]s[/imath] cannot equal [imath]0[/imath]....but i'm feeling [imath]1%[/imath] confident on my answer because the matrices was difficult to reduce to row echelon form Would appreciate any help, Thankyou! | 2699905 | For which values [imath]r, s \in \mathbb R[/imath] is the set [imath]\{u, v, w\}[/imath] linearly independent?
[imath]u = (r,1,1), \ v = (1,s,2s), \ w = (1,2,2)[/imath] I know to assume the set will be linearly dependent, equate them to the [imath]0[/imath] vector and find the values not equal to what [imath]r[/imath] and [imath]s[/imath] is. But I can't seem to reduce the matrix to echelon form. I've gotten to the point of: [imath] \begin{pmatrix} r & 1 & 1 \\ 1 & s & 2 \\ 1 & 2s & 2 \\ \end{pmatrix} [/imath] [imath]R_2 = R_2 - R_3[/imath] [imath] \begin{pmatrix} r & 1 & 1 \\ 1 & -s & 0 \\ 1 & 2s & 2 \\ \end{pmatrix} [/imath] [imath]R_3 = R_3 +2R_2[/imath] [imath] \begin{pmatrix} r & 1& 1 \\ 0 & -s & 0 \\ 1 & 0 & 2 \\ \end{pmatrix} [/imath] EDIT: Thank you @LordSharktheUnknown [imath]R_3 = r \times R_3 - R_1[/imath] [imath] \begin{pmatrix} r & 1 & 1 \\ 0 & -s & 0 \\ 0 & -1 & 2r-1 \\ \end{pmatrix} [/imath] And I'm stuck. I know [imath]s ≠ 0[/imath] for it to be linearly independent but I have no idea how to reduce the rows any further to get to [imath]r[/imath] by itself. Any help is appreciated. Thank you. |
2702331 | Calculating modular exponentiation
One of the question at HW in Number Theory was calculating [imath]147^{147}\mod 13[/imath] We didn't see much about the modular properties yet, so I am kind of stuck. Could someone give a hint or a direction how to approach it? | 1606780 | Modular arithmetic problem (mod [imath]13[/imath])
[imath]\large2017^{(2020^{2015})} \pmod{13}[/imath] I am practicing for my exam and I can solve almost all problem, but this type of problem is very hard to me. In this case, I have to compute this by modulo [imath]13[/imath]. |
2702328 | Let [imath]n \in \Bbb{N}[/imath]. Let [imath]x_1,x_2,\ldots,x_n[/imath] be positive real numbers such that [imath]\prod_{i=1}^n x_i = 1[/imath]. Prove that [imath]\sum_{i=1}^n x_i \ge n [/imath]
I'm trying to do this exercise: Let [imath]n \in \Bbb{N}[/imath]. Let [imath]x_1,x_2,\ldots,x_n[/imath] be positive real numbers such that [imath]\prod_{i=1}^n x_i = 1[/imath]. Prove that [imath]\sum_{i=1}^n x_i \ge n[/imath] Here is my attempt: Let [imath]S = \{n \in \Bbb{N} | \sum_{i=1}^n x_i < n \}[/imath] and [imath]S \ne \emptyset[/imath]. By the well ordering principle, S has a least element. If [imath] n = 1[/imath], we have: [imath]\prod_{i=1}^2 x_i = 1 \implies x_2 = \frac{1}{x_1} \implies \sum_{i=1}^2 x_i = \frac{x_1^2+1}{x_1}[/imath] and [imath]\frac{x_1^2+1}{x_1} \ge 2 \iff x_1^2+1 \ge 2x_1 \iff x_1^2 -2x_1 +1 \ge 0 \iff (x_1 - 1)^2 \ge 0[/imath] Therefore 1 is not the least element of S. Let k be the least element of S. We can conclude the following: [imath]k \ne 1 \implies k \ge 2 \implies (k-1) \in \Bbb{N} [/imath] (1) [imath]\sum_{i=1}^k x_i < k \implies \sum_{i=1}^{k-1} x_i < k - x_k[/imath] (2) If [imath]\prod_{i=1}^k x_i = 1[/imath] there are 3 cases: (i). [imath]\prod_{i=1}^{k-1} x_i = 1[/imath] and [imath]x_k = 1[/imath] (ii). [imath]\prod_{i=1}^{k-1} x_i < 1[/imath] and [imath]x_k > 1[/imath] (iii). [imath]\prod_{i=1}^{k-1} x_i > 1[/imath] and [imath]x_k < 1[/imath] Assuming cases (i) or (ii) would imply a contradiction with out assumption that k is the least element because we would have: [imath]\sum_{i=1}^k x_i < k \implies \sum_{i=1}^{k-1} x_i < k - x_k \le k-1[/imath] So [imath](k-1) \in S[/imath] and [imath]S = \emptyset[/imath] In case (iii) I'm stuck. Can we derive a contradiction assuming that [imath]\prod_{i=1}^{k-1} x_i > 1[/imath] and [imath]x_k < 1[/imath] ??? Or is there another approach to this problem? Any help will be appreciated. | 1553204 | Prove that [imath]a_1+\cdots+a_n \geq n[/imath] if [imath]a_1[/imath], [imath]a_2[/imath], ... [imath]a_n[/imath] are positive real numbers and their product is [imath]1[/imath]
Please give me the proof for the following: Let [imath]a_1,\,a_2,\,\dots\,a_n[/imath] be [imath]n[/imath] positive real numbers whose product is equal to [imath]1[/imath]. Prove that [imath]a_1+\cdots+a_n \geq n[/imath] and that the equality sign holds only if every [imath]a_k[/imath] is equal to [imath]1[/imath]. |
2702417 | Accuracy of A Equivalence about Absolutely Convex Sets
I remember I’ve seen a theorem but I’m not sure about remembering it truthfully or not. I cannot find this equivalence in anywhere. That is the reason of my doubt. Is it true or not? If it is true how should I show it via using definitions below? Could someone share exact theorem and proof or its link? If there is a mistake please fix me. Thanks a lot Let [imath]X[/imath] be a vector space over [imath]\mathcal F[/imath] field (it is real or complex numbers) and [imath]A \subseteq X[/imath] [imath]A[/imath] is absolutely convex if and only if A is convex and balanced Let [imath]\lambda , \beta \in \mathcal F[/imath] and [imath]x,y \in A[/imath] If [imath]|\lambda| + |\beta| \le 1[/imath] such that [imath]\lambda x+ \beta y \in A[/imath] then [imath]A[/imath] is an absolutely convex set If [imath]\lambda x \in A[/imath] such that [imath]|\lambda| \le 1[/imath] then A is balanced set If [imath]\lambda x + (1- \lambda )y \in A[/imath] such that [imath]|\lambda| \in [0,1] [/imath] then A is convex set | 1181597 | Seminorms: Absolute Convexity
This is a preparation for: Gauges vs. Cylinders Given a vector space [imath]\Omega[/imath]. One has the equivalence: [imath]U\text{ absolutely convex}\iff U\text{ balanced, convex}[/imath] Moreover, it holds the equivalence: [imath]\kappa U+\lambda U\subseteq U\quad(|\kappa|+|\lambda|\leq1)\iff\kappa U+\lambda U=(|\kappa|+|\lambda|)U[/imath] How to prove this from scratch? |
2702698 | Prove that [imath]\frac{x}{e^x}[/imath] tends to zero as [imath]x \to \infty [/imath]
As the title states, I want to prove [imath]\lim_{x \to \infty} \frac{x}{e^x} =0[/imath] Clearly, L'Hopital's rule easily solves this. However, I'm curious to see if there's another way to prove it, without involving some differential or integral calculus (that is, by algebraic means). What I'm really interested about, is to prove that [imath]\lim_{x \to \infty} \frac{x}{e^{x^2}}=0 [/imath] I assume that proving the first limit will provide a way to prove the second one, using the squeeze method. If you know a direct way to prove the second limit, it will be more than perfect. Thanks in advance! | 578133 | Some questions about the function [imath]xe^{-x}[/imath]
I want to prove that [imath]xe^{-x}[/imath] is bounded by some constant for all the [imath]x \in [0, \infty)[/imath]. So I take derivative of this guy, and found that on [imath]x \in [0, \infty)[/imath], its derivative is positive from [imath][0, 1)[/imath], [imath]0[/imath] at [imath]x=1[/imath], and negative from (1, in infinity). So [imath]xe^{-x}[/imath] achieves its maximum in [imath][0, \infty)[/imath] at [imath]x=1[/imath], which is [imath]1/e[/imath]. However, to prove that it's bounded by [imath]1/e[/imath], I need to prove that as [imath]x[/imath] goes to infinity, [imath]xe^{-x}[/imath] converges to [imath]0[/imath]. Because otherwise, if it went to some value below [imath](-1/e)[/imath], [imath]xe^{-x}[/imath] would not be bounded by [imath]1/e[/imath]. How can I prove that [imath]xe^{-x}[/imath] converges to [imath]0[/imath]? I mean, as [imath]x[/imath] goes to infinity, [imath]e^{-x}[/imath] converges to zero, but at the same time, x diverges to infinity. How can I formally prove that the product term converges to [imath]0[/imath]? Do I need to use the epsilon definition of convergence to prove it? Could you just show me the formal proof of [imath]xe^{-x}[/imath] converges to [imath]0[/imath]? Or is there's any other way to approach this question? My professor suggested that I should use the power series for [imath]e^{-x}[/imath]...? Your help is greatly appreciated! Thank you so much! I hope you enjoy your Saturday! |
2697905 | Composite functions, is this statement correct?
Let [imath]S = \{1,2,3,4\}[/imath]. Let [imath]f[/imath] be the sets of all functions from [imath]S[/imath] to [imath]S[/imath]. Now, is this statement correct? [imath]\forall f \in F[/imath] , [imath]\exists[/imath] [imath]g \in F[/imath] so that [imath](f\circ g) (1) = 2[/imath] I think this is false because suppose the only output [imath]f(x)[/imath] can produce is [imath]f(x)=3[/imath] and [imath]f(x) = 4[/imath] Now, there is no possibility of [imath](f\circ g)(1) = 2[/imath] because [imath]2[/imath] isn't an output of [imath]f[/imath]. Is that correct to say? | 2697948 | Am I on the right track to proving this statement about composite function?
Let [imath]S = \{1,2,3,4\}[/imath]. Let [imath]F[/imath] be the sets of all functions from [imath]S[/imath] to [imath]S[/imath]. Now I think this statement is true: [imath]\forall f \in F , \exists g\in F[/imath] so that [imath](g \circ f)(1) =2[/imath] I suppose [imath]f \in F[/imath], and I need to show that [imath]\exists g\in F[/imath] so that [imath](g \circ f)(1) = 2[/imath] Do I just make an example of [imath]g[/imath] so that [imath](g \circ f)(1) = 2[/imath] ? Like [imath]f(1) = y[/imath] and then $g (y)=2 ? |
2702668 | Prove that [imath]\frac{1}{2\pi}\int_0^{2\pi}{|f(re^{i\theta})|^2}d\theta=\sum_{n=0}^{\infty}{|a_n|^2r^{2n}}[/imath]
I don't know where to start with this one: Let [imath]f(z)=\sum_{n=0}^{\infty}{a_nz^n}[/imath] be a series with radius of convergence [imath]1[/imath]. a) Prove that [imath]\frac{1}{2\pi}\int_0^{2\pi}{|f(re^{i\theta})|^2}d\theta=\sum_{n=0}^{\infty}{|a_n|^2r^{2n}}[/imath] for all [imath]0<r<1[/imath]. If I plug the [imath]f(z)[/imath] expression into the integral, I have the square of a sum and I don't know if it's the easiest way to solve it. I'd appreciate any hint. Thanks for your time. | 451299 | Proving that [imath]\frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{it})|^2dt=\sum_{n=0}^{\infty}|a_n|^2r^{2n}[/imath]
Let [imath]f(z)=\sum_{n=0}^{\infty}a_nz^n[/imath] with radius of convergence equals to [imath]R[/imath]. Show that for every [imath]r<R[/imath]: [imath]\frac{1}{2\pi}\int_{0}^{2\pi}|f(re^{it})|^2dt=\sum_{n=0}^{\infty}|a_n|^2r^{2n}[/imath] I tried this: [imath]\int_{0}^{2\pi}|f(re^{it})|^2dt=\int_{0}^{2\pi}f(re^{it})\overline{f(re^{it})}dt=\int_{0}^{2\pi}\sum_{n=0}^{\infty}a_nr^ne^{int}\sum_{n=0}^{\infty}\overline{a_n}r^ne^{-int}dt= \int_{0}^{2\pi}{\sum_{n=0}^{\infty}\sum_{k=0}^{n}}a_k\overline{a_{n-k}}r^ne^{int}[/imath] But this is about it. Any ideas? Thanks! |
2703680 | Evaluate [imath]\arctan{\frac{1}{2}} + \arctan{\frac{1}{3}}[/imath]
The problem is finding the sum of the numbers: [imath]\arctan{\frac{1}{2}} + \arctan{\frac{1}{3}}[/imath] I've tried expressing [imath]\frac{1}{2}[/imath] and [imath]\frac{1}{3}[/imath] as tangent functions of some angles but I wasn't able to find a valid solution. Any help would be much appreciated. | 1456746 | Show that [imath]\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}[/imath]
Show that [imath]\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}[/imath]. Attempt: I've tried proving it but it's not equating to [imath]\frac{\pi}{4}[/imath]. Please someone should help try to prove it. Is anything wrong with the equation? If there is, please let me know. |
2703010 | Problem on divisibility (Fermat's Theorem?)
Let [imath]p, q[/imath] be prime numbers and [imath]n \in \mathbb{N}[/imath] such that [imath]p \nmid (n-1)[/imath]. If [imath]p \mid (n^q - 1)[/imath] then show that [imath]q \mid (p-1)[/imath]. Using Fermat's Theorem I got [imath]n^{p-1} \equiv n^q \, (\text{mod}\, p)[/imath]. How do I get to [imath]p \equiv 1 \, (\text{mod} \, q)[/imath]? Have I used Fermat's Theorem correctly or did I overlook some hypothesis? A detailed proof would be much appreciated. | 2699965 | How to show that that [imath]q \mid p-1[/imath]?
Let [imath]p, q[/imath] be prime numbers and [imath]n ∈ \mathbb{N}[/imath] such that [imath]p \not\mid n- 1[/imath]. If [imath]p \mid n^q -1[/imath] then show that [imath]q \mid p-1[/imath]. My ideas: I was thinking about the formula [imath]n^q -1 = (n-1)(n^{q-1} + n^{q-2} +\cdots+1),[/imath] but now here I do not know how to show that [imath]q \mid p-1[/imath]. Any hints or solution can be appreciated. |
2702973 | [imath]X_n[/imath] converges weakly to [imath]δ_p[/imath], then [imath]X_n \to p[/imath] in probability.
To prove that if [imath](S, d)[/imath] is a metric space, [imath]p \in S[/imath] and [imath]X_n[/imath] are random variables on a probability space [imath](\Omega, A, P)[/imath] such that the law of [imath]X_n[/imath] converges weakly to [imath]δ_p[/imath], then [imath]X_n \to p[/imath] in probability. Here [imath]\delta_p(S) = \begin{cases} 1 & \text{if } p \in S \\ 0 & \text{otherwise} \end{cases}[/imath]. Require some hints to proceed with the problem. | 206308 | If the laws of random variables weakly converge to a point mass at [imath]c[/imath], then the random variables converge in probability to [imath]c[/imath].
This is question 10.3.1 from Rosenthal's First Look at Rigorous Probability. Suppose [imath]\mathcal{L}(X_n) \Rightarrow \delta_c[/imath] for some [imath]c \in \mathbb{R}[/imath]. Prove that [imath]\{X_n\}[/imath] converges to [imath]c[/imath] in probability. Here [imath]\mathcal{L}(X_n)[/imath] refers to the distribution of [imath]X_n[/imath]. Is the following proof correct? Is a simpler proof available? Is there a simple proof using Skorohod's theorem? I attempted to prove the contrapositive. Suppose [imath]X_n \not \to c[/imath] in probability. This means that there is an [imath]\varepsilon > 0[/imath] so that for all [imath]N[/imath], there exists an [imath]m > N[/imath] such that \begin{equation} P(|X_m - c| \geq \varepsilon) > \eta, \text{ for some [imath]\eta > 0[/imath]}. \end{equation} Consider \begin{equation} B := \{x \in \mathbb{R} : |x - c| \leq \varepsilon\} \in \mathcal{B} \text{ the Borel sets on [imath]\mathbb{R}[/imath]}. \end{equation} Then [imath]B^c \in \mathcal{B}[/imath] and denoting the distribution of [imath]X_n[/imath] as [imath]\mu_n[/imath] we have \begin{equation} \mu_n(B^c) = P(X_n \in B^c) = P(|X_n - c| \geq \varepsilon) > \eta. \end{equation} For all [imath]N[/imath], there exists an [imath]m > N[/imath] such that the cumulative distribution function of [imath]X_m[/imath] is not the same as the cumulative distribution function of [imath]\delta_c[/imath]; in particular it's mass changes somewhere at least [imath]\varepsilon[/imath] distance from [imath]c[/imath]. Finally because [imath]\delta_c(\{x\}) = 0[/imath] for all [imath]x \neq c[/imath] this completes the proof. |
2704111 | Combinatorial proof for Stirling number of 1st kind's generating function
So I see this identity a lot but I can't find a detailed combinatorial proof for it. [imath]n(n+1)...(n+k-1)=\sum_{j=0}^k \begin{bmatrix}k \\j \end{bmatrix}n^j[/imath] Can someone please help? | 162151 | Stirling Numbers of the First Kind - a direct derivation
Usually, the Stirling numbers of the first kind are defined as the coefficients of the rising factorial: [imath](*) \prod_{i=0}^{n-1}(x+i) = \sum_{i=0}^{n} S(n,i) x^i[/imath]. With this definition, a recursive relation of [imath]S(n,i)[/imath] be derived, and it can be shown that is coincides with the recursive relation on the number of permutation on an n-set with i cycles and they have the same initial conditions, hence they coincide. 1) Is there any possibility to do it the other way around, i.e., define [imath]S(n,i)[/imath] combinatorially and then to show that [imath]\prod_{i=0}^{n-1}(x+i) = \sum_{i=0}^{n} S(n,i) x^i[/imath] holds for [imath]x \in \mathbb{N}[/imath] by some combinatorial argument, and thus it is a polynomial identity? (For Stirling numbers of the second kind, it is possible: it can be shown that [imath]n^k = \sum_{i=0}^{k} \binom{n}{i} i! S_2(k,i)[/imath] combinatorially ([imath]n^k[/imath] counts function from [imath][k][/imath] to [imath][n][/imath]).) 2) Additionally: equating coefficients in [imath](*)[/imath] shows that [imath]S(n,i)[/imath] is the elementary symmetric polynomial on [imath]n[/imath] variables of degree [imath]n-i[/imath] evaluated on [imath](0,1,\cdots ,n-1)[/imath]. Is there a combinatorial interpretation of this? |
1982680 | Why is rank [imath]uv^{T}[/imath] always equal to 1?
Let [imath]u \in \mathbb{R}^m[/imath] and [imath]v \in \mathbb{R}^n[/imath] and [imath]u,v \neq 0[/imath]. Since [imath]u[/imath] is an [imath]m\times1[/imath] matrix and [imath]v^T[/imath] is a [imath]1\times n[/imath] matrix, the product [imath]uv^{T}[/imath] is a [imath]m\times n[/imath] matrix. Then: [imath]\text{rank} (uv^{T}) =1[/imath] Just to illustrate: pick a random vector [imath]u[/imath] in [imath]\mathbb{R}^4[/imath], and [imath]v[/imath] in [imath]\mathbb{R}^5[/imath], for example: [imath]u= (1, 2, 3, 4),[/imath] and [imath]v= (1, 2, 3, 4, 5)[/imath]. The rank of the matrix [imath]m\times n[/imath] is [imath]1[/imath] which can be checked by row-reducing. Chosing arbitrary vectors [imath]u,v \neq 0[/imath], the rank of the product matrix mxn of [imath]uv^{T}[/imath] is always [imath]1[/imath]. Why is this? Any help would be greatly appreciated. | 1189989 | How to prove whether a matrix has rank [imath]1[/imath]
If [imath]u[/imath] [imath]∈ \mathbb R^{m \times 1}[/imath] and [imath]v ∈ \mathbb R^{n \times 1}[/imath] how do you show that the [imath](m \times n)[/imath] matrix [imath]uv^T[/imath] has rank [imath]1[/imath]? Would providing an example be sufficient to prove it? |
1070088 | If [imath]v_1,...,v_r[/imath] are eigenvectors that correspond to distinct eigenvalues, then they are linearly independent.
Prove: If [imath]v_1,...,v_r[/imath] are eigenvectors that correspond to distinct eigenvalues [imath]\lambda_1, ...,\lambda_r[/imath] of an [imath]n \times n[/imath] matrix [imath]A[/imath], then the set [imath]\{v_1,...,v_r\}[/imath] is linearly independent. Please give an example and tell me how this theorem works! | 2704394 | Prove three eigenvectors from three distinct eigenvalues are linearly independent
Here is the formal statement: Let [imath]\lambda_1, \lambda_2, \lambda_3[/imath] be distinct eigenvalues of [imath]n\times n[/imath] matrix [imath]A[/imath]. Let [imath]S=\{v_1, v_2, v_3\}[/imath], where [imath]Av_i = \lambda_i v_i[/imath] for [imath]1\leq i\leq 3[/imath]. Prove [imath]S[/imath] is linearly independent. Many resources online state the general proof or the proof for two eigenvectors. What is the proof for specifically 3? I tried to derive the 3 eigenvector proof from the 2 eigenvector proofs, but failed. |
2706223 | Subsequential limit and interval
Suppose that the sequence [imath](a_n)_{n \ge 1}[/imath] of real numbers is such that [imath]a_{n+1}− a_n \to 0[/imath]. Prove that the set of limits of its convergent subsequences is the interval with endpoints [imath]\liminf_{n \to \infty} a_n[/imath] and [imath]\limsup_{n \to \infty}a_n[/imath]. Can this be done this way that since the elements of the sequence are dense hence if you take a point then there are infinitely many elements around its neighbourhood and hence it acts as a subsequential limit now between lim sup and lim inf every point is a subsequential limit hence the proof follows Please help whether the above method is right or wrong | 2666730 | Let [imath]a = \liminf x_n[/imath], [imath]b = \limsup x_n[/imath]. If [imath]\lim (x_{n+1} - x_n) = 0[/imath] then every element of [imath](a,b)[/imath] is a subsequential limit of [imath](x_n)[/imath]
I'm fixing an arbitrary [imath]x[/imath] in [imath]a<x<b[/imath] and I must prove that [imath]x[/imath] is the limit of a subsequence of [imath](x_n)[/imath]. I have a theorem that says that [imath]x[/imath] is a subsequential limit of a sequence [imath](x_n)[/imath] iff every interval [imath](x-\epsilon, x+\epsilon)[/imath] containts infinitely many terms of [imath](x_n)[/imath]. I also know that [imath]a[/imath] and [imath]b[/imath] are the smallest and largest subsequential limits of [imath](x_n)[/imath]. I don't know where to go from here. |
2706391 | What let's us do stuff like [imath]F(a) = F(b)[/imath] for [imath]a=b[/imath]?
In Tao's Analysis vol 1 we have various proofs from properties and operations on natural numbers, as well as axioms. For example additive identity [imath]a+0=a[/imath]. But then in some proofs we apply these changes when the arguments are inside function calls. For example if [imath]a+0=a[/imath] then [imath]S(a+0)=S(a)[/imath]. What let's us do this? Maybe it's just so obvious it needs no further proof but what lets us say [imath]F(a) = F(b)[/imath] if [imath]a=b[/imath]? | 1965196 | Prove if a = b, then f(a) = f(b) for any function f (with natural deduction)
I want to be able to prove that for all functions f, that [imath]a = b \to f(a) = f(b)[/imath]. That this is true is obvious, but I'm not sure how to formally prove it using only the rules of inference in first-order logic (using natural deduction). I'm guessing I need to reference a formal definition for a function, where [imath]f \subset A \times B[/imath] and satisfies: [imath] (\forall x \in A)(\exists y \in B)((x,y) \in f \wedge (\forall z \in B)((x,z) \in f\to y=z))[/imath] However, I don't have any clue where to begin with this. Can anyone help? Please only give hints/answers using natural deduction as the deductive system. |
2706199 | The process behind completing the square
What is the process behind transforming [imath]x^2+x+1[/imath] into [imath](x+\frac12)^2+\frac34[/imath]? Does that same method works for every time I want to transform an expression like [imath]ax^2+bx+c[/imath] into a square like [imath](a+b)^2[/imath]? Thanks in advance. | 76772 | Formula for completing the square?
My math teacher said that this was the formula for completing the square. Original function: [imath]ax^2 + bx + c[/imath] Completed square: [imath]a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c[/imath] However, using this formula I'm not getting the same answers that I would get just by determining the stuff myself. Is this correct? |
2706555 | Why [imath]\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=e[/imath] doesn't imply [imath]\lim_{n\to \infty} \left(1+\frac 1n \right)≠1?[/imath]
I'm looking for the answer to this question. But I could not find the "satisfactory" answer. This is obvious, [imath]\lim_{n\to \infty} \left(1+\frac 1n \right)=1+0=1[/imath] and [imath]\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=e≈2.718281...[/imath] And also we can write, [imath]\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=\left(1+0 \right)^{\infty}=1^{\infty}≠1[/imath] The point, which that I can not understand is: [imath]\lim_{n\to \infty} 1^{\infty}=1^{\infty}=1[/imath] But, this is also true: [imath]\lim_{n\to \infty} \left (1+\frac 1n \right) =\lim_{n\to\infty}1=1[/imath] Why [imath]\lim_{n\to \infty} \left(1+ \frac 1n \right)^n=e[/imath] doesn't imply [imath]\lim_{n\to \infty} \left(1+\frac 1n \right)≠1?[/imath] | 1575719 | Does [imath]1^{\infty}=e[/imath] or [imath]1^{\infty}=1[/imath]?
In fact the real question is: Does [imath]\lim\limits_{n\to\infty}1^{n}=e[/imath]?. I know that [imath] \lim\limits_{n\to\infty}\left(1+\dfrac{1}{n}\right)^n=e, [/imath] So, can we say that [imath]1^\infty=e[/imath]? And, by logic, This product [imath]\underset{\text{infinity times}}{1\cdot1\cdot\ldots\cdot1},[/imath] gives [imath]1[/imath] not other value. |
2707187 | Does there exist a positive integer [imath]k[/imath] such that [imath]A = 2 ^ k + 3 ^ k[/imath] is a square number?
Does there exist a positive integer [imath]k[/imath] such that [imath]A = 2 ^ k + 3 ^ k[/imath] is a square number? I have tried from [imath]k = 1[/imath] to [imath]k = 10[/imath], seeing that [imath]A[/imath] is not a square number. Could you please answer in the general case? | 1794072 | Prove that [imath]2^n+3^n [/imath] is never a perfect square
My attempt : If [imath]n[/imath] is odd, then the square must be 2 (mod 3), which is not possible. Hence [imath]n =2m[/imath] [imath]2^{2m}+3^{2m}=(2^m+a)^2[/imath] [imath]a^2+2^{m+1}a=3^{2m}[/imath] [imath]a (a+2^{m+1})=3^{2m} [/imath] By fundamental theorem of arithmetic, [imath]a=3^x [/imath] [imath]3^x +2^{m+1}=3^y [/imath] [imath]2^{m+1}=3^x (3^{y-x}-1) [/imath] Which is not possible by Fundamental theorem of Arithmetic Is there a better method? I even have a nagging feeling this is wrong as only once is the equality of powers of 2 and 3 [imath](n) [/imath]are used. |
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