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2445746 | Prove that if [imath]f(1) = 0[/imath], then [imath]f(xy) = f(x) + f(y)[/imath] for all [imath]x, y > 0[/imath].
Suppose that [imath]f[/imath] is a function such that [imath]f'(x) = \frac{1}{x}[/imath] for all [imath]x > 0[/imath]. Prove that if [imath]f(1) = 0[/imath], then [imath]f(xy) = f(x) + f(y)[/imath] for all [imath]x, y > 0[/imath] I've tried applying the Mean Value Theorem, such that [imath]f'(x) = \frac {f(b) - f(a)}{b - a} = \frac {1}{x}[/imath], where [imath]y = f(x)[/imath] if continuous on [imath][a,b][/imath] and differentiable on [imath](a,b)[/imath] such that there exists at least one point [imath]c ∈ (a,b)[/imath]. I'm probably wrong, but i don't know how do i really approach this question and what they are really asking for (e.g. [imath]f(xy) = f(x) + f(y)[/imath]) | 1011467 | Suppose f is a function such that f'(x) = 1/x for all x > 0. Prove that if f(1) = 0, then f(xy) = f(x) + f(y) for all x,y > 0
Suppose f is a function such that [imath]f'(x) = \frac{1}{x}[/imath] for all [imath]x > 0[/imath]. Prove that if [imath]f(1) = 0, then f(xy) = f(x) + f(y)[/imath] for all [imath]x,y > 0[/imath] As stated above, how do i go about doing this question? Hi all, i would like to apologize if the question is missing context or other details. This question was part of my calculus tutorial. I was on the topic of application of differentiation when this question was given to me. Thus far i have learnt about intermediate value theorem, mean value theorem, rolle's theroem, Fermat's thereom. As of this point in my learning, i have not touched upon integration. The only reason why i accepted the answer with integration in it was because i understood it. Thus i believe i am suppose to use one the few theorems that i've stated to solve this question. But i have no idea where to start. At first i differentiated the equation [imath]f(xy)=f(x)+f(y)[/imath] with respect to x i believed the answer to be [imath]f'(xy)\frac{dy}{dx}=f'(x)+f'(y)\frac{dy}{dx}[/imath] which would be [imath]\frac{1}{xy}\frac{1}{x}=\frac{1}{x}+\frac{1}{y}\frac{1}{x}[/imath] At this point a few questions came to my mind 1) am i suppose to differentiate the eqn in the first place 2) what has [imath]f(1)=0[/imath] got to do with anything 3) what theorem am i suppose to use to solve this question This was my thought process throughout the question and i hope someone would point out my mistakes. Thank you |
2446228 | Bijection between homomorphisms [imath]\mathbb Z[i]\to R[/imath] and set [imath]x\in R[/imath] with [imath]x^2+1=0[/imath]
Let [imath]R[/imath] be a commutative ring. Show that there is a bijection between the set [imath]L[/imath] of ringhomomorphisms [imath]\mathbb Z[i]\to R[/imath] and the set [imath]X[/imath] of [imath]x\in R[/imath] with [imath]x^2+1=0[/imath]. I can show that [imath]\phi\colon L\to X\colon f\mapsto f(i)[/imath] is a well-defined ring homomorphism. Now I would either like to show directly that it is bijective, or I should show that its inverse exists; [imath]\phi^{-1}\colon X\to L\colon x\mapsto f:f(i)=x. [/imath] Now here I am a bit stuck, because I can't really make any progress either way. I also haven't used that [imath]R[/imath] is commutative, it seems. Could anyone give a hint on how to continue from this point? Btw: My course has just started dealing with rings and fields, so I barely know any theorem, except the isomorphism theorems. So I am looking for an elementary proof, if possible, at this point in my course. | 2445073 | Showing a bijection between the set of ringhomomorphisms [imath]\mathbb{Z}[i] \to R[/imath] and the set of [imath]x \in R[/imath] for which [imath]x^2+1=0[/imath]
I am studying basic ring theory and I am stuck on the following (possibly easy) question. Let [imath]R[/imath] be a commutative ring. Prove that there is a bijection between the set of ring homomorphisms [imath]\mathbb{Z}[i]\to R[/imath] and the set of [imath]x\in R[/imath] for which [imath]x^2+1=0[/imath]. Some things I remember and can prove: If [imath]f[/imath] is a ring homomorphism then [imath]f(1)=1, f(a+b)=f(a)+f(b)[/imath] and [imath]f(ab)=f(a)f(b)[/imath]. Also [imath]\mathbb{Z}[i]^* =\{1,i,-1,-i\}[/imath] and I can prove [imath]\mathbb{Z}[i] \cong \mathbb{Z}[X]/(X^2+1)[/imath]. My idea would be to construct a map from the set of ring homomorphisms to the set of [imath]x \in R[/imath] for which [imath]x^2+1=0[/imath] and show that it's bijective. The problem is that I don't know how to construct such a map (what does the set of ring homomorphisms look like?) and where to use that [imath]R[/imath] is commutative. Any answer/big hint would be appreciated. Thank you. |
2446457 | Proof by cases for sets
I have to prove by cases that for any two sets [imath]A[/imath] and [imath]B[/imath] [imath]A \subseteq B[/imath] if and only if [imath]A\cup(B\setminus A)=B[/imath] I don't get what cases should I consider, I am thinking of proving it by definition of sets, subsets, and union. Can someone help me understand what is needed from me? | 2111134 | For any sets [imath]A[/imath] and [imath]B[/imath], show that [imath](B\smallsetminus A)\cup A=B \iff A\subseteq B[/imath].
For any sets [imath]A[/imath] and [imath]B[/imath], show that [imath](B\smallsetminus A)\cup A=B \iff A\subseteq B[/imath]. I know I have to do the following: 1) Assume [imath](B\smallsetminus A) \cup A=B[/imath], prove [imath]A\subseteq B[/imath]. 2) Assume [imath]A\subseteq B[/imath], prove [imath](B\setminus A)\cup A=B[/imath]. This is what I have done so far: 1) Let [imath](B\smallsetminus A)\cup A=B[/imath] and [imath]x\in A[/imath]. Then [imath]x\in(B\smallsetminus A)\cup A[/imath]. Since [imath](B\smallsetminus A)\cup A=B[/imath], then [imath]x\in B\smallsetminus A[/imath]. Therefore, [imath]x\in B[/imath] by definition of difference. Thus, [imath]A\subseteq B[/imath]. 2) Let [imath]A\subseteq B[/imath] and [imath]x\in(B\smallsetminus A)\cup A[/imath]. Then [imath]x\in B\smallsetminus A[/imath] or [imath]x\in A[/imath] by definition of union. Then we have couple cases here: a) If [imath]x\in B\smallsetminus A[/imath], then we have proved the statement. b) Show that [imath]x\in A[/imath] and [imath]x\in B[/imath]. I have no idea if I am doing this correctly or not. Thus, any help is appreciated. |
2445657 | Solve initial-value quasilinear problem
I need help to solve initial-value quasilinear problem: [imath]xu_y - yu_x = u[/imath], [imath]u(x,0)=h(x)[/imath] Here what I did: The initial curve [imath]\Gamma: <x=s, y-0, z=h(s)> [/imath] The characteristic equation [imath]\frac{dx}{dt}=-y[/imath], [imath]\frac{dy}{dt}=x[/imath], [imath]\frac{dz}{dt}=z[/imath].. So [imath]x=-yt + s[/imath], [imath]y=xt[/imath] and [imath]z=h(s)e^t[/imath]. Solve for [imath]s[/imath] I get [imath]s=\frac{x^2 + y^2}{x}[/imath] then solution for this problem is [imath]u=h(\frac{x^2 + y^2}{x})e^{y/x}[/imath]. Is it right? I am not sure because the answer from text book, PDE 4th Edition by Fritz John, is [imath]u=h(\sqrt{x^2+y^2})e^{arctan(y/x)}[/imath] Please help. Thanks | 2080115 | Find the solution of the pde [imath]xu_y-yu_x=u[/imath].
I am trying to find the equation of characteristic curves and solution of the partial differential equation [imath]x\frac{\partial u}{\partial y}-y\frac{\partial u}{\partial x}=u[/imath] [imath]u(x,0)=\sin(\frac{\pi}{4}x)[/imath] . What I did: [imath]\frac{dx}{-y}=\frac{dy}{x}=\frac{du}{u}[/imath] with From first two equalities [imath]\frac{dx}{-y}=\frac{dy}{x}[/imath] by solving we get [imath]x^2+y^2=c_1[/imath] for some constant [imath]c_1[/imath]. Now I am stuck with the third and rest of the equalities to come up with a suitable manipulation to solve the problem. How can I do this? Any help would be great. Thanks. |
2446618 | If [imath]\frac{f'}{f}\left(\frac{1}{n}\right)=\frac{g'}{g}\left(\frac{1}{n}\right)[/imath] [imath]\forall n[/imath]
If [imath]f,g\in H(\mathbb{D})[/imath], and there are no zeroes in [imath]\mathbb{D}[/imath]. If [imath]\frac{f'}{f}\left(\frac{1}{n}\right)=\frac{g'}{g}\left(\frac{1}{n}\right)[/imath] [imath]\forall n[/imath], then [imath]f=kg[/imath] for some [imath]k\in \mathbb{C}[/imath]. I don't see what theorem should I use to prove it. Any hints to solve this problem? | 2437511 | If [imath]f'/f=g'/g[/imath] at every [imath]1/n[/imath] then [imath]f=kg[/imath] for some complex number [imath]k[/imath]
Suppose [imath]f(z)[/imath] and [imath]g(z)[/imath] are analytic in domain [imath]D[/imath], [imath]f(z)[/imath] and [imath]g(z)[/imath] never vanish at any [imath]z\in D[/imath] and that [imath] \frac{f'(z_n)}{f(z_n)}=\frac{g'(z_n)}{g(z_n)} [/imath] at a sequence of points [imath]\{z_n\}[/imath] converging to [imath]z_0\in D[/imath]. Show that [imath]f=Kg[/imath] for some [imath]K\in \mathbb{C}[/imath]. My Solution: As [imath]f,g[/imath] are analytic in [imath]D[/imath] and none of them have any zero in [imath]D[/imath] implies [imath]\frac{f'}{f}[/imath] and [imath]\frac{g'}{g}[/imath] are also analytic on [imath]D[/imath]. As [imath]\frac{f'(z_n)}{f(z_n)}=\frac{g'(z_n)}{g(z_n)}[/imath] and [imath]z_n\to z_0 \in D[/imath], using the identity theorem, [imath]\frac{f'}{f}\equiv \frac{g'}{g}[/imath] on [imath]D[/imath]. Now, \begin{align*} \frac{f'(z)}{f(z)}=\frac{g'(z)}{g(z)},\ \forall\ z\in D. \end{align*} Solving this differential equation gives me [imath]f=Kg[/imath] for some [imath]K\in \mathbb{C}[/imath]. Now my question is after doing the integration we will get [imath]\log(f)=\log(Kg)[/imath], is it valid in the complex logarithm? |
2446594 | prime factorization of number which is product of all divisors of another number
I need to calculate prime factorization of a number which is product of all divisors of a number represented as product of primes. For example [imath]2^1 \times 3^1 \times 5^1 = 30[/imath], and product of all divisors of 30 is [imath]2 \times 3 \times 5 \times 6 \times 10 \times 15 \times 30 = 2^x \times 3^y \times 5 ^ z[/imath] , I need to calculate [imath]x,y,z[/imath] which are [imath]4,4,4[/imath] in this example Is it possible to calculate [imath]x,[/imath]y,z without calculating actual product of divisors ? ( may be with divisor function ?) | 871064 | product of divisors of n as a power of n.
I am trying to prove that [imath]\prod_{d|n}{d}=n^\frac{d(n)}{2},[/imath] where [imath]d(n)[/imath] is the number of divisors of [imath]n[/imath]. My initial thought was to say let [imath]\{d_i\}_{i=1}^k[/imath] be the set of all divisors of [imath]n[/imath]. This means that [imath]d(n)=k[/imath]. But [imath]d(n)=(\alpha_1+1)(\alpha_2+1)...(\alpha_m+1)[/imath] when we define [imath]n=p_1^{\alpha_1}p_2^{\alpha_2}...p_m^{\alpha_m}[/imath]. So [imath]m<k[/imath]. Now the divisors of [imath]n[/imath] are just the product of prime powers with some combination of the primes/prime powers removed. For example, a divisor, [imath]d_i=\frac{n}{p_j^{\alpha_l}}[/imath]. The product of these divisors will yield [imath]n^k[/imath] in the numerator, and [imath]n^i[/imath] in the denominator with [imath]i<k[/imath] since the denominator will contain all [imath]p_j^{\alpha_j}[/imath]. But how to show that it is exactly [imath]n^\frac{d(n)}{2}[/imath] is escaping me. |
2446604 | Finding the Fibonacci term from the Fibonacci number
Given that the Fibonacci number is 165580141 Is it possible to find [imath]n[/imath] using the closed form. I tried simplifying the closed form but I get stuck at: [imath]\sqrt5F_n = (\frac{(1+\sqrt5}{2})^n - (\frac{(1+\sqrt5}{2})^n[/imath] [imath]\sqrt5(165580141) + (\frac{(1+\sqrt5}{2})^n = (\frac{(1+\sqrt5}{2})^n[/imath] ([imath]370248451 + (\frac{1+\sqrt5}{2})^n)^{1/n} = \frac{(1+\sqrt5)}{2}[/imath] I don't know how to proceed further. | 848691 | How to tell if a Fibonacci number has an even or odd index
Given only [imath]F_n[/imath], that is the [imath]n[/imath]th term of the Fibonacci sequence, how can you tell if [imath]n \equiv 1 \mod 2[/imath] or [imath]n \equiv 0 \mod 2[/imath]? I know you can use the Pisano period, however if [imath]n \equiv 1[/imath] or [imath]n \equiv 2[/imath] [imath]\mod \pi(k)[/imath], it can never be found, where [imath]k[/imath] is in [imath]\pi(k)[/imath] (The Pisano period). Also there is the fact that if [imath]\sqrt{5F_n^2+ 4}[/imath] is an integer then [imath]n \equiv 0 \mod 2[/imath], but is there a faster way? Lastly, because [imath]F_1 = F_2 = 1[/imath], that would have to be an exception for whatever rule/formula that would apply. |
2446111 | Removing path of a connected graph.
I bumped into this question in an old exercice sheet : Let [imath] G = (V,E) [/imath] be a connected graph with minimum degree [imath] \delta(G) = k \geq 2[/imath]. Prove that there exist a path [imath]P = x_1x_2...x_k[/imath] such that [imath]G \setminus P [/imath] is also connected [imath]G \setminus P[/imath] is the same as [imath]G[V\setminus (x_1,x_2,...x_k)][/imath]. I wanted to prove it by induction but even for the cas [imath]k=2[/imath] I didn't know where to begin. I thought maybe using the fact that in a graph with minimum degree k there always exist a cycle of length [imath]k+1[/imath] but then I have no idea how to show that it's possible to remove a part of the cycle. Thank you for your help | 2444351 | Graph Theory Path Problem
Let [imath]G[/imath] be a connected graph such that [imath]δ(G)≥k[/imath]. Proof there exists a path [imath]P[/imath] of length [imath]k[/imath] such that [imath]G-P[/imath] is connected. I am not sure how do I actually approach this question. Do I consider contradiction or inductive proof, if so how. Really appreciate it alot |
2428704 | How to show that the following relation is transitive (What is the idea?)
Suppose we consider the following set [imath] X = \big\{ (p,q) : p \in \mathbb{Z} \ \text{and} \ q \in \mathbb{N} \big\}[/imath] Define the following relation [imath](p_1,q_1) \sim (p_2,q_2) \iff p_1q_2 = p_2q_1[/imath] We can easily verify that this relation is both symmetric and reflexive. I am having trouble many with transitive. The issue that I am having is we can't really "divide". So what we assume is the following: Suppose that [imath](p_1,q_1) \sim (p_2,q_2)[/imath] and [imath](p_2,q_2) \sim (p_3,q_3)[/imath]. This means that the following equations are satisfied: [imath]p_1q_2 = p_2q_1[/imath] [imath]p_2q_3 = p_3q_2[/imath] We would like to show that the following equation is satisfied: [imath]p_1q_3 = p_3q_1[/imath] I am kinda lost in how to do that? | 2428368 | How to show that if [imath]x=(p,q)[/imath] and [imath]y=(p',q')[/imath] then [imath]x \sim y [/imath] if [imath]pq'=p'q[/imath] is an equivalence relation?
Suppose you have: [imath]x=(p,q)[/imath] and [imath]y=(p',q')[/imath] with [imath]p, q \in \mathbb{Z}[/imath] and [imath]q \not = 0[/imath] then [imath]x \sim y[/imath] if [imath]pq'=p'q[/imath] is an equivalence relation? So I managed to show that its reflexivity and symmetry, but I am unable to prove that it's transitive. |
2428910 | Given sequence converges to [imath]c \in \overline{\mathbb{R}}[/imath] extended reals, prove limit of average = c
Problem: Given [imath]a_n \to c \in \overline{\mathbb{R}}[/imath], prove that [imath]\displaystyle \lim_{n \to \infty} \frac{a_1 + a_2 + \cdots + a_n}{n} = c[/imath]. I have read many of the related posts on this topic: Convergence of the arithmetic mean Mean of a Convergent Sequence Two Limits Equal - Proof that limn→∞an=Llimn→∞an=L implies limn→∞∑n1akn=Llimn→∞∑1nakn=L Prove convergence of the sequence (z1+z2+⋯+zn)/n(z1+z2+⋯+zn)/n of Cesaro means These all attempt to split the summation in the numerator like so: [imath]\sum_{i=1}^N + \sum_{i=N}^n[/imath] and then bring in the c, so that eacn term is [imath]\frac{(a_1 - c) + (a_2 - c)}{n}[/imath]. However, all of these previous proofs assume that the sequence is convergent, which means that [imath]\frac{(a_1 - c) + (a_2 - c)}{n} = 0[/imath]. In my case, the sequence converges to a number in the extended reals and can thus diverge to [imath]-\infty[/imath] or [imath]\infty[/imath], meaning [imath]c[/imath] could be [imath]-\infty[/imath] or [imath]\infty[/imath], and we can't assume [imath]\lim_{n \to \infty}\frac{(a_1 - c) + (a_2 - c)}{n} = 0[/imath]. Are the methods in the above links still applicable? If not, how should I approach this proof? | 201274 | Cesàro mean for a divergent sequence
Given a real sequence [imath](a_n)_n[/imath] converging to a finite value [imath]a[/imath], a property of the Cesàro mean, defined as the arithmetic mean [imath] b_n=\frac{a_1+\ldots+a_n}{n}, [/imath] is [imath] \lim_{n\to\infty}b_n=a,\tag1 [/imath] so that, supposing [imath]a_n\neq0[/imath] [imath]\forall\,n[/imath] and [imath]a\neq0[/imath], we can also deduce [imath] \lim_{n\to\infty}\frac{b_n}{a_n}=1.\tag2 [/imath] Is result [imath](2)[/imath] also valid for [imath]a=0[/imath]? And are results [imath](1)[/imath] and/or [imath](2)[/imath] also valid for [imath]a=+\infty[/imath]? |
2447130 | [imath]\sum_{i=0}^n 2^{-i}[/imath]
This question might have been asked before but I have not been able to find it. How can I find: [imath]\sum_{i=0}^n 2^{-i}[/imath] Help? | 625394 | Sum with Exponent
I've got the following sum which I'm trying to figure out: [imath]\sum_{x=0}^n 2^{-x}[/imath] Wolframalpha tells me that it's equal to [imath]2 - 2^{-n}[/imath] but I am interested in figuring out why, and how to get that result by hand. Any help is greatly appreciated. Thanks. Also, as you can tell, I could really use some help with MathJax... |
2447646 | Counting ways for [imath]12[/imath] card-hands to contain [imath]4[/imath] cards in each of three suits
In a card deck of [imath]52[/imath], I want to count the ways for [imath]12[/imath] card-hands to contain 4 cards in each of three suits. I thought that it'd be: The ways of picking first suit [imath]\to 4[/imath] The ways of having 4 cards with this suit [imath]\to \binom{12}4[/imath] The ways of picking first suit [imath]\to 3[/imath] The ways of having 4 cards with this suit [imath]\to \binom{12}4[/imath] The ways of picking first suit [imath]\to 2[/imath] The ways of having 4 cards with this suit [imath]\to \binom{12}4[/imath] So total ways is: [imath]4\cdot 3\cdot 2 \cdot \binom{13}4 ^3[/imath]. My notes say this is incorrect but do not say why. | 2445773 | Deck Of Cards - 12 Card Hands
Problem: Find the number of twelve-card hands containing four cards in each of three suits. My solution: pick a suit : 4 ways. choose 4 cards from the 13 in that suit: [imath]{^{13}}{C}_{4}[/imath] ways. pick another suit: 3 ways. choose 4 cards from the 13 in the new suit: [imath]{^{13}}{C}_{4}[/imath] ways. pick the final suit: 2 ways. choose 4 cards from the 13 in the final suit: [imath]{^{13}}{C}_{4}[/imath] ways. Hence, total number of ways is [imath]4\cdot{^{13}}{C}_{4}\cdot3\cdot{^{13}}{C}_{4}\cdot2\cdot{^{13}}{C}_{4} = {^{4}}{P}_{3}\cdot\left({^{13}}{C}_{4}\right)^3 [/imath] total hands. This is the incorrect answer. Why? |
2445351 | Exact category and surjection
Let [imath]f:A\rightarrow B[/imath] be an epi morphism in an exact category and [imath]g:X\rightarrow B[/imath] be a morphism. Is there a morphism [imath]h:X\rightarrow A[/imath] such that [imath]g=f\circ h[/imath]? | 2201776 | Is this an equivalent formulation of "surjective" resp. "epimorphism"?
An [imath]A[/imath]-element [imath]x[/imath] of [imath]X[/imath], written [imath]x\in_A X[/imath], is a map [imath]x:A\to X[/imath]. If [imath]f[/imath] is a map with domain [imath]X[/imath], and [imath]x\in_A X[/imath] is an element, we write [imath]f(x)[/imath] to denote the composite of [imath]f[/imath] and [imath]x[/imath]. Now say that a map [imath]f:X\to Y[/imath] is cool if [imath]\forall A\quad \forall x, x'\in_A X\quad:\quad f(x) = f(x')\implies x = x'.[/imath] In the category of sets, a map is cool if and only if it is injective! In an arbitrary category, a map is cool if and only it is an monomorphism! Call a map [imath]f : X\to Y[/imath] fresh if for all [imath]A[/imath], and for all [imath]y\in_A Y[/imath], there is an [imath]x\in_A X[/imath] such that [imath]f(x) = y[/imath]. It would now be quite nice if, in the category of sets, a map is fresh if and only if it is surjective. 1. Is this true? 2. Also, in an arbitrary category, is a map fresh if and only if it is an epimorphism? |
2448117 | Can we imagine some strange "unit" [imath] S [/imath] such that [imath] S \times 0 = 1 [/imath]?
I'm not a professional mathematician (nor anything remotely close to that), but for sometimes I've wondered about this. Would it be possible to magicate some strange mathematical concept (lets call it [imath] S [/imath]), if that's even the correct word, such that: [imath] S \times 0 = 1 [/imath] Or in other words: [imath] S = \frac{1}{0} [/imath] I'm imagining something which at first looks absurd, such as the imaginary unit [imath] i = \sqrt{-1} [/imath], which nonetheless works very well in mathematics and in the real word. My sparse math skills can't take much further than some basic and sometimes bizarre properties such as the fact that [imath] S [/imath] seems to be completely indiferente to addition/subtraction, division and exponentiation for the natural numbers: [imath] n \in \mathbb{N} [/imath] [imath] S + n = \frac{1}{0} + n = \frac{1 + n \times 0}{0 \times 1} = \frac{1}{0} = S \Rightarrow S + n = S [/imath] [imath] S - n = \frac{1}{0} - n = \frac{1 - n \times 0}{0 \times 1} = \frac{1}{0} = S \Rightarrow S - n = S [/imath] [imath] \frac{S}{n} = \frac{1}{0} \times \frac{1}{n} = \frac{1 \times 1}{0 \times n} = \frac{1}{0} \Rightarrow \frac{S}{n} = S [/imath] [imath] S^n = \frac{1^n}{0^n} = \frac{1}{0} = S \Rightarrow S^n = S [/imath] For multiplication it's a bit more complicated, but the result is the same and it doesn't violate any of the above (I think). Illustrating for [imath] n = 2 [/imath]: [imath] 2 \times S = S + S = \frac{1}{0} + \frac{1}{0} = \frac{1 \times 0 + 1 \times 0}{0^2} = \frac{1/S}{1/S} = \frac{S}{S} \Rightarrow S = \frac{S}{S} - S = S \Bigl(\frac{1}{S} - 1\Big) = S (0 - 1) = S \times n = S \Rightarrow S \times n = S[/imath] Would this idea break math? I feel it's the most likely case, but I'm not skilled enough to pursue this much further than the above mentioned basics. Anyway, for me it's an interesting thing to think about, and I'd love the support of the community to come up with ways in which this idea fails or violates it's own rule-set in some manner. | 2436622 | Is it coherent to extend [imath]\mathbb{R}[/imath] with a reciprocal of [imath]0[/imath]?
I'm sure this is an obvious question, but I'm having trouble finding the right words to type into Google. I know that the definition of a ring allows that the additive identity not have a multiplicative inverse, but is this a requirement? Specifically, is something like [imath]\mathbb{R}\!\left[\frac{1}{0}\right][/imath] such that [imath]\frac{1}{0} \cdot 0 = 1[/imath] a ring, or does some contradiction arise from allowing the additive identity to have a multiplicative inverse? |
2446237 | Prove [imath]\exists[/imath] a sequence of polynomials which converges to [imath]f[/imath] on any compact subset of [imath]\mathbb{R}[/imath].
If [imath]f[/imath] is continuous function ([imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath]), then there exists a sequence of polynomials which converges to [imath]f[/imath] on any compact subset of [imath]\mathbb{R}[/imath]. Question: Since [imath]K[/imath] is compact, there exists some interval [imath][a,b][/imath], such that [imath]K[/imath] is contained in. By Weierstrass, for all [imath]\epsilon>0[/imath], there exits a sequence of polynomials of [imath]P_n[/imath], such that [imath]P_n[/imath] converges to [imath]f[/imath] uniformly, i.e. [imath]\exists N\in\mathbb{N}[/imath] [imath]\forall n\geq N[/imath], [imath]|P_n(x)-f(x)|<\epsilon[/imath], for all [imath]x\in [a,b][/imath], therefore, certainly in [imath]K[/imath]. So we are done. Correct? | 1074407 | Proving the existence of a sequence of polynomials convergent to a continuous function [imath]f[/imath].
I need to show that if [imath]f[/imath] is continuous function ([imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath]), then there exists a sequence of polynomials which converges to [imath]f[/imath] on any compact subset of [imath]\mathbb{R}[/imath]. I would appreciate any help/direction ... |
2448328 | Range of the function [imath]f(x) = \frac{2x}{x^{2}+1}[/imath]
Plotting the function [imath]f(x) = \frac{2x}{x^{2}+1}[/imath] we can see the range is [imath][-1,1][/imath]. Now I was told to do [imath]f(x)=y[/imath] and isolate [imath]x[/imath] doing this I have, using the quadratic formula [imath]x=\frac{1\pm \sqrt{1-y^2}}{y}[/imath] then [imath]range f(x)= domain \frac{1\pm \sqrt{1-y^2}}{y}[/imath]. But this domain does not include [imath]0[/imath] and [imath]0=f(0)[/imath] so [imath]0[/imath] must be in the range, what am I missing? | 2361415 | Show that [imath]y = \frac{2x}{x^2 +1}[/imath] lies between [imath]-1[/imath] and [imath]1[/imath] inclusive.
Prove, using an algebraic method,that [imath]y=\frac{2x}{x^2 +1}[/imath] lies between [imath]-1[/imath] and [imath]1[/imath] inclusive. Hence, determine the minimum and maximum points [imath]y=\frac{2x}{x^2 +1}[/imath] . What I tried: Firstly, I thought of using partial fractions but since [imath]x^2 +1 =(x-i)(x+i)[/imath], I don't think it is possible to show using partial fractions. Secondly, decided to use differentiation [imath]y=\frac{2x}{x^2 +1}[/imath] [imath]\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }[/imath] For stationary points: [imath]\frac {dy}{dx} = 0[/imath] [imath]\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0[/imath] [imath]x=-1[/imath] or [imath]x=1[/imath] When [imath]x=-1,y=-1[/imath] When [imath]x=1,y=1[/imath] Therefore, this implies that [imath]y=\frac{2x}{x^2 +1}[/imath] lies between [imath]-1[/imath] and [imath]1[/imath] inclusive. ^I wonder if this is the correct method or did I leave out something? The third way was using discriminant Assume that [imath]y=\frac{2x}{x^2 +1}[/imath] intersects with [imath]y=-1[/imath] and [imath]y=1[/imath] For [imath]\frac{2x}{x^2 +1} = 1[/imath], [imath]x^2 -2x+1 = 0[/imath] Discriminant = [imath] (-2)^2 -4(1)(1) = 0 [/imath] For [imath]\frac{2x}{x^2 +1} = -1[/imath], [imath]x^2 +2x+1 = 0[/imath] Discriminant = [imath] (2)^2 -4(1)(1) = 0 [/imath] So, since [imath]y=\frac{2x}{x^2 +1}[/imath] touches [imath]y=-1[/imath] and [imath]y=1[/imath], [imath]y=\frac{2x}{x^2 +1}[/imath] lies between [imath]-1[/imath] and [imath]1[/imath] inclusive. Is the methods listed correct?Is there any other ways to do it? |
2448735 | Evaluating [imath]\int_{0}^{2\pi}\frac{1}{4\cos^2 t + 9\sin^2 t} dt[/imath]
Evaluating [imath]\displaystyle \int_{0}^{2\pi}\frac{1}{4\cos^2 t + 9\sin^2 t} dt[/imath] Can someone show me a way using Complex Analysis methods? This is what I tried: Let [imath]C[/imath] be the contour defined by [imath]C(t)= 2\cos t + 3i\sin t[/imath], [imath]0 \leq t \leq 2\pi[/imath]. Considering the integral [imath]I= \int_{C}\frac{dz}{z}= 2\pi i[/imath] by Cauchy's Integral Formula, then also [imath]I = \int_{C}\frac{\overline{z}}{|z|^2}= \int_{0}^{2\pi}\frac{2\cos t - 3i\sin t}{4\cos^2 t + 9\sin^2 t}\cdot (-2\sin t + 3i\cos t) dt\\ = \int_{0}^{2\pi} \frac{5\sin t\cos t + 6i}{4\cos^2 t+ 9\sin^2 t}dt = 2\pi i.[/imath] I'm not sure where to go from here. | 455147 | closed form of [imath]\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}[/imath]
closed form of [imath]\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^n}[/imath] for [imath]a,b>0[/imath] n=1 we get [imath]\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^1}=\frac{2\pi}{ab}[/imath] n=2 we get [imath]\int_{0}^{2\pi}\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}=\frac{\pi(a^2+b^2)}{a^3b^3}[/imath] but what the integral for n ???? I hope to be there two different solution one by contour integration and the other by real analysis and thanks for all |
2446830 | Conics concerning Hyperbola. Tangent of ends of focal chord on hyperbola meet at directrix
How do you show that the tangents from the end points in a focal chord on a hyperbola meet at the directrix. Equation of hyperbola: [imath] \dfrac {x^2} {a^2}- \dfrac {y^2} {b^2}=1 [/imath] Original Question: Let [imath]P (a\sec(\theta),b\tan(\theta))[/imath] be a point on the hyperbola , with [imath]\tan(\theta)[/imath] not equal to [imath]0[/imath]. The tangent at [imath]P[/imath] meets the directrix at [imath]Q[/imath] and the [imath]S[/imath] is the corresponding focus. [imath]O[/imath] is the origin. Prove that [imath]SP[/imath] is perpendicular to [imath]SQ[/imath]. | 1117329 | Help with Conic: Hyperbola's chord of contact
please help with this proof. "Show that the tangents at the endpoints of a focal chord of the hyperbola [imath] \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 [/imath] meet on the corresponding directrix." This is a homework question with two part where the first part is to prove the converse of the above statement (namely prove that the chord of contact from a point on directrix is a focal chord) For the first part I was able to utilize the following equation for chord of contact from a point [imath]P(x_0, y_0)[/imath] \begin{align} \frac{x_0 x}{a^2} - \frac{y_0 y}{b^2} = 1 \end{align} Then substitute [imath]x_0[/imath] for [imath]\pm \frac{a}{e} [/imath] (x coordinate of directrix) and then show that [imath]S(\pm ae, 0) [/imath] (coordinate of focuses) satisfy the equation. But it seem I could not use the same method to show the second part. Please help. Thanks. |
2448772 | Prove [imath]\gcd(ka,kb)=k\cdot\gcd(a,b)[/imath]
This is my proof Let [imath]m[/imath] be [imath]\gcd(ka,kb)[/imath] then exists [imath]s[/imath] and [imath]t[/imath] such that [imath]ska+ tkb = m => k(sa+tb) = m [/imath] [imath]=> sa + tb = m/k[/imath] (1) since (1) is linear combination of [imath]a[/imath] and [imath]b[/imath] and [imath]m[/imath] is smallest positive linear combination of [imath]ka[/imath] and [imath]kb[/imath] => [imath]m/k[/imath] is smallest positive then [imath]\gcd(a,b) = m/k[/imath]. Is this proof is good enough to prove [imath]\gcd(ka,kb)=k\cdot \gcd(a,b)[/imath] | 477957 | Let [imath]a,k,m[/imath] be integers. Prove that [imath]\gcd(ka,km) = k\gcd(a,m)[/imath].
Here is what I got so far: Let [imath]\gcd(ka,km) = d[/imath] Then by the Euclidean Algorithm, we have integers [imath]s,t[/imath] such that: [imath]s(ka) + t(km) = d \implies k|d [/imath] Let [imath]d/k = g = sa + tm. \tag{1}[/imath] So now I need to show that [imath]g[/imath] is indeed = [imath]\gcd(a,m)[/imath]. Let [imath]\gcd(a,m) = h[/imath] Now since [imath]k|d \implies d|a , d|m \space [/imath] (since [imath]\gcd(ka,km) = d \implies g| \gcd(a,m) = h \\\implies h = dj ,[/imath] for some integer [imath]j[/imath]. Putting this into (1): [imath] h/jk = g = sa + tm [/imath] since [imath]h = dj \implies dj/j = kg \\ \implies d = kg. [/imath] So [imath]g[/imath] is the [imath]\gcd(a,m)[/imath]? Is this a correct proof? Or have I gone in some sort of circle? Many thanks for any hints or direction! |
2449122 | Prove [imath]BA-A^2B^2=I[/imath]
If [imath]A[/imath] and [imath]B[/imath] be two square matrices containing real elements and satisfying the following conditions [imath]AB-B^2A^2=I[/imath] [imath]A^3+B^3=O[/imath] then prove that [imath]BA-A^2B^2=I[/imath] My Attempt: If we are able to prove that if either of [imath]A[/imath] or [imath]B[/imath] is invertible then we are done. | 417508 | Prove [imath]BA - A^2B^2 = I_n[/imath].
I have a problem with this. Actually, still don't have the right way to start :/ Problem : Let [imath]A[/imath] and [imath]B[/imath] be [imath]n \times n[/imath] complex matrices such that [imath]AB - B^2A^2 = I_n[/imath]. Prove that if [imath]A^3 + B^3 = 0[/imath], then [imath]BA - A^2B^2 = I_n[/imath]. Thanks for any help. |
2449296 | Is [imath]x=0[/imath] if [imath]|x|<\varepsilon[/imath] for any [imath]\varepsilon>0[/imath] ??
Let [imath]x[/imath] be a real number. If [imath]|x|< \varepsilon[/imath] for any [imath]\varepsilon>0[/imath], then [imath]x=0[/imath]. That was a lemma in my lecture today. My Dr. proceeded as follows: Suppose that [imath]x≠0[/imath] Choose [imath]\varepsilon =|x|/2>0[/imath] We have [imath]|x|<|x|/2[/imath] (according to our assumption) Which is a contradiction Thus [imath]x=0[/imath] But I have a little question here: If I took [imath]\varepsilon=1>0[/imath] I have [imath]|x|<1[/imath] Then [imath]-1<x<1[/imath] Thus we can't assure the value [imath]x=0[/imath], it has more values, since [imath]x[/imath] is a real number in the interval [imath]]-1,1[[/imath] Is my Dr.'s solution accurate or am I missing something? | 2408066 | For any [imath]\epsilon[/imath], if [imath]\epsilon>0[/imath] and [imath]|x|<\epsilon[/imath], then [imath]x=0[/imath]
For any [imath]\epsilon[/imath], if [imath]\epsilon>0[/imath] and [imath]|x|<\epsilon[/imath], then [imath]x=0[/imath]. I understand that supposing [imath]\epsilon=\frac{x}{2}[/imath] will lead to a contradiction, but let’s take a correct case: Let [imath]\epsilon=3[/imath], then [imath]x[/imath] would have a whole set of values. Can you explain what is going on? |
2449492 | Explanation of [imath](x \notin A \setminus B)[/imath]
I don't understand [imath](x \notin A \setminus B)[/imath] Why is it not [imath]x \notin A \land x\in B[/imath] instead of [imath]x \notin A \lor x\in B[/imath] | 2142062 | If an element is not in the set difference [imath]A\setminus B \: \:?[/imath]
Given two sets[imath]A=\{1,3,5\}\quad,\quad B=\{1,3,8\}.[/imath] Then I compute the [imath]A\setminus B=\{5\}.[/imath] But my book[imath]^\dagger[/imath]said: ... . Also, observe that [imath]x\notin A\setminus B[/imath] does not mean that [imath]x\notin A\lor x\in B[/imath]. Why? I don't know how to explain the Why, please give me some example. [imath]^\dagger[/imath]a friendly introduction to analysis, second edition, page [imath]4[/imath]. EDIT: My intuition tells me that it does mean that. I guess there is some special case about null-set or about the restriction of the universal set. |
2449365 | If using L'Hopital's rule to show [imath]\lim\limits_{x\rightarrow0}\frac{\sin(x)}{x} = 1[/imath] is circular reasoning, then how do we justify it still applies?
I wrote out a proof using the definition of the derivative for the function [imath]f(x) = \sin(\theta)[/imath] to show that [imath]f'(x) = \cos(\theta)[/imath], in order to use L'H's rule to show that the limit, [imath]x \to 0[/imath], of [imath]\frac{\sin(x)}{x}[/imath] equals 1, but as one professor in my department warned me: this uses circular reasoning. There is a factor of [imath]\frac{\sin(h)}{h}[/imath] when using the definition [imath] f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h},[/imath] and we must show that [imath]\lim_{h \to 0} \frac{\sin(h)}{h} =1,[/imath] to continue the proof. So, how can we still justify using L'H's rule for [imath]\lim_{x \to 0} \frac{\sin(x)}{x}[/imath]? We're using the result to prove the result, which is absurd. Thanks, | 707469 | How to prove that [imath]d \sin(x)/dx = \cos(x)[/imath] without circular logic such as L'Hôpital's rule?
How do I prove that the derivative of [imath]\sin[/imath] is [imath]\cos[/imath] without resorting to L'Hôpital's rule (circular logic)? This part is easy: [imath] \begin{align*} \sin'(x) &= \lim_{\Delta x \to 0} \frac{\sin(x + \Delta x) - \sin(x)}{\Delta x} \\ \sin'(x) &= \lim_{\Delta x \to 0} \frac{\cos(x)\sin(\Delta x) + \sin(x) \cos(\Delta x) - \sin(x)}{\Delta x} \\ \sin'(x) &= \lim_{\Delta x \to 0} \left(\cos(x)\frac{\sin(\Delta x)}{\Delta x} + \sin(x)\frac{\cos(\Delta x) - 1}{\Delta x}\right) \\ \end{align*} [/imath] but where do I go from here? |
2449072 | Prove that exists [imath]\lim_{x\rightarrow \infty }\frac{f(x)}{x}[/imath] and determine its value
Let [imath]f:[0,\infty )\rightarrow \mathbb{R}[/imath] be an increasing function, such that [imath]\lim_{x\rightarrow \infty }\frac{1}{x^{2}}\cdot \int_{0}^{x}f(t)dt=1[/imath]. Prove that exists [imath]\lim_{x\rightarrow \infty }\frac{f(x)}{x}[/imath] and calculate this limit. If [imath]f[/imath] would be continuous, we'd have [imath]1=\lim_{x\rightarrow \infty }\frac{1}{x^{2}}\cdot \int_{0}^{x}f(t)dt=\lim_{x\rightarrow \infty }\frac{f(x)}{2x}[/imath], therefore [imath]\lim_{x\rightarrow \infty }\frac{f(x)}{x}=2.[/imath] But we don't know if [imath]f[/imath] is continuous or not, and I wasn't able to find any other idea. | 1323345 | Show that if [imath]\int_0^x f(y)dy \sim Ax^\alpha[/imath] then [imath]f(x)\sim \alpha Ax^{\alpha -1}[/imath]
Let [imath]f[/imath] be a real, continuous function defined on [imath][0,\infty)[/imath] such that [imath]xf(x)[/imath] is increasing for all sufficiently large values of [imath]x[/imath]. Show that if [imath]\int_0^x f(y)\,dy \sim Ax^\alpha \quad \left(\,x\to \infty\right)[/imath] for some positive constants [imath]A[/imath] and [imath]\alpha[/imath], then [imath]f(x)\sim \alpha Ax^{\alpha -1} \quad \left(\,x\to \infty\right).[/imath] Clearly, I have to use differentiation somewhere, but I don't know how to manipulate [imath]\lim_{x\to \infty}\frac{\int_0^x f(y)\,dy}{Ax^\alpha}=1[/imath] to get the desired result. From suggestions given, I know that by L'hospital's rule, it's enough to show that the limit [imath]\lim \frac{f(x)}{\alpha Ax^{\alpha -1}}[/imath] exists, and this limit equals [imath] \lim \frac{xf(x)}{\alpha Ax^{\alpha}}[/imath]. From here, I'll need to use the given assumption that [imath]xf(x)[/imath] is eventually increasing. But how can I show the existence of the limit based on these facts? I would greatly appreciate any solutions, hints or suggestions. |
2449980 | Modified pythagoras' theorem.
Given the equation: [imath]x^2 + y^2 = z^n [/imath] Prove that there are solutions in [imath]\mathbb N[/imath]. Provided that [imath]n\in\mathbb N.[/imath] Please don't give me the full answer, rather a way to attack the problem, I have no idea how to tackle this proof. | 1551756 | diophantine equation: [imath]x^2 +y^2 = z^n[/imath]
Prove that [imath]x^2+y^2=z^n[/imath] has a solution [imath](x, y, z)[/imath] in [imath]\mathbb{N}[/imath] for all [imath]n\in\mathbb{N}[/imath] I tried to prove this by induction, but couldn't. ( This was probably because the solution for some [imath]n[/imath] isn't necessarily related to the solution for [imath]n+1[/imath]) I can't seem to see any other way other than induction for proving the statement. Any help/hints on solving this problem, or any alternative approach will be highly appreciated :) |
2450080 | Prove that [imath]\lim\limits_{n \to \infty}\frac1n \sum\limits_{k=1}^nn^{1/k}=2[/imath]
Prove that [imath]\lim\limits_{n \to \infty}\frac1n \sum\limits_{k=1}^nn^{1/k}=2[/imath] How to estimate the sum? I tried to use Stolz's theorem but it is still a similar summation. | 2471532 | Limit of [imath](n+n^{1/2}+\cdots+n^{1/n})/n[/imath]
I'm pretty sure this has been asked here before. But perhaps due to high flexibility in formulating this problem I couldn't find any useful resources. Obviously a lower bound is [imath]2[/imath]. And, numerically testing it in WolframAlpha, I'm quite convinced the limit should just be [imath]2[/imath]. My first thought was to see it as a Riemann partial sum but it failed, at least in the equi-partition case. Could anybody help? Thanks. |
2450166 | Prove [imath]\lim_{x\to0} \sqrt {4 − x} = 2[/imath] using [imath]ε[/imath], [imath]δ[/imath]-definition.
Prove [imath]\lim_{x\to0} \sqrt {4 − x} = 2[/imath] using [imath]\varepsilon[/imath], [imath]\delta[/imath]-definition. Here's what I got: Given [imath]\varepsilon > 0[/imath], choose a particular [imath]\delta[/imath]. Then whenever [imath]0 < |x| < \delta[/imath], [imath]|\sqrt {4 − x} - 2|[/imath] has to equal [imath]\varepsilon[/imath] if we want to prove this limit. I know that [imath]4-x > 0[/imath]. How do I prove and present this question? Thanks. | 201476 | Prove [imath]\lim_{x\to0}\sqrt{4-x}=2[/imath] using the precise definition of limits. (Epsilon-Delta)
Prove [imath]\lim_{x\to0}\sqrt{4-x}=2[/imath] using the precise definition of limits. (Epsilon-Delta) I am not sure how to link [imath]0<\left |x \right |<\delta[/imath] with [imath]\left |\sqrt{4-x}-2\right |<\epsilon[/imath] . EDIT (Trying it out now) I worked till here, then I basically got stuck. |
2449533 | Equation of a plane by given 3 points
I've read of solutions, with let's say points [imath]P(1,2,3)[/imath], [imath]Q(3,1,2)[/imath], [imath]R(2,3,1)[/imath], where one subtracts point [imath]P-Q[/imath] and [imath]P-R[/imath] to get two vectors: [imath]\begin{pmatrix}1\\2\\3\end{pmatrix}-\begin{pmatrix}3\\1\\2\end{pmatrix}=\begin{pmatrix}-2\\1\\1\end{pmatrix}[/imath] [imath]\begin{pmatrix}1\\2\\3\end{pmatrix}-\begin{pmatrix}2\\3\\1\end{pmatrix}=\begin{pmatrix}-1\\-1\\2\end{pmatrix}[/imath] to now get the cross-product: [imath]\begin{pmatrix}-2\\1\\1\end{pmatrix}\times\begin{pmatrix}-1\\-1\\2\end{pmatrix}=\begin{pmatrix}2\\-1\\2\end{pmatrix}[/imath] hence the equation is [imath]2x-y+2z=d[/imath] by inserting, for example, point P we get [imath]2\times1-2+2\times3=d=6[/imath] and thus [imath]2x-y+2z-6=0[/imath] which can be transformed to [imath]z=-x+\frac12y+3[/imath] so when I put in any point it should work right? When I put in R though, I get: [imath]1=\frac52[/imath] So, what did I do wrong? | 1156983 | Find the equation of the plane knowing that it passes through 3 points
I have to find the equation of the plane that passes through [imath](0, 0, 0), (4, 0, -2), (0, 8, -6)[/imath]. I have done the following: The equation of the plane is of the form [imath]ax+by+cz+d=0[/imath] Since the points [imath](0, 0, 0), (4, 0, -2), (0, 8, -6)[/imath] are points o fthe plane, we have that [imath]d=0 \\ 4a-2c=0 \Rightarrow a=\frac{c}{2} \\ 8b-6c=0 \Rightarrow b=\frac{3c}{4}[/imath] So [imath]\frac{c}{2}x+\frac{3c}{4}y+cz=0 \Rightarrow \frac{1}{2}x+\frac{3}{4}y+z=0[/imath] Is it correct?? Is there also an other way to find the equation of the plane?? Maybe using the cross-product?? |
2450671 | How to calculate error function?
I am learning the normal (Or Gaussiance) distribution, I see a result for error function, that is: [imath]\frac{2}{\sqrt{\pi}}\int_{0}^{+\infty}e^{-t^2}dt=1[/imath] I thought so much about it, who can give me an explanation for this result? | 2161427 | Derivation of the gaussian integral
Can someone explain what's going on between the second equals sign, where things are converted to [imath]r[/imath]s? |
2450657 | homogeneous differential equations
In lectures , I learned that an ODE of the form : [imath]a_n(x)\frac{d^ny}{dx^n}+a_{n-1}(x) \frac{d^{n-1}y}{dx^{n-1}}+...+a_1(x)\frac{dy}{dx}+a_o(x)y=g(x) [/imath] is defined to be homogeneous if [imath]g(x)=0[/imath] but when solving this ODE for example [imath]x^3y'+8x^2yy'+4xy^2-y^3=0[/imath] we said it is homogeneous since [imath]f(tx,ty)=t^{3} f(x,y)[/imath] So I want to know when an ODE is said to be homogeneous ? do we have to check which condition? [imath]g(x)=0\ \ \ \ \ OR\ \ \ \ \ \ f(tx,ty)=t^{k} f(x,y)[/imath] | 1613456 | What is a homogeneous Differential Equation?
In first-order ODEs, we say that a differential equation in the form [imath]\frac{\mathrm d y}{\mathrm d x}=f(x,y)[/imath] is said to be homogeneous if the function [imath]f(x,y)[/imath] can be expressed in the form [imath]f\left(\displaystyle\frac{y}{x}\right)[/imath], and then solved by the substitution [imath]z=\displaystyle\frac{y}{x}[/imath]. In second-order ODEs, we say that a differential equation in the form [imath]a\frac{\mathrm d^2 y}{\mathrm d x^2}+b\frac{\mathrm d y}{\mathrm d x}+cy=f(x)[/imath] is said to be homogeneous if [imath]f(x)=0[/imath]. Is there a relation between these two? What does homogeneous mean? I thought it's when something [imath]=0[/imath], because in linear algebra, a system of [imath]n[/imath] equations is homogeneous if it is in the form [imath]\boldsymbol{\mathrm{Ax}}=\boldsymbol 0_{n\times 1}[/imath]; but this doesn't seem to be the case for first-order ODEs. |
2451119 | Confusing Integral Expression: [imath]\int{\frac{1}{x+\sqrt{x^2-x+1}}dx}[/imath]
I have this integral expression to be evaluated: [imath]\int{\frac{1}{x+\sqrt{x^2-x+1}}dx}[/imath] I've followed these steps: Completed the square in [imath]x[/imath] and then substituted [imath]u=x-\frac{1}{2}.[/imath] Trig substituted [imath]u = \frac{\sqrt{3}}{2}\tan t.[/imath] and then simplified the equation thereby giving: [imath]\int{\frac{\sqrt{3}\sec^2t}{\sqrt{3}\tan t+\sqrt{3}\sec t+1}dt}[/imath] I know up to this, but I'm unclear what to do next. Feel free to correct me if I'm wrong. | 2106937 | Evaluation of [imath]\int\frac{dx}{x+ \sqrt{x^2-x+1}}[/imath]
Evaluate : [imath]\frac{dx}{x+ \sqrt{x^2-x+1}}[/imath] After dividing and multiply with [imath]x-\sqrt{x^2-x+1}[/imath], I get [imath]x+\ln |x-1|-\int \frac{\sqrt{x^2-x+1}}{x-1}dx[/imath]. Is [imath]\int \frac{\sqrt{x^2-x+1}}{x-1}dx[/imath] is integrable in terms of elementary functions? |
2450136 | Quotient of complex polynomials ( alhfors complex analysis)
The Problem My Work We know that: [imath]\frac{Q'(z)}{Q(z)}=\sum\limits_{k=1}^{n}\frac{1}{(z-\alpha_k)}[/imath] Thus [imath]\frac{P(z)}{Q(z)}=\sum\limits_{k=1}^{n}\frac{P(z)}{Q'(z)(z-\alpha_k)}[/imath] This is about as far as I have traveled any ideas? Possible Ideas I get the feeling that I should use : But I am not sure of it's relevancy. | 2163293 | Show that [imath]\frac{P(z)}{Q(z)} = \sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}[/imath]
Here, [imath]Q[/imath] is a polynomial with distinct roots [imath]\alpha_1, \ldots, \alpha_n[/imath] and [imath]P[/imath] is a polynomial of degree [imath]<n[/imath]. Once again, the task is to show [imath]\frac{P(z)}{Q(z)} = \sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}[/imath] For reference, this is page 32 exercise 2 in Complex Analysis by Ahlfors. I'm having a ton of trouble making connections between what is presented in the text and this problem. The section is on rational functions and is incredibly concise. There was never a mention of derivatives and their relationships to rational functions, so I'm guessing here that it just means that [imath]Q'(a_k)[/imath] is a polynomial of degree [imath]n-1[/imath]. I don't even know where to start here, but I do have a couple question which might motivate the proof. What does [imath]P(\alpha_k)[/imath] mean? I don't see what a root of [imath]Q[/imath] has to do with a completely different polynomial. What is the relationship between the roots of [imath]Q[/imath] and the roots of [imath]Q'[/imath]? And finally, where does the derivative of [imath]Q[/imath] come in? |
2451565 | Prove [imath]\frac{x}{1+x^2} < \arctan x < x ,\quad x> 0[/imath]
So I should prove [imath]\frac{x}{1+x^2} < \arctan x < x ,\quad x> 0[/imath] I presume easiest way would be to: [imath]f(x)=\arctan x -\frac{x}{1+x^2}, \quad g(x)= x - \arctan x[/imath] then: [imath]f(x)>0[/imath], [imath]g(x)>0[/imath] But now I'm not really sure what I should utilize to prove those statements | 498532 | How to prove [imath]\frac{x}{1+x^2}<\arctan x for x>0?[/imath]
How to prove for [imath]x>0[/imath], [imath]\dfrac{x}{1+x^2}<\arctan x<x[/imath]? I started with saying [imath]0<\arctan x<\frac{\pi}{2}[/imath] when [imath]x>0[/imath] I'm just not sure how to proceed with this proof, it works when I take random values but I cant see how to prove that it works for any [imath]x[/imath]. |
2452095 | Let [imath]\{x_n\}[/imath] be a bounded sequence all of whose convergent proper subsequences converge to [imath]\ell[/imath]. Prove that [imath]\{x_n\}[/imath] converges to [imath]\ell[/imath].
How can i actually use the fact of the subsequences being proper. Are [imath]x_{2n+1}[/imath] and [imath]x_{2n}[/imath] proper subsequences? because being so solves this immediately. | 1326470 | Bounded sequence with subsequences all converging to the same limit means that the sequence itself converges to the same limit.
I have a question regarding one exercise in Stephen Abbotts' Understanding Analysis. The question is: Assume [imath](a_n)[/imath] is a bounded sequence with the property that every convergent subsequence of [imath](a_n)[/imath] converges to the same limit [imath]a \in \mathbb{R}[/imath]. Show that [imath](a_n)[/imath] must converge to [imath]a[/imath]. I was then presented this proof (from the solutions manual): Assume for contradiction that [imath](a_n)[/imath] does not converge to [imath]a[/imath], then from the negation of convergence we have (I'm not quite sure in this part): [imath]\exists \epsilon>0, \forall N\in \mathbb{N}, \exists n\in \mathbb{N} : n\geq N \wedge |a_n-a|\geq \epsilon. [/imath] Using this we can construct a subsequence [imath](a_{n_j})[/imath] that diverges from [imath]a[/imath] as follows: for arbitrary [imath]N\in \mathbb{N}[/imath], we can always find a, say, [imath]n_1[/imath] such that, [imath]n_1 \geq N[/imath] where [imath]|a_{n_1}-a|\geq \epsilon[/imath]. Since [imath]n_1[/imath] itself is in [imath]\mathbb{N}[/imath], then we can again find a [imath]n_2 \geq n_1[/imath] so that [imath]|a_{n_2}-a|\geq \epsilon[/imath]. And in general we can find a [imath]a_{n_{j+1}}[/imath] after choosing an appropriate [imath]a_{n_j}[/imath] so that [imath]|a_{n_{j+1}}-a|\geq \epsilon[/imath]. From the construction of such a subsequence, isn't the contradiction that [imath](a_{n_j})[/imath] does not converge to [imath]a[/imath] already proof that [imath](a_n)[/imath] should converge to [imath]a[/imath]? I am asking because the proof that I've read continues to say that the Bolzanno-Weierstrass theorem can be used to get another subsequence from the constructed [imath](a_{n_j})[/imath] which diverges from [imath]a[/imath], and that the proof ends there. I think I understand it, but I fail to see why it is necessary. Can somebody enlighten me on this? |
2451871 | Nilpotence of [imath]L =[/imath] [imath]b(n,F)[/imath]
Here by [imath]b(n,F)[/imath] we mean the Lie algebra of all [imath]n \times n[/imath] upper triangular matrices. So, the first step of the question asks to prove that [imath]L^m[/imath] has a basis consisting of all matrix units [imath]e_{ij}[/imath] with [imath]j - i > m[/imath]. So I thought I would try to compute [imath]L^m[/imath] for different [imath]m[/imath]. Now, [imath]e_{ij}[/imath] where [imath]i < j[/imath] form the basis for this Lie algebra. [imath][e_{ij}, e_{kl}][/imath] [imath]=[/imath] [imath]\delta_{jk}e_{il}[/imath] - [imath]\delta_{il}e_{kj}[/imath]. So we have two cases, [imath]j = k[/imath] or [imath]k = l[/imath], but not both. These two cases lead to [imath]L^1[/imath] being the span of [imath]e_{il}[/imath] and [imath]e_{kj}[/imath]. Now I don't know how to check if the claim in the question is true for [imath]L^1[/imath]. Can anyone help me out? | 2141517 | Strictly Upper Triangular [imath]n\times n[/imath] Matrices form a Nilpotent Lie Algebra over [imath]\mathbb{C}[/imath] for [imath]n\geq 2[/imath]
I have been attempting this problem for a while, it is an assignment problem so I don't want somebody to just post the answer, I'm just looking for hints. Let [imath]\mathfrak{u}(n,\mathbb{C})[/imath] be the Lie algebra of strictly upper triangular [imath]n\times n[/imath] matrices over [imath]\mathbb{C}[/imath]. I am trying to show that [imath]\mathfrak{u}(n,\mathbb{C})[/imath] is a nilpotent Lie algebra for [imath]n\geq 2[/imath]. This is what I have looked at so far. I have done some explicit calculations (I shall omit the bulk of them and just summarise below) for [imath]n=2,3,4[/imath] and have come up with some ideas. Let [imath]L=\mathfrak{u}(n, \mathbb{C})[/imath]. If [imath]n=2[/imath], [imath]L^1 = [L, L] = <0>_{\mathbb{C}}[/imath] So [imath]L[/imath] is nilpotent. If [imath]n=3[/imath], [imath]L^1 = [L, L] = <e_{13}>_{\mathbb{C}}[/imath] [imath]L^2 = [L, L^1] = <0>_{\mathbb{C}} = \left\{ 0 \right\}[/imath] So, [imath]L[/imath] is nilpotent. If [imath]n=3[/imath] [imath]L^1 = [L,L] = <e_{13},e_{14},e_{24}>_\mathbb{C}[/imath] [imath]L^2 = [L,L^1] = <e_{14}>_{\mathbb{C}}[/imath] [imath]L^3 = [L, L^2] = <0>_{\mathbb{C}} = \left\{ 0 \right\}[/imath] So, [imath]L[/imath] is nilpotent. So my hypothesis so far is that for an arbitrary [imath]m\in\mathbb{N}[/imath], [imath]L^{m-1}=\{0\}[/imath] My instinct is to do this by induction. My base case, [imath]n=2[/imath] has already been shown. Now I assume that [imath]\mathfrak{u}(n,\mathbb{C})[/imath] is nilpotent for [imath]n=k[/imath] ([imath]k\in\mathbb{N}[/imath]) and consider the case for n=k+1. However, I have no idea how to link these two cases together. I have read in Introduction to Lie Algebras and Representation Theory - J.E. Humphreys that [imath]L^k[/imath] should be spanned by basis vectors [imath]e_{ij}[/imath] where [imath]j-i=k+1[/imath] - however in my calculations for [imath]L^1[/imath], I obtain an [imath]e_{14}[/imath] which does not satisfy 4-1=2. Am I missing something completely obvious? Or have I made a mistake somewhere? (I have checked my calculations numerous times, but cannot find any errors!) Thanks, Andy. |
2453418 | Power series of [imath]f(x)=\frac{1}{1+x+x^2+x^3}[/imath]
Power series of [imath]f(x)=\frac{1}{1+x+x^2+x^3}[/imath] My try was: [imath]f(x)=\frac{1}{1+x+x^2+x^3}=\frac{1}{2}\frac{1}{1+x}-\frac{1(x-1)}{2}\frac{1}{1+x^2}[/imath] Now using the infinite geometric sum I got: [imath]\frac{1}{2}\sum_{n=0}^\infty(-x)^n-\frac{1}{2}(1-x)\sum_{n=0}^\infty(-x^2)^n=\frac{1}{2}\sum_{n=0}^\infty[(-1)^nx^n-(-1)^nx^{2n+1}+(-1)^nx^{2n}][/imath] The problem that I don't know if my moves are valid, and moreover, I don't know how to make it look as Power series. | 2164113 | Find series representation of a function [imath] y=\frac{1}{1+x+x^{2}+x^{3}} [/imath]
Let [imath]y = \frac{1}{1+x+x^{2}+x^{3}}[/imath]. And I want to find the series representation of this function. I've noticed that I can rewrite this like [imath]\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{1+x}*\frac{1}{1+x^{2}}[/imath] or I can rewite this like a sum of two fractions [imath]\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{2(1+x)}+\frac{1}{2(1+x^{2})}[/imath]. I know the series representations of these functions [imath](1+x)^{n}[/imath] and [imath]\frac{1}{1-x}[/imath], and I guess I should use one of them. But the result looks weird. Can you help me, please, may be I can't see some easy way to solve it. |
2453411 | How would I solve the following question involving elementary matrices?
[imath]A = \begin{pmatrix}0&5\\ 7&4\end{pmatrix}[/imath] (i) Write [imath]A[/imath] as a product of 4 elementary matrices. (ii) Write [imath]A^{-1}[/imath] as a product of 4 elementary matrices. I have managed to solve for [imath]A^{-1}[/imath], which is: [imath]\begin{pmatrix}-\frac{4}{35}&\frac{1}{7}\\ \frac{1}{5}&0\end{pmatrix}[/imath] But I am struggling to figure out how I would split each of these into 4 elementary matrices. I know that if I can solve the 4 elementary matrices of one of them, I just need to invert them to get the elementary matrices of the other. Any help? | 2452518 | How would I solve the following problem involving elementary matrices?
Let [imath]A=\begin{pmatrix}0&5\\ 7&4\end{pmatrix}[/imath]. 1) Write [imath]A[/imath] as a product of [imath]4[/imath] elementary matrices. 2) Write [imath]A^{-1}[/imath] as a product of [imath]4[/imath] elementary matrices. My work. I have managed to find [imath]A^{-1}[/imath], which came out to be this: \begin{pmatrix}-\frac{4}{35}&\frac{1}{7}\\ \frac{1}{5}&0\end{pmatrix} However, I am struggling to figure out how I would split each of these matrices into [imath]4[/imath] elementary matrices. Any help? |
2452457 | NECESSARY AND SUFFICIENT CONDITION for [imath]x[/imath] and [imath]y[/imath] lying in the same component of [imath]R^{\omega}[/imath] in uniform topology.
Consider [imath]R^{\omega}[/imath] in the uniform topology. Show that [imath]x[/imath] and [imath]y[/imath] lie in the same component of [imath]R^{\omega}[/imath] if and only if the sequence [imath]x - y = ({x_1 -y_1 , x_2 - y_2, .......})[/imath] is bounded. How to prove this? My Try : By uniform topology We can not have a separation in [imath]R^{\omega}[/imath]. Because we can not have separation on [imath]R[/imath]. I don't know I am wrong or right. Any help will be highly appreciated. Thank You. | 2391081 | connectedness of uniform topology
This question can also be found in this link, but because of dull brain I could not understood that One line proof. Let [imath]\mathbb{R}^{\omega}[/imath] be the set of all (infinite) sequences of real numbers. Determine whether [imath]\mathbb{R}^{\omega}[/imath] is connected or not in the uniform topology. Basis elements of uniform topology are [imath]B_{\rho}(x,\epsilon)=\{ y\mid \rho(x,y)<\epsilon \}[/imath], where [imath]\rho(x,y)=\sup\{ \min(d(x_\alpha,y_\alpha),1)\mid \alpha \in \mathbb{Z} \}[/imath] From the answer in the link, I understood that it is not connected. What I found: Note that [imath]B_{\rho}(x,1)=\overline{B_{\rho}(x,1)}[/imath]. So I need to find [imath]x_1,x_2\in \mathbb{R}^{\omega}[/imath] such that [imath]B_{\rho}(x_1,1)\cap B_{\rho}(x_2,1)=\phi[/imath] How to find such [imath]x_1[/imath] and [imath]x_2[/imath]? |
2452437 | Is there any domain [imath]\Omega\subset\mathbb{R}^3[/imath] with nontrivial homotopy group but with trivial first homology?
I'm using the word domain relatively loosely here. Ideally, I mean by a domain [imath]\Omega\subset\mathbb{R}^3[/imath] the interior of a connected smooth 3-manifold with corners in [imath]\mathbb{R}^3[/imath], though if this question has an answer for any connected open [imath]\Omega\subset\mathbb{R}^3[/imath], that would of course also be good. I know that for general manifolds [imath]M[/imath], it is possible that [imath]\pi_1(M)\ne 0[/imath] but [imath]H_1(M)=0[/imath], but how about if we only consider the case for domains [imath]\Omega\subset\mathbb{R}^3[/imath]? Does this remain true? | 679910 | Example of a domain in R^3, with trivial first homology but nontrivial fundamental group
Let [imath]\Omega \subset \mathbb{R}^3[/imath] be a domain. Is it true that if [imath]H_1(\Omega)[/imath] = 0, then [imath]\pi_1(\Omega) = 0[/imath]? For a counterexample, the group [imath]\pi_1(\Omega)[/imath] needs to be a perfect group and so I was trying with the smallest one i.e. [imath]A_5[/imath]. But I don't think the standard construction of the space from CW complexes, embeds into [imath]\mathbb{R}^3[/imath]. |
2452764 | Value of [imath]1/e[/imath] with the partial sum of [imath]e[/imath]
How to show [imath]1/e=1-1+\frac1{2!}-\frac1{3!}+\frac1{4!}+\dots[/imath] I have consider the product of the nth partial sums of the expansions for [imath]e[/imath] and [imath]1/e[/imath] but still can't prove there are equal to 1 with the definition [imath]e=1+1+\frac1{2!}+\frac1{3!}+\dots[/imath] | 2402730 | Why does the Taylor expansion of [imath]e^x[/imath] satisfy exponential properties?
Suppose I knew nothing about the function [imath]e^x[/imath]. If I wanted to find a power series that was its own derivative, I would logically start with the constant term, and first start by setting it to [imath]1[/imath]. Then, the next term should be the antiderivative of the first term, giving me [imath]x[/imath]. Doing this again would give me [imath]\frac{x^2}2[/imath]. Repeating this process over and over again, I would get [imath]1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots=\sum_{k=0}^{\infty}\frac{x^k}{k!}[/imath] Graphing a few terms of this, I might notice that this looks more and more like an exponential the more terms I graph. If I prove that this function satisfies the exponential relationship [imath]f(x+y) = f(x)f(y)[/imath], I would be able to prove that this series is an exponential function. How would I prove this? After this, how would I prove that the base of this exponential function is [imath]e[/imath], which is defined as [imath]\displaystyle{\lim_{n \to \infty}} (1+\frac1n)^n[/imath]? Edit: After expanding [imath](1+x)(1+y)[/imath] and [imath](1+x+\frac{x^2}2)(1+y+\frac{y^2}2)[/imath], I can see how extra terms get taken care of when the next degree is added. However, my second question still stands. |
2452567 | An olympiad Geometry problem involving triangles
In the triangle [imath]ABC\,[/imath] we denote by [imath]O[/imath] and [imath]I[/imath] the circumcenter and incenter respectively. The perpendicular bisectors of the line segments [imath]IA[/imath], [imath]IB[/imath] and [imath]IC[/imath] pairwise intersect thus defining the triangle [imath]A_1B_1C_1[/imath] Prove that [imath]\vec {OI}=\vec {OA_1}+\vec {OB_1}+\vec {OC_1}[/imath] Hi there. Im a high schooler new to olympiad geometry. Ive been unable to solve this problem despite repeated angle chasing could someone give a solution?Note I dont know stuff like projective geometry and hard lemmas so a simple proof would be most appreciated. | 2454334 | A Geometry Question utilizing triangles
In the triangle [imath]ABC\,[/imath] we denote by [imath]O[/imath] and [imath]I[/imath] the circumcenter and incenter respectively. The perpendicular bisectors of the line segments [imath]IA[/imath], [imath]IB[/imath] and [imath]IC[/imath] pairwise intersect thus defining the triangle [imath]A_1B_1C_1[/imath] Prove that [imath]\vec {OI}=\vec {OA_1}+\vec {OB_1}+\vec {OC_1}[/imath] Hi there. Im a high schooler new to olympiad geometry. Ive been unable to solve this problem despite repeated angle chasing could someone give a solution?Note I dont know stuff like projective geometry and hard lemmas so a simple proof would be most appreciated. |
2452937 | Proving this sequence is a Cauchy sequence
I have a positive infinite series [imath]a_n[/imath] and I know that $a_n[imath]a_{n+1}$ converges to 1 and also that $a_n[/imath]a_{n+2}$ converges to 1. How do I prove that [imath]a_n[/imath] is a Cauchy sequence (without using limit arithmetics)? How do I calculate [imath]lim_{n->inf}a_n[/imath] = ? Thanks. | 2448024 | how to prove that this positive sequence converges to 1
I have a positive infinite sequence [imath]a_n[/imath] and I know that $a_n[imath]a_{n+1}$ converges to 1 and also $a_n[/imath]a_{n+2}$ converges to 1. Why does it mean that [imath]a_n[/imath] also converges to 1? How can I prove it? Thanks. |
2453698 | Another Hard Inequality
Find the minimum of [imath]\frac{y^2+1}{x^2+z+1}+\frac{x^2+1}{y+z^2+1}+\frac{z^2+1}{x+y^2+1}[/imath] with [imath]x,y,z>-1[/imath]. The answer is 2, when [imath]x=y=z=1[/imath]. I am stuck. Here is what I have so far: [imath]\frac{y^2+1}{x^2+z+1}+\frac{x^2+1}{y+z^2+1}+\frac{z^2+1}{x+y^2+ 1}=3-\left(\frac{x^2-y^2+z}{x^2+z+1}+\frac{-x^2+y+z^2}{y+z^2+1}+\frac{x+y^2- z^2}{x+y^2+1}\right)\geq 3-\left(\frac{x^2-y^2+z}{2 x+z}+\frac{-x^2+y+z^2}{y+2 z}+\frac{x+y^2-z^2}{x+2 y}\right)=-\left(\frac{x^2-y^2+z}{2 x+z}+\frac{-x^2+y+z^2}{y+2 z}+\frac{x+y^2-z^2}{x+2 y}\right)+2+1=\left(\frac{(x-y) \left(x^2-y^2+z\right)}{(2 x+z) (x+y+z)}+\frac{(x-z) \left(x^2-y-z^2\right)}{(y+2 z) (x+y+z)}+\frac{(y-z) \left(x+y^2-z^2\right)}{(x+2 y) (x+y+z)}\right)+2[/imath] Source: https://brilliant.org/problems/minimum-11/?ref_id=1411033 (I did not write this question) | 1438053 | Prove [imath]\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2[/imath]
For positive integers Numbers [imath]a, b, c [/imath] prove that [imath]\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2[/imath] This inequality above take long time to prove it and still couldn't complete it. How I can prove this inequality? Any hint will help. Thanks |
2454240 | find the number of sequences of length [imath]n[/imath] over [imath]\{0,1,2,3,4\}[/imath] in which the difference between any two consecutive members is [imath]1[/imath] or [imath]-1[/imath]
Question: Find the number of series of length-[imath]n[/imath] over [imath]\{0,1,2,3,4\}[/imath] where the difference between any two consecutive members is [imath]1[/imath] or [imath]-1[/imath] Below are my workings. I'm stuck because I don't understand how I can make the recursion. Let [imath]a_n[/imath] be the number of [imath]n[/imath]-length series. Let [imath]a_n^k[/imath] be the number of [imath]n[/imath]-length series such that the first element is [imath]k[/imath]. We have that [imath]a_n=a_n^0+a_n^1+a_n^2+a_n^3+a_n^4[/imath] If the series starts with [imath]0[/imath], then the next element must be [imath]1[/imath], and we have that [imath]a_n^0=a_{n-1}^1[/imath] If the series starts with [imath]1[/imath], then the next element must be [imath]2[/imath] or [imath]0[/imath], and we have that [imath]a_n^1=a^2_{n-1}+a^0_{n-1}[/imath] If the series starts with [imath]2[/imath], then the next element must be [imath]3[/imath] or [imath]1[/imath], and we have that [imath]a_n^2=a^3_{n-1}+a^1_{n-1}[/imath] There is an similar answer to this but not in the same way that my college requires. | 192416 | In how many sequences of length [imath]n[/imath], the difference between every 2 adjacent elements is [imath]1[/imath] or [imath]-1[/imath]?
How do I find a recursion formula to solve the following question: In how many sequences of length [imath]n[/imath], with elements from [imath]\{{0,1,2,3,4}\}[/imath], the difference between every 2 adjacent elements is [imath]1[/imath] or [imath]-1[/imath]? I've been trying to solve the problem for the last 5 hours maybe, I'm sure that I'm missing something.. This is a problem in combinatorics, and the solution must be by finding a recursion formula for the problem. best solution I've come so far is that if we say that [imath]a_n[/imath] is the solution to the question, so let's count how many sequences of size [imath]n-1[/imath] there are that meet the requirement, and we complete the first element to make it a sequence of [imath]n[/imath] elements. so for [imath]n-1[/imath] there [imath]a_{n-1}[/imath] sequences, and for each sequence we have [imath]2[/imath] options to choose the first element (one option is to take a plus [imath]1[/imath] number from the number that start the [imath]n-1[/imath] sequence, and the other option to to take a minus [imath]1[/imath] number from the number that start the [imath]n-1[/imath] sequence), only problem as you figured out is that if the [imath]n-1[/imath] sequnce starts with [imath]0[/imath] or [imath]4[/imath] there is only [imath]1[/imath] option left for the first element in the sequnce.. so [imath]a_n=2a_{n-1}[/imath] won't work. I think that the solution may involve using [imath]2[/imath] or maybe [imath]3[/imath] recursion formulas. Hope I was clear enough. |
2454285 | Series converge or diverge?
Is this series converge ? [imath]\sum_{n=1}^{\infty} \frac{\tan(n)}{2^{n}}[/imath] My class disscussed this problem in some hours . Finally , my teacher used a lemma which isn't basic ( i.e : a result in a research paper ) . I hope someone can give me a more simpler proof . | 535386 | Converge or diverge? : [imath]\sum_{n=1}^{\infty}\frac{\tan{n}}{2^{n}}[/imath]
Determine this series converge or diverge, and if it converges, find its value. [imath]\sum_{n=1}^{\infty}\frac{\tan{n}}{2^{n}}[/imath] This was too hard for me, as unboundedness of [imath]\tan{x}[/imath] at infinite number of poles make harder to guess the behavior of value of tangent function at positive integers. Thanks. |
2455005 | Why is Integral from [imath]a[/imath] to [imath]b[/imath] equal to negative integral from [imath]b[/imath] to [imath]a[/imath]
Why is the integral from [imath]a[/imath] to [imath]b[/imath] equal to negative integral from [imath]b[/imath] to [imath]a[/imath]. According to my teacher this is a definition, but definitions still have some logic/reasoning behind it, so what is the logic/reasoning in this case? | 232455 | Is integration from [imath]a[/imath] to [imath]b[/imath] same or [imath]b[/imath] to [imath]a[/imath] or is negative?
The integration is generally area under the curve in [imath]\Bbb R^2 \to \Bbb R[/imath] Is integration in the range from [imath]a[/imath] to [imath]b[/imath] the same as [imath]b[/imath] to [imath]a[/imath] or is it negative? If it is negative, Is it merely notion of convention or is there some intuition for it? |
2451796 | [imath]q:X\to Y[/imath] continuous, show that [imath]X[/imath] is compact
I've been stuck on this problem for a very long time: Problem: Let [imath]X,Y[/imath] be topological spaces, [imath]Y[/imath] compact, and let [imath]q:X\to Y[/imath] be a continuous and closed map. Assume that [imath]q^{-1}(\{y\})[/imath] is compact for every [imath]y\in Y[/imath]. Show that [imath]X[/imath] is compact. What I have so far: [imath]q:X\to q(X)[/imath] is also continuous and closed, and [imath]q(X)[/imath] is compact (as a closed subset of a compact space). Thus, we could assume that [imath]q(X)=Y[/imath], or that [imath]q[/imath] is onto [imath]Y[/imath]. If [imath]q^{-1}(A)\subset X[/imath] is open, Then [imath](q^{-1}(A))^c=q^{-1}(A^c)[/imath] is closed, and so ([imath]q[/imath] is onto) [imath]A^c=q(q^{-1}(A^c))[/imath] is closed and [imath]A[/imath] is open. Thus [imath]q[/imath] is a quotient map. Thus, the quotient space [imath]X^*=\{q^{-1}(\{y\})|y\in Y\}[/imath] is compact, as it's homeomorphic to [imath]Y[/imath]. How do I carry on from here? | 902472 | Perfect Map [imath]p:\ X\to Y[/imath], [imath]Y[/imath] compact implies [imath]X[/imath] compact
I was assigned the following homework problem for an introductory course in topology: Let [imath]p:\ X\to Y[/imath] be a closed continuous surjective map such that [imath]p^{-1}(\{y\})[/imath] is compact for each [imath]y\in Y[/imath]. Show that if [imath]Y[/imath] is compact, then [imath]X[/imath] is compact. My question is, are the assumptions that [imath]p[/imath] be continuous and surjective even necessary for the result to hold? I've done the following: Let [imath]U[/imath] be an open set containing [imath]p^{-1}(\{y\})[/imath]. Since [imath]X-U[/imath] is closed, [imath]p(X-U)[/imath] is closed in [imath]Y[/imath]. Then [imath]W=Y-p(X-U)[/imath] is an open set that contains [imath]y[/imath]. Since [imath](X-U)\cap p^{-1}(W)=\varnothing[/imath], [imath]p^{-1}(\{y\})\subset p^{-1}(W) \subset U[/imath]. Let [imath]\{U_\alpha\}[/imath] be an open cover of [imath]X[/imath]. Then [imath]p^{-1}(\{y\})\subset \bigcup_{k=1}^{N}U_k[/imath]. For each [imath]y\in Y[/imath], define [imath]W_{y}[/imath] as above; then [imath]\{W_y\}[/imath] is an open cover of [imath]Y[/imath], so can be covered by [imath]W_{y_1},...,W_{y_m}[/imath], and [imath]p^{-1}(\{y_j\})\subset p^{-1}(W_{y_j})\subset \bigcup_{k=1}^{N}U_k[/imath]. Since [imath]X=p^{-1}(Y)[/imath], [imath]X=\bigcup_{j=1}^{n} p^{-1}(W_{y_j})[/imath]. Since each [imath]p^{-1}(W_{y_j})[/imath] is covered by finitely many [imath]U_k[/imath], so is [imath]X[/imath], so it is compact. If we dispense with surjectivity of [imath]p[/imath], the only thing that happens is that [imath]p^{-1}(\{y\})[/imath] may be empty, but then [imath]p^{-1}(\{y\})\subset p^{-1}(W)[/imath] holds vacuously, and the empty set is still compact. If [imath]p[/imath] is continuous, then [imath]p^{-1}(W)[/imath] is open, but I don't use the fact that it is open at all in my proof. I know that all 3 assumptions make a map "perfect" so that may be why they were mentioned, but can we dispense with continuity and surjectivity? Thanks. |
1729325 | How to prove that cardinality Borel [imath]\sigma[/imath]-algebra equals the cardinality of [imath]\mathbb R[/imath]?
My understanding at this point is that to assign a probability measure to a random variable defined on the real line, we need a Borel [imath]\mathscr{B}[/imath] sigma algebra, because otherwise we wouldn't be able to find a bijection from every subset of the power set ([imath]2^\Omega[/imath]) to the Lebesgue measure, due to the fact that the cardinality of [imath]2^\Omega[/imath] is not just greater, but much greater than the subsets of the Borel algebra. On the other hand, the Borel sigma algebra, even though it contains every number in the real line as a singleton set, has that same cardinality as [imath]\mathbb{R}[/imath]. How can this be proven? Is there a way to see this through the concept of standard topology? | 1954108 | cardinality of the Borel [imath]\sigma[/imath]-algebra of a second countable space
Second countability by itself doesn't restrict the cardinality of a topological space, since every set with the trivial topology is a second countable space, but it seems natural to ask whether second countability restricts the cardinality of the Borel [imath]\sigma[/imath]-algebra of the space. Can the cardinality of the Borel [imath]\sigma[/imath]-algebra of a second countable space be arbitrarily big? If this is the case is there a simple construction for a second countable space with an arbitrarily big Borel [imath]\sigma[/imath]-algebra? |
2455605 | Limit of sequence [imath] \left\{ \left(\frac{n^2+2}{2n^2+1}\right)^{n^2}\right\}_{n\geq 1}[/imath]
I trying to calculate [imath] \lim_{n\to\infty}\big(\frac{n^2+2}{2n^2+1}\big)^{n^2}.\tag{1}[/imath] Of course I cannot use theorem that if [imath]a_n\to\infty[/imath] then [imath]\lim_{n\to\infty}(1+\frac{1}{a_n})^{a_n}=e,[/imath] because in this particular situation we have [imath]\lim_{n\to\infty}\big(\frac{n^2+2}{2n^2+1}\big)^{n^2}=\lim_{n\to\infty}\big(1+\frac{1-n^2}{2n^2+1}\big)^{n^2}=\lim_{n\to\infty}\big(1+\frac{1}{\frac{(2n^2+1)}{(1-n^2)}}\big)^{n^2},[/imath] and [imath]\lim_{n\to\infty}\frac{(2n^2+1)}{(1-n^2)}=-2.[/imath] I would be grateful of any hints and comments. | 1995998 | Limit of [imath]\left(\frac{n^2+2}{2n^2+1}\right)^{n^2}[/imath]
The limit to solve is [imath]\lim_{n\to\infty}\left(\frac{n^2+2}{2n^2+1}\right)^{n^2}[/imath] I tried to use L'Hôpital's rule but the derivatives are quite complex. |
2455274 | Use Lagrange's theorem to prove that if [imath]G[/imath] is a finite group of order [imath]n[/imath] then the map [imath]x \to x^k[/imath] is surjective.
Use Lagrange's theorem to prove that if [imath]G[/imath] is a finite group of order [imath]n[/imath] then the map [imath]x \to x^k[/imath] is surjective. Edit: (n,k) = 1 which I completely blanked on mentioning. Also I truthfully have no idea how my post is a duplicate. My attempt: First prove [imath]x^k[/imath] is a subgroup of [imath]G[/imath]. It is a subset because is some element of [imath]x[/imath] of [imath]G[/imath] and [imath]x^k[/imath] was not in [imath]G[/imath], [imath]G[/imath] won't be closed under addition. This subset is non-empty since it contains the identity element because [imath]1^k=1[/imath]. It should be closed under multiplication because [imath]\varphi(ab) = a^k b^k = \varphi(a) \varphi(b)[/imath]. I am reasonably confident that I am correct through here. Let [imath]|G| = n[/imath] and [imath]|H| = m[/imath] for easier notation. Then by Lagrange's theorem [imath]\frac{n}{m} = c[/imath] with [imath]c \in \Bbb Z[/imath], [imath]1 \leq c \leq n[/imath]. Then maybe I could use [imath]n = c\cdot m[/imath] so [imath]\gcd(n,k) = 1 = \gcd(c\cdot m,k)[/imath] which implies [imath]\gcd(m,k) = 1[/imath]. But, if this reasoning is even correct, I don't know what to do now with it. | 522273 | If a group [imath]G[/imath] has odd order, then the square function is injective.
Suppose [imath]G[/imath] has odd order, show the function [imath]f:G\rightarrow G[/imath] defined by [imath]f(x)=x^2[/imath] is injective. This proposition is easily provable if we assume [imath]G[/imath] is Abelian, but I don't know how to start this without the assumption of being Abelian. |
2454243 | Minimum value of [imath]S=\frac{3a}{b+c}+\frac{4b}{a+c}+\frac{5c}{a+b}[/imath]
Find Minimum value of Minimum value of [imath]S=\frac{3a}{b+c}+\frac{4b}{a+c}+\frac{5c}{a+b}[/imath] My Try: we have [imath]S=3 \times \left (\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)+\frac{b}{a+c}+\frac{2c}{a+b}[/imath] By Standard [imath]AM[/imath] [imath]GM[/imath] inequality we know that [imath]\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \ge \frac{3}{2}[/imath] Is it possible to proceed here? | 949099 | Least value of an Expression?
Find the least value of [imath]\dfrac {3a}{b+c} + \dfrac{4b}{a+c}+ \dfrac{5c}{a+b}[/imath] for positive [imath]a, b, c[/imath]. I tried using the Cauchy-Schwarz inequality, but could not proceed after a bad equation which couldn't be further solved . |
2450074 | Perturbation theory and approximation - help!
I am having a bit of difficulties with expanding perturbation series. I need to find a first and second order perturbative approximation to the root of [imath]1+(x^2+\epsilon)^{1/2}=e^x[/imath]. I first tried the following: [imath]1+(x^2+\epsilon)^{1/2}=e^x[/imath] Set [imath]\epsilon = 0[/imath] [imath]1+(x^2)^{1/2}=e^x[/imath] [imath]1+x=e^x[/imath] [imath]x=0[/imath] But the derivative of [imath]1+x-e^x[/imath] evaluated at [imath]x=0[/imath] is [imath]1-e^0=0[/imath], so it is singular, and we can't use Taylor expansion. So, I tried, [imath]1+(x^2+\epsilon)^{1/2}=e^x[/imath] [imath](x^2+\epsilon)^{1/2}=e^x-1[/imath] [imath]x^2+\epsilon=(e^x-1)^2[/imath] [imath]x^2=(e^x-1)^2-\epsilon[/imath] Let [imath]y=e^x-1[/imath] [imath]e^x=y+1[/imath] So, [imath]x=ln(y+1)[/imath] Plugging it back to the above equation, [imath](ln(y+1))^2=y^2-\epsilon[/imath] Then I think I need to use the method of dominant balance and then go back to the very first equation written in terms of [imath]x[/imath] and [imath]\epsilon[/imath], but I am not quite sure how or if this is correct. Can someone please help? Thanks! | 2423383 | Find a two-term expansion for the root of 1+\sqrt(x^2+\epsilon)=e^x.
I am trying to find a two term expansion for the root of [imath]1+\sqrt{(x^2+\epsilon)}=e^x[/imath]. Since [imath]\epsilon << 1[/imath], I can tell that this equation behaves like [imath]1+x=e^x[/imath] which has a root close to zero. That gave me the idea that the root might look like [imath]x \approx x_0 + \epsilon^{\alpha} x_1 + \dots [/imath]. I used Taylor expansion for both functions [imath]\sqrt{(x^2+\epsilon)}[/imath] and [imath]e^x[/imath] and then plugged my guess into the Taylor expanded equation and balance of O(1) terms gave me [imath]x_0=0[/imath] as expected. My problem begins now. For the higher orders of [imath]\epsilon[/imath]. Long story short this expansion ends up with some inconsistencies. So I thought maybe [imath]x \approx x_0+\mu(\epsilon)[/imath]. But then again I get nowhere. I am getting frustrated and I would really appreciate some help. Thank you in advance. |
584228 | Exact probability of random graph being connected
The problem: I'm trying to find the probability of a random undirected graph being connected. I'm using the model [imath]G(n,p)[/imath], where there are at most [imath]n(n-1) \over 2[/imath] edges (no self-loops or duplicate edges) and each edge has a probability [imath]p[/imath] of existing. I found a simple formula online where [imath]f(n)[/imath] is the probability of [imath]G(n,p)[/imath] being connected. But apparently it's too trivial for the writer to explain the formula (it was just stated briefly). The desired formula: [imath]f(n) = 1-\sum\limits_{i=1}^{n-1}f(i){n-1 \choose i-1}(1-p)^{i(n-i)}[/imath] My method is: Consider any vertex [imath]v[/imath]. Then there's a probability [imath]{n-1 \choose i}p^i(1-p)^{n-1-i}[/imath] that it will have [imath]i[/imath] neighbours. After connecting [imath]v[/imath] to these [imath]i[/imath] neighbours, we contract them ([imath]v[/imath] and its neighbours) into a single connected component, so we are left with the problem of [imath]n-i[/imath] vertices (the connected component plus [imath]n-i-1[/imath] other "normal" vertices. Except that now the probability of the vertex representing connected component being connected to any other vertex is [imath]1-(1-p)^i[/imath]. So I introduced another parameter [imath]s[/imath] into the formula, giving us: [imath]g(n,s)=\sum\limits_{i=1}^{n-1}g(n-i,i){n-1 \choose i}q^i(1-q)^{n-1-i}[/imath], where [imath]q=1-(1-p)^s[/imath]. Then [imath]f(n)=g(n,1)[/imath]. But this is nowhere as simple as the mentioned formula, as it has an additional parameter... Can someone explain how the formula [imath]f(n)[/imath] is obtained? Thanks! | 2809016 | What is the probability of a random graph being connected?
Suppose [imath]G[/imath] is a random simple graph, that has [imath]n[/imath] vertices, and edges, that are present independently with probability [imath]p[/imath]. What is the probability of [imath]G[/imath] being connected? It is quite easy to calculate the probability for small [imath]n[/imath] (as for small [imath]n[/imath] we can classify all connected graphs with [imath]n[/imath] vertices). Thus for [imath]n = 1[/imath] the probability is [imath]1[/imath], for [imath]n = 2[/imath] it is [imath]p[/imath], and for [imath]n = 3[/imath] it is [imath]3p^2 - 2p^3[/imath]. However, I do not know, how to calculate it for arbitrary [imath]n[/imath]. Any help will be appreciated. |
2456137 | Show a group is solvable iff quotient by its center is solvable
Let [imath]G[/imath] be a group. Prove that [imath]G[/imath] is solvable [imath]\iff[/imath] [imath]G'[/imath] is solvable [imath]\iff[/imath] [imath]G/Z(G)[/imath] is solvable. [imath]G'[/imath] is the commutator subgroup. I've figured out the first of the stated equivalences but don't really know where to start on the second. | 39814 | For [imath]G[/imath] a group and [imath]H\unlhd G[/imath], then [imath]G[/imath] is solvable iff [imath]H[/imath] and [imath]G/H[/imath] are solvable?
I recently read the well known theorem that for a group [imath]G[/imath] and [imath]H[/imath] a normal subgroup of [imath]G[/imath], then [imath]G[/imath] is solvable if and only if [imath]H[/imath] and [imath]G/H[/imath] are solvable. In my book, only the fact that [imath]G[/imath] is solvable implies [imath]H[/imath] is solvable was proven. I was able to show that if [imath]H[/imath] and [imath]G/H[/imath] are solvable, then so is [imath]G[/imath], but I can't quite show that [imath]G[/imath] is solvable implies [imath]G/H[/imath] is solvable. My idea was this. Since [imath]G[/imath] is solvable, there exists a normal abelian tower [imath] G=G_0\supset G_1\supset\cdots\supset G_r=\{e\}. [/imath] I let [imath]K_i=G_i/(H\cap G_i)[/imath], in hopes of getting a sequence [imath] G/H\supset K_1\supset\cdots\supset K_r=\{e\}. [/imath] My hunch is that the above is also a normal abelian tower. However, I'm having trouble verifying that [imath]K_{i+1}\unlhd K_i[/imath] and that [imath]K_i/K_{i+1}[/imath] is abelian. Writing [imath]H_i=H\cap G_i[/imath], I take [imath]gH_{i+1}\in K_{i+1}[/imath] for some [imath]g\in G_{i+1}[/imath]. If [imath]g'H_i\in K_i[/imath], then I want to show [imath]g'H_igH_{i+1}g'^{-1}H_i[/imath] is still in [imath]K_{i+1}[/imath], but manipulating the cosets threw me off. I also tried to use either the second or third isomorphism theorems to show that [imath]K_i/K_{i+1}[/imath] is abelian, but I'm not clear on how to apply it exactly. I'd be grateful to see how this result comes through. Thank you. |
2456155 | Any example of an ideal of a ring which is not commutative
An ideal [imath]I[/imath] of a ring [imath]R[/imath] is often defined as follows: A subset [imath]I \subseteq R[/imath] is called an ideal if (1) [imath](I,+) \leq (R,+)[/imath] (2) [imath]IR \subseteq I[/imath] (3) [imath]RI \subseteq I[/imath] Here a ring does not necessarily contain [imath]1[/imath]. I know then every ideal is a subring, but I would also like to know if [imath]R[/imath] is always commutative or is there a ring which is not commutative but has a non-trivial ideal. | 1042144 | Example of non-commutative ring with exactly 2014 two sided-proper ideals.
Find a non-commutative ring with exactly 2014 two sided-proper ideals. Find a ring with exactly 2014 pairwise non-isomorphic irreducible modules. If it was the commutative ring i would have thought of the example [imath]Z/2^{2014}Z[/imath]. But what to do for non-commutative case? What is the strategy for constructing such examples? Will the same example work for having pairwise non-isomorphic irreducible modules?? |
2456228 | Solve this limit
Given that [imath]f(5x)=8x-f(3x)-\sin^2(2x),[/imath] find [imath]\lim _{ x\rightarrow 0 } \frac { f\left( x \right) -x }{ x^2 } =?\quad [/imath] Instead of putting f(x) I put f(5x) now i think it would be easier ? | 2455880 | What is [imath]\lim _{ x\rightarrow 0 } \frac { f(x) -x }{ x^2 }[/imath]?
Given that [imath]f(x)=8x-f(3x)-\sin^2(2x),[/imath] find [imath]\lim _{ x\rightarrow 0 } \frac { f\left( x \right) -x }{ x^2 }[/imath] |
2455932 | Counter-example: any algebraic extension is finite
I just proved that any finite extension of fields is an algebraic extension. That is, given fields [imath]F,K[/imath] such that [imath]K \subseteq F[/imath] is a subfield then if we can view [imath]F[/imath] as a vector space over [imath]K[/imath] with finite base, then any element of [imath]F[/imath] is the root of a polynomial with coefficients in [imath]K[/imath]. I was told that the converse is not true. Do you know any example where an extension is algebraic but not finite? | 552220 | Proving that [imath]\mathbb{A}=\{\alpha \in \mathbb{C}: \alpha \text{ is algebraic over } \mathbb{Q} \}[/imath] is not a finite extension
It is true that all finite field extensions are algebraic. It is not true however that all algebraic extensions are finite. In lectures we were given the example of the field extension [imath]\mathbb{A}=\{\alpha \in \mathbb{C}: \alpha \text{ is algebraic over } \mathbb{Q} \}[/imath] I want to prove to myself that this is indeed a non-finite algebraic extension. I have shown that [imath]\mathbb{A}[/imath] is a subfield of [imath]\mathbb{C}[/imath]. Now I consider roots of the equation [imath]x^n-2[/imath] for [imath]n \geq 1[/imath]. Clearly such roots are in [imath]\mathbb{A}[/imath] by definition. I think that if I show that for a given [imath]k[/imath], the root of [imath]x^k-2[/imath] (say [imath]\psi_k[/imath]) does not lie in [imath]\mathbb{Q}(\psi_1,...,\psi_{k-1})[/imath], then I have proven the extension is not finitie (since this will hold for all [imath]n[/imath]). Can someone please confirm this reasoning and suggest how the proof could be finished. |
2455440 | How many square numbers exist that have the length [imath]N[/imath] in the decimal system?
How many square numbers exist that have the length [imath]N[/imath] in the decimal system? E.g. for the length [imath]N=1[/imath] there exist 4 square numbers (0, 1, 4, 9). Thank you | 2437132 | What do you call the grouping of numbers based on the cardinality of digits in the square they produce?
I am trying to group numbers based on the count of digits in the square they produce, and below is the list containing the grouping that I've done so far. 1 - 3 // Numbers producing 1 digit when squared ... 4 - 9 // Numbers producing 2 digits ... 10 - 31 // Numbers producing 3 digits ... 32 - 99 // and so on ... 100 - 316 317 - 999 1000 - 3162 3163 - 9999 10000 - 31622 31623 - 99999 100000 - 316227 316228 - 999999 As one can observe, every range that produces odd digit count has a lower bound of [imath]10^{i-1}[/imath], while its upper bound has an observable pattern. I want to know if this can be represented by a formula or if the upper bound is a known constant such that we can use it as [imath]C \cdot 10^{i-1}[/imath]. |
2457412 | Be [imath]G[/imath] a group of order [imath]35,[/imath] Prove at least [imath]1[/imath] Element lies not in a subgroup of order [imath]7[/imath]
I am an exchange student here and have difficulties solving the first question of my group theory course tomorrow. The Question is: Let [imath]G[/imath] be a group of order [imath]35.[/imath] Prove that at least one element of [imath]G[/imath] does not lie in any subgroup of order [imath]7.[/imath] Hint: If [imath]H[/imath] and [imath]K[/imath] are distinct subgroups of [imath]G[/imath] such that both have order [imath]7,[/imath] what can you say about the order of the intersection of [imath]H[/imath] and [imath]K?[/imath] The only theorem we have learned is Lagranges Theorem, but I am not sure where it can help me. I have no idea how I should prove this, so please help me:( | 1761228 | Order of [imath]G[/imath] is 35, are there elements of order 5 and 7 in G?
I have a group of order 35 and I want to know if it contains elements of order 7 and 5. I know that it does and there is a proof that is much longer, but I wanted to know if the following worked to show that it does contain elements of order 5 and 7. My Approach: Let [imath]|G| = 35[/imath]. We know that any group of order 35 is cyclic so [imath]G[/imath] is isomorphic to [imath]C_7[/imath] x [imath]C_5[/imath] and [imath]C_7[/imath] contains an element of order 7 and [imath]C_5[/imath] contains an element of order 5. So, there are elements of order 5 and order 7 in [imath]G[/imath]. |
2457560 | Convergence of the real infinite series [imath]\sum \frac{n!e^{n}}{n^{n}}[/imath]
I am trying to check the convergence of the real infinite series [imath]\sum \frac{n!e^{n}}{n^{n}}[/imath]. My problem is i tried for it ratio and root test both but both failed as in both case limit comes to [imath]1[/imath]. Please give me a hint for its convergence.Thaks a lot. | 2162341 | Convergence of the series [imath]\sum_{n=1}^{\infty}\frac{e^n\,n!}{n^n}[/imath]
Is the following series convergent? [imath]\sum_{n=1}^{\infty}\frac{e^n\,n!}{n^n}[/imath] I treid the Ratio and Root tests, but both of them failed. |
2455392 | General solution of a wave equation
Consider the wave equation [imath]\frac{\partial^2u}{\partial t^2}-c^2\frac{\partial^2u}{\partial x^2}=0\tag{1}[/imath] which can be written as [imath]\Big(\frac{\partial}{\partial t}-c\frac{\partial}{\partial x}\Big)\Big(\frac{\partial}{\partial t}+c\frac{\partial}{\partial x}\Big)u=0\tag{2}.[/imath] The question here claims that (2) implies both [imath]\Big(\frac{\partial}{\partial t}+c\frac{\partial}{\partial x}\Big)u=0[/imath] as well as [imath]\Big(\frac{\partial}{\partial t}-c\frac{\partial}{\partial x}\Big)u=0[/imath]. I do not get this. Eq. (2) can be satisfied if [imath]\Big(\frac{\partial}{\partial t}+c\frac{\partial}{\partial x}\Big)u=v\neq 0[/imath] and [imath]\Big(\frac{\partial}{\partial t}+c\frac{\partial}{\partial x}\Big)v=0.[/imath] Am I missing something? | 833044 | Question about the solution of the wave equation
The solution of the wave equation [imath]u_{tt}-c^2u_{xx}=0[/imath] is given as [imath]u(x,t)=f(x+ct)+g(x-ct)[/imath] where [imath]f,g[/imath] are arbitrary functions of one variable. One way to prove this is the following: [imath](\partial_{tt}-c^2 \partial_{xx})u=0[/imath] [imath](\partial_t-c \partial_x)(\partial_t+c \partial_x)u=0[/imath] So solutions must satisfy [imath] (\partial_t-c \partial_x)u=0 \;\;\cup\;\; (\partial_t+c \partial_x)u=0 [/imath] [imath]\displaystyle{(\partial_t-c \partial_x)u=0 \Rightarrow u=f(x+ct)}[/imath] [imath]\displaystyle{(\partial_t+c \partial_x)u=0 \Rightarrow u=g(x-ct)}[/imath] Could you explain me why the last two equations stand?? |
2457830 | Let [imath]n[/imath] be a fixed integer and let [imath] H [/imath] = {[imath]x \in G : x^n = e[/imath]}. Prove that H is a subgroup of G.
Let G be an abelian group and let [imath]n[/imath] be a fixed integer and let [imath] H [/imath] = {[imath]x \in G : x^n = e[/imath]}. Prove that H is a subgroup of G. So firstly, [imath]e \in G[/imath] since the condition is that [imath]x^n : e[/imath]. I get confused about how to go further from here. If we suppose that [imath]a, b \in H[/imath] then [imath]a^n = e[/imath] and [imath]b^n=e[/imath]. But how do I get an expression for ab from here? Sorry, I think I am confusing myself. Any guidance would be much appreciated! | 2447190 | Show that H is a subgroup of G
Let n be a positive integer and let G be an Abelian group. Define [imath]H := \{x ∈ G : |x| \ \text{divides} \ n\}[/imath] where x denotes the order of |x|. Show that H is a subgroup of G. I know I need to prove that H is non-empty, contains the identity, e, and that it contains [imath]ab^{-1}[/imath]. I'm just not sure how to prove this, thank you for any help! |
2457419 | Find [imath]\lim a_n[/imath] if [imath]a_{n+1}=a_n+\frac{a_n^2}{n^2}[/imath]
[imath]0\lt a_1\lt 1[/imath], [imath]a_{n+1}=a_n+\dfrac{a_n^2}{n^2}[/imath], Find [imath]\lim\limits_{n\to\infty} a_n[/imath] Well, [imath]a_{n+1}\gt a_n[/imath], But how to show that there is an upper bound of [imath]\{a_n\}[/imath] ? What is the limit ? Thanks a lot | 680938 | Show that the sequence [imath]a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}[/imath] is upper bounded
Let [imath]\{a_{n}\}[/imath] be defined with [imath]a_{1}\in(0,1)[/imath], and [imath]a_{n+1}=a_{n}+\dfrac{a^2_{n}}{n^2}[/imath] for all [imath]n\gt 0[/imath]. Show that the sequence is upper bounded. My idea: since [imath]a_{n+1}=a_{n}\left(1+\dfrac{a_{n}}{n^2}\right)[/imath] then [imath]\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}}-\dfrac{1}{a_{n}+n^2}[/imath] then [imath]\dfrac{1}{a_{n}}-\dfrac{1}{a_{n+1}}=\dfrac{1}{a_{n}+n^2}[/imath] so [imath]\dfrac{1}{a_{1}}-\dfrac{1}{a_{n+1}}=\sum_{i=1}^{n}\dfrac{1}{a_{i}+i^2}[/imath] since [imath]a_{n+1}>a_{n}\Longrightarrow \dfrac{1}{a_{i}+i^2}<\dfrac{1}{a_{1}+i^2}[/imath] so [imath]\dfrac{1}{a_{n+1}}>\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)[/imath] But the RHS might be [imath]\lt0[/imath] for a sufficiently large starting value; for instance, with [imath]a_{1}=\dfrac{99}{100}[/imath] then [imath]\dfrac{1}{a_{1}}-\left(\dfrac{1}{1+a_{1}}+\dfrac{1}{2^2+a_{1}}+\cdots+\dfrac{1}{a_{1}+n^2}\right)<0,n\to\infty[/imath] see: so this method won't let me bound the series and I don't know what else to do. |
2457429 | Evaluate [imath] \int \frac{(\cos^3x)}{(\sin^2x)}dx [/imath]
How do you integrate the equation [imath] \int \frac{(\cos^3x)}{(\sin^2x)}dx [/imath] using U substitution? | 1646057 | Problem with Indefinite Integral [imath]\int\frac {\cos^4x}{\sin^3x} dx[/imath]
I'm stuck with this integral [imath]\int\frac {\cos^4x}{\sin^3x} dx[/imath] which I rewrote as [imath]\int \csc^3x \cos^4xdx[/imath] then after using the half angle formula twice for [imath]\cos^4x[/imath] I got this [imath]\frac 14\int \csc^3x (1+\cos(2x))(1+\cos(2x))dx[/imath] then after solving those products I got these integrals [imath]\frac 14 \{\int \csc^3xdx+2\int \csc^3x \cos(2x)dx + \int \csc^3x \cos^2(2x)dx\}[/imath] I do know how to solve the [imath]\int \csc^3xdx[/imath] one but I'm totally lost on the other ones, any tips/help/advice would be highly appreciated! thanks in advance guys! |
2458126 | non-zero Vector space with new scalar multiplication defined over finite field still vector space?
If [imath]V[/imath] is a vector space (not the zero vector space) over [imath]\mathbb{R}[/imath], and if [imath]F[/imath] is a finite field. how could I show that it is not possible to define a new scalar multiplication of [imath]F[/imath] on [imath]V[/imath], in a way that [imath]V[/imath] with this scalar multiplication and the usual addition becomes a vector space over [imath]F[/imath]? Thanks in advance. | 2458097 | Vector space with new scalar multiplication defined over finite field still vector space?
If [imath]V[/imath] is a vector space (not the zero vector space) over [imath]\mathbb{R}[/imath], and if [imath]F[/imath] is a finite field. how could I show that it is not possible to define a new scalar multiplication of [imath]F[/imath] on [imath]V[/imath], in a way that [imath]V[/imath] with this scalar multiplication and the usual addition becomes a vector space over [imath]F[/imath]? Thanks in advance. |
2458002 | Represent Integer as sum of at least two consecutive postive Integers
For the case that a natural number [imath]N =c\cdot k[/imath] where [imath]c[/imath] is odd, we can easily represent [imath]N[/imath] as following sum [imath]N = \sum_{n = k - \frac{c-1}{2}}^{k + \frac{c-1}{2}} n[/imath] This covers all integers except for powers of two: [imath]N=2^l[/imath]. Some experimenting suggests that we cannot decompose powers of two as sum of consecutive positive integers, but is there a way to prove this, or is there a counter example? | 2164482 | Proving any number of the form [imath]2^n[/imath] can't be written as the sum of k consecutive numbers.
I think i have proved that any number of the form [imath]2^n[/imath] can't be written as the sum of k consecutive positive integers, but i would be grateful if i could have some clarification as to whether this is a correct and valid proof or not. Let [imath]g=x+(x+1)+(x+2)+(x+3)+...+(x+(k-1))[/imath], hence g is the sum of k consecutive positive integers. [imath]g=kx+\sum_{d=1}^{k-1}d[/imath] Hence, [imath]g=kx+\frac{k(k-1)}{2}[/imath] Rearranging for x we find, [imath]x=\frac{g}{k}-\frac{(k-1)}{2}[/imath] So, for certain values of [imath]k[/imath], we can be sure that a positive integer value of [imath]x[/imath] can be found, and hence [imath]g[/imath] can be written as the sum of [imath]k[/imath] consecutive positive integers.There are two cases for [imath]k[/imath] to considor: Case 1) [imath]k[/imath] is odd. If [imath]x[/imath] exists, it must be the case [imath]k[/imath] | [imath]g[/imath], as for odd values of [imath]k[/imath], [imath]\frac{(k-1)}{2}[/imath] will alwyas be an integer, and hence [imath]x[/imath] will. Case 2) [imath]k[/imath] is even. [imath]k[/imath] does not divide [imath]g[/imath], but [imath]\frac{g}{k}[/imath] must be a multiple of [imath]\frac{1}{2}[/imath], and hence equals [imath]\frac{h}{2}[/imath], for some odd [imath]h[/imath]. Letting [imath]j=2^n[/imath], assume that we can write [imath]j[/imath] as the sum of [imath]k[/imath] consecutive positive integers. If [imath]k[/imath] is odd, then if a [imath]k[/imath] exists, it must divide [imath]j[/imath], by Case 1. But, [imath]j[/imath] has no odd factors as it is a power of two, hence [imath]k[/imath] could not be odd. If [imath]k[/imath] is even, by Case 2 it mut be the case [imath]k[/imath] does not divide [imath]j[/imath]. Also, [imath]\frac{j}{k}=\frac{h}{2}[/imath], for some odd [imath]h[/imath]. But, this implies [imath]2j[/imath] has an odd factor, [imath]h[/imath]. but, as [imath]2j[/imath] is a power of [imath]2[/imath], it can't have odd factors. So, [imath]k[/imath] can't be odd. Hence, [imath]k[/imath] is neither odd or even, and this is a contradiction to our assumption. Hence no [imath]x[/imath] exists, and hence and number of the form [imath]2^n[/imath] can't be written as the sum of [imath]k[/imath] consecutive positive integers. Any feedback on this proof would be appreciated. |
2457155 | [imath]N(ab) \geq N(a)[/imath] is not necessary for the ring to be euclidean
The definition of euclidean ring in my textbook is 1) [imath]N(ab) \geq N(a)[/imath] 2) Euclidean algorithm works. However there is a note that the first condition is not necessary. There is a hint that we can choose a following norm: [imath]N'(a) = \min_{b\sim a} N(b)[/imath]. I do not understand why it works. We need to check that euclidean algorithm still works with this norm. I tried to do the following. If [imath]a \sim b[/imath], then [imath]a = bx, x \in K^*[/imath]. Let [imath]N'(a)[/imath] be [imath]N(ax)[/imath]. Let us divide [imath]ax[/imath] by [imath]bx[/imath] instead of [imath]b[/imath]. We get [imath]ax = bxq + rx[/imath], [imath]N(rx) \leq N(bx)[/imath]. Here I am stuck and have no idea how to continue. | 2564283 | [imath]R[/imath] euclidean domain, show [imath]\tilde{N}(a)=\min_{0\neq x\in\left}N(x)[/imath] is a norm
Let [imath]R[/imath] be an euclidean domain with a norm [imath]N[/imath]. We define [imath]\tilde{N}(a) = \min_{0\neq x\in\left<a\right>}N(x)[/imath]. Show that [imath]\forall a,b\neq 0\quad \tilde{N}(ab) \geq \tilde{N}(a)[/imath] [imath]\tilde{N}(a)=\min_{0\neq x\in\left<a\right>}N(x)[/imath] is a norm I proved the first claim, but I'm struggling with the second one and would be thankful for some hint. |
2458080 | Showing that [imath]\mathsf{Cov}(f(X), g(Y)) \geq 0[/imath] for non-decreasing [imath]f[/imath] and [imath]g[/imath] taking values on [imath]\mathbb{R}[/imath].
We have two non-decreasing functions [imath]f[/imath] and [imath]g[/imath] and a random variable [imath]X[/imath]. We want to show that [imath]\mathsf {Cov}(f(X),g(X))\geqslant 0[/imath]. Intuitively, it makes sense, but how do I formalize this? | 74677 | covariance of increasing functions
Suppose [imath]f[/imath] and [imath]g[/imath] are monotonically increasing, and bounded, and let [imath]X[/imath] be a random variable. I want to show [imath]f[/imath], [imath]g[/imath] have positive covariance. I tried to compute it directly but I am not getting anything useful |
1700180 | Why [imath]c(a_1 \ a_2 \dots \ a_k)c^{-1}=(c(a_1) c(a_2)… c(a_k))[/imath]?
We investigate on an arbitrary [imath]a_i[/imath] : [imath]c(a_1 \ a_2 \dots \ a_k)c^{-1}(a_i)[/imath]. First step, [imath]c(a_i)=a_k[/imath]. Second step, [imath](a_1 \ a_2 \dots \ a_k)(a_k)=a_{k+1}[/imath], Third step, [imath]c^{−1}(a_{k+1})=? [/imath]. Any answer that I read in MSE was not helpful to understand. In the final step, they all imply [imath]c^{−1}(a_{k+1})=c(a_i)[/imath], but why [imath]c^{−1}(a_{k+1})=a_k=(a_1 \ a_2 \dots \ a_k)^{−1}(a_{k+1})[/imath]? which may imply [imath]c=(a_1 \ a_2 \dots \ a_k)[/imath] ! I would appreciate any simple clear detailed explanation. | 844279 | Conjugating a permutation
I am trying to see that two permutations are conjugate exactly when they have the same cycle decomposition. I fail to see that [imath]r(i_1,i_2,\dots,i_k)r^{−1}=(r(i_1),r(i_2),\dots,r(i_k))[/imath] for permutations [imath]r[/imath] in a set [imath]X\ni \{i_p\}^k_0[/imath] Any thoughts would be appreciated. p.s. I ask this question in a new post as suggested by Mr. Gerry Myerson. Still, I believe it belongs to the above thread. |
2458726 | [imath]C_0[/imath] is a closed subespace of [imath]l_\infty[/imath]?
Could anyone help me to prove is this statement is true or false? Thanks. | 680806 | [imath]c_0[/imath] is a closed subspace of [imath]l^{\infty}[/imath]
Put [imath] l^{\infty} = \{ (x_n) \subseteq \mathbb{C} : \forall j \; \;\ \;|x_j| \leq C(x)\} [/imath] I want to show that [imath]c_0[/imath], the space of all sequences of scalars that converges to [imath]0[/imath] is closed subspace of [imath]l^{\infty}[/imath]. MY atttemt: Take arbitrary sequence [imath](x_n) \in c_0[/imath]. We know [imath]x_n \to 0[/imath]. but [imath]0 \in c_0[/imath]. Therefore, every sequence in [imath]c_0[/imath] converges to an element in [imath]c_0[/imath]. Hence, [imath]c_0[/imath] must be closed Is this correct? thanks |
2458107 | Solving non linear ODE for y , x and y'
In a Math lecture , we solved the following DE [imath]yy'^2+2xy'-y=0[/imath] in 3 ways : First we solved for y' by taking y' in one side , making a substitution and integrating. We solved the DE again for x and one more time for y as following : We solved the DE for y , by taking y in one side and called y' as p , then we differentiated both sides w.r.t x to get finally a separable DE in p and x that can be solved easily. And we solved the DE for x by taking x in one side and called y' as p , then we differentiated both side w.r.t y to get finally a separable DE in p and y that can be solved easily. My problem is that I do not understand why can we need to solve a DE for x and y ?? solving the DE for y' gave us the solution we need . What do the solutions gotten from solving for x and y represent ?? and why do we differentiate w.r.t x when solving for y and differentiate w.r.t. y when solving for x ? ..Is this a standard method to solve non linear DE?? If you can provide me a reference name that introduces this method of solution ,I will be grateful! | 2016045 | How to solve the differential equation: [imath]yy'^2-2xy'+y=0[/imath]
Solve the differential equation: [imath]yy'^2-2xy'+y=0[/imath] (*) I have [imath](*)\Leftrightarrow (yy')^2-2xyy'+y^2=0\\ \Leftrightarrow z'^2-4xz+4z^2 =0 \ \ (z=y^2)[/imath] I can't know how to continue. Could you give me some hints. Thanks for helping. |
2458444 | Extremizing sum/difference of lengths using triangle inequality
We have three points [imath]P=(2,3)[/imath], [imath]Q=(4,-2)[/imath] and [imath]R=(\alpha,0)[/imath]. We need to find the value of [imath]\alpha[/imath] such that: Part 1: [imath]|PR+QR|[/imath] is maximized. Part 2: [imath]|PR-QR|[/imath] is minimized. I was given the solution for both. I have understood the solution for case 1 but not for case 2. Case 1's solution : https://imgur.com/a/ymUvM Case 2's Solution: https://imgur.com/a/lLK9k I have understood case 1 completely but in case 2 why was there a need to take the image of point Q? I fail to understand this. | 2458534 | Extremizing sum and difference of lengths using triangle inequality
We have three points [imath]P=(2,3)[/imath], [imath]Q=(4,-2)[/imath] and [imath]R=(\alpha,0)[/imath]. We need to find the value of [imath]\alpha[/imath] such that: Part 1: [imath]|PR+QR|[/imath] is maximized. Part 2: [imath]|PR-QR|[/imath] is minimized. I have the solution for both. I have understood the solution for case 1 but not for case 2. Case 1's solution : https://imgur.com/a/ymUvM Case 2's Solution: https://imgur.com/a/lLK9k I have understood case 1 completely but in case 2 why was there a need to take the image of point Q? I fail to understand this. |
2458603 | Rates Of Changes Calculus 1
The cost, in dollars, of producing [imath]x[/imath] units of a certain commodity is [imath]c(x)=920 + 2x - 0.02x^2 + 0.00007x^3[/imath] (a) find the marginal cost function. (b) find [imath]C'(100)[/imath]and explain its meaning (c) Compare [imath]C'(100)[/imath] with the cost of producing the [imath]101[/imath]st item | 814189 | Marginal cost function
Consider that the total cost to produce a [imath]x[/imath] units of a product is given by the function [imath]C(x) = 2000 + 3x + 0,01 x^2 + 0,0002 x^3[/imath] (a) Calculate the marginal cost at the production level [imath]x=100[/imath] (b) Calculate the cost to produce the product of number 101. Compare this result with the one obtained at item (a) I calculated the marginal as [imath]C'(100)[/imath] but I don't know how to obtain the cost of the product number 101. Can I have some help? thanks in advance! |
2459186 | How to use combinatorial proof instead of induction?
Let k ≥ 1 be an integer and consider a sequence [imath]n_1, n_2, . . . , n_k[/imath] of positive integers. Use a combinatorial proof to show that [imath]\binom{n_1}{2}[/imath] + [imath]\binom{n_2}{2}[/imath] +...+ [imath]\binom{n_k}{2}[/imath] [imath]\le[/imath] [imath]\binom{n_1 + n_2 + ...+ n_k}{2}[/imath] | 2451923 | How to relate combinatorial proof to bipartite graphs?
The question is - Let [imath]k \geq 1[/imath] be an integer and consider a sequence [imath]n_1,n_2,\cdots,n_k[/imath] of positive integers. Use a combinatorial proof to show that [imath]\binom{n_1}{2} + \binom{n_2}{2} +\cdots+ \binom{n_k}{2} \leq \binom{n_1+n_2+\cdots+n_k}{2}[/imath] How can i prove this without using induction? Additionally, For each [imath]i[/imath] with [imath]1 \leq i \leq k[/imath],consider the complete graph on [imath]n_i[/imath] vertices. How many edges does this graph have? I thought of making graphs and figuring out a relation but still wasnt able to. |
2459106 | Does this intermediate serie converge?
Does [imath]\sum_{n≥2}\frac{1}{n (\ln(\ln n))^a}[/imath] converge if a>1 ? I already know that [imath]\sum_{n≥2}\frac{1}{n (\ln n)^a}[/imath] and [imath]\sum_{n≥2}\frac{1}{n \ln n (\ln(\ln n))^a}[/imath] converge if a > 1. | 1825008 | Testing convergence of series [imath]\sum_{n=3}^\infty\frac{1}{n (\ln(n))^p(\ln\ln(n))^q}[/imath]
Lets have this problem. [imath]\sum_{n=3}^\infty\frac{1}{n (\ln(n))^p(\ln\ln(n))^q}[/imath] I have rewritten this to a form [imath]\sum\frac{1}{np'^{\ln\ln(n)}q'^{\ln\ln\ln(n)}}[/imath] For [imath]p,q\in\mathbb{R}[/imath]. Obviously, [imath]p',q'>0, p=\ln(p'),q=\ln(q')[/imath] |
2459355 | simply connected subset of [imath]\mathbb R^2[/imath] contractible?
I know that every simply connected open subset of [imath]\mathbb R^2[/imath] is homeomorphic with [imath]\mathbb R^2[/imath] . My question is : Is every simply connected subset of [imath]\mathbb R^2[/imath] contractible ? Remark : By simply connected I mean path connected with trivial fundamental group . | 548681 | Simply connected does not imply contractible. Is there a nice counter example in [imath]R^2[/imath]?
The standard counter example to the claim that a simply connected space might be contractible is a sphere [imath]S^n[/imath], with [imath]n > 1[/imath], which is simply connected but not contractible. Suppose that I were interested in a counter example in the plane - does anyone know of a subset of [imath]R^2[/imath] which is simply connected but not contractible? |
2459423 | The divisibility of the product of Pythagorean Theorem by 5
We know that there is a fundamental theorem that is: [imath]a^2+b^2=c^2[/imath] Then it is given that the product of a,b and c are divisible by 5 where [imath]a,b,c\in \mathbb{Z}[/imath]. Using numbers, I know for certain that this is in fact true. But by using [imath]\mod{5}[/imath], I am not certain how I can approach this question. Help is greatly appreciated. | 2457382 | Prove that if [imath]a^2 + b^2 = c^2[/imath], then 5 divides abc. a,b,c are positive integers.
My question is about proving that if [imath]a^2 + b^2 = c^2,[/imath] then 5 divides abc. a,b,c are positive integers. |
2459780 | Why do we divide by [imath]\sqrt{N}[/imath] to find the standard deviation?
The standard deviation of a population [imath][a_1, a_2, \dots a_N][/imath] is defined as follows: [imath] \sigma = \sqrt{\frac{\Sigma_{i=1}^{N} (a_i - \mu)^2}{N}} = \frac{\sqrt{\Sigma_{i=1}^{N} (a_i - \mu)^2}}{\sqrt{N}} [/imath] The numerator is the (Euclidean) length of the following vector: [imath] [a_1 - \mu, a_2 - \mu, \dots, a_n-\mu] [/imath] We may imagine that this vector lives in some [imath]N[/imath]-dimensional configuration space which describes the dispersion of the population. As we go into higher dimensions, this vector will naturally get longer, because we are adding more coordinates without changing the values of the existing coordinates. So clearly we have to divide by some scaling factor to normalize this between different population sizes. Why is that scaling factor the square root of [imath]N[/imath] instead of some other expression? Does it have anything to do with the [imath]\sqrt{m}[/imath] that appears in this problem? | 1418801 | Correct Intuition? Standard Deviation and distance in [imath]n[/imath] dimensional space.
Basic Question Is there an intuitive explanation of standard deviation in terms of Euclidean distance in [imath]n[/imath] dimensional space? Longer Version of Question To begin a more detailed sketch of my question, for simplicity let's just focus on the simple case of a discrete random variable that is uniformly distributed. In this case, the variance is given by the following formula, which I've abducted straight from Wikipedia: [imath] \frac1{n}\sum_{i =1}^n (x_i - \mu)^2[/imath] where [imath]\mu[/imath] is the mean. The standard deviation is then the square root of this. Now, I can't help noticing that the square root of the sum returns the euclidean distance from the vector [imath]X = (x_1, x_2, \dots, x_n)[/imath] to the vector [imath]\vec \mu = (\mu, \mu, \dots, \mu)[/imath]. That is, the standard deviation can be expressed as: [imath] \frac1{\sqrt{n}}|X - \vec \mu |[/imath] So I wonder, is there any significant conceptual relationship between this distance [imath]|X - \vec \mu |[/imath] and standard devation or is this just a coincidence? Even More Details... I have looked up many explanations of standard deviation and its cousin variance. Here are some that I've seen already, each sort of following from the previous one: We square the values before summing to get rid of the sign, which is obviously not important. This explanation is often criticised by hardcore statisticians and I can sort of see why: it doesn't explain why squaring beats taking the absolute value. We square the values so that we pay a greater price for greater deviations. This explains why squaring beats taking absolute values. But why not raise to the power of [imath]4[/imath], or [imath]6[/imath], or any other even power before summing? What is so special about [imath]2[/imath]? The thing that is so special about [imath]2[/imath] is that it's the second moment of intertia, whereas the mean is the first moment, so mechanically it makes sense. I don't follow this. My intuition is totally OK with the mean: the point where, if I put my finger, the weights on either side will balance. But the second moment is harder for me to imagine physically like this. Note, this is a question about intuition. I "understand" the mathematical formula at a shallow level: what all its terms mean, how to calculate it given a dataset. But I am not comfortable with my grasp on why this formula is "the best" one to use in so many applications e.g. the least squares method to fit data. I'm particularly confused as to why squaring is chosen as opposed to raising to some other even power e.g. [imath]9234324[/imath]. And this is where my intuition steps in and tries to provide an explanation that goes right back to the fundamental theorem of Pythagoras: euclidean distance. Here is my thought process: "The number [imath]2[/imath] is special. It's the unique power that makes Euclidean distance work. So maybe it's also the unique number that makes variance work." But then why the multiplying factor of [imath]\frac1{\sqrt{n}}[/imath]? Is it just simply a case of: swallow it up and accept the definition, or can this intuition be resolved somehow? |
2459200 | What is the Harish Chandra function?
I'm struggling to find a definition of the Harish Chandra function (usually denoted [imath]\Xi[/imath]) online. Could you tell me what it is? There is a definition on wikipedia that I asked about in this post, but I don't understand the notation used there. what does [imath]\cdot^\rho[/imath] mean in the Harish-Chandra function? | 2458390 | what does [imath]\cdot^\rho[/imath] mean in the Harish-Chandra function?
Wikipedia defines the Harish Chandra function on a semisimple Lie group to be [imath]g\mapsto \int_Ka(kg)^\rho dk[/imath] where [imath]K[/imath] is a maximal compact subgroup, [imath]a(g)[/imath] is the [imath]A[/imath] factor in the Iwasawa decomposition of [imath]g[/imath] and [imath]\rho[/imath] is the Weyl vector (half the sum of the positive roots). What does it mean to put that vector in the superscript? https://en.wikipedia.org/wiki/Harish-Chandra%27s_%CE%9E_function |
2458881 | Number of strings that start and end with the same letter or are palindromes or contain vowels only
In this question, we consider strings consisting of 26 characters, with each character being a uppercase letter. Determine the number of such strings that (a) start and end with the same letter, or (b) are palindromes, or (c) contain vowels only For this problem, I approached each scenario a, b, c separately. I then joined the 3 with an OR to satisfy the question in the final answer. a. [imath]26[/imath] characters, if the last letter must be the same as the first, we really only have [imath]25[/imath] characters to choose from, so part a = [imath]26^ {25}[/imath] b. [imath]26[/imath] characters, but if it is a palindrome, the length is cut in half, giving us [imath]26^{13}[/imath]. c. If the characters are vowels only, there are only A E I O U, [imath]5[/imath] characters to choose from. And we have a full 26 characters to pick. [imath]26^{5}[/imath] The three value then gets OR'd together [imath]26^ {25}[/imath] OR [imath]26^{13}[/imath] OR [imath]26^{5}[/imath] Is what I did correct? I'm working from practice problems and am not given an answer key so it's hard to really practice when I don't know if I'm doing it right or wrong. Thank you! | 2456599 | Counting the number of strings that start and end with the same letter, are palindromes, or contain vowels only
I have a question about this problem that I am trying to figure out: A string of letters is called a palindrome, if reading the string from left to right gives the same result as reading the string from right to left. For example, madam and racecar are palindromes. Recall that there are five vowels in the English alphabet: a, e, i, o, and u. In this question, we consider strings consisting of 28 characters, with each character being a lowercase letter. Determine the number of such strings that (i) start and end with the same letter, or (ii) are palindromes, or (iii) contain vowels only. My answer so far: [imath]A = 26^{27}[/imath], [imath]B = 26^{14}[/imath], [imath]C = 5^{28}[/imath] My question is, would I have to use a sum rule here or inclusion-exclusion rule? I am not sure about which method to use here. Any help would be appreciated. |
2105963 | How to prove that [imath]\sup \left( {{A_1} \cup {A_2}} \right) = \max \left( {\sup {A_1},\sup {A_2}} \right)[/imath]?
Suppose [imath]{{A_1}}[/imath]=[1,3] and [imath]{{A_2}}[/imath]=[2,4], then [imath]{{A_1} \cup {A_2}}[/imath]=[1,4] now [imath]\sup \left( {{A_1} \cup {A_2}} \right)[/imath] is clearly 4. so, [imath]\sup \left( {{A_1} \cup {A_2}} \right) = \max \left( {\sup {A_1},\sup {A_2}} \right)[/imath] is true. Confusion with definition: s is least upper bound for a set [imath]A \subseteq R[/imath] if two criterion are met (1) s is an upper bound for A (2) if b is any upper bound for A, then [imath]s \le b[/imath] In the proof if I take [imath]{s_1}[/imath] to be [imath]{\sup {A_1}}[/imath] and [imath]{s_2}[/imath] to be [imath]{\sup {A_2}}[/imath], then if I apply definition then least of [imath]{s_1}[/imath] and [imath]{s_2}[/imath] is [imath]\sup \left( {{A_1} \cup {A_2}} \right)[/imath], which is certainly not true. What is exactly am I missing here? Then once proved how can I extend it to [imath]\sup \left( { \cup _{k = 1}^n{A_k}} \right)[/imath] ? May be if i get clear with the base case then it will not be required. Edit: [imath]{{A_1}}[/imath] and [imath]{{A_2}}[/imath] are nonempty sets which are bounded above. | 921975 | Let [imath]A,B \subset \mathbb{R}[/imath]. Show that [imath]\sup(A \cup B) = \max\{\sup A, \sup B\}[/imath]
Let [imath]A,B \subset \mathbb{R}[/imath]. Show that [imath]\sup(A \cup B) = \max\{\sup A, \sup B\}[/imath] Here's what I did so far: Let [imath]\sup A, \sup B[/imath], and [imath]\sup (A \cup B)[/imath] denote upper bounds of [imath]A,B[/imath], and [imath]A \cup B[/imath], respectively. If [imath]x \in A[/imath], then [imath]x \le \sup A[/imath]. If [imath]x \in B[/imath], then [imath]x \le \sup B[/imath]. In either case, [imath]x \in (A \cup B)[/imath]. So either [imath]x \le \sup A[/imath] or [imath]x \le \sup B[/imath]. In either case, [imath]x \le \max \{\sup A, \sup B\}[/imath]. So [imath]\max \{\sup A, \sup B\}[/imath] is the upper bound of [imath]A \cup B[/imath], which means [imath]\sup (A \cup B) \le \max\{\sup A, \sup B\}[/imath]. I do not know if there's a better way to go at it, but all I did was show the inequality. Can I do this, and show the reverse inequality? If so, how can I go about doing that? |
2460193 | Show that for any constants [imath]k[/imath] and [imath]j[/imath], where [imath]j > 0[/imath], [imath](n+k)^j = \Theta(n^j)[/imath]
Does this solve the question: There exist [imath]c, n_0[/imath], such that [imath](n+k)^j = n^j[/imath] for all [imath]n>=n_0[/imath] I tired the above but couldn't find an answer.. | 1461909 | Show that for any real constants a and b, where b > 0, (n + a) b = Θ(n b ).
Show that for any real constants a and b, where b > 0, [imath](n + a)^b = Θ(n^b)[/imath]. So taking a stab at this.. I let [imath]a = 0[/imath], and [imath]b = 1[/imath] So I got [imath]f(n) = (n + a)^b \implies f(n) = (n+0)^1[/imath] which we know [imath]\implies n^1[/imath] or [imath]O(n) \implies \Theta(n^1)[/imath] ? I just took a guess. I'm not sure if that's even close to being right. |
2460211 | The closure of [imath](a;b)[/imath] in an order topology
Consider a linearly ordered set [imath](X; \prec)[/imath] with its order topology. Show that closure of [imath](a;b)[/imath] is a subset of [imath][a;b][/imath]. Under what conditions does equality hold? Give an example of a strict inclusion. I solves part 1: [imath][a,b]=X \setminus((−\infty,a)∪(b,+\infty))[/imath] is closed and contains [imath](a,b)[/imath] , so it contains the closure of [imath](a,b)[/imath]. Furthermore, it equals the closure iff both endpoints are limit points of the interval. But I cannot find example for strict inclusion. | 697912 | Topology - closures of intervals with the order topology
a question from my h.w.: Let [imath](X,<)[/imath] be a totally ordered set. Lets examine [imath]X[/imath] with the order topology. a. Prove that [imath]\overline{(a,b)} \subseteq [a,b][/imath] (where the overline denotes closure of the set). Give an example where the inclusion is strict. b. Give necessary and sufficient conditions such that [imath]a\in\overline{(a,b)}[/imath] and such that [imath]b\in\overline{(a,b)}[/imath] So I've managed a. easily enough. I think a proper example of strict inclusion would be a finite set - say [imath]\{1, 2, 3, 4, 5\}[/imath] with the obvious order. Then [imath](2,4)=\{3\}[/imath] so indeed [imath] \overline{(2,4)} = \{3\}[/imath] and not [imath][2,4][/imath]. I'm stumped about the conditions though. Can anyone give me some help? Thanks! |
2458576 | Finding the derived group
I need help to answer the folowing problem: Let [imath]F[/imath] be a field, [imath]n\ge 1[/imath] and [imath]G=F^n\bigoplus F^n\bigoplus Mat(n,F).[/imath] That is an element of [imath]G[/imath] has the form [imath](v,w,X)[/imath] where [imath]v[/imath] and [imath]w[/imath] are n-by-1 vectors and [imath]X[/imath] is [imath]n[/imath]-by[imath]-n[/imath] matrix, all with entires in [imath]F[/imath]. Define an operator on [imath]G[/imath] by: [imath](v_1,w_1,X_1)(v_2,w_2,X_2)=(v_1+v_2,w_1+w_2,X_1+X_2+v_1w_2^t)[/imath] Show that [imath]G'=\left\{(0,0,X), X\in Mat(n,F)\right\}[/imath]. Where [imath]G'[/imath] is the derived group of [imath]G[/imath] i.e [imath]G′=\left\{[a,b],a,b∈G\right\}[/imath] | 2457465 | Proving an inclusion for the derived group
I need help to answer the folowing problem: Let [imath]F[/imath] be a field, [imath]n\ge 1[/imath] and [imath]G=F^n\bigoplus F^n\bigoplus Mat(n,F).[/imath] That is an element of [imath]G[/imath] has the form [imath](v,w,X)[/imath] where [imath]v[/imath] and [imath]w[/imath] are n-by-1 vectors and [imath]X[/imath] is [imath]n[/imath]-by[imath]-n[/imath] matrix, all with entires in [imath]F[/imath]. Define an operator on [imath]G[/imath] by: [imath](v_1,w_1,X_1)(v_2,w_2,X_2)=(v_1+v_2,w_1+w_2,X_1+X_2+v_1w_2^t)[/imath] Show that [imath]G'=\left\{(0,0,X), X\in Mat(n,F)\right\}[/imath]. Where [imath]G'[/imath] is the derived group of [imath]G[/imath] i.e [imath]G′=\left\{[a,b],a,b∈G\right\}[/imath] I poved the following inclusion: we need to compute the commutator [imath][(v_1, w_1, X_1), (v_2, w_2, X_2)][/imath]. Before jumping into the computation it's clearer to first write out the inverses. We have: [imath](v_1, w_1, X_1)^{-1} = (-v_1, -w_1, -X_1 + v_1 w_1^t)[/imath] and [imath](v_2, w_2, X_2)^{-1} = (-v_2, -w_2, - X_2 + v_2 w_2^t)[/imath] Then [imath][(v_1, w_1, X_1), (v_2, w_2, X_2)][/imath] [imath]= (v_1, w_1, X_1)(v_2, w_2, X_2) (v_1, w_1, X_1)^{-1} (v_2, w_2, X_2)^{-1}[/imath] [imath]= (v_1, w_1, X_1)(v_2, w_2, X_2)(-v_1, -w_1, - X_1 + v_1 w_1^t) (-v_2, -w_2, - X_2 + v_2 w_2^t)[/imath] The [imath]v[/imath]'s and [imath]w[/imath]'s all cancel out nicely in the first two coordinates. The third coordinate is some [imath]n \times n[/imath] matrix. We have just proved that the set of commutators is a subset of the set of all elements of the form [imath](0, 0, X)[/imath]. So we have shown that [imath]G'\subset \{(0,0,X)\}[/imath]. Please help me to prove the other inclusion. Thanks in advance. |
2460276 | What is the dimension of the Euclidean motion groups?
The group of Euclidean motions of [imath]\mathbb R^n[/imath] is the semi-direct product [imath]G=\mathbb R^n\rtimes K[/imath] with [imath]K=SO(n, \mathbb R^n)[/imath]. Elements of G are written as pairs [imath]g=(x,k)[/imath]. I would like to know what is the dimension of [imath]G[/imath] ? Thank you in advance | 425995 | Dimension of isometry group of complete connected Riemannian manifold
Given an [imath]n[/imath]-dimensional geodesically complete connected Riemannian manifold [imath]M[/imath], we want to prove that the dimension of its isometry group is [imath]\dim {\rm ISO}(M) \leq \frac{n(n+1)}2.[/imath] Does it suffice to say that, since Euclidean space [imath]\mathbb{R}^n[/imath] is expected to be maximally symmetric, and the number above is the dimension of its isometry group, namely translations plus rotations, then the bound must be true for any other manifold? Do you know a more rigorous proof? |
2461530 | Show that [imath]x^2 + y^2 = z^n[/imath] has positive integer solutions for [imath]n = 1,2,3,...[/imath]
I am working through Kevin Houston's book How to think like a mathematician. The question I am currently stuck on is to show that [imath]x^2 + y^2 = z^n[/imath] has positive integer solutions for [imath]n = 1,2,3,\dots[/imath] My work thus far: [imath]f: \mathbb{N}^2 \rightarrow \mathbb{N}[/imath] defined by [imath]f(x,y) = x^2 + y^2[/imath] with image [imath]A = \{2,5,8,10,13,17,\dots\}[/imath] [imath]g: \mathbb{N} \rightarrow \mathbb{N}[/imath] defined by [imath]g(z) = z^n[/imath] with image [imath]B = \{1,2^n,3^n,\dots\}[/imath] If I can show that [imath]A \cap B \neq \varnothing[/imath] for [imath]n \in \mathbb{N}[/imath] then this shows that there are integer solutions for all [imath]n[/imath]. How can this be shown? | 2072371 | [imath]x^2+y^2=z^n[/imath]: Find solutions without Pythagoras!
I was presented with the following problem: Prove that there exist solutions to [imath]x^2+y^2=z^n[/imath] for all [imath]n[/imath], with [imath]x,y,z, n \in \mathbb{N}[/imath] I showed that by taking any Pythagorean triple [imath]x^2+y^2=z^2[/imath] and multiplying by [imath]z^{2(n-1)}[/imath] we get [imath](z^{n-1}x)^2+(z^{n-1}y)^2=(z^2)^n[/imath], which allows me to generate solutions easily for any value of [imath]n[/imath]. I managed to find several similar questions on this site, such as this one concerning the specific case [imath]n=3[/imath]. I notice that all of these questions take a similar approach and start with a Pythagorean triple and use it to generate general solutions. Is there a way to prove the statement (or better yet provide solutions to the equation) without first relying on Pythagoras? |
2461551 | Inductive proof of: [imath]3^n ≥ 1 + 2^n [/imath], every [imath]n\in\mathbb{N}[/imath].
How can I prove by induction that [imath]3^n ≥ 1 + 2^n[/imath] for every [imath]n\in\mathbb{N}[/imath]? | 966128 | Proof by Induction that [imath]3^n ≥ 1+2^n[/imath]
Use the PMI to prove the following for all natural numbers: [imath]3^n ≥ 1+2^n[/imath]. I have already verified the base case but am having trouble doing so with the inductive case. Thanks! |
1703602 | How does an almost complex structure on a manifold induce an orientation?
I have read that given a smooth even dimensional manifold [imath]M[/imath] with an almost complex structure [imath]J[/imath], then [imath]M[/imath] is orientable and there is a canonical choice of orientation. Why is this the case? How does one construct the orientation? Thanks! | 1817959 | Almost complex manifolds are orientable
I want to verify the fact that every almost complex manifold is orientable. By definition, an almost complex manifold is an even-dimensional smooth manifold [imath]M^{2n}[/imath] with a complex structure, i.e., a bundle isomorphism [imath]J\colon TM\to TM[/imath] such that [imath]J^2=-I[/imath], where [imath]I[/imath] is the identity. For every tangent space [imath]T_pM[/imath], there exists tangent vectors [imath]v_1,\cdots, v_n[/imath] such that [imath](v_1,\cdots, v_n,Jv_1,\cdots,Jv_n)[/imath] is a real ordered basis for [imath]T_pM[/imath]. I've checked that any two such bases are related by a matrix with positive determinant. So for each tangent space [imath]T_pM[/imath] I fix such a basis. Hence, I've obtained a possibly discontinuous global frame [imath]X_1,\cdots,X_n,JX_1,\cdots,JX_n[/imath]. To show this frame determine an orientation on [imath]M[/imath], I've been trying to construct for each [imath]p\in M[/imath] a coordinate neighborhood [imath](U,x^1,\cdots,x^{2n})[/imath] such that [imath](dx^1\wedge\cdots dx^{2n})(X_1,\cdots,X_n,JX_1,\cdots JX_n)>0[/imath] for every point in [imath]U[/imath]. How do I show that? Thanks! |
2461716 | Extension of Fundamental Theorem of Algebra
The problem states: Let [imath]p(x)[/imath] be a polynomial in [imath]x[/imath] of degree [imath]n[/imath] with [imath]n\ge2[/imath]. Recall that, according to the Fundamental Theorem of Algebra, [imath]p(x)[/imath] has [imath]n[/imath] number of roots in the complex number set. Suppose all roots of [imath]p(x)[/imath] are real and distinct. Prove that the roots of [imath]p'(x)[/imath] are all real. I know and kind of understand the proof of the Fundamental Theorem of Algebra, but I do not know how to extend it to [imath]p'(x)[/imath]. Any thoughts? Thanks! | 1371717 | Relation between real roots of a polynomial and real roots of its derivative
I have this question which popped in my mind while solving questions of maxima and minima. First Case:Let [imath]f(x)[/imath] be an [imath]n[/imath] degree polynomial which has [imath]r[/imath] real roots. Using this can we say anything about the number of real roots of [imath]f'(x)[/imath]? Second Case:Suppose, [imath]f(x)[/imath] has all [imath]n[/imath] real roots. Then will all of its derivatives also have all real roots? Also, if any of its derivatives do not have all real roots, then will [imath]f(x)[/imath] also have not all real roots? If the above is true then what about its converse? Comment Case:For the third case:Suppose f'(x) is a 5 degree polynomial with 3 real roots.Then f(x) will be a 6 degree polynomial.(correct me if I am wrong).What are the possible no. of roots that f(x) can have(3,4,5 etc.?).Basically I am asking for an example.Also it would be great if you follow all cases with an example like in the 4th case. |
2462102 | Find the determinant of addition of two matrices
Let [imath]x, y \in \mathrm{R^n}[/imath]. Show that det[imath](I - xy^T) = 1 - y^{T}x[/imath]. I test several cases for this problem, and it holds true when [imath]n = 2, 3[/imath], but when it comes to proof, I tried two ways and both failed: Write [imath]x, y[/imath] explicitly in the form of [imath]a_{ij}, b_{ij}[/imath], and tried to find the relationship between the left and right side. But the left side could not be reduced to a relatively clearer form and turns out to add more complexity. Try to transform [imath]I - xy^T[/imath] into something similar to the right side. I think the difficulty of this problem lies in the transformation from matrix [imath](I - xy^T)[/imath] into scalar [imath]1 - y^{T}x[/imath], and I do not have a good way to make such transformation. This question could be answered with matrix determinant lemma (the proof could be found here. Even though this is pretty easy to understand, it lacks the reasoning process, which I find very unsatisfying. | 1354007 | How to prove [imath]\det(I+uv^\intercal)=1+v^\intercal u[/imath]
Let be [imath]u,v\in\mathbb{R}^n[/imath], then [imath]\det(I+uv^\intercal)=1+v^\intercal u [/imath] where [imath]I[/imath] denotes the identity matrix of order [imath]n[/imath]. How to prove this? what I did: let be [imath]A=\{n\in\mathbb{N}: \forall u,v \in \mathbb{R^n}, \det(I+uv^\intercal)\neq1+v^\intercal u \} [/imath], and suppose [imath]A\neq \varnothing [/imath], then for the well-orderer-principle there exists [imath]n_0\in A[/imath] such [imath]n_0\leq n,\;\forall n\in A[/imath]. since [imath]1\notin A,\; n_0\neq 1 \rightarrow n_0-1\in\mathbb{N}\setminus A [/imath] so: [imath]\forall u,v\in\mathbb{R^{n_0-1}}: \det(I+uv^\intercal)=1+v^\intercal u [/imath] Then, let be [imath]u,v\in\mathbb{R}^{n_0}[/imath], so [imath]\det(I+uv^\intercal)= \displaystyle\sum_{j=1}^{n_0}(-)^{1+j}a_{1j}\det(A_{1j})[/imath] where [imath]a_{ij}=\begin{cases} u_iv_j+1 & i=j \\ u_iv_j & i\neq j \end{cases}[/imath], and [imath]A_{1j}[/imath] is the submatrix of [imath]I+uv^\intercal[/imath] that results deleting the i-file and j-column. Then I try to open the term for [imath]j=1[/imath] and try to use the relation for the new matrix of [imath]n_0-1[/imath] order with the determinant because it holds the same form but I have other terms which difficult me the work. Do you know other method?. Ps. here, in page 2 I've found something similar but I don't understand what it means PS. I'm taking a course of numerical analisys |
2461760 | Finding all subgroups of Direct product of groups
I'm having a hard time finding when I left a subgroup out. For example, lets take [imath]G=\mathbb{Z}_2 \times \mathbb{Z}_4.[/imath] Subgroups: $s_1:[imath]\langle(0,0)\rangle$[/imath] $s_2:\langle(0,1)\rangle=\{(0,0),(0,1),(0,2),(0,3)\}$ $s_3:[imath]\langle(0,2)\rangle=\{(0,0),(0,2)\}$[/imath] $s_4:\langle(0,3)\rangle=\{(0,0),(0,3),(0,2),(0,1)\}=s_2$ $s_5:[imath]\langle(1,0)\rangle=\{(0,0),(1,0)\}$[/imath] $s_6:\langle(1,1)\rangle=\{(0,0),(1,1),(0,2),(1,3)\}$ $s_7:[imath]\langle(1,2)\rangle=\{(0,0),(1,2)\}$[/imath] $s_8:\langle(1,3)\rangle=\{(0,0),(1,3),(0,2),(1,1)\}=s_6$ By looking at this I'd say all the subgroups are here, but for this [imath]G[/imath] there is also the subgroup [imath]\mathbb{Z}_2 \times 2\mathbb{Z}_4[/imath]. How should I know that? If I take some other examples like [imath]\mathbb{Z}_3 \times \mathbb{Z}_4[/imath] or [imath] \mathbb{Z}_3 \times \mathbb{Z}_3[/imath], the type of "list" I made is enough to determinate all the subgroups. | 361620 | How to find all subgroups of a direct product?
I am wondering how do we find all subgroups of a direct product? Is there a method to find it? For example, how can we find all the subgroups of [imath]\mathbb{Z}_2\times\mathbb{Z}_2[/imath]? There is the answer: [imath]\{(0, 0), (1, 0)\}, \{(0, 0), (0, 1)\}, \{(0, 0), (1, 1)\}[/imath], and the improper and trivial subgroup. But I do not know the method how to find it. All I can think about is the following: since the order of [imath]\mathbb{Z}_2\times\mathbb{Z}_2[/imath] is [imath]4[/imath], the order of a subgroup of it can be [imath]1,2[/imath] or [imath]4[/imath]. If order is [imath]1[/imath], then it is trivial, and if order is [imath]4[/imath], it is proper. Then, the order of a subgroup is [imath]2[/imath]. We know all the elements of [imath]\mathbb{Z}_2\times\mathbb{Z}_2[/imath], so we try all the pairs. Is that the method? But what if we are given a bigger order product, such as [imath]\mathbb{Z}_2\times\mathbb{Z}_8\times\mathbb{Z}_3\times\mathbb{Z}_9[/imath]? It looks very long to try all possibilities. Can anyone help? Thanks |
2461305 | Is it possible to apply Gram–Schmidt process to orthonormalise a matrix in Hilbert space?
Let [imath]V[/imath] be a separable infinite dimensional Hilbert space over [imath]\mathbb{C}[/imath] Let [imath]\{A_{i,j}\}_{i,j \in \mathbb{N}} \subset V[/imath] be a matrix of linearly independent vectors of [imath]V[/imath] My question is if is it possible to orthonormalise matrix [imath]A[/imath], getting the matrix [imath]B[/imath] such that [imath] \langle B_{i_1,j_1} , B_{i_2,j_2} \rangle = 0 [/imath] and such that [imath] \forall i \in \mathbb{N}: \operatorname{Span}(\operatorname{Row}_i(A))=\operatorname{Span}(\operatorname{Row}_i(B)) [/imath] I have to orthonormalise a matrix and not a sequence so the question is very different from Gram-Schmidt in Hilbert space? mainly for the second condition. I cannot see no one anything in common. thanks. | 620667 | Gram-Schmidt in Hilbert space?
EDIT: After some contemplation I decided to phrase the question better to avoid trivial answers. Consider a Hilbert space with a basis [imath]\{v_{i}\}[/imath] where [imath]i\in I[/imath] an index set, which could be uncountably infinite. We define a "Gram-Schmidt for infinite dimension" to consist of a total ordering [imath]\leq[/imath] on the index set [imath]I[/imath] and a set of orthonormal basis [imath]\{b_{i}\}[/imath] where [imath]i\in I[/imath], and the following condition is satisfied: for any [imath]i\in I[/imath], then [imath]b_{i}[/imath] can be written as a finite linear combination of [imath]v_{i}[/imath] and [imath]b_{j}[/imath] where [imath]j<i[/imath]. We define a "Gram-Schmidt for infinite dimension with series allowed" to be just like above, except that [imath]b_{i}[/imath] is allowed to be written as series instead of just finite linear combination. So the question is, can "Gram-Schmidt for infinite dimension" be done in a general Hilbert space starting with an arbitrary basis [imath]\{v_{i}\}[/imath]? What about "Gram-Schmidt for infinite dimension with series allowed"? Clearly, if [imath]I[/imath] is countable, the answer is trivial. The interesting case is when [imath]I[/imath] is uncountable. Thank you for your answer. =========== Original question: In Real Analysis for Graduate Student (Bass) chapter 19 (Hilbert space) page 188, wrote was "The Gram-Schmidt procedure from linear algebra also works in infinitely many dimensions". So how exactly would such process get carried out? Thank you for your answer. |
2462162 | Show that [imath]v[/imath] is harmonic and rotation invariant iff [imath]v(z)=A\log |z|+B[/imath] for some constants [imath]A[/imath] and [imath]B[/imath]
Let [imath]\Omega[/imath] be a domain in [imath]\mathbb{C}[/imath] and [imath]v[/imath] be a function on [imath]\Omega.[/imath] Suppose that [imath]v(z)[/imath] depends only on [imath]|z|[/imath] and not on arg[imath]z.[/imath] Show that [imath]v[/imath] is harmonic if and only if [imath]v(z)=A\log |z|+B[/imath] for some constants [imath]A[/imath] and [imath]B[/imath]. Any hint. Thanks in advance! | 1582569 | Find all possible functions, [imath]F(r)[/imath], harmonic in [imath]2[/imath] and [imath]3[/imath] dimensions,
Sources: this is an old advanced calculus exam question, which I think is asking for harmonic functions. The problem statement is: Suppose [imath]F(r)[/imath] is a smooth function of [imath]r[/imath] for [imath]r>0[/imath] . Define [imath]\Phi(x) = F(|x|)[/imath] with [imath]x = (x_1, ...,x_d) \in \mathbb{R}^d[/imath] and [imath]|x| = (\large \sum_i^dx_i^2)^{\frac{1}{2}}[/imath].For [imath]d=2[/imath] and [imath]d=3[/imath], find all possible functions [imath]F(r)[/imath] so that [imath]\sum_{i=1}^d (\frac {\partial}{\partial x_i})^2\Phi(x) = 0.[/imath] Any hints or comments are welcome. So, it appears that this problem statement is just a tricky way of asking to find all functions [imath]\Phi[/imath] such that [imath]\Phi_{x_1x_1} + \Phi_{x_2x_2} = 0 [/imath] and [imath]\Phi_{x_1x_1} + \Phi_{x_2x_2} + \Phi_{x_3x_3} = 0[/imath] I.e., we are looking for all functions that are harmonic. But what is weird about this question is that it doesn't really specify a domain in which we are to find these harmonic functions. Unless the domain is simply understood to be for all [imath]| x| \ne 0[/imath]. Thanks, |
2462467 | Topology on a set of matrices
Given a set [imath]M[/imath] of all [imath]n\times n[/imath] real matrices with the usual norm topology. Then, is the set of all symmetric positive definite matrices in [imath]M[/imath] connected? Also, is the set of all invertible matrices in [imath]M[/imath] compact? All I can infer so far is that the set [imath]M[/imath] is probably not compact as, if [imath]M[/imath] is positive definite then so is [imath]cM[/imath] for positive scalars [imath]c[/imath] and as [imath]c\rightarrow \infty[/imath], there is no limit. Thank You. | 196195 | Topology of matrices
1.Consider the set of all [imath]n×n[/imath] matrices with real entries as the space [imath]\mathbb R^{n^2}[/imath] . Which of the following sets are compact? (a) The set of all orthogonal matrices. (b) The set of all matrices with determinant equal to unity. (c) The set of all invertible matrices. 2.In the set of all [imath]n×n[/imath] matrices with real entries, considered as the space [imath]\mathbb R^{n^2}[/imath] , which of the following sets are connected? (a) The set of all orthogonal matrices. (b) The set of all matrices with trace equal to unity. (c) The set of all symmetric and positive definite matrices. FOR 1 (a) may be true as determinant mapping is continuous and it maps to the compact set{1,-1} but it is only a necessary condition.and (c) is not true as determinant mapping is continuous and it maps to a non compact set.do not know about (b).but i think it is not true. FOR2 (a) is not correct.do not know about (b) & (c) |
2463103 | About the proof of square root of 2 is irrational
I looked at the proof of [imath]\sqrt 2[/imath] is irrational. In first step we assume that [imath]\sqrt 2[/imath] is rational. Then we say it should be written as [imath]\frac ab[/imath] if it's rational.After that we assume gcd(a,b)=1 and end of the calculations we conclude that a and b are even numbers.So there is a common factor but we assumed that gcd(a,b)=1 so it's a contradiction , [imath]\sqrt 2[/imath] must be irrational.The thing that I don't understand why we assume that gcd(a,b)=1 ? We assume that [imath]\sqrt 2[/imath] is rational it's okay but why we need to assume gcd(a,b)=1.I don't think it's the need of being rational ? | 2136374 | Why no common factors in proving root 2 is irrational?
I'm taking 'Introduction to Mathematical Thinking' on Coursera platform and following proof steps are given : Proof of [imath]\sqrt{2}[/imath] is irrational. Assume [imath]\sqrt{2}[/imath] is rational. [imath]\sqrt{2}=p/q[/imath] p and q have no common factors. Why do p and q have no common factors? Is this a consequence of a property of the rational numbers? As p and q can be rational numbers we can set p = 6, q = 9 so p, q have common factors? |
2463220 | Let [imath]\{a_n\}[/imath] be bounded sequence, satisfied [imath]a_{n+1} \geq a_n - \displaystyle\frac{1}{2^n}, n \in \mathbb{N}[/imath]. Show that [imath]\{a_n\}[/imath] is convergent.
Problem: Let [imath]\{a_n\}[/imath] be bounded sequence, satisfied [imath]a_{n+1} \geq a_n - \displaystyle\frac{1}{2^n}, n \in \mathbb{N}[/imath]. Show that [imath]\{a_n\}[/imath] is convergent. My attempt: consider sequence [imath]\{x_n\} = \left\{ a_n - \displaystyle\frac{1}{2^{n-1}} \right\}[/imath]. Obviously [imath]\{x_n\}[/imath] is bounded. Then we have [imath]x_{n+1} - x_n = a_{n+1} - \displaystyle\frac{1}{2^n} - a_n + \displaystyle\frac{1}{2^{n-1}} = a_{n+1} - a_n + \displaystyle\frac{1}{2^n} \geq a_n - \displaystyle\frac{1}{2^n} - a_n + \displaystyle\frac{1}{2^n} = 0[/imath]. Hence, [imath]\{x_n\}[/imath] is a non-decreasing sequence, apply [imath]\{x_n\}[/imath] is a convergent sequence. Here, can we conclude that the sequence [imath]\{a_n\}[/imath] is convergent? If we can, please explain how can we do that. Sorry for my poor English. Thank all! | 705250 | How can we show that the following sequence converges?
[imath](a_n)[/imath] is a bounded sequence with the following condition [imath]a_{n+1}\geq a_n-\frac{1}{2^n}[/imath] The sequence converges, but how do we show it? |
2463557 | limsup [imath](n|a_{n}|)^{1/n}=[/imath]limsup [imath](|a_{n}|)^{1/n}[/imath]
I have troubles proving the next result: \begin{equation} \limsup\sqrt[n]{n|a_{n}|}=\limsup\sqrt[n]{|a_{n}|} \end{equation} where [imath]a_{n}[/imath] is a bounded sequence. My attempt: Since [imath]\limsup(s_{n}t_{n})\leq\limsup(s_{n})\limsup(t_{n})[/imath] and [imath]\lim\limits_{n\rightarrow\infty}\sqrt[n]{n}=1[/imath] I have the inequality [imath]\fbox{[/imath]\leq[imath]}[/imath]. But I can't prove the other inequality. I will appreciate any hint, thanks. | 596635 | Prove that [imath]\limsup\left(b_{n}a_{n}\right)=\limsup\left(a_{n}\right) [/imath] when [imath]\lim_{n\rightarrow\infty}\left(b_{n}\right)=1[/imath]
I'm having trouble with this homework question: "Let there be a sequence [imath]\left(b_{n}\right)_{n=1}^{\infty} [/imath] such that [imath]\lim_{n\rightarrow\infty}\left(b_{n}\right)=1[/imath]. Let there also be some bounded sequence [imath]\left(a_{n}\right)_{n=1}^{\infty}[/imath] Prove that the following statement is correct: [imath]\limsup\left(b_{n}a_{n}\right)=\limsup\left(a_{n}\right) [/imath]" Alright so I sat a good two hours on this question with no major breakthroughs as of yet. I have some semi-proofs all of which use some arbitrary assumptions which I can't make. Any hints? Thank you! |
2462945 | How to prove that [imath]\sin(x\sin x)[/imath] is NOT uniformly continuous on positive reals
How to prove that [imath]\sin(x \sin x)[/imath] is not uniformly continuous on [imath]\Bbb{R}^+[/imath]? My attempt: I know that [imath]\sin x[/imath] is uniformly continuous on that interval when [imath]\sin{x^2}[/imath] is not. Yes It is duplicate of one question. But the answer was not satisfactory. That's why I am posting it again. Any help will be highly appreciated. | 1919181 | Uniform continuity of [imath]\sin(x\sin x )[/imath] for all [imath]x\in (0,\infty)[/imath]
I want to prove or disprove that [imath]f(x)=\sin(x\sin x)[/imath] for all [imath]x\in (0,\infty)[/imath] is uniformly continuous. We know that [imath]g(x)=x\sin x[/imath] is not uniformly continuous on [imath](0,\infty)[/imath]. But what about [imath]f[/imath]? I am not getting any hint. Any help is appreciated. |
2463603 | How to find the closed form of [imath]f(n) = 9^k \times (-56) + f(n-1)[/imath]
I have to find the closed form for [imath]T(n) = \begin{cases} 2 , &\text{ if } n=0 \\ 9T(n-1) - 56n + 63, &\text{ if } n > 1 \end{cases}[/imath] I used the repeated substitution method and I found that the pattern for the coefficient of n is equal to the following: [imath]f(1) = -56[/imath] [imath]f(n) = 9^{n-1} \times (-56) + f(n-1)[/imath] I tried to find the closed form of [imath]f(n) = 9^{n-1} \times (-56) + f(n-1)[/imath], but it just got more and more confusing. I believe it may be a series of some sort. Is there a way to find a closed form for this? Thank you! | 2457918 | Solve the recurrence
[imath]T(n) = 2[/imath] if [imath]n=0[/imath] [imath]T(n) = 9T(n-1)-56n +63[/imath] if [imath]n>=1[/imath] Repeated substitution [imath]k=1[/imath] [imath]T(n) = 9T(n-1)-56n+63 [/imath] [imath]k =2[/imath] [imath]T(n) = 81T(n-2) -560n + 1134[/imath] [imath]k =3[/imath] [imath]T(n) = 729T(n-3) -5096n + 15309[/imath] I cant find the pattern for the n term and the integer For now i just have T(n) = [imath]9^k(n-k)[/imath] |
2460118 | Prove [imath]\mathbb{R}[X]/(ax^2 + bx +c) \cong \mathbb{C}[/imath] if [imath]b^2 - 4ac < 0[/imath]
I want to prove that [imath]\mathbb{R}[X]/(ax^2 + bx +c) \cong \mathbb{C}[/imath] if [imath]b^2 - 4ac < 0[/imath]. I belief I have all of the components needed for this exercise: Knowing the first isomorphic theorem [imath](2ax + b)^2 = b^2 - 4ac[/imath] [imath]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/imath] But I just don't see what the bigger picture is | 2463027 | Polynomial ring isomorphisms
Let [imath]f = aX^2 + bX + c \in \mathbb{R}[X][/imath]. I want to show that: [imath]\mathbb{R}[X]/(f) \cong \mathbb{C}[/imath] if [imath]b^2 - 4ac < 0[/imath], [imath]\mathbb{R}[X]/(f) \cong \mathbb{R}[\epsilon][/imath] if [imath]b^2 - 4ac = 0[/imath], [imath]\mathbb{R}[X]/(f) \cong \mathbb{R} \times \mathbb{R}[/imath] if [imath]b^2 - 4ac > 0[/imath]. My idea so far is to prove that in the first case, [imath](f) = (X^2 + 1)[/imath], in the second case [imath](f) = (X^2)[/imath] , in the third case [imath](f) = (X^2 - 1)[/imath], which would lead to the desired result. However, I don't know how to prove this or if I'm on the correct road at all. Does anyone have any ideas? |
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