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2463004 | Structural Induction Proof of Compositional Semantics
I need to show by structural induction on WFFs that every WFF, [imath]\theta[/imath], has the following property: For any PL (Propositional Logic) interpretations [imath]I[/imath] and [imath]I'[/imath] if [imath]I(\alpha) = I'(\alpha)[/imath] for every sentence letter [imath]\alpha[/imath] in [imath]\theta[/imath] then [imath]V_{I}(\theta) = V_{I'}(\theta).[/imath] I know that I need to show that: Base Step: Every wff [imath]\theta[/imath] is s.t. any PL (Propositional Logic) interpretations [imath]I[/imath] and [imath]I'[/imath] if [imath]I(\alpha) = I'(\alpha)[/imath] for every sentence letter [imath]\alpha[/imath] in [imath]\theta[/imath] then [imath]V_{I}(\theta) = V_{I'}(\theta).[/imath]; and Inductive Step: Modus Ponens preserves this property. But I am unclear about how to show the Base Step. I am unclear about how to start the proof because I'm having a hard time translating the stipulated property of [imath]\theta[/imath] into a conditional in English to go about giving a conditional proof. Can anyone help me translate the property [imath]\theta[/imath] above into an English conditional statement (or give me a hint as to how I should) so I can do a conditional proof for the Base Step? | 2462016 | Structural Induction: Metalogic Proof
I need to show that by structural induction that the truth value [imath]V[/imath] of a WFF depends only on the truth values of its sentence letters. In other words, I need to show show that for any WFF [imath]\theta[/imath] and any PL-interpretations [imath]I[/imath] and [imath]I'[/imath], if [imath]I(\alpha) = I'(\alpha)[/imath] for every sentence letter [imath]\alpha[/imath] in [imath]\theta[/imath], then [imath]V_I (\theta) = V_ I' (\theta).[/imath] I'm not sure where to begin because I'm not sure how to translate the former assertion into English. Does anyone have any tips about how to proceed with the structural inductive proof? |
2464012 | Possibilities of different numbers adding up to an integer
Let [imath]n_1 < n_2 < n_3 < n_4 < n_5[/imath] be positive integers such that [imath]n_1 + n_2 + n_3 + n_4 + n_5 = 20[/imath]. Then the number of such distinct arrangements [imath](n_1, n_2, n_3, n_4, n_5)[/imath] is My attempt First thing is that one would not be able to use stars and bars technique as the numbers cannot be equal. Hence I tried to count all the cases up to 20 which gave the answer as [imath]7[/imath] possibilities. [imath]n_1 + n_2 + n_3 + n_4 + n_5 = k[/imath] Or [imath]n_1 + n_2 + n_3 + n_4 + n_5 + n_6 + n_7 + \ldots + n_j = k[/imath] However , generalising it for an integer [imath]k[/imath] , counting each time would be tedious if [imath]k[/imath] is large. I was hoping someone could suggest an alternate way to solve such problems ? | 2401356 | How many solutions does the equation [imath]n_1 + n_2 + n_3 + n_4 + n_5 = 20[/imath] have in the positive integers if [imath]n_1 < n_2 < n_3 < n_4 < n_5[/imath]?
Let [imath]n_1 < n_2 < n_3 < n_4 < n_5[/imath] be positive integers such that [imath]n_1 + n_2 + n_3 + n_4 + n_5 = 20[/imath]. Then the number of such distinct arrangements [imath](n_1, n_2, n_3, n_4, n_5)[/imath] is...... I have no idea how to proceed. Manually, I have done it [imath]1+2+3+4+10[/imath] [imath]1+2+3+5+9[/imath] [imath]1+2+3+6+8[/imath] [imath]1+2+4+5+8[/imath] [imath]1+2+4+6+7[/imath] [imath]1+3+4+5+7[/imath] [imath]2+3+4+5+6[/imath] But is there any way I can do it by Permutation and Combination method? |
2464530 | Problem with [imath]0! = 1[/imath]
Okay, we know that [imath]4!=4 \cdot 3\cdot 2 \cdot 1=24[/imath] [imath]5!=5\cdot4\cdot3\cdot2\cdot1=120[/imath] [imath]6!=6\cdot5\cdot4\cdot3\cdot2\cdot1=720[/imath] This is can be written as [imath]n!=n\cdot(n-1)![/imath] We obtain [imath]6! = \frac {7!}{7} = 720[/imath] [imath]5! = \frac {6!}{6} = 120[/imath] [imath]4! = \frac {5!}{5} = 24[/imath] If we take [imath]0! = 1[/imath] [imath]0! = \frac {1!}{1} = \frac {1}{0}[/imath] Using other way [imath]0!=0\cdot0 = 0[/imath] Why is [imath]0![/imath] equal to [imath]1[/imath]? | 1169361 | Formula for factorial?
I need an equation that defines factorial without using factorial, that also works for [imath]0[/imath]. I have seen factorial defined like this: [imath]n! = 1\cdot2\cdot3\cdot4\cdots n[/imath] But if we plug [imath]0[/imath] into that, we end up with [imath]0[/imath], which doesn't work. So what would be the correct way to put it? |
2465407 | Evaluate the sum [imath]{1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}[/imath]
Evaluate the sum [imath]{1 - \frac{1}{2} {n \choose 1} + \frac{1}{3} {n \choose 2} + \ldots + (-1)^n \frac{1}{n+1} {n \choose n}}.[/imath] I have tried comparing this to the similar problem here. I believe I need to differentiate or integrate? But I'm not sure how that might work. Any ideas? Thanks. | 963882 | Alternating sum of binomial coefficients multiplied by (1/k+1)
I'm trying to prove that [imath]\sum_{k=0}^n {n \choose k} (-1)^k \frac{1}{k+1} = \frac{1}{n+1}[/imath] So far I've tried induction (which doesn't really work at all), using well known facts such as [imath]\sum_{k=0}^n {n \choose k} (-1)^k = 0[/imath] and trying to apply identities like [imath]{n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}[/imath] Would anyone be able to point me towards the right method? Should I be looking to apply an identity or is there a method I'm missing? |
2465842 | Prove [imath]a=(12345)[/imath] and [imath]b=(21345 )[/imath] are NOT conjugate in [imath]A_5[/imath]
I had to find two 5-cycles in [imath]A_5[/imath] that are not conjugate. I believe I found two, namely [imath]a=(12345)[/imath] and [imath]b=(21345)[/imath]. It wasn't part of the exercise, but I'd like to prove that they are not conjugate. But I'm not sure where to start. I think we have to start with that they are conjugate in [imath]S_5[/imath], so there exists a [imath]g \in S_5[/imath] such that [imath]gag^{-1}=b[/imath]. And then prove all the [imath]g's[/imath] for which that is true are [imath]\notin A_5[/imath]? Someone stated this as a duplicate, but I don't think it is, because they don't really ask for/ give the algebraic proof that I'm looking for. | 464582 | Show that not all 5 cycles in [imath]A_5[/imath] are conjugate in [imath]A_5[/imath]
This is a practice question (not assignment). One thing I am wondering is what this question exactly means. From my understanding, the conjugate of [imath]a \in A_5[/imath] is [imath]gag^{-1} \in A_5[/imath] where [imath]g \in A_5[/imath]. I was trying to look at a similar question here on MSE and someone mentioned [imath](12345) \in A_5[/imath] is not conjuagte in [imath]A_5[/imath]. So does that means there exists a [imath]g \in A_5[/imath] such that [imath]g(12345)g^{-1} \notin A_5[/imath] (I am guessing this is incorrect since [imath]A_5[/imath] is a group). I am also away of the property that [imath]g(12345)g^{-1} = (g(1)g(2)g(3)g(4)g(5))[/imath]. |
2464940 | How to integrate [imath]\int{xe^{x}\sin{x}}~dx[/imath]
How to integrate [imath]\int{xe^{x}\sin{x}}~dx[/imath] I tried taking [imath]v=xe^{x}[/imath] and using integration by parts, but that became lengthy and didn't work. | 1230290 | Loopy integral [imath]\int xe^x\sin(x)[/imath]
How to find integral of: [imath]\int xe^x\sin(x)[/imath] I know it should be repeated integration by parts, but i do not know how many times i should do this and when to stop. |
2466244 | Name of a special multivariate polynomial: each varaible appears alone
Is there a name for such multivariate polynomial where each varaible appears alone. For example: [imath]P(x,y,z)=x+x^2+3x^3+y+5y^2+z +2[/imath] So as it can be seen there is no terms in the polynomial where the variables are multiplied by each other like [imath]xy[/imath] or [imath]xz^2[/imath] for instance. | 151479 | Multivariate polynomial with no mixed terms
Is there a standard name for multivariate polynomials wherein each term consists of only one coordinate? That is, polynomials of this form: [imath]p(x_1, \ldots, x_n)= \sum_{i=1}^n \sum_{j=1}^n a_{i,j}x_i^{j}[/imath] where all of the [imath]a_{i,j}[/imath] are constants. |
2465652 | Find the equations of both of the tangent lines to the ellipse [imath]x^2+4y^2=36[/imath] that pass through the point [imath](12,3)[/imath].
Find the equations of both of the tangent lines to the ellipse [imath]x^2+4y^2=36[/imath] that pass through the point [imath](12,3)[/imath]. Finding Slope The derivative of [imath]x^2 + 4y^2 = 36[/imath] is [imath]y'= -\dfrac{x}{4y}[/imath]. Finding arbitrary point where tangent line is at If I arrange the equation [imath]x^2 + 4y^2 = 36[/imath], [imath]y= \pm\frac{\sqrt{x^2+36}}{2}[/imath] Let [imath](k, \pm\frac{\sqrt{k^2+36}}{2})[/imath] be the point where tangent line is at. Thus the tangent line slope at that point is [imath]-\frac{k}{4\left(\pm\frac{\sqrt{k^2+36}}{2}\right)}[/imath] Equation of tangent line is [imath]y-y_0 = m (x-x_0)[/imath] Let [imath](k, \pm\frac{\sqrt{k^2+36}}{2})[/imath] be [imath](x_0, y_0)[/imath]. Let [imath](12,3)[/imath] be [imath](x,y)[/imath] [imath]12-k = -\frac{k}{4\left(\pm\frac{\sqrt{k^2+36}}{2}\right)}\left(3- \pm \frac{\sqrt{k^2+36}}{2}\right)[/imath] And I'm stuck here. It gets really confusing? | 1778828 | Find the tangent lines to the graph of [imath]x^2+4y^2 = 36[/imath] that go through the point [imath]P=(12,3)[/imath]
I was solving a few problems from a textbook and I came across this one: Find the tangent lines to the graph of [imath]x^2+4y^2 = 36[/imath] that go through the point [imath]P=(12,3)[/imath] I could find the tangency points [imath] but my solution was completely messy (I had four equations that I had to go back-and-forth with.[/imath] Is there an elegant/simple solution to that? Tangency points: P_1 = (0,3)[imath] [/imath]P_2 = \left( \frac{24}{5} , 3-\frac{24}{5} \right)$$ |
2466546 | Abstract algebra-normal subgroups
I found the following problem is quite hard for me, and I tried a lot but something I can't bring in my mind. Let [imath]K[/imath] be a normal subgroup of a group [imath]G[/imath] ([imath]G[/imath] can be finte or infinite) such that [imath][G:K]=m[/imath]. If [imath]n[/imath] is a positive integer such that [imath]gcd(m,n)=1[/imath], then show [imath]K[/imath] contains all elements of order [imath]n[/imath] in [imath]G[/imath]. I start by assuming that there exists an element of order [imath]n[/imath] which is outside [imath]K[/imath], then tried for getting a contradiction. But the group isn't provided finite,so I can't reach too far. Kindly provide me some hints. | 2419761 | Normal subgroup of index m contains every element of order n where m and n are relatively prime
Let [imath]K[/imath] be a normal subgroup of [imath]G[/imath] of index [imath]m[/imath]. If [imath]n[/imath] is a positive integer such that [imath]\gcd(m,n)=1[/imath], show that all the elements of order [imath]n[/imath] are in [imath]K[/imath]. I could solve this if [imath]G[/imath] is finite as follows: Let [imath]g[/imath] be an element of order [imath]n[/imath]. [imath]|gK|[/imath] divides [imath]|G|[/imath] using canonical epimorphism, again [imath]|gK|[/imath] divides [imath]|G/K|[/imath]. Since [imath]m[/imath] and [imath]n[/imath] are relatively prime, we get order of [imath]gK[/imath] is [imath]1[/imath]. What if [imath]G[/imath] is infinite? |
2466041 | If [imath]G[/imath] is a group in which $(ab)^i = a^ib^i$ for two consecutive integers [imath]i[/imath], for all [imath]a,b \in G[/imath], can we conclude that [imath]G[/imath] is abelian?
Problem: If [imath]G[/imath] is a group in which [imath]$(ab)^i = a^ib^i$[/imath] for three consecutive integers [imath]i[/imath] for all [imath]a,b \in G[/imath], then [imath]G[/imath] is abelian. If we assume the relation [imath]$(ab)^i = a^ib^i$[/imath] for just two consecutive integers can we conclude that [imath]G[/imath] is abelian? Thank all! | 1199479 | Non-abelian group [imath]G[/imath] satisfying [imath](a \cdot b)^i=a^i \cdot b^i,[/imath] for two consecutive integers.
Given an example of a non-abelian group [imath]G[/imath] satisfying [imath](a \cdot b)^i=a^i \cdot b^i, \forall a, b \in G[/imath] for two consecutive integers. This is question 5 from Herstein Page 35. I have proved that conclusion holds for three positive integers.Thanks. |
2449282 | Constrained Optimization with an External Parameter
In theory, is it correct to pose a constrained optimization problem to minimize a nonlinear function of two variables with respect to three parameters, where the third parameter relates the two variables of the function? For example, suppose we aim to minimize an objective function [imath]\mathscr F(a,b)[/imath], for the design parameters [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath], such that [imath]c=a^2+b^2[/imath]. Is this mathematically correct? Please have a detailed and proven statement if possible. Thanks. | 2427092 | Optimization with Dependent Constraints
In the general theory of constrained optimization, is it possible to pose a minimization problem subject to some constraints that are dependent? For example, suppose we aim to minimize an objective function [imath]\mathscr F(x,y,z)[/imath], for the design parameters [imath]x[/imath], [imath]y[/imath], and [imath]z[/imath], such that [imath]z=x^2+y^2[/imath]. Is it correct to have such constraint, theoretically? EDIT [imath]z=x^2+y^2[/imath] |
2273603 | Solubility by Radicals
I am trying to show that [imath]f(t)=t^{5}-4t^{2}+2[/imath] is not soluble by radicals over [imath]\mathbb{Q}[/imath]. I think I need to show that [imath]\text{Gal}(f)[/imath] is not soluble. I think I need to construct an embedding [imath]\text{Gal}(f)\hookrightarrow A_{5}[/imath] which is not soluble since the commutator subgroup of [imath]A_{5}[/imath] is [imath]A_{5}[/imath]. So how do I do this? | 1541127 | Find splitting field of [imath]x^5-4x+2[/imath] over [imath]\mathbb{Q}[/imath]
I know how to get the splitting field from the roots but I just can't come up with any factorisation that gives me roots. Any suggestions? |
2466092 | Find the length of the curve of [imath]r^2 = a^2cos(2x)[/imath]
I am trying to find the arc length for the lemniscate [imath]r^2 = a^2cos(2x)[/imath] using the equation [imath]\int \sqrt{r^2 + (\frac{dr}{d\theta})^2} \, d\theta[/imath] However, I end up with the integration of [imath]\sqrt{sec(x)}[/imath] which is too difficult to integrate. Is there any different path that I can following using the same formula to find the length of the curve? Is there a way to make this easier? Any suggestion would be appreciated. | 1474317 | Find the arc length of lemniscate [imath]r=2(\cos(2\theta))^{1/2}[/imath]
I have to find the arc length of a lemniscate with polar equation [imath]r=2(\cos(2\theta))^{1/2}[/imath]. So far I got like [imath]\sqrt{4\cos(2\theta)+\left(-2\frac{\sin(2\theta)}{\sqrt{\cos(2\theta)}}\right)}[/imath]. I don't know how to proceed, I can't seem to figure out how to approach this. |
2017113 | Is the convex hull of a compact set compact?
Let [imath]V[/imath] be a normed vector space and [imath]K[/imath] a compact subset of [imath]V[/imath]. Is the convex hull of [imath]K[/imath] given by [imath]\langle K \rangle = \left\{\sum_{i=1}^n t_i x_i\mid x_i\in K, t_i≥0\text{ s.t. }\sum_i t_i=1\right\}[/imath] again compact? | 2548550 | In a normed space [imath]E[/imath], is the convex envelope of any compact set also compact?
In a normed space [imath]E[/imath]; One can show that, if [imath]E[/imath] is finte-dimensional, the convex envelope of any compact set is also compact. [imath](*[/imath]) Is (*) true if [imath]E[/imath] is no longer supposed finite-dimensional ? |
2466651 | Probability- Show that [imath] 1/6 \le P(A) \le 1/4[/imath]
I have [imath]A ∪ B ∪ C = Ω, P(B) = 2P(A), P(C) = 3P(A), P(A ∩ B) = P(B ∩ C)[/imath]. Pove that [imath] 1/6 \le P(A) \le 1/4[/imath] | 308386 | If [imath]P(A \cup B \cup C) = 1[/imath], [imath]P(B) = 2P(A) [/imath], [imath]P(C) = 3P(A) [/imath], [imath]P(A \cap B) = P(A \cap C) = P(B \cap C) [/imath], then [imath]P(A) \le \frac14[/imath]
We have ([imath]P[/imath] is probability): [imath]P(A \cup B \cup C) = 1[/imath] ; [imath]P(B) = 2P(A) [/imath] ; [imath]P(C) = 3P(A) [/imath] and [imath]P(A \cap B) = P(A \cap C) = P(B \cap C) [/imath]. Prove that [imath]P(A) \le \frac{1}{4} [/imath]. Well, I tried with the fact that [imath] 1 = P(A \cup B \cup C) = 6P(A) - 3P(A \cap B) + P(A \cap B \cap C) [/imath] but I got stuck... Could anyone help me, please? |
2466569 | Prove the following sequence always results in a perfect square.
There's a problem statement: For each [imath]m \in \mathbb{N}[/imath], we construct a sequence [imath]m_0[/imath], [imath]m_1[/imath], [imath]m_2,\dots[/imath] denoted [imath]S_m[/imath], recursively via [imath]m_0=m[/imath] and [imath]m_{i+1} = m_i + \left\lfloor \sqrt{m_i} \right\rfloor[/imath] for all [imath]i \ge 0[/imath]. Here, [imath]\lfloor x \rfloor[/imath] is the floor of [imath]x[/imath], the greatest integer less than or equal to [imath]x[/imath]. Hence, we have [imath]\left\lfloor \sqrt{10} \right\rfloor=3[/imath] and [imath]\left\lfloor \sqrt{29} \right\rfloor=5[/imath]. Show that for each positive integer [imath]m[/imath], the sequence [imath]S_m[/imath] contains the square of some integer. I'm pretty certain that this can be proved with induction. I am just not quite sure what to induct on. Examining examples shows that [imath]S_m[/imath] always results in a perfect square eventually, though I'm not sure how to prove it. | 2928771 | Prove that the sequence contains a perfect square for any natural number [imath]m [/imath] in the domain of [imath]f[/imath] .
The domain of a function [imath]f[/imath] is the set of natural numbers. The function is defined as follows: [imath]f(n)=n+\left\lfloor\sqrt{n}\right\rfloor[/imath] where [imath]\lfloor k\rfloor[/imath] denotes the nearest integer smaller than or equal to [imath]k[/imath] Prove that for every natural number [imath]m[/imath] the following sequence contains at least one perfect square [imath]m, f(m), f^2(m),f^3(m),\ldots[/imath]. The notation [imath]f^k[/imath] denotes the function obtained by composing [imath]f[/imath] with itself [imath]k[/imath] times e.g [imath]f^2 = f\circ f.[/imath] I tried this out using induction but I got stuck and couldn't make it. Please help me with this problem. |
2468258 | Probability that the given quadratic equation has real solutions
[imath]a[/imath] and [imath]b[/imath] are randomly chosen real numbers in the interval [imath][0, 1][/imath], that is both [imath]a[/imath] and [imath]b[/imath] are standard uniform random variables. Find the probability that the quadratic equation [imath]x^2 + ax + b = 0 [/imath] has real solutions? | 2272490 | Probability of [imath]x^2+bx+c=0[/imath] having real roots
Consider the quadratic equation [imath]x^2+bx+c=0[/imath], where [imath]b[/imath] and [imath]c[/imath] are [imath]Uni∼[0,1][/imath]. Let [imath]p(b)[/imath] represent the probability that the given equation has a real solution for a fixed value of[imath] b[/imath]. What is [imath]p(1/2)[/imath]? What is the probability that[imath] x^2+bx+c=0[/imath] has a real solution ? [imath]p(1/2)[/imath] = prob [imath]x^2+bx+c=0[/imath] has a real solution) now [imath]x^2+bx+c=0[/imath] has a real solution if discriminant [imath]≥0[/imath] i.e., [imath]1−16c≥0[/imath] which is same as [imath]c≤1/16[/imath]. since [imath]C∼UNI[0,1][/imath] the [imath]p(c)≤1/16=1/16[/imath] Is this correct and how to do the second part? |
1867041 | Problem 24 Chapter 2. Evans PDE 2nd edition
This is a Problem from Evans PDE 2nd Edition,Chapter 2 Problem 24. (Equipartition of energy). Let [imath]u[/imath] solve the initial-value problem for the wave equation in one dimension: [imath]\left \{ \begin{array}{ll} \ u_{ tt}-u_{ xx}=0, \text{ in } \Bbb R\times (0,\infty) \\ \ u=g,u_t=h,\text{ on } \Bbb R\times\{ t=0\}. \end{array} \right.[/imath] Suppose [imath]g,h[/imath] have compact support. The kinetic energy is $k(t):=\frac{\ 1}{2}[imath]\int_{-\infty}^{\infty}\ u_t^2(x,t)dx$ and the potential energy is $p(t):=\frac{\ 1}{2}[/imath]\int_{-\infty}^{\infty}\ u_x^2(x,t)dx$. Prove (a) [imath]k(t)+p(t)[/imath] is constant in [imath]t[/imath]. (b) [imath]k(t)=p(t)[/imath] for all large enough times [imath]t[/imath]. | 1081227 | Equipartition of energy
Let [imath]u[/imath] solve the initial-value problem or the wave equation in one dimension: [imath]\begin{cases}u_{tt}-u_{xx}=0 & \text{in } \mathbb{R} \times (0,\infty) \\ u = g, u_t = h & \text{on } \mathbb{R} \times \{t=0\}. \end{cases}[/imath] Suppose [imath]g,h[/imath] have compact support. The kinetic energy is [imath]k(t) := \frac 12 \int_{-\infty}^\infty u_t^2(x,t) \, dx[/imath] and the potential energy is [imath]p(t) := \frac 12 \int_{-\infty}^\infty u_x^2(x,t) \, dx[/imath]. Prove (a) [imath]k(t)+p(t)[/imath] is constant in [imath]t[/imath], (b) [imath]k(t)=p(t)[/imath] for all large enough times [imath]t[/imath]. This is Chapter 2, Exercise 24 of PDE Evans, 2nd edition. I am only doing part (a) right now; my work is shown below: Define [imath]e(t):=k(t)+p(t)=\int_{-\infty}^\infty u_t^2+u_x^2 \, dx.[/imath] Then [imath]e'(t)=\frac 12 \int_{-\infty}^\infty 2u_tu_{tt}+2u_{x}u_{xt} \, dx= \int_{-\infty}^\infty u_tu_{tt}+u_{x}u_{xt} \, dx.[/imath] Now, I want to get [imath]e'(t)=0[/imath] so that [imath]e(t)[/imath] is constant. How can I go about doing this? I do know that I can use [imath]u_{tt}-u_{xx}=0[/imath], if [imath]t > 0[/imath]. With [imath]t>0[/imath] into mind, do I have to consider the [imath]t=0[/imath] case separately? Or can I treat both cases together as [imath]t \ge 0[/imath], since [imath]g[/imath] and [imath]h[/imath] have compact support? |
2468404 | Show that there is an odd biholomorphic function [imath]g[/imath] on the unit disk such that [imath]g(z)^2=f(z^2).[/imath]
Let [imath]f[/imath] be biholomorphic (i.e. holomorphic and injective) on the unit disk with [imath]f(0)=0[/imath]. Show that there is a biholomorphic [imath]g[/imath] on the unit disk such that [imath]g(z)^2=f(z^2)[/imath]. Show that this function satisfies also [imath]g(-z)=-g(z)[/imath]. My attempt: Consider the power series expansion [imath]f(z)=f'(0)z+\cdots[/imath] so [imath]f(z^2)=f'(0)z^2+\cdots=z^2(f'(0)+\cdots)[/imath] Therefore, we have [imath] g(z)=z\cdot(f'(0)+\cdots)^{1/2}[/imath] defined on a neighborhood of [imath]0[/imath]. But how to extend the function to the whole unit disk? (Schwarz reflection principal?) And how to make the function to be an odd function? There seems to be a similar question here, but one important condition in that question does not hold here. So both ways do not work. | 2468336 | Prove the existence of an analytic function
Let [imath]\Delta=\{z\in \mathbb C: |z| < 1\}[/imath]. Let [imath]f: \Delta \rightarrow \mathbb C[/imath] be a one-to-one analytic function fixing the origin. Prove that there is a one-to-one analytic function [imath]g: \Delta \rightarrow \mathbb C[/imath] such that [imath][g(z)]^2=f(z^2)[/imath]. Further, show that such a function is odd. The only idea which comes to my mind is the following (not sure whether it is the right direction though). Let [imath]h: \Delta \rightarrow \Delta \rightarrow \mathbb C,\ z\mapsto z^2\mapsto f(z^2)[/imath]. Note that [imath]h'(0)=0,\ h''(z)=2f'(z^2)+4z^3f''(z^2)[/imath]; since [imath]f[/imath] is one-to-one and analytic, its derivative never vanishes, so [imath]h''(0)\ne 0[/imath]. So [imath]h[/imath] is an analytic function of [imath]\Delta[/imath] with a zero of order two at the origin. Then there is an analytic function [imath]g[/imath] defined in an open disc about the origin with a simple zero at the origin such that [imath][g(z)]^2=f(z^2)[/imath]. This is the only thing I came up with, but it seems to be a little bit irrelevant since I don't see any way of extending this [imath]g[/imath] to the whole [imath]\Delta[/imath], and also the injectivity and oddness of such a [imath]g[/imath] are unclear. |
2468155 | Proving that: [imath] (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)[/imath]
This problem is from Challenge and Thrill of Pre-College Mathematics: Prove that [imath] (a^3+b^3)^2\le (a^2+b^2)(a^4+b^4)[/imath] It would be really great if somebody could come up with a solution to this problem. | 2167148 | Inequalities help: [imath](a^3+b^3)^2\leq (a^2+b^2)(a^4+b^4)[/imath]
Prove: that [imath](a^3+b^3)^2\leq (a^2+b^2)(a^4+b^4)[/imath] for all real numbers [imath]a[/imath] and [imath]b[/imath]. My attempt: [imath](a^2+b^2)(a^4+b^4)=a^6+a^2 b^4+b^2 a^4+b^6\geq 4\cdot(a^6\cdot b^6\cdot a^2 b^4\cdot a^4 b^2)^{\frac{1}{4}}=4\cdot a^3\cdot b^3[/imath], by AM-GM. Now [imath](a^3+b^3)^2\geq 4a^3\cdot b^3[/imath], by AM-GM. putting this above gives:[imath](a^2+b^2)(a^4+b^4)\geq 4\cdot a^3\cdot b^3\leq (a^3+b^3)^2[/imath] But this don't gives good, please help, on what I did. How to improve solving Inequalities? |
2467541 | proving that a function from [imath]\Bbb R^2[/imath] to [imath]\Bbb R^2[/imath] is surjective
Let [imath]f: \Bbb R^2 \to \Bbb R^2[/imath] be a continuous function such that [imath]|f(x) - x| < 2017 \space[/imath] [imath]\forall[/imath] [imath]x\in \Bbb R^2[/imath]. prove that [imath]f[/imath] is surjective. please help me prove this statement. thank you very much. | 2107837 | [imath]|x-f(x)|<5777[/imath], Show that [imath]f[/imath] is surjective
Let [imath]f:\mathbb{R}^2\to\mathbb{R}^2[/imath] be a continuous map such that [imath]\forall \ x : |x-f(x)|<5777[/imath]. Show that [imath]f[/imath] is surjective. |
2468551 | Showing [imath]\{\sin(2^kx): k\in \mathbb{N}\}[/imath] is linearly independent
How would I go about showing that the set [imath]\{\sin(2^kx): k\in \mathbb{N}\}[/imath] is linearly independent without using any integration. I am really stuck on this and I don't know where to begin. | 2466459 | Proving something is a linearly independent subset
Let [imath]F(\mathbb{R},\mathbb{R})[/imath] be the vector space of all functions from [imath]\mathbb{R}[/imath] to [imath]\mathbb{R}[/imath]. Prove that for all [imath]n\in\mathbb{N}[/imath], the set of vectors [imath]\left\{sin(x), sin(2x),sin(2^2x),\dots,sin(2^nx)\right\}[/imath] form a linearly independent subset of [imath]F(\mathbb{R},\mathbb{R})[/imath]. I know that to form a linearly independent subset the set must be closed under addition, closed under scalar multiplication, and contain the neutral element. Showing that it contains the neutral element is relativley straight forward since when [imath]x=0[/imath] every element also equals [imath]0[/imath]. Therefore this subset contains the neutral element. However, I am having a hard time coming up with a way to approach the parts showing that it is closed under scalar multiplication and addition for all [imath]x[/imath]. Do you have a piece of advice or a hint on how to approach those part? |
1710010 | To calculate the integral [imath]\int_{0}^\pi e^{k \cos \theta} \cos(k \sin \theta)d\theta[/imath]
Question: let [imath]\Gamma[/imath] be the unit circle [imath]\{e^{i\theta}| 0\leq \theta \leq 2\pi\}[/imath]. Calculate the integral [imath]\int_{\Gamma} \frac{e^{kz}}{z} dz[/imath] and hence evaluate the integral [imath]\int_{0}^\pi e^{k \cos \theta} \cos(k \sin \theta) d\theta[/imath]. My answer: The first one is easy to evaluate you simply look at the function [imath]\frac{e^{kz}}{z}[/imath] and notice that it has a simple pole at 0 and the residue is 1 so the integral is [imath] 2\pi i[/imath]. The solution to the next part evades me. So clearly we have to convert the first integral into a form of the second but how? | 334652 | Calculating the following integral using complex analysis: [imath]\int_{0}^{\pi}e^{a\cos(\theta)}\cos(a\sin(\theta))\, d\theta[/imath]
I am trying to solve a question from my complex analysis test that I didn't manage to do during the test in order to practice for the next exam. The problem is as follows: Calculate [imath]\int_{0}^{\pi}e^{a\cos(\theta)}\cos(a\sin(\theta))\, d\theta[/imath] Where the method used should be using complex analysis. What I tried: I have noticed [imath]e^{a\cos(\theta)}\cos(a\sin(\theta))=Re(e^{acis(\theta)})[/imath] where [imath]cis(\theta):=\cos(\theta)+i\sin(\theta)[/imath] Hence the integral is [imath]\int_{0}^{\pi}Re(e^{a(\cos(\theta)+i\sin(\theta))}\, d\theta[/imath] I did the change of variables: [imath]z=ae^{i\theta}[/imath], [imath]dz=iae^{i\theta}\, d\theta\implies d\theta=\frac{dz}{iz}[/imath]. When [imath]\theta=0[/imath] we have that [imath]z=a[/imath] and when [imath]\theta=\pi[/imath] we get [imath]z=-ia[/imath] and so I wrote that the integral is [imath]\int_{a}^{-ia}Re(e^{z})\,\frac{dz}{iz}[/imath] Now I don't understand how I got [imath]e^{z}[/imath], which seems wrong to me, but what the checker actually marked in red and wrote a question mark by were the integration limits: [imath]a,-ia[/imath] . Can someone please help me understand what was wrong with what the integration limits, and how to actually solve this integral ? Any help is greatly appreciated! |
2468908 | In an inner product space, can we have [imath]\langle\sum_{n=1}^\infty x_n,y\rangle=\sum_{n=1}^\infty\langle x_n,y\rangle[/imath]?
In an inner product space [imath](X,\langle\cdot,\cdot\rangle)[/imath], we have [imath]\langle x_1+x_2,y\rangle=\langle x_1,y\rangle+\langle x_2,y\rangle[/imath]. If [imath]\sum_{n=1}^\infty x_n[/imath] is convergent in [imath]X[/imath], then can we have [imath]\langle\sum_{n=1}^\infty x_n,y\rangle=\sum_{n=1}^\infty\langle x_n,y\rangle[/imath]? Edit: If yes, how to prove it? | 35124 | Summation of inner products
I can't seem to find a way of asking a sub-question in relation to does linearity of inner product hold for infinite sum, which is in itself too generic a question for my purposes. Could someone please confirm or deny my understanding that the following: [imath]\left\langle \sum_{n=1}^\infty x_n, y\right\rangle =\sum_{n=1}^\infty \langle x_n, y\rangle [/imath] is always true in a Hilbert space equipped with the norm induced by its inner product, where the convergence on the LHS is convergence in the said norm and the convergence on the RHS is convergence in the norm of the scalar field. Thank you very much. |
2468734 | check that [imath]p(t)=\lambda e^{-\lambda t}[/imath] is a density function
I have to prove the following is a density function [imath]p(t)=\lambda e^{-\lambda t}[/imath] when [imath]t>=0[/imath]. [imath]p(t) = 0[/imath] for [imath]t<0[/imath] If I understand correctly, if solve the integral, I should get [imath]1[/imath], because this the definition of a density function. I tried the following... [imath]\int_{0}^{\infty}\int_{-\infty}^{\infty}\lambda e^{-\lambda t}[/imath] ...which resulted in... [imath]\int_{0}^{\infty}[(-\lambda e^{-\lambda t})/\lambda]^{\infty}_{-\infty}[/imath] I do not know how to proceed. If I just plug in the infinities then I [imath]0[/imath] I think. How do I check if the function is a density function? | 206050 | How do I tell if this function is a probability density function?
If I have [imath]f(x)=\sin(x\pi/10)\qquad\text{for}\;0\leq x\leq10.[/imath] How do I tell if it is a probability density function? And if it isn't how do I normalize it? |
2468975 | Prove the inequality [imath]2|xy|≤x^2+y^2[/imath]
As far as I know, there are no special conditions other than this is a space with only real numbers. I'm not sure how to move forward with the problem. I know that [imath]x^2 = |x|^2[/imath] but I don't know how to make use of that. [imath]2|xy|≤ x^2 + y^2[/imath] | 929199 | Prove that [imath]2|ab| \leq a^2 + b^2[/imath] and [imath]|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}[/imath]
I am having issues solving these problems: I tried using [imath](|a|+|b|)^2 \geq 0[/imath] and [imath](|a|-|b|)^2 \geq 0[/imath] but I am having problems constructing the proof. I need to gain some intuition on how to proceed. Show that for [imath]a,b\in (-\infty,+\infty)[/imath], the following inequalities apply: a) [imath]2|ab| \leq a^2 + b^2[/imath] b) [imath]|a|+|b| \leq \sqrt {2}(a^2+b^2)^{1/2}[/imath] |
1482075 | Convergence of cos, sin, tan functions
In Radian mode, continually pressing the [imath]\cos[/imath] function of a number causes the result to converge to [imath]x=0.739085133[/imath], a fixed point of [imath]\cos(x)[/imath]. Repeating this behavior with the [imath]\sin[/imath] function causes the result to converge to [imath]x=0[/imath], a fixed point of [imath]\sin(x)[/imath]. What happens if this is done using the tan function? It seems that [imath]x=0[/imath] is a repelling fixed point with no convergence in this case. What would be the best way to explain these results mathematically? | 2644481 | [imath]\cos(\cos(\cdots(\cos(x))\cdots))[/imath]
[imath]\forall n\in \mathbb N[/imath], Let [imath]f_n (x)=\underbrace{\cos(\cos(\cos(\cdots (\cos}_n(x\underbrace{))\cdots)}_n[/imath] It is known that [imath]\forall x\in\mathbb R,\space \lim_{n\rightarrow\infty}f_n(x)=0.7390851332\dots[/imath]Does the convergence uniform? |
2469060 | Why is the lim inf the union of intersections
For my statistics class we had elementary set theory. It was stated that: [imath]\inf_{k\geq n } A_k = \bigcap\limits_{k=n}^{\infty} A_k[/imath] and [imath]\sup_{k\geq n } A_k = \bigcup\limits_{k=n}^{\infty} A_k[/imath] From this was deduced that: [imath]\lim\limits_{n\to\infty} \inf A_k = \bigcup\limits_{n=1}^{\infty} \bigcap\limits_{k=n}^{\infty} A_k[/imath] and [imath]\lim\limits_{n\to\infty} \sup A_k = \bigcap\limits_{n=1}^{\infty} \bigcup\limits_{k=n}^{\infty} A_k[/imath] I absolutely have no idea why. Could someone explain it to me in the least technical way possible? I neither get why the intersection of Ak from n onwards should be the infimum nor why the union of all intersections should be the limit of that infimum. | 172167 | Interpretation of limsup-liminf of sets
What is an intuitive interpretation of the 'events' [imath]\limsup A_n:=\bigcap_{n=0}^{\infty}\bigcup_{k=n}^{\infty}A_k[/imath] and [imath]\liminf A_n:=\bigcup_{n=0}^{\infty}\bigcap_{k=n}^{\infty}A_k[/imath] when [imath]A_n[/imath] are subsets of a measured space [imath](\Omega, F,\mu)[/imath]. Of the first it should be that 'an infinite number of those events is verified', but I don't see how to explain (or interpret this). Thanks for any help! |
2469599 | What is [imath]GL_n(\mathbb{R}) /SL_n(\mathbb{R})[/imath]?
[imath]SL_n(\mathbb{R})[/imath] is the set of all real n by n matrices whose determinant is 1. [imath]GL_N(\mathbb{R})[/imath] is the set of all real n by n matrices whose determinant is non-zero. I have proved that [imath]SL_n[/imath] is normal in [imath]GL_n[/imath]. Then What is [imath]GL_N(\mathbb{R})/SL_n(\mathbb{R})[/imath]? I mean, can we find another expression for [imath]GL_n /SL_n=\{ASL_n \mid A \in GL_n \}[/imath]? My guess is it's the set of all real n by n matrices whose determinant is not 1. is my guess correct? or what should it be? | 1966320 | Prove [imath]GL_n(\Bbb{R})/SL_n(\Bbb{R}) \cong \Bbb{R}^\times[/imath]
[imath]GL_n(\Bbb{R})/SL_n(\Bbb{R}) \cong \Bbb{R}^\times[/imath] This is trivial to prove with the first isomorphism theorem - by using the determinant as a endomorphism, then as [imath]SL_n(\Bbb{R})[/imath] is [imath]1[/imath] under the determinant, it is the kernel, and by FIT, the above is proved. I was wondering how to prove this without the isomorphism theorem ? |
2469905 | For continuous functions, if they are equal on [imath]\mathbb{Q}[/imath] then they are equal on [imath]\mathbb{R}[/imath]
Let [imath]f[/imath] and [imath]g[/imath] be functions, and suppose [imath]f[/imath] and [imath]g[/imath] are continuous (in particular, this means [imath]D(f)=\mathbb{R}\; D(g)=\mathbb {R}[/imath]). Suppose that for every rational number [imath]x\in\mathbb{Q}[/imath], we have [imath]f(x)=g(x).[/imath] Prove that [imath]f(x)=g(x)[/imath] for [imath]{every}[/imath] number [imath]x\in\mathbb{R}[/imath]. Any suggestions on how I am supposed to even begin to approach this problem would help. | 590690 | Prove continuous functions [imath]f[/imath] and [imath]g[/imath] are equal
Suppose [imath]f,g[/imath]: [imath]\Bbb R[/imath] → [imath]\Bbb R[/imath] are continuous functions such that [imath]f(r) = g(r)[/imath] for all [imath]r[/imath] in [imath]\Bbb Q[/imath]. That is to say, [imath]f[/imath] and [imath]g[/imath] are equal on the rational numbers. Prove that [imath]f(x) = g(x)[/imath] for all [imath]x[/imath] in [imath]\Bbb R[/imath] |
2468418 | Linear independence in Jordan Normal Basis
I am trying to understand how do we form a Jordan Normal Basis and more specifically why the chains are linearly independent. I saw some relevant question, but still, I can't figure that out. We start with a highest order eigenvector and form its Jordan chain. Then, if we haven't had yet a basis, we take another eigenvector of highest possible order and form its Jordan chain and we continue like this. Why will two jordan chains we obtain be linearly independent? Thanks in advance edit:In this post, he says: The other method is to first determine the lengths of the chains (there are formulas you can use to find them). If the longest chains have length kk, then you find linearly independent vectors in [imath](T−\lambda I)^{k−1}[/imath] that are not in [imath](T−\lambda I)^{k−2}[/imath]. Then you apply [imath](T−\lambda I)[/imath] to them, and if necessary find linearly independent vectors from the ones you have so far that are in [imath](T−\lambda I)^{k−2}[/imath] but not in [imath](T−\lambda I)^{k−3}[/imath]. You continue this way, always selecting linearly independent vectors, until you are down to finding enough linearly independent vectors in [imath](T−\lambda I)[/imath] besides the ones you already have; you will be guaranteed they are linearly independent because at each stage we are always selecting vectors that are linearly independent. My question is actually, how do we know that after applying [imath](T−\lambda I)[/imath] to the vectors we already have selected, we get a linearly independent set of vectors? | 131819 | Are Jordan chains always linearly independent?
Firstly, I know this question might seem slightly vague, but I can't find any examples or specific questions to specify it. My question is in regards to finding the Jordan Canonical Form of a matrix, more specifically when you have two Jordan chain of length greater than [imath]1[/imath] for a particular eigenvalue. After you've found the eigenvectors, are these eigenvectors always linearly independent? Also, if they are, is it possible to find eigenvectors which form Jordan chains which are 'wrong' because they aren't linear independent? |
2469954 | Prove that if G is a graph with [imath]\chi[/imath](G-v-w)=[imath]\chi[/imath](G)-2 for every pair of vertices v and w in G, then G is complete.
Prove that if G is a graph with [imath]\chi[/imath](G-v-w)=[imath]\chi[/imath](G)-2 for every pair of vertices v and w in G, then G is complete. I think I can use contradiction method but I don't know how to do. | 1561592 | Prove that if [imath]\chi(G-u-v)=\chi(G)-2[/imath] for every vertices [imath]u, v[/imath] ([imath]u \ne v[/imath]) then G is complete graph.
I'm trying to prove this by contradiction: if [imath]G[/imath] isn't complete graph, there must exist vertices [imath]u,v \in V(G)[/imath] for which edge [imath]uv \notin E(G)[/imath]. Then [imath]u[/imath] and [imath]v[/imath] must have the same color in proper coloring [imath]C[/imath] and now I'd like to prove that by removing [imath]u, v[/imath] we have [imath]\chi(G-u-v) = \chi(G) - 1[/imath] (that would be the contradiction), but I'm not sure whether it's true. |
2470443 | Inverse of a continuos function
If [imath]f:I\subseteq \Bbb R \rightarrow \Bbb R[/imath] is injective and everywhere continuous in its domain then [imath]f^{-1}[/imath] is also continuous everywhere. Here [imath]I[/imath] cannot be discrete set and [imath]f[/imath] is single variable function The above statement is always correct or not? I thought it is always correct. Am I correct or not? | 2470282 | Inverse of function is continuous or not?
If [imath]f:I\subseteq \Bbb R \rightarrow \Bbb R[/imath] is injective and everywhere continuous in its domain then [imath]f^{-1}[/imath] is also continuous everywhere. Here [imath]I[/imath] cannot be discrete set and [imath]f[/imath] is single variable function The above statement is always correct or not? I thought it is always correct. Am I correct or not? |
2471173 | What is the convolution of [imath] t^m \ast t^n [/imath] where m and n are general positive integers?
I want to work out a general form of the convolution [imath] t^m \ast t^n [/imath]. I started with: [imath] t^m \ast t^n = \int_0^t (t - \tau)^m \tau^n \ d \tau [/imath] Then using the binomial expansion: [imath] \int_0^t \tau^n(\sum_{k=0}^{m} \binom{m}{k}t^{m-k}(-\tau)^k) \ d \tau [/imath] Integration by parts - let [imath] u = \sum (...)[/imath] and [imath] v' = \tau^n \ d\tau [/imath] : [imath] \frac{\tau ^{n+1}}{n+1}(\sum_{k=0}^{m} \binom{m}{k}t^{m-k}(-\tau )^k) - \int \tau^n(\frac{\mathrm{d} }{\mathrm{d} \tau } (\sum_{k=0}^{m} \binom{m}{k}t^{m-k}(-\tau )^k) ) [/imath] But I am unsure on how to differentiate the sum with the binomial in it. | 2146252 | Defined integral of [imath]x^{\alpha} (1-x)^{\beta-1}[/imath]
How can one prove the following equality? [imath]\int_{0}^{1} x^{\alpha} (1-x)^{\beta-1} \,dx = \frac{\Gamma(\alpha+1) \Gamma(\beta)}{\Gamma(\alpha + \beta + 1)}[/imath] |
182860 | What is the number of triangles with integer sides, given the length of the longest side?
Suppose [imath]a,b,c \in\mathbb N[/imath], and the value of [imath]c[/imath] is known and fixed, while [imath]a[/imath] and [imath]b[/imath] are unknown and are both smaller than [imath]c[/imath]. What is the total number of unique triangles possible with [imath]a, b[/imath] and [imath]c[/imath] as its sides? | 2527839 | The sides of a triangle are [imath]a,b,c[/imath], which are integers, and [imath]a\le b\le c[/imath]. If [imath]c[/imath] is given, find the number of triangles possible?
The sides of a triangle are [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath] inches, where [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath] are integers and [imath]a\le b\le c[/imath]. If [imath]c[/imath] is given, find the number of triangles possible. I don't know from where to start this question. Please provide the solution. |
920483 | Test the convergence of the series [imath]\sum \sin [\pi(\sqrt5+2)^n][/imath]
I have just approached the following series [imath]\sum_{n=1}^\infty \sin [\pi(\sqrt5+2)^n][/imath] And I already have a question. The [imath]\lim_{n \to \infty}\pi(\sqrt5+2)^n=+\infty[/imath]. And the [imath]\lim_{n\to \infty} \sin[\pi(\sqrt5+2)^n][/imath] does not exist. Can [imath]\sum a_n[/imath] converge if [imath]\lim_{n\to \infty}a_n[/imath] does not exist? I guess the answer is yes because I know that [imath]\sum a_n[/imath] cannot converge if [imath]\lim_{n\to \infty}a_n\neq 0[/imath]. Am I right? The second part of my question concerns the strategy to use to test convergence of such a series. I can test convergence of series with standard tests. But it seems I need to develop a more sophisticated approach in order to successfully deal with series like this. For example, here, should I use some sort of expansion? If yes, what kind of expansion? Thank you for your help. | 97548 | Does [imath]\sum_{n=1}^{\infty} \sin(\pi(2+\sqrt{3})^n)[/imath] converge? Converge absolutely?
Per the title, does [imath]\displaystyle\sum_{n=1}^{\infty} \sin(\pi(2+\sqrt{3})^n)[/imath] converge? Converge absolutely? I'm stuck on this question, not sure how to approach it. Thank you! |
2471390 | How can I prove that the following function is bounded?
How can I prove that the following function is bounded? [imath]f(x) = \frac{1-x^2}{4 + x^2},[/imath] And I wonder if the following 2 hints may help? Hints: [imath](x + 1)^2 \geq 0.[/imath] [imath](x - 1)^2 \geq 0.[/imath] thanks! | 2471368 | How can I prove that this function is bounded using the given 2 hints?
How can I prove that this function is bounded using the given 2 hints? [imath]f(x) = \frac{x}{x^2 + 1},[/imath] Hints: [imath](x + 1)^2 \geq 0.[/imath] [imath](x - 1)^2 \geq 0.[/imath] thanks! |
2090022 | Proof that [imath]\vert\int_a^bf(t)dt\vert\le\int_a^b\vert f(t)\vert dt [/imath] for complex integration
The following proof is given in my notes: Write the complex integral [imath]\int_a^bf(t)dt = Re^{i\theta}[/imath] where [imath] R=\int_a^bf(t)dt. \quad(1) [/imath] On the other hand, [imath] R=\int_a^be^{-i\theta}f(t)dt.\quad(2) [/imath] Write [imath]e^{-i\theta}f(t)=U(t)+iV(t)[/imath] where [imath]U(t)[/imath] and [imath]V(t) [/imath] are real-valued functions. Then because R is real, [imath] R=\int_z^bU(t)dt\quad(3) [/imath] But now, [imath] U(t)= Re\ e^{-i\theta}f(t) \le e^{-i\theta}f(t)=\vert f(t)\vert\quad(4) [/imath] Therefore, from the properties of real integrals, [imath] \int_a^bU(t)dt\le\int\vert f(t)\vert dt [/imath] which proves the desired result. I have trouble figuring out how each of the numbered equations actually hold. After writing the integral as [imath]\int_a^bf(t)dt = Re^{i\theta}[/imath] How can we say that (1) and (2) hold. (3) sort of makes sense as R has to be real. | 1610398 | property of integral of complex valued function of real variable
i have a question about this. if [imath]f(t):[a,b] \Rightarrow \mathbb{C}[/imath] is a continuous function, then is there a property in complex analysis that states, [imath] | \int_a^b f(t) dt | \leq \int_a^b |f(t)| dt [/imath] ?? I know there is one for real valued function, if yes, then is there a simple proof for it that use this property for real function to extend it to the complex case?? thank you. |
2471322 | Proof that [imath]\det(A^{-1})=\frac{1}{\det(A)}[/imath]
We can assume that A is [imath]n{\times}n[/imath] matrix. Prove that [imath]\det(A^{-1})=\frac{1}{\det(A)}[/imath] We know that [imath]AA^{-1}=I[/imath]. If someone could provide some insight on how to prove this that would be much appreciated Thanks, Tuki | 1857317 | Why [imath]\det(A^{-1}) = 1/\det(A)[/imath]?
I was hoping that someone could point me in the right direction on how I could start answering the following question: Based on the fact that [imath]\det(AB)=\det(A)\det(B)[/imath] for all square [imath]n \times n[/imath] matrices [imath]A,B[/imath], explain in every detail why: [imath]\det(A^{-1}) = 1/\det(A).[/imath] |
2471625 | The sum [imath]\sum_{k=1}^{\infty}(-1)^k \frac {\log(2k+1)}{(2k+1)}[/imath]
What can we say about [imath]\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^k\log (2k+1)}{2k+1}[/imath]? Clearly it takes a value of negative real number, and seems to be an analogy of [imath]\displaystyle \sum_{k=0}^{\infty}\frac{(-1)^k}{2k+1}=\frac{\pi}{4}[/imath], but I don't know the way of evaluating [imath]\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^k\log (2k+1)}{2k+1}[/imath]. | 329653 | Evaluating a slow sum
In my integration adventures, I came across this sum which I could not simplify: [imath]\sum_{n=1}^{\infty}\frac{(-1)^{n}\log(2n+1)}{2n+1}[/imath] Wolfram seems to believe the sum diverges and is not of much help here. Does a closed form for this sum exist? If not, can this sum be transformed nicely that has faster convergence? |
2471674 | Axiom of extensionality - why can't we replace [imath]\Rightarrow[/imath] with [imath]\iff[/imath]?
The axiom of extensionality says that [imath](\forall(x,y))\big(\forall(z)(z\in x\iff z\in y)\Rightarrow x=y\big)[/imath] Does it not work the other way around? [imath] (x=y)\Rightarrow\forall z(z\in x\iff z\in y) [/imath] If two sets are equal and [imath]z[/imath] is in one of them, then - by definiton - it has to be in the other one. If it is not true, could you show a counter-example? | 396748 | ZF Extensionality axiom
To familiarize myself with axiomatic set theory, I am reading Kenneth Kunen's The Foundations of Mathematics that presents ZF set theory. I haven't gotten really far since I am stuck at the axiom of Extensionality, stated as follows: [imath]\forall x,y \; [\forall z(z \in x \leftrightarrow z \in y) \rightarrow x = y][/imath] As far as I understood it, the purpose of this axiom is to state that every two sets that have exactly the same members are the same set. In terms of the above formula, when antecedent is true, the consequent needs to be true. What confuses me here is the case when the antecedent is false. Then, the consequent can be false or true, i.e., we can't say much about the consequent. But, don't we really want to say that in that case those two sets are not equal? More precisely, shouldn't we use equivalence instead of implication? If not, what is the reason? |
2471904 | Evaluating the integral:[imath]\int_0^\infty x\left(\log\left(\frac1{x^2}\right)\right)^n dx[/imath]
[imath]\int_0^\infty x\left(\log\left(\frac1{x^2}\right)\right)^n dx[/imath] put [imath]\log\left(\frac1{x^2}\right)=y^{1/n}[/imath] [imath]I=\frac{\ln10}{2n}\int_{-\infty}^\infty10^{-y^{1/n}}y^{1/n}dy[/imath] put [imath]10^{-y^{1/n}}=e^{-t}\implies[/imath] [imath]I=\frac1{(2\ln10)^n}\int_{-\infty}^0e^{-t}t^ndt+\frac1{(2\ln10)}\int_0^\infty e^{-t}t^ndt[/imath] [imath]I=\frac1{(2\ln10)^n}\int_{-\infty}^0e^{-t}t^ndt+\frac{\Gamma(n+1)}{(2\ln10)}[/imath] I want to know how can I evaluate the first integral in the last step of my turn I used some substitutions so i get two integrals one of them converted to the formula of gamma function but the second could not be so because the domain of integration is different ? [imath]n[/imath] is a natural number | 2471819 | Evaluate an proper integral:[imath]\int_0^\infty x\left(\log\left(\frac1{x^2}\right)\right)^n dx[/imath]
[imath]\int_0^\infty x\left(\log\left(\frac1{x^2}\right)\right)^n dx[/imath] put [imath]\log\left(\frac1{x^2}\right)=y^{1/n}[/imath] [imath]I=\frac{\ln10}{2n}\int_{-\infty}^\infty10^{-y^{1/n}}y^{1/n}dy[/imath] put [imath]10^{-y^{1/n}}=e^{-t}\implies[/imath] [imath]I=\frac1{(2\ln10)^n}\int_{-\infty}^0e^{-t}t^ndt+\frac1{(2\ln10)}\int_0^\infty e^{-t}t^ndt[/imath] [imath]I=\frac1{(2\ln10)^n}\int_{-\infty}^0e^{-t}t^ndt+\frac{\Gamma(n+1)}{(2\ln10)}[/imath] I want to know how can I evaluate the first integral in the last step of my turn I used some substitutions so i get two integrals one of them converted to the formula of gamma function but the second could not be so because the domain of integration is different ? [imath]n[/imath] is a natural number |
1126235 | Radius of convergence powers series s.t seriex [imath] \sum |a_n| [/imath]diverges
[UGC-CSIR-NET 2014 December] Let [imath]\{a_n:n\ge 1\}[/imath] be sequence of real numbers such that the series [imath]\sum a_n[/imath] converges and the series [imath]\sum |a_n|[/imath] diverges. Let [imath]R[/imath] be the radius of convergence of the power series [imath]\sum a_n x^n[/imath]; then which of the following is true. [imath]0\lt R \lt 1[/imath] [imath]1\lt R \lt \infty[/imath] [imath]R=1[/imath] [imath]R=\infty[/imath] In this question, I am a little confused regarding the role of the series [imath]\sum a_n[/imath] in its power series [imath]\sum a_n x^n[/imath] in earlier questions that I have faced, I find the radius of convergence of absolute version of coefficient of power series, where absolute version converges for those 'x' values origional power series also converges. Is that right? But as we see here absolute version diverges, so what does it imply we can't use ratio, root test etc to find radius of convergence here? | 1296307 | Radius of convergence of a power series with [imath]a_n[/imath] convergent
Let [imath]\{a_n:n \geq 1\}[/imath] be a sequence of real numbers such that [imath]\sum_{n=1}^{\infty} a_n[/imath] is convergent and [imath]\sum_{n=1}^{\infty} |a_n|[/imath] is divergent. Let [imath]R[/imath] be the radius of convergence of the power series [imath]\sum_{n=1}^{\infty} a_n x^n[/imath]. Then what is [imath]R[/imath] ? I dont know how to start, if the series is given we could use Radius of convergence formula, but how to solve this question?Pls help |
2472212 | How can I solve [imath]\int \sqrt{1+x^2}\,dx[/imath]?
[imath]\int \sqrt{1+x^2}\,dx[/imath] I've tried some step-by-step calculators, but what they give is way beyond my level. Is there any "simple" way to solve this? | 2125242 | How to integrate [imath]\int\sqrt{x^2+1}dx[/imath]
[imath]\int\sqrt{x^2+1}dx[/imath] I've been attempting this problem for days and I've made very little progress. I understand that I should substitute [imath]x=\tan(x)[/imath] and [imath]dx=\sec^2(x)dx[/imath] but beyond that I am very much lost. |
2472560 | Show that the series is convergent?
[imath]a_n=(1+\frac{1}{n})^n[/imath] and [imath]n\geq 1[/imath] I check first [imath]n\to \infty[/imath] then [imath]\lim_{n\to \infty}=(1+\frac{1}{\infty})^\infty[/imath] [imath]=(1+0)^\infty=1^\infty[/imath] I don't find the next step.Help me | 389793 | What is the most elementary proof that [imath]\lim_{n \to \infty} (1+1/n)^n[/imath] exists?
Here is my candidate for the most elementary proof that [imath]\lim_{n \to \infty}(1+1/n)^n [/imath] exists. I would be interested in seeing others. [imath]***[/imath] Added after some comments: I prove here by very elementary means that the limit exists. Calling the limit "[imath]e[/imath]" names it. [imath]***[/imath] It only needs Bernoulli's inequality (BI) in the form [imath](1+x)^n \ge 1+nx[/imath] for [imath]x > -1[/imath] and [imath]n[/imath] a positive integer, with equality only if [imath]x = 0[/imath] or [imath]n = 1[/imath]. This is easily proved by induction: It is true for [imath]n=1[/imath], and [imath](1+x)^{n+1} = (1+x)(1+x)^n \ge (1+x)(1+nx) = 1+(n+1)x+nx^2 \ge 1+(n+1)x [/imath]. (If [imath]-1 < x < 0[/imath], if [imath]1+nx \ge 0[/imath], the above proof goes through, and if [imath]1+nx < 0[/imath], [imath]1+mx < 0[/imath] for all [imath]m \ge n[/imath] so certainly [imath](1+x)^m > 1+mx[/imath].) This proof originally appeared in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563 and uses the arithmetic-geometric mean inequality (AGMI) in the form [imath]\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i[/imath] (all [imath]v_i[/imath] positive) with equality if and only if all the [imath]v_i[/imath] are equal. Let [imath]a_n = (1+1/n)^n[/imath] and [imath]b_n = (1+1/n)^{n+1}[/imath]. We will prove that [imath]a_n[/imath] is an increasing sequence and [imath]b_n[/imath] is an decreasing sequence. Since [imath]a_n < b_n[/imath], this implies, for any positive integers [imath]n[/imath] and [imath]m[/imath] with [imath]m < n[/imath] that [imath]a_m < a_n < b_n < b_m[/imath]. For [imath]a_n[/imath], consider n values of [imath]1+1/n[/imath] and [imath]1[/imath] value of [imath]1[/imath]. Since their sum is [imath]n+2[/imath] and their product is [imath](1+1/n)^n[/imath], by the AGMI, [imath]((n+2)/(n+1))^{n+1} > (1+1/n)^n[/imath], or [imath](1+1/(n+1))^{n+1} > (1+1/n)^n[/imath], or [imath]a_{n+1} > a_n[/imath]. For [imath]b_n[/imath], consider [imath]n[/imath] values of [imath]1-1/n[/imath] and [imath]1[/imath] value of [imath]1[/imath]. Since their sum is [imath]n[/imath] and their product is [imath](1-1/n)^n[/imath], by the AGMI, [imath](n/(n+1))^{n+1} > (1-1/n)^n[/imath] or [imath](1+1/n)^{n+1} < (1+1/(n-1))^n[/imath], or [imath]b_n > b_{n+1}[/imath]. Since [imath]b_n-a_n = (1+1/n)^{n+1} - (1+1/n)^n = (1+1/n)^n(1/n) =a_n/n [/imath] and every [imath]a_n[/imath] is less than any [imath]b_n[/imath] and [imath]b_3 = (1+1/3)^4 = 256/81 < 4[/imath], [imath]b_n-a_n < 4/n[/imath], so [imath]b_n[/imath] and [imath]a_n[/imath] converge to a common limit. These proofs do not seem to be really elementary, since they use the AGMI. However, they use a special form of the AGMI, where all but one of the values are the same, and this will now be shown to be implied by BI, and thus be truly elementary. Suppose we have [imath]n-1[/imath] values of [imath]u[/imath] and [imath]1[/imath] value of [imath]v[/imath] with [imath]u[/imath] and [imath]v[/imath] positive. The AGMI for these values is [imath](((n-1)u+v)/n)^n \ge u^{n-1}v[/imath] with equality if and only if [imath]u = v[/imath]. We will now show that this is implied by BI: [imath](((n-1)u+v)/n)^n \ge u^{n-1}v[/imath] is the same as [imath](u+(v-u)/n)^n \ge u^n(v/u)[/imath]. Dividing by [imath]u^n[/imath], this is equivalent to [imath](1+(v/u-1)/n)^n \ge v/u[/imath]. By BI, since [imath](v/u-1)/n > -1/n > -1[/imath], [imath](1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u[/imath] with equality only if [imath]n=1[/imath] or [imath]v/u-1 = 0[/imath]. Thus BI implies this version of the AGMI. |
2472488 | If [imath]A[/imath] is a linear map, then [imath]A(\alpha(t))'=A(\alpha'(t))[/imath]
Let [imath]A:\mathbb R^3\to \mathbb R^3[/imath] be a linear map. I need the following technical result to solve a problem I'm solving: [imath]A(\alpha(t))'=A(\alpha'(t))[/imath] Where [imath]\alpha:I\to \mathbb R^3[/imath] is a parametric curve. Intuitively, I know this is true, but I would like to prove it formally. | 153836 | Derivative of a linear transformation.
We define derivatives of functions as linear transformations of [imath]R^n \to R^m[/imath]. Now talking about the derivative of such linear transformation , as we know if [imath]x \in R^n[/imath] , then [imath]A(x+h)-A(x)=A(h)[/imath], because of linearity of [imath]A[/imath], which implies that [imath]A'(x)=A[/imath] where , [imath]A'[/imath] is derivative of [imath]A[/imath] . What does this mean? I am not getting the point I think. |
2472276 | if [imath]X[/imath] is independent of [imath]Y_1....Y_n[/imath], is it independent of [imath]f(Y_1,..Y_n)[/imath]
So we have the conclusion that if [imath]X[/imath] and [imath]Y[/imath] are independent, then [imath]g(X)[/imath] and [imath]h(Y)[/imath] are independent (g, h measurable?), is it still true that for general case [imath]X[/imath] independent of [imath]g(Y_1,..Y_n)[/imath], where [imath]Y_i[/imath] are themselves not independent to each other. I don't see how this is duplicate with that of showing [imath]g(X)[/imath] and [imath]h(Y)[/imath] are independent. If it is not generally true, can we at least show however, [imath]X[/imath] is independent of the sum of [imath]Y_i[/imath]? Let me add some more condition, what about [imath]X[/imath] and [imath]Y_i[/imath] are all gaussians? And the function [imath]g[/imath] [imath]h[/imath] are nice smooth functions. Essentially, I just want to show [imath]X[/imath] is independent of sum of [imath]Y_i[/imath]. | 8742 | Are functions of independent variables also independent?
It's a really simple question. However I didn't see it in books and I tried to find the answer on the web but failed. If I have two independent random variables, [imath]X_1[/imath] and [imath]X_2[/imath], then I define two other random variables [imath]Y_1[/imath] and [imath]Y_2[/imath], where [imath]Y_1[/imath] = [imath]f_1(X_1)[/imath] and [imath]Y_2[/imath] = [imath]f_2(X_2)[/imath]. Intuitively, [imath]Y_1[/imath] and [imath]Y_2[/imath] should be independent, and I can't find a counter example, but I am not sure. Could anyone tell me whether they are independent? Does it depend on some properties of [imath]f_1[/imath] and [imath]f_2[/imath]? Thank you. |
2472787 | Combinatorial Proofs with Summation
I've been trying to think up of two combinatorial proofs but I'm confused on the first and just completely stuck on where to begin for the second. [imath]\sum_{k=-m}^n \binom{m+k}{r}\binom{n-k}{s} = \binom{m+n+1}{r+s+1}[/imath] For the RHS, I understand that it is basically looking at the number of [imath](r+s+1)[/imath] subsets that can be formed from [imath](m+n+1)[/imath]. On expanding the LHS to yield [imath]\binom{0}{r}\binom{n+m}{s}+\binom{1}{r}\binom{n+m-1}{s}...\binom{m}{r}\binom{n}{s}+\binom{m+1}{r}\binom{n-1}{s}+...\binom{m+n}{r}\binom{0}{s}[/imath], I see that it is basically adding together the various ways of drawing a constant [imath]r[/imath] and [imath]s[/imath] from two changing populations that sum up to [imath]m+n[/imath]. However, I don't actually see where the [imath]+1[/imath] comes from. Is there anything wrong with my intepretation of the LHS? [imath]\sum_{i=1}^n (i-1)(n-i) = \binom{n}{3}[/imath] I know that RHS is counting the number of subsets of 3 that can be formed from [imath]n[/imath]. However, I do not have any idea of how to be interpreting the LHS. | 2462105 | Prove that [imath]\sum_{i = n - b}^{a - m} {a - i \choose m}{b + i \choose n} = {a + b + 1 \choose m + n + 1}[/imath]
Restrictions on the variables for this identity are as follows: [imath]a \geq m \geq 0[/imath] and [imath]n \geq b \geq 0[/imath]. I have not gotten far with this, but these are my very rudimentary thoughts: We need to define a special element, somehow, and then carve up the [imath]m + n[/imath] remaining elements to be chosen out of the remaining pool of [imath]a + b[/imath] elements by writing [imath]a + b = a - i + b + i[/imath]. From the [imath]a - i[/imath] pile, we choose [imath]m[/imath] things. From the [imath]b + i[/imath] pile, we choose [imath]n[/imath] things. My problem is, what is going on with this special element? How is it being defined so as to split things in this precise way? A hint would be great for the moment. |
2472754 | Topological embedding of real projective plane into [imath]\mathbb{R}^3[/imath]
I can prove that there is no [imath]C^2[/imath] embedding of the real projective plane into [imath]\mathbb{R}^3[/imath]. In fact, every closed [imath]C^2[/imath] surface of [imath]\mathbb{R}^3[/imath] is orientable (probably this is true for the [imath]C^1[/imath] category as well). Is there a topological embedding of [imath]\mathbb{R} \mathbb{P}^2[/imath] into [imath]\mathbb{R}^3[/imath]? An answer in this question tells to use the Alexander Duality Theorem, but I don't know how. | 585026 | A non orientable closed surface cannot be embedded into [imath]\mathbb{R}^3[/imath]
Can someone please remind me how this goes? Here's the idea of proof I'm trying to recall: let [imath]S[/imath] be a closed surface (connected, compact, without boundary) embedded in [imath]\mathbb{R}^3[/imath]. Then one can define the "outward-pointing normal unit vector" to [imath]S[/imath] at any point, and subsequently an orientation of the surface. One would like to define this vector by saying that it points towards exterior points to [imath]S[/imath]. So we need some kind of generalization of the Jordan curve theorem saying that the surface cuts [imath]\mathbb{R}^3[/imath] into two pieces (interior and exterior). What is this theorem exactly? Also, I apologize if this is silly, but is there an obvious argument that a piece of the surface cuts a small tubular neighborhood of it into interior and exterior points (this seems necessary to define the outward normal vector properly)? Is there a "cleaner" approach to prove this fact? Thanks in advance. |
1570366 | The minimum value of [imath] f(x) = | x - 1 | + | x - 2 | + | x - 3 | [/imath] is?
I don't get it why my solution is wrong : My solving : [imath] f(x) = | x - 1 | + | x - 2 | + | x - 3 | [/imath] When [imath] x\leq 1 [/imath] [imath] f(x) = | x - 1 | + | x - 2 | + | x - 3 | = 0 [/imath] = [imath] 6 -3x [/imath] since [imath] x\leq 1 [/imath] [imath] f(x)\leq 3 [/imath] also it clear [imath] f(x)\geq 0 [/imath] => The min value is 0 . I dont have to consider other cases, since 0 is the min value f(x) can take . Book has given the answer : [imath] 2 [/imath] . | 1758610 | Maximum and minimum of of [imath]f(x)=|x-1|+|x-2|+|x-3|[/imath]
I am trying to find the maximums or minimums of [imath]f(x)=|x-1|+|x-2|+|x-3|[/imath] (if there exist). My attempt: First I compute the derivative and tried to find critical point, i.e, [imath]f'(x) = \frac{x-1}{|x-1|} + \frac{x-2}{|x-2|} + \frac{x-3}{|x-3|}=0[/imath], first I noted that this derivative doesn't exist in 1,2,3 for the absolute value. And the are no critic point (Is that correct?) Then I stuck here because I don't know if there is a maximum, I think that it is not exist but how can I justified and I believe that there is a global minimum, but this occurs only if I found a critical point, some help pls. |
2008760 | Prove there exist [imath]0\le c_1 where \sum_{k=1}^n\frac1{f'(c_k)}=n[/imath]
A real valued function [imath]f(x)[/imath]is differentiable on [imath][0,1][/imath], where [imath]f(0)=0[/imath] and [imath]f(1)=1[/imath]. Prove that for any integer [imath]n[/imath] there exist [imath]0\le c_1<c_2<\dots<c_n\le1[/imath] where [imath]\sum_{k=1}^n\frac1{f'(c_k)}=n[/imath] This is a question a student asked me (I'm a teacher). I have no clue on how to prove it except that I guess it may use the mean-value theorem and the intermediate value theorem. Thanks in advance. | 1092299 | For f continuous on [imath][0,1][/imath], show that there exist points [imath]\alpha_k[/imath] such that [imath]\sum \limits_{k=1}^n \frac{1}{f'(\alpha_k)} = n [/imath]
Suppose that [imath]f[/imath] is continous on [imath][0,1][/imath] , differentiable on [imath](0,1)[/imath] , and [imath]f(0)=0[/imath] and [imath]f(1)=1[/imath].For every integer [imath]n[/imath] show that there must exist [imath]n[/imath] distinct points [imath]\alpha_1,\alpha_2\cdots,\alpha_n[/imath] in that interval so that [imath]\sum \limits_{k=1}^n \frac{1}{f'(\alpha_k)} = n [/imath] Is it theorem or i can show it with basic differentiation knowledge .Please give me hint . |
2473384 | Paradox in deriving the area of an ellipse
We have an ellipse [imath]x=a\cos(\theta)[/imath] and [imath]y=b\sin(\theta)[/imath]. We will find the area by adding sectors [imath]\frac12r^2\;d\theta[/imath]. Now, [imath] \begin{align} \int_0^{2\pi}\frac12r^2\;d\theta&=\frac12\int_0^{2\pi}\left(a^2\cos^2\theta+b^2\sin^2\theta\right)\;d\theta\\ &=\frac12\int_0^{2\pi}\left(\frac{a^2+b^2}2+\frac{a^2-b^2}2\cos2\theta\right)\;d\theta\\ &=\left.\frac12\left(\frac{a^2+b^2}2\theta+\frac{a^2-b^2}4\sin2\theta\right)\right|_0^{2\pi}\\ &=\frac\pi2\left(a^2+b^2\right) \end{align} [/imath] Why is it not [imath]\pi a b[/imath]? | 493104 | Evaluating [imath]\int_a^b \frac12 r^2\ \mathrm d\theta[/imath] to find the area of an ellipse
I'm finding the area of an ellipse given by [imath]\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1[/imath]. I know the answer should be [imath]\pi ab[/imath] (e.g. by Green's theorem). Since we can parameterize the ellipse as [imath]\vec{r}(\theta) = (a\cos{\theta}, b\sin{\theta})[/imath], we can write the polar equation of the ellipse as [imath]r = \sqrt{a^2 \cos^2{\theta}+ b^2\sin^2{\theta}}[/imath]. And we can find the area enclosed by a curve [imath]r(\theta)[/imath] by integrating [imath]\int_{\theta_1}^{\theta_2} \frac12 r^2 \ \mathrm d\theta.[/imath] So we should be able to find the area of the ellipse by [imath]\frac12 \int_0^{2\pi} a^2 \cos^2{\theta} + b^2 \sin^2{\theta} \ \mathrm d\theta[/imath] [imath]= \frac{a^2}{2} \int_0^{2\pi} \cos^2{\theta}\ \mathrm d\theta + \frac{b^2}{2} \int_0^{2\pi} \sin^2{\theta} \ \mathrm d\theta[/imath] [imath]= \frac{a^2}{4} \int_0^{2\pi} 1 + \cos{2\theta}\ \mathrm d\theta + \frac{b^2}{4} \int_0^{2\pi} 1- \cos{2\theta}\ \mathrm d\theta[/imath] [imath]= \frac{a^2 + b^2}{4} (2\pi) + \frac{a^2-b^2}{4} \underbrace{\int_0^{2\pi} \cos{2\theta} \ \mathrm d\theta}_{\text{This is $0$}}[/imath] [imath]=\pi\frac{a^2+b^2}{2}.[/imath] First of all, this is not the area of an ellipse. Second of all, when I plug in [imath]a=1[/imath], [imath]b=2[/imath], this is not even the right value of the integral, as Wolfram Alpha tells me. What am I doing wrong? |
2473039 | Limit of series/n convergence
Suppose we have bounded, monotonically decreasing [imath]\epsilon_i[/imath], such that as [imath]i \rightarrow \infty, \epsilon_i \rightarrow 0[/imath]. How can I show that [imath]\lim_{n \rightarrow \infty} \frac{\sum_{i=1}^\infty \epsilon_i}{n} = 0[/imath]? Is this even necessarily true? Tried Kronecker's lemma, as well as comparison tests to no avail. Seems like a pretty fundamental fact to prove. | 1713950 | How to compute the above limit
Let [imath]y_n[/imath] be a sequence such that [imath]\lim_{n\to \infty} y_n= y[/imath] then prove that [imath] \lim \dfrac{y_1+y_2+...+y_n}{n}= y[/imath] My try: As [imath]\lim_{n\to \infty} y_n= y[/imath] then given [imath]\epsilon>0[/imath] there exists [imath]N[/imath] such that [imath]\forall n\ge N ;||y_n-y||<\epsilon [/imath]. Now [imath]||\dfrac{y_1+y_2+...+y_n}{n}-y||\le n^{-1}\{||y_1-y||+||y_2-y||+...+||y_n-y||\}[/imath] How to make this limit tend to zero? |
1619879 | Proof of [imath]a^{\frac{p-1}{2}} \equiv \left(\frac{a}{p}\right)[/imath] mod [imath]p[/imath]
[imath]p[/imath] is a prime, odd integer. [imath]a[/imath] is an integer. we assume that [imath]p[/imath] does not divide [imath]a[/imath]. [imath]\left( \frac{a}{p} \right)[/imath] denotes the Legendre symbol. In order to prove [imath]a^{\frac{p-1}{2}} \equiv \left(\frac{a}{p}\right)[/imath] mod [imath]p[/imath] my course demonstrates the following statement first (Wilson) : [imath]\left( p-1 \right)! \equiv -\left(\frac{a}{p}\right) a^{\frac{p-1}{2}}[/imath] mod [imath]p[/imath] However it is not clear to me how this implies the first statement. | 401922 | Show that [imath]\left( \frac{q}{p} \right) \equiv q^{(p-1) / 2} \mod p[/imath], where [imath]\left( \frac{q}{p} \right)[/imath] is the Legendre Symbol
Show that if [imath]p[/imath] is any odd prime then [imath]\left( \frac{q}{p} \right) \equiv q^{\frac{p-1}{2}} \mod p.[/imath] stating any theory that you use. In particular, you may assume the existence of a primitive element in [imath]G_p[/imath]. Here [imath]\left( \frac{q}{p} \right)[/imath] is the Legendre Symbol and [imath]G_p[/imath] is the group of elements [imath]g \mod p[/imath] such that [imath]\gcd(g,p) = 1[/imath]. I said that for some [imath]a \in \mathbb{Z}[/imath], we have [imath]a \equiv q^{(p-1) / 2} \mod p \implies a^2 \equiv q^{(p-1)} \equiv 1 \mod p[/imath] by Fermat's little theorem. And so, by definition of the Legendre symbol, we have that if [imath]q[/imath] is a quadratic residue mod [imath]p[/imath] then [imath]\left( \frac{q}{p} \right) \equiv q^{(p-1) / 2} \mod p[/imath]. I'm now stuck on how to show that it is [imath]\equiv -1 [/imath] if it isn't a quadratic residue. Obviously the hint with primitive elements comes into play somehow, but I can't see how it does. Can someone help me please. |
1676011 | prove that maximal ideal in [imath]\mathbb{Z}[/imath] generated by a prime number
I am trying to prove [imath](a)[/imath] is a maximal ideal in [imath]\mathbb{Z}[/imath], if and only if [imath]a[/imath] is prime number. Now I wrote: assume [imath]a \in\mathbb{Z}[/imath], while it's not prime number we can write as [imath]a=xy[/imath], for some integers [imath]x[/imath] and [imath]y[/imath]. then [imath](a)\subset (x)[/imath] and [imath](a)\subset (y)[/imath] while if [imath](a)[/imath] is maximal ideal, it's not exist [imath](k)[/imath] such that [imath](a)\subset (k)\subset\mathbb{Z}[/imath]. so [imath](a)[/imath] cannot be maximal ideal. we can say by contradictory if [imath]a[/imath] is prime, [imath](a)[/imath] is maximal ideal. is this correct? and how to prove the other way. | 2809158 | Ring Theory, Prime Number
Consider the ring of integers [imath]\mathbb Z[/imath]. Prove that if [imath]p[/imath] is a prime number, then [imath]\langle p \rangle[/imath] is a maximal ideal. I know I need to show that there is no ideal [imath]J[/imath] such that [imath]\langle p \rangle[/imath] is a subset of [imath]J[/imath]. But I do not know how to do it. |
2473295 | How to show that [imath]\partial B(0,1)[/imath] is not weakly closed set?
Let X be an infinite dimensional normed [imath]\mathbb K[/imath]-linear space .Prove that [imath]\partial B(0,1)=\{x\in X: ||x||= 1\}[/imath] is not weakly closed in [imath]X[/imath]. We know that a convex subset of [imath]X[/imath] is weakly closed iff it is closed w.r.t. the normed topology. But here [imath]\partial B(0,1)[/imath] is not convex in X. (definition of weak topology): Let [imath]X[/imath] be a normed [imath]\mathbb K[/imath] -linear space . The weak topology of [imath]X[/imath] is the smallest topology [imath]J_{\omega}[/imath](which is a subset of the normed topology of X) such that [imath]f:(X,J_{\omega})\to\mathbb K[/imath] is continuous for all [imath]f \in X^*.[/imath] Please someone give some hints. Thank you. | 153889 | Prove: The weak closure of the unit sphere is the unit ball.
I want to prove that in an infinite dimensional normed space [imath]X[/imath], the weak closure of the unit sphere [imath]S=\{ x\in X : \| x \| = 1 \}[/imath] is the unit ball [imath]B=\{ x\in X : \| x \| \leq 1 \}[/imath]. [imath]\\[/imath] Here is my attempt with what I know: I know that the weak closure of [imath]S[/imath] is a subset of [imath]B[/imath] because [imath]B[/imath] is norm closed and convex, so it is weakly closed, and [imath]B[/imath] contains [imath]S[/imath]. But I need to show that [imath]B[/imath] is a subset of the weak closure of [imath]S[/imath]. [imath]\\[/imath] for small [imath]\epsilon > 0[/imath], and some [imath]x^*_1,...,x^*_n \in X^*[/imath], I let [imath]U=\{ x : \langle x, x^*_i \rangle < \epsilon , i = 1,...,n \} [/imath] then [imath]U[/imath] is a weak neighbourhood of [imath]0[/imath] What I think I need to show now is that [imath]U[/imath] intersects [imath]S[/imath], but I don't know how. |
2474474 | Prove that that the set [imath]P = \{p\mid \text{Prime}(p) \land p \equiv 3 \pmod{4}\}[/imath] is infinite
I was able to prove that there are infinitely many prime numbers but I'm not able prove that there are infinitely many primes that are equivalent to [imath]3 \pmod{4}[/imath]. How do I go about doing this? | 671440 | Proof that there are infinitely many primes congruent to 3 modulo 4
I'm having difficult proving this. As a hint the exercise to prove first, that if [imath]a\lneqq \pm 1[/imath] satisfies [imath]a \equiv 3 \pmod4[/imath], then exist [imath]p[/imath] prime, [imath]p \equiv 3 \pmod 4[/imath] such [imath]p\mid4[/imath]. But I'm not really getting for what purpose can this be used. |
2473975 | Metric mapping to sets other than [imath]\Bbb{R}[/imath]
A metric space is a set M together with a function [imath]d:M \times M \rightarrow \Bbb{R} [/imath], where [imath]d[/imath] satisfies: [imath]d(x,y)\ge 0[/imath] [imath]d(x,y)=0 \Leftrightarrow x=y[/imath] [imath]d(x,y)=d(y,x)[/imath] [imath]d(x,z) \le d(x,y)+ d(y,z)[/imath] [imath]\forall x,y,z \in M.[/imath] Naively it seems that [imath]\Bbb{R}[/imath] has too much structure than what is required of it to satisfy these axioms and [imath]d[/imath] could map to any ordered ring. So my question is why is [imath]\Bbb{R}[/imath] chosen and not a more general ring? Has there been any research on metrics that map to sets other than [imath]\Bbb{R}[/imath]? I have searched Google and not found anything useful. My question title seems similar to this one but after reading the full question text I believe they are asking different things. | 1315363 | Generalization of metric spaces?
Usually we define a metric on a space [imath]X[/imath] to be a map [imath]X\times X\to\mathbb{R}[/imath] that satisfies a few axioms. [imath]\mathbb{R}[/imath] has of course a total order. What if we instead have a metric [imath]X\times X\to A[/imath] where [imath]A[/imath] is a monoid with some kind of weaker order on it (I say monoid because the triangle inequality requires addition, and we'd also need a zero). It seems to me that if [imath]A[/imath] is a monoid and directed set, and if we define open sets as sets for which every point has a ball around it contained in the set, then this generalized metric induces a topology on [imath]X[/imath]. Is this a thing already? Which topological spaces can be endowed with this kind of metric? This feels natural because when going from metric spaces to topological spaces we have to talk about nets (maps from directed sets) rather than sequences (maps from the natural numbers). This is in a way analogous. Edit: Changed group to monoid. |
2475350 | [imath]\int_0^{\pi/4} (\ln\sin x)^2 \, dx[/imath]
Help me to integrate this, please. I tried to do this one for three days, but I have no idea. [imath] \int_0^{\pi/4} (\ln\sin x)^2\, dx [/imath] | 917154 | Looking for closed-forms of [imath]\int_0^{\pi/4}\ln^2(\sin x)\,dx[/imath] and [imath]\int_0^{\pi/4}\ln^2(\cos x)\,dx[/imath]
A few days ago, I posted the following problems Prove that \begin{equation} \int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}\\[20pt] -\int_0^{\pi/2}\ln^3(\cos x)\,dx=\frac{\pi}{2}\ln^3 2+\frac{\pi^3}{8}\ln 2 +\frac{3\pi}{4}\zeta(3) \end{equation} and the OP receives some good answers even I then could answer it. My next question is finding the closed-forms for \begin{align} \int_0^{\pi/4}\ln^2(\sin x)\,dx\tag1\\[20pt] \int_0^{\pi/4}\ln^2(\cos x)\,dx\tag2\\[20pt] \int_0^1\frac{\ln t~\ln\big(1+t^2\big)}{1+t^2}dt\tag3 \end{align} I have a strong feeling that the closed-forms exist because we have nice closed-forms for \begin{equation} \int_0^{\pi/4}\ln(\sin x)\ dx=-\frac12\left(C+\frac\pi2\ln2\right)\\ \text{and}\\ \int_0^{\pi/4}\ln(\cos x)\ dx=\frac12\left(C-\frac\pi2\ln2\right). \end{equation} The complete proofs can be found here. As shown by Mr. Lucian in his answer below, the three integrals are closely related, so finding the closed-form one of them will also find the other closed-forms. How to find the closed-forms of the integrals? Could anyone here please help me to find the closed-form, only one of them, preferably with elementary ways (high school methods)? If possible, please avoiding contour integration and double summation. Any help would be greatly appreciated. Thank you. |
2475146 | Find [imath] \lim_{x\to a^-}(x-a)\left\lfloor \frac{1}{x-a}\right\rfloor [/imath]
This problem is an updated version of this given here and I found it interesting for myself I would like no if the following limit exists or not. [imath] \lim_{x\to a^-}(x-a)\left\lfloor \frac{1}{x-a}\right\rfloor [/imath] Here the interesting fact is that we have the singularity at [imath]x =a[/imath] which not the case here Any idea,? | 1876062 | Does [imath]\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=a?[/imath]
Does [imath]\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=a?[/imath] I'm going to say this statement is false, and try to use the properties of limits [imath]\lim_{x \to 0+} \left(x\lfloor \frac{a}{x} \rfloor\right)=\lim_{x \to 0+} x.\lim_{x \to 0+}\lfloor \frac{a}{x}\rfloor[/imath] =[imath]0.\infty[/imath] which is undefined. Or is it [imath]\infty [/imath] because x ends up being a little bigger than 0? |
2475691 | Roots Of Polynomial With Integer-coefficients.
I have failed to prove the second part of the following problem in polynomials. Let [imath]P(x)[/imath] be a polynomial with integer coefficients and degree [imath]d>0[/imath]. Prove the following two statements: If [imath]α[/imath] and [imath]β[/imath] are two integers such that [imath]P(α)=1[/imath] and [imath]P(β)=-1[/imath], then prove that [imath]|α-β|[/imath] divides [imath]2.[/imath] Prove that the number of distinct roots of [imath](P(x))^2-1[/imath] is at most [imath]d+2.[/imath] | 1392860 | Find the number of distinct integer roots of [imath]P^2 (x)-1[/imath]
Let [imath]P(x)[/imath] be a polynomial with integer coefficients of degree [imath]d>0[/imath]. Prove that the number of distinct integer roots of [imath]P^2(x)-1[/imath] is at most [imath]d+2[/imath]. |
1969879 | Which is bigger? [imath]Ackermann(G_{64}, G_{64})[/imath] or [imath]G_{G_{64}}[/imath]
I have been playing around with the Ackermann function a bit and realized that it gets very big very fast. (Im going to use [imath]A[/imath] for [imath]Ackermann[/imath] throughout this question) Already [imath]A(5,1)[/imath] is (according to WolframAlpha) an integer too large to represent. It also presents me with a representation that looks like this: [imath] A(5,1) = 2 \uparrow \uparrow \uparrow 4 - 3 = 2 \uparrow^{3} 4 - 3 [/imath] After playing around a little I found out that it always represents it like this: [imath] A(n,m) = 2 \uparrow^{n-2} (m+3)-3 [/imath] After seeing this I started to wonder what would happen if you use Graham's number as the arguments. Since the value of it would uncomprehensible I tried to find something to compare it against. Remembering how the number is defined I asked myself if [imath]G_{G_{64}}[/imath] is bigger than [imath]A(G_{64},G_{64})[/imath]? Or in other words: [imath] A(G_{64},G_{64}) = 2 \uparrow^{G_{64} - 2} (G_{64} + 3) - 3 [/imath] vs [imath] G_{G_{64}} = 3 \uparrow^{G_{G_{64}} - 1} 3 [/imath] I personally supspect it is [imath]G_{G_{64}}[/imath] since it has a lot more arrows but I'm not entirely sure. | 832726 | If I call the Ackermann Function with Graham's number as both of its arguments will it be less than [imath]g_{65}[/imath]
In xkcd comic 207 it states that [xkcd] means calling the Ackermann function with Graham's number as the arguments just to horrify mathematicians. [imath]A(g_{64},g_{64})[/imath] In this explanation it states that Even simply making [imath]g_{65}[/imath] dwarfs the number [imath]A( A(g_{64}, g_{64}), A(g_{64}, g_{64}))[/imath] into insignificance. So: Is [imath]g_{65}[/imath] greater than [imath]A(g_{64}, g_{64})[/imath]? And is it greater than [imath]A( A(g_{64}, g_{64}), A(g_{64}, g_{64}))[/imath]? |
2476109 | Finding a sequence [imath]a_n[/imath] such that [imath]\sum a_n[/imath] converges but [imath]\sum a_n^3[/imath] diverges.
This is an exercise that was given to me, and I would like to emphasize that I am not looking for an answer, but rather a pointer or a hint in the right direction, since it appears I've reached a bit of a dead end. We are looking at sequences over the real numbers (at least that's my assumption, and that's where I've been looking). I've come with various results to help me try such a sequence: The ratio test has to be inconclusive, as does the root test. Since [imath]a_n[/imath] converges to [imath]0[/imath], so does [imath]a_n^3[/imath] The sequences [imath]|a_n|[/imath] cannot be monotone decreasing, otherwise [imath]\sum a_n^3[/imath] will converge. Other various things I've noticed: It's very easy to find [imath]a_n[/imath] such that [imath]a_n[/imath] converges but [imath]a_n^2[/imath] diverges. Is it possible to use that fact? Let [imath]u_n[/imath] be such a sequence. I tried to construct a sum of two sequences so that when we take the third power, [imath]u_n^2[/imath] appears in the binomial expansion. That "naive" approach didn't get me too far however. We can make the following observation: \begin{align} \sum a_n^3 &= a_1^3 + a_2^3+a_3^3+a_4^3\cdots \\ & = (a_1+a_2)((a_1-a_2)^2 + a_1a_2) + (a_3+a_4)((a_3-a_4)^2 + a_3a_4) +\cdots \end{align} This hasn't taken me anywhere. Other things I've thought about: I think a natural form for the sequence would be the product of two functions, [imath]f[/imath] and [imath]g[/imath], such that one converges to [imath]0[/imath] much faster than the other. By using the comparison test, we could try to require that [imath](fg)^3 > k_n[/imath] for all n for some sequence [imath]k_n[/imath] whose associated series is divergent. I think it is unlikely we could find such a product though, since if [imath]fg[/imath] plummets fast enough, then [imath](fg)^3[/imath] will plummet even faster. I've tried a few things for lack of better options, using exp, trig functions, polynomials, none of my tries have worked so far, almost always as a product of two functions. If someone could suggest a possible step forward, I would be very grateful! | 2216555 | General infinite series convergence or divergence.
Suppose the infinite series [imath]\sum a_n[/imath] converges ([imath]a_n \in \mathbb{R}[/imath] not all positive or negative). (1) Is it true that [imath]\sum a_n^2[/imath] converges? (2) Is it true that [imath]\sum a_n^3[/imath] converges? For (1), the answer is no. I found the counterexample [imath]a_n = (-1)^n / \sqrt{n}[/imath] which converges by alternating series, but [imath]\sum a_n^2 = \sum1/n [/imath] is divergent. I'm unsure about proving truth of (2) and can't seem to find a counterexample. Thanks for help. |
2471494 | Combinatorics with vowels and consonants
How many words of [imath]13[/imath] letters can be constructed from the English alphabet which contain [imath]4[/imath] vowels and [imath]9[/imath] different consonants (vowels can be the same). This is what I think: Pick the four vowels (does this arrange them as well?) = [imath]5^4[/imath]. Choose the consonants = [imath]\binom{13}9[/imath] Now then arrange these consonants in [imath]9![/imath] ways Total = [imath]5^4 \binom{13}9 9![/imath]. My friend got this: [imath]5^4 \binom{21}9 \binom{13}4[/imath] to arrange the four vowels (with the last binomial). | 2476169 | Combinatorics involving picking a specific amount of consonants without repeat and vowels (can repeat)
How many words of [imath]13[/imath] letters of the English Alphabet can we constructed which contain [imath]4[/imath] vowels and [imath]9[/imath] different consonants? This is what I did: Choose the consonants (diff.) in [imath]\binom{21}9[/imath] ways. Arrange these consonants in [imath]9![/imath] ways. Now we have [imath]4[/imath] empty spots remaining for the vowels, which can be chosen in [imath]5^4[/imath] ways. My question is: Is this the correct way to think about it? I'm unsure whether the [imath]5^4[/imath] arranges the vowels or not, |
2475473 | Prove the sequence is Cauchy and therefore Convergent
Let [imath](x_n)_{n\in\mathbb{N}}[/imath] be a sequence in [imath]\mathbb{R}[/imath]. Let [imath]r\in (0,1)[/imath] and suppose that [imath]|x_{n+1}-x_n|<r^n \forall n\in \mathbb{N}[/imath]. Show that [imath](x_n)_{n\in\mathbb{N}}[/imath] converges. I know that it is sufficient to show that the sequence is Cauchy. So, this is very similar to this past question (Please help prove that [imath](x_n)[/imath] is a Cauchy sequence if [imath]|x_{n+1} - x_n| \leq Cr^n[/imath]) but I don't completely understand the answers given so I'm hoping someone can help me out. | 378062 | Please help prove that [imath](x_n)[/imath] is a Cauchy sequence if [imath]|x_{n+1} - x_n| \leq Cr^n[/imath]
Past final exam question for an intro to Real Analysis course: Let [imath]C > 0[/imath], [imath]0<r<1[/imath] and suppose that [imath]\forall n\in \mathbb N, |x_{n+1} - x_n| \leq Cr^n[/imath]. Please help me prove that [imath](x_n)[/imath] is a Cauchy sequence. (We can assume the [imath]\lim\limits_{n\to\infty} r^n=0,[/imath] for [imath]0<r<1[/imath]) So, I know that a Cauchy series must satisfy [imath]|x_{n}-x_m| < \epsilon[/imath] for any [imath]\epsilon>0, \in \mathbb R[/imath] and for all [imath]n,m \gt H(\epsilon) \in \mathbb N[/imath] Note that there can't be any conditions on n and m (I saw somewhere else someone required [imath]m>n[/imath] which you can only do if you're proving its not at Cauchy sequence, right?) Another way of doing this is showing that it is contractive (and thus a Cauchy series) if there is a constant [imath]a[/imath] such that [imath]|x_{n+1}-x_n| \leq a|x_n-x_{n-1}|[/imath] Clearly I'm supposed to make use of [imath]\lim\limits_{n\to\infty} r^n=0,[/imath] for [imath]0<r<1[/imath]... But I don't even know how where to start with this. As I am bumbling through this problem, a more thorough answer would be much appreciated. Thanks! |
2476484 | For prime [imath]p\equiv 1(\operatorname{mod} 4)[/imath] , can [imath]\dfrac {p^p-1}{p-1}[/imath] be a prime?
Let [imath]p[/imath] be a prime of the form [imath]4k+1[/imath] , then can [imath]\dfrac {p^p-1}{p-1}[/imath] be a prime number ? | 1997613 | Prove that [imath]\frac{p^p-1}{p-1}[/imath] is not prime if [imath]p \equiv 1 \pmod 4[/imath]
Let [imath]p[/imath] be a prime number such that [imath]p \equiv 1 \pmod 4[/imath]. Prove that [imath]\frac{p^p-1}{p-1}[/imath] is not prime. We can rewrite [imath]\frac{p^p-1}{p-1}[/imath] as [imath]\dfrac{p^p-1}{p-1} = 1+p+p^2+\cdots+p^{p-1},[/imath]but how do we show this is not prime? |
1985612 | Polynomial of degree 5 such that [imath]P(x)-1[/imath] is divisible by [imath](x-1)^3[/imath] and [imath]P(x)[/imath] is divisible by [imath]x^3[/imath]
[imath]P(x)[/imath] is a polynomial of degree 5 such that [imath]P(x)-1[/imath] is divisible by [imath](x-1)^3[/imath] and [imath]P(x)[/imath] is divisible by [imath]x^3[/imath]. Find [imath]P(x)[/imath]. No idea where to start, would [imath]P(x)[/imath] be of the form [imath]x^3(Ax^2+Bx+C)[/imath]? | 1740767 | Find a polynomial f(x) of degree 5 such that 2 properties hold.
I have been trying to find a polynomial [imath]f(x)[/imath] such that these [imath]2[/imath] properties hold: [imath]f(x)-1[/imath] is divisible by [imath](x-1)^3[/imath] [imath]f(x)[/imath] is divisible by [imath]x^3[/imath] To start, I set [imath]f(x) =ax^5 + bx^4 + cx^3 + dx^2 + ex + f[/imath]. This is divisible by [imath]x^3[/imath], so [imath]d, e, f [/imath] must be [imath]0[/imath]. So polynomial [imath]f(x) = ax^5 + bx^4 + cx^3[/imath] This minus [imath]1[/imath] is divisible by [imath](x-1)^3[/imath]. So I used synthetic division and got that the remainder of [imath]ax^5 + bx^4 + cx^3[/imath] divided by [imath](x-1)[/imath] is [imath]a+b+c-1[/imath]. But it should divide out evenly, so [imath]a+b+c-1 = 0[/imath]. From here, I can't find any other equations involving the three variables. Could someone have any suggestions for how to continue? Thanks in advance. :) |
2476688 | How to show [imath]\vec{a} \cdot \vec{b}=\frac{1}{4}\left(\Vert{\vec{a}+\vec{b}\Vert^2}-\Vert{\vec{a}-\vec{b}\Vert^2}\right)[/imath]?
The Homework Exercise I am working on, is: Let [imath]\overrightarrow{a}, \overrightarrow{b}[/imath] be vectors. Show that [imath]\overrightarrow{a} \cdot \overrightarrow{b}=\frac{1}{4}\left(\Vert{\overrightarrow{a}+\overrightarrow{b}\Vert^2}-\Vert{\overrightarrow{a}-\overrightarrow{b}\Vert^2}\right)[/imath]. Things I tried: Using the property [imath]\Vert \overrightarrow{u}\Vert^2=\overrightarrow{u} \cdot \overrightarrow{u}[/imath], I tried to make [imath]\Vert{\overrightarrow{a}+\overrightarrow{b}\Vert^2} =\left(\overrightarrow{a}+\overrightarrow{b} \right) \cdot \left(\overrightarrow{a}+\overrightarrow{b} \right)[/imath] and [imath]\Vert{\overrightarrow{a}-\overrightarrow{b}\Vert^2} =\left(\overrightarrow{a}-\overrightarrow{b} \right) \cdot \left(\overrightarrow{a}-\overrightarrow{b} \right)[/imath] I'm not sure if that was a correct step, but then I substituted into the main equation to get [imath] \overrightarrow{a} \cdot \overrightarrow{b} = \frac{1}{4}\left( (\overrightarrow{a}+\overrightarrow{b} ) \cdot (\overrightarrow{a}+\overrightarrow{b} ) - (\overrightarrow{a}-\overrightarrow{b} ) \cdot (\overrightarrow{a}-\overrightarrow{b} ) \right) [/imath] Am not sure where to go from here, or if I'm even heading in the right direction... | 425173 | Derivation of the polarization identities?
For a real (or complex) inner product space [imath]V[/imath], the inner product can be expressed in terms of the norm as either [imath] \langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2) [/imath] or [imath] \langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2) [/imath] respectively. These identities are not hard to verify directly once they are given. I would like to know, is there a systematic way one would derive these from scratch without knowing them beforehand, or are both polarization identities just clever observations? |
2476864 | [imath]\sqrt[n]{ax^{n}+bx^{n-1}+...+c} ∼ \sqrt{a}|x+\frac{b}{na}| \ \ \ for:a>0 , n \in \mathbb{N} \ \ \text{even}[/imath]
Written in my textbook that : [imath]\sqrt[n]{ax^{n}+bx^{n-1}+...+c}_{x\to \infty} ∼ \sqrt{a}|x+\frac{b}{na}| \ \ \ for:a>0 , n \in \mathbb{N} \ \ \text{even}[/imath] And : [imath]\sqrt[n]{ax^{n}+bx^{n-1}+...+c} _{x\to \infty}∼ \sqrt{a}(x+\frac{b}{na}) \ \ \ for: , n \in \mathbb{N} \ \ \text{Odd number}[/imath] but I do not know why ? How to prove?please help me. | 2249973 | Substitution in determining limit of polynomials
In my calculus book has been written: "For finding limits sometimes we can change [imath]\sqrt[m]{a_m x^m + a_{m - 1} x^{m-1} + a_{m - 2} x^{m-2} + \cdots + a_0} [/imath] to [imath]\sqrt[m]{a_m}\left(x + \dfrac{a_{m - 1}}{ma_m}\right)[/imath] for odd [imath]m[/imath] and [imath]\sqrt[m]{a_m}\left|x + \dfrac{a_{m - 1}}{ma_m}\right|[/imath] for even [imath]m[/imath]." First of all I want to know how these formulas were built. After that , when I can use them without any mistake in value of limit. I searched in the internet but I didn't find these formulas. |
2474478 | What is the generalization of [imath]\Pr[ x_1 < X \leq x_2, y_1 < Y \leq y_2][/imath] to three random variables? To n random variables?
Given two random variables [imath]X,Y[/imath], it is well known that [imath]\Pr[ x_1 < X \leq x_2, y_1 < Y \leq y_2] = F_{X,Y}(x_2,y_2) - F_{X,Y}(x_1, y_2) - F_{X,Y}(x_2, x_1) + F_{X,Y}(x_1, y_1)[/imath] Where [imath]F_{XY}[/imath] is the joint CDF of [imath]X,Y[/imath]. Is there a generalization of the above to three random variables [imath]X,Y,Z[/imath]? To [imath]n[/imath] variables [imath]X_1, \ldots, X_n[/imath]? I have looked at a few probability references but I couldn't find any. | 2467903 | What is [imath]P([a_1,b_1]\times [a_2,b_2]\times...\times[a_k,b_k])[/imath]
How can I obtain an expression of [imath]P([a_1,b_1]\times [a_2,b_2]\times...\times[a_k,b_k])[/imath] knowing that [imath]F(x_1,...,x_k)=P((-\infty,x_1]\times (-\infty,x_2]\times...\times(-\infty,x_k])[/imath]? I solve the problem for [imath]k=2[/imath]. [imath]P([a_1,b_1]\times [a_2,b_2])=F(b_1,b_2)+F(a_1-,a_2-)-F(a_1-,b_2-)-F(b_1-,a_2-)[/imath]. But I dont know how to do it for any [imath]k[/imath]. Thanks a lot :) |
2477197 | A question about linear continous operators
Theorem: Let [imath]T:X\rightarrow Y[/imath] be linear operator. Show that if [imath]T[/imath] is continous on X, then it is bounded. The attempt: if [imath]T[/imath] is continous on X, it is contionous at origin. Then, [imath]\forall x\in X [/imath] we have [imath]\lVert T(x)\rVert<\epsilon [/imath] where [imath]\lVert x\rVert <\delta[/imath], we must show that there is a [imath]C>0[/imath] such that for [imath]\forall x\in X [/imath], [imath] \lVert A (x)\rVert <C \lVert x\rVert [/imath]. How? | 503433 | Prove that if a linear operator is continuous, then it is bounded.
I'm trying to prove that if a linear operator is continuous, then it is bounded. Let [imath]T:V\to W[/imath]. Let us assume it is continuous. Then for any [imath]\epsilon>0[/imath], [imath]\|T(x-x_0)\|<\epsilon[/imath] if [imath]\|x-x_0\|<\delta[/imath] for some [imath]\delta\in \Bbb{R}[/imath]. If [imath]T[/imath] is bounded, then [imath]\sup \frac{\|T(x-x_0)\|}{\|x-x_0\|}[/imath] exists, where [imath]x[/imath] and [imath]x_0[/imath] may be any vectors in [imath]V[/imath]. The proof in my book "Functional Analaysis" by Kreyszig (pg.97) proceeds by stating Let [imath]x=x_0+\delta\frac{y}{\|y\|}[/imath], where [imath]y[/imath] is any vector in [imath]V[/imath] Aren't we artificially restricting the possible values of [imath]x[/imath] in comparison with [imath]x_0[/imath]? Thanks in advance! |
2477064 | Proving that: [imath] | a + b | + |a-b| \ge|a| + |b|[/imath]
I am trying to prove this for nearly an hour now: [imath] \tag{$\forall a,b \in \mathbb{R}$}| a + b | + |a-b| \ge|a| + |b| [/imath] I'm lost, could you guys give me a tip from where to start, or maybe show a good resource for beginners in proofs ? Thanks in advance. | 986104 | Use the triangle inequality to show that [imath]|a|+|b| \leq |a+b|+|a-b|[/imath]
How would I go about proving that [imath]|a|+|b| \leq |a+b|+|a-b|[/imath] Using the triangle inequality? I tried squaring both sides, yielding: [imath]|a|^2 +|b|^2 +2|a||b| \leq 2|a|^2+2|b|^2+2||a|^2-|b|^2|[/imath] Is it correct to move the terms to the other side? I tried [imath]2|a||b| \leq |a|^2+|b|^2+2||a|^2-|b|^2|[/imath] Then it is obvious that [imath]0\leq |a|^2+|b|^2-2|a||b| +2||a|^2-|b|^2|[/imath] [imath]0\leq (|a|-|b|)^2 +2||a|^2-|b|^2|[/imath] [imath]0\leq ||a|-|b||^2 +2||a|^2-|b|^2|[/imath] But I have neither used the triangle inequality, nor it looks mathematically rigorous for my real analysis class. Any tips and tricks for solving these problems? |
2477240 | Can we convert infinite sums into infinite products?
Is there any method to convert an infinite sum (of any type) into an infinite product? That is, a method such that the equality [imath] \sum_{n = 0}^{\infty}a_n = \prod_{n = 0}^{\infty} f(a_n)[/imath] holds. | 1174647 | Is it possible to turn infinite sums into infinite products?
I am working on studying infinite products. I know that it is possible to convert an infinite product to an infinite sum using logarithms, where [imath]\log \prod s_n = \sum \log s_n[/imath] My question is, it always possible go from sum to product with [imath]e[/imath], where [imath]\exp \sum s_n = \prod e^{s_n}[/imath] Also, are there any other methods to convert sums into products? Thanks in advance. |
2476136 | Integral of 1/x- why does it behave this way?
First of all, I'm not entirely sure what to call polynomial functions that can have x raised to negative integers powers, so I will just call them polynomial type functions. The integral of every polynomial type function is another polynomial type function, unless, of course, our polynomial type function has [imath]\frac{1}{x}[/imath] in it. In that case, our integral is obviously [imath]\ln{x}[/imath]. In the case of every other function in the form [imath]x^{k}[/imath] for some integer [imath]k[/imath], we can use power rule to find the integral. With [imath]\frac{1}{x}[/imath], there is a problem, since our integral will have had a constant of [imath]0[/imath] multiplying the term, nullifying it. So my question is, why does it happen to be true that every single polynomial type function has a polynomial type integral, except [imath]\frac{1}{x}[/imath]? [imath]\frac{1}{x}[/imath] happens to be the only function where using power rule to evaluate the integral doesn't work, and it also happens to the be the only one without a polynomial type integral. This can't be a coincidence: the two special properties of [imath]\frac{1}{x}[/imath] must be related. What about [imath]\frac{1}{x}[/imath] makes it so special? | 498339 | Demystify integration of [imath]\int \frac{1}{x} \mathrm dx[/imath]
I've learned in my analysis class, that [imath] \int \frac{1}{x} \mathrm dx = \ln(x). [/imath] I can live with that, and it's what I use when solving equations like that. But how can I solve this, without knowing that beforehand. Assuming the standard rule for integration is [imath] \int x^a \, \mathrm dx = \frac{1}{a+1} \cdot x^{a+1} + C .[/imath] If I use that and apply this to [imath]\int \frac{1}{x} \,\mathrm dx[/imath]: [imath] \begin{align*} \int \frac{1}{x}\mathrm dx &= \int x^{-1} \,\mathrm dx \\ &= \frac{1}{-1+1} \cdot x^{-1+1} \\ &= \frac{x^0}{0} \end{align*} [/imath] Obviously, this doesn't work, as I get a division by [imath]0[/imath]. I don't really see, how I can end up with [imath]\ln(x)[/imath]. There seems to be something very fundamental that I'm missing. I study computer sciences, so, we usually omit things like in-depth math theory like that. We just learned that [imath]\int \frac{1}{x} dx = \ln(x)[/imath] and that's what we use. |
2476480 | Given that [imath]f[/imath] is continuous on [imath][0,1][/imath], differentiable on [imath](0,1)[/imath], and [imath]f(0)=0[/imath], [imath]f(1)=1[/imath]
Given that [imath]f[/imath] is continuous on [imath][0,1][/imath], differentiable on [imath](0,1)[/imath], and [imath]f(0)=0[/imath], [imath]f(1)=1[/imath], prove [imath]\text{There exists $c_1 \lt c_2$ such that $\frac{1}{f'(c_1)}+\frac{1}{f'(c_2)} =2$}[/imath] and [imath]\text{There exists $c_1 \lt c_2$ such that $f'(c_1)f'(c_2)=1$} [/imath] I have already proven that there exists [imath]c \in (0,1)[/imath] such that [imath]f'(c)=1[/imath] and there exist [imath]c_1 \lt c_2[/imath] such that [imath]f'(c_1)+f'(c_2)=2[/imath]. I tried using these to help to prove the existence of the above. I've just learned Mean Value Theorem, Rolle's Theorem and Intermediate Value Theorem, which I've used to answer the first two parts. Given the mark allocation for these two questions, they're supposed to be easy. However, I can't seem to find my way around. Please advise me on what I should do to simplify the problem. Thank you! | 2473014 | Prove that there are [imath]0 s.t. \frac{1}{g'(c_1)}+\frac{1}{g'(c_2)}=2 . The same for g'(c_1)g'(c_2)=1.[/imath]
Given [imath]g[/imath] is continuous on [imath][0,1][/imath] and differentiable on [imath](0,1)[/imath], [imath]g(0)=0[/imath], and [imath]g(1)=1[/imath]. 1) Prove that there exist [imath]0<c_1<c_2<1[/imath] such that [imath]\frac{1}{g'(c_1)}+\frac{1}{g'(c_2)}=2.[/imath] 2) Prove that there exist [imath]0<c_1<c_2<1[/imath] such that [imath]g'(c_1)g'(c_2)=1.[/imath] I am thinking if Mean Value Theorem would help, but Mean Value theorem could only prove that in interval (0,1) there exists [imath]g'(c)=1[/imath]. |
2477392 | Prove that the following polynomial has no multiple roots
Prove that the following polynomial [imath]p(x) = \sum_{k=0}^n x^k[/imath] has no multiple roots Hint: consider the polynomial [imath](x-1)\times \sum_{k=0}^n x^k[/imath] I have tried finding the derivative of the polynomial and trying to reach a contradiction by supposing there is a root such that p(x) = p'(x) = 0 but I can't reach any conclusions. The hint leads me nowhere, I don't know how to use that information Thanks in advance | 1284339 | Prove [imath]f=1+x+x^2+x^3+\cdots+x^n[/imath] has no multiple roots.
Prove [imath]f=1+x+x^2+x^3+\cdots+x^n[/imath] has no multiple roots. My attempt: Consider the polynomial [imath]g=(x-1)(1+x+x^2+x^3+\cdots+x^n)[/imath] As [imath]f\mid g, g[/imath] all the roots of [imath]f[/imath] are roots of [imath]g[/imath]. This means I have to prove the statement: If [imath]c[/imath] is a multiple root of [imath]f \Longrightarrow c[/imath] is multiple a root of [imath]g[/imath]. Equivalently: If [imath]c[/imath] is not a multiple root of [imath]g \Longrightarrow c[/imath] is not a multiple root of [imath]f[/imath]. I'll try to prove this last statement: Suppose [imath]\exists c[/imath] such that [imath]c[/imath] is a multiple root of [imath]g[/imath], then [imath]g(c)=0 \implies g'(c)=0[/imath] [imath] g=(x-1)(1+x+\cdots+x^n) \implies g'=nx^n [/imath] [imath] g'(c)=nc^n=0 \implies c=0 [/imath] But [imath] g(0)\neq 0 [/imath] And we've found the contradiction: [imath]\not \exists c: g(c)=g'(c)=0[/imath]. Then [imath]f[/imath] has no multiple roots. Is this correct? Also: Assuming this is correct, is it possible to prove [imath]\sum_{k=0}^n \frac {x^n}{n!} [/imath] has no multiple roots in a similar manner? |
2475362 | Solve identity: [imath]\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}[/imath]
[imath]\frac {\sin x}{1-\cos x} = \frac {1+\cos x}{\sin x}[/imath] The only way I can see of doing this is by cross multiplying but isn't that not allowed when trying to prove something? | 510016 | Prove the Trigonometric Identity: [imath]\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}[/imath]
I'm doing some math exercises but I got stuck on this problem. In the book of Bogoslavov Vene it says to prove that: [imath]\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}.[/imath] It is easy if we do it like this: [imath]\sin^2 x = (1-\cos x)(1+\cos x)=1-\cos^2 x[/imath]. But how to prove it just by going from the left part or from the right Can anybody help me? Thank you! |
2477632 | Calculate Ellipse axis length
Given an ellipse: [imath]ax^2 + 2xy + by^2 = 1[/imath] how can I calculate the length of the shorter and longer axis? | 1227369 | Calculating the length of the semi-major axis from the general equation of an ellipse
What is the most accurate way of solving the length of the semi-major axis of this ellipse? [imath]-0.21957597384315714 x^2 -0.029724573612439117 xy -0.35183249227660496 y^2 -0.9514941664721085 x + 0.1327709804087165 y+1 = 0[/imath] The answer should be extremely close to the correct value of the length of the semi-major axis which is equal to [imath]3.073400961177073[/imath] I already tried to rotate the graph so that the major axis will coincide with the x-axis making the xy term equal to zero, then I made it into standard form, in which I was able to calculate the length of semi-major axis. However, the result differs from the true value by about [imath]0.1[/imath] . This difference is not acceptable since this value will be used many times for the orbit propagation formulas and in our final computation, the result has about 5% error. We could not accept this 5% error since our goal is to have an error of at least 2%. Thank you in advance. |
2475771 | Solve the differential equation 3
[imath] y' = \frac{y}{\sin x} + \tan \frac {x}{2} [/imath] So from the beginning I do [imath]y' - \frac{y}{\sin x} = \tan \frac {x}{2} [/imath] Now I should equal it to 0, shouldn't I? [imath]y' - \frac{y}{\sin x} = 0 [/imath] [imath]\frac{dx}{dy} = \frac{y}{\sin x} [/imath] [imath] dx\sin x=ydy[/imath] [imath] \cos x=\frac{y^2}{2} [/imath] [imath]y=\sqrt{2\cos x}[/imath] And now i have problem with that derivative, is there any short way to do it? | 2475772 | Solve the differential equations:
[imath]y' = {y \over {\sin x}}+\tan{x \over 2}[/imath] I was trying to do this by substitution [imath]u=y/x [/imath] and it did not work and also with [imath]y' - {y \over {\sin x}} = 0[/imath] [imath]{dy \over dx} = {y \over {\sin x}} [/imath] [imath]{\ln(y) = \ln\left|\tan\frac{x}{2}\right|+c } [/imath] [imath]{y = c\cdot\tan\frac{x}{2} } [/imath] but then when im trying to calculate [imath]y'[/imath] I have a problem and I have too many equations. Is there some easier way or am I making some mistakes. |
2478106 | Is the unit circle homeomorphic to the unit square?
Let [imath]S_1:=\{(x,y\in\mathbb{R}^2:x^2+y^2=1\}[/imath] and [imath]S_2:=\{x,y\in\mathbb{R}^2:\max\{|x|,|y|\}=1\}[/imath] be endowed with subspace topologies induced by the usual topology on [imath]\mathbb{R}^2[/imath]. Then open sets in [imath]S_1[/imath] and [imath]S_2[/imath] are unions of open intervals in [imath]S_1[/imath] and [imath]S_2[/imath] respectively. Partition [imath]S_1[/imath] into four arbitrary non-empty disjoint open sets [imath]A,B,C,D[/imath] and four distinct points, [imath]a,b,c,d[/imath]. Similarly, partition [imath]S_2[/imath] into four disjoint open sets [imath]A',B',C',D'[/imath] and four points [imath](1,1), (1,-1), (-1,-1), (-1,1)[/imath], where [imath]A'[/imath] is the right vertical side of the square, [imath]B'[/imath] is the bottom side of the square, etc. Let [imath]f:S_1\rightarrow S_2[/imath] be a function such that [imath]f(A):=A',f(B):=B',f(C):=C',f(D):=D'[/imath], and [imath]f(a):=(1,1),f(b):=(1,1),f(c):=(1,1),f(d):=(1,1)[/imath]. It seems that [imath]f[/imath] is bijective and continuous, and [imath]f^{-1}[/imath] is also continuous, but I can't give a rigorous proof. | 103660 | Homeomorphism from square to unit circle
Can we find a homeomorphism from the square [imath]Q_2[/imath] of side length [imath]2[/imath] centered on the origin and the unit circle [imath]S^1[/imath]? We can easily define a map [imath]r:Q \longrightarrow S^1[/imath] by [imath](x,y) \mapsto \bigg(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}} \bigg)[/imath] which is a radial projection onto the unit circle, but how can we define [imath]r^{-1}[/imath] and show that it is continuous? Thoughts I think we may define the inverse map as [imath](x,y) \mapsto \bigg(\frac{x}{\sqrt2\max{│x│, │y│}} , \frac{y}{\sqrt2\max {│x│, │y│}}\bigg)[/imath]At least intuitively this maps to a square, and the [imath]\frac{1}{\sqrt2}[/imath] term scales appropriately. However, how can we demonstrate these are continuous maps, thus demonstrating that [imath]Q_2[/imath] and [imath]S^1[/imath] are homeomorphic? Any help would be appreciated. Regards, MM. |
2478338 | 3 questions on linear independence
Are the following linear independent? If S [imath]=[/imath] {[imath]a, b, c[/imath]} [imath]⊆ R^3[/imath], where [imath](2,3,4)[/imath] [imath]∉[/imath] span {[imath]a, b, c[/imath]}. If S [imath]=[/imath] {[imath]a, b, c[/imath]} [imath]⊆ R^3[/imath], where [imath](1,2,3)[/imath], [imath](1,3,2)[/imath], [imath](2,1,3)[/imath], and [imath](0,1,0)[/imath] [imath]ϵ[/imath] span {[imath]a, b, c[/imath]}. If [imath]S = (u + v, u - v, u - 2v + w) ⊆ R^n[/imath], where [imath](u, v, w) ⊆ R^n[/imath] is a linearly independent set. My thoughts: From what I know, linear independent means there is only one solution. As for question 1, what i understood by this [imath](2,3,4)[/imath] [imath]∉[/imath] span {[imath]a, b, c[/imath]} is that [imath](2,3,4) ≠ \lambda_{1} a + \lambda_{2} b + \lambda_{3} c [/imath] where [imath]\lambda[/imath] are real numbers. What then does this relation means? As for question 2, I was thinking along finding any form of linear combination between the four vectors and if it exists [imath]\implies[/imath] they are no linear independent. Am i right to say that? If so, what is the best way to find linear combinations. As for question 3, I was thinking of forming linear combinations as well, but the additional vector w makes it impossible for me to work out a solution. Please advice thanks. | 2475206 | Determine if the following sets of vectors are linearly independent
I need to determine if the following set of vectors are are linearly independent or not. 1) [imath]S=\{\mathbf u , \mathbf v , \mathbf w\}\subseteq \mathbb R^3[/imath], it is known that [imath](1,2,3) \notin \text{span}\{\mathbf u , \mathbf v , \mathbf w\}[/imath] 2) [imath]S=\{\mathbf u + \mathbf v,\mathbf u - \mathbf v , \mathbf u - 2\mathbf v +\mathbf w\}\subseteq \mathbb R^n[/imath] where [imath]\{\mathbf u , \mathbf v , \mathbf w\}\subseteq \mathbb R^n[/imath] is a linearly independent set. 3) [imath]S=\{\mathbf u , \mathbf v , \mathbf w\}\subseteq \mathbb R^3[/imath], where it is known that [imath](1,2,-3),(1,-3,2),(2,-1,5),(0,0,1)[/imath] all belong to [imath]\text{span}\{\mathbf u , \mathbf v , \mathbf w\}[/imath]. I honestly have no idea how to start this, all I know is that the if the vectors [imath]\mathbf u , \mathbf v , \mathbf w[/imath] are linearly independent, then [imath]a\mathbf u + b\mathbf v + c\mathbf w=0[/imath] has only the trivial solution. Any help here will be appreciated, many thanks! |
2478185 | Prove that every permutation is the product of disjoint cycles
I need to prove that every permutation [imath]\sigma \in S_n, \sigma \neq id[/imath] can be written as a distinct product of disjoint cycles and don't really know where to start. The statement is totally clear, but I cannot think of an elegant way to proof its correctness. | 622616 | Proving that any permutation in [imath]S_n[/imath] can be written as a product of disjoint cycles
I have attempted a proof of this, but upon looking at my notes, I feel it might be incorrect: it is noticeably simpler than the one in my notes. Proposition: any permutation in [imath]S_n[/imath] can be written as a product of disjoint cycles. Proof (attempt): [imath]s\in S_n[/imath]. Let [imath]\langle s\rangle[/imath] act on [imath]X=\{1,\cdots,n\}[/imath]. Define [imath]x\sim x'\iff x'\in\text{Orb}(x)[/imath] for [imath]x,x'\in X:[/imath] this is clearly an equivalent relation, hence the orbits partition [imath]X[/imath]. Each orbit corresponds a cycle in [imath]s[/imath], so we are done. Does this hold, or have I missed something vital? |
2478717 | Evaluate [imath] \int {(\tan x)}^{\frac{1}{3}} dx [/imath]
Evaluate the following indefinite integral. [imath] \int {(\tan x)}^{\frac{1}{3}} dx [/imath] How could i do this integral ? I tried multiplying the numerator by [imath]{(\sec x)}^2[/imath] and denominator by [imath]1+{(\tan x)}^2[/imath] but getting stuck after putting [imath]\tan x=t[/imath]. | 892087 | Evaluating [imath]\int \tan^{1/3}(\theta) d \theta[/imath]
I know [imath]\int \tan^{1/3}\theta d \theta[/imath] is a non integrable but wolfram alpha does it easily. And can anyone explain how a fuction is non integrable? My argument is that there has to be a function which represents the area under a graph. It's just that we do not know it. Please shed a bit light on it. Also, how can we solve [imath]\int \tan^{1/3}\theta d \theta[/imath]? |
2477983 | Prove the Integral of [imath]\frac{1}{x}[/imath] is not a Rational Function
The title is pretty self explanatory, I'd like to prove that \begin{align*} \int\frac{1}{x}\,dx \end{align*} cannot be a rational function. I have attempted a proof by contradiction, but it doesn't seem to lead anywhere. If it is assumed that [imath]F(x)=\frac{p(x)}{q(x)}[/imath], where [imath]p(x)[/imath] and [imath]q(x)[/imath] are polynomials, then using logarithmic differentiation, \begin{align*} \frac{1}{x}=\frac{p(x)}{q(x)}\left(\frac{p'(x)}{p(x)}-\frac{q'(x)}{q(x)}\right) \end{align*} I don't see how this leads to a contradiction. I get similar results using the quotient rule \begin{align*} \frac{1}{x}=\frac{p'(x)q(x)-p(x)q'(x)}{[q(x)]^2} \end{align*} Writing out the terms of [imath]p(x)[/imath] and [imath]q(x)[/imath] seems too messy. Any help or suggestions would be greatly appreciated. | 1820454 | What are some methods to show [imath]\log[/imath] is not a rational function?
It is easy to show [imath]\log[/imath] isn't a polynomial (no continuous extension to [imath]\mathbb{R}[/imath]). More challenging is showing it isn't rational. Suppose it were a rational function. Then write, the fraction in lowest terms [imath]\log x =\frac{G(x)}{Q(x)} \Longleftrightarrow\frac{G(x)}{\log x} = Q(x)[/imath] Clearly, as [imath]x \to 0[/imath], [imath]Q(x) \to 0[/imath]. However, [imath]Q(x)[/imath] then has [imath]x[/imath] as factor so that [imath]\frac{G(x)}{x\log x} = Q_2(x)[/imath] It is well known that [imath]x \log x \to 0[/imath] as [imath]x \to 0[/imath], so for [imath]Q_2(x)[/imath] to have a finite limit as [imath]x \to 0[/imath], which it must since it is a polynomial, [imath]G(x) \to 0[/imath] as [imath]x \to 0[/imath] so that [imath]x[/imath] is a factor of [imath]G(x)[/imath]. This contradicts the assumption that [imath]\frac{G}{Q}[/imath] was in lowest terms. If there is something wrong with this, please comment, but my main question is What are some other ways to prove that [imath]\log x[/imath] isn't a rational function? |
573856 | Find the prime factor decomposition of [imath]100![/imath] and determine how many zeros terminates the representation of that number.
Find the prime factor decomposition of [imath]100![/imath] and determine how many zeros terminates the representation of that number. Actually, I know a way to solve this, but even if it is very large and cumbersome, and would like to know if you have an easier way, or if I am applying wrong. Setting by [imath]\left[\frac{b}a \right][/imath] the quotient of [imath]b[/imath] with [imath]a[/imath], we have too [imath]E_p(m)[/imath] the largest exponent power [imath]p[/imath] dividing [imath]m[/imath], and found the demonstration of a theorem that says (that in my text says it was discovered that Lagendre) [imath]E_p(n!)=\left[\frac{n}p \right]+\left[\frac{n}{p^2} \right]+\left[\frac{n}{p^3} \right]+\;...[/imath]always remembering that there will be a number [imath]s[/imath] such that [imath]p^s\geq n![/imath] which tells us that [imath]\left[\frac{n!}{p^s} \right]=0[/imath] thus making the sum of a finite [imath]E_p(n!)[/imath]. So that I can address the first question I asked, really have to get all the cousins [imath](p_1,p_2,...,p_k)[/imath] and make all [imath]E_{p_1},\;E_{p_2},\;E_{p_3},\;...,\;E_{p_k}[/imath] with [imath]p[/imath] and cousin [imath]1<p<100[/imath]. And to find the zeros have to see how we both exponents in numbers 5 and 2. Example [imath]10!=2^83^45^27\\p<10\\E_2(10!)=5+2+1=8\\E_3(10!)=3+1=4\\E_5(10!)=2\\E_7(10!)=1\\[/imath] | 2175273 | How to find the prime factorization of [imath]100![/imath].
How can we find the prime factorization of [imath]100![/imath]. My only idea is to write it as, [imath](1)(2)(3)(\color{red}{2})(\color{red}2)...[/imath] And count the factors. But that would take way to long. Any ideas? |
2479994 | Evaluate [imath]\int_0^1(x^2-x^3)^{-\frac 1 3}dx.[/imath]
Evaluate this integral [imath]\int_0^1(x^2-x^3)^{-\frac 1 3}dx[/imath] by using a barbell-shaped contour with shrinking end surrounding 0 and 1, together with a large circle containing the barbell. The branch cut connects 0 and 1. Let the value of [imath]f(z)=z^{-\frac 2 3}(1-z)^{-\frac1 3}[/imath] over the branch cut be positive. When a point over the branch cut moves along the red curve to the corresponding point under the branch cut, [imath]\Delta \arg(z^{-\frac 2 3})=e^{-i{\frac {3\pi}{4}}}[/imath] and [imath]\Delta \arg\big((1-z)^{-\frac1 3}\big)=0[/imath]. So [imath]e^{-i\frac {4\pi} 3}[/imath] should be multiplied under the branch cut. The integral over the outer circle equals:[imath]-2\pi iRes(f(z),\infty)=2\pi iRes(f(\frac 1 t) \cdot \frac 1 {t^2},0)=2\pi iRes(\frac 1 t (t-1)^{-\frac 1 3})=2\pi i(-1)^{-\frac 1 3}.[/imath] Because of the branch we chose, [imath](-1)^{-\frac 1 3}=e^{-i{\frac \pi 3}}.[/imath] We get [imath](1-e^{-i{\frac {4\pi} 3}})\int_0^1(x^2-x^3)^{-\frac 1 3}dx + 2\pi ie^{-i{\frac \pi 3}}=0.[/imath] So,[imath]\int_0^1(x^2-x^3)^{-\frac 1 3}dx=\frac {2\pi} {\sqrt 3}.[/imath] | 748138 | Evaluating Contour Integral
How do I go about evaluating the following by contour integration? [imath] \int^1_0 \frac{dx}{(x^{2} - x^{3})^{1/3}} [/imath] The question does not fit in the standard form of : [imath]\int^{2\pi}_0[/imath] or [imath]\int^\infty_{-\infty}[/imath] etc. I tried substituting [imath] x = cos(\theta) [/imath] but got stuck. |
2472927 | Proving convergence of a sequence [imath]a_{n+1} = \frac{1}{1+a_{n}}[/imath] and finding the limit.
I am trying to prove the sequence (for n = 1,2,3...) [imath]a_{1} = 1,\,a_{n+1} = \frac{1}{1+a_{n}}[/imath] converges and then find its limit. I can find its limit easily but I cannot find a way to prove that it converges. I know there exists a theorem stating that a monotone increasing and bounded sequence has a limit. I do not think this sequence is increasing. Also, how can I apply the theorem [imath]\lim_{x\to\infty} q^{n}=0 \,{if} |q| < 1[/imath] to this question? I think it is irrelevant but it is good to know different approaches to prove convergence of a sequence. Thank you. | 38739 | Convergence of [imath]a_{n+1}=\frac{1}{1+a_n}[/imath]
We define [imath]a_{n+1}=\frac{1}{1+a_n}, a_0=c> 0[/imath] I stumbled across that sequence and Mathematica gives me that it converges against [imath]\frac{1}{2} \left(\sqrt{5}-1\right)[/imath] which doesn't even depend on [imath]a_0[/imath]. Normally I show that sequences converge by seeing that they are monotonic and then the limit can be easily found by setting [imath]a_n=a_{n+1}[/imath], but this one seems to be alternating. Also by checking OEIS I noticed that [imath]a_n=1/g(n)[/imath] where [imath]g(n)[/imath] gives the [imath]n+1[/imath]'th Golden Rectangle Number. I would be glad if someone can show me how to show that such sequences converge and how to find the limit, also maybe this is a well known sequence, as it has a quite simple form, too bad that google is very bad for looking for sequences and OEIS doesn't mention anything. |
2480391 | Perfect power of [imath]2[/imath] in whose decimal representation, each digit is also a non-negative power of [imath]2[/imath]
Let [imath]A:=\{n \in \mathbb N : 2^n=\sum_{j=0}^k 10^j2^{n_j} , [/imath] for some integers [imath] 0 \le n_j \le 3 , j=0,...,k [/imath] with [imath] k>1\}[/imath] ; then is the set [imath]A[/imath] finite or infinite ? I have been unable to find any number other than [imath]7[/imath] in [imath]A[/imath] . Please help . Thanks in advance . | 1057819 | Is 128 the only multi-digit power of 2 such that each of its digits is also a power of 2?
The number [imath]128[/imath] can be written as [imath]2^n[/imath] with integer [imath]n[/imath], and so can its every individual digit. Is this the only number with this property, apart from the one-digit numbers [imath]1[/imath], [imath]2[/imath], [imath]4[/imath] and [imath]8[/imath]? I have checked a lot, but I don't know how to prove or disprove it. |
2480458 | How to prove [imath]x_1 + x_2 +\ldots+x_n \le \left(\frac{1}{2}\right)\Rightarrow (1 - x_1)(1-x_2)\cdot \ldots \cdot(1-x_n)\ge \left(\frac{1}{2}\right)[/imath]?
Prove, that [imath]x_1 + x_2 +\ldots+x_n \le \left(\frac{1}{2}\right)\Rightarrow (1 - x_1)(1-x_2)\cdot \ldots \cdot(1-x_n)\ge \left(\frac{1}{2}\right)[/imath]. Value [imath]n[/imath] is a positive real number [imath](x_1, x_2, \ldots, x_n)[/imath] I tried using mathematical induction but didn't get too far with that. Any ideas how to solve it with induction? | 1118675 | Proof of an inequality by induction: [imath](1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n[/imath]
Let [imath]n \in \mathbb N^+[/imath]. Show that if [imath]x_1, x_2, ... , x_n[/imath] are [imath]n[/imath] real numbers such that [imath]-1 \le x_i \le 0[/imath] for each [imath]1 \le i \le n[/imath], then [imath](1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n[/imath] I am using a proof by induction, but I am unable to complete it. I would appreciate any tips on how I could finish the proof. Incomplete Proof Let [imath]P(n)[/imath] be the proposition that [imath](1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + \sum_{i = 1} ^ n x_i[/imath]. When [imath]n = 1[/imath], LHS = RHS = [imath]1 + x_i[/imath]. Clearly [imath]1 + x_i \ge 1 + x_i[/imath] and so [imath]P(1)[/imath] is true. Assume that [imath]P(k)[/imath] is true, i.e. [imath](1 + x_1)(1 + x_2)...(1 + x_k) \ge 1 + \sum_{i=1}^k x_i[/imath]. Case 1 Suppose [imath]x_{k + 1} = -1[/imath]. Then [imath]1 + x_{k + 1} = 0[/imath], hence LHS = [imath](1 + x_1)(1 + x_2)...(1 + x_k)(1 + x_{k+1}) = 0[/imath]. RHS = [imath]1 + x_1 + x_2 + ... + x_k - 1 = x_1 + x_2 + ... + x_k[/imath] which is always nonpositive since each [imath]x_i[/imath] is nonpositive. Hence RHS [imath]\le 0[/imath] and so LHS [imath]\ge[/imath] RHS. Case 2 Suppose [imath]x_{k+1} = 0[/imath]. Then [imath]1 + x_{k+1} = 1[/imath], hence multiplying it to LHS does not change anything. Also, adding [imath]x_{k+1}[/imath] to RHS does not change anything. Therefore, LHS [imath]\ge[/imath] RHS still holds. Case 3 Suppose [imath]-1 < x_{k+1} < 0[/imath]. Then [imath]0 < 1 + x_{k+1} < 1[/imath]. Adding [imath]x_{k+1}[/imath] to RHS makes the RHS smaller since [imath]x_{k+1}[/imath] is negative. If LHS [imath]= 0[/imath], then multiplying [imath]1 + x_{k+1}[/imath] to LHS does not change anything, and so LHS [imath]\ge[/imath] RHS holds. If LHS > [imath]0[/imath], then multiplying [imath]1 + x_{k+1}[/imath] to LHS makes the LHS smaller since [imath]0 < 1 + x_{k+1} < 1[/imath]. In this case LHS [imath]\ge[/imath] RHS does not necessarily hold? ... |
2480656 | Finding rank(a) as a varies
I computed the determinant to [imath]D= a^{3} - 27a-54[/imath]. When I solve [imath]D=0[/imath], I get [imath]a=6[/imath] and [imath]a=-3[/imath] Now when I plug in these values and bring the matrix to rref, I end up with the identity matrix, both for [imath]a=6[/imath] and [imath]a=-3[/imath]. That's confusing because that would mean the matrix [imath]A[/imath] has [imath]rank=3[/imath] for all values of [imath]a[/imath], and that can't be. So what is going on here? EDIT: I have already looked at similar questions so please don't mark this as duplicate. From those questions I have understood how to solve this, so that is not the problem. I took the route involving determinants but it gave me results that weren't expected. I have checked the algebra quite a few times and it always seems to come down to this, so I'm posting in the hope that someone can tell me what's wrong. | 2256440 | Finding rank of a matrix depending on parameter a
We have the matrix [imath]A = \begin{bmatrix}1 & a & 1\\-4 & 2 & a \\ a & -1 & -2 \\ 4 & 6 & 1\end{bmatrix}[/imath] and we want to find the rank of this matrix as a function of parameter [imath]a[/imath]. |
2345385 | Theorem of Primes in Arithmetic Progression
I'm am studying number theory and I came across this theorem by Dirichlet but the proof is not given and I do not know how to prove it Can anyone help me? [imath]Theorem[/imath] [imath]of[/imath] [imath]Primes[/imath] [imath]in[/imath] [imath]Arithmetic[/imath] [imath]Progression:[/imath] If [imath]a[/imath] and [imath]b[/imath] are integers, with [imath](a,b)=1[/imath] so the arithmetic progression [imath]an+b[/imath], [imath]n=1,2,3,...[/imath] has infinite primes. | 11330 | Dirichlet's theorem on primes in arithmetic progression
Is there a proof in the spirit of Euclid to prove Dirichlet's theorem on primes in arithmetic progression? (By the spirit of Euclid, I mean assuming finite number of primes we try to construct another number which has a prime factor which falls in the same equivalence class as the other primes but the number is not divisible by any of the primes we considered in the initial list.) I am aware of the proof using [imath]L[/imath] functions but I am curious to know if Euclid's "simple" idea can be extended to all other cases as well. I tried googling but am unable to find a proof other than the ones relying on [imath]L[/imath] functions. If it is not possible, to what cases can Euclid's idea be extended to? Any other proofs are welcome as well. |
373370 | Dirichlet's theorem on arithmetic progressions
The theorem can be found on Wikipedia. In the subsection "Proof" Wikipedia says that there is a proof for the case [imath]a=1[/imath] which uses no calculus, instead splitting behavior of primes in cyclotomic extensions. Could you help me proving this? Assumption. For every natural number [imath]n[/imath] there are infinitely many prime numbers [imath]p\equiv 1 \pmod n[/imath]. Proof: I assume there are only finitely many [imath]p_1,...,p_i[/imath], and let [imath]P=p_1\cdot...\cdot p_i[/imath]. The cyclotomic polynomial [imath]\phi_n(x):=\prod_{\gcd(k,n)=1,\ 1\le k<n}(x-\zeta_n^k)\;,\;\;\zeta_n=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n}[/imath] The hint in Neukirch books states that not all numbers [imath]\phi_n(xnP)[/imath] for [imath]x\in\mathbb Z[/imath] can equal [imath]1[/imath]. Why? Now let [imath]p \mid \phi_n(xnP)[/imath] for suitable [imath]x[/imath]. How can a contradiction be followed from this? | 677024 | arithmetic progression by Dirichlet
The arithmetic progression [imath]a_N=(p-1)N+1[/imath] contains infinitely many primes [imath]q[/imath] by Dirichlet. I have searched this part in wiki, but I din't get any relevant proof. Can any one prove it how [imath]a_N[/imath] contains infinitely many primes and what is Dirichlet proof? please explain. |
2480976 | Mean value Theorem with two values of d
[imath]f[/imath] is a continuous continuous on [imath][0,1][/imath], differentiable on [imath](0,1)[/imath], and [imath]f(0)=0, f(1)=1[/imath]. (a) There exists [imath]d_1 < d_2[/imath] in (0,1) such that [imath] \frac{1}{f'(d_1)}+\frac{1}{f'(d_2)}=2[/imath] (b) There exists [imath]d_1 < d_2[/imath] in (0,1) such that [imath]f'(d_1).f'(d_2) = 1[/imath] | 2473352 | Question on Application of Mean Value Theorem
Suppose that g is a function that is continuous on [imath][0,1][/imath] and differentiable on [imath](0,1)[/imath], [imath]g(0)=0[/imath] and [imath]g(1)=1[/imath]. Show that: 1) There exists [imath]a_1<a_2[/imath] in [imath](0,1)[/imath] such that [imath]g'(a_1)+g'(a_2)=2[/imath] 2) There exists [imath]a_1<a_2[/imath] in [imath](0,1)[/imath] such that [imath]\frac{1}{g'(a_1)}+ \frac{1}{g'(a_2)}=2[/imath] 3) There exists [imath]a_1<a_2[/imath] in [imath](0,1)[/imath] such that [imath]g'(a_1)g'(a_2)=1[/imath] I know by the Mean Value Theorem, there exists [imath]a[/imath] such that [imath]f'(a)=1[/imath], but I am not sure on how am I supposed to show the existence of two items satisfying the conditions. Many thanks for your help on this! |
2481194 | Let V be a finite set and let [imath]\Omega \subseteq 2^{V}[/imath] be a nested family of odd subsets of V. Show that [imath]|\Omega| \leq \frac{3}{2}\cdot|V|[/imath].
Let V be a finite set and let [imath]\Omega \subseteq 2^{V}[/imath] be a nested family of odd subsets of V, where subset [imath]A\subseteq V[/imath] is odd if |V(A)| is odd (if A has an odd amount of nodes). Question: How do I show that [imath]|\Omega| \leq \frac{3}{2}\cdot|V|[/imath]? I have no clue on how I should tackle this problem, thanks in advance! | 1993028 | Max family of odd subset of a nested set.
Hi I have to proof that for [imath]\Omega[/imath] a nested family of odd subsets of [imath]V[/imath] with [imath]V[/imath] finite, we have that [imath]|\Omega|\leq\frac{3}{2}|V|[/imath]. I can prove that for certain constructions of [imath]\Omega[/imath], it is satisfies [imath]|\Omega|\leq \frac{3}{2} |V|[/imath], the problem I encounter is that I have to proof that my construction of [imath]\Omega[/imath] is optimal. (Since I have to prove that this holds for every family of odd subsets) Can anyone give me a hint on how to proof that every [imath]\Omega[/imath] satisfies this equation. [imath]\Omega[/imath] is nested if for all [imath]A,B\in\Omega[/imath] either [imath]A\subseteq B[/imath], [imath]B\subseteq A[/imath], or [imath]A\cap B=\varnothing[/imath]; a set is odd if its cardinality is odd. |
2480967 | Notation for indefinite integrals
The notation [imath]\int{f(x)}\,dx[/imath] is commonly used to denote the set of ALL antiderivatives of the function [imath]f[/imath]. Is it wrong to use the notation [imath]\int{f(x)}\,dx[/imath] to denote a single antiderivative of the function [imath]f[/imath] without the constant of integration [imath]c[/imath]? I saw many instances where the notation [imath]\int{f(x)}\,dx[/imath] is used to denote a single antiderivative with the constant of integration omitted. For example to evaluate [imath]\int {xe^x}\,dx[/imath], some textbooks wrote [imath]v=\int\,dv=\int{e^x}\,dt=e^x[/imath], without the constant of integration. For example, to solve the differential equation [imath]y'+\frac{2}{t}y=e^t[/imath] by using an integrating factor, some textbooks wrote [imath]\mu(t)=\exp(\int{2/t}\,dt)=t^2[/imath], without a constant of integration. | 2174285 | Integrating Factor Method. Why Can We Set the Constant of Integration to Zero?
My textbook gives a demonstration of the use of the integrating factor method for solving first order ordinary differential equations: [imath]\dfrac{dx}{dt} + p(t)x = r(t)[/imath] [imath]\implies g(t)\dfrac{dx}{dt} + g(t)p(t)x = g(t)r(t)[/imath] [imath]\dfrac{d}{dt}[g(t)x][/imath] We use the product rule and compare the expression to the left-hand side of the above equation: [imath]g(t)\dfrac{dx}{dt} + \dfrac{dg}{dt}x = g(t)\dfrac{dx}{dt}+g(t)p(t)x[/imath] [imath]\therefore \dfrac{dg}{dt} = g(t)p(t)[/imath] The above equation is separable. [imath]\therefore \int\dfrac{dg}{g(t)} = \int p(t) dt[/imath] [imath]\implies ln[g(t)] = \int p(t)dt[/imath] My textbook then says that, "We can set the constant of integration to zero -- any function that satisfies [imath]\dfrac{dg}{dt} = g(t)p(t)[/imath] will do for our purposes". However, it gives no explanation as to why this is the case. Why can we set the constant of integration to zero? Why is any function that satisfied [imath]\dfrac{dg}{dt} = g(t)p(t)[/imath] appropriate for our purposes? I would greatly appreciate it if people could please take the time to elaborate on this. |
1872754 | Proving [imath](C_b(X,Y),d_{\infty})[/imath] is complete if [imath](Y,d_y)[/imath] is complete
Let [imath](X,d_X)[/imath] and [imath](Y,d_Y)[/imath] be metric spaces. We say a function [imath]f: X \rightarrow Y[/imath] is bounded if [imath]f(X)[/imath] is a bounded set of [imath]Y[/imath]. Consider [imath]C_b(X,Y) = \left\{f: X \rightarrow Y \mid f \ \text{is continuous and bounded} \right\} [/imath] with the metric [imath]d_{\infty}(f,g) = \sup\left\{d_y(f(x),g(x)) \mid x \in X \right\}. [/imath] Prove that [imath](C_b(X,Y),d_{\infty})[/imath] is complete if [imath](Y,d_Y)[/imath] is complete. Attempt: In my course, we already proved that if [imath](X,d_X)[/imath] is a metric space, then [imath]C_b(X, \mathbb{R}) = \left\{f: X \rightarrow \mathbb{R} \mid f \ \text{is continuous and bounded} \right\}[/imath] is a complete space for the [imath]d_{\infty}(f,g)= \sup\left\{|f(x) - g(x)| \mid x \in X \right\}[/imath] metric. I want to use this result now. I know that if [imath]Y[/imath] is a complete metric space, and [imath]X[/imath] is a bounded subspace of [imath]Y[/imath], then [imath]X[/imath] is also complete. So in this case I said [imath]C_b(X,Y)[/imath] is a subspace of the complete space [imath](C_b(X,\mathbb{R}),d_{\infty})[/imath]. So I need to prove that [imath]C_b(X,Y)[/imath] is closed in [imath]C_b(X, \mathbb{R})[/imath] ? Then let [imath]f_n \in C_b(X,Y)[/imath] be a convergent sequence. Denote the limit by [imath]f \in C_b(X,\mathbb{R})[/imath]. I need to prove that [imath]f \in C_b(X,Y)[/imath]. I'm not sure how to prove the continuity of [imath]f[/imath]. Also, I haven't used the fact yet that [imath](Y,d_Y)[/imath] is complete so I'm not sure if this is the right direction. Any help is appreciated. | 2480932 | Codomain of function complete implies function space complete
Let [imath](X,d_X)[/imath] and [imath](Y,d_Y)[/imath] be two metric spaces. Let [imath](C_b(X,Y), d_C)[/imath] be a function space with the metric [imath]d_C(f,g):=\sup\limits_{x\in X} d_Y(f(x), g(x)).[/imath] Prove that if [imath](Y,d_Y)[/imath] is complete then [imath](C_b(X,Y), d_C)[/imath] is complete. Unfortunately, I have no idea how to approach this problem. I can think of something like this: Let [imath](y_k)_k\subset Y[/imath] be a Cauchy sequence converging to [imath]y\in Y[/imath]. Then there exists some [imath]f\in C_b(X,Y)[/imath] such that [imath]f(x_k)=y_k[/imath] for all [imath]k\in \mathbb N[/imath]. Since [imath]f[/imath] is continuous, [imath]\lim\limits_{k\to\infty} d_Y(f(x_k), y)=0[/imath]. But how do I go from here? As far as I understand, what needs to be proved is that any Cauchy sequence of functions in [imath]C_b(X,Y)[/imath] converges in [imath]C_b(X,Y)[/imath], that is, that the limit is bounded and continuous. But how do we devise such a sequence of functions? |
2480023 | problem with calculation of conditional expectation
Let [imath]Y_1,Y_2,\dots , i.i.d, \ \mathbb Z \ [/imath]valuable and integrable random variables. [imath]Z_n:= \sum_{i=1}^n Y_i[/imath]. Derive [imath]E(Z_n \mid \sigma(Z_k)), k,n \in \mathbb N_0[/imath]. [imath]E(Z_n \mid \sigma(Z_k))=E(\sum_{i=1}^n Y_i \mid \sigma (Z_k))=n \cdot E(Y_1 \mid \sigma(Z_k))[/imath] by using the given properties. I am not sure how to continue now, should one consider two cases now ([imath]n \le k[/imath] and [imath]n>k[/imath])? Another point I am wondering is that if [imath]Y_1[/imath] is [imath]\sigma (Z_k)[/imath] measurable? But if so I could use the property that [imath]E(X \mid G)=X[/imath] if [imath]X[/imath] is [imath]G[/imath] measerable. I would appreciate any help! | 1039277 | Conditional expectation for random walks
The questions asks to [imath] E[X_1|S_n][/imath] where [imath] S_n = \sum_{[n]} X_i [/imath] with [imath]X_i[/imath] i.i.d. of finite expectation. My attempt was to consider an arbitrary Borel set, pull it back under [imath] S_n [/imath] to get a set in [imath]\sigma(S_n)[/imath], and then average [imath]X_1[/imath] over that pull back and normalize: [imath] E[X_1|S_n] = E[X_1 1_{S_n^{-1}(B)}]/P[S^{-1}_n(B)][/imath] for any Borel set [imath]B[/imath]. However, it just doesn't feel right - it seems just symbol pushing, without giving any intuition to what's happening. Can anyone explain if this is correct, of if I missing something deep? |
709873 | Proving that the unities of a ring form a group under multiplication
I am presented with the following task: Show that if [imath]U[/imath] is the collection of all units in a ring [imath]\langle R, +, \cdot\rangle[/imath] with unity, then [imath]\langle U, \cdot\rangle[/imath] is a group. I am still not confident with proofs in Abstract Algebra, and I'd like to receive some verification/slaughter/constructive critique on my reasoning here. I do not know if there are any mistakes, or how serious they are. We know that there is an identity element, since the unity itself is a unit, and therefore in the group. It follows directly from the definition of a unit that there exists an inverse for every element in the group, and associativity follows directly from the Ring-axioms. As for closure, we need to show that the product of two units is also a unit. Let [imath]a, b \in U[/imath] be two elements with the property that [imath]a\cdot b \ne e[/imath], where [imath]e[/imath] is the identity element in [imath]\langle U, \cdot \rangle[/imath]. We know that there exists elements [imath]a', b' \in U[/imath] with the property that [imath]a' \cdot a = e = a' \cdot a[/imath] and [imath]b' \cdot b = e = b \cdot b'[/imath]. Let [imath]a', b'[/imath] be two such elements with the property that [imath]a' \cdot b' \ne e[/imath]. We may now compute [imath](a\cdot b) \cdot (a' \cdot b')[/imath], and from the Ring-axioms we have associativity under multiplication, so we may write the expression as [imath]a \cdot (b \cdot a' \cdot b') = a \cdot (b \cdot b' \cdot a') = a \cdot (e \cdot a') = a \cdot a' = e[/imath]. Thus we have shown that the product of any two elements is a unity, which leads to closure. All the group-axioms are met and [imath]\langle U, \cdot \rangle[/imath] is a group. I have to apologize if you find the argument needlessly long; I've been having some trouble with assuming things I am not supposed to assume (like what I'm going to prove), so I wanted to stay clear of that. | 2481687 | Group of invertible elements make a multiplicative group
Working with a nontrivial ring with unity, the set S of all the invertible elements in a ring is a multiplicative group. So this is saying the invertible elements(or units) make a group that is closed under multiplication. So should I assume to have some invertible elements from my ring and show they are equal to 1? Then multiply the elements together to show the group is closed? Let [imath]a,b \in R[/imath], where [imath]R[/imath] is a nontrival ring. [imath]a[/imath] and [imath]b[/imath] are invertible, thus for some [imath]c[/imath], [imath]a\cdot c = c\cdot a =1[/imath] and [imath]b\cdot c =c\cdot b =1[/imath]. So [imath](a\cdot c) \cdot (b\cdot c) = 1\cdot 1 = 1[/imath] thus [imath]R[/imath] is closed? |
2438639 | A determinant expression
Supposing [imath]a\neq b \in \mathbb{R}[/imath], for [imath] A_n =\begin{bmatrix} p_1 &a &\ldots &a &a \\ b &p_2 &\ldots &a &a \\ \vdots &\vdots &\ddots &\vdots &\vdots \\ b &b &\ldots &p_{n-1} &a\\ b &b &\ldots &b &p_n \end{bmatrix} [/imath] (Other than on the diagonal line, entries are either [imath]a[/imath] or [imath]b[/imath]'s.) Show that [imath] \textrm{det}(A_n) =\frac{bf(a) -af(b)}{b-a} [/imath] where [imath] f(x) :=(p_1-x) (p_2-x)\dotsb (p_n-x) [/imath] I tried many values, and it seems to be correct. My obvious idea is to show this by mathematical induction, but things gets messy. And the expression seems to allow an application of MVT, but it is not clear how to proceed. | 1801627 | Prove that [imath]\det(A)=p_1p_2-ba={bf(a)-af(b)\over b-a}[/imath]
Let [imath]f(x)=(p_1-x)\cdots (p_n-x)[/imath] [imath]p_1,\ldots, p_n\in \mathbb R[/imath] and let [imath]a,b\in \mathbb R[/imath] such that [imath]a\neq b[/imath] Prove that [imath]\det A={bf(a)-af(b)\over b-a}[/imath] where [imath]A[/imath] is the matrix: [imath]\begin{pmatrix}p_1 & a & a & \cdots & a \\ b & p_2 & a & \cdots & a \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b & b & b & \cdots & p_n \end{pmatrix}[/imath] that is the entries [imath]k_{ij}=a[/imath] if [imath]i<j[/imath], [imath]k_{ij}=p_i[/imath] if [imath]i=j[/imath] and [imath]k_{ij}=b[/imath] if [imath]i>j[/imath] I tried to do it by induction over [imath]n[/imath]. The base case for [imath]n=2[/imath] is easy [imath]\det(A)=p_1p_2-ba={bf(a)-af(b)\over b-a}[/imath] The induction step is where I don´t know what to do. I tried to solve the dterminant by brute force(applying my induction hypothesis for n and prove it for n+1) but I don´t know how to reduce it. It gets horrible. I would really appreciate if you can help me with this problem. Any comments, suggestions or hints would be highly appreciated |
2482571 | Proof that there are infinite primes [imath]p[/imath] such that [imath]p+2[/imath] is not prime.
Let [imath]p\in\mathbb N[/imath] be a prime. I want to prove that there is an infinite amount of primes [imath]p[/imath] with the property [imath]p+2[/imath] is not a prime. I am stuck on how to do this, though. I have tried an alternation of Euclids proof of the infinitude of primes but I cannot seem to find a construction that implies the desired result. Any help would be appreciated. | 614209 | Infiniteness of non-twin primes.
Well, we all know the twin prime conjecture. There are infinitely many primes [imath]p[/imath], such that [imath]p+2[/imath] is also prime. Well, I actually got asked in a discrete mathematics course, to prove that there are infinitely many primes [imath]p[/imath] such that [imath]p + 2[/imath] is NOT prime. |
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