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2285551 | Independence of [imath]x = z_1^2 + z_2^2[/imath] and [imath]y = z_2 / z_1[/imath], with [imath]z_1, z_2[/imath] having a standart normal distribution
I'm trying to prove the independence of [imath]x = z_1^2 + z_2^2[/imath] and [imath]y = z_2 / z_1[/imath], with [imath]z_1, z_2[/imath] having a standart normal distribution. Knowing that [imath]z_1[/imath] and [imath]z_2[/imath] are independent from each other. I know [imath]x[/imath] is like the [imath]r^2[/imath] of a polar coordinate. And [imath]y[/imath] is like [imath]arctg(\alpha)[/imath]. But I still don't know how to use that to prove what I want. Things I'm trying todo since I posted the question: Trying to use Jacobian method to at least find [imath]f_{X,Y}[/imath], if I could then find [imath]f_{X}[/imath] and [imath]f_{Y}[/imath] I could prove the independence using the definition. Things I've done since first posted: I have found [imath]f_{Z_1,Z_2} (z_1,z_2) = f_{Z_1}(z_1)*f_{Z_2}(z_2) = \frac{1}{2\pi}*e^{\frac{-(z_1^2 + z_2^2)}{2}}[/imath] Some ideas: I'm stuck trying to find the inverse of the transformation, but if there was a way to find it I'd be closer to know the distribution of [imath]f_{X,Y}[/imath] If I do something along [imath]z_1 = r*cos(\theta), z_2 = r*sin(\theta)[/imath], then I have [imath]X = r^2, Y = tan(\theta)[/imath], maybe it's better to do two "easy" transformations than a hard one. That transformation leads to [imath]z_2/z_1 = tan(\theta)[/imath] [imath]\sqrt{z_1^2 + z_2 ^ 2} = r[/imath] Then going by the jacobian method, saying that g is the transformation. I have that [imath]f_{R,\theta}(r,\theta) = f_{z_1,z_2}(r*cos(\theta), r*sin(\theta))*J(g^-1(r,\theta)) = f_{z_1,z_2}(r*cos(\theta), r*sin(\theta))*r[/imath] [imath] = \frac{r}{2\pi}*e^{-r^2/2}[/imath] Then we have the other one, [imath]x = r^2[/imath] [imath]y = tan(\theta)[/imath] Once again I go with the jacobian method, [imath]|J(t^{-1}(X,Y)| = | \begin{bmatrix} \frac{\partial \sqrt{x}}{\partial x} & \frac{\partial \sqrt{x}}{\partial y} & \\ \frac{\partial \arctan(y)}{\partial x} & \frac{\partial arctan(y)}{\partial y} \end{bmatrix} | = | \begin{bmatrix} \frac{1}{2*x^{2}} & 0 & \\ 0 & \frac{1}{1 + y^2} \end{bmatrix} | = \frac{1}{2\sqrt(x)*(1+y^{2})}[/imath] [imath]f_{r,\theta}(\sqrt(x),arctan(\theta)) = \frac{1}{2\pi}*e^{-\frac{x}{2}}[/imath] [imath]f_{x,y} = \frac{1}{2\pi}*\sqrt{x}*e^{-\frac{x}{2}}*\frac{1}{2\sqrt{x}}*(1+y^{2}) = \frac{e^{-\frac{x}{2}}}{4\pi*(1+y^2)}[/imath] This is close to the answer, because X has an exponential distribution, and Y another one. But there is a mistake somewhere, because the probability of Y being any number is [imath]1/2[/imath] instead of [imath]1[/imath] | 2266214 | Independence of Gaussian variables
I want to solve the following problem: Suppose that [imath]X[/imath] and [imath]Y[/imath] are two independent, Gaussian variables, with mean zero and common variance [imath]\sigma^2[/imath]. Let [imath]R^2 = X^2+ Y^2[/imath] and [imath]Z = \arctan (X/Y)[/imath]. I know that in this case [imath]R^2[/imath] has an exponential distribution and [imath]Z[/imath] has a uniform distribution. However, are [imath]R[/imath] and [imath]Z[/imath] independent from each other? |
2286643 | When is [imath]G[/imath], a finite group, the sum of its Sylow [imath]p[/imath]-subgroups?
Questions like this have been asked before, but I just want some quick clarification on something. Suppose that [imath]G[/imath] is the sum of its Sylow [imath]p[/imath]-subgroups. Then each Sylow subgroup must be normal, and of course we have the other direction. If each Sylow subgroup is normal, then [imath]G[/imath] is the product of its Sylow subgroups. My question that I need clarified is, "if [imath]G[/imath] is a sum of its Sylow subgroups, when will this be abelian, non-abelian?" | 555269 | Direct product of Sylow subgroups
Proposition II.7.5 of Hungerford's Algebra goes Proposition 7.5. A finite group is nilpotent if and only if it is the direct product of its Sylow subgroups. Let [imath]G = (\mathbb{Z}_6, +)[/imath]. [imath]G[/imath] is abelian, so it is nilpotent and thus by the proposition is the direct product of its Sylow subgroups. The Sylow subgroups of [imath]G[/imath] are [imath]H = (\mathbb{Z}_3, +)[/imath] and [imath]K = (\mathbb{Z}_2, +)[/imath], so the proposition says [imath]G = H \times K[/imath]. In particular, [imath]0 \in H[/imath] and [imath]0 \in K[/imath], so [imath](0,0) \in H \times K = G[/imath]... but that's not true, since [imath]G = \{0,1,2,3,4,5\}[/imath]. Where is/are my misunderstanding(s)? |
2078777 | What is the remainder when [imath]40![/imath] is divided by [imath]1763[/imath]?
What is the remainder when [imath]40![/imath] is divided by [imath]1763[/imath]? My try : Factorizarion of [imath]1763 = 41 * 43[/imath] By using Wilson's theorem, I can say [imath]40! = -1 (mod 41) [/imath] and [imath]42.41.40! = -1(mod 43)[/imath] => [imath]40! = -22(mod 43)[/imath] How can I combine both the results ? | 1990448 | What is the remainder when [imath]40![/imath] is divided by [imath]1763[/imath]?
Need some assistance on answering this problem. For this question, there is no answer at the back of Kenneth Rosen's book. I initially began tackling this problem using Wilson's theorem, where [imath]40! \equiv -1[/imath] (mod [imath]41[/imath]) [imath]1763 = 41 * 43[/imath] gcd([imath]41,43[/imath]) [imath]= 1[/imath], hence these two numbers are relative primes. [imath]42! \equiv -1[/imath] (mod [imath]43[/imath]) [imath]\Rightarrow 42! = 40! * 41 * 42 = 40! * 2 \equiv -1[/imath] (mod [imath]43[/imath]). Using Euler's phi-function to find the inverse I obtained, [imath]40! \equiv -1 * 2^{41}[/imath] (mod [imath]43[/imath]) [imath]\equiv -22[/imath] (mod [imath]43[/imath]). Thus, [imath]40! \equiv 22[/imath] mod [imath]1763[/imath]). Have I done this correctly? If not, can you correct me and explain my mistakes. |
2286600 | Find the derivative of [imath]y=x^x[/imath].
Find the derivative of [imath]y=x^x[/imath]. My Attempt: [imath]y=x^x[/imath] Taking [imath]\textrm {ln}[/imath] on both sides, we get: [imath]\textrm {ln} y= \textrm {ln} x^x[/imath] [imath]\textrm {ln} y = x \textrm {ln} x[/imath] How do I procees further? | 146502 | Finding the derivative of [imath]x^x[/imath]
I am having trouble finding the derivative of the following: [imath]y = x^x+x^3+3^x+3^3[/imath] [imath]\frac{dy}{dx}= x \times x^{(x-1)}+3x^2+3^x\ln3+0[/imath] I think the [imath]x^x[/imath] part is wrong. Any help would be appreciated. |
2286885 | How to show [imath]M\subseteq J\left( R\right)[/imath]?
Let [imath]R[/imath] be a ring. Set [imath]M=\{ x\in R|\forall y\in R\Rightarrow 1-xy\;\; \text{unitary in}\; R\}[/imath] and [imath]J\left( R\right) =\cap \{ I|I\unlhd R, I \text{is a maximal ideal}\}[/imath]. How to show that [imath]M\subseteq J\left( R\right)[/imath]? | 1221256 | Two definitions of Jacobson Radical
I have in my notes that the Jacobson radical of a ring [imath]R[/imath] is: [imath]J(R) = \cap[/imath]{[imath]I[/imath] | [imath]I[/imath] primitive ideal of [imath]R[/imath]} [imath]= \cap[/imath] {[imath]Ann_R M[/imath] | [imath]M[/imath] simple [imath]R[/imath]-module}. I have now seen elsewhere that [imath]J(R) = \{x ∈ R: xy-1 ∈ R^\times \text{ for all } y ∈R\}[/imath] I was just wondering would anyone be able to provide an explanation as to how to get from one definition to the other? Thanks. |
2287071 | Standard deviation of a coin flipped 40 times
Let [imath]X[/imath] be the number of times that a fair coin that is flipped 40 times lands on heads. Find [imath]SD(X)[/imath] (the standard deviation). My attempt: Intuitively, [imath]E[X]=20[/imath] [imath]Var(X)=E[X^2]-(E[X])^2[/imath] and [imath]SD(X)=\sqrt{Var(X)}[/imath] [imath]\displaystyle E[X^2]=\sum_{x=0}^{40} x^2p(x)=\sum_{x=0}^{40}x^2\binom{40}{x}\left(\frac{1}{2}\right)^{40}[/imath] Is there an easier way of doing this? The answer given in my textbook is [imath]\sqrt{10}[/imath]. | 2268025 | How to calculate the expectation of the square of the number of heads in N tosses?
Suppose we have a coin that has a probability [imath]p[/imath] of landing head. Let H be the number of heads obtained in N tosses. How to calculate [imath]E(H^2)[/imath]? I need this to calculate some other things. But I can't figure out. I know I need to take [imath]\sum x^2p(x)[/imath], but I am unable to evaluate it to a nice expression. |
2286931 | Convex subset of a linear space
I am dealing with the following exercise A set [imath]\mathcal M[/imath] in a normed linear space [imath]\mathcal R[/imath] is said to be convex if [imath]\mathcal M[/imath] contains all elements of the form [imath]\alpha x + \beta y[/imath], where [imath]\alpha, \beta \geq 0[/imath], [imath]\alpha+\beta=1[/imath], provided that [imath]\mathcal M[/imath] contains x and y. Prove that the set of all elements [imath]x \in \mathcal R[/imath] satisfying the inequality [imath]\|x-x_0 \| \leq c[/imath], where [imath]x_0[/imath] is a fixed element of [imath]\mathcal R[/imath] and [imath]c > 0[/imath], is convex. Assume [imath]\|x-x_0 \|\leq c[/imath], and [imath]\|y-x_0 \| \leq c[/imath], we have to proof that [imath]\| \alpha x + (1-\alpha) y - x_0 \| \leq c.[/imath] By the triangular inequality, [imath]\begin{align*}\| \alpha x + (1-\alpha) y - x_0 \| & \leq \| \alpha x \| + \|(1-\alpha) y - x_0 \| \\ & \leq \| \alpha x \| + \|(1-\alpha) y \| +\|- x_0 \| \\ & \leq |\alpha| \| x \| + |(1-\alpha)| \| y \| +\| x_0 \| \\ & = c.\end{align*}[/imath] Is it enough for proof the theorem? If not, what path should I take? | 187502 | How to show convexity of a ball in metric space?
If [imath](X,\|\cdot\|)[/imath] is a normed linear space, then how to show any ball [imath]B(x,r)[/imath] is convex? I know that if [imath]x,y\in A\subset V[/imath] then [imath][x,y]\subset A[/imath], where [imath]A[/imath] is a convex subset of vector space [imath]V[/imath] and [imath][x,y]=\{(1-t)x+ty\mid 0\leq t \leq 1\}[/imath]. Please give me some hint. I tried the following: Claim. [imath][a,b] \subset B(x,r)[/imath] Let [imath]a,b \in B(x,r).[/imath] Then I get [imath]\|x-a\|<r[/imath] and [imath]\|x-b\|<r[/imath]. Is it right? |
2287344 | >Find the Equation of ellipse that is tangent to the lines [imath]x-4y=10[/imath] and [imath]x+y=5[/imath]
Find the Equation of Ellipse that is tangent to the lines [imath]x-4y=10 [/imath] and [imath]x+y=5[/imath] I don't know what is the idea of this problem , how can i begin Thank you for your help | 2287018 | Finding the equation of an ellipse tangent to two lines
Find the equation of the ellipse tangent to the two lines [imath]x-4y=10[/imath] and [imath]x+y=5[/imath] I know that the equation of ellipse in general is [imath] \frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2}=1[/imath] but how can continue? I substituted for [imath]x [/imath] from the equations of the two lines in the equation of ellipse but it still difficult to continue! is it true to take [imath](x_0,y_0)=(0,0)[/imath] ?? |
2281128 | Is there an intuitive explanation of [imath]n[/imath]-dimensional sphere's surface and volume being maximal in dimensions 5 and 6?
I was wondering if there is an intuitive explanation why the surface of an [imath]n[/imath]-dimensional sphere is maximal at [imath]n=6[/imath] and for the volume [imath]n=5[/imath]. I know that for the surface you have: [imath]A_n=\frac{2\pi^\frac{(n+1)}{2}}{\Gamma(\frac{n+1}{2})}[/imath] And for the volume [imath]V_n=\frac{A_{n-1}}{n}[/imath] Finally you get [imath]A_n(R)=A_nR^n[/imath] [imath]V_n(R)=V_nR^n[/imath] Where [imath]R[/imath] is the radius of the ball. Plugging in the dimensions it's easy to show that [imath]V_n[/imath] is max for [imath]n=5[/imath] and the surface for [imath]n=6[/imath] Is there an intuitive way of explaining this? | 15656 | Volumes of n-balls: what is so special about n=5?
The volume of an [imath]n[/imath]-dimensional ball of radius [imath]1[/imath] is given by the classical formula [imath]V_n=\frac{\pi^{n/2}}{\Gamma(n/2+1)}.[/imath] For small values of [imath]n[/imath], we have [imath]V_1=2\qquad[/imath] [imath]V_2\approx 3.14[/imath] [imath]V_3\approx 4.18[/imath] [imath]V_4\approx 4.93[/imath] [imath]V_5\approx 5.26[/imath] [imath]V_6\approx 5.16[/imath] [imath]V_7\approx 4.72[/imath] It is not difficult to prove that [imath]V_n[/imath] assumes its maximal value when [imath]n=5[/imath]. Question. Is there any non-analytic (i.e. geometric, probabilistic, combinatorial...) demonstration of this fact? What is so special about [imath]n=5[/imath]? I also have a similar question concerning the [imath]n[/imath]-dimensional volume [imath]S_n[/imath] ("surface area") of a unit [imath]n[/imath]-sphere. Why is the maximum of [imath]S_n[/imath] attained at [imath]n=7[/imath] from a geometric point of view? note: the question has also been asked on MathOverflow for those curious to other answers. |
2287579 | Finding number of positive integers using that [imath]p=2^{24036583} -1[/imath] is a prime number
We know [imath]p=2^{24036583} -1[/imath] is a prime number. Let [imath]s[/imath] be the number of positive integers [imath]c[/imath], such that the four quadratic equations [imath]x^2 ±px ±c =0[/imath] all have rational roots . Find [imath]s[/imath] modulo [imath]1000[/imath]. | 201056 | the number of the positive integer numbers [imath]k[/imath] that makes for the two quadratic equations [imath] \pm x^2 \pm px \pm k[/imath] rational roots.
If we assume that [imath]p=2^{24036583}-1[/imath] is the greatest prime number until now .How to find the number of the positive integer numbers [imath]k[/imath] that makes for the two quadratic equations [imath] \pm x^2 \pm px \pm k[/imath] rational roots. |
2287461 | [imath]\mathbb{Z}/n\mathbb{Z}[/imath] is a semisimple ring provided n is square-free
Show that [imath]\mathbb{Z}/n\mathbb{Z}[/imath] is a semisimple ring if n is square-free.. Some articles suggests that we can use Chinese remainder theorem. But I can't exactly figure out how to do that. Thanks in advance!!! | 1339842 | [imath]\mathbb Z_n[/imath] is semisimple iff [imath]n[/imath] is square free
[imath]\mathbb Z_n[/imath] is J-semisimple iff [imath]n[/imath] is square free. A ring [imath]R[/imath] is said to be [imath]J[/imath] semisimple if intersection of all maximal ideals of [imath]R[/imath] is [imath]\{0\}[/imath]. If [imath]n[/imath] is square free then [imath]n=p_1p_2...p_k[/imath] then [imath]\mathbb Z_n\cong \mathbb Z_{p_1}\times\mathbb Z_{p_2}\times...\mathbb Z_{p_k}[/imath] Since [imath]\mathbb Z_{n}[/imath] is a PID every prime ideal ideal is a maximal ideal and hence the maximal ideals will be [imath]\mathbb Z_{p_i}[/imath] for each [imath]i[/imath]; but intersection of all [imath]\mathbb Z_{p_i}[/imath]' s is [imath]\{0\}[/imath] since [imath]p_i[/imath] 's are distinct. How to do the converse? |
2287705 | The polynomial of a square matrix and its eigenvalues
For any polynomial [imath]p(x) = a_0 + a_1 x + \cdots + a_k x^k[/imath] and any square matrix [imath]A[/imath], polynomial [imath]p(A)[/imath] is defined as [imath]p(A) = a_0 I + a_1 A + \cdots + a_k A^k[/imath] Show that if [imath]v[/imath] is any eigenvector of [imath]A[/imath] and [imath]\chi_A(x)[/imath] is the characteristic polynomial of [imath]A[/imath], then [imath]\chi_A(A) v = 0[/imath]. Deduce that if [imath]A[/imath] is diagonalizable then [imath]\chi_A(A)[/imath] is the zero matrix. I don't get "if [imath]v[/imath] is any eigenvector of [imath]A[/imath] and [imath]\chi_A(x)[/imath] is the characteristic polynomial of [imath]A[/imath], then [imath]\chi_A(A)v = 0[/imath]". I have gotten [imath]p(\lambda)[/imath] is an eigenvalue of the matrix [imath]p(A)[/imath], but how do I continue? | 2284135 | Digonalizable Polynomial Matrix
For any polynomial [imath]p(x) = a_0 + a_1x + · · · + a^kx^k[/imath] and any square matrix A, [imath]p(A)[/imath] is defined as [imath]p(A) = a_0I + a_1A + · · · + a_kA^k[/imath]. Show that if v is any eigenvector of A and [imath]χ_A(x)[/imath] is the characteristic polynomial of A, then [imath]χ_A(A)v = 0[/imath], Deduce that if A is diagonalizable then [imath]χ_A(A)[/imath] is the zero matrix. I got that [imath]p(\lambda)[/imath] is the eigenvalue of [imath]p(A)[/imath] and if A is diagonalizable then [imath]χ_A(x)[/imath] is the zero matrix since p(A) is similar to its diagonal matrix p(D) [imath]p(D) = a_nD_n + · · · + a_1D + a_0I[/imath] = [imath]\begin{matrix} p(\lambda_1) &0&...&0\\0& p(\lambda_2)&...& 0 \\ ...&...&...&...\\0&0&... & p(\lambda_n) \end{matrix}[/imath] |
2285755 | Find a sufficient statistic for [imath]\theta[/imath] and show that a UMP test of [imath]H_0: \theta = 6[/imath] against [imath]H_1:\theta <6[/imath] is based on this statistic.
(20%) Let [imath]X_1, \dotsc, X_n[/imath] be an iid sample from a distribution with pdf [imath]f(x;\theta) = \theta x^{\theta-1}, 0< x< 1[/imath], zero elsewhere, where [imath]\theta >0[/imath]. Find a sufficient statistic for [imath]\theta[/imath] and show that a UMP test of [imath]H_0: \theta = 6[/imath] against [imath]H_1:\theta <6[/imath] is based on this statistic. I used the exponential family form to get that the summation of [imath]\ln(x)[/imath] is a sufficient statistic for [imath]\theta[/imath], but I do not know how to find a UMP Test based on this statistic. I believe it has something to do with likelihood ratios. Thanks in advance for your help. My question is different from that question because I do not already have a MP test, these are different numbers, the pdf is different, the significance level is not given, and the alternative hypothesis is an inequality. | 1328663 | (Uniformly) Most Powerful test
I'm having trouble to find a UMP test after finding a MP test. Consider one observation [imath]X[/imath] from CDF [imath]F_\theta(x) = x^\theta[/imath] where [imath]x \in [0, 1][/imath] and [imath]\theta > 0[/imath]. I found the MP test for testing [imath]H_0: \theta = 1[/imath] against [imath]H_1: \theta = 2[/imath] with significance level [imath]\alpha=0.05[/imath] using the Neyman Pearson lemma: [imath]\lambda_{\theta_0, \theta_1}(x) = \frac{f_{\theta_0}}{f_{\theta_1}}= \cdots = \frac{1}{2x}[/imath] Reject [imath]H_0[/imath] if [imath]\lambda_{\theta_0, \theta_1}(x)\leq\frac{1}{c}[/imath], hence if [imath]X\geq \tilde{c}[/imath], where [imath]\tilde{c} = 0.95[/imath]. Now I'm asked to find the UMP test for testing [imath]H_0: \theta \in [\frac{1}{2}, 1][/imath] against [imath]H_1: \theta=2[/imath] for significance level [imath]\alpha = 0.05[/imath]. How to proceed? |
2286149 | Prove that [imath]n^5+n^4+1[/imath] is composite for [imath]n>1.[/imath]
Prove that [imath]f(n)=n^5+n^4+1[/imath] is composite for [imath]n>1, n\in\mathbb{N}[/imath]. This problem appeared on a local mathematics competition, however it looks like there is no simple method to solve it. I tried multiplying it by [imath]n+1[/imath] or [imath]n-1[/imath] and then tried factorizing it, but it was too tough for me. Any help will be appreciated. | 2012344 | Prime factor of [imath]A=14^7+14^2+1[/imath]
Find a prime factor of [imath]A=14^7+14^2+1[/imath]. Obviously without just computing it. |
2287976 | The product of a sum of squares is a sum of squares
If [imath]x=a^2+b^2[/imath] and [imath]y=c^2+d^2[/imath] how can i prove that xy is also the sum of two rational squares? My teacher told me there are various methods to attack this problem but an easy way is to use the norms of guassian integers but I can't see how it would help. | 285563 | Product of sums of square is a sum of squares.
Given [imath]a,b,c,d \in \mathbb{Z}[/imath], there is [imath]x,y[/imath] such that [imath](a^2 + b^2)(c^2 + d^2) = x^2 + y^2[/imath] One can show this by considering the complex number [imath]a + bi[/imath] and [imath]c+ di[/imath], using complex properties to deduce that [imath]x = ac - bd, y = ad + bc[/imath] is a solution. However, given that either [imath]a^2, b^2[/imath] are distinct or [imath]c^2,d^2[/imath] are distinct, then how can one can find nonzero solutions [imath]x,y[/imath]? Sorry, I am assuming [imath]a,b,c,d[/imath] are nonzero. |
2288008 | Contour integration of [imath]\frac{\log( x)}{x^2+a^2}[/imath]
How do I intgrate [imath]\int_0^{\infty}\frac{\log( x)}{x^2+a^2} \,dx[/imath] using contour integration? Simple integration is easy but I can't understand contour integration. Please help! | 1826381 | Evaluate [imath]\int_0^{\infty} \frac{\log(x)dx}{x^2+a^2}[/imath] using contour integration
This question is Exercise 10 of Chapter 3 of Stein and Shakarchi's Complex Analysis. Show that if [imath]a>0[/imath], then [imath]\int_0^{\infty} \frac{\log(x)dx}{x^2+a^2}=\frac{\pi \log(a)}{2a}.[/imath] The hint is to integrate over the boundary of the domain [imath]\{z: \epsilon<|z|<R, \Im(z)>0\}.[/imath] I'm not sure what to do with the integral over the outer arc. |
2279971 | Identify the plane defined by [imath]|z-2i| = 2|z+3|[/imath]
I tried: [imath]|z-2i| = 2|z+3| \Leftrightarrow \\ |x+yi-2i|=2|x+yi+2|\Leftrightarrow \\ \sqrt{x^2+(y-2)^2}=\sqrt{4((x+2)^2+y^2)} \Leftrightarrow \\ \sqrt{x^2+y^2-4y+4} = \sqrt{4x^2+24x+36+4y^2} \Leftrightarrow \\ x^2+y^2-4y+4 = 4x^2+24x+36+4y^2 \Leftrightarrow \\ y^2-4y-4y^2=4x^2+24x+36+x^2 \Leftrightarrow \\ -3y^2-4y=5x^2+24x+26 \Leftrightarrow \\ ???[/imath] What do I do next? | 2289367 | Find the locus of [imath]|z-2i|=3|z+3|[/imath]
I got as far as [imath]|z-2i|=3|z+3| \Leftrightarrow \\ (\ldots) \Leftrightarrow \\ x^2-y^2+4y-4=9x^2+54x+81-9y^2 \Leftrightarrow \\ x^2-9x^2-y^2+9y^2+4y-4-54x-81=0\Leftrightarrow \\ -8x^2+8y^2+4y-85-54x=0 \Leftrightarrow \\ 8y^2+4y-8x^2-54x-85=0\Leftrightarrow \\ y^2+\frac{1}{2}y-x^2-\frac{27}{4}x=\frac{85}{8}[/imath] Then I tried to complete the square: [imath]y^2+\frac{1}{2}y = 0\Leftrightarrow (y+\frac{1}{4})^2-\frac{1}{16}[/imath] [imath]-x^2-\frac{27}{4}x=0 \Leftrightarrow -(x+\frac{27}{8})^2+\frac{729}{64} = 0 \Leftrightarrow (x+\frac{27}{8})^2-\frac{729}{64} = 0[/imath] And so the equation becomes: [imath](y+\frac{1}{4})^2-\frac{1}{16}+(x+\frac{27}{8})^2-\frac{729}{64}=\frac{85}{8} \Leftrightarrow \\ (y+\frac{1}{4})^2+(x+\frac{27}{8})^2=\frac{85}{8}+\frac{1}{16}+\frac{729}{64} [/imath] According to my book the center is [imath]-\frac{27}{8}-\frac{1}{4}i[/imath] which I got right but the radius is [imath]\frac{\sqrt{117}}{8}[/imath]. What went wrong? |
2288522 | How can I solve the infinite series 1+1/3+1/7+1/15+1/31+1/63+...
What is the summation of this infinite series [imath]1 + \frac{1}{3} + \frac{1}{7} + \frac{1}{15} + \frac{1}{31} + \frac{1}{63} + ... ?[/imath] | 266629 | Find the infinite sum [imath]\sum_{n=1}^{\infty}\frac{1}{2^n-1}[/imath]
How to evaluate this infinite sum? [imath]\sum_{n=1}^{\infty}\frac{1}{2^n-1}[/imath] |
2288757 | If [imath]\sum a_n[/imath] converges where [imath]a_n \ge 0 \; \forall \; n \in \Bbb N[/imath], then [imath]\{n a_n\}[/imath] converges or diverges?
I was asked this question in a PhD interview. They didn't say whether [imath]a_n[/imath] is non-increasing or strictly decreasing. I tried various examples such as [imath]a_n=\frac 1{n^p}[/imath] where [imath]p \gt 1[/imath], [imath]a_n=\frac 1{n^2 \ln n}[/imath]. Then [imath]n a_n \to 0[/imath]. I couldn't answer this question then. But when I saw Robert Israel's answer to this question, motivated from it, I defined a sequence [imath]a_n[/imath] as below : [imath] a_n = \begin{cases} \frac 1n, & \text{if [/imath]n[imath] is a power of 2} \\ 0, & \text{otherwise.} \end{cases}[/imath] Here, [imath]\sum a_n=\frac 12+ \frac 14+ \frac 18+ \cdots =\frac {\frac 12}{1 -\frac 12}=1[/imath] and [imath]\{n a_n\}[/imath] diverges, because there are infinitely many terms of [imath]0[/imath] and [imath]1[/imath] in it. i.e. [imath]n a_n = \begin{cases} 1, & \text{if [/imath]n[imath] is a power of 2} \\ 0, & \text{otherwise.} \end{cases}[/imath] So [imath]\{na_n\}[/imath] converges in some cases and diverges in some cases. Am I right? Note : I have gone through these posts, 1,2, 3 too but they have some extra conditions or specific details. Thanks. | 2084371 | Is it possible that [imath]a_n>0[/imath] and [imath]\sum a_n[/imath] converges then [imath]na_n \to 0[/imath]? (without assuming [imath]a_n[/imath] is decreasing)
Can it be proven that : If [imath]a_n>0[/imath] and [imath]\sum a_n[/imath] converges then [imath]na_n \to 0[/imath] (without assuming [imath]a_n[/imath] is decreasing) Please note This question does not require [imath]a_n[/imath] to be decreasing as condition, Is it possible to prove [imath]na_n \to 0[/imath] without requiring [imath]a_n[/imath] to be decreasing? I keep thinking that [imath]a_n>0[/imath] and [imath]\sum a_n[/imath] converging then that implies that [imath]a_n[/imath] must be decreasing anyway(not monotonically decreasing necessarily ) Is it possible that [imath]a_n>0[/imath] and [imath]\sum a_n[/imath] converges then [imath]na_n \not \to 0[/imath]? There are already answered questions that assume [imath]a_n[/imath] to be decreasing as a requirement: (this question does not assume [imath]a_n[/imath] is decreasing) If [imath](a_n)[/imath] is a decreasing sequence of strictly positive numbers and if [imath]\sum{a_n}[/imath] is convergent, show that [imath]\lim{na_n}=0[/imath] Prove that if [imath]\sum a_n[/imath] converges, then [imath]na_n \to 0[/imath]. This was the post that made me ask this question, a counter example of monotonoic decreasingness of [imath]a_n[/imath] was given in comments by the OP. |
2289008 | How to solve [imath]e^{x-2} = x[/imath]
A friend asked me what are the solutions to this equation. I know for sure it has 2 solutions (thanks to Desmos) - (0.159, 0.159) and (3.146, 3.146). I have no idea even where to start solving this problem. My knowledge in mathematics is sadly high school tier so please keep it as simple as possible. Thanks in advance! | 587070 | Solving [imath]x^2 - 1 = e^x[/imath]
Can someone help me solve the equation [imath]x^2 - 1 = e^x[/imath] ? I tried taking the natural logarithm of both sides but I don't know where to go from there.. I got: [imath]\ln(x^2 -1) = x[/imath] But I don't know how to solve it from here. Any help please? |
2288937 | Parametrized curve & differentiation and scalar product
Let [imath]\gamma\in C^2(I)[/imath] be a parametric continous curve with [imath]\parallel \ \dot\gamma(t)\parallel^2 \ = \ 1[/imath]. Show that [imath]\langle \ddot\gamma,\dot\gamma \rangle \equiv 0[/imath]. From that i know that we have to show that the vector of the velocity is perpendicular to the vector of acceleration. From [imath]\gamma \in C^2(I)[/imath] i know that [imath]\gamma[/imath] is exactly two times differentiable so [imath]\ddot\gamma[/imath] has to be constant. Then [imath]\dot\gamma[/imath] should have a constant slope which makes it a straight line. This straight line represents the velocity which should be 1. I think its not very far from here. Any tips? | 1725145 | Proving that second derivative is perpendicular to curve
How can I prove the following? [imath]\gamma (t)[/imath] is unit speed, [imath]\dot \gamma(t) \not= 0 \Rightarrow \ddot \gamma(t)[/imath] is perpendicular to [imath]\gamma(t)[/imath] I don't really see where a problem would arise when [imath]\dot \gamma(t)=0[/imath] would cause such a problem |
2288960 | Erroneous argument that every distribution function is left continuous.
I am trying to understand the distribution function, and read that it normally is only right continuous. But somehow I get left continuity as well, so could somebody please tell me what I am doing wrong: What is wrong with this 'proof'? Let X be a r.v. Take an arbitrary x and let [imath]x_n\to x[/imath] be monotonically increasing. Then [imath]\lim_{n\to\infty}F_X(x_n)=\lim_{n\to\infty}P^{X}((-\infty,x_n])=\lim_{n\to\infty}\int\mathbb{1}_{(-\infty,x_n]}dP^{X}[/imath] now monotone convergence gives [imath]=\int\mathbb{1}_{(-\infty, x]}dP^{X}=F_X(x)[/imath] hence [imath]F_X[/imath] is left continuous. | 522270 | Why left continuity does not hold in general for cumulative distribution functions?
Definition: The c.d.f. [imath]F[/imath] of a random variable [imath]X[/imath] is a function defined for each real number [imath]x[/imath] as follows:[imath]F(x)=\Pr(X\leq x) \text{ for } -\infty<x<\infty[/imath] Let [imath]F(x^-)=\lim_{y\rightarrow x,\,y<x}F(y)[/imath] and [imath]F(x^+)=\lim_{y\rightarrow x,\,y>x}F(y)[/imath] Property of cumulative distribution function: A c.d.f. is always continuous from the right; that is , [imath]F(x)=F(x^+)[/imath] at every point [imath]x[/imath]. Proof: Let [imath]y_1>y_2>\dots[/imath] be a sequence of numbers that are decreasing such that [imath]\lim_{n\rightarrow \infty}y_n=x.[/imath]Then the event [imath]\{X\leq x\}[/imath] is the intersection of all the events [imath]\{X\leq y_n\}[/imath] for [imath]n=1,2,\dots[/imath] .Hence, [imath]F(x)=\Pr(X\leq x)=\lim_{n\rightarrow \infty} \Pr(X\leq y_n)=F(x^+).[/imath] Now I think the left inequality can also be proved in the similar way as: Let [imath]y_1<y_2<\dots[/imath] be a sequence of numbers that are increasing such that [imath]\lim_{n\rightarrow \infty}y_n=x.[/imath]Then the event [imath]\{X\leq x\}[/imath] is the union of all the events [imath]\{X\leq y_n\}[/imath] for [imath]n=1,2,\dots[/imath] .Hence, [imath]F(x)=\Pr(X\leq x)=\lim_{n\rightarrow \infty}\Pr(X\leq y_n)=F(x^-).[/imath] Where am I wrong? |
2285143 | Shortest distance
Minimum distance between the curves [imath]y^2=x-1[/imath] and [imath]x^2=y-1[/imath] I just know that to go for common normal for such kind of problems but I am unable to go with it . It would be helpful if i can get exactly how to approach by this method. Any other methods are also welcomed as it can set an example for further persons who have same kind of difficulty. | 1209893 | Minimum distance between two parabolas
The shortest distance between the parabolas [imath]y^2=x-1[/imath] and [imath]x^2=y-1[/imath] is. Attempt: The shortest distance is along the common normal of the two curves. |
2290441 | Let [imath]x_1,x_2,x_3...[/imath] be all consecutive positive roots of the equation [imath]\tan x = x.[/imath] Find [imath]\lim_{n \rightarrow \infty}(x_n-x_{n-1}).[/imath]
Let [imath]x_1,x_2,x_3...[/imath] be all consecutive positive roots of the equation [imath]\tan x = x.[/imath] Find [imath]\lim_{n \rightarrow \infty}(x_n-x_{n-1}).[/imath] I feel like the answer is [imath]0[/imath], as the gap between [imath]x_n[/imath] and [imath]x_{n-1}[/imath] is getting smaller and smaller. My attempt: Note that [imath] n\pi\leq x_n \leq (n+1)\pi[/imath]. However, [imath]\pi \leq x_n - x_{n-1} \leq \pi[/imath] and I couldn't use Squeeze theorem to conclude. Any hint would be appreciated. | 364765 | [imath]x_n[/imath] is the [imath]n[/imath]'th positive solution to [imath]x=\tan(x)[/imath]. Find [imath]\lim_{n\to\infty}\left(x_n-x_{n-1}\right)[/imath]
[imath]x_n[/imath] is the [imath]n[/imath]'th positive solution to [imath]x=\tan(x)[/imath]. Find [imath]\lim_{n\to\infty}\left(x_n-x_{n-1}\right)[/imath]. |
2290250 | M contains all ordinals
Could you help me to prove the following: If [imath]M[/imath] is a transitive class such that [imath]Def(M)\subseteq M[/imath], then [imath]On^M =On[/imath] (only using this assumptions)? | 2287829 | [imath]M[/imath] contains all ordinals
I would like to prove the following claim: If M is a transitive class such that Def(M) [imath]\subseteq[/imath] M, then [imath]On^M[/imath]= [imath]On[/imath]. My own proof would start like this: Proof. Since "x is an ordinal " is absolute ([imath]\Delta_0[/imath]), we have [imath]On^M\subseteq On[/imath]. Suppose that [imath]On^M \neq On[/imath]. Let [imath]\alpha[/imath] be the least ordinal in [imath]On - On^L[/imath]. Then we will proof the following: Claim: [imath]\alpha = On^M[/imath] Could someone help me to finish the proof ? |
2289695 | 2D Integration With Quadratic Argument of Delta Function
The problem statement, all variables and given/known data. I have a 2D integral that contains a delta function: [imath]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp{-((x_2-x_1)^2+a x_2^2+b x_1^2-c x_2+d x_1+e)}\delta(p x_1^2-q x_2^2) dx_1 dx_2,[/imath] where [imath]x_1[/imath] and [imath]x_2[/imath] are variables, and [imath]a,b,c,d,e,p[/imath] and [imath]q[/imath] are some real constants. Relevant equations How can one integrate this without resulting in poles. Because integral over, for example, [imath]x_1[/imath] leads to a pole, given the property of delta function. The attempt at a solution If inside the delta function, the argument would be [imath]x_1^2+x_2^2[/imath], and not [imath]x_1^2-x_2^2[/imath], then one could go to polar coordinates, and get rid of the poles coming from the delta function integral. Sadly, this is not the case here. Further one can get rid of pole by going into hyperbolic coordinates [imath]x_1=RCosh(\theta)[/imath] and [imath]x_2=RSinh(\theta)[/imath]. The R integral can be done exactly because the poles cancel out due to Jacobian. However, the integral over ##\theta## then becomes unbounded. So this method also has a problem. Is there a smarter way to perform integral with quadratic argument of delta function without encountering poles? | 2287756 | 2D Integration With Quadratic Arg. of Delta Function
The problem statement, all variables and given/known data I have a 2D integral that contains a delta function: [imath]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp{-((x_2-x_1)^2)+(a x_2^2+b x_1^2-c x_2+d x_1+e))}\delta(p x_1^2-q x_2^2) dx_1 dx_2[/imath], where [imath]x_1[/imath] and [imath]x_2[/imath] are variables, and a,b,c,d,e,p and q are some real constants. Relevant equations How can one integrate this without resulting in poles. Because integral over, for example, [imath]x_1[/imath] leads to a pole, given the property of delta function. The attempt at a solution If inside the delta function, the argument would be [imath]x_1^2+x_2^2[/imath], and not [imath]x_1^2-x_2^2[/imath], then one could go to polar coordinates, and get rid of the poles coming from the delta function integral. Sadly, this is not the case here. Further one can get rid of pole by going into hyperbolic coordinates [imath]x_1=RCosh(\theta)[/imath] and [imath]x_2=RSinh(\theta)[/imath]. The R integral can be done exactly because the poles cancel out due to Jacobian. However, the integral over ##\theta## then becomes unbounded. So this method also has a problem. Is there a smarter way to perform integral with quadratic argument of delta function without encountering poles? |
2291319 | Find the value of [imath]\cos 1^\circ + \cos 2^\circ + \cos 3^\circ +.....+\cos 180^\circ [/imath]
Find the value of [imath]\cos 1^\circ + \cos 2^\circ + \cos 3^\circ +.....+\cos 180^\circ [/imath] My Attempt: [imath]=\cos 1^\circ + \cos 2^\circ + \cos 3^\circ +....+\cos 180^\circ [/imath] [imath]= \cos 1^\circ + \cos 2^\circ + \cos 3^\circ +....-1[/imath] How do I complete it? | 986835 | Finding the sum of [imath]\sin(0^\circ) + \sin(1^\circ) + \sin(2^\circ) + \cdots +\sin(180^\circ)[/imath]
I need help understanding the sum of [imath]\sin(0^\circ) + \sin(1^\circ) + \sin(2^\circ) + \cdots +\sin(180^\circ)[/imath] or [imath]\displaystyle \sum_{i=0}^{180} \sin(i)[/imath] This might be related to a formula to find the average voltage from a generator used to gauge waves: [imath]V_\text{avg} = 0.637 \times V_\text{peak}[/imath]. I am currently learning about AC circuits in the military. |
2291272 | Prove that the interval [imath](0,1)[/imath] and the Cartesian product [imath](0,1)\times(0,1)[/imath] have the same cardinality.
(Hint: Use the Schröder-Bernstein theorem.) I know I need to find two injective functions but I'm yet to find something that makes sense. | 2290136 | Prove that the interval [imath](0, 1)[/imath] and the Cartesian product [imath](0, 1) \times (0, 1)[/imath] have the same cardinality
Prove that the interval [imath](0, 1)[/imath] and the Cartesian product [imath](0, 1) \times (0, 1)[/imath] have the same cardinality using the SB theorem? Also how does one find a bijection on [imath]f:(0, \infty) \to (0,1)[/imath] such that they have the same cardiniality? |
2290763 | What is the integral of sin equal to?
I was wondering if the integral of [imath]\sin\left(\int(\sin(x)dx\right)[/imath] is equivalent to 0, if it had an answer, or if there was simply no answer? | 2048590 | How do we know what the integral of [imath]\sin (x)[/imath] is?
Since I started more or less formally learning the foundations of calculus, I naturally came across the Riemann definition of the integral. At first I considered this definition intuitive but not really useful since to obtain a closed form expression, one needed to add an infinite amount of values. Later on, an exercise prompted me to calculate a Riemann Integral, and from the definition and the expression for the sum of squares, I was able to calculate the limit with nothing more than I had learned at school. This was a revelation for me, since so far I had considered the definition a mere formalism. Now I knew how it gave results. The next integral I tried to calculate this way was, for obvious reasons [imath]\sqrt {1-x^2}[/imath]. Unfortunately, I found the sum intractable and gave up. I started to question the usefulness of the definition again. If it only works for simple functions like polynomials, how did we ever find out that the integral of [imath]\sin (x)[/imath] is [imath]-\cos (x)[/imath]? Did we use the Riemann definition or did we just say "the derivative of [imath]-\cos (x)[/imath] is [imath]\sin (x)[/imath] and therefore its integral must be [imath]-\cos (x)[/imath]? I would like to get some insight into the theory as well as the history that led to the tables of integrals we have today |
2291241 | Finding the generating pattern in an integer sequence
I got a problem to find the generating pattern of the following integer sequence: [imath]1,3,8,20,48,126,...[/imath] This page helped me in finding the pattern: https://oeis.org/search?q=1%2C3%2C8%2C20%2C48&language=english&go=Search What I am interested in is, how to find the generating pattern of an arbitrary infinite integer sequence on my own? The sequence can be monotonous or alternating. Is there a way instead of guessing it? By any means, correct me if I am using wrong terminology, English is not the language that my math professors teach on. | 1998900 | How to identify the pattern of a sequence
Are there some particular methods for identifying the following types of number series? [imath]6, 10, 19, 44, 93, \cdots[/imath] (Difference being prime no's square starting from 2) [imath]1, -2, 15, 52, -512, \cdots [/imath] ( [imath]^*2-4,\ ^*-6+3,\ ^*4-8,\ ^*-10+5[/imath], and so on) [imath]4, -2, -7, 25, 95,\cdots[/imath] ( [imath]^*-1+2,\ ^*2-3,\ ^*-3+4,\ ^*4-5[/imath], and so on) I mean they do not follow the arithmetic or geometric series nor do their common difference seem to follow any AM-GM pattern. So, is there any generalized mathematical theorems on these types of number series? Or we have to do it on a trial & error basis using intuition? |
2290734 | How to approach this problem of implicit differentiation?
I'm asked to find the derivative of [imath]\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}[/imath] using implicit differentiation. The question is why do I need to use implicit differentiation when I have a function defined only in terms of [imath]x[/imath]. | 2290590 | Derivation of inverse sine, what is wrong with this reasoning?
I'm trying to find the derivative of [imath]\sin^{-1}(x)[/imath]. I know the steps that lead to [imath]\frac{1}{\sqrt{1-x^2}}[/imath], however I don't understand why the following reasoning leads to a wrong answer. Because [imath]\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))} [/imath] If we plug in for [imath]f(x) = \sin(x)[/imath], and because [imath]\frac{d}{dx}\sin(x) = \cos(x)[/imath] we get [imath]\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\cos(\sin^{-1}(x))} [/imath] Since [imath]\sin(x) = \cos(x-\frac{\pi}{2})[/imath], we can state that [imath]\sin^{-1}(x) = \cos^{-1}(x-\frac{\pi}{2})[/imath]. (I suspect this is what is wrong) Thus, [imath]\frac{d}{dx}\sin^{-1}(x) = \frac{1}{\cos(\cos^{-1}(x-\frac{\pi}{2}))}[/imath] Then, by the definition of inverse function, we have [imath]\frac{d}{dx}\sin^{-1}(x) = \frac{1}{x - \frac{\pi}{2}}[/imath] |
2291776 | Evaluate [imath]\int_{0}^{\pi} x \ln(\sin x)dx.[/imath]
Evaluate [imath]\int_{0}^{\pi} x \ln(\sin x)dx.[/imath] My attempt, My first glance on this question I know that I'm going to use Integration by Parts. So, it equals to [imath]-\frac{1}{2}\int_{0}^{\pi} x^2 \cot x dx[/imath] I've tried every methods that I've learnt for integration for this and I ended up in nowhere. Hope someone could provide a solution and explain for it. Thanks in advance. Updated Obviously, my attempt is wrong as the integral does not converge. | 1562610 | Integral of [imath] x \ln( \sin (x))[/imath] from 0 to [imath] \pi [/imath]
[imath]\int_0^\pi x \ln(\sin (x))dx [/imath] I tried integrating this by parts but I end up getting integral that doesn't converge, which is this [imath] \int_0^\pi \dfrac{x^2\cos (x)}{\sin(x)} \ dx[/imath] So can anyone help me on this one? |
2291713 | Prove that [imath]\phi[/imath] is a partial relation for. Let [imath]\phi[/imath] is a relation on [imath]\mathbb{R}\times\mathbb{R}[/imath], [imath](a,b)\phi (x,y) \iff a\leq x[/imath] and [imath]b\leq y[/imath]
I'm stuck on how to prove this, most likely because I haven't really done an example like this before and I'm pretty new to partial order relations. It would be greatly appreciated if you could help me out, because I'm stressing over this, so any help would greatly be appreciated, thanks. Question Let [imath]\phi[/imath] be the relation on [imath]\mathbb{R}\times \mathbb{R}[/imath] in which [imath](a,b)\phi (x,y) \iff a\leq x[/imath] and [imath]b\leq y[/imath]. Prove that [imath]\phi[/imath] is a partial order relation. Additionally, is [imath]\phi[/imath] a total order relation? Explain why/why not. | 2291205 | Let [imath]σ[/imath] be the relation on [imath]R×R[/imath] in which [imath](a,b) σ (x,y)[/imath] if and only if [imath]a ≤ x[/imath] and [imath]b ≤ y[/imath]. Prove that [imath]σ[/imath] is a partial order relation.
I can prove it is reflexive and transitive but I cannot prove that it is antisymmetric. Help. |
2289019 | A linear operator [imath]T:V\rightarrow V[/imath] has a cyclic vector iff [imath]f_T=m_T[/imath] (minimal polynomial=characteristic polynomial)
I want to prove the following statement: Let linear operator [imath]T:V\rightarrow V[/imath] ([imath]V[/imath] is [imath]n[/imath]-dimentional). Then there exists a vector [imath]v[/imath] such that [imath]\left\{ v, Tv, ..., T^{n-1}v \right\}[/imath] form a basis of [imath]V[/imath], iff [imath]f_T=m_T[/imath]. (Actually, the [imath]\Rightarrow[/imath] direction is quite easy. I need to prove the other direction.) I've seen couple of answers to this problem in this site, all using the "Rational Canonial Form", which is something I wasn't tought in the course (but I am familiar with the Jordan Form). Moreover, the only proof I've found on the internet is this, which seemed to be promising, until I got to this sentence, which unfortunately seems to be a mistake: See that [imath]V_i[/imath] is a subspace of [imath]V[/imath], for all [imath]1 \leq i \leq m[/imath], and [imath]V=\bigcup_{i=1}^m V_i[/imath]. Therefore [imath]V=V_k[/imath], for some [imath]1 \leq k \leq m[/imath]. I don't know if that's true for infinite vector spaces, but for finite ones, such as [imath]\mathbb{F}_p^n[/imath] it's certainly isn't. Any step towards a proof would be appriciated (as well as an explanation for the suspicious claim stated above). Thank you! | 1035393 | Minimal polynomials and cyclic subspaces
I'm trying to make my way through two problems in Curtis's Linear Algebra, chapter 25. One of the two problems is this one, #5: Prove that [imath]V[/imath] is cyclic relative to a linear transformation [imath]T \in \mathcal{L}(V)[/imath] if and only if the minimal polynomial of [imath]T[/imath] is equal to the characteristic polynomial. Part of the issue is that I'm not really sure what it's asking. The statement that [imath]V[/imath] is cyclic seems to suggest that the entire vector space (say it has dimension [imath]n[/imath]) is generated by some vector [imath]v \in V[/imath], along with [imath]Tv, T^2v, \ldots T^{n-1}v[/imath]; Curtis's definition of a cyclic subspace (as opposed to a space, though I can't see why there should be a difference) seems to reinforce this: A [imath]T[/imath]-invariant subspace [imath]V_1[/imath] of [imath]V[/imath] is called cyclic relative to [imath]T[/imath] if [imath]V_1 \neq 0[/imath] and there exists a vector [imath]v_1 \in V_1[/imath] such that [imath]V_1[/imath] is generated by [imath]\{ v_1, Tv_1, T^2v_1, \ldots T^kv_1 \}[/imath] for some [imath]k[/imath]. However, I've seen several examples that seem to directly contradict this interpretation. For example, we recently had a take-home midterm that required we find elementary divisors, rational canonical form, etc., of a [imath]6\times 6[/imath] matrix; a classmate who received a perfect score on her test found that her matrix's characteristic polynomial was equal to its minimal polynomial — it was [imath](x+8)^2 (x+14)^2 (x^2+61x+80)[/imath] — and yet the vector space could be written as [imath]\left< v_6 \right> \oplus \left< v_4 \right> \oplus \left< -v_2+v_3+v_6 \right>[/imath] — i.e., as the direct sum of three cyclic subspaces, not as one cyclic subspace. So that suggests that "[imath]V[/imath] is cyclic" might mean the direct sum of cyclic subspaces. However, while working on the exercise that follows (#6), I came across this: [imath]S= \left(\begin{array}{ccc} -1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 2 \end{array}\right)[/imath] Here, the characteristic polynomial is not equal to the minimal polynomial, which is [imath]m(x) = (x-1)(x+1)(x+2)[/imath]. The eigenspace associated with [imath]\lambda = 1[/imath] has dimension [imath]2[/imath]. And so it seems as though, if [imath]V[/imath] has basis [imath]\{ v_1, v_2, v_3, v_4 \}[/imath], we can write [imath]V = \left< v_1 \right> \oplus \left< v_2 \right> \oplus \left< v_3 \right> \oplus \left< v_4 \right> [/imath]. This would seem to contradict my second interpretation of the original question. Obviously I'm missing something. Could someone fill me in on what it might be? (I've tried to be thorough, but I'm also trying not to make this ridiculously long; if I've left out necessary information, I'll be happy to append it.) |
2292104 | Test the convergence of [imath]\sum_{n=3}^{\infty}\frac{1}{(\log n)^{\log\log n}}[/imath]
I know that [imath]\sum_{n=3}^{\infty}\frac{1}{(\log n)^{\log n}}[/imath] converges and hence applying the Comparison test, I get the ratio [imath]\frac{a_n}{b_n}=\frac{(\log n)^{\log n}}{(\log n)^{\log\log n}}.[/imath] What I want to know is whether [imath]a_n/b_n[/imath] is bounded above or not. Or is there different approach to solving this problem? | 321738 | Does [imath]\sum_{n=3}^\infty \frac {1}{(\log n)^{\log(\log(n)}}[/imath] converge?
Does the following series converge: [imath]\sum_{n=3}^\infty \frac {1}{(\log n)^{\log(\log(n)}}[/imath] I've tried all test I know... Any ideas ? |
2283350 | Proving the bilinearity of a certain form
I have a question regarding a paper: let [imath]φ, ψ: ℝ^3 \to ℝ[/imath] two linear maps with [imath]φ(v)= 〈a,v[/imath]〉 and [imath]ψ(v)=〈b,v[/imath]〉with [imath]a,b ∈ ℝ^3[/imath]. We define the map with [imath]ℝ^3 \times ℝ^3 \to ℝ[/imath] by [imath]s(v,w):= φ(v) * ψ(w).[/imath] I need to show that s is a bilinear form and determine the transformation matrix of s regarding the canonical basis of [imath]ℝ^3[/imath]. I have no clue how to do this, so help would be much appreciated | 845932 | Simplifying the expression of a product of inner products
Let [imath]\mathbf{v}=(v_1,\cdots,v_n)^T, \mathbf{w}=(w_1,\cdots,w_n)^T, \mathbf{a}=(a_1,\cdots,a_n)^T \in\mathbb{R}^n[/imath], and let [imath] A = (\mathbf{w}\cdot\mathbf{a})(\mathbf{v}\cdot\mathbf{a})=\left(\sum_{i=1}^{n}w_ia_i\right)\left(\sum_{j=1}^{n}v_ja_j\right) [/imath] Is there any way of simplifying this quantity? Into one sum, maybe? |
2292342 | Two inequalities for holomorphic functions from [imath]\mathbb{D}[/imath] to [imath]\mathbb{D}[/imath]
Let [imath]f[/imath] be an holomorphic function from [imath]\mathbb{D}[/imath] to [imath]\mathbb{D}[/imath] and let [imath]z \in \mathbb{D}[/imath]. I have to prove that [imath] \frac{|f(0)| - |z|}{1 + |f(0)| |z|} \leq |f(z)| \leq \frac{|f(0)| + |z|}{1 - |f(0)| |z|} [/imath] and I have thought to use Schwarz-Pick theorem but I haven't obtain any result. Any help? Thank you very much. | 438875 | An inequality on holomorphic functions
Let [imath]D := \{z \in \mathbb{C}: |z| < 1\}[/imath] and [imath]f\colon D \rightarrow \mathbb{C}[/imath] be holomorphic. Suppose [imath]\lvert f(z)\rvert \leq 1[/imath] on [imath]D[/imath], show that [imath]\frac{|f(0)| - |z|}{1 + |f(0)||z|} \leq |f(z)| \leq \frac{|f(0)| + |z|}{1 - |f(0)||z|} \ \forall z \in D.[/imath] I have tried using Cauchy Integral Formula to [imath]f[/imath] and expanding [imath]f[/imath] at [imath]0[/imath], but I have no idea why [imath]f(0)[/imath] appears in the inequality in the former case. Both answers and hints are appreciated and thanks in advance! |
2292233 | Prove: [imath]\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq\sqrt{x}+\sqrt{y}[/imath]
[imath]\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq\sqrt{x}+\sqrt{y}[/imath] where [imath]x,y>0[/imath] How should I approach this? | 1851378 | Prove: [imath]\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}[/imath]
Prove: [imath]\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}\geq \sqrt{x}+\sqrt{y}[/imath] for all x, y positive [imath]\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\geq 0[/imath] [imath]\frac{x\sqrt{x}+y\sqrt{y}-x\sqrt{y}-y\sqrt{x}}{\sqrt{y}\sqrt{x}}\geq 0[/imath] [imath]\frac{x(\sqrt{x}-\sqrt{y})+y(\sqrt{y}-\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0[/imath] [imath]\frac{x(\sqrt{x}-\sqrt{y})-y(-\sqrt{y}+\sqrt{x})}{\sqrt{y}\sqrt{x}}\geq 0[/imath] [imath]\frac{(x-y)(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0[/imath] [imath]\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0[/imath] [imath]\frac{(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})}{\sqrt{y}\sqrt{x}}\geq 0[/imath] All the elements are positive and if [imath]\sqrt{x}=\sqrt{y}[/imath] we get [imath]0[/imath] Is the proof valid? |
2292930 | Prove the given set is closed
Let Y be an ordered set in the order topology. Let [imath]f,g: X \rightarrow Y[/imath] be two continuous function. Show that the set {[imath]x | f (x) \leq g (x) [/imath]} is closed in [imath]X[/imath]. Any help would be appriciated. Thanks. | 66335 | How to prove that this set is closed?
Suppose [imath]Y[/imath] be an ordered set in the order topology. Let [imath]f, g: X \to Y[/imath] be continuous. How to show that the set [imath]\{x| f(x) \leq g(x)\}[/imath] is closed? This is a excercise from munkres. Maybe trying to show that the set [imath]f(\{x| f(x) > g(x)\})[/imath] is open is suffice. But I could not figure out the connection between this set and the continuity of the two functions.Could you give a hint? Thanks. |
2292962 | Number of ways arranging 5 letter word from the word "Calculus"
How do I find out the amount of five letter word arrangements of the word "Calculus"? To clarify, my question is centered around the reason why that my methodology was incorrect. Clearly in this problem order does matter, And while it would be trivial to solve the problem there was only one letter that was repeated, the letter "u" "l" and "c" are being repeated My initial thoughts on the problem was to do the following: [imath]\frac{\binom{10}{5}}{3!\cdot 3!\cdot 2!}.[/imath] But upon delving into the problem further I discovered that it was wrong. I was wondering what would be the best method in solving this problem in which it has multiple letters repeating? | 2066257 | [imath]4[/imath] letter words taken from the letters of CONCENTRATIONS
How many words with or without meaning can be made from the letters of the word CONCENTRATIONS by taking [imath]4[/imath] letters at a time? There are [imath]14[/imath] letters total present in the word. [imath]4[/imath] letters can be picked in [imath]14\choose 4[/imath] ways. [imath]4[/imath] letters can be arranged in [imath]4![/imath] ways. So ideally answer has to be [imath]{14\choose 4} * 4![/imath]. But I know this answer is wrong because some letters are repeating. How do I solve this then? I appreciate any help. |
679842 | Find the last two digits of [imath]3^{45}[/imath]
I was wondering if there is a simpler way to find the last to digits of a power such as [imath]3^{45}[/imath]. I reduced it modulo 100 to get the answer, which is 43. But I was curious if there was a simpler, or more eloquent way to get this result. Here's my method: Using the fact that [imath]3^4=81\equiv -19 \pmod{100}[/imath] and [imath]19^5=2,476,099\equiv -1 \pmod{100}[/imath] [imath] 3^{45}\equiv 3(3^4)^{11}\equiv 3(-19)^{11}\equiv 3(-19)(19^5)^2\equiv 3(-19)(-1)^2 \equiv -57 \equiv 43 \pmod {100} [/imath] I just felt like this was sort of a messy way to find out that the last two digits are 43. | 2450158 | Find the last [imath]3[/imath] digits of [imath]3^{352}[/imath]?
Find the last [imath]3[/imath] digits of [imath]3^{352}[/imath] ? Apart from Carmichael Function, any other way of solving it ? |
2293419 | Positive integer solutions to [imath]\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=4[/imath]
I was wondering if anyone knew of any positive integer solutions to this Diophantine equation, or had a proof there are none. Integer solutions exist with negative values, such as (11,9,-5) and (4,11,-1), but checking positive integers up through 10,000 yielded nothing and I don't see a way to show there are none. | 2101118 | What are [imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] if [imath]\frac{x}{y + z} + \frac{y}{x + z} + \frac{z}{x + y} = 4[/imath] and [imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] are whole numbers?
What are [imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] if [imath]\dfrac{x}{y + z} + \dfrac{y}{x + z} + \dfrac{z}{x + y} = 4[/imath] and [imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] are whole numbers? MY ATTEMPT Let [imath]u = x + y + z[/imath]. Then the equation can be rewritten as [imath]\dfrac{x}{u - x} + \dfrac{y}{u - y} + \dfrac{z}{u - z} = 4[/imath] Suppose I set [imath]1 = \dfrac{x}{u - x} = \dfrac{y}{u - y}[/imath] and [imath]2 = \dfrac{z}{u - z}.[/imath] Then I get [imath]x = y + z[/imath] [imath]y = x + z[/imath] [imath]z = 2(x + y),[/imath] so that [imath]z = 0 = x + y,[/imath] which is impossible. Next, suppose I set [imath]\dfrac{4}{3} = \dfrac{x}{u - x} = \dfrac{y}{u - y} = \dfrac{z}{u - z}.[/imath] Then I get [imath]4(u - x) = 3x[/imath] [imath]4(u - y) = 3y[/imath] [imath]4(u - z) = 3z[/imath] so that [imath]12u - 4(x + y + z) = 3(x + y + z)[/imath] which implies that [imath]12u = 7(x + y + z) = 7u[/imath] from which it follows that [imath]u = 0.[/imath] This is, again, impossible. Alas, here is where I get stuck. Any hint(s) will be appreciated. |
2290471 | Surjection between free groups
Suppose we have two surjective homomorphisms [imath]f_1,f_2: F_{n} \rightarrow F_{m} [/imath] of free groups (so [imath]n\ge m[/imath]). One can always find homomorphism [imath]g:F_n \rightarrow F_n [/imath] giving a commuting triangle - when can it be enhanced to be an isomorphism? | 950835 | Epimorphisms from a free group onto a free group
Let [imath]f:F_n\to F_m[/imath] be an epimorphism ([imath]n\geq m[/imath]). Then it is true that there is a basis [imath]X=X_1\sqcup X_2[/imath] in [imath]F_n[/imath] such that [imath]f[/imath] maps [imath]\langle X_1\rangle[/imath] isomorphically onto [imath]F_m[/imath], and maps [imath]X_2[/imath] to identity. One can probably deduce this statement from the Grushko-Neumann theorem, but somehow I cannot get an elegant proof of it. I can see that by the Grushko-Neumann we can assume that [imath]m=1[/imath], and hence our epimorphism factors through the abelianization. And then an ugly linear algebra appears... Any suggestions how to make the proof slick? |
2293484 | Finite Definition of Span
Let S be a finite set of vectors [imath]S\subseteq V[/imath] with a vector space V over a field F. I have two alternate definitions of span(S): [imath]span(S)=\bigcap_{S\subseteq M\subseteq V} M [/imath] and [imath]span(S) = \left \{ \sum_{i=1}^{k}\lambda _{i}v_{i}: k\epsilon \mathbb{N}, v_{i}\epsilon S, \lambda _{i}\epsilon F \right \}[/imath]. Why are these two definitions equivalent? | 1264018 | The definition of span
In Linear Algebra by Friedberg, Insel and Spence, the definition of span (pg-[imath]30[/imath]) is given as: Let [imath]S[/imath] be a nonempty subset of a vector space [imath]V[/imath]. The span of [imath]S[/imath], denoted by span[imath](S)[/imath], is the set containing of all linear combinations of vectors in [imath]S[/imath]. For convenience, we define span[imath](\emptyset)=\{0\}[/imath]. In Linear Algebra by Hoffman and Kunze, the definition of span (pg-[imath]36[/imath]) is given as: Let [imath]S[/imath] be a set of vectors in a vector space [imath]V[/imath]. The subspace spanned by [imath]S[/imath] is defined to be intersection [imath]W[/imath] of all subspaces of [imath]V[/imath] which contain [imath]S[/imath]. When [imath]S[/imath] is finite set of vectors, [imath]S = \{\alpha_1, \alpha_2, ..., \alpha_n \}[/imath], we shall simply call [imath]W[/imath] the subspace spanned by the vectors [imath]\alpha_1, \alpha_2, ..., \alpha_n[/imath]. I am not able to understand the second definition completely. How do I relate "set of all linear combinations" and "intersection [imath]W[/imath] of all subspaces"? Please help. Thanks. |
2293456 | Let [imath]f:[0,\infty)\to \mathbb R[/imath] be Continuous and strictly decreasing function such that : [imath]\lim_{x\to \infty}f(x)=0[/imath]
Let [imath]f:[0,\infty)\to \mathbb R[/imath] be Continuous and strictly decreasing function such that : [imath]\lim_{x\to \infty}f(x)=0[/imath] prove [imath]\int_0^\infty \frac{f(x)-f(x+1)}{f(x)}[/imath] Diverges. I can not prove. please help . | 1141991 | Divergence of an improper integral
Let [imath]f:[0,\infty)\rightarrow \mathbb{R}[/imath] be a continuous and strictly decreasing function. If [imath]\displaystyle\lim_{x\rightarrow\infty}f(x)=0[/imath], prove that the following integral is divergent : [imath]\int_0^\infty \frac{f(x)-f(x+1)}{f(x)}dx[/imath] First, it's not hard to see that [imath]f(0)>0[/imath] and also [imath]f(x)>0[/imath] for each [imath]x[/imath]. I think we have to show that behavior of [imath]\frac{f(x)-f(x+1)}{f(x)}[/imath] is "similar" to [imath]\frac1{x+1}[/imath] or [imath]\frac{f(x+1)}{f(x)}[/imath] is "similar" to [imath]\frac{x}{x+1}[/imath] for big values of [imath]x[/imath]. But meaning of "similar" is not clear yet ! The most naive and rough approach is to show [imath]\frac{f(x+1)}{f(x)}<1-k[/imath] for some [imath]0<k<1[/imath] and therefore [imath]f(x)-f(x+1)>kf(x)[/imath]. One step forward, it's good to compare [imath]\frac{f(x+1)}{f(x)}[/imath] and a multiple of [imath]\frac{x}{x+1}[/imath] or equivalently to compare [imath]C (x+1)f(x+1)[/imath] and [imath]xf(x)[/imath]. But still I can't extend this idea . I tried more ways but I couldn't find a proper way to solve the problem. Any hint is appreciated ! |
2293814 | How can I find ALL complex roots of [imath]z^4 + 1 = 0[/imath]?
For [imath]z \in \mathbb{C}[/imath], how can I find all roots of the equation [imath] z^4 + 1 = 0 [/imath] Obviously, this equation implies [imath] z^4 = -1 = e^{i \pi} [/imath] and thus, one of our roots must be [imath] z_1 = e^{\frac{i \pi}{4}} [/imath] However, I am aware that this is not the only root. How can I find all other roots of this equation? | 497324 | Solve [imath]z^4+1=0[/imath] algebraically
I know the result and how to solve it using trigonometry and De Moivre. However, given that the complex number [imath]z[/imath] can be rewritten as [imath]a+bi[/imath], how can I solve it algebraically? |
2293807 | Every polynomial that takes square values is another polynomial squared
Let [imath]P(x)[/imath] be a polynomial with integer coefficients such that [imath]P(n)[/imath] is a perfect square for all integers [imath]n[/imath]. Prove or disprove: There exists a polynomial [imath]Q(x)[/imath] with integer coefficients such that [imath]P(x)=Q(x)^2[/imath]. | 428132 | Polynomial whose only values are squares
Given a polynomial [imath] P \in \Bbb Z [X] [/imath] such that, [imath] P (x)[/imath] is the square of an integer for all integers x, is [imath] P [/imath] necessarily of the form [imath] P (x)= Q (x)^2[/imath] with [imath] Q \in \Bbb Z [X][/imath]? |
2293345 | Analytic number theory question.
Let [imath]n[/imath] be a positive integer, and define [imath]f(n)[/imath] as [imath]n +\lfloor\sqrt{n}\rfloor[/imath], where [imath]\lfloor x\rfloor[/imath] is the greatest positive integer less than or equal to [imath]x[/imath]. Prove that the sequence [imath]n, f(n), f(f(n)), f(f(f(n))), \ldots[/imath] contains a perfect square. | 2288249 | Prove that the expression is a perfect square
Let [imath]m[/imath] be a natural number. Define [imath]f(m) = m + \lfloor\sqrt{m}\rfloor[/imath]. Prove that at least one of the number among [imath]m, f(m), f^2(m), \ldots[/imath] is a perfect square. Here [imath]f^k(m)[/imath] denotes the composition of [imath]f[/imath] over itself [imath]k[/imath] times. I tried the question, but the greatest integer along with square root is creating trouble. |
2294311 | Solving homogenous recurrence relation
Solve the recurrence relation [imath]f(n) = 2f(n - 1) + f(n - 2)[/imath] with initial conditions [imath]f(0) = a, f(1) = b[/imath]. (here a and b are fixed, arbitrary integers). I'm confused on how to approach this, a and b are fixed, arbitrary integers. | 2291375 | Solving recurrence relation. Recurrence
Solve the recurrence relation [imath]f(n) = 2f(n - 1) + f(n - 2)[/imath] with initial conditions [imath]f(0) = a, f(1) = b[/imath]. (here a and b are fixed, arbitrary integers). Can anyone show me how to solve this? a and b are fixed, arbitrary integers. Does it mean I can put any value for a and b? |
2293052 | Rational points on circles centered at the origin
How can I determine if [imath]x^2+y^2 = z[/imath] has infinitely many rational solutions [imath](x,y)[/imath] based on the value of [imath]z\in \Bbb{Q}[/imath]. I understand that [imath]a^2+b^2=1[/imath] has infinitely many rational solutions. As does [imath]a^2+b^2=2[/imath]. However, [imath]a^2+b^2=3[/imath] has no rational solutions. When [imath]z[/imath] is an integer I know how to determine if it has rational solutions. However when [imath]z[/imath] is not an integer it is not clear when it has rational solutions. For example, it is not immediately clear to me if the equation [imath]x^2+y^2 = \frac 43[/imath] has infinitely many rational solutions. | 1445058 | Which rationals can be written as the sum of two rational squares?
Which rational numbers can be written as the sum of two rational squares? That is, for which rational numbers [imath]a[/imath], are there rational numbers [imath]x[/imath] and [imath]y[/imath] such that [imath]a = x^2 + y^2[/imath]. It is a famous theorem that if an integer can be written as the sum of two rational squares then it can be written as the sum of two integral squares, and then the solution is the famous one by Fermat, but I didn't find anything about the general case. |
2294395 | Rubik's Cube Permutations
I know the formula for finding the number of permutations of the [imath]3[/imath]x[imath]3[/imath] cube, which is [imath]\frac{8! \times 3^8 \times 12! \times 2^{12}}{12}[/imath] But why must the division by [imath]12[/imath] be a part of the formula? From looking it up I know it has to do with some of the positions being impossible without disassembling the cube, but why specifically division by [imath]12[/imath] and not another number? | 256343 | Rubik's Cube Combination
Could anyone explain why the number of legal or reachable combinations of a [imath]3\times 3\times 3[/imath] Rubik's Cube is [imath]1/12\mbox{th}[/imath] of the total. I understood the logic behind the total number of combinations: [imath]8! \cdot 2^{12} + 3^{12} \cdot 12![/imath]. What I am not able to understand is why do we divide this quantity by [imath]2 \cdot 3\cdot 2[/imath]. I don't have any idea about parity or group/set theory. Is there another possible explanation for it? |
2290903 | Is the field [imath]\mathbb{Q}(3^{1/3},3^{1/4})[/imath] a Galois extension of [imath]\mathbb{Q}[/imath]?
My understanding is that for [imath]\mathbb{Q}(3^{1/3},3^{1/4})[/imath] to be a Galois extension of [imath]\mathbb{Q}[/imath] there must exist a polynomial over [imath]\mathbb{Q}[/imath] with roots [imath]3^{1/3},3^{1/4}[/imath] that splits in [imath]\mathbb{Q}(3^{1/3},3^{1/4})[/imath]. Is it true then that this field is not a Galois extension in this case as the field does not contain [imath]i[/imath] (since some roots of [imath]x^3-3[/imath] and [imath]x^4-3[/imath] are complex) or am I missing the point? Thanks. | 2287643 | Is [imath]\mathbb{Q}(\sqrt[3]{3}, \sqrt[4]{3})[/imath] a Galois extension of [imath]\mathbb{Q}[/imath]
From a previous question, we have that [imath]\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[4]{3})[/imath], and I am assuming that we would need to use this to justify the answer to the question. Would it be right to use some idea of a splitting field here? Or is it something to do with [imath]\mathbb{Q}(\sqrt[3]{3}, \sqrt[4]{3})[/imath] not containing a primitive 3rd root of unity? This is a past exam paper question worth 8 marks, so it doesn't seem like it's a few lined answer, but I am a little stuck on where to go. |
2294392 | A question about a bounded linear surjective operator
This is a question from Rudin's Functional Analysis. Let [imath]T\in \mathcal B(X,Y)[/imath] and [imath]T(X)=Y[/imath]. I have to prove that there exists [imath]\delta>0[/imath] such that [imath]\{S\in \mathcal B(X,Y):S(X)=Y\}\supset B_{\delta}(T)[/imath], where the ball is with respect to the metric generated by the operator norm. I could not proceed. Any help is appreciated. | 17087 | Why is the space of surjective operators open?
Suppose [imath]E[/imath] and [imath]F[/imath] are given Banach spaces. Let [imath]A[/imath] be a continuous surjective map. Why is there a small ball around [imath]A[/imath] in the operator topology, such that all elements in this ball are surjective? |
2294479 | How to solve the following recurrence relation question?
Solve the recurrence relation [imath]f(n) = f(n - 1) + f(n - 2)[/imath] with initial conditions [imath]f(0) = 2, f(1) = 1[/imath]. Give full details. The following is what I have done: - try [imath]f(n) = r^n[/imath], for some fixed [imath]r[/imath]. - [imath]r^n = r^{n - 1} + r^{n - 2}[/imath] - equation = [imath]r^2 - r - 1 = 0[/imath], if [imath]n = 2[/imath] | 2292707 | Fibonacci Recurrence Relations
Solve the recurrence relation [imath]f(n) = f(n − 1) + f(n − 2)[/imath] with initial conditions [imath]f(0)=1[/imath], [imath]f(1)=2.[/imath] So I understand that it grows exponentially so [imath]f(n) = r^n[/imath] for some fixed [imath]r[/imath]. This means substituting this [imath]r^n = r^n-1 + r^n-2[/imath] which gives the characteristic equation of [imath]r ^ 2 - r - 1 = 0[/imath]. I'm not 100% sure where to move on from here. |
2295060 | Equivalence relation for a group.
let [imath]H,K \subseteq G[/imath] all be groups. A relation ~ on [imath]G[/imath] is defined by x ~ y [imath]\iff x = hyk[/imath] for some [imath]h\in H[/imath] and [imath]k\in K[/imath]. I'm to prove that ~ is an equivalence relation on [imath]G[/imath]. I was just wondering if when I'm to prove reflexivity if this holds: x~x: [imath]x = hxk[/imath] [imath]h^{-1}xk^{-1} = h^{-1}hxkk^{-1} = x [/imath] , since [imath]h^{-1} \in H[/imath] and [imath]k^{-1} \in K[/imath] and so it's proven? Ie it doesn't have to be the same element of [imath]H[/imath] and the same element of [imath]K[/imath] right? | 930939 | Prove Equivalence Relation in G
Hei, guys! I'm having some trouble with the next problem: Let [imath]A[/imath] and [imath]B[/imath] be subgroups of [imath]G[/imath]. Show that [imath]\sim[/imath] is an equivalence relation when it is defined as follows: [imath]g\sim g'\Leftrightarrow g' = agb[/imath] for some [imath]a[/imath] from [imath]A[/imath], [imath]b[/imath] from [imath]B[/imath]. Should I prove this is valid for any a and b? It says for some elements a and b. But I might as well take the unit element. Could you give me a clue on how to solve it? Cheers! |
2293541 | Analytic function on unit disk, Schwartz lemma type inequality
Let [imath]f:\Bbb{D}\to\Bbb{C}[/imath] be an analytic function that satisfies the inequality [imath]|f(z)|\le\dfrac{1}{1-|z|}.[/imath] How can I show that [imath]|f'(0)|\le 4.[/imath] Also, I am looking for the equality case. Is this a sharp upper bound? | 896830 | Bound of a complex-valued function
This is a question from an old prelim in complex analysis. Show that if [imath]f[/imath] is an analytic function such that [imath]|f(z)| \leq \frac{1}{1-|z|}[/imath] for [imath]|z|<1[/imath], then [imath]|f'(0)| \leq 4[/imath]. I tried using both Cauchy's [imath]ML[/imath]-Estimates along with the Cauchy Integral Formula and even the Schwarz-Pick Lemma, but I still do not get the desired result. Any suggestions as to how best to proceed? |
2295359 | Understanding [imath]V=K\otimes_{\mathbb{F}_p}\mathbb{F}_{p^n}[/imath] where [imath]K[/imath] is the algebraic closure of [imath]\mathbb{F}_p[/imath]
So, let [imath]K[/imath] be an algebraic closure of [imath]\mathbb{F}_{p}[/imath] and consider the space [imath]V=K\otimes_{\mathbb{F}_p}\mathbb{F}_{p^n}[/imath] as a [imath]K[/imath] vector space. It seems pretty straight forward to show that this vector space is [imath]n[/imath] dimensional since for [imath]\{\alpha_1,\ldots\alpha_n\}[/imath] linearly independent elements of [imath]\mathbb{F}_{p^n}[/imath] (considered as itself a vector space over [imath]\mathbb{F}_p[/imath]) we have the linearly independant tensors [imath]1\otimes\alpha_i[/imath]. Then I think it is clear that any elementary tensor is a [imath]K[/imath] linear sum of these and so these must span [imath]V[/imath]. Anyway, what I don't understand is in what way this vector space is practically different from simply considering [imath]\mathbb{F}_{p^n}[/imath] as a [imath]\mathbb{F}_p[/imath] vector space. Studying for my algebra qualifying exam, I worked on a problem that asked to calculate the basis for [imath]V[/imath] for which the Froebenius automorphism is in Jordan canonical form. I don't see how this is different from doing it for [imath]\mathbb{F}_{p^n}[/imath]. Although, we know a priori that the form exists since the underlying field is complete algebraically closed, but isn't that true for [imath]\mathbb{F}_{p^n}[/imath]? Thanks. | 95887 | Frobenius Automorphism as a linear map
Let [imath]\phi(x) = x^p[/imath] be the Frobenius automorphism on [imath]\mathbb F_{p^n}[/imath]. We can view [imath]\mathbb F_{p^n}[/imath] as an [imath]n[/imath]-dimensional vector space over [imath]\mathbb F_p[/imath]. In this case, [imath]\phi[/imath] is a linear transformation. What are its characteristic and minimal polynomials? What are its Jordan and rational canonical forms? I know that [imath]\phi[/imath] satisfies [imath]x^n - 1[/imath], so its minimal and characteristic polynomials must divide this one. But, I can't seem to make much progress beyond this. Any suggestions? |
2295024 | Example of uniformly convergent sequence [imath](f_{n})[/imath] such that the sequence of cubes [imath](f^3_{n})[/imath] does not converge uniformly on the same set
So I know if the sequence [imath]f_{n}[/imath] has a pointwise limit [imath]f(z)=z[/imath] then the convergence is uniform on [imath]C[/imath] iff [imath]\sup\|(f_{n}(z)-f(z))|[/imath][imath]\rightarrow[/imath] 0 as n tends to infinity My trouble is in finding an example of a sequence is that is uniformly convergent and so is its sequence of cubes. Any help? Note: Sorry about my English! | 2285562 | Function Sequence Uniform Convergence
Just wondering if anyone has an example of a (complex) function sequence [imath](f_n)[/imath] which is uniformly convergent, but the sequence of cubes [imath](f_n^3)[/imath] is not uniformly convergent? |
1953218 | Is "The empty set is a subset of any set" a convention?
Recently I learned that for any set A, we have [imath]\varnothing\subset A[/imath]. I found some explanation of why it holds. [imath]\varnothing\subset A[/imath] means "for every object [imath]x[/imath], if [imath]x[/imath] belongs to the empty set, then [imath]x[/imath] also belongs to the set A". This is a vacuous truth, because the antecedent ([imath]x[/imath] belongs to the empty set) could never be true, so the conclusion always holds ([imath]x[/imath] also belongs to the set A). So [imath]\varnothing\subset A[/imath] holds. What confused me was that, the following expression was also a vacuous truth. For every object [imath]x[/imath], if [imath]x[/imath] belongs to the empty set, then [imath]x[/imath] doesn't belong to the set A. According to the definition of the vacuous truth, the conclusion ([imath]x[/imath] doesn't belong to the set A) holds, so [imath]\varnothing\not\subset A[/imath] would be true, too. Which one is correct? Or is it just a convention to let [imath]\varnothing\subset A[/imath]? | 2373457 | Confusion in understanding why empty set is a subset of every set
statement 1 : If [imath]x \in \emptyset[/imath] then [imath]x \in A[/imath]. statement 2 : If [imath]x \in \emptyset[/imath] then [imath]x \notin A[/imath]. I know that both statements are true since the hypothesis is false. But first statement says that [imath]\emptyset \subset A[/imath] while second statement says that [imath]\emptyset[/imath] is not a subset of [imath]A[/imath]. My question is why we prefer [imath]\emptyset \subset A[/imath] over the other implication that [imath]\emptyset[/imath] is not a subset of [imath]A[/imath]? Thanks. |
1656497 | Closed form of [imath]\sum_{k=1}^{2015}(-1)^{k(k+1)/2}k[/imath]
[imath]\sum_{k=1}^{2015}(-1)^\frac{k(k+1)}{2}\times k[/imath] How to solve this. Answer provided is [imath]0[/imath] | 2295814 | Find the sum of finite series [imath]S=\sum_{k=1}^{2015}{(-1)^{\frac{k(k+1)}{2}}}k[/imath]
Find the sum [imath]S=\sum_{k=1}^{2015}{(-1)^{\frac{k(k+1)}{2}}}k.[/imath] I partitioned k in two categories Either k is congruent to 0 , 3 mod(4) or congruent to 1,2 mod(4). But still I didn't get answer |
2296874 | Calculate [imath]\int_0^\pi{x\sin{x} \over 1+\cos^2x}dx[/imath]
I am trying to get started on [imath]\int_0^\pi{x\sin{x} \over 1+\cos^2x}dx[/imath] The usual trick I am familiar with would be to substitute [imath]y=\tan{x \over 2}[/imath]. This doesn't seem to work in this case. | 368064 | Integrate [imath]\int_0^\pi{{x\sin x}\over{1+\cos^2x}}dx[/imath].
Integrate [imath]\displaystyle \int \limits_0^\pi{{x\sin x}\over{1+\cos^2x}}dx[/imath]. I tried substituting [imath]t=\cos x[/imath], and then integrate with integration by parts. It got all messy... Thanks in advance for any help! |
2296988 | Bound on Commutator of Unitary Matrices
This is an exercise from Terry Tao's blog. If [imath]U[/imath] and [imath]V[/imath] are unitary [imath]n \times n[/imath] matrices, show that the commutator [imath][U,V] := UVU^{-1} V^{-1}[/imath] obeys the inequality [imath] \|[U,V]-I\|_{op} \leq 2 \|U-I\|_{op} \|V - I \|_{op} [/imath] We are provided the hint: (first control [imath]\|UV - VU\|_{op}[/imath].) | 2293450 | Prove that [imath]\|UVU^{-1}V^{-1}-I\|\leq 2\|U-I\|\|V-I\|[/imath]
[imath]U,V[/imath] are unitary [imath]n\times n[/imath] matrices, and the norm is the operator norm (so we can use [imath]\|UV\|\leq\|U\|\|V\|[/imath]). I've noticed that \begin{align} \|UVU^{-1}V^{-1}-I\|&= \|(UV-VU)U^{-1}V^{-1}\|\\ &\leq \|UV-VU\|\|U^{-1}V^{-1}\| \end{align} I can bound the first term by [imath]\|UV\|+\|VU\|[/imath], but I don't think this is useful. Hints (rather than complete answers) would be appreciated. The question comes from here (exercise 1) |
2297451 | Prove that [imath](a+b)^p \equiv a^p+b^p \mod p[/imath]
Prove that [imath](a+b)^p \equiv a^p+b^p \mod p[/imath] My try: I tried expanding this and then applying modulo to this expression but it did not work. | 1409017 | When is [imath](a+b)^n \equiv a^n+b^n[/imath]?
I remember a relation like [imath](a+b)^n \equiv a^n+b^n[/imath], but I don't remember mod which numbers this is true. Where can I learn more about this? |
2292004 | Help with System of nonlinear equations!
I have post once here about this question but i had some problems and i stopped with solving it. Now i started again and i think i`ve made it better. Here is where i need help. This is the system of nonlinear equations. f(x) = 0 where: [imath] \left\{ \begin{array}{c} x_1^5+x_2^3+x_3^4+1 \\ x_1^2*x_2* x_3 \\ x_3^4-1 \end{array} \right. [/imath] The right side of the equation is 0. A) Find manually all the zeroes of the system. B) Calculate the Jacobian J(X). (Notice that J(x) is singular for x3 = 0) C) Check this two starting solutions: 1) X0 = {-0.01, -0.01, -0.01} 2) X0 = {-0.1, -0.1, -0.1} D) Calculate the determinants |J(X0)| and |J^-1(X0)| for the two starting solutions. Notice that Jacobians are almost singular, altough the starting solutions are not so far from the real solutions. ------------------------------------------------------------------------------- This is the task that i need to solve. I started from here. From this system we can see that [imath] X_3 = \pm 1 [/imath] So going from this we can make 4 solutions. [imath]I:x_3=1, x_2=0, x_1=-1[/imath] [imath]II:x_3=1, x_2=-1, x_1=0[/imath] [imath]III:x_3=-1, x_2=0, x_1=-1[/imath] [imath]IV:x_3=-1, x_2=-1, x_1=0[/imath] After this i started with the Jacobian. I have made this for the Jacobian matrix. [imath] \left[ \begin{array}{ccc} 5x_1^4 & 3x_2^2x_3^4 &4x_3^3x_2^3\\ 2x_1 x_2 x_3 & 1*x_1^2 x_3 & 1*x_1^2 x_2\\ 0 & 0 & 4x_3^3 \end{array} \right] [/imath] After the jacobian i calculate the determinants for all 4 solutions and I have: I: the determinant is: 20. II: the determinant is: 0. III: the determinant is: 20. IV: the determinant is: 0. I stucked on this: When i need to see the two starting solutions C). And when i need to calculate their determinant. The first number that i calculate is 0.00000005 and i said okay its enough i am something wrong. | 2216245 | Problem with system of nonlinear equations
I need help for one math project. It's about a system of non-inear equations. The system go like this: [imath] f(x)=\left\{ \begin{array}{ll} x^5+y^3*z^4+1 \\ x^2*y*z \\ z^4-1 \end{array} \right. [/imath] The right side of the equation is 0. a) Calculate all the zeros in the system (manually) b) Find the Jacobian matrix, J(x). (Notice that J(x) is singular for z = 0) c) Consider independent two starting solutions: 1) x(0) = {-0.01, -0.01, -0.01}T 2) x(0) = {-0.1, -0.1, -0.1}T I have tried something but I don't know whether I am right... [imath] f(x)=\left\{ \begin{array}{ll} x^5+y^3*\pm1=-1 \\ x^2*y*\pm1=0 \\ z=\pm1 \end{array} \right. [/imath] So because in the middle line there is 0 in the right side some of X and Y must be 0. When i get for Y=0 the result is this: [imath]x= -1 , y= 0, z= \pm1[/imath] And for X=0 the result is this: [imath]x= 0 , y= -1, z= \pm1[/imath] I dont know where to go after this... Please help me. Thank you! |
2297615 | Find the eigenvectors and eigenvalues for this linear operator [imath]A:M_{n\times n }\mapsto M_{n\times n}[/imath] defined with [imath]A(X)=X^{T}[/imath]
Find the eigenvectors and eigenvalues for this linear operator [imath]A:M_{n\times n }\mapsto M_{n\times n}[/imath] defined with [imath]A(X)=X^{T}[/imath] How does the matrix of this linear operator look like? | 2117542 | The eigenvectors of the transpose operator
Define [imath]T : M_{n×n}(\mathbb{R}) → M_{n×n}(\mathbb{R})[/imath] by [imath]T(A) := A^t[/imath]. I know that the corresponding eigenvalues are [imath]+1[/imath] and [imath]-1[/imath], but I'm not sure how to find the eigenvectors of this transformation, in the case of a [imath]2\times 2[/imath] matrix it's simple, but not in the general case. |
2297504 | Find [imath]\lim_{x\to 1}\frac{p}{1-x^p}-\frac{q}{1-x^q}[/imath]
Find [imath]\lim_{x\to 1}(\frac{p}{1-x^p}-\frac{q}{1-x^q})[/imath] My attempt: I took LCM and applied lhospital but not getting the answer.Please help | 2349277 | Find [imath]\lim\limits_{x\to 1}\frac9{1-x^9}-\frac7{1-x^7}[/imath] without using L'Hopital
[imath]\lim_{x\to 1}\frac9{1-x^9}-\frac7{1-x^7}[/imath] I did this with l'Hôpital's rule, but how can we do this problem other than that? Any hints will be good. |
2297996 | Does the series [imath]\sum \sin{(\sin{(...\sin{(1)}...)})}[/imath] converge?
Let [imath]a_{n+1} = \sin{a_n}[/imath] and [imath]a_0 = 1[/imath]. Does the series [imath]\displaystyle \sum_{i=0}^\infty a_i[/imath] converge? Here is my solution. Let's prove by induction, that [imath]a_n > \frac{1}{n}[/imath]. We see that [imath]\sin{(1)} > \frac{1}{2}[/imath]. Suppose, that [imath]a_n > \frac{1}{n}[/imath]. Then: [imath]\underbrace{\sin{\sin{(...\sin{(1)}...)}} )}_\text{$n+1$ sines}\ ?\ \frac{1}{n+1}[/imath] [imath]\underbrace{\sin{\sin{(...\sin{(1)}...)}} )}_\text{$n$ sines}\ ?\ \mbox{arcsin}(\frac{1}{n+1})[/imath] If [imath]\mbox{arcsin}(\frac{1}{n+1}) < \frac{1}{n}[/imath] (becuase left part is bigger then it by induction hypothesis) then we are done. And it really is, because [imath]\sin{\frac{1}{n}} > \frac{1}{n+1} \Leftrightarrow \frac{1}{n} - \frac{1}{n^33!} > \frac{1}{n+1}[/imath] (here I used Teylor expansion for sine). As [imath]a_n > \frac{1}{n}[/imath] our series is divergent by comparison test. Is my reasoning correct? | 2293125 | Does the series [imath]\sum \sin^{(n)}(1)[/imath] converge, where [imath]\sin^{(n)}[/imath] denotes the [imath]n[/imath]-fold composition of [imath]\sin[/imath]?
I'm trying to solve the following task Sequence [imath]\{a_n\}[/imath] is given by the rule: [imath]a_1 = 1,\: a_{n+1} = \sin (a_n)[/imath]. Does the series [imath]\sum a_n[/imath] converge? Can you give me any hints how to solve it, cause i got totally stuck at the very beginning, please? |
2297898 | Geometric interpretation of inner product of two matrices
Consider space [imath]\mathcal{S}=\mathbb{R}^{n\times n}[/imath]. Let [imath]\mathbf{X},\mathbf{Y}\in\mathcal{S}[/imath]. The inner product between [imath]\mathbf{X},\mathbf{Y}[/imath] is defined as (https://en.wikipedia.org/wiki/Frobenius_inner_product) \begin{equation} <\mathbf{X},\mathbf{Y}> = trace(\mathbf{Y}^{T}\mathbf{X}^{})=\sum_{i=1}^{n}\sum_{j=1}^{n} \mathbf{X}_{i,j}\mathbf{Y}_{i,j} \end{equation} In [imath]\mathbb{R}^2,\mathbb{R}^3[/imath], "Inner Products" give us a notion of "angle" between the two vectors i-e \begin{equation} <\mathbf{x},\mathbf{y}> = \|\mathbf{x}\|\|\mathbf{y}\| cos(\theta) \end{equation} I am wondering how to extend this geometric interpretation to matrices? What does it mean for two matrices to be "aligned". Does it mean their singular vectors are aligned? | 1110603 | Geometrical or Physical significance (interpretation) of the inner-product [imath]\langle A,B \rangle := Trace (AB^t)[/imath] over [imath]M_n(\mathbb R)[/imath]
[imath]\langle A,B \rangle := Trace (AB^t)[/imath] is an inner product over the vector space [imath]M_n(\mathbb R)[/imath] of all real matrices of size [imath]n[/imath] , I would like to know whether this inner-product has any Geometrical or Physical significance (interpretation) or not ? Please shed some light . Thanks in advance |
2296620 | A number theoretical problem related to inverse sum of divisors
If [imath]n[/imath] is a perfect number then its sum of divisors is [imath]2n[/imath]. I would like to prove that [imath]\frac{1}{d_1}+\frac{1}{d_2}+\ldots+\frac{1}{d_{K}}=2[/imath] where [imath]d_k[/imath] are the divisors of [imath]n[/imath]. | 594280 | Sum of the reciprocals of divisors of a perfect number is [imath]2[/imath]?
How do I show that the sum of the reciprocals of divisors of a perfect number is [imath]2[/imath]? I tried [imath]d_i\mid n[/imath] with [imath]i\in\mathbb{N},\;d_i\leq n[/imath] then [imath]\frac{1}{d_1}+\frac{1}{d_2}+\frac{1}{d_3}+...+\frac{1}{d_i}=2[/imath] [imath]\sum_{d\mid n} \frac{1}{d}=2[/imath] So actually, I have to show this last equality, whereas [imath]n[/imath] is a perfect number. |
2293199 | Derivative of vector product with respect to a vector
I would like to derive the following expression with respect to a vector [imath]x \in \mathcal{R}^N[/imath]: [imath]a a^T b[/imath] where [imath]a, b[/imath] are vectors in [imath]\mathcal{R}^M[/imath]. My idea is to apply the following [imath] \frac{\partial a}{\partial x} a^T b + a (\frac{\partial a}{\partial x})^T b + a a^T \frac{\partial b}{\partial x}[/imath] but it is dimensionally wrong. In fact: [imath]\frac{\partial a}{\partial x} a^T b[/imath] has dimension: [imath](MxN) (NxM) (Mx1)[/imath] or [imath]a (\frac{\partial a}{\partial x})^T b[/imath] has dimension: [imath](Mx1) (NxM) (Mx1)[/imath] that are really dimensionally wrong. I'm assuming that the derivative of a vector [imath]a \in \mathcal{R}^M[/imath] with respect to another vector [imath]a \in \mathcal{R}^N[/imath] is a matrix [imath]a \in \mathcal{R}^{MxN}[/imath]. Moreover, even in the very simple example: I want to derive [imath]a^T b[/imath] with respect to [imath]x[/imath]: [imath](\frac{\partial a}{\partial x})^T b + a^T \frac{\partial b}{\partial x}[/imath], the first addendum seems to be a column vector in [imath]\mathcal{R}^N[/imath] while the second one is a row vector in [imath]\mathcal{R}^N[/imath]. What am I doing wrong? I know I can fix this by transposing the function [imath]a^T b[/imath] when partially deriving with respect to [imath]a[/imath] (then the first addendum becomes [imath]b^T \frac{\partial a}{\partial x}[/imath]) but I would like to know if there is a generic rule for that. Thank you to anyone that can provide to me some hints. Marco My question is a bit different with respect to this post since that is the differentiation about the transpose of a vector while here I'm questioning about the dimensionality of vector product derivative. Moreover, I add a dummy question that may help me in solving this: In case of a derivative of scalar-vector product, i.e. [imath]\frac{\partial (c a)}{\partial x}[/imath], with [imath]c \in \mathcal{R}, a \in \mathcal{R}^M[/imath] and [imath]x \in \mathcal{R}^N[/imath], I would say that the derivative is [imath]\frac{\partial (c a)}{\partial x} = \frac{\partial c}{\partial x} a + c \frac{\partial a}{\partial x}[/imath]. Now, the first addendum has a wrong dimensionality: (1xM) (Nx1). Now, I know that the things works if I do: [imath]\frac{\partial (c a)}{\partial x} = a \frac{\partial c}{\partial x} + c \frac{\partial a}{\partial x}[/imath], i.e., if I set the scalar on the right side before deriving with respect to [imath]c[/imath]. In this case I know that, for dimensionality, [imath]a c[/imath] is more correct than doing [imath]c a[/imath] but I would like to know if there are some rules that I'm aware of. Then my question is: why do I have to do this? | 20694 | Vector derivative w.r.t its transpose [imath]\frac{d(Ax)}{d(x^T)}[/imath]
Given a matrix [imath]A[/imath] and column vector [imath]x[/imath], what is the derivative of [imath]Ax[/imath] with respect to [imath]x^T[/imath] i.e. [imath]\frac{d(Ax)}{d(x^T)}[/imath], where [imath]x^T[/imath] is the transpose of [imath]x[/imath]? Side note - my goal is to get the known derivative formula [imath]\frac{d(x^TAx)}{dx} = x^T(A^T + A)[/imath] from the above rule and the chain rule. Thanks, Asaf |
2298446 | A ring homomorphism over rational numbers is the identity
Let [imath]f : \mathbb{Q} → \mathbb{Q}[/imath] be a ring homomorphism. Show that f is the identity. Im having trouble with this problem. I started considering two elements in [imath]\mathbb{Q}[/imath] and the definition of ring homomorphism: Let [imath]x, y \in \mathbb{Q}[/imath] such that [imath]x = \frac{m}{n}, y=\frac{m'}{n'}[/imath]with [imath]m,m',n,n'\in \mathbb{Z}[/imath] then: [imath]f(\frac{m}{n}*\frac{m'}{n'}) = f(\frac{m}{n})*f(\frac{m'}{n'})[/imath] and [imath]f(\frac{m}{n}+\frac{m'}{n'}) = f(\frac{mn'+m'n}{nn'})=f(\frac{m}{n})+f(\frac{m'}{n'})[/imath] Clearly, my goal is to show that: [imath]f(\frac{m}{n}*\frac{m'}{n'}) =\frac{m}{n}*\frac{m'}{n'}[/imath] or [imath]f(\frac{m}{n}+\frac{m'}{n'}) =\frac{m}{n}+\frac{m'}{n'}[/imath] I've been thinking for like an hour and still achieve nothing, any ideas/hints? | 175477 | If [imath]f: \mathbb Q\to \mathbb Q[/imath] is a homomorphism, prove that [imath]f(x)=0[/imath] for all [imath]x\in\mathbb Q[/imath] or [imath]f(x)=x[/imath] for all [imath]x[/imath] in [imath]\mathbb Q[/imath].
If [imath]f: \mathbb Q\to \mathbb Q[/imath] is a homomorphism, prove that [imath]f(x)=0[/imath] for all [imath]x\in\mathbb Q[/imath] or [imath]f(x)=x[/imath] for all [imath]x[/imath] in [imath]\mathbb Q[/imath]. I'm wondering if you can help me with this one? |
2299250 | Topology riddle - dense infinite subset
Made up this question while thinking about something in topology and thought I'd post it here for fun, to see how you guys might solve it: Give an example or disprove the existence of an infinite, topological Hausdorff space [imath](X,\tau)[/imath] in which every infinite subset is dense. | 1568343 | Prove [imath]\tau[/imath] is cofinite on [imath]X[/imath] iff [imath](X,\tau)[/imath] is [imath]T_1[/imath]-space and any infinite subset of [imath]X[/imath] is dense in [imath](X, \tau)[/imath]
I dont have a clue about the proof in the reverse implication on the iff statement. I will show what I have by now: 1) A cofinite topology implies that any infinite subset in [imath]X[/imath] is dense. Let [imath]U\in\tau[/imath] and is cofinite topology so [imath]U=X-\bigcup_{k=1}^{n}\{x_k\}[/imath]. Let [imath]V\in X[/imath] an infinite subset then [imath]V\cap U=V\cap(X-\bigcup_{k=1}^{n}\{x_k\})=(V\cap X)-(V\cap \bigcup_{k=1}^{n}\{x_k\})=V-(V\cap \bigcup_{k=1}^{n}\{x_k\})[/imath] and cause [imath]|V|\ge\aleph_0[/imath] and [imath]|\bigcup_{k=1}^{n}\{x_k\}|=n\in\Bbb N\,[/imath] we have that [imath]|V\cap \bigcup_{k=1}^{n}\{x_k\}|\le n[/imath]. And finally is obvious that [imath]V-(V\cap \bigcup_{k=1}^{n}\{x_k\})\ne\emptyset[/imath] so any infinite subset is dense on [imath]X[/imath]. 2) A cofinite topology implies a [imath]T_1[/imath]-space. Let [imath]x, y\in X[/imath] and [imath]U\in\tau: U=X-\{y\}[/imath] so [imath]x\in U\land y\notin U[/imath]. We can define too [imath]V\in\tau:V=X-\{x\}[/imath] so [imath]x\notin V\land y\in V[/imath] then a cofinite topology is a [imath]T_1[/imath]-space. But my problem is that I dont have a clue to prove the reverse implication i.e. if we have a [imath]T_1[/imath]-space where every infinite subset is dense then it is a cofinite topology. Some hint? Some advice or correction in the proof of above? Thank you in advance. I will complete the proof with the advice of @Cameron Buie. 1) A [imath]T_1[/imath]-space implies that any finite subset is closed. Let the singleton [imath]\{x\}\subset X[/imath]. Then [imath]\overline{\{x\}}=\{x\}\cup\{x\}'[/imath] where [imath]\{x\}'=\{y:(\forall U\in\tau : y\in U)\land x\in U\}[/imath], but because we are in a [imath]T_1[/imath]-space exist some [imath]U[/imath] for any points [imath]x,y\in X: x\in U\land y\notin U[/imath] so [imath]\{x\}'=\emptyset[/imath] then [imath]\overline{\{x\}}=\{x\}[/imath] and is closed. This means that for every [imath]T_1[/imath]-space it cofinite sets are open. 2) If any infinite subset is dense on [imath]X[/imath] it means that the closure of any infinite subset is [imath]X[/imath] so the only closed sets are finite (and [imath]X[/imath] itself), so the topology is cofinite. |
2298880 | Is the given function measurable?
Let [imath](X, A)[/imath] and [imath](Y,B)[/imath] be two measurable space. Let [imath]f \geq 0[/imath] be measurable w.r.t [imath]A \times B[/imath] (product [imath]\sigma[/imath] -algebra). Let [imath]g(x)=\sup_{y \in Y} f(x,y)[/imath] and suppose [imath]g(x)< \infty[/imath] for each [imath]x[/imath]. Is g necessarily measurable w.r.t [imath]A[/imath]? I am not sure whether the answer is no or yes. But, I haven't been able to produce a proof. Thanks in advance for any help! | 876544 | measurability of supremum of a class of functions
Let [imath]f:X\times Y \mapsto R[/imath] be a measurable function on product space [imath]X\times Y[/imath], where [imath]X[/imath] and [imath]Y[/imath] both are some metric spaces. Define [imath]g(x) := \sup_{y\in Y} f(x,y)[/imath]. [Q.] Is [imath]g[/imath] a measurable function on [imath]X[/imath]? If not, what is the sufficient condition to be measurable? |
2298630 | Is it possible to obtain a non-skewed coin from a skewed coin?
I was thinking about a puzzle, unfortunately I do not remember where I saw it, which asked about Given a skewed coin where the [imath]p_H \neq p_T[/imath] where [imath]p_H[/imath] and [imath]p_T[/imath] are probability of observing head in one coin flip and probability of observing tail in a single coin flip, respectively. Is it possible to obtain a non-skewed coin by flipping the given skewed coin? Even keywords for more search are greatly welcomed. | 1840180 | How to choose between two options with a biased coin
We would like to choose between theatre and cinema by tossing a coin. Unfortunately the only available coin we have has probapility of heads [imath]p\ \left(\dfrac{1}{2}<p<1\right)[/imath]. How could we use that coin to take a good decision so that the two options have equal possibilities? My solution: we will flip the coin twice, if it comes up heads first and tails second, then we will go to the theatre. If it comes up tails first and heads second, then we will go to the cinema. If the two flips are the same, we flip twice again, and repeat the process until we have a unbiased toss. Is it correct? Note added by joriki: As discussed in the comments, I'm reopening the question because of the aspect that the probability [imath]p[/imath] of heads is known and may be used to optimise the algorithm; this aspect was not present in the question Puzzle about technique of fair using of unfair coin as a duplicate of which this question had been closed. |
2299032 | fundamental group of SL(n,R)
I want to prove that [imath]SL(n,\mathbb{R})[/imath] is not simply connected. For this, How can I prove that the fundamental group of [imath]SL(n,\mathbb{R})[/imath] is [imath]\mathbb{Z}[/imath] for [imath]n=2[/imath] and [imath]\mathbb{Z_2}[/imath] for [imath]n>2[/imath]. | 214637 | fundamental group of [imath]GL^{+}_n(\mathbb{R})[/imath]
I would like to know whether the [imath]GL^{+}_n(\mathbb{R})[/imath] the set of all invertible matrices with positive determinant is simply connected or not? I guess it is not simply connected but that is just a guess only, I do not know how to prove that, i.e how to show that its fundamental group is non-trivial. Well, I can rigorously prove that this is connected and hense path connected as it is Lie Group. Could any one rigorously tell me how to approach this kind problem and solve them from the basic knowledge of fundamental group or some other way? So basically I need some result or tools by which I can compute fundamental groups of all known classical matrix Lie groups. |
2299389 | Bizarre equation as STQ
I just began filling out a survey for a contest that Walmart is hosting. Since I live in Canada there is a skill testing question which was as follows: [imath](4×2)+(6/3)?5=[/imath] The answer to the equation was omitted. What kind of mathematical significance does the question mark hold? I have tried Googling the equation for an explanation, but to no avail. Is the equation bunk? FWIW, the answer turned out to be 5... so obviously the question mark represented a subtraction symbol, but how was I ever supposed to induce that if the answer is/was omitted? | 2297302 | Question mark in SQT
I just began filling out a survey for a contest that Walmart is hosting. Since I live in Canada there is a skill testing question which was as follows: [imath](4×2)+(6/3) ? 5 =[/imath] The answer to the equation is/was omitted. What kind of mathematical significance does the question mark hold? I have tried Googling the equation for an explanation, but to no avail. Is the equation bunk? FWIW, the answer turned out to be 5... so obviously the question mark represented a subtraction symbol, but how was I ever supposed to induce that if the answer is/was omitted? |
2300051 | Does there exist a real square root for this diagonal matrix?
I am unable to find a real matrix [imath]A[/imath] that satisfies [imath]A^2=B[/imath] for the matrix given below by any methods I know, since it is already diagonalized. Does there exist such a real [imath]A[/imath] at all, and more importantly, is there any way to prove whether it does? [imath]B=\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}[/imath] | 2290094 | Does the given matrix have a square root?
Given the following matrix, I was asked to tell whether the matrix is the square of some matrix or not. [imath] \begin{pmatrix} 1 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{pmatrix} [/imath] Since this matrix is diagonal, a square root is [imath] \begin{pmatrix} 1 & 0 & 0\\ 0 & i & 0\\ 0 & 0 & i \end{pmatrix} [/imath] Am I right? |
2300357 | Chebyshev's inequality Problem
Let [imath]\mathrm{X}[/imath] be a random variable with [imath]\mathbb{E}[/imath][[imath]\mathrm{X}[/imath]]=0 and [imath]\mathbb{V}[/imath]ar[[imath]\mathrm{X}[/imath]]=[imath]\sigma^2[/imath], and [imath]\mathcal{a}[/imath] [imath]\gt[/imath] 0. Prove that P($\mathrm{X}[imath]\ge[/imath]\mathcal{a}$)$\le[imath]\frac{\sigma^2}{\sigma^2+\mathcal{a^2}}$[/imath] My attempts: I started with Chebyshev's inequality, which gave P($\vert$X$\vert\ge[imath]\mathcal{a}$)$\le[/imath]\frac{\sigma^2}{a^2}$, then I considered the random variable Y=X+t, and ended up with P([imath]\vert[/imath]X+t$\vert[imath]\ge[/imath]\mathcal{a}$+t)[imath]\le[/imath][imath]\frac{\sigma^2(1+2t)}{(a+t)^2}[/imath]. Any help is appreciated | 1949296 | One sided Chebyshev's inequality
How to prove the one-sided Chebyshev's inequality which states that if X has mean 0 and variance [imath]\sigma^2[/imath], then for any [imath]a > 0[/imath] [imath]P(X \geq a) \leq \frac{\sigma^2}{\sigma^2+a^2}[/imath] Solution. I know the Chebyshev's inequality which States that [imath]P(|X-\mu| \geq a) \leq \frac{Var(X)}{a^2}[/imath] If I first argue that for any [imath]b > 0[/imath] [imath]P(X \geq a) \leq P{[(X+b)^2 \geq (a+b)^2]}[/imath] [imath]P(X\geq a) \leq \frac{E(X+b)^2}{(a+b)^2}[/imath] [imath]P(X \geq a) \leq \frac{[E(X^2)+2E(X)b+b^2]}{(a+b)^2}[/imath] [imath]P(X \geq a) \leq \frac{\sigma^2+ b^2}{(a+b)^2}[/imath] I got the correct answer. |
2300399 | Help needed with this problem in algebra.
Let [imath]t[/imath] be a real number such that [imath]t^2 = at + b[/imath] for some positive integral values of [imath]a[/imath] and [imath]b[/imath], then [imath]t^3[/imath] can never be equal to: A. [imath]4t+3[/imath] B. [imath]8t+5[/imath] C. [imath]10t+3[/imath] D. [imath]6t+5[/imath] | 2300254 | Find the case when [imath]t^3[/imath] never equals to any of the given
Let [imath]t[/imath] be a real number such that [imath]t^2=at+b[/imath] for some positive integers [imath]a[/imath] and [imath]b[/imath], then for any choice of positive integers [imath]a[/imath] and [imath]b[/imath], [imath]t^3[/imath] never equals to a) [imath]4t+3[/imath] b) [imath]8t+5[/imath] c) [imath]10t+3[/imath] d) [imath]6t+5[/imath] I have recognized that if [imath]t=0[/imath] then all the answers are correct. But can not proceed in case of [imath]t\neq 0[/imath]. Please help to solve it. Thanks in advance. |
2241186 | prove that there are infinitely many primitive pythagorean triples [imath]x,y,z[/imath] such that [imath]y=x+1[/imath]
Here is my question: Prove that there are infinitely many primitive pythagorean triples [imath]x,y,z[/imath] such that [imath]y=x+1[/imath]. They gave me a hint that I should consider the triple [imath]3x+2z+1, 3x+2z+2, 4x+3z+2[/imath], but I honestly do not know how to start. Do I need contradiction? | 5540 | [imath]k^{2}+(k+1)^{2}[/imath] being a perfect square for infinitely many [imath]k[/imath]
Generally one can see that there are infinite number of solutions for this equation [imath]a^{2}+b^{2}=c^{2}[/imath] by taking multiples of the solution [imath]3,4[/imath] and [imath]5[/imath]. Can i use this as a fact to prove, that [imath]k^{2} + (k+1)^{2}[/imath] is a perfect square for infinitely many [imath]k \in \mathbb{N}[/imath]? Any hints, suggestions would be helpful. If not, then how do i prove this fact! |
2296050 | Degree of Division of polynomials [imath]\frac{p(x)}{q(x)}[/imath] when [imath]\deg(q(x)) > \deg(p(x))[/imath]
I'm not sure how to find the degree of the ratio [imath]\frac{p(x)}{q(x)}[/imath] of polynomials [imath]p(x)[/imath] and [imath]q(x)[/imath] when [imath]\deg(q(x)) > \deg(p(x))[/imath]. I couldn't find an answer. For example, [imath]\deg(p(x))=10[/imath] and [imath]\deg(q(x))=26[/imath]. Intuitively the degree of the ratio should be [imath]-\infty[/imath], and the degree of the remainder should be [imath]10[/imath]. Want to be sure if I'm right. Any help appreciated, thanks! | 1641780 | Degree of Rational Function
This might sound like a very trivial question but I found different answers on the web. Assume one has a rational function [imath]\frac{f(x)}{g(x)} ,[/imath] where [imath]f(x)[/imath] and [imath]g(x)[/imath] are polynomials. What is the degree of the rational function? Is it the maximum degree of [imath]f[/imath] and [imath]g[/imath]? Or is it [imath]\deg(f) – \deg(g)[/imath]? Thanks |
2300705 | Find all positive integers [imath]m[/imath] that satisfies: [imath]a^2 \equiv 0\mod{m} \Rightarrow a \equiv 0 \mod{m}[/imath], [imath]\forall a \in \mathbb{Z}[/imath]
I could prove that this is statement is true if [imath]m=1[/imath] or if [imath]m[/imath] is prime. However, I couldn't prove that the statement doesn't hold if [imath]m>1[/imath] is not prime. Any hints? | 1910194 | When is this true: If [imath]m^2[/imath] is a multiple of N then m is a multiple of N
I'm looking for a set of conditions and maybe a proof of said conditions for the thought proposed in the title. It seems to me that what was stated in the title always is true when N is not a perfect square and N < m, but I can find instances that work for perfect squares N, e.g. m = 12, N = 4. If it's not clear what I'm asking, I'm looking for a general case of what's asked in this question. Any help is appreciated! Thanks. Edit: m and N are positive integers. |
2300570 | how can I prove the value of correlation coefficient [imath]r[/imath] ranges between [imath]-1[/imath] and [imath]1[/imath]?
What is the proof for the claim that the value of correlation coefficient [imath]r[/imath] ranges between [imath]-1[/imath] and [imath]1[/imath]? | 158449 | Proving that the magnitude of the sample correlation coefficient is at most [imath]1[/imath]
How can you show that the magnitude of the sample correlation coefficient is at most [imath]1[/imath]? The formula is huge, I'm not even sure how to approach this. Can anyone point me in the right direction? Note that this is the sample correlation coefficient: [imath]r_{xy} = \dfrac{\displaystyle \sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{(n-1)s_xs_y} = \dfrac{\displaystyle \sum_{i=1}^n (x_i - \bar{x})(y_i - \bar{y})}{\sqrt{\displaystyle \sum_{i=1}^{n} (x_i - \bar{x})^2 \displaystyle \sum_{i=1}^{n} (y_i - \bar{y})^2}}[/imath] |
2300853 | Prove that [imath](3n)!/(3!)^n[/imath] is an integer
Prove that [imath]\frac{(3n)!}{(3!)^n}[/imath] is an integer where [imath]n[/imath] is a non-negative integer. I can prove it with mathematical induction. Is there any other method? | 496831 | Show that for any positive integer n, [imath](3n)!/(3!)^n[/imath] is an integer.
This is also a question on my exam paper that i proved by using mathematical induction. However, my tutor tells me that it can be proved without using mathematical induction. I really want to know how to deal with in another way. Show that for any positive integer n, [imath](3n)!/(3!)^n[/imath] is an integer. |
2295182 | Diophantine inequalities?
Do there exist naturaln numbers [imath]a, b, n \in \mathbb N[/imath] such that [imath]\sqrt{n}+\sqrt{n+1}<\sqrt{a}+\sqrt{b}<\sqrt{4n+2}?[/imath] One can easily see that both [imath]a, b[/imath] can't be perfect squares, because [imath](\sqrt{n} + \sqrt{n+1})^2 = 2n +1 + 2 \sqrt{n^2+n} < 4n+2,[/imath] which after trivial transformations is equivalent to [imath]4n^2+ 4n < 4n^2 + 4n +1.[/imath] | 820280 | Non-existence of natural numbers such that [imath]\sqrt{n} +\sqrt{n+1} <\sqrt{x} +\sqrt{y} <\sqrt{4n+2}[/imath]
Show that for any [imath]n\in\mathbb{N}[/imath] there does not exist natural numbers [imath]x,y[/imath] such that [imath]\sqrt{n} +\sqrt{n+1} <\sqrt{x} +\sqrt{y} <\sqrt{4n+2}.[/imath] |
2300827 | Let [imath]G[/imath] be a group of order [imath]p^2[/imath] then [imath]G[/imath] is isomorphic to [imath]\mathbb{Z}_{p^2}[/imath] or [imath] \mathbb{Z}_p \times \mathbb{Z}_p[/imath].
If [imath]|G|=p^2[/imath] then [imath]G[/imath] is abelian. Now, like [imath]G[/imath] is abelian for the Theorem of finitely generated abelian groups [imath]G[/imath] is isomorphic to [imath]\mathbb{Z}_p\times \mathbb{Z}_p[/imath], but I don't know if I can do this or what other way I have? | 1980824 | Group of Order [imath]p^2[/imath] Isomorphic to [imath]\mathbb{Z}_{p^2}[/imath] or [imath]\mathbb{Z}_{p}[/imath] [imath]\times[/imath] [imath]\mathbb{Z}_{p}[/imath]
I'm a bit lost on this problem: Show that a group of order [imath]p^2[/imath] is isomorphic to [imath]\mathbb{Z}_{p^2}[/imath] or [imath]\mathbb{Z}_{p}[/imath] [imath]\times[/imath] [imath]\mathbb{Z}_{p}[/imath]. The only thing I can think of that may relate to this problem is Lagrange's Theorem, where the order of a subgroup divides the order of a group. But I'm not sure if I can even use that for this problem... Any help would be tremendously helpful. Thank you! |
1566884 | [imath]f:\bf S^1 \to \bf R[/imath], there exist uncountably many pairs of distinct points [imath]x[/imath] and [imath]y[/imath] in [imath]\bf S^1[/imath] such that [imath]f(x)=f(y)[/imath]? (NBHM-2010)
Let [imath]\bf S^1[/imath] denote the unit circle in the plane [imath]\bf R^2[/imath]. True/False ? For every continuous function [imath]f:\bf S^1 \to \bf R[/imath], there exist uncountably many pairs of distinct points [imath]x[/imath] and [imath]y[/imath] in [imath]\bf S^1[/imath] such that [imath]f(x)=f(y)[/imath] Borsuk-Ulam or by taking the function [imath]g(x)=f(x)-f(-x)[/imath], IVT implies that there exist [imath]x[/imath] such that [imath]f(x)=f(-x)[/imath]. But I'm unable to show the existence of uncountably many pairs. I think the fact [imath]RP^1 \cong \bf S^1[/imath] may be helpful. Any ideas? | 639935 | For continuous function [imath] f:\mathbb S^1 \to \mathbb R[/imath] there exists uncountably many distinct points [imath]x,y[/imath] such that [imath]f(x)=f(y)[/imath]
Let [imath]\mathbb S^1[/imath] denote the unit circle in [imath]\mathbb R^2[/imath]. Then prove that for every continuous function [imath]f:\mathbb S^1 \to \mathbb R[/imath], there exist uncountably many pairs of distinct points [imath]x, y[/imath] in [imath]S^1[/imath], such that [imath]f(x)=f(y)[/imath]. |
2301406 | simplify [imath]\sum_{n \geq0 } \binom{2n}{n}\frac{1}{2^n}t^n[/imath]
How to simplify [imath]\sum_{n \geq0 } \binom{2n}{n}\frac{1}{2^n}t^n[/imath] to show it equals [imath](1-2t)^{-1/2}[/imath]? Here is a proof I came up with, but I use the fact that I know what the answer should be, so I am a bit dissatisfied with it. By the binomial theorem, the coefficient of [imath]t^n[/imath] in the [imath](1-2t)^{-1/2}[/imath] is [imath]\binom{-1/2}{n}(-2)^n[/imath], and we want to show this equals [imath]\binom{2n}{n}\frac{1}{2^n}[/imath]. This can be done via induction; it amounts to showing [imath] 1 \cdot 3 \cdot 5 \cdots (2n-1)\cdot 2^n = (2n)(2n-1)\cdots (n+1)[/imath] which can be easily proved via induction. I tried using [imath]\binom{2n}{n}=\sum_{i}\binom{n}{i}\binom{n}{n-i}[/imath] | 30343 | Summing the power series [imath]\sum\limits_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}\prod\limits_{k=1}^n\frac{2k-1}{2k} [/imath]
I'd like to determine the function corresponding to the following power series: [imath]x + \sum_{n=1}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n} \frac{x^{2n+1}}{2n+1}, [/imath] where [imath]|x|<1[/imath]. |
2302114 | Prove that [imath]\sum_{x\le{n}}\frac{1}{\varphi{(x)}} = \mathcal{O}(ln{}(n))[/imath]
So I've never proved anything like this before, as this is my first time working with big O notation. Upon being presented with it, I decided to use induction. From what I understand of this, the equation means that [imath]\exists k \mid \forall x \sum_{x\le{n}}\frac{1}{\varphi{(x)}} \le k\cdot\ln{}n[/imath]. So, I was going to show that this is true for [imath]n = 2[/imath], the base case here, and then assume its true for all [imath]n[/imath], then prove for [imath]n+1[/imath]. My work was as follows for the inductive step. [imath]\sum_{x\le{n+1}}\frac{1}{\varphi{(x)}} \le k\cdot\ln{}(n+1)[/imath] [imath]\sum_{x\le{n}}\frac{1}{\varphi{(x)}}+\frac{1}{\varphi{(n+1})} \le k\cdot\ln{}(n+1)[/imath] [imath]k*\ln{}(n+1)+\frac{1}{\varphi{(n+1})} \le k\cdot\ln{}(n+1)[/imath] At this point, we note that [imath]\varphi{(n)}[/imath] has a lower bound of [imath]\frac{n}{\ln{}\ln{}(n)}[/imath]. Hence, our left hand side will maximize with this value, so we can write [imath]k\cdot\ln{}(n+1)+\frac{\ln{}\ln{}(n+1)}{n+1} \le k\cdot\ln{}(n+1)[/imath] Raising e to each side of the equation and simplifying yields [imath]\sqrt[n+1]{\ln{}(n+1)} \le \left(\frac{n+1}{n}\right)^k[/imath] At this point, I don't know what to do. Am I going about this right, or am I completely off base? Edit: While the initial question is the same as the one this was tagged as a duplicate of, the question being asked here is whether induction is a valid approach to solving this problem without using the methods discussed in the other post. | 2299177 | If [imath]x \geq 2[/imath] prove that [imath]\sum_{ n \leq x} \frac{1}{\phi(n)} = \mathcal{O}(\log(x))[/imath]
If [imath]x \geq 2[/imath] prove that [imath]\sum_{ n \leq x} \frac{1}{\phi(n)} = \mathcal{O}(\log(x))[/imath]. This problem is from Apostol's analytic number theory book in the chapter 3 exercises. I am stuck on this problem as I am simply not sure how to do it. I know from the previous chapters that the following relation exists [imath]\phi(n) = \sum_{d \mid n}\mu(d) \frac{n}{d}[/imath] So I have [imath]\sum_{n \leq x} \phi(n) = \sum_{n \leq x} \sum_{d \mid n}\mu(d) \frac{n}{d}[/imath], where [imath]\mu(d)[/imath] is the Möbius function. Can anyone offer some information on how to proceed or use this relation? |
2301041 | Maximum number of subsets of [imath]\{1,\ldots,n\}[/imath] such that the intersection of every two of the subsets has cardinality [imath]\geq r[/imath].
Let [imath]S=\{1,2,\ldots,n\}[/imath] and [imath]\mathcal{F}=\{A_1,A_2,\ldots,A_k\}\subseteq\mathcal{P}(S)[/imath] such that [imath]|A_i\cap A_j|\geq r[/imath] for all [imath]1\leq i<j\leq k[/imath], where [imath]r\in\mathbb{N}_+[/imath] and [imath]1\leq r\leq n[/imath]. Find the maximum of [imath]k[/imath]. The special cases [imath]r=n-1,n[/imath] are trivial, where [imath]\max k=2^{n-r}[/imath]. The case [imath]r=1[/imath] is easy. We pair each [imath]A\subseteq S[/imath] with [imath]S\setminus A[/imath]. There are [imath]2^{n-1}[/imath] such pairs and the answer again is [imath]\max k=2^{n-r}[/imath] by the pigeonhole principle. However, [imath]k\leq 2^{n-r}[/imath] does not hold for all [imath]r[/imath]. An example is [imath]n=6[/imath], [imath]r=2[/imath] and [imath]\mathcal{F}[/imath] consists of all subsets of [imath]S[/imath] of cardinality [imath]\geq4[/imath], in which case [imath]k=22>2^4[/imath]. Is it possible to find the maximum for the general case and how do I proceed? | 833011 | maximum size of a [imath]k[/imath]-intersecting family of [imath][n][/imath]
What is the maximum size of a family of subsets of [imath][n]:=\{1,2,3,\dots,n\}[/imath] say [imath]\mathcal{A}[/imath] such that [imath]\mid A\cap B\mid \ge k[/imath] where [imath]A,B\in \mathcal{A}[/imath] and [imath]1\le k\le n-1[/imath]? This not Erdos-ko-rado theorem. In Erdos-ko-rado theorem, we place an extra restriction that subsets of [imath]\mathcal{A}[/imath] have to be of same size. My idea: There are [imath]2^{n-k}[/imath] subsets of [imath]\{k+1,k+2,\dots ,n\}[/imath]. Append [imath]\{1,2,\dots ,k\}[/imath] to each of these sets. Hence, [imath]2^{n-k}[/imath] is a lower bound. Is it possibly the maximum we are seeking? With some change: 2 Let [imath]C[/imath] be the set of subsets of [imath][n][/imath] such that size of each subset does not exceed [imath]r[/imath]. What is the maximum size of a family of subsets from [imath]C[/imath] (say [imath]\mathcal{B}[/imath]) such that [imath]\mid A\cap B\mid \ge k[/imath] where [imath]A,B\in \mathcal{B}[/imath] and [imath]1\le k\le r[/imath]? |
2301720 | Evaluate [imath]\sum \frac{2^k}{3^{2^k}+1}[/imath]
Evaluate the sum [imath]\sum_{k=0}^{\infty} \left ( \frac{2^k}{3^{2^k}+1} \right )[/imath] I first tried to see weather the sum actually converges, by ratio test it does as [imath]\lim_{n\to \infty}|\frac{a_{n+1}}{a_n}|<1[/imath] where [imath]a_n[/imath] is the nth term. Now I am left with finding the actual sum. I thought to write it so that it can telescope, but can't find any suitable representations. Any hints for the evaluation part? Any Taylor polynomials etc? Update: Wolfram says its [imath]1/2[/imath] | 1119245 | A closed form for the sum [imath]S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}[/imath] is ...
A closed form for the sum [imath]S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}[/imath] is [imath]1 - \frac{a^{n+b}}{3^{2^{n+c}}-1}[/imath], where [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath] are integers. Find [imath]a+b+c[/imath]. |
1785309 | Find [imath]\int_{\pi /4}^{65\pi /4} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})}[/imath]
Find the value of: [imath]I=\int_{\pi /4}^{65\pi /4} \frac{dx}{(1+2^{\cos x})(1+2^{\sin x})}[/imath] First, I rewrote the limits as the function goes from [imath]\frac{\pi}{4}[/imath] to [imath]\frac{9\pi}{4}[/imath]. Now the integral with the reduced limits has the value [imath]\frac{I}{8}[/imath]. I used standard properties of definite integrals like converting [imath]f(x)[/imath] to [imath]f(a+b-x)[/imath] where [imath]a,b[/imath] are the lower and upper limits of the integral. Considering symmetry of the trigonometric functions, I believe I can even change the limits to [imath]0[/imath] to [imath]2\pi[/imath]. However, this does not simplify the problem as much as I would like. Please advice. | 463881 | Evaluate [imath] \int_{25\pi/4}^{53\pi/4}\frac{1}{(1+2^{\sin x})(1+2^{\cos x})}dx [/imath]
How to evaluate the integral [imath]\int_{25\pi / 4}^{53\pi / 4}\frac{1}{(1+2^{\sin x})(1+2^{\cos x})}dx\ ?[/imath] |
2301612 | For [imath]n[/imath] sets with intersection of size at most [imath]1[/imath] sets we have [imath]\frac{1}{n}\sum\limits_{i=1}^{n}|S_i|=O(\sqrt n)[/imath]
Suppose [imath]S_1,\dots, S_n\subseteq[n][/imath] are such that [imath]|S_i\cap S_j|\leq 1[/imath] for all [imath]1\leq i\lt j\leq n[/imath]. Show that in this case [imath]\frac{1}{n}\sum\limits_{i=1}^{n}|S_i|=O(\sqrt n).[/imath] | 1750191 | Showing [imath]{1\over n}\sum|S_i|=O(\sqrt n)[/imath] for [imath]S_i\subset [n][/imath], [imath]|S_i\cap S_i|\le 1[/imath] for [imath]i\ne j[/imath]
Show [imath]{1\over n}\sum|S_i|=O(\sqrt n)[/imath] where [imath]S_i\subset [n][/imath], [imath]|S_i\cap S_i|\le 1[/imath] for [imath]i\ne j[/imath]. A previous question required showing [imath]|E|\le {1\over 2}(\sqrt{t-1}n^{3\over 2}+n)[/imath], for an [imath]n[/imath]-vertex graph [imath]G[/imath]. What I tried to do is placing [imath]S_1,...,S_n[/imath] in a row, and below it setting [imath]1,...,n[/imath] likewise. While I didn't mind the fact that as a graph, [imath]S_i[/imath] cannot be connected to [imath]S_j[/imath] as it will have no meaning here, I did pay attention to the restrictions I am given: I can't have a [imath]K_{2,2}[/imath] graph (also not a [imath]K_{2,t\ge 2}[/imath], but [imath]t=2[/imath] is tighter.). Is what I am doing admissible? If so, I am arriving at: [imath]|E|\le {1\over 2}(\sqrt 2n^{3\over 2}+2n)[/imath]. Acknowledge(or at least assuming) that the number of edges is simply the sum of [imath]|S_i|[/imath]'s, I get [imath]{1\over n}\sum|S_i|\le \sqrt2n^{1\over 2}+2[/imath] which seems [imath]O(\sqrt n)[/imath]. Is it graph theory acceptable arguing? I would really appreciate your insights. |
2302563 | Is the series [imath]\sum\limits_n\frac{(-1)^n}{\ln(n)+\cos(n)}[/imath] convergent?
What is the nature of the series [imath]\displaystyle \sum_n \dfrac{(-1)^n}{\ln(n)+\cos(n)}[/imath]? Since the sequence of general term [imath]\frac{1}{\ln(n)+\cos(n)}[/imath] does not decrease, the critera for alternated series cannot be applied. In these topics here and here however, we manage to show that its nature is the same as that of [imath]\sum_n \dfrac{\sin(2n+\frac12)}{(\ln(2n)+\cos(2n)) (\ln(2n+1)+cos(2n+1))}[/imath] but what next? An asymptotic expansion does not fit either. | 21175 | Convergence of the series [imath]\sum\limits_{n=2}^{\infty} \frac{ (-1)^n} { \ln(n) +\cos(n)}[/imath]
[imath]\sum_{n=2}^{\infty} \frac{ (-1)^n} { \ln(n) +\cos(n)}[/imath] |
2302305 | Functions involving real values
If [imath]f[/imath] is a real valued differentiable function satisfying [imath]|f(x)-f(y)|<(x-y)^2[/imath] for all real [imath]x[/imath] and [imath]y[/imath] and [imath]f(0)=0[/imath] then [imath]f(1)[/imath] equals: | 633081 | Prove that every function that verifies [imath]|f(x)-f(y)|\leq(x-y)^2[/imath] for all [imath]x,y[/imath] is constant.
Let [imath]f\colon \mathbb{R} \rightarrow \mathbb{R}[/imath] be a function. In order to prove that every function that verifies [imath]\lvert f(x)-f(y)\rvert \leq(x-y)^2[/imath] for all [imath]x,y[/imath], is constant, I have considered on applying the Lagrange's theorem. There exists some [imath]c\in[x,y][/imath] for which: [imath]\frac{\lvert f(x)-f(y)\rvert}{x-y}=f'(c)\leq (x-y)[/imath] Which would certainly lead to success if I'm able to prove that that is [imath]0[/imath], the problem is I don't know how to proceed with the proof. Any help appreciated. |
724350 | Munkres: Compact subsets of Hausdorff Space
Claim:If [imath]A,B[/imath] are compact disjoint subsets of the Hausdorff space [imath]X[/imath], then there exists disjoint open sets [imath]U,V[/imath] containing [imath]A,B[/imath] resp. Would I be on the right track in saying that since [imath]A,B[/imath] are compact subsets of [imath]X[/imath] then choose [imath]\{\mathbb{A}_\alpha \}[/imath], [imath]\{\mathbb{B}_ j \}[/imath] to be open covers for [imath]A,B[/imath] resp. Then since [imath]X[/imath] is Hausdorff we have that for each [imath]x[/imath] in [imath]\{\mathbb{A}_\alpha \}[/imath] and [imath]y \in \{\mathbb{B}_j \}[/imath] there exists disjoint open sets [imath]U,V[/imath]. Now if we take [imath]U = \cup_{x \in \mathbb{A}_\alpha} U_x[/imath] and [imath]V = \cup_{y \in \mathbb{B}_j} V_y[/imath] we have our disjoint open sets. Comment: However, I didn't uses the fact that [imath]A,B[/imath] were disjoint, or that they are closed, since every compact subset of a Hausdorff space is closed. Can someone give me some useful hint? This is hw, so I don't want and answer. Thanks in advance. | 18192 | Disjoint compact sets in a Hausdorff space can be separated
I want to show that any two disjoint compact sets in a Hausdorff space [imath]X[/imath] can be separated by disjoint open sets. Can you please let me know if the following is correct? Not for homework, just studying for a midterm. I'm trying to improve my writing too. My work: Let [imath]C[/imath],[imath]D[/imath] be disjoint compact sets in a Hausdorff space [imath]X[/imath]. Now fix [imath]y \in D[/imath] and for each [imath]x \in C[/imath] we can find (using Hausdorffness) disjoint open sets [imath]U_{x}(y)[/imath] and [imath]V_{x}(y)[/imath] such that [imath]x \in U_{x}(y)[/imath] and [imath]y \in V_{x}(y)[/imath]. Now the collection [imath]\{U_{x}: x \in C\}[/imath] covers [imath]C[/imath] so by compactness we can find some natural k such that [imath]C \subseteq \bigcup_{i=1}^{k} U_{x_{i}}(y)[/imath] Now for simplicity let [imath]U = \bigcup_{i=1}^{k} U_{x_{i}}(y)[/imath], then [imath]C \subseteq U[/imath] and let [imath]W(y) = \bigcap_{i=1}^{k} V_{x_{i}}(y)[/imath]. Then [imath]W(y)[/imath] is a neighborhood of [imath]y[/imath] and disjoint from [imath]U[/imath]. Now consider the collection [imath]\{W(y): y \in D\}[/imath], this covers D so by compactness we can find some natural q such that [imath]D \subseteq \bigcup_{j=1}^{q} W_{y_{j}}[/imath]. Finally set [imath]V = \bigcup_{j=1}^{q} W_{y_{j}}[/imath], then [imath]U[/imath] and [imath]V[/imath] are disjoint open sets containing [imath]C[/imath] and [imath]D[/imath] respectively. What do you think? |
2302802 | How to show that a subgroup is normal.
Let [imath]G[/imath] be the group of invertible [imath]2\times 2[/imath]-matrices over the real numbers. Let [imath]H \subseteq G[/imath] be a group consisting of matrices with determinant equal to 1. I'm to show that [imath]H[/imath] is a normal subgroup. I tried showing that [imath]ghg^{-1}[/imath] would end up with a determinant equal to 1, but that was just an awful lot of calculating. Then I thought perhaps I should find a group homomorphism [imath]\phi: G \to G[/imath] such that the kernel was the identity matrix. But I couldn't figure out how. Any tips would be awesome. | 359976 | Is [imath]SL_n(\mathbb{R})[/imath] a normal subgroup in [imath]GL_n(\mathbb{R})[/imath]?
Let [imath]N= GL_n(\mathbb{R})[/imath] and [imath]M=SL_n(\mathbb{R})[/imath]. Is [imath]M[/imath] normal subgroup in [imath]N[/imath]? Why or why not? I know how to do this with [imath]GL_2(\mathbb{R})[/imath] and [imath]SL_2(\mathbb{R})[/imath] but with [imath]N= GL_n(\mathbb{R})[/imath] and [imath]M=SL_n(\mathbb{R})[/imath] I don't even know all of their elements so I can't check the left and right cosets of [imath]M[/imath] in [imath]N[/imath]. |
2302726 | Show that [imath]\frac{a+b}{2} \geq \sqrt{ab} \geq \frac{2ab}{a+b}[/imath]
Question: Show that the harmonic mean is less than or equal to the arithmetic mean, and also less than or equal to the geometric mean, with equality if and only if [imath]a=b[/imath] ; ie., show that [imath]\frac{a+b}{2} \geq \sqrt{ab} \geq \frac{2ab}{a+b}[/imath] My attempt, I've shown the first equality which is [imath]\frac{a+b}{2}-\frac{2ab}{a+b}=(a-b)^2\geq0[/imath] My first question, should I write [imath]\frac{(a-b)^2}{2(a+b)}\geq0[/imath] or just leave as [imath](a-b)^2\geq0[/imath]? For the second, [imath]\sqrt{ab}-\frac{2ab}{a+b}=\frac{\sqrt{ab}(a+b)-2ab}{a+b}[/imath] [imath]=\frac{a^{\frac{3}{2}} \sqrt{b}+\sqrt{a} b^{\frac{3}{2}}-2ab}{a+b}[/imath] [imath]=\frac{\sqrt{ab}(a+b-2\sqrt{ab})}{a+b}[/imath] I stuck at here. Can anyone explain to me how to solve this question Thanks in advance. **I found other similar questions being asked in this site. But my question includes harmonic which is not a duplicate. | 1357549 | Prove that [imath] m_{a} \geq m_{g} \geq m_{h} [/imath] using strict inequalities unless [imath] a = b [/imath].
[imath] m_{a} = \frac{1}{2} (a + b) [/imath] [imath] m_{g} = \sqrt{ab} [/imath] [imath] \frac{1}{m_{h}} = \frac{1}{2}(\frac{1}{a} + \frac{1}{b})[/imath] Attempted Solution: I believe I have shown the first step, which is [imath] m_{a} \geq m_{g} [/imath] but i'm stuck on showing [imath] m_{g} \geq m_{h} [/imath] [imath] \frac{a + b}{2} > \sqrt{ab} [/imath] [imath] (\frac{a + b}{2})^2 > (\sqrt{ab})^2 [/imath] [imath]\frac{a^2 + 2ab + b^2}{4} > ab[/imath] [imath]a^2 + 2ab + b^2 > 4ab [/imath] [imath]a^2 -2ab + b^2 > 0 [/imath] [imath](a - b)^2 > 0 [/imath] so if [imath]a = b[/imath] we have [imath](a -b)^2 \geq 0[/imath] Therefore, [imath]m_{a} \geq m_{g} [/imath] Now, i'm not sure if that is correct or not, if not can you please guide me to the complete solution or if I am correct on this part, provide a solution for [imath]m_{g} \geq m_{h} [/imath] |
2303157 | Find the limit of f(x,y)
[imath]f(x,y) = \frac{xy^2}{x^2 + y^2},[/imath] determine the limit of [imath]f(x,y)[/imath] as [imath](x,y) \to (0,0)[/imath]. The answer for this would be [imath]0[/imath] right?? and can it be calculated from: [imath]f(x,0) = \frac{xy^2}{x^2 + y^2}= 0[/imath] and [imath]f(0,y) = \frac{xy^2}{x^2 + y^2}= 0[/imath] ?? | 415721 | How to determine the limit of [imath] f(x, y)=\frac{9xy}{x^2 + y^2}[/imath]?
Given [imath]f(x,y) = \frac{9xy}{x^2 + y^2},[/imath] determine the limit of [imath]f(x,y)[/imath] as [imath](x,y) \to (2,1)[/imath]. The answer for this would be [imath]\dfrac{18}{5}[/imath] right? |
2303065 | If [imath]f[/imath] and [imath]g[/imath] are analytic in a domain [imath]D[/imath] such that [imath]\overline{f}g[/imath] is also analytic, what is [imath]f[/imath] and [imath]g[/imath]?
I tried to solve it with Cauchy Riemann equations, let [imath]f = u+iv,g=a+ib[/imath] then we have [imath]u_x = v_y,u_y=-v_x,a_x=b_y,a_y=-b_x[/imath]. Consequently, since [imath]f^{'}g[/imath] is analytic also, we have Cauchy Riemann equation for this as well, [imath](au+bv)_x=(bu-av)_y~~~~~~(au+bv)_y= -(bu-av)_x[/imath] After simplifying, we have [imath]au_x = bu_y ~~~~~~~bu_x = -au_y[/imath] Then i am kind of stuck here. The answer to this question is [imath]f[/imath] is a constant in [imath]D[/imath] or [imath]g = 0[/imath] in [imath]D[/imath]. Any help will be appreciated. | 856521 | If [imath]f[/imath],[imath]g[/imath] and [imath]\overline{f}g[/imath] are holomorphic on [imath]\Omega[/imath], then [imath]g=0[/imath] or [imath]f[/imath] is constant
Show that if [imath]\Omega[/imath] is an open connected subset of [imath]\mathbb C[/imath], [imath]f[/imath] and [imath]g[/imath] are holomorphic on [imath]\Omega[/imath] and [imath]\overline{f}g[/imath] is holomorphic on [imath]\Omega[/imath], then [imath]g=0[/imath] or [imath]f[/imath] is constant. If [imath]f(x+iy)=u(x,y)+iv(x,y)[/imath] and [imath]g(x+iy)=u'(x,y)+iv'(x,y)[/imath], we have [imath]u_x=v_y \space ; u_y=-v_x,[/imath] [imath]u'_x=v'_y \space ; u'_y=-v'_x[/imath] The product [imath]\overline{f}g[/imath] is [imath]\overline{f}g=(u-iv)(u'+iv')[/imath] [imath]=uu'+vv'+i(uv'-vu')[/imath] By hypothesis, [imath]\overline{f}g[/imath] is holomorphic on [imath]\Omega[/imath], so by the Cauchy-Riemann equations we obtain [imath](1)\space u_xu'+uu'_x+v_xv'+vv'_x=u_yv'+uv'_y-(v_yu'+vu'_y),[/imath] [imath](2) \space u_yu'+uu'_y+v_yv'+vv'_y=-(u_xv'+uv'_x)+v_xu'+vu'_x[/imath] Using the Cauchy-Riemann equations for [imath]f[/imath] and [imath]g[/imath], we have [imath](1)[/imath] reduces to [imath](1)' \space v_yu'=v'u_y[/imath] [imath](2)[/imath] reduces to [imath](2)' \space u_xv'=v_xu'[/imath] I don't know how to arrive to the conclusion [imath]g=0[/imath] or [imath]f[/imath] is constant from here. I would appreciate any suggestions. |
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