Recent content by Artusartos

Let R be a local ring with maximal ideal J. Let M be a finitely generated R-module, and let V=M/JM. Then if \{x_1+JM,...,x_n+JM\} is a basis for V over R/J, then \{x_1, ... , x_n\} is a minimal set of generators for M. Proof Let N=\sum_{i=1}^n Rx_i. Since x_i + JM generate V=M/JM, we have...

If ##A_n = A_1## for any ##n>1##, then we automatically get a contradiction, because ##a_i \rightarrow b## implies that ##b## is in the closure of ##A_1##. However, ##\bar{A_1} = A_1## and ##b \not\in A_1##. A contradiction, right?

Homework Statement Let ##X## be a topological space. Let ##A_1 \supseteq A_2 \supseteq A_3...## be a sequence of closed subsets of ##X##. Suppose that ##a_i \in Ai## for all ##i## and that ##a_i \rightarrow b##. Prove that ##b \in \cap A_i##. Homework Equations The Attempt at a Solution...

Homework Statement Let ##X_1, ... , X_n## be iid normally distributed random variables ##N(\mu, \sigma^2)##, ##\mu \in \Bbb{R}##, ## \sigma^2 > 0##. a) Design a uniformly most powerful test with significance level ##\alpha## for testing ##H_0: \sigma^2 = \sigma^2_0## vs ##H_1: \sigma^2 >...

Thanks. But we do know that ##\frac{1-x}{x}## is convex...so we can come up with a similar result, right? (without taking the ln of both sides). Is that right?

Thanks a lot. We have, ##\frac{L(0)}{L(1)} = \frac{(1-x)^{10}}{x^{10}} = (\frac{1-x}{x})^{10}## So, ##(\frac{1-x}{x})^{10} \leq k## ## \implies (\frac{1-x}{x}) \leq k^{1/10}## ##\implies \ln (\frac{1-x}{x}) \leq \frac{1}{10} \ln(k) ## Let ##k' = \frac{1}{10} \ln(k)##...

Homework Statement Suppose a sample of size 10 is drawn from a distribution with probability density function ##f(x, \theta) = 2x^{\theta}(1-x)^{1-\theta}## if ##0<x<1## and ##0## otherwise, where ##\theta \in \{0,1\}##. Describe a best critical region of size ##\alpha## for testing ##H_0 ...

For finite sets, let's suppose that ##|X|=n## and ##|A|=m## where ##m > n##. Since ##A \subseteq X##, if ##a \in A## then ##a \in X##. So ##a_1 \in A \implies a_1 \in X##, ##a_2 \in A \implies a_2 \in X##, ... , a_m \in A \implies a_m \in X. So we have a bijection h from X to the set {1, ...

Homework Statement Let X be a set and let A be a subset of X. Suppose there is an injection $f: X \rightarrow A$. Show that the cardinalities of A and X are equal. Homework Equations The Attempt at a Solution I tried proving this for hours but couldn't really get anywhere. So...

Homework Statement I want to find the transition matrix for the rational canonical form of the matrix A below. Homework Equations The Attempt at a Solution Let ##A## be the 3x3 matrix ##\begin{bmatrix} 3 & 4 & 0 \\-1 & -3 & -2 \\ 1 & 2 & 1 \end{bmatrix}## The...

Thanks. So if ##\zeta_n^r+\zeta_n^s## is real, then ##r+s## must be a multiple of ##n##. So now I should prove that ##\zeta_n^r+\zeta_n^s = \bar{\zeta_n^r}+\bar{\zeta_n^s}##, right? But shouldn't we be proving the converse instead? That if ##r+s## is a multiple of ##n##, then...

Thanks a lot. :smile: But I'm not sure why it doesn't have to do with this problem. We want sine to be zero, because it is the complex term in ##\cos(2\pi/n) + i\sin(2\pi/n)##, right?

Oh I'm sorry. I was looking at ##\zeta_5## instead of ##\zeta_5^2## and ##\zeta_5^2## instead of ##\zeta_5^3##... So if I look at ##\zeta_5^2 + \zeta_5^3##, then it lies on the x-axis and so the sine is zero and it must be real, right?

Well, that looks like it's in the second quaderant, right?

When I draw a diagram, I need to go three times the angle of ##\zeta_5## for ##\zeta_5+\zeta_5^2##, and so I end up in quadrant three, right? So it is not real. But for ##\zeta_5^2 + \zeta_5^3## or ##\zeta_5+\zeta_5^4## we need to go 2+3=1+4=5 times the angle of ##\zeta_5##, and so we will be...