高一米唐 人
2022-03-22

You can still ask an expert for help

asked 2022-03-28

asked 2022-05-23

asked 2022-04-03

asked 2022-04-25

How to solve a cyclic quintic in radicals?

Galois theory tells us that

$\frac{{z}^{11}-1}{z-1}={z}^{10}+{z}^{9}+{z}^{8}+{z}^{7}+{z}^{6}+{z}^{5}+{z}^{4}+{z}^{3}+{z}^{2}+z+1$ can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:

Let the roots be $\zeta}^{1},{\zeta}^{2},\dots ,{\zeta}^{10$, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].

$\begin{array}{rl}{A}_{0}& ={x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}\\ {A}_{1}& ={x}_{1}+\zeta {x}_{2}+{\zeta}^{2}{x}_{3}+{\zeta}^{3}{x}_{4}+{\zeta}^{4}{x}_{5}\\ {A}_{2}& ={x}_{1}+{\zeta}^{2}{x}_{2}+{\zeta}^{4}{x}_{3}+\zeta {x}_{4}+{\zeta}^{3}{x}_{5}\\ {A}_{3}& ={x}_{1}+{\zeta}^{3}{x}_{2}+\zeta {x}_{3}+{\zeta}^{4}{x}_{4}+{\zeta}^{2}{x}_{5}\\ {A}_{4}& ={x}_{1}+{\zeta}^{4}{x}_{2}+{\zeta}^{3}{x}_{3}+{\zeta}^{2}{x}_{4}+\zeta {x}_{5}\end{array}$

Once one has $A}_{0},\dots ,{A}_{4$ one easily gets $x}_{1},\dots ,{x}_{5$. It's easy to find $A}_{0$. The point is that $\tau$ takes $A}_{j$ to $\zeta}^{-j}{A}_{j$ and so takes $A}_{j}^{5$ to $A}_{j}^{5$. Thus $A}_{j}^{5$ can be written down in terms of rationals (if that's your starting field) and powers of $\zeta$. Alas, here is where the algebra becomes difficult. The coefficients of powers of $\zeta$ in $A}_{1}^{5$ are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have $A}_{1$ as a fifth root of a certain explicit complex number. Then one can express the other $A}_{j$ in terms of $A}_{1$. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.

asked 2022-03-31

Find the length of the confidence interval given the following data

S=3 n=275 confidence level 95 %

asked 2022-04-25

Simple Linear Regression - Difference between predicting and estimating?

Here is what my notes say about estimation and prediction:

Estimating the conditional mean

"We need to estimate the conditional mean $\beta}_{0}+{\beta}_{1}{x}_{0$ at a value $x}_{0$, so we use $\hat{{Y}_{0}}=\hat{{\beta}_{0}}+\hat{{\beta}_{1}}{x}_{0}$ as a natural estimator." here we get

$\hat{{Y}_{0}}~N({\beta}_{0}+{\beta}_{1}{x}_{0},{\sigma}^{2}{h}_{00})\text{}\text{where}{h}_{00}=\frac{1}{n}+\frac{{({x}_{0}-\stackrel{\u2015}{x})}^{2}}{(n-1){s}_{x}^{2}}$

with a confidence interval for $E\left({Y}_{0}\right)={\beta}_{0}+{\beta}_{1}{x}_{0}$ is

$(\hat{{b}_{0}}+\hat{{b}_{1}}{x}_{0}-cs\sqrt{{h}_{00}},\hat{{b}_{0}}+\hat{{b}_{1}}{x}_{0}+cs\sqrt{{h}_{00}})$

where $c={t}_{n-2,1-\frac{\alpha}{2}}$ Where these results are found by looking at the shape of the distribution and at $E\left(\hat{{Y}_{0}}\right)$ and $var\left(\hat{{Y}_{0}}\right)$

Predicting observations

"We want to predict the observation $Y}_{0}={\beta}_{0}+{\beta}_{1}{x}_{0}+{\u03f5}_{0$ at a value $x}_{0$"

$E(\hat{{Y}_{0}}-{Y}_{0})=0\text{}\text{and}var(\hat{{Y}_{0}}-{Y}_{0})={\sigma}^{2}(1+{h}_{00})$

Hence a prediction interval is of the form

$(\hat{{b}_{0}}+\hat{{b}_{1}}{x}_{0}-cs\sqrt{{h}_{00}+1},\hat{{b}_{0}}+\hat{{b}_{1}}{x}_{0}+cs\sqrt{{h}_{00}+1})$

asked 2022-04-28

Sally has caught covid but doesn’t know it yet. She is testing herself with rapid antigen kits which have an 80% probability of returning a positive result for an infected person. For the purpose of this question you can assume that the results of repeated tests are independent.

a) If sally uses 3 test kits what is the probability that at least one will return a positive result?

b) In 3 tests, what is the expected number of positive results?

c) Sally has gotten her hands on more effective tests, these ones have a 90% probability of returning a positive result for an infected person. If she tested herself

twice with the new tests, how many positive results would she expect to see?