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Categories

## Sequence and permutations | AIME II, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.

## Sequence and permutations – AIME II, 2015

Call a permutation $a_1,a_2,….,a_n$ of the integers 1,2,…,n quasi increasing if $a_k \leq a_{k+1} +2$ for each $1 \leq k \leq n-1$, find the number of quasi increasing permutations of the integers 1,2,….,7.

• is 107
• is 486
• is 840
• cannot be determined from the given information

### Key Concepts

Sequence

Permutations

Integers

AIME II, 2015, Question 10

Elementary Number Theory by David Burton

## Try with Hints

While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end

Number of permutations with n elements is three times the number of permutations with n-1 elements

or, number of permutations for n elements=3 $\times$ number of permutations of (n-1) elements

or, number of permutations for n elements=$3^{2}$ number of permutations of (n-2) elements

……

or, number of permutations for n elements=$3^{n-2}$ number of permutations of {n-(n-2)} elements

or, number of permutations for n elements=2 $\times$ $3^{n-2}$

forming recurrence relation as the number of permutations =2 $\times$ $3^{n-2}$

for n=3 all six permutations taken and go up 18, 54, 162, 486

for n=7, here $2 \times 3^{5} =486.$

as

sds

Categories

## Numbers of positive integers | AIME I, 2012 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Numbers of positive integers.

## Numbers of positive integers – AIME 2012

Find the number of positive integers with three not necessarily distinct digits, $abc$, with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$.

• is 107
• is 40
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Number Theory

Algebra

AIME, 2012, Question 1.

Elementary Number Theory by David Burton .

## Try with Hints

Here a number divisible by 4 if a units with tens place digit is divisible by 4

Then case 1 for 10b+a and for 10b+c gives 0(mod4) with a pair of a and c for every b

[ since abc and cba divisible by 4 only when the last two digits is divisible by 4 that is 10b+c and 10b+a is divisible by 4]

and case II 2(mod4) with a pair of a and c for every b

Then combining both cases we get for every b gives a pair of a s and a pair of c s

So for 10 b’s with 2 a’s and 2 c’s for every b gives $10 \times 2 \times 2$

Then number of ways $10 \times 2 \times 2$ = 40 ways.

Categories

## Number of points and planes | AIME I, 1999 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.

## Number of points and planes – AIME I, 1999

Ten points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is $\frac{m}{n}$ where m and n are relatively prime positive integers, find m+n.

• is 107
• is 489
• is 840
• cannot be determined from the given information

### Key Concepts

Number of points

Plane

Probability

AIME I, 1999, Question 10

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

$10 \choose 3$ sets of 3 points which form triangles,

fourth distinct segment excluding 3 segments of triangles=45-3=42

Required probability=$\frac{{10 \choose 3} \times 42}{45 \choose 4}$

where ${45 \choose 4}$ is choosing 4 segments from 45 segments

=$\frac{16}{473}$ then m+n=16+473=489.

Categories

## Arithmetic Sequence Problem | AIME I, 2012 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.

## Arithmetic Sequence Problem – AIME 2012

The terms of an arithmetic sequence add to $715$. The first term of the sequence is increased by $1$, the second term is increased by $3$, the third term is increased by $5$, and in general, the $k$th term is increased by the $k$th odd positive integer. The terms of the new sequence add to $836$. Find the sum of the first, last, and middle terms of the original sequence.

• is 107
• is 195
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Number Theory

Algebra

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

## Try with Hints

After the adding of the odd numbers, the total of the sequence increases by $836 – 715 = 121 = 11^2$.

Since the sum of the first $n$ positive odd numbers is $n^2$, there must be $11$ terms in the sequence, so the mean of the sequence is $\frac{715}{11} = 65$.

Since the first, last, and middle terms are centered around the mean, then $65 \times 3 = 195$

Hence option B correct.

Categories

## Algebra and Positive Integer | AIME I, 1987 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Algebra and Positive Integer.

## Algebra and Positive Integer – AIME I, 1987

What is the largest positive integer n for which there is a unique integer k such that $\frac{8}{15} <\frac{n}{n+k}<\frac{7}{13}$?

• is 107
• is 112
• is 840
• cannot be determined from the given information

### Key Concepts

Digits

Algebra

Numbers

AIME I, 1987, Question 8

Elementary Number Theory by David Burton

## Try with Hints

Simplifying the inequality gives, 104(n+k)<195n<105(n+k)

or, 0<91n-104k<n+k

for 91n-104k<n+k, K>$\frac{6n}{7}$

and 0<91n-104k gives k<$\frac{7n}{8}$

so, 48n<56k<49n for 96<k<98 and k=97

thus largest value of n=112.

Categories

## Arithmetic Mean | AIME I, 2015 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Arithmetic Mean.

## Arithmetic Mean of Number Theory – AIME 2015

Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

• is 107
• is 431
• is 840
• cannot be determined from the given information

### Key Concepts

Inequalities

Algebra

Number Theory

AIME, 2015, Question 12

Elementary Number Theory by David Burton

## Try with Hints

Each 1000-element subset ${ a_1, a_2,a_3,…,a_{1000}}$ of ${1,2,3,…,2015}$ with $a_1<a_2<a_3<…<a_{1000}$ contributes $a_1$ to sum of least element of each subset and set ${a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$. $a_1$ ways to choose a positive integer $k$ such that $k<a_1+1<a_2+1,a_3+1<…<a_{1000}+1$ ($k$ can be anything from $1$ to $a_1$ inclusive

Thus, the number of ways to choose the set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is equal to the sum. But choosing a set ${k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}$ is same as choosing a 1001-element subset from ${1,2,3,…,2016}$!

average =$\frac{2016}{1001}$=$\frac{288}{143}$. Then $p+q=288+143={431}$

Categories

## Distance Time | AIME I, 2012 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Distance Time.

## Distance Time – AIME 2012

When they meet at the milepost, Sparky has been ridden for n miles total. Assume Butch rides Sparky for a miles, and Sundance rides for n-a miles. Thus, we can set up an equation, given that Sparky takes $\frac{1}{6}$ hours per mile, Butch takes $\frac{1}{4}$ hours per mile, and Sundance takes $\frac{2}{5}$ hours per mile.

• is 107
• is 279
• is 840
• cannot be determined from the given information

### Key Concepts

Time

Distance

Speed

AIME, 2012, Question 4

Problem Solving Strategies by Arther Engel

## Try with Hints

After meeting at milepost, Sparky for n miles. Let Butch with Sparky for a miles Sundance with Sparky for n-a miles.

Then
$\frac{a}{6} + \frac{n-a}{4}$ = $\frac{n-a}{6} + \frac{2a}{5}$ implies that $a = \frac{5}{19}n$

Then integral value of n is 19 and a = 5 and $t = \frac{13}{3}$ hours that is 260 minutes. Then $19 + 260 = {279}$.

Categories

## Sequence and fraction | AIME I, 2000 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.

## Sequence and fraction – AIME I, 2000

A sequence of numbers $x_1,x_2,….,x_{100}$ has the property that, for every integer k between 1 and 100, inclusive, the number $x_k$ is k less than the sum of the other 99 numbers, given that $x_{50}=\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 173
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

AIME I, 2000, Question 10

Elementary Number Theory by Sierpinsky

## Try with Hints

Let S be the sum of the sequence $x_k$

given that $x_k=S-x_k-k$ for any k

$100S-2(x_1+x_2+….+x_{100})=1+2+….+100$

$\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050$

$\Rightarrow S=\frac{2525}{49}$

for $k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}$

$\Rightarrow x_{50}=\frac{75}{98}$

$\Rightarrow m+n$=75+98

=173.

Categories

## Arithmetic and geometric mean | AIME I, 2000 Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Arithmetic and geometric mean with Algebra.

## Arithmetic and geometric mean with Algebra – AIME 2000

Find the number of ordered pairs (x,y) of integers is it true that $0 \lt y \lt 10^{6}$ and that the arithmetic mean of x and y is exactly 2 more than the geometric mean of x and y.

• is 107
• is 997
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Ordered pair

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

## Try with Hints

given that $\frac{x+y}{2}=2+({xy})^\frac{1}{2}$ then solving we have $y^\frac{1}{2}$-$x^\frac{1}{2}$=+2 and-2

given that $y \gt x$ then $y^\frac{1}{2}$-$x^\frac{1}{2}$=+2 and here maximum integer value of $y^\frac{1}{2}$=$10^{3}-1$=999 whose corresponding $x^\frac{1}{2}$=997 and decreases upto $y^\frac{1}{2}$=3 whose corresponding $x^\frac{1}{2}$=1

then number of pairs ($x^\frac{1}{2}$,$y^\frac{1}{2}$)=number of pairs of (x,y)=997.

.

Categories

## Logarithms and Equations | AIME I, 2000 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 2000 based on Logarithms and Equations.

## Logarithms and Equations – AIME I 2000

$log_{10}(2000xy)-log_{10}xlog_{10}y=4$ and $log_{10}(2yz)-(log_{10}y)(log_{10}z)=1$ and $log_{10}(zx)-(log_{10}z)(log_{10}x)=0$ has two solutions $(x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2})$ find $y_{1}+y_{2}$.

• is 905
• is 25
• is 840
• cannot be determined from the given information

### Key Concepts

Logarithms

Theory of Equations

Number Theory

AIME I, 2000, Question 9

Polynomials by Barbeau

## Try with Hints

Rearranging equations we get $-logxlogy+logx+logy-1=3-log2000$ and $-logylogz+logy+logz-1=-log2$ and $-logxlogz+logx+logz-1=-1$

taking p, q, r as logx, logy and logz, $(p-1)(q-1)=log2$ and $(q-1)(r-1)=log2$ and $(p-1)(r-1)=1$ which is first system of equations and multiplying the first three equations of the first system gives $(p-1)^{2}(q-1)^{2}(r-1)^{2}=(log 2)^{2}$ gives $(p-1)(q-1)(r-1)=+-(log2)$ which is second equation

from both equations (q-1)=+-(log2) gives (logy)+-(log2)=1 gives $y_{1}=20$,$y_{2}=5$ then $y_{1}+y_{2}=25$.


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# Are there any "simple and intuitive" models of hyperbolic geometry?

I've read many times that hyperbolic geometry is geometry on a negatively curved surface, but when I try to research it online, I usually get things like the Poincaré disk or Beltrami-Klein disk, which don't feel intuitive (especially on how distances are defined in both those models).

The current simplest model I can find is the Minkowski hyperboloid model (and both the Poincare disk and Beltrami-Klein disk models can be derived as azimuthal projections of the hyperboloid model), but there are two main difficulties I have with it:

• it relies on a "Minkowski metric" for deriving distance that is different from the Euclidean metric (and the paper doesn't seem to describe how one might discover this metric in the first place, except in the context of general relativity)
• it uses a hyperboloid of two sheets, which is a positively curved surface in 3D Euclidean space, not negatively curved

I also know of the pseudosphere model, which uses a negatively-curved surface but has complicated geodesics/lines as well as a video series called "Universal Hyperbolic Geometry" that looks promising in its simplicity but uses a completely different approach to hyperbolic geometry (and takes a very long to watch).

Because of this, is there a model of hyperbolic geometry that resolves all of these "problems" (i.e. it uses Euclidean metric, a negatively-curved surface, simple geodesics/lines, and a familiar geometric framework), or is it impossible?

• Are you familiar with the difference between intrinsic and extrinsic curvature, or for surfaces, the difference between the first and the second fundamental form? I am asking because the requirements from the last paragraph (Euclidean metric, negatively-curved surface, ...) seem to indicate some confusion about this... Apr 10 at 17:13
• In the last paragraph, I mean an extrinsically negatively-curved surface being used to described an intrinsically negatively-curved geometry. Apr 10 at 17:39
• Alright. Then see Somos' answer below. On a side note: you might also want to take a look at the half-plane model. I think it is quite intuitive and simple. Apr 10 at 17:56

format: plain


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# Recent questions tagged displacement-current

### A parallel plate capacitor with plate area $A$ and separation $d$ between the plates is charged by a constant current $i$. Consider a plane surface of area $A/4$ parallel to the plates and drawn symmetrically between the plates. Calculate the displacement current through this area.

To see more, click for the full list of questions or popular tags.


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Browse Questions

# If the function $f : R \rightarrow R$ be given by $f(x) = x^2+2$ and $g : R \rightarrow R$ be given by $g(x) = \large\frac{x}{x-1} , x \neq 1$, find fog and gof and hence find fog (2) and gof (-3)

Can you answer this question?


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Converting Between Linear & Rotational Video Lessons

Concept

Problem: What is the maximum speed of a point on the outside of the wheel 15 cm from the axle? Express your answer with the appropriate units.

FREE Expert Solution

ω = Δθ/Δt — angular velocity is slope of θ-t graph

Max ω is the steepest slope on the graph, here that's between 10 and 5 s

Linear velocity related to angular by v = ωr

81% (389 ratings)
Problem Details

What is the maximum speed of a point on the outside of the wheel 15 cm from the axle? Express your answer with the appropriate units.


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January 06, 2013

# Create a Web Applications with Symfony2 Framework Part 2 - Bundles and routing

9:00 am
2
In my previous article, I have explained about How to create a web application with Symfony 2 framework part 1. In this article, I'm going to explain about, How to bundle and routing with Symfony2 Framework.

If you have any doubt in installing and configuring the Symfony2 framework, Check my previous article "Installing and Configuring Symfony2 framework on Linux Ubuntu OS". If you need more details about, How to create a web application with Symfony2 framework, Check the article "Create a Web application with Symfony2 Framework - Part 1".

Step 1 : Create Bundle

At the beginning, Symfony 2 Framework makes sense that the main bundle "TestSymfonyBundle" create. Go to the cmd.exe the path to the root directory of the web application with php app/console it calls help. There you can read about the various basic commands and become better acquainted with the Symfony application consoles familiar.

With the following command we can create our example project finally the bundle : phpapp/console init: bundle Test\SymfonyBundle src/

* init: bundle is the command to initialize the bundle
* Test \ SymfonyBundle is the PHP 5.3 namespace
* src / corresponds to the folder in which the bundle is to be created

If everything goes well, an application of the consoles Symfony2 Framework already suggests the next steps. Additionally, you can look in the editor's choice created DefaultControllerFactory under / src / Test / SymfonyBundle / controller.

Step 2 : Bundle activate and integrate into the application

Symfony2 Framework is very economical in "guessing" of components or plugins, as this behavior costs a lot of performance and programmer unnecessarily squeezed into a programming style. Therefore, the developer must active the bundle tell the autoloader.

For this we open the first AppKernel (/app/AppKernel.php) and adds the newly created SymfonyBundle in the Function register bundles () added.
To do this simply extends the variable $bundles by the following lines: //Example Scandio new Test\SymfonyBundle\TestSymfonyBundle (), We also can use the namespace makes sense, we have to register it in the autoloader. Nothing easier than that: /// App/autoload.php$Loader -> register namespaces(array(
...
/ / Symfony
'Symfony' => __ DIR__. '/.. / src'
));

For our example, these settings are sufficient. As can be seen in the variety of configurations allow the developer Symfony2 very much freedom to design their applications as they wish.

Step 3 : Configuring Routing

An important component of Symfony2 is the routing. Routing is finding the way here the URL to the appropriate controller action. Unlike, for example CakePHP routes here are not automatically created (these are mostly based on conventions), but the developer the way is treated as a route before.

This turns - with the necessary programming discipline - to be very advantageous. As a developer, you can customize the URLs here with his wishes, without the risk of violating coding standards, and so produce unnecessary mistakes.

As already described in the first part of this series, we need for our application six basic routes ( index, add, edit, delete, markAsDone, markAsUndone ). The Referenzauf the routes are stored in the /app/config/routing.yml file. If that is not common or YML format to another format (ex.XML) is preferred, this can also use in Symfony2 Framework. The framework allows the programmer a lot of liberties here, as long as it is configured.

Based on the sample routes you can already see the Best Practices: The relevant route information should be stored in the bundle, in order to facilitate a transfer later. The following must be entered in the routing.yml so that the routes are detected by the application: #/app/config/routing.yml # Test Symfony2 tutorial test_symfony_tutorial: resource: "@TestSymfonyBundle/Resources/config/routing.yml"

Next the routes are entered into the currently referenced routing config. For this, I recommend reading the excellent again declared Symfony2 online manuals . There is explained clearly and in detail, what routes are and how they are created.

Maybe it will surprise some readers that, add/edit action in two different controller functions have been split. The point is that through this code is readable and understandable better. Additionally, one can wiedervewenden so features or actions directly. #index: Show all tasks ordered by priority test_symfony_index: pattern:/symfony defaults: {_controller: TestSymfonyBundle:Task:index}

test_symfony_view:
requirements:

requirements:
_method: get

requirements:
_method: post

# Edit: Show form with prevoius task data
test_symfony_edit_prepare:
requirements:
_method: get

# Edit: Process Form Data to edit task
: test_symfony_edit_prepare
pattern: / symfony / edit / {taskId}
requirements:
_method: post

# Delete: deletes an existing task
test_symfony_delete:
requirements:

# MarkAsDone: Mark an existing task as done
test_symfony_markAsDone:
requirements:

# MarkAsDone: Mark an existing task as undone
test_symfony_markAsUndone:
requirements:

If all routes were added successfully, we can test via console whether the required routes are recognized by Symfony2. The following output appears on successful route configuration:


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# Subcaption error

Im using TexStudio 2.12.22. When trying to make subfigures I get the following error:

! Illegal unit of measure (pt inserted).

! Missing number, treated as zero.

*
l.157 \begin{subfigure}{0.5*\textwidth}"

\usepackage{subcaption}
\begin{figure}[h]
\begin{subfigure}{0.5*\textwidth}
\includegraphics[width=\textwidth]{example-image-a}
\end{subfigure}
\hfill
\begin{subfigure}{0.5*\textwidth}
\includegraphics[width=\textwidth]{example-image-b}
\end{subfigure}
\end{figure}

Any ideas? Thanks

UPDATE Thanks to Leandris, the * was messing it up. replaced with {0.5\textwidth}

• Tried it with 0.2 and 0.01 but still get the same error. I also get:! Missing number, treated as zero. – potatasbravas Mar 15 at 13:33
• Remove the *. \begin{subfigure}{0.5\textwidth} should work but will result in an overfull box warning. – leandriis Mar 15 at 13:33
• Thanks it was the * that was messing things up. – potatasbravas Mar 15 at 13:37

Remove the *. \begin{subfigure}{0.5\textwidth} instead of \begin{subfigure}{0.5*\textwidth} works perfectly fine.

Placing two subfigures next to each other that both take up half the textwidth will result in an overfull box warning as there will be a small white spce between them. This can be overcome when removing the \hfill and when placing a %sign at the end of the first subfigure environment. Alternatively you could also decrease the width of both subfigures.

(Red lines indicate the start of the margins.)

\documentclass{article}
\usepackage{graphicx}
\usepackage{subcaption}
\begin{document}
\begin{figure}[h]
\begin{subfigure}{0.5\textwidth}
\includegraphics[width=\textwidth]{example-image-a}
\end{subfigure}
\hfill
\begin{subfigure}{0.5\textwidth}
\includegraphics[width=\textwidth]{example-image-b}
\end{subfigure}
\end{figure}

\begin{figure}[h]
\begin{subfigure}{0.5\textwidth}
\includegraphics[width=\textwidth]{example-image-a}
\end{subfigure}%
%    \hfill
\begin{subfigure}{0.5\textwidth}
\includegraphics[width=\textwidth]{example-image-b}
\end{subfigure}
\end{figure}

\begin{figure}[h]
\begin{subfigure}{0.45\textwidth}
\includegraphics[width=\textwidth]{example-image-a}
\end{subfigure}
\hfill
\begin{subfigure}{0.45\textwidth}
\includegraphics[width=\textwidth]{example-image-b}
\end{subfigure}
\end{figure}

\end{document}


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# Non-invertible Hessian Matrix

Published:

During my development of a novel high-dimensional Bayesian model for network data analysis (in my case is microbiome data), I ran into a problem that associates with Hessian Matrix. After using r optimizing() function to get the maximum likelihood estimates of the parameters, I needed to invert the Hessian matrix [TODO: add footnote “(or the estimate of a hessian matrix, depdening on the optimizing algorithm used. There are pirmarily three coming with rstan: Newton method, BFGS(quasi-newton algorithm), and LBFGS(quasi-newton algorithm). BFGS and LBFGS use a estimates of the Hessian matrix rather than calculate the Hessian matrix for the sake of computation speed.)”] to calculate the variance/standard error of the estimates.However, the task is not as trivial as it seems like. Even with a “converged” esimation [TODO: add footnote “when the algorithm reaches the default stoping criteria(not the one of maximum iterations, but the tolerance threshold)”] where the estimates are supposed to be the maximums, the Hessian matrix can still be un-invertible or non semi-positive definite. Both these two criteria are necessary to calculate vairance/standard error.

In order to move forward, I needed to locate the root cause of the problem, where I hypothesized couple before doing limited level of reading. 1) The estimation of Hessian matrix was bad. Since both BFGS and LBFGS do not calculate the Hessian matrix, but rather approximating the Hessian matrix in the algorithms, it is possible that the apporximation is bad, which leads to the invertible problem. Then I tried to run the code with Newton method (which I believed and still believe it calculates Hessian matrix). However, it is extremely slow, and seems that the Hessian matrix still is non-invertible (without seriously checking). 2) Another thought I had was increasing the training data sample size(thankfuly, it was a simulation where I could control the sample size.) In a test run, I increased my sample size from 100 to 500 where there were 330 variables where correlation structure exists. The Hessian matrix was invertible and gives variance/standard deivation. (Don’t be so happy, it was just one run.) Now I am re-runing the results with larger sample size on cluster with more simulation iterations.

Of course, I had to do more research on how to solve the problem completely or at least understanding why it is a problem, rather than relying on big sample or pure luck (refering to that one test). During my research on the topic, I came across this paper by J Gill and G King[TODO: insert citation], which fantastically explains what to do when Hessian matrix is not invertible, which is the title of the paper. The paper started with introducing the problem of Non-invertibel hessian and provided two alternative approaches (Generalized inverse and importance resample). These alternative is extremely different from the most common Internet solution, modifying the model specification. Please go ahead and read my takeaway from the paper.

The core idea of the paper about how to deal with non-invertable Hessian is to use some sampling method that is based on the posterior mode estimates and non-invertable hessian to approximate/create posterior denstities, which in turn gives the estimates of the mean and variance.

The sampling method is the sampling importance resampling. The prior distribution (in this sampling method) is the posterior mode estimates and some transformation of the non-invertable hessian. The transformation includes calcualting the generalized inverse of negative hessian, which is to deal with the non-invertability, and calculating the generalized Cholesky to calculate the pseudo-variance matrix (only if the generalized inverse is not positive definite, hence can’t be used as the variance matrix.)

 The asampling importance resampling method is a quiet delicate estimation algorithm. Refering back to the original sampling importance resampling algorithm, we have $\theta_1 = \beta$, $\theta_2 = \tilde \beta$ ($\tilde \beta$ is random draws from an approximation matrix), $f(\theta_1 X) = f(\beta y)$ the posterior distribution of interst, $f(\theta_2 \theta_1, X) = N(\tilde \beta \hat \beta, V’V)$ the normal distribution based on the MLE estimation and the pseudo-variance matrix generated by the generalized Cholesky decomposition, $\hat \beta$ is the maximum likelihood estimator,


text:
My Math Forum Interesting thoughts on square roots

 Number Theory Number Theory Math Forum

 October 28th, 2017, 12:32 AM #11 Senior Member   Joined: Nov 2015 From: USA Posts: 103 Thanks: 6 $\displaystyle R_4$ I have no idea what this is supposed to be. But seriously, I'm exploring possibilities that I think are awesome. Just looking at what everyone else has done and accepting that as the best or greatest or only valid way is just not my thing, and I believe that truth only comes from questioning these things, and indeed questioning everything, especially the most basic assumptions we don't even realize we make. Last edited by MystMage; October 28th, 2017 at 12:35 AM.
October 28th, 2017, 10:08 AM   #12
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Quote:
 Originally Posted by MystMage $\displaystyle R_4$ I have no idea what this is supposed to be. But seriously, I'm exploring possibilities that I think are awesome. Just looking at what everyone else has done and accepting that as the best or greatest or only valid way is just not my thing, and I believe that truth only comes from questioning these things, and indeed questioning everything, especially the most basic assumptions we don't even realize we make.
Consider $\displaystyle R_2$ for a moment. This is the group with members $\displaystyle \{ R(0), R( \pi ) \}$ where R is a rotation about the origin and multiplication is adding rotations. This group rotates a point by either no angle or 180 degrees. That is to say, given P = (x, y), $\displaystyle R(0) P = (x, y)$ and $\displaystyle R( \pi )P = (-x, -y)$. Setting $\displaystyle x = a^2, ~ y = 0$; then the square root of P is either $\displaystyle R(0)(a, 0) = (a, 0)$ or $\displaystyle R( \pi ) (a, 0) = (-a, 0)$. So the square root of $\displaystyle a^2$ is either (a, 0) or (-a, 0) ie. $\displaystyle \sqrt{a^2} = \pm a$.

This is pounding on a nail using an atom bomb.

But it's easy to generalize. I'll leave the details to you but the elements of $\displaystyle R_4$ are $\displaystyle \left \{ R(0), ~ R \left ( \frac{ \pi }{2} \right ), ~ R ( \pi ), R \left ( \frac{3 \pi }{2} \right ) \right \}$. (We are rotating by 90 degrees with this one.) This group takes a point P from (x, 0) to (x, 0), (0, x), (-x, 0), (0, -x) respectively. This means you can represent your operation as taking $\displaystyle P = (a^2, 0)$ to the four equivalent solutions (a, 0), (0, a), (-a, 0), (0, -a). I'll let you extend this to (x, y) because I don't know what you would want to do with it..

Reviewing old assumptions is a good idea. But you can't erase the Historical development of a well known and useful function with impunity. You are looking at the square root function and questioning it. Fine. But you can't just throw the baby out with the bath water. Either add a new aspect to the square root function or create a new, more general square root. You can't just throw it away just because it doesn't do what you want it to. (The same goes for your comments about the complex number system.)
-Dan

Last edited by topsquark; October 28th, 2017 at 10:18 AM.

 October 28th, 2017, 05:35 PM #13 Senior Member   Joined: Sep 2016 From: USA Posts: 417 Thanks: 231 Math Focus: Dynamical systems, analytic function theory, numerics I am not sure you understand what the word "dimension" means. I'm also curious how you can justify your definition of the square root. If we would follow your procedure, we would not have $\sqrt{x}\cdot \sqrt{x} = x$ which I personally feel is a fairly important property of the square root. What with it being the only important property and all. Thanks from topsquark
October 28th, 2017, 10:31 PM   #14
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Quote:
 Originally Posted by SDK I am not sure you understand what the word "dimension" means. I'm also curious how you can justify your definition of the square root. If we would follow your procedure, we would not have $\sqrt{x}\cdot \sqrt{x} = x$ which I personally feel is a fairly important property of the square root. What with it being the only important property and all.
What makes you think that what I proposed doesn't have this property?

The magnitude of the results remains unchanged.

When using powers you are basically "squeezing" the input space into half the output space. When using powers, two different inputs yield the same output, while with the inverse, roots have one input and yield two outputs, basically stretching the input space into twice the output space.

Basically, with normal roots, root 25 = +-5, a pair of solutions. To get root 5 * root 5 to equal 25, you assume the same solution from both of the root functions. If you don't make this assumption, you get four possibilities all with the same magnitude of 25, but two of which will be negative and two will be positive.

With my idea, the above remains true for rooting positive numbers. Now root -25 = +-5, also a pair of solutions, but to get root -5 * root-5 = -5, instead you assume opposing solutions from the root functions, else, without making an assumption about what solutions to use you would then get the same four possibilities for +-25 as a result.

In the end it comes out the exact same for rooting positive numbers as with current methods, but mine simply recognizes the assumptions being made and uses that to get simpler results.

Either way, an assumption is made. This idea probably fits better in a system that handles polarity independent of magnitude because then it would match up with the rest of the system instead of being unusual, but it still works.

If you really want to generalize though, then remove the assumptions.

Root X * Root X = Y.
Each Root X gives two solutions,
each of the two solutions multiplied by the two solutions of the other Root X gives four combinations and thus four solutions for Y.

In my method, all four of those solutions for Y, when rooted, get two of the solutions from the left side of the equation, but the current method of roots, adding in complex numbers, means that two of the solutions for Y can't be rooted back to two of the solutions on the left, unbalancing the equation.

Basically, when assumptions are removed, my method still loops all four solutions for rooting Y back to two of the four possibilities from left side, while current methods only equate two solutions and redefines the other two.

My way
root X = +x & -x
root X * root X = Y
(-x & +x) * (-x & +x) = Y
+x * +x = Y[1]
+x * -x = Y[2]
-x * +x = Y[3]
-x * -x = Y[4]
Y[1] = Y[4]
Y[2] = Y[3]
Y[1] = Y[2]*-1

Current way,
root X = +x & -x
root X * root X = Y
(-x & +x) * (-x & +x) = Y
+x * +x = Y[1]
-x * -x = Y[4]
Y[1] = Y[4]
Y[2] = Y[3]
Y[1] = Y[2]*-1
These are different from my method,
+x * -x =/= Y[2]
-x * +x =/= Y[3]
Y[2] = xi
Y[3] = xi ;Of course, now i needs to be defined somewhere

Does this show better why I think my method is simpler and cleaner?

P.S. How does everyone always get the fancy notation? I presume that you guys don't have fancy keyboards with a bunch special keys, so is there a reference somewhere for whatever code or whatever you guys use to do it?

October 29th, 2017, 04:40 AM   #15
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Quote:
 Originally Posted by MystMage How does everyone always get the fancy notation? I presume that you guys don't have fancy keyboards with a bunch special keys, so is there a reference somewhere for whatever code or whatever you guys use to do it?
It's called Latex. Enclose your code in $...$ tags. For a quick reference to some commonly used symbols, google "latex for beginners".

$\displaystyle \LaTeX: \frac{2}{3}$

To see the code I used to generate the above, quote this post.

October 29th, 2017, 05:15 AM   #16
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Quote:
 Originally Posted by MystMage Yea, and they seem rather ridiculous. All they are is the realization that a two dimensional number will result in two dimensional possibilities, and then did a rather clunky solution. Why have this ordered pair when you can simply have a solution that result in positive or negative numbers? The imaginary numbers thing is a construct. It is people deciding on certain rules then discovering the emergent consequences. If you've read my stuff before, you'll notice one of the things I like trying to show, is circumstances where the emergent consequences, like imaginary numbers, is purely the result of the rules chosen and that other mathematical systems will have different emergent results. This seems to me like a valid alternative to imaginary numbers. Though honestly, I'm curious how people might try to prove or disprove it's validity/usefulness, or just what people think of the concept. Sometimes I wonder if no one wants to discuss anything that questions the basic assumptions of math, or even question if there are assumptions (which is an assumption itself).
What 'stuff' have you 'shown' exactly? What is a valid alternative to imaginary numbers? Prove or disprove the validity of what?!

I suppose you don't approve of any mathematics at all since, it too, is a 'construct'.

Also, why the attitude? You seem to have some sort of chip on your shoulder about 'truthfulness' of mathematics. As though every mathematician is a zombie, mindlessly following rules and obeying orders without ever questioning things.

Ridiculous, indeed.

October 29th, 2017, 05:36 AM   #17
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Quote:
 Originally Posted by MystMage The imaginary numbers thing is a construct. It is people deciding on certain rules then discovering the emergent consequences.
This is literally the definition of mathematics.

Standard mathematics uses the rules that have proved most useful.

October 29th, 2017, 06:04 AM   #18
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Quote:
 Originally Posted by MystMage An analogy would be number bases. Are number bases somehow true and natural? Is base 10 somehow more true and natural than base 2? No. But a number in either base can easily be converted to the other. There is a significant difference though. Base 10 is a whole lot easier for us to handle and interpret than base 2, and different bases can be more difficult and imprecise in certain circumstances. Just like how 1/3 in base 10 is more easily represented accurately in the more complex form of a fraction in base ten, but swap to base 12 and a fraction is no longer required to get the same result with the same level of accuracy. The same applies to roots. Imaginary numbers is a more complicated way to show something, but that complication is a result of how we have defined our functions. Use a different definition and you can simplify the representation required. It might give us a whole new way to view the same thing we've been seeing all along, or it might do nought but simplify the set of symbols required to show it.
Mathematcians rarely use bases because they focus on the properties of numbers rather than the properties of representations of numbers. The differences between the bases are essentially trivial in most applications.

When people rail against a "more complicated way" of doing something, it's usually a tacit admission that they don't understand. In particular, they don't understand the power and benefits of what they rail against.

If one wishes to design a replacement for a tool, it is first necessary to understand what that tool does. Otherwise one has no hope of making sure that the replacement is even fit for purpose as a replacement, let alone better.

Of course, one can always design a new tool and then see what uses can be found for it, but then it isn't a replacement for anything else. It just is.

At a glance, one reason that your idea wouldn't replace complex numbers is the arbitrary nature of your pairing of signs (might one get different results for different pairings?). You talked about how the concept of number involves squeezing 2 dimensions into 1 and then proceeded to squeeze three dimensions into 2. By extension, your idea requires an arbitrarily large number of dimensions. And how will it handle $\sqrt[\pi]{x}$? You seem to need to introduce more and more arbitrary ideas.

The complex numbers, by contrast, introduce one single answer (not a technique) and the rest all falls into place more or less naturally.

 October 29th, 2017, 10:12 AM #19 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 Math Focus: Mainly analysis and algebra I could also add that the imaginary unit $i$ is perhaps the best example of one of the most powerful techniques in mathematics. What happens is that we simply say the suppose there were a number that satisfied the equation $x^2 + 1=0$. Let us call such a number $i$. What can we then determine about $i$? And what can we do with it as a result? In other words, the whole process starts not by decreeing that $i$ is a new type of number with various properties, but rather by assuming that the answer we are looking for exists and then finding what properties emerge. It is a massively more beautiful approach than attempting to rewrite a function in order to force it to give the result that you wish it had given before.
October 29th, 2017, 04:15 PM   #20
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Quote:
 Originally Posted by v8archie This is literally the definition of mathematics. Standard mathematics uses the rules that have proved most useful.
True, but that doesn't mean those rules are the only one worthwhile ones, nor that we shouldn't question them and see if entirely alternative rules sets have value in some cases.

Right now, most of the time everyone talks about math like it is some inherent aspect of nature that can't be changed or done differently, as though even the order of operations is natural and that no other order of operations could ever be possible.

Actually, this forum is the only place where anyone I've interacted with even acknowledges the fact that math is a system constructed by people.

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# Simplify p*5^2+p*5* square root of 5^2+24^2

Simplify each term.
Raise to the power of .
Move to the left of .
Move to the left of .
Pull terms out from under the radical, assuming positive real numbers.
Multiply by .
Raise to the power of .


text:
# Unified Theory of Garbage Collection

## Background

This week's paper, A Unified Theory of Garbage Collection (Bacon, Cheng, and Rajan (IBM)), appeared in OOPLSA 2004.

There are two classical algorithms for automatic garbage collection, tracing (also called mark and sweep), and reference counting, which were both first developed in the 1960s. Java's generational garbage collector is a typical example of a tracing collector, and CPython's garbage collector uses reference counting---but as this paper suggests, the reality is that these distinctions are certainly muddier in practice.

The two paradigms are associated important performance tradeoffs. In particular, the mark and sweep approach adds significant pauses to program progress, since in the simplest approach one must pause execution to "trace" all live objects before reclaiming unused space. Reference counting on the other hand does not have long pauses during execution, but rather pays for its overhead incrementally. Reference counting's major drawback is the difficulty presented by cycles appearing the link structure, which prevent objects' reference counts from ever reaching 0.

## Highlights

Bacon et al. begin by presenting the two fundamental garbage collection algorithms, reference counting and tracing, as dual views of the same underlying structure. This should hit you a bit like hearing that butter and I Can't Believe It's Not Butter have the same deep structure.

The deep comparison they make between the two paradigms is developed in a number of ways:

• Duality of operations: Tracing traverses the live objects on the heap in a technically analogous way to how reference counting traverses dead objects.

• Algorithm structure: They make this more concrete by giving pseudocode algorithms for each that really are strikingly similar.

• Fixed point forumation: They give a formal presentation of garbage collection from an abstract standpoint. In this abstract view, the goal of garbage collection is to compute a fixed-point of a system of references where nodes' reference counts accurately reflect in-edges from either (i) members of the root set, or (ii) objects whose reference count is nonzero. The fixed-point here could be imagined as the result of some iterate-to-convergence approach.

They go on to demonstrate that a whole continuum of hybrid approaches between the two not only make sense in principle but in fact really are already in use in practice. Here is a list of the the garbage collection strategies in rough order from most mark-n-sweepy to most-reference-county:

• T: Tracing. This is the classic tracing algorithm. The algorithm must traverse the link graph of the heap to 'mark' objects live. The authors point out that tail recursion can reduce the space cost of the traversal when singly linked structures are present.

• GT: Tracing Generational Collection. This form of collection uses a "nursery" to improve the time spent on collection. The idea is that recently allocated objects are more likely to be freed earlier, so we can spend our garbage-collection time more efficiently by looking at newer objects more often.

• Older First. (Sketched only.) In this approach, the oldest generation is traced first instead of the nursery.

• GCH: Generational with Counted Heap. Run normal tracing of the nursery, but do reference counting on the mature remaining space of the heap. The intuitive appeal here is twofold: (1) we expect the mutator is modifying nursery objects frequently, so we avoid the reference counting overhead of those actions, and (2) we don't expect to find a lot of garbage in the mature space doing tracing anyway, and the overhead is low to just reference count it.

• Train Algorithm. (Presented but not analyzed.) The idea here is generally that we can shorten pause times by developing a more subtle notion of "generation" in a multi-generational approach. (Remember, "short pause times" is the classic advantage of reference counting!) We divide the mature generation first into trains, and then into cars. Since only one car is collected during each pause, the cars can be made as small as needed to reduce collection pause times to the amount desired. To keep the scheme going, we need to find new cars for the live objects left over from a car collection, so we pick them based on the links themselves. The authors go on to suggest that the positional nature of the car choices, the "remembered sets" of links they use, and the order of collection ultimately serve as a sort of macro-reference counting scheme.

• CC: Reference Counting, Trial Deletion. This is mostly reference counting, but with some tracing of certain "candidates" mixed in to detect cycles (i.e., check if a node's reference count is artificially high due to circular references; evidently the case we care about is when a reference count is non-zero purely due to circular references, in which case its memory should be reclaimed).

• CD: Deferred Reference Counting. Do reference counting, but save overhead by not counting mutations of root references. Add items to a Zero Count Table (ZCT) when they are decremented to zero. References from the "root set" (a concept borrowed from tracing meaning "references into the heap from outside of the heap") can keep ZCT entries alive.

• CT: Reference Counting, Tracing Backup. Use reference counting without attempting to detect cycles. Occasionally do a full tracing collection to clean up cyclic garbage.

• C: Reference Counting. This is the classic reference counting algorithm. They give a really slick way to reduce space overhead for traversing the links by using the space guaranteed to exist in the dead objects themselves.

The paper concludes with a mathematical analysis of the costs of the above collection strategies. The authors point out that the analyses are not big-O, so these costs are actually "real", up to the accuracy of the constants for a particular system. It is not obvious to me that these particular constants even need to be constant on all systems. To the extent that these values might vary, certainly so would the validity of these results.

The analyses focus on computing the formulas for:

• $\phi$, the frequency of garbage collection
• $\kappa$, the cost of an individual collection, and
• $\tau$, the total cost of collection, which is generally $\phi \cdot \kappa$ plus the cost of mutation overhead

The computation of $\phi$ and $\kappa$ are kept separate, presumably because they quanitify the essence of the classic tradeoffs between tracing (large pauses) and reference counting (continuous overhead). Thus, by investigating the $\phi$ and $\kappa$ values of an approach one might be considering on the system one has in mind, one could evaluate where along the spectrum of hybrids the approach lies (on that system).

## Merits of the paper

The duality between tracing and reference counting itself is a both mind-bending and extremely well developed. Given that the paper is really just offering this insight about a nice way to categorize existing garbage collection methods, its organization is quite nice. That is, they give the deep comparison abstractly in several different ways, then completely flesh out how existing hybrids combine the two flavors.

The section of the paper that has the most potentional to be practically important is the cost analysis in section 7. Having never constructed a garbage collection system for a whole language, I do not know whether a compiler designer would actually use these formulas to tune a garbage collector (even with knowledge of their particular domain) since they would have to validate any decisions against benchmark performance anyway.

## Impact on state of the art

This paper does not technically offer any new approaches to storage reclamation (although to be sure, its authors have certainly done that in other works). On the other hand, it does seem like one of those papers that changes how one thinks about the topic forever. Of course, the extent to which it has done that for people in the field is certainly hard to know. It does seem, at least, that plenty of people spend a lot of time tuning the hybridization of their garbage collectors, so for them this paper gives a really nice framework for thinking about how to do that. (For those curious about more recent developments in this hybrid GC realm, there is Immix and its reference-counting-infused descendant, RCImmix).

They indicate their motivation for the cost analysis was actually to aid in development of dynamic construction of garbage collection algoirthms. That sounds like an interesting avenue for future work!

## Discussion Questions

1. The seminal papers they cite for reference counting and tracing were both from 1960. This one was 2004. Does it seem like this observation took a while? If so, why did it?
2. This paper doesn't give any new algorithms or approaches for garbage collection. In what ways does the mere observation of this duality have value to the community?
3. Can you give any situations or domains where using either (relatively) pure tracing or pure reference counting is advantageous over one of these "optimized hybrid approaches"?


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# Magnon

For other uses, see Magnon (disambiguation).

A magnon is a quasiparticle, a collective excitation of the electrons' spin structure in a crystal lattice. In the equivalent wave picture of quantum mechanics, a magnon can be viewed as a quantized spin wave. Magnons carry a fixed amount of energy and lattice momentum, and are spin-1, indicating they obey boson behavior.

## Brief history

The concept of a magnon was introduced in 1930 by Felix Bloch in order to explain the reduction of the spontaneous magnetization in a ferromagnet. At absolute zero temperature, a ferromagnet reaches the state of lowest energy, in which all of the atomic spins (and hence magnetic moments) point in the same direction. As the temperature increases, more and more spins deviate randomly from the common direction, thus increasing the internal energy and reducing the net magnetization. If one views the perfectly magnetized state at zero temperature as the vacuum state of the ferromagnet, the low-temperature state with a few spins out of alignment can be viewed as a gas of quasiparticles, in this case magnons. Each magnon reduces the total spin along the direction of magnetization by one unit of ħ and the magnetization by γħ, where γ is the gyromagnetic ratio.

The quantitative theory of magnons, quantized spin waves, was developed further by Theodore Holstein, Henry Primakoff (1940), and Freeman Dyson (1956). Using the second quantization formalism they showed magnons behave as weakly interacting quasiparticles obeying Bose–Einstein statistics (the bosons). A comprehensive treatment can be found in Kittel's textbook or the article by Van Kranendonk and Van Vleck.

Direct experimental detection of magnons by inelastic neutron scattering in ferrite was achieved in 1957 by Bertram Brockhouse. Since then magnons have been detected in ferromagnets, ferrimagnets, and antiferromagnets.

The fact that magnons obey the Bose–Einstein statistics was confirmed by the light scattering experiments done 1960s–1980s. Classical theory predicts equal intensity of Stokes and anti-Stokes lines. However, the scattering showed that if the magnon energy is comparable or smaller than the thermal energy: $\hbar \omega < k_B T$, the Stokes line becomes more intense, as follows from Bose–Einstein statistics. Bose–Einstein condensation of magnons was proven recently in an antiferromagnet at low temperatures by Nikuni et al. and in a ferrimagnet by Demokritov et al. at room temperature. See the news report by Schewe and Stein and the articles by Nikuni et al. and Demokritov et al. for more details.


text:
# Interpreting ANOVA results [duplicate]

I am having difficulty interpreting the results of a colleague. I don't know much about ANOVA other than it is a regression from classifications to floating point numbers.

From reading Gotelli's Ecological Statistics book, I have figured out more or less that the F-ratio is a step in getting the P value. The P value indicates the chance of obtaining this data if the null hypothesis were true.

I am stuck on matching that explanation with the data he provided me, shown below. I assume the Pr(>F) is the P value, but what is the hypothesis tested? Is it clear from just this?

fit <- with(divergence_slopes, aov(slopes~init_diversity + seed_population + init_diversity*seed_population))

summary(fit)
Df   Sum Sq   Mean Sq F value   Pr(>F)
init_diversity                   2 6.29e-14 3.146e-14   8.802 0.000228 ***
seed_population                  1 7.93e-14 7.925e-14  22.171 5.08e-06 ***
init_diversity:seed_population   2 8.34e-14 4.171e-14  11.667 1.76e-05 ***
Residuals                      174 6.22e-13 3.570e-15
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

• As y ~ x1 * x2 expands to y ~ x1 +x2 + x1:x2 the model could at least be simplified... – Rasmus Bååth Apr 2 '13 at 11:59
• "The P value indicates the chance of obtaining this data if the null hypothesis were true." --- no, it doesn't. It's the probability of obtaining a result at least as extreme as this one if the null were true. – Glen_b -Reinstate Monica Apr 2 '13 at 13:19
• Many very basic questions like this have been asked on the site before - this one was asked: Anova from R output interpretation and probably in several other places. Those who ask (and answer) such questions do the site a service by checking on that before posting, rather than allowing us to compile several answers to the same question in various places. – Macro Apr 2 '13 at 19:25

The null hypothesis being tested by the full model is that slopes are not related to init_diversity. seed_population, or the interaction. But that doesn't seem to be in what you've shown. You have individual F-values (and associated p-values) for each of three null hypotheses: 1. That there is no relation between slopes and init_diversity, after controlling for seed_population and the interaction

2, 3, similar, but for seed_pop and the interaction.

All three null hypotheses are rejected (p < 0.0001)

what is the hypothesis tested? Is it clear from just this?

There are three hypotheses being tested.

[Each p-value is effectively for that effect "fitted last", though this looks like it's probably a balanced ANOVA in which case the order makes no difference to the sum of squares (and hence to the p-values).]

Let's take some simpler cases first.

Imagine that only seed_population was considered to have any impact on the response; then the model would have been slopes ~ seed_population and the null hypothesis on seed_population (which I assume is a factor) would be that the population means for the two groups were equal.

Now, consider if you had a model like slopes~init_diversity (I realize this is probably not a very interesting model, but bear with me). This has 2 d.f., so it corresponds to three levels. The null hypothesis would be that the population mean of the 'slopes' variable for all three levels would be equal.

Now if you had say a main-effects model: slopes~init_diversity + seed+population
you'd be testing both those hypotheses I mentioned - each would get its own row of the ANOVA table and its own p-value.

Now to deal with your actual model. To those two terms we now add an interaction term between the two factors. There's now the additional null hypothesis that the population means for the interaction terms are both zero (which also corresponds to a particular linear combination of cell population means being zero). This hypothesis gets its own line (and its own p-value) in the table.

--

It's not common to formally test the main effects on finding a significant interaction.

Usually, you'd only concern yourself with the the interaction when it's significant - that alone is enough to tell you both variables are important in explaining variation in $y$ and that they interact.

• that's a good point; more generally. I notice someone downvoted my answer as well. But neither of us can tell why. – Peter Flom - Reinstate Monica Apr 2 '13 at 23:56
• @PeterFlom Likely the same person who downvoted the original question. I thought your answer was good and voted it up before. I don't mind downvotes - that's part of how we arrive at sorting out reputations after all - but I do want to understand what the problem is and try to fix it; downvotes are an opportunity to give a better answer, but it only works if we know what the issue is. – Glen_b -Reinstate Monica Apr 3 '13 at 0:16
• I thought it was obvious that I downvoted. The main reason was stated in my comment to the main question. It does no one any good to have the same introductory questions asked, and answered, again and again on this site. You two are veteran members of this site and know that answering such questions is a waste of time and undermines site policy against question duplication, yet you insist on doing it (for points I guess?). I didn't place a comment here before because it seemed unconstructive - I don't expect you two to stop answering duplicates as a result. Don't expect me to stop downvoting. – Macro Apr 3 '13 at 1:15
• Based purely on their merits as answers, I would downvote both regardless of anything else - glen_b, this answer is a comment, if anything, to the main question. I don't see how this answers the OP's question, in any way. Regarding @Peter's answer - "That there is no relation between slopes and init_diversity, after controlling for seed_population and the interaction" - no, that is not the null hypothesis that is being tested. – Macro Apr 3 '13 at 1:18
• @Macro Thanks for the explanation. My reasons for responding were simply that the question had been asked; if I'd seen a suitably similar one in the first few results returned I'd have pointed to it. Your complaint that my answer is more suitable as a comment ... yes, that's a reasonable criticism of it as an answer. I appreciate you taking the time. – Glen_b -Reinstate Monica Apr 3 '13 at 1:36


text:
# Finitely many embeddings of a finite extension in an algebraic closure

So I'm reading through Lang's Algebra, and he keeps saying something along the following lines:

"Let $K$ be a finite extension of a field $k$ and let $\sigma_1,\ldots,\sigma_r$ be the distinct embeddings of $K$ in an algebraic closure $k^\text{a}$ of $k$."

But the thing is that it doesn't seem clear to me that there need be only finitely many distinct embeddings, and I can't find anywhere where he proves it or figure out why it should be true or find its proof anywhere else. I feel like it should be obvious but it's not. So can anyone explain this to me?

EDIT: This all certainly makes more sense to me now. The only problem is that all of the answers make use of the minimal polynomial, which (as far as I know) only applies to algebraic finite extensions. I'm willing to believe that Lang is talking about algebraic extensions, but I do wonder if this is true of finite extensions in general.

EDIT 2: Nevermind Edit 1, I forgot what an algebraic closure is.

-
Re your edit: Every finite extension is algebraic. –  Prism Aug 12 '14 at 5:46
Following up on @Prism's comment, if $K$ is a finite extension of $k$, then by definition it is an $n$-dimensional $k$-vector space for some finite $n$. Given any $a \in K$, what can you say about the elements $1,a,a^2,\dots,a^n$? –  Morgan O Aug 12 '14 at 5:51
Oh haha I forgot that algebraic closures have to be algebraic –  desi Aug 12 '14 at 6:50

## 3 Answers

The key is that if you have a finite extension $K/F$, adding elements to $F$ will always raise the degree of the extension by some factor, so you get a tower of simple extensions $F \subseteq F(\alpha_1) \subseteq F(\alpha_1)(\alpha_2) \subseteq \cdots$. By finiteness of $K/F$, this process has to stop. Then you should recall the following :

If $\varphi : F \to F'$ is an isomorphism of fields, $f(X) \in F[X]$ and $f'(X) \overset{def}=\varphi(f)(X) \in F'[X]$ obtained by applying $\varphi$ on the coefficients of $f$, take $\alpha$ to be a root of $F$ and $\beta$ to be a root of $F'$ (in some extensions). Then there exists a unique isomorphism $\psi : F(\alpha) \to F'(\beta)$ such that $\psi|_F = \varphi$.

This will allow you to do the inductive argument. The main idea is that if you have a map $\psi : F(\alpha) \to K$ where $K$ is a field containing (an isomorphic copy of) $F$ and $\psi$ maps $F$ to this copy, then $\psi$ is entirely determined by where $\alpha$ maps to, and $\alpha$ must map to a root of its minimal polynomial, so it has finitely many options. I leave the details to you.

Hope that helps,

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If $K = k(a_1,\dots,a_n)$ is a finite field extension, then an embedding $\sigma: K \rightarrow k^{a}$ must send each $a_i$ to another root of its minimal polynomial over $k$. (Indeed, if $p(a_i)=0$ for $p \in k[x]$, then $p(\sigma(a_i))=0$. ) This shows there are only finitely many possibilities for the image of each $a_i$ under $\sigma$, and these determine $\sigma$. Hence there are only finitely many embeddings $K \hookrightarrow k^a$ over $k$.

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Try proving it for simple extensions of the form $k(a)/k$ first, then induct.

Hint: $k(a)\cong k[x]/(f)$ for a minpoly $f$, and embeddings $\hookrightarrow\overline{k}$ correspond to roots of $f(x)$.

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# If x€R then x^2 +2x +a/x^2 +4x+ 3a can take all real values if ? Answer is a€[0,1] Exerts I've already asked this and exert gave me link to.similar questions. But pls solve it. Pls solve it in the way that we take denominator to other side and multiply by y then write the new formed quadratic equation and then take D >or=0 and again get a quadratic equation and again D>or =0 Pls solve using this method Thanks

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