text:
## Monday, 25 April 2011

### It's official I've got Asperger's

Courtesy of Nilo it would appear according to this test I've got Asperger's syndrome

http://www.piepalace.ca/blog/asperger-test-aq-test/

I got 34 definitely in the Asperger's zone according to the test. A lot of my colleagues, joke at work that they have Asperger's, indeed one couldn't achieve anything in our work if we didn't have the ability to work on our own, and put up with a lot of boring tedious and repetive tasks and pay attention to detail. On a more general point if one wants to achieve anything intellectual, then one needs to spend a lot of time on one's own trying out ideas and  reading. If one has a full time job then it stands to reason that juggling the demands of OU study means that one's social life is bound to take a back burner.

As I commented on Nilo's blog it seems to me that the rise of this syndrome is partly due to the lack of appreciation of a person's intellectual abilities. Also it  reflects societies fear of people who are quite happy on their own and have an intellectual skill which they do not. It's interesting that it's scientist's mathematicians and engineers that are usually accused of having Asperger's syndrome. Whilst artist's, composer's and writer's who have the same lack of social skill's as mathematicians scientists and engineers are not considered Aspergic.

Our society seems to place a high value on superficial beauty, or wealth and a low value on things that really matter. Labelling people as Aspergic simply because they do not fit in with the norms of society does not help. As a final point muddying the waters between Asperger's and Autism will not help find a cure for Autism.

## Tuesday, 19 April 2011

### M208 M346 M337 I saw one go out and another Two come back in

Today 1 day late, I posted my off my first TMA for M346 linear statistical modelling. I felt it was fairly straightforward essentially being exercises in interpretation of output from GENSTAT and a review of some statistical tests. Essentially revision there were one or two points I hadn't come across before. So quietly confident of a reasonable grade although one can never be sure whether I've missed out something fundamental.

So thats one out. Also on my door step I found my returned TMA for complex variables slightly disappointed as I didn't make distinction due to some careless mistakes also some times I was quite rigorous in my arguments other times I missed out key justifications thus omitting to mention that a sequence tended to zero as it was a null sequence just stating that the sequence tended to zero.  I think it's a question of learning how to phrase arguments in analysis properly. Some times I'm guilty of making to many points other times I miss out key steps. Still one can only go on.

In contrast I seemed to have done really well with the TMA for M208 which I received earlier this week . I wonder if it will continue when I get to the analysis sections. Hopefully some of the lesson's I'm learning from M337 will help here.

As a final point talking with one of my mates Neil at the tutorial

http://neilanderson.freehostia.com/thoughts/degree/m208/

We both worried that in the chase to get good marks for the TMA's we were skimping on details and understanding. Still need's must Alan my tutor himself said that as far as the  first two units of linear algebra were concerned we should just concentrate on whether we have learnt the basic techniques and then concentrate on the more abstract stuff in the latter part of the block.

I think it's a question of concentrating on what you feel you need to know in detail for me this is group theory and analysis as far as M208 is concerned specifically learning how to deal with the dreaded

$$\epsilon - \delta$$ definitions of continuity and it's applications to pathological functions such as the so called blancmage function which is an iterated sequence of triangles. This function is every where continuous and no where differentiable. I'll speak more about this on another post.  Also most of M337 even though it is currently the most challenging part of pure mathematics I have tackled so far. Still as I said to Neil we are not here for an easy ride.

This means that M346 will have to take a back burner and I will be fairly pragmatic about the work I need to do for this course which means essentially working backwards from the TMA questions.

## Tuesday, 12 April 2011

### What do I want ??

A little time to reflect from the pressures of TMA deadlines made me think what is it I really want to achieve having taken the libertry of telling someone on the  M337 forum what makes a good PhD it's a question of 'physician heal thyself'. If one wants to achieve anything in life it's a question of gettting enough background knowledge to articulate a certiain insiight which no one else has thought of, but on the other hand not getting so bogged down in a lack of confidence of one's own abilities. Obviously this meams that one has to have a certain degree of self confidence, which I lack and also a degree of realism, as to what can be achieved, Thus it is highly unlikely that I am going to solve the Riemann hypothesis, or any of the Seven Millennia problems. It is highly unlikely that I am goiing to come up with a Grand unified theory of everyhiing and it is highlly unlikely that I am going to solve the problem of a coherent quantisiation of Einstein's theory of relativity. Given that better people have been there before me and have tried to solve the problem and have not achieved anything lasting.

Ok so what is it I think I can at least aritculate, it is, it is whether or not there is  a  connection between the philosophy of language and the rejection of bivalence as enshrined with Michael Dummet ideas
and the fact that the mathematical stucture of quantum mechanics seems to imply a similar rejection of bivalence as enshrined in the Kochen Specker paradox. Both would seem to imply a rejection of realism but how far does this go ?

See for example Dummett

http://www.iep.utm.edu/dummett/

http://en.wikipedia.org/wiki/Kochen%E2%80%93Specker_theorem

For a current interpreation of quantum mechanics which tries to circumvent this dilemma see Topos theory the simplest introduction to which I know of can be accessed here.

http://topos-physics.org/

It seems to me at least articulating what the possible connections between the three  issues namely whether

1) Does Micheal Dummet's assetion that realist theories of epsitemology or metaphysics are commited to bivalence hold water

2) Does the fact that quantum mechancs seems to imply an adherence to non realism via the Kochen Specker paradox spell the death knell for realist interpreations of quantum physics such as Bohm's intepretation or even worse the Many worlds interpretation ( I hope so).

3) How far do theories such as Topos theory circumvent the problem of realism and the implications of the Kochen Specker paradox . It seems to me that they go some way but not far enough.

would be worthy of an MSc by research in philosophy as I've not seen any full lengrh account of the issue, but it seems to be mentioned in passing in a number of papers.

So why don't I take a breather and write a research proposal for the Logic and language department of philosophy at Ediunbrgh University ?

http://www.ppls.ed.ac.uk/philosophy/groups/logic-language-edinburgh

and see whether they would take me on part time, and then get accepted by say Octobet this year or certainly October 2012 ? Instead of getting bogged down both time wise and financially in pursuing formal qualifications in mathematics and philosophy via the Open university or other sources. After all I do not need a knowledge of complex analyisis, Kant's critiique of pure reason or especially Heidegger's Being and Time or Nietszche's Genealogy of morals interesting as it would be to pursue these issues in some depth, to be able to pursue the question as  to whether there is a link between Dummet's rejection of Bivalence and realism and wheher or not quantum mechanics in it's present form implies the rejection of bivalence and as a consequence  lends some support to Dummet's ideas.  I do not need say for example the course in functional analysis offered by the OU MSc especially as it explicitly states that it does not deal with any concrete applications of the subject eg to say quantum physics.   I feel I can get such knowledge from say Kreyszig or even tackle Von Neumann directly now that being embroiled in M337 and M208 has given me a degree of confidence in handling pure mathematical concepts which I didn't have before.

Clearly I need to invesitgate these issues further the question is given time and finances whether or nor I can do this in a reasonable amount of time and financial cost ? Given that fee's are going to skyrocket over the next few years then it would seem reasonable to cut one's losses and just go for it.

On the other hand  I am handicapped by a lack of self confidence in putting this forward given the as yet lack of formal qualifications in both pure  Maths and philosophy. On the other hand waiting 5 years or so before I can gain  the qualifications to get me there isn't going to help me either. So as a compromse I wait a couple of years whilst I finish the degree in maths and my open degree in social sciences the Arts and philosophy supplemented by Geoffrey Klempner's pathway courses and his associateship. He is an expert in Dummet so at least from the philosophical side I should get some idea as to whether or not my ideas are on the right track or whether or niot I'm just talking rubbish.

http://www.philosophypathways.com/programs/pak2.html

After that just go for it rather than handicap myself with yet another financial and time burden just chasing after qualifications which whilst definitely worth it distract one from achieving one's real  goals. So yes I will complete my current courses and do the other pure maths course to complete the degree and also My open degree with the philosophy of mind course and Geoffrey Klempner;s pathways modules but then instead of spending circa £12000 on getting the MSc in maths and a correspoinding MA in continental philosophy spending what will be £6,000 on a parttime MSc by research (assuming I get accepted) will save me a lot of money and time to achieve the goal I really want.  So I'll finish the two degrees I'm aiming for but not bother with the MSc's and take my chances, sooner or later you have to break away, at least I am aware of what I should know and also what I want to explore.

## Sunday, 10 April 2011

### Amazing

Wow and I don't lknow how I did it but MathJAX now seems to work just as in the Forums of Moodle you have to mark off your LATEX input with two dollar signs so

$x^2$ just produces the crap you see here whereas enclosing it with two dolllar signs

where as $$x^2$$ now seems to work

Have amended my group theory post of March 6th to take into account the New LATEX facility.
Apologies if the blog takes longer to load that's the price you pay for sophistication.

### Math JAX

As some of you will have seen I've tried to incorporate math JAX on this blog however I can't get it to work
thus what should appear as a nice  x squared term comes out simply as $x^2$ or even $$x^2$$

Post 11/04/11 In light of the partial success I removed the rather defunct script but I stiil hate Computers !!

For reasons only known to the guru's of cypberspace Whilst Math JAX appears to work for the visible pages it doesn't seem to be able to update old posts even though when I preview the amendments it seems to incorporate the LATEX. As a result my old post on group theory is even more incomprehensible than ever.

On a brighter note finished both TMA's for M337 and M208, but still need to make inroads into the statistics. I shall have to take at least one day off next week.

Best wishes Chris

## Monday, 4 April 2011

### Some times you just have to cut your losses M337 TMA01

Well my somewhat disjointed attempt at the first TMA for M337 has been completed or at least what I've been able to do. I really found this a bit of a struggle and I feel like I'm walking through treacle. Sometimes what to do is quite straightforward other times it really is a struggle. For example we had to show that a function is conformal this is defined vaguely as a function which is angle preserving. One way of showing that a function is conformal is to calculate it's derivative and show that it is non zero and analytic at a point. OK but the question did not give a specific point neither did the course material give any general criteria for showing that a function is conformal.

Another really confusing point for me and this seemed to infect a large number of the questions was the domain of Log(z) which is continuous every where apart from the negative real axis. Ok fine but again there was a lot of ambiguity in specific questions.

I really will have to go over these blocks again. In the mean time I should have done enough to get a grade 2 pass for this TMA so as the deadline is Wednesday I just have to cut my losses and put it in the post first thing tomorrow.

In contrast M208 so far is proving a real joy and light relief to M337, No doubt in a years time (fingers crossed) I'll look back at M337 and wonder what the fuss is about. It should get better as we progress as the questions appear to be more calculational (which I prefer) rather than sketch this set, sketch that set, specify the domain of this function. Why doesn't this function have an inverse whereas another one has and What is the geometric effect of this transformation and so forth all of which there seems to be an undue emphasis on in the early part of M337.

Finish off writing the Group theory TMA for M208 tomorrow, then a few intensive days whilst I try and complete the first TMA for Statistics course. At least the statistics should be straightforward if tedious.
I'll let you know.


text:
CF758A Holiday Of Equality

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题目大意

今天是Berland的平等节。为了庆祝节日，国王决定用国库里的钱来使Berland所有公民的福利均等。

在Berland，有$N$个公民，每个人的福利都被估计为一个整数$a_{i}$ burle（burle是Berland的货币）。

你是皇家财政管理员，你需要计算国王为此要付出的最低费用。国王只能给公民们钱，他没有权力夺走任何钱财。

输入输出格式

输入格式

输入数据的第一行有一个整数$n$，表示Berland中公民的数量。

数据的第二行包括$n$个整数$a_{1},a_{2},...,a_{n}$，表示每个公民原有的福利。

输出格式

输出仅包括一个整数$S$，表示国王最少需要花费$S$ burle。

样例说明

第一个样例中，如果我们给第一个公民4 burles，第二个3 burles，第三个2 burles，第四个1 burle，那么所有公民的福利都等于4 burles。

第二个样例中，我们只需要给第三个公民1 burle就可以使得福利均等。

第三个样例中，我们需要分别给第一个和第三个公民2 burles，使得福利都等于3 burles。

第四个样例中，我们不需要给任何人钱因为所有人的福利都是12 burles。

数据范围

对于100%的数据，$1 \leq n \leq 100$，$0 \leq a_{i} \leq 10^{6}$。

翻译提供者：Simpson561

题目描述

In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.

Totally in Berland there are $n$ citizens, the welfare of each of them is estimated as the integer in $a_{i}$ burles (burle is the currency in Berland).

You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.

输入输出格式

输入格式：

The first line contains the integer $n$ ( $1<=n<=100$ ) — the number of citizens in the kingdom.

The second line contains $n$ integers $a_{1},a_{2},...,a_{n}$ , where $a_{i}$ ( $0<=a_{i}<=10^{6}$ ) — the welfare of the $i$ -th citizen.

输出格式：

In the only line print the integer $S$ — the minimum number of burles which are had to spend.

输入输出样例

输入样例#1： 复制
5
0 1 2 3 4

输出样例#1： 复制
10
输入样例#2： 复制
5
1 1 0 1 1

输出样例#2： 复制
1
输入样例#3： 复制
3
1 3 1

输出样例#3： 复制
4
输入样例#4： 复制
1
12

输出样例#4： 复制
0

说明

In the first example if we add to the first citizen $4$ burles, to the second $3$ , to the third $2$ and to the fourth $1$ , then the welfare of all citizens will equal $4$ .

In the second example it is enough to give one burle to the third citizen.

In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal $3$ .

In the fourth example it is possible to give nothing to everyone because all citizens have $12$ burles.

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# Existance of concave balanced function

by Eval   Last Updated May 15, 2019 18:20 PM

We know that the core of a balanced game is non-empty. The convexity ensures balancedness. However, I was wondering if a concave function too satisfies balancedness condition.

The balancedness is defined according to Bondareva–Shapley theorem as follows: Let the pair $$(N, v)$$ be a cooperative game in characteristic function form, where $$N$$ is the set of players and where the value function $$v:2^{N} \to \mathbb {R}$$ is defined on $$N$$'s power set (the set of all subsets of $$N$$).

The core of $$(N, v)$$ is non-empty if and only if for every function $$\alpha :2^{N \setminus \emptyset} \to [0,1]$$, where

$$\forall i\in N:\sum _{S\in 2^{N}:\;i\in S}\alpha (S)=1$$ the following condition holds:

$$\sum_{S\in 2^{N\setminus \emptyset}}\alpha (S)v(S)\leq v(N)$$

Tags :


text:
NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...

## JEE Main 2019 (Online) 10th January Morning Slot

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The total number of isotopes of hydrogen and number of radioactive isotopes amon...
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A process has $$\Delta$$H = 200 J mol–1 and $$\Delta$$S = 40 JK–1 m...
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The metal used for marking X–ray tube window is :
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text:
CGAL 5.0 - 2D and 3D Linear Geometry Kernel
CGAL::has_larger_distance_to_point()
CGAL::compare_distance_to_point()
CGAL::compare_signed_distance_to_line()
CGAL::compare_signed_distance_to_plane()
CGAL::has_larger_signed_distance_to_line()
CGAL::has_larger_signed_distance_to_plane()
CGAL::has_smaller_distance_to_point()
CGAL::has_smaller_signed_distance_to_line()
CGAL::has_smaller_signed_distance_to_plane()

## Functions

template<typename Kernel >
bool CGAL::has_larger_distance_to_point (const CGAL::Point_2< Kernel > &p, const CGAL::Point_2< Kernel > &q, const CGAL::Point_2< Kernel > &r)
returns true iff the distance between q and p is larger than the distance between r and p.

template<typename Kernel >
bool CGAL::has_larger_distance_to_point (const CGAL::Point_3< Kernel > &p, const CGAL::Point_3< Kernel > &q, const CGAL::Point_3< Kernel > &r)
returns true iff the distance between q and p is larger than the distance between r and p.


text:
Given a graph, suppose we have calculated shortest path from a source to all other vertices. If we modify the graph such that weights of all edges is becomes double of the original weight, then the shortest path remains same only the total weight of path changes.

The shortest path remains same


text:
## Square root of decimal number :[Procedure ,examples and practice problem}

In this post, we’re going to learn to “find the square root of decimal numbers using the long division method”.So, till now we have learned to find square root using factorization,one tens method, and long division method. If you didn’t read these stuff, check now  Today we’ll learn the complete concept of finding the square … Read more

## Finding Square root using long division method

In this post, we’re going to solve problems of the perfect square using the long division method. By the way, if you don’t know the basic procedure for finding square root by long division method. Check out this article: Square root using long division method  Q.1 Find the square root of 6084 using the long … Read more


text:
# How do you simplify ((5m)/(3z^3))^2 and write it using only positive exponents?

##### 1 Answer
Sep 4, 2016

$\frac{25 {m}^{2}}{9 {z}^{6}}$

#### Explanation:

Each factor in the bracket must be squared.

${\left(\frac{5 m}{3 {z}^{3}}\right)}^{2} = \frac{5 \times 5 \times m \times m}{3 \times 3 \times {z}^{3} \times {z}^{3}}$

$\frac{25 {m}^{2}}{9 {z}^{6}}$


text:
Kill Request¶

This extension may be referenced by the qualified name envoy.filters.http.kill_request

Note

This extension is intended to be robust against untrusted downstream traffic. It assumes that the upstream is trusted.

Kill Request configuration overview.

extensions.filters.http.kill_request.v3.KillRequest¶

[extensions.filters.http.kill_request.v3.KillRequest proto]

Configuration for KillRequest filter.

{
"probability": "{...}",
"kill_request_header": "..."
}

probability

(type.v3.FractionalPercent) The probability that a Kill request will be triggered.

kill_request_header

(string) The name of the kill request header. If this field is not empty, it will override the default header name. Otherwise the default header name will be used.


text:
## Loren on the Art of MATLABTurn ideas into MATLAB

Note

Loren on the Art of MATLAB has been archived and will not be updated.

# Creating C++ Shared Libraries and DLLs

Guest blogger Peter Webb returns with another in an occasional series of postings about application deployment.

### Contents

A previous post demonstrated how to use MATLAB Compiler to create standalone executables. In this article, I'll demonstrate how to use MATLAB Compiler to create C and C++ shared libraries or DLLs.

You create a shared library or DLL to add MATLAB-based functionality to an application you're developing. You select one or more MATLAB-based functions to include in the shared library, and MATLAB Compiler generates a binary file that you link against your application. Shared libraries generated by MATLAB Compiler are compatible with the Microsoft Visual Studio development environment on Windows and with the GNU C/C++ tool suite on most UNIX platforms. The list of supported compilers changes from one release to the next; be sure to check that list before starting your project.

### Building a Shared Library

To illustrate the process of creating and using shared libraries, I'll use a cryptographic algorithm, the Vigenere cipher. The program I've written consists of two parts:

• libvingenere: A shared library containing two MATLAB functions encrypt and decrypt.
• vigenere.cpp: A C++ main program that calls the functions in libvigenere.

The source code for the MATLAB functions and the main program is available on MATLAB Central. The download package also includes VigenereDetails.html, which describes the implementation of the Vigenere cipher in MATLAB.

The mcc command invokes MATLAB Compiler both from within interactive MATLAB and at the system (DOS or UNIX shell) prompt. Use mcc to build a shared library from the MATLAB encrypt and decrypt MATLAB functions:

 mcc -v -W cpplib:libvigenere -T link:lib encrypt decrypt

This mcc command breaks down into four parts:

• -v: Turn on verbose output.
• -W cpplib:libvigenere: Generate C++ wrapper code. Name the generated library libvigenere.
• -T link:lib: Invoke a C/C++ compiler to create a platform-specific binary shared library file from the generated code.
• encrypt decrypt: Place the encrypt and decrypt functions in the shared library. Generate C++ wrapper functions for each.

This command generates several files. Two of them are relevant here:

• libvigenere.dll: The platform-specific binary: the shared library itself. On most Unix systems, this file ends with a .so extension: libvigenere.so.
• libvigenere.h: Declarations of the C++ wrapper functions and C++ type conversion classes and utilities.

### The Generated Interface

MATLAB Compiler generates many different kinds of functions: initialization, termination, error and print handling and, of course, the functions you selected to compile into the library.

In our example, MATLAB Compiler generates a C++ entry point for encrypt and decrypt. Stripped of some bookkeeping annotation, these functions look very much like the corresponding MATLAB functions.

void encrypt(int nargout, mwArray& ciphertext,
const mwArray& plaintext, const mwArray& key);
void decrypt(int nargout, mwArray& plaintext,
const mwArray& ciphertext, const mwArray& key);

The generated functions differ from the MATLAB functions in two significant ways:

• The C++ functions explicitly declare the types of their arguments.
• The C++ functions return void, passing back results in output parameters provided in the argument list.

Let's look at the encrypt function in detail. For comparison, here's encrypt's MATLAB function signature:

 function ciphertext = encrypt(plaintext, key)

The encrypt MATLAB function has one output and two inputs. The encrypt C++ function has zero outputs (the void return type) and four inputs. The first C++ input indicates the number of outputs requested by the caller. As indicated by the MATLAB function, the number of outputs may be zero or one. Passing any other value will result in an error. The second C++ input is the output argument, passed by reference. encrypt will overwrite any data in this argument with the encrypted message. The third and fourth C++ input arguments are the function inputs, the plaintext and the encryption key. They are passed by constant reference which means the encrypt function cannot change their contents.

Unlike MATLAB, C++ requires all variables have declared, immutable types. Variables in MATLAB have types as well (matrix, cell array, structure, etc.), but the type of a MATLAB variable can change at any time. To accommodate this dynamic behavior in C++, the MATLAB Compiler provides the mwArray data type. All the functions generated by the MATLAB Compiler take mwArray inputs and return mwArray outputs. The mwArray API allows you to create mwArray objects from most C++ data types and to extract native C+++ data from a returned mwArray.

### Calling Functions in a Shared Library

Now that I've created a shared library, I need to write a program to call the library's public functions. The program performs six tasks:

• Includes the shared library's header file.
• Parses command line arguments.
• Initializes the MATLAB Compiler Runtime's global state and the shared library's state.
• Creates input arguments; convert native C++ data types to MATLAB data types.
• Invokes one or more functions from the shared library.
• Shuts down the library and the MATLAB Compiler Runtime.
• Below, I demonstrate how these steps translate into the code of the example's main program, vigenere.cpp.

Step 1: All programs that use a MATLAB Compiler-generated shared library must include the library's header file. The header file has the same base name as the compiled library, libvigenere in this case, and a .h extension.

// vingenere.cpp: Encrypt and decrypt using the Vigenere cipher.
#include "libvigenere.h"
#include <iostream>

Step 2: The main program parses the command line. The first argument, a switch, determines the type of action: -e means encrypt, -d decrypt. The second argument is the message, and the third, the key.

int main(int ac, const char *av[])
{
// Encrypt or decrypt? Determined by command line switch
bool enc = strcmp(av[1], "-e") == 0;

Step 3: Initialize the runtime and start the library before calling any functions exported from the runtime or the library. Failure to initialize will cause your program to crash. Always check for success (the initializers return false if they fail) and issue error messages as necessary.

    // Initialize the MATLAB Compiler Runtime global state
if (!mclInitializeApplication(NULL,0))
{
std::cerr << "Could not initialize the application properly."
<< std::endl;
return -1;
}
    // Initialize the Vigenere library
if( !libvigenereInitialize() )
{
std::cerr << "Could not initialize the library properly."
<< std::endl;
return -1;
}

Step 4: Convert the C++ strings from the command line (the message and the key) into MATLAB strings by creating mwArray objects. These declarations cannot appear before the initialization calls in Step 3.

    // Must declare all MATLAB data types after initializing the
// application and the library, or their constructors will fail.
mwArray text(av[2]);
mwArray key(av[3]);
mwArray result;

Step 5: Invoke the exported functions. Encrypt or decrypt as indicated by the command line switch. Note that the C++ functions have the same name as their MATLAB counterparts, and that all return values must be passed in by reference. The mwArray class defines

    // Initialization succeeded. Encrypt or decrypt.
    if (enc == true)
{
// Encrypt the plaintext text with the key.
// Request one output, pass in two inputs
encrypt(1, result, text, key);
}
else
{
// Decrypt the ciphertext text with the key.
// Request one output, pass in two inputs
decrypt(1, result, text, key);
}
std::cout << result << std::endl;

Step 6: Shut down the library and the runtime, in the opposite order of initialization (library first, then runtime).

    // Shut down the library and the application global state.
libvigenereTerminate();
mclTerminateApplication();
}

### Creating and Running the Application

MATLAB Compiler uses the mbuild function to compile the code it generates into a shared library. mbuild knows how to invoke the C/C++ compiler with the correct switches so the compiler can find the required include files and libraries. You can use mbuild to create your own executables and link them with MATLAB Compiler-generated shared libraries. On Windows, for example, issue this command:

mbuild vigenere.cpp libvigenere.lib

On UNIX, you link against a .so file instead of a .lib file:

mbuild vigenere.cpp libvigenere.so

In both cases, mbuild produces an executable called vigenere.

Encrypt the message with the secret key.

>> !vigenere -e "Algebra, the Music of the Reason" "MATLAB"
MLZPBSM LSEAYUKTCA FSDHFLRXLSPZ

The first argument is the message, the second argument the key. Note the leading exclamation point -- this command runs the vigenere executable from within MATLAB, using the system command. You can run these commands on the UNIX or DOS command line as well, but you have to configure your environment correctly.

Decrypt to verify the encryption worked:

>> !vigenere -d "MLZPBSM LSEAYUKTCA FSDHFLRXLSPZ" "MATLAB"
ALGEBRA THE MUSIC OF THE REASON

Because the alphabets in the Vigenere square only include letters and the space character, the decrypted message lacks punctuation marks. Extending the algorithm to handle punctuation is an exercise left to the reader.

And extra points if you can figure out why I chose that phrase (aside from its poetic merit, which is in the eye of the beholder).

### Next: Using Multiple Shared Libraries

I've shown you how to combine several MATLAB-based functions into a single shared library, and how to incorporate that library into a host application. In a later posting, I'll demonstrate how to use multiple shared libraries from a single host application.

What else can I tell you about shared libraries? Let me know here.

Published with MATLAB® 7.11

|


text:
# Prism : what if incident angle is greater than apex angle?

1. Oct 6, 2014

• Warning: Problem statement should include a description of the problem; Solution attempt must be included.
1. The problem statement, all variables and given/known data
incident angle : 60 degree
apex angle : 30 degree
prism refraction index : 1,6
deviation angle : ?

2. Relevant equations

3. The attempt at a solution

2. Oct 6, 2014

### ehild

You mean the angle of the prism? Why do you think it makes a problem if the angle of incidence is greater than that? What is the question?

ehild

3. Oct 6, 2014

if i solve this equation to find first refraction angle (r1) :
sin(I1) = np sin (r1), which (I1) is incident angle =60 degree, and np is prism refraction index = 1,6
i find r1 = 32,76 degree.
if beta = r1 + I2 , which beta is apex angle = 30 degree, and I2 is second incident angle,
i find I2 = -2,76 degree
i cant quite understand whats the meaning of (-) in my answer :/

4. Oct 6, 2014

### ehild

It means that the second angle of incidence falls to the opposite side of the normal. See picture. You still can use $|I_2|=|\beta-r_1|$. The second refracted ray will travel upward, in the direction of the apex.
I suggest to draw the rays always before applying a formula.

ehild

5. Oct 8, 2014

so i can still use
Deviation angle = i1+r2-beta to find second refraction angle? (r2)
thank you so much that enlightens me so much about prism.
many thanks man

6. Oct 8, 2014

### ehild

You find r2 by applying Snell's Law for I2. And I1 and r2 and beta determine the deviation angle.

ehild

7. Oct 8, 2014


text:
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# Common Core Standard HSN-RN.A.1 Questions

Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents. For example, we define 51/3 to be the cube root of 5 because we want (51/3)3 = 5(1/3)3 to hold, so (51/3)3 must equal 5.

You can create printable tests and worksheets from these questions on Common Core standard HSN-RN.A.1! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page.

Which of the following correctly explains why $(root(6){10})^6 = 10$ ? Choose all correct answers.
1. The 6's cancel out, leaving just 10.
2. $root(6){10}$ can be rewritten as $10^{1/6}$, and then exponent laws can be applied (namely, $(10^{1/6})^6 = 10^{6/6} = 10^{1}=10$).
3. If the left hand side is rewritten as $root(6){10^{6}}$, then the radical can be distributed to the 10 and the 6, and the rounded answer is about 10.
4. The 6th root of 10 is asking for the number that, if multiplied by itself 6 times, would equal 10. Therefore, this number raised to the power of 6, is equal to 10.
Starting with the variable x, raise x to the power of 4. Which of the following represents an inverse operation? Choose all correct answers.
1. Dividing the expression by 4.
2. Taking the base-4 logarithm of the expression.
3. Taking the 4th root.
4. Raising the entire expression to the power of 1/4.
Which of the following is a correct way to exactly rewrite the expression $2^{5/7}$? Choose all that apply.
1. $1.6$
2. $root(7)(2^5)$
3. $(root(7)(2))^5$
4. $2^(7/5)$
Which of the following is a correct way to rewrite the expression $4^(-5/3)$ ? Choose all that apply.
1. $root(3)(4^(-5))$
2. $-root(3)(4^5)$
3. $1 / (root(3)(4^5))$
4. $(root(3)(4))^(-5)$
Which of the following is a reason why, when dealing with an equation or expression, one would switch from using radical notation (square roots, cube roots, etc), to rational exponents? Choose all correct answers.
1. One can use exponent laws to simplify the expression or equation. For example, $root(3)(x^6) * sqrt(x^9)$ becomes $x^(6/3) * x^(9/2) = x^(2+9/2) = x^(4/2 + 9/2) = x^(13/2)$.
2. It avoids the issue of complex numbers. For example, $sqrt(-2)$ is complex, namely $2i$ where $i$ is the square root of -1, but changing this into a rational exponent, $-2^(1/2)$, it is no longer complex.
3. For an expression or equation with division, once the radical has been converted to a rational exponent, the factors can be more easily divided. For example, $sqrt(8x^7) / root(4)(4x^3)$ becomes $(8x^(7/2)) / (4x^(3/4)) = 2x^(7/2 - 3/4) = 2x^(14/4 - 3/4) = 2x^(11/4)$.
4. Using rational exponents can lead to easier forms for evaluation. For example, $root(3)(4^6)$ can be written as $4^(6/3) = 4^2$, which can then be easily evaluated.
Which of the following is a correct way to rewrite the expression $root(10)(3^5) ?$ Choose all that apply.
1. $3^(1/2)$
2. $3^(5/10)$
3. $(3^5)^(1/10)$
4. $(3^10)^(1/5)$
Which of the following is a correct way to rewrite the expression $root(3)(9^(-4)) ?$ Choose all that apply.
1. $(9^(-1/3))^(-4)$
2. $9^(4/3)$
3. $1/(9^(4/3))$
4. $(9^(-4))^(-3)$


text:
## Departmental PhD Thesis Exam – Justin Ko

Wednesday, August 25, 2020
11:00 a.m.

PhD Candidate:  Justin Ko
Supervisor:   Dmitry Panchenko
Thesis title: The Free Energy of Spherical Vector Spin Glasses

We study a class of vector spin models with configurations restricted to subsets of the sphere. We will prove a constrained free energy formula for these models. This formula defines a large deviations principle for the limiting distribution of the overlaps under the asymptotic Gibbs measure. The thesis builds on the mathematical results used to prove free energy formulas for the classical Sherrington–Kirkpatrick spin glass, spherical spin models, and vector spin glass models. The free energy formula proved in this thesis are true generalizations of the classical results, in the sense that these vector spin formulas restricted to one dimension coincide with the known results for classical models.

The first contribution of this thesis is a variational formula for contrained copies of classical spherical spin glasses sampled at different temperatures. The free energy for multiple systems of spherical spin glasses with constrained overlaps was first studied by Panchenko and Talagrand. They proved an upper bound of the constrained free energy using Guerra’s interpolation. In this thesis, we prove this upper bound is sharp. Our approach combines the ideas of the Aizenman–Sims–Starr scheme and the synchronization mechanism used in the vector spin models. We derive a vector version of the Aizenman–Sims–Starr scheme for spherical spin glass and use the synchronization property of arrays obeying the overlap-matrix form of the Ghirlanda–Guerra identities to prove the matching lower bound.

The second contribution of this thesis is the simplification of this variational formula to the form originally discovered for the classical spherical spin glass model by Crisanti and Sommers. In particular, we prove the analogue of the Crisanti–Sommers variational formula for spherical spin glasses with vector spins. This formula is derived from the discrete Parisi variational formula for the limit of the free energy of constrained copies of spherical spin glasses. In vector spin models, the variations of the functional order parameters must preserve the monotonicity of matrix paths which introduces a new challenge in contrast to the derivation of the classical Crisanti–Sommers formula.

A copy of the thesis can be found here: ut-thesis-Ko-updated


text:
# From Christoffel symbols to connection 1-forms

1. Mar 24, 2007

### cosmicstring1

I have Christoffel symbols for a metric and I want to find the connection 1-forms.

I have the relation:

w(^i j)=Chr(^i j k)*dx(^k)

w: conn. 1-form
Chr: Christoffel symbols

But Christoffel symbols do not share the symmetries of the conn. 1-forms. Do you know any way to make this possible?

2. Apr 11, 2007

### Chris Hillman

Have you studied the relevant chapter in Misner, Thorne, and Wheeler, Gravitation? Or the textbook by Harley Flanders, Differential forms, with applications to the physical sciences? These books offer clear introductions to computing the connection one-forms, as do many others.

3. Apr 12, 2007

### pmb_phy

I'm not aware of anything called the connection 1-forms. All I've heard are Chjristoffel symbols of the first and second kind. Are you sure that is not what you're refering to?

Pete

4. Apr 12, 2007

### garrett

You're going to have to pick a set of orthonormal basis vectors compatible with your metric, then either compute the connection from them, or use them with your Christoffel symbols to get the connection.

This is laid out in detail here:

http://deferentialgeometry.org/#[[Christoffel symbols]]

5. Apr 12, 2007

### coalquay404

Firstly, begin by taking $\{\hat{\theta}^i\}$ to be a non-coordinate basis. This non-coordinate basis is related to the coordinate basis $\{dx^i\}$ by

$$\hat{\theta}^i = \theta^i_{\phantom{i}j}dx^j$$

for some matrices $\theta^i_{\phantom{i}j}$. Then the relationship between the connection one-form and the Christoffel symbols is

$$\omega^i_{\phantom{i}j} = \Gamma^i_{\phantom{i}jk}\hat{\theta}^k$$

Thus, if you know the Christoffel symbols and the relationship between the non-coordinate and coordinate bases, the answer to your question is to simply read off the value of the components of the connection one form from the above.

He was perfectly clear when he said he was interested in the components of the connection one-form. This is a basic idea covered in, for example, the first few lectures of any decent differential geometry course.

6. Apr 13, 2007

### Hurkyl

Staff Emeritus
Consider the frame bundle, which is a principal SO(n)-bundle. (At least, it is for a Riemannian manifold)

If you have any curve, you can take a frame at one endpoint and invoke your connection to parallel transport it to become a frame at the other endpoint. In particular, we get an element of SO(n) detailing how the frame was rotated/twisted/whatever as it was parallel transported along the curve.

If we differentiate (i.e. if we parallel transport by an infinitessimal amount), then to each tangent vector at a point, we get an element of so(n). Since this gives an so(n) value to each tangent vector, and is a linear functional, we call this object a "so(n)-valued one-form".

This so(n)-valued one-form completely determines parallel transport (you integrate it along a curve to get an SO(n) value), thus the term "connection 1-form".

7. Apr 13, 2007

### SeReNiTy

Your relationship is correct, although i don't like to call them the christoffel symbols, i prefer to use the term connection co-efficients since they rely on the second half of the levi-civita connection to calculate.

your wijk = 1/2( g(ek,[ei,ej]) - g(ei,[ej,ek]) - g(ej, [ek,ei]) )

where ei, ej, ek are the othronormal vectors prescirbed by your metric.

8. Apr 14, 2007

### cosmicstring1

Yes, I read that section from MTW but it does not give the answer. I do'nt have Flanders' book.

9. Apr 14, 2007

### pmb_phy

What is a connection 1-form. Please defined the term for me.

Thank you

Pete

10. Apr 14, 2007

### coalquay404

Well, presumably you're familiar with the idea of using a coordinate basis $\{dx^i\}$ in differential geometry, right? Presumably you're also familiar with the idea that in many practical situations it's a far smarter idea to use a non-coordinate basis $\{\hat{\theta}^i\}$. Well, if you know about these things it's a simple step to define a matrix-valued one-form $[\omega^i_{\phantom{i}j}]$ as

$$\omega^i_{\phantom{i}j} \equiv\Gamma^i_{\phantom{i}jk}\hat{\theta}^k$$

That's really all there is to it. Once you have defined the matrix-valued connection one-form like this, it's a simple matter to go on and show that a neat and tidy way to express the curvature and torsion associated with a given metric is to use the Cartan structure equations.

11. Apr 14, 2007

### Chris Hillman

I suggested starting from a frame field, a la Cartan, as explained in MTW. I didn't answer the question you asked, I told you where to find the answer to the question you should have asked. You replied:

Sure it does--- look at the sections on computing curvature. This contains what you need to answer the question you asked, as well as showing you a much better way to do all this stuff.

I don't think that even makes sense!

But there is one small point concerning the symmetry properties of the matrix of connection one-forms which MTW slur over. Flanders doesn't cover it either because he only discusses Riemannian manifolds. Namely, the rule for adjoint is a bit different for Lorentzian vice Riemannian. You need to know this when you use the symmetries of the connection one-forms in order to read off the connection one-forms from the exterior derivatives of the basis one-forms, as suggested by Cartan.

By definition, if $A$ is some linear operator and $(\cdot , \cdot)$ is some inner product, the adjoint $A^\ast$ is the linear operator satisfying
$$(Ax, y) = (x, A^{\ast} y)$$
for all $x,y$. If you write this out in matrix notation, with the inner product written $(x,y) = x^T L y$, you obtain
$$x^T \, A^T \, L \, y = x^T \, L \, A^{\ast} \, y$$
This holds for all $x,y$, so
$$A^\ast = L^{-1} \, A^T \, L$$
Now apply this when $L = \operatorname{diag} (-1,1,1,1)$ to obtain the Lorentzian transpose. Now apply this to your matrix of one-forms when reading off the connection one-forms as suggested by Cartan.

In the past I have gone through all this stuff in great detail--- you can Google for old posts by myself in sci.physics.*

Last edited: Apr 14, 2007
12. Apr 14, 2007

### pmb_phy

I didn't say he wasn't clear. I mentioned my uncertainty about them once before but it now appears that such things exist, hence the new question, i.e. what are they? That was why I was asking him. No textbook that I have on this subject matter even mentions is. Can you please give me an example? Thank you.

Pete

13. Apr 14, 2007

### pmb_phy

Never mind. I finally found it defined in Lovelock and Rund. Thanks for the help folks.

Pete


text:
# Three shafts (spring constant k1, k2, k3) are connected in series such that they carries the same torque (T), then spring constant (k) for composite shaft will be

This question was previously asked in
RPSC ME Lecturer 2014: Official Paper
View all RPSC Lecturer Tech Edu Papers >
1. k = k1 + k2 + k3
2. k = (k1k2 + k2k3 + k3k1)½
3. $$\frac{1}{k} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}$$
4. $$k = {\left( {\frac{{{k_1}{k_2}{k_3}}}{{{k_1} + {k_2} + {k_3}}}} \right)^{\frac{1}{2}}}$$

## Answer (Detailed Solution Below)

Option 3 : $$\frac{1}{k} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \frac{1}{{{k_3}}}$$
Free
Building Material & Concrete Technology
15387
20 Questions 20 Marks 25 Mins

## Detailed Solution

Concept:

Stiffness of a coil spring is given by:

$$k = \frac{{G{d^4}}}{{64{R^3}n}}$$

G = Rigidity modulus, D = mean coil diameter, R = mean radius, d = wire diameter, n = number of coils

Stiffness is inversely proportional to number of coils

Equivalent stiffness of spring for series combination:

$$\frac{1}{{{k_{eq}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}+ \frac{1}{{{k_3}}}$$

Equivalent stiffness of spring for parallel combination:

keq = k1 + k2 + k3


text:
# Pseudopotentials

## Pseudopotentials in the Kleynman Bylander form¶

This page reports the basic definitions and equations needed to evaluate the nonlocal part of the Hamiltonian. We mainly focus on norm-conserving pseudopotentials. Providing a consistent introduction to the theory of pseudopotentials is beyond the purpose of this section, a more complete discussion of the topic can be found in specialized articles available in the literature [Hamann1979], [Bachelet1982], [Kleinman1982], [Hamann1989], [Troullier1991] [Gonze1991], [Fuchs1999]. For the generalization of norm conservation to multiple projectors (Optimized norm-conserving Vanderbilt pseudopotentials) we refer the reader to [Hamann2013]. The generation and validation of ONCVPSP pseudopotentials is described in the PseudoDojo paper [Setten2018].

For our purpose, we can summarize by saying that a pseudopotential is constructed in order to replace the atomic all-electron potential such that core states are eliminated and valence electrons are described by pseudo wavefunctions whose representation in the Fourier space decays rapidly.

Modern norm-conserving pseudopotentials are constructed such that the scattering properties of the all-electron atom are reproduced around the reference energy configuration up to first order in energy, following the procedure described in [Hamann1979], [Hamann1989], and [Troullier1991]. In modern ab initio codes, the fully separable KB form, proposed by Kleynman and Bylander, is usually employed as it drastically reduces the number of operations required for the application of the nonlocal part of the Hamiltonian as well as the memory required to store the operator [Kleinman1982]. In the KB form, the interaction between a valence electron and an ion of type $\aa$ is described by means of the operator:

$$\label{eq:KBsingleatom} \vv_\PS^\aa = \vloc^\aa(r) + \sum_\lm |\chi_\lm^\aa\ra\,E_l^\aa\,\la\chi_\lm^\aa|, % \vloc_\aa(\rr) + \sum_\lm |\Ylm\,u_{l\aa}^\KB\ra E_{l\aa}^\KB \lau_{l\aa}^\KB\,\Ylm|$$

where $\vloc^\aa(r)$ is a purely local potential with a Coulomb tail $\gamma/r$.

Note

For simplicity, the discussion is limited to pseudopotentials with a single projector per angular channel. The generalization to multi-projector pseudopotentials requires introducing an additional $n$ index in the KB energies i.e. $E_{nl}^\aa$ and in the projectors $\chi_{n\lm}^\aa(\rr)$.

The so-called Kleinman-Bylander energies, $E_l^\aa$, measure the strength of the nonlocal component with respect to the local part. The projectors $\chi_\lm^\aa(\rr)$ are short-ranged functions, expressed in terms of a complex spherical harmonic $\Ylm(\theta,\phi)$ multiplied by an $l$- and atom-dependent radial function:

$$\label{eq:KB_projectors} \chi_\lm^\aa(\rr) = \dfrac{1}{r}\,\chi_l^\aa(r)\,\Ylm(\theta,\phi).$$

The total nonlocal part of the Hamiltonian is obtained by summing the different atom-centered contributions. The final expression in real space reads:

$$\Vnl(\rr_1,\rr_2)= \sum_{\substack{\RR \\ \aa\tt_\aa}} \sum_\lm \la\rr_1-\RR-\tt_\aa|\chi_\lm^\aa\ra E_l^\aa\la\chi_\lm^\aa|\rr_2-\RR-\tt_\aa\ra,$$

where $\RR$ runs over the sites of the Bravais lattice, and $\tt_\aa$ over the positions of the atoms with the same type $\aa$ located inside the unit cell. Due to the nonlocality of the operator, its Fourier transform depends on two separate indices, $\kk+\GG_1$ and $\kk+\GG_2$, instead of the simple difference $\GG_1-\GG_2$. The shorthand notation $\KK = \kk + \GG$ is used in the following to simplify the derivation.

Using the Rayleigh expansion of planewaves in terms of spherical harmonics $\Ylm(\theta,\phi)$ and spherical Bessel functions $\jl(Kr)$:

$$\label{eq:PWinYlmPl} e^{i\KK\cdot\rr} = 4\pi\,\sum_\lm i^l \jl(Kr) \Ylm^\*(\KKhat)\,\Ylm(\rrhat),$$

the Fourier representation of the projector, defined by \ref{eq:KB_projectors}, can be expressed as:

$$\la\KK|\chi_\lm^\aa\ra = \frac{4\pi}{\Omega^{1/2}}\, (-1)^l\,\Ylm(\KKhat) \int_0^\infty\, r\jl(Kr)\,\chi_l^\aa(r)r\,\dd r = \frac{4\pi}{\Omega^{1/2}}\, (-1)^l\,\Ylm(\KKhat) F_l^\aa(K).$$

Finally, the expression for the total nonlocal operator in reciprocal space is:

$$\label{eq:VnlKBmatrixelements} \Vnl(\KK,\KKp)= \frac{(4\pi)^2}{\Omega} \sum_{\aa \tt_\aa} \sum_\lm \,e^{-i(\KK-\KKp)\cdot\tt_\aa}\, \Ylm(\KKhat)\Ylm^\*(\KKphat)\, E_l^\aa F_l^\aa(K) F_l^\aa(K').$$

Note

Abinit employs iterative eigenvalue solvers to solve the KS equations. This means that we only need to compute $\Vnl |\Psi\ra$, i.e., we only need to apply the Hamiltonian onto a set of trial eigenvectors. Therefore the full $\Vnl(\KK,\KKp)$ matrix is never constructed explicitly.