text:
• English only

# Lecture 2

## Manually proving an example

For software verification. Programs must be represented as logical formulas. These logical formulas can be used for proving invariants. This lecture explains the base theory about Propositional logic, First Order logic and Relations.

Recall our example from Lecture 1, with a loop invariant:

    public static int sum(int a0, int n0)
/*:
requires "a0 > 0 & n0 > 0"
ensures "result > 0"
*/
{
int res = 0, a = a0, n = n0;
while //: inv "a > 0 & res >= 0 & (n = n0 | res > 0)"
(n > 0)
{
a = 2*a;
res = res + a;
n = n - 1;
}
return res;
}

To prove ensured properties in the program, one first has to prove that this program will terminate or not, and then by showing that the program doesn't violate the required property prove it is complete. In case of loops in the program, an invariant of a loop in terms of a logical formula is needed such that after any number of iterations of the loop the invariant continues to hold. To prove the invariant for the above while loop, we have to show that after each iteration of the loop this invariant still holds. As we can observe the following logical formula shows that above invariant is true.

## Logical preliminaries

We used these constructs in mathematics all the time. Formalizing them is useful for

• foundation of mathematics
• proving formulas automatically: need to represent formulas and make correct reasoning steps on them

### Propositional logic

Basic propositional operations:

A formula of the propositional calculus is a string generated by the following grammar:

We can simplify this grammar and only use conjunction, disjuntion and negation by observing that the implication and equivalency relations can be expressed as:

Example formula:

We further define the interpretation of a formula. Let F be a propositional formula and let be the set of atoms appearing in F. An interpretation for F is a function , that is, e assigns one of the truth values true or false to each atom.

we can also write the following:

where is in fact a binary math function.

We next give the definitions of satisfiability and validity: a propositional formula F is satisfiable if its value is true in some intepretation. A satisfying interpretation is called a model for F. F is valid if its value is true in all interpretations – a valid propositional formula is also called a tautology. The notation is:

Is “false implies true” a valid formula?

Checking satisfiability is done by using a SAT solver. Now the question that arises is how to check for validity using a SAT solver ?

There are two ways of tackling the problem. One would be to notice that there are possibilities in total and just check for everyone of them if . The other is to convert to CNF or DNF and then use the existing techniques for solving.

SAT solvers: http://www.satlive.org/

#### Conjunctive Normal Form. Disjunctive Normal Form

We give the grammars for conjunctive normal form (CNF) and disjunctive normal form (DNF) as:

is the grammar for CNF, and

is the grammar for DNF.

### First-order logic (predicate calculus)

Predicate calculus is interesting because it supports quantifiers ( symbols). We define a predicate as a function from some arguments to a boolean value.

Example: “p divides q” is a binary predicate on p and q.

Predicate calculus is concerned with properties of predicates (regardless of what these predicates are).

Constructs in predicate calculus:

The syntax for first-order logic is given by:

Drinker's paradox (made more politically correct): in every restaurant there is a guest such that if this guest is having a dessert, then everyone in the restaurant is having a dessert.

Suppose predicate P(y) denotes that guest y is having a dessert. Then we can formally write the drinker's paradox as:

To prove the drinker's paradox formula holds true we have to notice that it can be reduced to a simple implication. Although implication is confusing in language, in logics we must note that P \rightarrow Q is false only if .

#### Semantics for First-order logics

for quantifiers we have:

Tarskian semantics for first-order logic.

Is there an algorithm for first-order logic?

Checking (and enumerating) proofs.

What is a formal proof?

### Sets and relations

A set can be thought of as any collection of distinct objects considered as a whole. For example,

We can define many operations over sets such as.

The Cartesian product is a direct product of sets. Specifically, the Cartesian product of two sets X and Y, denoted X × Y, is the set of all possible ordered pairs whose first component is a member of X and whose second component is a member of Y:

We will use binary relations on program states to represent the meaning of programs. A binary relation on set S is a subset of .

The diagonal relation is .

Because relation is a set, all set operations apply. In addition, we can introduce relation composition, denoted by circle

Then

Sets and predicates are interchangable.

turns set into predicate.

turns predicate into a set.

Here we present few properties of transition relation

Composition of a relation is associative:

Composition of a relation is not commutative

Composition of a relation is distributive over union

The property that is also true about composition of variables.

## Programs as relations

Now back to meaning of programs.

Simple programming language

x = exp
if (F) c1 else c2
c1 ; c2
while (F) c1

Consider a program that manipulates two integer variables.

What is the state - map from names to values.

What is the meaning of

• x=y+3
• x=x+y
• x=y+3 ; x=x+y

Observe how the following update commands are translated to binary relations:

If the above update commands appear one after another in a program then by relation composition they can be represented as the one relation:

## Programs as formulas

Can we represent relations symbolically?

In general? In our examples?

Accumulation of equalities.

## Guarded command language

Our simple language:

x = exp
if (F) c1 else c2
c1 ; c2
while (F) c1

Guarded command language:

assume(F)
c1 [] c2
c1 ; c2
assert(F)
havoc(x)
c1*

Relational semantics of guarded commands.

The following is the interpretation of special commands uses in Guarded Command Language.

Example of Guarded Command Language

Transforming language into guarded commands:

• conditionals:
if (F) c1 else c2

(assume(F); c1)  []
(assume(~F); c2)
• loops:
while (F) c1

(assume(F); c1)*;
assume(~F)
• loops, if we have loop invariants
while [inv I] (F) c1

assert(I);
havoc x1,...,xn;
assume(I);
(assume(~F) []
assume(F);
c1;
assert I;
assume(false));

Transformations such as these are used in Jahob, ESC/Java, Spec#.

## Weakest Preconditions

Weakest precondition of a set of states is the

What happens if r is deterministic (a function on states)?

Set of states - formula with state variables for one state.

Relation on states - formula with state variables for two states.

Important property: conjunctivity.

Weakest preconditions for guarded commands.

Weakest preconditions of assignments make wp very appealing.


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## Cryptology ePrint Archive: Report 2017/930

Four-state Non-malleable Codes with Explicit Constant Rate

Bhavana Kanukurthi and Sai Lakshmi Bhavana Obbattu and Sruthi Sekar

Abstract: Non-malleable codes (NMCs), introduced by Dziembowski, Pietrzak and Wichs (ITCS 2010), generalize the classical notion of error correcting codes by providing a powerful guarantee even in scenarios where error correcting codes cannot provide any guarantee: a decoded message is either the same or completely independent of the underlying message, regardless of the number of errors introduced into the codeword. Informally, NMCs are defined with respect to a family of tampering functions $F$ and guarantee that any tampered codeword either decodes to the same message or to an independent message, so long as it is tampered using a function $f \in F$. Nearly all known constructions of NMCs are for the $t$-split-state family, where the adversary tampers each of the $t$ blocks (also known as states), of a codeword, arbitrarily but independently. Cheraghchi and Guruswami (TCC 2014) obtain a Rate-1 non-malleable code for the case where $t = O(n)$ with $n$ being the codeword length and, in (ITCS 2014), show an upper bound of $1-1/t$ on the best achievable rate for any $t-$split state NMC. For $t=10$, Chattopadhyay and Zuckerman (FOCS 2014) achieve a constant rate construction where the constant is unknown. In summary, there is no known construction of an NMC with an explicit constant rate for any $t= o(n)$, let alone one that comes close to matching Cheraghchi and Guruswami's lowerbound!

In this work, we construct an efficient non-malleable code in the $t$-split-state model, for $t=4$, that achieves a constant rate of $\frac{1}{3+\zeta}$, for any constant $\zeta > 0$, and error $2^{-\Omega(\ell / log^{c+1} \ell)}$, where $\ell$ is the length of the message and $c > 0$ is a constant.

Category / Keywords: foundations / information theoretic cryptography, non-malleability

Original Publication (in the same form): IACR-TCC-2017

Date: received 17 Sep 2017, last revised 25 Sep 2017

Contact author: bhavana at iisc ac in, oslbhavana@gmail com, sruthi sekar1@gmail com

Available format(s): PDF | BibTeX Citation

Short URL: ia.cr/2017/930

[ Cryptology ePrint archive ]

format: plain


text:
#### Volume 17, issue 3 (2017)

 Recent Issues
 The Journal About the Journal Subscriptions Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Author Index To Appear ISSN (electronic): 1472-2739 ISSN (print): 1472-2747
The augmentation category map induced by exact Lagrangian cobordisms

### Yu Pan

Algebraic & Geometric Topology 17 (2017) 1813–1870
##### Abstract

To a Legendrian knot, one can associate an ${A}_{\infty }$ category, the augmentation category. An exact Lagrangian cobordism between two Legendrian knots gives a functor of the augmentation categories of the two knots. We study this functor and establish a long exact sequence relating the corresponding cohomology of morphisms of the two ends. As applications, we prove that the functor between augmentation categories is injective on the level of equivalence classes of objects and find new obstructions to the existence of exact Lagrangian cobordisms in terms of linearized contact homology and ruling polynomials.

##### Keywords
the augmentation category, Lagrangian cobordisms, wrapped Floer homology, Legendrian knots
##### Mathematical Subject Classification 2010
Primary: 53D42, 57R17
Secondary: 53D12, 57M50
##### Publication
Revised: 21 December 2016
Accepted: 1 January 2017
Published: 17 July 2017
##### Authors
 Yu Pan Mathematics Department Duke University Durham, NC United States http://sites.google.com/site/yupanduke/


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## Jackson's Theorem

Jackson's theorem is a statement about the error of the best uniform approximation to a Real Function on by Real Polynomials of degree at most . Let be of bounded variation in and let and denote the least upper bound of and the total variation of in , respectively. Given the function

 (1)

then the coefficients
 (2)

of its Legendre Series, where is a Legendre Polynomial, satisfy the inequalities
 (3)

Moreover, the Legendre Series of converges uniformly and absolutely to in .

Bernstein strengthened Jackson's theorem to

 (4)

A specific application of Jackson's theorem shows that if
 (5)

then
 (6)

References

Cheney, E. W. Introduction to Approximation Theory. New York: McGraw-Hill, 1966.

Jackson, D. The Theory of Approximation. New York: Amer. Math. Soc., p. 76, 1930.

Rivlin, T. J. An Introduction to the Approximation of Functions. New York: Dover, 1981.

Sansone, G. Orthogonal Functions, rev. English ed. New York: Dover, pp. 205-208, 1991.


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# How to delay the directory sync on startup in Confluence

#### Still need help?

The Atlassian Community is here for you.

## Purpose

In larger environments with over 1000 LDAP groups, or over 10,000 users, the LDAP sync on Confluence startup can take a long time to complete. This occurs on startup and in some cases can prevent users from logging into the product until the sync is complete.

This article explains how to delay this sync, which will allow for Confluence to complete startup in a reasonable amount of time and allow users to login. The sync can be scheduled to run during off-peak hours to minimize the impact.

This article will serve the purpose of a workaround until this feature request is implemented:  CONFSERVER-22326 - Getting issue details... STATUS

## Solution - Dynamic Sync Start Time

To delay the directory sync on startup of Confluence, set a value in milliseconds for which the sync should be delayed.

• Add the following jvm parameter to <confluence-install>/bin/setenv.sh (or setenv.bat in Windows), or your custom startup script:

-Dcrowd.polling.startdelay=<delay_value_in_milliseconds>
• For example, if you wish to delay the sync by 12 hours:

-Dcrowd.polling.startdelay=43200000
• Restart Confluence for the changes to take effect.

This will start the sync 12 hours after Confluence starts, effectively allowing Confluence to be fully up and running and not waiting on a sync to complete before becoming available to users.

## Solution - Fixed Sync Start Time

If you wish to use a fixed time, such as 3am to perform the syncs, you can define the number of milliseconds at startup dynamically by creating a variable:

• Add the following to your <confluence-install>/bin/setenv.sh file:

TIME_TO_3AM_IN_MS=$(($(($((86400+$(($(date -d "03:00" +%s)-$(date +%s)))))%86400))*1000))
echo "On startup, initial crowd sync will be delayed up to 3AM, which is in $TIME_TO_3AM_IN_MS ms." CATALINA_OPTS="-Dcrowd.polling.startdelay=$TIME_TO_3AM_IN_MS $CATALINA_OPTS" • Or, 7pm for example: TIME_TO_7PM_IN_MS=$(($(($((86400+$(($(date -d "19:00" +%s)-$(date +%s)))))%86400))*1000)) echo "On startup, initial crowd sync will be delayed up to 7PM, which is in$TIME_TO_7PM_IN_MS ms."

CATALINA_OPTS="-Dcrowd.polling.startdelay=$TIME_TO_7PM_IN_MS${CATALINA_OPTS}"
• Restart Confluence for the changes to take effect.

### Breakdown of each step above

# 3am in seconds
3AM_IN_SECONDS=$(date -d "03:00" +%s) # current time in seconds CURRENT_TIME_IN_SECONDS=$(date +%s)

# difference between the above
TIME_DIFF=$(($3AM_IN_SECONDS - $CURRENT_TIME_IN_SECONDS)) # add 1 day, modulo 1 day, in case the date command retrieved 3am in the past, causing a negative time difference # Note that modulo of a negative number *should* be positive according to mathematics but shell script doesn't do this correctly, which is why the day has to be added first ADJUSTED_TIME_DIFF=$(($((86400 +$TIME_DIFF)) % 86400))

# multiply by 1000 to get ms instead of seconds
TIME_TO_3AM_IN_MS=$(($ADJUSTED_TIME_DIFF * 1000))

### Notes

• While this KB article was written for Confluence, this JVM parameter can also be applied in other applications.
• Please note that this parameter is currently not working in newer versions of Crowd. See the following Crowd bug for more details CWD-5437 - Getting issue details... STATUS
• On Confluence Data Center the directory synchronization schedule works on different way and the parameter described in this KB doesn't work out of the box. See the following Confluence bug for more details CONFSERVER-60516 - Getting issue details... STATUS


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# What Are Integral Curves and Direction Fields Used for?

## Integral Curves

An integral curve—also known as a parametric curve—is the graph of a particular solution of a differential equation—that is, a solution where the constants are determined.

You can create these curves for particular solutions of both first and second order differential equations. If you have a collection of integral curves with different particular solutions, they are called a family of curves.

## Direction Fields

If you have a differential equation of the first order, you can find the tangent at any point $\left(x,y\right)$. You do this by moving terms around so that ${y}^{\prime }$ ends up alone on the left-hand side of the equation. This gives you a formula where you can enter $x$ and $y$ into the formula for ${y}^{\prime }$. If you do this for all the points in a region you get a direction field or slope field.

This direction field also shows which direction the integral curve would go for those specific points. You can say how $y$ develops from a point $\left(x,y\right)$ by looking at which direction the direction field shows at that point $\left(x,y\right)$.


text:
# -3dB Point Calculation

Discussion in 'Homework Help' started by simply_me, Feb 26, 2012.

1. ### simply_me Thread Starter Member

May 6, 2010
50
0
Hello,

In the question I suppose to find the -3dB point of the amplifier stages (current). I found an expression for the transfer function, and now I need to find the -3dB point. My transfer function expression is 1/(constant+Cs/gm). My idea to find the -3dB point is:
1. Assuming a low pass filter, I'll find the value of the transfer function when s=0, convert to dB, and subtract 3dB from it. Then convert back to decimal.
2. Then equate the obtained value to the general transfer function to find the value for s. But I'm not sure if I should square both sides (for magnitude) or not.
I'm not sure if this is the correct path, any thoughts?

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
You appear to have a first order transfer function of the form

G(s)=K/(s+a)

which has a -3dB point at ω=a [radians/sec]

So you need to re-cast your transfer into the form shown above and the rest is relatively simple.

3. ### simply_me Thread Starter Member

May 6, 2010
50
0
Hi t_n_k,

Sure I can change the form, but how do I show that the -3dB point is at omega=a (instead of stating it)?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
OK - your method will work.

Perhaps this can be done somewhat more efficiently as follows:

If

$G(s)=\frac{K}{s+a}$

Then

$|G(0)|=\frac{K}{a}$

$|G(j\omega)|=\frac{K}{\sqrt{\omega^2 +a^2}}$

and the ratio R is given

$R=\frac{|G(j\omega)|}{|G(0)|}$

or

$R=\frac{a}{ \sqrt{ \omega^2 + a^2}}$

One then solves

$20log(R)=-3 \ $dB$$

to find the ω value corresponding to the -3dB point.

The latter relationship (strictly speaking) reduces to

$R=\frac{1}{\sqrt{2}}$

Last edited: Feb 26, 2012
simply_me likes this.
5. ### simply_me Thread Starter Member

May 6, 2010
50
0
Thank you very much for the clarification.


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# Munkres’ Assignment Algorithm with RcppArmadillo

September 24, 2013
By

(This article was first published on Rcpp Gallery, and kindly contributed to R-bloggers)

Munkres’ Assignment Algorithm (Munkres (1957), also known as hungarian algorithm) is a well known algorithm in Operations Research solving the problem to optimally assign N jobs to N workers.

I needed to solve the Minimal Assignment Problem for a relabeling algorithm in MCMC sampling for finite mixture distributions, where I use a random permutation Gibbs sampler. For each sample an optimal labeling must be found, i.e. I have N parameter vectors and must assign each vector to one of the N components of the finite mixture under the restriction that each component gets a parameter vector. For the assignment of parameters to components Stephens (1997b) suggests an algorithm relying on the Minimal Assignment in regard to a loss matrix. The labeling with the smallest loss is then considered as optimal.

After an unsuccessful search for libraries implementing the algorithm easily for C++ or C, I made the decision to program it myself using RcppArmadillo for good performance. I found some guidance by the websites of Bob Pilgrim and TopCoder. These websites offer excellent tutorials to understand the algorithm and to convert it to code. The order of this implementation of Munkres’ algorithm is of an order N to the power of 4. There exists also a version of order N to the power of 3, but an order of N to the power of 4 works very good for me and coding time is as usual a critical factor for me.

In the following I walk through the different steps of Munkres’ algorithm and explain the main parts and their functionality.

Let’s begin. The first step in Munkres’ algorithm is to subtract the minimal element of each row from each element in this row.

#include<RcppArmadillo.h>

void step_one(unsigned int &step, arma::mat &cost,
const unsigned int &N)
{
for (unsigned int r = 0; r < N; ++r) {
cost.row(r) -= arma::min(cost.row(r));
}
step = 2;
}

Note, that we use references for all function arguments. As we have to switch between the steps of the algorithm continously, we always must be able to determine which step should be chosen next. Therefore we give a mutable unsigned integer step as an argument to each step function of the algorithm.

Inside the function we can easily access a whole row by Armadillo’s row() method for matrices. In the second step, we then search for a zero in the modified cost matrix of step one.

void step_two (unsigned int &step, const arma::mat &cost,
arma::umat &indM, arma::ivec &rcov,
arma::ivec &ccov, const unsigned int &N)
{
for (unsigned int r = 0; r < N; ++r) {
for (unsigned int c = 0; c < N; ++c) {
if (cost.at(r, c) == 0.0 && rcov.at(r) == 0 && ccov.at(c) == 0) {
indM.at(r, c)  = 1;
rcov.at(r)     = 1;
ccov.at(c)     = 1;
break;                                              // Only take the first
// zero in a row and column
}
}
}
/* for later reuse */
rcov.fill(0);
ccov.fill(0);
step = 3;
}

Only the first zero in a row is taken. Then, the indicator matrix indM indicates this zero by setting the corresponding element at (r, c) to 1. A unique zero - the only or first one in a column and row - is called starred zero. In step 2 we find such a starred zero.

Note, that we use here Armadillo’s element access via the method at(), which makes no bound checks and improves performance.

Note Bene: This code is thoroughly debugged - never do this for fresh written code!

In step 3 we cover each column with a starred zero. If already N columns are covered all starred zeros describe a complete assignment - so, go to step 7 and finish. Otherwise go to step 4.

void step_three(unsigned int &step, const arma::umat &indM,
arma::ivec &ccov, const unsigned int &N)
{
unsigned int colcount = 0;
for (unsigned int r = 0; r < N; ++r) {
for (unsigned int c = 0; c < N; ++c) {
if (indM.at(r, c) == 1) {
ccov.at(c) = 1;
}
}
}
for (unsigned int c = 0; c < N; ++c) {
if (ccov.at(c) == 1) {
++colcount;
}
}
if (colcount == N) {
step = 7;
} else {
step = 4;
}
}

We cover a column by looking for 1s in the indicator matrix indM (See step 2 for assuring that these are indeed only starred zeros).

Step 4 finds noncovered zeros and primes them. If there are zeros in a row and none of them is starred, prime them. For this task we program a helper function to keep the code more readable and reusable. The helper function searches for noncovered zeros.

void find_noncovered_zero(int &row, int &col,
const arma::mat &cost, const arma::ivec &rcov,
const arma::ivec &ccov, const unsigned int &N)
{
unsigned int r = 0;
unsigned int c;
bool done = false;
row = -1;
col = -1;
while (!done) {
c = 0;
while (true) {
if (cost.at(r, c) == 0.0 && rcov.at(r) == 0 && ccov.at(c) == 0) {
row = r;
col = c;
done = true;
}
++c;
if (c == N || done) {
break;
}
}
++r;
if (r == N) {
done = true;
}
}
}

We can detect noncovered zeros by checking if the cost matrix contains at row r and column c a zero and row and column are not covered yet, i.e. rcov(r) == 0, ccov(c) == 0. This loop breaks, if we have found our first uncovered zero or no uncovered zero at all.

In step 4, if no uncovered zero is found we go to step 6. If instead an uncovered zero has been found, we set the indicator matrix at its position to 2. We then have to search for a starred zero in the row with the uncovered zero, uncover the column with the starred zero and cover the row with the starred zero. To indicate a starred zero in a row and to find it we create again two helper functions.

bool star_in_row(int &row, const arma::umat &indM,
const unsigned int &N)
{
bool tmp = false;
for (unsigned int c = 0; c < N; ++c) {
if (indM.at(row, c) == 1) {
tmp = true;
break;
}
}
return tmp;
}

void find_star_in_row (const int &row, int &col,
const arma::umat &indM, const unsigned int &N)
{
col = -1;
for (unsigned int c = 0; c < N; ++c) {
if (indM.at(row, c) == 1) {
col = c;
}
}
}

We know that starred zeros are indicated by the indicator matrix containing an element equal to 1. Now, step 4.

void step_four (unsigned int &step, const arma::mat &cost,
arma::umat &indM, arma::ivec &rcov, arma::ivec &ccov,
int &rpath_0, int &cpath_0, const unsigned int &N)
{
int row = -1;
int col = -1;
bool done = false;
while(!done) {
find_noncovered_zero(row, col, cost, rcov,
ccov, N);

if (row == -1) {
done = true;
step = 6;
} else {
/* uncovered zero */
indM(row, col) = 2;
if (star_in_row(row, indM, N)) {
find_star_in_row(row, col, indM, N);
/* Cover the row with the starred zero
* and uncover the column with the starred
* zero.
*/
rcov.at(row) = 1;
ccov.at(col) = 0;
} else {
/* No starred zero in row with
* uncovered zero
*/
done = true;
step = 5;
rpath_0 = row;
cpath_0 = col;
}
}
}
}

Notice the rpath_0 and cpath_0 variables. These integer variables store the first vertex for an augmenting path in step 5. If zeros could be primed we go further to step 5.

Step 5 constructs a path beginning at an uncovered primed zero (this is actually graph theory - alternating and augmenting paths) and alternating between starred and primed zeros. This path is continued until a primed zero with no starred zero in its column is found. Then, all starred zeros in this path are unstarred and all primed zeros are starred. All primes in the indicator matrix are erased and all rows are uncovered. Then return to step 3 to cover again columns.

Step 5 needs several helper functions. First, we need a function to find starred zeros in columns.

void find_star_in_col (const int &col, int &row,
const arma::umat &indM, const unsigned int &N)
{
row = -1;
for (unsigned int r = 0; r < N; ++r) {
if (indM.at(r, col) == 1) {
row = r;
}
}
}

Then we need a function to find a primed zero in a row. Note, that these tasks are easily performed by searching the indicator matrix indM.

void find_prime_in_row (const int &row, int &col,
const arma::umat &indM, const unsigned int &N)
{
for (unsigned int c = 0; c < N; ++c) {
if (indM.at(row, c) == 2) {
col = c;
}
}
}

In addition we need a function to augment the path, one to clear the covers from rows and one to erase the primed zeros from the indicator matrix indM.

void augment_path (const int &path_count, arma::umat &indM,
const arma::imat &path)
{
for (unsigned int p = 0; p < path_count; ++p) {
if (indM.at(path(p, 0), path(p, 1)) == 1) {
indM.at(path(p, 0), path(p, 1)) = 0;
} else {
indM.at(path(p, 0), path(p, 1)) = 1;
}
}
}

void clear_covers (arma::ivec &rcov, arma::ivec &ccov)
{
rcov.fill(0);
ccov.fill(0);
}

void erase_primes(arma::umat &indM, const unsigned int &N)
{
for (unsigned int r = 0; r < N; ++r) {
for (unsigned int c = 0; c < N; ++c) {
if (indM.at(r, c) == 2) {
indM.at(r, c) = 0;
}
}
}
}

The function to augment the path gets an integer matrix path of dimension 2 * N x 2. In it all vertices between rows and columns are stored row-wise. Now, we can set the complete step 5:

void step_five (unsigned int &step,
arma::umat &indM, arma::ivec &rcov,
arma::ivec &ccov, arma::imat &path,
int &rpath_0, int &cpath_0,
const unsigned int &N)
{
bool done = false;
int row = -1;
int col = -1;
unsigned int path_count = 1;
path.at(path_count - 1, 0) = rpath_0;
path.at(path_count - 1, 1) = cpath_0;
while (!done) {
find_star_in_col(path.at(path_count - 1, 1), row,
indM, N);
if (row > -1) {
/* Starred zero in row 'row' */
++path_count;
path.at(path_count - 1, 0) = row;
path.at(path_count - 1, 1) = path.at(path_count - 2, 1);
} else {
done = true;
}
if (!done) {
/* If there is a starred zero find a primed
* zero in this row; write index to 'col' */
find_prime_in_row(path.at(path_count - 1, 0), col,
indM, N);
++path_count;
path.at(path_count - 1, 0) = path.at(path_count - 2, 0);
path.at(path_count - 1, 1) = col;
}
}
augment_path(path_count, indM, path);
clear_covers(rcov, ccov);
erase_primes(indM, N);
step = 3;
}

Recall, if step 4 was successfull in uncovering all columns and covering all rows with a primed zero, it then calls step 6. Step 6 takes the cover vectors rcov and ccov and looks in the uncovered region of the cost matrix for the smallest value. It then subtracts this value from each element in an uncovered column and adds it to each element in a covered row. After this transformation, the algorithm starts again at step 4. Our last helper function searches for the smallest value in the uncovered region of the cost matrix.

void find_smallest (double &minval, const arma::mat &cost,
const arma::ivec &rcov, const arma::ivec &ccov,
const unsigned int &N)
{
for (unsigned int r = 0; r < N; ++r) {
for (unsigned int c = 0; c < N; ++c) {
if (rcov.at(r) == 0 && ccov.at(c) == 0) {
if (minval > cost.at(r, c)) {
minval = cost.at(r, c);
}
}
}
}
}

Step 6 looks as follows:

void step_six (unsigned int &step, arma::mat &cost,
const arma::ivec &rcov, const arma::ivec &ccov,
const unsigned int &N)
{
double minval = DBL_MAX;
find_smallest(minval, cost, rcov, ccov, N);
for (unsigned int r = 0; r < N; ++r) {
for (unsigned int c = 0; c < N; ++c) {
if (rcov.at(r) == 1) {
cost.at(r, c) += minval;
}
if (ccov.at(c) == 0) {
cost.at(r, c) -= minval;
}
}
}
step = 4;
}

At last, we must create a function that enables us to jump around the different steps of the algorithm. The following code shows the main function of the algorithm. It defines also the important variables to be passed to the different steps.

arma::umat hungarian(const arma::mat &input_cost)
{
const unsigned int N = input_cost.n_rows;
unsigned int step = 1;
int cpath_0 = 0;
int rpath_0 = 0;
arma::mat cost(input_cost);
arma::umat indM(N, N);
arma::ivec rcov(N);
arma::ivec ccov(N);
arma::imat path(2 * N, 2);

indM = arma::zeros<arma::umat>(N, N);
bool done = false;
while (!done) {
switch (step) {
case 1:
step_one(step, cost, N);
break;
case 2:
step_two(step, cost, indM, rcov, ccov, N);
break;
case 3:
step_three(step, indM, ccov, N);
break;
case 4:
step_four(step, cost, indM, rcov, ccov,
rpath_0, cpath_0, N);
break;
case 5:
step_five(step, indM, rcov, ccov,
path, rpath_0, cpath_0, N);
break;
case 6:
step_six(step, cost, rcov, ccov, N);
break;
case 7:
done = true;
break;
}
}
return indM;
}

Note, this function takes the numeric cost matrix as an argument and returns the integer indicator matrix indM.

I chose to set the argument input_cost constant and copy it for reasons of reusability of the cost matrix in other C++ code. When working with rather huge cost matrices, it makes sense to make the argument mutable. Though, I used pass-by-reference for all the arguments in functions to avoid useless copying inside the functions.

To call the main function hungarian from R and to use our algorithm we construct an Rcpp Attribute:

// [[Rcpp::export]]

arma::imat hungarian_cc(Rcpp::NumericMatrix cost)
{
// Reuse memory from R
unsigned int N = cost.rows();
arma::mat arma_cost(cost.begin(), N, N, false, true);
// Call the C++-function 'hungarian'
arma::umat indM = hungarian(arma_cost);
//Convert the result to an Armadillo integer
//matrix - R does not know unsigned integers.
return arma::conv_to<arma::imat>::from(indM);
}

If we want to provide this function also to other users we should probably ensure, that the matrix is square (There exists also an extension to rectangular matrices, see Burgeois and Lasalle (1971)). This can be done easily with the exceptions provided by Rcpp passed over to R:

// [[Rcpp::export]]
arma::imat hungariansafe_cc(Rcpp::NumericMatrix cost)
{
unsigned int N = cost.rows();
unsigned int K = cost.cols();
if (N != K) {
throw Rcpp::exception("Matrix is not square.");
}
// Reuse memory from R
arma::mat arma_cost(cost.begin(), N, K, false, true);
// Call the C++-function 'hungarian'
arma::umat indM = hungarian(arma_cost);
//convert the result to an Armadillo integer
//matrix - R does not know unsigned integers.
return arma::conv_to<arma::imat>::from(indM);
}

Note, that it is also possible to use for the attribute directly an Armadillo matrix, but following the recent discussion on the Rcpp-devel list, a pass-by-reference of arguments is not yet possible. Romain Francois’ proposals seem promising, so maybe we can expect in some of the next releases shallow copies allowing pass-by-reference in Rcpp Attributes.

Let us begin now with a very easy example that makes clear what is going on.

# Check exception:
cost <- matrix(c(1:6), nrow = 3, ncol = 2, byrow = TRUE)
tryCatch(indM <- hungariansafe_cc(cost), error = function(e) {print(e)})

<Rcpp::exception: Matrix is not square.>

cost <- matrix(c(1, 2, 2, 4), nrow = 2, ncol = 2, byrow = TRUE)
cost

     [,1] [,2]
[1,]    1    2
[2,]    2    4

indM <- hungarian_cc(cost)
indM

     [,1] [,2]
[1,]    0    1
[2,]    1    0

min.cost <- sum(indM * cost)
min.cost

[1] 4

We can also compute a maximal assignment over a revenue matrix by simply considering the difference between a big value and this matrix as a cost matrix.

revenues <- matrix(seq(1, 4)) %*% seq(1, 4)
revenues

     [,1] [,2] [,3] [,4]
[1,]    1    2    3    4
[2,]    2    4    6    8
[3,]    3    6    9   12
[4,]    4    8   12   16

cost <- 100 - revenues
indM <- hungarian_cc(cost)
indM

     [,1] [,2] [,3] [,4]
[1,]    1    0    0    0
[2,]    0    1    0    0
[3,]    0    0    1    0
[4,]    0    0    0    1

max.revenue <- sum(indM * revenues)
max.revenue

[1] 30

CoCost matrices containing negative values work as well:

cost <- matrix(rnorm(100), ncol = 10, nrow = 10)
cost

          [,1]    [,2]    [,3]     [,4]     [,5]     [,6]    [,7]
[1,] -0.23018  0.3598 -0.2180 -0.29507 -0.20792 -0.02855 -0.5023
[2,]  1.55871  0.4008 -1.0260  0.89513 -1.26540 -0.04287 -0.3332
[3,]  0.07051  0.1107 -0.7289  0.87813  2.16896  1.36860 -1.0186
[4,]  0.12929 -0.5558 -0.6250  0.82158  1.20796 -0.22577 -1.0718
[5,]  1.71506  1.7869 -1.6867  0.68864 -1.12311  1.51647  0.3035
[6,]  0.46092  0.4979  0.8378  0.55392 -0.40288 -1.54875  0.4482
[7,] -1.26506 -1.9666  0.1534 -0.06191 -0.46666  0.58461  0.0530
[8,] -0.68685  0.7014 -1.1381 -0.30596  0.77997  0.12385  0.9223
[9,] -0.44566 -0.4728  1.2538 -0.38047 -0.08337  0.21594  2.0501
[10,]  1.22408 -1.0678  0.4265 -0.69471  0.25332  0.37964 -0.4910
[,8]    [,9]   [,10]
[1,] -2.309169  0.3853  0.5484
[2,]  1.005739 -0.3707  0.2387
[3,] -0.709201  0.6444 -0.6279
[4,] -0.688009 -0.2205  1.3607
[5,]  1.025571  0.3318 -0.6003
[6,] -0.284773  1.0968  2.1873
[7,] -1.220718  0.4352  1.5326
[8,]  0.181303 -0.3259 -0.2357
[9,] -0.138891  1.1488 -1.0264
[10,]  0.005764  0.9935 -0.7104

indM <- hungarian_cc(cost)
indM

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]    0    0    0    0    0    0    0    1    0     0
[2,]    0    0    0    0    1    0    0    0    0     0
[3,]    0    0    0    0    0    0    1    0    0     0
[4,]    0    0    0    0    0    0    0    0    1     0
[5,]    0    0    1    0    0    0    0    0    0     0
[6,]    0    0    0    0    0    1    0    0    0     0
[7,]    0    1    0    0    0    0    0    0    0     0
[8,]    1    0    0    0    0    0    0    0    0     0
[9,]    0    0    0    0    0    0    0    0    0     1
[10,]    0    0    0    1    0    0    0    0    0     0

min.cost <- sum(indM * cost)
min.cost

[1] -12.42

Finally let us make some benchmarking. We load the rbenchmark package and take now a more realistic cost matrix.

library(rbenchmark)
cost <- matrix(rpois(10000, 312), ncol = 100, nrow = 100)
benchmark(indM <- hungarian_cc(cost), columns=c('test','replications','elapsed','user.self','sys.self'), replications=1000)

                        test replications elapsed user.self sys.self
1 indM <- hungarian_cc(cost)         1000   10.53      10.5    0.004

But we also see, where the limitations of this algorithm lie - huge matrices:

cost <- matrix(rpois(250000, 312), ncol = 500, nrow = 500)
system.time(indM <- hungarian_cc(cost))

   user  system elapsed
3.656   0.008   3.674

Some last notes on the structure of the code. I prefer to put the code of the algorithm in an header file, e.g. hungarian.h. And use an attributes.cpp (attributes.cc) file to program the Rcpp Attributes. With this, I can easily reuse the algorithm in C++ code by simple inclusion (#include "hungarian.h") and have a complete overview on all the C++ functions I export to R.


text:
Marius - your [remarks on catalysis](https://forum.azimuthproject.org/discussion/comment/17825/#Comment_17825) are very interesting and important! One of the beauties of resource theory is that it lets us make the concept of "catalyst" very general and mathematical.

In [Lecture 20](https://forum.azimuthproject.org/discussion/2081/lecture-20-chapter-2-resource-theories), I talk about some reactions in manufacturing, like these:

$$\textrm{[processing chip]} + \textrm{[memory chip]} + 4 \textrm{[minute]} \to \textrm{[laptop]}$$

$$\textrm{[processing chip]} + 2 \textrm{[memory chip]} + 3 \textrm{[minute]} \to \textrm{[desktop]}$$

$$\textrm{[laptop]} \to 750\textrm{[profit]}$$

$$\textrm{[desktop]} \to 1000 \textrm{[profit]}$$

These are often studied using linear programming. Linear programming _ignores catalysis_ because it doesn't distinguish between a reaction, say

$$X + Y \to Z,$$

and a similar reaction that involves a catalyst:

$$X + Y + C \to Z + C.$$

The manufacturing reactions I listed don't involve catalysis, but if they did, linear programming would be somewhat inadequate to capture all the details! (At least the simple sort of linear programming I know about. Maybe there's a fancier version that handles catalysis.)


text:
# How do you solve ln x - ln (1/x) = 2?

Apr 27, 2018

$x = e$

#### Explanation:

As $\ln x - \ln \left(\frac{1}{x}\right) = 2$

we have $\ln \left(\frac{x}{\frac{1}{x}}\right) = 2$

or $\ln {x}^{2} = 2$

or $2 \ln x = 2$

or $\ln x = 1$

and $x = e$


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 Mat. Sb.: Year: Volume: Issue: Page: Find

 Mat. Sb. (N.S.), 1956, Volume 39(81), Number 1, Pages 51–148 (Mi msb5128)

The problem of Cauchy with operators as coefficients, the mixed boundary problem for systems of differential equations and an approximate method of their solution

M. I. Vishik

Full text: PDF file (8459 kB)

Bibliographic databases:

Citation: M. I. Vishik, “The problem of Cauchy with operators as coefficients, the mixed boundary problem for systems of differential equations and an approximate method of their solution”, Mat. Sb. (N.S.), 39(81):1 (1956), 51–148

Citation in format AMSBIB
\Bibitem{Vis56} \by M.~I.~Vishik \paper The problem of Cauchy with operators as coefficients, the mixed boundary problem for systems of differential equations and an approximate method of their solution \jour Mat. Sb. (N.S.) \yr 1956 \vol 39(81) \issue 1 \pages 51--148 \mathnet{http://mi.mathnet.ru/msb5128} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=80248} \zmath{https://zbmath.org/?q=an:0127.08002|0070.12202}

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• http://mi.mathnet.ru/eng/msb/v81/i1/p51

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This publication is cited in the following articles:
1. L. A. Kalyakin, “Galerkin's method for equations with a small parameter in the highest order derivatives”, Math. USSR-Sb., 14:4 (1971), 525–536
2. Uspenskii S. Demidenko G., “On Mixed Boundary-Value-Problems for One Class of Equations Unsolved with Respect to Higher Derivative”, 252, no. 3, 1980, 560–562
3. Yanov S., “The Behavior as T ->Infinity of the Solution of the First Exterior Mixed Boundary Value Problem for the Sobolev System”, Differ. Equ., 34:4 (1998), 554–558
4. S. V. Uspenskii, E. N. Vasil'eva, S. I. Yanov, “Differential Properties of the Solution to the First Mixed Boundary Value Problem for the Sobolev System”, Proc. Steklov Inst. Math., 227 (1999), 305–313
5. A. P. Oskolkov, “O nekotorykh psevdoparabolicheskikh sistemakh uravnenii s malym parametrom, voznikayuschikh pri chislennom analize uravnenii zhidkostei Kelvina–Foigta”, Vestnik ChelGU, 1999, no. 4, 155–173
6. G. A. Sviridyuk, V. E. Fedorov, “Polugruppy operatorov s yadrami”, Vestnik ChelGU, 2002, no. 6, 42–70
7. Prilepko A. Tkachenko D., “Fredholm Property of the Inverse Source Problem for Parabolic Systems”, Differ. Equ., 39:12 (2003), 1785–1793
8. V. N. Denisov, “On the behaviour of solutions of parabolic equations for large values of time”, Russian Math. Surveys, 60:4 (2005), 721–790
9. Gelman B.D., “Mnogoznachnye szhimayuschie otobrazheniya i ikh prilozheniya”, Vestn. Voronezhskogo gos. un-ta. Ser.: Fiz. Matem., 2009, no. 1, 74–86
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text:
# Integration tanx*tan2x*tan3x

1. Nov 21, 2003

### BlueDevil

I have been working on this integration problem for awhile now and am completely stuck:

Integral of tanx*tan2x*tan3x

I know I am suppose to use substituion and for that matter probably multiple substitions. But I am no sure what I should use for my first value of "u". I have changed the problem to be in terms of sin and cos, but I am still just ending up at a dead end. If some one can help me start off in the right step I would greatly appreciatte it. I DO NOT want the answer, I would like to solve it for myself, I just need help starting it off right. Thanks in advance.

2. Nov 21, 2003

### gnome

How about changing the form of your integrand using trig identities before attempting to integrate?

i.e.: tan(x+y) = ?

3. Nov 21, 2003

### BlueDevil

tan(x+y) = (tanx +tany)/(1-tanxtany) right?

But I dont see how that can help me. :-\

4. Nov 21, 2003

### sridhar_n

...

Try using the expansion of tan (x+y) to simplify tan 2x and tan 3x and try to bring all the three terms in terms of tanx only. Now substitute tanx=sinx/cosx and simplify the expression you get to see the below answer:

I = the integral = I1+I2;
where,
I1 = int[(4sin^3xcosx)dx]/(cos^2x-Sin^x) and

I2 = int[(2Sin^3xSecx)dx]/(cos^2x-Sin^x)

Put cosxdx = d(sinx) and then put, sinx = t and then I1 reduces to

int(4x^3dx/1-2X^2), solve this integral (can u???). Repeat a similar procedure for I2 and lo behold u have the answer with u...

Sridhar

5. Nov 21, 2003

### gnome

tan(2x) = tan(x + x)
tan(3x) = tan((x+x) + x)

So you end up with an expression in powers of tan(x), no tan(2x) and no tan(3x).

Then, integrate.

Last edited: Nov 21, 2003
6. Nov 21, 2003

### BlueDevil

I'm sorry but I'm just not gettin how you did that. How did you expand the integral of tan(x)*tan(2x)*tan(3x) to the integral you gave? Is there some trig identity I am not aware of?

7. Nov 21, 2003

### BlueDevil

Ok, I get that alittle better now, I'll start working on it like that and see where I get, thanks guys, Ill post again if I get lost again, which I probably will.

8. Nov 22, 2003

### BlueDevil

Ok I've been working on it for awhile now after changing the problem to:

integral tan(x)*tan(x+x)*tan(x+x+x)

I don't see how that is making the equation more favorable for me to solve. I still have to use substitution in the next step. I am just really confused at this point and am very desperate to get this done.

9. Nov 22, 2003

### sridhar_n

...

Your question has already been answered...Just scroll up the page a li'l bit and u have the answer there. I have given the step after using tan(a+b) and then the step after substituting tanx = sinx/cosx and then the method to solve the final integral...

I think u understand that cosx dx = d(sinx) {as d(sinx)/dx = cosx}
similarly, sinxdx = -d(cosx); These substitutions will help u solve the final integral that I have given u... That is what I have explained previously....

If u still do not understand follow these procedures:

1) Use tan(a+b) to bring the entire integral in terms of tanx alone. Note you might have to use tan2x = 2tanx/sec^2x.

2)After you have got an expression in terms of tanx:

integral{[(4tan^3x)/(sec^2x(Sec^2x-2tan^2x))]+ [(2tan^3x)/(sec^2x-2tan^2x)]}

3) now, substitute, tanx = sinx/cosx, you will get:

I (the main integral) = I1 + I2
where,

I1 = integral{[(4sin^3xcosxdx)/(cos^2x-Sin^2x)]}
I2 = integral{[(2sin^3xsecx)/(cos^2x-sin^2x)]}

4) Now consider I1 and I2 seperately and solve them seperately in the following way:

(i)To solve I1 - cos^2x = 1 - sin^2x and cosxdx = d(sinx). Now substitute sinx = t to get :

I1 = integral{[(4t^3)dt/(1-2t^2)]}. Can u solve this??? {to solve this put 1-2t^2 = y so that -4tdt = dy and then proceed}

(ii) To solve I2 - Proceed in a similar way as done for I1 to get:

I2 = integral{[(1-cos^2x)(sinxdx)/((cosx)(2cos^2x-1))]}
substitute cosx = p and sinxdx = -d(cosx) to get:

I2 = integral{[(1-p^2)dp/(p(2p^2-1))]} (u can solve this by adding and subtracting 2x^2 from the numerator to get 2 terms, u can easily solve the 2 terms to get I2)

5) Add the solved values of I1 and I2 to get the value of integral(tanxtan2xtan3x)...

Got it???????

Sridhar

10. Nov 22, 2003

### BlueDevil

How does tan2x = 2tanx/sec^2x? I do not see how this can be the same. They produce different y values when you use an x. Is there some kind of trig identitiy that states this? Once I understand this I will be bale to do the problem.

11. Nov 22, 2003

### Hurkyl

Staff Emeritus
He got the tangent identity wrong. It is

$$\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$$

so

$$\tan 2x = \frac{2\ tan x} {1 - \tan^2 x}$$

However, a general rule to use when confused about trig identities is to convert everything into sines and cosines. (reason: sines and cosines are, in general, much easier to manipulate) If you're doing an integral, you also want to convert everything to having the same argument, if possible. (reason: you're almost always going to have to do this anyways, so you might as well do it up front so it's easier to see the proper substitution to make) Applying this rule would have you do the conversion as follows:

$$\tan 2x = \frac{\sin 2x}{\cos 2x} = \frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x}$$

If all else fails, make the substitution

$$z = \tan \frac{x}{2}$$

Though this approach will tend to be far messier. You can show, using trig identities, that when using this substitution:

$$\sin x = \frac{2z}{1+z^2}$$
$$\cos x = \frac{1-z^2}{1+z^2}$$
$$dx = \frac{dz}{1+z^2}$$

So this will convert any rational function of trig functions into an ordinary (but possibly very ugly) rational function which you can do via partial fractions.

Last edited: Nov 22, 2003
12. Nov 22, 2003

### sridhar_n

...

Oops Sorry, As hurkyl said, I made a mistake in the identity. However the steps are the same. Bring every term to their sine and cosine forms and then proceed with the problem.

Sridhar

13. Nov 22, 2003

### BlueDevil

Ok, it makes more sense now. I really hate trig identities! I will most likely work on it tomorrow and I should be able to get an answer now. Thanks alot for your help guys, I really appreciatte it!

14. Nov 23, 2003

### BlueDevil

Ok a couple more things: First, how did you get those values of sinx, cosx, and dx from z=tan(x/2)? Secondly, I know the integral will be (sinx/cosx)(2sin^2(x)cos(x)/cos^3(x)-sin^2(x)*tan3x. But what does the tan3x equal in terms of sin and cos? I did some work and got confused again as to what value I should use for substituion. Sorry for being such a pain with this, but I do truely appreciatte you guys helping me out.

15. Nov 23, 2003

### Hurkyl

Staff Emeritus
You have a parenthesis error.

From the integral table in my CRC handbook.

They're not hard to do by hand, though. E.G.

$$\begin{equation*} \begin{split} \sin x &= \sin \left( 2 \frac{x}{2} \right) \\ &= 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) \\ &= 2 \frac{z}{\sqrt{1+z^2}} \frac{1}{\sqrt{1+z^2}} \\ &= \frac{2z}{1+z^2} \end{split} \end{equation*}$$

(If you don't see how to go from line 2 to line 3, try drawing a triangle with angle $$x/2$$; one such triangle has an opposite side with length $$z$$, and an adjacent side with length $$1$$. Note that the square root could be either positive or negative, but since it's being multiplied by itself, it doesn't matter)

Well, start by writing it in terms of sin and cos!

$$\tan 3x = \frac{\sin 3x}{\cos 3x} = \frac{\sin (x + 2x)}{\cos(x + 2x)} = \ldots$$

Can you post all of the work you did, so we can check to make sure it's right, and to show you the way to go about figuring out the next step?

Last edited: Nov 23, 2003
16. Nov 23, 2003

### BlueDevil

Ok, Ill try that. My work is wrong becasue I was using the wrong thing for tan3x. I had:

The integral of (sinx/cosx)(2sinxcosx/cos^2x-sin^2X) * the integral of (sinx/cosx)(2sinxcosx/cos^2x-sin^2X)

So then I distibuted the sinx and cosx and got:

The integral of 2sin^2xcosx/cos^3x-sin^2xcosx * The integral of 2sin^2xcosx/cos^3x-sin^2xcosx

So I then attempted to use substituion. I used a couple terms and did not get far at all. With all those trig function to the Nth powers I didn't know how to find the derivaitives. That is something we never learned.

17. Nov 23, 2003

### Hurkyl

Staff Emeritus
How did you get the product of two integrals?

Derivatives of powers of trig functions is easy; use the chain rule... why do you think such derivatives are necessary?

$$\frac{d}{dx}\left( (\sin x)^n \right) = n (\sin x)^{n-1} \cos x$$

Finally, you need to use parentheses better; 2sinxcosx/cos^2x-sin^2X means

$$\frac{2 \sin x \cos x}{\cos^2 x} - \sin^2 x$$

But 2 sin x cos x / (cos^2 x - sin^2 x) means

$$\frac{2 \sin x \cos x}{\cos^2 x - \sin^2 x}$$

18. Nov 25, 2003

### PrudensOptimus

$$\int tan(x)tan(2x)tan(3x) dx = \int tan(x)tan(2x)tan(2x + x) dx$$
$$=\int \frac{sin(x)}{cos(x)}(\frac{2tan(x)}{1-tan^2(x)})(\frac{tan(2x) + tan(x)}{1 - tan(2x)tan(x)}) dx$$
$$=\int \frac{sin(x)}{cos(x)}(\frac{sin(2x)}{cos(2x)})(...) dx$$

u = -cosx
du = sinx dx.

$$=\int \frac{sin(x)}{-u}(\frac{2sin(x)(-u)}{u^2 - (1 + u^2)})(...) \frac{du}{sin(x)}$$
$$=\int \frac{1}{-u}(u - \frac{1}{3u})(2u) du$$
$$= -2\int u - \frac{1}{3u} du$$
$$= \frac{2ln|-cosx|}{3} - cos^2(x) + C$$

Not sure if I got it right or not...

19. Nov 29, 2003

### BlueDevil

Ok I understand where your going, but you lost me on this step:
$$=\int \frac{sin(x)}{-u}(\frac{2sin(x)(-u)}{u^2 - (1 + u^2)})(...) \frac{du}{sin(x)}$$
to this step:
$$=\int \frac{1}{-u}(u - \frac{1}{3u})(2u) du$$

Plus, how did your make the cos(2x) into (u^2)-(1+u^2)?

20. Nov 29, 2003

### Hurkyl

Staff Emeritus
$$\begin{equation*} \begin{split} I &= \int \tan(x) \tan(2x) \tan(3x) \, dx \\ &= \int \frac{\sin(x)}{\cos(x)} \frac{\sin(2x)}{\cos(2x)} \frac{\sin(3x)}{\cos(3x)} \, dx \\ &= \int \frac{(\sin x) (2 \sin x \cos x) (\sin x (4 \cos^2 x - 1))} {(\cos x) (2 \cos^2 x - 1) (4 \cos^3 x - 3 \cos x)} \, dx \\ &= \int \frac{2 (\sin^3 x) (\cos x) (4 \cos^2 x - 1)} {(\cos^2 x) (2 \cos^2 x - 1) (4 \cos^2 x - 3)} \, dx \\ &= 2 \int \frac{(1 - \cos^2 x) (\cos x) (4 \cos^2 x - 1)} {(\cos^2 x) (2 \cos^2 x - 1) (4 \cos^2 x - 3)} (\sin x \, dx) \\ &= 2 \int \frac{(1 - u^2) u (4 u^2 - 1)}{u^2 (2 u^2 - 1) (4 u^2 - 3)} du \\ &= \ldots \end{split} \end{equation*}$$

Last edited: Nov 29, 2003


text:
# Probability to create an electron - hole pair inside Si

Given a slab of Silicon and a photon flux $\Phi[\frac{\gamma}{cm^2sec}]$ incident on the slab of thickness $h$, the incident power $I$ given by $\Phi$ will be absorbed following the Beer - Lambert law: $$I(x)=I_0\exp(-kx)$$ where $x$ is the vertical axis from the surface of the semiconductor. What is the probability for a photon $\gamma$ of wavelength $\lambda$ to create an electron - hole pair inside the slab? Thanks.

We will have to assume several things:

1) That your external Quantum Efficiency is unity (assuming you have a good AR coating).

- or conversely we are dealing only with the photons that make it into the substrate.

2) That your Si deep enough that all the photons are absorbed and that yo are concerned only with the photons absorbed within a certain thickness h.

- other wise you'll have to be concerned about reflections at the other interface too.

3) and we'll assume there is insignificant recombination

Since Si is an indirect bandgap material it's absorption coefficient does depend upon wavelength quite strongly.

In semi-conductor physics the coefficient is called alpha, here is a plot.

From the chart you pick your wavelength, which gives you your alpha. The absorption from the surface to a depth of h is simply: $$P=P_0(1-e^{-{\alpha(\lambda)}h})$$. This is the cumulative absorption in that depth of material (i.e. the integral).

Using the above data and assigning a 450 nm to "blue", 556 nm to "green", 630 nm to "red" and 850 nm to "NIR" we derive the following plot of internal QE. The total absorption is the area under the curve of each color. You will notice that blue gets absorbed closer to the surface. with NIR (Near IR) penetrates much deeper.

• You are assuming the quantum efficiency is equal to one. This means the probability for a photon to create an electron - hole pair is one inside the bulk of the slab. This is a strong assumption and I don't see how to agree. – Riccardo.Alestra Oct 11 '13 at 16:45
• This is wrong. Not every photon can create an electron-hole pair. The transition probability depends on the density of states, photon energy and matrix element for the transition. – mcodesmart Oct 11 '13 at 17:19
• Oh boy ... I am assuming the external QE ** is one. i.e. every photon **once ** it is the Si is subject to the probability of absorption as defined above. Since no external conditions are noted one simply needs to apply this to the photons that **do make it into the Si. – placeholder Oct 11 '13 at 18:10
• This is not wrong. Let me explain. The energy of the photon is the selected by the wavelength. The photon moving through the Si leaves a a probability trail of e/h pairs with a characteristic depth that \$e^{-1} \$ absorption. That formulate above is a probability. To check this out look at "Donati, Silvano". "Photodetectors: Devices, Circuits and Applications". Prentice Hall PTR, 1999. The graph above is from internal data that I've collected and verified. Do a little research. – placeholder Oct 11 '13 at 18:17
• Why are people down voting this good answer? @Riccardo.Alestra you define quantum efficiency (QE) incorrectly. QE is the fraction of electron hole pairs that decay radiatively. It has nothing to do with absorption. As pointed out above the transition probably depends on the density of states, however, it's convenient to use the macroscopic property of absorption coefficient as it encapsulates all of that information in a simple coefficient. A big +1 for your answer from me. – boyfarrell Oct 12 '13 at 1:06


text:
# Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?

Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$ ?

$$\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbf{Q}\}$$

$$\mathbf{Q}(\sqrt{2}+\sqrt{3}) = \lbrace a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \mathbf{Q} \rbrace$$

So if an element is in $\mathbf Q(\sqrt{2},\sqrt{3})$, then it is in $\mathbf{Q}(\sqrt{2}+\sqrt{3})$, because $\sqrt{6} = \sqrt{2}\sqrt{3}$.

How to conclude from there?

• $\mathbf{Q}(\sqrt{2}+\sqrt{3}) \not= \{a+b(\sqrt{2}+\sqrt{3})\ | a,b \in \mathbf{Q} \}$ because $\sqrt{2}+\sqrt{3}$ does not have degree 2 over $\mathbf{Q}$.
– lhf
Dec 22, 2011 at 14:28
• Hi lhf, then what is it? ? ? ? Dec 22, 2011 at 14:30
• $\alpha=\sqrt{2}+\sqrt{3}$ has degree 4 over $\mathbf{Q}$ and so a basis is $1,\alpha,\alpha^2,\alpha^3$.
– lhf
Dec 22, 2011 at 14:35
• @Tashi: This may just be an issue of notation, since your argument in the question seems to suggest that you thougt that $\{a+b(\sqrt{2}+\sqrt{3})\ | a,b \in \mathbf{Q} \}$ also includes $\sqrt2\sqrt3$. (It doesn't.) Dec 22, 2011 at 14:39

$\mathbb{Q}(\sqrt{2} + \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is clear.

Now note that $$(\sqrt{2} + \sqrt{3})^{-1} = \frac{1}{\sqrt{2} + \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{2 - 3} = \sqrt{3} - \sqrt{2}$$ hence $\sqrt{3} - \sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{2} + \sqrt{3} + \sqrt{3} - \sqrt{2} = 2 \sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\sqrt{3} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$. Note that by a similar argument you get $\sqrt{2} \in \mathbb{Q}(\sqrt{2} + \sqrt{3})$ and hence $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{Q}(\sqrt{2} + \sqrt{3})$.

To recap the notation: $\mathbb{Q}[x]$ denotes the ring of polynomials with rational coefficients. The square bracket notation $\mathbb{Q}[\sqrt{2}]$ means $\{p(\sqrt{2}) : p \in \mathbb{Q}[x]\}$. It's easy to show that $\mathbb{Q}[\sqrt{2}] = \{a+b\sqrt{2}:a,b,\in \mathbb{Q}\}.$

A really nice fact is that $\mathbb{Q}[\sqrt{2},\sqrt{3}] = \mathbb{Q}[\sqrt{2}][\sqrt{3}],$ where \begin{array}{ccc} \mathbb{Q}[\sqrt{2}][\sqrt{3}] &=& \{a+b\sqrt{3} : a,b \in \mathbb{Q}[\sqrt{2}] \} \\ \\ &=& \{p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6} : p,q,r,s \in \mathbb{Q}\}. \end{array} These all use square brackets because they are considered as rings. The round brackets give us the set of rational expressions, which are fields, e.g.

$$\mathbb{Q}(\sqrt{2},\sqrt{3}) = \left\{ \frac{\alpha}{\beta} : \alpha,\beta \in \mathbb{Q}[\sqrt{2},\sqrt{3}]\right\}$$

It turns out that, as sets, $\mathbb{Q}[\sqrt{2},\sqrt{3}] = \mathbb{Q}(\sqrt{2},\sqrt{3})$.

In turns of the representation of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ we have seen that, as a set, we have $\{p + q\sqrt{2} + r\sqrt{3} + s\sqrt{6}:p,q,r,s \in \mathbb{Q}\}$. There are many representations for this fiels, e.g. $\mathbb{Q}(1,\sqrt{2},\sqrt{3},\sqrt{6})$ or $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{6})$ or $\mathbb{Q}(1,\sqrt{2},\sqrt{3})$ or $\mathbb{Q}(\sqrt{2},\sqrt{3})$, etc. We can show that $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is also a representation of the same field too.

Think of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ as a $\mathbb{Q}$-vector space with $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ as a basis. Let $\gamma := \sqrt{2}+\sqrt{3}.$ We have $\gamma^2 = 5+2\sqrt{6},$ $\gamma^3 = 11\sqrt{2}+9\sqrt{3}$ and $\gamma^4 = 49 + 20\sqrt{6}$. Putting this together:

$$\left[\begin{array}{cccc} 0 & 1 & 1 & 0 \\ 5 & 0 & 0 & 2 \\ 0 & 11 & 9 & 0 \\ 49 & 0 & 0 & 20 \end{array}\right]\left[\begin{array}{c} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6} \end{array}\right] = \left[\begin{array}{c} \gamma \\ \gamma^2 \\ \gamma^3 \\ \gamma^4 \end{array}\right]$$

The 4-by-4 matrix on the left is non-singular, and so we can invert:

$$\left[\begin{array}{c} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6} \end{array}\right] = \frac{1}{2}\!\left[\begin{array}{cccc} 0 & 20 & 0 & -2 \\ -9 & 0 & 1 & 0 \\ 11 & 0 & -1 & 0 \\ 0 & -49 & 0 & 5 \end{array}\right]\left[\begin{array}{c} \gamma \\ \gamma^2 \\ \gamma^3 \\ \gamma^4 \end{array}\right]$$

This tells us that $1$, $\sqrt{2}$, $\sqrt{3}$ and $\sqrt{6}$ can all be expressed as rational polynomials in $\gamma = \sqrt{2}+\sqrt{3}$.

\begin{array}{ccc} 10\gamma^2-\gamma^4 &=& 1 \\ \tfrac{1}{2}(\gamma^3-9\gamma) &=& \sqrt{2} \\ \tfrac{1}{2}(11\gamma - \gamma^3) &=& \sqrt{3} \\ \tfrac{1}{2}(5\gamma^4-49\gamma^2) &=& \sqrt{6} \end{array}

It follows that $\mathbb{Q}(1,\sqrt{2},\sqrt{3},\sqrt{6}) \cong \mathbb{Q}(\gamma) = \mathbb{Q}(\sqrt{2}+\sqrt{3}).$

• I like this version, since it's clear how to extend to the situation where more roots are adjoined. May 28, 2013 at 19:23
• I'm a bit late to the party but this is a beautiful answer - If I could +2 I would, however, +1 anyway. Best, Bacon.
– user284001
Nov 3, 2015 at 9:26
• the best explanation for a general set up Apr 27, 2016 at 14:27
• How does it turn out that $\mathbb{Q}[\sqrt{2},\sqrt{3}] = \mathbb{Q}(\sqrt{2},\sqrt{3})$? How does one prove this? Sep 19, 2018 at 8:16
• @HansStricker They are the same as sets, i.e. they have the same elements. $\mathbb Q[\sqrt 2,\sqrt 3]$ contains $a+b\sqrt 2+c\sqrt 3 + d\sqrt 6$, while $\mathbb Q(\sqrt 2,\sqrt 3)$ contains all fractions $$\frac{a+b\sqrt 2+c\sqrt 3 + d\sqrt 6}{e + f\sqrt 2 + g\sqrt 3 + h\sqrt 6}$$ By "rationalising the denominator" we can write every element of $\mathbb Q(\sqrt 2,\sqrt 3)$ in the form $r+s\sqrt 2 + t\sqrt 3 + u\sqrt 6$, meaning $\mathbb Q(\sqrt 2,\sqrt 3) \subseteq \mathbb Q[\sqrt 2,\sqrt 3]$. Clearly, $\mathbb Q[\sqrt 2,\sqrt 3] \subseteq \mathbb Q(\sqrt 2,\sqrt 3)$. Sep 19, 2018 at 15:54

Hint $$\$$ If a field $$\rm F$$ has two $$\rm F$$-linear independent combinations of $$\rm\ \sqrt{a},\ \sqrt{b}\$$ then you can solve for $$\rm\ \sqrt{a},\ \sqrt{b}\$$ in $$\rm F$$. For example, the Primitive Element Theorem works that way, obtaining two such independent combinations by Pigeonholing the infinite set $$\rm\ F(\sqrt{a} + r\ \sqrt{b}),\ r \in F,\ |F| = \infty,\,$$ into the finitely many fields between $$\rm F$$ and $$\rm\ F(\sqrt{a}, \sqrt{b}),\,$$ e.g. see here or here.

In this case it's simpler to notice $$\rm\ E = \mathbb Q(\sqrt{a} + \sqrt{b})\$$ contains the independent $$\rm\ \sqrt{a} - \sqrt{b}\$$ since

$$\rm \sqrt{a}\ -\ \sqrt{b}\ =\ \dfrac{a-b}{\sqrt{a}+\sqrt{b}}\ \in\ E = \mathbb Q(\sqrt{a}+\sqrt{b})$$

To be explicit, notice that $$\rm\ u = \sqrt{a}+\sqrt{b},\ v = \sqrt{a}-\sqrt{b}\in E\$$ so solving the linear system for the roots yields $$\rm\ \sqrt{a}\ =\ (u+v)/2,\ \ \sqrt{b}\ =\ (v-u)/2,\$$ both of which are clearly $$\rm\in E,\:$$ since $$\rm\:u,\:v\in E\:$$ and $$\rm\:2\ne 0\:$$ in $$\rm\:E,\:$$ so $$\rm\:1/2\in E.\:$$ This works over any field where $$\rm\:2\ne 0,\:$$ i.e. where the determinant (here $$2$$) of the linear system is invertible, i.e. where the linear combinations $$\rm\:u,v\:$$ of the square-roots are linearly independent over the base field.

If it's allowed to use the Galois theory, it can be proved as following. Since the subgroup of the Galois group of the field extension $\mathbb{Q} (\sqrt2,\sqrt 3)$ over $\mathbb{Q}$ which the subfield $\mathbb{Q}(\sqrt 2+\sqrt 3)$ is trivial, therefor we know the result by the Galois theory. I admit it is not too trivial since one has to verify something as said above.

Also note, $\sqrt{2}=\frac{(\sqrt{2}+\sqrt{3})^3-9(\sqrt{2}+\sqrt{3})}{2}$ and $-\sqrt{3}=\frac{(\sqrt{2}+\sqrt{3})^3-11(\sqrt{2}+\sqrt{3})}{2}$ and so done.

Now suppose wants to show $\mathbb{Z}[\sqrt{2},\sqrt{3}]\neq \mathbb{Z}[\sqrt{2}+\sqrt{3}]$ then, note minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Z}$ is of degree $4$.

So we can write $\mathbb{Z}[\sqrt{2}+\sqrt{3}]=\{a_1+a_2x+a_3x^2+a_4x^3|x=\sqrt{2}+\sqrt{3},a_i\in \mathbb{Z}\}$ now simple case chase shows $\sqrt{2}$ not in $\mathbb{Z}[\sqrt{2}+\sqrt{3}]$

Another solution:

Clearly it has to be a subfield of $$\mathbf{Q}(\sqrt{2}, \sqrt{3})$$. Thus its degree has to be one of these numbers: $$1,2,4$$. We just have to rule out it is of degree $$2$$. If it is, then there is a minimal polynomial of degree two s.t $$\sqrt{2}+\sqrt{3}$$ is its root. However, if there was such polynomial, say $$x^2+bx+c$$ then $$5+c+2\sqrt{6}+b(\sqrt{3}+\sqrt{2})=0$$ which is impossible as $$1,\sqrt{2},\sqrt{3},\sqrt{6}$$ is a basis of $$\mathbf{Q}(\sqrt{2}, \sqrt{3})$$ and the equation implies it is linearly dependent!


text:
# $X:=\{z \in \mathbb C : |z|\le 1\}$ ; is $\{p(z,\bar z) | p(x,y) \in \mathbb R[x,y] \}$ dense in $C(X , \mathbb C)$ under sup metric?

Let $X:=\{z \in \mathbb C : |z|\le 1\}$ , is it true that any element in $C(X , \mathbb C)$ can be uniformly approximated by polynomials , in $z, \bar z$ , with real co-efficients ? If we wanted uniform approximation by polynomials in $z, \bar z$ , with complex co-efficients then I know it would be true by Complex version of Stone-Weierstrass theorem . But I have no idea what happens if we want real-coefficients ( I can't see whether the set $\{p(z,\bar z) | p(x,y) \in \mathbb R[x,y] \}$ is a subalgebra over $\mathbb C$ or not ) . Please help . Thanks in advance

• Unless I'm mistaken, the function $p(z, \bar{z}) = iz$ is a good example to have in mind. – Cameron Williams May 10 '17 at 11:39
• @CameronWilliams : what to do with that polynomial ? – user228168 May 10 '17 at 13:46
• Can you approximate that function? – Cameron Williams May 10 '17 at 14:09

To expand on the counterexample $p(z, \bar z)=iz$ given by Cameron Williams, note that when restricted to a subinterval of real line, such as $[-1,1]$, the set $\{p(z,\bar z) | p(x,y) \in \mathbb R[x,y] \}$ reduces to $\mathbb{R}[x]$. Clearly, any function one can approximate by the elements of $\mathbb{R}[x]$ must be real-valued on $[-1,1]$, and this does not include $iz$.


text:
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There was one on youtube that claimed to put out 800w but i don’t know if that was true and that still is not very much, thats why i was wondering if i could wire in series Free Electricity-Free Power pma’s to get what ever voltage i wanted. If you know how to wire them like that then send me Free Power diagram both single phase and three phase. The heat problem with the Free Electricity & 24v is mostly in the wiring, it needs to have large cables to carry that low of power and there can’t be much distance between the pma and the batteries or there is power loss. Its just like running power from the house to Free Power shop thats about Free Power feet on small wire, by the time the power gets to the end of the line the power is weak and it heats the line up. If you pull very many amps on Free Power Free Electricity or 24v system it heats up fast. Also, i don’t know the metric system. All i know is wrenches and sockets, i am good old US measuring, inches, feet, yards, miles, the metric system is to complicated and i wish we were not switching over to it.
Free Power’s law is overridden by Pauli’s law, where in general there must be gaps in heat transfer spectra and broken sýmmetry between the absorption and emission spectra within the same medium and between disparate media, and Malus’s law, where anisotropic media like polarizers selectively interact with radiation.

Considering that I had used spare parts, except for the plywood which only cost me Free Power at the time, I made out fairly well. Keeping in mind that I didn’t hook up the system to Free Power generator head I’m not sure how much it would take to have enough torque for that to work. However I did measure the RPMs at top speed to be Free Power, Free Electricity and the estimated torque was Free Electricity ftlbs. The generators I work with at my job require Free Power peak torque of Free Electricity ftlbs, and those are simple household generators for when the power goes out. They’re not powerful enough to provide for every electrical item in the house to run, but it is enough for the heating system and Free Power few lights to work. Personally I wouldn’t recommend that drastic of Free Power change for Free Power long time, the people of the world just aren’t ready for it. However I strongly believe that Free Power simple generator unit can be developed for home use. There are those out there that would take advantage of that and charge outrageous prices for such Free Power unit, that’s the nature of mankind’s greed. To Nittolo and Free Electricity ; You guys are absolutely hilarious. I have never laughed so hard reading Free Power serious set of postings. You should seriously write some of this down and send it to Hollywood. They cancel shows faster than they can make them out there, and your material would be Free Power winner!
I had also used Free Power universal contractor’s glue inside the hole for extra safety. You don’t need to worry about this on the outside sections. Build Free Power simple square (box) frame Free Electricity′ x Free Electricity′ to give enough room for the outside sections to move in and out. The “depth” or length of it will depend on how many wheels you have in it. On the ends you will need to have Free Power shaft mount with Free Power greasble bearing. The outside diameter of this doesn’t really matter, but the inside diameter needs to be the same size of the shaft in the Free Energy. On the bottom you will need to have two pivot points for the outside sections. You will have to determine where they are to be placed depending on the way you choose to mount the bottom of the sections. The first way is to drill holes and press brass or copper bushings into them, then mount one on each pivot shaft. (That is what I did and it worked well.) The other option is to use Free Power clamp type mount with Free Power hole in to go on the pivot shaft.
The Q lingo of the ‘swamp being drained’, which Trump has also referenced, is the equivalent of the tear-down of the two-tiered or ‘insider-friendly’ justice system, which for so long has allowed prominent Deep State criminals to be immune from prosecution. Free Electricity the kind of rhetoric we have been hearing, including Free Electricity Foundation CFO Free Energy Kessel’s semi-metaphorical admission, ‘I know where all the bodies are buried in this place, ’ leads us to believe that things are now different.
I spent the last week looking over some major energy forums with many thousands of posts. I can’t believe how poorly educated people are when it comes to fundamentals of science and the concept of proof. It has become cult like, where belief has overcome reason. Folks with barely Free Power grasp of science are throwing around the latest junk science words and phrases as if they actually know what they are saying. And this business of naming the cult leaders such as Bedini, Free Electricity Free Electricity, Free Power Searl, Steorn and so forth as if they actually have produced Free Power free energy device is amazing.
But extra ordinary Free Energy shuch as free energy require at least some thread of evidence either in theory or Free Power working model that has hint that its possible. Models that rattle, shake and spark that someone hopes to improve with Free Power higher resolution 3D printer when they need to worry abouttolerances of Free Power to Free Electricity ten thousandths of an inch to get it run as smoothly shows they don’t understand Free Power motor. The entire discussion shows Free Power real lack of under standing. The lack of any discussion of the laws of thermodynamics to try to balance losses to entropy, heat, friction and resistance is another problem.
Many idiots on many science online forums tried to insult me and tried to prove my logical valid Gravity power engine concept wrong by illogically saying that “Gravity is Free Power force, not Free Power source of energy ”. How foolish that idiot’s statement appears to be. Interesting posts, pro and con. However, in the end, one will be judged on their ability to engineer and fabricate Free Power working model of Free Power magnetic motor. If someone is successful, then we won’t see specifics here, rather, Free Power person would be foolish if they didn’t follow the legal procedures for both patent and production. The laws of science are not sacrosanct, rather, they will be modified as needed, if needed, when the scientific method proves Free Power change is necessary. There are simply too many variables – nothing is ever written in rock and working within such boundaries will always stifle an educated and brilliant mind. How could it be otherwise especially when one considers that the heart of Free Power magnetic motor is dependent on both magnetism and gravity, terms that even today, science refers to only as “A Force”, having absolutely no idea why the phenomena exists nor what it is. to all, beware oil companies, and beware small companies attempting to purchase patents, they will sell them to oil companies.
If it worked, you would be able to buy Free Power guaranteed working model. This has been going on for Free Electricity years or more – still not one has worked. Ignorance of the laws of physics, does not allow you to break those laws. Im not suppose to write here, but what you people here believe is possible, are true. The only problem is if one wants to create what we call “Magnetic Rotation”, one can not use the fields. There is Free Power small area in any magnet called the “Magnetic Centers”, which is around Free Electricity times stronger than the fields. The sequence is before pole center and after face center, and there for unlike other motors one must mesh the stationary centers and work the rotation from the inner of the center to the outer. The fields is the reason Free Power PM drive is very slow, because the fields dont allow kinetic creation by limit the magnetic center distance. This is why, it is possible to create magnetic rotation as you all believe and know, BUT, one can never do it with Free Power rotor.

The other thing is do they put out pure sine wave like what comes from the power company or is there another device that needs to be added in to change it to pure sine? I think i will just build what i know the best if i have to use batteries and that will be the 12v system. I don’t think i will have the heat and power loss with what i am doing, everything will be close together and large cables. Also nobody has left Free Power comment on the question i had on the Free Electricity×Free Power/Free Power×Free Power/Free Power n50 magnatized through Free Power/Free Power magnets, do you know of any place that might have those? Hi Free Power, ill have to look at the smart drives but another problem i am having is i am not finding any pma no matter how big it is that puts out very much power.
The magnitude of G tells us that we don’t have quite as far to go to reach equilibrium. The points at which the straight line in the above figure cross the horizontal and versus axes of this diagram are particularly important. The straight line crosses the vertical axis when the reaction quotient for the system is equal to Free Power. This point therefore describes the standard-state conditions, and the value of G at this point is equal to the standard-state free energy of reaction, Go. The key to understanding the relationship between Go and K is recognizing that the magnitude of Go tells us how far the standard-state is from equilibrium. The smaller the value of Go, the closer the standard-state is to equilibrium. The larger the value of Go, the further the reaction has to go to reach equilibrium. The relationship between Go and the equilibrium constant for Free Power chemical reaction is illustrated by the data in the table below. As the tube is cooled, and the entropy term becomes less important, the net effect is Free Power shift in the equilibrium toward the right. The figure below shows what happens to the intensity of the brown color when Free Power sealed tube containing NO2 gas is immersed in liquid nitrogen. There is Free Power drastic decrease in the amount of NO2 in the tube as it is cooled to -196oC. Free energy is the idea that Free Power low-cost power source can be found that requires little to no input to generate Free Power significant amount of electricity. Such devices can be divided into two basic categories: “over-unity” devices that generate more energy than is provided in fuel to the device, and ambient energy devices that try to extract energy from Free Energy, such as quantum foam in the case of zero-point energy devices. Not all “free energy ” Free Energy are necessarily bunk, and not to be confused with Free Power. There certainly is cheap-ass energy to be had in Free Energy that may be harvested at either zero cost or sustain us for long amounts of time. Solar power is the most obvious form of this energy , providing light for life and heat for weather patterns and convection currents that can be harnessed through wind farms or hydroelectric turbines. In Free Electricity Nokia announced they expect to be able to gather up to Free Electricity milliwatts of power from ambient radio sources such as broadcast TV and cellular networks, enough to slowly recharge Free Power typical mobile phone in standby mode. [Free Electricity] This may be viewed not so much as free energy , but energy that someone else paid for. Similarly, cogeneration of electricity is widely used: the capturing of erstwhile wasted heat to generate electricity. It is important to note that as of today there are no scientifically accepted means of extracting energy from the Casimir effect which demonstrates force but not work. Most such devices are generally found to be unworkable. Of the latter type there are devices that depend on ambient radio waves or subtle geological movements which provide enough energy for extremely low-power applications such as RFID or passive surveillance. [Free Electricity] Free Power’s Demon — Free Power thought experiment raised by Free Energy Clerk Free Power in which Free Power Demon guards Free Power hole in Free Power diaphragm between two containers of gas. Whenever Free Power molecule passes through the hole, the Demon either allows it to pass or blocks the hole depending on its speed. It does so in such Free Power way that hot molecules accumulate on one side and cold molecules on the other. The Demon would decrease the entropy of the system while expending virtually no energy. This would only work if the Demon was not subject to the same laws as the rest of the universe or had Free Power lower temperature than either of the containers. Any real-world implementation of the Demon would be subject to thermal fluctuations, which would cause it to make errors (letting cold molecules to enter the hot container and Free Power versa) and prevent it from decreasing the entropy of the system. In chemistry, Free Power spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly, because spontaneity is not related to kinetics or reaction rate. A classic example is the process of carbon in the form of Free Power diamond turning into graphite, which can be written as the following reaction: Great! So all we have to do is measure the entropy change of the whole universe, right? Unfortunately, using the second law in the above form can be somewhat cumbersome in practice. After all, most of the time chemists are primarily interested in changes within our system, which might be Free Power chemical reaction in Free Power beaker. Free Power we really have to investigate the whole universe, too? (Not that chemists are lazy or anything, but how would we even do that?) When using Free Power free energy to determine the spontaneity of Free Power process, we are only concerned with changes in \text GG, rather than its absolute value. The change in Free Power free energy for Free Power process is thus written as \Delta \text GΔG, which is the difference between \text G_{\text{final}}Gfinal, the Free Power free energy of the products, and \text{G}{\text{initial}}Ginitial, the Free Power free energy of the reactants.
This expression has commonly been interpreted to mean that work is extracted from the internal energy U while TS represents energy not available to perform work. However, this is incorrect. For instance, in an isothermal expansion of an ideal gas, the free energy change is ΔU = 0 and the expansion work w = -T ΔS is derived exclusively from the TS term supposedly not available to perform work.


text:
2 added two things to make it more clear, hopefully..

This property ( or rather the generalized version by Darij using (a,b)-shuffles ) means that f is what is called a "symmetral mould" in the context of Ecalle's theory of moulds. There is a related notion of "alternal mould" where the right hand side is 0 rather than a product of two f.

Here is just one reference among many : page 591 of

Jean Ecalle; Bruno Vallet The arborification-coarborification transform: analytic, combinatorial, and algebraic aspects

This may not be transparent when looking at this article. Maybe page 2 of my article

would be more clear, but it only defines "alternal moulds".

• The symmetral property is really a property of sequence of functions $f_n$, with $f_n$ a function of $n$ variables $x_1,\dots,x_n$.

• The notions of alternal and symmetral moulds, when considered under some specific point of view, turn into the notion of primitive and group-like element in a Hopf algebra.

1

This property ( or rather the generalized version by Darij using (a,b)-shuffles ) means that f is what is called a "symmetral mould" in the context of Ecalle's theory of moulds. There is a related notion of "alternal mould" where the right hand side is 0 rather than a product of two f.

Here is just one reference among many : page 591 of

Jean Ecalle; Bruno Vallet The arborification-coarborification transform: analytic, combinatorial, and algebraic aspects

This may not be transparent when looking at this article. Maybe page 2 of my article


text:
1. ## Integral

Evaluate the integral $\int_0^2(arctan(\pi x)-arctan (x)dx.$

Hint: Write the integrand as an integral.
This is what i've done so far:

$y=Aarctan(Bx) \Rightarrow \ tan \left( \frac{y}{A} \right)=B$

$\frac{dy}{dx}\frac{1}{cos^2 \left(\frac{y}{A} \right)}=B \Rightarrow \ \frac{dy}{dx}=Bcos^2 \left( \frac{y}{A} \right)$

This was to just get my bearings with what the integral should look like.

In double integral form, the initial integral is:

$\int_0^2 \int_x^{\pi x} cos^2 (y) \ dy \ dx$

Call the region we're integrating over R.
Let $R_1=\{ (x,y):\frac{y}{\pi} \leq x \leq y, \ 0 \leq y \leq 2 \}$ and $R_2=\{(x,y):\frac{y}{\pi} \leq x \leq 2, \ 2 \leq y \leq 2 \pi \}$.

$R= R_1 \cup R_2$.

This gives:

$\int_0^2 \int_{\frac{y}{\pi}}^y cos^2 y \ dx \ dy + \int_2^{2 \pi} \int_{\frac{y}{\pi}}^2 cos^2 y \ dx \ dy$

$\int_0^2 ycos^2 (y)-\frac{y}{\pi} cos^2(y)dy+\int_2^{2 \pi} 2cos^2(y)-\frac{y}{\pi}cos^2(y)dy$

$\int_0^{2 \pi} \left( y+2-\frac{2y}{\pi} \right) cos^2(y) \ dy$

Integrating by parts gives:

$\left[ arctan(y) \left( y+2-\frac{2y}{\pi} \right) \right]_0^{2 \pi}- \left( 1-\frac{2}{\pi} \right) \int_0^{2 \pi} arctan (y)\ dy$

$(2 \pi -2)arctan(2 \pi)- \left(1-\frac{2}{\pi} \right) \left[yarctan(y)-\frac{1}{2}ln (1+y^2) \right]_0^{2 \pi}$

$(2 \pi -2)arctan(2 \pi)-\left( 1-\frac{2}{\pi} \right) \left(2 \pi arctan (2 \pi) -\frac{1}{2}ln(1+4 \pi^2) \right)$

$2arctan (2 \pi)+\left( \frac{1}{2}-\frac{1}{\pi} \right) ln (1+4 \pi^2)$

Secondly, is there an easier way to do it than this??

2. Originally Posted by Showcase_22
This is what i've done so far:

$y=Aarctan(Bx) \Rightarrow \ tan \left( \frac{y}{A} \right)=B$

$\frac{dy}{dx}\frac{1}{cos^2 \left(\frac{y}{A} \right)}=B \Rightarrow \ \frac{dy}{dx}=Bcos^2 \left( \frac{y}{A} \right)$

This was to just get my bearings with what the integral should look like.

In double integral form, the initial integral is:

$\int_0^2 \int_x^{\pi x} cos^2 (y) \ dy \ dx$

Call the region we're integrating over R.
Let $R_1=\{ (x,y):\frac{y}{\pi} \leq x \leq y, \ 0 \leq y \leq 2 \}$ and $R_2=\{(x,y):\frac{y}{\pi} \leq x \leq 2, \ 2 \leq y \leq 2 \pi \}$.

$R= R_1 \cup R_2$.

This gives:

$\int_0^2 \int_{\frac{y}{\pi}}^y cos^2 y \ dx \ dy + \int_2^{2 \pi} \int_{\frac{y}{\pi}}^2 cos^2 y \ dx \ dy$

$\int_0^2 ycos^2 (y)-\frac{y}{\pi} cos^2(y)dy+\int_2^{2 \pi} 2cos^2(y)-\frac{y}{\pi}cos^2(y)dy$

$\int_0^{2 \pi} \left( y+2-\frac{2y}{\pi} \right) cos^2(y) \ dy$

Integrating by parts gives:

$\left[ arctan(y) \left( y+2-\frac{2y}{\pi} \right) \right]_0^{2 \pi}- \left( 1-\frac{2}{\pi} \right) \int_0^{2 \pi} arctan (y)\ dy$

$(2 \pi -2)arctan(2 \pi)- \left(1-\frac{2}{\pi} \right) \left[yarctan(y)-\frac{1}{2}ln (1+y^2) \right]_0^{2 \pi}$

$(2 \pi -2)arctan(2 \pi)-\left( 1-\frac{2}{\pi} \right) \left(2 \pi arctan (2 \pi) -\frac{1}{2}ln(1+4 \pi^2) \right)$

$2arctan (2 \pi)+\left( \frac{1}{2}-\frac{1}{\pi} \right) ln (1+4 \pi^2)$

Secondly, is there an easier way to do it than this??

$I = \int_{x = 0}^{x = 2} \int_{t = x}^{t = \pi x} \frac{1}{1 + t^2} \, dt \, dx$

and then reverse the order of integration (do the reversal carefully, it needs to be broken up into two seperate double integrals).

3. That's probably a better way of doing it

The reversal is the same as what I reversed previously so the integral is now $
\int_0^2 \int_{\frac{y}{\pi}}^y \frac{1}{1+y^2} \ dx \ dy + \int_2^{2 \pi} \int_{\frac{y}{\pi}}^2 \frac{1}{1+y^2} \ dx \ dy
$
(where y=t).

$\int_0^2 \left[ \frac{x}{1+y^2} \right]_{\frac{y}{\pi}}^y \ dy+\int_2^{2 \pi} \left[ \frac{x}{1+y^2} \right]_{\frac{y}{\pi}}^2 \ dy$

$\int_0^2 \frac{y}{1+y^2}-\frac{\frac{y}{\pi}}{1+y^2} \ dy+ \int_2^{2 \pi} \frac{2}{1+y^2}-\frac{\frac{y}{\pi}}{1+y^2} \ dy$

$\int_0^{2 \pi} \frac{y+2-\frac{2y}{\pi}}{1+y^2} \ dy$

Let $u=1+y^2 \Rightarrow \ \frac{du}{dy}=2y$

$\int_1^{4 \pi^2+1} \frac{y-\frac{2y}{\pi}}{u} \ \frac{du}{2y}+\int_0^{2\pi}\frac{2}{1+y^2} \ dy$

$\frac{1}{2}\int_1^{4 \pi ^2+1} \frac{1-\frac{2}{\pi}}{u} \ du+ \int_0^{2 \pi}\frac{2}{1+y^2} \ dy$

$\frac{1}{2}\left( 1-\frac{2}{\pi} \right) \left[ ln(u) \right]_1^{4 \pi^2+1}+2 \left[arctan(y) \right]_0^{2 \pi}$

$\frac{1}{2}\left( 1-\frac{2}{\pi} \right) ln(4 \pi^2+1)+2arctan( 2\pi)$

$\left( \frac{1}{2}-\frac{1}{\pi} \right) ln(4 \pi^2+1)+2arctan( 2\pi)$

Which is the same as before.

Well that was a good way of doing it, but can't you do the initial integral without splitting it up into two double integrals?

4. Why not integrate $\int_0^2 \tan^{-1}\pi x - \tan^{-1}x\, dx$ directly using parts?

5. I have no idea, I only did it this way because this is how the question wanted me to do it.

Perhaps there exist some integrals that can only be solved by turning them into double integrals. If you look at the second method, $tan^{-1}x$ is never integrated directly.

I can't actually think of any integrals that can only be solved by turning them into a double integral, so i'm only really speculating about where this method would be useful.

6. Originally Posted by Showcase_22

but can't you do the initial integral without splitting it up into two double integrals?
$\int_{0}^{2}{\left( \arctan \pi x-\arctan x \right)\,dx}=\int_{0}^{2}{\int_{1}^{\pi }{\frac{x}{1+t^{2}x^{2}}\,dt}\,dx}=\frac{1}{2}\int _{1}^{\pi }{\frac{\ln \left( 4t^{2}+1 \right)}{t^{2}}\,dt}.$

7. Originally Posted by Showcase_22
I have no idea, I only did it this way because this is how the question wanted me to do it.

Perhaps there exist some integrals that can only be solved by turning them into double integrals. If you look at the second method, $tan^{-1}x$ is never integrated directly.

I can't actually think of any integrals that can only be solved by turning them into a double integral, so i'm only really speculating about where this method would be useful.
Here's one would it would be beneficial (this is from Stewart's Calculus book)

$\int_0^{\infty} \frac{\tan^{-1}\pi x - \tan^{-1}x}{x}\, dx$

8. I was just trying Danny's post and there's something about Krizalid's post I don't quite understand.

$
\int_{0}^{2}{\left( \arctan \pi x-\arctan x \right)\,dx}=\int_{0}^{2}{\int_{1}^{\pi }{\frac{\color{red}{x}}{1+t^{2}x^{2}}\,dt}\,dx}=\f rac{1}{2}\int_{1}^{\pi }{\frac{\ln \left( 4t^{2}+1 \right)}{t^{2}}\,dt}.
$
Should that be an x or a 1?

$
\int_{0}^{2}{\int_{1}^{\pi }{\frac{x}{1+t^{2}x^{2}}\,dt}\,dx}
$

$
\int_{0}^{2}x{\int_{1}^{\pi }{\frac{1}{1+t^{2}x^{2}}\,dt}\,dx}
$

Let $tx=tan(u) \Rightarrow \frac{dt}{du}=sec^2(u)$

$\int_0^2 x \int_{arctan(x)}^{arctan(\pi x)} du \ dx$

$\int_0^2 x(arctan( \pi x)-arctan(x)) dx$

I just want to clarify this before I try the other integral.

9. Originally Posted by Showcase_22
I was just trying Danny's post and there's something about Krizalid's post I don't quite understand.

Should that be an x or a 1?

$
\int_{0}^{2}{\int_{1}^{\pi }{\frac{x}{1+t^{2}x^{2}}\,dt}\,dx}
$

$
\int_{0}^{2}x{\int_{1}^{\pi }{\frac{1}{1+t^{2}x^{2}}\,dt}\,dx}
$

Let $tx=tan(u) \Rightarrow \frac{dt}{du}=sec^2(u)$ where did your x go ?

$\int_0^2 x \int_{arctan(x)}^{arctan(\pi x)} du \ dx$

$\int_0^2 x(arctan( \pi x)-arctan(x)) dx$

I just want to clarify this before I try the other integral.
..

Bobak

10. I think I just did something a bit silly.

I thought that since the first integral is w.r.t t, then one is able to take the x (which is treated as a constant) out of the integral.

Can you do that?

11. Originally Posted by Showcase_22
I think I just did something a bit silly.

I thought that since the first integral is w.r.t t, then one is able to take the x (which is treated as a constant) out of the integral.

Can you do that?

you have $t =\frac{\tan u }{x}$ with x constant
$\frac{dt}{du} = \frac{sec^2 u }{x}$

12. Ah no, I see I was wrong (pretty much for the same reason I was defending myself with)

Anyway, i'll get cracking on the other integral which was posted.