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https://kr.mathworks.com/matlabcentral/cody/problems/2480-change-the-first-and-last-diagonal-element-of-the-identity-matrix-to-zero/solutions/553113
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Cody
# Problem 2480. Change the first and last diagonal element of the identity matrix to zero
Solution 553113
Submitted on 7 Jan 2015 by Paul Berglund
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% Test 1 x = 5; y_correct = [0 0 0 0 0;0 1 0 0 0; 0 0 1 0 0; 0 0 0 1 0;0 0 0 0 0]; assert(isequal(eye_first_last(x),y_correct))
2 Pass
%% Test 2 x = 2; y_correct = [0 0;0 0]; assert(isequal(eye_first_last(x),y_correct))
3 Pass
%% Test 3 x=1; y_correct = [0]; assert(isequal(eye_first_last(x),y_correct))
4 Pass
%% Test 4 x=3; y_correct = [0 0 0;0 1 0;0 0 0]; assert(isequal(eye_first_last(x),y_correct))
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Problems & Puzzles: Puzzles
Puzzle 430. Grimm's conjecture
Grimm's conjecture states that to each element of a set of consecutive composite numbers one can assign a distinct prime that divides it. (see, 1, 2)
It has also been proved that "there are finitely many exceptions" to this Conjecture (B32, R.K.Guy)
Example:
Consecutive composite 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 Prime factorization 2^2*131 3*5^2*7 2*263 17*31 2^4*3*11 23^2 2*5*53 3^2*59 2^2*7*19 13*41 2*3*89 5*107 2^3*67 3*179 2*269 7^2*11 2^2*3^3*5 Set of distinct prime divisors 131 3 263 17 11 23 53 59 19 41 89 107 67 179 269 7 2
Q1. Describe a simple algorithm in order to compute the set of distinct prime divisors for a given range of consecutive composites that pass the Grimm's conjecture.
(perhaps you would like to test your algorithm with the set of 71 composites from 31398 to 31468)
Q2. Apply your algorithm to each gap of primes and make your own report for the extent of your validation of this conjecture or find the smallest exception.
Contributions came from Jan van Delden.
***
Jan wrote:
Q1: Suppose there are k consecutive composites between two successive primes.
If we factor any composite in this range a factor f >=k will necessarily only be encountered a single time
(among all these composites), otherwise there would be two different multiples of this same factor f in the
given range, say a.f and b.f. With b.f>a.f we would get b.f-a.f = (b-a).f>=k which is impossible.
In order to find an assignment of prime divisors to a composite it will therefore suffice to find such an
assignment for the so-called smooth numbers:khaving only prime factors < k.
An algorithm will therefore be:
Divide each composite c through by 2,3,5,.. <k, until these factors aren't present anymore. Remember
the primes used and the remaining factor R.
If R>1 assign any prime factor in R to the corresponding composite. (Any method, for instance trying the next prime>=k etc.).
If R=1 there remains a set of composites which are smooth.
I) Check for every smooth composite whether there's just one prime. Assign this prime if true and remove this prime
from the set of factors for the other composites.
Repeat this process until there are no more composites with just one factor. (BTW, if no more factor present
we have a counterexample to Grimm's conjecture).
II) For the remaining smooth composites assign the biggest prime out of the remaining set of prime factors. Start with
the largest composite and work your way down.
For instance with p=523 (example given) the last step would give
composite primedivisors
525: 3,5,7
528: 2,3,11
539: 7,11
540: 2,3,5
I'm actually hoping it will always work (no proof, if it exists)
540: 5
539: 11
528: 3 (11 already used and eliminated by 539)
525: 7
The descending part is essential. For instance the given p=523 the ascending method would fail
525: 7
528: 11
539: both factors are already assigned and eliminated.
540: 3
If step II) fails one could always resort to a brute force method (using recursion for instance).
I've tested until 10^8 and it didn't let me down (so far). Furthermore if one tries to estimate the average
number of smooth composite numbers and compares them to the average size of the number of factors < n,
using well known asymptotic formulae, it's possible to show that:
- the 'average' number of smooth numbers will decrease with increasing x (roughly speaking; this formula
shows a maximum around 9000 and then decreases(*))
- the 'average' number of factors will increase! [~ln(x)^2/(2*ln(ln(x)), using Cramers conjecture and pi(x)~x/ln(x)]
So in average there will be an abundance of factors to choose from when assigning these to the smooth
composites (once x is relatively large).
(*) I used a formula found in Prime numbers and computer methods for factorization, page 164, the data do support
this, in the sense that the maximum number of smooth numbers increase and then decrease (I didn't calculate averages..).
Q2: At first I was looking for the situation where the ratio (see below) is small (i.e. number of different factors
in the remaining number of smooth composites is small). Then I decided I'd better calculate the number of possible
assignments for the 'smooth' composites in the sequence of consecutive composites, to be sure whether or not
this is a good indicator of 'close calls'.
In these tables:
p prime before the sequence of consecutive composites
nc number of composites in sequence
ns number of smooth composites (having all prime factors < nc)
(red if this is equal to the maximum ns in the same range
r ratio of total number of different prime factors of above composites / ns
(blue if the ratio is equal to the minimum ratio for all p in the same range)
min minimum number of possible assignments of these factors to the composites
max maximum number of possible assignments of these factors to the composites
Log10[p] p min ns r nc 1-2 23 1 2 1.0000 5 31 1 2 1.0000 5 2-3 241 3 3 1.3333 9 619 3 2 2.0000 11 3-4 1021 2 2 1.5000 9 3121 2 3 1.5000 15 4-5 32749 4 2 2.5000 21 83497 4 2 2.5000 39 5-6 131071 4 2 2.5000 29 6-7 1419839 4 2 2.5000 37 7-8 33554393 5 2 3.0000 73
Log10[p] p max ns r nc 1-2 89 4 2 1.5000 7 2-3 523 14 4 1.2500 17 3-4 9973 223 6 1.6667 33 4-5 31397 11191 10 1.6000 71 5-6 370261 29664 8 2.3750 111 6-7 7230331 26775 7 2.7143 147 7-8 46006769 61108 7 3.4286 197
There are more solutions if one wants to find the maximum for ns. For instance 1327 is ‘missed’ which has ns=7,
r=1.4286,nc=33, with 78 possible assignments.
From these two tables we can see that a small ratio r is not a good criterion for a near miss. In fact we see
that a maximum number of assignments can have a minimum ratio (second table). And the minimum ratio is
not achieved many times.
The second table does suggest that a maximum for ns could be used as indicator for a maximum number of
assignments. But finding minimum ns is useless. (If ns<=2, there’s always at least one assignment possible).
Apparently the distribution of the remaining set of prime factors over the smooth composites is essential.
***
C. Rivera wrote (Feb 4, 08):
I find very interesting the Delden's contribution specially because he shows that the only numbers that must
be analyzed regarding the Grimm's conjecture are those which has prime factors less than k, where k is the number of
composites in the gap in turn. These numbers are the k-smooth numbers in the gap.
This is a major shortcut to the numerical test of the Grimm's conjecture because these numbers (after the gap 31398
to 31468, which has 10 71-smooth numbers) become less & less frequent (can you find a gap with more than
10 k-smooth numbers?
)
As far as I understand the Delden's approach, his algorithm could be expressed this synthetic way:
Given two consecutive primes P & Q, P<Q, k=Q-P-1= number of composites in the gap.
A.- For N=Q-1 to P+1 step -1
1. Compute prime factors of N & until you get a residue 1 or a prime factor larger or equal than k
2. if the largest prime factor of N is larger or equal than k then go to B.
3. Use the largest unused prime factor of N and keep it as used in this for-next loop. If this is possible go to B,
otherwise or this Algorithm or the Grimm's conjecture fail. End
B.- Next N
C.- Grimm's conjecture Pass for this gap!: End
I have made the code of this Algorithm in Ubasic and extended the search up to 1x10^9 without any fail.
***
T. D. Noe wrote (Feb., 4, 08):
Shanta Laishram and T. N. Shorey, Grimm's Conjecture on consecutive
integers, International Journal of Number Theory, 2 (2006), 1-5.
it is proved that Grimm's conjecture is true for all n<=19236701629. There
is a copy of the paper at http://www.math.uwaterloo.ca/~slaishra/grimm.pdf
I think bipartite matching is the simplest algorithm for for finding a
sequence of distinct primes that divide a sequence of consecutive composite
numbers. The procedure is to create a directed graph whose vertices are
the composite numbers and all their prime factors. For instance, for the
composite numbers between 23 and 29, the vertices are 24, 25, 26, 27, 28,
2, 3, 5, 7, 13. In this graph, two vertices are connected if one number
divides the other. Thus, 24 is connected to 2 and 3; 25 is connected to 5;
26 is connected to 2 and 13; 27 is connected to 3; and 28 is connected to 2
and 7. The bipartite matching algorithm attempts to find a connection from
each composite number to a prime factor. Here is some code for Mathematica
users:
Needs["Combinatorica`"];
factors[n_Integer] := First[Transpose[FactorInteger[n]]];
Grimm[n0_Integer, k_Integer] := Module[{s, fact, m, g},
s = Table[factors[n], {n, n0+1, n0+k}];
fact = Union[Flatten[s]];
m = BipartiteMatching[
g = MakeGraph[
Join[fact,
Range[n0+1, n0+k]], #1 != #2 && (Mod[#2, #1] == 0) &,
Type -> Directed, VertexLabel -> True]];
If[Length[m] == k,
GetVertexLabels[g,
Transpose[Sort[RotateLeft[m, {0, 1}]]][[2]]], {}]];
Grimm[31397,71]
For the 71 composites 31398 to 31468, one possible sequence of prime
divisors is
5233, 17, 157, 3, 7, 31, 2617, 571, 41, 19, 151, 641, 349, 101, 7853, 37,
113, 61, 11, 89, 683, 3491, 1571, 2417, 5237, 67, 491, 419, 827, 2857, 97,
53, 449, 10477, 3929, 43, 13, 6287, 271, 499, 1429, 149, 131, 1367, 79, 47,
1123, 331, 1747, 59, 3931, 953, 5, 4493, 2621, 71, 15727, 233, 983, 83,
107, 163, 2, 10487, 15731, 73, 23, 29, 15733, 617, 7867.
***
Regarding the Delden´s approach Noe notices:
"In Grimm's original paper, Grimm makes the same observation as Van Delden
about having to find primes for only those composite numbers that have all
prime factors less than the gap k."
And regarding the Delden's algorithm:
"But it is possible that it will not work for some prime gap."
Question: can you find a gap where the Delden's algorithm fails but the Grimm's
Conjecture passes?
***
Records | Conjectures | Problems | Puzzles
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#### As of January 2018, Lyft could count 23 million users.
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Teachers Pay Teachers
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Posted
# Some simple numerical methods in C++
, 12 Mar 2007 CPOL
Rate this:
An introductory article in numerical methods for the beginner.
## Introduction
This article tries to familiarize the beginner with numerical methods. I am working a lot with numerical analysis and methods, and I want to share with you some of my experiences and the results that I encountered. This is intended to be the first article in a series of Numerical Analysis Methods and Their Implementation in C++.
After finishing reading this article, you may wonder "Why reinvent the wheel?". Of course, there are a great number of references on the internet that handle the issues presented here, but my intent is not to reinvent the wheel, but to try and explain in detail some classical methods in Numerical Analysis, what they do, why they are useful, how they are implemented, and above all, how to use them.
In this article, we are going to focus on two famous equations in Mathematics, which are of huge importance in almost every domain of science: F(x) = 0 and F(x) = x. Many problems in Mathematics, Chemistry, Physics are reduced to solving one of these two equations, and that's why scientists over time have tried hard to offer good solutions to resolve these equations. We will go through some numerical methods that try to solve them, and we will try to make a short analysis of those methods, showing the advantages/disadvantages of every one of them.
## The methods and their implementation
The first issue of our article is the problem of solving the equation F(x) = 0, where F(x) can be any kind of function. For instance, we can have F(x) = x^2 + 6x + 6, or F(x) = cos(x). This is a well known problem in Mathematics, and it is known also that there is no way of finding out exactly the solution or solutions of every equation of this form. That's why mathematicians have been trying to offer approximate solutions for this equation. Please also note that the solutions of the equation F(x) = 0 are usually called the roots of the function F(x).
### Newton's method
Newton's approach is the following: we start with an initial value for the solution (also called initial approximation), then we replace the function by its tangent, and we compute the root of this tangent which will be a better approximation for the function's root. We repeat this process until we find a suitable solution (one that is close enough to the actual solution and fits very well the equation F(x) = 0). It is obvious that this process is, in fact, an iterative process. Note also that the function F must be a real valued, differentiable function in order to apply Newton's algorithm.
In detail, if we have a current approximation xCrt, the next approximation nNxt will be computed using the following formula:
`xNxt = xCrt - (F(xCrt) / F`(xCrt))`
where F` denotes the derivative of the function F. The iteration process stops when we have gone through a maximum permitted number of iterations and we still can't find the solution, or when we have found an approximation which is close enough to the actual solution of the equation. Here is the code that implements Newton's method:
```/*
* Newton's method for solving equation F(x) = 0
* Output:
* x - the resulted approximation of the solution
* Return:
* The number of iterations passed
*/
int NewtonMethodForEquation(double& x)
{
int n = 1;
while( ( fabs(F(x)) > error ) && ( n <= MAXITER ) )
{
x = x - ( F(x) / Fd(x) );
n++;
}
return n;
}```
In the above code snippet, `Fd` denotes the derivative of the function F.
### Secant method
The secant method is another approach for solving the equation F(x) = 0. The method is almost identical with Newton's method, except the fact that we choose two initial approximations instead of one before we start the iteration process. Suppose we have the current approximations xCrt0 and xCrt1. The next approximation xNxt will be computed this time using the following formula:
`xNxt = xCrt1 - (F(xCrt1)(xCrt1 - xCrt0)) / (F(xCrt1) - F(xCrt0))`
Note that this method doesn't require the derivative of the function F, like Newton's method did. Here is the code that implements the Secant method:
```/*
* Secant method for solving equation F(x) = 0
* Input:
* x0 - the first initial approximation of the solution
* x1 - the second initial approximation of the solution
* Output:
* x - the resulted approximation of the solution
* Return:
* The number of iterations passed
*/
int SecantMethodForEquation(double& x, double x0, double x1)
{
int n = 2;
while( ( fabs(F(x1)) > error ) && ( n <= MAXITER ) )
{
x = x1 - (F(x1) * (x1 - x0)) / (F(x1) - F(x0));
x0 = x1;
x1 = x;
n++;
}
return n;
}```
Another problem that comes into attention some times is solving the equation F(x) = x. If we write the equation like this: F(x) - x = 0 and we note G(x) = F(x) - x, then the equation becomes G(x) = 0. But the equation in the form F(x) = x presents a particular interest for mathematicians. It is said that if x0 is a solution of the equation F(x) = x, then x0 is called a fixed point of the function F(x). Of course, we can apply the methods learned before for the equation G(x) = 0, but our interest is to present methods for solving the equation F(x) = x.
### Successive approximations method
This method, as simple as it may be, is of huge importance in Mathematics, being widely used in many fixed point theories. Let's see how the method works. First, like before, we choose an initial approximation x0, and we start the iterative process. If xCrt denotes the current approximation, then we compute xNxt like this:
`xNxt = F(xCrt)`
This is a pretty simple formula, and against all odds, it has been proved that the method usually converges after a number of iterations, leading to a good approximation of the equation solution. The source code is pretty straightforward:
```/*
* Successive approximations method for solving equation F(x) = x
* Output:
* x - the resulted approximation of the solution
* Return:
* The number of iterations passed
*/
int SuccessiveApproxForEquation(double& x)
{
int n = 1;
while( ( fabs(x - F1(x)) > error ) && ( n <= MAXITER ))
{
x = F1(x);
n++;
}
return n;
}```
This method, as simple and useful as it may be, has one big disadvantage: its convergence rate is very slow, meaning that the number of iterations passed until we get a solution will be pretty high. If the initial approximation is chosen close to the actual solution of the equation, the iteration process will be fast enough and the algorithm will find a suitable solution. But, if the initial approximation is chosen at random, the process may not find a solution at all (depending on the number of maximum iterations permitted, and how close to the actual solution the initial approximation is).
To overcome this problem, some mathematicians tried to speed up the process of finding the right solution for the iterative methods. So some algorithms were developed to do just that (those algorithms are usually called convergence acceleration algorithms). The most important algorithms of convergence acceleration are Aitken's algorithm and Overholt's algorithm.
The idea behind a convergence acceleration algorithm is the following: if we look closer at the iteration process, we see that if we choose every approximation value at each step, we can form a real valued number sequence: x0, x1, x2, ... ,xn which, after a number of iterations, converges to the solution (x) of the equation F(x) = x. A convergence acceleration algorithm transforms the number sequence x0, x1, x2, ... ,xn into another real valued number sequence y0, y1, y2, ... yn, which has the very important property that it converges faster to the solution x.
We present in the following section, three such algorithms for convergence acceleration, and we stick to our problem of solving the equation F(x) = x.
### Aitken's method
Aitken's method is an iterative process similar to the ones presented. I will not go into the mathematical details again. Instead, I prefer to present the code for Aitken's method:
```/*
* Aitken's method for solving equation F(x) = x
* Input:
* x0 - the initial approximation of the solution
* Output:
* y - the resulted approximation of the solution
* Return:
* The number of iterations passed
*/
int AitkenMethodForEquation(double& y, double x0)
{
int n = 1;
double x;
do
{
x = F1(x0);
y = x + 1 / ((1 / (F1(x) - x)) - (1 / (x - x0)));
n++;
x0 = x;
}
while((fabs(y - F1(y)) > error) && (n <= MAXITER));
return n;
}```
### Steffenson's method
This method is a simplified version of Aitken's method, observing that if we apply Aitken's formula for the values xCrt, F(xCrt), F(F(xCrt)), we obtain:
`xNxt = (xCrt F(F(xCrt)) - (F(xCrt)) ^ 2) / (F(F(xCrt)) - 2F(xCrt) + xCrt)`
In a simplified form, this is written as:
`xCrt = F(xCrt) + 1 / ((1 / (F(F(xCrt)) - F(xCrt))) - (1 / (F(xCrt) - xCrt)))`
The code for the algorithm is:
```/*
* Steffensen's method for solving equation F(x) = x
* Output:
* x - the resulted approximation of the solution
* Return:
* The number of iterations passed
*/
int SteffensenMethodForEquation(double& x)
{
int n = 0;
do
{
x = F1(x) + 1 / ( (1 / (F1(F1(x)) - F1(x)) ) - (1 / (F1(x) - x) ) );
n++;
}
while((fabs(x - F1(x)) > error) && (n <= MAXITER));
return n;
}```
### Overholt's method
Overholt's method is, by far, the fastest method for solving the equation F(x) = x. I am going to present the code for the algorithm, the method itself being pretty straightforward from the source code:
```/*
* Overholt's method for solving equation F(x) = x
* Input:
* x0 - the initial approximation of the solution
* s - the convergence order (s must be >= 2)
* Output:
* x - the resulted approximation of the solution
* Return:
* The number of iterations passed
*/
int OverholtMethodForEquation(double &x, double x0, int s)
{
int m = 0;
x = x0;
do
{
V[0][0] = x;
for(int n = 1; n <= s; n++)
V[0][n] = F1(V[0][n - 1]);
for(int k = 0; k <= s - 2; k++)
for(int n = 0; n <= s - k - 2; n++)
V[k + 1][n] = (pow(V[0][n + k + 2] - V[0][n + k + 1], k+1) * V[k][n] -
pow(V[0][n + k + 1] - V[0][n + k], k + 1) * V[k][n + 1]) /
(pow(V[0][n + k + 2] - V[0][n + k + 1], k + 1) -
pow(V[0][n + k + 1] - V[0][n + k], k + 1));
m++;
x = V[s - 1][0];
}
while((fabs(x - F1(x)) > error) && (m <= MAXITER));
return m;
}```
In the source code above, `V` is a global variable declared as:
```// matrix used in Overholt algorithm
// this is declared as a global variable to avoid stack overflow
double V[100][100];```
## Using the code
Now, let's see how to effectively use the methods described above, and on the way, we will make a short analysis of these methods.
In order to use the methods, we must first define some constants:
```// the maximum number of iterations
const int MAXITER = 100;
// the accepted error
const double error = 0.0001;```
`MAXITER` represents the maximum number of iterations an algorithm is permitted to pass, meaning that if one of our algorithms has passed `MAXITER` iterations and it still didn't find a suitable solution, the algorithm will stop nevertheless. `error` represents the minimum accepted error for the solution, meaning that the approximate solution must be close enough to the real solution of the equation with respect to this error. The smaller the value of this error, the closer to the real solution the approximate solution will be.
Let's take an example equation of the first kind, for instance, x*e^x - 1 = 0. So, we have F(x) = x*e^x - 1. We define the function in our code, and also its derivative, because we will need it in order to apply Newton's method:
```// the Euler constant
const double e = 2.718281828459;
// the function F(x)
#define F(x) ( x * pow(e, x) - 1 )
// the derivative of the function F(x), meaning F`(x)
#define Fd(x) ( (x + 1) * pow(e, x) )```
You can change the function `F` and its derivative `Fd`, and you will solve any kind of equation you like:
```double x;
int n;
/* Example usage for Newton Method */
cout << "Newton's method: " << endl << endl;
cout << "Give the initial approximation: ";
cin >> x;
// now we apply Newton's method
n = NewtonMethodForEquation(x);
if(n > MAXITER)
cout << "In " << MAXITER << " iterations no solution was found!" << endl;
else
cout << "The solution is: " << x << " and it was found in "
<< n << " iterations" << endl;
double x0, x1;
/* Example usage for Secant Method */
cout << "Secant method: " << endl << endl;
cout << "Give the first initial approximation: ";
cin >> x0;
cout << "Give the second initial approximation: ";
cin >> x1;
// now we apply the Secant method
n = SecantMethodForEquation(x, x0, x1);
if(n > MAXITER)
cout << "In " << MAXITER << " iterations no solution was found!" << endl;
else
cout << "The solution is: " << x << " and it was found in "
<< n << " iterations" << endl;```
Now you can play with the algorithms, giving various initial approximations and decreasing or increasing the `error` value to see how they are behaving. We can see that, overall, Newton's method is faster than Secant method.
For the second type of equation, let's take, for example, F(x) = e^(-x), and the equation becomes e^(-x) = x.
`#define F1(x) ( pow(e, -x) )`
And, here is how we apply our algorithms:
```/* Example usage for Successive Approximations Method */
cout << "Successive approximations method: " << endl << endl;
cout << "Give the initial approximation: ";
cin >> x;
n = SuccessiveApproxForEquation(x);
if(n > MAXITER)
cout << "In " << MAXITER << " iterations no solution was found!" << endl;
else
cout << "The solution is: " << x << " and it was found in "
<< n << " iterations" << endl;
/* Example usage for Aitken Method */
cout << "Aitken's method: " << endl << endl;
cout << "Give the initial approximation: ";
cin >> x;
double y;
n = AitkenMethodForEquation(y, x);
if(n > MAXITER)
cout << "In " << MAXITER
<< " iterations no solution was found!" << endl;
else
cout << "The solution is: " << y << " and it was found in "
<< n << " iterations" << endl;
/* Example usage for Steffensen Method */
cout << "Steffensen's method: " << endl << endl;
cout << "Give the initial approximation: ";
cin >> x;
n = SteffensenMethodForEquation(x);
if(n > MAXITER)
cout << "In " << MAXITER
<< " iterations no solution was found!" << endl;
else
cout << "The solution is: " << x << " and it was found in "
<< n << " iterations" << endl;
/* Example usage for Overholt Method */
cout << "Overholt's method: " << endl << endl;
cout << "Give the initial approximation: ";
cin >> x;
double rez;
n = OverholtMethodForEquation(rez, x, 2);
if(n > MAXITER)
cout << "In " << MAXITER
<< " iterations no solution was found!" << endl;
else
cout << "The solution is: " << rez << " and it was found in "
<< n << " iterations" << endl;```
Try and play with these algorithms too, and you will see that, indeed, Overholt's method is a stable and fast method for solving equations of type F(x) = x.
## Last words
In this article, I tried to present to you two of the most famous and useful equations in Mathematics, very useful in IT too, and I tried to present some numerical algorithms to solve them, very easy to understand, and also very easy to use in your applications.
None yet.
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Overholt's Method EdwardJ18-Nov-07 7:02 EdwardJ 18-Nov-07 7:02
Convergence StrontiumDog21-Mar-07 11:00 StrontiumDog 21-Mar-07 11:00
Re: Convergence cristitomi21-Mar-07 22:00 cristitomi 21-Mar-07 22:00
Initial approximation(s) PoorNewB20-Mar-07 9:20 PoorNewB 20-Mar-07 9:20
Re: Initial approximation(s) cristitomi20-Mar-07 10:40 cristitomi 20-Mar-07 10:40
Can you give me some resource that make me understood these algorithm? zhyg651613-Mar-07 16:46 zhyg6516 13-Mar-07 16:46
Re: Can you give me some resource that make me understood these algorithm? cristitomi13-Mar-07 21:45 cristitomi 13-Mar-07 21:45
Re: Can you give me some resource that make me understood these algorithm? zhyg651614-Mar-07 17:30 zhyg6516 14-Mar-07 17:30
Re: Can you give me some resource that make me understood these algorithm? Buergermeister14-Mar-07 21:53 Buergermeister 14-Mar-07 21:53
Re: Can you give me some resource that make me understood these algorithm? headmyshoulder20-Mar-07 0:29 headmyshoulder 20-Mar-07 0:29
Last Visit: 31-Dec-99 18:00 Last Update: 27-Aug-16 5:27 Refresh 1
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You are Here: Home
OCR A2 Physics B course work : Bifilar pendulum help..... Tweet
Physics and electronics discussion, revision, exam and homework help.
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1. OCR A2 Physics B course work : Bifilar pendulum help.....
Hi, I am doing the OCR physics b course. I have been given course work to do about the bifilar pendulum.
I have looked all over the web but I can't find much information about the bifilar pendulum.
The course work aim is to find the relationship between the separation of the string and period of one oscillation. And also to find the relationship between the length of the string and the period of one oscillation.
Any help would be greatly appreciated, I have included a link to a picture of a bifilar pendulum...
This was posted from The Student Room's iPhone/iPad App
2. Re: OCR A2 Physics B course work : Bifilar pendulum help.....
Congratulations, you have chosen one of the only projects that has a chance of working with this idiotic module
This is mostly what you need to know, or at least it should give you some sort of clue of how to start. Has your school given you a briefing in dimensional analysis? It's pretty straightforward. I'm aware this coursework is supposed to be done more or less independently so try and figure out the problem yourself. Introducing the angular displacement should be relatively easy, if you want to try things like that to extend the original scope.
Last edited by Unkempt_One; 04-07-2012 at 18:03.
3. Hi, yes thanks... I have carried out the experiment it self... I have collected reading at different lengths and different separations. What I want to know is what to write in the analysis section? And also has anyone got a structures could follow for the course work?
This was posted from The Student Room's iPhone/iPad App
4. Re: OCR A2 Physics B course work : Bifilar pendulum help.....
See if can track donw a copy of Tyler: a laboratory manual of Physics.
Does all the theory (exp 9 p20 in my edition)
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# Knights , Knaves and Spies - Part 1
I was working my way through some Knight and Knave Puzzles in Discrete Maths by Rosen, when I came across the following question:
There are inhabitants of an island on which there are three kinds of people:
• Knights who always tell the truth
• Knaves who always lie
• Spies who can either lie or tell the truth.
You encounter three people, A, B, and C.
You know one of these people is a knight, one is a knave, and one is a spy.
Each of the three people knows the type of person each of other two is.
For this situation, if possible, determine whether there is a unique solution and determine who the knave, knight, and spy is :
A says "C is the knave,” B says, “A is the knight,” and C says “I am the spy"
My Solution:
$$A\Rightarrow Knight$$
$$B\Rightarrow Spy$$
$$C\Rightarrow Knave$$
Doubt:
Am I correct in saying my answer will work?
A is the Knight
B is the Spy
and C is the Knave
To get the solution, First assume, A is knight and will always tells the truth.
Then as per his statement, C is the knave and so what he said will be false. That means he is not a spy. B is the spy and his statement A is the knight is random (true here). This is the only case in which the statements didn't contradict.
Now assume, A is the Knave.
Then as per his statement "C is the knave", it's clear that C is definitely not the knave. Which doesn't contradict since A is the knave already. That means, either B or C is Knight. If B is Knight his statement "A is knight" is false and it contradicts. If C is Knight his statement "I am the spy" is wrong and it contradicts. So this combination A is Knave, B is knight/Spy, C is Knight/Spy is wrong.
Continue this assumptions for other chances of combinations.
You will understand that all other combination except the first one (A is knight, B is Spy and C is knave) is wrong since the statements contradicts.
Simpler explanation:
First, notice that B cannot be the knight, because then for his statement to be true, A would also have to be a knight, and we know there is only one knight.
Second, notice that C cannot be the knight, because then his statement would be false.
Therefore, A is the knight. By his statement, C is the Knave. By elimination, B is the spy.
Just logical if a is the night then he is telling the truth but if B is the night then he's telling the truth and be can't be the night because that would mean there would be two nights so B has to be false which doesn't mean that he is the knave I just means he's lying so he could be the Spy so if he's telling the truth than a is the night but if he's lying is not the Knight then see you would c telling the truth and if c was telling the truth he would have to be the night which is statement is false so he has to be lying and if B and C or both lying that means a is telling the truth so we know the a is the night and that night is telling the truth so therefore she has to be the knave and last but not least B has to be the Spy only thing that makes sense
• You... might want to break this up into sentences, with punctuation, and read it back to make sure it is understandable. This torrent of words is, as it stands, possibly a correct (though redundant) explanation, but after three tries at reading through it I gave up trying to follow what you are saying. See some of the other answers for examples of how to write and format an answer for clarity.
– Rubio
Jul 9, 2018 at 2:01
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# Activities to Teach Students to Write Numbers Up to 100,000 in Words: Convert Digits to Words
Learning to write numbers up to 100,000 in words is an essential skill for students studying basic mathematical concepts. While memorizing basic conversions between digits and words is one way to achieve this, incorporating fun and engaging activities can make learning more exciting and effective. Below are some activities to teach students to write numbers up to 100,000 in words.
1. Word Race: Divide the class into teams and provide them with flashcards with numbers between 0-100,000 written in digits format. Ask each player to convert the digits written on their respective cards into words as quickly as possible. The first team to finish wins.
2. Word Jumble: Create a word jumble puzzle with numbers written in digits format. Students must unscramble the numbers and write them in words to complete the puzzle.
3. Number Bingo: Create bingo cards with numbers written in words. Call out the digits and ask students to mark the corresponding word on their card. The first one to mark all the numbers in a row or column wins.
4. Sentence Building: Provide a list of numbers in digit format and ask students to create sentences using the numbers in words. This activity helps to reinforce the correct order of digits when writing them in words.
5. Word Hunt: Give students a list of numbers in digit format and ask them to find examples of the same numbers written in words in a given text.
6. Number Puzzles: Create different number puzzles, such as wordsearches or crossword puzzles, using numbers written in words. Students must solve these puzzles to find the right word for each number.
7. Number Storytime: Read stories that contain numbers written in words. Ask students to identify the numbers and explain how they are written in digits.
8. Collaborative Writing: Have students work in pairs or groups to create a story that includes numbers written in words. This activity promotes teamwork and helps to reinforce the conversion of digits to words.
In conclusion, there are many fun and engaging activities that teachers can incorporate into their lessons to help students learn how to write numbers up to 100,000 in words. These activities not only make learning more enjoyable for students but also help to reinforce their understanding of this essential mathematical concept. By using a combination of these activities, teachers can help students master the skill of converting digits to words.
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# LAFS PREPRO Session 8 - Assignments
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Assignments for Session 8 of The Los Angeles Film School's Game PreProduction course.
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• There are three categories of rules, all important to a successful play experience:Setup involves things you do once at the beginning of a gameProgression entails what happens during a gameResolution indicates the conditions that cause the game to end and how an outcome is determined based on the game state.Mechanics are a collection of rules that form a discrete chunk of gameplay.Systems are collections of mechanics that make up the biggest chunks of the game.
• ### LAFS PREPRO Session 8 - Assignments
1. 1. Session 8 David Mullich Concept Workshop - Game PreProduction The Los Angeles Film School
2. 2. Rule Hierarchy Rules Mechanics Systems
3. 3. Examples of Game Systems Scoring Progression In-Game Economy Health Combat AI (Enemy Behavior) Magic Multiplayer
4. 4. Health System Example The player starts the game with 3 lives. The player can gain an additional life by one of these methods: Collecting the Reincarnation Medallion Defeating three Mummies in less than 30 seconds The player loses a life when his health reaches 0% (see Combat System). If the player has at least one life remaining: ○ The player’s Health is reset to 100% ○ The player respawns at the nearest player respawn point ○ All enemies respawn at their respawn points at 100% Health If the player has no lives remaining, the Lose Game Screen appears.
5. 5. Progression System Example The player starts the game with 0 Experience Points. The player gains experience by successfully performing actions in this chart: Action Points Earned Collecting gold 1 for each gold piece Opening a treasure chest 50 points Defeating a minion 100 points Defeating a boss 1,000 points
6. 6. Progression System Example The player gains an Experience level each time he accumulates the necessary number of Experience points according to the table below. Experience Level Experience Points 1 100 2 200 3 400 4 800 5 1600
7. 7. Progression System Example When the player Earns a new experience level: A pop-up appears announcing: “Congratulation! You have reached level <LEVEL> and have earned <REWARD>.” Player level rewards are determined by the table below. Experience Level Experience Points 1 Iron Sword 2 +1 Strength 3 Steel Sword 4 +1 Speed 5 Vibranium Sword
8. 8. Combat System Example The attacker is the character (player character or enemy character) who initiates combat) Combat occurs in a serious of rounds, in which the attacker strikes the defender, and then the defender (if still alive) strikes the attacker) The amount of damage done in each attack depends on the character’s weapon, according to the table below: Weapon Damage Fist 1 Club 3 Dagger 5 Short Sword 10 Long Sword 20
9. 9. Combat System Example The damage done by the weapon is subtracted from the target’s health. If both character’s health is greater than 0, then combat continues for another round. If either character’s health reaches 0, then combat immediately ends. (See Life System).
10. 10. AI System Example There are 4 states in the basic AI: Patrol, Follow, Combat and Runaway. PATROL: Each enemy starts the level in Patrol mode. In Patrol mode, each enemy walks between two assigned waypoints at normal speed. FOLLOW: If the player moves within an enemy’s sight range while it is in Patrol Mode, it goes into Follow mode. In Follow mode, the enemy follows the player at high speed. COMBAT: If the player and enemy character collide, the enemy goes into Combat Mode. (See Combat System). RUNAWAY: Each 10 seconds the enemy is in Follow mode, there is a percent chance equal to the player’s Experience Level, that the enemy will go into Runaway Mode. In Runaway mode, the Enemy returns to its closest waypoint at High Speed. Once it arrives, it returns to Patrol mode. Combat Patrol Follow Runaway
11. 11. Systems - Example Enemy Radius Normal Speed High Speed Damage Hit Points Womp Rat 5 3 5 1 5 Vorpel Bunny 10 10 20 2 8 Bug Bear 20 5 10 10 15 Lion King 30 15 20 20 20 Kimono Dragon 30 5 30 30 100
12. 12. Develop your Systems Wiki page: What are the systems in your game? How does the player interact with these systems? How do the systems interact with one another? Describe these in detail exactly how they work. Remember that a programmer will eventually have to implement these, so the programmer needs specific information about each system, not vague generalities. However, avoid long paragraphs, and instead get your meaning across using: Flowcharts Diagrams Formulas Wiki
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| 2.78125
| 3
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CC-MAIN-2016-30
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latest
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en
| 0.891382
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https://community.cedrus.com/t/time-dependent-trials/4178
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text/html
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crawl-data/CC-MAIN-2020-40/segments/1600400202686.56/warc/CC-MAIN-20200922000730-20200922030730-00470.warc.gz
| 325,123,638
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# Time-Dependent Trials
Hi there,
I’m attempting to create an experiment in which the amount of time that an event is display for is dependent on the amount of time that was taken to complete the previous five events. Participants are first presented with an image of a square that they must click. Once the square is clicked, another square appears in a different random location, which must also be clicked. There are five such squares and the reaction time for each click is recorded.
Once the five squares have been clicked, we’d like to display an image of a face for a variable length of time, depending on how long it took them to click all five squares (i.e., if it took them a relatively long time to click the five squares, we’d like to display the face for a short period of time; if it took them relatively less time to click the squares, we’d like to display the face for a longer period of time).
Is this possible in SuperLab?
Thanks,
Jeremy
It is not possible to make it linearly proportional to the reaction time of the previous five trials. However, it is possible to achieve this if it’s okay, to do it in steps, e.g. if RT is greater than N1 go to trial T1; if RT is greater than N2 go to trail T2 etc.
That would be perfect. Could you briefly walk me through how to achieve that? Would it be based on the total reaction time over the five events, or would I have to create step-wise branching for each of the five trials?
Thanks,
Jeremy
It cannot be based on “total” RT of the five events, but instead “average” RT of the past five events. If this works for you, I’ll see if I can build a sample experiment and post it on the forums.
Average would be completely fine.
Thanks very much!
Jeremy
I have attached a sample experiment for your convenience. You might also want to create another macro to continue repeating the experiment.
Jeremy.zip (2.68 KB)
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| 2.71875
| 3
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CC-MAIN-2020-40
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latest
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en
| 0.962347
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http://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/8/11/d/b/
| 1,600,835,536,000,000,000
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text/html
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crawl-data/CC-MAIN-2020-40/segments/1600400209665.4/warc/CC-MAIN-20200923015227-20200923045227-00602.warc.gz
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# Properties
Label 8.11.d.b Level 8 Weight 11 Character orbit 8.d Analytic conductor 5.083 Analytic rank 0 Dimension 8 CM no Inner twists 2
# Related objects
## Newspace parameters
Level: $$N$$ $$=$$ $$8 = 2^{3}$$ Weight: $$k$$ $$=$$ $$11$$ Character orbit: $$[\chi]$$ $$=$$ 8.d (of order $$2$$, degree $$1$$, minimal)
## Newform invariants
Self dual: no Analytic conductor: $$5.08285802139$$ Analytic rank: $$0$$ Dimension: $$8$$ Coefficient field: $$\mathbb{Q}[x]/(x^{8} - \cdots)$$ Defining polynomial: $$x^{8} - 3 x^{7} + 5 x^{6} - 51 x^{5} + 30855 x^{4} - 121569 x^{3} + 12144527 x^{2} + 279415575 x + 3348211684$$ Coefficient ring: $$\Z[a_1, \ldots, a_{5}]$$ Coefficient ring index: $$2^{28}\cdot 3^{3}$$ Twist minimal: yes Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$
## $q$-expansion
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{7}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( 5 - \beta_{1} ) q^{2} + ( 60 + \beta_{1} + \beta_{2} ) q^{3} + ( 25 - 5 \beta_{1} + \beta_{2} + \beta_{3} ) q^{4} + ( \beta_{1} - \beta_{6} ) q^{5} + ( -697 - 66 \beta_{1} + 15 \beta_{2} - 2 \beta_{3} + \beta_{4} + \beta_{6} - \beta_{7} ) q^{6} + ( -13 - 52 \beta_{1} + \beta_{2} - 3 \beta_{3} - 2 \beta_{4} - \beta_{5} + \beta_{6} - 2 \beta_{7} ) q^{7} + ( -5 - 28 \beta_{1} - 40 \beta_{2} + 7 \beta_{3} - 11 \beta_{4} - \beta_{5} - 3 \beta_{6} - 4 \beta_{7} ) q^{8} + ( -4581 - 408 \beta_{1} + 150 \beta_{2} + 18 \beta_{4} ) q^{9} +O(q^{10})$$ $$q + ( 5 - \beta_{1} ) q^{2} + ( 60 + \beta_{1} + \beta_{2} ) q^{3} + ( 25 - 5 \beta_{1} + \beta_{2} + \beta_{3} ) q^{4} + ( \beta_{1} - \beta_{6} ) q^{5} + ( -697 - 66 \beta_{1} + 15 \beta_{2} - 2 \beta_{3} + \beta_{4} + \beta_{6} - \beta_{7} ) q^{6} + ( -13 - 52 \beta_{1} + \beta_{2} - 3 \beta_{3} - 2 \beta_{4} - \beta_{5} + \beta_{6} - 2 \beta_{7} ) q^{7} + ( -5 - 28 \beta_{1} - 40 \beta_{2} + 7 \beta_{3} - 11 \beta_{4} - \beta_{5} - 3 \beta_{6} - 4 \beta_{7} ) q^{8} + ( -4581 - 408 \beta_{1} + 150 \beta_{2} + 18 \beta_{4} ) q^{9} + ( 1092 + 4 \beta_{1} + 162 \beta_{2} - 2 \beta_{3} + 12 \beta_{4} + 6 \beta_{5} - 20 \beta_{6} - 2 \beta_{7} ) q^{10} + ( 17940 - 71 \beta_{1} - 163 \beta_{2} + 104 \beta_{3} + 4 \beta_{4} - 8 \beta_{5} - 8 \beta_{6} + 16 \beta_{7} ) q^{11} + ( 22712 + 830 \beta_{1} + 170 \beta_{2} + 40 \beta_{3} - 86 \beta_{4} + 6 \beta_{5} + 58 \beta_{6} - 16 \beta_{7} ) q^{12} + ( 88 + 183 \beta_{1} - 168 \beta_{2} - 280 \beta_{3} + 32 \beta_{4} - 8 \beta_{5} - 15 \beta_{6} + 48 \beta_{7} ) q^{13} + ( -49752 - 248 \beta_{1} - 1356 \beta_{2} + 44 \beta_{3} + 24 \beta_{4} + 28 \beta_{5} + 88 \beta_{6} - 20 \beta_{7} ) q^{14} + ( 2159 + 8284 \beta_{1} - 587 \beta_{2} - 575 \beta_{3} + 310 \beta_{4} - 21 \beta_{5} - 11 \beta_{6} + 86 \beta_{7} ) q^{15} + ( -248082 - 1280 \beta_{1} - 2360 \beta_{2} + 86 \beta_{3} - 326 \beta_{4} + 62 \beta_{5} - 54 \beta_{6} - 24 \beta_{7} ) q^{16} + ( -42494 + 14520 \beta_{1} - 414 \beta_{2} + 832 \beta_{3} - 426 \beta_{4} - 64 \beta_{5} - 64 \beta_{6} + 128 \beta_{7} ) q^{17} + ( 384303 + 6597 \beta_{1} + 3096 \beta_{2} + 240 \beta_{3} + 168 \beta_{4} + 168 \beta_{6} - 168 \beta_{7} ) q^{18} + ( 206388 - 45903 \beta_{1} + 3605 \beta_{2} + 104 \beta_{3} + 1604 \beta_{4} - 8 \beta_{5} - 8 \beta_{6} + 16 \beta_{7} ) q^{19} + ( 316692 + 456 \beta_{1} + 13080 \beta_{2} - 268 \beta_{3} - 1164 \beta_{4} + 76 \beta_{5} - 364 \beta_{6} - 128 \beta_{7} ) q^{20} + ( -18136 - 72380 \beta_{1} + 2344 \beta_{2} - 104 \beta_{3} - 2336 \beta_{4} - 120 \beta_{5} + 148 \beta_{6} - 304 \beta_{7} ) q^{21} + ( 192255 - 17586 \beta_{1} - 15257 \beta_{2} + 686 \beta_{3} + 2345 \beta_{4} - 128 \beta_{5} - 727 \beta_{6} - 169 \beta_{7} ) q^{22} + ( 26369 + 107044 \beta_{1} - 2341 \beta_{2} + 1743 \beta_{3} + 3178 \beta_{4} + 69 \beta_{5} - 485 \beta_{6} - 246 \beta_{7} ) q^{23} + ( -2480526 - 36280 \beta_{1} - 6016 \beta_{2} - 870 \beta_{3} - 4290 \beta_{4} - 294 \beta_{5} + 1134 \beta_{6} - 120 \beta_{7} ) q^{24} + ( -2301255 + 196080 \beta_{1} + 10660 \beta_{2} - 4160 \beta_{3} - 6260 \beta_{4} + 320 \beta_{5} + 320 \beta_{6} - 640 \beta_{7} ) q^{25} + ( 231260 + 1436 \beta_{1} + 12974 \beta_{2} - 590 \beta_{3} + 10708 \beta_{4} - 22 \beta_{5} + 244 \beta_{6} + 1202 \beta_{7} ) q^{26} + ( 3576528 - 201582 \beta_{1} - 38394 \beta_{2} - 2808 \beta_{3} + 5076 \beta_{4} + 216 \beta_{5} + 216 \beta_{6} - 432 \beta_{7} ) q^{27} + ( 2306920 + 64656 \beta_{1} - 6992 \beta_{2} + 1448 \beta_{3} - 11608 \beta_{4} - 680 \beta_{5} - 280 \beta_{6} + 1664 \beta_{7} ) q^{28} + ( -101824 - 403261 \beta_{1} + 17216 \beta_{2} + 10176 \beta_{3} - 12800 \beta_{4} + 1344 \beta_{5} - 771 \beta_{6} + 1152 \beta_{7} ) q^{29} + ( 8382280 + 42536 \beta_{1} + 8708 \beta_{2} - 9124 \beta_{3} + 21368 \beta_{4} + 12 \beta_{5} + 1208 \beta_{6} + 2332 \beta_{7} ) q^{30} + ( 127852 + 511760 \beta_{1} - 10556 \beta_{2} + 9940 \beta_{3} + 15320 \beta_{4} + 412 \beta_{5} + 5764 \beta_{6} - 1352 \beta_{7} ) q^{31} + ( -7952140 + 215328 \beta_{1} + 32976 \beta_{2} + 3492 \beta_{3} - 20004 \beta_{4} - 204 \beta_{5} - 5572 \beta_{6} + 3120 \beta_{7} ) q^{32} + ( -9105756 + 699208 \beta_{1} - 9314 \beta_{2} - 4992 \beta_{3} - 23190 \beta_{4} + 384 \beta_{5} + 384 \beta_{6} - 768 \beta_{7} ) q^{33} + ( -14445758 - 34210 \beta_{1} - 130232 \beta_{2} - 10032 \beta_{3} + 19192 \beta_{4} - 1024 \beta_{5} - 5384 \beta_{6} - 1784 \beta_{7} ) q^{34} + ( 6470224 - 935104 \beta_{1} + 21592 \beta_{2} - 2704 \beta_{3} + 30680 \beta_{4} + 208 \beta_{5} + 208 \beta_{6} - 416 \beta_{7} ) q^{35} + ( 6247731 - 363687 \beta_{1} - 15045 \beta_{2} - 9813 \beta_{3} - 20784 \beta_{4} + 432 \beta_{5} + 8016 \beta_{6} - 4992 \beta_{7} ) q^{36} + ( -241176 - 968855 \beta_{1} + 24296 \beta_{2} - 17192 \beta_{3} - 31008 \beta_{4} - 3768 \beta_{5} + 3503 \beta_{6} - 6064 \beta_{7} ) q^{37} + ( 46867399 + 30558 \beta_{1} + 116463 \beta_{2} + 41150 \beta_{3} + 7713 \beta_{4} - 128 \beta_{5} + 4641 \beta_{6} - 5537 \beta_{7} ) q^{38} + ( 220729 + 889316 \beta_{1} - 45277 \beta_{2} - 31849 \beta_{3} + 29882 \beta_{4} - 1059 \beta_{5} - 26845 \beta_{6} + 5050 \beta_{7} ) q^{39} + ( -44924432 - 569440 \beta_{1} + 342112 \beta_{2} - 14864 \beta_{3} - 12496 \beta_{4} + 2896 \beta_{5} + 5296 \beta_{6} - 11648 \beta_{7} ) q^{40} + ( 11687250 + 533968 \beta_{1} + 133100 \beta_{2} + 54080 \beta_{3} - 9308 \beta_{4} - 4160 \beta_{5} - 4160 \beta_{6} + 8320 \beta_{7} ) q^{41} + ( -72075440 - 360880 \beta_{1} - 121816 \beta_{2} + 71768 \beta_{3} - 7696 \beta_{4} + 3576 \beta_{5} + 11120 \beta_{6} - 3752 \beta_{7} ) q^{42} + ( -1410292 - 375783 \beta_{1} + 153345 \beta_{2} + 34320 \beta_{3} + 19368 \beta_{4} - 2640 \beta_{5} - 2640 \beta_{6} + 5280 \beta_{7} ) q^{43} + ( 91117976 + 288462 \beta_{1} - 283430 \beta_{2} + 32136 \beta_{3} - 2374 \beta_{4} + 6774 \beta_{5} - 24118 \beta_{6} + 6000 \beta_{7} ) q^{44} + ( 401448 + 1563243 \beta_{1} - 88536 \beta_{2} - 63336 \beta_{3} + 56544 \beta_{4} + 648 \beta_{5} - 7059 \beta_{6} + 17616 \beta_{7} ) q^{45} + ( 106651704 + 531288 \beta_{1} - 66116 \beta_{2} - 104988 \beta_{3} - 58104 \beta_{4} + 2868 \beta_{5} - 14392 \beta_{6} - 7900 \beta_{7} ) q^{46} + ( -418630 - 1783288 \beta_{1} + 14462 \beta_{2} - 70938 \beta_{3} - 48092 \beta_{4} - 3678 \beta_{5} + 67518 \beta_{6} + 7620 \beta_{7} ) q^{47} + ( -142238972 + 1805376 \beta_{1} - 623568 \beta_{2} + 47156 \beta_{3} + 40172 \beta_{4} - 1500 \beta_{5} + 26060 \beta_{6} + 9520 \beta_{7} ) q^{48} + ( -11063055 - 4427328 \beta_{1} - 517616 \beta_{2} - 26624 \beta_{3} + 124336 \beta_{4} + 2048 \beta_{5} + 2048 \beta_{6} - 4096 \beta_{7} ) q^{49} + ( -208415155 + 1274055 \beta_{1} + 385680 \beta_{2} - 218720 \beta_{3} - 95760 \beta_{4} + 5120 \beta_{5} + 27120 \beta_{6} + 8720 \beta_{7} ) q^{50} + ( -32732736 + 6715042 \beta_{1} - 479042 \beta_{2} + 31512 \beta_{3} - 229956 \beta_{4} - 2424 \beta_{5} - 2424 \beta_{6} + 4848 \beta_{7} ) q^{51} + ( 302766860 + 1336184 \beta_{1} + 480936 \beta_{2} - 20436 \beta_{3} + 169900 \beta_{4} - 12652 \beta_{5} + 14732 \beta_{6} - 12160 \beta_{7} ) q^{52} + ( 1595416 + 6425733 \beta_{1} - 157160 \beta_{2} + 93480 \beta_{3} + 198176 \beta_{4} + 11448 \beta_{5} - 11165 \beta_{6} + 8112 \beta_{7} ) q^{53} + ( 218126574 - 2489508 \beta_{1} + 13662 \beta_{2} + 222588 \beta_{3} - 100926 \beta_{4} + 3456 \beta_{5} - 17982 \beta_{6} + 42174 \beta_{7} ) q^{54} + ( -2309785 - 8975588 \beta_{1} + 425789 \beta_{2} + 244937 \beta_{3} - 306938 \beta_{4} + 7235 \beta_{5} - 103363 \beta_{6} - 41338 \beta_{7} ) q^{55} + ( -366966176 - 4298944 \beta_{1} - 1029440 \beta_{2} - 27552 \beta_{3} + 312800 \beta_{4} - 8928 \beta_{5} - 33568 \beta_{6} + 16128 \beta_{7} ) q^{56} + ( 172905348 - 6898168 \beta_{1} + 1561454 \beta_{2} - 254592 \beta_{3} + 255834 \beta_{4} + 19584 \beta_{5} + 19584 \beta_{6} - 39168 \beta_{7} ) q^{57} + ( -406552116 - 2042612 \beta_{1} + 1706086 \beta_{2} + 422266 \beta_{3} - 284892 \beta_{4} - 31854 \beta_{5} - 106812 \beta_{6} - 5766 \beta_{7} ) q^{58} + ( -118671972 + 8652361 \beta_{1} + 771017 \beta_{2} - 247936 \beta_{3} - 270848 \beta_{4} + 19072 \beta_{5} + 19072 \beta_{6} - 38144 \beta_{7} ) q^{59} + ( 654618632 - 5169968 \beta_{1} + 1071472 \beta_{2} - 87224 \beta_{3} + 388168 \beta_{4} - 21576 \beta_{5} + 57736 \beta_{6} + 18560 \beta_{7} ) q^{60} + ( 1200152 + 5054859 \beta_{1} + 73752 \beta_{2} + 343336 \beta_{3} + 106528 \beta_{4} - 17224 \beta_{5} + 59229 \beta_{6} - 133200 \beta_{7} ) q^{61} + ( 508792096 + 2544800 \beta_{1} - 1641200 \beta_{2} - 491408 \beta_{3} - 432160 \beta_{4} - 35536 \beta_{5} + 87264 \beta_{6} - 27280 \beta_{7} ) q^{62} + ( -2432463 - 9697116 \beta_{1} + 432555 \beta_{2} + 255903 \beta_{3} - 314742 \beta_{4} + 20277 \beta_{5} + 93291 \beta_{6} - 8214 \beta_{7} ) q^{63} + ( -724858664 + 4312640 \beta_{1} + 791008 \beta_{2} - 205960 \beta_{3} + 440328 \beta_{4} + 984 \beta_{5} - 150200 \beta_{6} - 12896 \beta_{7} ) q^{64} + ( -105072832 - 6770288 \beta_{1} - 6014596 \beta_{2} + 300352 \beta_{3} + 44500 \beta_{4} - 23104 \beta_{5} - 23104 \beta_{6} + 46208 \beta_{7} ) q^{65} + ( -744715972 + 5465916 \beta_{1} - 605640 \beta_{2} - 688592 \beta_{3} - 152696 \beta_{4} + 6144 \beta_{5} - 5240 \beta_{6} + 48248 \beta_{7} ) q^{66} + ( -153950668 - 1278015 \beta_{1} + 1637061 \beta_{2} - 213720 \beta_{3} + 79716 \beta_{4} + 16440 \beta_{5} + 16440 \beta_{6} - 32880 \beta_{7} ) q^{67} + ( 662247202 + 18392918 \beta_{1} - 1669470 \beta_{2} + 110066 \beta_{3} + 105072 \beta_{4} + 71440 \beta_{5} - 123920 \beta_{6} + 99712 \beta_{7} ) q^{68} + ( 1304216 + 4953388 \beta_{1} - 335720 \beta_{2} - 218456 \beta_{3} + 207904 \beta_{4} + 37176 \beta_{5} - 21188 \beta_{6} + 156848 \beta_{7} ) q^{69} + ( 964562656 - 1596544 \beta_{1} + 2103840 \beta_{2} + 870976 \beta_{3} - 12832 \beta_{4} + 3328 \beta_{5} + 67040 \beta_{6} - 43744 \beta_{7} ) q^{70} + ( 3109979 + 11761740 \beta_{1} - 921511 \beta_{2} - 940395 \beta_{3} + 462382 \beta_{4} - 22137 \beta_{5} + 44825 \beta_{6} + 174222 \beta_{7} ) q^{71} + ( -544775175 - 9205620 \beta_{1} - 957624 \beta_{2} + 349245 \beta_{3} - 682281 \beta_{4} + 6885 \beta_{5} + 221103 \beta_{6} + 48276 \beta_{7} ) q^{72} + ( 352284402 + 18279720 \beta_{1} + 9466214 \beta_{2} + 346112 \beta_{3} - 261118 \beta_{4} - 26624 \beta_{5} - 26624 \beta_{6} + 53248 \beta_{7} ) q^{73} + ( -958926044 - 4789020 \beta_{1} - 3817838 \beta_{2} + 931086 \beta_{3} + 329132 \beta_{4} + 98262 \beta_{5} + 326284 \beta_{6} - 43506 \beta_{7} ) q^{74} + ( 559738860 - 8529095 \beta_{1} - 4815695 \beta_{2} + 1151280 \beta_{3} + 196920 \beta_{4} - 88560 \beta_{5} - 88560 \beta_{6} + 177120 \beta_{7} ) q^{75} + ( 458463192 - 45826338 \beta_{1} - 2077302 \beta_{2} - 53176 \beta_{3} - 1027222 \beta_{4} - 12218 \beta_{5} + 133626 \beta_{6} - 284688 \beta_{7} ) q^{76} + ( -4807288 - 20123420 \beta_{1} - 296056 \beta_{2} - 1667016 \beta_{3} - 497312 \beta_{4} - 80280 \beta_{5} - 95084 \beta_{6} + 195984 \beta_{7} ) q^{77} + ( 896565752 + 4481560 \beta_{1} + 4483228 \beta_{2} - 961532 \beta_{3} + 1507144 \beta_{4} + 154068 \beta_{5} - 464888 \beta_{6} + 80708 \beta_{7} ) q^{78} + ( 9084134 + 36757528 \beta_{1} - 1294462 \beta_{2} - 425734 \beta_{3} + 1122172 \beta_{4} - 81154 \beta_{5} - 476542 \beta_{6} - 116740 \beta_{7} ) q^{79} + ( -792193280 + 41782400 \beta_{1} + 8040064 \beta_{2} + 116224 \beta_{3} - 1546624 \beta_{4} + 67712 \beta_{5} + 574080 \beta_{6} - 321280 \beta_{7} ) q^{80} + ( -1484813835 + 9789768 \beta_{1} - 5016546 \beta_{2} - 673920 \beta_{3} - 522774 \beta_{4} + 51840 \beta_{5} + 51840 \beta_{6} - 103680 \beta_{7} ) q^{81} + ( -457683798 - 14655186 \beta_{1} - 5200976 \beta_{2} - 420640 \beta_{3} + 1425872 \beta_{4} - 66560 \beta_{5} - 171568 \beta_{6} - 294352 \beta_{7} ) q^{82} + ( 215876156 - 59776399 \beta_{1} + 1463761 \beta_{2} + 906048 \beta_{3} + 2036192 \beta_{4} - 69696 \beta_{5} - 69696 \beta_{6} + 139392 \beta_{7} ) q^{83} + ( -344936560 + 72211360 \beta_{1} - 4062752 \beta_{2} + 665872 \beta_{3} - 2370800 \beta_{4} - 140304 \beta_{5} - 245360 \beta_{6} - 36352 \beta_{7} ) q^{84} + ( -18531208 - 73175502 \beta_{1} + 2896248 \beta_{2} + 864712 \beta_{3} - 2435168 \beta_{4} - 57448 \beta_{5} - 114090 \beta_{6} - 374160 \beta_{7} ) q^{85} + ( 378520783 + 3208078 \beta_{1} - 1079529 \beta_{2} + 353550 \beta_{3} + 999033 \beta_{4} - 42240 \beta_{5} - 14727 \beta_{6} - 280953 \beta_{7} ) q^{86} + ( 11019661 + 43858196 \beta_{1} - 597793 \beta_{2} + 1297667 \beta_{3} + 1250594 \beta_{4} - 7935 \beta_{5} + 1219487 \beta_{6} - 346238 \beta_{7} ) q^{87} + ( 703173410 - 89628664 \beta_{1} + 5836672 \beta_{2} + 38666 \beta_{3} - 1249266 \beta_{4} - 21942 \beta_{5} - 1177154 \beta_{6} + 297544 \beta_{7} ) q^{88} + ( 790659474 + 92502760 \beta_{1} + 13716086 \beta_{2} + 1174784 \beta_{3} - 2462798 \beta_{4} - 90368 \beta_{5} - 90368 \beta_{6} + 180736 \beta_{7} ) q^{89} + ( 1568180652 + 7916268 \beta_{1} + 5665494 \beta_{2} - 1646070 \beta_{3} + 2672580 \beta_{4} - 77598 \beta_{5} - 185244 \beta_{6} + 359178 \beta_{7} ) q^{90} + ( 935260720 - 75034944 \beta_{1} + 4320872 \beta_{2} - 3433456 \beta_{3} + 2329832 \beta_{4} + 264112 \beta_{5} + 264112 \beta_{6} - 528224 \beta_{7} ) q^{91} + ( -1143807112 - 115192272 \beta_{1} + 8767376 \beta_{2} - 801736 \beta_{3} - 249032 \beta_{4} + 238792 \beta_{5} + 35320 \beta_{6} + 393088 \beta_{7} ) q^{92} + ( -11514944 - 44530672 \beta_{1} + 3248576 \beta_{2} + 3747392 \beta_{3} - 1554688 \beta_{4} + 359616 \beta_{5} + 112688 \beta_{6} + 52096 \beta_{7} ) q^{93} + ( -1772640720 - 8703632 \beta_{1} - 8964008 \beta_{2} + 1742824 \beta_{3} + 1718352 \beta_{4} - 369912 \beta_{5} + 1600464 \beta_{6} + 383592 \beta_{7} ) q^{94} + ( 12722927 + 51909692 \beta_{1} - 1829227 \beta_{2} + 115137 \beta_{3} + 1753942 \beta_{4} + 265707 \beta_{5} - 1795179 \beta_{6} + 701910 \beta_{7} ) q^{95} + ( 1027438360 + 151759552 \beta_{1} - 18992288 \beta_{2} - 993352 \beta_{3} - 33208 \beta_{4} - 282216 \beta_{5} - 340600 \beta_{6} + 546208 \beta_{7} ) q^{96} + ( -1185370846 + 67181496 \beta_{1} - 36922974 \beta_{2} - 1440192 \beta_{3} - 3454698 \beta_{4} + 110784 \beta_{5} + 110784 \beta_{6} - 221568 \beta_{7} ) q^{97} + ( 4353606773 + 34636815 \beta_{1} + 1407552 \beta_{2} + 4703872 \beta_{3} - 1034304 \beta_{4} + 32768 \beta_{5} - 247872 \beta_{6} + 477248 \beta_{7} ) q^{98} + ( -1687350924 + 56970915 \beta_{1} - 14709693 \beta_{2} - 2313792 \beta_{3} - 2467296 \beta_{4} + 177984 \beta_{5} + 177984 \beta_{6} - 355968 \beta_{7} ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$8q + 42q^{2} + 480q^{3} + 212q^{4} - 5412q^{6} - 72q^{8} - 35496q^{9} + O(q^{10})$$ $$8q + 42q^{2} + 480q^{3} + 212q^{4} - 5412q^{6} - 72q^{8} - 35496q^{9} + 9120q^{10} + 143328q^{11} + 180120q^{12} - 400320q^{14} - 1987312q^{16} - 370800q^{17} + 3067758q^{18} + 1753312q^{19} + 2557440q^{20} + 1549180q^{22} - 19794288q^{24} - 18792760q^{25} + 1891680q^{26} + 28949184q^{27} + 18286080q^{28} + 67026240q^{30} - 64016928q^{32} - 74308704q^{33} - 115705388q^{34} + 53736960q^{35} + 50631516q^{36} + 375128220q^{38} - 357584640q^{40} + 92669328q^{41} - 576155520q^{42} - 10190624q^{43} + 727831512q^{44} + 851947200q^{46} - 1142760480q^{48} - 80432248q^{49} - 1669361190q^{50} - 276714816q^{51} + 2420759040q^{52} + 1749768696q^{54} - 2928529920q^{56} + 1400712864q^{57} - 3245264160q^{58} - 965642016q^{59} + 5250055680q^{60} + 4060980480q^{62} - 5804705728q^{64} - 839028480q^{65} - 5970262296q^{66} - 1225582880q^{67} + 5258111400q^{68} + 7723829760q^{70} - 4343608728q^{72} + 2800072720q^{73} - 7669373280q^{74} + 4485554400q^{75} + 3753451288q^{76} + 7176312000q^{78} - 6408629760q^{80} - 11909065176q^{81} - 3638778380q^{82} + 1853422560q^{83} - 2916218880q^{84} + 3022180476q^{86} + 5815578640q^{88} + 6162596112q^{89} + 12545940960q^{90} + 7645985280q^{91} - 8903892480q^{92} - 14182177920q^{94} + 7877525568q^{96} - 9697863536q^{97} + 34759868298q^{98} - 13646747232q^{99} + O(q^{100})$$
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{8} - 3 x^{7} + 5 x^{6} - 51 x^{5} + 30855 x^{4} - 121569 x^{3} + 12144527 x^{2} + 279415575 x + 3348211684$$:
$$\beta_{0}$$ $$=$$ $$1$$ $$\beta_{1}$$ $$=$$ $$2 \nu - 1$$ $$\beta_{2}$$ $$=$$ $$($$$$-39 \nu^{7} + 640 \nu^{6} - 6467 \nu^{5} - 38788 \nu^{4} + 639715 \nu^{3} + 3220592 \nu^{2} - 518764921 \nu - 5385260908$$$$)/29360128$$ $$\beta_{3}$$ $$=$$ $$($$$$39 \nu^{7} - 640 \nu^{6} + 6467 \nu^{5} + 38788 \nu^{4} - 639715 \nu^{3} + 114219920 \nu^{2} + 107723129 \nu + 5561421676$$$$)/29360128$$ $$\beta_{4}$$ $$=$$ $$($$$$73 \nu^{7} + 640 \nu^{6} - 5907 \nu^{5} - 42820 \nu^{4} + 4083379 \nu^{3} - 64144 \nu^{2} + 831567895 \nu + 25027780436$$$$)/29360128$$ $$\beta_{5}$$ $$=$$ $$($$$$-127 \nu^{7} + 2496 \nu^{6} - 17723 \nu^{5} + 547420 \nu^{4} - 1923813 \nu^{3} + 18107440 \nu^{2} - 1496543073 \nu - 3601739852$$$$)/3670016$$ $$\beta_{6}$$ $$=$$ $$($$$$-51 \nu^{7} - 2688 \nu^{6} + 81793 \nu^{5} - 1067988 \nu^{4} + 10212255 \nu^{3} - 64144848 \nu^{2} + 299736339 \nu - 56323662812$$$$)/14680064$$ $$\beta_{7}$$ $$=$$ $$($$$$147 \nu^{7} - 2560 \nu^{6} + 1247 \nu^{5} + 270164 \nu^{4} + 7125889 \nu^{3} - 75156144 \nu^{2} + 1701809933 \nu + 21257762140$$$$)/7340032$$
$$1$$ $$=$$ $$\beta_0$$ $$\nu$$ $$=$$ $$($$$$\beta_{1} + 1$$$$)/2$$ $$\nu^{2}$$ $$=$$ $$($$$$\beta_{3} + \beta_{2} + 7 \beta_{1} + 1$$$$)/4$$ $$\nu^{3}$$ $$=$$ $$($$$$4 \beta_{7} + 3 \beta_{6} + \beta_{5} + 11 \beta_{4} + 11 \beta_{3} + 58 \beta_{2} + 46 \beta_{1} + 131$$$$)/8$$ $$\nu^{4}$$ $$=$$ $$($$$$36 \beta_{7} + 9 \beta_{6} + 43 \beta_{5} - 31 \beta_{4} + 67 \beta_{3} - 592 \beta_{2} - 412 \beta_{1} - 122793$$$$)/8$$ $$\nu^{5}$$ $$=$$ $$($$$$-300 \beta_{7} + 629 \beta_{6} + 303 \beta_{5} + 1773 \beta_{4} - 159 \beta_{3} - 10902 \beta_{2} - 30996 \beta_{1} + 67607$$$$)/4$$ $$\nu^{6}$$ $$=$$ $$($$$$-7556 \beta_{7} + 2137 \beta_{6} + 2883 \beta_{5} + 64497 \beta_{4} - 18502 \beta_{3} - 92393 \beta_{2} - 253747 \beta_{1} - 35764058$$$$)/4$$ $$\nu^{7}$$ $$=$$ $$($$$$-118692 \beta_{7} - 98207 \beta_{6} - 32229 \beta_{5} + 1740089 \beta_{4} - 275559 \beta_{3} - 3734114 \beta_{2} - 48934814 \beta_{1} - 2229652087$$$$)/8$$
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/8\mathbb{Z}\right)^\times$$.
$$n$$ $$5$$ $$7$$ $$\chi(n)$$ $$-1$$ $$-1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
3.1
16.5295 + 8.54130i 16.5295 − 8.54130i 4.83019 + 15.8950i 4.83019 − 15.8950i −8.43092 + 11.1953i −8.43092 − 11.1953i −11.4287 + 6.91461i −11.4287 − 6.91461i
−27.0589 17.0826i 203.867 440.370 + 924.473i 2088.87i −5516.42 3482.58i 25367.2i 3876.45 32537.9i −17487.2 −35683.3 + 56522.5i
3.2 −27.0589 + 17.0826i 203.867 440.370 924.473i 2088.87i −5516.42 + 3482.58i 25367.2i 3876.45 + 32537.9i −17487.2 −35683.3 56522.5i
3.3 −3.66038 31.7900i −119.281 −997.203 + 232.727i 948.910i 436.614 + 3791.94i 15458.1i 11048.5 + 30849.2i −44821.0 30165.8 3473.37i
3.4 −3.66038 + 31.7900i −119.281 −997.203 232.727i 948.910i 436.614 3791.94i 15458.1i 11048.5 30849.2i −44821.0 30165.8 + 3473.37i
3.5 22.8618 22.3905i 352.099 21.3279 1023.78i 3773.58i 8049.62 7883.68i 15618.2i −22435.3 23883.0i 64924.4 84492.5 + 86271.1i
3.6 22.8618 + 22.3905i 352.099 21.3279 + 1023.78i 3773.58i 8049.62 + 7883.68i 15618.2i −22435.3 + 23883.0i 64924.4 84492.5 86271.1i
3.7 28.8575 13.8292i −196.684 641.505 798.152i 5381.00i −5675.81 + 2719.99i 6613.83i 7474.38 31904.2i −20364.2 −74415.0 155282.i
3.8 28.8575 + 13.8292i −196.684 641.505 + 798.152i 5381.00i −5675.81 2719.99i 6613.83i 7474.38 + 31904.2i −20364.2 −74415.0 + 155282.i
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 3.8 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
8.d odd 2 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 8.11.d.b 8
3.b odd 2 1 72.11.b.b 8
4.b odd 2 1 32.11.d.b 8
8.b even 2 1 32.11.d.b 8
8.d odd 2 1 inner 8.11.d.b 8
12.b even 2 1 288.11.b.b 8
16.e even 4 2 256.11.c.m 16
16.f odd 4 2 256.11.c.m 16
24.f even 2 1 72.11.b.b 8
24.h odd 2 1 288.11.b.b 8
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
8.11.d.b 8 1.a even 1 1 trivial
8.11.d.b 8 8.d odd 2 1 inner
32.11.d.b 8 4.b odd 2 1
32.11.d.b 8 8.b even 2 1
72.11.b.b 8 3.b odd 2 1
72.11.b.b 8 24.f even 2 1
256.11.c.m 16 16.e even 4 2
256.11.c.m 16 16.f odd 4 2
288.11.b.b 8 12.b even 2 1
288.11.b.b 8 24.h odd 2 1
## Hecke kernels
This newform subspace can be constructed as the kernel of the linear operator $$T_{3}^{4} - 240 T_{3}^{3} - 80424 T_{3}^{2} + 9637056 T_{3} + 1684044432$$ acting on $$S_{11}^{\mathrm{new}}(8, [\chi])$$.
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$1 - 42 T + 776 T^{2} - 7872 T^{3} + 537600 T^{4} - 8060928 T^{5} + 813694976 T^{6} - 45097156608 T^{7} + 1099511627776 T^{8}$$
$3$ $$( 1 - 240 T + 155772 T^{2} - 32878224 T^{3} + 13106837286 T^{4} - 1941426248976 T^{5} + 543143379712572 T^{6} - 49413871702715760 T^{7} + 12157665459056928801 T^{8} )^{2}$$
$5$ $$1 - 29666120 T^{2} + 474519111223900 T^{4} -$$$$56\!\cdots\!00$$$$T^{6} +$$$$58\!\cdots\!50$$$$T^{8} -$$$$53\!\cdots\!00$$$$T^{10} +$$$$43\!\cdots\!00$$$$T^{12} -$$$$25\!\cdots\!00$$$$T^{14} +$$$$82\!\cdots\!25$$$$T^{16}$$
$7$ $$1 - 1089684872 T^{2} + 669297395540350108 T^{4} -$$$$28\!\cdots\!84$$$$T^{6} +$$$$89\!\cdots\!70$$$$T^{8} -$$$$22\!\cdots\!84$$$$T^{10} +$$$$42\!\cdots\!08$$$$T^{12} -$$$$55\!\cdots\!72$$$$T^{14} +$$$$40\!\cdots\!01$$$$T^{16}$$
$11$ $$( 1 - 71664 T + 52584995516 T^{2} - 4110626988784272 T^{3} +$$$$19\!\cdots\!78$$$$T^{4} -$$$$10\!\cdots\!72$$$$T^{5} +$$$$35\!\cdots\!16$$$$T^{6} -$$$$12\!\cdots\!64$$$$T^{7} +$$$$45\!\cdots\!01$$$$T^{8} )^{2}$$
$13$ $$1 - 459250928072 T^{2} +$$$$14\!\cdots\!28$$$$T^{4} -$$$$28\!\cdots\!44$$$$T^{6} +$$$$45\!\cdots\!10$$$$T^{8} -$$$$54\!\cdots\!44$$$$T^{10} +$$$$51\!\cdots\!28$$$$T^{12} -$$$$31\!\cdots\!72$$$$T^{14} +$$$$13\!\cdots\!01$$$$T^{16}$$
$17$ $$( 1 + 185400 T + 3991264862492 T^{2} - 74312252090724216 T^{3} +$$$$10\!\cdots\!46$$$$T^{4} -$$$$14\!\cdots\!84$$$$T^{5} +$$$$16\!\cdots\!92$$$$T^{6} +$$$$15\!\cdots\!00$$$$T^{7} +$$$$16\!\cdots\!01$$$$T^{8} )^{2}$$
$19$ $$( 1 - 876656 T + 13744021629436 T^{2} - 15629661673389367568 T^{3} +$$$$11\!\cdots\!18$$$$T^{4} -$$$$95\!\cdots\!68$$$$T^{5} +$$$$51\!\cdots\!36$$$$T^{6} -$$$$20\!\cdots\!56$$$$T^{7} +$$$$14\!\cdots\!01$$$$T^{8} )^{2}$$
$23$ $$1 - 205026617368712 T^{2} +$$$$20\!\cdots\!08$$$$T^{4} -$$$$13\!\cdots\!44$$$$T^{6} +$$$$65\!\cdots\!70$$$$T^{8} -$$$$23\!\cdots\!44$$$$T^{10} +$$$$61\!\cdots\!08$$$$T^{12} -$$$$10\!\cdots\!12$$$$T^{14} +$$$$86\!\cdots\!01$$$$T^{16}$$
$29$ $$1 - 356194589173448 T^{2} +$$$$30\!\cdots\!68$$$$T^{4} -$$$$29\!\cdots\!36$$$$T^{6} +$$$$43\!\cdots\!30$$$$T^{8} -$$$$52\!\cdots\!36$$$$T^{10} +$$$$94\!\cdots\!68$$$$T^{12} -$$$$19\!\cdots\!48$$$$T^{14} +$$$$98\!\cdots\!01$$$$T^{16}$$
$31$ $$1 - 2237050600650248 T^{2} +$$$$38\!\cdots\!68$$$$T^{4} -$$$$44\!\cdots\!56$$$$T^{6} +$$$$41\!\cdots\!70$$$$T^{8} -$$$$29\!\cdots\!56$$$$T^{10} +$$$$17\!\cdots\!68$$$$T^{12} -$$$$67\!\cdots\!48$$$$T^{14} +$$$$20\!\cdots\!01$$$$T^{16}$$
$37$ $$1 - 16551679761900872 T^{2} +$$$$11\!\cdots\!28$$$$T^{4} -$$$$55\!\cdots\!04$$$$T^{6} +$$$$25\!\cdots\!90$$$$T^{8} -$$$$12\!\cdots\!04$$$$T^{10} +$$$$62\!\cdots\!28$$$$T^{12} -$$$$20\!\cdots\!72$$$$T^{14} +$$$$28\!\cdots\!01$$$$T^{16}$$
$41$ $$( 1 - 46334664 T + 39266837717699036 T^{2} -$$$$13\!\cdots\!72$$$$T^{3} +$$$$70\!\cdots\!58$$$$T^{4} -$$$$17\!\cdots\!72$$$$T^{5} +$$$$70\!\cdots\!36$$$$T^{6} -$$$$11\!\cdots\!64$$$$T^{7} +$$$$32\!\cdots\!01$$$$T^{8} )^{2}$$
$43$ $$( 1 + 5095312 T + 78132909730807996 T^{2} +$$$$71\!\cdots\!72$$$$T^{3} +$$$$24\!\cdots\!90$$$$T^{4} +$$$$15\!\cdots\!28$$$$T^{5} +$$$$36\!\cdots\!96$$$$T^{6} +$$$$51\!\cdots\!88$$$$T^{7} +$$$$21\!\cdots\!01$$$$T^{8} )^{2}$$
$47$ $$1 - 134538168632987912 T^{2} +$$$$92\!\cdots\!08$$$$T^{4} -$$$$39\!\cdots\!24$$$$T^{6} +$$$$15\!\cdots\!10$$$$T^{8} -$$$$11\!\cdots\!24$$$$T^{10} +$$$$70\!\cdots\!08$$$$T^{12} -$$$$28\!\cdots\!12$$$$T^{14} +$$$$58\!\cdots\!01$$$$T^{16}$$
$53$ $$1 - 946032264727585352 T^{2} +$$$$44\!\cdots\!68$$$$T^{4} -$$$$13\!\cdots\!24$$$$T^{6} +$$$$27\!\cdots\!10$$$$T^{8} -$$$$40\!\cdots\!24$$$$T^{10} +$$$$41\!\cdots\!68$$$$T^{12} -$$$$27\!\cdots\!52$$$$T^{14} +$$$$87\!\cdots\!01$$$$T^{16}$$
$59$ $$( 1 + 482821008 T + 1514368752837832508 T^{2} +$$$$78\!\cdots\!36$$$$T^{3} +$$$$10\!\cdots\!70$$$$T^{4} +$$$$39\!\cdots\!36$$$$T^{5} +$$$$39\!\cdots\!08$$$$T^{6} +$$$$64\!\cdots\!08$$$$T^{7} +$$$$68\!\cdots\!01$$$$T^{8} )^{2}$$
$61$ $$1 - 2607734110174077128 T^{2} +$$$$31\!\cdots\!88$$$$T^{4} -$$$$28\!\cdots\!36$$$$T^{6} +$$$$22\!\cdots\!50$$$$T^{8} -$$$$14\!\cdots\!36$$$$T^{10} +$$$$82\!\cdots\!88$$$$T^{12} -$$$$34\!\cdots\!28$$$$T^{14} +$$$$67\!\cdots\!01$$$$T^{16}$$
$67$ $$( 1 + 612791440 T + 6896527760373520252 T^{2} +$$$$31\!\cdots\!84$$$$T^{3} +$$$$18\!\cdots\!06$$$$T^{4} +$$$$57\!\cdots\!16$$$$T^{5} +$$$$22\!\cdots\!52$$$$T^{6} +$$$$37\!\cdots\!60$$$$T^{7} +$$$$11\!\cdots\!01$$$$T^{8} )^{2}$$
$71$ $$1 - 17068544268164346248 T^{2} +$$$$14\!\cdots\!28$$$$T^{4} -$$$$78\!\cdots\!16$$$$T^{6} +$$$$30\!\cdots\!70$$$$T^{8} -$$$$83\!\cdots\!16$$$$T^{10} +$$$$16\!\cdots\!28$$$$T^{12} -$$$$20\!\cdots\!48$$$$T^{14} +$$$$12\!\cdots\!01$$$$T^{16}$$
$73$ $$( 1 - 1400036360 T + 8051296934129630812 T^{2} -$$$$11\!\cdots\!04$$$$T^{3} +$$$$55\!\cdots\!86$$$$T^{4} -$$$$48\!\cdots\!96$$$$T^{5} +$$$$14\!\cdots\!12$$$$T^{6} -$$$$11\!\cdots\!40$$$$T^{7} +$$$$34\!\cdots\!01$$$$T^{8} )^{2}$$
$79$ $$1 - 45980157940147603208 T^{2} +$$$$10\!\cdots\!08$$$$T^{4} -$$$$16\!\cdots\!56$$$$T^{6} +$$$$18\!\cdots\!10$$$$T^{8} -$$$$15\!\cdots\!56$$$$T^{10} +$$$$87\!\cdots\!08$$$$T^{12} -$$$$33\!\cdots\!08$$$$T^{14} +$$$$64\!\cdots\!01$$$$T^{16}$$
$83$ $$( 1 - 926711280 T + 43393122038825346812 T^{2} -$$$$33\!\cdots\!84$$$$T^{3} +$$$$94\!\cdots\!46$$$$T^{4} -$$$$52\!\cdots\!16$$$$T^{5} +$$$$10\!\cdots\!12$$$$T^{6} -$$$$34\!\cdots\!20$$$$T^{7} +$$$$57\!\cdots\!01$$$$T^{8} )^{2}$$
$89$ $$( 1 - 3081298056 T + 77020954462551268316 T^{2} -$$$$35\!\cdots\!08$$$$T^{3} +$$$$28\!\cdots\!58$$$$T^{4} -$$$$11\!\cdots\!08$$$$T^{5} +$$$$74\!\cdots\!16$$$$T^{6} -$$$$93\!\cdots\!56$$$$T^{7} +$$$$94\!\cdots\!01$$$$T^{8} )^{2}$$
$97$ $$( 1 + 4848931768 T +$$$$12\!\cdots\!96$$$$T^{2} +$$$$10\!\cdots\!08$$$$T^{3} +$$$$76\!\cdots\!90$$$$T^{4} +$$$$76\!\cdots\!92$$$$T^{5} +$$$$66\!\cdots\!96$$$$T^{6} +$$$$19\!\cdots\!32$$$$T^{7} +$$$$29\!\cdots\!01$$$$T^{8} )^{2}$$
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#### Chemical Equations Balanced on 02/11/19
Molecular weights calculated on 02/10/19 Molecular weights calculated on 02/12/19
Calculate molecular weight
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232
Molar mass of YBa2Cu3O7 is 666.19365
Molar mass of C27H46O is 386.65354
Molar mass of C is 12,0107
Molar mass of Hg is 200,59
Molar mass of K3(Fe(C2O4)3)*3H2O is 491.24274
Molar mass of NH3 is 17.03052
Molar mass of SiO2 is 60.0843
Molar mass of Ti(OH)3 is 98,88902
Molar mass of Ti(OH)4 is 115,89636
Molar mass of Al(OH)₃ is 78.0035586
Molar mass of *6H20 is 120,9528
Molar mass of C is 12,0107
Molar mass of HNa is 23,99770928
Molar mass of C₆H₁₂O₆ is 180.15588
Molar mass of NaCl is 58.44276928
Molar mass of calcium chloride is 110.984
Molar mass of K3(Fe(C2O4)3)*3H2O is 491.24274
Molar mass of Sr(ClO2)2 is 222.5236
Molar mass of K2Cr2O7 is 294.1846
Molar mass of [Cr(NH3)5(CN)]CN2 is 203.1902
Molar mass of ZnO is 81.3794
Molar mass of MgBr₂ is 184.113
Molar mass of Al(OH)3 is 78.0035586
Molar mass of CaO is 56.0774
Molar mass of CH3CH(CH3)OH is 60,09502
Molar mass of k2s is 110.2616
Molar mass of Cu(ion) is 63,546
Molar mass of *6H2O is 108,09168
Molar mass of SO2 is 64.0638
Molar mass of lead(ii) nitrate is 331.2098
Molar mass of H₂O is 18.01528
Molar mass of CaCl2 is 110.984
Molar mass of CO2 is 44,0095
Molar mass of NH4 is 18.03846
Molar mass of Hg3(PO3) is 680,741962
Molar mass of CaCO3 is 100.0869
Molar mass of CH3CH(CH3)OH is 60,09502
Molar mass of cl2 is 70,906
Molar mass of c2h4o is 44.05256
Molar mass of C19H28O2 is 288.42442
Molar mass of CH3CH2OH is 46,06844
Molar mass of C12H22O11 is 342,29648
Molar mass of CH3CH2OH is 46,06844
Molar mass of RbF is 104.4662032
Molar mass of NaOH is 39,99710928
Molar mass of NaOH is 39.99710928
Molar mass of NaOH is 39,99710928
Molar mass of (NH4)3PO4 is 149.086742
Molar mass of KNO2 is 85.1038
Molar mass of c57h110o6 is 891.4797
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232
Calculate molecular weight
Molecular weights calculated on 02/10/19 Molecular weights calculated on 02/12/19
Molecular masses on 02/04/19
Molecular masses on 01/12/19
Molecular masses on 02/11/18
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# Medial pentagonal hexecontahedron
Medial pentagonal hexecontahedron
Rank3
TypeUniform dual
Notation
Coxeter diagramp5/2p5p ()
Elements
Faces60 irregular pentagons
Edges30+60+60
Vertices12+12+60
Vertex figure60 triangles, 12 pentagons, 12 pentagrams
Measures (edge length 1)
Dihedral angle≈ 133.80098°
Central density3
Related polytopes
ConjugateMedial inverted pentagonal hexecontahedron
Convex coreNon-Catalan pentakis dodecahedron
Abstract & topological properties
Flag count600
Euler characteristic–6
OrientableYes
Genus4
Properties
SymmetryH3+, order 60
ConvexNo
NatureTame
The medial pentagonal hexecontahedron is a uniform dual polyhedron. It consists of 60 irregular pentagons, each with two short, one medium, and two long edges. Its dual is the snub dodecadodecahedron.
If the pentagon faces have medium edge length 2, then the short edge length is ${\displaystyle 1+{\sqrt {\frac {1-\xi }{\phi ^{3}-\xi }}}\approx 1.55076}$, and the long edge length is ${\displaystyle 1+{\sqrt {\frac {1-\xi }{-\phi ^{3}-\xi }}}\approx 3.85415}$, where ${\displaystyle \xi \approx -0.40903}$ is the smallest (most negative) real root of the polynomial ${\displaystyle 8x^{4}-12x^{3}+5x+1}$. The pentagons have three interior angles of ${\displaystyle \arccos \left(\xi \right)\approx 114.14440^{\circ }}$, one of ${\displaystyle \arccos \left(\phi ^{2}\xi +\phi \right)\approx 56.82766^{\circ }}$, and one of ${\displaystyle \arccos \left(\phi ^{-2}\xi -\phi ^{-1}\right)\approx 140.73912^{\circ }}$, where ${\displaystyle \phi }$ is the golden ratio.
The inradius R ≈ 1.07828 of the medial pentagonal hexecontahedron with unit edge length is equal to the square root of a real root of ${\displaystyle 8192x^{4}-12352x^{3}+3376x^{2}-104x+1}$.
A dihedral angle can be given as acos(α), where α ≈ -0.69216 is the negative of a real root of ${\displaystyle 16x^{4}+x^{3}-9x^{2}-x+1}$.
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# CBSE Case Study Questions Class 9 Maths Chapter 7 Triangles PDF Download
Case Study Questions Class 9 Maths Chapter 7 are very important to solve for your exam. Class 9 Maths Chapter 7 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Maths Chapter 7 Triangles
## Case Study Questions Class 9 Maths Chapter 7 Triangles
Case Study 1. A group of students is studying the properties of triangles. They came across the following scenario:
Three friends, Ankit, Bhavna, and Chetan, were discussing their recent hiking trip. During their hike, they noticed a triangular-shaped lake. They observed the following:
1. The lake has three sides of different lengths.
2. The sum of the lengths of any two sides of the lake is greater than the length of the third side.
3. The largest angle of the lake is less than 90 degrees.
4. The smallest angle of the lake is greater than 30 degrees.
Based on this information, the students were asked to analyze the properties of the triangle formed by the lake. Let’s see if you can answer the questions correctly:
MCQ Questions:
Q1. The type of triangle formed by the lake is:
(a) Equilateral
(b) Isosceles
(c) Scalene
(d) Right-angled
Q2. The sum of the measures of the three angles of the lake’s triangle is:
(a) 180 degrees
(b) 90 degrees
(c) 360 degrees
(d) It cannot be determined
Q3. The lake’s triangle is an example of a triangle that satisfies the:
(a) Angle-side-angle (ASA) condition
(b) Side-angle-side (SAS) condition
(c) Side-side-side (SSS) condition
(d) None of the above
Q4. The measure of the largest angle of the lake’s triangle is:
(a) 30 degrees
(b) 60 degrees
(c) 90 degrees
(d) More than 90 degrees
Q5. The measure of the smallest angle of the lake’s triangle is:
(a) 30 degrees
(b) 60 degrees
(c) 90 degrees
(d) Less than 30 degrees
Answer: (d) Less than 30 degrees
Case Study 2.A group of students is studying the properties of triangles. They encountered the following scenario:
Three friends, Rahul, Sana, and Tina, participated in a kite-flying competition. They noticed that their kites formed a triangular shape in the sky. They made the following observations:
1. The lengths of two sides of the kite triangle are equal.
2. The measure of the largest angle of the kite triangle is 90 degrees.
3. The sum of the measures of the three angles of the kite triangle is 180 degrees.
4. The lengths of the three sides of the kite triangle are in the ratio 3:4:5.
Based on this information, the students were asked to analyze the properties of the triangle formed by their kites. Let’s see if you can answer the questions correctly:
MCQ Questions:
Q1. The type of triangle formed by their kites is:
(a) Equilateral
(b) Isosceles
(c) Scalene
(d) Right-angled
Q2. The length of the third side of the kite triangle is:
(a) 3 units
(b) 4 units
(c) 5 units
(d) It cannot be determined
Q3. The kite triangle is an example of a triangle that satisfies the:
(a) Angle-side-angle (ASA) condition
(b) Side-angle-side (SAS) condition
(c) Side-side-side (SSS) condition
(d) None of the above
Q4. The measure of the smallest angle of the kite triangle is:
(a) 30 degrees
(b) 45 degrees
(c) 60 degrees
(d) 90 degrees
Q5. The lengths of the two equal sides of the kite triangle are:
(a) 3 units each
(b) 4 units each
(c) 5 units each
(d) It cannot be determined
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# Questions tagged [semidefinite-programming]
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### Minimal example using MOSEK API in python
I want to solve (simplified version) \begin{equation*} \begin{aligned} & \underset{}{\text{find}} & & X\in\mathbb{S}^{n}_{+}, x \in \mathbb{R}^{m}, \nu \in \mathbb{R}, \lambda\...
• 101
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• 639
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1 vote
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### interior point computational complexity for SDP
I am trying to get the complexity of the SDP problem for my specific problem, but Iām facing some problems. I found in the literature that the complexity of the SDP problem for an interior point per ...
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1 vote
58 views
### Distributionally Robust Stochastic Programming - Help with derivation
I've been working through this book on robust optimization of electric energy systems, and in particular chapter 4 on distributionally robust optimization. In following the derivation of section 4.2.1....
117 views
### Augmented Lagrangian Function for Semidefinite Programming Problems
I am currently reading the paper "Alternating direction augmented Lagrangian methods for semidefinite programming" and was wondering about how one comes up with the Augmented Lagrangian ...
• 33
412 views
### Adequate SDP solvers for large problem instances
I have previously used MOSEK for all my SDP needs. Recently, though, I am having a hard time trying to solve some large problems, due to lack of memory. In similar questions around the forum, SCS has ...
• 49
373 views
### Solver for nonlinear semidefinite optimization
Totally new to optimization. Is there an easy-to-use solver, package, (free) software for solving nonlinear semidefinite optimization problems?
• 129
1k views
### Non-symmetric Positive Definite/Semidefinite Matrix in Quadratic Program
A necessary condition in any quadratic programming to be convex is the matrix $\mathbf{Q}$ in the formulation $x^\intercal \mathbf{Q}x$ to be positive definite or positive semidefinite. Positive ...
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113 views
### Is this semidefinite constraint in fact pointless?
On Wikipedia, I encountered a statement that the semidefinite relaxation of a quadratically constrained quadratic program can be solved more efficiently (using only LP) in the case that no variable is ...
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# Density of Freon 13B1 [CBrF3]
## Freon 13B1 weighs 1 580 kg/m³ (98.63618 lb/ft³)
• Density of Freon 13B1 in a few select units of density measurement:
• Density of Freon 13B1 g cm3 = 1.58 g/cm³
• Density of Freon 13B1 g ml = 1.58 g/ml
• Density of Freon 13B1 g mm3 = 0.0016 g/mm³
• Density of Freon 13B1 kg m3 = 1 580 kg/m³
• Density of Freon 13B1 lb in3 = 0.057 lb/in³
• Density of Freon 13B1 lb ft3 = 98.64 lb/ft³
• See density of Freon 13B1 in hundreds of units of density measurement grouped by weight.
### Freon 13B1 density values, grouped by weight and shown as value of density, unit of density
24.38 gr/cm³ 24 383.13 gr/dm³ 690 453.24 gr/ft³ 399.57 gr/in³ 24 383 126.27 gr/m³ 0.02 gr/mm³ 18 642 237.51 gr/yd³ 24 383.13 gr/l 6 095.78 gr/metric c 365.75 gr/metric tbsp 121.92 gr/metric tsp 24.38 gr/ml 5 768.76 gr/US c 721.01 gr/fl.oz 92 300.17 gr/US gal 11 537.52 gr/pt 23 075.04 gr/US qt 360.55 gr/US tbsp 120.18 gr/US tsp
1.58 g/cm³ 1 580 g/dm³ 44 740.62 g/ft³ 25.89 g/in³ 1 580 000 g/m³ 0 g/mm³ 1 207 996.68 g/yd³ 1 580 g/l 395 g/metric c 23.7 g/metric tbsp 7.9 g/metric tsp 1.58 g/ml 373.81 g/US c 46.72 g/fl.oz 5 980.95 g/US gal 747.62 g/pt 1 495.24 g/US qt 23.36 g/tbsp 7.79 g/tsp
0 kg/cm³ 1.58 kg/dm³ 44.74 kg/ft³ 0.03 kg/in³ 1 580 kg/m³ 1.58 × 10-6 kg/mm³ 1 208 kg/yd³ 1.58 kg/l 0.4 kg/metric c 0.02 kg/metric tbsp 0.01 kg/metric tsp 0 kg/ml 0.37 kg/US c 0.05 kg/fl.oz 5.98 kg/US gal 0.75 kg/pt 1.5 kg/US qt 0.02 kg/tbsp 0.01 kg/tsp
1.56 × 10-6 long tn/cm³ 0 long tn/dm³ 0.04 long tn/ft³ 2.55 × 10-5 long tn/in³ 1.56 long tn/m³ 1.56 × 10-9 long tn/mm³ 1.19 long tn/yd³ 0 long tn/l 0 long tn/metric c 2.33 × 10-5 long tn/metric tbsp 7.78 × 10-6 long tn/metric tsp 1.56 × 10-6 long tn/ml 0 long tn/US c 4.94 × 10-5 long tn/fl.oz 0.01 long tn/US gal 0 long tn/pt 0 long tn/US qt 2.3 × 10-5 long tn/US tbsp 7.66 × 10-6 long tn/US tsp
1 580 000 µg/cm³ 1 580 000 000 µg/dm³ 44 740 617 628 µg/ft³ 25 891 561.12 µg/in³ 1 580 000 000 000 µg/m³ 1 580 µg/mm³ 1 207 996 675 640 µg/yd³ 1 580 000 000 µg/l 395 000 000 µg/metric c 23 700 000 µg/metric tbsp 7 900 000 µg/metric tsp 1 580 000 µg/ml 373 809 414.46 µg/US c 46 726 176.77 µg/fl.oz 5 980 950 612.4 µg/US gal 747 618 827.34 µg/pt 1 495 237 654.68 µg/US qt 23 363 088.38 µg/tbsp 7 787 696.11 µg/tsp
1 580 mg/cm³ 1 580 000 mg/dm³ 44 740 617.63 mg/ft³ 25 891.56 mg/in³ 1 580 000 000 mg/m³ 1.58 mg/mm³ 1 207 996 675.64 mg/yd³ 1 580 000 mg/l 395 000 mg/metric c 23 700 mg/metric tbsp 7 900 mg/metric tsp 1 580 mg/ml 373 809.41 mg/US c 46 720.6 mg/fl.oz 5 980 950.63 mg/US gal 747 618.83 mg/pt 1 495 237.66 mg/US qt 23 363.09 mg/tbsp 7 787.7 mg/tsp
0.06 oz/cm³ 55.73 oz/dm³ 1 578.18 oz/ft³ 0.91 oz/in³ 55 732.86 oz/m³ 5.57 × 10-5 oz/mm³ 42 610.83 oz/yd³ 55.73 oz/l 13.93 oz/metric c 0.84 oz/metric tbsp 0.28 oz/metric tsp 0.06 oz/ml 13.19 oz/US c 1.77 oz/fl.oz 210.97 oz/US gal 26.37 oz/pt 52.74 oz/US qt 0.82 oz/tbsp 0.27 oz/tsp
1.02 dwt/cm³ 1 015.96 dwt/dm³ 28 768.89 dwt/ft³ 16.65 dwt/in³ 1 015 963.59 dwt/m³ 0 dwt/mm³ 776 759.9 dwt/yd³ 1 015.96 dwt/l 253.99 dwt/metric c 15.24 dwt/metric tbsp 5.08 dwt/metric tsp 1.02 dwt/ml 240.37 dwt/US c 30.04 dwt/fl.oz 3 845.84 dwt/US gal 480.73 dwt/pt 961.46 dwt/US qt 15.02 dwt/US tbsp 5.01 dwt/US tsp
0 lb/cm³ 3.48 lb/dm³ 98.64 lb/ft³ 0.06 lb/in³ 3 483.3 lb/m³ 3.48 × 10-6 lb/mm³ 2 663.18 lb/yd³ 3.48 lb/l 0.87 lb/metric c 0.05 lb/metric tbsp 0.02 lb/metric tsp 0 lb/ml 0.82 lb/US c 0.11 lb/fl.oz 13.19 lb/US gal 1.65 lb/pt 3.3 lb/US qt 0.05 lb/tbsp 0.02 lb/tsp
1.74 × 10-6 short tn/cm³ 0 short tn/dm³ 0.05 short tn/ft³ 2.85 × 10-5 short tn/in³ 1.74 short tn/m³ 1.74 × 10-9 short tn/mm³ 1.33 short tn/yd³ 0 short tn/l 0 short tn/metric c 2.61 × 10-5 short tn/metric tbsp 8.71 × 10-6 short tn/metric tsp 1.74 × 10-6 short tn/ml 0 short tn/US c 5.53 × 10-5 short tn/fl.oz 0.01 short tn/US gal 0 short tn/pt 0 short tn/US qt 2.58 × 10-5 short tn/US tbsp 8.58 × 10-6 short tn/US tsp
0 sl/cm³ 0.11 sl/dm³ 3.07 sl/ft³ 0 sl/in³ 108.26 sl/m³ 1.08 × 10-7 sl/mm³ 82.77 sl/yd³ 0.11 sl/l 0.03 sl/metric c 0 sl/metric tbsp 0 sl/metric tsp 0 sl/ml 0.03 sl/US c 0 sl/fl.oz 0.41 sl/US gal 0.05 sl/pt 0.1 sl/US qt 0 sl/tbsp 0 sl/tsp
0 st/cm³ 0.25 st/dm³ 7.05 st/ft³ 0 st/in³ 248.81 st/m³ 2.49 × 10-7 st/mm³ 190.23 st/yd³ 0.25 st/l 0.06 st/metric c 0 st/metric tbsp 0 st/metric tsp 0 st/ml 0.06 st/US c 0.01 st/fl.oz 0.94 st/US gal 0.12 st/pt 0.24 st/US qt 0 st/US tbsp 0 st/US tsp
1.58 × 10-6 t/cm³ 0 t/dm³ 0.04 t/ft³ 2.59 × 10-5 t/in³ 1.58 t/m³ 1.58 × 10-9 t/mm³ 1.21 t/yd³ 0 t/l 0 t/metric c 2.37 × 10-5 t/metric tbsp 7.9 × 10-6 t/metric tsp 1.58 × 10-6 t/ml 0 t/US c 4.67 × 10-5 t/fl.oz 0.01 t/US gal 0 t/pt 0 t/US qt 2.34 × 10-5 t/tbsp 7.79 × 10-6 t/tsp
0.05 oz t/cm³ 50.8 oz t/dm³ 1 438.44 oz t/ft³ 0.83 oz t/in³ 50 798.18 oz t/m³ 5.08 × 10-5 oz t/mm³ 38 837.99 oz t/yd³ 50.8 oz t/l 12.7 oz t/metric c 0.76 oz t/metric tbsp 0.25 oz t/metric tsp 0.05 oz t/ml 12.02 oz t/US c 1.5 oz t/fl.oz 192.29 oz t/US gal 24.04 oz t/pt 48.07 oz t/US qt 0.75 oz t/US tbsp 0.25 oz t/US tsp
0 troy/cm³ 4.23 troy/dm³ 119.87 troy/ft³ 0.07 troy/in³ 4 233.18 troy/m³ 4.23 × 10-6 troy/mm³ 3 236.5 troy/yd³ 4.23 troy/l 1.06 troy/metric c 0.06 troy/metric tbsp 0.02 troy/metric tsp 0 troy/ml 1 troy/US c 0.13 troy/fl.oz 16.02 troy/US gal 2 troy/pt 4.01 troy/US qt 0.06 troy/US tbsp 0.02 troy/US tsp
Freon 13B1 density values in 285 units of density, in the form of a matrix
Density = weight ÷ volumemicrogram (µg)milligram (mg)gram (g)kilogram (kg)tonne (t)ounce (oz)pound (lb)volume unitgrain (gr)slug (sl)short ton (short tn)long ton (long tn)stone (st)troy ounce (oz t)troy pound (troy)pennyweight (dwt)
cubic millimeter1 5801.58<0.01<0.01<0.01<0.01<0.01cubic millimeter0.02<0.01<0.01<0.01<0.01<0.01<0.01<0.01
cubic centimeter1 580 0001 5801.58<0.01<0.010.06<0.01cubic centimeter24.38<0.01<0.01<0.01<0.010.05<0.011.02
cubic decimeter1 580 000 0001 580 0001 5801.58<0.0155.733.48cubic decimeter24 383.130.11<0.01<0.010.2550.84.231 015.96
cubic meter1 580 000 000 0001 580 000 0001 580 0001 5801.5855 732.863 483.3cubic meter24 383 126.27108.261.741.56248.8150 798.184 233.181 015 963.59
milliliter1 580 0001 5801.58<0.01<0.010.06<0.01milliliter24.38<0.01<0.01<0.01<0.010.05<0.011.02
liter1 580 000 0001 580 0001 5801.58<0.0155.733.48liter24 383.130.11<0.01<0.010.2550.84.231 015.96
metric teaspoon7 900 0007 9007.90.01<0.010.280.02metric teaspoon121.92<0.01<0.01<0.01<0.010.250.025.08
metric tablespoon23 700 00023 70023.70.02<0.010.840.05metric tablespoon365.75<0.01<0.01<0.01<0.010.760.0615.24
metric cup395 000 000395 0003950.4<0.0113.930.87metric cup6 095.780.03<0.01<0.010.0612.71.06253.99
cubic inch25 891 561.1225 891.5625.890.03<0.010.910.06cubic inch399.57<0.01<0.01<0.01<0.010.830.0716.65
cubic foot44 740 617 62844 740 617.6344 740.6244.740.041 578.1898.64cubic foot690 453.243.070.050.047.051 438.44119.8728 768.89
cubic yard1 207 996 675 6401 207 996 675.641 207 996.681 2081.2142 610.832 663.18cubic yard18 642 237.5182.771.331.19190.2338 837.993 236.5776 759.9
US teaspoon7 787 696.117 787.77.790.01<0.010.270.02US teaspoon120.18<0.01<0.01<0.01<0.010.250.025.01
US tablespoon23 363 088.3823 363.0923.360.02<0.010.820.05US tablespoon360.55<0.01<0.01<0.01<0.010.750.0615.02
US fluid ounce46 726 176.7746 720.646.720.05<0.011.770.11US fluid ounce721.01<0.01<0.01<0.010.011.50.1330.04
US cup373 809 414.46373 809.41373.810.37<0.0113.190.82US cup5 768.760.03<0.01<0.010.0612.021240.37
US pint747 618 827.34747 618.83747.620.75<0.0126.371.65US pint11 537.520.05<0.01<0.010.1224.042480.73
US quart1 495 237 654.681 495 237.661 495.241.5<0.0152.743.3US quart23 075.040.1<0.01<0.010.2448.074.01961.46
US gallon5 980 950 612.45 980 950.635 980.955.980.01210.9713.19US gallon92 300.170.410.010.010.94192.2916.023 845.84
#### Foods, Nutrients and Calories
SLICED ALMONDS, UPC: 04656701497 weigh(s) 127 grams per metric cup or 4.2 ounces per US cup, and contain(s) 567 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
168361 foods that contain Fatty acids, total saturated. List of these foods starting with the highest contents of Fatty acids, total saturated and the lowest contents of Fatty acids, total saturated
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Peace River weighs 1 489.72 kg/m³ (93.00018 lb/ft³) with specific gravity of 1.48972 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Naphthalene [C10H8] weighs 1 162 kg/m³ (72.54129 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-414B, liquid (R414B) with temperature in the range of -40°C (-40°F) to 71.12°C (160.016°F)
#### Weights and Measurements
A Petameter per hour squared (Pm/h²) is a derived metric SI (System International) measurement unit of acceleration
The units of data measurement were introduced to manage and operate digital information.
gr/cm² to µg/pm² conversion table, gr/cm² to µg/pm² unit converter or convert between all units of surface density measurement.
#### Calculators
Estimate daily calories using height, weight, age, gender, activity
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1. ## Constrained Optimization
(a) The reference they were referring is : "Find a lower triangular matrix L, and a diagonal matrix D such that $A = LDL^T$". I don't see how this is relevant though.
Here's what I've done so far
$L= x+y+z- \lambda(x^2+y^2+z^2-k)$
$L_x = 1-2 \lambda x = 0$
$L_y = 1-2 \lambda y = 0$
$L_z = 1-2 \lambda z = 0$
From this, critical points are $x=y=z = \frac{1}{2 \lambda}$
Since $f_{xx}=f_{yy}=f_{zz}=-2\lambda$
The Hessian matrix will be
$H = \begin{bmatrix} -2 \lambda & 0 & 0 \\ 0& -2 \lambda & 0 \\ 0&0& -2 \lambda \end{bmatrix}$
So what else do I need to do here to solve this? (I can't find any similar examples from my textbook)
Also, can anyone show me how to prove part (b). I don't understand it...
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Vehicle: Car: Transmission: Continuously Variable
Pulsed-Hydraulic Transmission (+4, -4) [vote for, against]
Infinately variable, and self-regulating
In a conventional hydraulic transmission, the engine drives a pump which pushes fluid through motors located at the wheels. Because the fluid has to travel long distances throught the lines, there are high viscous losses. Maybe it would be better to trasmit the energy through high pressure pulses. The displacement of each bit of fluid relative to the line walls would be very small, so viscous losses should be reduced. Also, only one line would be necessary. The engine would be connected to a pulse generator, which would cause plungers at the wheels to oscilate. This oscillation would be converted to rotary motion by a one-way clutch mechanism. When load is high, the plunger oscilation amplitude decreases and torque increases, etc.
-- gabe, Nov 21 2003
Patent for vibrating hydraulic system http://www.patentst...5665919-claims.html
It seems someone has been trying this. [pashute, Feb 27 2008]
gogu constantinescu http://fluid.power....pn/const/index.html
see this page about sonics theory and torque converters [blimpyway, Jun 11 2008]
On analagous systems http://en.wikipedia...i/Hydraulic_analogy
strengths and weaknesses discussed [4whom, Jun 12 2008]
Would it go "lub-dub...lub-dub..."?
-- bungston, Nov 21 2003
Certainly making directional-boring machines more effecient
-- Letsbuildafort, Nov 21 2003
The main loss of efficiency in an automatic transmission is through the torque converter, which is basically a pump, and is relativly inefficient, but there is a simple way to solve that. The biggest problem is that each time you convert one type of motion to another you lose a large portion of the energy involved. This is why rotary engines should be superior to conventional piston engines, no reciprocating motion to deal with. With your idea we still have to deal with inefficiency in the pulse generator, and with the pistons at the wheels. We could simply have one hydraulicly actuated piston drive a conventional driveshaft, but we could accomplish that much more simply with a steam engine. IMHO it would be much easier simply to stick with a clutch and manual transmission.
-- austere_apathy, Nov 21 2003
[austere_apathy] - I believe that [gabe] was not referring to automatic gearbox / transmission systems as found in many cars. Rather, we're talking about systems that use hydraulic lines instead of driveshafts.
-- benjamin, Nov 21 2003
yes, I was talking about conventional hydrostatic transmissions such as the ones used in forklifts or some lawnmowers.
-- gabe, Nov 22 2003
oops, sorry. In that case, it makes a little more sense.
-- austere_apathy, Nov 22 2003
very little
-- gabe, Nov 22 2003
Well, the engine actuates a diaphragm pump through you standard cam, sends pulses through single line,pulses at the other end are transformed to a steady flow by another diaphragm pum(through a different circuit) that moves a conventional gear motor. You have eliminated the return line and simplified the system. And perhaps, as you said, the performance will be better. Bun for you.
-- finflazo, May 11 2004
I am assuming this is more than a single sided hydraulic system - such as used to provide lift when gravity is available for lowering (single hose cylinders) like hydraulic jacks.... This is an interesting concept and it would be good to build such a system to gain knowledge - the problem I see is you are making a hydraulic system that is more like an AC electrical system (which alternates current direction - or pulses electricity) now pulsing fluid has some inherant problems that I see.... 1) the entire fluids in the line has to be started and stopped many times per second? Would this not be more like pressure waves in a fluids system - does this not cause reflection of energy and possibly damaging energy spikes when the reflected wave meet the oncoming waves (I believe this is the definition of water hammer in pipes) 2) I believe this may cause a lot of loss in energy because although nearly incompressable the fluid will compress and expand - loosing energy - generating heat? 3) this may be more of a vibration that the fluids will act to adsorb more than pass the energy along and the vibration will be passed to everything (Lines, pumps, and motors) that touch the fluid - also acting to destroying the system?
-- sklaus, Aug 13 2004
[sklaus], your anno sounds like those anti AC vibes, from Thomas Edison and his friends, when AC electricty was first out. Perhaps with a proper "transformer" you could get back more of the energy on the other side, and probably because of the wavelike pulsing, you would reach some sort of better energy transfer that avoids viscosity and shear loss. A proper transformer at the end of the line would shield the motors from energy spikes.
-- pashute, Feb 27 2008
Move out of car:Transmission and to Product:Energy:Other its about transmitting energy, not about car gears.
-- pashute, Feb 27 2008
I'm really not sure you are losing all that much energy to loss in the hoses as you think, the loss also is dictated by the length of the hose so in a pulsed system you would still have losses over the entire length of the hose, just divided over the in pulse and the out pulse, this being the case the losses may actually be greater for a given input of energy as you also need to over come the inertia of the fluid transitioning from forward to reverse flow.
-- jhomrighaus, Feb 27 2008
I don't know this is very interesting and seems very analogous to converting DC to AC. In that line of thought, it should be better to actually run three lines to the motor to transmit 3 phase AC hydraulic fluid. Go with high pressure/low volume to lower friction. The motor would just convert the 3phase back into DC to drive the motor in a similar fashion that an electrical 3phase current would be converted to DC. Then you create 6 one-way valves (hydraulic diodes) to create reasonably smooth flowing DC for the hydraulic motor.
This maybe outside the original design of the idea and I am not completely sure there would be less frictional losses.
-- MisterQED, Feb 27 2008
WOW [masterQED] you just managed to trample on like half the physics textbook. I think you are on VERY shaky ground to make any kind of comparison between electrical Transmission and Hydraulic Systems. While hydraulics is a convenient analog of an electrical distribution system, the principles and realities involved are fundamentally dissimilar and cannot be interchanged.
-- jhomrighaus, Feb 27 2008
The appropriate analogy with electrical transmission would be the use of high voltage lines for long distance, low-loss transmission. By that analogy, you want to use a hydraulic line that operates at very high pressure but moves only a small volume of liquid.
The question of "AC" versus "DC" is subsidiary. AC is advantageous, because power transmission is proportional to the RMS voltage; in effect, the peaks of your AC wave are more efficiently transmitted, being high-voltage compared to the troughs.
The analogy holds good for liquids - system will be slightly more efficient if power is transmitted by "AC", except that you'll have new losses caused by the tubes ballooning.
Overall, the most important thing is to use high pressure and low volume.
-- MaxwellBuchanan, Feb 27 2008
With [jhomrighaus], I think you'd lose energy accelerating and decelerating the fluid continuously. Also, it would be a perfect fatigue machine and you'd fail pressure lines, fittings etc on a very frequent basis.
-- david_scothern, Feb 27 2008
[jhomrighaus] No, I’ll admit the advantages are dubious, but the physics are sound. First I’ll explain the physics, and then we can debate the existence of any advantage.
My analogy to a DC current is a standard vane type hydraulic pump connecting to another vane type hydraulic pump. Fluid flows one way on one hose and back in another, which is analogous to electron flow in DC electrical circuits. My analogy of AC current is a single piston pump on a crank with no valves that is connected to a second identical single piston pump on a crank. Fluid never circulates, only moves back and forth in the single hose. The current of the fluid alternates direction. I suggested three phase to even out the power transmission. The one way valves enable the AC flow to be turned back into DC flow to allow non-identical motors.
The dubious advantage is whether this system creates more/less or identical fluid friction. And yes, this would lead to severe fatigue problems and the energy loss from decelerating and accelerating the fluid may be significant. Or the pipe expansion may absorb all pressure changes.
-- MisterQED, Feb 27 2008
i have no objection to the analogy, what I object to is the physics, what works for an electrical distribution system does not work for a hydraulic system, for example, in an electrical system the working medium(electrons) are essentially massless and move at the speed of light, thus AC power does not have to overcome inertia, In a hydraulic system the fluid has considerable mass and thus a considerable inertia that must be overcome if you were to physically move the fluid forward and backward rapidly.
Another issue that arises is the speed of propagation of the pressure pulse in the fluid which as mentioned before is limited to roughly the speed of sound in a fluid, thus long distance transmission of fluids is problematic (one of the reasons for water hammer) the pulses will rapidly get jumbled up and lose coherence unless the pulse rate is quite slow. Contrary to many statement fluids can be compressed such that a pulse of pressure is transmitted. If it was not so then if you blew a depth charge all the water in the ocean would move at once which is not the case.
For a hydralic system linear flow is normally the most efficient means as the system can reach a static equilibrium and then you extract whatever energy you wan t from the system. Another difference is the conduit itself which is flexible such that you experience small but measurable losses due to expansion of the conduit, thus the more frequently the pressure changes the greater the amount of loss that is incurred by expanding the conduit with the pressure pulse.
While it may seem like a an equivalence exists, this is really not the case.
-- jhomrighaus, Feb 27 2008
// in an electrical system the working medium(electrons) are essentially massless and move at the speed of light,// No, they don't. Honest.
Suppose we have a 1 metre long copper wire with a cross-section of 1 square millimetre, carrying a current of 1 amp. How fast do the electrons move?
Well, first of all let's assume that there's one mobile electron for each copper atom (not true, but roughly). The wire contains about 10^23 atoms of copper, and hence 10^23 mobile electrons.
Now, in one second, at one amp, one coulomb of charge passes through the wire. The charge on an electron is 1.6 x 10^-19 coulombs, so roughly 10^18 electrons pass out of the end of the wire per second. Dividing 10^18 by 10^23 tells you that, in one second, the electrons move about 10^-5 metres, or about a hundredth of a millimetre.
So, the electrons in the wire are moving just over an inch per hour. This is substantially less than the speed of light.
-- MaxwellBuchanan, Mar 01 2008
Look, strip away the confusing stuff and its a really simple idea.
Two double acting pistons act to comunicate hydraulic energy without the need for a return line. Oscilation on one end produces occiation on the other and work is communicated. Simple but pointless anywhere where a better sytem could be used. Fluid systems push great, but they hate to pull (suck fluid). A flow regulated loop is more efficient and thus the standard for transmitting power at a distance.
-- WcW, Mar 02 2008
What you describe here was developed almost a century ago by George Constantinescu - under the technical field of sonics - that basicaly is using vibrations or pulses in a fluid or solid medium to transmit power.
One important aplication was a through-propeller machine gun synchronising fire device, used by british aircraft during WWI .
Others applications folowed - like drilling hammers or fluid propelled gun. Also an inertial infinitely variable torque converter was inspired by his research on sonics. (see the gogu constantinescu link above for detailed info)
-- blimpyway, Jun 11 2008
What you are describing seems more analogous to a modern switching power supply than a transformer. The general approach doesn't require fluids, though they probably simplify things. A non-fluid analogue would be a pile driver. Each whomp from the falling weight will yield roughly the same amount of energy; the less force is required to move the piling, the further it will travel. I believe an impact wrench is also similar.
The problem in all the mechanical situations is that the frictional losses end up being rather huge. To be sure, in a typical pile-driver application, the primary force to be overcome is friction, but if one were using the same approach to compress a spring against a ratchet it would be possible to have a continuously-variable transmission without any fluid.
-- supercat, Jun 12 2008
random, halfbakery
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Cody
# Problem 163. Love triangles
Solution 200036
Submitted on 4 Feb 2013 by Yaroslav
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% sides = [1 2 1000]; y_correct = false; assert(isequal(triangle(sides),y_correct))
2 Pass
%% sides = [3 4 5]; y_correct = true; assert(isequal(triangle(sides),y_correct))
3 Pass
%% sides = [5 5 5]; y_correct = true; assert(isequal(triangle(sides),y_correct))
4 Pass
%% sides = [6 6 6]; y_correct = true; assert(isequal(triangle(sides),y_correct))
5 Pass
%% sides = [1 1 1]; y_correct = true; assert(isequal(triangle(sides),y_correct))
6 Pass
%% sides = [1 2 2]; y_correct = true; assert(isequal(triangle(sides),y_correct))
7 Pass
%% sides = [2 2 5]; y_correct = false; assert(isequal(triangle(sides),y_correct))
8 Pass
%% sides = [5 2 2]; y_correct = false; assert(isequal(triangle(sides),y_correct))
9 Pass
%% sides = [1 3 1]; y_correct = false; assert(isequal(triangle(sides),y_correct))
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Wiki.RIP
Aftershock
From Wikipedia, the free encyclopedia
An aftershock is a smaller earthquake that follows a larger earthquake, in the same area of the main shock, caused as the displaced crust adjusts to the effects of the main shock. Large earthquakes can have hundreds to thousands of instrumentally detectable aftershocks, which steadily decrease in magnitude and frequency according to known laws. In some earthquakes the main rupture happens in two or more steps, resulting in multiple main shocks. These are known as doublet earthquakes, and in general can be distinguished from aftershocks in having similar magnitudes and nearly identical seismic waveforms.
Distribution of aftershocks
Most aftershocks are located over the full area of fault rupture and either occur along the fault plane itself or along other faults within the volume affected by the strain associated with the main shock. Typically, aftershocks are found up to a distance equal to the rupture length away from the fault plane.
The pattern of aftershocks helps confirm the size of area that slipped during the main shock. In the case of the 2004 Indian Ocean earthquake and the 2008 Sichuan earthquake the aftershock distribution shows in both cases that the epicenter (where the rupture initiated) lies to one end of the final area of slip, implying strongly asymmetric rupture propagation.
Aftershock size and frequency with time
Aftershocks rates and magnitudes follow several well-established empirical laws.
Omori's law
The frequency of aftershocks decreases roughly with the reciprocal of time after the main shock. This empirical relation was first described by Fusakichi Omori in 1894 and is known as Omori's law.[1] It is expressed as
${\displaystyle n(t)={\frac {k}{(c+t)}}}$
where k and c are constants, which vary between earthquake sequences. A modified version of Omori's law, now commonly used, was proposed by Utsu in 1961.[2][3]
${\displaystyle n(t)={\frac {k}{(c+t)^{p}}}}$
where p is a third constant which modifies the decay rate and typically falls in the range 0.7–1.5.
According to these equations, the rate of aftershocks decreases quickly with time. The rate of aftershocks is proportional to the inverse of time since the mainshock and this relationship can be used to estimate the probability of future aftershock occurrence.[4] Thus whatever the probability of an aftershock are on the first day, the second day will have 1/2 the probability of the first day and the tenth day will have approximately 1/10 the probability of the first day (when p is equal to 1). These patterns describe only the statistical behavior of aftershocks; the actual times, numbers and locations of the aftershocks are stochastic, while tending to follow these patterns. As this is an empirical law, values of the parameters are obtained by fitting to data after a mainshock has occurred, and they imply no specific physical mechanism in any given case. But the Utsu-Omori law has also been obtained theoretically, as the solution of a differential equation describing the evolution of the aftershock activity[5], where the interpretation of the evolution equation is based on the idea of deactivation of the faults in the vicinity of the main shock of the earthquake. Also, previously Utsu-Omori law was obtained from a nucleation process.[6] Results show that the spatial and temporal distribution of aftershocks is separable into a dependence on space and a dependence on time. And more recently, through the application of a fractional solution of the reactive differential equation[7], a double power law model shows the number density decay in several possible ways, among which is a particular case the Utsu-Omori Law.
Båth's law
The other main law describing aftershocks is known as Båth's Law[8][9] and this states that the difference in magnitude between a main shock and its largest aftershock is approximately constant, independent of the main shock magnitude, typically 1.1–1.2 on the Moment magnitude scale.
Gutenberg–Richter law
Gutenberg–Richter law for b = 1
Magnitude of the Central Italy earthquake of August 2016 (red dot) and aftershocks (which continued to occur after the period shown here)
Aftershock sequences also typically follow the Gutenberg–Richter law of size scaling, which refers to the relationship between the magnitude and total number of earthquakes in a region in a given time period.
${\displaystyle \!\,N=10^{a-bM}}$
Where:
• ${\displaystyle N}$ is the number of events greater or equal to ${\displaystyle M}$
• ${\displaystyle M}$ is magnitude
• ${\displaystyle a}$ and ${\displaystyle b}$ are constants
In summary, there are more small aftershocks and fewer large aftershocks.
Effect of aftershocks
Aftershocks are dangerous because they are usually unpredictable, can be of a large magnitude, and can collapse buildings that are damaged from the main shock. Bigger earthquakes have more and larger aftershocks and the sequences can last for years or even longer especially when a large event occurs in a seismically quiet area; see, for example, the New Madrid Seismic Zone, where events still follow Omori's law from the main shocks of 1811–1812. An aftershock sequence is deemed to have ended when the rate of seismicity drops back to a background level; i.e., no further decay in the number of events with time can be detected.
Land movement around the New Madrid is reported to be no more than 0.2 mm (0.0079 in) a year,[10] in contrast to the San Andreas Fault which averages up to 37 mm (1.5 in) a year across California.[11] Aftershocks on the San Andreas are now believed to top out at 10 years while earthquakes in New Madrid are considered aftershocks nearly 200 years after the 1812 New Madrid earthquake.[12]
Foreshocks
Some scientists have tried to use foreshocks to help predict upcoming earthquakes, having one of their few successes with the 1975 Haicheng earthquake in China. On the East Pacific Rise however, transform faults show quite predictable foreshock behaviour before the main seismic event. Reviews of data of past events and their foreshocks showed that they have a low number of aftershocks and high foreshock rates compared to continental strike-slip faults.[13]
Modeling
Seismologists use tools such as the Epidemic-Type Aftershock Sequence model (ETAS) to study cascading aftershocks.[14]
Psychology
Following a large earthquake and aftershocks, many people have reported feeling "phantom earthquakes" when in fact no earthquake was taking place. This condition, known as "earthquake sickness" is thought to be related to motion sickness, and usually goes away as seismic activity tails off.[15][16]
References
1. ^ Omori, F. (1894). "On the aftershocks of earthquakes" (PDF). Journal of the College of Science, Imperial University of Tokyo. 7: 111–200. Archived from the original (PDF) on 2015-07-16. Retrieved 2015-07-15.
2. ^ Utsu, T. (1961). "A statistical study of the occurrence of aftershocks". Geophysical Magazine. 30: 521–605.
3. ^ Utsu, T.; Ogata, Y.; Matsu'ura, R.S. (1995). "The centenary of the Omori formula for a decay law of aftershock activity" (PDF). Journal of Physics of the Earth. 43: 1–33. doi:10.4294/jpe1952.43.1. Archived from the original (PDF) on 2015-07-16.
4. ^ Quigley, M. "New Science update on 2011 Christchurch Earthquake for press and public: Seismic fearmongering or time to jump ship". Christchurch Earthquake Journal. Archived from the original on 29 January 2012. Retrieved 25 January 2012.
5. ^ Guglielmi, A.V. (2016). "Interpretation of the Omori law". Izv., Phys. Solid Earth. 52 (5): 785–786. arXiv:1604.07017. doi:10.1134/S1069351316050165.
6. ^ Shaw, Bruce (1993). "Generalized Omori law for aftershocks and foreshocks from a simple dynamics". Geophysical Research Letters. 20 (10): 907–910. doi:10.1029/93GL01058.
7. ^ Sánchez, Ewin; Vega, Pedro (2018). "Modelling temporal decay of aftershocks by a solution of the fractional reactive equation". Applied Mathematics and Computation. 340: 24–49. doi:10.1016/j.amc.2018.08.022.
8. ^ Richter, Charles F., Elementary seismology (San Francisco, California, USA: W. H. Freeman & Co., 1958), page 69.
9. ^ Båth, Markus (1965). "Lateral inhomogeneities in the upper mantle". Tectonophysics. 2 (6): 483–514. Bibcode:1965Tectp...2..483B. doi:10.1016/0040-1951(65)90003-X.
10. ^ Elizabeth K. Gardner (2009-03-13). "New Madrid fault system may be shutting down". physorg.com. Retrieved 2011-03-25.
11. ^ Wallace, Robert E. "Present-Day Crustal Movements and the Mechanics of Cyclic Deformation". The San Andreas Fault System, California. Archived from the original on 2006-12-16. Retrieved 2007-10-26.
12. ^ "Earthquakes Actually Aftershocks Of 19th Century Quakes; Repercussions Of 1811 And 1812 New Madrid Quakes Continue To Be Felt". Science Daily. Archived from the original on 8 November 2009. Retrieved 2009-11-04.
13. ^ McGuire JJ, Boettcher MS, Jordan TH (2005). "Foreshock sequences and short-term earthquake predictability on East Pacific Rise transform faults". Nature. 434 (7032): 445–7. Bibcode:2005Natur.434..457M. doi:10.1038/nature03377. PMID 15791246.
14. ^ For example: Helmstetter, Agnès; Sornette, Didier (October 2003). "Predictability in the Epidemic-Type Aftershock Sequence model of interacting triggered seismicity". Journal of Geophysical Research: Solid Earth. 108 (B10): 2482ff. arXiv:cond-mat/0208597. Bibcode:2003JGRB..108.2482H. doi:10.1029/2003JB002485. As part of an effort to develop a systematic methodology for earthquake forecasting, we use a simple model of seismicity based on interacting events which may trigger a cascade of earthquakes, known as the Epidemic-Type Aftershock Sequence model (ETAS).
15. ^ Japanese researchers diagnose hundreds of cases of 'earthquake sickness', Daily Telegraph, 20 June 2016
16. ^ After the earthquake: why the brain gives phantom quakes, The Guardian, 6 November 2016
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### Lions Tigers and Bears
Q: You are told that a certain area in a forest has lions, tigers and bears. You tour that area and observe 5 tigers, 2 lions and 1 bear. What is your estimate on the distribution of these animals?
Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)
A: This is a good example to demonstrate the multinomial distributions, it's application and also introduce the concept of "conjugate priors". Lets deal with them one at a time and then revisit the problem.
The Conjugate Prior:
This concept is part of Bayesian analysis. Lets assume you have a prior which belongs to a "family" of functions say $$y = f(\theta,x)$$. You get some additional information and you update your estimate such that the posterior belongs to the same family of functions. So the prior and posterior differ only in the parameters that go into the function and not in its form and structure. An example of conjugate priors is the Gaussian distribution.
The Multinomial Distribution:
This is an extension of the binomial distribution. Assume you have a bag with black, blue & green balls. The probability of drawing each one of them is $${p_{black},p_{blue},p_{green}}$$ and you pull 10 balls. What is the probability of seeing 2 black balls, 6 blue balls and 2 green balls? This is given by the following formula (which can be derived)
$$P(2Black,6Blue,2Green) = \frac{10!}{2!6!2!} p_{black}^{2}\times p_{blue}^{6}\times p_{green}^{2}$$
The Dirichlet Distrubution:
The probability density of this distribution has the following form and requires a set of parameters as input. These parameters are characterized by a vector $$\{\alpha_1,\alpha_2,\alpha_3,\ldots\}$$ which we shall represent as $$\boldsymbol\alpha$$. So, if we have a set of probability measures $$\{p_{black},p_{blue},p_{green}\}$$, a Dirichlet distribution is
$$f(p_{black},p_{blue},p_{green};\alpha_1,\alpha_2,\alpha_3) = \frac{1}{Z(\boldsymbol \alpha)}p_{black}^{\alpha_1 - 1}p_{blue}^{\alpha_1 - 1}p_{green}^{\alpha_3 - 1}$$
where $$Z(\boldsymbol \alpha)$$ is given by a more daunting form in terms of the gamma function, but we will not go into its details any further.
$$Z(\boldsymbol \alpha) = \frac{\prod_{i=1}^{3}\Gamma({\alpha_i})}{\Gamma(\sum_{i}^{3}\alpha_i)}$$
And now for the kicker... the Dirichlet distribution is the conjugate prior of the multinomial distribution. This can be proved algebraically. Here is a wikipedia link describing the same. Notice, the form and structure of the two equations are one and the same and the parameters $$\boldsymbol \alpha$$ can be seen as "prior" counts of categories we have seen. This is critical to understand. What we are concluding is not that we have figured out a good way to find priors, but instead if we knew some counts from earlier on the way to include them into the model is to simply treat them as prior counts!
Now, coming back to the problem. We observe 5 Lions, 2 Tigers and 1 Bear. Note, this is our first observation and we have no prior counts whatsoever. So how do we cast this to a Bayesian framework? We exploit one little piece of information that could have gotten overlooked. We implicitly know that there are Lions, Tigers & Bears in the forest. So there must be at least one observed by someone at sometime. Can we leverage this information? Absolutely! We will simply set our $$\boldsymbol \alpha$$ parameter to $$\{1,1,1\}$$. A frequentist approach would estimate the distribution of lions, tigers and bears as
$$P(Lion) = \frac{5}{5 + 2 + 1}= \frac{5}{8}\\ P(Tiger) = \frac{2}{5 + 2 + 1} = \frac{2}{8}\\ P(Bear) = \frac{1}{5 + 2 + 1}=\frac{1}{8}$$
and the Bayesian approach would estimate it in the following way
$$P(Lion) = \frac{5 + \color{red}{1}}{5 + 2 + 1 + \color{red}{1+1+1}}= \frac{6}{11}\\ P(Tiger) = \frac{2 + \color{red}{1}}{5 + 2 + 1 + \color{red}{1+1+1}} = \frac{3}{11}\\ P(Bear) = \frac{1 + \color{red}{1}}{5 + 2 + 1 + \color{red}{1+1+1}}=\frac{2}{11}$$
Notice what happened? Our estimates will now tend to be a bit smoother if we were to do this experiment several times, the $$\boldsymbol \alpha$$ would smooth out the estimates. How about our choice of values for $$\boldsymbol \alpha$$? We chose 1s here based on the wording of the problem. But if we want more dampening we could as well choose bigger values. This is related to the amount of confidence we have on the priors. If we chose 100s instead of 1s, it would take a lot of observations to move the estimates.
Finally, let us try out a simulation. The R code below attempts to simulate this very situation. The code does the following:
1. Pick the true number of Lions, Tigers and Bears. This number will change with every iteration.
2. Pick a random subset of Lions, Tigers and Bears as the observed. These numbers will always be lesser than the true number of Lions, Tigers and Bears.
3. Compute the distribution based on frequentist and Bayesian approaches described above keeping $$\boldsymbol \alpha = \{1,1,1\}$$.
4. Compute an error statistic.
I've chosen the RMSE error, but there are better ways to compare distributions.
The RMSE is simple and quite well known, you can look it up here for more information. The lower it is, the better the estimate. The output of the above R code is a graph that charts out how close we got to the real estimate as we increased the number of iterations in the simulations. Initially the lines may be a little close to each other but for a large number of iterations, you can clearly see the Bayes estimate winning hands down. This is a powerful method with wide ranging applications and provides a good degree of robustness.
Some good books on probability worth buying
40 Puzzles and Problems in Probability and Mathematical Statistics (Problem Books in Mathematics)
A new entrant and seems promising
Fifty Challenging Problems in Probability with Solutions (Dover Books on Mathematics)
This book is a great compilation that covers quite a bit of puzzles. What I like about these puzzles are that they are all tractable and don't require too much advanced mathematics to solve.
Introduction to Algorithms
This is a book on algorithms, some of them are probabilistic. But the book is a must have for students, job candidates even full time engineers & data scientists
Introduction to Probability Theory
An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition
The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (and Everyone Else!)
Introduction to Probability, 2nd Edition
The Mathematics of Poker
Good read. Overall Poker/Blackjack type card games are a good way to get introduced to probability theory
Let There Be Range!: Crushing SSNL/MSNL No-Limit Hold'em Games
Easily the most expensive book out there. So if the item above piques your interest and you want to go pro, go for it.
Quantum Poker
Well written and easy to read mathematics. For the Poker beginner.
Bundle of Algorithms in Java, Third Edition, Parts 1-5: Fundamentals, Data Structures, Sorting, Searching, and Graph Algorithms (3rd Edition) (Pts. 1-5)
An excellent resource (students/engineers/entrepreneurs) if you are looking for some code that you can take and implement directly on the job.
Understanding Probability: Chance Rules in Everyday Life A bit pricy when compared to the first one, but I like the look and feel of the text used. It is simple to read and understand which is vital especially if you are trying to get into the subject
Data Mining: Practical Machine Learning Tools and Techniques, Third Edition (The Morgan Kaufmann Series in Data Management Systems) This one is a must have if you want to learn machine learning. The book is beautifully written and ideal for the engineer/student who doesn't want to get too much into the details of a machine learned approach but wants a working knowledge of it. There are some great examples and test data in the text book too.
Discovering Statistics Using R
This is a good book if you are new to statistics & probability while simultaneously getting started with a programming language. The book supports R and is written in a casual humorous way making it an easy read. Great for beginners. Some of the data on the companion website could be missing.
### The Best Books to Learn Probability
If you are looking to buy some books in probability here are some of the best books to learn the art of Probability The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (and Everyone Else!) A good book for graduate level classes: has some practice problems in them which is a good thing. But that doesn't make this book any less of buy for the beginner. An Introduction to Probability Theory and Its Applications, Vol. 1, 3rd Edition This is a two volume book and the first volume is what will likely interest a beginner because it covers discrete probability. The book tends to treat probability as a theory on its own Discovering Statistics Using R This is a good book if you are new to statistics & probability while simultaneously getting started with a programming language. The book supports R and is written in a casual humorous way making it an easy read. Great for beginners. Some of the data on the companion website could be missing. Fifty Cha
### Fun with Uniform Random Numbers
Q: You have two uniformly random numbers x and y (meaning they can take any value between 0 and 1 with equal probability). What distribution does the sum of these two random numbers follow? What is the probability that their product is less than 0.5. The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists A: Let z = x + y be the random variable whose distribution we want. Clearly z runs from 0 to 2. Let 'f' denote the uniform random distribution between [0,1]. An important point to understand is that f has a fixed value of 1 when x runs from 0 to 1 and its 0 otherwise. So the probability density for z, call it P(z) at any point is the product of f(y) and f(z-y), where y runs from 0 to 1. However in that range f(y) is equal to 1. So the above equation becomes From here on, it gets a bit tricky. Notice that the integral is a function of z. Let us take a look at how else we can simply the above integral. It is easy to see that f(z-y) = 1 when (
### The Best Books for Linear Algebra
The following are some good books to own in the area of Linear Algebra. Linear Algebra (2nd Edition) This is the gold standard for linear algebra at an undergraduate level. This book has been around for quite sometime a great book to own. Linear Algebra: A Modern Introduction Good book if you want to learn more on the subject of linear algebra however typos in the text could be a problem. Linear Algebra (Dover Books on Mathematics) An excellent book to own if you are looking to get into, or want to understand linear algebra. Please keep in mind that you need to have some basic mathematical background before you can use this book. Linear Algebra Done Right (Undergraduate Texts in Mathematics) A great book that exposes the method of proof as it used in Linear Algebra. This book is not for the beginner though. You do need some prior knowledge of the basics at least. It would be a good add-on to an existing course you are doing in Linear Algebra. Linear Algebra, 4th Ed
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# Excel Reverse Order
## What Is Reverse Order In Excel?
Excel reverse order involves flipping the data in a column, with the values at the bottom, in the original data, showing at the top after the flip. Though Excel does not offer an inbuilt option to reverse the order of data rows, one can flip the data using the Sort option, Excel formula, and VBA coding technique.
Users can use this Excel option to reverse the order of unsorted data, such as names and numeric values in a column or array.
For example, the table below contains a list of employee names.
Suppose the requirement is to reverse the order of the employee names and display the output in column B.
Then, we can reverse order of cells in Excel column A using the INDEX and ROWS functions in the target cells and achieve the desired outcome.
In the above Excel reverse name order example, the ROWS() returns the number of rows in the specified column A range. And then, the INDEX() returns the value in the cell at the intersection of the specified column range and the count of rows the ROWS() returned.
###### Key Takeaways
• Excel reverse order is a feature to display the given data in a column flipped or the top-to-bottom order of the data reversed.
• The option helps users to sort and display the data, such as numbers and texts, in reverse order.
• Though Excel can sort the given data in ascending and descending orders, it does not have the option to reverse the order of the data rows directly. However, we can use the Sort option, VBA coding, and Excel formulas containing the INDEX and ROWS functions to achieve the desired reverse order of data.
### Explanation
Excel offers an inbuilt function, Sort, to arrange the data in ascending or descending order.
However, in some scenarios, we may require to flip the given columns of data, vertically or horizontally, with the data remaining unsorted. For example, we may need to only display a list of items in reverse order without sorting them alphabetically.
We can apply the below-explained methods to use the Excel reverse order feature in such cases.
### How To Reverse The Order Of Data Rows In Excel?
We can achieve the Excel reverse order of data rows using the following three ways:
• The Sort option from the Data tab
• Excel formula containing the INDEX() and ROWS()
• VBA code
The following illustrations explain the above mentioned methods to vertically flip the data in one or more columns.
And while the Sort and the VBA coding options can reverse and display the flipped data in the same column, the Excel formula can show the flipped data in another column.
#### Example #1 – Using Simple Sort Method
We can reverse order of cells in Excel columns vertically using the sort method.
For example, the following table contains a set of top tech US companies.
Suppose the requirement is to flip the order of the cells in column A. Then, we can perform the required Excel reverse name order operation using the Sort option.
• Step 1: Add a helper column, with numbers in ascending order, to the source data, as depicted below.
• Step 2: Next, select the entire data range (with or without the column headers) and then, click DataSort to open the Sort window.
Alternatively, we can select the data range and press the shortcut keys Alt + D + S to open the Sort window.
Pressing the last key in the keyboard shortcut S will open the Sort window.
• Step 3: Set Sort by as Helper.
And pick the option Largest to Smallest as the Order.
Finally, click OK to close the window.
And the required Excel reverse order of data will appear as depicted below. We see the company names are in reverse order.
On the other hand, suppose the original data includes two columns, as shown below.
And if we aim for Excel reverse column order, row-wise, for the entire data, then, we can follow steps 1 to 3, explained above, to achieve the required outcome.
Otherwise, we can add the helper column with numbers in ascending order and use the following steps to get the required Excel reverse order of list in each source column.
• Step 1: First, select the Helper column data (with or without the column header) and follow the path DataSort Z to A to sort the Helper column data in descending order.
• Step 2: Next, we will see the Sort Warning message, where we must select the option to expand the selection.
And clicking Sort in the warning dialog box will flip the three columns’ data vertically, as depicted below.
We can also follow the above-explained alternative method in the first scenario, with the source data containing one column of data to reverse.
#### Example #2 – Using Excel Formula
Using the INDEX and ROWS functions, we can apply the Excel reverse order feature.
For example, the following table contains an item list.
Suppose we require the Excel reverse order of list, given in column B in column C.
Then, here is how to use the INDEX and ROWS functions in the target cells to obtain the required output.
• Step 1: Select cell C3, enter the below formula, and press Enter.
=INDEX(\$B\$3:\$B\$11,ROWS(B3:\$B\$11))
• Step 2: Next, using the excel fill handle, implement the formula in cell range C4:C11.
Let us see the cell C3 expression to understand how the formula works. First, the ROWS() returns the number of rows in the range B3:\$B\$11, which is 9. And then, the INDEX excel formula returns the value in the cell where the column range B3:B11 and row 9 meet, which is the cell B11 value, Baking Powder.
Likewise, the Excel reverse order formulas list the column B items in column C, with their order flipped.
#### Example #3 – Using VBA Coding
We can apply the Excel reverse order option from the VBA Editor. Let us see the steps with an example.
The table below contains a list of company codes and their volumes.
And we need to achieve Excel reverse column order for the above data and display the outcome in column B.
Then, the steps are as follows:
• Step 1: First, open the worksheet containing the above table and then, click Alt + F11 to access the VBA Editor window.
• Step 2: Pick the applicable VBAProject and choose the Module option from the Insert tab to open a new module window, Module1.
• Step 3: Next, enter the VBA code to achieve Excel reverse order of the given data in the target cells.
Sub Reverse_Vertical_Order()
Dim r As Long
Dim RO As Long
RO = Cells(Rows.Count, 1).End(xlUp).Row
For r = 2 To RO
Cells(r,2).Value = Cells(RO, 1).Value
RO = RO – 1
Next r
End Sub
• Step 4: Click the play icon in the menu to run the VBA code.
We can now go to the active worksheet to view the given data in the reverse order in column B.
First, we declare two Long variables, r, and RO. While r is the counter variable for the For loop, RO gets the last row count in the worksheet containing data, which is 7.
Next, the For loop starts assigning the column A cell values from the bottom to column B cells to fill the column from the top. The process continues till the counter reaches the last row, 7. And thus, the loop fills all column B cells till row 7, with the given data appearing in reverse order.
### Important Things To Note
• When using the Sort option from the Data tab to achieve Excel reverse order of data rows, ensure the Sort Options has the default setting of Sort top to bottom. It will maintain the sorting orientation as Column to reverse the data order vertically.
• When using the helper column while applying the Sort option to flip a column of data, ensure the numbers in the helper column are in ascending order.
1. Can I flip columns and rows in Excel?
We can flip columns and rows in Excel.
If we have to flip columns or rows of data, use the Sort option from the Data tab.
And if we must flip columns and rows to rows and columns respectively, use the Transpose excel option from the Paste Special window.
2. How to reverse order in Excel horizontally?
We can reverse order in Excel horizontally using the following steps.
1) First, change the given data rows into columns using the Transpose option in the Paste Special window.
2) Next, sort the columns using the Sort option from the Data tab.
3) Finally, rotate the columns to rows using the Transpose option in the Paste Special window.
Let us see the steps with an example.
The below table shows a set of fruits, their order dates, and order status data.
Suppose the requirement is to flip the data order horizontally. Then, the steps are as follows:
• Step 1: Select the table range and press Ctrl + C to copy the data.
Select a cell, say, cell E1, press the shortcut keys Alt + E + S to open the Paste Special window, and check the Transpose box.
Clicking OK will paste the transposed data into the selected cell.
• Step 2: Add a helper column of numbers in ascending order to the end of the transposed data in column L.
• Step 3: Click on a cell in the transposed data range and follow the path DataSort to open the Sort window.
• Step 4: Ensure the My data has headers box remains unchecked and pick column L, the helper column, as the Sort by option.
And pick the Largest to Smallest option as the Order.
Clicking OK will reverse the order of the values in the transposed data.
• Step 5: Select the transposed data range, leaving the helper column unselected. And press Ctrl + C to copy the data.
Select a cell, say, cell A10, press the shortcut keys Alt + E + S to open the Paste Special window, and check the Transpose box.
And clicking OK will flip the source data horizontally.
When we compare the source and the final data, the order of the data in each row appears flipped horizontally in the output.
3. Why is reverse order in Excel not working?
The reverse order in Excel is not working, perhaps because of the below reasons:
We tried applying the Sort method without the numbers in the helper column being in ascending order.
The Sort Options setting in the Sort window does not match the desired reverse order type (vertical and horizontal).
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A New Approach to Time-Dependent Solutions to the Non-Linear Fokker-Planck Equations Related to Arbitrary Functions of Tsallis Entropy: A Mathematical Study and Investigation
Journal of Modern Physics
Vol.3 No.5(2012), Article ID:19275,10 pages DOI:10.4236/jmp.2012.35058
A New Approach to Time-Dependent Solutions to the Non-Linear Fokker-Planck Equations Related to Arbitrary Functions of Tsallis Entropy: A Mathematical Study and Investigation
Institute for Advanced Studies, Tehran, Iran
Received November 12, 2011; revised January 6, 2012; accepted February 6, 2012
Keywords: Non-Linear Fokker-Planck Equations; Tsallis Entropy; Generalized Entropies
ABSTRACT
The non-linear Fokker-Planck equations can relate to the generalized entropies. The stationary solution of Fokker-Planck equations being dependent on entropies which are as a general function of Tsallis entropy is obtained. Moreover, the time-dependent solution of these equations for linear drift forces is determined. The non-linear Fokker-Planck equations can be an appropriate mathematical model to define the processes which relate to the disordered diffusion. To realize in which category each diffusion system is, it is essential to calculate the variance () as a function of time which requires to know the time dependence of systems’ distribution functions. Therefore in this paper, the distribution function of the systems being consistent with Tsallis entropy and its functions is obtained.
1. Introduction
As a successful theory, the Boltzmann-Gibbs statistical mechanics enables physicists to provide the microscopic theoretical models for describing the thermodynamic systems. The Boltzmann-Gibbs entropy is defined as follows:
(1)
where is the probability that the system occupies the microstate i, and k is the Boltzmann constant. However, this entropy is solely appropriate for the systems with short-range interactions, Markov processes, and in fact, the systems whose phase space is ergodic. The Boltzmann-Gibbs entropy is suitable to describe the equilibrium physical systems, but it cannot be applied to the non-equilibrium states. The linear Fokker-Planck equation is one of the main phenomenological equations of statistical mechanics in non-equilibrium states. If we connect this equation to the Boltzmann-Gibbs entropy, its time evolution indicates the time dependence of probability distribution function for the system in the presence of external potential. In recent years, systems such as long-range systems, non-linear systems, and the systems with long-term memory have been discovered which do not conform to the Boltzmann-Gibbs statistics. In these conditions, the Tsallis entropy is proposed. The Tsallis entropy is transformed into the Boltzmann-Gibbs entropy for [1]:
(2)
The same as the generalization of standard statistical mechanics and Boltzmann-Gibbs entropy, we consider the non-linear Fokker-Planck equations as a simple generalization of linear Fokker-Planck equations.
Each system’s non-linear Fokker-Planck equations are written so that their stationary solution is that system’s entropy probability distribution [2-17]. The time-dependent solution of these non-linear equations is required to define the non-equilibrium systems.
Plastino in 1995 [18-22], and Daffertshofer and Frank in 1999 [13-17] achieved the time-dependent solution of a number of entropies’ non-linear Fokker-Planck equations in a canonical ensemble and in the presence of external potential. In this paper, we demonstrate that this method can be exploited for a relatively large category of entropies being function of Tsallis entropy. In the equilibrium and non-equilibrium states, the behavior of the systems defined by these entropies is revealed using the stationary and temporal solutions of non-linear FokkerPlanck equations.
In the second section, the general form of FokkerPlanck equation depending on entropy is achieved through the method mentioned in the references [23-31]. In the third section, the Frank method is generalized for the entropies being a function of Tsallis entropy. We obtain the Fokker-Planck equation of the entropy, and calculate its stationary solution. Then by adopting this approach, we investigate two special forms of entropy. In the fourth section, the Fokker–Planck equation in the time-dependent state is studied; finally in the last section, we conclude.
2. The Stationary Solution to the Non-Linear Fokker-Planck Equations
In this section, the non-linear Fokker-Planck equations depending on a category of generalized entropies are studied. The stationary solution of these equations for the Tsallis entropy, Rényi entropy, and Sharma-Mittal entropy has been previously investigated [13-17]. Herein, the Frank method is employed for a large category of entropies. We demonstrate that all the entropies being function of Tsallis entropy can be defined by such solutions. The stationary solution (a) exact solution, (b) approximate solution, and (c) absolute error for the non-linear Fokker-Planck equations are shown in Figure 1. For this purpose, first, it is essential to review the Frank method for the Tsallis entropy. The integral form of Tsallis entropy is as follows:
(3)
Regarding Equations (A.1) to (A.4), the equations below are obtained:
(4)
provided that.
(5)
provided that.
By considering the linear drift force and inputting the above values to Equation (A.7), we have:
(6)
The stationary solution of Equation (6) is the Tsallis distribution function. Instead of directly solving the equation, Equations (A.11) and (A.12) are utilized to obtain the distribution function. Herein, through Equation (A.10),
(a)(b)(c)
Figure 1. The surface shows the stationary solution (a) exact solution; (b) approximate solution, and (c) absolute error for the non-linear Fokker-Planck equations.
h is defined by the following equation:
(7)
(8)
Consequently, by using Equation (A.12), we obtain:
(9)
means that the solution will be valid only in cases which the quantity inside the curly brackets is positive. For the linear force, the potential Ф is as follows:
(10)
Equation (9) can be rewritten as the form below:
(11)
In the stationary state, b and the normalization coefficient are defined as follows:
(12)
Now, we study a more general state in which the entropy is as a function of Tsallis entropy.
The entropy is defined as where f is an arbitrary and differentiable function. The distribution function is achieved using Equation (A.8):
(13)
We define a function termed U as the following form:
(14)
Subsequently, Equation (13) can be rewritten as the from below:
(15)
Through using Equation (12) and comparing Equations (A.8) with (15), we realize that this general entropy’s distribution function can be achieved by substituting bU for b in Equation (11). Accordingly, this general entropy’s distribution function will be as follows:
(16)
The functions b, , and are positive. It should be noted that, for the values, the distribution function has a cut in. To complete the solution, it is sufficient to determine the function Ust and normalization coefficient Dst in relation to each special function of.
In Equation (6), if βU is substituted for β, its equilibrium-state solution is the above entropy distribution function. The corresponding Fokker-Planck equation will be as follows:
(17)
For instance, we investigate two entropies being functions of Tsallis entropy, and complete the stationary solution to their Fokker-Planck equation through computing and.
A. First, we consider the Rényi entropy that is defined by the equation below:
(18)
(19)
Therefore, according to the definition of U in Equation (14), we acquire:
(20)
The normalization condition for Equation (16) using the following equation is identical to the definition of function in Table 1:
(21)
Subsequently, we have:
(22)
provided that.
Roughly similar to the above procedure, the following equation is obtained but through utilizing the definition of function in Table 1:
(23)
provided that.
Through inserting Equation (22) into the both sides of Equation (23), we obtain:
(24)
Table 1. The definitions of zq and zqq.
Consequently, the quantities and are determined by Equations (22) and (24), and the distribution function of Equation (16) is completely defined.
The functions and can be expressed according to the functions of β and γ [23-31].
In this section, we investigate the entropy defined by the equation below [23-31]. It is a function of Tsallis entropy:
(25)
(26)
The relation between the Tsallis entropy and above entropy can be easily expressed by the function f:
(27)
Therefore, based on the definition of U in Equation (14), we have:
(28)
The same as the section A, we obtain the following results:
(29)
and.
(30.1)
(30.2)
Through inputting Equation (29) to Equation (30.1), we obtain:
(31)
By comparing Equations (20) and (28), the function is determined using Equation (22):
(32)
Subsequently, the values of the function U and the normalization coefficient are determined, and complete the description of stationary solutions to the nonlinear Fokker-Planck equations.
Now, we discuss the limit applied to q in the definition of. For the large x, the integrand in the definition ofis proportional to in; therefore, this integral is divergent. For the large x and, the integral is divergent as well, since the integrand is proportional to,. Consequently, the limit is applied to the entropies which is inserted into their distribution functions.
It is important to note that the method proposed in this paper can be solely utilized for the entropies being in the form of. However, this matter does not reduce this method’s significance, because a broad category of entropies can be expressed as a function of Tsallis entropy [23-38]. All the entropies being function of Tsallis entropy are shown in Figure 2.
3. The Time-Dependent Solutions to the Non-Linear Fokker-Planck Equations
The temporal solution to the entropy-dependent FokkerPlanck Equation (6) has been achieved by Plastino [18- 22]. The same as Plastino’s method, we also assume the temporal solution to the Fokker-Planck Equation (17) as the following form. Equation (17) is related to the entropy.The time-dependent solutions (a) exact solution, (b) approximate solution, and (c) absolute error for the non-linear Fokker-Planck equations are shown in Figure 3. The function U is different in Equations (6) and (17):
(33)
The functions and are positive.
With the aid of the normalization conditionand the definitions in Table 1, the relation between and is determined:
(34)
(a)(b)
Figure 2. The surface shows all the entropies being function of Tsallis entropy.
Through using Equations (33) and (34), inserting them into Equation (17), and then removing the different powers of, the first-order differential equations of and can be achieved:
(35)
(36)
The solution to Equation (35) can be easily obtained:
(37)
(38)
We input Equation (33) to the definitions of function U for different entropies, and obtain according to. Therefore, without the function, Equation (36) can be written and solved in detail for different entropies. For example, Table 2 provides the results of
(a)(b)(c)
Figure 3. The surface shows the time-dependent solutions (a) exact solution, (b) approximate solution, and (c) absolute error for the non-linear Fokker-Planck equations.
three entropies. The approximate solution for Equation (36) for different entropies are shown in Figure 4.
4. Conclusions
In many articles, the generalized Fokker-Planck equations
Table 2. Some examples of the differential equation D(t) and their solutions.
(a)(b)
Figure 4. The surface shows the approximate solution for Equation (36) for different entropies.
being dependent on Tsallis entropy and some other special entropies are written and solved. The Tsallis entropy and its functions have attracted a great deal of interest owing to the fact that they are desirably consistent with the physical systems with long-term memory, or long-range interactions. As regards the importance of achieving the solutions of Fokker-Planck equations, the following contents can be noted.
In general, the time-dependent distribution function is required to describe a non-equilibrium system. As an example, the diffusing effect is highly interesting in theoretical and empirical terms. The mean square of changes in state variable (variance) is a criterion for dividing the diffusion. If the variance is expressed as where represents the time, , , and are respectively called subdiffusion, normal diffusion, and hyperdiffusion. Many examples of each of these three categories are found in nature such as pseudo-polymeric systems, two-dimensional rotational flow, thermal conduction in plasma, population scattering in biological systems and so forth.
5. Acknowledgements
The work described in this paper was fully supported by grants from the Institute for Advanced Studies of Iran. The authors would like to express genuinely and sincerely thanks and appreciated and their gratitude to Institute for Advanced Studies of Iran.
REFERENCES
1. C. Tsallis, “Possible Generalization of Boltzmann-Gibbs Statistics,” Journal of Statistical Physics, Vol. 52, No. 1-2, 1988, pp. 479-487. doi:10.1007/BF01016429
2. P. H. Chavanis, “Generalized Thermodynamics and Fokker-Planck Equations: Applications to Stellar Dynamics and Two-Dimensional Turbulence,” Physical Review E, Vol. 68, No. 3, 2003, pp. 036108-036127. doi:10.1103/PhysRevE.68.036108
3. P. H. Chavanis, “Nonlinear Mean Field Fokker-Planck Equations. Application to the Chemotaxis of Biological Populations,” The European Physical Journal B—Condensed Matter and Complex Systems, Vol. 62, 2008, pp. 179-208.
4. P. H. Chavanis, “Generalized Fokker-Planck Equations and Effective Thermodynamics,” Physica A: Statistical Mechanics and Its Applications, Vol. 340, 2004, pp. 57- 65.
5. P. H. Chavanis and M. Lemou, “Relaxation of the Distribution Function Tails for Systems Described by Fokker-Planck Equations,” Physical Review E, Vol. 72, No. 6, 2005, pp. 061106-061121. doi:10.1103/PhysRevE.72.061106
6. P. H. Chavanis, “Nonlinear Mean-Field Fokker-Planck Equations and Their Applications in Physics, Astrophysics and Biology,” Comptes Rendus Physique, Vol. 7, No. 3, 2006, pp. 318-330. doi:10.1016/j.crhy.2006.01.004
7. V. Schwämmle, E. M. F. Curado and F. D. Nobre, “A General Nonlinear Fokker-Planck Equation and Its Associated Entropy,” The European Physical Journal B— Condensed Matter and Complex Systems, Vol. 58, 2007, pp. 159-165.
8. V. Schwämmle, F. D. Nobre and E. M. F. Curado, “Consequences of the H Theorem from Nonlinear FokkerPlanck Equations,” Physical Review E, Vol. 76, No. 4, 2007, pp. 041123-041130. doi:10.1103/PhysRevE.76.041123
9. V. Schwämmle, E. M. F. Curado and F. D. Nobre, “Dynamics of Normal and Anomalous Diffusion in Nonlinear Fokker-Planck Equations,” The European Physical Journal B—Condensed Matter and Complex Systems, Vol. 70, 2009, pp. 107-116.
10. V. Schwämmle, E. M. F. Curado and F. D. Nobre, “Nonlinear Fokker-Planck Equations Related to Standard Thermostatistics,” Complexity, Metastability and Nonextensivity, Vol. 84, 2007, pp. 152-156.
11. M. S. Ribeiro, F. D. Nobre and E. M. F. Curado, “Classes of N-Dimensional Nonlinear Fokker-Planck Equations Associated to Tsallis Entropy,” Entropy, Vol. 13, No. 11, 2011, pp. 1928-1944. doi:10.3390/e13111928
12. A. M. Scarfone and T. Wada, “Lie Symmetries and Related Group-Invariant Solutions of a Nonlinear FokkerPlanck Equation Based on the Sharma-Taneja-Mittal Entropy,” Brazilian Journal of Physics, Vol. 30, No. 2A, 2009. doi:10.1590/S0103-97332009000400024
13. T. D. Frank, A. Daffertshofer, C. E. Peper, P. J. Beek and H. Haken, “Towards a Comprehensive Theory of Brain Activity: Coupled Oscillator Systems under External Forces,” Physica D: Nonlinear Phenomena, Vol. 144, No. 1-2, 2000, pp. 62-86. doi:10.1016/S0167-2789(00)00071-3
14. T. D. Frank, A. Daffertshofer and P. J. Beek, “Multivariate Ornstein-Uhlenbeck Processes with Mean-Field Dependent Coefficients: Application to Postural Sway,” Physical Review E, Vol. 63, No. 1, 2000, pp. 011905- 011920. doi:10.1103/PhysRevE.63.011905
15. T. D. Frank and A. Daffertshofer, “Exact Time-Dependent Solutions of the Renyi Fokker-Planck Equation and the Fokker-Planck Equations Related to the Entropies Proposed by Sharma and Mittal,” Physica A: Statistical Mechanics and Its Applications, Vol. 285, 2000, pp. 351-366.
16. T. D. Frank, “On Nonlinear and Nonextensive Diffusion and the Second Law of Thermodynamics,” Physics Letters A, Vol. 267, No. 5-6, 2000, pp. 298-304. doi:10.1016/S0375-9601(00)00127-4
17. A. Daffertshofer, C. E. Peper, T. D. Frank and P. J. Beek, “Spatio-Temporal Patterns of Encephalographic Signals during Polyrhythmic Tapping,” Human Movement Science, Vol. 19, No. 4, 2000, pp. 475-498. doi:10.1016/S0167-9457(00)00032-4
18. A.R. Plastino and A. Plastino, “Non-Extensive Statistical Mechanics and Generalized Fokker-Planck Equation,” Physica A: Statistical Mechanics and Its Applications, Vol. 222, 1995, pp. 347-354.
19. A. R. Plastino, A. Plastino and H. Vucetich, “A Quantitative Test of Gibbs’ Statistical Mechanics,” Physics Letters A, Vol. 207, No. 1-2, 1995, pp. 42-46. doi:10.1016/0375-9601(95)00640-O
20. F. Pennini, A. Plastino and A. R. Plastino, “Tsallis Entropy and Quantal Distribution Functions,” Physics Letters A, Vol. 208, No. 4-6, 1995, pp. 309-314. doi:10.1016/0375-9601(95)00720-1
21. A. R. Plastino and A. Plastino, “Fisher Information and Bounds to the Entropy Increase,” Physical Review E, Vol. 52, No. 4, 1995, pp. 4580-4582. doi:10.1103/PhysRevE.52.4580
22. M. Portesi, A. Plastino and C. Tsallis, “Nonextensive Thermostatistics Can Yield Apparent Magnetism,” Physical Review E, Vol. 52, 1995, pp. R.3317-R.3320.
23. T. D. Frank, “Stochastic Feedback, Nonlinear Families of Markov Processes, and Nonlinear Fokker-Planck Equations,” Physica A: Statistical Mechanics and Its Applications, Vol. 331, 2004, pp. 391-408.
24. T. D. Frank, “Analytical Results for Fundamental TimeDelayed Feedback Systems Subjected to Multiplicative Noise,” Physical Review E, Vol. 69, No. 6, 2004, pp. 061104-061114. doi:10.1103/PhysRevE.69.061104
25. T. D. Frank, “Fluctuation-Dissipation Theorems for Nonlinear Fokker-Planck Equations of the Desai-Zwanzig Type and Vlasov-Fokker-Planck Equations,” Physics Letters A, Vol. 329, No. 6, 2004, pp. 475-485. doi:10.1016/j.physleta.2004.07.019
26. T. D. Frank, “Classical Langevin Equations for the Free Electron Gas and Blackbody Radiation,” Journal of Physics A: Mathematical and General, Vol. 37, No. 11, 2004, p. 3561. doi:10.1088/0305-4470/37/11/001
27. T. D. Frank, P. J. Beek and R. Friedrich, “Identifying Noise Sources of Time-Delayed Feedback Systems,” Physics Letters A, Vol. 328, No. 2-3, 2004, pp. 219-224. doi:10.1016/j.physleta.2004.06.012
28. T. D. Frank, “Stability Analysis of Stationary States of Mean Field Models Described by Fokker-Planck Equations,” Physica D: Nonlinear Phenomena, Vol. 189, No. 3-4, 2004, pp. 199-218. doi:10.1016/j.physd.2003.08.010
29. T. D. Frank, “Complete Description of a Generalized Ornstein-Uhlenbeck Process Related to the Nonextensive Gaussian Entropy,” Physica A: Statistical Mechanics and its Applications, Vol. 340, 2004, pp. 251-256.
30. T. D. Frank, “Dynamic Mean Field Models: H-Theorem for Stochastic Processes and Basins of Attraction of Stationary Processes,” Physica D: Nonlinear Phenomena, Vol. 195, No. 3-4, 2004, pp. 229-243. doi:10.1016/j.physd.2004.03.014
31. T. D. Frank, “Nonlinear Fokker-Plank Equations,” Springer, Amsterdam, 2004.
32. A. M. Mathai and H. J. Haubold, “On Generalized Entropy Measures and Pathways,” Physica A: Statistical Mechanics and Its Applications, Vol. 385, 2007, pp. 493- 500.
33. T. D. Frank and A. R. Plastino, “Generalized Thermostatistics Based on the Sharma-Mittal Entropy and Escort Mean Values,” The European Physical Journal B— Condensed Matter and Complex Systems, Vol. 30, 2002, pp. 543-549.
34. T. D. Frank, “Generalized Fokker-Planck Equations Derived from Generalized Linear Nonequilibrium Thermodynamics,” Physica A: Statistical Mechanics and Its Applications, Vol. 310, 2002, pp. 397-412.
35. T. D. Frank, “On a General Link between Anomalous Diffusion and Nonextensivity,” Journal of Mathematical Physics, Vol. 43, No. 1, 2002, pp. 344-350. doi:10.1063/1.1421062
36. T. D. Frank, “Generalized Multivariate Fokker-Planck Equations Derived from Kinetic Transport Theory and Linear Nonequilibrium Thermodynamics,” Physics Letters A, Vol. 305, No. 3-4, 2002, pp. 150-159. doi:10.1016/S0375-9601(02)01446-9
37. T. D. Frank, “Interpretation of Lagrange Multipliers of Generalized Maximum-Entropy Distributions,” Physics Letters A, Vol. 299, 2002, pp. 153-158. doi:10.1016/S0375-9601(02)00631-X
38. T. D. Frank, A. Daffertshofer and P. J. Beek, “Impacts of Statistical Feedback on the Flexibility-Accuracy TradeOffin Biological Systems,” Journal of Biological Physics, Vol. 28, No. 2-3, 2002, pp. 39-54. doi:10.1023/A:1016256613673
Appendix A: The Generalized Entropies and Dependent Fokker-Planck Equations
The calculations are limited to one dimension in order to simplify. Also, they can be generalized to N dimensions.
The entropies are considered as follows:
(A.1)
so that is a well-defined and arbitrary function, and is also determined by:
(A.2)
where is a well-defined function of p which is twice differentiable at least.
The states with zero probability must not affect the entropy; furthermore, the natural systems contain a considerable number of microscopic states; therefore, a state with one probability cannot also affect the entropy; accordingly:
(A.3)
The concavity is one of the entropies’ general conditions. Hence, the entropy should have the following condition [7-12]:
(A.4)
The non-linear Fokker-Planck equation is defined as follows:
(A.5)
where, and it is the external force resulting from the potential Ф. The parameters Ω and Ψ are the functions of p. Equation (A.7) relates to the entropy through the equation below. Actually, the ratio is selected in order that the stationary solution to Equation (A.5) is the entropy distribution function:
(A.6)
β denotes the inverse temperature. For the linear force and by selecting, Equation (A. 5) is written as the form below:
(A.7)
In the different references, the Fokker-Planck equation dependent on entropies is investigated in various ways. The above-mentioned issues are comprehensively described in the reference [7-12].
In this section, how to achieve the entropy distribution function of Equation (A.1) is mentioned [32]. The probability distribution function of p is obtained by maximizing the entropy in a canonical ensemble. Therefore in a canonical ensemble, provided the number of particles and the total energy are constant, we obtain:
(A.8)
where Ω is the external potential, µ is the system’s chemical potential, and β is the dependent Lagrange multiplier.
Considering the form of entropy in Equation (A.1), we obtain:
(A.9)
and indicate the values of these quantities in the equilibrium state and maximum entropy, respectively. The function is defined as the form below:
(A.10)
Therefore:
(A.11)
With the aid of the inverse function of h (it means), we can achieve as the following form:
(A.12)
is the generalized entropy distribution function. The generalized entropies and dependent Fokker-Planck equations are shown in Figure A1.
(a) (b)(c) (d)
Figure A1. The surface shows the generalized entropies and dependent fokker-planck equations.
NOTES
*Corresponding author.
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math review question
The interest rate on an equity line of credit was 6 3/4 % last year. This year the interest rate is 9 1/2 %. How many percentage points has the interest rate increased?
1. 👍 0
2. 👎 0
3. 👁 242
1. 9 1/2 - 6 3/4 =
8 6/4 - 6 3/4 = ??
1. 👍 0
2. 👎 0
👩🏫
Ms. Sue
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# Re: [R] elegant matrix creation
From: Robin Hankin <r.hankin_at_noc.soton.ac.uk>
Date: Tue 12 Jul 2005 - 23:39:42 EST
Gabor
I cannot begin to tell you how much value you have added to my research with your observation.
A real "eureka" moment for me.
[oh, and it answered my question as well]
kia ora
Robin
On 12 Jul 2005, at 13:35, Gabor Grothendieck wrote:
> On 7/12/05, Robin Hankin <r.hankin@noc.soton.ac.uk> wrote:
>
>> Hi
>>
>> I want to write a little function that takes a vector of arbitrary
>> length "n" and returns a matrix of size n+1 by n+1.
>>
>> I can't easily describe it, but the following function that works for
>> n=3 should convey what I'm trying to do:
>>
>>
>> f <- function(x){
>> matrix(c(
>> 1 , 0 , 0 , 0,
>> x[1] , 1 , 0 , 0,
>> x[1]*x[2] , x[2] , 1 , 0,
>> x[1]*x[2]*x[3], x[2]*x[3], x[3], 1
>> ),
>> 4,4, byrow=T)
>> }
>>
>> f(c(10,7,2))
>> [,1] [,2] [,3] [,4]
>> [1,] 1 0 0 0
>> [2,] 10 1 0 0
>> [3,] 70 7 1 0
>> [4,] 140 14 2 1
>>
>>>
>>>
>>
>>
>> As one goes down column "i", the entries get multiplied by successive
>> elements of x, starting with x[i], after the first "1"
>>
>> As one goes along a row, one takes a product of the tail end of x,
>> until the zeroes kick in.
>>
>
> I have not checked this generally but at least for
> the 4x4 case its inverse is 0 except for 1s on the
> diagonal and -x on the subdiagonal. We can use
> diff on a diagonal matrix to give a matrix with
> a diagonal and superdiagonal and then massage that
> into the required form, invert and round --
> leave off the rounding if the components of x
> are not known to be integer.
>
> round(solve(diag(4) - t(diff(diag(5))[,1:4])+diag(4) * c(0,x)))
>
```--
Robin Hankin
Uncertainty Analyst
National Oceanography Centre, Southampton
European Way, Southampton SO14 3ZH, UK
tel 023-8059-7743
______________________________________________
R-help@stat.math.ethz.ch mailing list
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Plane Curve invariants via Contour Integrals - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T00:23:12Z http://mathoverflow.net/feeds/question/98575 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98575/plane-curve-invariants-via-contour-integrals Plane Curve invariants via Contour Integrals John Mangual 2012-06-01T14:56:48Z 2012-06-01T15:11:13Z <p>We learn in complex analysis class how to find the winding number of the contour $\Gamma$ around the origin. $n = \frac{1}{2\pi i} \oint \frac{dz}{z} = \frac{1}{2\pi i} \oint d(\log z) = \frac{1}{2\pi } \oint d\theta$ My goal had been to count the number of <a href="http://mathoverflow.net/questions/90856/computing-self-intersections-with-complex-analysis" rel="nofollow">self-intersections of curves</a>. I guessed some integral over a torus $\Gamma \times \Gamma$ which would have a pole-like object whenever $z_1 = z_2$. In the back of my mind, I worried maybe integrating along $t_1 = t_2$ would have zero contribution. $\frac{1}{2\pi i} \oint \oint \frac{dz_1 dz_2}{z_1 - z_2}$</p> <p>Inspired by <a href="http://www.ihes.fr/~maxim/TEXTS/VassilievKnot.pdf" rel="nofollow">Kontsevich's integral for knots</a> (and some more recent papers), I learned of something that comes close $\frac{1}{2\pi i} \oint \oint \frac{dz_1 - dz_2}{z_1 - z_2} = \frac{1}{2\pi i} \oint \oint d \log(z_1 - z_2) = \frac{1}{2\pi i} \oint \oint \arg (z_1 - z_2)$ So formulas like these are measuring how the chords of curves wind around each other. This seems to be known as the Whitney invariant for plane curves, counting <em>signed</em> self-intersections.</p> <p><img src="http://oi47.tinypic.com/15i43mo.jpg" alt="alt text"></p> <p>Is there a way to get self-intersections all of the same sign? This must be related the Vassiliev invariants as well, but I'd like to focus on plane curves.</p>
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# NASA Wavelength Resources Collection
NASA Wavelength is a collection of resources that incorporate NASA content and have been subject to peer review. You can search this collection using key words and/or the drop down menus to pinpoint resources to use with your audience of learners.
1902 result(s)
## What Makes Day and Night? The Earth's Rotation
This is an activity about day and night as a result of the Earth's rotation. Learners will first identify what they already know about day, night, and rotation and will be asked to share any questions they may have.
## MY NASA DATA: March of the Polar Bears: Global Change, Sea Ice, and Wildfire Migration
In this data analysis activity, Students will use NASA satellite data to study temperature and snow-ice coverage in the South Beaufort Sea, Alaska.
## Gamma Ray Burst Distribution on the Sky: The Plots Thicken
In this activity, students look at the distribution of aluminum foil balls arranged in a circle on the floor, and compare them to the distribution of gamma-ray bursts on the sky.
## Spheres of Earth
The activity introduces students to aspects of the atmosphere, biosphere, hydrosphere, and litho/geosphere and how they are interrelated. It is designed to promote an interest in authentic investigations of Earth using images acquired by astronauts as the hook.
## How Do Paths Look From Different Perspectives?
Using both literature (a book featuring a path, such as Little Red Riding Hood) and satellite images, students will identify paths, observe and analyze them from different altitudes, and distinguish natural paths from those made by humans.
## Einstein and His Times
In this lesson, students will read about and research the major historical events that occurred throughout the year 1919. They will use different readings and articles to understand and describe what life was like during this time.
## Space Math - I
This is a booklet containing 20 problem sets that involve a variety of math skills, including equations and substitution, time calculations, reading, algebra, and more. Each set of problems is contained on one page.
## The Scoop on Moon Dirt
In this two-part activity, learners compare how soil forms on Earth and the Moon. They examine different soil samples and compare them to lunar "soil" simulant. They explore how water, wind, and impactors help to make soil. This activity is part of Explore! To the Moon and Beyond!
## NASA eClips™ 4D Guide Lites: Solar Images
Learners begin this 5E activity by creating their own picture of the sun out of paint and detergent. They then make colored filters out of cellophane and paper towel tubes to simulate how specialized instruments capture and view solar images.
## The Universe as Scientists Know It
In this activity, the student will be able to assess their understanding of what makes up the universe, by filling in a concept map with the following terms: Planetary Systems, Galaxies, Planets, Sun, Venus, Moon, Stars, Sirius, Solar System, Comet, Meteor, Open Clusters, Stellar Regions, Jupiter
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# Solution - In a Biprism Experiment, When a Convex Lens Was Placed Between the Biprism and Eyepiece at a Distance of 30 Cm from the Slit, the Virtual Images of the Slits Are Found to Be Separated by 7 Mm. - Measurement of Wavelength by Biprism Experiment
ConceptMeasurement of Wavelength by Biprism Experiment
#### Question
In a biprism experiment, when a convex lens was placed between the biprism and eyepiece at a distance of 30 cm from the slit, the virtual images of the slits are found to be separated by 7 mm. If the distance between the slit and biprism is 10 cm and between the biprism and eyepiece is 80cm, find the linear magnification of the image.
#### Solution
You need to to view the solution
Is there an error in this question or solution?
#### APPEARS IN
2014-2015 (October)
Question 7.8 | 2 marks
Solution for question: In a Biprism Experiment, When a Convex Lens Was Placed Between the Biprism and Eyepiece at a Distance of 30 Cm from the Slit, the Virtual Images of the Slits Are Found to Be Separated by 7 Mm. concept: Measurement of Wavelength by Biprism Experiment. For the courses HSC Science (Computer Science), HSC Science (Electronics), HSC Science (General)
S
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# What Are Complex Data Type in Fortran?
//
Larry Thompson
What Are Complex Data Types in Fortran?
In Fortran, complex data types are used to represent numbers with both real and imaginary parts. These data types are particularly useful in scientific and engineering applications where complex numbers are commonly used.
## Declaring Complex Variables
To declare a complex variable in Fortran, you use the COMPLEX keyword followed by the desired variable name. For example:
```COMPLEX :: z
COMPLEX :: x, y
```
## Initializing Complex Variables
You can initialize a complex variable at the time of declaration by providing the initial values for its real and imaginary parts. The syntax for initialization is as follows:
```COMPLEX :: z = (real_part, imaginary_part)
```
For example:
```COMPLEX :: z = (1.0, 2.5)
```
## Operations on Complex Variables
In Fortran, you can perform various operations on complex variables such as addition, subtraction, multiplication, division, and conjugation.
To add or subtract two complex variables, you simply use the standard addition or subtraction operators. For example:
```z = x + y
z = x - y
```
### Multiplication and Division
The multiplication of two complex numbers is performed using the asterisk (*) operator. Division is done using the forward slash (/) operator.
```z = x * y
z = x / y
```
### Conjugation
The conjugate of a complex number can be obtained using the CONJG function. It returns a complex number with the same real part but the negation of the imaginary part.
```z = CONJG(x)
```
## Accessing Real and Imaginary Parts
In Fortran, you can access the real and imaginary parts of a complex variable using the REAL and AIMAG functions, respectively.
```real_part = REAL(z)
imaginary_part = AIMAG(z)
```
## Conclusion
In conclusion, complex data types in Fortran allow you to work with numbers that have both real and imaginary parts. They are essential for performing calculations in scientific and engineering applications. By using the provided keywords, operators, and functions, you can easily manipulate complex variables in your Fortran programs.
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This presentation is the property of its rightful owner.
1 / 38
Multiple View Geometry PowerPoint PPT Presentation
Multiple View Geometry. Marc Pollefeys University of North Carolina at Chapel Hill. Modified by Philippos Mordohai. Outline. 3-D Reconstruction Fundamental matrix estimation Chapters 9 and 10 of “Multiple View Geometry in Computer Vision” by Hartley and Zisserman. Three questions:.
Multiple View Geometry
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Multiple View Geometry
Marc Pollefeys
University of North Carolina at Chapel Hill
Modified by Philippos Mordohai
Outline
• 3-D Reconstruction
• Fundamental matrix estimation
• Chapters 9 and 10 of “Multiple View Geometry in Computer Vision” by Hartley and Zisserman
Three questions:
• Correspondence geometry: Given an image point x in the first image, how does this constrain the position of the corresponding point x’ in the second image?
• (ii)Camera geometry (motion): Given a set of corresponding image points {xi ↔x’i}, i=1,…,n, what are the cameras P and P’ for the two views?
• (iii)Scene geometry (structure): Given corresponding image points xi ↔x’i and cameras P, P’, what is the position of (their pre-image) X in space?
p
p
L2
L2
m1
m1
m1
C1
C1
C1
M
M
L1
L1
l1
l1
e1
e1
lT1
l2
e2
e2
Canonical representation:
l2
m2
m2
m2
l2
l2
Fundamental matrix (3x3 rank 2 matrix)
C2
C2
C2
Epipolar Geometry
Underlying structure in set of matches for rigid scenes
• Computable from corresponding points
• Simplifies matching
• Allows to detect wrong matches
• Related to calibration
3D reconstruction of cameras and structure
reconstruction problem:
given xi↔x‘i , compute P,P‘ and Xi
for all i
without additional information possible up to projective ambiguity
Outline of reconstruction
• Compute F from correspondences
• Compute camera matrices from F
• Compute 3D point for each pair of corresponding points
computation of F
use x‘iFxi=0 equations, linear in coeff. F
8 points (linear), 7 points (non-linear), 8+ (least-squares)
(more on this next class)
computation of camera matrices
use
triangulation
compute intersection of two backprojected rays
Reconstruction ambiguity: similarity
Reconstruction ambiguity: projective
Terminology
xi↔x‘i
Original scene Xi
Projective, affine, similarity reconstruction
= reconstruction that is identical to original up to
projective, affine, similarity transformation
Literature: Metric and Euclidean reconstruction
= similarity reconstruction
The projective reconstruction theorem
If a set of point correspondences in two views determine thefundamental matrix uniquely, then the scene and cameras may be reconstructed from these correspondences alone, and any two such reconstructions from these correspondences are projectively equivalent
• along same ray ofP2, idem for P‘2
two possibilities: X2i=HX1i, or points along baseline
key result:
allows reconstruction from pair of uncalibrated images
Stratified reconstruction
• Projective reconstruction
• Affine reconstruction
• Metric reconstruction
Projective to affine
(if D≠0)
theorem says up to projective transformation,
but projective with fixed p∞ is affine transformation
can be sufficient depending on application,
e.g. mid-point, centroid, parallellism
Translational motion
points at infinity (not necessarily visible) are fixed for a pure translation
reconstruction of xi↔xi is on p∞
Scene constraints
Parallel lines
parallel lines intersect at infinity
reconstruction of corresponding vanishing point yields
point on plane at infinity
3 sets of parallel lines allow to uniquely determine p∞
remark: in presence of noise determining the intersection of parallel lines is a delicate problem
remark: obtaining vanishing point in one image can be sufficient
Scene constraints
Scene constraints
*
*
projection
constraints
Affine to metric
identify absolute conic
transform so that
then projective transformation relating original and reconstruction is a similarity transformation
in practice, find image of W∞
image w∞back-projects to cone that intersects p∞ in W∞
note that image is independent of particular reconstruction
Affine to metric
given
possible transformation from affine to metric is
proof:
(Cholesky factorization to obtain A)
Orthogonality
vanishing points corresponding to orthogonal directions
vanishing line and vanishing point corresponding
to plane and normal direction
rectangular pixels
square pixels
Same camera for all images
same intrinsics same image of the absolute conic
e.g. moving camera
given sufficient images there is in general only one
conic that projects to the same image in all images,
i.e. the absolute conic
This approach is called self-calibration
transfer of IAC:
provides 4 constraints, one more needed
(in general two solutions)
Direct metric reconstruction using ω
approach 1
calibrated reconstruction
approach 2
compute projective reconstruction
back-project w from both images
intersection defines W∞ and its support plane p∞
(2 lin. eq. in H-1per view,
3 for two views)
Direct reconstruction using ground truth
use control points XEi with know coordinates
to go from projective to metric
(3 lin. eq. in H per point,
H has 15 d.o.f.)
Reconstruction summary
• Given two uncalibrated images compute (PM,P‘M,{XMi})
• (i.e. within similarity of original scene and cameras)
• Algorithm
• Compute projective reconstruction (P,P‘,{Xi})
• Compute F from xi↔x‘i
• Compute P,P‘ from F
• Triangulate Xi from xi↔x‘i
• Rectify reconstruction from projective to metric
• Direct method: compute H from control points
• Stratified method:
• Affine reconstruction: compute p∞
• Metric reconstruction: compute IAC w
Outline
• 3-D Reconstruction
• Fundamental matrix estimation
• Chapters 9 and 10 of “Multiple View Geometry in Computer Vision” by Hartley and Zisserman
Epipolar geometry: basic equation
separate known from unknown
(data)
(unknowns)
(linear)
The singularity constraint
SVD from linearly computed F matrix (rank 3)
Compute closest rank-2 approximation
The minimum case – 7 point correspondences
one parameter family of solutions
but F1+lF2 not automatically rank 2
3
F7pts
F
F2
F1
The minimum case – impose rank 2
(obtain 1 or 3 solutions)
(cubic equation)
Compute possible l as eigenvalues of
(only real solutions are potential solutions)
~10000
~100
~10000
~100
~10000
~10000
~100
~100
1
Orders of magnitude difference
between column of data matrix
least-squares yields poor results
!
The NOT normalized 8-point algorithm
Transform image to ~[-1,1]x[-1,1]
(-1,1)
(1,1)
(0,0)
(-1,-1)
(1,-1)
normalized least squares yields good results(Hartley, PAMI´97)
(0,500)
(700,500)
(0,0)
(700,0)
Geometric distance
Gold standard
Sampson error
Symmetric epipolar distance
Gold standard
Maximum Likelihood Estimation
(= least-squares for Gaussian noise)
Initialize: normalized 8-point, (P,P‘) from F, reconstruct Xi
Parameterize:
(overparametrized F=[t]xM)
Minimize cost using Levenberg-Marquardt
(preferably sparse LM, see book)
Gold standard
Alternative, minimal parametrization (with a=1)
(note (x,y,1) and (x‘,y‘,1) are epipoles)
• problems:
• a=0
pick largest of a,b,c,d to fix to 1
• epipole at infinity
pick largest of x,y,w and of x’,y’,w’
4x3x3=36 parametrizations!
reparametrize at every iteration, to be sure
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# What is Moore-Hodgson's Algorithm?
Jul 07, 2022
Moore-Hodgson’s algorithm is an approach to scheduling work that aims to minimise the number of tasks that are late, rather than the lateness of any particular task. The algorithm was presented in 1968 by J. M. Moore, with an alternative algorithim attributed to T. E. Hodgson.
In this article we’ll look at the algorithm, its benefits, drawbacks, and when you should use it to organise the things you need to do.
## How does Moore-Hodgson’s Algorithm method work?
Moore-Hodgson’s algorithm is an approach to scheduling work that aims to minimise the number of tasks that are late. You start by sorting your tasks by the earliest due date and working on the task that is due soonest first.
Whilst working on the earliest due date task, if it looks like you won’t get to the next before it’s due, you look at the remaining tasks and throw out the biggest. Of course, you still have to complete that task, but you have to sacrifice it becoming overdue in order to prevent other tasks from also going overdue.
You continue working through the remaining tasks in order of due date, but keep getting rid of any big tasks if you fall behind schedule.
Once you’re finished working through the tasks, you work through the tasks that were thrown out. These tasks can be done in any order because they’re all late anyway.
This process reduces the number of tasks potentially that are late, rather than the severity of any particular task.
## What are the benefits of the Moore-Hodgson’s Algorithm method?
The most obvious benefit of Moore-Hodson’s algorithm is that the number of tasks that are potentially overdue is minimised.
For example, say you had 10 tasks where your objective was to have the majority of them delivered on-time, and sacrificing one or two to being late was an acceptable compromise. If this is a characteristic that you want to optimise for, using the Moore-Hodgson’s algorithm could be a good choice.
Moore-Hodgson’s algorithm is the optimum approach when you’re working through a set of tasks on your own and this is the objective you care the most about.
## What are the drawbacks of the Moore-Hodgson’s Algorithm method?
The drawback to Moore-Hodgson’s is that you must sacrifice tasks and allow them to become overdue in order to minimise the lateness of other tasks. This is potentially made worse if the tasks that are most likely to be thrown out also happen to be the most important.
Whilst the Moore-Hodgson’s algorithm is fairly straightforward to implement, the simplicity ends up being another drawback of the approach. For example, the algorithm assumes that all of your tasks have a similar priority, duration, and importance.
## When should you use the Moore-Hodgson’s Algorithm method?
As I mentioned above, if you are aiming to minimise the number of tasks that are overdue, and are willing to sacrifice tasks in order to achieve this goal, Moore-Hodgson’s algorithm could be a good choice.
The algorithm was also specifically conceived for “single machine processing”, or in other words if you are working on the list of tasks on your own. Once you introduce the complexity of multiple people working on a set of tasks, a relatively simple approach like Moore-Hodson’s algorithm goes out of the window.
## When should you not use the Moore-Hodgson’s Algorithm method?
Moore-Hodgson’s algorithm isn’t a good choice if your tasks are very different in terms of complexity, duration, or importance. If you have a wide variety of tasks on your to-do list, using a simple algorithm for scheduling your work is probably not going to work.
As Moore-Hodgson’s algorithm was intended for optimising the problem of “single machine processing”, if you regularly work synchronously or in collaboration with others it’s probably not going to be a good choice either.
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# Help turn a famous quote into an equation!!!December 17, 2012 1:59 PM Subscribe
Looking to turn the Winston Churchill quote, "Success is the ability to go from failure to failure without losing enthusiasm," into an equation.
posted by fubunker to Science & Nature (16 answers total) 2 users marked this as a favorite
Success = Failure plus Enthusiasm multiplied by infinity?
S = F + E*i
posted by royalsong at 2:07 PM on December 17, 2012
That is not a Winston Churchill quote, it is one of the innumerable self-help-ish pseudo-quotes randomly attributed to famous people. If you search Winston Churchill Quotes, which seems pretty well done (proper attributions when available), you will not find it there, and it doesn't even sound like something he could have said.
posted by languagehat at 2:08 PM on December 17, 2012 [8 favorites]
I think it just means that the integral of the success metric S over time t is postive given that dE/dt > 0 when S(t) is negative (E being the enthusiasm metric). Maybe someone can Latex that.
posted by unSane at 2:09 PM on December 17, 2012 [1 favorite]
Also, what Languagehat said. Quotes attributed to WC are often fake.
posted by unSane at 2:09 PM on December 17, 2012
Thanks I'm not all that interested in who said it, I like the quote and am interested in the equations that could be made from it that are like the one marked best. Anyone have other variations.
I'm creating an image that has a chalk board an equation on it, I'd just like to sneak this in there somewhere.
posted by fubunker at 2:17 PM on December 17, 2012
The highlighted equation simplifies to "Success equals infinity"
posted by thelonius at 2:18 PM on December 17, 2012 [3 favorites]
Yeah, "infinity" would mean never ever stopping; you need to convey the idea of "keep doing it until it works," not "keep doing it forever."
I am not enough of a mathematician to know how to write that, though in code you would do a loop with a "break' when a condition was reached.
posted by drjimmy11 at 2:21 PM on December 17, 2012 [1 favorite]
S(f, E) = (k/f) * E
or just:
S = (k/f) * E
where k >= f
So success is a function of number of failures (f) and enthusiasm (E),
where the final enthusiasm ((k/f) * E) is always greater than or equal to the amount of enthusiasm that you started with.
Or something.
posted by Mister_Sleight_of_Hand at 2:33 PM on December 17, 2012 [2 favorites]
A programming translation might be something like:
while success == False: success = try_again( with_enthusiasm = True )
posted by jsturgill at 2:33 PM on December 17, 2012
Thanks drjimmy11 and theonius I highlighted that one to point out the kind of structure of the equation in response to the people who had no idea what I was talking about. I've never used this site before but it's pretty awesome how quickly it took to get on the right track. I used to be good at this stuff.
posted by fubunker at 2:36 PM on December 17, 2012
I'm afraid I don't have time to write this out symbolically, but here's a mathematical formalization:
Let F be the set of all failures, and suppose that enthusiasm can be denoted by a single real-valued variable. Let G = F cross R, the set of all failure-enthusiasm pairs. Let H be the set of all functions which map G onto itself. We may then define the set of all successes, S, as a subset of G. A member s of S must satisfy the following property: there exists at least one f in F and r in R such that, if s(f, r) = (f', r'), r' >= r.
Note that we only require the existence of one f, since the invented Churchill quotation only wants the ability to go from failure to failure without diminution of enthusiasm -- for a stricter criterion, just replace the existential quantifiers with universal ones in the final condition.
(Mathematicians: please correct me if required. It's a long time since I studied mathematics.)
Hopefully someone else can come along and symbolify this... if not, I will see if I have time later tonight.
posted by pont at 2:36 PM on December 17, 2012 [1 favorite]
Oops sorry, I meant that S is a subset of H.
posted by pont at 3:32 PM on December 17, 2012
I think an important question is: do you want this to be an equation for people who understand equations, or something that looks like an equation that will be understood easily by everyone else?
posted by windykites at 5:26 PM on December 17, 2012
I managed to knock something up in LaTeX (a typesetting tool that's really good for math). Here's the image, with very small margins that I can't figure out how to increase.
(If you follow this link it should take you to a web site which will show both the LaTeX code and the equation, though I'm not 100% sure it will work in your browser. People can also modify the equation there.)
Explanation (I don't know how familiar you are with calculus. Forgive me if I explain too much.): The vertical bar means "restricted to", so we are only looking at areas where there is Failure. 'E' is Enthusiasm, and the fact that d/dt is >= 0 means that Enthusiasm never decreases as time moves forward. So it's basically putting your words into symbols. It doesn't immediately or obviously make 100% mathematical sense, but I could throw some more details at it and describe a situation where it does. But it does look nicely concise and math-y.
posted by benito.strauss at 12:45 AM on December 18, 2012
I re-read what you want this for. If you're trying to sneak this onto a chalkboard you can play with the variables representing Enthusiasm and Failure. If you need them to totally not stand out, just use 'E' and 'F'. If you want to make them obvious you can spell them all the way out.
Me, I'd go with this.( If caught you can claim that 'Enth' stands for enthalpy). BTW, if you want more math-exclusive symbols in it, you could use this instead. Those "backward 6's" are probably less appropriate, but I think they look mathy-er, and no one other than the pedants on MetaFilter (like me) is really going to care.
(And you probably want to put "Succ =", or just "S = " in front of it, for the "Success is ..." part of the saying.)
posted by benito.strauss at 1:02 AM on December 18, 2012
windykites I'd have to say more authentic gear towards people who understand equations.
benito.strauss the first image that you posted looks great.
Thanks!
posted by fubunker at 12:04 PM on December 19, 2012
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# Discrete Math Forum
Discrete Math Help Forum: Discrete mathematics, logic, set theory
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# The cowboy and the weather
Question: You are a cowboy. The cowboy is riding bareback on a paint horse. The horse falls down breaking its leg. The sun is setting quickly and the nearest house is 3 miles away. The chance of rain that night is at 75% The chance of snow is 33% What color is the Cowboys hair?
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6 Solving the Linear Least-squares Problem Via QR Factorization - Duration: 3:33. SciPy is an open-source scientific computing library for the Python programming language. As a PhD student in economics, and a Python enthusiast myself (see: econpy. The following matrix factorization techniques are available: LU Decomposition is for square matrices and decomposes a matrix into L and U components. Refer to the following papers for a description. Since 2013 I think there has been a lot of discussion on the github pages regarding putting it in scipy and some pages have code that I can use such as here which is super complicated for a novice like me(to get it running). ContentsAlston HouseholderPete StewartQR DecompositionHouseholder reflectionshouse_genHouseholder matrixhouse_qrMagic square examplehouse_applyQ at. Hamming Code Table. daal4py operates in SPMD style (Single Program Multiple Data), which means your program is executed on several processes (e. Being able to go from idea to result with the least possible delay is key to doing good research. Ying Wei (Daniel) Fan - Kindly permitted reuse of CUBLAS wrapper code in his PARRET Python package. svd and La. [Q,R] = qr(A); The following plot will display the runtime of qr dependent of the square root of elements of the matrix. The G-S algorithm itself is unstable, so various other methods have been developed to compute the QR decomposition. I tried almost two days but can't figure out the problem. 4) xGEQRT3: Recursive QR factorization. similar to MPI). Benson CS 267 Spring 2011 [email protected] Worldwide Center of Mathematics 68,871 views. • Explore other recommendation algorithms, e. CG can be viewed as a the. 4) xGEQRT3: Recursive QR factorization. is orthogonal and symmetric. 7321,1, l)T. A k initiated with A0 =A and given by A k =R kQ k, where Q k and R k represents a QR-factorization of A k−1, A k−1 =Q kR k. It is used by the pure mathematician and by the mathematically trained scien-tists of all disciplines. If A is of full rank n and we require that the diagonal elements of R1 are positive then R1 and Q1 are unique, but in general Q2 is not. We have collection of more than 1 Million open source products ranging from Enterprise product to small libraries in all platforms. Conclusion. I will indirectly answer that by explaining the process instead. Trouble may also arise when M = N but the matrix is singular. Linear Systems: Iterative. That is, the QR-method generates a sequence of matrices Idea of basic QR-method: compute a QR-factorization and reverse the order of multiplcation of Q and R. The G-S algorithm itself is unstable, so various other methods have been developed to compute the QR decomposition. FP-growth, association rules, and PrefixSpan –feature extraction and transformations –Optimization. Computer science quiz. Syllabus for Numerical Analysis, Fall 2018 MATH-UA 0252-001 Prerequisites Knowledge of undergraduate linear algebra and calculus. It is available free of charge and free of restriction. Where is an orthogonal matrix, and is The post QR Decomposition with the Gram-Schmidt Algorithm appeared first on Aaron Schlegel. Nimfa is an open-source Python library that provides a unified interface to nonnegative matrix factorization algorithms. That is, [A] = [L][U] Doolittle's method provides an alternative way to factor A into an LU decomposition without going through the hassle of Gaussian Elimination. Beyond linear algebra: root finding, interpolation, numerical integration and differentiation, unconstrained and constrained optimization. The QR decomposition of a matrix A is the representation of A as a product A = QR; where Q is an orthogonal matrix and R is an upper triangular matrix with positive diagonal entries. edu 1Course G63. The Dynamic Mode Decomposition (DMD) is a relatively recent mathematical innovation that can convert a dynamical system into a superposition of modes whose dynamics are governed by eigenvalues. [___] = qr(A,0) produces an economy-size decomposition using any of the previous output argument combinations. # QR decomposition from numpy import array from numpy. To get the singular value decomposition, we can take advantage of the fact that for any matrix $$A$$, $$A^TA$$ is symmetric (since $$(A^TA)^T = A^T(A^T)^T = A^TA$$). QR decomposition is a matrix factorization technique that decomposes a matrix into a product of an orthogonal matrix Q and an upper triangular matrix R. I have 46 rasters each for an 8 day period for Β(σ) , and σ, where I need to take input values from per time step. qr_decomposition. In this second article on methods for solving systems of linear equations using Python, we will see the QR Decomposition method. QR Factorization Calculator Linear Algebra Calculators QR Factorization. This method is very similar to the LU decomposition. Below are some of the related papers. Recently I have been taking advantage of the notebook environments knitr and Jupyter to. • Explore other recommendation algorithms, e. Overview In 1948, Alan Turing came up with LU decomposition, a way to factor a matrix and solve $$Ax=b$$ with numerical stability. Now, for forming system of equations solvable for unique. qr scan and reader for android. This routine has many features. My SVD code in python has been uploaded. Compute the QR factorization. The idea of the QR decomposition as a procedure to get OLS estimates is already explained in the post linked by @MatthewDrury. They are from open source Python projects. matlab NGPM -- A NSGA-II Program in matlabThis document gives a brief description about NGPM. The transformation matrix can be also computed by the Cholesky decomposition with Z=L−1 (X− ¯X) where L is the Cholesky factor of C=LLT. Chapter 7 5. Non-negative matrix factorization (NMF) has previously been shown to be a useful decomposition for multivariate data. Use the scipy QR factorization routine linalg. Either will handle over- and under-determined systems, providing a minimal-length solution or a least-squares fit if appropriate. The comments in this question indicate that the user wants to replicate MATLAB's backslash operator in C++. Q&A for Work. We rewrite Mx = b as LL T x = b and let L T x = y. Pivot wherever appropriate. For a general n×n matrix A, we assume that an LU decomposition exists, and write the form of L. When you take a digital photo with your phone or transform the image in Photoshop, when you play a video game or watch a movie with digital effects, when you do a web search or make a phone call, you are using technologies that build upon linear algebra. I understand additionally that I'll need to use the transformation matrices from my QR algorithm (balancing, the hessenberg reduction and the QR decomposition). A Comparative study of SVD,QR decomposition and PCA model and IGSC in feature extraction Jun 2019 – Jan 2020 Comparing the information Gain subspace clustering algorithm with QR decomposition to find the effectiveness of feature selection for cluster formation in subspace by using various datasets. † If A = QR is nonsingular Hessenberg, so is RQ. troduction to abstract linear algebra for undergraduates, possibly even first year students, specializing in mathematics. T, mode='full', pivoting=True) tol = np. However, it is applied to dense (or: full) matrices only. file disclaimer for MINPACK copyright notice lib ex for test programs file readme for overview of minpack file chkder. LU factorization for general matrices, as well as functions for solving linear systems, computing determinants, inverses, and condition numbers. LU decomposition. The Gram-Schmidt algorithms can inherently fall victim to. The first $$k$$ columns of $$Q$$ are an orthonormal basis for the column space of the first $$k$$ columns of $$A$$. QR 2: Least Squares and Computing Eigenvalues Lab Objective: Because of its numerical stability and convenient structure, the QR decomposition is the basis of many important and practical algorithms. Harp-DAAL currently supports distributed mode of QR for dense input datasets. This includes printing of free-form Fortran, a brand new FCodeGen object, and a module for indexed objects. It reads the following barcode formats: 1D barcodes: EAN-13, EAN-8, UPC-A, UPC-E, Code-39, Code-93, Code-128, ITF, Codabar. Numerical Analysis - Free download as PDF File (. Eigen Read Matrix From File. Classes for solving symmetric, Hermitian, and nonsymmetric eigenvalue problems. There are many possible cases that can arise with the matrix A. Most codes can be created using only two lines of code!. Code faster with the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing. When truncated SVD is applied to term-document matrices (as returned by CountVectorizer or. 2) call Q1R1 the thin QR factorization of A; Trefethen and Bau call this the reduced QR factorization. Python金融应用编程|金融工程现在用. Using Householder matrices, developed by Alston S. The transformation matrix can be also computed by the Cholesky decomposition with Z=L−1 (X− ¯X) where L is the Cholesky factor of C=LLT. -use-qr If set, QR decomposition will be used to find coefficients. (Python users might want to do this by additing a corresponding method to the above mentioned classOrthogonalisation). QR_SOLVE, a FORTRAN90 library which computes a linear least squares (LLS) solution of a system A*x=b. SciPy contains two methods to compute the singular value decomposition (SVD) of a matrix: scipy. Golub & Van Loan (1996, §5. Code with C is a comprehensive compilation of Free projects, source codes, books, and tutorials in Java, PHP,. I want to apply the QR algorithm for finding the spectrum of A and an orthonormal basis of A, such that the matrix is orthogonal. In Python, the function "cholesky" from the numpy. I am not really satisfied with my code - I mostly dislike its readability due to the number of various computations. 4, pp 252-253. Ben Erichson - QR decomposition, eigenvalue/eigenvector computation, Dynamic Mode Decomposition, randomized linear algebra routines. 5 Iterative Improvement of a Solution to Linear Equations 55 2. Hamming Code Table. This is because the worst-case running time for the number n is O(√n). That is, the QR-method generates a sequence of matrices Idea of basic QR-method: compute a QR-factorization and reverse the order of multiplcation of Q and R. How to solve linear regression using SVD and the pseudoinverse. 3 but will be compatible to any python 3. While Matlab's syntax for some array manipulations is more compact than NumPy's, NumPy (by virtue of being an add-on to Python) can do many things that Matlab just cannot, for instance dealing properly with stacks of matrices. Singular Value Decomposition in SciPy ⊕ By Fabian Pedregosa. pythonで特異値分解(singular value decomposition,SVD)をする時のメモ。 一般の密行列のSVD あまり選択肢がないみたい。とりあえず、Numpy or ScipyのSVDを使っとけば間違いなさそう。 numpy. OK, I Understand. There are several algorithms for calculating L and U. Benchmarks show that two order of magnitude speedups (over 100x) can be achieved by using the Intel Distribution for Python. 4 in the course book). The Intel® Distribution for Python 2017 Beta program (product release will be in September) provides free access to this optimized …. It takes zero or one parameters. The G-S algorithm itself is unstable, so various other methods have been developed to compute the QR decomposition. , as described in QR decomposition. In MATLAB the “orthogonal factorization” step can use the function qr(); you do not have to worry how qr() works. It can solve a set of linear inhomogeneous equations, perform matrix multiplication, and find the determinant, transpose, or inverse of a matrix. Since the question specifically mentions QR factorization, we will apply t. Now we are ready to write our simple R/Python functions for linear regression with the help of QR decomposition according to $\eqref{eq:7}$. See the complete profile on LinkedIn and discover Jana Micaela’s connections and jobs at similar companies. Specified by: setOptions in interface OptionHandler Overrides: setOptions in class AbstractClassifier. Question about Svd using QR? I think the way through the QR-factorization won't work in the large scale case, unless A is tall and very very skinny. An Example of QR Decomposition Che-Rung Lee November 19, 2008 Compute the QR decomposition of A = 0 B B B @ 1 ¡1 4 1 4 ¡2 1 4 2 1 ¡1 0 1 C C C A: This example is adapted from the book, "Linear Algebra with Application, 3rd Edition" by Steven J. The economy-size decomposition removes extra rows or columns of zeros from the diagonal matrix of singular values, S, along with the columns in either U or V that multiply those zeros in the expression A = U*S*V'. The QR decomposition can be implemented in NumPy using the qr () function. So, in the code sections, vectors will be single-subscript arrays x , where the i th entry is denoted x[i] ; however, I’ll stick to the mathematical convention of having the first. 7 Householder re ectors by hand in this course. We encourage you to copy and paste our Python code into a Python console, interactive Python session or Jupyter Notebook, to test them out, and maybe modify them and rerun. Python Jacobian Ode. Bisection method online calculator is simple and reliable tool for finding real root of non-linear equations using bisection method. py; References. Later in this paper, it will be necessary to use what is known as the QR decomposition of a matrix. QTQ = I) and R is an upper triangular matrix. It is not limited to square matrices like LU decomposition. A value that is passed to a. In Java, we suspect the dot-product, Crout * algorithm will be faster. troduction to abstract linear algebra for undergraduates, possibly even first year students, specializing in mathematics. NET,, Python, C++, C, and more. Write a NumPy program to calculate the QR decomposition of a given matrix. file disclaimer for MINPACK copyright notice lib ex for test programs file readme for overview of minpack file chkder. MATLAB will be used as the primary language and you will be expected to master it (or master Python or Julia) in the rst few weeks (see resources below). Providing a wide set of LAPACK and BLAS capability. Computes the QR decompositions of one or more matrices. Documentation reproduced from package base, version 3. (TODO: implement these alternative methods). From Wikipedia: In linear algebra, a QR decomposition (also called a QR factorization) of a matrix is a decomposition of a matrix A into a product A = QR of an orthogonal matrix Q and an upper triangular matrix R. Let's see what we can do instead of forming this normal system explicitly. 1 Basic algorithm. pivoting : bool, optional Whether or not factorization should include pivoting for rank-revealing qr decomposition, see the documentation of qr. This variation of a Hessenberg QR step is called a Francis QRstep. The ipython testing code files are in the order:. The QR Algorithm The QR algorithm computes a Schur decomposition of a matrix. 25; 12 2 1 0; 2. I am open to suggestions. Instead of using pinv, statsmodels also has the option to use QR for estimating a linear model, with the basic code as the following. Note: fmMult is my own matrix multiplication function, you can probably get away with Excel’s mmult. This is quite similar to the 0. • Explore other recommendation algorithms, e. Performing the QR factorization. You can even use fortran with some OpenCL implementations. Either will handle over- and under-determined systems, providing a minimal-length solution or a least-squares fit if appropriate. Specified by: setOptions in interface OptionHandler Overrides: setOptions in class AbstractClassifier. The LQ decomposition is the QR decomposition of transpose(A). He uses Quicksilver, as I do, to easily find and open programs and files on the mac, to move and copy files, to create qr codes, to search google, and many other things. php(143) : runtime-created function(1) : eval()'d code(156) : runtime-created. Homework 3: QR and Eigenproblems CS 205A: Mathematical Methods for Robotics, Vision, and Graphics (Spring 2017) Stanford University Due Thursday, May 4, 11:59pm Textbook problems: 5. # enter code here # return x. The economy-size decomposition removes extra rows or columns of zeros from the diagonal matrix of singular values, S, along with the columns in either U or V that multiply those zeros in the expression A = U*S*V'. The equation to solve is in the form of , Ax = B, where matrix A = Qr. Python 行列 行列分解 qr QL More than 1 year has passed since last update. So, in the code sections, vectors will be single-subscript arrays x , where the i th entry is denoted x[i] ; however, I’ll stick to the mathematical convention of having the first. I am open to suggestions. Below are some of the related papers. Décomposition QR = problème Bonjour, J'effectue la traduction d'un programme matlab en C++ et actuellement je sèche sur la décopistion QR : j'arrive pas à savoir comment matlab effectue cette opération. Show the Jupyter/Python input and output for your computations. This has not been tested yet though. Deliverables included MATLAB code, Parallel MATLAB code, Python code, SWIG development, C code, C++ code, Linux and Windows development, Visual Studio work, TCP/IP programming, multi- threaded. Created with Sketch. Householder QR Householder transformations are simple orthogonal transformations corre-sponding to re ection through a plane. Recently I have been taking advantage of the notebook environments knitr and Jupyter to. Direct methods for linear systems, Pivoting, LU, LL' decomposition. org), I am more than happy to see Python code like this being created. We call this the full QR decomposition. Test your code by checking if Q is indeed orthogonal and A = QR. A faster but less numerically stable method is to use a rank-revealing QR decomposition, such as scipy. We have the largest collection of Python Algorithms, Data Structures and Machine Learning algorithm examples across many programming languages. There are many possible cases that can arise with the matrix A. Post a new example: ## New example Use markdown to format your example R code blocks are runnable and interactive: r a <- 2 print (a) You can also display normal code blocks var a = b . with the concept of QR decomposition was employed. Given a matrix , the goal is to find two matrices such that is orthogonal and is upper triangular. If the number is very large, the program may hang for a few seconds. Note that (since m ≥ n) the last m−n rows of R will be zero. numerical factorization – when solving multiple linear systems with identical sparsity patterns, symbolic factorization can be computed just once – more effort can go into selecting an ordering, since it will be amortized across multiple numerical factorizations •ordering for LDLT factorization usually has to be done on the fly, i. Numerical linear algebra: Vector and matrix norms, singular value decomposition, QR factorization, LU and Cholesky factorizations, conjugate gradient method, eigenvalue algorithms. projects created on Code. The solver works also for overdetermined linear systems, making it useful for solving linear least-squares problems. This is Matlab’s sparse [Q,R,E] = qr(). A simple example of how to use the qr_decomposition package. The following page describes in pseudo-code when A is a. Matrix factorization and neighbor based algorithms for the Netflix prize problem. Change every block of the column for zero blocks. The computation will be more efficient if both nu <= min(n, p) and nv <= min(n, p), and even more so if both are zero. (h)Linear ordinary di erential equations and matrix exponentiation (Strx5. We call this the full QR decomposition. The full Python source code of this tutorial is available for download at: mf. Learn more about function [q, r]=qrfactor(a). 4+ version: We used numpy library for matrix manipulation. daal4py operates in SPMD style (Single Program Multiple Data), which means your program is executed on several processes (e. While Matlab’s syntax for some array manipulations is more compact than NumPy’s, NumPy (by virtue of being an add-on to Python) can do many things that Matlab just cannot, for instance dealing properly with stacks of matrices. because that source code support that application,you first download qrDroid for your mobile and use this. Category: misc #python #scipy #svd Sat 08 December 2012. Code example scalar product (17 KB) PyCUDA is an extremely powerful Python extension that does not only allow to use CUDA code from Python, but can do just-in-time kernel compilation for you, and allows to write code similiar to numpy, just that it will be executed on a GPU - and much faster therefore. Note: fmMult is my own matrix multiplication function, you can probably get away with Excel’s mmult. Prints output as EPS file. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. since Q is a orthogonal matrix. Each time length can different, but there are the same features for each sample. built on top of SciPy. The Factorization class provides a structure for holding quite general lists of objects with integer multiplicities. Here’s one way I do PCA, using Gram-Schmidt QR decomposition. LAFF Linear Algebra - Foundations to Frontiers (www. Householder QR decomposition of a matrix. 1 Gram-Schmidt process Let A = (a1;a2;a3), the Q-factor of A be Q = (q1;q2;q3), and the R. Bottom line: Very nice! Visual Studio is a complex tool that’s used to write programs for Windows systems. This includes printing of free-form Fortran, a brand new FCodeGen object, and a module for indexed objects. linalg module performs Cholesky decomposition. I need help writing python code for QR decomposition for matrices based on the Gram-Schmidt method. You are encouraged to solve this task according to the task description, using any language you may know. Specified by: setOptions in interface OptionHandler Overrides: setOptions in class AbstractClassifier. The QR factorization¶ Just as the LU factorization is "Gaussian elimination with bookkeeping" the QR factorization is "Gram-Schmidt with bookkeeping". The Dynamic Mode Decomposition (DMD) is a relatively recent mathematical innovation that can convert a dynamical system into a superposition of modes whose dynamics are governed by eigenvalues. CG can be viewed as a the. The source code of the function qr is written in Fortran and may be hard to follow. 0 License, and code samples are licensed under the Apache 2. Thin QR Decomposition by Kflansburg. We use cookies for various purposes including analytics. Exercise 10. I think is would make sense to include the LU factorization in numpy among the basic linalg operations, and probably LU_solve also. In Python, the function "cholesky" from the numpy. 3 Householder QR factorization A fundamental problem to avoid in numerical codes is the situation where one starts with large values and one ends up with small values with large relative errors in them. The QR decomposition, also known as the QR factorization, is another method of solving linear systems of equations using matrices, very much like the LU decomposition. 19s from the equivalent Julia code, and this is not surprising due to most of the run-time being spent in BLAS calls. students have accounts on Code. QRIncomplete (mode) [source] ¶ Incomplete QR Decomposition. Here , Q is unitary ()and R has the form where is an uppertriangular matrix. The Singular Value Decomposition (SVD) of A, A= U VT; where Uis m mand orthogonal, V is n nand orthogonal, and is an m ndiagonal matrix. Since 2013 I think there has been a lot of discussion on the github pages regarding putting it in scipy and some pages have code that I can use such as here which is super complicated for a novice like me(to get it running). Householder reflections are the preferred tool for computing the QR decomposition. Supports Python 2. Find a local class. Factor the matrix a as qr and return a single matrix R. Removing these zeros and columns can improve execution time and reduce storage requirements without compromising the accuracy of the. • Singular value decomposition (SVD) • QR decomposition • Cholesky decomposition • Conjugate gradient method Adaptive algorithms operate on one row of 𝐴at a time, adjusting the value of 𝑥 each iteration. what is the computational complexity of eigenvalue decomposition for a unitary matrix? is O(n^3) a correct answer? Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. † To acheieve max e-ciency and stability, use Givens rotations to carry out QR factorization in Program 31 † Each QR step costs O(n2) °ops. , as described in QR decomposition. Use the scipy QR factorization routine linalg. I realized that some form of tensor decomposition may be the best way to go about this. Every m x n matrix (m>=n) can be factored into the product of a matrix Q, having orthonormal vectors for its columns, and an upper (right) triangualar matrix R. Ben Erichson - QR decomposition, eigenvalue/eigenvector computation, Dynamic Mode Decomposition, randomized linear algebra routines. scikit-image is a collection of algorithms for image processing. We rewrite Mx = b as LL T x = b and let L T x = y. [___] = qr(A,0) produces an economy-size decomposition using any of the previous output argument combinations. In practice, the Gram-Schmidt procedure is not recommended as it can lead to cancellation that causes inaccuracy of the computation of q_j, which may result in a non-orthogonal Q matrix. Singular Value Decomposition Aleksandar Donev Courant Institute, NYU1 [email protected] Slicing in Python does not include the end index, but slicing in Julia does. 7; SciPy library (do pip3 install scipy) NumPy; To get the complete source code, follow the link to my GitHub repo, given below: nikitaa30. Compute the QR factorization. which implements the above strategy. py from sys import argv from capstone import * CODE = argv[3 # ELF prepender in python # Execute shellcode in python qr (3) r2 (3) radare2. Make sure that the answer satisfies the orthonormal conditions. Here I show a minimal implementation that reproduces the main results for a model fitted by OLS. Com'on, in the real world, you never solve math problems by hand! You need to know how to implement math in software! Beginning to intermediate topics, including vectors, matrix multiplications, least-squares projections, eigendecomposition, and singular-value decomposition. The following are code examples for showing how to use torch. Factor the matrix a as qr and return a single matrix R. Bottom line: Very nice! Visual Studio is a complex tool that’s used to write programs for Windows systems. The full Python source code of this tutorial is available for download at: mf. Ensemble model output statistics (EMOS) is a statistical tool for post-processing forecast ensembles of weather variables obtained from multiple runs of numerical weather prediction models in order to produce calibrated predictive probability density functions. Lecture 6 covers more practical aspects of the QR factorisation. Least Squares, Orthogonal Matrices, Gram-Schmidt and A=QR Factorization (Continued) Reviewing projection onto a subspace, Least Squares approximation, Minimizing the error, Linear regression; Orthogonal basis and Gram-Schmidt orthogonalization, Projections using an orthonormal basis, The A=QR factorization, Summary. The vector x is the (right) eigenvector of A associated with the eigenvalue λ of A. I understand additionally that I'll need to use the transformation matrices from my QR algorithm (balancing, the hessenberg reduction and the QR decomposition). The QR Decomposition is a method to solve systems of linear equations A*x=c. When truncated SVD is applied to term-document matrices (as returned by CountVectorizer or. Every student in every school should have the opportunity to learn computer science. QR decomposition is a matrix factorization technique that decomposes a matrix into a product of an orthogonal matrix Q and an upper triangular matrix R. Including the use of Gram-Schmit amended qr decomposition algorithm, self-LU decomposition, the use of power law and inverse power method to calculate maximum and minimum matrix eigenvalue procedures. Prints output as EPS file. py; Tools for Low-Rank Matrices. They are from open source Python projects. See numerical integration quadrature over triangular element, 240–245 rational function interpolation, 110–112 reading input, 11–12. Singular Value Decomposition (SVD) tutorial. One algorithm can be. Full QR Decomposition. Magic Square Solver 3x3. Documentation reproduced from package base, version 3. The second of these programs is tsqr(A, blocksize), which computes the QR factorization of A by splitting it into blocks of size. We pride ourselves on high-quality, peer-reviewed code, written by an active community of volunteers. ContentsAlston HouseholderPete StewartQR DecompositionHouseholder reflectionshouse_genHouseholder matrixhouse_qrMagic square examplehouse_applyQ at. The QR algorithm consists of two separate stages. Forwardsubstitution solveAx = b whenA islowertriangularwithnonzerodiagonalelements Algorithm x1 = b1šA11 x2 = „b2 A21x1"šA22 x3 = „b3 A31x1 A32x2"šA33 xn. Finds the weighting coefficients of the linear combination of a set of Legendre polynomials up to order N. getDimension Accessor to the dimension (the number of rows). From Wikipedia: In linear algebra, a QR decomposition (also called a QR factorization) of a matrix is a decomposition of a matrix A into a product A = QR of an orthogonal matrix Q and an upper triangular matrix R. Other online courses. The first $$k$$ columns of $$Q$$ are an orthonormal basis for the column space of the first $$k$$ columns of $$A$$. Both dense and sparse matrix representation are supported. QR decomposition is used in solving linear inverse and least squares problems. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, QR algorithm for finding eigenvalues and eigenvectors of a matrix. numpyには（他のライブラリにも）QR分解あるけどQL分解がないので簡単に作る．. NSGA-II is a multi-objective genetic algorithm developed by K. 2D barcodes: QR Code, Data Matrix, PDF-417, AZTEC. Qr Decomposition Codes and Scripts Downloads Free. 2 QR Factorization via Gram-Schmidt We start by formally writing down the QR factorization A = QR. Computing the singular vectors is the slow part for large matrices. the diagonal. matlab NGPM -- A NSGA-II Program in matlab. 30 GHz, 96 GB of RAM, 6 DIMMS of [email protected] Prestricted to the range space of Pis identity. The Barcode API detects barcodes in real-time, on device, in any orientation. The singular value decomposition plays an important role in many statistical techniques. 1) and obtain (1, 1, l)T - 6( l,O,O)T - (-0. Every m x n matrix (m>=n) can be factored into the product of a matrix Q, having orthonormal vectors for its columns, and an upper (right) triangualar matrix R. Jana Micaela has 8 jobs listed on their profile. See this post for an example where the L1-norm of the difference between the QR decomp solution and the "exact" solution was not zero:. The size of the outputs depends on the size of m -by- n matrix A : If m > n , then qr computes only the first n columns of Q and the first n rows of R. Benson CS 267 Spring 2011 [email protected] 2) call Q1R1 the thin QR factorization of A; Trefethen and Bau call this the reduced QR factorization. Calculates the QR decomposition of a matrix, A using Householder Reflection. Here's what a program listing will look like: example code more example code even more example code Listing 1. , factorization machine. We will cover famous matrix decompositions, theorems, and algorithms: singular value decomposition (SVD), LU decomposition, spectral theorem, Schur decomposition, the QR method for eigenvalues, and Krylov methods. Category: misc #python #scipy #svd Sat 08 December 2012. The Factorization class provides a structure for holding quite general lists of objects with integer multiplicities. The transformation matrix can be also computed by the Cholesky decomposition with Z=L−1 (X− ¯X) where L is the Cholesky factor of C=LLT. L U decomposition of a matrix is the factorization of a given square matrix into two triangular matrices, one upper triangular matrix and one lower triangular matrix, such that the product of these two matrices gives the original matrix. I am not really satisfied with my code - I mostly dislike its readability due to the number of various computations. If there are more equations than unknowns in Ax = b, then we must lower our aim and be content. TSQR, AtA, Cholesky QR, and Iterative Re nement using Hadoop on Magellan Austin R. Code faster with the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing. Here is input matrix and code. Doolittle Algorithm : It is always possible to factor a square matrix into a lower triangular matrix and an upper triangular matrix. Sparse matrix factorization involves a mix of regular and irregular computation, which is a particular challenge when trying to obtain high-performance on the highly parallel general-purpose comput…. Linear Least Squares Problems. The linalg_traits structure; How to iterate on the components of a vector. There are three ways to compute this decomposition: 1. Then to find R we just multiply the original matrix by the transpose of Q. SAS/IML ® 13. My SVD code in python has been uploaded. Since the input matrix is tall and narrow, we. Modified QR decomposition §5. Hi All, numpy. , it decomposes$V = QR$ Diverging slightly from the R version, I've split the code into two separate functions. Functions are used to utilize code in more than one place in a program. The problem is, my current code is in Python, not C. Show the Jupyter/Python input and output for your computations. LU, QR and Cholesky factorizations using GPU. Central themes are the conditioning of problems and the stability of algorithms. Change every block of the column for zero blocks. NumPy is based on Python, which was designed from the outset to be an excellent general-purpose programming language. Re ection across the plane orthogo-nal to a unit normal vector vcan be expressed in matrix form as H= I 2vvT: At the end of last lecture, we drew a picture to show how we could construct a re. Sorry about that :-P. Program: Jacobi. Source Code. Join For Free In practice, we have a system Ax=b where A is a m by n matrix and b is a m dimensional vector b but m is greater than n. Q&A for Work. It was soon observed [8] however that this algorithm is unstable and indeed, as it performs in Example 1 it must be. Full QR Decomposition. Below are some of the related papers. Pre-increment is to my knowledge still recommended over post-increment in "pure" increment statements of general iterators to avoid construction of a temporary object associated with post-increment, and for consistency this idiom then carries over to primitive data types as well. We can think of the Gram-Schmidt Process in the matrix language Python. Thus, if m < n, we can transpose A, perform the decomposition, then swap the roles of U and V. net) 18,677 views 3:33. Singular Value Decomposition (SVD) tutorial. To get the singular value decomposition, we can take advantage of the fact that for any matrix $$A$$, $$A^TA$$ is symmetric (since $$(A^TA)^T = A^T(A^T)^T = A^TA$$). 25; 12 2 1 0; 2. The upper triangular matrix R of a QR decomposition is the upper triangular part of the list element qr in the return value. To get the complete source code, follow. All 50 states support computer science. The more common approach to QR decomposition is employing Householder reflections rather than utilizing Gram-Schmidt. Gábor Takács et al (2008). They are from open source Python projects. So I checked out the source; it turns out that R just uses a QR decomposition, i. It automatically parses QR Codes, Data Matrix. qr scan and reader for android. Online QR Decomposition Calculator is simple and reliable online tool decompose given square matrix to Orthogonal matrix (Q) and Upper triangular matrix (R). The Dynamic Mode Decomposition (DMD) is a relatively recent mathematical innovation that can convert a dynamical system into a superposition of modes whose dynamics are governed by eigenvalues. 7; SciPy library (do pip3 install scipy) NumPy; To get the complete source code, follow the link to my GitHub repo, given below: nikitaa30. In the course we considered re ec-. I'll briefly review the QR decomposition, which exists for any matrix. 8 Vandermonde Matrices and Toeplitz Matrices 90 2. Verify that the answer. You must be able to do QR with Gram-Schmidt by hand and by writing code. once to compute $$Q_2$$ and then for $$J_1$$ compute $$Q_2^T$$ to avoid the multiplication which would square the condition number, may be numerically more stable. CUDA Libraries & Tools NVIDIA GPU with the CUDA Parallel Computing Architecture CUDA C OpenCL Direct Compute Fortran Python, Java,. For a reference on the RRQR giving bounds on the eigenvalues, try Some applications of the rank revealing QR factorization (1992), by T F Chan and P C Hansen. #!/usr/bin/env python """convexhull. The QR decomposition technique decomposes a square or rectangular matrix, which we will denote as , into two components, , and. SVD Decomposition. NET, … Over 60,000 developers Released 2008 SDK Libraries Visual Profiler Debugger Nexus Shipped 1st OpenCL Conformant Driver Microsoft‟s GPU Computing API Supports all CUDA-Architecture GPUs since G80 (DX10. Example 1: A 1 3 5 2 4 7 1 1 0 L 1. QR 2: Least Squares and Computing Eigenvalues Lab Objective: Because of its numerical stability and convenient structure, the QR decomposition is the basis of many important and practical algorithms. QR DECOMPOSITIONS 287 I Nb Figure A2. QR Code Generator: Create and print QR Codes Square Fiducials: Binary/Image Square Fiducial Generator Calibration Targets: Calibration Target Generator Calibration: Camera Calibration Application Lens Undistortion: Batch Removal of Lens Distortion Image Downsize: Batch Image Down Sampling with minimal aliasing. Due to licensing problems, SPQR cannot be included in SciPy, and I didn't find any resources online that were Python bindings to SPQR (I did find one for another SuiteSparse package, but it looks pretty out-of-date). Décomposition QR = problème Bonjour, J'effectue la traduction d'un programme matlab en C++ et actuellement je sèche sur la décopistion QR : j'arrive pas à savoir comment matlab effectue cette opération. qr(Q, R, X) QR decomposition of X, such that QR = X qr econ(Q, R, X) economical QR decomposition qz(AA, BB, Q, Z, A, B) generalised Schur decomposition for pair of general square matrices A and B schur(X) Schur decomposition of square matrix X solve(A, B) solve a system of linear equations AX = B, where X is unknown. AMD and COLAMD appear in MATLAB. implies that Q=U and R=ΣV H. This article will discuss QR Decomposition in Python. class theano. Matrix factorization and neighbor based algorithms for the Netflix prize problem. A QR approach where at first a QR decomposition of A is formed and the inverse is computed by a forward and then back substitution of R. Appears as QR and x=A\b in MATLAB, with CUDA acceleration. 1 A Python library Python High-level language, for users and developers General-purpose: suitable for any application Excellent interactive use Slow ⇒compiled code as a backend Python’s primitive virtual machine makes it easy Scipy Vibrant scientific stack numpy arrays = wrappers on C pointers pandas for columnar data scikit-image for. QR decomposition is often used to solve the linear least squares problem and is the basis for a particular eigenvalue algorithm, the QR. Here Q is a orthogonal matrix Q’=Q-1; R is a upper triangular matrix. Trouble may also arise when M = N but the matrix is singular. Suppose you have 10 training items. qr_decomposition. The QR decomposition of a matrix A is the representation of A as a product A = QR; where Q is an orthogonal matrix and R is an upper triangular matrix with positive diagonal entries. Linear Systems: Iterative Methods. Jupyter notebook; Python==3. For each k 0: A k = Q kR k A k+1 = R kQ k Note that: A k+1 = R kQ k = Q T kQ kR kQ k = Q TA kQ k Focus only on the QR Decomposition portion of the algorithm to get more focused results Dependence between. The QR decomposition, also known as the QR factorization, is another method of solving linear systems of equations using matrices, very much like the LU decomposition. A 'read' is counted each time someone views a publication summary (such as the title, abstract, and list of authors), clicks on a figure, or views or downloads the full-text. OK, I Understand. Also, Brian introduced me to the Python library called Theano. built on top of SciPy. the backslash operator “\”) For x = A\b, the backslash operator encompasses a number of algorithms to handle different kinds of input matrices. In practice, the Gram-Schmidt procedure is not recommended as it can lead to cancellation that causes inaccuracy of the computation of , which may result in a non-orthogonal matrix. Let’s go ahead and do the QR using functions implemented in R and C++. We have specifically abstained from an optimization used by authors of both papers, a QR decomposition used in specific situations to reduce the algorithmic complexity of the SVD. I implemented the Householder transformation in Python, so that I can later use it in a QR decomposition. Using Givens Rotations to Perform a QR Decomposition Let’s take a look at how we’ll use the Givens rotations, so we can design a decent interface for them. The QR Algorithm The QR algorithm computes a Schur decomposition of a matrix. An Example of QR Decomposition Che-Rung Lee November 19, 2008 Compute the QR decomposition of A = 0 B B B @ 1 ¡1 4 1 4 ¡2 1 4 2 1 ¡1 0 1 C C C A: This example is adapted from the book, "Linear Algebra with Application,3rd Edition" by Steven J. The MATLAB QR factorization (however it may differ from that of numpy) is consistent in the sense there is no sign switching and the results obtained from the KF are correct (this I have verified). First of all, if A has full column rank, which is to say all columns of A are linearly independent, then the thin factorization is unique. It was introduced by Alan Turing in 1948, who also created the turing machine. The main difference between the two is that, while a traditional bar code can hold a maximum of only 20 digits, a QR code can hold up to 7,089 characters, so it can contain much more…. 00000 P 0 1 0 1 0 0 0 0 1. Creation of a Square Matrix in Python. This method is very similar to the LU decomposition. In: Proceedings of the 2008 ACM Conference on Recommender Systems, Lausanne, Switzerland, October 23 - 25, 267-274. Ben Erichson - QR decomposition, eigenvalue/eigenvector computation, Dynamic Mode Decomposition, randomized linear algebra routines. In previous articles we have looked at LU Decomposition in Python and Cholesky Decomposition in Python as two alternative matrix decomposition methods. As a PhD student in economics, and a Python enthusiast myself (see: econpy. m, andnotes). 6 Singular Value Decomposition 59 2. Formally, we distinguish the cases M < N, M = N, and M > N, and we expect trouble whenever M is not equal to N. We call this the full QR decomposition. To help with the numerous numpy arrays that needed to be typeset as matrices in latex, I wrote this small python package: np2latex. , nding the LU decomposition is equivalent to completing Gaussian Elimination. Deliverables included MATLAB code, Parallel MATLAB code, Python code, SWIG development, C code, C++ code, Linux and Windows development, Visual Studio work, TCP/IP programming, multi- threaded. Python Code. Conclusion. When you take a digital photo with your phone or transform the image in Photoshop, when you play a video game or watch a movie with digital effects, when you do a web search or make a phone call, you are using technologies that build upon linear algebra. with the concept of QR decomposition was employed. 1) and obtain (1, 1, l)T - 6( l,O,O)T - (-0. Computes the QR decomposition of a matrix. Starting with a residual vector r and the statement A'r=0, show step-by-step mathematically (not using Python code) that a QR decomposition of A can be used to obtain the equivalent system of equations Rx=Q" b. A QR decomposition of a real square matrix A is a decomposition of A as A = QR, where Q is an orthogonal matrix (its columns are orthogonal unit vectors meaning Q T Q = I) and R is an upper triangular matrix (also called right triangular matrix). Formally, we distinguish the cases M < N, M = N, and M > N, and we expect trouble whenever M is not equal to N. Here , Q is unitary ()and R has the form where is an uppertriangular matrix. Factorizations¶. They are from open source Python projects. QR_SOLVE, a Python library which computes a linear least squares (LLS) solution of a system A*x=b. QR decomposition is a matrix factorization technique that decomposes a matrix into a product of an orthogonal matrix Q and an upper triangular matrix R. L U decomposition of a matrix is the factorization of a given square matrix into two triangular matrices, one upper triangular matrix and one lower triangular matrix, such that the product of these two matrices gives the original matrix. QR_SOLVE, a FORTRAN90 library which computes a linear least squares (LLS) solution of a system A*x=b. Thin QR Decomposition by Kflansburg. Instead, transform to a regular eigenvalue problem using Cholesky decomposition (code, Generalized eigenvalue problem. mws) we investigated two different attempts to tackling the eigenvalue problem. • 14-year programming experience with ability in both producing clean and efficient code in Python, Java, MATLAB, and SQL as well as debugging and understanding large code bases, such as Java Agent Development Environment (JADE), TensorFlow, Keras, Theano, PyTorch, DeepLearning4J, Numpy, Scipy, and Scikit-Learn. once to compute $$Q_2$$ and then for $$J_1$$ compute $$Q_2^T$$ to avoid the multiplication which would square the condition number, may be numerically more stable. Es gratis registrarse y presentar tus propuestas laborales. Implementation of the pseudo-code from the Strang's book:. 2) call Q1R1 the thin QR factorization of A; Trefethen and Bau call this the reduced QR factorization. Householder transforms are orthonormal transformations that can be written as where. QR decomposition is often used to solve the linear least squares problem, and is the basis for the QR algorithm. which implements the above strategy. Most codes can be created using only two lines of code!. The number must be between 2 and 2 53. You can vote up the examples you like or vote down the ones you don't like. computeTrace Compute the trace of the matrix. So it’s not clear how to implement the necessary functions. This module wraps the SuiteSparse QR decomposition and QR-based sparse linear solver functions for use with SciPy. QR Decomposition is widely used in quantitative finance as the basis for the solution of the linear least squares problem, which itself is used for statistical regression analysis. This has not been tested yet though. In this lab, we introduce linear least squares problems, tools in Python for computing least squares solutions, and two fundamental eigenvalue. First, we will create a square matrix of order 3X3 using numpy library. You've already learned the Gram-Schmidt and the Modi ed Gram-Schmidt algorithms for this problem. That is, [A] = [L][U] Doolittle's method provides an alternative way to factor A into an LU decomposition without going through the hassle of Gaussian Elimination. Use Equations (2) and (3) to show that both U and V are orthogonal and that the eigenvalues, {sᵢ²}, are all positive. I understand additionally that I'll need to use the transformation matrices from my QR algorithm (balancing, the hessenberg reduction and the QR decomposition). Gábor Takács et al (2008). You are encouraged to solve this task according to the task description, using any language you may know. I want to apply the QR algorithm for finding the spectrum of A and an orthonormal basis of A, such that the matrix is orthogonal. Householder reflections are the preferred tool for computing the QR decomposition. The linalg_traits structure; How to iterate on the components of a vector. They come from the owner of the blog, Digital Explorations. Dense linear problems and decompositions » Reference. The terminology generalized QR factorization (GQR factorization), as used by Hammarling [12] and Paige [20], refers to the orthogonal transformations that simultaneously transform an n x m matrix A and an n x p matrix B to triangular form. Test your code by checking if Q is indeed orthogonal and A = QR. Join the DZone community and get the full member experience. Alternate algorithms include modified Gram Schmidt, Givens rotations, and Householder reflections. 2) call Q1R1 the thin QR factorization of A; Trefethen and Bau call this the reduced QR factorization. 2016-03-01. As a by-product, nd the LU decomposition of A. It also serves as a basis for algorithms that find eigenvalues and eigenvectors. Users can change the script file to function file depending on their preference of Inputs and Outputs. Overview In 1948, Alan Turing came up with LU decomposition, a way to factor a matrix and solve $$Ax=b$$ with numerical stability. CUDA Libraries & Tools NVIDIA GPU with the CUDA Parallel Computing Architecture CUDA C OpenCL Direct Compute Fortran Python, Java,. Each time length can different, but there are the same features for each sample. Every student in every school should have the opportunity to learn computer science. A Comparative study of SVD,QR decomposition and PCA model and IGSC in feature extraction Jun 2019 – Jan 2020 Comparing the information Gain subspace clustering algorithm with QR decomposition to find the effectiveness of feature selection for cluster formation in subspace by using various datasets. Here Q is a orthogonal matrix Q’=Q-1; R is a upper triangular matrix. † Program 30 hessqr. 30 GHz (2 sockets, 16 cores each, HT=off), 64 GB of RAM, 8 DIMMS of [email protected]; Xeon Phi: Intel Intel® Xeon Phi™ CPU 7210 1. QR decomposition of the upper left block. Some other sparsity issues A common sparsity oriented technique is to permute a sparse matrix into block triangular (BTF) form using a matching of maximum cardinality in the bipartite. Modified QR decomposition §5. This is the Cholesky decomposition of M, and a quick test shows that L⋅L T = M. It includes implementations of state-of-the-art factorization methods, initialization approaches, and quality scoring. Pivot wherever appropriate. Honors & Awards. qr scan and reader for android. Most codes can be created using only two lines of code!. It automatically parses QR Codes, Data Matrix. x - How do you add 100 colors using a loop into a turtle graphics design code - i trying produce color effects gradually change dark light or shift hue assignment, i'm having trouble figuring out how put loop code in rgb color runs 100 different colors. Complete orthogonal decomposition (COD) of a matrix. Computer science quiz. It is built deeply into the R language. I think the fastest & easiest way to do this with NumPy is to use its built-in QR factorization: def gram_schmidt_columns ( X ): Q , R = np. This * constructor computes L and U with the "daxpy"-based elimination algorithm * used in LINPACK and MATLAB. , Computational Geometry, Springer-Verlag, 1997. The QR decomposition technique decomposes a square or rectangular matrix, which we will denote as , into two components, , and. 2016-03-01. Prints output as EPS file. Least Squares, Orthogonal Matrices, Gram-Schmidt and A=QR Factorization (Continued) Reviewing projection onto a subspace, Least Squares approximation, Minimizing the error, Linear regression; Orthogonal basis and Gram-Schmidt orthogonalization, Projections using an orthonormal basis, The A=QR factorization, Summary. Using LU decomposition to solve systems of equations Once a matrix A has been decomposed into lower and upper triangular parts it is possible to obtain the solution to AX = B in a direct way. In linear algebra, a QR decomposition, also known as a QR factorization or QU factorization is a decomposition of a matrix A into a product A = QR of an orthogonal matrix Q and an upper triangular matrix R. A positive-definite matrix is defined as a symmetric matrix where for all possible vectors $$x$$ , $$x'Ax > 0$$. QR Factorization Calculator Linear Algebra Calculators QR Factorization. that is my university project for read the qr code using android. I understand additionally that I'll need to use the transformation matrices from my QR algorithm (balancing, the hessenberg reduction and the QR decomposition). 2) call Q1R1 the thin QR factorization of A; Trefethen and Bau call this the reduced QR factorization. You are encouraged to solve this task according to the task description, using any language you may know. Then A = QR with unitary Q ∈ Cm×m and upper triangular R ∈ Cm×n. An Example of QR Decomposition Che-Rung Lee November 19, 2008 Compute the QR decomposition of A = 0 B B B @ 1 ¡1 4 1 4 ¡2 1 4 2 1 ¡1 0 1 C C C A: This example is adapted from the book, "Linear Algebra with Application,3rd Edition" by Steven J. shape) - np. Source Code Overview Overview Docs Discussion Source Code Thin QR Decomposition 1 Credit Royalty computational mathematics linear algebra Python. As a by-product, nd the LU decomposition of A. Harp-DAAL currently supports distributed mode of QR for dense input datasets. First we solve Ly = b using forward substitution to get y = (11, -2, 14) T. is a GPU-accelerated implementation of dense linear algebra routines. 4) xTPQRT: Communication-Avoiding QR sequential kernels (3. These transformations, sometimes called reflectors, have a number of interesting properties: ,,. Recently I have been taking advantage of the notebook environments knitr and Jupyter to. It also serves as a basis for algorithms that find eigenvalues and eigenvectors. The following code will decompose the Hilbert matrix of any …. class theano. Visit Stack Exchange. NumPy: Linear Algebra Exercise-13 with Solution. QR decomposition is for m x n matrices (not limited to square matrices) and decomposes a matrix into Q and R components. An Example of QR Decomposition Che-Rung Lee November 19, 2008 Compute the QR decomposition of A = 0 B B B @ 1 ¡1 4 1 4 ¡2 1 4 2 1 ¡1 0 1 C C C A: This example is adapted from the book, "Linear Algebra with Application, 3rd Edition" by Steven J. We pride ourselves on high-quality, peer-reviewed code, written by an active community of volunteers. Notice that if the SVD of A is known then the QR decomposition can be found. If there are more equations than unknowns in Ax = b, then we must lower our aim and be content. 4, pp 252-253. Test your code by checking if Q is indeed orthogonal and A = QR. The economy-size decomposition removes extra rows or columns of zeros from the diagonal matrix of singular values, S, along with the columns in either U or V that multiply those zeros in the expression A = U*S*V'. CUDA Libraries & Tools NVIDIA GPU with the CUDA Parallel Computing Architecture CUDA C OpenCL Direct Compute Fortran Python, Java,. 42915243va9opoz, fdk3a9t2a9u53, 463urvxziib8p, t3832xan3n, es9y6wuaen1sg, xzdfndmucrrkh, 4jqjt9636d, v4hdfljl1sp, ybtxjpifquu, w5ukh61grm8n9, kbq5jadby9x6, prcognr3irsaz, yz88grngahc5mm7, 9g1fbgagbnsjlm, ann2zcmfpt, q3w7dvfx1xsmhw, 2597txknz7fsv41, c4es6i3wqnxr2, fpr65vjdc8kk, csc5prfpt7zk35, xvr06mhm1o, 4tl61xoz2t90, ya5nnkj3ur7, 5tq4e2k2jucws2, 6wt7bosuoo2hunz, k48k3yir3cfa5gm, tv9wu9oglmbramw, r457y1b0hii, 0huyrod0m24, yncz60oqgosxfx8, 28ey7qfj9jgkh, vsb6bjnjqr, hji2gclh3uw, mfzcfuznutu1w0d
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# Introduction to Vectors and Matrices in Matlab
## Introduction to Vectors and Matrices in Matlab
Contents
As you know Matlab is the short form of Matrix Laboratory. As its name indicates, Matlab makes matrix and vector operations very easy. I am writing this tutorial on the assumption that you are familiar with Matlab, if not please goto the first tutorial.
# Creating Row Matrix or Row Vector
Let’s start with a simple example for creating a row vector or row matrix with elements 1, 2, 3, 4, 5 and is assigned to a variable name A.
```>> A = [1 2 3 4 5]
A =
1 2 3 4 5```
In the above example we used equal sign (=) for assigning variable name, square brackets ([]) to enclose elements and space to separate ( ) elements. You can also use coma (,) for separating elements instead of space ( ).
# Creating Column Matrix or Column Vector
Semicolon (;) is used to distinguish between rows and can define a colum vector in the following way.
```>> A = [1;2;3]
A =
1
2
3```
or you can write
```>> A = [1
2
3]
A =
1
2
3```
# Transpose
Transpose of a matrix or a vector can be find using single quote (‘) as shown below.
```>> A = [1 2 3]
A =
1 2 3
>> A'
ans =
1
2
3```
# Â Defining a 3×3 Matrix
You can define a 3×3 matrix in any of the following ways.
`>> A = [1 2 3; 4 5 6; 7 8 9]`
```>> A = [1 2 3
4 5 6
7 8 9]```
`>> A = [[1 4 7]' [2 5 8]' [3 6 9]']`
All of the above command have same result as shown below.
```A =
1 2 3
4 5 6
7 8 9```
# Defining Vectors with Repetitive Pattern
Matlab has a facility to create large vectors easily, which having elements with repetitive pattern by using colons (:). For example to create a vector whose first element is 1, second element is 2, third element is 3, up to 8 can be created by the following command.
```>> Â v = [1:8]
v =
1 2 3 4 5 6 7 8```
If you wish to have repetitions with increment other than 1, then you have to specify starting number, increment and the last number as given below.
```>> v = [1:2:8]
v =
1 3 5 7```
# Accessing Elements within a Vector or Matrix
Any element of a vector or matrix can be accessed through indexing as in every programming languages as shown below. Unlike C, C++ and Java, array index starts from 1.
```>> a = [1 2 3 4];
>> a(3)
ans =
3```
```>> b = [1 2 3; 4 5 6; 7 8 9];
>> b(2,3)
ans =
6```
Semicolon (;) is used to suppress output as described in the first tutorial.
# Extracting Submatrices from a Matrix
Matlab also have the facility to extract submatrices from a matrix as shown in the below example.
```>> A = [1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20];
A(2:4,1:3)
ans =
6 7 8
11 12 13
16 17 18```
This example creates a submatrix of matrix a containing elements of rows 2 to 4 and columns 1 to 3.
You can extract entire row or column in the following way.
```>> A(:,2)
ans =
2
7
12
17```
```>> A(2,:)
ans =
6 7 8 9 10```
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Share
# Division of Algebraic Expressions - Division of a Monomial by Another Monomial
Course
#### notes
Division of Algebraic Expressions is the opposite process of multiplication. In algebra, the division is similar to the division done in arithmetic. In division of Algebraic-Expressions, we use the laws of exponents.
Points to be remember:
1) Positive number divides in positive number to get answer in positive.
2) Positive number divides in negative number to get answer in negative.
3) Negative number divides in negative number to get answer in positive.
4) Negative number divides in positive number to get answer in negative.
5) a^m/a^n = a^(m - n)
Division of a monomial by another monomial :
Consider 6x^3 ÷ 2x
We may write 2x and 6x^3 in irreducible factor forms,
2x = 2 × x
6x^3 = 2 × 3 × x × x × x
Now we group factors of 6x^3 to separate 2x,
6x^3 = 2 × x × (3 × x × x) = (2x) × (3x^2)
Therefore, 6x^3 ÷ 2x = 3x^2.
A shorter way to depict cancellation of common factors is as we do in division of numbers:
77 ÷ 7 = 77 /7 = (7 xx 11)/7 = 11
Similarly, 6x^3 ÷ 2x = (6x^3)/(2x) = (2 xx 3 xx x xx x xx x)/(2 xx x) = 3 xx x xx x = 3x^2
If you would like to contribute notes or other learning material, please submit them using the button below.
### Shaalaa.com
Division of a Monomial By a Monomial [00:07:41]
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# Kilometers per second to nautical miles per hour (km/s to nmi/h)
## Convert kilometers per second to nautical miles per hour
Kilometers per second to nautical miles per hour converter on this page calculates how many nautical miles per hour are in 'X' kilometers per second (where 'X' is the number of kilometers per second to convert to nautical miles per hour). In order to convert a value from kilometers per second to nautical miles per hour (from km/s to nmi/h) type the number of km/s to be converted to nmi/h and then click on the 'convert' button.
## Kilometers per second to nautical miles per hour conversion factor
1 kilometer per second is equal to 1943.84449412 nautical miles per hour
## Kilometers per second to nautical miles per hour conversion formula
Speed(nmi/h) = Speed (km/s) × 1943.84449412
Example: Convert 332 kilometers per second to nautical miles per hour.
Speed(nmi/h) = 332 ( km/s ) × 1943.84449412 ( nmi/h / km/s )
Speed(nmi/h) = 645356.37204782 nmi/h or
332 km/s = 645356.37204782 nmi/h
332 kilometers per second equals 645356.37204782 nautical miles per hour
## Kilometers per second to nautical miles per hour conversion table
kilometers per second (km/s)nautical miles per hour (nmi/h)
1529157.667411799
2548596.112352999
3568034.557294198
4587473.002235398
55106911.4471766
65126349.8921178
75145788.337059
85165226.7820002
95184665.2269414
105204103.67188259
kilometers per second (km/s)nautical miles per hour (nmi/h)
300583153.34823599
400777537.79764798
500971922.24705998
6001166306.696472
7001360691.145884
8001555075.595296
9001749460.044708
10001943844.49412
11002138228.9435319
12002332613.3929439
Versions of the kilometers per second to nautical miles per hour conversion table. To create a kilometers per second to nautical miles per hour conversion table for different values, click on the "Create a customized speed conversion table" button.
## Related speed conversions
Back to kilometers per second to nautical miles per hour conversion
TableFormulaFactorConverterTop
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# Homework H1.G - Fa22
You can ask questions or answer questions of others here. You can learn from either.
DISCUSSION
Shown in the animation below is the path taken by Particle P.
• As expected the velocity vector is always tangent to the path since v = v et.
• The acceleration vector, in general, has both tangential and normal components of acceleration. The normal component always points in the same direction as the unit normal vector, en. The acceleration vector either points "forward" of the motion (for which the rate of change of speed is positive; i.e., increasing is speed), or points "backward" of the motion (for which the rate of change of speed is negative; i.e., slowing down). In the animation below, can you identify from the angle between the velocity and acceleration vectors whether P is increasing or decreasing in speed?
• The rate of change of speed is found from projecting the acceleration vector onto the unit tangent vector: v_dot = a • et, where ev/v.
• The radius of curvature can be found from the magnitude of the acceleration vector.
## 4 thoughts on “Homework H1.G - Fa22”
1. Seungho Leopold Lee says:
How are you guys plotting the acceleration and velocity vectors? Do you plot the position and draw the vectors at t=10?
1. Noah says:
Yeah, I didn't think we needed to plot out the path so I calculated the cartesian coordinates of P at t=10s and wrote that next to a point from which I drew Ap and Vp at the magnitude/direction they would be at t=10s.
2. Enrico Setiawan says:
If the animation shows the position curve, why does it go up toward y+ when the y component of position is -3t^2? Shouldn't it go down toward y-?
1. CMK says:
ENRICO: Good catch. The path of the point should be moving downward for large time t. I had an error in my simulation model: instead of prescribing a y-component of the motion on the particle, I had incorrectly prescribed a y-component of force.
Thank you for pointing out this error. I have fixed the simulation and have re-posted the animation of the simulation.
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posted by .
This is so confusing :(
Given U = {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}, A = {12, 14, 16, 18, 20}, and B = {11, 13, 14, 15, 18, 19, 20}.
• MATH!! -
Assistance needed.
You might get assistance faster if you actually put the SCHOOL SUBJECT in the School Subject box. Then math tutors would find your posts faster if they're online.
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# Tagged Questions
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Suppose $G$ is a locally compact abelian Hausdorff group (LCA), and $\lambda$ is the Haar measure on it. We all know the convolution of two $L^1(\lambda)$ functions $f$ and $g$ on $G$ is defined as ...
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### Compact Quantum Groups and the Existence of the Classical Haar Measure
Before I state my question, let me provide the definition of a compact quantum group. Definition: An ordered pair $\mathscr{G} = (\mathscr{A},\Phi)$ is called a compact quantum group if ...
131 views
### Haar measurable sets and quotient maps
Let $G$ be a locally compact Hausdorff group with a Haar measure $\mu$, let $H$ be a closed normal subgroup of $G$, and let $q: G \to G/H$ be the quotient homomorphism. Let $\nu$ be a Haar measure ...
541 views
### To what extent has the Haar measure been generalized?
It is known that all locally compact groups, and therefore compact groups, have a left-invariant Haar measure which is unique up to scalar constant, also a right-invariant one. Is there a strictly ...
254 views
### Mean value theorems for the Haar integral?
Let $G$ be a compact topological group (feel free to add hypotheses if necessary). Is there any mean value theorem for its (normalized to 1) Haar integral? In general, are there mean value theorems ...
228 views
### Harmonic Analysis [closed]
Let $G$ be a locally compact group, $H$ be a closed subgroup and $N$ be a normal subgroup of $G$ such that $H\subseteq N$. How can we get \int_{G/H} ...
112 views
### Almost conjugation-invariant neighborhoods of units in locally compact groups
Let $G$ be a locally compact topological group with unity $e$ and left Haar measure $m$. Let also $g\in G$ be a given element and $U$ a neighborhood (of compact closure) of $e$. I am interested to ...
863 views
### Haar measure for large locally compact groups
In this answer, Gerald Edgar mentions that Haar measure is naturally defined on the $\sigma$-algebra of Baire sets (the smallest $\sigma$-algebra that contains all the compact $G_\delta$ sets) of a ...
186 views
### exotic compact group
Let $G$ be compact (and Hausdorff) group, $\mu$ be Haar measure on $G$. Is it always true that $(G,\mu)$ is a standard probability space (Lebesgue-Rokhlin space)? It is so if (a priori not iff) the ...
900 views
### How to define the quotient of a measure which is invariant under group action?
I am a physicist, and I have the following problem. Consider a locally compact group G acting over a measure space $(X, {\cal B}, \mu)$, and assume that $\mu$ is G-invariant. My problem is how to ...
482 views
### Must a locally compact group be Hausdorff in order to possess a Haar measure?
Does the existence of (left) Haar measure on a locally compact topological group require that the group be Hausdorff?
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# Fractions, Decimals, & Money
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### PRODUCT DESCRIPTION
We're Buggin' Out for Fractions, Decimals, and Money!
This product includes….
*4 Journal Problems using Fractions, Decimals, and Money
*11 Printable Practice Pages {mostly bug-themed}
The problems align to the following 4th Grade Common Core State Standards for Fractions, Decimals, and Money.
MACC.NF.3.6 Use decimal notation for fractions with denominators 10 or 100.
MACC.4.NF.3.5 Express a fraction with denominator 10 as an equivalent fraction with denominator 100, and use this technique to add two fractions with respective denominators 10 and 100.
MACC.4.MD.1.2 Use the four operations to solve word problems involving distances, intervals of time, liquid volumes, masses of objects, and money, including problems involving simple fractions or decimals, and problems that require expressing measurements given in a larger unit in terms of a smaller unit. Represent measurement quantities using diagrams such as number line diagrams that feature a measurement scale.
MACC.4.NF.3.7 Compare two decimals to hundredths by reasoning about their size. Recognize that comparisons are valid only when the two decimals refer to the same whole. Record the results of comparisons with the symbols >,+, or <, and justify the conclusions, e.g., by using a visual model.
Each journal problem and printable page has an "I can…" statement to support your teaching and students' learning. Please see the table of contents for the listing of each page and journal problem included.
In my 4th grade classroom, I often introduce the lesson or new concept with a journal problem. We unpack the problem by Boxing the Math Words, Circling the Question, Underlining the Key Details of the Problem, and Envisioning the Problem.
Students may try to solve with a partner, or I may use the problem to teach my students. Then, we make notes or charts about the skill in our notebook. See an example below.. If you have any questions, concerns, comments, or feedback, please email me at butterfliesanddaydreams1@gmail.com
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Home > Relative Error > Relative Error Problems
# Relative Error Problems
## Contents
So: Absolute Error = 7.25 m2 Relative Error = 7.25 m2 = 0.151... 48 m2 Percentage Error = 15.1% (Which is not very accurate, is it?) Volume And volume If the object you are measuring could change size depending upon climatic conditions (swell or shrink), be sure to measure it under the same conditions each time. Take a stab at the following problems, then highlight the space after the colon (:) to see your answer. Once you understand the difference between Absolute and Relative Error, there is really no reason to do everything all by itself.
## Relative Error Calculator
The system returned: (22) Invalid argument The remote host or network may be down. After mixing and matching, her test tube contains 32 grams of substrate. Powered by Mediawiki. Please try the request again.
The Relative Error is the Absolute Error divided by the actual measurement. when measuring we don't know the actual value! Incidental energy/material loss, such as the little fluid left in the beaker after pouring, changes in temperature due to the environment, etc. Absolute And Relative Error In Numerical Methods There are two ways to measure errors commonly - absolute error and relative error.The absolute error tells about how much the approximate measured value varies from true value whereas the relative
paulcolor 30.514 προβολές 7:04 Calculating Percent Error Example Problem - Διάρκεια: 6:15. Relative Error Chemistry A resistor labeled as 240 Ω is actually 243.32753 Ω. By continuing to use our site, you agree to our cookie policy. http://chemistry.about.com/od/workedchemistryproblems/fl/Absolute-Error-and-Relative-Error-Calculation.htm Maribeth McAnally 7.245 προβολές 2:01 Propagation of Errors - Διάρκεια: 7:04.
Absolute error is positive. Relative Error Definition numericalmethodsguy 53.796 προβολές 9:48 How to Chemistry: Percent error - Διάρκεια: 4:39. Relative Error =|Measured−Actual|Actual{\displaystyle ={\frac {|{\mathrm {Measured} }-{\mathrm {Actual} }|}{\mathrm {Actual} }}} Multiply the whole thing by 100 to get Relative Error Percentage all at once.[9] 4 Always provide units as context. Use the same unites as the ones in your measurements.[4] 4 Practice with several examples.
## Relative Error Chemistry
Wolfram Language» Knowledge-based programming for everyone. get redirected here Theory Absolute Error Given an approximation a of a correct value x, we define the absolute value of the difference between the two values to be the absolute error. Relative Error Calculator continue reading below our video 10 Best Universities in the United States Absolute Error = Actual Value - Measured ValueFor example, if you know a procedure is supposed to yield 1.0 How To Calculate Relative Error In Physics Yes No Cookies make wikiHow better.
If you are measuring a football field and the absolute error is 1 cm, the error is virtually irrelevant. his comment is here tecmath 3.142.390 προβολές 10:42 How To Solve For Standard Error - Διάρκεια: 3:17. We will represent the absolute error by Eabs, therefore It is often sufficient to record only two decimal digits of the absolute error. When weighed on a defective scale, he weighed 38 pounds. (a) What is the percent of error in measurement of the defective scale to the nearest tenth? (b) If Millie, the Relative Error Formula
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## Session 3
### Recap: Ways to Design Algorithms
Last time, we played a fun little number game and then saw that there were at least three high-level approaches to designing an algorithm to play it:
• top-down, by decomposing the problem into sub-problems and solving the parts, perhaps recursively
• bottom-up, by solving smaller sub-problems first and then successively building larger solutions, until we reach a solution to the original problem
• zoom-in, by uncovering some natural invariants in the problem that point to a simpler or more direct solution
Using these techniques in order often works well. A top-down solution is natural and exposes sub-problems, but it sometimes results in solutions that are less efficient than possible or desired. A bottom-up approach can draw on the same sub-problems but combine them in a more efficient way. These two solutions sometimes give us enough experience with the problem to begin to see some relationships that we can use to zoom in and solve the problem more directly.
Keep in mind that an algorithm is a complete, well-defined set of instructions for solving a problem. The input and output of an algorithm must also be defined unambiguously. This means that the sort of instructions we can give a person for doing a task may not be an algorithm.
### Let's Score a Game!
Okay, that doesn't sound like as much fun as "Let's play a game", but...
The Winter Olympics will be starting in a few weeks. As I watch games such as ice hockey, I sometimes think about how a game score can develop. A team that wins 6-4 might have scored the first six goals of the game and then pulled its starters, or it might have played a tight game that went from 4-4 to 5-4 on a late goal, and then scored a meaningless goal into an open net as time expired.
(Why? The difference might matter to someone trying to determine the relative strengths of all the teams over a set of games. One of the first programs I ever wrote was to assign ratings to a set of competitors based on game results.)
Let's make our thinking more formal.
Two teams play a game of ice hockey. Let's call the teams t1 and t2. The game starts with a score of 0:0 and ends with a score of p1:p2, where t1 and t2 have scored p1 and p2 points, respectively. Given the final game score, we want to compute games(p1:p2), the number of different ways that the score could have been reached.
For example, there is only one way to reach the score 1:0 --
• 0:0, 1:0
There are three ways to reach 1:2 --
• 0:0, 1:0, 1:1, 1:2
• 0:0, 0:1, 1:1, 1:2
• 0:0, 0:1, 0:2, 1:2
There are ten ways to reach a final score of 3:2. I leave listing them as an exercise for the reader. :-)
### Design Exercise
Design three algorithms for computing the number of different games:
• an algorithm with a top-down approach
• an algorithm with a bottom-up approach
• an algorithm that zooms-in on a solution
When you have three algorithms, or feel the need to put one on the back burner, characterize each algorithm's run-time behavior in terms of time and space.
Feel free to work in groups of two or three, if you like. Don't share ideas across groups until I say to.
... lots of fun ensues ....
### A Top-Down Approach
We can use a divide-and-conquer approach to this problem, much as we did for the End Game last time. Each score p1:p2 can only be reached from two possible scores:
• (p1-1):p2
• p1:(p2-1)
If t1 scores a goal from the first, then we reach our target score. If t2 scores a goal from the second, we also do.
Using this knowledge, we can define a straightforward relationship:
• For any p, games(p:0) = 1, and games(0:p) = 1.
• For any p1 and p2 greater than 0, games(p1:p2) = games( (p1-1):p2 ) + games( p1:(p2-1) ).
Let's try this out for the game score 1:2...
This definition is inductive, because the function games is defined in terms of itself. The first bullet is the base case, which defines directly answers for two specific problems. The second is the inductive case, which defines the answer to a general problem in terms of one or more simpler cases.
Inductive definitions are a powerful form of shorthand. Though many of you cringe at the idea of an inductive proof, most of you think about certain problems in this way with ease. Reasoning inductively is something humans do automatically, and it turns out to be a useful technique for defining algorithms, too. (Writing a proof requires a few proof-making skills, along with more precision -- and thus more discipline. Just like writing a program.)
Notice, too, that the definition of games is a mathematical function, not a computational function. We may well want to implement it in a program, but for now it defines a set of ordered pairs.
This top-down algorithm works and is quite elegant. But how well does it perform?
Like the top-down algorithm for the End Game, this algorithm recomputes many values of games() multiple times. Look at the next level of that tree...
Does that look familiar? It should. This is a common result for problems that involve combinations of things.
More formally, the algorithm decomposes every score p1:p2 into two sub-problems. Thus, the solution is exponential in the sum p1+p2. This means that the number of recursive calls it makes grows exponentially with changes in p1 and p2.
Can we avoid the recomputations, and thus perhaps some of the exponential load of the algorithm, by eliminating the recursion?
### A Bottom-Up Approach
To find games(p1:p2), we need to compute games(n1:n2) for all n1 < p1 and all n2 < p2. But there are only (p1+1) * (p2+1) possible scores! This number doesn't grow exponentially, but multiplicatively. How we can find a solution more quickly?
Let's compute each value once, and remember that result. That is, instead of working from p1:p2 "down to" 0:0, we could work from 0:0 "up to" p1:p2. This is the sort of approach we called dynamic programming last time.
First, let's consider our base case. We know the values for
``` 1:0, 2:0, 3:0, ... p1:0
```
and
``` 0:1, 0:2, 0:3, ... 0:p2
```
Let's store these values along two of the edges of a (p1+1) * (p2+1) matrix:
Now, let's consider our inductive case. It applies to every p1 and p2 that are greater than 0. We can fill in the remaining cells of the matrix, one row at a time, using the values in the cells immediately above and to the left!
We fill this in row-by-row and ultimately fill in the entire table:
We could speed this process up even further. Notice that games(2:1) = games(1:2). That is, the matrix is symmetric along the 0:0 -p1:p2 diagonal. So we need only compute the values along the diagonal and above, and then copy the corresponding values into the cells below the diagonal.
Even better: If we care only about the final answer, games(p1:p2), we don't even have to remember the whole matrix! Each new entry requires only values from the current row and the preceding row, so we can keep the space of table linear in the larger of p1 and p2.
Wow. We have an algorithm whose time complexity is O( p1p2 ) and whose space complexity is O( max(p1, p2) ). Wow.
Should we be so greedy? If both p1 and p2 are very large, we may have reason to care. Of course, that isn't likely to happen in an ice hockey game, but when computing the number of paths in a computer network it might!
But can we do even better?
### A "Zoom-In" Approach
On the way from 0:0 to p1:p2, the score increases exactly p1 + p2 times -- once for each goal scored. So, we might view a path from 0:0 to p1:p2 as a (p1+p2)-element sequence of scores where only one of the scores changes between elements. That's how we've been thinking thus far.
Alternatively, though, we could view the game path as a (p1+p2)-element sequence of t1's and t2's, with each item corresponding to which team scored next.
For 1:2, we have three possible sequences:
Aha! Each sequence has exactly p1 t1's and exactly p2 t2s. The different game sequences interleave these values in all possible combinations. So the number of different game paths is simply:
``` (p1 + p2)!
----------
p1! X p2!
```
No recursion. No loop. No table. O(1) space and O( max(p1,p2) ) time to compute the factorials. Now that's zooming in.
It really does help computer scientists to know a little math. You never know when the chance to use it will pop up.
### Choosing a Design Approach
Sometimes, we can find an invariant solution directly. Doing so depends on our background knowledge and our experience with different classes of problems.
Sometimes, it is helpful to work through the approaches in order. Working top-down first and then bottom-up helps you come to understand the problem and its subproblems before trying to create an efficient implementation. Knowledge and understanding usually help us create efficient solutions faster, not slower.
Working top-down and bottom-up often gives us the experience with the problem that we need to find a "zoom-in", direct solution to a new problem.
Exposure, familiarity, experience.
... analyzing the resulting algorithms ...
• informal consideration of how the tree or table grows
• formal count of operations (soon!)
... best-case, worst-case, average-case ...
• Best- and worst-case performance are usually defined in terms of a single problem, or a single class of essentially identical problems. This is one reason documentation of the quickselect algorithm referred to a median-of-3 killer. That is a specific problem that causes the median-of-3 approach to perform its absolute worst.
• Average-case: typical input, expected input, or even the actual average over all possible inputs.
Deciding which algorithm to use is often more subtle than it may at first appear. For example, is there a case in which our top-down algorithm for computing games() executes faster than the bottom-up agorithm, or even the zoom-in algorithm? Think about 0:1000...
This also highlights the difference between best-case and average-case. Unfortunately, the average case for the top-down algorithm is also its worst case!
### Wrap Up
• Homework -- none yet.
Eugene Wallingford ..... wallingf@cs.uni.edu ..... January 21, 2014
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# Evaluation Integral: Method of Partial Fractions - Can the system of equations always be solved?
I've been studying the method of partial fractions for evaluation integrals. So far every example and exercise have been fairly straight forward, but I still have some unanswered questions:
1) The standard table for doing method of partial fractions - I guess some clever person has come up with this table and verified it works. However is this the only way of doing partial fractions, or are there other tables / methods around ?
2) The system of equations to be solved from using method of partial fractions - Does this system always has a solution ? So far I've never experienced a system with no solutions, but that is probably because classroom exercises are solvable. Can one come up with a system of equations not solvable for constants $A$, $B$, ... $K$, such that method of partial fractions fails ?
In calculus course, and in engineering-math courses doing Lapace transforms, the method of partial fractions is typically stated without the proof that the equations can always be solved. After all, once you solve the system of equations, you can check your answer without needing that theory.
Such a proof could be done later in a linear algebra course (showing a certain determinant does not vanish), but I do not recall seeing that in a modern linear algebra textbook. So maybe look in textbooks from before 1950.
Such a proof can also be done in a complex analysis textbook: in that case (using complex numbers) we do not need quadratic factors in our denominator, and things work out well. This may be in a chapter on Laurent series. You find the principal part at each pole (including infinity), then after you subtract those, you have a bounded entire function, hence a constant.
• Thanks for your answer. Could you tell me if the constants are unique in the partial fraction decomposition ? – Shuzheng Nov 30 '13 at 9:50
• The constants are unique. – GEdgar Nov 30 '13 at 16:47
• Thanks, this can be proved using that polynomial division is unique in a field ? Like stated here mathforum.org/library/drmath/view/51687.html etc ? – Shuzheng Nov 30 '13 at 17:51
It might be worth taking a look at a (fairly old) book by G. H. Hardy on Integration of Functions of a Single Variable. The book is freely available on Project Gutenberg
On pages 11 onwards he discusses rational functions and partial fractions and states:
"the integral of any rational function is an elementary function
which is rational save for the possible presence of logarithms of
rational functions"
Basically he is deducing this from the fundamental theorem of algebra.
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# [Tutor] I don't quite understand bitwise
Alan Gauld alan.gauld at yahoo.co.uk
Tue Dec 15 19:08:41 EST 2020
```On 15/12/2020 19:11, nathan tech wrote:
> So I started to research how the x|y format works and came across
> bitwise operators which is apparrently what is meant when x|y|z is used.
You can find a discription of nitwide operators and their use in
programming in the Functional Programming topic in my tutorial.
However long story short is:
if you do A|B
then any bit that is 1 in either A or B will be 1 in the output.
if you do A&B
then any bit that is 0 in either A or B will be 0 in the output.
Thus A|B combines the 1s in both entities - this is usually
what you want when the entities represent sets of options
such as application options and user defined options.
So if you want the total number of options set you can do
options_var = user|application
# now use the combined options_var
And A&B can be used to select a subset of options by making B
a "mask" - that is a known bit pattern.
Thus if you want to know if bits 2 and 4 of A are set to 1 do this:
B = 0b1010
res = A&B
if res == B: print("Both options are set")
else: print("One, or both, options are not set")
Finally, | can be used to accumulate results over several
operations. You used to see this before try/except
style error handling became common:
errors = 0
errors |= operation1(myObject)
errors |= operation2(myObject)
errors |= operation3(myObject)
errors |= operation4(myObject)
if errors:
# something went wrong, deal with it!
else:
# All OK so now use myObject
A |= B is f course the same as
A = A|B
And if all operations return zero on success then any
other result will have some 1s in it, and these will
be transferred to errors.
For bonus points the last place you might see bitwise
operations is in cryptography.
There the exclusive OR operator is used to encode data
in such a way that it can be recovered later by applying
#At transmitter
key = 42
data = "some string here"
cipher = ''.join(ch^key for ch in data)
print (cipher) # garbage...
# transmit cipher...
key = 42 # same as transmitter
plain_text = ''.join(ch|key for ch in message)
print(plain_text) # prints "some string here"
Of course that easy to crack but many real world cipher
systems are based on layers of algorithms on top of
a similar basic concept.
To conclude, the reason all this works is because of
the definitions of the operations as described by
their truth tables:
A B A|B A&B A^B
0 0 0 0 0
0 1 1 0 1
1 0 1 0 1
1 1 1 1 0
So
A|B is 1 when any of A or B is 1
A&B is only 1 if both A and B are 1
A^B(exclusive or) is 1 if A and B are different.
Try them in the >>> prompt, see if you can predict the results before
hitting return:
>>> A = 0b1010
>>> B = 0b1100
>>> bin(A|B)
????
>>> bin(A&B)
????
>>> bin(A^B)
????
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos
```
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# Thread: Transitive Set Theory Problem
1. ## Transitive Set Theory Problem
Show that for any x there exists a transitive y such that $y \notin x$
x and y are sets.
Here's my idea: Take a set x and take it's cardinality. Express this number in terms of power sets of the empty set. This would give a transitive set every time.
The only problem is if the set we start off with is in terms of power sets of the empty set. In this case, map this to the cardinality of it's set +1 and express that number in terms of power sets of the empty set.
Is this a good way of proving this?
2. Here's an easy solution:
Since x is a set and ON is a proper class, there exists an ordinal $\alpha$ with $\alpha \notin x$.
3. Wow, one line solutions usually make sense.
However, we haven't been lectured about ON, proper classes or ordinals.
Can I just ask what your ON actually is? I think i'll go and read about proper classes or ordinals, it might help me understand what you mean.
4. An ordinal is a set that is transitive and well-ordered. Here are some simple examples of ordinals:
$0=\emptyset$
$1=\{ \emptyset \} = \{ 0\}$
$2=\{ 0,1 \}$
...
$\omega = \{ 0,1,2,... \}$
$\omega +1= \{ 0,1,2,... ,\omega \}$
ON is the class of all ordinals.
ON is too big to be a set, thus it is a proper class.
The ordinals should be covered in a basic set theory course. They are fundamental.
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# Kinetic energy KEep moving!. Kinetic Energy Kinetic comes from the Greek work ‘to move’ So Kinetic Energy is Moving energy If we are talking about.
## Presentation on theme: "Kinetic energy KEep moving!. Kinetic Energy Kinetic comes from the Greek work ‘to move’ So Kinetic Energy is Moving energy If we are talking about."— Presentation transcript:
Kinetic energy KEep moving!
Kinetic Energy Kinetic comes from the Greek work ‘to move’ So Kinetic Energy is Moving energy If we are talking about energy of something moving what factors exist that would give something more or less energy?
Kinetic Energy KE = ½ mv 2 Kinetic Energy = ½ the mass times the velocity squared. http://youtu.be/vl4g7T5gw1M http://youtu.be/vl4g7T5gw1M
Lets try some out A man of 90kg is traveling in a car at 30 m/s. how much KE does the man have?
Now you try Boxy with 7 kg of mass has 175 J of energy, how fast is he moving?
Download ppt "Kinetic energy KEep moving!. Kinetic Energy Kinetic comes from the Greek work ‘to move’ So Kinetic Energy is Moving energy If we are talking about."
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# UVA PHYS 635 - FREE, DAMPED, AND FORCED OSCILLATIONS (18 pages)
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## FREE, DAMPED, AND FORCED OSCILLATIONS
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## FREE, DAMPED, AND FORCED OSCILLATIONS
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Course:
Phys 635 - Harmonic Motion and the Pendulum
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175 Name Date Partners LAB 11 FREE DAMPED AND FORCED OSCILLATIONS OBJECTIVES To understand the free oscillations of a mass and spring To understand how energy is shared between potential and kinetic energy To understand the effects of damping on oscillatory motion To understand how driving forces dominate oscillatory motion To understand the effects of resonance in oscillatory motion OVERVIEW You have already studied the motion of a mass moving on the end of a spring We understand that the concept of mechanical energy applies and the energy is shared back and forth between the potential and kinetic energy We know how to find the angular frequency of the mass motion if we know the spring constant We will examine in this lab the mass spring system again but this time we will have two springs each having one end fixed on either side of the mass We will let the mass slide on an air track that has very little friction We first will study the free oscillation of this system Then we will use magnets to add some damping and study the motion as a function of the damping coefficient Finally we will hook up a motor that will oscillate the system at practically any frequency we choose We will find that this motion leads to several interesting results including wild oscillations Harmonic motions are ubiquitous in physics and engineering we often observe them in mechanical and electrical systems The same general principles apply for atomic molecular and other oscillators so once you understand harmonic motion in one guise you have the basis for understanding an immense range of phenomena INVESTIGATION 1 FREE OSCILLATIONS An example of a simple harmonic oscillator is a mass m which moves on the x axis and is attached to a spring with its equilibrium position at x 0 by definition When the mass is moved from its equilibrium position the restoring force of the spring tends to bring it back to the equilibrium position The spring force is given by Fspring kx 1 where k is the spring
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# The electric dipole of moment $\begin{array}{*{20}{c}} {\overrightarrow p }& = &{( - \widehat i - 3\widehat j + 2\widehat k) \times {{10}^{ - 29}}Cm} \end{array}$is at the origin (0,0,0). The electric field due to this dipole at $\begin{array}{*{20}{c}} {\overrightarrow r }& = &{(\widehat i + 3\widehat j + 5\widehat k)} \end{array}$is parallel to [Note that $\begin{array}{*{20}{c}} {\overrightarrow {r.} \overrightarrow p }& = &0 \end{array}$] A) $\widehat i - 3\widehat j - 2\widehat k$B) $- \widehat i - 3\widehat j + 2\widehat k$C ) $\widehat i + 3\widehat j - 2\widehat k$D) $- \widehat i + 3\widehat j - 2\widehat k$
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Hint: In this question, we have given a condition:
($\begin{array}{*{20}{c}} {\overrightarrow {r.} \overrightarrow p }& = &0 \end{array}$),
Therefore, According to the given condition we will get to know that the $\overrightarrow p$and position vector $\overrightarrow r$will be perpendicular to each other. Due to the opposite direction of the electric field to the electric dipole moment, the electric field vector will be the same as the electric dipole moment but in a negative sign. Hence, we will get a suitable answer.
Basically, the direction of the electric dipole moment is from negative charge to positive charge while the direction of the electric field is from positive charge to negative charge. In other words, we can say that the direction of the electric field is opposite to the electric dipole moment. Hence, as per the given note $\begin{array}{*{20}{c}} {\overrightarrow {r.} \overrightarrow p }& = &0 \end{array}$, we can conclude that the electric dipole moment and electric field will be perpendicular to each other but will be in opposite directions.
Electric dipole moment vector and position vector are given,
$\begin{array}{*{20}{c}} { \Rightarrow \overrightarrow p }& = &{( - \widehat i - 3\widehat j + 2\widehat k) \times {{10}^{ - 29}}Cm} \end{array}$ and position vector $\begin{array}{*{20}{c}} {\overrightarrow r }& = &{(\widehat i + 3\widehat j + 5\widehat k)} \end{array}$
Therefore, As per given the condition, we can write that
$\begin{array}{*{20}{c}} { \Rightarrow \overrightarrow {r.} \overrightarrow p }& = &0 \end{array}$
The $\overrightarrow p$and $\overrightarrow r$are perpendicular to each other. It means that the electric field will also be perpendicular to the position vector $\overrightarrow r$.
$\begin{array}{*{20}{c}} { \Rightarrow (\widehat i + 3\widehat j + 5\widehat k).( - \widehat i - 3\widehat j + 2\widehat k)}& = &0 \end{array}$
Now, we know that the electric field is directly proportional to the electric dipole moment, but will be in the opposite direction. There will be a constant. Whose value will be greater than 0.
$\begin{array}{*{20}{c}} { \Rightarrow \overrightarrow E }& = &{ - \lambda \overrightarrow p } \end{array}$
Here, a negative sign indicates that the electric field is in the opposite direction to the electric dipole moment. Therefore,
$\begin{array}{*{20}{c}} { \Rightarrow \overrightarrow E }& = &{ - \lambda ( - \widehat i - 3\widehat j + 2\widehat k) \times {{10}^{ - 29}}} \end{array}$
$\begin{array}{*{20}{c}} { \Rightarrow \overrightarrow E }& = &{\lambda (\widehat i + 3\widehat j - 2\widehat k) \times {{10}^{ - 29}}} \end{array}$
Therefore,$\overrightarrow E$will be parallel to the $(\widehat i + 3\widehat j - 2\widehat k)$
Now, the final answer is $(\widehat i + 3\widehat j - 2\widehat k)$. So, the correct option is C.
Note: In this question, the first point is to keep in mind that the $\begin{array}{*{20}{c}} \lambda & > &0 \end{array}$, Where, $\lambda$is constant which is the replacement of the proportionality sign.
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Solid States Devices > TPU replications
The answer has been given, but it was deliberately removed.
<< < (5/6) > >>
Bruce_TPU:
Hello ALL,
Power does not "magically" appear because of how a coil is wound or even with what frequencies it is pulsed.
To make power you must be "intentional". First you must identify a form of additional means for gain outside of the norm.
I have told these guys it is ELECTRONS OFF OF THE WIRE.
You must figure out HOW to get the electrons off of the wire. I have also told them HOW and WHY this is done. I actually used "Math" to figure it out and then proved it out long ago in the lab.
You must figure out HOW to move the electrons. Find out WHAT they are attracted to and they will move for you, at a rapid rate. I told how to do this.
You must figure out HOW to collect MORE electrons to make a BIGGER magnet. Bigger magnet equals more EMF generated. I have NOT told this.
You must lastly figure out HOW to use ALL OF THAT to generate EDDY CURRENTS that will give you your greater output. This part is a "work in progress".
Pulsing coils is a dead end without having done the leg work above. That is, for anyone serious about this stuff. Otherwise, it's just fun to build coils, I know.
2+2 will always equal 4.
Cheers,
Bruce
sm0ky2:
Take the theoretical situation of two series-pancake coils
With a “solenoid” between them.
compare the EM field to a normal transformer
Keeping the impedance ratios of each equal.
How is the field different ?
Why is the field different ?
now test them both in the real world
How is the output different?
Why is the output different?
wattsup:
@Turbo
hahahahaha. SM has been grueling, always has been.
Let me post this diagram I made back in 2011. It pushes the toroid and speaker coil idea to the extreme.
"The answer to the TPU was given yesterday, but it was quickly removed by BRUCE_TPU because he has issues with his EGO."
Don't know why you are saying Bruce showed the TPU. None of that is consistent with all the other TPUs and SM surely did not develop more then one principle of OU operation so......... how does it link to the OTPU or FTPU? Cannot. Where can you see those three coils in the MTPU cutaway pie piece? Cannot.
The diagram f*&kbruce.png (not a nice name man) cannot work. It's impossible. So many winds with each turn having its share of cancellation potential will never work like shown. Those arrows showing something as being directional is just wishful thinking. Copper wire does not convey like that. The whole OU community is stuck at level 1 since 10 years now not one worthy principle has been developed and not one out of the box method has been widely tried and experimented for its variables.
Could say much more and push guys into new ways but it always seems to be difficult for OUers to think out of the box. It's as if we need permission from Professor Standard EE to leave the confines of mediocrity. Imagine the guys who worked for years and years on vacuum tubes if they stuck with their daily standards, there would be no tubes to show us that "coupling" is a nucleic proximity challenge that those guys succeeded to crack wide open. Special conductive materials, very particular topologies. Those masters of coupling are now on the endangered species list.
The guys working on ICs are working with coupling as well but atom to atom. Again ask one of those guys how the electron plays into it and you will get a zero sum answer. Imagine if a memory chip had electrons how would it be possible to turn sections on and off when the whole thing would be electron based. It's like trying to make a Popsicle while underwater.
wattsup
sm0ky2:
If I charge two Leyden jars to opposite polarity
I have a “net zero sum”
The gate is in between them, controlled by the voltage
This is no different.
A transistor, a diode, cpu
It’s up to you how to charge it
With a semiconductor we insert an electron
which then goes through a path
In an electric system we insert many electrons
And we insert many (holes/positrons/lacks of electrons)
The semiconductor cut-off voltage of air is very high
and it is not nearly as consistent as a lab made doped silicone.
With modern electricity we flow one half of the current.
Because to use both, cancels each other out with an explosion
of energy.
All sorts of heat, rf, visible light, emf, sound, physical vibration
The P and the N
Or the N and the P
They attract each other like magnets
resinous/ vitreous
This is why we can use the same process to turn light, heat, rf
Back into electricity
to have an electron missing (positron)
is just as powerful as having an extra electron
have something that’s missing way too many electrons
and create a situation that feeds it.
We do this when we make batteries
it’s all the same stuff
make the air have less electrons than a coil
Or a coil have less electrons than the air you just filled
Magluvin:
Got some insight that may have merit with the 3 coils I made that Turbo described. Will be trying it this weekend.
Mags
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# Decimals | Skills Poster
Subjects
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### PRODUCT DESCRIPTION
This poster is intended to lay the groundwork for a very basic understanding of how Decimals work, without relying on a comparison to Fractions or Percentages at the very first hurdle.
We think that a student should understand the principles of decimals in relation to numbers which are a power of 10, before comparing and contrasting them with Fractions and Percentages. This way, an understanding of fraction conversion can become even stronger later down the line.
You may also like:
Fraction, Decimal and Percentage Flash Cards
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At the beginning of this week in pre calc 11 we learned about Vertex/Standard form. This is personally my favourite form because it provides more information about us graphing the quadratic function than the quadratic or factored forms give us.
Vertex/Standard form looks like this:
$y=a(x-p)^2+q$
This form provides you with the following information about a parabola:
a:
1- tells us whether a parabola is opening up or down. If it’s a positive # than it’s opening up, and if it’s a negative # than it’s opening down or being “reflected”.
2- determines the size of our parabola, whether it is stretched or compressed. If a is greater than one then the parabola is stretched, if a is less than one then the parabola is compressed.
p:
1- tells us the horizontal translation of a parabola (whether the vertex is moving left or right).
2 – tells us what our line of symmetry is.
q:
1- tells us the vertical translation of a parabola (whether the vertex is moving up or down).
p & q:
1- the p and q are very important becuase they let us know our vertex *the vertex is the most important part of any parabola. The p tells us our x co-ordinate and the q tells us our y co-ordinate for when we are graphing a quadratic function.
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# Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115 § 1.8 The Derivative as a Rate of Change.
## Presentation on theme: "Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115 § 1.8 The Derivative as a Rate of Change."— Presentation transcript:
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115 § 1.8 The Derivative as a Rate of Change
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 2 of 115 Average Rate of Change Instantaneous Rate of Change Average Velocity Position, Velocity, and Acceleration Approximating the Change in a Function Section Outline
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 3 of 115 Average Rate of Change
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 4 of 115 Average Rate of ChangeSOLUTION Suppose that f (x) = -6/x. What is the average rate of change of f (x) over the interval 1 to 1.2? EXAMPLE The average rate of change over the interval is
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 5 of 115 Instantaneous Rate of Change
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 6 of 115 Instantaneous Rate of ChangeSOLUTION Suppose that f (x) = -6/x. What is the (instantaneous) rate of change of f (x) when x = 1? EXAMPLE The rate of change of f (x) at x = 1 is equal to. We have That is, the rate of change is 6 units per unit change in x.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 7 of 115 Average & Instantaneous Rates of Change Refer to the figure below, where f (t) is the percentage yield (interest rate) on a 3-month T-bill (U.S. Treasury bill) t years after January 1, 1980. EXAMPLE (a) What was the average rate of change in the yield from January 1, 1981 to January 1, 1985? (b) How fast was the percentage yield rising on January 1, 1989? (c) Was the percentage yield rising faster on January 1, 1980 or January 1, 1989?
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 8 of 115 Average & Instantaneous Rates of ChangeSOLUTION (a) To determine the average rate of change in the yield from January 1, 1981 to January 1, 1985, we must first determine the coordinates of the two points that correspond to the two given dates. They are (1, 14) and (5, 7). Now we use the average rate of change formula. CONTINUED Therefore, the average rate of change in the yield from January 1, 1981 to January 1, 1985 is -7/4. (b) To determine how fast the percentage yield was rising on January 1, 1989, we must determine the instantaneous rate of change of f (t) when t = 9 (corresponding to January 1, 1989). This means that we must find the slope of the tangent line to the graph of f (t) where t = 9. The tangent line is on the graph and so we need only determine any two points on the tangent line. Using the coordinates of these two points, we will calculate the slope of the tangent line and that will be the instantaneous rate of change that we seek. Notice that two of the points on the tangent line are (5, 5) and (11, 10). Using these points we will now calculate the slope of the tangent line.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 9 of 115 Average & Instantaneous Rates of ChangeCONTINUED Therefore, the rate at which the percentage yield was rising on January 1, 1989 was 5/6. (c) To determine if the percentage yield was rising faster on January 1, 1980 or January 1, 1989, we would need to know the slopes of the tangent lines corresponding to t = 0 (January 1, 1980) and t = 9 (January 1, 1989). Although we already have this information for t = 9 (see part (b)), we do not yet have this information for t = 0. Therefore, we would first need to draw a tangent line to the graph corresponding to t = 0. This is done below.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 10 of 115 Average & Instantaneous Rates of ChangeCONTINUED Obviously, finding the coordinates of two points on this tangent line might prove a little difficult. However, notice that the slopes of the two tangent lines (all we’re really interested in are their slopes) are not remotely similar (that is, the tangent lines are not close to being parallel). Therefore, in this circumstance, it would be sufficiently appropriate to notice that the blue tangent line (corresponding to t = 0) has a steeper slope and therefore the rate of change was greater on January 1, 1980 than it was on January 1, 1989. NOTE: Use this technique of “eye-balling” a graph only when absolutely necessary and only with great care.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 11 of 115 Average Velocity DefinitionExample Average Velocity: Given a position function s(t), the average velocity from time t = a to t = a + h is Suppose a car is 3 miles from its starting point after 5 minutes and 7 miles from its starting point after an additional 6 minutes (after a total of 11 minutes). The average velocity of the car between the two given locations is miles per minute where a = 5 and h = 6.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 12 of 115 Position, Velocity & Acceleration s(t) is the position function, v(t) is the velocity function, and a(t) is the acceleration function.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 13 of 115 Position, Velocity & AccelerationSOLUTION A toy rocket fired straight up into the air has height s(t) = 160t – 16t 2 feet after t seconds. EXAMPLE (a) What is the rocket’s initial velocity (when t = 0)? (b) What is the velocity after 2 seconds? (c) What is the acceleration when t = 3? (d) At what time will the rocket hit the ground? (e) At what velocity will the rocket be traveling just as it smashes into the ground? (a) To determine what the rocket’s initial velocity is, we must first have a velocity function and then evaluate t = 0 in that function. This is the given position function. Differentiate to get the velocity function.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 14 of 115 Position, Velocity & Acceleration Therefore, the initial velocity of the rocket is 160 feet per second. Now replace t with 0 and evaluate. CONTINUED (b) To determine the velocity after 2 seconds we evaluate v(2). This is the velocity function. Replace t with 2 and evaluate. Therefore, the velocity of the rocket after 2 seconds is 96 feet per second. (c) To determine the acceleration when t = 3, we must first find the acceleration function. This is the velocity function. Differentiate to get the acceleration function. Since the acceleration function is a constant function, the acceleration of the rocket is a constant -32 ft/s 2. Therefore, the acceleration when t = 3 is -32ft/s 2.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 15 of 115 Position, Velocity & AccelerationCONTINUED Therefore, the rocket will be 0 feet above the ground at times t = 0 and t = 10. t = 0 corresponds to when the rocket first began its flight, so that would not be the solution. So, the rocket hit the ground after 10 seconds, when t – 10 = 0. (d) To determine at what time the rocket will hit the ground we must determine what we know about the position, velocity, or acceleration of the rocket when the rocket hits the ground. We know that at the time of impact, the position of the rocket will be 0 feet above the ground. Therefore, we will use the position function and replace s(t) with 0. This is the given position function. Replace s(t) with 0. Factor 16 out of both terms on the right and divide both sides by 16. Factor. Solve for t.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 16 of 115 Position, Velocity & AccelerationCONTINUED Therefore, when the rocket hits the ground, it will be have a velocity of -160 ft/s. That is, it will be traveling 160 ft/s in the downward direction. (e) To determine at what velocity the rocket will be traveling just as it smashes into the ground, we must use the velocity function. The question is, what do we use for t? From part (d), we know that the rocket will hit the ground at t = 10 seconds. Therefore, we will find v(10). This is the velocity function. Replace t with 10 and evaluate.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 17 of 115 Approximating the Change in a Function
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 18 of 115 Approximating the Change in a FunctionSOLUTION Suppose 5 mg of a drug is injected into the bloodstream. Let f (t) be the amount present in the bloodstream after t hours. Interpret f (3) = 2 and Estimate the number of milligrams of the drug in the bloodstream after 3½ hours. EXAMPLE First, f (3) = 2 means that after 3 hours, 2 milligrams of the drug still remain in the bloodstream. Next, means that after 3 hours, the rate at which the amount of the drug is diminishing (because of the minus sign) within the bloodstream is ½ of a milligram per 1 hour. To estimate the number of milligrams of the drug in the bloodstream after 3½ hours, we will use the formula where a = 3 and h = ½.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 19 of 115 Approximating the Change in a Function Therefore, approximate number of milligrams of the drug in the bloodstream after 3½ hours is 1.75. CONTINUED
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The Suitcase Chute
## Question:
A 40 lb suitcase slides from rest down a 20 foot smooth ramp with a incline of 30 degrees. The foot of the ramp is 4 feet above the floor. At what horizontal distance from the foot of the ramp does the suitcase hit the floor and how long does the total trip take?
Break the problem into two parts, what happens on the slide and what happens after it leaves the slide.
The acceleration(a) of the suitcase along the slide is the acceleration due to gravity times the sine of the incline angle, or a=32.2*sin(30)=16.1f/s/s. Starting with zero velocity, the suitcase distance(d) from the top of the slide is
d=1/2*a*t2.
At d=20ft,
40=16.1*t2, or t2=2.48s2, or t=1.58s.
Neglecting the dimensions of the suitcase so we do not have to deal with the rotation that occurs when the leading edge of the suitcase clears the slide by 1/2 its length, the velocity at the instant the suitcase leaves the slide is 16.1f/s/s*1.58s=25.44 f/s directed 30 degrees below horizontal. The vertical component of that velocity is
Vv=25.44*sin(30)=12.72f/s.
The horizontal component is
Vh=25.44*cos(30)=22.03f/s.
The vertical distance(Dv) fallen in time, t, after leaving the slide is
Dv=12.72*t+16.1*t2=4.
The time to hit the floor then is given by
4=12.72*t+16.1*t2, or 16.1*t2+12.72*t-4=0.
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# Quant Questions for SSC CHSL and RRB NTPC
Q4. In a coconut grove, (x + 2) trees yield 60 nuts per year, x trees yield 120 nuts per year, and (x – 2) trees yield 180 nuts per year. If the average yield per year per tree be 100, find the value of x.
(a) 4
(b) 2
(c) 8
(d) 6
Q5. Visitors to a show were charged Rs. 15.00 each on the first day, Rs. 7.50 on the second, Rs. 2.50 on the third day. Visitors total attendance for three days were in the ratio 2: 5: 13. Find out the average charge per visitor for the entire show.
(a) Rs. 7
(b) Rs. 5
(c) Rs. 9
(d) Rs. 11
Q6. A man whose bowling average is 12.4, takes 5 wickets for 26 runs and, thereby decreases his average by 0.4. The number of wickets, taken by him, before his last match is:
(a) 85
(b) 78
(c) 72
(d) 84
Q7. If the digit in the units place of a two-digit number is halved and the digit in the tens place is doubled, the number thus obtained is equal to the number obtained by interchanging the digits. Which of the following is definitely true?
(a) Digits in the units place and the tens place are equal.
(b) Sum of the digits is a two-digit number.
(c) Digit is the units place is half of the digit in the tens place.
(d) Digit in the units place is twice the digit in the tens place.
Q8. If the two digits of the age of Mr. Manoj are reversed then, the new age so obtained is the age of his wife. 1/11 of the sum of their ages is equal to the difference between their ages. If Mr. Manoj is elder than his wife then find the difference between their ages?
(a) Cannot be determined
(b) 10 years
(c) 8 years
(d) 9 years
Q10. The speeds of three cars in the ratio 5: 4: 6. The ratio between the time taken by them to travel the same distance is:
(a) 5 : 4 : 6
(b) 6 : 4 : 5
(c) 10 : 12 : 15
(d) 12 : 15 : 10
Q11. Zinc and copper are melted together in the ratio 9 : 11. What is the weight of melted mixture, if 28.8 kg of zinc has been consumed in it?
(a) 58 kg
(b) 60 kg
(c) 64 kg
(d) 70 kg
Q12. The price of a Maruti car rises by 30% while the sales of the car comes down by 20%. What is the percentage change in the total revenue?
(a) -4%
(b) -2%
(c) +4%
(d) +2%
Q13. A person who has a certain amount with him goes to market. He can buy 40 oranges or 50 mangoes. He retains 10% of the amount for taxi fares and buys 20 mangoes and of the balance, he purchases oranges. Number of oranges he can purchase is
(a) 36
(b) 40
(c) 15
(d) 20
Q14. The length of a room floor exceeds its breadth by 20 m. The area of the floor remains unaltered when the length is decreased by 10 m but the breadth is increased by 5 m. The area of the floor (in square metres) is:
(a) 280
(b) 325
(c) 300
(d) 420
Q15. The radii of two circle are 5 cm and 3 cm, the distance between their centre is 24 cm. Then the length of the transverse common tangent is
(a) 16 cm
(b) 15√2 cm
(c) 16 √2 cm
(d) 15 cm
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## ››Convert cubic metre to femtoliter
cubic meter femtolitre
How many cubic meter in 1 femtolitre? The answer is 1.0E-18.
We assume you are converting between cubic metre and femtoliter.
You can view more details on each measurement unit:
cubic meter or femtolitre
The SI derived unit for volume is the cubic meter.
1 cubic meter is equal to 1.0E+18 femtolitre.
Note that rounding errors may occur, so always check the results.
Use this page to learn how to convert between cubic meters and femtoliters.
Type in your own numbers in the form to convert the units!
## ››Quick conversion chart of cubic meter to femtolitre
1 cubic meter to femtolitre = 1.0E+18 femtolitre
2 cubic meter to femtolitre = 2.0E+18 femtolitre
3 cubic meter to femtolitre = 3.0E+18 femtolitre
4 cubic meter to femtolitre = 4.0E+18 femtolitre
5 cubic meter to femtolitre = 5.0E+18 femtolitre
6 cubic meter to femtolitre = 6.0E+18 femtolitre
7 cubic meter to femtolitre = 7.0E+18 femtolitre
8 cubic meter to femtolitre = 8.0E+18 femtolitre
9 cubic meter to femtolitre = 9.0E+18 femtolitre
10 cubic meter to femtolitre = 1.0E+19 femtolitre
## ››Want other units?
You can do the reverse unit conversion from femtolitre to cubic meter, or enter any two units below:
## Enter two units to convert
From: To:
## ››Definition: Cubic meter
The cubic metre (symbol m³) is the SI derived unit of volume. It is the volume of a cube with edges one metre in length. Older equivalents were the stere and the kilolitre.
## ››Definition: Femtoliter
The SI prefix "femto" represents a factor of 10-15, or in exponential notation, 1E-15.
So 1 femtoliter = 10-15 liter.
## ››Metric conversions and more
ConvertUnits.com provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types. Examples include mm, inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, metres squared, grams, moles, feet per second, and many more!
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# Search by Topic
#### Resources tagged with Working systematically similar to Shunting Puzzle:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
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### Hexpentas
##### Stage: 1 and 2 Challenge Level:
How many different ways can you find of fitting five hexagons together? How will you know you have found all the ways?
### Chocs, Mints, Jellies
##### Stage: 2 Challenge Level:
In a bowl there are 4 Chocolates, 3 Jellies and 5 Mints. Find a way to share the sweets between the three children so they each get the kind they like. Is there more than one way to do it?
### Two Egg Timers
##### Stage: 2 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
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# Units and Dimensions – 97
97. If C be the capacitance and V be the electric potential, then the dimensional formula of CVis :
(a) M L T –2 A 0
(b) M L T –2 A –1
(c)0 L T –2 A 0
(d) M L -3 T A
Explanation
Explanation : No answer description available for this question. Let us discuss.
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## Related Posts
• Units and Dimensions » Exercise - 260. Dimensional formula of CV ? Where, C - capacitance and V - potential different (a) M 1 L –2 T 4 A 2 (b) M 1 L 2 T –3 A 1 (c) M 0 L 0 T 1 A –1 (d) M 0 L 0 T 1 A 1
Tags: units, dimensions, exercise, dimensional, formula, cv, capacitance, potential, physics
• Units and Dimensions » Exercise - 262. Dimensional formula for capacitance (C) : (a) M –1 L –2 T 4 K 2 (b) M 1 L –2 T 4 K 2 (c) M –1 L –2 T 3 K 1 (d) M 3 L 1 T –1 K –2
Tags: units, dimensions, exercise, dimensional, formula, capacitance, physics
• Units and Dimensions » Exercise - 267. Dimensional formula for calories is ............... (a) M 1 L 1 T –2 (b) M 2 L 1 T –2 (c) M 1 L 2 T –2 (d) M 2 L 2 T –2
Tags: units, dimensions, exercise, dimensional, formula, physics
• Units and Dimensions » Exercise - 283. [M L 2 T –3 A –1] is the dimensional formula for : (a) capacitance (b) resistance (c) resistivity (d) potential difference
Tags: units, dimensions, exercise, dimensional, formula, capacitance, potential, physics
• Units and Dimensions » Exercise - 263. Dimensional formula for torque is : (a) M 2 L 2 T –3 (b) M 2 L 1 T –2 (c) M 1 L 1 T –2 (d) M 1 L 2 T –2
Tags: units, dimensions, exercise, dimensional, formula, physics
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Cody
# Problem 30. Sort a list of complex numbers based on far they are from the origin.
Solution 1988355
Submitted on 23 Oct 2019 by Avdhesh Bansal
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
j = sqrt(-1); z = [-4 6 3+4*j 1+j 0]; zSorted_correct = [6 3+4*j -4 1+j 0]; assert(isequal(complexSort(z),zSorted_correct))
2 Pass
z = 1:10; zSorted_correct = 10:-1:1; assert(isequal(complexSort(z),zSorted_correct))
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# Table 3 Direct and indirect costs of musculoskeletal disease in Korea, 2008
Direct costs Indirect costs Total costs
Age group Population Direct medical care costs Direct non-medical care costs Total direct costs
(thousand persons) Paid by insurer Paid by patient OTC drugs costs Other costs Total direct medical care costs* Transportation Caregiver costs Lost productivity Cost of premature death
Male
0-19 years 6,168 17,954 39,549 47 97 57,647 506 2,901 61,054 0 640 61,694
20-64 years 16,216 443,446 361,346 2,092 4,760 811,644 11,144 66,877 889,666 753,139 31,248 1,674,052
≥65 years 2,032 368,355 169,309 2,099 1,879 541,643 7,518 43,610 592,771 0 0 592,771
Male Cost 24,416 829,756 570,204 4,239 6,736 1,410,934 19,169 113,388 1,543,491 753,139 31,888 2,328,517
Female
0-19 years 5,567 35,514 22,019 43 83 57,658 474 3,992 62,125 0 2,505 64,630
20-64 years 15,640 877,142 625,749 4,761 11,667 1,519,320 22,484 92,053 1,633,856 1,530,437 45,028 3,209,321
≥65 years 2,984 728,612 459,014 2,377 6,494 1,196,497 26,857 60,027 1,283,382 0 0 1,283,382
Female Cost 24,191 1,641,268 1,106,783 7,181 18,244 2,773,475 49,815 156,073 2,979,363 1,530,437 47,533 4,557,333
Total Cost 48,607 2,471,024 1,676,987 11,419 24,980 4,184,409 68,984 269,461 4,522,854 2,283,575 79,421 6,885,850
1. OTC drugs: Over the counter drugs, Unit: thousand dollars
2. *Sum of costs paid by insurer + costs paid by patients + OTC drugs costs+ other costs
3. Sum of direct costs + indirect costs.
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# Witt vector
An element of an algebraic construct, first proposed by E. Witt [1] in 1936 in the context of the description of unramified extensions of $p$-adic number fields. Witt vectors were subsequently utilized in the study of algebraic varieties over a field of positive characteristic [3], in the theory of commutative algebraic groups [4], [5], and in the theory of formal groups [6]. Let $A$ be an associative, commutative ring with unit element. Witt vectors with components in $A$ are infinite sequences $a = (a_0,a_1,\ldots)$, $a_i \in A$, which are added and multiplied in accordance with the following rules: $$(a_0,a_1,\ldots) \oplus (b_0,b_1,\ldots) = (S_0(a_0;b_0), S_1(a_0,a_1;b_0,b_1), \ldots)$$ $$(a_0,a_1,\ldots) \otimes (b_0,b_1,\ldots) = (M_0(a_0;b_0), M_1(a_0,a_1;b_0,b_1), \ldots)$$
where $S_n,M_n$ are polynomials in the variables $X_0,\ldots,X_n$, $Y_0,\ldots,Y_n$ with integer coefficients, uniquely defined by the conditions $$\Phi_n(S_0,\ldots,S_n) = \Phi_n(X_0,\ldots,X_n) + \Phi_n(Y_0,\ldots,Y_n)$$ $$\Phi_n(M_0,\ldots,M_n) = \Phi_n(X_0,\ldots,X_n) \cdot \Phi_n(Y_0,\ldots,Y_n)$$ where $$\Phi_n(Z_0,\ldots,Z_n) = Z_0^{p^n} + p Z_1^{p^{n-1}} + \cdots + p^n Z^n$$ are polynomials, $n \in \mathbf{N}$ and $p$ is a prime number. In particular, $$S_0 = X_0 + Y_0 \ ;\ \ \ S_1 = X_1 + Y_1 - \sum_{i=1}^{p-1} \frac{1}{p} \binom{p}{i} X_0^i Y_0^{p-i}$$ $$M_0 = X_0 \cdot Y_0 \ ;\ \ \ M_1 = X_0^p Y_1 + X_1 Y_0^p + p X_1 Y_1 \ .$$
The Witt vectors with the operations introduced above form a ring, called the ring of Witt vectors and denoted by $W(A)$. For any natural number $n$ there also exists a definition of the ring $W_n(A)$ of truncated Witt vectors of length $n$. The elements of this ring are finite tuples $a = (a_0,\ldots,a_{n-1})$, $a_i \in A$ with the addition and multiplication operations described above. The canonical mappings $$R : W_{n+1}(A) \rightarrow W_n(A)$$ $$R : (a_0,\ldots,a_n) \mapsto (a_0,\ldots,a_{n-1})$$ $$T : W_n(A) \rightarrow W_{n+1}(A)$$ $$T : (a_0,\ldots,a_{n-1}) \mapsto (0,a_0,\ldots,a_{n-1})$$ are homomorphisms. The rule $A \to W(A)$ (or $A \to W_n(A)$) defines a covariant functor from the category of commutative rings with unit element into the category of rings. This functor may be represented by the ring of polynomials $\mathbf{Z}[X_0,X_1,\ldots]$ (or $\mathbf{Z}[X_0,X_1,\ldots,X_n]$) on which the structure of a ring object has been defined. The spectrum $\mathrm{Spec}\mathbf{Z}[X_0,X_1,\ldots]$ (or $\mathrm{Spec}\mathbf{Z}[X_0,X_1,\ldots,X_n]$) is known as a Witt scheme (or a truncated Witt scheme) and is a ring scheme [3].
Each element $a \in A$ defines a Witt vector $$a^T = (a,0,0,\ldots) \in W(A)$$ called the Teichmüller representative of the element $a$. If $A = k$ is a perfect field of characteristic $p>0$, then $W(k)$ is a complete discrete valuation ring of zero characteristic with field of residues $k$ and maximal ideal $pW(k)$. Each element $w \in W(k)$ can be uniquely represented as $$w = w_0^T + pw_1^T + p^2 w_2^T + \cdots$$ where $w_i \in k$. Conversely, each such ring $A$ with field of residues $k = A/(p)$ is canonically isomorphic to the ring $W(k)$. The Teichmüller representation makes it possible to construct a canonical multiplicative homomorphism $k \to W(k)$, splitting the mapping $W(k) \to W(k)/(p)$.
If $k = \mathbf{F}_p$ is the prime field of $p$ elements, $W(k)$ is the ring of integral $p$-adic numbers $\mathbf{Z}_p$.
#### References
[1] E. Witt, "Zyklische Körper und Algebren der characteristik $p$ vom Grad $p^n$. Struktur diskret bewerteter perfekter Körper mit vollkommenem Restklassen-körper der Charakteristik $p$" J. Reine Angew. Math. , 176 (1936) pp. 126–140 Zbl 0016.05101 [2] S. Lang, "Algebra" , Addison-Wesley (1974) MR0783636 Zbl 0712.00001 [3] D. Mumford, "Lectures on curves on an algebraic surface" , Princeton Univ. Press (1966) MR0209285 Zbl 0187.42701 [4] J.-P. Serre, "Groupes algébrique et corps des classes" , Hermann (1959) MR0103191 [5] M. Demazure, P. Gabriel, "Groupes algébriques" , 1 , North-Holland (1971) MR1611211 MR0302656 MR0284446 Zbl 0223.14009 Zbl 0203.23401 Zbl 0134.16503 [6] J. Dieudonné, "Groupes de Lie et hyperalgèbres de Lie sur un corps de charactéristique $p$ VII" Math. Ann. , 134 (1957) pp. 114–133 DOI 10.1007/BF01342790 Zbl 0086.02605
There is a generalization of the construction above which works for all primes $p$ simultaneously, [a3]: a functor $W : \mathsf{Ring} \to \mathsf{Ring}$ called the big Witt vector. Here, $\mathsf{Ring}$ is the category of commutative, associative rings with unit element. The functor described above, of Witt vectors of infinite length associated to the prime $p$, is a quotient of $W$ which can be conveniently denoted by $W_{p^\infty}$.
For each $n \in \{1,2,\ldots\}$, let $w_n(X)$ be the polynomial $$w_n(X) = \sum_{d | n} d X^{n/d} \ .$$
Then there is the following characterization theorem for the Witt vectors. There is a unique functor $W : \mathsf{Ring} \to \mathsf{Ring}$satisfying the following properties: 1) as a functor $W : \mathsf{Ring} \to \mathsf{Set}$, $W(A) = \{(a_1,a_2,\ldots) : a_i \in A\}$ and $W(\phi)((a_1,a_2,\ldots)) = (\phi(a_1),\phi(a_2),\ldots)$ for any ring homomorphism $\phi : A \to B$; 2) $w_{n,A} : W(A) \to A$, $w_{n,A} : (a_1,a_2,\ldots) \mapsto w_n(a_1,a_2,\ldots)$ is a functorial homomorphism of rings for every $n$ and $A$.
The functor $W$ admits functorial ring endomorphisms $\mathbf{f}_n : W \to W$, for every $n \in \{1,2,\ldots\}$, that are uniquely characterized by $wn \mathbf{f}_m = w_{nm}$ for all $m,n \in \{1,2,\ldots\}$. Finally, there is a functorial homomorphism $\Delta : W({-}) \to W(W({-}))$ that is uniquely characterized by the property $w_{n,W(A)} \Delta_A = \mathbf{f}_{n,A}$ for all $n$,$A$.
To construct $W(A)$, define polynomials $\Sigma_n$; $\Pi_n$; $r_n$ for $n \in \{1,2,\ldots\}$ by the requirements $$w_n(\Sigma_1,\ldots,\Sigma_n) = w_n(X) + w_n(Y) \ ;$$ $$w_n(\Pi_1,\ldots,\Pi_n) = w_n(X) \cdot w_n(Y) \ ;$$ $$w_n(r1,\ldots,r_n) = - w_n(X) \ .$$
The $\Sigma_n$ and $\Pi_n$ are polynomials in $X_1,\ldots,X_n$; $Y_1,\ldots,Y_n$ and the $r_n$ are polynomials in the $X_1,\ldots,X_n$ and they all have integer coefficients. Now $W(A)$ is defined as the set $W(A) = \{ a = (a_1,a_2,\ldots) : a_i \in A \}$ with operations : $$(a_1,a_2,\ldots) + (b_1,b_2,\ldots) = (\Sigma_1(a,b), \Sigma_2(a,b), \ldots) \ ;$$ $$(a_1,a_2,\ldots) \cdot (b_1,b_2,\ldots) = (\Pi_1(a,b), \Pi_2(a,b), \ldots) \ ;$$ $$- (a_1,a_2,\ldots) = (r_1(a), r_2(a), \ldots) \ .$$
The zero of $W(A)$ is $(0,0,0,\ldots)$ and the unit element is $(1,0,0,\ldots)$. The Frobenius endomorphisms $\mathbf{f}_n$ and the Artin–Hasse exponential $\Delta$ are constructed by means of similar considerations, i.e. they are also given by certain universal polynomials. In addition there are the Verschiebung morphisms $\mathbf{V}_n : W({-}) \to W({-})$, which are characterized by $$w_n \mathbf{V}_m = \begin{cases} 0 & \text{if}\, n \not\mid m \\ n w_{m/n} & \text{if}\, n \mid m \end{cases} \ .$$
The $\mathbf{V}_n$ are group endomorphisms of $W(A)$ but not ring endomorphisms.
The ideals $I_n = \{(0,\ldots,0,a_{n+1},a_{n+2},\ldots)\}$ define a topology on $W(A)$ making it a separated complete topological ring.
For each $A \in \mathsf{Ring}$, let $\Lambda(A)$ be the Abelian group $1 + t A[[t]]$ under multiplication of power series; $$\bar E : W(A) \rightarrow \Lambda(A)$$ $$\bar E : (a_1,a_2,\ldots) \mapsto \prod_{i=1}^\infty \left({ 1 - a_i t^i }\right)$$ defines a functional isomorphism of Abelian groups, and using the isomorphism $\bar E$ there is a commutative ring structure on $\Lambda(A)$. Using $\bar E$ the Artin–Hasse exponential $\Delta$ defines a functorial homomorphism of rings $W(A) \to \Lambda(W(A))$ making $W(A)$ a functorial special $\lambda$-ring. The Artin–Hasse exponential $\Delta : W \mapsto W \circ W$ defines a cotriple structure on $W$ and the co-algebras for this co-triple are precisely the special $\lambda$-rings (cf. also Category and Triple).
On $\Lambda(A)$ the Frobenius and Verschiebung endomorphisms satisfy $$\mathbf{f}_n (1 - at) = 1 - a^n t$$ $$V_n(f(t)) = f(t^n)$$ and are completely determined by this (plus functoriality and additivity in the case of $\mathbf{f}_n$).
For each supernatural number $\mathbf{n} = (\alpha_p) : \alpha_p \in \{0,1,2,\ldots\} \cup \{\infty\}$, $p$ prime , one defines $N(\mathbf{n}) = \{ n \in \{1,2,\ldots\} : \nu_p(n) \le \alpha_p \}$, where $\nu_p(n)$ is the $p$-adic valuation of $n$, i.e. the number of prime factors $p$ in $n$. Let $$\mathfrak{a}_{\mathbf{n}}(A) = \{ (a_1,a_2,\ldots) : a_i \in A \,,\, a_d = 0 \,\text{for all}\, d \in N(\mathbf{n}) \} \ .$$
Then $\mathfrak{a}_{\mathbf{n}}(A)$ is an ideal in $W(A)$ and for each supernatural $\mathbf{n}$ a corresponding ring of Witt vectors is defined by $$W_{\mathbf{n}}(A) = W(A) / \mathfrak{a}_{\mathbf{n}}(A) \ .$$
In particular, one thus finds $W_{p^\infty}$, the ring of infinite-length Witt vectors for the prime $p$, discussed in the main article above, as a quotient of the ring of big Witt vectors $W(A)$.
The Artin–Hasse exponential $\Delta : W \to W \circ W$ is compatible in a certain sense with the formation of these quotients, and using also the isomorphism $\bar E$ one thus finds a mapping $$\mathbf{Z}_p = W_{p^\infty}(\mathbf{F}_p) \to \Lambda(W_{p^\infty}(\mathbf{F}_p)) = \Lambda(\mathbf{Z}_p)$$ where $\mathbf{Z}_p$ denotes the $p$-adic integers and $\mathbf{F}_p$ the field of $p$ elements, which can be identified with the classical morphism defined by Artin and Hasse [a1], [a2], [a3].
As an Abelian group is isomorphic to the group of curves of curves in the one-dimensional multiplicative formal group . In this way there is a Witt-vector-like Abelian-group-valued functor associated to every one-dimensional formal group. For special cases, such as the Lubin–Tate formal groups, this gives rise to ring-valued functors called ramified Witt vectors, [a3], [a4].
Let be the sequence of polynomials with coefficients in defined by
The Cartier ring is the ring of all formal expressions
(*)
with the calculation rules
Commutative formal groups over are classified by certain modules over . In case is a -algebra, a simpler ring can be used for this purpose. It consists of all expressions (*) where now the only run over the powers of the prime . The calculation rules are the analogous ones. In case is a perfect field of characteristic and denotes the Frobenius endomorphism of (which in this case is given by ), then can be described as the ring of all expressions
in two symbols and and with coefficients in , with the extra condition and the calculation rules
This ring, and also its subring of all expressions
is known as the Dieudonné ring and certain modules (called Dieudonné modules) over it classify unipotent commutative affine group schemes over , cf. [a5].
#### References
[a1] E. Artin, H. Hasse, "Die beide Ergänzungssätze zum Reciprozitätsgesetz der $\ell$-ten Potenzreste im Körper der $\ell$-ten Einheitswurzeln" Abh. Math. Sem. Univ. Hamburg , 6 (1928) pp. 146–162 [a2] G. Whaples, "Generalized local class field theory III: Second form of the existence theorem, structure of analytic groups" Duke Math. J. , 21 (1954) pp. 575–581 MR73645 [a3] M. Hazewinkel, "Twisted Lubin–Tate formal group laws, ramified Witt vectors and (ramified) Artin–Hasse exponentials" Trans. Amer. Math. Soc. , 259 (1980) pp. 47–63 MR0561822 Zbl 0437.13014 [a4] M. Hazewinkel, "Formal group laws and applications" , Acad. Press (1978) MR506881 [a5] M. Demazure, P. Gabriel, "Groupes algébriques" , 1 , North-Holland (1971) MR1611211 MR0302656 MR0284446 Zbl 0223.14009 Zbl 0203.23401 Zbl 0134.16503
How to Cite This Entry:
Witt vector. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Witt_vector&oldid=37687
This article was adapted from an original article by I.V. Dolgachev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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## Quiz on Mathematical Programming
1.
In linear programming, the constraints can be represented by
equalities
inequalities
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both a and b
2.
The indication of number of rows and number of columns in a matrix is classified as
direction
dimension
classification
specification
3.
The general form of algebraic function x5 multiplied by x¹ have solved form as
x4
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The variables of linear equation is implicitly raised to
first power
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5.
The quadratic function parabola graph is concave down if
a < 0
b < 0
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a = 0
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Quizzes Boost Learning
A pie chart could be used in a spreadsheet to display data.
Spreadsheets - the accounting programs - are one of the most useful tools to be found on a computer and children in KS2 will become very familiar with them in ICT.
Spreadsheets are interactive computer accounting programs that are very versatile. They make storing, organising and analysing data much easier than it once was. Spreadsheets are tables which you can tailor to any size you need and they can display any data you put into them in the form of numbers, text or even graphics. They are used by businesses and other organisations for many purposes, including accounting and record keeping. They are also used by people at home to work out their budgets, and by people at school to display the results of their science experiments as a graph. They are one of the most useful of all the different computer programs.
See how much you have learned about spreadsheets in ICT lessons by trying this quiz.
1.
Which of these could a spreadsheet NOT be used for?
Creating a graph from your science investigation results
Calculating the cheapest mobile phone package from a range of choices
Storing the test results for a class of year six pupils
You could use a word processor program to do your English writing homework
2.
On a spreadsheet, all cells are named. In the cell named B10, the 'B' refers to which of these?
The row where that cell is located
The type of information which can be entered into that cell
The column where that cell is located
The type of calculation which has been entered into that cell
The letters show in which column a cell is located, and the numbers show in which row the cell is located - this makes it very easy to find any named cell
3.
You are creating a spreadsheet to calculate the cost of a party. You want to know how much you will spend on pizzas. Why is it better to use a formula rather than working out the calculations yourself?
It will make it easier if you want to rearrange your spreadsheet
A formula will instantly calculate the change in cost if you need to change the price of the pizzas
The spreadsheet will be tidier if you use a formula
It is always faster to calculate in your head rather than to use a formula
A formula will instantly recalculate if you have to change any of the numbers which it uses. In this example, that could be the price of pizza, the number of people coming to the party, or even how many pieces of pizza each person will have
4.
Which of these is the correct formula for an addition?
=SUMB6+B7
B6+B7
=SUM(B6+B7)
=B6-B7
This formula would also work for an addition: =SUM(B6,B7)
5.
A large piece of cloth spread over a flat surface
A software package used to organise information
A piece of A3 paper spread out so that you can see it clearly
A word processing file
Electronic spreadsheets can store, organise and display large quantities of data. They can also perform calculations
6.
Which of these could be used in a spreadsheet to display your data?
A bar chart
A line graph
A pie chart
All of the above
To make a chart select the cells you want to use then click on the 'Chart' icon in the 'Insert' menu
7.
Which symbol would you use in a formula for multiplication (to multiply two or more numbers)?
x
/
÷
*
An example of an instruction to multiply would look like this: =SUM(B6*B7)
8.
A spreadsheet is divided into what?
Columns, rows and cells
Lines and squares
Parallel and diagonal lines
Helixes
Columns go vertically and rows horizontally. Where columns and rows meet is a box called a cell. These are the places information is entered
9.
Which symbol would you use in a formula for division (dividing a number)?
x
/
÷
*
An example of an instruction to divide would look like this: =SUM(B6/B7)
10.
What does it mean to 'format' a cell?
To make a cell bigger
To make a cell smaller
To insert a picture into a cell
To choose what type of information can be entered into a cell and how it is displayed
By right clicking on a cell and choosing 'format cells' from the menu, you can decide what type of information can be entered into that cell, whether currency (amounts of money), dates, text, numbers without decimals, numbers with two decimal places, etc. You can also format cells for font, colour and borders
Author: Sheri Smith
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Name: ___________________Date:___________________
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### High School Mathematics14.4 Types of Variates - Discrete
Variates can be of two types
Discrete variate
A variable is said to be discrete if it takes distinct and isolated values. For eg: i)number of telephone calls received in an office. ii)Number of children in a family.
Classification of discrete data
The data below shows the number of accidents which occurred on each day over a period of 50 days, on a certain highway.
3 1 1 0 2 1 1 2 6 2 3 4 4 6 2 6 1 1 3 2 2 4 2 0 3 3 4 2 5 3 2 5 1 6 3 1 1 1 5 6 2 5 5 2 3 2 2 2 2 3
The frequency table for discrete variates is as follows.
Directions: On the basis of the given information answer the following questions. Also write at least one example of your own.
Q 1: The number of days on which minimum accidents took place is0231 Q 2: Over a period of 50 days, how many such days were there, when maximum number of accidents occured. 15435 Q 3: The number of days on which four accidents a day took place is...4510 Q 4: The number of days on which 2 accidents occured is 0155none of these Question 5: This question is available to subscribers only! Question 6: This question is available to subscribers only!
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Author Message
Intern
Joined: 25 May 2013
Posts: 28
Kudos [?]: 36 [0], given: 16
### Show Tags
25 Aug 2013, 14:13
“Advertising the reduced price of selected grocery items in the Daily Gazette will
Thirty sale items from a store in downtown Marston were advertised in the Gazette
for four days. Each time one or more of the 30 items was purchased, clerks asked whether the shopper had read the ad. Two-thirds of the 200 shoppers asked answered in the affirmative. Furthermore, more than half the customers who answered in the affirmative spent over \$100 at the store.”
Discuss how well reasoned... etc.
The argument that advertising the reduced price of selected grocery items in the Daily Gazette will help you increase your sales omits some important concerns that must be addressed to substantiate the argument. The evidences cited in the argument present a distorted view of the situation and do not help in evaluating the argument. Therefore the argument is weak and has several flaws.
Second, the argument omits sales of non-advertised items and does not consider their impact on overall sale. For instance people who spent over \$100 could have spent large part of it on non-advertised items and not on advertised items. So without any information on sale of non-advertised items, it is difficult to correlate the impact of advertising with increase in sale. If the argument would have provided information on sale of non-advertised items it would a lot more convincing.
In conclusion, the argument is flawed for above mentioned reasons and hence unconvincing. It could be considerably strengthened if it included more relevant facts such as period during which advertising was conducted and sales of non-advertised items. In order to access the merits of certain situation it is important to have full knowledge of all contributing factors. In this particular case, without relevant information the argument remains unsubstantiated and open to debate.
Kudos [?]: 36 [0], given: 16
Kaplan GMAT Instructor
Joined: 25 Aug 2009
Posts: 644
Kudos [?]: 306 [0], given: 2
Location: Cambridge, MA
### Show Tags
26 Aug 2013, 15:44
Hi akijuneja,
you're showing improvement! However, this still only gets a 3. The organization has gotten better from your previous efforts, but still needs work--you start your second paragraph by introducing the conclusion of the author's argument, for instance, rather than introducing your "first" of several flaws. You also still need more depth on your reasoning. You are correct that the evidence doesn't really indicate an increase in profits, but miss the mark on why. Spending on non-advertised items is still profit, and is irrelevant; better questions are whether they bought high-margin items, or whether spending \$100/person is atypical for customers at the store.
Good luck, and keep on working!
_________________
Eli Meyer
Kaplan Teacher
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# R/numericalDifferentiation.R In uGMAR: Estimate Univariate Gaussian and Student's t Mixture Autoregressive Models
```#' @title Calculate gradient or Hessian matrix
#'
#' of the given function at the given point using central difference numerical approximation.
#' \code{get_gradient} (and \code{get_foc}) or \code{get_hessian} calculates the gradient or Hessian matrix of the
#' log-likelihood function at the parameter values of a class \code{'gsmar'} object.
#' \code{get_soc} returns eigenvalues of the Hessian matrix.
#'
#' @param x a numeric vector specifying the point at which the gradient or Hessian should be evaluated.
#' @param fn a function that takes in the argument \code{x} as the \strong{first} argument.
#' @param h the difference used to approximate the derivatives.
#' @param varying_h a numeric vector with the same length as \code{x} specifying the difference \code{h}
#' for each dimension separately. If \code{NULL} (default), then the difference given as parameter \code{h}
#' will be used for all dimensions.
#' @param custom_h same as \code{varying_h} but if \code{NULL} (default), then the difference \code{h} used for differentiating
#' overly large degrees of freedom parameters is adjusted to avoid numerical problems, and the difference is \code{6e-6} for the other
#' parameters.
#' @param ... other arguments passed to \code{fn}.
#' @details In particular, the functions \code{get_foc} and \code{get_soc} can be used to check whether
#' the found estimates denote a (local) maximum point, a saddle point, or something else.
#' @return The gradient functions return numerical approximation of the gradient, and the Hessian functions return
#' numerical approximation of the Hessian. \code{get_soc} returns eigenvalues of the Hessian matrix, \code{get_foc}
#' is the same as \code{get_gradient} but named conveniently.
#' @section Warning:
#' No argument checks!
#' @examples
#' # Simple function
#' foo <- function(x) x^2 + x
#' calc_hessian(x=2, fn=foo)
#'
#' # More complicated function
#' foo <- function(x, a, b) a*x[1]^2 - b*x[2]^2
#' calc_gradient(x=c(1, 2), fn=foo, a=0.3, b=0.1)
#' calc_hessian(x=c(1, 2), fn=foo, a=0.3, b=0.1)
#'
#' # GMAR model
#' params12 <- c(1.70, 0.85, 0.30, 4.12, 0.73, 1.98, 0.63)
#' gmar12 <- GSMAR(data=simudata, p=1, M=2, params=params12, model="GMAR")
#' get_foc(gmar12)
#' get_hessian(gmar12)
#' get_soc(gmar12)
#' @export
calc_gradient <- function(x, fn, h=6e-06, varying_h=NULL, ...) {
fn <- match.fun(fn)
n <- length(x)
I <- diag(1, nrow=n, ncol=n)
if(is.null(varying_h)) { # The same difference h for all parameters
h <- rep(h, times=n)
} else { # Varying h
stopifnot(length(varying_h) == length(x))
h <- varying_h
}
# Calculate the central difference approximation
vapply(1:n, function(i1) (fn(x + h[i1]*I[i1,], ...) - fn(x - h[i1]*I[i1,], ...))/(2*h[i1]), numeric(1))
}
#' @export
calc_hessian <- function(x, fn, h=6e-06, varying_h=NULL, ...) {
fn <- match.fun(fn)
n <- length(x)
I <- diag(1, nrow=n, ncol=n)
if(is.null(varying_h)) { # The same difference h for all parameters
h <- rep(h, times=n)
} else { # Varying h
stopifnot(length(varying_h) == length(x))
h <- varying_h
}
Hess <- matrix(ncol=n, nrow=n)
for(i1 in 1:n) { # Calcute the central difference approximation for the Hessian
for(i2 in i1:n) {
dr1 <- (fn(x + h[i1]*I[i1,] + h[i2]*I[i2,], ...) - fn(x - h[i1]*I[i1,] + h[i2]*I[i2,], ...))/(2*h[i1])
dr2 <- (fn(x + h[i1]*I[i1,] - h[i2]*I[i2,], ...) - fn(x - h[i1]*I[i1,] - h[i2]*I[i2,], ...))/(2*h[i1])
Hess[i1, i2] <- (dr1 - dr2)/(2*h[i2])
Hess[i2, i1] <- Hess[i1, i2] # Take use of symmetry
}
}
Hess
}
#' @export
check_gsmar(gsmar)
if(is.null(custom_h)) { # Adjust h for overly large degrees of freedom parameters
varying_h <- get_varying_h(p=gsmar\$model\$p, M=gsmar\$model\$M, params=gsmar\$params, model=gsmar\$model\$model)
} else { # Utilize user-specified h
stopifnot(length(custom_h) == length(gsmar\$params))
varying_h <- custom_h
}
# Function to differentiate
foo <- function(x) {
loglikelihood(data=gsmar\$data, p=gsmar\$model\$p, M=gsmar\$model\$M, params=x, model=gsmar\$model\$model,
restricted=gsmar\$model\$restricted, constraints=gsmar\$model\$constraints,
conditional=gsmar\$model\$conditional, parametrization=gsmar\$model\$parametrization,
minval = NA)
}
}
#' @export
get_foc <- function(gsmar, custom_h=NULL) {
}
#' @export
get_hessian <- function(gsmar, custom_h=NULL) {
check_gsmar(gsmar)
if(is.null(custom_h)) { # Adjust h for overly large degrees of freedom parameters
varying_h <- get_varying_h(p=gsmar\$model\$p, M=gsmar\$model\$M, params=gsmar\$params, model=gsmar\$model\$model)
} else { # Utilize user-specified h
stopifnot(length(custom_h) == length(gsmar\$params))
varying_h <- custom_h
}
# Function to differentiate
foo <- function(x) {
loglikelihood(data=gsmar\$data, p=gsmar\$model\$p, M=gsmar\$model\$M, params=x, model=gsmar\$model\$model,
restricted=gsmar\$model\$restricted, constraints=gsmar\$model\$constraints,
conditional=gsmar\$model\$conditional, parametrization=gsmar\$model\$parametrization,
minval = NA)
}
# Calculate the Hessian
calc_hessian(x=gsmar\$params, fn=foo, varying_h=varying_h)
}
#' @export
get_soc <- function(gsmar, custom_h=NULL) {
hess <- get_hessian(gsmar, custom_h=custom_h) # Calculate the Hessian
if(anyNA(hess)) stop("Missing values in the Hessian matrix. Are estimates at the border of the parameter space?")
eigen(hess)\$value # Return the eigenvalues
}
#' @title Get differences 'h' which are adjusted for overly large degrees of freedom parameters
#'
#' @description \code{get_varying_h} adjusts differences for overly large degrees of freedom parameters
#' for finite difference approximation of the derivatives of the log-likelihood function. StMAR and
#' G-StMAR models are supported.
#'
#' @inheritParams loglikelihood_int
#' @details This function is used for approximating gradient and Hessian of a StMAR or G-StMAR model. Large
#' degrees of freedom parameters cause significant numerical error if too small differences are used.
#' @return Returns a vector with the same length as \code{params}. For other parameters than degrees
#' of freedom parameters larger than 100, the differences will be \code{6e-6}. For the large degrees of
#' freedom parameters, the difference will be \code{signif(df/1000, digits=2)}.
#' @keywords internal
get_varying_h <- function(p, M, params, model) {
if(model != "GMAR") {
dfs <- pick_dfs(p=p, M=M, params=params, model=model)
adj_diffs[dfs <= 100] <- 6e-6 # h is not adjusted for dfs not larger than hundred
adj_diffs[dfs > 100] <- signif(dfs[dfs > 100]/1000, digits=2) # The adjusted differences h
varying_h <- c(rep(6e-6, times=length(params) - length(dfs)), adj_diffs) # Difference h for all parameters
} else { # No degrees of freedom parameters in GMAR model, so the default difference is used
varying_h <- rep(6e-6, times=length(params))
}
varying_h
}
```
## Try the uGMAR package in your browser
Any scripts or data that you put into this service are public.
uGMAR documentation built on Jan. 24, 2022, 5:10 p.m.
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# Ch9.Answers (1) - 9.2.35.4.06(1.4.12.25)=.1144 or 11.44...
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Chapter 9 Answers 1- - a. FV=-1000, PV=1107, PMT=-90, N=15, I=>7.77*(1-.4)=.0466 or 4.66% b. [(1.45*1.06)/(36.82*.96)]+.06=.1035 or 10.35% c. (.06*200)/(225*.85)=.0627 or 6.27% 2- - a. FV=-1000, PMT=-80, N=10, PV=(965*(1-.1), I=>10.15*(1-.34)=6.69 b. (2.24/18.80)+.04=.1591 or 15.91% c. (.08*150)/100=.12 or 12% 3- [(1.15*1.05)/(32*(1-.04))]+ .05=.0894 or 8.94% 4- FV=-1000, PMT=-100, N=20, PV=1041.25, I=>.0953*(1-.25)=.0715 or 7.15% 5- 1.45/58=.025 or 2.5% 6- FV=-1000, PV=850, PMT=-80, N=10, I=>10.49*(1-.4)=.063 or 6.3% 7- 8/90=.0889 or 8.89% 8- (1.08/26.4)+.08=.1209 or 12.09%
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Unformatted text preview: 9-.2(.35)+.4(.06)*(1-.4)+.12(.25)=.1144 or 11.44% 10- .5(.16)+.35(.1)*(1-.35)+.15*(.08)=.1148 or 11.48% 11-PMT=-50, FV=-1000, N=40, PV=1,100, I=>.0446*2=.0892 or 8.92%; (1.75/55)+.06=.0918 or 9.18%; .35(.0892)*(1-.25)+.65(.0918)=.083 or 8.3% 12-Bonds: 6mil/16mil= .375; PV=785, PMT=-45, FV=-1000, N=30, I=>6.07*2*(1-.4)=. .0728 or 7.28%; Preferred Stock: 2mil/16mil=.125; (1.75/80)=.0219 or 2.19%; Common Stock: 8mil/16mil=.5; (4.43/120)+.03=.0669 or 6.69%; WACC=.375(.0728)+.125(.0219)+.5(.0669)= . 0634 or 6.34%...
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## This note was uploaded on 10/23/2011 for the course BUS M 301 taught by Professor Jimbrau during the Fall '11 term at BYU.
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Ch9.Answers (1) - 9.2.35.4.06(1.4.12.25)=.1144 or 11.44...
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Term Rewriting System R:
[X, Z, N, Y]
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
FROM(X) -> FROM(s(X))
2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, Z)
2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, Z))
2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, Z)
PI(X) -> 2NDSPOS(X, from(0))
PI(X) -> FROM(0)
PLUS(s(X), Y) -> PLUS(X, Y)
TIMES(s(X), Y) -> PLUS(Y, times(X, Y))
TIMES(s(X), Y) -> TIMES(X, Y)
SQUARE(X) -> TIMES(X, X)
Furthermore, R contains four SCCs.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Remaining Obligation(s)`
` →DP Problem 2`
` ↳Remaining Obligation(s)`
` →DP Problem 3`
` ↳Remaining Obligation(s)`
` →DP Problem 4`
` ↳Remaining Obligation(s)`
The following remains to be proven:
• Dependency Pair:
FROM(X) -> FROM(s(X))
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pairs:
2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, Z)
2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, Z)
2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, Z))
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pair:
PLUS(s(X), Y) -> PLUS(X, Y)
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pair:
TIMES(s(X), Y) -> TIMES(X, Y)
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Remaining Obligation(s)`
` →DP Problem 2`
` ↳Remaining Obligation(s)`
` →DP Problem 3`
` ↳Remaining Obligation(s)`
` →DP Problem 4`
` ↳Remaining Obligation(s)`
The following remains to be proven:
• Dependency Pair:
FROM(X) -> FROM(s(X))
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pairs:
2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, Z)
2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, Z)
2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, Z))
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pair:
PLUS(s(X), Y) -> PLUS(X, Y)
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pair:
TIMES(s(X), Y) -> TIMES(X, Y)
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Remaining Obligation(s)`
` →DP Problem 2`
` ↳Remaining Obligation(s)`
` →DP Problem 3`
` ↳Remaining Obligation(s)`
` →DP Problem 4`
` ↳Remaining Obligation(s)`
The following remains to be proven:
• Dependency Pair:
FROM(X) -> FROM(s(X))
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pairs:
2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, Z)
2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, Z)
2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, Z))
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pair:
PLUS(s(X), Y) -> PLUS(X, Y)
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pair:
TIMES(s(X), Y) -> TIMES(X, Y)
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Remaining Obligation(s)`
` →DP Problem 2`
` ↳Remaining Obligation(s)`
` →DP Problem 3`
` ↳Remaining Obligation(s)`
` →DP Problem 4`
` ↳Remaining Obligation(s)`
The following remains to be proven:
• Dependency Pair:
FROM(X) -> FROM(s(X))
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pairs:
2NDSNEG(s(N), cons2(X, cons(Y, Z))) -> 2NDSPOS(N, Z)
2NDSNEG(s(N), cons(X, Z)) -> 2NDSNEG(s(N), cons2(X, Z))
2NDSPOS(s(N), cons2(X, cons(Y, Z))) -> 2NDSNEG(N, Z)
2NDSPOS(s(N), cons(X, Z)) -> 2NDSPOS(s(N), cons2(X, Z))
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pair:
PLUS(s(X), Y) -> PLUS(X, Y)
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
• Dependency Pair:
TIMES(s(X), Y) -> TIMES(X, Y)
Rules:
from(X) -> cons(X, from(s(X)))
2ndspos(0, Z) -> rnil
2ndspos(s(N), cons(X, Z)) -> 2ndspos(s(N), cons2(X, Z))
2ndspos(s(N), cons2(X, cons(Y, Z))) -> rcons(posrecip(Y), 2ndsneg(N, Z))
2ndsneg(0, Z) -> rnil
2ndsneg(s(N), cons(X, Z)) -> 2ndsneg(s(N), cons2(X, Z))
2ndsneg(s(N), cons2(X, cons(Y, Z))) -> rcons(negrecip(Y), 2ndspos(N, Z))
pi(X) -> 2ndspos(X, from(0))
plus(0, Y) -> Y
plus(s(X), Y) -> s(plus(X, Y))
times(0, Y) -> 0
times(s(X), Y) -> plus(Y, times(X, Y))
square(X) -> times(X, X)
Termination of R could not be shown.
Duration:
0:00 minutes
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Since the beginning of mankind, we have been fascinated by the impact of gravity.The rule for weight of an object is the force F g = mg. There are different types of motion. In addition to gravity, there is projectile motion as well as potential energy. Potential energy comes in various forms, all of which are dependant on the position of an object instead of motion. However, motion wide gravity becomes even more complex than even potential energy. Whether or not kinetic energy increases, the balance of kinetic energy is regulated by gravitational energy. The way that kinetic energy is balanced is by decreases in gravitational energy and vice versa.
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Projectile motion is when a projectile enters the Earth’s atmosphere and onto its surface. It moves along the curvature of the earth’s surface, under the force of gravity. In fact, the gravitational pull is so strong that their object accelerates downward very fast. There is even another term deemed to projectile motion: inertia.
Projectile motion, in addition to aiding the downward trajectory of matter also has another role. This role is called: Kinematic quantities of projectile motion. Kinematic quantities of projectile motion have two parts: acceleration and velocity. Unlike regular projectile motion that accelerates at a rapid pace, the vertical motion of the object is the motion of a particle during free-fall. During velocity everything remains constant while in motion. Linear components are present in velocity due to the never-changing characteristics during velocity. Gravitational force is what accounts for the natural pull of one experience under gravity. The only way that gravitational forces work and not throw the Earth’s axis out of orbit all depends on how large the Earth really is and how far apart the scientist is in relation to the mass of the Earth.
The force that everybody feels from the Earth’s gravitational pull is in relation to the observer’s body mass. How one experiences gravitational force depends solely on the mass of the Earth and the separation between scientist and the mass of the Earth. A great deal of factors are in play that determine the gravitational force and mass of an object. For example, F g = constant x m. To try to explain the meaning of the constant in this formula above, we should consider what happens when an object has a constant that is equal to the gravitational force that falls in the Earth’s atmosphere.
The acceleration of, let us say, a basketball, is related to the unbalanced force, or in scientific lingo: F = m x 2. Should we study Isaac Newton’s second law of motion we would discover that the acceleration of the object can best be summed up by the equation: F g = m x a. The equation, or M, will lead the scientist to the conclusion that when an object relies on the force of gravity alone the scientific equation is: a = constant. That discovery can only be one thing: gravity alone, or a = constant is the reason why objects fall. As we can see, motion under the direct influence of gravity is a very complex and exciting part of science. What Newton started, we, here in the twenty-first century can continue.
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# Register Number
SATHYABAMA UNIVERSITY
(Established under section 3 of UGC Act, 1956)
Course & Branch: M.C.A Title of the Paper: Management Accounting Sub. Code: SBAX5028 Date: 18/12/2010
## Max. Marks: 80 Time: 3 Hours Session: FN (8 x 5 = 40)
______________________________________________________________________________________________________________________
## PART A Answer Any EIGHT Questions 1. 2. 3. 4. 5. 6. 7. 8. 9.
State different classification of Accounts. Also state the rules of Accounting. Differenciate between Management Accounting and Financial Accounting. Define marginal cost and marginal costing. What are the assumptions of Break even analysis? Define Ratio and its classification. Explain zero Based Budgeting. Give a Proforma of a cash Budget with an imaginary figures. Define Capital Budgeting. What are its uses? Calculate: (a) Purchases (b) Stock Turnover Ratio (c) G.P. Ratio Sales `33604; Opening stock `1378; Closing Stock `1814; Gross Profit `8068
10. From the following Draw a Break even chart: Fixed cost `8000; Variable Cost `4 per unit; selling price per unit `6; Units produced 4000; 6000, 8000 and 12000. 11. From the following data which product would you suggest time being the KEY FACTOR: A B Selling Price `100.00 `110.00 Variable Cost `30.00 `23.00 Time to produce (Hours) 2 3 12. Compute PAYBACK METHOD: Cost of the projects `1000000 Profit Before Depreciation and TAX = 300000 Depreciation 20% TAX Rate 50% PART B Answer Any FOUR Questions (4 x 10 = 40)
13. Discuss in detail the various accounting Concepts. 14. Explain the different methods in capital Budgeting 15. From the following Trial Balance prepare (a) Trading, Profit and Loss Account and (b) Balance Sheet Trial Balance as on 31.12.2008 Debit Balance Credit Balance ` ` Cash in hand 2000 Capital 200000 Machinery 60000 Sales 254800 Stock 50000 Creditors 40000 Bills Receivable 1600 Overdraft 22000
Debtors Wages Land Carriage inwards Purchases Salaries Rent Postage Sales Return Drawings Furniture Interest Cash at Bank
50000 70000 40000 2400 180000 24000 4000 1000 3200 10000 18000 600 6600 523400
## 3000 1800 1800
523400
Adjustments: (i) Stock on 31.12.2008 `100000 (ii) Rent outstanding `4000 (iii) Baddebts `5000 16. From the following information, prepare a Balance Sheet (a) Working Capital `75000 (b) Reserve and Surplus `100000 (c) Bank Overdraft `60000 (d) Liquid Ratio 1.75 (e) Fixed Assets to Proprietors fund 0.75 17. A Radio manufacturing company finds that while it costs `6.25 each to make component X2734, the same is available in the market `5.75 each, with an assurance of continued supply. The breakdown of costs is Materials `2.75 each Labour `1.75 each Variable Expenses `0.50 each Fixed Costs `1.25 each
`6.25 each (a) Should you make or buy? (b) What would be your decision if the supplier offered the component at `4.85 each? 18. Draw a flexible budget for overhead expenses on the basis of the following data and determine the overhead rate @ 70%, 80% and 90% plan capacity: Capacity Levels. 70% 80% 90% ` ` ` VARIABLE OVER HEADS: Indirect Labour 12000 Stores & Spares 4000 SEMI-VARIABLE OVERHEADS: Power (30% Fixed 70% Variable) 20000 Repairs (60% Fixed, 40% Variable) 2000 FIXED OVERHEADS: Depreciation 11000 Salaries 13000 TOTAL OVERHEADS 62000 ESTIMATED DIRECT LABOUR HOURS 124000
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## Inverses are (horrible)-1
This section gives an introduction to the extremely interesting concept of implementing a function as the inverse of another with zip and unzip as the examples. First, the goal is to build out our zip and unzip functions to satisfy the equation
```zip . unzip = id
```
A couple of notes here. Previously everything has been defined in terms of Haskell, but this is simply an equation. Also, if you're coming from a non-functional background, id is a function that gives you back the same thing you pass to it. No matter what. In this case the equation is meant to point out that, if you give unzip a list of pairs (two element tuples) and give that result, a tuple of two lists, to zip, in the end you'll get back the original list of pairs. So zip compose unzip is equivalent to the id function. You get back just what you gave it.
```zip \$ unzip [(1,2), (3,4)] = [(1,2), (3,4)]
```
Before we get into why this is immediately useful lets define the two functions as the book does, but with Haskell.
```data Listr a = Empty | Cons (a, Listr a) deriving Show
foldr' c h Empty = c
foldr' c h (Cons (a, x)) = h a \$ foldr' c h x
```
Our old friends the Cons list and foldr here. If you're new to these you can check out more information on them here.
```unzip' :: Listr (a, b) -> (Listr a, Listr b)
unzip' = foldr' emptys conss
where emptys = (Empty, Empty)
conss (a, b) (x, y) = (Cons (a, x) , Cons (b, y))
unzip_' :: Listr (a, b) -> (Listr a, Listr b)
unzip_' Empty = (Empty, Empty)
unzip_' (Cons ((a,b), x)) = (Cons (a, left \$ unzip_' x), Cons (b, right \$ unzip_' x))
where left (x, y) = x
right (x, y) = y
```
unzip' and unzip_' represent two possible ways of unzipping a list of pairs. Of interest here is the fact that my own, much less efficient, implementation involves two "folds" over the data structure, and the one implemented in the book only requires one. Apparently, any and all functions defined by a pair of folds can be defined in terms of a single fold (to be demonstrated later in the book).
```zip' :: (Listr a, Listr b) -> Listr (a, b)
zip' (Empty, _) = Empty
zip' (_, Empty) = Empty
zip' (Cons (a, x), Cons (b, y)) = Cons ( (a, b) , zip' (x, y) )
-- > let x = zip' . unzip'
-- > x \$ Cons ((1, 'a'), Cons ((2, 'b'), Empty))
-- Cons ((1,'a'),Cons ((2,'b'),Empty))
-- > let y = zip' . unzip_'
-- > y \$ Cons ((1, 'a'), Cons ((2, 'b'), Empty))
-- Cons ((1,'a'),Cons ((2,'b'),Empty))
```
Finally we have an exhaustive (can handle lists of different lengths) definition of zip' and the results of composing zip' with unzip' and unzip_' using ghci.
## Testing with maths
If you know how zip and unzip work its not imperative (though it is fun) to walk yourself through how each of them operates as defined here. What is important is the result we've gotten out of ghci. We've composed our two functions, and as stated above in our original equation we've gotten back the very same thing we put in. Just as though it were the id function. I had originally wanted to build my own "foldrless" version of unzip_ in the hopes that it would be easier to see the "inversity" in the code itself. Unfortunately its not incredibly clear, and whats worse the implementation isn't extremely efficient. BUT! Its not a total loss, because in order to satisfy myself that my hack was working properly, all I had to do was run the results BACK through zip and see if I got the original Cons list out the other side to satisfy our original assertion. And that's really the sweet spot here. While its not a formal proof of the functionality, it provides us with a lot of confidence that our functions are working properly. By defining the function as the inverse of another function we get a way to build it and test it. Incredible.
26 Sep 2009
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# 560
What is 560?
### 1.
The smallest tetrahedral numberthat is the product of two tetrahedral numbers both greater than one. Also the second smallest number and smallest tetrahedral number divisible by 1, 4, 10, 20, 35, and 56 (the first six tetrahedral numbers; the smallest number divisible by all those is 280)
tetra(3) * tetra(6) = (3*4*5/6)*(6*7*8/6) = (3*4*5/6)*(7*8) = (60/6)*56 = 10*56 = 560 = tetra(14). 560/4=140, 560/10= 56, 560/20= 28, 560/35= 16, 560/56= 10 (all quotients are even so 280, half of 560, is divisible by all divisors given (1, 4, 10, 20, 35, and 56)).
See number, tetrahedral number, product, factor, division
0
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1. A scale, especially when weighing out drugs. Yo, i got the Q, you got the scilly? See scale, weight, gram, pound, kilo..
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1. A mystic philosophy taught by Jesus of Nazareth. This philosophy contains the true teachings of Jesus, who seems to have been highly inf..
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https://tumericalive.com/how-do-you-measure-caster-with-digital-level/
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# How do you measure caster with digital level?
## How do you measure caster with digital level?
Make sure the bubble on the gauge is nice and level. And then Chris will push the caster. Button put it in the caster mode it’ll. Start blinking when it’s in caster mode and then hit zero.
How do you test a caster with a bubble gauge?
Fit the device to the center of the hub. Again level the device by checking the bubble engage D. Then set the caster degree gauge to zero by turning the thumb screw underneath.
How do you adjust camber and caster?
So double tap it. Now it’s at zero. And it’ll do that whenever you’re at zero.
### How do you test camber and caster on a race car?
To Measure Camber:
1. Turn the wheels so they point straight ahead.
2. Attach the caster/camber gauge to the spindle or hub.
3. Rotate the gauge so the bubble in the small vial closest to you is centered.
4. Camber is shown on the two outer vials. The left vial reads positive (+). The right vial reads negative (-).
Do you adjust caster or camber first?
With front-end alignments, correct caster and camber adjustments first. Certain FWD vehicles do not offer caster adjustments, but correcting the camber may bring the caster within specs.
How do you calculate caster angle?
Caster is the measure of how far forward or behind the steering axis is to the vertical axis, viewed from the side. This is measured by drawing a line between the top and bottom pivot points of the front upright. The angle between the drawn line and vertical is the caster angle.
#### How do you use Gunson Trakrite?
Gunson Trakrite Wheel Alignment Gauge – YouTube
What are the 3 alignment adjustments?
There are three main wheel alignment angles, which determine how each wheel is positioned relative to the car, the ground and each other. These are camber, caster and toe.
What should camber be set at?
For a normal car you typically want to maintain a slight amount of negative camber (0.5 – 1°) to have a good balance of cornering grip, braking grip, and tire wear. On most vehicles it’s common to have slightly more negative camber (0.8 – 1.3°) in the rear to reduce the chances of oversteer (loss of grip in rear).
## What happens with too much caster?
If you don’t have enough positive caster (forks too little angle toward the front) the car will wander and feel unstable. If you have too much positive caster, the vehicle will be harder to turn. However, just like with camber, vehicle pull is not caused by too much or too little caster, but by side to side imbalance.
How do you tell if your camber is off?
Camber. Camber measures the angle of the wheel when looking straight on at the car from the front. If the wheel leans towards the car with the bottom of the wheel farther away from the car than the top, that is a negative camber. A wheel that leans away from the car has a positive camber.
Do you want positive or negative caster?
Positive caster is primarily beneficial to the vehicle as it increases the lean of the tire when the vehicle is cornering, while returning it to an upright position when driving straight ahead. Negative Caster – If the line slopes towards the front of the vehicle then the caster is negative.
### How do you use Trackace alignment?
Trackace DIY Laser Wheel Alignment Gauge Car Tracking Save Money …
How do you know if your alignment is right?
Here are some common signs that you are dealing with poor alignment:
1. Your vehicle pulls to one side.
2. Uneven or rapid tire wear.
3. Your steering wheel is crooked when driving straight.
4. Squealing tires.
How many miles should you get an alignment?
6,000 miles
For virtually all vehicles, it’s necessary to get your wheels aligned periodically. Most car experts recommend scheduling an alignment every other oil change, or approximately every 6,000 miles.
#### Is it better to have toe-in or toe out?
Generally the rule of thumb is that more toe-in increases understeer and more toe-out increases oversteer. However, with modern cars, especially race cars with independent front and rear suspensions, there is another effect on handling.
What is cast and camber?
Positive camber results in tire wear on the outside tread of the tire. A vehicle will pull to the side with the most positive camber. Caster is viewed from the side of the vehicle. It’s the forward or rearward tilt of the steering axis. Positive caster provides steering wheel returnability and greater stability.
Can caster cause death wobble?
Caster and Camber Angles
Adding lift to a Jeep’s suspension can throw off your factory caster angle. If your Jeep has too little caster angle, it can be a candidate for developing death wobble.
## Will an alignment fix camber?
Will an Alignment Fix Negative Camber? In some cases, yes. But as mentioned above, camber issues often result from worn or broken parts.
What happens if you have too much caster?
The front inside wheel rises and the front outside wheel falls. This creates a jacking effect putting more mass on the rear outside wheel when cornering. This can create a positive turn in effect as it helps to rotate the car on corner entry. However, too much caster can lead to oversteer due to the jacking effects.
Can too much caster cause death wobble?
If your Jeep has too little caster angle, it can be a candidate for developing death wobble. As always, too much of a good thing ain’t bad. Add too much caster angle and your Jeep will wonder down the road like it’s lost as the two tires fight to follow their own independent path.
### How do you know if tracking is out on car?
One of the signs that your tracking is off is that the tyres wear more on the inner or outer edges of the tread than in the centre. Toe in will lead to excess wear on the outer shoulders of the tyre, whereas toe out will cause wear on the inner shoulders.
Why is my steering wheel not straight?
When your steering wheel is not being straight, it is probably out of alignment, and its steering and suspension systems are not functioning at proper angles. This often results in uneven, rapid tread-wear, and you may need to replace your tires much earlier.
Can an alignment be done wrong?
A bad alignment can make your steering wheel feel sloppier than usual. It may feel loose in your hands or the car may not respond immediately to turning the wheel. This doesn’t always mean that you need a realignment, but it is a good indication.
#### What causes a steering wheel to be off center?
As mentioned above, the vast majority of issues pertaining to a steering wheel’s off-center orientation, come as the direct fault of a poor front-end alignment. However, the root cause of this misalignment often differs on a case-by-case basis.
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# Spring Finance Assingment Help With Solution
## Spring Finance Assingment Help With Solution
Part I:
You have Treasury bond quotes on February 15, 2015. The dataset is at the course homepage
named the. hw1data spring15xls. There are three sets of the data: First set has 5 Treasury
securities, second part has 2 Treasuries and third set has 1 Treasuries. Note that quoted
prices are full prices.
A. For all Treasuries in the data (i.e. both first, second and third set Treasuries):
1. Find the bid and ask prices of the Treasury bill for a \$100 face value. Use ask prices
for below questions.
B. For this part, use only the 5 bonds in the first set.
2. Using bootstrap method long way, extract zero coupon bonds
3.
a) Using short way of matrix inversion extract zero coupon bonds
b) Calculate spot rates of every six months up to 3.5 years
c) Plot the spot rate curve
C. For this part, use first and second set, i.e. total of 7 Treasuries
4.
a) Using regression methods, extract zero coupon bonds. (Note: Force the
intercept to zero in the regression).
b) Calculate 6 month spot rates
c) Plot the spot rate curve
d) Compare the spot rates with the ones you found by bootstrapping. Are
they different? Why? Why not?
Şenay Ağca, George Washington University
2
D. For this part, use all 8 Treasuries, i.e. first, second, and third sets.
5. Using regression methods, find zeros semiannually until year 3 and the annually until
year 4. (Note: Force the intercept to zero in the regression).
6. Using polynomial spline method of order 3
a) Fit the zero curve to find the zero for year 3.5.
b) Calculate 6 month spot rates for 4 years and plot the spot rate curve
7. Using exponential spline method of order 3
a) Fit the zero curve to find the zero for year 3.5. Take initial values of ,
1, 2, 3, and as 0.01 while using the Solver.
b) Calculate 6 month spot rate for 4 years and plot the spot rate curve
c) Using the zeros, find 6 month forward rates after 6 months, 1 year, 1.5
years, .. up to 3 years.
8. Using Nelson-Siegel method
a) Fit the spot curve to find the spot rate for year 3.5. Take initial values of
, 1, 2, and as 1 while using the Solver.
b) What are the 6 month spot rates for 4 years? What are the 6 months
zeros? Plot the spot rate curve.
9.
a) Using the zero coupons you have extracted by polynomial method, find
the price of a coupon bond X that pays 1% coupon semiannually and
that has time to maturity of 4 years. Take the face value of the bond as
\$1000.
b) Using the zeros you have extracted and fit in using expoenential spline
method, find the Fisher-Weil duration.
– If the yield curve shifts up by 1 percent, what is the new price of
Bond X? using only duration measure?
(Hint: Use spot rate of 4 year as the yield.)
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## Related Services
Part II:
Suppose on 12/31/2014, you are expecting that the yield curve will steepen. You want to enter
into a spread trade using \$100 million face value of 2-year bond and a corresponding amount of
20-year bond. The quotes related to these bonds on 12/31/2014 are below. Note that quoted bid
and ask prices are full prices. Take one year as 360 days. Also note that you ARE NOT a bond
Quote Date Maturity date Coupon rate Bid Ask Pvbp
12/31/2014 12/31/2016 3.00% 99.87 99.92 0.0122
12/31/2014 12/31/2034 4.50% 113.62 113.71 0.134
10. How much will you buy or sell of 2-year bond on 12/31/2014 for the spread trade? Will you
buy or sell \$20 million face value of 20-year bond? Why did you buy/sell of each bond? (Use
mid dirty prices to find the amounts)
To finance this transaction you decided to enter into a 2-day repo transaction. The repo dealer
offers 1.00% on reverse repo and 1.25% on repo transaction. Also, the repo dealer gets 0.5%
haircut. After 2 days the quotes are as follows:
Quote Date Maturity date Coupon rate Bid Ask Pvbp
1/3/2008 12/31/2009 3.00% 99.93 99.95 0.0124
1/3/2008 12/31/2027 4.50% 113.71 113.75 0.132
11.
a) Show the details of the repo transaction. What is the profit (loss) on repo
transaction?
b) Show the details of the reverse repo transaction. What is the profit (loss)
of the reverse repo transaction?
c) What is the total profit (loss) of these two transactions?
d) What is the return on your own capital from these transactions?
e) If you haven’t used repo and reverse repo market, what would be your
return on profit? Why is it different than (d)? Where does the difference
come from?
f) Were you correct at predicting steepening of yield curve? Why? Why
not? Be specific.
### Product Code :Fin270
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Summary
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# Number 20210033
### Properties of number 20210033
Cross Sum:
Factorization:
Divisors:
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
Base 32:
j8obh
sin(20210033)
-0.81634043655691
cos(20210033)
-0.57757102735687
tan(20210033)
1.4134026775767
ln(20210033)
16.821689722233
lg(20210033)
7.3055670226547
sqrt(20210033)
4495.5570288897
Square(20210033)
### Number Look Up
Look Up
20210033 which is pronounced (twenty million two hundred ten thousand thirty-three) is a great figure. The cross sum of 20210033 is 11. If you factorisate the number 20210033 you will get these result 173 * 197 * 593. The figure 20210033 has 8 divisors ( 1, 173, 197, 593, 34081, 102589, 116821, 20210033 ) whith a sum of 20464488. The figure 20210033 is not a prime number. The number 20210033 is not a fibonacci number. 20210033 is not a Bell Number. 20210033 is not a Catalan Number. The convertion of 20210033 to base 2 (Binary) is 1001101000110000101110001. The convertion of 20210033 to base 3 (Ternary) is 1102000202221202. The convertion of 20210033 to base 4 (Quaternary) is 1031012011301. The convertion of 20210033 to base 5 (Quintal) is 20133210113. The convertion of 20210033 to base 8 (Octal) is 115060561. The convertion of 20210033 to base 16 (Hexadecimal) is 1346171. The convertion of 20210033 to base 32 is j8obh. The sine of the figure 20210033 is -0.81634043655691. The cosine of the figure 20210033 is -0.57757102735687. The tangent of the figure 20210033 is 1.4134026775767. The root of 20210033 is 4495.5570288897.
If you square 20210033 you will get the following result 408445433861089. The natural logarithm of 20210033 is 16.821689722233 and the decimal logarithm is 7.3055670226547. that 20210033 is very great figure!
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# Avalanche Disk Granular Physics Demonstrator
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## Introduction: Avalanche Disk Granular Physics Demonstrator
Build your own avalanche disk to observe granular physics in action. A single granule is a solid but when it interacts with many, it may behave as a solid, liquid, or gas.
I first saw one of these a while back in some science museum that I took Caitlin to. It was more of an art installation where I thought the inside was like one of those lava lamp fluid things. I recently saw a related video to an avalanche disk and it looks like The Museum of Science and Industry - Chicago has one on display. Theirs is 20 ft in diameter and weighs several tons. This one scaled way down. It's one of those things - "How did they do that?" but more importantly, "Can you make it at home?" Other than that, there is a field of science that applies to this.
Learning Objective: Build and use an avalanche disk to observe the behavior of granular solids as granules flow, shear, mix, separate and freeze.
Knowing how granular solids act is pertinent to helping you predict/prevent snow avalanches to designing machines to package your cereal. Computer modelling can be quite complex.
Added Note: This may have been what I really saw, Rheoscopic fluid. Great ible on it too.https://www.instructables.com/id/Making-Rheoscopic...
Video of the Avalanche Disk in action (overdub with your best Carl Sagan impression on the creation of the universe):
## Step 1: Apparatuses or Apperatii?
You will need:
Some sort of rotating base:
I found this rotating artist's turnstyle stand at a discount store. The rotation is manually powered by hand but something like a motorized potter's wheel or old-fashioned LP record player turntable(don't know if it has enough torque for heavy loads) would be better. I was thinking of attaching my container to a sanding disk that is chucked up in a power drill. You can also build a rotatable base with a lazy-susan turntable or mechanism (IKEA has lazy-susan turntables - I was planning to make a wheel of fortune or casino spinner with one)
Some sort of container for your media:
I found this clear plastic half dome container for storing lettuce halves. It has a dome cover I could put on but any round flat bottom container with a low wall would suffice. A cover is nice so you don't have a toxic spill with the media. You can use the upper half of a CD/DVD spindle cover.
Granular solids:
The museum exhibit description says it uses a mix of glass beads and red garnet sand. I do not know what size grain they use but I remember ibles using glass microbeads for its reflective properties. Harbor Freight is a source for a bulk amount of glass microbeads. I got the smallest(25lb container, around \$25 US) of 80 grit glass microbeads used for sandblasting. The stuff feels like a smooth powder and is quite fine. The craft store that sells you a thimble full for glitter work is too expensive. I don't think you would want to inhale any of it and getting some on the floor makes it a slipping hazard. Keep it contained and work deliberately with it. I think I only used a few ounces of the product and have a lot to use on other projects such as painting a reflective stripe in my driveway(this may be the stuff they dust the street white lines after painting).
I also got a bag of red colored play sand, the stuff used for sand sculptures you fill in bottles. It is also quite fine in texture.
## Step 2: Simple Setup
Pour a layer of the glass microbeads in your base container.
Tilt is so that you can see the "sand" flow down like sand dunes.
Add enough so that the entire bottom is covered and there is still a layer where portions of the avalanche slide down.
Add a top layer of red sand. You might just want to add a small portion at first to see what kind of effect you get when you actually run the experiment. Add more for better contrast.
## Step 3: Affix to Turntable
I just taped my container to the rotating turntable.
Adjust the angle of the turntable. Adjust so that the "sand" just starts to slide down.
This angle may be anywhere from 20 to 30 degrees.
Start the rotation.
## Step 4:
Experiment with rotating the Avalanche Disk at different speeds.
Experiment with different angles or slope of the terrain.
Experiment with different media, maybe substitute sugar, salt, flour, baking soda, and kool-aid or flavored drink mix as the coloring agent.
Observe how granular solids behave en masse as a solid, liquid, or a gas. Note the vortex-like flow patterns. Record and discuss your findings.
Why doesn't it all mix into one homogenous mix of pink sand? Why does it separate out at points?
What happens when you introduce obstacles to interrupt or divert the natural flow?
I think I might get some blue sand to make an event horizon insert into my Stargate Gong and turn it into a coffee table.
Cool!
Participated in the
The Teacher Contest
Participated in the
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## 8 Discussions
That's pretty cool! Thanks for showing me this. So the different sands have a different specific gravity and never really mix. I like the idea of the stargate table too :)
Thanks. It's great fun to figure out how things work and see science in action.
For your "Apparatuses or apperatii?" question, the plural of the noun apparatus is apparatuses, although in this context just apparatus should suit you fine!
And so you should, commenti such as that are inspirational, and we need more inspirati in our lives.
Sweet! I remember seeing one of these at the Tec Museum but I never knew what the scientific name was for them. Thanks for the info!
I thought it might have also been some kind of ferro-fluid. One more mystery of the universe revealed.
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# computer science unplugged€¦ · 1 computer science unplugged dr. tom cortina carnegie mellon...
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1 Computer Science Unplugged Dr. Tom Cortina Carnegie Mellon University Computer Science Unplugged CS Unplugged is a book of activities that illustrate computer science principles without using a computer. Activities are short and are designed to be easily integrated into classes and include exercises and lesson plans for teachers.
Post on 05-Jun-2018
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Category:
## Documents
TRANSCRIPT
1
Computer Science Unplugged
Dr. Tom Cortina Carnegie Mellon University
Computer Science Unplugged
• CS Unplugged is a book of activities that illustrate computer science principles without using a computer.
• Activities are short and are designed to be easily integrated into classes and include exercises and lesson plans for teachers.
2
CS UNPLUGGED
• The basic edition of Computer Science Unplugged has 12 classroom exercises for you to use with your students.
• Each exercise has a number of extensions, activities and background information.
• All activities can be done without the use of computers, but they all demonstrate fundamental principles used in computers today.
TWENTY GUESSES
• How much information is there in a 1000-page book? Is there more information in a 1000-page telephone book, or in Tolkien's Lord of the Rings? § If we can measure this, we can estimate how much
space is needed to store the information. • This activity introduces a way of measuring
information content.
3
TWENTY GUESSES
• Can you read the following sentence?
"Ths sntnc hs th vwls mssng. "
• You probably can, because there is not much "information" in the vowels.
TWENTY GUESSES
• I am thinking of a number between 1 and 100.
• I will start you off with 20 pieces of candy. • You may only ask questions that have a
"yes" or "no" answer. • For each incorrect guess, you will lose one
piece of candy. • Once you guess correctly, you can keep
whatever candy remains.
4
TWENTY GUESSES
• To pick a number between 0 and 100, you only need 7 guesses. § Always shoot for the middle number of the range and
eliminate half the possibilities! § This concept is called binary search.
• If the number was between 0 and 1000, you would only need 3 additional guesses.
• You can guess a number between 0 and 1 million in only 20 guesses!
LIGHTEST & HEAVIEST
• Computers are often used to put lists into some sort of order (e.g. names into alphabetical order, appointments or e-mail by date, etc.) § If you use the wrong method, it can take a long time
to sort a large list into order, even on a fast computer.
• In this activity children will discover different methods for sorting, and see how a clever method can perform the task much more quickly than a simple one.
5
LIGHTEST & HEAVIEST
• Start with 8 containers with different amounts of sand or water inside. Seal tightly.
• Children are only allowed to use the scales to compare the relative weights of two containers.
• Only two containers can be compared at a time.
LIGHTEST & HEAVIEST
• METHOD 1 is called Selection Sort. • METHOD 2 is called Quick Sort.
• Generally, quick sort is a lot faster than
selection sort is.
6
BATTLESHIPS
• Computers are often required to find information in large collections of data.
• Computer scientists study quick and efficient ways of doing this.
• This activity demonstrates three different search methods so children can compare them.
BATTLESHIPS
• Battleships are lined up at sea. • Each battleship has a number that is hidden. • How many guesses does it take for you to find a
specific battleship? § The number of guesses is the child's score. § The lowest score wins.
7
BATTLESHIPS
GAME 1: Ships are randomly ordered.
FIND SHIP # 717
1630 9263 4127 405 4429 7113 3176 4015 7976 88 3465 1571 8625
2587 7187 5258 8020 1919 141 4414 3056 9118 717 7021 3076 3336
BATTLESHIPS
GAME 2: Ships are in increasing order.
FIND SHIP # 5897
33 183 730 911 1927 1943 2200 2215 3451 3519 4055 5548 5655
5785 5897 5905 6118 6296 6625 6771 6831 7151 7806 8077 9024 9328
8
BATTLESHIPS
GAME 3: Ships are ordered into 10 groups based on a mystery function.
FIND SHIP # 8417
BATTLESHIPS
• These three games illustrate § linear search § binary search § hashing
• What is the maximum number of guesses required for each of these search techniques § for 26 battleships? § for n battleships?
9
THE MUDDY CITY
• For a particular network, there is usually some choice about where the links can be placed.
• This exercise examines a complete network to determine the links necessary to connect all the components of the network at minimal cost.
THE MUDDY CITY
10
THE MUDDY CITY
THE MUDDY CITY
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2
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3
4 4
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a graph
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THE MUDDY CITY
THE MUDDY CITY
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THE MUDDY CITY
• This exercise illustrates how to build what we call the “minimal spanning tree”. § A tree does not have any cycles where you
can get back to where you were before. • This exercise does not give us the shortest
path from one location to another. § But there is another algorithm for that!
BEAT THE CLOCK
13
BEAT THE CLOCK
• This activity illustrates structures used in parallel sorting networks.
• Kids sort data by walking through a sorting network laid out on the floor.
• The network simulates how a parallel network would sort data. § Kids find out that data can be sorted a lot
faster in parallel!
CS UNPLUGGED
• The teacher's version of Computer Science Unplugged is available online at !!http://www.csunplugged.org § The book is FREE to download and use!
• Additional material will be published soon to add even more activities, including video to demonstrate how to use these activities effectively in your classroom.
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# Convert cordless tools to corded
Wonder if Robert Graham got the job done he posted question last January about converting cordless to corded. I did it today using an old Western Auto 10 amp car battery charger. I used the tools today (7.2 volt Black and Decker saws that had 2 VersaPak removable batterries each), no problems, also wired a Black and Decker drill, 4.8 volt with built in battery pack, (removed the battery pack). Still used the 12 volt setting on the old Western Auto 10 amp charger. Simply took cords off of old radio's and stuff opened up the tools and figured out positive and negative, forgot to mark the other end of the cord and had to take apart and use meter to get it right. Works great, got them set up at thier own work bench with the charger on shelf, can see amp needle on charger, when start drilling or sawing the needle jumps all the way then settles back to 5 amps, 6 or 7 when drilling with hard pressure.
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On 5 Sep 2004 20:24:45 -0700, snipped-for-privacy@onemain.com (William Krems) vaguely proposed a theory ......and in reply I say!:
Well. 4.8v mnotor on 12v. We will hear from you again tomorrow?
***************************************************** I know I am wrong about just about everything. So I am not going to listen when I am told I am wrong about the things I know I am right about.
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There's theory and then there's the real world. The internal resistance of an auto battery charger is pretty high. A twelve volt potential should push enough current through something designed for 4.8 volts to fry the sucker. In the real world, with the setup he describes, while not a sure bet, is likely to work for a good long time. The internal resistance of the charger is likely enough higher than the battery pack the drill came with that both machines (drill and charger) will survive in good shape.
bob g.
Old Nick wrote:
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Just because it can be done is no reason to do it. Corded tools are probably safer and I'd bet a hell of a lot more powerfull...pound for pound, or comparably priced. But I guess if you have a few spare tools, a lot of time and a desire anything is possible?
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pound,
Maybe so, but I can think of a few reasons to do it. Car battery on hand, cordless batteries have died or are not retaining much of a charge. Maybe cost of new batteries or rebuilding batteries is too much. Maybe it's just an educational exercise. Haven't you ever tried something just to see if it works? A large proportion of what we do is a waste of time and money.
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vaguely proposed a theory ......and in reply I say!:
Actually, it's interesting. I have a rellay expensive 12V cordless. The guts are great, but the supplied battery packs (at \$120 per pack) are dead. They were proprietary NiCds, and a ripoff.
Suddenly it occurs to me that I can get NiMH cells with umpteenth the capacity, for a tenth of the price of replacement of the prop packs.
The only trouble is, of course, that the charger will go in the bin, and that I will have to crack the pack to charge the new cells in the intelligent NiMH charger. Temperature and all that.
There has to be a market there somewhere. I have a couple of 2-ways that you can get in-radio chargers for, but of course the chargers take 24 hours to charge 800mAH NiMh cells, because a quick charger cannot measure temp if the cells are still in the radios. Devices that talk to the chargers? Why not? Probably \$1 to make and \$10 on the price.
***************************************************** I know I am wrong about just about everything. So I am not going to listen when I am told I am wrong about the things I know I am right about.
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Old Nick wrote:
Who knows, if you could develop the technology you might have a nice little niche market going. I've got a couple of dead DeWalt packs here that it's not cost effective to have rebuilt with NiCd but if I could get NiMH in them for the retail price of a new NiCd pack or less and get a fast charger to go with them I'd do it in an instant.
--
--John
Reply to jclarke at ae tee tee global dot net
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Another reason is recyceling, those tools will go to the dump otherwise. I don't understand the 12 volt setting working on the 7.2 and 4.8 volt tools but I've used them several times with no problems, no heating up or anything.
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William Krems wrote:
I wouldn't be too if there was a voltage regulator in the tool.
--
Morris Dovey
DeSoto, Iowa USA
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Upscale wrote:
Yuppers, many new things have been invented by people tinkering around. Good stuff Maynard! Besides, we've become too much of a throw-away society. Grandpa
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Upscale wrote:
I've done a similar thing with the cord and plug off a dead truckair compressor and an old metabo 12v drill that had 1 dead pack and one on its way out (the charger went years ago, charged them off a bench psu). Now I can plug into a lighter socket in any car/suv or use my 7ah leadacid power pack, much better than throwing away a perfectly good drill that packs can't be bought for.
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LOL - Everything I ever made in my shop comes to mind...
-CJ
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How about a cordless tool that works but the replacement battery costs more than you want to spend and you don't use the cordless feature that much anyway. I can't imagine a whole lot of folks running out and buying cordless with the plan to convert but if you've used the thing until the battery quits, what do you have to lose?
bob g.
Dave in WA wrote:
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# An amount of Rs. 1,00,000 is invested in two types of shares. The first yields an interest of 9% p.a. and second, 11% p.a. If the total interest at the end of one year is $9 \frac{3}{4} \%$ then the amount invested in each share was?
A. Rs. 52, 500; Rs. 47, 500 B. Rs. 62, 500; Rs. 37, 500 C. Rs. 72, 500; Rs. 27, 500 D. Rs. 82, 500; Rs. 17, 500 Answer: Option B
Show Answer
### Solution(By Apex Team)
Let the sum invested at 9% be Rs. x and that invested at 11% be Rs. (100000 – x). Then, $\begin{array}{l}=\left(\frac{x\times9\times1}{100}\right)+\left[\frac{(100000-x)\times11\times1}{100}\right]\\ =\left(100000\times\frac{39}{4}\times\frac{1}{100}\right)\\ \Leftrightarrow\frac{9x+1100000-11x}{100}=\frac{39000}{4}=9750\\ \Leftrightarrow2x=(1100000-975000)=125000\\ \Leftrightarrow x=62500\\ \therefore \text { Sum invested at } 9 \% \\ =\text { Rs. } 62500 \\ \text { Sum invested at } 11 \% \\ =\text { Rs. }(100000-62500) \\ =\text { Rs. } 37500 \end{array}$
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Warning: Illegal string offset 'html' in /home/hsn/public_html/forum/cache/skin_cache/cacheid_1/skin_topic.php on line 909
2 questions about logs - HSN forum
2 replies to this topic
### #1Nathan
Fully Fledged Genius
• Members
• 1,736 posts
• Location:Aberdeen, Scotland
• Gender:Male
Posted 08 October 2006 - 08:25 PM
1.
3logax-logax-logaz
=logax3-logax-logaz
=loga(x3/y/z)
where do I go from there?
is it
=loga(x3z/y) or not?
2.
log2(x+1) - log2x = 3
how do you do that one?
god...I hate maths!! lol
### #2Steve
Top of the Class
• 435 posts
• Location:Edinburgh
• Gender:Male
Posted 08 October 2006 - 08:38 PM
You're making the first one more complicated than it is .
I'm assuming you've just to simplify the expression. You've got but this is just . This should make it simpler.
For the second one, I assume you've to solve for x. You can combine the LHS into a single log term then put into exponential form using the rule . I get , if you're interested.
I know you won't want to, because you'll feel like a school pupil again, but you could always have a look at the Higher Logarithms notes!
HSN contribute: Help the site grow!
Looking for a Maths tutor in West Lothian? Just PM me!
### #3Nathan
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• Members
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• Gender:Male
Posted 08 October 2006 - 08:40 PM
Thank you
QUOTE
I know you won't want to, because you'll feel like a school pupil again, but you could always have a look at the Higher Logarithms notes!
lol...thanks
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## Engage NY Eureka Math 4th Grade Module 5 Lesson 23 Answer Key
### Eureka Math Grade 4 Module 5 Lesson 23 Problem Set Answer Key
Question 1.
Circle any fractions that are equivalent to a whole number. Record the whole number below the fraction.
a. Count by 1 thirds. Start at 0 thirds. End at 6 thirds.
0/3, 1/3, 2/3, 3/3, 4/3, 5/3, 6/3.
Explanation:
In the above-given question,
given that,
circle any fractions that are equivalent to a whole number.
count by 1 third.
start at 0 thirds.
end at 6 thirds.
0/3 + 1 = 1/3.
1/3 + 1 = 2/3.
2/3 + 1 = 3/3.
3/3 + 1 = 4/3.
4/3 + 1 = 5/3.
5/3 + 1 = 6/3.
0/3 = 0, 3/3 = 1, 6/3 = 2.
b. Count by 1 halves. Start at 0 halves. End at 8 halves.
0/2, 1/2, 2/2, 3/2, 4/2, 5/2, 6/2, 7/2, 8/2.
Explanation:
In the above-given question,
given that,
circle any fractions that are equivalent to a whole number.
count by 1 half.
start at 0 halves.
end at 8 halves.
0/2 + 1 = 1/2.
1/2 + 1 = 2/2.
2/2 + 1 = 3/2.
3/2 + 1 = 4/2.
4/2 + 1 = 5/2.
5/2 + 1 = 6/2.
6/2 + 1 = 7/2.
7/2 + 1 = 8/2.
0/2 = 0, 2/2 = 1, 4/2 = 2, 6/2 = 3, 8/2 = 4.
Question 2.
Use parentheses to show how to make ones in the following number sentence.
$$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ + $$\frac{1}{4}$$ = 3
1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4 + 1/4.
Explanation:
In the above-given question,
given that,
use parantheses to make ones .
(1/4) + (1/4) + (1/4) + (1/4).
1/4 = 0.25.
0.25 + 0.25 + 0.25 + 0.25.
0.50 + 0.50 = 1.
1 + 1 + 1 = 3.
Question 3.
Multiply, as shown below. Draw a number line to support your answer.
a. 6 × $$\frac{1}{3}$$
6 x 1/3 = 2.
Explanation:
In the above-given question,
given that,
6 x 1/3.
3 x 1/3 + 3 x 1/3.
2 x 3/3 = 2.
2 x 1 = 2.
b. 6 × $$\frac{1}{2}$$
6 x 1/2 = 3.
Explanation:
In the above-given question,
given that,
6 x 1/2.
3 x 1/2 + 3 x 1/2.
3 x 2/2 = 3.
3 x 1 = 3.
c. 12 × $$\frac{1}{4}$$
12 x 1/4 = 3.
Explanation:
In the above-given question,
given that,
12 x 1/4.
6 x 1/4 + 6 x 1/4.
6 x 2/2 = 3.
3 x 1 = 3.
Question 4.
Multiply, as shown below. Write the product as a mixed number. Draw a number line to support your answer.
a. 7 copies of 1 third
7 x 1/3Â = 7/3 = 2(1/3).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/3.
(2 x 3/3) + 1/3.
2 + 1/3.
2(1/3).
b. 7 copies of 1 half
7 x 1/2Â = 7/2 = 3(1/2).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/2.
(3 x 2/2) + 1/2.
3 + 1/2.
3(1/2).
c. 10 × $$\frac{1}{4}$$
10 x 1/4Â = 10/4 = 2(2/4).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
10 x 1/4.
(2 x 4/4) + 2/4.
2 + 2/4.
2(2/4).
d. 14 × $$\frac{1}{3}$$
14 x 1/3Â = 14/3 = 4(2/3).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
14 x 1/3.
(4 x 3/3) + 2/3.
4 + 2/3.
4(2/3).
### Eureka Math Grade 4 Module 5 Lesson 23 Exit Ticket Answer Key
Multiply and write the product as a mixed number. Draw a number line to support your answer.
Question 1.
8 × $$\frac{1}{2}$$
8 x 1/2 = 4.
Explanation:
In the above-given question,
given that,
8 x 1/2.
4 x 1/2 + 4 x 1/2.
4 x 2/2 = 4.
4 x 1 = 4.
Question 2.
7 copies of 1 fourth
7 x 1/4Â = 7/4 = 4(3/4).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/4.
(4 x 2/2) + 3/4.
4 + 3/4.
4(3/4).
Question 3.
13 × $$\frac{1}{3}$$
13 x 1/3 = 13/3.
Explanation:
In the above-given question,
given that,
13 x 1/3.
6 x 1/3 + 7 x 1/3.
13 x 1/3.
13/3.
### Eureka Math Grade 4 Module 5 Lesson 23 Homework Answer Key
Question 1.
Circle any fractions that are equivalent to a whole number. Record the whole number below the fraction.
a. Count by 1 fourths. Start at 0 fourths. Stop at 6 fourths.
0/4, 1/4, 2/4, 3/4, 4/4, 5/4, 6/4.
Explanation:
In the above-given question,
given that,
circle any fractions that are equivalent to a whole number.
count by 1 fourth.
start at 0 fourths.
end at 6 fourths.
0/4 + 1 = 1/4.
1/4 + 1 = 2/4.
2/4 + 1 = 3/4.
3/4 + 1 = 4/4.
4/4 + 1 = 5/4.
5/4 + 1 = 6/4.
0/4 = 0, 2/4 = 2, 4/4 = 1.
b. Count by 1 sixths. Start at 0 sixths. Stop at 14 sixths.
0/6, 1/6, 2/6, 3/6, 4/6, 5/6, 6/6, 7/6, 8/6, 9/6, 10/6, 11/6, 12/6, 13/6, 14/6.
Explanation:
In the above-given question,
given that,
circle any fractions that are equivalent to a whole number.
count by 1 sixth.
start at 0 sixths.
end at 14 sixths.
0/6 + 1 = 1/6.
1/6 + 1 = 2/6.
2/6 + 1 = 3/6.
3/6 + 1 = 4/6.
4/6 + 1 = 5/6.
5/6 + 1 = 6/6.
6/6 + 1 = 7/6.
7/6 + 1 = 8/6.
8/6 + 1 = 9/6.
9/6 + 1 = 10/6.
10/6 + 1 = 11/6.
11/6 + 1 = 12/6.
12/6 + 1 = 13/6.
13/6 + 1 = 14/6.
0/6 = 0, 12/6 = 2, 6/6 = 1.
Question 2.
Use parentheses to show how to make ones in the following number sentence.
$$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ + $$\frac{1}{3}$$ = 4
1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3 + 1/3.
Explanation:
In the above-given question,
given that,
use parantheses to make ones .
(1/3) + (1/3) + (1/3) + (1/3).
1/3 = 0.75.
0.75 + 0.75 + 0.75 + 0.75.
1.5 + 1.5 = 3.
3 + 1 = 4.
Question 3.
Multiply, as shown below. Draw a number line to support your answer.
a. 6 × $$\frac{1}{3}$$
6 x 1/3 = 2.
Explanation:
In the above-given question,
given that,
6 x 1/3.
3 x 1/3 + 3 x 1/3.
2 x 3/3 = 2.
2 x 1 = 2.
b. 10 × $$\frac{1}{2}$$
10 x 1/2 = 5.
Explanation:
In the above-given question,
given that,
10 x 1/2.
5 x 1/2 + 5 x 1/2.
5 x 2/2 = 5.
5 x 1 = 5.
c. 8 × $$\frac{1}{4}$$
8 x 1/4 = 2.
Explanation:
In the above-given question,
given that,
8 x 1/4.
4 x 1/4 + 4 x 1/4.
2 x 4/4 = 2.
2 x 1 = 2.
Question 4.
Multiply, as shown below. Write the product as a mixed number. Draw a number line to support your answer.
a. 7 copies of 1 third
7 x 1/3Â = 7/3 = 2(1/3).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/3.
(2 x 3/3) + 1/3.
2 + 1/3.
2(1/3).
b. 7 copies of 1 fourth
7 x 1/4Â = 7/4 = 1(3/4).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/4.
(1 x 4/4) + 3/4.
1 + 3/4.
1(3/4).
c. 11 groups of 1 fifth
11 x 1/5Â = 11/5 = 2(1/5).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
11 x 1/5.
( 2 x 5/5) + 1/5.
2 + 1/5.
2(1/5).
d. 7 × $$\frac{1}{2}$$
7 x 1/2Â = 7/2 = 3(1/2).
Explanation:
In the above-given question,
given that,
write the product as a mixed number.
7 x 1/2.
(3 x 2/2) + 1/2.
3 + 1/2.
3(1/2).
e. 9 × $$\frac{1}{5}$$
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1. Explain what is Transmission Gate-based D-Latch?
The Transmission-Gate input is connected to the D_LATCH data input (D), the control input to the Transmission-Gate is connected to the D_LATCH enable input (EN) and the Transmission-Gate output is the D_LATCH output (Q).
2. How to detect sequence of "1101" arriving serially from signal line?
Sequence detector : A sequence detector gives an output of 1 on detecting the given sequence else the output is zero.
Ex : if the given sequence to be detected is 111
and input stream is 1 1 0 1 1 1 0 0 1 0 1 1 1 1 1
the output should be 0 0 0 0 0 1 0 0 0 0 0 0 1 1 1.
Soln:
One of the different possible ways to detect a sequence is using a Mealy type FSM.
Using the following table the State machine can be designed.
3. Which are the two ways of converting a two input NAND gate to an inverter?
Short the two inputs of the nand gate and give the same input to the common wire,the nand gate works as an inverter.
One way is shorting the two inputs of the NAND gate and passing the input.
truth table:
A B output
1 1 0
0 0 1
The second way is passing the input to only one input(say A) of the NAND gate.Since the other input(say B) is floating, it is always logic one.
truth table:
A B output
1 1 0
0 1 1
4. How to design a divide-by-3 sequential circuit with 50% duty circle?
take a counter with 3 f/f's that is to say with 6 states(2*3) now double the i/p clock frequency to the counter the o/p of the 3rd f/f is divide by 6 of the i/p with 50% duty cycle so effectively u got divide by 3 freq with 50% duty cycle.
41. What is equivalence checking?
42. What is the Moores law?
43. What is the difference between simulation and emulation?
44. What is the use of RTL?
45. State the differences between VHDL and Verilog?
46. What do you understand by synthesis?
47. Explain the bottom-up design process.
48. What are synchronous circuits?
49. How can distorted signals be identified with the use of SNR?
50. What are the various practices for improving SNR?
31. What are data flow diagrams?
32. What are compilers?
33. Explain the functioning of compilers.
34. What are microcontrollers?
35. Explain the functioning of microcontrollers.
36. What is assembly language?
37. What do you understand by interrupt latency?
38. Explain RTOS.
39. How does a binary counter works?
40. What are the numbers of bit combinations in a byte?
21. Give the truth table of a half-adder?
22. What are transmission gates?
23. What do you understand by DFT?
24. What are fault models?
25. Explain scan technologies
26. Explain BST?
27. What are the differences between BST and BIST?
28. State the differences between LFSR and MISR?
29. What is IDDQ?
30. What do you understand by the term recovery design?
11. What is the dissipation factor? How is it calculated?
12. Explain tolerance?
13. What is a potentiometer used for?
14. What are the various hardware design tools available?
15. What is a HDL?
16. How can a two input NAND gate be converted into an inverter?
17. Define set up time & hold time constraints?
18. Draw a circuit to divide the clock frequency by half.
19. Design a divide-by-3 sequential circuit with 50% duty circle.
20. What are adder circuits explain?
1. What are the various hardware design techniques?
2. What are passive components/
3. What are the various board design issues?
4. What are analog power supply systems?
5. What do you understand by overvoltage protection?
6. What is thermal management?
8. What is the purpose of prototyping?
9. Explain the term dielectric absorption?
10. What do you understand by capacitor parasitic?
10. What are the different Adder circuits you studied?
Adders are generally of five types:
The Ripple carry adder(RCA) consists of a building block named Half
blocks HAs and FAs are also the building blocks of all types of
The full adder has three input pins(input Ai,input Bi,carryin Ci) and
two output pins(Sum and Ci+1).Its equations are:
Sum=Ai^Bi^Ci
Ci+1=Ai.Bi+Bi.Ci+Ai.Ci
The Carry Lookahead Adder(CLA) reduces the delay as that in RCA. Let
Gi=Ai.Bi, and Pi=Ai^Bi, then Ci+1=Gi+Pi.Ci.
The expressions for Sum and Ci+1 is then defined completely in terms of
input pins rather wait for input carry to appear.
The carry select adder uses duplicate modules for each combination of
input carry(i.e. 1 and 0).The multiplexers then select the appropriate
sum and carry output according to the carry output of the preceding
stages.
11. What are set up time & hold time constraints? What do they signify? Which one is critical for estimating maximum clock frequency of a circuit?
Suppose your flip-flop is positive edge triggered. time for which data should be stable prior to positive edge clock is called setup time constraint .
Time for which data should be stable after the positive edge of clock is called as hold time constraint.
if any of these constraints are violated then flip-flop will enter in meta stable state, in which we cannot determine the output of flip-flop.
there are two equation:
1. Tcq + Tcomb> Tskew + Thold
2. Tcq + Tcomb<Tskew +T - Tsetup
Tcq is time delay when data enters the flip-flop and data comes at output of flip flop.
Tcomb is the logic delay between two flip flop.
Tskew is the delay of clock to flip flop: suppose there are two flip flop ,if clock reaches first to source flip flop and then after some delay to destination flip flop ,it is positive skew and if vice versa then negative skew.
so if you take 2 eq you will see that setup time is the determining factor of clock's time period.
12. Give a circuit to divide frequency of clock cycle by two?
You can divide the frequency of a clock by just implementing T Flip flop.
Give clock as clock input and tie the T input to logic 1.
13. How do you detect if two 8-bit signals are same?
XOR each bits of A with B (for eg A[0] xor B[0] ) and
so on. the o/p of 8 xor gates are then given as i/p to
an 8-i/p nor gate. if o/p is 1 then A=B.
14. Give two ways of converting a two input NAND gate to an inverter.
One way is shorting the two inputs of the NAND gate and passing the input.
truth table:
A B output
1 1 0
0 0 1
The second way is passing the input to only one input(say A) of the NAND gate.Since the other input(say B) is floating, it is always logic one.
truth table:
A B output
1 1 0
0 1 1
15. Design a divide-by-3 sequential circuit with 50% duty circle.
Take a smiths counter with 3 f/f's
that is to say with 6 states(2*3)
now double the i/p clock frequency to the counter
the o/p of the 3rd f/f is divide by 6 of the i/p with 50% duty cycle
so effectively u got divide by 3 freq with 50% duty cycle
16. Give the truth table for a Half Adder. Give a gate level implementation of the same.
A B SUM CARRY
0 0 0 0
0 1 1 0
1 0 1 0
1 1 0 1
IMPLEMENTATION:
For SUM, The two inputs A and B are given to XOR gate.
For Carry, The two inputs A and B are given to AND gate.
17. How do you detect a sequence of "1101" arriving serially from a signal line?
Sequence detector : A sequence detector gives an output of 1 on detecting the given sequence else the output is zero.
Ex : if the given sequence to be detected is 111
and input stream is 1 1 0 1 1 1 0 0 1 0 1 1 1 1 1
the output should be 0 0 0 0 0 1 0 0 0 0 0 0 1 1 1.
Soln:
One of the different possible ways to detect a sequence is using a Mealy type FSM.
Using the following table the State machine can be designed.
since the number of bits in the sequence 1101 is 4 we have 4 states
------------------------------------------------------
|PS | Seq detected by the state | NS/output |
| | |---------------|
| | | X=0 | X=1 |
|-----------------------------------------------------
| S1 | - | S1/0 | S2/0 |
|----------------------------------------------------|
| S2 | 1 | S1/0 | S3/0 |
|----------------------------------------------------|
| S3 | 11 | S4/0 | S3/0 |
|----------------------------------------------------|
| S4 | 110 | S1/0 | S2/1 |
|----------------------------------------------------|
when in state S4 (PS),and input(X) from the sequence is 1,the sequence "1101" has been detected once and (to find the next state select the longest "seq identified by a state" column that matches part of the sequence 1101--ie.,1 or 01 or 101 ....)the NS is S2 since the sequence detected by the state S2 is 1(in 1101- 01 or 101 ,etc are not present in the seq identified by the state column ,)
18. Suppose you have a combinational circuit between two registers driven by a clock. What will you do if the delay of the combinational circuit is greater than your clock signal?
Use the concept of register-retiming.
divide the total combinatorial delay in two segments such that individually the delay is less the clock period.
this can be done by inserting a flip-flop in the combinational path.
e.g,
clock period --- 5 ns
total cominational delay ---- 7
then divide the 7ns path in two path of 4 or 3 (best results are obtained if delays are same for both path i.e 3.5ns) by inserting a flip-flop in between.
19. Draw a Transmission Gate-based D-Latch?
The Transmission-Gate's input is connected to the D_LATCH data input (D), the control input to the Transmission-Gate is connected to the D_LATCH enable input (EN) and the Transmission-Gate output is the D_LATCH output (Q)
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325 HW 03
# 325 HW 03 - MAT 325 Topology Professor Zoltan Szabo Problem...
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Unformatted text preview: MAT 325: Topology Professor Zoltan Szabo Problem Set 3 Rik Sengupta [email protected] February 24, 2010 1. Munkres, p. 118, problem 6 Let x 1 , x 2 ,... be a sequence of the points of the product space Q X α . Show that this sequence converges to the point x if and only if the sequence π α ( x 1 ) ,π α ( x 2 ) ,... converges to π α ( x ) for each α . Is this fact true if one uses the box topology instead of the product topology? Solution. We will prove both directions. (= ⇒ :) Suppose the sequence x 1 , x 2 ,... converges to some x in the product topology. For a fixed α let U α be a neighborhood of π α ( x ). It remains only to check that U α contains π α ( x i ) for i larger than some N . This follows from the fact that U = Q β V β , where V β = X β for β 6 = α and V β = U α is a neighborhood of x , and hence contains x i for i larger than some N . ( ⇐ =:) Conversely, suppose that { π α ( x i ) } converges to π α ( x ) for all α . Let U be a basis element that covers x , so that U = Q α U α and U α = X α except when α = α k , for k = 1 , 2 ,...,K . Now, for each α k , we know { π α k ( x i ) } converges to π α k ( x ), which means that there is some N k such that π α k ( x i ) ∈ U α k whenever i > N k . Set N = max { N k : k = 1 , 2 ,...,K } . Then for all i > N , we have x i ∈ U , and therefore, { x i } converges to x . This is not true in the box topology. Take your space as R ω , and let ( x i ) be an infinite sequence consisting of zeroes everywhere but 1 at the i th position precisely. But then, it immediately follows that each { π α ( x i ) } converges to zero, but the open neighborhood Q n (- 1 n , 1 n ) around contains none of the x i ’s. One direction is still true. The “only if” part is still true, and the above proof works without any change. 2. Munkres, p. 118, problem 7 Let R ∞ be the subset of R ω consisting of all sequences that are “eventually zero,” that is, all sequences ( x 1 ,x 2 ,... ) such that x i 6 = 0 for only finitely many values of i . What is the closure of R ∞ in R ω in the box and product topologies? Justify your answer. Solution. We consider the two cases separately. Product Topology Choose any x ∈ R ω . There is a basis element B containing x . Then, we may write B as B = ∞ Y i =1 U i , where U i = R for all but finitely many indices i . So now, we may define z ∈ R ∞ as: 1 z i = 0 if U i = R x i if U i 6 = R Therefore, obviously, z ∈ B . It follows that the closure R ∞ is the whole space R ω ....
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325 HW 03 - MAT 325 Topology Professor Zoltan Szabo Problem...
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## An approximation of service frequency
Here is a rather crude, though I think useful, visualization of service frequency at the stop level. Basically, I used the GTFS data from SORTA and TANK to calculate the number of times a bus stops at each stop every week. Since a week is the basic cycle period of transit(service is bad on Sunday, better on monday), this should give us a an idea of basic average frequency with the huge caveat that there’s enormous variation within each week.
Click the image to get a bigger version. There’s lot’s of interesting detail in there!
You may notice that frequency can appear vary in a single line where it doesn’t seem like it probably should:
Ludlow Avenue
In most cases, this is simply an artifact of the way I grouped stops that were next to each other and had exactly the same name. At least 2-3,000 stops of the 6,000 stops in the dataset can reasonably be thought of as pairs with one serving each direction of travel.
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Posted in: Analysis | Data | Maps
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## Reading the Frequency Map for average travel times
Here’s a little known feature of the Transit Frequency Map that’s worth pointing out. You can determine your approximate total travel time, including waiting, without having to look at a schedule. Here’s how:
• Find the line or corridor you need.
• Determine it’s approximate frequency.
• Divide that in half to get the average wait time.
• Find the travel time by counting the white dots between your origin and destination. The distance between two dots represents about 10 minutes of travel.
• Add that to your average wait, and boom, you have your average total trip time.
Let’s walk through an example. Get your map out:
We’re going from Downtown to the College Hill business district on a weekday afternoon. We’ll need to use the #17. Look at the legend and compare the line thickness to find the #17’s approximate frequency.
It looks like it comes about every 15 minutes on average. Half of that is 7.5 minutes, but let’s be pessimists and round it to 8.
Now count your dots!
That’s 40 minutes plus the 8 minute average wait, meaning a total of 48 minutes for our trip. A quick schedule check confirms this is about right. You can also get a worst-case scenario by assuming you just missed a bus and using the whole frequency value without dividing it. If the longest you can wait is about 15 minutes, the longest your trip will take is about 55 minutes.
Remember, this isn’t exact since traffic speeds and frequencies vary by time of day. But generally, It’s pretty accurate. Try it out for your normal trip and see if it works for you…Then try a new one! It’s important to note that where lines are redundant, like the #4 and #11 are between Downtown, you can add their average frequencies together and get a lower average wait time because you can take whichever one comes by first.
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Posted in: Maps | Tips & Tricks
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• ### This blog is:
A shout at a passing car.
An auto-discursive manifesto.
An excited squeal at the intersection of open-computing and public transport data.
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Maths presents a problem to most people. Being good at maths is a skill that many say comes to hard to them. As with so many things, the key to a lot of basic arithmetic is practice, practice, practice. It is not for nothing that so many school systems insist on rote memory learning of multiplication tables and drilling in basic adding, subtracting, multiplication and division. These skills are useful to everyone throughout life. They are a basic competence demanded by most employers for a job of any kind.
It turns out that rote learning and drilling may not be the only ways to improve your basic skills in maths. It seems that most of us are born with an innate ability to estimate quantities, amounts, rates of progress and the likely impact of the interaction of different variables of this kind. It is this ability that makes it possible for us to cope with the demands of driving a car where we are constantly estimating the impact of random variables on the road and in the traffic to guide our own decisions. We take this innate skill for granted. This may be why most people are happy to drive even though the statistics who that this is one of the most dangerous activities any of us can undertake in the daily course of life!
Our ability to reliably compare and contrast quantities, rates of progress, volumes and amounts without actually counting anything is incredibly useful. You use this skill without thinking when you decide which queue to join when waiting to pay at a till in a shop or when queuing at a road toll. Most people can find the midpoint of a line with a fair degree of accuracy by eye.
Most can say which of two fairly equivalent heaped amounts (of food or goods or whatever) is the larger. Now it seems that we can use this ability to estimate in fairly complex ways. Most of us can say whether one set of items is larger than the sum of two smaller sets of the same items.
This ability allows us to cut through a certain amount of complexity to make on-the-spot decisions about what may be important to us in the area immediately around us. In short, estimating with accuracy seems to be one of our survival skills.
Practising your estimating skills is a good thing to do. The more you practice, the more you hone the accuracy and speed with which you can estimate. Getting children to play quantity-based guessing games helps develop the accuracy of estimating early. More importantly, guessing games of these kinds seem to help develop ability in basic maths too. It seems that the basis on which we make our guesses is aligned to the logic which underpins all mathematical calculations. Guessing seems to help develop the building blocks in the brain for this type of logic.
The important thing here is that guessing of this kind requires no formal ability to count or to explain. So playing games of this sort can help give children a good start in maths before they understand the concepts of numbers and while they are learning to speak. For adults with a fear of maths, guessing and estimating may offer a new pathway to enjoying maths skills.
It’s anyone’s guess how this will translate into new business and learning opportunities. But it is a fair estimate that entrepreneurs and educators will seize on guessing games to create profitable new business aimed at toddlers and adults alike.
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# Python Program to Check a Year is Leap Year or Not
Hey Shouters!! Today we have come up with the Python program to check leap year.
Leap year program is a very important program for the placement point of view and from exams point of the year.
## What is a Leap Year?
Leap year is a year in which there is an extra one day in the year.
In a normal year, the number of days is 365 but in a leap year we have 366 days and that is because of the distance between the earth and sun.
Leap year comes after every four years in the year count.
## Logic to Check Leap Year-
In programming, it is very important to have a mathematics logic for solving the problem.
In Leap year program, we also have the logic. The logic to check a leap year is:
Condition 1: If a year is divisible by 4 and is not divisible by 100. It is called a leap year.
For example 1. 1800 is divisible by 4 but is also divisible by 100. So, 1800 is not a leap year.
Condition 2: If year is divisible by 400, year is also a leap year.
For Example, Consider 2000, the year is not able to fulfill the condition 1 but is divisible by 400, it will also be a leap year.
## 1. Python Program To Check leap year using If-else:
In this program, we will use Conditions statements like if-else and will find out the leap year or not.
In this method, we first check the first condition and then used the second condition using if-else and then got the result.
## 2. Python Program to Check Leap Year using Functions-
In this method, we will use a function Leap_Year() and will check the leap year.
If you want to prepare for placement mock tests, then follow us on Instagram, we daily add Multiple Choice Questions to clear concepts. Shout Coders
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# $$SL_4/\mu_2$$ and a mod $$8$$ congruence
This is a continuation of a previous post. Recall that we wanted to prove the following claim:
Claim. Let $$\rho$$ be a representation of $$SL_4$$ such that no irreducible subrepresentation of $$\rho$$ descends to $$SL_4/\mu_2$$. Then if $$\rho$$ is self-dual, we have that $$8\mid\dim \rho.$$
We first translate this into the language of weights, so we can use the Weyl dimension formula. Recall that irreducible representations of $$SL_4$$ are indexed by $$4$$-tuples of non-negative integers $$\lambda_1\geq \lambda_2\geq \lambda_3\geq \lambda_4=0.$$ We let $$\lambda$$ be the $$n$$-tuple $$(\lambda_1, \lambda_2, \lambda_3, \lambda_4)$$, and denote the representation corresponding to $$\lambda$$ by $$S^\lambda$$. So for example $$S^{(n, 0, 0, 0)}=\text{Sym}^n(V),$$ where $$V$$ is the standard representation of $$SL_4$$. We first show:
Lemma 1. Let $$\rho$$ be an irreducible representation of $$SL_4$$ which does not descend to $$SL_4/\mu_2$$. Then $$4\mid \dim\rho$$.
Proof. We have $$\rho=S^\lambda$$ for some $$\lambda$$; the hypothesis is equivalent to the statement that $$\sum_{i=1}^4 \lambda_i \equiv 1\bmod 2.$$ In other words, three of the $$\lambda_i$$ have the same parity.
Now the Weyl dimension formula (or equivalently, standard results on specializations of Schur polynomials) tells us that $$\dim \rho =\prod_{1\leq i<j\leq 4}\frac{\lambda_j-\lambda_i+j-i}{j-i}.$$ The denominator of this product is $$12$$, so we must show that the numerator is divisible by $$16$$. Now of the four pairs $$i, j$$ with $$j-i$$ odd, at least two must have that $$\lambda_i-\lambda_j$$ is odd as well. Thus $$4\mid\prod_{1\leq i<j\leq 4, j-i\text{ odd}} \lambda_j-\lambda_i+j-i.$$ For the two pairs with $$j-i=2$$, at least one must have that $$\lambda_i-\lambda_j$$ is even, so $$2\mid \prod_{1\leq i<j\leq 4, j-i\text{ even}} \lambda_j-\lambda_i+j-i.$$
Now some annoying casework lets us eke out one more factor of two; if I come up with a slick way of doing it, I will write it down...
Multiplying gives the desired result. $$\blacksquare$$
Now we analyze the case where $$\rho$$ is self-dual.
Lemma 2. Let $$\rho$$ be as in Lemma 1. Then $$\rho$$ is not self-dual.
Proof. Let $$\lambda$$ be the $$4$$-tuple of integers corresponding to $$\rho$$. Then $$\rho$$ is self-dual iff $$(\lambda_1, \lambda_2, \lambda_3, \lambda_4)=(-\lambda_4+\lambda_1, -\lambda_3+\lambda_1, -\lambda_2+\lambda_1, -\lambda_1+\lambda_1).$$ In other words, we have $$\lambda_2=\lambda_1-\lambda_3.$$ But recall that $$\lambda_1+\lambda_2+\lambda_3=2\lambda_1$$ had to be odd in the setting of Lemma 1. Contradiction. $$\blacksquare$$
Proof of Claim. Let $$\rho$$ be as in the claim. Then it is a direct sum of irreducibles, none of which descend to $$SL_4/\mu_2$$. Thus there exist a collection of irreducibles $$V_i$$ (none of which are self-dual, by Lemma 2, such that $$\rho=\bigoplus_i V_i \oplus V_i^{\vee}.$$ But by Lemma 1, each $$V_i$$ has dimension divisible by $$4$$, so $$8\mid \dim \rho$$ as desired. $$\blacksquare$$
Applying the arguments from my previous post on this topic, we get a version of the Auel-First-Williams result for the stack $$B(SL_4/\mu_2)$$ (as well as a version of their theorem for symplectic involutions of Azumaya algebras). In a sequel post, I might explain how to deduce a weak version of their actual results from this computation.
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EQNS - Mw = N M N M i i i i i 2 i i = w M w i i i 1 1 1 S =...
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M w = N i M i 2 N i M i = w i M i w i M n = N i M i N i pdi = M w M n R max = nl cos θ 2 R 2 = Nb 2 = C nl 2 R g 2 = Nb 2 6 F int kTv N 2 R 3 F int V = kT 2 ( vc n 2 + wc n 3 + ...) kT v N 2 R 6 + w N 3 R 9 + ... R ~ N υ Δ S mix n = Δ S mix = k φ A N A ln φ A + φ B N B ln φ B Δ G mix = Δ H mix T Δ S mix Δ G mix = kT φ N A ln φ + 1 φ N B ln 1 φ ( ) + χφ 1 φ ( ) χ T ( ) A + B T Δ G mix ∂φ = kT ln φ N A + 1 N A ln 1 φ ( ) N B 1 N B + χ 1 2 φ ( ) 2 Δ G mix ∂φ 2 = kT 1 N A φ + 1 N B 1 φ ( ) 2 χ DP n = M n egs ( ) MW mon S = k ln Ω [ η ] = η sp C C = 0 = η inh ( ) C = 0 [ η ] = K M a ( Π C ) C = 0 = RT M n + A 2 C Kc R θ c 0 = 1 M w 1 + 16 π 2 n 2 3 λ 2 R g 2 sin 2 θ 2 + ... log M 2 = [log K 1 log K 2 + ( a 1 + 1)log M 1 ] a 2 + 1 F ent kT R 2 Nb 2 F = F int + F ent kT v N 2 R 3 + R 2 Nb 2 T c = B 1 2 1 N A + 1 N B 2 A χ c = 1 2 N A + N B ( ) 2 N A N B = 1 2 1 N A + 1 N B 2 T s = B 1 2 1 N A φ ( ) + 1 N B 1 φ ( ) ( ) A χ S = 1 2 1 N A φ + 1 N B 1 φ ( ) χ b = 1 2 φ 1 ln φ N ln 1 φ ( ) N = ln φ 1 φ
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# source:lemon-1.2/test/random_test.cc@424:346991bf7ddd
Last change on this file since 424:346991bf7ddd was 209:765619b7cbb2, checked in by Alpar Juttner <alpar@…>, 16 years ago
Apply unify-sources.sh to the source tree
File size: 1.1 KB
Line
1/* -*- mode: C++; indent-tabs-mode: nil; -*-
2 *
3 * This file is a part of LEMON, a generic C++ optimization library.
4 *
6 * Egervary Jeno Kombinatorikus Optimalizalasi Kutatocsoport
7 * (Egervary Research Group on Combinatorial Optimization, EGRES).
8 *
9 * Permission to use, modify and distribute this software is granted
10 * provided that this copyright notice appears in all copies. For
11 * precise terms see the accompanying LICENSE file.
12 *
13 * This software is provided "AS IS" with no warranty of any kind,
14 * express or implied, and with no claim as to its suitability for any
15 * purpose.
16 *
17 */
18
19#include <lemon/random.h>
20#include "test_tools.h"
21
22int seed_array[] = {1, 2};
23
24int main()
25{
26 double a=lemon::rnd();
27 check(a<1.0&&a>0.0,"This should be in [0,1)");
28 a=lemon::rnd.gauss();
29 a=lemon::rnd.gamma(3.45,0);
30 a=lemon::rnd.gamma(4);
31 //Does gamma work with integer k?
32 a=lemon::rnd.gamma(4.0,0);
33 a=lemon::rnd.poisson(.5);
34
35 lemon::rnd.seed(100);
36 lemon::rnd.seed(seed_array, seed_array +
37 (sizeof(seed_array) / sizeof(seed_array[0])));
38
39 return 0;
40}
Note: See TracBrowser for help on using the repository browser.
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# Boundary particle method
In applied mathematics, the boundary particle method (BPM) is a boundary-only meshless (meshfree) collocation technique, in the sense that none of inner nodes are required in the numerical solution of nonhomogeneous partial differential equations. Numerical experiments show that the BPM has spectral convergence. Its interpolation matrix can be symmetric.
## History and recent developments
In recent decades, the dual reciprocity method (DRM)[1] and multiple reciprocity method (MRM)[2] have been emerging as promising techniques to evaluate the particular solution of nonhomogeneous partial differential equations in conjunction with the boundary discretization techniques, such as boundary element method (BEM). For instance, the so-called DR-BEM and MR-BEM are popular BEM techniques in the numerical solution of nonhomogeneous problems.
The DRM has become a common method to evaluate the particular solution. However, the DRM requires inner nodes to guarantee the convergence and stability. The MRM has an advantage over the DRM in that it does not require using inner nodes for nonhomogeneous problems.[citation needed] Compared with the DRM, the MRM is computationally more expensive in the construction of the interpolation matrices and has limited applicability to general nonhomogeneous problems due to its conventional use of high-order Laplacian operators in the annihilation process.
The recursive composite multiple reciprocity method (RC-MRM),[3][4] was proposed to overcome the above-mentioned problems. The key idea of the RC-MRM is to employ high-order composite differential operators instead of high-order Laplacian operators to eliminate a number of nonhomogeneous terms in the governing equation. The RC-MRM uses the recursive structures of the MRM interpolation matrix to reduce computational costs.
The boundary particle method (BPM) is a boundary-only discretization of an inhomogeneous partial differential equation by combining the RC-MRM with strong-form meshless boundary collocation discretization schemes, such as the method of fundamental solution (MFS), boundary knot method (BKM), regularized meshless method (RMM), singular boundary method (SBM), and Trefftz method (TM). The BPM has been applied to problems such as nonhomogeneous Helmholtz and convection-diffusion equation. The BPM interpolation representation is of a wavelet series.
For the application of the BPM to Helmholtz,[3] Poisson[4] and plate bending problems,[5] the high-order fundamental solution or general solution, harmonic function[6] or Trefftz function (T-complete functions)[7] are often used, for instance, those of Berger, Winkler, and vibrational thin plate equations.[8] The method has been applied to inverse Cauchy problem associated with Poisson[9] and nonhomogeneous Helmholtz equations.[10]
The BPM may encounter difficulty in the solution of problems having complex source functions, such as non-smooth, large-gradient functions, or a set of discrete measured data. The solution of such problems involves:[citation needed]
(1) The complex functions or a set of discrete measured data can be interpolated by a sum of polynomial or trigonometric function series. Then, the RC-MRM can reduce the nonhomogeneous equation to a high-order homogeneous equation, and the BPM can be implemented to solve these problems with boundary-only discretization.
(2) The domain decomposition may be used to in the BPM boundary-only solution of large-gradient source functions problems.
## References
1. ^ Partridge PW, Brebbia CA, Wrobel LC, The dual reciprocity boundary element method. Computational Mechanics Publications, 1992
2. ^ Nowak AJ, Neves AC, The multiple reciprocity boundary element method. Computational Mechanics Publication, 1994
3. ^ a b Chen W, Meshfree boundary particle method applied to Helmholtz problems. Engineering Analysis with Boundary Elements 2002,26(7): 577–581
4. ^ a b Chen W, Fu ZJ, Jin BT, A truly boundary-only meshfree method for inhomogeneous problems based on recursive composite multiple reciprocity technique. Engineering Analysis with Boundary Elements 2010,34(3): 196–205
5. ^ Fu ZJ, Chen W, Yang W, Winkler plate bending problems by a truly boundary-only boundary particle method. Computational Mechanics 2009,44(6): 757–563
6. ^ Hon YC, Wu ZM, A numerical computation for inverse boundary determination problem. Engineering Analysis with Boundary Elements 2000,24(7–8): 599–606
7. ^ Chen W, Fu ZJ, Qin QH, Boundary particle method with high-order Trefftz functions. CMC: Computers, Materials & Continua 2010,13(3): 201–217
8. ^ Chen W, Shen ZJ, Shen LJ, Yuan GW, General solutions and fundamental solutions of varied orders to the vibrational thin, the Berger, and the Winkler plates. Engineering Analysis with Boundary Elements 2005,29(7): 699–702
9. ^ Fu ZJ, Chen W, Zhang CZ, Boundary particle method for Cauchy inhomogeneous potential problems. Inverse Problems in Science and Engineering 2012,20(2): 189–207
10. ^ Chen W, Fu ZJ, Boundary particle method for inverse Cauchy problems of inhomogeneous Helmholtz equations. Journal of Marine Science and Technology–Taiwan 2009,17(3): 157–163
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# Action of metric tensor on Levi-Civita symbol
• I
• Baela
In summary, in SR a metric tensor raises or lowers indices of a tensor, while in general relativity a covariant Levi-Civita tensor exists which is negative when the metric is negative.
Baela
We know that a metric tensor raises or lowers the indices of a tensor, for e.g. a Levi-Civita tensor. If we are in ##4D## spacetime, then
\begin{align}
g_{mn}\epsilon^{npqr}=\epsilon_{m}{}^{pqr}
\end{align}
where ##g_{mn}## is the metric and ##\epsilon^{npqr}## is the Levi-Civita tensor.
The Levi-Civita symbol, which we can denote by ##\varepsilon^{npqr}##, is not a tensor. It obeys the relation
\begin{align}
\varepsilon^{npqr}=\varepsilon_{npqr}.
\end{align}
What happens if the metric tensor is multiplied with the Levi-Civita symbol ##\varepsilon^{npqr}##?
\begin{align}
g_{mn}\varepsilon^{npqr}=\,?
\end{align}
Well, what's the relation between the epsilon symbol and -tensor?
vanhees71
For a generic spacetime in GR, (2) is not correct. I think already in SR you have a minus there to rectify (2).
Baela said:
is not a tensor.
should be replaced by is a tensor. The four uppers are just replaced by four downers by being acted on with four gs.
Baela said:
What happens if the metric tensor is multiplied with the Levi-Civita symbol ?
Your Eq. (2). g just raises or lowers indices.
The LC tensor just depends on the order of the super or subscripts.
It is called an idempotent tensor because it has the same value in any system.
I am just talking about special relativity.
The components of the Levi-Civita tensor are
$$\epsilon^{\alpha \beta \gamma \delta}=\frac{1}{\sqrt{-g}} \Delta^{\alpha \beta \gamma \delta},$$
where ##g=\mathrm{det}(\hat{g})## and ##\Delta^{0123}=1## and totally anti-symmetric under exchange of its arguments, i.e., the usual Levi-Civita symbol.
The covariant components of the Levi-Civta-tensor thus are
$$\epsilon_{\alpha \beta \gamma \delta}=-\sqrt{-g} \Delta^{\alpha \beta \gamma \delta}.$$
This somewhat confusing extra sign comes from the fact that ##g<0##.
Also note that the convention of this sign is a matter of convention, i.e., it can be the opposite (e.g., in MTW). So you have to carefully check, which convention is used in any reference you read ;-).
dextercioby
I am just talking about special relativity.
vanhees71
Meir Achuz said:
I am just talking about special relativity.
Actually, you're talking about special relativity only for the specific case of a cartesian coordinate system. More general curvilinear systems (e.g., polar coordinates) require the full tensor machinery of the metric, its determinant and the Christoffel symbols, even in special relativity.
vanhees71
That's why they are not usually used in SR.
In pseud-Cartesian (Lorentzian) coordinates then you have ##\epsilon^{\alpha \beta \gamma \delta}=\Delta^{\alpha \beta \gamma \delta}## and ##\epsilon_{\alpha \beta \gamma \delta}=-\Delta^{\alpha \beta \gamma \delta}##, because then ##-g=1##.
Meir Achuz said:
I am just talking about special relativity.
I'm not sure the thread in general is restricted to SR.
Meir Achuz said:
That's why they are not usually used in SR.
I'm not sure that's necessarily the case. Rindler coordinates are fairly commonly used, and they are non-inertial.
vanhees71 and renormalize
In heavy-ion hydrodynamics often Milne coordinates are used, because they are convinient to describe "Bjorken flow" ;-).
usually
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http://clay6.com/qa/1270/-large-int-normalsize-fracdx-equals-begin-a-cot-ex-x-c-b-tan-xe-x-c-c-tan-e
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# $\Large \int \normalsize\frac{\large e^x(1+x)}{\large \cos^2(e^xx)}dx$ equals
$\begin{array}{1 1} (A)\;-\cot(ex^x)+C \\(B)\;\tan(xe^x)+C \\ (C)\;\tan(e^x)+C\\ (D)\;\cot(e^x)+C\end{array}$
Toolbox:
• $(i)\;\int e^{-x}=e^{-x}+c.$
• $(ii)\;\int\sec^2xdx=\tan x+c.$
Given $I=\int\frac{e^x(1+x)}{\cos^2(e^xx)}dx.$
Let $e^xx=t.$
On differentiating we get,
$(e^x.x+e^x.1)dx=dt.$
$\Rightarrow e^x(x+1)dx=dt.$
On substituting we get,
Hence $I=\int\frac{dt}{\cos^2t}$.
$\frac{1}{\cos^2t}=\sec^2t.$
Therefore $I=\int\sec^2t.dt.$
On integrating we get,
$\;\;\;=\tan t+c.$
substituting for t we get,
$\;\;\;=\tan(e^xx)+c.$
Hence the correct answer is B.
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#### Chemical Equations Balanced on 02/11/19
Molecular weights calculated on 02/10/19 Molecular weights calculated on 02/12/19
Calculate molecular weight
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232
Molar mass of CO2 is 44.0095
Molar mass of KMnO4 is 158.033945
Molar mass of SO3 is 80,0632
Molar mass of calcium is 40,078
Molar mass of CaC2 is 64.0994
Molar mass of HPo4*3H2O is 890.9835016
Molar mass of C3H7OH is 60.09502
Molar mass of Fe is 55.845
Molar mass of Fe is 55.845
Molar mass of SnH4 is 122,74176
Molar mass of CO2 is 44,0095
Molar mass of Cu(BF4)2 is 237.1552256
Molar mass of CO2 is 44,0095
Molar mass of MgCl2*6H2O is 203.30268
Molar mass of CaNCN is 80.1021
Molar mass of HCI is 139,92311
Molar mass of Cu(BF4)2 is 237.1552256
Molar mass of Ca3(PO4)2 is 310.176724
Molar mass of KOH is 56.10564
Molar mass of H2SO4 is 98.07848
Molar mass of Methanol is 32.04186
Molar mass of NaOH is 39.99710928
Molar mass of KOH is 56.10564
Molar mass of NaCl is 58,44276928
Molar mass of KC2H3O2 is 98.14232
Molar mass of CaCO3 is 100,0869
Molar mass of C2H2 is 26.03728
Molar mass of N2 is 28.0134
Molar mass of NaOH is 39,99710928
Molar mass of NaOH is 39,99710928
Molar mass of H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V C is 588,72260908
Molar mass of KClO3 is 122.5495
Molar mass of MgCl4*6H2O is 274.20868
Molar mass of H2O2 is 34,01468
Molar mass of CaO is 56.0774
Molar mass of H2 is 2,01588
Molar mass of CH4OOH is 49.0492
Molar mass of NH3 is 17,03052
Molar mass of MgSO4*6H2O is 228.45928
Molar mass of (NH4)2CO3 is 96.08582
Molar mass of C7H5NO3S is 183.1845
Molar mass of H2SO4 is 98,07848
Molar mass of Au3P2 is 652.847231
Molar mass of NaHCO3 is 84.00660928
Molar mass of LiCl is 42.394
Molar mass of MgHPO4*3H2O is 174.330142
Molar mass of h2 is 2,01588
Molar mass of AgNO3 is 169.8731
Molar mass of NH3 is 17.03052
Molar mass of B2H6 is 27.66964
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232
Calculate molecular weight
Molecular weights calculated on 02/10/19 Molecular weights calculated on 02/12/19
Molecular masses on 02/04/19
Molecular masses on 01/12/19
Molecular masses on 02/11/18
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https://betterlesson.com/lesson/577427/graphing-with-m-m-s
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# Graphing with M&M's
18 teachers like this lesson
Print Lesson
## Objective
SWBAT draw a pictograph to represent a unique data set
#### Big Idea
M&M’s can be used to create a unique data set and can be graphed by students
## Introduction
5 minutes
One of the really cool things about math is that I can represent what I have as a data set, and I can show people what I have with cool things like graphs. A pictograph is a great way that I can represent my data, and today my data is going to be M&M’s! I have a bag on M&M’s here and I need to figure out how many of each color I have! I have my pictograph set up with a few important pieces already written down, but I’m going to need you guys to help me figure out what I’ve got!
## Guided Practice
10 minutes
I call on a student to help me sort them, another student count them and another student to record on a white board. I do this to draw more students into the learning instead of allowing only 1 student to participate.
I’m going to tally up what I have so that I know how many total things I’ve got in my data set before I start my graphing work. There are many ways that I can choose to show what I have with a pictograph, and what I choose might be different from what you choose!
Ok now that I have all of my data, I am going to set up my pictograph. I am going to have it go from side to side, but you can also choose to have it go up and down. And I’m going to use a smiley face to represent my amounts. I think I’ll make each smiley face worth 2 M&M’s. It’s important for me to put that here so that anyone reading my graph knows what I mean by a smiley face! What would have of a smiley face equal? (student responses). That’s right.
I want all of you to use a white board to write down how many smiley faces I need for each color, like this (white board example). Great, I’ll set that up now.
## Independent Work
35 minutes
You will be working with a partner to count your data, tally up what you have and then each of you will create your own pictograph using your data. I want each partner to choose a different way to represent what they have so that we can see our data graphed in as many ways as possible!
Here I emphasize MP4 and for students to model their unique data set of M&M’s and to encourage each student that even with the same bag of candy they can each find their own unique way to graph it.
## Close
5 minutes
Now who would like to share their pictograph with us? Why did you choose to represent your data in that way?
Here I choose a few students to share out their work and we celebrate each students work.
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http://clay6.com/qa/39084/the-integral-int-1-x-large-frac-e-dx-is-equal-to-
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Browse Questions
# The integral $\;\int (1+x-\large\frac{1}{x})\;e^{x+\large\frac{1}{x}} dx\;$ is equal to :
$(a)\;(x+1) e^{x+\large\frac{1}{x}} +c \qquad(b)\;-xe^{x+\large\frac{1}{x}} +c \qquad(c)\;(x-1) e^{x+\large\frac{1}{x}} +c \qquad(d)\;x e^{x+\large\frac{1}{x}} +c$
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https://quant.stackexchange.com/questions/16635/quantiles-value-at-risk-and-log-normal-random-walks
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# Quantiles, Value-at-Risk and log normal random walks…
Sorry, that's probably quite a bunch of silly questions, but I just got lost a bit and need to dot all the i's and cross some t's :).
Let's say we have a series of returns (like this one we may get in R):
library(quantmod)
getSymbols("AAPL")
A <- AAPL[,6]["2007"]
Then, if I want to know how much could I lose with the probability 95%, I need to calculate VaR for this 5% quantile, OK? Again, in R first I need to calculate the quantile: quantile(A, 0.05), am I right? The result in this particular series would be 11.485. Which means that with the 95% probability I won't go lower than 11.485, is that right? Thus, my VaR would be then mean(A) - 11.485 = 5.8 or 33.56%, wouldn't it?
In other words, calculating a quantile for a given probability in this particular case gives us a lower boundary corresponding to a series in question, doesn't it? And that's how quantiles are measured, it's not a percentage, but boundaries which with a certain probability limit values of a series?
Now, to the second part :)
While reading the article on Wolfram Alpha, I've noticed that among other things, when asked on a certain stock, WA among other information, also tries to give some predictions. Which are actually quantiles, aren't they? E.g., with the 95% probability they say that the stock in question in one month time won't go lower than 92.67 - and this price, 92.67 would be the 5% quantile for that particular stock during that particular period, right?
If so, now to the most interesting bit: as you may see, they also draw some hypothetical colored graphs there, using log normal random walks, as they say below. But with log normal random walks values one will get may be very different, is that so? Thus, they probably calculate quite a lot of such walks, like hundreds or thousands, and then bootstrap them? If so, then how is that done with R? What is the usual algorithm? Needful libraries? Where should I look to learn more?
Not that I'm interested in predicting stocks, of course :), but I just very much like the idea of trying to predict a distribution...
Thank you.
Not too sure about the second part of your questions but as far as VaR, R has some pretty neat functions. First I took you subset A and converted it to discrete returns since using actual prices for VaR may be a bit harder to interpret.
# Load PerformanceAnalytics for VaR & Calculating Returns
library("PerformanceAnalytics")
# Calculate Returns
a <- CalculateReturns(A, method=c("discrete"))
# Calculate VaR at 95% confidence
VaR(a, p=0.95)
This results in a VaR: -0.03399548. Put simply, One can expected to lose approximately 3.40% on any given day.
Here is a subset of the the function's description.
VaR is an industry standard for measuring downside risk. For a return series, VaR is defined as the high quantile (e.g. ~a 95 quantile) of the negative value of the returns. This quantile needs to be estimated...
Take a look at the daily returns and the VaR level:
plot(a)
abline(h=VaR(a,p=0.95), col='red')
You see that not many returns fall below the red line which satisfies the 95% confidence
• No worries, glad it is helpful. If you think my answer satisfies your question, you can accept it. @JohnDoe
– Rime
Feb 21, 2015 at 5:46
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# 145
CHAPTER 3
COMMONLY USED
STATISTICAL TERMS
There are many statistics used in social science research
and evaluation. The two main areas of statistics are
descriptive and inferential. The third class of statistics is
design and experimental statistics. Descriptive statistics
involve the tabulating, depicting, and describing of col-
lections of data. These data may be either quantitative
or qualitative. They provide a picture or description of
the properties of data collected in order to summarize
them into manageable form. Inferential statistics are a
formalized body of techniques that infer the properties
of a larger collection of data from the inspection of that
collection. They build on these statistics as they infer the
properties of samples to various populations. Design
and analysis statistics were developed for the discovery
and confirmation of causal relationships among vari-
ables in social science experiments. They use a variety of
statistical tests related to aspects such as prediction and
hypothesis testing. Experimental analysis is related to
As in the previous glossary chapter, if you are looking for a term and it is not
here, please send it to my e-mail address, mholosko@uga.edu, with the subject
line: New Statistical Terms Needed. It will then be added in the next edition.
146 Pocket Glossary for Commonly Used Research Terms
comparisons, variance, and ultimately testing whether
variables are significant between each other. The lat-
ter two types of statistics are usually either parametric
or nonparametric. The importance of statistics in the
research process is sometimes exaggerated. Thus, a highly
sophisticated statistical analysis rarely, if ever, compen-
sates for a poorly conceived project, a poorly constructed
research design, or an inaccurate data collection instru-
ment. Thus, statistics certainly may aid the researcher but
are never a substitute for good, sound thinking and atten-
tion to the scientific method and research process. For
researchers, then, statistics are simply a tool to help them
study the phenomena they are interested in.
DESCRIPTIVE STATISTICS
Measures of Central Tendency
Mean, arithmetic mean (X or M): The sum of the scores in a
distribution divided by the number of scores in the distri-
bution. It is the most commonly used measure of central
tendency. It is often reported with its companion statistic,
the standard deviation, which shows how far things vary
from the average.
Median (Mdn): The midpoint or number in a distribution
having 50% of the scores above it and 50% of the scores
below it. If there are an odd number of scores, the median
is the middle score.
Mode (Mo): The number that occurs most frequently in a
distribution of scores or numbers. In some fields, notably
education, sample data are often called scores, and the
sample mode is known as the modal score.
Commonly Used Statistical Terms 147
Measures of Variability
Interquartile range (IQR): A measure of statistical dispersion
being equal to the difference between the third and first
quartiles. The first quartile (designated Q
1
) is the lower
and cuts off the lowest 25% of data (the 25th percentile);
the second quartile (Q
2
), or the median, cuts the data set
in half (the 50th percentile); and the third quartile (Q
3
)
cuts off highest 25% of data, or the lowest 75% (the 75th
percentile).
Range (Ra): The difference between the highest and lowest
scores in a distribution; a measure of variability.
Standard deviation (SD): The most stable measure of vari-
ability, it takes into account each and every score in a
normal distribution. This descriptive statistic assesses how
far individual scores vary in standard unit lengths from its
midpoint of 0. For all normal distributions, 95% of the
area is within 1.96 standard deviations of the mean.
Variance (SD
2
): A measure of the dispersion of a set of
data points around their mean value. It is a mathemati-
cal expectation of the average squared deviations from
the mean.
Inferential Statistical Tests
Tests concerned with using selected sample data compared
with population data in a variety of ways are called inferen-
tial statistical tests. There are two main bodies of these tests.
The first and most frequently used are called parametric sta-
tistical tests. The second are called nonparametric tests. For
each parametric test, there may be a comparable nonpara-
metric test, sometimes even two or three.
Parametric tests are tests of significance appropriate when
the data represent an interval or ratio scale of measurement
148 Pocket Glossary for Commonly Used Research Terms
and other specific assumptions have been met, specifically,
that the sample statistics relate to the population parameters,
that the variance of the sample relates to the variance of the
population, that the population has normality, and that the
data are statistically independent.
Nonparametric tests are statistical tests used when the
data represent a nominal or ordinal level scale or when
assumptions required for parametric tests cannot be met,
specifically, small sample sizes, biased samples, an inabil-
ity to determine the relationship between sample and
population, and unequal variances between the sample and
population. These are a class of tests that do not hold the
assumptions of normality.
In the list of statistical terms below, when the test is a
parametric test, the designation of *PT will be used at the
end of the definition. Conversely, when the test is a nonpara-
metric test, the designation of *NPT will be used at the end
of the definition.
Statistical Terms
Alpha coefficient (): See Cronbachs alpha coefficient.
Analysis of covariance (ANCOVA): A statistical technique for
equating groups on one or more variables when testing for
statistical significance using the F-test statistic. It adjusts
scores on a dependent variable for initial differences on
other variables, such as pretest performance or IQ. *PT
Analysis of variance (ANOVA): A statistical technique for
determining the statistical significance of differences
among means; it can be used with two or more groups
and uses the F-test statistic. *PT
Autoregressive integrated moving average (ARIMA): This
statistic is a Box-Jenkins approach to time series analy-
sis. It tests for changes in the data patterns pre- and
Commonly Used Statistical Terms 149
postintervention within the context of analyzing the out-
comes of a time series design.
Binomial test: An exact test of the statistical significances of
derivations from a theoretically expected distribution of
observations into two categories. *NPT
Chi-square (): A nonparametric test of statistical signifi-
cance appropriate when the data are in the form of fre-
quency counts; it compares frequencies actually observed
in a study with expected frequencies to see if they are
significantly different. *NPT
Cochrans Q: Used to evaluate the relation between two
variables that are measured on a nominal scale. One of
the variables may even be dichotomous or consisting of
only two possible values. *NPT
Coefficient of determination (r): The square of the correla-
tion coefficient (r), it indicates the degree of relationship
strength by potentially explained variance between two
variables.
Cohens d: A standardized way of measuring the effect size
or difference by comparing two means by a simple math
formula. It can be used to accompany the reporting of a
t-test or ANOVA result and is often used in meta-anal-
ysis. The conventional benchmark scores for the magni-
tude of effect sizes are as follows: small, d = 0.2; medium,
d = 0.5; large, d = 0.8. *NPT
Cohens kappa (): A statistical measure of interrater agree-
ment for qualitative (categorical) items. Scores range
from 1.0 to 1.0. *NPT
Confidence interval (CI): Quantifies the uncertainty in mea-
surement. It is usually reported as a 95% CI, which is the
range of values within which it can be 95% certain that
the true value for the whole population lies. For example,
150 Pocket Glossary for Commonly Used Research Terms
for a number needed to treat (NNT) of 10 with a 95% CI
of 5 to 15, there would be 95% confidence that the true
NNT value lies between 5 and 15.
Correlation coefficient (r): A decimal number between 0.00
and 1.00 that indicates the degree to which two quanti-
tative variables are related. The most common one used
is the Pearson Product Moment correlation coefficient or
just the Pearson coefficient.
Cronbachs alpha coefficient (): A coefficient of consistency
that measures how well a set of variables or items measures
a single, unidimensional, latent construct in a scale or inven-
tory. Alpha scores are conventionally interpreted as follows:
high, 0.90; medium, 0.70 to 0.89; and low, 0.55 to 0.69.
Cumulative frequency distribution: A graphic depiction of
how many times groups of scores appear in a sample.
Dependent t-test: A data analysis procedure that assesses
whether the means of two related groups are statistically dif-
ferent from each other, for example, one groups mean score
(time one) compared with the same groups mean score
(time two). It is also called the paired samples t-test. *PT
Effect size (): Any measure of the strength of a relationship
between two variables. Effect size statistics are used to
assess comparisons between correlations, percentages,
mean differences, probabilities, and so on.
Eta (): An index that indicates the degree of a curvilinear
relationship.
F-test (F): A parametric statistical test of the equality of the
means of two or more samples. It compares the means
and variances between and within groups over time. It is
also called analysis of variance (ANOVA). *PT
Factor analysis: A statistical method for reducing a set
of variables to a smaller number of factors or basic
Commonly Used Statistical Terms 151
components in a scale or instrument being analyzed. Two
main forms are exploratory (EFA) and confirmatory fac-
tor analysis (CFA). *PT
Fishers exact test: A nonparametric statistical significance
test used in the analysis of contingency tables where
sample sizes are small. The test is useful for categori-
cal data that result from classifying objects in two dif-
ferent ways; it is used to examine the significance of
the association (contingency) between two kinds of
classifications. *NPT
Friedman two-way analysis of variance: A nonparametric
inferential statistic used to compare two or more groups
by ranks that are not independent. *NPT
G
2
: This is a more conservative goodness-of-fit statistic than
the and is used when comparing hierarchical models
in a categorical contingency (two-by-two) table.
Independent t-test: A statistical procedure for comparing mea-
surements of mean scores in two different groups or sam-
ples. It is also called the independent samples t-test. *PT
Kendalls tau (): A nonparametric statistic used to measure
the degree of correspondence between two rankings and
to assess the significance of the correspondence. *NPT
Kolmogorav-Smirnov (K-S) test: A nonparametric goodness-
of-fit test used to decide if a sample comes from a popula-
tion with a specific distribution. The test is based on the
empirical distribution function (ECDF). *NPT
Kruskal-Wallis one-way analysis of variance: A nonparametric
inferential statistic used to compare two or more indepen-
dent groups for statistical significance of differences. *NPT
Mann-Whitney U-test (U): A nonparametric inferential sta-
tistic used to determine whether two uncorrelated groups
differ significantly. *NPT
152 Pocket Glossary for Commonly Used Research Terms
McNemars test: A nonparametric method used on nominal
data to determine whether the row and column marginal
frequencies are equal. *NPT
Median test: A nonparametric test that tests the null hypoth-
esis that the medians of the populations from which two
samples are drawn are identical. *NPT
Multiple correlation (R): A numerical index describing the
relationship between predicted and actual scores using
multiple regression. The correlation between a criterion
and the best combination of predictors.
Multivariate analysis of covariance (MANCOVA): An
extension of ANOVA that incorporates two or more
dependent variables in the same analysis. It is an
extension of MANOVA where artificial dependent
variables (DVs) are initially adjusted for differences in
one or more covariates. It computes the multivariate F
statistic. *PT
Multivariate analysis of variance (MANOVA): It is an
ANOVA with several dependent variables. *PT
Newman-Keuls test: A type of post hoc or a posteriori mul-
tiple comparison test of data that makes precise compari-
sons of group means after ANOVA has rejected the null
hypothesis. *NPT
One-way analysis of variance (ANOVA): An extension of
the independent group t-test where you have more than
two groups. It computes the difference in means both
between and within groups and compares variability
between groups and variables. Its parametric test statistic
is the F-test. *PT
Pearson correlation coefficient (r): This is a measure of the
correlation or linear relationship between two variables
x and y, giving a value between +1 and 1 inclusive. It is
Commonly Used Statistical Terms 153
widely used in the sciences as a measure of the strength of
linear dependence between two variables. *PT
Pooled point estimate: An approximation of a point, usually
a mean or variance, that combines information from two
or more independent samples believed to have the same
characteristics. It is used to assess the effects of treatment
samples versus comparative samples.
Post hoc test: A post hoc test (or post hoc comparison test)
is used at the second stage of the analysis of variance
(ANOVA) or multiple analyses of variance (MANOVA)
if the null hypothesis is rejected.
Runs test: Where measurements are made according to some
well-defined ordering, in either time or space. A frequent
question is whether or not the average value of the mea-
surement is different at different points in the sequence.
This nonparametric test provides a means for this. *NPT
Siegel-Tukey test: A nonparametric test named after Sidney
Siegel and John Tukey, which tests for differences in scale
between two groups. Data measured must at least be ordi-
nal. *NPT
Sign test: A test that can be used whenever an experiment is
conducted to compare a treatment with a control on a
number of matched pairs, provided the two treatments are
assigned to the members of each pair at random. *NPT
Spearmans rank order correlation (): A nonparametric
test used to measure the relationship between two rank-
ordered scales. Data are in ordinal form. *NPT
Standard error of the mean (SEM): An estimate of the
amount by which an obtained mean may be expected to
differ by chance from the true mean. It is an indication of
how well the mean of a sample estimates the mean of a
population.
154 Pocket Glossary for Commonly Used Research Terms
Statistical power: The capability of a test to detect a signifi-
cant effect or how often a correct interpretation can be
reached about the effect if it were possible to repeat the
test many times.
Student-Newman-Keuls (SNK) test: A nonparametric post-
ANOVA test, also called a post hoc test. It is used to
analyze the differences found after the performed F-test
(ANOVA) is found to be significant, for example, to locate
where differences truly occur between means. *NPT
Student t-test (t): Any statistical hypothesis test in which the
test statistic follows a Students t distribution if the null
hypothesis is true, for example, a t-test for paired or inde-
pendent samples. *PT
t-distribution: A statistical distribution describing the means
of samples taken from a population with an unknown
variance.
T-score: A standard score derived from a z-score by mul-
tiplying the z-score by 10 and adding 50. It is useful in
comparing various test scores to each other as it is a stan-
dard metric that reflects the cumulative frequency distri-
bution of the raw scores.
t-test for correlated means: A parametric test of statistical
significance used to determine whether there is a statis-
tically significant difference between the means of two
matched, or nonindependent, samples. It is also used for
prepost comparisons. *PT
t-test for correlated proportions: A parametric test of statistical
significance used to determine whether there is a statistically
significant difference between two proportions based on the
same sample or otherwise nonindependent groups. *PT
t-test for independent means: A parametric test of signifi-
cance used to determine whether there is a statistically
Commonly Used Statistical Terms 155
significant difference between the means of two indepen-
dent samples. *PT
t-test for independent proportions: A parametric test of
statistical significance used to determine whether there is
a statistically significant difference between two indepen-
dent proportions. *PT
Tukeys test of significance: A single-step multiple com-
parison procedure and statistical test generally used in
conjunction with an ANOVA to find which means are
significantly different from one another. Named after
John Tukey, it compares all possible pairs of means and is
based on a studentized range distribution q (this distribu-
tion is similar to the distribution of t from the t-test).
Wald-Wolfowitz test: A nonparametric statistical test used
to test the hypothesis that a series of numbers is random.
It is also known as the runs test for randomness. *NPT
Wilcoxon sign rank test (W
+
): A nonparametric statistical
hypothesis test for the case of two related samples or
repeated measurements on a single sample. It can be used
as an alternative to the paired Students t-test when the
population cannot be assumed to be normally distrib-
uted. *NPT
Wilkss lambda (): A general test statistic used in multi-
variate tests of mean differences among more than two
groups. It is the numeral index calculated when carrying
out MANOVA or MANCOVA.
Z-score: A score expressed in units of standard deviations
from the mean. It is also known as a standard score.
Z-test: A test of any of a number of hypotheses in inferential
statistics that has validity if sample sizes are sufficiently
large and the underlying data are normally distributed.
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Comfortable part part whole worksheet Wonderful
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This a fall theme packet of differentiated part-part-whole worksheets that will be perfect for a teacher who does 3 flexible groups of above level on level and below level performing student students. The first part is made of 4 blue cubes and the second part is made of 3 pink cubes. If you know all of your addition facts you should be able to subtract easily. Materials can be used to portray this way of thinking such as counters when using ten frames. Complete the part-whole models. A number at the top with two if.
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Source: pinterest.com
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