text:
# Physical world simulation

Original author: Jamie Wong
• Transfer

How would you approach the simulation of rain, or of any other long-lasting physical process?

Simulation, be it rain, air flow over the wing of an airplane, or falling on the steps of a slink (remember a spring toy from a childhood rainbow?), Can be imagined if you know the following:

1. The state of everything at the start of the simulation.
2. How does this state change from one point in time to another?

By “state of everything” I mean any variable data that either determines what the environment looks like or the changes that occur over time. As an example of a condition, one can cite the position of a raindrop, the direction of the wind, or the speed of each part of a slink spring.

If we assume that our entire state of environment is one big vector, then we can formulate the data we need, indicated above, into the following:

1. Find value satisfying ?
2. Find function  such that .

Why do we need to store the state in just one vector, I will explain a little later. This is one of those cases when it seems that we are overdoing it with a generalization of the problem, but I promise that this approach has its own interests.

Now let's see how you can store all the information about the environment in one vector using a simple example. Let's say we have 2 objects in a 2D simulation. Each object is determined by its position. and speed .

So, to get the vector we need to combine the vectors together  into one, consisting of 8 components, like this:

If you are confused, why do we want to find , not the initial definition , then the meaning is that we need the derivative as an expression that depends only on the current state, , constants and time itself. If this is not possible, then for certain any status information has not been taken into account.

## Initial state

To determine the initial state of the simulation, you need to define a vector . Thus, if the initial state of our two-object example looks something like this:

Then in vector form it can be represented as follows:

If we combine all this into one vector, we will get what we need :

## Derived function

determines the initial state and now all we need is to find a way to transition from the initial state to the state that occurs the instant after it, and from the received state a little further in time and so on.

To do this, solve the equation for . Find the derivative of:

Wow! The formula turned out high! But you can bring it into a more readable form if we break our vector back to the component parts.

 and  bound by similar rules, as well as  from , so despite the bunch of expressions above, all we really need to find are the following 2 things:

The quality of the simulation depends on the definition of these two derivatives, it is in them all the strength. And so that the simulation does not resemble a program where everything happens in a chaotic way, you can turn to physics for inspiration.

## Kinematics and Dynamics

Kinematics and dynamics are essential ingredients to create an interesting simulation. Let's start with the very basics in our example.

Responsible for the position in space, and the first derivative of the point’s position in time is its velocity . In turn, the derivative of speed over time is acceleration,.

It may seem that we have already found the function we needbecause we already know the following:

And in fact, we brilliantly dealt with because  this is part of our state vector , but we need to make out the second formula a little more, because with not so simple.

Here Newton’s second law will help us:. If we assume that the mass of our objects is known, then rearranging the variables in the expression, we get:

So wait a minute  and  are not part , so it’s hard to call promotion (remember, we need a derivative function that depends only on  and ) Nevertheless, we moved forward, because we found all the useful formulas that determine the behavior of objects in our physical world.

Suppose that in our simple example, the only force that acts on objects is gravitational attraction. In this case, we can determineusing Newton’s law of gravity:

Where  this is the gravitational constant , a  and these are the masses of our objects (which, we assume, are constants).

To create the simulation itself, we also need a direction and somehow indicate through vector components . Assuming that is the force acting on the first object, then we can do it as follows:

To summarize. The change of states in our system of two objects is fully expressed through variables. And the changes are:

Now we have everything that distinguishes our simulation from all other simulations:  and .

But how, having a strictly defined simulation, turn it into a beautiful animation?

If you had experience writing a simulation or a game, then maybe you can suggest something like this:

x += v * delta_t
v += F/m * delta_t

But let's stop a bit and see why this works.

## Differential equations

Before we begin implementation, we need to decide what information we already have and what we need. We have a meaningwhich satisfies there is also satisfying . And we need a function that will give us the state of the system at any time. Mathematically, we need a function.

Bearing this in mind and looking closely at, you can see that this equation relates  with its derivative . This means that our equation is differential! Ordinary differential equation of the first order, to be precise. If we solve it, then we will find a function.

The task of finding according to  and called the Cauchy problem .

## Numerical integration

For some examples of Cauchy problems, you can easily find the answer by the analytical method, but in complex simulations the analytical approach can be very difficult. Therefore, we will try to find a way to find an approximate solution to the problem.

For example, take the simple Cauchy problem.
Given: and . Find an approximate solution for.

Consider the problem from a geometric point of view and look at the value and tangent at a point. From what is given to us, we have and

We don’t know what it looks like but we know that near the point , the value is close to the tangent. Now try to calculate for small value using the tangent. Let's try for a start.

If you paint, then we approximate the value  in the following way:

So, for . Now we can continue to be calculated for other points. Although, of course, we did not find the exact value

but if our approximate value is very close to exact, then the approximated tangent will also be very close to real!

display \ begin \left\{aligned\right\} f \left(t, y \left(t\right)\right) & = y \left(t\right) \\ f \left(0.5,1.5\right) & = 1.5 \ end \left\{aligned\right\} display

Next, let's move on  units to the right tangent.

Repeat the process and get the tangent slope :

The procedure can be performed recursively and for this we derive the formula:

This numerical method for solving differential equations is called the Euler method. For the general case, a step x += v * delta_t.

In our particular case, a step-by-step solution looks like this:

Using this method, it is convenient to present the results in a table:

It turns out that our problem has a beautiful analytical solution :

What do you think will happen if the step is reduced in the Euler method?

The difference between approximated and exact solutions decreases with decreasing ! In addition, in addition to reducing the step, other methods of numerical integration can be used, which can lead to a better result, such as the middle rectangle method , the Runge-Kutta method, and the Adams method .

## It is time to code!

With the same success as we deduced the mathematical representation of the description of the simulation, we can write the implementation of the simulation programmatically.

Because I am most familiar with JavaScript, and I like the clarity that annotations add to the code, all examples will be written in TypeScript .

And we will start with the version, which implied that it is a one-dimensional array of numbers, just like in our mathematical model.

function runSimulation(
// y(0) = y0
y0: number[],
// dy/dt(t) = f(t, y(t))
f: (t: number, y: number[]) => number[],
// показывает текущее состояние симуляции
render: (y: number[]) => void
) {
// Шаг вперёд на 1/60 секунды за тик
// Если анимация будет 60fps то это приведёт к симуляции в рельном времени
const h = 1 / 60.0;
function simulationStep(ti: number, yi: T) {
render(yi)
requestAnimationFrame(function() {
const fi = f(ti, yi)
// t_{i+1} = t_i + h
const tNext = ti + h
// y_{i+1} = y_i + h f(t_i, y_i)
const yNext = []
for (let j = 0; j < y.length; j++) {
yNext.push(yi[j] + h * fi[j]);
}
simulationStep(tNext, yNext)
}
}
simulationStep(0, y0)
}

It is not always convenient to operate with one-dimensional arrays; you can abstract the functions of adding and multiplying the simulation process into an interface and get a brief generalized simulation implementation using TypeScript Generics .

interface Numeric {
plus(other: T): T
times(scalar: number): T
}
function runSimulation>(
y0: T,
f: (t: number, y: T) => T,
render: (y: T) => void
) {
const h = 1 / 60.0;
function simulationStep(ti: number, yi: T) {
render(yi)
requestAnimationFrame(function() {
// t_{i+1} = t_i + h
const tNext = ti + h
// y_{i+1} = y_i + h f(t_i, y_i)
const yNext = yi.plus(f(ti, yi).times(h))
simulationStep(yNext, tNext)
})
}
simulationStep(y0, 0.0)
}

The positive side of this approach is the ability to concentrate on the basis of the simulation: what exactly distinguishes this simulation from any other. We use the simulation example with the two objects mentioned above:

Code simulation of two objects
// Состояние симуляции двух объектов в один тик времени
class TwoParticles implements Numeric {
constructor(
) { }
plus(other: TwoParticles) {
return new TwoParticles(
this.x1.plus(other.x1), this.v1.plus(other.v1),
this.x2.plus(other.x2), this.v2.plus(other.v2)
);
}
times(scalar: number) {
return new TwoParticles(
this.x1.times(scalar), this.v1.times(scalar),
this.x2.times(scalar), this.v2.times(scalar)
)
}
}
// dy/dt (t) = f(t, y(t))
function f(t: number, y: TwoParticles) {
const { x1, v1, x2, v2 } = y;
return new TwoParticles(
// dx1/dt = v1
v1,
// dv1/dt = G*m2*(x2-x1)/|x2-x1|^3
x2.minus(x1).times(G * m2 / Math.pow(x2.minus(x1).length(), 3)),
// dx2/dt = v2
v2,
// dv2/dt = G*m1*(x1-x1)/|x1-x2|^3
x1.minus(x2).times(G * m1 / Math.pow(x1.minus(x2).length(), 3))
)
}
// y(0) = y0
const y0 = new TwoParticles(
/* x1 */ new Vec2(2, 3),
/* v1 */ new Vec2(1, 0),
/* x2 */ new Vec2(4, 1),
/* v2 */ new Vec2(-1, 0)
)
const canvas = document.createElement("canvas")
canvas.width = 400;
canvas.height = 400;
const ctx = canvas.getContext("2d")!;
document.body.appendChild(canvas);
// Текущее состояние симуляции
function render(y: TwoParticles) {
const { x1, x2 } = y;
ctx.fillStyle = "white";
ctx.fillRect(0, 0, 400, 400);
ctx.fillStyle = "black";
ctx.beginPath();
ctx.ellipse(x1.x*50 + 200, x1.y*50 + 200, 15, 15, 0, 0, 2 * Math.PI);
ctx.fill();
ctx.fillStyle = "red";
ctx.beginPath();
ctx.ellipse(x2.x*50 + 200, x2.y*50 + 200, 30, 30, 0, 0, 2 * Math.PI);
ctx.fill();
}
// Запускаем!
runSimulation(y0, f, render)

If you play with numbers, you can get a simulation of the orbit of the moon!

Simulation of the orbit of the moon, 1 pix. = 2500 km. 1 sec Simulation is 1 day on Earth. The proportion of the moon to the earth is increased 10 times

## Collisions and restrictions

The above mathematical model actually simulates the physical world, but in some cases, the method of numerical integration, unfortunately, breaks down.

Imagine a simulation of a ball bouncing on the surface.

The simulation state can be described as follows:

Where  this is the height of the ball above the surface, and his speed. If you release the ball from a height of 0.8 meters, we get:

If you plot then we get something like the following:

Derivative function during ball fall  easy to compute:

With the acceleration of gravity, .

But what happens when the ball touches the surface? That the ball has reached the surface we can find out by. But with numerical integration, at one point in time, the ball can be above the surface, and already at the next below it:.

This problem could be solved by determining the moment of collision. But even if this moment is found, how to determine the accelerationso that it changes in the opposite direction.

You can, of course, determine the collision in a limited period of time and apply another force for this length of time , but it’s much easier to define a discrete constant limiting simulation.

And in order to reduce the amount of penetration by the ball of the surface, it is possible to calculate several simulation steps in one tick at once. Together with this, the code of our simulation will change like this:

function runSimulation>(
y0: T,
f: (t: number, y: T) => T,
applyConstraints: (y: T) => T,
iterationsPerFrame: number,
render: (y: T) => void
) {
const frameTime = 1 / 60.0
const h = frameTime / iterationsPerFrame
function simulationStep(yi: T, ti: number) {
render(yi)
requestAnimationFrame(function () {
for (let i = 0; i < iterationsPerFrame; i++) {
yi = yi.plus(f(ti, yi).times(h))
yi = applyConstraints(yi)
ti = ti + h
}
simulationStep(yi, ti)
})
}
simulationStep(y0, 0.0)
}

And now you can write the code of our bouncing ball:

Bouncing ball code

const g = -9.8; // m / s^2
const r = 0.2; // m
class Ball implements Numeric {
plus(other: Ball) { return new Ball(this.x + other.x, this.v + other.v) }
times(scalar: number) { return new Ball(this.x * scalar, this.v * scalar) }
}
function f(t: number, y: Ball) {
const { x, v } = y
return new Ball(v, g)
}
function applyConstraints(y: Ball): Ball {
const { x, v } = y
if (x <= 0 && v < 0) {
return new Ball(x, -v)
}
return y
}
const y0 = new Ball(
/* x */ 0.8,
/* v */ 0
)
function render(y: Ball) {
ctx.clearRect(0, 0, 400, 400)
ctx.fillStyle = '#EB5757'
ctx.beginPath()
ctx.ellipse(200, 400 - ((y.x + r) * 300), r * 300, r * 300, 0, 0, 2 * Math.PI)
ctx.fill()
}
runSimulation(y0, f, applyConstraints, 30, render)

## Attention to the developers!

Although this model has its advantages, it does not always lead to productive simulations. For me, such a framework is useful for representing the behavior of a simulation, even if a lot of unnecessary things happen in it.

For example, the rain simulation at the beginning of this article [ approx. In the original article, a beautiful interactive rain simulation was inserted at the beginning, I recommend that you look first-hand] was not created using the template described in the article. It was an experiment using Entity – component – system . The simulation source can be found here: rain simulation on GitHub .

## See you later!

I find the intersection of mathematics, physics, and programming something really impressive. Creating a working simulation, running it and rendering it is a special kind of something out of nothing .

All of the above was inspired by the materials of the SIGGRAPH lecture, just like in the fluid simulation . If you want to find more comprehensive information about the above, then take a look at the materials of the SIGGRAPH 2001 course "Introduction to Physical Modeling" . I give a link to the 1997 course, as Pixar seems to have removed version 2001.

I want to thank Maggie Cai for the wonderful illustration of the couple under the umbrella and for their patience with painstaking selection of colors, while I can not distinguish between blue and gray.

And if you are interested, then the illustrations were created in Figma .

Only registered users can participate in the survey. Please come in.

## Whether to translate the article “fluid simulation” (the link is given in the conclusion of the article)

• 96.8% Yes 154
• 3.1% No 5


text:
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

## Add and subtract decimals by lining up the decimal points.

%
Progress
Progress
%

Stephen went shopping to buy some new school supplies. He bought a backpack that cost $28.67 and a scientific calculator for$34.88. How much money did Stephen spend altogether?

When you add decimals you are adding whole numbers and like fraction parts. To make this process simpler, you should add decimal numbers using the vertical alignment method. The decimal points must be kept directly under each other and the digits in the same place value must be kept in line with each other. If the decimal numbers are signed numbers, the rules for adding integers are applied to the problem.

#### \begin{align*}87.296+48.6\end{align*}

Begin by writing the question using the vertical alignment method.

\begin{align*} 87.296 & \\ \underline{+48.6\phantom{00}}\end{align*}

Don't forget that the decimal points must be kept directly under one another. To ensure that the digits are aligned correctly, add zeros to 48.6.

\begin{align*} 87.296 & \\ \underline{+48.6{\color{blue}00}} & \end{align*}

\begin{align*} 87.296 & \\ \underline{+ 48.6{\color{blue}00}} & \\ 135.896 & \end{align*}

#### Now, let's try adding positive and negative decimals:

\begin{align*}(97.38)+(-45.17)\end{align*}

The first step is to write the problem using the vertical alignment method. The two decimal numbers that are being added have opposite signs. Apply the same rule that you used when adding integers that had opposite signs – subtract the numbers and use the sign of the larger number in the answer.

\begin{align*} 97.38 & \\ \underline{- 45.17} & \\ \quad 52.21 & \end{align*}

The larger number is 97.38 and it has a positive sign. This means that the sign of the answer will also be a positive value.

#### Finally, let's add two negative decimals together:

\begin{align*}(-168.8)+(-217.4536)\end{align*}

The first step is to write the problem using the vertical alignment method.

\begin{align*} -168.8\phantom{000} & \\ \underline{+-217.4536} & \end{align*}

To ensure that the digits are aligned correctly, add zeros to 168.8.

\begin{align*} -168.8{\color{blue}000} & \\ \underline{+-217.4536} & \end{align*}

\begin{align*} -168.8{\color{blue}000} & \\ \underline{+-217.4536} & \\ -386.2536 & \end{align*}

The numbers being added each had negative signs. This means that the sign of the answer is also a negative value.

### Examples

#### Example 1

Earlier, you were told that Stephen went shopping to buy some new school supplies and he bought a backpack that cost $28.67 and a scientific calculator for$34.88.

Stephen bought two items. To determine the total amount of money he spent, add the prices of the items.

The numbers and the decimal points have been correctly aligned. Now add the numbers.

Stephen spent 63.55 total. #### Example 2 Add the decimals: \begin{align*}45.36+15+137.692+32.8=?\end{align*} You can add all three numbers at once. \begin{align*}& \overset{2 1 \ \ 1}{\quad 45.36{\color{blue}0}}\\ & \quad 15.{\color{blue}000}\\ & \ 137.692\\ & \underline{+ 32.8{\color{blue}00}}\\ & \ 230.852\end{align*} #### Example 3 Add the decimals: \begin{align*}(53.69)+(-33.7)+(6.298)=?\end{align*} Add the two positive numbers together: \begin{align*} 53.69{\color{blue}0} & \\ \underline{+ \ \ 6.298} & \\ 59.988 & \end{align*} Subtract the negative number from the result: \begin{align*} 59.988 & \\ \underline{-33.7{\color{blue}00}} & \\ 26.288 & \end{align*} #### Example 4 Add the decimals: \begin{align*}14.68+39.217=?\end{align*} \begin{align*}14.68+39.217=53.897\end{align*} ### Review Add the following numbers. 1. \begin{align*}14.36+9.42\end{align*} 2. \begin{align*}52.72+27.163\end{align*} 3. \begin{align*}0.26+4.5+1.137\end{align*} 4. \begin{align*}37.231+14.567\end{align*} 5. \begin{align*}78.32+6.2+19.46\end{align*} 6. \begin{align*}65.23+12.75\end{align*} 7. \begin{align*}148.067+53.78+6.9\end{align*} 8. \begin{align*}56.75+14.9294+17.854\end{align*} 9. \begin{align*}18+26.87+65.358\end{align*} 10. \begin{align*}23.067+268.93+9.4\end{align*} 11. \begin{align*}(-24.69)+(-39.87)\end{align*} 12. \begin{align*}(76.35)+(-36.68)\end{align*} 13. \begin{align*}(-12.5)+(47.97)+(-21.653)\end{align*} 14. \begin{align*}(62.462)+(254.69)+(-427.9)\end{align*} 15. \begin{align*}(-37.76)+(-45.8)+(53.92)\end{align*} Determine the answer to the following problems. 1. When the owners of the Finest Fixer Co. completed a small construction job, they found that the following expenses had been incurred: labour,975.75; gravel, $88.79; sand,$43.51; cement, $284.96; and bricks$2214.85. What bill should they give the customer if they want to make a profit of $225 for the job? 2. A tile setter purchases the following supplies for the day: • One bag of thin-set mortar - $5.67 per bag • 44 sq ft of tile -$107.80 for 44 sq ft of tile • One gallon of grout - $17.97 per gallon • One container of grout sealer -$32.77 per container • 3 containers of grout and tile cleaner - $5.99 per container • 4 scrub pads -$2.78 each • One trowel - @$3.95 each
• 2 packages of tile spacers - $2.27 each • One grout bag -$2.79 each
• One grout float - $10.45 each What is the cost of these items before tax is added? 3. The four employees of the Broken Body Shop earned the following amounts last week: 815.86, 789.21, 804.18 and 888.35. What is the average weekly pay for the employees? 4. Jennifer bought the following school supplies: • 1000 sheets of paper -$14.67
• 36 pencils - @ $6.55 • 1 binder - @$18.48
• 1 backpack - @ $22.74 • 1 lunch bag - @$4.64

How much did Jennifer spend on these supplies before taxes?

5. A local seamstress needs to purchase fabric to sew curtains for the local theatre. She needs 123.75 yd. of black cotton for a backdrop, 217.4 yd. of white linen for stage curtains, 75 yd. for accessory curtains and 98.5 yd. for costumes. How many yards of fabric must be purchased to fill this order?

To see the Review answers, open this PDF file and look for section 1.3.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English

Decimal point

A decimal point is a period that separates the complete units from the fractional parts in a decimal number.

Rounding

Rounding is reducing the number of non-zero digits in a number while keeping the overall value of the number similar.

format: plain


text:
## Make {x^2 – 2xy + y^2, x^2 + 3xz + z^2, y^2 – 5yz + z^2} squares … Part 2

Find three positive integers   $x, y, z$   such that   $(x, y, z) = 1$   and

$x^2 \; - \; 2 \, x \, y \; + \; y^2 \; = \; A^2$
$x^2 \; + \; 3 \, x \, z \; + \; z^2 \; = \; B^2$
$y^2 \; - \; 5 \, y \, z \; + \; z^2 \; = \; C^2$

Here are the first few solutions:

$\{x, \; y, \; z, \; A, \; B, \; C \}$

$\{ 112, \; 259, \; 48, \; 147, \; 176, \; 85 \}$
$\{ 456, \; 888, \; 95, \; 432, \; 589, \; 613 \}$
$\{ 1488, \; 3216, \; 527, \; 1728, \; 2201, \; 1465 \}$
$\{ 4720, \; 9520, \; 1239, \; 4800, \; 6431, \; 5761 \}$
$\{ 11088, \; 22995, \; 3344, \; 11907, \; 15664, \; 12469 \}$
$\{ 23352, \; 44520, \; 4031, \; 21168, \; 29051, \; 33181 \}$
$\{ 30096, \; 59499, \; 6992, \; 29403, \; 39824, \; 38845 \}$
$\{ 42784, \; 80416, \; 6303, \; 37632, \; 51761, \; 63025 \}$
$\{ 66352, \; 127699, \; 12528, \; 61347, \; 83984, \; 92005 \}$
$\{ 127920, \; 241995, \; 20336, \; 114075, \; 156784, \; 185389 \}$

Find the next few solutions.

Advertisements

## About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.


text:
# Pre Bonsai Murasaki Kiyhome 2- 3 yr #213

\$48.99

This Murasaki Kiyohime Acer palmatum Japanese maple is ideal for Bonsai since the grafting scars are minimal and very low. The Muraski Kiyohime is a great plant for bonsai because it has small leaves and is a dwarf maple.  Leaves emerge with a tinge of red on the edges in spring. Summertime foliage is a pleasing green. This maple is a 2-3 yrs old, 1 gallon plant. Muraski kiyohime is a dwarf that if planted out in the landscape will reach about 4- 5′.

Out of stock

Join the waitlist to be emailed when this product becomes available


text:
## 9 Group Title Find an equation of tangent line .... one year ago one year ago

1. tennisgirl5621 Group Title

WHO '

2. 9 Group Title

of $y(x)=\int\limits_{2}^{x} \cos ( \pi t ^{3} dt$ at x = 2

$\huge{y'(x)=F(x)=\cos(\pi x^3)}$


text:
# Is possible to perform algebric operation with bitwise operations in PDM domain?

I have to compute the some operations (say sum and subtraction) between 2 signals. I have data in PDM domain: x1_pdm(t)=[0101011....]; and x2_pdm(t)=[0101011....]; What I do now is: convert x1_pdm and x2_pdm in PCM domain (low pass filter and decimation factor M=64) and that compute the sum and difference (or any other operation). Is it possible to perform this sum and sub in PDM domani with bitwise operation ? and then compute one pdm to pcm conversion... thank you mattia

• What on earth is meant by x1_pdm(t)=[0101011....]; and x2_pdm(t)=[0101011....];? Are the bits to be interpreted as a binary integer that tells us how wide the PDM pulse is? – Dilip Sarwate Sep 2 '14 at 13:22
• x1_pdm(t)=[0101011....] is the pdm stream out out the sigma delta modulator. – user3246711 Sep 15 '14 at 20:30


text:
# How to form multiple clusters of given size from a set of points?

Let us consider a 2 dimensional space. There are 100 points in it. So, each point has its $$x$$ and $$y$$ coordinates. I want to form 10 clusters. Each cluster has 10 points.

How can I form such clusters.

• – Edmund Oct 25 '18 at 22:18
• Your "definition" of a cluster needs some clarification. For a start: When are two points in the point cloud "adjacent"? – Henrik Schumacher Oct 25 '18 at 22:18
• @HenrikSchumacher, I have x and y-axis coordinates of 100 points. I want to perform clustring over these points. Lets say I want to generate 10 clusters, each cluster with exactly 10 elements. – dipak narayanan Oct 26 '18 at 16:20
• The result is depend by distance of points to cluster center. Consider this situation: cluster 1-points[1-11]-center1, cluster2-ponits[12-20]-center2. there are 11 points in cluster1, 9 points in cluster2, we can move point 11 to cluster2? But distance[center1, point11]< distance[center2, point11], is this valid? – HyperGroups Oct 28 '18 at 5:05

You need an exact definition of "adjacent" as Henrik already said, because it is the bread and butter of finding clusters of similar elements.

pts = RandomReal[1, {100, 2}];
ListPlot[FindClusters[pts, 10], PlotStyle -> PointSize[0.05]]

• Thanks. How do you gurantee that each cluster has 10 points? – dipak narayanan Oct 26 '18 at 11:20

Here an example for constructing two clusters, which can easily expanded to n clusters:

cstart = {1, 2};

Now this forms the "start"-cluster

   clusterStart = (# + cstart) & /@
RandomVariate[NormalDistribution[0, 0.5], {10^3, 2}];

Next a bunch of clusters, somewhat drifting apart from the start cluster

clusterList =
Table[(# + cstart + delta*{1, 1}) & /@
RandomVariate[NormalDistribution[0, 0.5], {10^3, 2}], {delta, 0.5,
4, 0.3}];

Lastly, forming the clusters:

clusterList = {clusterStart, #} & /@ clusterList;

Finally plotting the resulting point sets:

With[{out =
Graphics[{{Red, Point[#[[1]]]}, {Green, Point[#[[2]]]}},
Frame -> True] & /@ clusterList},
GraphicsGrid[Partition[out, 3], ImageSize -> Large]
]

this delivers:


text:
# Is there “intuition” as to why the Continuum Hypothesis is independent of most large cardinal axioms?

I could not find a question that seemed to answer my specific query, despite lots of material on the Continuum Hypothesis (CH) on MSE and MO. If there is already a question on this, I'd greatly appreciate a link.

My question has to do with the resilience of CH to being resolved. I realize that it is independent of ZFC and "most" large cardinal axioms, and that CH can fail as "badly as you want" (see this question).

My question is: why is this the case, and is there intuition as to why "ZFC + any large cardinal axiom" cannot seem to anchor this problem? I realize the "multiverse" approach is simply to study CH in whatever universe you are in. But my question whether there is intuition as to why CH has such an indeterminate status no matter what axioms you throw at it. Thanks!

-
There are axioms that determine the truth of $\mathsf{CH}$. For example, $\diamondsuit$ and $\mathsf{PFA}$. –  William Aug 20 '14 at 0:19
@William: Sure, as well as CH, GCH, and Woodin's work. But my question is about the context within large cardinal axioms. Since large cardinals don't settle CH, some set theorists are comfortable saying that this unsettled state is the natural order (multiverse). I'm interested in the intuition as to why we should accept the unsettled large cardinal state of affairs. –  trb456 Aug 20 '14 at 0:25
@trb456: There's a lot about history, tradition, the fact that we are interested in both CH holding and failing since both have interesting consequences, and of course, we aim to have a minimal axiomatic setting. So once the dust settled and $\sf ZFC$ seemed like a good candidate, it's hard to add more axioms to the "hardcore of mathematics" without a good reason. –  Asaf Karagila Aug 20 '14 at 0:57
Large cardinal axioms are not accepted as "correct" either. They are popular because rich consequences can be derived from them with current methods. Before that forcing axioms were popular, some still are. Before that GCH itself was popular, etc. The reason why they weren't permanently adopted is not specific to CH: none of them are needed in most of mathematical practice. –  Conifold Aug 20 '14 at 1:08
@zibadaw: No, that's not quite right. The continuum hypothesis and its negations have implications to operator theory and they continue from there. All the way to physics. To say that no one but set theorists care for its value is plain false; in fact I don't think that many set theorists care for the value of the continuum, and perhaps the emerging school of multiverse approach to set theory (rather than formalist or Platonist) will end up as the dominant one in the future. Precisely because there are so many conflicting interests. –  Asaf Karagila Aug 20 '14 at 1:47

Let me address the question as stated first.

Why does $\sf CH$ has no determinate provability from any of the axioms we throw at it? This is false. As remarked in the comments. Plenty of axioms prove $\sf CH$ or disprove it. Things like $V=L$ or $\lozenge$ imply $\sf CH$ whereas things like $\sf PFA$ and similar forcing axioms imply its negation (these in particular imply that $2^{\aleph_0}=\aleph_2$ of all values).

Specifically you might be asking about large cardinals. Well, this is because large cardinals are large, and the continuum is small. Historically, I believe, the motivation for the thought that large cardinals might settle the continuum hypothesis came from the fact that we can prove the continuum hypothesis for Borel sets. Namely every uncountable Borel set has size continuum, and we can push this to analytic sets, but not further.

However, if there exists a measurable cardinal then we can prove the continuum hypothesis for co-analytic sets as well. Which is a push forward. So perhaps by having "enough" large cardinals, or a strong enough large cardinal axiom we can push this proof and go through all the possible sets of reals?

Well, no, not really. While these things definitely can occur (in the presence of Woodin cardinals we can push the continuum hypothesis higher up the projective hierarchy), there are more sets of real numbers than that. And at some point the large cardinals exhaust their power.

In particular we have the Levy-Solovay phenomenon:

Suppose that $\kappa$ is an inaccessible cardinal, and $\Bbb P$ is a forcing such that $|\Bbb P|<\kappa$. Then $\Vdash_\Bbb P\check\kappa\text{ is inaccessible}$.

Namely, small forcings cannot change the largeness of an inaccessible cardinal. The same is true for many other types of large cardinals, e.g. weakly compact cardinals, measurable cardinals, Woodin cardinals, etc.

On the other hand we always have a forcing of size $\aleph_2$ which violates the continuum hypothesis: adding $\aleph_2$ Cohen reals; and a forcing of size continuum which restores the continuum hypothesis: Levy collapse of $2^{\aleph_0}$ to $\aleph_1$ (and it doesn't add real numbers too!)

So with particularly small forcings we can switch the truth value of the continuum hypothesis on and off as we like. Therefore large cardinals, which are unaffected by small forcings, cannot possibly decide the truth value of $\sf CH$.

(That been said, there are large cardinals properties which can be destroyed by small forcings, but so far these were shown consistent with $\sf CH$ and with its negation by various means, e.g. by showing that we can perhaps destroy the large property, but then we can have another forcing which restores it and doesn't change the value of the continuum.)

-
Therefore large cardinals, which are unaffected by small forcings, cannot possibly decide the truth value of CH. I think this is the key statement. The fact that we cannot decide which combination of large cardinal and small forcing is "correct" is a major factor in not being able to resolve CH. –  trb456 Aug 20 '14 at 1:44
It's like you don't even read what I wrote in all those comments and this long answer. :-) We're not even trying to decide what is "correct". You asked why large cardinals didn't decide the continuum hypothesis, this is why. Moreover saying that a "combination of large cardinal and small forcing is "correct" is a major factor ..." shows complete lack of understanding what forcing is used for and how it is used for. –  Asaf Karagila Aug 20 '14 at 1:50
We use forcing to introduce new objects to a model of set theory, and thus show that a certain statement is consistent. We don't use it (often) to decide a particular model of set theory is "the one true model of set theory". Philosophically this is hardly makes any sense. If you insist that $\sf CH$ is necessary true, then you can't really use any type of forcing which changes the value of the continuum. In fact, just adding real numbers seems morally objectionable once you decide that $\sf CH$ is true. Forcing is a force majeure in how little we work towards settling the truth value of CH. –  Asaf Karagila Aug 20 '14 at 1:53
Look, I'm not a professional mathematician, and I did read what you wrote, so please don't be insulting. I have an advanced, but not professional, background in mathematics. And I highly regard your work! I'm trying to narrow down to why we cannot resolve CH beyond that it eliminates lots of interesting set theory. As you noted, we could decide all groups are abelian, but we don't, for good reasons. So given that there is lots of interesting set theoretical exploration, why does none of this settle the problem? And this is where I acknowledge that there may be no good intuition as to why. –  trb456 Aug 20 '14 at 2:04
I'm sorry if it came off as insulting. I even put a little smiley there to make sure it reads with a smile, as a joke. As for the rest of the comment, "can't prove it from the axioms we regard as standard + either truth value has interesting results + the technique which revolutionized modern set theory will be handicapped" are three excellent reasons not to want to settle the problem "once and for all". –  Asaf Karagila Aug 20 '14 at 2:20


text:
## 0213 Group Title Could someone walk me through this question and explain to me what happens. I have been working on it for 7 hours. I don't understand my prof and I might fail the class and I want to cry. Its solving a quadratic equation that had complex numbers 9 months ago 9 months ago

1. 0213

$2x^2-ix-2+i=0$

2. DragonFlame

:/ Sorry... I can't help..

3. 0213

:'(

4. DragonFlame

@jdoe0001

5. jdoe0001

tried using the quadratic formula yet?

6. DragonFlame

Phew! GO JDOE!

7. 0213

I tried it, i solved it but i got the wrong answers. I tried it over and over again. I looked up videos, still wrong answer

8. jdoe0001

$$\large \begin{array}{llll} {\color{red}{ 2}}x^2&{\color{blue}{ -i}}x&{\color{green}{ -2+i}}=0\\ \uparrow &\ \uparrow &\quad \uparrow \\ {\color{red}{ a}}&\ {\color{blue}{ b}}&\quad {\color{green}{ c}} \end{array}\\ \quad \\ \bf x=\cfrac{i\pm\sqrt{i^2-4(2)(-2+i)}}{2(2)}\implies x=\cfrac{i\pm\sqrt{-1-8(-2+i)}}{2(2)}\\ \quad \\ x=\cfrac{i\pm\sqrt{-1+16-8i}}{2(2)}\implies x=\cfrac{i\pm\sqrt{15-8i}}{4}$$

9. 0213

ok..got that

10. jdoe0001

well.. that's.... as far as I can see it going... what does your answer say?

11. 0213

x=1 and x=-1+0.5i

12. jdoe0001

hmmm I don't see it simplifying ... to that

13. 0213

:(

14. RaphaelFilgueiras

|dw:1389574709566:dw|

15. RaphaelFilgueiras
16. 0213

how did u do that

17. RaphaelFilgueiras

that is an algorithm

18. 0213

19. RaphaelFilgueiras

|dw:1389575164382:dw|

20. 0213

wait...is this like synthetic division

21. RaphaelFilgueiras

|dw:1389575421091:dw|

22. RaphaelFilgueiras

this is the briot-ruffini algorithmm, i don't what is synthetic division

23. 0213

so in my question the root is 2?

24. RaphaelFilgueiras

no the root is one, the first coef in your polynomial is 2

25. 0213

how do i know what the root is?

26. RaphaelFilgueiras

one, was a easy root to see in that problem

27. 0213

oh. so if it had all 4...then it will be 2?

28. 0213

could u do it step by step and explain what you were doing at each step

29. RaphaelFilgueiras

|dw:1389575930097:dw|


text:
# Math Help - Expected number of questions to win a game

1. ## Expected number of questions to win a game

Let, a person is taking part in a quiz competition. For each questions, there are 3 answers, and for each correct answer he gets 1 point.
When he gets 5 points, he wins the game.
If he gives 2 consecutive wrong answers, then his points resets to zero (i.e. if his score is now 4 and he gives 2 wrong answers, then his score resets to 0).

My question is, on an average how much questions he needs to answer to win the game?

2. ## Re: Expected number of questions to win a game

Originally Posted by concept
Let, a person is taking part in a quiz competition. For each questions, there are 3 answers, and for each correct answer he gets 1 point. When he gets 5 points, he wins the game.
If he gives 2 consecutive wrong answers, then his points resets to zero (i.e. if his score is now 4 and he gives 2 wrong answers, then his score resets to 0). On an average how much questions he needs to answer to win the game?
This seems to be quite complicated. The difficulty will in modeling the outcomes.

This person can win on the $5^{th},~6^{th},~\cdots~10^{th}$ game.
If the person gets to play the $10^{th}$ and does not win, then the score must reset.

Now that is not to say that the score does not reset before the tenth game.
In fact, it could reset after the second game.

Have you worked with conditional expectation?

3. ## Re: Expected number of questions to win a game

Yes, I know about conditional probability, but I can not figure out how to solve this problem

4. ## Re: Expected number of questions to win a game

Originally Posted by concept
Yes, I know about conditional probability, but I can not figure out how to solve this problem
No it is not conditional probability, it is conditional expectation.

5. ## Re: Expected number of questions to win a game

Originally Posted by Plato
No it is not conditional probability, it is conditional expectation.
I did not work with conditional expectation. I searched in "Probability and Statistics- walpole" but could not find. Then found in wikipedia, but could not understand properly.
Could you please give hints how to solve the problem using conditional expectation or without using conditional expectation?

Thank you

6. ## Re: Expected number of questions to win a game

Originally Posted by concept
I did not work with conditional expectation. I searched in "Probability and Statistics- walpole" but could not find. Then found in wikipedia, but could not understand properly. Could you please give hints how to solve the problem using conditional expectation or without using conditional expectation?

In that case I suppose you are not to use it.
Do you understand what I have said about the setup?

Now I assume the person is simply doing random guessing.
The probability that the person wins on the eight game is $\binom{5}{3}{\left( {\frac{2}{3}} \right)^3}{\left( {\frac{1}{3}} \right)^5}$.

The reasoning behind this is that the fifth win is gotten on the eighth game requires that there be four wins and three losses none of which are together.

Now $\binom{5}{k-5}{\left( {\frac{2}{3}} \right)^{k-5}}{\left( {\frac{1}{3}} \right)^5}$ is the probability if winning on the $k^{th}$ game, where $k=5,6,7,8,9,10$.

The points reset to 0 for the eleventh game. And the next chance to win is on the fifteenth game.
All of that assumes no-resets.

The computations are quite tedious. I am not prepared to do more.

7. ## Re: Expected number of questions to win a game

Originally Posted by concept
Let, a person is taking part in a quiz competition. For each questions, there are 3 answers, and for each correct answer he gets 1 point.
When he gets 5 points, he wins the game.
If he gives 2 consecutive wrong answers, then his points resets to zero (i.e. if his score is now 4 and he gives 2 wrong answers, then his score resets to 0).

My question is, on an average how much questions he needs to answer to win the game?
Is the person randomly guessing answers? If yes, suppose we start from the beginning or, from a "reset to zero".

Let n = the number of questions answered to win, l = the # of losses, w = the # of wins

n = l + w, and as mentioned $n \geq 5$. The probability of winning in 5 questions is simple (1/3)^5. Note that the winning string of letters must consist of the substrings W and LW (any combinations of these does not result in a "Reset"). e.g. n = 6 = LW W LW W (as just one possibility). Now, P(W) = 1/3 and P(LW)=2/9, and we can find the probability of winning after n questions by:

$P(n=5) = {5 \choose 0} \left(\frac{1}{3}\right)^5 \left(\frac{2}{9}\right)^0 = 1/243$
$P(n=6) = {5 \choose 1} \left(\frac{1}{3}\right)^4 \left(\frac{2}{9}\right)^1 = 10/729$ i.e. LW W W W W
$P(n=7) = {5 \choose 2} \left(\frac{1}{3}\right)^3 \left(\frac{2}{9}\right)^2 = etc.$ i.e. LW LW W W W
$P(n=8) = {5 \choose 3} \left(\frac{1}{3}\right)^2 \left(\frac{2}{9}\right)^3$ i.e. LW LW LW W W
$P(n=9) = {5 \choose 4} \left(\frac{1}{3}\right)^1 \left(\frac{2}{9}\right)^4$ i.e. LW LW LW LW W
$P(n=10) = {5 \choose 5} \left(\frac{1}{3}\right)^0 \left(\frac{2}{9}\right)^5$ i.e. LW LW LW LW LW

The one thing I don't have time to help you with is the conditional probability prior to having won (according to the above possibilities). So, you must calculate the probability of seeing a string that combines LW's and W's and then LL at the end (so, in this case, you can't have more than 4 wins, or else you've won already!). This part is quite complicated, because you could have a large number of "resets to zero", and these particular strings would be in various sizes. Seems like it would take less time to do this via a computer program to play this game thousands of times to estimate the answer.

8. ## Re: Expected number of questions to win a game

Originally Posted by majamin
$P(n=7) = {5 \choose 2} \left(\frac{1}{3}\right)^3 \left(\frac{2}{9}\right)^2 = etc.$ i.e. LW LW W W W
$P(n=8) = {5 \choose 3} \left(\frac{1}{3}\right)^2 \left(\frac{2}{9}\right)^3$ i.e. LW LW LW W W

There is a problem with those calculations without further assumptions.
They are correct with no resets.
But $n=7$ is possible with a reset on $n=2$.

Thus $P(n=7) = {5 \choose 2} \left(\frac{1}{3}\right)^3 \left(\frac{2}{9}\right)^2 +\left(\frac{2}{3}\right)^2P(n=5)$
That is no resets plus one reset.

Now for $n=8$ we could have a reset on $n=2\text{ or }n=3$ or none at all.

9. ## Re: Expected number of questions to win a game

Originally Posted by Plato
There is a problem with those calculations without further assumptions.
They are correct with no resets.
But $n=7$ is possible with a reset on $n=2$.

Thus $P(n=7) = {5 \choose 2} \left(\frac{1}{3}\right)^3 \left(\frac{2}{9}\right)^2 +\left(\frac{2}{3}\right)^2P(n=5)$
That is no resets plus one reset.

Now for $n=8$ we could have a reset on $n=2\text{ or }n=3$ or none at all.
Yes, that was my intent, to only calculate a "winning" string. I agree with your addition to the problem, and I stand by my original comment that it would be difficult to include all possibilities (especially for large n) without the aid of some kind of computer assisted code. My hunch is that a "nice", explicit solution is out of reach.


text:
# Solve the slope…. pls

###### Question:

solve the slope…. pls

### 3(3x - 8) - 5(3x - 8) = 4(x - 2) - 6(x - 2) please help

3(3x - 8) - 5(3x - 8) = 4(x - 2) - 6(x - 2) please help...

### Member of red velvet

member of red velvet...

### What does hola señor mean in english?

what does hola señor mean in english?...

### Which answer choice it is? A 44 B 30 C 49 D 41

Which answer choice it is? A 44 B 30 C 49 D 41...

### Which is one theme of this passage "The True Story of My Life" by Hans Christian Anderson? A) A bird in a cage cannot succeed. B) Ridicule often pushes one to fail. C) The road to success is not always easy. D) Self-esteem is not connected to success.

Which is one theme of this passage "The True Story of My Life" by Hans Christian Anderson? A) A bird in a cage cannot succeed. B) Ridicule often pushes one to fail. C) The road to success is not always easy. D) Self-esteem is not connected to success....

### Fill in the blank with the correct form of the verb *ir*. ¿Tú al concierto mañana? Mi hermana y yo no . Yo no porque tengo mucha tarea. Pero mi amiga con su novio. Ellos a un restuarante antes de escuchar la música. Ellos a estar muy cansados en clase mañana.

Fill in the blank with the correct form of the verb *ir*. ¿Tú al concierto mañana? Mi hermana y yo no . Yo no porque tengo mucha tarea. Pero mi amiga con su novio. Ellos a un restuarante antes de escuchar la música. Ellos a estar muy cansados en clase mañana....

### Where is genetic information stored? DNA Proteins RNA

Where is genetic information stored? DNA Proteins RNA...

### Which expression is equivalent to 7^5/7^2

Which expression is equivalent to 7^5/7^2...

### Identity each container a. element or compound

Identity each container a. element or compound...

### Using the function in example 4, what is the y value (output) when x= -3?

Using the function in example 4, what is the y value (output) when x= -3?...

### What is the Molarity of a solution containing 2.3 moles of KCL in 0.45 L of solution

What is the Molarity of a solution containing 2.3 moles of KCL in 0.45 L of solution...

### Help help help! This is 40% of my grade and i am stuck. FIND THE MEASURES OF THE INTERIOR ANGLES

Help help help! This is 40% of my grade and i am stuck. FIND THE MEASURES OF THE INTERIOR ANGLES...

### A machine is supplied energy at a rate of 4000 watts and does useful work at a rate of 3761 watts what is the efficiency of the machine

A machine is supplied energy at a rate of 4000 watts and does useful work at a rate of 3761 watts what is the efficiency of the machine...

### The wire netting has been installed to prevent loss of property or life resulting from

The wire netting has been installed to prevent loss of property or life resulting from...

### Which of the following places is most similar to an internment camp? a jail a resort an apartment complex the army

Which of the following places is most similar to an internment camp? a jail a resort an apartment complex the army...


text:
Ideal rootkit elimination device for a server?

0
2019-05-05 12:09:23
Source Share

If your system has a rootkit, you need to :

1. Collect any kind of details concerning running procedures on your equipment.
2. Make a duplicate of RAM and also your harddrive.
4. Install a new system/restore your backup.

The first 2 factors serve, if you intend to explore because concern. Possibly it is additionally valuable to do not touch the system till your examination has actually finished.

In instance of a rootkit a few other individual had possibly complete accessibility of your computer system. So it is necessary to entirely remove the old system. Thatswhy you need to reformat your drive. If your fortunate and also have a current backup, you need to recover it and also you're done. Otherwise, you need to re-install the system. This is the only means to firmly remove the rootkit and also ahead up with a tidy system.

0
2019-05-08 19:43:40
Source

There are no computerized rootkit elimination tools for Ubuntu, just tools to look for rootkits.

chkrootkit and also rkhunter are rather durable tools when it involves identifying rootkits, yet they're just like their regulations. Additionally check into tripwire, which checks essential apply for adjustments.

You need to have every one of the above run consistently using cron.

0
2019-05-08 09:54:43
Source


text:
# Measurement in quantum mechanics

A measurement always causes the system to jump into an eigenstate of the dynamical variable that is being measured, the eigenvalue this eigenstate belongs to being equal to the result of the measurement.

P.A.M. Dirac (1958) in "The Principles of Quantum Mechanics" p. 36

The framework of quantum mechanics requires a careful definition of measurement. The issue of measurement lies at the heart of the problem of the interpretation of quantum mechanics, for which there is currently no consensus.

## Measurement from a practical point of view

Measurement plays an important role in quantum mechanics, and it is viewed in different ways among various interpretations of quantum mechanics. In spite of considerable philosophical differences, different views of measurement almost universally agree on the practical question of what results from a routine quantum-physics laboratory measurement. To understand this, the Copenhagen interpretation, which has been commonly used,[1] is employed in this article.

### Qualitative overview

In classical mechanics, a simple system consisting of only one single particle is fully described by the position $\vec{x} (t)$ and momentum $\vec{p} (t)$ of the particle. As an analogue, in quantum mechanics a system is described by its quantum state, which contains the probabilities of possible positions and momenta. In mathematical language, all possible pure states of a system form an abstract vector space called Hilbert space, which is typically infinite-dimensional. A pure state is represented by a state vector in the Hilbert space.

Once a quantum system has been prepared in laboratory, some measurable quantity such as position or energy is measured. For pedagogic reasons, the measurement is usually assumed to be ideally accurate. The state of a system after measurement is assumed to "collapse" into an eigenstate of the operator corresponding to the measurement. Repeating the same measurement without any evolution of the quantum state will lead to the same result. If the preparation is repeated, subsequent measurements will likely lead to different results.

The predicted values of the measurement are described by a probability distribution, or an "average" (or "expectation") of the measurement operator based on the quantum state of the prepared system.[2] The probability distribution is either continuous (such as position and momentum) or discrete (such as spin), depending on the quantity being measured.

The measurement process is often considered as random and indeterministic. Nonetheless, there is considerable dispute over this issue. In some interpretations of quantum mechanics, the result merely appears random and indeterministic, whereas in other interpretations the indeterminism is core and irreducible. A significant element in this disagreement is the issue of "collapse of the wavefunction" associated with the change in state following measurement. There are many philosophical issues and stances (and some mathematical variations) taken—and near universal agreement that we do not yet fully understand quantum reality. In any case, our descriptions of dynamics involve probabilities, not certainties.

### Quantitative details

The mathematical relationship between the quantum state and the probability distribution is, again, widely accepted among physicists, and has been experimentally confirmed countless times. This section summarizes this relationship, which is stated in terms of the mathematical formulation of quantum mechanics.

#### Measurable quantities ("observables") as operators

Main article: Observable

It is a postulate of quantum mechanics that all measurements have an associated operator (called an observable operator, or just an observable), with the following properties:

1. The observable is a self-adjoint operator mapping a Hilbert space (namely, the state space, which consists of all possible quantum states) into itself.
2. Thus, the observable's eigenvectors (called an eigenbasis) form an orthonormal basis that span the state space in which that observable exists. Any quantum state can be represented as a superposition of the eigenstates of an observable.
3. Hermitian operators' eigenvalues are real. The possible outcomes of a measurement are precisely the eigenvalues of the given observable.
4. For each eigenvalue there are one or more corresponding eigenvectors (eigenstates). A measurement results in the system being in the eigenstate corresponding to the eigenvalue result of the measurement. If the eigenvalue determined from the measurement corresponds to more than one eigenstate ("degeneracy"), instead of being in a definite state, the system is in a sub-space of the measurement operator corresponding to all the states having that eigenvalue.

Important examples of observables are:

• The Hamiltonian operator $\hat{H}$, which represents the total energy of the system. In nonrelativistic quantum mechanics the nonrelativistic Hamiltonian operator is given by ${\hat H} = \hat T + \hat V ={\hat p^2 \over 2m} + V( \hat x )$.
• The momentum operator ${\hat p}$ is given by ${\hat p} = -i\hbar {\partial \over \partial x}$ (in the position basis), or ${\hat p} = p$ (in the momentum basis).
• The position operator ${\hat x}$ is given by ${\hat x} = x$ (in the position basis), or ${\hat x}= i\hbar {\partial \over \partial p}$ (in the momentum basis).

Operators can be noncommuting. Two Hermitian operators commute if (and only if) there is at least one basis of vectors, each of which is an eigenvector of both operators (this is sometimes called a simultaneous eigenbasis). Noncommuting observables are said to be incompatible and cannot in general be measured simultaneously. In fact, they are related by an uncertainty principle as discovered by Werner Heisenberg.

#### Measurement probabilities and wavefunction collapse

There are a few possible ways to mathematically describe the measurement process (both the probability distribution and the collapsed wavefunction). The most convenient description depends on the spectrum (i.e., set of eigenvalues) of the observable.

##### Discrete, nondegenerate spectrum

Let $\hat{O}$ be an observable. By assumption, $\hat{O}$ has discrete eigenstates $|1 \rang, |2 \rang, |3 \rang,...$ with corresponding distinct eigenvalues $O_1, O_2, O_3,...$. That is, the states are nondegenerate.

Consider a system prepared in state $|\psi \rang$. Since the eigenstates of the observable $\hat{O}$ form a complete basis called eigenbasis, the state vector $|\psi \rang$ can be written in terms of the eigenstates as

$|\psi\rang = c_1 | 1 \rang + c_2 | 2 \rang + c_3 | 3 \rang + \cdots = \sum_n c_n | n \rang$,

where $c_1,c_2,\ldots$ are complex numbers in general. The eigenvalues $O_1, O_2, O_3,...$ are all possible values of the measurement. The corresponding probabilities are given by

$\Pr( O_n ) = \frac{ | \lang n | \psi \rang |^2 }{ |\lang \psi | \psi\rang |} = \frac{ | c_n |^2 }{\sum_k | c_k |^2}$

Usually $|\psi\rang$ is assumed to be normalized, i.e. $\lang \psi | \psi\rang=1$. Therefore, the expression above is reduced to

$\Pr( O_n ) = | \lang n | \psi \rang |^2 = | c_n |^2$

If the result of the measurement is $O_n$, then the system (after measurement) is in pure state $|n\rang$. That is,

$| \psi' \rang = | n \rang$

so any repeated measurement of ${\hat O}$ will yield the same result $O_n$. When there is a discontinuous change in state due to a measurement that involves discrete eigenvalues, that is called wavefunction collapse. For some, this is simply a description of a reasonably accurate discontinuous change in a mathematical representation of physical reality; for others, depending on philosophical orientation, this is a fundamentally serious problem with quantum theory.

##### Continuous, nondegenerate spectrum

Let $\hat{O}$ be an observable. By assumption, $\hat{O}$ has continuous eigenstate $|x\rang$, with corresponding distinct eigenvalue $x$. The eigenvalue forms a continuous spectrum filling the interval (a,b).

Consider a system prepared in state $|\psi\rang$. Since the eigenstates of the observable $\hat{O}$ form a complete basis called eigenbasis, the state vector $|\psi \rang$ can be written in terms of the eigenstates as

$|\psi\rang = \int_a^b c(x) | x \rang \, dx$,

where $c(x)$ is a complex-valued function. The eigenvalue that fills up the interval $(a,b)$ is the possible value of measurement. The corresponding probability is described by a probability function given by

$\Pr( d

where $(d,e)\subseteq(a,b)$. Usually $|\psi\rang$ is assumed to be normalized, i.e. $\int_a^b\lang\psi|\psi\rang\, dx=1$. Therefore, the expression above is reduced to

$\Pr( d

If the result of the measurement is $x$, then the system (after measurement) is in pure state $|x\rang$. That is,

$|\psi'\rang = |x\rang.$

Alternatively, it is often possible and convenient to analyze a continuous-spectrum measurement by taking it to be the limit of a different measurement with a discrete spectrum. For example, an analysis of scattering involves a continuous spectrum of energies, but by adding a "box" potential (which bounds the volume in which the particle can be found), the spectrum becomes discrete. By considering larger and larger boxes, this approach need not involve any approximation, but rather can be regarded as an equally valid formalism in which this problem can be analyzed.

##### Degenerate spectra

If there are multiple eigenstates with the same eigenvalue (called degeneracies), the analysis is a bit less simple to state, but not essentially different. In the discrete case, for example, instead of finding a complete eigenbasis, it is a bit more convenient to write the Hilbert space as a direct sum of eigenspaces. The probability of measuring a particular eigenvalue is the squared component of the state vector in the corresponding eigenspace, and the new state after measurement is the projection of the original state vector into the appropriate eigenspace.

##### Density matrix formulation
Main article: Density matrix

Instead of performing quantum-mechanics computations in terms of wavefunctions (kets), it is sometimes necessary to describe a quantum-mechanical system in terms of a density matrix. The analysis in this case is formally slightly different, but the physical content is the same, and indeed this case can be derived from the wavefunction formulation above. The result for the discrete, degenerate case, for example, is as follows:

Let ${\hat O}$ be an observable, and suppose that it has discrete eigenvalues $O_1,O_2,O_3,\ldots$, associated with eigenspaces $V_1,V_2,\ldots$ respectively. Let $P_n$ be the projection operator into the space $V_n$.

Assume the system is prepared in the state described by the density matrix ρ. Then measuring ${\hat O}$ can yield any of the results $O_1, O_2, O_3, \ldots$, with corresponding probabilities given by

$\Pr( O_n ) = \mathrm{Tr}(P_n \rho)$

where $\mathrm{Tr}$ denotes trace. If the result of the measurement is n, then the new density matrix will be

$\rho' = \frac{P_n \rho P_n}{\mathrm{Tr}(P_n \rho)}$

Alternatively, one can say that the measurement process results in the new density matrix

$\rho'' = \sum_n P_n \rho P_n$

where the difference is that $\rho''$ is the density matrix describing the entire ensemble, whereas $\rho'$ is the density matrix describing the sub-ensemble whose measurement result was $n$.

#### Statistics of measurement

As detailed above, the result of measuring a quantum-mechanical system is described by a probability distribution. Some properties of this distribution are as follows:

Suppose we take a measurement corresponding to observable $\hat O$, on a state whose quantum state is $|\psi\rang$.

$\lang \psi | \hat O | \psi \rang$.
$\lang \psi | \hat O^2 | \psi \rang - (\lang \psi | \hat O | \psi \rang)^2$
$\sqrt{\lang \psi | \hat O^2 | \psi \rang - (\lang \psi | \hat O | \psi \rang)^2}$

These are direct consequences of the above formulas for measurement probabilities.

#### Example

Suppose that we have a particle in a 1-dimensional box, set up initially in the ground state $|\psi_1\rang$. As can be computed from the time-independent Schrödinger equation, the energy of this state is $E_1=\frac{\pi^2\hbar^2}{2mL^2}$ (where m is the particle's mass and L is the box length), and the spatial wavefunction is $\lang x|\psi_1\rang = \sqrt{ \frac{2}{L} }~{\rm sin}\left(\frac{\pi x}{L}\right)$. If the energy is now measured, the result will always certainly be $E_1$, and this measurement will not affect the wavefunction.

Next suppose that the particle's position is measured. The position x will be measured with probability density

$\Pr(S

If the measurement result was x=S, then the wavefunction after measurement will be the position eigenstate $|x=S\rang$. If the particle's position is immediately measured again, the same position will be obtained.

The new wavefunction $|x=S\rang$ can, like any wavefunction, be written as a superposition of eigenstates of any observable. In particular, using energy eigenstates, $| \psi_n\rang$, we have

$|x=S\rang = \sum_n | \psi_n \rangle \left\langle \psi_n | x=S \right\rangle = \sum_n | \psi_n \rangle \sqrt{ \frac{2}{L} }~{\rm sin}\left(\frac{n \pi S}{L}\right)$

If we now leave this state alone, it will smoothly evolve in time according to the Schrödinger equation. But suppose instead that an energy measurement is immediately taken. Then the possible energy values $E_n$ will be measured with relative probabilities:

$\Pr(E_n) = |\lang \psi_n | S \rang|^2 = \frac{2}{L}~{\rm sin}^2\left(\frac{n \pi S}{L}\right)$

and moreover if the measurement result is $E_n$, then the new state will be the energy eigenstate $|\psi_n\rang$.

So in this example, due to the process of wavefunction collapse, a particle initially in the ground state can end up in any energy level, after just two subsequent non-commuting measurements are made.

## Wavefunction collapse

The process in which a quantum state becomes one of the eigenstates of the operator corresponding to the measured observable is called "collapse", or "wavefunction collapse". The final eigenstate appears randomly with a probability equal to the square of its overlap with the original state.[2] The process of collapse has been studied in many experiments, most famously in the double-slit experiment. The wavefunction collapse raises serious questions regarding "the measurement problem",[3] as well as questions of determinism and locality, as demonstrated in the EPR paradox and later in GHZ entanglement. (See below.)

In the last few decades, major advances have been made toward a theoretical understanding of the collapse process. This new theoretical framework, called quantum decoherence, supersedes previous notions of instantaneous collapse and provides an explanation for the absence of quantum coherence after measurement. Decoherence correctly predicts the form and probability distribution of the final eigenstates, and explains the apparent randomness of the choice of final state in terms of einselection.[4]

### von Neumann measurement scheme

The von Neumann measurement scheme, the ancestor of quantum decoherence theory, describes measurements by taking into account the measuring apparatus which is also treated as a quantum object.

#### "Measurement" of the first kind — premeasurement without detection

Let the quantum state be in the superposition $\scriptstyle |\psi\rang = \sum_n c_n |\psi_n\rang$, where $\scriptstyle |\psi_n\rang$ are eigenstates of the operator for the so-called "measurement" prior to von Neumann's second apparatus. In order to make the "measurement", the system described by $\scriptstyle |\psi\rang$ needs to interact with the measuring apparatus described by the quantum state $\scriptstyle |\phi\rang$, so that the total wave function before the measurement and interaction with the second apparatus is $\scriptstyle |\psi\rang |\phi\rang$. During the interaction of object and measuring instrument the unitary evolution is supposed to realize the following transition from the initial to the final total wave function:

$|\psi\rang |\phi\rang \rightarrow \sum_n c_n |\psi_n\rang |\phi_n\rang \quad \text{(measurement of the first kind),}$

where $\scriptstyle |\phi_n\rang$ are orthonormal states of the measuring apparatus. The unitary evolution above is referred to as premeasurement. The relation with wave function collapse is established by calculating the final density operator of the object $\scriptstyle \sum_n |c_n|^2 |\psi_n\rang\lang \psi_n|$ from the final total wave function. This density operator is interpreted by von Neumann as describing an ensemble of objects being after the measurement with probability $\scriptstyle |c_n|^2$ in the state $\scriptstyle |\psi_n\rang.$

The transition

$|\psi\rang \rightarrow \sum_n |c_n|^2 |\psi_n\rang \lang \psi_n|$

is often referred to as weak von Neumann projection, the wave function collapse or strong von Neumann projection

$|\psi\rang \rightarrow \sum_n |c_n|^2 |\psi_n\rang \lang \psi_n| \rightarrow |\psi_n\rang$

being thought to correspond to an additional selection of a subensemble by means of observation.

In case the measured observable has a degenerate spectrum, weak von Neumann projection is generalized to Lüders projection

$|\psi\rang \rightarrow \sum_n |c_n|^2 P_n,\; P_n = \sum_i |\psi_{ni}\rang \lang \psi_{ni}|,$

in which the vectors $\scriptstyle |\psi_{ni}\rang$ for fixed n are the degenerate eigenvectors of the measured observable. For an arbitrary state described by a density operator $\scriptstyle \rho$ Lüders projection is given by

$\rho \rightarrow \sum_n P_n \rho P_n.$

#### Measurement of the second kind — with irreversible detection

In a measurement of the second kind the unitary evolution during the interaction of object and measuring instrument is supposed to be given by

$|\psi\rang |\phi\rang \rightarrow \sum_n c_n |\chi_n\rang |\phi_n\rang,$

in which the states $\scriptstyle |\chi_n\rang$ of the object are determined by specific properties of the interaction between object and measuring instrument. They are normalized but not necessarily mutually orthogonal. The relation with wave function collapse is analogous to that obtained for measurements of the first kind, the final state of the object now being $\scriptstyle |\chi_n\rang$ with probability $\scriptstyle |c_n|^2.$ Note that many present-day measurement procedures are measurements of the second kind, some even functioning correctly only as a consequence of being of the second kind. For instance, a photon counter, detecting a photon by absorbing and hence annihilating it, thus ideally leaving the electromagnetic field in the vacuum state rather than in the state corresponding to the number of detected photons; also the Stern–Gerlach experiment would not function at all if it really were a measurement of the first kind.[5]

### Decoherence in quantum measurement

One can also introduce the interaction with the environment $\scriptstyle |e\rang$, so that, in a measurement of the first kind, after the interaction the total wave function takes a form

$\sum_n c_n |\psi_n\rang |\phi_n\rang |e_n \rang,$

which is related to the phenomenon of decoherence.

The above is completely described by the Schrödinger equation and there are not any interpretational problems with this. Now the problematic wavefunction collapse does not need to be understood as a process $\scriptstyle |\psi\rangle \rightarrow |\psi_n\rang$ on the level of the measured system, but can also be understood as a process $\scriptstyle |\phi\rangle \rightarrow |\phi_n\rang$ on the level of the measuring apparatus, or as a process $\scriptstyle |e\rangle \rightarrow |e_n\rang$ on the level of the environment. Studying these processes provides considerable insight into the measurement problem by avoiding the arbitrary boundary between the quantum and classical worlds, though it does not explain the presence of randomness in the choice of final eigenstate. If the set of states

$\{ |\psi_n\rang\}$, $\{ |\phi_n\rang\}$, or $\{ |e_n\rang\}$

represents a set of states that do not overlap in space, the appearance of collapse can be generated by either the Bohm interpretation or the Everett interpretation which both deny the reality of wavefunction collapse. Both of these are stated to predict the same probabilities for collapses to various states as the conventional interpretation by their supporters. The Bohm interpretation is held to be correct only by a small minority of physicists, since there are difficulties with the generalization for use with relativistic quantum field theory. However, there is no proof that the Bohm interpretation is inconsistent with quantum field theory, and work to reconcile the two is ongoing. The Everett interpretation easily accommodates relativistic quantum field theory.

## Philosophical problems of quantum measurements

### What physical interaction constitutes a measurement?

Until the advent of quantum decoherence theory in the late 20th century, a major conceptual problem of quantum mechanics and especially the Copenhagen interpretation was the lack of a distinctive criterion for a given physical interaction to qualify as "a measurement" and cause a wavefunction to collapse. This is best illustrated by the Schrödinger's cat paradox. Certain aspects of this question are now well understood in the framework of quantum decoherence theory, such as an understanding of weak measurements, and quantifying what measurements or interactions are sufficient to destroy quantum coherence. Nevertheless, there remains less than universal agreement among physicists on some aspects of the question of what constitutes a measurement.

### Does measurement actually determine the state?

The question of whether (and in what sense) a measurement actually determines the state is one which differs among the different interpretations of quantum mechanics. (It is also closely related to the understanding of wavefunction collapse.) For example, in most versions of the Copenhagen interpretation, the measurement determines the state, and after measurement the state is definitely what was measured. But according to the many-worlds interpretation, measurement determines the state in a more restricted sense: In other "worlds", other measurement results were obtained, and the other possible states still exist.

### Is the measurement process random or deterministic?

As described above, there is universal agreement that quantum mechanics appears random, in the sense that all experimental results yet uncovered can be predicted and understood in the framework of quantum mechanics measurements being fundamentally random. Nevertheless, it is not settled[6] whether this is true, fundamental randomness, or merely "emergent" randomness resulting from underlying hidden variables which deterministically cause measurement results to happen a certain way each time. This continues to be an area of active research.[7]

If there are hidden variables, they would have to be "nonlocal".

### Does the measurement process violate locality?

In physics, the Principle of locality is the concept that information cannot travel faster than the speed of light (also see special relativity). It is known experimentally (see Bell's theorem, which is related to the EPR paradox) that if quantum mechanics is deterministic (due to hidden variables, as described above), then it is nonlocal (i.e. violates the principle of locality). Nevertheless, there is not universal agreement among physicists on whether quantum mechanics is nondeterministic, nonlocal, or both.[6]


text:
find the Delaunay triangulation for a given list of points - Maple Help

Home : Support : Online Help : Mathematics : Discrete Mathematics : Graph Theory : GraphTheory Package : GraphTheory/DelaunayTriangulation

GraphTheory[DelaunayTriangulation] - find the Delaunay triangulation for a given list of points

 Calling Sequence DelaunayTriangulation(points)

Parameters

 points - list of 2 element lists representing n coordinates

Description

 • This function computes a Delaunay triangulation of the input points returning a sequence of two outputs: the set of triangles and an edge adjacency Matrix.
 • The Delaunay triangulation is a triangulation that maximizes the smallest angle of the triangles in the triangulation (i.e. it avoids thin triangles).
 • The triangles are returned in a set containing the three vertices of each triangle as integer references to the input points list.
 • The adjacency matrix is an n by n symmetric Matrix that contains a $1$ in entry $i,j$ if one of the triangle edges is from ${\mathrm{points}}_{i}$ to ${\mathrm{points}}_{j}$, and $0$ otherwise.


text:
# Thread: Finding exact zeros and tangent lines.

1. ## Finding exact zeros and tangent lines.

So, here's the problem:

1) Let f be the function defined by $\displaystyle f(x) = -A + ln(x^2)$, where A is the number of letters in your last name. (For me, that would be 6).
(a) Find the exact zeros of f (using algebra, not your calculator).
(b) Write an equation for the tangent line to the graph of f at x = -1.

Can someone explain to me how to get started solving these questions?

2. To begin with, $\displaystyle x$ is a zero of $\displaystyle f(x)$ when

$\displaystyle -6 + \ln x^2=0.$

Solving this equation gives us the zeros of $\displaystyle f(x)$. To find the equation of the tangent line at $\displaystyle x=-1$, we differentiate $\displaystyle f(x)$ to find its slope at $\displaystyle x=-1$, remembering that the equation for a line is

$\displaystyle y-y_1=m(x-x_1).$

3. Originally Posted by Rumor
So, here's the problem:

1) Let f be the function defined by $\displaystyle f(x) = -A + ln(x^2)$, where A is the number of letters in your last name. (For me, that would be 6).
(a) Find the exact zeros of f (using algebra, not your calculator).
(b) Write an equation for the tangent line to the graph of f at x = -1.

Can someone explain to me how to get started solving these questions?

for a)

$\displaystyle -A + \ln (x^2) =0$

$\displaystyle A = \ln (x^2)$

$\displaystyle e^{A} = x^2$

$\displaystyle x = \mp e^{\frac{A}{2}}$

for 2)

$\displaystyle f(x) = -A + \ln (x^2)$

$\displaystyle f'(x) = 0 + \frac{2x}{x^2}$

$\displaystyle f'(-1) = \frac{-2}{1} = -2$ the slope of the function at x=-1

the equation of the tangent line when x=-1 , first find y when x=-1

$\displaystyle f(-1) = -A + \ln (x^2) = -A + 0 =-A$

the equation

$\displaystyle y-(-A ) = slope (x-(-1))$

$\displaystyle y+A = -2(x+1)$

4. Thank you to the both of you! That was really helpful.


text:
# Properties

 Label 3.13.ap_dy_aqx Base Field $\F_{13}$ Dimension $3$ Ordinary Yes $p$-rank $3$ Principally polarizable Yes Contains a Jacobian No

## Invariants

 Base field: $\F_{13}$ Dimension: $3$ L-polynomial: $( 1 - x + 13 x^{2} )( 1 - 7 x + 13 x^{2} )^{2}$ Frobenius angles: $\pm0.0772104791556$, $\pm0.0772104791556$, $\pm0.455715642762$ Angle rank: $2$ (numerical)

This isogeny class is not simple.

## Newton polygon

This isogeny class is ordinary.

 $p$-rank: $3$ Slopes: $[0, 0, 0, 1, 1, 1]$

## Point counts

This isogeny class is principally polarizable, but does not contain a Jacobian.

 $r$ 1 2 3 4 5 6 7 8 9 10 $A(\F_{q^r})$ 637 4213755 10125466624 22758933199275 50961221941086457 112500615380537180160 247131956265713730738709 542811441823314176232085275 1192537920497238107784957242368 2620019459632019140624088299414275

 $r$ 1 2 3 4 5 6 7 8 9 10 $C(\F_{q^r})$ -1 149 2096 27893 369659 4828748 62765639 815746757 10604540528 137859744989

## Decomposition and endomorphism algebra

Endomorphism algebra over $\F_{13}$
 The isogeny class factors as 1.13.ah 2 $\times$ 1.13.ab and its endomorphism algebra is a direct product of the endomorphism algebras for each isotypic factor. The endomorphism algebra for each factor is: 1.13.ah 2 : $\mathrm{M}_{2}($$$\Q(\sqrt{-3})$$$)$ 1.13.ab : $$\Q(\sqrt{-51})$$.
All geometric endomorphisms are defined over $\F_{13}$.

## Base change

This is a primitive isogeny class.

## Twists

Below are some of the twists of this isogeny class.
 Twist Extension Degree Common base change 3.13.an_cw_ald $2$ (not in LMFDB) 3.13.ab_ak_x $2$ (not in LMFDB) 3.13.b_ak_ax $2$ (not in LMFDB) 3.13.n_cw_ld $2$ (not in LMFDB) 3.13.p_dy_qx $2$ (not in LMFDB) 3.13.ag_be_afm $3$ (not in LMFDB) 3.13.ad_g_abr $3$ (not in LMFDB) 3.13.d_bn_cw $3$ (not in LMFDB) 3.13.g_bq_fq $3$ (not in LMFDB) 3.13.j_cc_ib $3$ (not in LMFDB)
Below is a list of all twists of this isogeny class.
 Twist Extension Degree Common base change 3.13.an_cw_ald $2$ (not in LMFDB) 3.13.ab_ak_x $2$ (not in LMFDB) 3.13.b_ak_ax $2$ (not in LMFDB) 3.13.n_cw_ld $2$ (not in LMFDB) 3.13.p_dy_qx $2$ (not in LMFDB) 3.13.ag_be_afm $3$ (not in LMFDB) 3.13.ad_g_abr $3$ (not in LMFDB) 3.13.d_bn_cw $3$ (not in LMFDB) 3.13.g_bq_fq $3$ (not in LMFDB) 3.13.j_cc_ib $3$ (not in LMFDB) 3.13.ab_bk_ax $4$ (not in LMFDB) 3.13.b_bk_x $4$ (not in LMFDB) 3.13.an_di_aoj $6$ (not in LMFDB) 3.13.al_ck_ajr $6$ (not in LMFDB) 3.13.al_cw_alz $6$ (not in LMFDB) 3.13.ak_ck_ako $6$ (not in LMFDB) 3.13.aj_cc_aib $6$ (not in LMFDB) 3.13.ai_bs_ahm $6$ (not in LMFDB) 3.13.ai_ce_aik $6$ (not in LMFDB) 3.13.ag_bq_afq $6$ (not in LMFDB) 3.13.af_bv_afe $6$ (not in LMFDB) 3.13.ae_u_aeo $6$ (not in LMFDB) 3.13.ae_bg_adq $6$ (not in LMFDB) 3.13.ad_bn_acw $6$ (not in LMFDB) 3.13.ac_ba_ack $6$ (not in LMFDB) 3.13.ab_c_acj $6$ (not in LMFDB) 3.13.ab_o_ab $6$ (not in LMFDB) 3.13.ab_bj_aw $6$ (not in LMFDB) 3.13.b_c_cj $6$ (not in LMFDB) 3.13.b_o_b $6$ (not in LMFDB) 3.13.b_bj_w $6$ (not in LMFDB) 3.13.c_ba_ck $6$ (not in LMFDB) 3.13.d_g_br $6$ (not in LMFDB) 3.13.e_u_eo $6$ (not in LMFDB) 3.13.e_bg_dq $6$ (not in LMFDB) 3.13.f_bv_fe $6$ (not in LMFDB) 3.13.g_be_fm $6$ (not in LMFDB) 3.13.i_bs_hm $6$ (not in LMFDB) 3.13.i_ce_ik $6$ (not in LMFDB) 3.13.k_ck_ko $6$ (not in LMFDB) 3.13.l_ck_jr $6$ (not in LMFDB) 3.13.l_cw_lz $6$ (not in LMFDB) 3.13.n_di_oj $6$ (not in LMFDB) 3.13.ab_aj_w $12$ (not in LMFDB) 3.13.ab_m_b $12$ (not in LMFDB) 3.13.b_aj_aw $12$ (not in LMFDB) 3.13.b_m_ab $12$ (not in LMFDB)


text:
Use implicit differentiation to find an equation of the tangent line to the curve $5xy^{3}+4xy = 54$ at the point $\left(6,1\right)$.

The equation defines the tangent line to the curve at the point $\left(6,1\right)$.

Your overall score for this problem is


text:
## The study of Borromean halo nuclei by the neutron wall with simulation    [PDF]

Liu Longxiang, Sun Zhiyu, Yue Ke, Xiao Guoqing, Chen Ximeng, Yu Yuhong, Zhang Xuehong, Wang Shitao, Tang Shuwen, Zhou Yong, Yan Duo, Fang Fang
The model of three-body Borromean halo nuclei breakup was described by using standard phase space distributions and the Monte Carlo simulation method was established to resolve the detection problem of two neutrons produced from breakup reaction on the neutron wall detector. For $^{6}$He case, overall resolution $\sigma_{E_{k}}$ for the Gaussian part of the detector response and the detection efficiency including solid angle acceptance with regard to the excitation energy $E_{k}$ are obtained by the simulation of two neutrons from $^{6}$He breakup into the neutron wall. The effects of the algorithm on the angular and energy correlations of the fragments are briefly discussed.
View original: http://arxiv.org/abs/1305.0675


text:
# How do you find (dy)/(dx) given -2y^2+3=x^3?

Jun 10, 2018

$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = - \setminus \frac{3 {x}^{2}}{4 y}$

#### Explanation:

Differentiate both sides as such:
$\setminus \frac{d}{\mathrm{dx}} \left(- 2 {y}^{2} + 3\right) = \setminus \frac{d}{\mathrm{dx}} \left({x}^{3}\right)$
$- 2 \setminus \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 3 {x}^{2}$
$- 2 \left(2 y \setminus \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 3 {x}^{2}$
rearrange to make $\setminus \frac{\mathrm{dy}}{\mathrm{dx}}$ the factor to get:
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = - \setminus \frac{3 {x}^{2}}{4 y}$


text:
# Generalization of Fatou's Lemma Without the Monotone Convergence Theorem?

Prove the following generalization of Fatou's Lemma: if $\{f_n\}$ is a sequence of nonnegative measurable functions on $E$, then $\int_E \lim \inf f_n \le \lim \inf \int_E f_n$.

I found this, but I don't understand how or why the Monotone Convergence Theorem is of any help. Here's my proof with just relies on Fatou's Lemma.

By definition, $\inf_{k \ge n} f_k$ converges pointwise to $\lim \inf f_n$, so by Fatou's lemma we get

$$\int_E \lim \inf f_n \le \lim \inf \int_E \inf_{k \ge n} f_k$$ But $\inf_{k \ge n} f_k \le f_n$ for every $n \in \Bbb{N}$ which implies $\int_E \inf_{k \ge n} f_k \le \int_E f_n$ for every $n \in \Bbb{N}$ and finally

$$\lim \inf \int_E \inf_{k \ge n} f_k \le \lim \inf \int_E f_n$$ The above two inequalities yield

$$\int_E \lim \inf f_n \le \lim \inf \int_E f_n,$$ which concludes the proof.

Have I made a mistake that is presently eluding me?

• In fact DCT is not necessary for this version of Fatou's lemma (and some authors acrually call this one "Fatou's lemma"). I think I recall hearing during a lecture that Fatou (this one), Beppo Levi and DCT are three theorems that each author decides to prove in whatever order he prefers (don't quote me on that). For instance, Royden proves this Fatou and then Beppo Levi and DCT.
– user228113
Mar 22, 2018 at 18:07
• @G.Sassatelli So the way I've proven it is okay? Mar 22, 2018 at 19:08
• Yes, the proof is okay. This generalization is usually called 'Fatou's Lemma', see e.g. Wikipedia. Mar 24, 2018 at 8:39


text:
Thread: Homework Help!! exact value, -pi < θ ≤ pi

1. Homework Help!! exact value, -pi < θ ≤ pi

Hi All,

Could somebody please explain (step by step) how to find the exact value of sinx= -1/√2 when -pi < θ ≤ pi? Been trying to google but the only example is between 0 < θ ≤ 2pi..

Many thanks

2. Re: Homework Help!! exact value, -pi < θ ≤ pi

Get the solutions for the easy region, then add multiples of 2pi until you have all the solutions in the region you are given.

3. Re: Homework Help!! exact value, -pi < θ ≤ pi

$\text{Disregard the minus sign temporarily and put} \\ \\ sin^{-1} \frac{1}{\sqrt{2}} \\ \\ \text{in your calculator, this will give you the reference angle of 45 or} \ \frac{\pi}{4} \\ \\ \text{The sine is negative in 3rd and 4th quadrants so your 2 solutions will be} \\ \\ - \pi \ + \ \frac{\pi}{4} \ = \ - \frac{3 \pi}{4} \\ \\ and \\ \\ - \frac{\pi}{4}$

4. Re: Homework Help!! exact value, -pi < θ ≤ pi

Originally Posted by agentmulder
$\text{Disregard the minus sign temporarily and put} \\ \\ sin^{-1} \frac{1}{\sqrt{2}} \\ \\ \text{in your calculator, this will give you the reference angle of 45 or} \ \frac{\pi}{4} \\ \\ \text{The sine is negative in 3rd and 4th quadrants so your 2 solutions will be} \\ \\ - \pi \ + \ \frac{\pi}{4} \ = \ -3 \frac{\pi}{4} \\ \\ and \\ \\ - \frac{\pi}{4}$
I'm sure you mean $\displaystyle -\frac{3\pi}{4}$...

5. Re: Homework Help!! exact value, -pi < θ ≤ pi

Yes, i guess the way i wrote it the fraction looked like a mixed number, i edited the post.

6. Re: Homework Help!! exact value, -pi < θ ≤ pi

Keep in mind that the sin(-x) = -sinx

Howard Heller
InteractiveMathTutor.com

7. Re: Homework Help!! exact value, -pi < θ ≤ pi

Originally Posted by IMTinstructor
Keep in mind that the sin(-x) = -sinx

Howard Heller
InteractiveMathTutor.com
Anyone stupid enough to fall for you scam deserves the outcome.
I can find no entry for you at MathGenealogy Project

So what equips you to tutor?

8. Re: Homework Help!! exact value, -pi < θ ≤ pi

What equips me to tutor? First of all, equips is not english to you must be from another country. I though am from the United States. Lets see... Math major, expert in math, actuary science major, mba, etc etc.