text:
# Induced emf in various rectangular loops

## Homework Statement

[/B]
Three conducting loops, all with the same resistance ##R## move towards a uniform and constant magnetic field, all with the same velocity ##v##. Their relative sizes can be identified by the grid. As the loops move into the magnetic field an induced current begins to flow.

a) Will the induced current be the same [in the loops] or not? If they differ- in which loop is the current the highest. Motivate your answer. [see picture below]

## Homework Equations

Magnetix flux: ##Φ_B=∫B⋅dA####=BAcosθ##
Ohm's law: ##V=RI##

## The Attempt at a Solution

So I take it that it's the moment as the loops just enter the magnetic field. Time or orientation is not specified. Since the B-field and the dA is parallel ##(cos(0)=1)## we have that the magnetic flux ##Φ_B## is ##B*dA##.

If we take the height (in the y-direction) to be ##l## and the width (in the x-direction) to be ##x## then we can write the area ##A## as ##x*l## and since ##l## is constant the area depends on how far the loop has entered the magnetic field. We take the amount in the x-direction that has entered the field to be ##dx##.

Now we have ##Φ_B=B*dA=B*dx*l##. And when we take the derivative with respect to time to calculate the Emf we get: ##ε=(-dΦ_B/dt)=B*l*(dx/dt)##

We note that dx/dt is just velocity and we can rewrite ##ε=(-dΦ_B/dt)=Blv##

This means that the length of the wire doesn't matter, only the height ##l##. This would mean that the induced current is greater in ##C## and lower but equally so in ##A## and ##B##.

Due to Lenz's law the current flows counterclockwise in the loops as they're drawn to create a magnetic field that opposes the B-field in the drawing.

------------------------

Now here's where I get worried if I missed out on anything important.

When the loops have moved JUST over one whole grid unit into the magnetic field there's no change in magnetix flux through two of the loops so there's no induced emf in them. Loop ##B## still has one grid unit to move before this happens so at this moment the current ##I## in ##A## and ##C## is 0. But not in ##B##.

Since the height in ##C## is ##2l## (compared to the other two loops) i take it that when all three loops have moved one grid unit into the field (just at the moment where there's still change in magnetic flux in all three loops) the current in ##C## is twice as large as in both ##A## and ##B##.

Did I do anything crazy here? Did I miss out on any details or cases which makes this false or did I get it right?

BiGyElLoWhAt
Gold Member
Looks good. C, however, is twice as large from the moment they enter to the moment just before C is fully inside the field. Assuming v is constant, the emf is constant of the domain of your emf function. (Your function is only valid from just after the loops enter the field to just before they are fully in, and the same with exiting [ - sign])

You shoul use Faraday's law: $$ε=-\frac{d∅}{dt}$$ => In the same Δt, the loop sweeps an area which is the biggest, this loop has the highest ε. So the current the highest.

BiGyElLoWhAt
Gold Member
You shoul use Faraday's law: $$ε=-\frac{d∅}{dt}$$
Isn't that what he did?

Looks good. C, however, is twice as large from the moment they enter to the moment just before C is fully inside the field. Assuming v is constant, the emf is constant of the domain of your emf function. (Your function is only valid from just after the loops enter the field to just before they are fully in, and the same with exiting [ - sign])

Cool, thanks! Yeah so the emf in ##C## is twice that of ##A## and ##B## during the whole period that all the loops still are subject to an induced emf, right? =)

BiGyElLoWhAt
BiGyElLoWhAt
Gold Member
Si, senor(/ita).


text:
## teeny Group Title calculate the mass of a cube of copper, each side of which is 15 cm one year ago one year ago

1. Khushal_Shah

so you can determine the volume, its just 15*15*15=3375 m^3

2. Azteck

$\huge V_{cube}=(side)^3$

3. Khushal_Shah

after that, you need the density of copper. Multiply the Density by the Volume, and you'll have your solution :)

4. teeny

okay thanks :). this was physics and i was wondering how to actually get the answer, when i was only given one piece of information. So thanks :)

5. Khushal_Shah

Anytime @teeny :)

6. Syamantak

$Mass of Cube = \left( side of cube \right)^{3}$ So the Mass of the copper cube would be (15*15*15)cm = 3,375 $cm ^{3}$

7. UnkleRhaukus

$\rho=\frac mV$$\rho_{\text{Cu}}=8.96 \left[\text {g·cm}^{−3}\right]$

format: plain


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# Left Endpoint Approximation Calculator for a Function

An online calculator for approximating the definite integral using the left endpoints (the left Riemann sum), with steps shown.

Related calculator: Left Endpoint Approximation Calculator for a Table

If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.

Approximate the integral $\int\limits_{0}^{4} \sqrt{\cos^{4}{\left(x \right)} + 2}\, dx$ with $n = 5$ using the left endpoint approximation.

## Solution

The left Riemann sum (also known as the left endpoint approximation) uses the left endpoints of a subinterval:

$\int\limits_{a}^{b} f{\left(x \right)}\, dx\approx \Delta x \left(f{\left(x_{0} \right)} + f{\left(x_{1} \right)} + f{\left(x_{2} \right)}+\dots+f{\left(x_{n-2} \right)} + f{\left(x_{n-1} \right)}\right)$

where $\Delta x = \frac{b - a}{n}$.

We have that $f{\left(x \right)} = \sqrt{\cos^{4}{\left(x \right)} + 2}$, $a = 0$, $b = 4$, and $n = 5$.

Therefore, $\Delta x = \frac{4 - 0}{5} = \frac{4}{5}$.

Divide the interval $\left[0, 4\right]$ into $n = 5$ subintervals of the length $\Delta x = \frac{4}{5}$ with the following endpoints: $a = 0$, $\frac{4}{5}$, $\frac{8}{5}$, $\frac{12}{5}$, $\frac{16}{5}$, $4 = b$.

Now, just evaluate the function at the left endpoints of the subintervals.

$f{\left(x_{0} \right)} = f{\left(0 \right)} = \sqrt{3}\approx 1.732050807568877$

$f{\left(x_{1} \right)} = f{\left(\frac{4}{5} \right)} = \sqrt{\cos^{4}{\left(\frac{4}{5} \right)} + 2}\approx 1.495196773630485$

$f{\left(x_{2} \right)} = f{\left(\frac{8}{5} \right)} = \sqrt{\cos^{4}{\left(\frac{8}{5} \right)} + 2}\approx 1.414213819387789$

$f{\left(x_{3} \right)} = f{\left(\frac{12}{5} \right)} = \sqrt{\cos^{4}{\left(\frac{12}{5} \right)} + 2}\approx 1.515144715776502$

$f{\left(x_{4} \right)} = f{\left(\frac{16}{5} \right)} = \sqrt{\cos^{4}{\left(\frac{16}{5} \right)} + 2}\approx 1.730085700215823$

Finally, just sum up the above values and multiply by $\Delta x = \frac{4}{5}$: $\frac{4}{5} \left(1.732050807568877 + 1.495196773630485 + 1.414213819387789 + 1.515144715776502 + 1.730085700215823\right) = 6.309353453263581.$

$\int\limits_{0}^{4} \sqrt{\cos^{4}{\left(x \right)} + 2}\, dx\approx 6.309353453263581$A


text:
in

# How To Find Unit Rate With Fractions

Remember back to how to solve complex fraction: You can conclude that the per minute lap unit rate is 0.4/1.

Divide Unit Fractions and Whole Numbers Math charts

### For example, if 50km is covered in 5.5 hours, the unit rate will be 50km/5.5 hours = 9.09 km/hour.

How to find unit rate with fractions. = today we will be finding unit rates by using complex fractions just like these! Unit rate vocabulary ratios compare two or more types of. 120 problems in 5 hours, find the unit rate for:

In the last example, we calculated that bob was driving at a rate of $\frac{\text{175 miles}}{\text{3 hours}}$. To divide fractions, multiply the fraction in the numerator by the reciprocal of the fraction in the denominator. The calculator gives explanation of the working steps involved and simplifies the result using the greatest common denominator.

To master this lesson, you must understand how ratios and fractions will help to find the unit rate based off of a higher ratio of values. Did meredith simplify the complex fraction correctly? Explain how the fraction 5/6 was obtained.

Finding unit rates by simplifying complex fractions this video demonstrates two methods of find a unit rate: To find the turtle’s unit rate, meredith wrote and simplified the following complex fraction. 122 patients in 4 weeks unit rates complex fractions

In this lesson you will learn how to find unit rates in situations involving fractions by using division. The total laps goes in the numerator. Kylie can run 0.4 laps per minute (0.4 laps per 1 minute = 0.4/1).

A turtle walks 7/8 of a mile in 50 minutes. _____ examples for the following examples find the unit rate for each. Use this popular fractions calculator to add, subtract, multiply and divide fractions, including mixed number fractions.

In the main program, all problems are. Write the value that goes with each quantity in the numerator and denominator (e. 1/4 mi in numerator, 1/20 hr in denominator).

Then, find the unit rate by simplifying the fractions. Compute unit rates associated with ratios of fractions lesson 1 of. After 15 minutes, pool a has leaked 2/3 gallon.

We saw this yesterday when working with unit fractions. Unit rate on these printable worksheets, student will learn the concept of a unit rate, and they will practice solving word problems related to linear measurement, constant speed, and pricing. Find the unit rate for:

What is the unit rate expressed in miles per hour? This helped us to see that if the 2nd term of a rate is a unit fraction, when we scale the rate by the denominator of the 2nd term we find the unit rate (ie 1/4 * 4 = 4/4 or 1; For example, i find the unit rate when you walk 1/2 mile in 1/4 hour.

Mrs math geek 7th grade math anchor charts this lady is so. Which pool is leaking faster ? Then, i solve a word problem using the unit rate.

49 points in 7 games, find the unit rate for: To find the unit rate of change, simply divide the total amount of change in a quantity by the amount of time it took to achieve that change. We can use unit rates to simplify rates and ratios that appear complicated, such as those containing fractions in both the numerator and denominator.

Determine the unit rate in each problem. To calculate a unit rate, you want to figure out how much of one item exists for every 1 unit of a second item. Create your free account teacher student.

To make this point, and to tie the concept to a variety of fractions other than unit. Unit rateword problemswith complex fractions i can compute unit rates as complex fractions. So the fraction is 12/30.

These math activities align closely with the following common core standards: We want to know the laps per minute unit so we set up a ratio with minutes in the denominator. 1/5 * 5 equals 5/5 or 1).

Some of the worksheets displayed are unit rates with fractions, ratios rates unit rates, lesson 11 ratios of fractions and their unit rates, grades mmaise salt lake city, p 7 unit rates, unit rate, ratio and rates, gr6 8 savvy shopping unit pricing work student version. If given miles driven in an hour, you can determine. Unit rates with complex fractions notes and practice 7 rp 1.

12 ÷ 30 = 0.4. We usually want the rate to reflect the number of miles in one hour. How to find the unit rate including with fractions pre algebra.

So, to find unit rate, divide the denominator with the numerator in a way that the denominator becomes 1. First, rewrite your data as a division problem, where the numerator is the amount you’re trying to calculate and the denominator is the unit. References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program.

After 20 minutes, pool b has leaked 3/4 gallon. Answers (1) jerry greer 12 february, 23:37. \$112 in 8 hours, find the unit rate for:

This is correct, but not very useful. This tells us that every three hours, bob will travel $175$ miles. Think back what is a unit rate?

Some of the worksheets for this concept are unit rates with fractions, ratios rates unit rates, lesson 11 ratios of fractions and their unit rates, grades mmaise salt lake city, p 7 unit rates, unit rate, ratio and rates, gr6 8 savvy shopping unit pricing work student version. Create a new teacher account for learnzillion. In this practice worksheet, learners are presented with word problems that contain two different quantities, and they will need to find the unit rate and write it as a proper fraction.

How do you find unit rate of two fractions? This skill directly address ccss.math.content.7.rp.a.1 compute unit rates associated with ratios of fractions, including ratios of lengths, areas and other quantities measured in like or different units.

3rd Grade Fractions Unit Equivalent Fractions

Fraction Unit Equivalent Fractions Worksheet Fractions

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Divide Unit Fractions and Whole Numbers Fractions

4.md.5 anchor chart. Unit fraction angle measures Using

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Pin by Cassie Kennedy on 7th Grade Compacted Math Common

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Multiplying Fractions Unit (5th Grade CC Aligned) Math

Fractions Unit for Grade 2 (Ontario Curriculum) Ontario


text:
# Is L={<M>|M is a TM and L(M) is uncountable} decidable?

Is $L=\{\langle M\rangle\mid \text{$M$is a Turing machine and$L(M)$is uncountable}\}$ decidable?

My intuition is that it is not, but I'm not sure if Rice's Theorem applies in this case. If it is not decidable, how can I prove that using reducibility?


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# How to draw small sprite bigger?

I have created an image with the resolution of 25px(width)*34px(height) (I know it's unusual, however it doesn't bother me now). The problem is that when I try to display it by creating a bitmap, it becomes really really small on my Galaxy S4. It should look a little pixelated, kinda like a flappy bird sprite. I checked those sprites as well and they are even smaller in comparison to mine however it is displayed much bigger than my sprite on the screen. I want to know if I can make it look larger without scaling it. Is there a way to make my sprite look bigger?

@Override
protected void onDraw(Canvas canvas) {
Log.i(TAG, "onDraw called");
BitmapFactory.Options op = new BitmapFactory.Options();
op.inPreferredConfig = Bitmap.Config.ALPHA_8;
bm.setDensity(DisplayMetrics.DENSITY_DEFAULT);
bm = Bitmap.createBitmap(BitmapFactory.decodeResource(getResources(),R.drawable.giraffe_default,op));
canvas.drawBitmap(bm,getWidth()/2,getHeight()/2,null);  //get from res, set coords to 10,10
}

This is my onDraw method where I draw my canvas and bitmap. As I said, my image appears to be so small, I can barely see it.

Draw the bitmap using a Matrix.

Matrix matrix;
matrix.reset();
matrix.postScale(2.0f, 2.0f);
matrix.postTranslate(getWidth()/2.0f, getHeight()/2.0f);
canvas.drawBitmap(bm, matrix, null);

EDIT: you probably should tell it where to draw too. My bad.

EDIT2: In response to you blurry comment, you can try turning off anti-aliasing, and bitmap filtering.

Paint paint = new Paint();
paint.setAntiAlias(false);
paint.setFilterBitmap(false);

canvas.drawBitmap(bm, matrix, paint);

If that doesn't give you the results you are after you can create the larger image by using blocks of color (i.e. 3x3, or 10x10) to represent each pixel of the original image. Also, consider searching this site, there have been similar or related questions already asked and answered.

Angry Birds Scaling

• I added that snippet in my onDraw, however everytime i enter this gamesurface, the app crashes – user3624245 Jan 7 '15 at 21:38
• My logcat tell me it has something to do with the thread. public void run(){ Canvas canvas; long frames = 0; while(isRunning){ canvas = null; frames++; try{ canvas = this.sHolder.lockCanvas(); synchronized(sHolder){ this.gamesurface.draw(canvas); } }finally{ if(canvas != null){ sHolder.unlockCanvasAndPost(canvas); } } } Log.d(TAG,"Gameloop executed: "+frames+" times"); } } – user3624245 Jan 7 '15 at 21:50
• I now added matrix.postTranslate(getWidth()/2.0f, getHeight()/2.0f); but it still crashes :/ – user3624245 Jan 8 '15 at 21:30
• Try using the debugger and see if you can find exactly what line it is puking on. – Aholio Jan 9 '15 at 0:45
• Alright so i got it working. The problem is that it's really blurry, as if it was zoomed in. I can't figure how to display my image larger without increasing the resoultion since i want it to look pixelated just like in flappy bird for instance. – user3624245 Jan 9 '15 at 1:12


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# Change the location of macros

Hello, I would like to change the path to the macros at LibreOffice.

The macros are currently in the path .. C: \ Users \ [USER] \ AppData \ Roaming \ LibreOffice \ 4 \ user \ basic ..saved.

I would like to have the macros permanently saved elsewhere. I know that I can export the modules and macros, but that doesn't change the storage location.

Thank you very much

edit retag close merge delete

( 2020-02-04 09:43:49 +0100 )edit

Sort by » oldest newest most voted

The macro libraries are stored in the user profile. There is a way to create and use a profile stored in a different place: see -env command line switch.

The default profile location is defined in bootstrap.ini file located in your LibreOffice installation directory (usually C:\Program Files\LibreOffice\program). The value there is \$SYSUSERCONFIG/LibreOffice/4, which means the profile created and used by default (i.e., when LO is started using normal shortcuts without any switches) is in user's AppData\Roaming. I don't recommend to modify the setting in bootstrap.ini, since that will be reverted on upgrade anyway.

I don't think there's a way to move libraries path independently from the whole profile.

more


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CCO '03 P3 - Cube

View as PDF

Points: 20 (partial)
Time limit: 1.0s
Memory limit: 16M

Problem type
2003 Canadian Computing Competition, Stage 2

Imagine a cube formed from solid interlocking pieces of various shapes. If the pieces are sufficiently entwined, the only way to separate them would be to cut some of them. We can ask the question: "is the cube stable?" That is, is it physically impossible to separate the cube into or more fragments without deforming or cutting any individual piece?

Your program must answer this question for a variety of such cubes.

The pieces that make up a cube will be specified as follows: divide the cube into a grid of miniature cubes, each labeled by a capital letter. Two adjacent (face-sharing) minicubes are joined together if and only if they are labeled by the same letter. For instance, the first example cube given consists of solid pieces.

Input Specification

Your program will be given the specification of up to 10 different cubes. The first two lines of each specification will consist of the size of that cube, , and a blank line. The remaining lines will specify the horizontal layers of the cube from bottom to top. Each layer specification consists of an square showing the labels for each minicube on that layer, followed by a blank line. There will be no spaces in the input. The input will be terminated by the number on a line by itself.

Output Specification

For each cube given, in the order specified, print Yes if that cube is stable, and No if it is not.

Sample Input

2

AB
AB

BB
BA

3

AAA
BBB
AAA

AAA
ABA
AAA

ABA
ABA
ABA

0

Sample Output

No
Yes


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G13 Chapter Contents
G13 Chapter Introduction
NAG Library Manual

# NAG Library Routine DocumentG13AAF

Note:  before using this routine, please read the Users' Note for your implementation to check the interpretation of bold italicised terms and other implementation-dependent details.

## 1  Purpose

G13AAF carries out non-seasonal and seasonal differencing on a time series. Information which allows the original series to be reconstituted from the differenced series is also produced. This information is required in time series forecasting.

## 2  Specification

 SUBROUTINE G13AAF ( X, NX, ND, NDS, NS, XD, NXD, IFAIL)
 INTEGER NX, ND, NDS, NS, NXD, IFAIL REAL (KIND=nag_wp) X(NX), XD(NX)

## 3  Description

Let ${\nabla }^{d}{\nabla }_{s}^{D}{x}_{i}$ be the $i$th value of a time series ${x}_{i}$, for $i=1,2,\dots ,n$ after non-seasonal differencing of order $d$ and seasonal differencing of order $D$ (with period or seasonality $s$). In general,
 ${\nabla }^{d}{\nabla }_{s}^{D}{x}_{i}$ $=$ ${\nabla }^{d-1}{\nabla }_{s}^{D}{x}_{i+1}-{\nabla }^{d-1}{\nabla }_{s}^{D}{x}_{i}$ $d>0$ ${\nabla }^{d}{\nabla }_{s}^{D}{x}_{i}$ $=$ ${\nabla }^{d}{\nabla }_{s}^{D-1}{x}_{i+s}-{\nabla }^{d}{\nabla }_{s}^{D-1}{x}_{i}$ $D>0$
Non-seasonal differencing up to the required order $d$ is obtained using
 ${\nabla }^{1}{x}_{i}$ $=$ ${x}_{i+1}-{x}_{i}$ for $i=1,2,\dots ,\left(n-1\right)$ ${\nabla }^{2}{x}_{i}$ $=$ ${\nabla }^{1}{x}_{i+1}-{\nabla }^{1}{x}_{i}$ for $i=1,2,\dots ,\left(n-2\right)$ $\text{ }⋮$ ${\nabla }^{d}{x}_{i}$ $=$ ${\nabla }^{d-1}{x}_{i+1}-{\nabla }^{d-1}{x}_{i}$ for $i=1,2,\dots ,\left(n-d\right)$
Seasonal differencing up to the required order $D$ is then obtained using
 ${\nabla }^{d}{\nabla }_{s}^{1}{x}_{i}$ $=$ ${\nabla }^{d}{x}_{i+s}-{\nabla }^{d}{x}_{i}$ for $i=1,2,\dots ,\left(n-d-s\right)$ ${\nabla }^{d}{\nabla }_{s}^{2}{x}_{i}$ $=$ ${\nabla }^{d}{\nabla }_{s}^{1}{x}_{i+s}-{\nabla }^{d}{\nabla }_{s}^{1}{x}_{i}$ for $i=1,2,\dots ,\left(n-d-2s\right)$ $\text{ }⋮$ ${\nabla }^{d}{\nabla }_{s}^{D}{x}_{i}$ $=$ ${\nabla }^{d}{\nabla }_{s}^{D-1}{x}_{i+s}-{\nabla }^{d}{\nabla }_{s}^{D-1}{x}_{i}$ for $i=1,2,\dots ,\left(n-d-D×s\right)$
Mathematically, the sequence in which the differencing operations are performed does not affect the final resulting series of $m=n-d-D×s$ values.

None.

## 5  Parameters

1:     X(NX) – REAL (KIND=nag_wp) arrayInput
On entry: the undifferenced time series, ${x}_{\mathit{i}}$, for $\mathit{i}=1,2,\dots ,n$.
2:     NX – INTEGERInput
On entry: $n$, the number of values in the undifferenced time series.
Constraint: ${\mathbf{NX}}>{\mathbf{ND}}+\left({\mathbf{NDS}}×{\mathbf{NS}}\right)$.
3:     ND – INTEGERInput
On entry: $d$, the order of non-seasonal differencing.
Constraint: ${\mathbf{ND}}\ge 0$.
4:     NDS – INTEGERInput
On entry: $D$, the order of seasonal differencing.
Constraint: ${\mathbf{NDS}}\ge 0$.
5:     NS – INTEGERInput
On entry: $s$, the seasonality.
Constraints:
• if ${\mathbf{NDS}}>0$, ${\mathbf{NS}}>0$;
• if ${\mathbf{NDS}}=0$, ${\mathbf{NS}}\ge 0$.
6:     XD(NX) – REAL (KIND=nag_wp) arrayOutput
On exit: the differenced values in elements $1$ to ${\mathbf{NXD}}$, and reconstitution data in the remainder of the array.
7:     NXD – INTEGEROutput
On exit: the number of differenced values in the array XD.
8:     IFAIL – INTEGERInput/Output
On entry: IFAIL must be set to $0$, $-1\text{ or }1$. If you are unfamiliar with this parameter you should refer to Section 3.3 in the Essential Introduction for details.
For environments where it might be inappropriate to halt program execution when an error is detected, the value $-1\text{ or }1$ is recommended. If the output of error messages is undesirable, then the value $1$ is recommended. Otherwise, if you are not familiar with this parameter, the recommended value is $0$. When the value $-\mathbf{1}\text{ or }\mathbf{1}$ is used it is essential to test the value of IFAIL on exit.
On exit: ${\mathbf{IFAIL}}={\mathbf{0}}$ unless the routine detects an error or a warning has been flagged (see Section 6).

## 6  Error Indicators and Warnings

If on entry ${\mathbf{IFAIL}}={\mathbf{0}}$ or $-{\mathbf{1}}$, explanatory error messages are output on the current error message unit (as defined by X04AAF).
Errors or warnings detected by the routine:
${\mathbf{IFAIL}}=1$
 On entry, ${\mathbf{ND}}<0$, or ${\mathbf{NDS}}<0$, or ${\mathbf{NS}}<0$, or ${\mathbf{NS}}=0$ when ${\mathbf{NDS}}>0$.
${\mathbf{IFAIL}}=2$
 On entry, ${\mathbf{NX}}\le {\mathbf{ND}}+\left({\mathbf{NDS}}×{\mathbf{NS}}\right)$.

## 7  Accuracy

The computations are believed to be stable.

The time taken by G13AAF is approximately proportional to $\left({\mathbf{ND}}+{\mathbf{NDS}}\right)×{\mathbf{NX}}$.

## 9  Example

This example reads in a set of data consisting of $20$ observations from a time series. Non-seasonal differencing of order $2$ and seasonal differencing of order $1$ (with seasonality of $4$) are applied to the input data, giving an output array holding $14$ differenced values and $6$ values which can be used to reconstitute the output array.

### 9.1  Program Text

Program Text (g13aafe.f90)

### 9.2  Program Data

Program Data (g13aafe.d)

### 9.3  Program Results

Program Results (g13aafe.r)


text:
# They're watching your every move

There are lots of fun ways to play with your phone: Angry Birds, Snapchat, Facebook, Face-swapping. Maybe you like to play with sensors? Anyone? Anyone? Well, in this post I will share some Python code and a video of how you can stream your Android's accelerometer data to your laptop and then visualize it. In fact, you can do this with any of the sensors in the unit, the accelerometer, gyroscope or the magnetometer.

An accelerometer is a sensor that every phone has. It measures the acceleration felt along three orthogonal axes. The left-right axis, the back-and-forth axis and the up-down axis. This forms a basis with which we can describe any acceleration felt in a three-dimensional space. Since gravity is always present, an accelerometer at rest will only measure the acceleration due to gravity. The figure below shows the grid present on my Android phone, and I assume an iPhone as well.

In this configuration the readings of the accelerometer will be

X = 0.00, Z = 0.00, Y = 9.81

This is because gravity is pulling down on the y-axis with an acceleration of -9.81 m/s$^2$, so, in order to be still, the positive y-axis reading must be equal of strength and opposite of sign.

### Applications

Most of use accelerometers every day and we don't know it. An accelerometer is the sensor in a phone that tells the display when to switch orientations from normal to landscape.

Accelerometers are used in internal navigation of vehicles and airplanes. They help with autopilot of airplanes and determining the position of cars. They are also used to determine when to deploy an airbag by sensing a dramatic change in acceleration. In some washing machines they are used to detect load imbalances and are used to prevent your load from spinning out of control.

There are many applications in medicine.  Recently, I came across a paper where the authors used an accelerometer in a catheter to try and determine urological diagnosis and treatment. The ideas was to "see" how the urethra is moving when there is activity. Be warned:

In fact, your phone, and it's sensors may be quite insidious. Using your phone and it's sensors, it is possible to recreate your body's physical movements while you drive. Texting and driving? They can also be used to see through walls and potentially tell where you are in your own home. Here is a story where an accelerometer is used to deduce your ATM pin code. Pretty amazing!

### Real-Time streaming of a phone's sensor

Using the android app Sensorstream, an app that streams your phone's data over WiFi, Python can be used to capture the data and visualize it. Python's socket package captures the data while matplotlib and numpy help us visualize it.

 ### make connection with phone
host = ''
port = 5555

s.bind((host, port))

On the app use the port settings need to match and the IP address needs to be the IP address your machine is logged onto. If you only want the Stream and no logs check the "UDP Stream." Once you switch it on the socket s will be capturing the data.

Once we capture the data we simply need to visualize the data. Python's Matplotlib is maybe not the best solution for this but we make it work 😉

The full code can be found at my GitHub page: https://github.com/theholymath/acceleromter-filter-testing/blob/master/sensorstream_live_stream_acceleration.py.

The relevant code to stream with is this:

while 1:
messageString = message.decode("utf-8")
Acc = messageString.split(',')[2:5]
Acc = [float(Acc[i])/10.0 for i in range(3)]

Matplotlib's ability to plot fast streaming data is limited unless there are a few tweaks. Most of what I did came from this source. The code I have posted displays three graphs: the raw accelerometer data, the filtered accelerometer data, and the tilt angle. animated_accelerometer_phone_tilt_angles

Feel free to download and improve the code. Accelerometers are fun to play with and this is a great toy problem to learn sensor fusion and digital filtering. Enjoy!

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text:
# Are my questions really this bad?

I just noticed this "recommended" badge in my profile and I'm rather confused.

I have $$20$$ out of the $$5$$ needed questions for this badge, but apparently they are "poor" questions so they don't actually count towards the count? Kinda bummed to see that, I've been quite diligent and working hard to make my questions as best I can.

• FYI, see Asking days badges, and the Meta FAQ answer. Also, somewhat related is Misleading description of new Curious badge. Feb 18 at 19:54
• How many questions have you asked and later deleted? Feb 21 at 18:18
• I have deleted none of my questions Feb 21 at 18:50
• (How is a "positive question record calculated?")[meta.stackexchange.com/questions/290545/… Do you perhaps have a few old downvoted, closed, and deleted questions? It would only take 5 to give you a negative question record with 18 positive questions and 2 closed ones. Edit: I see you haven't deleted any questions. Its worth noting that unanswered downvoted questions will eventually be deleted automatically. Maybe this is the cause? I don't have enough reputation to check. Barring these scenarios this is baffling. Feb 22 at 16:24

## 1 Answer

I am confused as well. I read the criteria described in Asking days badges, and as far as I can tell, Clyde has met the criteria.

Here are seven questions asked by Clyde. For each question, Clyde asked no other questions that day. Each is well-received, because each has a positive score, and is neither deleted nor closed. Finally, Clyde's question record should be impeccable, since they have asked $$38$$ questions, and have no negative questions, and only two closed questions.

Since seven > five, shouldn't this qualify?

Prove $f(r, \theta) = (\cos\theta, \sin\theta)$ is continuous

Is $C([0,1])$ complete if $\int_{0}^{1} f(x)g(x)dx$ is the inner product?

If $(f_n)$ converges pointwise to $f$, then $f$ is uniformly continuous

Find intersection points of $x^2 - 3xy+ 2y^2 - x + 1 = 0$ and $y = \alpha x + \beta$

What is the difference between an $FG$-module and a group algebra?

In $\triangle ABC$, $DE$ parallel to $BC$, $F$ midpoint of $DE$, $AF$ meets $BC$ at $G$. Prove $G$ is the midpoint of $BC$

• Thanks for looking into this Mike, I'm glad that someone else understands why this is confusing. Feb 19 at 23:38
• <Joke> I think you could use MathJAX when Comparing 5 & 7 . . . </Joke>
– Prem
Feb 22 at 10:16


text:
# A reflection group is Euclidean if and only if it has an invariant of degree 2

Let $$G \subset GL(V)$$ be a FINITE IRREDUCIBLE complex reflection group acting on its natural representation $$V$$. Then $$G$$ acts on the dual representation $$V^*$$ contragrediently which induces an action on the polynomial ring $$k[V]$$, the subalgebra, $$G$$-invariant polynomials are finitely-generated. The claim is that the natural representation of $$G$$ can be realized over the real, that is $$G$$ is a Euclidean reflection group if and only if it has an invariant of degree 2.

Now suppose $$G$$ is a Euclidean reflection group, then it leaves the norm polynomial invariant, since it is a subgroup of the orthogonal group.

The converse have been a headache for me, what I'm thinking is that to derive a invariant quadratic form from this invariant of degree 2, and somehow to show that the representation of this group can be define over the real numbers.

Can someone help me with this problem please? Thanks

A Hint [From 'Unitary Reflection Group' by Lehrer and Taylor], use the invariant of degree $$2$$ to construct a non-degenerate $$G$$-invariant quadratic form. From the quadratic form and the hermitian inner product to construct a semi-linear map $$f:V \to V$$ and prove that there is a positive real number $$\lambda$$ such that $$f^2(v) = \lambda v$$ for all $$v\in V$$. Show that the set $$V_0$$ of fixed points of a suitable multiple of $$f$$ is a real vector space such that $$V = \mathbb{C} \otimes V_0$$.

• Are you assuming $G$ is finite? – user10354138 Aug 1 at 12:40
• @user10354138 yes I'm assuming $G$ is finite – kkkkk Aug 1 at 12:51
• There seems to be some missing irreducibility condition here. Take any complex reflection group on $V$ and embed it as a complex reflection group on $W=\mathbb{C}\oplus V$ by doing nothing to the $\mathbb{C}$, and we would have an invariant of degree 2. – user10354138 Aug 1 at 14:20
• @user10354138 Yes you are right, we require the reflection group to be irreducible. – kkkkk Aug 1 at 16:51
• So can you get the quadratic form from the invariant? That's a very good start. – David A. Craven Aug 1 at 20:03


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# Multivariate Taylor Series Derivation (2D)

I am trying to understand the derivation here

kiwi.atmos.colostate.edu/group/dave/pdf/TaylorSeries.pdf

I understand how first, second total differentials are derived. I do not understand how they are plugged into a form that is compatible with the single variable Taylor Series, which is

$f(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a) (x-a)^2 + ...$

The coefficients of multivariate expansion are exactly the same as the single variable version. Sure given the total differential, $(x-a)$ is substituted for $dx$, $(y-b)$ for $dy$, but it's not shown why, or how the rest of $f(x,y)$ expansion should mirror the single variable form based on the total differential. Expansion for $f(x,y)$ is below

$f(x,y) = f(a,b) + \bigg[(x-a)\frac{\partial}{\partial x} + (y-b)\frac{\partial}{\partial x}\bigg]f + \frac{1}{2!} \bigg[(x-a)\frac{\partial}{\partial x} + (y-b)\frac{\partial}{\partial x}\bigg]^2 f + ...$

If you know a better 2D Taylor Series derivation, that would be welcome as well.

Thanks,

-
This is not a rigorous proof: Suppose that one can write $f(x,y)=f(a,b)+A(x-a)+B(y-b)+C(x-a)^2+D(x-a)(y-b)+E(y-b)^2+\cdots$, how can the values $A, B, C, D, E, \dots$ be determined? –  pharmine Sep 27 '11 at 11:08
I think I came across a similar idea here physics.miami.edu/~nearing/mathmethods/series.pdf pg 7. Thanks. –  BB_ML Sep 27 '11 at 11:26

So we want to express $f(a+u,b+v)$ in terms of $f(a,b)$, the increment $(u,v)$ and the partial derivatives of $f$ at $(a,b)$. To this end we look at the auxiliary function $$\phi(t):=f(a+t u, b+t v)\qquad(0\leq t\leq 1)\ .$$ By the one-variable Taylor theorem we have $$f(a+u,b+v)=\phi(1)=\sum_{r=0}^n{1\over r!}\phi^{(r)}(0)+{\rm error\ term}\ .\qquad(*)$$ Therefore we have to compute the successive derivatives of $\phi$ at $t=0$.

Obviously $$\phi^{(0)}(t)=\phi(t)=f(a+tu,b+tv)$$ and therefore $\phi^{(0)}(0)=f(a,b)$. To compute $\phi'(t)$ we need the chain rule: $$\phi'(t)={d\over dt}\phi(t)={d\over dt}f(a+t u,b+ t v)=f_1(a+t u,b+ tv)u+f_2(a+ tu, b+ tv) v\ ,$$ whence $\phi'(0)=f_1(a,b)u+f_2(a,b)v$. Proceeding in this manner we get \eqalign{\phi''(t)&={d\over dt}\bigl(f_1(a+t u,b+ tv)u+f_2(a+ tu, b+ tv) v\bigr) \cr &=f_{11}(a+tu,b+tv)u^2+2f_{12}(a+tu,b+tv) uv +f_{22}(a+tu,b+tv)v^2\cr} and therefore $\phi''(0)=f_{11}(a,b)u^2 + 2f_{12}(a,b)uv+f_{22}(a,b) v^2\ .$ And so on for the higher derivatives of $\phi$. The values $\phi(0)$, $\phi'(0)$, $\phi''(0)$, $\ldots$ so obtained have to be plugged into the equation $(*)$

Of course all of this could be written up more elegantly using vector notation.

-
Hi Christian, with $1$ in $f_1$ did you mean the first block of variables, which are $a+tu$? –  BB_ML Sep 28 '11 at 6:01
–  BB_ML Sep 28 '11 at 6:02
@user6786: $f$ is a function of two real inputs, whatever their name. By $f_1$ I mean the partial derivative with respect to the first input. It seemed strange to write $f_x$ when there is nowhere an $x$ in the argument. –  Christian Blatter Sep 29 '11 at 13:26

If you are looking for a physical intuition behind multivariate Taylor series, remember that $T_x = \partial_x$ is the generator of translations, as for suitable class of functions: $$(T_x f)(x) = \lim_{\epsilon \to 0} \frac{ f(x+\epsilon) - f(x)}{\epsilon}$$ Thus $\mathcal{T}_a = \mathrm{e}^{a T_x}$ is the operator of translation by $a$: $$(\mathcal{T}_a f)(x) = f(x+a)$$ Writing defining series of exponential give univariate Taylor expansion: $$f(x+a) = (1 + a T_x + \frac{a^2}{2!} T_x \circ T_x + \ldots ) \circ f(x)$$ Taking this to mulrivariate, the translation by vector $\vec{u}$ becomes $$\mathcal{T}_\vec{u} = \exp( \sum_{i=1}^n u_i T_{x_i} ) = \circ_{i=1}^n \mathcal{T}_{u_i}$$ You can now develop each univariate translation operator into a series, giving the multivariate Taylor expansion formula or, alternatively, you could expand the combined exponential:

$$f(\vec{x} + \vec{u}) = \sum_{n=0}^\infty \frac{1}{n!} \left( \langle \vec{u}, \mathcal{T}_{\vec{x}} \rangle \right)^n \circ f(\vec{x})$$

Hope this helps.

-


text:
# Projection onto the span of linearly independent vectors in Hilbert spaces

Let $H$ be a Hilbert space and $S\subseteq H$ a closed subspace. Moreover, let $\{s_{n}\}_{n=1}^{\infty}\subseteq S$ a complete and linear independent sequence in $S$, i.e.

• $S=\overline{\text{Span}\big(\{s_{n}\}_{n=1}^{\infty}\big)}$ and
• for $N\geq 1$ and $\lambda\in\mathbb{R}^{N}$, $\sum_{n=1}^{N}\lambda_{n}\,s_{n}=0$ implies $\lambda=0$.

Denote by $\mathcal{P}$ the orthogonal projection from $H$ onto $S$ and by $\mathcal{P}_{N}$ the orthogonal projection from $H$ onto $\text{Span}\big(\{s_{n}\}_{n=1}^{N}\big)$.

Is it true that, for all $x\in H$, $$\mathcal{P}_{N}(x) \quad\rightarrow\quad \mathcal{P}(x)$$ for $N\rightarrow\infty$ ?

This is used in a paper without proof or comment and I am wondering how to show this rigorously.

Any help or comment is highly appreciated! Thanks.

Apply Gram-Schmidt process on the sequence $\{s_n\}_{n=1}^\infty$ to obtain an orthonormal sequence $\{e_n\}_{n=1}^\infty$ in $S$ such that $\operatorname{span} \{e_1, \ldots, e_n\} = \operatorname{span} \{s_1, \ldots, s_n\}, \forall n \in \mathbb{N}$ and $\operatorname{span} \{e_n\}_{n=1}^\infty = \operatorname{span} \{s_n\}_{n=1}^\infty$.
Then clearly $\{e_n\}_{n=1}^\infty$ is an orthonormal basis for $S$ so
$$P_nx = \sum_{k=1}^n \langle x, e_k\rangle e_k \xrightarrow{n\to\infty} \sum_{k=1}^\infty \langle x, e_k\rangle e_k = Px$$
for all $x \in H$.


text:
# mars.tensor.special.gammasgn#

mars.tensor.special.gammasgn(x, **kwargs)[source]#

Sign of the gamma function.

It is defined as

$\begin{split}\text{gammasgn}(x) = \begin{cases} +1 & \Gamma(x) > 0 \\ -1 & \Gamma(x) < 0 \end{cases}\end{split}$

where $$\Gamma$$ is the gamma function; see gamma. This definition is complete since the gamma function is never zero; see the discussion after [dlmf].

Parameters

x (array_like) – Real argument

Returns

Sign of the gamma function

Return type

scalar or ndarray

Notes

The gamma function can be computed as gammasgn(x) * np.exp(gammaln(x)).

gamma

the gamma function

gammaln

log of the absolute value of the gamma function

loggamma

analytic continuation of the log of the gamma function

References

dlmf

NIST Digital Library of Mathematical Functions https://dlmf.nist.gov/5.2#E1


text:
Tumor-associated antibodies labeled with131 I and <tex-math>${}^{90}{\rm Y}$</tex-math> have been used in the treatment of malignant disease with some success. The use of α-particle-emitting radionuclides as radiolabels offers potential advantages over β-particle sources. The short range in tissue (<100 μm) and the high linear energy transfer associated with α-particle emitters will result in a more concentrated deposition of energy at the site of radionuclide decay. Thus, if radiolabeled antibodies can be bound to malignant cells specifically, a high differential cell killing can be achieved between the malignant and the normal cells. However, the energy deposition pattern will be strongly dependent upon the configuration of α-particle sources relative to the cells, and will consequently impact upon the dose-response characteristics. The purpose of this paper is to study distributions of energy deposition from α-particle-emitting radioimmunoconjugates distributed uniformly and nonuniformly around cells through theoretical modeling. Energy deposition spectra for cell nuclei are calculated and used to estimate the survival fraction by a simple biological model. We show that survival curves resulting from nonuniform distributions of α-particle-emitting radiolabeled antibodies can depart significantly from the classical exponential survival model applied to external α-particle beams. The survival curves may have initial slopes much steeper than those produced by a uniform distribution of sources, and they may also depart from linearity. Furthermore, the results of the modeling indicate how survival curves are dependent on the cell and radiolabel spacing. The results from our model compare reasonably well with published experimental data and can be used to facilitate the design and interpretation of radiobiological experiments.

This content is only available as a PDF.


text:
## Calculus 8th Edition

$R_s(1,2)=32, R_t(1,2)=-39$
We will use the Chain Rule to find $R_s(s,t)$ and $R_t(s,t)$ and then we will find it for $s=1$ and $t=2$: $$R_s(s,t)=\frac{\partial}{\partial s}(G(u(s,t),v(s,t)))=G_u(u(s,t),v(s,t))u_s(s,t)+G_v(u(s,t),v(s,t))v_s(s,t)$$ $$R_t(s,t)=\frac{\partial}{\partial t}(G(u(s,t),v(s,t)))=G_u(u(s,t),v(s,t))u_t(s,t)+G_v(u(s,t),v(s,t))v_t(s,t)$$ So we have: $$R_s(1,2)=G_u(u(1,2),v(1,2))u_s(1,2)+G_v(u(1,2),v(1,2))v_s(1,2)= G_u(5,7)\cdot4+G_v(5,7)\cdot2=9\cdot4-2\cdot2=32$$ $$R_t(1,2)=G_u(u(1,2),v(1,2))u_t(1,2)+G_v(u(1,2),v(1,2))v_t(1,2)= G_u(5,7)\cdot(-3)+G_v(5,7)\cdot6=9\cdot(-3)-2\cdot6=-39$$


text:
• anonymous
A( $$z_A$$ ) , B ( $$z_B$$ ) , C ( $$z_C$$ ) are vertices of right angled triangle, $$z_c$$ being the orthocentre. A circle is described on AC as diameter. Find the point of intersection of the circle with hypotenuse.
Mathematics

Looking for something else?

Not the answer you are looking for? Search for more explanations.


text:
# How do you solve (x^2 - 7)/(x - 4) = 0?

Apr 8, 2016

$x = \sqrt{7}$

#### Explanation:

color(blue)((x^2-7)/(x-4)=0

Multiply both sides by $x - 4$

$\rightarrow \frac{{x}^{2} - 7}{\cancel{x - 4}} \cdot \cancel{x - 4} = 0 \cdot x - 4$

$\rightarrow {x}^{2} - 7 = 0$

Add $7$ both sides

$\rightarrow {x}^{2} - 7 + 7 = 0 + 7$

$\rightarrow {x}^{2} = 7$

Take the square root of both sides

$\rightarrow \sqrt{{x}^{2}} = \sqrt{7}$

color(green)(rArrx=sqrt7


text:
# Do We Need Pythagoras Here?

Geometry Level 1

If lines $$l$$ and $$m$$ are parallel, then what is the value of angle $$x$$ (in degrees)?

×


text:
Comment
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Q)

# A black and a red dice are rolled. Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.

This is a multi part question answered separately on Clay6.com

Comment
A)
Toolbox:
• In any problem that involves tossing a coin or a die and determing the outcomes, we can write down the sample space and cout the number of favorable outcomes. Then we find P(E), P(F), P(E$\cap$F) using the set of outcomes.
• Given P(E), P(F), P(E $\cap$ F), P(E/F) $= \large \frac{P(E \;\cap \;F)}{P(F)}$
• Given P(E), P(F), P(E $\cap$ F), P(E $\cup$ F) = P(E) + P(F) - P(E $\cap$ F)
Given one black die and one red die are rolled. The sample space of equally likely events = 6 $\times$ 6 = 36.
Let E: set of events where the sum is greater than 9. E: (6,4), (4,6), (5,5), (5,6), (6,5), (6,6). The total possible outcomes = 6.
$\Rightarrow P(E) = \large \frac{\text{Number of favorable outcomes in E}}{\text{Total number of outcomes in S}} = \frac{6}{36} = \frac{1}{6}$
Let F be set of events where the black die rolled a 5. F: (1,5), (2,5), (3,5), (4,5), (5,5), (6,5). The total possible outcomes = 6.
$\Rightarrow P(F) = \large \frac{\text{Number of favorable outcomes in F}}{\text{Total number of outcomes in S}} = \frac{6}{36} = \frac{1}{6}$
For our set of events $E \cap F =(5,5), (6,6). Total number of outcomes = 2.$\Rightarrow P(E\;\cap\;F) = \large \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes in S}} = \frac{2}{36}$. Given P(E), P(F), P(E$\cap$F), P(E/F)$= \large \frac{P(E \;\cap \;F)}{P(F)}\Rightarrow P(E/F) = \Large\frac {\Large \frac{2}{36}}{\Large\frac{1}{6}}$=$\large\frac{1}{3}\$


text:
Do mirages really occur due to total internal reflection?

Mirages occur in areas such as deserts due to difference in the refractive indices of layers of air. As the light ray travels from denser to rarer layers of air, slowly bending, it approaches the critical angle and after refracting, it goes along with the boundry and the angle becomes 90 degrees. Now this light ray becomes the incident ray and if this ray undergoes total internal reflection, its angle of reflection would be 90degrees which means it would never bend, but in case of mirages, the ray bends in the upward direction. Does this mean that mirages dont occur due to total internal reflection?

$$`$$

An alternative explanation would be "index gradient reflection". Normally, TIR refers to reflection at an interface between a high index material and a low index material. "index gradient reflection" occurs without need for an interface: instead, it requires only a refractive index that changes gradually. The light rays never need to encounter an interface where the index changes abruptly.

A good example is a gradient index rod, in which the refractive index decreases from the axis outwards. In the case of a mirage, usually the air is warmer and less dense, and thus has lower refractive index, near the ground.

Yes, a situation comes, when the ray has bent so much that it's angle of incidence at new layer becomes greater than critical angle. Assuming ray comes from above, it gets refracted by rarer layers below, so its angle of incidence keeps on increasing; it becomes more horizontal. So a point comes when TIR occurs, and ray bounces upwards.

YES, reason for mirage is not Total internal Reflection.

You said correct at lowest most point when the ray bounce upward, angle of incident is 90° so if TIR is the reason angle of reflection is 90°, hence, the ray should continue move in straight line but we know this is not happen.

So lets consider refractive index of surrounding increases gradually upward as μ = μ0 + αy , (say, α is small) and assume origin at point of minima (convergence).

Now, using snell's law , μ sinθ = constant

(μ0 + αy)sinθ = μ0 {at bottom}

sinθ = μ0 / μ0 + αy

now , use dy/dx = tanθ

to find equation of trajectory,

taking α to be small, equation of trajectory is parabolic in nature whose minima is at origin , And that explain why rays is bouncing upward.

so its not TIR but equation of trajectory (snell's law) is the reason.


text:
# Test Power

## Priori power

The probability that the test will reject an incorrect H0, for a predefined effect size

While the significance level (α) represents the probability of rejecting a correct H0 , it doesn't mean that H0 is correct. It only indicates that the probability of rejecting a correct H0 will be bigger than α.
This a risk we are not willing to take. Usually α is 0.05 meaning there is a 5% chance of the above error (Type I Error).

We plan the study and determine the required priori test power before running the study.
The statistical power represents the probability of the test to reject an incorrect H0, but it depends on the H1 value. In some cases we don't know the H1 value as it just proposes that H0 is incorrect. In such cases we need to define the “expected effect” that the test is expected to discover. The smaller the expected effect, the weaker the statistical power. Let d be “absolute effect” needed to be discovered. .
Commonly, a priori power of 0.8 is being used.

statistical power = 1 - β .

β is the probability of a type II error, meaning the test won't reject an incorrect H0.

## Observed power (post-hoc power)

Be careful not to confuse the priori power with the observed power, they are different measurements.
The priori power is calculated before collecting data, while the observed power is calculated after collecting data.
The observed power represents the probability of the test to reject H0 based on the observed effect. It is directly correlated to the significance level (α). When accepting H0, the observed effect will be small and the observed power will be weak, on the contrary, when rejecting H0, the observed effect will be big and the observed power will be strong.

## z-test power

In this tool I prefer to treat the expect effect as a proportion from the H0 value

### Two tailed

H0: μ = μ0
H1: μ ≠ μ0

The test need to reject H0 when μ = μ1 = μ0 ± d
Since z is symmetrical it get the same result when calculating for μ0 + d or μ0 - d

#### Calculate reject area

$$p(z < Z_1) = \alpha/2 \:,\: p(z < Z_2) = 1-\alpha/2$$ H0 The test's accepted range for x is: [R1 , R2]. Reject H0 when x< R1 or x > R2

• $$p(z < \frac { R_1 - \mu_0} {\frac {\sigma }{\sqrt{n}}}) = \alpha/2 \:\rightarrow\: R_1= \mu_0 + Z_{\alpha/2}* \frac {\sigma }{\sqrt{n}}$$
• $$p(z < \frac { R_2 - \mu_0} {\frac {\sigma }{\sqrt{n}}}) = 1-\alpha/2 \:\rightarrow\: R_2= \mu_0 + Z_{1-\alpha/2}* \frac {\sigma }{\sqrt{n}}$$
• Now we can calculate the probability to reject H0 when we know the mean is μ1 instead of μ0, $$\overline{X}\:distribution\:N(\mu_1, \frac {\sigma }{\sqrt{n}})$$
• $$power = p(\overline{x} < R_1 |\mu_=\mu_1 ) + p(\overline{x} > R_2 |\mu_=\mu_1)$$
• $$power = p(z < \frac { R_1 - \mu_1} {\frac {\sigma }{\sqrt{n}}}) + 1 - p(z < \frac { R_2 - \mu_1} {\frac {\sigma }{\sqrt{n}}})$$

β - type II error
power

### Left tail

Since z is symmetrical, for the same d, left tail test has the same power as right tail test. H0: μ ≥ μ0
H1: μ < μ0

The test need to reject H0 when μ = μ1 = μ0 - d

#### Calculate reject area

$$p(z < Z_1) = \alpha$$ H0 The test's accepted range for x is: [R1 , ∞ ]. Reject H0 when x < R1

• $$p(z < \frac { R_1 - \mu_0} {\frac {\sigma }{\sqrt{n}}}) = \alpha \:\rightarrow\: R_1= \mu_0 + Z_{\alpha}* \frac {\sigma }{\sqrt{n}}$$
• Now we can calculate the probability to reject H0 when we know the mean is μ1 instead of μ0, $$\overline{X}\:distribution\:N(\mu_1, \frac {\sigma }{\sqrt{n}})$$
• $$power = p(\overline{x} < R_1 |\mu_=\mu_1 )$$
• $$power = p(z < \frac { R_1 - \mu_1} {\frac {\sigma }{\sqrt{n}}})$$

β - type II error
power

### Right tail

H0: μ ≤ μ0
H1: μ > μ0

The test need to reject H0 when μ = μ1 = μ0 + d

#### Calculate reject area

$$p(z < Z_2) = 1-\alpha$$ H0 The test's accepted range for x is: [-∞ , R2]. We reject H0 when x > R2

• $$p(z < \frac { R_2 - \mu_0} {\frac {\sigma }{\sqrt{n}}}) = 1-\alpha \:\rightarrow\: R_2= \mu_0 + Z_{1-\alpha}* \frac {\sigma }{\sqrt{n}}$$
• Now we can calculate the probability to reject H0 when we know the mean is μ1 instead of μ0, $$\overline{X}\:distribution\:N(\mu_1, \frac {\sigma }{\sqrt{n}})$$
• $$power = p(\overline{x} > R_2 |\mu_=\mu_1)$$
• $$power = 1 - p(z < \frac { R_2 - \mu_1} {\frac {\sigma }{\sqrt{n}}})$$

β - type II error
power

## Effect size

A measurement of the size of a statistical phenomenon, for example a mean difference, correlation, etc. there are different ways to measure the effect size.

### Effect size (unstandardized)

the pure effect as it, for example if needs to identify change of 1 mm in the size of mechanical part, the effect size is 1mm

### Standardizes effect size

Used when there is no clear cut definition regard the required effect, or the scale is arbitrary,or to compare between different researches with different scales
this site uses the Cohen's d as Standardizes effect size

Expected Cohen's d $$\:d=\frac{|\mu_1-\mu_0|}{\sigma}$$ Observed Cohen's d $$\:d=\frac{|\overline{x}-\mu_0|}{\sigma}$$ The Cohen's standardized effect was also named as following:

• 0.2 - small effect
• 0.5 - medium effect
• 0.8 - large effect

### Effect ratio

Used when there is no clear cut definition regard the required effect, and compare the effect size to the current expected value.
Calculate the effect size as a ratio of the expected value


text:
#### Volume 6, issue 5 (2006)

 Recent Issues
 The Journal About the Journal Editorial Board Editorial Interests Subscriptions Submission Guidelines Submission Page Policies for Authors Ethics Statement ISSN (electronic): 1472-2739 ISSN (print): 1472-2747 Author Index To Appear Other MSP Journals
Knots with unknotting number 1 and essential Conway spheres

### Cameron McA Gordon and John Luecke

Algebraic & Geometric Topology 6 (2006) 2051–2116
 arXiv: math.GT/0601265
##### Abstract

For a knot $K$ in ${S}^{3}$, let $T\left(K\right)$ be the characteristic toric sub-orbifold of the orbifold $\left({S}^{3},K\right)$ as defined by Bonahon–Siebenmann. If $K$ has unknotting number one, we show that an unknotting arc for $K$ can always be found which is disjoint from $T\left(K\right)$, unless either $K$ is an EM–knot (of Eudave-Muñoz) or $\left({S}^{3},K\right)$ contains an EM–tangle after cutting along $T\left(K\right)$. As a consequence, we describe exactly which large algebraic knots (ie, algebraic in the sense of Conway and containing an essential Conway sphere) have unknotting number one and give a practical procedure for deciding this (as well as determining an unknotting crossing). Among the knots up to 11 crossings in Conway’s table which are obviously large algebraic by virtue of their description in the Conway notation, we determine which have unknotting number one. Combined with the work of Ozsváth–Szabó, this determines the knots with 10 or fewer crossings that have unknotting number one. We show that an alternating, large algebraic knot with unknotting number one can always be unknotted in an alternating diagram.

As part of the above work, we determine the hyperbolic knots in a solid torus which admit a non-integral, toroidal Dehn surgery. Finally, we show that having unknotting number one is invariant under mutation.

##### Keywords
unknotting number 1, Conway spheres, algebraic knots, mutation
Primary: 57N10
Secondary: 57M25


text:
# Calculate the number of moles in 360 g of this gas sample. [Given, atomic masses of C = 12 u, O = 16 u, NA = 6.022 × 10^23 per mole]

93 views

edited

Calculate the number of moles in 360 g of this gas sample. [Given, atomic masses of C = 12 u, O = 16 u, NA = 6.022 × 1023 per mole]

+1 vote
by (58.0k points)

Given:

Mass of gas = 360g

Molar mass = 44g/mol

To calculate the number of moles, we apply the formula:

Number of moles = $\frac{Mass\,of\,CO_2}{Molar\,mass\,of\,CO_2}$

⇒ Number of moles $\frac{360g}{44g/mol}$

⇒ Number of moles = 8.1mol

Thus, the number of moles in 360g of the sample is 8.1


text:
Search
Search

Hi Guys,

Im using SAP XI version 7...im just a beginner using XI..

In runtime workbench...the comm channel for sender and receiver is correctly configured and started but in status its has an error.

This is the actual path of my comm channel.

*file sender path

f:\XI\Sender - <where f: is the local drive in xi server>

*file archive path

f:\XI\Sender\archive

if you double check it under Start-> Run

xidev\f\XI\Sender <hostname><local drive><shared folder>

Is there an issue regarding shared folder in XI Server? do i need to map drive with $sign of the folder like this? xidev\f$\xi\sender

##### Former Memberreplied

hi ,

i think there is no issue with your shared folder in xi.

if u r using NFS protocol than just replace "\" with the "/" in your comm channel source and target folder path .

it will work

regards,

navneet

0 View this answer in context
View more on this topic or


text:
# A charge of 8 C is passing through points A and B on a circuit. If the charge's electric potential changes from 64 J to 8 J, what is the voltage between points A and B?

Jan 26, 2016

Electric potential is $\left(\text{Energy")/("Charge}\right)$, and the potential difference is the change in potential.
$V = \frac{E}{Q}$
$\Delta V = \frac{\Delta E}{Q}$
$\Delta V = \frac{8 J - 64 J}{8 C}$
$\Delta V = - 7 V$


text:
# Hard disks and the liquid-solid transition in two dimensions_Questions

To edit the page, you have to become a member of the wiki by clicking on the Join link (top-right of the page), or by sending an email to one of us.
You can insert math formulæ using LaTeX code: e.g. typing
[[math]]
\pi_C(t+1)=\sum_{C'} \pi_{C'}(t) p_{C' \to C}
[[math]]
yields:
$\pi_C(t+1)=\sum_{C'} \pi_{C'}(t) p_{C' \to C}$

## Questions and Remarks:

• Q: in the first variant we should replace a=choice(L) - actually there are two line which read "a = choice(L)": one two choose a random disc to propose a move, and one to calculate the pair correlation. Which lines should be replaced? the first, the second or both?
• A: you have to replace both instances.

• Q: in the question about the two variants of the MC algorithm, the second variant includes both the replacements or only the second one?
• A: the second variant include only the replacement for b.

• Q: In the question 2 of the very first section ("why can we forget about the velocities?"), I'm not sure what the question is about: should we disscuss the fact that we are sampling equilibrium, distribution or compare the algorithms to molecular dynamics?
• A: The dynamics of hard spheres is described by their individual uniform motion punctuated by collisions. It occurs in phase space, whose coordinates consists of the set of their positions and velocities. As discussed in the lectures and in the class sessions, according to the ergodic hypothesis, their stationnary distribution is uniform in phase space (which includes positions and velocities). However, in the proposed programs, the sampling is performed uniquely on positions. We ask to justify why we can forget about the velocities.

• Q: In the questions about the algorithm A, what is exactly the property ? I read it as "the probability to find a disk at position (x,y) is uniform in space" but this is clearly false (as the histogram of the x-coordinate of the disks shows), so I do not understand the following questions.
• A: The hard spheres are distributed with the microcanonical distribution: their probability density in phase space is uniform. The property that we ask you to decide whether it is true of false indeed reads: "the probability to find the center of a disk at spatial position (x,y) is uniform". In question 2, you have to justify theoretically why you think this property is true or false. In the following questions (3 and 4), you have to test your choice numerically.
Besides, we cannot comment on the rest of your remark « but this is clearly false (as the histogram of the x-coordinate of the disks shows) », as it would partially answer to the question. However, we advice you to clearly identify and rethink of the premises and implications of your reasoning.

• Q: I think %1 is missing in the last question (second variant).
• A: Yes. The question is now reformulated in a clearer way.


text:
# Homework Help: Laplace Transforms

1. Jun 27, 2010

### soulk

1. The problem statement, all variables and given/known data

Okay, i'm pretty lost in my differential equations class and I'm not sure how to transform the functions

2. Relevant equations

f(t)= 1 / (s+2)^2
f(t)= 1 / (s+3)^3
f(t)= 1 / (s^2-3^2)

They seem pretty simple, but i have no clue if whether i'm suppose to use 2 formulas or separate them?

I'd look at the solutions in the back of my book, but they're even problems :(
Thanks, I'd appreciate any help!

Last edited: Jun 27, 2010
2. Jun 27, 2010

### Zryn

Are you familiar with Laplace transform tables or do you need to show calculations, as per:

$$\int e^{-st} f(t) dt$$

3. Jun 27, 2010

### soulk

Yes, i am familiar with the tables, but I'm not sure how to use them, lol.
I don't need the calculations, just the solution. Thanks again.

4. Jun 27, 2010

### Zryn

So looking at the tables, you have a list of time domain functions and a list of equivalent Laplace domain functions. It doesnt matter which you start with, you can work from either side back to the other.

Your question gives you Laplace domain functions and wants the time domain equivalent.

Can you transform $$\frac{1}{s}$$ and $$\frac{1}{s^{2}}$$ into their time domain equivalents?

5. Jun 27, 2010

### soulk

Yes I can, 1 and t

6. Jun 27, 2010

### Zryn

Yep, spot on.

How about $$\frac{2}{s^{3}}$$ and $$\frac{1}{s+2}$$ ?

7. Jun 27, 2010

### soulk

Hm, I'm not sure about 2/s^3 but 1/s+2 is e^-2t, right?

8. Jun 27, 2010

### Zryn

Ah, I see the problem then.

There are a billion and one different Laplace transform tables, some more comprehensive than others.

If for example, you took a look at:

http://math.fullerton.edu/mathews/c2003/laplacetransform/LaplaceTransformMod/Images/Table.12.2.jpg [Broken]

Check out transform #2 and #5.

* Yes $$\frac{1}{s+2}$$ becomes $$e^{-2t}$$ by the way, how about, $$\frac{2}{s^{3}}$$ ?

Last edited by a moderator: May 4, 2017
9. Jun 27, 2010

### soulk

^is it t^2?
Ah, I'm still not really sure, could you please show me the solution to
f(t)= 1 / (s+2)^2
That will probably make it much clearer to me. Thanks

10. Jun 27, 2010

### Zryn

Using #5 on that table:

$$\frac{n!}{(s-a)^{n+1}}$$ becomes $$t^{n}$$$$e^{at}$$

Therefore:

$$\frac{1!}{(s- -2)^{1+1}}$$ turns into ...

* Yes, $$\frac{2}{s^{3}}$$ turns into $$t^{2}$$

11. Jun 27, 2010

### soulk

te^-2t?

12. Jun 27, 2010

### soulk

So f(t)= 1 / (s+3)^3 would be te^-3t?

13. Jun 27, 2010

### Zryn

Bingo!

Now the second and third equation of yours require a small amount of manipulation. You may notice that the second equation is almost the same as the first, except you would need a 2 as the numerator for the transform to work.

When this happens, its just a matter of being tricksy.

Keep in mind that$$\frac{1}{2}$$*2 = 1. Make the numerator of your equation a 2 (multiply by 2) instead of a 1 so the transform works, then multiply the answer after you finish transforming by $$\frac{1}{2}$$ so that everything works out.

* You type faster than me, and you're fairly close but to get that answer you cant do the transform with the 1 as the numerator. Much like with 2/s^3 = t^2, you cant have 1/s^3 = t^2, you need to make (2*)1/s^3=(1/2*)t^2.

14. Jun 27, 2010

### soulk

I'm sorry, I'm not really sure of what you mean?

15. Jun 27, 2010

### Zryn

No problems!

When you do the intgral of 2x, you bring the 2 out the front of the integral sign, integrate x, then multiply the answer by 2, to get x^2. You can confirm this by then differentiating x^2 to get 2x.

When you do Laplace transforms, you can do the same thing.

If you have 2/s, you actually have 2* 1/s. 1/s Transforms from Laplace domain to time domain to 1, and then you multiply by 2 for a final answer of 2.

If you had 4/s^3 you actually have 2 * 2/s^3. You transformed 2/s^3 earlier to t^2, so you then multiply the transformed answer by 2 and the final answer is 2t^2.

If you have 1/(s+3)^3, you actually have 1/2 * 2/(s+3)^3. You need the 2 in the numerator to be able to do the transform, which then becomes t^2e^-3t, multiplying by the 1/2 gives a final answer of 1/2*t^2e^-3t.

Does that make more sense?

16. Jun 27, 2010

### soulk

Sort of, but I'm still kind of lost..
What is the solution to f(t)= 1 / (s+3)^3?
I'll understand if I see the solution.

17. Jun 27, 2010

### Zryn

Based on Line 6 in my previous post. Repetition (+Bold) for emphesis!

If you have 1/(s+3)^3, you actually have 1/2 * 2 * 1/(s+3)^3 (since 1/2 * 2 = 1, and you can just about always multiply something by 1 if youre in the mood to do so). You need the 2 in the numerator to be able to do the transform, and you take the 1/2 out the front of the transform (like you would take 2 out the front of the integral example I mentioned earlier) which then becomes t^2e^-3t, multiplying by the 1/2 gives a final answer of 1/2*t^2e^-3t.

* Theres only two problems with learning Laplace theory, the first is knowing how to use a transform table, and the second (and much harder) is being able to see what you need to do to an equation to get it into the form of one of the transforms on the transform table. If you can do both of those you will be able to solve all of your class problems however! Just takes a bit of practice.

Last edited: Jun 27, 2010
18. Jun 27, 2010

### soulk

So is it 2t^2e^-3t?

19. Jun 27, 2010

### Zryn

Not quite.

Starting on the time domain side of the table, what does te^-3t transform into? (using transform #5 on the laplace table). Suffer through my continuous questions if you can, theres a point and a goal we're heading towards ;)

20. Jun 27, 2010

1/(s+3)^2?


text:
# Vectors Problem

Printable View

• Dec 11th 2008, 02:34 PM
Ataxia
Vectors Problem
http://i87.photobucket.com/albums/k1...ectorsprob.gif

I'm not quite sure what I'm being asked to prove here. I presumably have to show that P satisfies some equation that shows it is a plane and not some other kind of surface but I don't know what that equation is.
• Dec 11th 2008, 03:55 PM
Skalkaz
Dear Ataxia,

If you use a vector formalism, the equation of the plane is same in 3 or 4 or other dimension.

Let n is a normal, so it is perpendicular to the plane, therefore their inner product (or scalar product) is null. So the equation is:
$\vec n \cdot \vec r = 0$


text:
# Following mentioned are the Data Sufficiency Questions for GMAT

Time Duration: 30 minutes.

Directions to solve: Each of the following given problems consist of a question and two statements which are marked as (1) and (2). Kindly use the data mentioned in (1) and (2) together with the other given information like the number of hours in a day, the definition of clockwise, mathematical facts, etc to figure out whether the statements are sufficient to answer the question or not.

All the numbers that have been used in this section are real numbers. A figure for a problem is basically given to provide information consistent with that in the question, not essentially with the additional information quoted in the statements.

1. How many degrees Celsius is 100° Fahrenheit? Degrees Celsius = 5/9 (degrees Fahrenheit) -32 Degrees Fahrenheit = 9/5 (degrees Celsius) + 32

1. if you can get the answer from (1) alone but not from (2) alone

2. if you can get the answer from (2) alone but not from (1) alone

3. if you can get the answer from (1) and (2) together, although neither statement by itself suffices

4. if statement (1) alone suffices and statement (2) alone suffices

5. if you cannot get the answer form statements (1) and (2) together, but need even more data.

2. What is the area of the shaded part of the circle? O is the center of the circle. The radius of the circle is 4. x is 60.

1. if you can get the answer from (1) alone but not from (2) alone

2. if you can get the answer from (2) alone but not from (1) alone

3. if you can get the answer from (1) and (2) together, although neither statement by itself suffices

4. if statement (1) alone suffices and statement (2) alone suffices

5. if you cannot get the answer form statements (1) and (2) together, but need even more data.

3. What was Mr. Kliman's income in 1970? His total income for the year 1968, 1969, and 1970 was $41, 000. He made 20% more in 1969 than he did in 1968. 1. if you can get the answer from (1) alone but not from (2) alone 2. if you can get the answer from (2) alone but not from (1) alone 3. if you can get the answer from (1) and (2) together, although neither statement by itself suffices 4. if statement (1) alone suffices and statement (2) alone suffices 5. if you cannot get the answer form statements (1) and (2) together, but need even more data. 4. If I and J are straight lines, Then find y. x = 100 & z = 80 1. if you can get the answer from (1) alone but not from (2) alone 2. if you can get the answer from (2) alone but not from (1) alone 3. if you can get the answer from (1) and (2) together, although neither statement by itself suffices 4. if statement (1) alone suffices and statement (2) alone suffices 5. if you cannot get the answer form statements (1) and (2) together, but need even more data. 5. Is X > Y (X + Y)? 2 > 0 X is positive 1. if you can get the answer from (1) alone but not from (2) alone 2. if you can get the answer from (2) alone but not from (1) alone 3. if you can get the answer from (1) and (2) together, although neither statement by itself suffices 4. if statement (1) alone suffices and statement (2) alone suffices 5. if you cannot get the answer form statements (1) and (2) together, but need even more data. 6. How long will it take to travel from A to B? It takes 4 hours to travel from A to B and back to A, It takes 25% more time to travel from A to B than it does to travel from B to A. C is midway between A and B, and it takes 2 hours to travel from A to C and back to A 1. if you can get the answer from (1) alone but not from (2) alone 2. if you can get the answer from (2) alone but not from (1) alone 3. if you can get the answer from (1) and (2) together, although neither statement by itself suffices 4. if statement (1) alone suffices and statement (2) alone suffices 5. if you cannot get the answer form statements (1) and (2) together, but need even more data. 7. Did ABC Company make a profit in 1980? ABC Company made a profit in 1979. ABC Company made a profit in 1981. 1. if you can get the answer from (1) alone but not from (2) alone 2. if you can get the answer from (2) alone but not from (1) alone 3. if you can get the answer from (1) and (2) together, although neither statement by itself suffices 4. if statement (1) alone suffices and statement (2) alone suffices 5. if you cannot get the answer form statements (1) and (2) together, but need even more data. 8. Did the price of a bushel of soybeans increase during every week of 1980? The price of a bushel of soybeans was$2 on January 1, 1980. The price of a bushel of soybeans was $4 on January 1, 1981. 1. if you can get the answer from (1) alone but not from (2) alone 2. if you can get the answer from (2) alone but not from (1) alone 3. if you can get the answer from (1) and (2) together, although neither statement by itself suffices 4. if statement (1) alone suffices and statement (2) alone suffices 5. if you cannot get the answer form statements (1) and (2) together, but need even more data. 9. Fifty students have signed up for at least one of the courses German I and English I, How many of the 50 students are taking German I but not English I? 16 students are taking German I and English I. The number of students taking English I but not German I is the same as the number taking German I but not English I. 1. if you can get the answer from (1) alone but not from (2) alone 2. if you can get the answer from (2) alone but not from (1) alone 3. if you can get the answer from (1) and (2) together, although neither statement by itself suffices 4. if statement (1) alone suffices and statement (2) alone suffices 5. if you cannot get the answer form statements (1) and (2) together, but need even more data. 10. The XYZ Corporation has 7, 000 employees. What is the average yearly wage of an employee of the XYZ Corporation? 4, 000 of the employees are executives. The total amount the company pays in wages each year is$77, 000, 000.

1. if you can get the answer from (1) alone but not from (2) alone

2. if you can get the answer from (2) alone but not from (1) alone

3. if you can get the answer from (1) and (2) together, although neither statement by itself suffices

4. if statement (1) alone suffices and statement (2) alone suffices

5. if you cannot get the answer form statements (1) and (2) together, but need even more data.


text:
Drawing a complete graph of an ego network

I have an ego network (a network of a person (ego) and his friends). It is drawn with the ego at the center surrounded by the friends. It might be possible that all the friends are connected to each other, in which case one can have a complete graph. Mathematica however draws a complete graph with no focal node. How can I make Mathematica draw a complete graph with a center node in it? How can I remove some connections so that it would still be an ego network but not a complete graph? Thanks.

-
Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the FAQs! 3) When you see good Q&A, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. ALSO, remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign –  Vitaliy Kaurov Apr 24 '13 at 0:43

---------- Update -------------

Now because of your comment I understand your problem better. Then it is basically a one-liner:

RandomGraph[{11, 25}, GraphLayout -> {"StarEmbedding", "Center" -> 2},
GraphStyle -> "SmallNetwork"]

---------- Older versions -------------

Start from {"BalloonEmbedding", "RootVertex" -> k} or {"RadialEmbedding", "RootVertex" -> k} where k is the root vertex (focus) - see this and this for the references:

g = CompleteGraph[12,
GraphLayout -> {"BalloonEmbedding", "RootVertex" -> 3},
GraphStyle -> "SmallNetwork"]

You got 55 edges in this case

el = EdgeList[g]; el // Length
55

Now remove some of them but preserve embeding:

Manipulate[Graph[EdgeList[g][[1 ;; m]], VertexCoordinates -> GraphEmbedding[g],
GraphStyle -> "SmallNetwork"], {m, 10, 55, 1}]

-
Could you please link to the doc page of BalloonEmbedding? :) –  belisarius Apr 23 '13 at 23:56
@belisarius Here. The graph related functionality leaves some to be desired both in documentation and quality though ... –  Szabolcs Apr 24 '13 at 0:06
@belisarius I put the link in ;) –  Vitaliy Kaurov Apr 24 '13 at 0:08
Pfff and I was doing everything manually. Well, you can never know all the functionality :). –  Jacob Akkerboom Apr 24 '13 at 0:37
Thanks Vitaly!You have the things I need...(In your manipulate graph though, my case would be 3 (instead of 1) would be connected to everyone else and its the connections among the friends that would be changing). Thanks a lot! –  newbie py Apr 24 '13 at 1:05

Ideally, it would be possible to fix the coordinates of one vertex, and let the layout algorithm take care of the rest. I couldn't figure out how to do this with Graph, but it is possible with GraphPlot:

GraphPlot[CompleteGraph[10], VertexCoordinateRules -> {1 -> {0, 0}}]

-
Thanks a lot Szabolcs! This gives me lots of options! –  newbie py Apr 24 '13 at 1:07
nFriends = 10;
coords = Prepend[#, {0, 0}] &[{Sin@#,
Cos@#} & /@ (Range[nFriends]/nFriends*2 Pi)];
CompleteGraph[nFriends+1, VertexCoordinates -> coords]

Will give you the complete graph with a vertex in the center.

func = Function[xxxx, UndirectedEdge[#, Last[xxxx]] & /@ Most[xxxx]];
connections = Array[Sequence @@ func@Range[0, #] &, nFriends];

Graph[Range[0, nFriends], connections, VertexCoordinates -> coords]

will give you the same thing, but now the connections are given in the variable connections so that you can now edit them. You can then do

connections = DeleteCases[connections,
UndirectedEdge[int1, int2]]
`

to delete the connection between friend int1 and friend int2 (which have to satisfy int1 < int2). Note that the ego is numbered 0 and he is at the center at position {0,0}.

-
Thanks Jacob for the feedback. –  newbie py Apr 24 '13 at 1:06


text:
Volume 380 - Particles and Nuclei International Conference 2021 (PANIC2021) - Standard model physics at the TeV scale
Double Higgs production at the LHC
L. Portales*  on behalf of the ATLAS and CMS Collaborations
Full text: pdf
Pre-published on: March 02, 2022
Published on: May 24, 2022
Abstract
An overview of the latest results on resonant and non-resonant Higgs boson pair production, using up to 140~fb$^{-1}$ of $\sqrt{s}=13$~TeV proton-proton collision data recorded during Run~2 of the LHC by the ATLAS and CMS Collaborations, is presented. A wide range of decay modes is covered, with one Higgs boson decaying to $b\bar{b}$, and the second to $b\bar{b}$, $W^+W^-$, $ZZ$, $\tau^+\tau^-$ or $\gamma\gamma$, as well as different topologies, with studies of both resolved and boosted signals and for both gluon-gluon and vector-boson fusion production modes. Analysis improvements, and newly introduced techniques with respect to previous studies, are highlighted.
DOI: https://doi.org/10.22323/1.380.0417
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


text:
# Cancel the leading terms of two polynomials

## Definition

• cancel_leads p q: the polynomial formed by multiplying p and q by monomials so that they have the same leading term, and then subtracting.

## Main Results

The degree of cancel_leads is less than that of the larger of the two polynomials being cancelled. Thus it is useful for induction or minimal-degree arguments.

def polynomial.cancel_leads {R : Type u_1} [comm_ring R] (p q : polynomial R) :

cancel_leads p q is formed by multiplying p and q by monomials so that they have the same leading term, and then subtracting.

Equations
@[simp]
theorem polynomial.neg_cancel_leads {R : Type u_1} [comm_ring R] {p q : polynomial R} :

theorem polynomial.dvd_cancel_leads_of_dvd_of_dvd {R : Type u_1} [comm_ring R] {p q r : polynomial R} (pq : p q) (pr : p r) :


text: Let $M$ be a reasonable commutative monoid written multiplicatively and $R$ a commutative ring. Then the natural numbers act on the monoid algebra $R[M]$, with $n$ acting by raising elements of $M$ to the $n-th$ power. Consider a functor $E$ from rings to spectra; then the natural numbers act on $E(R[M])/E(R)$ by the induced maps, which we call dilations. Gubeladze conjectured that if $R$ is regular and $E = K$ is algebraic $K-theory$, then this action is nilpotent. He proved his conjecture whenever $R$ contains the rational numbers. A different proof of this result was later given by the speaker jointly with Cortinas, Walker and Weibel and they adapted their proof to all regular $R$ containing any field. I will discuss this proof and outline how the approach can be used to prove the general case of the conjecture.


text:
### 4.2 Cauchy's theorem for microstructural contact actions

We assume that τ is a function of the point x and the normal n to $∂b$ at all points where the normal itself is uniquely defined and at every instant. In other words, we presume the validity of Cauchy's postulate for the microstructural contact actions. A question is then whether we can prove the Cauchy theorem for τ.

A proof is given in Capriz and Virga (1990), but there the microstructure is represented in a manifold embedded into a linear space, with undoubted advantages. Here we want to maintain the representation of the microstructure in a manifold as abstract as possible, avoiding ...

Get Advances in Applied Mechanics now with the O’Reilly learning platform.

O’Reilly members experience live online training, plus books, videos, and digital content from nearly 200 publishers.


text:
# The jump set of $SBV$ function for different value of parameter in image denoising problem

The classical Mumford-Shah image denoisng problem study the minimizer of the following functional, for each $\alpha>0$ where $\Omega\subset \mathbb R^2$ is open bounded with sommth boundary, $$u_\alpha:=\operatorname{argmin}_{u\in SBV(\Omega)}\left\{\int_\Omega\left|{u-u_0}\right|^2dx+\alpha\left(\int_\Omega\left|{\nabla u}\right|^2dx+\mathcal H^1(S_u)\right)\right\}$$ where $u_0\in SBV(\Omega)$ and $\nabla u$ denotes the absolute continuous part of $Du$ which is the weak derivative of $u$ in the sense of Radon measure. $S_u$ denote the jump set of $u$.

Clearly, for $\alpha=0$, $u_0=u_0$, and for $\alpha=+\infty$, $S_\infty$ would be empty.

My question: Would it be possible to prove that, if $\alpha_1<\alpha_2$,we have $$\mathcal H^1(S_{u_{\alpha_2}}\setminus S_{u_{\alpha_1}})=0 \tag 1$$ hold?

Moreover, do we even know that, for any $\alpha>0$, $$\mathcal H^1(S_{u_{\alpha}}\setminus S_{u_{0}})=0 \tag 2$$ hold?

Please advise. If you know some past study about a similar problem, please direct me there.

Thank you!

• I would look at "Singular Sets of Minimizers for the Mumford-Shah Functional" by Guy David. – Dirk Jun 27 '16 at 20:59
• @Dirk I know this book sir. I think it just discussd the properties of $S_{u_\alpha}$ a lot, but it does not discuss the relation between $S_{u_0}$ and $S_{u_\alpha}$... – JumpJump Jun 27 '16 at 21:05
• Ah, OK. It's been some time since I had this book in my hands. – Dirk Jun 27 '16 at 21:41
• You can also take a look at this paper: tuomov.iki.fi/mathematics/jumpset.pdf – Christian Clason Jun 28 '16 at 7:04
• @ChristianClason Thank you sir. This paper discusss my problem $(1)$ in $L^1$ setting, i.e., for ROF not MS, but it very useful. Thank you! Btw, do you have an idea that is there some paper disscuss about $(2)$? – JumpJump Jul 5 '16 at 14:30


text:
# NCERT Solutions for Class 6 Maths Decimals Chapter 7 Exercise 8.2 and 8.3

In this page we have NCERT Solutions for Class 6 Maths Decimals Chapter 7 Exercise 8.2 and 8.3. Hope you like them and do not forget to like , social share and comment at the end of the page.

## Exercise   8.2

Question 1

Complete the table with the help of these boxes and use decimals to write the number.

(a)

(b)

(c)

 Ones Tenths Hundredths Number (a) - - - - (b) - - - - (c) - - - -

First counting the boxes in each figure we get

a) There are two tenths and 6 hundredths. So, the decimal is 0.26

b)      One whole or ones, 3 tenths and 8 hundredths. Decimal is 1.38

c)      One whole or ones, two tenths and 8 hundredths. Decimal is 1.28

 Ones Tenths Hundredths Number (a) 0 2 6 .26 (b) 1 3 8 1.38 (c) 1 2 8 1.28

Question 2

Write the numbers given in the following place value table in decimal form.
 Hundreds 100 Tens 10 Ones 1 Tenths Hundredths Thousandths (a) 0 0 3 2 5 0 (b) 1 0 2 6 3 0 (c) 0 3 0 0 2 5 (d) 2 1 1 9 0 2 (e) 0 1 2 2 4 1

 a) $0+0+3+\frac {2}{10}+\frac {5}{100}+ \frac {0}{1000}=3+ .2+ .05=3.25$ b) $100+0+2+ \frac {6}{10} + \frac {3}{100} + \frac {0}{1000}= 100+2+.6+.03= 102.63$ c) $0+30+0+0+ \frac {2}{100} + \frac {5}{1000}=30+ +.02+.005=30.025$ d) $200+ 10+1+ \frac {9}{10} + \frac {0}{100} + \frac {2}{10000}= 211+.9+.002=211.902$ e) $0+ 10+2+ \frac {2}{10} + \frac {4}{100} + \frac {1}{1000}= 12+.2+.04+.001= 12.241$

Question 3

Write the following decimals in the place value table.

(a) 0.29

(b) 2.08

(c) 19.60

(d) 148.32

(e) 200.812

 Hundreds 100 Tens 10 Ones 1 Tenths Hundredths Thousandths (a) 0 0 0 2 9 0 (b) 0 0 2 0 8 0 (c) 0 1 9 6 0 0 (d) 1 4 8 3 2 0 (e) 2 0 0 8 1 2

Question 4

Write each of the following decimals.

(a)$20 + 9 + \frac {4}{10} + \frac {1}{100}$

(b)$137 + \frac {5}{100}$

(c)$\frac {7}{10} + \frac {6}{100} + \frac {4}{1000}$

(d)$23 + \frac {2}{10} + \frac {6}{1000}$

(e)$700 + 20 + 5 + \frac {9}{100}$

 a) $20 + 9 + \frac {4}{10} + \frac {1}{100}$ 29+.4+.01=29.41 b) $137 + \frac {5}{100}$ 137+.05= 137.05 c) $\frac {7}{10} + \frac {6}{100} + \frac {4}{1000}$ .7+.06+.004=.764 d) $23 + \frac {2}{10} + \frac {6}{1000}$ 23+.2+.006=23.206 e) $700 + 20 + 5 + \frac {9}{100}$ 725+.09=725.09

Question 5

Write each of the following decimals in words.

(a) 0.03

(b) 1.20

(c) 108.56

(d) 10.07

(e) 0.032

(f) 5.008

a)      0.03 = zero point zero three

b)      1.20 = one point two zero

c)      108.56 = one hundred eight point five six.

d)     10.07 = ten point zero seven.

e)      0.032 = zero point zero three two

f)       5.008 = five point zero zero eight

Question 6

Between which two numbers in tenths place on the number line does each of the given number lie?

(a) 0.06

(b) 0.45

(c) 0.19

(d) 0.66

(e) 0.92

(f) 0.57

1.      0.06 lies between 0 and 0.1

2.      0.45 lies between 0.4 and 0.5

3.      0.19 lies between 0.1 and 0.2

4.      0.66 lies between 0.6 and 0.7

5.      0.92 lies between 0.9 and 1

6.      0.57 lies between 0.5 and 0.6

Question 7

Write as fractions in lowest terms.

(a) 0.60

(b) 0.05

(c) 0.75

(d) 0.18

(e) 0.25

(f) 0.125

(g) 0.066

a) .60= 60/100=3/5
b)  .05= 5/100=1/20
c) .75= 75/100= 3/4
d) .18= 18/100=9/50
e) .25= 25/100=1/4
f) .125= 125/1000=1/8
g) .066= 66/1000=      33/500

# Exercise 8.3

Question 1

Which is greater?

(a) 0.3 or 0.4

(b) 0.07 or 0.02

(c) 3 or 0.8

(d) 0.5 or 0.05

(e) 1.23 or 1.2

(f) 0.099 or 0.19

(g) 1.5 or 1.50

(h) 1.431 or 1.490

(i) 3.3 or 3.300

(j) 5.64 or 5.603

To find larger number, we can follow these steps

1) When comparing two positive decimals, here are the tips
 a) If two numbers have unequal number of digits on the whole part, then the number with the greater number of digits is greater. b) If two numbers have equal number of digits on the whole part then, the number with greater valued digit on the extreme left is greater. If the digits on extreme left of the numbers are equal, then the digits to the right of the extreme left digits are compared and so on. c)  if the whole part is same, then the number with a greater value digit at tenth part is greater, if the digits on tenth part are equal, then the digit on hundredth part compared and so on

2)  When comparing two negative decimals, we can remove the sign and compare them as positive integer given above, the smallest of them would be the greater when we put the sign back

3)We comparing two different sign decimals, the number with positive sign is larger

With that in mind, lets solve the above questions

a)      0.3 and 0.4

Apply the same principle as given above

The whole number part is the same. It can be seen that the tenth part (3) of 0.3 is less than the tenth part (4) of 0.4.

So, 0.4 is greater.

b)      0.07 and 0.02

Apply the same principle as given above

The whole number part is the same. It can be seen that the hundredth part (7) of 0.07 is greater than the hundredth part (2) of 0.02.

So, 0.07 is greater.

c)      3 or 0.8

Apply the same principle as given above

The whole part 3 is greater than the decimal 0.8

So, 3 is greater

d)     0.5 or 0.05

Apply the same principle as given above

The whole part is the same. Tenth part of 0.5 is 5 and the tenth part of 0.05 is 0. So, we see that the tenth part of 0.5 is greater than the tenth part of 0.05. So, 0.5 is greater.

e)      1.23 or 1.2

Apply the same principle as given above

Whole part is 1 and is the same. The tenth part is also the same. The hundredth part of 1.23 is 3 and the hundredth part of 1.2 is 0. So, we see that 1.23 I greater.

f)       0.099 or 0.19

Apply the same principle as given above

Whole number part is the same. The tenth part of 0.099 is 0 and that of 0.19 is 1. So, the tenth part of 0.19 is greater than 0.099. Hence, 0.19 is greater.

g)      1.5 or 1.50

Apply the same principle as given above

Whole number is the same. The tenth part is the same. So, the numbers are equal.

h)      1.431 or 1.490

Apply the same principle as given above

The whole number is the same. The number in the tenth part is also the same. The hundredth part of 1.431 is 3 and that of 1.491 is 9. We see that 3 < 9 and hence 1.431 is lesser than 1.490.

Hence, 1.490 is greater.

I)        3.3 or 3.300

Apply the same principle as given above

The whole number part and the tenth part are the same. When there are no digits in the hundredths and the thousandths place of the number, we know that the digits are zero.

So, the numbers are the same.

j)        5.64 or 5.603

Apply the same principle as given above

The whole number and the tenth place digits are the same. As 4 is greater than 0, we know that 5.64 is greater than 5.603.

So, 5.64 is greater.

### Practice Question

Question 1 What is $\frac {1}{2} + \frac {3}{4}$ ?
A)$\frac {5}{4}$
B)$\frac {1}{4}$
C)$1$
D)$\frac {4}{5}$
Question 2 Pinhole camera produces an ?
A)An erect and small image
B)an Inverted and small image
C)An inverted and enlarged image
D)None of the above


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# Symmetric Smoothing Filters from Global Consistency Constraints

We introduce a simple principle of consistency in image denoising that argues that the relative similarities between the pixels as imputed by an averaging matrix should be preserved in the filtered output and hence results in a consistency filter that has the theoretically desirable properties of being symmetric, stable and belongs to the family of generalized doubly stochastic matrices. Our consistency filter has an interpretation as a specific form of Laplacian regularization unifying two strands of image denoising methods, i.e. symmetric smoothing filters and spectral graph theory.

#### Observation Model

Given the image noising observation model

$$\mathbf{y}=\mathbf{z}+\mathbf{n}$$,

where $$\mathbf{y}$$ and $$\mathbf{z}$$ are $$N_p$$-dimensional observed and true image vectors respectively and $$\mathbf{n} \sim \mathcal{N}(\mathbf{0},\sigma^2\mathbf{I})$$, in a denoising problem, we wish to recover $$\mathbf{z}$$.

#### Non-parametric Regression

In a non-parametric regression setting, one uses the following filtering technique to estimate $$\mathbf{z}$$ given as

$$\hat{\mathbf{z}} = \mathbf{W}{\mathbf{y}} = \mathbf{D}^{-1}\mathbf{K} {\mathbf{y}}$$

where $$\hat{\mathbf{z}}$$ is the estimate, $$K_{ij}$$ are defined by the choice of a kernel function that represents the notion of similarity we wish to capture and $$\mathbf{D}$$ is diagonal with $$i$$ -th entry of $${\sum}_{j} K(i,j)$$.

#### Notion of Consistency

##### Strict Consistency

Our notion of consistency implies

$$\hat{\mathbf{z}} = \mathbf{W}(\hat{\mathbf{z}})\hat{\mathbf{z}} = \mathbf{W}\hat{\mathbf{z}}$$

i.e

$$\left(\mathbf{I}-\mathbf{W}\right)\hat{\mathbf{z}} = \mathcal{L}\hat{\mathbf{z}} = \mathbf{0}$$

where $$\mathcal{L} =\mathbf{I}-\mathbf{W}$$ is the normalized Laplacian. The consistency constraint tells us that the filtered image has the same intensity relationships as encoded by $$\mathbf{W}$$, and no amount of further filtering will change this condition.

##### Relaxed Consistency

Strict consistency leads to the solution of piecewise constant images. However, this is excessively restrictive for our problem of denoising natural images which are far from being piecewise constant. Hence, we replace our strict consistency criterion of $$\mathbf{z} = \mathbf{W} \mathbf{z}$$ by a more general consistency model

$$\mathbf{z} = \mathbf{W} \mathbf{z} + \mathbf{\epsilon}(\mathbf{z})$$

where $$\mathbf{\epsilon}(\mathbf{z})$$ signifies the degree to which any estimated image deviates from the consistency requirement specified by $$\mathbf{W}$$ and the argument $$\mathbf{z}$$ emphasizes that this measure of inconsistency is dependent on $$\mathbf{z}$$.

#### Our Solution is the $$\mathbf{C}$$ filter

Our denoising framework minimizes the following cost function.

$$\hat{\mathbf{z}} = \displaystyle{\textrm{argmin}_{\mathbf{z}}}\, ||\mathbf{z}-\mathbf{y}||^2 +\lambda ||\mathbf{\Lambda}\left(\mathcal{L}\mathbf{z}\right)||^2$$

or,

$$\hat{\mathbf{z}} = \displaystyle{\textrm{argmin}_{\mathbf{z}}}\, ||\mathbf{z}-\mathbf{y}||^2 +\lambda ||\mathbf{L}\mathbf{z}||^2$$

which gives $$\mathbf{C}=(\mathbf{I}+\lambda\mathbf{L}^{T}\mathbf{L})^{-1}$$.

#### Properties of the $$\mathbf{C}$$ filter

• A stable consistent smoothing filter
• Belongs to the family of generalized doubly stochastic matrices
• A Weiner filter in the eigenspace of the Laplacian

#### Comparison with the usual Graph Regularization

Figure 1.a shows the adjacency matrix visualization of the regular K-NN + bilateral graph. Figure 1.b shows that of our enriched equivalent graph.

Figure 1.a The regular Graph Adjacency matrix
Figure 1.b Our modified Graph Adjacency matrix

#### Results

We compare the performance of our C filter with the vanilla filter W constructed from regular K-NN + bilateral graph, its Sinkhorn symmetrized version S(W), the usual graph-regularized R = (I + λL)-1, K-SVD and BM3D respectively below in Table 1.

 Image Noise Level W = D-1K S(W) R=(I + λL)-1 K-SVD BM3D Our C=(I + λLTL)-1 σ = 20 28.69 28.48 28.84 30.33 30.92 30.44 σ = 40 23.24 22.74 26.00 26.37 27.43 26.82 σ = 20 27.27 27.11 28.26 29.41 29.66 28.61 σ = 40 22.42 21.97 24.61 25.56 26.25 25.73 σ = 20 24.78 24.89 25.54 26.53 26.79 26.18 σ = 40 22.16 21.96 21.96 22.77 23.23 23.08 σ = 20 24.89 25.29 26.24 26.97 27.56 27.35 σ = 40 21.83 21.73 21.37 21.96 22.70 22.39 σ = 20 28.38 27.91 31.39 33.03 33.64 33.12 σ = 40 22.68 22.13 27.88 29.20 30.66 29.55 σ = 20 28.26 27.98 29.09 30.94 31.43 30.67 σ = 40 22.93 22.41 26.10 27.01 27.93 27.33 σ = 20 28.20 27.93 29.84 30.96 31.44 30.53 σ = 40 22.95 22.43 26.17 27.34 28.18 27.23

Table 1. Denoising performance comparison (PSNR in dB) of our method with various state-of-the-art techniques.

## Publications

1. Symmetric Smoothing Filters from Global Consistency Constraints (Sk. Mohammadul Haque, Gautam Pai, Venu Madhav Govindu), In IEEE Transactions on Image Processing, volume 24, 2015. [bibtex] [pdf]


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# Trying to understand neural network's performance

I am trying to build a RNN for classification and below is the layout of the network

import tensorflow as tf
model  = tf.keras.Sequential()

from sklearn.utils.class_weight import compute_class_weight
class_weight = compute_class_weight(class_weight='balanced',classes =np.unique(y),y=y)
class_weight = {i : class_weight[i] for i in range(len(class_weight))}

and fitting using

model.fit(X,y,epochs=10,validation_split=0.4,class_weight=class_weight)

My X has structure :

array([[[  0.29615385,   0.9375    ,   0.52268041, ...,   0.        ,
0.13050997,   0.15842043],
[  0.29615385,   0.9375    ,   0.52268041, ...,   0.        ,
0.13050997,   0.15842043],
[  0.26538462,   0.9875    ,   0.52268041, ...,   0.        ,
0.09115686,   0.12374666],
...,
[-100.        , -100.        , -100.        , ..., -100.        ,
-100.        , -100.        ],
[-100.        , -100.        , -100.        , ..., -100.        ,
-100.        , -100.        ],
[-100.        , -100.        , -100.        , ..., -100.        ,
-100.        , -100.        ]],

:
:
:

The -100 is the padding done to make those inner 2d arrays of same shape

However the model is not learning as I expect it to be i.e increasing accuracy

The model's accuracy does not increase more than this, it decreases on more epochs. Is there anything wrong with my Neural Network structure or I need to experiment with more number of hidden layers (which I already tried)

757/757 [=============>] - ETA: 10:08 - loss: 0.6800 - accuracy: 0.5290 - auc: 0.6303