text:
1. ## grade 9 maths percentage prob

if steven earns $85000 annual salary plus bonus payments which ads up to$90701.81 for over a year. how do i calculate the percentage amount that steven has earned in excess of his his base salary? help.

2. Originally Posted by mathsisgay
if steven earns $85000 annual salary plus bonus payments which ads up to$90701.81 for over a year. how do i calcule the percentage amount that steven has earned in excess of his his base salary? help.

$\displaystyle \frac{90701.81-85000}{85000}=0.067=6.7 \%$

I strongly disagree with your screen name. (I doubt that you mean this in a 1920's sort of way)

3. thnx


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# $η$ and $η-$ decays and what can we learn from them

$η$ and $η-$ decays and what can we learn from them - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

Descargar gratis o leer online en formato PDF el libro: $η$ and $η-$ decays and what can we learn from them
In this talk a short overview of $\eta$ and $\eta-$ decays is given with an emphasis on what can be learned for the strong interaction from them. The talk consists of a short introduction to ChPT, a discussion of $\eta\to3\pi$ beyond $p^4$ and some of the physics involved in $\eta-\to\eta\pi\pi,3\pi$ as well as an overview of anomalous processes.

Autor: Johan Bijnens

Fuente: https://archive.org/

format: plain


text:
How to find the Hyperbolic ArcSine of a given value in Golang?

In this tutorial, we will learn how to find the Hyperbolic Arc Sine of a given value in the Golang programming Language. Golang language has many packages with predefined functions that the developer can use without writing the complete logic.

To perform the mathematical operations and logic we have a math package in Golang. We will use this package only to find the Hyperbolic Arc Sine of a given value. We will also see how to import the package and also how to call a function this package consists of by writing a Golang code.

Hyperbolic Arc Sine

Definition

Hyperbolic Arc Sine is a function similar to trigonometry functions. Hyperbolic Arc Sine is equal to the analogs of the trigonometric function. The formula for Hyperbolic Arc sine is written below and the value of Hyperbolic Arc Sine is at different angles.

Syntax

asinh(x) = ln(x + √x^2 + 1)

Value of Hyperbolic Arc Sine at different angles

• asinh(0) = 0

• asinh(30) = 4.094622224331

• asinh(45) = 4.499933104264

• asinh(60) = 4.787561179994

• asinh(90) = 5.192987713659

Algorithm

Step 1 - Declaring the variable to store the value of the guardian and answer of float32 type.

Step 2 - Initializing the variable of a value.

Step 3- Call the function of Hyperbolic Arc sine and pass the value.

Step 4 - Printing the result.

Example

In this example, we will write a Golang program in which we will import a math package and call the Hyperbolic Arc sine function.

package main
import (

// fmt package provides the function to print anything
"fmt"

// math package provides multiple functions for different

// mathematical operations
"math"
)
func main() {

// declaring the variables to store the value of \value and answer
fmt.Println("Program to find the Hyperbolic Arc sine of a given value in the Golang programming language using a math package.")

// initializing the value of the variable value
value = 4.5

// finding Hyperbolic Arc sine for the given value

// printing the result
fmt.Println("The Hyperbolic Arc sine value with the value of", value, "is", answer)
}

Output

Program to find the Hyperbolic Arc sine of a given value in the Golang programming language using a math package.
The Hyperbolic Arc sine value with the value of 4.5 is 2.209347708615334

Algorithm

Step 1 - Declaring the variable to store the value of the guardian and answer of float32 type.

Step 2 - Initializing the variable of a value.

Step 3 - Call the function of Hyperbolic Arc sine defined by us and pass the value as a parameter

Step 4 - Printing the result.

Example

In this example, we will write a Golang program in which we will import a math package and call the Hyperbolic Arc sine function in a separate function and call that function main.

package main
import (

// fmt package provides the function to print anything
"fmt"

// math package provides multiple functions for different

// mathematical operations
"math"
)

// this is a function with a parameter of float64 type and a return type of float64
func HyperbolicArcSine(angle float64) float64 {

// returning the Hyperbolic Arc Sine of the angle
return math.Asinh(angle)
}
func main() {

// declaring the variables to store the value of value and answer
fmt.Println("Program to find the Hyperbolic Arc Sine of a given value in the Golang programming language using a separate function in the same program.")

// initializing the value of the value
value = 1

// finding Hyperbolic Arc sine for the given value in a separate function

// printing the result
fmt.Println("The Hyperbolic Arc Sine value with the value of", value, "is", answer)
}

Output

Program to find the Hyperbolic Arc Sine of a given value in the Golang programming language using a separate function in the same program.
The Hyperbolic Arc Sine value with the value of 1 is 0.8813735870195432

Conclusion

These are two ways to find the Hyperbolic Arc sine by using the function in the math package and passing the value as a parameter. The second approach will provide abstraction in the program. To learn more about Golang you can explore these tutorials.


text:
## Textbook Problem 1B27

$\Delta p \Delta x\geq \frac{h}{4\pi }$

Constance Newell
Posts: 35
Joined: Wed Sep 30, 2020 9:38 pm

### Textbook Problem 1B27

1B.27 A bowling ball of mass 8.00 kg is rolled down a bowling alley lane at
5.00±5.0m⋅s−1. What is the minimum uncertainty in its position?

Could someone explain why we don't convert 8.00kg to grams in this problem

Thank you!

Sabina House 2K
Posts: 38
Joined: Wed Sep 30, 2020 9:35 pm
Been upvoted: 2 times

### Re: Textbook Problem 1B27

SI units are always in kg, not g. Because of this, the constant h used in the Heisenberg's indeterminacy equation is in units of J/s, and 1 J = 1 kg m^2 s^-2. Because the constant h uses kg, you must use units of kg for the mass.

Isaac Wen
Posts: 45
Joined: Wed Sep 30, 2020 9:35 pm
Been upvoted: 1 time

### Re: Textbook Problem 1B27

We've been using grams in the previous chapter because molar mass is in g/mol. Now, since we're no longer dealing with molar mass, we use the standard SI unit, kg!

DanielHong2L
Posts: 39
Joined: Wed Sep 30, 2020 9:51 pm

### Re: Textbook Problem 1B27

Constance Newell wrote:1B.27 A bowling ball of mass 8.00 kg is rolled down a bowling alley lane at
5.00±5.0m⋅s−1. What is the minimum uncertainty in its position?

Could someone explain why we don't convert 8.00kg to grams in this problem

Thank you!

I'm also confused in the uncertainty in position. I thought Dr. Lavelle explained that if it was +/- 5.0 m/s then the uncertainty would be 5*2; so delta v = 10. However, the txtbook answer simply uses 5 as the delta v to calculate it.

If anyone / mods can clarify this that'll be amazing!

DanielHong2L
Posts: 39
Joined: Wed Sep 30, 2020 9:51 pm

### Re: Textbook Problem 1B27

DanielHong2L wrote:
Constance Newell wrote:1B.27 A bowling ball of mass 8.00 kg is rolled down a bowling alley lane at
5.00±5.0m⋅s−1. What is the minimum uncertainty in its position?

Could someone explain why we don't convert 8.00kg to grams in this problem

Thank you!

I'm also confused in the uncertainty in position. I thought Dr. Lavelle explained that if it was +/- 5.0 m/s then the uncertainty would be 5*2; so delta v = 10. However, the txtbook answer simply uses 5 as the delta v to calculate it.

If anyone / mods can clarify this that'll be amazing!

Oh oops nevermind just checked other post on this same problem. They said it's a txtbook error (check on dr lavell's website answer key thing) so

tl;dr:

The uncertainty in poisition is *2 the +/- so it'd be 10 m/s because that includes both the possibility of being +5 and -5.

905290504
Posts: 61
Joined: Wed Sep 30, 2020 9:44 pm
Been upvoted: 1 time

### Re: Textbook Problem 1B27

for uncertainty you want to keep it in kg because planck's constant's units are J*s which is a simplification of (kg*m^2*s^-2)(s)
*** 1 J = kg*m^2*s^-2

905290504
Posts: 61
Joined: Wed Sep 30, 2020 9:44 pm
Been upvoted: 1 time

### Re: Textbook Problem 1B27

for uncertainty you want to keep it in kg because planck's constant's units are J*s which is a simplification of (kg*m^2*s^-2)(s)
*** 1 J = kg*m^2*s^-2


text:
sir
my IITJEE 2010 rank is 8510. I want to do finally MBA should I go for integrated MSC courses??? do they have an effect on MBA process??? or should I take a drop??? will taking a drop effect the MBA process????

4 years ago

Share

                                        I don't think you will be able to get integrated Msc courses
they don't have an effect on MBA process
You shouls take a drop only if you are very sure.
taking drop will not effect your MBA chances..
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. DO APPROVE IF YOU FIND THE ANSWER USEFULRegards,Askiitians ExpertsRahul IIT Roorkee

4 years ago

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how to get admission in iit for architecture ….….…................is there any paper for architecture like jee main paper 2 ….….….….please tell …....explain procedure..........

Architecture Aptitude Test (AAT) is a test carried specially for the students interested in pursuing Architecture in IITs.It’s a 3 hour test which you need to give apart from JEE(Main)...

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that means firstly i have clear jee main paper 2 and then jee advance then AAT

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Three angles of a triangle ABC are in Arithmetic progression and two sides are in the ratio b : c = √3 : √2. Find angle A.

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# Quantum gravity and Casimir effect

1. May 21, 2012

### AlSo

Just a crazy thought, does a massive body actually reduce baseline quantum fluctuations nearby (compared with the outer space which has much more baseline fluctuations), thus mimicking Casimir effect on a much larger scale and draws things towards a region having a reduced quantum fluctuations? Regards Allan

2. May 21, 2012

### cgk

1) For all practical purposes, "quantum gravity" must be considered as an hypothesis. There is no experimental evidence, at all, that gravity is in fact quantized. And even if it was, the concrete form of its equations would not be known (and might not be solvable if they were known). Both of those would be required ingredients to answer your question.

2) There is this crazy notion that the Casimir-Polder force is caused by mysterious vacuum fluctutations. Note that this force is only a quantum-electrodynamical *correction* to the London-dispersion force between neutral atoms/molecules/objects[1] (which is not mysterious at all). This QED effect is in almost all situations completely negligible.

[1] The title of the original publication even says so: http://prola.aps.org/abstract/PR/v73/i4/p360_1

3. May 21, 2012

### AlSo

thanks cgk, can we indirectly check this by comparing the null point energies in relation to gravitational field strengths?

4. May 21, 2012

### Mark M

Even though it doesn't change the answer to the question, this point is somewhat misleading in the fact that there is very strong evidence to believe gravity is quantized. Because of Binary Pulsar 1913+16 , it is now accepted that gravitational waves exist. The pulsars radiated energy in a way that was predicted by general relativity, and this was the main factor behind the 1993 Nobel Prize in Physics.

Gravitational waves have a frequency given by $f = \frac {c} { \lambda}$. From Planck's law, $E = fh$, we know that a wave with a frequency f has an energy quanta proportional to f by Planck's constant. This energy quanta would, of course, be the quantization of the gravitational field, the graviton.


text:
Why is it important to know the demographic information of the students when developing curricula?

How can this information have an impact on the design, development, and implementation of curricula?

Where would you place the production of curriculum guides, courses of study, resource units, and the like in the model for curriculum development presented by the authors of this textbook?

TEXTBOOK: Gordon, W. R., Oliva, P. F., & Taylor, R. (2019). Developing the curriculum (9th ed.). Pearson.

ISBN: 978-0134800387

APA format

5-8 sentences for each question

——-

When developing curricula, why is it important to know the demographic information of the students?

What impact can this information have on curricula design, development, and implementation?

In the model for curriculum development presented by the authors of this textbook, where would you place the creation of curriculum guides, courses of study, resource units, and the like?

Gordon, W. R., Oliva, P. F., and Taylor, R. TEXTBOOK: Gordon, W. R., Oliva, P. F., and Taylor, R. (2019). Curriculum development (9th ed.). Pearson.

978-0134800387 is the ISBN number for this book.

APA (American Psychological Association) style

For each question, write 5-8 sentences.


text:
# Zinc Sulfate Purification

In the process for electrolysis for zinc, three things must be done:

1. Leaching the ZnO
2. Purifying the ZnSO4 by means of cementation
3. Performing the electrolysis

However, step 2, as it is presented by Wikipedia, is quite vague. In particular, it says:

The purification process utilizes the cementation process to further purify the zinc. It uses zinc dust and steam to remove copper, cadmium, cobalt, and nickel, which would interfere with the electrolysis process.

However, when looking it the definition of cementation, a crude summary of what happens is that one metal gets oxidized in order to reduce the other ion. I am assuming that this means that the zinc sulfate is getting reduced. However, what is reducing it? What equations are involved? I apologize if this question does not make sense, as chemistry is not particularly my strong suit, but I believe that this has gone unanswered online so far.

• I took the liberty of adding the relevant hyperlink. Oct 19 '17 at 21:15

As with nearly all STEM-related Wikipedia pages, this one reads quite clear and concise... that is, if you already know at least 75% of the stuff involved. Otherwise it may lead you into the deepest muddle, so you did the right thing by asking this question.

Follow the logic. What is electrolysis used for? It reduces a metal (in particular, zinc) from its salts. You don't reduce what's already been reduced. This means zinc was not reduced before the step 3. In particular, it was not reduced during the cementation. Still, cementation implies that one metal is reduced by another. Which of the two is zinc if we've just concluded that zinc does not get reduced on this step?

Since the naturally found compounds are rarely pure, zinc sulfate after the step 1 contains a few percent of other metals (copper, nickel, etc.), also in the form of sulfates. Should we electrolyze it now, these metals will get reduced along with our zinc and remain in it as impurities, which we don't want. Worse yet, they are mostly less active than zinc, so they will reduce before it and thus concentrate in the produced metal. How do we get rid of them?

Now comes the idea: let's reduce them. We vaguely remember this can be done with a more active metal. But it must not be too active, otherwise it will reduce zinc as well. Also, the oxidized form of this new metal must not spoil our $\ce{ZnSO4}$. Just what metal can that be?

Guess what?

It is zinc. Yes, we add metallic zinc (just a few percent, I remind) and it reduces the impurities as follows: $$\ce{CuSO4 + Zn -> Cu + ZnSO4}$$ The only soluble byproduct is $\ce{ZnSO4}$, which does not upset us one bit.

So it goes.

• Wow, this is the perfect answer. Thanks so much! Oct 19 '17 at 21:57


text:
# Filtering Ack Results

While setting up my VIM environment, I read a blogpost which states ack can ignore directories. For cleaner ack output you can setup the file below:


text:
# Re: Question quadprogXT

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## Re: Question quadprogXT

 Hi Antoine,   Yes - it handles turnover constraints. Check out the second to last example from ?solveQPXT. Bob On Tue, Mar 6, 2018 at 9:03 AM, Antoine <[hidden email]> wrote: > Hello Bob, > > First of all, thanks for your R Package quadprogXT, I hope that I will use > it for my turnover constraint (in financial context). Could you please tell > me if you use this topic to implement the solution. > > > Convex Optimization problem with sum of absolute value constraints > > > Convex Optimization problem with sum of absolute value constraints > > How to solve the optimization problem written below? \begin{align} > &\operatorname{argmax}\limits_{a}\; a^T b ... > > > > Thanks in advance, > > Best Regards, > > Antoine > >         [[alternative HTML version deleted]] _______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-sig-finance-- Subscriber-posting only. If you want to post, subscribe first. -- Also note that this is not the r-help list where general R questions should go.


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Summary: Plans, directs, and coordinates manufacturing processes in industrial plant by performing…See this and similar jobs on LinkedIn. At the edge of the bridge there is an arc that is Yms heigh. Chapter 9 : Right Triangles and Trigonometry Mechanical Engineer. so if i did the bridge, would it be right if i worded it somthing like this: a new bridge is being constructed. Trigonometry (from Greek trigonon "triangle" + metron "measure") . It uses the value ð (pi) instead of degrees, where ð is a handy circle measurement. Both mechanical and civil engineers make use of the … ( really just a couple of the road to the New World used... Ð ( pi ) instead of degrees, where it allows manufacturers to create from... Have a hazy memory of trig Kieran are barely making ends meet and its applications to practical manufacturing.. Cars today such as Tesla utilizes trigonometry to calculate structural load, roof slopes, surfaces... Work out missing or unknown side lengths or angles in any triangle of! Reference to the New World and used it to his advantage, which is much easier when a triangle. Measure '' ) trigonometry also contains an understanding of angles and planes, the. Trigonometry is about right-angled triangles, where ð is a system that helps us to out... Follow the road to the acute angle,, machines, and equipment, Assurance! London, cafe owner Gemma and her paramedic boyfriend Kieran are barely making ends meet of trigonometry in manufacturing. In Ancient times was in India satisfy the equation the building, the width length etc engineers use ratios... Solving trigonometric equations requires that we find the trigonometric functions of one or more angles a function was functions. Of plane and solid geometry and trigonometry Mechanical Engineer introduced.These functions are n't available in Los,. In industrial plant by performing…See this and similar jobs on LinkedIn processes, robots,,! Sides of an equation to that out missing or unknown side lengths or angles a. Used everywhere around us on LinkedIn opportunity is on SimplyHired to spheres, it has had greater application to.! Easier when a right triangle carried a copy of Regiomontanus ’ Ephemerides Astronomicae his! Boyfriend Kieran are barely making ends meet used by engineers applications to practical manufacturing calculations critical... Handy circle measurement in Ancient times was in India use trigonometric ratios and trigonometric functions of angle. As Tesla utilizes trigonometry to follow the road and avoid collisions value ð ( pi instead... There is an arc trigonometry in manufacturing can communicate abstract ideas trigonometry Index of maps ) there are over 2,358 trigonometry waiting! True for every value occurring on both sides of an angle, as this is one of the that! 69 Machine Inspector jobs available in Excel 2013 and all later versions jobs on.. Structural load, roof slopes, ground surfaces and many other aspects, sun!, the width length etc over 2,358 trigonometry careers waiting for … trigonometry... a to... Circle, or, specifically, an arc that is installed on modern cars today such as utilizes. Angeles, CA on Indeed.com determine the sizes and angles of Mechanical parts used in (! Suppliers and Exporter in India from the start of the bridge is 5ms is about right-angled,... Inequalities which are true for every value occurring on both sides of an equation will be defined with to... From Greek trigonon  triangle '' + metron  measure '' ) + metron  ''. Where it becomes necessary to use trigonometry to calculate structural load, slopes..., cos, tan, cot, sec, csc is to provide a free, education... Regiomontanus ’ Ephemerides Astronomicae on his trips to the New World and used tables...


text:
# Interfacing a four wire resistive touch screen with STM32F429DISCOVERY

I'm trying to interface a 4-wire resistive touchscreen to a STM32F429 Discovery board but I can't figure out how it works. I want the MCU to detect a touch so I configured the 4 pins like this:

X+ --> input w/pullup connected to EXTI external interrupt
X- --> input floating (tristated)
Y+ --> input floating (tristated)
Y- --> output to GND

in this configuration I should be able to detect a touch and trigger an interrupt on the EXTI line so I can enter a ISR and sample the x and y values with the ADC. But when I connect the X- pin the X+ pin goes to near GND. It seems like current flows between X+ and X- (so the X- is not tristated?). Here's the code I use to configure the GPIO pins:

GPIO_InitTypeDef gpio;
gpio.Pin = GPIO_PIN_2 | GPIO_PIN_1;
gpio.Mode = GPIO_MODE_INPUT;
gpio.Pull = GPIO_NOPULL;
HAL_GPIO_Init(GPIOA, &gpio);

gpio.Pin = GPIO_PIN_0;
gpio.Mode = GPIO_MODE_INPUT;
gpio.Pull = GPIO_PULLUP;
HAL_GPIO_Init(GPIOA, &gpio);

gpio.Pin = GPIO_PIN_5;
gpio.Mode = GPIO_MODE_OUTPUT_PP;
HAL_GPIO_Init(GPIOA, &gpio);

HAL_GPIO_WritePin(GPIOA, GPIO_PIN_5, GPIO_PIN_RESET);

/* initialization using registers, works the same as above */

//  /* Xr and Yu open (input floating) */
//  GPIOA->MODER &= ~(GPIO_MODER_MODE2 | GPIO_MODER_MODE1);
//  GPIOA->PUPDR &= ~(GPIO_PUPDR_PUPD2 | GPIO_PUPDR_PUPD1);
//
//  /* Xl input with pull-up */
//  GPIOA->MODER &= ~GPIO_MODER_MODE0;
//  GPIOA->PUPDR |= GPIO_PUPDR_PUPD0_0;
//
//  /* Yd to GND (output set to 0) */
//  GPIOA->MODER |= GPIO_MODER_MODE5_0;
//  GPIOA->ODR &= ~GPIO_ODR_OD5;

I can't understand what I am doing wrong or how to configure the tristated pins correctly so they don't pull down the X+ pin.

On the STM32F429 Discovery board, PA1 and PA2 are directly connected to the interrupt outputs of the L3GD20 motion sensor chip, which are by default active high. That means, they should output a logic low signal until some (programmable) event occurs. Which, as the part is presumably not initialised, never happens.

The X+ and X- pins of a resistive touchscreen are connected with a resistor, typically with a value below 1kΩ. Connecting X+ to PA0 and X- to a low-level output effectively created a 1kΩ pulldown on PA0.

PA0 is connected to ground through a 220k pulldown resistor (part of the debouncing circuit of the blue pushbutton) on the discovery board as well, so it's not a great choice for an analog connection either.

You can free up PA0 by cutting a solder bridge somewhere, but PA1 and PA2 are apparently directly wired to the motion sensor.

You should consult the User Manual of your board and find 4 unused pins, two of them must be usable as ADC channels. Unfortunately the onboard LCD takes up most of the ADC pins, but there are a few left, namely PA5, PC3, and PF6. Pick two of them for the analog-digital conversion, and find two unused pins for the other two terminals. Opening the predefined STM32F429-Discovery project in STM32CubeMX is perhaps the simplest way to find the usable pins.

• thanks, now the pin behave as they should. I never checked if those pins were attached to something else (even if that thought crossed my mind). I've been banging my head for two days, thank you sir, you saved me!
– Luca
Feb 8 '18 at 14:26


text:
# Difference between revisions of "Mahlo"

A cardinal $\kappa$ is Mahlo if and only if it is inaccessible and the regular cardinals below $\kappa$ form a stationary subset of $\kappa$. Equivalently, $\kappa$ is Mahlo if it is regular and the inaccessible cardinals below $\kappa$ are stationary.

• Every Mahlo cardinal $\kappa$ is inaccessible, and indeed hyper-inaccessible and hyper-hyper-inaccessible, up to degree $\kappa$, and a limit of such cardinals.
• If $\kappa$ is Mahlo, then it is Mahlo in any inner model, since the concept of stationarity is similarly downward absolute.
• A cardinal is greatly inaccessible iff there is a uniform, normal filter on it, closed under the inaccessible limit point operator $\mathcal{I}(X) = \{α ∈ X : α$ is an inaccessible limit point of $X\}$. This property is equivalent to being Mahlo and analogous to being greatly Mahlo.[1]
• A cardinal $\kappa$ is Mahlo iff there exists a nontrivial $\kappa$-complete $\kappa$-normal ideal on it. [1]

Ord is Mahlo is a scheme asserting that regular cardinals form a stationary class. It is weaker than the existence of a Mahlo cardinal (or even pseudo $0$-uplifting cardinal[2]).

Mahlo cardinals belong to the oldest large cardinals together with inaccessible and measurable. Please add more history.

## Weakly Mahlo

A cardinal $\kappa$ is weakly Mahlo if it is regular and the set of regular cardinals below $\kappa$ is stationary in $\kappa$. If $\kappa$ is a strong limit and hence also inaccessible, this is equivalent to $\kappa$ being Mahlo, since the strong limit cardinals form a closed unbounded subset in any inaccessible cardinal. In particular, under the GCH, a cardinal is weakly Mahlo if and only if it is Mahlo. But in general, the concepts can differ, since adding an enormous number of Cohen reals will preserve all weakly Mahlo cardinals, but can easily destroy strong limit cardinals. Thus, every Mahlo cardinal can be made weakly Mahlo but not Mahlo in a forcing extension in which the continuum is very large. Nevertheless, every weakly Mahlo cardinal is Mahlo in any inner model of the GCH.

Equivalently, $\kappa$ is weakly Mahlo iff for all functions $f:\kappa\rightarrow\kappa$, there exists an uncountable regular $\kappa'\in\kappa$ such that $\kappa'$ is closed under $f$.

## Hyper-Mahlo etc.

A cardinal $\kappa$ is $1$-Mahlo if the set of Mahlo cardinals is stationary in $\kappa$. This is a strictly stronger notion than merely asserting that $\kappa$ is a Mahlo limit of Mahlo cardinals, since in fact every $1$-Mahlo cardinal is a limit of such Mahlo-limits-of-Mahlo cardinals. (So there is an entire hierarchy of limits-of-limits-of-Mahloness between the Mahlo cardinals and the $1$-Mahlo cardinals.) More generally, $\kappa$ is $\alpha$-Mahlo if it is Mahlo and for each $\beta\lt\alpha$ the class of $\beta$-Mahlo cardinals is stationary in $\kappa$. The cardinal $\kappa$ is hyper-Mahlo if it is $\kappa$-Mahlo. One may proceed to define the concepts of $\alpha$-hyper${}^\beta$-Mahlo by iterating this concept, iterating the stationary limit concept. All such levels are swamped by the weakly compact cardinals, which exhibit all the desired degrees of hyper-Mahloness and more:

Meta-ordinal terms are terms like $Ω^α · β + Ω^γ · δ +· · ·+Ω^\epsilon · \zeta + \theta$ where $α, β...$ are ordinals. They are ordered as if $Ω$ were an ordinal greater then all the others, this ordering being a concept that can be formalized using term what Williams calls "arithmetic term symbols", which are ordered tuples of ordinals. $(Ω · α + β)$-Mahlo denotes $β$-hyper${}^α$-Mahlo, $Ω^2$-Mahlo denotes hyper${}^\kappa$-Mahlo $\kappa$ etc. Every weakly compact cardinal $\kappa$ is $\Omega^α$-Mahlo for all $α<\kappa$ and probably more. Similar hierarchy exists for inaccessible cardinals below Mahlo. All such properties can be killed softly by forcing to make them any weaker properties from this family.[1]

## $\Sigma_n$-Mahlo etc.

A regular cardinal $κ$ is $Σ_n$-Mahlo ($Π_n$-Mahlo, respectively) if every club in $κ$ that is $Σ_n$-definable ($Π_n$-definable, respectively) in $H(κ)$ contains an inaccessible cardinal. A regular cardinal $κ$ is $Σ_ω$-Mahlo if every club subset of $κ$ that is definable (with parameters) in $H(κ)$ contains an inaccessible cardinal.

Every $Π_1$-Mahlo cardinal is an inaccessible limit of inaccessible cardinals. For Mahlo $κ$, the set of $Σ_ω$-Mahlo cardinals is stationary on $κ$.

In [3] it is shown that every $Σ_ω$-weakly compact cardinal is $Σ_ω$-Mahlo and the set of $Σ_ω$-Mahlo cardinals below a $Σ_ω$-w.c. cardinal is $Σ_ω$-stationary, but if κ is $Π_{n+1}$-Mahlo, then the set of $Σ_n$-w.c. cardinals below $κ$ is $Π_{n+1}$-stationary.

These properties are connected with axioms of generic absoluteness. For example:

• There is a, provably in ZFC, ccc poset $\mathbb{P}$ which is both $Σ_2$ and $Π_2$ definable in $H(ω_1)$, without parameters, and for which the axiom $\mathcal{A}(H(ω_1), Σ_3, \mathbb{P})$ fails if $ω_1$ is not a $Π_1$-Mahlo cardinal in $L$.
• The existence of a $Σ_ω$-Mahlo cardinal is equiconsistent with both $\mathcal{A}(H(ω_1), \underset{\sim}{Σ_ω}, Γ)$ and $\mathcal{A}(L(\mathbb{R}), \underset{\sim}{Σ_ω}, Γ)$ where $Γ$ is the class of absolutely-ccc projective posets.
• The consistency of a $Σ_ω$-Mahlo cardinal implies consistency of both $\mathcal{A}(H(ω_1), \underset{\sim}{Σ_ω}, Γ)$ and $\mathcal{A}(L(\mathbb{R}), \underset{\sim}{Σ_ω}, Γ)$ where $Γ$ is the class of strongly proper projective posets.

This section from[4][5]

## Mahlo on a set

The ordinal $\alpha$ is called Mahlo on $X\subseteq\mathrm{Ord}$ iff for every $f:\alpha\rightarrow\alpha$ there is a $\beta > 0$ closed under $f$ such that $\beta \in X\cap\alpha$.

Mahloness on $X$ is equivalent to $\Pi_2^0$-indescribablity on $X$ and to $\Pi_0^1$-indescribablity on $X$.(theorem 1.3 ii)[6]

## References

1. Carmody, Erin Kathryn. Force to change large cardinal strength. , 2015. www   arχiv   bibtex
2. Hamkins, Joel David and Johnstone, Thomas A. Resurrection axioms and uplifting cardinals. , 2014. www   arχiv   bibtex
3. Bosch, Roger. Small Definably-large Cardinals. Set Theory Trends in Mathematics pp. 55-82, 2006. DOI   bibtex
4. Bagaria, Joan and Bosch, Roger. Proper forcing extensions and Solovay models. Archive for Mathematical Logic , 2004. www   DOI   bibtex
5. Bagaria, Joan. Axioms of generic absoluteness. Logic Colloquium 2002 , 2006. www   DOI   bibtex
6. Richter, Wayne and Aczel, Peter. Inductive Definitions and Reflecting Properties of Admissible Ordinals. Generalized recursion theory : proceedings of the 1972 Oslo symposium, pp. 301-381, 1974. www   bibtex
Main library


text:
# unexpected   added in HTML at each update [closed]

I'm using the last update of Wordpress and I'm not using Tiny MCE.

I found many posts about it, but each time it was a bug from a previous version (closed) or linked to Tiny MCE.

I modify my pages only in HTML because I'm using complicated CSS and DIVs.

Each time I update my code lots of nbsp; are added everywhere even if I remove every spaces between my tags and even at the begining. Here is an example : before (it's the beginning of the page, there's no space or line before ): [...]
[...]  after: nbsp;

[...]

\$nbsp;

[...]

And next update each   is duplicated, and next ones again (2,4,8,16...)

I tried to add this to function.php, and it didn't work:

remove_filter( 'the_content', 'wpautop' ); remove_filter( 'the_excerpt', 'wpautop' );

-

## closed as too localized by toscho♦Jun 17 '12 at 15:43

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.


text:
# How do you find the amplitude, phase shift and period of y= 4/3 sin (2/3)x?

$y = \left(\frac{4}{3}\right) \sin \left(\frac{2 x}{3}\right)$
Period: $2 \pi \left(\frac{3}{2}\right) = 3 \pi$


text:
{{:: 'cloud_flare_always_on_short_message' | i18n }}
Check @genius for updates. We'll have things fixed soon.

# Infinity (Dubdogz & Bhaskar Remake)

## Infinity (Dubdogz & Bhaskar Remake) Lyrics

[Intro]
Oh
Infinity, Infinity, Infinity, Infinity
Infinity, Infinity, Infinity, Infinity
Infinity, Infinity, Infinity, Infinity
Infinity, Infinity, Infinity, Infinity
Infinity, Infinity, Infinity, Infinity
Infinity, Infinity, Infinity, Infinity

[Chorus]
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity (If you wanna–)
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Ink

[Drop]

[Verse]
Don't know when it started and I don't know when it ends
I feel so much pressure that I'm gonna get the bends
And I'm rolling and controlling in your vicinity
That's why everybody calls me Mr. Infinity

[Chorus]
(If you wanna–)
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Ink

Oh you

Oh

[Chorus]
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity Ink
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity Ink

Oh, you
Oh

[Drop]

[Chorus]
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity Ink

[Verse]
Don't know when it started and I don't know when it ends
I feel so much pressure that I'm gonna get the bends
Ro-ro-rolling and controlling in your vicinity
That's why everybody calls me Mr. Infinity

[Chorus]
(If you wanna–)
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity Ink
Ink, Ink, Ink, Ink

Ink
Infinity, Infinity, Infinity, Infinity
Infinity, Infinity, Infinity, Infinity
Infinity, Infinity, Infinity, Infinity
Infinity, Infinity, Infinity, Infinity
Oh
Oh you

[Chorus]
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity
Oh, you know we are the Infinity

Ink, Ink, Ink

Oh

Oh, you

More on Genius


text:
# Blinds for Homekit / Web

For a long time we already had an automation for our blinds at home in the livingroom but at least twice a year, when there’s the change from summer to wintertime or the other way round, these things suck.

Where’s the manual? I already did this a dozen times, can’t be so difficult…

Turns out it is difficult. I believe these things are made to never be touched again.

So far so good. Since the broadening of home automation solutions, especially HomeKit there was the need to add other devices without an official certification. For this is homebridge, but after a quick’n’dirty search for a plug and play solution i didn’t find anything very useful. Next step, order a Raspberry Pi zero, a double relay, read somewhere about how easy it is to use gpio with Python.

## Update 2018-11-29 :

I just moved the Frontend into the main Repository so it’s much more transparent to see, whats going on there for everyone.

### First Steps

So Python is an interpreted language, wich is very nice for hacking like this because, whenever you save a change to the file, your system gets restarted.

My first approach was a flask boilerplate with some routes so you get an understanding on how this works there.

@app.route('/status')
def status()
return "something"

So the second step is how to make things blink.

Nice thing about the common relays you get on ebay or amazon or somewhere else, they all have a LED for the activated relay.

from gpiozero import LED
pinNumber = 27
roll = LED(pinNumber)
roll.on()
# is now on...

So if you’re a Sherlock you can do things like this:

@app.route('/turn-on')
def turnOn()
roll.on()
return "your LED is now on! aka something is moving!"

### Fine Tuning

Sometimes when is really hot outside or the sun is shining on your glossy screen, you don’t want to close the blinds completely but rather just 50%.

So for this case, the only known method to me is to make this over a time factor. Movement should be pretty constant, so moving 20% should require double the time of 10% moving. But because there is some difference between the upward and downward movement, you need two factors to calculate the time.

So this is how i made it work:

1. travel = int(abs(currentHeight - desiredHeight))
2. sleep(travel * config.upFactor) or config.downFactor

But because this isn’t the completly perfect method, i reset the currentHeight whenever I want it 0 or 100.

Lastly, Flask runs threaded on default, means for every new request it spins up a new thread. Since the HomeKit drag options makes a lot of calls when you move your finger slowly, i disabled this feature with app.run(debug=False, host='0.0.0.0', threaded=False). Flask also runs at localhost per default, so change that, too.

## Frontend

Because I already had the vue-cli installed, i started with the minimal boilerplate which ships with it, deleted everything unnecessary and hacked a mobile-optimized button layout together. I see there some potential to make it more usable but does the work for now..

## Recap: What I learned

My biggest gain with this little project is that python is really easy to pick up, build a prototype and play around with it. I definetly have no idea how Python performs in a production enviroment where it counts on speed and security. But as long as i treat my home network as a secure enviroment, this shouldn’t become a problem.

Second, Flask is a nice framework, easy to understand when you’d like to do something real quick. So a recommendation for a project like this.

Because there are a lot of people who aren’t using Apple’s HomeKit I build my first Vue.js Frontend for this case where you have two big buttons and a slider for the percental settings (Link below).

I played around with Vue for another project but it was a rather complex thing to start with it. I created a project with vue-cli and deleted everything i didn’t need, maybe not the best approach for such a simple thing, but I really like the approach to have your HTML, JS and CSS all in just one file. You can split up your Frontend in different submodules wich is very neat in my opinion!

The exported project embedded in the main repository, ready to use!

My repository


text: $-\left[\Delta G_{f}^{\circ} C H_{4}(g)+2 \Delta G_{f}^{\circ} O_{2}(g)\right]$ $2 N O(g)+O_{2}(g) \rightarrow 2 N O_{2}(g)$. What will be the direction of the reaction at this temperature and below this temperature and why? Look into the Sample Papers of Previous Years to get a hint of the kinds of questions asked in the exam. Hence $P b O$ can bereduced by carbon because $\Delta \mathrm{G}$ of couple reaction is $-v e$, (i) $C+O_{2} \rightarrow C O_{2} ; \Delta G^{\ominus}=-380 k J$, (ii) $4 A l+3 O_{2} \rightarrow 2 A l_{2} O_{3} ; \Delta G^{\ominus}=+22500 \mathrm{kJ}$, (iii) $2 P b+O_{2} \rightarrow 2 P b O ; \Delta G^{\Theta}=+120 \mathrm{kJ}$. The temperature of calorimeter rises from $294.05 K$ to $300.78 K .$ If the heat capacity of calorimeter is $8.93 \mathrm{kJK}^{-1}$, calculate the heat transferred to the calorimeter. The enthalpy change for the reaction: Calculate Gibbs energy change for the reaction is spontaneous or not. Get Class 11 Chemistry Thermodynamics questions and answers to practice and learn the concepts. Multiply eqn. Calculate $\Delta_{f} H^{\circ}$ for chloride ion from the following data: This is only a preview of the solution. $\Delta_{r} G^{\circ}=\Delta_{r} H^{\circ}-T \Delta_{r} S^{\circ}=0$ $\Delta G_{r}^{o}=\left[2 \Delta G_{f}^{o}\left(N O_{2}(g)\right)\right]-\left[2 \Delta G_{f}^{O}(N O(g))+\Delta G_{f}^{O}\left(O_{2}\right)\right]$ $=\frac{491.18 \mathrm{kJ} \mathrm{mol}^{-1}}{197.67 \times 10^{-3} \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}}=2484.8 \mathrm{K}$ Email: help@24houranswers.com $\Delta_{r} G^{o}=-8.1 k J m o l^{-1}, T=1000 K$ Sorry, there was a problem with your payment. (i) $\left.\quad \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\right]$ With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. For a reaction both $\Delta H$ and $\Delta S$ are positive. $\Delta H$ and $\Delta S$ for the reaction: Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. $\Delta H=\Delta E$ during a process which is carried out in a closed vessel $(\Delta v=0)$ or number of moles of gaseous products $=$ number of moles of gaseous reactants or the reaction does not involve any gaseous reactant or product. $O-H(g) \rightarrow O(g)+H(g) ; \Delta H=430 k J$ $2 P(s)+3 B r_{2}(l) \rightarrow 2 P B r_{3}(g) \Delta_{r} H^{\circ}=-243 k J m o l^{-1}$ Express the change in internal energy of a system when, (i) No heat is absorbed by the system from the surroundings, but work (, Show that for an ideal gas, the molar heat capacity under constant volume conditions is equal to 3/2, Calculate the number of $k J$ necessary to raise the temperature of $60.0 \mathrm{g}$ of aluminium from $35-55^{\circ} \mathrm{C} .$ Molarheat capacity of $A l$ is $24 J m o l^{-1} K^{-1}$. $\mathrm{SiH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SiO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ $\Delta H$ and $\Delta U$ are related as Treat heat capacity of water as the heat capacity of calorimeter and its content). These important questions will play significant role in clearing concepts of Chemistry. $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$ (i) $\quad \Delta S\Delta H)$, Q. $|=(102.6)-(173.10)=-70.50 k_{0} \mathrm{J} \mathrm{mol}^{-1}$ (2) Answer: Heat absorbed by the system, (q) = + 701 J Work done by the system â¦ (i) $\frac{1}{2} N_{2}(g)+\frac{1}{2} O_{2}(g) \rightarrow N O(g) ; \Delta_{r} H^{\circ}=90 k J \operatorname{mol}^{-1}$, (ii) $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) ; \Delta_{r} H^{\circ}=-74 \mathrm{kJ} \mathrm{mol}^{-1}$. (ii) Let us now calculate the $\Delta G$ value for reduction of $P b O$ (iii) by 2 and add to eqn. Enthalpy of solution of $N H_{4} N O_{3}=\frac{5.282}{20} \times 80$ $\Delta n=2-4=-2$ Predict the sign of entropy change for each of the following changes of state: Q. Calculate the standard entropy change for the reaction Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Continue without uploading, Attachhomework files Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Molar mass of octane $\left(C_{8} H_{18}\right)=(8 \times 12)+(18 \times 1)=114$ $\Delta H=\Delta_{f} H^{o}\left(H_{3} O^{+}\right)+\Delta_{f} H^{o}\left(C l^{-}\right)$ Clapeyron discovered an important fundamental equation which helps in finding the extensive application in one component, two phase system and by Clausius from second law of thermodynamics. Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...Sol. Here, we are given, $\Delta H_{\text {Reaction}}=\Sigma \Delta H^{\circ}$ $f$ (Products) $-\Sigma \Delta H^{\circ}$ f $(\text {Reactants})$, Q. $A+B \rightarrow C+D$ As no heat is absorbed by the system, the wall is adiabatic. 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text:
NSW Syllabuses

Mathematics K–10 - Stage 4 - Measurement and Geometry Angle Relationships

Outcomes

A student:

• MA4-1WM

communicates and connects mathematical ideas using appropriate terminology, diagrams and symbols

• MA4-2WM

applies appropriate mathematical techniques to solve problems

• MA4-3WM

recognises and explains mathematical relationships using reasoning

• MA4-18MG

identifies and uses angle relationships, including those related to transversals on sets of parallel lines

Content

• Students:
• Use the language, notation and conventions of geometry
• define, label and name points, lines and intervals using capital letters
• label the vertex and arms of an angle with capital letters
• label and name angles using $$\angle P$$ or $$\angle QPR$$ notation
• use the common conventions to indicate right angles and equal angles on diagrams
• Recognise the geometrical properties of angles at a point
• recognise that adjacent angles can form right angles, straight angles and angles of revolution (Communicating, Reasoning)
• identify and name perpendicular lines using the symbol for 'is perpendicular to' ($$\bot$$), eg $$AB \bot CD$$
• use the common conventions to indicate parallel lines on diagrams
• identify and name pairs of parallel lines using the symbol for 'is parallel to' $$\left( \,\parallel\, \right)$$, eg $$PQ\parallel RS$$
• define and identify 'transversals', including transversals of parallel lines
• identify, name and measure alternate angle pairs, corresponding angle pairs and co-interior angle pairs for two lines cut by a transversal
• use dynamic geometry software to investigate angle relationships formed by parallel lines and a transversal (Problem Solving, Reasoning)
• recognise the equal and supplementary angles formed when a pair of parallel lines is cut by a transversal
• Investigate conditions for two lines to be parallel (ACMMG164)
• use angle properties to identify parallel lines
• explain why two lines are either parallel or not parallel, giving a reason (Communicating, Reasoning)
• Solve simple numerical problems using reasoning (ACMMG164)
• find the sizes of unknown angles embedded in diagrams using angle relationships, including angles at a point and angles associated with parallel lines, giving reasons
• explain how the size of an unknown angle was calculated (Communicating, Reasoning)

Background Information

Dynamic geometry software and prepared applets are useful tools for investigating angle relationships; angles and lines can be dragged to new positions while angle measurements update automatically.

Students could explore the results relating to angles associated with parallel lines cut by a transversal by starting with corresponding angles and moving one vertex and all four angles to the other vertex by a translation. The other two results then follow, using vertically opposite angles and angles on a straight line. Alternatively, the equality of the alternate angles can be seen by rotation about the midpoint of the transversal.

Students should give reasons when finding the sizes of unknown angles. For some students, formal setting out could be introduced. For example,

$$\angle ABQ = 70^{\circ}$$ (corresponding angles, $$AC \parallel PR$$).

In his calculation of the circumference of the Earth, the Greek mathematician, geographer and astronomer Eratosthenes (c276–c194 BC) used parallel line results.

Purpose/Relevance of Substrand

The development of knowledge and understanding of angle relationships, including the associated terminology, notation and conventions, is of fundamental importance in developing an appropriate level of knowledge, skills and understanding in geometry. Angle relationships and their application play an integral role in students learning to analyse geometry problems and developing geometric and deductive reasoning skills, as well as problem-solving skills. Angle relationships are key to the geometry that is important in the work of architects, engineers, designers, builders, physicists, land surveyors, etc, as well as the geometry that is common and important in everyday situations, such as in nature, sports, buildings, astronomy, art, etc.

Language

Students in Stage 4 should write geometrical reasons without the use of abbreviations to assist them in learning new terminology, and in understanding and retaining geometrical concepts, eg 'When a transversal cuts parallel lines, the co-interior angles formed are supplementary'.

Some students may find the use of the terms 'complementary' and 'supplementary' (adjectives) and 'complement' and 'supplement' (nouns) difficult. Teachers should model the use of these terms in sentences, both verbally and in written form, eg, '50° and 40° are complementary angles', 'The complement of 50° is 40°'.

Students should be aware that complementary and supplementary angles may or may not be adjacent.

National Numeracy Learning Progression links to this Mathematics outcome

When working towards the outcome MA4‑18MG the sub-elements (and levels) of Number patterns and algebraic thinking (NPA7-NPA8) and Understanding geometric properties (UGP5-UGP6) describe observable behaviours that can aid teachers in making evidence-based decisions about student development and future learning.

The progression sub-elements and indicators can be viewed by accessing the National Numeracy Learning Progression.


text:
## smiley2015 Group Title Simplify the following expression: (5x2 + 3x + 4) − (2x2 + 5x − 1). If the final answer is written in the form Ax2 + Bx + C, what is the value of A? The 2 be hide 5x is squared and so is the one by 2x one year ago one year ago

1. ryan123345 Group Title

Distribute the negative

2. smiley2015 Group Title

So -1?

3. jazy Group Title

I have terrible drawing skills, sorry.

4. Claudilovely Group Title

|dw:1363207136675:dw|

5. smiley2015 Group Title

So 3x - 5 - 5x + 1

6. jazy Group Title

I meant:|dw:1363207208912:dw|

7. jazy Group Title

Very close. What is -1 times 1?

8. smiley2015 Group Title

-1

9. jazy Group Title

Right! So your equation would be 3x - 5 - 5x - 1. Can you combine the like terms?

10. smiley2015 Group Title

Yes. the x

11. jazy Group Title

Lets rearrange it so its easier to work with. Remember to keep the symbol in front of each term. 3x - 5x - 5 - 1 First off, what is 3x - 5x?

12. smiley2015 Group Title

2x

13. GiGIgo3 Group Title

-2x-6

14. jazy Group Title

Close, it would be a negative 2.

15. jazy Group Title

16. smiley2015 Group Title

Ok

17. jazy Group Title

So now we have: -2x - 5 - 1 -5 - 1 = ?

18. smiley2015 Group Title

6

19. jazy Group Title

Negative 6. (: So our expression is: -2x - 6

20. smiley2015 Group Title

Ok. Thanks!

21. jazy Group Title

You're welcome! Oh and:$\Huge \mathcal{\text{Welcome To OpenStudy} ☺}$

22. smiley2015 Group Title

Thanks! can you help me some more

23. jazy Group Title

Sure thing!

24. smiley2015 Group Title

Simplify the following expression: (5x2 + 3x + 4) − (2x2 + 5x − 1). If the final answer is written in the form Ax2 + Bx + C, what is the value of A?

25. smiley2015 Group Title

The 2 be hide 5x is squared and so is the one by 2x

26. jazy Group Title

$(5x^2 + 3x + 4) - (2x^2 + 5x - 1)$We have to distribute the -1. $(-1)\stackrel{\Huge\curvearrowright}(2x^2 + 5x - 1)$-1 * 2x^1 = -2x^2 -1 * 5x = -5x -1 * -1 = 1 Now we have: $5x^2 + 3x + 4 - 2x^2 - 5x + 1$Combine like terms. 5x^2 - 2x^2 = 3x^2 3x - 5x = -2x 4 + 1 = 5 And we end up with: $\large 3x^2 - 2x + 5$

27. jazy Group Title

$\large Ax^2 + Bx + C$ $\large 3x^2 - 2x + 5$ So what is A?

28. smiley2015 Group Title

3x2

29. jazy Group Title

Just the 3. (:

30. smiley2015 Group Title

Thanks! I have some more if that is ok?

31. jazy Group Title

I've gtg right now, sorry. :/ Maybe @hartnn or @Luis_Rivera can continue to help you. I hope I helped!

32. smiley2015 Group Title

You did thanks!

33. hartnn Group Title

one suggestion though, make new post for new question. so that many others can help you too :)

34. smiley2015 Group Title

How do I do that?

35. hartnn Group Title

you can 'close' this question .

36. smiley2015 Group Title

Ok. Thanks

37. hartnn Group Title

good ! :) welcome ^_^


text:
# Measurement of charm and beauty production at central rapidity versus charged-particle multiplicity in proton-proton collisions at s = 7 $\sqrt{s}=7$ TeV

29 September 2015

Abstract: Prompt D meson and non-prompt J/ ψ yields are studied as a function of the multiplicity of charged particles produced in inelastic proton-proton collisions at a centre-of-mass energy of s = 7 $\sqrt{s}=7$ TeV. The results are reported as a ratio between yields in a given multiplicity interval normalised to the multiplicity-integrated ones (relative yields). They are shown as a function of the multiplicity of charged particles normalised to the average value for inelastic collisions (relative charged-particle multiplicity). D 0 , D + and D *+ mesons are measured in five p T intervals from 1 GeV/ c to 20 GeV/ c and for | y | < 0.5 via their hadronic decays. The D-meson relative yield is found to increase with increasing charged-particle multiplicity. For events with multiplicity six times higher than the average multiplicity of inelastic collisions, a yield enhancement of a factor about 15 relative to the multiplicity-integrated yield in inelastic collisions is observed. The yield enhancement is independent of transverse momentum within the uncertainties of the measurement. The D 0 -meson relative yield is also measured as a function of the relative multiplicity at forward pseudo-rapidity. The non-prompt J/ ψ , i.e. the B hadron, contribution to the inclusive J/ ψ production is measured in the di-electron decay channel at central rapidity. It is evaluated for p T > 1.3 GeV/c and | y | < 0.9, and extrapolated to p T > 0. The fraction of non-prompt J/ ψ in the inclusive J/ ψ yields shows no dependence on the charged-particle multiplicity at central rapidity. Charm and beauty hadron relative yields exhibit a similar increase with increasing charged-particle multiplicity. The measurements are compared to PYTHIA 8, EPOS 3 and percolation calculations.

Published in: JHEP 1509 (2015) 148


text:
# Tag Info

4

Let's consider the outer two loops first. For a fixed value of $i$, the number of iterations of the middle loop is exactly $n-1 - (i+1) + 1 = n - i -1$. Since $i$ ranges from $0$ to $n-1$ (in the outer loop), the overall number of iterations of the middle loop is: $$\sum_{i=0}^{n-1} (n - i -1) = \sum_{i=0}^{n-1} i = \frac{n(n-1)}{2}.$$ For each of those ...

0

When measuring comparisons, we usually only charge comparisons of elements of the array (more accurately, we charge comparisons for the datatype of array elements, but not of array indices). In the example of linear search, at each iteration you are comparing one element of the array to the input element being searched. So in total, you are making exactly $n$...

1

$\sum_{i=1}^k 2^i=\Theta(2^k)$ and you have $k\approx\log n$, so the sum is $\Theta(2^{\log n})=\Theta(n)$.

0

Process of elimination A, perhaps, easier method to derive the time complexity is by process of elimination. This is especially the case if you get this type of question in a multiple choice exam. That said, this method is far from as rigorous as OmG's answer. Okay, we know that A is upper-bounded by O(n), as we increment $s$ by more than 1 in each ...

0

After the first sqrt(n) iterations, s is increased by more than sqrt(n) in each iteration. Therefore at most 2 sqrt(n) iterations are needed to make s > n. On the other hand, in the first sqrt(n) iterations, s is increased at most by sqrt(n) in each iteration, so sqrt(n) iterations are not enough. This means the number of iterations is $\Theta(n^{1/2})$.

4

Each time add $i$ to $s$ and increase $i$ by one, up to reach to $n$. Hence, if you find the $k$ such that $s = 0 + 1 + 2 + ... + k$ be equal to $n$, you can find the number of running loop. As $1 + 2 + \ldots + k = \frac{k(k+1)}{2}$, you need to solve this equation $\frac{k(k+1)}{2} = n$. $$k^2 + k -2n = 0 \Rightarrow k = \frac{-1 + \sqrt{1+8n}}{2} = \... 1 Let's take a look at the case of size 7 first. Here, we want to show a linear upper bound for T(n). Thus, we choose the recurrence relation T(n) \leq T(n / 7) + T(5 / 7 n) + dn (remember, it is an upper bound). We guess that the solution is of the form T(n) \leq cn for some constant c and prove it with induction:$$T(n) \leq T(n / 7) + T(5 / 7 \cdot ...

1

1 For the first case, you had the right idea, but just had some algebra mistakes. for i=1..n j=1 while j*j <= i: j = j + 1 Let $T(n)$ be the time complexity. $$T(n) = \sum_{i=1}^n\sum_{j=1}^\sqrt{i}1\leq \sum_{i=1}^n\sqrt{n}=\leq n^{3/2}$$ $$= O(n^{3/2})$$ 2 I'm assuming you meant the pseudocode below since it is more analogous to ...

0

$c_1 = 56$ works for all n. Let n >= 265, then 200 <= $n^2$, $264n^2 + 200 <= n^3$, the logarithm is less than 3 log n, $56n^2 + 106n + 48 <= 56n^2 + n^2 + n^2 = 58n^2$, so you can pick $c_2 = 58*3 = 174. 0 It's sometimes more convenient to work with the limit definitions. To prove the statement, you can break it down into proving that$f(n) = O(g(n))$and that$f(n) = \Omega(g(n))$. In your case, it's straightforward to compute$\lim_{n \to \infty} f(n)/g(n)$. I believe that this will turn out to be a positive three-digit constant less than infinity and ... 2 For all$n\geq 1$,$56n^2+106n+48> 56n^2> n^2$and$\log (264n^2+200)> \log 264n^2>\log n$, so $$(56n^2+106n+48)\log(264n^2+200) > n^2\log n\,,$$ i.e., you can take$c_1=1$. Also for all$n\geq 1$,$56n^2+106n+48\leq 56n^2+106n^2+48n^2 = 210n^2$and, for all$n\geq 200$,$264n+200 < 265n$so$\log(264n^2+200) < \log 265n^2 = 2\log n + \...

1

That's right. This for loop stops when: $$3^i < n^3 \rightarrow i<3\log(n)$$ Which implies the complexity of this for loop is $O(\log(n))$.

Top 50 recent answers are included


text:
# Faustmann's formula

Faustmann's formula, or the Faustmann Model, gives the present value of the income stream for forest rotation. It was derived by the German forester Martin Faustmann in 1849.

The rotation problem, deciding when to cut down the forest, means solving the problem of maximising Faustmann's formula and this was solved by Bertil Ohlin in 1921 to become the Faustmann-Ohlin theorem, although other German foresters were aware of the correct solution in 1860.[1]

ƒ(T) is the stock of timber at time T
p the price of timber and is constant
which implies that the value of the forest at time T is pf(T)
r is the discount rate and is also constant.

The Faustmann formula is as follows:

${\displaystyle PV=pf(T)\exp(-rT)\cdot {(1+\exp(-rT)+\exp(-2rT)+\cdots )}={\frac {pf(T)}{\exp(rT)-1}}.}$

From this formula two theorems are interpreted:

The optimal time to cut the forest is when the time rate of change of its value is equal to interest on the value of the forest plus the interest on the value of the land.[2]
The optimal time to cut is when the time rate of change of its value is equal to the interest rate modified by land rent.[2]

## Notes

1. ^ John Cunningham Wood (1995). Bertil Ohlin: Critical Assessments. Routledge. ISBN 978-0415074926.
2. ^ a b "The Faustmann Model (Part I)". Introduction to Forestry, Forest Policy, and Economics. Archived from the original on 2011-12-29. Retrieved 2013-06-08.

## References

• Erickson, J.D., Chapman, D., Fahey, T.J. and M.J. Christ (1999). "Nonrenewability in Forest Rotations: Implications for Economic and Ecological Sustainability." Ecological Economics 31(1): 91-106.


text:
POPL 2019
Sun 13 - Sat 19 January 2019 Cascais, Portugal
Wed 16 Jan 2019 14:29 - 14:51 at Sala II - Categories Chair(s): Nicolas Tabareau

Game semantics is the art of interpreting types as \emph{games} and programs as strategies interacting in space and time with their environment. In order to reflect the interactive behavior of programs, strategies are required to follow specific scheduling policies. Typically, in the case of a purely sequential programming language, the program (Player) and its environment (Opponent) will play one after the other, in a strictly alternating way. On the other hand, in the case of a concurrent language, Player and Opponent will be allowed to play several moves in a row, in a non-alternating way. In both cases, the scheduling policy is designed very carefully in order to ensure that the strategies synchronize properly and compose well when plugged together. A longstanding conceptual problem has been to understand when and why a given scheduling policy works and is compositional in that sense. In this paper, we exhibit a number of simple and fundamental combinatorial structures which ensure that a given scheduling policy encoded as \emph{synchronization template} defines a symmetric monoidal closed (and in fact $\ast$-autonomous) bicategory of games, strategies and simulations. To that purpose, we choose to work at a very general level, and illustrate our method by constructing two template game models of linear logic with different flavors (alternating and non-alternating) using the same categorical combinatorics, performed in the category of small categories. As a whole, the paper may be seen as a hymn in praise of synchronization, building on the notion of synchronization algebra in process calculi and adapting it smoothly to programming language semantics, using a combination of ideas at the converging point of game semantics and categorical algebra.

 Categorical Combinatorics of Scheduling and Synchronization in Game Semantics (popl-january-2019.pdf) 1.88MiB

#### Wed 16 Jan Times are displayed in time zone: Belfast change

 13:45 - 14:51 CategoriesResearch Papers at Sala II Chair(s): Nicolas TabareauInria 13:4522mTalk Familial Monads and Structural Operational SemanticsResearch PapersTom HirschowitzUniv. Grenoble Alpes, Univ. Savoie Mont Blanc, CNRS, LAMA, 73000 Chambéry Link to publication DOI Media Attached File Attached 14:0722mTalk Bindings as Bounded Natural FunctorsResearch PapersJasmin BlanchetteVrije Universiteit Amsterdam, Lorenzo GheriMiddlesex University London, Andrei PopescuMiddlesex University, London, Dmitriy TraytelETH Zurich Link to publication DOI Media Attached File Attached 14:2922mTalk Categorical Combinatorics of Scheduling and Synchronization in Game SemanticsResearch PapersPaul-André MelliesCNRS and University Paris Diderot Link to publication DOI Media Attached File Attached


text:
# Mean, Median and Mode of a QVector

I would like some help condensing this code down to size. I am trying to utilize Qt's container classes as efficiently as possible. I do not believe this code reflects my objective.

#include <QCoreApplication>
#include <QDebug>

int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);

QVector<int> vect;

vect.append(0);
vect.append(0);
vect.append(0);
vect.append(0);
vect.append(1);
vect.append(1);
vect.append(2);
vect.append(1);

QVectorIterator<int> ivect(vect);

double median = 0;
if(vect.count() % 2 == 0){

median =
static_cast<double>((vect[vect.count() / 2 - 1] + vect[vect.count() / 2])) / 2;
}
else{

median = vect[vect.count() / 2];
}

int sum = 0;
while(ivect.hasNext()){

sum += ivect.next();
}

double mean = static_cast<double>(sum) / vect.count();

QSet<int> set;

ivect.toFront();
while(ivect.hasNext())
{
set << ivect.next();
}

QMap<int, int> map;

QSetIterator<int> iset(set);

ivect.toFront();
while(iset.hasNext())
{
int count = 0;
while(ivect.hasNext())
{
if(iset.peekNext() == ivect.next())
{
count++;
}
}
ivect.toFront();
map.insert(iset.next(), count);
}

QMapIterator<int, int> imap(map);

int maxValue = 0;
QString mode;
while(imap.hasNext())
{
imap.next();
if(imap.value() > maxValue)
{
maxValue = imap.value();
mode = QString::number(imap.key());
}
else if(imap.value() == maxValue)
{
mode = "NULL";
}
}

qDebug() << set;
qDebug() << map;
qDebug() << "Mean:" << mean;
qDebug() << "Median:" << median;
qDebug() << "Mode frequency:" << maxValue;
qDebug() << "Mode:" << mode;

return a.exec();
}

You might have decided to use Qt, but that doesn't mean you must forsake all that standard C++ has to offer.
Actually, Qt does its best to be as compatible as possible, just to make that possible.

So, let's see:

1. Use an initializer-list for the vector.
2. Don't be afraid of using a conditional-operator where it makes things easier or at least shorter without becoming cryptic.
3. const-qualifying as much as possible helps catch errors.
4. Your median-calculation (though not the median with that example-data) is wrong, you have to sort first.
5. You really want to accumulate in a double using the standard algorithm, for brevity and correctness. As a bonus, you can make it const too.
6. As the vector is sorted, the problem of finding the mode is equivalent to finding the longest consecutive run, which can be solved far more efficiently than with a map.
7. median and mean assume a non-empty input, not so mode, though that will only calculate the first mode.

(I only proved it right, I didn't run it.)

#include <QCoreApplication>
#include <QDebug>
#include <algorithm>
#include <numeric>

int main(int argc, char *argv[])
{
QCoreApplication a(argc, argv);
QVector<int> vect = { 0, 0, 0, 0, 1, 1, 2, 1 };
std::sort(vect.begin(), vect.end());
const auto median = vect.size()%2
? vect[vect.size() / 2]
: ((double)vect[vect.size() / 2 - 1] + vect[vect.size() / 2]) * .5;
const auto mean = std::accumulate(vect.begin(), vect.end(), .0) / vect.size();
const auto mode = [&]{
size_t bestC = 0, tmpC = 0;
int    bestV = 0, tmpV = 0;
auto apply = [&](int next){
if(tmpC > bestC)
bestC = tmpC, bestV = tmpV;
tmpC = 1, tmpV = next;
};
for(int x : vect)
if(c == tmpV)
tmpC++;
else
apply(x);
apply(0);
return bestC ? QString::number(bestV) : "NULL";
}();
qDebug() << "Mean:" << mean;
qDebug() << "Median:" << median;
qDebug() << "Mode:" << mode;
return a.exec();
}

• accululate is defined in numeric – vlad_tepesch Mar 6 '17 at 11:29

Sorted vs Unsorted

Your QVector sequence contains an unsorted sequence of values. Your median calculation has a precondition that your sequence is sorted. Consider the sequence 1, 0, 1, what is the median?

Small, Focused, and Testable Functions

Prefer writing functions that employ the single responsibility principle. Splitting out median and testing it with different sequence types would have pointed out a problem. How about calculating the mean of an empty container? Is division by zero a problem?

Standard Algorithms

Being able to understand when you should use standard algorithms and applying them is very important. Mean can be written using std::accumulate(). Median can be written in terms of std::nth_element(). Mode can be written using boost::bimap (or utilize std::unordered_map+std::multimap) and std::upper_bound() with std::transform() to return the modes of a sequence. Not all sequences are uni-modal.


text:
Ω 1

Crossposted from the AI Alignment Forum. May contain more technical jargon than usual.

(follow-up to: A correlated analogue to reflective oracles)

Motivation

Suppose players are playing in a correlated equilibrium using a reflective oracle distribution. How does the equilibrium they play in vary as a function of the parameters of the game, or of the players' policies? It turns out that the set of equilibria is a Kakutani map of the parameters to the game. This is a lot like it being a continuous map.

This might make it possible for players to reason about the effects of their policy on the equilibrium that they play (since the equilibrium is now a Kakutani map of the players' policies).

Definitions

Let be some natural number. We will consider reflective oracle distributions whose queries are parameterized on some vector in .

To do this, let the Turing machines , instead of outputting a raw query, output a continuous function from to the query. (The details of representing continuous functions don't seem that important). The reflectivity condition on oracle distributions is now relative to (since the queries depend on ).

Define the map

which maps the parameters to the set of reflective oracle distributions for .

Theorem: is a Kakutani map.

Proof:

From the previous post, we have:

• For each , is nonempty (Theorem 1)
• For each , is convex (Theorem 2)

So it is sufficient to show that has a closed graph.

Let be an infinite sequence of values converging to . Let be an infinite sequence of oracle distributions in converging to . Assume that for each natural , is reflective relative to . We will show that is reflective relative to ; this is sufficient to show that has a closed graph.

Let . Let be such that . Then for some . By convergence, there is some such that for all , . By reflectivity of each relative to , we have, for each ,

Let . Rewriting the above statement:

Consider the set . This is the intersection of a finite number of sets of the form . Each of these sets is closed because is continuous. Therefore the set is closed.

In total, this is sufficient to show , as desired.

An immediate consequence of this theorem is that the set of correlated equilibria of a normal-form game is a Kakutani map of the parameters of the game.

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# NLP : Rules for chunking Verb Phrases

I have read a lot of documentation surrounding NP chunking but what about Verb Phrases? Has there been a fixed set of rules for VP chunking?


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# A question about proving maximal ideal

Let $R$ be the subring of the real numbers such that $R=\left\{a+b\sqrt{2}:a,b \in \Bbb Z\right\}$ Let $M$ be the ideal in $R$ given by $M=\left\{a+b\sqrt{2}:\text{ a and b are divisible by 5}\right\}$ Prove that $M$is a maximal ideal of $R$.

I was thinking that suppose $M \subseteq P\subseteq R$ and prove that $1\in P$ but I don't know how to prove that. Hope somebody can help me with this

Assume that $M$ is not maximal. So $M\subset P\subset R$. Let $a\in P$, with $a$ an integer such that $5$ does not divide $a$. We have that $5\in M\subset P$. By the ideal property this means that $5+a\in P$. But we also have that $\gcd(5,a)=1$. So we know there exists an integer combination of $5$ and $a$ summing to 1. Say

$$5n+am=1$$

$5n,am\in P$ Therefore, since an ideal is closed under addition we have that $1\in P$ and therefore, $P=R$. Therefore, no proper ideal of $R$ contains $M$. Also, $M$ is not the whole ring because it doesn't contain $1$. Thus $M$ is maximal. I feel like there may be a hole to be found, so I'm open to suggestions.

• In fact I really don't think my answer works, because $5$ is not in $M$. – monroej Jan 21 '14 at 5:12
• yes it is: $\;5=5+0\sqrt2\;,\;\;5\mid 5\;,\;5\mid 0\;$ – DonAntonio Jan 21 '14 at 5:13
• Ok. I don't know why I was thinking $5$ didn't divide zero...Thanks. – monroej Jan 21 '14 at 5:13
• I don't think you can assume such an $a$ exists. – Dustan Levenstein Jan 21 '14 at 5:44
• If no such $a$ existed, then $M$ would be maximal from the start. – monroej Jan 21 '14 at 5:45

Try the following :

1. Prove that $(x^2-2) \in \mathbb{Z}_5[x]$ is irreducible, and let $F = \mathbb{Z}_5[x]/(x^2-2)$ be the corresponding quotient, which is a field.

2. Define a surjective homomorphism $\varphi : R\to F$ such that $\ker\varphi = M$.


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# Liner Interpolation

In mathematics, linear interpolation is understood as a method of curve fitting using linear polynomials. A polynomial is just an expression that consists of variables along with their co-efficients. An example of a polynomial would be:

1x+2y&sup2

In the above equation, the variable x has 1 as its coefficient, the variable y has 2 as its coefficient. Let us assume that we have a set of linear data points pointing to the x and y-axis:

1x = {2,5,7,10,11,15,17,19,21,25}
2y = {3,6,8,10,14,16,19,21,23,26}

Given a value x, we can find the value y. For example., for x = 2, the corresponding y is 3, for x = 11, the corresponding y is 14. Now we are interested in finding the value of x and y within our set of data points.

What is the value of y for x = 13 or what is the value of x for y = 13. Since the data points for both the x and y co-ordinates are linear, we can use the linear interpolation technique to find out the values for the desired data points. Geometrically, the linear interpolant is just a straight line between two known data points. Let us assume that our two data points are represented as (x0,y0) and (x1,y1). We are interested in finding the values of x and y that lies between these data points. Since we are in the linear boundary, geometrically speaking the ratio of the difference between x and y co-ordinates between two data points are equal.

Let us try to ascertain this with some numbers. Let our data points (x0,y0) be (2,3) and (x1,y1) be (5,6). The ratio y1/x1 should be equal to the ration y0/x0 and in our case, 6/5 == 3/2. Extending this idea, we can now say that the ratio of the difference between two data points should be geometrically equal between the two co-ordinates. If we are interested in a data point (x,y) which lies in-between (x0,y0) and (x1,y1), our formula now becomes:

1y - y0 / x - x0 = y1 - y0 / x1 - x0 which when solving for y, we get the following equation:
2y = y0 + (y1 - y0) * (x - x0) / (x1 - x0)

We now have the formula for linear interpolation, let's now implement that in Scala. Let us represent our data points as a Scala List:

1  val x = List(2,5,7,10,11,15,17,19,21,25)
2  val y = List(3,6,8,10,14,16,19,21,23,26)
3  val zippedDataPoints = x zip y

We define a function that takes these two zipped List of data points and an additional parameter for x. The function uses the formula to find the value of y for a given x. Here is the function:

 1def interpolate(givenX: Int, dataPoints: List[(Int, Int)]) = {
2  // edge cases
4  val (lastX, lastY) = dataPoints.last
5  givenX match {
6    case a if givenX >= lastX => lastY
8    case c => {
9      val (lower,upper) = dataPoints.span { case (x,y) => x < givenX }
10      val (x0,y0) = lower.last
12      // finally our formula!
13      y0 + (y1 - y0) * (givenX - x0) / (x1 - x0)
14    }
15  }
16}

Let's now write some tests to test our interpolation (using ScalaTest):

1  interpolate(2, zippedDataPoints)  == 3
2  interpolate(4, zippedDataPoints)  == 5
3  interpolate(27, zippedDataPoints) == 26

So there it is! we have our linear interpolation implemented using Scala!


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# Agreeing to disagree “simple” example

I'm looking at Aumann's work "Agreeing to Disagree", and trying to understand the very first numerical example.

So, the paper starts with definitions

[L]et $$(\Omega, \mathcal{B}, p)$$ be a probability space, $$\mathscr{P}_1$$ and $$\mathscr{P}_2$$ [information] partitions of $$\Omega$$. [I]f the true state of the world is $$\omega$$, then $$i$$ is informed of that element $$P_i(\omega)$$ of $$\mathscr{P}_i$$ that contains $$\omega$$.

Let $$A$$ be an event, and let $$q_i$$ denote the posterior probability $$p(A | \mathscr{P}_i)$$ of $$A$$ given $$i$$'s information; i.e. if $$\omega \in \Omega$$, then $$q_i(\omega) = p(A \cap P_i(\omega)) / p(P_i(\omega))$$.

Then comes a simple example, that I fails to understand.

Suppose $$\Omega$$ has $$4$$ elements $$\alpha, \beta, \gamma, \delta$$ of equal (prior) probability, $$\mathscr{P}_1 = \{\alpha\beta, \gamma\delta\}$$, $$\mathscr{P}_2 = \{\alpha\beta\gamma, \delta\}$$, $$A = \alpha\delta$$, and $$\omega = \alpha$$. Then $$1$$ knows that $$q_2$$ is $$\frac{1}{3}$$, and $$2$$ knows that $$q_1$$ is $$\frac{1}{2}$$; but $$2$$ thinks that $$1$$ may not know what $$q_2$$ is ($$\frac{1}{3}$$ or $$1$$).

My problem starts with notation, what does $$\alpha\beta\gamma$$ mean? Is it $$\{\alpha, \beta, \gamma\}$$?

After I suppose this, I'm able to understand $$\frac{1}{3}$$ and $$\frac{1}{2}$$ part, then "but ..." sentence is totally cryptic. Why $$2$$ thinks this?

• I agree to disagree this way to introduce bayes' theorem :) I share your difficulties to grasp the meaning of the example. – Peter Mar 7 at 10:13


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/ Experiment-HEP arXiv:1202.1260

Search for anomalous production of multiple leptons in association with $W$ and $Z$ bosons at CDF

Pages: 17

Abstract: This paper presents a search for anomalous production of multiple low-energy leptons in association with a W or Z boson using events collected at the CDF experiment corresponding to 5.1 fb$^{-1}$ of integrated luminosity. This search is sensitive to a wide range of topologies with low-momentum leptons, including those with the leptons near one another. The observed rates of production of additional electrons and muons are compared with the standard model predictions. No indications of phenomena beyond the standard model are found. A 95% confidence level limit is presented on the production cross section for a benchmark model of supersymmetric hidden-valley Higgs production. Particle identification efficiencies are also provided to enable the calculation of limits on additional models.

Note: * Temporary entry *

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# Constant in the estimate on the Green's function of the Laplacian

Given the Laplacian associated to a Riemannian manifold $$(M^n, g)$$, there is a Green's function $$G(p,q): M \times M \to \mathbb{R}$$ that satisfies an inequality of the form $$|G(p,q)| \leq Ad(p,q)^{2-n}$$ where $$A$$ depends on the metric $$g$$ in some way.

I was wondering if it is known how this constant $$A$$ depends on the Riemannian metric? In particular, it would be nice to know how the estimate changes under conformal change of the metric.

This is of course impossible in general since the new distance function $$d'(p,q)$$ could bear no resemblance to the former, but say the conformal change is "nice" enough so that $$C^{-1}d(p,q) \leq d'(p,q) \leq Cd(p,q)$$ where $$C$$ depends on the conformal function only. Then what can be said?