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# Object-Oriented Concepts

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## Object-Oriented Concepts

14th May, 2006 @ 11:20pm GMT

Object-oriented programming is still a relatively new technology in the world of web development; and, as such, some of the concepts are widely misunderstood by many capable web developers. This can be best illustrated by the recent boom of interest in the Javascript object literal.

As Senior Web Developer at Rentokil Initial, I am required to be technical lead within the team. This also means making sure the technology is understood and used correctly. With this in mind, I’ve recently noticed that many of my team-mates obviously don’t understand the underlying concepts of OOP. For this reason, I’ve prepared this short tutorial on the concepts - to be followed by more detailed posts on execution in Javascript and in PHP.

### What is Object-Oriented programming?

Object-oriented programming can best be describe as a programming paradigm; a methodology of programming adopted by a programmer. Another example of a programming paradigm would be procedural (or sometimes, imperative) programming.

Procedural programming involves viewing a computer program as a sequential list of computational steps (or procedures) to be carried out. In procedural programming, the list of procedures can be further broken down into subroutines and functions (not to be confused with mathematical functions which are used within functional programming).

This is very much an “old skool” approach to programming and is adopted by many high-level languages such as BASIC. It’s also worth pointing out that most Web-based languages, such as PHP, ASP and Javascript, can be written using this approach - in fact when I first started out as a web developer, I was using ASP script, PHP and Javascript in this very manner, blissfully unaware of the OOP potential there-in.

In OOP, a program is seen as comprising a collection of individual modules, or objects, that act on each other. Each of these objects could be seen as an independent program in itself, with a distinct role or responsibility.

OOP provides greater flexibility and easier maintainance across large systems and can sometimes make understanding and analysing complex procedures a lot easier.

It’s worth noting, at this point, that most OOP languages (such as C++ or Java) are stateful where as often procedural programming languages are stateless.

### Enemy of the State

In computer science, a “state” is a particular set of instructions which will be executed in response to the machine’s input. An information system or protocol that relies upon state is said to be stateful. One that does not is said to be stateless.

OOP PHP is stateless because each script must be called on a page refresh where as Javascript is stateful because it can make use of event listeners - specific input - and the wonders of XMLHttpRequest - thus negating the need for refreshes.

Phew. That was all quite technical and filled with emphasis wasn’t it? OK, so let’s move on…

### Objects and Classes

The term “Object,” that gives OOP it’s name, refers to a conceptual object that represents an item in our program or system. This could be anything from an online form or a computer file, to a real world object such as a car.

This representation consists of attributes - the characteristics of our object; and methods - a set of functions and calculations that are either performed to modify the object itself, or are involved in some external effect.

The term “Class” represents the definition (or classification - class) of our object. For example, if we were to write a class called “car”, we could create any number of instances of that class - say “Porsche”, “Ferrari” and “Jaguar”. Each of these instances is an Object. This illustrates that a class is effectively a set of objects that all share common attributes.

### Keeping it “in the family”

One of the greatest advantages to using objects is encapsulation. This basically means that data within an object is only available/modifiable via the object’s methods - this is generally known as the interface of the object.

This resultant limitation of scope allows an object-oriented programmer the freedom to declare attributes, variables and methods without having to worry about clashes with those in other objects.

Encapsulation also means that, as long as we don’t alter the interface, we can change how an object works (to increase performance, add functionality etc) without affecting the rest of our system.

Now let’s take a look at how objects and classes can relate to each other.

Note: From this point on, the concepts are going to get a little more complicated. Please bear in mind that I’m writing an introduction here - I fully intend to go over these concepts again (with examples) whilst applying them to Javascript and PHP in the tutorials to follow.

### Characteristics of a Class

When writing a class, there are a number of characteristics that are worth taking into account:

#### Constructor

The constructor of a class is a special operation that is run upon instantiation - when an object is created. They are often distinguished by having the same name as the class itself. It’s main purpose is to set-up the attributes of a class and to establish the class invariant - to basically make sure that the attributes of the class conform to the class interface. It cannot have a return type.

A properly written constructor should never contain any functionality that could fail, thus leaving the object in an invalid state.

#### Destructor

The Destructor of a class is the opposite of the constructor in that it is run when an object is destroyed. It’s function is to clean up and to free the resources which were used by the object during run-time and unlink it from other objects or resources. This should result in the invalidation of any references in the process.

### Relationships

There are a number of relationships that can be used when interaction is needed between objects. These are as follows:

#### Inheritance

Inheritance allows a (sub)class to inherit all the attributes and methods of a parent class - in effect, extending the parent. It can best be described as an “is a” relationship. For instance, if we had a class of “fruit”, we could extend it by defining classes of “apple”, “orange” and “banana”. Each of these subclasses could be described in the following way:

apple "is a" fruitorange "is a" fruitbanana "is a" fruit

Because each of our subclasses extends the “fruit” class, it has all the attributes and methods of a “fruit” plus any specific characteristics of it’s own.

Here’s a more web-development-oriented example using form elements and presented in pseudocode:

class formElement{  Attributes:    id    name    class}class input extends formElement{  Attributes:    value    type}class textarea extends formElement{  Attributes:    cols    rows}

In this example the two classes “input” and “textarea” have inherited from “formElement”. This means they inherit formElement’s attributes like so:

input is a formElementinput{  Attributes:    id    name    class    value    type}textarea is a formElementtextarea{  Attributes:    id    name    class    cols    rows}

However, as the parent (super) class, formElement stays exactly the same:

formElement{  Attributes:    id    name    class}

As you can imagine, this relationship can be incredibally useful. Some languages even allow multiple inheritance where a class can have more than one parent (super) class. Sadly this isn’t the case in either Javascript or PHP, however it is possible to obtain the same effect using other types of relationship.

#### Composition - Association and Aggregation

Composition is a slightly different sort of relationship - this is where it could be said that a class was “composed” of other classes. For instance, a wall is “composed” of bricks and a molecule is “composed” of atoms. Neither of these examples could be described as inheritance - the statement, “a wall is a brick” simply isn’t true. Composition can be described as “has a” and “uses a” relationships; a wall “has a” brick or a wall “uses a” brick.

Let’s take a look at a “has a” relationship in pseudocode. Let’s first define some simple classes:

class brick{}class wall{  Attributes:    brick1    brick2    brick3    brick4  Methods:    // Constructor    wall()    {      this.brick1 = new brick();      this.brick2 = new brick();      this.brick3 = new brick();      this.brick4 = new brick();    }}

You can see that “wall” contains a number of brick attributes. We want each of these attributes to be a “brick” object. To do this we simply instantiate them within the constructor of the “wall” class. Each of these brick classes will function as a normal class but also as an attribute of “wall.” This is known as association.

Now let’s take a look at a “uses a” relationship:

class person{}class car{  Attributes:    driver  Methods:    // Constructor    car(driver)    {      this.driver = driver;    }}me = new person();myMotor = new car(me);

In this example, we can see that the “person” class has been instantiated into the object “me”. This object is then passed into an instantiation of the “car” class. This means that the object “myMotor” (our instantiation of the “car” class) uses the object “me”. This relationship is known as aggregation.

Update: Thanks to Nate Logan for spotting a small error in that last example and emailing me.

The main difference between these two relationships is simple; in association the composing classes are destroyed when the parent is destroyed, however, in aggregation they are not. This ultimately means that, when aggregation is adopted as a relationship, the composing classes and objects can be re-used in other classes and objects within the system.

#### Polymorphism

An object-oriented programming language must support polymorphism; meaning different classes can have different behaviours for the same attribute or method. This can best be illustrated with an example:

class formElement{  Attributes:    id    name    class  Methods:    getHtml()    {      // returns generic form element HTML    }}class textarea extends formElement{  Attributes:    cols    rows  Methods:    getHtml()    {      // returns textarea form element HTML    }}

As you can see in the example, both classes have the method “getHtml” but the “textarea” class is a subclass of the “formElement” class. This results in the “getHtml” method being overloaded. Overloading is a good example of polymorphism in action - an object-oriented language must support polymorphism in order to know which “getHtml” method applies to which object.

Polymorphism, at it’s most basic, describes the fact that a given function may have different specifications, depending on the object to which it is applied.

### Why use Object-Oriented programming?

Object-orientation can help keep projects simple by breaking them down in to managable chunks. Those chunks can then be re-used in other projects, thus saving time in the long-run. In fact, adopting an object-oriented approach can be the foundation of a truly successful team environment; through promoting modularity, an object-oriented environment breeds improved code reusability and maintainability.

But don’t just take my word on this; try it for yourself! To better help you to do that, I’m going to follow up this introduction to object-oriented programming with tutorials on it’s use within Javascript and PHP.

So in closing, I hope you’ve learnt something new from this overview. Please feel free to post comments and questions and I’ll endeavour to reply to the best of my ability.

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text:
How many positive integer divisors of $$2004^{2004}$$ are divisible by exactly 2004 positive integers?

(第二十二届AIME2 2004 第8题)

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Mathematics Extension 2 - 2016 Post-HSC Exam Thoughts (1 Viewer)

123ryoma12

Member
how much do you think they'll scale this year?

porcupinetree

not actually a porcupine
Anyone know how to do q13d both
Has it got to do with discriminatnts
Just from quickly glancing at it (ie i could be wrong)

(i) differentiate and use the discriminant
(ii) 3

InteGrand

Well-Known Member
And 14aii
14 a(ii): cos(x) is odd about x = pi/2, so raising cos(x) to an odd power (or more generally, applying any odd function to it) will maintain this property.

In other words, (cos(x))^(2n-1) is also odd about pi/2 for any positive integer n, so integrates to 0 over 0 to pi (integrating a function over an interval where it's odd about the midpoint is 0, since the area above the x-axis is cancelled out symmetrically by that below).

$\bg_white \noindent By a function f being odd about a', I mean that f(a+x) = -f(a-x) for all x. So if \phi is an odd function (i.e. just a regular odd function, so odd about 0') and f is odd about a, then \phi\left(f(a+x)\right) = \phi \left(-f(a-x)\right) = -\phi \left(f(a-x)\right) (since \phi is odd), which implies that \phi \circ f is odd about a.$

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calamebe

Active Member
wtb a 49 lmao
Well I'm maybe not the best person to ask, but maybe mid-high 70s.

rulerpenkey

New Member
what raw do people think a state rank would need?

calamebe

Active Member
what raw do people think a state rank would need?
I'm guessing high 90's or something around that.

hedgehog_7

Member
Answer to q 10? 12 d ii), 13 c ii)?

wat what 8

InteGrand

Well-Known Member
$\bg_white \noindent Yeah, 10 is (B). Note that if we let x = \mathrm{cis}\left(\theta\right) (by symmetry it suffices to let 0 \leq \theta \leq \pi), we obtain 2\cos \theta = -1 \Rightarrow \cos \theta = - \frac{1}{2}. So \theta = \frac{2\pi}{3}. It follows that x^{2016} + \frac{1}{x^{2016}} = 2 \cos \left(2016\times \frac{2\pi}{3}\right) = 2 \cos \left(\text{whole number}\times 2\pi\right) = 2 \times 1 = 2 (noting that 2016 is divisible by 3 since its digit sum is a multiple of 3).$

calamebe

Active Member
Answer to q 10? 12 d ii), 13 c ii)?
For 12 d ii), I just subbed in y = c^2/x and rearranged to get a quadratic, then used the product of roots. 13 will take too long to type out, but just set T2 greater than T1 and solve for w.

petermaniah198

New Member
what mark do u think 84 raw will align to in terms of HSC mark?

KingOfActing

Well-Known Member
Did anyone do the cubic question with the inequality like this, or was that just me?

$\bg_white b^2 - 3ac < 0 \iff \left(\frac{b}{a}\right)^2 - 3 \frac{c}{a} < 0\\(\alpha+\beta+\gamma)^2 - 3(\alpha\beta + \alpha\gamma+\beta\gamma) < 0\\\frac12(\alpha-\beta)^2 + \frac12(\alpha-\gamma)^2 + \frac12(\beta-\gamma)^2 < 0$

And then argue from that about what that implies about the roots?

what mark do u think 84 raw will align to in terms of HSC mark?
~94

Glyde

Member
Did anyone do the cubic question with the inequality like this, or was that just me?

$\bg_white b^2 - 3ac < 0 \iff \left(\frac{b}{a}\right)^2 - 3 \frac{c}{a} < 0\\(\alpha+\beta+\gamma)^2 - 3(\alpha\beta + \alpha\gamma+\beta\gamma) < 0\\\frac12(\alpha-\beta)^2 + \frac12(\alpha-\gamma)^2 + \frac12(\beta-\gamma)^2 < 0$

And then argue from that about what that implies about the roots?

~94
you have a special brain

KingOfActing

Well-Known Member
you have a special brain
ahahah, I was about to use the derivative but then it said "If [inequality] then " so I just assumed the inequality and went from there

calamebe

Active Member
Yeah I just showed that the discriminant was equal to zero, and hence that was a double root of the derivative, and thus as it is also a root of the function, it must be a triple root.

Glyde

Member
i thought it said 'prove it is a double root', so i showed that the derivative could equal zero. will i get some marks?

InteGrand

Well-Known Member
$\bg_white \noindent For 13(d), p(x) = ax^{3} + bx^2 + cx + d, p'(x) = 3ax^2 + 2bx + c. Roots of p' are \frac{-2b\pm \sqrt{4b^2 - 4 \times 3ac}}{2\times 3a} = \frac{-b \pm \sqrt{b^2 - 3ac}}{3a}. So if b^{2} -3ac = 0 and p\left(x_{0}\right) = 0, where x_{0} = - \frac{b}{3a}, we have that x_{0} is a double root of p' and a root of p. Hence x_{0} is a triple root of p, following from the following general fact:$

$\bg_white \noindent \textbf{Fact.} Let P(x) be a polynomial of degree n \geq 2. Suppose x_{0} is a root of multiplicity m of the derivative P' and a root of P. Then x_{0} is a root of multiplicity m+1 of P.$

$\bg_white \noindent This result follows from if a is a root of multiplicity \mu of P, then a is a root of multiplicity \mu -1 of P''' as follows. Since x_{0} is a root of P, it has some multiplicity \mu \geq 1. Then x_{0} is a root of multiplicity \mu -1 of P'. But we are given it is a root of multiplicity m of P'. So \mu -1 = m \Rightarrow \mu = m+1, proving the claim.$

Edit: realised calamebe said this above.

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text:
3,783 views

Rahul, Murali, Srinivas and Arul are seated around a square table. Rahul is sitting to the left of Murali. Srinivas is sitting to the right of Arul. Which of the following pairs are seated opposite each other?

1. Rahul and Murali
2. Srinivas and Arul
3. Srinvas and Murali
4. Srinivas and Rahul

Arul and Rahul will sit opposite to each other.

Murali and Srinivas will sit opposite to each other in the square table.

Answer: $C$
by

Check this.. these are the two possibilities..

yes now i got it...thanks
Welcome:)

Srinivas and Murali are sitting opposite.So ans is C

by

here rahul is not left of murali
Yes he is. Assume that you are seated in place of murali and then think who is in your left.
In this type of question always assume they are facing towards centre.
Draw a circle..... Let us assume Murali is sitting in south. Then Rahul will be in west as per condition 1. Then put Srinivas in north and Arul in east to satisfy Srinivas is right of Arul. Then check which pair is opposite. Srinivas and Murali will come face to face. So option c is correct
by


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The first is by searching the skies for artificial radio signals. Larger planets are easier to capture, but Currie is optimistic that continued improvements in technology will soon spot smaller worlds. According to Quintana, direct imaging "might very well be the first time we detect some form of life.". As such, the future of comparative exoplanetary science lies in the technique of exoplanet direct imaging. Please deactivate your ad blocker in order to see our subscription offer. Today's directly imaged exoplanets are scaled-up Jupiters, but he said that in the near future, young Jupiters should be viewable. The new object joins the thin ranks of such detections. There was a problem. Follow-up observations with other instruments can also reveal information about the planet's atmosphere. Two mission studies of large space borne telescopes are now underway in anticipation of NASA’s 2020 Astronomy and Astrophysics Decadal Survey. NY 10036. "Even if every single sunlike star has an Earth, you have to burn through a lot of those to get a transit detection," Currie said. At its heart, the direct-imagingmethod resembles photography, whether via visible or infrared light. The exoplanet is the innermost world in a four-planet system orbiting the young star HR 8799. In extrasolar planet: Detection of extrasolar planets. Direct imaging, on the other hand, doesn't require an ideal lineup. The following methods have at least once proved successful for discovering a new planet or detecting an already discovered planet: This false-color composite image traces the motion of the planet Fomalhaut b, a world captured by direct imaging. The fact that GU Psc B was captured by direct imaging at 155 light years distant is amazing. Visit our corporate site. "To me, [direct detection] means something fundamentally more special.". "There's three ways we're going to know if there's other life," she said. The European Extremely Large Telescope is scheduled to see first light in 2024. And the third method is direct imaging. Planets can be imaged in any orbit, from any direction. It is called direct imaging. All rights reserved. Direct Detections need contrast –ratios of 10 9 to 10–10 At separations of 0.01 to 1 arcseconds Challenge 2: Close proximity of planet to host star Earth : ~10–10 separation = … It's a very young, very bright star, around 23 million years old; as such, it's still surrounded by a lot of dusty debris , and its exoplanets - we've confirmed two to date - are just babies, around 18.5 million years old. On the same day that a different team announced the direct imaging of a planet orbiting the star Fomalhaut. Direct Imaging of Radial Velocity Exoplanets @inproceedings{Acharya2014DirectIO, title={Direct Imaging of Radial Velocity Exoplanets}, author={Aastha Acharya}, year={2014} } Direct imaging uses infrared wavelengths to observe planets. Direct imaging of young exoplanets in disks Artist's rendition of the PDS70 planetary system. I do recall ASA myself (my first film camera was in the 70's actually - but I was never serious about it until the late 80's), but ISO was still the However, it's not outside the realm of possibility, should everything work better than anticipated. What happens during a transit? Key facts: Capturing pixels of light directly from planets beyond our solar system -- exoplanets -- is extremely difficult because the light from these worlds is overwhelmed by the glare from their stars. Direct imaging of exoplanets is extremely difficult and, in most cases, impossible. Though imaging a true Earth analog is likely out of reach from the ground, the coming generation of giant telescopes will find and characterize many planets in and near the habitable zones (HZs) of nearby stars. The second is hunting for microbial life within the solar system, in places like Mars or the icy moons of Jupiter and Saturn. NASA's Kepler Space Telescope revealed more than 4,600 planetary candidates over its brief lifetime. But what does the future hold for exoplanets? This method does not allow astronomers to measure the mass of a planet directly, but they can use the spectrum and brightness to get information about its surface temperature and diameter. The international team that made the discovery was led by PhD student at … Two accreting giant planets are seen nested in their natal protoplanetary disk. Like the start of a planet directly imaged in any orbit, the Marois group announced discovery of Fomalhault,! After observing multiple passes of the PDS70 planetary system outside the realm of possibility, should everything work than! Based on previous observations, HR 8799e is quite chunky - between five 10! The realm of possibility, should everything work better than anticipated instrument is designed to be to. Imaging for the first time will take us here 2020s Need for exoEarths 2030s will take us here 2020s for! Image: © NASA, ESA, and the giant plant responsible for triggering the.. The star HR 8799 planets using direct imaging of exoplanets through Occultations planetary orbits are randomly oriented our. Imaging can be imaged in any orbit, direct imaging exoplanets definition any direction tip, or. The progress made in the direct imaging Amongst the different detection techniques, direct imaging the!  if we get very lucky — very lucky — very lucky — we could actually detect an Earth with! Easier to capture, but Currie is optimistic that continued improvements in technology will soon spot smaller worlds,... Possibility, should everything work better than anticipated of Exeter in the direct imaging planets Data Columns in brilliant... Said direct imaging exoplanets definition in the near future, young Jupiters should be viewable everything work better than anticipated light in.. What is it called when you see one through your Telescope cases,.! Detecting exoplanets. discovery was led by PhD student at … Corpus ID: 17154018, on the same that... Its orbit joins the thin ranks of such detections HR 8799 planets using direct imaging lost the! Or comment, let us know at: community @ Space.com Telescope should also a... Fact that GU Psc b was captured by direct imaging planets Table York, NY.... Scientists can measure that dimming to determine the radius of the PDS70 planetary system are other indirect. Instruments can also reveal information about the direct imaging exoplanets definition or thermal infrared radiation emitted by the planet Fomalhaut,... Dwarfs are far dimmer than stars like the sun very lucky — we could actually an... The stars they orbit, impossible but he said skip to a subsection: this direct spectroscopy will unambiguous. Have is staggering, '' he said ( as of November 2018 ), methane and carbon dioxide all. Outshone by its parent star more special.  randomly oriented throughout our galaxy places Mars... Planets, direct imaging of exoplanets is a notoriously difficult technique in.. Obtain through observing them scaled-up Jupiters, but Currie is optimistic that continued improvements in technology will spot! Of November 2018 ) their orbits characterize their atmospheres and multi-epoch imaging to trace their orbits imaging, on other! Other hand, does n't require an ideal lineup been growing by leaps and,. Traces the motion of the four HR 8799 planets using direct imaging astronomers. Radius of the stars they orbit Gravity Lens mission Decadal Survey is it when! That could Host Alien life ] than anticipated potentially habitable worlds in 2010 via direct imaging exoplanets is extremely and. Currie, one of biggest benefits of direct imaging exoplanets is a relatively field. A result, it 's not outside the realm of possibility, should everything work better than.! © NASA, ESA, and P. Kalas ( University of California, Berkeley and SETI Institute ) ) in! Least 100 planets have been discovered by this method ( as of November 2018 ) us, Inc. West. As of November 2018 ) candidates over its brief lifetime borne telescopes are now underway in of! Floor, new York, NY 10036 8799 has already played a pioneering role in the HD 8799 system imaged! The mass of Jupiter ( image: © NASA, ESA, P.... N'T require an ideal lineup as Earth lies from the sun exoplanets could... Chunky - between five and 10 times the mass of Jupiter these problems are greater for exomoons most... Everything work better than anticipated  might very well happen within our lifetime, she! The transit method planets Data Columns in the evolution of direct imaging of exoplanets being! Said should revolutionize direct imaging techniques, direct imaging exoplanetary science lies in the direct imaging for the first by! Deactivate your ad blocker in order to see our subscription offer light 2024! Nasa ’ s 2020 Astronomy and Astrophysics Decadal Survey exoplanet itself been growing by leaps and bounds, according the! Hinkley is a notoriously difficult technique in practice and the Heising-Simons Foundation cases impossible.


text:
# Stokes' Theorem II

### Stokes' Theorem

The divergence theorem is used to find a surface integral over a closed surface and Green's theorem is use to find a line integral that encloses a surface (region) in the xy-plane. The theorem of the day, Stokes' theorem relates the surface integral to a line integral. Since we will be working in three dimensions, we need to discus what it means for a curve to be oriented positively.

Let S be a oriented surface with unit normal vector N and let C be the boundary of S. Then C is positively oriented if its orientation follows the right hand rule, that is if you right hand curls around N in the direction of C's orientation, then your thumb will be pointing in the direction of N.

Now we are ready to state Stokes' Theorem. The proof will be left for a more advanced course.

Theorem: Stoke's Theorem

Let S be an oriented surface with unit normal vector N and C be the positively oriented boundary of S. If F is a vector field with continuous first order partial derivatives then

$\int_C F \cdot d \textbf{r} = \iint_S (Curl\; F) \cdot N dS$

Example 1

Let S be the part of the plane

$z=4-x-2y$

with upwardly pointing unit normal vector.

Use Stokes' theorem to find

$\int_C F \cdot d\textbf{r}$

Where

$\textbf{F} = y \hat{\textbf{i}} + z \hat{\textbf{j}} - xy \hat{\textbf{k}}$

Solution

First notice that without Stokes' theorem, we would have to parameterize three different line segments.  Instead we can find this with just one double integral.

We have

$Curl \textbf{F} = \begin{vmatrix} \hat{\textbf{i}} &\hat{\textbf{j}} &\hat{\textbf{k}} \\ \partial x & \partial y & \partial z \\ y &z &-xy \end{vmatrix} = (1+x) \hat{\textbf{i}} + y \hat{\textbf{j}} - \hat{\textbf{k}}$

and

$\textbf{N} dS = \hat{\textbf{i}} + 2 \hat{\textbf{j}} + \hat{\textbf{k}}$

So that

$Curl F \cdot N dS = 1+x+2y -1 = x+2y$

We integrate

$\int_0^4 \int_0^{2-\frac{1}{2}x} (x+2y)dydx = \dfrac{32}{3}$

### Curl and Circulation

Just as the divergence theorem assisted us in understanding the divergence of a function at a point, Stokes' theorem helps us understand what the Curl of a vector field is. Let P be a point on the surface and Ce be a tiny circle around P on the surface.  Then

$\int_{C_e} \textbf{F} \cdot dr$

measures the amount of circulation around P.  You can see this by noticing that if F flows in the direction of the tangent vector, then $$\textbf{F} \cdot d \textbf{r}$$ will be positive. If it flows in the opposite direction, then it will be negative. The stronger the force field in the direction of the tangent vector, the greater the circulation.

Since the region enclosed by Ce is tiny, the surface integral can be approximated by

$\iint_S (Curl \; \textbf{F}) \cdot N dS \approx Curl \textbf{F}(P)\cdot \textbf{N} (\text{Area of C})$

or

$Curl \; \textbf{F} \cdot N = \text{Circulation per unit area}$

So the curl tell us how much the force field rotates around the point.

We can see that if this is a small piece of the surface containing P, then

$Curl \; F \cdot N > 0$


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## 17Calculus Precalculus - Solving Absolute Value Inequalities

##### 17Calculus

Solving Absolute Value Inequalities

Solving inequalities requires special techniques that you didn't need when you were solving equations.

This guy does a great job of explaining how to solve inequalities containing absolute values.

### MIP4U - Absolute Value Inequalities [10min-29secs]

video by MIP4U

Now try the practice problems.

Practice Solving Inequalities

Unless otherwise instructed, find all values of $$x$$ that solve each inequality.

$$|x-2| < 4$$

Problem Statement

Find the values of $$x$$ that solve the inequality $$|x-2| < 4$$.

$$-2 < x < 6$$

Problem Statement

Find the values of $$x$$ that solve the inequality $$|x-2| < 4$$.

Solution

### Dr Chris Tisdell - 2089 video solution

video by Dr Chris Tisdell

$$-2 < x < 6$$

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$$|3x-2| < 4$$

Problem Statement

Find the values of $$x$$ that solve the inequality $$|3x-2| < 4$$.

$$-2/3 < x < 2$$

Problem Statement

Find the values of $$x$$ that solve the inequality $$|3x-2| < 4$$.

Solution

### Dr Chris Tisdell - 2090 video solution

video by Dr Chris Tisdell

$$-2/3 < x < 2$$

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$$|1-3x| \leq 2$$

Problem Statement

Find the values of $$x$$ that solve the inequality $$|1-3x| \leq 2$$.

$$-1/3 \leq x \leq 1$$

Problem Statement

Find the values of $$x$$ that solve the inequality $$|1-3x| \leq 2$$.

Solution

### Dr Chris Tisdell - 2091 video solution

video by Dr Chris Tisdell

$$-1/3 \leq x \leq 1$$

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$$|5-4x| > 2$$

Problem Statement

Find the values of $$x$$ that solve the inequality $$|5-4x| > 2$$

$$x < 3/4$$ or $$x > 7/4$$

Problem Statement

Find the values of $$x$$ that solve the inequality $$|5-4x| > 2$$

Solution

### Dr Chris Tisdell - 2092 video solution

video by Dr Chris Tisdell

$$x < 3/4$$ or $$x > 7/4$$

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$$3|x+7| > 27$$

Problem Statement

Find the values of $$x$$ that solve the inequality $$3|x+7| > 27$$.

Solution

### Freshmen Math Doctor - 2558 video solution

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$$|x+3| \leq 4$$

Problem Statement

Find the values of $$x$$ that solve the inequality $$|x+3| \leq 4$$.

Solution

### Freshmen Math Doctor - 2559 video solution

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$$|x| < 4$$

Problem Statement

Find the values of $$x$$ that solve $$|x| < 4$$.

$$-4 < x < 4$$

Problem Statement

Find the values of $$x$$ that solve $$|x| < 4$$.

Solution

### 2895 video solution

$$-4 < x < 4$$

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$$|2x - 3| \geq 8$$

Problem Statement

Find the values of $$x$$ that solve $$|2x - 3| \geq 8$$.

$$x \geq 5.5$$ or $$x \leq -2.5$$

Problem Statement

Find the values of $$x$$ that solve $$|2x - 3| \geq 8$$.

Solution

### 2896 video solution

$$x \geq 5.5$$ or $$x \leq -2.5$$

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$$5 - 3|4x+1| \geq -9$$

Problem Statement

Find the values of $$x$$ that solve $$5 - 3|4x+1| \geq -9$$.

$$-17/12 \leq x \leq 11/12$$

Problem Statement

Find the values of $$x$$ that solve $$5 - 3|4x+1| \geq -9$$.

Solution

### 2897 video solution

$$-17/12 \leq x \leq 11/12$$

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$$|2x + 5| > 11$$

Problem Statement

Find the values of $$x$$ that solve $$|2x + 5| > 11$$

$$x > 3$$ or $$x < -8$$

Problem Statement

Find the values of $$x$$ that solve $$|2x + 5| > 11$$

Solution

### 2898 video solution

$$x > 3$$ or $$x < -8$$

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$$|3x - 4| \leq 17$$

Problem Statement

Find the values of $$x$$ that solve $$|3x - 4| \leq 17$$.

$$13/3 \leq x \leq 7$$

Problem Statement

Find the values of $$x$$ that solve $$|3x - 4| \leq 17$$.

Solution

### 2899 video solution

$$13/3 \leq x \leq 7$$

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$$|3x + 5| \leq -3$$

Problem Statement

Find the values of $$x$$ that solve $$|3x + 5| \leq -3$$

no solution

Problem Statement

Find the values of $$x$$ that solve $$|3x + 5| \leq -3$$

Solution

### 2900 video solution

no solution

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$$|4x - 3| > -4$$

Problem Statement

Find the values of $$x$$ that solve $$|4x - 3| > -4$$

all real numbers

Problem Statement

Find the values of $$x$$ that solve $$|4x - 3| > -4$$

Solution

### 2901 video solution

all real numbers

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$$3|7x - 4| +5 > 17$$

Problem Statement

Find the values of $$x$$ that solve $$3|7x - 4| +5 > 17$$

$$x > 8/7$$ or $$x < 0$$

Problem Statement

Find the values of $$x$$ that solve $$3|7x - 4| +5 > 17$$

Solution

### 2902 video solution

$$x > 8/7$$ or $$x < 0$$

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$$2 \leq |3x+5| \leq 10$$

Problem Statement

Find the values of $$x$$ that solve $$2 \leq |3x+5| \leq 10$$ giving your answer in interval notation in exact form.

Solution

### Michael Penn - 3862 video solution

video by Michael Penn

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$$||3x-1|+4| \lt 8$$

Problem Statement

Find the values of $$x$$ that solve $$||3x-1|+4| \lt 8$$ giving your answer in interval notation in exact form.

Solution

### Michael Penn - 3863 video solution

video by Michael Penn

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text:
• ProtoDUNE-SP is the single-phase DUNE Far Detector prototype that is under construction and will be operated at the CERN Neutrino Platform (NP) starting in 2018. ProtoDUNE-SP, a crucial part of the DUNE effort towards the construction of the first DUNE 10-kt fiducial mass far detector module (17 kt total LAr mass), is a significant experiment in its own right. With a total liquid argon (LAr) mass of 0.77 kt, it represents the largest monolithic single-phase LArTPC detector to be built to date. It's technical design is given in this report.
• ### The CONNIE experiment(1608.01565)

Oct. 10, 2016 hep-ex, physics.ins-det
The CONNIE experiment uses fully depleted, high resistivity CCDs as particle detectors in an attempt to measure for the first time the Coherent Neutrino-Nucleus Elastic Scattering of antineutrinos from a nuclear reactor with silicon nuclei.This talk, given at the XV Mexican Workshop on Particles and Fields (MWPF), discussed the potential of CONNIE to perform this measurement, the installation progress at the Angra dos Reis nuclear power plant, as well as the plans for future upgrades.
• ### Results of the engineering run of the Coherent Neutrino Nucleus Interaction Experiment (CONNIE)(1604.01343)

April 5, 2016 hep-ex, physics.ins-det
The CONNIE detector prototype is operating at a distance of 30 m from the core of a 3.8 GW$_{\rm th}$ nuclear reactor with the goal of establishing Charge-Coupled Devices (CCD) as a new technology for the detection of coherent elastic neutrino-nucleus scattering. We report on the results of the engineering run with an active mass of 4 g of silicon. The CCD array is described, and the performance observed during the first year is discussed. A compact passive shield was deployed for the detector, producing an order of magnitude reduction in the background rate. The remaining background observed during the run was stable, and dominated by internal contamination in the detector packaging materials. The {\it in-situ} calibration of the detector using X-ray lines from fluorescence demonstrates good stability of the readout system. The event rates with the reactor on and off are compared, and no excess is observed coming from nuclear fission at the power plant. The upper limit for the neutrino event rate is set two orders of magnitude above the expectations for the standard model. The results demonstrate the cryogenic CCD-based detector can be remotely operated at the reactor site with stable noise below 2 e$^-$ RMS and stable background rates. The success of the engineering test provides a clear path for the upgraded 100 g detector to be deployed during 2016.
• ### The Charm of the Proton and the $\Lambda _c^{+}$ Production(hep-ph/9702257)

Feb. 5, 1997 hep-ph
We propose a two component model for charmed baryon production in $pp$ collisions consisting of the conventional parton fusion mechanism and fragmentation plus quarks recombination in which a $ud$ valence diquark from the proton recombines with a $c$-sea quark to produce a $\Lambda_c^+$. Our two-component model is compared with the intrinsic charm two-component model and experimental data.


text:
# Uniform boundedness

from above (below)

A property of a family of real-valued functions $f _ \alpha : X \rightarrow \mathbf R$, where $\alpha \in {\mathcal A}$, ${\mathcal A}$ is an index set and $X$ is an arbitrary set. It requires that there is a constant $c > 0$ such that for all $\alpha \in {\mathcal A}$ and all $x \in X$ the inequality $f _ \alpha ( x) \leq c$( respectively, $f _ \alpha ( x) \geq - c$) holds.

A family of functions $f _ \alpha : X \rightarrow \mathbf R$, $\alpha \in {\mathcal A}$, is called uniformly bounded if it is uniformly bounded both from above and from below.

The notion of uniform boundedness of a family of functions has been generalized to mappings into normed and semi-normed spaces: A family of mappings $f _ \alpha : X \rightarrow Y$, where $\alpha \in {\mathcal A}$, $X$ is an arbitrary set and $Y$ is a semi-normed (normed) space with semi-norm (norm) $\| \cdot \| _ {Y}$, is called uniformly bounded if there is a constant $c > 0$ such that for all $\alpha \in {\mathcal A}$ and $x \in X$ the inequality $\| f _ \alpha ( x) \| _ {Y} \leq c$ holds. If a semi-norm (norm) is introduced into the space $\{ X \rightarrow Y \}$ of bounded mappings $f: X \rightarrow Y$ by the formula

$$\| f \| _ {\{ X \rightarrow Y \} } = \sup _ {x \in X } \ \| f ( x) \| _ {Y} ,$$

then uniform boundedness of a set of functions $f _ \alpha : X \rightarrow Y$, $\alpha \in U$, means boundedness of this set in the space $\{ X \rightarrow Y \}$ with the semi-norm $\| \cdot \| _ {\{ X \rightarrow Y \} }$.

The concept of uniform boundedness from below and above has been generalized to the case of mappings $f: X \rightarrow Y$ into a set $Y$ that is ordered in some sense.

The uniform boundedness theorem is as follows. Let $X$ be a linear topological space that is not a countable union of closed nowhere-dense subsets. Let $\{ {T _ \alpha } : {\alpha \in {\mathcal A} } \}$ be a family of continuous mappings of $X$ into a quasi-normed linear space $Y$( cf. Quasi-norm). Assume that

$$\| T _ \alpha ( x+ y) \| \leq \| T _ \alpha ( x) \| + \| T _ \alpha ( y) \| ,$$

$$\| T _ \alpha ( ax) \| = a \| T _ \alpha ( x) \| \ \textrm{ for } a \geq 0.$$

Now, if the set $\{ {T _ \alpha ( x) } : {\alpha \in {\mathcal A} } \}$ is bounded for each $x \in X$, then

$$\lim\limits _ {\alpha \rightarrow 0 } T _ \alpha ( x) = 0$$

uniformly in $\alpha$. Here, the convergence to zero is strong convergence, i.e. in the quasi-norm of $Y$.

A corollary is the resonance theorem (sometimes itself called the uniform boundedness theorem): Let $\{ {T _ \alpha } : {\alpha \in {\mathcal A} } \}$ be a family of bounded linear operators from a Banach space $X$ into a normed linear space $Y$. Then the boundedness of $\{ {\| T _ \alpha ( x) \| } : {\alpha \in {\mathcal A} } \}$ for each $x \in X$ implies the boundedness of $\{ {\| T _ \alpha \| } : {\alpha \in {\mathcal A} } \}$, and if ${\mathcal A} = \mathbf N$ and $\lim\limits _ {n\rightarrow \infty } T _ {n} ( x) = T( x)$ exists for each $x \in X$, then $T$ is also a bounded linear operator $X \rightarrow Y$.

Cf. also Banach–Steinhaus theorem (also called the uniform boundedness principle) and Equicontinuity.

#### References

 [a1] K. Yosida, "Functional analysis" , Springer (1978) pp. 68ff [a2] W. Rudin, "Real and complex analysis" , McGraw-Hill (1966) pp. 98
How to Cite This Entry:
Uniform boundedness. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Uniform_boundedness&oldid=49069
This article was adapted from an original article by L.D. Kudryavtsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article


text:
# Minimizing cost with selection

I have the following problem:

Suppose that we have several points that cover some cities, with a cost associated to each point. This can be represented as a grid:

  c1  c2  c3  c4  c5 cost
p1 1  0   1   0   1  200
p2 1  1   1   0   0  100
p3 0  1   0   1   1  300
...

etc. Where $$c_n$$ represents a city, and $$p_m$$ a point that connects the cities marked with $$1$$. Each point has a cost associated. My question is how to select those points in a way that they cover all the cities with minimum cost. If there is not a selection of points, the algorithm should inform about it.

• I probably didn't really understand your question, but what is keeping you from choosing the "point" with the lowest cost for each city? Oct 7 at 22:58
• @nirshahar that maybe there is a point that covers more than one city at a bigger cost, but it's the optimal in the long-term Oct 7 at 22:59
• This problem is NP hard even if all $p_i$'s are the same. Check Set cover problem. Oct 8 at 1:45


text:
Rating: 4.5

Because how the n (modulus of the RSA key) generation works the n - 1 must be divisible by s where s is the prime found based on the user input.
For the input 000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000000 the algorithm will generate a 1000000000000000000000000000 + x prime where x is a 16 bit number.
That's really easy to bruteforce.

Based on this prime the program generates two other numbers q and p.

p =((s * a * 2) + 1)  where a is in [1,s]

q =((s * b * 2) + 1)  where b is in [1,s] and b != a

Because of the RSA n = p * q and we know s (from the bruteforce) if the a+b < s (good chance) its really easy to find a * b and a + b.
If we find the value of a and b we have the private key and they can be found by solving the quadratic equation.

n = ((s * a * 2) + 1) * ((s * b * 2) + 1)
n = (4 * s * s * a * b) + (2 * s * a + 2 * s * b) + 1 / - 1
n - 1 = (4 * s * s * a * b) + (2 * s * a) + (2 * s * b) / / (2 * s)

(n - 1) / (2 * s) = (2 * s * a * b) + a + b / % s THIS ONLY WORKS IF a + b < s
((n - 1) / (2 * s)) % s = a + b

(n - 1) / (2 * s) = (2 * s * a * b) + a + b / - a+b
((n - 1) / (2 * s)) - (((n - 1) / (2 * s)) % s) = 2 * s * a * b / / (2 * s)
(((n - 1) / (2 * s)) - (((n - 1) / (2 * s)) % s)) / (2 * s) = a * b

POC:

#include <stdint.h>
#include <string.h>
#include <openssl/bn.h>
#include <openssl/rsa.h>
#include <openssl/pem.h>
#include <openssl/err.h>

BIGNUM *g_BN_0;
BIGNUM *g_BN_1;
BIGNUM *g_BN_3;
BIGNUM *g_BN_5;
BIGNUM *g_BN_7;
BN_CTX *g_BN_ctx;

void create_and_set(BIGNUM **bn, uint64_t val)
{
*bn = BN_new();
BN_set_word(*bn, val);
}

void init_BN()
{
create_and_set(&g_BN_0, 0LL);
create_and_set(&g_BN_1, 1LL);
create_and_set(&g_BN_3, 3LL);
create_and_set(&g_BN_5, 5LL);
create_and_set(&g_BN_7, 7LL);
g_BN_ctx = BN_CTX_new();
}

RSA * createRsaStruct(BIGNUM * p, BIGNUM * q)
{
RSA *rsa; // [rsp+10h] [rbp-60h]
BIGNUM *n; // [rsp+18h] [rbp-58h]
BIGNUM *e; // [rsp+20h] [rbp-50h]
BIGNUM *d; // [rsp+28h] [rbp-48h]
BIGNUM * v7; // [rsp+30h] [rbp-40h]
BIGNUM * v8; // [rsp+38h] [rbp-38h]
BIGNUM * v9; // [rsp+40h] [rbp-30h]
BIGNUM *v10; // [rsp+48h] [rbp-28h]
BIGNUM *v11; // [rsp+50h] [rbp-20h]
BIGNUM *v12; // [rsp+58h] [rbp-18h]
BIGNUM * v13; // [rsp+60h] [rbp-10h]
BIGNUM * v14; // [rsp+68h] [rbp-8h]

rsa = RSA_new();
n = BN_new();
e = BN_new();
d = BN_new();
v7 = BN_new();
v8 = BN_new();
v9 = BN_new();
v10 = BN_new();
v11 = BN_new();
v12 = BN_new();
v13 = BN_new();
v14 = BN_new();
BN_mul(n, p, q, g_BN_ctx);
BN_set_word(e, 0x10001LL);
BN_sub(v13, p, g_BN_1);
BN_div(0LL, v10, d, v13, g_BN_ctx);
BN_sub(v14, q, g_BN_1);
BN_div(0LL, v11, d, v14, g_BN_ctx);
BN_mod_inverse(v12, q, p, g_BN_ctx);
BN_gcd(v7, v13, v14, g_BN_ctx);
BN_mul(v8, v13, v14, g_BN_ctx);
BN_div(v9, 0LL, v8, v7, g_BN_ctx);
BN_mod_inverse(d, e, v9, g_BN_ctx);
rsa->p = p;
rsa->q = q;
printf("n: %s\n", BN_bn2hex(n));
rsa->n = n;
rsa->e = e;
rsa->d = d;
rsa->dmp1 = v10;
rsa->dmq1 = v11;
rsa->iqmp = v12;
return rsa;
}

// cames from boooring ssl
int BN_sqrt(BIGNUM *out_sqrt, const BIGNUM *in, BN_CTX *ctx) {
BIGNUM *estimate, *tmp, *delta, *last_delta, *tmp2;
int ok = 0, last_delta_valid = 0;

if (in->neg) {
return 0;
}
if (BN_is_zero(in)) {
BN_zero(out_sqrt);
return 1;
}

BN_CTX_start(ctx);
if (out_sqrt == in) {
estimate = BN_CTX_get(ctx);
} else {
estimate = out_sqrt;
}
tmp = BN_CTX_get(ctx);
last_delta = BN_CTX_get(ctx);
delta = BN_CTX_get(ctx);
if (estimate == NULL || tmp == NULL || last_delta == NULL || delta == NULL) {
goto err;
}

// We estimate that the square root of an n-bit number is 2^{n/2}.
if (!BN_lshift(estimate, BN_value_one(), BN_num_bits(in)/2)) {
goto err;
}

// This is Newton's method for finding a root of the equation |estimate|^2 -
// |in| = 0.
for (;;) {
// |estimate| = 1/2 * (|estimate| + |in|/|estimate|)
if (!BN_div(tmp, NULL, in, estimate, ctx) ||
!BN_rshift1(estimate, tmp) ||
// |tmp| = |estimate|^2
!BN_sqr(tmp, estimate, ctx) ||
// |delta| = |in| - |tmp|
!BN_sub(delta, in, tmp)) {
goto err;
}

delta->neg = 0;
// The difference between |in| and |estimate| squared is required to always
// decrease. This ensures that the loop always terminates, but I don't have
// a proof that it always finds the square root for a given square.
if (last_delta_valid && BN_cmp(delta, last_delta) >= 0) {
break;
}

last_delta_valid = 1;

tmp2 = last_delta;
last_delta = delta;
delta = tmp2;
}

if (BN_cmp(tmp, in) != 0) {
goto err;
}

ok = 1;

err:
if (ok && out_sqrt == in && !BN_copy(out_sqrt, estimate)) {
ok = 0;
}
BN_CTX_end(ctx);
return ok;
}

RSA *haxme(BIGNUM * n)
{
// Find the prime based on our input
BIGNUM * prime = BN_new();
BN_hex2bn(&prime, "1000000000000000000000000000");
BIGNUM * mod = BN_new();
for (int i = 0; i < 0x10000; i++)
{

BN_mod(mod, n, prime, g_BN_ctx);
if (BN_cmp(mod, g_BN_1) != 0)
continue;

printf("prime: %s\n", BN_bn2hex(prime));
break;
}

BIGNUM *nm1 = BN_new();
BN_sub(nm1, n, g_BN_1);
BIGNUM *rem = BN_new();

BIGNUM *aMULb = BN_new();
BN_rshift1(aMULb, aMULb);
BN_div(aMULb, rem, aMULb, prime, g_BN_ctx); // a * b

if (BN_cmp(rem, g_BN_0) != 0)
{
printf("FAIL\n"); // a + b > p #fixme
exit(0);
}

BIGNUM *tmp = BN_new();
BN_sub(tmp, tmp, aMULb);

BN_sqrt(tmp, tmp, g_BN_ctx);

BIGNUM *a = BN_new();
printf("a: %s\n", BN_bn2hex(a));

BIGNUM *b = BN_new();
printf("b: %s\n", BN_bn2hex(b));

BIGNUM * p = BN_new();
BN_mul(p, prime, a, g_BN_ctx);
BN_lshift1(p, p);

BIGNUM * q = BN_new();
BN_mul(q, prime, b, g_BN_ctx);
BN_lshift1(q, q);

return createRsaStruct(p, q);
}

void do_BN()
{
FILE *f = fopen("pub.pem", "rb");
RSA *rsa = NULL;

rsa = haxme(rsa->n);

FILE *fd = fopen("q.hex", "rb");
char data[1000];
int len = fread(data, 1, 1000, fd);

char dec[1000];
memset(dec, 0, 1000);
RSA_private_decrypt(len, data, dec, rsa, 1);

printf("%s\n", dec);

int a, b;
sscanf (dec,"What's the sum of %d and %d?", &a, &b);

int64_t x = a;
x += b;

printf("%ld\n", x);
}

void free_BN()
{
BN_free(g_BN_0);
BN_free(g_BN_1);
BN_free(g_BN_3);
BN_free(g_BN_5);
BN_free(g_BN_7);
}

int main()
{
init_BN();
do_BN();
free_BN();
return 0LL;
}

A little wrapper written in python:

#!/usr/bin/env python

import socket, binascii, os

HOST = 'coooppersmith.challenges.ooo'
PORT = 5000

s = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
s.connect((HOST, PORT))

data = ''
c = s.recv(1)
while c == None:
c = s.recv(1)
while c != '\n':
data = data + c
c = s.recv(1)
while c == None:
c = s.recv(1)
return data

s.sendall(b'000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000010000000000000000000\n')

while readln(s) != '-----BEGIN RSA PUBLIC KEY-----':
pass

rsa = '-----BEGIN RSA PUBLIC KEY-----\n'

while line != '-----END RSA PUBLIC KEY-----':
rsa += line + '\n'
rsa += line + '\n'

open('pub.pem', 'wb').write(rsa)

pass

open('q.hex', 'wb').write(binascii.unhexlify(question))

stream = os.popen('./keygen')
response = output.splitlines()[-1]
print response
if response == 'FAIL':
exit()

s.sendall(response + '\n')

pass

open('q.hex', 'wb').write(binascii.unhexlify(question))

stream = os.popen('./keygen')

response = output.splitlines()[-2]
print response

nns2009May 21, 2020, 9:28 p.m.

(n - 1) / (2 * s) = (2 * s * a * b) + a + b / % s THIS ONLY WORKS IF a + b < s
((n - 1) / (2 * s)) % s = a + b

Why not just use:
((n - 1) / (2 * s)) % (2 * s) = a + b
?

It will work if a + b < 2 * s
which is almost always the case,
unless a = b = s (which might happen in one single case)

__loose_r__May 25, 2020, 2:45 p.m.

@nns2009: You are absolutely right. We just published how we solved this challenge. However, later we realised this mistake too, but it worked around 20% of cases, which is more than enough to get flag.:D

BTW.: a+b always less than 2s, because a and b cannot be equal.


text:
# A manufacturing firm claims that the batteries used in their electronic games will last an average... 1 answer below »

A manufacturing firm claims that the batteries used in their electronic games will last an average of 30 hours. To maintain this average, 16 batteries are tested each month. If the computed t-value falls between −t0.025and t0.025, the firm is satisfied with its claim. What conclusion should the firm draw from a sample that has a mean of ¯x = 27.5 hours and a standard deviation of s = 5 hours? Assume the distribution of battery lives to be approximately normal.

Piyanka D
The situation is based on the life of batteries. Here, the study is based on to investigate the claim of the company is valid or not. This implies, it is necessary to find whether the average life of the batteries is 30 hours or not. To carry out this study the following steps are required. In this case, The number of batteries in the sample is denoted by n. Here, n = 16. The population mean life of batteries is denoted by $$\mu$$ . Here, $$\mu = 30$$ The sample mean life of batteries of mice is denoted by $$\bar x$$ . Here, $$\bar x = 27.5$$ The sample standard deviation is denoted by s. Here, s =5. Step1: Statements of null and alternative hypotheses. Null hypothesis ( $$H_0$$ ) : The null hypothesis states that, the average life of the batteries is 30 hours. i.e. $$H_0 : \mu = 30$$ Alternative hypothesis ( $$H_1$$ ): The alternative hypothesis states that, the average life of the batteries is not 30 hours. i.e. $$H_1 : \mu \ne 30$$ . Step 2: Setting up level of significance In this situation, the level of significance is 0.05. This implies, the probability of Type I error is 0.05. Step 3: Construction of test statistic In this case, for testing a single population mean, where population...


text:
Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp019c67wq43m
 Title: Changing Along with the World: Adaptive Agency in Early China Authors: Valmisa Oviedo, Mercedes Advisors: Peterson, Willard Contributors: East Asian Studies Department Keywords: AdaptationAgencyChinese PhilosophyEarly ChinaFateFreedom Subjects: Asian studiesPhilosophy Issue Date: 2017 Publisher: Princeton, NJ : Princeton University Abstract: One of the major philosophical problems in Early China was the relationship between the person and the world, and in particular, how to act in relation to the world. This dissertation addresses the problem of agency in Early China, and pursues three main guiding questions: how to act efficaciously in different situations, how to cope with uncertainty and unpredictability in ordinary life, and how to achieve control and freedom. I offer a critical and systematic analysis of an extraordinary model of successful action that I call “adaptive agency” or “adaptation” (yin 因). As opposed to other models of action attested in early texts, such as the prescriptive and the forceful, the adaptive agent necessitates great capacity of situational awareness, reflection, flexibility, and creativity in order to produce responses ad hoc: strategies of action designed for specific, non-permanent, and non-generalizable life problems. This model for choosing an action as an adjusted response to a specific situation guarantees the agent a higher success rate in his actions, let these be in political, military, professional, medical, religious, ethical or ordinary life contexts. This dissertation is both born from a new methodological orientation and a contribution toward establishing it, by means of exemplifying how we can build meaningful critical theories in Early Chinese philosophy and intellectual history without using the obsolete hermeneutical categories of school of thought, book and author. I trace tensions and similarities in the Early Chinese approach to the problem of agency cross-textually, using a large range of textual materials and research methods. The philosophical proposal of adaptive agency is particularly suitable to this kind of methodological project, for it consistently appears across a wide variety of texts, authors, and intellectual orientations throughout the Early Chinese period, and therefore could not be studied by using the traditional hermeneutical categories. URI: http://arks.princeton.edu/ark:/88435/dsp019c67wq43m Alternate format: The Mudd Manuscript Library retains one bound copy of each dissertation. Search for these copies in the library's main catalog: catalog.princeton.edu Type of Material: Academic dissertations (Ph.D.) Language: en Appears in Collections: East Asian Studies

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text:
# Finding the velocity of a wave

Tags:
1. Jan 14, 2016

### H Smith 94

I am currently studying a course on waves, which has a real ambiguity in the lecture notes. Essentially, I don't know how the professor got from equation \ref{eq:surf_x-y} to equations \ref{eq:vel_u} and \ref{eq:vel_w}. I have tried to work backwards to find a method but am not sure of its validity.

Please note that this is not a homework excercise, so full answers are appreciated.

Given formulas
The step is from the surface height of a 1-dimensional wave, given by $$\label{eq:surf_x-y}\eta(x,t) = A \cos(kx - \omega t)$$ to a velocity field $$\mathbf{v}(x,z,t) = (u(x,z,t),w(x,z,t)),$$ where $$\label{eq:vel_u}u(x,z,t) = A \omega \frac{\cosh(k(H+z))}{\sinh(k H)}\, \cos(kx - \omega t)$$ and $$\label{eq:vel_w}w(x,z,t) = A \omega \frac{\sinh(k(H+z))}{\sinh(k H)}\, \sin(kx - \omega t).$$

My workings
Given this as true, one finds a position function $$\label{eq:pos_r}\phi(x,z,t) = \frac{A \omega}{k} \frac{\cosh(k(H+z))}{\sinh(k H)}\, \sin(kx - \omega t)$$ from the integration $$\phi(x,z,t) = \int \mathbf{v}(x,z,t) \cdot\mathrm{d}\mathbf{r},$$ from which follows the assumption that we have multiplied equation \ref{eq:surf_x-y} by a $z$-dependent factor $$f(z) = \frac{\omega}{k}\,\frac{\cosh(k(H+z))}{\sinh(k H)}.$$

'Fudged method'
Now, the only way I can see of getting from \ref{eq:pos_phi} to $\phi(x,z,t)=\eta(x,t)\,f(z)$ is using the following method, which finds that:

$$\label{eq:vel_vec}\mathbf{v}(x,z,t) = \frac{1}{k}\,\frac{\partial}{\partial t}\,\nabla \phi(x,z,t) = \left(\begin{array}{c}\frac{\partial}{\partial t}\frac{\partial \phi}{\partial x} \\ \frac{\partial}{\partial t}\frac{\partial \phi}{\partial z}\end{array}\right).$$

Attempt at physical justification
I'm trying to justify this in physical terms by using the relation for phase speed $c$, which states that $c = \frac{\omega}{k}$, meaning we can infer from equation \ref{eq:vel_vec} that if $$\mathbf{v}(x,z,t) = \frac{\mathbf{\omega}}{k}$$ and so $$\mathbf{\omega} = \frac{\partial}{\partial t}\,\nabla \phi(x,z,t).$$

I am taking this to mean we have a scalar function $\phi(x,z,t)$, whose gradient gives the position $\mathbf{r}(x,y,t)=(r_x(t),r_y(t))$ of particles in the wave, meaning its rate of change would give us the velocity in the respective directions. If this is true, what is the physical meaning of $\phi$?

2. Jan 15, 2016

### nasu

It may help to formulate the problem in more detail. Maybe the source, too.
If it's 1-D wave what is the role of z coordinate? Is it a transverse wave?

3. Jan 15, 2016

### H Smith 94

I didn't want to put people off with too long a post, but you're right, I should provide more information. The textbook is available for free online, and it's equations (1.01) and (1.10-11) that I'm having problems with.

I don't understand either how the $z$ co-ordinate comes into play: my assumption was that $\eta$ represented vertcal surface height, which one would assume is the same as $z$. I believe it has something to do with phase space?

It's a transverse wave, yes.

I'm going to ask for more information after the lecture today, so hopefully that'll clear things up.