text:
Theory of relativity/General relativity/Introduction

A great many attempts have been made to explain the basic concepts of general relativity to non-experts. These can range from museum exhibits that roll a ball around on a curved surface, to treatments that are mathematically quite daunting. This article will attempt to explain it at the level of undergraduate, or ambitious high-school, mathematics and physics.

The general relativity formulation of gravity states that gravity arises from the curvature of spacetime, and, in analogy with the classical notion that massive objects create a gravitational field attracting other objects, matter causes spacetime to curve. In the words of physicist John Wheeler, "Space tells matter how to move, matter tells space how to curve."

Coordinate Systems and Spacetime Diagrams
Figure 1 shows a "spacetime diagram", with my house, my neighbour's house, and my neighbour walking from his house to mine.

 * This is the same kind of diagram that is used in explanations of special relativity. The "spacetime" is sometimes called "Minkowski space".  Spacetime is actually four-dimensional, but we can only show two dimensions, so we leave out y and z.  The single x spatial coordinate is good enough for our purposes, so the diagram has x going from left to right, and t (time) going upward.  For the purposes of this explanation, don't worry about the considerations of special relativity such as the speed of light, the Lorentz transform, or light cones.  None of that is important just now.

The diagram shows the calibration, in space (that is, x) and time. These measurements are made with respect to my (stationary) frame of reference. My house is at x=0, and my neighbour's house is at x=5 (all units are arbitrary). My neighbour walks at 1 unit per minute.

The diagram shows some "events"—my house, now; my house, 5 minutes from now; and my neighbour's house now and 5 minutes from now. The green diagonal line depicts my neighbour walking from his house to mine, arriving 5 minutes from now. That line is called his world line. The red line going straight up in my house is my own world line (I'm sitting at home.)

A car is driving down the street, from left to right. Figure 2 shows the same four events and two world lines, but with different calibration—the car's own coordinate system. The car is driving at 1 unit per minute, but in the opposite direction from my neighbour's walking. The event of my neighbour's arrival at my house is now at x=-5. It's way behind the car, though the car was directly adjacent to my house at t=0. The car passes my neighbour (that is, my neighbour is observed to be at x=0 in this coordinate system) at t=2.5.

Because the car's frame of reference is in motion, the calibration lines in figure 2 are not perpendicular. The formerly vertical lines are now slanted. But there is something very important to notice about both this coordinate system and the previous one: They are both flat. The flatness comes from the fact that the calibration lines are straight and parallel. The boxes created by the lines are parallelograms. But note that the lines don't have to be perpendicular, and the boxes don't have to be rectangles. Straight parallel lines and parallelograms are all that is required.

These two flat coordinate systems have a very important physical property: Neither I, sitting at home, nor a passenger in the car, experience any "fictitious forces" (this concept is described below.)  That is, people in the car don't feel any recoil from acceleration, or centrifugal force, or Coriolis force. These frames of reference are said to be inertial. This leads to an important principle of geometrical physics:

 * Inertial frames of reference have flat coordinate systems. Flat coordinate systems lead to an absence of fictitious forces.

Now consider figure 3. The coordinate system is once again that of the car, but the car is accelerating, starting at a standstill in front of my house at t=0. Its world line is curved. Once again, it crosses paths with my neighbour. This case is very different from the other two. The calibration lines are curved, and the boxes that they create are not parallelograms. This coordinate system is curved. Another thing to notice is that people in the car will feel a fictitious force—a "recoil" force agains the back of the seat. This frame of reference is not inertial.

 * Accelerating frames of reference have curved coordinate systems. Curved coordinate systems lead to fictitious forces.

The notion of flat and curved coordinate systems is actually quite familiar from analytic geometry. Figures 4 and 5 show two common coordinate systems (or "ways of drawing the calibration lines on graph paper".) Figure 4 shows Cartesian coordinates, which are flat, and figure 5 shows polar co-ordinates, which are curved.

Geodesics and Fictitious Forces
A geodesic is a line that is "objectively straight". What we mean by that is that a line's straightness transcends different coordinate systems. Different coordinate systems may have different equations for a line, but whether a line is a geodesic is independent of that.

We have drawn the figures above (and those below) so that geodesics are straight lines on your computer screen. The various colored world lines of Figures 1, 2, and 3 are all geodesics except for the blue line of Figure 3, for the accelerating car.

In a flat coordinate system, the equation of a geodesic is a linear equation. For example, the equation of the green line in figure 2 is $$x = 5 - t$$.

The equation of the blue line in Figure 3 is $$x = 0$$, but that doesn't make it a geodesic, because the coordinate system is not flat.

In a curved coordinate system, the formula for a geodesic is much more complicated. It will be explained below that, in the presence of gravitational fields, the space itself is curved. This means that all coordinate systems are curved, which makes the calculation of geodesics very complicated. This is one of the reasons that general relativity is so difficult.

Everyone except the passengers in the accelerating car (blue line, Figure 3), are undergoing inertial motion, and feel no fictitious forces:
 * Observers following geodesics in spacetime are undergoing inertial motion, and feel no fictitious forces.

An observer following a geodesic is said to be in "free fall". Planets are in free fall around the Sun. People falling down elevator shafts are also in free fall. (We consider the various observers in Figures 1 and 2 to be in free fall, because we are ignoring gravity for now. They feel no fictitious forces other than gravity.)

How about observers whose world lines are not geodesics? They are being accelerated by some external force, which is what makes their world line deviate from straightness. If they are "at rest", or moving uniformly, in their own coordinate system, the force making their world line curve may not seem real, but their recoil against that force will seem real. That recoil is a fictitious force.

For example, passengers in the car of Figure 3 feel a recoil force against the back of the car seat. People in various rotating amusement park rides feel a "centrifugal force", another fictitious force, seeming to push them outward. An observer standing on the ground can see that the centrifugal force is fictitious—what is really happening is that the ride mechanism is pushing them into circular motion. There is another fictitious force associated with rotation: the Coriolis force.

Gravity as a Fictitious Force
Finally, we come to gravity itself.
 * Gravity is a fictitious force, just like all the others. It arises from particles not following geodesics.

Figure 6 shows a spacetime diagram of the coordinate system of a person falling down an elevator shaft. His world line is shown in red. The left-to-right axis, which was the "x" axis in the earlier figures, is now the "z" axis, that is, height. The coordinate system travels down the shaft with him, and he thinks he is at rest in it. He does not feel any fictitious forces, including gravity. The coordinate system is flat and inertial.

Figure 7 shows the same person in the coordinate system of "stationary" people standing on the top floor of the building. The world line of the falling person is still red, and that of the stationary people is blue. The stationary coordinate system is curved; the stationary people are not following geodesics. The floor is pushing up against them, making them deviate from geodesics. Their recoil against this is the fictitious force that they call "gravity". The curved coordinate lines are the lines of "stationary" constant height. Think of them as being like the floors of the building. Note that the unfortunate victim is passing downward through them.

The aspect of gravity that indicates that it is a fictitious force is that it shares an important property with the standard fictitious forces: it is exactly proportional to the object's mass. For the standard fictitious forces, it is easy to see why this is so—they are actually the result of accelerations. By Newton's law $$F = ma$$, the force is proportional to the mass for a given acceleration.

This connection is the basis for Einstein's equivalence principle—one can't tell the difference between gravity and a traditional fictitious force. The thought experiment making this plausible was the "elevator experiment". A person in a closed elevator can't tell whether it is stationary in a building on Earth, or is out in space being dragged, at 9.8 meters per second squared, by a cable attached to a rocket.

For ordinary fictitious forces, we can readily explain things by transforming away from the curved coordinate system to some flat coordinate system. For example, we can "transform away" the centrifugal force of a rotating amusement park ride by viewing the action from a stationary observer on the ground.

For gravity, the equivalent transformation is to a coordinate system that is in free fall, such as that of the person falling down an elevator shaft. This leads to the following not-very-satisfactory explanation of the Earth's gravity: The gravity observed on the Earth's surface arises because the Earth is flat, and there are rocket engines on the underside, accelerating everything upward at 9.8 meters per second squared. This would explain everything, at least locally, but it has one fatal flaw: The Earth is round, with gravity going inward everywhere. So the surface of the Earth would have to be accelerating outward in all directions. This would require that the Earth continually get bigger, which it does not do.

Curvature of Spacetime
The problem with the not-very-satisfactory explanation of gravity given above was that, while we can transform the curved coordinate system of a person standing on the ground to the coordinate system of a person falling down an elevator shaft locally, we can't do it everywhere. People falling down elevator shafts on opposite sides of the Earth will not have consistent coordinate systems. The problem is that spacetime itself is curved.

Up until now, we have been dealing with curved and flat coordinate systems on spaces (the technical term is "manifold") that are themselves flat. We put coordinate systems on them as though we were drawing lines on a flat piece of graph paper. Now we have to deal with curved manifolds.

The curvature of a 2-dimensional manifold embedded in 3-dimensional space can be visualized by looking at it. It's just what you think it means. A typical curved 2-dimensional manifold is the surface of a sphere. We can place a coordinate system on it that is nearly flat over a tiny region. For example, ordinary latitude/longitude coordinates will work near the equator. But this coordinate system doesn't work everywhere. As one moves away from the equator, it measures distances incorrectly, and at the North and South poles it turns into polar coordinates, which are clearly not flat. (One could use a different coordinate system that looks flat at the North pole, but it wouldn't be flat at the equator.) This coordinate system that is flat at the equator is analogous to the coordinate system of the person falling down the elevator shaft—it can't be extended around the world.

Whether a manifold is curved or not is determined by the Riemann tensor, also called the curvature tensor. This is a mathematical object that has an "objective" or "intrinsic" existence, independent of coordinate system. (Tensors are defined to have this property. Doing this correctly is what makes tensor calculus so difficult.)  All of the spaces we have discussed so far, even those with curved coordinate systems, are flat. In fact, they are 2-dimensional Euclidean spaces.

Riemann's tensor on the 2-dimensional surface of a sphere is nonzero.

For the case of a 2-dimensional surface, the curvature is determined by a single number K, the Gaussian curvature. Figure 8 shows some surfaces with negative, zero, and positive Gaussian curvature.

The figure on the left, with negative Gaussian curvature, is often referred to as being (locally) "saddle-shaped". The figure in the middle has zero curvature. Only "intrinsic" curvature counts; rolling up a sheet of paper does not.

The fact that we can't make a flat coordinate system globally around a non-microscopic region of the Earth's surface tells us that in the vicinity of the Earth, spacetime itself is curved.

It is important to note that the curvature of 4-dimensional spacetime is vastly more complicated than the Gaussian curvature of a surface. It has 20 components instead of 1. Attempts to visualize it as a (perhaps saddle-shaped) surface are not correct; they merely suggest the meaning of curvature. The actual analysis of the curvature, and how it gives rise to gravity, requires careful analysis of tensor calculus.

 * This point needs to be emphasized: Visualizations of gravitational curvature in terms of an ordinary curved surface, with a ball rolling across it and being deflected, such as one sees in science museum exhibits, do not do justice to the actual curvature of general relativity.  They simulate a classical gravity well, such as exists in the Newtonian formulation of gravity.

The curvature of spacetime is manifested in the metric tensor, which has 10 non-trivial parameters, from which one derives the curvature itself in Riemann's tensor (20 parameters), from which one further derives Ricci's tensor and Einstein's tensor (10 parameters each.)

The Schwarzschild Solution
Under general relativity, the metric tensor describing the curved spacetime in the vicinity of the Earth (or the Sun, or any spherically symmetric gravitating body) is the celebrated Schwarzschild solution. Exact closed-form solutions to the gravity equations are notoriously difficult to obtain, and the Schwarzschild solution is one of very few such solutions. (The Kerr solution, which takes the body's rotation into account, is another.) In the Schwarzschild solution, the strength of an object's gravity is determined by a parameter, with the dimensions of length, called the Schwarzschild radius. When the equations are solved, the instantaneous downward acceleration of a particle initially at rest (for example, the instant someone jumps into the elevator shaft) is
 * $$\frac{d^2 r}{dt^2} = - \frac{c^2\,a\ (r - a)}{2\,r^3}$$

where $$a$$ is the Schwarzschild radius. Classically, the acceleration would be
 * $$\frac{d^2 r}{dt^2} = - \frac{G M}{r^2}$$

where $$G$$ is Newton's constant of gravitation, and $$M$$ is the mass of the gravitating body.

Setting these equal, under the assumption that $$a\,$$ is very small in comparison to $$r\,$$, we get
 * $$a = \frac{2\,G M}{c^2}$$

For the Earth, $$a$$ is about 1 centimeter. For the Sun it is about 3 kilometers.

But, as will be seen below, this equation only applies in the absence of gravitating matter, that is, outside of the Earth. The Earth would only have a Schwarzschild radius if its mass were all concentrated inside a sphere of 1 centimeter. If that were the case, it would be a black hole, with an event horizon at the Schwarzschild radius.

The Schwarzschild solution is just one simple case (gravity in the vacuum surrounding a spherically symmetric massive object) of general relativity, but it provides many of the observed phenomena:
 * bending of light (both around the Sun during an eclipse and around distant galaxies with gravitational lensing)
 * the anomalous precession of the perihelion of Mercury
 * gravitational redshift
 * gravitational time dilation
 * gravitational delay (Shapiro effect) of spacecraft signals past planets

Nonzero Gravitating Mass
We have seen how the curvature of spacetime tells matter how to move. Now we will examine how matter tells space how to curve.

The "gravitational field", which we might denote by a vector $$V\,$$, arises from the curvature of spacetime. Two tensor fields—the Ricci tensor $$R_{\mu\nu}\,$$ and the closely related Einstein tensor $$G_{\mu\nu}\,$$ govern this. The divergence of $$V\,$$ is just the 00 component of Ricci's tensor:
 * $$\nabla \cdot V = R_{00}$$

Just as the statement that an electric field is spherically symmetric and has zero divergence leads to an inverse square law, the statement that a coordinate system is spherically symmetric and has $$R_{\mu\nu} = 0\,$$ leads to an inverse square law (in the classical approximation) for the "gravitational field". (The Schwarzschild solution was just the exact solution to $$R_{\mu\nu} = 0\,$$.)

Now the actual law for electrostatics is Coulomb's law:
 * $$E = \frac{Q}{4 \pi \epsilon r^2}$$

where $$Q\,$$ is the charge and $$\epsilon\,$$ is the physical constant governing the force. This is equivalent to Maxwell's first equation:
 * $$\nabla \cdot E = \frac{\rho}{\epsilon}$$

where $$\rho\,$$ is the charge density.

Similarly, the equation for classical gravity is:
 * $$V = \frac{G M}{r^2}$$

where $$M\,$$ is the gravitating object's mass and $$G\,$$ is Newton's constant of gravitation. This is equivalent to:
 * $$\nabla \cdot V = 4 \pi G \rho$$

where $$\rho\,$$ is the matter density.

Therefore we have:
 * $$R_{00} = 4 \pi G \rho\,$$

(We had $$R_{00} = 0\,$$ before because we were calculating the field in a vacuum.)

The actual tensor of interest is Einstein's tensor $$G_{\mu\nu}\,$$, which has $$G_{00} = 2 R_{00}\,$$ in this simple case, so:
 * $$G_{00} = 8 \pi G \rho\,$$

Specifically:
 * $$G_{\mu\nu} = \begin{bmatrix} 8 \pi G \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

The other components relate to the motion of the gravitating matter—when the matter moves, things become even more complicated.

The stress-energy-momentum tensor $$T_{\mu\nu}\,$$, describes the configuration of matter and other things (e.g. electrical energy) that create gravity. For motionless matter, we have:
 * $$T_{\mu\nu} = \begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

So the equation of gravity, Einstein's equation, is:
 * $$G_{\mu\nu} = 8 \pi G\ T_{\mu\nu}\,$$


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User talk:Howickmoor86

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User talk:Clare M

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User talk:SamBennet

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User talk:Msafiri~enwikiversity

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Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Msafiri. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Msafiri~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

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User:Techeplo

Course name:	Technology exploitation

Term:	T1, 2010/2011

Course Code: :	922_507

E-mail:	techeplo@gmail.com (homework)

Appointment Times:	Per request

Course Overview:	To understand the critical roles and functions of technology within a global enterprise and be able to evaluate and select cost performance effective technology system for a global organization. Through simulations students will experience the decision making process behind analyzing and selecting enterprise wide and specific technology solutions.

title: User:Techeplo


text:
User:Techexplo

Term:	T1, 2010/2011
On-line Supplement to the Course

Course Code:	922_507

E-mail:	techexplo@gmail.com

Appointment Times:	Per request

Course Overview:	To understand the critical roles and functions of technology within a global enterprise and be able to evaluate and select cost performance effective technology system for a global organization. Through simulations students will experience the decision making process behind analyzing and selecting enterprise wide and specific technology solutions.

 * /Syllabus/
 * /Technology Exploitation Reading/
 * /Schedule/
 * /Homework/
 * /Homework Posting/

In-class
/In-class Work/
 * /Case Study 1/
 * /Case Study 2/
 * /Case Study 3/

Help

 * 1) Adding Content in Wikiversity
 * 2) Editing
 * 3) Picture Tutorial
 * 4) Embedding images
 * 5) Image Policy
 * 6) Categorizing
 * 7) Learning by Doing

title: User:Techexplo


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User:DarkKarma

I'm a young South African man and interested in computer programming. I can program in Object Pascal and a little bit of C/C++. Unfortunately I don't have any tertiary qualifications, though I hope on studying in 2010. I'm also a hardened Linux user and developing for the OS is important for me, but I still use Windows for development as well.

title: User:DarkKarma


text:
File talk:25142.070113235749.djvu

If it's on public domain, why do you publish as image rather than PDF so it will make copy and use easier?

title: File talk:25142.070113235749.djvu


text: Category:New Zealand

title: Category:New Zealand


text:
Conservative Judaism

What is Conservative Judaism?
In general, unlike Orthodox Judaism, the Conservative branch of the Jewish religion is based in the Conservative movement. The movement is not centralized, and hence force it is divided into local branches (For example, Israel and England's movement's are called “Masorti”, which means traditional in Hebrew), that can hold different halachic opinions.

Essentially, the conservative movement differs from orthodox Judaism by believing in changing halacha to suite the modern world, even at the cost of making drastic changes that do not take after most rabbis over the ages. On the other hand, it differs from Reform Judaism by accepting halacha as the structure of the religion, and working within it's framework.

title: Conservative Judaism


text:
User:Techexplo/Technology Exploitation Reading

Course name:	Technology exploitation Term:	T1, 2010/2011

Course Code:	922_507

E-mail:	techexplo@gmail.com

Reading Assignments

 * 1) Intro to Technology Exploitation: Emerging Technology, Technology Convergence,Technology Definition, List of Technologies
 * 2) * Business Models, Making Money by Technology: Business Models, Business Models (Michael Rappa NCSU,
 * 3) * Spreadsheets: Spreadsheet Concepts
 * 4) U-Commerce:
 * 5) * M-Commerce: m-Commerce
 * 6) * Case Study: ABLBoats
 * 7) Technology Scanning: Research Universities Ranking by NSF, Wiki Technology Portal, Research_and_Development
 * 8) * Using Internet for scanning: RSS Feeds
 * 9) New ERP: Enterprise Resource Planning
 * 10) * CRM for Small Companies: Customer Relations Manaement, e-CRM
 * 11) Open Society:
 * 12) * Open Source Software: Open Sources Software
 * 13) Developing Technology: Feasibility Study
 * 14) * Request for Proposals: RFP
 * 15) Preparing for Organizational Change: Change Management
 * 16) Implement Technology: SDLC,
 * 17) * Other System Development: System Development

 * 1) Midterm

 * 1) Social Marketing: Digital Marketing
 * 2) Supply Chain Management: SCM
 * 3) Cloud Computing: Cloud Computing
 * 4) * XML: XML
 * 5) Data Mashup: Mashup
 * 6) * Personal Portal: Web Portal
 * 7) * Cascading Style Sheets: CSS
 * 8) Smart Object: Smart Objects
 * 9) * Software Agents: Software Agents

 * 1) Presentations
 * 2) Final

Other Links

 * EDGAR US Securities & Exchange listings
 * MITRE Emerging Technologies
 * NIST Available Technologies
 * How Stuff Works
 * R&D Magazine
 * Wiki Technology Portal
 * Technology Review (MIT)
 * Wiki 101 Instructions
 * Yale Tomorrow


text:
User:Etabojr

Darlingtony Etabo Ekuwom student of Moi University-Kenya a major of Community Development and Philosophy.

title: User:Etabojr


text:
User talk:Redcamarocruiser

 Hello Redcamarocruiser, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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title: User talk:Redcamarocruiser


text: User:Techexplo/Schedule

title: User:Techexplo/Schedule


text:
Talk:Astrophysics

I've done some work to this. Well? --StalinUpstaged 08:52, 29 November 2009 (UTC)

title: Talk:Astrophysics


text:
User:Egm6321.f09.Team1.andy/HW7

= Problem 1: Odd and Even Solutions of Legendre DE =

From Lecture Slide 37-1.

Given
From lecture slide 31-3, general expression for Legendre Polynomial is given by,

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{1 \cdot 3 \dots \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * },
 * },

and from lecture slide 37-1, the general expression for the second solution of Legendre Differential Equation is given as,

 * {| style="width:100%" border="0" align="left"

Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3,5...}^{J} \frac{2n-2j+1}{(2n-j+1)j} P_{n-j}(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * },
 * },

where $$\displaystyle J $$ is given by,

 * {| style="width:100%" border="0" align="left"

J := 1 + 2 \left\lfloor \frac{n-1}{2} \right\rfloor $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }.
 * }.

Find
Using Equation (2), to show when $$\displaystyle Q_n $$ is even and odd, depending on whether "$$\displaystyle n $$" is even or odd.

Solution
We know, from Odd and Even property of $$\displaystyle P_n $$, that $$\displaystyle P_n $$ is odd, when $$\displaystyle n $$ is odd and even, when $$\displaystyle n $$ is even and the fact that

 * {| style="width:100%" border="0" align="left"

tanh^{-1}(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * },
 * },

is odd.

''' (i). When $$\displaystyle n $$ is odd.'''

When $$\displaystyle n $$ is odd, $$\displaystyle P_n $$ is odd and so,

 * {| style="width:100%" border="0" align="left"

P_n(x)tanh^{-1}(x) $$
 * $$\displaystyle
 * $$\displaystyle
 * },
 * },

is even.

 * {| style="width:100%" border="0" align="left"

So, when $$\displaystyle n $$ is odd, first term in Equation (2) is even.
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

When $$\displaystyle n $$ is odd, value of $$\displaystyle n - j $$ is even since $$\displaystyle j $$ values are odd,  and so, $$\displaystyle P_{n-j} $$ is even.

 * {| style="width:100%" border="0" align="left"

So, when $$\displaystyle n $$ is odd, all the terms in the summation of Equation (2) are even.
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

And so,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Rightarrow $$ So, when $$\displaystyle n $$ is odd, $$\displaystyle Q_n $$ is even.
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

''' (ii). When $$\displaystyle n $$ is even.'''

When $$\displaystyle n $$ is even, $$\displaystyle P_n $$ is even and so,

 * {| style="width:100%" border="0" align="left"

P_n(x)tanh^{-1}(x) $$
 * $$\displaystyle
 * $$\displaystyle
 * },
 * },

is odd.

 * {| style="width:100%" border="0" align="left"

So, when $$\displaystyle n $$ is even, first term in Equation (2) is odd.
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

When $$\displaystyle n $$ is even, value of $$\displaystyle n - j $$ is odd since $$\displaystyle j $$ values are odd, and so, $$\displaystyle P_{n-j} $$ is odd.

 * {| style="width:100%" border="0" align="left"

So, when $$\displaystyle n $$ is even, all the terms in the summation of Equation (2) are odd.
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

And so,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Rightarrow $$ So, when $$\displaystyle n $$ is even, $$\displaystyle Q_n $$ is odd.
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }.
 * }.

= Problem 2: Legendre Solutions Plot =

From Lecture Slide 37-1.

Given
From lecture slide 31-3, general expression for Legendre Polynomial is given by,

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{1 \cdot 3 \dots \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * },
 * },

and from lecture slide 37-1, the general expression for the second solution of Legendre Differential Equation is given as,

 * {| style="width:100%" border="0" align="left"

Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3,5...}^{J} \frac{2n-2j+1}{(2n-j+1)j} P_{n-j}(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * },
 * },

where $$\displaystyle J $$ is given by,

 * {| style="width:100%" border="0" align="left"

J := 1 + 2 \left\lfloor \frac{n-1}{2} \right\rfloor $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }.
 * }.

Find
Plot {$$\displaystyle P_0, P_1, P_2, P_3, P_4$$ } and {$$\displaystyle Q_0, Q_1, Q_2, Q_3, Q_4$$ } using Matlab.

Solution
Please refer Problem 6 Solution of HW6, for $$\displaystyle P_0(x),\ P_1(x),\ P_2(x),\ P_3(x),\ P_4(x) $$.

Please refer, p. 33, for expressions of $$\displaystyle Q_0(x), Q_1(x), Q_2(x), Q_3(x)$$.

The expression for $$\displaystyle Q_4(x)$$ is given by,

 * {| style="width:100%" border="0" align="left"

Q_4= P_4(x) tanh^{-1}(x) - 2 \sum_{j=1,3}^{} \frac{2n-2j+1}{(2n-j+1)j} P_{n-j}(x) $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

 * {| style="width:100%" border="0" align="left"

\Rightarrow Q_4= \left[\frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8}\right] tanh^{-1}(x) - 2 \left[ \frac{2(4)-2+1}{(2(4)-1+1)1} P_{3}(x) \right] - 2 \left[ \frac{2(4)-2(3)+1}{(2(4)-3+1)3} P_{1}(x) \right] $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

 * {| style="width:100%" border="0" align="left"

\Rightarrow Q_4= \left[\frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8}\right] tanh^{-1}(x) - \frac{7}{8} \left[ 5 x ^ 3 - 3 x\right] - \frac{1}{3} x $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

 Matlab Code:

 Plots:

The plots for $$\displaystyle P_n(x) $$ with $$\displaystyle n = 0,1,2,3,4 $$ are shown below:

.

The plots for $$\displaystyle Q_n(x) $$ with $$\displaystyle n = 0,1,2,3,4 $$ are shown below:

.

= Problem 3: Orthogonality of Legendre Solutions =

From Lecture Slide 37-2.

Given
From lecture slide 31-3, general expression for Legendre Polynomial is given by,

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{1 \cdot 3 \dots \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * },
 * },

and from lecture slide 37-1, the general expression for the second solution of Legendre Differential Equation is given as,

 * {| style="width:100%" border="0" align="left"

Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3,5...}^{J} \frac{2n-2j+1}{(2n-j+1)j} P_{n-j}(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * },
 * },

where $$\displaystyle J $$ is given by,

 * {| style="width:100%" border="0" align="left"

J := 1 + 2 \left\lfloor \frac{n-1}{2} \right\rfloor $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }.
 * }.

Find
Show that, $$\displaystyle < P_n, Q_n > = 0$$.

Solution
From lecture slide 33-1, the inner (scalar) product is given by,

 * {| style="width:100%" border="0" align="left"

 = \int_{-1}^{+1} G(x)F(x)dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)

 * },
 * },

and Equation (4) becomes zero, when $$\displaystyle G(x) $$ is even and $$\displaystyle F(x) $$ is odd or when $$\displaystyle G(x) $$ is odd and $$\displaystyle F(x) $$ is even.

For $$\displaystyle P_n(x) $$ and $$\displaystyle Q_n(x) $$ using Equation (4),

 * {| style="width:100%" border="0" align="left"

 = \int_{-1}^{+1} P_n(x)Q_n(x)dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)

 * }.
 * }.

From Solution to Problem 1 above, $$\displaystyle Q_n(x) $$ is odd and $$\displaystyle P_n(x) $$ is even, when $$\displaystyle n $$ is even. When $$\displaystyle n $$ is odd, $$\displaystyle Q_n(x) $$ is even and $$\displaystyle P_n(x) $$ is odd.

So, for all values of $$\displaystyle n $$, $$\displaystyle Q_n(x) $$ is opposite to $$\displaystyle P_n(x) $$ in oddness and evenness, which means,

 * {| style="width:100%" border="0" align="left"

\int_{-1}^{+1} P_n(x)Q_n(x)dx = 0 $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

And so,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle  = 0 $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }.
 * }.

= Problem 4: Integration by Parts: Legendre Solutions = From lecture slide 37-3.

Given
From lecture slide 37-2,

 * {| style="width:100%" border="0" align="left"

\alpha = \int_{-1}^{+1} L_m \left[ (1-x^2) L'_n \right]' dx $$ $$ where $$\displaystyle L_n$$ and $$\displaystyle L_m$$ are solutions to the Legendre differential equation.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
Show that
 * {| style="width:100%" border="0" align="left"

\alpha = -\int_{-1}^{+1} (1-x^2) L'_n L'_m dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * },
 * },

after integration by parts of $$\displaystyle \alpha $$ in Equation (1).

Solution
Let,

 * {| style="width:100%" border="0" align="left"

u = L_m \Rightarrow du = L'_m dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * },
 * },

and,

 * {| style="width:100%" border="0" align="left"

dv = \left[ (1-x^2) L'_n \right]' dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }.
 * }.

 * {| style="width:100%" border="0" align="left"

\Rightarrow \int dv = \int\left[ (1-x^2) L'_n \right]' dx \Rightarrow v = \left[ (1-x^2) L'_n \right] $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

We know,

 * {| style="width:100%" border="0" align="left"

\int_{a}^{b} udv = \left[uv \right]_{a}^{b} - \int_{a}^{b} vdu $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * },
 * },

and so, with $$\displaystyle a = -1 $$ and $$\displaystyle b = +1 $$

 * {| style="width:100%" border="0" align="left"

\alpha = \int_{-1}^1 L_m \left[ (1-x^2) L'_n \right]' dx = \left[ L_m(1-x^2) L'_n\right]_{-1}^{+1} - \int_{-1}^{+1} (1-x^2) L'_n L'_m dx $$
 * $$\displaystyle
 * $$\displaystyle

$$
 *
$$\displaystyle (Eq. 6)

 * },
 * },

For values of $$\displaystyle x = -1 $$ and $$\displaystyle x = +1 $$,

 * {| style="width:100%" border="0" align="left"

(1-x^2) = 0 $$
 * $$\displaystyle
 * $$\displaystyle

$$
 *
$$\displaystyle (Eq. 7)

 * }.
 * }.

Substituting Equation (7) in first term of Equation (6), first term becomes zero.

 * {| style="width:100%" border="0" align="left"

\alpha = \cancelto{0}{\left[ L_m(1-x^2) L'_n\right]_{-1}^{+1}} - \int_{-1}^{+1} (1-x^2) L'_n L'_m dx $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

So,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \alpha = - \int_{-1}^{+1} (1-x^2) L'_n L'_m dx $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }.
 * }.

= Problem 5: Attraction of Spheres = From lecture slide 38-2.

Given
From lecture slide 38-2,

 * {| style="width:100%" border="0" align="left"

(r_{PQ})^2 = \sum_{i=1}^{3} \left(x_Q^i - x_P^i\right)^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * },
 * },

where,

$$\displaystyle x_P^1 = x_P,\ x_P^2 = y_P,\ x_P^3 = z_P,\ $$

and,

$$\displaystyle x_Q^1 = x_Q,\ x_Q^2 = y_Q,\ x_Q^3 = z_Q,\ $$

given by,

 * {| style="width:100%" border="0" align="left"

x_P = r_P cos(\theta_P) cos(\psi_P);\ y_P = r_P cos(\theta_P) sin(\psi_P);\ z_P = r_P sin(\theta_P) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * },
 * },

and,

 * {| style="width:100%" border="0" align="left"

x_Q = r_Q cos(\theta_Q) cos(\psi_Q);\ y_Q = r_Q cos(\theta_Q) sin(\psi_Q);\ z_Q = r_Q sin(\theta_Q) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }.
 * }.

Find
Show that,

 * {| style="width:100%" border="0" align="left"

(r_{PQ})^2 = (r_P)^2 + (r_Q)^2 - 2 (r_P) (r_Q) cos\gamma $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * },
 * },

where,

 * {| style="width:100%" border="0" align="left"

cos\gamma = cos(\theta_Q) cos(\theta_P) cos(\psi_Q-\psi_P) + sin(\theta_Q)sin(\theta_P) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Solution
Equation (1) can be written as,

 * {| style="width:100%" border="0" align="left"

(r_{PQ})^2 = \left(x_Q - x_P\right)^2 + \left(y_Q - y_P\right)^2 + \left(z_Q - z_P\right)^2 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }.
 * }.

Substituting Equations (2) and (3) in (6),

 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow (r_{PQ})^2 = & \left[r_Q cos(\theta_Q) cos(\psi_Q) - r_P cos(\theta_P) cos(\psi_P)\right]^2 + \\ & \left[r_Q cos(\theta_Q) sin(\psi_Q) - r_P cos(\theta_P) sin(\psi_P)\right]^2 + \\ & \left[r_Q sin(\theta_Q) - r_P sin(\theta_P)\right]^2 \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow (r_{PQ})^2 = & \left[r_Q^2 cos^2(\theta_Q) cos^2(\psi_Q) + r_P^2 cos^2(\theta_P) cos^2(\psi_P) - 2 r_P r_Q cos(\theta_Q) cos(\psi_Q) cos(\theta_P) cos(\psi_P) \right] + \\ & \left[r_Q^2 cos^2(\theta_Q) sin^2(\psi_Q) + r_P^2 cos^2(\theta_P) sin^2(\psi_P) - 2 r_P r_Q cos(\theta_Q) sin(\psi_Q) cos(\theta_P) sin(\psi_P)\right] + \\ & \left[r_Q^2 sin^2(\theta_Q) + r_P^2 sin^2(\theta_P) - 2 r_P r_Q sin(\theta_Q) sin(\theta_P)\right] \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

 * }.
 * }.

 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow (r_{PQ})^2 = &\ r_Q^2 cos^2(\theta_Q) \cancelto{1}{\left[ cos^2(\psi_Q) + sin^2(\psi_Q) \right]} + r_Q^2 sin^2(\theta_Q) -  2 r_P r_Q cos(\theta_Q) cos(\theta_P) \left [cos(\psi_Q) cos(\psi_P) \right] + \\ &\ r_P^2 cos^2(\theta_P) \cancelto{1}{\left[ cos^2(\psi_P) + sin^2(\psi_P) \right]} + r_P^2 sin^2(\theta_P) - 2 r_P r_Q cos(\theta_Q)cos(\theta_P) \left [sin(\psi_Q) sin(\psi_P)\right] - 2 r_P r_Q sin(\theta_Q) sin(\theta_P)
 * $$\displaystyle
 * $$\displaystyle

\end{align} $$

 * }.
 * }.

 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow (r_{PQ})^2 = &\ r_Q^2 \cancelto{1}{\left[cos^2(\theta_Q) + sin^2(\theta_Q) \right]} -  2 r_P r_Q cos(\theta_Q) cos(\theta_P) \underbrace{\left [cos(\psi_Q) cos(\psi_P) + sin(\psi_Q) sin(\psi_P)\right]}_{=\ cos(\psi_Q-\psi_P)} + \\ &\ r_P^2 \cancelto{1}{\left[ cos^2(\theta_P) + sin^2(\theta_P) \right]} - 2 r_P r_Q sin(\theta_Q) sin(\theta_P)
 * $$\displaystyle
 * $$\displaystyle

\end{align} $$

 * }.
 * }.

 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow (r_{PQ})^2 = &\ r_Q^2 + r_P^2 -  2 r_P r_Q \left[\underbrace{cos(\theta_Q) cos(\theta_P) {cos(\psi_Q-\psi_P)} + sin(\theta_Q)sin(\theta_P)}_{=\ cos\gamma} \right]
 * $$\displaystyle
 * $$\displaystyle

\end{align} $$

 * }
 * }

And so,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle (r_{PQ})^2 = (r_P)^2 + (r_Q)^2 - 2 (r_P) (r_Q) cos\gamma $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }.
 * }.

= Notes and References =

title: User:Egm6321.f09.Team1.andy/HW7


text:
User talk:Techexplo

Students, please feel free to make comments on class, assignments in this area --Techexplo 09:09, 29 November 2009 (UTC)

title: User talk:Techexplo


text:
User:Egm6321.f09.Team1/HW7

= Problem 1: Odd and Even Solutions of Legendre DE =

From Lecture Slide 37-1.

Given
From lecture slide 31-3, general expression for Legendre Polynomial is given by,

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{1 \cdot 3 \dots \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i}, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

and from lecture slide 37-1, the general expression for the second solution of Legendre Differential Equation is given as,

 * {| style="width:100%" border="0" align="left"

Q_n(x) = P_n(x) \tanh^{-1}(x) - 2 \sum_{j=1,3,5...}^{J} \frac{2n-2j+1}{(2n-j+1)j} P_{n-j}(x), $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

where $$\displaystyle J $$ is given by,

 * {| style="width:100%" border="0" align="left"

J := 1 + 2 \left\lfloor \frac{n-1}{2} \right\rfloor. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Find
Using Equation (2), to show when $$\displaystyle Q_n $$ is even or odd, depending on whether "$$\displaystyle n $$" is even or odd.

Solution
We know, from Odd and Even property of $$\displaystyle P_n $$, that $$\displaystyle P_n $$ is odd, when $$\displaystyle n $$ is odd and even, when $$\displaystyle n $$ is even and the fact that

 * {| style="width:100%" border="0" align="left"

\tanh^{-1}(x), $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

is odd.

''' (i). When $$\displaystyle n $$ is odd.'''

When $$\displaystyle n $$ is odd, $$\displaystyle P_n $$ is odd and so,

 * {| style="width:100%" border="0" align="left"

P_n(x)\tanh^{-1}(x), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

is even.

 * {| style="width:100%" border="0" align="left"

So, when $$\displaystyle n $$ is odd, first term in Equation (2) is even.
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

When $$\displaystyle n $$ is odd, value of $$\displaystyle n - j $$ is even since $$\displaystyle j $$ values are odd,  and so, $$\displaystyle P_{n-j} $$ is even.

 * {| style="width:100%" border="0" align="left"

So, when $$\displaystyle n $$ is odd, all the terms in the summation of Equation (2) are even.
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

And so,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Rightarrow $$ So, when $$\displaystyle n $$ is odd, $$\displaystyle Q_n $$ is even.
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

''' (ii). When $$\displaystyle n $$ is even.'''

When $$\displaystyle n $$ is even, $$\displaystyle P_n $$ is even and so,

 * {| style="width:100%" border="0" align="left"

P_n(x)\tanh^{-1}(x), $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

is odd.

 * {| style="width:100%" border="0" align="left"

So, when $$\displaystyle n $$ is even, first term in Equation (2) is odd.
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

When $$\displaystyle n $$ is even, value of $$\displaystyle n - j $$ is odd since $$\displaystyle j $$ values are odd, and so, $$\displaystyle P_{n-j} $$ is odd.

 * {| style="width:100%" border="0" align="left"

So, when $$\displaystyle n $$ is even, all the terms in the summation of Equation (2) are odd.
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * style="width:50%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

And so,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \Rightarrow $$ So, when $$\displaystyle n $$ is even, $$\displaystyle Q_n $$ is odd.
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style= |
 * }
 * }

= Problem 2: Legendre Solutions Plot =

From Lecture Slide 37-1.

Given
From lecture slide 31-3, general expression for Legendre Polynomial is given by,

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{1 \cdot 3 \dots \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

and from lecture slide 37-1, the general expression for the second solution of Legendre Differential Equation is given as,

 * {| style="width:100%" border="0" align="left"

Q_n(x) = P_n(x) \tanh^{-1}(x) - 2 \sum_{j=1,3,5...}^{J} \frac{2n-2j+1}{(2n-j+1)j} P_{n-j}(x), $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

where $$\displaystyle J $$ is given by,

 * {| style="width:100%" border="0" align="left"

J := 1 + 2 \left\lfloor \frac{n-1}{2} \right\rfloor. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Find
Plot {$$\displaystyle P_0, P_1, P_2, P_3, P_4$$ } and {$$\displaystyle Q_0, Q_1, Q_2, Q_3, Q_4$$ } using Matlab.

Solution
Please refer Problem 6 Solution of HW6, for $$\displaystyle P_0(x),\ P_1(x),\ P_2(x),\ P_3(x),\ P_4(x) $$.

Please refer, p. 33, for expressions of $$\displaystyle Q_0(x), Q_1(x), Q_2(x), Q_3(x)$$.

The expression for $$\displaystyle Q_4(x)$$ is given by,

 * {| style="width:100%" border="0" align="left"

Q_4= P_4(x) \tanh^{-1}(x) - 2 \sum_{j=1,3}^{} \frac{2n-2j+1}{(2n-j+1)j} P_{n-j}(x) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\Rightarrow Q_4= \left[\frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8}\right] \tanh^{-1}(x) - 2 \left[ \frac{2(4)-2+1}{(2(4)-1+1)1} P_{3}(x) \right] - 2 \left[ \frac{2(4)-2(3)+1}{(2(4)-3+1)3} P_{1}(x) \right] $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\Rightarrow Q_4= \left[\frac{35}{8} x ^ 4 - \frac{15}{4} x ^ 2 + \frac{3}{8}\right] \tanh^{-1}(x) - \frac{7}{8} \left[ 5 x ^ 3 - 3 x\right] - \frac{1}{3} x $$
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

 Matlab Code:

 Plots:

The plots for $$\displaystyle P_n(x) $$ with $$\displaystyle n = 0,1,2,3,4 $$ are shown below:

.

The plots for $$\displaystyle Q_n(x) $$ with $$\displaystyle n = 0,1,2,3,4 $$ are shown below:

.

= Problem 3: Orthogonality of Legendre Solutions =

From Lecture Slide 37-2.

Given
From lecture slide 31-3, general expression for Legendre Polynomial is given by,

 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{\lfloor n/2 \rfloor} \frac{1 \cdot 3 \dots \left(2n - 2i - 1\right)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

and from lecture slide 37-1, the general expression for the second solution of Legendre Differential Equation is given as,

 * {| style="width:100%" border="0" align="left"

Q_n(x) = P_n(x) \tanh^{-1}(x) - 2 \sum_{j=1,3,5...}^{J} \frac{2n-2j+1}{(2n-j+1)j} P_{n-j}(x), $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

where $$\displaystyle J $$ is given by,

 * {| style="width:100%" border="0" align="left"

J := 1 + 2 \left\lfloor \frac{n-1}{2} \right\rfloor $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Find
Show that, $$\displaystyle < P_n, Q_n > = 0$$.

Solution
From lecture slide 33-1, the inner (scalar) product is given by,

 * {| style="width:100%" border="0" align="left"

 = \int_{-1}^{+1} G(x)F(x)dx, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)

 * }
 * }

and Equation (4) becomes zero, when $$\displaystyle G(x) $$ is even and $$\displaystyle F(x) $$ is odd or when $$\displaystyle G(x) $$ is odd and $$\displaystyle F(x) $$ is even.

For $$\displaystyle P_n(x) $$ and $$\displaystyle Q_n(x) $$ using Equation (4),

 * {| style="width:100%" border="0" align="left"

 = \int_{-1}^{+1} P_n(x)Q_n(x)dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
 * }
 * }

From Solution to Problem 1 above, $$\displaystyle Q_n(x) $$ is odd and $$\displaystyle P_n(x) $$ is even, when $$\displaystyle n $$ is even. When $$\displaystyle n $$ is odd, $$\displaystyle Q_n(x) $$ is even and $$\displaystyle P_n(x) $$ is odd.

So, for all values of $$\displaystyle n $$, $$\displaystyle Q_n(x) $$ is opposite to $$\displaystyle P_n(x) $$ in oddness and evenness, which means,

 * {| style="width:100%" border="0" align="left"

\int_{-1}^{+1} P_n(x)Q_n(x)dx = 0. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

And so,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle  = 0 $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

= Problem 4: Integration by Parts: Legendre Solutions = From lecture slide 37-3.

Given
From lecture slide 37-2,

 * {| style="width:100%" border="0" align="left"

\alpha = \int_{-1}^{+1} L_m \left[ (1-x^2) L'_n \right]' dx $$ $$ where $$\displaystyle L_n$$ and $$\displaystyle L_m$$ are solutions to the Legendre differential equation.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Find
Show that

 * {| style="width:100%" border="0" align="left"

\alpha = -\int_{-1}^{+1} (1-x^2) L'_n L'_m dx, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

after integration by parts of $$\displaystyle \alpha $$ in Equation (1).

Solution
Let,

 * {| style="width:100%" border="0" align="left"

u = L_m \Rightarrow du = L'_m dx, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

and,

 * {| style="width:100%" border="0" align="left"

dv = \left[ (1-x^2) L'_n \right]' dx. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\Rightarrow \int dv = \int\left[ (1-x^2) L'_n \right]' dx \Rightarrow v = \left[ (1-x^2) L'_n \right]. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

We know,

 * {| style="width:100%" border="0" align="left"

\int_{a}^{b} udv = \left[uv \right]_{a}^{b} - \int_{a}^{b} vdu , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

and so, with $$\displaystyle a = -1 $$ and $$\displaystyle b = +1 $$

 * {| style="width:100%" border="0" align="left"

\alpha = \int_{-1}^1 L_m \left[ (1-x^2) L'_n \right]' dx = \left[ L_m(1-x^2) L'_n\right]_{-1}^{+1} - \int_{-1}^{+1} (1-x^2) L'_n L'_m dx , $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

For values of $$\displaystyle x = -1 $$ and $$\displaystyle x = +1 $$,

 * {| style="width:100%" border="0" align="left"

(1-x^2) = 0. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }

Substituting Equation (7) in first term of Equation (6), first term becomes zero.

 * {| style="width:100%" border="0" align="left"

\alpha = \cancelto{0}{\left[ L_m(1-x^2) L'_n\right]_{-1}^{+1}} - \int_{-1}^{+1} (1-x^2) L'_n L'_m dx. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

So,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle \alpha = - \int_{-1}^{+1} (1-x^2) L'_n L'_m dx $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

= Problem 5: Attraction of Spheres = From lecture slide 38-2.

Given
From lecture slide 38-2,

 * {| style="width:100%" border="0" align="left"

(r_{PQ})^2 = \sum_{i=1}^{3} \left(x_Q^i - x_P^i\right)^2, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }

where,

$$\displaystyle x_P^1 = x_P,\ x_P^2 = y_P,\ x_P^3 = z_P,\ $$

and,

$$\displaystyle x_Q^1 = x_Q,\ x_Q^2 = y_Q,\ x_Q^3 = z_Q,\ $$

given by,

 * {| style="width:100%" border="0" align="left"

x_P = r_P \cos(\theta_P) \cos(\psi_P);\ y_P = r_P \cos(\theta_P) \sin(\psi_P);\ z_P = r_P \sin(\theta_P), $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

and,

 * {| style="width:100%" border="0" align="left"

x_Q = r_Q \cos(\theta_Q) \cos(\psi_Q);\ y_Q = r_Q \cos(\theta_Q) \sin(\psi_Q);\ z_Q = r_Q \sin(\theta_Q). $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }

Find
Show that,

 * {| style="width:100%" border="0" align="left"

(r_{PQ})^2 = (r_P)^2 + (r_Q)^2 - 2 (r_P) (r_Q) \cos\gamma, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

where,

 * {| style="width:100%" border="0" align="left"

\cos\gamma = \cos(\theta_Q) \cos(\theta_P) \cos(\psi_Q-\psi_P) + \sin(\theta_Q)\sin(\theta_P) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Solution
Equation (1) can be written as,

 * {| style="width:100%" border="0" align="left"

(r_{PQ})^2 = \left(x_Q - x_P\right)^2 + \left(y_Q - y_P\right)^2 + \left(z_Q - z_P\right)^2. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }

Substituting Equations (2) and (3) in (6),

 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow (r_{PQ})^2 = & \left[r_Q \cos(\theta_Q) \cos(\psi_Q) - r_P \cos(\theta_P) \cos(\psi_P)\right]^2 + \\ & \left[r_Q \cos(\theta_Q) \sin(\psi_Q) - r_P \cos(\theta_P) \sin(\psi_P)\right]^2 + \\ & \left[r_Q \sin(\theta_Q) - r_P \sin(\theta_P)\right]^2 \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow (r_{PQ})^2 = & \left[r_Q^2 \cos^2(\theta_Q) \cos^2(\psi_Q) + r_P^2 \cos^2(\theta_P) \cos^2(\psi_P) - 2 r_P r_Q \cos(\theta_Q) \cos(\psi_Q) \cos(\theta_P) \cos(\psi_P) \right] + \\ & \left[r_Q^2 \cos^2(\theta_Q) \sin^2(\psi_Q) + r_P^2 \cos^2(\theta_P) \sin^2(\psi_P) - 2 r_P r_Q \cos(\theta_Q) \sin(\psi_Q) \cos(\theta_P) \sin(\psi_P)\right] + \\ & \left[r_Q^2 \sin^2(\theta_Q) + r_P^2 \sin^2(\theta_P) - 2 r_P r_Q \sin(\theta_Q) \sin(\theta_P)\right] \end{align} $$
 * $$\displaystyle
 * $$\displaystyle

 * }
 * }

 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow (r_{PQ})^2 = &\ r_Q^2 \cos^2(\theta_Q) \cancelto{1}{\left[ \cos^2(\psi_Q) + \sin^2(\psi_Q) \right]} + r_Q^2 \sin^2(\theta_Q) -  2 r_P r_Q \cos(\theta_Q) \cos(\theta_P) \left [\cos(\psi_Q) \cos(\psi_P) \right] + \\ &\ r_P^2 \cos^2(\theta_P) \cancelto{1}{\left[ \cos^2(\psi_P) + \sin^2(\psi_P) \right]} + r_P^2 \sin^2(\theta_P) - 2 r_P r_Q \cos(\theta_Q)\cos(\theta_P) \left [\sin(\psi_Q) \sin(\psi_P)\right] - 2 r_P r_Q \sin(\theta_Q) \sin(\theta_P) \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow (r_{PQ})^2 = &\ r_Q^2 \cancelto{1}{\left[\cos^2(\theta_Q) + \sin^2(\theta_Q) \right]} -  2 r_P r_Q \cos(\theta_Q) \cos(\theta_P) \underbrace{\left [\cos(\psi_Q) \cos(\psi_P) + \sin(\psi_Q) \sin(\psi_P)\right]}_{=\ \cos(\psi_Q-\psi_P)} + \\ &\ r_P^2 \cancelto{1}{\left[ \cos^2(\theta_P) + \sin^2(\theta_P) \right]} - 2 r_P r_Q \sin(\theta_Q) \sin(\theta_P)
 * $$\displaystyle
 * $$\displaystyle

\end{align} $$
 * }
 * }

 * {| style="width:100%" border="0" align="left"

\begin{align} \Rightarrow (r_{PQ})^2 = &\ r_Q^2 + r_P^2 -  2 r_P r_Q \left[\underbrace{\cos(\theta_Q) \cos(\theta_P) {\cos(\psi_Q-\psi_P)} + \sin(\theta_Q)\sin(\theta_P)}_{=\ \cos\gamma} \right] \end{align} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

And so,

 * {| style="width:100%" border="0" align="left"

$$\displaystyle (r_{PQ})^2 = (r_P)^2 + (r_Q)^2 - 2 (r_P) (r_Q) \cos\gamma $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

= Problem 6: Binomial Series = From lecture note slide 38-4.

Given
The binomial series expansion is
 * {| style="width:100%" border="0" align="left"

(x+y)^r = \sum_{k=0}^\infty \begin{pmatrix} r \\ k \end{pmatrix} x^{r-k} y^k $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\begin{pmatrix} r \\ k \end{pmatrix} = \frac{r(r-1) \cdot\cdot\cdot (r-k+1)}{k!} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }

Find
Use Eqs. 1 and 2 to show that
 * {| style="width:100%" border="0" align="left"

(1-x)^{-1/2} = \sum_{i=0}^\infty \alpha_i x^i $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\alpha_i = \frac{1 \cdot 3 \cdot ... \cdot (2i - 1)}{2 \cdot 4 \cdot ... \cdot (2i)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Solution
Using Eq. 3, we can rewrite the LHS of Eq. 1 as
 * {| style="width:100%" border="0" align="left"

(1-x)^{-1/2} = \sum_{k=0}^\infty \begin{pmatrix} -1/2 \\ k \end{pmatrix} 1^{-1/2-k} (-x)^k $$ $$ We can expand this
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(1-x)^{-1/2} = \sum_{k=0}^\infty \frac{(-1/2)(-3/2)\cdot\cdot\cdot(1/2-k)}{k!} (-x)^k $$ $$ The number of factors in the numerator is $$\displaystyle k$$, hence we can rewrite this as
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(1-x)^{-1/2} = \sum_{k=0}^\infty \frac{(1)(3)\cdot\cdot\cdot(2k-1)}{(-2)^k k!} (-x)^k $$ $$ Change $$\displaystyle k \rightarrow i$$, and cancel two minus signs
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(1-x)^{-1/2} = \sum_{i=0}^\infty \frac{(1)(3)\cdot\cdot\cdot(2i-1)}{2^i i!} x^i $$ $$ The denominator can then be expanded
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(1-x)^{-1/2} = \sum_{i=0}^\infty \underbrace{\frac{1 \cdot 3\cdot\cdot\cdot(2i-1)}{2 \cdot 4 \cdot \cdot \cdot (2i)}}_{\alpha_i} x^i $$ $$ So that
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (1-x)^{-1/2} = \sum_{i=0}^\infty \alpha_i x^i $$ $$ where
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \alpha_i = \frac{1 \cdot 3\cdot\cdot\cdot(2i-1)}{2 \cdot 4 \cdot \cdot \cdot (2i)} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 11)
 * }
 * }

= Problem 7: Generating Functon for Legendre Polynomials = From lecture note slide 39-2.

Given
We have that
 * {| style="width:100%" border="0" align="left"

A(\mu, \rho) := 1 - 2 \mu \rho + \rho^2 $$ $$ and the binomial expansion
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

(1-x)^{-1/2} = \sum_{i=0}^\infty \alpha_i x^i $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\alpha_i = \frac{1 \cdot 3 \cdot ... \cdot (2i - 1)}{2 \cdot 4 \cdot ... \cdot (2i)} $$ $$ Using Eqs. 1 and 2, we can expand the expression
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

[A(\mu,\rho)]^{-1/2} = \alpha_0 + \alpha_1(2\mu \rho- \rho^2) + \alpha_2(2\mu \rho- \rho^2)^2 + ... $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

= \underbrace{\alpha_0}_{P_0(\mu)} + \underbrace{2 \mu \alpha_1}_{P_1(\mu)} \rho + \underbrace{(- \alpha_1 + 4 \mu^2 \alpha_2)}_{P_2(\mu)} \rho^2 + ... $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 4)
 * }
 * }

Find
Continue the expansion given by Eq. 4 to yield $$\displaystyle P_3$$, $$\displaystyle P_4$$, and $$\displaystyle P_5$$, and compare the result to that obtained by Eq. 7 on lecture note slide 31-3
 * {| style="width:100%" border="0" align="left"

P_n(x) = \sum_{i=0}^{[n/2]} = \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 5)
 * }
 * }

Solution
We need the first 6 terms of the expansion
 * {| style="width:100%" border="0" align="left"

[A(\mu,\rho)]^{-1/2} = \alpha_0 + \alpha_1(2\mu \rho- \rho^2) + \alpha_2(2\mu \rho- \rho^2)^2 + \alpha_3(2\mu \rho- \rho^2)^3 + \alpha_4(2\mu \rho- \rho^2)^4 + \alpha_5(2\mu \rho- \rho^2)^5 + ... $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

= \alpha_0 $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ \alpha_1(2\mu \rho - \rho^2) $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ \alpha_2(4\mu^2 \rho^2 - 4\mu \rho^3 + \rho^4) $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ \alpha_3 (8\mu^3 \rho^3 - 12\mu^2 \rho^4 + 6\mu \rho^5 -  \rho^6) $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ \alpha_4 (16\mu^4 \rho^4 - 32\mu^3 \rho^5 +  24\mu^2 \rho^6 - 8 \mu \rho^7 +  \rho^8) $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ \alpha_5 (32\mu^5 \rho^5 - 80 \mu^4 \rho^6 + 80 \mu^3 \rho^7 - 40 \mu^2 \rho^8 +  10 \mu \rho^9 -  \rho^{10}) + ... $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

= \alpha_0 + \alpha_1 \rho $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ (-\alpha_1 + 4 \alpha_2 \mu^2) \rho^2 $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ (- 4 \alpha_2 \mu + 8 \alpha_3 \mu^3) \rho^3 $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ (\alpha_2 - 12 \alpha_3 \mu^2 + 16 \alpha_4 \mu^4) \rho^4 $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ (6 \alpha_3 \mu - 32 \alpha_4 \mu^3 + 32 \alpha_5 \mu^5) \rho^5 + ... $$ $$ Using Eq. 3, we have that
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 6)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

\alpha_0 = 1 \quad \alpha_1 = \frac{1}{2} \quad \alpha_2 = \frac{3}{8} \quad \alpha_3 = \frac{15}{48} \quad \alpha_4 = \frac{105}{384} \quad \alpha_5 = \frac{945}{3840} \quad $$ we can substitute this into Eq. 6
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * }
 * {| style="width:100%" border="0" align="left"

= 1 + \frac{1}{2} \rho $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ \left(-\frac{1}{2} + \frac{3}{2} \mu^2\right) \rho^2 $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ \left(- \frac{3}{2} \mu + \frac{5}{2} \mu^3\right) \rho^3 $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ \left(\frac{3}{8} - \frac{15}{4} \mu^2 + \frac{35}{8} \mu^4\right) \rho^4 $$
 * $$\displaystyle
 * $$\displaystyle
 *

 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

+ \left(\frac{15}{8} \mu - \frac{35}{4} \mu^3 + \frac{63}{8} \mu^5\right) \rho^5 + ... $$ $$ From the above, we can extract
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 7)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_3(\mu) = - \frac{3}{2} \mu + \frac{5}{2} \mu^3 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 8)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_4(\mu) = \frac{3}{8} - \frac{15}{4} \mu^2 + \frac{35}{8} \mu^4 $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 9)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_5(\mu) = \frac{15}{8} \mu - \frac{35}{4} \mu^3 + \frac{63}{8} \mu^5 $$ $$ Using Eq. 5, we get
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 10)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

P_3(x) = \sum_{i=0}^1 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 11)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_3(x) = \frac{1 \cdot 3 \cdot 5}{ 3!} x^3 - \frac{1 \cdot 3 }{2} x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 12)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_3(x) = \frac{5}{ 2} x^3 - \frac{3}{2} x $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 13)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_4(x) = \sum_{i=0}^2 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 14)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_4(x) = \frac{1 \cdot 3 \cdot ... \cdot 7}{4!} x^4 - \frac{1 \cdot 3 \cdot 5}{4} x^2 + \frac{1 \cdot 3 }{8} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 15)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_4(x) = \frac{35}{8} x^4 - \frac{15}{4} x^2 + \frac{3}{8} $$ $$
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 16)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_5(x) = \sum_{i=0}^2 \frac{1 \cdot 3 \cdot ... \cdot (2n - 2i - 1)}{2^i i! (n-2i)!} (-1)^i x^{n-2i} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 17)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

P_5(x) = \frac{1 \cdot 3 \cdot ... \cdot 9}{5!} x^5 - \frac{1 \cdot 3 \cdot ... \cdot 7}{2 \cdot 3!} x^3 + \frac{1 \cdot 3 \cdot 5}{8} x $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 18)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$\displaystyle P_5(x) = \frac{63}{8} x^5 - \frac{35}{4} x^3 + \frac{15}{8} x $$ $$ We have confirmed that the methods are equivalent, at least for terms up to $$\displaystyle P_5$$.
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 * style="width:2%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle (Eq. 19)
 * }
 * }

= Notes and References =

= Contributing Team Members =

Egm6321.f09.Team1.andy 12:03, 29 November 2009 (UTC)

Egm6321.f09.Team1.sallstrom 18:48, 2 December 2009 (UTC)

Egm6321.f09.Team1.vasquez 05:09, 5 December 2009 (UTC)

Egm6321.f09.Team1.AH 15:59, 8 December 2009 (UTC)

title: User:Egm6321.f09.Team1/HW7


text:
User:Egm6321.f09/Writing mediawiki articles with google doc

Almost-WYSiWYG wiki writing (Dr. Vu-Quoc)


text:
Political Science Construct

Overview
The Political Science Construct is the conceptualization of how political science is presented on Wikiversity. The Wikiversity School of Political Science builds student knowledge upon four pillars: Political Theory, Government & Politics, Comparative Politics and International Relations. Each of these concepts provides a unique perspective on the study of political science and serves as the core for more advanced and nuanced studies such as the Congress and U.S. Foreign Policy or African E.U. Policy. The -100 level of instruction provides an introduction to these four pillars with an emphasis on the major concepts within the pillar. More detailed instruction in the pillars occurs at the higher course levels. The School does offer a 200 level of instruction which focuses on research, a key takeaway from undergraduate instruction for those students interested in graduate studies. Of note, while graduate studies is not currently offered by the School of Political Science, this course of study will be added to future iterations of the School.

Political Theory
This is the fundamental pillar of Political Science as it provides the "why?" that describes the "what?", "where?" and "how?" in our course of study. This pillar explores the various political philosophers that have given rise to the field of political science and the intellectual contributions that the philosophers have made that make the subject matter distinct from other schools of study. According to Aristotle "[...] man is a political animal [...]" and thus humanity has developed, distinctive political modalities in the four corners of the planet which allow the study of political theory to be geographically diffuse. In other words, as a student broadens their understanding of the political theories of one side of the planet in addition to the other, their perspective broadens and may change but ultimately provides them with a much more nuanced understanding of global political dimensions.

Government & Politics
The Government & Politics pillar answers the "what?" question. This a descriptive pillar that seeks to deepen student knowledge about what states, nations and other organizations due to govern and how they politic among themselves. This is an extremely important pillar as it builds our knowledge base of what works and what doesn't work in terms of government and political structures. This pillar explores things like democracy, socialism, and autocracy as well the political parties within a particular political structure such as the political parties of Germany. Individuals that deepen their knowledge in this pillar will be experts in describing political structures for particular countries and organizations such as for Japan or the United Nations. This pillar is partially supported by the theories that may have affected the development of the governmental and political structure in a particular geographic area and serves as a partial support for the study of comparative politics and international relations.

title: Political Science Construct


text:
Comparative Teaching of Old Greek and Latin/Lesson 04 Part 1

Comparative Teaching of Old Greek and Latin Lesson 04 Part 1

Translated from the Greek Wikipedia|Wikiversity: "Συγκριτική διδασκαλία των κλασικών γλωσσών / (Comparative Teaching of the Classical Languages)"

'''LESSON 04  PART 1. The verb. Regular verbs. Present, imperfect, future indicative, active voice.'''

4.1.1. Present, imperfect, future indicative in active voice, of the Old Greek verbs of the first conjugation in -ω.

Verbs are the declinable words that show what makes or what suffers or in what situation the subject is found. Indicative is the mood of the real, the basic mood that is used mainly in main clauses as well as in subordinate clauses denoting real statements. It usually expresses something which the speaking person considers real or opposite to real or possible in the past or repeated in the past or a wish that cannot be realised etc. Old Greek Language also maintained both the thematic conjugation, i.e. the one that uses, in forming  verbal forms, the thematic vowels -ε-, -ο- (λύ-ο-μεν, λύ-ε-τε) of the  Indoeuropean Language, and the not thematic conjugation, the verb forms of which are formed, by adding the endings directly on the verb stem (ἵ-στα-μεν, τι-θέ-ασι).

Declension of the indicative in the active voice

Persons	/  Indoeuropean   /   Indicative present   /   Indicative imperfect   /  Indicative future

1st singular   /   *ag-o1,2,5,6	/         ἄγ-ω	         /         ἦγ-ον	           /   ἄξ-ω

2nd singular	/  *ag-e-si	        /         ἄγ-εις         /         ἦγ-ες	           /   ἄξ-εις

3rd singular	/  *ag-e-ti	        /         ἄγ-ει	         /         ἦγ-ε(ν)	   /   ἄξ-ει

1st plural	/  *ag-o-mes	        /         ἄγ-ο-μεν	 /         ἤγ-ομεν	   /   ἄξ-ομεν

2nd plural	/  *ag-e-te	        /         ἄγ-ε-τε	 /         ἤγ-ε-τε	   /   ἄξ-ε-τε

3rd plural     /   *ag-o-nti	        /  ἄγ-ουσι < ἄγ-ο-ντι	 /         ἦγ-ον           /    ἄξ-ουσι

2nd plural	/	 -             /        ἄγ-ε-τον        /	   ἤγ-ε-τον 	    /   ἄξε-τον

3rd plural	/	 -             /        ἄγ-ε-τον	 /         ἠγ-έτην	    /   ἄξε-τον

The thematic vowels -ε-, -ο- of the Indoeuropean Language were short. The thematic vowel -o- is used in front of nasals (μ,ν), while in the other cases it is used the thematic vowel -ε-. The present and future tenses use the endings of the initial tenses (-ω, -εις, -ει-, -ομεν, -ετε, -ουσι), the imperfect uses the endings of the historical tenses (-ον, -ες, -ε, -ομεν, -ετε, -ον). The imperfect tense, in order to denote the past, apart from the endings of the historical tenses, it also takes  the syllable augment (λύ-ω > ἔ-λυ-ον) or the time augment (ἄγ-ω > ἦγ-ον). The forms of the dual number will be analyzed in the texts.

4.1.2. Present, imperfect, future indicative, active voice, of the Latin verbs.

Latin kept only the thematic conjugation of the Indoeuropean Language which it developed in further four conjugations, the first one having as thematic vowel a long -a- (am-a'-re), the second one with thematic vowel a long -e- (del-e'-re), the third one with thematic vowel a short -i- (le'g-e-re) or, according to other linguists, without thematic vowel and the fourth one with thematic vowel a long -i- (aud-i'-re).

Declension of the indicative in the active voice.

Persons  /  Imperfect of the Indoeuropean	/  Indicative present	/   Indicative imperfect   /  Indicative future

1st singular /	*e-bher-o-m3,4,7	/       am-o	        /         ama’-ba-m	/       ama’-bo

2nd singular /	*e-bher-e-s	        /       am-a-s	        /         ama’-ba-s	/       ama’-bi-s

3rd singular /	*e-bher-e-t	        /       am-a-t	        /         ama’-ba-t	/       ama’-bi-t

1st plural   /	*e-bher-o-me	        /       ama’-mus	/         ama-ba’-mus	/       ama’bi-mus

2nd plural   /	*e-bher-e-te	        /       ama-ti-s	/         ama-ba’-tis	/       ama’-bi-tis

3rd plural   /	*e-bher-o-nt	        /       ama-nt	        /         ama’-ba-nt	/       ama’-bunt

The present and future tenses use the endings of the initial tenses (-o,  -s, -t, -mus, -tis, -nt), the imperfect the endings of the historical tenses (-m, -s, -t,  -mus, -tis, -nt), that are same with the endings of the initial tenses with difference in the first singular person. The imperfect, in order to denote the past, apart from the endings of the historical tenses, takes between the verb stem and the ending the suffix -ba-. The future takes the suffix -bi-, having a short vowel, which is contracted in the 1st singular person in -bo and in the 3rd plural person the short-voweled  -bi- becomes    -bu- in front of the ending  -nt.

The second conjugation is formed and conjugated precisely as the first one but it has as thematic vowel a long -e-, instead of the -a- of the first conjugation, and does not contract, in the first singular person of the present indicative, the thematic vowel with the ending -o (del'-e-o, del-e'-bam, del-e'-bo). The third conjugation in the present indicative uses as thematic vowel a short -i-, which becomes -u- in front of the ending -nt (leg-o, leg-i-s, leg-i-t, le'g-i-mus, le'g-i-tis, leg-u-nt). The third conjugation forms the imperfect indicative with the long thematic vowel -e-, precisely as the second one (leg-e'-bam), and the future  indicative with the long thematic vowel -a- for the first singular person  and  -e- for the other persons (leg-a-m, leg-e-s, leg-e-t, leg-e'-mus, leg-e'-tis leg-e-nt).

Finally the fourth conjugation forms the present indicative with a long thematic vowel -i- (fi'n-i-o, fin-i-s, fin-i-t, fin-i'-mus, fin-i'-tis, fi'n-i-unt), which maintains in the first singular and third plural person together with the ending (fin-i-o, fin-i-unt). The imperfect and the  future  are formed with the thematic vowels of the third conjugation, but they keep before the endings the thematic vowel -i- of the fourth conjugation (fin-i-e-ba-m, fin-i-a-m).

4.1.3. Peculiarities, deflections and  completions:

(These elements are studied in second and in third phase, i.e. after it has been completed the study of the regular course.)

1. Character of the verbal stem  is the last sound of the stem. Enclitic vowels are those vowels that show the mood of the verb, e.g. the Old Greek verb in the optative mood has the enclitic vowel -ι- (λύ-ο-ι-μι) and the Latin subjunctive of the second conjugation has  -ea- (de'l-ea-m). Tense suffix is the syllable or the sound that is added to the verbal stem, in order to be formed the stem of some verb tense (βαλ-j-ω, λυ-θή-σ-ο-μαι). 2. The stem of the present tense is same with the verbal stem (λύ-ω) or is formed with reduplication (γί-γν-ο-μαι) or is formed by the verbal stem with addition of present tense suffixes (βαλ- j - ω, ἄρ- J - ω,  εὑρ- ίσκ - ω,  ἁμαρτ- άν - ω,  κόπ- τ - ω,  φυλάκ- j - ω, φυλάσσω,  πλάθ- J - ω, πλάσσω, πλάττω).

3. Verbs ending in -ίω, -ύω have usually long  the vowel  -ι- or -υ- of the ending (λύω, λῦε). Exceptions are the verbs:ἐσθίω, ἑλκύω, μεθύω, ἐπαΐω, ἀνύω, ἀρύω. The verb φύω has the vowel -υ- short  before a vowel and long before a consonant. The verbs ending in -ζω, -ττω and -πτω usually, not always, have, the vowel of the penultimate short.

4. Greek imperfect and future tense are generally formed with the present tense stem, the imperfect also taking the syllable augment or the time augment (ἔ-λυ-ον) and the future taking the tense suffix -σ- between the stem and the ending (λύ-σ-ω). The future tense endings are the same with the present tense endings. 5. Latin imperfect indicative of the 4th conjugation (aud-i-re) was formed, in older days, with the ending  -i'bam instead of the ending  -ie'bam (audi'bam instead of audie'bam) and the future tense with the ending -i'bo instead of the ending -iam (aud-i'-bo instead of au'd-i-am).

To continue look at: ../Lesson 04 Part 2/ To see the Introduction look at: ../Introduction/


text:
User:ScienceWhiz2000

ScienceWhiz2000
Reach for the stars!

My pages

 * Astronomy
 * Inventing
 * Filmmaking
 * Photography
 * Spanish
 * French
 * Biology
 * Nature journal

title: User:ScienceWhiz2000


text:
User talk:ScienceWhiz2000

 Hello ScienceWhiz2000, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Trinity507 20:08, 29 November 2009 (UTC)

title: User talk:ScienceWhiz2000


text:
User:ScienceWhiz2000/French

Bonjour, je mapelle ScienceWhiz! This page is for homework and other links outside of Wikiversity that I am using to learn French.

Learning the basics of French

 * Bonjour.com

title: User:ScienceWhiz2000/French


text:
Category:ScienceWhiz2000

Pages in ScienceWhiz2000's user space.

title: Category:ScienceWhiz2000


title: User:ScienceWhiz2000/Astronomy


text:
User:Akohnert/Proportional-integral controller in a flow system

Consider a problem where a small tank is connected to a large pressure reservoir. The tank has a release valve to the atmosphere which is governed by a proportional-integral controller. The time behavior of this system is nearly impossible to solve analytically, but very simple to solve numerically. Given the gain (G) and reset (T) parameters of the controller, we can easily determine its output signal using the formula for a proportional integral system,

$$\mathrm{Output}=G \frac{p-p_{set}}{p_{set}}+\frac{G}{T}\int{\frac{p-p_{set}}{p_{set}}}$$

This will give us a target value for the valve position. We can interpret a signal of one to mean full open and a signal of zero to mean fully closed. We will assume a linear dependence between the valve position and flow coefficient. The valve coefficient ($$c_v$$) is used in the Bernoulli equation to determine the flow rate out of the tank.

$$Q=c_v\sqrt{\frac{P_1-P_2}{\gamma}}$$

We have to be careful to limit the valve coefficient to 1 (fully open) even if the output signal rises above 1. The same equation can be used with the known inlet valve coefficient to determine the inlet flow from the pressure reservoir. Using the two flow rates, we can find the change in tank pressure using a "tank capacitance factor". This is a constant that relates the difference in inlet and outlet flows to the differential pressure change in the tank,

$$\frac{\delta P}{\delta t} = C \left( Q_i-Q_o \right)$$

This is all the requisite information to begin writing the program. We need to know the pressure as time progresses, so the program will move through small increments in time. The structure will be as follows:
 * Calculate the flow rates based on existing value of tank pressure
 * Calculate the change in tank pressure due to flow rate differences
 * Using the new tank pressure, calculate the output signal
 * Adjust the valve position to match the output signal
 * Increment the time
 * Tabulate time, valve position, and tank pressure
 * Repeat

An example of this code is provided below for Mathematica. The first block sets the initial conditions, namely a reservoir pressure of 100 psi, a target pressure of 80 psi, a tank capacitance of 0.5 psi per gpm per second,and a time step of 0.05 seconds. cvo = 0; pi = 100; po = 0; psp = 80; G = 2; T = 1; $$\Delta$$ t = .05; $$\epsilon$$ = {}; Op = 0; cvi = 1.484; P = {100}; t = {0}; Qi = 0; Qo = 0; c = .5; $$\gamma$$ = 62.4; currentP= 0; cvolist = {0};

The second block represents the steps in the loop outlined above. Notice that the rate of change in the valve position has been capped to reflect real limits on the speed of opening and closing valves. For[i = 1, i <= 200$$\Delta$$t, i = i + 1, {  (*Select most recent pressure*) currentP = P[[i] ];  (*Determine the change in pressure tank capacitance factor and current flow rates*)   dp=$$\Delta$$t/c*(cvi*Sqrt[(pi-currentP)/$$\gamma$$]-cvo*Sqrt[(currentP-po)/$$\gamma$$]);   (*Tabulate values for error, new pressure, and time*)   AppendTo[$$\epsilon$$, (currentP - psp)/psp];   AppendTo[P, currentP + dp];   AppendTo[t, $$\Delta$$t*(i)];   (*Determine the controller output signal using the given formula*)      Op = G*$$\epsilon$$[[i] ] + G $$\Delta$$t/T*Total[$$\epsilon$$];   (*Adjust valve position in accordance with the output signal*)   If[Op > cvo + .01, cvo = cvo + .25 $$\Delta$$t];   If[Op < cvo - .01, cvo = cvo - .25 $$\Delta$$t];   (*Ensure real limits on valve position*)   If[cvo > 1, cvo = 1];   If[cvo < 0, cvo = 0];   (*tablulate valve position*)   AppendTo[cvolist, cvo];  };]


text:
User:Jwood/codes

This page will be for running through some codes.

title: User:Jwood/codes


text:
User:Inviiere

Learn!

title: User:Inviiere


text:
User:Fetorg

Hello My name is Christopher Bryant, and I'm the founder of the Free Education and Training Organization, an organization dedicated to providing accredited tertiary education to impoverished people and developing nations free or nearly free of charge.

The Free Education & Training Organization Website

title: User:Fetorg


text:
User talk:Fetorg

 Hello Fetorg, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. Jtneill - Talk - c 04:04, 30 November 2009 (UTC)

 Hi , I am Percy. Nice to meet you. I love the beach, swimming, fishing, and I am currently doing some research at Wikiversity on oceanography. I am also learning to play piano. What do you like? Can I be your mascot?

FET
Hi Chris - be great to hear more about FET and explore how Wikiversity could help/collaborate. -- Jtneill - Talk - c 04:04, 30 November 2009 (UTC)

title: User talk:Fetorg