text:
User:EML5526.S11.Team3.vnarayanan/Homework6

=Problem 6.4(a) Verify the Divergence Theorem=

Problem Statement
Given a vector field $$\;\;q_{x} = -y^2\;;\; q_{y} = -2xy \;\;$$ on the domain shown below. Verify the divergence theorem.

Solution
$$ \vec{q} = q_{x}\vec{i}+ q_{y}\vec{j}$$

$$\therefore\;\; \vec{q} = -y^2\vec{i} - 2xy\vec{j}$$

Divergence of $$\vec{q} = \vec{\triangledown }. \vec{q} = = \frac{\partial q_{x}}{\partial x}+\frac{\partial q_{y}}{\partial y} = = 0 - 2x = -2x$$ The divergence theorem states,

$$\int_{\Omega }\vec{\triangledown}.\vec{q} = \oint_{\Gamma }\vec{q}.\vec{n}$$

$$\oint_{\Gamma }\vec{q}.\vec{n} = \int_{AB}\vec{q}.\vec{n^{1}}\;d\Gamma_{AB}+\int_{BC}\vec{q}.\vec{n^{2}}\;d\Gamma_{BC}+\int_{CD}\vec{q}.\vec{n^{3}}\;d\Gamma_{CD}+\int_{DA}\vec{q}.\vec{n^{4}}\;d\Gamma_{DA}$$

$$\vec{q}.\vec{n^{1}} = 2xy\;;\vec{q}.\vec{n^{2}} = -y^2\;;\vec{q}.\vec{n^{3}} = -2xy\;;\vec{q}.\vec{n^{4}} = y^2$$

$$d\Gamma_{AB} = dx;\;d\Gamma_{BC} = dy;\;d\Gamma_{CD} = -dx;\;d\Gamma_{DA} = -dy$$

From equations (1) & (2), we have

$$\int_{\Omega }\vec{\triangledown}.\vec{q} = \oint_{\Gamma }\vec{q}.\vec{n}$$

Hence, Divergence theorem is verified

=Problem 6.4(b) Verify the Divergence Theorem=

Problem Statement
Given a vector field $$\;\;q_{x} = 3x^2y+y^3\;;\; q_{y} = 3x+y^3 \;\;$$ on the domain shown below. Verify the divergence theorem.The curved boundary of the domain is a parabola.

Solution
$$ \vec{q} = q_{x}\vec{i}+ q_{y}\vec{j}$$

$$\therefore\;\; \vec{q} = (3x^2y+y^3)\vec{i}+(3x+y^3) \vec{j}$$

Divergence of $$\vec{q} = \vec{\triangledown }. \vec{q} = \frac{\partial q_{x}}{\partial x}+\frac{\partial q_{y}}{\partial y} = 6xy+3y^2$$

The divergence theorem states,

$$\int_{\Omega }\vec{\triangledown}.\vec{q} = \oint_{\Gamma }\vec{q}.\vec{n}$$

The equation of the given parabola is $$x^2 = 4-y \;\;\Rightarrow \;\; y = 4-x^2$$

$$\int_{\Omega }\vec{\triangledown}.\vec{q} = \int_{-2}^{2}\left ( \int_{0}^{4-x^2}(6xy+3y^2)\;dy\right )dx = \int_{-2}^{2}\left (  \int_{0}^{4-x^2}(6x(4-x^2)+3(4-x^2)^2)\;dy\right )dx = \int_{-2}^{2}\left ( -(x^2-4)^2(x^2-3x-4)\right )dx$$

$$\oint_{\Gamma }\vec{q}.\vec{n} = \int_{AB}\vec{q}.\vec{n^{1}}\;d\Gamma_{AB}+\int_{BCA}\vec{q}.\vec{n^{2}}\;d\Gamma_{BCA}$$

$$\vec{q}.\vec{n^{1}} = -(3x+y^3)\;;\vec{q}.\vec{n^{2}} = 3x+y^3$$

$$d\Gamma_{AB} = dx;\;d\Gamma_{BCA} = dy$$

Equation of line AB is $$y = 0$$

$$\therefore\;\;\int_{AB}\vec{q}.\vec{n^{1}}\;d\Gamma_{AB} = \int_{-2}^{2}-3xdx = 0$$

$$\int_{BCA}\vec{q}.\vec{n^{2}}\;d\Gamma_{BCA} = \int_{-2}^{2}3x+y^3$$

Equation of parabola is $$y = 4 - x^2$$

, using [| Wolframalpha]

From equations (1) & (2), we have

$$\int_{\Omega }\vec{\triangledown}.\vec{q} = \oint_{\Gamma }\vec{q}.\vec{n}$$

Hence, Divergence theorem is verified

=Problem 6.4(c) Use Divergence Theorem=

Problem Statement
Using the divergence theorem prove

$$\oint_{\Gamma}n\;\;\mathrm{d}\Gamma = 0$$

Solution
Let, $$\vec{q} = \vec{i}+\vec{j}\;\;\; and \;\; \vec{n} = \vec{n_{1}}+\vec{n_{2}}$$

Divergence Theorem states,

$$\int_{\Omega }\vec{\triangledown}.\vec{q}\mathrm{d}\Omega = \oint_{\Gamma }\vec{q}.\vec{n}\mathrm{d}\Gamma$$

$$\oint_{\Gamma }\vec{q}.\vec{n}\;\mathrm{d}\Gamma = \oint_{\Gamma }(n_{1}+n_{2})\;\mathrm{d}\Gamma = \oint_{\Gamma }n\;\mathrm{d}\Gamma$$

$$\therefore\;\;\oint_{\Gamma }n\;\mathrm{d}\Gamma = \int_{\Omega }\vec{\triangledown}.\vec{q}\mathrm{d}\Omega $$

$$\vec{\triangledown}.\vec{q} = \frac{\partial (1)}{\partial x}+\frac{\partial (1)}{\partial y} = 0$$

$$\therefore\; \int_{\Omega }\vec{\triangledown}.\vec{q}\mathrm{d}\Omega = \int_{\Omega }0\;\mathrm{d}\Omega = 0$$

$$\Rightarrow\;\;\oint_{\Gamma}n\;\;\mathrm{d}\Gamma = 0$$

=Problem 6.8.2(a) Gauss Quadrature=

Problem Statement
Use Gauss quadrature to obtain exact values for the following integrals. Verify by analytical integration:

$$(a).\;\int_{0}^{4}(x^2+1)\mathrm{d}x$$

$$(b).\;\int_{-1}^{1}(\xi^4+2\xi^2)\mathrm{d}\xi$$

Solution
(a).

$$I =\int_{0}^{4}(x^2+1)\mathrm{d}x$$

As $$n_{gp} = 2\; (two-point \;\;quadrature),\; p = 2n_{gp} - 1 = 3$$

$$ \begin{bmatrix} 1 &1 \\ \xi_{1}&\xi_{2} \\ \xi_{1}^2&\xi_{2}^2 \\ \xi_{1}^3&\xi_{2}^3 \\ \end{bmatrix} \begin{bmatrix} W_{1}\\ W_{2}\\ \end{bmatrix} = \begin{bmatrix} 2\\ 0\\ \frac{2}{3}\\ 0\\ \end{bmatrix} $$

Solving the above system of 4 equations and 4 unknowns, we get

$$W_{1} = W_{2} = 1\;\; and \; \xi_{1} = \frac{1}{\sqrt3}\;\;\xi_{2} = \frac{-1}{\sqrt3}$$

We have, $$a = 0\;\;;b = 4$$

$$x = \frac{1}{2}(a+b)+\frac{1}{2}\xi(b-a) = 2+2\xi$$

The value of the function $$f(\xi) = (2+2\xi)^2+1$$

$$I = \frac{l}{2}\int_{-1}^{1}((2+2\xi)^2+1)\; \mathrm{d}\xi$$

$$I = \frac{l}{2}\left[W_{1}((2+2\xi_{1})^2+1)\; + \;W_{2}((2+2\xi_{2})^2+1)\right] \;and \; l = b - a = 4 - 0 = 4$$

$$= 2\left[10.953+1.715\right] = 25.334$$

Analytical Method

$$I =\int_{0}^{4}(x^2+1)\mathrm{d}x$$

$$= \left[\frac{x^3}{3}+x\right]_{0}^{4} = \frac{4^3}{3}+4 = 25.333$$

The analytical method also yields exactly the same result as the Gauss Quadrature.

(b). $$I =\int_{-1}^{1}(\xi^4+2\xi^2)\mathrm{d}\xi$$

$$n_{gp} = 4\; (fo-point \;\;quadrature),\; p = 2n_{gp} - 1 = 7$$

$$ \begin{bmatrix} 1 &1 \\ \xi_{1}&\xi_{2} \\ \xi_{1}^2&\xi_{2}^2 \\ \xi_{1}^3&\xi_{2}^3 \\ \xi_{1}^4&\xi_{2}^4 \\ \xi_{1}^5&\xi_{2}^5 \\ \xi_{1}^6&\xi_{2}^6 \\ \xi_{1}^7&\xi_{2}^7 \\ \end{bmatrix} \begin{bmatrix} W_{1}\\ W_{2}\\ W_{3}\\ W_{4}\\ \end{bmatrix} = \begin{bmatrix} 2\\ 0\\ \frac{2}{3}\\ 0\\ \frac{2}{5}\\ 0\\ \frac{2}{7}\\ 0\\ \end{bmatrix} $$

Solving the above system of 8 equations and 8 unknowns, we get

$$W_{1} = W_{3} = 0.348\;W_{2} = W_{4} = 0.652\; and \; \xi_{1} = 0.861\;\;\xi_{2} = 0.339\;\;\xi_{3} = -0.861\;\;\xi_{4} = -0.339$$

We have, $$a = -1\;\;;b = 1$$

$$x = \frac{1}{2}(a+b)+\frac{1}{2}\xi(b-a) = \xi$$

The value of the function $$f(\xi) = (\xi^4+2\xi^2)$$

$$I = \frac{l}{2}\int_{-1}^{1}(\xi^4+2\xi^2)\; \mathrm{d}\xi$$

$$I = \frac{l}{2}\left[W_{1}(\xi_{1}^4+2\xi_{1}^2)\; + \;W_{2}(\xi_{2}^4+2\xi_{2}^2)\;+ (\xi_{3}^4+2\xi_{3}^2)\;+ (\xi_{4}^4+2\xi_{4}^2)\;\right] \;and \; l = b - a = 4 - 0 = 4$$

$$= 1.4144+0.3169 = 1.731$$

Analytical Method

$$I =\int_{-1}^{1}(\xi^4+2\xi^2)\mathrm{d}\xi$$

$$= \left[\frac{\xi^5}{5}+\frac{2\xi^3}{3}\right]_{-1}^{1} = 2\left[\frac{1}{5}+\frac{2}{3}\right] = 1.731$$

The analytical method also yields exactly the same result as the Gauss Quadrature.

=Problem 6.8.2(b)Three-point Gauss quadrature=

Problem Statement
Use three-point Gauss quadrature to evaluate the following integrals. Compare to the analytical integral.

(a).$$ \int_{-1}^{1}\frac{\xi}{\xi^2+1}\;\mathrm{d}x $$

(b). $$\int_{-1}^{1}cos^2\pi\zeta\mathrm{d\xi}$$

Solution
(a).

$$I =\int_{-1}^{1}\frac{\xi}{\xi^2+1}\;\mathrm{d}x $$

As $$n_{gp} = 3\; (three-point \;\;quadrature),\; p = 2n_{gp} - 1 = 5$$

$$ \begin{bmatrix} 1 &1 \\ \xi_{1}&\xi_{2} \\ \xi_{1}^2&\xi_{2}^2 \\ \xi_{1}^3&\xi_{2}^3 \\ \xi_{1}^4&\xi_{2}^4 \\ \xi_{1}^5&\xi_{2}^5 \\ \end{bmatrix} \begin{bmatrix} W_{1}\\ W_{2}\\ W_{3}\\ \end{bmatrix} = \begin{bmatrix} 2\\ 0\\ \frac{2}{3}\\ 0\\ \frac{2}{5}\\ 0\\ \end{bmatrix} $$

Solving the above system of 6 equations and 6 unknowns, we get

$$W_{1} = W_{3} = 0.557\;W_{2} = 0.889\; and \; \xi_{1} = 0.7745\;\;\xi_{2} = 0\;\;\xi_{3} = -0.7745$$

We have, $$a = -1\;\;;b = 1$$

$$x = \frac{1}{2}(a+b)+\frac{1}{2}\xi(b-a) = \xi$$

$$\therefore\;\; \mathrm{dx} = \mathrm{d\xi}$$

The value of the function $$f(\xi) = \frac{\xi}{\xi^2+1}$$

$$I = \frac{l}{2}\int_{-1}^{1}(\frac{\xi}{\xi^2+1})\; \mathrm{d}\xi$$

$$I = \frac{l}{2}\left[W_{1}(\frac{\xi_{1}}{\xi_{1}^2+1})\; + \;W_{2}(\frac{\xi_{2}}{\xi_{2}^2+1})\;+ W_{3}(\frac{\xi_{3}}{\xi_{3}^2+1})\right] \;and \; l = b - a = 1 - (-1) = 2$$

$$= 0.556\left(\frac{0.7745}{0.7745^2+1}\right)+0.556\left(\frac{-0.7745}{0.7745^2+1}\right) = 0$$

Analytical Method

$$I =\int_{-1}^{1}(\frac{\xi}{\xi^2+1})\mathrm{d}\xi = 0,$$, using [Wolframalpha]

The analytical method also yields exactly the same result as the Gauss Quadrature.

(b). $$\int_{-1}^{1}cos^2\pi\zeta\mathrm{d\xi}$$

(b).

$$I =\int_{-1}^{1}cos^2\pi\zeta\mathrm{d\xi} $$

As $$n_{gp} = 3\; (three-point \;\;quadrature),\; p = 2n_{gp} - 1 = 5$$

$$ \begin{bmatrix} 1 &1 \\ \xi_{1}&\xi_{2} \\ \xi_{1}^2&\xi_{2}^2 \\ \xi_{1}^3&\xi_{2}^3 \\ \xi_{1}^4&\xi_{2}^4 \\ \xi_{1}^5&\xi_{2}^5 \\ \end{bmatrix} \begin{bmatrix} W_{1}\\ W_{2}\\ W_{3}\\ \end{bmatrix} = \begin{bmatrix} 2\\ 0\\ \frac{2}{3}\\ 0\\ \frac{2}{5}\\ 0\\ \end{bmatrix} $$

Solving the above system of 6 equations and 6 unknowns, we get

$$W_{1} = W_{3} = 0.557\;W_{2} = 0.889\; and \; \xi_{1} = 0.7745\;\;\xi_{2} = 0\;\;\xi_{3} = -0.7745$$

We have, $$a = -1\;\;;b = 1$$

$$\zeta = \frac{1}{2}(a+b)+\frac{1}{2}\xi(b-a) = \xi$$

The value of the function $$f(\xi) = cos^2\pi\xi$$

$$I = \frac{l}{2}\int_{-1}^{1}cos^2\pi\xi\mathrm{d\xi}$$

$$I = \frac{l}{2}\left[W_{1}(cos^2\pi\xi_{1})\; + \;W_{2}(cos^2\pi\xi_{2})\;+W_{3}(cos^2\pi\xi_{3})\right] \;and \; l = b - a = 1 - (-1) = 2$$

$$I = 0.556(cos^2\pi(0.7745))+0.889(cos^2\pi(0))+0.556(cos^2\pi(0.7745)) = 1.53$$

Analytical Method

$$I =\int_{-1}^{1}cos^2\pi\zeta\mathrm{d\xi} = 1,$$, using [Wolframalpha]

The value from the analytical method is different from the value evaluated from Gauss Quadrature

=Problem 6.8.2(c) Check for accuracy in Gauss Quadrature=

Problem Statement
The integral $$\int_{-1}^{1}(3\xi^3+2)\;\mathrm{d\xi}$$ can be integrated exactly using two-point Gauss quadrature. How is the accuracy affected if a. one-point quadrature is employed; b. three-point quadrature is employed.

Solution
$$I =\int_{-1}^{1}(3\xi^3+2)\;\mathrm{d\xi}$$

As $$n_{gp} = 2\; (two-point \;\;quadrature),\; p = 2n_{gp} - 1 = 3$$

$$W_{1} = W_{2} = 1\;\; and \; \xi_{1} =0.5774\;\;\xi_{2} = -0.5774$$

$$f(\xi) = (3\xi^3+2)$$

$$I = \frac{l}{2}\int_{-1}^{1}(3\xi_{1}^3+2)\; \mathrm{d}\xi$$

$$I = \frac{l}{2}\left[W_{1}(3\xi_{1}^3+2)\; + \;W_{2}(3\xi_{2}^3+2)\right] \;and \; l = b - a = 1 - (-1) = 2$$

$$= 2+2 = 4$$

(a).One - Point Quadrature: $$W = 2\;\; and \; \xi =0$$

$$f(\xi) = (3\xi^3+2)$$

$$I = \frac{l}{2}\int_{-1}^{1}(3\xi^3+2)\; \mathrm{d}\xi$$

$$I = \frac{l}{2}\left[W(3\xi^3+2)\right]\;and\;\; l = b - a = 1 - (-1) = 2$$

$$= 2(2) = 4$$

(b).Three - Point Quadrature:

$$W_{1} = W_{3} = 0.556\;\; and \; \xi_{1} =0.7745\;\;\xi_{2} = 0\;\;\xi_{3} =-0.7745$$

$$f(\xi) = (3\xi^3+2)$$

$$I = \frac{l}{2}\int_{-1}^{1}(3\xi_{1}^3+2)\; \mathrm{d}\xi$$

$$I = \frac{l}{2}\left[W_{1}(3\xi_{1}^3+2)\; + \;W_{2}(3\xi_{2}^3+2)+\;\;W_{3}(3\xi_{3}^3+2)\right] \;and \; l = b - a = 1 - (-1) = 2$$

$$= 2(0.556)(2)+(0.889)(2) = 4$$

'''The value of the Integral is same when evaluated with 0ne -,Two - and Three- point Quadratures. Hence all the three are accurate.'''

=Problem 6.8 Verify the given table=

Problem Statement
Verify the table of {(wi,xi),i=1,2,..,5} against NIST handbook and also with table on textbook page 89

Solution
The table from Lecture 36-3 is given below:

Nodes and weights for the 5-point Gauss–Legendre formula from | NIST Handbook are given below:

The values of the Nodes and Weights in the given table in Lecture 36-3 are calculated and the values are tabulated:

The table of Nodes and Weights with the calculated values from the text book Introduction to Finite Element Analysis by Fish and Belytschko is as below:

From the above tables, it can be concluded that the values of Nodes and Weights from the table in Lecture 36-3, NIST handbiik and the text book agree with each other.


license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: File:Fe1.s11.team1.HW6.fig2.1.pdf


text: File:Nieuw frame.pdf

title: File:Nieuw frame.pdf


title: File:Fe1.s11.team1.HW6.fig2.2.pdf


title: File:Fe1.s11.team1.HW6.fig2.3.pdf


text:
File:Table Points.png

Licensing
for school project

title: File:Table Points.png


text:
User:Rudzko

REFLECTIVE PRACTICE ANALYSIS
The aim of this assignment is to analyse several political simulations and to provide the reader with the potential benefits these simulations may have. The analysis will argue that playing board games could be not only entertaining, but extremely useful as well, in the sense that “they help you think about the practical issues and engage with the theory” (Matthew Reisz; 2011, online). It will aim to prove that board games, especially those of political nature could be of great aid to anyone who wishes to simulate the experience of any given (political) field. The analysis will be separated into four parts – each depicting a different political simulation, the chief points of which will be summarized in the conclusion.

==== * MONOPOLY ====

Among the great number of board games, it could be argued that Monopoly is by far the most popular. Created in the first half of the 20th century (est. 1935) – although the game’s development dates back to 1903, Monopoly has undeniably dominated the world of board games. Furthermore, it could be also argued that Monopoly has been the matrix for, and has facilitated the creation of countless other board games. Taking into account the popularity of the game, Monopoly has been adapted to fit individual countries, by changing various properties of the board. The fact that Monopoly is easy to play and entertaining (even among children) has further contributed to the game’s popularity. Every few years, national champions from all over the world meet for the MONOPOLY World Championship tournament. World Champions have emerged from 10 different countries, including: the United States, Ireland, Singapore, Italy, New Zealand, United Kingdom, the Netherlands, Hong Kong, Japan and Spain (Hasbro; 2011, online).

Monopoly tends to be very realistic, as it engulfs the players in a world of capitalism, where greed, risk-taking and self interest are the key to victory. Additionally, it enables players to conspire together in order to diminish the chances of other players, but ultimately only one can achieve victory. As Monopoly is a dice-based board game – luck is a core factor; this facilitates new, inexperienced players to win, despite them being new, and not entirely familiar with the rules of the game. Taking into account that the stage in Monopoly is based on actual places (Atlantic City, Islington, ect.) it is easy for players to relate to them, and feel like being a part of the capitalist machine. All players start at the same stage with an equal opportunity to win. “Money and property ownership symbolise success in the 1980’s and the MONOPOLY brand seems to match up. Localisations, licenses, and spin-offs of MONOPOLY allow people all over the world to live the dream of owning it all” (Hasbro; 2011, online).

The political nature of the game could be argued to be somewhat unsettled, and open for debate. The board game involves obtaining and selling properties, and ultimately winning at all costs, putting personal interest before everything else in order to own it all – which at first glance could be seen as entirely capitalist in nature. Furthermore, during the Cold War era, both Russia and America claimed that Monopoly is a capitalist game. Nevertheless, the argument that Monopoly is in fact a socialist game possesses an equal amount of validity. The board game suggests that a free market would eventually lead to a monopoly, which in its turn may results in the bankruptcy of all the rest. Monopoly teaches players to take heed when dealing with money in order to avoid becoming bankrupt – which is the opposite of the (risk-taking) capitalist way.

Recently, Monopoly has undergone various modifications to certain platforms, such as the introduction of debit cards taking the place of cash (in the original board game) – which is more suitable for the contemporary players; with more innovation in the near future to come.

==== * KINGMAKER ====

Kingmaker is a historical strategy board game (first created by Andrew McNeil and PhilMar in 1974) based in the times of struggle and conflict in medieval England (more accurately 1455 – 1487) known as the Wars of the Roses. The players will undertake the military and political struggles between the houses of Lancaster and York to decide the rightful heir to the English throne. Every single one of the players aims to build and takes control of a faction of nobles that, through skirmishes, politics and diplomacy, attempts to eliminate the factions of all the other players (Vintage Toys and Games; 2011, online).

What generates further interest in the game is that although “Kingmaker” is more or less historically accurate, the players are in no way bound to abide by past historical events, and are free to alter the historical outcome of England. The fact that “Kingmaker” is based in medieval times, the game helps players to understand the way society functioned in the past. Politically “Kingmaker” could be regarded as an excellent recreation of the balance of power between the barons and royals in the Wars of the Roses. Furthermore, playing the game is an entertaining way to understand English politics in medieval times. Additionally, upon playing the game, a player starts identifying with the barons and other historical characters who took part in the Wars of the Roses. Realism is another key feature of the board game (especially considering that luck was an important aspect in medieval warfare); for example – the plague card, a card drawn by a player by sheer chance, could entirely cripple the said player’s efforts in finishing the game – not unlike the actual plague, which was a major issue which could alter various events it the times of medieval England; by playing “Kingmaker” a player learns that despite being well on his way to victory – all can be lost in an instant. Subsequently the game’s realism is further highlighted by the way it simulates the struggle between baronial families rather than royal claimants as in other Wars of the Roses games – additionally – the game underlines the importance of the church in medieval times. The necessity to be in control of one archbishop or two bishops in order to crown a monarch in a cathedral town is a superb imitation of the connection between the church and the state in that era. The board could be regarded as an excellent lesson in the political and economic geography of medieval England as the events take place around towns and forts and castles which bore a great amount of significance at the time, but are no longer.

Perhaps the only shortcomings of the game, which may require some attention, are large amount of rules, which render new players to stand little if any chance against an experienced veteran of the game. Certain simplifications in the rules of the game could greatly facilitate new players to learn how to play, while at the same time avoid altering the entertaining gameplay of “Kingmaker”.

==== * ORIGINS OF WORLD WAR II ====

The Origins of World War II is a board game first published in 1971 by Avalon Hill, and was based on A. J. P Taylor’s views on the origins of the Second World War. The action of the game takes place in the years leading up to World War II (more precisely 1933-1939). Players assume the roles of some of the chief participants of World War II, including, Germany, the USSR, Great Britain, France and the United States. It could be argued that the action of the game is highly abstract. Participating players have to place political factor markers on their countries, their opponents' countries, as well as the countries which are non-playable (like Italy for example). Players put up to five factors in any given country which can create an understanding or a control power in that country, given that a country is allowed by the rules to have that power. The Origins of World War II tends to be a highly realistic game as well, as it is based on the actual events of the pre-World War II period. In a standard game, the United States for example is not allowed to have either understanding or control of any other country, reflecting isolationist sentiments and a general lack of any significant influence in European politics. Furthermore, it is highly difficult to win if playing with Nazi Germany (despite the fact that Germany starts at the highest position) as all the other countries will form an alliance, and team up against it. The amount of political factors chosen to the country varies, depending on the player’s country (the US has the least and Germany the most, respectively) and the stage of the game in progress (in the later year, all players receive more political factors, reflecting heightening tensions). Victory and defeat are based on an individualized scoring system that gives the various countries points for either having a control or understanding in a specific country, or preventing other countries from having said controls or understandings.

The highly abstract nature of Origins of World War II board game is perhaps its greatest asset as well as (arguably) its most severe design flaw. Understanding the dynamic of how exactly countries get involved in wars is a rather problematical task. Consequently, attempting to recreate a simulation of these mentioned dynamics that could be easily learned and applied within a game is perhaps an even greater and more difficult feat to achieve. In an attempt to make the game play simple and easily understandable, the game uses political factors which seem to combine military invasion, treaties, and informal alliances. “To direct the flow of the game play, Origins of World War II creates game objectives that force players into conflict. While these models succeed, to a certain extent, in re-creating the unfolding of history, they seem to do so only by hard-wiring historical outcomes into the model of play itself. Ultimately, the game does a nice job of walking a player through the tangle of politics that preceded World War II, but a poor job of explaining why World War II was structurally inevitable” (Wargames; ND, online). Nevertheless, for anyone interested in the historical events prior to World War II, the game provides an enjoyable way of exploring the Origins of World War II, and offers an accurate simulation of pre-war political behaviour.

==== * THE GAME OF WAR ====

“The Game of War” is a military strategy board game which is based on classic eighteen century warfare. The game puts the players in control of the manoeuvres of two armies of identical force, each one of them aiming to crush their opponent via superior strategy and well calculated skirmishes. “The Game of War” could be argued to be a highly demanding game, requiring a great amount of dedication to master. New players may face some difficulty at the beginning. Unlike many other board games, “The Game of War” excludes the factor of luck, making the player carefully analyse possible moves, in order to avoid mistakes – and ultimately, defeat. Furthermore, risk taking is a part of the game as well, as a well calculated risk may result in victory. The core feature of the game is strategy – the strategic aspects of the game are realistic (maintaining a single strong army, for instance tends to be much more beneficial to a player than separating his forces). In addition, a very (if not most) crucial aspect of “The Game of War” are the lines of communication, which must be guarded throughout the continuation of the game – needless to say, keeping armies in constant contact is critical in times of war. Some aspects of the game however, are less-so realistic – for example weather conditions and fog of war are ignored within the game – whereas said conditions could have turned the tide of any battle in the eighteenth century (It was in fact the severe weather conditions which ultimately saved Russia from permanent French occupation, greatly crippling Napoleon’s army). Not unlike chess – “The Game of War” requires the participants to think ahead of their current moves – and teaches new players how to do so. Furthermore, in order to achieve victory, the board game demands from the player to attempt to predict the future moves of their opponents as well.

“The Game of War” teaches players strategic thinking and the cost of making mistakes in warfare-related conditions. By playing the game, players undertake the full weight of being in command of an army in the eighteenth century during times of war. As the game often requires teammates to exchange confidential plans of action, participants can really experience the importance of planning and strategy.

Despite being challenging and perhaps time-consuming, “The Game of War” tends to be rather fun, and is possibly the ultimate in strategy simulation board games, where every next decision could result in the potential victory or defeat of the player. The rules of the game become clear to new participants with further play, and the players could develop various intellectual skills as they play.

IN CONCLUSION, it is important to state that board games are much more than just a way to pass time or a means of entertainment. Not only are simulations, such as “Kingmaker” or “Origins of World War II” an invaluable source for the understanding and practicing of political theory – they also provide the players with various historic facts. Finally, simulations facilitate the players to experience scenarios (political or otherwise) which would have been impossible in real life.

==== '''BIBLIOGRAPHY ''' ====

Class Wargames (ND) ‘The Game of War’. Online, available from http://www.classwargames.net/?page_id=829, accessed on 23 March 2011.

Hasbro (2011) ‘(Monopoly) Did you know?’ Online, available from http://www.hasbro.com/monopoly/en_US/discover/75-Years-Young.cfm, accessed on 22 March 2011.

Kingmaker Board Game (2011). Online, available from http://www.youtube.com/results?search_query=kingmaker+board+game&aq=f, accessed on 23 March 2011.

Reisz Matthew (2011) ‘Tales of the unexpected’. Online, available from http://www.timeshighereducation.co.uk/story.asp?sectioncode=26&storycode=415486&c=2, accessed on 21 March 2011.

Vintage Toys and Games (2011) ‘Kingmaker, board Game’. Online, available from http://www.vintagetoysgames.co.uk/kingmaker_1974.html, accessed on 22 March 2011.

Wargames (ND) ‘The Origins of World War II’. Online, available from http://www.faculty.virginia.edu/setear/students/wargames/page3a.htm, accessed on 23 March 2011.

title: User:Rudzko


text:
User:Eml5526.s11.team3.ushnish12/Homework 6

Problem statement
Please refer to lecture note 37-2 and 38 for more details.

Solve the heat conduction problem in 2D with boundary convection.

1> Construct the weak form for heat conduction in 2D with boundary convection.

2> Develop semi-discrete equations (ODEs) similar to (3) 23-3 Give detailed expression of all quantities.

3>Solve G2DM2.0/D1 using 2D LIBF till $$\displaystyle 10^{-6} $$ accuracy at the center. Plot solution in 3D with contour.

Developing the Weak Form of Newton's law of Cooling
i> From Lecture 33-2, the PDE of the problem is given by:-

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\frac{\partial }\left( {{K}_{ij}}\frac{\partial u} \right)+f\left( x,t \right)=\rho c\frac{\partial u}{\partial t}

$$ (Eq )<1>
 *
 * }

From the PDE the weak form can be written as:-

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\int{w\left( \frac{\partial }{\partial {{x}_{i}}}\left( {{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)+f\left( x,t \right)-\rho c\frac{\partial u}{\partial t} \right)}d\text{ }\!\!\Omega\!\!\text{ =0}

$$ (Eq )<2>
 *
 * }

Now from the theory of Integration by parts, we can write

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

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

\int{w\left( \frac{\partial }{\partial {{x}_{i}}}\left( {{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right) \right)}d\text{ }\!\!\Omega\!\!\text{ =}\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial }{\partial {{x}_{i}}}\left( w{{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)}d\text{ }\!\!\Omega\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial w}{\partial {{x}_{i}}}{{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}}d}\text{ }\!\!\Omega\!\!\text{ }

$$
 * }

Hence equation 2 can be now written as:-

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial }{\partial {{x}_{i}}}\left( w{{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)}d\text{ }\!\!\Omega\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial w}{\partial {{x}_{i}}}\left( {{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)}d\text{ }\!\!\Omega\!\!\text{ +}\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{wf\left( x,t \right)}d\text{ }\!\!\Omega\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{w\rho c\frac{\partial u}{\partial t}}d\text{ }\!\!\Omega\!\!\text{ }=0

$$ (Eq )<3>
 *
 * }

Now we can write $$\displaystyle d\Omega $$ as $$\displaystyle d\Omega ={{\Gamma }_{g}}\cup {{\Gamma }_{h}}\cup {{\Gamma }_{H}}$$ , where $$\displaystyle {{\Gamma }_{g}}$$ = essential boundary condition. $$\displaystyle {{\Gamma }_{h}}$$ = natural boundary condition. $$\displaystyle {{\Gamma }_{H}}$$ = Mixed boundary condition.

So by applying the Gauss theorem on the first term we obtain,

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}w{{K}_{ij}}\frac{\partial u}{{n}_{i}}d\text{ }\!\!\Gamma\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial w}{\partial {{x}_{i}}}\left( {{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)}d\text{ }\!\!\Omega\!\!\text{ +}\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{wf\left( x,t \right)}d\text{ }\!\!\Omega\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{w\rho c\frac{\partial u}{\partial t}}d\text{ }\!\!\Omega\!\!\text{ }=0

$$ (Eq )<4>
 *
 * }

Now the heat flux is given by:-

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

{{q}_{i}}=-{{\text{ }\!\!\Kappa\!\!\text{ }}_{ij}}\frac{\partial u}{\partial {{x}_{j}}}

$$ (Eq )<5>
 *
 * }

So by rearranging Equation 5, we can write

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{-w{{q}_{i}}{n}_{i}}d\text{ }\!\!\Gamma\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial w}{\partial {{x}_{i}}}\left( {{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)}d\text{ }\!\!\Omega\!\!\text{ +}\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{wf\left( x,t \right)}d\text{ }\!\!\Omega\!\!\text{ }-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{w\rho c\frac{\partial u}{\partial t}}d\text{ }\!\!\Omega\!\!\text{ }=0

$$ (Eq )<6>
 *
 * }

We can write the first term as-

$$\displaystyle \int\limits_{\Gamma }{-w{{q}_{i}}{{n}_{i}}d\Gamma =}\int\limits_{-w{{q}_{i}}{{n}_{i}}d{{\Gamma }_{g}}}-\int\limits_{-w{{q}_{i}}{{n}_{i}}d{{\Gamma }_{h}}}-\int\limits_{-w{{q}_{i}}{{n}_{i}}d{{\Gamma }_{H}}}$$

We select $$\displaystyle w $$ such that $$\displaystyle w=0 $$ on $$\displaystyle {{\Gamma }_{g}}$$

On $$\displaystyle {{\Gamma }_{h}},\quad  q(x,t)n(x)=h(x,t) ,\forall x\in {{\Gamma }_{h}}$$

$$\displaystyle {{\Gamma }_{H}},\quad q(x,t)n(x)=H[u(x,t)-{u}_{\infty}] ,\forall x\in {{\Gamma }_{H}}$$ So we can write Equation 6 as :-

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

-\int\limits_{whd{{\text{ }\!\!\Gamma\!\!\text{ }}_{h}}}-\int\limits_{wH(u-{{u}_{\infty }})d{{\text{ }\!\!\Gamma\!\!\text{ }}_{H}}}-\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\frac{\partial w}{\partial {{x}_{i}}}\left( {{K}_{ij}}\frac{\partial u}{\partial {{x}_{j}}} \right)d\text{ }\!\!\Omega\!\!\text{ }}+\int\limits_{\Omega }{wf\left( x,t \right)d\text{ }\!\!\Omega\!\!\text{ }-}\int\limits_{\Omega }{w\rho c\frac{\partial u}{\partial t}d\text{ }\!\!\Omega\!\!\text{ }=0}

$$ (Eq )<7> This can be re-arranged in the form:-
 *
 * }

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{w\rho c\frac{\partial u}{\partial t}d\text{ }\!\!\Omega\!\!\text{ +}}\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\nabla w\left( {{K}_{ij}}\nabla u \right)d\text{ }\!\!\Omega\!\!\text{ }}+\int\limits_{wHu}d{{\text{ }\!\!\Gamma\!\!\text{ }}_{H}}=-\int\limits_{whd{{\text{ }\!\!\Gamma\!\!\text{ }}_{h}}}+\int\limits_{wH{{u}_{\infty }}d{{\text{ }\!\!\Gamma\!\!\text{ }}_{H}}}+\int\limits_{\Omega }{wf\left( x,t \right)d\text{ }\!\!\Omega\!\!\text{ }}

$$ (Eq )<8>
 *
 * }

This Equation is of the form:-

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #000000" |
 * style="width:92%; padding:10px; border:2px solid #000000" |

\tilde{m}(w,u)+\tilde{k}(w,u)=\tilde{f}(w) \quad \forall w $$ such that $$\displaystyle w=0 \quad on \quad {{\Gamma }_{g}}

$$ where,
 * }

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\tilde{m}(w,u)=\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{w\rho c\frac{\partial u}{\partial t}d\text{ }\!\!\Omega\!\!\text{ }}$$ $$\displaystyle\tilde{k}(w,u)=\int\limits_{\text{ }\!\!\Omega\!\!\text{ }}{\nabla w\left( {{K}_{ij}}\nabla u \right)d\text{ }\!\!\Omega\!\!\text{ }}+\int\limits_{wHu}d{{\text{ }\!\!\Gamma\!\!\text{ }}_{H}}$$ $$\displaystyle\tilde{f}(w)=-\int\limits_{whd{{\text{ }\!\!\Gamma\!\!\text{ }}_{h}}}+\int\limits_{wH{{u}_{\infty }}d{{\text{ }\!\!\Gamma\!\!\text{ }}_{H}}}+\int\limits_{\Omega }{wf\left( x,t \right)d\text{ }\!\!\Omega\!\!\text{ }}

$$ (Eq )<9>
 *
 * }

Developing Semi-Discrete equations

ii>

Let $$\displaystyle u=\left[ N \right]{{u}_{e}}\quad and \quad w=\left[ N \right]{{w}_{e}},$$

where $$\displaystyle [N] $$ is the vector of interpolation functions (Lagrange polynomial) and $$\displaystyle {u}_{e}\quad and \quad {w}_{e} $$ are the nodal values of $$\displaystyle u $$ and $$\displaystyle w $$ respectively.

Now by taking the derivative of basis functions with respect to $$\displaystyle {x}_{1}$$ and $$\displaystyle {x}_{2}$$, we can write:-

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

\left[ \begin{align} & \frac{dN}{d{{x}_{1}}} \\ & \frac{dN}{d{{x}_{2}}} \\ \end{align} \right]=\left[ B \right]

$$ (Eq )<10>
 *
 * }

Substituting the Lagrangian forms in Equation 9, we obtain:-

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

$$\displaystyle \int\limits_{\Omega }{{{w}_{e}}^{T}\rho c{{[N]}^{T}}[N]\frac{\partial u}{\partial t}d\Omega }+\int\limits_{\Omega }{{{w}_{e}}^{T}{{[B]}^{T}}[B]{{u}_{e}}d\Omega }+\int\limits_{{{w}_{e}}^{T}H{{[N]}^{T}}[N]{{u}_{e}}d{{\Gamma }_{h}}=-}\int\limits_{{{w}_{e}}^{T}h{{[N]}^{T}}d{{\Gamma }_{h}}}+\int\limits_{{{w}_{e}}^{T}H{{[N]}^{T}}{{u}_{\infty }}d{{\Gamma }_{h}}}+\int\limits_{\Omega }{w_{e}^{T}{{[N]}^{T}}f(x,t)d\Omega } $$ (Eq )<11>
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 *
 * }

From the above equation the conductivity matrix, the heat source vector and the capacitance matrix can be described as:- Capacitance matrix

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #000000" |
 * style="width:92%; padding:10px; border:2px solid #000000" |

\tilde{m}(w,u)= \int\limits_{\Omega }{\rho c{{[N]}^{T}}[N]\frac{\partial u}{\partial t}d\Omega } $$
 * }

Conductivity matrix

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #000000" |
 * style="width:92%; padding:10px; border:2px solid #000000" |

\tilde{k}(w,u)= \int\limits_{\Omega }{{{[B]}^{T}}[B]d\Omega }+\int\limits_{H{{[N]}^{T}}[N]d{{\Gamma }_{h}}} $$
 * }

Heat source matrix

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #000000" |
 * style="width:92%; padding:10px; border:2px solid #000000" |

\tilde{f}(w)= -\int\limits_{h{{[N]}^{T}}d{{\Gamma }_{h}}}+\int\limits_{H{{[N]}^{T}}{{u}_{\infty }}d{{\Gamma }_{h}}}+\int\limits_{\Omega }{{{[N]}^{T}}f(x,t)d\Omega } $$
 * }

plotting
consider the case of n=m=3 the number of nodes is 9 The trial solution is
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:95%" |
 * style="width:95%" |

u_{p}^{h}(x)={{d}_{1}}*N_{1}+{{d}_{2}}N_{2}+{{d}_{3}}N_{3}+\cdots +{{d}_{9}}N_{9},_ – ^ – j=0,1,2,\cdots ,9

$$
 * }

Since the trial solution must satisfy the essential boundary condition,, we should have  $${{d}_{1}}=2,{{d}_{2}}=2,{{d}_{3}}=2,{{d}_{6}}=2,{{d}_{9}}=2$$. , because it is specified that through out the boundary the temperature is 2 i.e for all the nodes on the boundary temperature is 2

In order to solve for the rest coefficient $${{d}_{j}},_ – ^ – j=1,2,\cdots $$, we construct the Conductance matrix, Heat Source matrix and Capacitance matrix as given by Equations 12,13 and 14.

Using Matlab to construct above matrices until satisfying the convergence criterion, we finally have
 * {| style="width:100%" border="0"

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

{{K}_{FF}}=\left( \begin{matrix} 151/200 &  -1/5 &  -1/30 & -133/1000 & -16/45 &    1/9 & -33/500 &    1/9 &  -1/45\\ -1/5 &  88/45 &   -1/5 & -16/45 & -16/15 & -16/45 & 1/9 & 0 & 1/9\\ -1/30 & -1/5 & 28/45 & 1/9 & -16/45 & -1/5 &  -1/45 & 1/9 & -1/30\\ -133/100 & -16/45 & 1/9 & 112/45 & -16/15 & 0 & -133/1000 & -16/45 & 1/9\\ -16/45 & -16/15 & -16/45 & -16/15  & 256/45 & -16/15 & -16/45 & -16/15 & -16/45\\ 1/9 & -16/45 & -1/5 & 0 & -16/15 & 88/45 & 1/9 & -16/45 & -1/5\\ -33/500 & 1/9 & -1/45 & -133/1000 & -16/45 & 1/9 & 1511/2000 & -1/5 & -1/30\\ 1/9 & 0 & 1/9 & -16/45 & -16/15 & -16/45 & -1/5 & 88/45 & -1/5\\ -1/45 & 1/9 & -1/30 & 1/9 & -16/45 & -1/5 & -1/30 & -1/5 & 28/45\\ \end{matrix} \right)

$$
 * }

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

{F}=\left( \begin{matrix} 166/1000\\ 0\\ 0\\ 33/500\\ 0\\ 0\\ -59/60\\ -4\\ -1\\ \end{matrix} \right)

$$
 * }

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

$$\displaystyle
 * style="width:92%; padding:10px; border:2px solid #ff0000" |
 * style="width:92%; padding:10px; border:2px solid #ff0000" |

{d}=\left( \begin{matrix} 2.000\\ 2.000\\ 2.000\\ 0.335\\ 0.847\\ 2.000\\ -1.460\\ -1.331\\ 2.000\\ \end{matrix} \right)

$$
 * }

The trial solution using 2D LIBF obtained is as follows:-

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

$$\displaystyle \begin{align} &U_{F}^{h}=(x)0.50*x*y*(x - 1)*(y - 1) - 1*x*(1.0*x + 1.0)*(1.0*y + 1.0)*(y - 1)\\& \quad + 0.665*y*(1.0*x + 1.0)*(1.0*y + 1.0)*(x - 1) - 0.1678*x*(1.0*y + 1.0)*(x - 1)*(y - 1)\\& \quad - 1.000*y*(1.0*x + 1.0)*(x - 1)*(y - 1) + 0.847*(1.0*x + 1.0)*(1.0*y + 1.0)*(x - 1)*(y - 1) \\& \quad + 0.500*x*y*(1.0*x + 1.0)*(1.0*y + 1.0) + 0.500*x*y*(1.0*x + 1.0)*(y - 1) - 0.365*x*y*(1.0*y + 1.0)*(x - 1)\\ \end{align} $$ (Eq )<15>
 * style="width:92%; padding:10px; border:2px solid #000000" |
 * style="width:92%; padding:10px; border:2px solid #000000" |
 *
 * }


text: File:Two elements.jpg

title: File:Two elements.jpg


title: File:Fe1.s11.team3.HW6.9.3.fig1.jpg


title: File:Fe1.s11.team3.HW6.9.3.fig2.jpg


title: File:Fe1.s11.team3.HW6.9.3.fig3.jpg


title: File:Fe1.s11.team3.HW6.9.3.fig4.jpg


title: File:Fe1.s11.team3.HW6.9.3.fig5.jpg


title: File:Fe1.s11.team3.HW6.9.3.fig6.jpg


title: File:Fe1.s11.team3.HW6.9.3.fig7.jpg


title: File:Fe1.s11.team3.HW6.9.3.fig8.jpg


title: File:Fe1.s11.team3.HW6.9.3.fig9.jpg


text:
File:Table.pdf

Licensing
for school project

title: File:Table.pdf


title: File:Fe1.s11.tm7.HW6.1nel2.JPG


title: File:Fe1.s11.tm7.HW6.1nel4.JPG


title: File:Fe1.s11.tm7.HW6.1nel6.JPG


title: File:Fe1.s11.tm7.HW6.1nel8.JPG


text:
User:Pkogut

Reflective Practice Analysis
Games are designed to model reality, or are they? Just like beauty, reality is in the eye of the beholder. If one agrees with this cliché, one will find oneself in a situation where reality may have multiple meanings, in effect becoming contingent upon the beholder’s interpretation. Navigating away from this view, this paper suggests that games are designed to model conflict. Conflict enlivens games just like alcohol enlivens perfume. Human history is one long conflict. Even when ‘rule of law’ replaced ‘law of the jungle,’ conflict still occupied a centre stage of human existence. When physical survival became less of a concern, social struggle for a better place ‘under the sun’ came to the fore. What spurs conflict is up to debate, but as far as this paper is concerned, the author will assume that it is usually an interplay of ideational, material and social factors. Drawing on these assumptions, this paper will analyse the accuracy of reflection of conflict in the following games: Liberte, Origins of WWII, Vietnam 1955, Comrade Koba and The Settlers of Catan. A caveat needs to be inserted at this point to inform the reader of the subjective nature of the analysis to follow. What the reader will find is not a detailed discussion of the structure and rules of the games, but rather a subjective analysis of the degree to which their game mechanics accurately model conflict as well as their strengths and weaknesses. Each game will also be accompanied by comments and suggestions for improvement.

Liberte features conflict so well that you can actually feel the chill down the spine as ‘children’ of the French Revolution lose their heads on the guillotine. On a more serious note, however, the game does not model conflict ‘all the way down.’ Three aspects of conflict – ideational, material and social – are illustrated to different degrees, with ideational being the lowest, material and social being the highest. Ideational factor is not fully acknowledged in the game, but those familiar with French or European history would know that 1789 Revolution shattered the foundation of monarchical rule and prepared the soil for future sprouts of liberal democracy. Although ideational face-off between forces of liberalism and absolutism are not duly acknowledged in Liberte, material and social factors make up for this absence. Conflict over material factors exemplifies geographic areas e.g. cities, provinces and adds a realist touch to the game. Social rivalry, on the other hand, is illustrated by three factions – Royalists, Moderates and Radicals – and is portrayed vividly in the game. It is perhaps through this mechanism that conflict is best modelled in Liberte.

Having said that, a player is prevented from experiencing rivalry to its fullest as there is no alignment with a particular party. This deprives players of feeling identified with one particular faction. On balance, though, Liberte creates a sense of conflict throughout the gameplay. Players are compelled to make rational choices and weight their decisions as they anticipate regional elections. Because Liberte is a sequential game, the last player enjoys an information advantage over other players as their plans unfold before his turn. This places the last player in a competitive position by enabling him to take advantage of the moves of previous players. Introducing some simultaneous moves to the game can definitely improve the game by making it more balanced as well as simplifying the rules to increase the pace of gameplay and players’ engagement.

Origins of WWII, as the name suggests, spans the time of the 20th century when Great Powers were preparing for the second ‘big fight.’ A game that models the Second World War is anything but peaceful. Conflict prevails throughout the game as players operate within strict time constraints – only six rounds to go before the outcome of continental rivalry is revealed. Origins of WWII features ideational and material conflict, leaving domestic social dimension out of sight. The game conveys the battle of ideas that characterises the inter-war period. Three competing ideologies i.e. Nazism, Capitalism and Communism clashed at the crossroads of history. The game captures this conflict rather accurately and simulates this tension to a high degree during the gameplay. Material conflict is exemplified in the game through territorial conquest. As Great Powers build their material capabilities, they expand their territories at the expense of small states, or ‘pawns,’ demonstrating insatiable desire for world domination.

Origins of WWII is almost a perfect information game, except when it comes to political factors that can be concealed to confuse other players. Other than that, information on outcomes, victory conditions and miscellaneous rules is revealed to all players, rendering information imbalance almost impossible. Another unique feature of the Origins is that it can be modified to be played as a simultaneous move game thereby preventing first several players from feeling disadvantaged in relation to the last one. Because of the disposition of political factors, the game encourages players to ‘gang up’ on the most powerful actors, mainly Germany and the Soviet Union. Origins, being very interesting and sociable to play, could be improved by having more historical actors such as Italy and Japan. Ability to form alliances that increase member states’ political factors could be a useful add-on as well as an increase in the number of rounds from six to ten, for example. Chance or community cards that simulate natural disasters, defection of top officials, desertion, and discovery of new weapons of mass destruction can provide a value-added experience to the enjoyment of all players.

In a similar vein, Vietnam 1955 models a certain time in history when Great Powers were dividing the world into spheres of influence to swing the balance of power in their favour. Like the Origins, Vietnam models conflict by juxtaposing ideational and material factors, mirroring the established post-Yalta World Order with a certain degree of realism. The uniqueness of Vietnam lies in its game mechanics. Unlike other board games discussed in this paper, Vietnam utilizes role-playing to simulate conflict in the Far-East Asia a little more than half a century ago. Players are divided into competing groups and take on the roles of actors involved: the US, the USSR, the UK, France, French Indo-China and Vietminh. Moderated by the narrator, Vietnam becomes a hot spot of debates, accusations, conspiracy and secret negotiations. Victory conditions are designed so that players’ outcomes are interwoven.

Having said that, during the gameplay this aspect was miscommunicated and players interpreted it as they saw fit. It is absolutely imperative to have clear guidance from the narrator in any role-playing game to avoid confusion and misunderstanding. Nevertheless, microcosm that Vietnam scenario simulated imbued the gameplay with a spirit of diplomacy, especially when it came to the Geneva Peace conference. Although arguing for and against certain outcomes during the conference lacked concrete reasoning and understanding of the subject, many found it enjoyable just for the sake of public speaking. The game could benefit from having more visual material to deliver the spirit of the era. For example, a detailed map of the area the size of A2 could prove useful in identifying key locations of the conflict and would also enable players to refer to them during secret and open negotiations. Clear delegation of responsibilities and victory conditions would lock players in purpose-specific roles and enable them to improve their argument when negotiating with other players. Lack of communication from the narrator can deprive players of strategic and tactical thinking – something that needs to be addressed during future game-testing. On balance, Vietnam delivered fun, tension and excitement – three components of a successful role-playing game.

Comrade Koba depicts conflict mainly from a psychological perspective of an individual and does not account for the greater socio-historic context of the era. Comrade Koba presents conflict as a struggle of fears, characterised by one’s paranoid suspicion and mistrust. Notwithstanding this one-sided view of the conflict, the game confirms the zero-sum logic of great leaders and the absurdity it can sometimes lead to. The game offers players to choose from a variety of Stalin’s cronies and experience the whims of the great dictator as their lives are ‘shuffled’ along the way. Random kill and real historical figures used in the game convey realism of the Great Purges to a high degree. Negligence of human life sets an eerie tone for the forthcoming gameplay as players aim to keep their characters as further away as possible from Stalin to achieve victory. Being a game of not only political, but also physical survival, Comrade Koba elevates the principle of ‘all against all’ above other moral considerations, mainly friendship and loyalty. ‘A stab in the back’ is expected every time new player takes his turn. If anything, this game confirms the common knowledge that ‘absolute power corrupts absolutely.’

Like many games, Comrade Koba is not a perfect information game in that players keep their chosen characters in secret. Neither is it a simultaneous game that eliminates information disadvantage. One’s success also depends on chance as Koba dice rolls determine the future outcome of a character’s life. A great advantage of this game is that its rules are very simply to follow and easy to remember, which allows to engage players from the start. Comrade Koba is anything but time-consuming and in less than half an hour the game has its first victor. Time and chance are probably two factors that make this game enjoyable and sociable to play. Comrade Koba, although it is classified as a strategy game, shares few characteristics with the Settlers of Catan, also a strategy game, but one in which thinking several steps ahead is a necessity, not a choice. In terms of improvement, the game would benefit from some background information to accompany its main characters. Especially death stories can prove very useful as they inform players of the peculiarities of Stalin Purges, on the one hand, and amuse the team, on the other.

The Settlers of Catan falls short of a political simulation as it over-stresses the role of trade and exploration of natural resources in a given environment. This game models development, city-planning and infrastructure, leaving political element somewhat out of the equation. One understands that all of the aforementioned activities cannot be initiated without a centralized authority, but politics is more than just city-planning and trade. Nevertheless, the game manages to instil competitive spirit in all players as they vie for victory points. The Settlers of Catan features material factors that spur conflict i.e. the struggle for natural resources (ore, timber, stock etc). The problem of scarce resources becomes acute after a few rounds of gameplay and keeps players on the edge of their seats, least they run out of timber and ore – two crucial components for building settlements. This feature is integral not least for encouraging players to leave disagreements behind and start negotiating.

The Settlers of Catan, therefore, like no other game, combines cooperation and competition by proving that these two options are not mutually exclusive. If anything, it teaches players to compromise and make deals with their rivals. A major drawback of this game is its limited distribution of resources and territory. One needs to be rather quick to occupy best lands and start erecting settlements. In addition, chance determines the outcome of many events, be it having a Robber placed near your territory or giving some of your resources to a competitor as a result of a chance card, which sometimes makes strategic decision making futile. Hence, it may prove difficult for players to follow strategies as they are constantly hampered by bad luck. Having said that, the game models difficulties associated with uninhabitable lands or barren soil – a reality of some parts of the world – rather accurately. A country with scarce natural resources has no other option but to trade with its neighbours in order to survive. The game could be improved by introducing power and destruction abilities. For example, special circumstances need to be introduced to allow players to take over their rivals’ lands, settlements and roads – something that current version of the game does not offer. Adding a power component to the game would definitely shift it from trade to political dimension. Finally, an extended map with added resources would help eliminate frustration among those who were out of luck when it was time to conquer lands and are now left to enjoy only a handful of resources.

To conclude, this paper analysed the degree to which five chosen games simulate conflict. Liberte, for example, simulates rivalry among different social groups during the French Revolution. Origins of WWII depicts conflict as a battle of ideas and a battle for power on the continental Europe in the first half of the 20th century. Vietnam 1955 is similar to Origins in that it focuses on the post-WWII World Order in which Great Powers compete for spheres of influence. Comrade Koba personifies conflict on an individual level and offers players to enter the psychic world of Joseph Stalin. Finally, The Settlers of Catan, although it elevates trade over politics, simulates conflict over material factors, namely natural resources, in a realistic fashion. After reading this analysis, the reader is encouraged to take part in at least one of these games. The five games discussed in this paper appeal to different public as there is something to be found for students of politics and international relations (Origins, Vietnam), history (Liberte) and just admirers of Stalin (Comrade Koba). All games have proved to be very stimulating and sociable to play, to say the least.

title: User:Pkogut


text: File:Eml5526.s11.team2.reiss.hw6.8.1.png


text: File:Eml5526.s11.team2.reiss.hw6.8.2.png


text: File:Eml5526.s11.team2.reiss.hw6.8.3.png


text: File:FE HW 6.9 pseudocode.pdf

title: File:FE HW 6.9 pseudocode.pdf


text: File:Tm7 ex 8 4 temp.jpg

title: File:Tm7 ex 8 4 temp.jpg


title: File:Fe1.s11.team1.hw6.fig6.9.1.jpg


text:
User:Egm6341.s11.team4.kurth/hw6

=Problem 6.5: Using the Recurrence Relation to Calculate Polynomials $$p_{2i}(t)$$ and $$p_{2i+1}(t)$$=

Objective
Use the recurrence relation, equation [[media:Nm1.s11.mtg32.djvu|(1)p.32-2]]

to find $$\displaystyle \;\; (p_2,p_3),\;(p_4,p_5),\;(p_6,p_7),\;(p_8,p_9)$$.

Solution
From equations [[media:Nm1.s11.mtg32.djvu|(1)p.32-1]] and [[media:Nm1.s11.mtg32.djvu|(3)-(4)p.32-1]] the polynomials $$\displaystyle p_{2i}(t)$$ and $$\displaystyle p_{2i+1}(t)$$ take the following form:

Find $$(p_2,p_3)$$
Here, $$\displaystyle i=1$$. Equation (5.2) becomes

Using equations (5.3) and (5.4) to reconstruct $$\displaystyle p_2$$ and $$\displaystyle p_3$$

Find $$(p_4,p_5)$$
Now, $$\displaystyle i=2$$. Equation (5.2) becomes

Using equations (5.3) and (5.4) to reconstruct $$\displaystyle p_4$$ and $$\displaystyle p_5$$

Find $$(p_6,p_7)$$
Now, $$\displaystyle i=3$$. Equation (5.2) becomes

Using equations (5.3) and (5.4) to reconstruct $$\displaystyle p_6$$ and $$\displaystyle p_7$$

Find $$(p_8,p_9)$$
Now, $$\displaystyle i=4$$. Equation (5.2) becomes

Using equations (5.3) and (5.4) to reconstruct $$\displaystyle p_8$$ and $$\displaystyle p_9$$

title: User:Egm6341.s11.team4.kurth/hw6


text: File:Egm5526.s11.team2.oztekin.Hw6.10.2.png


text: File:Egm5526.s11.team2.oztekin.Hw6.10.1.png


title: File:Fe1.s11.team1.hw6.fig6.9.2.jpg


title: File:Fe1.s11.tm7.HW6.1error.JPG


title: File:Fe1.s11.team1.hw6.fig6.9.3.jpg


title: File:Fe1.s11.team1.hw6.fig6.9.4.jpg


text:
Psycholinguistics/Development of Speech Production answers

Learning Exercise Answers
1. a)
 * video 1 - Integration (Jargon stage)
 * His vocalizations are similar to the English language in their prosodic structure, syllable structure and vowel-consonant combinations. You can also see that he is using gestures as if he were having a conversation..

 * video 2 - Canonical babbling(reduplicated babbling)
 * CC produces the sounds "badada" showing that he is able to produce consonants and vowels, and he can control the sounds to produce the ones he wants.

 * video 3 - Vocal play
 * Here we see a range of sounds, that seem vowel-like, however none seem to be repeated ( and thus are not canonical). Furthermore, at one point, he seems to raise his voice (yells) showing an attempt to control or display his vocal range.

 * video 4 - Reflexive vocalization
 * We hear very little vocalization besides a sneeze towards the end of the video.

 * video 5 - cooing
 * CC is making vowel-like ("a") sounds in responds to parental attention and is this attempting at interaction. We almost see a smile too!

b) This vocalization is called a quasi-word as it is used consistently in the presence of a "bottle". It occurs during integration (stage 5) as the child is bridging the gap between babbling and words, and beginning to understand the connection between vocalization patterns and their associated meaning.

2. a) b)
 * 1.Possesive morphemes:
 * CHI: that Kathy.
 * MOT:yes # that's Kathy.
 * MOT:	Eve's letter.
 * CHI:	Mommy letter.
 * MOT:	there's Mommy's letter.
 * CHI:	Eve letter.
 * 2.in, on:
 * MOT:	you have tapioca on your finger.
 * CHI:	tapioca finger . [+ IMIT]
 * 3.present progressive (-ing)
 * CHI:	Kathy cry.
 * MOT:	yes # Kathy was crying.

c) Although this child appears to have used "a" in her utterance "a fly", she does not use "a" consistently throughout the sample. For example, soon after that utterance, the child repeats a similar utterance "fly" without the "a" (MOT: you go in the room and kill a fly. CHI: fly. ). Other examples that should include "a" are the following: The child shows no evidence of using "the."
 * MOT: what is that ?
 * CHI: letter.
 * MOT: oh # what's that ?
 * CHI: hat.

3.

4.
 * (a) Vocabulary spurt. There are three theories on how this may have developed. This process may begin from the naming insight, where children begin to understand that referents can be labeled. Second, children would need to expand their vocabulary to label categories, as this this period seems to coincide with Piaget’s sensorimotor stage  (where children are expanding their understanding of concepts and objects).  Finally, leveraged learning may facilitate the vocabulary explosion where the child learns words slowly (one at a time) and then each word can act as a "leverage" to learn the next word and this facilitates word learning.
 * (b) The Noun Bias.
 * (c) Overextension.

5.
 * (a) lexical innovations - The child has not yet learned a word associated with the meaning they are trying to express, or they simply cannot retrieve it properly. It is sometimes easier to remember this word than the traditional word.
 * (b) over-extension - The child is attempting to fill the lexical gap because they do not have a current label for that particular item, they don’t know the target word sufficiently and thus have difficulty determining which label the target should receive, or they temporarily cannot retrieve the word.
 * (c) over-regularization - The child does not have a complete understanding of the word meaning and thus incorrectly selects it, or the child is improperly categorizing the word.
 * (d) cluster reduction (leave out second stop in a consonant cluster) - difficulty producing multiple consonants in a row
 * (e) under-extension - The child is attempting to fill the lexical gap because they do not have a current label for that particular item, they don’t know the target word sufficiently and thus have difficulty determining which label the target should receive, or they temporarily cannot retrieve the word.
 * (f) skip unstressed syllable  the child fails to attend to the unstressed syllable
 * (g) assimilation - another phoneme is influencing the production of the target phoneme
 * (h) omission of function words - The child may focus on words that have referents, the child recognizes only the content words, which have greater stress and emphasis, or the child's production system is not mature enough to produce all the various parts of speech.


text:
Psycholinguistics/Development of Speech Production example

Learning Exercise Example
Example: video 1 video 2 video 3
 * In terms of phonology, this child is able to produce most sounds appropriately. She does produce a consonant cluster reduction (kechup - rather than ketchup).
 * Her semantic development is good as she understands that ketchup refers to the sauce in the red bottle and mustard in the yellow bottle.
 * Her syntactic speech production shows evidence of her still developing negative phrase structure. She says “I no like ketchup” which is an incorrect syntactic sentence. It should be “I don’t like ketchup”
 * In this example, we see that this child may not yet understand the meaning of the word “gift” as her mother asked her if it’s a gift for her, and the girl replies “ no it’s a mirror”
 * In this video, the child can combie multiple utterances, but seems to make some overregularization errors (morphological development). She says “it didn’t ate him” and it and he falled down”


title: File:Fe1.s11.tm7.HW6.8gauss.JPG


text: File:Fe1 s11 team3 HW6 1 fig1.jpg

title: File:Fe1 s11 team3 HW6 1 fig1.jpg


text: File:Tm7 ex 8 4 mesh.jpg

title: File:Tm7 ex 8 4 mesh.jpg


text:
User:Egm6341.s11.team4.yang/hw6

=HW 6.7 problem: compute circumference bifolium and ellilpse =

Refer to lecture slide [[media:nm1.s11.mtg33.djvu|mtg-33]] for the problem statement.

Objective
Compute circumference of : 1).bifolium; 2). ellipse using the following methods: a). composite trap b). Romberg c). clenshaw-curtis quad d). Fast CC winckel

1.bifolium
$$r(t)=2\sin (t)\cos {{(t)}^{2}},t\in [0,\pi ]$$ The circumference of bifolium can be computed by: $$C=\int\limits_{t=0}^{\pi }{\sqrt{{{r}^{2}}+{{(\frac{dr}{dt})}^{2}}}}dt$$

Four methods to compute the integral: a).Composite trap Use error estimate for Composite Trapezoidal Rule to find n, such that the error is to the order of $$\displaystyle O(10^{-8}) $$.

Use successive numerical Integration results as a stopping criterion, i.e,

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

$$ $$
 * $$\displaystyle
 * I_{n+2} - I_{n}| < 10^{-8}
 * I_{n+2} - I_{n}| < 10^{-8}
 * $$\displaystyle
 * }.
 * }.

 Matlab Code:  b).Romberg The first column of Romberg table is calculated by composite trap, and then use the following equation to continue. $$\displaystyle T_{k}(2n) = \frac{2^{2k}T_{k-1}(2n) - T_{k-1}(n)}{2^{2k}-1}. $$  Matlab Code:  c).Clenshaw-Curtis Using Clenshaw-Curtis matlab code to compute the result.  Matlab Code:  d).Fast CC Using fast Clenshaw-Curtis matlab code to compute the result.  Matlab Code:  The results from the different methods are compared here:

2.ellipse-A

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

$$\displaystyle r(\theta) = \frac{a(1-e^2)}{1-e \, cos(\theta)} $$ $$
 * $$\displaystyle
 * }.
 * }.

Here, $$\displaystyle a=10,b=3 $$ and $$\displaystyle e=\sqrt{1-{{\left( \frac{b}{a} \right)}^{2}}} $$

Arc Length for an elliptical curve with $$\displaystyle \theta$$ varying from 0 to $$\displaystyle 2\pi $$ is given by,

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

$$\displaystyle

I=\int_{0}^{2\pi }{\sqrt{{{r}^{2}}+{{\left( \frac{dr}{d\theta } \right)}^{2}}}}d\theta

$$ $$ Four methods to compute the integral: a).Composite trap Use error estimate for Composite Trapezoidal Rule to find n, such that the error is to the order of $$\displaystyle O(10^{-10}) $$.
 * $$\displaystyle
 * }.
 * }.

Use successive numerical Integration results as a stopping criterion, i.e,

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

$$ $$
 * $$\displaystyle
 * I_{n+2} - I_{n}| < 10^{-10}
 * I_{n+2} - I_{n}| < 10^{-10}
 * $$\displaystyle
 * }.
 * }.

b).Romberg The first column of Romberg table is calculated by composite trap, and then use the following equation to continue. $$\displaystyle T_{k}(2n) = \frac{2^{2k}T_{k-1}(2n) - T_{k-1}(n)}{2^{2k}-1}. $$

c).Clenshaw-Curtis Using Clenshaw-Curtis matlab code to compute the result. d).Fast CC Using fast Clenshaw-Curtis matlab code to compute the result.

The results from the different methods are compared here:

3.ellipse-B
The elliptic integral of the second kind is given as:

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

$$\displaystyle I=4aE(e)=4a\int_{0}^{\pi /2}{\sqrt{1-{{e}^{2}}{{\sin }^{2}}\theta }}\ d\theta $$ $$ Four methods to compute the integral: a).Composite trap Use error estimate for Composite Trapezoidal Rule to find n, such that the error is to the order of $$\displaystyle O(10^{-10}) $$.
 * $$\displaystyle
 * }.
 * }.

Use successive numerical Integration results as a stopping criterion, i.e,

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

$$ $$
 * $$\displaystyle
 * I_{n+2} - I_{n}| < 10^{-10}
 * I_{n+2} - I_{n}| < 10^{-10}
 * $$\displaystyle
 * }.
 * }.

b).Romberg The first column of Romberg table is calculated by composite trap, and then use the following equation to continue. $$\displaystyle T_{k}(2n) = \frac{2^{2k}T_{k-1}(2n) - T_{k-1}(n)}{2^{2k}-1}. $$

c).Clenshaw-Curtis Using Clenshaw-Curtis matlab code to compute the result. d).Fast CC Using fast Clenshaw-Curtis matlab code to compute the result.

The results from the different methods are compared here:

Observations
1.For all the problem, Romberg method always need less time than Composite trap method. 2.The second kind of ellips (ellip-B) need less time than ellip-A. 3.The fast CC method always performs better than CC method. 4.For the problem of ellip-B, the computation time is similar for the four methods. This problem was solved by shengfeng yang

title: User:Egm6341.s11.team4.yang/hw6


text: File:Fe1 s11 team3 HW6 1 fig3.jpg

title: File:Fe1 s11 team3 HW6 1 fig3.jpg


text: File:Fe1 s11 team3 HW6 1 fig4.jpg

title: File:Fe1 s11 team3 HW6 1 fig4.jpg


text: File:Fe1 s11 team3 HW6 1 fig2.jpg

title: File:Fe1 s11 team3 HW6 1 fig2.jpg


title: File:Eml.5526.s11.tm7.6.3.1.Pic3.PNG


title: File:Eml.5526.s11.tm7.6.3.1.Pic4.PNG


text: File:Hw6 1 Uh vs Ue nel=14.jpg

title: File:Hw6 1 Uh vs Ue nel=14.jpg


title: File:Eml.5526.s11.tm7.6.3.1.Pic5.PNG


title: File:Eml.5526.s11.tm7.6.3.1.Pic6.PNG


title: File:Eml.5526.s11.tm7.6.3.1.Pic7.PNG


title: File:Eml.5526.s11.tm7.6.3.1.Pic8.PNG


title: File:Eml.5526.s11.tm7.6.3.1.Pic11.PNG


title: File:Eml.5526.s11.tm7.6.3.1.Pic12.PNG


text:
File:Media Table.png

Licensing
for school project

title: File:Media Table.png


text: File:Hw6 1 Error.jpg

title: File:Hw6 1 Error.jpg


title: File:Eml.5526.s11.tm7.6.3.1.Pic2.PNG


title: File:Eml.5526s11.tm7.6.3.2.Pic1.PNG


title: File:Eml.5526s11.tm7.6.3.2.Pic2.PNG


title: File:Peml.5526s11.tm7.6.3.2.Pic3.PNG


title: File:Eml.5526s11.tm7.6.3.2.Pic4.PNG


title: File:Eml.5526s11.tm7.6.3.2.Pic7.PNG


title: File:Eml.5526s11.tm7.6.3.2.Pic9.PNG


title: File:Eml.5526s11.tm7.6.3.2.Pic10.PNG


title: File:Eml5526.s11.team2.4elements.png


title: File:Seml.5526.sll.tm7.6.3.3.Pic1.PNG


title: File:Seml.5526.sll.tm7.6.3.3.Pic2.PNG


title: File:Seml.5526.sll.tm7.6.3.3.Pic3.PNG


text: File:Pic5.PNGseml.5526.sll.tm7.6.3.3.Pic5.PNG


title: File:Eml.5526.s11.tm7.6.4.4.Pic1.PNG


title: File:Eml.5526.s11.tm7.6.4.4.Pic3.PNG


title: File:Eml.5526.s11.tm7.6.4.4.Pic4.PNG


title: File:Eml.5526.s11.tm7.6.4.4.Pic5.PNG


text: File:Eml5526.s11.team7.hw6.6 1.jpg

title: File:Eml5526.s11.team7.hw6.6 1.jpg


text:
User:Anand ST14

=Problem 7.4-Solution to a general 2-D problem (Heat Transfer)with Lagrange Interpolation Basis Functions(LIBF) =

Given
Problem statement initiated under problem 6.10 (Re-Done here) in mtgs - [[media:Fe1.s11.mtg37.djvu|37-2]] and [[media:Fe1.s11.mtg37.djvu|38-2]]

2D space with boundary convection
Consider a heat conduction problem in 2-dimensional space with boundary convection as shown in the below Figure 6.10.1.

Boundary Conditions
Above 2D space has temperature $$ T(x,y)$$ as space unknown to solve for. We can state it as a boundary value problem with following boundary conditions specified at 3 different kind of boundaries as : $${\Gamma _g}$$, $${\Gamma _h}$$ and $${\Gamma _H}$$

Essential BC on $${\Gamma _g}$$
$$\displaystyle (Eq. 7.4.1) $$

Note : $$T (x,y)$$ can also can be defined as a varying quantity along $${\Gamma _g}$$

Natural BC on $${\Gamma _h}$$
Refer to the Figure 7.4.2 below. We specify the heat flux on boundary $${\Gamma _h}$$ as $${q_n} = {q_0}$$.

This is as we have : $$ {q_n} = \overline q .\overline n $$ with $$\overline n $$ as the unit normal vector on $${\Gamma _h}$$ Hence we can also write : $${q_n} = ( - k.\overline \nabla T).\overline n  =  - k(\overline \nabla  T.\overline n )$$ OR we then have :

$$\displaystyle (Eq. 6.10.2) $$

Note- $$ \nabla T$$ is positive as explained in the - Assumption- subsection below

We are to consider only the normal component of $$\overline q $$, i.e. $$ {q_n}$$, for heat transactions as that is the only component causing heat flow in or out of the element. The tangential component of $$\overline q $$, i.e. $$ {q_t}$$, is just circulating over the boundary.

Assumption
$$\overline n  $$ is unit vector pointing in the outward direction. This implies that $$ \nabla T$$ is positive and temperature increases as we go out away from the surface towards inside of the body. Also we note that $$\overline n  $$ is a vector quantity normal to the $$ T(x,y)$$ contours in 2D space. As $$ T(x,y)$$ increases from inside to outside, heat flux $${q_n}$$ goes into the volume from outside of the boundary $${\Gamma _h}$$ This implies that when $${q_0}$$ is positive, heat comes into the system

Convection BC on $${\Gamma _H}$$
We have as the convection boundary where heat transfer occurs via heat convection Hence we can say that the normal component $${q_n}$$ of the heat flux $$\overline q $$ equals the heat transfer due to heat convection at $${\Gamma _H}$$. Hence we write-

$${q_n} = h(T - Ta)$$ where, we have - $$T$$ : Temperature at the surface of $${\Gamma _H}$$ $${T_a} = {T_\infty } =$$ Say, Ambient temperature $$h$$ :Coefficient of heat convection at $${\Gamma _H}$$ Thus :

$$\displaystyle (Eq. 7.4.3) $$

Given domain with 2-dimensional model data set 2
The following data set needs to be solved using 2D LIBFs -

Solution
Referenced from Section 6.3 (pg 142 )and Section 8.1 (pg 189 ) of text book - ' A first course in finite element method' by authors Fish & Belytschco

Deriving the Equilibrium Equation for 2D- heat transfer
We first derive the equilibrium equation using elemental plane surface as shown in Figure- 7.4.3 below. The elemental surface has length as 'dx' and height as 'dy'. The inflow and outflow of heat is shown symbolically with arrows at various sections.

Note- Heat flux ' $$ \underline q $$ ' is a vector quantity in real 2D space defined per unit surface area and also $$f = $$ Heat generation per unit volume for the given elemental surface We assume unit dimension along the direction perpendicular to the screen or paper

Now the equilibrium of heat can be written in following manner -

Now we use | Taylor Series Expansion technique and can then write -

$$ {({q_x})_{x + dx}} = {({q_x})_x} + {\left( {\frac} \right)_x}dx + higher\_order\_terms$$ OR we can say with the higher order terms neglected :

Similarly we can write for y- direction -

And now $$ Eq.7.4.4 $$ can be modified as below - Canceling out $$dxdy $$ throughout as we have used it for elemental surface and it has to be true over the entire 2D space,

$$ Eq. 7.4.8 $$ can be written with different notation in the form of Divergence of $$\overline q $$ as - Note that $$ Eq. 7.4.9 $$ is valid for 2D as well as 3D heat transfer with $$\overline q $$ as 2D or 3D vector respectively in real 2D or 3D space. Here we are solving for temperature $$T(x,y)$$ so we use 2D space with $$T(x,y)$$

Constitutive Equation for 2D- heat transfer
Here we use the Fourier Law for 2D space which is stated as : $$\overline q  =  - \left[ k \right]\overline \nabla  T $$ It can be written as - Note- Usually, $$=0 $$ and $$=0 $$ for most of the materials as we do not see flow of heat perpendicular to the direction of the temperature gradient line. Also, we have : $$= $$ for isotropic materials and $$ \ne  $$ for anisotropic materials We consider isotropic materials here hence we can assume : $$= $$. Hence $$ Eq. 7.4.10 $$ now becomes - Hence we write then :

Governing Equation for 2D- heat transfer
Using $$ Eq.7.4.9 $$ and $$ Eq. 7.4.10 $$ we can say that :

Thus the strong form for 2D - Heat Transfer can be written as follows-

$$\displaystyle (Eq. 7.4.11) $$

Deriving the Weak Form for 2D- heat transfer with boundary convection
Let us select the weighting function $$ w(x,y) $$ such that $$ w(x,y)=0$$ at boundary $$ {{\Gamma }_{g}} $$ where the essential boundary condition is applied. This is actually obvious to say as the essential boundary condition is specified at $$ {{\Gamma }_{g}} $$, the value of $$ T (x,y)$$ will be constant there and will no longer require any weighting function multiplication to approximate it in the given domain. Also this will eliminate the unknown flux at the boundary $${{\Gamma }_{g}} $$

Integrating over volume (for general 3D - case) after multiplying with the weighting function w(x), in the form of weighted residual, we get - Note- For 2D- case : Put $$ dz $$ = $$1$$ which will give us the Area Integral Now we use Green's Theorem to perform the integration by parts. Our aim to get rid of the derivative on $$ \overline{\nabla }.({{K}_{ij}}\overline{\nabla }T)$$ and a derivative should appear on $$ w(x) $$ We state the general Green's Theorem for scalar fields $$ \varphi $$, T - is as follows -

Here, $${{\nabla }^{2}} $$ serves as Laplacian or Divergence of the gradient. Also, $$\overline{n} $$ is the unit normal at boundary and is variable at each point on the boundary. Thus ,we have a volume integral converted to a volume plus a surface integral Hence, $$ Eq. 6.10.12 $$ can now be modified as follows - This is general equation for 3D as well as 2D case and can be converted to specific case depending upon whether the gradient is 2D or 3D Now again we note that in the weak form the natural boundary conditions get pulled in automatically, for example- the surface integral in above equation will be split into integral over the 3 boundaries - $${{\Gamma }_{g}} $$, $${{\Gamma }_{H}} $$ , and$${{\Gamma }_{h}} $$ , as follows :

Note- In the above Equation, the first term on the RHS will vanish as the weighting function $$ w (x) $$ vanishes on $$ {{\Gamma }_{g}}$$ as explained. Also here we flipped the signs for convection boundary condition on boundary on $${{\Gamma }_{g}} $$ Now, we put $$ Eq. 7.4.15 $$ into $$ eq. 6.10.14$$ and keep the terms required to get the stiffness matrix $$ \left[ K \right]$$ along with the terms involving temperature on LHS and all the terms with known values on the RHS.

$$\displaystyle (Eq. 7.4.16) $$

$$ Eq. 7.4.16 $$ is the weak form for the given 2D Heat Transfer domain with convection boundary specified as $$ {{\Gamma }_{H}}$$

This equation is of the following form -

It is the continuous form of the weak equation and we need to convert it into a discrete form using the method of Desensitization in FEM. Also note that - 1. In $$ Eq. 7.4.16 $$, temperature is a continuous function of field parameters (x,y). 2. This equation is good for 2D as well as 3D domains. 3. We can handle anisotropic materials by converting conductivity 'k' into matrix form as explained.

Developing the semi-discrete equations
Now the temperature distribution in 2D space can be written as follows with subscript 'e' denoting to an element i.e. temperature at a point (x,y) in the 2D space is a sum over that of the elemental nodes. Also here $$ N_i $$ denotes shape function and $$ T_i $$ denotes nodal temperature values.

This can be written in matrix form as follows where the nodal values matrix is shown by $$\left\{ {{X}_{e}} \right\} $$

Also applying the same for the weighting function, we can write : This can be written in matrix form as follows -

We note that we have got gradient terms in the weak form as seen in $$Eq. 7.4.16 $$. The $$\overline{\nabla }T $$ can be written as : Here $$ \left[ B \right]$$ denotes the matrix of derivatives of the shape functions as shown in below $$ Eq. 7.4.22 $$.

Similar equation can be written for the weighting function as follows -

Now the LHS of the weak form in $$ Eq. 7.4.16 $$ can be written as for a single element -

Hence, LHS becomes for $$ Eq. 7.4.16$$ as :

Depending upon various boundary conditions the RHS of $$ Eq. 7.4.16 $$ can be written as $$\left[ {{F}_{e}} \right]{{\left\{ \partial {{X}_{e}} \right\}}^{T}} $$ Hence the discrete form Equation for first term of the stiffness matrix becomes -

where, $$ \left[ {{K}_{e1}} \right]=\int\limits_{V}{{{\left[ B \right]}^{T}}\left[ K \right]}\left[ B \right]dv$$

The second term of the stiffness can be written in discrete form as follows-

Now the mass matrix can be written in discretized form from the first term of the continuous weak form of $$ Eq. 7.4.16 $$ as follows -

Now the first term on RHS can be discretized as -

Now the second term on RHS can be discretized as -

And the last term on RHS of the weak form can be discretized as -

Capacitance matrix

Conductivity matrix

Heat source matrix

Semi- discrete equations form the given data set G2DM2.0/D1
Referring to mtg - [[media:Fe1.s11.mtg37.djvu|41-5]], the discrete form equations for the given data set as explained in the 'Given' section above, can be written as follows -

Note- In all the matrices below, we have overlap at Node 13 of the domain as shown in Figure 7.4.5 below - Now the conduction contribution from $${{\Gamma }_{H}} $$ can be written as follows -

Solve the weak form
The following Matlab codes are used to solve the weak form given above using LLEBF.

compare results for different numbers of elements
We use uniform elements and the results of different numbers of elements are as following

From the above results, we can find these results are identical to each other, which means that our codes are correct to achieve the same results even with different numbers of elements.

We can find "negative temperature" at the intersection corner of natural boundary condition and mixed boundary condition. Although it's quite unphysical to have "negative temperature", we try to adjust the boundary condition to find out the reason why the "negative temperature" exists.

compare results for different essential boundary conditions
First we compare the results of different essential boundary conditions with g=2,4,10,300

From the above results, we can find that we won't have "negative temperature" if g>4. From the 2D temperature distribution, we can find that whatever the essential boundary is, the distribution is similar with each other but the magnitude of maximum and minimum are different. However, the distance between maximum and minimum temperature is identical to each other, which means that the essential boundary can "shift" the results. Therefore maybe the "negative temperature" exists because the essential boundary is a little low for this case.

compare results for different natural boundary conditions
Then we compare the results of different natural boundary conditions with h=6,3,2,1

From the above results, we can find that the lower outward heat flux is, the minimum temperature is, because with lower outward heat flux, the temperature will dissipate slowly. When h = 1, the "negative temperature" disappears because less and less temperature will flow out of the area.

Why do we have "negative temperature"?
It's true that the "negative temperature" exist without any physical meaning, because in the universe the minimum temperature is zero K where "kelvin" is a unit of measurement for temperature. But if we suppose the unit of the temperature is Celsius scale, which is another unit scale for thermal temperature, we can have negative temperature just like what we have in winter. In other word, the actual governing equation we solved in this problem is a Laplace equation which can not guarantee that all the results on all the nodes are not negative, unless we can give it the "true" boundary condition which are physical in real world.

And also from the above procedures that we compare results from different boundary conditions, we can find that the boundary conditions do dominate the temperature distribution and "negative temperature" disappears with certain boundary conditions.

Therefore, We believe that we can have physical temperature distribution with physical boundary conditions, even sometimes "negative temperature" can exist with using different aspects of measure unit.

title: User:Anand ST14


title: File:Eml5526.s11.team2.6.2error.png


title: File:Eml5526.s11.team2.6.2sol.png


title: File:Fe1.s11.team3.HW6.9.2.fig1.jpg


title: File:Fe1.s11.team3.HW6.9.2.fig2.jpg


title: File:Fe1.s11.team3.HW6.9.2.fig3.jpg


title: File:Fe1.s11.team3.HW6.9.2.fig4.jpg


title: File:Fe1.s11.team3.HW6.9.2.fig5.jpg


title: File:Fe1.s11.team3.HW6.9.2.fig6.jpg


text:
Statistics/Bayesian Statistics

The difference between a Bayesian and a frequentist
Bayesian statistics assumes a fundamentally different model of the universe from the one of frequentist statistics. This difference can sometimes divide the two “camps” of statistics along philosophical lines. Some problems are more naturally suited to one or the other approach, or some can be viewed from either camp.

Bayesian statistics takes its name from Thomas Bayes, an English Presbyterian minister and amateur mathematician. This is why Bayesian is capitalized.

Philosophy of Bayesian statistics
The basic Bayesian philosophy can be summarized as follows: Everyone has information, or opinions, about the world. When we encounter new information, we take that information and combine it with what we already know to create new opinions. The goal of Bayesian statistics is to do this in a coherent way. Whereas a frequentist assumes that there is an "exact truth" out there, which can only be measured with measurement error, the Bayesian regards measurements as exact and the "underlying universe" as the thing subject to uncertainty.

What do we mean by coherency? The basic idea is that nobody should be able to make a Dutch book against you: it should not be possible to get you to agree to a bet where you lose no matter what. This seems like a very simple principle, but lies at the foundations of modern statistics.

This philosophy has some interesting implications: most fundamental of all, my opinions may not be the same as your opinions. If we are sitting in a room together, you might think there is a 40% chance that it will rain tomorrow, and I might think there is a 20% chance. There are many reasons we might disagree: for example, you might have a bad knee that hurts when it will rain, and your knee is hurting. This may also have an impact on our reaction to new information. If we turn on the weather report, and the weatherman says there will be brief showers, you may say there is a 90% chance it will rain, but I may only say there is a 30% chance, because I distrust weathermen.

This might sound like there is no way Bayesians can ever agree on anything, and we might as well all go home now. However, the principle of coherency has an interesting implication: if we can manage to agree on our current opinions, and we observe the same piece of new information, our updated opinions should agree.

How does this compare to the frequentist philosophy?
The frequentist philosophy says that the likelihood of something should be equal to the long-run chance that it happens. For example, if I have a fair coin, and I could flip my fair coin exactly the same way a very large number of times, then I should get heads approximately half the time, and the larger the number of flips, the closer the proportion should be to a half. The study of probability and statistics started in gambling, in which games are (theoretically) played over and over under very similar conditions, and the idea of fairness is paramount: this is one of the motivations for the frequentist philosophy. Under the frequentist philosophy, all events have a true and fixed probability of occurring.

This is fundamentally different from the Bayesian viewpoint, in which a probability is viewed as a subjective phenomenon, according to the information available at the time.

Terminology
Bayesian statistics is full of strange terms that may not make sense at first. Bayesians talk a lot about priors, posteriors, and likelihoods. These are all very simple terms that we can define in terms of the above example.

 * A prior is the opinion you hold before you observe a new piece of information. In the above example, this would be our opinions of how likely it is to rain.  In addition, we should figure out how strongly we hold that opinion: this is often called the prior information.
 * A likelihood refers to a mathematical statement about the probability of something happening. This is a complex and delicate thing, and is one of the reasons Bayesian statistics is usually studied after frequentist statistics, where likelihoods are more natural creatures to study.
 * A posterior is the updated opinion you hold after you get a new piece of information. Because there is additional information coming in (in our example, the weatherman's report) we should have more information than we started with.  This is a natural concept.

Likelihoods
The complex part of the above has to do with the likelihood. What, you might ask, is a likelihood? We can think of a likelihood as being like a statement about long-run odds, but what it really is is a statement about the mathematical properties of any information we get about what we're interested in.

In our example, the thing in question---the piece of information we've gotten---is the weatherman's report. In reality, a weatherman might say any number of things, but suppose we simplify it down to predicting rain, or predicting sunshine. This makes it a yes-no question. Yes-no questions are often dealt with by using the Bernoulli likelihood, in which something either ‘succeeds‘ or ‘fails‘. We could agree, for example, that we will call it a ‘success‘ if the weatherman predicts rain, and a ‘failure‘ if he doesn't (it doesn't matter which one we pick to be a success: the maths comes out the same).

Most likelihoods that people actually work with are extremely simple versions of the world. In reality, weathermen usually tell us a lot more information, like showing us forecast maps, giving us an idea of the probability that it will rain, and telling us something about the current conditions that lead them to believe what they're saying. But here, we've simplified it down to a yes or no question: Does the weatherman predict rain, or sun?

One term that comes up a lot in dealing with both priors and likelihoods is parameters. A parameter is any unknown quantity in a mathematical model of something. Here we have one obvious unknown quantity: the probability that it will rain.

Priors
Priors are also mathematical models. Rather than being mathematical models of something we observe, however, they are mathematical models about something we think.

While priors are more intuitive to think about having, they are somewhat less intuitive to come up with, most of the time. The difference is that while a likelihood asks us to come up with a mathematical model for whatever it is we're talking about (in this case, whether or not it will be raining), priors ask us to come up with a mathematical model for the probability of whatever parameters we used in the likelihood.

When stated that way, this seems simple: you have a model for what is going to happen, and now you have to talk about the probability of things that drive that model. In practicality, however, this gets complicated fast. What is the probability that the probability of rain tomorrow is 30%? The process of figuring out what your prior actually is is called eliciting a prior, and is an area of research in and of itself.

Conjugate priors
Eliciting a prior can be simplified sometimes because many likelihoods have a mathematical form that suggests a mathematical form for the prior. For example, if we use a Bernoulli likelihood to talk about whether or not it will be raining, there is a natural choice of prior: the Beta distribution. This is because when we go to update our prior to produce the posterior, we get another Beta distribution back out.

When this happens, we call the prior a conjugate prior for that likelihood. Conjugate priors are often used because they are easy to deal with, and make it easy to add more information of the same type later on. In our example, if we saw a second weather report, we would have yet another updated Beta distribution.

What is probability?
To a frequentist, probability is the long run chances of something happening. For example, if I flip a coin twice, I might get two heads, or two tails, or one head and one tail. If I flip a coin twice another time, I might get something different. If I repeat this experiment many many times, an infinite number of times, I should get two heads 25% of the time. This is the "true chance" of getting two heads on a flip of two coins.

To a Bayesian, however, probability is subjective. We say that your personal probability for something happening is equal to the point at which you'd take a bet either way. To understand this a little better, let's take a moment to talk about odds.

Odds versus probability versus chance
Lots of books and lots of people use "chance" to mean something like "things that happen at random" or "the chances of" to mean something like "probability". I did it myself two paragraphs above. People do this because "chance" is not a very scary word, unlike "probability" or even worse "stochastic". However, it is not very specific. From here on out, we will try to avoid this word when we have a better one to use. This means we need to define a couple of words that are more specific.

In a betting context, odds are probably the easiest thing to understand---but the hardest to understand outside that context! Imagine that you have a very large urn filled with some colored marbles (and pretend all the marbles are clearly colored: there's no multicolored or in-between shades). Maybe they are red or blue marbles. The odds of drawing a red marble are simple: the number of red marbles to the number of not-red (blue) marbles. If we call the number of red marbles R and the number of blue marbles B, then the odds in favor of red are R to B, or R:B. The odds against red are B:R, read as B to R. We can see that odds are symmetric in this way: the odds against red are the same as the odds for blue, if one of the two has to happen.

Probability is related to the odds. The probability of drawing a red marble is just the number of red marbles over the total number of marbles. If we think about it as a frequentist, this makes sense: if I draw a marble, then put it back and mix up the marbles really well, then repeat this infinitely often, in the long run, I should end up with a red marble R times out of R + B. So the probability of red is R / (R+B).

Fair bets and personal probability
Now imagine that you are going to make a bet based on this urn full of marbles. You know how many red marbles and how many blue marbles are in the bin. You have to pay $1 to play the game. How much money has to be offered before this becomes a fair bet?

Imagine there are no red marbles at all. If you bet on red, you will lose. If you bet on blue, you will win. There is no question about it at all. How much money should you be offered in order to take the bet? Logically, no payoff in the world would be big enough to get you to take a bet on red. But conversely, they should only have to offer you your $1 back in order to get you to bet on blue. It is a sure thing: you don't need any extra money for it to be fair, because you know you'll get your $1 back.

Now suppose there are two marbles, one red and one blue. If you bet on red, one-half of the time you will get the red payoff. If you bet on blue, one-half of the time you will get the blue payoff. How much money should they offer you for this to be a fair game? The logical answer is $2 in either case: if you made this bet a very large number of times, you would, in the long run, break even.

title: Statistics/Bayesian Statistics


text:
Talk:Statistics/Bayesian Statistics

I am a PhD student in statistics and noticed there was no Bayesian course. I just joined this website today and I have no idea what I'm doing, but hopefully this is sort of helpful to someone.

I tried to write at a level that would be accessible to introductory students.

What I've written is just a start, but RL calls. Kalany 05:02, 6 April 2011 (UTC)

title: Talk:Statistics/Bayesian Statistics


text:
User:Kalany

Hello. I am a PhD student in statistics.

title: User:Kalany


text:
User talk:Kalany

I'm new, and I have no idea what I'm doing. I did make one page already! Introduction to Bayesian Statistics and I hope I did it right.

 * Welcome aboard, Kalany. The best way to learn around here is to do exactly what you are doing - have a go and be open to suggestions . Some officialish welcome below. Let me know if I can help out. Sincerely, James. -- Jtneill - Talk - c 10:01, 6 April 2011 (UTC)

 Hello Kalany, 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. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

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.
 * Explore Wikiversity with the links to your left.

 * 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.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! Jtneill - Talk - c 10:01, 6 April 2011 (UTC)

 Hi! My name is WikiversityJack. I am a pumpkin and a Wikiversity mascot. My aunt told me that fun + learning = Wikiversity. Let me know if I can help you out.

title: User talk:Kalany


title: File:Preprocess1.png


text: File:Eml s11 team2 694 1.jpg

title: File:Eml s11 team2 694 1.jpg


text: File:Eml s11 team2 6916 1.jpg

title: File:Eml s11 team2 6916 1.jpg


text: File:Eml s11 team2 6964 1.jpg

title: File:Eml s11 team2 6964 1.jpg


text: File:Eml s11 team2 694 2.jpg

title: File:Eml s11 team2 694 2.jpg


text: File:Eml s11 team2 6916 2.jpg

title: File:Eml s11 team2 6916 2.jpg


text: File:Eml s11 team2 6964 2.jpg

title: File:Eml s11 team2 6964 2.jpg


text: File:Eml s11 team2 694 3.jpg

title: File:Eml s11 team2 694 3.jpg


text: File:Eml s11 team2 6916 3.jpg

title: File:Eml s11 team2 6916 3.jpg


text: File:Eml s11 team2 6964 3.jpg

title: File:Eml s11 team2 6964 3.jpg


title: File:Preprocess2.png


title: File:Preprocess3.png


title: File:Preprocess4.png


title: File:Preprocess5.png


title: File:Preprocess6.png


title: File:Preprocess7.png


title: File:Preprocess8.png


title: File:Preprocess10.png


title: File:Preprocess11.png


title: File:Preprocess12.png


text: File:Solver.png

title: File:Solver.png


text: File:Graph 0.png

title: File:Graph 0.png


title: File:PostProcess1.png


title: File:PostProcess2.png


title: File:PostProcess3.png


title: File:PostProcess4.png


title: File:PostProcess5.png


title: File:PostProcess6.png


title: File:PostProcess7.png


title: File:PostProcess8.png


title: File:PostProcess9.png


text: File:Cantibeamani.gif

title: File:Cantibeamani.gif


text: File:Eml5526.s11.team2.brenner.hw6.1d.png


title: File:Eml.5526.s11.tm7.6.10.Slide1.jpg