text:
User talk:STANELY

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

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title: User talk:STANELY


text:
User:Egm6341.s10.Team4.roni/HW6

HW6.4

HW6.10

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

title: User:Egm6341.s10.Team4.roni/HW6


text:
User:Egm6341.s10.Team4.roni/HW6/6.4

= Problem 4: Matrix Inversion for Hermite - Simpson Algorithm =

Given
Matrix A [[media:Egm6341.s10.mtg35.djvu|35-3]], which is:

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

$$
 * $$\displaystyle
 * $$\displaystyle

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 1 & 1 & 1 & 1\\ 0 & 1 & 2 & 3 \end{bmatrix}
 * $$\displaystyle A= \begin{bmatrix}

$$
 * style= |
 * }
 * }

Show
That the Inversion of this Matrix A Gives the following Matrix: on slide [[media:Egm6341.s10.mtg35.djvu|35-4]]

$$\displaystyle $$

$$\displaystyle Inv( A ) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ -3 & -2 & 3 & -1\\ 2 & 1 & -2 & 1 \end{bmatrix}

$$

Solution
Using MatLab:

 Matlab Code:

Giving Result,

A_Inv =

1    0     0     0     0     1     0     0    -3    -2     3    -1     2     1    -2     1

Author
Solved and typed by - --Egm6341.s10.Team4.roni 01:15, 7 April 2010 (UTC) Reviewed by -

title: User:Egm6341.s10.Team4.roni/HW6/6.4


text:
User:Kirgintseva

Hello and welcome to my user page!

title: User:Kirgintseva


text:
User talk:Kirgintseva

Welcome
Hello Kirgintseva, 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 with " "; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Hillgentleman | //\\ |Talk 07:51, 6 April 2010 (UTC)

title: User talk:Kirgintseva


text:
User talk:Dxgriffiths

Welcome
Hello Dxgriffiths, 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 with " "; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Hillgentleman | //\\ |Talk 07:54, 6 April 2010 (UTC)

page move

 * To see what is in your userspace click here

Here are two wikitext scripts that you can use to keep track of subpages in your userspace.

creates a page called My subpages

creates a list of all subpages in your userspace.

Keep in mind that these two scripts here on this talk page, and on your user page User:. They won't work on most Wikiversity pages.--Guy vandegrift (discuss • contribs) 05:09, 2 January 2024 (UTC)

title: User talk:Dxgriffiths


text: Category:Woodworking

title: Category:Woodworking


text:
Learning tools

This page was created, but so far, little content has been added. Everyone is invited to help expand and create educational content for Wikiversity. If you need help learning how to add content, see the editing tutorial and the MediaWiki syntax reference.

To help you get started with content, we have automatically added references below to other Wikimedia Foundation projects. This will help you find materials such as information, media and quotations on which to base the development of "Reference Materials" as an educational resource. However, please do not simply copy-and-paste large chunks from other projects. You can also use the links in the blue box to help you classify this page by subject, educational level and resource type.

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Some essential reading materials and references:

Art of Digital Video, Third Edition by John Watkinson Art of Digital Audio, Third Edition by John Watkinson Television Fundamentals by John Watkinson Convergence in Broadcast and Communications Media by John Watkinson The MPEG Handbook, Second Edition by John Watkinson

title: Learning tools


text:
User:Dxgriffiths/Clojure

Welcome to the Wikiversity project for Clojure, part of the Division of Computer Programming. The aim of this course is to get you fluent in Clojure programming by completing modular coding drills. Each lesson has a different technique with some explanation about what you're doing, followed by the drills.

A major goal of this course to have a smooth learning curve, so we take care not to mix advanced terms and topics with basic or intermediate ones. If we need to refer to concepts from other languages to explain why something happens in Clojure, you'll get plenty of notice and the chance to read a nutshell version.

As much as possible, this is a jargon-free course. If you come across a word or phrase you don't know, or see a word used in a weird way, please look it up on Wikipedia to see what you're missing. Attempting to push forward when something is unclear assumes that the unclear part is unimportant, and that's almost never the case.

Note to contributors: This course has been started with the intent to satisfy Open Study Tech requirements, which means that contributors must:

 * 1) do everything reasonable to ensure that text can be fully understood, including replacing jargon terms with plain English.
 * 2) encourage readers to type out example code themselves, so they can verify that it works and gain first-hand experience.
 * 3) ensure that all topics are presented in a logically-determined order so readers can progress comfortably to the next level.

Prerequisites
Clojure is a unique programming system designed to run on the JVM platform (Java Virtual Machine). Clojure mixes elements from several language families (notably ALGOL and LISP), and can be best described as an object-oriented / list processing hybrid. The Clojure language itself is not intended to be used in an object-oriented way, but Clojure is nonetheless written in Java and that underlying foundation does poke through from time to time. Because of this, Clojure is best not considered as a LISP dialect so much as a Java-based LISP simulator.

Clojure programming requires getting comfortable with a challenging paradox. Its goal is to simulate an interactive, dynamic programming environment via the use of a virtual machine. The nature of virtual machines is that they adopt a specific personality for as long as they run, and that personality is set by a pre-compiled, fixed main program. Virtual machines are typically special-purpose and limited to a particular task, done in a particular way. This mirrors the days of computing before operating systems were invented, when engineers literally had to rewire a machine between restarts to make it do different things. Clojure is therefore a bit like a leap into the operating system era; it allows a single VM to change its behavior and abilities even after it has been started.

In a way, Clojure can be seen as a souped-up calculator that has a command line instead of buttons, and can run code in batches if needed. This is a drastic oversimplification of course but it's an approachable metaphor to begin with.

Properly understanding Clojure requires some knowledge of several topics including: operating systems, virtual machines, Java, LISP, lambda notation, object-oriented programming, functional programming, and the history of programming language families. Clojure is exciting because it allows for fun, instant-feedback noodling as well as the development of powerful, professionally-deployed software. It has the benefit of running on a popular platform, the JVM, so it can be used in business or consumer settings without special requirements.

However, because of Clojure's mixed heritage it's only fair to say that it does require more technical knowledge than is usually necessary.

Wikibooks textbook and Wikipedia articles
Wikiversity participants should use and help develop the the book Programming Clojure at Wikibooks and make use of Wikipedia.

Wikiversity learning resources should not take the form of a general encyclopedia article, textbook or manual: Wikiversity is for other types of learning resources. Wikiversity has a "learn by doing" approach to e-learning, so learning projects are particularly useful. Design activities and projects for Wikiversity participants and attempt to establish a collaborative learning environment.

title: User:Dxgriffiths/Clojure


text:
Physics/Essays/Anonymous/Solar planets as gravitational resonators

Solar planets as gravitational resonators form some law like the s.c. Titzius-Bode law for the planets mass-radius characteristics.

History
The theory of the quantum gravitational resonator (QGR) is based on the Maxwell-like gravitational equations and similar in many relation to the theory of quantum electromagnetic resonator (QER), therefore the QGR history is close connected with the QER history.

Gravitational resonators
Due to McDonald first who used Maxwell equations to describe gravity was Oliver Heaviside The point is that in the weak gravitational field the standard theory of gravity could be written in the form of Maxwell equations

In the 90-ties Kraus first introduced the gravitational characteristic impedance of free space, which was detaled later by Kiefer, and now Raymond Y. Chiao    who is developing the ways of experimental determination of the gravitational waves.

Velocity circulation quantum
First the VCQ was proposed in the early 50-th for the quantum superfluids in the general form by R.Feynman, :
 * $$\oint_{L} \mathbf{v}\cdot\,d\mathbf{l} = n\frac{h}{m}, \ $$

where $$n $$ could be integer or fractional in the general case.

Further developments this approach was made by Yakymakha (1994) for inversion layers in MOSFETs.

General resonator characteristics
Geometrical properties of a planet define the following resonance frequency:
 * $$\omega_{pl} = \frac{c}{R_{pl}}, \ $$

where $$c $$ is velocity of light, and $$R_{pl} $$ is the planet radius. This frequency could be connected with the "minimal mass conseption":
 * $$m_{pl} = \frac{\hbar \omega_{pl}}{c^2} = \frac{\hbar}{cR_{pl}}, $$

where $$\hbar $$ is the reduced Planck constant. Considering that total planet mass $$M_{pl} $$ is replaced on the resonator surface:
 * $$M_{pl} \rightarrow S_{pl} = 4\pi R_{pl}^2, \ $$

and therefore the "minimal mass" should be placed on the minimal surface:
 * $$m_{pl} \rightarrow s_{pl} = 4\pi r_{pl}^2. \ $$

Thus, the minimal radius will be:
 * $$r_{pl} = \sqrt{\frac{m_{pl}}{M_{pl}}}. \ $$

Velocity circulation quantum approach
In the general case the velocity circulation quantum is defined as:
 * $$\omega \cdot S = n_x \frac{h}{m}, \ $$

where $$n_x $$ is integer number. This equation could be rewritten in the "mass form":
 * $$m\cdot S = \frac{2\pi n_x}{c^2}\frac{\hbar^2}{m}. \ $$

For $$n_x = 2 $$ this equation defines the minimal mass as:
 * $$m = \frac{\hbar}{cR}. \ $$

Note that this definition is compatible with the gravitational resonato approach presented in the above section.

Solar system gravitational characteristics
The full sets of the planetary data are presented in the Table 1.

Note that all planetary data were taken from the textbook .


text:
User:Moonriddengirl

My primary user page is located at Wikipedia:User:Moonriddengirl. This is a new environment for me; please forgive any errors caused by my inexperience. I certainly welcome constructive feedback, since it seems that things are done quite differently here!

I am e-mail enabled and have requested notification if my talk page is changed, but we all know that e-mail isn't infallible. As I may not log in here often or frequently check my discussion page, if you need me quickly, my wikipedia talk page is probably faster.

title: User:Moonriddengirl


text:
User talk:Moonriddengirl

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 13:22, 6 April 2010 (UTC)

 * Thanks! I'm sure the links will be very useful; as I said on my userpage, this environment is quite alien. Congratulations, by the way, on your recent DYK. Lovely article. --Moonriddengirl 13:38, 6 April 2010 (UTC)
 * Thanks. If there is anything you need help with, the Colloquium is the fastest way to get responses. There is also an IRC room that can be entered by clicking here (it is java). There is always at least two editors around at any given time. Ottava Rima (talk) 13:51, 6 April 2010 (UTC)

title: User talk:Moonriddengirl


text: User:Stephan Schulz

title: User:Stephan Schulz


title: User:Flyguy649


text:
User:Screamaniac

All credit goes to http://ucspace.canberra.edu.au/display/7125/Lecture+Introduction

Introduction to Social Psychology

Defining Social psychology
Social psychology is defined as a joint function of personal and situational influences. (Baumeister & Bushman, 2008)

''' There are 3 main facets of Socila psychology; the ABCs. '''
 * 1) Affect(feelings)
 * 2) Behaviour(action)
 * 3) Cognition(thoughts)

Social psychology can be between 2 individuals, an individual and a group and among groups.

''' How does Social psychology differ from Sociology? '''

Simply, Sociology focuses on groups while Social psychology focuses both on groups and individuals

There are 3 aspects of Social psychology. Applications of Social Psychology Social psychology can be applied in Business when sales assistants make deals with their customers, health when peers influence us to drink or smoke, Education when we learn more about ourselves and others, Law and environment when we engage in prosocial behaviour.
 * 1) Social perception-How we interpret social objects
 * 2) Social influence-How others influence our attitudes & behaviours
 * 3) Social interaction-How we interact with others in the social world

History of Social psychology

 * volkerpsychologie(folk psych)
 * crwod psychology (group mind)
 * influences of WWII
 * Gordon Allport
 * Gestalt theory
 * Post WWII-explaining atrocities with experiments like Milgram (obedience) and Zimbardo (roles)

Research

 * state problem
 * create hypothesis
 * design study/experiment
 * conduct study, collect data & results
 * data analysis
 * conclusion and discussion

Research methods
 * 1) experimental (DV, IV) +control, low flexibility, low ecological validiy vs non experimental (more naturalistic, ethical, higher construct validity, low control)

 * 1) quantitative vs qualitative

Research ethics
 * informed consent
 * harm, discomfort, risk, benefits
 * deception
 * confidentiality
 * debrief

Culture and nature
 * psyche-mind-
 * evolution-natural selection: survival, mutation, reproduction

title: User:Screamaniac


text:
User talk:Screamaniac

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 15:32, 6 April 2010 (UTC)

title: User talk:Screamaniac


text:
User talk:Flyguy649

 Hello Flyguy649, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 15:33, 6 April 2010 (UTC)

title: User talk:Flyguy649


text:
User talk:Stephan Schulz

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 15:33, 6 April 2010 (UTC)

title: User talk:Stephan Schulz


text:
User:Egm6341.s10.team3.ynahn81/HW6-1

= (1) Computation of arc length an ellipse $$\displaystyle \overset{\frown}{PQ} $$ using successive numerical integration = Ref: Lecture notes [[media:Egm6341.s10.mtg32.djvu|p.32-1]]

Problem Statement
Compute arc length of ellipse $$\displaystyle \overset{\frown}{PQ} $$ to $$\displaystyle O(10^{-10}) $$ from $$\displaystyle \theta_{P} = 0 $$ to $$\displaystyle \theta_{Q} = \pi/3 $$. Successive numerical integration results should be used as stopping criterion, i.e. $$\displaystyle |I_{2n} - I_{n}| < O(10^{-10}) $$.

i) Use error estimate for composite trapezoidal rule
 * {| style="width:100%" border="0" align="left"

$$ $$ to find $$\displaystyle n $$.
 * $$\displaystyle
 * E_{n}| \leq \frac{(b-a)h^2}{12}M_{2}
 * E_{n}| \leq \frac{(b-a)h^2}{12}M_{2}
 * $$\displaystyle (Eq. 1)
 * }
 * }

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

\overset{\frown}{PQ} = \int_{\theta_{P}}^{\theta_{Q}} dl $$ $$ where $$\displaystyle dl = d\theta\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}} $$ and $$\displaystyle r(\theta) = \frac{(1-e^2)}{1-e\cos\theta} $$ to obtain the arc length from composite trapezoidal rule, Romberg table, clencurt in Matlab and sum in Chebfun.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

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

\overset{\frown}{PQ} = \int_{\theta_{P}}^{\theta_{Q}} \left[1-e^2\sin^2\theta\right]^{\frac{1}{2}}d\theta. $$ $$ to obtain the arc length from composite trapezoidal rule, Romberg table, clencurt in Matlab and sum in Chebfun.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Solution
i) Find $$\displaystyle n $$ In (Eq. 1), recall that $$\displaystyle h = \frac{b-a}{n} $$ and $$\displaystyle M_{2} = \max_{\zeta \in [a,b]}|f^{(2)}(\zeta)| $$. Here, $$\displaystyle a = 0 $$ and $$\displaystyle b = \pi/3 $$.

Also, rearrange (Eq. 1) to get the following equation.
 * {| style="width:100%" border="0" align="left"

n^2 \leq \frac{(b-a)^3}{12}\frac{\max_{\zeta \in [a,b]}|f^{(2)}(\zeta)|}{|E_{n}|}. $$ $$   Now, put the below values into (Eq. 4).  $$\displaystyle a = 0 $$  <br\> $$\displaystyle b = \pi/3 $$ <br\> <br\> $$\displaystyle |E_{n}| = 10^{-11} (\because |I_{2n} - I_{n}| < O(10^{-11}) $$ <br\> <br\> $$\displaystyle M_{2} = \max_{\zeta \in [a,b]}|f^{(2)}(\zeta)| = \max_{\theta \in [0,\frac{\pi}{3}]}|dl^{(2)}(\theta)| = 0.2934 $$ for (Eq. 2) and <br\> <br\> $$\displaystyle M_{2} = \max_{\zeta \in [a,b]}|f^{(2)}(\zeta)| = \max_{\theta \in [0,\frac{\pi}{3}]}|\left(\sqrt{1-e^2\sin^2(\theta)}\right)^{(2)}| = 0.0670 $$ for (Eq. 3) <br\> <br\> <br\>
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
 * }
 * }

Then, $$\displaystyle n^2 \leq 2.8077e+09 \quad \Rightarrow \quad n \leq 52988.59 $$ for (Eq. 2) and<br\> <br\> $$\displaystyle n^2 \leq 6.4105e+08 \quad \Rightarrow \quad n \leq 25319.00 $$ for (Eq. 3).<br\> <br\> <br\>

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

$$\displaystyle n \leq 53988 $$ for (Eq. 2) and
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle n \leq 25319 $$ for (Eq. 3). <br\> <br\> <br\>
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

ii) Based on (Eq. 2) In what follows, we are using the follow matlab function as a given function $$\displaystyle dl(\theta) $$.

<br\> <br\>

 (a) Composite trapezoidal rule  <br\> Recall the formula of composite trapezoidal rule.

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

T_{0}(n) = h\left[\frac{1}{2}f(x_{0}) + f(x_{1}) + f(x_{2}) + \cdot\cdot\cdot + f(x_{n-1}) + \frac{1}{2}f(x_{n}) \right] $$ where $$\displaystyle h = \frac{b-a}{n} $$ and $$\displaystyle x_{0} = a,\quad x_{n} = b $$.
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Also, realize the fact that if we want to double the number of sub-intervals, we don't have to calculate the functions at even point, i.e. $$\displaystyle f(x_{0}), f(x_{2}), \cdot\cdot\cdot, f(x_{n-2}), f(x_{n}) $$ because we already calculated the functions at these even point at the previous step. Therefore, all we have to do is evaluating the functions only at the odd points and combine them with the calculated values at the previous step. As a result, we can use the following inductive formula.
 * {| style="width:100%" border="0" align="left"

T_{0}(2n) = T_{0}(n) + h\sum_{i=1}^{n}f(x_{2i-1}). $$ <br\> <br\> <br\> <br\>
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

<br\> The obtained number of nodes to achieve $$\displaystyle O(10^{-10}) $$ is $$\displaystyle n = 32768 $$ which satisfies the result of part i) $$\displaystyle n \leq 53988 $$. <br\> The integration result, final error and computational time are summarized in Table at the last part of this problem. <br\> <br\>

 (b) Romberg Table  <br\>

Since the first column of Romberg table corresponds to Trap. rule, we can use the below formulas in order to obtain the first column of Romberg talbe.

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

T_{0}(n) = h\left[\frac{1}{2}f(x_{0}) + f(x_{1}) + f(x_{2}) + \cdot\cdot\cdot + f(x_{n-1}) + \frac{1}{2}f(x_{n}) \right] $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

T_{0}(2n) = T_{0}(n) + h\sum_{i=1}^{n}f(x_{2i-1}). $$ where $$\displaystyle h = \frac{b-a}{n} $$ and $$\displaystyle x_{0} = a,\quad x_{n} = b $$. <br\>
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Also, we gradually set up the Romberg table in parallel, as soon as the result for $$\displaystyle T_0(2^j) $$ becomes available using the below formula.

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

T_{k}(2n) = \frac{2^{2k}T_{k-1}(2n) - T_{k-1}(n)}{2^{2k}-1}. $$ <br\> <br\> <br\>
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Result

<br\>

The integration result, final error and computational time are summarized in Table at the last part of this problem.

<br\> <br\>

 (c) clencurt  <br\>

To achieve the desired tolerance of error, we use the following Matlab code. <br\>

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem. <br\> <br\>

 (d) chebfun sum command  <br\> The 'sum' command in chebfun performs integration of a given function up to the machine accuracy. Therefore, we can achieve the tolerance of error of numerical integration for the given function $$\displaystyle dl $$ up to the machine accuracy by using the following Matlab code.

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem. <br\> <br\>

 (e) Summary  <br\>

In the previous HW, recall that composite trapezoidal rule was a very efficient method to perform integration due to the periodicity of the given function. In this HW, however, note that composite trapezoidal rule requires a very large number of nodes and long computation time. The reason is that the given function does not show a periodicity within a given range $$\displaystyle \theta \in [0, \pi/3] $$  even though it shows a periodicity for the range of $$\displaystyle \theta \in [0, 2\pi] $$.

As shown above, both Romberg table and clencurt require only 32 nodes to achieve the desired tolerance of error and the computation times of the both methods are very short. Therefore, both Romberg Table and clencurt are efficient methods to integrate the given function. <br\> <br\> <br\>

iii) Based on (Eq. 3) In what follows, we are using the follow matlab function as a given function $$\displaystyle dl(\theta) $$.

<br\> <br\>

 (a) Composite trapezoidal rule  <br\> We use the same Matlab code with that of part ii).

The obtained number of nodes to achieve $$\displaystyle O(10^{-10}) $$ is $$\displaystyle n = 16384 $$ which satisfies the result of part i) $$\displaystyle n \leq 25319 $$. <br\> The integration result, final error and computational time are summarized in Table at the last part of this problem. <br\> <br\>

 (b) Romberg Table  <br\> We use the same Matlab code with that of part ii).

<br\>

The integration result, final error and computational time are summarized in Table at the last part of this problem.

<br\> <br\>

 (c) clencurt  <br\> We use the same Matlab code with that of part ii).

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem. <br\> <br\>

 (d) chebfun sum command  <br\> The 'sum' command in chebfun performs integration of a given function up to the machine accuracy. Therefore, we can achieve the tolerance of error of numerical integration for the given function $$\displaystyle sqrt(1 - e^2\sin^2\theta) $$ up to the machine accuracy by using the following Matlab code.

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem. <br\> <br\>

 (e) Summary  <br\>

As we already explained in summary of part i), composite trapezoidal rule is not a efficient method to integrate the given function due to the same reason (i.e. The given function does not show a periodicity within a given range $$\displaystyle \theta \in [0, \pi/3] $$). Romberg table and clencurt require only 16 nodes to achieve the desired tolerance of error and the computation times of the both methods are very short. Therefore, both Romberg Table and clencurt are efficient methods to integrate the given function.

In addition, note that the integration result of part iii) is different from the integration result of part ii) although the given ranges of the integrations are with each other (i.e. $$\displaystyle \theta \in [0, \pi/3] $$). As the figures show in problem statement, the definition of $$\displaystyle \theta $$ of part ii) is different from that of part iii). Therefore, it is very clear that the integration results of the both parts are not same with each other even though the range of $$\displaystyle \theta $$ is same.

title: User:Egm6341.s10.team3.ynahn81/HW6-1


text:
User:Egm6341.s10.team3.sa/HW6

= Computation of arc length an ellipse $$\displaystyle \overset{\frown}{PQ} $$ using successive numerical integration = Ref: Lecture notes [[media:Egm6341.s10.mtg32.djvu|p.32-1]]

Problem Statement
Compute arc length of ellipse $$\displaystyle \overset{\frown}{PQ} $$ to $$\displaystyle O(10^{-10}) $$ from $$\displaystyle \theta_{P} = 0 $$ to $$\displaystyle \theta_{Q} = \pi/3 $$. Successive numerical integration results should be used as stopping criterion, i.e. $$\displaystyle |I_{2n} - I_{n}| < O(10^{-10}) $$.

i) Use error estimate for composite trapezoidal rule
 * {| style="width:100%" border="0" align="left"

$$ $$ to find $$\displaystyle n $$.
 * $$\displaystyle
 * E_{n}| \leq \frac{(b-a)h^2}{12}M_{2}
 * E_{n}| \leq \frac{(b-a)h^2}{12}M_{2}
 * $$\displaystyle (Eq. 1)
 * }
 * }

ii)

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

\overset{\frown}{PQ} = \int_{\theta_{P}}^{\theta_{Q}} dl $$ $$ where $$\displaystyle dl = d\theta\left[r^2 + \left(\frac{dr}{d\theta}\right)^2\right]^{\frac{1}{2}} $$ and $$\displaystyle r(\theta) = \frac{(1-e^2)}{1-e\cos\theta} $$ to obtain the arc length from composite trapezoidal rule, Romberg table, clencurt in Matlab and sum in Chebfun.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

iii)

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

\overset{\frown}{PQ} = \int_{\theta_{P}}^{\theta_{Q}} \left[1-e^2\sin^2\theta\right]^{\frac{1}{2}}d\theta. $$ $$ to obtain the arc length from composite trapezoidal rule, Romberg table, clencurt in Matlab and sum in Chebfun.
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Solution
i) Find $$\displaystyle n $$ In (Eq. 1), recall that $$\displaystyle h = \frac{b-a}{n} $$ and $$\displaystyle M_{2} = \max_{\zeta \in [a,b]}|f^{(2)}(\zeta)| $$. Here, $$\displaystyle a = 0 $$ and $$\displaystyle b = \pi/3 $$.

Also, rearrange (Eq. 1) to get the following equation.
 * {| style="width:100%" border="0" align="left"

n^2 \leq \frac{(b-a)^3}{12}\frac{\max_{\zeta \in [a,b]}|f^{(2)}(\zeta)|}{|E_{n}|}. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Now, put the below values into (Eq. 4). $$\displaystyle a = 0 $$

$$\displaystyle b = \pi/3 $$

$$\displaystyle |E_{n}| = 10^{-11} (\because |I_{2n} - I_{n}| < O(10^{-10}) $$

$$\displaystyle M_{2} = \max_{\zeta \in [a,b]}|f^{(2)}(\zeta)| = \max_{\theta \in [0,\frac{\pi}{3}]}|dl^{(2)}(\theta)| = 0.2934 $$ for (Eq. 2) and

$$\displaystyle M_{2} = \max_{\zeta \in [a,b]}|f^{(2)}(\zeta)| = \max_{\theta \in [0,\frac{\pi}{3}]}|\left(\sqrt{1-e^2\sin^2(\theta)}\right)^{(2)}| = 0.0670 $$ for (Eq. 3)

Then, $$\displaystyle n^2 \leq 2.8077e+09 \quad \Rightarrow \quad n \leq 52988.59 $$ for (Eq. 2) and

$$\displaystyle n^2 \leq 6.4105e+08 \quad \Rightarrow \quad n \leq 25319.00 $$ for (Eq. 3).

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

$$\displaystyle n \leq 53988 $$ for (Eq. 2) and
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle n \leq 25319 $$ for (Eq. 3).
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

ii) Based on (Eq. 2) In what follows, we are using the follow matlab function as a given function $$\displaystyle dl(\theta) $$.

 (a) Composite trapezoidal rule

Recall the formula of composite trapezoidal rule.

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

T_{0}(n) = h\left[\frac{1}{2}f(x_{0}) + f(x_{1}) + f(x_{2}) + \cdot\cdot\cdot + f(x_{n-1}) + \frac{1}{2}f(x_{n}) \right] $$ where $$\displaystyle h = \frac{b-a}{n} $$ and $$\displaystyle x_{0} = a,\quad x_{n} = b $$.
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Also, realize the fact that if we want to double the number of sub-intervals, we don't have to calculate the functions at even point, i.e. $$\displaystyle f(x_{0}), f(x_{2}), \cdot\cdot\cdot, f(x_{n-2}), f(x_{n}) $$ because we already calculated the functions at these even point at the previous step. Therefore, all we have to do is evaluating the functions only at the odd points and combine them with the calculated values at the previous step. As a result, we can use the following inductive formula.
 * {| style="width:100%" border="0" align="left"

T_{0}(2n) = T_{0}(n) + h\sum_{i=1}^{n}f(x_{2i-1}). $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

The obtained number of nodes to achieve $$\displaystyle O(10^{-10}) $$ is $$\displaystyle n = 32768 $$ which satisfies the result of part i) $$\displaystyle n \leq 53988 $$. The integration result, final error and computational time are summarized in Table at the last part of this problem.

 (b) Romberg Table

Since the first column of Romberg table corresponds to Trap. rule, we can use the below formulas in order to obtain the first column of Romberg talbe.

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

T_{0}(n) = h\left[\frac{1}{2}f(x_{0}) + f(x_{1}) + f(x_{2}) + \cdot\cdot\cdot + f(x_{n-1}) + \frac{1}{2}f(x_{n}) \right] $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

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

T_{0}(2n) = T_{0}(n) + h\sum_{i=1}^{n}f(x_{2i-1}). $$ where $$\displaystyle h = \frac{b-a}{n} $$ and $$\displaystyle x_{0} = a,\quad x_{n} = b $$.
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Also, we gradually set up the Romberg table in parallel, as soon as the result for $$\displaystyle T_0(2^j) $$ becomes available using the below formula.

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

T_{k}(2n) = \frac{2^{2k}T_{k-1}(2n) - T_{k-1}(n)}{2^{2k}-1}. $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Result

The integration result, final error and computational time are summarized in Table at the last part of this problem.

 (c) clencurt

To achieve the desired tolerance of error, we use the following Matlab code.

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem.

 (d) chebfun sum command

The 'sum' command in chebfun performs integration of a given function up to the machine accuracy. Therefore, we can achieve the tolerance of error of numerical integration for the given function $$\displaystyle dl $$ up to the machine accuracy by using the following Matlab code.

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem.

 (e) Summary

In the previous HW, recall that composite trapezoidal rule was a very efficient method to perform integration due to the periodicity of the given function. In this HW, however, note that composite trapezoidal rule requires a very large number of nodes and long computation time. The reason is that the given function does not show a periodicity within a given range $$\displaystyle \theta \in [0, \pi/3] $$  even though it shows a periodicity for the range of $$\displaystyle \theta \in [0, 2\pi] $$.

As shown above, both Romberg table and clencurt require only 32 nodes to achieve the desired tolerance of error and the computation times of the both methods are very short. Therefore, both Romberg Table and clencurt are efficient methods to integrate the given function.

iii) Based on (Eq. 3) In what follows, we are using the follow matlab function as a given function $$\displaystyle dl(\theta) $$.

 (a) Composite trapezoidal rule

We use the same Matlab code with that of part ii).

The obtained number of nodes to achieve $$\displaystyle O(10^{-10}) $$ is $$\displaystyle n = 16384 $$ which satisfies the result of part i) $$\displaystyle n \leq 25319 $$. The integration result, final error and computational time are summarized in Table at the last part of this problem.

 (b) Romberg Table

We use the same Matlab code with that of part ii).

The integration result, final error and computational time are summarized in Table at the last part of this problem.

 (c) clencurt

We use the same Matlab code with that of part ii).

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem.

 (d) chebfun sum command

The 'sum' command in chebfun performs integration of a given function up to the machine accuracy. Therefore, we can achieve the tolerance of error of numerical integration for the given function $$\displaystyle \sqrt{(1 - e^2\sin^2\theta)} $$ up to the machine accuracy by using the following Matlab code.

The integration result, final error, final number of nodes and computational time are summarized in Table at the last part of this problem.

 (e) Summary

As we already explained in summary of part i), composite trapezoidal rule is not a efficient method to integrate the given function due to the same reason (i.e. The given function does not show a periodicity within a given range $$\displaystyle \theta \in [0, \pi/3] $$). Romberg table and clencurt require only 16 nodes to achieve the desired tolerance of error and the computation times of the both methods are very short. Therefore, both Romberg Table and clencurt are efficient methods to integrate the given function.

In addition, note that the integration result of part iii) is different from the integration result of part ii) although the given ranges of the integrations are same with each other (i.e. $$\displaystyle \theta \in [0, \pi/3] $$). As the figures show in problem statement, the definition of $$\displaystyle \theta $$ of part ii) is different from that of part iii). Therefore, it is very clear that the integration results of the both parts are not same with each other even though the range of $$\displaystyle \theta $$ is same.

Yong Nam Ahn 02:11, 7 April 2010 (UTC) Abhishekksingh 09:13, 7 April 2010 (UTC)

=Show that $$ dp_y=mVdr$$ = Ref: Lecture Notes [[media:Egm6341.s10.mtg33.djvu|p.33-2]]

Problem Statement
At t+dt, consider V+dV Show that $$ dp_{\overline{y}}=mVdr$$

Solution

At t+dt, V+dV,see the above figure
 * {| style="width:100%" border="0" align="left"

$$ By neglecting higher order terms, mdVdr, We have: At t+dt, V+dV,see the above figure
 * $$\displaystyle dp_{\overline{y}}=m(V+dV)dr=mVdr+mdVdr
 * $$\displaystyle dp_{\overline{y}}=m(V+dV)dr=mVdr+mdVdr
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ End the proof
 * $$\displaystyle dp_{\overline{y}}=mVdr
 * $$\displaystyle dp_{\overline{y}}=mVdr
 * }
 * }

Min Zhong 12:00, 7 Arip 2010 (UTC)

= Evaluate the $$z_{i+1}$$ and $$z'_{i+1}$$ in terms of $$c_0,c_1,c_2,c_3$$ and represent it in the coefficient matrix = Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.35-3]]

Problem Statement
Evaluate the $$z_{i+1}$$ and z'_{i+1} in terms of $$c_0,c_1,c_2,c_3$$ and represent it in the coefficient matrix i.e, Evaluate the remaining coefficient of Matrix by using degrees of Freedom

Solution
We have

$$ Z(s)=\sum_{i=0}^{3}c_is^i=\sum_{i=0}^{3}N_i(s)d_i $$

such that

$$ d_1=Z_i$$

$$ d_2=\dot{Z_i}$$

$$ d_3=Z_{i+1}$$

$$ d_4={\dot{Z}_{i+1}}$$

where

$$ \dot{Z}=\frac{dZ}{dt}={\frac{dZ}{ds}} \frac{ds}{dt} $$

such that $$ \frac{ds}{dt} = \frac {1}{h} $$

We know the coefficient of matrix for first two rows from lecture notes   [[media:Egm6341.s10.mtg32.djvu|p.35-3]]

$$\begin{bmatrix} 1 &0 &0  &0 \\  0&1  &0  &0 \end{bmatrix}$$

Using the equations above we have

$$  d_3=Z_{i+1}=Z(s=1)=c_0+c_1+c_2+c_3 $$

$$  d_4=\dot{Z}_{i+1}=\frac{1}{h}Z'(s=1)=c_1+2c_2+3c_3 $$

Putting the results in matrix form we obtain

$$\begin{bmatrix} 1 &0 &0  &0 \\ 0 &1  &0  &0 \\ 1 &1  &1  &1 \\ 0 &1  &2  &3 \end{bmatrix}\begin{Bmatrix} c_0\\ c_1\\ c_2\\

c_3\end{Bmatrix}= \begin{Bmatrix} Z_i\\ Z'_i\\ Z_{i+1}\\ Z'_{i+1} \end{Bmatrix}$$

Abhishekksingh 09:13, 7 April 2010 (UTC)

= Verify the inverse of matrix using Matlab = Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.35-4]]

Problem Statement
Find the inverse of given Matrix

A =

1    0     0     0     0     1     0     0     1     1     1     1     0     1     2     3

Solution
which is same as the one given on [[media:Egm6341.s10.mtg32.djvu|p.35-4]]

Hence Verified

Abhishekksingh 09:13, 7 April 2010 (UTC)

= Identify basis functions and plot them=

Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.35-4]]

Problem Statement
Identify the basis functions

$$ N_i (s) $$

where $$ i = 1,2,3,4$$

Solution
We have

$$ z(s) = \sum_{i=0}^3 c^i s^i = \sum_{i=1}^4 N_i(s) d_i $$

Expanding above we obtain

$$ N_1 d_1 + N_2 d_2 + N_3 d_3 + N_4 d_4 = C_0 s^0 + C_1 s^1 + C_2 s^2 + C_3 s^3$$

$$                                      = C_0  + C_1 s^1 + C_2 s^2 + C_3 s^3$$

$$d_1 = z_i = z(s=0) = C_0$$

$$d_2 = \dot z_i = \dot z_i (s=0) = C_1$$

$$d_3 = z_{i+1}= z_{i+1} (s=1) = C_0 + C_1 + C_2  + C_3 $$

$$ d_4 = \dot z_{i+1} =C_1 + 2 C_2 + 3 C_3 $$

Inserting above values in first eq we obtain

$$(N_1 + N_3) C_0 + (N_2+N_3+N_4) C_1 + (N_3+2 N_4)C_2 + (N_3 + 3 N_4) C_3 = C_0 + C_1 s + C_2 s^2 + C_3 s^3$$

Comparing both LHS and RHS we obtain

$$N_1 + N_3 = 1$$

$$N_2+N_3+N_4 = s$$

$$N_3+2 N_4=s^2$$

$$N_3 + 3 N_4 = s^3$$

Solving above we obtain basis functions

$$N_1 = 1- 3 s^2 +2 s^3$$

$$N_2 = s^3 - 2 s^2 + s $$

$$N_3 = 3 s^2 - 2 s^3$$

Below is the plot of above basis functions

$$ N_4 = s^3 - s^2$$

Abhishekksingh 09:13, 7 April 2010 (UTC)

= Express $$ t(s) $$ in terms of $$ s(t) $$=

Ref Lecture Notes [[media:Egm6341.s10.mtg32.djvu|p.36-1]]

Problem Statement
We have to show that s is the function of t (s = s(t) )

Solution
We have (from [[media:Egm6341.s10.mtg32.djvu|p.35-1]] eq (1))

$$ t(s)= (1-s) t_i + s t_{i+1} $$

$$    = t_i + s (t_{i+1} - t_i ) $$

$$ but \frac {ds}{dt} = \frac{1}{h} $$

$$ or \frac {dt}{ds} = h $$

so $$ t_{i+1} - t_i = \frac{t_{i+1} - t_i}{1-0}= h $$

$$ t(s) = t_i +h s $$ $$ h s = t - t_i $$ $$ s = \frac {(t - t_i)} {h} $$ $$ s = s(t) $$

Hence Proved

Abhishekksingh 09:13, 7 April 2010 (UTC)

= Show that $$ Z(s= 1/2)= (1/2)(Z_i-Z_{i+1})+ (h/8)(f_i-f_{i+1}) $$  = Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]

Problem Statement
Proof that $$ Z(\varsigma=2)= \frac{1}{2}(Z_i-Z_{i+1})+ \frac{h}{8}(f_i-f_{i+1}) $$

In Lecture Notes [[media:Egm6341.s10.mtg35.djvu|p.35-2]](2), We have:
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle Z(\varsigma)= \sum_{n=0}^{3}C_i{\varsigma}^i
 * $$\displaystyle Z(\varsigma)= \sum_{n=0}^{3}C_i{\varsigma}^i
 * $$\displaystyle (Eq. 7-1)
 * }
 * }

In lecture notes [[media:Egm6341.s10.mtg35.djvu|p.35-4]](1), we have:

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

\begin{Bmatrix} C_0 \\ C_1 \\ C_2 \\ C_3 \\ \end{Bmatrix}= \begin{bmatrix} 1 & 0 & 0&0 \\ 0 & 1 & 0&0 \\  -3 & -2 & 3&-1 \\  2 & 1 & -2&1\\ \end{bmatrix} \begin{Bmatrix} Z_i \\ Z_i^\prime \\ Z_{i+1} \\ Z_{i+1}^\prime \\ \end{Bmatrix}
 * $$\displaystyle
 * $$\displaystyle

$$ $$ In lecture notes [[media:Egm6341.s10.mtg35.djvu|p.36-1]](1), we have:
 * $$\displaystyle (Eq. 7-2)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle Z^\prime= h\dot{Z}=hf
 * $$\displaystyle Z^\prime= h\dot{Z}=hf
 * $$\displaystyle (Eq. 7-3)
 * }
 * }

Solution
We start the proof from Equation 1.
 * {| style="width:100%" border="0" align="left"

\begin{align} Z(\varsigma= \frac{1}{2})& = \sum_{n=0}^{3}C_i{ \frac{1}{2}}^i\\ &=C_0+ \frac{1}{2}C_1+ \frac{1}{4}C_2+ \frac{1}{8}C_3\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7-4)
 * }
 * }

Next, we will find the expression of Ci by Zi and fi From the matrix Eq. 7-2, we have:
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle C_0=Z_i
 * $$\displaystyle C_0=Z_i
 * $$\displaystyle (Eq. 7-5)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle C_1=Z_i^\prime
 * $$\displaystyle C_1=Z_i^\prime
 * $$\displaystyle (Eq. 7-6)
 * }
 * }

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

$$ $$
 * $$\displaystyle C_2=-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime
 * $$\displaystyle C_2=-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime
 * $$\displaystyle (Eq. 7-7)
 * }
 * }

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

$$ $$
 * $$\displaystyle C_3=2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime
 * $$\displaystyle C_3=2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime
 * $$\displaystyle (Eq. 7-8)
 * }
 * }

Plug Eq.7-5,Eq.7-6,Eq.7-7,Eq.7-8, into Eq.7-4,We can get:
 * {| style="width:100%" border="0" align="left"

\begin{align} Z(\varsigma= \frac{1}{2})& =Z_i+ \frac{1}{2}Z_i^\prime + \frac{1}{4}(-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime)+ \frac{1}{8}(2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime)\\ &=\frac{1}{2}Z_i+\frac{1}{2}Z_{i+1}+\frac{1}{8}Z_i^\prime-\frac{1}{8}Z_{i+1}^\prime\\ &=\frac{1}{2}(Z_i+Z_{i+1})+\frac{1}{8}(Z_i^\prime-Z_{i+1}^\prime)\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7-9)
 * }
 * }

Plug Eq 3 into above equation, we have

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

$$\displaystyle Z(\varsigma=\frac{1}{2})=\frac{1}{2}(Z_i+Z_{i+1})+\frac{h}{8}(f_i-f_{i+1}) $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Min Zhong 12:00, 7 Apri 2010 (UTC)

= Show that $$ Z(s=1/2)'=-(3/2)(Z_i-Z_{i+1})-(1/4)(Z'_i-Z'_{i+1})$$ = Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]

Problem Statement
Proof that $$\displaystyle Z(\varsigma=\frac{1}{2})^\prime=-\frac{3}{2}(Z_i-Z_{i+1})-\frac{1}{4}(Z_i^\prime-Z_{i+1}^\prime) $$ From Lecture Notes [[media:Egm6341.s10.mtg35.djvu|p.35-1]](2)

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

$$ $$
 * $$\displaystyle Z(\varsigma)=C_0+ C_1\varsigma+ C_2{\varsigma}^2 + C_3{\varsigma}^3
 * $$\displaystyle Z(\varsigma)=C_0+ C_1\varsigma+ C_2{\varsigma}^2 + C_3{\varsigma}^3
 * $$\displaystyle (Eq. 8-1)
 * }
 * }

Solution
We start the proof from Eq.8-1,

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

$$ $$
 * $$\displaystyle Z(\varsigma)^\prime=C_1+ 2C_2\varsigma+ 3C_3{\varsigma}^2
 * $$\displaystyle Z(\varsigma)^\prime=C_1+ 2C_2\varsigma+ 3C_3{\varsigma}^2
 * $$\displaystyle (Eq. 8-2)
 * }
 * }

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

\begin{align} Z(\varsigma={\frac{1}{2}}^\prime)&=C_1+ 2C_2\frac{1}{2}+ 3C_3{\frac{1}{2}}^2\\ &=C_1+ C_2 +\frac{3}{4}C_3\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8-3)
 * }
 * }

Plug Eq.7-5,Eq.7-6,Eq.7-7,Eq.7-8, into Eq.8-3, we get:

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

\begin{align} Z(\varsigma=\frac{1}{2})^\prime & = Z_i^\prime-3Z_i-2Z_i^\prime+3Z_{i+1}-Z_{i+1}^\prime+\frac{1}{2}(2Z_i+Z_i^\prime-2Z_{i+1}+Z_{i+1}^\prime) \\ &=-\frac{3}{2}Z_i+\frac{3}{2}Z_{i+1}-\frac{1}{4}Z_i^\prime-\frac{1}{4}Z_{i+1}^\prime\\ &=-\frac{3}{2}(Z_i-Z_{i+1})-\frac{1}{4}(Z_i^\prime+Z_{i+1}^\prime)\\ \end{align}
 * $$\displaystyle
 * $$\displaystyle

$$ Min Zhong 12:00, 7 Apri 2010 (UTC)
 * }
 * }

=Show that $$ Z_{i+1}=Z_i+(h/2)/(3)[f_i+4f_{i+1/2}+f_{i+1}]$$ = Ref: Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-3]]

Problem Statement
Proof that $$ Z_{i+1}=Z_i+\frac{h/2 }{3}[f_i+4f_{i+\frac{1}{2}}+f_{i+1}]$$

From Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-3]](1)

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

$$ $$
 * $$\displaystyle \triangle=\dot Z_{i+ \frac{1}{2}}-f_{i+ \frac{1}{2}}
 * $$\displaystyle \triangle=\dot Z_{i+ \frac{1}{2}}-f_{i+ \frac{1}{2}}
 *
$$\displaystyle (Eq. 9-1)
 * }
 * }

From Lecture Notes [[media:Egm6341.s10.mtg36.djvu|p.36-2]]

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

\dot Z_{i+ \frac{1}{2}}=- \frac{3}{2h}(Z_i-Z_{i+1})- \frac{1}{4}(f_i+f_{i+1}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-2)
 * }
 * }

Solution
When $$\triangle=0$$, from Eq.9-1, we have :
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle \dot Z_{i+ \frac{1}{2}}=f_{i+ \frac{1}{2}}
 * $$\displaystyle \dot Z_{i+ \frac{1}{2}}=f_{i+ \frac{1}{2}}
 *
$$\displaystyle (Eq. 9-3)
 * }
 * }

Plug Eq. 9-3 into Eq, 9-2, we can get:
 * {| style="width:100%" border="0" align="left"

f_{i+ \frac{1}{2}}=- \frac{3}{2h}(Z_i-Z_{i+1})- \frac{1}{4}(f_i+f_{i+1}) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-4)
 * }
 * }

Rearrange Eq. 9-4 we have:

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

\begin{align} Z_{i+1}&=Z_i+ \frac{h}{6}(f_i+4f_{i+\frac{1}{2}}+f_{i+1})\\ &=Z_i+ \frac{h/2}{3}(f_i+4f_{i+\frac{1}{2}}+f_{i+1})\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 9-2)
 * }
 * }

= Kessler's code =

Problem Statement
1. Run Kessler's code to reproduce the table of constants and the values of polynomials at t=1 obtained by Kessler. 2. Explain Kessler's code by giving comments for every line. 3. Obtain $$ \displaystyle (p_2,p_3) \; (p_4,p_5) \; (p_6,p_7)$$ by understanding Kessler's code line by line.

Ref: Lecture Notes [[media:Egm6341.s10.mtg30.djvu|p.30-2]] and [[media:Egm6341.s10.mtg37.djvu|p.37-1]]

Ref: Trapezoidal rule error

Solution
Part I : Generation of the Table To generate the table given in Kessler's paper, the following modifications are to be carried out.

1. The first half of the code is the main code. Give that a name called kessler.m

2. The second half of the code is a subroutine with the name fracsum. This should not be part of the file kessler.m   We need to create a new file called fracsum.m in the same folder that contains the file kessler.m

3. Set n=8. This is to ensure that we get the constants $$ c_1 \; through \; c_{17} $$ and polynomials $$ p_2(1) \; through\; p_{16}(1) $$.

4. The most important part is that the subroutine fracsum should end with the line, dsum=dsum/div;

5. The following lines have been appended to kessler.m to generate the table.

Result

Part II: Line by Line Explanation of Kessler's MATLAB Code - Adding Comments

Part III : Generating $$\displaystyle (p_2,p_3),(p_4,p_5),(p_6,p_7)$$
 * {| style="width:100%" border="0" align="left"

 * Recall the following:
 * }
 * }

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

$$
 * $$\displaystyle p_1(t) = c_1 t $$
 *
$$\displaystyle (Eq. 1)
 *
$$\displaystyle (Eq. 1)
 * }

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

 * $$\displaystyle p_2(t) = c_1 \frac{t^2}{2!} + c_3 $$
 *
$$\displaystyle (Eq. 2) $$
 * }
 * }

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

 * $$\displaystyle p_3(t) = c_1 \frac{t^3}{3!} + c_3 t $$
 *
$$\displaystyle (Eq. 3) $$
 * }
 * }

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

 * $$\displaystyle p_4(t) = c_1 \frac{t^4}{4!} + c_3 \frac{t^2}{2!} + c_5 $$
 *
$$\displaystyle (Eq. 4) $$
 * }
 * }

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

 * $$\displaystyle p_5(t) = c_1 \frac{t^5}{5!} + c_3 \frac{t^3}{3!} + c_5 t $$
 *
$$\displaystyle (Eq. 5) $$
 * }
 * }

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

 * $$\displaystyle p_6(t) = c_1 \frac{t^6}{6!} + c_3 \frac{t^4}{4!} + c_5 \frac{t^2}{2!} + c_7 $$
 *
$$\displaystyle (Eq. 6) $$
 * }
 * }

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

 * $$\displaystyle p_7(t) = c_1 \frac{t^7}{7!} + c_3 \frac{t^5}{5!} + c_5 \frac{t^3}{3!} + c_7 t $$
 *
$$\displaystyle (Eq. 7) $$
 * }
 * }

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

$$\displaystyle c_1, c_3, c_5 \;and\; c_7 $$
 * Hence we need to find out the constants,
 * Hence we need to find out the constants,
 * }

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

And then we need to have n=3 ,because,
 * So we start off with setting $$ c_1 = -1 $$
 * So we start off with setting $$ c_1 = -1 $$
 * }

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

 * $$\displaystyle n=1 \;will\; yield\; c_3 $$
 * }
 * }

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

 * $$\displaystyle n=2 \;will\; yield\; c_5 $$
 * }
 * }

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

 * $$\displaystyle n=3 \;will\; yield\; c_7 $$
 * }
 * }

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

$$\displaystyle f=1 \,\,\, g=2$$ $$\displaystyle n=3; $$
 * $$\displaystyle c_1 = -1 $$
 * $$\displaystyle c_1 = -1 $$
 * }

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

 * Consider the first iteration: i.e, k=1
 * }
 * }

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

f= 1*2*3 =6
 * $$ \displaystyle f=f.*g.*(g+1) $$ yields
 * $$ \displaystyle f=f.*g.*(g+1) $$ yields
 * }

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

$$\displaystyle f=6 $$
 * Hence,
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

 * $$\displaystyle [newcn,newcd]=fracsum(-1*cn,cd.*f);$$
 * }
 * }

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

fracsum(-1*1, 1*6) i.e, fracsum(-1,6)
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * }

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

$$\displaystyle fracsum(-1,6) $$
 * Hence,we have
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

$$\displaystyle div=gcd(round(n),round(d))$$
 * Now looking into the fracsum(n,d) function,
 * Now looking into the fracsum(n,d) function,
 * }

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

because, round(-1) = -1 round(6) = 6 and hence, gcd(-1,6) = 1.
 * will yield div = 1.
 * will yield div = 1.
 * }

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

$$n=round(n./div);$$ will give n=1 becasue round(1/1) =1
 * Now,
 * Now,
 * }

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

will give d=6 becasue round(6/1) =6
 * $$\displaystyle d=round(d./div);$$
 * $$\displaystyle d=round(d./div);$$
 * }

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

$$\displaystyle n=1\; and\; d=6 $$
 * Thus,
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

$$\displaystyle for k=1:length(d) $$ $$\displaystyle dsum=lcm(dsum,d(k)); $$ $$\displaystyle end $$
 * $$\displaystyle dsum=1; $$
 * $$\displaystyle dsum=1; $$
 * }
 * }

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

and lcm (1,6) = 6, we have, dsum =6.
 * Since length(d) = 1
 * Since length(d) = 1
 * }
 * }

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

$$\displaystyle dsum = 6 $$
 * Thus,
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

$$ \displaystyle div=gcd(round(nsum),round(dsum)); $$
 * $$ \displaystyle nsum=dsum*sum(n./d);$$
 * $$ \displaystyle nsum=dsum*sum(n./d);$$
 * }
 * }

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

and
 * nsum = 6* sum(1/6) = 1
 * nsum = 6* sum(1/6) = 1
 * }

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

since, round(nsum) = round(1) = 1 round(dsum) = round(6) = 6 gcd(1,6) =1 Now,
 * div = 1
 * div = 1
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle dsum=dsum/div; $$
 * $$ \displaystyle nsum=nsum/div; $$
 * $$ \displaystyle nsum=nsum/div; $$
 * }

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

dsum = 6
 * nsum = 1
 * nsum = 1
 * }

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

$$\displaystyle [newcn \; newcd] = [1 \; 6] $$
 * Now going back,
 * Now going back,
 * }

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

$$\displaystyle \therefore c_3 = \frac{1}{6} $$
 * Thus,
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

$$\displaystyle cn=[cn;newcn]; cd=[cd;newcd] $$ $$\displaystyle f=[f;1]; $$
 * Next, we have,
 * Next, we have,
 * }

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

cn = [-1;1]; cd=[1;6]
 * Hence,
 * Hence,
 * }

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

 * and,
 * and,

So, f=[6;1]
 * }

This is followed by,

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

which means, g = [2+2, 2] = [4;2]
 * $$\displaystyle g=[2+g(1);g] $$ ;
 * $$\displaystyle g=[2+g(1);g] $$ ;
 * }

Now consider the second iteration - i.e, k=2

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

 * Consider the second iteration: i.e, k=2
 * }
 * }

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

f= (6*4*5; 1*2*3)
 * $$ \displaystyle f=f.*g.*(g+1) $$ yields
 * $$ \displaystyle f=f.*g.*(g+1) $$ yields
 * }

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

$$\displaystyle f=(120;6) $$
 * Hence,
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

 * $$\displaystyle [newcn,newcd]=fracsum(-1*cn,cd.*f);$$
 * }
 * }

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

fracsum(-1*-1, 1*120) i.e, fracsum(1,120)
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * }

and,

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

fracsum(-1*1, 6*6) i.e, fracsum(-1,36)
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * $$ \displaystyle fracsum(-1*cn,cd.*f) $$ yields
 * }

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

$$\displaystyle fracsum(1,120)\; and\; fracsum(-1,36) $$
 * Hence,we have
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

$$\displaystyle div=gcd(round(n),round(d))$$
 * Now looking into the fracsum(1,120) function,
 * Now looking into the fracsum(1,120) function,
 * }

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

because, round(1) = 1 round(120) = 120 and hence, gcd(1,120) = 1.
 * will yield div = 1.
 * will yield div = 1.
 * }

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

$$n=round(n./div);$$ will give n=1 becasue round(1/1) =1
 * Now,
 * Now,
 * }

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

will give d=120 becasue round(36/1) =120
 * $$\displaystyle d=round(d./div);$$
 * $$\displaystyle d=round(d./div);$$
 * }and,

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

$$\displaystyle div=gcd(round(n),round(d))$$
 * Now looking into the fracsum(-1,36) function,
 * Now looking into the fracsum(-1,36) function,
 * }

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

because, round(-1) = -1 round(36) = 36 and hence, gcd(-1,36) = 1.
 * will yield div = 1.
 * will yield div = 1.
 * }

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

$$n=round(n./div);$$ will give n=-1 becasue round(-1/1) =-1
 * Now,
 * Now,
 * }

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

will give d=6 becasue round(36/1) =36
 * $$\displaystyle d=round(d./div);$$
 * $$\displaystyle d=round(d./div);$$
 * }

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

$$\displaystyle n=[1;-1]\; and\; d=[120;36] $$
 * Thus,
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style="width:35%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

$$\displaystyle for k=1:length(d) $$ $$\displaystyle dsum=lcm(dsum,d(k)); $$ $$\displaystyle end $$
 * $$\displaystyle dsum=1; $$
 * $$\displaystyle dsum=1; $$
 * }
 * }

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

and For d(1), dsum = lcm (1,120) = 120, and once again for d(2), dsum = lcm (120,36) = 360 Finally we have, dsum = 360.
 * Since length(d) = 2
 * Since length(d) = 2
 * }
 * }

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

$$\displaystyle dsum = 360$$
 * Thus,
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style="width:15%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

$$ \displaystyle div=gcd(round(nsum),round(dsum)); $$
 * $$ \displaystyle nsum=dsum*sum(n./d);$$
 * $$ \displaystyle nsum=dsum*sum(n./d);$$
 * }
 * }

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

and
 * nsum = 360* sum(1/120 - 1/36) = -7
 * nsum = 360* sum(1/120 - 1/36) = -7
 * }

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

since, round(nsum) = round(-7) = -7 round(dsum) = round(360) = 360 gcd(-7,360) =1 Now,
 * div = 1
 * div = 1
 * }
 * {| style="width:100%" border="0" align="left"

$$ \displaystyle dsum=dsum/div; $$
 * $$ \displaystyle nsum=nsum/div; $$
 * $$ \displaystyle nsum=nsum/div; $$
 * }

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

dsum = 360
 * nsum = -7
 * nsum = -7
 * }

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

$$\displaystyle [newcn\; newcd] = [-7 \;360] $$
 * Now going back,
 * Now going back,
 * }

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

$$\displaystyle \therefore c_5 = -\frac{7}{360} $$
 * Thus,
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

$$\displaystyle cn=[cn;newcn]; cd=[cd;newcd] $$ $$\displaystyle f=[f;1]; $$
 * Next, we have,
 * Next, we have,
 * }

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

cn = [-1;1;7]; cd=[1;6;360]
 * Hence,
 * Hence,
 * }

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

 * and,
 * and,

So, f=[120;6;1]
 * }

This is followed by,

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

which means, g = [2+4;4; 2] = [6;4;2]
 * $$\displaystyle g=[2+g(1);g] $$ ;
 * $$\displaystyle g=[2+g(1);g] $$ ;
 * }

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

we will obtain,
 * Again following the same procedure for Iteration 3, i.e for k=3
 * Again following the same procedure for Iteration 3, i.e for k=3
 * }

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

$$\displaystyle \therefore c_7 = \frac{31}{15120} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

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

 * On substituting, the values of
 * }
 * }

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

in (Eq.1) through (Eq.7) we obtain,
 * $$\displaystyle c_1\;c_3\;c_5\;and\;c_7$$
 * $$\displaystyle c_1\;c_3\;c_5\;and\;c_7$$
 * }

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

 * $$\displaystyle p_1(t) = - t $$
 * }
 * }

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

 * $$\displaystyle p_2(t) = - \frac{t^2}{2!} + \frac{1}{6} $$
 * }
 * }

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

 * $$\displaystyle p_3(t) = - \frac{t^3}{3!} +\frac{t}{6}$$
 * }
 * }

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

 * $$\displaystyle p_4(t) = - \frac{t^4}{4!} + \frac{t^2}{12} -\frac{7}{360} $$
 * }
 * }

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

 * $$\displaystyle p_5(t) = - \frac{t^5}{5!} + \frac{t^3}{36} -\frac{7t}{360} $$
 * }
 * }

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

 * $$\displaystyle p_6(t) = - \frac{t^6}{6!} + \frac{t^4}{144} - \frac{7 t^2}{720} + \frac{31}{15120}$$
 * }
 * }

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

 * $$\displaystyle p_7(t) = - \frac{t^7}{7!} + \frac{t^5}{720} - \frac{7 t^3}{2160} + \frac{31t}{15120} $$
 * }
 * }

Subramanian Annamalai 15:31, 7 April 2010 (UTC)

=  Verification of results for the question 10 of HW 5 =

Problem Statement
$$ \displaystyle Compare\, results\, of\, Team 3\, to\, Team4\, $$

$$ \displaystyle i)\, Integrate\, the\, ellipse\, arc\, length\, by\, Quad.,\, Comp.\, Trap.,\, Romberg,\, Clencurt\, and\, the\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

$$ \displaystyle ii)\, Evaluate\, the\, given\, circumference\, formular\, by\, Quad.,\, Comp. Trap.,\, Romberg\, and\, Clencurt\, $$

$$ \int_{\theta=0}^{\theta=2\pi} dl = C $$

$$ \displaystyle Where\,\,\,\, \begin{align} dl &=d\theta [r^2+(\frac{dr}{d\theta})^2]^(\frac{1}{2})\\ r(\theta) &= \frac{1-e^2}{1-ecos(\theta)} \\ e &= sin(\frac{\pi}{12}) \\\end{align}\ $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg25.djvu|p.25-2]] and [[media:Egm6341.s10.mtg30.djvu|p.30-3]]

$$ \displaystyle iii)\, Evaluate\, the\, complete\, elliptic\, integral\, of\, the\, second\, kind\, method\, $$

$$ \displaystyle C = 4a E (e) $$

$$ \displaystyle E(e) = \int_{0}^{\pi /2} [1-e^2sin^2(\theta)]^{(\frac{1}{2})} d\theta $$

Ref: Team 4 in HW 4 [|problem_11] and Wikipedia[|Elliptic_integral]

$$ \displaystyle iiii)\, Compare\, and\, analyze\, the\, results\, $$

Solution
$$ \displaystyle i)\, Summary\, of\, Results\, given\, by\, Team4\, and\, Team3\, $$

$$ \displaystyle \dagger\, Note:\,\, In\, the\, result\, of\, Clencurt,\, X\, means\, ocillating\, $$

$$ \displaystyle ii)\, Comparision\, $$

$$ \displaystyle a)\, The\, big\, difference\, was\, caused\, by\, the\, calculation\, of\, error.\, $$

$$ \displaystyle :\, Team 4\, computed\, I_n\, to\, the\, error\, 10^{-10}\, b/w\, the\, exact\, I\, and\, I_n.\, $$

$$ \displaystyle \,\, However,\, Team 3\, computed\, I_n\, to\, the\, error\, 10^{-10}\, b/w\, I_{n-1}\, and\, I_n.\, $$

$$ \displaystyle \,\, In\, order\, to\, clearly\, know\, the\, exact\, I,\,\, Team 3\, directly\, computed\, I\, with\, Quad\, again\, $$

$$ \displaystyle b)\, Team 4\, exactly\, found\, I_n\, with\, Clencurt\, but\, the\, result\, of\, Team 3\, is\, oscillating\, on\, the\, place\, for\, 10^{-15}\, $$

$$ \displaystyle c)\, Correction\, for\, Clencurt\, and\, 2nd\, method\, $$

$$ \displaystyle :\, Since\, the\, Clencurt's\, result\, of\, Team 3\, is\, going\, in\, busy\, mode\, and\, the\, input\, of\, 2nd\, method\, was\, not\, correctly\, uploaded,\, it\, is\, not\, easy\, to\, clearly\, see\, the\, results.\, $$

$$ \displaystyle Thus,\, the\, inputs\, for\, above\, two\, are\, attached\, below\, again.\, $$

$$ \displaystyle c-1)\, Clencurt\, $$

$$ \displaystyle c-2)\, 2nd\, method\, $$

$$ \displaystyle iii)\, Analyzing\, the\, Results\, of\, Clencurt\, $$

$$ \displaystyle The\, oscillating\, on\, the\, place\, for\, 10^{-15}\, is\, caused\, by\, the\, difference\, b/w\, I_{exact}\, and\, I_{cl}\, in\, the\, result\, of\, team 3.\, $$

$$ \displaystyle Thus,\, we\, verified\, the\, Clencurt's\, result\, with\, a\, different\, order\, of\, errors\, $$

$$ \displaystyle a)\, Correctional\, Clencurt\, $$

$$ \displaystyle b)\, The\, difference\, when\, I_{exact}\, -\, I_{cl}\, <\, 10^{-8}\, is\, 7.591901329817574*10^{-9}.\, $$

$$ \displaystyle This\, value\, is\, pretty\, small\, and\, can\, be\, thought\, that\, the\, Clencurt\, integration\, close\, up\, to\, the\, I_{exact},\, the\, Quad\, integration.\, $$

--Heejun Chung 13:07, 7 April 2010 (UTC)

=Contributing Team Members=

1.Abhishekksingh 09:16, 7 April 2010 (UTC) Authored 3,4,5 and 6 Proof read 2,7,8 and 9 2.Yong Nam Ahn 14:12, 7 April 2010 (UTC) Authored 1, Proof read 11 3.Min 14:20, 7 April 2010 (UTC) Authored 2,7,8,9, Proof read 3,4,5 and 6 4.Heejun Chung 14:53, 7 April 2010 (UTC) Authored 11 Proof read 10 5.Subramanian Annamalai 15:10, 7 April 2010 (UTC) Authored 10 Proof read 1

=References= 1. Introduction to Numerical Methods,2nd Edition by Kendall E Atkinson Second Edition 2. Numerical Methods for Engineers,5th Edition by Steven C Chapra and Raymond P Canale

title: User:Egm6341.s10.team3.sa/HW6


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User:Rjorjorjo/HW6

HW 6 Problem Set  =Find the equation of motion $$\mathbf{Z}$$ when local coordinate vector $$ s = \frac{1}{2}$$=

Problem Statement
Prove $$\begin{align} \mathbf{Z}_{i+\frac{1}{2}}& = \ \mathbf{Z}(s = \frac{1}{2}) & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1}+\frac{h}{8}(f_{i}-f_{i+1}))\end{align}$$

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

C_{i}s^{i}$$
 * (a): (2) [[media:Egm6341.s10.mtg35.djvu|p.35-2]] $$\mathbf{Z}(s) = \ \sum_{i=0}^{3}
 * (a): (2) [[media:Egm6341.s10.mtg35.djvu|p.35-2]] $$\mathbf{Z}(s) = \ \sum_{i=0}^{3}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

C_{0} \\ C_{1} \\ C_{2} \\ C_{3} \\ \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ -3&-2&3&-1 \\ 2&1&-2&1 \\ \end{bmatrix} \cdot \!\, \begin{bmatrix} \mathbf{Z}_{i} \\ \mathbf{Z}_{i}^{'} \\ \mathbf{Z}_{i+1} \\ \mathbf{Z}_{i+1}^{'} \\ \end{bmatrix} $$
 * (b): (1) [[media:Egm6341.s10.mtg35.djvu|p.35-4]] $$\begin{bmatrix}
 * (b): (1) [[media:Egm6341.s10.mtg35.djvu|p.35-4]] $$\begin{bmatrix}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

 * (c): (1) [[media:Egm6341.s10.mtg36.djvu|p.36-1]] $$ \mathbf{Z}^{'} = h \dot{\mathbf{Z}}$$
 * }
 * }
 * }

Ref: Lecture Notes [[media:Egm6341.s10.mtg35.djvu|p.35-2&4]]  [[media:Egm6341.s10.mtg36.djvu|p.36-1]]

Solution
$$\begin{align} \mathbf{Z}(s = \frac{1}{2})& = \ \sum_{i=0}^{3}C_{i}s^{i} \\ & =\ C_{0}(\frac{1}{2})^{0}+C_{1}(\frac{1}{2})^{1}+C_{2}(\frac{1}{2})^{2}+C_{3}(\frac{1}{2})^{3} \\ & =\ \mathbf{Z}_{i}+\mathbf{Z}_{i}^{'}(\frac{1}{2})+(3\mathbf{Z}_{i+1}-3\mathbf{Z}_{i}-2\mathbf{Z}_{i}^{'}-\mathbf{Z}_{i+1}^{'})(\frac{1}{4})+(\mathbf{Z}_{i+1}^{'}-2\mathbf{Z}_{i+1}+2\mathbf{Z}_{i}+\mathbf{Z}_{i}^{'})(\frac{1}{8}) \\ & = \ (1-\frac{3}{4}+\frac{1}{4})\mathbf{Z}_{i}+(\frac{1}{2}-\frac{1}{2}+\frac{1}{8})\mathbf{Z}_{i}^{'}+(\frac{3}{4}-\frac{1}{4})\mathbf{Z}_{i+1}+(-\frac{1}{4}+\frac{1}{8})\mathbf{Z}_{i+1}^{'} \\ & = \ \frac{1}{2} \mathbf{Z}_{i}+\frac{1}{8}\mathbf{Z}_{i}^{'}+\frac{1}{2}\mathbf{Z}_{i+1}-\frac{1}{8}\mathbf{Z}_{i+1}^{'} \\ & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{1}{8}(\mathbf{Z}_{i}^{'}-\mathbf{Z}_{i+1}^{'}) \\ & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{h}{8}(\frac{1}{h}\mathbf{Z}_{i}^{'}-\frac{1}{h}\mathbf{Z}_{i+1}^{'}) \\ & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{h}{8}(\dot{\mathbf{Z}_{i}}-\dot{\mathbf{Z}_{i+1}}) \\ & =\ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{h}{8}(f_{i}-f_{i+1}) \end{align}$$

=Run the circumference formula with the complete elliptic integral codes of Team 3, and compare to the results of our own code=

Problem Statement
verify the results circumference formula with the complete elliptic integral codes

Solution
1Composite Trapezoidal Rule (a)team 3's codes :

subfunction is

(b)Ours

2Use Clenshaw-Curtis quadrature :

(a)Team 3

(b)Team1(ours)

3Use Romberg Table:

(a)Team 3

(b)Team 1 (ours)

EGM6341.s10.Team1.Ya-Chiao Chang 02:53, 7 April 2010 (UTC)

title: User:Rjorjorjo/HW6


text:
User:Egm6341.s10.Team4.roni/HW6/6.10

= Problem 10: Trapezoidal Rule Error Computation =

Given
Kessler's Paper, MatLab code and Result Table.

Patch Kessler Code for Trapezoidal Rule Error 5-9 problem given in Page [[media:Egm6341.s10.mtg30.djvu|30-2]]

 Matlab Code:

Find
Run Kessler's Code line by line to reproduce his table. Complete HW 5-9 Refer Lecture slide [[media:Egm6341.s10.mtg30.djvu|30-2]] Produce (P2,P3) (P4,P5) (P6,P7)and put comments in his code

Solution
but denominator is obviously wrong this denominator should have turned to be 3 newpd =  360; pd =     6   360 (All wrong! Algorithm does not work)

Table 1 shows the results from running Kessler's Code

Table 2 shows the results from running Team4 Code

 Team4 Trapezoidal rule Error (alternative to to Kessler code)

 Matlab Code:

Discussion:

Under Kessler's code, the program computes P(2k) for n=1 to any n number selected. From the paper we can see that (P3=P5=P7=0 at t=1). So we can use this equality to compute the C coefficients.

Kessler's code starts missing the correct C coefficients from C6 as shown in Table 1. Although the decimal value of C6-C9 is shown correctly in Table 1, The program can't not compute it correctly since it is derived from the numerator and the incorrect denominator. Team4 algorithm is based on decimal calculation of the C coefficients and therefore all coefficients are accurate to C9. Kessler's Algorithm start breaking down due to inaccuracies when calculating the C coefficients as fraction. Team4 algorithm allows the user to convert the decimal to fraction after the computation of the coefficients. The loss of accuracy in the fraction calculation from C7 in Team4 code is due to numerical errors of using single precision variables and functions.

Kessler's code in the paper took 0.102152 seconds to compute coefficients for n=1 to 8 The code developed by Team4 is much faster.021921 seconds for n=2 to 9 and does the work accurately for the decimal terms. Fraction terms are accurate up to C6 and P10. Those are shown in table 2 and could be improved if Double Precision is used to find the C coefficients and the factorial terms.

Author
Solved and typed by - --Egm6341.s10.Team4.roni 04:08, 7 April 2010 (UTC)

Proof read by -

title: User:Egm6341.s10.Team4.roni/HW6/6.10


title: File:Arclength1.png


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User:Egm6341.s2010.Team1/HW6

HW 6 Problem Set  =Problem 1: Compute the arc length of an ellipse on $$[0^{\circ},60^{\circ}]$$=

Statement
An arc is denoted $$\widehat{PQ}$$ and lies on an ellipse defined by,
 * $$r(\theta)=\frac{a(1-e^2)}{1-e \cos{(\theta)}}$$

where $$a=1$$ and $$e=\sin{(\frac{\pi}{12})}$$. The equation for the arc length is,
 * $$PQ=\int_a^bdl=\int_{\theta_1}^{\theta_2}d\theta \sqrt{r^2+\left (\frac{dr}{d\theta} \right )^2}$$

1.) Estimate the number of nodes required for the composite trapezoidal estimate of the integral to have an error $$O(10^{-10})$$ 2.)  Using successive numerical integrations, find the actual number of nodes required for the trapezoidal technique by considering the stopping criterion $$\left |I_{2n}-I_n \right |<10^{-10}$$ 3.) How many nodes are required for the Romberg table (verify vs trapezoidal), clencurt.m (verify with a similar stopping criterion), and sum (chebfun - compare to trapezoidal).

Solution
The integrating function is,
 * $$F=\sqrt{r^2+\left (\frac{dr}{d\theta} \right )^2}=\frac{1-e^2}{(1-e\cos{\theta})^2}\sqrt{1-2e\cos{\theta}+e^2}$$

1.) An estimate for the error in the composite trapezoidal technique is,
 * $$\left | E_n \right |\leq \frac{(b-a)^3}{12n^2}M_2$$

Rearranging for the number of nodes, $$n$$, gives,
 * $$n=\sqrt{\frac{(b-a)^3}{12(10^{-10})}M_2}$$

where $$M_2=max\left | F^{(2)}(\zeta) \right |, \zeta\in [a,b]$$ and $$[a,b]=[0,60\frac{\pi}{180}]$$. The following MATLAB code solves for $$M_2$$ and determines $$n<16,755$$

2.) Numerically integrating $$F$$ using the trapezoidal rule detemines that at $$n=16,384$$ we get $$\left | I_{16,384}-I_{8,192} \right | < 10^{-10}$$. The following MATLAB code is used.  Notice that in step sizes of 2n, the previous summation can be utilized as shown in the code.

3.) Numerically integrate using the Romberg, clencurt, and sum (chebfun).

Romberg gives $$O(10^{-10})$$ compared to itself at $$n=2^7=128$$, see code.

For the sum (chebfun) the code is shown next. Comparing this to the composite trapezoidal result shows a difference $$O(10^{-11})$$ so is considered accurate enough.

And finally for clencurt.m is shown last. The result is compared to itself and at n=10 has an error less than $$O(10^{-10})$$

Problem solved by EGM6341.s10.Team1.Kumanchik 22:52, 7 April 2010 (UTC)

=Problem 2-Consider v+dv,show lin.mom.compo. =

Find
At $$ t+dt\,$$,consider $$ v+dv\,$$, show $$ dp_{\bar{y}}=mvd\gamma\,$$ by neglecting hot.

Given
Refer Lecture slide (1) [[media:Egm6341.s10.mtg33.djvu|p.33-2]] for problem statement

[Figure 1]

$$ dp_{\bar{y}}=mvd\gamma\,$$

Solution

[Figure 2]

From Fig.2,it can be seen that

$$ dp_{\bar{y}}=m(v+dv)d\gamma\,$$

$$ =mvd\gamma+ \cancelto{}{mdvd\gamma}\,$$

By neglecting higher order term$$ dvd\gamma\,$$,just retention the lowest term

It can be derived that

$$ dp_{\bar{y}}=mvd\gamma\,$$

Egm6341.s10.team1.lei15:00, 7 April 2010 (UTC)

= Problem 3: Determine 3rd and 4th rows of the Matrix =

Given
The matrix is given to be:
 * {| style="width:70%" border="0" align="center"

 * $$\underbrace{\begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \end{bmatrix}.}_{A} \underbrace{\begin{bmatrix}

C_0 \\ C_1 \\ C_2 \\ C_3 \end{bmatrix}}_{C_i} = \begin{bmatrix}

z_i \\ z'_i \\ z_{i+1} \\ z'_{i+1} \end{bmatrix} $$
 * }.
 * }.

Find
Need to determine the 3rd and 4th row of the matrix using

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \bar{d_3}=z_{i+1}=z(s=1)
 * }.
 * }.

and

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \bar{d_4}=z'_{i+1}=z'(s=1)
 * }.
 * }.

Solution
We know that

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z(s)=\sum_{i=0}^{3} C_i s^i= C_0+ C_1s+ C_2s^2+ C_3s^3
 * }.
 * }.

and

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z'(s)= \sum_{i=1}^{3} iC_i s^{i-1}= C_1+ 2C_2s+ 3C_3s^2
 * }.
 * }.

thus using the above relations where $$s=1$$ in both cases we then have

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z_{i+1}= C_0+ C_1+ C_2+ C_3
 * }.
 * }.

and

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z'_{i+1}= C_1+ 2C_2+ 3C_3
 * }.
 * }.

Therefore from here it is easy to surmise that the 3rd and 4th rows are $$[1 1 1 1]$$ and $$[0 1 2 3]$$ respectively.

Egm6341.s10.team1.toddmock 19:13, 7 April 2010 (UTC)

= Problem 4: verify the inverse of the Matrix =

Given
From the matrix

 * {| style="width:70%" border="0" align="center"

 * $$A=\begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \end{bmatrix} $$
 * }.
 * }.

Find
Use Matlab to show that the inverse is

 * {| style="width:70%" border="0" align="center"

 * $$A^{-1}=\begin{bmatrix}

1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ -3 & -2 & 3 & -1 \\ 2 & 1 & -2 & 1 \end{bmatrix} $$
 * }.
 * }.

Solution
This yields the result

Which is exactly what we were hoping for.

Egm6341.s10.team1.toddmock 19:13, 7 April 2010 (UTC)

=Problem 5: Identify the basis functions $${N_i(s)}\,$$, $$i=1,2,3,4\,$$. Plot $$\bar{N_i}\,$$'s=

Solution
$$ \sum_{i=0}^{3} C_is^i = \sum_{i=1}^{4} N_i(s)d_i\,$$ $$ C_0+C_1s+C_2s^2+C_3s^3=N_1(s)d_1+N_2(s)d_2+N_3(s)d_3+N_4(s)d_4\,$$ $$ C_0=z_i \,$$ $$ C_1=z_i' \,$$ $$ C_2=-3z_i-2z_i'+3z_{i+1}-z_{i+1}' \,$$ $$ C_3=2z_i+z_i'-2z_{i+1}+z_{i+1}' \,$$ $$ d_1=z_i\,$$ $$ d_2=\dot{z_i}=z_i'\frac{1}{h}\,$$ $$ d_3=z_{i+1}\,$$ $$ d_4=\dot{z_{i+1}}=z_{i+1}'\frac{1}{h} \,$$ $$ \Rightarrow z_i+z_i's+(-3z_i-2z_i'+3z_{i+1}-z_{i+1}')s^2+(2z_i+z_i'-2z_{i+1}+z_{i+1}')s^3=N_1(s)z_i+N_2(s)z_i'\frac{1}{h}+N_3(s)z_{i+1}+N_4(s)z_{i+1}'\frac{1}{h} \,$$ $$ \Rightarrow z_i(1-3s^2+2s^3)+z_i'(s-2s^2+s^3)+z_{i+1}(3s^2-2s^3)+z_{i+1}'(-s^2+s^3)=z_i(N_1(s))+z_i'(N_2(s)\frac{1}{h})+z_{i+1}(N_3(s))+z_{i+1}'(N_4(s)\frac{1}{h}) \,$$

$$ \bar{N_1}(s) = 1-3s^2+2s^3 \,$$ $$ \bar{N_2}(s) = s-2s^2+s^3 \,$$ $$ \bar{N_3}(s) = 3s^2-2s^3 \,$$ $$ \bar{N_4}(s) = -s^2+s^3 \,$$

Note for verification that: $$ \bar{N_1}(0)=1, \dot{\bar{N_1}}(0)=0, \bar{N_1}(1)=0, \dot{\bar{N_1}}(1)=0 \,$$ $$ \bar{N_2}(0)=0, \dot{\bar{N_2}}(0)=1, \bar{N_2}(1)=0, \dot{\bar{N_2}}(1)=0 \,$$ $$ \bar{N_3}(0)=0, \dot{\bar{N_3}}(0)=0, \bar{N_3}(1)=1, \dot{\bar{N_3}}(1)=0 \,$$ $$ \bar{N_4}(0)=0, \dot{\bar{N_4}}(0)=0, \bar{N_4}(1)=0, \dot{\bar{N_4}}(1)=1 \,$$

Egm6341.s10.team1.andrewdugan 21:27, 7 April 2010 (UTC)

=Problem 6- Find s=s(t) =

Find
Find $$ s=s(t)\,$$ such that $$ Z(t)=\sum_{i=1}^{4}N_{i}(t)d_{i}\,$$

Given
(1) [[media:Egm6341.s10.mtg35.djvu|p.35-1]]

Local coord. for $$ [t_{i},t_{i+1}]:\,$$

$$ t(s)=(1-s)t_{i}+st_{i+1}\, $$

Solution
Convert

$$ t(s)=(1-s)t_{i}+st_{i+1}\, $$

We can get

$$ s=\frac{t-t_{i}}{t_{i+1}-t_{i}}\,$$

thus

$$ s=s(t)=\frac{t-t_{i}}{t_{i+1}-t_{i}}\,$$

plug$$ s=s(t)\,$$ into

$$ Z(s)=\sum_{i=1}^{4}N_{i}(s)d_{i}\,$$

$$ \Rightarrow Z(T)=\sum_{i=1}^{4}N_{i}(T)d_{i}\,$$

Egm6341.s10.team1.lei15:00, 7 April 2010 (UTC)

=Problem 7: Find the equation of motion $$\mathbf{Z}$$ when local coordinate vector $$ s = \frac{1}{2}$$=

Problem Statement
Prove $$\begin{align} \mathbf{Z}_{i+\frac{1}{2}}& = \ \mathbf{Z}(s = \frac{1}{2}) & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1}+\frac{h}{8}(f_{i}-f_{i+1}))\end{align}$$ Given :
 * {| style="width:100%" border="0" align="left"

C_{i}s^{i}$$
 * (a): (2) [[media:Egm6341.s10.mtg35.djvu|p.35-2]] $$\mathbf{Z}(s) = \ \sum_{i=0}^{3}
 * (a): (2) [[media:Egm6341.s10.mtg35.djvu|p.35-2]] $$\mathbf{Z}(s) = \ \sum_{i=0}^{3}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

C_{0} \\ C_{1} \\ C_{2} \\ C_{3} \\ \end{bmatrix} = \begin{bmatrix} 1&0&0&0 \\ 0&1&0&0 \\ -3&-2&3&-1 \\ 2&1&-2&1 \\ \end{bmatrix} \cdot \!\, \begin{bmatrix} \mathbf{Z}_{i} \\ \mathbf{Z}_{i}^{'} \\ \mathbf{Z}_{i+1} \\ \mathbf{Z}_{i+1}^{'} \\ \end{bmatrix} $$
 * (b): (1) [[media:Egm6341.s10.mtg35.djvu|p.35-4]] $$\begin{bmatrix}
 * (b): (1) [[media:Egm6341.s10.mtg35.djvu|p.35-4]] $$\begin{bmatrix}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

 * (c): (1) [[media:Egm6341.s10.mtg36.djvu|p.36-1]] $$ \mathbf{Z}^{'} = h \dot{\mathbf{Z}}$$
 * }
 * }
 * }

Ref: Lecture Notes [[media:Egm6341.s10.mtg35.djvu|p.35-2&4]]  [[media:Egm6341.s10.mtg36.djvu|p.36-1]]

Solution
$$\begin{align} \mathbf{Z}(s = \frac{1}{2})& = \ \sum_{i=0}^{3}C_{i}s^{i} \\ & =\ C_{0}(\frac{1}{2})^{0}+C_{1}(\frac{1}{2})^{1}+C_{2}(\frac{1}{2})^{2}+C_{3}(\frac{1}{2})^{3} \\ & =\ \mathbf{Z}_{i}+\mathbf{Z}_{i}^{'}(\frac{1}{2})+(3\mathbf{Z}_{i+1}-3\mathbf{Z}_{i}-2\mathbf{Z}_{i}^{'}-\mathbf{Z}_{i+1}^{'})(\frac{1}{4})+(\mathbf{Z}_{i+1}^{'}-2\mathbf{Z}_{i+1}+2\mathbf{Z}_{i}+\mathbf{Z}_{i}^{'})(\frac{1}{8}) \\ & = \ (1-\frac{3}{4}+\frac{1}{4})\mathbf{Z}_{i}+(\frac{1}{2}-\frac{1}{2}+\frac{1}{8})\mathbf{Z}_{i}^{'}+(\frac{3}{4}-\frac{1}{4})\mathbf{Z}_{i+1}+(-\frac{1}{4}+\frac{1}{8})\mathbf{Z}_{i+1}^{'} \\ & = \ \frac{1}{2} \mathbf{Z}_{i}+\frac{1}{8}\mathbf{Z}_{i}^{'}+\frac{1}{2}\mathbf{Z}_{i+1}-\frac{1}{8}\mathbf{Z}_{i+1}^{'} \\ & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{1}{8}(\mathbf{Z}_{i}^{'}-\mathbf{Z}_{i+1}^{'}) \\ & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{h}{8}(\frac{1}{h}\mathbf{Z}_{i}^{'}-\frac{1}{h}\mathbf{Z}_{i+1}^{'}) \\ & = \ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{h}{8}(\dot{\mathbf{Z}_{i}}-\dot{\mathbf{Z}_{i+1}}) \\ & =\ \frac{1}{2}(\mathbf{Z}_{i}+\mathbf{Z}_{i+1})+\frac{h}{8}(f_{i}-f_{i+1}) \end{align}$$

EGM6341.s10.Team1.Ya-Chiao Chang 02:57, 7 April 2010 (UTC)

=Problem 8: Verify the expression for $$ z_{i+\frac{1}{2}}' \,$$=

Find
Verify that $$ z_{i+\frac{1}{2}}'=-\frac{3}{2}(z_i-z_{i+1})-\frac{1}{4}(z_i'+z_{i+1}')\,$$

Solution
$$ z_{i+\frac{1}{2}}'=z'(s=\frac{1}{2}) \,$$ Recall that $$ z'(s)=\sum_{i=1}^{3}iC_is^{i-1} \,$$ Thus $$ z'(s=\frac{1}{2})= (1)(C_1)(\frac{1}{2})^{1-1}+(2)(C_2)(\frac{1}{2})^{2-1}+(3)(C_3)(\frac{1}{2})^{3-1} \,$$ $$ z'(s=\frac{1}{2}) = (1)(C_1)(1)+(2)(C_2)(\frac{1}{2})+(3)(C_3)(\frac{1}{4}) = C_1+C_2+\frac{3}{4}C_3 \,$$ Recall that $$ C_1=z_i', C_2=-3z_i-2z_i'+3z_{i+1}-z_{i+1}', C_3=2z_i+z_i'-2z_{i+1}+z_{i+1}' \,$$ Then, $$ z'(s=\frac{1}{2}) = z_i'+(-3z_i-2z_i'+3z_{i+1}-z_{i+1}')+\frac{3}{4}(2z_i+z_i'-2z_{i+1}+z_{i+1}') \,$$ $$ z'(s=\frac{1}{2}) = z_i(-3+\frac{3}{2})+z_i'(1-2+\frac{3}{4})+z_{i+1}(3-\frac{3}{2})+z_{i+1}'(-1+\frac{3}{4}) \,$$ $$ z'(s=\frac{1}{2}) = -\frac{3}{2}z_i-\frac{1}{4}z_i'+\frac{3}{2}z_{i+1}-\frac{1}{4}z_{i+1}' \,$$

Egm6341.s10.team1.andrewdugan 21:33, 7 April 2010 (UTC)

= Problem 9: Proof of $$z_{i+1}$$ =

Given
We know that

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dot{z}_{i+1/2}=\dfrac {-3}{2h}(z_i-z_{i+1})-\dfrac {1}{4}(f_i+f_{i+1})
 * }.
 * }.

and we want to use collocation at $$t_{i+1/2}$$ which gives the relation

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \triangle = 0 = \dot{z}_{i+1/2}-f_{i+1/2}
 * }.
 * }.

Find
The mission here is to find a relation for $$z_{i+1}$$

Solution
Therefore using the collocation we have
 * {| style="width:100%" border="0" align="left"

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle f_{i+1/2}= -\dfrac {3}{2h}(z_i-z_{i+1})- \dfrac {1}{4}(f_i+ f_{i+1})
 * }.
 * }.

which can be rearranged to give

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dfrac{1}{4}f_i+\dfrac{1}{4}f_{i+1}+f_{i+1/2}=\dfrac{3}{2h}(z_{i+1}-z_i)
 * }.
 * }.

or

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle \dfrac{1}{4}(f_i+4f_{i+1/2}+f_{i+1})=\dfrac{3}{2h}(z_{i+1}-z_i)
 * }.
 * }.

rearranging this gives

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z_{i+1}-z_i=\dfrac{h}{6}(f_i+4f_{i+1/2}+f_{i+1})
 * }.
 * }.

here it should be noted that the right hand side is Simpson's rule, proceeding further we obtain the expression

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

$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle z_{i+1}=z_i+\dfrac{h/2}{3}(f_i+4f_{i+1/2}+f_{i+1})
 * }.
 * }.

Egm6341.s10.team1.toddmock 19:14, 7 April 2010 (UTC)

=Problem 10: Use Kessler's code to generate his table=

Statement
Use Kessler's code to generate his table and then follow the code line-by-line to get $$(P_2,P_3)(P_4,P_5)(P_6,P_6)$$. Comment on the code.

Solution
Kessler's table was generated with the code below for n=8:

The code computes the Trapezoidal error polynomials evaluated at t=1. The polynomial equation is,
 * $$p_{2k}(t)=c_1\frac{t^{2k}}{(2k)!}+c_3\frac{t^{2k-2}}{(2k-2)!}+c_5\frac{t^{2k-4}}{(2k-4)!}+...+c_{2k-3}\frac{t^4}{4!}+c_{2k-1}\frac{t^2}{2!}+c_{2k+1}\frac{t^0}{0!}$$

When evaluated at t=1, the equation is,
 * $$p_{2k}(1)=\frac{c_1}{(2k)!}+\frac{c_3}{(2k-2)!}+\frac{c_5}{(2k-4)!}+...+\frac{c_{2k-3}}{4!}+\frac{c_{2k-1}}{2!}+\frac{c_{2k+1}}{0!}$$

The coefficients, $$c$$, are computed by solving the below equation using $$c_1=-1$$ as a starting point,
 * $$\frac{c_1}{(2k-1)!}+\frac{c_3}{(2k-3)!}+\frac{c_5}{(2k-5)!}+...+\frac{c_{2k-5}}{5!}+\frac{c_{2k-3}}{3!}+\frac{c_{2k-1}}{1!}=0$$

For illustration, the first few coefficients and polynomial equations are shown below.

Notice that the factorials in the denominators of the polynomials differ by 1 from those in the coefficients. Also, to solve for the next coefficient, the preceding coefficients only need to be summed and then moved to the other side of the equal sign, i.e $$c_7=-(\frac{c_1}{7!}+\frac{c_3}{5!}+\frac{c_5}{3!})$$.

The following code will be commented line-by-line to explain the algorithm of Kessler:

Finally, the code will be followed to generate $$(P_2,P_3)(P_4,P_5)(P_6,P_6)$$: The polynomials are written as,
 * $$p_2(t)=c_1\frac{t^2}{2!}+c_3$$
 * $$p_3(t)=c_1\frac{t^3}{3!}+c_3t$$
 * $$p_4(t)=c_1\frac{t^4}{4!}+c_3\frac{t^2}{2!}+c_5$$
 * $$p_5(t)=c_1\frac{t^5}{5!}+c_3\frac{t^3}{3!}+c_5t$$
 * $$p_6(t)=c_1\frac{t^6}{6!}+c_3\frac{t^4}{4!}+c_5\frac{t^2}{2!}+c_7$$
 * $$p_7(t)=c_1\frac{t^7}{7!}+c_3\frac{t^5}{5!}+c_5\frac{t^3}{3!}+c_7t$$

Following the code to get $$c_{1-7}$$, n=3 because 3 loops will generate all 7 coefficients f=1, g=2, cn=-1, cd=1 entering loop which runs 3 times f=1*2*3 (3! = 6) enter fracsum with (1,6) and will sum the fractions w/o leaving a decimal div = gcd(1,6) = 1 n = 1 d = 6 dsum = 1 enter loop (runs once) dsum = 6 leave loop nsum = 6*(1/6)= 1 div = gcd (1,6) = 1 nsum = 1 dsum = 6 leave fracsum with (1,6) cn = [-1; 1] cd = [1; 6] f = [6;1] g = [4;2] enter fracsum with ([-3;1],[6;6]) div = [3;1] n=[-1;1] d=[2;6] enter loop dsum = 2 then 6 leave loop nsum=6*-1/2+6*1/6=-3+1=-2; div=2 nsum=-1; dsum=3; leave fracsum with (-1,3) pn(1,1)=-1; pd(1,1)=3; move on to next loop k=2

At this point it is apparent that fracsum will sum up fractions without reducing them to decimals Also, the second fracsum call does not affect cn so we will skip calculating it And, cn=[-1;1] cd=[1;6] so we have c1 and c3

now k=2 f=[6;1].*[4;2].*[5;3] = [24;2].*[5;3] = [120;6] enter fracsum with (-[-1;1],[1;6].*[120;6])=([1;-1],[120;36]) this is like 1/120-1/36 = -7/360 leave fracsum cn = [-1;1;-7]; cd = [1;6;360]; f = [120;6;1] g = [2+4;[4;2]] = [6;4;2] skip to k=3 f=[120;6;1].*[6;4;2].*[7;5;3] = [720;24;2].*[7;5;3] = [5040;120;6] enter fracsum with ([1;-1;7],[1;6;360].*[5040;120;6]) = ([1;-1;7],[5040;720;2160]) this is 1/5040-1/720+6/2160 = 31/15120 leave fracsum cn = [-1;1;-7;31]; cd = [1;6;360;15120]

The result is,
 * $$c_1=-1 \!$$
 * $$c_3=-\frac{1}{6}$$
 * $$c_5=-\frac{-7}{360}$$
 * $$c_7=-\frac{31}{15120}$$

Plugging into the expressions for $$p_1(t) -> p_7(t)$$ generates the correct result.

Problem solved by EGM6341.s10.Team1.Kumanchik 22:53, 7 April 2010 (UTC)

=Problem 11: Comparison of the elliptic integral codes of Team 3 with those of ours=

Problem Statement
verify the results circumference formula with the complete elliptic integral codes

Solution
1Composite Trapezoidal Rule (a)team 3's codes : subfunction is (b)Ours

2Use Clenshaw-Curtis quadrature : (a)Team 3 (b)Team1(ours) 3Use Romberg Table: (a)Team 3 (b)Team 1 (ours) EGM6341.s10.Team1.Ya-Chiao Chang 02:53, 7 April 2010 (UTC)

=Contributing Members= Egm6341.s10.team1.lei15:03, 7 April 2010 (UTC) Author of problem 2 & 6 EGM6341.s10.Team1.Ya-Chiao Chang 15:23, 7 April 2010 (UTC) Author of problem 7 & 11 Egm6341.s10.team1.toddmock 19:24, 7 April 2010 (UTC) Author of problem 3, 4, & 9 Egm6341.s10.team1.andrewdugan 21:35, 7 April 2010 (UTC) Author of problems 5 & 8

title: User:Egm6341.s2010.Team1/HW6


text: File:KesslerTable.png

title: File:KesslerTable.png


text: File:My Table.png

title: File:My Table.png