text:
User talk:Gorn~enwikiversity

 Hello Gorn, 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.

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 * Visit a (kind of) random project.
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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! --Ottava Rima (talk) 15:05, 4 February 2011 (UTC)

Your account will be renamed
Hello,

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

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

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:37, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:09, 19 April 2015 (UTC)

title: User talk:Gorn~enwikiversity


text:
Survey research and design in psychology/Materials/Software

Software
To complete the tutorials and survey research project for Survey research and design in psychology access will be needed to statistical software, including IBM SPSS and spreadsheet software such as Microsoft Excel (part of Microsoft Office). This software can be accessed on University of Canberra campus computers at no additional cost by those enrolled.

You can also use versions of the software on other computers. Options for accessing SPSS include:
 * 1) Download and install a 14-day trial version.
 * 2) Purchase a SPSS Standard Grad Pack 6 or 12 month license via SPSS buy now - see bottom right hand corner for links to student version resellers

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


text:
Talk:University of Canberra/RCC2011/What key facilitation capacities help create successful wikis?

Is Wikipedia a "successful wiki?"
It's clear that Wikipedia has been successful in certain ways. If I want to look up the meaning and some explanation for some scientific term, usually I get a decent explanation. But I've discovered that if I'm familiar with a topic, and happen to look, I can often see that the article is quite defective, sometimes seriously. I used to think, "Just fix it!" However, sometimes -- certainly not always -- there will be editors who will resist the fix, sometimes based on technicalities, but sometimes obviously based on either some point of view, or a kind of default point of view: "I'm right and it looks like you disagree with me, so you are wrong." If this editor has friends who are administrators, forget about it. It's impossible, unless you also have those friends or are expert in using the structure and willing to put in the considerable effort it can take. Just to make one little change to an article. Horribly inefficient. And you can go through all this, and look at the article a year later, and it backslid. Sisyphus.

Whenever there is real-world controversy over a topic, Wikipedia can be grossly unreliable. The basic mission was to create a compendium of "all human knowledge," but that was an overstatement. Down in the nuts and bolts, "all" meant "all notable," which, in the end, by policy -- but not necessarily in practice -- meant "all knowledge covered in reliable sources so that it's verifiable." If you know a thing, your friends know a thing, and you are the world's experts on a thing, it means nothing, though it certainly is human knowledge. Yet, Wikipedia is full of unverified information, because people believe the advertising: all human knowledge, and no practical structure has been developed to ensure that information actually is verified, and, perhaps even more important, that coverage is neutral and balanced.

Another fundamental aspect of the mission was "neutrality." Wikipedia articles are supposed to adhere to NPOV, or "Neutral Point of View." Great idea. It's subjective, however. "Neutral" is a relative term. It's easy for cherry-picked text to appear neutral to those who don't know the subject. It's easy for sources to be quoted or used out-of-context, so that what the text says seems verifiable, but any subject expert would see the problem in a flash. Someone who favors a point of view may see material that confirms that point of view as "neutral," and text that might seem to contradict the POV is, of course, "POV." Even if equally verifiable. And, indeed, it might be. But it might be needed for balance, properly framed and attributed.

One of the common phenomena, in articles under dispute, is "cherry-picking." Evidence is picked and inserted that appears to support one point of view, and if one looks at the editor's contributions, a curious phenomenon can be seen. Even with a complex controversy, with mounds of evidence on both sides (and sometimes there are more than two sides!), the editor is always inserting or restoring evidence on one side, and never inserts "inconvenient evidence." Does this mean that the editor is biased? Almost certainly. However, is there something wrong with this? Not necessarily. It depends on how the editor goes about it.

There is only one way to approach neutrality, and that is to maximize true consensus. Jimbo Wales understood the principle; if text is truly neutral, all sides will see that their point of view is fairly represented. Factions only get exercised when their point of view is unfairly denigrated, but with proper balance and completion of coverage, according to sourcing standards, all factions should be able to sign on. There can still be "die-hards" who won't accept the insertion of "inconvenient fact," rather than accepting it with due balance, but these are, in my experience, rare, and they have generally been trained to be intransigent by experience with the success of intransigence on Wikipedia. That happens when the user's point of view is MPOV, i.e., aligns with the "majority point of view," which means "majority among Wikipedia editors, especially administrators," not necessarily "majority" as in "majority of experts."

Wikipedia did not develop, and apply widely, decent consensus process, designed to maximize consensus. Instead, the project went toward maximizing the number of articles and the bulking up of content. Consensus process is generally time-consuming. It requires detailed discussion. Wikipedia is famous for harassing and banning experts, because experts typically love their subject, as well as knowing about it, and they will often assume that other editors will be similarly interested. They are called "SPAs," single purpose accounts, and are given short shrift if there is a conflict. An editor with wide contributions and an editor who is, say, a professor in a field, who only contributes in that field, are no match in a conflict. As the administrative structure consists of users who aren't SPAs, their sympathies are naturally with the wide-focus editor, the "content contributor," and surely banning this editor who only works on one article won't be a big loss!

I used consensus process in a few cases I came across of conflict between an expert and a general-purpose editor, and successfully. It's not that difficult. But as soon as I became expert in a field myself, I was practically dead meat, and in my field of (new) expertise, the Wikipedia article is atrocious, heavily biased to what became the "scientific consensus" twenty years ago, but which, in recent years, flipped. In spite of recent peer-reviewed literature being 100% in one direction, and that is hundreds of papers published in mainstream journals, including 19 positive reviews of the field, over the last six years, the Wikipedia article excludes almost all the recent review and work, and often presents the topic with radical misinformation, clearly false -- which can be sourced from newspapers and other popular sources -- and because experts have been banned from the article (at least three), and non-experts who were aware of the current science were marginalized, edits reverted consistently, the article bias continues, and has been, in fact, getting worse.

'''When administrators intervene and see an expert as "biased, POV-pushing," they tend to warp the process. Experts are, in fact, biased.''' They tend to believe that they know the field they have studied more than those who have not, or who have only looked at cherry-picked sources. Whenever consensus process was followed on Wikipedia, with respect to material in this field, article content improved. But whenever experts were banned, and those who had tried to negotiate consensus were banned -- because they were fostering a lot of discussion, which is somehow considered disruptive -- article content declined.

Am I biased about the field? Perhaps. I started out with the general opinion, the "skeptical position." But then I read the sources, and I actually bought the books (on all sides of the controversy) and, it turns out, those who read the sources tend to have a different perspective than those who don't read them, who assume that anything that contradicts their opinions must be bogus, that peer-reviewers have been duped, that thousands of scientists are deluded, wishful thinking. Skeptics in the field have been unable to be published under peer review in relevant mainstream journals for at least five years. ("Skeptic" here means those who reject the entire field, not those who are normally skeptical about this or that result.)

Now, I know how to go to ArbComm, and I was successful going to ArbComm over the two issues I took there. But ArbComm isn't sophisticated, and tends to try to protect the majority. I.e., if there are two dozen editors yelling at one (me!), and even if ArbComm agrees with the point of the filing (what I was asking for), and implement it, they are worried that they will offend the two dozen, who include "valuable contributors." And, at the time of my last case, they had not clearly idenfied the two dozen as a "cabal," a faction that operated more or less coherently to exclude certain POV and reinforce their own. They did later, and that particular faction has been "defanged," more or less, but the operating principles haven't changed. ArbComm can, and will, shoot the messenger, and only a minority of arbitrators are inclined to stop it.

'''So I could go to ArbComm, but doing this in the presence of an opposed faction is quite time-consuming, and, eventually I concluded, just too much work. A single article isn't worth it!''' Now, repeat this over thousands of articles on controversial topics, with different experts or informed and interested users....

Wikipedia is failing at its basic principle of neutrality, because inadequate consensus structure was built, and such structure threatens the existing oligarchy, so it can't be built there. It's still possible on Wikiverity, I believe, or I wouldn't be active here. --Abd 16:35, 4 February 2011 (UTC)


text:
User talk:Tirtha banerjee

 Hello Tirtha banerjee, 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.

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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! --Ottava Rima (talk) 17:53, 4 February 2011 (UTC)

title: User talk:Tirtha banerjee


text:
Dental Anatomy, Histology and Embryology

Welcome to the stage 1 of dental studies... -

Department of Dental Anatomy [DA], Dental Histology [DH] and Embryology

*) Dental Anatomy [DA]

*) Dental Histology [DH]

*) Embryology


text:
User talk:Abymathews

 Hello Abymathews, 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.

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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! --Ottava Rima (talk) 19:00, 4 February 2011 (UTC)

title: User talk:Abymathews


text:
User talk:Celestine6

 Hello Celestine6, 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.

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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! --Ottava Rima (talk) 20:29, 4 February 2011 (UTC)

title: User talk:Celestine6


text:
User talk:Bk314159

 Hello Bk314159, 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.

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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! --Ottava Rima (talk) 21:00, 4 February 2011 (UTC)

title: User talk:Bk314159


text:
User talk:Boud

 Hello Boud, 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.

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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! --Ottava Rima (talk) 21:00, 4 February 2011 (UTC)

Thanks for the invitation
It's true that I've neglected Wikiversity. I should note that physics isn't my area of specialization, and despite some amount of conversation on the topic I haven't actually done general relativity calculations, so I would make a much better student than teacher for your course. I have a few very general comments, and I could be far off base on them: Wnt 20:29, 21 June 2011 (UTC)

 * Student level should be fine. Boud 13:08, 23 June 2011 (UTC)

 * As an instructor at a bricks-and-telescope university, your primary commitment is to your paying students. So my thought is that, however much you may organize notes here, your lectures to your students will be the best work you can do.  Therefore, I should ask: would you be willing to videotape them and contribute them here (or on Wikimedia Commons) as multimedia resources?  Likewise, would you be willing to scan the transparencies you draw, post your recommended reading list, give ISBNs and other details for the textbooks you use?  That way the larger community has an actual course to seed their organization, and can help in getting them organized into a form more easily read online. Wnt 20:29, 21 June 2011 (UTC)
 * i see university education as a human right under Article 26 of the Universal declaration of human rights, so IMHO the creeping McDonaldisation of the university system in Europe - allegedly related to the Bologna Process - should not affect my commitment to teaching them. Regular day students who do not fail subjects still (so far) do not have to pay fees AFAIK, though i admit that i have not followed the details of this human rights struggle. Anyway, to your concrete suggestions. This semester's course is finished, so it's too late to videotape. i can see arguments for and against videotaping next year (northern spring/summer 2012, most likely), but probably we can shift that discussion to the talk page once i start the main course page. i have some (brief) notes on my user page, i'll copy them over soon to the course page, but briefly, the pdf presentation (beamer presentation like powerpoint(TM) but with LaTeX + postscript source files) and WIMS exercises/exam constitute the main part of the course. The pdf extensively links to Wikipedia for in-depth reading. Anyway, this will make more sense once i have an approximate first draft in place. Boud 13:08, 23 June 2011 (UTC)

 * I do a lot of answering at w:Wikipedia:Reference desk/Science. I think that in general Wikiversity should try to make this resource as much a part of it as it is of Wikipedia.  Thus, people with questions about the course you design should eventually (once it's pretty well established) post their questions to the Wikipedia refdesk, including a link back to your course so that these questions can be found.  The course should point students in that direction, and give them instructions on how to tag their questions, and provide an index of questions by your students which have been answered there (together with instructions how to add their own to this list).  Of course, I'm confident you personally can answer questions as they arise, but I don't know if you'll always have the time, and multiple perspectives might help. Wnt 20:29, 21 June 2011 (UTC)
 * Connecting with the WP Reference desk is a good idea. Feel free to add a link if i forget :). As for me answering questions, best to redirect to the community - at the moment at Wikipedia IMHO there are a huge number of people competent to answer questions at this elementary level. Boud 13:08, 23 June 2011 (UTC)

 * My feeling is that Wikiversity needs much stronger teacher-student relationships, and that no course can ever be certified as complete unless it has actual students going through it from beginning to end who are in communication with the people writing the course. It isn't enough to have a nice design for a set of detailed lessons - Wikiversity has to have people actually doing their homework, taking their quizzes, and graduating the class, in order to become a reality.  (I know this apparently contradicts the previous point, but having both informal study groups and formal question periods seems necessary for education) Wnt 20:29, 21 June 2011 (UTC)
 * i suspect that you're right. In my case, next northern summer semester (Feb-June 2012, presumably), communicating with Wikiversity students "simultaneously" (except for students flying past in rockets at nearly unity speeds) with local students would probably not take up much extra time for me (though i don't have great expectations for local students participating wikiwise). Once you have a look at WIMS exercises, you'll hopefully agree that this sort of exercise is potentially vastly superior to multiple choice quizzes, though the actual exercises i've prepared so far significantly underuse what is available in WIMS (partly because i'm using an old version). Boud 13:08, 23 June 2011 (UTC)

I wish you success in all of your teaching endeavors.

Wnt 20:29, 21 June 2011 (UTC)

 * Thanks :). Let's get to work... Boud 13:08, 23 June 2011 (UTC)

title: User talk:Boud


text:
User talk:Richarderay

 Hello Richarderay, 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.

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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! --Leutha 21:05, 4 February 2011 (UTC)

title: User talk:Richarderay


text:
User talk:Sreenadh~enwikiversity

 Hello Sreenadh, 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.

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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
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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! --Leutha 21:05, 4 February 2011 (UTC)

Your account will be renamed
Hello,

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

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

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 00:03, 18 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:23, 19 April 2015 (UTC)

title: User talk:Sreenadh~enwikiversity


text:
User talk:Valentine731

 Hello Valentine731, 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.

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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! --Leutha 21:06, 4 February 2011 (UTC)

title: User talk:Valentine731


text:
User talk:Djhickey1

 Hello Djhickey1, 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.

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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! --Leutha 21:06, 4 February 2011 (UTC)

title: User talk:Djhickey1


text:
User talk:Jguindon~enwikiversity

 Hello Jguindon, 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.

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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! --Leutha 21:06, 4 February 2011 (UTC)

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title: User talk:Jguindon~enwikiversity


text:
User talk:D289999

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title: User talk:D289999


text:
User:Egm6341.s11.team3/sub3

= 3.1 Taylor Series vs. Lagrange approximation error =

Given
Given a function

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

$$ f(x) = sin(x) \quad $$ in interval $$ x \in [0, \pi] \quad $$, $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.1)
 * }

that is to be approximated by Lagrange polynomial, $$ f_4^L(x) \quad $$, at 5 nodes. Nodes were chosen to be equidistant $$ x_0 = 0, x_1 = \frac{\pi}{4}, x_2 = \frac{\pi}{2}, x_3 = \frac{3\pi}{4}, x_4 = \pi $$.

Moreover, the same function $$ f(x) \quad $$ was approximated with Taylor series, $$ f_n^T(x) \quad $$, around a point $$ \frac{3\pi}{8} \quad $$.

Find
At a chosen point $$ x = \frac{7\pi}{8} $$, find what degree 'n' of Taylor polynomial approximation would be needed to produce an error, lesser then Lagrange polynomial at 5 nodes.

or

Find 'n' such that $$ |e_n^T(\frac{7\pi}{8})| \le |e_4^L(\frac{7\pi}{8})| $$

Solution
We start with an error estimation for Lagrange interpolation at a point $$ x = \frac{7\pi}{8} $$.

From we see that

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

$$ f(x) - f_n^L(x) = e_n^L(x) =\frac{q_{n+1}(x)}{(n+1)!}f^{n+1}(\xi) $$ $$ where
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.2)
 * }
 * {| style="width:100%" border="0"

$$ q_{n+1}(x) = \prod_{j=0}^{n}(x-x_j) = (x-x_0)(x-x_1)(x-x_2).......(x-x_n) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.3)
 * }

for $$ n = 4 \quad $$, nodes of $$ x_0 = 0, x_1 = \frac{\pi}{4}, x_2 = \frac{\pi}{2}, x_3 = \frac{3\pi}{4}, x_4 = \pi $$, and $$ f(x) = sin(x) \quad $$

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

$$ e_4^L(x) =\frac{q_{5}(x)}{5!}f^{5}(\xi) = \frac{(x-0)(x-\frac{\pi}{4})(x-\frac{\pi}{2})(x-\frac{3\pi}{4})(x-\pi)}{5!}cos(\xi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle\!
 * }

since $$ \xi \in [x_0, x_n] = [x_0, x_5] = [0, \pi] \Rightarrow cos(\xi) \in [0, 1] \quad $$.

Assuming error to be maximum

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

$$ max(e_4^L(x)) \Rightarrow cos(\xi) = 1 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.4)
 * }

Therefore,

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

$$ e_4^L(x) = \frac{(x-0)(x-\frac{\pi}{4})(x-\frac{\pi}{2})(x-\frac{3\pi}{4})(x-\pi)}{5!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.5)
 * }

and at a point of an interest $$ x = \frac{7\pi}{8} $$

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

$$ e_4^L(x) = \frac{(\frac{7\pi}{8}-0)(\frac{7\pi}{8}-\frac{\pi}{4})(\frac{7\pi}{8}-\frac{\pi}{2})(\frac{7\pi}{8}-\frac{3\pi}{4})(\frac{7\pi}{8}-\pi)}{5!} = -0.0082 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle\!
 * }

Therefore,

$$ |e_4^L(x)| = 0.0082 \quad $$

Consequently, we start solving for error for Taylor Series. Error in Taylor Series is expressed as reminder $$ R \quad $$

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

$$ R_{n+1}(x) := \frac{(x-\hat x)^{n+1}}{(n+1)!}f^{n+1}(\xi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 3.1.6)
 * }

as it was mentioned before, since $$ \xi \in [x_0, x_n] = [x_0, x_5] = [0, \pi] \quad $$ and $$ |f^{(n)}(x)| \quad $$ is either $$ cos(x) \quad $$ or $$ sin(x) \Rightarrow |f^{(n+1)}(\xi)| \in [0, 1] \quad $$

Again, we assume a maximum error, that is $$ |f^{(n+1)}(\xi)| = 1 \quad $$

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

$$ e_n^T = \frac{(x-\hat x)^{n+1}}{(n+1)!} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.1.7)
 * }

Since we know the point around witch we are computing Taylor series from a given parameters $$ (\hat x = \frac{3\pi}{8} )$$ and the point that we want to find an approximation for $$( x = \frac{7\pi}{8} )$$, we have

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

$$ e_n^T = \frac{(\frac{7\pi}{8}-\frac{3\pi}{8})^{n+1}}{(n+1)!} = \frac{(\frac{\pi}{2})^{n+1}}{(n+1)!}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.1.8)
 * }

Then we equate the errors to find the order of Taylor Series polynomial that we need in order to produce results at least as good as 5 node Lagrange polynomial.

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

$$ |e_n^T(\frac{7\pi}{8})| \le |e_4^L(\frac{7\pi}{8})| $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ \Rightarrow \frac{(\frac{\pi}{2})^{n+1}}{(n+1)!} \le 0.0082 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.1.9)
 * }

Since this can't be solved by analytical methods, we used Matlab to find 'n' value for us

$$ n = 6, \quad |e_n^T(\frac{7\pi}{8})|=0.0047 $$

Therefore, you need Taylor Series of $$ 6^{th} \quad $$ order $$ (n = 6) \quad $$ to approximate a function at least as good as Lagrange Series at $$ 4^{th} \quad $$ order $$ (n = 4) \quad $$.

--Egm6341.s11.team3.JA 22:29, 9 February 2011 (UTC)

= 3.2 Error Analysis of Lagrange Interpolation at nodes =

Given

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

$$ G(x) = q_{n+1}(x) d(x) \frac{}{}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.2.1)
 * }

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

$$ q_{n+1}(x) = \prod_{j=0}^{n}(x-x_j) = (x-x_0)(x-x_1)(x-x_2).......(x-x_n) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.2.2)
 * }

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

$$ d(x) = \frac{e(x)}{q_{n+1}(x)} - \frac{e(t)}{q_{n+1}(t)}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.2.3)
 * }

Find

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

$$ G(x_i) \frac{}{}$$ for $$ i=0,1,...,n \frac{}{}$$
 * style="width:95%" |
 * style="width:95%" |
 * }

Solution
Substituting the point of interest into Equation 3.2.2 yields:

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

$$ q_{n+1}(x_i) = \prod_{j=0}^{n}(x_i-x_j) = (x_i-x_0)(x_i-x_1)(x_i-x_2).......(x_i-x_n) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.2.4)
 * }

When variable i is substituted into Equation 3.2.4, the result becomes:

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

$$ q_{n+1}(x_0) = \prod_{j=0}^{n}(x_0-x_j) = (0)(x_0-x_1)(x_0-x_2).......(x_0-x_n) = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.2.4.1)
 * }

i=1:
 * {| style="width:100%" border="0"

$$ q_{n+1}(x_1) = \prod_{j=0}^{n}(x_1-x_j) = (x_1 - x_0)(0)(x_1-x_2).......(x_1-x_n) = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.2.4.2)
 * }

Until finally i=n:
 * {| style="width:100%" border="0"

$$ q_{n+1}(x_n) = \prod_{j=0}^{n}(x_n-x_j) = (x_n - x_0)(x_n - x_1)(x_n-x_2).......(0) = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.2.4.n)
 * }

Thus, for each value of variable i Equation 3.2.4 becomes:
 * {| style="width:100%" border="0"

$$ q_{n+1}(x_i) = \prod_{j=0}^{n}(x_i-x_j) = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.2.5)
 * }

Substituting the point of interest into Equation 3.2.1 yields:

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

$$ G(x_i) = q_{n+1}(x_i) d(x_i) \frac{}{}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.2.6)
 * }

Substituting Equation 3.2.5 into Equation 3.2.6 yields the final desired answer:
 * {| style="width:100%" border="0"

$$ G(x_i) = (0) d(x_i) = 0 \frac{}{}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.2.7)
 * }

$$\therefore G(x_i) = 0$$

Egm6341.s11.team3.russo 01:43, 15 February 2011 (UTC)

= 3.3.1 Plotting figures similar to proof of LIET =

Given
Given a function

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

$$ f(x) = \frac{exp(x)-1}{x} \quad $$ in interval $$ x \in [-1, 1] \quad $$, $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.3.1)
 * }

Find
Find by Lagrange polynomial,weights and $$ q(n+1) \quad $$ for, $$ L_2,4^L(x) \quad $$, at 5 nodes. Nodes were chosen to be equidistant $$ x_0 = -1, x_1 = \frac{-1}{2}, x_2 = \frac 0, x_3 = \frac{1}{2}, x_4 = 1 $$.

also find errors at points $$ x = 0.75 \quad $$ and $$ t = 5 \quad $$

Solution
Consider a Lagrange method on approximating exact function:

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

f_n^L\left( x \right)=\,\sum_{i=0}^{n}l_{i,n}\left ( x \right)f\left ( x_i \right)$$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3.3.2)
 * }
 * }

For given 5 nodes at $$ n = 4 \rightarrow x_{0} = -1, x_{1} = -0.5, x_{2} = 0, x_{3} = 0.5, x_{4} = 1 $$.

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

f_{4}^L=\sum_{i=0}^{4}l_{i,4}(x)f(x_{i})=l_{0,4}(x)f(x_{0})+l_{1,4}(x)f(x_{1})+l_{2,4}(x)f(x_{2})+l_{3,4}(x)f(x_{3})+l_{4,4}(x)f(x_{4}) $$ $$ where
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle\!
 * }
 * }
 * {| style="width:100%" border="0" align="left"

l_{i,n}=\prod_{j=0,j\neq 0}^{4}\frac{x-x_{j}}{x_{i}-x_{j}} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 3.3.3)
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{0,4}=\prod_{j=0,j\neq 0}^{4}\frac{x-x_{j}}{x_{0}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{1,4}=\prod_{j=0,j\neq 1}^{4}\frac{x-x_{j}}{x_{1}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{2,4}=\prod_{j=0,j\neq 2}^{4}\frac{x-x_{j}}{x_{2}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{3,4}=\prod_{j=0,j\neq 3}^{4}\frac{x-x_{j}}{x_{3}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

l_{4,4}=\prod_{j=0,j\neq 4}^{4}\frac{x-x_{j}}{x_{4}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

For example:

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

l_{0,4}=\prod_{j=0,j\neq 0}^{4}\frac{x-x_{j}}{x_{0}-x_{j}} = \frac{x-x_1}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}}\frac{x-x_{3}}{x_{0}-x_{3}}\frac{x-x_{4}}{x_{0}-x_{4}} = \left(\frac{x-(-0.5)}{(-1)-(-0.5)} \right) \left(\frac{x-0}{(-1)-0} \right ) \left( \frac{x-0.5}{(-1)-0.5} \right) \left (\frac{x-1}{(-1)-1} \right ) $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Since matlab does not form the matrices/vectors for the symbolic and numeric values, we did a sample run for $$ l_{0,4}= \quad $$ to check the output for the expected numerical values above. Note, since we used loops to populate for values of $$ l \quad $$ we denoted $$ x_j = x_i \quad $$ case to be $$ 1 \quad $$ so that we could run the loops smoothly and we would not get the answer of $$ \frac{x-x_j}{\inf} $$.

These values did came out to be the same as the expected values from the equation above

Noting that $$ q_{n+1} \quad $$ is

$$\displaystyle q_{n+1}(x) := (x - x_0)(x - x_1)(x - x_2)(x - x_3)\cdots(x - x_n) $$
$$\displaystyle (Eq. 3.3.4) $$

and that error is

$$\displaystyle e_T^L = f(x) - f^L(x) \quad $$
$$\displaystyle (Eq. 3.3.5) $$

we generated matlab code to calculate and plot parameters such as $$ q_{n+1}, \ e_T^L(x), \  e_T^L(t), l_{i,n}, f^L \quad $$

as you may see, the answers came out to be

it is clear that q part hits all roots at specified nodes. Also, it may be seen that true and approximate values are really close to each other and overlap. From that image is hard to see if the approximation does cross the exact function at the nodes (note, we used the same function for n=8 where more oscillations are present and it clearly shows approximate function passing through the nodes).

We attached another figure showing the numerical values along with approximate values, just to show that the 2 graphs do follow 1 another really close

Errors at x = 0.75 and t = 5 came out to be:

$$ e_4^L(x = 0.75) = -1.6274*10^{-4} \quad $$

$$ e_4^L(t = 5) = 11.0239 \quad $$

It can be clearly seen that error inside the bound interval is really small. Therefore, functions are close to each other. On the other hand for t= 5 (outside the function bounds), error was relatively huge (note the function maximum value was only around 1.8). This suggests that Lagrange interpolation is a poor tool to extrapolate the function.

--Egm6341.s11.team3.JA 20:47, 16 February 2011 (UTC)

= 3.3.2 Runge-Phenomenon =

Given
Given a function

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

$$ f(x) = \frac{1}{1+4x^2}\quad $$ in interval $$ x \in [-5,5] \quad $$, $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.3.2)
 * }

Find
Find the error estimate for the function given above and plot, $$ L_3,8^L(x) \quad $$, for n=8 at equidistant nodes.

Solution
For $$ n = 8 \quad $$ nodes are:

$$x_{0} = -5, x_{1} = -3.75, x_{2} = -2.5, x_{3} = -1.25, x_{4} = 0, x_{5} = 1.25, x_{6} = 2.5, x_{7} = 3.75, x_{8} = 5 \quad $$.

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

 * $$\displaystyle
 * $$\displaystyle

f_{8}^L=\sum_{i=0}^{8}l_{i,8}(x)f(x_{i})=l_{0,8}(x)f(x_{0})+l_{1,8}(x)f(x_{1})+l_{2,8}(x)f(x_{2})+l_{3,8}(x)f(x_{3})+l_{4,8}(x)f(x_{4})+l_{5,8}(x)f(x_{5})+l_{6,8}(x)f(x_{6})+l_{7,8}(x)f(x_{7})+l_{8,8}(x)f(x_{8}) $$ $$
 *
$$\displaystyle (Eq.3.3.2.1)
 * }
 * }

where,

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

l_{3,8}=\prod_{j=0,j\neq 3}^{8}\frac{x-x_{j}}{x_{3}-x_{j}} $$
 * $$\displaystyle
 * $$\displaystyle
 * }
 * }

Using the same basic equations as in part 1 $$( Eq. 3.3.2 - 5 )  \quad $$ and relying on previous matlab code, we generated another matlab code

As you may see, high degree oscillations are apparent in this plot and oscillations do increase in frequency towards the endpoints of the function interval as well as increase in amplitude. This phenomenon (increase in oscillations towards the endpoints as number of members/nodes is increased) is known as Runge's phenomenon

Also note from the plot in 1st part and 2nd part(this part), that at whatever node we choose to plot $$ l_{i,n} \quad $$, for example $$ l_{3,8} \quad $$, as in this case, all the nodes except for $$ l_{3,8} \quad $$ node hit zero while the node $$ l_{3,8} \quad $$ hits value of $$ 1 \quad $$. The same was for $$ l_{2,4} \quad $$ node in 1st part of Problem 3.3.

$$ e_8^L(x) = -0.3786 \quad $$

$$ e_8^L(t) = 1.7840 \quad $$

--Egm6341.s11.team3.JA 20:48, 16 February 2011 (UTC)

= 3.4 Area of Biofolium =

Given
Cartesian coordinate

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

$$((x^2+y^2)^2)-4axy^2=0 \quad $$, $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle\!
 * }
 * }

when $$ a= 0.5 \quad $$

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

$${({x}^{2}+{y}^{2})}^{2}=2x{y}^{2} \quad $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\! $$
 * }

Polar Coordination

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

$$ x=rcos(t) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$y=rsin(t) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$r=2cos(t){sin(t)}^{2} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

Find
1)Do literature search to find history and application of this classic curve

2)Find area in one leaf to $${10}^{-6}$$ accuracy

2.1) Trapozoial Rule

2.2) Simpson's Rule

2.3) Summation of Triangles

Solution
1) Refer the Web below 2) Area of Leaf

From Wolframalpha

Way to Calculation by Trapozoial Rule and Simpson's Rule

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

$${({x}^{2}+{y}^{2})}^{2}=2x{y}^{2} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4.1)
 * }

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

$$y=\sqrt{{-x}^{2}+\sqrt{{x}^{2}-2{x}^{3}}+x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4.2)
 * }

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

$$y=\sqrt{{-x}^{2}-\sqrt{{-x}^{2}(2x-1)}+x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4.3)
 * }

2.1)Trapozoial Rule

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

$${I}_{1}=\frac{b-a}{2}[f(a)+f(b)]$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4.4)
 * }

by (Eq. 3.4.2)

$$f(0)=0, f(0.5)=0.5\quad$$

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

$${I}_{1}=\frac{0.5-0}{2}[0+0.5]=0.125$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4.5)
 * }

by (Eq 3.4.3)

$$f(0)=0, f(0.5)=0.5\quad$$

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

$${I}_{1}=\frac{0.5-0}{2}[0+0.5]=0.125$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4.6)
 * }

Area of One Leaf by Trapozoial Rule $$0.125-0.125=0\quad$$

$$
 *
 $$ \displaystyle\!

2.2) Simpson's Rule

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

$${I}_{2}=\frac{h}{3}[{f}_{0}+4{f}_{1}+{f}_{2}]$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4.7)
 * }

$${x}_{0}= a = 0, {x}_{1}= \frac{{x}_{2}+{x}_{0}}{2}=0.25, {x}_{2}= b = 0.5\quad$$

$$h=\frac{b-a}{2}=0.25$$

by Eq(3.4.2)

$$f({x}_{0})= 0, f({x}_{1})= 0.603553, f({x}_{2})= 0.5 \quad$$

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

$${I}_{2}=\frac{0.25}{3}[0+4(0.603553)+0.5] = 0.242851 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4.8)
 * }

by (Eq 3.4.3)

$$f({x}_{0})= 0, f({x}_{1})= 0.103553, f({x}_{2})= 0.5 \quad$$

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

$${I}_{2}=\frac{0.25}{3}[0+4(0.103553)+0.5] = 0.076184 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4.9)
 * }

Area of One Leaf by Simpson's Rule

$$ 0.242851-0.076184 = 0.166667 \quad$$ $$
 *
 $$ \displaystyle\!

2.3)Summation of Triangles

$${x}_{0}=0,{x}_{1}=0.125,{x}_{2}=0.25,{x}_{3}=0.375,{x}_{4}=0.5\quad$$

by (Eq 3.4.2)

$$f(0)=0, f(0.125)=0.466506, f(0.25)=0.603553, f(0.375)=0.649519, f(0.5)=0.500000 \quad$$

Length between two points

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

$$\sqrt{{({x}_{1}-{x}_{0})}^{2}+{({y}_{1}-{y}_{0})}^{2}}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4.10)
 * }

$$\overline{ox_1}=0.482962$$

$$\overline{ox_2}=0.653281$$

$$\overline{ox_3}=0.749999$$

$$\overline{ox_4}=0.707106$$

$$\overline{{x}_{1}{x}_{2}}=0.185490$$

$$\overline{{x}_{2}{x}_{3}}=0.133183$$

$$\overline{{x}_{3}{x}_{4}}=0.194886$$

Heron's formula

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

$$ A=\sqrt{s(s-a)(s-b)(s-c)}$$
 * style="width:95%" |
 * style="width:95%" |

$$ s=\frac{a+b+c}{2}$$

$$ upperline area
 *
 $$ \displaystyle (Eq. 3.4.11)
 * }

$$ \overline{o{x}_{1}{x}_{2}}= 0.020590$$

$$ \overline{o{x}_{2}{x}_{3}}= 0.031976 $$

$$ \overline{o{x}_{3}{x}_{4}}= 0.068629 $$

Summation of the Triangles of Upper Parts

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

$$Area = 0.020590 + 0.031976 + 0.068629=0.121195\quad$$
 * style="width:95%" |
 * style="width:95%" |

$$
 *
 $$ \displaystyle (Eq. 3.4.12)
 * }

by (Eq 3.4.3)

$$f(0)=0, f(0.125)=0.033493, f(0.25)=0.0.03553, f(0.375)=0.216506, f(0.5)=0.500000\quad$$

Length between two points

by (Eq 3.4.10)

$$\overline{o{x}_{1}}=0.129409$$

$$\overline{o{x}_{2}}=0.270597$$

$$\overline{o{x}_{3}}=0.433012$$

$$\overline{o{x}_{4}}=0.707106$$

$$\overline{{x}_{1}{x}_{2}}=0.143294$$

$$\overline{{x}_{2}{x}_{3}}=0.168473$$

$$\overline{{x}_{3}{x}_{4}}=0.309828$$

by (Eq 3.4.11)

$$ \overline{o{x}_{1}{x}_{2}}= 0.002285$$

$$ \overline{o{x}_{2}{x}_{3}}= 0.007646 $$

$$ \overline{o{x}_{3}{x}_{4}}= 0.039623 $$

Summation of the Triangles of lower Parts

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

$$Area = 0.002285 + 0.007646 + 0.039623 = 0.049554\quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.4.13)
 * }

Sum of (Eq 3.4.12) and (Eq 3.4.13)

Total Area by Sum of the Triangles

Area of Leaf = 0.121195 + 0.049554 = 0.170749 $$
 *
 $$ \displaystyle\!

Results

Area by Trapozoial Rule = 0

Area by Simpson's Rule = 0.166667

Area by Sum of Triangles = 0.170749

$$
 *
 $$ \displaystyle\!

– Hard to get exact area of leaf by Trapozoial Rule –

Egm6341.s11.team3.sungsikkim 03:39, 14 February 2011 (UTC) Egm6341.s11.team3.sungsikkim 07:07, 15 February 2011 (UTC)

= 3.5 Deriving the n+1th order derivative of En  =

Given
Given a function

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

$$
 * style="width:95%" |$$E_{n}=I(F)-I_{n}(F) \quad$$,
 *
 $$ \displaystyle\!
 *
 $$ \displaystyle\!
 * }

Find
Show that

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

$$
 * style="width:95%" |$$E_{n}^{n+1}\left ( x \right )=F^{n+1}\left ( x \right)-0$$,
 *
 $$ \displaystyle\!
 *
 $$ \displaystyle\!
 * }

Solution

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

$$ E_{n}^{n+1}\left ( x \right )=F^{n+1} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.1)
 * }

We start with an general equation for error estimation

From we learn that

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

$$E_{n} = I-I_{n} \quad $$ $$ the first derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1)
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.2)
 * }
 * {| style="width:100%" border="0"

$$E_n^{1}=\int_{0}^{1}f^{1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{1}\left ( x \right )dx\, \quad $$ $$ the second derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1)
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.3)
 * }
 * {| style="width:100%" border="0"

$$E_n^{2}=\int_{0}^{1}f^{2}\left ( x \right )dx-\int_{0}^{1}f_{n}^{2}\left ( x \right )dx\, \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.4)
 * }

the third derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1)

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

$$E_n^{3}=\int_{0}^{b}f^{3}\left ( x \right )dx-\int_{0}^{b}f_{n}^{3}\left ( x \right )dx\, \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.5)
 * }

the fourth derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1)

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

$$E_n^{4}=\int_{0}^{1}f^{4}\left ( x \right )dx-\int_{0}^{1}f_{n}^{4}\left ( x \right )dx\, \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.6)
 * }

the nth derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1)

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

$$E_n^{n}=\int_{0}^{1}f^{n}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n}\left ( x \right )dx\, \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.7)
 * }

the n+1 th derivative of $$E_{n}\,$$ would yield us the following result for an interval (0,1)

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

$$E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n+1}\left ( x \right )dx\, \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.8)
 * }

we have learned that

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

 * style="width:95%" |
 * style="width:95%" |

$$f_{n}\left ( x \right )=\sum _{i=0}^{n}l_{i,n}\left ( x \right ) f\left ( x_i \right)\,$$ $$
 *
 $$ \displaystyle (Eq. 3.5.9)
 * }

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

 * style="width:95%" |
 * style="width:95%" |

where $$l_{i,n}\left ( x \right)= \prod_{j=0,i \neq j}^{n}\frac{x-x_{j}}{x_{i}-x_{j}}\,$$ $$
 *
 $$ \displaystyle (Eq. 3.5.10)
 * }

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

$$ \begin{align} \\n=1, \quad & l_{i,1} (x)=\frac{x-x_{1}}{x_{0}-x_{1}}  l_{0}=c_1x+c_0 \\ \\n=2, \quad & l_{i,1} (x)=\frac{x-x_{1}}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}}  l_{0}=c_2x^2 + c_1x + c_0
 * style="width:95%" |
 * style="width:95%" |

\\.\, \\.\, \\n=n, \quad & l_{i,n} ( x )=\frac{x-x_{1}}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}}...\frac{x-x_{n-1}}{x_{0}-x_{n-1}}\frac{x-x_{n}}{x_{0}-x_{n}}  l_{0}= c_nx^n+c_{n-1}x^{n-1}++c_{n-2}x^{n-2}+......+c_{1}x + c_0 \end{align} $$ $$
 *
 $$ \displaystyle\!
 * }

where $$ c_n, c_{n-1}, c_{n-2},.... c_2, c_1, c_0 \quad $$ are constants

since $$f\left( x_{i} \right)\,$$ is constant, hence forth  $$f_n\left( x \right)\,$$ is is of highest order of $$x^{n}\,$$

Consequently,

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

$$f_n^{n+1}\left ( x \right )= \frac{d}{dx^{n+1}}(c_nx^n+c_{n-1}x^{n-1}++c_{n-2}x^{n-2}+......+c_{1}x + c_0) = 0\,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.11)
 * }

Finally,

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

$$E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n+1}\left ( x \right )dx\,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.12)
 * }

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

$$E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1} 0 dx\,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.13)
 * }

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

$$ E_n^{n+1}=\!F^{n+1}\left ( x \right )dx- 0 \,$$ $$ Therefore,
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.14)
 * }

$$ E_n^{n+1}=\!F^{n+1}\left ( x \right )dx- 0 \,$$

Egm6341.s11.team3.rakesh 20:32, 11 February 2011 (UTC)

= 3.6 Deriving the n+1th order derivative of Q n+1th order =

Find
Prove that $$q_{n+1}^{(n+1)}\left ( x \right )={(n+1)}!$$

Given
$$q_{n+1}\left ( x \right )=(x-x_{0})(x-x_{1})(x-x_{2})...(x-x_{n}) $$

Solution
We know from our lectures

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

$$q_{n+1}\left ( x \right )=(x-x_{0})(x-x_{1})(x-x_{2})...(x-x_{n})$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.1)
 * }

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

$$n=0 \,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$q_{1}\left ( x \right )=(x-x_{0})=x^1+ a_0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.2)
 * }

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

$$q_{1}^{(1)}\left ( x \right )=1$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.3)
 * }

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

$$n=1 \,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$q_{2}\left ( x \right )=(x-x_{0})(x-x_{1})= x^2 + a_1x^1+ a_0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.4)
 * }

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

$$q_{2}^{(2)}\left ( x \right )=2\times 1=2!$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.5)
 * }

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

$$n=2 \,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$q_{3}\left ( x \right )=(x-x_{0})(x-x_{1})(x-x_{2})= x^3 + a_2x^2 + a_1x^1+ a_0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.6)
 * }

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

$$q_{3}^{(3)}\left ( x \right )=3\times 2\times 1=3!$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.7)
 * }

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

$$n=3 \,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$q_{4}\left ( x \right )=(x-x_{0})(x-x_{1})(x-x_{2})(x-x_{3})$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.8)
 * }

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

$$q_4 = (x-x0)(x-x1)(x-x3)(x-x4) = x^4 + a_3x^3 + a_2x^2 + a_1x^1+ a_0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.9)
 * }
 * {| style="width:100%" border="0"

$$q_{4}^{(4)}\left ( x \right )=4\times 3\times 2\times 1=4!$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.10)
 * }

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

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

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

$$n=n \,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$q_{n+1}\left ( x \right ):=(x-x_{0})(x-x_{1})(x-x_{2})(x-x_{3}...(x-x_{n}) = x^n + a_{n-1}x^{n-1} + ....+ a_2x^2 + a_1x^1+ a_0 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.11)
 * }

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

$$q_{n+1}^{(n+1)}\left ( x \right )=n\times (n-1)\times (n-2)...\times 1=(n+1)!$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.6.12)
 * }

where constants $$ a_0, a_1, a_2,....., a_n \quad $$ are combination of constants $$ x_0, x_1, x_2,....., x_n \quad $$

Therefore,

$$ P_{n+1}^{(n+1)}\left ( x \right )={(n+1)}!$$

Egm6341.s11.team3.rakesh 20:17, 11 February 2011 (UTC)

= 3.7 Integration Derivation for Error of Trap. rule =

Given
Give the Integration for Error of Trap. rule:
 * {| style="width:100%" border="0"

$$ \frac{M_2}{2!} \int_{a}^{b} \left | (x-a)(x-b) \right | dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.7.1)
 * }

Find
Derive the value of the integration in terms of $$ a \quad $$, $$ b \quad $$ and $$ M_2 \quad $$.

Solution

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

$$ \frac{M_2}{2!} \int_{a}^{b} \left | (x-a)(x-b) \right | dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{2!} \int_{a}^{b} (x-a)(b-x) dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{2} \int_{a}^{b} [-x^2+(a+b)x-ab] dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{2} [-\frac{1}{3}x^3+\frac{1}{2}(a+b)x^2-abx]\mid _{a}^{b}  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{2} [-\frac{1}{3}(b^3-a^3)+\frac{1}{2}(a+b)(b^2-a^2)-ab(b-a)] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{12} (b^3-3b^2a+3a^2b-a^3) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_2}{12} (b-a)^3 $$ $$ Therefore,
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

$$ \frac{M_2}{2!} \int_{a}^{b} \left | (x-a)(x-b) \right | dx = \frac {(b-a)^3}{12} M_2 $$

Egm6341.s11.team3.Xia 19:00, 11 February 2011 (UTC)

= 3.8 Integration Derivation for Error of Simple Simpson's rule =

Given
Give the Integration for Error of Simple Simpson's rule:
 * {| style="width:100%" border="0"

$$ \frac{M_3}{3!} \int_{a}^{b} \left | (x-a)(x-\frac{a+b}{2})(x-b) \right | dx = \frac {(b-a)^4}{192} M_3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.8.1)
 * }

Find
Derive the value of the integration in terms of $$ a \quad $$, $$ b \quad $$ and $$ M_3 \quad $$.

Solution

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

$$ \frac{M_3}{3!} \int_{a}^{b} \left | (x-a)(x-\frac{a+b}{2})(x-b) \right | dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{3!} [\int_{a}^{\frac{a+b}{2}} (x-a)(x-\frac{a+b}{2})(x-b) dx + \int_{\frac{a+b}{2}}^{b} (x-a)(x-\frac{a+b}{2})(b-x) dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{3!} (\int_{a}^{\frac{a+b}{2}} - \int_{\frac{a+b}{2}}^{b}) [x^3-\frac{3}{2}(a+b)x^2+\frac{a^2+4ab+b^2}{2}x- \frac{a^2b+b^2a}{2}] dx $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{3!} [\frac{1}{4}x^4-\frac{1}{2}(a+b)x^3+\frac{a^2+4ab+b^2}{4}x^2- \frac{a^2b+b^2a}{2}x] (\mid_{a}^{\frac{a+b}{2}} + \mid_{b}^{\frac{a+b}{2}}) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{2M_3}{6} [\frac{1}{64}(a+b)^4-\frac{1}{16}(a+b)(a+b)^3+\frac{a^2+4ab+b^2}{16}(a+b)^2- \frac{a^2b+b^2a}{4}(a+b)] $$ $$ - \frac{M_3}{6} [\frac{1}{4}a^4-\frac{1}{2}(a+b)a^3+\frac{a^2+4ab+b^2}{4}a^2- \frac{a^3b+b^2a^2}{2}] $$ $$ - \frac{M_3}{6} [\frac{1}{4}b^4-\frac{1}{2}(a+b)b^3+\frac{a^2+4ab+b^2}{4}b^2- \frac{a^2b^2+b^3a}{2}] $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{192} (a^4-4a^3b-10a^2b^2-4ab^3+b^4) - \frac{M_3}{6} (-\frac{a^2b^2}{2}) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{192} (a^4-4a^3b+6a^2b^2-4ab^3+b^4) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

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

$$ = \frac{M_3}{192} (b-a)^4 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle\!
 * }

Therefore,

$$ \frac{M_3}{3!} \int_{a}^{b} \left | (x-a)(x-\frac{a+b}{2})(x-b) \right | dx = \frac {(b-a)^4}{192} M_3 $$

Egm6341.s11.team3.Xia 20:00, 11 February 2011 (UTC)

= References =

= Contributing Members =

--Egm6341.s11.team3.rakesh 15:18, 14 February 2011 (UTC)

--Egm6341.s11.team3.sungsikkim 17:12, 14 February 2011 (UTC)

--Egm6341.s11.team3.ren 14:22, 15 February 2011 (UTC)

--Egm6341.s11.team3.Xia 18:35, 15 February 2011 (UTC)

--Egm6341.s11.team3.elango 20:00, 15 January 2011 (UTC)

--Egm6341.s11.team3.JA 20:48, 16 February 2011 (UTC)

--Egm6341.s11.team3.russo 20:57, 16 February 2011 (UTC)

title: User:Egm6341.s11.team3/sub3