text:
User talk:Mu301/Archive 2011

Recreation
Yet another recreation. Ottava Rima (talk) 22:07, 6 January 2011 (UTC)
 * and more. Ottava Rima (talk) 22:08, 6 January 2011 (UTC)
 * Cool. Now, what do we do? Range blocks? You know where that road goes. Endless maintenance. I'm going to do something with these IP-created pages, and we'll see what happens. I'll yank the speedy tags as part of this. I want to see if I can snag these users and encourage them to participate. Feel free to copy what I did, you can use your user space or mine. --Abd 03:10, 7 January 2011 (UTC)

 * I'm covering this, now that I can revision-delete. It may take some time for these (kids?) to get it, I'm willing to be patient for a while. After all, I have seven children and five grandchildren, I've seen a bit in my time. --Abd 21:12, 12 January 2011 (UTC)

Seriously?
. Ottava Rima (talk) 19:13, 9 January 2011 (UTC)

Request for you to step down
Please see. Since there is no Bureaucratship policy, no one can make up statements about a Crat being required to remove a Crat, and seeing as how Bureaucratship requires a super majority of support, it seems highly unlikely that you will be allowed to continue. You can end unnecessary drama by resigning your Bureaucratship now. Ottava Rima (talk) 15:54, 30 January 2011 (UTC)

Cold fusion categorization
I noticed that you added and removed category Cold fusion from a subpage under that topic. Do you think that all resources under Cold fusion should have the category? You removed the category in this case, I think, because it was a Talk page, and that's correct, I believe.

I'm motivated to consider, here, what is appropriate as to categorization. A category for subpages is a bit redundant, but would still have some use, I suppose. Is this worth the maintenance labor?

Anyway, thanks for supporting this resource with routine maintenance, as well as by occasional discussion. --Abd 16:36, 6 January 2011 (UTC)

 * Yes, I had intended to tag the page, and not the talk page. My feeling is that the more methods of helping people find resources the better off we are. I would not consider a category of subpages to be redundant, as some people might use different methods to search for information. I seem to recall that there is also a dynamic page list feature to automatically generate a list of subpages. --mikeu talk 20:52, 6 January 2011 (UTC)

Anyway, Cold fusion will be, my hope, a demonstration of how Wikiversity can address real-world controversies, using and developing and exploring them, in depth. As with any academic process, this might even eventually help to resolve some controversies, because certain controversies are maintained by lack of such detailed exploration. At the very least, we may develop an understanding of the basis for controversy, why people are lined up on this side or that, with others scratching their heads. --Abd 16:40, 6 January 2011 (UTC)
 * It's also a good example of issues arising in the Sociology of Science. —Caprice 22:20, 6 January 2011 (UTC)

On moving to project space
I'm fine with moving User talk:SB Johnny/CR CR to the project space. I worry a little bit that it will be a magnet for tl;dr post once it goes there. Followed by tl;dr arguments about how tl;dr is a good thing, and tl;dr rants about abuse if the tl;dr comments are removed.

Just sayin', but I suppose there's no way to find out aside from finding out :-). --SB_Johnny talk 19:59, 3 February 2011 (UTC)

 * tl;dr --mikeu talk 20:42, 3 February 2011 (UTC)

 * I'm in no hurry to move it. Perhaps we should let folks chime in as they notice it for a short while, and then move it later? WV:NBD --mikeu talk 20:42, 3 February 2011 (UTC)
 * Don't move it there yet. See how it develops, work on the process, managing it as "chair." Consider that you have formed a CR CR committee. If I don't like your committee, I can form my own damn committee! But as long as you allow me to participate in yours, why should I bother?
 * Your CR CR can be announced on the Colloquium, being explicit that it's not binding on anything until moved to CR subspace and announced in the site message (and you should not personally put it there!). If there are any polls there, pending, they are just "committee polls," to make temporary decisions about process, subject to your approval. What becomes irrelevant can be archived to page history or moved to page Talk. --Abd 22:10, 3 February 2011 (UTC)

We've already had one person getting angry, it seems :-). The beauty of userspace is that the user can set the rules, or so I've been told. --SB_Johnny talk 22:30, 3 February 2011 (UTC)
 * Who got angry? Yes. The beauty of user space. I've used it on Wikipedia, I ran a "self-RfC" there, it was effective to resolve a dispute, it was used to then negotiate with Jehochman, and he caved and apologized (without admitting bad faith, which I wasn't alleging) and we became good friends, even met off-wiki. Of course, there were the usual suspects screaming about how crazy it was to have a user manage his own RfC, and they tried to delete it, but it didn't fly. The RfC was presented as being for the purpose of advising me. Not binding on anyone else. --Abd 23:50, 3 February 2011 (UTC)

Abuse of ops
This is a clear and egregious abuse of ops.

You gave SB Johnny rights without consensus. The community responded and shot that down. You have no right to try and use ops to fight against the community.

Policy is very clear and states that to retain them he must have consensus in support. This is clearly not what he had, and you put an archive box on a discussion which policy does not allow for and protect a page in clear defiance of Wikiversity tradition.

You have violated just about every possible policy we have. Ottava Rima (talk) 17:16, 8 February 2011 (UTC)

Privacy
See here.

Mabuabsdd
If you didn't notice, Mabuabsdd is part of the Aaqib group, the one we use to just block over the past two years instead of allowing them to spread there nonsense into dozens of pages. Now that Abd no longer has Custodian rights, can we clean up the "play area" and stop the kids from running amok? Ottava Rima (talk) 15:55, 1 February 2011 (UTC)
 * Well, not having ops does make it more difficult for me to police that "group." This is apparently a number of elementary school kids, though it's a bit possible they are middle school. Ottava Rima should use Request custodian action like any other mortal, and not solicit individual administrator action based on private argument.

 * There has been some (limited) success in encouraging these kids to communicate. See, however, Playspace for the proposal, and User:Abd/Playspace for my personal implementation. These kids are, if nothing else, learning to use wikitext. The largest problem is with Privacy, as these are apparently minors and may sometimes use real names. How would we know? However, massive intervention to prevent this will involve blocking school IP, which is not exactly a good course to pursue if we want to encourage schools to participate in Wikiversity. I've thought of contacting the school, but would want our community consensus before doing so.

 * No attention was paid in the recent CR on me to what I'd actually been doing as a custodian.... only to the very visible flap Ottava was creating. I was pursuing a graduated and cautious, friendly but firm, approach to dealing with these kids, encouraging them to keep their "games" out of mainspace, shoving it into user space where it was less disruptive and required a bit less attention. I will be responsible for pages in User:Abd/Playspace, but I can no longer revision delete, necessary to handle revelation of names. I'd appreciate guidance on how to handle this, now that I don't have ops. I suppose I can blank, use speedy deletion tags. This is not quite the emergency that "outing" represents, we are not obligated to instant response, my opinion, and revdel need not be instant. It may be enough that names are confined to history, for a short time.

 * I'll ask how successful blocking the Aaquib group was. That procedure requires constant sysop attention. If my approach fails, it won't make things worse, merely return them to the prior status quo. Ottava has a very different philosophy about Wikiversity, much more limited to a no-original-research, all-knowledge-is-officially approved, approach. Lousy with kids nowadays. I do have seven kids and six grandchildren.... At one point I had five teenagers, proof of my obvious insanity. --Abd 17:10, 1 February 2011 (UTC)

Deletion riquired! www.@mdj.com
Do not acting like this! But thanks for the deletion of Sponsor Episode. Draubb--Draubb 17:44, 3 February 2011 (UTC)

 * My suspicion is strong that the user is creating many accounts. These are almost certainly elementary school kids, and I'm not sure how much they understand what they are doing and what is being said to them. I don't wonder that their stuff was routinely deleted before. However, carrot and stick. I think we should leave the carrot in place, allow user space pages, within reason. I'm unclear on whether or not we have one user here, or many, the appearance is that of a number of kids going to the same school, but also creating more accounts than one each. They are not particularly communicative with the adults here. I'm trying to keep the door open for positive or at least harmless play, while shutting it to continual need for custodial attention. Maybe I'll fail. But I'm willing to continue trying for a while. Without the tools, it's harder, but that's the breaks. The above made no sense to me. --Abd 19:54, 3 February 2011 (UTC)

What next then?
Greetings. Peter Damian 21:55, 11 February 2011 (UTC)

 * Thrilled you are here. Can you convince Milton Roe to join us, too?  —Moulton 22:00, 11 February 2011 (UTC)

To answer a question and relate it to the above, we need to write a pile of policies so that we can ignore them. Welcome, I assume that Mike won't mind my welcoming you here. You can come on over to User talk:Abd if you need any doorstops. --Abd 02:12, 12 February 2011 (UTC)

Resolutions
Hi Mike. I think we're making some progress with reforming Community Review to make it more useful and approachable. I've created a "resolutions" subpage in the interests of adopting the things we might agree upon, and leaving the things we still need to discuss for later. My hope is that by getting these reforms through we will be able to pursue further reforms, since that's pretty much what CRs should do.

With that in mind, please comment on Community Review/CR process discussion/Resolutions.

I'm leaving this note because you've already commented. If you comment on nothing else, please let your opinion be clear on using the sitenotice (part of resolution #4). --SB_Johnny talk 23:35, 11 February 2011 (UTC)

Notice of an amended resolution
Hi Mike. I'd like to make sure that you're aware of an amended version of a CR resolution you commented upon. Please let the community know if you're satisfied with the text of the new resolution.

On a personal note, please have a gander at my scratch page, and let me know your thoughts on that! --SB_Johnny talk 23:24, 14 February 2011 (UTC)

Moulton on User talk:Ottava Rima
User talk:Ottava Rima Thanks for removing that, supporting what I'd been doing. However, you did, then, leave the use of a personal name in place, when Ottava has previously objected to that. It's pretty certain that that message would also be unwanted. --Abd 23:18, 24 February 2011 (UTC)

 * Please see Wikiversity talk:Community Review/Pseudonymity and external correspondence where a discussion has started on how to clarify wv policy on the use of names. --mikeu talk 23:53, 24 February 2011 (UTC)

Moulton flagrantly violating your warning re outing
you warned Moulton.]

See Request_custodian_action, which requests oversight of two edits. In this edit, Moulton edits a prior comment of his to show her real name. This is pure trolling with no possible academic justification.

In response, Moulton repeated the name, including in edit summaries, with and, and with the last edit, establishes clearly that his intention is to harass.

With, Moulton also outs another WP user.

There is no privacy policy if it is not enforced.

As this is clear harassment, with apparent intention to harm, and besides revision deletion of outing edits, it is essential that Moulton be promptly blocked, until and unless it is clear that the risk of continued harassment has abated. This is not just about Wikiversity, there are major cross-wiki issues. Accordingly, I am discussing, with some involved, should this not be resolved locally, going to meta to request relinking of the WV Moulton account with the global account, so that the global lock is again effective here, as well as the addition of Caprice. (It could then be whitelisted if any admin wants to allow Moulton to edit. Apparently the whitelist works now.) --Abd 03:08, 2 March 2011 (UTC)

RCA request re-opened due to continued disruption by Moulton
Request_custodian_action. Moulton is also pushing the edge with outing, see. I have re-opened the request at meta due to continued and gratuitous disruption. --Abd 15:03, 4 March 2011 (UTC) Ummm, would you consider it "outing" for me to post this link: Cold_fusion/Experts/Abd_ul-Rahman_Lomax? --mikeu talk 15:19, 4 March 2011 (UTC)
 * Gosh this is exciting. —Montana Mouse 15:15, 4 March 2011 (UTC)
 * No. There is a reason for that page, it discloses possible conflict of interest. I pointed to Moulton's edit because he is using the name offensively, and, when one looks at what he's posted elsewhere, he's being fairly explicit about what he's hinted at with others: threatened legal action for libel. The use of the name, with me as with others, is to underscore personal responsibility. From a practical standpoint, apparent outing should be interdicted because investigating each situation is to cumbersome. I'm not suggesting that Moulton should be blocked for outing, but that he will push every edge, creating more and more disruption, until he finally manages to get himself blocked again. The core of this complaint is "Death Eater Bitch," and SBJ's apparent willingness to tolerate this. There are plenty of other signs, as well, that Moulton intends to be absolutely as disruptive as possible. He should not have been unblocked without assurances. --Abd 15:52, 4 March 2011 (UTC)

 * What I wanna know is why Mr. Lomax didn't complete his studies and earn a degree from CalTech. But I suppose I'm not entitled to know the true reason, on account of privacy of academic records.  —Moulton 15:29, 4 March 2011 (UTC)
 * And the reason you want to know this? You would not find the answer in the records. I left in good standing, I could have returned at will. The bottom line is that I decided to study something else. --Abd 15:54, 4 March 2011 (UTC)
 * You decided to study Cold Fusion, something that Richard Feynman would have derided, as an example of Cargo-Cult Science? —Moulton 16:58, 4 March 2011 (UTC)
 * Eh? I left Cal Tech in 1964 or so. CF was discovered, probably, in 1989. I knew Feynman. You didn't. What he'd have derided would have been your pseudoskepticism, the rejection of experimental evidence in favor of attachment to personal opinion. I.e., what you do every day. --Abd 18:28, 4 March 2011 (UTC)

 * You made your position clear at meta, Mike. Thanks for revealing it. It will take some time to sort this. --Abd 16:43, 4 March 2011 (UTC)
 * Take all the time you need. —Moulton 17:00, 4 March 2011 (UTC)

Privacy violation
Thank you for the recent oversights; however you seem to have missed. I am also a little confused as to why I have seen no actions other than cleanup/removal/oversight, not even a warning, regarding this behavior. Is this standard practice here? KillerChihuahua 13:06, 18 February 2011 (UTC)

 * Thank you for contributing to our research project on discovering best practices for dealing with intimidating practices. —Gastrin Bombesin (talk) 13:24, 18 February 2011 (UTC)
 * I'm not voluntarily contributing to that "research" at all. I don't know why you're harassing me and trying to intimidate me. I have asked you to stop. KillerChihuahua 15:32, 18 February 2011 (UTC)
 * Are you feeling harassed and intimidated? —Montana Mouse (talk) 16:06, 18 February 2011 (UTC)

--mikeu talk 16:14, 18 February 2011 (UTC)

Further discussion on WV:Privacy is being conducted here: Wikiversity_talk:Privacy_policy --mikeu talk 17:49, 18 February 2011 (UTC)

Thanks for archiving on Requests for deletion.
There are some very old discussions there, and some recent ones that might have an apparent close, but my tentative "close while involved" was rejected, repeatedly, in one case. My view is that RfDs should not hang around forever, the existence of many on the page has an effect of suppressing discussion, and the argument that leaving discussions open will lead to improvements is very incorrect; many people will not improve a page that has a deletion tag on it, I learned, the hard way, to avoid that on wikipedia. If I wanted to improve a marginal page, I'd wait for deletion process to complete and then, if it was deleted, arrange for the page to be userfied, so that work could proceed without disruption.

Do you think that you could review some of the discussions and close them? Either with a keep/delete conclusion, or accepting userification, or other result, or as "no consensus", the latter judging that enough time has passed without a clear conclusion?

("No consensus" is always "without prejudice," except for possibly inhibiting immediate renomination. For simplicity, we should establish some standards, which will matter if the WV scale increases.

Thanks for considering this. --Abd 18:29, 5 March 2011 (UTC)

Is this an outing in the desert?
Is this outing content? The purported real name of a contributor to Wikipedia is in there. (I also asked SBJ about this.) --Abd 23:52, 5 March 2011 (UTC)

 * Who the hell knows if it's his real name? He writes under six different names. "Paul Mitchell" is the name he writes under on the WMF Mailing List. It's just one of the many pseuds used by the author who, here on Wikiversity, wrote under at least two registered names plus an IP or two. You certainly had no qualms about listing all the pseuds and IPs under which I post content on WMF sites. —Caprice 00:00, 6 March 2011 (UTC)

 * I've responded to this at User talk:SB Johnny and probably should have simply cited that page instead of copying my notice. Sorry. You may remove this section, as far as I'm concerned, it's enough that you be aware of what's going on, so that you may act if needed. Thanks. --Abd 00:11, 6 March 2011 (UTC)

 * Are you aware that Mike really doesn't care for Lunatic Psychodrama? —Caprice 00:15, 6 March 2011 (UTC)

 * Uh, reread what I wrote above. Mike is free to disregard this, though he's the one who warned you about the outing thing, so he's a proper admin to notify that You Are Still Doing It. If he doesn't care for Lunatic Psychodrama, what in the world was he thinking when he accepted (and continued accepting) being a Wikiversity admin? That's a question, not an accusation. He is not obligated to answer! --Abd 01:19, 6 March 2011 (UTC)
 * I reckon he was thinking that Wikiversity was an authentic learning community and not a Post-Modern Theater of the Absurd to be frequented by a curious cast of characters dressed up as Eight-Line Rhyme Schemes, Narcissistic Centaurs, Doubtful Salmon, Killer Canines, and Gamboling Goats going for a musical outing in the desert. —Caprice 04:28, 6 March 2011 (UTC)

Category:Charles Moore 2010 WWA etc
Category:Charles Moore 2010 WWA Hi mikeu

Here is my reply to you concerning the titular matter. Cheers! -- KYPark [T] 09:12, 26 March 2011 (UTC)

Welcome
Hey, thanks for welcoming me to Wikiversity. Well, I've been looking at things here for the last days, and I liked a few things. For example, I signed up for a Breton course (School of Language and Literature under Faculty for Humanities. I'll come back here and post if I need help - though I hope that won't be necessary. Will that be okay? --User:Scorpio March 18, 2011 5:41 PM (IST)

science communication
Folling up on what we discussed the other day, this podcast has a lot of starting points: http://www.abc.net.au/rn/ockhamsrazor/stories/2011/3089284.htm

--SB_Johnny talk 17:58, 14 February 2011 (UTC)

Re: Welcome Message
Thanks a lot mike, It was really nice to meet you. The information shared by you is really useful. Lets learn a lot about meteors. --Venkyzealous 16:20, 5 March 2011 (UTC)

Philly astronomy night
Thought you'd get a kick out of this. --SB_Johnny talk 08:38, 19 April 2011 (UTC)

Slide show
Thanks for your suggestion of Commons:User:Dschwen/Slideshow. For the reason you mention, and because I'd rather keep the slide show to a fraction of the screen, however, I'd like to keep looking.

dark lama  has offered me an implementation that seems likely to meet my needs. I need to experiment with it to see how far I can push it.

Thanks for your help and interest so far.

BillBell 13:06, 9 September 2011 (UTC)

Great, yeah Darklama is very adept at working out solutions to these kinds of things. Let me know if there is anything else that I can help with, though. --mikeu talk 15:43, 9 September 2011 (UTC)

Biblical RfD
Hi Mike, I reopened an RfD you commented on previously: Requests_for_Deletion. –SJ + > 03:54, 13 July 2011 (UTC)

Belated Thanks plus Question
First, Mike, belated thanks for your welcome!

Second, I would like to be able to put a slide show within a topic on wikiversity using, say, |deck.js, and perhaps some other javascript libraries. Would you please tell me where the rules for using such things are outlined--if indeed they are permitted at all?

BillBell 18:55, 5 September 2011 (UTC)

 * Thanks for your advice so far.
 * BillBell 15:17, 8 September 2011 (UTC)
 * BillBell 15:17, 8 September 2011 (UTC)

Changing username
Mu301, I see you made an edit today. If you have time, would you mind looking at Changing username? I've clerked the page to make it easier, and have made recommendations. If any of my recommendations are improper, I'd love to know. It should now be practically one-button to handle good rename requests, and two-button for usurpations that are ready. Let me know what you think, okay? Thanks. --Abd 16:03, 8 September 2011 (UTC)

User:Marshallsumter
User:Marshallsumter Mu301, I'm contacting you because you're a bureaucrat here. We've had some problems over at en.wikipedia due to the activities of Marshallsumter, who has been creating very problematic articles as part of a "research project" he's been writing about over here - see Dominant group. Although Marshallsumter is now blocked at en.wikipedia, and is likely to receive a ban, he continues to contribute here. All of his contributions on Wikiversity relate to his own original research on creating meaningless pages about combinations of words on Wikipedia, and he's now using Wikiversity to host versions of these pages, which have since been deleted at en.wikipedia. I am unfamiliar with most Wikiversity policies, but I don't think it's acceptable for a user to coordinate disruption of another Wikimedia project from here. Would you please look into a way of preventing this activity ? Thanks. --S Larctia 10:59, 10 September 2011 (UTC)
 * I was aware that the user had been active at Wikipedia with "problematic articles." However, original research is allowed at Wikiversity. "Coordinating disruption" would not be allowed. Can you explain what you mean by that? What activity is it that you wish to prevent? We have no control over what users do on other projects. --Abd 13:57, 10 September 2011 (UTC)
 * If you read through Dominant group, it is clear that Marshallsumter's "research" here involves creating large numbers of problematic articles on Wikipedia to see how the Wikipedia community responds. All of his Wikiversity activity relates to creating the said disruptive articles. I'd like Marshallsumter to be prevented from writing about this particular "research" on Wikiversity and using Wikiversity as a place to store drafts and copies of deleted Wikipedia articles (you can find them as subpages of Dominant group). This is clearly an abuse of his privilege to edit Wikiversity. Thanks. --S Larctia 14:55, 10 September 2011 (UTC)
 * Larctia, it seems you are trying to prevent a user from writing material that may be allowed here, not being satisfied with arranging his block on enwiki. If there is a specific problem where Wikiversity inclusion policy is being violated, please point it out; however, "deleted Wikipedia articles" are not prohibited content here, per se, depending on details and how they are placed. Frankly, I'd prefer to arrange for transwiki for deleted articles that may have some educational value, to preserve contribution history, but if he's the only major contributor, that's not a problem. The user can surely "store drafts" off-line, but he can share them here. There are possible concerns, to be sure.

 * Do you have the idea that material not appropriate for Wikipedia is therefore not appropriate for any WMF wiki? In any case, Mu301 is quite inactive. I'd suggest we not continue to discuss this here. There is your user Talk page and mine, if you have any specific concerns, or there is Request custodian action, if you believe a user is being disruptive here, and we have speedy deletion tags and Requests for deletion. --Abd 16:43, 10 September 2011 (UTC)

 * Much as it's sort of funny watching you try to convince this person, Abd, I'm pretty sure you've met him before ;-). --SB_Johnny talk 20:06, 10 September 2011 (UTC)
 * You may be pretty sure, but I'm not. Maybe you know or have noticed something I haven't, beyond "deletionist," and the one that would come to mind for me, I haven't seen specific signs. I did comment on the user's Talk page. To put it politely, this is a "highly experienced returning user, not disclosing prior account." As to "convincing" him or her, I'm giving the person an opportunity. I'm not relying on him or her accepting it. --Abd 01:55, 11 September 2011 (UTC)

You are invited to register for the Wikiversity Assembly

 * The Assembly has been established as a technique for developing reports on topics of import for Wikiversity administration. The Assembly is not a decision-making body, per se. Rather, it is designed to create or discover or estimate consensus, through focused, facilitated, thorough deliberation. Assembly reports may be referenced in regular Wikiversity discussions, but will not directly control outcomes. Where full consensus is not found, minority reports may be issued.
 * I invite you to register for the Assembly by adding your user name to the Delegable proxy/Table.

 * Registering for the Assembly creates no specific obligation, but does consent to direct communication as the Assembly may determine is appropriate. You may opt out of such direct communication by adding "no messages" to the Table when you register, in the user comment field, but it is unlikely that the default (communication allowed) will create burdensome traffic for you.

You are invited to name a proxy

 * When you register for the Assembly, you may optionally designate a "proxy."

 * I suggest that you nominate, as a proxy, the user whom you most trust to participate positively in a Wikiversity discussion if you are unable to participate yourself. The proxy will not be voting for you in any process. Rather, the proxy will be considered to loosely represent you, as a means of estimating probable large-scale consensus based on small-scale participation, in the event that you do not personally participate.

 * If you name a proxy, you will be consenting to direct communication with you by that proxy. If a named proxy accepts the proxy, you become, as long as you maintain the nomination (you may change it at any time), the "client" of the proxy, and by accepting, a proxy has consented to direct communication from the client.

 * See Delegable proxy for details.

Comments
This is a device for allowing thorough deliberative process while not requiring massive participation, which, without attention to process details, can become enormously inefficient and confusing. The proposals may look complicated, but they will, in practice, be simple. I hope you will consider this. Thanks. --Abd 19:41, 11 September 2011 (UTC)

FLOSS4Science
FLOSS4Science Hi :-)

A couple of years ago on my talk page:

I hadn't noticed Portal:FLOSS4Science until just now. This is a great idea, and I would like to see this developed. --mikeu talk 11:46, 6 September 2009 (UTC)

and I responded

Great! I would love to see a community grow around this project. Please go for it (i.e. do whatever you want to take it further) and let me know if there is anything specific I can do to help. K 20:30, 6 September 2009 (UTC)

Someone seems to have taken the idea further (www.floss4science.com).

This is a good thing ... just wish they had not decided to impose NC-ND on all the content on their web site (scroll down to the foot of the front page) - breaks the spirit in which the knowledge was shared.

Of course it is entirely possible they came up with the idea independently.

- K 12:19, 30 September 2011 (UTC)

Closures needed
Hi Mike. I have offered to mentor a new arrival. Are you available for button pushing?

Also, Abd's confirmation is on it's 5th day now. I'm too involved to close (as is James, I would imagine).

Thanks! --SB_Johnny talk 11:20, 17 September 2011 (UTC)
 * That was handled by agreement. --Abd 17:53, 30 September 2011 (UTC)

Bot learning resources
I am interested in your plans! Keep planning! This bot WILL used to generate examples for the creation of learning resources, if I have any say in the matter! Ray Calvin Baker 21:09, 1 August 2011 (UTC) (P. S. I got here trying to find out when the sandbox was invented. I would like to nominate those involved for a Nobel Peace Prize!)

Mu301Bot at Astronomy
Hi Mu301!

Your Bot has kindly stopped by the Astronomy article that I have been contributing to pointing out apparently problematic links. I wanted to update you on these.
 * Dead link - has been deleted.
 * Dead link 2 - "Reference checked and link works fine. No clue why bot's having a problem!"
 * Dead link 3 - "Section of article is a cited quote from Wikipedia article "Andromeda Galaxy", which contains the broken link even now."

I cannot leave messages at WP, but perhaps you can if you wish to verify and notify accordingly.

Just FYI Marshallsumter 20:42, 13 November 2011 (UTC)

 * Hi, thanks for the note. I'll take a look at the bot logs to see what is up. --mikeu talk 14:29, 14 November 2011 (UTC)

 * I just changed link 2 from https to http which the bot seems to like better. Andromeda Galaxy already has that link tagged as broken, but I'll need to look around to find a reliable source as a replacement. Let me know if the bot is incorrectly reporting any other broken links. --mikeu talk 15:18, 14 November 2011 (UTC)

Hi Mu301!

Your Bot has again kindly stopped by the Dominant group/Astronomy article to report the following link to be dead:
 * http://www.sciencedirect.com/science/article/pii/S0168822708003355.

When I click on the link it takes me to the Abstract on ScienceDirect, which I believe should be okay. I've added a pmid from Pubmed to this same link. No idea if this is a problem with the bot. Marshallsumter 17:39, 15 January 2012 (UTC)
 * http://www.sciencedirect.com/science/article/pii/0012821X80900047, does the same thing. Is this a problem? Marshallsumter 17:54, 15 January 2012 (UTC)
 * It seems like the problem occurs when the bot tries to check an url for a website that does redirects, ie. when the wiki page has a link to http://www.sciencedirect.com/science/article/pii/S0168822708003355 but the Science Direct webserver rewrites the url as http://www.diabetesresearchclinicalpractice.com/article/S0168-8227%2808%2900335-5/abstract One solution would be to try to the second url. Another option is to try the  10.1016/j.diabres.2008.07.017  syntax which gives a link to the same article that looks like this: 10.1016/j.diabres.2008.07.017 or this: Global complication rates of type 2 diabetes in Indigenous peoples: A comprehensive review. Also, please let me know if the bot is annoying ;) I can tell it to ignore specific pages. --mikeu talk 00:10, 16 January 2012 (UTC)

title: User talk:Mu301/Archive 2011


text:
User:Egm6341.s11.team5.reiss

Homework #2 Homework #3 Homework #5

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

title: User:Egm6341.s11.team5.reiss


text:
User:Egm6341.s11.team5.reiss/HW2

Given
The expression to find the Legendre polynomial of degree n is given by: $${{\rm P}_n}(x) = \sum\limits_{i = 0}^{\left[ {\frac{n}{2}} \right]} {{{( - 1)}^i}\frac}$$ where $$\left[ \frac{n}{2}\right]$$ is the integer value of the fraction.

Find
A) Use the Legendre Polynomial expression to find $${{\rm P}_5}(x)$$ and $${{\rm P}_{10}}(x)$$ B) Find and plot the roots of both polynomials and comment on the location of the roots near the end points of the interval $$\left[ {-1,1}\right]$$

Solution
For $${{\rm P}_5}(x)$$ the equation is as follows: $${{\rm P}_5}(x) = \sum\limits_{i = 0}^{\left[ {\frac{5}{2}} \right]} {{{( - 1)}^i}\frac} $$ for $$i=0$$ $$ \Rightarrow \frac = \frac{8}{x^5}$$ for $$i = 1$$ $$ \Rightarrow - \frac =  - \frac{4}{x^3}$$ for $$i = 2$$ $$ \Rightarrow \frac = \frac{8}{x^{}}$$ $$ \Rightarrow {{\rm P}_5}(x)=\frac{8}{x^5} - \frac{4}{x^3} + \frac{8}x$$ For $${{\rm P}_{10}}(x)$$ the equation is as follows: $${{\rm P}_{10}}(x) = \sum\limits_{i = 0}^{\left[ {\frac{10}{2}} \right]} {{{( - 1)}^i}\frac} $$ for $$i=0$$ $$ \Rightarrow \frac = \frac{x^{10}}$$ for $$i=1$$ $$\Rightarrow - \frac =  - \frac{x^8}$$ for $$i=2$$ $$\Rightarrow \frac = \frac{x^6}$$ for $$i=3$$ $$\Rightarrow - \frac =  - \frac{x^4}$$ for $$i=4$$ $$ \Rightarrow \frac = \frac{x^2}$$ for $$i=5$$ $$ \Rightarrow - \frac =  - \frac$$ $$ \Rightarrow {{\rm P}_{10}}(x) = \frac{x^{10}} - \frac{x^8} + \frac{x^6} - \frac{x^4} + \frac{x^2} - \frac$$

title: User:Egm6341.s11.team5.reiss/HW2


title: File:Nm1.s11.team5.HW2.P6.fig1.png


title: File:Nm1.s11.team5.HW2.3.fig1.jpg


text:
User:Egm6341.s11.team4/HW2

Homework 2 =Problem 2.1: Taylor Series Expansion Proof=

As assigned in lecture 6

Given
Given the remainder term after the first integration by parts upon a function

 * $$ \int_{x_{0}}^{x} (x-t) f^{(2)}(t) dt   $$    (1.1)

Objectives
1.    Repeat the integration by parts method used for 3 more degrees of the Taylor Series Approximation

 * $$ \frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0}) +   \frac{(x-x_{0})^{3}}{3!}   f^{(3)}(x_{0})   +     \frac{(x-x_{0})^{4}}{4!}   \!$$  (1.2)

2. Use the Intermediate Mean Value Theorem to express the remainder in terms of

 * $$ f^{(5)}(\xi) \!$$ (1.3)
 * $$ \xi \epsilon [x_{0},x] \!$$

3. Do integration by parts a fourth time on the equation to verify the Taylor Series remainder equations for (n+1) expansion with $$ R_{n+2}(x) $$

4. Use the Intermediate Mean Value Theorem on the Remainder term to show

 * $$ R_{n+1}(x) = \frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}(\xi) \!$$ (1.4)

Solution
1. First we will start with the "remainder" term from the Taylor series expansion as derived in class lecture [].

 * $$ R_{1}=\int_{x_{0}}^{x}(x-t)f^{(2)}(t)dt \!$$   (1.5)

The same integration by parts method used in class will be used on this remainder to further expand the Taylor series.

 * $$\int v(t) u'(t)\, dt = v(t) u(t) - \int v'(t) u(t)\, dt\!$$ (1.6)

Here we will now define the parts in our integration by parts scheme.

 * $$ u^{\prime}=(x-t) \!$$ (1.7)

 * $$ v=f^{(2)}(t) \!$$ (1.8)

 * $$  u=(xt-\frac{t^{2}}{2})       \!$$  (1.9)

Now, substituting into our integration by parts equation.

 * $$  R_{1}=\left [ (xt-\frac{t^{2}}{2}) f^{(2)}(t)\right ]_{t=x_{0}}^{t=x}-\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt       \!$$  (1.10)

Expanding out the term in brackets we obtain

 * $$ R_{1}=(\frac{1}{2}x^{2})f^{(2)}(x)-(xx_{0}-\frac{1}{2}x_{0}^{2})f^{(2)}(x_{0}) -\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt  \!$$

 * $$+ [ (\frac{1}{2}x^{2})f^{(2)}(x_{0})-(\frac{1}{2}x^{2})f^{(2)}(x_{0}) ]     \!$$  (1.11)

The second line in this equation sums to zero and is added so that we can rearrange the resulting equation into an integral term and a non-integral term.

 * $$     R_{1}=[ \frac{1}{2}x^{2}f^{(2)}(x)-\frac{1}{2}x^{2}f^{(2)}(x_{0}) ] +(\frac{1}{2}x^{2}-xx_{0}+\frac{1}{2}x_{0}^{2})f^{(2)}(x_{0})  -\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt                     \!$$  (1.12)

The bracketed term can be re-expressed as an integral. Additionally, the term $$ (\frac{1}{2}x^{2}-xx_{0}+\frac{1}{2}x_{0}^{2}) \!$$ can be expressed as $$ \frac{1}{2}(x-x_{0})^{2} \!$$. This gives us the following.

 * $$   R_{1}= \int_{x_{0}}^{x}\frac{1}{2}x^{2}f^{(3)}(t)+\frac{1}{2}(x-x_{0})^{2}f^{(2)}(x_{0})-\int_{x_{0}}^{x}(xt-\frac{t^{2}}{2}) f^{(3)}(t)dt                                 \!$$  (1.13)

Now regrouping the integral terms under one integral we obtain the following.

 * $$     R_{1}=  \int_{x_{0}}^{x}\frac{1}{2}(x-t)^{2}f^{(3)}(t)+\frac{1}{2}(x-x_{0})^{2}f^{(2)}(x_{0})                       \!$$  (1.14)

This can be grouped back into our Taylor series expansion to give us the following.

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

$$\displaystyle f(x)=f(x_{0})+\frac{(x-x_{0})^{1}}{1!}f^{(1)}(x_{0})+\frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0})+\int_{x_{0}}^{x}\frac{1}{2}(x-t)^{2}f^{(3)}(t) $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }. (1.15)
 * }.
 (1.15)

Our new remainder term, the term in the integral, will now be further expanded.

 * $$      R_{2} =   \int_{x_{0}}^{x}\frac{1}{2}(x-t)^{2}f^{(3)}(t)                       \!$$
  (1.16)

Our integration by parts terms are now the following.
 * $$          u'=  \frac{1}{2}(x-t)^{2}                  \!$$
  (1.17)
 * $$             v=f^{(3)}(t)                  \!$$
  (1.18)
 * $$             u=\frac{1}{6}t^{3}-.5t^{2}x+.5tx^{2}                 \!$$
  (1.19)

Substituting into our integration by parts equation we obtain the following.

 * $$ R_{2} =   [(\frac{1}{6}t^{3}-.5t^{2}x+.5tx^{2})f^{(3)}(t)]_{t=x_{0}}^{t=x} - \int_{x_{0}}^{x}  (\frac{1}{6}t^{3}-.5t^{2}x+.5tx^{2} )  f^{(4)}(t) dt            \!$$
  (1.20)

For clarity, the term in brackets will now be expressed as $$ \alpha \!$$. Expanding the terms in bracket we obtain the following.
 * $$  R_{2} =(\frac{1}{6}x^{3}) f^{(4)}(x)- (\frac{1}{6}x_{0}^{3}-\frac{1}{2}x_{0}^{2}x+\frac{1}{2}x_{0}x^{2}) f^{(4)}(x_{0}) - \alpha             \!$$

 * $$ +(\frac{1}{6}x^{3}) f^{(4)}(x_{0}) - (\frac{1}{6}x^{3}) f^{(4)}(x_{0})                    \!$$
  (1.21)

The second line of the equation adds to zero as before and is added to help rearrange the expressions into an integral later on.

Now we can group together the terms appropriately so that we can simplify the equation.

 * $$ R_{2} =  \int_{x_{0}}^{x} \frac{1}{6}x^{3} f^{(4)} dt   + (\frac{1}{6}x^{3} - \frac{1}{6}x_{0}^{3}-\frac{1}{2}x_{0}^{2}x+\frac{1}{2}x_{0}x^{2})f^{(3)}(x_{0}) - \alpha                 \!$$
  (1.22)

Pulling out the fractional coefficients of the terms outside of the integrals.

 * $$ R_{2} =  \int_{x_{0}}^{x} \frac{1}{6}x^{3} f^{(4)} dt +\frac{1}{3!}(x^{3}-3x_{0}^{2}x+3x_{0}x^{2}-x_{0}^{3})  f^{(3)}(x_{0}) -\alpha                           \!$$
  (1.23)

We can further simplify the term outside of the integral to be

 * $$ R_{2} =  \int_{x_{0}}^{x} \frac{1}{6}x^{3} f^{(4)} dt +\frac{1}{3!} (x-x_{0})  f^{(3)}(x_{0}) -\alpha                           \!$$
  (1.24)

Now combining the terms under the integrals to eliminate our $$ \alpha \!$$ expression.

 * $$  R_{2} =  \int_{x_{0}}^{x} (\frac{1}{3!}x^{3}-\frac{1}{6}t^{3}-\frac{1}{2}t^{2}x+\frac{1}{2}tx^{2})f^{(4)} + \frac{1}{3!} (x-x_{0})  f^{(3)}(x_{0})                           \!$$
  (1.25)

Further simplifying, this becomes

 * $$ R_{2} =  \frac{1}{3!} (x-x_{0})  f^{(3)}(x_{0})   +\int_{x_{0}}^{x} (\frac{1}{3!}(x-t)^{3})f^{(4)}(t)dt                         \!$$
  (1.26)

Substituting this back into our original equation we obtain the following.

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

$$\displaystyle f(x)=f(x_{0})+\frac{(x-x_{0})^{1}}{1!}f^{(1)}(x_{0})+\frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0})+  \frac{(x-x_{0})^{3}}{3!}   f^{(3)}(x_{0})   +\int_{x_{0}}^{x} (\frac{1}{3!}(x-t)^{3})f^{(4)}(t)dt                                    $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle
 * style= |
 * }.
 (1.27)
 * }.
 (1.27)

Now we can use the same method for the next term in the Taylor series.

 * $$       \int_{x_{0}}^{x} \frac{1}{3!}(x-t)^{3}f^{(4)}(t)dt                             \!$$
  (1.28)
 * $$      u'=     \frac{1}{3!}(x-t)^{3}                   \!$$
  (1.29)
 * $$       v= f^{(4)}(t)                       \!$$
  (1.30)
 * $$       u=(-\frac{t^{4}}{24}+\frac{1}{6}t^{3}x-\frac{1}{4}t^{2}x^{2}+\frac{1}{6}tx^{3} )                      \!$$
  (1.31)

Now substituting into our integration by parts equation
 * $$   R_{3}= [(-\frac{t^{4}}{24}+\frac{1}{6}t^{3}x-\frac{1}{4}t^{2}x^{2}+\frac{1}{6}tx^{3} ) f^{(4)}(t)  ]_{x_{0}}^{x} -\int_{x_{0}}^{x}  (-\frac{t^{4}}{24}+\frac{1}{6}t^{3}x-\frac{1}{4}t^{2}x^{2}+\frac{1}{6}tx^{3} )  f^{(5)}(t)  dt                           \!$$
  (1.32)

Expanding the bracket and substituting $$ \alpha \!$$ for the integral term, we obtain the following equation.

 * $$ R_{3}= (-\frac{x^{4}}{24}+\frac{1}{6}x^{4}-\frac{1}{4}x^{4}+\frac{1}{6}x^{4} )f^{(4)}(x)  -(-\frac{x_{0}^{4}}{24}+\frac{1}{6}x_{0}^{3}x-\frac{1}{4}x_{0}^{2}x^{2}+\frac{1}{6}x_{0}x^{3} )f^{(4)}(x_{0}) -\alpha                            \!$$
  (1.33)
 * $$ +\frac{1}{24}x^{4}  f^{(4)}(x_{0})-    \frac{1}{24}x^{4}   f^{(4)}(x_{0})                         \!$$
  (1.34)

The second line sums to zero and is added in order to re-arrange the terms into an integral.

Now we can express the equation as the following.

 * $$ R_{3}=   \int_{x_{0}}^{x}   \frac{1}{24}x^{4}f^{(5)}(t)dt -\alpha +[\frac{1}{24}x^{4} +\frac{x_{0}^{4}}{24}-\frac{1}{6}x_{0}^{3}x+\frac{1}{4}x_{0}^{2}x^{2}-\frac{1}{6}x_{0}x^{3}  ]             \!$$
  (1.35)

Now, combining the integral term with our $$ \alpha \!$$ term and then simplifying the terms in brackets we can express the equation as follows

 * $$R_{3}=  \frac{(x-x_{0})^{4}}{4!}   +  \int_{x_{0}}^{x}   \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt  \!$$
  (1.36)

Substituting this into our Taylor series polynomial expression, we obtain

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

$$\displaystyle f(x)=f(x_{0})+\frac{(x-x_{0})^{1}}{1!}f^{(1)}(x_{0})+\frac{(x-x_{0})^{2}}{2!}f^{(2)}(x_{0})        \!$$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$ +  \frac{(x-x_{0})^{3}}{3!}   f^{(3)}(x_{0})   +     \frac{(x-x_{0})^{4}}{4!}   +  \int_{x_{0}}^{x}   \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt  $$

$$ 2.
 *
$$\displaystyle
 * style= |
 * }.
 (1.37)
 * }.
 (1.37)

First we state the Intermediate Mean Value Theorem (IMVT)
 * $$ \int_{a}^{b} w(t)f(t)dt = f(\xi) \int_{a}^{b} w(t)dt \!$$
  (1.38)

This was previously proven in assignment 1 [] Now, we know the remainder for our equation is the following
 * $$ R_{4}=\int_{x_{0}}^{x}  \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt \!$$
  (1.39)

Now, if we identify w(t) and f(t) we can use the IMVT for our problem
 * $$ W(t)= \frac{1}{4!}(x-t) \!$$
 (1.40)
 * $$ f(t) = f^{(5)}(t) \!$$
 (1.41)

Since we are integrating from our constants of integration x and $$ x_{0} $$ we can use the IMVT, substituting our equations, to obtain the following.

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

$$\displaystyle \int_{x_{0}}^{x}   \frac{1}{4!}(x-t)^{4}f^{(5)}(t)dt = f^{(5)}(\xi)\int_{x_{0}}^{x}  \frac{1}{4!}(x-t)^{4} dt    $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle
 * style= |
 * }.
 (1.42)
 * }.
 (1.42)

Where

$$ \xi \in [x_{0},x] \!$$
 (1.43)

3.

Here, we must do integration by parts for a fourth time to verify equations 3 and 4 on lecture slide 3-3 for (n+1) expansion with $$ R_{n+2}(x) $$. In essence, this requires us to see if

 * $$ R_{n+2}(x) = \frac{1}{(n+1)!} \int_{x_{0}}^{x} (x-t)^{(n+1)} f^{n+2}(t)dt \!$$
 (1.44)

Where, in our case, n=4.

We will start with the remainder we obtained in part 1.
 * $$ R_{5}(x)=\frac{1}{4!}\int_{x_{0}}^{x} (x-t)^{4} f^{(5)}(t)dt \!$$
  (1.45)

Now identifying our integration by parts terms.

 * $$  u'=\frac{(x-t)^{4}}{4!}       \!$$
  (1.46)
 * $$   v=f^{(5)}(t)      \!$$
  (1.47)
 * $$    u=(\frac{t^{5}}{5!}-\frac{t^{4}x}{4!}+\frac{t^{3}x^{2}}{12}-\frac{t^{2}x^{3}}{12}+\frac{tx^{4}}{4!})     \!$$
  (1.48)

Substituting into our integration by parts equation.

 * $$ R_{5}(x)= [(\frac{t^{5}}{5!}-\frac{t^{4}x}{4!}+\frac{t^{3}x^{2}}{12}-\frac{t^{2}x^{3}}{12}+\frac{tx^{4}}{4!}) (f^{(5)}(t) )]_{x_{0}}^{x} - \int_{x_{0}}^{x}  (\frac{t^{5}}{5!}-\frac{t^{4}x}{4!}+\frac{t^{3}x^{2}}{12}-\frac{t^{2}x^{3}}{12}+\frac{tx^{4}}{4!}) f^{(6)}(t) dt     \!$$
  (1.49)

Now referring to the term to the integral term as $$ \alpha $$ as in the previous several iterations, expanding the bracketed term as before, and adding two terms that sum to zero in order to aid in re-arranging the equation

 * $$ R_{5}(x)=(\frac{x^{5}}{5!}-\frac{x^{5}}{4!}+\frac{x^{5}}{12}-\frac{x^{5}}{12}+\frac{x^{5}}{4!}) f^{(5)}(x)   - (\frac{x_{0}^{5}}{5!}-\frac{x_{0}^{4}x}{4!}+\frac{x_{0}^{3}x^{2}}{12}-\frac{x_{0}^{2}x^{3}}{12}+\frac{x_{0}x^{4}}{4!}) f^{(5)}(x_{0})   -\alpha \!$$
 * $$ +\frac{x^{5}}{5!}f^{(5)}(x_{0}) - \frac{x^{5}}{5!}f^{(5)}(x_{0})        \!$$
  (1.50)

The second line of the above equation sums to zero and is added so that we can rearrange the equation into a simpler form.

 * $$  R_{5}(x)= [\frac{x^{5}}{5!}f^{(5)}(x) -\frac{x^{5}}{5!}f^{(5)}(x_{0}) ]_{x_{0}}^{x} -(\frac{x_{0}^{5}}{5!}-\frac{x_{0}^{4}x}{4!}+\frac{x_{0}^{3}x^{2}}{12}-\frac{x_{0}^{2}x^{3}}{12}+\frac{x_{0}^{4}x}{4!}-\frac{x^{5}}{5!}) f^{(5)}(x)  -\alpha    \!$$
  (1.51)

Now we can express the bracketed term as an integral. Meanwhile, the complex polynomial simplifies to $$ \frac{(x-x_{0})^{5}}{5!} $$.
 * $$  R_{5}(x)= \int_{x_{0}}^{x}\frac{x^{5}}{5!}f^{(6)}(t)dt+\frac{(x-x_{0})^{5}}{5!}f^{(5)}(x_{0})-\alpha      \!$$
  (1.52)

Now, combining the integral term with the $$ \alpha $$ term we end up with the following equation

 * $$ R_{6}= \frac{(x-x_{0})^{5}}{5!}f^{(5)}(x_{0}) +  \int_{x_{0}}^{x}\frac{(x-t)^{5}}{5!}f^{(6)}(t)dt     \!$$
  (1.53)

Since n=4 in our case, this verifies the expansion with $$ R_{n+2}(x) $$

4.

We can use the IMVT on the Remainder term, given as equation 4 lecture slide 3-3  in order to show equation 5 lecture slide 3-3. The IMVT can be stated as follows.

 * $$  \int_{a}^{b}w(t)g(t)dt=g(\xi)\int_{a}^{b}w(t)dt       \!$$
  (1.54)

When applied to equation 4, the Taylor series remainder, we can evaluate the individual terms of the IMVT to be
 * $$  a=x_{0}          \!$$
  (1.55)
 * $$  b=x        \!$$
  (1.56)
 * $$  w(t)= \frac{(x-t)^{n}}{n!}       \!$$
  (1.57)
 * $$     g(t)=f^{(n+1)}(t)       \!$$
  (1.58)

Simple substitution of these equations into the IMVT gives us the following

 * $$  \int_{x_{0}}^{x}\frac{(x-t)^{n}}{n!}f^{(n+1)}(t)dt=f^{(n+1)}(\xi)  \int_{x_{0}}^{x}  \frac{(x-t)^{n}}{n!} dt        \!$$
  (1.59)

Now, evaluating the integral present on the right hand side, we can further simplify the remainder to the following
 * $$   R_{n+1}(x) = \frac{(x-x_{0})^{(n+1)}}{(n+1)!}f^{(n+1)}(\xi)      \!$$
  (1.60)

And this result is simply equation 5, so we have effectively shown that equation 5 arises naturally out of equation 4 in conjunction with the Intermediate Mean Value Theorem.

Problem solved by Brendan Mahon.

Given
The function f(x)
 * $$ f(x)=sin(x) \!$$

and the span of x
 * $$ x \epsilon [0, \pi] \!$$

Objective
Evaluate the Taylor Series about the point
 * $$ x_{0}=\frac{3\pi}{8} \!$$

For Taylor Series degrees of
 * $$ n=0,1,2,...,10 \!$$

We must then plot these series for each value of n in order to visually see how the Taylor Series improves with each degree of approximation.

Next we must estimate the error in our solution at the point $$ x= \frac{3\pi}{4} \!$$ for each degree of Taylor Series approximation.

Solution
Evaluating the Taylor Series of the sin(x) is a simple matter of following the general Taylor Series expansion formula for the case where f(x) is sin(x).

The general Taylor Series expansion as presented in the lecture slides is

 * $$ P_{n}(x) = f(x_{0}) + \frac{(x-x_{0})}{1!} f^{(1)}(x_{0}) +...+\frac{(x-x_{0})^{n}}{n!}f^{(n)}(x_{0}) \!$$
  (2.1)

For the case of sin(x), the equation then becomes

 * $$ f(x) = sin(x_{0}) + \frac{(sin(x)-sin(x_{0}))^{1}}{1!}cos(x_{0}) - \frac{(sin(x)-sin(x_{0}))^{2}}{2!}sin(x_{0}) - ...\!$$
 (2.2)

The periodicity of the sine function shows up in our evaluation of the Taylor Series and can be capitalized upon. Since the derivative of sine is the sine function itself every 4 derivatives evaluating the nth derivative is a simple matter of determining what the remainder is when n is divided by 4 and matching the derivative with the appropriate trigonometric function based on what the remainder is.

This method is implemented in code about the point $$ x_{0} = \frac{3\pi}{8}$$ up to degree n=10 and then plotted. The result is a plot that graphically shows how the Taylor Series comes closer to the true function it evaluates with each successive degree evaluated.

The maximum error of the Taylor Series can be expressed mathematically using equation 5 lecture slide 3-3
 * $$ \left | R_{n+1}(x) \right | \leq \frac{(x-x_{0})^{n+1}}{(n+1)!}(\alpha ) \!$$
 (2.3)

Where $$ \alpha \!$$ represents the maximum of the maximum value of the n+1 derivative over the relevant range.

 * $$ \alpha = max \left | f^{(n+1)}(t) \right | \!$$
 * $$ t \epsilon [x_{0},x] \!$$
 (2.4)

Now we will apply these equations to our specific function; the sine function and an evaluation point of $$ x_{0} = \frac{3\pi}{8} $$ and measurement point of $$ x = \frac{3\pi}{4} $$. First, we can say that

 * $$ \alpha = max \left | f^{(n+1)}(t) \right | = 1 \!$$
 * $$ t \epsilon [x_{0},x] \!$$
 (2.5)

This arises because of the nature of the sine function. Every nth order derivative of sine is going to be positive or negative sine or cosine. Sine and cosine functions are bounded above and below by 1 and -1. This means that the maximum absolute value of any derivative of sine is 1.

Now we can use this result to simplify our remainder equation.

 * $$ \left | R_{n+1}(x) \right | \leq \frac{(x-x_{0})^{n+1}}{(n+1)!} \!$$
 (2.6)

Since, for the problem we are evaluating $$ x_{0} = \frac{3\pi}{8} $$ and $$ x = \frac{3\pi}{4} $$ we can say that $$ (x-x_{0})=\frac{3\pi}{8} $$

 * $$ \left | R_{n+1}(\frac{3\pi}{4}) \right | \leq \frac{(\frac{3\pi}{8})^{n+1}}{(n+1)!} \!$$
 (2.7)

Now we have an upper bound on the error for any degree Taylor Series approximation of our function. This equation is included in the code used to plot the sine function and evaluated for all degree Taylor Series approximations from 0 to 10.

Matlab Code:

The code as printed above gives us several outputs, the plot of the Taylor Series and the error for each degree of the Taylor series about the point $$ \frac{3\pi}{4} $$.

Plots:

As can be seen, the higher order Taylor Series functions do an excellent job of approximating the sine function. In fact, every approximation above the 5th degree approximation cannot even be seen in the plot since they coincide with and are obscured by the plotting of the true value of sin(x). Thus, we would expect to see a very small error associated with these higher order approximations. The code used above gives us the following error estimations.

 * $$ \left | R_{1}(\frac{3\pi}{4}) \right | \leq 1.1781 \!$$
  (2.8)
 * $$ \left | R_{2}(\frac{3\pi}{4}) \right | \leq 0.6940 \!$$
  (2.9)
 * $$ \left | R_{3}(\frac{3\pi}{4}) \right | \leq 0.2725 \!$$
  (2.10)
 * $$ \left | R_{4}(\frac{3\pi}{4}) \right | \leq 0.0803 \!$$
  (2.11)
 * $$ \left | R_{5}(\frac{3\pi}{4}) \right | \leq 0.0189 \!$$
  (2.12)
 * $$ \left | R_{6}(\frac{3\pi}{4}) \right | \leq 0.0037 \!$$
  (2.13)
 * $$ \left | R_{7}(\frac{3\pi}{4}) \right | \leq 6.2494*10^{-4} \!$$
  (2.14)
 * $$ \left | R_{8}(\frac{3\pi}{4}) \right | \leq 9.2030*10^{-5} \!$$
  (2.15)
 * $$ \left | R_{9}(\frac{3\pi}{4}) \right | \leq 1.2047*10^{-5} \!$$
 (2.16)
 * $$ \left | R_{10}(\frac{3\pi}{4}) \right | \leq 1.4192*10^{-6} \!$$
  (2.17)
 * $$ \left | R_{11}(\frac{3\pi}{4}) \right | \leq 1.5200*10^{-7} \!$$
  (2.18)

As expected, the higher order approximations from about degree six onward produce an almost negligible error.

Problem solved by Brendan Mahon.

Given

 * $$f(x)=\frac{e^{x}-1}{x}$$

Objectives
To examine the Taylor series expansions of both $$\displaystyle e^x$$ and $$\displaystyle f(x)$$ and their remainders to verify that:

 * $$R_{n+1}[f(x);x] = \frac{(x-0)^{n+1}}{x(n+1)!}e^\xi = \frac{R_{n}[e^{x};x]}{n+1}$$

Solution
For this problem we will examine Taylor series expansions about the point $$ x_0 = 0$$.

In general, functions for which the (n+1)th derivative exists and is continuous, can be expressed in the form

where $$p_{n}(x)$$ is a polynomial of order n and $$R_{n+1}(x)$$ is the corresponding remainder.

During lecture 3, it was shown that these polynomials and remainders take a specific form for Taylor series expansions.

Equation 3 from lecture slide [[media:nm1.s11.mtg3.djvu|3-3]]:

where $$\displaystyle f^{(n)}(x_0)$$ denotes the n-th derivative of f evaluated at $$\displaystyle x_0$$.

Equation 5 from lecture slide [[media:nm1.s11.mtg3.djvu|3-3]]:

For the Taylor series expansion of $$\displaystyle e^x$$ about $$\displaystyle x_0=0$$, all derivatives of $$\displaystyle e^x$$ are also $$\displaystyle e^x$$ and are simplified to 1 when evaluated at $$\displaystyle x_0=0$$.

The result of (3.4) will be manipulated to solve for the expansion and remainder of $$\displaystyle f(x)$$:

(3.8) verifies equation 2 from lecture slide [[media:nm1.s11.mtg6.djvu|6-5]]. Comparing (3.5) and (3.8), it is obvious that the remainders of $$e^x$$ and $$f(x)$$ are similar. If the Taylor series expansion for $$e^x$$ were expanded to (n-1) terms rather than n, then the remainders of both $$e^x$$ and $$f(x)$$ would be of the same degree.

(3.10) confirms equation 1 from lecture slide [[media:nm1.s11.mtg7.djvu|7-1]]!

Problem solved by William Kurth.

=Problem 2.4: Comparing Methods of Numerical Integration= Refer to lecture slide [[media:nm1.s11.mtg7.djvu|7-3]] for problem statement.

Objectives
Plot f(x). Then, integrate f(x) using the following methods:
 * (1) Taylor series expansion
 * (2) Composite trapezoidal rule
 * (3) Composite Simpson rule
 * (4) Gauss-Legendre quadrature

for $$n=2,4,8...$$, until the error reaches the order $$10^{-6}$$.

Solution
The following MATLAB code was used to plot f(x) from -1 to 1.

(1) Integration by Taylor Series Expansion

To evaluate (4.1) by Taylor series expansion, the function f(x) will be approximated by an expansion about the point $$\displaystyle x_0 = 0$$. In homework problem 1.2, this expansion was shown to be

To solve for n, we must first examine the nature of the error.

On lecture slide [[media:nm1.s11.mtg7.djvu|7-1]], this error was shown to have the form that follows (adjusted for different interval):

When observing $$\displaystyle e^{\xi(\alpha)}$$ over the interval $$\displaystyle\alpha\;\in\;[-1,1]$$ it is seen that the function is bounded on both the top and bottom by

This can be extended to say that the error, $$\displaystyle E_n(f)$$, is also bounded as follows:

Now we will calculate the integration of f(x) and the upper bound of the error for increasing values of n until the error reaches the order of $$10^{-6}$$. The following table shows these values.

(2) Integration by Composite Trapezoidal Rule

Equation (1) from lecture slide [[media:nm1.s11.mtg7.djvu|7-4]]:

where $$\displaystyle h=\frac{b-a}{n}$$ and  $$\displaystyle f_n=f(x_n)$$.

The error associated with the composite trapezoidal rule is (Atkinson, p.253)

For this problem (4.10) becomes

Values of In and En are calculated from equations (4.9) and (4.13), respectively, for increasing values of n and shown in the table below.

(3) Integration by Composite Simpson Rule

Equation (4) from lecture slide [[media:nm1.s11.mtg7.djvu|7-4]]:

where $$\displaystyle h=\frac{b-a}{n}$$ and  $$\displaystyle f_n=f(x_n)$$ and n must be even.

The error associated with the composite trapezoidal rule is (Atkinson, p.257)

For this problem (4.15) becomes

Values of In and En are calculated from equations (4.14) and (4.18), respectively, for increasing values of n and shown in the table below.

(4)Integration by Gauss-Legendre Quadrature

Equation (1) from lecture slide [[media:nm1.s11.mtg7.djvu|7-5]]:

where $$\displaystyle \{x_i, i=1,2,3,...,n\}$$ are roots of the Legendre polynomial of order n, $$\displaystyle \mathcal{P}_n(x)$$, and $$\displaystyle w_i$$ are their corresponding weights. Some of these values are shown in the table below supplied in lecture slide [[media:nm1.s11.mtg7.djvu|7-5]].

The error associated with using this method for approximating the value of an integral is given by equation (3) on lecture slide [[media:nm1.s11.mtg8.djvu|8-2]].

Equations (4.19) and (4.20) were solved for increasing values of n until the error reached the order of 10-6 and shown in the table below.

Solved by William Kurth.

=Problem 2.5: Verifying Legendre Polynomial Roots and Weights= Refer to lecture slide [[media:nm1.s11.mtg7.djvu|mtg-7]] for the problem statement.

Given
Gauss-legendre quadrature:

Objectives
Verify this table against NIST handbook (lecture plan).

Solution
According to the Legendre polynomial in lecture note [[media:nm1.s11.mtg8.djvu|mtg-8]], {  Xi, i=1….n }  is the root of Legendre Polynomial of order n, and the associated weight with Xi is :
 * {| style="width:100%" border="0"

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

{W_i} = \frac

$$     (Eq 5.1 ) where the Legendre Polynomial is:
 *

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

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

{P_n}(x) = \sum\limits_{i = 0}^{[n/2]} \frac

$$     (Eq 5.2 )
 *

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

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

{P_1}(x) = x

$$     (Eq 5.3 )
 *

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

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

{P_2}(x) = \frac{1}{2}(3{x^2} - 1)

$$     (Eq 5.4 )
 *

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

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

{P_3}(x) = \frac{1}{2}(5{x^3} - 3x)

$$     (Eq 5.5)
 *

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

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

{P_4}(x) = \frac{8}{x^4} - \frac{4}{x^2} + \frac{3}{8}

$$     (Eq 5.6)
 *

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

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

{P_5}(x) = \frac{8}{x^5} - \frac{8}{x^3} + \frac{8}x

$$     (Eq 5.7 )
 *

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

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

{P_6}(x) = \frac{x^6} - \frac{x^4} + \frac{x^2} - \frac{5}

$$     (Eq 5.8 )
 *

 * }

1). When n=1, the root of P1(x)=0 is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{1}} = 0

$$     (Eq 5.9 )
 *

 * }

And the weight is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{1}}=\frac{-2}{(1+1)[P_{1}^{'}({{x}_{1}}){{P}_{2}}({{x}_{1}})]}=\frac{-2}{(1+1)[1\centerdot (3{{x}_{1}}^{2}-1)/2]}=\frac{-2}{(1+1)[1\centerdot (3\centerdot 0-1)/2]}=1

$$     (Eq 5.10 )
 *

 * }

2). When n=2, the root of P2(x)=0 is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{1,2}} = \pm 1/\sqrt 3

$$     (Eq 5.11 )
 *

 * }

And the weight is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{1}}=\frac{-2}{(2+1)[P_{2}^{'}({{x}_{1}}){{P}_{3}}({{x}_{1}})]}=\frac{-2}{(2+1)[(3{{x}_{1}})\centerdot (5{{x}_{1}}^{3}-3{{x}_{1}})/2]}=1

$$     (Eq 5.12 )
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{2}}=\frac{-2}{(2+1)[P_{2}^{'}({{x}_{2}}){{P}_{3}}({{x}_{2}})]}=\frac{-2}{(2+1)[(3{{x}_{2}})\centerdot (5{{x}_{2}}^{3}-3{{x}_{2}})/2]}=1

$$     (Eq 5.13 )
 *

 * }

3). When n=3, the root of P3(x)=0 is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_1} = 0,\begin{array}{ccccccccccccccc} {}&{{x_{2,3}} = \pm \sqrt {15} /5} \end{array}

$$     (Eq 5.14 )
 *

 * }

And the weight is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{1}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{1}}){{P}_{4}}({{x}_{1}})]}=\frac{-2}{(3+1)[(15{{x}_{1}}^{2}-3)/2\centerdot (\frac{35}{8}{{x}_{1}}^{4}-\frac{15}{4}{{x}_{1}}^{2}+\frac{3}{8})]}=\frac{8}{9}

$$     (Eq 5.15 )
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{2}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{2}}){{P}_{4}}({{x}_{2}})]}=\frac{-2}{(3+1)[(15{{x}_{2}}^{2}-3)/2\centerdot (\frac{35}{8}{{x}_{2}}^{4}-\frac{15}{4}{{x}_{2}}^{2}+\frac{3}{8})]}=\frac{5}{9}

$$     (Eq 5.16)
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{3}}=\frac{-2}{(3+1)[P_{3}^{'}({{x}_{3}}){{P}_{4}}({{x}_{3}})]}=\frac{-2}{(3+1)[(15{{x}_{3}}^{2}-3)/2\centerdot (\frac{35}{8}{{x}_{3}}^{4}-\frac{15}{4}{{x}_{3}}^{2}+\frac{3}{8})]}=\frac{5}{9}

$$     (Eq 5.17)
 *

 * }

4). When n=4, the root of P4(x)=0 is:
 * {| style="width:100%" border="0"

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

\frac{8}{x^4} - \frac{4}{x^2} + \frac{3}{8} = 0

$$     (Eq 5.18)
 *

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

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

{x^2} = (3 \pm 2\sqrt {6/5} )/7

$$     (Eq 5.19)
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{1,2}} = \pm \sqrt {(3 + 2\sqrt {6/5} )/7}

$$     (Eq 5.20 )
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{3,4}} = \pm \sqrt {(3 - 2\sqrt {6/5} )/7}

$$     (Eq 5.21 )
 *

 * }

And the weight is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{1}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{1}}){{P}_{5}}({{x}_{1}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{1}}^{3}-\frac{15}{2}{{x}_{1}})\centerdot (\frac{63}{8}{{x}_{1}}^{5}-\frac{70}{8}{{x}_{1}}^{3}+\frac{15}{8}{{x}_{1}})]}=(18-\sqrt{30})/36

$$     (Eq 5.22 )
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{2}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{2}}){{P}_{5}}({{x}_{2}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{2}}^{3}-\frac{15}{2}{{x}_{2}})\centerdot (\frac{63}{8}{{x}_{2}}^{5}-\frac{70}{8}{{x}_{2}}^{3}+\frac{15}{8}{{x}_{2}})]}=(18-\sqrt{30})/36

$$     (Eq 5.23 )
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{3}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{3}}){{P}_{5}}({{x}_{3}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{3}}^{3}-\frac{15}{2}{{x}_{3}})\centerdot (\frac{63}{8}{{x}_{3}}^{5}-\frac{70}{8}{{x}_{3}}^{3}+\frac{15}{8}{{x}_{3}})]}=(18+\sqrt{30})/36

$$     (Eq 5.24)
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{W}_{4}}=\frac{-2}{(4+1)[P_{4}^{'}({{x}_{4}}){{P}_{5}}({{x}_{4}})]}=\frac{-2}{(4+1)[(\frac{35}{2}{{x}_{4}}^{3}-\frac{15}{2}{{x}_{4}})\centerdot (\frac{63}{8}{{x}_{4}}^{5}-\frac{70}{8}{{x}_{4}}^{3}+\frac{15}{8}{{x}_{4}})]}=(18+\sqrt{30})/36

$$     (Eq 5.25)
 *

 * }

5). When n=5, the root of P5(x)=0 is:
 * {| style="width:100%" border="0"

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

{P_5}(x) = \frac{8}{x^5} - \frac{8}{x^3} + \frac{8}x

$$     (Eq 5.26)
 *

 * }

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

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

{x_1} = 0,\begin{array}{ccccccccccccccc} {}&{{{({x_{2,3,4,5}})}^2} = (5 \pm 2\sqrt {10/7} )/9} \end{array}

$$     (Eq 5.27)
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{2,3}} = \pm \sqrt {(5 + 2\sqrt {10/7} )} /3

$$     (Eq 5.28)
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{x_{4,5}} = \pm \sqrt {(5 - 2\sqrt {10/7} )} /3

$$     (Eq 5.29)
 *

 * }

And the weight is:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{W}_{1}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{1}}){{P}_{6}}({{x}_{1}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{1}}^{4}-\frac{210}{8}{{x}_{1}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{1}}^{6}-\frac{315}{16}{{x}_{1}}^{4}+\frac{105}{16}{{x}_{1}}^{2}-\frac{5}{16})]} \\ & =128/225 \\ \end{align}

$$     (Eq 5.30)
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{W}_{2}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{2}}){{P}_{6}}({{x}_{2}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{2}}^{4}-\frac{210}{8}{{x}_{2}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{2}}^{6}-\frac{315}{16}{{x}_{2}}^{4}+\frac{105}{16}{{x}_{2}}^{2}-\frac{5}{16})]} \\ & =(322-13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.31)
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{W}_{3}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{3}}){{P}_{6}}({{x}_{3}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{3}}^{4}-\frac{210}{8}{{x}_{3}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{3}}^{6}-\frac{315}{16}{{x}_{3}}^{4}+\frac{105}{16}{{x}_{3}}^{2}-\frac{5}{16})]} \\ & =(322-13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.32)
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{W}_{4}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{4}}){{P}_{6}}({{x}_{4}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{4}}^{4}-\frac{210}{8}{{x}_{4}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{4}}^{6}-\frac{315}{16}{{x}_{4}}^{4}+\frac{105}{16}{{x}_{4}}^{2}-\frac{5}{16})]} \\ & =(322+13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.33 )
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

\begin{align} & {{W}_{5}}=\frac{-2}{(5+1)[P_{5}^{'}({{x}_{5}}){{P}_{6}}({{x}_{5}})]}=\frac{-2}{(5+1)[(\frac{315}{8}{{x}_{5}}^{4}-\frac{210}{8}{{x}_{5}}^{2}+\frac{15}{8})\centerdot (\frac{231}{16}{{x}_{5}}^{6}-\frac{315}{16}{{x}_{5}}^{4}+\frac{105}{16}{{x}_{5}}^{2}-\frac{5}{16})]} \\ & =(322+13\sqrt{70})/900 \\ \end{align}

$$     (Eq 5.34)
 *

 * }

Problem solved by Shengfeng Yang.

=Problem 2.6: Verifying Legendre Polynomials=

Refer to lecture slide [[media:nm1.s11.mtg8.djvu|mtg-8]] for the problem statement.

Given
Legendre polynomial is given by:

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

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

{{P}_{n}}(x)=\sum\limits_{i=0}^{[n/2]}\frac{(2n-2i)!{{x}^{n-2i}}}{{{2}^{n}}i!(n-2i)!(n-i)!}

$$     (Eq 6.1)
 *

 * }

Where [ n/2 ] denotes the integer part of n/2, e.g. n=5, [ n/2 ] =2.
 * {| style="width:100%" border="0"

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

{{P}_{0}}(x)=1

$$
 *

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

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

{{P}_{1}}(x)=x

$$
 *

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

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

{{P}_{2}}(x)=\frac{1}{2}(3{{x}^{2}}-1)

$$
 *

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

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

{{P}_{3}}(x)=\frac{1}{2}(5{{x}^{3}}-3x)

$$
 *

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

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

{P_4}(x) = \frac{8}{x^4} - \frac{4}{x^2} + \frac{3}{8}

$$
 *

 * }

Objectives
Verify P0(x) – P4(x) using the Legendre polynomial.

Solutions
1). When n=0, [ n/2 ] =0, the Equation (6-1) becomes:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{P}_{0}}(x)={{(-1)}^{0}}\frac{0!{{x}^{0}}}{{{2}^{0}}\centerdot 0!\centerdot 0!}=1

$$     (Eq 6.2 ) 2). When n=1, [ n/2 ] =0, the Equation (6-1) becomes:
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{P}_{1}}(x)={{(-1)}^{0}}\frac{2!{{x}^{1}}}{{{2}^{1}}\centerdot 0!\centerdot 0!}=x

$$     (Eq 6.3) 3). When n=2, [ n/2 ] =1, the Equation (6-1) becomes:
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{P}_{2}}(x)={{(-1)}^{0}}\frac{4!{{x}^{2}}}{{{2}^{2}}\centerdot 2!\centerdot 2!}+{{(-1)}^{1}}\frac{2!{{x}^{0}}}{{{2}^{2}}\centerdot 1!\centerdot 0!}=\frac{1}{2}(3{{x}^{2}}-1)

$$     (Eq 6.4) 4). When n=3, [ n/2 ] =1, the Equation (6-1) becomes:
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{P}_{3}}(x)={{(-1)}^{0}}\frac{6!{{x}^{3}}}{{{2}^{3}}\centerdot 3!\centerdot 3!}+{{(-1)}^{1}}\frac{4!{{x}^{0}}}{{{2}^{3}}\centerdot 1!\centerdot 1!\centerdot 2!}=\frac{1}{2}(5{{x}^{3}}-3x)

$$     (Eq 6.5) 5). When n=4, [ n/2 ] =2, the Equation (6-1) becomes:
 *

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

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

{{P}_{4}}(x)={{(-1)}^{0}}\frac{8!{{x}^{4}}}{{{2}^{4}}\centerdot 4!\centerdot 4!}+{{(-1)}^{1}}\frac{6!{{x}^{2}}}{{{2}^{4}}\centerdot 1!\centerdot 2!\centerdot 3!}+{{(-1)}^{2}}\frac{4!{{x}^{0}}}{{{2}^{4}}\centerdot 2!\centerdot 0!\centerdot 2!}=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8}

$$     (Eq 6.6)
 *

 * }

Problem solved by Shengfeng Yang.

=Problem 2.7: Constructing the Gram Matrix= Refer to lecture slide [[media:nm1.s11.mtg9.djvu|9-1]] for the problem statement.

Given
Legendre polynomials: Where $$[\frac{n}{2}]=$$ integer part of $$\frac{n}{2}$$.

Objectives
Construct (Gram matrix) $$\underline{\Gamma }$$; show $$\underline{\Gamma }$$ is diagonal or not; and compute $$ \det (\underline{\Gamma })$$.

Construct (Gram matrix) $$\underline{\Gamma }$$
By definition of Legendre polynomials: We can obtain: For $$n=0$$, For $$n=1$$, For $$n=2$$, For $$n=3$$, For $$n=4$$, For $$n=5$$, By definition of Gram matrix, we have

Hence, Check Result: Wolframe|Alpha. Check Result: Wolframe|Alpha. Check Result: Wolframe|Alpha. Check Result: Wolframe|Alpha. For $$[a,b]=[-1,1]$$, we have

compute $$ \det (\underline{\Gamma })$$
Problem solved by Hailong Chen.

Given
Newton-Cotes formula: WhereLagrange interpolating functions and $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$. Simple trapezoidal rule: Where $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$.

Objectives
Show when $$n=1$$, Newton-Cotes method implies Simple trapezoidal rule.

Solution
By definition of Lagrange interpolating functions, we can obtain: For $$n=1 $$,$$ {{x}_{0}}=a$$,$$ {{x}_{1}}=b$$ Hence, by definition of Newton-Cotes formula, we consequently have Thus, We get Simple trapezoidal rule from Newton-Cotes method when $$n=1$$. Problem solved by Hailong Chen.

Given

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

(9.1)
 * $$\displaystyle P_n (x) = \sum_{i=0}^{[n/2]} (-1)^i \frac{(2n-2i)! x^{n-2i}}{2^n i! (n-i)! (n-2i)!}$$
 *

 * }
 * }

Objectives
a.) Use 9.1 to generate P5(x) and compute the roots of P5(x) to check values in the table on lecture slide [[media:nm1.s11.mtg8.djvu|7-5]]. Plot the roots on [-1,1].

b.) Repear steps in a. for P10(x). Observe the location of the roots near endpoints -1 and +1.

Solution
a.)Using Wolfram Alpha to compute P5(x) we get:

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

(9.2)
 * $$\displaystyle P_5 (x) = \frac{63 x^5}{8} - \frac{35 x^3}{4} + \frac{15 x}{8}$$
 *

 * }
 * }

Matlab code to produce the roots of P5(x) and the plot of the roots:

>> C = [63/8 0 -35/4 0 15/8 0]; >> x = roots(C)

x =

0 -0.906179845938664  -0.538469310105683   0.906179845938664   0.538469310105683 >> y = [0;0;0;0;0]; >> plot(x,y,'.','markersize',15)

These roots compare well with the values in the table in lecture [[media:nm1.s11.mtg8.djvu|7-5]].

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

(9.3)
 * $$\displaystyle 0 = 0$$
 *

 * }
 * }

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

(9.4)
 * $$\displaystyle \pm0.538469310105683 \approx \pm \frac{1}{3} \sqrt{5 - 2\sqrt{10/7}}$$
 *

 * }
 * }

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

(9.5)
 * $$\displaystyle \pm0.906179845938664 \approx \pm \frac{1}{3} \sqrt{5 - 2\sqrt{10/7}}$$
 *

 * }
 * }

Below is a plot of the roots of P5(x):

b.)Using Wolfram Alpha to compute P10(x) we get:

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

(9.6)
 * $$\displaystyle P_{10} (x) = \frac{46189 x^{10}}{256} - \frac{109395 x^8}{256} + \frac{45045 x^6}{128} - \frac{15015 x^4}{128} + \frac{3465 x^2}{256} - \frac{63}{256}$$
 *

 * }
 * }

Matlab code to produce the roots of P10(x) and the plot of the roots:

>> C = [46189/256 0 -109395/256 0 45045/128 0 -15015/128 0 3465/256 0 -63/256]; >> x = roots(C)

x =

-0.973906528517167 -0.865063366688989  -0.679409568299026   0.973906528517167   0.865063366688989   0.679409568299024  -0.433395394129247   0.433395394129247  -0.148874338981631   0.148874338981631

>> y = [0;0;0;0;0;0;0;0;0;0]; >> plot(x,y,'.','markersize',15)

Below is a plot of the roots of P10(x):

We note that the roots are located disproportionately near the edges of our integration interval. Runge's Phenomenon is defined as a problem of oscillation near the end points of an interval. The more roots we have near the edges of our interval, the more severe this problem will be.

Problem solved by Erle Fields.

=Problem 2.10: Expression for {Ci}=

Refer to lecture slides [[media:nm1.s11.mtg10.djvu|10-2 and 10-4]] for the problem statement.

Given
Equations (2) and (3) on lecture slide [[media:nm1.s11.mtg10.djvu|10-2]]:

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

(10.1)
 * $$\displaystyle P_2 (x) = c_2x^2 + c_1x + c_0 $$
 *

 * }
 * }

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

(10.2)
 * $$\displaystyle P_2 (x_i) = f(x_i) \, for \, i = 0, 1, 2 $$
 *

 * }
 * }

Objectives
Use equations (10.1) and (10.2) to find an expression for {Ci} in terms of (xi,f(xi)) for i = 0, 1, 2.

Solution
From equation (4) on lecture slide [[media:nm1.s11.mtg10.djvu|10-2]] we know:

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

(10.3)
 * $$\displaystyle P_2 (x) = \sum_{i=0}^{n=2}l_{i,2}(x)f(x_i) $$
 *

 * }
 * }

Where, from equation (1) on lecture slide [[media:nm1.s11.mtg9.djvu|9-2]]

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

(10.4)
 * $$\displaystyle l_{i,2}(x) = \prod_{j=0,j \ne i}^{n=2}\frac{x-x_j}{x_i-x_j} $$
 *

 * }
 * }

Now we solve for $$ l_{0,2}(x),$$  $$l_{1,2}(x)$$  and  $$l_{2,2}(x) $$

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

(10.5)
 * $$\displaystyle l_{0,2}(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} $$
 *

 * }
 * }

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

(10.6)
 * $$\displaystyle l_{1,2}(x) = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} $$
 *

 * }
 * }

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

(10.7)
 * $$\displaystyle l_{2,2}(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} $$
 *

 * }
 * }

Using equation (10.3) we get:

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

(10.8)
 * $$\displaystyle P_2 (x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}f(x_0) + \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}f(x_1) + \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}f(x_2) $$
 *

 * }
 * }

Combining terms of $$ x^2 $$  and  $$ x $$ we get:

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

(10.9)
 * $$\displaystyle P_2 (x) = \Big(\frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)}\Big)x^2 $$
 *

 * }
 * }

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

 * $$\displaystyle - \Big(\frac{(x_1+x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0+x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0+x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)}\Big)x $$

 * }
 * }

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

 * $$\displaystyle + \frac{(x_1x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)} $$

 * }
 * }

After setting equation (10.1) equal to (10.9) we get:

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

(10.10)
 * $$\displaystyle C_2 = \frac{f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{f(x_2)}{(x_2-x_0)(x_2-x_1)} $$
 *

 * }
 * }

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

(10.11)
 * $$\displaystyle C_1 = - \Big(\frac{(x_1+x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0+x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0+x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)}\Big) $$
 *

 * }
 * }

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

(10.12)
 * $$\displaystyle C_0 = \frac{(x_1x_2)f(x_0)}{(x_0-x_1)(x_0-x_2)} + \frac{(x_0x_2)f(x_1)}{(x_1-x_0)(x_1-x_2)} + \frac{(x_0x_1)f(x_2)}{(x_2-x_0)(x_2-x_1)} $$
 *

 * }
 * }

Problem solved by Erle Fields.

Given
Equation (4) on lecture slide [[media:nm1.s11.mtg10.djvu|10-2]]:

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

(11.1)
 * $$\displaystyle P_2 (x) = \sum_{i=0}^{2} l_{i,2}(x) f(x_i)$$
 *

 * }
 * }

Simple Simpson's Rule given in equations (2) and (3) on lecture slide [[media:nm1.s11.mtg7.djvu|7-4]]:

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

(11.2)
 * $$\displaystyle \int_a^bf(x)dx \approx I_2 = \int_a^bP_2(x)dx = \frac{h}{3}[f(x_0) + 4f(x_1) +f(x_2)]$$
 *

 * }
 * }

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

(11.3)
 * $$\displaystyle h = \frac{b-a}{2}$$
 *

 * }
 * }

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

 * $$\displaystyle x_0 = a$$

 * }
 * }

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

 * $$\displaystyle x_1 = \frac{a+b}{2}$$

 * }
 * }

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

 * $$\displaystyle x_2 = b$$

 * }
 * }

Objectives
Use equation (11.1) to derive the Simple Simpson's Rule (11.2).

Solution
From equation (2) on lecture slide [[media:nm1.s11.mtg9.djvu|9-2]]:

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

(11.4)
 * $$\displaystyle \int_a^bP_2(x)dx = \sum_{i=0}^{2} \int_a^bl_{i,2}(x)dx f(x_i)$$
 *

 * }
 * }

From equation (1) on lecture slide [[media:nm1.s11.mtg9.djvu|9-2]]:

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

(11.5)
 * $$\displaystyle l_{i,2}(x) = \prod_{j=0,j \ne i}^{n=2}\frac{x-x_j}{x_i-x_j} $$
 *

 * }
 * }

Now we solve for $$ l_{0,2}(x),$$  $$l_{1,2}(x)$$  and  $$l_{2,2}(x) $$

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

(11.6)
 * $$\displaystyle l_{0,2}(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)} $$
 *

 * }
 * }

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

(11.7)
 * $$\displaystyle l_{1,2}(x) = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)} $$
 *

 * }
 * }

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

(11.8)
 * $$\displaystyle l_{2,2}(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)} $$
 *

 * }
 * }

Using Wolfram Alpha to solve the following integrals we get:

WA

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

(11.9)
 * $$\displaystyle \int_a^b l_{0,2}(x)dx = \int_{a}^{b} \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}dx = \frac{b-a}{6}$$
 *

 * }
 * }

WA

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

(11.10)
 * $$\displaystyle \int_a^b l_{1,2}(x)dx = \int_{a}^{b} \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}dx = -\frac{2}{3}(a-b)$$
 *

 * }
 * }

WA

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

(11.11)
 * $$\displaystyle \int_a^b l_{2,2}(x)dx = \int_{a}^{b} \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}dx = \frac{b-a}{6}$$
 *

 * }
 * }

Using Equation (11.4) and the results of equations (11.9), (11.10) and (11.11) we get:

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

(11.12)
 * $$\displaystyle \int_a^bP_2(x)dx = \frac{b-a}{6}f(x_0) + \frac{2}{3}(b-a)f(x_1) + \frac{b-a}{6}f(x_2)$$
 *

 * }
 * }

Simplifying equation (11.12) leaves us with the Simple Simpson's Rule:

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

(11.13)
 * $$\displaystyle I_2 = \int_a^bP_2(x)dx = \frac{b-a}{6}[f(x_0) + 4f(x_1) +f(x_2)] = \frac{h}{3}[f(x_0) + 4f(x_1) +f(x_2)]$$
 *

 * }
 * }

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

 * $$\displaystyle h = \frac{b-a}{2}$$

 * }
 * }

Problem solved by Erle Fields.

=Problem 2.12: Estimation by Lagrange Interpolation Functions=

Given
$$ f(x) = \frac{e^x-1}{x}, [-1,1], x_0 = -1, x_n = 1, n= 1,2,4,8,16 $$

Objectives
2-12-1) Construct

2-12-2) Plot

Solution
2-12-1) Plot

Matlab Code:

2-12-2) Plot Matlab Code:

Given
Trapezoidal rule: Simple rule: Composite rule: Where $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$. $$a={{x}_{0}}<{{x}_{1}}<\cdots <{{x}_{n-1}}<{{x}_{n}}=b$$. The distance among consecutive nodes is $$h=\frac{b-a}{n}$$. Simpson’s rule: Simple rule: Composite rule: Where $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$. >. $$a={{x}_{0}}<{{x}_{1}}<\cdots <{{x}_{n-1}}<{{x}_{n}}=b$$. The distance among consecutive nodes is $$h=\frac{b-a}{n}$$.

Objectives
Derive each composite rule from the simple rule.

Trapezoidal rule
Divide the interval into $$n$$ subintervals with equal length, hence $$h=\frac{b-a}{n}$$. Break the integral into $$n$$ subintegrals, Approximate each subintegrals using simple rule, then Hence, we get the composite trapezoidal rule.

Simpson’s rule
Using Newton-Cotes formula, we can get simple Simpson’s rule as follow: Where $$ c=\frac{b+a}{2}$$. In order to evaluate above integral directly, we use $$h=\frac{b-a}{2}$$ to make it  easy to solve. Then The complete evaluation of equation $$4$$ yields Like what we do in deriving composite trapezoidal rule, dividing interval $$[a,b]$$ into $$ n$$ subintervals, each containing three nodes. Thus Approximating each subintegrals by equation $$8$$ yields Hence, we get the composite Simpson’s rule. Problem solved by Hailong Chen.

Objective
Give details of how to obtain (14.1)

Solution
Let $$\displaystyle x=x(y)$$ be a function of y that satisfies (14.1). To satisfy the boundary conditions x(y) must satisfy the following conditions:

The solution to these conditions lies upon a line passing through the two ordered pairs.

Taking the partial derivative of each side yields an expression for dx:

Plugging these expression back into the middle term of (14.1) yields

Now, let

Combining equations (14.6) and (14.7) proves (14.1):

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

$$\displaystyle \therefore I(f) = \int_{a}^{b}f(x)dx = \int_{-1}^{1}\bar{f}(y)dy $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 *
$$\displaystyle
 * style= |
 * }
 * }

Solved by William Kurth and Brendan Mahon.

title: User:Egm6341.s11.team4/HW2


text:
Computer Architecture Lab/SS2011

Project page for the course in summer 2011.

Group 1 AID
Nicolas Ingemann Borup

Jonas Andersen Dindorp

Cristian Pandele

Instruction Set 1
Summary and comparison between instruction sets of:

 * 1) MIPS
 * 2) ARM7
 * 3) LC3
 * 4) Zilog Z8

Document: Assignment1

Architecture implementing decimal (AID)
This page contains information about the implementation of a decimal processor in the course Advanced computer architecture on DTU

 * [Report]

Instruction set architecture
We have several formats for the instruction set, as the operations needs different inputs

Type 1:

Type 2:

Type 3:

Type 4:

Type 5:

Group 2 FooBarBaz fighters
Mark Ruvald Pedersen

Laust Brock-Nannestad

Jaspur Højgaard

Rasmus Bo Sørensen

Assignment I
Summary and comparison between instruction sets of:

 * 1) PIC16F84 microcontroller
 * 2) ARM
 * 3) Intel 80x86

 * [Download pdf]

ISA / CPU Design
Document (Assignment II):
 * [Initial project description]

 * [Final report]

 * [Our SVN repo]

UART Exercise (Assignment III)

 * [Source code]

Group 3 [Newbies]
Ioannis Chionidis

Georgios Papanikas

Xia Hua Henning Dietmar Weiss

Document:
 * [Assignment 1 pdf]
 * [ISA Reference pdf]

Group 4
Andreas Erik Hindborg

Niklas Quarfot Nielsen

Tobias Jeppe

Kristoffer Lyder Andersen

Assignment 1
Summary of:
 * 1) ARMv7-M
 * 2) PIC
 * 3) x86
 * 4) POWER7

Document: https://www.campusnet.dtu.dk/cnnet/filesharing/ViewFileHistory.aspx?ElementId=347459&FolderId=567165&FileId=2700363

Group ??
Caroline Johannsen

Assignment I
...

Leros
Martin Schoeberl (no group - this is just a template ;-)

An FPGA optimized tiny processor core for utility functions (e.g., SW UART). The challenge is to get the resources below 500 LC and use just 2 RAM blocks. The processor is named after the Greek island Leros where the architecture was designed. Details of the design can be found at the Leros processor page.

Perhaps I will do this design along the course to see how much work the assignment actually is ;-)

Group 6
Dimitrios Kosmadakis

Assignment I
Summary of:
 * 1) Intel 4004
 * 2) Intersil 9100
 * 3) TMS 9900

title: Computer Architecture Lab/SS2011


title: File:Nm1.s11.team5.HW2.P9.fig1.png


title: File:Nm1.s11.team1.HW2.fig1.jpg


title: File:Nm1.s11.team1.HW2.fig2.jpg


title: File:Nm1.s11.team1.HW2.fig3.jpg


title: File:Nm1.s11.team1.HW2.fig4.jpg


text:
User:Egm6341.s11.team4.hylon/hw2

=Homework 2.7= Refer to lecture slide [[media:nm1.s11.mtg9.djvu|9-1]] for the problem statement.

Given
Legendre polynomials: Where $$[\frac{n}{2}]=$$ integer part of $$\frac{n}{2}$$.

Objectives
Construct (Gram matrix) $$\underline{\Gamma }$$; show $$\underline{\Gamma }$$ is diagonal or not; and compute $$ \det (\underline{\Gamma })$$.

Construct (Gram matrix) $$\underline{\Gamma }$$
By definition of Legendre polynomials: We can obtain: For $$n=0$$, For $$n=1$$, For $$n=2$$, For $$n=3$$, For $$n=4$$, For $$n=5$$, By definition of Gram matrix, we have

Hence, Check Result: Wolframe|Alpha. Check Result: Wolframe|Alpha. Check Result: Wolframe|Alpha. Check Result: Wolframe|Alpha. For $$[a,b]=[-1,1]$$, we have

compute $$ \det (\underline{\Gamma })$$
Problem solved by Hailong Chen.

Given
Newton-Cotes formula: WhereLagrange interpolating functions and $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$. Simple trapezoidal rule: Where $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$.

Objectives
Show when $$n=1$$, Newton-Cotes method implies Simple trapezoidal rule.

Solution
By definition of Lagrange interpolating functions, we can obtain: For $$n=1 $$,$$ {{x}_{0}}=a$$,$$ {{x}_{1}}=b$$ Hence, by definition of Newton-Cotes formula, we consequently have Thus, We get Simple trapezoidal rule from Newton-Cotes method when $$n=1$$. Problem solved by Hailong Chen.

Given
Trapezoidal rule: Simple rule: Composite rule: Where $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$. $$a={{x}_{0}}<{{x}_{1}}<\cdots <{{x}_{n-1}}<{{x}_{n}}=b$$. The distance among consecutive nodes is $$h=\frac{b-a}{n}$$. Simpson’s rule: Simple rule: Composite rule: Where $$f(x)$$ is the original function to be interpolated, with $$x\in [a,b]$$. >. $$a={{x}_{0}}<{{x}_{1}}<\cdots <{{x}_{n-1}}<{{x}_{n}}=b$$. The distance among consecutive nodes is $$h=\frac{b-a}{n}$$.

Objectives
Derive each composite rule from the simple rule.

Trapezoidal rule
Divide the interval into $$n$$ subintervals with equal length, hence $$h=\frac{b-a}{n}$$. Break the integral into $$n$$ subintegrals, Approximate each subintegrals using simple rule, then Hence, we get the composite trapezoidal rule.

Simpson’s rule
Using Newton-Cotes formula, we can get simple Simpson’s rule as follow: Where $$ c=\frac{b+a}{2}$$. In order to evaluate above integral directly, we use $$h=\frac{b-a}{2}$$ to make it  easy to solve. Then The complete evaluation of equation $$4$$ yields Like what we do in deriving composite trapezoidal rule, dividing interval $$[a,b]$$ into $$ n$$ subintervals, each containing three nodes. Thus Approximating each subintegrals by equation $$8$$ yields Hence, we get the composite Simpson’s rule. Problem solved by Hailong Chen.

title: User:Egm6341.s11.team4.hylon/hw2


title: File:Octave2-5.jpg


text:
User:Egm6341.s11.team3.ren

HW 2.13
(a) Trapezoidal rule Simple rule: $$I_1=\dfrac{b-a}{2}\left[f(a)+f(b)\right] $$

Composite rule:

$$I_n=\dfrac{x_1-x_0}{2}[f(x_0)+f(x_1)]+\dfrac{x_2-x_1}{2}[f(x_1)+f(x_2)]+\ \cdot\cdot\cdot\ +\dfrac{x_n-x_{n-1}}{2}[f(x_{n-1})+f(x_n)]$$

$$=\sum^n_{j=0}\dfrac{x_j-x_{j-1}}{2}[f(x_{j-1})+f(x_j)]$$

Using uniform intervals, $$h=x_j-x_{j-1} (j=0,1,2,\cdot\cdot\cdot,n)$$, the integral becomes

$$I_n=\frac{h}{2}\sum^n_{j=0}[f(x_{j-1})+f(x_j)]=\frac{h}{2}[f(x_0)+2f(x_1)+2f(x_2)+\ \cdot\cdot\cdot\ +2f(x_{n-1})+f(x_n)]$$ $$=h[\frac{1}{2}f(x_1)+f(x_2)+\ \cdot\cdot\cdot\ +f(x_{n-1})+\frac{1}{2}f(x_n)]$$

(b) Simpson's rule

Simple rule: $$I_2=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]$$

Composite rule:

Using uniform intervals, $$h=(b-a)/n\ (n=2k, k==0,1,2,\cdot\cdot\cdot)$$, the integral can be written as

$$I_n=\frac{h}{3}\sum^{n-1}_{j=1,3,5,\cdot\cdot\cdot}[f(x_{j-1})+4f(x_j)+f(x_{j+1})]$$

$$=\dfrac{h}{3}[f(x_0)+4f(x_1)+f(x_2)]+\dfrac{h}{3}[f(x_2)+4f(x_3)+f(x_4)]+\ \cdot\cdot\cdot\ +\dfrac{h}{3}[f(x_{n-2})+4f(x_n)+f(x_n)]$$

$$=\dfrac{h}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4)+ \cdot\cdot\cdot\ +2f(x_{n-2})+4f(x_n)+f(x_n)]$$

Egm6341.s11.team3.ren 22:55, 30 January 2011 (UTC)

HW 2.14

Assume y is related to x in a simple form $$y=g(x)=C_1x+C_2$$. We want y=-1 as x=a, and y=1 as x=b, which is

$$g(a)=aC_1+C_2=-1$$

$$g(b)=bC_1+C_2=1$$

Solving for $$C_1$$ and $$C_2$$, we have

$$C_1=\frac{2}{b-a}$$

$$C_2=\frac{a+b}{a-b}$$

$$\Rightarrow\ y=\frac{2}{b-a}x+\frac{a+b}{a-b}\ \Rightarrow\ x=\frac{b-a}{2}y-\frac{1}{2}\ \Rightarrow\ dx=\frac{b-a}{2}dy$$

$$\Rightarrow\ \int^{b}_{a}f(x)dx=\int^{1}_{-1}f(\frac{b-a}{2}y-\frac{1}{2})\frac{b-a}{2}dy=\int^{1}_{-1}\bar f(y)dy$$

Therefore,

$$\bar f(y)=\frac{b-a}{2}f(\frac{b-a}{2}y-\frac{1}{2})$$

Given
An ideal mass-spring-damper system with mass m, spring constant k and viscous damper of damping coefficient c is subject to an constant force F=u.

Find
1) Derive equation of motion in terms of d,c,k,m,u.

2) Let $$ x=\left\{ {\begin{array}{*{20}{c}} d \\  \dot d  \\ \end{array}} \right\} =\left\{ {\begin{array}{*{20}{c}} x_1  \\  x_2  \\ \end{array}} \right\}  \ $$, find $$\left(F,G\right)$$

3) Find $$c_{cr}$$ in terms of k,m such that system is critically damped.

4) Let $$k=1,m=1/2,x_0=[0.8,-0.4]^T$$

a) For u=0, plot $$x_k$$ for $$c=\frac{1}{2}c_{cr}, c_{cr}, \frac{3}{2}c_{cr}$$.  b) For u=0.5 Gaussian noise, and $$c=\frac{3}{2}c_{cr}$$, plot $$x_k$$ c) For u=0.5 Cauchy noise, and $$c=\frac{3}{2}c_{cr}$$, plot $$x_k$$

Solution
1)

For the spring k
 * {| style="width:100%" border="0"

$$ f_k=kd \ $$ $$ (Eq.5.4.1) \ $$
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 *
 * }

For the damping c
 * {| style="width:100%" border="0"

$$ f_c=c\cdot \frac{d}{dt}d(t)=c\dot d \ $$ $$ (Eq.5.4.2) \ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

For the mass m
 * {| style="width:100%" border="0"

$$ m\frac{d^2}{dt^2}d(t) =u-f_k-f_c  \ $$ $$ (Eq.5.4.3) \ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$ \Rightarrow\quad m\ddot d = u - c\dot d-kd \quad \Rightarrow\quad  m\ddot d + c\dot d+kd= u  \ $$ $$ (Eq.5.4.4) \ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

2)

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

$$ x=\left\{ {\begin{array}{*{20}{c}} d \\  \dot d  \\ \end{array}} \right\} =\left\{ {\begin{array}{*{20}{c}} x_1  \\  x_2  \\ \end{array}} \right\}  \ $$ $$ (Eq.5.4.5) \ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

From (Eq.5.4.4), we have

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

$$ \left\{ {\begin{array}{*{20}{c}} \dot x_1 = x_2 \\  \dot x_2 = u - \frac{c}{m}x_2-\frac{k}{m}x_1\\ \end{array}} \right. \ $$   $$ (Eq.5.4.6) \ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

that is

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

$$ \left\{ {\begin{array}{*{20}{c}} \dot x_1 \\ \dot x_2 \\ \end{array}} \right\} = \left[ {\begin{array}{*{20}{c}} 0 & 1 \\  -\frac{k}{m} & -\frac{c}{m} \\ \end{array}} \right] \left\{ {\begin{array}{*{20}{c}} x_1 \\ x_2 \\ \end{array}} \right\} + \left[ {\begin{array}{*{20}{c}} 0 & 0 \\  1 & 0 \\ \end{array}} \right] \left\{ {\begin{array}{*{20}{c}} u \\ 0 \\ \end{array}} \right\} $$   $$ (Eq.5.4.7) \ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

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

$$ \Rightarrow \quad A = \left[ {\begin{array}{*{20}{c}} 0 & 1 \\  -\frac{k}{m} & -\frac{c}{m} \\ \end{array}} \right] \quad B= \left[ {\begin{array}{*{20}{c}} 0 & 0 \\  1 & 0 \\ \end{array}} \right]$$ $$ (Eq.5.4.8) \ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 * }

According to lecture notes

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

$$ \Rightarrow \quad F=I+\Delta t A = \left[ {\begin{array}{*{20}{c}} 1 & \Delta t \\ -\frac{k\Delta t}{m} & 1-\frac{c\Delta t}{m} \\ \end{array}} \right] \quad G=\Delta t B= \left[ {\begin{array}{*{20}{c}} 0 & 0 \\  \Delta t & 0 \\ \end{array}} \right]$$ $$ (Eq.5.4.9) \ $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

3) For critical damping system, the governing equation can be written as
 * {| style="width:100%" border="0"

$$ \ddot d + 2\omega_0\dot d+\omega^2 d=const. \ $$   $$ (Eq.5.4.10) \ $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

Comparing (Eq.5.4.10) with (Eq.5.4.4), we know

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

$$ \frac{k}{m} = \left(\frac{c_{cr}}{2m}\right)^2 \quad\Rightarrow\quad c_{cr}=2\sqrt{mk}\ $$ $$ (Eq.5.4.11) \ $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

4) The governing equation can be integrated according to

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

$$ x_{k+1}=Fx_k+Gu_{k+1} \quad\ $$ $$ (Eq.5.4.11) \ $$ with  $$\quad x_0=[0.8,-0.4]^T \ $$
 * style="width:95%" |
 * style="width:95%" |
 *

 * }

title: User:Egm6341.s11.team3.ren


text:
SPIR608 Political Simulations and Gaming/2011/Week 4

Friday 11th February

 * Week 4 Discussion

Links

 * Photos of Week 4 game playing.
 * War on Terror - the board game.
 * Origins of World War II on BoardGameGeek.
 * A.J.P. Taylor and the origins of the Second World War controversy.
 * Game Theory .net - educational resources about Game Theory.
 * Benjamin Polak, Yale Video Course on Game Theory.
 * BBC & Open University, Interactive Prisoner's Dilemma - requires Shockwave plug-in.


text:
SPIR608 Political Simulations and Gaming/2011/Week 5

Friday 18th February

 * Week 5 Discussion

Links

 * Photos of Week 5 game playing.
 * Comrade Koba on BoardGameGeek.
 * Modern Society on BoardGameGeek.
 * Markku Eskelinen, The Gaming Situation.
 * Ruth Catlow, Rethinking Wargames - pawns join together to defend world peace.
 * Molly Hankwitz, Re-thinking Wargames: a chance to remaster conflict - interview with Ruth Catlow.
 * Andy Cameron, Dissimulations.
 * Tetsuya Mizguchi, Rez.