text:
User:Thanin89

Each student answer the questions on his or her wikiversity User page:

How do you open the Select Pane in PowerPoint 2007? In PowerPoint 2007, how do you change the maximum number of Undoes to 30? How do you save a PowerPoint 2007 file as a web page?

office button/save as/go to save as type:/select "web page"/save. How do you change the color of a SmartArt object in PowerPoint 2007?

right click on the Smart art/ Format Shape/color/ select color/close. In PowerPoint 2007, how do you include slide numbers on every slide?

View/Presentation views/slide master/insert/header&footer/slide number. How do you change the slide orientation of a PowerPoint 2007 presentation to portrait? How do you modify a textbox so that its text shrinks to fill the space in the shape in PowerPoint 2007? How do you add a circle shape to a slide in PowerPoint 2007? On the Ribbon in PowerPoint 2007, how do you change text in the textbox to WordArt? How do you open a current presentation in PowerPoint 2007? How do you turn on the grids in PowerPoint 2007? What will allow you to quickly make a PowerPoint presentation as quickly as you can type? How do you go to Handout Master in PowerPoint 2007#What does the Contrast function do in PowerPoint 2007? In PowerPoint 2007, how do you print out the slide you are currently looking at? Retrieved from "http://en.wikiversity.org/wiki/User:Icns141/In-Class_Active_Quiz/February_12,_Section_2"

title: User:Thanin89


text:
User talk:Trips303

 Hello Trips303, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 12 February 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 12 February 2010 (UTC)

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

title: User talk:Trips303


text:
User talk:Nlibucha

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

 * Take a guided tour and learn to edit.
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 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 12 February 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 12 February 2010 (UTC)

title: User talk:Nlibucha


text:
User talk:Swestgate 5060

 Hello Swestgate 5060, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 12 February 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 12 February 2010 (UTC)

title: User talk:Swestgate 5060


text:
User talk:Sasfireandsecurity

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

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 12 February 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 12 February 2010 (UTC)

title: User talk:Sasfireandsecurity


text:
The Near Death Experience as Possible Evidence for an Afterlife

Those who study the Near Death Experience tend to remain convinced about whatever they believed before. Those who believed in an afterlife before tend to see strong evidence for afterlife through the NDE while sceptics tend to remain skeptical. Below is a link to a sceptical website that tried to refute common arguments for an afterlife based on the NDE. Hallucinatory Near-Death Experiences. The author of this article is a weak atheist or agnostic and sees no compelling evidence that the NDE is more than the product of a dying brain or a brain that is near death, very much afraid. Different aspects will be discussed.

The life review
There is disagreement about how frequent the life review as a learning experience is though it certainly happens in at least fifteen to twenty percent of cases and other NDE experiencers may have repressed the disturbing memory.

Judgements appear human
"Carter Mills saw himself when he was a child and killed a mother bird with a sling shot. At the time, he was so proud of that shot; but during his life review, he felt the pain that the mother bird's babies went through when they starved to death. (Carter Mills)" That all looks a bit harsh for a wrong done at the age of eight, It seems an adult who did anything similar would not suffer harder.

Why do people who eat factory farmed produce, meat, milk, eggs not report suffering the physical pain, boredom and other stress suffered by factory farmed animals? Why do meat eaters from developed countries not report suffering the pain animals go through during slaughter? Third World People sometimes need to use any protein source they can get to keep themselves and their children alive but people from developed countries can survive successfully as vegetarians. It is well known that factory farmed animals live terrible lives and those who do not know the details are wilfully ignorant. Further many people eat far more meat than they need to stay healthy (even without plant protein to suplement their diet) and damage their health by eating too much of the wrong type of meat. As far as this author knows no NDE experiencers report suffering for the unnecessary stress they cause farm animals when they eat more meat than they need.

Basically this is consistent with our dying brains or brains near death judging us. We appear to judge ourselves more impartially than we normally do during life but there is still that, ‘If everyone else is doing what I do then it can’t be very bad.’ feel to the judgement and it looks more like a human judgement than the judgement of a divine omniscient being.

Why NDE experiencers may feel pleasure and pain they have caused others
Reports of the life review frequently include account of having felt pleasant or painful feelings that the NDE experiencer has caused to other people or even to animals as shown above. These pleasant or painful experiences are reported to be so vivid that the experiencer feels exactly as if the pleasant or painful act were being done to himself/herself personally.

Feeling the pleasure and pain of other people, even of animals is a normal part of being human outside the NDE experience. We experience something like another sentient being’s emotions when we see a person or animal feeling a particular way, when we see a film, a photograph, a drawing, hear or read an account. Further when we are kind or cruel to another we conceptualize the other’s feelings by understanding similar feelings that we would have in that situation. We use this conceptualization of the other’s feelings to fine tune our actions and get the result we want whether we are being ethical or unethical.

During ordinary life outside the NDE such feelings are nowhere near as strong as similar feelings would be if we were on the receiving end of any kind or unkind act. If our sympathies with other sentient beings were too vivid this would be disabling and might preoccupy us to the point where we could not run our lives. Further we are genetically programmed to look after our selfish genes and the genes of relatives preferentially. If during ordinary life we truly felt the pleasure and pain of others the way we feel our own pleasure and pain we would be forced to look after all sentient beings the way we look after ourselves and our relatives. Evolution would never allow us to become like that.

During ordinary life either our sympathetic emotions when we do things to others are naturally weaker than similar emotions are when others do things to us or there exist inhibitory mechanisms that prevent sympathetic emotions being so powerful that they become disabling or both apply. The author is not a neuroscientist and knows of no such inhibitory mechanism, none the less if there is an inhibitory mechanism explaining vivid sympathetic emotion during the life review is easy. All that is needed is to postulate that the inhibitory mechanism breaks down and supernatural explanations are not needed.

Expectations also influence what people experience during the NDE, see Cultural Differences, many cultures teach us to expect some type of judgment at death, therefore the NDE frequently includes some kind of judgement. That can predispose NDE experiencers to feel vivid pleasure over past actions that they believe were ethical and to feel vivid pain over past actions that make them feel guilty.

Neurobiological hypotheses
Most of this section is by a different author.
 * "During this process of disintegration of neural connections and the neural network, reminiscences, memories and building stones of the personal identity are ‘released’ and strengthen a feeling of time travel and life review." Heaven can wait -or down to earth in real time: Near-death experience revisited by C. van Tellingen

Related resources

 * Science Fiction Challenge


text:
User:Alexander Cherkasky

I am biologist with Masters degree and inventor. My inventions include new anticancer and anti-inflammatory treatments, novel multiparallel diagnostic systems, novel fast genome sequencing systems, new lubricants and new materials. I have 34 publications in patent literature including German patents, US patent applications, international PCT patent applications and German Patent applications. I got 7 German patents and have also 4 published US Patent Applications. I invent since early youth and as I was student at Goethe-Gymnasium in Dusseldorf, I presented my inventions of Anti-Alzheimer's disease therapy and anticancer therapy on scientific youth competitions "Youth researches" (Yugend forscht) in the years 1999 and 2000. In 1999 I have presented my invention of biomolecules that specifically bind and cleave pathogenic proteins especially so-called beta-amyloid plaques. In 2000, I have presented fusion proteins that selectively inhibit division of target cancer cells. It evoked interest of journalists. About me and inventions of mine reported known newspapers and magazines including Stern(The Star), Rheinische Post (The Rhein Post), Westdeutsche Zeitung (Westgerman Newspaper) and other media. I am also the inventor of ordered DNA winding around the bobbins (my German Patent DE 19929530). In the year 2003, I was awarded the Bronze Medal "For Outstanding Achievements" and the Honorary Certificate of the International Exhibition and Competition of Inventors "Ideas-Inventions-New Products" IENA 2003 in Nuremberg. IENA is the largest international inventors exhibition and competition in Germany already over 60 years. My name was also placed on the Young Inventors International Hall of Fame. I was only one inventor from Germany whose name was placed on this Hall of Fame. My invention of novel lubricants was selected and published by the largest Chinese data provider Wanfang Data, which is affiliated with the Chinese Ministry of Science and Technology and customers of Wanfang include Governments, Library of Congress, Yale and Oxford Universities. Wanfang scanned patent literature worldwide, and recognized in this my lubricant invention a very big commercial potential.

Paradoxally, certain of my inventions were confirmed experimentally with success and were published both in scientific and US patent literature but although my name was not mentioned, my priority and my authorship are documented clearly. The examples include the following:

1. My invention of a medicine comprising protease region for cleavage, combined with the beta-amyloid region for selective recognition and binding in order to cleave specifically the beta-amyloid plaques by Alzheimer’s disease and other amyloid-associated diseases was presented by myself during youth scientific competition “Youth researches” (“Jugend forscht”) in1999 and was highlighted in the most popular magazine in Germany Stern (The Star) (March 25, 1999). In that article my invention was described and named revolution. My patent application DE19822406 for this invention, especially for protease fusion proteins specifically for cleaving amyloid proteins in order to treat amyloid diseases causally, was filed with the German Patent and Trademark Office on May 19,1998 and was published on November 25, 1999. The group of Roger Nitsch, Johannes Pohlner, Axel Wollmer and Hasan Mohajeri which are employed at Evotec NeuroSciences GmbH in Germany and at the University of Zurich, tried to get patents for “Methods and compounds for treating brain amyloidosis”, which is invention of mine. They filed the European patent application EP00120449.4 on September 19, 2000 and then international patent application WO0225279, PCT/EP01/10816, as well as two US patent applications US 20040038302 (filed on July 29, 2003 in the US) and US20060142194 (filed on November 16, 2005). Thus I have the priority. So in addition to the fact of my priority and authorship of my invention, it is well documented and confirmed that my inventions deserve to be implemented commercially. The following are the links to my German patent application DE 19822406 and to two US patent applications US20040038302 and US 20060142194.

http://v3.espacenet.com/publicationDetails/biblio?DB=EPODOC&adjacent=true&locale=en_V3&FT=D&date=19991125&CC=DE&NR=19822406A1&KC=A1

http://appft.uspto.gov/netacgi/nph-Parser?Sect1=PTO1&Sect2=HITOFF&d=PG01&p=1&u=%2Fnetahtml%2FPTO%2Fsrchnum.html&r=1&f=G&l=50&s1=%2220040038302%22.PGNR.&OS=DN/20040038302&RS=DN/20040038302

http://appft.uspto.gov/netacgi/nph-Parser?Sect1=PTO1&Sect2=HITOFF&d=PG01&p=1&u=%2Fnetahtml%2FPTO%2Fsrchnum.html&r=1&f=G&l=50&s1=%2220060142194%22.PGNR.&OS=DN/20060142194&RS=DN/20060142194

2. My invention of socalled Fc-Autoantigen fusion proteins for causal treatment of autoimmune diseases such as multiple sclerosis, was described in my published German patent application DE10160248, filed on December 07,2001 and published on June 26, 2003. The German company Micromet AG, with quarters in Munich and Bethesda, MD, USA, filed later (February 13, 2002) the European patent application EP02003332, and international patent application PCT/EP03/01389, published as WO03/068822 on August 21,2003. Micromet Inc. is on the stock exchange NASDAQ and its ticker symbol is MITI. The examiner with the United States Patent and Trademark Office (USPTO) Prema Mertz missed my published German patent application and granted to Micromet the US Patent 7323440 on January 29, 2008. There are surprisingly and obviously many similarities in claims and description and examples between my earlier patent application and their patent so that the patent to them was issued by an error of the examiner. Below are the links to my patent application DE 10160248 and the US patent 7322440.

http://v3.espacenet.com/publicationDetails/biblio?DB=EPODOC&adjacent=true&locale=en_V3&FT=D&date=20030626&CC=DE&NR=10160248A1&KC=A1

http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO1&Sect2=HITOFF&d=PALL&p=1&u=%2Fnetahtml%2FPTO%2Fsrchnum.htm&r=1&f=G&l=50&s1=7323440.PN.&OS=PN/7323440&RS=PN/7323440

3. A Japanese group of Kyohei Terao, Masao Washizu and Hidehiro Oana experimentally confirmed my inventions for sequencing and manipulation of giant DNA sequences described in two my German patents DE19929530B4 and DE19937512B4 filed in 1999 and published in 2001. In my German patent DE19929530B4 is described my invention of ordered winding DNA around the bobbins and this work was carried out by the above-mentioned Japanese group and claimed as sensation. Their paper “On-site manipulation of single chromosomal DNA molecules by using optically driven mictostructures” was published on June 23, 2008 in the journal Lab on a Chip of the Royal Society of Chemistry in the UK. Their work was highlighted in Chemical Science and in New Scientist. Thereby my German patent DE 19937512 was mentioned by American inventor Rouvain Bension in his US patent 7163658. There are links to the patents DE 199 29530, DE 199 37512 and US 7163658 as well as to certain related articles.

http://v3.espacenet.com/publicationDetails/biblio?DB=EPODOC&adjacent=true&locale=en_V3&FT=D&date=20010104&CC=DE&NR=19929530A1&KC=A1

http://v3.espacenet.com/publicationDetails/biblio?DB=EPODOC&adjacent=true&locale=en_V3&FT=D&date=20010215&CC=DE&NR=19937512A1&KC=A1

http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO1&Sect2=HITOFF&d=PALL&p=1&u=%2Fnetahtml%2FPTO%2Fsrchnum.htm&r=1&f=G&l=50&s1=7163658.PN.&OS=PN/7163658&RS=PN/7163658

http://www.rsc.org/Publishing/Journals/LC/article.asp?doi=b803753a

http://www.rsc.org/Publishing/ChemScience/Volume/2008/07/Miniaturised_sewing_machines.asp

http://www.internetchemie.info/news/2008/jul08/dna-sewing-machine.html

4. The Nobel Laureate Joshua Lederberg tried to get priority for the invention for which I have well documented priority. This invention concerns genetically engineered medicine carriers especially bacteriophage-antibody and bacteriophage-receptor complexes against bacterial diseases and cancer. His US Patent Application 20040033584 was filed as a PCT international application on December 21, 2000, it means later than my earlier German patent application DE19818938A1, filed on April 28, 1998 and published in November 4, 1999. Lederberg was a famous scientist and is considered as the father of molecular biotechnology and his experiments were basis for the experiments of Stanley N. Cohen and Herbert Boyer, inventors of recombinant DNA technology, of genetic engineering. Interestingly, Lederberg’s last invention was my first invention. Please see below the links to DE 19818938 and US 20040033584.

http://v3.espacenet.com/publicationDetails/biblio?DB=EPODOC&adjacent=true&locale=en_V3&FT=D&date=19991104&CC=DE&NR=19818938A1&KC=A1

http://appft.uspto.gov/netacgi/nph-Parser?Sect1=PTO1&Sect2=HITOFF&d=PG01&p=1&u=%2Fnetahtml%2FPTO%2Fsrchnum.html&r=1&f=G&l=50&s1=%2220040033584%22.PGNR.&OS=DN/20040033584&RS=DN/20040033584

These facts are also confirmations for sponsors, investors and potential business partners that my inventions work and are worthy to be implemented commercially.

In conclusion, that the invention implementation process is the key process for economic growth and development I propose the new economic model based upon the renewal of patent law and patent system as well as the creation and establishing of the invention implementation system in order to motivate inventors and more efficiently implement inventions of new therapies and technologies:

1. Condition-Dependent or Condition-Responsive Patents (CoDe Patents or CoRe Patents)

The patent protection will be switched in dependence of or in response to defined conditions (Condition-Responsive Protection Switching). Preferably three levels of protection (Condition Dependent or Condition-Responsive Protection Levels) may be distinguished:

1.1. If an invention does not bring profits to inventor or company (no profit condition) and nobody tried to obtain license, the invention should remain protected both free-of-charge and unlimited in time.

1.2. If invention will be entered to market and begins to bring profits, than patent costs may arise and time of protection may be limited to 20 years from entry to market. Thus, my Cherkasky Theory of Living or Switching Patent System is conform with Anti-Trust Law.

1.3. As soon as an invention pays off initial investments and continues to bring profits to inventor or company, government can be directly shared in these profits in addition to taxes from these profits as compensation for free of charge protection in level 1.1. Switch from 1.1 to 1.2 may be reversible in certain exceptions, but switch from 1.2 to 1.3 is irreversible. The futuristic legislative act can be named Condition-Responsive Patent Act. Thus, the patent system will be made living. The Switching Patent System or Condition-Dependent Switching Patent System issuing Condition-Dependent Patents can be established on the basis of existing patent system. It can be governmentally funded and refunded by incomes from increasing number of successful implementations and commercialisation of more new inventions. The legislative act for establishing such system or for improvement of current patent system can be named Switching Patent System Act or Switching Patent Act.

2. Free of Charge Protection of Inventions

Each inventor should have legal possibility to protect his invention free of charge but for share of government.

3. Expansion of Patent Office Network

For dynamic consideration of increasingly number of filed patent applications, the Patent Office Network can be expanded and contain regional and local offices.

4. Publication after Filing

Patent applications should be published directly after filing and than searched and examined. After-filing-publications will increase transparency and allow secure searches and reduction of investment risks.

5. Invention Implementation Office or Invention Implementation System

Billions are invested in seeking new solutions for healing diseases such as cancer. The futuristic Invention Implementation Act can build the basis for establishing the Invention Implementation System and allow to solve problems dynamically and to maximize innovation, because inventions as problem solutions must be evaluated and implemented independently from status of inventor and high values of inventions may not be reduced or destroyed. I propose to establish Laboratories for Implementation of Inventions especially of Biotechnological Inventions in order to openly constructively evaluate and implement inventions independently from status of inventor. The inventions can be selected in current idea competition. Experimental scientists working for implementation must be rewarded additionally for successful implementations. Thus, more inventions will be implemented, more problems will be dynamically solved and after successful proofs, investment risks will be reduced, managers can be found easier, more new companies can be established with share of government, investors and inventors. Thus, more people will get more new causal therapies.

6. Office for Promotion of Inventors and Inventions

I propose to establish the Office for Promotion of Inventors and Inventions in order to evaluate constructively and scale inventions according to their potential, find producers and managers for joint ventures with inventors and find companies interested in obtaining licences as well as organize significantly rewarded competitions in order to find and promote solutions of problems.

7. Patent Exchange and New Patent Market

Strong patent protection for high values will open new possibilities not only for inventors, but also for investors and for new managers, wishing to make new companies and seeking for suitable ideas.

Parties involved will profit from the suggestions: Government will obtain new sources of income and investors will get more investment options and reduction of risk. Inventors and companies will be more motivated to create in a secure legal medium. More new cures will be created and entered to market as well as more new careers will be possible.

title: User:Alexander Cherkasky


text:
User:5a(

I lost the game

Jokes:

We are sorry, but the phone number you have dialed is imaginary. Please turn your phone 90 degrees and try again.

There are 10 types of people in the world. Those who understand binary, and those who don't.

Why doesn't DOS ever say "EXCELLENT Command or Filename"?

Your /dev/null is 98% full.

Hello, Mr. 5a(. Jobjams 01:01, 28 February 2010 (UTC)

title: User:5a(


text:
User:Gaby nicolas habis

i just done it, on february 12 , 2010

title: User:Gaby nicolas habis


text:
User:Brinpegasus/Literature for teens

Units

 * /The Absolutely True Diary of a Part-Time Indian

title: User:Brinpegasus/Literature for teens


text: Category:Literature for teens

title: Category:Literature for teens


text: File:French Verb - être.ogg

title: File:French Verb - être.ogg


text:
User:Egm6341.s10.team3.sa/HW3

HW Poblem Set 3 = Proof the error bound for Simple Simpson' Rule = Ref: Lecture Notes [[media:Egm6341.s10.mtg17.djvu|p.17-1]]

Problem Statement
 Part 1  Replicate the proof of tighter error bound for Simple Simpson for two cases:
 * {| style="width:100%" border="0" align="left"

a) G(t) = e(t) - t^{4}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

b) G(t) = e(t) - t^{6}e(1) $$ Point out where proof breaks down  Part 2  $$ \displaystyle G(t) = e(t) - t^{6}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Find $$ \displaystyle G^{(3)}(0) $$ and follow same steps in proof and see what happens

Solution
 Part 1

a) For case 1
 * {| style="width:100%" border="0" align="left"

G(t) = e(t) - t^{4}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

From (3) in lecture[[media:Egm6341.s10.mtg14.pdf|p.14-2]]we have
 * {| style="width:100%" border="0" align="left"

e(t) = \int_{+t}^{-t}F(t)\, dt-\frac{t}{3}[F(-t)+4f(0)+F(t)] $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

e(0)= \int_{0}^{0}f(x)\, dx -\frac{0}{3}[F(-0)+4f(0)+F(0)]=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

G(0) = e(0) - 0^{4}e(1)=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G(1) = e(1) - 1^{4}e(1)= 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

According to Rolle' thm,
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _1 \ni ]0,1[,  G^{(1)}(\zeta _1) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G^{(1)}(t) = e^{(1)}(t) - 4t^{3}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

e^{(1)}(t)& = F(-t)+F(t)- \frac{1}{3}[F(-t)+4F(0)+F(t)]-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)]\\ & = \frac{2}{3}[F(-t)+F(t)] - \frac{4}{3}F(0)-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)]\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }

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

e^{(1)}(0) &= \frac{2}{3}[F(0)+F(0)] - \frac{4}{3}F(0)-\frac{0}{3}[-F^{(1)}(-0)+F^{(1)}(0)]\\ &=\frac{4}{3}-\frac{4}{3}F(0)-0\\ &=0\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }

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

G^{(1)}(0) = e^{(1)}(0) - 4(0^{3})e(1)= 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G^{(1)}(\zeta _1) = 0  $$ Thus According to Rolle' thm,
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _2 \ni ]0,\zeta _1[,  G^{(2)}(\zeta _2) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G^{(2)}(t) = e^{(2)}(t) - 12t^{2}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G^{(2)}(0) = e^{(2)}(0) - 12(0^{2})e(1) = e^{(2)}(0)  $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

e^{(2)}(t) &= \frac{2}{3}[-F^{(1)}(-t)+F^{(1)}(t)] - \frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]-\frac{1}{3}[F^{(2)}(-t)+F^{(2)}(t)]\\ &=\frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]-\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

e^{(2)}(0) = \frac{1}{3}[-F^{(1)}(-0)+F^{(1)}(0)]-\frac{0}{3}[F^{(2)}(-0)+F^{(2)}(0)]=0 $$ Thus
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

G^{(2)}(0) = e^{(2)}(0)=0   $$ Recall
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

G^{(2)}(\zeta _2) = 0  $$ Thus According to Rolle' thm,
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _3 \ni ]0,\zeta _2[,  G^{(3)}(\zeta _3) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G^{(3)}(t) = e^{(3)}(t) - 24te(1) $$ Where
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

e^{(3)}(t) = -\frac{t}{3}[F^{(3)}(t)+F^{(3)}(-t)] $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * }

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

G^{(3)}(\zeta_3) & = -\frac{\zeta_3}{3}\underbrace{[F^{(3)}(\zeta_3)+F^{(3)}(-zeta_3)]}_{use DMVT}- 24\zeta_3e(1)\\ &=-\frac{\zeta_3}{3}[2\zeta_3F^{(4)}(\zeta_4)]- 24\zeta_3e(1)\\ &=0 \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}

$$
 * $$\displaystyle (Eq. 1)
 * }
 * }

Where $$\zeta _{4} \in [-\zeta _{3}, \zeta _{3}]$$ Since $$\zeta _{3} \neq 0, $$

Solve for

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

$$\displaystyle e(1)= -\frac{\zeta_3}{36}F^{(4)}(\zeta_4) $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

From the relation ship between $$\zeta_4 and \xi $$ See HW2, No 15 for details

$$ \displaystyle F(t)=f(x(t)) $$

$$ \displaystyle x(t)=x_{1}+ht $$

$$ \displaystyle F^{(4)}(t) = \frac{d\left(F(t)^{(3)}\right)}{dt} = \frac{d\left(h^{3}f(x)^{(3)}\right)}{dx}\frac{dx}{dt} = h^{3}f^{(4)}(x) \times h = h^{4}f^{(4)}(x) $$

Let

$$ \displaystyle \xi = x_{1}+h \zeta_{4} $$

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

$$\displaystyle e(1)= -\frac{\zeta_3}{36}h^{4}f^{(4)}(\xi )=-\frac{\zeta_3(b-a)^{4}}{576}f^{(4)}(\xi )
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$ We know the Error for Simple Simpson's rule from lecture[[media:Egm6341.s10.mtg14.pdf|p.14-2]] $$ \displaystyle
 * style = |
 * }
 * }

E_{2} = I - I_2 = \frac{b-a}{2}e(1) = h e(1) $$

Thus the error is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle E_{2} = h e(1)=-\frac{\zeta_3(b-a)^{5}}{1152}f^{(4)}(\xi )
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$
 * style = |
 * }
 * }

Sum: the proof breaks though at Eq 1, where we can not cancel $${\zeta_3}^2$$, because the third derivative of G(t) at $$\zeta_3$$ gives $$\zeta_3 e(1)$$ instead of $${\zeta_3}^2 e(1)$$

b) For case 2
 * {| style="width:100%" border="0" align="left"

G(t) = e(t) - t^{6}e(1) $$ The proof procedure is similar to case 1 <\br> From (3) in lecture[[media:Egm6341.s10.mtg14.pdf|p.14-2]]we have
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

e(t) = \int_{+t}^{-t}F(t)\, dt-\frac{t}{3}[F(-t)+4f(0)+F(t)] $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

e(0)= \int_{0}^{0}f(x)\, dx -\frac{0}{3}[F(-0)+4f(0)+F(0)]=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

G(0) = e(0) - 0^{6}e(1)=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G(1) = e(1) - 1^{6}e(1)= 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

According to Rolle' thm,
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _1 \ni ]0,1[,  G^{(1)}(\zeta _1) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G^{(1)}(t) = e^{(1)}(t) - 6t^{5}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

e^{(1)}(t)& = F(-t)+F(t)- \frac{1}{3}[F(-t)+4F(0)+F(t)]-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)]\\ & = \frac{2}{3}[F(-t)+F(t)] - \frac{4}{3}F(0)-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)]\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }

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

e^{(1)}(0) &= \frac{2}{3}[F(0)+F(0)] - \frac{4}{3}F(0)-\frac{0}{3}[-F^{(1)}(-0)+F^{(1)}(0)]\\ &=\frac{4}{3}-\frac{4}{3}F(0)-0\\ &=0\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }

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

G^{(1)}(0) = e^{(1)}(0) - 6(0^{5})e(1)= 0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G^{(1)}(\zeta _1) = 0  $$ Thus According to Rolle' thm,
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _2 \ni ]0,\zeta _1[,  G^{(2)}(\zeta _2) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G^{(2)}(t) = e^{(2)}(t) - 30t^{3}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G^{(2)}(0) = e^{(2)}(0) - 30(0^{4})e(1) = e^{(2)}(0)  $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

e^{(2)}(t) &= \frac{2}{3}[-F^{(1)}(-t)+F^{(1)}(t)] - \frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]-\frac{1}{3}[F^{(2)}(-t)+F^{(2)}(t)]\\ &=\frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]-\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * {| style="width:100%" border="0" align="left"
 * {| style="width:100%" border="0" align="left"

e^{(2)}(0) = \frac{1}{3}[-F^{(1)}(-0)+F^{(1)}(0)]-\frac{0}{3}[F^{(2)}(-0)+F^{(2)}(0)]=0 $$ Thus
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * }
 * {| style="width:100%" border="0" align="left"

G^{(2)}(0) = e^{(2)}(0)=0   $$ Recall
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

G^{(2)}(\zeta _2) = 0  $$ Thus According to Rolle' thm,
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _3 \ni ]0,\zeta _2[,  G^{(3)}(\zeta _3) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

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

G^{(3)}(t) = e^{(3)}(t) - 120t^{3}e(1) $$ Where
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

e^{(3)}(t) = -\frac{t}{3}[F^{(3)}(t)+F^{(3)}(-t)] $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * }

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

G^{(3)}(\zeta_3) & = -\frac{\zeta_3}{3}\underbrace{[F^{(3)}(\zeta_3)+F^{(3)}(-\zeta_3)]}_{use DMVT}- 120{\zeta_3}^{3}e(1)\\ &=-\frac{\zeta_3}{3}[2\zeta_3F^{(4)}(\zeta_4)]- 120{\zeta_3}^{3}e(1)\\ &=0 \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}

$$
 * $$\displaystyle (Eq. 2)
 * }
 * }

Where $$\zeta _{4} \in [-\zeta _{3}, \zeta _{3}]$$ Since $$\zeta _{3} \neq 0, $$

Solve for

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

$$\displaystyle e(1)= -\frac{1}{180\zeta_3}F^{(4)}(\zeta_4) $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

From the relation ship between $$\zeta_4 and \xi $$ See HW2, No 15 for details

$$ \displaystyle F(t)=f(x(t)) $$

$$ \displaystyle x(t)=x_{1}+ht $$

$$ \displaystyle F^{(4)}(t) = \frac{d\left(F(t)^{(3)}\right)}{dt} = \frac{d\left(h^{3}f(x)^{(3)}\right)}{dx}\frac{dx}{dt} = h^{3}f^{(4)}(x) \times h = h^{4}f^{(4)}(x) $$

Let

$$ \displaystyle \xi = x_{1}+h \zeta_{4} $$

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

$$\displaystyle e(1)= -\frac{1}{180\zeta_3}h^4F^{(4)}(\xi)= -\frac{(b-a)^4}{2880\zeta_3}F^{(4)}(\xi)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$
 * style = |
 * }
 * }

We know the Error for Simple Simpson's rule from lecture[[media:Egm6341.s10.mtg14.pdf|p.14-2]] $$ \displaystyle

E_{2} = I - I_2 = \frac{b-a}{2}e(1) = h e(1) $$

Thus the error is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle E_{2} = h e(1)= -\frac{(b-a)^4}{2880\zeta_3}hF^{(4)}(\xi)=-\frac{(b-a)^5}{5760\zeta_3}F^{(4)}(\xi)
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$
 * style = |
 * }
 * }

Sum: the proof breaks though at Eq 2, where we can not cancel $${\zeta_3}^2$$, but the third derivative of G(t) at $$\zeta_3$$ gives $${\zeta_3}^3 e(1)$$ instead of $${\zeta_3}^2 e(1)$$

 Part 2  Continue the proof from lecture [[media:Egm6341.s10.mtg15.djvu|Lecture p.15-2]] where we have
 * {| style="width:100%" border="0" align="left"

G^{(3)}(t) = e^{(3)}(t) - 60t^{2}e(1) $$
 * $$ \displaystyle
 * $$ \displaystyle

$$
 * $$\displaystyle (Eq. 3)
 * }
 * }

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

G^{(3)}(\zeta _3)=0 $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

e^{(3)}= \frac{t}{3}[F^{(3)}(t)-F^{(3)}(-t)] $$ $$
 * $$ \displaystyle
 * $$ \displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

From Eq 3. and Eq 4, we have

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

G^{(3)}(t) = \frac{t}{3}[F^{(3)}(t)-F^{(3)}(-t)] - 60t^{2}e(1) $$ Thus
 * $$ \displaystyle
 * $$ \displaystyle
 * }
 * {| style="width:100%" border="0" align="left"

$$\displaystyle G^{(3)}(0) = \frac{0}{3}[F^{(3)}(0)-F^{(3)}(-0)] - 60(0)^{2}e(1) = 0 $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Again, according to Rolle's thm,
 * {| style="width:100%" border="0" align="left"

\exists  \zeta _4 \ni ]0,\zeta _3[,  G^{(4)}(\zeta _4) = 0   $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

Next is to find the 4th derivative of G(t), we calculate the 4th derivative of e(t) first
 * {| style="width:100%" border="0" align="left"

e^{(4)}= -\frac{1}{3}[F^{(3)}(t)-F^{(3)}(-t)]-\frac{t}{3}[F^{(4)}(t)+F^{(4)}(-t)] $$
 * $$ \displaystyle
 * $$ \displaystyle
 * }

The 4th derivative of G(t) is given as:
 * {| style="width:100%" border="0" align="left"

G^{(4)}(t) & = e^{(4)}(t) - 120te(1) \\ & = -\frac{1}{3}[F^{(3)}(t)-F^{(3)}(-t)]-\frac{t}{3}[F^{(4)}(t)+F^{(4)}(-t)] - 120te(1)\\ \end{align} $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }

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

G^{(4)}(\zeta _{4}) &= -\frac{1}{3}\underbrace{[F^{(3)}(\zeta _{4})-F^{(3)}(-\zeta _{4})]}_{use DMVT}-\frac{\zeta _{4}}{3}[F^{(4)}(\xi_4)+F^{(4)}(-\xi_4)] - 120\xi_4e(1)\\ &=-\frac{1}{3}[2\zeta _{4}F^{(4)}(\xi_5)]-\frac{\zeta _{4}}{3}[F^{(4)}(\xi_4)+F^{(4)}(-\xi_4)]-120\zeta _{4}e(1)\\ &=0 \end{align}$$ Where $$\zeta _{5} \in [-\zeta _{4}, \zeta _{4}]$$ Since $$\zeta _{4} \neq 0, $$
 * $$ \displaystyle \begin{align}
 * $$ \displaystyle \begin{align}
 * }
 * }

Solve for

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

$$\displaystyle e(1)= -\frac{1}{360}[2F^{(4)}(\zeta _{5})+F^{(4)}(\zeta _{4})+F^{(4)}(-\zeta _{4})] $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

Since

$$ \displaystyle F(t)=f(x(t)) $$

$$ \displaystyle x(t)=x_{1}+ht $$

$$ \displaystyle F^{(4)}(t) = \frac{d\left(F(t)^{(3)}\right)}{dt} = \frac{d\left(h^{3}f(x)^{(3)}\right)}{dx}\frac{dx}{dt} = h^{3}f^{(4)}(x) \times h = h^{4}f^{(4)}(x) $$

See HW2, No 15 for details

Let

$$ \displaystyle \xi_1 = x_{1}+h \zeta_{4} $$

$$ \displaystyle \xi_2 = x_{1}+h \zeta_{5} $$

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

$$\displaystyle e(1)= -\frac{1}{360}[2h^{4}f^{(4)}(\xi_2 )+h^{4}f^{(4)}(\xi_1 )+h^{4}f^{(4)}(-\xi_1 )] =-\frac{(b-a)^4}{5760}[2f^{(4)}(\xi_2 )+f^{(4)}(\xi_1 )+f^{(4)}(-\xi_1 )] $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

We know the Error for Simple Simpson's rule from lecture[[media:Egm6341.s10.mtg14.pdf|p.14-2]] $$ \displaystyle

E_{2} = I - I_2 = \frac{b-a}{2}e(1) = h e(1) $$

Thus the error is
 * {| style="width:100%" border="0" align="left"

$$\displaystyle E_{2} = h e(1)= -\frac{(b-a)^5}{11520}[2f^{(4)}(\xi_2 )+f^{(4)}(\xi_1 )+f^{(4)}(-\xi_1 )]
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |

$$
 * style = |
 * }
 * }

Sum: Simple simpson is exact for poly deg less than 3. Because $$f ^4(t) = 0$$, thus $$ E_{2} = 0 $$

-- By Min Zhong 12:43, 17 February 2010 (UTC)

= Error bound for Composite Simpson's Rule=

Problem Statement
$$ \displaystyle Show\, the\, error\, for\, Composite\, Simpson's\, Rule,\, $$

$$ \displaystyle \left| {E}_{n}^{2} \right| \leq \frac{(b-a)^{5}}{2880 n^{4}} M_{4} = \frac{(b-a)h^{4}}{2880} M_{4} $$

$$ \displaystyle Where\,\, M_{4} := max \left| f^{4}(\xi) \right| \,\, and \,\, \xi \in [a,b]$$

Ref: Lecture Notes [[media:Egm6341.s10.mtg17.djvu|p.17-2]]

Solution
Ref: Lecture Notes [[media:Egm6341.s10.mtg16.pdf|p.16-2&3]]

$$ \begin{align} \displaystyle E_{n}^{2} &:= I - I_{n} \\ &=\int_{a}^{b}f(x)dx - \frac{h}{3}[f(x_{0})+4f(x_{1})+2f(x_{2})+...+2f(x_{n-2})+4f(x_{n-1})+f(x_{n})] \\ &=\sum_{i=1}^{n} [ \int_{x_{i-1}}^{x_{i}}f(x)dx - \frac{h}{3}[f(x_{i-1}) + 4f(\frac{(x_{i-1}+x_{i})}{2}) + f(x_{i})] \\\end{align}\ $$

$$ \left| E_{n}^{2} \right| \leq \left| \frac{h^{5}}{90} \underbrace {\sum_{n}^{i=1}max\left(F^{(4)}(\xi )\right)}_{Call \bar{M}_{4}} \right| $$

$$ \displaystyle \xi \in [x_{i-1},x_{i}] $$

Ref: Lecture Notes [[media:Egm6341.s10.mtg14.pdf|p.14-1]]

$$ \Rightarrow M_{4} := max \left| f^{(4)}(\xi)\right| $$

$$ \displaystyle \xi \in [a,b] $$

$$ \bar{M}_{4} \leq n \times M_{4} $$

$$ \Rightarrow \left| E_{n}^{2} \right| \leq \left| \frac{(b-a)^{5}}{2880n^{5}}nM_{4} \right| = \left| \frac{(b-a)(b-a)^{4}}{2880n^{4}} \right| = \frac{(b-a)h^{4}}{2880}M_{4} $$

where $$ \displaystyle h= \frac{(b-a)}{n} $$

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

$$\displaystyle \displaystyle \left| {E}_{n}^{2} \right| \leq \frac{(b-a)^{5}}{2880 n^{4}} M_{4} = \frac{(b-a)h^{4}}{2880} M_{4} $$
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style="width:10%; padding:10px; border:2px solid #8888aa" |
 * style = |
 * }
 * }

--Heejun Chung 18:15, 17 February 2010 (UTC)

= Use Error estimate for Taylor series, composite trapezoidal and composite Simpsons rule =

Problem Statement
Use Error estimate for Taylor series, composite trapezoidal and composite Simpson's rule to find n such that $$ E_n = I - I_n = O(10^{-6}) $$ and compare to numerical results.

Solution

 * $$I=\int_{0}^{1}\frac{e^{x}-1}{x}dx$$

Taylor Series

 * $$e^{x}=\sum _{j=0}^{\infty }\frac{x^j}{j!}\Rightarrow \frac{e^{x}-1}{x}=\sum _{j=1}^{\infty }\frac{x^{j-1}}{j!}$$
 * $$I_n=\int_{0}^{1}f_n(x)dx=\int_{0}^{1}\sum _{j=1}^{\infty} \frac{x^{j-1}}{j!}dx=\sum _{j=1}^{n}\frac{x^j}{j!j}|_0^1=\sum _{j=1}^n \frac{1}{j!j}$$
 * $$f(x)-f_n(x)=R_n(x)=\frac{(x-x_0)^n}{(n+1)!}f^{(n+1)}(\xi)$$ with $$\xi \in [0,x] \,$$ and $$x_0=0 \,$$
 * $$\Rightarrow f(x)-f_n(x)=\frac{x^n}{(n+1)!}f^{(n+1)}(\xi)$$
 * $$E_n=I-I_n=\int_{0}^{1}\left [ f(x)-f_n(x) \right ]dx=\int_{0}^{1}\underbrace{\frac{x^n}{(n+1)!}}_{w(x)}\underbrace{f^{(n+1)}\left (\xi(x) \right )}_{g(x)}dx$$
 * $$\Rightarrow g(\alpha)\int_{0}^{1}w(x)dx$$ for $$\alpha \in [0,1]$$
 * $$E_n =\,$$
 * $$\Rightarrow max = \frac{g(\alpha)}{(n+1)!(n+1)}$$, $$\alpha= \frac{e}{(n+1)!(n+1)} \,$$
 * $$\Rightarrow min = \frac{g(\alpha)}{(n+1)!(n+1)}$$, $$\alpha= \frac{1}{(n+1)!(n+1)}\,$$
 * $$n=2, E_2 \leq .151016 \,$$
 * $$n=4, E_5 \leq .00453 \,$$
 * $$ n=6, E_6 \leq 6.2792 X 10^{-4} \,$$
 * $$ n=8, E_8 \leq 8.42 X 10^{-6} \,$$

Below are the values from Numerical Analysis of Taylor series from HW_1

 * $$n=2, E_2 \leq .151016 \,$$
 * $$n=4, E_5 \leq .00453 \,$$
 * $$ n=6, E_6 \leq 6.2792 X 10^{-4} \,$$
 * $$ n=8, E_8 \leq 8.42 X 10^{-6} \,$$

We are getting same values from both analysis for Taylor series

Trapazoidal Rule

Error for Composite Trapazoidal rule is given by

$$  \displaystyle \left | E_n^1 \right |\leq \frac{(b-a)^3}{12n^2}M_2 $$

where $$M_2 = max \left | f^{(2)} (\zeta )\right |$$ for $$\zeta \varepsilon [a,b]$$

For the given function $$f(x)= \frac{e^x - 1}{x}$$, we have

$$  \displaystyle f^{(2)}(x)= \frac{e^x [x^2-2x+2]-2}{x^3} $$

For the given interval [0,1] the maximum value of function $$f^{(2)}(x)$$ is achieved at $$x=1$$

$$  \displaystyle M_2= f^{(2)}(x=1)=\frac{e[1+2-2]-2}{1}=0.71828182 $$

$$ E_n^1 \leq \frac{(1-0)^3}{12n^2}X 0.71828182 $$

$$ E_2^1 \leq .014964204 $$

$$ E_4^1 \leq 3.74105 X 10^{-3} $$

$$ E_6^1 \leq 1.662689 X 10^{-3} $$

$$ E_8^1 \leq 9.3526278 X 10^{-4}$$

$$ E_{16}^1 \leq 2.3381569 X 10 ^{-4}$$

$$ E_{32}^1 \leq 5.845392 X 10 ^{-5}$$

$$ E_{64}^1 \leq 1.4613481 X 10^{-5}$$

$$ E_{128}^1 \leq 3.65337026 X 10^{-6} $$

$$ E_{256}^1 \leq 9.1334256 X 10^{-7}$$

Below are the results from Numerical analysis from HW 1

$$ E_2^1 \leq 0.010389576$$

$$ E_4^1 \leq 0.002602468 $$

$$ E_8^1 \leq 0.650935 X 10^{-3}$$

$$ E_{16}^1 \leq 162753 X 10^{-4}$$

$$ E_{32}^1 \leq 4.06892 X 10 ^{-5}$$

$$ E_{64}^1 \leq 1.0172 X 10^{-5}$$

$$ E_{128}^1 \leq 2.54263 X 10^{-6} $$

$$ E_{256}^1 \leq 6.3528 X 10^{-7}$$

We can see that we are getting order $$ 10^{-6}$$ at n=128 from both the analysis.

Composite Simpsons Rule

The error estimate of the Composite Simpson's rule is given as

$$   E_n^2 \leq \frac{(b-a)^5}{2880n^4}M_4 $$

where $$M_4 = max \left | f^{(4)} (\zeta )\right |$$ for $$\zeta \varepsilon [a,b]$$

For the given function $$f(x)= \frac{e^x - 1}{x}$$, we have

$$  \displaystyle f^{(4)}(x)= \frac{e^x [x^4-4x^3+12x^2-24x]-24}{x^5} $$

The function $$f^{4}(x)$$ has maximum value at $$x=1$$

$$  \displaystyle M_4= \frac{e[1-4+12-24+24]-24}{1}=0.46453645

$$

$$ E_n^1 \leq \frac{(1-0)^5}{2880n^4}X 0.46453645 $$

$$ E_2^2 \leq 1.00810 X 10^{-5} $$

$$ E_4^2 \leq 6.300678 X 10^{-7} $$

Below are the results from Numerical analysis from HW 1

$$ E_2^2 \leq 1.06514 X 10^{-4} $$

$$ E_4^2 \leq 6.76473 X 10^{-6} $$

We can see that we are getting order $$ 10^{-6}$$ at n=4 from both the analysis.

Numerical determination of power of h

Below are the plots for Taylor series, Composite Trapazoidal rule and Composite Simpson's rule.

Taylor Series

Above is a log Error vs Log h plot(we have used the numerical data from HW1-Problem 8 .We can see that slope of the line is 6.56 which is close to the exact value of 7. Composite Trapazoidal Rule

Above is a log Error vs Log h plot(we have used the numerical data from HW1-Problem 8).We are getting slope of 1.9997 which is very close of exact value 2.

Composite Simpson's rule

Above is a log Error vs Log h plot(we have used the numerical data from HW1-Problem 8).We are getting slope of 3.9987 which is very close to the exact value 4 of the slope for Simpson's rule.

=Asymptotic Errors and Peano Kernel error estimates=

Asymptotic Error: Composite Trapezoidal rule
Ref: Lecture Notes [[media:Egm6341.s10.mtg17.djvu|p.17-3]]

Problem Statement
$$ \displaystyle I=\int_{0}^{\pi}e^{x}cos(x)dx,\,\, $$

$$\displaystyle Evaluate\, I_{n}, E_{n}, Ratio \,\, and \,\, \bar{E}_{n}\, by\, Trapezoidal\, rule.\,\,\, where\,\,\, n=1,2...,9 \,$$

Solution
=Matlab Code for Trapezoidal Rule= MATLAB Code

Results: Trapezoidal Rule

Formula

a) Comp. Trap. rule

$$ \int_{a}^{b}f(x)dx=\frac{b-a}{2n}*[f(x_{0})+2f(x_{1})+2f(x_{2})+.........+2f(x_{n-1})+f(x_{n})] $$

b) Corrected Trap. rule

$$ \int_{a}^{b}f(x)dx=\frac{b-a}{2n}*[f(x_{0})+2f(x_{1})+2f(x_{2})+.........+2f(x_{n-1})+f(x_{n})]-\frac{h^2}{12}*[f^{'}(b)-f^{'}(a)] $$

c) Asymptotic error

$$ \bar{E}_{n} \equiv -\frac{h^{2}}{12}[f^{'}(b)-f^{'}(a)] $$

--Heejun Chung 01:31, 17 February 2010 (UTC)

Asymptotic errors: Composite Simpson rule
Ref: Lecture notes [[media:Egm6341.s10.mtg17.djvu|p.17-3]]

Problem Statement

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

$$\displaystyle Evaluate\, \int_{a}^{b} f(x)\,dx \,using\, Composite\,  Simpson\, rule\, with\,\,  n\;\; =2,4,8,16,32,64\, and\,128.$$ $$\displaystyle Also\, evaluate\,\,  asymptotic\,  errors\,.$$
 * $$\displaystyle f(x) = e^{x}cos(x) \,\,, x\in [a,b] = [0,\pi]$$
 * $$\displaystyle f(x) = e^{x}cos(x) \,\,, x\in [a,b] = [0,\pi]$$
 * }
 * }

Solution

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

 * The asymptotic error for the composite Simpson rule is given by,
 * }
 * }
 * }

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

 * $$\displaystyle \bar{E_n}(f) = \frac{-h^4}{180}\big[ f^{(3)}(b) - f^{(3)}(a)\big]$$
 * }
 * }
 * }

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

i.e, $$\displaystyle h:= \frac{b-a}{n} $$,
 * where,h is the step size
 * where,h is the step size
 * }
 * }

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

 * The Composite Simpson rule is given by,
 * }
 * }
 * }

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

where n = 2k  and k = 1,2,3,4.....
 * $$ \displaystyle I_{\text{n}} = \frac {h}{3} [f_{\text{o}}+4 f_1+2f_2+4f_{\text{3}}+2f_4+4f_5+f_6.......+2f_{\text{n-2}}+4 f_{\text{n-1}} +f_{\text{n}}] $$
 * $$ \displaystyle I_{\text{n}} = \frac {h}{3} [f_{\text{o}}+4 f_1+2f_2+4f_{\text{3}}+2f_4+4f_5+f_6.......+2f_{\text{n-2}}+4 f_{\text{n-1}} +f_{\text{n}}] $$
 * }
 * }

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

 * The above integrals are calculated using MATLAB codes and the errors are tabulated.
 * }
 * }
 * }

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

 * }
 * }
 * }
 * }

Simpson Rule : Asymptotic error

Subramanian Annamalai 18:35, 17 February 2010 (UTC)

Application: Peano Kernel errors estimates : $$ f^{(4)}(x)$$ is singular at x=0
Ref: Lecture notes [[media:Egm6341.s10.mtg18.djvu|p.18-1]]

Problem Statement

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

$$\displaystyle Find\, the\,  Peano\, Kernel\, error\, estimates\, for\, Composite\,  Trapezoidal\, $$ $$\displaystyle and\, Composite\,  Simpson\,  rules\, with\,  n\;\; =2,4,8,16,32,64\, and\,128.$$ $$\,Comment.$$
 * $$\displaystyle f(x) = x^3 \sqrt{x} \,\,, x\in [a,b] = [0,1]$$
 * $$\displaystyle f(x) = x^3 \sqrt{x} \,\,, x\in [a,b] = [0,1]$$
 * }
 * }

Solution

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

 * The Peano Kernel error estimate for the composite Trapezoidal rule is given by,
 * }
 * }
 * }

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

 * $$\displaystyle E_n(f) = \int_{a}^{b} K(t) f ^{(2)}(t)\,dt$$
 * }
 * }
 * }

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

 * where,the Peano Kernel error estimate,
 * }
 * }
 * }

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

 * $$\displaystyle K(t) = \frac{1}{2}(t-t_{j-1})(t-t_j) \;\;\; t_{j-1}<t<t_j \;\;\;\;\; j=1,2,....,n.$$
 * }
 * }
 * }

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

$$\displaystyle t_{j-1} < t <t_j  \;\;\;\;\; j=1,2,....,n$$ represents the intervals.
 * Here
 * Here
 * }
 * }

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

 * The Peano Kernel error estimate for the composite Simpson rule is given by,
 * }
 * }
 * }

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

 * $$\displaystyle E_n(f) = \int_{a}^{b} K(t) f ^{(4)}(t)\,dt$$
 * }
 * }
 * }

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

 * where,the Peano Kernel error estimate,
 * }
 * }
 * }

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

$$\displaystyle K(t) = \begin{cases} \frac{1}{72} (t-t_{j-2})^3 (3t - t_{j-2} - 2t_j),\;\;\;\;  t_{j-2} \leq t \leq \frac{t_{j-2}+t_j}{2} \\ \frac{1}{72} (t_{j}-t)^3 (t_{j} + 2t_{j-2} -3t), \;\;\;\;\;  \frac {t_{j-2}+t_j}{2}  \leq t \leq {t_j} \end{cases} $$

 * }
 * }

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

 * The above integrals are calculates using MATLAB codes and the corresponding errors are tabulated.
 * }
 * }
 * }

Observation: It can be observed that: $$\displaystyle f^{(4)}(x) \rightarrow \infty\, as\, x\rightarrow 0$$. Here the actual error formula for the Simpson's rule cannot be used as it involves $$f^{(4)}(x)$$.

Subramanian Annamalai 18:35, 17 February 2010 (UTC)

Peano Kernel errors estimates -The ratio of errors need not necessarily be 4 and 16 for the Trapezoidal and Simpson's rules respectively on halfing the step size
Ref: Lecture notes [[media:Egm6341.s10.mtg18.djvu|p.18-1]]

Problem Statement

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

$$\displaystyle Find\, the\,  Peano\, Kernel\, error\, estimates\, for\, Composite\,  Trapezoidal\, $$ $$\displaystyle and\, Composite\,  Simpson\,  rules\, with\,  n\;\; =2,4,8,16,32,64\, and\,128. $$ $$\,Comment.$$
 * $$\displaystyle f(x) = \frac{1}{1+(x-\pi)^2} \,\,, x\in [a,b] = [0,5]$$
 * $$\displaystyle f(x) = \frac{1}{1+(x-\pi)^2} \,\,, x\in [a,b] = [0,5]$$
 * }
 * }

Solution

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

 * The Peano Kernel error estimate for the composite Trapezoidal rule is given by,
 * }
 * }
 * }

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

 * $$\displaystyle E_n(f) = \int_{a}^{b} K(t) f ^{(2)}(t)\,dt$$
 * }
 * }
 * }

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

 * where,the Peano Kernel error estimate,
 * }
 * }
 * }

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

 * $$\displaystyle K(t) = \frac{1}{2}(t-t_{j-1})(t-t_j) \;\;\; t_{j-1}<t<t_j \;\;\;\;\; j=1,2,....,n.$$
 * }
 * }
 * }

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

$$\displaystyle t_{j-1} < t <t_j  \;\;\;\;\; j=1,2,....,n$$ represents the intervals.
 * Here
 * Here
 * }
 * }

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

 * The Peano Kernel error estimate for the composite Simpson rule is given by,
 * }
 * }
 * }

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

 * $$\displaystyle E_n(f) = \int_{a}^{b} K(t) f ^{(4)}(t)\,dt$$
 * }
 * }
 * }

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

 * where,the Peano Kernel error estimate,
 * }
 * }
 * }

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

$$\displaystyle K(t) = \begin{cases} \frac{1}{72} (t-t_{j-2})^3 (3t - t_{j-2} - 2t_j),\;\;\;\;  t_{j-2} \leq t \leq \frac{t_{j-2}+t_j}{2} \\ \frac{1}{72} (t_{j}-t)^3 (t_{j} + 2t_{j-2} -3t), \;\;\;\;\;  \frac {t_{j-2}+t_j}{2}  \leq t \leq {t_j} \end{cases} $$

 * }
 * }

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

 * The above integrals are calculates using MATLAB codes and the corresponding errors are tabulated.
 * }
 * }
 * }

Observation: In this case when 'n' is doubled the the error ratio is doesn't reduce by 4 times (Trapezoidal rule) and by 16 times (for Simpson's rule). This is because the function is peaked and hence takes time to converge.

Subramanian Annamalai 18:37, 17 February 2010 (UTC)

Application of Peano Kernel errors estimates : $$f^{(1)}(x)$$ tends to infinity as x tends to zero
Ref: Lecture notes [[media:Egm6341.s10.mtg18.djvu|p.18-1]]

Problem Statement

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

$$\displaystyle Find\, the\,  Peano\, Kernel\, error\, estimates\, for\, Composite\,  Trapezoidal\, $$ $$\displaystyle and\, Composite\,  Simpson\,  rules\, with\,  n\;\; =2,4,8,16,32,64\, and\,128. $$ $$\,Comment.$$
 * $$\displaystyle f(x) = \sqrt{x} \,\,, x\in [a,b] = [0,1]$$
 * $$\displaystyle f(x) = \sqrt{x} \,\,, x\in [a,b] = [0,1]$$
 * }
 * }

Solution

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

 * The Peano Kernel error estimate for the composite Trapezoidal rule is given by,
 * }
 * }
 * }

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

 * $$\displaystyle E_n(f) = \int_{a}^{b} K(t) f ^{(2)}(t)\,dt$$
 * }
 * }
 * }

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

 * where,the Peano Kernel error estimate,
 * }
 * }
 * }

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

 * $$\displaystyle K(t) = \frac{1}{2}(t-t_{j-1})(t-t_j) \;\;\; t_{j-1}<t<t_j \;\;\;\;\; j=1,2,....,n.$$
 * }
 * }
 * }

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

$$\displaystyle t_{j-1} < t <t_j  \;\;\;\;\; j=1,2,....,n$$ represents the intervals.
 * Here
 * Here
 * }
 * }

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

 * The Peano Kernel error estimate for the composite Simpson rule is given by,
 * }
 * }
 * }

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

 * $$\displaystyle E_n(f) = \int_{a}^{b} K(t) f ^{(4)}(t)\,dt$$
 * }
 * }
 * }

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

 * where,the Peano Kernel error estimate,
 * }
 * }
 * }

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

$$\displaystyle K(t) = \begin{cases} \frac{1}{72} (t-t_{j-2})^3 (3t - t_{j-2} - 2t_j),\;\;\;\;  t_{j-2} \leq t \leq \frac{t_{j-2}+t_j}{2} \\ \frac{1}{72} (t_{j}-t)^3 (t_{j} + 2t_{j-2} -3t), \;\;\;\;\;  \frac {t_{j-2}+t_j}{2}  \leq t \leq {t_j} \end{cases} $$

 * }
 * }

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

 * The above integrals are calculates using MATLAB codes and the corresponding errors are tabulated.
 * }
 * }
 * }

Observation: It should be noted that: $$\displaystyle f^{(4)}(x) \rightarrow \infty as x\rightarrow 0$$. So the Peano curve error estimate is helpful in such cases. The errors reduce by the same ratio in both Trapezoidal and Simpson's rule.

Subramanian Annamalai 18:36, 17 February 2010 (UTC)

Peano Kernel errors estimates -Trapezoidal and Simpson rules - Periodic functions
Ref: Lecture notes [[media:Egm6341.s10.mtg18.djvu|p.18-1]]

Problem Statement

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

$$\displaystyle Find\, the\,  Peano\, Kernel\, error\, estimates\, for\, Composite\,  Trapezoidal\, $$ $$\displaystyle and\, Composite\,  Simpson\,  rules\, with\,  n\;\; =2,4,8,16,32,64\, and\,128. $$ $$\,Comment.$$
 * $$\displaystyle f(x) = e^{cos(x)} \,\,, x\in [a,b] = [0,2\pi]$$
 * $$\displaystyle f(x) = e^{cos(x)} \,\,, x\in [a,b] = [0,2\pi]$$
 * }
 * }

Solution

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

 * The Peano Kernel error estimate for the composite Trapezoidal rule is given by,
 * }
 * }
 * }

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

 * $$\displaystyle E_n(f) = \int_{a}^{b} K(t) f ^{(2)}(t)\,dt$$
 * }
 * }
 * }

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

 * where,the Peano Kernel error estimate,
 * }
 * }
 * }

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

 * $$\displaystyle K(t) = \frac{1}{2}(t-t_{j-1})(t-t_j) \;\;\; t_{j-1}<t<t_j \;\;\;\;\; j=1,2,....,n.$$
 * }
 * }
 * }

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

$$\displaystyle t_{j-1} < t <t_j  \;\;\;\;\; j=1,2,....,n$$ represents the intervals.
 * Here
 * Here
 * }
 * }

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

 * The Peano Kernel error estimate for the composite Simpson rule is given by,
 * }
 * }
 * }

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

 * $$\displaystyle E_n(f) = \int_{a}^{b} K(t) f ^{(4)}(t)\,dt$$
 * }
 * }
 * }

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

 * where,the Peano Kernel error estimate,
 * }
 * }
 * }

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

$$\displaystyle K(t) = \begin{cases} \frac{1}{72} (t-t_{j-2})^3 (3t - t_{j-2} - 2t_j),\;\;\;\;  t_{j-2} \leq t \leq \frac{t_{j-2}+t_j}{2} \\ \frac{1}{72} (t_{j}-t)^3 (t_{j} + 2t_{j-2} -3t), \;\;\;\;\;  \frac {t_{j-2}+t_j}{2}  \leq t \leq {t_j} \end{cases} $$

 * }
 * }

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

 * The above integrals are calculates using MATLAB codes and the corresponding errors are tabulated.
 * }
 * }
 * }

Observation: It can be seen that when we have periodic functions the composite trapezoidal and composite Simpson's rules converge rapidly. And moreover the trapezoidal rule proves itself superior to that of Simpson's.

Subramanian Annamalai 18:36, 17 February 2010 (UTC)

= Comparison of Romberg table with Taylor, Trap and Simpson's Rules = Ref: Lecture notes [[media:Egm6341.s10.mtg19.djvu|p.19-2]]

Problem Statement
This problem is continued from the HW on [[media:Egm6341.s10.mtg6.pdf|p.6-5]]

i) Modify matlab code to make the computation of $$\displaystyle T_{0}(2^{j}) $$ efficient, i.e. $$\displaystyle T_{0}(2^{j}) = T_{0}(2^{(j-1)}) + \cdot\cdot\cdot $$.

ii) Construct Romberg table and compare to previous results, i.e. compare Romberg table to Taylor, Trap. and Simpson's rules.

Solution
i)

Recall the formula of comp. Trap. rule.

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

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

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

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

In the following table, we summarize $$\displaystyle T_{0}(n) $$ for $$\displaystyle n = 1, 2, 4, 8, 16, 32, 64, 128 $$. The Matlab code for this calculation is provided in the end of this document.

ii)

In order to construct Romberg table, recall that the following inductive formula from the lecture note [[media:Egm6341.s10.mtg19.djvu|p.19-2]],

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

T_{k}(2n) = \frac{2^{2k}T_{k-1}(2n) - T_{k-1}(n)}{2^{2k}-1}. $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }

Since we already obtain $$\displaystyle T_{0}(n) $$ in part i), the following Romberg table can be constructed using Eq.(3).  Matlab code to make this table is provided in the end part of this document.

'''Note the fact that the first and second columns of the above table correspond to Trap. and Simpson's rules respectively.'''

In order to compare the Romberg integration table to Taylor, Trap. and Simpson's rules, let's evaluate errors of the Romberg integration table.

In the previous HW (i.e. HW on [[media:Egm6341.s10.mtg6.pdf|p.6-5]] of the lecture note), we already found that we can achieve $$\displaystyle E_{n} = O(10^{-7})$$ when $$\displaystyle n = 8$$ in Taylor series expansion. As we shows the first and second column of above table, to achieve $$\displaystyle E_{n} = O(10^{-7})$$, $$\displaystyle n > 128$$ for trap. rule and $$\displaystyle n = 8$$ for Simpson's rule. On the other hand, we can achieve the same order of error when $$\displaystyle n = 4$$ using Romberg table (see. the third column of above table).

Therefore, Romberg integration table is the most efficient and accurate integration method.

Matlab code for part i) and ii)

Yong Nam Ahn 15:18, 17 February 2010 (UTC)

= Contributing Team Members = 1. Subramanian Annamalai 18:52, 17 February 2010 (UTC) Authored: 4.2,4.3,4.4,4.5 and 4.6  Proof-read: 4.1 and 5

2. Heejun Chung 19:38, 17 February 2010 (UTC) Authored: 2 and 4.1  Proof-read: 4.2,4.3,4.4,4.5 and 4.6

3. Min Zhong 19:56, 17 February 2010 (UTC) Authored: 1   Proof-read: 2

4. Yong Nam Ahn 21:03, 17 February 2010 (UTC) Authored: 5   Proof-read: 2 and 4.1

5. Abhishekksingh 21:03, 17 February 2010 (UTC) Authored: 3   Proof-read: 1 and 4.1

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

title: User:Egm6341.s10.team3.sa/HW3


text:
Talk:The Near Death Experience as Possible Evidence for an Afterlife

Like the article that Schmidt linked to I suspected a connection with Savantism. Savantism happens when damage to some parts of the brain release special abilities in other parts, similarly I felt the breakdown in parts of the brain during the near death process could release abilities in other parts. This is a good resource on Savantism that I haven’t had time to read yet. Savbant Syndrome, Islands of Genius. Proxima Centauri 05:15, 13 February 2010 (UTC)

My own NDE
When I was maybe 8 I was swimming in the ocean when a large wave and the subsequent rip-tide pulled me under and away from shore. I didn't know which way was up or have any sense of how much time was passing. I remember thinking how beautiful and peaceful everything was. Then another wave deposited me on the shore, completely disoriented. I wandered the beach for some time before I began to see familiar landmarks. So, was this the "beauty and peace of God" I experienced ? It's hard to say. StuRat 02:35, 15 February 2010 (UTC)

 * It was part of the NDE,endorphins flooded your brain preventing you from feeling fear. Near-death experience gives a good summary. Proxima Centauri 08:01, 15 February 2010 (UTC)

Tunnel of light
A common explanation for the tunnel of light vision is that low blood pressure causes tunnel vision. Should this be included on the main page ? StuRat 02:35, 15 February 2010 (UTC)

 * I'm not an expert but if you have good sources for that suggestion please put it into the article, the experts can review what you write later. I hope I know enough to evaluate sources. Proxima Centauri 08:14, 15 February 2010 (UTC)

Judgement
It only addresses the judgement because that’s as far as I’ve got so far, I’ll write more later. Proxima Centauri 12:57, 15 February 2010 (UTC)


text:
User talk:Jkinsey~enwikiversity

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Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:12, 19 April 2015 (UTC)

title: User talk:Jkinsey~enwikiversity


text:
User talk:Ewin~enwikiversity

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 13 February 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 13 February 2010 (UTC)

Your account will be renamed
Hello,

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

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

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:35, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:08, 19 April 2015 (UTC)

title: User talk:Ewin~enwikiversity


text:
User talk:Krinn 8023

 Hello Krinn 8023, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 13 February 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 13 February 2010 (UTC)

title: User talk:Krinn 8023


text:
User talk:Lobabah

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 13 February 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 13 February 2010 (UTC)

title: User talk:Lobabah


text:
User talk:Tjunderh

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 13 February 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 13 February 2010 (UTC)

Academy of Special Education
Academy of Special Education in Warsaw (Akademia Pedagogiki Specjalnej im. Marii Grzegorzewskiej w Warszawie)

This is a very long title. Perhaps "Academy of Special Education in Warsaw" would be more appropriate as the full title is already within the page? I am going to move it to the shorter title so it is easier for people to link to it. Ottava Rima (talk) 19:35, 17 February 2010 (UTC)

title: User talk:Tjunderh


text:
User talk:Dscott

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Mu301Bot 08:18, 13 February 2010 (UTC)

Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 08:18, 13 February 2010 (UTC)

title: User talk:Dscott


text:
Talk:Java

what is the difference bitween public class members and private class memebers?

should a link to JavaScript be added?
javascript is not Java, but Java is part of the javascript name.... any preferences? limitless peace Michael Ten (discuss • contribs) 01:51, 28 October 2022 (UTC)

 * They are very different in actual use. When in doubt, I use Wikipedia as a guide. JavaScript is not referenced in Java, except at the top to make it clear that they are not related. -- Dave Braunschweig (discuss • contribs) 01:52, 31 October 2022 (UTC)
 * I hear that. Java (disambiguation) lists JavaScript as a see also. So I might do that here also (with note), unless I misunderstood. Michael Ten (discuss • contribs) 02:52, 31 October 2022 (UTC)

title: Talk:Java