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User:Sorenhogberg/Intersektionalitet 1 SH

Intersektionalitet Jag möter aldrig i boken Intersektionalitet, Kritiska reflektioner över (o)jämlikhetens landskap skriven av Paulina de los Reyes och Diana Mulinari (2005) någon formulerad definition av begreppet intersektionalitet. Därmed inte sagt att detta kan utläsas via de reflektioner som författarna presenterar. Jag menar ändå att det kunde vara på sin plats eftersom det finns en språklig motsättning i hur begreppet används t.ex. i meningen: Ett intersektionellt perspektiv ställer frågor om hur makt och ojämlikhet vävs in i uppfattningen om vithet, manlighet, könstillhörighet, heterosexualitet, klasstillhörighet m.m. genom ett ständigt (åter)skapande av nya markörer som gör skillnaden mellan ”vi” och ”dem” till sociala koder. (s.9-10) Begreppen intersektionalitet och perspektiv ger uttryck för skilda ambitioner. Att välja att betrakta ett fenomen ur ett valt perspektiv är att avgränsa studieobjektet, vilket i sig utgör själva grundproblematiken till varför begreppet intersektionalitet införs. Intersektionalitetens strävan är att involvera flera perspektiv i betraktandet av någon form av samhällsföreteelse, dvs. att frångå ett alltför ensidigt betraktande av studieobjektet. Begreppen intersektionalitet och perspektiv drar således språkligt sätt åt två olika håll. Med begreppet intersektionalitet avses enligt den förståelse jag finner i de los Reyes och Mulinaris text en utvidgning av perspektivseendet till att omfatta ett flertal samverkande perspektiv. Det är just denna samverkan som öppnar upp för ny förståelse av vad som blir oavsiktliga (men kanske även avsiktliga) följder av ett kunskapande ur ett avgränsat perspektiv. Ett intersektionellt perspektiv, uttryckt i singularis, kan därför ge intryck av att vara något annat än de olika uppfattningar som lyfts fram i den ovan citerade meningen, dvs. något annat som inte definierats. Mer begripligt framstår det när författarna uttrycker sig på följande sätt: Ett intersektionellt maktperspektiv ser vi inte enbart som synliggörande av den simultana verkan av skilda och interagerande strukturer av förtyck, vare sig det handlar om kön, klass eller etnicitet (s.8). I denna mening får begreppet intersektionalitet stå för ett förtydligande av på vilket sätt ett maktperspektiv är betjänt av att betraktas via flera olika utgångspunkter. Det är således makten som ska analyseras genom att tänka intersektionellt. Begreppet intersektionalitet blir därför starkt knutet till frågor om makt och avsaknad av jämlikhet men begreppet, såsom jag uppfattar det, har mer en metodologisk innebörd än en innehållslig eftersom intersektionen aldrig är begränsad till särskilda fördefinierade områden. Tvärtom finns, såsom jag uppfattar det, en ambition att ständigt finna nya utgångspunkter för att dekonstruera talet om den andre. Detta tal som diskursivt skapar ojämlikheter människor emellan. Författarna tar inte någon explicit diskussion om möjligheten av synliggöra förhållanden via andra icke nämnda perspektiv men alla deras reflektioner är i sig en presentation av allehanda olika möjligheter att intersektionellt använda olika perspektiv i betraktandet av diverse olika samhällsföreteelser. Framträdande och återkommande är dock perspektiven klass, kön och etnicitet men även sexualitet, ålder, tid och rum lyfts fram och påvisar vad som är möjligt att upptäcka när olika perspektiv samverkar. Det författarna vill framföra är att ”ett intersektionalietsperspektiv” möjliggör analysresultat som påvisar och förankrar maktrelationer i samhället utan att förfalla i ”en ontologisk rangordning om ojämlikhetens primat” (s.11). Med detta som syfte skulle även andra aspekter som exempelvis bristande kroppsfunktioner eller olika kroppskonstitutioner kunna samverka med andra perspektiv för att synliggöra den diskurs som konstruerar och vidmakthåller skilda kulturella och materiella villkor för människor. Intersektionaliteten har dock sitt ursprung ur en antirasistisk kritik av feminismen som trots sitt emanciperande perspektiv bidragit med en exkluderande praktik gentemot ”icke-vita” kvinnor, ett i sig paradoxalt resultat. Men fokus på könsmaktordning i en västerländsk kontext har osynliggjort kvinnors villkor i andra sammanhang, så även ”icke-vita” kvinnor levande i västvärlden. Deras situation har försvunnit i en allmän uppfattning om ”de andra” i bemärkelsen de ”icke-vita” som en grupp andra där frågan om kön blivit till en icke-fråga. Detta osynliggörande har snarare befäst rådande maktordning grundat på etnicitet. Det är ur detta som poängen med ett intersektionellt perspektiv växer fram, nämligen att synliggörandet av maktordning ur ett perspektiv kan paradoxalt leda till ett osynliggörande ur ett annat. Den vita feminismens hegemoni är ett exempel på detta vilket synliggjorts av ”black feminism” som förenar fler perspektiv i analysen av ojämlik fördelning av materiella och symboliska resurser. Ett annat exempel som de los Reyas och Mulinari lyfter fram är den diskurs som i Sverige förts med välfärdsstaten som kristalliseringspunkt. Den politiska stävan att utjämna ojämlikheter ur ett klassperspektiv har samtidigt tagit sin utgångspunkt i att se nationer som en självklar indelningskategori av världen. En indelning som i sig skapat förutsättningar för rasifiering, vilket ur ett etnicitetsperspektiv snarare har bidragit till att cementera ojämlikheter ur detta perspektiv. Framväxten av idealbilden ”det svenska folkhemmet” skapade samtidigt ur etnicitetssynpunkt en vit svensk hegemoni. I förhållande till motsvarande problematik faller det postkoloniala perspektivet väl in i intersektionalitetens strävan att synliggöra maktstrukturer. Intressanta och viktiga analysresultat är kolonialismens och imperialismens nära koppling till framväxten av upplysningens värderingar. Talet om mänsklig frihet, jämlikhet och broderskap har sin hemvist i ett västerländskt perspektiv som tagit sin utgångspunkt i klasskillnader och osynliggjort både könsmaktordning och rasistiska orättvisor. Att inte synliggöra olika perspektiv och ställa dessa i relation till varandra leder således till ett osynliggörande enligt ett intersektionellt maktperspektiv. Detta kan också bli följden av ett postmodernistiskt synsätt där talet om alltings icke-essentiella position också gör ojämlikheter till en icke-fråga. Ett postkolonialt perspektiv kan därför inte reduceras till ett postmodernistiskt teoretiskt projekt. Den postkoloniala teorin strävar också, till skillnad mot postmodernismens subjektssyn, mot (i bemärkelsen för) ett hävdande och bejakande av förvägrad subjektivitet. De strukturer av över- och underordningar som existerar i samhället kan därför också synliggöras genom uppmärksammandet av identifierbara motståndsdiskurser. Av ovanstående återgivning av de los Reyes och Mulinaris text framgår tydligt att intersektionalitet placerar sig inom traditionen kritisk forskning, vilket också uttrycks av författarna (s.11). Jag ser också stark koppling till den kritiska diskursanalysen såsom den presenterats av Jörgensen och Phillips (1999) i och med att intersektionalitetsperspektivet står i ett dialektiskt förhållande till diskurs som konstituerande och diskurs som konstituerad. de los Reyes och Mulinari återkommer gång på gång till ett diskursivt angreppssätt parat med analysen av materiella resurser för att ett intersektionellt maktperspektiv ska kunna ge ökad rättvisa i synliggörandet av ”(o)jämlikhetens landskap”.

Referenser de los Reyes, Paulina & Mulinari, Diana (2005) Intersektionalitet, Kritiska reflektioner över (o)jämlikhetens landskap Malmö: Liber.

Winther Jørgensen Marianne & Phillips Louise (2000) Diskursanalys som teori och metod. Lund: Studentlitteratur.


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User:Egm6341.s11.team4.kurth

EGM 6341, Numerical Methods for Engineering Analysis, Spring 2011 this is a work in progress

Problem 1.4

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User talk:Egm6341.s11.team4.kurth

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Reflections on intersectionality

De los Reyes, Paulina & Mulinari, Diana (2005) Intersektionalitet: kritiska reflektioner over (o)jämlikhetens landskap. Malmö: Liber
In their book, Intersektionalitet, de los Reyes and Mulinari are interested in “finding new ways to think intersectionally” (8). By providing a wide theoretical background and some examples from different empirical studies, the authors aim to deepen the dialogue about power and power relations in order to understand how different kinds of disempowerments are linked and are mutually strengthened by the intersection. An interesting point that the authors are keen to make in their work is that the question on whether feminism as an issue of gender is constitutive of power relations: “Bland såväl neomarxister som poststrukturalister finns en ambition att utveckla teoretiska modeller som kan fånga nutida sociala relationer, identitetsformeringar och materiell ojämlikhet bortom endimensionella maktanalyser” (8). This means that if dynamics of oppression and disempowerment are to be found beyond the boundaries of gender, intersectionality can no longer affirm to belong to feminism theory only: “Makt och ojämlikhet konstrueras även genom den vetenskapliga internalisering av en atomiserad världsbild som essentialiserar vår uppfattning av sociala processer och mänskligt handlande” (9). In other words, what de los Reyes and Mulinari want to do is also to analyze how knowledge, politics and social life influence the intersection of different power relations in institutional settings and how, from an intersectional perspective, it is possible to define social differences and variations in terms of class, ethnicity, sexuality etc. (and the nature of this ‘etc’ is also in need of a further discussion according to some researchers such as Alice Ludvig (2006)) as labels that allows us to create an “us” in contraposition to “them”. Furthermore, another important aim of the book is to analyze how metaphors of time and space have been used: “med utgångspunkt i postkoloniala teorier och teorier om globarisering vill vi problematisera tidens och rummets roll som självklara analytiska redskap” (11). De los Reyes and Molina thus criticise the view of power relations as a well define system, and they aim to create tools that allows the analysis of these relations as the result of the uneven distribution of what they call ‘material and symbolic resources’ (11).

“Globaliseringens motsägefulla karaktär” – The contradictory nature of globalization
In the heart of Italy, in Toscana, there is a rather small town, Prato, the inhabitants of which always boast about their status of ‘true’ Italians, stigmatizing what ‘the outside world’, the ‘others’ consider to be true Italian characteristics. Prato is also very famous for its textile industry that supplies for the Italian high fashion industry, or the phenomenon of the ‘made in Italy’. Due to globalization and the equation ‘cheap workers – cheap products’, thousands of immigrants from China have been pouring to Prato to work in its industries. Most of them are illegal and work in barely human conditions and with low wages. Chinese have taken over the textile manufacture in Prato, being able to provide Italian industry and market with low price products, still ‘Made in Italy’. The result is that the Chinese community in Prato is one of the largest in the country and the interesting aspects here is that they are not integrated but very much considered as “others than Italian” by the local community. The Chinatown in Prato, with all its Chinese restaurants and shops, is the proof of the segregation in which the population is actually living in one of Italy’s ‘most Italian’ cities. I thought this was yet an example of what de los Reyes and Mulinari define as “globaliseringen motsägelselfulla karaktär” (20) i.e. globalization creates and recreates diversity between the different regions in the world in spite of the increasing speed with which people are moving, looking for jobs, an education and also the incredible possibility we are entitled to today, being able to interact with other people from anywhere in the world through the internet, comfortably sitting in our homes. De los Reyes and Mulinari do not mention the aspect of how social media have influenced power relationships and identity formation from an intersectional perspective. Being online and using the computer as an interface, we are able to show or hide identity markers others than ours and this can be especially interesting when the context are institutional settings that provide their services online. The authors underline the fact that globalization implies a move back and forth of individuals, disturbing what they call a ‘geopolitical inequality’: “De andra” finns nu mitt ibland “oss” och ställer krav på en plats i den rika världen” (22) and this is true if considered from the other way around as well. Wealthy people from Europe living in ‘poor countries’ are ‘the others’ craving a place in the poor world, being rich.

The analysis of ‘the doing of’, rather than ‘talking about’ identities
Drawing from the work of Althusser (how the subject is created through language and naming) and Foucault (how the subject is created in relation to the notion of what is the norm and what is not in society), power and power relations become such in social interaction by individuals as opposed to the Marxist tradition where a structural view of class differences is undertaken. The authors consider Tilly’s position in this debate as an attempt to formulate a model of analysis that allows the merge of structural relations and institutional processes: “Aktörernas handlingar och institutionella arrangemang formas inom ramen för en strukturell ordning, men det finns även utrymme för individuella val och strategier” (37). Nevertheless Tully does not wish for an analysis of social differences from an identity perspective but rather an analysis of the social interaction where identity formations are embedded in. According to de lo Reyes and Mulinari, to analyze power relations from an intersectional perspective, does not mean to focus on the different categories from which power can be generated, but rather to put into question the upcoming of those categories, trying to understand what they actually mean. The focus should thus lie on the analysis of ‘the doing of’ a category, not on the existence of the category itself, and how this doing of different categories, in their turn, influences other categories. This could prevent the analysis and later on the production of knowledge to be fragmented. The critic that de los Reyes and Mulinari address to Ambjörsson’s work arises from this starting point: “Invandrarna” är kulisserna i vithetens iscensättning. Vi får aldrig veta om och hur kön konstrueras utanför vitheten och inte heller vad det innebär att tillskrivas en ”avvikande” icke vit kvinnlighet […] Fanny Ambjörnsson erbjuder läsaren en bild som redan är filtrerad av andra blickar och förmedlad av andras röster” (93). This does not mean that structural power relations can be ignored (this, according to the authors, would lead to a stigmatization of those very structures) in the analysis. De los Reyes and Mulinari consider that an intersectional perspective can give individuals a voice and the possibility to become subjects of their own story. According to Bagga-Gupta: “the agency of human subjects-in-context and the playing out of diversity at the realm of praxis-institutions is a complex enterprise” (6: 2010). I think this is an interesting starting point for further discussion and it continues in the same direction as my previews reflection about the emic perspective and the role of scholars in institutional interactional spaces where diversity and identity formation are the main focus.

Literature (besides de los Reyes and Mulinari)
Bagga-Gupta, S. (2010) A common education-for-all & life-long learning? Reflections on inclusion, equity and integration. In Hansen, B. (Ed.) Theory and methodology in international comparative classroom studies. Volume 2, December 2010

Ludvig, A. (2006) Differences Between Women? Intersecting voices in a Female Narrative. European Journal of Women’s Studies. Volume 13 (3) s. 244-258

Giulia Messina Dahlberg 16:47, 24 January 2011 (UTC) [Original author of this resource]

title: Reflections on intersectionality


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TESOL/Describing objects - elementary

What does it look like?

 * It's plain white.
 * It's made of cloth.

title: TESOL/Describing objects - elementary


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User:Aeffanga

About me
Political Simulations and Gaming 2011

Reflective Analysis

The purpose of this essay is to compare and contrast the various games, while identifying political practices within the game itself. According to Hannah Arendt, politics is created through our interactions with others, identifying political practices requires games with multiple players. The essay will be structured, beginning with an initial introduction to the module by an analysis of monopoly. The chosen games within which political practices would be analyzed include, WWII,' Kingmaker’ and Red Guard.' The initial game which we were subject to dissect was monopoly. A game which the majority of people are familiar with. My previous experience of playing the game did two things, firstly it dismissed all political relevance and secondly made me identify the game as a ‘fun.' Playing the game under critical circumstances I developed two concepts. The concept of ‘gaming the game’ and the concept of ‘strategic movement which is a political contract.' Gaming the game within monopoly involved an understanding of the game itself. With limited number of houses available in the bank, buying them all without converting to hotels or buying as many as possible before anyone else, has several significances to the other players. Their development is halted, as they cannot buy houses or landing on a street with several houses could potentially bankrupt them therefore rendering them out of the game. Strategic movement involves calculated risks; the presence of chance over strategy is unfavorable in politics. Monopoly, based on the ‘Landlords game’ by Elizabeth Maggie, was based on a resentment of capitalism. Playing monopoly there is no initial political reference but careful analysis after the game allows for political enlightenment. Guy Debord a situationist and Marxist based his ‘Game of War on Prussian general Carl von Clausewitz’s. A fascination and in-depth strategy as to how a relatively small Prussian army was able to defeat a much more reputable French army. The Game of War is able to capture certain aspects of realism during battle through the movement of game pieces on the board. It incorporates the politics of war and strategic placement, based on calculated risks and limited movement. However, in attempting to create a sense of realism, the Game of War constrains a player’s ability to make instant decisions. In a battle-like scenario, the ability to react or act fast often leads to victory. The Game of War does however take on a realist perspective and leaves little down to luck, seemingly your advantage down to your understanding of the rules, or rather you opponents lack of understanding.

Kingmaker refers to the medieval war of roses, and the meddling of power by influential Baron’s in their attempt to crown a favorable Royal Family. With a prevalent political connotation, politics within Kingmaker is more visible within the game. The individuals within the game attempt to gain political legitimacy first, in the crowning of their chosen Royal house, which also provides the house with a monopoly of power. Andrew McNeil, in the mechanics of the game, provides the ability for players to increase influence through military power and tittles. The historical reference of the game places power within the land owned by Barons, the dispute within the Baronial factions acted not as a conquest for land but rather influence and untimely power over the chosen royal. Equality is promoted regardless of affiliation, and the game becomes a race to establish a ‘first among equals’. Realistic practices within the game was further promoted by assention through offices, at the base a nobleman, with opportunities to become, Dukes, Earls, Chancellors and gain control over the church. The assention through offices for the players, established a self awareness of positions held, with more offices comes more power, the players also were limited from holding conflicting offices, in prevention of a monopoly of power which would lead to simplification of objectives due to offices held.

Historically, the Kingmakers role involves a support for a chosen leader. McNeil’s Kingmaker was replicated a struggle between Baronial factions rather than the Houses themselves. The use of cards promoted an accurate version of the War of Roses. The cards acted an Achilles heels and provided the element of luck to the game. ‘The fall of the dice may signify and determine the divine workings; by it we may move the gods as efficiently as by any other form of contest’ (Huizinga 2002). However the element of luck present throughout the game can act as distraction from the objectives and lead to the concept of ‘Gaming the game’. Through memory, a player is able to establish and control their chances at luck, I.e. events such as plagues and uprisings only take place once in each of the towns, with memory of where these have taken place, a player is then able to strategically move to that position and has an advantage in the game. These information and reasoning were not available during the war.

Origins of WWII followed. The game was presented, with a game board and game pieces. The most simplified game as of yet. Playing as either, France, Britain, the USA, Germany, or the USSR. Like the game of war, and kingmaker, moves were made on the board using the individuals given pieces. Avalon hill, attempt to recreate the 1935-39 international diplomacy, which led up to WW11. It attempts to simulate real life situations through transparency of state strengths and weaknesses which lead up to formation of Alliances, usually alliances to Stop Germany. Geographically accurate, the game succeeds as a political educational tool in its allocation of points. Germany had far more political factors then any other player and could win the game within the second of 6 rounds and alone, the USA stood no chance of winning, not even gaining control over a territory. Without prior knowledge of the crisis, a player is able to take up a historically proportionate position within the game. I.e. the USA has very little in the way of political factions due to their lack of interest and unwillingness to commit within Europe.

Eventually the game presents itself as everyone Vs Germany; alliances are formed to prevent a European domination. The opportunity to ‘appease like Neville Chamberlin’ was also incorporated into the game. Political factors were able to be placed within states and with enough opposing factions, the state was able to displace the factors and therefore break alliances formed, which is historically accurate. However without involvement from Austria or Italy it was less realistic. While shifting the object of the game to ‘stop Germany’ focuses drift and the opportunity emerges for another player to win. The ability to move blind, added a sense of realism to the game, with appropriate appeasement and diplomacy, a player is able to convince or deceive another about their influence within a certain territory.

Origins of WWII compared to the previous games, involved more of the political. Decisions, Movement, appeasement through diplomacy, creating and breaking alliances, are all within a player’s control. While focusing on self interest, players through a desire for victory or Stooping Germany are able to negotiate and create strategic blockades. With a sense of the not too distant history and universal knowledge of the conflict prompted individuals to take a stand as to what state they represent before the game began. The game, forced players to create alliances, without alliances no other player stand an adequate chance of victory, when compared to Germany. In creating alliances, players were forced to make politically beneficial moves by putting aside their initial ambitions for victory. With proper understanding of the rules, the accuracy of the game is also politically beneficial to the play, in recreating WWII. Germany is limited from establishing peace with certain states. The number of rounds also, adds to the politicalization of the game, with 6 rounds, players have to negotiate from the get go and Germany was left to make the last move. With alliances formed from the start, Germany's chances of success are decreased. Similar to the game of war, there is no element of luck. Every move potentially counts even the negotiations; negotiations were best established with players who possessed significant numbers of political factors able to stand a chance at defeating or deceiving Germany.

Finally ‘Red Guard'. General Mao’s revolution depicted into aboard game by Brian Train, attempts to recreate the Cultural Revolution through collective up to 6 players attempting to lead the Chinese Cultural Revolution. General Mao, the much loved Chinese revolutionary is the key to influence “Whatever policy originated from Chairman Mao, we must continue to support," and "Whatever directions were given to us from Chairman Mao, we must continue to follow”. With Mao, presenting the game, the faction which possessed Mao’s influence was unable to be eliminated. The Red Guards, sought out to defend Mao as their supreme leader, dysfunction’s within factions as to who truly represented Mao, is the mechanics in which the game basis itself on.

The game is represented by cards, a die, objective pieces and small cardboard pieces to represent you military might. Players have secret objectives in which they could claim victory. The objectives were motivation for attacks on other players in order to gain necessary institutions to satisfy victory. With no previous in-depth understanding of the Chinese Cultural Revolution, the game provided basic background of the period in concern. However the game failed in its attempt to create a realistic environment. The secret objectives allow for inconsistencies and provide no real room for political leverage. Negotiations and alliances can be formed, but a lack of trust and skepticism present no middle ground for the any player as victory is achievable by any player and is the only means of ending the game. The political aspect of the game represents the realist conception of valuing individual goals over goals of others. Alliances could be forms but the knowledge that they could be broken at any point drove players to pursue self interests as primary.

The mechanics of the game differ significantly. In the game of war, the players were limited to moving accordingly to the capabilities of their pieces, the movement was strictly between the pieces and the board. The mechanics within origins of WWII involve simple and easy placements of the political factors within the chosen country. Kingmaker, was unique in as in every move, a player was entitles 5 moves, and as to where they placed those moves was all within the players control. The final game Red Guard was shown through the movement of combat pieces. When attacking or purging, a player had to separate allocated combat pieces from defense pieces. The movement was theoretical and the combat results were decided by the role of a die. The stronger the attacking force the more likely victory was guaranteed. The very idea of politics is a compromise between individuals. For politics to occur, more than one individual has to be present. The process of establishing individual ideals, discussing them in order to influence or come to a common good is politics. Playing the game alone has no political implication.

In delivering the various political messages, the games differ. The Game of War, as a game failed to present itself as ‘interesting’. The manner in which it was laid out and the fact that the pieces have to be self made, limits the audience it could reach out to. Its historical and political relevance is not initially inferable by way of presentation. It is in during play that the player is able to ‘make political decisions’. Debord succeeds in placing power in the hands of the individual but due to the limitations of accessibility, the aspect of political relevance is reduces to one of nostalgic historical interests. Andrew McNeil’s Kingmaker, by way of mechanics, promotes the political. He incorporates accurately the aims of the game with historically relevant events. Without knowledge of the War on Roses, players are able to grasp the objective connotation of the game aims. Kingmaker’s design and availability make it an accurate political educational tool, however the historical period in which it is relevant presents the game as a historical tool, rather than a political, although the political is at play during the game.

Origins of WWII for the subject of politics, presents compared to Kingmaker or The Game of War, presents the most accurate and realistic political educational tool. It fuses, history, with state interest (through victory objectives), with history and it promotes alliances, in the desire to create a balance of power. Players have limited abilities within their grasp, due to historical simulation, to maximise abilities in order to satisfy victory, the players must negotiate to prevent Germany’s domination by either diplomatically appeasement with Germany or form ‘realist’(self interested) alliances with others. The strategy is signified in the alliances, the interests are what drives the need for alliances and the allocation of political factors (although exaggerated) attempts to replicate historical strength. The simplistic nature of the game, promotes diplomacy on the subject of politics. Diplomacy is central to international politics and states ability to promote diplomatic alliances, can determines its standing within the international community. Avalon Hill’s game succeeds in promoting the political, in game form, where ‘The Game of War’ failed. Red Guard bears similarities between Kingmaker and Origins of WWII. However the possibility at winning is unequal as each player has different victory points. The element of luck, in way of a die and a random selection of reinforcements after each round takes away from historical accuracy. As a political instrument, it models the Cultural Revolution through a separation of factions. The emphasis on the element of luck however is the game is limited in the political. The Game of War had no element of luck and therefore little opportunity in being able to ‘game the game’. Kingmaker on the other hand, presented the element of luck, but luck which was mathematically reversible and appropriately fused strategy with game aims. Similarly to The Game of War, Origins of WWII presented little, in the way of luck but placed great emphasis on player’s negotiation skills. ‘Politics is a game of minds’ and to be political is to understand how to ‘play the game’.

All the above games bar ‘The Game of War’ encourage socialising, the Game of war, demands concentration. The rules on the whole, for all the games were east to follow after a test round; Origins of WWII presented the easiest rules, as the rules were printed on the game packaging. An improvement is needed for the game of war as there are not enough instructions available, the single a4 sheet simply demonstrated the point’s value of the game and for a beginner he rules were not clear. All the games attempted to promote realistic environments for the various aims, Origins of WWII succeed in promoting these while The Game of War presented the most abstract scenario. With no background, similar to a chess board, the game is simply a game of minds, with players reacting to each other’s movements. All games placed different levels of power in the hand of players. Origins of WWII placed the least amount of power into reach of the players, due to the unfavourable point system. Players were forced in alliances. The Game of War placed the most power into within players as the element of luck was inexistent. Simply a two sided game, there was no opportunity for alliances and each player possessed the same abilities, the power of the players was within their decisions. And due to this the decision making process is accurately replicated. Kingmaker presented the most inaccurate opportunity at decision making. A mathematical strategy, entails taking calculated risks which for the most part, are not decisions made through desire. The game subject to the most improvement would be The Game of War. A lack of understanding of the games purpose, due to limited rules and no universal game design, make the game more of a ‘hobby’. Marking are also needed to make it aesthetically appealing, even a board of chess looks interesting. Upon first glance, the Game of War resembles contemporary art and not a game. The game needs to be made to look like a game. I would improve the game by creating a universally accepted game board, ratify all the rules, state clear aims in the rules and make it more available. In improving Kingmaker, the board and playing pieces need to be more legible. Origins of WWII is in need of inclusion of Countries such as Italy and Austria need to be included as active players in the game, to increase the realistic representation of the game. To improve Red Guard, the playing pieces need to be made larger or simply replaces with figures and the allocation of pieces at the end of each round needs to be made equal to all players. A reduction in the way of luck will aid in the games political replication.

As an insight to politics, the various games present a non-traditional approach to the study of ideology. Traditional materials such as textbooks and articles are able to provide an insight into politics but games present ways in which theories can be tested and perfected. Games have been used increasingly in the simulation of real life situations and understanding the purpose of ideology is best demonstrated by games.

Bibliography
 * Huizinga, Johan. Homo Ludens, Routledge.2002

 * In, Qiu (1999). The Culture of Power: Lin Biao and the Cultural

 * Revolution. Stanford, California: Stanford University Press

 * Jiaqi Yan, Gao Gao, Danny Wynn Ye Kwok, Turbulent decade: a history of the cultural revolution, Honolulu Univ. of Hawai'i Press

 * Ladwig, E (2006). Cycle of Life and development in Beijing.

 * Kathleen Bawn. Politics, Strategy and Game Theory. UCLA lecture

Background Paper

The acronym BRIC is derived from Jim O’Neill’s, Chairman of Goldman Sachs Global Assets Management, document entitled ‘Building Better Global Economic BRICS’. In this paper he explores the rapid socio-economic rise of: Brazil, Russia, India and China; all seemingly in competition to exert their regional dominance through trade and politics. It was from this hierarchy of international states that the idea to develop a game based on socio-economic gain came about. What remains a vital factor in winning the game is the necessary need for multilateral trade, as is with global situation. Although each BRIC nation is rich in its specific natural or social resource; becoming a highly advanced country is founded on a level of interdependence.

Each player will receive one of the four developing states card: Brazil, Russia, India or China detailing a specific key attribute e.g.: Brazil-Natural resources. At the beginning of the game each player will start with 200 points. The points received are used to negotiate trading resources needed in achieving infrastructure objectives. Once the required resources have been reached they are deposited into the bank, in return, the corresponding infrastructure piece is received. In developing this game out initial inspiration was drawn from the property-developing game Monopoly. Just as is with Monopoly there is the possibility to trade, bribe or manipulate a player(s) in order to obstruct the goal(s) of another. For example, should a player choose to over-price a commodity then other players have the opportunity to under-cut the market price. It was not until later in the module that it became apparent that our trading process strongly resembled Catan, in that trading natural resources was fundamental to reaching one’s objective. Where BRIC differs from Catan is in returning the resources back to their states after achieving the matching infrastructure piece.

As with Catan the mechanics of the game focus in trade; where the players exchange their natural resource for an agreed upon price. For every infrastructural improvement gained the state becomes stronger, an essential factor in order to protect themselves from chance cars. Upon understanding the basic mechanisms integral to BRIC very little play-testing was initially required. Nonetheless, it soon became apparent that the numerial aspect would prove difficult in over-coming. By limiting the number of points, natural resources and requirements for infrastructure objectives it simplified the mathematical process, by working within a defined boundary.

title: User:Aeffanga


text:
User:Egm6341.s11.team4/HW1

Homework 1 = Problem 1.1: Limits and Taylor Series=

Refer to lecture slide [[media:nm1.s11.mtg3.djvu|3-1]] for the problem statement.

Given

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

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

 * }.
 * }.

Objectives
1. Find the limit of the function as x approaches zero.

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

\lim_{x \rightarrow 0}f(x)=\frac{e^{x}-1}{x} $$
 * $$\displaystyle

 * }.
 * }.

2. Plot the function over the range from zero to one inclusive
 * {| style="width:70%" border="0" align="center"

f(x)=\frac{e^{x}-1}{x},\ x\in[0,1] $$
 * $$\displaystyle

 * }.

Solution
1. Function Limit

Before employing the Taylor series method of evaluating the limit for the function as x approaches zero, we will use the traditional L'Hopital's rule.

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

\lim_{x \rightarrow 0}\frac{e^{x}-1}{x} $$ Now, seeing that both the numerator and the denominator trend towards zero as x approaches zero, we will take the derivatives of both the numerator and the denominator.
 * $$\displaystyle
 * $$\displaystyle
 * }.
 * }.

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

\lim_{x \rightarrow 0}\frac{e^{x}}{1} $$
 * $$\displaystyle
 * $$\displaystyle

Evaluating the numerator at x equals zero we find
 * }.
 * }.

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

$$\displaystyle \frac{1}{1}=1 $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

L'Hopital's rule clearly tells us that the limit of the function approaches one. We will now seek a similar answer using the Taylor series expansion method as demonstrated in class.

We will begin with the Taylor series expansion for $$\displaystyle e^x $$ about zero.

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

\begin{align} e^x \approx \sum_{i=0}^n \frac{x^i}{i!} \\ \end{align} $$
 * $$\displaystyle

 * }.
 * }.

This Taylor series expansion to the n-th order can be displayed so simply since $$\displaystyle e^x $$ evaluated at zero is one and every derivative of the function $$\displaystyle e^x $$ is the function itself. As n goes to infinity this function becomes an equality.

Furthermore, we can evaluate the first term in the summation to get

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

\begin{align} e^x \approx 1+\sum_{i=1}^n \frac{x^i}{i!} \\ \end{align} $$
 * $$\displaystyle

 * }.
 * }.

We can algebraically rearrange this equation to get

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

\begin{align} e^x -1 \approx \sum_{i=1}^n \frac{x^i}{i!} \\ \end{align} $$
 * $$\displaystyle

 * }.
 * }.

Now we have the Taylor expansion of our numerator. To find the Taylor series expansion of our original function we simply need to divide both sides of the above equation by x.

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

\begin{align} \frac{e^x -1}{x} \approx \sum_{i=1}^n \frac{x^{i-1}}{i!} \\ \end{align} $$
 * $$\displaystyle

 * }.
 * }.

The left hand side of this equation is our function $$\displaystyle f(x)$$. The right hand side is a Taylor series approximation of the function. For n=$$\infty$$ this can be expressed as

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

\begin{align} \frac{e^x -1}{x} &= \sum_{i=1}^\infty \frac{x^{i-1}}{i!} \\ \end{align} $$
 * $$\displaystyle

 * }.
 * }.

Inspecting the Taylor series approximation we can see that, when evaluated at $$x=0$$, the numerator will be zero for every i except i=1. Mathematically, this can be expressed by pulling out the term i=1 from our summation, leaving the summation from 2 to infinity.

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

\begin{align} \lim_{x \rightarrow 0} \frac{e^x -1}{x} &= \frac{0^{0}}{0!}+\cancelto{0}{\sum_{i=2}^\infty \frac{0^{i-1}}{i!} }
 * $$\displaystyle

\end{align} $$

 * }.
 * }.

This leads us to our final conclusion (Equation 2), consistent with our expectations via L'Hopital's rule, that

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

$$\displaystyle \lim_{x \rightarrow 0} f(x) = \frac{e^x - 1}{x}=1 $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

2. Function Plot

The function f(x) is then plotted over the span [0,1] to graphically make clear how the function approaches 1 as x approaches 0. Two functions are plotted for this purpose. First, we will plot the function itself. This function, due to a discontinuity, is not evaluated at 0. Next, we will plot our Taylor series approximation using a value of n=30. Though this value is chosen arbitrarily to minimize computational expense, it is shown in problem 2 that this selection is sufficiently large to make the approximation error negligible. The Taylor series expansion, expressing no discontinuities, is evaluated over the entirety of the span [0,1].

Matlab Code:

Plots:

Both the Taylor series approximation and the function itself clearly show that the value of the function as x approaches zero is one.

Solved by Brendan Mahon.

Given
1. $$\displaystyle f(x)=e^x$$

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

Objectives
Find $$\displaystyle p_{x}(x)$$ and $$\displaystyle R_{n+1}(x)$$ of both the given functions at the point $$ x_{0}=0$$.

Solution
According to the Taylor series from the lecture [[media:nm1.s11.mtg3.djvu|Mtg 3]], the Taylor polynomials is as following:

$${{p}_{n}}(x)=f({{x}_{0}})+\frac{(x-{{x}_{0}})}{1!}{{f}^{(1)}}({{x}_{0}})+\frac{2!}{{f}^{(2)}}({{x}_{0}})....+\frac{n!}{{f}^{(n)}}({{x}_{0}})$$

$${{R}_{n+1}}(x)=\frac{1}{n!}\int\limits_^{x}{{{(x-t)}^{n}}{{f}^{(n+1)}}(t)}dt=\frac{(n+1)!}{{f}^{(n+1)}}(\xi )\begin{matrix} , & \xi \in \left[ {{x}_{0}},x \right] \\ \end{matrix}$$

1. $$\displaystyle f(x)=e^x$$

For $$\displaystyle f(x)={{e}^{x}}$$, take the i-th order derivative:

$$\displaystyle  {{f}^{(1)}}(x)={{f}^{(2)}}(x)=...={{f}^{(n)}}(x)={{e}^{x}}$$.

At point $$\displaystyle {{x}_{0}}=0$$,

$$\displaystyle  {{f}^{(1)}}(0)={{f}^{(2)}}(0)=...={{f}^{(n)}}(0)=1$$.

Thus, the n-th order polynomial and residual is:
 * {| style="width:70%" border="0" align="center"

$$\displaystyle {{p}_{n}}(x)=1+\frac{x}{1!}+\frac{2!}+...+\frac{n!} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

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

$$\displaystyle {{R}_{n+1}}(x)=\frac{1}{n!}\int\limits_{0}^{x}dt=\frac{(n+1)!}{{e}^{\xi }}\begin{matrix} , & \xi \in \left[ 0,x \right] \\ \end{matrix} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

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

For $$\displaystyle {{f}_{2}}\left( x \right)=\frac{{{e}^{x}}-1}{x}$$, based on the result of 1): $$\displaystyle {{f}_{1}}(x)={{p}_{n}}(x)+{{R}_{n+1}}(x)$$ , we obtain:

$$\displaystyle \begin{align} & {{f}_{2}}^ – (x)=\frac{{{f}_{1}}^ – (x)-1}{x} \\ & \begin{matrix} {} & \begin{matrix} {} & =\frac{{{p}_{n}}(x)+{{R}_{n+1}}(x)-1}{x} \\ \end{matrix} \\ \end{matrix} \\ & \begin{matrix} {} & \begin{matrix} {} & = \\ \end{matrix}  \\ \end{matrix}\frac{1+\frac{x}{1!}+\frac{2!}+...+\frac{n!}+\frac{(n+1)!}{{e}^{\xi }}-1}{x} \\ & \begin{matrix} {} & \begin{matrix} {} & = \\ \end{matrix}  \\ \end{matrix}\frac{1}{1!}+\frac{x}{2!}+...+\frac{n!}+\frac{(n+1)!}{{e}^{\xi }} \\ \end{align}$$

Thus, At point $$\displaystyle {{x}_{0}}=0$$ ,the n-th order polynomial and residual for $$\displaystyle {{f}_{2}}\left( x \right)=\frac{{{e}^{x}}-1}{x}$$ is:

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

$$\displaystyle {{p}^{'}}_{n}(x)=\frac{1}{1!}+\frac{x}{2!}+...+\frac{n!} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

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

$$\displaystyle {{R}^{'}}_{n+1}(x)=\frac{(n+1)!}{{e}^{\xi }}\begin{matrix} , & \xi \in \left[ 0,x \right] \\ \end{matrix} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

Solved by Shenfeng Yang and Hailong Chen.

Given
IMVT: $$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx}\,\!$$

Objectives
1. Prove the Integral Mean Value Theorem (IMVT) for the non-negative function $$w(\cdot)$$

2. Prove the IMVT for $$w(x) \neq 0 \,\,\,\, \forall \,x \in [a,b]$$

Solution
1.  $$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx}\,\!$$

Where w(x) and f(x) are continuous on [a,b], w(x) is non-negative and ξ is a point on [a,b].

For the case where w(x) = 0, both integrals = 0 and we are left with 0 = 0

The rest of this proof is for w(x) > 0

Since w(x) > 0, $$\int_{a}^{b}{w(x)dx}  >  0$$

Let $$\,m := min f(x) \,\,\forall x \in [a,b]$$

Let $$\,M := max f(x) \,\,\forall x \in [a,b]$$

It follows that:  $$\,m\,\le  \,f(x) \,\le\, M\,\, \forall x \in [a,b]$$

We now multiply through by w(x) to get $$\,m w(x)\,\le \,f(x) w(x) \,\le\, M w(x)\,\, \forall x \in [a,b]$$

Which leads to: $$\,m \int_{a}^{b}{w(x)dx}\,\le  \, \int_{a}^{b}{f(x)}{w(x)dx}\,\le\, M \int_{a}^{b}{w(x)dx}\,\, \forall x \in [a,b]$$

We can now divide through by $$\int_{a}^{b}{w(x)dx}$$

To get $$\,m \,\le \, \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}\,\le\, M \, \forall x \in [a,b]$$

Since f(x) is continuous, there must be at least one value ξ on the interval [a,b]; such that $$f(\xi) = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}$$

Plugging this value into the original statement $$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx} = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\cancel{\int_{a}^{b}{w(x)dx}}}\cancel{\int_{a}^{b}{w(x)dx}}  =  \int_{a}^{b}{w(x)f(x)dx}$$

proves the theorem.

2. The proof for w(x) > 0 is contained in part a. above.

What follows is the proof for the IMVT with w(x) < 0.

$$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx}\,\!$$

Where w(x) and f(x) are continuous on [a,b], w(x) < 0 and ξ is a point on [a,b].

Since w(x) < 0, $$\int_{a}^{b}{w(x)dx}  <  0$$

Let $$\,m := min f(x) \,\,\forall x \in [a,b]$$

Let $$\,M := max f(x) \,\,\forall x \in [a,b]$$

It follows that:  $$\,m\,\le  \,f(x) \,\le\, M\,\, \forall x \in [a,b]$$

We now multiply through by w(x) to get $$\,m w(x)\,\ge \,f(x) w(x) \,\ge\, M w(x)\,\, \forall x \in [a,b]$$

Which leads to: $$\,m \int_{a}^{b}{w(x)dx}\,\ge  \, \int_{a}^{b}{f(x)}{w(x)dx}\,\ge\, M \int_{a}^{b}{w(x)dx}\,\, \forall x \in [a,b]$$

We can now divide through by $$\int_{a}^{b}{w(x)dx}$$

To get $$\,m \,\le \, \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}\,\le\, M \, \forall x \in [a,b]$$

Since f(x) is continuous, there must be at least one value ξ on the interval [a,b]; such that $$f(\xi) = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\int_{a}^{b}{w(x)dx}}$$

Plugging this value into the original statement $$\int_{a}^{b}{w(x)f(x)dx} = f(\xi)\int_{a}^{b}{w(x)dx} = \frac{\int_{a}^{b}{f(x)}{w(x)dx}}{\cancel{\int_{a}^{b}{w(x)dx}}}\cancel{\int_{a}^{b}{w(x)dx}}  =  \int_{a}^{b}{w(x)f(x)dx}$$

proves the theorem.

Solved by Erle Fields and Taeho Kim.

Given

 * $$\displaystyle f(x)=3x-2x^3, \; x \in [0,1]$$

 * $$\displaystyle g(x)=\sinh(x), \; x \in [0,1]$$

Objectives

 * 1. Plot $$\displaystyle f(x)$$ and $$\displaystyle g(x)$$.

 * 2. Find $$||f(x)||_\infty$$, $$||g(x)||_\infty$$ , and $$||f(x)-g(x)||_\infty$$.

Solution
1. Plotting the functions

The following MATLAB code was used to plot f(x) from x=0 to x=1:

Similarly for g(x):

2. Evaluating the Infinity Norms

 * $$||f(x)||_\infty = max(|f(x)|)$$

As seen in Figure 3 above f(x) has a local maxima within the interval $$ x \in [0,1]$$, and will be determined by inspection of the derivative of f(x):

 * $$ \frac{d}{dx}f(x) = 3-6x^2$$

Solving for the x-value for which f(x) attains its local maxima:

 * $$ \frac{d}{dx}f(x) = 3-6x^2=0$$
 * $$\Rightarrow x = \frac{\sqrt{2}}{2}$$

Then solving for the local maximum of f(x):

 * $$\displaystyle f(\frac{\sqrt{2}}{2})=3*\frac{\sqrt{2}}{2}-2*(\frac{\sqrt{2}}{2})^{3} = \sqrt{2}$$

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

$$\displaystyle \Rightarrow ||f(x)||_\infty=\sqrt{2} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

 * $$||g(x)||_\infty = max(|g(x)|)$$

As can be seen in figure 4 above, g(x) on the interval $$x \in [0,1]$$ attains its maximum at x=1.

 * $$max(|g(x)|) = g(1) = sinh(1) \approx 1.1752$$

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

$$\displaystyle \Rightarrow ||g(x)||_\infty \approx 1.1752 $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

 * $$||f(x)-g(x)||_\infty = max(|f(x)-g(x)|)$$

The following MATLAB code was used to calculate this infinity norm:

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

$$\displaystyle \Rightarrow ||f(x)-g(x)||_\infty \approx 0.7392 $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

Solved by William Kurth.

=Contributing Team Members= The following members have contributed to this homework assignment and have reviewed all of the presented material:
 * Egm6341.s11.team4.fields 03:43, 26 January 2011 (UTC)
 * Egm6341.s11.team4.kurth 10:03, 26 January 2011 (UTC)
 * Egm6341.s11.team4.BM 11:39, 26 January 2011 (UTC)
 * Egm6341.s11.team4.yang 08:49, 26 January 2011 (UTC)
 * Egm6341.s11.team4.hylon 10:30, 26 January 2011 (UTC)
 * Egm6341.s11.team4.KTH 20:36, 26 January 2011 (UTC)

Report formatted and compiled by William Kurth.

title: User:Egm6341.s11.team4/HW1


text:
User talk:Yewande

 Hello Yewande, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

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

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 * Give feedback about your initial observations
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You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 17:58, 24 January 2011 (UTC)

Hi
Hi Yewande, thanks for registering for the course. See you on Friday. All the best Leutha 11:08, 25 January 2011 (UTC)

title: User talk:Yewande


text:
User:Eml5526.s11.team3.ushnish12

Refer to meeting 6.1 for the figure.
 * User:Eml5526.s11.team3.ushnish12/Homework 1
 * User:Eml5526.s11.team3.ushnish12/Homework 2
 * User:Eml5526.s11.team3.ushnish12/Homework 3
 * User:Eml5526.s11.team3.ushnish12/Homework 4
 * User:Eml5526.s11.team3.ushnish12/Homework 5
 * User:Eml5526.s11.team3.ushnish12/Homework 6
 * User:Eml5526.s11.team3.ushnish12/Homework 7

title: User:Eml5526.s11.team3.ushnish12


text:
User:EML5526.S11.Team3.risher/homework2

Problem 2.1
From notes 6-2: Derive heat problem in 1-D, i.e, (1) p.4-2

$$ \frac{\partial}{\partial x} \left[ A(x)k(x)\frac{\partial u}{\partial x} \right] + f(x,t) = A \rho c \frac{\partial u}{\partial x} $$

$$ m = \rho A $$

Problem 2.2
From notes 7-3 and 7-4:

1.) Find $$det \left[ b_{jk} \right]$$

2.) Find $$ \underline{ \Gamma } (\underline{b}_{1},\underline{b}_{2},\underline{b}_{3}) = \underline{K}, det \underline{ \Gamma }$$

3.) $$ \underline{F} = \left[ F_{i} \right] = \left[ \underline{b}_{i} * \underline{v} \right]   $$

4.) Solve (5) on p. 7-2 for $$ \underline{d} = \left[ v_{j} \right] $$

(5) on p. 7-2 : $$ \underline{K} \underline{d} = \underline{F} $$

5.) Use $$ \underline{w}_{i} * \underline{P} (\underline{v}) = 0 $$, $$ \forall_{i} = 1, ..., n $$ to find $$ \bar { \underline{K}} \underline{d} = \bar{ \underline{F}} $$

What is $$ \bar{ \underline{K} } $$ and $$ \bar{ \underline{F} } $$ ? Note: $$ \underline{d} = \left[ v_{j} \right] $$

6.) Solve for $$ \underline{d} $$; compare to $$ \underline{d} $$ in part 4.)

7.) Observe symetric properties of $$ \underline{K} $$ and $$ \bar{ \underline{K} } $$

Discuss pros and cons of the 2 methods

Note: (5) p. 7-2 → Bubnov-Galerkin

$$ \underline{K} \underline{d} = \underline{F} $$

Note: (2) p. 7-4 → Petrov-Galerkin

$$ \bar { \underline{K}} \underline{d} = \bar{ \underline{F}} $$

Problem 2.3
From notes 8-3:

Using $$ \left[ \underline{a}_{i} \right] $$ (basis), let $$ \underline{w} = \Sigma_{i} \beta_{i} \underline{a}_{i} $$

Then (1) p. 8-1 is equivalent to:

$$ \underline{w} * \underline{P} (\underline{v}) = 0 $$

$$ \forall \left[ \beta_{1}, ..., \beta_{n} \right] \in \mathbb{R}^n$$ at $$ \underline{w} = \Sigma_{i} \beta_{i} \underline{a}_{u} $$

Show that $$ \underline{w} * \underline{P} (\underline{v}) = 0 $$ is equivalent to:

$$ \underline{a}_{i} * \underline{P} (\underline{v}) = 0 $$

$$ \forall_{i} = 1, ..., n $$

Problem 2.4
From notes 9-2 and 10-1:

Show $$ \int_{} \frac{x^2}{1+x}dx = \frac{x^2}{2} - x +log(1+x) + k $$

1.) $$ \int_{} log(x)dx = xlog(x) - x $$ (HINT: Integration by parts)

2.) $$ \int_{} xlog(x)dx = \frac{1}{2}x^2 \left[ log(x) -\frac{1}{2} \right] $$

3.) Find $$ \int_{} \frac{x^2 dx}{1+cx} $$ (in particular c=1)

4.) find $$ \int_{} \frac{x^2 dx}{a+bx} $$

5.) Find exact solution u(x) for Pb. (3) p. 9-2

$$ \frac{\partial}{\partial x} \left[ (2+3x)\frac{\partial u}{\partial x} + 5x = 0 \right] $$ $$ \forall x \in ] 0,1 [ $$ $$ u(1)=4, -\frac{\partial u(x=0)}{\partial x} =6 $$

6.) Plot u(x)

Note: Use WolframAlpha to help check and solve some problems (very helpful tool) http://www.wolframalpha.com/

e.g., (debt usa)/(gdp usa)

integrate log(x), etc

Problem 2.5
From notes 10-1:

Complete part B of proof of (3) p. 8-2, i.e., show (2) p. 7-2 → (2) p. 8-2

(2) p. 7-2: $$ \underline{b}_{i} * \underline{P} (\underline{v}) = 0 $$ $$ i=1, ..., n $$

(2) p. 8-2:

$$ \underline{w} * \underline{P} (\underline{v}) = 0 $$ $$ \forall \left[ \alpha_{1}, ..., \alpha_{n} \right] \in \mathbb{R}^n$$ at $$ \underline{w} = \Sigma_{i} \alpha_{i} \underline{b}_{u} $$

Problem 2.6
From notes 10-2 and 10-3

Consider $$ F = \left[ 1, cos(iwx), sin(iwx) \right] $$ on interval [0,T], i.e., i = 1,2

$$ \left[ 1, cos(wx), cos(2wx), sin(wx), sin(wx), sin(2wx) \right] = \left[ b_{1}(x), b_{2}(x), b_{3}(x), b_{4}(x), b_{5}(x) \right] $$

1.) Constr. $$ \underline{\Gamma}(F) $$; obs. prop. of $$ \underline{\Gamma} $$

2.) Find $$ det \underline{\Gamma}(F) $$

3.) Conclude F is orthogonal basis, i.e., $$ \Gamma_{ij} = {<}b_{i},b_{j}{>} =\delta_{ij} $$ $$ \delta_{ij} = 1  $$ for  i = j   $$ \delta_{ij} = 0  $$ for  i (not equal to) j

Problem 2.7
From notes 10-3

Similarly for $$ F = [1, x , x^2 , x^3 , x^4] $$

$$ \Omega = \left[ 0,1 \right]$$ in Part 3.) of Problem 2.6

Is F orthog. family?

Problem 2.8
From notes 11-1:

Show (1) p. 10-4 ←→ (2) p. 10-4

Problem 2.9
From notes 12-1

My Solutions
Solutions for homework #2 by Risher

Risher's WikiHome
Go to Risher's WikiHome

title: User:EML5526.S11.Team3.risher/homework2


text:
User talk:Eml5526.s11.team3.ushnish12

 Hello Eml5526.s11.team3.ushnish12, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

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

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twiter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --Ottava Rima (talk) 18:41, 24 January 2011 (UTC)

title: User talk:Eml5526.s11.team3.ushnish12


text:
User talk:Eml5526.s11.team7.jin/HW1

Jin,

Equation 1.2 does not match with snooks equation.

Lalith

title: User talk:Eml5526.s11.team7.jin/HW1


text:
User talk:Fem.tm7/HW

My comments: 1. The force balance equation does not match with jin's solution equation 1.2. 2. For Part 2 would it possible to put new figure in place of the image cut from the lecture.

lalith

title: User talk:Fem.tm7/HW


text:
User:Sorenhogberg/Intersectionality 2 SH

Reflections on the following articles

•	European Journal of Women’s Studies (2006) Volume 13 (3) s. 187-290, 103 s. Sage Journals, Online.

•	McCall, Leslie (2005): The complexity of intersectionality. Signs: Journal of Women in Culture and Society. Volume 3 (3) s. 1771-1800, 29 s.

•	Staunaes, Dorthe (2003): Where have all the subjects gone? Bringing the concepts of intersectionality and subjectification, s. 101-110, 9 s. NORA Volume 11 (2)

Reading the articles mentioned above has certainly broaden my understanding of intersectionality as theory. Taking it’s point of departure from a critical perspective upon the fact that inequalities seem to last in spite of convincing results along with previous as well as today’s research that develop critical results on the one and at the same time create conservative contributions on the other hand, as Bredström shows according to race within the context of people having HIV/AIDS, has opened for a reflexive attitude to research methods in general. The knowledge emerged out of black feminism has led to a richer and more complex ontology and a continuously ongoing process of questioning methods of intersectionality itself.

There is probably a number of ways that these articles could be read but my way of approaching them will be to seek for the ways the authors claim the necessity to expand ways of understanding intersectionality. Yuval-Davis and others question the additive way of understanding inequalities. When so is the case research seem to remain only on one level of analysis, the experiential. This becomes a problem since results then will be mainly understood as different kinds of identities. Instead by looking on the relations between discourse and peoples way of identify themselves it’s possible to understand how hegemonies works in society. That’s why Yuval-Davis argues for the need to scrutinize inequalities on several levels. What creates identity is a complex process that involves differences relating to different powering positions but also how these are interpreted according to cultural differences. I also find it interesting when existing political values are implemented into the understanding of how identity epistemology is formed.

Several authors (Prins, Verloo, Yuval-Davis and others) argue that mistakes are being made if different kinds of categories as race, gender etc are managed in similar ways. Verloo explains this in a convincing way by taking a closer look at the complexity of multiple inequalities. When doing this she also makes a distinction between structural intersectionality and political intersectionality showing that differences between categories are obvious when we take a close look. She then addresses a number of strategies to avoid this mistake. If the differences between categories are not taken into consideration there is a risk that political intentions, when dealing with these kinds of questions, will lead to unwished consequences.

Another debate within the field on intersection has to do with the dimension of power according to the process of positioning the subject in relation to the agency of the subject. The question touches the arguments held by Yuval-Davis but is more in detail elaborated by Buitelaar who shows how articulations of I-positions are made in order to make a stand. The agency conducted by the individual can be understood as an exploration of how identity can be co-constructed and reconstructed, in relation to different categories, for a certain purpose. Looking upon sub-cultures that take a resisting stand by redefining their status of position also give evidence to the fact that individuals are not absolutely determined to identify themselves with what at first seems to be a definite subject position. In fact Staunaes shows us how inequality and exclusion in one category can create an agency for strengthening positions in another category.

Futhermore Prins makes a distinction between systematic and constructionist interpretations of intersectionality. The point, hereby made, links to the question of categories which has a big place in the debate. Prins argues for a constructionist perspective since we otherwise will overlook possible ways of understanding inequality. In the study a new category, origins, was found that could explain how different subject positions were constructed. If Prins would have taken his point of departure from already accepted categories, the individuals’ way of identifying themselves would have been making no sense.

Well then, are categories good or bad? Leslie McCall put this question in new light when arguing for a possibility to use categories deliberately. By discussing a continuum from (inter)categorical complexity to anticategorical complexity where an intracategorical complexity is placed somewhere in-between, she clarifies that intersectionality is neither a positivist nor a postmodern science. I found this discussion very important since it places intersectionality in another position relating to theory of science than I first expected. I believe that McCall:s paper contributes to understand intersectionality as a theory of methodology. Perhaps postmodern anticategorical complexity studies have given me, and perhaps others, the idea that intersectionality belongs in that field. When McCall argues for at more positivist approach, she talks about critical realism, using categories and looking for quantitative data, she convinces me that intersectionality is a possible approach also when taking her point of departure from an angel of categorical complexity. But maybe my main belief has been that of placing intersectionality in a position constructed by a combination of discourse theory and Marxist theory. In any case, McCall clarifies intersectionality by taking an unusual position.

The intracategorical category could also be understood as a category in-between. An act of balance where focus alternate from the ways of how structural powers in society are creating demarcation, which leads to exclusion and inequality to how individuals handle the position offered (se the discussion above). An intracategorical complexity will then try to reveal and understand differences within a group. This is what Ludvig accomplishes in her work looking for differences between women.

Finally I want to give voice to Staunaes who broadens the object of intersectionality from “the otherness” to also include the “firstness”. Maybe this has been taken for granted but I think Staunaes has got a point when trying to find the “firstness” on same conditions as the “otherness” instead of just creating a notion of the “firstness” based upon the “otherness”.

- A question to be discussed according to the discussion above is: If intersectionality is a theory of methodology, or if focus is upon “firstness”, is it possible to understand intersectionality as not a member of critical research?


text:
Category:Egm6341.s11.team2

Team 2

title: Category:Egm6341.s11.team2


text:
User:Egm6341.s11.team3.rakesh

http://en.wikiversity.org/wiki/User:Egm6341.s11.team3.rakesh/hw2

= HW 1.1 =

Problem
Find the limit of function: $$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac {e^x - 1}{x} $$

Find
$$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac {e^x - 1}{x} $$,

Solution
Take the limit of the function:

$$ \lim_{x\to 0} f(x) = \frac {e^0 - 1}{0} = \frac {0}{0}$$

Hence we can apply L-Hopital Rule:

$$ \lim_{x\to 0} f(n) = \frac {f'(n)}{g'(n)} \qquad (1)$$

By above rule:

$$ \lim_{x\to 0} f(x) = \lim_{x\to 0} \frac {e^x - 1}{x} =\lim_{x\to 0} \frac {f(x)}{g(x)} \Rightarrow

\begin{matrix} f'(x) = e^x -0 \\ g'(x) = 1 \end{matrix} $$

$$ \Rightarrow \lim_{x\to 0} f(0) = \frac {f'(0)}{g'(0)} = \frac {e^0 - 0}{1} = 1 \ $$

$$ Ans.: \lim_{x\to 0} f(x) = \frac {e^x - 1}{x} = 1 $$

= HW 2.1 =

Problem
Do the integration by parts to reveal three more terms  :  $$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\int\limits_{x}^{x_0}(x-t)f''(t)dt } $$

Problem 2.1.2
use IMVT theorem to express the remainder $$ f^5(\S)for (x,x_0) $$

Problem
use IMVT theorem to express the remainder $$ f^5(\S)for (x,x_0) $$

Solution
a) Given That
 * {| style="width:100%" border="0"

$$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\int\limits_{x}^{x_0}(x-t)f''(t)dt } $$|| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}(x-t)f(t)dt = [-\frac {(x-t)^2}{2}]_{t=x_0}^{t=x} f(t)-\int\limits_{x}^{x_0}\frac{-(x-t)^2}{2}f'''(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}(x-t)f(t)dt = \frac {(x-x_0)^2}{2} f(x_0)+\int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f'''(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f(t)dt = -[\frac {(x-t)^3}{2*3}]_{t=x_0}^{t=x} f(t)-\int\limits_{x}^{x_0}\frac {-(x-t)^3}{2*3}f(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^2}{2}f(t)dt = \frac {(x-x_0)^3}{3!} f(t)+\int\limits_{x}^{x_0}\frac {(x-t)^3}{3!}f(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^3}{3}f'(t)dt = -[\frac {(x-t)^4}{4*3!}]_{t=x_0}^{t=x} f'(t)-\int\limits_{x}^{x_0}-\frac {(x-t)^4}{4!}f'(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^3}{3}f'(t)dt = \frac {(x-x_0)^4}{4!} f'(t)+\int\limits_{x}^{x_0}\frac {(x-t)^4}{4!}f(t)dt $$| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ {f(x)=f(x_0)+\frac {x-x_0}{1!} f(x_0)+\frac {(x-x_0)^2}{2} f(x_0)+ \frac {(x-x_0)^3}{3!} f'(t)+\frac {(x-x_0)^4}{4!} f'(x_0)+\int\limits_{x}^{x_0}\frac {(x-t)^4}{4!}f'(t)dt} $$|| $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$ \int\limits_{x}^{x_0}\frac {(x-t)^4}{4}f^(4)(t)dt =  f^5(\xi)\frac {(x-x_0)^5}{5!} $$, $$   \xi  \in (x,x_0) $$ By IMVT THEROEM
 * style="width:95%" |
 * style="width:95%" |
 * }

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

 * style="width:95%" |Integrate by parts of $$R_{n+1}\left ( x \right )$$
 * $$\frac{1}{n!}\int_{x_{0}}^{x}\left ( x-t \right )^{n}f^{(n+1)}\left ( t \right )dt=\frac{1}{n!}\left \{ \left [ -\frac{(x-t)^{n}}{(n+1)}f^{(n+1)}\left ( t \right ) \right ]_{x_{0}}^{x} - \int_{x_{0}}^{x}\frac{-(x-t)^{n+1}}{(n+1)}f^{(n+1)}\left ( t \right )dt\right \}$$||
$$\displaystyle
 * $$\frac{1}{n!}\int_{x_{0}}^{x}\left ( x-t \right )^{n}f^{(n+1)}\left ( t \right )dt=\frac{1}{n!}\left \{ \left [ -\frac{(x-t)^{n}}{(n+1)}f^{(n+1)}\left ( t \right ) \right ]_{x_{0}}^{x} - \int_{x_{0}}^{x}\frac{-(x-t)^{n+1}}{(n+1)}f^{(n+1)}\left ( t \right )dt\right \}$$||
$$\displaystyle

$$
 * }

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

 * style="width:95%" |
 * $$ \frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}\left ( x_{0} \right )+\int_{x_{0}}^{x}\frac{(x-t)^{n+1}}{(n+1)!}f^{(n+2)}dt$$
$$\displaystyle
 * $$ \frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}\left ( x_{0} \right )+\int_{x_{0}}^{x}\frac{(x-t)^{n+1}}{(n+1)!}f^{(n+2)}dt$$
$$\displaystyle

$$
 * }

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

 * style="width:95%" |
 * $$f\left ( x \right )=P_{n+1}\left ( x \right )+R_{n+2}\left ( x \right )$$||
$$\displaystyle
 * $$f\left ( x \right )=P_{n+1}\left ( x \right )+R_{n+2}\left ( x \right )$$||
$$\displaystyle

$$
 * }

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

$$ P_{n+1}\left ( x \right )=f\left ( x_{0} \right )+\frac{(x-x_{0})}{1!}f^{(1)}\left ( x_{0} \right )+...+\frac{(x-x_{0})^{n+1}}{(n+1)!}f^{(n+1)}\left ( x_{0} \right ) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle
 * }

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

$$ R_{n+2}\left ( x \right )=\frac{1}{(n+1)!}\int_{x_{0}}^{x}(x-t)^{n+1}f^{(n+2)}dt $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$\displaystyle
 * }

APPLYING IMVT TO EQUATION

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

$$\int\limits_{x}^{x_0}\frac{(x-t)^n}{n!}f^{n+1}(t)dt = f^{n+1}(\xi)\frac {(x-x_0)^n}{n+1!} \xi \in (x,x_0)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

= HW 2.2 =

 REMARK: We did the solution on our own

HW 2.2.a
1. Make sure to leave spaces in between the code like I have, b'fore finishing the stuff. 2. The way I do is I make extra space, then copy there the code and when I finished doing that, I copy/paste the formulas to the code that I put above them and then delete the plain formulas.

 * Example*

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

$$ F = force \quad (N) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x) = \sin x, x\in(0,\pi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x) = \sin x \quad at \quad x_0=\frac{\pi}{4} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x) = cos(x) \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

As for font, it seems that wiki has it's own.. though process or whatever...

use "\quad" command at the end to up the font. Quad is nothing more then saying "skip space" but it does a job.

 * example*

//

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

$$ f'(x) = \cos(x) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x) = \cos(x) \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

 * end of example

!!!! also if u can do cos(x) in place of \cos x and similarly for sin

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

$$f(x) = \sin x, x\in(0,\pi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x) = \sin x \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$f(x) = \sin x, x\in(0,\pi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x) = \sin x  \quad $$ $$ x\in(0,\pi)$$   |$$ x_0=\frac{\pi}{4}$$ $$
 * style="width:95%"  |
 * style="width:95%"  |
 *
 $$ \displaystyle
 * }

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

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

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

$$
 * style="width:95%" |$$  f'(x) = \cos x \quad$$
 *
$$ \displaystyle
 *
$$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$  f''(x) = -\sin x \quad $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$ f'''(x) = -\cos x \quad $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

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

$$
 * $$ f(x) =  \sin x \quad $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

we know from taylor series
 * {| style="width:100%" border="0"

$$ f(x) = P_n + R_n+1 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

where as

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

$$P_n =f(x_0)+\frac {x-x_0}{1!} f(x_0)+\frac {(x-x_0)^2}{2} f(x_0)+\frac {(x-x_0)^3}{3!} f'(t)+.....+\frac {(x-x_0)^n}{n!} f^n(t) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_n+1 = \frac {(x-x_0)^n+1}{n+1!} f^{n+1}(\S) $$ $$ when N =0
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$  f(x) = P_0+ R_1 \quad $$
 *
$$ \displaystyle
 *
$$ \displaystyle
 * }

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

$$P_0 = f(x_0) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_1= (x-\frac{\pi}{4})$$   $$ \,f(\S)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x_0) = \frac{1}{\sqrt{2}}+(x-\frac{\pi}{4}) \,f(\S) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ f(x) = P_1+ R_2 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$P_0 = f(x_0) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$P_1 = f(x_1) = \frac{x-\frac{\pi}{4}}{1!}\cos\frac{\pi}{4}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_2= \frac{(x-\frac{\pi}{4})^2}{2!} \,f^2(\S)$$$$ \displaystyle $$ $$P_1 + R_2= \frac{1}{\sqrt{2}}+\frac{x-\frac{\pi}{4}}{1!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^2}{2!} \,f^2(\S) $$ $$ when n=2
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ f(x) = P_2+ R_3 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}+\frac{x-\frac{\pi}{4}}{1!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^3}{3!} \,f^3(\S)  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

when n=3

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

$$ f(x) = P_3+ R_4 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}+\frac{x-\frac{\pi}{4}}{1!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^3}{3!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^4}{4!} \,f^4(\S)  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

when n=4

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

$$ f(x) = P_4+ R_5 \quad$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}+\frac{x-\frac{\pi}{4}}{1!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^3}{3!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{\pi}{4})}^4}{4!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^5}{5!} \,f^5(\S)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

so on and when n=10
 * {| style="width:100%" border="0"

$$ f(x) = P_{10}+ R_{11}\quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}+\frac{x-\frac{\pi}{4}}{1!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}-\frac{({x-\frac{\pi}{4})}^3}{3!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{\pi}{4})}^4}{4!}\frac{1}{\sqrt{2}}+.......-\frac{({x-\frac{\pi}{4})}^{10}}{10!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^{11}}{11!} \,f^{11}(\S)  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

finally the maximum can be computed by

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

$$R_{n+1} = \frac {(x-x_0)^{n+1}}{n+1!} f^{n+1}(\S)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$R_{n+1}\le\frac {(x-x_0)^{n+1}}{n+1!} |maxf^{n+1}(\S)|\le\frac {(x-x_0)^{n+1}}{n+1!}=\frac {(\pi/4)^{n+1}}{n+1!}\le(\pi/4)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

HW 2.2.b

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

$$ f(x) = \sin x, x\in(0,\pi) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x) = \sin x \quad at \quad x_0=\frac{3\pi}{4} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f'(x) = COS(X) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f'(x) = COS(X) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$  f''(x) = -\sin x  $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

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

$$
 * style="width:95%" |$$ f'''(x) = -\cos x  $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

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

$$
 * $$ f(x) =  \sin x  $$
 *
 $$ \displaystyle
 *
 $$ \displaystyle
 * }

we know from taylor series
 * {| style="width:100%" border="0"

$$ f(x) = P_n + R_n+1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

where as

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

$$P_n =f(x_0)+\frac {x-x_0}{1!} f(x_0)+\frac {(x-x_0)^2}{2} f(x_0)+\frac {(x-x_0)^3}{3!} f'(t)+.....+\frac {(x-x_0)^n}{n!} f^n(t) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_n+1 = \frac {(x-x_0)^n+1}{n+1!} f^{n+1}(\S) $$ $$ when N =0
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$P_0 = f(x_0) = \sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_1= (x-\frac{3\pi}{4})$$   $$ \,f(\S)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ f(x_0) = \frac{1}{\sqrt{2}}+(x-\frac{3\pi}{4}) \,f(\S) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ f(x) = P_1+ R_2 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$R_2= \frac{(x-\frac{3\pi}{4})^2}{2!} \,f^2(\S)$$$$ \displaystyle $$ $$f(x)= \frac{1}{\sqrt{2}}-\frac{x-\frac{3\pi}{4}}{1!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{3\pi}{4})^2}{2!} \,f^2(\S) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

when n=2

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

$$ f(x) = P_2+ R_3 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}-\frac{x-\frac{3\pi}{4}}{1!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{3\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{3\pi}{4})^3}{3!} \,f^3(\S)  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

When n=3

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

$$ f(x) = P_3+ R_4 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}-\frac{x-\frac{3\pi}{4}}{1!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{3\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{3\pi}{4})}^3}{3!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{3\pi}{4})^4}{4!} \,f^4(\S)  $$ $$ ...so on When n=10
 * style="width:95%" |
 * style="width:95%" |
 *
$$ \displaystyle
 * }
 * {| style="width:100%" border="0"

$$ f(x) = P_{10}+ R_{11} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$ =\frac{1}{\sqrt{2}}-\frac{x-\frac{3\pi}{4}}{1!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{3\pi}{4})}^2}{2!}\frac{1}{\sqrt{2}}+\frac{({x-\frac{3\pi}{4})}^3}{3!}\frac{1}{\sqrt{2}}++.......-\frac{({x-\frac{3\pi}{4})}^{10}}{10!}\frac{1}{\sqrt{2}}+\frac{(x-\frac{\pi}{4})^{11}}{11!} \,f^{11}(\S)  $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

finally the maximum can be computed by

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

$$R_{n+1} = \frac {(x-x_0)^{n+1}}{n+1!} f^{n+1}(\S)$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

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

$$R_{n+1}\le\frac {(x-x_0)^{n+1}}{n+1!} |maxf^{n+1}(\S)|\le\frac {(x-x_0)^{n+1}}{n+1!}=\frac {(\pi/2)^{n+1}}{n+1!}\le(\pi/2)$$ $$ = 3.5 = $$E_{n}^{n+1}\left ( x \right )=F^{n+1}\left ( x \right)-0$$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

Solution

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

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

We start with an general equation for error estimation

From we learn that

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(i) where $$l_{i,n}\left ( x \right)=l_{i}\left ( x \right)=\prod_{j=0,i \neq j}^{n}\frac{x-x_{j}}{x_{0}-x_{j}}\,$$ $$E_n^{n+1}=\int_{0}^{1}f^{n+1}\left ( x \right )dx-\int_{0}^{1}f_{n}^{n+1}\left ( x \right )dx\, \quad $$ $$
 *
 $$ \displaystyle (Eq. 3.5.10)
 * }

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

$$ \begin{align} \\n=1, & l_i\left ( x \right)=\frac{x-x_{1}}{x_{0}-x_{1}}  l_{0}=l\left ( x^{1}\right)\;  equation \mbox{ }of\mbox{ } degree\mbox{ } 1 \\n=2, & l_i\left ( x \right)=\frac{x-x_{1}}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}}  l_{0}=l\left ( x^{2}\right)\; equation \mbox{ }of \mbox{ }degree \mbox{ } 2 \\.\, \\.\, \\.\, \\n=n, & l_i\left ( x \right)=\frac{x-x_{1}}{x_{0}-x_{1}}\frac{x-x_{2}}{x_{0}-x_{2}}...\frac{x-x_{n-1}}{x_{0}-x_{n-1}}\frac{x-x_{n}}{x_{0}-x_{n}}  l_{0}=l\left ( x^{n}\right)\, equation \mbox{ }of\mbox{ } degree\mbox{ } n \end{align} $$||
 $$ \displaystyle $$
 * style="width:95%" |
 * style="width:95%" |
 * }

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

$$f\left( x_{i} \right)\,$$ is constant, Hence forth  $$f_n\left( x \right)\,$$ is $$x^{n}\,$$'s function $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle
 * }

which is inferred from our professors lecture

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

$$f_n^{n+1}\left ( x \right )=0\,$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *
 $$ \displaystyle (Eq. 3.5.11)
 * }

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

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

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

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

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

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

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

title: User:Egm6341.s11.team3.rakesh


text:
User:Egm6341.s11.team4.yang/hw1

=HW 1.2 problem=

Find $${{p}_{n}}(x)$$ and $${{R}_{n+1}}(x)$$ of

1).$${{e}^{x}}$$

2).$${{f}_{2}}\left( x \right)=\frac{{{e}^{x}}-1}{x}$$

at point $${{x}_{0}}=0$$.

=Answer:=

According to the Taylor series from the lecture [[media:nm1.s11.mtg3.djvu|Mtg 3 (d)]], the Taylor polynomials is as following:

$${{p}_{n}}(x)=f({{x}_{0}})+\frac{(x-{{x}_{0}})}{1!}{{f}^{(1)}}({{x}_{0}})+\frac{2!}{{f}^{(2)}}({{x}_{0}})....+\frac{n!}{{f}^{(n)}}({{x}_{0}})$$

$${{R}_{n+1}}(x)=\frac{1}{n!}\int\limits_^{x}{{{(x-t)}^{n}}{{f}^{(n+1)}}(t)}dt=\frac{(n+1)!}{{f}^{(n+1)}}(\xi )\begin{matrix} , & \xi \in \left[ {{x}_{0}},x \right] \\ \end{matrix}$$

1). For $$f(x)={{e}^{x}}$$, take the i-th order derivative:

$$ {{f}^{(1)}}(x)={{f}^{(2)}}(x)=...={{f}^{(n)}}(x)={{e}^{x}}$$.

At point $${{x}_{0}}=0$$,

$$ {{f}^{(1)}}(0)={{f}^{(2)}}(0)=...={{f}^{(n)}}(0)=1$$.

Thus, the n-th order polynomial and residual is:
 * {| style="width:70%" border="0" align="center"

$$\displaystyle {{p}_{n}}(x)=1+\frac{x}{1!}+\frac{2!}+...+\frac{n!} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

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

$$\displaystyle {{R}_{n+1}}(x)=\frac{1}{n!}\int\limits_{0}^{x}dt=\frac{(n+1)!}{{e}^{\xi }}\begin{matrix} , & \xi \in \left[ 0,x \right] \\ \end{matrix} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

2). For $${{f}_{2}}\left( x \right)=\frac{{{e}^{x}}-1}{x}$$, based on the result of 1): $${{f}_{1}}(x)={{p}_{n}}(x)+{{R}_{n+1}}(x)$$ , we obtain:

$$\begin{align} & {{f}_{2}}^ – (x)=\frac{{{f}_{1}}^ – (x)-1}{x} \\ & \begin{matrix} {} & \begin{matrix} {} & =\frac{{{p}_{n}}(x)+{{R}_{n+1}}(x)-1}{x} \\ \end{matrix} \\ \end{matrix} \\ & \begin{matrix} {} & \begin{matrix} {} & = \\ \end{matrix}  \\ \end{matrix}\frac{1+\frac{x}{1!}+\frac{2!}+...+\frac{n!}+\frac{(n+1)!}{{e}^{\xi }}-1}{x} \\ & \begin{matrix} {} & \begin{matrix} {} & = \\ \end{matrix}  \\ \end{matrix}\frac{1}{1!}+\frac{x}{2!}+...+\frac{n!}+\frac{(n+1)!}{{e}^{\xi }} \\ \end{align}$$

Thus, At point $${{x}_{0}}=0$$ ,the n-th order polynomial and residual for $${{f}_{2}}\left( x \right)=\frac{{{e}^{x}}-1}{x}$$ is:

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

$$\displaystyle {{p}^{'}}_{n}(x)=\frac{1}{1!}+\frac{x}{2!}+...+\frac{n!} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

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

$$\displaystyle {{R}^{'}}_{n+1}(x)=\frac{(n+1)!}{{e}^{\xi }}\begin{matrix} , & \xi \in \left[ 0,x \right] \\ \end{matrix} $$ $$
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * style="width:20%; padding:10px; border:2px solid #8888aa" |
 * $$\displaystyle
 * style= |
 * }.
 * }.

title: User:Egm6341.s11.team4.yang/hw1


text:
User:Eml5526.s11.team4.yang

=Homework=

Homework-1 team4

 * homework 1 team4

Homework-2

 * Homework 2.3
 * Homework 2.6

Homework-3

 * Homework 3

Homework-4

 * Homework 4

Homework-5

 * Homework 5

Homework-6

 * Homework 6

=Links for this course=
 * EML5526-FEA-Team4
 * EML5526-FEA-Course Website
 * EML5526-FEA-Course Lecture Plan

=The other course: EGM6341-NM1=
 * EGM6341-NM1-Team4-yang
 * EGM6341-NM1-Team4
 * EGM6341-NM1-Course Website
 * EGM6341-NM1-Course Lecture Plan

title: User:Eml5526.s11.team4.yang